An Introduction to Finite Tight Frames

Shayne F. D. Waldron

– Monograph –

July 26, 2017

Springer

To my wife Catherine Ann Etheredge.

Preface

This book gives a unified introduction to the rapidly developing area of finite tight frames. Fifteen years ago, the existence of equal–norm tight frames of n > d vectors for Rd and Cd was not widely known. Now equal–norm tight frames are known to be common, and those with optimal cross–correlation and symmetry properties are being constructed and classified. The impetus behind these rapid developments are applications to areas as diverse as signal processing, quantum information theory, multivariate orthogonal polynomials and splines, and compressed sensing. It can be thought of as an extension of the first chapter of Ole Christensen’s book: An introduction to Frames and Riesz Bases (in this series), which deals mostly with the infinite dimensional case. For finite dimensional Hilbert spaces the technicalities of Riesz bases disappear (though infinite frames are still of interest), and, with some work, usually a nice tight frame can be constructed explicitly. Hence the focus is on finite tight frames, which are the most intuitive generalisation of orthonormal bases. In addition to analogues of familiar ideas from the infinite dimensional setting such as group frames, there is a special geometry, e.g., the variational characterisation and the frame force. The book is structured into chapters, with the first paragraph intended to give a feeling for its content. These give a logical development, while being as independent as possible. For example, one could jump to those giving the important examples of harmonic frames, equiangular frames and SICs, referring back to the motivating chapters on symmetries and group frames as desired. Similarly, in the text we give forward references to such nice examples. Grey boxes are used to emphasize or paraphrase some key ideas, and may help readers to navigate this book. Each chapter has Notes, which primarily give a brief description of my source material and suggestions for further reading. The interdisciplinary nature of the subject makes it difficult (and often senseless) to give exact attribution for many results, e.g., the discrete form of Na˘ımark’s theorem, and so I have attempted to do so only in some cases. This introductory book is deliberately as short as possible, and so its scope and list of references is far from definitive. If only for this reason, some contributions may not be noted explicitly.

vii viii Preface

The chapters conclude with Exercises, which contain more details and examples. Brief solutions are given, since some are parts of proofs, and in the interests of being both introductory and self contained. Those marked with a m are suitable for using a computer algebra package such as matlab, maple and magma. To get a real sense for finite tight frames the reader is encouraged construct the various examples given in the book numerically or algebraically. In particular, the harmonic frames, MUBs and Weyl–Heisenberg SICs for d 100 (or higher), which really are quite remarkable. ≤ This book can be used for an introductory (graduate or undergraduate) course in finite tight frames: chapters 1, 2, 5, 6, 7, 8, 11, together with some examples of interest, e.g., group frames (chapter 10), equiangular tight frames (chapter 12) and SICs (chapter 14). The latter chapters rely heavily on parts of classical algebra, such as graph theory (real equiangular tight frames), character theory (harmonic frames), representation theory (group frames), and Galois theory (SICs). Examples from these could be included in a graduate algebra course, e.g., Theorem 10.8 as an orthogonal decomposition of FG–modules. It could also act as the theoretical background for a course involving applications of tight frames, e.g., some of the topics treated in Finite frames Theory and Applications (in this series). It is also suitable for self study, because of the complete proofs and solutions for exercises. How does one come to write such a book? My initial interest in finite tight frames was accidental. While investigating the eigenstructure of the Bernstein operator, I encountered deficiencies in the existing orthogonal and biorthogonal expansions for the Jacobi polynomials on a triangle. Having heard of frames while a graduate student in Madison, I supposed that a finite tight frame might just be what was required. This eventually proved to be so (see 15), and along the way I found that finite frames are a fascinating and very active area§ of research. When I started this book, there was just enough material for a short book on finite tight frames, a small part of frame theory, at the time. Since then, there has been an explosion of research on finite tight frames, which continues apace. I have called time on the project: the book contains results from last week (Lucas-Fibonnacci SICs), but not next week (maybe a proof of Zauner’s conjecture). There are chapters that could have been written, e.g., ones on erasures and data transmission. I maintain a strong interest in finite tight frames. On my home page, there are various links of relevance to the book (lists of SICs, harmonic frames, typos, etc). It is my hope that this book conveys the basic theory of finite tight frames, in a friendly way, being mindful of its connections with existing areas and continuing development. The applications given are just the tip of an iceberg: an invitation to use finite tight frames in any area where there is a natural inner product on a finite dimensional space.

Shayne F. D. Waldron Honeymoon Valley, Far North Aotearoa (New Zealand) July 2017 Acknowledgements

There are many people to thank. Catherine for her support. My mentors Carl de Boor (Orcas island) and Len Bos (Verona). My colleagues at home: Tom ter Elst, Sione Ma’u and Warren Moors, and in Italy: Marco Vianello and Stefano Di Marchi. My students: Irine Peng, Richard Vale, Simon Marshall, Nick Hay, Helen Broome, Tan Do, Jennifer Bramwell, Xiaoyang Li, Paul Harris, Tuan Chien and Daniel Hughes. Many in the frame community: John Benedetto and Pete Casazza for their ongoing encouragement with this book, Joe Renes for introducing me to SICs, and others for their careful and insightful reading of various chapters of the book. Algebraists that I have bothered: Charles Leedham-Green, Mark Lewis, Martin Liebeck, Gabriel Verret, and those that I will, now that I have time.

Contents

1 Introduction ...... 1 1.1 Somehistory ...... 2 1.2 Desirableproperties ...... 3 Notes ...... 4 Exercises ...... 4

2 Tight frames ...... 7 2.1 Normalisedtightframes ...... 7 2.2 Unitarily equivalent ﬁnite tight frames ...... 10 2.3 Projective and complex conjugate equivalences...... 11 2.4 The analysis, synthesis and frame operators ...... 12 2.5 TheGramian ...... 13 2.6 Tight frames as orthogonal projections ...... 15 2.7 The construction of tight frames from orthogonal projections...... 17 2.8 Complementarytightframes ...... 18 2.9 Partitionframes...... 20 2.10 Real and complex tight frames...... 21 2.11 SICsandMUBs...... 22 Notes ...... 23 Exercises ...... 24

3 Frames ...... 31 3.1 Motivation ...... 31 3.2 Aframeanditsdual ...... 32 3.3 Thecanonicaltightframe...... 35 3.4 Unitarilyequivalentframes ...... 40 3.5 Similar frames and orthogonal frames ...... 41 3.6 Frames as orthogonal projections ...... 43 3.7 Condition numbers and the frame bounds ...... 46 3.8 Normalising frames and the distances between them ...... 47 3.9 Approximate inverses of the frame operator...... 53

xi xii Contents

3.10 Alternateduals...... 54 3.11 Obliqueduals...... 57 Notes ...... 60 Exercises ...... 60

4 Canonical coordinates for vector spaces and affine spaces ...... 71 4.1 The canonical Gramian of a spanning sequence ...... 72 4.2 The canonical coordinates of a spanning sequence ...... 74 4.3 A characterisation of the canonical coordinates ...... 76 4.4 Properties of the canonical coordinates ...... 78 4.5 The canonical inner product for a spanning sequence ...... 80 4.6 Canonical coordinates for cyclotomic fields ...... 81 4.7 Generalised barycentric coordinates for affine spaces ...... 87 4.8 TheBernsteinframe ...... 92 4.9 PropertiesoftheBernsteinframe ...... 93 4.10 The generalised Bernstein operator ...... 94 Notes ...... 98 Exercises ...... 98

5 Combining and decomposing frames ...... 99 5.1 Unions ...... 99 5.2 Directsums ...... 100 5.3 Lifting...... 104 5.4 Complements...... 105 5.5 Sums...... 106 5.6 Tensorproducts...... 108 5.7 Decompositions...... 110 Notes ...... 110 Exercises ...... 110

6 Variational characterisations of tight frames ...... 113 6.1 Welch bound equality sequences ...... 113 6.2 The variational characterisation ...... 114 6.3 Theframepotential...... 115 6.4 Spherical t-designsandtheWaringformula...... 116 6.5 Other characterisations of spherical t–designs...... 117 6.6 The existence of cubature formulas ...... 119 6.7 Tightframesofsymmetrictensors ...... 120 6.8 Cubature on the real and complex spheres ...... 122 6.9 Spherical (t,t)–designs...... 126 6.10 Weighted (t,t)–designs...... 128 6.11 Complex projective t–designs ...... 129 6.12 Isometricembeddings...... 131 6.13 Weighted (2,2)–designs of orthonormal bases...... 132 6.14 Theframeforce...... 135 Contents xiii

6.15 Local and global minimisers of the frame potential...... 136 6.16 The numerical construction of spherical (t,t)–designs ...... 139 Notes ...... 142 Exercises ...... 143

7 The algebraic variety of tight frames ...... 151 7.1 The real algebraic variety of normalised tight frames ...... 152 7.2 The algebraic variety of equal–norm tight frames ...... 153 7.3 Thedensityofrationaltightframes...... 155 7.4 The existence of tight frames with given norms ...... 157 7.5 The construction of tight frames with given norms ...... 158 Notes ...... 162 Exercises ...... 163

8 Projective unitary equivalence and fusion frames ...... 165 8.1 Projective unitary equivalence ...... 166 8.2 The m–products...... 167 8.3 Theframegraph ...... 168 8.4 Characterisation of projective unitary equivalence ...... 169 8.5 Reconstruction from the m–products...... 172 8.6 Triple products, equiangular lines, SICs and MUBs ...... 174 8.7 Projective similarity and canonical m–products for vector spaces . . . 176 8.8 Fusionframes ...... 179 8.9 Signedframes ...... 181 8.10 Scaling the vectors of a frame to obtain a tight frame ...... 182 Notes ...... 186 Exercises ...... 186

9 Symmetries of tight frames ...... 189 9.1 The symmetries of a sequence of vectors ...... 190 9.2 The symmetry group of a sequence of vectors ...... 191 9.3 The projective symmetry group of a sequence of vectors ...... 195 9.4 Symmetries of combinations of frames ...... 197 9.5 Maximallysymmetrictightframes ...... 198 9.6 Algorithmsandexamples...... 200 9.7 Case study: 16 equiangular lines in R6 ...... 205 9.8 Case study: A spherical (4,4)–design of 12 lines in C2 ...... 206 Notes ...... 206 Exercises ...... 207

10 Group frames ...... 209 10.1 Representations and G-frames...... 210 10.2 The frame operator of a G–frame ...... 211 10.3 Group matrices and the Gramian of a G–frame...... 213 10.4 Identifying G–framesfromtheirGramian ...... 216 10.5 Irreducible G–frames ...... 218 xiv Contents

10.6 TheverticesofthePlatonicsolids...... 222 10.7 Irreducible G–frames from vertex–transitive graphs ...... 224 10.8 Maschke’s theorem and homogeneous G–frames ...... 225 10.9 The characterisation of all tight G–frames ...... 227 10.10G–frames of multivariate orthogonal polynomials ...... 231 10.11G–invariantframes ...... 234 10.12Frames invariant under the action of an abelian group ...... 236 10.13The minimal number of generators for a G–invariant frame...... 238 10.14The coinvariants of a ﬁnite reﬂection group ...... 240 Notes ...... 241 Exercises ...... 241

11 Harmonic frames ...... 245 11.1 Introduction ...... 245 11.2 Charactertables...... 246 11.3 Harmonicframes...... 247 11.4 Harmonic frames with distinct vectors and with real vectors ...... 249 11.5 Combining and decomposing harmonic frames ...... 250 11.6 Realharmonicframes...... 251 11.7 Unitary equivalence preserving the group structure ...... 253 11.8 Noncyclicharmonicframes ...... 256 11.9 Unitary equivalence not preserving the group structure ...... 257 11.10The number of cyclic harmonic frames ...... 258 11.11Projective unitary equivalence of harmonic frames ...... 260 11.12The projective symmetry group of a harmonic frame ...... 262 Notes ...... 263 Exercises ...... 263

12 Equiangular and Grassmannian frames ...... 265 12.1 Equiangular lines and frames ...... 266 12.2 Grassmannianframes ...... 268 12.3 Equiangular harmonic frames and difference sets ...... 269 12.4 Equiangular tight frames from block designs ...... 275 12.5 Tremain equiangular tight frames ...... 278 12.6 Equiangular frames and their signature matrices ...... 279 12.7 The reduced signature matrix and projective unitary equivalence . . . 284 12.8 The relative bound on the number of real equiangular lines...... 287 12.9 The connection with algebraic graph theory...... 289 12.10Real equiangular tight frames and strongly regular graphs ...... 291 12.11Conditions for the existence of real equiangular tight frames ...... 295 12.12A list of real equiangular tight frames ...... 297 12.13Nontight real equiangular frames from graphs ...... 299 12.14Spherical two–distance tight frames ...... 304 12.15Two–distance tight frames and partial difference sets...... 307 12.16The standard m–distancetightframe...... 309 Contents xv

12.17Complex equiangular tight frames...... 310 12.18Algebraic equations for tight complex equiangular lines ...... 316 12.19Mutually unbiased bases and s–angular tight frames...... 318 12.20Mutually unbiased bases and Hadamard matrices ...... 320 12.21ExamplesofMUBs...... 324 Notes ...... 326 Exercises ...... 326

13 Tight frames generated by nonabelian groups ...... 331 13.1 The identification of the G–matrices with the group algebra ...... 332 13.2 Tight G–frames as idempotents of the group algebra ...... 333 13.3 Characters of nonabelian groups ...... 334 13.4 Central G–frames ...... 336 13.5 The classification of central tight G–frames...... 337 13.6 The idempotents and the homogeneous decomposition ...... 339 13.7 Anillustrativeexample...... 340 13.8 Thehighlysymmetrictightframes ...... 341 13.9 The construction of highly symmetric tight frames ...... 342 13.10Complex polytopes and finite reflection groups ...... 343 13.10.1Imprimitivegroups(ST1-3) ...... 345 13.10.2Primitive reflection groups (ST 4-37) ...... 347 13.11Projective representations ...... 351 13.12The canonical abstract (error) group ...... 352 13.13Niceerrorframes...... 353 13.14Tensor products of nice error frames ...... 355 13.15Computing all nice error frames ...... 356 13.16Examples of nice error frames ...... 357 Notes ...... 359 Exercises ...... 359

14 Weyl–Heisenberg SICs ...... 361 14.1 Maximal sets of complex equiangular lines and vectors ...... 362 14.2 SICs ...... 363 14.3 Quantummeasurements ...... 364 14.4 Groupcovariance ...... 365 14.5 The discrete Heisenberg group and Weyl–Heisenberg SICs...... 366 14.6 TheHoggarlines...... 367 14.7 The Clifford group (normaliser of the Heisenberg group) ...... 368 14.8 GeneratorsfortheCliffordgroup ...... 372 14.9 Indexing the Clifford operations ...... 373 14.10Applebyindexing ...... 374 14.11Symplecticunitaries ...... 378 14.12Permutationmatrices ...... 380 14.13Calculating a symplectic matrix from its symplectic index ...... 382 14.14TheZaunermatrix...... 384 xvi Contents

14.15The Scott–Grassl numerical SICs ...... 385 14.16Symmetries of the Weyl–Heisenberg SICs ...... 389 14.17Monomial representations of the Clifford group ...... 391 14.18The Clifford trace and symplectic unitaries of order 3 ...... 392 14.19Conjugates of the canonical order 3 symplectic unitaries...... 398 14.20The SIC field of a Weyl–Heisenberg SIC ...... 400 14.21The Galois group of a generic SIC ...... 401 14.22The Galois action on a fiducial projector ...... 402 14.23 Constructing exact SICs from numerical SICs ...... 404 14.24The Galois action on a centred fiducial ...... 404 14.25Minimal and maximal SIC fields ...... 410 14.26Rayclassfields ...... 412 14.27Equivalent equations for SIC fiducial vectors ...... 414 Notes ...... 418 Exercises ...... 419

15 Tight frames of orthogonal polynomials on the simplex ...... 429 15.1 Jacobi polynomials and their Bernstein coefﬁcients ...... 430 15.2 The Bernstein–Durrmeyer operator ...... 434 15.3 Tight frames of Jacobi polynomials with symmetries ...... 435 15.4 The orthogonal polynomials of Appell and Proriol ...... 437 Notes ...... 438 Exercises ...... 439

16 Continuous tight frames for ﬁnite dimensional spaces ...... 441 16.1 Continuous and discrete frames ...... 441 16.2 The analysis and synthesis operators ...... 443 16.3 Reproducingkernels ...... 444 16.4 Zonalharmonics ...... 445 16.5 Homogeneous polynomials ...... 448 16.6 Orthogonal polynomials for a radially symmetric weight...... 451 16.6.1 The O(d)–invariant subspaces of Pn ...... 451 16.6.2 Continuous tight frames for Pn ...... 454 16.6.3 The reproducing kernel for Pn ...... 457 16.6.4 Finite tight frames for Pn ...... 459 16.7 Functions on the complex sphere...... 461 16.8 G–framesforinﬁnitegroups ...... 463 Notes ...... 465 Exercises ...... 466

Solutions ...... 471

References ...... 563 References...... 563

Index ...... 575 Chapter 1 Introduction

The prototypical example of a ﬁnite tight frame is three equally spaced unit 2 vectors u1,u2,u3 in R , which provide the following redundant decomposition

3 2 2 f = ∑ f ,u j u j, f R . (1.1) 3 j=1 ∀ ∈

Such generalisations of an orthogonal expansion have been used extensively for infinite dimensional function spaces. Most notably, in the area of wavelets, where they allow expansions in terms of functions which have nicer properties, such as good time–frequency localisation and a simple description, than is possible for an orthogonal expansion. Despite the fact that in these methods, ultimately a numerical approximation is computed in a finite dimensional subspace, until recently little attention has been paid to frames for finite dimensional spaces. Over the last decade, it has become increasingly apparent that tight frames for finite dimensional spaces are useful for similar reasons. They can have desirable properties, such as good time–frequency localisation, and share symmetries of the space, which may be impossible for an orthonormal basis. In addition, there are computational advantages of stability and robustness to erasures. There is also a special geometry (different for real and complex spaces) which has no analogue in the infinite dimensional setting.

1 2 1 Introduction 1.1 Some history

Like many great ideas in mathematics, frames (which have been crucial to the development of wavelets) have, with hindsight, been around in some form or other for quite a while. For example, in 1937, Schonhardt¨ [Sch37] proved the generalisation 2 of (1.1) to n equally spaced unit vectors u1,...,un R , i.e., ∈ n 2 2 f = ∑ f ,u j u j, f R . (1.2) n j=1 ∀ ∈

This idea received some attention, with Brauer and Coxeter [BC40] extending the result to the orbit of any irreducible group of orthogonal matrices (also see [Had40]). They also mention the possibility of extending the result to the orbit of a continuous group, e.g., taking the group of rotations gives the continuous version of (1.2)

2π 2 2 cosθ f = f ,uθ uθ dθ, f R , uθ := , (1.3) 2π 0 ∀ ∈ sinθ which is an example of a continuous tight frame. I don’t doubt that there are even earlier instances. In 1952, Dufﬁn and Schaeffer set out the modern theory of frames in their seminal paper [DS52], which included the deﬁnition in terms of frame bounds

2 2 2 A f ∑ f , f j B f , f . ≤ j | | ≤ ∀

They were interested in Fourier type series for functions in L2[ π,π] involving iλ t − functions f j : t e j , for frequencies λ j R, which might not be integers (see Young [You01]→ for an excellent account). ∈ From the late 1980’s onwards came the wavelet era (see, e.g., [Dau92], [Kai94]). d Here frames were used to obtain Fourier expansions for L2(R ) in terms of functions with both a simple description and good time–frequency localisation. At the risk of over simpliﬁcation, this was done by taking a single (wavelet) function, and obtaining the others from it by applying the operations of

d translation: (Ta f )(x) := f (x a), a R , (1.4) − ∈ 2πib x d modulation: (Mb f )(x) := e f (x), b R , (1.5) ∈ d dilation: (Dc f )(x) := c 2 f (cx), c > 0. (1.6)

The parameters a,b,c may be chosen to be either discrete or continuous. The theory has two strands: when the operations form a group (Gabor systems), and when they don’t (wavelet systems). In the former case, the group allows a description of the dual frame as the orbit of a single function, and in the latter the method of multiresolution analysis yields a suitable function. 1.2 Desirable properties 3 1.2 Desirable properties

Surely, three equally spaced unit vectors (aka the Mercedes–Benz frame) give the “nicest” possible tight frame of three vectors for R2. In this example, one can see many desirable properties that one might hope for more generally in a tight frame. To illustrate these properties, many of which do extend to quite general situations, we also consider a second prototypical example: the four unit vectors Φ :=(v,Sv,Ωv,SΩv) in C2, given by

1 3 + √3 0 1 1 0 v : π S : Ω : (1.7) = i , = , = , √6 e 4 3 √3 1 0 0 1 − − which form a tight frame for C2, i.e., 2 f = ∑ f ,φ φ, f C2. (1.8) 4 φ Φ ∀ ∈ ∈ The operations S and Ω are discrete analogues of translation and modulation, and the tight frame Φ, which will be termed a SIC, is a discrete analogue of a Gabor system (Weyl–Heisenberg frame). In addition to being tight, we might hope that a ﬁnite frame have some of the following properties. Equal norms. As in both examples, the vectors in the tight frame have equal • norms. Equivalently, the decomposition of the form (1.1) and (1.8) is a sum of one–dimensional projections with equal weightings. Symmetries. The frame be invariant under some group G of symmetries, such as • the three equally spaced vectors (which are invariant under the group of order 3 generated by rotation through 2π/3). Equivalently, the frame is the G–orbit of one, or a small number of vectors. Equiangularity. The (equal norm) vectors in both frames have the cross–correlation • φ,ψ constant for all vectors φ = ψ. For vectors in Rd, this is equivalent to the vectors| | having equal angles between each other. Robustness to erasures. In both examples, each pair of frame vectors spans a • 2–dimensional space. Hence if all but two of the coefﬁcients f ,φ (φ a frame vector) are lost, then f can be reconstructed from these values.

Stability. Suppose the three coefﬁcients in (1.1) are perturbed, say to f ,u j +a j. • Then the error in the computed f is a1u1 +a2u2 +a3u3, which is bounded by the norm of a =(a j). This error might even be zero, for a nonzero perturbation (take a1 = a2 = a3), a phenomenon which can not occur for an orthogonal expansion. A further property, not shared by these examples, is that of sparseness, i.e., the frame vectors having many zeros. This is the discrete analogue of having small or compact support, and is of importance in algorithms for compressed sensing and data compression. 4 1 Introduction Notes

In some sense, the theory of frames is dual to that of compressed sensing (sparse sampling). In compressed sensing, the (sparseness) structure of some vectors allow them to be represented by using fewer vectors than in a basis, whilst in ﬁnite frame theory more vectors than are needed for a basis are used in a redundant expansion which has more desirable properties than would be possible by using just a basis.

Exercises

1.1. Prove (1.1) holds, by using the fact f ,u j = u f , to write it in the matrix form ∗j 2 1 0 VV = I, V :=[u ,u ,u ], I := . 3 ∗ 1 2 3 0 1 Hint. Any two sets of three equally spaced unit vectors are rotations of each other.

1.2. Robustness to erasures. 2 (a) Suppose one of the coefficients in (1.1) is lost, say f ,u3 . Show that f R can be reconstructed from the remaining two, by giving an explicit formula for∈ this, i.e., 2 find the basis for R dual to the functionals f f ,u j , j = 1,2. → (b) Now suppose a coefficient is changed, say f ,u3 , can this be detected? (c) What if two of the coefficients are changed?

1.3. Equiangularity/equispacing. (a) Verify that the three unit vectors in C2 given by

1 1 1 1 1 1 2πi v1 := , v2 := , v3 := , ω := e 3 √2 1 √2 ω √2 ω2 2 2 ∑3 2 form a tight frame for C , i.e., f = 3 j=1 f ,v j v j, for all f C . (b) Show these three vectors, and those of Exer. 1.1 are equia∈ngular, i.e., 1 v j,vk = u j,uk = , j = k. | | | | 2

√3 v j vk = , u j uk = √3, j = k. − √2 −

(d) How can the distances between the real vectors (u j) be larger than those for the 2 2 complex vectors (v j)? Isn’t there more space in C than in R ! 1.2 Desirable properties 5

1.4. Gabor and Wavelet systems. (a) Show that translation and modulation satisfy the commutation relation

2πia b TaMb = e− MbTa.

(b) Use this to conclude if A and B are nonzero subgroups of (R,+), then

2πia b G := cTaMb : a A,b B,c C , C := e : a A,b B . { ∈ ∈ ∈ } { ∈ ∈ } d is a group, which is abelian only if R = R. The G–orbit TaMbφ a A,b B (up to d { } ∈ ∈ scalar multipliers) of a suitable φ L2(R ) is called a Gabor system/frame. ∈ (c) For a suitable function ψ L2(R), a wavelet system/frame is given by the functions ∈ j j ψ jk := 2 2 ψ(2 k)= D j Tkψ, j,k Z. − 2 ∈ D T Show 2 j k j,k Z is not a group, and so wavelet systems are not group orbits. { } ∈

Chapter 2 Tight frames

Decompositions like those in our two prototypical examples, i.e.,

f = ∑ f , f j f j, f H , (2.1) j J ∀ ∈ ∈ will come from what is called a tight frame ( f j) j J. The basic ingredients are ∈ H – a real or complex Hilbert space (for us usually finite dimensional) • J – an index set (often with a group structure) • ( f j) j J – a sequence (set, or multiset) of vectors in H • ∈ ∑ j J – a sum (for us usually finite, but sometimes continuous) • ∈ The emphasis here, is on the possible redundancy (over completeness) of the vectors ( f j) in the expansion, i.e., the case when (2.1) is not an orthogonal expansion. In the first instance, you are encouraged to consider H as Rd or Cd, with the usual (Euclidean) inner product, and to think in familiar matrix terms.

2.1 Normalised tight frames

The polarisation identity (see Exer. 2.1) implies that (2.1) is equivalent to

2 2 f = ∑ f , f j , f H , j J | | ∀ ∈ ∈ which explains the following deﬁnition.

Deﬁnition 2.1. A countable sequence ( f j) j J in a Hilbert space H is said to be a tight frame (for H ) if there exists a (frame∈ bound) A > 0, such that

2 2 A f = ∑ f , f j , f H . (2.2) j J | | ∀ ∈ ∈

Further ( f j) j J is normalised if A = 1, and ﬁnite if J is ﬁnite. ∈

7 8 2 Tight frames

The Bessel identity (2.2) is equivalent (see Exer. 2.2) to either of the identities 1 Parseval: f = ∑ f , f j f j, f H , (2.3) A j J ∀ ∈ ∈ 1 Plancherel: f ,g = ∑ f , f j f j,g , f ,g H . (2.4) A j J ∀ ∈ ∈

For A = 1, (2.2) says that f ( f , f j ) j J is an isometry, and so normalised tight → ∈ frames for H are equivalent to isometries H ℓ2(J). The maps taking normalised tight frames to normalised tight frames are the→partial isometries (Exer. 2.7).

We prefer the term normalised tight frame9 to Parseval frame (which is also used), as it emphasizes the fact the frame bound A > 0 is simply a normalising factor, i.e., if ( f j) is a tight frame, then ( f j/√A) is the unique positive scalar multiple of it which is a normalised tight frame. We will soon see that this normalised version of a tight frame is convenient in many situations.

We have deﬁned a tight frame to be a sequence, which is standard, but not universal. By contrast with a basis (which can be a set or a sequence), a tight frame can have repeated vectors. At times, e.g., when the vectors in a frame are all distinct or the indexing is unimportant, it can be convenient to think of them as a (multi)set. We will not labour this point, making statements such as the set f j j J is a tight frame, without further explanation. { } ∈

Fig. 2.1: Examples of normalised tight frames of n = 4,5,...,11 vectors for R2.

1 This term dates back to [HL00]. Just to confuse matters, the term normalised tight frame has also been used for a tight frame with f j = 1, j J (we call these unit–norm tight frames). ∀ ∈ 2.1 Normalised tight frames 9

Example 2.1. (Exer. 2.4) An orthonormal basis is a normalised tight frame. These are the only normalised tight frames in which all the vectors have unit length (all vectors in a normalised tight frame have length 1). ≤ Example 2.2. (Exer. 2.5) The unitary image of a normalised tight frame is again a normalised tight frame, and the only invertible linear maps which map a normalised tight frame to a normalised tight frame are the unitary maps. Example 2.3. (Exer. 2.6) The orthogonal projection of a normalised tight frame is again a normalised tight frame (for its span). In particular, if U is an n n unitary matrix, then the columns of any d n submatrix is a normalised tight× frame of n vectors for Cd. This is effectively× the projection of the orthonormal basis for Cn given by the columns of U onto the d–dimensional subspace of vectors which are zero in some ﬁxed n d coordinates. − We say that ( f j) j J is an equal–norm tight frame if f j = fk , j,k J, and ∈ ∀ ∈ is a unit–norm tight frame if f j = 1, j J. ∀ ∈ Example 2.4. (Exer. 2.8) Equal–norm tight frames of n vectors for Fd can be obtained from an n n unitary matrix U with entries of constant modulus, by taking the columns of any× d n submatrix. Examples of such U include the Hadamard matrices (real entries)× and the Fourier (transform) matrix

1 1 1 1 2 n 1 1 ω ω ω − 1  ω2 ω4 ω2(n 1)  2πi 1 ω n U = Fn = − , := e . (2.5) √n . . . .  . . . .    1 ωn 1 ω2(n 1) ω(n 1)(n 1)  − − − −    Equal–norm tight frames which come from the Fourier matrix in this way will be known as harmonic tight frames (see Chapter 11). In 2.6, we will show that every normalised tight frame can be obtained as the orthogonal§ projection of an orthonormal basis (in a larger space).

2 n Example 2.5. (Exer. 2.9) The tight frames for R . A sequence of vectors (v j) j=1, 2 2 v j =(x j,y j) R is a tight frame for R if and only if the diagram vectors which ∈ 2 are deﬁned by w j :=(x j + iy j) C, 1 j n, sum to zero (in C). ∈ ≤ ≤ In applications, the interest in (2.3) is usually that it gives a decomposition of the identity into a weighted sum of projections (see Chapter 8), i.e.,

2 f j f , f j I = ∑ c jPj, c j := , Pj f := f j, (2.6) j J A f j, f j ∈ where the particular (unit modulus) scalar multiple of f j that is used to deﬁne the orthogonal projection Pj is unimportant. The pair (Pj),(c j) above is a fusion frame (see 8.8). When taking this point of view, we will use the epithet projective. § 10 2 Tight frames 2.2 Unitarily equivalent ﬁnite tight frames

Before giving any further concrete examples of ﬁnite tight frames, we deﬁne an equivalence, under which any set of three equally spaced vectors with the same norm in R2 would be considered equivalent.

Deﬁnition 2.2. We say that two normalised tight frames ( f j) j J for H and (g j) j J for K , with the same index set J, are (unitarily) equivalent∈ if there is a unitary∈ transformation U : H K , such that g j = Uf j, j J. → ∀ ∈ Since unitary transformation preserve inner products, unitarily equivalent tight frames have the same inner products (angles) between their vectors. Furthermore, these inner products uniquely determine the equivalence classes (see 2.5). This equivalence is dependent on the indexing, which is appropriate§ when set J has some natural (e.g., group) structure. The normalised tight frames of two vectors (e1,0) and (0,e1) for the one–dimensional space H = span e1 are not equivalent, { } since there is no unitary map e1 0 (or 0 e1). For such cases, where it is useful to consider these as equivalent, we→ extend our→ deﬁnition of equivalence as follows.

Deﬁnition 2.3. We say that two ﬁnite normalised tight frames ( f j) j J for H and ∈ (g j) j K for K , are (unitarily) equivalent up to reordering if there is a bijection ∈ σ : J K for which ( f j) j J and (gσ j) j J are unitarily equivalent. → ∈ ∈ We will say that tight frames are unitarily equivalent (up to reordering) if after normalisation they are, in which case we say they are equal up to unitary equivalence (and reordering). 2 Example 2.6. Let u1,u2,u3 be equally spaced unit vectors in R , and Rθ be rotation through an angle θ. Then each of the sets of six vectors π u1,u2,u3,Rθ u1,Rθ u2,Rθ u3 , 0 < θ { } ≤ 3 forms a tight frame. Since unitary maps preserve angles, none of these are unitarily equivalent (up to reordering).

θ π π π Fig. 2.2: The unitarily inequivalent tight frames obtained for = 12 , 6 , 3 . 2.3 Projective and complex conjugate equivalences 11 2.3 Projective and complex conjugate equivalences

There are other equivalences which appear in the frame literature. Most notably, the normalised tight frames of Deﬁnition 2.2 are projectively (unitarily) equivalent if

g j = α jUf j, j J, ∀ ∈ where U is unitary, and α j = 1, j. | | ∀ All tight frames (α j f j), α j = 1, j, obtained from a given tight frame ( f j), are projectively unitarily equivalent,| | but∀ are not unitarily equivalent, in general.

Example 2.7. For tight frames of n nonzero vectors in R2 the equivalence classes for projective equivalence up to reordering are in 1–1 correspondence with convex polygons with n sides (see Exer. 2.10).

Fig. 2.3: Equal–norm tight frames of three vectors for R2 which are projectively equivalent, but are not unitarily equivalent.

The complex conjugation map on Cd is the antilinear map

d d C C : v =(v j) v :=(v j). → → d Since v,w = v,w , this maps a tight frame ( f j) for C to a tight frame ( f j), and these are said to be complex conjugate (or anti) equivalent. The conjugation map C : H H extends these ideas to H (see Exer. 2.11). These→ basic types of equivalences can be combined, in the obvious way, to obtain others, e.g., the tight frames ( f j) j J and (gk)k K for H and K would be anti projectively unitarily equivalent up to∈ reordering if∈

gσ j = α jUf j, j J, ∀ ∈

for σ : J K a bijection, α j = 1, j, and U : H K unitary. → | | ∀ → 12 2 Tight frames 2.4 The analysis, synthesis and frame operators

The Parseval identity (2.3) consists of an extraction of “coordinates”

c j = f , f j , j J ∈ for the vector f (analysis), and a reconstruction of f from these (synthesis). Many important properties of a tight frame follow from this factorisation. For simplicity of presentation, we suppose J is ﬁnite, write F for R or C, ℓ2(J) J for F , with the usual inner product, and I = IH for the identity on H .

Deﬁnition 2.4. For a ﬁnite sequence ( f j) j J in H the synthesis operator (reconstruction operator or pre-frame operator∈) is the linear map

V :=[ f j] j J : ℓ2(J) H : a ∑ a j f j, ∈ → → j J ∈ and its dual is the analysis operator (or frame transform operator)

V ∗ : H ℓ2(J) : f ( f , f j ) j J. → → ∈

It is convenient to make little distinction between the sequence ( f j) j J and the ∈ linear map V =[ f j] j J, which we will say has j–th column f j. ∈ The product S := VV ∗ : H H is known as the frame operator. A simple calculation (see Exer. 2.12) shows→ the trace of S and S2 are given by

1 2 2 2 2 2 trace(S)= S 2 F = ∑ f j , trace(S )= S F = ∑ ∑ f j, fk . (2.7) j J j J k J | | ∈ ∈ ∈

Proposition 2.1. A ﬁnite sequence ( f j) f J in H is a tight frame for H (with frame bound A) if and only if ∈

S = VV ∗ = AIH , V :=[ f j] f J. (2.8) ∈ In particular, a tight frame satisﬁes

2 ∑ f j = dA, d := dim(H ), (2.9) j J ∈ and 2 2 1 ∑ ∑ f j, fk = ∑ f j, f j . (2.10) j J k J | | d j J ∈ ∈ ∈ Proof. Since Sf = VV ∗ f = ∑ f , f j f j, f H , j ∀ ∈ the Parseval identity (2.3) implies the condition (2.2) is equivalent to (2.8). Taking the trace of (2.8) and its square gives 2.5 The Gramian 13

2 ∑ f j = trace(S)= trace(AIH )= dA, j 2 2 2 2 1 2 1 ∑∑ f j, fk = trace(S )= trace(A IH )= (Ad) = ∑ f j, f j , j k | | d d j which are (2.9) and (2.10). ⊓⊔ The equations (2.3) and (2.8) will be referred to as the Parseval identity, (2.9) as the trace formula, and (2.10) as the variational formula.

For H = Fd and J = n, V is a d n matrix, and the condition (2.8) says that the columns of V are| | orthogonal and× of length √A, i.e., V/√A is a coisometry, equivalently, V ∗/√A is an isometry.

In 6.2 we show that the variational formula characterises tight frames for ﬁnite dimensional§ spaces. There is no inﬁnite dimensional counterpart of this result.

2.5 The Gramian

Unitary equivalence has the advantage (over projective unitary equivalence) that it preserves the inner product between vectors, and hence the Gramian matrix. Indeed, we will show that the Gramian characterises the equivalence class.

2 Deﬁnition 2.5. For a ﬁnite sequence of n vectors ( f j) j J in H , the Gramian or Gram matrix is the n n Hermitian matrix ∈ ×

Gram(( f j) j J) :=[ fk, f j ] j,k J. ∈ ∈

This is the matrix representing the linear map V V : ℓ2(J) ℓ2(J) with respect ∗ → to the standard orthonormal basis e j j J. The possible Gramian matrices{ are} precisely∈ the orthogonal projections:

Theorem 2.1. An n n matrix P =[p jk] j,k J is the Gramian matrix of a normalised × ∈ tight frame ( f j) j J for the space H := span f j j J if and only if it is an orthogonal ∈ 2 { } ∈ projection matrix, i.e., P = P∗ = P . Moreover,

2 d = dim(H )= rank(P)= trace(P)= ∑ f j . (2.11) j J ∈

Proof. (= ) Let Φ =( f j) j J be a normalised tight frame, and P = Gram(Φ). Take ⇒ ∈ f = fℓ in (2.3) to get fℓ = ∑ j J fℓ, f j f j, and take the inner product of this with fk to obtain ∈

2 Note the ( j,k)–entry of the Gramian is fk, f j = f j∗ fk (so it factors V ∗V), not f j, fk , which is sometimes used to deﬁne the Gramian. 14 2 Tight frames

2 fk, fℓ = ∑ f j, fℓ fk, f j pℓk = ∑ pℓ j p jk P = P . j J ⇐⇒ j J ⇐⇒ ∈ ∈

But P is Hermitian, since p jk = φk,φ j = φ j,φk = pkj, and so is an orthogonal projection. ( =) Suppose that P is an n n matrix, such that P = P = P2. The columns of ⇐ × ∗ P are f j := Pe j, j J, where e j j J is the standard orthonormal basis of ℓ2(J). Fix ∈ n { } ∈ f H := span f j ℓ2(J). Then f = Pf , so that ∈ { } j=1 ⊂

f = P ∑ Pf ,e j e j = ∑ f ,Pe j Pe j = ∑ f , f j f j, j J j J j J ∈ ∈ ∈ n H i.e., ( f j) j=1 is a normalised tight frame for , with Gramian P. Finally, taking the trace of P gives (2.11). ⊓⊔ The condition that P = Gram(Φ) be an orthogonal projection is equivalent to it having exactly d nonzero eigenvalues all equal to 1 (see Exer. 2.17). Corollary 2.1. (Characterisation of unitary equivalence) Normalised tight frames are unitarily equivalent if and only if their Gramians are equal.

Proof. Let Φ =( f j) j J, Ψ =(g j) j J be normalised tight frames for H and K . ∈ ∈ (= ) If Φ and Ψ are unitarily equivalent, i.e., g j = Uf j, j, for some unitary U : H⇒ K , then their Gramians are equal since ∀ →

g j,gk = Uf j,Ufk = f j, fk .

( =) Suppose the Gramians of Φ and Ψ are equal, i.e., g j,gk = f j, fk , j,k. ⇐ ∀ Then, by Exer. 2.19, there is a unitary U : H K with g j = Uf j, j. Hence Φ and Ψ are unitarily equivalent. → ∀ ⊓⊔ In other words:

The properties of a tight frame (up to unitary equivalence) are determined by its Gramian.

Example 2.8. Equal–norm tight frames of three vectors for C2 are given by

1 ω ω2 1 1 1 2πi Φ :=( , , ), Ψ :=( , , ), ω := e 3 . 1 ω2 ω 1 ω ω2 These harmonic frames are not unitarily equivalent since their Gramians

2 1 1 2 1 + ω 1 + ω2 Gram(Φ)= 1− 2 −1 , Gram(Ψ)= 1 + ω2 2 1 + ω − −    1 1 2 1 + ω 1 + ω2 2 − −     are different. They are however projectively unitarily equivalent (see Exer. 2.21). 2.6 Tight frames as orthogonal projections 15 2.6 Tight frames as orthogonal projections

We have seen (Exer. 2.6) that the orthogonal projection of an orthonormal basis is a normalised tight frame (for its span). The converse is also true.

Theorem 2.2. (Na˘ımark) Every ﬁnite normalised tight frame Φ =( f j) j J for H is ∈ the orthogonal projection of an orthonormal basis for ℓ2(J). Indeed, the orthogonal projection P = Gram(Φ) of the standard orthonormal basis (e j) j J (onto the ∈ column space of the Gramian) is unitarily equivalent to Φ via f j Pe j, i.e., →

Pe j,Pek = f j, fk H , j,k J. ℓ2(J) ∀ ∈ Proof. By Theorem 2.1, P = Gram(Φ) is an orthogonal projection, and so

Pe j,Pek = Pe j,ek =(k, j)–entry of P = f j, fk .

⊓⊔ When Ψ and Φ are unitarily equivalent, then we will say that Ψ is a copy of Φ. With this terminology, Na˘ımark’s theorem says:

A canonical copy of a tight frame Φ is given by the columns of Gram(Φ).

Fig. 2.4: The normalised tight frames 0,e1,e2 and three equally spaced vectors obtained as the orthogonal projection of an orthonormal{ basis for} R3 onto R2.

This is one of those often rediscovered theorems, which can be considered as a special case of Na˘ımark’s theorem (see [AG63] and Exer. 2.26). Hadwiger [Had40] n d showed that ( f j) j=1 in R is a coordinate star (normalised tight frame) if and only if it is a Pohlke normal star (projection of an orthonormal basis). In signal processing this method of obtaining tight frames is called seeding. 16 2 Tight frames

Fig. 2.5: Examples of normalised tight frames of three vectors for R2 obtained as the orthogonal projection of an orthonormal basis for R3.

Example 2.9. The Gramian of the three equally spaced vectors in R2 is

2 1 1 3 3 3 1 1 1 −2 − 1 2 1 2 2 P = V ∗V = 3 3 3 , V =[v1,v2,v3]= 3 −√3 −√3 . (2.12) − 1 1 −2  0 2 − 2 − 3 − 3 3   2 The (particular choice of) vectors is normalised v j = (see Exer. 2.15), so that 3 (v j) is a normalised tight frame, and hence P is an orthogonal projection. The columns (Pe j) of P give a canonical copy of this normalised tight frame (up to unitary equivalence), e.g.,

2 2 2 1 3 3 3 3 1 1 2 1 −2 1 Pe1,Pe1 = 3 , 3 = , Pe1,Pe2 = 3 , 3 = . − 1  − 1  3 − 1   1  −3 − 3 − 3 − 3 − 3         Example 2.10. A cross in Rn is the set obtained by taking an orthonormal basis and its negatives e1,..., en , and the orthogonal projection of a cross onto a d– dimensional subspace{± V±is a} called a eutactic star (see Coxeter [Cox73]). In view of Theorem 2.2, a eutactic star is precisely a tight frame of the form a1,..., an {± ± } for V, i.e., the union of a tight frame a1,...,an and the equivalent frame obtained { } by taking its negative. When the vectors ai all have the same length one obtains a so called normalised eutactic star. Since equal–norm tight frames always exist (see Chapters 7 and 11), so do normalised eutactic stars in Rd for every n d. ≥ 2.7 The construction of tight frames from orthogonal projections 17 2.7 The construction of tight frames from orthogonal projections

Φ n The Gramian of a normalised tight frame =(v j) j=1 for a d–dimensional space is an orthogonal projection P. By Theorem 2.2, the columns (Pe j) of P give a canonical copy of the frame (up to unitary equivalence) as a d–dimensional subspace of Fn. To obtain a copy of Φ in Fd, we consider the rows of P = Gram(Φ).

n n Theorem 2.3. (Row construction). Let P C × be an orthogonal projection matrix ∈ d n of rank d. The columns of V =[v1,...,vn] C × are a normalised tight frame for Cd with Gramian P if and only if the rows∈ of V are an orthonormal basis for the row space of P. In particular, such a V can always be obtained by applying the Gram–Schmidt process to the rows of P.

Proof. (= ) Suppose the columns of V are a normalised tight frame for Cd with ⇒ Gramian P, i.e., VV ∗ = I (the rows of V are orthonormal) and P = V ∗V. Then

row(P)= row(V ∗V) row(V)= row(VV ∗V) row(V ∗V)= row(P), ⊂ ⊂ so that row(P)= row(V), and the rows of V are an orthonormal basis for row(P). ( =) Suppose the rows of V are an orthonormal basis for row(P). Then VV ∗ = I, and⇐ we have 2 (V ∗V) = V ∗(VV ∗)V = V ∗V, so that V ∗V is an orthogonal projection matrix with the same row space (and hence column space) as P. Thus V V = P. ∗ ⊓⊔ In other words:

A frame V =[v1,...,vn] is a copy of a normalised tight frame Φ if and only if the rows of V are an orthonormal basis for the row space of Gram(Φ).

Example 2.11. For the three equally spaced vectors of Example 2.9, applying the Gram–Schmidt process to the ﬁrst two rows of the Gramian P gives

2 1 1 3 3 3 1 1 1 −2 − 1 2 1 2 2 P = 3 3 3 , V =[v1,v2,v3]= 3 −√3 −√3 . (2.13) − 1 1 −2  −→ 0 2 − 2 − 3 − 3 3   Φ 1 Φ Example 2.12. For a tight frame with frame bound A, the matrix P = A Gram( ) is an orthogonal projection, and so a copy of Φ is given by an orthogonal basis for the row space of Q = Gram(Φ) consisting of vectors of length √A. For the four equally spaced unit vectors of (1.7), we have A = 2, and applying the Gram–Schmidt process to the ﬁrst two rows of the Gramian Q gives 18 2 Tight frames

1 1 1 i √3 √3 − √3 1 1 i 1 1 1 1 i Q =  √3 − √3 √3  V = √3 √3 − √3 . 1 i 1 1 √2 i 1 1 i √ √ √ −→ 0 +  3 3 − 3  √3 − √2 − √2√3 √2 √2√3  i 1 1 1   √3 √3 − √3    Here col(Q) = row(Q), and applying Gram–Schmidt to the columns of Q (instead of the rows) does not give a copy of Φ.

2.8 Complementary tight frames

Tight frames are determined (up to unitary equivalence) by their Gramian matrix P, which is an orthogonal projection matrix (when the frame is normalised), and all orthogonal projection matrices correspond to normalised tight frames. Thus there is normalised tight frame with Gramian given by the complementary projection I P. − Definition 2.6. Given a finite normalised tight frame Φ with Gramian P, we call any normalised tight frame with Gramian I P its complement. More generally, we say that two tight frames are complements− of each other, if after normalisation the sum of their Gramians is the identity I. The complement of a finite tight frame is unique up to unitary equivalence (and normalisation), the complement of the complement is the frame itself, and a tight frame is equiangular (or equal–norm) if and only if its complement is.

Fig. 2.6: Projecting an orthonormal basis for R3 onto a two–dimensional subspace and its orthogonal complement, thereby obtaining a normalised tight frame of three vectors for R2 and the complementary tight frame of three vectors for R. 2.8 Complementary tight frames 19

Fig. 2.7: Examples of normalised tight frames of four vectors for R2 and the complementary frames for R2 (below). A tight frame and its complement can never be unitarily equivalent.

In view of (2.11), the complement of a tight frame of n vectors for a space of dimension d is a tight frame of n vectors for a space of dimension n d. A tight frame and its complement can never be unitarily or projectively− unitarily equivalent (Exer. 2.23), though they do have the same symmetries (see 9.2, 9.3). § § Example 2.13. The complement of an orthonormal basis for Cd is the frame for the zero vector space given by d zero vectors.

Example 2.14. By (2.12), the Gramian of the complementary frame to the three equally spaces vectors in R2 is

2 1 1 1 1 1 1 0 0 3 3 3 3 3 3 1 −2 − 1 1 1 1 Q = I P = 0 1 0 3 3 3 = 3 3 3 . − 0 0 1 − − 1 1 −2   1 1 1  − 3 − 3 3 3 3 3       By the row construction (Theorem 2.3). this is the Gramian of normalised tight frame 1 , 1 , 1 of three repeated vectors for R. The Gramian of an arbitrary { √3 √3 √3 } normalised tight frame of three vectors for C2 is considered in Exer. 2.22.

We call the tight frame of n = d + 1 vectors for Rd which is the complement of 1 ,..., 1 , the vertices of the (regular) simplex in Rd. This has Gramian { √d+1 √d+1 }

1 d−+1 , j = k; P =[p jk], p jk := d (2.14) d+1 , j = k.

To ﬁnd a copy in Rd, one can apply the method of Theorem 2.3 to the Gramian. This example can be generalised to obtain partition frames. 20 2 Tight frames 2.9 Partition frames

k Deﬁnition 2.7. Let α =(α1,...,αk) Z be a partition of n, i.e., ∈

n = α1 + + αk, 1 α1 α2 αk. ≤ ≤ ≤≤ The α–partition frame for Rd, d = n k, is the complement of the normalised tight frame of n vectors for Rk given by − e e e e 1 ,..., 1 ,..., k ,..., k . (2.15) √α1 √α1 √αk √αk α 1 times αk times It is said to be proper if α j 2, j. ≥ ∀ The Gramian of the α–partition frame is the block diagonal n n matrix × α j 1 1 1 1 α− −α −α −α B j j j j 1 α 1 j 1 1 1 . − .  −α j α j −α j −α j   .  α 1 1 j 1 1 − P = B j , B :=  −α −α α −α  (2.16)   j  j j j j   .   . . .   ..   . . .. 1  −α j    α   Bk  1 1 1 1 j 1     −α −α −α −α α−     j j j j j    where the above B j is a α j α j orthogonal projection matrix of rank α j 1. Since each normalised tight frame× is unitarily equivalent to the columns of its− Gramian, it follows that the vectors in a proper partition frame are distinct and nonzero. If α j = 1, then the corresponding partition frame vector is zero. Example 2.15. (Simplex) For n = d + 1, the trivial partition α =(d + 1) gives the vertices of the simplex in Rd.

Fig. 2.8: The proper α–partition frames in R3 for α =(4),(2,3) and (2,2,2), respectively. These are the vertices of the tetrahedron, trigonal bipyramid and octagon. In four dimensions the possible choices for α are (5), (2,4), (3,3) and (2,2,3). 2.10 Real and complex tight frames 21

Table 2.1: The proper partition frames in R2 and R3. Here G is the order of their symmetry group G = Sym(Φ) (see Chapter 9, Exer. 9.5). | | Partition n Description of partition frame G | | (3) 3 three equally spaced vectors in R2 6 (2,2) 4 four equally spaced vectors in R2 8 (4) 4 vertices of the tetrahedron in R3 24 (2,3) 5 vertices of the trigonal bipyramid in R3 12 (2,2,2) 6 vertices of the octahedron in R3 48

2.10 Real and complex tight frames

A tight frame for a real Hilbert space is a tight frame for its complexiﬁcation (see Exer. 2.28). We will call frames that come in this way real tight frames.

Deﬁnition 2.8. We say that a tight frame ( f j) j J is real if its Gramian is a real matrix, and otherwise it is complex. ∈

By Theorem 2.3 (row construction), a tight frame for a space H of dimension d d d is real if and only if there is a unitary matrix U : H F for which Uf j R , j. Moreover, a frame is complex if and only if its complementary→ frame is. ∈ ∀

Example 2.16. The vertices of a simplex (or partition frame) are a real frame, by deﬁnition. The second prototypical example (1.7), and the example (a) of Exer. 1.3 are complex frames, since their Gramians (after normalisation) are

1 1 1 i − 2 2√3 2√3 2√3 2 ω2 ω 1 1 i 1 3 −3 3 − − 2  2√3 2 2√3 2√3 , ω 2 ω . (2.17) 1 i 1 1  −3 3 3  2√3 2√3 2 2−√3 ω2 ω −2  i 1 1 1  −3 3 3  −   −   2√3 2√3 2√3 2      There are intrinsic differences between the classes of real and complex frames, e.g., see Exer. 1.3, 6.8, 12.1, 12.10 and 12.17. The real algebraic§ variety§ of§ real and complex§ normalised ﬁnite tight frames (and unit–norm tight frames) is considered in Chapter 7.

Remark 2.1. One could extend the Deﬁnition 2.8 to other ﬁelds, e.g., say that the three equally spaced unit vectors in R2 are a rational tight frame, since their Gramian has rational entries. In this case, the columns of the Gramian give a copy of this frame in a rational inner product space (Example 2.9), but the row construction (Example 2.11) does not give a copy in Q2 (with the Euclidean inner product). These ideas are explored in [CFW15]. 22 2 Tight frames 2.11 SICs and MUBs

There are many interesting and useful examples of equal–norm tight frames (v j), e.g., group frames (see 10). Those for which the cross–correlation v j,vk , j = k, takes a small number of§ values are of particular interest (especially for| applications).| We briefly mention two such classes of frames: the SICs and the MUBs. These are simple to describe, and come with some intriguing conjectures, which are still unproven (despite considerable work on them). Indeed, the construction of SICs (see 14) and maximal sets of MUBs (see 14.2, 8.6) are two central problems in the theory§ of finite tight frames. § § 2 d Definition 2.9. A tight frame of d unit vectors (v j) for C is a SIC if

2 1 v j,vk = , j = k. | | d + 1 SICs can be viewed as maximal sets of complex equiangular lines. It follows from the bounds of Theorem 12.2 on such lines, that SICs are complex frames. Their origins as quantum measurements, and the known constructions are detailed in Chapter 14. The conjecture that SICs exist in every dimension d is known as Zauner’s conjecture or the SIC problem. Example 2.17. The second prototypical example (1.7) is a SIC for C2. The case d = 3 seems to be an exception for SICs. Here the SICs form a a continuous family, while for d = 3 there is currently only a finite number of SICs for Cd known. Definition 2.10. A tight frame consisting of m orthonormal bases B1,...,Bm is said to be a MUB (or a set of m MUBs) for Cd if the bases are mutually unbiased, i.e., 2 1 v,w = , v B j, w Bk, j = k. | | d ∈ ∈ Mutually unbiased bases have similar uses in quantum state determination as SICs do. The maximal number M (d) of MUBs for Cd is bounded above by d + 1 (Proposition 12.12). This bound is attained for d a prime power. Beyond this not much is known [BWB10], e.g., for d = 6 (the first d which is not a prime power), it is only known that 3 M (6) 7. ≤ ≤ The MUB problem is to say anything more, e.g., to show that M (6)= 3 (as is commonly believed). Example 2.18. Three mutually unbiased bases in C2 are given by

1 0 1 1 1 1 1 1 1 1 B1 = , , B2 = , , B3 = , . 0 1 √2 1 √2 1 √2 i √2 i − − The ﬁrst two are real MUBs. The question on how many real MUBs there are and its connection with association schemes is of interest (see [LMO10]). Three MUBs for R4 can be obtained by choosing a subset of the vertices of the 24–cell in R4. 2.11 SICs and MUBs 23 Notes

The key idea (not to be underestimated) of this section is:

Tight frames are best understood via their Gramian.

Φ n Indeed, a tight frame =( f j) j=1 is determined up to unitary equivalence (and normalisation) by its Gramian P = PΦ , which is an orthogonal projection matrix. The columns of PΦ give a (canonical) copy of Φ, and so the kernel of PΦ is the space of linear dependencies between the vectors in Φ, i.e.,

n n ker(PΦ )= a F : Pa = ∑a jPe j = 0 = a F : ∑a j f j = 0 =: dep(Φ). { ∈ j } { ∈ j }

Since PΦ is determined by ker(PΦ ), this observation allows the theory of tight frames to be extended to any finite dimensional vector space over a subfield of C which is closed under conjugation (see Chapter 4). Many notions of equivalence of tight frames appear in the literature (see [Bal99], [HL00], [GKK01], [Fic01], [HP04]). Here we use a descriptive terminology (from which all of these can be described). For finite tight frames viewed as sequences of vectors, unitary equivalence is the natural equivalence, and when viewed as (weighted) projections (fusion frames) projective unitary equivalence is natural. Unitary equivalence is determined by the Gramian (Corollary 2.1) and projective unitary unitary equivalence is determined by certain m–products (see Chapter 8). It is implicit in the Definition 2.1 of a tight frame that H be separable, i.e., have a countable orthonormal basis. The theory extends, in the obvious way, to nonseparable spaces, with J now an uncountable index set. In these cases, it turns out that all tight frames for H (with nonzero vectors) have the same infinite cardinality, i.e., the Hilbert dimension of H . By way of contrast, if H has finite dimension d, then there exist tight frames for H with any countable cardinality greater than or equal to d. We will have good reason to consider representations such as (1.3), where the sum ∑ j J is replaced by a continuous sum (with respect to some measure). This generalisation∈ (see Chapter 16) will be called a continuous tight frame, with the special case of Definition 2.1 referred to as a (discrete) tight frame. The book [HKLW07] covers the material of this section. It has a section on frames in R2 (for tight frames in R3 see [Fic01]). The popular article [KC07a], [KC07b] advocates the use of tight frames in a number of engineering applications. It outlines standard terminology for frames (resulting from an e-mail discussion within the frame community), which we adopt, except for our preference of normalised tight frame over Parseval frame. In this parlance a ENPTF is a equal–norm Parseval tight frame, and similarly. 24 2 Tight frames Exercises

2.1. By expanding, or otherwise, verify the polarisation identity for an inner product space H , i.e., f ,g H that ∀ ∈ 1 ℜ f ,g = ( f + g 2 f g 2), 4 − − 1 ℑ f ,g = ( f + ig 2 f ig 2), (for H complex). 4 − − 2.2. Use the polarisation identity to show that the following conditions are equivalent to being a ﬁnite tight frame 1 Parseval: f = ∑ f , f j f j, f H , A j J ∀ ∈ ∈ 1 Plancherel: f ,g = ∑ f , f j f j,g , f ,g H . A j J ∀ ∈ ∈

2.3. Orthogonal projection formula. Let ( f j) j J be a ﬁnite tight frame (with frame bound A) for a subspace K H . Show that∈ P the orthogonal projection onto this subspace is given by ⊂ 1 1 P = VV ∗ : f ∑ f , f j f j, V :=[ f j] j J. A → A j J ∈ ∈

2.4. Orthogonal bases and tight frames. (a) Show that an orthogonal basis ( f j) j J for H is a tight frame if and only if all its vectors have the same norm, and that it∈ is a normalised tight frame if and only if it is an orthonormal basis. (b) Show that if ( f j) is a normalised tight frame, then f j 1, j J, and ≤ ∀ ∈

f j = 1 f j fk, k = j. ⇐⇒ ⊥ ∀ In particular, the only normalised tight frames whose vectors all have unit length are the orthonormal bases.

2.5. Unitary images of tight frames. (a) Show that the image of a tight frame ( f j) j J under a unitary map U is a tight frame with the same frame bound. ∈ (b) Show that if ( f j) is a ﬁnite normalised tight frame for H , and T is a linear map for which (Tf j) is also, then T is a unitary map.

2.6. Projections of normalised tight frames are normalised tight frames. A linear map P : H H on a Hilbert space is an orthogonal projection if P2 = P → 2.11 SICs and MUBs 25 and P∗ = P. Show that if ( f j) j J is a normalised tight frame for a Hilbert space ∈ K and P is an orthogonal projection onto a subspace H K , then (Pf j) is a normalised tight frame for H . (This is obvious in the context⊂ of Theorem 2.2.)

2.7. Partial isometries map normalised tight frames to normalised tight frames. A linear map L : H K between Hilbert spaces is an isometry if L∗L = IH , i.e., it is norm preserving:→ Lx = x , x H . ∀ ∈ It is a coisometry if L∗ : K H is an isometry, i.e., L∗ is norm preserving. Let Φ be a ﬁnite normalised→ tight frame for H , and Q : H K be a linear map. Show that the following are equivalent → (a) Q is a partial isometry, i.e., its restriction to (kerQ)⊥ = ran(Q∗) is an isometry. (b) QQ∗ is an orthogonal projection. (c) Q∗Q is an orthogonal projection. (d) QΦ is a normalised tight frame (for its span). Remark. Since unitary maps and orthogonal projections are partial isometries, this generalises Exercises 2.5 and 2.6. It appears as a special case in Exer. 3.5.

2.8.m If U is an n n unitary matrix (or a scalar multiple of one) with entries of constant modulus, then× an equal–norm tight frame for Fd is given by the columns of the d n submatrix obtained from it by selecting any d of its rows. (a) When×F = R, such U, with entries 1, are called Hadamard matrices. Use the ± n n matlab function hadamard(n) (deﬁned for n, 12 or 20 a power of 2) to construct equal–norm tight frames of n vectors in Rd. Remark: It can be shown that if a Hadamard matrix exists, then n = 1,2 or n is divisible by 4. The Hadamard conjecture is that there exists a Hadamard matrix of size n = 4k, for every k. The smallest open case (in 2010) is n = 668. π 1 jk 2 i (b) Show that the Fourier matrix F = [ω ]0 j,kFourier transform) to construct F, and hence equal–norm tight frames of n vectors in Cd. Remark: It is always possible to obtain a real frame in this way.

2.9. Tight frames for R2. n 2 2 (a) Show the vectors (v j) j=1, v j =(x j,y j) R are a tight frame for R if and only 2 ∈ if the diagram vectors w j :=(x j + iy j) C sum to zero (in C). (b) Show that two tight frames for R2 are∈ projectively unitarily equivalent if and only if their diagram vectors are scalar multiples of each other. (c) Show that up to projective unitary equivalence the only equal–norm tight frame of three vectors for R2 is three equally spaced unit vectors. (d) Show that all unit–norm tight frames of four vectors for R2 are the union of two orthonormal bases. This gives a one–parameter family of projectively unitarily inequivalent unit–norm tight frames of four vectors for R2. (e) Show the tight frames of ﬁve unit vectors for R2 with diagram vectors 26 2 Tight frames π iθ iθ i(θ+ψ) i(θ+ψ) 1 e ,e− ,e ,e− , 1 , 0 < θ < , cos2θ + cos(2θ + 2ψ)= { − } 2 2 are projectively unitarily inequivalent, and that none is the union of an orthonormal basis and three equally spaced vectors.

2.10. Projective unitary equivalence in R2. (a) For unit–norm tight frames of n vectors for R2 show that the equivalence classes for projective unitary equivalence up to reordering are in 1–1 correspondence with convex n–gons with sides of unit length (given by a sum of diagram vectors). (b) What do subsets of orthonormal vectors correspond to on the polygon? (c) What is the n–gon corresponding to the tight frame for R2 given by n equally spaced unit vectors? (d) Does every ﬁnite tight frames for R2 correspond to some convex polygon? 2.11. The complex conjugate of H is the Hilbert space H of all formal complex conjugates with addition, scalar multiplication and inner product given by

v + w = v + w, αv = αv, v,w = v,w . (2.18) (a) Show that the conjugation map C : H H : v v is antilinear. → → (b) Suppose that Φ =( f j) is a sequence of vectors in H and Φ :=( f j) H . Show that the frame operator and Gramian satisfy ⊂

1 SΦ = CSΦC− , Gram(Φ)= Gram(Φ).

Hence C maps tight frames to a tight frames (with the same bound A). (c) Suppose that H = V, with V a subspace of Cd. Show that H is isomorphic to d the subspace V := v : v V of C , where v = (v j) :=(v j). { ∈ } 2.12. Show the frame operator S for a sequence of vectors f1,..., fn satisﬁes: 2 (a) trace(S)= ∑ j f j . 2 2 2 (b) trace(S )=( S F ) = ∑ j ∑k f j, fk . Hint: The trace operator satisﬁes| trace(AB| )= trace(BA).

2.13. Trace formula. Show that if ( f j) j J is a ﬁnite normalised tight frame for H and L : H H is a linear transformation,∈ then its trace is given by →

trace(L)= ∑ Lf j, f j . j J ∈ In particular, when L is the identity map we obtain the trace formula (2.9). 2.14. Let H be have finite dimension d 1. Show that ∞ ≥H (a) There exists a tight frame ( f j) j=1 for , with infinitely many nonzero vectors. (b) For any such tight frame, f j 0 as j ∞. (c) There are no equal–norm tight frames→ for→H with infinitely many vectors. Remark: In contrast, the continuous tight frame (uθ ) for R2 of (1.3) has uncountably many equal–norm vectors. 2.11 SICs and MUBs 27

2.15. Equal–norms. Show that if ( f j) is an equal–norm tight frame of n vectors (with frame bound A) for a space H of dimension d, then

dA f j = , j. n ∀ 1 d In particular, if ( f j) is unit–norm, i.e., f j = 1, j, then = , ∀ A n

2.16. Equiangularity. Show that if ( f j) is an equiangular tight frame of n > 1 vectors (with frame bound A) for a space H of dimension d, then its Gramian satisﬁes

dA A d(n d) f j, f j = , j, f j, fk = − , j = k. n ∀ | | n n 1 −

2.17. Let Φ =( f j) j J be a ﬁnite sequence of vectors in H , where d = dim(H ), ∈ and V :=[ f j] j J. Show Φ is a normalised tight frame for H if and only if ∈ (a) Gram(Φ)= V ∗V has exactly d nonzero eigenvalues all equal to 1. (b) The frame operator SΦ = VV ∗ has all its eigenvalues equal to 1. (c) The synthesis operator V has d singular values equal to 1. (d) The analysis operator V ∗ has d singular values equal to 1.

2.18. Isometries. Let Φ =( f j) j J H , and V ∗ : H ℓ2(J) : f ( f , f j ) j J be the analysis operator. Show that∈ the⊂ following are equivale→ nt → ∈ (a) Φ is a normalised tight frame for H . (b) V ∗ is inner product preserving, i.e., V ∗ f ,V ∗g = f ,g , f ,g H . (c) V is an isometry, i.e., V f = f , f H . ∀ ∈ ∗ ∗ ∀ ∈

Φ n Ψ n 2.19. Suppose that =( f j) j=1 and =(g j) j=1 are sequences of vectors, with H = span(Φ) and K = span(Ψ). Show there is a unitary map U : H K with → g j = Uf j, j if and only if f j, fk = g j,gk , j,k, i.e., Gram(Φ)= Gram(Ψ). ∀ ∀ 2.20. (a) Express unitary equivalence up to reordering in terms of the Gramian. (b) Express projective unitary equivalence up to reordering in terms of the Gramian. (c) Show that a necessary, but not sufﬁcient, condition for normalised tight frames ( f j) j J and (g j) j K to be projectively equivalent up to reordering is that there is ∈ ∈ a permutation σ : J K with gσ j,gσk = f j, fk , j,k J. In particular, the → | | | | ∀ ∈ multisets f j, fk j,k J and g j, fk j,k J must be equal. {| |} ∈ {| |} ∈ 2.21. (a) Show that normalised tight frames Φ and Ψ are projectively unitarily equivalent up to reordering if and only if their complements are. (b) Show that all equal–norm tight frames of n = d +1 vectors in Fd are projectively unitarily equivalent, and hence are equiangular. (c) For the unitarily inequivalent equal–norm frames of three vectors for C2 given in Example 2.8, ﬁnd a unitary matrix U (and scalars α j) which give the projective unitary equivalence g j = α jUf j, where Φ =( f j) and Ψ =(g j). 28 2 Tight frames

2.22. Find all possible normalised tight frames of three vectors for C2 up to unitary equivalence. 2.23. Show that no tight frame can be unitarily equivalent to its complement. Can a tight frame be projectively unitarily equivalent to its complement? 2.24.m Write a matlab function for the complementary tight frame using null. 2.25.m (a) By using an inductive argument based on complements, prove that an equal–norm tight frame of n vectors for Fd can be constructed, for all n > d. (b) Write a function ENTF(n,d) to construct such equal–norm tight frames. (c) Construct an equal–norm tight frame of 8 vectors for R3.

2.26. M. A. Na˘ımark’s theorem. An orthogonal resolution of the identity for a Hilbert space H is a one parameter family (Et )t R of orthogonal projections on H , for which t Et is left continuous, and ∈ → lim Et = 0, lim Et = IH , EsEt = Emin s,t . t ∞ t ∞ { } →− → A generalised resolution of the identity is a family (Ft )t R, for which the differ- ∈ ences Ft Fs, s < t are bounded positive operators, t Ft is left continuous, and − →

lim Ft = 0, lim Ft = IH . t ∞ t ∞ →− → Na˘ımark’s theorem (see, e.g., [AG63]) says that every generalised resolution of the identity for H is the orthogonal projection onto H of an orthogonal resolution of the identity for some larger Hilbert space K H . n ⊃ (a) Let ( f j) j=1 be a ﬁnite normalised tight frame for which none of the vectors are zero. Show that a generalised resolution of the identity is given by

Ft f := ∑ f , f j f j, f H . j t ∀ ∈ ≤ (b) By Na˘ımark’s theorem, there is a Hilbert space K H , and an orthogonal ⊃ resolution of the identity (Et ) for K , such that Ft = PEt , where P is the orthogonal projection of K onto H . Conclude that

n n IH = ∑(Fj Fj 1)= ∑ PQ j, Q j := E j E j 1 j=1 − − j=1 − − where Q j is an orthogonal projection, and Q j Qk, k = j. (c) Show that K can be taken to be n dimensional.⊥ d (d) Prove Na˘ımark’s theorem for H = F by taking V =[ f1,..., fn] which has orthonormal rows, and extend it to obtain a unitary matrix. n d 2.27. Suppose that (u j + iv j) j=1 is a normalised tight frame of n vectors for C , d where u j,v j R . Prove that (u1,...,un,v1,...,vn) is a normalised tight frame of 2n vectors for∈Rd. 2.11 SICs and MUBs 29

2.28. Show that a tight frame for Rd is tight frame for Cd.

2.29. Normalised tight frames and linear mappings. Let ( f j) j J and (gk)k K be ﬁnite normalised tight frames for H and K . Denote the vector space∈ of all linear∈ maps H K by L (H ,K ). (a) Show that the Hilbert–Schmidt →inner product on L (H ,K ) satisﬁes

L,M HS := trace(M∗L)= ∑ Lf j,Mf j = ∑ M∗gk,L∗gk . j J k K ∈ ∈ Remark: Taking M = IH gives the trace formula of Exer. 2.13. H (b) Let f j∗ be F : f f , f j . Show that (gk f j∗) j J,k K is a normalised tight frame (of rank one→ maps) for→ L (H ,K ) with the Hilbert-Schmidt∈ ∈ inner product.

2.30. Matrices with respect to a normalised tight frame. Normalised tight frames can be used to represent vectors and linear maps in much the same way as orthonormal bases. Suppose that ( f j) j J and (gk)k K are ﬁnite ∈ ∈ normalised tight frames for H and K , and let V =[ f j] j J, W =[gk]k K. Then the ∈ ∈ coordinates x of f H with respect to ( f j), and the matrix A representing a linear ∈ map L : H K with respect to ( f j) and (gk) are → J J J x =[ f ] := V ∗ f F , A =[L] := W ∗LV F × . ∈ ∈

(a) Show that [Lf ]= Ax, and f , L can be recovered via f = Vx, L = WAV ∗. (b) Show that [αL + βM]= α[L]+ β[M], α,β F, and [L∗]=[L]∗. (c) Show that the composition of linear maps satisﬁes∈ [ML]=[M][L]. (d) Suppose L : H H , and W = V. Show that f is an eigenvector of L for the eigenvalue λ if and only→ if Ax = λx, i.e., eigenvectors of L correspond to the eigenvectors of A that are in the range of V ∗. (e) Show L and A have the same singular values, and hence the same rank.

Chapter 3 Frames

n H A ﬁnite (normalised) tight frame is a spanning set ( f j) j=1 for , for which

n f = ∑ f , f j f j, f H . j=1 ∀ ∈

This expansion can be further generalised, by replacing the rank one orthogonal projections f f , f j f j, by rank one projections f f ,g j f j, to obtain what is called a (nontight)→ frame expansion. → This elegant theory includes orthogonal and biorthogonal expansions as special cases. It will be mostly used as a route to obtain the so called canonical tight frame. This is a normalised tight frame naturally associated with a given frame, which is as close as possible to it.

3.1 Motivation

n Suppose that ( f j) spans H , so that each f H can be reconstructed j=1 ∈ n n f = ∑ c j f j, Vc = f , V :=[ f j] j=1, j=1 ⇐⇒ for some coefﬁcients c j = c j( f ) F, which are unique if and only if ( f j) is a basis, i.e., n = dim(H ). However, there∈ is always a unique least squares solution (one 2 minimising ∑ c j ) given by j | | † 1 1 c = V f = V ∗S− f c j = f ,S− f j , ⇐⇒ † 1 n where S := VV ∗, and V = V ∗(VV ∗)− : H F is the pseudoinverse of V (see → 1 Exercises 3.2 and 3.3), The coefﬁcients c j = f ,S− f j are linear functions of f , whose Riesz representers 1 g j = f˜j := S− f j will be called the dual frame.

31 32 3 Frames 3.2 A frame and its dual

A ﬁnite frame ( f j) j J for H is simply a spanning set (see Exer. 3.6), i.e., ∈

V =[ f j] j J : ℓ2(J) H is onto H , ∈ →

which implies that S = VV ∗ is (boundedly) invertible. For H infinite dimensional (or J infinite), the following condition ensures that V and S can be defined, such that V is onto and S has a bounded inverse.

Deﬁnition 3.1. A countable sequence ( f j) j J in a Hilbert space is said to be a frame (for H ) if there exists (frame bounds) A,B∈> 0, such that

2 2 2 A f ∑ f , f j B f , f H . (3.1) ≤ j J | | ≤ ∀ ∈ ∈ The best possible A,B are called the (optimal) frame bounds.

A ﬁnite frame for H is precisely a spanning sequence for H .

Given a frame Φ =( f j) for H , we recall the frame operator S = SΦ : H H → is the self adjoint operator deﬁned by S := VV ∗, i.e.,

Sf := ∑ f , f j f j, f H , j J ∀ ∈ ∈

where V :=[ f j] j J is the synthesis operator. The frame bounds (3.1) can be written ∈

Af , f Sf , f Bf , f , f H AIH S BIH . (3.2) ≤ ≤ ∀ ∈ ⇐⇒ ≤ ≤ From this, it follows that S is positive, with a bounded inverse satisfying

1 1 1 IH S− IH . B ≤ ≤ A and the optimal frame bounds are (cf Exer. 3.7)

A = AΦ := smallest eigenvalue of S, B = BΦ := largest eigenvalue of S.

We can therefore deﬁne the dual frame, which is indeed a frame.

Deﬁnition 3.2. Given a frame Φ = ( f j) j J for H , with frame operator S, the 1 ∈ (canonical) dual frame Φ˜ =(g j) j J for H is deﬁned by ∈ 1 g j = f˜j := S− f j.

1 Sometimes the dual frame is called the canonical dual frame to distinguish it from a so called alternate dual frame (see §3.10). 3.2 A frame and its dual 33

1 The synthesis operator for the dual frame Φ˜ is W =[ f˜j]= S− V, and so

1 1 1 1 SΦ˜ = WW ∗ =(S− V)V ∗S− = S− = SΦ− . (3.3)

Thus the dual frame Ψ = Φ˜ is a frame, with Ψ˜ = Φ, and optimal frame bounds 1 1 AΦ˜ = , BΦ˜ = . BΦ AΦ The Gramian of a frame and its the dual are pseudoinverses of each other (Exer. 3.3)

Gram(Φ˜ )= Gram(Φ)†, Gram(Φ)= Gram(Φ˜ )†. (3.4)

Proposition 3.1. Let ( f j) j J be a ﬁnite frame for H , i.e., a spanning set, and let 1 ∈ f˜j := S− f j be the dual frame. Then we have the frame expansion

f = ∑ f , f˜j f j = ∑ f , f j f˜j, f H . (3.5) j J j J ∀ ∈ ∈ ∈

Moreover, suppose that f = ∑ j c j f j, for some coefﬁcients c j, then

2 2 2 ∑ c j = ∑ f , f˜j + ∑ c j f , f˜j . (3.6) j J | | j J | | j J | − | ∈ ∈ ∈ 1 Proof. Since [ f˜j] j J = S− V, V :=[ f j] j J, the equation (3.5) can be written ∈ ∈ 1 1 IH = V(S− V)∗ =(S− V)V ∗, S := VV ∗, which clearly holds. Suppose f = Vc, and write c as the least squares solution, plus the error 1 1 c = V ∗S− f +(c V ∗S− f ). − These vectors are orthogonal (see Exer. 3.2) since

1 1 1 1 1 V ∗S− f ,c V ∗S− f = S− f ,Vc VV ∗S− f = S− f , f f = 0, − − − 2 1 2 1 2 and so, by Pythagoras, c = V ∗S− f + c V ∗S− f , which gives (3.6), 1 1 − since (V S f ) j = ((S V) f ) j = f , f˜j . ∗ − − ∗ ⊓⊔ It follows from Proposition 3.1 that given a ﬁnite spanning sequence ( f j) for a vector space (over R or C), for each vector f there are unique coefﬁcients (c j) with 2 f = ∑ j c j f j and ∑ j c j minimal. These c j are linear functions of f and are called the canonical coordinates| | of f (see 4). § Example 3.1. For ( f j) a basis for H , (3.5) implies that ( , f˜j ) is the corresponding dual basis (coordinate functionals), i.e., ( f j) and ( f˜j) form a biorthogonal system

1, j = k; f j, f˜k = 0, j = k. 34 3 Frames

Example 3.2. If ( f j) is a tight frame, then (3.2) reduces to S = AIH , and so the dual ˜ 1 1 frame is given by f j = S− f j = A f j.

2 2 2 Example 3.3. Consider the frame Φ =(e1,αe2,βe2), α + β > 0, for R . Then

1 0 0 1 0 1 1 0 0 V = , S = , W = S V = α β , 0 α β 0 α2 + β 2 − 0 α2+β 2 α2+β 2 β so the dual frame is Φ˜ e α e e , and (3.5) gives the expansion =( 1, α2+β 2 2, α2+β 2 2) α β x1 x2 x2 2 x = = x1e1 + (αe2)+ (βe2), x R . x2 α2 + β 2 α2 + β 2 ∀ ∈ 1 1 Other choices for Φ˜ that satisfy (3.5) include (e1, e2,0) and (e1,0, e2) (α,β = 0), α β which give the (orthogonal) expansions

x2 x2 2 x = x1e1 + (αe2)+ 0(βe2)= x1e1 + 0(αe2)+ (βe2), x R . α β ∀ ∈

Observe that these coefﬁcients have greater ℓ2–norm than those for the dual frame. This illustrates the tendency of a frame to distribute information about a function evenly over the coefﬁcients, rather than concentrate it on a few. Indeed, since S is invertible, it follows that f j = 0 if and only if f˜j = 0. Example 3.4. If ( f j) is a normalised tight frame for H and T : H H is an 1 → invertible linear map, then the frames (T ∗ f j) and (T − f j) are dual (see Exer. 3.4).

Fig. 3.1: A frame (solid heads) and its dual, for 5, 10 and 15 random vectors in R2.

Example 3.5. (Exer. 3.5) The image Ψ = QΦ of a frame Φ under a linear map Q : H K is again a frame (for its span), with frame bounds → † 2 2 AΨ AΦ Q − , BΨ BΦ Q . ≥ ≤ 3.3 The canonical tight frame 35 3.3 The canonical tight frame

Given a frame (or its dual), there is a naturally associated tight frame, which inherits properties of the original (see, e.g., Theorem 10.1).

Deﬁnition 3.3. If Φ =( f j) is a frame for H , with frame operator S = SΦ , then the Φcan can corresponding canonical tight frame =( f j ) is given by

can 1 g j = f := S− 2 f j, j. j ∀ 1 To ﬁnd the canonical tight frame, one must calculate S− 2 , the positive square root of the inverse of the frame operator S. In practice, this may be a difﬁcult numerical or analytical calculation (see 3.9). If V = U1ΣU is a singular value decomposition § 2∗ of V =[ f j], then (see Exer. 3.9) the canonical tight frame is given by

can [ f j ]= U1 I 0 U2∗ = U1U2∗. (3.7)

Theorem 3.1. Let Φ =( f j) j J be a ﬁnite frame for H . Then the canonical tight 1 ∈ can 2 frame f j := SΦ− f j is a normalised tight frame, i.e., can can H f = ∑ f , f j f j , f . (3.8) j J ∀ ∈ ∈ Moreover, (Φ˜ )can = Φcan, and the Gramian is the orthogonal projection matrix with the same kernel as the synthesis operator V =[ f j], and satisﬁes

Gram(Φcan)= Gram(Φ)Gram(Φ˜ )= Gram(Φ˜ )Gram(Φ). (3.9)

can 1 can Proof. Let U =[ f j ] j J = S− 2 V be the synthesis operator of ( f j ). Then ∈

1 1 1 1 UU∗ =(S− 2 V)(V ∗S− 2 )= S− 2 SS− 2 = IH , which is (3.8). The other observations follow similarly (see Exer. 3.10). ⊓⊔

Example 3.6. If ( f j) is a basis for H , then the canonical tight frame is an orthonormal basis known as the symmetric or Lowdin¨ orthogonalisation of ( f j). It was ﬁrst obtained via a symmetric version of the Gram–Schmidt algorithm, and is the closest orthonormal basis to ( f j) (see Corollary 3.3).

Example 3.7. If ( f j) is an equal–norm frame for H , then the dual frame and the canonical tight frame need not have vectors of equal length, e.g., the canonical dual can α β of the frame of Example 3.3 is Φ =(e1, e2, e2) (take α = β = 1). √α2+β 2 √α2+β 2 They do satisfy (see Exer. 3.13)

˜ ˜ can 2 ˜ H f j, f j = f j, f j = f j , j = ∑ f j, f j = d = dim( ). ∀ ⇒ j J ∈ 36 3 Frames

Example 3.8. (Exer. 3.16) If Φ =( f j) is a frame for H , and Q : H H is an → invertible linear map, then Ψ := QΦ =(Qf j) is a frame for H , and

Ψ can = UΦcan,

where U is unitary. Equivalently, Gram(Ψ can)= Gram(Φcan).

The synthesis operator of the canonical tight frame for Φ =( f j) j J ∈ 1 can 2 H U =[ f j ] j J = S− V : ℓ2(J) ∈ →

is a partial isometry, which appears in the polar decomposition of V =[ f j] j J. ∈ Corollary 3.1. (Polar decomposition). Let Φ =( f j) j J be a ﬁnite frame for H , and can ∈ U =[ f j ] j J. Then the polar decomposition of the synthesis operator V =[ f j] j J into a product∈ of a partial isometry and a positive operator is ∈

1 1 V = UG 2 = S 2 U, G = Gram(Φ)= V ∗V, S = SΦ = VV ∗. (3.10) Σ Proof. Let V = U1 U2∗ be a singular value decomposition of V. Then (3.10) is the polar decomposition of A = V, where

1 1 1 1 U = U1U2∗, P =(V ∗V) 2 = G 2 , Q =(VV ∗) 2 = S 2 ,

and we recognise that U is [ f can], by (3.7). j ⊓⊔ The Gramian of the canonical tight frame is an orthogonal projection

can 1 P = Gram(Φ )= V ∗S− V : ℓ2(J) ℓ2(J), (3.11) →

which gives the coefﬁcients c with f = ∑ j c j f j of minimal ℓ2–norm.

Corollary 3.2. Let Φ = ( f j) j J be a ﬁnite frame for H , and V = [ f j] j J. If ∈ ∈ f = Va, then the coefﬁcients c of minimal ℓ2–norm with f = Vc are given by c = Gram(Φcan)a. In particular, V can be expanded

V = V Gram(Φcan), (3.12)

where can ran(V ∗)= ker(V)⊥ = ran(Gram(Φ )). (3.13)

Proof. By Proposition 3.1, the coefﬁcients c of minimal ℓ2–norm are given by

1 1 1 1 can c = V ∗S− f = V ∗S− Va =(S− 2 V)∗(S− 2 V)a = Gram(Φ )a. (3.14)

1 can Left multiplying c = V ∗S− Va = Gram(Φ )a by V gives

1 can can VV ∗S− Va = Va = V Gram(Φ )a, a = V = V Gram(Φ ). ∀ ⇒ Finally, since S is invertible and V is onto, we obtain (3.13) from (3.11) ⊓⊔ 3.3 The canonical tight frame 37

Fig. 3.2: A frame of n randomly chosen unit vectors (solid arrowheads), its dual, and the canonical tight frame (circles), for n = 8 and n = 16.

Thus, if V =[ f j] is the synthesis operator of a ﬁnite frame Φ =( f j), then

1 1 can 2 V = V Gram(Φ )= U Gram(Φ) 2 = SΦU,

Φcan can where Gram( ) is an orthogonal projection, U :=[ f j ] is a coisometry, Gram(Φ) is positive semideﬁnite, and SΦ is positive deﬁnite.

n For a given a frame ( f j) j=1, the canonical tight frame is the closest tight frame n (g j) j=1 in the sense of minimising the least squares error

n 2 n n 2 ∑ f j g j = [ f j] j=1 [g j] j=1 F . j=1 − − Φ n H λ λ Theorem 3.2. Let =( f j) j=1 be a frame for , and 1,..., d be the eigenvalues Ψ n H of SΦ . If =(g j) j=1 is a tight frame for , with frame bound A, then

n d 2 2 ∑ f j g j ∑( λk √A) , (3.15) j=1 − ≥ k=1 − with equality if and only if Ψ = √AΦcan. n n Proof. With V =[ f j] j=1, W =[g j] j=1, we compute

2 ∑ f j g j = trace((V W)∗(V W)) j − − − = trace(V ∗V)+ trace(W ∗W) trace(W ∗V) trace(V ∗W) − − = trace(VV ∗)+ trace(WW ∗) trace(W ∗V) trace(W V) − − ∗ = trace(SΦ )+ trace(SΨ ) 2ℜtrace(W ∗V) − = ∑ λk + ∑ A 2ℜtrace(W V). k k − ∗ 38 3 Frames

Hence we must prove that

ℜ(trace(W ∗V)) = ℜ∑ f j,g j ∑ λk√A. (3.16) j ≤ k Let u1,...,ud an orthonormal basis of eigenvectors for SΦ corresponding to the eigenvalues λ1,...,λd. Then (3.16) can be expanded

n n d ℜ ∑ f j,g j = ℜ ∑ ∑ f j,uk uk,g j j=1 j=1 k=1 d n d = ∑ ℜ ∑ f j,uk g j,uk ∑ λk√A. k=1 j=1 ≤ k=1 It would therefore sufﬁce to prove that

n ℜ ∑ f j,uk g j,uk λk√A, k j=1 ≤ ∀ which we now do. Firstly,

n n ℜ ∑ f j,uk g j,uk ∑ f j,uk g j,uk , ≤ j=1 j=1 with equality if and only if

n n ∑ f j,uk g j,uk = ∑ f j,uk g j,uk 0. (3.17) ≥ j=1 j=1 By the Cauchy–Schwarz inequality,

n 1 1 2 2 2 2 ∑ f j,uk g j,uk ∑ f j,uk ∑ g j,uk j=1 ≤ j | | j | | = SΦ uk,uk SΨ uk,uk = λk√A, with equality if and only if

f j,uk = ck g j,uk , j, ckA = λk√A. (3.18) ∀ | | Thus we obtain the desired inequality (3.15). There is equality in (3.15) if and only if (3.17) and (3.18) hold. These together imply ck = λk/√A, and hence 3.3 The canonical tight frame 39

d d √A √A d 1 1 g = ∑ g ,u u = ∑ f ,u u = ∑ S− 2 f ,S 2 u u j j k k λ j k k λ Φ j Φ k k k=1 k=1 k k k=1 √A d 1 1 = ∑ S− 2 f , λ u u = √AS− 2 f = √Af can, λ Φ j k k k Φ j j k k=1 as claimed. ⊓⊔ Φ n H λ λ Corollary 3.3. Let =( f j) j=1 be a frame for , and 1,..., d be the eigenvalues of the frame operator S = SΦ . Then

n d d 2 2 1 min ∑ f j g j : Ψ =(g j) is a tight frame for H = ∑ λk ∑ λk , { j=1 − } k=1 − d k=1 which is attained if and only if

d can 1 Ψ = √AΦ , √A := ∑ λk. d k=1 2 Proof. The minimum of √A ∑ ( λk √A) occurs when its derivative is zero: → k − d 1 d 2 ∑( λk √A)= 0 = √A = ∑ λk, − k=1 − ⇒ d k=1 2 and this minimum value ∑ ( λk √A) can be simpliﬁed as claimed. k − ⊓⊔ In terms of matrices (see Exer. 3.17) this result says that the minimum

min V W F , rank(V)= d W Fd n, A>0 − ∈ × WW∗=AI is uniquely attained for

1 1 trace((VV ∗) 2 ) 1 trace(S 2 ) 1 W = (VV ) 2 V = S 2 V. d ∗ − d −

d Let V =[ f j] be a ﬁnite frame for F . Then there is a unique tight frame closest to it given by can 1 √A( f j ), √A = trace((VV ∗) 2 )/d. 40 3 Frames 3.4 Unitarily equivalent frames

As for tight frames, we say that ﬁnite frames Φ =( f j) j J and Ψ =(g j) j J, with the same index set J, are unitarily equivalent if there is a∈c > 0 and a unitary∈ U for which g j = cU f j, j J. ∀ ∈ The dual and canonical tight frames are related as follows (see Exer. 3.16)

1 can can g˜ j = U f˜j, g = Uf , j J. c j j ∀ ∈ Moreover, the Gramians satisfy Gram(Ψ)= c2 Gram(Φ), which gives the following generalisation of Corollary 2.1.

Unitarily equivalent frames are uniquely determined by their Gramians (up to a positive scalar multiplier).

The Gramian of a normalised tight frames is a projection matrix, and vice a versa. Any positive semideﬁnite matrix is the Gramian of some frame.

Theorem 3.3. An n n matrix M =[m jk] j,k J is the Gramian matrix of a frame × ∈ Φ =( f j) j J for the space H := span f j j J if and only if it is positive semideﬁnite, i.e., M = M∈ and Mf , f 0, f H{ .} Moreover,∈ ∗ ≥ ∀ ∈ d = dim(H )= rank(M), (3.19) and a copy of the frame is given by the columns of any matrix L for which L∗L = M 1 (e.g., the positive square root L = M 2 , or a Cholesky factor L = R).

Proof. Since the Gramian can be factorised as Gram(Φ)= V ∗V, V =[ f j] j J, it is ∈ positive semideﬁnite: V ∗Vf , f = Vf ,Vf 0. Conversely, suppose that M is a positive≥ semideﬁnite n n matrix. Then M can be factored M = L L for some m n matrix. For example,× let m = n, and take ∗ × L = R from a Cholesky factorisation M = R∗R (which exists, though it may not be numerically stable to calculate), or with M = UDU∗ a unitary diagonalisation of M, 1 1 take the positive square root L = M 2 := D 2 U∗. The columns of L are a frame with Gramian M, since

Le j,Lek = L∗Le j,ek = Me j,ek =(k, j)–entry of M.

⊓⊔ This result, i.e., that every positive semideﬁnite matrix is the Gramian of some sequence of vectors (which is a frame for its span) is well known (cf [Ros97]). 3.5 Similar frames and orthogonal frames 41

Φ n d Example 3.9. (Exer. 3.19) A sequence of n unit vectors =(v j) j=1 in R , d > 1, is said to be θ–isogonal2 if

v j,vk = a := cosθ, j = k. ∀ The eigenvalues of the Gramian M of such a sequence are na a + 1 and 1 a −1 − (multiplicity n 1). Since the Gramian is positive semideﬁnite, n− 1 a 1, and its − 1 − ≤ ≤ rank is either 1 (a = 1), n 1(a = − ), or n. Thus the only isogonal conﬁgurations − n 1 up to unitary equivalence are the d +−1 vertices of of the regular simplex in Rd, or a d 1 unique set of d vectors in R with n− 1 < a < 1. In the latter case − b c c c b c † c a Gram(Φ˜ )= M = . . ., = − , . .. . b na 2a + 1   − c c b     and so the dual frame Φ˜ is isogonal, and it is unitarily equivalent to Φ if and only if a = 0, i.e., Φ is an orthonormal basis.

Remark 3.1. As for tight frames, a frame is said to be real if its Gramian is real, and to be complex otherwise. A frame is real if and only if its dual is, in which case the canonical tight frame is real also. However, if the canonical tight frame is real, then the frame itself need not be real (see Exer. 3.20).

3.5 Similar frames and orthogonal frames

We deﬁne a second equivalence relation on frames, for which each equivalence class contains a unique tight frame (up to unitary equivalence).

Deﬁnition 3.4. Frames Φ =( f j) j J and Ψ =(g j) j J for H and K , with the same index set J, are said to be similar∈if there is an invertible∈ linear map Q : H K such that → g j = Qf j, j J. ∀ ∈ For frames which are not tight, this equivalence relation is weaker than unitary equivalence. Indeed, a frame Φ =( f j), its dual frame, and canonical tight frame are all similar, since 1 ˜ 1 can 2 f j = S− f j, f j = S− f j, but they are unitarily equivalent if and only if Φ is tight.

Example 3.10. Every basis is similar to an orthonormal basis.

2 Isogonal vectors appear in the structure of soap ﬁlms and bubbles, see [Mur93]. 42 3 Frames

Theorem 3.4. There is exactly one tight frame (up to unitary equivalence) in each equivalence class of similar frames, namely the canonical tight frame.

Proof. Suppose Φ =( f j) and Ψ =(g j) are similar, via Q, and S = SΦ . Then

1 1 can g j = Qf j = QS 2 S− 2 f j = Tf , j, j ∀ 1 Ψ can where T = QS 2 is invertible. Hence =(Tf j ) is tight if and only if

can can 2 SΨ = T[ f j ][ f j ]∗T ∗ = TT ∗ = c I, c > 0,

i.e., T=cU, with U unitary, and Ψ is unitarily equivalent to Φcan. ⊓⊔ In other words, the study of frames up to similarity reduces to the study of tight frames up to unitary equivalence.

Finite frames are similar if and only if their canonical Gramians are equal.

Since orthogonal projections, such as P = Gram(Φcan), are uniquely determined by their range (or kernel), Theorem 3.4 implies the equivalence classes of similar frames (or of unitarily equivalent tight frames) of n vectors for a d–dimensional space are in 1–1 correspondence with the d–dimensional subspaces of Fn, i.e.. points n on the Grassmannian Grd(F ). These considerations lead to many conditions equivalent to similarity.

Proposition 3.2. (Similarity) Let Φ =( f j) j J and Ψ =(g j) j J be ﬁnite frames with synthesis operators V and W. Then the following∈ are equivale∈ nt (a) Φ and Ψ are similar. can can 1 1 (b) Gram(Φ )= Gram(Ψ ), i.e.,V ∗(VV ∗)− V = W ∗(WW ∗)− W. (c) ran(V ∗)= ran(W ∗), or, equivalently, ker(V)= ker(W). Furthermore, these imply the equivalent conditions (d) WV ∗ is invertible. (e) ∑ f , f j g j = 0, f = 0. j ∀ can Proof. Recall, from (3.13), that ran(V ∗)= ran(Gram(Φ )). (a) (b) Theorem 3.4 and Corollary 2.1. (b)⇐⇒(c) The orthogonal projections Gram(Φcan) and Gram(Ψ can) (which are de- ⇐⇒ termined by their ranges) are equal if and only if ran(V ∗)= ran(W ∗). (c)= (d) If ran(V )= ran(W ), then ran(WV )= ran(WW )= ran(SΨ ), so that ⇒ ∗ ∗ ∗ ∗ WV ∗ onto, and hence invertible. (d) (e) This is follows immediately since WV f = ∑ f , f j g j. ⇐⇒ ∗ j ⊓⊔ At the other extreme, we consider frames which are far from being similar.

Deﬁnition 3.5. Frames Φ =( f j) j J and Ψ =(g j) j J for H and K , with the same index set J, are said to be orthogonal∈ (or strongly∈ disjoint) if

ran(V ∗) ran(W ∗), V =[ f j], W =[g j]. ⊥ 3.6 Frames as orthogonal projections 43

Orthogonal frames cannot be similar (similar frames have ran(V ∗)= ran(W ∗)). There are many conditions equivalent to orthogonality (see Lemma 5.1), including

∑ f , f j g j = 0, f H WV ∗ = 0. (3.20) j ∀ ∈ ⇐⇒

This can be written suggestively as

H ∑g j f j∗ = 0, f j∗ : F : f f , f j . (3.21) j → →

Example 3.11. (1–dimensional frames) For H = K = F, the condition (3.21) is orthogonality of vectors. Hence the rows of any unitary matrix, e.g., the Fourier matrix (2.5), are orthogonal frames for C.

Example 3.12. (Projections) Let ( f j) be a normalised tight frame for H , and P, Q be orthogonal projections onto subspaces H1, H2, with H1 H2, i.e., PQ = 0. Then ⊥ Φ =(Pf j), Ψ =(Qf j) are orthogonal normalised tight frames for H1 and H2, since

WV ∗ = QUU∗P∗ = QP = 0, U =[ f j], V =[Pf j]= PU, W =[Qf j]= QU.

In 5.2 we will see that effectively all orthogonal frames appear in this way. §

3.6 Frames as orthogonal projections

We have seen (in 2.6) that every ﬁnite normalised tight frame is the orthogonal projection of an orthonormal§ basis (tight frame without redundancy). Thus, it is natural to ask whether every frame and its dual is the projection of a biorthogonal system with the same frame bounds (cf Exer. 3.5). Let ( f j) and (g j) be a biorthogonal system for K , i.e., ( f j) be a basis for K , with (g j) the dual basis uniquely determined by

f j,gk = δ jk VW ∗ = WV ∗ = I, V :=[ f j], W :=[g j], ⇐⇒ and P be the orthogonal projection onto a subspace H . Then (see Exer. 3.12)

f = ∑ f ,Pg j Pf j = ∑ f ,Pf j Pg j, f H , j j ∀ ∈ but the frames Φ =(Pf j) and Ψ =(Pg j) for H need not be the (canonical) duals of each other (they are alternate duals in the sense of 3.10). They are dual if and only if the synthesis operators of Φ˜ and Ψ are equal, i.e.,§

1 1 (PV(PV)∗)− PV = PW = P(V ∗)− , 44 3 Frames which equivalent to

PS = PSP SP = PSP SP = PS, S := VV ∗. ⇐⇒ ⇐⇒ In other words (see Exer. 3.21):

Let ( f j) and (g j) be a biorthogonal system for K , S = VV ∗, V =[ f j], and P be the orthogonal projection of K onto a subspace H . Then (Pf j) and (Pg j) are (canonically) dual frames for H if and only if

H is an invariant subspace of S, i.e., SH H . ⊂

Therefore, the answer to the above question takes the following form. Theorem 3.5. Every ﬁnite frame and its dual is the orthogonal projection of a biorthogonal system (onto an invariant subspace of frame operator). Moreover, the two bases making up the biorthogonal system can be taken to have the same frame bounds as their projections, i.e., the frame and its dual.

Proof. Let (φ j) j J be a frame for H , with frame bounds A and B. This will be the ∈ projection of a basis (φ j + ψ j) j J in a larger Hilbert space H L . To see what this might be, consider∈ the condition that H be⊕ invariant under the φ ψ frame operator S = S(φ j+ψ j) of ( j + j), i.e.,

Sf = ∑ f ,φ j + ψ j (φ j + ψ j)= ∑ f ,φ j φ j + ∑ f ,φ j ψ j H , f H . j j j ∈ ∀ ∈

This is precisely the condition that the frames be orthogonal (Deﬁnition 3.5), i.e.,

∑ f ,φ j ψ j = 0, f H ∑ g,ψ j φ j = 0, g L , (3.22) j ∀ ∈ ⇐⇒ j ∀ ∈ which implies (cf Theorem 5.1)

S φ ψ ( f + g)= S φ ( f )+ S ψ (g), f H , g L . ( j+ j) ( j) ( j) ∀ ∈ ∈

Hence the dual frame to ( f j), f j := φ j + ψ j, is

1 1 1 f˜j = S− (φ j + ψ j)= S− φ j + S− ψ j = φ˜j + ψ˜ j, (φ j+ψ j) (φ j) (ψ j) and the orthogonal projections of f j and f˜j onto H are φ j and φ˜j. Further, (3.22) implies

2 2 2 ∑ f + g,φ j + ψ j = ∑ f ,φ j + ∑ g,ψ j , f H , g L , j | | j | | j | | ∀ ∈ ∈ so that (φ j + ψ j) will have the desired frame bounds (A and B) provided that 3.6 Frames as orthogonal projections 45

2 2 2 A g ∑ g,ψ j B g , g L . (3.23) ≤ j | | ≤ ∀ ∈

Let V :=[φ j]. The bijection [φ j + ψ j] : ℓ2(J) H L maps the kernel of V onto L (by a dimension count). Hence, we take → ⊕

L := ker(V)= ran(V ∗)⊥, and it therefore remains only to ﬁnd ψ j L = ker(V) satisfying (3.22) and (3.23). ∈ Let Q : L ℓ2(J) be positive (self adjoint), with range L , and eigenvalues between √A and→√B, e.g., Qg := λg, g, where √A λ √B. Then ∀ ≤ ≤

ψ j := Q∗e j, where (e j) j J is the standard basis for ℓ2(J), satisﬁes ∈

∑ g,ψ j φ j = ∑ g,Q∗e j φ j = ∑ Qg,e j φ j = V(Qg)= 0, j j j

2 2 2 2 2 2 A g ∑ g,ψ j = ∑ g,Q∗e j = ∑ Qg,e j = Qg B g , ≤ j | | j | | j | | ≤ which completes the proof. ⊓⊔ The sum (φ j + ψ j) is the motivating example of a direct sum (see 5.2). The above argument holds with L a proper subspace of ker(V), with the only§ difference being that (φ j + ψ j) is not a basis. It also extends to frames for inﬁnite dimensional H , where the role of a basis (and dual basis) is replaced by that of a Riesz basis (and dual Riesz basis). We give the relevant facts (cf [Chr03]).

Deﬁnition 3.6. A sequence of vectors ( f j) is a Riesz basis for a Hilbert space H if it is a perturbation of an orthonormal basis (e j) for H , i.e., there is a bounded invertible operator T : H H , such that f j = Te j, j. → ∀ Equivalently, ( f j) is a Riesz basis if it is an unconditional basis with

0 < inf f j sup f j < ∞. j ≤ j

From the orthogonal expansion, we have

1 1 1 f = T(T − f )= T ∑ T − f ,e j e j = ∑ f ,(T − )∗e j Te j, j j

1 and similarly for T replaced by (T − )∗. This gives the biorthogonal expansion

f = ∑ f ,g j f j = ∑ f , f j g j, f H , j j ∀ ∈

1 1 where g j :=(T ∗)− e j =(T − )∗e j is called the dual Riesz basis (cf Example 3.4). 46 3 Frames 3.7 Condition numbers and the frame bounds

The frame bounds for ( f j) imply, and are equivalent to, a number of similar bounds for the various maps obtainable from V =[ f j] (see Exer. 3.24).

Proposition 3.3. Let Φ =( f j) j J be a ﬁnite sequence in H , and V =[ f j], S = VV ∗. The frame bounds (3.1) are equivalent∈ to the inequalities

A f Sf B f , f H , (3.24) ≤ ≤ ∀ ∈

√A f V ∗ f √B f , f H , (3.25) ≤ ≤ ∀ ∈ A c Gram(Φ)c B c , c ran(Gram(Φ)), (3.26) ≤ ≤ ∀ ∈ √A c Vc √B c , c ran(V ∗), (3.27) ≤ ≤ ∀ ∈ which are sharp if and only if A and B are the optimal frame bounds, respectively.

The frame expansion (3.5) of a function f H involves the representation ∈ 1 f = Vc = ∑ c j f j, c = V ∗S− f ran(V ∗). j J ∈ ∈ When f is constructed from c in this way, there will be rounding errors which yield a perturbed vector c+δc. The perturbation will have a component δa in ran(V ∗), and a component δb ker(V)= ran(V ) . Hence, we will obtain a perturbed function ∈ ∗ ⊥

f + δ f = V(c + δc)= V(c + δa + δb)= Vc +V(δa), δa ran(V ∗). ∈ By (3.27), the relative error in the computed f satisﬁes

√A δa δ f √B δa √B δc . (3.28) √B c ≤ f ≤ √A c ≤ √A c where A and B are the frame bounds. This estimate motivates the following.

Deﬁnition 3.7. The condition number (of the frame expansion) of a frame Φ, with frame bounds A and B, is B cond(Φ) := 1. A ≥ We observe that a frame has condition number 1 if and only if it is tight, and that a frame and its dual have the same condition number, since

1/√A √B cond(Φ˜ )= = = cond(Φ). 1/√B √A

In view of (3.28), the relative error in the computed value f +δ f of f is bounded by the condition number times the relative error in the frame coefﬁcients c. If Φ is 3.8 Normalising frames and the distances between them 47

1 a basis, then cond(Φ)= V V − , the usual condition number of a basis, and V has kernel 0, so that δa =δc, and we obtain the lower estimate on the relative error 1 δc δ f . cond(Φ) c ≤ f In contrast, for a frame that is not a basis, a nonzero perturbation δc ker(V) in the coefﬁcients leads to no error in the constructed f . ∈ Another measure of how errors propagate in calculations with a frame (and its dual) is the condition number of the frame operator S = SΦ , i.e.,

1 B 2 cond(SΦ ) := S S− = = cond(Φ) 1. A ≥

Thus, either cond(Φ) or cond(SΦ ), and its distance from 1, can be used as a measure of how well conditioned calculations with the frame Φ are.

Both the redundancy and the tightness (as measured by the condition number) of a frame make it better conditioned than a basis.

3.8 Normalising frames and the distances between them

Here we consider whether or not there is a natural normalisation of a frame, which extends that of a tight frame. Since the canonical tight frame of a tight frame is its normalised version (which is its own dual), we would hope that a normalised frame and its dual are both close, in some sense, to the canonical tight frame. There does seem to be one normalisation which captures all the properties one might reasonably hope for, and so we proceed directly to it. We then investigate how close this normalised frame and its dual are to the canonical tight frame for the following metrics

dist(Φ,Ψ)= Gram(Φ) Gram(Ψ) (frames up to unitary equivalence) − 1 distB(Φ,QΦ)= log(max( I Q , I Q− )+ 1) (for similar frames). − − If a frame Φ =( f j) with frame bounds A and B is multiplied by a scalar c > 0, Ψ Φ Ψ˜ 1 Φ˜ then the resulting frame = c =(cf j) and its dual = c have frame bounds 2 2 1 1 c A,c B and c2B , c2A , respectively. Thus there is a unique scaling which ensures that a frame and its dual have the same optimal frame bounds, i.e., 1 1 1 c2A = , c2B = c = . c2B c2A ⇐⇒ √4 AB 48 3 Frames

Deﬁnition 3.8. We say that a frame Φ is normalised if its optimal frame bounds A and B satisfy AB = 1. We emphasize this is simply a normalising factor:

If Φ =( f j) is a frame, with optimal frame bounds A and B, then 1 1 Φ =( f j) √4 AB √4 AB is the unique positive scalar multiple of it which is a normalised frame.

After this normalisation (used in Figures 3.1 and 3.2), the frame bounds are

√A 1 √B = , = cond(Φ). (3.29) √B cond(Φ) √A Hence being normalised is equivalent to either of

SΦ = SΦ˜ , Gram(Φ) = Gram(Φ˜ ) . A natural distance on the unitarily equivalent frames (where the non unit scalar multiples are not identiﬁed) is given by

dist(Φ,Ψ) := Gram(Φ) Gram(Ψ) , − where is the induced (spectral) norm. Let λ1,...,λd > 0 be the eigenvalues of the frame operator SΦ , Then (see Exer. 3.25)

Φ 1Φ˜ 2λ 1 λ 1 dist(c , )= ScΦ S 1 Φ˜ = max c j j− , (3.30) c − c 1 j d | − c2 | ≤ ≤ can 2 dist(cΦ,Φ )= ScΦ SΦcan = max c λ j 1 , (3.31) − 1 j d | − | ≤ ≤ 1Φ˜ Φcan 1 λ 1 dist( , )= S 1 Φ˜ SΦcan = max j− 1 . (3.32) c c − 1 j d |c2 − | ≤ ≤ The following inequality shows that (3.30) is minimised by a unique c > 0. Lemma 3.1. The following inequality holds 1 1 1 B A max A , λ , B − , 0 < A λ B, { − A − λ − B }≥ √AB ≤ ≤ with equality if and only if AB = 1. 1 Proof. Since λ λ λ is increasing for λ > 0, and zero at λ = 1, the maximum 1 → − 1 B A is either A or B , which is bounded below by − (see Exer. 3.26). | − A | | − B | √AB ⊓⊔ 3.8 Normalising frames and the distances between them 49

Theorem 3.6. Let Φ be a ﬁnite frame with optimal frame bounds A and B, then B A dist(Φ,Φ˜ )= Gram(Φ) Gram(Φ˜ ) = SΦ SΦ˜ − , − − ≥ √AB with equality if and only if Φ is normalised.

Proof. Let A = λ1 λd = B be the eigenvalues of SΦ . Then by Exer. 3.25 (with α = 1, β = ≤≤1, γ = 0) and Lemma 3.1, we have − 1 B A Gram(Φ) Gram(Φ˜ ) = SΦ SΦ˜ = max λ j − , − − 1 j d − λ j ≥ √AB ≤ ≤ with equality if and only if AB = 1, i.e., when Φ is normalised. ⊓⊔ B We note that any function of the ratio A of the optimal frame bounds, such as the condition number, and B A B A − = , √AB A − B is invariant under any scaling of the frame. We now investigate the minima of the distances (3.31) and (3.32) of cΦ and its 1 Φ˜ Φcan dual c , from the canonical tight frame . They are the same 1 B A mindist(cΦ,Φcan)= mindist( Φ˜ ,Φcan)= − < 1, c>0 c>0 c A + B

2 2 2 A+B which occur for c = A+B and c = 2AB , respectively (see Exer. 3.27). These values for c are equal if and only if A = B, i.e., the frame is tight. Hence, when Φ is not tight, the scaling which makes Φ closest to Φcan does not make Φ˜ closest to Φcan. Thus we seek to minimise the maximum of these distances. The minimum is B 1, which occurs once (see Exer. 3.27), when A − 2 A + B 1 c4 = = cΦ and 1 Φ˜ are normalised. A + B 2AB AB ⇐⇒ c In this case, and only this case, the distances are equal:

Theorem 3.7. Let Φ be a ﬁnite frame with optimal frame bounds A and B, then

Gram(Φ) Gram(Φcan) = Gram(Φ˜ ) Gram(Φcan) , − − which is equivalent to SΦ I = SΦ˜ I , − − if and only if Φ is normalised. In this case, the common distance is

B dist(Φ,Φcan)= dist(Φ˜ ,Φcan)= cond(Φ) 1 = 1. (3.33) − A − 50 3 Frames

Proof. Let A = λ1 λd = B be the eigenvalues of SΦ . For λ > 0, the functions 1≤≤1 λ λ 1 and λ λ 1 decrease to 0 at λ = 1 and then increase, and so take → | − | → | − | 1 1 their maxima over [A,B] and [ B , A ] at an endpoint. Thus (see Exer. 3.25), we obtain can Gram(Φ) Gram(Φ ) = SΦ I = max λ j 1 = max A 1 , B 1 , − − j | − | | − | | − | can 1 1 1 Gram(Φ˜ ) Gram(Φ ) = SΦ˜ I = max 1 = max 1 , 1 . − − j |λ j − | A − B − By Exer. 3.28, these are equal if and only if AB = 1, i.e., Φ is normalised. The frame bounds of Φ after normalisation are given by (3.29), and so the common distance is

√A √B √B √A √B √A √B max 1 , 1 = max − , − = 1. √B − √A − √B √A √A − ⊓⊔ We now summarise our results so far:

If Φ is a ﬁnite frame with optimal frame bounds A and B, then the following are equivalent 1. Φ is normalised, i.e., AB = 1. 2. Φ and Φ˜ have the same frame bounds. 3. SΦ = SΦ˜ , i.e., Gram(Φ) = Gram(Φ˜ ) . B A B A 4. SΦ SΦ˜ = − , i.e., Gram(Φ) Gram(Φ˜ ) = − . − √AB − √AB 5. SΦ I = SΦ˜ I . 6. Gram− (Φ) Gram− (Φcan) = Gram(Φ˜ ) Gram(Φcan) . − −

There is a unique closest tight frame to a normalised frame Φ and its dual Φ˜ (individually and simultaneously) amongst all tight frames similar to Φ.

Theorem 3.8. Let Φ be a ﬁnite frame which is normalised, i.e., its optimal frame 1 bounds satisfy B = A , then

1 A2 min max dist(Φ,Ψ),dist(Φ˜ ,Ψ) = − . Ψ tight { } 2A Ψ=QΦ

√A2 1 This is attained for Ψ = + Φcan (up to unitary equivalence), for which √2A

1 A2 1 A dist(Φ,Ψ)= dist(Φ˜ ,Ψ)= − − = dist(Φ,Φcan)= dist(Φ˜ ,Φcan), 2A ≤ A with equality in the above if and only if Φ is tight.

Proof. If Ψ is tight and is similar to Φ, then it is similar to Φcan, and so 3.8 Normalising frames and the distances between them 51

Ψ = cUΦcan, c > 0, U unitary,

(see Exer. 2.5). For this Ψ, Gram(Ψ)= c2 Gram(Φcan). Hence (see Exer. 3.27) 1 dist(Φ,Ψ)= dist(Φ˜ ,Ψ)= max A c2 , c2 . | − | A − 1 A2 This function of c has a minimum value of −2A , which occurs precisely when

A + 1 A2 + 1 c2 = A = . 2 2A

(the minimum must occur when A c2 = 1 c2 ). By (3.33), we have | − | | A − | 1 A2 √B 1 A − dist(Φ,Φcan)= dist(Φ˜ ,Φcan)= 1 = − . 2A ≤ √A − A

1 A 1 A2 (A 1)2 Since − − = − , this this inequality is sharp, unless Φ is tight. A − 2A 2A ⊓⊔ Another notion of distance between (certain) frames is based on the theory of perturbations of orthonormal bases (called Riesz bases). We follow [Bal99]. A frame Ψ =(g j) j J is close to a frame Φ =( f j) j J if there exists a λ 0 such that ∈ ∈ ≥

∑ c j(g j f j) λ ∑ c j f j , c ℓ2(J), j J − ≤ j J ∀ ∈ ∈ ∈ with the inﬁmum over such λ called the closeness bound, and denoted by cl(Ψ,Φ). We say that Φ and Ψ are near if Φ is close to Ψ and Ψ is close to Φ. This is an equivalence relation, and a metric distB called the quadratic distance can be deﬁned on all frames which are near each other by

distB(Φ,Ψ) := log(max cl(Φ,Ψ),cl(Ψ,Φ) + 1). { } Frames Φ and Ψ are near if and only if they are similar (see Exer. 3.30), i.e., Φ = QΨ, for some invertible Q, in which case

1 distB(Φ,Ψ) := log(max Q I , Q− I + 1). (3.34) { − − }

Unlike dist(Φ,Ψ)= Gram(Φ) Gram(Ψ) , distB is scaling invariant, i.e., −

distB(cΦ,cΨ)= distB(Φ,Ψ), c = 0. ∀ A frame, its dual, and canonical tight frame are all near. By (3.34), we have

1 2 1 1 distB(cΦ, Φ˜ )= log(max c SΦ I , SΦ− I + 1), c { − c2 − }

1 1 can 1 can 2 1 2 distB(cΦ,Φ )= distB( Φ˜ ,Φ )= log(max cSΦ I , SΦ− I + 1). c { − c − } 52 3 Frames

By minimising these distances over all c > 0, we obtain the following analogue of Theorems 3.6 and 3.7. Theorem 3.9. Let Φ be a ﬁnite frame with optimal frame bounds A and B, then 1 distB(Φ,Φ˜ ) (logB logA), (3.35) ≥ 2 −

can can 1 distB(Φ,Φ )= distB(Φ˜ ,Φ ) (logB logA), (3.36) ≥ 4 − with equality in each if and only if Φ is normalised. Proof. Since log is strictly increasing, it sufﬁces to minimise the maxima in these distances. By Exer. 3.25 (and a slight variation of Exer. 3.27), these are

2 1 1 2 2 1 1 max c SΦ I , SΦ− I = max c A 1 , c B 1 , 1 , 1 , { − c2 − } | − | | − | c2A − c2B − 1 1 2 1 2 1 1 max cSΦ I , SΦ− I = max c√A 1 , c√B 1 , 1 , 1 . { − c − } | − | | − | c√A − c√B − By Exer. 3.27 (with appropriate changes of variables) the respective minima of these 1 B 1 B are B/A 1 and √B/√A 1, giving the distB distances log( ) and log( ), − − 2 A 4 A which are attained if and only if c2 = 1 , i.e., cΦ and 1 Φ˜ are normalised. √AB c ⊓⊔ The analogue of Theorem 3.8 is as follows. Theorem 3.10. Let Φ be a ﬁnite frame with optimal frame bounds A and B, then 1 min distB(Φ,Ψ)= (logB logA), Ψ tight 4 − Ψ=QΦ

with equality if Ψ = √4 ABΦcan (and possibly other unitarily equivalent frames). Thus, if Φ is normalised, then Φcan is a best tight frame approximation to it.

Proof. Since Ψ is tight and similar to Φcan, we have Ψ = cUΦcan, where c > 0 and U is unitary. Hence

1 1 2 2 1 distB(Φ,Ψ)= log(max cUSΦ− I , (cUSΦ− )− I + 1). { − − } By Exer. 3.23, the maximum above is only made smaller by choosing U = I, and 1 1 2 1 2 so, effectively, we need to minimise max cSΦ− I , SΦ I , i.e., { − c − } c c √A √B max 1 , 1 , 1 , 1 . { √A − √B − c − c − 4 1 By Exer. 3.27, this is minimised if and only if c = √AB , with minimum (B/A) 4 1, can 1 B − which gives the minimal distance distB(Φ,cΦ )= log( ). 4 A ⊓⊔ 3.9 Approximate inverses of the frame operator 53 3.9 Approximate inverses of the frame operator

Calculations with a frame Φ require the inverse of the frame operator SΦ , e.g., to determine the dual frame. This can be done numerically using standard iterative algorithms for the calculating the inverse based on an approximate inverse.

Deﬁnition 3.9. An approximate (left) inverse of bounded linear map S on H is a bounded linear map M on H for which

IH MS < 1. − If M is an approximate left inverse of S, then MS is boundedly invertible. Hence 1 1 1 if S is invertible, then S− can be calculated as S− =(MS)− M (see Exer. 3.31). Let A and B be known, but possibly not optimal frame bounds for a frame Φ. 2 Φ Then A+B IH is an approximate inverse of S = S , with 2 B A I S − < 1. − A + B ≤ A + B This leads to the standard ﬁxed point iteration method for ﬁnding the solution of Sg = h (see Exer. 3.32) 2 2 g0 := 0, gk 1 := gk Sgk + h, k = 0,1,2,..., + − A + B A + B with gn converging to g at the rate

n n 1 B A h B A gn g S− h − − . − ≤ A + B ≤ A A + B 1 In particular, taking h = f gives a sequence converging to S− f , and the choice h = Sf , i.e., g = f , leads to the so called frame algorithm 2 g0 := 0, gk 1 := gk + S( f gk), k = 0,1,2,.... (3.37) + A + B − The frame algorithm, which requires some estimate for the frame bounds, allows f to be reconstructed from the coefﬁcients f , f j without calculating the dual frame. It can be accelerated by standard techniques, including the Chebyshev method and the conjugate gradient method (see [Gro93] for a detailed analysis). 1 It is also possible to calculate S− 2 (and hence the canonical tight frame) numerically in terms of S, via the absolutely convergent series expansion

∞ 1 2 (2 j)! 2 j 2 S− = ∑ 2 j 2 I S A + B j=0 2 ( j!) − A + B (see Exer. 3.33 for details). 54 3 Frames 3.10 Alternate duals

The notion of a basis and its dual functionals can be further generalised as follows.

Deﬁnition 3.10. Finite frames ( f j) and (g j) for H , with synthesis operators V and W, are said to be dual if VW ∗ = I, i.e., they give the reconstruction formula

f = ∑ f ,g j f j = ∑ f , f j g j, f H . (3.38) j j ∀ ∈

Since (VW ∗)∗ = WV ∗, being dual does not depend on the order of the frames. The set A = AΦ of all frames dual to a given ﬁnite frame Φ =( f j) j J for H is an afﬁne subspace of H J, which contains the (canonical) dual frame Φ˜∈. The elements of A are commonly referred to as dual frames, and those of A Φ˜ as alternate (or noncanonical) dual frames of the frame Φ. \

Example 3.13. Let ( f j) and (g j) be a biorthogonal system for K , and P be the orthogonal projection onto a subspace H of K , then (Pf j) and (Pg j) are dual frames for H . Moreover, (Pg j) is the canonical dual of (Pf j) if and only if H is an invariant subspace of the frame operator for ( f j). See Exer. 3.21 for details.

A ﬁnite frame ( f j) is dual to (g j), where we write g j = f˜j + h j, if and only if

[ f j][g j]∗ =[ f j][ f˜j]∗ +[ f j][h j]∗ = I +[ f j][h j]∗ = I [ f j][h j]∗ = 0, ⇐⇒

i.e., the frame (h j) is orthogonal to ( f j), in the sense of Deﬁnition 3.5, see (3.20). Some conditions equivalent to being dual include the following.

Proposition 3.4. Let Φ =( f j) j J and Ψ =(g j) j J be a ﬁnite frames for H , with ∈ ∈ 1 synthesis operators V and W. Let S = VV ∗, and PΦ = V ∗S− V be the canonical Gramian of Φ. Then the following are equivalent (a) Φ and Ψ are dual. (b) W is a left inverse of V ∗, i.e.,WV ∗ = I. (c) W ∗ is a right inverse of V, i.e., VW ∗ = I. (d) Ψ Φ˜ is orthogonal to Φ, i.e., g j = f˜j + h j, with (h j) orthogonal to ( f j). − 1 (e) ran((W S− V)∗) ran(V ∗). −1 ⊥ (f) W = S− V + L(I PΦ ), where L : ℓ2(J) H is a linear operator. 1 − → (g) W ∗ = V ∗S− +(I PΦ )R, where R : H ℓ2(J) is a linear operator. − 2 → (h) Q = V ∗W is a projection, i.e., Q = Q.

Proof. (a) (b) (c) Use VW ∗ f = ∑ j f ,g j f j and (VW ∗)∗ =(WV ∗). ⇐⇒ ⇐⇒ 1 (d),(e) (a) As observed (Ψ Φ˜ and Φ have synthesis maps W S− V and V). (b)= ⇐⇒(f) Suppose that WV =−I, and take L = W. Then we have − ⇒ ∗ 1 1 1 S− V + L(I PΦ )= S− V +W (WV ∗)S− V = W. − − 1 1 (f)= (b) We have WV ∗ =(S− V + L(I V ∗S− V))V ∗ = I + L(V ∗ V ∗)= I. (f) ⇒ (g) Take adjoints. − − ⇐⇒ 3.10 Alternate duals 55

2 (b)= (h) If WV ∗ = I, then (V ∗W) = V ∗(WV ∗)W = V ∗W. ⇒ 2 (h)= (b) If (V ∗W) = V ∗W, then V(V ∗WV ∗W)W ∗ = V(V ∗W)W. Since VV ∗ and WW⇒are invertible (Φ and Ψ are frames), they cancel to give WV = I. ∗ ∗ ⊓⊔ Corollary 3.4. Let Φ be a frame of n vectors for H , where d = dim(H ). Then the afﬁne subspace A of all frames dual to Φ has dimension d(n d). In particular, there exist alternate dual frames if and only if Φ is not a basis. −

Proof. By (3.13), we have that ker(I PΦ )= ran(PΦ )= ran(V ). Thus by (f), the − ∗ dual frames are in 1–1 correspondence with linear maps L K : K H , where | → K = ran(V ∗)⊥ = ker(V), dim(K )= n d. The space of linear maps K H is isomorphic to the subspace A S 1V, and− so A has dimension d(n d). → − − − ⊓⊔

Let Φ =( f j) be a ﬁnite frame for H , V =[ f j], with canonical Gramian PΦ . Then all frames (g j) that are dual to ( f j) are given by

W =[g j]=[ f˜j]+ L(I PΦ ), (3.39) − where L : ker(V) H and PΦ = V (VV ) 1V. → ∗ ∗ −

Example 3.14. (Exer. 3.34) Let Φ =( f1, f2, f3) be the tight frame of three equally spaced unit vectors in R2. Then the afﬁne space AΦ of all duals of Φ has dimension d(n d)= 2(3 1)= 2, and is given by − − 2 ( f˜1 + w, f˜2 + w, f˜3 + w), w R . ∈

Fig. 3.3: The tight frame of three equally spaced vectors for R2, and some alternate duals. Here the vector w of Example 3.14 is depicted with a hollow arrow head .

If (g j) is dual to a ﬁnite frame ( f j) for H , then taking c j = f ,g j in (3.6) gives 2 2 2 ∑ f ,g j = ∑ f , f˜j + ∑ f ,g j f˜j , f H . j | | j | | j | − | ∀ ∈

2 Thus the canonical dual is characterised by minimising ∑ f ,g j , f H . j | | ∀ ∈ 56 3 Frames

In a similar vein, taking the Frobenius norm: A 2 := trace(AA ) of the formula F ∗ (3.39) for a dual frame (g j) of ( f j), using (3.12), i.e., V = VPΦ , gives

2 2 1 1 ∑ g j = W F = trace (S− VPΦ + L(I PΦ ))(PΦV ∗S− +(I PΦ )L∗) j − − 1 2 2 ˜ 2 2 = S− V F + L(I PΦ ) F = ∑ f j + L(I PΦ ) F . − j − (3.40)

Thus the canonical dual frame to ( f j) is the unique dual frame (g j) of ( f j) 2 which minimises ∑ g j . j

Example 3.15. (Exer. 3.35) Let Φ =( f j) be a ﬁnite frame for H , V =[ f j], and Q : ℓ2(J) ℓ2(J) be an invertible linear map. Then → 1 W =[g j]=(VQ∗QV ∗)− VQ∗Q

1 gives a frame (g j) dual to ( f j), which uniquely minimises WQ F , W =[g j]. For − D an invertible ﬁrst–order difference operator, the (alternate) dual frame (g j) which r minimises the Sobolev–type norm W(D )∗ F is called the r–th order Sobolev dual. These are motivated by Σ∆–quantisation (see [BLPY10]). The canonical tight frame can also be characterised in terms of similarity: Proposition 3.5. No two distinct duals of a ﬁnite frame Φ are similar to each other. Thus the canonical dual Φ˜ is the unique dual of Φ which is similar to Φ.

Proof. Suppose that (g j) and (Qg j) are similar frames, which are duals of Φ =( f j). Then Q∗ f = ∑ Q∗ f ,g j f j = ∑ f ,Qg j f j = f , f H , j j ∀ ∈ so that Q = I. Hence Φ˜ = S 1Φ is the only dual which is similar to Φ. − ⊓⊔ For ﬁnite frames ( f j) and (g j) for H , with synthesis operators V and W, the condition VW ∗ = I for being dual can be weakened as follows: ( f j) and (g j) are approximately dual frames if VW I < 1, ∗ − ( f j) and (g j) are pseudo–dual frames if VW ∗ is invertible. It is easy to verify that these notions do not depend on the frame order, and that the pairs of frames satisfying the different dualities satisfy the inclusions:

canonically dual dual approximately dual pseudo–dual. ⊂ ⊂ ⊂ Example 3.16. The set of all frames approximately dual to a given ﬁnite frame Φ is a convex set. If Φ has optimal frame bounds A and B, then cΦ is approximately Φ 2 dual to for any scalar 0 < c < B (see Exer. 3.37). 3.11 Oblique duals 57

Example 3.17. If two frames are similar, then their synthesis operators are related W = QV, where Q is invertible. Since VW ∗ =(VV ∗)Q∗ is invertible, it follows that similar frames (for the same space) are pseudo–duals. They may not necessarily approximate duals (see Example 3.16).

3.11 Oblique duals

The dual frame expansion (3.38) can be generalised by allowing the coefﬁcients c j = f ,g j to be given by vectors g j from outside the space H . For example, in signal processing one may try and choose the g j to be outside some subspace for which measurements are known to be corrupted by noise.

Deﬁnition 3.11. Let ( f j) and (g j) be ﬁnite frames for subspaces V and W of H , 3 and V :=[ f j], W :=[g j]. Then (g j) is an oblique dual of ( f j) if VW V = IV , i.e., ∗|

f = ∑ f ,g j f j, f V . j ∀ ∈

Example 3.18. Let Vk(µ) be the space of multivariate orthogonal polynomials of degree k for a measure µ (see 10.10), and Q be the orthogonal projection onto it. § Let ( f j) be a frame for Vk(µ), and g j be leading term of f˜j (its homogeneous term of degree k), so that (g j) is a frame for the homogeneous polynomials of degree k. Since g j = Q f˜j and Qf = f , f Vk(µ), we have ∈

∑ f ,g j f j = ∑ f ,Q f˜j f j = ∑ f , f˜j f j = f , f Vk(µ), j j j ∀ ∈

i.e., (g j) is an oblique dual of ( f j).

Example 3.19. Suppose ( f j) is a ﬁnite frame for V , and (gˆ j) V is a dual frame. ⊂ Let Q be the orthogonal projection onto V . For h j V = H V , deﬁne ∈ ⊥ ⊖

g j := gˆ j + h j, W := span g j . { }

Then (g j) is an oblique dual of ( f j), since Qg j = gˆ j gives

f = ∑ f ,gˆ j f j = ∑ f ,Qg j f j = ∑ Qf ,g j f j = ∑ f ,g j f j, f V . j j j j ∀ ∈

Similarly, if (g j) W is an oblique dual, then (Qg j) is a dual frame for ( f j). It ⊂ 2 3 may be that dim(W ) > dim(V ). For example, let V = R 0 R , and ( f j) be the × ⊂ normalised tight frame for V given by three equally spaced vectors. Then ( f˜j +ce3), 3 c = 0, e3 =(0,0,1) is an oblique dual where W = R .

3 The term pseudodual is also used in the literature, e.g., see [LO04]. 58 3 Frames

Let Φ =( f j) be a ﬁnite frame for V , V =[ f j], with canonical Gramian PΦ . Then all frames (g j) that are oblique duals of ( f j) are given by

W =[g j]=[ f˜j]+ L(I PΦ )+[h j], (3.41) − 1 where L : ker(V) H , PΦ = V (VV ) V, and h j V . → ∗ ∗ − ∈ ⊥

The condition of being an oblique dual is not symmetric, i.e., if (g j) is an oblique dual of ( f j), then ( f j) may or may not be an oblique dual of (g j) (see Example 3.19). For (g j) a frame for W to be an oblique dual of a frame ( f j) for V , we must have

V W ⊥ = 0. ∩ Otherwise 0 = f V W ⊥ = f = ∑ f ,g j f j = 0. ∈ ∩ ⇒ j Thus a necessary condition to ensure being an oblique dual is symmetric is that

V W ⊥ = W V ⊥ = 0, ∩ ∩ which is equivalent to the algebraic direct sums

H = V a W ⊥ = W a V ⊥, ⊕ ⊕ and implies that dim(V )= dim(W ) (see Exer. 3.38). We now show this condition is sufﬁcient. For an algebraic direct sum H = V a W the oblique projection of ⊕ ⊥ H onto V along W ⊥ is the linear map P = PV W on H given by , ⊥

P V = IV , P(W ⊥)= 0. |

Proposition 3.6. Let ( f j) and (g j) be finite frames for subspaces V and W of H , with V W = W V = 0. Then the following are equivalent ∩ ⊥ ∩ ⊥ 1. The oblique projection PV W of H onto V is given by PV W f = ∑ j f ,g j f j. , ⊥ , ⊥ 2. The oblique projection PW V of H onto W is given by PW V f = ∑ j f , f j g j. , ⊥ , ⊥ 3. (g j) is an oblique dual of ( f j), i.e., f = ∑ f ,g j f j, f V . j ∀ ∈ 4. ( f j) is an oblique dual of (g j), i.e., f = ∑ f , f j g j, f W . j ∀ ∈ Proof. The condition V W ⊥ = W V ⊥ = 0 ensures PV W and PW V are well ∩ ∩ , ⊥ , ⊥ defined, and satisfy (PV W )∗ = PW V (see Exer. 3.38). Let V =[ f j] and W =[g j]. , ⊥ , ⊥ 1. 2. These conditions are PV W = VW ∗, PW V = WV ∗, which are equivalent. ⇐⇒ , ⊥ , ⊥ 1.= 3. If PV W = VW ∗, then f = VW ∗ f = f ,g j f j, f V . ⇒ , ⊥ ∀ ∈ 3.= 1. Define P on H by Pf := ∑ f ,g j f j, then Pf = f , f V and Pf = 0, ⇒ j ∀ ∈ f W ⊥ (since then f ,g j = 0), so that P = PV W . ∀ ∈ , ⊥ 2. 4. Interchange ( f j) and (g j) in the argument for 1. 3.. ⇐⇒ ⇐⇒ ⊓⊔ 3.11 Oblique duals 59

For a given ﬁnite frame for V there is a canonical oblique dual from W . Theorem 3.11. Suppose that V and W are subspaces of H , with

V W ⊥ = W V ⊥ = 0, ∩ ∩ and ( f j) is a ﬁnite frame for V . Then the unique coefﬁcients c j = c j( f ) satisfying

PV W f = ∑c j f j, f H (3.42) , ⊥ j ∈ which have minimal ℓ2–norm are given by

† c =(W ∗V) W ∗ f , V :=[ f j], (3.43) where W is the synthesis operator of any frame for W . † Proof. It is easy to verify (take Λ = W ∗ in Exer. 3.40) that PV W = V(W ∗V) W ∗, , ⊥ and so we seek a minimum norm solution c to

† Vc = V(W ∗V) W ∗ f .

† † † The unique such c is given by c = V V(W ∗V) W ∗ f . Since V V is the orthogonal † projector onto the range of V ∗, and ran((W ∗V) )= ran(V ∗W) ran(V ∗), we can simplify this to c =(W V)†W f . ⊂ ∗ ∗ ⊓⊔ For the c j( f )= f ,g j of (3.43), the frame (g j) for W , which is given by † † [g j]=((W ∗V) W ∗)∗ = W(V ∗W) , (3.44) is called the canonical oblique dual of ( f j) in W . As would be hoped, the canonical oblique dual of this (g j) in V is ( f j) (see Exer. 3.39). Example 3.20. If V = W = H and W = V, then (3.44) gives (cf Exer. 3.9)

† † [ f˜j]= V(V ∗V) = V Gram(Φ) , Φ =( f j).

Example 3.21. In signal processing, (3.42) can be used as follows (see [Eld03]). Let P be the orthogonal projection onto W . Then c j( f )= f ,Pg j = Pf ,g j , so that the sampling of a signal f can be done by ﬁrst projecting it onto thesampling space W (where signals can be accurately measured), then measuring the projected signal Pf .

Corollary 3.5. Suppose that Φ =( f j) is a ﬁnite sequence in H , with V :=[ f j], and S := VV : H H (which may not be invertible). Then ( f j) is a ﬁnite frame for ∗ → V := span f j , with the canonical dual frame ( f˜j) given by { } † † [ f˜j]= V Gram(Φ) = S V.

Proof. Take W = V in (3.44), and use V Gram(Φ)† = S†V (see Exer. 3.11). ⊓⊔ 60 3 Frames Notes

An early appearance of the canonical tight frame was in Lowdin’s¨ well known work in Quantum Chemistry in the late 1940’s, where he constructed “orthonormalised atomic orbitals” from a basis of orbitals (see [Low70],¨ [AEG80]). This also known as Schweinler–Wigner orthogonalisation (see [SW70]). Symmetric Gram–Schmidt methods for inﬁnite dimensional spaces are explored in [FPT02]. The material of this chapter extends to inﬁnite dimensional spaces, see [HL00], [Chr03]. Source material and further reading includes: dual frames [Li95], oblique dual frames [Eld03], [CE04], [LO04], approximate dual frames [CL10], [LY09], and optimal dual frames for erasures [LH10].

Exercises

3.1. Rank one projections. Let A j : H H : f f ,g j f j, where f j,g j H . → → ∈ (a) Show that trace(A j)= f j,g j . (b) Show that A = A j is a scalar multiple of a projection, i.e., A = cP, where c F 2 ∈ and P = P, provided f j,g j = 0. (c) Show P is orthogonal if and only if f j and g j are multiples of each other. (d) Show that Hilbert–Schmidt (Frobenius) inner product between them is

trace(A jA∗)= f j, fk gk,g j . k

3.2. Least squares solution. Let A := c ℓ2(J) : Vc = f be all possible sequences of coefficients from which { ∈ } f H can be reconstructed f = ∑ j c jv j, where ( f j) spans H . (a)∈ Show A is an affine subspace, i.e., λa +(1 λ)b A , a,b A , λ R. 1 − 1 ∈ ∀ ∈ ∈ (b) Show that c = V ∗S− f A . Here c j = f ,S− f j , where S = VV ∗. (c) Since affine subspaces are∈ translates of linear subspace s, (a) and (b) give

1 A = V ∗S− f + ker(V).

1 Show that c = V ∗S− f is the unique solution to f = Vc of minimal ℓ2–norm.

3.3. Pseudoinverse. Take the deﬁnition of the pseudoinverse of A : H K to be the unique linear map A† : K H satisfying → → AA†, A†A are Hermitian, AA†A = A and A†AA† = A†.

† 1 (a) Show that if A is onto, then A = A∗(AA∗)− . In particular, for V =[ f j] j J the synthesis operator of a ﬁnite spanning set for H ∈

† 1 1 V = V ∗(VV ∗)− = V ∗S− , S := VV ∗. 3.11 Oblique duals 61

(b) Show that if P is an orthogonal projection P, then P† = P. (c) Show that the Gramian of a frame and its dual are pseudoinverses, i.e.,

Gram(Φ˜ )= Gram(Φ)†.

(d) Let Φ =( f j) be a ﬁnite frame with synthesis operator V =[ f j]. Show that

Gram(Φcan)= V †V = Gram(Φ)Gram(Φ)†.

3.4. Perturbation of a normalised tight frame. Let ( f j) j J be a normalised tight frame for H , and T : H H be invertible. ∈ 1 → Show that the frames (T ∗ f j) and (T − f j) for H are dual, i.e.,

1 1 f = ∑ f ,T ∗ f j T − f j = ∑ f ,T − f j T ∗ f j, f H . j J j J ∀ ∈ ∈ ∈

3.5. If Φ =( f j) is a ﬁnite frame for H , then its image Ψ =(Qf j) under a linear map Q : H K is a frame for its span, with frame bounds satisfying → † 2 2 AΦ Q − AΨ BΨ BΦ Q . ≤ ≤ ≤ In particular, if Q is a partial isometry, e.g., an orthogonal projection or unitary map, then AΦ AΨ BΨ BΦ , ≤ ≤ ≤ and so partial isometries map tight frames to tight frames (cf Exer. 2.7).

n H n 3.6. Let ( f j) j=1 be a ﬁnite sequence in , with synthesis operator V =[ f j] j=1. Prove the following are equivalent: 2 n f , f j ∑ | | (a) A := inf f =0 j=1 f 2 > 0. (b) ( f j) is a frame for H . (c) ( f j) spans H . (d) V is onto. (e) V ∗ is 1–1. 2 2 (f) Sf , f = V f ,V f = V f = ∑ f , f j > 0, f = 0. ∗ ∗ ∗ j | | ∀

3.7. Suppose that ( f j) is a ﬁnite frame, and V =[ f j]. Let λ j be the eigenvalues of { } S = VV ∗ (so σ j = λ j are the nonzero singular values of V). Show that (a) The optimal frame bounds are A = min λ and B = max λ . j j j j (b) The frame bounds (3.1) are equivalent to A min j λ j and B max j λ j. ≤ ≥

3.8. Let ( f j) be a ﬁnite frame with optimal frame bounds A and B. Show that 2 (a) f j B, j. 2 ≤ ∀ (b) f j = B if and only if f j spank= j fk. 2 ⊥ (c) f j < A implies f j spank= j fk. ∈ 62 3 Frames

3.9. Let Φ =( f j) be a ﬁnite frame for H , and Σ Σ σ σ V = U1 U2∗, = diag( 1, 2,...)

be a singular value decomposition of the synthesis operator V =[ f j]. Show that ˜ Φ † σ σ (a) [ f j]= V Gram( ) = U1 diag(1/ 1,1/ 2,...)U2∗. 1 can Φ † 2 (b) [ f j ]= V(Gram( ) ) = U1 diag(1,1,...)U2∗.

3.10. Canonical tight frame. Let Φ =( f j) j J be a frame for H . Show that (a) (Φ˜ )can = Φcan. ∈ (b) Gram(Φcan)= Gram(Φ)Gram(Φ˜ )= Gram(Φ˜ )Gram(Φ). 1 1 1 1 can can 2 2 2 2 (c) f = f˜ = SΦ− f j = S f j = S− f˜j = SΦ f˜j, j J. j j Φ˜ Φ˜ ∀ ∈

3.11. Commutativity of the synthesis, frame and Gramian operators. Let Φ =( f j) j J be a ﬁnite sequence in H , with synthesis operator V =[ f j] j J, ∈ ∈ frame operator S = VV ∗ and Gramian G = V ∗V. Show that these satisfy (a) S jV = VG j, j = 1,2,.... 1 1 (b) S 2 V = VG 2 . (c) S†V = VG†.

3.12. Orthogonal projection formula (generalises Exer. 2.3). Suppose ( f j) j J is a ﬁnite frame for a subspace K H , and let V :=[ f j] j J, ∈ ⊂ ∈ W :=[ f˜j] j J. Show the orthogonal projection onto this subspace is given by ∈

P = VW ∗ = WV ∗ : f ∑ f , f˜j f j = ∑ f , f j f˜j. → j J j J ∈ ∈

3.13. Let ( f j) be ﬁnite frame for H . Prove the analog of the trace formula (2.9)

˜ ˜ can 2 ˜ H f j, f j = f j, f j = f j , j = ∑ f j, f j = d = dim( ). ∀ ⇒ j J ∈

3.14. Show that if V =[ f1,..., fn] maps onto a proper subspace K of H , so that S = VV : H H is not invertible. Then the dual frame and canonical tight frame ∗ → for ( f j) are given by

1 1 ˜ † can † 2 2 † f j = S f j, f j =(S ) f j =(S ) f j. Φ n Φ 3.15. If the Gramian of a frame =( f j) j=1 can be factored Gram( )= L∗L, where L =[v1,...,vn] is an m n matrix, then the columns of L give a copy of Φ as a subspace of Fm (cf Th. 3.3).× Show that the dual frame and canonical tight frame of † † 1 this copy are given by the columns of (LL∗) L and ((LL∗) ) 2 L. 3.11 Oblique duals 63

3.16. Let Φ =( f j) be a ﬁnite frame for H with synthesis operator V =[ f j], and Ψ =(g j) := QΦ =(Qf j), where Q : H H is invertible. Show that (a) Ψ is a frame for H . → 1 (b) Ψ˜ =(Q∗)− Φ˜ . can can 1 1 (c) Ψ = UΦ , U :=(QVV ∗Q∗)− 2 Q(VV ∗) 2 , where U is unitary. (d) If Q = cU, with c > 0 and U unitary, then this U is the U of part (c).

3.17. Let V be a d n matrix, n d, with full rank, i.e., rank(V)= d. Use the × ≥ Σ singular value decomposition V = U1 U2∗ to show that the minimum

min V W F W Fd n,A>0 − ∈ × WW∗=AI is uniquely attained for

1 1 1 1 1 2 2 1 2 2 W = d trace((VV ∗) )(VV ∗)− V = d trace(S )S− V. Remark: This generalises the problem of ﬁnding the unitary matrix W which best approximates a square matrix V (cf [HJ90], Problem 3 of 7.4). § 3.18. Modify the argument of Theorem 3.2 to show that for any sequences

n d d d d 2 ∑ f j g j ∑ λk + ∑ µk 2 ∑ ∑ λk √µk uk ,vk , − ≥ − 1 2 | 1 2 | j=1 k=1 k=1 k1=1 k2=1

where (λk), (µk) are the eigenvalues of the frame operators for ( f j), (g j), and (uk), (vk) are corresponding orthonormal bases of eigenvectors. Φ n 3.19. Isogonal conﬁgurations (Example 3.9). Let =(u j) j=1 be a sequence of isogonal of unit vectors in Rd, d > 1, i.e., one with Gramian

1, j = k; M = Gram(Φ)= a, j = k. (a) Show the eigenvalues of M are na a + 1 and 1 a (of multiplicity n 1). (b) Determine the condition on a that− ensures M is− positive semideﬁnite,− and the corresponding rank of M (which can be 1, n 1 or n). − (c) When rank(M)= n 1, conclude (u j) are the vertices of the simplex. − (d) When rank(M)= n, show that the dual frame (u˜ j) is an isogonal conﬁguration.

3.20. Real and complex frames. Let Φ be a ﬁnite frame. (a) Show that Φ is real if and only if its dual Φ˜ is real. (b) Show that if Φ is real, then canonical tight frame Φcan is real. (c) If the canonical tight frame is real, then does it follow that Φ and Φ˜ are? 64 3 Frames

3.21. Suppose that ( f j) and (g j) form a biorthogonal system for K , and P is the orthogonal projection onto a subspace H . Let S be the frame operator for ( f j). Show that the dual frame of (Pf j) (for H ) being equal to (Pg j) is equivalent to (a) PS = PSP. (b) SP = PSP. (c) SP = PS. (d) H is invariant under S, i.e., SH H . ⊂ 3.22. Let L : H K be a linear map between ﬁnite–dimensional Hilbert spaces, → with singular values σ1,σ2, ,σm. Show that

Lx maxσ j x , x H , ≤ ∀ ∈ with equality for x = 0 if and only if x (kerL) is a right–singular vector for the ∈ ⊥ largest singular value (which implies L = max j σ j), and

min σ j x Lx , x (kerL)⊥, σ j=0 ≤ ∀ ∈ with equality for x = 0 if and only if x is a right–singular vector for the smallest nonzero singular value.

3.23. Let L be a self adjoint invertible map on a ﬁnite–dimensional Hilbert spaces. Show that if U is unitary, then

L U , LU I , UL I L I . − − − ≥ −

3.24. Suppose that Φ =( f j) is a ﬁnite frame for H with (possibly not optimal) frame bounds A and B. Let V =[ f j]. Show that the frame bounds (3.1) are equivalent to any of the bounds

√A c Vc √B c , c ran(V ∗), ≤ ≤ ∀ ∈

√A f V ∗ f √B f , f H , ≤ ≤ ∀ ∈ A f Sf B f , f H , ≤ ≤ ∀ ∈ A c Gram(Φ)c B c , c ran(Gram(Φ)), ≤ ≤ ∀ ∈ and these are sharp if and only if A and B are the optimal frame bounds.

3.25. Suppose that Φ =( f j) is a ﬁnite frame. Let λ1,...,λd be the eigenvalues of its frame operator SΦ . Use a singular value decomposition for the synthesis operator V =[ f j], to show that for any scalars α,β,γ F one has ∈ can αSΦ + βSΦ˜ + γI = α Gram(Φ)+ β Gram(Φ˜ )+ γ Gram(Φ ) αλ βλ 1 γ = max j + j− + . 1 j d ≤ ≤ 3.11 Oblique duals 65

2 1 In particular, since ScΦ = c SΦ and S 1 Φ˜ = 2 SΦ˜ , we obtain (3.30) (3.31), (3.32), by c c taking (α,β,γ) to be (c2, 1 ,0), (c2,0, 1), (0, 1 , 1), respectively. We also have c2 − c2 − 2 SΦ = Gram(Φ) = V = maxλ j = BΦ . j

3.26. Show that the inequality 1 1 B A max A , B − , 0 < A B − A − B ≥ √AB ≤ holds, with equality if and only if AB = 1. Φ Φ 1 Φ˜ 3.27. Let be a ﬁnite frame. Here we investigate how close c , its dual c , and Φcan can be for the metric dist(Φ,Ψ) := Gram(Φ) Gram(Ψ) . (a) Fix 0 < A B. Show that − ≤ 1 1 B A minmax tA 1 , tB 1 = minmax 1 , 1 = − , t>0 {| − | | − |} t>0 { tA − tB − } A + B 2 A+B which are attained if and only if t = A+B and t = 2AB , respectively. (b) Let A and B be the optimal frame bounds for Φ. Use (a) to show that 1 B A mindist(cΦ,Φcan)= mindist( Φ˜ ,Φcan)= − < 1, c>0 c>0 c A + B

2 2 2 A+B which are attained for c = A+B and c = 2AB , respectively. (c) Fix 0 < A B. Show that ≤ 1 1 B minmax tA 1 , tB 1 , 1 , 1 = 1, t>0 | − | | − | tA − tB − A − which is attained if and only if t = 1/√AB. (d) Let A and B be the optimal frame bounds for Φ. Show that

1 B minmax dist(cΦ,Φcan),dist( Φ˜ ,Φcan) = 1, c>0 c A − 4 1 which is attained if and only if c = AB . (e) Suppose Φ is normalised, i.e., AB = 1. Let c > 0 and U be unitary. Show that 1 dist(Φ,cUΦcan)= dist(Φ˜ ,cUΦcan)= max A c2 , c2 , | − | A − and 1 1 A2 minmax A c2 , c2 = − , c>0 | − | A − 2A 2 1 1 A2+1 with the minimum occuring for c = 2 (A+ A )= 2A . 66 3 Frames

3.28. Let 0 < A B. Show that ≤ 1 1 max A 1 , B 1 = max 1 , 1 AB = 1. | − | | − | A − B − ⇐⇒ 3.29. Prove the following one–sided version of Proposition 3.2. If Φ =( f j) and Ψ =(g j) are ﬁnite frames, with the same index set, and synthesis operators V and W, then the following are equivalent (a) Ψ = QΦ for some linear map Q (possibly not invertible). (b) Gram(Φcan)Gram(Ψ can)= Gram(Ψ can). (c) ran(W ∗) ran(V ∗), or, equivalently, ker(V) ker(W). Furthermore,⊂ these imply the equivalent conditions⊂ (d) VW ∗ is 1–1. (e) ∑ g,g j f j = 0, g. j ∀ 3.30. Suppose that Ψ =(g j) j J is close to a ﬁnite frame Φ =( f j) j J for H , i.e., ∈ ∈

∑ c j(g j f j) λ ∑ c j f j , c ℓ2(J). j J − ≤ j J ∀ ∈ ∈ ∈ Use Exer. 3.29 to show (a) Ψ = QΦ for some linear map Q. (b) The closeness bound (smallest λ) is cl(Ψ,Φ)= cl(QΦ,Φ)= Q I . λ Ψ Φ Φ Ψ Φ Ψ λ − (c) If := cl( , ) < 1, then is close to , with cl( , ) < 1 λ . In particular, the Q of (b) is invertible (since Q I < 1). − −

3.31. Let S : X Y be a bounded linear map between normed linear spaces, and M : Y X be an→ approximate left inverse of S, i.e., a bounded linear map with → I MS < 1. − (a) Show that S is invertible. (b) Let k X. Show that the map F : X X given by ∈ → F(g) :=(I MS)g + k − is a contraction map with constant κ = I MS , and its ﬁxed point g satisﬁes − MSg = k, i.e., MS is invertible. 1 (c) Show that (MS)− is bounded.

3.32. Let A, B be (possibly not optimal) frame bounds for a ﬁnite frame Φ for H . 2 Φ (a) Show that M := A+B I is an approximate left inverse for S = S , i.e., 2 B A I S − . − A + B ≤ A + B 3.11 Oblique duals 67

(b) Consider the contraction mapping F : H H given by → F(g) :=(I MS)g + Mh, − 1 which has ﬁxed point g satisfying MSg = Mh, i.e., g = S− h (see Exer. 3.31). Estimate the error in the ﬁxed point iteration method 2 2 g0 := 0, gk 1 :=(I MS)gk + Mh = gk Sgk + h, + − − A + B A + B

1 for ﬁnding g = S− h as limgn. 3.33. Let A and B be (possibly not optimal) frame bounds for a ﬁnite frame Φ, 1 with frame operator S = SΦ . Show that S− 2 can be calculated via the absolutely convergent series

∞ 1 2 (2 j)! 2 j 2 S− = ∑ 2 j 2 I S . A + B j=0 2 ( j!) − A + B 2 3.34. Let ( f j) be the tight frame for R given by the three equally spaced vectors 2π j 2π j 2 f j =(cos ,sin ), 0 j 2, and ( f˜j), f˜j = f j, be the canonical dual frame. 3 3 ≤ ≤ 3 (a) Show that all dual frames (g j) of ( f j) are given by

2 g j = f˜j + w, w R . ∈ 2 (b) From (a), it follows that there is a dual (g j) of Φ with g1 R arbitrary. Show 2 ∈ that there is a pseudo–dual (g j) of Φ with g1,g2 R , g1 = g2 arbitrary. ∈ 3.35. Let Φ =( f j) j J be a finite frame for H , with synthesis operator V =[ f j], ∈ and Q : ℓ2(J) ℓ2(J) be an invertible linear map. Define a norm on the linear maps → ℓ2(J) H by → 1 W Q := WQ− F (Frobenius norm). (a) Show that frames with synthesis operators V and W are dual if and only if the 1 frames with synthesis operators VQ∗ and U = WQ− are dual. (b) Show that there is a unique dual frame (g j) of Φ minimising W Q, W =[g j], given by 1 W =(VQ∗QV ∗)− VQ∗Q. 1 r Remark. For J = 1,...,n and Q− =(D∗) , r 1, where D is the first–order difference operator{ } ≥ 1 1 −1 1  −. .  D = .. .. ,    1 1  −   1    the dual frames of part (b) are called r–th order Sobolev duals of Φ (see [BLPY10]). 68 3 Frames

3.36. Let Φ =( f j) be a ﬁnite frame with frame bounds A and B, and Φˆ =( fˆj) be a perturbation satisfying

2 2 ∑ f , fˆj f j R f , f . j | − | ≤ ∀

(a) Show that if R < A, then Φˆ is a frame with bounds (√A √R)2 and (√B+√R)2. A Φˆ − Φ (b) Show that if R < 4 , then the canonical dual of is an approximate dual of .

3.37. Show that a ﬁnite frame Φ with optimal frame bounds A and B is approxi- Φ 2 mately dual to the scalar multiple c if and only if 0 < c < B , and that VW ∗ is 2 minimised by the choice c = A+B . 3.38. Suppose that V and W are ﬁnite dimensional subspaces of H . (a) Show that the following are equivalent (i) V W = W V = 0. ∩ ⊥ ∩ ⊥ (ii) H = V a W ⊥ = W a V ⊥ (algebraic direct sums). (b) Show that if⊕V W = W⊕ V = 0, then the oblique projections satisfy ∩ ⊥ ∩ ⊥

dim(V )= dim(W ), (PV W )∗ = PW V . , ⊥ , ⊥

3.39. Let ( f j) be a finite frame for V and (g j) be its canonical oblique dual in W . Show that the canonical oblique dual of (g j) in V is ( f j). n λ m 3.40. Suppose ( f j) j=1 is a finite sequence in a linear space X, and ( k)k=1 is a finite n m seqence of linear functionals X F. Let V :=[ f j] : F X and Λ ′ :=[λk] : F Λ λ m → → → X′, i.e., ( f ) :=( k( f ))k=1, and

† P := V(ΛV) Λ, V := span( f j) X, L := span(λk) X′. ⊂ ⊂ (a) Show that P : X X is a projection, i.e., P2 = P. (b) Show that Λ(Pf→)= Λ( f ), f V = ran(V). ∀ ∈ (c) Show that rank(P)= rank(ΛV)= rank(Λ V )= dim(L V ). | | (d) Show that if dim(V ) dim(L V ), then P projects onto V , i.e., PV = V. ≤ | (e) Show that if dim(L ) dim(L V ), then P interpolates L , i.e., ΛP = Λ. ≤ | (f) What is the formula for P if X is a Hilbert space, and so λk( f )= f ,gk . 3.41.mOrthogonal polynomials. For a suitable (nonnegative) weight function w on (a,b), an inner product can be deﬁned on the univariate polynomials by

b f ,g := f (t)g(t)w(t)dt. a If the Gram–Schmidt algorithm is applied to the ﬁrst n + 1 monomials, or any other sequence of polynomials with degrees 0,1,...,n, then polynomials obtained are (for various weights) the classical orthogonal polynomials. 3.11 Oblique duals 69

Fix some standard weight, e.g., the Legendre weight w = 1 (a = 1,b = 1). Investigate the orthonormal basis obtained by taking the canonical tight− frame for the monomials (Lowdin¨ orthogonalisation). What happens for other bases?

Chapter 4 Canonical coordinates for vector spaces and afﬁne spaces

If ( f j) j J is a finite frame for H , i.e., a spanning sequence for H , then each f H can be∈ written ∈ f = ∑c j f j, j for some choice of coefficients c =(c j). The unique coefficients which minimise

2 2 c = ∑ c j j | | are c j = c j( f ) := f , f˜j , where ( f˜j) is the canonical dual frame. We observe that

The linear functionals f c j( f ) do not depend on the inner product on H . →

In this way, it is possible to extend the frame expansion (and other elements of frame theory) to any finite spanning sequence for a vector space over any subfield of the complex numbers which is closed under conjugation. The linear functionals f c j( f ) will be called the canonical coordinates for f → with respect to the spanning sequence ( f j) for the vector space X = span f j . They depend only on the vector space structure of X, though they can be calculated{ } via the canonical dual frame if X is endowed with an inner product. They generalise the dual basis (the case when the vectors are linearly independent), and are characterised by the fact that the associated Gramian matrix is an orthogonal projection. The unique inner product for which a finite spanning sequence is a normalised tight frame is given by the Euclidean inner product between the canonical coordinates of vectors. The canonical coordinates are ideally suited to situations where there is a natural spanning set for a vector space (which is not a basis), e.g., the n–th roots of unity in cyclotomic field (as vector space over the rationals). In such cases, computations can be done directly with the canonical coordinates, in an efficient and stable way, which preserves the geometry of the spanning sequence.

71 72 4 Canonical coordinates for vector spaces and afﬁne spaces 4.1 The canonical Gramian of a spanning sequence

Throughout, let X be a ﬁnite dimensional vector space over a subﬁeld F of C. To be 2 J able to calculate ∑ j c j for a vector c =(c j) F , it is necessary that F be closed under conjugation. Therefore,| | from now on: ∈

We assume that F is closed under complex conjugation.

Recall (Theorem 2.2) that if Φ =( f j) j J is ﬁnite normalised tight frame with ∈ P = Gram(Φ) (an orthogonal projection), then Φ is unitarily equivalent to (Pe j), the columns of P. In particular, they have the same (linear) dependencies, i.e.,

J J dep(Φ) := c F : ∑c j f j = 0 = c F : ∑c jPe j = Pc = 0 = ker(P). { ∈ j } { ∈ j }

Thus P is the orthogonal projection onto dep(Φ)⊥. We now show that the orthogonal decomposition of FJ extends to the case when F is not R or C, e.g., F = Q.

Lemma 4.1. (Orthogonal projections in FJ). Suppose F = F, and W is a subspace of FJ (J ﬁnite). Then there is the orthogonal direct sum decomposition

J J F = W W ⊥, W ⊥ := x F : x,a = 0, a W , ⊕ { ∈ ∀ ∈ } and matrices Q,P FJ J giving the orthogonal projections onto W and W . ∈ × ⊥ Proof. It suffices to show there exists matrix Q giving the orthogonal projection onto W (then take P to be the complementary orthogonal projection P = I Q). We can apply Gram–Schmidt orthogonalisation, without normalising,− to any J spanning set for W to obtain an orthogonal basis of vectors w1,...,wr in F (it may not be possible to normalise and stay within the field F).{ The j–th column} of the matrix Q is then defined by

r e j,vk J Qe j = ∑ vk F . vk,vk ∈ k=1

By construction, this is the orthogonal projection of e j on W . ⊓⊔

Deﬁnition 4.1. Let Φ =( f j) j J be a ﬁnite sequence in a vector space X over F (with ∈ F = F). The canonical Gramian of Φ, denoted by PΦ FJ J, is the orthogonal ∈ × projection onto dep(Φ)⊥.

By Lemma 4.1, PΦ is well deﬁned, and can be calculated by

r vkvk∗ PΦ = I ∑ , (4.1) − vk,vk k=1 4.1 The canonical Gramian of a spanning sequence 73 where (vk) is any orthogonal basis for dep(Φ). Such a (vk) can be obtained by applying Gram–Schmidt (without normalising) to any spanning set for dep(Φ). We can generalise Proposition 3.2 as follows.

Proposition 4.1. (Similarity) Let Φ =( f j) j J and Ψ =(g j) j J be ﬁnite spanning sequences. Then the following are equivalent∈ ∈

1. Φ and Ψ are similar, i.e., there is a invertible linear map Q : f j g j. → 2. PΦ = PΨ . 3. dep(Φ)= dep(Ψ). In other words:

Every spanning sequence Φ corresponds to a unique normalised tight frame determined by dep(Φ) (with Gramian PΦ ).

Example 4.1. (Basis) If Φ =( f j) is a basis, then dep(Φ)= 0, and so PΦ = I. Φ n Example 4.2. (Simplex) Suppose that =( f j) j=1 has just one dependency

f1 + f2 + + fn = 0. 1 Then 1 =(1,...,1) spans dep(Φ), so that PΦ = I 1∗1, d = dim(X)= n 1, i.e., − n − d , j = k; Φ d+1 (P ) jk = 1 − , j = k. d+1

Example 4.3. (Frames) Let ( f j) j J be a ﬁnite frame, with synthesis map V =[ f j]. ∈ Then PΦ can be calculated as above from a spanning sequence for dep(Φ)= ker(V), or by 1 PΦ = V ∗S− V, S := VV ∗. (4.2) 1 This follows, since V ∗S− V is clearly an orthogonal projection, and its kernel is dep(Φ)= ker(V) (since V is onto, V ∗ is 1–1). The ( j,k)–entry of PΦ is

1 1 1 (PΦ ) jk =(S− Ve j)∗S− (Vek)=(S− f j)∗ fk = fk, f˜j 1 1 can can =(S− 2 f j)∗S− 2 fk = f , f . (4.3) k j In particular, PΦ is the Gramian of the canonical tight frame associated with Φ. Example 4.4. (Equiangular tight frames) We can use the identiﬁcation of a spanning sequence Φ with the unique normalised tight frame with Gramian PΦ , to extend elements of frame theory to spanning sequences. For example, we say a spanning sequence Φ is equiangular if the tight frame given by the columns of PΦ =[p jk] is, i.e., PΦ has a constant diagonal and p jk = C, j = k. In this way, an equiangular tight frame could be given by a spanning| sequence| for a vector space X (which is not Cd), as in Example 4.2. 74 4 Canonical coordinates for vector spaces and afﬁne spaces 4.2 The canonical coordinates of a spanning sequence

We now deﬁne the canonical coordinates via the canonical Gramian.

Theorem 4.1. Let Φ =( f j) j J be a ﬁnite spanning sequence for a vector space X ∈ J over F (with F = F), and f = ∑ j a j f j X, a F . Then there are unique coefﬁcients Φ J ∈ ∈ c = c ( f ) F of minimal ℓ2–norm with ∈

f = ∑c j( f ) f j, (4.4) j which are given by Φ c ( f )= PΦ a, (4.5) where PΦ is the canonical Gramian of Φ.

Proof. Let V =[ f j] (the synthesis map for Φ). Since I PΦ is the orthogonal projection onto dep(Φ)= ker(V), we have − V = V PΦ +(I PΦ ) = VPΦ . (4.6) − Φ Thus, f = Va = V(PΦ a)= V(c ( f )) = ∑ j c j( f ) f j. Φ Finally, we show the choice c = c has minimal ℓ2–norm. For any choice of coefﬁcients c FJ, we have the orthogonal decomposition ∈

c =(c PΦ c)+ PΦ c dep(Φ) dep(Φ)⊥, − ∈ ⊕ and so Pythagoras gives

2 2 2 c = c PΦ c + PΦ c . − Φ 2 Φ 2 If f = ∑ c j f j, then c = PΦ c, and we obtain c c , with equality if and j ≥ only if c = cΦ . ⊓⊔ It follows from (4.5), that f cΦ ( f ), j J, are a linear functionals, and → j ∈ Φ PΦ =[c j ( fk)] j,k J. (4.7) ∈ Definition 4.2. The canonical coordinates (or canonical dual functionals) for a finite spanning sequence Φ =( f j) for an F–vector space X are the linear functionals Φ Φ c =(c j ) given by Φ c ( f ) := PΦ a, f = ∑a j f j, j where PΦ is the canonical Gramian of Φ. We call (4.4) the canonical expansion, and (4.6) the canonical factorisation of the synthesis map V =[ f j]. By Theorem 4.1, the canonical coordinates cΦ are well defined. They are the minimal ℓ2–norm coefficients (c j) for which f = ∑ j c j f j, and 4.2 The canonical coordinates of a spanning sequence 75

The linear map

Φ LΦ : X ran(PΦ )= dep(Φ)⊥ : f c ( f ) → → Φ is a vector space isomorphism, with LΦ ( f j)= c ( f j)= PΦ e j. In particular,

Φ c ( f )= 0 f = 0. ⇐⇒

Example 4.5. (Dual basis) If Φ =( f j) is a basis, then PΦ = I, and (4.7) implies that the canonical coordinates are the dual basis. The canonical coordinates therefore generalise the dual functionals to the case when ( f j) is not a basis.

Example 4.6. (Frames) If Φ =( f j) is a frame, then (4.3) and (4.7) give

Φ c ( f )= f , f˜j , f , j ∀ ˜ Φ i.e., f j is the Riesz representer of the linear functional c j . Example 4.7. (Matrices) We can deﬁne matrices with respect to spanning sequences in the usual way. The (canonical) matrix representing a linear map L : X Y with Φ n Ψ m → respect to spanning sequences =( f j) j=1 and =(gk)k=1 for X and Y is the m n A = AL F given by ∈ × Ψ j–th column of A = Ae j = c (Lf j).

Ψ Φ i.e., c (Lf )= A(c ( f )), f X. The map L AL is linear, and L can be recovered ∀ ∈ → from A = AL via Φ L = WAc , W =[gk].

The canonical coordinates for a Φ which spans X can be computed from any spanning sequence (λk) for the algebraic dual space X′, as follows. m m Proposition 4.2. Let Φ =( f1,..., fn) span X. If Λ =(λk) : X F is any 1–1 k=1 → linear map, i.e., λ1,...,λm span X′, then the canonical coordinates for Φ are given by Φ + c ( f )=(ΛV) Λ f , V =[ f1,..., fn], (4.8) where (ΛV)+ is the pseudoinverse the matrix ΛV Fm n. ∈ × Proof. Let a = c( f ) Fn be a solution to (4.4), i.e., to Va = f . Since the linear ∈ functionals λ1,...,λm span X , Va = f is equivalent to λk(Va)= λk( f ), k, i.e., ′ ∀ ΛVa = Λ f ,

m n n where ΛV F × and Λ f F . This (possibly) underdetermined linear system has ∈ ∈ Φ a unique minimal ℓ2–norm (least squares) solution a = c ( f ) given by (4.8). ⊓⊔ 76 4 Canonical coordinates for vector spaces and afﬁne spaces 4.3 A characterisation of the canonical coordinates

Let Φ =( f j) j J span a vector space X, and Ψ =(λ j) j J be linear functionals on X. We say that Φ∈and Ψ are dual sequences (see Exer. 4.1)∈ if

f = ∑λ j( f ) f j, f X. (4.9) j ∀ ∈

This implies ( f j) spans X and (λ j) spans X′, and is equivalent to

λ = ∑λ( f )λ j, λ X′. j ∀ ∈

The matrix G =[λ j( fk)] is called the Gramian of Φ and Ψ. We now show that:

(λ j) are the canonical coordinates (canonical dual functionals) for ( f j) if and only if G =[λ j( fk)] is an orthogonal projection of rank d = dim(span f j ). { }

Recall that the canonical isomorphism between X and its bidual X′′ is

ˆ: X X′′ : f fˆ, fˆ(λ) := λ( f ), λ X′. → → ∀ ∈ Theorem 4.2. (Characterisation). Suppose X is an F–vector space (with F = F), Φ =( f j) in X and Ψ =(λ j) in X′ are dual, i.e., satisfy (4.9), and G is the Gramian

J J G = ΛV =[λ j( fk)], V :=[ f j] : F X, Λ =(λ j) : X F . → → Then the following are equivalent 1. cΦ = Ψ. 2. cΨ = Φˆ . 3. G = G∗, i.e., G is an orthogonal projection. 4. PΦ = G. 5. PΨ = GT . T 6. PΨ = PΦ . 7. dep(Ψ)= dep(Φ).

Proof. First, observe that (4.9) can be written as VΛ = IX , and so

G2 = Λ(VΛV)= ΛV = G, i.e., G is a projection of rank d = dim(X). Ψ Φ Ψ Φ ˆ 1.= 2. Suppose that = c . Then =(c j ) spans X′ and ( f j) spans X′′, and so Proposition⇒ 4.2 gives

Ψ + Φ J J c =(LW) L, W :=[c ] : F X′, L =( fˆj) : X′ F . j → → 4.3 A characterisation of the canonical coordinates 77

ˆ Φ Φ T + Now LW =[ f j(ck )]=[ck ( f j)] = PΦ is an orthogonal projection, so (LW) = LW, and we obtain Ψ λ T λ T λ Φ ˆ λ Φ λ c j ( )=(PΦ L ) j = ∑(PΦ ) jk(L )k = ∑ck ( f j) fk( )= ∑ck ( f j) ( fk) k k k λ Φ λ ˆ λ λ Ψ ˆ = ∑ck ( f j) fk = ( f j)= f j( ), X′ = c j = f j. k ∀ ∈ ⇒

2.= 5. If ( fˆj) are the canonical coordinates for Ψ, then (4.7) gives ⇒ T PΨ =[ fˆj(λk)]=[λk( f j)] = G .

5.= 4. If PΨ = GT , then G is an orthogonal projection matrix of rank d = dim(X). Moreover,⇒ Φ Φ Φ PΦ G =(c V)(ΛV)= c (VΛ)V = c V = PΦ , and so G = PΦ . 4.= 3. Immediate (PΦ is an orthogonal projection). ⇒ 3.= 7. Suppose G = G , i.e., λ j( fk)= λk( f j), k. Then ⇒ ∗ ∀

a dep(Ψ) ∑a jλ j( fk)= 0, k ∑a jλk( f j)= 0, k ∈ ⇐⇒ j ∀ ⇐⇒ j ∀

∑a jλk( f j)= 0, k λk(∑a j f j)= 0, k ⇐⇒ j ∀ ⇐⇒ j ∀

∑a j f j = 0 a dep(Φ). ⇐⇒ j ⇐⇒ ∈

7.= 6. Suppose that dep(Ψ)= dep(Φ). Let (vk) be an orthogonal basis for dep(Φ), ⇒ so (vk) is an orthogonal basis for dep(Ψ), and from (4.1), we obtain

vk(vk)∗ vkvk∗ T PΨ = I ∑ = I ∑ = PΦ = PΦ . − vk,vk − vk,vk k k 6.= 1. For each f X, we have the orthogonal decomposition ⇒ ∈ Φ Ψ( f )=(λ j( f )) = c ( f )+ ρ( f ) ran(PΦ ) ker(PΦ ). ∈ ⊕ For a ker(PΦ ), we have ∈

ρ( f ),a = Ψ( f ),a = ∑a jλ j( f ). j

T T Now suppose PΦ = PΨ , so that a ker(PΦ )= ker(PΨ ), and we obtain ∈ ρ( f ),a = 0, a = ρ( f )= 0, ∀ ⇒ i.e., Ψ = cΦ . ⊓⊔ 78 4 Canonical coordinates for vector spaces and afﬁne spaces

Example 4.8. The second equivalence shows that any spanning sequence Ψ for X′ is the canonical coordinates for some Φ, i.e., the Φ given by Φˆ = cΨ . For example, 2 let X = Π1 the (three dimensional) space of linear polynomials on R , and Ψ be the point evaluations

Ψ = δ ,δ ,δ ,δ , δx : Π1 R : f f (x). (0,0) (1,0) (0,1) (a,b) → → These are the canonical coordinates for Φconsisting of the linear polynomials (with the obvious indexing)

(ab 1 a b2)x +(ab 1 b a2)y + 1 + a2 + b2 f (x,y)= − − − − − − , (0,0) (a + b)2 +(a 1)2 +(b 1)2 − − (2 + ab a + 2b2 2b)x +(a a2 2ab)y + ab + a2 a f (x,y)= − − − − − , (1,0) (a + b)2 +(a 1)2 +(b 1)2 − − (b b2 2ab)x +(2 + ab b + 2a2 2a)y + ab + b2 b f (x,y)= − − − − − , (0,1) (a + b)2 +(a 1)2 +(b 1)2 − − (2a + b 1)x +(a + 2b 1)y + 1 a b f (x,y)= − − − − . (4.10) (a,b) (a + b)2 +(a 1)2 +(b 1)2 − − Note that these polynomials are continuous functions of (a,b).

4.4 Properties of the canonical coordinates

Φ Because PΦ =[c j ( fk)] is an orthogonal projection, the canonical coordinates share many properties of the coordinates for a basis (where PΦ = I). We write the sequence obtained by removing the vector f j from Φ =( f1,..., fn) as

Φ f j :=( f1,..., f j 1, f j+1,..., fn). \ − Φ Proposition 4.3. (Properties). The canonical coordinates c =(c j) for Φ satisfy

1. c j( fk)= ck( f j). 2. c j( fk) 1 with c j( fk) = 1 if and only if k = j and f j span(Φ f j), in which | |≤ | | ∈ \ case c j( f j)= 1 and c j = 0 on span(Φ f j). \ 3. c j( f j) 0, with c j( f j)= 0 if and only if f j = 0. ≥ 4. ∑ j c j( f j)= d = dim(X). 5. c j = αck, α F if and only if f j = α fk. ∈ Proof. Let P =[p jk]=[c j( fk)]. These properties follow from those of orthogonal projection matrices, e.g., P = P∗, gives c j( fk)= p jk = pkj = ck( f j). Orthogonal projections P have the property Px x with equality if and only if Px = x, so that ≤ c j( fk) = p jk Pek ek = 1, | | | | ≤ ≤ 4.4 Properties of the canonical coordinates 79 with equality if and only if Pek = ek and p jk = 1, i.e., j = k, c j( f j)= 1, and | | c j( fℓ)= 0, ℓ = j. The condition Pe j = e j is equivalent to the j–th column of P not being in the span of the others, and since the columns of P and the vectors of Φ have the same linear dependencies, this is the same as f j span(Φ f j). Similarly, ∈ \ 2 c j( f j)= e j,Pe j = Pe j,Pe j = Pe j 0, ≥ Φ with equality if and only if Pe j = 0, i.e., c ( f j)= 0, which gives f j = 0. We have ∑c j( f j)= trace(P)= rank(P)= d. j Finally,

f j = α fk c ( f j)= c (α fk)= αc ( fk), ℓ ⇐⇒ ℓ ℓ ℓ ∀ c j( f )= αck( f ), ℓ ⇐⇒ ℓ ℓ ∀ ∑c j( fℓ) fℓ = α ∑ck( fℓ) fℓ f j = α fk. ⇐⇒ ℓ ℓ ⇐⇒

⊓⊔

n Example 4.9. If Φ =( f j) satisﬁes ∑ f j = 0, i.e., (1,...,1) dep(Φ), and Q is j=1 j ∈ the orthogonal projection onto (1,...,1)⊥, then

1 1 n 1 1 c ( f ) PΦ e Qe = e ∑e = (1 )2 + − = 1 . j k k k k n ℓ n n2 n | | ≤ ≤ − ℓ − − The canonical coordinates transform naturally under the action of a linear map.

Proposition 4.4. (Linear maps). Suppose that Φ =( f j) spans the F–vector space X, L : X Y is an invertible F–linear map, and Ψ = LΦ =(Lf j). Then the canonical coordinates→ for Φ and Ψ satisfy

Φ Φ cL (Lf )= c ( f ), f X. (4.11) j j ∀ ∈ Proof. Choose Λ as in (4.8), so that

Φ + c ( f )=(ΛV) Λ f , V =[ f j].

1 m Then ΛL : Y F is 1–1, and so, with W =[Lf j]= LV, we have − → Ψ 1 + 1 + Φ c (Lf )=(ΛL− W) ΛL− (Lf )=(ΛV) Λ f = c ( f ).

⊓⊔ The canonical coordinates cΦ have the same symmetries as Φ (see Example 9.5 of 9.2 for details). § 80 4 Canonical coordinates for vector spaces and afﬁne spaces 4.5 The canonical inner product for a spanning sequence

We observe that if F = F, then inner products can be deﬁned on F–vector spaces in the usual way. In this case, there is a unique inner product for which a spanning sequence is a normalised tight frame.

Theorem 4.3. Let Φ =( f j) j J be a ﬁnite spanning sequence for an F–vector space X (with F = F). Then there∈ exists a unique inner product on X for which Φ is a normalised tight frame, namely

Φ Φ f ,g Φ := c ( f ),c (g) . (4.12) Proof. The linear map X FJ : f cΦ ( f ) is 1–1, and so (4.12) deﬁnes an inner product on X. For this, → →

fk, f j Φ = PΦ ek,PΦ e j = PΦ ek,e j =(PΦ ) jk, so that Gram(Φ)= PΦ , and Φ is a normalised tight frame (Theorem 2.1). Conversely, for any inner product f ,g X on X for which Φ is a normalised tight frame Gram(Φ)= PΦ , and hence (since Φ is a spanning set)

fk, f j Φ = fk, f j X , j,k = f ,g Φ = f ,g X , f ,g X. ∀ ⇒ ∀ ∈ This proves the uniqueness. ⊓⊔

Deﬁnition 4.3. The canonical inner product , Φ for a spanning sequence Φ for an F–vector space X is the unique inner product on X for which Φ is a normalised tight frame, i.e., the inner product given by (4.12).

Example 4.10. (Frames) If Φ is a ﬁnite frame for H , with frame operater S = VV ∗, then 1 1 1 f ,g Φ = S− 2 f ,S− 2 g = f ,S− g , f ,g H . (4.13) ∀ ∈ Φ 1 1 1 This follows from c ( f )= f ,S f j = S 2 f ,S 2 f j (see Exer. 4.2). j − − − In summary:

Whenever there is a natural spanning sequence for a vector space, it can be viewed as normalised tight frame (in a unique) way, and computations can be done directly with it, in an efﬁcient and stable way. This avoids the need to obtain a basis by thinning which may destroy the inherent geometry.

We now illustrate this principle for the cyclotomic fields (as Q–vector spaces). 4.6 Canonical coordinates for cyclotomic fields 81 4.6 Canonical coordinates for cyclotomic fields

Let ω be the primitive n–th root of unity

2πi ω := e n .

The cyclotomic ﬁeld Q(ω) is a Q–vector space of dimension d = ϕ(n), where ϕ is the Euler phi (totient) function. The number of primitive n–th roots is ϕ(n), but they do not form a basis for Q(ω) in general, e.g., the primitive 4–th roots are i, which are Q–linearly dependent. For n square free, the primitive n–th roots are± a basis. When the primitive roots are not a basis, bases with additional properties can be constructed in a noncanonical way. Most prominently used are the integral bases (each element of the ring of integers has its coefﬁcients in Z), and the power bases 2 d 1 (these have the form 1,z,z , ,z − ). A natural spanning{ sequence for Q(}ω) is given by the n–th roots themselves, i.e., Φ ω j ω ω2 ωn 1 =( ) j Zn =(1, , , , − ). ∈ We now consider the corresponding canonical coordinates c = cΦ . These naturally inherit the geometry of Q(ω), e.g., if

n 1 z = a0 + a1ω + + an 1ω − , a0,...,an 1 Q, − − ∈ then

2 n n 1 ωz = a0ω + a1ω + + an 1ω , z = a0 + a1ω − + + an 1ω, − − and so the canonical coordinates satisfy

c j(ωz)= c j+1(z), (4.14)

c j(z)= c j(z), (4.15) − i.e., multiplication by ω corresponds to a forward cyclic shift of coordinates, and complex conjugation to the permutation j j of the indices. Let µ be the Mobius¨ function → −

1, n = 1; µ(n) := ( 1)n, n is square free;  − 0, otherwise which satisﬁes  ∑ ω j = µ(n). (4.16) j Z ∈ ∗n ω j Here Zn∗ is the group of units in Zn (the primitive n–th roots are , j Zn∗). The canonical coordinates can be computed as follows. ∈ 82 4 Canonical coordinates for vector spaces and afﬁne spaces

Φ ω j Proposition 4.5. (Calculation). The canonical Gramian for =( ) j Zn is ∈ 1 χ χ χ ω j ω2 j ω(n 1) j T PΦ = ∑ j ∗j , j :=(1, , ,..., − ) , (4.17) n j Z ∈ ∗n which has entries

1 a( j k) 1 n (PΦ ) jk = ∑ ω − = ϕ(g)µ , g := gcd( j k,n). n n g − a Z∗n ∈ Proof. The vectors χ j are the characters of Zn, and hence are orthogonal. Suppose that a dep(Φ), i.e., ∈ χ ω ωn 1 ∗ 1a = a0 + a1 + an 1 − = 0, − − then applying the Galois action ω 1 ω j, j Z , which ﬁxes Q, gives − → ∈ n∗ χ ω j ω (n 1) ∗j a = a0 + a1 − + + an 1 − − = 0, − and so χ j dep(Φ) , j Zn. A dimension count shows that χ j : j Z is an ∈ ⊥ ∈ { ∈ n∗} orthogonal basis for dep(Φ), and hence PΦ is given by (4.17). Evaluating the entries of PΦ gives the Ramanujan sum

1 χ χ 1 ω ak χ 1 ωa( j k) (PΦ ) jk = e∗j ∑ a a∗ek = ∑ − e∗j a = ∑ − . n a Z n a Z n a Z ∈ ∗n ∈ ∗n ∈ ∗n ϕ ϕ ϕ n Using (4.16), and (n)= (g) ( g ), this can be simpliﬁed to

1 j k 1 1 1 n ∑ ωag −g = ∑ ωag = ϕ(g) ∑ (ωg)b = ϕ(g)µ . n n n n g a Z∗n a Z∗n b Z∗ ∈ ∈ ∈ n/g

⊓⊔

Example 4.11. The canonical Gramians PΦ for n = 3,4,5 are

4 1 1 1 1 2 0 2 0 − − − − 2 1 1 − 1 4 1 1 1 1 − − 1 0 2 0 2 1 − − − −  1 2 1 ,  − , 1 1 4 1 1 . 3 − −  4 20 2 0 5 − − − − 1 1 2 −  1 1 1 4 1 − −  0 2 0 2  − − − −     −   1 1 1 1 4    − − − −    Example 4.12. For n = 2k, the canonical expansion of ω j is

j 1 j 1 j n ω = ω ω + 2 . 2 − 2 More generally, for n = pk, p a prime, the canonical expansion of ω j is 4.6 Canonical coordinates for cyclotomic ﬁelds 83

j 1 j j+ n j+2 n j+(p 1) n ω = (p 1)ω ω p ω p ω − p . p − − − −− These canonical coordinates have norm

j 1 2 1 c(ω ) 2 = (p 1) +(p 1)= 1 < 1. p − − − p Multiplication in Q(ω) corresponds to convolution of the canonical coordinates: Φ Φ ω j Proposition 4.6. The canonical coordinates c = c for =( ) j Zn satisfy ∈ c(αx + βy)= αc(x)+ βc(y), α,β Q, x,y Q(ω), (4.18) ∈ ∈ c(xy)= c(x) c(y), x,y Q(ω), (4.19) ∗ ∈ where a b is the cyclic convolution of a and b over Zn, which is given by ∗ n 1 − (a b)k := ∑ a jbk j. ∗ j=0 −

Proof. The ﬁrst is just the fact f c( f ) is linear. For the second, observe that if M → is a circulant matrix (such as PΦ ), then

M(a b)=(Ma) b = a (Mb). ∗ ∗ ∗ Let a,b Qn be coordinates for x,y, then ∈ s r s+r k k xy = ∑asω ∑brω = ∑∑asbrω = ∑∑a jbk jω = ∑(a b)kω , s r s r k j − k ∗ so that a b are coordinates for xy, and we have ∗ c(xy)= PΦ (a b)=(PΦ a) b = c(x) b = c(x) c(y), ∗ ∗ ∗ ∗ where for the last equality, we made the particular choice b = c(y). ⊓⊔ The canonical inner product has the following properties.

Proposition 4.7. (normalised tight frame) The unique Q–inner product on Q(ω) for Φ ω j which =( ) j Zn is a normalised tight frame satisﬁes ∈ xy,z Φ = y,xz Φ , x,y,z Q(ω) (4.20) ∀ ∈ ϕ (n) 2 z,z Φ = z , whenever z 2 Q. (4.21) n | | | | ∈ In particular, multiplication by any z Q(ω) of unit modulus is a unitary operation. ∈ 84 4 Canonical coordinates for vector spaces and afﬁne spaces

Proof. In view of (4.15) and (4.19), the ﬁrst amounts to showing

a b,w = b,a˜ w , a˜ =(a j), ∗ ∗ − where a = c(x), b = c(y) and w = c(z). This holds for all a,b,w by direct calculation:

a b,w = ∑ ∑a jbk j wk = ∑∑a jbk jwk, ∗ k j − j k −

b,a˜ w = ∑bk ∑a jwk j = ∑∑a jbkwk+ j = ∑∑a jbk jwk. ∗ k j − − j k j k − Finally, when z 2 Q, we have | | ∈ 2 2 z,z Φ = zz,1 Φ = z 1,1 Φ = z 1,1 Φ . | | | | In particular, all the n–roots have the same norm, and so

dim(Q(ω)) = ϕ(n)= rank(PΦ )= trace(PΦ )= ∑ ω j,ω j Φ = n 1,1 Φ . j Combining these gives (4.21). ⊓⊔ Example 4.13. This inner product is different from the one induced by viewing the n–th roots of unity as vectors in R2 (with the Euclidean inner product), which gives a tight frame for R2. For example, when n = 5, we have

1 1 cos 2π 2π 1,ω Φ = , , 5π = cos Q. −5 0 sin 2 5 ∈ 5 2 n n Here, the coordinates for 1 which minimise ∑ c j over c R , and over c Q j | | ∈ ∈ (the canonical coordinates), and those given by 1 = 1 1 + 0 ω + + 0 ωn 1 are − 2 4 1 5 π 5 2 cos 2 1 0 5 5π Φ − 5  2 cos 4 , c (1)=  1 , 0, 5 5π − 5  2 cos 6   1  0  5 5π  − 5     2 cos 8   1  0  5 5   5      −    respectively, which have norms

π π 2 1 + 2cos2 2 + 2cos2 4 0.63246, 2 0.89442, 2 0.89442, 1. 5 5 5 ≈ √5 ≈ √5 ≈ Example 4.14. We observe that n ω j is orthogonal to ωk µ )= 0, g = gcd( j k,n), ⇐⇒ g − and so two n–th roots cannot be orthogonal if n is square free. 4.6 Canonical coordinates for cyclotomic ﬁelds 85

2πi Example 4.15. (Subﬁelds) Let n = 8. Then ω = √i = e 8 , and Q(ω) is 4–dimensional, with cyclotomic subﬁelds Q Q(i) Q(√i). ⊂ ⊂ The canonical coordinates of 1,ω,ω2,ω3 are

1 2 0 0 0 1 0 2 0 0      1    0 0 2 0 0 0 0  1  c(1)=  , c(√i)=  , c(i)=  , c(i√i)=  2 .  1  0 0 0  2        0  1  0 0    2      0 0  1  0      2    0 0 0  1         2          Therefore one can determine what is the smallest cyclotomic subﬁeld of Q(√i) a given element lies in by considering which of its canonical coordinates are zero. Φ ω j By viewing =( ) j Zn as a normalised tight frame (for the canonical inner ∈ product), we can easily show that the natural action of the cyclic group Cn = a on the Q–vector space Q(ω) given by a ω j = ω j+1 is irreducible. Theorem 4.4. (Irreducibility) Let , be the canonical inner product on Q(ω) Φ ω j ω ω ωn 1 given by =( ) j Zn . For any nonzero z Q( ), the vectors (z, z,..., − z) are an equal norm tight∈ frame for Q(ω), i.e.,∈ 1 x = ∑ x,ω jz ω jz, x Q(ω). (4.22) z 2 ∀ ∈ j Zn | | ∈ ω j Proof. Using (4.20) and the fact ( ) j Zn is a normalised tight frame, we calculate ∈ ∑ x,ω jz ω jz = ∑ zx,ω j ω j z =(zx)z = z 2x. j j | |

Thus every Cn–orbit of z = 0 spans Q(ω), i.e., the action is irreducible. ⊓⊔ The (forward) cyclic shift operator S on Qn, which is given by

Se j := e j 1, j Zn, + ∈ n and deﬁnes a natural action of Cn = a on Q via a v = Sv. By (4.14) the canonical coordinates c = cΦ for Φ =(ω j) satisfy

c(ωkz)= Skc(z). (4.23)

n Thus it follows that dep(Φ)⊥ is an irreducible shift invariant subspace of Q . 86 4 Canonical coordinates for vector spaces and afﬁne spaces

Corollary 4.1. (Irreducibility) The shifts of any nonzero b dep(Φ) are an equal ∈ ⊥ norm tight frame for ran(PΦ )= dep(Φ)⊥ (with the Euclidean inner product), i.e., ϕ (n) 1 j j a = ∑ a,S b S b, a dep(Φ)⊥. n b,b ∀ ∈ j Zn ∈ Proof. Expanding the canonical inner product in (4.22), using (4.23), gives 1 x = ∑ c(x),S jc(z) ω jz, x Q(ω). z 2 ∀ ∈ j Zn | | ∈ Applying c to this, and letting a = c(x), b = c(z), gives the result. ⊓⊔ The shift invariant subspace dep(Φ) of Qn is not irreducible when n is not a prime. In this case there is a proper 1–dimensional shift invariant subspace spanned by (1,1,...,1). Nevertheless, we are able to give an single linear dependence aΦ whose shifts give a tight frame for dep(Φ).

Theorem 4.5. (Shift invariant tight frame) Let aΦ Zn be n times the ﬁrst column ∈ of I PΦ , i.e., − aΦ = ∑ χ j. j Z ∈ ∗n Then the shifts of aΦ are an equal norm tight frame for dep(Φ), i.e.,

1 j j x = ∑ x,S aΦ S aΦ , x dep(Φ). (4.24) n2 ∀ ∈ j Zn ∈

Proof. By (4.17) and the fact the characters χ j are orthogonal, I Pφ is the circulant matrix given by −

1 χ χ 1 2 n 1 ∑ j ∗j = [aΦ ,SaΦ ,S aΦ ,...,S − aΦ ]. n j Z n ∈ ∗n The columns of the orthogonal projection matrix I PΦ are a normalised tight frame for its range dep(Φ), and the corresponding frame− expansion is (4.24). ⊓⊔ Example 4.16. For n = 6, Z = 1,5 , and 6∗ { } 4 11 2 1 1 4 1− 4 11 2− 1 1 1 −1 1− 4 1 1 2  −1  I PΦ = − − , aΦ = χ0 + χ2 + χ3 + χ4 = . − 6  2 1 1 4 1 1   2   − −     1 2 1 1 4 1  1   − −     11 2 1 1 4   1 − −  −      Thus the linear dependencies between the 6–th roots can be expressed as 4.7 Generalised barycentric coordinates for afﬁne spaces 87

4ω j ω j+1 + ω j+2 + 2ω j+3 + ω j+4 ω j+5 = 0, 0 j < 6. − − ≤ Here the 4–dimensional subspace dep(Φ) can be decomposed into two 1–dimensional and one 2–dimensional orthogonal shift invariant subspaces, which are generated by

(1,1,1,1,1,1)T , (1, 1,1, 1,1, 1)T , (0,1, 1,0,1, 1)T , − − − − − respectively.

4.7 Generalised barycentric coordinates for afﬁne spaces

We now give the analogue of the canonical coordinates (for a vector space) for affine spaces. An affine space X is, in effect, a vector space for which there is no distinguished point that plays the role of the origin in a vector space (or, equivalently, the translation of a vector subspace). As such, we can take affine combinations of “points” in X, i.e., linear combinations where the sum of the coefficients is 1, and differences of points to obtain “vectors”. Let X be an affine space with dimension d, i.e., d + 1 is the number of points in affinely independent affine spanning set for X. A sequence v1,...,vn of n = d + 1 points in X is affinely independent if and only if each point x X can be written uniquely as an affine combination of them, i.e., ∈

x = ∑ξ j(x)v j, ∑ξ j(x)= 1. (4.25) j j

The afﬁne functions ξ j, so deﬁned, are the barycentric coordinates for Θ =(v j).

Definition 4.4. Let X be an affine space over F, with F = F, and Θ =(v1,...,vn) be points with affine span X. Then the (affine) generalised barycentric coordinates1 Θ n n ξ (x)=(ξ j(x)) F for Θ of a point x X are the (unique) coefficients of j=1 ∈ ∈ minimal ℓ2–norm for which x is an affine combination of Θ, i.e., (4.25) holds. n These are well defined, since the set of vectors a F satisfying x = ∑ j a jv j and n ∈ ∑ j a j = 1 form a nonempty affine subspace of F . ξΘ ξ n Proposition 4.8. (Calculation) The generalised barycentric coordinates =( j) j=1 Θ n 1 ∑n for points =(v j) j=1, with barycentre b = bΦ = n j=1 v j, are given by

Θ Θ bΘ 1 ξ (x)= c − (x bΘ )+ , (4.26) j j − n

Θ bΘ where c − =(c j) are the canonical coordinates of Θ bΘ =( f j), f j := v j bΘ . In particular, each x ξΘ (x) is an affine function. − − → j 1 There are many generalisations of barycentric coordinates used in geometric modeling (computer graphics) for nonsimplical polytopes. These are the only ones where each ξ j is an affine function. 88 4 Canonical coordinates for vector spaces and affine spaces

∑ ξ 2 ξ 1 Proof. We seek to minimise j j(x) subject to (4.25). Write j(x)= a j(x)+ n . ξ | | Then ∑ j j(x)= 1 is equivalent to ∑ j a j(x)= 0, and so

ξ 2 2 1 1 1 2 1 ∑ j(x) = ∑ a j(x) + a j(x)+ a j(x)+ 2 = ∑ a j(x) + . j | | j | | n n n j | | n

Since ∑ f j = ∑ (v j b)= ∑ v j nb = 0, expanding gives j j − j − 1 x = ∑ξ j(x)v j = ∑ξ j(x) f j + ∑ξ j(x)b = ∑ a j(x)+ f j + b = ∑a j(x) f j + b. j j j j { n} j

2 Thus we must minimise ∑ a j(x) , subject to the constraints j | |

x b = ∑a j(x) f j, ∑a j(x)= 0. − j j

Θ bΦ The minimiser subject to just the ﬁrst constraint is a j(x)= c − (x bΘ ). But j − ∑ j f j = 0 implies the dependency ∑ j c j = 0 (by Theorem 4.2), and so the second constraint is also satisﬁed by this choice for a j(x). ⊓⊔ The generalised barycentric coordinates have similar properties to those of the canonical coordinates (see Proposition 4.3).

Θ Proposition 4.9. (Properties) The generalised barycentric coordinates ξ =(ξ j) Θ n for points =(v j) j=1 with afﬁne span X satisfy

1. ξ j(vk)= ξk(v j). 2. ξ j(vk) 1 with ξ j(vk) = 1 if and only if k = j and v j aff(Θ v j), in which | |≤ | | ∈ \ case ξ j(v j)= 1 and ξ j = 0 on aff(Θ v j). 1 ξ ξ 1 \ Θ 3. n j(v j) 1, with j(v j)= n if and only if v j = bΘ (the barycentre of ). ≤ξ ≤ 4. ∑ j j(v j)= d + 1. 5. ξ j = ξk if and only if v j = vk.

Proof. By Proposition 4.8, 1 1 ξ j(vk)= c j( fk)+ , fk := vk bΘ , bθ := ∑v j, n − n j and so, by Proposition 4.3, we immediately obtain 1, the lower bound in 3, 4 and 5. Since vk can be written as the affine combination vk = 1vk + ∑ j=k 0v j, we have 2 2 2 2 ξ j(vk) ∑ ξ j(vk) 1 + ∑ 0 = 1 = ξ j(vk) 1, | | ≤ j | | ≤ j=k ⇒ | |≤ with equality if and only if k = j and ξℓ(v j)= 0, ℓ = j, i.e., ξ j = 0 on aff(Θ v j). If ξ Θ Θ ∀ \ j = 0 on aff( v j), then v j aff( v j). Otherwise, v j = ∑k= j akvk, ∑k= j ak = 1, \ ∈ \ and v j can be written as an affine combination 4.7 Generalised barycentric coordinates for affine spaces 89

v j =(1 t)v j + ∑ takvk, t R, − k= j ∈ where the sum of the squares of the coefﬁcients above is

2 2 2 2 2 (1 t) + ∑ t ξk = 1 2t +t 1 + ∑ ξk , − k= j | | − k= j | | which is strictly less than 1 for t > 0 sufﬁciently small, and so λ j(v j) < 1. ⊓⊔ From the formula (4.26), we also observe that ξ 1 The coordinates of the barycentre b = bΘ are j(b)= n , j. • 1 ∀ ξ j is constant (equal to ) if and only if v j is the barycentre c. • n These imply that the set of points where the generalised barycentric coordinates are nonnegative Θ N = NΘ := x X : ξ (x) 0, j (4.27) { ∈ j ≥ ∀ } is a convex polytope, with the barycentre as an interior point. For the purpose of illustration, we also deﬁne the ellipsoid with centre the barycentre ξΘ 2 E = EΘ := x X : ∑ j (x) = 1 , { ∈ j | | } inside which the points have generalised barycentric coordinates with ℓ2–norm < 1.

Example 4.17. (Four points in R2) (See Figs. 4.1 and 4.2) Suppose, without loss of generality, that 0 1 0 a Θ = , , , , 0 0 1 b where there are no restrictions on (a,b). The generalised barycentric coordinates (indexed by the points) are given by the linear polynomials of (4.10). We observe that these coordinates depend continuously on (a,b) R2. If the convex hull of the points is a quadrilateral Q∈, i.e., a,b > 0 and a + b > 1, then the polytope NΘ (which depends continuously on Θ) has four vertices, one of which lies on the edge from (0,0) to (0,1), namely

ξ ξ a+b 1 (0,1)(x,y)= (a,b)(x,y)= 0 (x,y)= 2a+b− 1 ,0 . ⇐⇒ − Thus we conclude that NΘ circumscribes the boundary of Q, with one point on each 1 edge. The barycentre of the vertices of NΘ is not bΘ = 4 (a + 1,b + 1) in general, since the vertices are

a+b 1 ab a2+1 2a b− 1 0 b+1 + , , 2 , a+1 . 0 − a+b 1 b +1 ab a+2b− 1 b+1 a+1 − 90 4 Canonical coordinates for vector spaces and afﬁne spaces

Θ 2 1 8 4 3 3 Fig. 4.1: The points , EΘ ,NΘ ,cΘ for Example 4.17 with (a,b)=( 3 , 3 ),( 9 , 9 ),(1,1),( 2 , 4 ).

Θ 1 1 4 4 1 4 Fig. 4.2: The points , EΘ ,NΘ ,cΘ for Example 4.17 with (a,b)=(0,0),( 3 , 3 ),( 9 , 9 ),( 5 , 5 ).

Example 4.18. (The vertices of a regular polygon) (see Fig. 4.3). Let (v j) be n equally spaced unit vectors in R2, say

cos 2π j v = nπ , j = 1,...,n. j sin 2 j n Then 2 1 ξ j(x)= x, v j + , n n π so that NΘ is the n–sided regular polygon (inscribing a circle of radius 1/(2cos n )) given by

2π 1 1 cos n ( j + 2 ) NΘ = conv w j : j = 1,...,n , w j := π π , { } −2cos sin 2 ( j + 1 ) n n 2 n 1 ξ 3 and EΘ is the disc of radius r = −2 centred at 0. Here j(v j)= n . By writing the expansion as 1 n 1 n x = ∑(nξ j(x)) p j, ∑(nξ j(x)) = 1, n j=1 n j=1 we can obtain the limiting case (of points on the unit circle)

1 2π 1 2π x = ξθ (x)vθ dθ, ξθ (x)dθ = 1, 2π 2π 0 0 where 4.7 Generalised barycentric coordinates for afﬁne spaces 91

cosθ ξθ := vθ , vθ := , ξθ (x) := 2 x,vθ + 1, 0 θ 2π. sinθ ≤ ≤ ξ 1 Here the coordinates θ are nonnegative on the disc with centre 0 and radius 2 .

Fig. 4.3: The n = 3,4,5,6 equally spaced points Θ of Example 4.18, with NΘ and EΘ .

We have the following analogue of Proposition 4.4.

Proposition 4.10. (Affine maps). Let A : X Y be an invertible affine map between → affine spaces X and Y, and Θ =(v1,...,vn) be points in X with affine span X. Then the canonical barycentric coordinates for Θ and AΘ =(Av1,...,Avn) satisfy Θ Θ ξ A (Ax)= ξ (x), x X. ∀ ∈

Proof. Write Ax = L(x bΘ )+a, where L is a linear map (on the vectors in X), and 1 ∑ − Θ Θ bΘ := n j v j is the barycentre of . Then the barycentre of A is 1 1 bAΘ = ∑ L(v j bΘ )+ a = L ∑(v j bΘ ) + a = a, n j − n j − Φ n Θ Θ and so Ax bAΘ = L(x bΘ ). Let =(v j bΘ ) j=1, then A bAΘ = L( bθ ), and using− (4.11), we obtain− − − −

Φ AΘ AΘ bAΘ 1 L( bΘ ) 1 ξ (Ax)= c − (Ax bAΘ )+ = c − (L(x bΘ ))+ j j − n j − n Φ bΘ 1 Θ = c − (x bΘ )+ = ξ (x). j − n j

⊓⊔ A similar argument shows that Θ and ξΘ have same symmetry group (see 9.2). § 92 4 Canonical coordinates for vector spaces and afﬁne spaces 4.8 The Bernstein frame

Here we index the generalised barycentric coordinates for m d +1 points Θ, with d ≥ Θ afﬁne span X = R , by the points that they correspond to, i.e., ξ =(ξv)v Θ . Let d d ∈ Πn(R ) denote the polynomials R R of degree n. Since → ≤

x = ∑ ξv(x)v, ∑ ξv(x)= 1, (4.28) v Θ v Θ ∈ ∈ d it follows that (ξv)v Θ spans Π1(R ). More generally, for any n 1, the polynomials ∈ ≥ α α Θ ξ ξ v α α α = ∏ v , = ∑ v = n ( Z+) v | | v ∈

d Θ span Πn(R ). Here the α Z are multi-indices. By the multinomial theorem, ∈ + n ξ n ξ α n n! ∑ v = ∑ α = 1, α := α . v α =n ! | | n+m 1 Π d Thus the m −1 polynomials (which span n(R )) − Θ α α Bα = Bα := | | ξ , α = n, (4.29) α | | form a partition of unity, which is nonnegative on the region NΘ given by (4.27).

Deﬁnition 4.5. We call the polynomials (Bα ) α =n of (4.29) the Bernstein frame d | | (of degree n) for Πn(R ) given by the points Θ. By construction (and a dimension count):

Π d The Bernstein frame (Bα ) α =n is a frame (spanning sequence) for n(R ). It | | is a basis if and only if Θ consists of d + 1 afﬁnely independent points in Rd.

Example 4.19. (Bernstein basis) When the Bernstein frame is a basis, i.e., Θ is d +1 afﬁnely independent points in Rd, then it is the usual Bernstein basis. For example, when d = 1, Θ =(0,1), we have ξ0(x)1 x, ξ1 = x, which gives the univariate Bernstein basis − n j j B(n j, j)(x)=(1 x) − x , 0 j n, − − ≤ ≤ and when d = 2, Θ = (0,0),(1,0),(0,1) , we have

ξ (x,y)= 1 x y, ξ (x,y)= x, ξ (x,y)= y, (0,0) − − (1,0) (0,1) which gives the bivariate Bernstein basis

n j k j k B(n j k, j,k)(x)=(1 x y) − − x y , 0 j,k j + k n. − − − − ≤ ≤ ≤ 4.9 Properties of the Bernstein frame 93

Example 4.20. (Vertices of the square) When Θ is the vertices (0,0),(1,0),(0,1),(1,1) of the unit square in R2, then a = b = 1 in Example 4.17, gives 1 1 ξ = ( 2x 2y + 3), ξ = (2x 2y + 1), (0,0) 4 − − (1,0) 4 − 1 1 ξ = ( 2x + 2y + 1), ξ = (2x + 2y 1). (0,1) 4 − (1,1) 4 − The corresponding Bernstein frame for the bivariate quadratics has 10 polynomials (a basis has 6). However, the Bernstein frame shares the symmetries of the square, whereas a Bernstein basis cannot.

4.9 Properties of the Bernstein frame

The Bernstein frame shares the following well known and useful properties of the Bernstein basis. Let ev be the multi-index which is 1 at v, and 0 at all other points.

Proposition 4.11. The Bernstein frame (Bα ) α =n can be calculated recursively via | | ξ Bα = ∑ vBα ev , B0 = 1, (4.30) v Θ − ∈ and expressed in terms of the Bernstein frame for polynomials of degree n + 1 via

αv + 1 Bα ∑ Bα (4.31) = α +ev . v V + 1 ∈ | | Proof. We calculate

α 1 α αv α α ∑ ξ Bα ∑ ξ ∑ ξ Bα v ev = |α |− = α |α| = , v V − v V ev v V ∈ ∈ − ∈ | | ξ and, using ∑v v = 1, that α α α α ! α v + 1 + ev ! α+e v + 1 Bα Bα ∑ ξ ∑ ξ ξ ∑ ξ v ∑ Bα = v = |α| v = α |α | = α +ev . v V v V ! v V + 1 ( + ev)! v V + 1 ∈ ∈ ∈ | | ∈ | |

⊓⊔ d Let Dv f denote the directional derivative of f in the direction v R . We recall ∈ that v Dv f is linear. → Proposition 4.12. (Differentiation). For u,v,w V, we have ∈

Dv wξu = ξu(v) ξu(w)= ξv(u) ξw(u). − − − Thus the Bernstein frame satisﬁes 94 4 Canonical coordinates for vector spaces and afﬁne spaces

α ξ ξ Dv wBα = ∑ u(v) u(w) Bα eu . − | | u Θ − − ∈ Proof. Since ξu is afﬁne

ξu(x +t(v w)) ξu(x) (Dv wξu)(x)= lim − − − t 0 t → ξu(x)+tξu(v) tξu(w) ξu(x) = lim − − = ξu(v) ξu(w). t 0 t − → By the product and chain rules, we have

α ! α α ! α α α ξ u α ξ u 1 ξ ξ ξ ueu Dv wBα = |α| Dv w ∏ u = |α| ∑ u u − u(v) u(w) − − ! − u V ! u V − ∈ ∈ ( α 1)! α α ξ ξ ξ eu = ∑ u(v) u(w) |α |− − | | u V − ( eu)! ∈ − α ξ ξ = ∑ u(v) u(w) Bα eu . | | u V − − ∈ ⊓⊔

4.10 The generalised Bernstein operator

Θ d For a Bernstein frame (Bα ) α =n given by points in R , with convex hull | | T = conv(Θ),

d we deﬁne the (generalised) Bernstein operator Bn = Bn,Θ : C(T) Πn(R ) of degree n 1 by the usual formula → ≥ αv Bn( f ) := ∑ Bα f (vα ), vα := ∑ α v, (4.32) α =n v Θ | | ∈ | | which is equivalent to

v1 + + vn Bn( f )= ∑ ∑ f ξv ξv . n 1 n v Θ vn Θ 1∈ ∈ This has the positivity property

f 0 on T = conv(Θ) = Bn f 0 on NΘ , (4.33) ≥ ⇒ ≥ and reproduces the linear polynomials (see Example 4.21). We now show that the generalised Bernstein operator is degree reducing, i.e., 4.10 The generalised Bernstein operator 95

Fig. 4.4: The points vα α =n used in the deﬁnition of Bn,V f , where n = 7 and V is the vertices of a triangle, square, pentagon{ }| | and hexagon (respectively).

d Bn( f ) Πk(R ), f Πk (k = 0,1,...). ∈ ∀ ∈ Deﬁne the univariate and multivariate (falling) shifted factorials by

n β βv [x] := x(x 1) (x n + 1), [α] := ∏[αv] , − − v V ∈ and the multivariate Stirling numbers of the second kind by

S(τ,β) := ∏ S(τv,βv), v V ∈ where S(τv,βv) are the Stirling numbers of the second kind. We note that

S(τ,β)= 0, β τ, (4.34) ≤ and define α | | := 0, α 0. (4.35) α ≥ These are related by τ β α = ∑ S(τ,β)[α] . (4.36) β τ ≤ Lemma 4.2. For any τ and n, we have α ατ ξ α τ β β ξ β ∑ |α| = ∑ S( , )[n]| | . (4.37) α =n β τ | | ≤ β α β α β Proof. Since [ α ] | − | = | | [α] (without restriction on α and β), (4.36) | | | | α β α gives − α β α α τ β α β τ β α β ατ ∑ S( , )[ ]| | |α − β| = |α| ∑ S( , )[ ] = |α| . β τ | | β τ ≤ − ≤ Thus, we calculate 96 4 Canonical coordinates for vector spaces and affine spaces α α β ατ ξ α τ β α β ξ α ∑ |α| = ∑ ∑ S( , )[ ]| | |α − β| α =n α =n β τ | | | | | | ≤ − α β τ β β ξ β ξ α β τ β β ξ β = ∑ S( , )[n]| | ∑ |α − β| − = ∑ S( , )[n]| | , β τ α =n β τ ≤ |α | β − ≤ ≥ with the last equality given by the multinomial identity. ⊓⊔ Observe that (4.28) is equivalent to the reproduction formula for affine functions

d f = ∑ f (v)ξv, f Π1(R ). (4.38) v Θ ∀ ∈ ∈

Theorem 4.6. (Degree reducing). The generalised Bernstein operator Bn is degree reducing. More precisely,

β γ ξ β [n]| | ξ β [n]| | γ β ξ γ Bn( )= β + ∑ β a( , ) , (4.39) n| | 0< γ < β n| | | | | | where w1,...,wm is the sequence of points in V, and β β w1 τ1 wm τm a(γ,β) := ∑ ∑ ξ (w1) ξ (wm)S(τ1 + + τm,γ). τ1 τm τ =βw τm =βw | 1| 1 | | m (4.40)

Proof. Since each ξw is an afﬁne function, and ξw(v)= ξv(w), we have α α α ξ ξ v v ξ v ξ w(vα )= w ∑ α v = ∑ α w(v)= ∑ α v(w), v V v V v V ∈ | | ∈ | | ∈ | | and the multinomial identity gives

β τ β αv w βw α τ (ξ )(vα )= ∏ ∑ ξv(w) = ∏ ∑ ξ (w) α τ α βw w V v V w V τ =βw ∈ ∈ | | ∈ | | | | τ + +τ β β τ τ α 1 m w1 wm ξ 1 ξ m = ∑ ∑ (w1) (wm) β . τ1 τm α τ =βw τm =βw | | | 1| 1 | | m | | Thus, by rearranging (4.32) and Lemma 4.2, we have

τ + +τ β β τ β τ α 1 m n α ξ w1 ξ 1 wm ξ m ξ Bn( )= ∑ ∑ (w1) (wm) ∑ β τ1 τm n α τ =βw τm =βw α =n | | | 1| 1 | | m | | β β w1 τ1 wm τm = ∑ ∑ ξ (w1) ξ (wm) τ1 τm τ =βw τm =βw | 1| 1 | | m γ [n]| | τ τ γ ξ γ ∑ β S( 1 + + m, ) . × γ τ + +τm n| | ≤ 1 4.10 The generalised Bernstein operator 97

β Here Bn(ξ ) is written as a polynomial in ξ of degree β , so that Bn is degree reducing. The terms of degree β can be simpliﬁed using≤| the| multinomial identity, ξ ξ | | v(w j)= w j (v), and (4.38), as follows

Bn(ξv)= ξv, v V, (4.41) ∀ ∈ d i.e., Bn reproduces the linear polynomials Π1(R )= span ξv . This is equivalent to { } d x = ∑ Bα (x)vα , ∑ Bα (x)= 1, x R . (4.42) α =n α =n ∈ | | | | Example 4.22. (Quadratics) For β = 2, we recall S(1,0)= 0, S(2,1)= 1, so that | | ξ 2 ξ 2 ξ ξ ξ ξ a(eu,2ew)= u (v)= v (u), a(eu,ev + ew)= u(v) u(w)=( v w)(u), v = w, and we obtain β 1 β 1 β Bn(ξ )= 1 ξ + ∑ ξ (u)ξu, β = 2. (4.43) − n n u V | | ∈ Corollary 4.2. (Convergence) For all polynomials f , Bn( f ) f , as n ∞. → → Proof. It sufﬁces to consider f = ξ β . For n β , (4.39) gives ≥| | β γ ξ β ξ β [n]| | ξ β [n]| | γ β ξ γ Bn( ) = β 1 ∑ β a( , ) , − n| | − − γ < β n| | | | | | β γ [n]| | [n]| | 1 ∞ where β 1, β = O( n ), as n . n| | − n| | → ⊓⊔ 98 4 Canonical coordinates for vector spaces and afﬁne spaces

The spectral structure of Bn can easily be deduced Theorem 4.6 (see [DW15]).

Notes

The canonical coordinates were introduced in [Wal11]. They allow the theory of frames to be extended to vector spaces (without an inner product) in a natural way.

Exercises

4.1. Dual sequences. Let ( f j) j J and (λ j) j J be spanning sequences for a vector ∈ ∈ space X and its algebraic dual X′, respectively, with Gramian

J J G :=[λ j( fk)] = ΛV, V :=[ f j] : F X, Λ =(λ j) : X F . → → Show that the following are equivalent (a) VΛ = IX . (b) f = ∑ λ j( f ) f j, f X. j ∀ ∈ (c) λ = ∑ λ( f j)λ j, λ X . j ∀ ∈ ′ (d) G2 = G. (e) V = VG. (f) Λ = GΛ.

4.2. Show that if Φ is a ﬁnite frame for H , with frame operater S = VV ∗, then then the canonical inner product is given by

1 1 1 f ,g Φ = S− 2 f ,S− 2 g = f ,S− g , f ,g H . ∀ ∈ Φ 1 Hint: Use c ( f )= f ,S f j . j − 4.3. Let Θ be a set of d + 1 afﬁnely independent points with afﬁne span X, and ℓ =(ℓv)v Θ be the corresponding barycentric coordinates. Suppose that Ψ =(ψ j) ∈ is a sequence of points in Θ, with each v Θ appearing with multiplicity mv 1. ∈ ≥ Show that the canonical barycentric coordinates (ξ j) for Ψ are 1 ξ j = ℓv when ψ j = v. mv Chapter 5 Combining and decomposing frames

We now give a list of ways in which two or more frames can be combined to obtain a new frame, for which the frame and its dual are related in a natural way to those of its constituent parts. Given a frame, these methods offer a way of decomposing it into simpler parts. Sometimes this can be done in several different ways, and at present there is no coherent decomposition theory for frames (or tight frames). The inner product on the orthogonal direct sum H1 H2 and the tensor product ⊕ H1 H2 are given by ⊗

( f1,g1),( f2,g2) := f1, f2 + g1,g2 , ( f1,g1),( f2,g2) H1 H2, ∀ ∈ ⊕

f1 g1, f2 g2 := f1, f2 g1,g2 , f1 g1, f2 g2 H1 H2. ⊗ ⊗ ∀ ⊗ ⊗ ∈ ⊗ If no confusion arises, then one can identify H1 and H2 as subspaces of H1 H2, ⊕ and write f j or f j + 0 in place of ( f j,0), etc.

5.1 Unions

The (disjoint) union of ﬁnite frames Φ =( f j) j J and Ψ =(gk)k K for H1 and H2 ∈ ∈ f 0 Φ Ψ := j , (5.1) ∪ 0 gk j J,k K ∈ ∈ is a frame for the orthogonal direct sum H1 H2 (which is indexed by J K), with dual frame Φ˜ Ψ˜ , and ⊕ ∪ ∪ Gram(Φ) 0 Gram(Φ Ψ)= Gram(Φ) Gram(Ψ) := . (5.2) ∪ ⊕ 0 Gram(Ψ) Conversely, if a frame for H has a Gramian matrix which can be block diagonalised as in (5.2), for some partitioning Φ Ψ of its vectors, then Φ and Ψ are frames for ∪

99 100 5 Combining and decomposing frames their spans H1 and H2, which are orthogonal complements, i.e., H = H1 H2, and the dual frames of Φ and Ψ are given by the corresponding dual vectors⊕ of the original frame (see Exer. 5.2).

Example 5.1. The frame bounds satisfy (see Exer. 5.4)

AΦ Ψ = min AΦ ,AΨ , BΦ Ψ = max BΦ ,BΨ , ∪ { } ∪ { } and so a union of normalised tight frames is again a normalised tight frame.

Example 5.2. In the (extreme) case when Gram(Φ) is diagonal, we have

H φ φ˜ 1 φ = span , = φ φ , φ Φ { } , ∈ and Φ is an orthogonal basis.

The union is a natural ﬁrst candidate for decomposing a frame.

Each frame can be uniquely decomposed into a union of frames for orthogonal subspaces (each corresponds to a component of the frame graph of 8.3). §

If Φ =( f j) and Ψ =(gk) are finite tight frames for H , then so is the frame consisting their union as a subset of H . In [LMO14] a finite tight frame for H is said to be divisible if it can be partitioned into two tight frames for H , i.e., it has a proper subset which is a tight frame for H , and otherwise it is prime. Clearly, every finite tight frame can be partitioned into prime tight frames (though not uniquely).

5.2 Direct sums

We consider two notions of the sum of frames: the direct sum, and sum (see 5.5). The ﬁrst of these requires that each summand have the same index set, and§ the second has no counterpart for inﬁnite dimensional spaces.

Deﬁnition 5.1. Let Φ =(φ j) j J and Ψ =(ψ j) j J be frames for H1 and H2, with the same index set J. Then their∈ (inner) direct sum∈ is

Φ Ψ =(φ j + ψ j) j J H1 H2. ⊕ ∈ ⊂ ⊕

The direct sum may not be a frame for H1 H2, e.g., when ⊕

dim(H1 H2)= dim(H1)+ dim(H2) > J . ⊕ | | A necessary and sufﬁcient condition (see Exer. 5.3) is disjointness, i.e, 5.2 Direct sums 101

ran(V ∗) ran(W ∗)= 0 , V :=[φ j] j J, W :=[ψ j] j J. (5.3) ∩ { } ∈ ∈ If Φ Ψ is a frame, i.e., Φ and Ψ are disjoint, then we say ⊕ Φ Ψ is an orthogonal dilation of Φ and Ψ, ⊕ Φ, Ψ are orthogonal compressions of Φ Ψ. ⊕ The direct sum of three (or more) mutually disjoint frames is not well deﬁned, in general. For example, the frames

Φ1 =(e1,0,e2,0,), Φ2 =(0,e1,0,e2), Φ3 =(e1,e1,e2,e2) are mutually disjoint frames for R2, so the direct sum of any two is a basis for R4, which cannot be disjoint from the remaining frame (by a dimension count). The stronger condition of orthogonality (see Deﬁnition 3.5), i.e.,

ran(V ∗) ran(W ∗), ⊥ leads to a direct sum for which the dual frame is the direct sum of the duals. The direct sum jΦ j = Φ1 Φn of mutually orthogonal frames for H1,...,Hn can be deﬁned in the⊕ obvious way,⊕⊕ and is associative, since

ran([φ j + ψ j]∗)= ran([φ j]∗)+ ran([ψ j]∗).

We observe by deﬁnition (and Proposition 3.2) that:

Disjointness and orthogonality depend only on the frames up to similarity.

Lemma 5.1. (Orthogonality) Let Φ =(φ j) j J and Ψ =(ψ j) j J be ﬁnite frames for ∈ ∈ H1 and H2, with V =[φ j] and W =[ψ j]. Then the following are equivalent 1. Φ and Ψ are orthogonal (strongly disjoint), i.e., ran(V ) ran(W ). ∗ ⊥ ∗ 2. The canonical coordinates of Φ and Ψ are orthogonal, i.e., ran(PΦ ) ran(PΨ ). 3. Gram(Φcan)Gram(Ψ can)= 0. ⊥ 4. WV = 0, i.e., ∑ f ,φ j ψ j = 0, f H1. ∗ j ∀ ∈ 5. VW = 0, i.e., ∑ g,ψ j φ j = 0, g H2. ∗ j ∀ ∈ Proof. 1. 2. We observe that ran(V )= ker(V) = dep(Φ) = ran(PΦ ), and, ⇐⇒ ∗ ⊥ ⊥ similarly, ran(W ∗)= ran(PΨ ). 2. 3. Since PΦ = Gram(Φcan) and PΨ = Gram(Ψ can), this follows by the fact that⇐⇒ two subspaces are orthogonal if and only if the product of the orthogonal projections onto them is zero. 3.= 4. By the factorisation (3.12), we have ⇒

∑ f ,φ j ψ j = WV ∗ f = WPΨ (VPΦ )∗ f = W(PΨ PΦ )V ∗ f = 0. j 102 5 Combining and decomposing frames

4.= 5. We have VW ∗ =(WV ∗)∗ = 0∗ = 0. 5.=⇒1. If WV = 0, i.e., W(V f )= 0, f , then ⇒ ∗ ∗ ∀

ran(V ∗) ker(W)=(ran(W ∗))⊥ = ran(V ∗) ran(W ∗). ⊂ ⇒ ⊥

⊓⊔

Theorem 5.1. The (inner) direct sum Φ Ψ is a frame for H1 H2, with ⊕ ⊕

SΦ Ψ ( f + g)= SΦ ( f )+ SΨ (g), f H1, g H2, (5.4) ⊕ ∀ ∈ ∀ ∈ or, equivalently, dual frame given by

(Φ Ψ)˜= Φ˜ Ψ˜ (5.5) ⊕ ⊕ if and only if Φ and Ψ are orthogonal. In this case, the frame bounds are

AΦ Ψ = min AΦ ,AΨ , BΦ Ψ = max BΦ ,BΨ , (5.6) ⊕ { } ⊕ { } and so a direct sum of normalised tight frames is a normalised tight frame.

Proof. Expanding gives

SΦ Ψ ( f + g)= SΦ ( f )+ SΨ (g)+ ∑ f ,φ j ψ j + ∑ g,ψ j φ j, ⊕ j j with the last two sums depending only on f and g, respectively. Thus Φ and Ψ are orthogonal if and only if (5.4) holds. In this case (see Exer. 5.4), Φ Ψ is a frame ⊕ for H1 H2, with frame bounds (5.6). The condition (5.4) is equivalent to (5.15) for the spanning⊕ sequence Φ Ψ, i.e., ⊕ 1 1 1 (φ j + 0)˜= SΦ− Ψ (φ j + 0)= SΦ− (φ j)+ SΨ− (0)= φ˜j, ⊕ and similarly (0 + ψ j)˜= ψ˜ j, which gives (5.5). ⊓⊔ Example 5.3. (Gramian) The Gramian and canonical Gramian of a direct sum satisfy

Gram(Φ Ψ)= Gram(Φ) Gram(Ψ), ⊕ ⊕ ran(PΦ Ψ )= ran(PΦ )+ ran(PΨ ). ⊕ Therefore, when Φ and Ψ are orthogonal, we have

PΦ Ψ = PΦ + PΨ , ⊕ and the canonical coordinates satisfy

Φ Ψ Φ Ψ c ⊕ ( f + g)= c ( f )+ c (g). 5.2 Direct sums 103

Theorem 3.5 can be restated in terms of direct sums as follows.

Theorem 5.2. Given a ﬁnite frame Φ =(φ j) j J for H , there exists an orthogonal ∈ frame Ψ =(ψ j) j J for some K , such that Φ Ψ is a basis for H K , with the same frame bounds∈ as Φ. In particular, Φ and⊕ its dual are the orthogonal⊕ projection of the biorthogonal system given by Φ Ψ and its dual. ⊕ The above Ψ is an example of what will be called a complement of Φ.

Example 5.4. (Decomposition) Let Φ =( f j) j J be a ﬁnite frame, with V =[ f j], and J J ∈ P : F F be any orthogonal projection onto a subspace of ran(V ∗). Then Φ is a direct sum→ of the orthogonal frames with synthesis operators VP and V(I P), since −

ran((VP)∗)= ran(PV ∗) ran((V(I P))∗)= ran((I P)V ∗). ⊥ − − Thus a frame can be decomposed into a direct sum of orthogonal one–dimensional frames in many ways. Therefore the usefulness of direct sums for decompositions seems very limited, unless some additional structure is present (see Example 11.7).

Example 5.5. Consider the frame of four equally spaced unit vectors in R2

1 0 1 0 Φ = , , , , 0 1 −0 1 − 1 and the frames Ψ1 and Ψ2 for R given by 1 1 1 1 1 1 1 1 Ψ1 =( , , , ), Ψ2 =( , , , ). √2 √2 √2 √2 √2 −√2 √2 −√2 These are mutually orthogonal tight frames, e.g.,

1 0 1 0 0 VW = 1 1 1 1 = , ∗ 0 1− 0 1 √2 √2 √2 √2 0 − 3 with the same frame bound. The tight frames Φ Ψ1 and Φ Ψ2 for R are the lifted four equally spaced vectors (see Example 5.9)⊕ and the vertic⊕ es of the tetrahedron. 2 The direct sum Ψ1 Ψ2 consists of two copies of an orthonormal basis for R , and ⊕ 4 Φ Ψ1 Ψ2 is an orthogonal basis for R . ⊕ ⊕ Example 5.6. Let Φ =(1,1,1) which is a (tight) frame for F. Both of the frames

Ψ1 =(1, 1,0), Ψ2 =(2, 1, 1) − − − are orthogonal to Φ, but neither Ψ1 and Ψ2 are similar, nor are Φ Ψ1 and Φ Ψ2. ⊕ ⊕ We now consider two special cases of the direct sum: when a summand is for a space of 1–dimension (lifting) and when the direct sum is a basis (complements). 104 5 Combining and decomposing frames 5.3 Lifting

The idea of lifting is to take a frame, and add an additional component to each of its vectors, so as to obtain a frame for a space of dimension one higher, i.e., take the direct sum with a frame for a one–dimensional space.

Deﬁnition 5.2. Let Φ =(φ j) j J be frame for H , and Ψ =(α jψ) j J be a frame for K = span ψ , ψ = 0. Then ∈Φ Ψ is a lift of Φ to H K (by∈Ψ) if Φ and Ψ are orthogonal,{ } i.e., ⊕ ⊕ ∑α¯ jφ j = 0, (5.7) j and is a simple lift when all the α j are equal. Since the lift of a frame is an inner direct sum, Theorem 5.1 gives that the dual frame of the lift of Φ by Ψ =(α jψ) j J is the lift of Φ˜ by ∈ Ψ˜ α ψ 1 =(c j ) j J, c := 2 . ∈ ψ,ψ ∑ α j j | | φ φ A ﬁnite frame ( j) is said to be balanced if ∑ j j = 0. Clearly, a balanced frame cannot be a basis, and a frame is balanced if and only if its dual frame is balanced.

Example 5.7. (Simple lifts) A frame has a simple lift if and only if it is balanced, and (see Exer. 5.5) a frame Φ =( f j) is a simple lift if and only if

∑ f j = 0, ∑ f j, fk = C, k. j j ∀

Φ φ n d Example 5.8. (Repeated lifts) If =( j) j=1 is in F , then condition (5.7) is that α =(α j) is a nonzero vector orthogonal to the rows of the matrix [φ1,...,φn]. Thus a ﬁnite frame can be lifted if and only if it is not a basis. Moreover, since

2 2 AΨ = BΨ = ψ ∑ α j , Ψ =(α jψ), j | | the lift by Ψ has the same frame bounds as Φ provided

2 2 AΦ ψ ∑ α j BΦ . ≤ j | | ≤

In this way, one can successively lift Φ until a basis is obtained, which gives a simple proof of Theorem 3.5 (also see Proposition 5.1). For example, the frame

1 1 1 1/3 1/3 1/3 Φ = , , , Φ˜ = , , , 0 1 1 0 1/2 1/2 − − for R2 has frame bounds AΦ = 2, BΦ = 3. The vector α =(2c, c, c) is orthogonal to rows of the matrix with these columns, and so lifts Φ to a basis− − for R3, i.e., 5.4 Complements 105

1 1 1 1/3 1/3 1/3 Ψ = 0 , 1 , 1 , Ψ˜ = 0 , 1/2 , 1/2 . 2c −c  c 1/(3¯c)  −1/(6¯c)  1/(6¯c) − − − −             2 2 1 1 This has the same frame bounds as Φ if 2 ∑ α j = 6 c 3, i.e., c . ≤ i | | | | ≤ √3 ≤| |≤ √2 n Example 5.9. (Lifted roots of unity) Let Φ =(u j) be n 3 equally spaced unit j=1 ≥ vectors in R2. Since these vectors sum to zero, they can be lifted by adding a third coordinate, say u j lifts to v j =(u j,α), α > 0. The condition on α which ensures n 2 1 that (v j) is tight is AΦ = = n α , i.e., α = . Thus we obtain the equal norm 2 | | √2 tight frame 2π j 2π j 1 (cos ,sin , ) : j = 1,...,n { n n √2 } which we call the lifted n–th roots of unity (or the lifted equally spaced vectors).

Fig. 5.1: The lifted n equally spaced unit vector in R2 for n = 3,6,9.

5.4 Complements

Every normalised tight frame has a unique complement, and each frame is similar to a unique normalised tight frame. Thus we can deﬁne a complement (which is unique up to similarity) for an arbitrary frame. This can be thought of as a maximal orthogonal direct summand, or as the direct sum of a sequence of lifts which take the frame to a basis.

Definition 5.3. We say that finite frames Φ =(φ j) j J and Ψ =(ψ j) j J for H1 and ∈ ∈ H2, with the same index set J, are complements of each other if they are disjoint and H1 H2 ℓ2(J), i.e., dim(H1)+ dim(H2)= J . ⊕ ≈ | | For normalised tight frames this is equivalent to Definition 2.6. We recall (see 2.9) that the α–partition frame is the normalised tight frame of nonzero vectors §which is the complement of e e e e 1 ,..., 1 ,..., k ,..., k . √α1 √α1 √αk √αk α 1 times αk times 106 5 Combining and decomposing frames

Disjoint frames cannot be similar, and so, in particular, no normalised tight frame is equal to its complement. By way of contrast, this is not true in inﬁnite dimensions, e.g., [Cas98] shows that every frame for an inﬁnite dimensional Hilbert space can be written as the direct sum of three orthonormal bases. Here are some equivalences that follow easily from the previous discussion.

Proposition 5.1. Let Φ =(φ j) j J, Ψ =(ψ j) j J be a frames for H1, H2, with index ∈ ∈ set J, and synthesis maps V =[φ j], W =[ψ j]. Then the following are equivalent 1. Φ and Ψ are complements. 2. Φcan and Ψ can are complements. 3. Gram(Φcan)+ Gram(Ψ can)= I. 4. ℓ2(J)= ran(PΦ ) ran(PΨ ). ⊕ 5. dim(H1)+ dim(H2)= J and VW ∗ = 0. 6. Φ Ψ is a basis and (Φ| | Ψ)˜= Φ˜ Ψ˜ . 7. Ψ ⊕= Q(I Gram(Φcan)),⊕ where Q is⊕ 1–1 on ran(I Gram(Φcan)). − − 8. Ψ = Ψ1 Ψk, k = J dim(H1), where Ψj lifts Φ Ψ1 Ψj 1. ⊕⊕ | |− ⊕ ⊕⊕ −

5.5 Sums

The sum of frames requires that at least one of them is balanced. Φ φ n1 Ψ ψ n2 H H Deﬁnition 5.4. Let =( j) j=1 and =( k)k=1 be frames for 1 and 2, at least one of which is balanced, i.e., has vectors which sum to zero. Then their sum is 1 1 Φ +ˆ Ψ := φ j + ψk . (5.8) 1 j n1 √n2 √n1 1≤k≤n ≤ ≤ 2

This is a frame of n1n2 elements for H1 H2, which is tight if the summands are tight with the same frame bound. ⊕

Theorem 5.3. The sum Φ +ˆ Ψ deﬁned by (5.8) is a frame of n1n2 vectors for H1 ⊕ H2, which is balanced if and only if both Φ and Ψ are balanced, with

SΦ ˆ Ψ ( f + g)= SΦ f + SΨ g, f H1, g H2, + ∀ ∈ ∀ ∈ or, equivalently, dual frame given by

(Φ +ˆ Ψ)˜= Φ˜ +ˆ Ψ˜ .

The frame bounds are

AΦ ˆ Ψ = min AΦ ,AΨ , BΦ ˆ Ψ = max BΦ ,BΨ . (5.9) + { } + { } In particular, a sum of normalised tight frames is a normalised tight frame. 5.5 Sums 107

Proof. Let S be the frame operator for (αφ j + βψk), with α,β scalars, then

S( f + g)= ∑∑ f + g,αφ j + βψk (αφ j + βψk) j k 2 ¯ 2 = α n2SΦ f + αβ¯ f ,∑φ j ∑ψk + αβ g,∑ψk ∑φ j + β n1SΨ g. | | j k k j | |

If Φ or Ψ is balanced, and we choose α = 1 , β = 1 , then this reduces to √n2 √n1

S( f + g)= SΦ f + SΨ g.

By Exer. 5.4, Φ +ˆ Ψ is a frame for H1 H2, with the asserted properties. ⊕ ⊓⊔ The sum +ˆ also satisﬁes the following rules

(Φ1 +ˆ Φ2) +ˆ Φ3 = Φ1 +(ˆ Φ2 +ˆ Φ3), α(Φ1 +ˆ Φ2)=(αΦ1) +(ˆ αΦ2).

Example 5.10. Three equally spaced vectors Φ in R2 are a balanced tight frame. Taking the sum of Φ with itself gives the following tight frame of 9 vectors for R4

2π cos 3 j 2π Φ Φ sin 3 j +ˆ :=  2π  : 1 j,k 3 . cos 3 k ≤ ≤ sin 2π k  3    Example 5.11. (Roots of unity) The n–th roots of unity sum to zero, and so form 2πi/n 2πi/n a balanced tight frame for C. Let ω := e 1 , µ := e 2 be n1,n2 2 roots of unity. Then Φ :=( 1 ω j)n1 and Ψ :=( 1 µk)n2 are balanced normalised≥ tight √n1 j=1 √n2 k=1 frames, and so their sum

1 ω j Φ +ˆ Ψ := µk 1 j n1 √n1n2 1≤k≤n ≤ ≤ 2 2 is an equal–norm balanced normalised tight frame of n1n2 vectors for C .

Example 5.12. Taking d summands of Φ =( 1 , 1 ) gives the vertices of the √2 − √2 cube in Rd, i.e.,

( 1) j1 1 − Φ +ˆ +ˆ Φ = . √ d  j  1 j1,..., jd 2 2 ( 1) d ≤ ≤ −   Example 5.13. (Equal–norm tight frames) Clearly the sum of equal–norm frames is an equal–norm frame, and by (5.9), the sum of equal–norm normalised tight frames (one of them balanced) is an equal–norm normalised tight frame. 108 5 Combining and decomposing frames 5.6 Tensor products

The tensor product of vector spaces V and W is the (abstract) vector space V W with the property that any bilinear map from V W factors uniquely through V ⊗W. There is an associated bilinear map × ⊗

: V W V W : ( f ,g) f g ⊗ × → ⊗ → ⊗ whose images f g are called irreducible tensors, and span V W. ⊗ ⊗ If v1,...,vm and w1 ...,wn are bases for V and W, then V W can be realised as the{ mn–dimensional} { vector space} with a basis given by (the⊗ formal symbols) v j wk, 1 j m, 1 k n, which satisfy the distributive law ⊗ ≤ ≤ ≤ ≤

∑α jv j ∑βkwk = ∑∑α jβ j(v j wk), α j,βk F. j ⊗ k j k ⊗ ∀ ∈

For Hilbert spaces H1 and H2, the tensor product Hilbert space H1 H2 is ⊗ obtained by taking the inner product on H1 H2 given by ⊗

f g,φ ψ H H := f ,φ H g,ψ H , f ,φ H1, g,ψ H2. ⊗ ⊗ 1⊗ 2 1 2 ∀ ∈ ∀ ∈

Deﬁnition 5.5. The tensor product of frames Φ and Ψ for H1 and H2 is

Φ Ψ :=(φ ψ)φ Φ,ψ Ψ H1 H2. ⊗ ⊗ ∈ ∈ ⊂ ⊗ The tensor product of frames is a frame.

Theorem 5.4. If Φ and Ψ are frames for H1 and H2, then Φ Ψ is a frame for the ⊗ tensor product Hilbert space H1 H2, with frame operator ⊗ SΦ Ψ = SΦ SΨ , (5.10) ⊗ ⊗ frame bounds AΦ Ψ = AΦ AΨ , BΦ Ψ = BΦ BΨ , (5.11) ⊗ ⊗ and dual frame given by (φ ψ)˜= Φ˜ Ψ˜ . (5.12) ⊗ ⊗ Proof. For irreducible tensors f g H1 H2, we have ⊗ ∈ ⊗ SΦ Ψ ( f g)= ∑ ∑ f g,φ ψ φ ψ = ∑ ∑ ( f ,φ φ) ( g,ψ ψ) ⊗ ⊗ φ Φ ψ Ψ ⊗ ⊗ ⊗ φ Φ ψ Ψ ⊗ ∈ ∈ ∈ ∈ = ∑ f ,φ φ ∑ g,ψ ψ =(SΦ f ) (SΨ g), φ Φ ⊗ ψ Ψ ⊗ ∈ ∈ and so we obtain SΦ Ψ = SΦ SΨ by linearity. Since the eigenvalues of a tensor product of operators are⊗ the products⊗ of eigenvalues of the operators, we obtain the asserted frame bounds. 5.6 Tensor products 109

1 1 1 Finally, the dual frame is SΦ− Ψ (φ ψ)=(SΦ− φ) (SΨ− ψ)= φ˜ ψ˜ . ⊗ ⊗ ⊗ ⊗ ⊓⊔

Tensor products of frames with various properties inherit those properties.

Corollary 5.1. . Let Φ and Ψ be frames. Then 1. Φ Ψ is a tight frame if and only its factors are. 2. Φ ⊗Ψ is an equal–norm tight frame if and only its factors are. 3. Φ ⊗Ψ is an orthonormal basis if and only its factors are. 4. Φ ⊗Ψ is an orthogonal basis if and only its factors are. 5. Φ ⊗Ψ is a real frame if and only if its factors are. ⊗ Proof. Let Φ =(φ j) and Ψ =(ψk). 1. Use (5.11) and A B. ≤ 2. Use φ j ψk = φ j ψk . ⊗φ ψ φ ψ φ φ ψ ψ 3., 4. Use j1 k1 , j2 k2 = j1 , j2 k1 , k2 . 5. If Φ Ψ is⊗ real (and Ψ⊗has a nonzero vector), then ⊗ 2 φ j ,φ j ψk = φ j ψk,φ j ψk R = φ j ,φ j R, 1 2 1 ⊗ 2 ⊗ ∈ ⇒ 1 2 ∈ and similarly ψk ,ψk R. If Φ and Ψ are real, then clearly Φ Ψ is real. 1 2 ∈ ⊗ ⊓⊔

m n Example 5.14. (Equally spaced vectors) Let (v j) j=1 and (wk)k=1 be the tight frames of m and n equally spaced unit vectors in R2, given by 2π j 2π j 2πk 2πk v := cos e + sin e , w := cos e + sin e , j m 1 m 2 k n 1 n 2 where e j is the standard basis. By taking their tensor product { } 2π j 2πk 2π j 2πk v j wk := cos cos e1 e1 + cos sin e1 e2 ⊗ m n ⊗ m n ⊗ 2π j 2πk 2π j 2πk + sin cos e2 e1 + sin sin e2 e2, m n ⊗ m n ⊗

4 we obtain the tight frame Φ =(φ jk) of mn unit vectors for R , given by

2π j 2πk cos m cos n 2π j 2πk cos m sin n j = 1,...,m, φ jk :=  π π , sin 2 j cos 2 k k = 1,...,n. mπ n  sin 2 j sin 2πk   m n    Example 5.15. (Orthogonal polynomials) Tensor products of orthogonal polynomials are orthogonal polynomials for the tensor product weight. This can be used to construct frames of orthogonal polynomials in several variables (see [DX01]). 110 5 Combining and decomposing frames 5.7 Decompositions

In general, there is not a unique way to decompose a frame.

Example 5.16. (Nonuniqueness) The tight frame of four equally spaced unit vectors in R2 (the vertices of the square) can be written as a disjoint union, direct sum (lift), sum and tensor product, e.g.,

1 1 0 0 (1, 1) (1, 1)= , − , , , − ∪ − 0 0 1 1 − 1 1 0 0 (1, 1,0,0)+(0,0,1, 1)= , − , , , − − 0 0 1 1 − 1 1 1 1 ( 1,1)+(ˆ 1,1)= , − , − , , − − 1 1 1 1 − − ( 1,1) (e1,e2)=(e1,e2, e1, e2). − ⊗ − − This tight frame is also divisible, e.g., it can be partitioned into two prime tight frames e1,e2 e1, e2 . { }∪{− − }

Notes

The direct sum, and the associated notions of a lift and complement can be found throughout the frame literature. There can be some variation in terminology (when named), e.g., the term lift is used for the simple lift in [BF03].

Exercises

5.1. Show that a frame ( f j) is a disjoint union of tight frames if and only if each f j is an eigenvector of the frame operator S (such a frame is said to be semicritical).

5.2. Unions. Let Φ =(φ j) and Ψ =(ψk) be frames for H1 and H2. (a) Show that Φ Ψ, as a sequence in H1 H2, has frame operator S, with ∪ ⊕

S( f + g)= SΦ ( f )+ SΨ (g), f H1, g H2. ∀ ∈ ∀ ∈ (b) Show that the dual frame is Φ˜ Ψ˜ , by using Exer. 5.4. (c) Show the Gramian is block diagonal,∪ i.e., 5.7 Decompositions 111

Gram(Φ) 0 Gram(Φ Ψ)= . ∪ 0 Gram(Ψ)

5.3. Direct sums. Let Φ =(φ j) j J and Ψ =(ψ j) j J be ﬁnite frames for H1 and H2, ∈ ∈ with V =[φ j], W =[ψ j]. Show that (φ j + ψ j) is a frame for H1 H2 if and only if ⊕

ran(V ∗) ran(W ∗)= 0, ∩ in which case

ran([φ j + ψ j]∗)= ran(V ∗)+ ran(W ∗) (algebraic direct sum).

5.4. Suppose Ξ is a sequence in H1 H2 for which S = SΞ can be decomposed ⊕

S( f + g)= SΦ ( f )+ SΨ (g), f H1, g H2, (5.13) ∀ ∈ ∀ ∈ where Φ and Ψ are frames for H1 and H2, e.g., the sequences

Φ Ψ, Φ Ψ, Φ +ˆ Ψ. (5.14) ∪ ⊕

(a) Show that Ξ is a frame for H1 H2, with frame bounds ⊕ A = min AΦ ,AΨ , B = max BΦ ,BΨ . { } { } In particular, this implies that unions, direct sums and sums of normalised tight frames are again normalised tight frames. (b) Show that (5.13) is equivalent to

1 1 1 S− ( f + g)= SΦ− ( f )+ SΨ− (g), f + g Ξ. (5.15) ∀ ∈ For the choices (5.14), this gives

(Φ Ψ)˜= Φ˜ Ψ˜ , (Φ Ψ)˜= Φ˜ Ψ˜ , (Φ +ˆ Ψ)˜= Φ˜ +ˆ Ψ˜ . ∪ ∪ ⊕ ⊕

5.5. Show that a ﬁnite frame ( f j) j J is a simple lift if and only if ∈

∑ f j = 0, ∑ f j, fk = C, k. j j ∀

Chapter 6 Variational characterisations of tight frames

If ( f j) j J is a finite tight frame for H , then (see Proposition 2.1) ∈ 2 2 1 ∑ ∑ f j, fk = ∑ f j, f j , d = dim(H ). (6.1) j J k J | | d j J ∈ ∈ ∈ For a general finite frame, this becomes an inequality, with equality if and only if the frame is tight. Thus (6.1) characterises finite tight frames. Important instances (for unit vectors) include Welch bound equality sequences, minimisers of the frame potential, spherical half–designs of order 2, and spherical (1,1)–designs.

6.1 Welch bound equality sequences

d Let f1,..., fn be n d unit vectors in C (signals of unit energy). Then ≥ n n 2 2 n F ( f1,..., fn) := ∑ ∑ f j, fk , (6.2) j=1 k=1 | | ≥ d which is known as the Welch bound, after [Wel74], which used (6.2) to prove

2 2 n /d n n d max f j, fk − = − . (6.3) j=k | | ≥ n2 n d(n 1) − − Each gives a lower bound on how small the cross–correlation of a set of signals of unit energy can be, i.e., how “spread out” the signals are. Unit vectors f1,..., fn which give equality in (6.2) are called WBE sequences (Welch bound equality sequences), see, e.g., [MM93] where they are used for CDMA (code division multiple access) systems. Equality in the Welch bound is the same as the equality (6.1).

113 114 6 Variational characterisations of tight frames 6.2 The variational characterisation

The following extends the Welch bound and frame potential results (see 6.3) to the case where the vectors may have arbitrary lengths (also see Exer. 6.1). §

Theorem 6.1. Let f1,..., fn be vectors in H , not all zero, and d = dim(H ). Then

n n n 2 2 1 2 ∑ ∑ f j, fk ∑ f j , (6.4) j=1 k 1 | | ≥ d j=1 = n H with equality if and only if ( f j) j=1 is a tight frame for .

Proof. Let V =[ f j]. We recall from (2.7) that the frame operator S = VV ∗ satisﬁes

2 2 2 trace(S)= ∑ f j , trace(S )= ∑∑ f j, fk . j j k | |

Since S = VV ∗ is positive deﬁnite, it is unitarily diagonalisable with eigenvalues λ1,...,λd 0. By the Cauchy–Schwarz inequality ≥ 2 λ 2 λ 2 2 λ 2 λ 2 2 trace(S) =(∑ j) = (1),( j) (1) ( j) = d ∑ j = d trace(S ), j ≤ j which is (6.4), with equality if and only if λ j = A, j, A > 0 i.e., ∀

S = AIH ( f j) is a tight frame for H . ⇐⇒

Note above, since one of the vectors ( f j) is nonzero, S = 0, and so A = 0. ⊓⊔ Example 6.1. If all the vectors f j have unit norm, then (6.1) reduces to the Welch bound (6.2), i.e., n n n 2 2 2 1 2 n ∑ ∑ f j, fk ∑ 1 = . j=1 k 1 | | ≥ d j=1 d = Example 6.2. The corresponding generalisation of the Welch bound (6.3) is

2 2 4 2 (∑ℓ fℓ ) /d ∑ℓ fℓ max f j, fk − > 0. j=k | | ≥ n2 n − The tightness of a ﬁnite frame can be determined from the absolute values of the entries of its Gramian:

n H H The vectors ( f j) j=1 are a tight frame for , d = dim( ), if and only if

n n n 2 2 1 2 ∑ ∑ f j, fk = ∑ f j > 0. (6.5) j=1 k 1 | | d j=1 = 6.3 The frame potential 115 6.3 The frame potential

Let S = f H : f = 1 be the unit sphere in H , d = dim(H ). The function { ∈ } n n n ∞ n F 2 FP : S [0, ) : ( f j) j=1 ( f1,..., fn)= ∑ ∑ f j, fk (6.6) → → j=1 k=1 | | of (6.2) was called the frame potential by Fickus (see [Fic01]) who derived it from a frame force (see 6.14). We recall that unit norm tight frames exist (Example 2.4). Since a tight frame§ for H is a spanning set (which must have at least d elements), n2 Theorem 6.1 implies that for n d the minimum of the frame force is d , which is attained precisely for unit norm≥ tight frames For n < d the minimum of n is attained when ( f j) is a (nonspanning) orthonormal sequence (see Exer. 6.5). In summary: Φ n H H Theorem 6.2. ([BF03]). Let =( f j) j=1 be n unit vectors in , d = dim( ). Then the frame potential is bounded below by

n2 FP(Φ) max ,n ≥ { d } with equality (a minimum) if and only if either (a) Φ is a unit norm tight frame (and hence n d), or (b) Φ is a nonspanning orthonormal sequence≥ (and hence n < d). A careful analysis of the Lagrange multiplier equations (see 6.15) shows that all local minimisers of the frame potential give the global minimum.§ This is in contrast with many other well known potentials, e.g., the Coulomb electrostatic potential, which do not have this property. In view of (6.4), a normalised frame potential FPˆ : H n 0 [0,∞) can be deﬁned (see Exer. 6.6) on sequences of vectors which are not\ all { zero} → by

∑n ∑n 2 1 j=1 k=1 f j, fk FPˆ ( f1,..., fn) := | | 1. (6.7) n 2 2 d ≤ ∑ f j ≤ j=1 This leads to the following extension of Theorem 6.2.

Theorem 6.3. ([Wal03]). For n d = dim(H ) the normalised frame potential ˆ ≥ 1 n FP( f1,..., fn) attains its minimum of d if and only if ( f j) j=1 is a tight frame for H n , and its maximum of 1 if and only if span( f j) j=1 is 1–dimensional.

1 Thus the distance of the normalised frame potential from d gives a scaling independent way of measuring how far from being tight it is (also see the discussion of 3.8). In particular, sequences of vectors which do not span Fd are far from tight, since§ 1 FPˆ ( f1,..., fn) . ≥ dim(span1 j n f j ) ≤ ≤ { } 116 6 Variational characterisations of tight frames 6.4 Spherical t-designs and the Waring formula

We now outline some of the basic results about cubature on the sphere in Rd, and its relationship to tight frames. These are proved, together with the analogous results d for the complex sphere, in 6.8. Let Hom j(R ) be the homogeneous polynomials of total degree j in d real variables.§ Definition 6.1. A finite subset Φ of the unit sphere S in Rd is a (real) spherical t–design if the normalised surface integral satisfies

1 d f dσ = ∑ f (φ), f Hom j(R ), 0 j t, S Φ φ Φ ∀ ∈ ≤ ≤ | | ∈ anditisa(real) spherical half–design1 of order t if

1 d f dσ = ∑ f (φ), f Homt (R ). S Φ φ Φ ∀ ∈ | | ∈ Spherical designs have been studied since the 1970’s (see [DGS77] and [BB09]). Their existence for every t (and d 2) was proved by [SZ84] (see Theorem 6.5). ≥ Example 6.3. The 12 vertices of the regular icosahedron give a spherical 5–design. Example 6.4. The n equally spaced vectors in R2 are a spherical (n 1)–design, and the n = t + 1 equally spaced lines in R2 give a spherical half design− of order 2t. The following simple observations are very useful (and illuminating). d 2 If f Hom j(R ), then g = f Hom j+2, and the restriction of f and g to • the sphere∈ S are equal. Thus if Φ is∈ a spherical half–design of order t, then it is also a spherical half–design of order t 2,t 4,.... d − − For j odd, every f Hom j(R ) is an odd function, and so has zero integral. Thus • if Φ is centrally symmetric∈ , i.e, Φ = Φ, then it is a spherical half–design of order 1,3,5,.... − d For t even, a set Φ = φ1,...,φn of unit vectors in R is a spherical half–design of order t if and only if it{ satisﬁes the} so called Waring formula

n t/2 d(d + 2) (d +t 2) 1 t d x,x = − ∑ x,φ j , x R , (6.8) 1 3 5 (t 1) n ∀ ∈ − j=1 or, equivalently (see [Sid74], [GS79], [Sei01])

n n t 2 1 3 5 (t 1) ∑ ∑ φ j,φk = n − . (6.9) d(d + 2) (d +t 2) j=1 k=1 − 1 There is some variation of terminology in the literature, e.g., Seidel [Sei01] refers to a spherical half–design of order t as a “spherical t–design”. Spherical t–designs whose number of vectors satisfy the lower bounds of [DGS77] are said to be tight (see [BBHS10] for a classiﬁcation). 6.5 Other characterisations of spherical t–designs 117

For t = 2, these become

n n n 2 2 d 2 d 2 n x = ∑ x,φ j , x R , ∑ ∑ φ j,φk = , (6.10) n j=1 | | ∀ ∈ j=1 k=1 d i.e., Φ is a tight frame by deﬁnition, and by equality in (6.4), respectively.

A spherical half–design of order 2 is precisely a tight frame of distinct unit vectors in Rd (equivalently, a WBE sequence of distinct vectors).

We now consider the relationship between tight frames and spherical 2–designs. d Proposition 6.1. A set Φ := φ1,...φn of unit vectors in R is a spherical 2–design if and only if it is a balanced{ tight frame,} i.e.,

n n n 2 2 n ∑ φ j = 0, ∑ ∑ φ j,φk = . (6.11) j=1 j=1 k=1 d

Moreover, such a Φ is also a 3–design if it is centrally symmetric. Proof. The second condition in (6.11) ensures that Φ is spherical half–design of order 2 (and hence 0). It therefore remains to add to this a condition which ensures Φ is is spherical half–design of order 1. Since the homogeneous polynomials of d degree 1 have zero integral, and are spanned by py := ,y , y R , the condition is ∈ n n n 1 1 1 d 0 = py dσ = ∑ py(φ j)= ∑ φ j,y = ∑ φ j,y , y R , n n n ∀ ∈ j=1 j=1 j=1 φ Φ i.e., ∑ j j = 0. We already observed that if is centrally symmetric, then it is a spherical half design of order t = 1,3,5,.... ⊓⊔

6.5 Other characterisations of spherical t–designs

The description of homogeneous polynomials on the sphere (and their integrals) is intimately related to the harmonic polynomials (see 16.5 for details). From this connection come various characterisations of spherical§t–designs, which we brieﬂy d Π d consider. Each homogeneous polynomial f Homt (R )= t◦(R ) can be written uniquely ∈

2 j d f (x)= ∑ x ft 2 j(x), ft 2 j Harmt 2 j(R ), (6.12) − − − 0 j t ∈ ≤ ≤ 2 d d where Harmk(R )= Hk(R ) are the harmonic polynomials of degree k. Since the harmonic polynomials of different degrees are orthogonal with respect to the inner 118 6 Variational characterisations of tight frames product given by the normalised surface measure σ on S, integrating f gives

0, t is odd; f dσ = ∑ ft 2 j dσ = S S − f t is even 0 j t 0, . ≤ ≤ 2

Therefore Φ = φ j is spherical t–design if and only if { } d ∑ f (φ j)= 0, f Harm (R ), 1 ℓ t, (6.13) j ∀ ∈ ℓ ≤ ≤ and is a spherical half–design of order t if and only if

d ∑ f (φ j)= 0, f Harm (R ), ℓ = t,t 2,... (ℓ 1). (6.14) j ∀ ∈ ℓ − ≥

Φ φ d A set satisfying (6.14) for for ℓ = t, i.e., ∑ j f ( j)= 0, f Harmt (R ), is called a spherical design of harmonic index t (see [BOT15]). ∈ By using basic properties of reproducing kernels (see 16.3, 16.4 for details), [SW09] have converted the condition (6.13) into a variationa§ l characterisation§ of spherical t–designs similar to that of Theorem 6.2.

(d) Theorem 6.4. ([SW09]) For each ℓ, let Kℓ ( x,y ) be a positive scalar multiple of d n n the reproducing kernel for Harm (R ), Φ = φ j , and deﬁne At : S R by ℓ { } j=1 → t n n Φ (d) φ φ At ( ) := ∑ ∑ ∑ Kℓ ( j, k ). (6.15) ℓ=1 j=1 k=1

Then At (Φ) 0 and Φ is a spherical t–design if and only if At (Φ)= 0. ≥ (ℓ) d Proof. Let (Ys ) be an orthonormal basis for Hℓ(R ). Then Proposition 16.3 gives

(d) (ℓ) (ℓ) Kℓ ( x,y )= cℓ ∑Ys (x)Ys (y), cℓ > 0, s and we compute

t n n t n n (ℓ) (ℓ) (ℓ) (ℓ) At (Φ)= ∑ ∑ ∑ cℓ ∑Ys (φ j)Ys (φk)= ∑ cℓ ∑ ∑ Ys (φ j) ∑ Ys (φk) ℓ=1 j=1 k=1 s ℓ=1 s j=1 k=1 2 t n (ℓ) = ∑ cℓ ∑ ∑ Ys (φ j) 0. ℓ=1 s j=1 ≥

Clearly, there is equality above if and only if

(ℓ) ∑ Ys (φ j)= 0, s, 1 ℓ t, j ∀ ≤ ≤ 6.6 The existence of cubature formulas 119 i.e., by (6.13), Φ is a spherical t–design. ⊓⊔ d 2 (d) ( −2 ) Example 6.5. By (16.22), Kℓ is a multiple of the Gegenbauer polynomial Cℓ . By (16.40), we may take K(d)(z)= z and K(d)(z)= dz2 1, which gives 1 2 − 2 A2(Φ)= ∑∑ φ j,φk + ∑∑ d φ j,φk 1 j k j k − 2 2 2 n = ∑φ j + d ∑∑ φ j,φk . j j k − d

This is clearly minimised, i.e., gives a spherical 2–design, if and only if (6.11) holds. Example 6.6. Restricting the sum in (6.15) to ℓ = t,t 2,... (ℓ 1) and to ℓ = t gives − ≥ a nonnegative functional Bt (Φ) for which Bt (Φ)= 0 if and only if Φ is a spherical half–design of order t and a spherical design of harmonic index t, respectively.

6.6 The existence of cubature formulas

Φ φ n Spherical t–designs = j j=1 and their variants are examples of cubature rules. Their existence (without any{ } estimate of n) is guaranteed by the following result: Theorem 6.5. ([SZ84]) Let X be a path–connected topological space, and µ be a finite (positive) measure on X, defined on the open sets, with full support, i.e., µ(U) > 0 for every nonempty open set U X. For a continuous integrable function f : X Rm, there exists a finite set of samples⊂ A X for which → ⊂ 1 µ 1 µ f d = ∑ f (a). (X) X A a A | | ∈ Here A , the size of A, can be any number with a finite number of exceptions. | | This is a generalisation of the integral form of the mean value theorem.

Example 6.7. Let f =( f1,..., fm). For µ = σ the normalised surface area on S, d choosing f1,..., fm to be a spanning set for Homt (R ) gives the existence of a { } Π d spherical half–design of order t (equal weight cubature rule for t◦(R )). Similarly, by choosing a spanning set for the space of functions integrated by the cubature rule, one has

Spherical t–designs, spherical half–designs of order t and spherical designs of harmonic index t exist, for all values of t.

The construction of such designs (cubature rules) with small numbers of points (and estimates the smallest number of points) is a subject of ongoing interest. 120 6 Variational characterisations of tight frames 6.7 Tight frames of symmetric tensors

Here we extend (6.4) to an inequality, for which equality gives a tight frame for the symmetric tensors in t H := H H (t times), and a cubature rule for integration of certain homogeneous⊗ polynomials⊗⊗ on the sphere (see 6.8). For simplicity, we follow the development of 6.2 as closely as§ possible. Thus we deﬁne the symmetric tensors of rank t to be the§ subspace of t H given by ⊗ t t t Sym (H ) := span v⊗ : v H , v⊗ := v v (t times). { ∈ } ⊗⊗ This Hilbert space has dimension

t + d 1 dim(Symt (H )) = − , d = dim(H ), (6.16) t and we recall (see 5.6) that its inner product satisﬁes § t t t v⊗ ,w⊗ = v,w , v,w H . (6.17) ∀ ∈ t t t The dual space (Sym (H ))∗ = Sym (H ∗) contains ,v ⊗ , v H , and its inner product is given by ∈

t t t ,v ⊗ , ,w ⊗ = w,v , v,w H . (6.18) ∀ ∈ t There is a vector space isomorphism between Sym (H ∗) and the space Lt (H ,F) of symmetric t–linear maps from H t F which is given by → t λ ⊗ L, (λ H ∗) L(v1,...,vt ) := λ(v1) λ(vt ). → ∈ We deﬁne the space of homogeneous polynomials on H of degree t to be

Π ◦(H ) := Lˆ : L Lt (H ,F) , Lˆ : H F, Lˆ(v) := L(v,...,v). t { ∈ } → ˆ L H Π H The map L L above gives a vector space isomorphism t ( ,F) t◦( ). → Π H t H → The inner product on t◦( ) induced from that on (Sym ( ))∗ via the above isomorphisms is the apolar (or Bombieri or Fisher) inner product, which is given by t t t t t ,v , ,w := ,v ⊗ , ,w ⊗ = w,v . (6.19) ◦ It follows from (6.19) that the apolar inner product satisﬁes

t Π H H p, ,w = p(w), p t◦( ), w , (6.20) ◦ ∀ ∈ ∀ ∈ i.e., ,w t is the Riesz representer of point evaluation at w. Withg ˜(z) := g(z), the apolar inner product is given by (see Exer. 6.17) 6.7 Tight frames of symmetric tensors 121 α 1 1 D f (0) α Π d f ,g = ( f (D)g˜)(0)= ∑ D g(0), f ,g t◦(F ). (6.21) ◦ α t! t! α =t ! ∀ ∈ | | α α α α! In particular, the monomials z α =t are orthogonal with z ,z = α ! . { }| | ◦ | |

Theorem 6.6. Fix t 1,2,... . Let f1,..., fn be vectors in H , not all zero. Then ∈ { } n n n 2 ∑ ∑ 2t 1 ∑ 2t f j, fk t+d 1 f j , (6.22) j=1 k 1 | | ≥ − j=1 = t with equality precisely when any of the equivalent conditions holds t n t H (a) ( f j⊗ ) j=1 is a tight frame for the symmetric tensors Sym ( ). t n t H t H (b) ( , f j ⊗ ) j=1 is a tight frame for (Sym ( ))∗ = Sym ( ∗). t n (c) ( , f j ) is a tight frame for Π (H ) with the apolar inner product (6.19). j=1 t◦ t n t H Proof. Firstly, we observe that ( f j⊗ ) j=1 is a sequence of vectors in Sym ( ) which t are not all zero, since v⊗ is zero if and only if v = 0. Thus we may apply Theorem 6.1, using (6.16), to obtain

n n n 2 ∑ ∑ t t 2 1 ∑ t 2 f j⊗ , fk⊗ t+d 1 f j⊗ , j=1 k 1 | | ≥ − j=1 = t with equality if and only if (a) holds. By (6.17), the equation above equals (6.22). A similar argument, using (6.18) and (6.19) in place of (6.17), gives (b) and (c), respectively. ⊓⊔ A simple calculation (Exer. 6.15) shows that equality in (6.22) is also equivalent to the following generalised Plancherel and Bessel identities (also see Theorem 6.7)

d+t 1 n t t− t t x,y = ∑ x, f j f j,y , x,y H , (6.23) ∑n f 2t ∀ ∈ ℓ=1 ℓ j=1

d+t 1 n 2t t− 2t x = ∑ x, f j , x H . (6.24) ∑n f 2t | | ∀ ∈ ℓ=1 ℓ j=1

n H Example 6.8. For t = 1, Theorem 6.22 says that ( f j) j=1 is a tight frame for if n H and only if ( , f j ) j=1 is a tight frame for its dual ∗, or, equivalently, for the Π H homogeneous linear polynomials 1◦( ) (with the apolar inner product), and that each of these are equivalent to the variational condition (6.5).

For t > 1, the existence vectors ( f j) giving equality in (6.22) is not immediately obvious (the frames given Theorem 6.6 have a special form). We now investigate this question by showing that (6.22) is equivalent to certain cubature rules. 122 6 Variational characterisations of tight frames 6.8 Cubature on the real and complex spheres

Here we will show that equality in (6.22) is closely related to cubature formulas for the integrals of certain homogeneous polynomials over the complex sphere. Further, if H is a real Hilbert space and t > 1, then it turns out that equality in (6.22) can never be attained. In this case, there is a sharper inequality (Theorem 6.7) for which equality corresponds to cubature formulas for the real sphere. For simplicity, we suppose that H = Fd, and use standard multi-index notation, e.g., (x)α denotes the multivariate Pochhammer symbol. Let σ be the normalised surface area measure on the real or complex unit sphere S. For H = Cd the integrals of the monomials in z =(z1,...,zd) and z =(z1,...,zd) over the unit sphere are

α β α (d 1)!α! z z dσ(z)= 0, α = β, z 2 dσ(z)= − . (6.25) S S | | (d 1 + α )! − | | d For H = R the integrals of the monomials in x =(x1,...,xd) are

1 ( )α α σ α d 2α σ 2 x d (x)= 0, (2Z) , x d (x)= d . (6.26) S ∈ S ( 2 ) α | | Of interest here is the space of polynomials Fd F → d α β Π ◦ (F )= Hom(t,t) := span z z z : α = β = t , (6.27) t,t { → | | | | } which are homogeneous of degree t in z and in z. Equivalently (see Exer. 6.17)

d 2t d Π ◦ (F )= span z z,v : v F . (6.28) t,t { → | | ∈ } Π d Π d We note t◦,t (R )= 2◦t (R ). Recall that a homogeneous polynomial f of degree 2t is uniquely determined by its values on S by f (x)= x 2t f (x/ x ), x = 0. n H Deﬁnition 6.2. A sequence ( f j) j=1 of vectors in is a cubature rule for a space Π H P of homogeneous polynomials of degree 2t, such as t◦,t ( ), if

n n 2t 1 f j f j p(x)dσ(x)= ∑ p( f j)= ∑ p( ), p P. ∑ f 2t ∑ f 2t f ∀ ∈ S k k j=1 j=1 k k j f =0 j

A cubature rule for which the vectors ( f j) have equal norms gives an unweighted cubature rule for the integration of P over S. By Theorem 6.5, these exist (take the coordinates of f to be the real and imaginary parts of a basis for P). 2t t Let ct = ct (d,F) denote the integral of the monomial z z1 =(z1z1) over S. From (6.25) and (6.26), we have → | |

1 1 3 5 (2t 1) c (d,C)= , c (d,R)= − . t d+t 1 t d(d + 2) (d + 2(t 1)) t− − 6.8 Cubature on the real and complex spheres 123

The invariance of surface area measure under unitary maps implies

2t 2t x,y dσ(y)= x ct (d,F), x H . (6.29) | | ∀ ∈ S Denote the restriction of a polynomial space P to the unit sphere by P(S).

d Theorem 6.7. Fix t 1,2,... . Let f1,..., fn be vectors in H = F , not all zero. Then ∈ { } n n n 2 2t 2t ∑ ∑ f j, fk ct (d,F) ∑ fℓ , (6.30) j=1 k 1 | | ≥ 1 = ℓ= with equality when any of the following equivalent conditions hold (a) The generalised Bessel identity

n 2t 1 2t ct (d,F) x = ∑ x, f j , x H . (6.31) ∑n f 2t | | ∀ ∈ ℓ=1 ℓ j=1 (b) The generalised Plancherel identity

n t 1 t t d ct (d,F) x,y = ∑ x, f j f j,y , x,y F . (6.32) ∑n f 2t ∀ ∈ ℓ=1 ℓ j=1

n σ 1 Π H p(x)d (x)= n 2t ∑ p( f j), p t◦,t ( ), (6.33) S ∑ f ∀ ∈ ℓ=1 ℓ j=1

or, equivalently, for Πt,t (S)

n 2t f j f j p(x)dσ(x)= ∑ p( ), p Π ◦ (S). (6.34) ∑n f 2t f ∀ ∈ t,t S j=1 ℓ=1 ℓ j f =0 j (d) The tensor product integration formula

n t t 1 t t σ ⊗ x⊗ x⊗ d (x)= n 2t ∑ f j⊗ f j . (6.35) S ⊗ ∑ f ⊗ ℓ=1 ℓ j=1 (e) The integration formula

n t t σ 1 t t ,x⊗ x⊗ d (x)= n 2t ∑ , f j⊗ f j⊗ . (6.36) S ∑ f ℓ=1 ℓ j=1

(f) For all univariate polynomials g Πt (R), we have ∈ 124 6 Variational characterisations of tight frames

n n 2t 2t f j fk f j f g x,y 2 dσ(y)dσ(x)= ∑ ∑ g , k 2 . | | (∑ f 2t )2 | f f | S S j=1 k=1 ℓ ℓ j k f =0 f =0 j k (6.37) ∑n 2t ξ t H t H Proof. Let C := ℓ=1 fℓ . Deﬁne a tensor Sym ( ) Sym ( ) and a self adjoint operator Q on Symt (H ) by ∈ ⊗

n t t 1 t t ξ := x⊗ x⊗ dσ(x) ∑ f ⊗ f j⊗ , ⊗ − C j ⊗ S j=1

n t t 1 t t Q := ,x⊗ x⊗ dσ(x) ∑ , f ⊗ f ⊗ . − C j j S j=1 Equip Symt (H ) Symt (H ) with the apolar inner product, and the space of linear operators on Sym⊗t (H ) with the Frobenius inner product. Then (see Exer. 6.16) a simple calculation using (6.17) and (6.29) shows that

ξ ξ 1 2t , = Q,Q F = 2 ∑∑ f j, fk ct (d,F) 0, ◦ C j k | | − ≥ which is (6.30). Moreover, equality in (6.30) is equivalent to (d) or to (e). By the polarisation identity and (6.18), (a) and (b) are equivalent. We now complete the proof by showing

(d)= (c)= (a),(f)= equality in (6.30). ⇒ ⇒ ⇒ (d)= (c): Expand x t x t in terms of the coordinates of x. Since ⇒ ⊗ ⊗ ⊗ d d t η x⊗ = ∑ xk1 ek1 ∑ xkt ekt = ∑ pk(x) k, ⊗⊗ t k1=1 kt =1 k 1,...,d ∈{ }

pk(x) := xk xk xk , ηk := ek ek ek , 1 2 t 1 ⊗ 2 ⊗⊗ t we obtain t t x⊗ x⊗ = ∑ pk(x)pℓ(x)ηk ηℓ. ⊗ k,ℓ ⊗ Thus (d) can be written as

1 n ∑ pk(x)p (x)ηk η dσ(x)= ∑ ∑ pk( f j)p ( f j)ηk η . ℓ ⊗ ℓ C ℓ ⊗ ℓ S k,ℓ j=1 k,ℓ

Since the tensors ηk η are linearly independent, equating their coefﬁcients gives ⊗ ℓ the cubature rule for all the polynomials x pk(x)p (x), and hence for Π (H ). → ℓ t◦,t (c)= (a): Let p = x, 2t Π (Fd) in (6.33) and use (6.29) to obtain ⇒ | | ∈ t◦,t 6.8 Cubature on the real and complex spheres 125

2t 2t 1 2t ct (d,F) x = x,y dσ(y)= ∑ x, f j . | | C | | S j

(c)= (f): Let p = 2(t s) x, 2s Π , 0 s t in (6.34) to get ⇒ − | | ∈ t◦,t ≤ ≤ 2t fk fk x,y 2s dσ(y)= ∑ x, 2s. S | | C | fk | k

fk 2s 2(t s) fk 2s For x S, x, f = x − x, f , and so using (6.34) again gives ∈ | k | | k | 2t 2t f j fk f j fk x,y 2s dσ(y)dσ(x)= ∑ ∑ , 2s. S S | | C C | f j fk | j k s Thus (6.37) holds for the monomials ( ) , 0 s t, and hence for Πt (R). ≤ ≤ (a)= equality in (6.30): Take x = fk in (a) then sum over k to obtain the required equality⇒ 2t 1 2t ct (d,F) fk = ∑ fk, f j , C j | |

2t 1 2t ct (d,F)C = ct (d,F)∑ fk = ∑∑ fk, f j . k C k j | | (f)= equality in (6.30): Take g =( )t in (f) to obtain the desired equality ⇒ n n 2t 2t 2t f j fk f j fk 2t ct (d,F)= x,y dσ(y)dσ(x)= ∑ ∑ , | | (∑ f 2t )2 | f f | S S j=1 k=1 ℓ ℓ j k f =0 f =0 j k 1 2t = 2 ∑∑ f j, fk . C j k | |

⊓⊔ d Example 6.9. For unit vectors ( f j) in R , (c) reduces to the deﬁnition for a spherical half–design of order 2t. The condition (a) is the Waring formula (6.8), and equality in (6.30) is the condition (6.9).

Since ct (d,R) ct (d,C), with strict inequality for t,d > 1 (Exer. 6.9), we have: ≥

d Let f1,..., fn be vectors in H = R , not all zero. Then

n n n 2 2t 1 3 5 (2t 1) 2t ∑ ∑ f j, fk − ∑ f j , (6.38) j=1 k 1 | | ≥ d(d + 2) (d + 2(t 1)) j=1 = − which is a sharper bound than (6.22) when t,d > 1. A sequence of unit vectors giving equality in (6.38) is precisely a spherical half–design of order 2t. 126 6 Variational characterisations of tight frames 6.9 Spherical (t,t)–designs

We now give examples of sequences giving equality in Theorem 6.7 (also see 6.13 and the numerical study of 6.16). § § d Deﬁnition 6.3. A nonzero sequence ( f j) in F giving equality in Theorem 6.7, i.e.,

n n n 2 2t 2t ∑ ∑ f j, fk = ct (d,F) ∑ fℓ , (6.39) j=1 k 1 | | 1 = ℓ= is called a (spherical) (t,t)–design2 for Fd. Theorem 6.5 implies the (unweighted) cubature rules in Theorem 6.7 exist, i.e.,

For each t 1, unit–norm spherical (t,t)–designs for Fd always exist. The art is in≥ constructing those with a small number of vectors.

d Example 6.10. By Theorem 6.6, a unit–norm (t,t)–design (v j) for C (which always t t d exists) gives a unit–norm tight frame (v⊗j ) for Sym (C ). t t d Example 6.11. For d 2 and t > 1, there is no tight frame (v⊗j ) for Sym (R ). If there was, then Theorem≥ 6.1 gives

2 ∑∑ t t 2 ∑∑ 2t 1 ∑ 2t v⊗j ,vk⊗ = v j,vk = t+d 1 v j , j k | | j k | | − j t which violates the sharpened Welch bound (6.38). Example 6.12. The unit–norm (t,t)–designs for Rd are the spherical half–designs of order 2t (see 6.4 and Example 6.9). § Example 6.13. In view of (6.31), a (1,1)–design is precisely a tight frame for Fd. Example 6.14. Three equally spaced unit vectors in R2 are a (1,1)–design for F2. Further, they are a (2,2)–design for R2, but not for C2, since

2 4 1 4 27 4 2 ∑∑ f j, fk = 3 + 6( 2 ) = , ∑ fℓ = 3 = 9, j k | | − 8 ℓ 1 3 3 1 1 c (2,R)= = , c (2,C)= = . 2 2 4 8 2 2+2 1 3 2− Example 6.15. There is a (3,3)–design of 40 unit vectors for C4 given by a highly symmetric tight frame (see Exer. 6.12). It has the property that each given vector is orthogonal to 12 others, and makes an angle 1 with 27 others. √3

2 For H = Cd , these are also known as complex t–designs 6.9 Spherical (t,t)–designs 127

2 d Example 6.16. A SIC (see 2.11, 14.1), i.e., a set of d unit vectors ( f j) in C with § § 2 1 f j, fk = , j = k, | | d + 1 is a (2,2)–design of d2 unit vectors for Cd. Example 6.17. A set of d + 1 MUBs (mutually unbiased bases) for Cd (see 2.11), i.e., orthogonal bases with § 1 f ,g = , f and g in different bases, | | √d gives a (2,2)–design of d(d + 1) unit vectors for Cd [KR04]. Example 6.18. Three MUBs for C2 give a (3,3)–design of six vectors for C2, e.g., one can take 1 0 1 1 1 1 1 1 1 1 , , , , , . 0 1 √2 1 √2 1 √2 i √2 i − − For a given (t,t)–design ( f j), the cubature rule (6.34) can be written

n p(x)dσ(x)= ∑ w j p(φ j), p Π ◦ (S), (6.40) ∀ ∈ t,t S j=1 f =0 j where 2t φ f j f j j = , w j = 2t . (6.41) f j ∑ f ℓ ℓ Since 2(t r)q Π (Fd), for q Π (Fd), 1 r t, we have − ∈ t◦,t ∈ r◦,r ≤ ≤

Π ◦ (S) Π ◦ (S), 0 r t. (6.42) r,r ⊂ t,t ≤ ≤ Combining these observations gives: n d t/r 1 Proposition 6.2. Fix t 1. If ( f j) is a (t,t)–design for F , then ( f j f j) is ≥ j=1 − an (r,r)–design for Fd, 1 r t, i.e., ≤ ≤ n n n 2 2r 2(t r) 2(t r) 2t ∑ ∑ f j, fk f j − fk − = cr(d,F) ∑ fℓ . (6.43) j=1 k 1 | | 1 = ℓ= t/r 1 2(t r) Proof. Let g j := f j f j, q Π (S). Since p := q Π (S), we have − ∈ r◦,r − ∈ t◦,t n 2r n 2t g j g j f j f j ∑ q( )= ∑ p( )= pdσ = qdσ, ∑n g 2r g ∑n f 2t f j=1 ℓ=1 ℓ j j=1 ℓ=1 ℓ j S S g =0 f =0 j j and so, by (6.34), (g j) is an (r,r)–design. Substituting into (6.39) gives (6.43). ⊓⊔ 128 6 Variational characterisations of tight frames

In particular, we have:

A unit norm (t,t)–design for Fd is an (r,r)–design, 1 r t. ≤ ≤

t 1 If ( f j) is a (t,t)–design, t 1, then ( f j f j) is tight frame. ≥ −

If the norms of ( f j) are not all equal, then the properties (6.43) for 1 r t of a (t,t)–design and the corresponding equivalent conditions given by Theorem≤ ≤ 6.7 are most naturally described in terms of weighted (t,t)-designs.

6.10 Weighted (t,t)–designs

Φ φ n d n Deﬁnition 6.4. Suppose that =( j) j=1 are unit vectors in F , and w =(w j) j=1 3 satisfy w j 0, ∑ w j = 1. Then (Φ,w) is a weighted (spherical) (t,t)–design if ≥ j n n 2t ∑ ∑ w jwk φ j,φk = ct (d,F). (6.44) j=1 k=1 | |

There is a 1–1 correspondence between the (t,t)–designs ( f j) and the weighted (t,t)–designs (Φ,w) given by (6.41).

In this terminology, Theorem 6.7 becomes: Φ φ n Corollary 6.1. (Weighted version) Let =( j) j=1 be a sequence of unit vectors in d n F , and w =(w j) be nonnegative weights, i.e., w j 0, ∑ w j = 1. Then j=1 ≥ j n n 2t ∑ ∑ w jwk φ j,φk ct (d,F), (6.45) j=1 k=1 | | ≥ with equality if and only if (Φ,w) is a weighted (t,t)–design, or, equivalently,

n p(x)dσ(x)= ∑ w j p(φ j), p Π ◦ (S). (6.46) ∀ ∈ t,t S j=1 If (Φ,w) is a weighted (t,t)–design, then it is a weighted (r,r)–design, 0 r t. ≤ ≤ Proof. Make the substitution (6.41) in Theorem 6.7, and observe that (c) can be written as (6.46). The last assertion follows from this and (6.42). ⊓⊔ 3 These are also known as weighted spherical half–designs of order t when F = R (see [KP11]). 6.11 Complex projective t–designs 129

Example 6.19. A weighted (t,t)–design (Φ,w) satisﬁes

n n 2r ∑ ∑ w jwk φ j,φk = cr(d,F), 1 r t, j=1 k=1 | | ≤ ≤ which is the weighted version of (6.43). Substituting (6.41) into Theorem 6.7 gives a weighted version of each of the equivalent conditions (see Exer. 6.18), e.g., condition (a) becomes

n 2t 2t d ct (d,F) x = ∑ w j x,φ j , x F , j=1 | | ∀ ∈ or, equivalently n 2t ∑ w j x,φ j = ct (d,F), x S. (6.47) j=1 | | ∀ ∈

6.11 Complex projective t–designs

The equality (6.39) deﬁning (t,t)–designs is invariant under multiplying the vectors by unit scalars, and so (t,t)–designs can be extended to a projective setting. This has been done not only for R and C, but also the quaternians H and the d 1 octonians O (see [Hog82]). The complex projective sphere CP − can be viewed variously as The complex sphere S(Cd) with points z and az, a = 1 identiﬁed. • The 1–dimensional subspaces of Cd (the complex| lines| through 0). • The rank 1 orthogonal projections on Cd. • The polynomials on S(Cd) which carry over to this space, i.e., those with

p(z)= p(az), z, a F, α = 1 ∀ ∀ ∈ | | Π d Π d Π d are precisely those in 0◦,0(F ) 1◦,1(F ) 2◦,2(F ) . There is no notion of odd polynomials on this space (see⊕ Exer. 6.25 for⊕ details). d 1 We will take the elements of CP − to be rank one orthogonal projections. There d 1 is a unique unitarily invariant probability measure µ on FP − induced from the area measure σ on the sphere S(Fd), via

f (P)dµ(P)= f (Px)dσ(x), (6.48) d 1 d FP − F(C ) where Px = ,x x denotes the rank one orthogonal projection onto span x , x = 1. The Frobenius inner product between rank one orthogonal projections{ is}

2 Px,Py = trace(PxPy)= x,y R. (6.49) | | ∈ 130 6 Variational characterisations of tight frames

P n d Deﬁnition 6.5. Let =(Pj) j=1 be rank one orthogonal projections on F , and n w =(w j) satisfy w j 0, ∑ w j = 1. We say (P,w) isa(weighted) projective j=1 ≥ j (t,t)–design4 if n n t ∑ ∑ w jwk Pj,Pk = ct (d,F). j=1 k=1

The (t,t)–designs Φ =( f j) (up to multiplication by unit scalars) are in 1–1 correspondence with the projective (t,t)–designs (P,w), via

2t 1 f j Pj = 2 , f j f j, w j = 2t . (6.50) f j ∑ f ℓ ℓ

This gives the projective version of Theorem 6.7 (see [RS07], cf. Corollary 6.1). P n Corollary 6.2. (Projective version) Let =(Pj) j=1 be a sequence of rank one d n orthogonal projections in F , and w =(w j) satisfy w j 0, ∑ w j = 1. Then j=1 ≥ j n n t ∑ ∑ w jwk Pj,Pk ct (d,F), (6.51) j=1 k=1 ≥ with equality if and only if (P,w) is a projective (t,t)–design. A projective (t,t)–design is a projective (r,r)–design, 1 r t. ≤ ≤ Proof. With Pj and w j given by (6.50), (6.49) gives

2t 2t 2 t 2t t f j fk f j, fk f j, fk w jwk Pj,Pk = 2t 2 | 2 | 2 = | |2t 2 . (∑ f ) f j fk (∑ f ) ℓ ℓ ℓ ℓ Thus, making the substitution (6.50) in Theorem 6.7 gives (6.51), with equality for projective (t,t)–designs. The last part follows from Corollary 6.1 and (6.50). ⊓⊔ Other conditions equivalent to being a projective (t,t)–design can be obtained by substituting (6.50) into Theorem 6.7, e.g., by using (6.49), condition (a) becomes

n t d 1 ct (d,F)= ∑ w j Q,Pj , Q FP − . (6.52) j=1 ∀ ∈

The condition (e) gives Levenshtein’s deﬁnition [Lev98] of a weighted t–design

n n g P,Q dµ(P)dµ(Q)= ∑ ∑ w jwk g Pj,Pk , g Πt (R). Pd 1 Pd 1 ∀ ∈ F − F − j=1 k=1 (6.53)

4 Other terms such as weighted or quantum t–design are also commonly used. 6.12 Isometric embeddings 131

The condition (d) becomes

n t t P⊗ dµ(P)= ∑ w j P⊗ . (6.54) d 1 j FP − j=1 For F = C, Schur’s lemma implies that the left–hand side of (6.54) is

t µ 1 Π (t) P⊗ d (P)= d+t 1 sym, CPd 1 − t−

(t) t d where Πsym is the orthogonal projection of the tensors C onto the symmetric tensors Symt (Cd). ⊗ A naive numerical search (see [Bra11]) for (t,t)–designs in Cd suggests that in some cases, e.g., t = 4, d = 4 and t = 3, d = 3, those with the minimal number of vectors do not have constant weights. There are few known constructions of such weighted (t,t)–designs. A very general construction is given in 6.13. §

6.12 Isometric embeddings

The generalised Bessel identity (6.31) says that the linear map

d n n H = ℓ2(F ) ℓ2t (F ) : x ( x, f j ) → → j=1 is an isometric embedding, i.e., there is a constant C > 0 with

n d ( x, f j ) = C x , x ℓ2(F ). (6.55) j=1ℓ2t ℓ2 ∀ ∈ Conversely, any such embedding corresponds (via the Riesz representation) to a n sequence of vectors ( f j) j=1 giving equality in (6.55). Further, isometric embeddings d n ℓ2(F ) ℓp(F ) can exist only when p = 2t (see [LS04] which considers F = R,C and also→ the quaternians H). Collecting these observations gives:

d n There is an isometric embedding ℓ2(F ) ℓp(F ) if and only if p = 2t and there is a weighted (t,t)–design for Fd. →

d n We give a little more detail. Suppose that g =(g1,...,gn) : ℓ2(F ) ℓ2t (F ) is → d an isometric embedding, i.e., a linear map with g(x) = C x , x ℓ2(F ). By ℓ2t ∀ ∈ the Riesz representation, the linear maps x g j(x) have the form g j(x)= x, f j , d → where f j F . We compute ∈ n ∑ x f 2t g x 2t C x 2t c x 2t , j = ( ) ℓ2t =( ) = . j=1 | | 132 6 Variational characterisations of tight frames

2t Integrating the above over x S, using (6.29), gives ct (d,F)∑ f = c, and so ∈ ℓ ℓ ( f j) is a (t,t)–design by virtue of satisfying the generalised Bessel identity (6.31).

Example 6.20. The three equally spaced unit vectors (1,0),( 1 , √3 ),( 1 , √3 ) are − 2 2 − 2 2 a (2,2)–design (4–design) for R2 and a (1,1)–design (tight frame) for F2, which gives the isometric embeddings

2 3 1 √3 1 √3 g : ℓ2(R ) ℓ4(R ), g(x,y)= x, x + y, x y , → −2 2 −2 − 2 2 3 1 √3 1 √3 g : ℓ2(F ) ℓ2(F ), g(z1,z2)= z1, z1 + z2, z1 z2 . → −2 2 −2 − 2 The corresponding generalised Bessel identities are

4 4 4 1 √3 1 √3 9 2 2 2 9 x + x + y + x y = (x + y ) , = 3c2(2,R), −2 2 −2 − 2 8 8 2 2 2 1 √3 1 √3 3 2 2 3 z1 + z1 + z2 + z1 z2 = ( z1 + z2 ), = 3c1(2,F). | | −2 2 −2 − 2 2 | | | | 2 Example 6.21. The E. Lucas identity (1876)

4 2 2 4 4 6 ∑ x j = ∑ (x j + xk) + ∑ (x j xk) , j=1 1 j

e j ek : 1 j < k 4 . { ± ≤ ≤ } 4 12 Equivalently, it provides an isometric embedding ℓ2(R ) ℓ4(R ). →

6.13 Weighted (2,2)–designs of orthonormal bases

We now give a very general construction of weighted (2,2)–designs (Theorem 6.8). A special case is a construction of d + 1 MUBs in Cd for d a prime power. Let f : G H be a function between ﬁnite abelian groups with G H . The equation → | |≤| | f (x + a) f (x)= b (6.56) − has G solutions for (a,b)=(0,0). If f is linear, i.e., f (x)= mx, m Z, then it has G solutions| | for b = f (a). In view of this, we have the following notion∈ of a highly |nonlinear| function. 6.13 Weighted (2,2)–designs of orthonormal bases 133

Deﬁnition 6.6. We say a function f : G H between ﬁnite abelian groups with G H is (differentially) 1–uniform5 →if | |≤| | f (x + a) f (x)= b − has at most one solution for each (a,b) =(0,0). 2 Example 6.22. The functions f : Z5 Z6, g : Z5 Z5 : x x given by → → → f (0)= 0, f (1)= 1, f (2)= 0, f (3)= 2, f (4)= 2,

g(0)= 0, g(1)= 1, g(2)= 4, g(3)= 4, g(4)= 1, are 1–uniform.

We require some basic facts about the character group Gˆ (see 11). These include: §

Gˆ is the group homomorphisms G C 0 under pointwise multiplication. • Different characters are orthogonal→ (as vectors\ { } in CG). • Gˆ is isomorphic to G. • The Pontryagin duality map (11.7) gives a canonical isomorphism G Gˆ. • → Theorem 6.8. ([RS07]) Let f : G H be 1–uniform, G = d. Then there exists a weighted (2,2)–design for Cd = C→G given by the H +|1 orthonormal| bases | | 1 B0 =(eg)g G, w0 = , ∈ d(d + 1) B ψ 1 ψ ˆ ψ =(eg )g G, wψ = w1 = , H ∈ H (d + 1) ∈ | | where g χg is an isomorphism G G,ˆ and → → ψ 1 χ ψ eg := ∑ g(k) ( f (k))ek. √d k G ∈

Proof. We verify that (6.44) holds for t = 2. For φ j,φk B0, we have ∈

2 4 d 1 ∑ ∑ w0 ea,eb = 2 2 = 2 , a G b G | | d (d + 1) d(d + 1) ∈ ∈ and for φ j B0, φk Bψ and vice versa, we have ∈ ∈

ψ 4 2 H 1 1 2 2 ∑ w0w1 ∑ ∑ ea,e = | | ∑∑ = . | b | d(d + 1) H (d + 1) d2 d(d + 1)2 ψ Hˆ a G b G k g ∈ ∈ ∈ | |

5 The terms perfect nonlinear and maximally nonlinear are also used when H = G. 134 6 Variational characterisations of tight frames

ψ ξ We now consider the last case φ j Bψ , φk Bξ . Expanding ea ,e gives ∈ ∈ b

1 1 1 ∑ χa(k)ψ( f (k))ek, ∑ χb(ℓ)ξ( f (ℓ))eℓ = ∑ χa b(k)(ψξ − )( f (k)), d k G ℓ G d k G − ∈ ∈ ∈ 1 since ξ = ξ − (inverse in the character group). Thus

ψ ξ 4 1 χ ψξ 1 ea ,eb = 4 ∑ a b(w + x y z)( − )( f (w)+ f (x) f (y) f (z)), | | d w,x,y,z G − − − − − ∈ ∑ ∑ ψ ξ 4 and so b G ξ Hˆ ea ,eb equals ∈ ∈ | | 1 ∑ ∑ χ(w + x y z) ∑ ψ( f (w)+ f (x) f (y) f (z)). d4 − − − − w,x,y,z G χ Gˆ ψ Hˆ ∈ ∈ ∈ By Pontryagin duality, and the orthogonality of characters, the two inner sums are nonzero if and only if w+x y z and f (w)+ f (x) f (y) f (z)= 0, which occurs d(2d 1) times (see Exer. 6.27),− − giving d H each time.− Thus− − | |

4 1 2 ∑∑w jwk f j, fk = 2 1 + 2 +(2d 1) = = c2(d,C), j k | | d(d + 1) − d(d + 1) as claimed. ⊓⊔ n Example 6.23. (Maximal MUBs) Let G = H = Fd the finite field of order d = p , p odd, e.g., Fp = Zp. The following functions f : G G are 1–uniform (see [YCD06]) → 1. f (x)= x2, k 2. f (x)= xp +1, where n/gcd(n,k) is odd, k 3. f (x)= x(3 +1)/2, where p = 3, k is odd, gcd(n,k)= 1, 4. f (x)= x10 ux6 u2x2, where p = 3, n is odd, u = 0. − − These give d +1 MUBs (mutually unbiased bases) for Cd. The first (which is special case of the second) gives the d + 1 MUBs for Cd discussed in 12.19. § Example 6.24. Let d +1 be a prime power, and y be a generator for Fd∗+1 (the multiplicative group for the field of order d +1). Then the function f : Zd Zd+1 defined by → f ( j) := y j, is 1–uniform (see Exer. 6.26), and we obtain a weighted (2,2)–design for Cd that consists of d + 2 orthonormal bases, i.e., has n = d(d + 2) elements. 6.14 The frame force 135 6.14 The frame force

Motivated by well distributed points on the sphere, e.g., the n equally spaced vectors in R2 and the vertices of the Platonic solids in R3, Fickus [Fic01] gave a central force for which the equilibrium conﬁgurations are tight frames. He deﬁned the frame force between unit vectors in Rd to be

FF(a,b) := a,b (a b), a,b Rd, a = b = 1. − ∈ a b 2 Since a,b = 1 − for a = b = 1, − 2 x2 FF(a,b)= f ( a b )(a b), f (x) := 1 , (6.57) − − − 2 which gives a so called central force on Rd . The frame force between equal vectors is zero, whereas many physical forces are only deﬁned for distinct vectors, e.g., the Coulomb force between unit charges at a = b, given by

(a b) d FC(a,b) := − , a,b R . a b 3 ∈ − For a central force on Rd such as (6.57), the potential between a and b is

P(a,b) := p( a b ), where p′(x)= xf (x), − − so that (∇P( ,b))(a)= f ( a b )(a b), − − − n d and the total potential for a sequence of vectors (x j) j=1 in R is

∑ ∑ P(x j,xk). (6.58) j k= j 4 2 For the frame force, the choice p(x)= x x gives the potential 8 − 2 a b 4 a b 2 P(a,b) := − − 8 − 2 1 = ( a,b 2 1) (for a = b = 1), 2 −

d and hence the total potential for unit vectors x1,...,xn in R

1 2 n ∑ ∑ P(x j,xk)= ∑∑ x j,xk . j k= j 2 j k − 2 In this way, Fickus arrived at the frame potential (6.6) for unit vectors in Fd. 136 6 Variational characterisations of tight frames 6.15 Local and global minimisers of the frame potential

The local minimisers of the frame potential (6.6) give a set of unit vectors which are at equilibrium with respect to the frame force (see Exer. 6.28). Moreover, it turns out these local minimisers are in fact global minimisers. d For f : C R with f (x1 + iy1,...,xd + iyd) a differentiable function of the real → d d variables x1,y1,...xd,yd R, deﬁne a gradient ∇ f = 2(∂ 1 f ,...,∂ d f ) : C C by ∈ → ∂ ∂ ∇ d f := f (x1 + iy1,...,xd + iyd)+ i f (x1 + iy1,...,xd + iyd) j=1. (6.59) ∂x j ∂y j Then for both Rd and Cd, we have

∇( 2)(a)= 2a, ∇( ,b 2)(a)= 2 a,b b. (6.60) | | The minimisation of the frame potential for unit vectors in Fd = Rd or Cd can be viewed as a constrained optimisation problem:

2 minimise ∑∑ v j,vk subject to v1 = = vn = 1. j k | |

If (a1,...,an) is a local minimiser, then each aℓ is a local minimiser of

2 2 f (vℓ) := ∑ vℓ,ak subject to g(vℓ) := vℓ = 1. k=ℓ | | The critical points of this constrained optimisation of n or 2n real variables are given by Lagrange multipliers: ∇ f (aℓ)= λ∇g(aℓ), which by (6.60) becomes

∑ aℓ,ak ak = λaℓ, λ = λℓ R. k=ℓ ∈

Thus, each aℓ is an eigenvector of the frame operator S for (a j), i.e.,

n 2 Saℓ = ∑ aℓ,ak ak =(λℓ + aℓ )aℓ, ℓ = 1,...,n. (6.61) k=1 Using this, we now show that local minimisers are global minimisers. Theorem 6.9. ([BF03]). Let S = S(H ) be the unit sphere in H , d = dim(H ). The local minimisers of the frame potential

n n 2 FP(v1,...,vn) := ∑ ∑ v j,vk , v1,...,vn S j=1 k=1 | | ∈ are global minimisers, which in turn are tight frames for H (n d), or nonspanning orthonormal sequences (n < d). In particular, there exist equal≥ norm tight frames of n vectors in Rd and Cd, for all values of n d. ≥ 6.15 Local and global minimisers of the frame potential 137

Proof. Let (a j) be a local minimiser of the frame potential on S, and S be its frame operator. We have seen, (6.61), that the Lagrange multiplier equations imply each a j is an eigenvector of S. Moreover, the eigenvalue λ is 1, since ≥ 2 Sa j = λa j = λ = λ a j,a j = Sa j,a j a j,a j = 1. ⇒ ≥| |

Let λmax be the largest eigenvalue of S. If λmax = 1, i.e., it is the only nonzero eigenvalue, then S = I on span(a j), and (a j) is an orthonormal basis for span(a j) (see Exer. 2.4). If λmax is the only eigenvalue, i.e., S = λmaxI, then (a j) is a tight frame. Thus it reduces to showing that S cannot have a second nonzero eigenvalue λ2 < λmax. We suppose that it does, and show that (a j) can be perturbed to obtain ε (a j ) with a smaller frame potential, thereby giving a contradiction. Let J be the subset of indices j for which a j is a λmax–eigenvector, and u be a λ2–eigenvector. For ε R sufficiently small, define ∈ 2 2 ε 1 ε β j a j + εβ ju, j J; a j := − | | ∈ a j, j J ∈ ε ∞ where the β j C are to be chosen later. Then FP (a ) is a C –function of ε for ∈ j ε < 1/max( β j ). Thus, we consider the asymptotic expansion given by the first few| | terms of the| Taylor| series. By definition,

ε ε 2 2 a ,a = a j,ak , j J,k J. (6.62) | j k | | | ∈ ∈ Since S is Hermitian, its eigenspaces are orthogonal to each other, so that

a j,ak = 0, j J, k J, a j,u = 0, j J. ∈ ∈ ∈ and we obtain

ε ε 2 2 2 2 2 a ,a = a j,ak + ε β j u,ak , j J, k J. (6.63) | j k | | | | | | | ∈ ∈

ε ε 2 2 2 2 2 a ,a = 1 ε β j 1 ε βk a j,ak + ε β jβk, j J, k J. j k − | | − | | ∈ ∈ Using the Taylor expansion

2 2 1 2 2 4 1 ε β j = 1 ε β j + O(ε ), ε 0, − | | − 2 | | → we expand the expression for j J, k J, to obtain ∈ ∈ ε ε 1 2 2 2 2 4 a ,a =(1 ε ( β j + βk )) a j,ak + ε β jβk + O(ε ), j k − 2 | | | | which gives 138 6 Variational characterisations of tight frames

ε ε 2 2 2 2 2 2 a ,a =(1 ε ( β j + βk )) a j,ak + 2ε ℜ( a j,ak β jβk) | j k | − | | | | | | + O(ε4), ε 0, j J, k J. (6.64) → ∈ ∈ Adding (6.62), (6.63) and (6.64) gives ε ε2 β 2 β 2 2 FP (a j ) = FP (a j) ∑ ∑( j + k )) a j,ak − j J k J | | | | | | ∈ ∈ 2 2 2 2 4 + 2ε ∑ ∑ ℜ( a j,ak β jβk)+ 2ε ∑ ∑ β j u,ak + O(ε ). j J k J j J k J | | | | ∈ ∈ ∈ ∈ Since the eigenspaces of S are orthogonal, we have

2 2 ∑ a j,ak = Sa j,a j = λmax, j J, ∑ u,ak = Su,u = λ2, k J | | ∈ k J | | ∈ ∈ and so obtain ε ε2 β 2 λ λ ε2 β 2 ε4 FP (a j ) = FP (a j) 2 ∑ j ( max 2)+ 2 ∑ ja j + O( ). − j J | | − j J ∈ ∈ Since S = λmaxI on the λmax–eigenspace E = span a j : j J , the vectors (a j) j J { ∈ } ∈ are a unit–norm tight frame for E with frame bound λmax, and so J λmax = | | > λ2 1. dim(E) ≥

Hence the vectors (a j) j J are linearly dependent, and there is (β j) j J = 0 with β ∈ β ∈ ∑ j J ja j = 0, i.e., (a j) j J and ( j) j J are mutually orthogonal. Thus ∈ ∈ ∈ ε 2 4 FP (a ) = FP (a j) cε + O(ε ), ε 0, j − → 2 where c = 2(∑ β j )(λmax λ2) > 0, which gives the desired contradiction. j | | − ⊓⊔ In the motivating case H = Rd, these minimisers are at equilibrium with respect to the frame force. Indeed, deﬁne the effective frame force of b on a to be the component of the frame force FF(a,b) of b on a which is orthogonal to a, i.e.,

EFF(a,b)= a,b (a a,b b), − and so the total effective frame force on each point a j is zero (see Exer. 6.28), i.e.,

∑EFF(a j,ak)= 0. k

Example 6.25. A sequence (a j) of points on the sphere is said be FF–critical (frame force critical) if each a j is an eigenvector of the frame operator of (a j). The minimisers of the frame force are FF–critical. A sequence is FF–critical if and only if it is a union of equal–norm tight frames for orthogonal subspaces. 6.16 The numerical construction of spherical (t,t)–designs 139 6.16 The numerical construction of spherical (t,t)–designs

d n Let V =[vαβ ]=[v1,...,vn], and p,g : F × R be the homogeneous polynomials given by → 2t 2t p(V) := ∑∑ v j,vk , g(V) := ∑ vℓ . (6.65) j k | | ℓ

The spherical (t,t)–designs for Fd are the nontrivial zeros of the nonnegative homogeneous polynomial

d 2 f (V) := p(V) ct (d,F )g(V) (6.66) − d n of degree 4t in the real (and imaginary) parts of entries of V =[vαβ ] F . ∈ ×

The minimisers of p(V) 0 with g(V) ﬁxed, e.g., V =[v j] a unit norm sequence, satisfy the Lagrange equations:≥ ∇p(V)= λ∇g(V). Moreover, the ones that give spherical (t,t)–designs are minima of f , and so satisfy ∇ f (V)= 0, i.e.,

∇p(V)= 2ct (d,F)g(V)∇g(V). (6.67)

Thus we obtain the following condition for the existence of spherical (t,t)–designs. Theorem 6.10. Let t 1 and f : Fd n R be the nonnegative function given by ≥ × → n n n 2t 2t 2 f ([v1,...,vn]) := ∑ ∑ v j,vk ct (d,F) ∑ vℓ . j=1 k=1 | | − ℓ=1 Then the critical points of f satisfy

2(t 1) 2t 2(t 1) ∑ v j,vβ − vβ ,v j v j = ct (d,F) ∑ vℓ vβ − vβ , 1 β n. j | | ≤ ≤ ℓ In particular, for t = 1 the nonzero critical points of f are the tight frames for Fd, which are all global minima. Proof. The critical point of f are given by (6.67), where ∇ f is the gradient of f viewed as a function of real variables, as in (6.59). A calculation (see Exer. 6.29) show that the β–columns of ∇p(V) and ∇g(V) are

2(t 1) 2(t 1) 4t ∑ v j,vβ − vβ ,v j v j, 2t vβ − vβ . j | |

Substituting this into (6.67) gives the desired condition. For t = 1, the V = 0 which are critical points of f (V) satisfy

1 2 ∑ vβ ,v j v j = ∑ vℓ vβ , 1 β n, j d ≤ ≤ ℓ 140 6 Variational characterisations of tight frames

d and so, by linearity, (v j) is tight frame for H := span vβ 1 β n F , with frame 1 2 { } ≤ ≤ ⊂ bound A = ∑ v . The trace condition (2.9) gives dim(H )= d, so that (v j) is d ℓ ℓ a tight frame for Fd. Thus the nonzero critical points of f (V) are precisely the tight frames for Fd. ⊓⊔ Spherical (t,t)–designs can be found numerically, by minimising f (V), with g(V) ﬁxed. This can be done by an iterative algorithm which starts a random V0, and chooses Vk+1 = Vk +Wk, where Wk is such that f (Vk+1)= f (Vk +Wk) < f (Vk). The direction Wk can be random (of an appropriate size) [Bra11], or in the direction of maximal decrease [Hug16] (which is more effective close to a minimum).

The maximal decrease of f at V is in the direction W = ∇ f (V), where − 2(t 1) d 2t 2(t 1) (∇ f (V))αβ = 4t ∑ v j,vβ − vβ ,v j vα j 4tct (d,F ) ∑ vℓ vβ − vαβ . j | | − ℓ

A summary of these numerical results (which have motivated various analytic constructions) is given in Tables 6.1 and 6.2.

Table 6.1: The minimum numbers nw and n of vectors in a weighted and in a unit–norm spherical (t,t)–design for Rd (spherical half–design of order 2t) as calculated by Daniel Hughes [Hug16].

t d nw ne Comments 1 d d d orthonormal bases in Rd t 2 t + 1 t + 1 equally spaced lines in R2 2 3 6 6 equiangular lines in R3 2 4 11 12 no structure repeated angles 2 5 16 20 group structure no structure 2 6 22 24 group structure work in progress 2 7 28 28 equiangular lines in R7 2 8 45 >45 no structure 3 3 11 16 no structure possible group structure 3 4 23 >23 group structure 3 5 41 >41 group structure 4 3 16 25 group structure no structure 4 4 43 >43 work in progress 5 3 24 35 no structure no structure 6.16 The numerical construction of spherical (t,t)–designs 141

Table 6.2: The minimum numbers nw and n of vectors in a weighted and in a unit–norm spherical (t,t)–design for Cd as calculated by Jennifer Bramwell [Bra11].

t d nw ne Comments 1 d d d orthonormal bases in Rd 2 d d2 d2 SICs 3 2 6 6 three MUBs for C2 3 3 22 27 some structure 3 4 40 40 highly symmetric tight frame (Example 6.15) 3 5 >100 4 2 10 12 two orbits, see §9.8 4 3 47 >47 4 4 >85 >85 5 2 12 12 group frame, see Exer. 10.12 6 2 18 24 some structure 7 2 22 24 some structure 8 2 37 >37 some structure 9 2 44 >44 some structure

It is also possible to calculate (numerically) the Hessian (second derivative) of f and p at V to investigate the nature of the critical points of f (see Exer. 6.30).

For the real case V = X Rd n, the formulas for the Hessian simplify to ∈ × ∂ 2 p 2(t 1) 2t 1 (V)= 4t(2t 1) vb,vβ − vαbvaβ + 4tδaα vb,vβ − ∂xab∂xαβ − 2(t 1) + 4t(2t 1)δbβ ∑ v j,vβ − vα jvaj, − j ∂ 2 2 (g ) 2(t 1) 2(t 2) (V)= 2g(V) 2tδaα δbβ vβ − + 4t(t 1)δbβ vβ − vαβ vaβ ∂xab∂xαβ − 2(t 1) 2(t 1) + 2 2t vb − vab 2t vβ − vαβ . 142 6 Variational characterisations of tight frames Notes

Neil Sloane has a webpage of putatively optimal real spherical t–designs http://neilsloane.com/sphdesigns/ (see [HS96]) and the author has a similar list for real and complex spherical (t,t)–designs. There is recent interest in complex spherical (t,t)–designs (weighted complex projective t–designs), see, e.g., [KR05], [RS07], [RS14]. Our uniﬁed treatment of real spherical half–designs of order 2t and complex (t,t)–designs in Theorem 6.7 was adapted from [Kon99],¨ [Wal16] (also see [DHC12], [BH15]). There are other equivalences. Those involving the evaluation of Gegenbauer polynomials can be used to estimate the minimum number of vectors in a (t,t)–design (see [DGS77], [Hog89] for spherical t–designs). Some (t,t)–designs meeting these bounds (for spherical 2t–designs) are termed tight (this is not related to being a tight frame). Thanks to Aidan Roy, Andreas Klappenecker and Wei–Hsuan Yu for insightful discussions about this chapter. 6.16 The numerical construction of spherical (t,t)–designs 143 Exercises

2 6.1. Let A n n be Hermitian, A 0. Show that rank A trace(A) with equality C × = ( ) trace(A2) ∈ n r ≥ if and only if A = cUU , U =[u1,...,ur] C , with orthonormal columns. ∗ ∈ × Hint. If λ1,...,λr are the nonzero eigenvalues of A, then Cauchy–Schwarz gives

n r 2 λ 2 λ 2 2 (trace(A)) =(∑ j) r ∑ j = r trace(A ). j=1 ≤ j=1

6.2. Use the generalised Welch bound (6.4) to prove Example 6.2, i.e., that

∑ 2 ∑ 4 2 ( j f j ) /d j f j max f j, fk − > 0. j=k | | ≥ n2 n − 6.3.m Do a numerical investigation of the variational inequality (6.4). Does one get close to equality for large numbers of random (unit or otherwise) vectors?

6.4. We may write the variational characterisation (6.4) for tight frames (v j) j J as ∈ 4 4 4 4 2 v j + vk 1 2 2 v j + vk ∑ 2 v j,vk + = ∑ 2 v j vk + . j,k J | | n 1 d j,k J n 1 { }⊂ − { }⊂ − n d We say that a frame (v j) j=1 for C is perfectly tight if equality holds for all pairs. (a) Show that for d = 1, every tight frame is perfectly tight. (b) Show that for d 2, every perfectly tight frame has nonzero vectors. (c) Describe the equal–norm≥ perfectly tight frames. (d) Do there exist perfectly tight frames for d 2 which do not have equal norms? ≥ 6.5. Let ( f j) be a sequence of n d unit vectors in a H , where dim(H )= d. Show that ≤ 2 FP( f1,..., fn) := ∑∑ f j, fk j k | | has a minimum value of n, which is attained if and only if ( f j) is orthogonal. 6.6. Show the normalised frame potential satisﬁes (6.7), i.e., 1 FPˆ ( f1,..., fn) 1, d ≤ ≤ and that ˆ 1 (a) FP( f1,..., fn) equals d if and only if ( f j) is a tight frame. (b) FPˆ ( f1,..., fn) equals 1 if and only if span( f j) is 1–dimensional. 2 Remark: For unit norm vectors FP = n FPˆ n , and so this extends Theorem 6.2. |S d 6.7. Real spherical 2–designs. Let Φ = φ1,...,φn be unit vectors in R . Φ { } d Π d (a) Suppose that is a real spherical 2–design. For y R , let py 2◦(R ) be given by ∈ ∈ 144 6 Variational characterisations of tight frames

2 2 py(x) := y,x =( y,x ) . | | 2 Show that the integral of py over the unit sphere S is c y , with c > 0 independent of y, and hence conclude that Φ is a tight frame. (b) Now suppose that Φ is a tight frame. By considering the integral of py above, or otherwise, show that it is a real spherical 2–design. Remark: This is a special case of the key arguments of 6.8. § 6.8. Let d 2. Show that there exists a real spherical 2–design of n points for Rd unless n ≥d or n = d + 2 and n is odd (the existence part requires a construction). In particular,≤ there is no real spherical 2–design of ﬁve points for R3. Hint: For the construction use harmonic frames (see Chapter 11).

6.9. Real spherical designs and Waring type formulas. Here we consider the case of equality in Theorem 6.7 for H = Rd. (a) Show that ct (d,R) ct (d,C), with strict inequality when t > 1, d > 1. ≥ d (b) A sequence ( f j) of unit vectors in R satisfying (6.33) is, by deﬁnition, a real spherical half–design of order 2t. By substituting t for 2t, write down the equivalent conditions for being a real spherical half–design of order t (for t even) given by equality in (6.30), and (6.31), (6.35). n/2 (c) Show that if ( f j) is centrally symmetric, i.e., of the form ( f j) j=1, then the cubature rule (6.33) holds for all odd polynomials, i.e., p Π ±Π Π . ∈ 1◦ ⊕ 3◦ ⊕ 5◦ ⊕ 6.10. The n equally spaced (unit) vectors in R2 are 2π 2π Φ =(v j)= cos j,sin j : j = 0,...,n 1 , n n − and the n equally spaced lines in R2 are π π Ψ =(w j)= cos j,sin j : j = 0,...,n 1 . n n − (a) Show that the n equally spaced vectors in R2 are a spherical (n 1)–design. (b) Show that the n = t +1 equally spaced lines in R2 are a spherical− half–design of order 2t, i.e., a (t,t)–design. Hint: Use the integrals of (6.25) and (6.26).

6.11. A SIC consists of d2 equiangular unit vectors in Cd, with a common angle 2 1 d f j, fk = , j = k, and m MUBs are m orthonormal bases for C , with the | | √d+1 property that f ,g = 1 for f and g from different bases (see 2.11). | | √d § (a) Show that a SIC is a (2,2)–design. (b) Show that m MUBs in Cd form a (2,2)–design if and only if m = d + 1. (c) Show that d + 1 MUBs in Cd form a (3,3)–design if and only if d = 2.

6.12. There is a highly symmetric tight frame of 240 vectors for C4 given as an orbit of the Shephard Todd group 32 (see 13.8), which gives 40 lines, since the group contains scalar multiplication by the§ 6–th roots of unity. Take a vector from each 6.16 The numerical construction of spherical (t,t)–designs 145 line. This set Φ of 40 vectors has the property that each is orthogonal to 12 others and makes an angle 1 with 27 others. Show that Φ is a (3,3)–design for C4. √3

d 6.13. Show that an equiangular tight frame Φ =(v j) of n unit vectors for R is a d 1 spherical (2,2)–design for R if and only if n = 2 d(d + 1). Remark: Such equiangular lines are known to exist for d = 2,3,7,23 (see 12.1). § 6.14. Let Φ be the set of 240 vectors v R8 of with v 2 = 2, and the form ∈ 1 type 1: v j and v has an even number of positive entries, ∈ {± 2 } type 2: v j 0, 1 (such v have two nonzero entries). ∈ { ± }

Since Φ is centrally symmetric, it can be written Φ = Φ0 Φ0. ∪− 8 (a) Show that Φ0 is a spherical (3,3)–design of 120 vectors for R . (b) Show that Φ is a spherical 7–design of 240 vectors R8. Remark: This Φ (minimal vectors of the Korkin–Zolotarev lattice) is due to [KP11]. 6.15. Show that equality in (6.22) is equivalent to (a) The generalised Plancherel identity

d+t 1 n t t− t t x,y = ∑ x, f j f j,y , x,y H . ∑n f 2t ∀ ∈ ℓ=1 ℓ j=1 (b) The generalised Bessel identity

d+t 1 n 2t t− 2t x = ∑ x, f j , x H . ∑n f 2t | | ∀ ∈ ℓ=1 ℓ j=1 6.16. Deﬁne ξ Symt (H ) Symt (H ) and Q : Symt (H ) Symt (H ) by ∈ ⊗ → n t t 1 t t ξ := x⊗ x⊗ dσ(x) ∑ f ⊗ f j⊗ , ⊗ − C j ⊗ S j=1 n t t 1 t t Q := ,x⊗ x⊗ dσ(x) ∑ , f ⊗ f ⊗ , − C j j S j=1 where C := ∑ f 2t . Show that ℓ ℓ ξ ξ 1 2t , = Q,Q F = 2 ∑∑ f j, fk ct (d,F), ◦ C j k | | − where the apolar and Frobenius inner products are used, respectively. Π d d 6.17. We consider the vector space t◦,r(C ) of polynomials C C which are homogeneous of degree t in z and of degree r in z, i.e., →

d α β Π ◦ (C ) := span z z z : α = t, β = r . (6.68) t,r { → | | | | } 146 6 Variational characterisations of tight frames

This absolutely irreducible U –invariant space is denoted by H(t,r) in 16.7. There § is a natural identiﬁcation Symt (H ) Symr(H ∗) Π (H ) given by ∗ ⊗ → t◦,r t r t r t r ,v ⊗ ,x ⊗ ,v ,x = ,v x, . ⊗ → For polynomials p : Cd C, we deﬁne an associated differential operator p(∂) by → α β α β p(∂) := ∑ cαβ ∂ ∂ , where p(z)= ∑ cαβ z z , (6.69) (α,β) (α,β) where ∂ and ∂ are the Wirtinger complex differential operators given by

∂ 1 ∂ ∂ ∂ 1 ∂ ∂ ∂ j = = i , ∂ j = = + i . ∂z 2 ∂x − ∂y ∂z 2 ∂x ∂y j j j j j j (a) Show that the monomials z zα zβ in (6.68) are linearly independent, and so → d +t 1 d + r 1 dim(Π (Fd)) = − − . (6.70) t◦,r t r (b) By taking the apolar inner product on Symt (H ) Symr(H ∗), show that ∗ ⊗ ,v t x, r, ,w t y, r := w,v t x,y r (6.71) ◦ Π d deﬁnes an inner product on t◦,r(C ). (c) Show the Riesz representer of point evaluation at w is ,w t w, r, i.e., t r Π d d p, ,w w, = p(w), p t◦,r(C ), w C . ◦ ∀ ∈ ∀ ∈ Π d (d) Use this to conclude that t◦,r(C ) is spanned by ridge functions, i.e.,

d t r d Π ◦ (C )= P := span z z,v v,z : v C . (6.72) t,r { → ∈ } Π d In particular, t◦,t (C ) is spanned by ridge functions (plane waves), i.e.,

2t d Π ◦ (Ω)= span z z,v : v Ω , where Ω = C or S . t,t { → | | ∈ } C (e) With p(∂) given by (6.69) andq ˜(z) := q(z), show that

1 ∂ Π d p,q = p( )q˜(0), p,q t◦,r(C ). ◦ t!r! ∀ ∈ In particular, the monomials z zα zβ in (6.68) form an orthogonal basis. → Π d Π d Remark: It follows from (6.25), that for t◦,0(C ) and 0◦,t (C ) one has

t + d 1 p,q = − p(z)q(z)dσ(z). ◦ t d S(C ) 6.16 The numerical construction of spherical (t,t)–designs 147

Π d T T Cubature rules which integrate t◦,r(C ), (t,r) for some set of indices are studied in [RS14], where they are called spherical∈ T –designs. 6.18. Make the substitution (6.41) in Theorem 6.7 to obtain the weighted versions of the conditions (a)–(e). 6.19. Let ∆ be the Laplacian for functions Fd F, i.e., for F equal R and C → d ∂ 2 d ∂ 2 d ∂ 2 d ∆ ∆ ∂ ∂ = ∑ ∂ , = ∑ ∂ + ∑ ∂ = 4 ∑ j j. j=1 x j j=1 x j j=1 y j j=1 (a) Take the Laplacian with respect to x Rd to get ∈ 2t 2t 2 2t 2t 2 2 ∆( x )= 2t(d + 2t 2) x − , ∆( x,v )= 2t(2t 1) x,v − v . − − (b) Take the Laplacian with respect to z Rd to get ∈ 2t 2t 2 2t 2 2(t 1) 2 ∆( z )= 4t(d +t 1) z − , ∆( z,v )= 4t z,v − v . − | | | | (c) Using (a) and (b), apply the Laplacian to the Bessel identity (6.31). d 6.20. Show that if (v j) and (wk) are spherical (t,t)–designs for F , with

2t 2t ∑ v j = ∑ wk , j k

d then their union (v j) (wk) is a spherical (t,t)–design for F . ∪ 6.21. Use the generalised Bessel identity (6.31) to show that the minimal number n of vectors in a weighted (t,t)–design satisﬁes

d+t 1 2 d t− , F = C 2t n dim(Π ◦ (F )) = = O(d ), d ∞. ≤ t,t d+2t 1 → 2t− , F = R 6.22. Use the generalised Plancherel identity (6.32) to show that the number n of vectors in a weighted (t,t)–design for Fd satisﬁes

d t + d 1 t n dim Π ◦(F ) = − = O(d ), t ∞. ≥ t d 1 → − d 6.23. Suppose that ( f j) is a tight frame for F , i.e., is a (1,1)–design. Show the condition which ensures it comes from a (t,t)–design, as per Proposition 6.2, is that

n n 2t n 2 f j, fk 2 ∑ ∑ |2t 2 | 2t 2 = ct (d,F) ∑ fℓ . j=1 k 1 f j − fk − 1 = ℓ= 6.24. The lines Fx : x = 0 in H = Fd are in 1–1 correspondence with the rank { } d 1 one orthogonal projections, i.e., points in the projective space FP − , via 148 6 Variational characterisations of tight frames

,x Fx Px := x. ←→ x,x (a) Show that the Frobenius inner product between orthogonal projections given by unit vectors is 2 Px,Py = x,y = Py,Px . | | d 1 (b) Show that the metric on FP − given by the Frobenius inner product is

d 1 ρ(P,Q)= √2 1 P,Q , P,Q FP − . − ∈ (c) Show that in terms of lines this metric is

x y 2 ρ(Fx,Fy)= √2 1 , . − x y (d) Show that the set of lines can also be embedded into the real vector space of traceless Hermitian matrices (with the Frobenius norm), via 1 Fx Px I. → − d

6.25. Polynomials on projective spaces. Determine the vector space of polynomials p : Fd F whose value at each z = 0 depends only on the 1–dimensional subspace given→ by z, i.e.,

p(z)= p(az), z Fd, a F, a = 1. ∀ ∈ ∀ ∈ | |

6.26. Let Fd+1 be the ﬁeld of order d +1, where d +1 is a prime power. Suppose that j y is a generator for the multiplicative group Fd∗+1. Show that f : Zd Zd+1 : y y is a 1–uniform function. → →

6.27. Let f : G H be a map between ﬁnite abelian groups, with G = d. Show that if f is 1–uniform,→ then | |

w + x y z = 0, f (w)+ f (x) f (y) f (z)= 0 − − − − has exactly d(2d + 1) solutions in (w,x,y,z) G4. ∈ 6.28. Equilibrium with respect to the frame force in Rd and Cd. For unit vectors a,b Cd, the frame force of b on a can be extended ∈ FF(a,b) := a,b (a b), − though this is no longer a central force. The frame force between orthogonal vectors and between coincident vectors is zero. The effective frame force EFF(a,b) of b on a is the component of the frame force v = FF(a,b) which is orthogonal to a. (a) Calculate EFF(a,b) for a,b S. ∈ 6.16 The numerical construction of spherical (t,t)–designs 149

n (b) Show that if (a j) j=1 is a minimiser of the frame potential, then the total effective frame force on each a j is zero, i.e.,

∑EFF(a j,ak)= ∑ EFF(a j,ak)= 0. k k= j d n 6.29. Let V =[vαβ ]=[v1,...,vn], and p,g : F R be given by × → 2t 2t p(V) := ∑∑ v j,vk , g(V) := ∑ vℓ . j k | | ℓ

With ∇ f given by (6.59) when F = C, show the β–columns of ∇p(V) and ∇g(V) are 2(t 1) 2(t 1) 4t ∑ v j,vβ − vβ ,v j v j, 2t vβ − vβ . j | | d n 6.30. Here we calculate the Hessian matrix Hf of f of the functions p,g : F × R given by (6.65). Let X be the real variables with some ordering, i.e., →

X = xαβ yαβ for F = C, X = xαβ for F = R. { } ∪ { } { }

Then Hf (V) is the X X real symmetric matrix with (r,s)–entry given by × ∂ 2 f H (V) = (V). f rs ∂r∂s Find the Hessian matrix of p, g, and hence the function f given by (6.66).

n d 6.31. A sequence ( f j) j=1 is a ﬁnite tight frame for F if and only if

n d 2 1 g(x) := 2 ∑ x, f j 2 = 1, x = 0. ∑ fk | | x ∀ k j=1 (a) Show for a general frame that g can take values which are > 1 and < 1. Thus the obvious generalisation of Bessel’s inequality does not hold. There are various generalisations in the literature. We now develop a few. 2 2 2 (b) By Cauchy–Schwarz, ∑ j c j x, f j = x,∑ j c j f j x ∑ j c j f j . Use the triangle and Cauchy-Schwarz| inequalities | | to show Pe|cariˇ ≤c’s´ inequality

n 2 n n 2 2 ∑ c j x, f j x ∑ c j ∑ f j, fk . ≤ | | | | j=1 j=1 k=1 (c) From this deduce Selberg’s inequality

n 2 x, f j 2 ∑ n| | x . ∑ f , f j ≤ j=1 ℓ=1 | ℓ | (d) From Selberg’s inequality, deduce Bombieri’s inequality 150 6 Variational characterisations of tight frames

n n 2 2 ∑ x, f j x max ∑ fℓ, f j . | | ≤ 1 ℓ n | | j=1 ≤ ≤ j=1 (e) Vary the argument of of (b) to show the inequality.

n n n 1 2 2 2 2 ∑ x, f j x ∑ ∑ f j, fk . j=1 | | ≤ j=1 k 1 | | = Chapter 7 The algebraic variety of tight frames

d Let V =[v1,...,vn] be the synthesis operator of a normalised tight frame for F , i.e., a d n matrix with VV ∗ = I (Proposition 2.1). Since VV ∗ = I, the collection of normalised× tight frames of n vectors for a space of dimension d can be viewed as an d n algebraic variety (in F × ), as can other classes of frames, such as the equal–norm tight frames. Here we consider some geometry of this algebraic variety, including:

What norms the vectors v j can have. • The dimension of the variety, and how to describe points on it. • The fact the rational points are dense on the variety, i.e., every normalised tight • frame can be arbitrarily well approximated by a normalised tight frame consisting of vectors with rational entries. Our treatment is based on the following simple observations. If U is unitary, then

(VU)(VU)∗ = V(UU∗)V ∗ = VV ∗ = I, so that W = VU is (the synthesis operator of) a normalised tight frame. Conversely, if V and W are normalised tight frames of n vectors for Fd, then we may choose complementary normalised tight frames Vc and Wc. Since the matrix with rows given by the rows of a normalised tight frame and a complement is unitary, we have

W W ∗ V V ∗ W V V ∗ W = I = = = U, U := .               Wc Wc Vc Vc ⇒ Wc Vc Vc Wc               Thus W = VU, where U = V W +V Wc is an n n unitary matrix. In other words: ∗ c∗ ×

Let V =[v1,...,vn] be a normalised tight frame for H . Then all normalised tight frames of n vectors for H have the form W = VU, where U is an n n unitary matrix. ×

151 152 7 The algebraic variety of tight frames 7.1 The real algebraic variety of normalised tight frames

d n Two useful descriptions of the normalised tight frames V =[v1,...,vn] F × are the orthogonality of rows of V and the variational characterisation (see ∈6), i.e., the systems of equations § VV ∗ = I, n n n n 2 1 2 2 ∑ ∑ v j,vk = ∑ v j,v j , ∑ v j = d. (7.1) j=1 k=1 | | d j=1 j=1 Since these involve the entries of V and their complex conjugates, it follows that:

The normalised tight frames of n vectors in Fd are a real algebraic variety, N denoted by n,Fd . For F = R it is in dn variables (the entries of V), and for F = C itisin2dn variables (the real and imaginary parts of the entries of V).

n N Since the action of right multiplication by the real Lie group U(F ) on n,Fd N is transitive, it follows that n,Fd is irreducible and smooth. Since the stabiliser of V =[I,0] under this action is all n n unitary matrices of the form ×

I 0 n d , U U(F − ), 0 U ∈ the dimension of this variety is dim(U(Fn)) dim(U(Fn d)), which gives − − 1 dim(N d )= d(2n d 1), dim(N d )= d(2n d). n,R 2 − − n,C − We have observed that the normalised tight frames of n vectors for Fd can be indexed by the n n unitary matrices (for n > d this is not 1–1). In particular, by choosing V to be the× normalised tight frame given by the standard orthonormal basis (and zero vectors), we see (Exer. 7.1) that all normalised tight frames have the form

U1 n W =[w1,...,wn]=[I,0]U = U1, U = U(F ), (7.2)   U2 ∈   i.e., are the d n submatrices of the n n unitary matrices. If n = d, i.e., U2 = 0, then the normalised× tight frames are given× by the unitary matrices (orthonormal bases), N d i.e., d,Fd = U(F ). For n > d, a row of the submatrix U2 can be multiplied by a scalar, so that U has determinant 1. Hence, for n > d the normalised tight frames W can be indexed by elements of the special unitary group SU(Fn). Since SU(Fn) and U(Cn) are path–connected, and SU(Rn) is not, it follows that:

N The set of normalised tight frames n,Fd is path–connected, except for when n = d and F = R. 7.2 The algebraic variety of equal–norm tight frames 153 7.2 The algebraic variety of equal–norm tight frames

F N We can deﬁne the subvariety n,Fd of n,Fd consisting of all equal–norm tight frames of n vectors in Fd by replacing the norm equation in (7.1), by the n equations

2 d v j = , j = 1,...,n. n For n = d (equal–norm orthogonal bases) these spaces are the same, and for n > d the dimension reduces by n 1 (see [CMS13]), i.e., − d dim(F d )=(n 1)(d 1), dim(F d )= d(n d)+n(d 1)+1, n > d. n,R − 2 − − n,C − −

Example 7.1. The dimensions of the algebraic varieties of equal–norm real tight frames of n vectors for R2 and R3 are

dim(F 2 )= 1, dim(F 2 )= n 2, n > 2, 2,R n,R −

dim(F 3 )= 3, dim(F 3 )= 2n 5, n > 3. 3,R n,R − For R2 the points on the algebraic variety are determined by the ﬁrst n 2 diagram − vectors (see Exer. 2.9), since these are equal–norm complex numbers (w j) with 2 3 wn 1,wn determined by wn 1 + wn = (w1 + + wn 2). For C and C , we have{ − } − − − dim(F 2 )= 4, dim(F 2 )= 3n 3, n > 2, 2,R n,R − dim(F 3 )= 9, dim(F 3 )= 5n 8, n > 3. 3,R n,R − F The variety n,Fd has been studied extensively (see [DS06], [Str11], [CMS13]). F Classical results of [Whi57] imply that n,Fd is a union of ﬁnitely many manifolds, F and for n and d relatively prime n,Fd is a manifold [DS06]. We now consider its connectivity (the frame homotopy problem). F For a V n,Fd its Gramian P = V ∗V is an n n orthogonal projection matrix ∈ G× with diagonal entries d/n (and hence rank d). Let n,Fd be the set of such orthogonal n projections (the Grassmann manifold of d–planes in F ). Since every P G d is ∈ n,F the Gramian of some equal–norm tight frame for Fd, it follows that:

The map F d G d : V V V is onto and preserves path–connectivity. • n,F → n,F → ∗ For d 2, the map G d G n d : P I P (taking the Gramian to the n,F n,F − • Gramian≥ of the complementary→ equal–norm→ tight− frame) is a homeomorphism. Since tight frames are deﬁned up to unitary equivalence by their Gramians G F d (Corollary 2.1), n,Fd can also be described as the orbit space n,Fd /U(F ) for d F the action of U(F ) on n,Fd given by right multiplication. In [DS06] it is shown G G that 4,R2 is homeomorphic to a graph with 12 vertices and 24 edges, and 5,R2 is the the orientable surface of genus 25. 154 7 The algebraic variety of tight frames

G Example 7.2. For n = d, the space d,Fd has a single point (the identity matrix) and F d F F d,Fd = U(F ). Thus d,Cd is path–connected and d,Rd is not. G G Example 7.3. For n = d + 1 (and d = 1), we can calculate d+1,Fd = I d+1,F1 . F − The variety n,F1 consists of all V of the form

1 V = v11 v12 v1 n , v1 j = , j. , | | n ∀ F n Therefore n,R1 consists of 2 isolated points, and so is not path–connected. The G n 1 space n,R1 has 2 − isolated points (and so is not path–connected for n = d + 1). F F Hence d+1,Rd and n,R1 are not path–connected. For V,W F 1 , an explicit path from V to W in F 1 is given by ∈ n,C n,C

γ F w11 t w1n t : [0,1] n,C1 : t v11 ... v1,d+1 . → → v11 v1n F F Thus n,C1 and d+1,Cd (by taking complements) are path–connected. It turns out, that the only examples of path–disconnectedness are those above. Theorem 7.1. ([CMS13]) The variety of equal–norm tight frames satisﬁes: F 1. n,Rd is path–connected if and only if n d + 2 and d 2. F ≥ ≥ 2. n,Cd is path–connected. We will only give an indication of the proof given in [CMS13], which is involved. (k) (k) For a V =[v1,...,vn] F d let λ λ be the eigenvalues of the k–th ∈ n,R 1 ≤≤ d partial sum of the frame operator v1v + + vkv . These satisfy 1∗ k∗ (i) λ (0) = 0, j. j ∀ (ii) λ (n) = 1, j. j ∀ (iii) λ (k+1) λ (k) λ (k+1), λ (k) λ (k+1) λ (k) , 1 j d 2, 1 k n 1. j ≤ j+1 ≤ j+2 j ≤ j+1 ≤ j+2 ≤ ≤ − ≤ ≤ − (iv) ∑n λ (k) + d = ∑n λ (k+1), 1 k n 1. j=1 j n j=1 j ≤ ≤ − λ λ (k) The sequences = ( j )1 j d,0 k n satisfying the above conditions are called F ≤Λ≤ ≤ ≤ eigensteps for n,Fd . The set n,d of all eigensteps is a convex polytope, and hence it and its interior int(Λn,d) are path–connected. It is shown

The map F d Λn d : V λ = λV of a frame to its eigensteps is onto. • n,F → , → For any frame V F d with eigensteps Λ(V) int(Λn d), there is a continuous • ∈ n,F ∈ , map θ = θV : int(Λn d) F d with θ(Λ(V)) = V and Λθ = I on int(Λn d). , → n,F , For n > d, F d is path–connected if and only if F n d is, and this extends to n,F n,F − • the subsets of nonorthodecomposible frames (those that can’t be partitioned into two orthogonal subsets). Λ F These results are used to lift paths in n,d to paths in n,Fd . The case F = C exploits the fact that U(Cd) is connected. The case F = R is more technical and is proved using induction on n and d, and some special cases. 7.3 The density of rational tight frames 155 7.3 The density of rational tight frames

N d We have seen that the normalised tight frames V n,Fd of n vectors for C can be indexed by the n n unitary matrices, e.g., (7.2)∈ gives × n V =[Id,0]U, U U(F ), (7.3) ∈ where the index U is not unique for n > d. Here we use (7.3) to describe points on N n the variety n,Fd . There are various parametrisations of the unitary group U(F ), e.g., factorising its elements into Givens rotations or Householder transformations. We now consider the description in terms of the Cayley transform. This allows us to N F n n show that the rational points are dense on the n,Fd (but not on n,Fd ). Let A C × be a skew Hermitian matrix, i.e., A = A. Then A + I is invertible, and ∈ ∗ − I A U := − (7.4) I + A is a unitary matrix, called the Cayley transform of A. If U is unitary, and does not have 1 as an eigenvalue (so that I +U is invertible), then − I U A := − (7.5) I +U is a skew Hermitian matrix. These maps are the inverses of each other, and so

The unitary matrices (without eigenvalue 1) can be parametrised by the skew Hermitian matrices. −

Cayley’s original presentation (1846) was in the real case, where the Cayley transform (restricted to real matrices) maps into SO(Rd). A complex number x + iy is a (Gaussian) rational if x,y Q. We now show: ∈ N The rational points are dense in the variety n,Fd of normalised tight frames.

d d Theorem 7.2. ([CFW15]) Every tight frame V =[v1,...,vn] for C or R can be approximated arbitrary closely by one with vectors in (Q+iQ)d or Qd, respectively. Proof. Suppose, without loss of generality, that V is normalised. For n > d, we can choose an index U U(Fn) in (7.3) for which U does not have eigenvalue 1 (by multiplying the last∈ row of U by a suitable scalar in F). Thus, for n > d, the truncated− Cayley transform I A V =[I ,0] − (7.6) d I + A N maps the skew Hermitian matrices onto n,Fd . Thus the normalised tight frames V N d can be parametrised by the entries which determine the skew symmetric ∈ n,F matrices A, i.e., the 1 n(n 1) strictly upper triangular entries and n purely imaginary 2 − diagonal entries (for real matrices this reduces to 1 n(n 1) real parameters). 2 − 156 7 The algebraic variety of tight frames

Taking the truncated Cayley transform of such a parametrised matrix A gives a N V n,Fd with entries in the same ﬁeld as the parameters. Thus we can approximate the∈ parameters as closely as desired by elements in Q + iQ (which is dense in C) or Qd, and the truncated Cayley transform of the skew Hermitian matrix given by these approximate parameters will approximate V as closely as desired. For n = d, scale the last row of V to obtain a U U(Fd) which does not have 1 as an eigenvalue, approximate as above, and then∈ unscale the row. This can be done− since the rational points on the unit circle are dense (this classical result is the special case d = 1 and n = 1, incidently). ⊓⊔ Example 7.4. (Three vectors in R2) The 3 3 skew symmetric matrices A have three real parameters × 0 a b A =  a 0 c, a,b,c R. − ∈    b c 0 − −  The Cayley transform is the orthogonal  matrix 1 a2 b2+c2 2(a+bc) 2(ac b) − − 1+−a2+−b2+c2 1+a2+b2+c2 1+a2+b2+c2 I A 2(a bc) 1 a2+b2 c2 2(c+ab) U = − =  − − . I + A 1+a2+b2+c2 1+−a2+b2+−c2 1+a2+b2+c2  2(ac+b) 2(ab c) 1+a2 b2 c2   2 2 2 − 2 −2 2 2− 2− 2   1+a +b +c 1+a +b +c 1+a +b +c    The truncated Cayley transform gives the following indexing of V N d ∈ 3,R 1 a2 b2+c2 2(a+bc) 2(ac b) − − − − V = 1+a2+b2+c2 1+a2+b2+c2 1+a2+b2+c2 , a,b,c R. 2(a bc) 1 a2+b2 c2 2(c+ab)  − −  ∈ 1+a2+b2+c2 1+−a2+b2+−c2 1+a2+b2+c2 The normalised tight frame of three equally spaced equal–no rm vectors is given by 2 1 1 1 a = 2 √3 + √2√3 + √2, V = − 2 − 2 , − − √ √ 3 0 3 3  b = √3 √2, c = √2 1. 2 − 2 − − We can approximate these parameters to 5 decimal places by rationals a˜ 13165 , b˜ 31784 , c˜ 41421 . ≈ 100000 ≈ 100000 ≈ 100000 The corresponding truncated Cayley transform

5266079680 2633025064 2633092535 V˜ = 6449619561 − 6449619561 − 6449619561 ,  25064 4560603095 4560536360  − 6449619561 6449619561 − 6449619561 gives a normalised tight frame which approximates the equally spaced vectors of V to 5 decimal places. It is not an equal–norm frame. Theorem 7.2 does not imply that F F there is a dense set of rational points in n,Fd (there are none on 3,R2 ). 7.4 The existence of tight frames with given norms 157 7.4 The existence of tight frames with given norms

d If [v1,...,vn] is a normalised tight frame for H = F , then its norms must satisfy

2 2 2 v1 , v2 ,..., vn 1, v1 + v2 + + vn = d, (7.7) ≤ since it is the projection of an orthonormal basis (Theorem 2.2) and by (2.9). It is natural to ask whether there exists a normalised tight frame with some given norms which satisfy this condition, e.g., an equal–norm frame. We will show there is always such a frame. First we show the existence of such a tight frame follows from the Schur–Horn majorisation theorem, and then give a simple constructive proof. A vector β Rn is said to majorise a vector α Rn if after reordering so that their entries are∈ increasing, one has ∈

α1 +α2 + +αk β1 +β2 + +βk, 1 k n, α1 + +αn = β1 + +βn. ≤ ≤ ≤ Theorem 7.3. (Schur–Horn) Let a Rn be a vector which majorises λ Rn, then there is positive semideﬁnite real n ∈n matrix A with diagonal a and eigenvalues∈ λ. × Example 7.5. Let a Rn be a vector with ∈

0 a1,a2,...,an 1, a1 + a2 + + an = d, ≤ ≤ and λ = en d+1 + + en =(0,...,0,1,...,1) (n d zeros and d ones). Then a majorises λ−, by the calculation −

k k ∑ a j ∑ λ j = 0, 1 k n d, j=1 ≥ j=1 ≤ ≤ −

k n k ∑ a j = d ∑ a j d (n k)= d n + k = ∑ λ j, n d + 1 k n. j=1 − j=k+1 ≥ − − − j=1 − ≤ ≤

Thus there exists a semideﬁnite matrix A with eigenvalues λ, i.e., an orthogonal projection of rank d, and diagonal a. This matrix is the Gramian of a normalised d 2 tight frame V =[v1,...,vn] for R with v j = a j, 1 j n. ≤ ≤ Thus, we have:

Corollary 7.1. (Existence) There is a normalised tight frame V =[v1,...,vn] for d 2 H = F with norms v j = a j, 1 j n, if and only if ≤ ≤

0 a1,a2,...,an 1, a1 + a2 + + an = d. (7.8) ≤ ≤ d Example 7.6. (Equal–norm tight frames). By taking a j = , 1 j n, we conclude n ≤ ≤ that an equal–norm tight frame of n d vectors for Fd exists (for every n and d), F ≥ i.e., n,Fd is a nontrivial variety. Explicit constructions are given by Example 2.4, or, more generally, group frames (see 10). § 158 7 The algebraic variety of tight frames 7.5 The construction of tight frames with given norms

N All normalised tight frames are the orbit of any given one V =[v1,...,vn] n,Fd under the action right multiplication by the n n unitary matrices. It is therefore∈ N × natural to try and move along the variety n,Fd by using unitary matrices U that only make small (hence controlled) changes. Here we consider U of the form Uαβ = δαβ , α,β j,k , i.e., ∈ { } 1 .  ..     u jj u jk  j    . ← U =  ..  (7.9)       k  ukj ukk   ←  ..   .       1     These have the effect of ﬁxing all vectors of [v1,...,vn] except for v j and vk, which transform to iθ v′ = av j + bvk, v′ = e (bv j avk), (7.10) j k − − where VU =[v1′ ,...,vn′ ] and u u a eiθ b jj jk 2 2 θ U j k 2 = = − , a + b + 1, R. (7.11) | ,    iθ  | | | | ∈ { } ukj ukk b e a     Since (VU)∗(VU)= V ∗(U∗U)V = V ∗V, we have the U conservation of norms

2 2 2 2 2 2 v′ + v′ + + v′ = v1 + v2 + + vn , 1 2 n which for matrices of the form (7.9) gives

2 2 2 2 v′ + v′ = v j + vk . (7.12) j k

We now investigate precisely what norms the v′j (and hence vk′ ) of (7.10) can take.

Lemma 7.1. Let v,w Fd. Then for a,b F with a 2 + b 2 = 1, we have ∈ ∈ | | | | 1 1 av + bw 2 ( v 2 + w 2) ( v 2 w 2)2 + 4 v,w 2, (7.13) − 2 ≤ 2 − | | where the maximum and minimum of av + bw over a 2 + b 2 = 1 give equality. | | | | Proof. Without loss of generality, we suppose that a = t, b = σ 1 t2, 0 t 1, σ F, σ = 1. − ≤ ≤ ∈ | | 7.5 The construction of tight frames with given norms 159

It therefore suffices to find the maxima and minima of f = fσ : [0,1] R given by → f (t) := tv + σ 1 t2w 2 = t2 v 2 +(1 t2) w 2 + 2t 1 t2α, − − − where α = ασ := ℜ(σ v,w ), and then to optimise these values over σ = 1. These calculations (see Exer. 7.2) give the result. | | ⊓⊔ Example 7.7. The interval (7.13) for the norms av+bw , a 2 + b 2 = 1, is smallest when v and w are orthogonal, i.e., v,w = 0, which gives | | | | min v , w av + bw max v , w , (7.14) { } ≤ ≤ { } and largest when v and w are linearly dependent, i.e., v,w = v w , which gives | | 0 av + bw 2 v 2 + w 2. (7.15) ≤ ≤ Without any knowledge of v,w , it is only possible to construct vectors in the interval (7.14). This is sufficient for our purposes.

d Lemma 7.2. If V =[v1,...,vn] is a normalised tight frame for F , and

min v j , vk r max v j , vk , (7.16) { } ≤ ≤ { } then there exists a real n n unitary matrix U of the form (7.9) with ×

2 2 2 v′ = r, v′ = v j + vk r , j k − where [v1′ ,...,vn′ ] := VU. Proof. By Lemma 7.1 there are a,b F with av + bw = r, a 2 + b 2 = 1. Since ∈ | | | | 2 g(t) := tv j 1 t vk ± − is a continuous function of t with g(0)= vk , g(1)= v j , we can choose such a,b R, and determine them by solving g(t)= r. A unitary matrix U with the desired∈ properties is then given by (7.11). ⊓⊔ Example 7.8. One can take the U above to be a Givens rotation, by the choice

θ a = cosψ, b = sinψ, ei = 1, − − or to be a Householder transformation I 2wwT , by the choice − b 1 a , a = 1; √2(1 a) w j = − , wk = − − wℓ = 0, ℓ = j,k. 2 1, a = 1 160 7 The algebraic variety of tight frames

We now use Lemma 7.2 to give a constructive proof of Corollary 7.1, i.e., to 2 construct a normalised tight frame V =[v1,...,vn] with given norms v j = a j, n 1 j n, where a R satisﬁes (7.8). Since v′j is an average of v j and vk , one≤ must≤ take some∈ care. We now describe the algorithm of [FWW06], which starts with a normalised tight frame with large intervals (7.16), and moves it closer to one with the desired norms. Another algorithm based on the factorisation of elements of U(Rd) into Givens rotations was given earlier by [Cas04], [CL06]. Algorithm (for constructing a tight frame with given norms): Suppose that

1 a1 a2 an 0, a1 + a2 + + an = d. ≥ ≥ ≥≥ ≥ (0) (k) (k) (k) Starting with V , we construct iterates V =[v ,...,vn ] N d with 1 ∈ n,F (k) 2 (i) v = a j, 1 j k. j ≤ ≤ (k) 2 (k) (ii) v ak 1 or v = 0, k + 2 j n. j ≥ + j ≤ ≤ (n 1) 2 In view of (7.12), V has norms v j = a j, j, and so gives the desired frame. − ∀ (0) (0) Let V =[B,0] with B unitary, e.g., V =[e1,...,ed,0,...,0]. • Suppose that V (k), 0 k n 2, has been constructed. There are three cases: • ≤ ≤ − (k) 2 (k+1) (k) 1. If v = ak 1, then we can take V = V . k+1 + (k) 2 (k) 2 2. If v < ak 1, then ak 1 v , for some k + 2 j0 n. k+1 + + ≤ j0 ≤ ≤ (k) 2 (k) 2 (k) 2 3. If v > ak 1, then ak 1 v or v = 0, for some k+2 j0 n. k+1 + + ≥ j0 j0 ≤ ≤ (see Exer. 7.3) Interchange the vectors v(k) and v(k) in the cases 2 and 3. k+2 j0 (k) (k) Since ak 1 is in the interval (7.16) given by v and v , there is a unitary • + k+1 k+2 matrix U of the form (7.9) for which V (k+1) := V (k)U satisﬁes

(k+1) (k) (k+1) 2 v = v , j = k + 1,k + 2, v = ak 1. j j k+1 + (k+1) Since ak 1 ak 2, it follows that V N d satisfies properties (i) and (ii). + ≥ + ∈ n,F Example 7.9. Here we construct an equal–norm tight frame of three vectors for R2. (0) We start with V =[e1,e2,0]. Since the first two columns are orthonormal, every (0) (0) 2 2 vector av1 + bv2 , a + b = 1, has norm 1 (the interval of Lemma 7.1 is a single point). Thus we take a linear combination of the first and third columns given by a Givens rotation

cosθ 0 sinθ 1 0 0 cosθ 0 sinθ V (0)U = , U :=  0 1 0 . 0 1 0 0 1 0   θ θ  sin 0 cos     −    7.5 The construction of tight frames with given norms 161

Clearly the first column can take any norm between 0 and 1. We want it to be 2/3, and so take cosθ = 2/3, sinθ = 1/3, to obtain 2 0 1 V (1) = 3 3 . 0 1 0    Applying a rotation with fixes the first column (which has the desired norm) gives

1 0 0 2 1 ψ 1 ψ 3 3 sin 3 cos V (1)U = − , U := 0 cosψ sinψ . 0 cosψ sinψ   ψ ψ 0 sin cos     −    1 The squared norm of the second column above can take any value between 3 and 1. Choosing cosψ = 1/√2, sinψ = 1/√2 gives the equal–norm tight frame − 2 1 1 1 V (2) = − 2 − 2 . √ √ 3 0 3 3  − 2 2   N The algorithm presented moves between certain elements of the variety n,Fd , i.e., those satisfying (i) and (ii), by using Lemma 7.2. The next example indicates how one can move on the variety by using the more general Lemma 7.1.

Example 7.10. Consider the tight frame of three equally spaced unit vectors

1 1 1 2 2 V =[v1,v2,v3]= − − . 0 √3 √3  2 − 2   We will transform this to a tight frame with a two orthogonal vectors and the zero vector, by right multiplication by unitary matrices. Since the vectors v j have unit 1 norm, Lemma 7.2 cannot be applied. Since v j,vk = 2 , j = k, by Lemma 7.1 we 2 2 2| | 1 3 can choose a + b = 1 so that av j + bvk is as small as 2 and as large as 2 . The choice 1 0 0 1 0 1 V (1) := VU = − √2 , U = 0 1 1 , 0 √3 0  √2 √2 √2  1 1  − 0     − √2 √2    gives squared norms 1, 3 , 1 and v(1),v(1) = 1 . Thus 0 av(1) + bv(1) 2 3 , 2 2 | 1 3 | √2 ≤ 1 3 ≤ 2 we obtain 162 7 The algebraic variety of tight frames

√2 0 1 √3 0 0 √3 √3 V (2) = V (1)U = √2 , U =  0 1 0 .  0 √3 0 √2  1 √2  −  0    − √3 √3    Since every unitary matrix is a product of matrices of the form (7.9), it is not N difﬁcult to imagine that one could move from any element of n,Fd to any other by right multiplying by a ﬁnite sequence of such matrices.

Notes

The existence of equal–norm tight frames was not widely known until recently. This question was raised at Bommerholz in September 2000 (see [BF03]), and was “settled” in various ways: retrospectively [GVT98], by explicit constructions [Zim01],[RW02] and by minimisation of the frame potential [BF03]. The connection with results such as the Schur–Horn majorisation theorem are now well known and there are sophisticated algorithms [Str12],[CFM12], [CFM+13],[FMP16] for N F moving over the varieties n,Fd and n,Fd . The corresponding algebraic varieties of spherical (t,t)–designs for t = 1 (see 6.9) are far less studied. § 7.5 The construction of tight frames with given norms 163 Exercises

n 7.1. Let U(F ), the real Lie group of n n unitary matrices over F, act on N d × n,F (the normalised tight frames of n vectors for Fd) via right multiplication. (a) Show that the stabiliser of V =[I,0] N d is ∈ n,F

I 0 n d Stab(V)= : U U(F − . {0 U ∈ }   N n (b) Since the action is irreducible, it follows that n,Fd is isomorphic to U(F )/Stab(V). Use this to calculate its dimension. Hint: dim(U(Cn)) = n2, dim(U(Rn)) = 1 n(n 1). 2 − 7.2. For v,w Fd and σ F, σ = 1, define f = fσ : [0,1] R by ∈ ∈ | | → f (t) := tv + σ 1 t2w 2 = t2 v 2 +(1 t2) w 2 + 2t 1 t2α, − − − where α = ασ := ℜ(σ v,w ). (a) For a fixed σ, find a possible local maximum and minimum of f = fσ over [0,1]. (b) Optimise the possible local maximum and minimum from (a) over all σ = 1. | | (c) By considering the end points t = 0,1 find the maximum and minimum of fσ (t) over t and σ.

7.3. Suppose that 1 a1 a2 an 0, a1 + a2 + + an = d, and there is a ≥ (≥k) ≥≥(k) (k≥) normalised tight frame V =[v ,...,vn ] N d with 1 ∈ n,F (k) 2 (i) v = a j, 1 j k. j ≤ ≤ (k) 2 (k) (ii) v ak 1 or v = 0, k + 2 j n. j ≥ + j ≤ ≤ Show that (k) 2 (k) 2 (a) If v < ak 1, then ak 1 v , for some k + 2 j0 n. k+1 + + ≤ j0 ≤ ≤ (k) 2 (k) 2 (k) 2 (b) If v > ak 1, then ak 1 v or v = 0, for some k + 2 j0 n. k+1 + + ≥ j0 j0 ≤ ≤

Chapter 8 Projective unitary equivalence and fusion frames

Two ﬁnite sequences of vectors Φ =(v j) and Ψ =(w j) in inner product spaces are unitarily equivalent if and only if their respective inner products (Gramian matrices) are equal (Corollary 2.1, 3.4). For projective unitary equivalence, i.e., §

v j = α jUw j, j, ∀ where α j = 1, j, and U is unitary, the inner products are not projective unitary invariants,| | since ∀

v j,vk = α jUw j,αkUwk = α jαk w j,wk . Obvious projective invariants are

2 v j,v j = v j , 2 v j,vk v j,vk = v j,vk , | | but these don’t characterise projective unitary equivalence, unless (v j) is orthogonal. A projective unitary invariant is given by

v j,vk vk,v v ,v j . ℓ ℓ These “triple products” do characterise projective unitary equivalence when none of the inner products v j,vk are zero, e.g., for equiangular frames, but not in general. Here we show that ﬁnite sequences of vectors (lines) in inner product spaces are projectively unitarily equivalent if and only if certain projective unitary invariants (called m–products) are equal (Theorem 8.1). This is proved by giving an algorithm to recover a sequence of vectors (up to projective unitary equivalence) from a small subset of these projective invariants, which are determined by a spanning tree for the “frame graph”. We also extend our results to the projective similarity of vectors in F–vector spaces (where F = F).

165 166 8 Projective unitary equivalence and fusion frames 8.1 Projective unitary equivalence

As in 2.3 and 3.4, ﬁnite sequences of vectors Φ =(v j) and Ψ =(w j) in (real or § § complex) inner product spaces H1 and H2 are projectively unitarily equivalent if there is a unitary map U : H1 H2 and unit scalars α j, such that →

w j = α jUv j, j, (8.1) ∀ or, equivalently, w jw∗ = U(v jv∗)U∗, j. j j ∀ The study of lines in Rd and Cd, in particular, equiangular lines (Chapter 12), is effectively the study of conﬁgurations of unit vectors up to projective unitary equivalence. Many applications, such as signal analysis, depend only on frames up to projective unitary equivalence. For example, if the vectors are multiplied by unit modulus scalars, say w j = α jv j, then the frame operator is unchanged, and so the frame expansion is essentially unchanged, i.e.,

f = ∑ f ,w˜ j w j = ∑ f ,α jv˜j α jv j = ∑ f ,v˜j v j, f . j j j ∀

The condition (8.1), can be written as w j = U(α jv j), i.e., (w j) and (α jv j) are unitarily equivalent, which by the Gramian condition (see 3.4) is equivalent to §

wk,w j = αkα j vk,v j , j,k. (8.2) ∀ Equivalently, in terms of the Gramian:

Frames Φ and Ψ are projectively unitarily equivalent if and only if

Gram(Ψ)= C∗ Gram(Φ)C, (8.3) where C is the diagonal matrix with diagonal entries α j.

This is a practicable method for determining projective unitary equivalence only when the inner product space is real. In this case α j = 1, and so there are only ﬁnitely many possible matrices C. ± In view of (8.2), the inner products between vectors are not projective unitary invariants (in general). However, some products of them are, e.g.,

w j,wk wk,w w ,w j = α jUv j,αkUvk αkUvk,α Uv α Uv ,α jUv j ℓ ℓ ℓ ℓ ℓ ℓ = α jαk Uv j,Uvk αkα Uvk,Uv α α j Uv ,Uv j ℓ ℓ ℓ ℓ = v j,vk vk,v v ,v j . (8.4) ℓ ℓ This projective unitary invariant generalises in the obvious way. 8.2 The m–products 167 8.2 The m–products

Deﬁnition 8.1. Let Φ =(v j) be a sequence of n vectors (in a Hilbert space). Then the m–products (or the m–vertex Bargmann invariants) of Φ are

∆(v j ,v j ,...,v j ) := v j ,v j v j ,v j v j ,v j , 1 j1,..., jm n. (8.5) 1 2 m 1 2 2 3 m 1 ≤ ≤ The 3–products will also be called triple products. We observe that there are only ﬁnitely many m–products, and by the argument of (8.4), we have

The m–products of Φ are projective unitary invariants, i.e., if Φ and Ψ are projectively unitarily equivalent, then they have the same m–products.

The main result of this chapter (Theorem 8.1) is the converse of this, i.e., that if Φ and Ψ have the same m–products then they are projectively unitarily equivalent.

Example 8.1. The 1–products, 2–products, and 3–products of Φ =(v j) are

2 ∆(v j)= v j,v j = v j , 2 ∆(v j,vk)= v j,vk v j,vk = v j,vk , | | ∆(v j,vk,v )= v j,vk vk,v v ,v j . ℓ ℓ ℓ The 1–products and 2–products can be deduced from the 3–products, since

3 2 ∆(v j,v j,v j)= v j,v j , ∆(v j,v j,vk)= v j,v j v j,vk . (8.6) | | Example 8.2. (Conjugation) The m–products are closed under complex conjugation, i.e., ∆ ∆ (v j1 ,v j2 ,...,v jm )= (v jm ,...,v j2 ,v j1 ). (8.7) Example 8.3. (Decompositions) We observe that in some cases m–products can be decomposed into products of smaller ones, e.g.,

∆(v1,v2,...,vn 1)∆(v1,vn 1,vn) ∆(v1,v2,...,vn)= − − , (8.8) ∆(v1,vn 1) − provided ∆(v1,vn 1) = 0 and n 2. − ≥ We now deﬁne the notion of a “generating set” for the m–products. Deﬁnition 8.2. A subset of the m–products of Φ (or the corresponding indices/cycles) is a determining set if all the m–products can be determined from them. Example 8.4. If all the inner products between the vectors in Φ are nonzero, then (8.8) implies that the triple products are determining set for the m-products of Φ. 168 8 Projective unitary equivalence and fusion frames 8.3 The frame graph

Here we give examples which show that orthogonality (zero inner products) between vectors in a frame (v j) affect which m–products determine it up to projective unitary equivalence. This motivates the following deﬁnition (see [Str11], [AN13]). Deﬁnition 8.3. The frame graph (or correlation network) of a sequence of vectors (v j) is the graph with vertices v j (or the indices j themselves) and { }

an edge between v j and vk, j = k v j,vk = 0. ⇐⇒ Clearly, the frame graph is determined by the 2–products, and so projectively unitarily equivalent frames have the same frame graph.

Edges in the frame graph correspond to inner products which are nonzero. • m–cycles in the frame graph correspond to m–products which are nonzero. •

Example 8.5. (Empty graph) The frame graph of (v j) is empty (edgeless) if and only if all the inner products between different vectors are zero, i.e., the nonzero vectors are orthogonal. In this case, the Gramian is diagonal, and so by (8.3) all projectively unitarily equivalent frames have the same Gramian, i.e., the frame (v j) is determined up to projective unitary equivalence by its 2–products. Example 8.6. (Complete graph) We will see (Example 8.11) that if a frame has a complete frame graph, then it is determined up to projective unitary equivalence by its triple products. n Example 8.7. (n–cycle) Let (e j) be the standard basis vectors in C . Fix z = 1, and let | | e j + e j+1, 1 j < n, v j := ≤ en + ze1, j = n.

Then the frame graph of (v j) is the n–cycle (v1,...,vn), and so the only nonzero m–products for distinct vectors are

2 ∆(v j)= v j = 2, 1 j n, (8.9) ≤ ≤ 2 ∆(v j,v j 1)= v j,v j 1 = 1, 1 j < n, (8.10) + | + | ≤ ∆(v1,v2,...,vn)= z, (8.11) and their complex conjugates. Therefore, different choices for z give projectively unitarily inequivalent frames. Thus, for n > 3, the vectors (v j) are not deﬁned up to projective unitary equivalence by their triple products.

Example 8.8. (Connected components) The vectors in a connected component of the frame graph are orthogonal to all the other vectors of the frame, and hence a frame is union (see 5.1) of the frames given by the vertices of each connected component. § 8.4 Characterisation of projective unitary equivalence 169 8.4 Characterisation of projective unitary equivalence

We now show that a sequence of n vectors is determined up to projective unitary equivalence by its m–products for 1 m n (or a determining set). ≤ ≤ This is done by using the m–products of Φ =(v j) to construct all the possible Gramians G =[ wk,w j ] given by sequences of vectors (w j) which are projectively unitarily equivalent to Φ (and so have the same m–products as Φ). 3 We motivate the proof with an example. Let Φ =(v j) be the frame for R given 1 1 1 0 − 2 − 2 Φ = 0, √3 , √3 ,0 , 2 − 2         0  0   0  1         i.e., three equally spaced unit vectors  in a 2–dimensional    subspace together with a unit vector orthogonal to them all. The frame graph of Φ has an edge between each pair of the points v1,v2,v3, and v4 as an isolated point. By (8.3), all the possible Gramians of frames Ψ =(w j) which are projectively equivalent to Φ are given by

1 1 1 α1α2 α1α3 0 − 2 − 2  1 α α 1 α α  2 2 1 1 2 2 3 0 G =[ wk,w j ]= C Gram(Φ)C∗ = − − , α j = 1.  1 α α 1 α α  | |  3 1 3 2 1 0 − 2 − 2     0 0 01     Here we assume that only the m–products of Φ are known, so above we only know Φ 1 the modulus of the inner products between vectors in (this happens to be 2 for vectors which are not orthogonal). Clearly, any of the nonzero inner products (which correspond to edges) is a free variable of the form w j,wk = v j,vk a, a = 1. We 1 | | | | suppose that w1,w2 = a, a = 1 (we could choose any edge). Effectively, we 2 | | have ﬁxed α1 and α2 (without knowing v1,v2 ). We now consider an edge from one of the points v1,v2 (which have already been scaled) to an unscaled point, say α 1 v2 to v3. Since v3 has not been scaled, we can choose 3, so that w2,w3 = 2 b, b = 1, is a second free variable. We now have a spanning tree for the connected | | component of the frame graph which involves the vertices v1,v2,v3. The remaining edge (from v3 to v1) no longer corresponds to a free variable (α1,α2,α3 have been ﬁxed). This edge is in a cycle (v1,v2,v3), where the inner products corresponding to the other edges of the cycle are free variables (i.e., belong to the spanning tree), and so the inner product w3,w1 given by this edge is determined by the m–product given by the cycle

1 1 1 3 ∆(w1,w2,w3)= ∆(v1,v2,v3) = a b w3,w1 = ⇒ 2 2 −2 1 = w1,w3 = ab. ⇒ −2 170 8 Projective unitary equivalence and fusion frames

This example illustrates the main points of the proof below:

The edges in a spanning tree for a connected component of the frame graph of Φ correspond to inner products which can be taken as free variables. Once these are chosen (with the appropriate moduli), then there is a unique choice for the others (determined by the m–products of Φ) which gives the Gramian of a frame which is projectively unitarily equivalent to Φ.

We recall the following facts: Every ﬁnite graph Γ has a spanning tree (forest) T . • For each edge e Γ T , there is a unique cycle in e T called the fundamental • cycle (corresponding∈ \ to e). ∪

Theorem 8.1. (Characterisation) Sequences Φ =(v j) and Ψ =(w j) of n vectors are projectively unitarily equivalent if and only if their m–products are equal, i.e.,

∆(v j ,v j ,...,v j )= ∆(w j ,w j ,...,w j ), 1 j1,..., jm n, 1 m n. 1 2 m 1 2 m ≤ ≤ ≤ ≤ Proof. We have already observed that projectively unitarily equivalent sequences have the same m–products. We therefore suppose that Φ and Ψ have the same m–products, and will show that we can choose α1,...,αn so that (8.2) holds. The Gramians of Φ and Ψ are block diagonal (with entries having the same moduli), with blocks given by the vertices of the connected components of the common frame graph. We therefore assume without loss of generality that there is a single block, i.e., the frame graph Γ is connected. Spanning tree argument. Find a spanning tree T of Γ with root vertex r. By working outwards from the root r, we can multiply the vertices v Γ r by unit ∈ \ { } scalars αv so that for an edge v j,vk T , (8.2) holds, i.e., { }∈

wk,w j = αkα j vk,v j .

In this way, we can choose α1,...,αn so that (8.2) holds for all edges v j,vk T . Completing cycles. It remains only to show that (8.2) also holds{ for all}∈ edges e = v j,vk Γ T . Let (v j,vk,v ,...,v ) be the fundamental cycle given by the { }∈ \ ℓ1 ℓr edge e = v j,vk . Since the m–products are equal, and the other edges in this cycle belong to{T , we} obtain

∆(w j,wk,w ,...,w )= w j,wk wk,w w ,w w ,w j ℓ1 ℓr ℓ1 ℓ1 ℓ2 ℓr = w j,wk αkα vk,v α α v ,v α α j v ,v j ℓ1 ℓ1 ℓ1 ℓ2 ℓ1 ℓ2 ℓr ℓr =(αkα j w j,wk ) vk,v v ,v v ,v j ℓ1 ℓ1 ℓ2 ℓr = v j,vk vk,v v ,v v ,v j ℓ1 ℓ1 ℓ2 ℓr ∆ = (v j,vk,vℓ1 ,...,vℓr ), and cancellation gives (8.2) for the edge v j,vk Γ T . { }∈ \ ⊓⊔ 8.4 Characterisation of projective unitary equivalence 171

Above we associated the (directed) m–cycle (v j1 ,...,v jm ) in the frame graph with ∆ the nonzero m–product (v j1 ,...,v jm ). For m 3, all m–products can be calculated from those corresponding to simple cycles, since≥ if a cycle crosses at a, we have

∆(v1,...,vs,a,w1,...,wt ,a)= ∆(v1,...,vs,a)∆(w1,...,wt ,a). (8.12)

The cycle space of a ﬁnite graph Γ is the set of its Eulerian subgraphs (those with vertices of even degree). This can be viewed as a Z2–vector space, where the addition is the symmetric difference of sets. From this, it follows that the cycle space is spanned by the simple cycles (and its elements are disjoint unions of cycles). If the sum of two simple cycles is a simple cycle (in the frame graph), then corresponding m–product can be determined from those of the summands

∆(v1,...,vs,e1,...,er)∆(w1,...,wt ,er,...,e1)

= ∆(e1,e2) ∆(er 1,er)∆(v1,...,vs,e1,w1,...,wt ,er). − Combining these observations, we have:

A determining set for the m–products of Φ is given by the 2–products and the m–products corresponding to a basis for the cycle space of the frame graph.

The fundamental cycles corresponding to a spanning tree (forest) of a ﬁnite graph form a basis for the cycle space, called a fundamental cycle basis. We therefore have the following strengthening of Theorem 8.1.

Corollary 8.1. A finite frame Φ, with frame graph Γ , is determined up to projective unitary equivalence by a determining set for the m–products, e.g., 1. The 2–products. 2. The m–products, 3 m n, corresponding to a fundamental cycle basis (for the cycle space of Γ≤) formed≤ from a spanning tree (forest) T for Γ . In particular, if M is the number of edges of Γ T , then it is sufficient to know all of the 2–products, and M of the m–products, 3 \ m n. ≤ ≤ Proof. It suffices to verify the condition of Theorem 8.1 for a determining set. ⊓⊔

Example 8.9. Let Φ =(v j) be four equiangular vectors with C > 0. The frame graph of Φ is complete, and M = 6 3 = 3. Spanning trees (see Figure 8.1) include −

Tp := the path v1,v2,v3,v4,

Ts := the star graph with internal vertex v1 and leaves v2,v3,v4.

For Tp, the fundamental cycles given by the edges v1,v4 , v1,v3 , v2,v4 are { } { } { }

(v1,v2,v3,v4), (v1,v2,v3), (v2,v3,v4). 172 8 Projective unitary equivalence and fusion frames

v1 v1

v2 v4 v2 v4

v3 v3

Fig. 8.1: The spanning trees Tp and Ts (and cycle completions) of Example 8.9.

For Ts, the fundamental cycles given by the edges v2,v3 , v2,v4 , v3,v4 are { } { } { }

(v1,v2,v3), (v1,v2,v4), (v1,v3,v4).

Thus Φ is determined up to projective unitary equivalence by its 2–products, and the either of the following sets of m–products

∆(v1,v2,v3,v4), ∆(v1,v2,v3), ∆(v2,v3,v4),

∆(v1,v2,v3), ∆(v1,v2,v4), ∆(v1,v3,v4).

8.5 Reconstruction from the m–products

We now state the characterisation in way which summarises how all projectively unitarily frames can be constructed from a small determining set of m–products.

Theorem 8.2. (Reconstruction) Suppose Φ =(v j) is a frame of n vectors. Let Γ be the frame graph of Φ, T be a spanning tree (forest) for Γ , and

N = the number of edges in T , M = the number of edges in Γ T . \

Then the collection of all Gramians G =[ wk,w j ] of frames Ψ =(w j) which are unitarily projectively equivalent to Φ can be parameterised by N free variables. More precisely, for each of the N edges v j,vk T (choose an order) we have a free variable { }∈ w j,wk = v j,vk a , a = 1, | | ( j,k) | ( j,k)| and for the remaining M edges e = v j,vk Γ T , w j,wk is uniquely determined by equality of the m–products of Φ{and Ψ}∈for the\ fundamental cycle given by e. 8.5 Reconstruction from the m–products 173

2 Example 8.10. Let Φ =(v j) be the “two mutually unbiased bases” (see 8.6) for C given by §

1 0 1 1 √2 √2 1 0 1 1  0 1 1 1  Φ = , , √2 , √2 , Gram(Φ)= √2 − √2 .      1   1   1 1  0 1  1 0  √2 − √2  √2 √2           1 1   0 1   √2 − √2    The frame graph Γ of Φ is the 4–cycle (v1,v3,v2,v4). A spanning tree T is given by the path v1,v3,v2,v4. Corresponding to the three edges of T , we have three free variables

a b c w1,w3 = , w3,w2 = , w2,w4 = . √2 √2 √2

1 The remaining inner product w4,w1 = z is determined the fundamental cycle √2 given by v1,v4 , i.e., by completing the 4–cycle { }

w1,w3 w3,w2 w2,w4 w4,w1 = v1,v3 v3,v2 v2,v4 v4,v1 , which gives abcz = 1. − Thus all the Gramians of Ψ which are projectively unitarily equivalent to Φ are given by 1 0 a abc √2 − √2  0 1 b c  G = √2 √2 , a = b = c = 1.  a b  | | | | | |  1 0   √2 √2   abc c   0 1  − √2 √2  This particular Φ is in fact determined up to projective unitary equivalence by just its 2–products. This is because Sylvester’s criterion for G (as function of a,b,c,z) to be positive semideﬁnite gives

1 (bz + ac)2 1 bz 2 bz ac det(G)= = + 1 0 = +1 = 0 = z = . −4 abcz −4 ac ≥ ⇒ ac ⇒ − b In contrast, the (v j) of Example 8.7 forn = 4 also has frame graph a 4–cycle, but it is not determined up to projective unitary equivalence by its 2–products (and triple products).

We now consider those frames which are determined up to projective unitary equivalence by their 2–products and triple products. 174 8 Projective unitary equivalence and fusion frames 8.6 Triple products, equiangular lines, SICs and MUBs

The following special case of Corollary 8.1 is often useful.

Corollary 8.2. (Triple products) A ﬁnite frame Φ is determined up to projective unitary equivalence by its triple products (3–products) if the cycle space of its frame graph is spanned by 3–cycles (and so the cycle space has a basis of 3–cycles).

Proof. The 2–products can be deduced from the triple products by (8.6). ⊓⊔ Example 8.11. (Chordal graphs) A graph is said to be chordal (or triangulated) if each of its cycles of four or more vertices has a chord, and so the cycle space is spanned by the 3–cycles. Hence a frame is determined by its triple products if its frame graph is chordal. The extreme cases are the empty graph (orthogonal bases) where there are no cycles, and the complete graph where all subsets of three vectors lie on a 3-cycle (equiangular lines).

A set of equiangular lines given by a frame Φ is determined up to projective unitary equivalence by the triple products of Φ.

We now give an example (Corollary 8.3) where the cycle space of the frame graph has a basis of 3–cycles, but the frame graph is not chordal.

d Deﬁnition 8.4. A family of orthonormal bases B1,B2,...,Bk for C is said to be mutually unbiased if for r = j 2 1 v,w = , v Br, w B j. | | d ∈ ∈

We call B1,...,Bk a sequence of k MUBs (mutually unbiased bases). The maximal number of MUBs is a question of considerable interest (see 2.11). The frame graph of two or more MUBs (d > 1) is not chordal, because there§ is a 4–cycle (v1,w1,v2,w2), v1,v2 Br, w1,w2 Bs not containing a chord. We now show for three or∈ more MUBs∈ the cycle space of the frame graph is spanned by the 3–cycles. This is not case for two MUBs (see Example 8.10).

Corollary 8.3. (MUBs) A frame Φ consisting of three or more MUBs in Cd , d 2, is determined up to projective unitary equivalence by its triple products. ≥

Proof. It sufﬁces to show that the cycle space of the frame graph Γ of Φ has a d basis of 3–cycles. To this end, let B j, j = 1,...,k, be the MUBs for C , so that Γ is a complete k–partite graph (with partite sets B j). Fix v1 B1 and v2 B2. A ∈ ∈ spanning tree T for Γ is given by taking an edge from v1 to each vertex of B j, j = 1, and an edge from v2 to each vertex of B1 v1. Each of the remaining edges of Γ T gives a fundamental cycle. These have two\ types (see Figure 8.2): \ 8.6 Triple products, equiangular lines, SICs and MUBs 175

1 2 1. d (k 1)(k 2) edges between vertices in Br and Bs, r,s = 1, which give 2 − − fundamental 3–cycles (involving v1). 2. (d 1)((k 1)d 1) edges between vertices u B1 v1 and w j=1B j v2, − − − ∈ \ ∈∪ \ which give fundamental 4–cycles (u,w,v1,v2). These can be written as a sum (symmetric difference) of the 3–cycles (u,w,v2) and (v1,v2,w). Thus the cycle space is spanned by 3–cycles. ⊓⊔

v2 u

v1 v1

3 Fig. 8.2: Details from the proof of Corollary 8.3 for MUBs B1,B2,B3 in C . The frame graph Γ , the spanning tree T , and fundamental cycles of type 1 and 2.

There exist graphs which are not chordal, with every edge on a 3–cycle (as is the case for the frame graph of three or more MUBs), but for which the cycle space is not spanned by 3–cycles (see Figure 8.3).

Fig. 8.3: A nonchordal graph for which each edge is on a 3–cycle. 176 8 Projective unitary equivalence and fusion frames 8.7 Projective similarity and canonical m–products for vector spaces

We now use the previous results to characterise projective similarity. Let Φ =(v j) j J and Ψ =(w j) j J be ﬁnite sequences of vectors which span vector spaces X and∈ Y over a subﬁeld∈ F of C. We say that Φ and Ψ are similar if there is an invertible linear map Q : X Y with →

w j = Qv j, j, ∀ and are projectively similar if there unit scalars α j and Q invertible, with

w j = α jQv j, j. ∀ J J Assume that F = F. Then the canonical Gramian PΦ F × is deﬁned (see ), and ∈ § LΦ : X ran(PΦ ) : v j Pe j → → is an invertible linear map. Now (PΦ e j) and (PΨ e j) are projectively similar if and only if (α jPΦ e j) and (PΨ e j) are similar (for α j as above). But (α jPΦ e j) and (PΨ e j) are normalised tight frames, and so are similar if and only if they are unitarily equivalent (see Exer. 2.5). Combining these observations gives:

Φ =(v j) and Ψ =(w j) are projectively similiar

(PΦ e j) and (PΨ e j) are projectively similiar (8.13) ⇐⇒ (PΦ e j) and (PΨ e j) are projectively unitarily equivalent. ⇐⇒ This motivates the deﬁnition:

Deﬁnition 8.5. Let Φ =(v j) be a ﬁnite sequence of vectors in an F–vector space, with F = F. Then the canonical m–products of Φ are the m–products of (PΦ e j), which we denote by

∆C(v j ,...,v j ) := ∆(PΦ e j ,...,PΦ e j )= p j j p j j p j j , (8.14) 1 m 1 m 1 2 2 3 m 1 where PΦ =[pkj]. Φ These depend on as well as v j1 ,...,v jm , (unlike the usual m–products), and ∆ Φ ∆ one could use notation such as C = C the emphasize this. In this way, we may apply Theorem 8.1.

Theorem 8.3. (Characterisation) Let Φ =(v j) and Ψ =(w j) be ﬁnite sequences of vectors in vector spaces over a subﬁeld F of C with F = F. Then

1. Φ and Ψ are similar if and only if PΦ = PΨ (the canonical Gramians are equal). 2. Φ and Ψ are projectively similar if and only if their canonical m–products (for a determining set) are equal. 8.7 Projective similarity and canonical m–products for vector spaces 177

Proof. The ﬁrst follows from Proposition 4.1, and second from the observation (8.13) and Theorem 8.1. ⊓⊔ For projective similarity, one can calculate the α j and Q in w j = α jQv j explicitly:

Corollary 8.4. (Construction) Suppose that Φ =(v j) and Ψ =(w j) are projectively similar, i.e., w j = α jQv j, j, and Γ is the frame graph of (PΦ e j). Then the unit ∀ scalars α j are unique up to multiplication of those corresponding to a component of Γ by a unit scalar. All possible choices for (α j) can be constructed as follows

1. Fix the α j corresponding to the root(s) of a spanning tree (forest) for Γ . 2. Determine the remaining α j by the applying the spanning tree argument to

PΨ = C∗PΦC, C = diag(α j).

The invertible linear map Q is then deﬁned by Q(α jv j)= w j, j. ∀ We illustrate this with a simple example (also see 9.7). § Example 8.12. Suppose that Φ =(v j) and Ψ =(w j) span 2–dimensional spaces, say, for simplicity,

2 2 2 a1v1 + a2v2 + a3v3 = 0, a1 + a2 + a3 = 1, | | | | | | (8.15) 2 2 2 b1w1 + b2w2 + b3w3 = 0, b1 + b2 + b3 = 1. | | | | | | T For a general ﬁeld F, it may not be possible to normalise the vector a =(a1,a2,a3) , which spans dep(Φ), in which case one can modify the argument below. We have

2 1 a1 a1a2 a1a3 −| | − −  2  PΦ = I aa∗ = a1a2 1 a2 a2a3 . (8.16) − − −| | −  2  a1a3 a2a3 1 a3   − − −| |    The canonical 2–products are uniquely determined by the a j , since | | 2 2 2 2 2 ∆C(v j,v j)=(1 a j ) , ∆C(v j,vk)= a jak = a j ak , j = k, −| | |− | | | | | as are the canonical 3–products corresponding to a 3–cycle

1. Φ and Ψ are similar if and only if a jak = b jbk, j,k. ∀ 2. Φ and Ψ are projectively similar if and only if a j = b j , j. | | | | ∀ We now suppose that Φ and Ψ are projectively similar, i.e., w j = α jQv j, j, and ∀ calculate (α j) and Q from 178 8 Projective unitary equivalence and fusion frames

2 1 a1 α1α2( a1a2) α1α3( a1a3) −| | − −  2  PΨ = C∗PΦC = α2α1( a1a2) 1 a2 α2α3( a2a3) , (8.17) − −| | − α α α α 2   3 1( a1a3) 3 2( a2a3) 1 a3   − − −| |    and the frame graph Γ of (PΦ e j). In view of (8.15) are (8.16), there only are three possibilities for Γ (up to graph isomorphism). Γ is complete, i.e., a1,a2,a3 = 0. A spanning tree for Γ is given by the path PΦ e1,PΦ e2,PΦ e3. Fix α1 (corresponding to to the root), then α2,α3 are determined by the entries of (8.17) corresponding to the edges, i.e.,

α1α2( a1a2)= b1b2, α2α3( a2a3)= b2b3. − − − − Solving these gives

b1b2 b2b3 b2b3 b1b2 b1b3 α2 = α1, α3 = α2 = α1 = α1. a1a2 a2a3 a2a3 a1a2 a1a3

Thus, with α1 ﬁxed, say α1 = 1, Q is given by

b1b2 Q(v1)= w1, Q v2 = w2. a1a2 Suppose Γ is not complete, say a2a3 = 0 with a3 = 0, then (8.17) reduces to − 2 1 a1 α1α2( a1a2) 0 −| | −  2  PΨ = C∗PΦC = α2α1( a1a2) 1 a2 0 , − −| |  2  0 0 1 a3   −| |    where one of a1,a2 must be nonzero, say a1 = 0. This gives the remaining two cases. Γ has one edge, i.e., a1,a2 = 0, a3 = 0. The edge PΦ e1,PΦ e2 together with PΦe3 is a spanning forest. Fix α1,α3 (corresponding to the roots). Then α2 is given by

b1b2 α1α2( a1a2)= b1b2 = α2 = α1. − − ⇒ a1a2

Thus, with α1,α3 ﬁxed, say α1,α3 = 1, Q is given by

Q(v1)= w1, Q(v3)= w3.

Γ has no edges, i.e., a1 = 0, a2,a3 = 0. The vertices PΦ e1,PΦ e2,PΦ e3 are a spanning forest. We can make any choice for α1,α2,α3, and then Q is determined by Q(α jv j)= w j. Here a1v1 = 0, so v1 = 0, and so the j = 1 equation is vacuous. Thus, for α2 = α3 = 1, Q is given by

Q(v2)= w2, Q(v3)= w3. 8.8 Fusion frames 179 8.8 Fusion frames

As observed in (2.6), the frame expansion for a ﬁnite tight frame ( f j) for H can be written 1 H f = ∑ f , f j f j = ∑c jPWj f , f , A j j ∀ ∈

1 2 where c j = A f j 0, and PWj is the orthogonal projection onto Wj := span f j . This expansion is projectively≥ invariant, i.e., it only depends on the vectors up{ to} multiplication by unit modulus scalars. It can be generalised by letting the Wj be subspaces with any possible dimension, to obtain a tight fusion frame. Let (Wj) j J be subspaces of H and (c j) j J be nonnegative weights. Then the ∈ ∈ collection of pairs (Wj,c j) is a tight fusion frame for H if for some A > 0 { } 1 H f = ∑ c jPWj f , f . (8.18) A j J ∀ ∈ ∈ Many (projectively invariant) results for tight frames extend to tight fusion frames, e.g., taking the trace of the linear operators in (8.18) generalises (2.9) to

∑c j dim(Wj)= dA, d := dim(H ). (8.19) j

A tight fusion frame (Wj),(c j) for H is normalised if A = 1, i.e., ∑ j c j = dim(H ).

The ﬁnite normalised tight frames ( f j) for H (up to projective equivalence) are equivalent to the tight fusion frames (Wj),(c j) for H with

dim(Wj)= 1, c j = 0, ∑c j = dim(H ), j

2 via c j = f j , Wj = span f j . { }

Example 8.13. A trivial tight fusion frame is given by taking a single subspace, i.e., W1 = H , c1 = 1.

Example 8.14. If (Wj,c j) is a normalised tight fusion frame for H , then (8.18) and (8.19) give { }

c j ∑c jPW = I = IH = ∑c j(I PW )=(d 1)I = ∑ P = I. j ⇒ − j − ⇒ d 1 Wj⊥ j j j − c j H H Thus (Wj⊥, d 1 ) is a normalised tight fusion frame for , d = dim( ) > 1. { − } When (Wj,c j) is a tight frame, this fusion frame expresses the identity as a weighted{ sum of} orthogonal projections onto hyperplanes. 180 8 Projective unitary equivalence and fusion frames

Example 8.15. There is no tight fusion frame for C3 consisting two 2–dimensional subspaces. If there were, then the construction of Example 8.14 would give a tight frame of two vectors for C3.

The study of nontight fusion frames (frames of subspaces) was initiated by [AK05], [CKL08]. It parallels theory of frames. We now give a few of the salient deﬁnitions and details. The fusion frame operator of the subspace and nonnegative weight pairs (Wj,c j) j J is the positive semideﬁnite operator S : H H given by { } ∈ → H Sf := ∑c jPWj f , f . j ∀ ∈

The collection (Wj,c j) is a fusion frame for H if S is boundedly invertible, i.e., { } AIH S BIH (Loewner order), ≤ ≤ which is equivalent to

2 2 2 H A f Sf , f = ∑c j PWj f B f , f . ≤ j ≤ ∀ ∈

To calculate the orthogonal projections PWj in the fusion frame operator S one can F introduce a local frame j =( f jk)k Kj for each Wj, so that (see Exer. 3.12) ∈

PW f = ∑ f , f˜jk f jk = ∑ f , f jk f˜jk, f H , j ∀ ∈ k Kj k Kj ∈ ∈ 1 where f˜jk = SF− f jk. The collection of triples (Wj,c j,F j) is known as a fusion j { } frame system, and each F j as a local frame (for Wj). The calculation of Sf can be distributed by using a fusion frame system, which is natural for sensor networks can can 1/2 (see [CKL08]). Let ( f jk )k Kj , f jk := SF− f jk be the canonical tight frame for a ∈ j local frame F j for Wj. Then the fusion frame operator can be written as

can can can can Sf = ∑c j ∑ f , f f = ∑ ∑ f ,√c j f √c j f , f H , jk jk jk jk ∀ ∈ j k Kj j k Kj ∈ ∈ so that the fusion frame operator can be viewed as the frame operator of the frame can H 1 ( f jk ) j J,k Kj for , which has frame bounds A and B as above. Thus S− f can be calculated∈ ∈ using the frame algorithm (see 3.9) and the distributed calculation (parallel processing) of Sf outlined above. The§fusion frame expansion is

1 1 1 f = S− Sf = ∑c jS− PW f = ∑c j ∑ f , f˜jk S− f jk, f H . j ∀ ∈ j j k Kj ∈ The frame operator of a frame can similarly be calculated by a parallel algorithm. 8.9 Signed frames 181 8.9 Signed frames

The tight frame expansion (8.8) can be generalised by letting the scalars c j 0 take (possibly negative) real values. We call such a conﬁguration: ≥

σ H f = ∑ j f , f j f j = ∑c jPWj f , f , (8.20) j j ∀ ∈

2 σ j 1 , f j H , c j := σ j f j , Wj := span f j , (8.21) ∈ {± } ∈ { } a tight signed frame for H with signature σ =(σ j). By the polarisation identity, the condition (8.20) is equivalent to

2 2 f = ∑σ j f , f j , f H . (8.22) j | | ∀ ∈

Using this presentation, one can develop a theory of signed frames along the lines of that for frames (see [PW02], Exercises 8.1, 8.2).

Example 8.16. The tight signed frames with positive signature σ =(1) are precisely the normalised tight frames.

Example 8.17. Take any three unit vectors in R2 none of which are multiples of each other. Then there are unique c j that give a tight signed frame, which are given by

cos(β α) c = − , j sinα sinβ where π/2 α < β π/2 are the (acute) angles from the subspace spanned by this vector− to≤ those spanned≤ by the other two. This is negative if α < 0, β > 0, β α < π/2, i.e., the subspace generated by the vector lies in the region between the− acute angle made by the other two.

+ + +

+ − −

+ + +

Fig. 8.4: Tight signed frames of three vectors in R2 with the signature indicated.

Example 8.18. There exist tight signed frames of n vectors for Fd for any signature which takes the value +1 at least d times (see Exer. 8.3). 182 8 Projective unitary equivalence and fusion frames 8.10 Scaling the vectors of a frame to obtain a tight frame

1 We say a ﬁnite frame ( f j) of (nonzero) vectors for H can be scaled to a tight frame if there are scalars c j 0, such that ≥ f j I = IH = ∑c jPj, where Pj f := f ,u j u j, u j := , (8.23) f j j i.e., (√c j f j) is a normalised tight frame for H . Similarly, we say that ( f j) can be scaled to a tight signed frame if (8.23) holds for some choice of c j R. We now use tight signed frames to consider the question: ∈

When can a ﬁnite frame of n vectors for H = Fd be scaled to a tight frame?

It turns out (Corollary 8.5), that for the particular value of n

1 H 2 d(d + 1), real; n = 2 (8.24) d , H complex almost all sequences of n vectors have a unique scaling to a tight signed frame for H = Fd. In the generic situation, there is no scaling for less than n vectors, and inﬁnitely many for more than n vectors. We ﬁrst consider the geometry of the set of best possible scalings.

Proposition 8.1. (Best approximation) Let u1,...,un be unit vectors in H . Then n n the coefﬁcients c =(c j) F which minimise the Frobenius (matrix) norm j=1 ∈ n I ∑ c jPj F , Pj f := f ,u j u j, (8.25) − j=1 are the solutions of the n n linear system × 2 n Ac =[1], A :=[ uk,u j ] . (8.26) | | j,k=1

In particular, (u j) can be scaled to a tight frame if and only if this minimum is zero and there is a solution with c j 0. ≥ 2 Proof. We ﬁrst recall (see Exer. 3.1), that Pj,Pk F = u j,uk , I,Pk F = 1. The | | minimum (least squares solution) of (8.25) occurs when (the error) I ∑ c jPj is − j orthogonal to all the Pk, i.e., k ∀ 2 I ∑ c jPj Pk ∑ c j Pj,Pk F = I,Pk F ∑ c j u j,uk = 1. − j ⊥ ⇐⇒ j ⇐⇒ j | |

⊓⊔ 1 The term scalable is used in [KOPT13]. 8.10 Scaling the vectors of a frame to obtain a tight frame 183

We now seek a condition on the vectors (u j) which ensures the normal equations (8.26) have a unique solution, i.e., there is a unique scaling to a tight signed frame. Let be the Hadamard (pointwise) product of matrices: (S T) jk := s jkt jk. Then the matrix◦ A in the normal equations is the Hadamard product◦

2 A :=[ uk,u j ]= B B, B :=[ uk,u j ], (8.27) | | ◦ where B is the Gramian of (u j). We will use the Schur product theorem (cf [HJ91]).

Theorem 8.4. (Schur product) If A and B are positive semidefinite, then so is A B. If, in addition, B is positive definite and A has no diagonal entry equal to zero, then◦ A B is positive definite. ◦ 2 The following Lemma gives a condition which ensures that [ uk,u j ] (and other matrices) is invertible. It uses Lebesgue measure on Fd | Fd,| and the fact that the zero set of a nonzero polynomial has measure zero.××

d Lemma 8.1. For almost every v1,...,vn F ∈

r d + r 1 rank([ vk,v j ]) = min n, − , r 0. { r } ≥ d For almost every v1,...,vn C ∈

r s d + r 1 d + s 1 rank([ vk,v j vk,v j ]) = min n, − − , r,s 0. { r s } ≥

Proof. Let B = V ∗V be the Gramian of V =[v1,...,vn]. The matrices above are Hadamard products of B and B, i.e., respectively

r A :=[ vk,v j ]= B B B, ◦ ◦◦ r times r s A :=[ vk,v j vk,v j ]= B B B B B B. ◦ ◦◦ ◦ ◦ ◦◦ r times s times

Since B (and hence B) is positive semidefinite, it follows from the Schur product n theorem that A is also. Almost every choice of (v j) j=1 is in general position, and so we may assume without loss of generality that they are chosen to be so. First suppose that n d. Then the (v j) are linearly independent, so that B is positive definite, and by≤ the Schur product theorem A is positive definite, giving rank(A)= n, as asserted. Hence it suffices to suppose that n > d. Clearly, rank(A) n. Since B and V have ≤ the same kernel, and rank(V)= d, the positive semidefinite matrix B = V ∗V has rank d. Thus B can be written d B = ∑ u ju∗j , j=1 184 8 Projective unitary equivalence and fusion frames where u j,...,u j is an orthogonal basis for the range of B. We{ now consider} only the second case (the first is similar, following from the algebra for s = 0). Expanding (see Exer. 8.4) gives

r s A =[ vk,v j vk,v j ]= B B B B B B ◦ ◦◦ ◦ ◦ ◦◦ r times s times d d d d = ∑ ∑ ∑ ∑ (uk uk u j u j )(uk uk u j u j )∗, 1 ◦◦ r ◦ 1 ◦◦ s 1 ◦◦ r ◦ 1 ◦◦ s k1=1 kr=1 j1=1 js=1

d+r 1 d+s 1 a sum of at most − − rank one matrices ( is commutative), giving r s ◦ d + r 1 d + s 1 rank(A) − − . ≤ r s Thus, it sufﬁces to show that rank(A)= n, where

d + r 1 d + s 1 n = − − , r s for some choice of nonzero vectors (v j). Since det(A) is a polynomial in v1,...,vn and v1,...,vn, this then implies that det(A) will be nonzero for almost every choice of (v j), which gives the result. The existence of vectors (v j) for which A is invertible follows from the fact that s Π (Cd) has a basis of ridge polynomials z z,v r z,v (see Exer. 8.5). r◦,s → ⊓⊔

Theorem 8.5. (Equivalence) Let u1,...,un be unit vectors in a Hilbert space H of dimension d, where 1 H 2 d(d + 1), real; n = 2 d , H complex. 2 Let A :=[ uk,u j ]. Then the following are equivalent | | (a) The n n positive semideﬁnite matrix A is invertible. × (b) The vectors u1,...,un have a unique scaling which gives a tight signed frame, 1 with the c of (8.23) given by c = A− [1]. (c) The Hermitian forms on H have a basis given by

( f ,g) f ,u j u j,g , j = 1,...,n. → (d) The Hermitian operators on H have a basis given by the rank 1 orthogonal projections Pj : f f ,u j u j, j = 1,...,n. → Proof. (a) (b) As discussed in Proposition 8.1, A is the matrix giving the normal ⇐⇒ equations for ﬁnding a best scaling. If A is invertible, then the system ∑ j c jPj = I 1 has a unique solution given by c =(c j)= A− [1]. 8.10 Scaling the vectors of a frame to obtain a tight frame 185

We now consider the case H = Cd. The case H = Rd is similar, and easier. It Π d d uses the indentiﬁcation of 2◦(R ) with the symmetric bilinear forms on R (real Hermitian forms). (a) (c) By taking r = s = 1 in Exer. 8.5, the invertibility of A is equivalent ⇐⇒ d to the polynomials z z,u j u j,z , 1 j n, being a basis for Π (C ). Using → ≤ ≤ 1◦,1 the indentiﬁcation of Π (Cd) with the Hermitian forms (see 6.7), we conclude 1◦,1 § that the Hermitian forms ( f ,g) f ,u j u j,g , 1 j n, are a basis for the (real vector space) of Hermitian forms→ on Cd. ≤ ≤ (c) (d) We observe the Hermitian operator corresponding to the Hermitian ⇐⇒ form ( f ,g) f ,u j u j,g is P = u ju . → ∗j ⊓⊔ Corollary 8.5. (Scaling to a tight signed frame). Let H be a Hilbert space of dimension d, and 1 H 2 d(d + 1), real; n = 2 d , H complex.

Then for almost every choice of unit vectors (u1,...,un) in H there is a unique scaling that gives a tight signed frame, with the constants c j in (8.23) given by

1 2 c = A− [1], A :=[ uk,u j ]. (8.28) | |

Proof. We observe that det(A) is a polynomial in u1,...,un (and u1,...,un for H complex). By Lemma 8.1, this polynomial is nonzero for almost every choice for u1,...,un. When it is nonzero, i.e., A is invertible, Theorem 8.5 implies that there is unique scaling of (u j) to a tight signed frame given by (8.28). ⊓⊔

Since the set of scalings c of (u j) which give a tight signed frame is an afﬁne subspace (by Proposition 8.1), the number n above is a cut off. Almost all sequences of less than n vectors don’t have a scaling to a tight signed frame, and almost all sequences of more than n vectors have inﬁnitely many such scalings.

Example 8.19. (2 dimensions). Almost every set of three vectors in R2 can be uniquely scaled to a tight signed frame. See Example 8.17 (and Figure 8.4) for a description on when this scaling is a tight frame. Two vectors in R2 can be scaled to a tight signed frame if and only if they are orthogonal. This shows directly that almost every set of two vectors in R2 cannot be scaled to a tight signed frame. Almost every set of four vectors in C2 can be uniquely scaled to a tight signed frame for C2. The possible signatures are ++++ (a tight frame), +++ , and ++ (see Exer. 8.3). − −− 186 8 Projective unitary equivalence and fusion frames Notes

The characterisation of SICs up to projective unitary equivalence by their triple products was given by [AFF11]. This work was adapted to the general case (which includes MUBs) by [CW16] (see Theorems 8.1 and 8.2). The results of this chapter allow projective objects such as spherical (t,t)–designs and frames viewed as fusion frames to be classiﬁed (up to projective unitary equivalence) and their projective symmetries to be determined (see 9.3). There is ongoing interest in fusion§ frames, e.g., see [BE15], the Fusion frame page of the Frame Research Centre, and www.fusionframe.org. Tight signed frames were introduced in [PW02], where their relationship to the question of scaling to a tight frame (as presented here) was studied. The scaling question was also addressed in [KOPT13], who gave geometric descriptions of when a frame can be scaled to a tight frame.

Exercises

φ n H 8.1. Let ( j) j=1 be a sequence of vectors in , and c j F be scalars. (a) Show that there exists a representation of the form ∈

f = ∑c j f ,φ j φ j, f H , (8.29) j ∀ ∈ if and only if 2 2 f = ∑c j f ,φ j , f H . (8.30) j | | ∀ ∈

(b) Suppose that (8.29) holds. Show that there is a unique choice for (c j) which 2 minimises ∑ j c j , and that this satisﬁes c j R, j. Prove the analogue of (2.9), i.e., | | ∈ ∀ 2 ∑c j φ j = dim(H ). j

n H σ σ σ 8.2. Let ( f j) j=1 be vectors in , and =( j), j 1 . We say that ( f j) is a signed frame with signature σ for H if there exist (signed∈ {± } frame bounds) A,B > 0 with 2 2 2 A f ∑σ j f , f j B f , f H . (8.31) ≤ j | | ≤ ∀ ∈

The signed frame operator S;H H of a vector, signature pair ( f j), (σ j) is given by → Sf := ∑σ j f , f j f j, f H . j ∀ ∈ (a) Show that the frame operator S of a signed frame with bounds A,B is invertible, 1 with (1/B)IH S (1/A)IH . ≤ − ≤ 8.10 Scaling the vectors of a frame to obtain a tight frame 187

(b) For a signed frame ( f j) with signature σ and frame operator S, deﬁne the dual 1 signed frame to be ( f˜j) with signature σ, where f˜j := S− f j. Show that the dual 1 signed frame is a signed frame with frame operator S− , and one has the expansion

f = ∑σ j f , f˜j f j = ∑σ j f , f j f˜j, f H . j j ∀ ∈

can σ Deﬁne the canonical tight signed frame to be ( f j ) with signature , where can 1/2 f j := S− f j, and show that this is a tight signed frame.

8.3. Show that there is a tight signed frame of n vectors for Fd with signature σ if and only if σ takes the value +1 at least d times.

8.4. Show that the Hadamard product satisﬁes

d (aa∗) (bb∗)=(a b)(a b)∗, a,b F . ◦ ◦ ◦ ∀ ∈ Π d 8.5. Here we consider the space r◦,s(C ) of Exer. 6.17, which has dimension

d + r 1 d + s 1 n = − − . r s d Let v1,...,vn C . Show that the following are equivalent ∈ r s Π d (a) The polynomials p j : z z,v j v j,z are a basis for r◦,s(C ). δ→ Π d (b) The point evaluations j : f f (v j) are a basis for dual space r◦,s(C )′. → r s (c) The n n positive semideﬁnite matrix A =[ v j,vk v j,vk ] is invertible. Remark: Since× Π (Cd) has a basis of ridge polynomials z z,v r v,z s, it follows r◦,s → that A is invertible for some choices of (v j).

Chapter 9 Symmetries of tight frames

The angle preserving transformations of R2 form the real orthogonal group

2 2 T O(2) := A R × : A A = I , { ∈ } which can be thought of as the symmetries of the inner product space H = R2. One might reasonably hope that an expansion for this space would reﬂect this structure as much as possible. To understand the issues involved here, consider an orthonormal basis, and the tight frame of three equally spaced vectors.

The first is invariant (mapped to itself) under a reflection (which generates a group of order 2), and the second is invariant under the dihedral group (of order 6) generated 2π by a rotation (through 3 ) and a reflection. Thus the three equally spaced vectors have more of the symmetry of the space R2 than an orthonormal basis does. Further, this “large” symmetry group is closely related to the vectors of the frame being equiangular, which is desirable. It is even possible to have an expansion with all of the symmetry of the space, as in (1.3), but necessarily this requires one to use an (uncountably) infinite set of vectors. We now make these (intuitively obvious) ideas precise by defining the symmetry group and projective symmetry group of a frame (which can be calculated from a small set of invariants). For simplicity, we suppose that the frame is finite, though the theory extends to infinite frames without any complications (see Chapter 16).

189 190 9 Symmetries of tight frames 9.1 The symmetries of a sequence of vectors

We deﬁne the symmetry group and the projective symmetry group of a ﬁnite frame in a very general setting (which includes versions allowing antilinear symmetries). Each of these “symmetry groups” has the following key features:

It is defined for all finite frames as a group of permutations on the index set. • It is simple to calculate from a small set of invariants. • The symmetry group of a frame and all similar frames are equal. In particular, a • frame, its dual frame and canonical tight frame have the same symmetry group. The symmetry group of various combinations of frames, such as tensor products • and direct sums, are related to those of the constituent frames in a natural way. The symmetry group of a frame and its complementary frame are equal. • Frames with a large symmetry group have a simple structure (which embodies underlying symmetries of the space). Often the symmetry group can be very useful in constructing the frame itself. A special case is when the action of the symmetry group is transitive, which gives a group frame (see Chapter 10). If the symmetry group is abelian, then the group frames are the harmonic frames (see Chapter 11), all of which can be constructed from the abstract abelian groups. All the known maximal sets of complex equiangular lines come as the orbit of the projective action of an abelian group (see Chapter 14). Another example is multivariate orthogonal polynomials, where tight frames sharing the symmetries of the weight function can be constructed (see Chapter 15). Let SJ denote the symmetric group on the set J, i.e., the group of all bijections J J (called permutations) under composition. →Throughout, let H be a finite dimensional vector space over F, where F = F. Any finite sequence Φ =( f j) j J in H can be thought of as normalised tight frame for an appropriate inner product∈ on H (see 4.5). Thus, one can suppose that Φ is a tight frame (and thereby bypass Chapter 4).§ Let C : v v be the complex conjugation map (see 2.3). A product of C and a linear (unitary)→ map is called an antilinear (antiunitary§ map). In this way, we can extend the linear and unitary maps:

EGL(H ) := L,LC : L GL(H ) (Extended general linear group), { ∈ } EU(H ) := U,UC : L U(H ) (Extended unitary group). { ∈ } Of course these are groups, with GL(H ) EGL(H ), U(H ) EU(H ), where there is strict inclusion if and only if F R⊂(i.e., C = I). ⊂ ⊂ It is very convenient to have the possibility of a symmetry of Φ =( f j) j J which ∈ takes f j to fk where j = k, but f j = fk, i.e., nonidentity symmetries which map repeated vectors to themselves. For this reason, we deﬁne the symmetry groups to be permutations of the indices (given by a possibly unfaithful action). 9.2 The symmetry group of a sequence of vectors 191 9.2 The symmetry group of a sequence of vectors

Deﬁnition 9.1. Let Φ =( f j) j J be a ﬁnite sequence of vectors which spans H . Then the symmetry group and∈ the extended symmetry group of Φ are the groups

Sym(Φ) := σ SJ : Lσ GL(H ) with Lσ f j = fσ j, j J , { ∈ ∃ ∈ ∀ ∈ }

SymE(Φ) := σ SJ : Lσ EGL(H ) with Lσ f j = fσ j, j J . { ∈ ∃ ∈ ∀ ∈ } In other words, σ Sym(Φ) if ( f j) and ( fσ j) are similar. If ( f j) is a tight frame, ∈ i.e., SΦ =[ f j][ f j]∗ = AI, A > 0, then this similarity becomes unitary equivalence, since 1 1 1 1 Lσ Lσ = Lσ [ f ][ f ] Lσ = [Lσ f ][Lσ f ] = [ fσ ][ fσ ] = (AI)= I. ∗ A j j ∗ ∗ A j j ∗ A j j ∗ A It is easy to check that these symmetry groups are indeed groups, with

Sym(Φ) SymE(Φ) SJ. ⊂ ⊂ Since linear and antilinear maps are determined by their action on a spanning set, it follows that if σ is a symmetry, then there is a unique linear or antilinear map Lσ : H H with → Lσ f j = fσ j, j J. ∀ ∈ and the linear map πΦ : Sym(Φ) GL(H ) : σ Lσ (9.1) → → is a group homomorphism, i.e., a linear representation of G = Sym(Φ) on H . We will refer to both σ and Lσ (the action of σ) as a symmetry of Φ when Lσ is linear, and as an antisymmetry when Lσ is antilinear. With this understanding, we have:

If Φ is tight frame, then its symmetry group consists of unitary maps, i.e., the action of Sym(Φ) is unitary (and its extended symmetry group consists of unitary and antiunitary maps).

If the vectors in Φ are distinct, then πΦ is injective, i.e., the representation is faithful (see Exer. 9.2). In this case Sym(Φ) can be identiﬁed with its image.1

Proposition 9.1. (Similarity) If ﬁnite sequences Φ and Ψ are similar, then

Sym(Ψ)= Sym(Φ), SymE(Ψ)= SymE(Φ).

In particular, if Φ is a frame, then its dual and canonical tight frame have the same symmetry group, i.e., Sym(Φ)= Sym(Φ˜ )= Sym(Φcan).

1 The group of linear maps πΦ (Sym(Φ)) is sometimes deﬁned to be the “symmetry group” of Φ. 192 9 Symmetries of tight frames

Proof. Suppose that Φ =( f j) is similar to Ψ, say Ψ =(Qf j) for some invertible linear map Q. If σ Sym(Φ), then ∈ 1 Lσ f j = fσ j, j = (QLσ Q− )Qf j = Qfσ j, j, ∀ ⇒ ∀ so that σ Sym(Ψ), and Sym(Φ) Sym(Ψ). The reverse inclusion follows since ∈ ⊂ Ψ is similar to Φ. A similar argument shows that SymE(Ψ)= SymE(Φ). ⊓⊔ In other words:

The (extended) symmetry group of Φ depends only on its similarity class.

Example 9.1. (Bases) If Φ =( f j) j J is a basis, then for each σ SJ, Lσ f j := fσ j ∈ ∈ deﬁnes a linear map, and hence Sym(Φ)= SJ.

Example 9.2. (Vertices of a simplex) Suppose that Φ =( f j) j J is the vertices of a ∈ simplex, i.e., the vectors have a single linear dependence ∑ j f j = 0. Fix an index k, then for each σ SJ, Lσ f j := fσ j, j = k deﬁnes a linear map, with ∈

Lσ ( fk)= Lσ ∑ f j = ∑ fσ j = fσk, − j=k − j=k and hence Sym(Φ)= SJ.

Example 9.3. (see Figure 9.1) Let Φ =(v1,v2,v3) be the tight frame of three equally 2 spaced unit vectors for R , and Ψ =(v1,v2, v3). Then − Φ Ψ Sym( )= S3 = S 1,2,3 (order 6), Sym( )= 1,(12) (order 2). { } { }

Fig. 9.1: The frames Φ and Ψ of Example 9.3, which have Sym(Φ) = 6 and Sym(Ψ) = 2. | | | |

Φ n Example 9.4. The symmetry group of the tight frame =(v j) j=1 consisting of n equally spaced unit vectors, say 9.2 The symmetry group of a sequence of vectors 193

2π j sin n 2 v j = R ,  2π j  ∈ cos n   n 2 1 1 is the dihedral group of order 2n, i.e., Dn := a,b : a = 1,b = 1,b− ab = a− , π where a =(12 n) acts as rotation through 2 , and b as a reﬂection. n We now explain how the symmetry group of Φ can be calculated directly from its canonical Gramian PΦ (see 4.1). There is a bijection between the permutations § σ SJ and the so called J J permutation matrices, given by σ Pσ , where ∈ × →

Pσ e j := eσ j.

Let V =[ f j], and fσ j = Lσ f j, then VPσ =[Veσ j]=[ fσ j]=[Lσ f j]= LσV, so that

σ Sym(Φ) VPσ = LσV, for some Lσ GL(H ). (9.2) ∈ ⇐⇒ ∈

Lemma 9.1. (Calculation) Suppose that Φ =( f j) is a ﬁnite sequence of vectors, with canonical Gramian PΦ . For Φ a frame, PΦ = Gram(Φcan). Then

σ Sym(Φ) Pσ∗PΦ Pσ = PΦ . (9.3) ∈ ⇐⇒

Proof. By Propositions 4.1 and 9.1, we can suppose Φ =( f j) is the normalised tight frame given by the columns of PΦ (an orthogonal projection), so that V =[ f j]= PΦ . (= ) If σ Sym(Φ), then Lσ is unitary, and we have ⇒ ∈

Pσ∗PΦ Pσ =(PΦ Pσ )∗PΦ Pσ =(Lσ PΦ )∗Lσ PΦ = PΦ∗ (Lσ∗ Lσ )PΦ = PΦ .

( =) If Pσ PΦ Pσ = PΦ , then by choosing Lσ := Pσ , and writing PΦ = V, we have ⇐ ∗ VPσ = PΦ Pσ = Pσ PΦ = LσV, and so, by (9.2), we have that σ Sym(Φ). ∈ ⊓⊔ If Φ is tight frame, then PΦ is a scalar multiple of Gram(Φ), and hence we have:

σ Sym(Φ) Pσ∗ Gram(Φ)Pσ = Gram(Φ) ∈ ⇐⇒ fσ j, fσk = f j, fk , j,k. ⇐⇒ ∀

The condition for σ to be an extended symmetry (for R F) is that ⊂ T Pσ∗PΦ Pσ = PΦ , (9.4) which for a tight frame becomes

fσ j, fσk = f j, fk = fk, f j , j,k. ∀ 194 9 Symmetries of tight frames

Example 9.5. (Canonical coordinates) A sequence of vectors Φ and its canonical coordinates Ψ = cΦ (see 4.2) have the same symmetries, i.e., Sym(Φ)= Sym(cΦ ). § T T This follows since PcΦ = PΦ (Theorem 4.2) and Pσ = Pσ∗ give

T T σ Sym(Φ) Pσ∗PΦ Pσ = PΦ Pσ∗PΨ Pσ = PΨ ∈ ⇐⇒ ⇐⇒ Pσ∗PΨ Pσ = PΨ σ Sym(Ψ). ⇐⇒ ⇐⇒ ∈ We now show that a frame and its complement have the same symmetry group. Such a result is not possible if the symmetry group is deﬁned to be a group of linear transformations (as in [VW05]). Theorem 9.1. (Complements) Suppose that Φ is a ﬁnite sequence of vectors, and Ψ is a complementary sequence, i.e., PΦ + PΨ = I. Then

Sym(Φ)= Sym(Ψ), SymE(Φ)= SymE(Ψ).

Proof. By Lemma 9.1 and Pσ∗Pσ = I, we have

σ Sym(Φ) Pσ∗PΦ Pσ = PΦ Pσ∗(I PΨ )Pσ = I PΨ ∈ ⇐⇒ ⇐⇒ − − Pσ∗PΨ Pσ = PΨ σ Sym(Ψ). ⇐⇒ ⇐⇒ ∈

For SymE(Φ) a similar argument using (9.4) gives the result. ⊓⊔ Example 9.6. We consider the equal–norm tight frames Φ of four vectors for C3 with nontrivial symmetries, i.e., Sym(Φ) > 1. A complement Ψ consists of four equal–norm vectors for C. The| symmetries| of Ψ (and hence of Φ) are given by those permutations of Ψ which can be realised by multiplication by a unit modulus complex number, e.g., the permutation σ =(12)(34) is a symmetry of Ψ =([1],[ 1],[1],[ 1]) corresponding to multiplication by 1. Therefore, the only possibilities− for these− complementary frames (up to similarity)− are

([1],[1],[1],[z]), z = 1, ([1],[1],[z],[z]), z = 1, ([1],[1],[z],[w]), z = w, z,w = 1, ± ([1],[1],[1],[1]), ([1],[ 1],[1],[ 1]), ([1],[i],[ 1],[ i]). − − − − The corresponding symmetry groups are (up to group isomorphism)

S3, S2 S2, S2, S4, D4 (dihedral group of order 8), C4. × The Gramian matrices for the last three (which are harmonic frames) are 3 1 1 1 3 1 1 1 3 i 1 i − − − − − 1  1 3 1 1 1  1 3 1 1 1  i 3 i 1  − − − , − , − . 4   4   4    1 1 3 1  11 3 1   1 i 3 i − − −  −   −         1 1 1 3   1 1 1 3   i 1 i 3  − − −   −  −        This example can be generalised to give all possible symmetry groups for a tight n 1 frame of n vectors in C − (see [VW10]). 9.3 The projective symmetry group of a sequence of vectors 195 9.3 The projective symmetry group of a sequence of vectors

We now deﬁne projective symmetries of a sequence of vectors Φ =( f j), and give a parallel theory to that for (nonprojective) symmetries (here representations become projective representations, inner products are replaced by m–products, etc).

Deﬁnition 9.2. Let Φ =( f j) j J be a ﬁnite sequence of vectors which spans H . The projective symmetry group ∈and extended projective symmetry group of Φ are

SymP(Φ) := σ SJ : Lσ GL(H ), α j = 1 with Lσ f j = α j fσ j, j J , { ∈ ∃ ∈ | | ∀ ∈ }

SymEP(Φ) := σ SJ : Lσ EGL(H ), α j = 1 with Lσ f j = α j fσ j, j J . { ∈ ∃ ∈ | | ∀ ∈ } In other words, a permutation σ SymP(Φ) if ( f j) and ( fσ j) are projectively similar (equivalently, projectively unitarily∈ equivalent, when Φ is a tight frame). These projective symmetry groups are groups, which contain the corresponding symmetry groups, i.e.,

Sym(Φ) SymP(Φ), SymE(Φ) SymEP(Φ). ⊂ ⊂ 2 Example 9.7. Let Φ =(v1,v2,v3) and Ψ =(v1,v2, v3) be the tight frames for R of Example 9.3. These clearly have the same projective− symmetries, so that

S3 = Sym(Φ) SymP(Φ)= SymP(Ψ) S3. ⊂ ⊂

Hence SymP(Ψ)= S3 (of order 6) properly contains Sym(Ψ) (of order 2).

It is possible to associate with σ SymP(Φ) the projective linear map induced by ∈ Lσ , thereby obtaining a projective linear representation. We won’t labour this point, but do observe that the representation gives a projective unitary action of SymP(Φ) (on the lines of H ) when Φ is a tight frame. We have the projective analogue of Proposition 9.1. Proposition 9.2. (Projective similarity) If ﬁnite sequences Φ and Ψ are projectively similar, then

SymP(Ψ)= SymP(Φ), SymEP(Ψ)= SymEP(Φ).

In particular, if Φ is a frame, then its dual and canonical tight frame have the same (extended) projective symmetry group. In other words:

SymP(Φ) and SymEP(Φ) depend only on Φ up to projective similarity.

We now show how the projective symmetry group of Φ can be calculated directly from its canonical m–products (see 8.7). § 196 9 Symmetries of tight frames

Lemma 9.2. (Calculation) Suppose that Φ =( f j) is a ﬁnite sequence of vectors, with canonical m–products ∆C( f j ,..., f j ). Then σ SymP(Φ) if and only if 1 m ∈ ∆ ∆ C( f j1 ,..., f jm )= C( fσ j1 ,..., fσ jm ), (9.5) for a determining set of canonical m–products.

Proof. The condition σ SymP(Φ) is that ( f j) and ( fσ j) are projectively similar, and so we can apply Theorem∈ 8.3. ⊓⊔ For σ SymEP(Φ) (an antisymmetry), the condition (9.5) is replaced by ∈ ∆ ∆ ∆ C( f j1 ,..., f jm )= C( fσ j1 ,..., fσ jm )= C( fσ jm ,..., fσ j1 ). (9.6) For Φ a tight frame, the canonical m–products are a nonzero scalar multiple of the m–products, so that:

If Φ =( f j) is tight frame, then σ SymP(Φ) if and only if ∈ ∆ ∆ ( f j1 ,..., f jm )= ( fσ j1 ,..., fσ jm ), (9.7) for a determining set of m–products for Φ.

A frame and its complement have the same projective symmetry group.

Theorem 9.2. (Complements) If Φ =(v j) j J is a ﬁnite sequence of vectors, and ∈ Ψ =(w j) j J is a complement up to projective similarity, i.e., PΦ + CPΨC∗ = I, ∈ where C = diag(α j) is a unitary diagonal matrix, then

SymP(Ψ)= SymP(Φ), SymEP(Ψ)= SymEP(Φ).

Proof. Let PΦ =[pkj], PΨ =[qkj]. Then the canonical inner products, which are given by (8.14), satisfy

Φ ∆ (v j ,...,v j )= p j j p j j p j j C 1 m 1 2 2 3 m 1 α α α α α α =( j1 j2 q j1 j2 )( j2 j3 q j2 j3 ) ( jm j1 q jm j1 ) − m − − =( 1) q j j q j j q j j − 1 2 2 3 m 1 m Ψ =( 1) ∆ (w j ,...,w j ). − C 1 m The result then follows from Lemma 9.2 and (9.6). ⊓⊔

d Example 9.8. Let Φ =(v j) be an equal–norm tight frame of d + 1 vectors for C , e.g., the vertices of the regular simplex, and Ψ =(w j) be the complementary tight frame for C1. Since PΦ has a constant diagonal, the vectors of Ψ are equal–norm, say w j =(a j), a j = r > 0. Therefore, the m–products of Ψ are | | 9.4 Symmetries of combinations of frames 197

2m ∆(w j ,...,w j )=(a j a j )(a j a j ) (a j a j )= r . 1 m 1 2 2 3 m 1 d Thus all equal–norm tight frames Φ =(v j) of d + 1 vectors in C are projectively similar (to the vertices of the simplex), with SymP(Φ)= SJ.

9.4 Symmetries of combinations of frames

We now consider how the symmetry groups of a combination of frames Φ and Ψ (see Chapter 5) is related to their symmetry groups. In view of Theorem 4.1, these results also hold for spanning sequences of vectors. Let Φ =(φ j) j J and Ψ =(ψk)k K be ﬁnite frames for H1 and H2. The inner ∈ ∈ products on the orthogonal direct sum H1 H2 and tensor product H1 H2 are given by ⊕ ⊗

( f1,g1),( f2,g2) := f1, f2 + g1,g2 , ( f1,g1),( f2,g2) H1 H2, ∀ ∈ ⊕

f1 g1, f2 g2 := f1, f2 g1,g2 , f1 g1, f2 g2 H1 H2. ⊗ ⊗ ∀ ⊗ ⊗ ∈ ⊗ For σ Sym(Φ), τ Sym(Ψ), with corresponding Lσ GL(H1), Lτ GL(H2), ∈ ∈ ∈ ∈ let L σ τ = Lσ Lτ GL(H1 H2), i.e., ( , ) ⊕ ∈ ⊕

f Lσ f L(σ,τ) := , f H1, g H2. (9.8) g Lτ g ∀ ∈ ∀ ∈     We interpret (σ,τ) as a permutation on J K in the obvious way. This induces symmetry on the union and sum of Φ and Ψ∪, via

φ j Lσ φ j φσ j 0 0 0 L σ τ = = , L σ τ = = , ( , )       ( , )       0 0 0 ψk Lτ ψk ψτk             1 1 1 φ j Lσ φ j φσ j √n2 √n2 √n2 L(σ,τ) = = . 1 1 1  ψ   Lτ ψ   ψτ  √n1 k √n1 k √n1 k In this way, we have      

Sym(Φ) Sym(Ψ) Sym(Φ Ψ), Sym(Φ) Sym(Ψ) Sym(Φ +ˆ Ψ). × ⊂ ∪ × ⊂ For the direct sum, where J = K, we have

φ j Lσ φ j φσ j L σ τ = = , ( , )       ψk Lτ ψk ψτk       198 9 Symmetries of tight frames which is a permutation of the direct sum provided σ = τ. In this way, we have

Sym(Φ) Sym(Ψ) Sym(Φ Ψ). ∩ ⊂ ⊕

For the tensor product, deﬁne L σ τ GL(H1 H2) by L σ τ = Lσ Lτ . Then ( , ) ∈ ⊗ ( , ) ⊗

L σ τ (φ j ψk)=(Lσ φ j) (Lτ ψk)= φσ j ψτk ( , ) ⊗ ⊗ ⊗ and so we obtain Sym(Φ) Sym(Ψ) Sym(Φ Ψ). × ⊂ ⊗ In summary, we have: Proposition 9.3. The symmetry group of a ﬁnite frame satisﬁes

Sym(Φ) Sym(Ψ) Sym(Φ Ψ), × ⊂ ∪ Sym(Φ) Sym(Ψ) Sym(Φ +ˆ Ψ), × ⊂ Sym(Φ) Sym(Ψ) Sym(Φ Ψ), × ⊂ ⊗ Sym(Φ) Sym(Ψ) Sym(Φ Ψ). ∩ ⊂ ⊕ Moreover, these inclusions also hold for the other symmetry groups. Each of these inclusions can be strict (see Exer. 9.1).

9.5 Maximally symmetric tight frames

If Φ is a frame of n vectors, then

Sym(Φ) Sn = Sym(Φ) n! = Sym(Φ) n!. ⊂ ⇒ | | ⇒ | |≤ Thus, there are maximally symmetric frames in any class of such frames. Deﬁnition 9.3. Let C be a class of frames of n vectors, e.g., the tight frames or equal–norm frames in Fd. We say that Φ C is maximally symmetric if ∈ Sym(Φ) = max Sym(Ψ) . | | Ψ C | | ∈ This deﬁnition should be treated with a little caution for frames with repeated 2 vectors. For example, the frame of n vectors for R consisting e1 repeated n 1 − times and e2 has symmetry group of order (n 1)!, whilst that of the n equally spaced unit vectors has order 2n. − Example 9.9. The only cases when a frame Φ of n vectors for Fd (d > 1) can have maximal symmetry by virtue of Sym(Φ)= Sn is when n = d, i.e., Φ is a basis, or when n = d +1, i.e., Φ is the vertices of the simplex (see Example 9.8). This follows since the canonical 3–products (for distinct vectors) are all equal (see Exer. 9.4). 9.5 Maximally symmetric tight frames 199

Example 9.10. The tight frame of n equally spaced unit vectors in R2 is a maximally symmetric tight frame of n distinct vectors for R2. This is because the unitary transformations on R2 are products of rotations and reﬂections.

Example 9.11. The n equally spaced unit vectors in R2 are not always maximally symmetric tight frames of n distinct vectors in C2. For n even, the harmonic frame

2 3 4 n 2 n 1 1 ω ω ω ω ω − ω − 2πi , , , , ,... , , ω := e n 1  ω ω2  ω3 ω4 ωn 2  ωn 1 − − − − −               1 2 has a symmetry group of order 2 n .

Example 9.12. (Five vectors in C3). We consider the maximally symmetric tight frames Φ of ﬁve vectors in C3, by considering the complementary tight frames Ψ (which have the same symmetry group). We note Sym(Φ) = Sym(Ψ) divides 5! | | | | Since zero vectors are ﬁxed by a symmetry, the most symmetric Φ with a zero vector is given by the (1,4)–partition frame corresponding to

1 0 0 0 0 Ψ = , , , , , Sym(Φ) = 4! = 24, 0  1   1   1   1  | | 2 2 2 2           which is the vertices of the tetrahedron and a zero vector. If Sym(Φ) does not have an element of order 5, then the next most symmetric is the (2,3)–partition frame given by

1 1 0 0 0 Ψ = √2 , √2 , , , , Sym(Φ) = 2!3! = 12,  0   0   1   1   1  | | √3 √3 √3           which is the vertices of the trigonal bipyramid, followed by

1 1 0 0 0 Ψ = √2 , √2 , , , , Sym(Φ) = 8,  0   0   1   1  0 | | √2 √2           which is four equally spaced vectors and one orthogonal. If Sym(Φ) has an element of order 5, then Ψ must be a harmonic frame with the largest possible symmetry group (see Chapter 11), i.e., ﬁve equally spaced vectors

π cos 2 j Ψ 2 5 Φ = π : j = 1,...,5 Sym( ) = D5 = 10, 5 sin 2 j  | | | | 5   and Φ is the lifted ﬁve equally spaced vectors. We therefore conclude: 200 9 Symmetries of tight frames

The most symmetric tight frame of ﬁve (nonzero) vectors in C3 is the vertices of the trigonal bipyramid (the solution of Tammes’ problem).

Fig. 9.2: The most symmetric tight frames of ﬁve distinct nonzero vectors in R3. The trigonal bipyramid (12 symmetries), ﬁve equally spaced vectors lifted (10 symmetries), and four equally spaced vectors and one orthogonal (8 symmetries).

The known examples of maximally symmetric tight frames (such as those above) suggest a close relationship with group frames (see Chapter 10): Conjecture 9.1. A maximally symmetric tight frame is a union of group frames.

9.6 Algorithms and examples

To calculate the symmetry groups of a finite frame Φ =(v j) of n vectors, one must determine which of the n! permutations σ give a frame (vσ j) which is (extended) (projectively) unitarily equivalent to Φ. This can be checked (in theory) by applying Theorem 8.3 To make this feasible (for large n) requires an algorithm which checks the inner product or m–product condition efficiently, i.e., for many permutations at a time. We mention two such algorithms. The projective symmetry group of a frame can be viewed as the stabiliser (under the action of the symmetric group) of the frame graph with its m–cycles labelled by the corresponding m–products. It is not necessary to fully label the frame graph, e.g., the triple products suffice when the cycle space of the frame graph is spanned by the 3–cycles (Corollary 8.2). Algorithms are being developed (with Markus Grassl) which are efficient, and will be made publicly available. We now give a simple algorithm, which is suited to hand calculations, and cases where the symmetry group is small. For frames Φ =(v j) j J and Ψ =(w j) j J of ∈ ∈ n vectors, the algorithm determines the set of σ SJ for which Φ and (wσ j) are projectively similar, i.e., ∈ 9.6 Algorithms and examples 201

∆ ∆ C(v j1 ,...,v jm )= C(wσ j1 ,...,wσ jm ), (9.9) for all cycles ( j1,..., jm) from a determining set for Φ. In particular, for Ψ = Φ it calculates SymP(Φ), and if there is some σ then Φ and Ψ are projectively similar. There are two cases:

1. SymP(Φ) is large, i.e., the m–products take few different values. 2. SymP(Φ) is small, i.e., the m–products take many different values. An extreme example of the first is the vertices of a d–simplex (see Example 9.8), where d d+1 , j = k; (PΦ ) jk = SymP(Φ)= SJ. 1 , j = k, − d+1 Here the m–products are all equal (for fixed m), and so it is easy to check that each σ SJ is a projective symmetry. ∈ Our algorithm is best suited to the second case: when SymP(Φ) is small, and the m–products take many different values. This is the generic situation. Indeed, if the diagonal entries of PΦ (the 1–products) are distinct, then SymP(Φ) = 1. For an index set J of size n, we define a k–flag f to be| an ordering| of k distinct elements of J f =( j1, j2,..., jk). For a given fixed n–flag fb =( j1,..., jn), we can represent the permutation σ : jℓ σ jℓ (giving a projective similarity or symmetry) by the n–flag → fσ =(σ j1,...,σ jn).

Determining whether Φ =(v j) and Ψ =(wσ j) are projectively similar is equivalent to determining which of the n! permutations σ, i.e., n–flags fσ , satisfy (9.9). We think of each possible n–flag fσ =(σ j1,...,σ jn) as being built up from the 0–flag () by successively adding entries

0 1 2 n fσ = (), fσ =(σ j1), fσ =(σ j1,σ j2), ... fσ =(σ j1,σ j2,...,σ jn).

We will call the operation of going from a set Fk 1 of (k 1)–flags to a set Fk of k–flags as growing. At the k–th stage there are n − k + 1 choices− for the next entry, so that − Fk =(n k + 1) Fk 1 . | | − | − | k If Sym(Φ) < n!, then, at some stage, not all fσ Fk will be extendable to an n–flag| satisfying| (9.9). A necessary condition for such∈ an extension to exist is that (9.9) hold for all cycles (of length k) on the first k indices of the fixed flag fb = ≤ F ( j1,..., jn) from a determining set for (v j1 ,...,v jk ). Removing elements from k because they fail this condition (either in full or in part) will be called pruning. When the full condition is imposed we have a full pruning, otherwise a partial 202 9 Symmetries of tight frames pruning. In these terms, our algorithm for finding the set Fn of n–flags fσ giving the permutations σ that Φ and (wσ j) are projectively similar is:

Algorithm (to determine the n–ﬂags Fn giving a projective similarity).

Let F0 := () consist of the empty flag { } for k from 1 to n do

Grow Fk 1 to Fk − Prune Fk end for

Fully prune Fn, if necessary.

The art is in balancing the cost of pruning, with that of growing the set of possible k–ﬂags overly large. One can do this on a case by case basis, or by using an adaptive algorithm. The algorithm can easily be parallelised: simply partition Fk in any way, at any stage k, and apply the algorithm to each subset. We now illustrate our algorithm with a couple of examples, where Ψ = Φ. As a pruning rule we ask that a k–ﬂag ( j1,..., jk) match ∆ ∆ C(v j1 ,...,v jk )= C(wσ j1 ,...,wσ jk ). Thus at each stage we check only one new m–product, which is easily calculated.

Example 9.13. (SICs) Consider the equiangular tight frame Φ :=(v,Sv,Ωv,SΩv) of four vectors for C2 (the second prototypical example), where

3 + √3 0 1 1 1 Ω v := π , S := , := . √6 e 4i 3 √3 1 0  1 − −       We have 1 1 1 i √3 √3 − √3  1 i 1  1 √ 1 √− √ PΦ = 3 3 3 . 2  1 i 1   1   √3 √3 − √3   i 1 1   1   √3 √3 − √3    Take the base flag to be (1,2,3,4). The empty flag (0–flag) F0 = () grows to the set of 1-flags { } F1 = (1),(2),(3),(4) . { }

The pruning rule is that v1,v1 = wσ1,wσ1 , i.e., the norm is preserved, and so there is no pruning. Growing gives

F2 = (1,2),(1,3),(1,4),(2,1),(2,3),(2,4),(3,1),(3,2),(3,4),(4,1),(4,2),(4,3) , { } 9.6 Algorithms and examples 203 and pruning gives no reduction since Φ is equiangular. We now consider growing the 2–flag (3,2), the others being similar. This grows to the 3–flags (3,2,1), (3,2,4). Since i i i ∆C(v1,v2,v3)= , ∆C(w3,w2,w1)= , ∆C(w3,w2,w4)= , 24√3 −24√3 24√3 the 3–flag (3,2,1) is pruned. Continuing in this way gives

F3 = (1,2,3),(1,3,4),(1,4,2),(2,1,4),(2,3,1),(2,4,3), { (3,1,2),(3,2,4),(3,4,1),(4,1,3),(4,2,1),(4,3,2) . }

The ﬁnal stage k = n, growing does not increase the size of Fn 1, and in this case nothing gets pruned, by the rule used, or a full prune. Thus we have−

SymP(Φ)= F4 = (1,2,3,4),(1,3,4,2),(1,4,2,3),(2,1,4,3),(2,3,1,4),(2,4,3,1), { (3,1,2,4),(3,2,4,1),(3,4,1,2),(4,1,3,2),(4,2,1,3),(4,3,2,1) . }

This is the alternating group A4. Example 9.14. (Anti projective symmetries) Applying the full pruning algorithm to ∆ the previous example, with the m–products C(v j1 ,...,v jm ) replaced by their conjugates, and base ﬂag (1,2,3,4) gives the following anti projective symmetries

F4 = (1,2,4,3),(1,3,2,4),(1,4,3,2),(2,1,3,4),(2,3,4,1),(2,4,1,3), { (3,1,4,2),(3,2,1,4),(3,4,2,1),(4,1,2,3),(4,2,3,1),(4,3,1,2) . } Hence, we have SymP(Φ)= A4 SymEP(Φ)= S4. ⊂ In general, since the product of two anti projective symmetries is a projective symmetry, SymEP(Φ) is generated by SymP(Φ) together with any anti projective symmetry (if they exist). 2 Example 9.15. (MUBS) Let Φ =(v j) be the following two MUBs in C (see 8.6) § 1 1 v1 = e1, v2 = e2, v3 = (e1 + e2), v4 = (e1 e2). √2 √2 − Here 1 0 1 1 √2 − √2  1 1  1 0 1 √ √ PΦ = 2 2 . 2  1 1   1 0   √2 √2   1 1   0 1  − √2 √2    We arrive at the same F2 as in Example 9.13, without pruning. The pruning rule says that modulus of the inner product between v1 and v2 must be preserved. Since 204 9 Symmetries of tight frames this is zero, the index pairs in F2 must correspond to pairs of orthogonal vectors, which leads to the pruning

F2 = (1,2),(2,1),(3,4),(4,3) . { } Growing this gives

F3 = (1,2,3),(1,2,4),(2,1,3),(2,1,4),(3,4,1),(3,4,2),(4,3,1),(4,3,2) . { } All 3–products for distinct vectors are zero, and so there is no pruning at this stage. Growing, then full pruning leads to

SymP(Φ)= F4 = (1,2,3,4),(1,2,4,3),(2,1,3,4),(2,1,4,3), { (3,4,1,2),(3,4,2,1),(4,3,1,2),(4,3,2,1) . }

This group is the dihedral group of order 8 (the only subgroup of S4 of order 8), which is generated by the following permutations

(1324) (rotation through 90 degrees), (34) (reﬂection in the x–axis).

For d a prime power, d +1 MUBs for Cd can be constructed from the columns of elements RℓF from the Clifford group (see Theorem 12.22, 14.7). The projective symmetry groups, as calculated by our algorithm, for the ﬁrs§ t few d are given in Table 9.1.

Table 9.1: The projective symmetry groups SymP(Φ) and SymEP(Φ) for Φ the tight frame of n d vectors given by d + 1 MUBs in C , including the transitive subgroups of SymP(Φ).

d n SymP(Φ) SymEP(Φ) transitive subgroups of SymP(Φ) 2 6 < 24,12 > < 48,48 > < 6,1 >,< 12,3 > 3 12 < 216,153 > < 432,734 > < 72,41 > 4 20 1920 3840 < 20,3 >,< 60,5 >,< 80,49 >,< 120,34 >, < 160,234 >,< 320,1635 >,< 960,11357 > 5 30 3000 6000 < 600,150 >

The projective symmetry group of harmonic frames is considered in 11.12. § 9.7 Case study: 16 equiangular lines in R6 205 9.7 Case study: 16 equiangular lines in R6

We consider the 16 equiangular lines in R6 as presented by Janet Tremain ([Tre08])

1 1 1 1 1 11111111111 − − − − − 1 11 1 1 1 1 1 1 1111111  − − − − −   1 1 11 1 1 11 1 1 1 1 11 1 1  V =[v ]=  − − − − −  j   1 1 1 11 1 1 1 1 1 1 1 1 1 1 1   − − − − −    11 1 1 11 1 1 1 1 1 1 1 1 1 1  − − − − −    11 1 1 1 11 1 1 1 1 1 1 1 1 1  − − − − −    The projective symmetry group has order 11520 = 28 3 5, and is generated by the two permutations

a =(1,2,6,11,16,9)(3,12,8)(4,14,7)(5,10), b =(1,12,15,16,5,10)(2,11,4)(3,6,14,8,13,9).

We now seek a unitary matrix Lσ , σ a,b , with ∈ { }

Lσ v j = α jvσ j, j. ∀ By Corollary 8.4, Q = Lσ is unique up to a scalar 1, and can be calculated from its ± action on a basis of the v j, once a suitable choice of the scalars α j = 1 is known. ± We suppose that α1 = 1, so that

vσ1,vσ j v1,v j = Lσ v1,Lσ v j = α1vσ1,α jvσ j = α j = , j. ⇒ v1,v j ∀ In this way, we obtain

0 0 00 1 0 0 0 001 0 − 0 100 0 0 0 0 100 0  −     000100 0 0 000 1 L =  , L =  − . a   b   001000 1 0 000 0          000001 0 1000 0     −      100000 0 0 010 0          The group G = La,Lb is irreducible and has order 23040 (it contains the scalar matrix I). Since the projective symmetry group is transitive, it follows (see 10.7) − § that the 16 equiangular lines v j are an irreducible G–frame. {± } 206 9 Symmetries of tight frames 9.8 Case study: A spherical (4,4)–design of 12 lines in C2

Several unit–norm spherical (4,4)–designs of 12 vectors (lines) in C2 were computed numerically using the techniques of 6.16. The projective symmetry group for each was calculated (as described in 9.6)§ to be the dihedral group of order 10, with the projective action giving two orbits:§ one of size 2 (with the vectors orthogonal), and one of size 10. This suggests there is a (4,4)–design of the form

2 3 4 2 3 4 Φ =(v,av,a v,a v,a v,bv,abv,a bv,a bv,a bv) (u1,u2), (9.10) ∪ where v C2 is a unit vector, a (a rotation) and b (a reﬂection) are generators of the ∈ dihedral group and u1,u2 is an orthonormal basis. Taking { }

ω 0 2πi 0 1 1 0 a = , ω := e 5 , b = , u1 = , u2 = , (9.11)  0 ω 1 0 0 1         and optimising over v to obtain a (4,4)–design numerically suggested that the ratio √5+1 of the components of a suitable v was the golden ratio 2 , i.e.,

1 (1 + √5)ζ v := , ζ = 1. (9.12) 10 + 2√5  2  | |   An elementary calculation (see Exer. 9.7) shows that (9.10), (9.11), (9.12) deﬁne a spherical (4,4)–design Φ of 12 unit vectors for C2. The general method used here is known as precision bumping (see 14.20, 14.24). § §

Notes

The symmetry group of a finite tight frame (which is a natural notion) was first studied in [VW05], where it was considered as a group of unitary matrices. The definition given here (a group of permutations acting on the frame) was introduced in [VW10]. The projective symmetry group has been studied for SICs [AFF11], [Zhu12], and by [WC14] for a general finite frame. The calculation of the projective symmetry group as the stabiliser of a suitably labelled frame graph was initiated by Markus Grassl. 9.8 Case study: A spherical (4,4)–design of 12 lines in C2 207 Exercises

9.1. Show that following inclusions can be strict (a) Sym(Φ) Sym(Ψ) Sym(Φ Ψ). (b) Sym(Φ)× Sym(Ψ) ⊂ Sym(Φ ∪+ˆ Ψ). (c) Sym(Φ) ×Sym(Ψ) ⊂ Sym(Φ Ψ). (d) Sym(Φ)×Sym(Ψ) ⊂ Sym(Φ ⊗Ψ). ∩ ⊂ ⊕ 9.2. Let Φ be a frame of n vectors for H , dim(H )= d. (a) Show that Sym(Φ) divides n!. | | (b) Let πΦ be the representation σ Lσ of (9.1). Show that if Φ has distinct vectors, → then πΦ is faithful, and

πΦ (Sym(Φ)) m(m 1) (m d + 1), | |≤ − − where m is the number of distinct vectors in Φ.

9.3. Here we consider the map πΦ : Sym(Φ) GL(H ) : σ Lσ given by (9.1). → → (a) Show that πΦ is a group homomorphism, i.e., is a linear representation (action) of the group G = Sym(Φ) on H . (b) Let SΦ be the frame operator of Φ. Show that if g GL(H ), then ∈ 1 SΦ (gf )=(g∗)− Sg Φ ( f ), f H . ∗ ∀ ∈ (c) Show that if Φ is tight, then the action of Sym(Φ) on H commutes with SΦ , i.e., SΦ (σ f )= σSΦ ( f ), σ Sym(Φ), f H . ∀ ∈ ∀ ∈ 9.4. Let Φ be a sequence of n vectors. Show that if Sym(Φ)= Sn, then either Φ is a basis, or the vertices of the simplex (up to projective unitary equivalence). Remark: Therefore the symmetry group of θ–isogonal conﬁguration of n vectors (see Example 3.9) is Sn. 9.5. Describe the symmetry group of an α–partition frame for Rd (see 2.9). § 9.6. Find the symmetry groups and their action on C2 of the tight frames

2 1 ω ω 1 1 1 2πi Φ :=( , , ), Ψ :=( , , ), ω := e 3 , 1 ω2  ω  1 ω ω2             of Example 2.8. 9.7. Show that

2 3 4 2 3 4 Φ =(v,av,a v,a v,a v,bv,abv,a bv,a bv,a bv) (u1,u2), ∪ as given by (9.11) and (9.12) is a spherical (4,4)–design of 12 unit vectors for C2.

Chapter 10 Group frames

Here we introduce an important example of a structured frame, i.e., one in which the frame vectors can be obtained from the index set in a simple way. Prototypical examples of such frames include Gabor and wavelet systems. We consider the analogue of a Gabor system, where the index set is a ﬁnite group G, and the vectors are an orbit under a unitary action of G. This is equivalent to a frame whose symmetry group acts transitively on its vectors (Theorem 10.4). A useful example to keep in mind is the n equally spaced unit vectors in R2

or the vertices of a Platonic solid in R3. A group frame Φ for H will be, effectively, a frame of the form

Φ =(gv)g G, v H , (10.1) ∈ ∈ where G GL(H ) is a ﬁnite group of linear transformations, i.e., one which is the orbit of⊂ a single vector v under the linear action of G. Many important frames come in this way, e.g., the vertices of the Platonic solids, the harmonic frames, many equiangular tight frames (including all the known SICs), and all the known MUBs. By using representation theory, we will consider increasingly more general cases. We start with the case when the G–orbit of all vectors v = 0 gives a frame for H , and ﬁnish with a complete characterisation of the G–invariant frames (Theorem 10.9).

209 210 10 Group frames 10.1 Representations and G-frames

Suppose that G is a ﬁnite abstract group. A representation or linear action of G on a (ﬁnite dimensional) vector space H (over F) is a group homomorphism

ρH : G GL(H ). → The linear action of G on H given by ρ = ρH will often be written

gv := ρ(g)(v), g G, v H . ∈ ∈ Equivalent terminology is that H is an FG–module, or just a G–module (if the ﬁeld F and action of G is clear from the context). This is because H is a module over the group algebra FG (the F–vector space with a basis given by the elements of G and multiplication given by extending the multiplication of G linearly). A subspace (or set) V H is said to be G–invariant if gv V, v V, and so ⊂ ∈ ∀ ∈

The G–invariant subspaces of H are precisely the FG–submodules of H .

Two representations (H ,ρH ) and (K ,ρK ) are said to be equivalent if there is an invertible linear map T : H K such that → 1 ρK (g)= TρH (g)T − , g G. ∀ ∈ Any representation of G is equivalent to one in which H is a Hilbert space, and all the ρH (g) are unitary transformations (see Exer. 10.4). It therefore sufﬁces to consider frames of the form (10.1) where the action is unitary (see Corollary 10.1).

Deﬁnition 10.1. Let G be a ﬁnite group. We say that a frame (φg)g G for H is a group frame or G–frame if there exists a unitary representation ρ ∈: G U (H ) such that → gφh := ρ(g)φh = φgh, g,h G. ∀ ∈

Example 10.1. Let Cn = a be the cyclic group of order n. Then the n equally spaced 2 unit vectors are a Cn–frame (gv)g Cn for R given by the unitary group action ∈

2π 2π j j j cos n sin n a v := ρ 2 (a )v = ρ 2 (a) v, ρ 2 (a) := − . (10.2) R R R  2π 2π  sin n cos n   Since the representation (10.2) is faithful, i.e., injective, we can take G to be the ρ ρ abstract group Cn, or the isomorphic copy R2 (G)= R2 (a) . In contrast, ([1])g Cn j ∈ is a Cn–frame for R, via the trivial representation (a v := v, v R), but here G ∀ ∈ cannot be taken to be ρR(G)= [1] U (R) (the trivial group). This “repetition” has certain technical advantages,{ e.g.,}⊂ for taking combinations (see Theorem 10.2), or decomposing a G–frame into its constitute parts (see Theorem 10.7). 10.2 The frame operator of a G–frame 211 10.2 The frame operator of a G–frame

Since unitary maps preserve lengths, the deﬁnition implies that

A G–frame is an equal–norm frame.

The unitary action of G in a G–frame gives:

Lemma 10.1. The frame operator S of a G–frame commutes with (the unitary action of) G, i.e., S(hf )= hS( f ), h G, f H . ∀ ∈ ∀ ∈ Proof. Let Φ =(φg)g G be a G–frame for H , with frame operator S = SΦ . Then 1 ∈ 1 ρ(h)∗ = ρ(h)− = ρ(h− ) gives

1 1 S(hf )= ∑ hf ,φg φg = h ∑ f ,h− φg h− φg g G g G ∈ ∈ = h ∑ f ,φ 1 φ 1 = hS( f ), h− g h− g g G ∈ as supposed. ⊓⊔ The G–frame structure carries over to the dual and canonical tight frames.

Theorem 10.1. If Φ is G–frame for H , then so is The dual frame Φ˜ . • The canonical tight frame Φcan. • Any unitarily equivalent frame Ψ. • In particular, Φcan is an equal–norm tight frame.

Proof. Let Φ =(φg)g G be a G–frame for H . By Lemma 10.1, the frame operator ∈ 1 1 S = SΦ commutes with G, and hence with S− and S− 2 (as these can be written as power series in S). Thus the dual frame Φ˜ =(φ˜g)g G satisﬁes ∈ 1 1 1 gφ˜h = gS− φh = S− gφh = S− φgh = φ˜gh, g,h G, ∀ ∈ 1 can and so is a G–frame. A similar argument (with S− 2 ) shows that Φ is a G–frame. Suppose that Ψ is unitarily equivalent to Φ, say Ψ = cUΦ, c > 0, U U (H ). ∈ Let ρΦ be the representation with ρΦ (g)φh = φgh, g,h, and deﬁne a representation ∀ ρΨ : G U (H ) by ρΨ (g) := UρΦU 1. Then ρΨ is unitary, and → − 1 ρΨ (g)ψh = UρΦ (g)U− cUφh = cUρΦ (g)φh = cUφgh = ψgh, g,h G, ∀ ∈ so that Ψ is a G–frame. ⊓⊔ 212 10 Group frames

θ d d Example 10.2. A –isogonal conﬁguration (x j) j=1 of d unit vectors for R , i.e., one with 1 x j,xk = cosθ = , j = k −d 1 − is a group frame (Exer. 10.13), and hence so is its dual. Its dual is also isogonal (see Exer. 3.19). The canonical tight frame, an orthonormal basis, is also a group frame. A frame given by a linear action ρ : G GL(H ) is similar to one given by a unitary action: →

Corollary 10.1. Let G be a ﬁnite group, and Φ =(φg)g G be a frame for H . Then the following are equivalent ∈

1. Φ =(ρ(g)v)g G for ρ : G GL(H ) a group homomorphism and v H . 2. Φcan is a (tight)∈ G–frame. → ∈

Proof. 1= 2. If Φ =(ρ(g)v)g G is a frame, then it is similar to a tight G–frame Ψ can (see Exer.⇒ 10.4). Since similar∈ frames have unitarily equivalent canonical tight frames (Theorem 3.4), Φcan is unitarily equivalent to Ψ can, and hence is a G–frame. can 2= 1. Suppose that Φ =(τ(g)v)g G is a G–frame, and let S be the frame operator⇒ for Φ. Then ∈

1 can 1 1 1 Φ = S 2 Φ =(S 2 τ(g)S− 2 w)g G, w := S 2 v, ∈ 1 1 where G GL(H ) : g S 2 τ(g)S 2 is a group homomorphism. → → − ⊓⊔ For many of the methods of combining frames (Section 5), a combination of group frames is again a group frame. Theorem 10.2. Group frames can be combined as follows. 1. The direct sum of disjoint G–frames is a G–frame. 2. The sum of a G1–frame and a G2–frame is a G1 G2–frame. × 3. The tensor product of G1–frame with a G2–frame is a G1 G2–frame. × Proof. Use the notation x y and x y for the elements of H1 H2 and H1 H2. ⊕ ⊗ ⊕ ⊗ 1. If Φ =(gv)g G and Ψ =(gw)g G are disjoint G–frames, then their direct sum ∈ ∈ is Φ Ψ =(g(v w))g G, where g(x y) := gx gy is a unitary action. ⊕ Φ⊕ ∈ ⊕ ⊕ 2 and 3. Let j =(g jv j)g j G j be G j–frames. Then ∈

(g1,g2)(x y) := g1x g2y, (g1,g2)(x y) := g1x g2y ⊕ ⊕ ⊗ ⊗

deﬁne a unitary action of G1 G2 on H1 H2 and H1 H2, respectively. Thus × ⊕ ⊗ 1 1 Φ1+ˆ Φ2 = (g1,g2)( v1 v2) , (g1,g2) G1 G2 G2 ⊕ G1 ∈ × | | | | Φ Φ 1 2 = (g1,g2)(v1 v2) (g ,g) G G , ⊗ ⊗ 1 2 ∈ 1× 2 and so the sum and tensor product of group frames is again a group frame. ⊓⊔ 10.3 Group matrices and the Gramian of a G–frame 213 10.3 Group matrices and the Gramian of a G–frame

If Φ =(φg)g G is a G–frame for H , then the unitary action of G gives ∈ 1 1 1 g∗v := ρ(g)∗v = ρ(g)− v = ρ(g− )v = g− v, f H , ∀ ∈ and so the entries of its Gramian [ φh,φg ]g,h G have the special form ∈ 1 φh,φg = hφ1,gφ1 = g− hφ1,φ1 . (10.3) This is an example of what is called a group matrix or G–matrix.

Definition 10.2. Let G be a finite group. A matrix A =[agh]g,h G is a G–matrix (or group matrix1) if there exists a function ν : G C such that ∈ → 1 agh = ν(g− h), g,h G. ∀ ∈ It is natural to think of an n n matrix A as a G–matrix for some G of order n if there is an indexing of its entries× by G which yields a G–matrix. However, we do not take this as our definition, as it complicates the description of the algebraic properties of group matrices.

Example 10.3. Let G = D3 ∼= S3 be the dihedral group of order 6, i.e., 3 2 1 1 G = D3 = a,b : a = 1,b = 1,b− ab = a− , (10.4) and order its elements 1,a,a2,b,ab,a2b. Then each G–matrix has the form

1 a a2 b ab a2b 1 ν(1) ν(a) ν(a2) ν(b) ν(ab) ν(a2b) a  ν(a2) ν(1) ν(a) ν(a2b) ν(b) ν(ab)  2  ν ν 2 ν ν ν 2 ν  a  (a) (a ) (1) (ab) (a b) (b) . (10.5)    ν ν 2 ν ν ν 2 ν  b  (b) (a b) (ab) (1) (a ) (a)     ν ν ν 2 ν ν ν 2  ab  (ab) (b) (a b) (a) (1) (a )    2 ν 2 ν ν ν 2 ν ν  a b  (a b) (ab) (b) (a ) (a) (1)     

Example 10.4. If G is a cyclic group, say Zn with its elements ordered 0,1,...,n 1, then a G–matrix is a circulant matrix (see [Dav79]) −

1 Recently there has been a revival of interest in group matrices, see, e.g., [BR04] and [Joh07]. 1 Some authors write the (g,h)–entry as ν(gh− ), and variations thereof. 214 10 Group frames

c0 c1 cn 2 cn 1 − − cn 1 c0 c1 cn 2 − −  . .. .  ν A =  . cn 1 c0 . . , a jk = ck j := (k j).  −  − −  . .   c .. .. c   2 1     c1 c2 cn 1 c0   −    We now show:

A frame is a G–frame if and only if its Gramian is a G–matrix.

Theorem 10.3. (Characterisation) Let G be a ﬁnite group. Then Φ =(φg)g G is a G–frame (for its span H ) if and only if its Gramian is a G–matrix. ∈ Proof. If Φ is a G–frame, then (10.3) implies that its Gramian is a G–matrix. Conversely, suppose that the Gramian of a frame Φ for H is a G–matrix. Let Φ˜ =(φ˜g)g G be the dual frame, so that ∈

f = ∑ f ,φ˜g φg, f H . (10.6) g G ∀ ∈ ∈

For each g G, deﬁne a linear operator Ug : H H by ∈ → φ˜ φ H Ug( f ) := ∑ f , h1 gh1 , f . h G ∀ ∈ 1∈

Since Gram(Φ)=[ φh,φg ]g,h G is a G–matrix, we have ∈ 1 1 φgh ,φgh = ν((gh2)− gh1)= ν(h− h1)= φh ,φh . (10.7) 1 2 2 1 2

It follows from (10.6) and (10.7) that Ug is unitary, by the calculation φ˜ φ φ˜ φ Ug( f1),Ug( f2) = ∑ f1, h1 gh1 , ∑ f2, h2 gh2 h G h G 1∈ 2∈ φ˜ φ˜ φ φ = ∑ ∑ f1, h1 f2, h2 gh1 , gh2 h G h G 1∈ 2∈ φ˜ φ˜ φ φ = ∑ ∑ f1, h1 f2, h2 h1 , h2 h G h G 1∈ 2∈ φ˜ φ φ˜ φ = ∑ f1, h1 h1 , ∑ f2, h2 h2 = f1, f2 . h G h G 1∈ 2∈ Similarly, we have φ φ φ˜ φ φ φ˜ φ φ Ug h = ∑ h, h1 gh1 = ∑ gh, gh1 gh1 = gh. h G h G 1∈ 1∈ 10.3 Group matrices and the Gramian of a G–frame 215

This implies that ρ : G U (H ) : g Ug is a group homomorphism, since → → φ φ φ φ H φ Ug1g2 h = g g h = Ug1 g h = Ug1Ug2 h, = span h h G. 1 2 2 { } ∈ Thus ρ is a unitary representation of G on H , with

ρ(g)φh = φgh, g,h G, ∀ ∈ i.e., Φ is a G–frame for H . ⊓⊔ In particular, we can characterise normalised tight G–frames:

Corollary 10.2. Let G be a ﬁnite group. Then Φ =(φg)g G is a normalised tight G–frame (for its span H ) if and only if its Gramian P is∈ a G–matrix which is a projection, i.e., P2 = P.

Proof. A ﬁnite frame is a normalised tight frame if and only if its Gramian P is 2 an orthogonal projection matrix (Theorem 2.1), i.e., P = P and P = P∗. The result then follows, since every Gramian satisﬁes P = P . ∗ ⊓⊔ This extends to spanning sequences via the canonical coordinates (Chapter 4).

Corollary 10.3. A spanning sequence Φ =(φg)g G for an F–vector space (with ∈ F = F) is similar to a G–frame if and only if the canonical Gramian P = PΦ is a G–matrix. In this case, we can take Ψ =(PΦ eg)g G with the unitary action given by ∈

g(Peh)= Pegh.

We will see in 13.1 that the G–matrices form an algebra, which is isomorphic to the group algebra§ CG, via the correspondence

1 A =[ν(g− h)]g,h G ∑ ν(g)g CG. (10.8) ∈ ⇐⇒ g G ∈ ∈ Thus G–frames (up to similarity) can also be identiﬁed with elements of CG. We now show that if Φ is similar to a G–frame, then its complement is similar to a G–frame.

Corollary 10.4. The complement of a tight G–frame is a tight G–frame.

Proof. Let Ψ be the complement of a normalised tight G–frame Φ =(φg)g G. Since Gram(Φ)+ Gram(Ψ) is the identity matrix, we have ∈

φ φ 1 φ φ ψ ψ h, g = g− h 1, 1 , h = g; h, g = − − 1 1 φh,φg = 1 g hφ1,φ1 , h = g − − − i.e., Ψ is a G–frame. ⊓⊔ 216 10 Group frames 10.4 Identifying G–frames from their Gramian

The previous results imply that we can determine whether a frame Φ is similar to a G–frame by considering its (canonical) Gramian PΦ . This gives simple necessary conditions for Φ to be similar to a G–frame, e.g.,

The diagonal entries of PΦ must be constant. • The entries of every row/column of PΦ must be the same (up to reordering). • By investigating the symmetry group Sym(Φ) (which can be calculated), one can determine whether a frame Φ is similar to a G–frame (possibly with repetitions).

Theorem 10.4. A frame Φ with n vectors is similar to a G–frame if and only if its symmetry group Sym(Φ) has a transitive subgroup G. In this case, Φ with each of its vectors repeated G /n times is similar to the tight G–frame | |

(σPΦ e j)σ G =(PΦ eσ j)σ G ( j ﬁxed). ∈ ∈ n Proof. Let Φ =(φ j) . The action of σ Sym(Φ) on ran(PΦ ) given by j=1 ∈

σ(PΦ e j) := PΦ eσ j

n σ is unitary (since (PΦ e j) j=1 is a tight frame). Hence, for j ﬁxed, ( PΦ e j)σ G is a ∈ tight G–frame, which is similar to Ψ =(φσ j)σ G. If G is transitive, i.e., for any j,k 1,...,n there is a permutation σ G mapping∈ j to k, then Ψ consists of G /∈n {copies of}Φ. ∈ | |Conversely, if Sym(Φ) is not transitive, then Φ is not a group frame. ⊓⊔ Φ 3 2 Example 10.5. Let = (u j) j=1 be the three equally spaced unit vectors in R (Mercedes–Benz frame), which has symmetry group Sym(Φ)= S3 (Exer. 9.2). The transitive group S3 has one transitive proper subgroup, i.e., the cyclic group

C3 = a , a =(123). The canonical Gramian of Φ, which determines Sym(Φ), is

1 1 1 − 2 − 2 2 1 1 2 PΦ =  1  = [v1,v2,v3]. 3 − 2 − 2 3  1 1   1  − 2 − 2    Thus (by fixing j = 1) Φ can be thought of as a S3–frame (vσ1)σ S of six vectors, ∈ 3 or as a C3–frame (vσ1)σ C3 of three vectors. ∈ Φ 2π Suppose the action of a on is rotation through 3 , and b =(23) acts as the reflection which fixes u1, i.e., bu1 = u1, bu2 = u3, bu3 = u2. Then the S3–frame (vσ1)σ S is similar to the S3–frame (σu1)σ S (with the above unitary action). As ∈ 3 ∈ 3 10.4 Identifying G–frames from their Gramian 217 these two frames are tight, they have the same Gramian (up to a scalar), which is an S3–matrix. This matrix, with the same indexing as in (10.5), is given by

1 a a2 b ab a2b 1 1 1 1 1 1 1 − 2 − 2 − 2 − 2 a  1 1 1 1 1 1  − 2 − 2 − 2 − 2 2  1 1 1 1  a  1 1 . (10.9) − 2 − 2 − 2 − 2   1 1 1 1  b  1 1   − 2 − 2 − 2 − 2   1 1 1 1  ab  1 1  − 2 − 2 − 2 − 2  2  1 1 1 1  a b  1 1  − 2 − 2 − 2 − 2    This can be calculated via

1 1 2 ν(g− h) := hu1,gu1 = g− hu1,u1 = v 1 ,v1 , 3 g− h1 e.g., 2 2 2 3 1 ν(ab)= v ,v1 ,= v2,v1 = = . 3 (123)(23)1 3 3 −4 −2 It is easy to check that this matrix is a rank 2 orthogonal projection (up to a scalar). The element of the group algebra CS3 that it corresponds to via (10.8) is 1 1 1 1 1 1 ∑ ν(g)g = 1 a a2 + b ab a2b =(1 + b)(1 a a2). g S − 2 − 2 − 2 − 2 − 2 − 2 ∈ 3 Example 10.6. (Repeated vectors) If Φ is a frame of n vectors, then it may be that Sym(Φ) is transitive, but doesn’t contain a transitive subgroup G of order n. Thus it may be necessary to the repeat the vectors in Φ in order to view it as a group frame. For example, let Φ be the tight frame given by the 20 vertices of dodecahedron (see 10.6). With the labelling of Figure 10.2, §

Sym(Φ)= a,b,r = A5 Z2, Sym(Φ) = 120, ∼ × | | where a and b act as rotations through 72◦, c as a reﬂection, and are given by

a =(1,2,3,4,5)(6,8,10,12,14)(7,9,11,13,15)(16,17,18,19,20), b =(1,6,7,8,2)(3,5,15,16,9)(4,14,20,17,10)(12,13,19,18,11), c =(2,5)(3,4)(7,15)(8,14)(9,13)(10,12)(16,20)(17,19).

This group has one subgroup of order 20, which is not transitive, no subgroup of order 40, and unique and transitive subgroups of orders 60 and 120. Thus the vertices of the dodecahedron can be viewed as an A5–frame (vectors repeated three times) and as an A5 Z2–frame (vectors repeated six times). × 218 10 Group frames 10.5 Irreducible G–frames

Given a unitary action of a ﬁnite group G on H , and some v H , one has ∈

Φ =(gv)g G is a G–frame for its span. ∈ In 10.9 we will answer the questions: § When is Φ a frame for H ? • When is Φ tight? • The answer is particularly simple, and instructive, in the following situation.

Deﬁnition 10.3. A linear action (or representation) of a group G on H = 0 is said to be irreducible if the only G–invariant subspaces of H are 0 and H , i.e., { }

span gv g G = H , v = 0, v H . { } ∈ ∀ ∈ A G–frame given by such a unitary action is called an irreducible G–frame.

Theorem 10.5. (Irreducible G–frames). Suppose a unitary action of a group G on H is irreducible. Then (gv)g G is a tight G–frame for H for any v = 0, i.e., ∈ H dim( ) 1 H f = 2 ∑ f ,gv gv, f . G v g G ∀ ∈ | | ∈

Proof. Fix a nonzero vector v, and let S be the frame operator of (gv)g G. Since S is positive, it has an eigenvalue λ > 0 with eigenvector w. By Lemma∈ 10.1, S commutes with (the action of) G, and so gw is also an eigenvector for λ, for any g G, by the calculation ∈ S(gw)= gS(w)= g(λw)= λ(gw).

But gw g G spans H , and so S = λIH , i.e., (gv)g G is a tight frame. Taking traces gives{ } ∈ ∈ 2 2 trace(S)= ∑ gv = G v = λ dim(H )= trace(λIH ), g | | which determines λ. ⊓⊔ Corollary 10.5. All irreducible G–frames are tight.

Example 10.7. (Equally spaced vectors) The n equally spaced unit vectors in R2 (vertices of an n sided regular polygon) are an irreducible (tight) Cn–frame. They are an orbit of the cyclic group of order n acting via rotations as in (10.2), which is clearly irreducible. By way of comparison, to show this frame is tight by direct calculation would require the identities 10.5 Irreducible G–frames 219

n 2π j 2 n 2π j 2 n n 2π j 2π j ∑ cos = ∑ sin = , ∑ cos sin = 0, n n 2 n n j=1 j=1 j=1 which can now be viewed as a consequence of Theorem 10.5.

We note that all irreducible representations of abelian groups (such as Cn) over complex vector spaces are one–dimensional. From this it follows (see Chapter 11) that there are only ﬁnitely many G–frames for G abelian, i.e., the harmonic frames. We now show:

If G is nonabelian, then there are uncountably many unitarily inequivalent irreducible G–frames.

Proposition 10.1. Let G be a ﬁnite nonabelian group with an irreducible unitary action on Cd, d 2 (such an action always exists). Then there are uncountably many irreducible≥ G–frames for Cd (up to projective unitary equivalence).

Proof. Since all irreducible G–frames are tight, they are unitarily equivalent if and only if their Gramians are equal (when the generating vectors have equal norms). Let ρ be the representation. Then some ρ(g) has two distinct eigenvalues λ1 = λ2, otherwise each ρ(g) would be a scalar matrix, and

d span gv g G = span v g G = C , { } ∈ { } ∈ contradicting the irreducibility of ρ. Let u1 u2 be corresponding unit eigenvectors, ⊥ and vα be the unit vector

2 vα := αu1 + 1 α u2, α 1. −| | | |≤ The (1,g)–entry of the Gramian of the irreducible G–frame Φα =(gvα )g G is ∈

2 2 2 gvα ,vα = αλ1u1 + 1 α λ2u2,αu1 + 1 α u2 = αλ1 + 1 α λ2, −| | −| | −| | and so for different α, the frames Φα are unitarily inequivalent. ⊓⊔ Example 10.8. The above argument extends to some irreducible actions ρ on Rd. For example, if there is an element ρ(g) = I of order 2, then it has eigenvalues 1 (with orthogonal eigenvectors in Rd). An− example of this is the action of the ± 2 dihedral group Dn on R (as reﬂections and rotations) given in Example 9.4, which 2 gives uncountably many Dn–frames for R (see Figure 2.2 for the case n = 3).

Example 10.9. Let G be the nonabelian group of orthogonal matrices generated by a the rotation through 2π/3 and b the reﬂection in the y–axis, i.e., 220 10 Group frames

1 1 √3 1 a = − − , b = − , 2 √3 1   1 −     2 which is the dihedral group D3. This action of D3 on R is irreducible. Thus

2 2 Φ :=(gv)g D =(v,av,a v,bv,abv,a bv) ∈ 3 2 is an irreducible D3–frame for every nonzero vector v R . For v =(x,y) a unit ∈ vector, the Gramian of Φ is the D3–matrix

1 1/2 1/2 α + β α β − − − −  1/2 1 1/2 β α + β α  − − − −    1/2 1/2 1 α β α + β Gram(Φ)= − − − − , α β β α   + 1 1/2 1/2  − − − −   α α β β   + 1/2 1 1/2  − − − −   β α α β   + 1/2 1/2 1   − − − −    where 1 1 1 1 α := x2 √3xy y2, β := x2 + √3xy y2, α + β = y2 x2. − 2 − − 2 − 2 − 2 −

We observe that this gives uncountably many unitarily inequivalent D3–frames (see Figure 2.2). By considering the leading principal minor of order 3, we can see that the irreducible unitary action of the abelian subgroup C3 = a gives just one irreducible C3–frame up to unitary equivalence (the Mercedes–Benz frame).

d d d For irreducible actions on H = C there is an associated tight frame for C × (with the Frobenius inner product).

Theorem 10.6. Suppose that there is a unitary action ρ of a ﬁnite group G on Cd. Then the following are equivalent 1. The action of ρ is irreducible. 2. If v,w = 0, then (gv)g G and (gw)g G are dual tight G–frames, i.e., ∈ ∈ d 1 f = ∑ f ,gw gv, f Cd. (10.10) G v,w g G ∀ ∈ | | ∈

3. The unitary maps (ρ(g))g G are a tight G–frame for the d d matrices, i.e., ∈ ×

d d d A = ∑ A,ρ(g) ρ(g), A C × . (10.11) G g G ∀ ∈ | | ∈ 10.5 Irreducible G–frames 221

Proof. Let S be the operator Cd Cd given by → Sf := ∑ f ,gw gv, f Cd. g G ∈ ∈ This commutes with the action of G on Cd (see Exer. 10.1). 1= 2. If ρ is irreducible, then it is absolutely irreducible, and Schur’s Lemma (see Lemma⇒ 10.4) implies that S = λI. Take the trace of this, using Exer. 3.1 and the fact the action of G is unitary, to get

λd = trace(λI)= trace(S)= ∑ gv,gw = ∑ v,w = G v,w . g G g G | | ∈ ∈

2= 3. We have f ,gw = trace( f (ρ(g)w)∗)= trace( f w∗ρ(g)∗)= f w∗,ρ(g) , v,w ⇒= w v, and so (10.10) can be written as ∗

d d f w∗v = ∑ f w∗,ρ(g) ρ(g)v, f ,w,v C G ∀ ∈ | | g d d f w∗ = ∑ f w∗,ρ(g) ρ(g), f ,w C ⇐⇒ G ∀ ∈ | | g d ρ ρ d d A = ∑ A, (g) (g), A C × = span e jek∗ 1 j,k d. ⇐⇒ G ∀ ∈ { } ≤ ≤ | | g 3= 1. Take A = f v , and use f v ,ρ(g) = f ,ρ(g)v , v v = v 2 to get ⇒ ∗ ∗ ∗

d 2 d f v∗ = ∑ f v∗,ρ(g) ρ(g) = f v = ∑ f ,ρ(g)v ρ(g)v, G ⇒ G | | g | | g i.e., f span gv g G, v = 0, and so the action is irreducible. ∈ { } ∈ ∀ ⊓⊔ Example 10.10. The real unitary matrices

0 1 0 1 0 0 0 0 1, 0 1 0  −     1 0 0 0 0 1    −      3 generate a group G of order 12 (isomorphic to A4) whose action on C is irreducible. 3 3 Thus the 12 matrices in G form a tight frame for the 9–dimensional space R × .

Example 10.11. (Nice error bases) If ρ is an irreducible unitary action of G on Cd, then, by Schur’s lemma, the centre Z(ρ(G)) of the matrix group ρ(G) consists of scalar matrices, and ρ gives a projective representation of H = ρ(G)/Z(ρ(G)). When H has order d2, i.e., the d2 matrices of ρ(G) (up to scalar multiples) are d d an orthogonal basis for C × , the matrices are known as a nice error basis with index group H (see 13.13). § 222 10 Group frames 10.6 The vertices of the Platonic solids

The unitary action of the symmetry groups of the ﬁve Platonic solids (with centre of gravity at the origin) on R3 is irreducible (see Exer. 10.8), and so the vertices of the Platonic solids are irreducible G–frames. Similarly, the vertices of the truncated icosahedron (aka the ‘soccer ball’, ‘bucky ball’) form a tight frame for R3.

Fig. 10.1: The ﬁve Platonic solids: tetrahedron, cube, octahedron, dodecahedron and icosahedron.

We may apply Theorem 10.4 to determine for which groups G the vertices of a given Platonic solid are a (possibly not irreducible) G–frame (see Table 10.1). These calculations are outlined in Figure 10.2, which were adapted from [Mor04] (who kindly allowed us to reproduce his ﬁgures here).

Table 10.1: The groups G for which the vertices Φ of a Platonic solid are a G–frame.

Platonic solid vertices faces transitive subgroups G of Sym(Φ)

Tetrahedron 4 4 S4 (order 24), A4 (order 12),

D4 (order 8), Z2 Z2, Z4 (order 4). × Cube 8 6 S4 Z2 (order 48), × S4, A4 Z2 (order 24), D4 Z2 (order 16), × × Z2 Z2 Z2, Z2 Z4, D4 (order 8). × × × Octahedron 6 8 S4 Z2 (order 48), S4, A4 Z2 (order 24), × × A4, D6 (order 12), S3, Z6 (order 6).

Dodecahedron 20 12 A5 Z2 (order 120), A5 (order 60). × Icosahedron 12 20 A5 Z2 (order 120), A5 (order 60). × A4 Z2 (order 24), A4 (order 12). ×

Let G be the symmetry group of a Platonic solid acting on R3 (the order of G can 3 be G = 24,48,120). For a generic vector v R , the irreducible G–frame (gv)g G has| G| distinct vectors (none are repeated),∈ and there are uncountably many such∈ unitarily| | inequivalent frames. The vertices of the Platonic solids have the additional 10.6 The vertices of the Platonic solids 223 property that they are stabilised by a proper subgroup of G. Generalising this leads to the ﬁnite class of highly symmetric tight frames (see 13.8). §

Tetrahedron

a :=(1 2 3), b :=(4 2 3), c :=(1 2), a,b,c = S4 (tetrahedral group). ∼

Cube

a :=(1234)(5678), b :=(1485)(2376), c :=(2 4)(6 8), a,b,c = S4 Z2 (octohedral group). ∼ ×

Octahedron

a :=(1234), b :=(5 1 2)(6 3 4), c :=(1 2)(3 4), a,b,c = S4 Z2 (octohedral group). ∼ ×

Dodecahedron

a :=(12345)(6 8 10 12 14)(7 9 11 13 15)(16 17 18 19 20), b :=(16782)(3 5 15 16 9)(4 14 20 17 10)(12 13 19 18 11), c :=(2 5)(3 4)(7 15)(8 14)(9 13)(10 12)(16 20)(17 19), a,b,c = A5 Z2 (icosohedral group). ∼ ×

Icosahedron

a :=(12345)(7 8 9 10 11), b :=(1 2 6)(3 5 7)(4 11 8)(9 10 12), c :=(1 2)(3 5)(8 11)(9 10), a,b,c = A5 Z2 (icosohedral group). ∼ ×

Fig. 10.2: Generators for the symmetry groups of the vertices of the Platonic solids. 224 10 Group frames 10.7 Irreducible G–frames from vertex–transitive graphs

Let Γ be a graph on n vertices. There is a unitary action of the automorphism group Aut(Γ ) of Γ on Rn (where the vertices are labelled 1,2,...,n) given by

σe j := eσ j, j. ∀

Let A =[a jk] be the (0,1)–adjacency matrix of the graph Γ (a jk = 1 if and only if there is an edge from j to k). For σ Aut(Γ ), aσ j σk = a jk, and so we have ∈ ,

σ(Ae j)= σ ∑arjer = ∑arjσer = ∑arjeσr = ∑aσ 1 es − s, j r r r s = ∑as,σ jes = Aeσ j = A(σe j), (10.12) s i.e., the action of Aut(Γ ) commutes with the action A. Now suppose that Γ is a vertex–transitive graph, i.e., a graph for which there is an automorphism taking any vertex to any other. Let G be any subgroup of Aut(Γ ) whose action on the vertices of Γ is transitive. Then by (10.12), the columns of A are a G–frame (σAe j)σ G =(Aeσ j)σ G (where each column is repeated G /n ∈ ∈ | | times). Since A is symmetric, it is unitarily diagonisable. Let Eλ be the orthogonal projection onto the λ–eigenspace. The Eλ are called the primitive idempotents. By (10.12),

σ ∑λEλ e j = ∑Eλ eσ j = σEλ e j = Eλ eσ j. λ λ ⇒ The unitary action of G on the λ–eigenspace is irreducible (this is true for any distance–transitive graph), and so the G–orbit of any λ–eigenvector v is an irreducible G–frame for the λ–eigenspace (Theorem 10.5). Taking the choice v = Eλ e j and letting G be any transitive subgroup of Aut(Γ ) gives a tight G–frame

(σEλ e j)σ G =(Eλ eσ j)σ G ∈ ∈ for the λ–eigenspace, which has at most n distinct vectors. The Gramian of the n normalised tight frame (Eλ e j) j=1 is Eλ . From these observations, we obtain: Proposition 10.2. Let Γ be a vertex–transitive graph, with adjacency matrix A, and primitive idempotents Eλ . For each choice αλ 0,1 (not all αλ = 0), deﬁne an orthogonal projection matrix ∈ { }

Pα := ∑αλ Eλ , rank(Pα )= ∑αλ dim(Eλ ). λ λ

Then the n columns of Pα can be viewed as a normalised tight G–frame (Pα eσ j)σ G, where G is any transitive subgroup of Aut(Γ ). ∈

n By construction, the tight frame Φ =(Pα e j) has Aut(Γ ) Sym(Φ). j=1 ⊂ 10.8 Maschke’s theorem and homogeneous G–frames 225 10.8 Maschke’s theorem and homogeneous G–frames

To understand the G–frames for unitary actions which are not irreducible, we need to decompose a space into G–invariant subspaces (FG–submodules). This is often presented in the language of FG–modules (see Exer. 10.9, [JL01]). We will use the following terminology:

Deﬁnition 10.4. Suppose that there is a linear action of a group G on F–vector spaces V and W. Then a linear map σ : V W is an FG–homomorphismif it commutes with the action of G, i.e., →

σ(gf )= g(σ f ), g G, f H . ∀ ∈ ∀ ∈ A bijective FG–homomorphism is an FG–isomorphism, written V ∼= W. An FG–homomorphism is also called a G–morphism, a G–equivariant map, or a G–map, where the ﬁeld F is understood from the context.

Example 10.12. The frame operator S of a G–frame is an FG–isomorphism (Lemma 10.1), and if there is a unitary action of G on V and W, then

Sf := ∑ f ,gv gw, f V, g G ∀ ∈ ∈ deﬁnes an FG–homomorphism S : V W for any v V, w W (see Exer. 10.1). → ∈ ∈ We now give a version of Maschke’s theorem, where the unitary action ensures that the direct sum is orthogonal.

Lemma 10.2. (Maschke). Suppose that there is unitary action of ﬁnite group G on V = H . Then V can be written as an orthogonal (internal) direct sum

V = V1 V2 Vm (10.13) ⊕ ⊕⊕

of irreducible G–invariant subspaces Vj, where the Vj are unique up to ordering and FG–isomorphism, and the homogeneous components

HV (W) := ∑ X = Vj (10.14) X V V W X ⊂W j∼= ∼= corresponding to an irreducible W are unique.

Proof. Use strong induction on d = dim(H ). The case d = 0 is trivial. For d > 0, let V1 = 0 be an irreducible G–invariant subspace. Then V = V1 X, with X the ⊕ orthogonal complement of V1. We now show that X is G–invariant, which gives (10.13). Since the action is unitary, for g G and x X, we have ∈ ∈ 1 gx,v = x,g− v = 0, v V1 = gx V ⊥ = X. ∀ ∈ ⇒ ∈ 1 226 10 Group frames

The usual arguments to show that the Vj are unique up to FG–isomorphism (Jordan–Holder),¨ and HV (W)= V W Vj, apply without modiﬁcation. j∼= ⊓⊔ Thus there is a unique orthogonal decomposition of the space H = V into its homogeneous components

V = HV (W), W W ∈ where W is the collection of different irreducible G–invariant subspaces of H (up to FG–isomorphism). The HV (W) can be calculated by Theorem 13.2.

It is easy to check that a G–invariant subspace of some HW (V) is again a sum of irreducible G–invariant subspaces which are FG–isomorphic to W.

Deﬁnition 10.5. A G–frame for a space H with just one nonzero homogeneous component is called a homogeneous G–frame.

Example 10.13. An irreducible G–frame is a homogeneous G–frame. For G nonabelian, there are uncountably many homogeneous G–frames (see Proposition 10.1).

To decompose a G–frame into homogeneous G–frames, we need:

Lemma 10.3. (Schur). Suppose that A : V W is a linear map between irreducible G–invariant subspaces of H which commutes→ with the linear action of G, i.e, an FG–homomorphism. Then either 1. A = 0. 2. A is invertible, i.e., V and W are FG–isomorphic.

Proof. Suppose that A = 0. Then ker(A) = V and ran(A) = 0. Since the map A is an FG–homomorphism, its kernel and range are G–invariant subspaces of V and W. Hence, by irreducibility, ker(A)= 0, ran(A)= W, i.e., A is a bijection. ⊓⊔ Theorem 10.7. (Homogeneous decomposition). Suppose there is a unitary action of a ﬁnite group G on V = H = HV (W). W W ∈ Φ If v = ∑W vW , vW HV (W), then the G–frame =(gv)g G (for its span) can be decomposed as an∈ orthogonal direct sum of homogeneous G–fra∈ mes, i.e.,

Φ = ΦW , ΦW :=(gvW )g G. ∈ W W ∈ In particular, G–frames in different homogeneous components are orthogonal. 10.9 The characterisation of all tight G–frames 227

Φ Proof. Since g(∑W vW )= ∑W gvW , it follows immediately that is the direct sum of the homogeneous G–frames ΦW for span(ΦW ) HW (V). It remains only to show the orthogonality of this direct sum, i.e., that the maps⊂ Φ Φ B : span( W1 ) span( W2 ) : f ∑ f ,gvW1 gvW2 , W1 = W2 → → g G ∈

are zero. By Maschke’s theorem, it sufﬁces to show that A := PV B V = 0, where 2 | 1 Vj span(ΦW ) HV (Wj), Vj = Wj, are irreducible G–invariant subspaces, and ⊂ j ⊂ ∼ PV is the orthogonal projection onto Vj. The map A : V1 V2 commutes with the j → action of G, since PV and A V do (the latter by the argument of Lemma 10.1). By 2 | 1 Schur’s Lemma (Lemma 10.3), we have A = 0, as otherwise V1 and V2 would be FG–isomorphic, contradicting Vj = Wj. ∼ ⊓⊔ In other words:

A G–frame is the orthogonal direct sum of homogeneous G–frames.

10.9 The characterisation of all tight G–frames

If there is a G–frame for H , then there is one which is tight: Proposition 10.3. Suppose that there is a unitary action of a ﬁnite group G on H . Then the following are equivalent

1. For some v H , gv g G spans H . 2. There is a G–frame∈ { for} H∈ . 3. There is a tight G–frame for H . Proof. Since ﬁnite frames are simply spanning sequences, Theorem 10.1 gives

span gv g G = H Φ =(gv)g G is a G–frame for H { } ∈ ⇐⇒ ∈ Φcan is a tight G–frame for H . ⇐⇒

⊓⊔ This gives a simple necessary condition for the existence of a tight G–frame for H . We now give a constructive characterisation of which G–orbits under a unitary action are tight G–frames. We recall that a real vector space H can be complexified H C = H iH , and C ∼ ⊕ a linear/unitary map U on H extends to H via U(v1 iv2) :=(Uv1) (iUv2). The complexification of H is H C when F = R, and H⊕when F = C. ⊕ In this way, an FG–module V can always be thought of as a CG–module. Definition 10.6. A linear action of a finite group G on H (or representation, or FG–module) is absolutely irreducible if the action on H C is irreducible. 228 10 Group frames

Example 10.14. For F = C irreducibility and absolute irreducibility coincide. The 2 unitary action (10.2) of the cyclic group Cn on R is irreducible, but is not absolutely ρ irreducible, since the eigenspaces of R2 (a) as a C–linear map are G–invariant.

Schur’s lemma (Lemma 10.3) implies that the F–vector space HomFG(V,W) of the FG–homomorphisms between the FG–isomorphic irreducibles V and W is a division ring. When F = C (or V and W are absolutely irreducible), then this division ring is one–dimensional. This leads to the following.

Lemma 10.4. (Schur) Suppose that S : Vj Vk is an FG–homomorphism between → absolutely irreducible FG–modules, which are FG–isomorphic via σ : Vj Vk. Then → S = cσ, for some c F. ∈ Proof. Since Vj and Vk are absolutely irreducible, HomFG(Vj,Vk) is one–dimensional, and so spanned by σ. This gives the result (with c possibly zero). ⊓⊔ We now show that every tight G–frame is a direct sum of irreducible G–frames:

Theorem 10.8. (Characterisation). Let there be a unitary action of a ﬁnite group G on H = V1 V2 Vm, an orthogonal direct sum of irreducible G–invariant subspaces. Then⊕ ⊕⊕

(gv)g G, v = v1 + + vm, v j Vj ∈ ∈ is a tight G–frame for H if and only if

2 v j dim(Vj) v j = 0, j, 2 = , j = k, (10.15) ∀ vk dim(Vk) and when Vj = Vk are FG–isomorphic, (gv j)g G and (gvk)g G are orthogonal, i.e., ∈ ∈

∑ v j,gv j gvk = 0. (10.16) g G ∈

Moreover, if Vj is absolutely irreducible, then (10.16) can be replaced by

σv j,vk = 0, (10.17) where σ : Vj Vk is any FG–isomorphism. → Proof. Φ =(gv)g G is a tight frame for H if and only if there exists a λ > 0 with ∈ SΦ ( f )= ∑ f ,gv gv = λ f , f H . g G ∀ ∈ ∈

By linearity, it sufﬁces to show this for f j Vj, 1 j m, i.e., ∈ ≤ ≤

∑ f j,gv gv = ∑ f j,gv j gv j + ∑ ∑ f j,gv j gvk = λ f j, (10.18) g G g G g G k= j ∈ ∈ ∈ 10.9 The characterisation of all tight G–frames 229 since gv = ∑ j gv j. By equating the Vk components, (10.18) holds if and only if

∑ f j,gv j gv j = λ f j, ∑ f j,gv j gvk = 0, k = j. (10.19) g G g G ∈ ∈

By Theorem 10.5, the ﬁrst condition will hold for all f j Vj provided v j = 0, with ∈ a λ = λ j > 0, which depends on j, given by

2 G v j λ j = | | . dim(Vj)

Therefore λ j is independent of j if and only if the second part of (10.15) is satisﬁed. By Theorem 10.7, the second condition in 10.19 automatically holds if Vj = Vk. ∼ Since hv j h G spans Vj, the second condition in 10.19 can be rewritten { } ∈ 1 1 ∑ hv j,gv j gvk = h ∑ v j,h− gv j h− gvk = h ∑ v j,gv j gvk = 0, k = j, g G g G g G ∈ ∈ ∈ which gives (10.16), since h maps Vk Vk bijectively. → Finally, if Vj is absolutely irreducible, then

S : Vj Vk : f j ∑ f j,gv j gvk → → g G ∈ is a nonzero FG–homomorphism (Exer. 10.1), so Lemma 10.4 gives S = cσ, c F. We now determine c. Since the action is unitary, using Theorem 10.5, we calcula∈te

Sv j,σv j = ∑ v j,gv j gvk,σv j = ∑ v j,gv j gvk,σv j g G g G ∈ ∈ 1 1 1 1 = ∑ g− v j,v j vk,σg− v j = vk,σ ∑ v j,g− v j g− v j g G g G ∈ ∈ 2 2 G v j G v j = vk,σ | | v j = | | vk,σv j . dim(Vj) dim(Vj)

2 Since Sv j,σv j = cσv j,σv j = c σv j , we have 2 G v j Sf = ∑ f ,gv gv = v ,σv σ f , j k | | σ 2 k j g G dim(Vj) v j ∈ and the condition (10.16), i.e., Sv j = 0, is equivalent to (10.17). ⊓⊔ Example 10.15. (One summand). For m = 1 this reduces to Theorem 10.5.

Example 10.16. (Harmonic frames) If G is an abelian group, then all the absolutely irreducible FG–modules are one–dimensional, with the action of G given by

gv j = ξ j(g)v j, v j Vj, ∈ 230 10 Group frames where ξ j is a character of G (see 11.2). The condition (10.16) becomes § 2 ∑ v j,ξ j(g)v j ξk(g)vk = ∑ ξk(g)ξ j(g) v j vk = 0 = ∑ ξk(g)ξ j(g)= 0, g G g G ⇒ g G ∈ ∈ ∈ i.e., the characters are orthogonal, and so each character can be taken at most once. These harmonic frames are studied in Chapter 11. Example 10.17. (Dimension of a G–frame) For an absolutely irreducible W W ∈ (for which the action of G is unitary) and σ j : Vj W any FG–isomorphism, the condition (10.17) can be written (see Exer. 10.2) as→

σ jv j,σkvk = 0, j = k. Therefore the maximal number of summands which can be FG–isomorphic to W is dim(W). When the number of summands in H = jVj that are FG–isomorphic to a given W is zero or dim(W), then one obtains a central⊕ G–frame (see xx). Taking each absolutely irreducible W W as a summand dim(W)§times gives rise to tight G–frame, so that ∈

∑ dim(W)2 G . (10.20) W W ≤| | ∈

On the other hand, taking P = I in Corollary 10.3 gives the tight G–frame (eg)g G for CG, so there is equality. More generally, all of the possible dimensions for∈ a G–frame are ∑ aW dim(W), 0 aW dim(W), W W ≤ ≤ ∈ where aW is the multiplicity of the absolutely irreducible W as a summand. Example 10.18. Suppose that there is an irreducible unitary action of G on W = Cd. This induces a unitary action ρ on d d matrices via matrix multiplication × d ρ(g)[u1,...,ud]=[gu1,...,gud], u j C . ∈ d d Let Vj be the G–invariant subspace of C × consisting of the matrices which are zero in all but the j–th column (and the zero matrix),

d v j =[0,...,u j,...,0], σ j : Vj C : v j u j. → →

Then σ j is a CG–isomorphism, and so the G–orbit of U =[u1,...,ud]= v1 + +vd d d is a tight G–frame for C if and only if v j = u j is constant (for all j) and ×

σ jv j,σkvk = u j,uk = 0, j = k, i.e., U is a unitary matrix (up to a scalar). Thus, with U unitary, (10.11) generalises to d d d A = ∑ A,ρ(g)U ρ(g)U, A C × . G g G ∀ ∈ | | ∈ 10.10 G–frames of multivariate orthogonal polynomials 231 10.10 G–frames of multivariate orthogonal polynomials

Here we apply Theorem 10.8 to construct tight frames of orthogonal polynomials of several variables which share the symmetries of the weight (also see 15 and 16). d § d § Let Π := Π(R ) be the polynomials in d real variables, and Πk := Πk(R ) those of degree k. Let µ be a measure on Rd, for which ≤

f1, f2 µ := f1 f2 dµ deﬁnes an inner product on Π. The space of orthogonal polynomials of degree k with respect to the measure µ is

Vk(µ) := f Πk : f , p µ = 0, p Πk 1 . { ∈ ∀ ∈ − } This has dimension k + d 1 dim(V (µ)) = − . k d 1 − The symmetry group of the measure µ is

G = Sym(µ) := g Aff(Rd) : f gdµ = f dµ, f Π , { ∈ ◦ ∀ ∈ } where Aff(Rd) denotes the group of afﬁne transformations on Rd. This acts on Π via g f := f g 1, with each g Sym(µ) inducing a unitary transformation ◦ − ∈ 1 1 1 g f1,g f2 µ = ( f1 g− )( f2 g− )dµ = ( f1 f2) g− dµ = f1 f2 dµ = f1, f2 µ , ◦ ◦ ◦ which maps Vk(µ) onto itself since g Πk 1 = Πk 1, i.e., Vk(µ) is a G–invariant − − subspace. It is therefore natural to seek a G–invariant tight frame for Vk(µ), which has a small number of vectors.

Example 10.19. ( Legendre polynomials on a square) Let P2 be the 3–dimensional space of quadratic Legendre polynomials on the square [ 1,1]2 R2. Here − ⊂ 1 f1, f2 = f1 f2, 4 [ 1,1]2 − so that 1 = 1, and G = D4 acts as the symmetries of the square. The orthogonal decomposition of P2 into its homogeneous components consists of three 1–dimensional G–invariant subspaces, which are given by

2 2 2 2 2 p1(x,y)= x + y , p2(x,y)= xy, p3(x,y)= x y . − 3 −

None of these spaces are RG–isomorphic. To see this, observe that D4 acts as the identity on p1, and the symmetry (x,y) (y,x) ﬁxes p2 but not p3. → 232 10 Group frames

By Theorem 10.8, any f which is a sum of vectors of equal length from these 2√2 1 subspaces gives a G–frame (gf )g G for P2. Since p1 = p3 = , p2 = , ∈ 3√5 3 we can take

√5 2 √5 1 f (x,y)= (x2 + y2 )+ √2xy + (x2 y2)= √5(x2 )+ √2xy. 2 − 3 2 − − 3

This gives a D4–invariant tight frame of four vectors for P2. Since the size of an orbit divides the order of the group, there is no D4–invariant basis for P2.

Fig. 10.3: Contour plots of the f of Example 10.19, and its orbit (showing the square symmetry).

Example 10.20. (Legendre polynomials on a triangle). Let P2 be the 3–dimensional space of quadratic Legendre polynomials on a triangle T with vertices V. Here 1 f1, f2 = f1 f2, Area(T) T and G = SV = D3 acts as the symmetries of the triangle, i.e., for σ SV ∈ σ σ ∑ avv = ∑ av v, (where ∑v av = 1). v V v V ∈ ∈ The action of G on the barycentric coordinates ξ =(ξv)v V for V (see 4.7) is ∈ §

σ ξv = ξσv, σ SV . (10.21) ∀ ∈ The inner product between powers of the barycentric coordinates (which are linear polynomials) is given by the special case d = 2 of the formula

α β (α + β)! ξ ,ξ = , (d + 1) α + β | | | | for (normalised) integration over a simplex in Rd with vertices V. Let V = u,v,w . { } 10.10 G–frames of multivariate orthogonal polynomials 233

In view of (10.21), the 1–dimensional subspace V1 of P2 spanned by

2 2 2 1 2 1 f1 := ξ + ξ + ξ , f1 = u v w − 2 60 is G–invariant. Further, its G–invariant orthogonal complement V2 := V = P2 V1 1⊥ ⊖ is irreducible. This follows since the action of G on P2 is faithful, and so if V2 was not irreducible, then linear maps P2 P2 induced by G would be simultaneously → ξ 2 diagonalisable, and so G ∼= S3 would be abelian. The orthogonal projection of v onto V2 is given by

2 4 1 1 2 1 f2 := ξ ξv + f1, f2 = . v − 5 10 − 3 675 By Theorem 10.8, the G–orbit of the unit vector

f1 1 f2 2 ξ 2 4ξ 1 f := =(2√5 5√2) f1 15√2 v v + , f1 √3 ± f2 3 ∓ ± − 5 10 is a tight frame for the 3–dimensional space P2. The polynomial f is ﬁxed by any permutation which ﬁxes v, and so this orbit has three distinct vectors. These give an orthonormal basis for P2 (the quadratic Legendre polynomials on the triangle), which is invariant under the symmetries of the triangle (see Figure 10.4).

Fig. 10.4: Contour plots of the f of Example 10.20, and its orbit (showing the triangular symmetry).

Since the orders of the symmetry groups G of the square and triangle are 8 and 6, and the size of a G–frame must divide G , it is not possible to ﬁnd a tight G–frame for the space of orthogonal polynomials| of| degree k on the square and triangle (or for any weight with a ﬁnite symmetry group) as soon as its dimension is greater than G . In these cases, it is natural to seek tight frame which is the G–orbit of a small |number| of vectors (see 10.11 and Chapter 15). § 234 10 Group frames 10.11 G–invariant frames

Suppose there is unitary action of a finite group G on H . If there is no G–frame (gv)g G for H , then it is natural to seek a G–invariant (tight) frame for H , which is the∈ orbit of a small number of vectors. Here we give a complete characterisation of such G–invariant tight frames (Theorem 10.9). This allows one to calculate the minimal number of generators for a G–invariant frame ( xx). § Definition 10.7. The number of generators of a G–invariant frame is the number of orbits under the action of G on the set of its vectors. We now prove the main result: a characterisation of which G–orbits of vectors w1,...,wr give tight frames. Theorem 10.9. (Characterisation). Let H be a Hilbert space over F = C or R. Suppose there is a unitary action of a finite group G on H = V1 V2 Vm, ⊕ ⊕⊕ an orthogonal direct sum of irreducible G–invariant subspaces. Let Pj = PVj be the orthogonal projection of H onto Vj. Then

Φ =(gws)g G,1 s r, w1,...,wr H , ∈ ≤ ≤ ∈ is a tight G–invariant frame for H if and only if

r ∑r 2 2 s=1 Pjws dim(Vj) ∑ Pjws = 0, j, r 2 = , j = k, (10.22) ∀ ∑ Pkws dim(Vk) s=1 s=1 and when Vj = Vk are FG–isomorphic

∑ ∑ v j,gPjws gPkws = 0, (10.23) s g G ∈ for any (and hence all) nonzero v j Vj. Moreover, if Vj is absolutely irreducible, then (10.23) can be replaced by ∈

r ∑ σPjws,Pkws = 0, (10.24) s=1 where σ : Vj Vk is any choice of FG–isomorphism. → Proof. Φ is a tight frame for H if and only if there exists a λ > 0 with

SΦ ( f )= ∑ ∑ f ,gws gws = λ f , f H . s g G ∀ ∈ ∈

By linearity, it sufﬁces to show this for f j Vj, 1 j m, i.e. to show that there exists λ (independent of j) such that, ∈ ≤ ≤

∑ ∑ f j,gws gws = ∑ ∑ ∑ f j,gPjws gPkws = λ f j, (10.25) s g G s g G k ∈ ∈ 10.11 G–invariant frames 235 since ws = ∑k Pkws. By equating the Vk components, (10.25) holds if and only if

∑ ∑ f j,gPjws gPjws = λ f j, ∑ ∑ f j,gPjws gPkws = 0, k = j. (10.26) s g G s g G ∈ ∈

By Theorem 10.5, the ﬁrst part of (10.26) will hold for all f j Vj provided that 2 ∈ λ λ some ws has a nonzero Vj–component, i.e., ∑s Pjws = 0, with a = j > 0, which depends on j, given by

G 2 λ j = | | ∑ Pjws . dim(Vj) s

This λ j is independent of j if and only if (10.22) holds. By Theorem 10.7, the second part of (10.26) automatically holds if Vj ≇ Vk, and so reduces to

∑ ∑ f j,gPjws gPkws = 0, f j Vj, k = j, Vj ∼= Vk. (10.27) s g G ∀ ∈ ∈

By Lemma 10.3, this holds if and only if it holds for f j some nonzero v j Vj. ∈ We now seek to simplify (10.27) in the case that Vj is absolutely irreducible. Let τ : Vj Vk be the FG–homomorphism (see Exer. 10.1) →

τ f := ∑∑ f ,gPjws gPkws. s g

Then for σ : Vj Vk an FG–isomorphism, we calculate →

τv j,σv j = ∑∑ v j,gPjws gPkws,σv j s g

= ∑∑ v j,gPjws gPkws,σv j s g 1 1 = ∑∑ g− v j,Pjws Pkws,σg− v j s g 1 1 = ∑ Pkws,σ ∑ Pjws,g− v j g− v j s g 2 v j G = ∑ Pkws,σ | |Pjws , s dim(Vj) with the last equality given by Theorem 10.5. Since Vj is absolutely irreducible, Lemma 10.4 implies that τ = cσ, for some c F (possibly zero). Substituting τ = cσ into the above gives ∈

2 τ G v j σ σ = | | 2 ∑ Pkws, Pjws . dim(Vj) σv j s Thus (10.27), which is equivalent to τ = 0, holds if and only if (10.24) does. ⊓⊔ For a single generator (r = 1), Theorem 10.9 reduces to Theorem 10.8. 236 10 Group frames 10.12 Frames invariant under the action of an abelian group

We now consider Theorem 10.9 when G is abelian. In this case, all the absolutely irreducible G–invariant subspaces Vj are 1–dimensional, with the action of G given by gv = ξ(g)v, v Vj, ∈ where ξ is a character of G, i.e., a homomorphism G C (see 11.2). For one generator, there are a ﬁnite number of tight→ G–frames§ for Cd, the so called harmonic frames (see Example 10.16 and Chapter 11). We now use Theorem 10.9 to describe the situation for two or more generators. Corollary 10.6. (G abelian) Suppose that there is a unitary action of a ﬁnite abelian group G on Cd, and, without loss of generality, that the irreducible G–invariant subspaces are Vj = span e j , with the action on Vj given by ge j = ξ j(g)e j, where { } d ξ j : G C is a character of G. Let w1,...,wr C . Then → ∈

Φ =(gws)1 s r,g G ≤ ≤ ∈ is a tight frame for Cd if and only if

1. The matrix W =[w1,...,wr] has rows of equal norm. 2. The rows of W corresponding to the same character are orthogonal. In particular, there is a G–invariant tight frame for Cd with r generators if and only if d r G . ≤ | | Proof. Since Pjws =(ws) je j, we have

2 2 2 ∑ Pjws = ∑ (ws) j =(norm of the j–th row of W) , s s | | and so (10.22) reduces to 1. If Vj and Vk are CG–isomorphic, i.e., correspond to the same character, then σ : Vj Vk : e j ek is a CG–isomorphism, since → →

σ(ge j)= σ(ξ j(g)e j)= ξ j(g)σ(e j)= ξk(g)ek = g(σe j).

In this case, σPjws = σ(ws) je j = e(ws) jek, and so (10.24) becomes

∑ σPjws,Pkws = ∑ (ws) jek,(ws)kek = ∑(ws) j(ws)k = 0, s s s i.e., the j and k rows of W are orthogonal, which is condition 2. ⊓⊔ Example 10.21. For G = a the cyclic group of order 2, deﬁne a unitary action of G on C3 by

1 0 0 v1     av := 0 1 0 v = v2 ,     0 0 1  v3  −  −      10.12 Frames invariant under the action of an abelian group 237 i.e., take the trivial representation on V1 = Ce1, V2 = Ce2, and the sign representation 3 on V3 = Ce3. There is no G–frame (gv)g G for C . However, there are many choices ∈ 3 for w1,w2 so that (gw j)1 j 2,g G is a tight G–invariant frame for C , e.g., ≤ ≤ ∈

1 0

W =[w1,w2]= 0 1 , u 1. | |≤  2 u 1 u   −| |    2 Here w1,w2 = u 1 u , and so inﬁnitely many unitarily inequivalent G–invariant tight frames for C3 can−| be| constructed in this way. 2 2 2 Given some w1 =(x,y,z), a suitable w2 can be chosen provided z x + y , 2 2 2 2 | | ≤| | | | since we can take w2 =(y, x,u) where u = x + y z . For example, if − | | | | | | −| | w1 =(1,2,2), then choosing w2 =(2, 1,1) gives the G–invariant tight frame − 1 2 1 2

(w1,w2,aw1,aw2)= 2, 1, 2 , 1 . − −         2  1   2  1     −  −          Example 10.22. Let G = a be the cyclic group of order n. An irreducible unitary action of G on R2 is given by (10.2), i.e.,

cos 2π sin 2π av = Av, A := n − n .  2π 2π  sin n cos n   Suppose that G acts on R4 = R2 R2 componentwise, i.e., a(v,w)=(Av,Aw). Then the conditions for the G–orbit of×(v,w) to be a tight frame for R4 are

n n j j j j 2 v = w = 0, ∑ f1,A v A w = ∑ A wv∗A− f1 = 0, f1 R . j=1 j=1 ∀ ∈ A calculation shows that this is not possible, and so no G–orbit is a tight frame. This can also be seen by appealing to Theorem 10.9, as follows. The action of G on R2 given by av = Av is not absolutely irreducible. On the complexification C2 there are two orthogonal G–invariant subspaces: the eigenspaces of A corresponding 2πi to the eigenvalues λ = ω,ω, ω := e n . For n > 2, ω = ω, and so these subspaces are not CG–isomorphic. Thus the complexification C4 of R2 R2 decomposes as the sum of four 1–dimensional G–invariant subspaces, with two× pairs CG–isomorphic to each other. By Theorem 10.9, it is therefore not possible to find a G–frame for C4, and hence neither for R4. 238 10 Group frames 10.13 The minimal number of generators for a G–invariant frame

Suppose that there is a unitary action of a ﬁnite group G on a Hilbert space H . Let W be a complete set of non-isomorphic irreducible CG–modules, i.e., each irreducible CG–module occurs once in W up to CG-isomorphism. We denote the direct sum of k copies of an FG–module V by V k. Example 10.17 shows that there is a tight G–frame for H if and only if H is CG–isomorphic to an orthogonal direct sum

aW W , 0 aW dim(W). (10.28) W W ≤ ≤ ∈ On the other hand, Wedderburn’s Theorem states that CG is CG–isomorphic to

W dim(W). W W ∈ Combining these observations gives the following.

Proposition 10.4. The following are equivalent:

1. There is a tight frame (gv)g G for H . ∈ 2. There is a frame (spanning set) (gv)g G for H . 3. H is CG–isomorphic to a submodule∈ of CG.

In other words:

The existence of a tight G–frame for H depends only on the CG–module structure of H . It does not depend on the inner product on H .

We now generalise this. Denote the number of times an irreducible CG–module appears in a direct sum decomposition of H into irreducible CG–modules by

mult(W,H )= dim(HomCG(W,H )) = dim(HomCG(H ,W)).

Theorem 10.10. (Minimal number of generators) The following are equivalent.

1. There is a G–invariant tight frame (gv j)g G,1 j r for H with r generators. ∈ ≤ ≤ 2. There is a G–invariant frame (gv j)g G,1 j r for H with r generators. 3. H is CG–isomorphic to a submodule∈ of≤(≤CG)r, i.e.,

mult(W,H ) r dim(W), (10.29) ≤ for every irreducible CG–module W. 10.13 The minimal number of generators for a G–invariant frame 239

Proof. (1 2) This follows from a variation of the argument for Theorem 10.1, ⇐⇒ i.e., if Φ = (gw j)g G,1 j r spans H and S is the frame operator of Φ, then 1 ∈ ≤ ≤ (gS− 2 w j)g G,1 j r is a tight frame for H . ∈ ≤ ≤ (2 3) First, suppose that (gw j)g G,1 j r spans H . Then H is a quotient of ⇐⇒ ∈ ≤ ≤ j Wj, where Wj := span gv j g G. By Proposition 10.4, each Wj is CG–isomorphic to a submodule of CG, and{ hence} ∈ H is CG–isomorphic to a submodule of (CG)r. Conversely, now suppose that mult(W,H ) r dim(W) for every irreducible ≤ CG–module W. Then we can write H = Z1 Zr, where each submodule Z j ⊕⊕ has mult(W,Z j) dim(W), and so is CG–isomorphic to a submodule of CG. We ≤ can choose vectors w j Z j for which (gw j)g G spans Z j (by Proposition 10.4). Thus ∈ ∈ (gw j)g G,1 j r spans H . ∈ ≤ ≤ ⊓⊔ We now briefly indicate how Theorems 10.9 and 10.10 can be used to construct tight frames of multivariate orthogonal polynomials (see 10.10) which are invariant under the symmetries of the weight and have a minimal number§ of generators. Recall from 10.10, that the symmetry group G of a measure µ has a unitary § d 1 action on the d–variate polynomials Π = Πk(R ) given by g f := f g . We ◦ − are interested in finding G–invariant tight frames for space Vk(µ) of orthogonal polynomials of degree k for this measure. Π Π d We denote the space of homogeneous polynomials of degree k by k◦ = k◦(R ). d This is a G–invariant subspace of Π(R ), indeed it is FG–isomorphic to Vk(µ). Proposition 10.5. Let G be a finite subgroup of the symmetry group of a measure µ. V µ Π Then k( ) is FG–isomorphic to k◦.

Proof. Let PV be the orthogonal projection onto a subspace V. If V is G–invariant, then PV commutes with the action of G. Thus

Π V µ Π k◦ k( ) : f f P k 1 f = PVk(µ)( f ) → → − − is an FG–homomorphism onto Vk(µ). Moreover, this is an FG–isomorphism since dim(Π )= dim(Vk(µ)). k◦ ⊓⊔ Thus the problem of ﬁnding G–invariant frames for Vk(µ) therefore reduces to Π understanding the CG–module structure of k◦. To this end, denote the G–invariant polynomials by Π G := f Π : g f = f , g G . { ∈ ∀ ∈ }

Example 10.23. (Trivial representation). Let W be the trivial irreducible CG–module. This is one–dimensional, with the G ﬁxing every vector in W. Thus (10.29) becomes

mult(W,Π 0)= dim(Π G Π 0) r dim(W)= r, k ∩ k ≤ G 0 and so the minimal number of generators r for Vk(µ) satisfies r dim(Π Π ). ≥ ∩ k Π When G is a finite reflection group, there are techniques to decompose k◦ into irreducible CG-submodules. We now briefly outline these. 240 10 Group frames 10.14 The coinvariants of a finite reflection group

A linear transformation on a complex vector space is a complex reflection if it has finite order and fixes a hyperplane. A group G generated by complex reflections is called a complex reflection group. A complete classification of the finite irreducible complex reflection groups was given by Shephard and Todd. Π G I Let + be the G–invariant polynomials with zero constant term, and G be the Π G Π ideal generated by + in . The ring of coinvariants (or coinvariant space) is the quotient ring Π ΠG := . IG This is naturally graded with respect to degree and inherits the action of G (since IG is a homogeneous ideal which is G–invariant). We now suppose that G is a finite reflection group. This implies Π G is a ring (this characterises reflection groups). • Π G ∼= CG (CG–isomorphism). • There is a decomposition of Π as a tensor product of graded CG–modules: • G Π = Π ΠG. (10.30) ∼ ⊗C

We identify ΠG as a CG–submodule of Π. Then (10.30) gives the CG–module decomposition k Π Π G Π Π Π k◦ = ( j◦) C ( G k◦ j). ∩ ⊗ ∩ − j=0 Π G Π Π G Π We observe that j◦ consists of dim( j◦) copies of the trivial module. Hence, for W an irreducible∩ CG–module, ∩

k Π Π G Π Π Π mult(W, k◦)= ∑ dim( j◦)mult(W, G k◦ j). j=0 ∩ ∩ − where ∑ mult(W,ΠG Π ) mult(W,ΠG)= mult(W,CG)= dim(W). This gives: j ∩ j◦ ≤

For G Sym(µ) a ﬁnite reﬂection group, determining the minimal number ⊂ of generators r for a G–invariant tight frame for Vk(µ) (via Theorem 10.10) amounts to calculating mult(W,ΠG Π ), 0 j k, for each irreducible W. ∩ j◦ ≤ ≤

These calculations are done for the orthogonal polynomials on a regular polygon in R2 and the cube in R3 (see [VW16]). The discussion above also gives the following estimate for r Π G Π r max dim( j◦). ≤ 0 j k ∩ ≤ ≤ 10.14 The coinvariants of a ﬁnite reﬂection group 241 Notes

The basic theory of groups frame was given in [VW05], [Han07] (one generator), and [VW16] (multiple generators). Thanks to Patrick Morandi for the graphics used in Figures 10.1 and 10.2.

Exercises

10.1. Suppose that there are unitary actions of a ﬁnite group G on Vj and Vj. For any v j Vj, vk Vk, show that S : Vj Vk given by ∈ ∈ →

Sf := ∑ f ,gv j gvk, f Vj, g G ∀ ∈ ∈ is an FG–homomorphism, i.e., commutes with the action of G. 10.2. Show that if W is absolutely irreducible (with the action of G unitary) and σ j : Vj W is an FG–isomorphism, then the condition (10.17) can be written as →

σ jv j,σkvk = 0. 8 10.3. Let H be the subspace of R consisting of vectors x with ∑ j x j = 0. H ρ σ 8 (a) Show there is unitary action of G = S8 on given by ( )x :=(xσ j) j=1. (b) Show that there are 28 distinct vectors in the G–frame (gv)g G for H given by the vector v =(3,3, 1, 1, 1, 1, 1, 1). ∈ Remark: These 28 vectors− − are− an− equiangular− − tight frame for R7 (see Example 12.3). They can be thought of as a G–frame (apply the permutation matrices to the vectors in R8 and calculate the size of the stabiliser).

10.4. Let Φ =(gv)g G =(ρ(g)v)g G be a ﬁnite frame for the vector space H , where ρ : G GL(H∈ ) is a representation∈ (group homomorphism). Choose some → inner product , on H , and let A∗ be the Hermitian transpose with respect to it. (a) Show that a second inner product on H can be deﬁned by 1 1 x,y ρ := ∑ ρ(g)x,ρ(g)y = Ax,y , A = Aρ = ∑ ρ(g)∗ρ(g). G g G G g G | | ∈ | | ∈

(b) Show that each ρ(g) is unitary with respect to , ρ , i.e., it is G–invariant ρ(h)x,ρ(h)y ρ = x,y ρ , h G. ∀ ∈ 1 (c) Let B = A 2 be the positive square root of the positive deﬁnite A above. Show that ρ˜(g) := Bρ(g)B 1 deﬁnes a representation ρ˜ : G U (H ). − → (d) Show that Φ is similar to the G–frame Ψ =(ρ˜(g)Bv)g G, and hence is similar to the tight G–frame Ψ can. ∈ 242 10 Group frames

10.5. Let , ρ be the G–invariant inner product of Exer. 10.4. Show that x,y := Mx,y ρ , M positive deﬁnite with respect to , ρ gives a G–invariant inner product on H if and only if M commutes with the action of G, i.e., ρ(g)M = Mρ(g), g G. In particular, if ρ is absolutely irreducible, then there is a unique G–invariant∀ inner∈ product.

10.6. Let (gv)g G be a G–frame for H . (a) Show that the∈ variational condition (6.4) for tightness becomes

dim(H ) ∑ v,gv 2 v 4. G g G | | ≥ | | ∈

(b) Show that (gv)g G is a (t,t)–design if and only if ∈

1 2t 4t ∑ v,gv = ct (d,F) v . G g G | | | | ∈ 10.7. Prove the assertions of Theorem 10.2, i.e., (a) The direct sum of disjoint G–frames is a G–frame. (b) The sum of a G1–frame and a G2–frame is a G1 G2–frame. × (c) The tensor product of G1–frame with a G2–frame is a G1 G2–frame. (d) The complement of a tight G–frame is a tight G–frame. × 10.8. Let G be the symmetry group of a Platonic solid T R3 (with centre of gravity the origin) acting as unitary transformations. Use the fact⊂ that ﬁnite subgroups of U (R2) are cyclic or dihedral to prove this action is irreducible. 10.9. FG–modules. Let G be a group. An action of the group G on a set X is a map G X X : (g,x) gx satisfying g1(g2x)=(g1g2)x, 1x = x. A vector space V over×F is→ an FG–module→ if there is a multiplication gv, g G, v V for which (g,v) gv is an action and v gv is a linear map, g G. ∈ ∈ (a) Show→ that V is an FG–module→ if and only if ρ :∀G ∈ GL(V), ρ(g)v := gv is a (linear) representation/action. → (b) Show that FG–submodules are the same as G–invariant subspaces. Remark: Other terminology carries over in the obvious way, e.g., the action of G on V = 0 given by some representation is irreducible if and only if the only the FG–module V is simple, i.e., has no FG–submodules other than 0 and V.

10.10. Suppose that Vj and Vk are absolutely irreducible G–invariant subspaces of H , and σ, τ are FG–isomorphisms Vj Vk. Prove that → σ = λτ, for some λ C. ∈

10.11. Let Φ =(gv)g G be a G–frame for H given by ρ, Ψ be the set of vectors in Φ, and H := g G :∈gv = v be the stabiliser of v. (a) Show the vectors{ ∈ in Ψ can} be indexed by the set of left cosets of H in G 10.14 The coinvariants of a ﬁnite reﬂection group 243

Ψ =(gv)gH C , C := gH : g G . (10.31) ∈ { ∈ } (b) Suppose H is normal, so that C becomes the group G/H. Show that (10.31) gives a G/H–frame if and only H = N, where N is the kernel of ρ. (c) Suppose G is abelian. Show that H = N, which is independent of v, and hence Ψ is a G/N–frame of distinct vectors.

10.12. Let G = a,b be the binary icosahedral group of order 120 generated by 1 1 t− ti 1 i 0 a = − , b = − , 2  1 t 1 +ti  0 i − −     1 √ where t = 2 (1 + 5) is the golden ratio. 2 (a) Show that for every unit vector v C , the G–frame (gv)g G is a (5,5)–design. (b) Show that there is (5,5)–design of∈ 12 vectors for C2. ∈

10.13. Show that a θ–isogonal conﬁguration (see Example 10.2) is a group frame by determining the group G and its unitary action.

Chapter 11 Harmonic frames

11.1 Introduction

Here we consider the tight G–frames for G abelian. We will see that: There are finitely many such frames (they will be called the harmonic frames). • Each is given by a subset of the characters of G. • Each is given by a subset of G. • We first motivate the definitions to come, by considering the Example 2.4 for n = 3. The character table (Fourier matrix) of the cyclic group G = C3 of order 3 is

1 1 1 1 ω ω2, ω = 1, ω3 = 1,  ω2 ω  1      which has orthogonal columns of equal length. Since the orthogonal projection of an orthonormal basis is a tight frame, removing rows (i.e., characters) from this character table gives a submatrix whose columns are an equal–norm tight frame, e.g., removing rows 1 or 2 gives the equal–norm tight frames for C2

1 ω ω2 1 1 1 ( , , ), ( , , ). (11.1) 1 ω2  ω  1 ω ω2             Equivalently, one could remove columns (i.e., elements of G). This amounts to restricting the characters to a subset J of G, which is the most convenient way to describe harmonic frames (we take this as the deﬁnition). The tight frames of (11.1) are clearly G–frames under the respective unitary actions of G = a given by ω 1 ρ(a)= ω2 , ρ(a)=( ω ).

245 246 11 Harmonic frames 11.2 Character tables

We first outline the basics of character theory for finite abelian groups (cf. [Rud90]). Let G be a finite abelian group. The (irreducible) characters of G are the group homomorphisms ξ : G C 0 , where C 0 is a group under multiplication. Here we think of them as→ vectors\ { }ξ CG (with\ { the} Euclidean inner product), which satisfy ∈ ξ(gh)= ξ(g)ξ(h), g,h G. (11.2) ∀ ∈ The set of irreducible characters of the abelian group G is denoted by Gˆ. The characters Gˆ form group under the multiplication (ξη)(g) := ξ(g)η(g), which is called the character group. The character group Gˆ is isomorphic to G. For χ Gˆ, (11.2) implies that χ(g) is a G –th root of unity, and so the inverse of χ satisfies∈ | | 1 1 χ− (g)= = χ(g). (11.3) χ(g) The square matrix with the irreducible characters of G as rows is referred to as the character table of G. For example, if G = a is the cyclic group of order n, n 1 with its elements ordered 1,a,...,a − , then its character table is

1 1 1 1 2 n 1 1 ω ω ω −  2πi  ω2 ω4 ω2(n 1)  ω n 1 − , := e . (11.4)   . . . .  . . . .     n 1 2(n 1) (n 1)(n 1) 1 ω − ω − ω − −      1 This (and the scalar multiple by √n ) is also known as the Fourier matrix. Example 11.1. If all elements of G have order 2, i.e., G is an elementary abelian 2–group Z2 Z2, then the entries of the character table of G are 1. ×× ± The rows and columns of the character table are orthogonal, i.e.,

ξ,η := ∑ ξ(g)η(g)= 0, ξ = η, (11.5) g G ∈ ∑ χ(g)χ(h)= 0, g = h. (11.6) χ Gˆ ∈ These are referred to as the orthogonality and column orthogonality relations. The Pontryagin duality map (canonical group isomorphism) is given by

G Gˆ : g gˆ, gˆ(χ) := χ(g), χ Gˆ, g G. (11.7) → → ∀ ∈ ∈ 11.3 Harmonic frames 247 11.3 Harmonic frames

For G abelian, the absolutely irreducible FG–modules are one–dimensional, with the action of G given by

gv j = ξ j(g)v j, ξ j Gˆ, (11.8) ∈ where ξ j Gˆ. As outlined in Example 10.16, the condition (10.16) in Theorem ∈ 10.8 (the characterisation of tight G–frames) is that the characters ξ j in (11.8) be orthogonal, and hence can be taken at most once. By choosing v j = e j, we therefore conclude that all tight G–frames for Cd (up to unitary equivalence) are given by Ψ ψ ψ ξ d d =( g)g G, g :=( j(g)) j=1 C , (11.9) ∈ ∈ where ξ1,...,ξd Gˆ are distinct. In view of (11.2), the action of G on Ψ =(ψg) is ∈

ξ1(g) ξ1(h) ξ1(gh) ψ . . . ψ g h :=  ..  .  =  .  = gh.       ξd(g)ξd(h) ξd(gh)           The construction (11.9) amounts to taking d rows of the character table of G, i.e., a subset of d elements of Gˆ. Equivalently (since G and Gˆ are isomorphic), one can select columns of the character table, i.e., a subset J G. The latter is the most convenient (cf. Theorem 11.2), and so we take it as our⊂ deﬁnition.

Deﬁnition 11.1. Let G be a ﬁnite abelian group of order n. A tight frame for Cd which is unitarily equivalent to Φ ξ J d J =( J)ξ Gˆ C C , (11.10) | ∈ ⊂ ≈ where J G, J = d, is called a harmonic frame (given by J G). If G can be a cyclic group,⊂ then| | we say that the harmonic frame is cyclic1. ⊂

Our observations lead to the following:

Theorem 11.1. (Characterisation of harmonic frames) Let Φ be an equal–norm tight frame for H Cd. Then the following are equivalent: ≈ 1. Φ is a G–frame, where G is abelian. 2. Φ is given by a submatrix of d rows of the character table of an abelian G. 3. Φ is harmonic, given by some J G, J = d. 4. The symmetry group Sym(Φ) has⊂ a transitive| | abelian subgroup G. For each Φ, G can be taken to be the same in 1,2,3 and 4, but it need not be unique.

1 Harmonic frames are also called geometrically uniform frames [EB03] (orbits of abelian matrix groups), and the term harmonic frame is also used for what we call a cyclic harmonic frame. 248 11 Harmonic frames

Proof. The equivalence of 1 and 2 follows from Theorem 10.8, as already discussed. The equivalence of 1 and 4 follows from Theorem 10.4. (3= 1) The ΦJ given by (11.10) is a Gˆ–frame, via the unitary action ⇒

χ (ξ J) :=(χ J)(ξ J)=(χξ) J. | | | |

Since Gˆ and G are isomorphic, it therefore follows that ΦJ is a G–frame. (2= 3) We use the Pontryagin duality map (11.7). Suppose that Φ is given by a submatrix⇒ of the character table of G. Since Gˆ and G are isomorphic Φ, is given by a submatrix of the character table of Gˆ, say ˆ ξ ξ ξ [gˆ( )]g J,ξ Gˆ =[ (g)]g J,ξ Gˆ =[ J]ξ Gˆ , J G, ∈ ∈ ∈ ∈ | ∈ ⊂ Φ Φ ξ and so is given by the harmonic frame J =( J)ξ Gˆ . For the possible nonuniqueness of G, see Examples| ∈ 11.2, 11.3 and 11.4. ⊓⊔ Early applications of cyclic harmonic frames include robust signal transmission with quantization and erasures [GVT98], [GKK01], [CK03] (which introduces the term harmonic tight frame), and multiple–antenna code design [HMR+00].

Example 11.2. (Nonuniqueness of G) Let Φ =(e1,e2, e1, e2) be the equal–norm tight frame of four equally spaced unit vectors for R2.− This− has symmetry group the 2 dihedral group D8 = a,b (see Exercise 9.4), where the action of a and b on R is given by 0 1 0 1 a = − , b = . 1 0  1 0 The frame Φ is the G–orbit of the nonisomorphic  abelian subgroups

2 a ,b C2 C2, a C4, ≈ × ≈ and hence the group G in Theorem 11.1 need not be unique. Example 11.3. (Orthonormal bases) Since the columns of a character table have equal norm and are orthogonal, any orthonormal basis for Rn (or Cn)isa G–frame for any abelian group G of order n.

Example 11.4. (Simplex) By removing the ﬁrst row from (11.4), i.e., the trivial character χ = 1, one obtains the n = d + 1 vertices of the regular simplex in Rd, which is therefore a tight Cd+1–frame. Similarly, the vertices of the simplex are a G–frame for any abelian group G of order n.

Example 11.5. (Noncyclic harmononic frames) Consider the harmonic frame given 3 by the eight vertices of the cube [ 1,1] . This is a both a Z2 Z2 Z2–frame and − × × a Z2 Z4–frame, but not a cyclic harmonic frame since its symmetry group (the × octohedral group S4 Z2) contains no elements of order 8. × 11.4 Harmonic frames with distinct vectors and with real vectors 249 11.4 Harmonic frames with distinct vectors and with real vectors

The condition for a harmonic frame to have distinct vectors, be real, or lifted are: Φ Φ ξ Theorem 11.2. Let G be an abelian group of order n, and = J =( J)ξ Gˆ be | ∈ the harmonic frame of n vectors for Cd given by a choice J G, J = d. Then ⊂ | | 1. Φ has distinct vectors if and only if J generates G. 2. Φ is a real frame if and only J is closed under taking inverses. 3. Φ is a lifted frame if and only if the identity is an element of J. Proof. We write G additively, and observe that ξ( j)= ξ( j), ξ Gˆ, j G. 1. Let H be the subgroup of G generated by J. Then−Φ has∀ distinct∈ ∀ vectors∈ if J and only if the composition of maps Gˆ Hˆ C : ξ ξ H ξ J is 1–1. Since each h H can be written as a sum of elements→ → in J, and→ ξ| is→ a character,| ξ(h) is ∈ determined by ξ J, and so ξ H ξ J is 1–1. Hence ξ ξ J is 1–1 if and only if | | → | → | the group homomorphism given by Gˆ Hˆ : ξ ξ H is 1–1, i.e., Gˆ = Hˆ , and so G = H = J . → → | 1 1 2. The frame Φ is real if and only if χ J,η J = (χη− ) J,1 J R, χη− Gˆ i.e., | | | | ∈ ∀ ∈ ˆ ψ := ∑ jˆ RG. (11.11) j J ∈ ∈ First, suppose that J is closed under taking inverses, and j J. Then either j is its own inverse, so ξ( j)= ξ( j)= ξ( j) R, or the pair j∈, j J contributes − ∈ { − }⊂ ξ( j)+ ξ( j)= ξ( j)+ ξ( j) R to the sum above. Thus we conclude each inner product is− real. ∈ Conversely, suppose the inner products are real, i.e., (11.11) holds, and ψ = ψ. Let ζ,τ be the Euclidean inner product on CGˆ normalised so that the characters of Gˆare orthonormal. Then

j J ψ, jˆ = 1 ψ, jˆ = ψ,( j)ˆ = 1 j J. ∈ ⇐⇒ ⇐⇒ − ⇐⇒ − ∈ 3. By the column orthogonality relation (11.6) for characters, Φ is unlifted if and only if

∑ ξ J = 0 ∑ ξ( j)= ∑ ξ( j)ξ(0)= 0, j J 0 J. | ⇐⇒ ∀ ∈ ⇐⇒ ∈ ξ Gˆ ξ Gˆ ξ Gˆ ∈ ∈ ∈

⊓⊔ It sufﬁces to consider harmonic frames with distinct vectors: Φ ξ Corollary 11.1. Let J =( J)ξ Gˆ be a harmonic frame, and H be the subgroup of | ∈ Φ χ G generated by H. Then the H distinct vectors of J are an H–frame ( J)χ Hˆ . | | | ∈ Proof. For ξ Gˆ, χ = ξ H is a character of H with the property that χ J appears ∈ | | exactly G / H times in ΦJ. | | | | ⊓⊔ 250 11 Harmonic frames 11.5 Combining and decomposing harmonic frames

We ﬁrst observe that Theorem 11.2 guarantees the existence of harmonic frames of any number of distinct vectors for Cd. Corollary 11.2. (Existence) Let G be a ﬁnite abelian group, with minimal number d of generators d∗, Then there is a G–frame of distinct vectors for C if and only if

d∗ d G . ≤ ≤| |

Example 11.6. Let G be an elementary abelian p–group (p a prime), i.e.,

G = Zp Zp (k times), ×× Then G gives a harmonic frame of distinct vectors for Cd only for k d pk. ≤ ≤ Since harmonic frames are G–frames, we have a special case of Theorem 10.2. Theorem 11.3. Harmonic frames can be combined as follows. The direct sum of disjoint harmonic frames is a harmonic frame. • The sum of harmonic frames a harmonic frame. • The tensor product of harmonic frames is a harmonic frame. • The complement of a harmonic frame is a harmonic frame. • Proof. For the ﬁrst three, use Theorem 10.2, and observe that products of abelian groups are abelian. For the last use Corollary 10.4. ⊓⊔

Example 11.7. (Direct sums) The orthogonality of the irreducible characters ξ j Gˆ 1 ∈ of an abelian group G is equivalent to the harmonic frames (ξ j(g))g G for C being orthogonal (see Example 3.11). Thus the construction of harmonic frames∈ by taking rows ξ j of the character table can be interpreted as taking a direct sum of these harmonic frames for C1. The decomposition of a harmonic frame as a direct sum of irreducible characters given above is unique (for G given). Its decomposition into sums and tensor products may not be unique: Example 11.8. The four equally spaced vectors in R2 (the vertices of the square) can be written as a sum 1,1 +ˆ 1,1 , and as a tensor product 1,1 e1,e2 , {− } {− } {− } ⊗ { } where e1,e2 is an orthonormal basis. { } Example 11.9. (Symmetries) The symmetry group of a harmonic frame which is combination of harmonic frames can be larger than that guaranteed by Proposition 9.3. For example, the harmonic frame of 9 vectors for R4 given by the tensor product (see Example 5.14) and by the sum of the three equally spaced unit vectors in R2 with themselves has symmetry group of order 3!33 = 162. These two harmonic frames are unitarily equivalent (up to a reordering). The projective symmetry group of a harmonic frame is considered in 11.12. § 11.6 Real harmonic frames 251 11.6 Real harmonic frames

Most harmonic frames are complex (see [WH06], 11.10), but real harmonic frames can always be constructed by using Theorem 11.2.§

Corollary 11.3. (Existence of real cyclic harmonic frames). For all d 2 and n 2, there exists a cyclic harmonic frame of n distinct vectors for Rd. ≥ ≥

Proof. Let G = Zn. By choosing J Zn, J = d, to be a union of the disjoint sets ⊂ | | 0 , j, j , 1 j n 1, n (for n even), { } { − } ≤ ≤ 2 − { 2 } we obtain a real cyclic harmonic frame for Rd (they all come in this way). We can ensure that this frame has distinct vectors by making J a generating set for Zn, e.g., by choosing 1, 1 J. { − }⊂ ⊓⊔ The construction of Corollary 11.3 gives all real cyclic harmonic frames. It is equivalent to selecting real rows and complex conjugate pairs of rows from the Fourier matrix (11.4). The real rows corresponding to 0 and n (n even) are { } { 2 } 1111 1 1 , 1 1 1 1 1 1 . − − − To obtain a copy of the frame explicitly in Rd, one can apply the unitary map

1 1 1 z ℜz U := , U = √2 , √2  i i z ℑz −       to the complex conjugate pairs of rows corresponding to j, j to obtain real rows { − } j 2 j (n 1) j j 2 j (n 1) j 1 ω ω ω − 1 cos(2π ) cos(2π ) cos(2π − ) U = √2 n n n . j 2 j (n 1) j j 2 j (n 1) j 1 ω ω ω  1 sin(2π ) sin(2π ) sin(2π − ) − − − − n n n     The orthogonality of the rows of V above can be viewed as orthogonality of the corresponding trigonometric polynomials with respect to the discrete inner product

n 1 2πk 2πk f ,g := ∑− f g . k 0 n n = Example 11.10. (Equally spaced vectors) The n equally spaced unit vectors in R2 a cyclic harmonic frame, since they are a Zn–frame (Example 10.1). By the above calculation, they are given by J = 1, 1 Zn (or j, j for j a generator of Zn). { − }⊂ { − } Example 11.11. For the special case p = 2 in Example 11.6, all nonzero elements k of G = Z2 have order 2, and so are equal to their inverse. Thus all harmonic frames given by G = Zk are real. These are noncyclic harmonic frames when k 3 and 2 ≥ 252 11 Harmonic frames the vectors are distinct (cf. Example 11.5). The character table of the elementary abelian 2–groups Z2 Z2 can be calculated by taking Kronecker product of ×× that for Z2, e.g., 11 1 1 1 1 1 1 1 1 1 1 = − − . (11.12)   ⊗     1 1 1 1 1 1 1 1 − −  − −        1 1 1 1   − −    Example 11.12. The vertices of the Platonic solids (see 10.6) give tight frames Φ for R3 (with a high degree of symmetry). By Theorem 11.1,§ these are harmonic if and only if Sym(Φ) has has a transitive abelian subgroup G. By Table 10.1, we have that the tetrahedron, the cube and the octahedron are harmonic frames, and the icosahedron (12 vertices) and dodecahedron (20 vertices) are not harmonic frames. 3 The eight vertices of the cube give a noncyclic harmonic frame for R (since Z8 is not a transitive subgroup of their symmetry group). See Table 11.1. We observe that the complex unlifted cyclic harmonic frame of 12 vectors given by 1,5,9 Z12 has a symmetry group of order 384, and the complex lifted and { }⊂ unlifted cyclic harmonic frames of 20 vectors given by 0,1,11 , 1,10,11 Z20 have a symmetry group of order 200 (see Prob. 11.2). { } { }⊂

Example 11.13. There exist real harmonic frames with “large” symmetry groups, e.g, the real cyclic harmonic frames of 14 vectors for C5 given by J = 0, 1, 6 and J = 1, 6,7 have symmetry groups of order 392. The remaining{ 334± cyclic± } harmonic{± frames± of} 14 vectors for C5 have symmetry group orders 98,42,28,14.

n d d = 1 d = 2 d = 3 d = 4 d = 5 d = 6 d = 7 \ n = 4 0(0) 1(0) 2(0) 1(0) n = 5 0 1 1 1 1 n = 6 013321 n = 7 0111111 n = 8 0(0) 1(0) 3(1) 5(2) 5(1) 3(0) 2(0) n = 9 0(0) 1(0) 1(0) 3(1) 3(1) 2(0) 2(0) n = 10 0135764 n = 11 0112222 n = 12 0 1(0) 3(1) 9(2) 15(3) 17(4) 17(3)

Table 11.1: The numbers of real harmonic frames of n 12 distinct vectors for Rd , 1 d 7 (up to unitary equivalence and reordering), with the number≤ which are not cyclic in brackets.≤ ≤ 11.7 Unitary equivalence preserving the group structure 253 11.7 Unitary equivalence preserving the group structure

Here we outline techniques for efﬁciently determining whether harmonic frames are unitarily equivalent to each other. Since a harmonic frame Φ =(φg)g G is a G–frame, its Gramian is a G–matrix (Theorem 10.3). Thus each row and column∈ of the Gramian has the same entries. We call this multiset minus the diagonal entry the angle multiset of Φ

Ang(Φ) := multiset of off diagonal entries of any row/column of Gram(Φ).

The angle multiset of the cyclic harmonic frame given by J = j1,..., jd Zn is { }⊂ 2π aj1 aj2 aj Ang(ΦJ)= ω + ω + + ω d : 1 a n 1 , ω = e n . (11.13) { ≤ ≤ − } Since the Gramian determines a frame up to unitary equivalence, we have:

Harmonic frames which are unitarily equivalent up to a reordering must have the same angle multiset.

Fig. 11.1: The angle multisets of the unitarily inequivalent harmonic frames of 7 vectors for C3. Note that one is real, and three are equiangular.

Example 11.14. (7 vectors in C3) The angle multiset for the 7 unitarily inequivalent (cyclic) harmonic frames of 7 vectors for C3 are depicted in Figure 11.1. Observe that one is real (Corollary 11.3) and three are equiangular. 254 11 Harmonic frames

Most, but not all, reindexings (reorderings) of a given harmonic frame which make it unitarily equivalent to another are automorphisms. Let Aut(G) denote the group of automorphisms of G, i.e., isomorphisms σ : G G. → Deﬁnition 11.2. We say G–frames (vg)g G and (wg)g G are unitarily equivalent ∈ ∈ via an automorphism σ Aut(G) if (vg)g G and (wσg)g G are unitarily equivalent. ∈ ∈ ∈ Subsets J and K of G are multiplicatively equivalent if there is a σ Aut(G) 1 ∈ for which K = σJ. In this case, σˆ : Gˆ Gˆ : χ χ σ − is an automorphism of Gˆ, and → → ◦ ξ J,η J = σξˆ K,σηˆ K , | | | | i.e., ΦJ and ΦK are unitarily equivalent after reindexing by the automorphism σˆ . For G = Zn, each σ Aut(G) has the form g ag, with a Z a unit, and hence ∈ → ∈ n∗ J and K are multiplicatively equivalent if and only if K = aJ for some a Zn∗. We now give a simple condition which ensures harmonic frames are∈ unitarily equivalent via an automorphism.

Theorem 11.4. (Multiplicative equivalence) Let G be a ﬁnite abelian group, and ΦJ and ΦK be the harmonic frames for J,K G. Then the following are equivalent: ⊂ 1. The harmonic frames ΦJ and ΦK are unitarily equivalent via an automorphism. 2. The subsets J and K are multiplicatively equivalent.

Proof. Let σ Aut(G). Using the Pontryagin duality map (11.7), and the fact that the characters∈ are a basis (they are orthogonal), we calculate:

ΦσJ and ΦK are unitarily equivalent

ξ σJ,η σJ = ξ K,η K , ξ,η Gˆ ⇐⇒ | | | | ∀ ∈ ∑ ξ(σ j)η(σ j)= ∑ ξ(k)η(k), ξ,η Gˆ ⇐⇒ j J k K ∀ ∈ ∈ ∈ 1 1 ∑(ξη− )(σ j)= ∑ (ξη− )(k), ξ,η Gˆ [by (11.3)] ⇐⇒ j J k K ∀ ∈ ∈ ∈ 1 ∑ χ(σ j)= ∑ χ(k), χ Gˆ (take χ = ξη− ) ⇐⇒ j J k K ∀ ∈ ∈ ∈ ∑(σ j)ˆ= ∑ kˆ (by Pontryagin duality) ⇐⇒ j J k K ∈ ∈ σJ = σ j : j J = K (the characters of Gˆ are a basis) ⇐⇒ { ∈ } i.e., J and K are multiplicatively equivalent. ⊓⊔ Multiplicative equivalence is an equivalence relation, with the equivalence classes being the orbits of the natural action of Aut(G) on the d–element subsets of G. The number of multiplicative equivalence classes of d–element subsets of a group G which generate G is essentially Hall’s Eulerian function Φd(G), which counts the ordered d–element generating subsets of G. 11.7 Unitary equivalence preserving the group structure 255

In view of Theorem 11.4, the question of determining all the harmonic frames (for a given group) up to unitary equivalence (and reordering) reduces to determining whether harmonic frames corresponding to different multipicative equivalence classes are unitarily equivalent. This can most often be done by a simple calculation, e.g., comparing their angle multisets. Examples where there is unitary equivalence via a permutation which is not an automorphism are considered in 11.2. § 2 Example 11.15. (Four vectors in C ) First consider G = Z4. The automorphism σ group Aut(G) has order 2, and is generated by : g 3g (Z4∗ = 1,3 ). Thus the multiplicative equivalence classes of 2–element subse→ts of G are { }

0,1 , 0,3 , 1,2 , 2,3 , 1,3 , 0,2 . {{ } { }} {{ } { }} {{ }} {{ }} The ﬁrst three give cyclic harmonic frames with distinct vectors (1 generates G), while the last does not. None are unitarily equivalent, since their angle multisets are

i + 1,0,i + 1 , 0, i 1,i 1 , 0,0, 2 , 0,0,2 . {− } { − − − } { − } { }

Now consider G = Z2 Z2, which is generated by any two of its three elements a,b,a+b of order 2. The× automorphism group of G has order 6, with an automorphism{ corresponding} to each permutation of a,b,a + b . Thus the multiplicative equivalence classes are { }

a,b , a,a + b , b,a + b , 0,a , 0,b , 0,a + b . {{ } { } { }} {{ } { } { }} Only the ﬁrst gives a harmonic frame with distinct vectors. This frame is unitarily equivalent to the cyclic harmonic frame of four equally spaced unit vectors for R2. 3 Example 11.16. (Seven vectors in C ) For G = Z7, there are seven multiplicative equivalence classes of 3–element subsets, with representatives

1,2,6 , 1,2,3 , 0,1,2 , 0,1,3 , 1,2,5 (size 6) { } { } { } { } { } 0,1,6 (size 3) 1,2,4 (size 2). { } { } Each gives a cyclic harmonic frame with distinct vectors (nonzero elements generate G). None are unitarily equivalent since their angle multisets differ (see Fig. 11.1). There is just one harmonic frame of n distinct vectors for C1. Example 11.17. (Cyclic harmonic frames for C1) There is a unique harmonic frame of n distinct vectors for C1, namely the cyclic harmonic frame given by the n–th roots of unity. This follows since such a frame must be given by J = g , where g generates G (which is therefore cyclic). There is a automorphism of G{ }taking any generator to any other.

Example 11.18. (Cyclic harmonic frames for C2) The cyclic harmonic frames of n distinct vectors for C2 are unitarily equivalent (up to reordering) if and only if the subsets of Zn that give them are multiplicatively equivalent. This follows by considering the angle multisets (see [CW11] for details). 256 11 Harmonic frames 11.8 Noncyclic harmonic frames

The eight vertices of cube are a noncyclic harmonic frame for R3 (Example 11.12).

Example 11.19. (Noncyclic harmonic frame for C2). There is a noncyclic complex 2 harmonic frame of eight vectors for C given by J = (0,1),(1,0) G = Z4 Z2, i.e., { }⊂ × 1 1 i i 1 1 i i , , , , − , − , − , − . 1  1 1  1  1   1  1   1 − − − −                 A ﬁnite abelian group G can be written as a direct sum of p–groups

Gp = Z e1 Z e2 Zpem p ⊕ p ⊕⊕ where p are the prime divisors of G . The automorphism group of Gp has order | | m m m d k 1 e j m d j ei 1 m ci+1 Aut(Gp) = ∏(p k p − ) ∏(p ) − ∏(p − ) − , (11.14) | | k=1 − j=1 i=1 where ck := min r : er = ek k, dk := max r : er = ek k, and so the order of Aut(G) is the{ product of these}≤ orders (see [HR07]).{ In effect,}≥ the less cyclic an abelian group is, the larger its automorphism group becomes.

d = 2 d = 3 d = 4 n non cyc harm n non cyc harm n non cyc harm 4 0 3 3 4 0 3 3 4 0 1 1 8 1 7 8 8 5 16 21 8 8 21 29 9 1 6 7 9 3 15 18 9 5 23 28 12 2 13 15 12 11 57 68 12 30 141 171 16 4 13 17 16 28 74 102 16 139 228 367 18 2 18 20 18 19 121 140 18 80 494 574 20 3 19 22 20 29 137 166 20 154 622 776 24 6 27 33 24 89 241 330 24 604 1349 1953 25 1 15 16 25 8 115 123 25 37 636 673 27 3 18 21 27 33 159 192 27 202 973 1175 28 4 25 29 28 57 255 312 28 443 1697 2140 32 9 25 34 32 158 278 436 32 1379 2152 3531

Table 11.2: The numbers of noncyclic, cyclic harmonic frames of n 35 distinct vectors for Cd , d = 2,3,4 (up to unitary equivalence and reordering) when there is a≤ noncyclic abelian group. 11.9 Unitary equivalence not preserving the group structure 257

It was observed in [WH06] that most harmonic frames seem to be cyclic, with increasingly fewer as G becomes less cyclic. A heuristic explanation for this is that: as the group G becomes less cyclic, its automorphism group becomes larger (so the number of multiplicative equivalence classes becomes smaller), and the orders of its elements become smaller (so that J G is less likely to generate G, and hence give a harmonic frame with distinct vectors).⊂

11.9 Unitary equivalence not preserving the group structure

By considering the angle multisets (see [CW11]), it can be shown that:

Theorem 11.5. Cyclic harmonic frames of n distinct vectors for C2 are unitarily equivalent (up to a reordering) if and only if the subsets of Zn that give them are multiplicatively equivalent. Theorem 11.5 implies that unitary equivalence (up to reordering) and multiplicative equivalence are the same for cyclic harmonic frames for C3, except if both frames are unlifted. There do exist examples of unlifted cyclic harmonic frames which are unitarily equivalent after a reordering which is not an automorphism.

Example 11.20. For Z8 there are 17 multiplicative equivalence classes of 3–element subsets which generate it. Only two of these give frames with the same angles, namely 1,2,5 , 3,6,7 , 1,5,6 , 2,3,7 . { } { } { } { } The common angle multiset is 1,i,i, i, i, 2i 1,2i 1 . These two frames are unitarily equivalent up to a reordering.{− − Since− − they− are g−iven} by multiplicatively inequivalent subsets, this reordering cannot be an automorphism (see Table 11.3). Three inﬁnite families of such “exceptional cases” where unitary equivalence (up to reordering) does not imply multiplicative equivalence are given in in [Chi10]. It is not known whether or not these are all of them. If n is square free, i.e., is a product of distinct primes, then the primitive n–th 2πi roots of unity are a basis for the cyclotomic ﬁeld Q(ω), ω = e n . This leads to: Theorem 11.6. Let n be square free. Then the cyclic harmonic frames of n vectors d for C given by J,K Zn∗ (the units) are unitarily equivalent (up to a reordering) if and only if J and K are⊂ multiplicatively equivalent.

Proof. Suppose that ΦJ and ΦK are unitarily equivalent (after reordering), but are ω j Φ Φ not multiplicatively equivalent. Then the angle ∑ j J of J is an angle of K, ω j ωak ∈ so that ∑ j J = ∑k K , where a Zn∗. Since gcd(ak,n)= gcd(a,n), k K, ∈ ∈ ∈ ∀ ∈ it follows that ∑ ωak = ∑ c ωb, where each c Z has the same sign (see k K b Z∗n b b ∈ ∈ ω j ∈ ωak [Chi10] for details). Thus it follows that ∑ j J = ∑k K (since n is not even), which is a contradiction. ∈ ∈ ⊓⊔ 258 11 Harmonic frames

Computations of [Chi10] suggest that Theorem 11.6 also holds when n is not square free, i.e., all examples of cyclic harmonic frames for which multiplicative equivalence and unitary equivalence are not equivalent (such as Example 11.20) involve cyclic harmonic frames ΦJ for which J contains a nonunit.

11.10 The number of cyclic harmonic frames

Computations of [WH06] suggest that the number of harmonic frames of n distinct d d 1 vectors for C grows like n − (for d ﬁxed), and most harmonic frames are cyclic. In [MW04], it was established that the number hn,d of cyclic harmonic frames of n distinct vectors for Cd up to unitary equivalence (and reordering) grows like

d n d 1 hn d n − , n ∞, , ≈ ϕ(n) ≥ → where ϕ(n) is Euler’s totient function. The proof uses a correspondence between unitarily equivalent cyclic harmonic frames and points on the torus T2d, in which multiplicative equivalence gives a torsion point. Here we consider the particularly simple case of when n is a prime, for which all harmonic frames are cyclic and there is an explicit formula for hn,d (see [MW04], [Hir10]). Let Zn∗ (as the automorphisms of Zn) act on the d–element subsets of Zn, and S be invariant under this action. Then, by Burnside’s counting lemma, the number of multiplicative equivalence classes in S is 1 S/Zn∗ = ϕ ∑ Fix(a) , (11.15) | | (n) a Z | | ∈ ∗n where Fix(a) is the set of elements of S ﬁxed by a Z . ∈ n∗ u l Theorem 11.7. Let p be a prime, and hp,d and hp,d be the numbers of unlifted and lifted (cyclic) harmonic frames of p d distinct vectors for Cd, d > 1. Then ≥ p 1 u 1 −j d 1 h = ∑ ϕ( j) p − , p ∞, p,d p 1 d ≈ → j gcd(p 1,d) j − | − p 1 l 1 −j d 2 h = ∑ ϕ( j) p − , p ∞. p,d p 1 d 1 ≈ → j gcd(p 1,d 1) −j − | − − u l d 1 In particular, the number of harmonic frames hp d = h + h p , p ∞. , p,d p,d ≈ − → Proof. Let a Z be one of the ϕ( j) elements of order j. The orbit of a nonzero ∈ n∗ element of Zn under the action of a has size j, and the orbit of zero is a singleton. Thus if a d–element subset J Zn is ﬁxed by a, then either ⊂ 11.10 The number of cyclic harmonic frames 259

d J is unlifted and is the union of j of the a –orbits of nonzero singletons. • d 1 J is lifted and is the union of 0 and − of the a –orbits of nonzero singletons. • { } j Since d > 1 and p is prime, all the harmonic frames have distinct vectors. Now Z∗p is cyclic of order ϕ(p)= p 1 (so j divides p 1), and (11.15) gives the result. − − ⊓⊔ 1 Example 11.21. For d = 2, we have hp,2 = 2 (p + 1), since

p 1 1 p 1 − 1 hu = hl = 1, hu = hl = − + 2 = (p 1). p,1 p,2 p,2 p,3 p 1 2 1 2 − − For d = 3, we have

p 1 u l 1 −3 , p 1 (mod 3); h = h = p 1 ≡ p,3 p,4 p 1 − p 1 − + 2 3 , p 1 (mod 3). − 3 1 ≡ Hence 1 2 6 (p 2p + 3), p 1 (mod 3); hp,3 = − ≡ 1 (p2 2p + 7), p 1 (mod 3). 6 − ≡ As indicated above, formulas for hp,d depending on p modulo d and d 1 can be always be constructed. It is also possible to modify the proof of Theorem− 11.7 to count the real harmonic frames: Proposition 11.1. Let p be an odd prime. For d even, the number of real harmonic (unlifted) frames of p vectors for Rd (up to unitary equivalence) is

p 1 p 1 1 − − hR = ∑ j ϕ( j)+ ∑ 2 j ϕ( j) . p,d p 1 d d j gcd(p 1,d) j j gcd(p 1, d ) 2 j − | j even− | − 2 j odd

For d odd, the number of real harmonic (lifted) frames of p vectors for Rd is

p 1 p 1 1 − − hR = ∑ j ϕ( j)+ ∑ 2 j ϕ( j) . p,d p 1 d 1 d 1 j gcd(p 1,d 1) −j j gcd(p 1, d 1 ) 2−j − | j even− − | − −2 j odd

Proof. See Exer. 11.11 for details. ⊓⊔ Example 11.22. For d 3, there is a single real harmonic frame of p vectors, i.e., ≤ R R R hp,1 = hp,2 = hp,3 = 1. For d even, d 4, we have the estimate ≥ R R d 1 h = h p 2 − , p ∞. p,d p,d+1 ≈ → 260 11 Harmonic frames 11.11 Projective unitary equivalence of harmonic frames

We have seen that multiplicatively equivalent subsets J G give harmonic frames which are unitarilly equivalent up to a reindexing (Theorem⊂ 11.4). We now show that the translates of J gives projectively unitarily equivalent harmonic frames. The translation of a ﬁnite abelian group G by b is the bijection

τb : G G : j j + b, b G. → → ∈

We say that K is a translate of J if K = J + b, i.e., K = τbJ. Theorem 11.8. (Translates) Let G be a ﬁnite abelian group. If K is a translateofJ, then the harmonic frames ΦJ and ΦK are projectively unitarily equivalent. Φ ξ Proof. Suppose that K = J + b. Since J =( J)ξ Gˆ , we need to show | ∈

ξ K = cξU(ξ J), ξ Gˆ, | | ∈ J K J K where U : C C is unitary. Let Ub = C C be the unitary map given by → → (Ubv)(k) := v(k b), k K. Since ξ is a character, we have − ∈

(Ubξ J)(k)= ξ J(k b)= ξ(k b)= ξ(k)ξ( b)= ξ(b)ξ K(k), | | − − − | and so we can take U = Ub and cξ = ξ(b). ⊓⊔ The converse: that projective unitary equivalence (without reordering) implies that J and K are translates of each other, appears to be true.

Example 11.23. (Seven vectors in C3) We now revisit Example 11.16. The seven multiplicative equivalence classes of 3–element subsets of G = Z7 give rise to seven (cyclic) harmonic frames up to unitary equivalence and reordering. By translating a given subset, we obtain a projectively unitarily equivalent harmonic frame. Since the representatives of the multiplicative equivalence classes are related by

0,1,3 = 6 1,2,6 + 2, 1,2,4 = 6 1,2,6 + 3, { } { } { } { } 0,1,2 = 1,2,3 + 6, 1,2,5 = 3 1,2,3 + 6, 0,1,6 = 1,2,3 + 5, { } { } { } { } { } { } we conclude that there are exactly two harmonic tight frames of seven vectors for C3 up to projective unitary equivalence and reordering. By considering the angle multisets of Figure 11.1, we have that one is real, and the other is equiangular.

Example 11.24. Let p > 2 be a prime. Then all harmonic frames of p vectors for C2 are projectively unitarily equivalent up to reindexing to the p equally spaced unit vectors in R2. This follows since there is a unique afﬁne map, taking a sequence of two distinct elements of Zp to any other. In particular the two harmonic frames of three vectors in C2 which are unitarily inequivalent (one is real, one is complex) are projectively unitarily equivalent up to a reordering. 11.11 Projective unitary equivalence of harmonic frames 261

We define the affine group of G to be the group of bijections θ : G G generated by the translations and automorphisms of G, i.e., the G Aut(G) maps→ of the form | || | θ(g)= σ(g)+ b, σ Aut(G), b G. ∈ ∈ If K = θJ, for some θ in the affine group, we say J and K are affinely equivalent, i.e., they are in the same orbit under the natural action of the affine group of G on the subsets of G. Combining Theorems 11.4 and 11.8 gives the following:

Theorem 11.9. (Affine equivalence) Let G be a finite abelian group. If J and K are affinely equivalent subsets of G, then the harmonic frames ΦJ and ΦK that they give are projectively unitarily equivalent up to reindexing by an automorphism.

In view of Theorem 11.9, the question of determining the harmonic frames up to projective unitary equivalence (and reordering) reduces to determining whether frames given by afﬁnely inequivalent subsets are projectively unitarily equivalent. In all cases considered (see, e.g., Table 11.3) this can be done by using a simple test, such as comparing the multiset of m–products.

d = 2 d = 3 d = 4 d = 5 d = 6

n uni proj n uni proj n uni proj n uni proj n uni proj 2 1 1 3 1 1 5 2 1 5 1 1 6 1 1 3 2 1 4 3 1 6 9 3 6 4 1 7 2 1 4 3 2 5 3 1 7 7 2 7 4 1 8 11 3 5 3 1 6 11 3 8 21 6 8 19 4 9 16 3 6 6 3 7 7 2 23 5 20 10 55 9 7 4 1 8 16 4 9 23 4 9 23 4 56 8 7 3 17 24 24 11 48 6 9 6 2 9 15 3 10 53 9 10 67 9 10 9 3 10 29 4 54 11 48 6 11 6 1 11 17 2 11 34 4 12 13 5 12 56 9 12 138 21 13 7 1 57 141 14 12 3 13 25 3 15 13 3

Table 11.3: The number of unitary and projective unitary equivalence classes (up to reindexing) of cyclic harmonic frames of n vectors for Cd (d 6). When the group theoretic estimate given by Theorem 11.9 is larger (i.e., there are reindexings≤ which are not automorphisms), then it is given in the row below. 262 11 Harmonic frames 11.12 The projective symmetry group of a harmonic frame

The projective symmetry group of various harmonic frames was calculated in [WC14] (see Table 11.4). We say that a frame has projectively repeated vectors if some vectors are scalar multiples of each other (in this case the projective symmetry group is not given). When a harmonic frame is complex, the extended projective symmetry group is given also. The number of erasures of a frame for Cd is the maximum number of vectors that can be removed from it so that those remaining still span Cd.

d n real orth reps SymP(Φ) SymEP(Φ) J erasures 3 3 R y <6,1> 0,1,2 0 { } 3 4 R <24,12> 1,2,3 1 { } 3 5 R <10,1> 0,1,3 2 { } 3 6 <18,3> <36,10> 0,1,4 2 R y <12,4> {1,2,3} 3 R y y {1,3,5} 1 { } 3 7 <21,1> <42,1> 1,2,6 4 R <14,1> {1,3,5} 4 { } 3 8 <16,8> <32,43> 1,3,4 5 <32,11> <64,134> {0,1,4} 3 R <16,7> {0,1,2} 5 R y {1,3,5} 3 { } 3 9 <9,1> <18,1> 1,4,6 5 R y <18,1> {0,1,2} 6 R y y {1,4,7} 2 { } 3 10 <50,3> <100,13> 0,1,5 4 R <20,4> {0,1,9} 7 <10,2> <20,4> {0,1,8} 7 R y {1,5,7} 5 { } 3 11 <11,1> <22,1> 0,1,3 8 R <22,1> {1,2,3} 8 { } 3 12 <12,2> <24,6> 1,2,11 8 R y <24,6> { 1,2,3 } 9 R y {1,4,10} 5 <12,2> <24,6> { 0,3,4 } 7 y {1,5,7} 5 y <24,5> <48,38> {0,1,8} 7 R y y {1,3,5} 7 <72,30> <144,154> {2,3,8} 5 R y y {1,5,9} 3 { } 3 13 R <26,1> 0,1,12 10 <13,1> <26,1> { 0,1,3 } 10 <39,1> <78,1> {1,2,11} 10 { } Table 11.4: The cyclic harmonic frames of n vectors for C3. If the frame doesn’t have projectively repeated vectors, then its projective symmetry group is given, and when it is complex, then its extended projective symmetry group is given. We also indicate if the frame has orthogonal vectors, and its number of erasures. 11.12 The projective symmetry group of a harmonic frame 263 Notes

It has been known (at least) since [Zim01] that there exists at least one real harmonic frame of n d vectors for Cd. ≥ d 1 n − pn d , ≈ ϕ(n)

Exercises

d 11.1. An equal–norm frame Φ =(v j) for F is said to be equispaced if the distances v j vk are constant for j = k (this is equivalent to ℜ v j,vk being constant). (a) − Show that the real and complex harmonic frames of three vectors for C2 are equispaced (and also equiangular), but that the spacings are different in each case. 2 (b) Let Φ =(v j) be the cyclic harmonic frame for C given by 1,2 Z5, i.e., { }⊂ 2 3 4 1 ω ω ω ω π Φ = , , , , , ω := e2 i/5. 1 ω2 ω4 ω ω3 Show that Φ is equispaced. (c) The equispaced harmonic frames of (a) and (b) can be generalised as follows. d Let Φ = (v j) be the cyclic harmonic frame of 2d 1 vectors for C given by − 0,1,...,d 1 Z2d 1, and Ψ =(w j) be the cyclic harmonic frame of 2d + 1 { −d }⊂ − vectors for C given by 1,2,...,d Z2d 1. Show that Φ and Ψ are equispaced. { }⊂ + 11.2.m Write a function to compute the symmetry group of a frame from its Gramian. Apply it in the following cases: 3 (a) The unlifted cyclic harmonic frame of 12 vectors for C given by 1,5,9 Z12. 3 { }⊂ (b) The cyclic frames of 20 vectors for C given by 0,1,11 , 1,10,11 Z20. { } { }⊂ d j n 1 11.3. Let U be a unitary matrix and v F . Show that (U v) − is an equal–norm ∈ j=0 tight frame for Fd if and only if it is projectively unitarily equivalent to a cyclic harmonic frame, i.e., Un = cI.

11.4. Show that the set of n–th roots of unity is the unique harmonic frame of n distinct vectors for C1.

11.5. Show that the only real harmonic frame of n distinct vectors for C2 is the n equally spaced unit vectors in R2.

11.6. Find all the (unitarily inequivalent) harmonic frames of six vectors for C2. Show that none are complex conjugates of each other, by computing the distances between their vectors, or otherwise. 264 11 Harmonic frames

d 11.7. Suppose that Φ =(gv)g G is a frame for F generated by a ﬁnite abelian group of matrices G GL(H ∈). Show that Φ has distinct vectors. Does this hold for nonabelian groups?⊂

11.8. Show that if Φ is a harmonic frame given by J G, then the complementary tight frame is the harmonic frame given by G J (the⊂ complement of J). \ 11.9. Suppose that Φ is an unlifted harmonic frame with Sym(Φ) = m. Show that Ψ := Φ+ˆ +ˆ Φ (d summands) has a symmetry group| of order at| least d!md. Remark: The vertices of the cube [ 1,1]d in Cd (see Example 5.12) are a balanced tight frame of 2d vectors for Rd with− symmetry group of order d!2d.

11.10. Let G be a ﬁnite abelian group. (a) Show that a harmonic frame (vg)g G satisﬁes ∈

v j,vk = v j a,vk a , j,k,a G. + + ∀ ∈

(b) Suppose that harmonic frames (vg)g G and (wg)g G are unitarily equivalent up to reindexing. Show that for all j,k,b ∈G there exists∈ some a G with ∈ ∈

v j,vk = wa,wb .

(c) Suppose that the harmonic frames given by ξ1,...ξd , η1,...,ηd Gˆ are unitarily equivalent up to reindexing. Show that { } { }⊂

d d ∑ ξ j(1)= ∑ η j(a), j=1 j=1 for some a G. ∈ 11.11. Let p be an odd prime. (a) Show that if J Zp, J = d gives a real cyclic harmonic frame, then J = K K when d is even,⊂ and J |=| 0 K K when d is odd (disjoint unions), where∪− { }∪ ∪− K Zp 0 generates Zp (for d > 1). (b)⊂ Use\ Burnside { } orbit counting to show that the number of real (cyclic) harmonic frames of p vectors for Rd is given by

p 1 p 1 1 − − hR = ∑ j ϕ( j)+ ∑ 2 j ϕ( j) (d even), p,d p 1 d d j gcd(p 1,d) j j gcd(p 1, d ) 2 j − | j even− | − 2 j odd

p 1 p 1 1 − − hR = ∑ j ϕ( j)+ ∑ 2 j ϕ( j) , (d odd). p,d p 1 d 1 d 1 j gcd(p 1,d 1) −j j gcd(p 1, d 1 ) 2−j − | j even− − | − −2 j odd Chapter 12 Equiangular and Grassmannian frames

d A set of n unit vectors f j in F is said to deﬁne a set of equiangular lines if the angles between the subspaces{ } (lines) they determine are equal, i.e., C 0 with ∃ ≥

f j, fk = C, j = k. (12.1) | | ∀ An illuminating example is three isogonal vectors in R3 (see Example 3.9),

where one can choose 0 C < 1, or even C = 1 (in which case the vectors are equal). By the variational≤ characterisation of tight frames (Theorem 6.1), it follows that n d C − , (12.2) ≥ d(n 1) − d with equality in (12.2) if and only if f j is a tight frame for F . Moreover, the maximum C can be is 1, when the vectors{ } lie in a common line. Here we consider those sets of unit vectors that define a set of equiangular lines, which, in addition, give equality in (12.2), i.e., are a finite tight frame for Fd. The corresponding sets of tight equiangular lines can be thought as being spread out in Fd as much as is possible. This extra structure gives sets of equiangular lines with special properties, e.g., the Gramian of the vectors defining the lines is an orthogonal projection matrix, and so a complementary set of equiangular lines exists.

265 266 12 Equiangular and Grassmannian frames 12.1 Equiangular lines and frames

d Sets f j of n unit vectors in F for which the maximum cross–correlation { }

M := max f j, fk [0,1] j=k | | ∈ is small have applications to wireless communication and coding theory [SH03]. These correspond to sets of lines L j = cf j : c F (through 0 and f j) for which the minimal angle between them { ∈ }

1 1 θ := mincos− ( f j, fk ) [0, π] j=k | | ∈ 2 is large. The lower bound (6.3) for M is attained as follows.

Deﬁnition 12.1. A set f j of equal–norm vectors in H is equiangular if C 0 with { } ∃ ≥ f j, fk = C, j = k. | | ∀ For a set of equiangular vectors of unit length, the constant is often written

C = α = cosθ, α [0,1], θ [0,π/2], ∈ ∈ with both C = α and θ referred to as the angle between the vectors (or lines). n H H Theorem 12.1. Let ( f j) j=1 be a sequence of n unit vectors in , d = dim( ). Then n d M = max f j, fk − , (12.3) j=k | | ≥ d(n 1) − or, equivalently,

1 1 n d θ = mincos− f j, fk cos− − , (12.4) j=k | | ≤ d(n 1) − n with equality if and only if ( f j) j=1 is an equiangular tight frame. Proof. We consider the case of equality in Exer. 6.2. Firstly, observe that

2 1 2 1 n n d ∑∑ f j, fk n n = − , n2 n | | − ≥ n2 n d − d(n 1) − j k − − 12.1 Equiangular lines and frames 267 with equality if and only if ( f j) is a tight frame. The two inequalities above give (12.3), and the fact that cos 1 is strictly decreasing on [0, 1 π] gives (12.4). − 2 ⊓⊔ Example 12.1. The d vectors of an orthonormal basis in Fd and the d +1 vertices of d 1 the regular simplex in R give real equiangular tight frames with C = 0 and C = d . These exist in every dimension d, and are the unique equiangular tight frames of d and d + 1 vectors in Cd up to projective unitary equivalence (see Example 9.8). The equiangularity condition on a set of lines ensures that the corresponding set of one-dimensional orthogonal projections is linearly independent, which in turn gives a bound on the possible number of equiangular lines in Fd. d Theorem 12.2. Suppose d > 1. Let ( f j) be a sequence of n unit vectors in F giving a set of n equiangular lines, then the orthogonal projections

Pj : f f , f j f j, j = 1,...,n → are linearly independent, and hence

1 2 d(d + 1), F = R; n 2 (12.5) ≤ d , F = C

n with equality if and only if Pj is a basis for the Hermitian matrices. { } j=1 Proof. Since d > 1, the equiangularity constant C is less than 1. By Exer. 3.1

2 2 trace(PjP∗)= f j, fk = C , j = k, k | | and so the Frobenius norm of the linear combination ∑ j c jPj is

2 ∑c jPj F = trace(∑c jPj ∑ckPk∗)= ∑∑c jck trace(PjPk∗) j j k j k 2 2 = ∑∑c jckC + ∑c jc j(1 C ) j k j − 2 2 2 2 = C ∑c j +(1 C )∑ c j , | j | − j | | which is zero only for the trivial linear combination. The projections Pj belong to the real vector space of Hermitian matrices, which has dimension (see{ Exer.} 12.1) given by the right hand side of (12.5). ⊓⊔ The right hand side of (12.5) is the number n of (8.24) which ensures that a generic sequence of n vectors in Fd has a unique scaling to a tight signed frame. d If the upper bound (12.5) is attained, then ( f j) is a unit norm tight frame for F (see Exer. 12.2), i.e., the identity matrix can be written

d n I = ∑ Pj. n j=1 268 12 Equiangular and Grassmannian frames

The upper bound (12.5) for the maximum number of equiangular lines in Rd is called the absolute bound (or Gerzon bound). It is rarely attained: only the cases d = 2,3,7,23 are known (see Example 12.35). Many maximal set of equiangular lines turn out to be tight frames (see Table 12.3). By way of contrast, the upper bound (12.5) for complex equiangular lines is conjectured to always be attained (Zauner’s conjecture), by a tight frame known as a SIC (see Chapter 14). Example 12.2. There are sets of two and three equiangular lines in R2. Theorem 12.2 conﬁrms the familiar geometric fact that there are not four (or more) equiangular lines in R2, and hence no equiangular tight frame of four or more vectors for R2. There is a tight frame giving four equiangular lines in C2 (see Example 2.16). Example 12.3. An equiangular tight frame of 28 vectors for R7 can be constructed as follows. Let ( f j) be the 28 unit vectors of the form 1 (x1,x2,...,x8), x j 1,3 , ∑x j = 0. (12.6) √24 ∈ {− } j

These vectors span a 7–dimensional subspace of R8, and give an equiangular tight 1 frame (see Exer. 12.4, Example 12.56), which gives equality n = 2 d(d +1) in (12.5).

12.2 Grassmannian frames

If an equiangular tight frame does not exist for a given n d, then a good substitute is a Grassmannian frame (see [SH03], [BK06]) or a variation≥ thereof. Φ n H Definition 12.2. A frame =( f j) j=1 of n unit vectors for is a Grassmannian frame if it minimises M∞(Φ) := max f j, fk , (12.7) j=k | | and is optimal if it gives equality in (12.3), i.e., is an equiangular tight frame. Grassmannian frames exist for all n d, for H real or complex, since the the set of n element unit norm frames is a compact≥ subset of H n. By the same reasoning, there exist equal–norm normalised tight frames which minimise (12.7). These were frames were called 2–optimal frames by [HP04], who proved that they are optimal for the 2–erasure problem. The Grassmannian space G (H ,ℓ)= G (d,ℓ) is the set of all ℓ–dimensional subspaces of the d–dimensional space H = Fd (usually real). The Grassmannian packing problem is to find the best packing of n subspaces of dimension ℓ in H , so that the angle between any two is as large as possible (see [CHS96], [DHST08]). Clearly, for ℓ = 1 this equivalent to finding a Grassmannian frame. Example 12.4. The first n vertices of the regular 2n–gon (multiplied by 1), i.e., the n equally spaced lines in R2 (see Exer. 6.10) are the unique Grassmannian± frame of n vectors for R2 (see Exer. 12.5). This is tight, and it is equiangular only for n = 3. 12.3 Equiangular harmonic frames and difference sets 269

Example 12.5. By Theorem 12.1, an equiangular tight frame is Grassmannian.

Example 12.6. The only known example of a nontight Grassmannian frame is that of 5 vectors in R3 which lie on the diagonals of the regular icosahedron, which is the optimal Grassmannian line packing (see [CHS96], Exer. 12.7). This is the nontight equiangular frame given by the graph of the 5–cycle (see Example 12.44).

There is an extensive literature on the closely related problem of finding the dis- tributions of points on a sphere that minimise a given potential (see [SK97]). A classical example is Tammes’ problem of determining how to place n points on the unit sphere in R3 so as to maximise the minimum distance between them. For five vectors the solution to Tammes’ problem is the north and south poles together with three equally spaced points on the equator. This set of five vectors is not a tight frame, or a Grassmannian frame for R3.

12.3 Equiangular harmonic frames and difference sets

Φ ξ Here we show that a harmonic frame J =( J)ξ Gˆ is equiangular if and only if J G is a difference set for the abelian group G.| This∈ leads to some inﬁnite families of⊂ equiangular tight frames.

Deﬁnition 12.3. A d element subset J of a ﬁnite group G of order n is said to be a (n,d,λ)–difference set if every nonidentity element of G can be written as a 1 λ difference j1 j2− of two elements j1, j2 J in exactly ways. The difference set is said to be abelian, cyclic, etc, when the∈ group G is.

The complement G J of a (n,d,λ)–difference set J is a \ (n,n d,n 2d + λ)–difference set − − and so lists of difference sets (Tables 12.1 and 12.2) usually assume 2d n, λ > 0. Since there are exactly d2 d pairs of elements from J whose difference≤ is not the identity, a (n,d,λ)–difference− set must satisfy

d(d 1)=(n 1)λ. (12.8) − − The set of all translates of a difference set J gives a symmetric block design (not all symmetric block designs appear in this way). In particular, the parameters of a difference set must satisfy the Bruck–Ryser–Chowla theorem, which gives If n is even, then its order d λ is a square. • If n is odd, the Diophantine− equation • 2 2 (n 1)/2 2 x +(d λ)y ( 1) − λz = 0 − − − has a nontrivial solution (x,y,z). 270 12 Equiangular and Grassmannian frames

We assume the basic theory of harmonic frames (and characters) as detailed in Chapter 11. The orthogonality between a character χ = 1 and the trivial character 1 gives ∑ χ(g)= 0 = ∑ χ(g)= 1, χ = 1. (12.9) g G ⇒ g=0 − ∈ The characters of an abelian group G sum to the so called regular character

G , g = 1; ∑ χ = χreg, χreg(g) := | | (12.10) 0 g 1 χ Gˆ , = . ∈

Theorem 12.3. Let G be an abelian group of order n, and ΦJ =(ξ J)ξ ˆ be the | G harmonic frame given by J G, J = d. Then the following are equivalent∈ ⊂ | | d 1. ΦJ is an equiangular tight frame of n vectors for C . 2. J is a (n,d,λ)–difference set for G. ∑ χ 2 d(n d) χ χ ˆ 3. j J ( j) = n −1 , for all = 1, G. | ∈ | − ∈ Proof. (1 3) Suppose that ξ = η, i.e., χ := ξη 1 = 1. We calculate ⇐⇒ − 1 ξ J,η J = ∑ ξ( j)η( j)= ∑(ξη− )( j)= ∑ χ( j), | | j J j J j J ∈ ∈ ∈ and so ΦJ is equiangular if and only if

ξ J η J 1 n d | , | = ∑ χ( j) = − , √d √d d j J d(n 1) ∈ − which gives the equivalence of 1 and 3. We now show the equivalence of 1 and 2. Since χ(k)= χ( k), we have that − 2 ξ J,η J = ∑ ∑ χ( j k)= ∑ αgχ(g), χ = 1, (12.11) | | | | j J k J − g G ∈ ∈ ∈ α α α 2 where g := # ( j,k) J J : j k = g . Observe that 0 = d, ∑g=0 g = d d. { ∈ × − } − (2 =1) Suppose that J is a difference set, i.e., αg = λ, g = 0. Then (12.11), together⇐ with (12.9), gives

2 ξ J,η J = λ ∑ χ(g)+ d = λ + d, | | | | g=0 − i.e., ΦJ is equiangular. (1= 2) Suppose that ΦJ is equiangular. Then (12.11) gives ⇒ 2 ξ J,η J = ∑ αgχ(g)+ d = λ + d, χ = 1, | | | | g=0 − for some ﬁxed λ. By Pontryagin duality, this can be written as 12.3 Equiangular harmonic frames and difference sets 271

∑ αggˆ(χ)= λ, χ = 1. g=0 − Moreover, by counting the number of differences

ˆ 2 ∑ αg0ˆ(χ)= ∑ αg = d d. g=0 g=0 − These combine to give the regular character of Gˆ as a sum of characters

2 λ α ˆ λ ˆ d d + χGˆ ∑ ggˆ + 0 = − reg. g=0 n Since the (irreducible) characters of Gˆ are linearly independent, (12.10) implies that αg = λ, g = 0, i.e., J is a (n,d,λ)–difference set for G. ∀ ⊓⊔ The condition 3 for difference sets is well known [Tur65], [LM90]. However, its interpretation in terms of equiangular lines (harmonic frames) seems to be more recent [Kon99],¨ [XZG05], [Kal06]. In [XZG05] equiangular tight frames are called MWBE (maximum–Welch–bound–equality) codebooks. Example 12.7. The cyclic harmonic frame of n vectors for Cd given by

j 2 j (n 1) j 1 ω 1 ω 1 ω − 1 . . . . 2πi Φ . . . . ω n J = ., . , . ,..., .  , := e 1 ω jd ω2 jd ω(n 1) jd        −          is equiangular if and only if J = j1, j2,..., jd is a (n,d,λ)–difference set for Zn. The cyclic difference sets for d {10 are given} in Table 12.1. ≤ Example 12.8. (Orthonormal bases) The choice J = G is an (n,n,n)–difference set which gives an orthonormal basis. Example 12.9. (Simplex) There is a unique difference set with λ = 0, d 1, i.e., the (n,1,0)–difference set consisting of the identity element of G. The complementary≥ (n,n 1,n 2)–difference set (of the nonidentity elements) gives the vertices of the simplex.− Both− of these difference sets (and those with d = 0,n) are said to be trivial.

Difference sets J1 and J2 for a group G are said to be equivalent if there is an automorphism σ of G and some g G with ∈

σJ1 = gJ2, for some g G. ∈ By Theorem 11.8, it follows that

Equivalent difference sets give projectively unitarily equivalent harmonic frames (up to reordering by an automorphism). 272 12 Equiangular and Grassmannian frames

Table 12.1: List of the cyclic (n,d,λ)–difference sets J for d 10, λ > 0 (including complements). These give the cyclic equiangular harmonic frames of n > d≤+ 1 vectors for Cd .

n d λ J Zn n d λ J Zn ⊂ ⊂ 7 3 1 1,2,4 15 8 4 3,6,7,9,11,12,13,14 { } { } 7 4 2 0,3,5,6 57 8 1 0,1,6,15,22,26,45,55 { } { } 13 4 1 0,1,3,9 13 9 6 2,4,5,6,7,8,10,11,12 { } { } 11 5 2 1,3,4,5,9 19 9 4 1,4,5,6,7,9,11,16,17 { } { } 21 5 1 3,6,7,12,14 37 9 2 1,7,9,10,12,16,26,33,34 { } { } 11 6 3 0,2,6,7,8,10 73 9 1 0,1,12,20,26,30,33,35,57 { } { } 31 6 1 1,5,11,24,25,27 19 10 5 0,2,3,8,10,12,13,14,15,18 { } { } 15 7 3 0,1,2,4,5,8,10 91 10 1 0,2,6,7,18,21,31,54,63,71 { } { }

Table 12.2: The abelian (n,d,λ)–difference sets for d 50, n 2d, λ > 0, where G has invariant ≤ ≥ d factors [a1,...,am]. These give the equiangular harmonic frames of n > d + 1 vectors for C .

n d λ G n d λ G n d λ G 7 3 1 [7] 31 15 7 [31] 757 28 1 [757] 13 4 1 [13] 273 17 1 [273] 109 28 7 [109] 21 5 1 [21] 35 17 8 [35] 59 29 14 [59] 11 5 2 [11] 307 18 1 [307] 175 30 5 [5,35] 16 6 2 [4,4] 96 20 4 [2,4,12] 871 30 1 [871] 16 6 2 [2,8] 96 20 4 [2,2,2,12] 156 31 6 [156] 16 6 2 [2,2,4] 96 20 4 [2,2,2,2,6] 63 31 15 [63] 16 6 2 [2,2,2,2] 381 20 1 [381] 63 31 15 [3,21] 31 6 1 [31] 85 21 5 [85] 993 32 1 [993] 15 7 3 [15] 43 21 10 [43] 133 33 8 [133] 57 8 1 [57] 47 23 11 [47] 67 33 16 [67] 73 9 1 [73] 553 24 1 [553] 71 35 17 [71] 37 9 2 [37] 101 25 6 [101] 1407 38 1 [1407] 19 9 4 [19] 651 26 1 [651] 79 39 19 [79] 91 10 1 [91] 64 28 12 [2,2,16] 121 40 13 [121] 23 11 5 [23] 64 28 12 [2,4,8] 121 40 13 [11,11] 45 12 3 [3,15] 64 28 12 [4,4,4] 83 41 20 [83] 133 12 1 [133] 64 28 12 [2,2,4,4] 1723 42 1 [1723] 27 13 6 [3,3,3] 64 28 12 [8,8] 1893 44 1 [1893] 40 13 4 [40] 64 28 12 [4,16] 2257 48 1 [2257] 183 14 1 [183] 64 28 12 [2,2,2,2,2,2] 99 49 24 [3,33] 36 15 6 [3,12] 64 28 12 [2,2,2,2,4] 197 49 12 [197] 36 15 6 [6,6] 64 28 12 [2,2,2,8] 2451 50 1 [2451] 12.3 Equiangular harmonic frames and difference sets 273

By (12.8), the number n of vectors in an equiangular harmonic frame for Cd is 1 n = (d2 d)+ 1 d2 d + 1. (12.12) λ − ≤ − This is largest when λ = 1, i.e., J is a planar difference set. Here the corresponding symmetric block design gives a projective plane (the blocks are the points on a line, and so each pair of distinct lines intersects in a single point). For example, the (7,3,1)–difference set J = 1,2,4 Z7 gives the Fano plane (7 points, 7 lines, 3 points on each line), whose{ lines are}⊂ the translates of J, i.e.,

1,2,4 , 2,3,5 , 3,4,6 , 4,5,0 , 5,6,1 , 6,0,2 , 0,1,3 . { } { } { } { } { } { } { } There is considerable work on difference sets (for both abelian and nonabelian groups). The online La Jolla Difference Set Repository maintained by Dan Gordon is an excellent resource. From this, we obtained the Table 12.2 of all abelian difference sets for n 50, i.e., all equiangular harmonic frames of n 50 vectors (for brevity, the complementary≤ difference sets are not included). ≤ There are several inﬁnite families of difference sets. These are often classiﬁed via their parameters, e.g., difference sets with parameters (4m 1,2m 1,m 1) are of Paley–type1, and those with parameters (4m2,2m2 m,m−2 m)−are Hadamard− difference sets (see [DJ96]). We now give a few examples.− −

Example 12.10. (Singer) Let PGm(q) be the m–dimensional projective geometry over Fq. The intersection of two hyperplanes is an (m 2)–dimensional subspace m 2 − q − 1 containing q 1− projective points, and so one obtains a symmetric block design where the blocks− are the projective points on a hyperplane. A classical result of Singer shows that there is a cyclic subgroup of the projective linear transformations which acts regularly on the points and hyperplanes of PGm(q). From this, it follows m m 1 m 2 q 1 q − 1 q − 1 that there exists a cyclic ( q −1 , q 1− , q 1− )–difference set. In particular, by tak- − − − ing m = 3, and letting d = q + 1, one obtains a cyclic (d2 d + 1,d,1)–difference set, i.e., an equiangular cyclic harmonic frame of n = d2 d−+1 vectors for Cd. The Prime Power Conjecture is that every abelian planar difference− set (such as these) has d 1 a prime power. This has been veriﬁed for d 2,000,000 [BG04]. − ≤ Example 12.11. (Paley) Let q = 4m 1 be a prime power (q 3 mod 4). Then the set of all nonzero squares in (GF(q)−,+) is a (4m 1,2m 1,≡m 1)–difference set. − − − Example 12.12. (Twin prime powers) Let q and q + 2 be twin prime powers, i.e., a pair of odd integers, each of which is a prime power, e.g, q = 7,q+2 = 9 = 32. Then q(q+2) 1 q(q+2) 3 there exists a (q(q + 2), 2 − , 4 − )–difference set. The last two examples, and the q = 2 Singer difference sets are of Paley–type. Example 12.13. (n = 2d + 1, d odd) For a difference set of Paley–type, by letting d 1 d = 2m 1(d is odd), the parameters become (2d + 1,d, − ). Hence, Paley–type − 2 1 The term Hadamard type difference set is also used. 274 12 Equiangular and Grassmannian frames difference sets give equiangular harmonic frames of n = 2d + 1 vectors for Cd. These exist when 2d + 2 is a power of 2, or 2d + 1 is a prime power congruent to 3 mod 4, or 2d +1 is a product of twin prime powers. The ﬁrst odd d where there is not a an equiangular harmonic frame of 2d + 1 vectors for Cd is d = 19 (2d + 1 = 39).

Example 12.14. (n = 2d 1, d even) The complement of a Paley–type difference set is a (4m 1,2m,m)–difference− set. By letting d = 2m (d is even), these parameters become −(2d 1,d, d ), and so Paley–type difference sets give rise to equiangular − 2 harmonic frames of n = 2d 1 vectors for Cd. − Example 12.15. (Reversible difference sets) A differences set J is reversible if it is closed under taking inverses, i.e., the corresponding equiangular harmonic frame is real (Theorem 11.2). A reversible difference set for G = Z4 Z4 is given by × J = (1,0),(2,0),(3,0),(0,1),(0,2),(0,3) , { } and so there exists a real equiangular harmonic frame of 16 vectors for C6. There 4 is also a (16,6,2)–difference set for G = Z2, which must be reversible (as are all 4 difference sets for groups of exponent 2), since each element of Z2 is its own inverse. The set J = (x,0),(0,x),(x,x) : x = 0 Z6 Z6 { }⊂ × is a reversible (36,15,6)–difference set for G = Z6 Z6, and so there exists a real equiangular harmonic frame of 36 vectors for R15. ×

The bound n d2 d + 1 of (12.12) for equiangular We say that a set of unit vectors in Cd (or≤ the lines− they determine) is ﬂat if each vector entry has constant modulus (of 1 ). Harmonic frames are ﬂat. √d

Proposition 12.1. ([GR09]) There can be at most d2 d + 1 ﬂat equiangular lines in Cd, and any such set must be tight. −

d Proof. Suppose that the unit vectors Φ =(v j) give n ﬂat equiangular lines in C , i.e., v j,vk = α, j = k. The Gram matrix of the rank one orthogonal projections | | v1v1∗,...,vnvn∗ and e1e1∗,...,dded∗ (with the Frobenius inner product) is

α2 α2 1 J +(1 )In d J A = 1 − , J =[1]. d J Id

2 d d 2 Since these vectors lie in the d dimensional space C × , we have rank(A) d . A calculation (see Exer. 12.9) shows that rank(A)= n + d + 1 and Φ is≤ tight, or rank(A)= n + d. Thus n + d 1 rank(A) d2, which gives the result. − ≤ ≤ ⊓⊔ Example 12.16. For 2 d 6 there are d2 d + 1 ﬂat equiangular lines for Cd, given by an equiangular≤ harmonic≤ frame corresponding− to a cyclic difference set (see Table 12.1). 12.4 Equiangular tight frames from block designs 275 12.4 Equiangular tight frames from block designs

d We call an equiangular tight frame (v j) of n unit vectors for C subsimplicial (of dimension r) if 1 v j,vk = , j = k, for some integer r, | | r i.e., if every r + 1 element subset of (v j) is an r–simplex, up to projective unitary equivalence. The possible d,n for a subsimplicial frame are restricted by the fact that d(n 1) r = − Z. (12.13) n d ∈ − A single d–simplex is a trivial example of a subsimplicial frame.

Example 12.17. (Complementary frames) If d,n satisfy the necessary condition (12.13) for being a subsimplicial frame, then the complementary frame may not. For example, for d = 8, n = 15 there is a harmonic equiangular tight frame given by the difference set 3,6,7,9,11,12,13,14 Z15 which is subsimplicial with r = 4, { }⊂ 7 but the complementary tight frame is not subsimplicial (r = 2 ). Example 12.18. (Real equiangular lines) A set of real equiangular lines meeting the 1 absolute bound n = 2 d(d + 1) is subsimplicial (with r = √d + 2) if and only if d + 2 is a square. Many other maximal sets of tight real equiangular lines are also subsimplicial, e.g., all those for d = 3 listed in Table 12.3 (including the conjectured sets of lines in grey).

Example 12.19. (SICs) A set of complex equiangular lines meeting the absolute bound n = d2 is subsimplicial (with r = √d + 1) if and only if d + 1 is a square. The ﬁrst two examples d = 3,8 are exceptional cases for SICs (see Chapter 14).

We now outline a construction of [FMT12] which uses Steiner systems (a type of block design) to construct subsimplicial equiangular tight frames directly in Cd. A Steiner system S(t,k,ν) is a collection B of k–element subsets (called blocks) of an ν element set V , with the property that each element of V is in exactly r blocks and each t–element subset of V is contained in exactly one block. A count shows

ν 1 t −1 r = k−1 . t −1 − B V ν 1 We consider Steiner systems on with t = 2 (so r = k−1 ), i.e., the property that every pair of points lies in exactly one block. We illust−rate the construction using an S(2,2,4) system (all two element subsets of a four element set). Each B can be represented by a B V matrix AT with a 1 in the (β,a) entry if the point a is in the block β, and a 0 otherwise× (A is the incidence matrix of the system). For each point a V , choose a Hadamard matrix of size r + 1, i.e., ∈ 276 12 Equiangular and Grassmannian frames

(a) h0 (a) hβ (a)  1  (a) (a) (a) (a) H = . , hβ = hβ,1 hβ,2 hβ,r+1 ,  .     (a) hβ   r    (a) where β1,...,βr B are the r blocks containing a. The r + 1 rows of H are orthogonal, and of∈ length √r + 1. Let V be the B ν(r + 1) block matrix obtained × by replacing the nonzero (β,a)–entries of AT by the 1 (r + 1) blocks h(a), and × β replacing the zero entries by the 1 (r + 1) zero matrix. × blocks (1) (2) h 1,2 h 1,2 0 0 1100 1,2 { } { } { } h(1) 0 h(3) 0 1010 1,3  1,3 1,3    { } {(1) } { } (4) h 0 0 h T 1001 1,4  1,4 1,4  A = { } V =  { } { }. 0110 2,3 −→  0 h(2) h(3) 0    { }  2,3 2,3     {2 } { } 4  0101 2,4  0 h( ) 0 h( )    { }  2,4 2,4  0011 3,4  { } (3) ({4) }   { }  0 0 h h     3,4 3,4   { } { } B ν(ν 1) The columns of V =[va, j]a V ,1 j r+1 are the vectors in C , B = − , given ∈ ≤ ≤ | | k(k 1) by − (a) β hβ, j, a ; va, j(β)= ∈ (12.14) 0, a β ∈ Recall that the rows and columns of a Hadamard matrix are orthogonal. Thus, the rows of V are orthogonal, and so (va, j) is an equal norm tight frame. The inner product between different vectors indexed by the same point a is the inner product between the corresponding columns of H(a) with the ﬁrst entries removed, giving

(a) (a) va j,va k = h h , j = k. (12.15) , , − 0, j 0,k Vectors indexed by different points a and b are both nonzero only in the β–entry, where β is the unique block containing both points, which gives

(a) (b) va j,vb k = h h , a = b. (12.16) , , β, j β,k We have just proved the following result. ν B ν 1 Theorem 12.4. (Steiner systems). For a (2,k, )–Steiner system , r = k−1 , and − one can construct an equiangular tight frame (va, j) of n = ν(r + 1) vectors for a ν(ν 1) space of dimension d = k(k −1) , via (12.14). Moreover, the inner products (scaled to have unit modulus) are given− by (12.15) and (12.16). 12.4 Equiangular tight frames from block designs 277

The frame (va, j) above is called a Steiner equiangular tight frame (constructed from r–simplices). It is subsimplicial of dimension r. A natural choice for the Hadamard matrices H(a) in the construction above is the Fourier matrix for an abelian group of order r + 1. In particular, if p1,..., p j are the (a) prime factors of r + 1 and m := p1 p2 p j, then one can choose H to have m–th roots of unity entries. The scaled inner products of the resulting Steiner equiangular tight frame are then m–th roots of unity and their negatives.

Example 12.20. (Real case) There exists a real Steiner equiangular tight frame if and only if there is a real Hadamard matrix of size r + 1, e.g., when r + 1 = 2 j.

Example 12.21. For the Steiner system S(2,2,4) of our worked example r = 3, and H(a) can be chosen to be one of

11 1 1 11 1 1 1 1 1 1 1 1 1 1 1 i 1 i =  − − ,  − − . 1 1 ⊗ 1 1 1 1 1 1 1 1 1 1 − −  − −   − −  1 1 1 1  1 i 1 i   − −   − −      The ﬁrst choice gives a real Steiner equiangular tight frame of 16 vectors in R6, i.e.,

1 1 1 1 1 1 1 100 0 000 0 0 − − − − 1 1 1 100 0 01 1 1 100 0 0  − − − −  1 1 1100 0 000 0 01 1 1 1 V =  − − − − . 00 0 011 1 1 1 1 1 100 0 0   − − − −  00 0 01 1 1100 0 011 1 1  − − − −  00 0 000 0 01 1 1 1 1 1 1 1     − − − −  Example 12.22. The collection B of all 2-element subsets of 1,...,ν , ν 2, is an S(2,2ν) Steiner system, with r + 1 = ν. The corresponding{ Steiner equiangular} ≥ ν2 1 ν ν ν tight frames have n = vectors in a space of dimension d = 2 ( 1). For = 3, taking the Fourier matrix gives −

1 1 0 1 1 1 1 ω ω2 1 ω ω2 0 0 0 AT = 1 0 1 , H(a) = 1 ω ω2 , V = 1 ω2 ω 00 01 ω ω2 ,     −→   0 1 1 1 ω2 ω 00 01 ω2 ω 1 ω2 ω             2πi where ω = e 3 .

Example 12.23. There are eight inﬁnite families of Steiner equiangular tight frames arising from known inﬁnite families of Steiner systems S(2,k,ν) (see [FMT12]).

We now use certain Steiner equiangular tight frames to obtain equiangular tight frames with more vectors in a higher dimension. 278 12 Equiangular and Grassmannian frames 12.5 Tremain equiangular tight frames

Let (va, j) be the Steiner equiangular tight frame given by a Steiner triple system B on ν points V (blocks of size three) and Hadamard matrices H(a), a V . We now outline a construction of [Tre08] (see [FJMP16]) which gives an equiangular∈ tight 1 ν ν 1 ν ν frame of 2 ( + 1)( + 2) vectors for a space of dimension 6 ( + 2)( + 3). β For the vectors va j C , a V , 1 j r + 1, given by (12.14), let , ∈ ∈ ≤ ≤ (a) B V v˜a j :=(va j,√2h ea,0) C C C. (12.17) , , 0, j ∈ ⊕ ⊕

α1 α2 αν+1 ν For a (complex) Hadamard matrix , α j C, w j C , let w1 w2 wν 1 ∈ ∈ +

1 3 B V w˜ℓ :=(0, wℓ, αℓ) C C C. (12.18) √2 2 ∈ ⊕ ⊕ 2 Since wℓ,wm = αℓαm, ℓ = m, wℓ = ν, the vectors (wℓ)1 ℓ ν+1 are said to be − ν ≤ ≤ the vertices of a unimodular simplex in C with complement (α j).

Theorem 12.5. (Tremain frames) Let (va, j) be the Steiner equiangular tight frame given by a Steiner triple system B on ν points V and Hadamard matrices H(a), and ν (wℓ) be the vertices of a unimodular simplex for C , with complement (αℓ). Deﬁne v˜a, j and w˜ℓ by (12.17) and (12.18). Then

(v˜a, j)a V ,1 j r+1 (w˜ℓ)1 ℓ ν+1 (12.19) ∈ ≤ ≤ ∪ ≤ ≤ 1 ν ν 1 ν ν 6 ( +2)( +3) is an equiangular tight frame of 2 ( + 1)( + 2) vectors for C , which is 1 ν subsimplicial of dimension r + 2 = 2 ( + 3). Proof. A calculation shows that the vectors are equiangular and tight, with angle α = 1 (see Exer. 12.10). r+2 ⊓⊔ Subsimplicial equiangular tight frames of the form (12.19) are known as Tremain (equiangular tight) frames. Steiner triple systems on ν points exist if and only if ν 1,3 (mod 6), ν 3. Thus (since Hadamard matrices of all sizes exist), we have ≡ ≥

1 ν ν There exists a Tremain equiangular tight frame of 2 ( +1)( +2) vectors for 1 ν 2 ν 3 C 6 ( + )( + ) if and only if ν 1,3 (mod 6), ν 3. ≡ ≥

Example 12.24. The construction gives a real equiangular tight frame provided there 1 ν ν exists a real Hadamard matrix of size r+1 = 2 ( +1) (and hence one of size +1). 1 ν ν Thus there is a Tremain equiangular tight frame of 2 ( + 1)( + 2) vectors for 1 ν ν 6 ( +2)( +3) 1 ν R if and only if there is a real Hadamard matrix of size 2 ( + 1), where 1 (ν + 1) 1,2 (mod 3), e.g., there are 820 equiangular lines in R287 (ν = 39). 2 ≡ 12.6 Equiangular frames and their signature matrices 279 12.6 Equiangular frames and their signature matrices

Since frames are determined up to unitary equivalence by their Gramian matrices, the Gramian of an equiangular frame Φ with C > 0 has the form

1 Cz12 Cz13 Cz1n Cz12 1 Cz23 Cz2n   Q = Cz13 Cz23 1 = I +CΣ, z jk = 1.   | |  . . ..   . . .    Cz Cz 1   1n 2n    We call any n n Hermitian matrix Σ of the above form, i.e., with zero diagonal and off diagonal× entries of modulus 1 a signature matrix. Since signature matrices are nonzero Hermitian matrices with zero trace, they have at least two eigenvalues: one negative and one positive. Let F stand for R or C. The n n signature matrices with entries from F are in 1–1 correspondence with the equiangular× frames of n vectors (which are not orthogonal bases). Theorem 12.6. (Signature matrices) Let Σ be an n n signature matrix (over F), with smallest eigenvalue λ of multiplicity n d, d × 1, then − − ≥ 1 1 Q := I + Σ = (Σ ( λ)I), r > 0 λ λ − − is the Gramian matrix of an equiangular frame of n vectors for Fd, and every Gramian of an equiangular frame of n > d vectors for Fd can be constructed in this way. Further, the frame is tight if and only if Σ has (exactly) two eigenvalues λ1,λ2, in which case

d(n 1) (n d)(n 1) λ1 = λ = − , λ2 = − − . (12.20) − − n d d − Proof. By construction, the matrix Q is positive semideﬁnite of rank d > 0, and so 1 2 has a positive square root B = Q 2 . Since Q = B = B∗B, Q is the Gramian matrix of the frame given by the columns of B (which span a d–dimensional subspace). Conversely, a frame of n > d vectors is tight if only if its Gramian has a zero eigenvalue of multiplicity n d and exactly one nonzero eigenvalue (see Exer. 2.17). − Hence an equiangular frame ( f j) of n > d vectors for a d–dimensional space is tight if and only if its signature matrix has exactly two eigenvalues. Moreover, by Exercise 2.16, we have

dA 1 A d(n d) 1 = f j, f j = , j, = f j, fk = − , j = k, n ∀ λ | | n n 1 − which gives the formula for λ = λ1. Since trace(Σ)=(n d)λ1 + dλ2 = 0, we have − − 280 12 Equiangular and Grassmannian frames

n d (n d)(n 1) λ2 = − λ1 = − − . − d d Thus we have (12.20). ⊓⊔ 1 1 Theorem 12.6 gives rise to a system of 2 n(n 1) equations in the 2 n(n 1) entries of Σ (see [BP05]). − −

Corollary 12.1. Let (z jk)1 j

0 z12 z13 z1n z12 0 z23 z2n   Σ = z13 z23 0    . . ..   . . .    z z 0   1n 2n    gives an equiangular tight frame of n > d vectors for Fd if and only if

2 n 1 Σ (λ1 + λ2)Σ (n 1)I = 0, λ1 + λ2 :=(n 2d) − , (12.21) − − − − d(n d) − which is equivalent to

j 1 k 1 n n 1 − − (n 2d) − z jk = ∑ z jz k + ∑ z j z k + ∑ z j zk , 1 j < k n. − d(n d) ℓ ℓ ℓ ℓ ℓ ℓ ≤ ≤ − ℓ=1 ℓ= j+1 ℓ=k+1 (12.22)

Proof. The signature matrix has two eigenvalues λ1 = λ (with multiplicity n d) − − and λ2 if and only if it satisﬁes the minimal polynomial

2 Σ (λ1 + λ2)Σ + λ1λ2I = 0, (12.23) − where λ1λ2 = n 1. From the entries of the matrix equation (12.23), we therefore 2 − obtain n equations in the z jk, with real coefﬁcients depending only on n and d. Those from the diagonal entries hold automatically, and since the ( j,k) and (k, j) entries are complex conjugates, we obtain the equivalent system

2 (λ1 + λ2)z jk =(Σ ) jk, 1 j < k n, ≤ ≤ which can be written as (12.22). ⊓⊔

Example 12.25. For n = 4, d = 2, (12.22) gives 6 equations. Let z12 = a, z13 = b, z14 = c. Then the ( j,k)=(1,2) and (1,3) equations are z13z23 + z14z24 = 0, z13z23 + z14z34 = 0 = z24 = bcz23, z34 = acz23. ⇒ − −

Making the above substitutions for z24 and z34 reduces the other 4 equations to one 12.6 Equiangular frames and their signature matrices 281

2 (abz23) = 1 = z23 = iab. − ⇒ ± Hence there is a three parameter family of unitarily inequivalent equiangular tight frames of four vectors for C2 given by the signature matrices

0 a b c a 0 iab iac Σ =  ± ∓ , a = b = c = 1. (12.24) b iab 0 ibc | | | | | |  ∓ ±  c iac ibc 0   ± ∓    If an n n signature matrix Σ satisﬁes Σ 2 = µΣ + τI for some µ,τ, then µ is real and τ ×= n 1 (by considering entries), so that Σ has two eigenvalues with sum µ. Therefore, Corollary− 12.1 can be stated in the following convenient form. Theorem 12.7. (Characterisation) Let Σ be an n n signature matrix. Then the following are equivalent × 1. Σ is the signature matrix of an equiangular tight frame for Cd. 2. Σ 2 =(n 1)I + µΣ for some necessarily real µ. 3. Σ has exactly− two eigenvalues (with sum µ). Further, when these hold, we have n nµ d = . (12.25) 2 − 2 4(n 1)+ µ2 − Proof. Solving n 1 µ2 =(n 2d)2 − − d(n d) − for d gives n nµ d = . 2 ± 2 4(n 1)+ µ2 − Substituting this formula for d back into

n 1 µ =(n 2d) − , − d(n d) − shows that the choice must be made. − ⊓⊔

Example 12.26. (Hermitian complex Hadamard matrices) If µ = λ1 + λ2 = 2, so that Σ 2 + 2Σ =(n 1)I, then Σ + I is a Hermitian complex Hadamard matrix,− via the calculation −

2 2 (Σ + I)(Σ + I)∗ =(Σ + I) = Σ + 2Σ + I =(n 1)I + I = nI. − This example generalises to a correspondence between the signature matrices with µ 2 and the complex Hadamard matrices with constant diagonal [Szo13].¨ | |≤ 282 12 Equiangular and Grassmannian frames

Corollary 12.2. (Complex Hadamard matrices) Let Σ be an n n signature matrix. Then the following are equivalent ×

1. Σ gives an equiangular tight frame for Cd, where n √n d n + √n . 2 − 2 ≤ ≤ 2 2 2. Σ 2 =(n 1)I + µΣ, where 2 µ 2. 3. Σ + ζI is− a complex Hadamard− ≤ matrix≤ for ζ = 1 ( µ 4 µ 2i). 2 − ± −| | Proof. The equivalence of the ﬁrst two follows from the observation that

n 1 n 2 n µ2 =(n 2d)2 − 4 d . − d(n d) ≤ ⇐⇒ − 2 ≤ 4 − (2 = 3) Suppose that 2 holds, then ζ = 1 ( µ 4 µ 2i) is a unit modulus ⇒ 2 − ± −| | complex number, with ζ + ζ = µ, and − 2 (Σ + ζI)∗(Σ + ζI)=(Σ + ζI)(Σ + ζI)= Σ +(ζ + ζ)Σ + I = ((n 1)I + µΣ) µΣ + I = nI, − − i.e., Σ + ζI is a complex Hadamard matrix. (3 = 2) Suppose that 3 holds, then ζ = 1 ( µ 4 µ 2i) has unit modulus, ⇒ 2 − ± −| | and 2 µ = (ζ + ζ)= 2ℜ( ζ) 2, which gives − ≤ − − ≤ 2 Σ =(Σ + ζI)∗(Σ + ζI) (ζ + ζ)Σ I = nI + µΣ I =(n 1)I + µΣ. − − − −

⊓⊔ Example 12.27. For a SIC, i.e., d2 equiangular lines in Cd, the signature matrix satisﬁes Σ 2 = I when d = 2(µ = 0), and Σ 2 = 2I + 2Σ when d = 3(µ = 2).

Example 12.28. Let Σ be the signature matrix for the equiangular harmonic frame of 7 vectors for C3 given by the cyclic difference set J = 1,2,4 . Here µ = 1 , { } √2 and we obtain the constant diagonal Hadamard matrix

ζζζ ζ ζ ζ ζ ζζζζ ζ ζ ζ   ζ ζζζζ ζ ζ   1 √7 Σ + ζI = ζ ζ ζζζζ ζ, ζ := + i,   − √ √ ζ ζ ζ ζζζζ 2 2 2 2   ζ ζ ζ ζ ζζζ     ζ ζ ζ ζ ζ ζ ζ     1 2πi where ζ = (ω + ω2 + ω4), ω = e 7 . √2 For V a complex Hadamard matrix, we now deﬁne a variant of the Kronecker product of V and V ∗ which yields Hadamard matrices with constant diagonal. 12.6 Equiangular frames and their signature matrices 283

Theorem 12.8. Let V =[v1,...,vm] be a complex Hadamard matrix of order m. Then v1v1∗ vmv1∗ . . K =  . .  (12.26)

v1vm∗ vmvm∗      is a Hermitian complex Hadamard matrix of order m2, with 1’s on its diagonal. Thus Σ = (K I) is the signature matrix of equiangular tight frame of n = m2 vectors for C±d, where− d = 1 m(m 1). 2 ± Proof. By construction K is Hermitian with entries of unit modulus, and 1’s on its diagonal. The ( j,k)–block of K2 is

m2I, j = k; ∑vℓv∗vkv∗ =(v∗vk)∑vℓv∗ =(v∗vk)VV ∗ =(v∗vk)mI = j ℓ j ℓ j j 0, j = k ℓ ℓ i.e., K = Σ + I is a complex Hadamard matrix. Corollary 12.2 implies that Σ is the signature matrix of an equiangular tight frame of n = m2 vectors for Cd, where µ = 2, and, by (12.25), − n nµ n √n 1 d = = + = m(m + 1). 2 − 2 4(n 1)+ µ2 2 2 2 − 2 1 1 The complementary equiangular tight frame for d = m 2 m(m + 1)= 2 m(m 1) dimensions has signature matrix (K I). − − − − ⊓⊔ Example 12.29. (Cube root signature matrices) Taking V to be the 3 3 Hadamard matrix given by the Fourier matrix gives ×

111111111 1 1 1 ω ω ωω2 ω2 ω2   1 1 1 ω2 ω2 ω2 ω ω ω  2 2 2 1 1 1 1 ω ω 1 ω ω 1 ω ω  2πi   V = 1 ω ω2 , ω = e 3 K = 1 ω2 ω ω2 1 ω ω ω2 1    −→   ω2 ω  ω2 ω ω ω2 ω2 ω  1 1 1 1     ω ω2 ω2 ω ω2 ω    1 1 1    1 ω ω2 ω2 ω 1 ω 1 ω2   1 ω ω2 ω 1 ω2 ω2 ω 1      with K I the signature matrix for an equiangular tight frame of 9 vectors in C6, and K−+ I the signature matrix for an equiangular tight frame of 9 vectors in C3. − 284 12 Equiangular and Grassmannian frames 12.7 The reduced signature matrix and projective unitary equivalence

Recall from 2.3 (and 8.1) that frames Ψ are Φ are projectively unitarily equivalent if and only if§ their Gramians§ satisfy

Gram(Ψ)= Λ ∗ Gram(Φ)Λ, Λ = diag(α j), where α j = 1, j. For equiangular frames, this condition can be expressed in terms of the signature| | ∀ matrices ΣΨ = Λ ∗ΣΦΛ. Hence every equiangular frame is projectively unitarily equivalent to one with a signature matrix of the form

ΛΣΛ 0 1∗ Λ ∗ = Σˆ , := diag(1,z12,z13,...,z1n), 1 :=(1,1,...,1)∗. 1 (12.27) We will call the above matrix Σˆ the reduced signature matrix of the equiangular frame (and its projective unitary equivalence class).

An equiangular frame is uniquely determined by its reduced signature matrix.

Theorem 12.9. (Triple products) An equiangular frame is uniquely determined up to projective unitary equivalence by its reduced signature matrix.

Proof. Each equiangular frame Φ has a reduced signature matrix Σˆ . The matrix

Σ 0 1∗ := Σˆ 1 is the signature matrix of an equiangular frame Ψ which is projectively unitarily equivalent to Φ. It therefore sufﬁces to show that the entries of the above Σ are projectively unitarily invariant. Without loss of generality, assume that Φ =(v j) has be scaled so that v j,vk = Σkj. Then

v1,v j = 1, j. ∀ Thus the (k, j)–entry of Σ can be written as

v j,vk = v1,v j v j,vk vk,v1 .

Clearly, the “triple products” on the right hand side depend only on (v j) up to projective unitary equivalence (see for the general theory). § ⊓⊔ 12.7 The reduced signature matrix and projective unitary equivalence 285

Example 12.30. The reduced signature matrix for the Σ of (12.24) is

0 i i ± ∓ Σˆ = i 0 i . (12.28) ∓ ±  i i 0 ± ∓    Thus there are just two equiangular tight frames of four vectors in C2 up to projective unitary equivalence (and reordering). Moreover, these can be obtained from each other by taking the entrywise complex conjugate of the reduced signature matrix.

Example 12.31. The complement of a tight equiangular frame is a tight equiangular frame. If its signature and reduced signature matrices are Σ and Σˆ , then those of the complementary equiangular tight frame are Σ and Σˆ . − − By Theorem 12.6, the construction of equiangular frames with large numbers of vectors (compared to the dimension) is equivalent to ﬁnding signature matrices Σ whose smallest eigenvalue has a large multiplicity. We now seek to translate this to the reduced signature matrix Σˆ .

Proposition 12.2. (Spectral structure of Σˆ ) Let Σˆ be the reduced signature matrix of a Σ satisfying (12.27), and v be a λ–eigenvector of Σˆ . Then

(0,v) is a λ–eigenvector of ΛΣΛ ∗ if and only if v 1. • (a,1) is an eigenvector of ΛΣΛ if and only if 1 is⊥ a β–eigenvector of Σˆ and • ∗ 1 a = β β 2 + 4(n 1) . (12.29) 2 − ± − Proof. The ﬁrst follows since

0 1∗ 0 1∗v 1∗v Σˆ = Σˆ = λ . 1 v v v For the second, observe that

0 1∗ a 1∗1 n 1 Σˆ = Σˆ = −Σˆ . 1 1 a1 + 1 a1 + 1

Thus (a,1) can be can be an eigenvector only if a1 + Σˆ 1 is a multiple of 1, i.e., 1 is a β–eigenvector of Σˆ . In this case, (a,1) is a α–eigenvector if and only if

αa = n 1, (a + β)= α, − which gives (12.29), where α = a + β. ⊓⊔ 286 12 Equiangular and Grassmannian frames

Since the eigenspaces of the Hermitian matrix Σˆ are orthogonal, we obtain a correspondence between the eigenstructures of Σ and Σˆ when 1 is a µ–eigenvector of Σˆ . Example 12.32. The reduced signature matrix Σˆ of (12.28) has eigenvector 1 for eigenvalue β = 0, and the other eigenvalues are √3. Thus the eigenvalues of any Σ (of size n = 4) with this reduced signature matrix± are 1 α = a + β = β β 2 + 4(n 1) = √3, 2 ± − ± together with √3 (which happen to be the same). ± The factorisation (12.27) allows the number of variables in (12.22) to be reduced by n 1 to 1 (n 1)(n 2). We now express these equations in terms of the reduced − 2 − − signature matrix Σˆ , and show that 1 must be an eigenvector of Σˆ .

Proposition 12.3. There exists an equiangular tight frame of n > d vectors for Fd with reduced signature matrix Σˆ if and only if

2 n 1 (λ1 + λ2)Σˆ = Σˆ + J (n 1)I, λ1 + λ2 :=(n 2d) − , J := 11∗, − − − d(n d) − and 1 is an eigenvector of Σˆ for the eigenvalue λ1 + λ2.

Σ 0 1∗ Proof. Substitute = Σˆ into (12.23) 1

n 1 1 Σˆ 0 1 1 0 ∗ λ λ ∗ n 1 0 Σ−ˆ Σˆ 2 ( 1 + 2) Σˆ ( ) = , 1 J + − 1 − − 0 I and equate the blocks. ⊓⊔ Example 12.33. (The simplex) Up to projective unitary equivalence and reordering, there is a unique equiangular tight frame of n = d + 1 vectors for Rd or Cd, which is given by the vertices of the regular simplex. For n = d + 1, the d d reduced × signature matrix Σˆ has 1 as an eigenvector for λ1 + λ2 = (d 1), and so all its off diagonal entries must be 1. In particular, there is a unique− equiangular− tight frame of three vectors in C2, which− is a real frame (the Mercedes–Benz frame). In summary:

Each equiangular frame is uniquely determined up to unitary equivalence by its signature matrix, and it is uniquely determined up to projective unitary equivalence by its reduced signature matrix. 12.8 The relative bound on the number of real equiangular lines 287 12.8 The relative bound on the number of real equiangular lines

d Equiangular lines in R are given by a set of unit vectors f j with { }

f j, f j = α = cosθ, j = k. | | The variational characterisation of tight frames (see Theorem 6.1) immediately gives the following estimate on the maximum number of equiangular lines.

Theorem 12.10. (Relative bound) Suppose that f j gives n equiangular lines in { } Rd with angle α = cosθ. If α < 1/√d, then

d dα2 n − , (12.30) ≤ 1 dα2 − d with equality if and only if f j is a tight frame for R . { } Proof. By the variational characterisation, we have

2 2 2 2 1 2 1 2 ∑∑ f j, fk = n +(n n)α n = ∑ f j , j k | | − ≥ d d j with equality if and only if f j is a tight frame. The above inequality is equivalent to { } 1 n( α2) 1 α2. d − ≤ − Hence if α < 1/√d, then we can divide (without changing the sign) to obtain the result. ⊓⊔ Example 12.34. The isogonal vectors of Example 3.9 give d equiangular lines in Rd with 1 α < 1. √d ≤ We observe that the relative bound is not a generalisation of the absolute bound (Theorem 12.2). Indeed, it also holds (by the proof) for complex equiangular lines. In the algebraic graph theory literature (12.30) is called the relative bound, and the tight frame condition is usually stated as the orthogonal projections Pj := f j f j∗ satisfy n n ∑ Pj = I. j=1 d

Thus tightness (equality) in (12.30) occurs if and only the frame f j is tight. { }

We say that a set of equiangular lines in Rd is tight if the vectors deﬁning them are a tight frame, equivalently, there is equality in the relative bound.

The relative bound is very useful when used in conjunction with the following. 288 12 Equiangular and Grassmannian frames

Theorem 12.11. Suppose there are n equiangular lines in Rd with angle α = cosθ. If n > 2d, then 1/α is an odd integer. Proof. Let Q = I +αΣ be the Gramian of n unit vectors giving the equiangular lines d in R . Since these vectors span a subspace of dimension d′ d, by Theorem 12.6, the signature matrix ≤ 1 Σ = Q I α − has smallest eigenvalue 1/α, with multiplicity m = n d′ n d. Since Σ is an integer matrix, its eigenvalues− are algebraic integers. Th−e eigenvalue≥ − 1/α must be rational, and hence is an integer. Otherwise, the conjugate of 1/α− would be a second eigenvalue with multiplicity m, so n 2m 2(n d), which− contradicts n > 2d. ≥ ≥ − To see 1/α is odd, consider the integer matrix 1 B = Σ + 11 + I . 2 ∗ Since Σ has a 1/α eigenspace of dimension m n d > 2d d 1 and 11∗ has a − ≥ β− 1 1− ≥ 0 eigenspace of dimension n 1, the matrix B has a = 2 α +0+1 eigenvalue. Since β is a rational algebraic− integer, it is an integer, and− so 1/α = 2β + 1 is odd. − ⊓⊔ Unlike the relative bound, Theorem 12.11 does not hold for complex equiangular lines in Cd, e.g., for n = d2 such lines 1/α = √d + 1. Example 12.35. If the absolute and relative bounds hold for n > 2d, then

1 2 d = 2, where 1 is an odd integer, α − α i.e., d = 7,23,47,79,.... Thus the absolute bound (Theorem 12.2) for the maximal number of equiangular lines in Rd can only hold when d = 2,3 or d +2 is the square of an odd integer. The maximum size M(d) of a set of n real equiangular lines in Rd has been investigated since [Haa48] showed that M(2)= 3 (the Mercedes–Benz frame) and M(3)= 6 (lines through the opposite vertices of an icosahedron). As of [BY14], [GKMS16], the best estimates of M(d), d 47 are given in Table 12.3. We observe (see Example 12.41) that the≤ ﬁrst instance where M(d) is known to be not attained by a tight frame (in a possibly lower dimension) is d = 16. 2 2 d A nontight set of 9 (d +1) equiangular lines for C was constructed by de Caen [dC00] for d = 3 4t 1, t 1. By using the existence of a set of d + 1 mutually 2 − ≥ 2 unbiased bases in Rd for d = 4t , t 1, [GKMS16] adapted this example to construct nontight equiangular lines in Cd giving≥ the following lower bound

32d2 + 328d + 296 (d + 2)2 M(d) > . ≥ 1089 72 12.9 The connection with algebraic graph theory 289

Table 12.3: The maximum number M(d) of equiangular lines in Rd , d 47. The last column gives the number of vectors in tight frame in Rd given by a strongly regular≤ graph (see §12.10). When the existence of the graph (and hence the equiangular lines) is still unknown, then it is in grey.

1 1 d M(d) α tight d M(d) α tight 2 3 2 3 19 72 75 5 − 3 6 √5 6 20 90 95 5 − 4 6 √5,3 21 126 5 126 5 10 3 10 22 176 5 176 6 16 3 16 23 41 276 5 276 − 7 13 28 3 28 42 276 288 5,7 288 − − 14 28 29 3,5 43 344 7 344 − 15 36 5 36 44 344 422 7 − 16 40 41 5 45 344 540 7 540 − − 17 48 50 5 46 344 736 7 736 − − 18 54 60 5 47 344 1127 7 − −

12.9 The connection with algebraic graph theory

The Seidel matrix Σ = Seid(Γ ) of a graph Γ with n vertices is the n n matrix with a 1 in the ( j,k)–entry if the j and k vertices are adjacent (connected× by an edge), a− 1 if they are nonadjacent, and 0 diagonal entries. Clearly, Seidel matrices are signature matrices over R, and vice versa. For F = R, there are finitely many possible n n signature matrices, and hence finitely many real equiangular frames of n vectors,× each in 1–1 correspondence with a graph on n vertices, namely the graph whose Seidel matrix is its signature matrix. This correspondence between vectors defining a set of real equiangular lines and a graph goes back to the foundation of algebraic graph theory (see [GR01]). d The set of n vectors f j defining a set of equiangular lines in R can each { } be multiplied by σ j = 1 to obtain a set defining the same equiangular lines, but possibly with different Gramian± matrices

Gram( σ j f j )= Λ Gram( f j )Λ, Λ := diag(σ1,...,σn), (12.31) { } { } and hence possibly different corresponding graphs. Any two graphs related in this way (for some ordering of their points) are said to be switching equivalent (by switching on the vertices j : σ j = 1 ). This is an equivalence relation. Switching on a vertex j of a graph{ entails changing− } all edges from j to nonedges, and all nonedges from j to edges. The set of all graphs switching equivalent to Γ is called 290 12 Equiangular and Grassmannian frames

2 1 the switching class of Γ In view of (12.31), Seid( σ j f j )= Λ − Seid( f j )Λ, and so the eigenvalues of the Seidel matrices of any two{ switc}hing equivalent{ graphs} are the same (reindexing a set of vectors does not change the eigenvalues of their Gramian). The eigenvalues of the Seidel matrix are called the Seidel spectrum of the graph (or its switching class).

Each real equiangular frame which is not a basis is uniquely determined up to projective unitary equivalence and reordering by a switching class of graphs (the one containing a graph whose Seidel matrix is its signature matrix). It is tight if and only if its Seidel matrix has exactly two eigenvalues.

Example 12.36. (The simplex) Let Γ be the complete graph Kn on n 2 vertices. Then its Seidel matrix Σ has two eigenvalues: (n 1) of multiplicity≥ 1, and 1. The n −1 − corresponding tight frame of n vectors in R − (cf. Example 12.33) is the vertices of a regular simplex. Similarly, its complement, the empty graph on n vertices, gives the equiangular tight frame for R1 consisting of a nonzero vector repeated n times. By switching on a vertex of Kn on obtains the graph consisting of a point together with Kn 1 For n = 3, Γ is the 3–cycle (see Figure 12.1). Its Seidel matrix S, and those obtained− by switching on the ﬁrst, second and third vertices are

0 1 1 0 1 1 0 1 1 0 1 1 − − − − S =  1 0 1, 1 0 1,  1 0 1 ,  1 0 1. − − − −          1 1 0  1 1 0   1 1 0   1 1 0 − −   −  −           

Fig. 12.1: The graphs in the switching class of the complete graph K3, and the corresponding equiangular (tight) frames of three vectors in R2.

Example 12.37. For n = 4 the eleven graphs lie in three switching classes. The one containing the complete graph gives the vertices of the regular tetrahedron, the one containing the empty graph gives equiangular lines in R1. The third switching class, which contains the path graph, gives four lines at an angle of cos 1( 1 ) 63.4 , − √5 ≈ ◦ which we recognise as diagonals of the regular icosahedron (see Example 12.44).

2 The switching class is also known as a two–graph. 12.10 Real equiangular tight frames and strongly regular graphs 291 12.10 Real equiangular tight frames and strongly regular graphs

Let Σ be the Seidel matrix of a graph Γ , and Σˆ given by (12.27) be its reduced Seidel matrix. The condition that Σ have two eigenvalues, and hence give a real equiangular tight frame, is most easily expressed in terms of the graph with Seidel matrix Σˆ (which can depend on the ordering of the points of Γ ). This graph must be strongly regular, with particular parameters (Theorem 12.12). A regular graph of degree k with ν vertices is said to be strongly regular, or a srg(ν,k,λ, µ), if there are integers λ and µ such that Every two adjacent vertices have λ common neighbours. • Every two non-adjacent vertices have µ common neighbours. • The adjacency matrix A = Adj(Γ ) (which has a 1 for adjacency, and a 0 otherwise) of a strongly regular graph Γ which is not complete or empty is characterised by

AJ = kJ, A2 +(µ λ)A +(µ k)I = µJ, (12.32) − − where J = Jν is the ν ν matrix of all 1’s and I = Iν is the identity. The eigenvalues of× the adjacency matrix A are

λ µ + √∆ λ µ √∆ θ = − , τ = − − , ∆ :=(λ µ)2 + 4(k µ), (12.33) 2 2 − − with multiplicities

1 2k +(ν 1)(λ µ) 1 2k +(ν 1)(λ µ) mθ = ν 1 − − , mτ = ν 1+ − − , 2 − − √∆ 2 − √∆ (12.34) and k with eigenvector 1 (see [GR01]). Since the trace of A is zero, its smallest eigenvalue is τ < 0. We note that the adjacency and Seidel matrices A = Adj(Γ ) Σ Γ Γ 1 Σ Σ and = Seid( ) of a graph are related by A = 2 (J I ), = J I 2A. The spectral structure of one can be deduced from the other− − when Γ is k−–regular,− i.e., 1 =(1,...,1) is an eigenvector of A for eigenvalue k (cf. Proposition 12.2). Theorem 12.12. Let Σ be the Seidel matrix of a graph Γ on n vertices which is not switching equivalent to the complete or empty graph, and Σˆ be given by (12.27). Then Σ has two eigenvalues (and so corresponds with an equiangular tight frame of n > d + 1 vectors for Rd) if and only if Σˆ is the Seidel matrix of a strongly regular graph Γˆ of the type 3k n k srg(n 1,k,λ, µ), λ = − , µ = . (12.35) − 2 2 The n, k, d above are related as follows

1 1 n(n 2k 2) 1 n d(n 1) d = n − − > 1, k = n 1+ 1 − . (12.36) 2 − 2 (n 2k)2 + 8k 2 − − 2d n d − − 292 12 Equiangular and Grassmannian frames

Proof. By Proposition 12.3, Σ has two eigenvalues (i.e., gives rise to an equiangular d tight frame for R ), if and only if 1 is an eigenvector of Σˆ for the eigenvalue λ1 +λ2, and 2 (λ1 + λ2)Σˆ = Σˆ + J (n 1)I, J := 11∗. (12.37) − − 1 Σˆ Γˆ Thus 1 is an eigenvector of the adjacency matrix A = 2 (J I ) of for the eigenvalue − −

(n 1) 1 (λ1 + λ2) 1 n d(n 1) k = − − − = n 1 + 1 − , (12.38) 2 2 − − 2d n d − with k a positive integer (since the nonzero entries of A are 1). Hence Γˆ is a regular graph of degree k, which is not the complete or empty graph (by our assumption). Using Σˆ = J I 2A, AJ = JA = kJ and (12.38), we can rewrite (12.37) as − − n k k 3k n k k A2 +( k)A I = A2 +( − )A +( k)I = J, 2 − − 2 2 − 2 2 − 2 Γˆ λ µ λ 1 µ k which is equivalent to being a srg(n 1,k, , ), = 2 (3k n), = 2 . Finally, solving (12.38) for d gives − −

1 1 n(n 2k 2) d = n − − , 2 ± 2 (n 2k)2 + 8k − with the choice of sign determined by the multiplicities of the eigenvalues of Σ. More precisely, d is the multiplicity of the largest eigenvalue of Σ, which by Propo- 1 Σˆ sition 12.2 and A = 2 (J I ) is 1 plus the multiplicity of the smallest eigenvalue of A, i.e., − − 1 2k +(n 2)( 1 (2k n)) 1 + n 2 + − 2 − , 2 − 1 (2k n)2 + 4 1 k 4 − 2 which simpliﬁes to the formula for d in (12.36). ⊓⊔ The linear equations of (12.35) can be solved for n and k, giving

n = 2λ + 6µ, k = 2µ. − In this way, the equiangular tight frames be indexed by the pair of integers (λ, µ). Example 12.38. (Complements) Let Σˆ be the reduced signature matrix of an equiangular tight frame of n > d +1 vectors for Rd, which corresponds to a strongly regular graph with the parameters (12.35). The complementary equiangular tight frame n d of n vectors for R − has reduced signature matrix Σˆ (see Exer. 12.14), and so corresponds to the complementary strongly regular graph,− which has the parameters

2n 3k 6 n k 2 srg(n 1,n k 2, − − , − − ). − − − 2 2 where k is given by (12.36). 12.10 Real equiangular tight frames and strongly regular graphs 293

By Theorem 12.12, the existence and construction of equiangular tight frames in Rd can be expressed in terms of strongly regular graphs with certain parameters. Strongly regular graphs which are the union m complete graphs on w vertices, and their complements, are considered boring. The strongly regular graphs which are not boring are said to be primitive. These graphs have parameters

srg(mw,w 1,w 2,0), srg(mw,(m 1)w,(m 2)w,(m 1)w), − − − − − and so give rise to equiangular tight frames (as above) if and only if m = w = 1.

Corollary 12.3. There exists an equiangular tight frame of n > d +1 vectors for Rd if and only if there exists a strongly regular graph Γˆ, with Seidel matrix Σˆ , of the type

3k n k 1 n d(n 1) srg(n 1,k, − , ), k := n 1 + 1 − . − 2 2 2 − − 2d n d − Moreover, all graphs Γ giving an equiangular tight frame of n > d + 1 vectors for Rd have Seidel matrices of the form

1 1 0 1∗ P− Λ − ΛP, Λ = diag(σ1,...,σn), σ j = 1, (12.39) 1 Σˆ  ±   where P is a permutation matrix and Σˆ is as above. In particular, we can take Γ to be Γˆ together with an isolated point.

Example 12.39. The switching class of the graph Γ obtained by adding an isolated point to the unique srg(5,2,0,1), i.e., the 5–cycle, is given in the Figure 12.2. The corresponding tight frame of n = 6 vectors for R3 consists of vectors which are on the six diagonals of the regular icosahedron.

Fig. 12.2: The switching class of the graph Γ consisting of a 5–cycle and an isolated point.

We observe that graph Γ consisting of a strongly regular graph Γˆ together with an isolated point is switching equivalent to Γˆ together with a point which is adjacent to all points of Γˆ. 294 12 Equiangular and Grassmannian frames

The switching class of a graph Γ giving an equiangular tight frame of n vectors is called a regular two–graph. It may contain graphs consisting of an isolated point (one with no neighbours) together various nonisomorphic strongly regular graphs on n 1 points (called neighbourhoods). This is sometimes used as a method to construct− strongly regular graphs. Thus reordering the vectors of an equiangular tight frame may yield a reduced signature matrix for a different strongly regular graph (with the same parameters). Since the unique equiangular tight frame of n = d + 1 vectors in Rd (the vertices of the simplex) is given by the boring strongly regular graph

Γˆ = Kn 1 = srg(n 1,n 2,n 3,0) (the complete graph), − − − − we can summarise our results as follows.

Up to projective unitary equivalence and reordering, each equiangular tight frame (which isn’t an orthonormal basis) corresponds to a collection of strongly regular graphs. These strongly regular graphs have the property that the graphs obtained by adding an isolated point are switching equivalent.

A list of the known real equiangular tight frames (for d 50) obtained by this correspondence is given in Table 12.5. ≤ The construction of Theorem 12.12 also gives nontight equiangular frames.

Corollary 12.4. Let Σˆ be the Seidel matrix of a strongly regular graph Γˆ of the type 3k n srg(n 1,k,λ, µ), λ = − . (12.40) − 2 Then there exists a nontight equiangular tight frame of n > d +1 vectors for Rd with reduced signature matrix Σˆ if and only if 1 1 λ + µ √∆ < n 2 2k (n 2 2k)2 + 4(n 1) , − − − 2 − − − − − − where ∆ :=(λ µ)2 + 4(k µ), and − − 1 2k +(n 2)(λ µ) d = n + 2 + − − . 2 √∆ Proof. Let ν = n 1, then by (12.33), (12.34) and Proposition 12.2, the eigenvalues of Σ are 1 2θ−= 1 λ + µ √∆, 1 2τ = 1 λ + µ + √∆, and − − − − − − − − − 1 β + β β 2 + 4(n 1) , β := n 2 2k = ν 1 2k, 2 − ± − − − − − with multiplicities mθ ,mτ ,1,1. The condition for 1 2θ to be the smallest eigenvalue is − − 12.11 Conditions for the existence of real equiangular tight frames 295 1 1 2θ = 1 λ + µ √∆ β β 2 + 4(n 1) . − − − − − ≤ 2 − − If there is equality, then the frame is tight, and if the inequality doesn’t hold, then Σ has rank n 1, and so corresponds to the n = d + 1 vertices of the simplex. For − strict inequality n d = mθ , which completes the proof. − ⊓⊔

Table 12.4: Selected examples of nontight equiangular frames of n > d +1 vectors for Rd given by Corollary 12.4, including the Seidel spectrum of Σ.

n d srg parameters eigenvalues multiplicities 11 6 (10,3,0,1) 3,3, 2,5 5,4,1,1 − − 17 7 (16,5,0,2) 3,5, 1 (1 √89) 10,5,1,1 − 2 ± 17 11 (16,6,2,2) 5,3 1 (3 √73) 6,9,1,1 − 2 ± 27 14 (26,10,3,4) 5,5, 1 (5 √129) 13,12,1,1 − 2 ± 37 14 (36,14,4,6) 5,7, 1 (7 √193) 21,14,1,1 − 2 ± 41 17 (40,12,2,4) 5,7, 1 (15 √385) 24,15,1,1 − 2 ± 65 20 (64,18,2,6) 5,11, 1 (27 √985) 45,18,1,1 − 2 ±

12.11 Conditions for the existence of real equiangular tight frames

For tight real equiangular lines, we reﬁne the absolute bound and Theorem 12.11. Theorem 12.13. Suppose d > 1. Then a necessary condition for an equiangular tight frame of n > d + 1 vectors for Rd to exist is that n be even, and 1 1 n min d(d + 1), (n d)(n d + 1) , (12.41) ≤ {2 2 − − } or, equivalently,

2d + 1 + √8d + 1 1 d + 2 < n d(d + 1). (12.42) 2 ≤ ≤ 2 Moreover, for n = 2d, one must have λ d(n 1) λ (n d)(n 1) Σ The eigenvalues 1 = n −d , 2 = − d − of the signature matrix • − − are odd integers. n 1 is odd, but not a prime. • 1 −n2(n 1) 1 (n 2d)2(n 1) 4 d(n −d) and 4 −d(n d)− are perfect squares. • − − 296 12 Equiangular and Grassmannian frames

Proof. Since the complement of an equiangular tight frame of n vectors for Rd is n d one for R − , from the bound (12.5) of Theorem 12.2, we get (12.41). Now 1 n (n d)(n d + 1) f (n) := n2 (2d + 1)n + d2 d 0. ≤ 2 − − ⇐⇒ − − ≥ The quadratic f is non-negative if and only if n is less than its smallest root or greater than its largest, i.e., 1 1 1 1 n d + √8d + 1 < d, n d + + √8d + 1, ≤ 2 − 2 ≥ 2 2 which gives (12.42). The eigenvalues λ1,λ2 of Σ are given by (12.20), and are real algebraic integers with multiplicities n d,d 2. By Theorem 12.12, Proposition 12.2 and the spectral − ≥ structure of strongly regular graphs, λ1,λ2 are eigenvalues of Σˆ with multiplicities n d 1,d 1 1, and the corresponding eigenvectors are orthogonal to 1. Hence, − − − ≥ Σˆ 1 Σˆ the adjacency matrix of the graph given by , i.e., A := 2 (J I ) has eigenvalues 1 λ 1 λ − − 2 ( 1 1), 2 ( 1 2) with multiplicities n d 1,d 1 1. −For−n = 2d,− the− pairs of multiplicities above− are− not− equal,≥ and so the pairs of eigenvalues are algebraic integers which are not algebraic conjugates, A simple calculation shows λ1 = 1, λ2 = 1. The three conditions then follow from the calculations − 1 λ j λ j = 2− − 1, λ1λ2 = (n 1), − 2 − − − λ λ 2 2 1 1 1 2 1 2 1 n (n 1) − − − − = (λ2 λ1) = − , 2 − 2 4 − 4 d(n d) − 2 1 λ1 1 λ2 2 1 1 (n 2d) (n 1) − − + − − = (λ + λ )2 = − − . 2 2 4 1 2 4 d(n d) − ⊓⊔ For n = 2d, the eigenvalues of Σ are √2d 1. In this case, we obtain an inﬁnite family of equiangular lines, corresponding± to the− conference graphs. An n n matrix C is a conference matrix if its diagonal entries are 0, its off × diagonal entries are 1, and C∗C =(n 1)I. The reduced signature matrix Σˆ of a symmetric conference± matrix C = Σ (for− which n must be even) is a strongly regular graph n 2 n 6 n 2 srg n 1, − , − , − , − 2 4 4 called a conference graph. Equivalently, conference graphs are the strongly regular graphs srg(ν,k,λ, µ) with

2k +(ν 1)(λ µ)= 0, − − i.e., the multiplicities (12.34) of the eigenvalues θ,τ of the adjacency matrix equal. We now characterise equiangular tight frames of n = 2d vectors for Rd. 12.12 A list of real equiangular tight frames 297

Proposition 12.4. For d 3, the following are equivalent ≥ 1. There is an equiangular tight frame of 2d vectors for Rd. 2. There exists a 2d 2d symmetric conference matrix. 3. There exists a conference× graph with 2d 1 vertices, i.e., a − d 3 d 1 srg(2d 1,d 1, − , − ). − − 2 2 For these to hold it is necessary that d be odd, and 2d 1 be a sum of two squares. − Proof. Suppose that n = 2d, d 3. Let Σˆ be the reduced signature matrix of an n n signature matrix Σ. By Corollary≥ 12.1 and Proposition 12.3, the following conditions× are equivalent to Σ giving an equiangular tight frame of n = 2d vectors for Rd Σ 2 (n 1)I = 0, − − Σˆ 1 = 0, Σˆ 2 + J (n 1)I = 0. − − The ﬁrst says that Σ is a symmetric conference matrix. Writing the second in terms of the adjacency matrix A = 1 (J I Σˆ ) gives 2 − − AJ = 1 (n 2)J, A2 + A + 1 (2 n)I = 1 (n 2)J. 2 − 4 − 4 − By the characterisation (12.32), this says that Σˆ gives a srg(2d 1,k,λ, µ) with 1 λ 1 1 µ 1 d−1 k = 2 (n 2)= d 1, = 4 (n 6)= 2 (d 3), = 4 (n 2)= −2 . A well− known− necessary condition− for an−n n conference− matrix to exist is that n 2 (mod 4) and n 1 be a sum of squares.× This completes the proof. ≡ − ⊓⊔ An inﬁnite family of conference graphs are given by the Paley graphs.

Example 12.40. (Paley graphs) Let q = pm be a prime power with q 1 (mod 4). ≡ This implies Fq, the unique ﬁnite ﬁeld of order q, has a square root of 1 and so b a is a square if and only if b a is a square. The Paley graph of order−n 1 = q − − − (or q–Paley) is the graph with vertices Fq, and an edge between a and b if a b is a square (aka quadratic residue mod p if q = p). When q is prime its Paley− graph is a Hamiltonian circulant graph. The Paley graph on n 1 = q vertices is a self complementary conference graph. −

12.12 A list of real equiangular tight frames

There may or may not be a real equiangular tight frame of n d > 1 vectors for Rd. For example, for ≥ n = d one has orthonormal bases. • n = d + 1 one has the vertices of the regular simplex. • n = d + 2 there are no equiangular tight frames, by Theorem 12.13. • 298 12 Equiangular and Grassmannian frames

If there did exist an equiangular tight frame of n = d + 2 vectors for Rd, then its complement would be a real equiangular tight frame of n > 3 vectors for R2. In Table 12.5, we list all equiangular tight frames of n > d + 1 vectors known to exist for d 50. This was constructed using the literature on strongly regular graphs (see Brouwer≤ [Bro07] and the associated internet page). We also include (in grey) those which are conjectured to exist, e.g., the longstanding open question in algebraic graph theory:

Does there exist a conference graph srg(65,32,15,16), i.e., an equiangular tight frame of 66 vectors for R33?

Table 12.5: The equiangular tight frames of n > d + 1 vectors for Rd (d 50), for n = 2d and n = 2d (conference graphs). Here # gives the number of associated strongly≤ regular graphs Gˆ (with + indicating at least one exists). The existence of those in grey are open problems.

n d # λ1 λ2 Gˆ n d # Γˆ 2d 1 16 6 1 3 5 srg(15,6,1,3) − − 6 3 1 Paley(5) 12 + 22 16 10 1 5 3 srg(15,8,4,4) − 10 5 1 Paley(9) 02 + 32 28 7 1 3 9 srg(27,10,1,5) − 14 7 1 Paley(13) 22 + 32 28 21 1 9 3 srg(27,16,10,8) − 18 9 1 Paley(17) 12 + 42 36 15 3854 5 7 srg(35,16,6,8) − 26 13 15 Paley(25) 32 + 42 36 21 + 7 5 srg(35,18,9,9) − 30 15 41 Paley(29) 22 + 52 126 21 + 5 25 srg(125,52,15,26) − 38 19 6760 Paley(37) 12 + 62 176 22 + 5 35 srg(175,72,20,36) − 42 21 + Paley(41) 42 + 52 276 23 1 5 55 srg(275,112,30,56) − 46 23 + Conference 32 + 62 64 28 + 7 9 srg(63,30,13,15) − 50 25 + Paley(49) 02 + 72 64 36 + 9 7 srg(63,32,16,16) − 54 27 + Paley(53) 22 + 72 120 35 + 7 17 srg(119,54,21,27) − 62 31 + Paley(61) 52 + 62 148 37 ? 7 21 srg(147,66,25,33) − 66 33 ? Conference 12 + 82 246 41 ? 7 35 srg(245,108,39,54) − 74 37 + Paley(73) 32 + 82 288 42 ? 7 41 srg(287,126,45,63) − 82 41 + Paley(81) 02 + 92 344 43 + 7 49 srg(343,150,53,75) − 86 43 ? Conference 22 + 92 100 45 + 9 11 srg(99,48,22,24) − 90 45 + Paley(89) 52 + 82 540 45 ? 7 77 srg(539,234,81,117) − 98 49 + Paley(97) 42 + 92 736 46 ? 7 105 srg(735,318,109,159) −

Example 12.41. There is no equiangular tight frame of 40 or 41 vectors for R16. Therefore (by Table 12.3) the maximal number of equiangular lines in R16 is given by a nontight equiangular frame. This is currently the only known case where the maximum number of equiangular lines is not given by a tight frame. 12.13 Nontight real equiangular frames from graphs 299 12.13 Nontight real equiangular frames from graphs

We now consider some nontight real equiangular frames and the graphs that they come from. We recall Theorem 12.6 as it applies:

Proposition 12.5. Let Γ be a graph (with two or more vertices) whose n n Seidel matrix Σ has smallest eigenvalue λ of multiplicity n d. Then Σ is the× signature matrix of an equiangular frame of− n > d vectors for Rd−, which is tight if and only if Σ has only two eigenvalues.

Example 12.42. The construction of Corollary 12.4 (see Table 12.4) gives examples of nontight frames whose associated graphs have four Seidel eigenvalues.

Example 12.43. The lower bounds on the maximum number of real equiangular lines in Rd given by Table 12.3 are attained for d = 16,17,19,20 by graphs with three (but not two) eigenvalues. See Table 12.6 (and [GKMS16] for details).

Table 12.6: The graphs corresponding to the nontight equiangular frames giving the maximum number n of equiangular lines in Rd currently known (see [GKMS16]).

n d Origin of graph eigenvalues multiplicities 40 16 srg(40,12,2,4) 5,7,12 24,15,1 − 48 17 Netto triples 5,7,11 31,8,9 − 72 19 Witt design 5,13,19 53,16,3 − 90 20 Taylor [Tay72] 5,15,19,25 70,9,10,1 −

Example 12.44. Let Φ be the nontight equiangular frame of ﬁve vectors for R3 given by Γ the 5–cycle, whose Seidel matrix Σ has eigenvalues √5, √5,0,√5,√5. This a nontight Grassmannian frame (see Example 12.6). The− dual−Φ˜ and canonical tight frame Φcan are equal–norm frames. The minimal angles for Φ, Φ˜ and Φcan are

1 1 1 3 + √5 1 1 + √5 cos− 63.4◦, cos− 49.1◦, cos− 57.4◦. √5 ≈ 8 ≈ 6 ≈ It is easy to verify that Φ consists of vectors that lie in ﬁve of the six diagonals of the regular icosahedron, and that the tight frame Φcan is the harmonic frame given by the lifted ﬁfth roots of unity (see Exer. 12.7).

We now generalise Example 12.44 to when Γ is a strongly regular graph (these have adjacency matrices with three eigenvalues). We say that an equal norm frame Φ =( f j) has two distances (cf 12.14) if § f j, fk , j = k takes two values. 300 12 Equiangular and Grassmannian frames

Lemma 12.1. Let A be the adjacency matrix of a srg(ν,k,λ, µ), and Pθ ,Pτ ,Pk be the orthogonal projections onto its θ,τ,k eigenspaces, where θ,τ are given by (12.33). Then J 1 J J Pk = , Pθ = A τI (k τ) , Pτ = I Pθ . (12.43) ν θ τ − − − ν − − ν − In particular, the diagonal entries of Pθ and Pτ are constant, and their off diagonal entries take only two values.

Proof. We have, Pk =(1/√ν)(1/√ν)∗ = J/ν, and I = Pk + Pθ + Pτ , so it sufﬁces to prove the formula for Pθ . From the spectral decomposition of A, we have

θ τ J τ τ τ τ J A = Pθ + Pτ + k ν , I = Pθ + Pτ + ν .

Eliminating τPτ , and solving for Pθ gives the desired formula for Pθ . ⊓⊔ Proposition 12.6. Let Γ be a strongly regular graph srg(ν,k,λ, µ), and Φ be the equiangular frame of ν vectors for Rd that it determines. Then Φ is tight if and only if λ µ 2k + ν = (λ µ)2 + 4(k µ). (12.44) − − ± − − Otherwise, either

(λ µ)2 + 4(k µ) > µ λ + 2k ν, (12.45) − − − − and the frame, its dual, and canonical tight frame are equal–norm frames with two distances, where

1 2k +(ν 1)(λ µ) d = ν + 1 + − − , (12.46) 2 (λ µ)2 + 4(k µ) − − or d = ν 1. − Proof. By (12.33), (12.34), the eigenvalues of the Seidel matrix Σ of Γ are

1 2θ, 1 2τ, ν 1 2k, − − − − − − with multiplicities mθ ,mτ ,1. These are distinct, unless 1 2θ or 1 2τ equals − − − − ν 1 2k, which is equivalent to (12.44), and Φ is tight, with d = ν (mθ + 1) or − − − d = ν mθ , respectively (by Theorem 12.6). − Otherwise, the minimal eigenvalue of Σ is (1 + 2θ) (with multiplicity mθ ) when 1 2θ < ν 1 k, i.e., (12.45) holds. Here,− the spectral decomposition of the symmetric− − matrix−Σ−is J Σ = (1 + 2θ)Pθ +( 1 2τ)Pτ +(ν 1 2k) , − − − − − ν 12.13 Nontight real equiangular frames from graphs 301 where Pθ and Pτ are the orthogonal projections onto the θ and τ eigenspaces of A. Since Pθ = I Pτ J/ν, the Gramian of the associated equiangular frame for Rd, − − d = ν mθ , has the form − 1 J 2θ 2τ 2θ + ν 2k I + Σ = αPτ + β , α = − , β = − . 1 + 2θ ν 1 + 2θ 1 + 2θ The dual and canonical tight frames have Gramians

1 Σ † 1 1 J 1 Σ † 1 Σ J (I + λ ) = α Pτ + β ν , (I + λ ) (I + λ )= Pτ + ν .

By Lemma 12.1, Pτ has a constant diagonal, and off diagonal entries taking two possible values. Hence Φ, its dual and canonical tight frame are equal–norm frames with two distances (all the entries of J are 1). The only remaining case is when ν 1 k < 1 2θ, in which case the minimal eigenvalue ν 1 k has multiplicity one,− − and so−d −= ν 1. − − − ⊓⊔

Fig. 12.3: Three switching equivalent graphs that give 10 lines in R5: The Paley graph on 9 vertices and a point, the Petersen graph, and the triangular graph T5.

Example 12.45. For n 50 there are 28 equal–norm tight frames with two angles that can be constructed≤ by Theorem 12.6 (see Table 12.7). There are also many equiangular tight frames that can be constructed in this way, e.g., for n = 10 the unique graphs srg(10,3,0,1) (the Petersen graph) and srg(10,6,3,4) (the triangular 5 graph T5) give a set of 10 equiangular vectors in R . Since there is a unique such frame up to switching equivalence of the graphs, it follows these two graphs are switching equivalent to that obtained by taking the Paley graph on nine vertices and adding an isolated vertex (see Figure 12.3).

Example 12.46. The strongly regular graph srg(40,12,2,4) gives 40 equiangular lines in R16, which are not tight. There are no strongly regular graphs giving tight frames of 40 or 41 vectors in R16 (see Table 12.5). Therefore the maximal set of equiangular lines in 16 dimensions is not tight (see Table 12.3). 302 12 Equiangular and Grassmannian frames

Table 12.7: The list of all equiangular frames of n 50 vectors for Rd constructed from strongly regular graphs Γ . Here the type refers to the three cases≤ in Theorem 12.6.

n d type Γ n d type Γ 5 3 nontight srg(5,2,0,1) 36 26 nontight srg(36,10,4,2) 9 5 nontight srg(9,4,1,2) 36 35 – srg(36,25,16,20) 10 5 tight srg(10,3,0,1) 36 15 tight srg(36,14,4,6) 10 5 tight srg(10,6,3,4) 36 21 tight srg(36,21,12,12) 13 7 nontight srg(13,6,2,3) 36 28 nontight srg(36,14,7,4) 15 6 nontight srg(15,6,1,3) 36 35 – srg(36,21,10,15) 15 10 nontight srg(15,8,4,4) 36 21 tight srg(36,15,6,6) 16 6 tight srg(16,5,0,2) 36 15 tight srg(36,20,10,12) 16 10 tight srg(16,10,6,6) 37 19 nontight srg(37,18,8,9) 16 10 tight srg(16,6,2,2) 40 16 nontight srg(40,12,2,4) 16 6 tight srg(16,9,4,6) 40 39 – srg(40,27,18,18) 17 9 nontight srg(17,8,3,4) 41 21 nontight srg(41,20,9,10) 21 7 nontight srg(21,10,3,6) 45 25 nontight srg(45,12,3,3) 21 15 nontight srg(21,10,5,4) 45 44 – srg(45,32,22,24) 25 17 nontight srg(25,8,3,2) 45 36 nontight srg(45,16,8,4) 25 24 – srg(25,16,9,12) 45 44 – srg(45,28,15,21) 25 13 nontight srg(25,12,5,6) 45 23 nontight srg(45,22,10,11) 26 13 tight srg(26,10,3,4) 49 37 nontight srg(49,12,5,2) 26 13 tight srg(26,15,8,9) 49 48 – srg(49,36,25,30) 27 7 nontight srg(27,10,1,5) 49 31 nontight srg(49,18,7,6) 27 21 nontight srg(27,16,10,8) 49 48 – srg(49,30,17,20) 28 21 tight srg(28,12,6,4) 49 25 nontight srg(49,24,11,12) 28 7 tight srg(28,15,6,10) 50 22 nontight srg(50,7,0,1) 29 15 nontight srg(29,14,6,7) 50 49 – srg(50,42,35,36) 35 15 nontight srg(35,16,6,8) 50 25 nontight srg(50,21,8,9) 35 21 nontight srg(35,18,9,9) 50 25 tight srg(50,28,15,16)

Example 12.47. The strongly regular graphs srg(76,30,8,14) and srg(76,40,18,24) give equiangular tight frames of 76 vectors for C19, i.e., a srg(75,32,10,16). Since there are no other strongly regular graphs on 75 or 76 vertices (other than the complements) [Hae93], it follows that there is a strongly regular graph on 75 points if and only if there is a strongly regular graph on 76 points. In [BPR14] it was shown there is no srg(76,30,8,14)), and in [AM15] that there is no srg(75,32,10,16). It therefore follows that there is no srg(76,40,18,24) (and hence no strongly regular graphs on 75 or 76 points), and no equiangular tight frames of 76 vectors for C19. 12.13 Nontight real equiangular frames from graphs 303

Example 12.48. In [AM16], it was shown that no srg(95,40,12,20) exists, and so there are not 96 equiangular lines in R20. From Proposition 12.6, it follows that no srg(96,38,10,18) exists (see [Deg07]) and no srg(96,50,22,30) exists.

The next open problem of the above type is the following.

Does there exist 148 equiangular lines in R37, i.e., a srg(147,66,2533)? The existence of such an equiangular tight frame implies the existence of a srg(148,63,22,30) or a srg(148,77,36,44), and the nonexistence implies that there is no strongly regular graph with 148 vertices.

The 6–cycle is a regular, but not strongly regular graph. Its Seidel matrix has three eigenvalues, and gives an equiangular frame of six vectors for R4. The dual and canonical tight frames have equal norms (and more than two angles). This is a consequence of the 6–cycle being a circulant graph. Let C be a subset of Zn which is closed under taking additive inverses, i.e, c C, c C. Then the circulant graph G with connection set C is the graph with− vertices∈ ∀ ∈ Zn and an edge from j to k if j k C. The choice C = 1,1 gives the n–cycle. − ∈ {− } Proposition 12.7. (Circulant graphs) . Let Γ be a circulant graph, and Φ be the real equiangular frame that it determines. Then the dual frame Φ˜ and canonical tight frame Φcan are equal–norm frames.

Proof. Since Γ is a circulant graph, the Gramian of Φ is an n n circulant matrix, and hence is diagonalised by the Fourier matrix F, i.e., ×

1 1 2πi 1 Σ λ λ ω jk ω n F− (I + )F = diag( 1,..., n), F := [ ] j,k Zn , := e . λ √n ∈ Since F is unitary, we can write this spectral decomposition as 1 1 I Σ ∑λ P P : f f f : ω jk + λ = j j, j = j j∗, j = ( )k Zn . j √n ∈

1 The rank one projection matrices Pj have constant diagonal entries (equal to n ), and so the dual and canoncial tight frames have equal–norms (their Gramians are 1 ∑ j λ Pj and ∑ j Pj). j ⊓⊔

Example 12.49. (n–cycle graph) The Seidel matrix of the n–cycle has a minimal 2π eigenvalue 1 4cos n of multiplicity 2 with corresponding eigenvectors f 1, f1. − − n 2 − The corresponding equiangular frame is of n vectors for R − , and its complement is the tight frame of n equally spaced vectors for R2 (up to similarity).

Example 12.50. (Paley graphs) For n a prime congruent to 1 (mod 4), the Paley graph is circulant, with connection set C given by the quadratic residues modulo n. 304 12 Equiangular and Grassmannian frames 12.14 Spherical two–distance tight frames

d A set Φ =(x j) of n unit vectors in R is called a (spherical) two–distance set if the inner products between distinct vectors take two values a and b, i.e.,

x j,xk : j = k = a,b , a = b. { } { } 1 The spherical distance between points x j and xk is cos− x j,xk , and so these sets have precisely two spherical distances between their points. The maximum size g(d) of a spherical two–distance set has been studied for several decades [BY13]. When b = a, a spherical two–distance set is equivalent to a real equiangular frame (see 12.9).− We therefore primarily consider the case when a + b = 0. § d+1 1 Let e1,...,ed+1 be the standard basis for R . Then the n = 2 d(d + 1) points e j +ek, j = k give a spherical two–distance set (after scaling), as does the orthogonal projection of these points onto the d–dimensional subspace given by the orthogonal complement of e1 + + ed 1 (see 12.16). We therefore have the lower bound + § 1 g(d) d(d + 1). ≥ 2 Delsarte, Goethals and Seidel [DGS77] proved the “harmonic” upper bound 1 g(d) d(d + 3). ≤ 2 Various improvements and refinements of this have been made, e.g., Musin [Mus09] proved the upper bound 1 g(d) d(d + 1), when a + b 0. ≤ 2 ≥ This generalises the estimate (12.2) for the maximal number of of real equiangular lines (the case a + b = 0). As of the improvements of [BY13], the best estimates of g(d) for d 93 are as follows. ≤ Theorem 12.14. The maximum size g(d) of a spherical two–distance set in Rd satisfies g(2)= 5, g(3)= 6, g(4)= 10, g(5)= 16, g(6)= 27, 1 g(d)= d(d + 1), 7 d 93, d = 22,46,78, 2 ≤ ≤ g(22)= 275, 1 1 d(d + 1) g(d) d(d + 3) 1, d = 46,78. 2 ≤ ≤ 2 − All the known maximal configurations are tight.

Given a spherical two–distance set Φ =(x j), one can associate with it the graph Γa with points x j and an edge from x j to xk if x j,xk = a. The following result { } 12.14 Spherical two–distance tight frames 305 of [BGOY14] shows that if Φ is tight and b = a (Φ is not equiangular), then Γa is a strongly regular graph, and that all tight nonequiangular − two–distance frames can be constructed in this way.

Theorem 12.15. Let Φ =(x j) be a two–distance set, with a = b and associated − graph Γa. Then Φ is tight if and only if Γa is a strongly regular graph srg(ν,k,λ, µ), in which case (after normalisation) its Gramian is one of J J Pθ , Pθ + ν , Pτ , Pτ + ν , (12.47)

J where Pθ ,Pτ , ν are the orthogonal projections onto the eigenspaces of the adjacency matrix A of Γa. Further, all the projections of (12.47) give tight two distance sets. Φ ν d Proof. Suppose that =(x j) j=1 is a tight two–distance frame for R . Let Na be the number of points a distance a from a given point x j. Since Φ is a ν unit–norm tight frame (with A = d ), the norm of the j–th column of its Gramian is ν 2 2 1 + Naa +(ν 1 Na)b = . − − d

2 2 Since a = b, a b = 0, and so the above can be solved for Na, which is therefore − − independent of j, i.e., Γa is a regular graph. Fix a pair of indices k,ℓ with xk,x = a. Let Ca be the number of indices j = k,ℓ ℓ for which the distances between x j and xk,xl are both a. Since Φ is tight, Parseval gives ν ν xk,xℓ = ∑ xk,x j x j,xℓ , d j=1 from which we obtain ν 2 2 a = 2a + 2(Na Ca 1)ab +Caa +(ν 2Na +Ca)b d − − − 2 2 = 2a + 2(Na 1)ab +(ν 2Na)b +(a b) Ca. − − −

Since a = b, this gives a unique solution for Ca, i.e., every two adjacent vertices have λ = Ca common neighbours. Similarly, every two nonadjacent vertices has the same number of adjacent vertices, and so Γa is strongly regular. The (normalised) Gram matrix of Φ is d P = (1 b)I +(a b)A + bJ . ν − − The spectral decomposition of A (see Lemma 12.1) gives

J θ τ J I = Pθ + Pτ + ν , A = Pθ + Pτ + k ν , and so P can be written in the form 306 12 Equiangular and Grassmannian frames

α β γ J P = Pθ + Pτ + ν .

The condition P2 = P implies that α2 = α, β 2 = β, γ2 = γ, and so α,β,γ must be 0 J J or 1. The choices P = Pθ +Pτ + ν = I and P = Pθ +Pτ = I ν give an orthonormal basis and the vertices of the simplex, which are not two–dist− ance sets. The other J choices give two–distance sets. For example, when P = Pθ + ν (which has constant diagonal), we have d = rank(Pθ )+ rank(J)= mθ + 1, and, by Lemma 12.1,

mθ + 1 1 J J P = (1 b)I +(a b)A + bJ = A τI (k τ) + . ν − − θ τ − − − ν ν − By taking an entry with A jk = 1, we obtain

mθ + 1 1 1 1 ν k + θ a = 1 (k τ) + = a = − . ν θ τ − − ν ν ⇒ (mθ + 1)(θ τ) − − Similarly, taking an off diagonal entry with A jk = 0 gives a formula for b. The results of these calculations for the four possible choices are given in Table 12.8. We observe that, since ν = 0, each of these choices gives a two–distance set. ⊓⊔

d Table 12.8: The parameters of the tight two–distance sets for R given by Γa a srg(ν,k,λ, µ). Here the Gramian matrix (after normalisation) is the orthogonal projection matrix P, and Na is the number of neighbours of a. The condition for these to be equiangular is given in the last column.

Gramian matrix P d Na a b a = b holds − ν k+τ k+τ Pθ mθ k ν 2 k τ mθ−(θ τ) mθ−(θ τ) = ( ) − − − J ν k+θ k+θ Pθ mθ 1 k ν 2 k θ + ν + (mθ +−1)(θ τ) (mθ +−1)(θ τ) = ( ) − − − J ν+k θ k θ Pτ I Pθ ν mθ 1 k ν 2 k θ = ν (ν −mθ 1−)(θ τ) (ν mθ −1)(θ τ) = ( ) − − − − − − − − − − − J ν+k τ k τ Pτ I Pθ ν mθ k ν 2 k τ + ν = (ν −mθ )(−θ τ) (ν mθ−)(θ τ) = ( ) − − − − − − −

By combining Corollary 12.3 and Theorem 12.15, we obtain a characterisation of all spherical two–distance tight frames in terms of the strongly regular graphs.

Corollary 12.5. All spherical two–distance tight frames of n > d +1 vectors for Rd can be constructed from the strongly regular graphs as follows

3k n k 1 n d(n 1) 1. (Equiangular) From the srg(n 1,k, − , ), k := n 1 + 1 − . − 2 2 2 − − 2d n d 2. (Nonequiangular) From the srg(ν,k,λ, µ), ν = n. − Each complementary pair of strongly regular graphs gives four nonequiangular spherical two–distance tight frames, unless ν = 2(k τ) or ν = 2(k θ), in which case there are only two (the other two being equiangular).− − 12.15 Two–distance tight frames and partial difference sets 307

Example 12.51. For a srg(10,3,0,1), we have θ = 1, τ = 2, so that − ν = 10 = 2(3 ( 2)) = 2(k τ). − − − J Hence the spherical two–distance tight frames with Gramians Pθ and Pτ + ν are J equiangular, and those with Gramians Pθ + ν and Pτ are nonequiangular. Here Pθ is J the tight frame constructed by Proposition 12.6, and Pτ + ν is the one constructed from the complementary graph. Example 12.52. Here, we consider the nonprimitive strongly regular graph given by the union of m complete graphs on w vertices, which is a srg(mw,w 1,w 2,0). − − We have θ = w 1, mθ = m 1, τ = 1, mτ = m(w 1), and − − − − 1 J Pθ = (A + I) . w − ν In this case, there are only two eigenvalues, since k = w 1 = θ. The corresponding spherical two–distance tight frames are described in Table− 12.9.

Table 12.9: The parameters of the two–distance tight frames for Rd given by the nonprimitive Γa = srg(mw,w 1,w 2,0) , which consists of m copies of a complete graph on w points. − −

Gramian matrix P d Na a b description 1 J 1 Q = w (A + I) ν m 1 w 1 1 m 1 w copies of m–simplex − − − − − J Q + m w 1 1 0 w copies of e1,...,em ν − { } J 1 R = I (Q + ν ) n m w 1 w 1 0 complement of second − − − − − J m 1 1 R + ν = I Q n m + 1 w 1 n m−+1 n m+1 complement of ﬁrst − − − − − −

12.15 Two–distance tight frames and partial difference sets

We now show that nonequiangular two–distance tight frames which are harmonic can be constructed from partial difference sets (cf. Theorem 12.3). Deﬁnition 12.4. A d element subset J of a ﬁnite group G of order n is said to be a (n,d,λ, µ)–partial difference set if the multiset of (nonzero) differences

1 j2 j− : j1, j2 J, j1 = j2 { 2 ∈ } contains each nonidentity element of J exactly λ times and each nonidentity element of G J exactly µ times. If λ = µ then one has a difference set. \ 308 12 Equiangular and Grassmannian frames

Partial difference sets with λ = µ are are reversible (i.e., closed under taking inverses). A partial difference set J for which 1 J is said to be regular. It sufﬁces to study the regular partial difference sets, since∈ if J is a partial difference set with 1 J, then J 1 is a (regular) partial difference set. ∈ \ { } Proposition 12.8. Let G be an abelian group of order n, and ΦJ =(ξ J)ξ ˆ be the | G harmonic frame given by J G, J = d. Then the following are equivalent∈ ⊂ | | 1. J is a (n,d,λ, µ)–partial difference set for G. 2. For all χ = 1, χ G,ˆ ∈ λ µ √(λ µ)2+4(d λ) − ± − − , 0 J; ∑ χ( j)= 2 ∈ (12.48)  λ µ √(λ µ)2+4(d µ) j J − ± − − , 0 J. ∈  2 ∈

In particular, if these hold with λ= µ, then ΦJ is a nonequiangular two–distance tight frame of n vectors for Rd. Proof. The equivalence of 1 and 2 is a standard characterisation of partial difference sets [LM90],[Ma94]. By (12.48), the harmonic frame given by a partial difference set is real, with two–distances, and is equiangular if and only if λ = µ. ⊓⊔

Example 12.53. The set J = 1,4 Z5 (of nonzero quadratic residues) is a regular cyclic (5,2,0,1)–partial difference{ }⊂ set. The regular (9,4,1,2)–partial difference set

J = (0,1),(0,2),(1,0),(2,0) Z3 Z3. { }⊂ × is a noncyclic example. Table 12.10 lists the regular abelian partial difference sets for d 21. ≤

Table 12.10: The regular (n,d,λ, µ)–partial difference sets for d 21, where 0 λ k 1, 1 n 1 ≤ ≤ ≤ − ≤ µ k 1, k − . This was adapted from [Ma94] (which lists those with d 100). ≤ − ≤ 2 ≤ n d λ µ n d λ µ n d λ µ n d λ µ 5 2 0 1 17 8 3 4 29 14 6 7 37 18 8 9 9 4 1 2 36 10 4 2 36 15 6 6 81 20 1 6 16 5 0 2 49 12 5 2 81 16 7 2 121 20 9 2 16 6 2 2 25 12 5 6 64 18 2 6 41 20 9 10 13 6 2 3 36 14 4 6 49 18 7 6 64 21 8 6 25 8 3 2 64 14 6 2 100 18 8 2 243 22 1 2

Example 12.54. The set J = 0,1,4 Z5 is a cyclic (5,2,0,1)–partial difference set. It is not regular, since it contains{ }⊂ the identity 0. 12.16 The standard m–distance tight frame 309

In view of Theorem 12.15, the nonequiangular two–distance harmonic frames must be associated with strongly regular graphs. These turn out to be Cayley graphs. For J a generating subset of a group G with 1 J, the (uncoloured) Cayley graph ∈ has vertices G and a directed edge from g1 to g2 if g2 = g1 j for some j J (J is 1 ∈ called the connection set). For J with J− = J (e.g., a partial difference set with λ = µ), the Cayley graph given by J is undirected. Proposition 12.9. If J is a regular (n,d,λ, µ)–partial difference set for a group G with λ = µ, then the (undirected) Cayley graph given by J is a srg(n,d,λ, µ). There is a converse of Proposition 12.9 (the condition λ = µ is replaced by the condition that the partial difference set be reversible) [Ma94].

12.16 The standard m–distance tight frame

d We say that a set Φ =(x j) of n unit vectors in R isa(spherical) m–distance set if the inner products between distinct vectors take m values. This generalisation of the isogonal conﬁgurations of Example 3.9 (m = 1) and spherical two–distance sets (m = 2) are not well studied. However, we can provide one nice example. d+1 d+1 Let e1,...,ed+1 be the standard basis for R . Then the n = m points

e j + + e j , 1 j1 < < jm d + 1 1 m ≤ ≤ give an m–distance set (after scaling). The orthogonal projection of these points onto d (e1 + + ed 1) gives an m–distance tight frame for R . + ⊥ Proposition 12.10. Suppose 1 m d. For J = j1,..., jm 1,2,...,d +1 , let ≤ ≤ { } ⊂ { } m wJ := e j + e j + + e j (e1 + ed 1). 1 2 m − d + 1 +

Then (wJ) J =m is a tight frame for (e1 + + ed+1)⊥, with Gramian given by | | m2 wJ,wK = r , J K = r. − d + 1 | ∩ | Proof. This frame is the highly symmetric tight frame of Example xx. In particular, it is tight, since it is the orbit of an irreducible unitary action. Writing m WJ = ∑ e j + ∑ e j (e1 + ed+1), j J K j J K − d + 1 ∈ ∩ ∈ \ and expanding the inner product gives

m m m m2 wJ,wK = r m m (d + 1) = r , J K = r. − d + 1 − d + 1 − d + 1 − d + 1 | ∩ | 310 12 Equiangular and Grassmannian frames

⊓⊔ m d+1 m The number of r entries in a given column of the Gramian is r m −r , and so for there to be at least one r entry, one must have − d + 1 m m r d + 1 2m r. − ≥ − ⇐⇒ ≥ − In particular, for d + 1 2m, ≥

Φ d + 1 =(cwJ) J =m, c := | | m(d + 1 m) − d+1 d is an m–distance tight frame of m vectors for R , with distances

d + 1 m2 αr := r , r = 0,1,...,m 1, m(d + 1 m) − d + 1 − − which we call the standard m–distance tight frame.

Example 12.55. For d 3, the standard 2–distance tight frame has ≥ d 3 2 a = α1 = − , b = α0 = , (12.49) 2(d 1) −d 1 − − and corresponds (via Pθ in Theorem 12.15) to Γa = Td+1, the triangular graph, 1 which is a srg 2 d(d + 1),2(d 1),d 1,4 . The triangular graph is the unique graph with these parameters, except− for− when d + 1 = 8, in which case there are three other srg(28,12,6,4) (the Chang graphs).

Example 12.56. (28 equiangular lines in R7) The standard 2–distance tight frame gives a set of equiangular lines if and only if a = b in (12.49), i.e., d = 7, which gives the 28 lines of (12.6)

12.17 Complex equiangular tight frames

We ﬁrst observe that the complex analogue of Theorem 12.13 holds.

Proposition 12.11. Suppose d > 1. A necessary condition for an equiangular tight frame of n > d + 1 vectors for Cd to exist is that

n min d2,(n d)2 , (12.50) ≤ { − } or, equivalently,

1 + √4d + 1 2d + 1 + √4d + 1 d + = n d2. (12.51) 2 2 ≤ ≤ 12.17 Complex equiangular tight frames 311

Example 12.57. There is no equiangular tight frame of n = d + 2 vectors for Cd when d 3. ≥ The existence and construction of equiangular tight frames of n = d2 vectors for Cd is a compelling subject (see Chapter 14). There have been attempts to adapt the methods of constructing real equiangular tight frames to complex ones. Most notably, by restricting the entries of the Gramian (or the vectors) to be m–th roots of unity: thereby giving a ﬁnite set of possible n vector equiangular tight frames for Cd. When the entries of the vectors are roots of unity (as for harmonic frames), then n d2 d + 1 (Theorem 12.16). When the entries of the signature matrix are roots≤ of unity− (generalising real equiangular tight frames), then there exist some intriguing examples, including a maximal set of n = d2 equiangular lines for d = 3,8. We now give some details on the known constructions.

Theorem 12.16. Suppose that Φ =( f j) is an equiangular tight frame of n (unit) vectors for Cd, for which the entries of d Gram(Φ) are contained in a subring A of the algebraic integers. Then

A d(n d) 1. has elements of modulus n −1 . d(n d) − 2. − A Z. n 1 ∈ ∩ 3. n =− 1 d(d 1)+ 1, for some positive integer λ. In particular, n d2 d + 1. λ − ≤ − Proof. By our assumption a jk := d f j, fk A , j = k. Since Φ is equiangular, ∈ n d d(n d) a jk = d f j, fk = d − = − . | | | | d(n 1) n 1 − − For any ﬁxed j = k, d(n d) 2 − = a jk = a jkakj A , n 1 | | ∈ − d(n d) A so that n −1 is a rational algebraic integer, and hence is an integer (in ). Further, − d(n d) d(d 1) λ = d − = − − n 1 n 1 − − is a rational algebraic integer, and hence is a positive integer. ⊓⊔ Example 12.58. (Unital frames) If the n vectors of Φ have the form

v1 1 . v =  . , v1,...,vd A , √d ∈ v   d   312 12 Equiangular and Grassmannian frames where A is a subring of the algebraic integers which is closed under conjugation, then d Gram(Φ) has entries in A , and so n d2 d + 1. ≤ − A special case is when the entries v j are m–th roots of unity, which is known as a unital equiangular tight frame (of degree m). These are a generalisation of the equiangular harmonic frames (see 12.3). The only know example of a unital equiangular tight frame which is not harmonic§ is one of 576 vectors for R276, given by a Kirkman frame [JMF14]. For example, [STDH07] found a unital equiangular tight frame of 27 vectors for C13 with m = 3 via a computer search. This corresponds 4 to a (27,13,6)–difference set for Z3.

Theorem 12.17. (m–th root signature matrices) Let Σ be the signature matrix of an equiangular tight frame of n vectors for Cd. Then

n 1 λ1 + λ2 =(n 2d) − Z[Σ] (the ring generated by the entries of Σ). − d(n d) ∈ − If n = 2d, and the entries of Σ are algebraic integers, then eigenvalues of Σ d(n 1) (n d)(n 1) λ1 = − , λ2 = − − − n d d − are algebraic integers in the subﬁeld of C generated by the entries of Σ. Thus, if the 2πi entries of Σ are powers of the primitive m–th root of unity ω = e m , then

2π λ1,λ2 Z[ω + ω]= Z[cos ]. ∈ m

Proof. Multiply (12.22) by zkj = z jk to obtain λ1 +λ2 Z[Σ]. If the entries of Σ are algebraic integers, then so are the coefficients of the chara∈ cteristic polynomial of Σ, whose roots λ1,λ2 are therefore algebraic integers. For n = 2d, solving n 1 λ1 + λ2 =(n 2d) − Z[Σ], (n d)λ1 + dλ2 = 0, − d(n d) ∈ − − shows that λ1,λ2 are in the field generated by the entries of Σ. Finally, when Σ consists of m–th roots of unity, we have that λ1,λ2 are in the cyclotomic field Q(ω). Since λ1,λ2 are also real algebraic integers, they are in Z(ω + ω) (the intersection of Q(ω) and the real algebraic integers). ⊓⊔ Example 12.59. (SICs) For n = d2, we have

λ1 + λ2 =(d 2)√d + 1. − Thus, for d = 2, √d + 1 must be in the ﬁeld generated by the entries of Σ. Example 12.60. (Cube roots) If the entries of a Σ giving an equiangular tight frame are third roots of unity then λ1,λ2 must be integers. It can be shown [BPT09] that 12.17 Complex equiangular tight frames 313

λ1,λ2 2 (mod 3). ≡ 2Πi In particular, since λ1λ2 = n 1, one must have n 0 (mod 3). Let ω = e 3 . For n = 9, d = 6 there is a cube− root signature matrix≡ Σ given by the Kronecker product type construction of Example 12.29 (taking the product again gives n = 81, d = 45). This example can also be obtained by associating Σ with a directed graph (1 corresponds to a nonedge, and ω,ω2 to directed edges) [BPT09]. The choice n = 33, d = 11 gives λ1 = 4, λ2 = 8. It is unknown whether or not there is a cube root signature matrix with these− parameters.

The above condition on λ1,λ2 has been generalised by [BE10] as follows. Theorem 12.18. Let p > 2 be prime. If the entries of Σ are p–th roots of unity and Σ is the signature matrix of an equiangular tight frame of n vectors for Cd, then

1. λ1,λ2 are integers, with λ1,λ2 p 1 (mod p). 2 ≡ − 2 2. 4(n 1)+(λ1 λ2) is a perfect square, which is 0 modulo p . − − Many of the known p–th root signature matrices for equiangular tight frames are given by Theorem 12.8 applied to Butson type complex Hadamard matrices V, i.e., those with p–th root entries. Corollary 12.6. Let V H(p,m), i.e., be a complex Hadamard matrix of order m with p–th root entries, and∈ K H(p,m2) be given by (12.26). Then ∈ 1. K I is a p–th root signature matrix for an equiangular tight frame of n = m2 − 1 vectors for a 2 m(m + 1) dimensional space. 2. I K isa 2p–th root (or a p–th root, when p is even) signature matrix for an equiangular− tight frame of n = m2 vectors for a 1 m(m 1) dimensional space. 2 − Proof. By construction, the entries of K I are p–th roots of unity, and those of I K are the negatives of p–th roots of unity− (which are 2p–th roots when p is odd, and− are p–th roots when p is even). ⊓⊔ The equiangular tight frames of Corollary 12.6 are subsimplicial with r = m 1. ± Example 12.61. (Butson) Let p be a prime. There are complex Hadamard matrices V of order m = 2 j pk, 0 j k, whose entries are p–th roots of unity [But62]. Thus, with ≤ ≤ n =(2 j pk)2, 0 j k. ≤ ≤ there is a p–th root signature matrix of size n giving an equiangular tight frame of n 1 j k j k vectors in d = 2 2 p (2 p + 1) dimensions. The complementary equiangular tight frame in 1 2 j pk(2 j pk 1) dimensions has a 2p–th (or p–th) root signature matrix. 2 − Example 12.62. (n = 36) There exist 6 6 complex Hadamard matrices with cube root entries and with fourth root entries.× Thus there is a p–th root signature matrix for an equiangular tight frame of 36 vectors in C21 for p = 2,3,4, and one in C15 for p = 2,6,4 (the complement). The real equiangular tight frames (p = 2) can be obtained from a reversible difference set (see Example 12.15). 314 12 Equiangular and Grassmannian frames

Example 12.63. (Conference matrices) Let C = CT be an n n skew–symmetric conference matrix (n must be a multiple of 4 when− n = 2). Then× Σ = iC is the signature matrix (with entries i) of an equiangular tight frame of n = 2d vectors for Cd. This follows by Theeorem± 12.7, via the calculation (see Exer. 12.11)

Σ 2 = ( CT )C =(n 1)I. − − − These frames are not subsimplicial, since r = √2d 1 implies that d is odd, and so n = 2d is not a multiple of 4. Given an n n skew–symmetric− conference matrix C, one of size 2n is given by × C C I − . C + I C For example, −

0 i i i − − − 0 1 C C I i 0 i i C = − Σ = i − =  − , −→ 1 0 C + I C i i 0 i  −  −  i i i 0   −    which is the signature matrix of an equiangular tight frame of four vectors for C2. Fourth root Seidel matrices are studied in [DHS10]. A suitable choice for H(a) in Theorem 12.4 gives p–th root signature matrices. Corollary 12.7. For a (2,k,ν)–Steiner system B, let m be the product of the prime ν 1 factors of r + 1 = k−1 + 1. Then there exists a Steiner equiangular tight frame of ν − ν(ν 1) n = (r + 1) vectors for a space of dimension d = k(k −1) whose signature matrix has nonzero entries the m–th roots of unity and their negativ− es. Proof. In the construction of Theorem 12.4 take each H(a) to be a Fourier matrix with m–th root entries. ⊓⊔ Example 12.64. There exist Steiner equiangular tight frames of n vectors for Cd with m–th root signature matrices in cases where such frames can not be constructed by the methods already considered, e.g., when (n,d) is (45,12) (r = 4, m = 10), (65,13) (r = 4, m = 10), (65,13) (r = 5, m = 6), (65,13) (r = 6, m = 14). Example 12.65. The inner products between the vectors in a Tremain equiangular tight frame are products of the entries of the Hadamard matrices that are used in their construction (see Exer. 12.10). The Hadamard matrices H(a) can be taken as in Corollary 12.7, and the Hadamard matrix giving the the unimodular simplex in Cν can be taken to be one of these tensored by a 2 2 real Hadamard matrix. In this way, Tremain equiangular tight frames can be construct× ed with with signature matrix have nonzero entries given by the m–th roots of unity and their negatives. 12.17 Complex equiangular tight frames 315

Motivated by Theorem 12.18, in Table 12.11 we list all n,d (d < 30) for which the eigenvalues λ1,λ2 of the signature matrix Σ of an equiangular tight frame of n > d + 1 vectors for Cd are integers. In view of (12.13), this condition is equivalent to both the frame and its complementary frame being subsimplicial. We indicate when an m–th root a signature matrix of such a frame is known to exist, by Corollaries 12.3, 12.6 and 12.7, or the Hoggar lines (n = 64, d = 8 see 14.6). §

Table 12.11: The n,d (d < 30) for which the eigenvalues λ1,λ2 of the signature matrix Σ of an equiangular tight frame of n > d +1 vectors for Cd are integers. We indicate when such a signature matrix is known to exist with entries which are m–th roots, and use grey when it is unknown.

n d λ1 λ2 exists n d λ1 λ2 exists n d λ1 λ2 exists 9 3 2 4 m = 6 65 13 4 16 m 10 126 21 5 25 m = 2 − − − − 10 5 3 3 m = 2 105 14 4 26 m 3 33 22 8 4 m 4 − − ≥ − ≥ 9 6 4 2 m = 3 25 15 6 4 m = 5 55 22 6 9 m 4 − − − ≥ 16 6 3 5 m = 2 36 15 5 7 m = 2 176 22 5 35 m = 2 − − − 28 7 3 9 m = 2 225 15 4 56 m 2 276 23 5 55 m = 2 − − ≥ − 64 8 3 21 m = 4 51 17 5 10 m 4 576 24 5 115 m 4 − − ≥ − ≥ 16 10 5 3 m = 2 76 19 5 15 m 2 50 25 7 7 m = 2 − − ≥ − 25 10 4 6 m = 10 96 20 5 19 m = 6 91 26 6 15 m = 14 − − − 33 11 4 8 m 3 28 21 9 3 m = 2 49 28 8 6 m = 7 − ≥ − − 45 12 4 11 m = 10 36 21 7 5 m = 2 64 28 7 9 m = 2 − − − 26 13 5 5 m = 2 49 21 6 8 m = 14 145 29 6 24 m 4 − − − ≥

Example 12.66. (SICs) For n = d2 and d = j2 1, j = 2,3,..., we have − 2 λ1 = j, λ2 = j( j 1), √d + 1 = j. − − The ﬁrst two cases are: d = 3, where there is a cube root signature matrix, • d = 8, where there is a fourth root signature matrix (the Hoggar lines). • Both are exceptional cases for SICs. In view of this, one can speculate about the existence of a SIC for d = 15,24,35,... with a simple form, which is not a Heisenberg SIC (see 14.5). § 316 12 Equiangular and Grassmannian frames 12.18 Algebraic equations for tight complex equiangular lines

The equations (12.22) determining the signature matrix Σ of an equiangular tight d frame of n vectors for C each involve variables z jk from each row/column of Σ. The number of these variables can be reduced by considering a reduced signature matrix Σˆ (Proposition 12.3). Here we consider necessary equations that involve only variables from principal submatrices of Σ =[z jk]. These have been used to good effect by Szoll¨ osi˝ [Szo14]¨ to investigate the existence of complex equiangular tight frames with small numbers of vectors. We now consider these ideas. The Gramian of an equiangular tight frame of n unit vectors for Cd, n > d, has the form

1 αz12 αz13 αz1n αz12 1 αz23 αz2n αΣ α α  α n d Q = I + =  z13 z23 1 , = − , z jk = 1,   d(n 1) | |  . . ..  −  . . .      αz1n αz2n 1      (12.52) where P := d Q is a rank d orthogonal projection matrix (see 12.6). In particular, n §

2 (i) The (d + 1) (d + 1) minors of I + αΣ are zero, giving n equations. × d+1 From P2 = P, i.e., dQ2 nQ = 0, a block matrix calculation (see Exer. 12.15) gives −

αΣ n (ii) If Qn r is a principal submatrix of Q = I + of size n r, 0 r 2 , then − − ≤ ≤ 2 rank(dQn r nQn r) r. − − − ≤ n 2 For a given r, this gives r+1 equations. Less obvious equations involving principal submatrices of Σ of size n 2, can be obtained from the equations (12.22) which characterise equiangular tight− frames, by using the following identity for the triple product of three ﬂat vectors in C2.

2 Lemma 12.2. Let x,y,z C , with x1 = x2 = y1 = y2 = z1 = z2 = 1, then ∈ | | | | | | | | | | | | x,y y,z z,x = x,y 2 + y,z 2 + z,x 2 4. | | | | | | − Proof. By direct calculation, using the fact that ξ = 1/ξ, when ξ = 1. | | ⊓⊔ 12.18 Algebraic equations for tight complex equiangular lines 317

With z jj := 0, the equations (12.21), which characterise equiangular tight frames, can be written

n 1 n (n 2d) − zab = ∑ zaszsb (n 1)δab, 1 a,b n. − d(n d) s=1 − − ≤ ≤ − s=a,b Thus, for 1 a,b n 2, we have ≤ ≤ − n 2 n 1 − cab :=(n 2d) − zab ∑ zaszsb +(n 1)δab = va,vb , (12.53) − d(n d) − s=1 − − s=a,b 2 where va :=(za,n 1,za,n) C . Since each va is flat, we can apply Lemma 12.2 to obtain equations− which∈ depend only on the principal submatrix of the signature matrix Σ =[z jk] of size n 2. − Theorem 12.19. ([Szo14])¨ Let n 5 and Σ =[z jk] be the signature matrix of an ≥ d equiangular tight frame of n vectors for C , n > d (here z jj = 0,z jk = zkj, j = k). For 1 a,b n 2, a = b, define ≤ ≤ − n 2 n 1 − cab :=(n 2d) − zab ∑ zaszsb, a = b, caa = 2. (12.54) − d(n d) − s=1 − s=a,b Then Σ satisfies

2 2 2 c jkck c j = c jk + ck + c j 4, 1 j,k,ℓ n 2. (12.55) ℓ ℓ | | | ℓ| | ℓ | − ≤ ≤ −

Proof. By Lemma 12.2, with x = v j, y = vk, z = vℓ, we have (12.55), where the formula (12.53) deﬁning cab can be written as (12.54). ⊓⊔ In [Szo14],¨ the equations (i), (ii) and (12.55) for an (n 2) (n 2) principal submatrix of the Gramian was used to give a description− of all×possible− sets of n tight complex equiangular lines in C3. We now outline what they are.

Example 12.67. There exists a unique set of n tight equiangular lines in C3 (up to projective unitary equivalence) for n = 3 (orthonormal basis) and n = 4 (vertices of the tetrahedron). For n = 5 there are no tight equiangular lines (Example 12.57). For n = 6 there are tight real equiangular lines given by the diagonals of the regular icosahedron (Example 12.39). Further, a calculation (using Groebner bases) of the possible 4 4 principal submatrices, shows that all the Gramian matrices for equiangular tight× frames six vectors in C3 have the (reduced signature) form 318 12 Equiangular and Grassmannian frames

√51 1 1 1 1  1 √5 a a 1 1  − −  √  (1) 1  1 a 5 1 a 1  G (a)=  − − , a C, a = 1. 6 √   ∈ | | 5  1 a 1 √5 a 1   − −     1 1 a a √5 1   − −     1 1 1 1 1 √5   − −    For n = 7, there is a unique set of 7 equiangular lines (given by the harmonic frame for the (7,3,1)–difference set). For n = 8, there are no tight equiangular lines in C3 (the equations have no solution). For n = 9, there is a one parameter family of SICs (see [BW07], [Zhu12], [Szo14]).¨

Example 12.68. For C4, sets of n tight equiangular lines are known for n = 4,5 (orthonormal basis, simplex), n = 7,13 (harmonic frames given by difference sets), n = 8 (Example 12.63), and n = 16 (a SIC). There are no tight equiangular lines for n = 6 (Example 12.57), and so we can pose the following elementary open problem:

Is there an equiangular tight frame of 9 vectors for C4?

12.19 Mutually unbiased bases and s–angular tight frames

For an equal–norm frame Φ (or set of lines), its set of angles is

Ang(Φ) := v,w : v,w Φ are are not scalar multiples . {| | ∈ } It is said to be s–angular (or has s angles) if Ang(Φ) has s elements. The 1–angular frames/lines are precisely the equiangular frames/lines, and MUBs are 2–angular tight frames. We now generalise Theorem 12.2.

Theorem 12.20. Let Φ be a set of n unit vectors in Cd (d > 1) giving a system of n (distinct) lines with s angles A = Ang(Φ), and

2 α k x,v x,x 1, 0 A; pv(x) := x,v ∏ | | −α , k := ∈ α A 0 1 0, 0 A. ∈ \ − ∈

Then the polynomials pv v Φ are linearly independent, and hence { } ∈ d+s 1 2 − , 0 A; n s ∈ (12.56) ≤ d+s 1 d+s 2 s− s −1 , 0 A. − ∈ 12.19 Mutually unbiased bases and s–angular tight frames 319

Proof. Each pv is a polynomial which is homogeneous of degree s in x1,...,xd and Π d is homogeneous of degree s k in x1,...,xd. The space s◦,s k(C ) of such polyno- − s+d 1 s k+d 1 − mials has complex dimension d −1 −d 1− (see Exer. 6.17). By construction, − − δw(pv)= pv(w)= δv w, v,w Φ, , ∈ so the pv v Φ are linearly independent, and we obtain (12.56). { } ∈ ⊓⊔ Remark 12.1. The same argument gives the following bound for vectors giving systems of lines in Rd 2s+d 1 d 1− , 0 A; n 2s+−d 2 ∈ (12.57) ≤ d 1− , 0 A. − ∈ Example 12.69. For nonorthogonal equiangular lines (s = 1, 0 A), (12.56) and (12.57) give Theorem 12.2. ∈ Example 12.70. Spherical two–distance sets (and tight frames) are 2–angular frames for Rd (see 12.14). The estimate of the maximum size of a spherical two–distance set in Rd given§ by (12.57) has a higher order of growth in d than that given by Theorem 12.14 (though it is exact for d = 2). Example 12.71. The highly symmetric tight frames given by ﬁnite reﬂection groups (see 13.8, 13.10) are s–angular tight frames with s small (see Tables 13.1, 13.2). § § For the 2–angular frame give by m mutually unbiased bases for Cd (see 2.11), the bound (12.56) gives §

d + 1 d 1 1 md = d2(d + 1) = m d(d + 1). ≤ 2 1 2 ⇒ ≤ 2 The following argument of [WF89] gives the sharper bound m d + 1. ≤ Proposition 12.12. Let B be m mutually unbiased bases for Cd. Then m d + 1. ≤ d I Proof. Let Pv be the orthogonal projection onto v C , so that Pv is a traceless ∈ − d Hermitian operator. Since B B is an orthonormal basis, ∑v B Pv = I, so that the I ∈ ∈ matrices Pv , v B are linearly dependent and span a space VB of dimension − d ∈ d 1. The spaces VB, B B are orthogonal in the Frobenius norm, since for v ≤and w−in mutually unbiased∈ bases, we have I I 1 1 1 Pv ,Pw = trace(PvPw) trace(Pv) trace(Pw) trace(I) − d − d − d − d − d2 1 2 1 = + = 0. d − d d The dimension of the real vector space of traceless of Hermitian operators is d2 1. 2 − Thus a dimension count gives dim( B BVB m(d 1) d 1 =(d + 1)(d 1), and cancelling d 1 gives the result.⊕ ∈ ≤ − ≤ − − − ⊓⊔ 320 12 Equiangular and Grassmannian frames 12.20 Mutually unbiased bases and Hadamard matrices

We now investigate MUBs in more detail. First consider the case of two MUBs (v j) d and (w j) for C . The synthesis operator of this tight frame (v j) (w j) is [V,W], ∪ where V =[v j], W =[w j], and so its Gramian has the form

V ∗V V ∗W I V ∗W [V,W]∗[V,W]= = .     W ∗V W ∗W W ∗V I     Thus, to understand two MUBs up to unitary equivalence, it sufﬁces to consider the d d matrix V W. This leads to the following connection with Hadamard matrices. × ∗ d Lemma 12.3. Two orthonormal bases (v j) and (w j) for C are mutually unbiased if and only if

w1,v1 wd,v1 . . . H := √dV ∗W = √d  . .. .  (12.58)    w1,vd wd,vd      is a (complex) Hadamard matrix (of size d).

Proof. If V =[v j] and W =[w j] are the synthesis operators of orthonormal bases d for C , i.e., V ∗V = I and WW ∗ = I, then H := √dV ∗W satisﬁes

H∗H = d(W ∗V)(V ∗W)= dW ∗(VV ∗)W = dWW ∗ = dI.

Thus (v j) and (w j) are mutually unbiased if and only if H has entries of modulus 1, i.e., H is a Hadamard matrix. ⊓⊔ We will refer to the H of (12.58) as the Hadamard matrix of (v j) and (w j). Conversely, if H is a (complex) Hadamard matrix of size d, then

I 1 H Q = √d (12.59)  1 H I  √d ∗   is the Gramian of two mutually unbiased bases for Cd. This follows since Q has the d 1 correct form, and is a tight frame for C , since P = 2 Q is an orthogonal projection of rank d, by the calculation

1 1 1 1 1 I + HH∗ H + H 1 I H P2 = d √d √d = √d = P. 4  1 H + 1 H 1 HH + I  2  1 H I  √d ∗ √d ∗ d ∗ √d ∗     12.20 Mutually unbiased bases and Hadamard matrices 321

Two MUBs for Cd are determined up to unitary equivalence by their Hadamard matrix, and each Hadamard matrix corresponds to two MUBs.

The vectors (v j) and (w j) can be multiplied by unit scalars (α j) and (β j) so that the Hadamard matrix of (12.58) has each entry of it ﬁrst row and column 1. This is called a dephased Hadamard matrix. In this way, each pair of MUBs corresponds to a unique dephased Hadamard matrix: once α1 is chosen the remaining scalars are determined by

βkwk,α1v1 = 1, 1 k d, β1w1,α jv j = 1, 2 j d, (12.60) ≤ ≤ ≤ ≤ with each choice of α1 giving the same dephased Hadamard matrix, which we call the dephased Hadamard matrix of (v j) and (wk). Proposition 12.13. Two mutually unbiased bases for Cd are uniquely determined up to projective unitary equivalence by their dephased Hadamard matrix.

d Proof. Let (v j) and (w j) be mutually unbiased bases for C . We already observed that their dephased Hadamard matrix is projectively unitarily invariant. Indeed, by solving (12.60), we see that the ( j,k)–entry of the dephased Hadamard matrix is the 4–product βkwk,α jv j = wk,v j v j,w1 w1,v1 v1,wk . It therefore sufﬁces to show that (v j) and (w j) are determined (up to projective unitary equivalence) by their dephased Hadamard matrix H. This follows from the formula (12.59), which gives the Gramian Q of a frame that is projectively unitarily equivalent to (v j) (w j). ∪ ⊓⊔ Hadamard matrices H1 and H2 are said to be equivalent if there are unitary diagonal matrices Λ1 = diag(α j), Λ2 = diag(β j) and permutation matrices Pσ , Pτ , with Λ 1 1 Λ H1 = 1− Pσ− H2Pτ 2

In particular, For the Hadamard matrix H = √d[ wk,v j ] of (12.58) is equivalent to 1 1 1 Λ − Pσ− HPτΛ2 = Λ − [ wτk,vσ j ]τΛ2 =[ βkwτk,α jvσ j ]. 1 1 Thus we have:

Two MUBs for Cd are determined up to projective unitary equivalence by their dephased Hadamard matrix H, and are determined to projective unitary equivalence after a reordering by H up to Hadamard matrix equivalence.

A catalogue of complex Hadamard matrices was given in [TZ06a], and an online version (for 2 d 16) is maintained by Bruzda, Tadej and Zyczkowski.˙ ≤ ≤ 322 12 Equiangular and Grassmannian frames

Example 12.72. For d = 2,3,5 there is a unique complex Hadamard matrix of size d up to equivalence, which is given by the Fourier matrix (see [Haa97], which uses the language of maximal abelian –algebras). For d = 4, there is one–parameter family of inequivalent Hadamard matrices∗ given by

11 1 1 1 ieia 1 ieia F(1)(a)= − − , 0 a < π. 4   ≤ 1 1 1 1   − −   ia ia  1 ie 1 ie   − −    Motivated by this example, it was conjectured that there is a unique Hadamard matrix of size d (up to equivalence), given by the Fourier matrix, when d is a prime. However, this is not the case for primes p 7 (see the discussion of [Szo10]).¨ ≥ Example 12.73. The Hadamard matrices of size d = 6 have not been fully classiﬁed. There is evidence that their variety has dimension 4 (see [SNS09], [Szo12]).¨ Let H be d d matrix with complex entries of unit modulus. Then then equations ×

H∗H = dI characterise when H is a Hadamard matrix. These are the analog of the equations 2 Σ (λ1 +λ2)Σ (n 1)I = 0 for the signature matrix of a equiangular tight frame (Corollary− 12.1).− Deﬁne− the dephased form of H as for when it is a Hadamard matrix

1 1∗ t d 1 (d 1) (d 1) , 1 =(1,...,1) R − , Hˆ C − × − . 1 Hˆ  ∈ ∈   We call Hˆ (which has unit modulus entries) the reduced Hadamard matrix of H. The analogue of the equations of Proposition 12.3 is as follows. Proposition 12.14. Let Hˆ be the reduced Hadamard matrix of a d d matrix with entries of unit modulus. Then Hˆ gives a d d Hadamard matrix if and× only if × 1. Hˆ ∗Hˆ = dI J, J := 1∗1. 2. 1 is an eigenvector− of Hˆ and of Hˆ for eigenvalue 1. ∗ − Proof. Block multiplication of the condition for being a Hadamard matrix gives

∗ 1 1∗ 1 1∗ d 1∗ + 1∗Hˆ d 0 = = dI = .         1 Hˆ 1 Hˆ 1 + Hˆ ∗1 11∗ + Hˆ ∗H 0 dId 1 −         Equating the blocks gives 1, and that 1 is an eigenvector of Hˆ for eigenvalue 1. ∗ − By considering the condition HH∗ = dI instead, we conclude that 1 must also be an eigenvector of Hˆ (for eigenvalue 1). − ⊓⊔ 12.20 Mutually unbiased bases and Hadamard matrices 323

By using Lemma 12.2 (as in Theorem 12.19), one can obtain necessary equations for being a Hadamard matrix that depend only of the submatrices of H of size d 2 (see [Haa97]). We now brieﬂy consider how to go from two mutually unbiased bases− (a Hadamard matrix) to three or more. Theorem 12.21. There are m mutually unbiased bases for Cd if and only if there 1 are m 1 Hadamard matrices H2,...,Hm of size d, for which the 2 (m 1)(m 2) matrices− − − 1 H∗Hk, 2 j < k m, √d j ≤ ≤ have entries of modulus of 1.

Proof. Let V1,...,Vm be the synthesis maps for m mutually unbiased bases B1,...,Bm d B B for C . Let Hj be the Hadamard matrix for 1 and m, i.e., Hm = √dV1∗Vm. Then the Hadamard matrix for B j and Bk is 1 1 Hjk := √dV ∗Vk = √dV ∗(V1V ∗)Vk = (√dV ∗Vj)∗(√dV ∗Vk)= H∗Hk. j j 1 √d 1 1 √d j

Thus all the Hadamard matrices for the B1,...,Bm are determined by H2,...,Hm by the above formula for 2 j < k m. These matrices satisfy ≤ ≤ 1 1 1 H H = (H H )(H H )= H (dI)H = H (dI)H = dI, ∗jk jk d ∗j k k∗ j d ∗j j d ∗j j and so are Hadamard (giving mutually unbiased bases) if and only if they have entries of modulus 1. ⊓⊔ The condition that H∗j Hk have entries of constant modulus, gives the following concrete construction for the mutually unbiased bases.

If H2,...,Hm are the Hadamard matrices of Theorem 12.21, then the columns of the d d matrices × 1 1 I, H2,..., Hm √d √d give m mutually unbiased bases for Cd.

Example 12.74. For the three MUBs for C2 of Example 2.18, i.e.,

1 0 1 1 1 1 1 1 1 1 B1 = , , B2 = , , B3 = , , 0 1 √2 1 √2  1 √2 i √2  i − −             the Hadamard matrices of Theorem 12.21 (which are all equivalent) are

1 1 1 1 1 1 1 + i 1 i H1 = , H2 = , with H23 = H2∗H3 = − . 1 1 i i √2 √2 1 i 1 + i − − −       324 12 Equiangular and Grassmannian frames 12.21 Examples of MUBs

Let F be the d d Fourier matrix as deﬁned by (2.5), i.e., × π jk 2 i Fjk := ω , j,k Zd, ω := e d . ∈ The columns of F are an orthonormal basis for Cd. Since F has entries of constant modulus, this basis is mutually unbiased to the standard basis (e j). This still holds if the rows/columns of F are multiplied by unit scalars. In this way, we seek further mutually unbiased bases. Suitable scalars can be described in terms of the diagonal matrix R of 14.7, which is given by § π j( j+d) 2 i R jk := µ δ jk, j,k Zd, µ := e 2d . (12.61) ∈ This matrix plays a key role in the description of the known SICs (see Chapter 14).

Theorem 12.22. Let F be the d d Fourier matrix, R be the d d diagonal matrix given by (12.61), and deﬁne orthonormal× bases by ×

ℓ E := e j , B := R Fe j . { } ℓ { }

Then for ℓ,m Zd, the following pairs are mutually unbiased bases ∈ 1. B ,E . { ℓ } 2. B ,Bm , ℓ m Z . { ℓ } − ∈ d∗ In particular, for d a prime, E ,B0,...,Bd 1 are d + 1 mutually unbiased bases. −

Proof. We have already observed that Bℓ and E are mutually unbiased, since

ℓ ℓ 1 R Fek,e j = (R F) jk = . | | | | √d

To show that Bℓ and Bk are mutually unbiased, we require

ℓ m 1 ℓ m 1 R Fek,R Fe j = (F− R − F) jk = , a := ℓ m Zd, | | | | √d − ∈

1 a i.e., the matrix F− R F, a Zd, has entries of constant modulus (for a = 0). In 1 ∈a other words, Ha := √dF− R F is a Hadamard matrix. This can be done by a direct calculation using Gauss sums (see Exercise 12.19), or as follows. A calculation (see Exer. 12.19) with Gauss sum shows that

1 a a(d 1) a+ j k 1 a (F− R F) jk = µ − ω − (F− R F) j,k a. (12.62) − 1 a Since F− R F is circulant (it is diagonalised by the Fourier matrix), this implies that it has entries of constant modulus when a is a unit (and so generates Zd). ⊓⊔ 12.21 Examples of MUBs 325

2πi Example 12.75. For d = 3, the diagonal matrix is R = diag(1,ω2,ω2), ω = e 3 . Thus four mutually unbiased bases for C3 are given by the standard basis, together with the columns of the matrices

1 1 1 1 1 1 1 1 1 1 1 1 F = 1 ω ω2, RF = ω2 1 ω, R2F = ω ω2 1 . √3 √3 √3  ω2 ω  ω2 ω  ω ω2 1   1   1              Example 12.76. For d = 6 (which is not a prime power), any pair of (cyclically) consecutive bases B0,...,B5 are mutually unbiased. Therefore E together with 6 any consecutive pair B j,B j+1 gives three mutually unbiased bases for C . A long standing open question is the MUB problem:

Do there exist more than three mutually unbiased bases in C6?

There are various constructions of d + 1 MUBs for Cd when d is a prime, or a prime power (see [GR09]). All of these constructions are cases of a construction of [CCKS97] based on symplectic spreads and Z4–Kerdock codes, which gives many projectively unitarily inequivalent sets of d + 1 MUBs for Cd, when d is an odd power of two. We now give the construction of [WF89] (also see [KR04]).

n Example 12.77. Let d = p be an odd prime power, Fd be the Galois field of order d (viewed as an extension of Zp), with field trace tr : Fd Zp. Define d d matrices → × 1 ωtr(aj2+ jk) Va = j,k F , a Fd. √d ∈ d ∈ d Then I, Va, a Fd, are the synthesis maps of d + 1 mutually unbiased bases for C . ∈ For d = p > 2, the field trace is the identity, and so the matrices Va are given by

2a Va = R F, a Zd, ∈ which are the bases of Theorem 12.22 (since 2 is a unit). Example 12.78. Let d = 2n be an even prime power, and R be the Galois ring 2 GR(2 ,n) with Teichmuller¨ set Tn and trace tr : R Z4. Deﬁne d d matrices by → × 1 tr(aj2+2 jk) Wa = i , j,k Tn √d ∈ d Then I, Wa, a Tn, are the synthesis maps of d +1 mutually unbiased bases for C . ∈ The examples of this section can be viewed as the eigenvectors of orthogonal (in the Frobenius inner product) commuting (up to a scalar) unitary matrices, e.g., see [BBRV02], [KR04], [GR09], Exer. 14.17 and Theorem 12.22) 326 12 Equiangular and Grassmannian frames Notes

There is currently considerable activity on determining estimates of the maximal number M(d) of real equiangular lines in Rd (see 12.8 and Table 12.3), e.g., [Gre16] reduced the bound for equiangular lines in R§18 from 48 61 to 48 60, then [Szo17]¨ improved it to 54 60, [BY14] give bounds on M−(d) obtained− by semideﬁnite programming, and [Buk16],− [BDKS16] give asymptotic estimates in d for when the angle α is ﬁxed. Maximal relative projection constants and equiangular tight frames are studied in [FS17]. There are various internet sites with lists of equiangular tight frames (also see [FM15]), and the associated geometric structures, such as difference sets, Steiner systems, strongly regular graphs and Hadamard matrices. An excellent resource is the talks and follow up preprints on the webpage of the workshop Systems of Lines: Applications of Algebraic Combinatorics organised by Bill Martin. Thanks to Alexander Barg, Peter Cameron, Simon Foucart, Gary Greaves, John Jasper, Ferenc Szoll¨ osi˝ and Wei-Hsuan Yu for many insightful discussions.

Exercises

12.1. Show that the dimension of the real vector space of d d Hermitian matrices 1 2 × is 2 d(d + 1) when F = R, and is d when F = C.

n d 12.2. Suppose that (Pj) j=1 are the projections onto a set of equiangular lines in F , α ∑n d > 1, with constant C = , and the identity can be written as I = j=1 c jPj. d (a) Show that c j = n , j. (b) Show that ∀ n(1 dα2)= d(1 α2), − − α2 1 and hence < d ,

d(1 α2) n d n = − , α2 = − . 1 α2 d(n 1) − − d 12.3. Suppose that ( f j) is a ﬁnite normalised tight frame of nonzero vectors for F (d 2) which satisfy the equiangularity condition ≥ f j f , k = C, j = k. f j fk Show that ( f j) is an equal–norm frame, i.e., it is an equiangular tight frame.

12.4. Show that the 28 unit vectors of (12.6) are equiangular.

2 12.5. Let Φ =( f j) be n unit vectors in R . Show that (12.3) can be sharpened to 12.21 Examples of MUBs 327

π n 2 M∞(Φ) := max f j, fk cos > − , n > 3 j=k | | ≥ n 2(n 1) − with equality if and only if the f j are the ﬁrst n vertices of the regular 2n–gon (up to multiplication by 1), i.e., they give n equally spaced lines in R2. ± d 12.6. Show that if Φ =( f j) is a unit–norm tight frame of n vectors for F which minimises M∞(Φ) := max f j, fk , j=k | | n d then so is the complementary tight frame Ψ =(g j) for F − (scaled appropriately).

12.7. Let Φ be the nontight equiangular frame of five vectors for R3 given by Γ the 5–cycle (see 12.13). (a) Calculate§ the minimal angle between the five lines given by Φ, Φ˜ , and Φcan. (b) Show the vectors in Φ lie on five of the six diagonals of the regular icosahedron. Remark: Since the solution to the Grassmannian packing problem for five lines in R3 is given by five diagonals of the regular icosahedron (see [CHS96]), we conclude that Φ is a nontight Grassmannian frame. (c) Show that Φcan is the harmonic frame given by the fifth roots of unity. Hint: The vertices of the regular icosahedron are given by the cyclic permutations of the vector (0, 1, c), where c = 1+√5 is the golden ratio. ± ± 2 12.8. If a (nontight) sequence of unit vectors defines a set of equiangular lines, then does the dual frame define a set of equiangular lines?

12.9. Suppose that (v j) is a set of n ﬂat equiangular lines (see Proposition 12.1). Let A be the Gram matrix of the rank one orthogonal projections v1v1∗,...,vnvn∗ and e1e1∗,...,dded∗ (with the Frobenius inner product), i.e.,

2 2 1 α J +(1 α )In J A = − d , J =[1].  1  d J Id   (a) Row reduce A to an upper triangular matrix B (to determine its rank). (b) Show that n+d 1 rank(A) n+d, and Φ is tight when rank(A)= n+d 1. − ≤ ≤ − 12.10. Let (va, j) be the Steiner equiangular tight frame of Theorem 12.4 given by a (2,3,ν)–Steiner system B on points V (a Steiner triple system), and Hadamard matrices H(a), a V . Show that the vectors ∈

(v˜a, j)a V ,1 j r+1 (w˜ℓ)1 ℓ ν+1 ∈ ≤ ≤ ∪ ≤ ≤ given by (12.17) and (12.18) form an equiangular tight frame for Cβ CV C, with α 1 ⊕ ⊕ angle = r+2 . 328 12 Equiangular and Grassmannian frames

12.11. (Conference matrices) Let Σ = iC be an n n complex signature matrix with off diagonal entries i. Here C has zero diagonal× and off diagonal entries 1. (a) Show that C = ±CT , i.e., C must be skew–symmetric. ± (b) Show that Σ gives− an equiangular tight frame of n vectors for Cd if and only if CTC =(n 1)I, i.e., C is skew–symmetric conference matrix, where n = 2d. − 12.12. Let C be a skew–symmetric conference matrix of size n + 1 in the standard form 0 1∗ C = , 1 :=(1,1,...,1).  1 A  −   These exist for n = 2k 1 (Example 12.63) and n = pm an odd prime power with n 3 (mod 4) (with A−the Seidel adjacency matrix (12.64) of the Paley digraph). ≡ 2 (a) Show that A satisﬁes A = J nI and AJ = JA = 0, where J = 11∗. (b) Show that the n n matrix − × 1 Σ := (J I)+ i√nA (12.63) √n + 1 ± − is the signature matrix of an equiangular tight frame of n vectors for Cd, where n 1 d = ± (n = 2d 1). 2 ∓ (c) Show that Σ + ζI, ζ = 1 (√ni 1), is an n n complex Hadamard matrix. √n+1 ± × 12.13. Let q = pm = 4m 1 be a (necessarily odd) prime power. Let S be the set of all nonzero squares in the− Galois ﬁeld GF(q), i.e.,

S := x2 : x GF(q),x = 0 . { ∈ } m This is a (4m 1,2m 1,m 1)–difference set for G :=(GF(q),+) ∼= Zp , i.e., gives − − − 2m 1 an equiangular harmonic frame of 4m 1 vectors for C − (by Theorem 12.3). Denote the nonsquares in G by N := G− (S 0 ). The q q matrix given by \ ∪ { } × 0, j = k; A jk := 1, j k S; (12.64)  − ∈  1, j k N, − − ∈ is the Seidel adjacency matrix of the Paley digraph. (a) Show that S 0 is a (4m 1,2m,m)–difference set for G. (b) Show that N∪and { }N 0 are− difference sets for G. ∪ { } 2m 1 (c) Show that there are equiangular tight frames of 4m 1 vectors for C − and C2m for which the inner products between their vectors have− a constant real part. (d) Show that there is an equiangular tight frame of 4m vectors for C2m.

12.14. Let Σˆ be the reduced signature matrix of an equiangular tight frame of n > d + 1 vectors for Rd, which corresponds (by Theorem 12.12) to a strongly regular graph 12.21 Examples of MUBs 329

3k n k 1 n d(n 1) srg n 1,k, − , , k := n 1 + 1 − . − 2 2 2 − − 2d n d − n d Show that the complementary equiangular tight frame of n vectors for R − has reduced signature matrix Σˆ , and so is given by the complementary strongly regular graph, which has the parameters−

2n 3k 6 n k 2 srg(n 1,n k 2, − − , − − ). − − − 2 2 12.15. The Gramian matrix Q of an equiangular tight frame of n unit vectors for Cd satisﬁes dQ2 nQ = 0. Show that this implies − 2 rank(dQn r nQn r) r, − − − ≤ where Qn r is any principal submatrix of Q of size n r, 0 r < n. − − ≤ 12.16. Let G be the Gramian of an equiangular tight frame of n unit vectors for Cd, and U be the n n matrix × d U := I 2 G. − n d (a) Show that U is Hermitian, unitary, with constant diagonal entries 1 2 n and constant modulus off diagonal entries. − (b) Suppose that U is an n n Hermitian unitary matrix with constant diagonal entries λ = 1 2 d [ 1,1]×and constant modulus off diagonal entries. Show that − n ∈ − n G := (I U) 2d − is the Gramian of an equiangular tight frame of n unit vectors for Cd. 2 Remark: A matrix B is unistochastic if it has the form B jk = Ujk , for U unitary. Such a B is bistochastic (its rows and columns sum to 1). The above| | correspondence between equiangular tight frames and unistochastic matrices with constant diagonal and constant modulus off diagonal entries is considered by [GT16].

d 12.17. Let (v j) be a unit norm tight frame of n vectors for F . (a) For 0 < p < 2, show that

2 p n 2 p ( d n) ∑∑ v j,vk − p + n, 2 1 j | | ≤ (n n) 2 k − − with equality if and only if (v j) is an equiangular tight frame. (b) For 2 < p < ∞, show that

2 p n 2 p ( d n) ∑∑ v j,vk − p + n, 2 1 j | | ≥ (n n) 2 k − − 330 12 Equiangular and Grassmannian frames with equality if and only if (v j) is an equiangular tight frame. (c) Show that for p = 2 equality holds in the inequalities of parts (a) and (b).

12.18. Let H1 and H2 be d d (complex) Hadamard matrices. By (12.58), these × d are given by pairs of MUBs for C with synthesis operators [V1,W1] and [V2,W2]. Suppose that these two pairs of MUBs are projectively unitarily equivalent up to a reordering which maps the respective bases to each other, i.e., there is a unitary map U, unitary diagonal matrices Λ1 = diag(α j), Λ2 = diag(β j), and permutation matrices Pσ , Pτ , with

Pσ Λ1 U[V1,W1]=[V2,W2] . (12.65)   PτΛ2   (a) Show that the condition (12.65) is equivalent to H1 and H2 being Hadamard matrix equivalent. (b) Use the fact that a pair of MUBs for Cd is determined up to projective unitary equivalence by its frame graph and the nonzero 4–products (the 4–cycles span the frame graph) to show that H1 and H2 are Hadamard matrix equivalent if and only if

(H1) jk(H1)kℓ(H1)ℓm(H1)mj =(H2)σ j,τk(H2)σk,τℓ(H2)σℓ,τm(H2)σm,τ j, for some permutations σ,τ Sd. In particular, the multiset ∈

HjkHk H mHmj : 1 j,k,ℓ,m d { ℓ ℓ ≤ ≤ } must be the same for all equivalent Hadamard matrices H.

12.19. Gauss sums. Let F and R be given by (2.5) and (12.61), and a Zd∗ be a unit. 1 a ∈ Since F− R F is diagonalised by the Fourier matrix F, it is a circulant matrix. 1 a (a) Show that the entries of F− R F are generalised Gauss sums, i.e.,

c 1 π 1 a 1 − 2 i (an2+bn) (F− R F) jk = G(a,ad + 2(k j),2d), G(a,b,c) := ∑ e c . 2d − n=0

(b) Let b be the multiplicative inverse of a unit b Z . Show that ∈ d∗ d 1 a k j 2 2ω −2 ( ) G(2a,0,d), d odd; G(a,ad + 2(k j),2d)= − µ a( d +a(k j))2 − − 2 − G(a,0,2d), d even.

G(2a,0,d) = √d, d odd, G(a,0,2d) = √2√2d, d even. | | | | 1 a Use this to prove that the entries of F− R F have constant modulus. Chapter 13 Tight frames generated by nonabelian groups

If G is a ﬁnite abelian group, then there are a ﬁnite number of tight G–frames, i.e., the harmonic frames (see 11). If G is nonabelian, then there is an uncountable number of unitarily inequivalent§ G–frames (see Proposition 10.1). To illustrate this, consider the smallest nonabelian group G = D3 = a,b ∼= S3, the dihedral group of order 6, acting on R2 as unitary transformations via:

2π a = rotation through 3 , b = reﬂection in the x–axis. (13.1) π Then (see Fig. 13.1), for each of the unit vectors vθ :=(cosθ,sinθ), 0 θ , the ≤ ≤ 6 tight D3–frames (gvθ )g D are unitarily inequivalent (since their angles differ). ∈ 3

θ π π Fig. 13.1: The unitarily inequivalent tight D3–frames given by vθ , for = 0, 12 , 6 .

Here we study the tight G–frames (gv)g G for G nonabelian by: ∈ Showing that (gv)g G corresponds to an element of the group algebra CG. • ∈ Putting additional restrictions on (gv)g G to obtain a ﬁnite set of G–frames, e.g., • the central G–frames and the highly symmetric∈ G–frames. Investigating nonabelian groups G which come as projective representations of • nice groups. These give interesting tight frames, e.g., all the known SICs are G–frames for a projective representation of an abelian group (see X). §

331 332 13 Tight frames generated by nonabelian groups 13.1 The identiﬁcation of the G–matrices with the group algebra

Let G be a finite group. By Corollary 10.2, (φg)g G is a normalised tight G–frame if and only if its Gramian is a G–matrix which is∈ a projection. To understand such projections, we now consider the structure of the algebra of G–matrices. Definition 13.1. Given a function ν : G C, let M(ν) be the G–matrix → 1 M(ν) :=[ν(g− h)]g,h G. (13.2) ∈ The group algebra CG of G, is the algebra obtained from the complex vector space with basis the elements of G, and the multiplication given by extending the multiplication in G linearly. Proposition 13.1. (G–matrix algebra) Let G be a finite group. Then the G–matrices form an algebra, i.e., the sum and product of G–matrices is again a G–matrix. This algebra is isomorphic to the group algebra CG, via the map

π : M(ν) ∑ ν(g)g. (13.3) → g G ∈

Proof. Clearly, M is an injective linear map from CG onto the G–matrices, which therefore form a vector space. The product of G–matrices is a G–matrix since

1 M(ν)M(µ)= M(ν µ), (ν µ)(g) := ∑ ν(gh)µ(h− ), (13.4) ∗ ∗ h G ∈ and so the G–matrices form an algebra. G Let eg : G C, g G be the standard basis vectors for C . In view of the natural vector space isomorphism→ ∈ between CG and CG, it follows that

g M(eg), g G (13.5) → ∈ gives a vector space isomorphism between CG and the G–matrices. Further, this is an isomorphism of algebras since M(eg )M(eg )= M(eg eg ) where 1 2 1 ∗ 2 1 1 (eg1 eg2 )(g)= ∑ eg1 (gh)eg2 (h− )= eg1 (gg2− )= eg1g2 (g) = eg1 eg2 = eg1g2 . ∗ h G ⇒ ∗ ∈ Finally, we observe that the inverse of this isomorphism (13.5) is (13.3). ⊓⊔ The corresponding element of the group algebra for the three equally spaced unit vectors viewed as an S3–frame was calculated in Example 10.5. We observe that the Hermitian transpose of a G–matrix is given by the formula

ν ν ν ν 1 M( )∗ = M( ˜ ), ˜ (g) := (g− ). (13.6) Next we characterise the Gramians of the normalised tight G–frames when viewed as elements of the group algebra via (13.3). 13.2 Tight G–frames as idempotents of the group algebra 333 13.2 Tight G–frames as idempotents of the group algebra

We now show that:

The normalised tight G–frames, i.e., the G–matrices which are orthogonal projections, correspond to idempotents of the group algebra CG.

Proposition 13.2. (Characterisation) The normalised tight G–frames Φ =(φg)g G ∈ are in a 1–1 correspondence with the elements p = ∑g cgg of the group algebra CG 2 satisfying c 1 = cg, g G, and p = p, i.e., g− ∀ ∈

∑ chc 1 = cg, g G, (13.7) h− g h G ∀ ∈ ∈ given by P = Gram(Φ)= M(ν) p = ∑ν(g)g, (13.8) ←→ g where ν(g)= φg,φ1 = gφ1,φ1 . Proof. Here (13.8) is the correspondence p = πP of Proposition 13.1. Since Φ is a normalised tight frame if and only if its Gramian is an orthogonal projection matrix (Corollary 10.2), it sufﬁces to determine the conditions on p = ∑g cgg CG which 2 ∈ ensure that P is an orthogonal projection, i.e., P∗ = P, P = P. By (13.6), the ﬁrst condition is that cg = c 1 . Since the π of (13.3) is an isomorphism of algebras, the g− second condition is that

2 p = ∑ ∑ ch1 ch2 h1h2 = ∑ cgg = p. h G h G g G 1∈ 2∈ ∈ This can be rewritten as (13.7). ⊓⊔

Example 13.1. Let G = C3 = a be the cyclic group of order 3. The ﬁrst condition gives c1 R (this always the case) and c 2 = ca. The second condition (13.7) is that ∈ a 2 2 2 c1 + caca2 + ca2 ca = c1, c1ca + cac1 + ca2 = ca, c1ca2 + ca + ca2 c1 = ca2 . Solving these equations gives the following six choices for p 1 1 1 1 0, (1+a+a2), (1+ωa+ω2a2), (2 ω2a ωa2), (2 a a2), 1. 3 3 3 − − 3 − − The ranks of the corresponding orthogonal projections P are 0,1,1,2,2,3. Example 13.2. The SIC of (1.7) viewed as a G–frame for G = S,Ω corresponds to ∑ν(g)g = 1 √3I + S + Ω iSΩ √3( I) ( S) ( Ω)+ i( Sω) . 4√3 g − − − − − − − − 334 13 Tight frames generated by nonabelian groups 13.3 Characters of nonabelian groups

The (linear) characters of an abelian group generalise as follows. The character of a representation ρ : G GL(H ) of a ﬁnite group G (or the → FG–module H ) is the map χ = χρ : G C deﬁned by → χ(g) := trace(ρ(g)).

The degree of χ is deg(χ) := dim(H ). A character is said to be irreducible if the corresponding representation is irreducible. The character χ of G satisﬁes χ is constant on the conjugacy classes of G. • 1 χ(g− )= χ(g). • χ(1)= deg(χ)= dim(H ). • Characters are important in study of CG–modules, in particular CG–modules are CG–isomorphic if and only if they have the same character. • If χ is a character of G, then χ is a character of G. • The following example motivates the class of central G–frames (see 13.4). § Proposition 13.3. Let W be an irreducible CG–module of dimension d. Suppose H d that there is a unitary action of G on ∼= W = W W (CG-isomorphism). Then all normalised tight G–frames Φ for H are unitarily⊕⊕ equivalent, with χ 1 (1) P = Pχ := Gram(Φ)=[ν(g− h)]g,h G, ν(g) := χ(g), (13.9) ∈ G | | where χ is the character of W.

Proof. Without loss of generality, assume the action ρ on W is unitary, and that the unitary action on V = W d is given by

g (w1,...,wd) :=(ρ(g)w1,...,ρ(g)wd).

Let Vj be the absolutely irreducible subspace of vectors

v j =(0,...,w j,...,0), w j W, ∈ which are zero in all but the j–th coordinate (and the zero vector). Clearly, these are orthogonal, and σ j : Vj W : v j w j is a CG–isomorphism. → → Φ Let v = ∑ j v j jVj. By Theorem 10.8, =(gv)g G is a normalised tight frame for H if and only∈⊕ if ∈

2 2 d v j = w j = , σ jv j,σkvk = w j,wk = 0, j = k. G | | Thus the ν : G C deﬁning the Gramian P of the G–frame Φ is given by → 13.3 Characters of nonabelian groups 335 ρ ν d (g)w j,w j (g) := gv,v = g∑v j,∑vk = ∑ gv j,v j = ∑ 2 G w j j k j | | j dim(W) χ(1) = trace(ρ(g)) = χ(g). G G | | | | ⊓⊔ By Theorem 10.6 (or from the proof of Proposition 13.3), it follows that the Φ H d unique normalised tight G-frame for ∼= W , W an irreducible CG–module of dimension d and character χ, can be realised by

χ(1) Φ = Φχ := ρ(g) , A,B = trace(AB∗), (13.10) G g G | | ∈ where ρ : G U (Cd) is a unitary representation equivalent to W. → Example 13.3. Let G = D3 = S3 be the dihedral group of order 6 (see Example 10.3), ∼2 2 and order its elements 1,a,a ,b,ab,a b. An irreducible representation of D3

2 2 2 4 ρ : D3 U (C ) C × C → ⊂ ≈ with character χ =(2, 1, 1,0,0,0) is given by − − 1 ω ω2 1 0 0 ω 0  0  ω2 0  0  ρ(1)= , ρ(a)= , ρ(a2)= ,   ≈    ω2 ≈    ω ≈   0 1 0 0  0  0  0              ω2    ω  1                 0 0 0 0 1 1 0 ω  ω  0 ω2 ω2 ρ(b)= , ρ(ab)= , ρ(a2b)= ,   ≈   ω2  ≈ ω2 ω  ≈  ω  1 0 1 0   0                     0  0   0        2πi       where ω := e 3 . Thus from (13.10) we obtain the normalised tight D3–frame

1 ω ω2 0 0 0 1 0  0   0  1  ω  ω2 Φ = , , , , , √          2   3 0  0   0  1 ω   ω                ω2  ω        1     0  0   0                          4 Φ 1 χ 1 for C , which has P = Gram( )=[ 3 (g− h)]g,h D3 . ∈ 336 13 Tight frames generated by nonabelian groups 13.4 Central G–frames

1 The G–frame of Proposition 13.3 with Gramian P =[ν(g− h)]g,h G has the property that ν : G C is a class function, i.e., is constant on the conjugacy∈ classes of G. The irreducible→ characters form a basis for the vector space of class functions on G, and ν is a class function if and only if

∑ ν(g)g Z(CG), (13.11) g G ∈ ∈ where Z(CG) denotes the centre of the group algebra CG.

Definition 13.2. A G–frame Φ =(φg)g G is said to be central if ν : G C defined by ∈ → ν(g) := φg,φ1 = gφ1,φ1 is a class function. In view of Proposition 13.3 and (13.11), each of the following conditions on a 1 G–frame Φ with Gram(Φ)=[ν(g− h)]g,h H are equivalent to it being central ∈ ν is a class function. • ν ∑g G (g)g Z(CG). • Gram∈ (Φ) is∈ in the centre of the algebra of group matrices. • The symmetry condition gv,hv = gw,hw , g,h G, v,w Φ. • ∀ ∈ ∀ ∈ For G abelian, all G–frames are central (the conjugacy classes are singletons), and so the central G–frames are a generalistion of the harmonic frames. We will show (Theorem 13.1) that there are a finite number of tight central G– frames (for a given G). To this end, consider the homogeneous normalised tight central G–frame of Proposition 13.3, which has Gramian

χ(1) χ 1 χ(1) χ Pχ :=[ G (g− h)]g,h G = G M( ), (13.12) | | ∈ | | where χ is an irreducible character of G. Since χ is a class function, the idempotent pχ CG that it corresponds to (via Proposition 13.2) is in the centre of the group algebra,∈ i.e., χ(1) pχ := ∑ χ(g)g Z(CG). (13.13) G g G ∈ | | ∈ Moreover, for different characters, these homogeneous G–frames are orthogonal (Theorem 10.7), and so the product of their Gramians is zero (Lemma 5.1), which gives pχ pχ = 0, χ j = χk. (13.14) j k

χ χ Thus, if 1,..., r are the irreducible characters of G, then pχ j 1 j r is a basis of (orthogonal) idempotents for Z(CG). { } ≤ ≤ 13.5 The classiﬁcation of central tight G–frames 337 13.5 The classiﬁcation of central tight G–frames

The central tight G–frames can be characterised in terms of the Gramian.

Theorem 13.1. (Classification) Let G be a finite group with irreducible characters χ1,..., χr. Then Φ is a central normalised tight G–frame if and only if its Gramian is given by χ Φ j(1) χ Gram( )= ∑ Pχ j = M ∑ j , (13.15) j J j J G ∈ ∈ | | for some J 1,...,r , where Pχ is defined by (13.12), and M by (13.2). ⊂ { } Φ Proof. The two formulas given for Gram( ) are equal. Since Pχ j Pχk = 0, j = k, the first gives the Gramian of a normalised tight G–frame, and this is central (by the second formula). Thus, it suffices to assume that Φ is a central normalised tight G–frame, and to show that P := Gram(Φ)= M(ν) is given by (13.15). Φ ν ν Since is central, is a class function, and so the idempotent p = ∑g (g)g corresponding to P is in Z(CG). Write p in terms of the basis p j 1 j r, p j := pχ j for CG { } ≤ ≤ p = ∑α j p j, α j C. j ∈ Since p is an idempotent, (13.14) gives 2 α α α2 α α2 α p = ∑∑ j k p j pk = ∑ j p j = p = ∑ j p j = j = j. j k j j ⇒ α α Hence j 0,1 , and p = ∑ j J p j, where J := j : j = 1 . We therefore have ∈ { } ∈ { } π 1 π 1 P = − (p)= − ∑ pχ j = ∑ Pχ j . j J j J ∈ ∈ ⊓⊔ 4 The G–frame Φ of Example 13.3 is a central normalised tight D3–frame for C (by construction). Excluding the frame (0)g G, there are six others. More generally: ∈

For a given ﬁnite group G with r distinct irreducible characters there are 2r 1 nontrivial central normalised tight G–frames (up to unitary equivalence). −

Theorem 13.1 leads to the following count.

Corollary 13.1. Suppose there is a unitary action of G on the complex space H . Then either 1. There is no G–frame for H (H is not CG–isomorphic to a submodule of CG). 2. There is one tight G–frame for H (which is central) 3. There are uncountably many tight G–frames for H (none of which are central). 338 13 Tight frames generated by nonabelian groups

a Proof. In terms of the the homogeneous decomposition H = W W W W of (10.28), the three cases are ⊕ ∈

1. aW > dim(W), for some W (apply Proposition 10.4). 2. aW 0,dim(W) , for all W (apply Proposition 13.3). ∈ { } 3. 1 < aW < dim(W), for some W (apply Proposition 10.1).

In the second case, J = j : χ j = char(W) and aW = dim(W), W W in (13.15), where char(W) is the character{ of W. ∈ } ⊓⊔ The character of H can be determined from any G–orbit which spans H .

Proposition 13.4. Let P = M(ν) be the canonical Gramian of a G–frame for H . Then the character χ of the representation is given by

1 χ(g)= ∑ ν(h− gh). h G ∈ In particular, if the G–frame is central, then χ = G ν. | | Proof. By Corollary 10.3, we can assume that the G–frame is (Peg)g G, with ∈

g(Peh)= egh, H = ran(P).

Since (Peh)h G is a normalised tight frame, the trace formula (Exercise 2.13) gives ∈ 1 χ(g)= ∑ gPeh,Peh = ∑ Pegh,eh = ∑ Ph,gh = ∑ ν(h− gh). h G h G h G h G ∈ ∈ ∈ ∈ If ν is a class function, then ν(h 1gh)= ν(g), and we get χ(g)= G ν(g). − | | ⊓⊔ For the orthogonal projections Pχ of (13.12) given by irreducible characters χ, 2 the condition Pχ = Pχ , and the orthogonality relation

Pχ Pχ = 0, χ j = χk, j k can be expressed using (13.4) as

G χ χ = | | χ, (13.16) ∗ χ(1)

1 (χ j χk)(g)= ∑ χ j(gh)χk(h− )= 0, g G (χ j = χk). (13.17) ∗ h G ∀ ∈ ∈ These formulas for the convolution of irreducible characters are well known. The special case g = 1 in (13.17) gives the orthogonality of characters (11.5), i.e., 1 χ j, χk := ∑ χ j(h)χk(h)= 0, = δ jk. (13.18) G h G | | ∈ 13.6 The idempotents and the homogeneous decomposition 339 13.6 The idempotents and the homogeneous decomposition

The (orthogonal) idempotents pχ of (13.13), or, more precisely, the idempotents χ (1) 1 qχ := pχ = ∑ χ(g− )g Z(CG). (13.19) G g G ∈ | | ∈ play a very special role in the homogeneous decomposition of a CG–module V (Lemma 10.2). For W an irreducible CG–module with character χ, we recall the homogeneous component of V is

HV (χ)= HV (W) := ∑ X. X V X ⊂W ∼= Let Gˆ denote the irreducible characters of G. We now state (and prove) the well known formula for the homogeneous components.

Theorem 13.2. Let G be a ﬁnite group, and V be a CG–module. Then the direct sum decomposition of V into it homogeneous components is given by

V = qχV, (13.20) χ Gˆ ∈ i.e., HV (χ)= qχV, where the sum is orthogonal if the action of G on V is unitary.

Proof. Let W be a d–dimensional irreducible with character χ. We will ﬁrst show that qχ w = w, w W. (13.21) ∀ ∈ For this, we can assume that the action of G is unitary (Corollary 10.1). As in the proof of Proposition 13.3, choose w1,...,wd W so that (gv)g G, v =(w1,...,wd) ∈ d∈ χ(1) is a central normalised tight G–frame for W , i.e., gv,v = χ(g). Here w1 can G be any nonzero element of W, up to a scalar multiple. We calculate| | χ (1) 1 1 qχ v = ∑ χ(g− )gv = ∑ g− v,v gv = ∑ v,gv gv = v. g G G g G g G ∈ | | ∈ ∈

Taking the ﬁrst component gives qχ w1 = w1, which gives (13.21). As in Lemma 10.2, write V as a direct sum

V = V1 V2 Vm, ⊕ ⊕⊕ of irreducible G–invariant subspaces (which are orthogonal if the action is unitary). 2 χ χ χ χ χ ∑ χ Using the properties qχ = q and q j q k = 0, j = k it follows that χ Gˆ q = 1 ∈ (apply ∑χ qχ to the irreducible submodules W of CG), which gives (13.20). ⊓⊔ 340 13 Tight frames generated by nonabelian groups 13.7 An illustrative example

We now return to our motivating example of the nonabelian group G = D3 = a,b of order 6 acting on R2 via (13.1), i.e.,

1 1 √3 1 0 a := − 2 2 , b := . (13.22)  1 √3 1  0 1 − 2 − 2 −     The conjugacy classes of G are 1 , a,a2 , b,ab,a2b . Since the action of G is { } { } { } irreducible, the orbit Φ =(gw)g G of any vector ∈ 1 x w := , x2 + y2 = 1 3 y is a normalised tight frame, with Gramian the G–matrix P = M(ν), given by

ν(1)= 1 , ν(a)= 1 , ν(a2)= 1 , ν(b)= 1 x2 1 y2, 3 − 6 − 6 3 − 3 ν(ab)= 1 x2 1 xy + 1 y2, ν(a2b)= 1 x2 + 1 xy + 1 y2. − 6 − √3 6 − 6 √3 6 It is easy to verify (13.7) holds, e.g., for g = 1, we have 1 ν(1)2 +ν(a)ν(a2)+ν(a2)ν(a)+ν(b)2 +ν(ab)2 +ν(a2b)2 = (x2 +y2)2 = ν(1). 3 From Proposition 13.4, we can determine the character χ of the representation

χ(1)= 6ν(1)= 2, χ(a)= χ(a2)= 3ν(a)+ 3ν(a2)= 1, − χ(b)= χ(ab)= χ(a2b)= 2ν(b)+ 2ν(ab)+ 2ν(a2b)= 0. With the order 1,a,a2,b,ab,a2b, the characters of G are

1 1 2 1 1 1     −  1 1 1 χ1 =  , χ2  , χ3 = − . (13.23) 1  1  0          −    1  1  0    −    1  1  0    −          χ χ Thus we can deduce from P that the action of G is irreducible with = 3. With qχ j the idempotents of (13.19), Theorem 13.2 gives qχ1V = qχ2V = 0, qχ3V = V, i.e.,

I + a + a2 (b + ab + a2b)= 0, 2 (2I a a2 + 0(b + ab + a2b)) = I. ± 6 − − where a and b are deﬁned by (13.22). 13.8 The highly symmetric tight frames 341 13.8 The highly symmetric tight frames

2 For the irreducible action of G = D3 on R given by (13.1) there are uncountably many inequivalent G–frames (see Figure 13.1). None of these are central, since the 2 only central D3–frame for R is the one obtained by taking the characters χ1 and χ2 of (13.23), which results in three copies of an orthogonal basis. We would like to θ π think of the six equally spaced unit vectors ( = 6 ), which has a larger symmetry group than the others, as being a G–frame worth singling out. This leads to the notion of a highly symmetric frame. Definition 13.3. A finite frame Φ of distinct vectors is highly symmetric if the action of its symmetry group Sym(Φ) is irreducible, transitive, and the stabiliser of any one vector (and hence all) is a nontrivial subgroup which fixes a subspace of dimension exactly one. We recall that the action of the symmetry group of a finite frame is given by (9.1). If Φ is a highly symmetric frame, then (by the orbit size theorem)

Sym(Φ) > Φ . | | | | As defined, a highly symmetric frame has distinct vectors, and so it may not be a group frame (Theorem 10.4) unless the vectors are repeated some fixed number of times, e.g., it is naturally a Sym(Φ)–frame. Since a frame is highly symmetric if and only if the canonical tight frame is, it suffices to consider only the highly symmetric tight frames. The key features of the class of highly symmetric tight frames are:

There is a finite number of highly symmetric tight frames of n vectors for Cd. • They can be computed from the representations of abstract groups. • It is possible to determine whether or not a given tight frame is highly symmetric. • Some harmonic frames are highly symmetric tight frames. • The vertices of the regular complex polytopes are highly symmetric tight frames. • All finite reflection groups give highly symmetric tight frames. • There are no highly symmetric frames for C1 (by definition), so we let d > 1. Example 13.4. (Equally spaced vectors) The three equally spaced unit vectors in R2 are a highly symmetric tight C3–frame (each vector is fixed by the reflection through the line it lies on). The six equally spaced unit vectors are a highly symmetric tight D3–frame, as discussed above. More generally, the n equally space unit vectors are a highly symmetric tight Cn–frame, and Dn/2–frame also, when n is even. d Example 13.5. (Harmonic frames) The standard orthonormal basis e j for F is { } not a highly symmetric tight frame, since its symmetry group fixes e1 + + ed (and so its action is not irreducible). On the other hand, the vertices of the regular d–simplex always are (the three equally spaced vectors is the case d = 2). Since both of these frames are harmonic, we conclude that a highly symmetric tight frame may or may not be harmonic. 342 13 Tight frames generated by nonabelian groups 13.9 The construction of highly symmetric tight frames

We now show that up to similarity:

There is a ﬁnite number of highly symmetric frames of n vectors for Fd.

Theorem 13.3. (Finiteness) Fix n d. There is a ﬁnite number of highly symmetric normalised tight frames of n vectors≥ for Fd (up to unitary equivalence).

Proof. Suppose Φ is a highly symmetric normalised tight frame of n vectors for Fd. Then it is determined, up to unitary equivalence, by the representation induced by Sym(Φ), and a subgroup H which fixes only the one–dimensional subspace spanned by some vector in Φ. There is a finite number of choices for Sym(Φ) since its order is at most n(n 1) (n d +1) (Exercise 9.2), and hence (by Maschke’s theorem) a finite number− of possible − irreducible representations. As there is only a finite number of choices for H, it follows that the class of such frames is finite. ⊓⊔ This yields the following algorithm, which can be implemented in a symbolic algebra package such as Magma (see the worked example that follows). We denote the stabiliser of v by Stab(v)= StabG(v) := g G : gv = v . { ∈ } Algorithm: To construct all highly symmetric tight frames Φ of n vectors in Fd.

1. Start with an abstract group G. This corresponds to Sym(Φ) or an appropriate subgroup, so that G divides n! and n < G n(n 1) (n d + 1). | | | |≤ − − 2. Take all faithful irreducible representations ρ : G GLd(F). There is a ﬁnite number of these, and they can be→ computed. 3. Find (up to conjugacy) all subgroups H of ρ(G) which ﬁx a subspace span v , { } v = 0. Then gv g G is a highly symmetric tight frame of G /Stab(v) vectors. No other subgroups{ } ∈ of Stab(v) need be considered. | | 4. Determine which of the highly symmetric tight frames obtained are unitarily equivalent (up to a reordering).

Example 13.6. (Magma calculation) Let G be the solvable group <18,3>, for which Magma gives the presentation

2 3 3 1 2 G = g1,g2,g3 : g = g = g = 1,g− g3g1 = g . 1 2 3 1 3 The representations of G over C can be computed: G:=SmallGroup(18,3); r:=AbsolutelyIrreducibleModules(G,Rationals()); There are six of dimension 1, and three of dimension 2, the first given by rho:=Representation(r[7]); rG:=ActionGroup(r[7]); a:=rG.1=rho(G.1); b:=rG.2; c:=rG.3; sg:=Subgroups(rG); 13.10 Complex polytopes and finite reflection groups 343

0 1 ω2 0 ω2 0 a = ρ(g1)= , b = ρ(g2)= , c = ρ(g3)= , 1 0  0 ω2  0 ω       2πi where ω := e 3 . The subspace ﬁxed by a (nontrivial) subgroup H given by sg can be found by the command NullspaceMatrix(M-Id), where M is a block matrix of generators for H and Id is the corresponding identity block matrix. Thus, we obtain two highly symmetric tight frames:

6 vectors: v = v1 =(1,0), Stab(v1)= bc , 9 vectors: v = v2 =(1,1), Stab(v2)= a , which are a cross and a cube (see Example 13.10). These are the only highly symmetric tight frames we obtain, since the eighth representation is not faithful, and ρ(G) is the same for the seventh and ninth.

Example 13.7. There are no highly symmetric tight frames of ﬁve vectors in C3. Such a tight frame would have a symmetry group of order a multiple of 5, which is at most 5 4 3 = 60. A computer search over all groups in this range shows there is no such frame. By way of contrast, the tight frame of ﬁve vectors in C3 with the largest symmetry group is the vertices of a trigonal bipyramid, which has symmetry group of order 12 (see Example 9.12).

13.10 Complex polytopes and ﬁnite reﬂection groups

The n equally spaced unit vectors can be viewed as the vertices of the regular n–gon. We now investigate the highly symmetric tight frames which come as the vertices of (regular) complex polytopes (which include the n–gon and the Platonic solids). The main idea is that by imposing enough regularity (symmetries must map flags to flags), the symmetry group is generated by (complex) reflections, which leads to a complete classification via the symmetry group. A transformation g GLd(F) is a (complex) reflection (or pseudoreflection) if it has finite order m and∈ rank(g I)= 1, i.e., g fixes a hyperplane H, and maps some v ωv where v H is nonzero− and ω is a primitive m–th root of unity. The terminology→ and geometric∈ motivation comes from Rd with ω = 1. A finite − subgroup of GLd(F) is a reflection group if it is generated by its reflections. Frames are sequences of vectors (points), whereas polytopes, such as the Platonic solids, have points, lines (through points), and faces, etc. The technical definition (to follow), specifies these j-faces ( j = 0,1,...) as affine subspaces of Fd, together with some combinatorial properties motivated by the case R3. Of course, such a face is the affine hull of the vertices it contains, and it is convenient to move between the two. For complex spaces, a line (1–face) may contain more than two points, which challenges one’s intuition. 344 13 Tight frames generated by nonabelian groups

Definition 13.4. (see [Sch04]) A d–polytope–configuration is a finite family P of affine subspaces of Fd of dimensions j = 1,0,1,...,d, called elements or j–faces, ordered by inclusion , which form lattice− with the properties ⊂ (i) If Fj 1 Fj+1 are j 1 and j + 1 faces, then there are at least two j–faces contained− ⊂ between them.− (Modified diamond condition) (ii) If F G are faces, then there is a sequence of faces F = H0 H1 Hk = ⊂ ⊂ ⊂⊂ G with dim(Hj)= dim(F)+ j, j. (Connectedness) ∀ For brevity, we call such a P a complex polytope. We now follow the usual practice and translate P so that the barycentre (average of the vertices) is zero. This allows the vertices to be thought of as vectors, and ensures that the affine maps of the vertices to themselves are linear (and ultimately unitary). Definition 13.5. The symmetry group Sym(P) of a d–polytope–configuration P (with barycentre 0) is the group of g GLd(F) which map the elements of P to themselves. ∈

In particular, if ΦP is the points (vectors) of P, then Sym(P) is a subgroup of d Sym(ΦP ) (viewed as linear transformations of F ). Definition 13.6. A flag of d–polytope–configuration P is a sequence F of faces with

F =(F 1,F0,F1,...,Fd), F 1 F0 F1 Fd, dim(Fj)= j, j, − − ⊂ ⊂ ⊂⊂ ∀ and P is regular if Sym(P) is transitive on the flags of P. Shephard [She52], [She53] showed the symmetry group of a regular complex polytope is an irreducible reflection group, and classified all such polytopes via their symmetry groups. More precisely, let F be a flag of a regular complex polytope P, and c j be the centre of the j–face Fj, i.e., the average of its vertices. Then there are generating reflections r0,...,rd 1 for Sym(P) where r j fixes − c0,...,c j 1,c j+1,...,cd and maps Fj to another j–face, i.e., r j maps F to a flag which differs− only in the j–face. A parabolic subgroup of a finite reflection group G GLd(F) is the pointwise stabiliser of a subset V Fd. Steinberg’s fixed point theorem⊂ ([Ste64]) says that a parabolic subgroup is a⊂ finite reflection subgroup. Theorem 13.4. The vertices of the regular complex polytopes are highly symmetric tight frames. In particular, the vertices of the regular complex polytopes can be constructed from their abstract symmetry groups (which contains the corresponding reflection group).

Proof. Let P be a regular complex polytope, and Φ = ΦP be its vertices. View G = Sym(Φ) as a subgroup of GLd(F). Then H = Sym(P) is a subgroup of G, which is irreducible and transitive on the flags, and in particular is transitive on the vertices Φ. Thus, Φ will be a highly symmetric (tight) frame provided that StabH (v) StabG(v) fixes a space of dimension exactly one for each v Φ. ⊂ ∈ 13.10 Complex polytopes and finite reflection groups 345

Fix a vertex v Φ. Since H is a reflection group, Steinberg’s fixed point theorem ∈ implies that StabH (v) is the group generated by all the reflections which fix v. If F is a flag with F0 = v , then the d 1 generating reflections r1,...,rd 1 fix v, and so the subspace fixed{ } by them all is− one-dimensional (and equal to span− v ). Thus { } StabG(v) fixes only span v . { } ⊓⊔

Corollary 13.2. If G GLd(F) is an irreducible ﬁnite reﬂection group, then (gv)g G is a highly symmetric⊂ tight frame for Fd if and only if H = Stab(v) is a maximal∈ proper parabolic subgroup.

Proof. Since the parabolic subgroups are generated by reflections, and reflections fix a hyperplane, the set V fixed by a maximal proper parabolic subgroup must be a one–dimensional subspace V = span v , v = 0. { } ⊓⊔ In [ST54] (cf. [LT09]) all finite reflection groups were classified. Essentially, they appear as the symmetry groups of “semi–regular” complex polytopes. In the next subsections we outline the highly symmetric tight frames which can be obtained from the (imprimitive and primitive) finite reflection groups.

13.10.1 Imprimitive groups (ST 1-3)

d d d A representation of G on F is imprimitive if F is a direct sum F = V1 Vm d ⊕⊕ of nonzero subspaces, such that the action of G on F permutes the Vj, otherwise it is primitive. The Shephard–Todd classification of the imprimitive irreducible complex reflection groups consists of three infinite families (ST 1–3) given by the groups G(m, p,d), where m > 1, p m, and | G(m, p,d) = mdd!/p. | | These are available in Magma via ImprimitiveReflectionGroup(m,p,d), and can be constructed (cf. [LT09]) as a group of unitary transformations

G(1,1,d)= r1,r2,...,rd 1 , − G(m,m,d)= s,r1,r2,...,rd 1 , − G(m,1,d)= t,r1,r2,...,rd 1 , − p G(m, p,d)= s,t ,r1,r2,...,rd 1 , 1 < p < m, p m − | π 2 i 1 where r j swaps e j and e j+1, t is the reﬂection e1 ωe1, ω = e m , and s = t− r1t, i.e., → 0 1 ω 0 ω

r1 = 1 0 , t =  1 , s = ω 0 , (13.24)        I  I  I             346 13 Tight frames generated by nonabelian groups where I is the identity matrix of size d 2. The three inﬁnite families are: − ST 1: G(1,1,d + 1) = Sd+1 acting on the d–dimensional subspace of vectors in d+1 ∼ F which are orthogonal to e1 + + ed+1. ST 2: G(m, p,d), m,d > 1, p m, (m, p,d)=(2,2,2) acting on Cd. | ST 3: G(m,1,1) ∼= Zm acting on C. There are no highly symmetric tight frames for C, so only ST 2 and ST 3 can give highly symmetric tight frames. We now give some indicative examples.

Example 13.8. (m–distance tight frame) Let G = G(1,1,d + 1) = Sd+1 act on the d+1 ∼ d–dimensonal subspace H =(e1 + + ed 1) of F via σe j = eσ j, and + ⊥ m wm := e1 + + em (em 1 + + ed 1), 1 m d. − d + 1 m + + ≤ ≤ − Then Stab(wm) = m!(d +1 m)!, so Φm := gwm g G is a highly symmetric tight | d+1 | − { } ∈ frame of vectors for the d–dimensional space H , with Sd 1 Sym(Φm). m + ⊂ These are the only possibilities. This frame Φm is the standard m–distance tight frame of 12.16. For m = 1, we obtain the simplex with vertices given by Φ1. For § the other cases, the vectors of Φm are the barycentres of the (m 1)–faces of this − simplex, and so Φd is also a simplex. The special case m = 2, d = 3 gives the six vertices of the octahedron (which have symmetry group S4 Z2). × Example 13.9. (28 equiangular lines in R7) The special case of Example 13.8 where G = G(1,1,8) acts on the vector

v = 3w2 =(3,3, 1, 1, 1, 1, 1, 1). − − − − − − gives an orbit of 28 vectors in R7 which are an equiangular tight frame.

Example 13.10. (The generalised cross and cube) Let G = G(m,1,d), G = d!md, and | | vk := e1 + + ek, 1 k d. ≤ ≤ d k Then vk has Stab(vk) = k!(d k)!m − , and so its orbit gives a highly symmetric | d k | − d tight frame of k m vectors for R . These are the only possibilities. The extreme cases are the (generalised) cross (k = 1) and cube (k = d), which are regular complex polytopes. These terms originate from the case m = 2, d = 3, where we have the octahedron (6 vertices), the cuboctahedron (12 vertices), and the cube (8 vertices), respectively, and G = G(2,1,3) is Oh (the full octahedral group). For m = 2, d = 4 (see Figure 13.2), the polytopes are the hexadecachoron (16–cell) (8 vertices), octaplex (24–cell) (24 vertices), rectiﬁed tesseract (32 vertices), and tesseract (16 vertices). The cross and cube are harmonic frames, generated by the cyclic subgroups

r1r2 rd 1t , q1, ,qd , − 1 where q j =(r1r2 r j 1)− t(r1r2 r j 1) is the reflection e j ωe j. − − → 13.10 Complex polytopes and finite reflection groups 347

Fig. 13.2: Symbolic projections of the cross (hexadecachoron) and cube (tesseract) in R4.

The imprimitive reflection groups of the ST 2 family can be nested, (see [LT09]), e.g., G(m, p,d) ⊳ G(m,1,d), G(m, p,2) ⊳ G(2m,2,2). Hence, a highly symmetric tight frame obtained from an imprimitive reflection group G may be a subset of one obtained for a larger imprimitive reflection group. Example 13.11. (Nested irreducible reflection groups) Let

d 1 G = G(2,2,d), d > 2, G = 2 − d! (Coxeter group Dd). | |

There are highly symmetric tight frames given by the orbits of e1 and e1 + + ed. The ﬁrst of these is the cross, which has a symmetry group larger than G, namely G(2,1,d). The second is the demicube, a subset of half the vertices of the cube, which has symmetry group G(2,1,d).

13.10.2 Primitive reﬂection groups (ST 4-37)

There are 34 (exceptional) finite reflection groups in the Shephard–Todd classification. Their numbers and rank (the dimension of the space they act on) are

ST 4–22 (rank 2), ST 23–27 (rank 3), ST 28–32 (rank 4), ST 33 (rank 5), ST 34–35 (rank 6), ST 36 (rank 7), ST 37 (rank 8).

Magma calculations (see Tables 13.1 and 13.2) indicate the following behaviour: There are highly symmetric tight frames given by each primitive reﬂection group • (Theorem 13.4). Some are not the vertices of a regular complex polytope. These highly symmetric tight frames are not harmonic. • They may or may not be G–frames (of distinct vectors). • They have a small number of angle moduli. • We now highlight a few examples (with indicative Magma code). 348 13 Tight frames generated by nonabelian groups

Example 13.12. (ST 23). All highly symmetric tight frames obtained from rank 2 reﬂection groups are group frames (of distinct vectors). This is not the case in higher dimensions. Let G be the Shephard–Todd group 23, G = 120, for which Magma gives the generators | |

1 0 0 1 1 (√5 + 1) 0 1 0 0 − 2 1 g1 =  (√5 + 1) 1 0, g2 = 0 1 0, g3 = 0 1 1 , 2 −        0 0 1 0 1 1 0 0 1      −        which are not unitary matrices. We obtain three highly symmetric tight frames:

12 vectors: v =(√5 1,0,2), − 20 vectors: v =(√5 + 3,0,2), 30 vectors: v =(1,1,1), which are the vertices of the icosahedron, dodecahedron, and icosidodecahedron. The ﬁrst of these is a group frame (for <12,3>), and the other two are not.

Example 13.13. (24 vectors in C2). There are five regular complex polygons with 24 vertices. Their (flag to flag) symmetry groups are

ShephardTodd(6)=<48,33>, ShephardTodd(6)=<48,33>, ShephardTodd(5)=<72,25>, ShephardTodd(8)=<96,67>, ImprimitiveReflectionGroup(12,1,2)=<288,239>.

The four obtained from the primitive groups are not harmonic. The ﬁfth frame is a generalised cross, which is harmonic. In addition to these, there is a highly symmetric tight frame of 24 vectors (which is not a polygon) that can be obtained from the group

G:=ShephardTodd(12)=<48,29>, G = g1,g2,g3 , 1 ω3 ω ω3 + ω 1 ω3 ω ω3 ω 0 ω g1 := − − , g2 := − − , g3 := − . 2  ω3 + ω ω3 + ω 2 ω3 ω ω3 + ω ω3 0  − − − −   2πi     and the vector v =(1,ω3), where ω = e 8 . Similarly, this frame is not harmonic.

Let n be the number of vectors in a frame Φ with s angles Ang(Φ), and k be the order of the group of scalar matrices which map Φ to Φ. Then the estimate of Theorem 12.20 implies that

d+s 1 d+s 2 − − , 0 Ang(Φ); s s 1 ∈ n b := k d+s 1 2 − (13.25) ≤ − , 0 Ang(Φ). s ∈ 13.10 Complex polytopes and ﬁnite reﬂection groups 349

The following tables list the highly symmetric tight frames of n vectors for Cd given by the complex reﬂection groups ST 4–37, as calculated in [BW13]. In these we list the “small group library” number (when possible), the bound b of (13.25) on the number of vectors (here (ℓ) denotes the number of lines when it is sharp), the number of angles s, and the groups for which it is group frame with distinct vectors (when possible).

Table 13.1: The highly symmetric tight frames of n vectors in C2 given by the primitive reﬂection groups ST 4–22. Here (P) denotes a non-starry regular complex polytope. ST d order n b s group frame 4 2 24,3 8 (P) 8 (4) 1 8,4 5 72,25 24 (P) 24 (4) 1 24,3 , 24,11 6 48,33 16 (P) 16 (4) 1 16,13 24 (P) 24 (6) 2 24,3 7 144,157 48 48 (4) 1 48,47 , 48,33 72 72 (6) 2 72,25 8 96,67 24 (P) 24 (6) 2 24,3 , 24,1 9 192,963 48 (P) 48 (6) 2 48,4 , 48,28 , 48,29 96 (P) 160 4 96,67 , 96,74 10 288,400 72 (P) 72 (6) 2 72,12 , 72,25 96 (P) 144 3 96,54 , 96,67 11 576,5472 144 144 (6) 2 144,69 , 144,121 , 144,122 192 288 3 192,876 , 192,963 288 480 4 288,400 , 288,638 12 48,29 24 40 4 24,3 13 96,192 48 80 4 48,28 , 48,29 48 48 (6) 2 48,28 , 48,33 14 144,122 48 (P) 72 3 48,26 , 48,29 72 (P) 120 4 72,25 15 288,903 96 144 3 96,182 , 96,192 144 240 4 144,121 , 144,122 144 144 (6) 2 144,121 , 144,157 16 600,54 120 (P) 120 (12) 3 120,5 , 120,15 17 1200,483 240 (P) 240 (12) 3 240,93 , 240,154 600 (P) 1440 8 600,54 18 1800,328 360 (P) 360 (12) 3 360,51 , 360,89 600 (P) 900 5 600,54 19 3600, 720 720 (12) 3 720,420 , 720,708 ∗ 1200 1800 5 1200,483 1800 4320 8 1800,328 20 360,51 120 (P) 180 5 120,5 21 720,420 240 (P) 360 5 240,93 360 (P) 864 8 360,51 22 240,93 120 288 8 120,5 350 13 Tight frames generated by nonabelian groups

Table 13.2: The highly symmetric tight frames of n vectors in Cd , 3 d 8, given by the primitive reﬂection groups ST 23–37. Here (P) denotes a non-starry regular complex≤ ≤ polytope. ST d order n b s group frame 23 3 120 12 (P) 18 1 12,3 20 (P) 72 2 - 30 300 4 - 24 336 42 120 3 42,2 56 450 4 - 25 648 27 (P) 27 (9) 1 27,3 , 27,4 72 108 2 - 26 1296 54 (P) 54 (9) 1 54,8 , 54,10 , 54,11 72 (P) 108 2 - 216 1350 4 216,88 27 2160 216 1350 4 - 270 1890 5 - 360 9720 8 - 28 4 1152 24 (P) 80 2 24,1 , 24,3 , 24,11 96 9408 6 96,67 , 96,201 , 96,204 29 7680 80 160 2 80,30 160 800 3 - 320 7840 5 320,1581 , 320,1586 640 251680 10 - 30 14400 120 (P) 1400 4 120,5 , 120,15 600 (P) 1109760 15 600,54 720 3032400 18 - 1200 78330560 32 - 31 46080 240 800 3 - 1920 145200 9 1920, ∗ 3840 3162816 16 - 32 155520 240 (P) 240 (40) 2 - 2160 28224 6 - 33 5 51840 80 450 2 - 270 450 2 - 432 31752 5 - 1080 138600 7 - 34 6 39191040 756 * * * * * ** 35 51840 27 441 2 27,3 , 27,4 72 252 2 - 216 213444 6 216,86 , 216,88 720 232848 6 - 36 7 2903040 126 * * * * * ** 37 8 696729600 240 * * * * * ** 13.11 Projective representations 351 13.11 Projective representations

The four equiangular vectors Φ =(v,Sv,Ωv,SΩv) in C2 of (1.7) are not the orbit of a group of order four. Indeed, if this was so, then they would be a harmonic frame (as all groups of order four are abelian), and hence have inner products in Q(i). Nevertheless, since ΩS = SΩ, the group H generated by S and Ω contains the scalar matrix I, and so is− H = I, Ω, S, SΩ . Thus, we can think of Φ as a H–frame (with− vectors repeated),{± or± as the± orbit± of} the projective action of G = H/ I ∼= Z2 Z2 (for which the vectors are only defined up to a unit scalar multiple).− The natural× way to describe these equivalent viewpoints is via projective representations of the index group G. We will see that: For a given projective representation of G there is a canonical choice for H. • A projective representation of G can be calculated from the representations of H. • Many sets of equiangular lines come as projective orbits of irreducible actions. • Let G be a finite abstract group. A projective representation of G on a finite dimensional vector space V (over F) is a group homomorphism

GL(V ) ρP : G PGL(V ) := , C := c F : c = 0 . (13.26) → C { ∈ } Here PGL(V ) is the projective linear group, i.e., the group invertible linear transformations up to a nonzero scalar. The theory of representations (see 10.1) extends in the obvious way. In particular, by an appropriate choice of inner product§ on V , the action of a projective representation can be taken to unitary. Example 13.14. The matrices of (1.7) give a projective unitary representation of Z2 Z2 via × 2 j k G = Z2 Z2 PU(C ) : ( j,k) [S Ω ], × → → where PU(Fd) is the projective unitary group, i.e., the quotient of the unitary group by the subgroup C of unit modulus scalar matrices, and [S jΩ k]= S jΩ kC. We now explain how a projective representation of a finite group G on Fd can be d associated with a finite group H GL(F ). Let Eg be any matrix in ρP(g) We recall ∈ that ρP(g) is the set of all nonzero scalar multiples of E. The fact ρP(g) has finite k order, i.e., Eg = cI for some nonzero scalar c, does not imply that Eg has finite order. This can be rectified by using the key observation:

d There are exactly d scalings of Eg GL(C ) which have determinant 1, i.e., ∈ ω j Eˆ = E , j = 0,1,...,d 1, (13.27) g 1/d g det(Eg) −

π 1/d 2 i where det(Eg) is any ﬁxed d–th root of det(Eg), and ω := e d . 352 13 Tight frames generated by nonabelian groups 13.12 The canonical abstract (error) group

Let Eˆg denote any of the d scalings of (13.27), so that

det(Eˆg)= 1, g G. ∀ ∈ j Then the d G matrices H := ω Eˆg : j = 0,...,d 1,g G are distinct. Moreover, they form a| group,| since { − ∈ }

EˆgEˆh = wˆg hEˆgh. g,h G, , ∀ ∈ for somew ˆg h C, and taking determinants of this gives , ∈ d 2 d 1 1 = wˆ = wˆg h 1,ω,ω ,...ω − , g,h G. g,h ⇒ , ∈ { } ∀ ∈ Thus, we arrive at the following deﬁnition.

d Deﬁnition 13.7. Let ρP be an irreducible projective representation of G on C , and Eg ρP(g), g G. Then associated canonical error group is ∈ ∈ j H := ω Eˆg : j = 0,...,d 1, g G , { − ∈ } and the abstract version of this group is called the canonical abstract error group. We will call G the index group of H, and d is rank.

Since conjugation preserves the determinant, it follows that equivalent projective representations have the same canonical abstract error group. The action ρ of the canonical error group H on Cd is irreducible. Hence if this action is taken to be unitary, then (by Theorem 10.6) the unitary matrices (Eg)g G ∈ are an equal–norm tight frame for Md(C), i.e., d A = ∑ A,Eg Eg, A Md(C). (13.28) G g G ∀ ∈ | | ∈

Moreover, by (13.10), the tight frame (ρ(h))h H is a central group frame. The canonical error group H is (particular)∈central extension of G. It has centre

Z(H)= ωI = Zd, ∼ since if a matrix commutes with the spanning sequence (Eg)g G for Md(C), then it ∈ commutes with all of Md(C), and therefore is a scalar matrix (Schur’s lemma). Thus a group H can be a canonical (abstract) error group for at most one dimension d. Further, the index group G of a canonical (abstract) error group H is given by H G = . Z(H) 13.13 Nice error frames 353 13.13 Nice error frames

In applications, it is convenient to describe a projective representation ρP of G in terms of unitary matrices (Eg)g G, where Eg ρP(g). A sequence (Eg)g G of unitary matrices gives a faithful unitary∈ projective representati∈ on of G if and only∈ if

1. E1 is a scalar multiple of the identity I, and no other Eg is. 2. EgEh = wg hEgh, g,h G, where wg h C. , ∀ ∈ , ∈ Moreover, by Theorem 10.6, this projective representation is irreducible if and only if (13.28) holds. By Theorem 6.1, this irreducibilty condition can be written as 2 3. ∑ trace(Eg) = G . g G | | | | ∈ 2 For G a group (of order d ), a sequence (Eg)g G of unitary matrices in Md(C) is a nice (unitary) error frame≥ with index group G∈ if it satisﬁes 1,2 and 3 (above). Our discussion gives:

Proposition 13.5. Let G be a group and (Eg)g G be unitary matrices in Md(C). Then the following are equivalent ∈

1. (Eg)g G is a nice error frame for Md(C). ∈ 2. g Eg is a faithful irreducible projective representation of G of degree d. → In this case, the action of the canonical error group H on Cd is a special unitary faithful irreducible ordinary representation of the canonical abstract error group of degree d.

Above (and to follow) we write the projective representation ρP of (13.26) as g Eg, with the understanding that Eg ρP(g). → ∈ Example 13.15. (Nice error bases) When G = d2, the condition 3 of a nice error frame reduces to | |

3. trace(Eg)= 0, g = 1, g G. ∈ i.e., (13.28) is an orthogonal expansion. In the ﬁeld of quantum error correcting codes, an orthonormal expansion for Md(C) is called a error operator basis and a nice (unitary) error basis, if the matrices come from a projective representation (and are unitary) [Kni96a]. These were then generalised to nice error frames.

Example 13.16. The Pauli matrices (used to study spin in quantum mechanics)

0 1 0 i 1 0 σ1 = σx := , σ2 = σy := − , σ3 = σz := , (13.29) 1 0 i 0 0 1 − together with the identity, are a nice error basis for the 2 2 matrices, with index × group Z2 Z2. They have determinant 1, and generate the group <16,13> of × − order 16. The group generated by just the reﬂections σ1 and σ3 contains iσ2, and ± 354 13 Tight frames generated by nonabelian groups is the dihedral group <8,3>. The canonical error group for the nice error basis I,σ1,σ2,σ3 is { } H = iσ1,iσ2,iσ3 , which is the quaternian group <8,4>.

We now specify which nice error frames are considered to be “equivalent”.

Deﬁnition 13.8. Nice error frames (Eg)g G and (Fh)h H for Md(C) are equivalent ∈ ∈ if there is bijection σ : G H between their index groups, scalars (cg)g G and an → ∈ invertible T Md(C), such that ∈ 1 Fσg = cgT − EgT, g G. (13.30) ∀ ∈ This is more general than the equivalence of projective representations, where G = H, and reindexing of the elements of (Eg)g G is not allowed. ∈ Proposition 13.6. Equivalent nice error frames have the same canonical abstract error group, and (in particular) the same index group.

Proof. Suppose nice error frames (Eg)g G and (Fh)h H for Md(C) are equivalent. Then (13.30) scales to ∈ ∈

1 Fˆσg = T − (cˆgEg)T, g G, ∀ ∈ 2 d 1 wherec ˆg 1,ω,ω ,...,ω − (by considering determinants). Thus the canonical error groups∈ { are conjugate via }T, and so are isomorphic. Since the index group is the abstract error group factored by its centre, the nice error frames also have the same index groups. ⊓⊔ Example 13.17. For d = 1, the only canonical abstract error group is H = 1.

Example 13.18. A nice error basis (projective representation of Zd Zd) is given by × j k G = Zd Zd Md(C) : ( j,k) E = S Ω , × → → ( j,k) where S is the cyclic shift matrix, and is Ω the modulation matrix, given by

π j 2 i (S) jk := δ j,k+1, (Ω) jk = ω δ j,k, ω := e d .

This is the only nice error basis (up to equivalence) for Md(C) with index group G = Zd Zd (cf. [BK73]). This nice error basis (which generalises the Pauli matrices) plays× a key role in the construction of SICs (see Chapter 14). It is known as the Heisenberg nice error basis.

In 13.15, we outline how all nice error frames can be constructed using Magma. § 13.14 Tensor products of nice error frames 355 13.14 Tensor products of nice error frames

Nice error frames can be constructed (and also deconstructed) is via tensor products.

Proposition 13.7. Let (Eg1 )g1 G1 ,(Fg2 )g2 G2 be nice error frames for Md1 (C),Md2 (C). Then their tensor product ∈ ∈

(Eg1 Fg2 )(g ,g ) G G ⊗ 1 2 ∈ 1× 2 is a nice error frame for Md1d2 (C). In particular, a product of index groups is an index group. Moreover, the canonical error group is

π j 2 i H = ω (h1 h2) : 0 j < d 1,h1 H1,h2 H2 , ω := e d , d := d1d2. { ⊗ ≤ − ∈ ∈ } where H1,H2 are the canonical error groups of the nice error frames. Proof. In view of Proposition 13.5, the first part follows from the basic theory of (projective) representations. Alternatively, it can be verified directly, e.g., the tensor product satisfies condition 3 of the definition of a nice error basis since

2 2 ∑ trace(Eg Fg ) = ∑∑ trace(Eg )trace(Fg ) | 1 ⊗ 2 | | 1 2 | (g ,g ) G G g1 g2 1 2 ∈ 1× 2 2 2 = ∑ trace(Eg1 ) ∑ trace(Fg2 ) = G H = G H . g | | g | | | || | | × | 1 2 The tensor product group H1 H2 consists of scalar multiples of each Eg Eg , ⊗ 1 ⊗ 2 with determinant 1, but may not contain all d–roots of unity (if d1 and d2 are not coprime), and so we add these. ⊓⊔ Corollary 13.3. A product of index groups is an index group, and in particular, a product of index groups for nice error bases is an index group for a nice error basis. Example 13.19. Let K be a ﬁnite abelian group of order d. Since K is a product of cyclic groups, it follows by taking tensor products of the Heisenberg nice error basis (Example 13.18) that G = K K is the index group of a nice error basis for Md(C). × It can be shown (Theorem 13.5) that:

A nice error frame can have an abelian index group only if it is a nice error basis.

Example 13.20. Taking the tensor product of the two nonabelian index groups for d = 4, with the (abelian) index group for d = 2, gives two nonabelian index groups for d = 8, i.e.,

<16,3> <4,2> = <64,193>, <16,11> <4,2> = <64,261>. × × 356 13 Tight frames generated by nonabelian groups 13.15 Computing all nice error frames

Finding the centre of a ﬁnite group H and its irreducible representations are fast calculations, and a representation can always be made unitary. Thus the following characterisation of abstract error groups gives rise to a practical algorithm for their calculation, and hence that of the nice error frames they correspond to. Proposition 13.8. A group H is a canonical abstract error group if and only if 1. Its centre Z(H) is cyclic of order d. 2. It has a faithful irreducible ordinary representation ρ of degree d, which is special, i.e., det(h)= 1, h H. ∀ ∈ In particular, for d > 1 all canonical abstract error groups are nonabelian.

The nice error frame given by such a representation is (Eg)g G, where ∈ H G := , Eg ρ(g). Z(H) ∈ It remains only to determine which of these are equivalent. In this regard, we have:

Proposition 13.9. (Equivalence) If ρ : H Md(C) is a faithful irreducible special unitary ordinary representation of H, then→ so is

ρσ : h ρ(σh), σ Aut(H), → ∈ where Aut(H) denotes the automorphisms of H. These give equivalent nice error frames, even though the representations may not be equivalent when σ is an outer automorphism. Proof. Since any given automorphism σ of H ﬁxes the centre Z(H), it induces an automorphism σG Aut(G) on the index group G = H/Z(H). Thus a nice error ∈ frame (Fg)g G for ρσ is reindexing of one for ρ, since ∈

Fg := ρσ (g)= ρ(σg)= ρ(σG(g)), g G. ∀ ∈ 1 If σ is an inner automorphism, i.e., σh = k− hk, for some k H, then ρ and ρσ are equivalent ordinary representations of H, since ∈

1 1 ρσ (h)= ρ(k− hk)) = ρ(k)− ρ(h)ρ(k).

⊓⊔ In practice, the action groups ρ(H) of the ordinary representations ρ of a given H calculated in magma with the command

AbsolutelyIrreducibleModules(H,Rationals()); are often the same monomial group. 13.16 Examples of nice error frames 357 13.16 Examples of nice error frames

In [WC15], lists of nice error frames were produced using Proposition 13.8, i.e., Algorithm: To construct all rank d abstract error groups H of a given order. 1. Find all groups H (of the given order) with a cyclic centre of order d. 2. Find the faithful irreducible (ordinary) representations ρ of H of degree d. 3. Determine whether ρ is special, i.e., det(h)= 1, h H. ∀ ∈

Table 13.3: The canonical abstract error groups H and index groups G for the ﬁrst few nice error frames which are not bases, 2 d 4. ≤ ≤ d = 2 d = 3 d = 4

H G H G H G 12,1 6,1 36,11 12,3 80,28 20,3 16,9 8,3 54,8 18,4 96,157 24,8 20,1 10,1 63,3 21,1 96,215 24,14 24,3 12,3 72,42 24,12 128,523 32,27 24,4 12,4 81,9 27,3 128,545 32,24 28,1 14,1 108,15 36,9 128,749 32,34 32,20 16,7 108,22 36,11 128,782 32,31 36,1 18,1 117,3 39,1 128,864 32,6 40,4 20,4 144,68 48,3 128,880 32,9 44,1 22,1 162,14 54,5 128,1750 32,27 48,8 24,6 171,4 57,1 128,1799 32,28 48,28 24,12 189,8 63,3 128,2146 32,39

Example 13.21. (d = 2) In view of Proposition 13.5, all canonical abstract error groups for d = 2 are given by the ADE classiﬁcation of the ﬁnite subgroups of SL2(C) [Ste85]. The generalised quaternian group or dicyclic group of order 4n (n > 1), which is generated by the matrices

ω2n 0 0 1 2πi , − , ω2n := e 2n , 0 ω 1 1 0 2−n gives an inﬁnite family of rank 2 of abstract canonical error groups H. These account for all rank the 2 abstract canonical error groups in Table 13.3, except for

H = <24,3>, G = <12,3>, H = <48,28>, G = <24,12>. which come from the Shephard–Todd reﬂection groups with numbers 4 and 8.

From Table 13.3, we observe that index groups may be repeated in different dimensions. 358 13 Tight frames generated by nonabelian groups

Example 13.22. (Repeated index groups) A group G may be the index group for nice error frames in more than one dimension d, e.g., G = <12,3> is the index group for a nice error frame for M2(C) (H = <24,3>), and also one for M3(C) (H = <36,11>).

None of the index groups in Table 13.3 are abelian. Indeed, basic results from character theory imply the following.

Theorem 13.5. (Abelian index group) A nice error frame can have an abelian index group only if it is a nice error basis.

Proof. Suppose that H is the canonical abstract error group of a nice error frame, and that χ : H C is the character of a faithful irreducible representation with deg(χ)= χ(1)=→d. Recall that the centre of a character χ : H C is the subgroup → Z(χ) := h H : χ(h) = χ(1) , { ∈ | | } and that if χ is irreducible then

Z(χ) H = Z , ker(χ) := h H : χ(h)= χ(1) . ker(χ) ker(χ) { ∈ } Since the representation is faithful, ker(χ)= 1, and so this becomes

Z(χ)= Z(H).

Thus the index group is G = H/Z(χ), by Theorem 2.13 of [Isa06], if G is abelian, then G =[H : Z(χ)] = χ(1)2 = d2. | | ⊓⊔ The nice error bases (see Tables 13.4 and 14.1) play a prominent role in the construction of SICs. These may or may not have an abelian index group. It follows from Example 13.19, that G = K K is the index group of a nice error basis for every abelian group K. ×

Example 13.23. (Inequivalent nice error bases) For d = 8, there are 47 canonical abstract error groups, and only 42 index groups (see Table 14.1). In particular, there are three canonical abstract error groups for G = <64,67>, and hence at least three inequivalent nice error bases with this index group. Moreover, two of these give rise to SICs, and one does not. 13.16 Examples of nice error frames 359

Table 13.4: Nice error bases for d < 14, d = 8. Here H is the canonical abstract error group, G is the index group, and sic indicates that a SIC exists numerically.

d H G d H G

1 1,1 1,1 = Z1 sic 12 1728,1294 144,68 sic 2 8,4 4,2 = Z2 sic 1728,2011 144,92 2 3 27,3 9,2 = Z2 sic 1728,2079 144,101 = Z2 sic 3 12 4 64,19 16,2 = Z2 sic 1728,2983 144,132 4 64,94 16,3 1728,10718 144,95 64,256 16,11 1728,10926 144,100 64,266 16,14 1728,11061 144,102 5 125 3 25 2 2 sic 1728,13457 144,136 , , = Z5 1728,20393 144,170 6 216,42 36,11 = Z3 A4 sic × 1728,20436 144,172 216,66 36,13 2 1728,20556 144,177 216,80 36,14 = Z6 sic 2 1728,20771 144,179 7 343,3 49,2 = Z7 sic 2 1728,30353 144,184 9 729,24 81,2 = Z9 sic 1728,30562 144,189 729,30 81,4 1728,30928 144,193 729,405 81,9 sic 1728,30953 144,194 729,489 81,12 2 1728,31061 144,196 729,503 81,15 =(Z3 Z3) × 1728 31093 144 197 2 10 1000,70 100,15 , , =(Z2 Z6) 2 × 1000,84 100,16 = Z2 sic 13 2197,3 169,2 = Z13 sic 10 11 1331,3 121,2 = Z2 sic 11

Notes

Relaxing the condition of irreducibility gives a stable isogon (see [MVW16]).

Exercises

Chapter 14 Weyl–Heisenberg SICs

The maximal number of equiangular vectors (lines) in Cd is less than or equal to d2 1 (Theorem 12.2). The corresponding bound of 2 d(d + 1) for real equiangular lines in Rd is rarely attained (only for d = 2,3,7,23). There is compelling evidence for

Zauner’s conjecture (1999): There are d2 equiangular lines in Cd (for all d).

This was given in Zauner’s 1999 thesis (see [Zau10]), and is often stated in stronger forms which give a specific structure for the lines (as a group frame). These lines will be called SICs (see 14.2), and their existence is also known as the SIC problem. The evidence for this§ conjecture (which is steadily increasing) includes SICs have been constructed numerically for all d 121 (to 8000 digit precision), • and for a handful of other dimensions up to d = 323.≤ SICs have been constructed analytically for d = 2,...,16,19,24,28,35,48 [SG10], • and most recently for d = 17,18,20,21,30,31,37,39,43 [ACFW16]. We now outline the detailed structure of SICs, using the following road map: SICs are group covariant, i.e., are G–frames for a discrete Heisenberg group. • This reduces their construction to finding a fiducial vector (generating vector). The Clifford group (normaliser of the Heisenberg group) maps fiducial vectors to • fiducial vectors. Fiducial vectors have certain symmetries, in particular they are eigenvectors of • the Zauner matrix Z of order 3 (or M1 in some special cases). The field given by the triple products defining a SIC has a solvable Galois group. • It induces Galois symmetries of SICs (this is part proved and part conjecture). Before reading the remainder of this chapter, please be warned:

The SIC problem is very addictive.

361 362 14 Weyl–Heisenberg SICs 14.1 Maximal sets of complex equiangular lines and vectors

2 d Suppose that (v j) is sequence of n = d unit vectors in C which gives equality in the lower bound (6.22) for the second frame potential (t = 2), i.e.,

3 3 2 2 4 2d 4 2d 2 d (d 1) ∑∑ v j,vk = , or ∑ v j,vk = d = − 2 . j k | | d + 1 j=k | | d + 1 − (d + 1) By the variational characterisation (Theorem 6.1),

2 1 2 2 2 3 2 ∑ v j,vk (d ) d = d d , (14.1) j=k | | ≥ d − − with equality if and only if (v j) is a tight frame. By Cauchy–Schwarz,

2 2 2 2 2 4 d (d 1) 3 2 ∑ v j,vk d (d 1) ∑ v j,vk = − = d d , j=k | | ≤ − j=k | | d + 1 − 2 so by (14.1) there must be equality, i.e., v j,vk = C, where C > 0 is constant. | d | Thus (v j) is an equiangular tight frame for C with the maximal possible number of vectors (see Theorem 12.2). This condition on the vectors v j can be expressed in terms of the associated lines Cv j, and the rank one orthogonal projections Pj = v jv∗j , which satisfy

2 Pj,Pk = trace(PjP∗)= trace(v jv∗vkv∗)= trace(v∗vkv∗v j)= v j,vk . k j k j k | | We now give some of these characterisations. 2 d Proposition 14.1. Suppose that (v j) is d unit vectors in C , and Pj := v jv∗j . Then the following are equivalent d 1. (v j) is a equiangular tight frame for C . 2. The lines (Cv j) are equiangular. 2 1 3. v j,vk = d+1 , j = k. | | 1 4. Pj,Pk = , j = k. d+1 4 2d3 5. ∑ ∑ v j,vk = . j k | | d+1 2 2d3 6. ∑ ∑ Pj,Pk = . j k d+1 7. (v j) is a complex (2,2)–design. 8. (Pj) is a complex projective (2,2)–design. 2 Proof. If (v j) is an equiangular tight frame with v j,vk = C, j = k, then (2.10) gives | | 1 (d4 d2)C = d3 d2) = C = . − − ⇒ d + 1 Thus the equivalences follow from the previous discussion, and the deﬁnition of (2,2)–designs (see 6.9). § ⊓⊔ 14.2 SICs 363 14.2 SICs

2 The “rank one quantum measurements” (Pj) corresponding to d equiangular lines in Cd have applications in quantum mechanics, e.g., in quantum state tomography. Deﬁnition 14.1. A SIC1, SIC-POVM, or symmetric informationally complete positive operator valued measure for Cd is d2 rank one orthogonal projections d (Pj) on C , which satisfy the condition 1 Pj,Pk = trace(PjP∗)= , j = k. (14.2) k d + 1 For an explanation of the quantum physic underlying this deﬁnition see 14.3. The space of d d matrices (linear operators on the Hilbert space H = Cd) can§ be expressed in terms× of these rank one Hermitian matrices: d Proposition 14.2. If (Pj) is a SIC for C , then

1. (Pj) is a basis for the real vector space of Hermitian matrices. d d 2. (Pj) is a basis for C × . 1 3. (Pj d I) is an equiangular tight frame for the traceless Hermitian matrices. − 1 d d 4. (Pj I) is an equiangular tight frame for the traceless matrices in C . − d × 2 Proof. By Theorem 12.2 the d matrices (Pj) are linearly independent (over C and d d hence R), and so form a basis for the Hermitian matrices and C × . The matrices 1 1 1 1 Pj I are traceless, with Pj I,Pk I = Pj,Pk , and so are equiangular. − d − d − d − d In addition, they are a tight frame for the (d2 1)–dimensional spaces of traceless matrices asserted, as the variational characterisation− (6.4) for being a tight frame holds, via the calculation

1 2 1 1 2 (d 1)d2 1 1 2 d2 1 +(d4 d2) = − = d2(1 ) . − d − d + 1 − d d + 1 d2 1 − d − 1 Here we used Pj,Pk = , j = k and Pj,Pj = 1. d+1 ⊓⊔ 1 2 Remark 14.1. The equiangular tight frame (Pj I) above is the d vertices of a − d regular simplex in a (d2 1)–dimensional real space. Indeed (see Exer. 14.2), the − d rank one orthogonal projections vv∗ (or unit vectors v C ) giving a SIC are mapped to the traceless Hermitian matrices with unit norm (called∈ the Bloch sphere when d = 2) by the map d 1 vv∗ (vv∗ I). (14.3) → d 1 − d − Since their image (the vertices of a regular simplex) has symmetry group Sd2 , it is expected that SICs should also have symmetries. This is borne out by the known constructions. Unfortunately, SICs can only be found by taking the preimage of certain regular simplices under (14.3), as it is not onto for d > 2.

1 Vectors (v j) satisfying 1,2,3,5,7 of Proposition 14.1 are also known as SICs. 364 14 Weyl–Heisenberg SICs 14.3 Quantum measurements

Much of the literature on SICs comes from quantum mechanics (which uses Dirac notation). I now outline the physical interpretation of SICs (as kindly explained to me by Marcus Appleby). The Dirac or bra–ket notation consists of the bra w and the ket v which combine to give the bra–ket w v . We illustrate this notation: | | |

mathematics w = ,w v v,w = w v vw e j ∗ ∗ ∗ bra–ket w v w v =( w )( v ) v w e j = j | | | | | | | | | In particular, our inner products are linear in the first variable, and the bra–ket is linear in the second variable. The state of a quantum system of dimension d is described by a d d positive semidefinite matrix ρ with trace(ρ)= 1 called a density matrix. A measurement× is described by positive semidefinite matrices E1,...,En with ∑ j E j = I. The (E j) are called a positive operator valued measure (POVM). The probability distribution for obtaining the measurement outcome j for a state ρ is trace(ρE j).

For many measurements (E j), e.g. spin and energy, there are infinitely many • states which give some fixed probability distribution. There are some measurements for which the probability distribution fixes the • state – these measurements are said to be informationally complete. Informationally complete measurements can be used to infer the state statistically • (quantum tomography). An informationally complete POVM must have at least d2 operators. It is said to • be minimally informationally complete if it has exactly d2 operators. d If (Pj) is a SIC for C , say Pj = v jv∗j , then (v j) is an equiangular tight frame for Cd, and the frame expansion can be written as 1 I = ∑Pj. d j

Thus the operators 1 E j := d Pj are a

Symmetric Informationally Complete Positive Operator Valued Measure, where the term symmetric refers to the fact that E j,Ek takes just two values, i.e., 1 j k; d2(d+1) , = E j,Ek = trace(E jEk)= 1 d2 , j = k. This explains the origin of the term SIC-POVM (SIC). 14.4 Group covariance 365 14.4 Group covariance

All known SICs for Cd have a group covariance property, i.e., they are an H–frame. The group H can be the discrete (Weyl–)Heisenberg group, except when d = 8, • where it can also be a product of such groups for the Hoggar lines (see 14.6). § The Heisenberg group is a projective representation of G = Zd Zd, i.e., a nice • error basis with index group G. Other nice error bases (with possibly× nonabelian index groups) give rise to SICs, but these SICs are group covariant with respect to the Heisenberg group (or are the Hoggar lines). For d prime, the only group covariant SICs are for the Heisenberg group [Zhu10] • (there might be SICs which are not group covariant). The Clifford group (normaliser of the Heisenberg group) maps Weyl–Heisenberg • SICs to Weyl–Heisenberg SICs. By deﬁnition, a SIC or set of equiangular lines is a projective object. To describe the SICs for Cd which are the projective orbit of a given vector (line) under

ρ : G PU(Cd) (projective unitary group), → it is often convenient to deal with a generating unit vector v Cd (any in the line) ∈ and unitary matrices Eg ρ(G), i.e., the corresponding nice error basis (see 13.13). ∈ § The matrices (Eg)g G can be replaced by a ﬁnite group H which consists of unit scalar multiples of them,∈ e.g., the canonical abstract error group, and the vector v by the orthogonal projection Pv = vv∗ onto the line it deﬁnes. In this way, a SIC can be equivalently presented as

(Egv)g G (each line appears once) • ∈ An H–frame (hv)h H (the lines are repeated) • ∈ 1 A projective G–frame (g Pv)g G, where g A := gA˜ g˜− ,g ˜ ρ(g). • ∈ ∈ The vector v and the rank one orthogonal projection Pv = vv∗ are said to be ﬁducial. By Corollary 8.2, the equiangular lines given by (hv)h H are determined up to projective unitary equivalence by their triple products ∈

1 1 1 1 h1v,h2v h2v,h3v h3v,h1v = gv,v hv,v g− h− v,v , g := h− h1, h := h− h2. 2 3 The field associated with a given covariant SIC is the extension of Q by these triple products, in which the fiducial projector Pv = vv∗ lies. The structure of this field plays a crucial role in the construction of Weyl–Heisenberg SICs (see 14.20). § Example 14.1. The SIC (v,Sv,Ωv,SΩv) of (1.7) is covariant with respect to Z2 Z2 (see Example 13.14). The fiducial vector and fiducial projector are ×

1 3 + √3 1 3 + √3 √3 √3i v π P vv = i , v = ∗ = − . √6 e 4 3 √3 6 √3 + √3i 3 √3 − − We observe that here the ﬁeld required to present v is larger than that for Pv = vv∗. 366 14 Weyl–Heisenberg SICs 14.5 The discrete Heisenberg group and Weyl–Heisenberg SICs

We now generalise the Pauli matrices and the group (nice error basis) they generate to higher dimensions (see Examples 13.16 and 13.18). Throughout, let ω and µ be the primitive d–th and 2d–th roots of unity

2πi 2πi ω := e d , µ := e 2d ,

d d d and take the indices for elements of C and C × from Zd = 0,1,...,d 1 . Let d d d d { − } S C × be the cyclic shift matrix, and Ω C × be the modulation matrix given∈ by ∈ j (S) jk := δ j,k+1, (Ω) jk := ω δ j,k. (14.4)

For d = 2, these are the Pauli matrices σ1 = σx and σ3 = σz, and for d = 3 they are

0 0 1 1 S = 1 0 0 , Ω = ω .     0 1 0 ω2     These have order d, and satisfy the commutativity relation

Ω kS j = ω jkS jΩ k. (14.5)

Thus the group generated by the unitary matrices S and Ω is

r j k H := S,Ω = ω S Ω : r, j,k Zd . (14.6) { ∈ } 2 This is called the Heisenberg group (for Zd), as is the group

Hˆ := ch : c T,h H U (Cd), T := c C : c = 1 . (14.7) { ∈ ∈ }⊂ { ∈ | | } Since trace(S jΩ k)= 0, ( j,k) =(0,0), we have (see Proposition 13.5) jΩ d d The unitary matrices (S )( j,k) Zd Zd are a nice error basis for C × . • j k ∈ × ( j,k) S Ω is a faithful irreducible projective representation of Zd Zd. • → × (see Example 13.18). In particular, the unitary action of H on Cd is irreducible, and j k d so (hv)h H and (S Ω v) j,k Z are tight frames for C for any v = 0 (Theorem 10.5). ∈ ∈ d Deﬁnition 14.2. A SIC (equiangular tight frame of d2 vectors) for Cd of the form j k Φv =(S Ω v) j,k Z issaidtobea(Weyl-)Heisenberg SIC (or Heisenberg frame). ∈ d A Heisenberg frame for Cd is generated from a single vector v by applying S (translation) and Ω (frequency shift). Thus, it is a discrete analogue of a Gabor system (Weyl–Heisenberg frame), and has good time–frequency localisation. In this analogy the ﬁducial vector v corresponds to the mother wavelet.

2 It is also known as the generalised Pauli or Weyl–Heisenberg group. 14.6 The Hoggar lines 367 14.6 The Hoggar lines

All the known SICs are group covariant, and all are Weyl–Heisenberg SICs, except for the Hoggar lines (d = 8), which we now describe. The original presentation of these lines [Hog98] was in the 4–dimensional space H4 over the quaternians H. We follow the description of [GR09]. Consider the nice error basis given by the tensor product (see 13.14) of the Heisenberg nice error basis for C2 (see Example 13.18) with itself three§ times, i.e.,

j1 Ω j2 k1 Ω k2 ℓ1 Ω ℓ2 Eg g (Z2)3 = S ) (S ) (S ) ( j,k,ℓ) (Z2)3 , { } ∈ 2 { ⊗ ⊗ } ∈ 2 0 1 1 0 where S = σ = σ = , Ω = σ = σ = . Then the Hoggar lines are 1 x 1 0 3 z 0 1 8 − the SIC of 64 (equiangular) lines in C given by Egv g (Z2)3 , where { } ∈ 2 v =(0,0,1 + i,1 i,1 + i, 1 i,0,2). − − − The Hoggar lines are coinvariant with respect to other groups (see Table 14.1), many of which can be obtained as a subgroup of the Clifford group (see 14.7). §

Table 14.1: The nice error bases for d = 8. Those which occur as subgroups of the Clifford group are labelled with an *, and those which give rise to SICs are labelled SIC. All these SICs are the 2 Hoggar lines (up to projective unitary equivalence), except for H = <512,451>, G = Z8.

H G H G 2 512,451 64,2 = Z8 SIC* 512,400443 64,123 * 512,452 64,3 SIC* 512,401215 64,91 SIC* 512,35969 64,8 SIC* 512,402896 64,128 * 512,36083 64,10 512,402951 64,138 SIC 512,59117 64,34 * 512,402963 64,138 512,59133 64,35 * 512,403139 64,162 * 512,260804 64,58 * 512,406850 64,174 * 512,261506 64,67 SIC* 512,406879 64,167 * 512,261511 64,67 SIC* 512,406902 64,179 * 2 512,261518 64,67 * 512,6276980 64,192 =(Z2 Z4) * 512,262018 64,60 SIC* 512,6277027 64,193 SIC* × 512,262052 64,62 SIC* 512,6278298 64,195 SIC* 512,265618 64,69 SIC* 512,6279917 64,202 SIC 512,265839 64,68 SIC* 512,6279938 64,202 SIC 512,265911 64,71 SIC* 512,6280116 64,203 512,266014 64,72 * 512,6291080 64,226 512,266267 64,73 512,6339777 64,211 512,266357 64,75 SIC 512,6339869 64,207 512,266373 64,74 SIC 512,6375318 64,236 512,266477 64,78 SIC 512,6376278 64,216 512,266583 64,77 SIC 512,7421157 64,242 512,266616 64,82 512,10481364 64,261 3 2 512,400195 64,90 SIC * 512,10494180 64,267 =(Z2) SIC 512,400223 64,90 SIC 368 14 Weyl–Heisenberg SICs 14.7 The Clifford group (normaliser of the Heisenberg group)

Here we study the Clifford group (normaliser of the Heisenberg group), which plays a key role in the construction (and counting) of Heisenberg SICs.

Proposition 14.3. If U is a unitary matrix which normalises Hˆ of (14.7), i.e.,

j k 1 US Ω U− Hˆ , j,k Zd, ∈ ∀ ∈ j k and Φv :=(S Ω v) j,k Z is a SIC (i.e., is equiangular), then so is ΦUv. ∈ d Proof. Since S j1 Ω k1 v,S j2 Ω k2 v = S j1 j2 Ω k1 k2 v,v , the angles between distinct − − vectors in Φv are S jΩ kv,v = r, ( j,k) =(0,0). | | Since U 1S jΩ kU = ch, c T, h H, with h not a scalar for ( j,k) =(0,0), − ∈ ∈ j k 1 j k S Ω (Uv),Uv = U− S Ω Uv,v = chv,v = r, ( j,k) =(0,0), | | | | | | and so the angles in the second frame ΦUv are equal. ⊓⊔ Let [U] := cU : c T = eitU : t R , so that [I] is the unitary scalar matrices. { ∈ } { ∈ } Deﬁnition 14.3. The normaliser of the Heisenberg group Hˆ in the group of unitary matrices is called the Clifford group, and it is denoted by C(d). The projective Clifford group is PC(d) := C(d)/[I] (its elements are called Clifford operations).

Since H C(d), the action of C(d) on Cd is irreducible, and Z(C(d))=[I]. Since ⊂ T 1 1 T S∗ = S = S− , Ω ∗ = Ω − , Ω = Ω, (14.8) the Heisenberg group and Clifford group are closed under taking the transpose and T Hermitian transpose, and hence also entrywise conjugation A =(A∗) , we have

Entrywise conjugation maps a given Heisenberg SIC ﬁducial to another.

This motivates the following.

Deﬁnition 14.4. The normaliser of the Heisenberg group Hˆ in EU(Cd) (the group of unitary and antiunitary maps C C) is the extended Clifford group EC(d). The extended projective Clifford group→ is PEC(d) := EC(d)/[I].

Combining the above observation with Proposition 14.3 gives:

The extended Clifford group maps a given Heisenberg SIC ﬁducial to another. 14.7 The Clifford group (normaliser of the Heisenberg group) 369

Some elements of the Clifford group include the Fourier matrix3 F, the diagonal matrix R, M = RF, and the permutation matrices Pσ , σ Z , which are given by ∈ d∗

1 jk (F) jk := ω , (14.9) √d j( j+d) (R) jk := µ δ jk, (14.10)

1 j( j+d)+2 jk (M) jk := µ , (14.11) √d (Pσ ) jk := δ j σk, σ Z∗. (14.12) , ∈ d We observe that R, M are well deﬁned, i.e., the value of j( j + d) depends only on the integer j mod d. The entry µ j( j+d) has many alternative descriptions, e.g.,

2 2 2 2 2 µ j( j+d) = µ j ( 1) j = µ j ( 1) j =( µ) j = µ(d+1) j . − − − Elementary computations (see Exer. 14.5) give:

Lemma 14.1. The unitary matrices F,R,M,Pa normalise H.ˆ Indeed

j k 1 jk k j F(S Ω )F− = ω− S− Ω , (14.13) j k 1 j( j+d) j j+k R(S Ω )R− = µ S Ω , (14.14) j k 1 k(k 2 j+d) k j k M(S Ω )M− = µ − S− Ω − , (14.15) j k 1 σ j σ 1k Pσ (S Ω )Pσ− = S Ω − , (14.16)

σ 1 σ where − is the multiplicative inverse of Zd∗. If d is odd, then all the powers of µ above are even, and so can be expressed as∈ powers of ω. We will see (Theorem 14.1) that R (or M) along with F and H generate the Clifford group. The appearance of R (and hence M) can be explained. It appears in a direct search for diagonal matrices in the normaliser of Hˆ . Proposition 14.4. The group of diagonal unitary matrices which normalise the Heisenberg group Hˆ is generated by the scalar matrices, Ω, and the matrix R given by j( j+d) R jk := µ δ jk.

Proof. Suppose that Λ = diag(λ j) normalises Hˆ , and λd := λ0. Then

λ 0 0 d λd 1 − λ1 λ 0 0 λ Λ Λ 1  0  Ω k j+1 ω jk S − = λ2 = cS , i.e., = c , j, 0 λ 0 λ j ∀  1   . .   . .   . .    3 1 The matrix F∗ = F− = F is also commonly referred to as the Fourier matrix, or as the DFT (discrete Fourier transform) matrix. 370 14 Weyl–Heisenberg SICs where c C, k Z. Solving this recurrence gives ∈ ∈ j 1 j( j 1)k j j( j 1)k j( j+d)k k(d+1) j λ j = λ0c ω 2 − = λ0c µ − = λ0µ (cµ− ) .

Since λd = λ0, this gives

k(d+1) d k(d+1) 0 k(d+1) m (cµ− ) =(cµ− ) = 1 = cµ− = ω , ⇒ k m and so Λ = λ0R Ω . ⊓⊔ We now consider the structure of C(d). Let

U := S jΩ k, ( j,k) Z2. (14.17) ( j,k) ∈ d If a C(d), then ∈ 1 2 aUλ a− = za(λ)Uψ λ , λ Z , (14.18) a( ) ∀ ∈ d ψ 2 2 2 which deﬁnes functions a : Zd Zd and za : Zd T, since no Uλ is a scalar multiple of another. For example,→ (14.13), (14.14) give→

j k ψ z j k ω jk (14.19) F k = −j , F ( , )= − , j j ψ = , z ( j,k)= µ j( j+d). (14.20) R k j + k R We now show the elements of the Clifford group factored by Hˆ can be indexed by the elements of SL2(Zd) (the 2 2 matrices over Zd with determinant 1). × For a 2 2 matrix A, we deﬁne a symmetric matrix σA by × αγ βγ α β σ A := βγ βδ , A = γ δ . (14.21)

Lemma 14.2. Let ψa and za be given by (14.18). Then the map

ψ : C(d) SL2(Zd) : a ψa (14.22) → → is a group homomorphism with kernel H,ˆ and za satisﬁes

T p σAq 2 za(p + q)= ω za(p)za(q), p,q Z (14.23) ∈ d where A = ψa and σA is given by (14.21).

p q Proof. By (14.5), we have UpUq = ω 2 1Up+q. and so

p2q1 1 1 1 1 ω (aUp+qa− )= aUpUqa− =(aUpa− )(aUqa− ), which gives 14.7 The Clifford group (normaliser of the Heisenberg group) 371

ω p2q1 za(p + q)Uψa(p+q) = za(p)Uψa(p)za(q)Uψa(q) ψ ψ ω a(p)2 a(q)1 = za(p)za(q) Uψa(p)+ψa(q). and hence ψa(p + q)= ψa(p)+ ψa(q), (14.24)

p2q1 ψa(p)2ψa(q)1 ω za(p + q)= za(p)za(q)ω . (14.25) 2 For p = p1e1 + p2e2 Z , from (14.24) we obtain ∈ d

ψa(p)= p1ψa(e1)+ p2ψa(e2)=[ψa(e1),ψa(e2)]p, i.e., ψa can be represented by the 2 2 matrix [ψa(e1),ψa(e2)]. × Let [p′,q′]=[ψa(p),ψa(q)] = ψ[p,q], so that det([p′,q′]) = det(ψa)det([p,q]). Since the quotient za(p)za(q)/za(p + q) is symmetric in p and q, (14.25) gives

p2q1 p q q2 p1 q p ω − 2′ 1′ = ω − 2′ 1′ = p′ q′ q′ p′ = p1q2 q1 p2 ⇒ 1 2 − 1 2 − = det([p′,q′]) = det([p,q]), ⇒ = det(ψa)= 1, (14.26) ⇒ i.e., ψa SL2(Zd). Using this (see Exer. 14.8), (14.25) can be written as (14.23). ∈ 1 1 1 Since (ab)Uλ (ab)− = a(bUλ b− )a− , we have λ λ 1 λ ψ λ zab( )Uψab(λ) = a(zb( )Uψb(λ))a− = zb( )za( b( ))Uψa(ψb(λ)), (14.27) so that ψab(λ)= ψa(ψb(λ)), i.e., a ψa is a homomorphism. → We now determine the kernel of ψ. By (14.5), Hˆ kerψ. Suppose ψa = I, so 1 1 ⊂d d that aSa− = za(1,0)S and aΩa− = za(0,1)Ω. Since S = Ω = I, this implies that za(1,0) and za(0,1) are d–th roots of unity, say

1 α 1 β aSa− = ω S, aΩa− = ω Ω. (14.28)

β α If a Hˆ , then (14.28) implies that a is a scalar multiple of S− Ω . Hence, we ∈ β α 1 consider the unitary matrix b =(S− Ω )− a. By (14.28) and repeated application of (14.5), we have that

j k 1 α β 1 j 1 k β α b(S Ω )b− = Ω − S (aSa− ) (aΩa− ) S− Ω α β α j β k β α j k = Ω − S (ω S) (ω Ω) S− Ω = S Ω .

j k d d Since b commutes with the basis (S Ω ) j,k Z for C × , Schur’s Lemma implies ∈ d that b must be a (unit) scalar matrix cI, and hence a = cS β Ω α Hˆ . − ∈ ⊓⊔

Example 14.2. From Lemma 14.1, we have the following ψa SL2(Zd), ∈ σ 1 ψ 0 1 ψ 1 0 ψ 0 1 ψ − F = 1− 0 , R = 1 1 , M = 1 −1 , Pσ = σ . − 372 14 Weyl–Heisenberg SICs 14.8 Generators for the Clifford group

The subgroup of the Clifford group C(d) generated by F,R (and the scalars) is the symplectic unitaries4 CSp(d) := F,R,[I] . The elements of CSp(d)/[I] are called symplectic operations, and elements of the Heisenberg group Hˆ (or Hˆ /[I]) are referred to as Heisenberg operations, (Weyl) displacements or time–frequency shifts. For d even (see Exer. 14.6),

d d d 1 d 1 d d d 1 d d Ω 2 = R , S 2 = F− Ω 2 F = F− R F, S 2 Ω 2 = F− R FR . (14.29)

Thus there are nontrivial symplectic operations which are also displacements.

It turns out that (14.29) are the only cases (see Corollary 14.2). This makes the description of the Clifford group more technical for d even (here R has order 2d). We now show that C(d) is generated by the normal subgroup Hˆ and F,R, i.e.,

Every Clifford operation is the product of a displacement operation and a symplectic operation.

Theorem 14.1. (Clifford group generators) The homomorphism

ψ : C(d) SL2(Zd) : a ψa → → maps F and R to generators for SL2(Zd), and hence is onto. Therefore C(d) is generated by the unitary scalar matrices, and

S, Ω, F, R.

The Clifford group C(d) is closed under taking the transpose, Hermitian transpose, T and entrywise conjugation A =(A∗) .

Proof. By Lemma 14.2, the kernel of a ψa is Hˆ . Since Hˆ is generated by the → unitary scalar matrices and S,Ω, it sufﬁces to show that SL2(Zd) is generated by

0 1 1 0 ψ ψ (14.30) F = 1− 0 , R = 1 1 .

It is well known these matrices generate SL2(Z). Since the map of taking the entries of A SL2(Z) modulo d is a homomorphism onto SL2(Zd), they generate SL2(Zd). We observed∈ the closure properties are a consequence of (14.8). Alternatively, for R and F, we have R = R = R 1, RT = R, F = F = F 1, FT = F. ∗ − ∗ − ⊓⊔ 4 This is because their action on Hˆ is given by a symplectic matrix (see Remark 14.2). 14.9 Indexing the Clifford operations 373

The order of SL2(Zd) is known (see [Gun62] Theorem 3, Chapter I)

3 1 SL2(Zd) = d ∏ 1 2 , (p the prime factors of d). | | p d − p | Hence, by Theorem 14.1, the number of Clifford operations is

C(d) Hˆ C(d) 1 = = d2 SL (Z ) = d5 ∏ 1 . ˆ 2 d 2 [I] [I] H | | p d − p | The extended Clifford group EC(d) is generated by the antiunitary map C : v v of conjugation together with any set of generators for C(d) (see Corollary 14.3).→

14.9 Indexing the Clifford operations

We now show each Clifford operation is uniquely determined by the pair (ψa,za). Z2 Deﬁne the semidirect product SL2(Zd) ⋉T d via the multiplication

(A,zA)(B,zB) :=(AB,(zA B)zB), (14.31) ◦ where functions Z2 T are multiplied pointwise. d → Corollary 14.1. With the multiplication (14.31), the map

Z2 C(d) SL2(Zd) ⋉T d : a (ψa,za) (14.32) → → is a homomorphism with kernel [I]. Thus every Clifford operation [a] = C(d)/[I] ∈ has a unique index (ψa,za), and these satisfy

ψab = ψaψb, zab =(za ψb)zb, (14.33) ◦ 1 1 1 ψa = ψ 1 = ψ− , za = z 1 = z− ψ− = za ψa , (14.34) ∗ a− a ∗ a− A ◦ a ◦ ∗ Further, if ψa = ψb, then za/zb is a character.

Z2 Proof. It is easy to check (see Exer. 14.7) that SL2(Zd) ⋉T d is a group with the 1 1 1 1 multiplication (14.31), identity (I,1), and inverse (A,zA)− =(A− ,zA− A− ) By (14.27), we have ◦ ψab = ψaψb, zab =(za ψb)zb, ◦ i.e., the map a (ψa,za) is a homomorphism. Thus (14.33) holds, as does (14.34) → 1 1 1 1 by the calculation (ψ 1 ,z 1 )=(ψa,za)− =(ψ− ,z− ψ− ). a− a− a a ◦ a Now suppose that a is in the kernel, i.e., ψa = I, za = 1. By Lemma 14.2, we have a = cS jΩ k Hˆ . Using (14.5), we therefore obtain (see Exer. 14.4) ∈ p p 1 j k p p k j kp jp p p aS 1 Ω 2 a− = S Ω S 1 Ω 2 Ω − S− = ω 1− 2 S 1 Ω 2 , 374 14 Weyl–Heisenberg SICs

kp jp 2 so that za(p)= ω 1 2 = 1, p Z . Thus j = k = 0 and a = cI [I], as supposed. − ∀ ∈ d ∈ For ψa = ψb = A, it follows from (14.23) or (14.25) that za/zb is a character. ⊓⊔ Z2 We call the subgroup of SL2(Zd) ⋉T d given by

Ind(d) := (ψa,za) : a C(d) { ∈ } the index group of the Clifford operations, and the index map is the isomorphism

C(d)/[I] Ind(d) : [a] (ψa,za). (14.35) → → Example 14.3. For M = RF, from (14.19) and (14.20), we calculate

ψ ψ ψ 1 0 0 1 0 1 M = R F = 1 1 1− 0 = 1 −1 , − ( k)( k+d) jk k(k+d)+2 jk zM =(zR ψF )zF = zM( j,k)= µ − − ω− = µ . ◦ ⇒

The Clifford group C(d) and the index map a (ψa,za) can easily be set up in a computer algebra package such as magma→by taking the matrix group generated by the matrices S,Ω,F,R (deﬁned over a suitable cyclotomic ﬁeld).

1 By Theorem 14.1, ψ− (A)= ha : h Hˆ , A SL2(Zd), for any a with ψa = A. Hence to describe the elements of{ the Clifford∈ } group∈ C(d), it sufﬁces to know an a 1 in each coset ψ− (A). We now show this representative can be a symplectic matrix.

14.10 Appleby indexing

d+1 If d is odd, then µ = ω 2 , and it follows from (14.23) that − T p σA p 2 za(p)=( µ) zâ(p), p Z , (14.36) − ∀ ∈ d where A = ψa, andz â is a character (see Exer. 14.8). T σ If d is even, then the factor ( µ)p A p above is not well defined. To obtain an − 2 analogue of (14.36), it is necessary “lift” A to a B SL2(Z2d). This “doubling” works, but the corresponding (Appleby) index [B, χ] is∈ not unique. We now give the details as described by [App05]. Define displacement operators by

p1 p2 p1 p2 2 Dˆ p :=( µ) S Ω , p Z . (14.37) − ∈

These satisfy det(Dp)= 1, and (see Exer. 14.15)

ˆ 1 ˆ ˆ ˆ µ p,q ˆ ω p,q ˆ ˆ D−p = D p, DpDq =( ) Dp+q = DqDp, (14.38) − − 14.10 Appleby indexing 375 and ˆ ˆ Dp, d odd; Dp+dq = p,q (14.39) ( 1) Dˆ p, d even, − where , is the symplectic form

T 0 1 p,q := p2q1 p1q2 = p − q. − 1 0

It follows from (14.39) that Dˆ p depends only on p mod d′, where

d, d odd; d′ := 2d, d even.

We observe that p,q has the property Ap,Aq = det(A) p,q , p,q. (14.40) ∀ We now generalise (14.36), to show that for each [a] C(d)/[I] there exists a 2 ∈ B SL2(Z ) and χ Z , such that ∈ d′ ∈ d 1 χ,Bp 2 aDˆ pa− = ω Dˆ Bp, p Z . ∀ ∈ d′ Here χ,Bp is interpreted as χ,Ap , A := B mod d, when d is even. We will write the pair(B, χ) as [B, χ], and call it an Appleby index.

Theorem 14.2. Deﬁne the semidirect product SL (Z ) ⋉Z2 via the multiplication 2 d′ d

[B1, χ1][B2, χ2] :=[B1B2, χ1 + A1χ2], A1 := B1 mod d. (14.41)

Then there is a unique surjective homomorphism onto the Clifford operations

2 f : SL2(Z ) ⋉Z C(d)/[I], (14.42) d′ d → with the property that for [a]= f ([B, χ])

1 χ,Bp 2 aDˆ pa− = ω Dˆ Bp, p Z , (14.43) ∀ ∈ d′ i.e.,

T χ,Ap p σB p 2 A := ψa = B mod d, za(p)= ω ( µ) , p Z . (14.44) − ∀ ∈ d

This f is an isomorphism for d odd (i.e., d′ = d), and for d even it has kernel

1 + rd sd s d ker f = , 2 : r,s,t 0,1 . (14.45) td 1 + rd d t 2 ∈ { } 376 14 Weyl–Heisenberg SICs

Proof. If a C(d) satisfies (14.43), then (14.44) follows (see Exer. 14.16). In view of the isomorphism∈ (14.35), f is uniquely defined, and it suffices to show that

2 θ : SL2(Z ) ⋉Z Ind(d) : [B, χ] (A,za), d′ d → → given by (14.44) is a surjective homomorphism. Z2 We ﬁrst show it is a homomorphism (as a map to SL2(Zd) ⋉T d ). Now θ χ χ θ χ χ [B1, 1][B2, 2] = [B1B2, 1 + A1 2] =(A1A2,za1a2 ),

T σ χ1+A1χ2,A1A2 p p B B p A j := B j mod d, za a (p) := ω ( µ) 1 2 , 1 2 − and

θ [B1, χ1] θ [B2, χ2] =(A1,za )(A2,za )=(A1A2,(za A2)za ), 1 2 1 ◦ 2 T σ T σ χ1,A1A2 p (B2 p) B B2 p χ2,A2 p p B p (za A2)za (p)= ω ( µ) 1 ω ( µ) 2 , 1 ◦ 2 − − so that θ is a homomorphism provided that

χ1 + A1χ2,A1A2 p = χ1,A1A2 p + χ2,A2 p , T σ T σ T σ p B1B2 p =(B2 p) B1 B2 p + p B2 p. The ﬁrst follows since (14.40) gives

χ2,A2 p = A1χ2,A1A2 p , and the second follows by the identity σ T σ σ B1B2 = B2 B1 B2 + det(B1) B2 . We calculate (as in Example 14.4)

0 1 1 0 θ 0 ψ z θ 0 ψ z (14.46) [ 1− 0 , ] =( F , F ), [ 1 1 , ] =( R, R), and α θ ψ [I, β ] =(zSα Ω β , Sα Ω β ), so that θ maps generators for SL (Z ) ⋉Z2 to generators for Ind(d), and hence is 2 d′ d a surjective homomorphism. Finally, we determine ker f = kerθ. By (14.44), we have [B, χ] ker f if ∈ T χ,Ap p σB p A := ψa = B mod d = I, za(p)= ω ( µ) = 1, p. − ∀ χ p χ p For d odd, d′ = d, and so B = I and za(p)= ω 2 1− 1 2 = 1, p. Thus [B, χ]=[I,0], and f is an isomorphism. For d even, B mod d = I gives ∀ 14.10 Appleby indexing 377

1 + rd sd B = , r,s,t,u 0,1 , td 1 + ud ∈ { } and the condition det(B)= 1 gives

2 det(B)=(1 + rd)(1 + ud) std 1 +(r + u)d mod d′ = r = u, − ≡ ⇒ so that

1 + rd sd td(1 + rd) tdsd td B = , σB = mod d′. td 1 + rd sdtd sd(1 + rd) ≡ sd χ,p pT σ p χ p χ p tdp2+sdp2 Hence za(p)= ω ( µ) B = ω 2 1 1 2 ( µ) 1 2 = 1, which gives − − − χ p χ p tp2+sp2 tp +sp d (tp +sp ) ω 1 2− 2 1 =( 1) 1 2 =( 1) 1 2 ω 2 1 2 , p. − − ∀ d d d Thus, χ1 = s, χ2 = t = t, and we obtain (14.45). 2 − 2 2 ⊓⊔ In other words:

2 Each Clifford operation has an Appleby index [B, χ] SL2(Z ) Z . ∈ d′ × d This is unique for d odd. • There are eight choices (each differing by an element of ker f ) for d even. •

a ψa za( j,k) [B, χ] α α Ω β ωβ j αk S I − [I, β ] 0 1 ω jk 0 1 F 1− 0 − [ 1− 0 ,0] 1 0 µ j( j+d) 1 0 R 1 1 [ 1 1 ,0]

Table 14.2: The index (ψa,za) and Appleby index [B, χ] for generators of the Clifford operations.

0 1 0 1 Example 14.4. For the Appleby index B χ 0 , we have σ , [ , ]=[ 1− 0 , ] B = 1− 0 −

T χ,Bp p σB p 2 jk jk j ω ( µ) =( µ)− = ω− = zF ( j,k), p = , − − k 0 1 0 1 so that θ 0 ψ z and 0 is an Appleby index for F . [ 1− 0 , ] =( F , F ) [ 1− 0 , ] [ ] 378 14 Weyl–Heisenberg SICs 14.11 Symplectic unitaries

By (14.29), for d even, there are nontrivial symplectic unitaries (those generated by F, R and the scalars) which are in the Heisenberg group. We now characterise these. Let md be the surjective homomorphism

md : SL2(Z ) SL2(Zd) : B A := B (mod d), d′ → → which is the identity for d odd, and for d even has kernel (see Theorem 14.2)

1 + rd sd K := : r,s,t 0,1 , K = 8. (14.47) td 1 + rd ∈ { } | | Corollary 14.2. A matrix a C(d) is a symplectic unitary if and only if it has an Appleby index of the form [B∈,0]. Indeed, the map

α : SL2(Z ) CSp(d)/[I] : B f ([B,0]) (14.48) d′ → → is a surjective homomorphism, which is an isomorphism for d odd. For d even, kerα = I,(d + 1)I , and so the only nontrivial Heisenberg operations which are symplectic{ are given} by

d d d d S 2 , Ω 2 , S 2 Ω 2 (d even).

Proof. By (14.41), we have

[B1B2,0]=[B1,0][B2,0], and so α is a homomorphism. It is onto, since by (14.46), its image contains

0 1 1 0 α F α R (14.49) 1− 0 =[ ], 1 1 =[ ], which are generators for CSp(d)/[I]. Since ψ has kernel Hˆ (Lemma 14.2), it induces a well deﬁned homomorphism ψˆ : CSp(d)/[I] SL2(Zd), with → 0 1 1 0 ψˆ F ψ F ψˆ R ψ R (14.50) ([ ]) = ( )= 1− 0 , ([ ]) = ( )= 1 1 . By (14.49) and (14.50), we conclude that

md = ψˆ α, ◦ since it holds for the generators (14.30) of SL (Z ). The kernel of ψˆ consists of the 2 d′ symplectic operations which are also Heisenberg operations, i.e.,

kerψˆ = CSp(d)/[I] Hˆ /[I]. ∩ 14.11 Symplectic unitaries 379

For d odd, md is an isomorphism, so that kerψˆ = [I] . For d even, { }

md = ψˆ α = kerψˆ kerα = kermd = K = 8, ◦ ⇒ | || | | | | | and (14.39) gives

p,p Dˆ =( 1) Dˆ p = Dˆ Ip = (d + 1)I kerα = kerα 2. (d+1)Ip − ⇒ ∈ ⇒ | |≥ In view of (14.29), we must have

d d d d kerα = I,(d + 1)I , kerψˆ = [I],[S 2 ],[Ω 2 ],[S 2 Ω 2 ] , { } { } as claimed. ⊓⊔ In other words:

Each symplectic operation [a] has an Appleby index of the form [B,0]. This is unique for d odd. • There are two choices ([B,0] and [(d + 1)B,0]) for d even. • We call B SL2(Z ) a symplectic index for [a]. ∈ d′

The following commutative diagram summarises Corollary 14.2. α SL (Z ) -- (d)/[I] 2 d′ CSp

ψ m ˆ (14.51) d -- ? SL2(Zd) In particular, we have the following 1–1 indexing of the symplectic operations

CSp(d) SL2(Zd), d odd; = ∼ SL2(Z2d ) [I] (d+1)I , d even. Remark 14.2. Matrices in SL (Z ) are said to be symplectic (see Exer. 14.10). If a 2 d′ is a symplectic unitary, with symplectic index B, then (14.43) gives

1 aDˆ pa− = Dˆ Bp, p( Z ), ∀ ∈ d′ i.e., the conjugation action of a on the displacement Dˆ p is given by multiplication of p by the symplectic matrix B. This is the origin of the term symplectic unitary.

The group CSp(d) of symplectic unitaries is not irreducible for d > 2, since its 2 centre contains the nondiagonal matrix P 1 = F . Calculations in CSp(d)/[I] can be done in the group generated by F and R,− which is ﬁnite (see Exer. 14.19 for details). 380 14 Weyl–Heisenberg SICs 14.12 Permutation matrices

The permutation matrices are a subgroup of the symplectic unitaries.

Proposition 14.5. The permutation matrices Pb, b Zd∗, are symplectic. Indeed, with 1 b < d, we have ∈ ≤ 1 1 1 b− b b− Pb =(cb,d)− R FR FR F, (14.52)

1 where b− is the inverse of b in Z∗ , and cb,d = c 1 is the Gauss sum d′ b− ,d

1 bj( j+d) 1 cb,d := ∑ µ = G b(d + 1),2d . √d 2√d j Zd ∈ b 0 Proof. Let B = 1 SL2(Zd ). Then σB = 0, and so (14.16) gives 0 b− ∈ ′ T 0,Ap p σB p 2 A := ψP = B mod d, zP (p)= 1 = ω ( µ) , p Z . b b − ∀ ∈ d

By Theorem 14.2, this implies that [B,0] is an Appleby index for Pb , which is therefore a symplectic unitary, with symplectic index B. Now B can be factored

b 0 1 0 1 0 1 B = 1 = − 1 − − 1 . (14.53) b− 1 b− 1 b 1 b− − − − In view of (14.46), a symplectic index for RbF is given by

b 0 1 1 0 0 1 1 −b = 1 1 1− 0 , − and so applying the homomorphism α of Corollary 14.2 to (14.53) gives (14.52), for some scalar cb,d, to be determined. From (14.9) and (14.10), we have

b 1 bj( j+d)+2 jk (R F) jk = µ . (14.54) √d

b 1 1 b b 1 Hence, equating the (0,0)–entries of cb,dPb(R − F)− = R FR − F, gives

1 1 1 1 b b− b− j( j+d) cb,d = ∑ (R F)0 j(R F) j0 = ∑ µ . √d j Z d j Z ∈ d ∈ d 2 We recall that µ j( j+d) depends only on j modulo d, and µ jd = µdj , so that 14.12 Permutation matrices 381

2d 1 2d 1 1 1 1 1 1 2 1 − b− j( j+d) − b− (d+1) j 1 cb,d = ∑ µ = ∑ µ = G b− (d + 1),2d . 2 √d j=0 2√d j=0 2√d Evaluating the (0,0)–entries of (14.52), using (14.54), gives

1 bj( j+d)+2 jk+b 1k(k+d) cb d = ∑ ∑ µ − = c 1 . , b− ,d d√d j Z k Z ∈ d ∈ d

⊓⊔ The formulas for evaluating Gauss sums imply that cb,d is an 8–th root of unity, 1 d e.g., if b has odd order, then cb,d =(√i) − (see Exer. 14.11). Example 14.5. When b = 1, (14.52) gives

π 3 3 2 i (d 1) M =(RF) = c1,dP1 = e− 8 − I.

The map Z C(d) : σ Pσ is a group homomorphism, since d∗ → → δ δ σ (Pσ1σ2 ) jk = ∑(Pσ1 ) jr(Pσ2 )rk = j,σ1r r,σ2k = j,σ1σ2k =(Pσ1σ2 ) jk. r From Corollary 14.2, we observe that

If a1 and a2 are symplectic matrices with symplectic indices B1 and B2, then B1B2 is a symplectic index for the symplectic matrix a1a2.

Using the symplectic index for Pβ given by Proposition 14.52, and those for F and R given by (14.49), one obtains the Table 14.3.

Symplectic matrix A = ψa SL(Zd ) a f ([B,0]) B SL(Z∈ ) z ( j,k) index description d′ a ∈ β∈ 0 Pβ 1 1 diagonal 0 β − β α 0 µ αβ 2 j2 µαβ 2 j( j+d) R Pβ αβ β 1 ( ) = lower triangular − − β 1 αβ 1 α − α µ αβ 2k2 F− R Pβ F − , = 0 ( )− upper triangular 0 β − βγ β α 1 γ µ γ(αβ 2γ 1) j2+2(αβ 2γ 1) jk+αβ 2k2 R Pβ F− R αβγ β 1 αβ ( ) − − b12 is a unit − − αβ β− 1 αβγ 1 α 1 γ − µ αβ 2 j2+2(αβ 2γ 1) jk γ(αβ 2γ 1)k2 F− R Pβ F− R F − ( )− − − − b21 is a unit β βγ − −

Table 14.3: Some symplectic unitaries [a] CSp(d), with an Appleby index [B,0] (B is a symplectic index), the index (ψ ,z ). Here β is a unit∈ in Z (and hence in Z ). For the cases when a is a a d d′ antiunitary, or the off diagonal entries of A = ψa are nonzero and not units, see §14.13. 382 14 Weyl–Heisenberg SICs 14.13 Calculating a symplectic matrix from its symplectic index

Using Table 14.3, a symplectic matrix a can be determined from its symplectic index B (or ψa = B mod d), except for when the off diagonal entries of B are nonzero and not units. We now consider this case, and also when a is antiunitary. The antilinear map C : Cd Cd;z z → → of entrywise complex conjugation is in the extended Clifford group EC(d), i.e., it normalises Hˆ , since j k 1 j k C(S Ω )C− = S Ω − . (14.55) Thus the antiunitary elements of EC(d) are precisely aC, a C(d). Let ∈ 1 0 ψ : J : C = = 0 1 . − We note det(J)= 1. With Uλ deﬁned by (14.17), from (14.55) we obtain − 1 1 1 1 ψ λ (aC)Uλ (aC)− = a(CUλC− )a− = a(UψC(λ))a− = za( C( ))UψaψC(λ).

In this way, we can extend the index (za,ψa) to a EC(d), where ∈

ψa ESL2(Zd) := A SL2(Zd);det(A)= 1 = SL2(Zd) AJ : A SL2(Zd) , ∈ { ∈ ± } ∪ { ∈ } and zaC = za ψC, ψaC = ψaψC, a E(d). ◦ ∈ Similarly, the Appleby indexing extends. With Dˆ p deﬁned by (14.37), (14.55) gives

p1 p2 p1 p2 p1 p2 p1 p2 CDˆ pCv = C( µ) S Ω v =( µ)− S Ω − v − − (Jp)1(Jp)2 (Jp)1 (Jp)2 =( µ) S Ω = Dˆ Jpv, − so that if a C(d) has Appleby index [B, χ], then ∈ 1 1 χ,BJp 2 (aC)Dˆ p(aC)− = a(Dˆ Jp)a− = ω Dˆ BJp, p Z , (14.56) ∀ ∈ d′ i.e., aC has extended Appleby index [BJ, χ]. Thus, the surjective homomorphism f of Theorem 14.2 extends to

2 fE : ESL2(Z ) ⋉Z EC(d)/[I]. (14.57) d′ d → We say that an antiunitary element aC EC(d) is symplectic if it has an (extended) Appleby index of the form [BJ,0], B ∈SL (Z ). 2 d′ We now illustrate how a symplectic∈ Clifford operation can be constructed from its (extended) Appleby index. This allows a matrix a EC(d) to be constructed with ∈ 2 any given extended Clifford index [B, χ], B SL2(Zd ), χ Z . ∈ ′ ∈ d′ 14.13 Calculating a symplectic matrix from its symplectic index 383

Example 14.6. (Antiunitary symplectic) For d = 8, consider the symplectic index

6 11 ESL2(Z16) 5 1 ∈ from the Table 14.4. This matrix has determinant 1, and so given an antiunitary symplectic operation aC, a C(d), where a symplectic− index of a is given by ∈ 6 11 6 11 1 0 6 5 B J 1 = 5 1 − = 5 1 0 1 = 5 15 . −

Since 5 is a unit in Z16 with inverse 13, we have

1 0 5 0 0 1 1 0 B = 3 1 0 13 1 0 14 1 , − 3 1 14 and so we can take aC = R P5F− R C (see the fourth row of Table 14.3).

Lemma 14.3. Let B SL2(Z ), then for x Z , we have ∈ d′ ∈ d′ α β 1 0 0 1 γ + xα δ + xβ B (14.58) = γ δ = x 1 1− 0 α β , − − − where x can be chosen so that δ + xβ Z∗ . One such choice for x is the product of d′ the primes which divide d but not δ :=∈ b Z . In this way, a symplectic matrix 22 d′ a C(d) with symplectic index B can be constructed∈ by using Table 14.3. ∈ Proof. The formula (14.58) holds by multiplying out. Since det(B)= αδ βγ = 1, − a prime p dividing d (and hence d′) cannot divide both δ and β. We have two cases

p δ = p ∤ β, p ∤ x = p ∤ δ + xβ, | ⇒ ⇒ p ∤ δ = p x = p ∤ δ + xβ. (14.59) ⇒ | ⇒ In both of these cases, p ∤ δ + xβ, and so δ + xβ is a unit in Z . d′ ⊓⊔

Example 14.7. For d = 6, d′ = 12, let B be the order 4 matrix

3 2 B := SL2(Z12). 4 3 ∈ The prime divisors of d are p = 2,3 and δ = 3, so we can take x = 2, and (14.58) gives 3 2 1 0 0 1 10 7 B = 4 3 = 2 1 1− 0 3 2 . − − − 2 2 1 2 Thus a = R− FR− F− R− is a symplectic unitary, with Appleby index [B,0]. This matrix a has three zero entries in each column and satisﬁes a4 = I. 384 14 Weyl–Heisenberg SICs 14.14 The Zauner matrix

It follows from (14.46) that M := RF has the symplectic index

1 1 0 0 1 − 0 1 BM = = − SL2(Zd ). (14.60) 1 1 1 0 1 1 ∈ ′ − − 3 From BM = I, or Example 14.5, it follows the symplectic operation [M] has order 3. This also follows from the Gauss sum

d 1 d 1 2d 1 − − 2 1 − 2πi 2 1 √d j( j+d) j 2d (d+1) j ∑ µ = ∑ ( µ) = ∑ e = G(d + 1,2d)= π , 2 2 2 i (d 1) j=0 j=0 − j=0 e 8 − (14.61) by a direct calculation

2 1 j2+2 jr r2+2rk 1 (r+ j+k)2 k2 2kj (M ) jk = ∑( µ) ( µ) = ∑( µ) − − d r − − d r −

1 √d 2 2πi µ k 2kj 8 (d 1) 1 = π ( )− − =(e )− − (M− ) jk. (14.62) d 2 i (d 1) e 8 − − Thus the Zauner matrix5 (see [Zau99],[Zau10]) given by the normalisation

π d 1 d 1 2 i Z := ζ − M = ζ − RF, ζ := e 24 , (14.63)

2 has order 3, as do Z2,Z,Z C(d). By (14.15), ∈ j k 1 k(k 2 j+d) k j k Z(S Ω )Z− = µ − S− Ω − , (14.64) and by (14.33), we have

0 1 1 1 0 1 1 1 ψ = , ψ 2 = , ψ = , ψ 2 = . Z 1 −1 Z −1 0 Z 1 1 Z −1− 0 − − − − The Zauner matrix satisﬁes

1 2 1 2 R− ZR = Z , R− Z R = Z.

We have the following corollary of Theorem 14.1. Corollary 14.3. The extended Clifford group is generated by H,ˆ and

C (order 2), Z (order 3), F (order 4).

Proof. The extended Clifford group is generated by C, and a set of generators for C(d). Here, the generator R of Theorem 14.1 is replaced by Z = ζ d 1RF. − ⊓⊔ 5 d 1 2 1 1 The Z = ζ − FG of [Zau99] is Z , with F = F− , G = R, U = Ω, V = S− . 14.15 The Scott–Grassl numerical SICs 385 14.15 The Scott–Grassl numerical SICs

A stronger form of Zauner’s conjecture asserts the existence of a Weyl–Heisenberg SIC with a speciﬁc structure, i.e.,

Zauner’s conjecture (Strong): For every dimension d, a Weyl–Heisenberg SIC (set of d2 equiangular vectors) in Cd of the form

j k (S Ω v) j,k Z , ∈ d can be constructed, where v is an eigenvector of the Zauner matrix Z.

Following from the work of Renes, et al [RBKSC04], an extensive (and ongoing) search led by Andrew Scott6 [SG10] found numerical Weyl–Heisenberg SICs (for d 67) and counted and indexed them (for d 50). This has now been extended to a putative≤ list of all Weyl–Heisenberg SICs with≤ certain symmetries for d 90 and at least one for d 120 and d = 124,143,147,168,172,195,199,228,259≤,323. These (Scott–Grassl≤ ) numerical SICs were obtained by finding unit vectors v d j k ∈ C that minimise the second frame potential of (S Ω v) 2 , i.e., by Proposition ( j,k) Zd j k k j jk j k ∈ 14.1 and (S Ω )∗ = Ω − S− = ω S− Ω − , 1 2d ∑ S jΩ kv,v 4 = ∑∑ Sp1 Ω p2 v,Sq1 Ω q2 v 4 , | | Z2 | | ≥ d + 1 ( j,k) Z2 d p q ∈ d | | with equality if and only v is a fiducial vector for a Weyl–Heisenberg SIC. These SIC fiducials can be presented to high accuracy, e.g., 1000 decimal places (see [Chi15]). The numerical SICs given by a fiducial vector v have been invaluable in the study and analytic construction of SICs [ACFW17]. We now detail some of their properties which are summarised in Table 14.4 (kindly provided by Scott and Grassl). The Clifford action (of the extended Clifford group) on the projectors Pv = vv∗ 1 given by fiducial vectors v is g Pv =(gv)(gv) = gPvg , which gives orbits ∗ − 1 orb(v) := g(vv∗)g− g EC(d). { } ∈ The symmetries of a fiducial vector v is the stabiliser of the induced action of the projective group PEC(d), i.e.,

S = S(v) := [a] PEC(d) : av = λv for some (unit) scalar λ C . (14.65) { ∈ ∈ } The symmetries S = S(v) simplify the construction of a ﬁducial vector v, e.g., if v is an eigenvector of the Zauner matrix Z, then v lies in a subspace of Cd of dimension approximately d/3 (see Table 14.7). The symmetries of numerical SICs all appear

6 Using Andrew Scott’s code, C. A. Fuchs, M. C. Hoang and B. C. Stacey have found numerical Weyl–Heisenberg SICs for d 151 (and counting) using a Chimera supercomputer. ≤ 386 14 Weyl–Heisenberg SICs to be symplectic operations, which give a cyclic group of order a multiple of three for d > 3. These include the operations with symplectic indices Fz,Fa,Fb,Fc,Fd,Fe (see 14.18). The number of SICs modulo the action of the extended projective Clifford§ group PEC(d) is given in the # column of Table 14.4.

Table 14.4: The Weyl-Heisenberg covariant numerical SIC-POVMs of [Sco17]. Gaps in the classiﬁcation are marked by ?’s, which either indicate an unexplored dimension (to be ﬁlled in later) or note the likely presence of an unknown general symmetry.

PEC(d) orbits d stabiliser # labels S S notes | | 0 1 2 1 6 1− 0 ,Fz a ∞ −0 1 6 1− 0 ,Fz a 3 0− 1 1 12 1− 0 , Fz b − − 1 48 ESL2(Z3) c 4 1 6 FcFz = Fc Fz a 5 1 3 Fz a 6 1 3 Fz a 1 3 Fz a 7 1 6 FcFz = Fc Fz b 1 3 Fz a 8 6 11 3 6 1 12 = Fz Fb ? b 5 1 10 9 ∋ 9 2 3 Fz a,b 10 1 3 Fz a 11 3 3 Fz a–c 1 3 Fz a 12 0 17 0 17 1 6 Fa Fe b 17 15 ∋ 17 15 ∼ 13 2 3 Fz a,b 14 2 3 Fz a,b 3 3 Fz a–c 15 1 6 FbFz = Fb Fz d 16 2 3 Fz a,b 17 3 3 Fz a–c 18 2 3 Fz a,b 3 3 Fz a–c 19 1 6 FcFz = Fc Fz d 3 12 7 14 7 14 1 18 = Fc Fz Fd e 7 15 5 2 ∋ 5 2 ∼ 20 2 3 Fz a,b 4 3 Fz a–d 21 1 3 Fa e 22 1 3 Fz a 23 6 3 Fz a– f 2 3 Fz a,b 24 1 6 FbFz = Fb Fz c 25 2 3 Fz a,b 14.15 The Scott–Grassl numerical SICs 387

PEC(d) orbits d stabiliser # labels S S notes | | 26 4 3 Fz a–d 27 6 3 Fz a– f 2 3 Fz a,b 28 1 6 FcFz = Fc Fz c 29 4 3 Fz a–d 3 3 Fz a–c 30 1 3 Fa d 31 7 3 Fz a–g 32 2 3 Fz a,b 33 4 3 Fz a–d 34 2 3 Fz a,b 8 3 Fz a–h 35 1 6 FbFz = Fb Fz i 15 3 3 15 1 12 = Fz Fb ? j 32 18 20 18 ∋ 36 4 3 Fz a–d 37 4 3 Fz a–d 38 4 3 Fz a–d 6 3 Fz a– f 39 2 3 Fa g,h 0 7 0 7 2 6 Fa Fe i, j 28 6 ∋ 28 6 ∼ 40 2 3 Fz a,b 41 8 3 Fz a–h 42 4 3 Fz a–d 43 6 3 Fz a– f 44 6 3 Fz a– f 45 4 3 Fz a–d 46 3 3 Fz a–c 47 8 3 Fz a–h 4 3 Fz a–d 1 3 Fa e 48 1 6 FbFz = Fb Fz f 4 37 1 24 Fa,Fb ? g 25 63 ∋ 49 7 3 Fz a–g 50 2 3 Fz a,b 51 14 3 Fz a–n 3 3 Fz a–c 52 1 6 FcFz = Fc Fz d 9 3 Fz a–i 53 7 21 7 21 1 9 Fz Fd j 32 28 ∋ 32 28 ∼ 54 4 3 Fz a–d 55 6 3 Fz a– f 56 6 3 Fz a– f 6 3 Fz a– f 57 2 3 Fa g,h 58 4 3 Fz a–d 388 14 Weyl–Heisenberg SICs

PEC(d) orbits d stabiliser # labels S S notes | | 59 12 3 Fz a–l 60 4 3 Fz a–d 61 6 3 Fz a– f 62 5 3 Fz a–e 14 3 Fz a,d–p 63 2 6 FbFz = Fb Fz b,c 64 4 3 Fz a–d 65 8 3 Fz a–h 6 3 Fz d–i 66 3 3 Fa a–c 7 3 Fz c–i 67 2 6 FcFz = Fc Fz a,b 68 4 3 Fz a–d 69 8 3 Fz a–h 70 5 3 Fz a–e 71 18 3 Fz a–r 72 4 3 Fz a–d 73 4 3 Fz a–d 74 7 3 Fz a–g 12 3 Fz a–l 75 3 3 Fa m–o 76 6 3 Fz a– f 77 8 3 Fz a–h 78 7 3 Fz a–g 79 14 3 Fz a–n 8 3 Fz a–h 80 1 6 FbFz = Fb Fz i 81 12 3 Fz a–l 82 3 3 Fz a–c 83 16 3 Fz a–p 6 3 Fz a– f 84 2 3 Fa g,h 2 2 6 Fe Fe Fa i, j ∼ 85 4 3 Fz a–d 86 10 3 Fz a– j 87 12 3 Fz a–l 88 4 3 Fz a–d 89 10 3 Fz a– j 90 4 3 Fz a–d 91-121 1 3 ? Fz a ≥ ≥ ∋ 99 3 6 ? Fz,Fb b–d ≥ ≥ ∋ 3 111 1 9 ? Fd Fd Fa ≥ ≥ ∋ ∼ 1 6 ? Fz,Fb 120 ≥ ≥ ∋ 1 6 ? Fa,Fb ≥ ≥ ∋ 124 1 6 ? Fz,Fc a ≥ ≥ ∋ 14.16 Symmetries of the Weyl–Heisenberg SICs 389

PEC(d) orbits d stabiliser # labels S S notes | | 143 1 6 ? Fz,Fb a ≥ ≥ ∋ 2 147 1 6 ? Fe Fe Fa a ≥ ≥ ∋ ∼ 168 1 6 ? Fz,Fb a ≥ ≥ ∋ 172 1 6 ? Fz,Fc a ≥ ≥ ∋ 195 1 6 ? Fz,Fb a ≥ ≥ ∋ 199 1 9 ? Fd a ≥ ≥ ∋ 2 228 1 6 ? Fe Fe Fa a ≥ ≥ ∋ ∼ 259 1 6 ? Fz,Fc a ≥ ≥ ∋ 323 1 9 ? Fd a ≥ ≥ ∋

14.16 Symmetries of the Weyl–Heisenberg SICs

The symmetries of a ﬁducial v (under the Clifford group action) given in Table 14.4 include the symplectic (anti)unitaries with symplectic indices (see Exer. 14.16)

0 d 1 0 1 Fz := − =(d + 1) − , d 2, d + 1 d 1 1 1 ≥ − − 1 d + 3 d + 1 3 Fa := 4d 3 =(d + 1) d 3 , d 3 mod 9, d = 3, − d 2 − d 2 ≡ 3 − 3 − β d 2 Fb := − , d = β 1, β 3, d d β − ≥ − κ d 2κ κ d + 2κ 1 0 d =(3k 1)2 + 3, Fc := − = , ± k 0, d + 2κ d κ d + 2κ d + κ 0 1 κ = 3k2 k + 1, ≥ − − ± 0 1 2 Fd := , d = k (k + 3) 1, 1 k(k + 3) − − − 0 1 0 1 1 0 F := = , d = 9k2 + 3. e 1 d + 3k 1 d− 3k 0 1 − − − The ﬁrst two are order three symplectic unitaries, namely the Zauner matrix Z, and

d 1 d 1 1 3 d M1 = aF :=( 1) − R − F− R− FR 3 , d 3 mod 9, d = 3. (14.66) a − ≡ These have symplectic indices with trace 1 (see 14.18), and (see Table 14.4) − § For every dimension d there is a ﬁducial with the Zauner symmetry. • In addition, for d 3 mod 9, d = 3, there are exceptional ﬁducials which are • ≡ eigenvectors of M1 (but not of a conjugate of Z), namely those with the labels

12b, 21e, 30d, 39ghi j, 48eg, 57gh, 66abc, 75mno, 84ghi j, ... 390 14 Weyl–Heisenberg SICs

For all known Weyl–Heisenberg numerical SICs, the symmetries S(v) of the ﬁducial vector v is a cyclic group of order a multiple of three for d > 3.

In addition to a symmetry Z or M1, the tabulated SICs have symmetries: The order 2 symplectic unitary • 1 d(d+1) 1 d(d+1) 2 aF = S 2 Ω 2 P k, d = k 1, b − − for d = 8b, 15d, 24c, 35ij, 48 f , 63bc, 80i, 99, 120, 143a, 168a, 195a, 224,... The order 2 symplectic antiunitary • 1 1 κ(2κ d) 1 κd 2 aF = R− F− R − P3κ FR − C, d =(3k 1) + 3, c ± where κ = 3k2 k + 1, for d = 4a, 7b, 19de, 28c, 52d, 67ab, 103, 124a, 172a, 199, 259a, 292,...± The order 9 unitary • k(k+3) 1 2 aF = R− F− , d = k (k + 3) 3, d − for d = 19e, 53 j, 111, 199a, 323a, 489,... The order 6 antiunitary • d+3k 2 aFe = R FC, d = 9k + 3,

for d = 12b, 39ij, 84ij, 147a, 228a, 327,...

Example 14.8. (d = 4) The Scott–Grassl SIC 4a for C4 is in the 1–eigenspace of Z 1 1 2 3 (dimension 2) and has antiunitary symmetry aC, a := R− F− R− P3FR− , where

√i √i √i √i 1 i 0 1 i 0 1 i 1 i 1 1 −0 1 i − 0− i 1 Z = − − , a = − − − , 2  √i √i √i √i 2  1 i 0 1 i 0  − i 1 −i 1 − 0− i 1− 0 1 i    − − −   − −    It therefore satisfies7 Zv = v, av = 1 (1 i)v. (14.67) √2 − d Each eigenspace of Z and M1 (of order 3) has dimension 3 , see Table 14.7 and (14.109). Equations such as (14.67) aid the search for exact≈ fiducials. Further simplifications of the equations that determine a Weyl–Heisenberg SIC (see 14.27) can be obtained by choosing a conjugate of Z (or other symmetries) to be monomial§ (see 14.17 and 14.18). § § 7 Here v is the numerical SIC 4a, which has the (unnormalised) analytic form given by (14.83). The SIC chosen in (14.68) does not have the symmetry av = cv, c T. ∈ 14.17 Monomial representations of the Clifford group 391 14.17 Monomial representations of the Clifford group

If d is a square, say d = n2, then the commutativity relation (14.5) gives

2 Ω nSn = ωn SnΩ n = SnΩ n, so the Weyl–Heisenberg group has an abelian subgroup generated by Sn and Ω n. By diagonalising this subgroup, [ABB+12] were led to a monomial representation of the Clifford group, i.e., one in which the matrices have exactly one nonzero entry in each row and column. We now summarise (without proof) this phase-permutation d representation. Let (ek)k Zd is the standard basis for C . d∈ In the basis (u j) for C given by

n 1 − ω nt j1 2 u j := ∑ − ent+ j2 , j =( j1, j2) Zn, t=0 ∈ the Clifford group elements a transform to monomial matrices

1 aˆ := U− aU, U :=[u j], where

e , j2 + 1 = 0; 2πi ˆ ( j1, j2+1) Ωˆ ω j2 σ n Se j := e j := e( j 1, j ), := e . σ j1 1− 2 e( j1,0) j2 + 1 = 0,

Example 14.9. (d = 4) For n = 2, with the order (0,0),(0,1),(1,0),(1,1) on the 2 indices from Z2, we have 010 0 0010 101 0 100 0 0 0 0 i 010 1 Sˆ Ωˆ U = 0 0 0 1, = 1000, = 1 0 1 0 , 001− 0 0 i 0 0 0 1− 0 1      −        100 0 0 0 1 0 0 √i 0 0 001 0 0 √i 0 0 0 0 i 0 Fˆ = , Rˆ = , Zˆ = . 010 0  1− 0 0 0  √i 0− 0 0 0 0 0 i 0 0 0 √i 0 0 01  −         −    1 In the basis (u j) the condition Zv = v of (14.67) can be written as Zwˆ = w, w =U− v, i.e., √iw2 = w1, iw3 = w2 and √iw1 = w3. Thus we seek a ﬁducial of the form − w =( i√i, i,1,a), a C. It is easy to check that a = 2 + √5√i gives a ﬁducial vector−v, which− is a 1–eigenvector∈ of Z, namely 1 i√i − 1 i + 2 + √5√i 1 v = − , √i = (1 + i). (14.68) √ 1 i√i √2 10 + 2 5 −−  √ √   i 2 + 5 i − −  392 14 Weyl–Heisenberg SICs 14.18 The Clifford trace and symplectic unitaries of order 3

Each known Weyl–Heisenberg SIC has an order 3 symmetry Z or M1 (see 14.16). The symplectic indices of these symplectic unitaries satisfy §

trace(Fz)= trace(Fa)= 1. − We now classify such matrices up to conjugation by a symplectic operation. The Clifford trace is the map

trC : C(d) Zd : a trace(ψa). → →

Since a ψa is a homomorphism with kernel Hˆ (Lemma 14.2), this satisﬁes →

trC(ab)= trC(ba), a,b C(d), (14.69) ∀ ∈

trC(ah)= trC(a), a C(d), h Hˆ . (14.70) ∀ ∈ ∀ ∈ In particular, the Clifford trace of any conjugate of Z or Z 1 = Z2 is 1, e.g., − − 1 1 trC(gZg− )= trC(Zg− g)= trC(Z)= trace(ψZ)= 1, − and the Clifford trace is well deﬁned on the Clifford operations, i.e.,

trC([a]) := trC(a), [a] PC(d). ∀ ∈

The order of a Clifford operation is related to its Clifford trace, since

2 A = trace(A)A I, A SL2(Zd). (14.71) − ∀ ∈

Lemma 14.4. A nonidentity extended Clifford operation [a] EC(d)/[I] with index ∈ (A,za) and Clifford trace t = trace(A) has order 3 if and only if

T 2 (t+1)p MA p 2 (t 1)A =(t + 1)I, za((t + 1)Ap)= ω , p Z , (14.72) − ∀ ∈ d γ α3 2α2δ αδ 2 2α δ βγ α δ 1 α δ 1 α β where M = ( + + ) ( + )( + + ) , A = . A βγ(α + δ 1)(α +−δ +−1) β(δ 3 + 2αδ 2−+ α2δ 2δ α) γ δ − − − Proof. Since a product of three antiunitaries is a unitary matrix, we have a C(d). In view of the isomorphism (14.35), [a] has order 3 if and only if ∈

3 3 2 (A,za) =(A ,(za A )(za A)za)=(I,1). ◦ ◦ From (14.71), we obtain

A3 = A(tA I)= t(tA I) A =(t2 1)A tI, − − − − − 14.18 The Clifford trace and symplectic unitaries of order 3 393 so that the condition A3 = I can be written as the ﬁrst condition of (14.72). 2 We now consider the condition (za A )(za A)za = 1. By (14.23), we calculate ◦ ◦ T 2 p σA(Ap) 2 za(p)za(Ap)za(A p)= ω− za(p + Ap)za(A p) T T 2 p σA(Ap) (p+Ap) σA(A p) 2 = ω− ω− za(p + Ap + A p) T 2 T 2 p (σAA+σAA +A σAA )p 2 = ω− za(p + Ap + A p).

By (14.71), we have

I + A + A2 = I + A +tA I =(1 +t)A. − Using det(A)= αδ βγ = 1, a calculation gives − 2 T 2 σAA + σAA + A σAA γ(α3 + 2α2δ + αδ 2 2α δ) βγ(α + δ 1)(α + δ + 1) =(α + δ + 1) − − − . βγ(α + δ 1)(α + δ + 1) β(δ 3 + 2αδ 2 + α2δ 2δ α) − − − 2 Thus we may rewrite the condition (za A )(za A)za = 1, to obtain the result. ◦ ◦ ⊓⊔ Since za(0)= 1, a EC(d), and trC(I)= 2 = 1 if and only if d = 3, we have ∀ ∈ −

If a C(d) has Clifford trace 1 and d = 3, then [a] has order 3. ∈ −

Further, by taking the trace of the condition (t2 1)A =(t + 1)I, we have −

The Clifford trace t of a Clifford operation of order 3 satisﬁes

(t 2)(t + 1)2 = 0. (14.73) −

For d prime, Clifford operators of order 3 must have Clifford trace 1. − Proposition 14.6. Suppose that d = 3 and a C(d). Then ∈ 1. If a has Clifford trace 1, then [a] has order 3. 2. If d is prime, then [a] has− order 3 if and only if a has Clifford trace 1. − Proof. Since we have already proved 1, it sufﬁces to prove for d = 3 prime and [a] of order 3 that the Clifford trace t = trC(a) is 1. We recall that t is a root of (14.73). − If t = 1, then t + 1 is a unit (all nonzero elements of Zd are units for d prime), − so that t = 2. But, if t = 2, then (14.72) gives 3A = 3I, and hence A = I (3 Zd∗ for d = 3 prime), so that a Hˆ (by Lemma 14.2). Since (S jΩ k)3 = ω3SrjΩ rk (see∈ Exer. 14.4) and S, Ω have order∈ d, the order of [a] cannot be 3 (since 3 does not divide d). Thus t = trC(a) 1 (when [a] has order 3 and d = 3 is prime). − ⊓⊔ 394 14 Weyl–Heisenberg SICs

A Clifford operation of order 3 is said to be canonical order 3 if it has Clifford trace 1 (see [App05]), e.g., the Zauner matrix Z and M1 are canonical order 3. − Example 14.10. It follows from (14.69) and (14.70) that left or right multiplication of a canonical order 3 Clifford operation by a displacement operation gives another canonical order 3 operation, e.g., [h1Zh2] is canonical order 3 for any h1,h2 Hˆ . ∈ There are Clifford operations of order 3 with Clifford trace 2.

d′ Example 14.11. If 3 divides d, then the symplectic unitary R 3 (and its inverse) has order 3 and Clifford trace

d′ 1 0 tr (R 3 )= trace = 2, C d′ 1 3 d d d d as do the Weyl displacement operators S 3 ,Ω 3 ,S 3 Ω 3 . There are Clifford operations of order 3 with Clifford trace t = 1,2, i.e., for − which (14.73) holds with t 2 and t + 1 not units in Zd. − Example 14.12. For d = 10, SL2(Z10) has a single conjugacy class of elements of order 3 and trace 4 and 7. These have representatives

3 2 6 5 A (trace 4) B (trace 7) = 6 1 , = 5 1 . These can be lifted to symplectic indices which give symplectic unitaries of order 3 and Clifford trace 4 and 7 (see 14.13), e.g., a = R10FR8F 1R6, b = R10FR5F 1R15. § − − We need the following technical lemma of [BW17]. Lemma 14.5. Suppose that d 2, and let ≥ 0 1 1 1 z : ψ z2 (14.74) = Z = 1 −1 , = −1 0 , − − 1 3 m1 := d 3 , d 3 mod 9, (14.75) − 2 ≡ 3 − 1 3 m2 := 2d 3 , d 6 mod 9. (14.76) − 2 ≡ 3 −

Then the conjugacy classes of elements of order 3 and trace 1 in SL2(Zd) have representatives −

z , d 0 mod 3, (14.77) { } ≡ z,z2 , d 0 mod 9 or d = 3, (14.78) { } ≡ 2 z,z ,m1 , d 3 mod 9, d = 3, (14.79) { } ≡ 2 z,z ,m2 , d 6 mod 9. (14.80) { } ≡ 14.18 The Clifford trace and symplectic unitaries of order 3 395

Lemma 14.6. Let ϕ : G H be a homomorphism of G onto H, with kerϕ = 2k. If h H has order 3, then→ there is an element g G of order 3 with ϕ(|g)= h.| ∈ ∈ Proof. By the ﬁrst isomorphism theorem for groups, we may assume that H = G/K, where K = kerϕ. Suppose that h = aK G/K has order 3, i.e., a3 = x K, where a K. By Bezout’s´ identity (the Euclidean∈ algorithm) choose integers∈α,β with 1 =∈ 3α + 2kβ. Let g = axα a . Then ϕ(g)= axα K = aK, and − ∈ α α α kβ g3 =(ax )3 = a3x3 = x3 +1 = x2 = 1.

⊓⊔

Lemma 14.7. For d even, SL2(Z2d) has no elements of of order 3 and trace d 1. − Proof. If A SL2(Z2d) has order 3, and t = trace(A), then, by (14.71), we have ∈ A3 = A(tA I)= t(tA I) A =(t2 1)A tI = I = (t2 1)A =(t + 1)I. − − − − − ⇒ − For t = d 1, this gives (d2 2d)A = 0 = dI (mod 2d), which not possible. − − ⊓⊔ We now characterise all symplectic unitaries of canonical order 3. Theorem 14.3. (Characterisation) The symplectic operations of canonical order 3 are conjugate in CSp(d)/[I] to [a], where a CSp(d) is one of the following ∈ Z , d 0 mod 3, { } ≡ Z,Z2 , d 0 mod 9 or d = 3, { } ≡ 2 Z,Z ,W1 , d 3 mod 9,d = 3, { } ≡ 2 Z,Z ,W2 , d 6 mod 9, { } ≡ where

π 2 i (d 1) 1 Z := e 24 − RF− , d 1 2d a 1 3 Wa :=( 1) − R 3 F− R FR, (14.81) − have order 3 in CSp(d). Proof. The key idea is to apply the fact that group homomorphisms map conjugacy classes to conjugacy classes to the commutative diagram (14.51) of 14.11, i.e., § α SL (Z ) -- (d)/[I] 2 d′ CSp

ψ m ˆ d -- ? SL2(Zd) We observe that 396 14 Weyl–Heisenberg SICs

Since the kernel of Ψˆ has order 1 or 4 (d odd or even), the conjugacy classes of • elements of order 3 and Clifford trace 1 in CSp(d)/[I] map onto the conjugacy − classes of elements of order 3 and trace 1 in SL2(Zd) (by Lemma 14.6). Since the kernel of α has 1 or 2 (d odd− or even), each conjugacy class of an • element of order 3 and Clifford trace 1 in CSp(d)/[I] is the image under α of the conjugacy class of an element of order− 3 in SL (Z ) (by Lemma 14.6) and 2 d′ of trace 1 (by Lemma 14.7). − Thus the conjugacy classes of elements of order 3 and trace 1 in SL2(Z ) map − d′ onto the conjugacy classes of elements of canonical order 3 in CSp(d)/[I], which in turn map onto the conjugacy classes of elements of order 3 and trace 1 in SL2(Zd). A count of the conjugacy classes in SL (Z ) and SL (Z ) (for d even)− shows that 2 d′ 2 d these maps are 1–1, i.e., representatives of the conjugacy classes of elements of order 3 and trace 1 in SL (Z ) give symplectic indices for representatives of the 2 d′ conjugacy classes− of the symplectic operations of canonical order 3. We now use Lemma 14.5 to calculate these symplectic indices (and show the injectivity asserted above) for the various cases. For d 0 (mod 3), we have 2d 0 (mod 3), and so there is a single conjugacy class with≡ symplectic index z. ≡ For d 0 (mod 9), d = 3, we have 2d 0 (mod 9), and so there is are two ≡ ≡ 2 conjugacy classes given by the symplectic indices z,z . For d = 3, we have d′ = d, and there are two conjugacy classes given by the symplectic indices z,z2. For d 3 (mod 9), d = 3, we have 2d 6 (mod 9), so that there are three conjugacy≡ classes given by the symplectic indices≡ z,z2, and

1 3 1 3 d 3 SL2(Zd) (d odd), 2(2d) 3 SL2(Z2d) (d even). − 2 ∈ − 2 ∈ 3 − 3 − The second formula gives the ﬁrst for d odd, and so works in both cases. For d 6 (mod 9), we have 2d 3 (mod 9), so that there are three conjugacy classes given≡ by the symplectic indices≡ z,z2, and

1 3 1 3 2d 3 SL2(Zd) (d odd), 2d 3 SL2(Z2d) (d even). − 2 ∈ − 2 ∈ 3 − 3 − In the last two cases, the third conjugacy class is given by the symplectic indices m1 and m2 (respectively), where 1 3 m j := 4dj 3 SL2(Zd ). − 2 ∈ ′ 3 − 2 and m j is conjugate to m j (since otherwise there would be four conjugacy classes). For convenience of presentation, we take the representative with symplectic index

2d j 1 3 2 3 1 0 3 0 1 − 1 0 0 1 1 0 w := m2 = − − = . j j 1 + 2d j 1 1 1 1− 0 1 1 1− 0 1 1 3 14.18 The Clifford trace and symplectic unitaries of order 3 397

2 By taking the symplectic operations corresponding to the representatives z,z ,w1,w2 2 in the above conjugacy classes, i.e., Z,Z ,W1,W2, we obtain representatives for the conjugacy classes of canonical order 3 symplectic operations. The normalisation of Wa in its deﬁnition (14.81) ensures that it has order 3 (see Exer. 14.14). ⊓⊔ From the above proof, we have:

The conjugacy classes of order 3 and trace 1 elements in SL(Z ) are in 1–1 d′ correspondence with the conjugacy classes− of canonical order 3 symplectic operations.

The matrices W0,W1,W2 given by (14.81) are deﬁned for d 3, and are sparse, i.e., |

µ 1 ( j k)2+ 2d aj2+k2 3 1 3d ( )− 3 − 3 , j k 0 mod 3; (Wa) jk = ( √i) − − − ≡ d − 0, j k 0 mod 3. − ≡ 1 d 1 1 1 Further W0 is deﬁned for all d, and is conjugate to Z via τ − W0 =(R− F)− Z(R− F).

1 d 1 1 1 τ − W0 =(R− F)− Z(R− F).

Example 14.13. The canonical order 3 unitary given by the symplectic index

1 d + 1 3 1 0 − 2 1 0 2 1 3 (d + 1)Fa = d 3 = w1 , w1 = 4d 3 − d 2 d 1 d 1 − 2 3 − 3 − 2 1 is conjugate to W1 = W1− , and so is given by the order 3 symplectic unitary

d 1 d d 1 d 1 1 3 d aF = R W − R =( 1) − R − F− R− FR 3 . a 1 −

All the known Weyl–Heisenberg SICs appear as eigenvectors of the canonical order 3 symplectic unitaries Z or M1 = aFa (equivalently W1). As yet, no SIC ﬁducials have been found which are eigenvectors of W2 (for d 6 mod 9). ≡

Since the symmetries appear to form a cyclic group, it is natural to look for roots of the symplectic indices for z and m1 as the symplectic indices of extra symmetries. Example 14.14. For d = 8, the matrix

2 11 13 6 11 b = = SL2(Z16) 3 8 5 1 ∈ is a square root of (d + 1)z. It is the symplectic index for the square root of Z given 17 8 1 7 by B = ζ R P13F− R , which is a symmetry of the SIC 8(c) of Table 14.4. 398 14 Weyl–Heisenberg SICs 14.19 Conjugates of the canonical order 3 symplectic unitaries

For ﬁducial vectors which are eigenvectors of Z or M1 (all those known do date) the equations characterising the SIC can be simpliﬁed (as in the Example 14.9) by conjugating Z (or M1) by an element of EC(d) to obtain a monomial matrix (the conjugates of S jΩ k will continue to be monomial matrices). We now use Theorem 14.3 to determine when the conjugate of Z or M1 (or M2 α for that matter) by a symplectic operation is a monomial matrix of the form R Pσ . Since the permutation matrices in C(d) are symplectic (see 14.12), there exists § a permutation matrix Pσ C(d), σ Z of canonical order 3 if and only if ∈ ∈ d∗ 3 1 Pσ = Pσ 3 = I, trC(Pσ )= σ + σ − = 1, − i.e., the existence of an integer σ (for d = 3) with σ 3 1 mod d, σ 2 + σ + 1 0 mod d. (14.82) ≡ ≡ σ 3 σ 3 σ 0 0 1 0 For such a , we have 1 = 2 3 = , so that: α σ − α(1 + σ + σ ) σ 0 1 −

α If σ satisﬁes (14.82), then [R Pσ ] is a canonical order 3 symplectic operation.

By the Chinese remainder theorem, it follows (see [App05]) that the condition (14.82) is equivalent to d satisfying: (i) d has at least one prime divisor 1 mod 3. (ii) d has no prime divisors 2 mod≡ 3 (so that d is odd). (iii) d is not divisible by 9. ≡ The ﬁrst few such d are

d = 7,13,19,21,31,37,39,43,49,57,61,67,73,79,91,93,97,...

α By Theorem 14.3, the monomial operation [R Pσ ] is conjugate (via a symplectic 2 operation) to one of [Z], [Z] , [M1]=[W1], [W2]. For d not a multiple of 3 (d 0 mod 3), i.e., • ≡ d = 7,13,19,31,37,43,49,61,67,73,79,91,97,...

α there is single conjugacy class, and so all [R Pσ ] are conjugate to [Z]. For d a multiple of 3, i.e., • d = 21,39,57,93,111,129,147,183,201,219,237,...

d we have 3 1 mod 3, i.e., d 3 mod 9, and so the conjugacy classes are given 2 ≡ ≡ by [Z],[Z] ,[M1]. 14.19 Conjugates of the canonical order 3 symplectic unitaries 399

For a σ satisfying σ 3 = 1, 1 + σ + σ 2, the symplectic index calculations

σ 0 σ 2 0 1 σ 0 1 gzg 1 = , gz2g 1 = , g := , z = , − 1 σ 2 − 1 σ 0 1 1 −1 − − give the following:

For any d, if σ satisﬁes (14.82), then

1. The monomial operation [RPσ ] is a symplectic conjugate of [Z]. 1 2 2. The monomial operation [R− Pσ ] is a symplectic conjugate of [Z ].

Whenever d is a multiple of 3, i.e., d 3 mod 9, it appears (for the d listed above) ≡ that [Pσ ] is always a symplectic conjugate of [M1].

Example 14.15. For d = 21, no symplectic conjugate of Z is a permutation matrix, but many conjugates are monomial, e.g., the symplectic index calculation

1 1 4 0 1 1 4 − 4 0 0 1 1 −1 0 1 = 1 16 , − together with Table 14.3, gives

1 7 16 1 4 gZg− = ω R P4, g := F− R− F.

Example 14.16. For d = 19, Appleby [App05] constructed an exact SIC ﬁducial which was an eigenvector of the order 18 antiunitary P10C with symplectic index

5 9 0 9 0 3 12 5 1 J g 1 g g : −0 2 = −0 2 = − 7 15 , = 3 8 . − In view of Table 14.4, this is the SIC 19e. The exact ﬁducial is

18 θ iℓr 5+9√5 10 √5 θ 1 √5 1 v = b0e0 + ∑ b1e er, b0 = 95 , b1 = 190− , = cos− 8− , r=1 r where ℓr =( ) 1,1 is the Legendre symbol. 19 ∈ {− } The extra symmetries given by P10C immediately imply that the ﬁducial vector v has the very simple structural form above. At the time of its construction, the only known exact SICs were for d = 2,...,7 and d = 8 (the Hoggar lines). By using similar techniques, the exact Scott–Grassl SICs 24c, 35 j and 48g were constructed [SG10]. In subsequent constructions of exact SICs, the symmetries deduced from numerical SICs have been exploited. 400 14 Weyl–Heisenberg SICs 14.20 The SIC ﬁeld of a Weyl–Heisenberg SIC

The Weyl–Heisenberg SICs for d = 3 that have been found exactly, e.g., for d = 4 8 (√10 + √2 2√5 2) √5 + 1 4 i +(√10 + √2) √5 + 1 + 4√2 4 − − − −  (8√2 8)i  − −  √10 √2 2√5 2 √5 1 4 i √10 √2 √5 1 4√2 4  ( + ) + + ( + ) + +  − − − − (14.83)−  have the following features: They are very complicated to present (in general), e.g., the three exact fiducial • vectors for d = 11 given in [SG10] take up 22 pages. They are expressible by radicals (nested roots). • Since the equations defining a SIC involve quartic polynomials in several variables (see 14.27), it does not follow (from the fact a univariate quadratic can be solved by radicals)§ that the components of a SIC should be expressible by radicals. We now consider the field defined by a Weyl–Heisenberg SIC. We will see that some of its Galois automorphisms map SICs to SICs. This leads to methods for the construction of exact SICs from numerical SICs. The presentation (to follow) assumes a basic familiarity with Galois theory. In theory, the natural field in which to define a SIC is that generated by the triple products of its vectors (see 8.6). Since the Weyl–Heisenberg SICs are given by the § orbit of a fiducial projector Π = vv∗ under the action of the Heisenberg group, and the Clifford group maps SIC fiducials to SIC fiducials (see 14.7), it is convenient and effective to (possibly) enlarge8 the field to contain the entries§ of Π and µ.

Definition 14.5. The SIC field E of a Weyl–Heisenberg SIC with fiducial Π = vv∗ πi is the smallest extension of Q containing the entries of Π and µ = e d .

The inclusion of µ ensures (see Exercises 14.22, 14.23) that:

The SIC ﬁeld E = Q(Π, µ) depends only on the extended Clifford orbit of Π.

T T 1 Since Π =(Π ∗) = Π and µ = µ− , it follows

The SIC ﬁeld is closed under complex conjugation (denoted by gc).

We have the (inclusion reversing) Galois correspondence between subfields of E and subgroups of the Galois group G (of automorphisms of the field E which fix Q).

8 In earlier work [AYAZ13], the SIC field E was defined to be the smallest normal extension of Q containing the entries of Π, √d, and µ. Here we use the more recent definition of [AFMY17]. 14.21 The Galois group of a generic SIC 401 14.21 The Galois group of a generic SIC

For d > 1, a SIC is said to be generic if it is a Weyl–Heisenberg SIC for d 4, otherwise it is sporadic SIC. The sporadic SICs have special properties not shared≥ by the generic SICs (see [Zhu15], [Sta17], [AFMY17], Exer. 14.21), e.g.,

The symmetry groups of the known sporadic SICs, i.e., the Hoggar lines and the d = 2,3 Weyl–Heisenberg SICs are nonabelian (and doubly transitive on the SIC ﬁducial projectors [Zhu15]), whereas the known generic SICs have cyclic symmetry groups.

The generic SICs seem to have many special properties (in addition the Clifford group action), that were observed in the early constructions. These were formalised as a set of conjectures about the SIC ﬁeld E (and the action of its Galois group) in [AYAZ13], and a reﬁned version of these facts is given in [AFMY17]. We now outline these facts (conjectures) about the generic SICs. From now on, we assume that SICs are generic, and the known SICs are those reported in [ACFW16].

Fact 1. In every known case, the SIC ﬁeld E is normal over Q.

We recall E is normal over Q means that every irreducible polynomial over Q which has a root in E splits over E (i.e., all its roots are in E).

Fact 2. In every known case, E is an extension of K := Q( (d 3)(d + 1)). − Here (d 3)(d + 1) is never an integer. − Fact 3.In every known case, Gal(E/K) is a ﬁnite abelian group. In particular, the Galois group of E over Q is solvable, and so a generic SIC is expressible by radicals.

We recall the Kronecker–Weber theorem, that the finite abelian extensions of Q are subfields of some cyclotomic field Q(e2πi/n). Finding a similar characterisation for the finite abelian extensions of a quadratic field (such as the extension E of K above) is an instance of Hilbert’s 12–th problem. This was solved for imaginary quadratic fields Q(i√n), n a positive integer, where such extensions are a subfield of a field generated by the torsion points of certain elliptic curves. For abelian extensions of Q(√n), such as the conjectured E, there is currently no such construction. Even if the SIC field E is known (there are conjectures that it is a ray class field over K), then it is not immediately obvious how to go from a Scott–Grassl numerical SIC (of high precision) to an exact SIC with entries in E (since E is dense in C). This is part of the intrigue of the SIC problem, i.e., given the high precision numerical SICs, it is tempting to imagine that exact SICs cannot be too far away. 402 14 Weyl–Heisenberg SICs 14.22 The Galois action on a fiducial projector

The extended Clifford group permutes the SIC ﬁducials via the action (Exer. 14.22)

1 a Π :=(av)(av)∗ = aΠa− , Π = vv∗, a EC(d). ∈ There is a natural action of the Galois group G = Gal(E/Q) of the SIC ﬁeld E on matrices over E given by

m n g(A)= g([a jk]) :=[g(a jk)], g G , A E × . ∈ ∈ One might hope that this Galois action also maps SIC fiducials to SIC fiducials. Indeed we have already seen ( 14.7) that this is the case for complex conjugation, § which we now denote by gc G . We will show that this is true for a large (index 2) ∈ subgroup Gc (consisting of the elements which commute with gc). If g G = Gal(E/Q) and Π =[a jk] is a fiducial projector, then a necessary ∈ condition for g(Π) to be a fiducial projector is g(Π)∗ = g(Π)= g(Π ∗), i.e.,

g(akj)= g(akj) gcg(akj)= ggc(akj). ⇐⇒ Since the automorphism g maps µ to another 2d–th root of unity, this implies that g G must commute with complex conjugation gc G (see Exer. 14.20). ∈ ∈

For g G = Gal(E/Q) and Π a SIC ﬁducial, g(Π) is a rank one orthogonal ∈ projection if and only if g is in the centraliser of complex conjugation gc G . ∈

Let Gc be the centraliser of gc in G . We now show that Gc maps SIC ﬁducials to Π jΩ k SIC ﬁducials. Let = vv∗ be a rank one orthogonal projection. Since (S ) j,k Zd is an orthogonal (nice error) basis, we have ∈ 1 Π = ∑ Π,S jΩ k S jΩ k, d j,k Z ∈ d i.e., Π is determined by its overlaps (or scalar multiples of them)

j k j k j k Π,S Ω = trace(vv∗(S Ω )∗)= v,S Ω v ( j,k) =(0,0), and Π is a SIC fiducial if and only if these have constant modulus. We define the overlaps χΠ := trace(ΠDˆ ), p Z , where the Dˆ are the displacement operators p p d′ p of (14.37). We note that Π is a SIC∈ fiducial if and only if its overlaps satisfy

Π 1, p 0 mod d; χ := trace(ΠDˆ p)= iθp ≡ p Z . (14.84) p e , p 0 mod d, ∈ d′ √d ≡

We refer to the θp as the overlap phases of the SIC ﬁducial Π. 14.22 The Galois action on a ﬁducial projector 403

Let g G = Gal(E/Q). Since ( µ) is a primitive d –th root of unity, there is a ∈ − ′ unique integer 0 kg < d , with ≤ ′ kg g( µ)=( µ) , kg Z∗ . (14.85) − − ∈ d′ so that

p1 p2 p1 p2 kg p1 p2 p1 kg p2 g(Dˆ p)= g(( µ) S Ω )=( µ) S Ω = Dˆ H p, (14.86) − − g where 1 0 Hg := GL2(Zd ), g G . (14.87) 0 kg ∈ ′ ∈ Lemma 14.8. Let g Gc. If Π = vv is a ﬁducial projector with SIC ﬁeld E, then so ∈ ∗ is g(Π), i.e., Gc maps SICs to SICs (possibly on different extended Clifford orbits).

Proof. Since g Gc, it follows g(Π) is a rank one orthogonal projection. By (14.86), ∈ χΠ Π ˆ Π ˆ χg(Π) 2 g( p )= g(trace( Dp)) = trace(g( )DHg p)= , p Z . Hg p ∀ ∈ d′ 2 2 Since g Gc, i.e., it commutes with conjugation, we have g( z )= g(z) , z E, and so ∈ | | | | ∈ g(Π) 2 Π 2 Π 2 χp = g(χ 1 ) = χ 1 . | | | Hg− p | | Hg− p| In view of (14.84), we conclude that g(Π) gives a SIC, with SIC ﬁeld E. ⊓⊔

The subgroup Gc of Gal(E/Q) maps SICs to SICs (with the same SIC ﬁeld), with the Galois action on the overlaps given by

χΠ χg(Π) 2 g( p )= , p Z . (14.88) Hg p ∀ ∈ d′

It appears that Gc is a large (index 2) subgroup of G .

Fact 4. In every known case,

Gal(E/K)= Gc = the centraliser of complex conjugation.

By (14.88), the Galois action of Gc maps SICs to SICs with the same SIC field, but on possibly different extended Clifford orbits. We will call the extended Clifford orbits with the same SIC field a Galois multiplet, with the number of orbits being its size, e.g., 4a is a Galois singlet, 9ab is a Galois doublet, 30abc is a Galois triplet, and 21abcd is a Galois quartet. It appears that orbit of a SIC fiducial under the combined Galois and Clifford actions is (all of) a multiplet. 404 14 Weyl–Heisenberg SICs 14.23 Constructing exact SICs from numerical SICs

Before giving more detail about the Galois action on SICs, we explain why it has been so pivotal in the construction of exact SICs from numerical SICs. Suppose now, for a moment, that

The Gc–orbits O j of the overlaps of a SIC ﬁducial are known. • { } K = Q( (d 3)(d + 1)) is ﬁxed by Gc (Fact 4). • − Gc is abelian (Facts 3 and 4). • Then the Gc–invariant polynomials

f j(x) := ∏ (x χ), − χ O j ∈ would have coefficients in the field K. By using the Scott–Grassl numerical SICs, these coefficients could be calculated to a high degree of accuracy. If it were then possible to “guess” these coefficients in K exactly, then one could find the overlaps exactly by factoring the exact f j (Gc is solvable), and so convert a numerical SIC into an exact SIC. This general method is called precision bumping (also see 9.8), and has been successful in constructing many exact SICs (see [Chi15], [ACFW16]),§ by using the PSLQ algorithm or LLL (Lenstra–Lenstra–Lovasz)´ algorithm to “guess” elements of K (or some small extension of it, such as the E0 of 14.24). §

14.24 The Galois action on a centred ﬁducial

The precision bumping algorithm of 14.23 for constructing exact SIC fiducials § supposes that one know the Gc–orbit of the overlaps. At this point, only the action of the subgroup G0 which maps a SIC to one on the same extended Clifford orbit is understood. We now summarise these results (see [AYAZ13], [AFMY17]). Let G0 be the group which fixes the extended Clifford orbit of a fiducial Π, i.e., Π G0 = G := g Gc : g(Π) is on the same extended Clifford orbit as Π , 0 { ∈ } Π G G G and E0 = E0 be the fixed field of 0. Clearly, 0 is a subgroup of c, which depends only on the extended Clifford orbit of Π (or it seems the multiplet).

G G Π Π Fact 5. In every known case, 0 = 0 depends only on the multiplet of , and the index [E0 : K] is the size of the multiplet.

The group S(v) of the symmetries (14.65) of a SIC ﬁducial vector v, are the stabiliser of the projector Π = vv∗ under the Clifford action (see Exer. 14.22), since

S(v)= [a] PEC(d) : (av)(av)∗ = vv∗ = [a] PEC(d) : a Π = Π =: SΠ . { ∈ } { ∈ } 14.24 The Galois action on a centred ﬁducial 405

By a simple calculation, 1 S[a] Π =[a]SΠ [a]− , Hence it is natural to seek a ﬁducial [a] Π on the extended Clifford orbit of Π for 1 which the symmetries [a]SΠ [a]− have a simple form. Now (see Exer. 14.24)

The Clifford action of an [a] PEC(d), with extended Appleby index [B,b], on the overlaps is given by ∈

χΠ ω b,det(B)Bp χ[a] Π 2 p = , p Z . (14.89) det(B)Bp ∀ ∈ d′

1 In particular, if [a] is symplectic, i.e., b = 0, then the overlaps of Π and aΠa− are a permutation of each other. We say that a ﬁducial Π is centred if its symmetries SΠ are (extended) symplectic operations. In particular, all the Scott-Grassl SICs are centred (see Table 14.4). In every known case, the symmetry group SΠ of a SIC ﬁducial can always be conjugated to be symplectic, i.e.,

On every extended Clifford orbit there is a centred SIC ﬁducial.

For a given centred SIC ﬁducial Π, we denote by S0(Π) the group of (extended) symplectic indices for SΠ , i.e., with fE given by (14.57),

S0(Π) := B ESL2(Z ) : αE (B) SΠ , αE (B) := fE ([B,0]). { ∈ d′ ∈ } Since [a] Π = Π for [a] SΠ , it follows from (14.89) that ∈

If Π is a centred SIC ﬁducial, then

Π Π B S0(Π) χ = χ . ∈ ⇐⇒ p det(B)Bp

This clearly reduces the number of overlaps that must be found to construct a SIC. We now consider a further refinement of the SIC field E. Let D be the square–free part of (d 3)(d + 1), so that K := Q( (d 3)(d + 1)) = Q(√D). − − Let g Gal(E/Q) be any element with g(√D)= √D (these exist, by the Galois ∈ 1 − correspondence). Then g1 := ggcg− is an element of order 2, which fixes K, since

g1(√D)= g( √D)= g( √D)= √D, − − and is independent of the choice of g, since

1 1 1 1 ggcg− = gg˜ cg˜− g− g˜ Gc = Gal(E/K) g g˜ fixes K, ⇐⇒ ∈ ⇐⇒ − 1 1 with the last equivalence by: g− g˜(√D)= g− ( √D)= √D. In every known case, Π − g1 G , i.e., g1 maps fiducials to fiducials on the same extended Clifford orbit. ∈ 0 406 14 Weyl–Heisenberg SICs

1 From (14.85), we have g 1( µ)=( µ)kg− , so that − − −

1 1 k kg 1 1 g1( µ)= ggcg− ( µ)=(( µ)− g− ) =( µ)− = Hg = J := . − − − − ⇒ 1 1 −

Let E1 be the ﬁxed ﬁeld of the order 2 subgroup G1 = g1 of Gc. By the Galois correspondence, Q K E0 E1 E, with the index [E : E1]= 2. ⊂ ⊂ ⊂ ⊂

Fact 6. In every known case, E = E1(i√d′).

We now seek to understand the extension of E0 to E1.

Fact 7. On every known extended Clifford orbit, there is a ﬁducial Π with its χΠ overlaps p in E1, i.e., by (14.88),

Π g1(Π) Π g1(χ )= χ = χ , p Z . (14.90) p Jp p ∀ ∈ d′

A centred SIC fiducial satisfying (14.90) is said to be strongly centred. For d 0 mod 3, every centred fiducial is strongly centred (see Exer. 14.27). Whilst for d ≡ 0 mod 3, some, but not all, centred fiducials are strongly centred. Hence ≡

On every extended Clifford orbit there is a strongly centred SIC ﬁducial.

We now assume that the SIC ﬁducial Π is strongly centred. Let O1,...,Oℓ be the extended Clifford orbits of the multiplet for Π = Π1, and Π j be a strongly centred ﬁducial on O j. Then for every g Gc, there is a symplectic operation a = aB ∈ ∈ EC(d), with symplectic index B = Fg j ESL2(Z ), for which , ∈ d′ 1 g(Π j)=[a] Πk = aΠka− , [a]= fE ([Fg j,0]). ,

The fact that an a EC(d) with g(Π j)=[a] Πk can be chosen to be symplectic is ∈ a consequence of Π j and Πk being strongly centred (see Exer. 14.27). By (14.88) and (14.89), we calculate

Π Π Π Π Π χ j χg( j) χ[a] k χ k χ k 1 g( p )= H p = H p = 1 = G p, Gg, j := det(Fg, j)Fg−, j Hg. g g det(B)B− Hg p g, j (14.91) For g G0, we have j = k, and, in particular, for g1, we have (see Exer. 14.27) ∈ 1 Fg j = J, Gg j = det( J)( J)− J = I. 1, − 1, − −

For a given strongly centred Π = Π j, the matrix Fg j ESL2(Z ) is not unique, , ∈ d′ as it may be replaced by any element of the left coset Fg, jS0(Π). Correspondingly, the matrix Gg j GL2(Z ) may be replaced by any element of the form , ∈ d′ 14.24 The Galois action on a centred ﬁducial 407

1 1 1 1 1 det(Fg jA− )(Fg jA− )− Hg = det(A− )Adet(Fg j)F− Hg, A S0(Π). , , , g, j ∈ 1 Since det(A− )= det(A), we may write this set as the right coset S(Π)Gg, j, for the subgroup S(Π) := det(A)A : A S0(Π) . { ∈ } We now investigate whether the mapping of g to these cosets is a homomorphism. In view of the homomorphism αE , SΠ is a homomorphic image of S0(Π), i.e.,

SΠ = S0(Π) (d odd), SΠ = S0(Π)/ dI (d even). ∼ ∼ It is easy to see that the map

Θ : GL2(Z ) GL2(Z ) : A det(A)A (14.92) d′ → d′ → is a homomorphism, which maps S0(Π) onto S(Π), and has kernel

3 kerΘ = cI : c = 1,c Z∗ . { ∈ d′ } 2 If cI S0(Π) has order 3, then from det(cI)= c = 1, we have that c = 1, i.e., ∈ ±

The groups S0(Π) and S(Π) are isomorphic (via θ).

Let N(S0(Π)) and N(S(Π)) be the normaliser of S0(Π) and S(Π) in GL2(Zd ). 1 1 ′ If gMg− = L, then det(M)= det(gMg− )= det(L), and so

1 1 gMg− = L g(det(M)M)g− = det(L)L. ⇐⇒

Hence S (Π) and S(Π) have the same normaliser in GL (Z ), which we 0 2 d′ denote by N(Π)= N(S0(Π)) = N(S(Π)).

Since g(Π)=[a] Π, a = aF , we have (see Exer. 14.27) g, j 1 Sg(Π) = S[a] Π = g(SΠ )=[a] SΠ = aSΠ a− . ⇒

Hence, for [aL]= αE (L) SΠ there is [aM]= αE (M) SΠ with (see Exer. 14.26) ∈ ∈ 1 g([aL])=[a 1 ]=[aaMa− ]=[aFg, j aMa 1 ], HgLHg− Fg−, j

1 1 i.e., HgLHg− = Fg, jMFg−, j , up to equivalence of symplectic indices, and we have Π 1 Π 1 HgS0( )Hg− = Fg, jS0( )Fg−, j . (14.93)

We observe that g kg and g Hg of (14.85) and (14.86) are homomorphisms. → → 408 14 Weyl–Heisenberg SICs

Lemma 14.9. Let Π be a strongly centred ﬁducial. Then for a ﬁxed j, the maps Π G N( ) 1 Π 0 : g Fg−, j HgS0( ), (14.94) → S0(Π) → Π N( ) 1 G0 : g Gg jS(Π), Gg j := det(Fg j)F− Hg, (14.95) → S(Π) → , , , g, j

are homomorphisms, which have kernels gc and g1 , respectively. 1 Π 1 1 Π Proof. Let Fg = Fg, j, Gg = Gg, j. By (14.93), Fg− HgS0( )(Fg− Hg)− = S0( ), so 1 Π d d that Fg− Hg N(S0( )), and the ﬁrst map is well deﬁned. With aB chosen in E × , we calculate∈ Π Π 1 g1g2( )= aFg g a− , 1 2 Fg1g2 1 1 1 g1g2(Π)= g1(aF Πa− )= a 1 aF Πa− a− . g2 Fg Hg Fg Hg− g1 Fg 1 2 1 2 1 1 Hg1 Fg2 Hg−1 By equating, rearranging, and taking symplectic indices, we obtain

1 1 1 1 1 1 1 F− Hg g (F− Hg )− (F− Hg )− = F− Hg Fg H− Fg S0(Π), (14.96) g1g2 1 2 g2 2 g1 1 g1g2 1 2 g1 1 ∈

i.e., the ﬁrst map is a group homomorphism. Above we used Hg1g2 = Hg1 Hg2 , For the second map, we apply the homomorphism Θ of (14.92), for which Θ Π Π 1 (S0( )) = S( ). It is well deﬁned, since det(Hg)det(Hg− )= 1, gives

1 1 Π 1 1 1 Π det(Fg− )Fg− HgS( )(det(Fg− )Fg− Hg)− = S( ). Θ Applying to (14.96), and using det(Hg1g2 )= det(Hg1 )det(Hg2 ) to simplify, gives G G 1G 1 S Π , and so it is a homomorphism. g1g2 g−2 g−1 ( ) For the kernel∈ of the ﬁrst homomorphism, we have

1 1 F− Hg S0(Π) = det(F− )det(Hg)= 1 = kg = det(Hg)= 1. g ∈ ⇒ g ± ⇒ ±

For kg = 1, we have g ﬁxes µ and Hg = I, so that

Fg S0(Π) = g(Π)= Π = g ﬁxes E = Q(Π, µ) = g = I. ∈ ⇒ ⇒ ⇒ 1 For kg = 1, we have g(µ)= µ = gc(µ) and Hg = J, so that − −

Fg JS0(Π) = g(Π)=[C] Π = gc(Π) = g = gc on E = g = gc. ∈ ⇒ ⇒ ⇒ For the kernel of the second, we have a similar argument. As before

1 det(Fg)F− Hg S(Π) = kg = det(Hg)= 1. g ∈ ⇒ ± 1 1 Π Π For kg = 1, we have det(Fg− )Fg− S( ) so that Fg S0( ) and g = I. For kg = 1, we have ∈ ∈ −

1 1 1 1 det(F− )F− J = det( JF− )( JF− ) S(Π) = JFg S0(Π), g g − g − g ∈ ⇒ − ∈ 14.24 The Galois action on a centred ﬁducial 409 so that Fg JS0(Π). Hence ∈− 1 g(Π)=[a J] Π = g1(Π), g(µ)= µ− = g1(µ) = g = g1. − ⇒ ⊓⊔

By (14.91), the action of g G0 on the overlaps of a strongly centred ﬁducial ∈ Π = Π j is given by

χΠ χΠ g( p )= Gg p, Gg = Gg, j. (14.97) Thus the second homomorphism (14.95) provides an explicit way to compute the G0–orbits of the overlaps, given that the image of G0 in N(Π)/S(Π) is known. This makes the precision bumping method of 14.23 feasible. §

Factoring out (14.95) by its kernel G1 := g1 gives an injective homomorphism N(Π) G0/G1 = Gal(E1/E0) : g g1 GgS(Π). (14.98) → S(Π) → In view of Fact 3, the image of this map must be an abelian group.

Key fact. In every known case, the map g GgS(Π) of (14.98) deﬁnes an isomorphism → Π Π Gal(E1/E0) ∼= M( )/S( ), (14.99) where M(Π) is a maximal abelian subgroup of GL (Z ) containing S(Π), 2 d′ and g permutes the overlaps via g(χΠ )= χΠ , p. p Gg p ∀

Let C(X) denote the centraliser in GL (Z ) of a set X of matrices. 2 d′ Example 14.17. (Type-z orbits) For a strongly centred ﬁducial Π which is given by an eigenvector of the Zauner matrix (or a conjugate of it) with symplectic index B, a calculation (see Exer. 14.25) shows the centraliser of B in GL (Z ) is abelian, so 2 d′ that Π Π Π Π Gal(E1/E0) ∼= C( )/S( ), C( ) := C(S( )) = C(B). This is the original conjecture of [AYAZ13].

Example 14.18. (Type-a orbits) For a strongly centred ﬁducial Π which given by an eigenvector of the canonical order 3 operation with symplectic index Fa, the centraliser of F in GL (Z ) is not abelian, and there are three maximal abelian a 2 d′ subgroups containing S(Π) (see [ACFW16]). In this case, the isomorphism (14.99) holds for an appropriate choice of M(Π) in the known cases: 12b, 21e, 48g. 410 14 Weyl–Heisenberg SICs 14.25 Minimal and maximal SIC ﬁelds

The facts and their consequences for the SIC ﬁeld of 14.24 can be summarised: §

The SIC field E of a generic strongly centred fiducial Π which is an n-let (multiplet of size n) is conjectured to have the Galois correspondence ⊳ 1 ⊳ G1 G0 ⊳ Gc ⊳ G ⊲ ⊲ ⊲ ⊲ E E1 Π E0 K Q (14.100) 2 M( ) n 2 | S(Π) | where Gc is abelian, K = Q( (d 3)(d + 1)), and all of the subgroups and subfields shown are normal (with degree− of the field extension given below).

Under these assumptions the precision bumping algorithm outlined in 14.20 G G Π Π § proceeds as follows (see [ACFW16]). Let 0/ 1 ∼= M( )/S( ) act on the overlaps of a centred ﬁducial via (14.97) to obtain orbits O j . If d 0 mod 3, then Π may not be strongly centred, and it is convenient to{ replace} the≡ overlaps by their third powers (which are permuted by G0/G1). If Π is an n–let, then the polynomials

f j(x) := ∏ (x χ), (14.101) − χ O j ∈ have coefficients in the field E0 (a degree n extension of K). For a singlet (n = 1), i.e., Gc = G0, the algorithm proceeds as already outlined. For an n–let, the Gc–orbit of an overlap will be the union of n of the G0/G1–orbits of overlaps from centred fiducials lying on different extended Clifford orbits. Either the Gc–orbit can be guessed in this way (e.g., if n = 2 and gc G0 then one could conjugate the Clifford orbit), or ∈ the polynomials f j can be dealt with directly. In practice, so far, it has been possible to “guess” the next to leading coefficient of f j exactly (since E0 is a small degree extension of K), thereby determining E0, and applying the algorithm as before. For a given SIC multiplet, there is a SIC field E (the SIC field of the multiplet). For different multiplets, the SIC field (and the Galois group G ) might be different. Calculations of [ACFW16] (for fixed d) suggest that

Each SIC multiplet has a different SIC field. • There is minimal SIC field Emin, which is contained in all SIC fields. • There is maximal SIC field Emax, which contains in all SIC fields. •

A SIC field which is not minimal or maximal is called an intermediate SIC field. We say that a multiplet (or any SIC in it) is minimal, intermediate or maximal if its SIC field is. See Figure 14.1 and Table 14.5 for the a summary of the SIC fields obtained in [ACFW16]. 14.25 Minimal and maximal SIC fields 411

15ac 19bc 35bcdg

4 4 2 2 2

15b 19d 35e 35af 35h

2 2 4 2 15d 19a 35i 4 3 2

19e 35 j

39acde 48abcd

2 3 3 8

39bf 39gh 48e 48 f

3 2 8 3 39ij 48g

Fig. 14.1: The SIC ﬁeld inclusions (with Emin at the bottom) for cases where there are one or more intermediate ﬁelds, together with the degrees of the extensions.

minimal intermediate maximal d [E : E ] multiplet multiplet(s) multiplet max min 15 15d 15b 15ac 4 17 17c 17ab 2 18 18ab 19 19e 19a, 19d 19bc 12 20 20ab 21 21e 21abcd 3 24 24c 24ab 4 28 28c 28ab 4 30 30d 30abc 3 35 35 j 35i, 35e, 35h, 35af 35bcdg 16 39 39ij 39bf , 39gh 39acde 6 48 48g 48e, 48 f 48abcd 24

Table 14.5: Minimal, maximal and intermediate multiplets 412 14 Weyl–Heisenberg SICs 14.26 Ray class ﬁelds

A ray class field is an abelian extension of a global field (such as the algebraic number field K = Q(√D)) associated with a ray class group. Every finite abelian extension of a number field is contained in one of its ray class fields. The motivating example is the abelian extensions of Q. Here the ray class field over Q associated with the ray class group Zn∗ is the field generated by the n–th roots of unity. Every finite abelian extension of Q is a subfield of such a ray class (cyclotomic) field (the Kronecker–Weber theorem). In general, the role of n above is is played by the conductor of a ray class field. Most importantly, there are algorithms for calculating in ray class fields. Calculations (see [AFMY16], [ACFW16]) suggest

Ray class conjecture: The minimal SIC field E = Emin is the ray class field over K = Q( (d 3)(d + 1)= Q(√D) with conductor d′, and ramification allowed at both infinite− places9.

Given this, the minimal SIC field (and extensions of it) can be constructed in Magma using RayClassField(m). This makes calculations in the precision bumping algorithm, such as factoring the polynomials f j of (14.101), faster than when the ray class field is constructed as a tower of extensions. There are further conjectures about the SIC field (see [ACFW16]), e.g.,

Let E = Emin be the minimal SIC field in dimension d. Then in every known case the tower of fields (14.100) satisfies:

E and E1 are ray class fields over Q(√D) for which the finite part of the • conductor is d′. E1 is the class field with ramification only allowed at the infinite place • taking √D to a positive real number. E0 is the Hilbert class field over Q(√D). •

We recall that D is the square–free part of (d 3)(d + 1). If D is a square–free positive integer, then it is the square–free part of (−d 3)(d + 1) if and only if − (d 3)(d + 1)= m2D (d 1)2 m2D = 4. − ⇐⇒ − − for some positive integer m. This is a modified version of Pell’s equation, and has infinitely many solutions d1 < d2 < d3 < (see [AFMY17]), given by d1 1 d = 1 + 2T − , j > 1, j j 2 9 There are four ray class fields over K = Q( (d 3)(d + 1) with conductor d′. The SIC field E is the largest of these, and the three others, which include− E , are subfields of E. 1 14.26 Ray class fields 413 where the Tj are the Chebyshev polynomials of the first kind. If Zauner’s conjecture holds with the SIC field structure conjectured, then

For every square–free positive integer D, there are infinitely many minimal SIC fields which are ray class fields over Q(√D).

Example 14.19. The ﬁrst few sequences of dimensions with D ﬁxed are

d = 4, 8, 19, 48, 124, 323, 844, 2208, 5799, 15128, ... D = 5, d = 5, 15, 53, 195, 725, 2703, 10085, 37635, 140453,... D = 3, d = 6, 24, 111, 528, 2526, 12099, 57966, 277728, ... D = 21, d = 7, 35, 199, 1155, 6727, 39203, 228487, 1331715,... D = 2, d = 9, 63, 489, 3843, 30249, 238143, 1874889, 14760963, ... D = 15.

It can be shown [AFMY16] that for each one of these sequences, there are inﬁnitely many distinct subsequences for which each dimension divides the following one. These subsequences are called dimension towers (the corresponding minimal SIC ﬁelds are nested).

For the D = 5 sequence (d j)=(4,8.19,48,124,...), it has been conjectured in [GS17] that there is a SIC with an antiunitary symmetry of order 6 j given by the symplectic index

0 1 0 1 1 0 F : AJ f = 1 1 = 1 −1 0 1 = , − − i.e., the symplectic operation ZC. Such a SIC is called a Lucas–Fibonacci SIC, on account of the fact that the entries of the powers of A = Ff J are the Fibonacci numbers. By directly solving (14.102) using Groebner basis methods, [GS17] have constructed an exact Lucas–Fibonacci SIC for d = 124 (labelled 124a), and have found numerical SICs for d6 = 323, d7 = 844, by searching in eigenspaces of ZC. It is natural to speculate about similar infinite families of SICs for other values of D. In this regard, we list the first few dimensions for D 19 ≤ D 2 3 5 6 7 10 11 13 14 15 17 19 d1 7 5 4 11 17 39 21 12 31 9 67 341 d2 35 15 8 99 255 1443 399 120 899 63 4355 115599 d3 199 53 19 971 4049 54759 7941 1299 26911 489 287299 39302981 and observe from the symmetries given in Table 14.4, that for D = 2, there appears to be a family with S = 6 j, d = d j, and for D = 3 a family with S = 3 j, d = d j. There much on going| | work in this direction. | | 414 14 Weyl–Heisenberg SICs 14.27 Equivalent equations for SIC fiducial vectors

The vectors v Cd giving SIC fiducials are determined by the d2 quartic equations ∈ 1 S jΩ kv,v 2 = , ( j,k) =(0,0), v 2 = 1, (14.102) | | d + 1 in v1,...,vd, v1,...,vd with coefficients from Q(ω). The variational characterisation of (2,2)–designs (see Exer. 10.6) gives two equations (the first has degree 8) 1 2 ∑ S jΩ kv,v 4 = v 4, v 2 = 1. (14.103) d2 | | d(d + 1) ( j,k) Z2 ∈ d The algebraic variety of SIC fiducials (see 7) can be viewed as being real (by § taking real variables ℜv j and ℑv j and polynomials with complex coefficients) or complex. In any case, it appears that:

The algebraic variety of SIC ﬁducials is zero dimensional (and nonempty), except for d = 3, where it is one dimensional.

The direct solution of equations such as above led to the earliest exact SICs, e.g., [SG10] used Groebner basis methods on (14.102), with v in an eigenspace of Z. We observe that

j k j k kr S Ω v,v =(S− v)∗Ω v = ∑vr+ jvrω , (14.104) r ω which is a polynomial in , with coefficients Uj =(vrvr+ j)r Zd . By using finite Fourier methods, this leads to a third set of simplified equat∈ions (see [ADF14], [BW07], [Kha08]), which we now present, and solve in a couple of cases.

Theorem 14.4. A vector v Cd is a Weyl–Heisenberg SIC ﬁducial if and only if ∈ 0, s,t = 0; t 1 Us,S− Us = ∑ vrvr+svr+t vr+s+t = , s = 0,t = 0, s = 0,t = 0;  d+1 r Zd  2 ∈  d+1 , (s,t)=(0,0). (14.105)  Proof. Using (14.104), we may write S jΩ kv,v 2 as a polynomial of degree d 1 evaluated at a d–th root of unity | | −

j k 2 kr ks S Ω v,v = ∑vrvr+ jω ∑vsvs+ jω− | | r s kt k = ∑∑vrvr+ jvr t vr t+ jω =: f j(ω ). t r − − 14.27 Equivalent equations for SIC ﬁducial vectors 415

1 For (14.102) to hold for j = 0, we must have f j be constant, equal to d+1 (since f j is uniquely determined by its values at the d–th roots of unity), i.e.,

0, t = 0; ∑vrvr+ jvr t vr t+ j = j = 0, − − 1 r d+1 , t = 0, which (after a change of variables) gives the ﬁrst two equations in (14.105). Given that this holds, we have

d 1 k 1 − kt 4 f0(ω )= ∑ ω + ∑ vr . d + 1 t=1 r | | For (14.102) to hold for j = 0, k = 0, we must have k 1 4 1 4 2 f0(ω )= + ∑ vr = ∑ vr = ∑vrvrvrvr = , −d + 1 r | | d + 1 ⇐⇒ r | | r d + 1 i.e., the third equation in (14.105) must hold. This then ensures (14.102) holds for j = 0 and k = 0, i.e., d 1 2 f (1)= − + = 1. 0 d + 1 d + 1 Thus for v to be fiducial it is necessary and sufficient that the subset of the equations (14.105) considered above holds. Given the symmetry in s and t of the left hand side, it follows that they all hold for a fiducial. ⊓⊔ From the proof, it is sufficient to require that (14.105) hold for 0 s t d . ≤ ≤ ≤⌊ 2 ⌋ Example 14.20. With r j := v j , the last two conditions in (14.105) become | |

2 2 1 4 2 ∑r j r j+s = , s = 0, ∑r j = . j d + 1 j d + 1

θ Example 14.21. For d = 2 and v j := r je j , the equations of (14.105) are 1 2 v2v 2 + v 2v2 = 0, 2r2r2 = , r4 + r4 = . 0 1 0 1 0 1 3 0 1 3

2 2 Solving the last two (using r0 + r1 = 1 to simplify the algebra) gives

3 √3 3 √3 r2 = ± , r2 = ∓ . 0 6 1 6

The ﬁrst then reduces to cos(2(θ1 θ0)) = 0, which has solution − π π iθ1 iθ0 i n 2(θ1 θ0)= + πn e = e e 4 i , n = 0,1,2,3. − 2 ⇐⇒ This gives the unique SIC for C2 (up to projective unitary equivalence). 416 14 Weyl–Heisenberg SICs

Example 14.22. We consider the sporadic SICs for d = 3, which is the only known case where there are inﬁnitely many inequivalent SICs. Here (14.105) gives 1 1 v v v2 + v v v2 + v v v2 = 0, r2r2 + r2r2 + r2r2 = , r4 + r4 + r4 = . 2 1 0 0 2 1 1 0 2 2 1 0 2 1 0 4 0 1 2 2

4 4 4 2 2 2 2 2 2 From r0 + r1 + r2 = 1/2 = 2(1/4)= 2(r0r1 + r1r2 + r2r0), we obtain

(r2 (r2 + r2))2 = r4 +(r4 + 2r2r2 + r4) 2r2(r2 + r2)= 4r2r2. 0 − 1 2 0 1 1 2 2 − 0 1 2 1 2 If r2 r2 + r2, then taking square roots gives 0 ≥ 1 2 2 2 2 2 2 2 2 r (r +r )= 2r1r2 = r = r +2r1r2 +r =(r1 +r2) = r0 = r1 +r2, 0 − 1 2 ⇒ 0 1 2 ⇒ 2 2 2 and if r0 < r1 + r2, then

2 2 2 2 2 2 2 r (r + r )= 2r1r2 = r = r 2r1r2 + r =(r1 r2) , 0 − 1 2 − ⇒ 0 1 − 2 − so that r0 = r1 r2 for r1 r2, and r0 = r2 r1 for r2 r1. Thus we have three cases − ≥ − ≥

r0 = r1 + r2, r1 = r0 + r2, r2 = r0 + r1.

2 2 2 Since r0 + r1 + r2 = 1 and r j 0, these three cases describe the three sides of a spherical triangle in the ﬁrst octant≥ on the unit sphere. The vertices of this triangle are given by the intersections of the three great circles, e.g.,

1 1 r0 = r1 + r2 and r1 = r0 + r2 = (r0,r1,r2)=( , ,0), ⇒ √2 √2 with the other vertices being ( 1 ,0, 1 ) and (0, 1 , 1 ). √2 √2 √2 √2 It remains only to satisfy the ﬁrst equation, i.e.,

2 2 2 v2v1v0 + v0v2v1 + v1v0v2 = 0. (14.106)

We consider a representative case when r =(r0,r1,r2) is on a vertex, and on an edge of the spherical triangle. The other fiducials come from a symmetry of the triangle, i.e., by permuting the entries of v. 1 1 The vertex (r0,r1,r2)=( , ,0). Since r2 = v2 = 0, the equation (14.106) is √2 √2 | | trivially satisfied, and as there are no other conditions, we have fiducial vectors

iθ 1 e v = eiφ , θ,φ R. (14.107) √2  0  ∈   2 The triangle edge r0 = r1 +r2, r1,r2 > 0. For convenience, let z j := v j /v j, so that z j = v j = r j. Then dividing (14.106) by v0v1v2 yields the equivalent equation | | | |

z0 + z1 + z2 = 0. 14.27 Equivalent equations for SIC ﬁducial vectors 417

Since z0 = z1 z2 and z0 = r0 = r1 + r2 = z1 + z2 , the complex numbers − − | | |− | |− | z0, z1, z2 must have the same argument, i.e., − − iφ iφ iφ z0 = r0e , z1 = r1e , z2 = r2e , φ R. − − ∈ iθ 3iθ Hence, writing v j = r je j , so that z j = r je j , we calculate

φ 3iθ0 iφ 2π r0e = r0e = θ0 = + k0, k0 = 0,1,2, ⇒ 3 3 θ φ φ π π 3i 1 i θ 2 r1e = r1e = 1 = 3 + 3 + 3 k1, k1 = 0,1,2, − ⇒ φ 3iθ2 iφ π 2π r2e = r2e = θ2 = + + k2, k2 = 0,1,2. − ⇒ 3 3 3

After the change of variables θ :=(φ +2πk0)/3, we can describe the ﬁducial vectors corresponding to the point (r0,r1,r2) on this edge of the triangle by

r0 iθ j v = e r1µω 1 , θ R, j1, j2 0,1,2 ,  j  ∈ ∈ { } r2µω 2   2πi where µ = e 6 , ω = µ2. We can give a more precise description of this edge by 2 2 2 solving r0 + r1 + r2 = 1 and r0 = r1 + r2 for r1,r2 in terms of r0, which gives

2 2 r0 2 3r0 r0 2 3r0 r = ± − and r = ∓ − . 1 2 2 2

Here the half of the edge where r1 r2 is given by the ‘top’ choice in the formulas, ≥ and the other choice gives the half with r1 r2. For these to give r1,r2 > 0, we must have ≤ 2 2 2 2 2 1 2 3r 0, r0 2 3r > 0 r , r > . − 0 ≥ − − 0 ⇐⇒ 0 ≤ 3 0 2 1 2 2 Hence < r along this edge, with r1 and r2 given by the formulas above. 2 0 ≤ 3 For the vertex SIC ﬁducial given by (14.107), we have the triple product

∆(v,Sv,S2v)= v,Sv Sv,S2v S2v,v = 1 eit , t = 3(φ θ). 8 − and so there are uncountably many projectively unitarily inequivalent vertex SIC ﬁducials parameterised by t = 3(φ θ). In 8.5 of [Zhu12], the symmetries S(v) of the SICs in three dimensions were calculated.− § For these, S(v) can be 6 (the inﬁnite family 3a), 12, or 28 (types 3b and 3c). | |

Example 14.23. Direct solution of the equations (14.105) shows that there exist real ﬁducial vectors v Rd for d = 3,7,19 (see [BW07], [Kha08]). ∈ The solution of systems of equations such as those presented in this section, still play a vital role in the analytic construction of exact SICs. 418 14 Weyl–Heisenberg SICs Notes

Many of the Gauss sums that appear when multiplying elements of the Clifford group can be calculated by using the following quadratic reciprocity law

For a,b,c Z, with ac = 0 and ac + b even, one has the quadratic reciprocity ∈ c 1 a 1 | |− π 2 c π 2 | |− π 2 ∑ e i(aj +bj)/c = | | e i( ac b )/4ac ∑ e i(ck +bk)/a. (14.108) a | |− − j=0 | | k=0

The Clifford group can be viewed from the point of view of Gabor analyis for the finite abelian group Zd [FKL09], where the Clifford operations are referred to as metaplectic operations [FHK+08]. There is currently much activity on Zauner’s conjecture, due to The recent conjectures about the structure of the SIC field and the action of its • Galois group on fiducials, e.g., the ray class conjecture. The recent construction of many more exact SICs, and consequently SIC fields. • The recent construction of many more numerical SICs. • Efficienct methods for calculating the symmetry group from numerical SICs, and • consequently conjectures about families of SICs with large symmetry groups, such as the Lucas–Fibonnacci SICs. Some history and an extensive collection of references is given in [FHS17]. I wish to thank Marcus Appleby, Len Bos, Tuan Chien, Steve Flammia, Chris Fuchs, Gary Mcconnell, Markus Grassl, Andrew Scott, Jon Yard, Blake Stacey and Hjuangjun Zhu for sharing their results and insights on the SIC problem with me. 14.27 Equivalent equations for SIC fiducial vectors 419 Exercises

d d d 14.1. Let (Pj) be a SIC for C , and C × have the Frobenius inner product. (a) Determine the condition on c1,c2 R which ensures that Pj c1I and Pk c2I are orthogonal for j = k. ∈ − − (b) Find the dual basis to (Pj). 1 (c) Show that (Pj cI) is a basis if and only if c = d . − 1 1 (d) Show that (Pj cI), c = (1 ) is orthogonal. − d ± √d+1 (e) Find the projection of (Pj cI), c R onto the traceless matrices. − ∈ 14.2. The Bloch sphere. Each unit vector v C2 can be scaled so that ∈ θ cos 2 v = θ φ , 0 θ π, 0 φ < 2π. sin ei ≤ ≤ ≤ 2

(a) Express the rank one orthogonal projection matrix vv∗ in terms of

a = sinθ cosφ, b = sinθ sinφ, c = cosθ.

Here (a,b,c) R3, with a2 + b2 + c2 = 1. (b) Show that∈ the map 1 vv∗ √2 vv∗ I → − 2 maps the rank one orthogonal projections on C2 (equivalently, unit vectors in C2) onto the traceless Hermitian matrices with unit Frobenius norm. Remark: This identiﬁcation of the rank one orthogonal projections (pure states of a two-level quantum mechanical system) with the unit sphere in the 2 2 traceless Hermitian matrices is called the Bloch sphere, with (a,b,c) a Bloch vector× . (c) Show that the map

d 1 vv∗ A = vv∗ I → d 1 − d − maps the rank one orthogonal projections on Cd (unit vectors in Cd) to the traceless Hermitian matrices with unit Frobenius norm, but is not onto for d > 2. 3 2 14.3. The Bloch vector (a,b,c) R for a rank one orthogonal projection vv∗ on C is given by ∈ c a ib 2vv∗ I = − . − a + ib c − (a) Calculate the the Bloch vectors of the SIC for C2 given by (1.7). (b) The vectors ( 1,0, 1 ), (0, 1, 1 ) are the vertices of a regular tetrahedron. ± − √2 ± √2 Find the SIC corresponding to the Bloch vectors these give when normalised. 14.4. Show that the displacement operators S jΩ kgiven by (14.4) satisfy (a) Ω kS j = ω jkS jΩ k. 420 14 Weyl–Heisenberg SICs

j k r 1 r(r 1) jk rj rk (b) (S Ω ) = ω 2 − S Ω . (c) For h = cSaΩ b Hˆ , one has hS jΩ kh 1 = ωbj akS jΩ k. ∈ − − 14.5. Show that the matrices F,R,M,Pσ of 14.5 satisfy j k 1 jk k j § (a) F(S Ω )F− = ω− S− Ω . j k 1 j( j+d) j j+k (b) R(S Ω )R− = µ S Ω . j k 1 k(k 2 j+d) k j k (c) M(S Ω )M− = µ − S− Ω − . j k 1 σ j σ 1k (d) Pσ (S Ω )Pσ− = S Ω − .

d d 14.6. For d even, show that S 2 and Ω 2 are symplectic unitaries, i.e., belong to the subgroup of the Clifford group generated by F and R, with

d d d 1 d 1 d Ω 2 = R , S 2 = F− Ω 2 F = F− R F.

Z2 14.7. Show that SL2(Zd) T d is a group with the multiplication ×

(A,zA)(B,zB) :=(AB,(zA B)zB), ◦ where functions Z2 T are multiplied pointwise. d → 2 14.8. The function za : Z T of (14.18) satisﬁes (14.25), i.e., d →

cA(p,q) za(p + q)= ω za(p)za(q),

2 2 where the symmetric function cA : Z Z Zd is given by d × d →

cA(p,q) :=(Ap)2(Aq)1 p2q1, A := ψa. −

(a) Show that cA(p,q) can be written as a quadratic form αγ βγ α β T σ σ ψ cA(p,q)= p Aq, A := βγ βδ , A = a = γ δ . d+1 d+1 p (b) Suppose that d is odd. Then µ = µ = ω 2 , so that Zd T : p ( µ) 2− → → − is well defined. Show thatz â : Z T given by d → T cA(p,p) p σA p p1 p2 (Ap)1(Ap)2 zâ(p) :=( µ)− za(p)=( µ)− za(p)=( µ) − za(p), − − − is a character, i.e.,z â(p + q)= zâ(p)zâ(q), p,q, and that ∀

zˆab =(zˆa ψb)zˆb, ◦ ωkp1 jp2 zˆF = zˆR = zˆM = 1, zˆS jΩ k (p)= zS jΩ k (p)= − .

(c) Suppose that d is even. Choose a B SL2(Z2d) with B = A mod d (there are T σ ∈ eight choices). Show that ( µ)p B p depends only on p mod 2d, so that − T 2 p σB p zâ B : Z T : p ( µ)− za(p) , d → → − 14.27 Equivalent equations for SIC fiducial vectors 421 is well defined. Show thatz â,B is a character, and so has the form

χ,Ap 2 zâ B(p)= ω , where χ Z . , ∈ d 14.9. Here we consider how the indexing of 14.9 depends on the normalisation of S and Ω. Let § Sˆ := c1S, Ωˆ := c2Ω, c1,c2 T, ∈ j k 2 and Uˆ := Sˆ Ωˆ . As in (14.18), define a correspondingz â : Z T by ( j,k) d → cp1 cp2 ˆ 1 ˆ 1 2 aUpa− = zâ(p)Uψ (p) = zâ(p)= ψ ψ za(p). a ⇒ a(p)1 a(p)2 c1 c2

(a) Show thatz ˆa satisﬁes

bj ak zâb = zˆb(zâ ψb), zˆ aΩ b ( j,k)= ω − , ◦ S and so Corollary 14.1 holds with za replaced byz â. (b) For the choice c1 = c2 = µ, show that det(Uˆ p)= 1 andz â(p) is a power of ω. − zâ( j,k) is a power of ω, and calculatez â( j,k) explicitly for a = F,R,M,Pσ

14.10. A 2n 2n matrix M over a ﬁeld F is said to be symplectic if × 0 I MT AM A A : = , = I 0 . − The symplectic matrices form a group Sp(n)= Sp(2n,F). (a) Show that for n = 1 the symplectic matrices are the matrices with determinant 1, i.e., Sp(2)= SL2(F). (b) Show SL2(Z) is a subgroup of SL2(R). (c) Show that SL2(Zd) is a group. Remark: The elements of SL2(Zd) are called symplectic matrices.

14.11. Show that if b Z∗ has odd order, then ∈ d′

1 bj( j+d) 1 d cb,d := ∑ µ =(√i) − . √d j Z ∈ d

14.12. Let A be a normal matrix with eigenvalues λ1,...,λk, and Pj be the projection onto the λ j–eigenspace. Since normal matrices are unitarily diagonalisable, we have

j j j j A = λ P1 + λ P2 + + λ Pk, j = 0,1,2,.... 1 2 k 2 k 1 (a) Show that Pj can be written as a linear combination of I,A,A ,...,A − . π 2 k 1 2 i (b) Suppose A has order k, so that its eigenvalues are 1,ω,ω ,...,ω − , ω := e k (possibly with zero multiplicity). Show that the projection onto the λ–eigenspace is 422 14 Weyl–Heisenberg SICs

1 2 k 1 Pλ = I +(λA)+(λA) + +(λA) − . k (c) Show that if A is unitary and of order k, then the multiplicities of its eigenvalues ℓ k (dimension of it eigenspaces) can be determined from trace(A ), 0 ℓ 2 . (d) Calculate the multiplicities of the eigenvalues of the Fourier matrix≤ ≤F, i.e.,

d 1 i 1 i 4m m + 1 m −m m− 1 4m + 1 m + 1 m m m− 4m + 2 m + 1 m m + 1 m 4m + 3 m + 1 m + 1 m + 1 m

Table 14.6: The multiplicities of the eigenvalues 1,i,i2,i3 of F.

(e) Calculate the multiplicities of the eigenvalues of the the Zauner matrix Z, i.e.,

d 1 τ τ2 3m m + 1 m m 1 3m + 1 m + 1 m m− 3m + 2 m + 1 m + 1 m

2 Table 14.7: The multiplicities of the eigenvalues 1,τ,τ2 of Z and Z .

2πi 14.13. Suppose that ℓ d. Let √i = e 8 . ℓ | 1 ℓ 1 ℓ d′ (a) Show that R , and hence F− R F =(F− RF) , has order ℓ . 1 ℓ (b) For d odd, and for d even and ℓ odd, show that F− R F is sparse, i.e.,

j kµ 1 ( j k)2 1 ℓ ℓ 1 ℓd ( 1) − − ℓ − , j k 0 mod ℓ; (F− R F) jk = (√i) − − − ≡ d 0, j k 0 mod ℓ. − ≡ 1 ℓ (c) For d even and ℓ even, show that F− R F is sparse, i.e.,

j k 1 ( j k)2 d ℓ ( 1) µ ℓ , j k mod ℓ; 1 ℓ √ 1 ℓd − − − 2 (F− R F) jk = ( i) − − − ≡ d d 0, j k mod ℓ. − ≡ 2 2πi 2πi 14.14. Suppose that 3 d. Let τ = e 3 , √i = e 8 , and | d 1 2d a 1 3 Wa :=( 1) − R 3 F− R FR, a = 0,1,2. −

(a) Show that Wa is sparse, i.e.,

µ 1 ( j k)2+ 2d aj2+k2 3 1 3d ( )− 3 − 3 , j k 0 mod 3; (Wa) jk = ( √i) − − − ≡ d − 0, j k 0 mod 3. − ≡ 14.27 Equivalent equations for SIC ﬁducial vectors 423

(b) Show that Wa has order 3. (c) Show that i d a d trace(Wa)= 1 + 2τ 3 + 2τ (1 τ 3 ) . −√3 − (d) Find the multiplicities of the eigenvalues 1,τ,τ2 of

(i) W1 for d 3 mod 9, d = 3. ≡ (ii) W2 for d 6 mod 9. ≡ 2 Remark: The ﬁrst implies that the multiplicities of the eigenvalues mλ of M1 = W1 are d+6 λ 1 2 3 , = 1; mλ = d + 3(λ + λ) = (14.109) d 3 λ τ τ2 3 −3 , = , . (e) The matrix W0 is well deﬁned for all d. Show that it is conjugate to the Zauner matrix via 1 d 1 1 1 τ − W0 =(R− F)− Z(R− F).

p p p p 2 14.15. Show that the displacement operators Dˆ p =( µ) 1 2 S 1 Ω 2 , p Z , satisfy − ∈ ˆ 1 ˆ ˆ ˆ µ p,q ˆ ω p,q ˆ ˆ D−p = D p, DpDq =( ) Dp+q = DqDp, − − and ˆ ˆ Dp, d odd; Dp+dq = p,q ( 1) Dˆ p, d even, − where , is the symplectic form p,q := p2q1 p1q2. − 14.16. Appleby indexing. We consider some details from 14.10. 2 § (a) Suppose that B SL2(Z ) and χ Z . Show that if a C(d) satisﬁes ∈ d′ ∈ d ∈ 1 χ,Bp 2 aDˆ pa− = ω Dˆ Bp, p Z , ∀ ∈ d′ then the unique index (ψa,za) is given by

T χ,Ap p σB p ψa = A := B mod d, za(p)= ω ( µ) . − (b) Calculate the kernel of the Appleby homomorphism (14.42). (c) Find the index (ψa,za) for the Clifford operations with Appleby index 0 d 1 0 1 (i) [F ,0], F := =(d + 1) . z z d + 1 d − 1 1 −1 − − 1 d + 3 d + 1 3 (ii) [Fa,0], Fa := 4d 3 =(d + 1) d 3 , d 3 mod 9, d = 3. − d 2 − d 2 ≡ 3 − 3 − √d + 1 d (iii) [Fb,0], Fb := − , √d + 1 Z, d 8. d d √d + 1 ∈ ≥ κ −κ d 2 2 2 (iv) [Fc,0], Fc := − , d =(3k 1) + 3, κ = 3k k + 1, k 0. d + 2κ d κ ± ± ≥ − 424 14 Weyl–Heisenberg SICs and determine the Clifford operation [a] if it is obvious.

ℓ 14.17. We show the eigenvectors of SΩ , ℓ Zd, give the MUBs of Theorem 12.22. Ω ℓ ∈ (a) Show that the matrices S ℓ Zd commute up to a scalar. Ω ℓ Ω{m } ∈ (b) Show that S and S are (Frobenius) orthogonal when ℓ m Zd∗. ℓ ℓ − ∈ ℓ (c) Show that the columns Bℓ := R Fe j of R F are eigenvectors of SΩ . (d) For any d, the d2 1 matrices{ S jΩ k}: 0 j = k < d can be partitioned into d + 1 subsets of d 1− matrices { ≤ } − 2 d 1 ℓ 2 2ℓ d 1 (d 1)ℓ Ω,Ω ,...,Ω − , M := SΩ ,S Ω ,...,S − Ω − , ℓ Zd. { } ℓ { } ∈

Show that matrices in the same set have the same eigenvectors, i.e., E = e j and { } Bℓ, respectively. 14.18. Here we consider the canonical abstract error group given by Clifford group. (a) Find the determinant of the generators S, Ω, F, R for C(d). Hint: First find the determinant of Z. (b) Explain how each of the generators M above can be replaced by a scalar multiple 1 Mˆ = c− M with determinant 1. (c) Use Magma to investigate the canonical abstract error groups which appear as subgroups of Sˆ,Ωˆ ,Fˆ,Rˆ for d = 2,3,4. 14.19. Let G = F,R be the group generated by F and R. (a) Show that Gis finite, i.e., it contains only finitely many unitary scalar matrices. 2 (b) Show that for d > 2, the nondiagonal matrix P 1 = F is in the centre of G. − Remark: By Schur’s lemma, this implies that F,R and CSp(d) are not irreducible. m n 14.20. Let A =[a jk] C × with E := Q(A)= Q( a jk ), and g Gal(E/Q). (a) Show that if A is∈ Hermitian, then g(A) is Hermitian{ } if and only∈ if g commutes with complex conjugation. (b) Show that if A = vv∗ is a rank one orthogonal projection, then g(A) is also if and only if g commutes with complex conjugation. (c) Now let E be the SIC field of a fiducial Π. Show that if g(Π) is a SIC fiducial for some g Gal(E/Q), then g must commute with complex conjugation on E. ∈ 14.21. The symmetry group S(v) of (14.65) for a generic SIC fiducial is abelian. Here we show that it is nonabelian for the SICs for d = 2,3 (listed in Table 14.4). 1 (a) Show that the order two antiunitary symplectic operation b = F− C does not commute with the Zauner matrix Z. (b) Show that the SICs for d = 2,3 have nonabelian symmetry groups. 1 (c) Show that if b = F− C is a symmetry of a SIC (with a symmetry Z or M1), then its symmetry group is nonabelian. Remark: The Hoggar lines are the only other known SIC with a nonabelian (in fact noncyclic) symmetry group. Because of this (and other reasons), [Sta17] refers to the Hoggar lines and d = 2,3 SICs as sporadic SICs, and all other SICs as generic. 14.27 Equivalent equations for SIC fiducial vectors 425

14.22. If v is a Weyl–Heisenberg SIC fiducial, then so is av, a EC(d), (see 14.7). Show that ∈ § [a] Π :=(av)(av)∗, Π = vv∗, [a] PEC(d) ∈ defines an action of the extended Clifford group on the SIC fiducial projectors.

14.23. Here we show the SIC field E = Q(Π, µ) of a Weyl–Heisenberg SIC fiducial projector Π = vv∗ depends only its extended Clifford orbit. (a) Show that E is closed under complex conjugation. (b) Let Π =(av)(av) , [a] PEC(d), be a another fiducial in the extended Clifford ′ ∗ ∈ orbit of Π. Show that Q(Π ′, µ)= Q(Π, µ), i.e., the SIC field E = Q(Π, µ) depends only on the orbit of Π, and not the particular choice of the fiducial projector. χΠ Π 14.24. We consider the extended Clifford action on the overlaps p of a SIC . (a) Show that if a C(d) has Appleby index [B,b], then ∈ χΠ ω b,Bp χ[a] Π 2 p = , p Z . Bp ∀ ∈ d′ (b) Show that the overlaps of Π = Π T satisfy

χΠ χΠ 2 1 p = Jp, p Zd , J = − . − ∀ ∈ ′ − 1 (c) Show that if [a] PEC(d) has (extended) Appleby index [B,b], then ∈ χΠ ω b,det(B)Bp χ[a] Π 2 p = , p Z . det(B)Bp ∀ ∈ d′ Π Π Π 14.25. Let be a centred ﬁducial for which S0( ) ∼= S( ) is abelian (all the known cases). Show that if B S0(Π) is a conjugate of Fz, then there is a unique maximal abelian subgroup of GL∈ (Z ) containing S (Π) and S(Π), which is given by 2 d′ 0

C(Π) := C(S(Π)) = C(S0(Π)) = C(B)= αI + βB : α,β Z GL2(Z ), { ∈ d′ }∩ d′ where C(X) denotes the centraliser of X in GL (Z ). 2 d′ 14.26. Let G = Gal(E/Q) be the Galois group of the SIC ﬁeld E of a ﬁducial Π, and Hg be given by (14.87). It can be shown (see []) that

d 1 √i − Q(µ). √d ∈

d d (a) Show that for every Clifford operation [a], we can choose a E × . d d ∈ (b) Show that if a C(d) E × and g Gc then g(a) C(d). (c) Show that ∈ ∩ ∈ ∈ d d g([a]) :=[g(a)], g Gc, a E × , ∈ ∈ deﬁnes an action of the group Gc on the Clifford operations. (d) Show that action of Gc on PC(d) in terms of Appleby indexes is given by 426 14 Weyl–Heisenberg SICs

1 g( f ([B,b])) = f ([HgBHg− ,Hgb]).

In particular, g Gc maps the symplectic operation with symplectic index B to the ∈ 1 symplectic operation with symplectic index HgBHg− .

14.27. Let Π be a SIC fiducial. We recall that g(Π), g Gc, is a fiducial (on the ∈ same multiplet), with g(Π) on the same extended Clifford orbit when g G0. (a) Show that the Clifford action commutes with the symmetries of a fiducial,∈ i.e.,

1 S[a] Π =[a] SΠ := aSΠ a− , [a] PEC(d). ∀ ∈ (b) Show that the Galois action commutes with symmetries of a ﬁducial, i.e.,

S Π = g(SΠ ), g Gc. g( ) ∀ ∈

(c) Let O1,...,Oℓ be the extended Clifford orbits of a SIC ﬁducial multiplet, and suppose that on every orbit O j there is a centred ﬁducial Π j. Fix Π j. Show that for every g Gc there is a Πk and [a] PEC(d) with Appleby index [B,q], such that ∈ ∈

Π Π 0 mod d, d is odd; g( j)=[a] k, 3q = d 0 mod 2 , d is even.

(d) Show that if d 0 mod 3, then we can chose q = 0 above, so that a is symplectic, ≡ and Gc maps centred fiducials to centred fiducials (on the same multiplet). (e) Show that if d 0 mod 3, then all centred fiducials are strongly centred. ≡ (f) Show that if all the Π j above are strongly centred, then we can take [a] to be symplectic, and in this case

g1(Π j)= P 1Π jP 1. − − 14.28. Here we consider the group generated by the Clifford operations and the Galois symmetries of the SIC ﬁeld E, and its action on SIC ﬁducials. For g G , and a C(d) Ed d, we write ag for the map Ed Ed given by ∈ ∈ ∩ × → (ag)v := a(g(v)), v Ed. ∀ ∈ d d (a) Show that X = ag : a C(d) E × ,g Gc is a group under composition, i.e., the multiplication { ∈ ∩ ∈ } (a1g1)(a2g2)= a1g1(a2)g1g2, 1 1 1 1 with inverse (ag)− = g− (a− )g− . Remark: The map ag is said to be a g–unitary, since a is unitary and

(ag)(αv + βw)= g(α)(ag)(v)+ g(β)(ag)(v), α,β E, v,w Ed. ∈ ∈

A 1–unitary map is unitary, and a gc–unitary map is antiunitary. (b) Show that PC(d) Gc = [a]g : [a] PC(d),g Gc is group with the induced multiplication × { ∈ ∈ } 14.27 Equivalent equations for SIC ﬁducial vectors 427

d d ([a1]g1)([a2]g2) :=[a1g1(a2)]g1g2, a1,a2 C(d) E × , ∈ ∩ which contains PC(d), Gc and PEC(d) as subgroups. We call [a]g a g–unitary. (c) Show there is a (natural) action of PC(d) Gc = on ﬁducials Π = vv given by × ∗ 1 d d [a]g Π :=(agv)(agv)∗ = ag(Π)a− =[a] g(Π), a C(d) E × . ∈ ∩ d d Show that if [a]g, a C(d) E × , stabilises Π, then v is an “eigenvector” of ag, i.e., ∈ ∩ (ag)v = λv, λ E, λ = 1. ∃ ∈ | | Remark: The stabiliser of Π = vv in the subgroup PEC(d)= PC(d) gc is the ∗ × group SΠ = S(v). The stabiliser is in general larger, since for a g G0, we can choose [a] PEC(d), with g(Π)=[a] Π, and so the g–unitary [a 1]g∈stabilises Π. ∈ − 428 14 Weyl–Heisenberg SICs

(d) For a g–unitary [a]g PC(d) Gc, one can deﬁne an extended index (ψa,za,g), or an extended Appleby∈ index [B,×b,g], [a]= f ([B,b]). Show that

2 fG : SL2(Z ) Z Gc PC(d) Gc : [B,b,g] f ([B,b])g, c d′ × d × → × → 2 is a homomorphism, where SL2(Z ) Z Gc is equipped with the multiplication d′ × d × B b g B b g B H B H 1 b B H b g g [ 1, 1, 1][ 2, 2, 1]=[ 1 g1 2 g−1 , 1 + 1 g1 2, 1 2].

2 (e) The subgroup SL2(Z ) Z 1,gc gives the extended Appleby triple indices d′ × d ×{ } for PEC(d). Let J = Hgc . Show that the map

2 2 j j Θ : SL2(Z ) Z 1,gc ESL2(Z ) Z : [B,b,g ] [BJ ,b] d′ × d × { } → d′ × d c → is an isomorphism between indices. 1 2 Remark: The map fG Θ : ESL2(Z ) Z PEC(d) is the fE of (14.57). c ◦ − d′ × d → Chapter 15 Tight frames of orthogonal polynomials on the simplex

The orthogonal polynomials of degree k on the triangle are a ﬁnite dimensional inner product space which is invariant under the unitary action of the symmetry group G of the triangle (the dihedral group of order 6) given by

1 g f = f g− , ◦ i.e., is a G–invariant space (see 10.10). It is natural to seek an orthogonal expansion for this space which is invariant§ under these symmetries. The polynomials in such an orthogonal expansion have a simple form, i.e., they consist of a small number of polynomials, together with those obtained by the changes of variables given by the action of G. In a few cases it is possible to find a G–invariant orthonormal basis, e.g., for the 3–dimensional space of quadratic Legendre polynomials (Example 10.20), but in general it is not. Here we present a natural G–invariant tight frame for this space (and more generally the Jacobi polynomials on the simplex). This is great illustration of the usefulness of tight frames – the redundancy of a tight frame allows us to find an expansion with the symmetries of the space, which is not possible for a basis. Key aspects of this construction include: The Jacobi polynomials in the tight frame are given explicitly in terms of the • Bernstein basis by using a multivariate generalisation of the 2F1 hypergeometric function (the Lauricella function of type A). The tight frame considerably improves upon the previously known expansions • (Appell’s biorthogonal system which has some of the symmetries, and Proriol’s orthogonal basis which has no symmetries and is given by a recursive formula). A polynomial is a Jacobi polynomial if and only if its Bernstein basis coefficients • give a Hahn polynomial. The Bernstein coefficients are characterised by certain linear dependencies which can be expressed in terms of the adjoint of the degree elevation operator (for the Bernstein form). The proof that the frame is tight uses the fact that the Jacobi polynomials are • eigenspaces of the Bernstein–Durrmeyer operator.

429 430 15 Tight frames of orthogonal polynomials on the simplex 15.1 Jacobi polynomials and their Bernstein coefﬁcients

Throughout, let ξ =(ξ0,ξ1,...,ξd) be the barycentric coordinates of a d–simplex d T R with vertices V = v0,v1,...,vd , and volume vold(T). We recall (see 4.7) that⊂ the barycentric coordinates{ are the unique} linear polynomials with §

d x = ∑ξ j(x)v j, ∑ξ j(x)= 1, x R . j j ∀ ∈

In particular, the barycentric coordinate ξ j is1at v j and zero at all the other vertices. It is sometimes convenient to index the barycentric coordinates by the vertices that they correspond to (rather than labels for them). We will use standard multiindex notation as outlined in 4.9, e.g., Γ (ν)= ∏ Γ (ν j), where Γ is the Gamma function. § j Example 15.1. The barycentric coordinates of the interval (1–simplex) [ 1,1] with vertices 1,1 are − {− } 1 x 1 + x ξ (x)= − , ξ (x)= , (15.1) 0 2 1 2 and for the triangle (2–simplex) with vertices 0,e1,e2 they are { }

ξ0(x,y)= 1 x y, ξ1(x,y)= x, ξ2(x,y)= y. − − The univariate Jacobi polynomials are the orthogonal polynomials given by the inner product

Γ ν ν 1 ν 1 ν 1 ( 0 + 1) 1 1 x 0− 1 + x 1− f ,g ν := f (x)g(x) − dx, ν0,ν1 > 0. Γ (ν0)Γ (ν1) 2 1 2 2 − The normalisation above is chosen so 1,1 ν = 1. The parameters are usually written as ν =(ν0,ν1)=(α +1,β +1), with the Jacobi polynomial of degree n denoted by (α,β) Pn . The condition ν j > 0 ensures that the weight function is integrable.

Example 15.2. Well known Jacobi polynomials include the Legendre polynomials ν ν 1 ( j = 1) and the Chebyshev polynomials ( j = 2 ), which are given by 1 1 1 1 dx f ,g (1,1) = f (x)g(x)dx, f ,g ( 1 , 1 ) = f (x)g(x) . 2 1 2 2 π 1 √1 x2 − − − In view of (15.1), the weight function on the interval [ 1,1] used to deﬁne the Jacobi polynomials is a product of powers of the barycentric− coordinates of [ 1,1]. By replacing [ 1,1] by a d–simplex T (a triangle for d = 2), we obtain the following− multivariate generalistion− of the univariate Jacobi inner product.

Deﬁnition 15.1. Let T be a d–simplex in Rd with barycentric coordinates ξ. Then (multivariate) Jacobi inner product on T with parameters ν is given by 15.1 Jacobi polynomials and their Bernstein coefﬁcients 431 Γ ν ( ) 1 ν 1 d+1 f ,g ν := | | f gξ − , ν R , ν j > 0. (15.2) Γ (ν) d!vol (T) T ∈ d ν 1 We call the Lebesgue measure on T weighted by ξ − the Jacobi measure.

A simple calculation shows that the condition ν j > 0, j, ensures the weight ν 1 ∀ function ξ − is integrable on T, so the Jacobi inner product is well deﬁned, and that (ν)α β ξ α ξ β + α β d+1 , ν = ν , , Z+ , (15.3) ( ) α + β ∈ | | | | | | ν ∏ ν where ( )α := j( j)α j , with the Pochhammer symbol (x)n given by

(x)n := x(x + 1) (x + n 1). − The orthogonal polynomials of degree k (see 10.10) for the Jacobi measure are § Pν called the (multivariate) Jacobi polynomials, and we denote them by k , i.e., Pν Π d Π d k := f k(R ) : f ,h ν = 0, h k 1(R ) , { ∈ ∀ ∈ − } d d where Πn(R ) is the space of polynomials of degree n on R . Since ≤ ν k + d 1 dim(P )= − , k d 1 − Pν the space k has dimension greater than 1 when k > 0 and d > 1. d Each polynomial f Πn(R ) can be expressed in terms of the Bernstein basis ∈ f = ∑ cα ( f )Bα = ∑ cα Bα , α =n α =n | | | | where the Bernstein polynomials of degree n are defined by α α α ! α n! α d+1 Bα := | | ξ = | | ξ = ξ , α = n, α Z . α α! α! | | ∈ + d This basis for Πn(R ) is well suited to representing polynomials on the simplex T (see 4.9 for a generalisation to a frame for other polytopes). The coefficients § n c( f )= c ( f )= c =(cα ) α =n | | are referred to as the Bernstein(–Bezier)´ coefficients. By the multinomial theorem

d j j f = ∑ cα Bα = ∑ cα Bα ∑ ξi = ∑ (R c)α Bα , α =n α =n i=0 α =n+ j | | | | | | where the powers of the degree raising operator R are given by 432 15 Tight frames of orthogonal polynomials on the simplex α j j ( )γ (R c)α = ∑ − cα γ , j = 0,1,2,.... (15.4) γ α − γ = j ( ) j | | −| | Here we think of the Bernstein coefﬁcients as a function c : α cα deﬁned on the simplex points →

d+1 d ∆n := α Z : α = n , #∆n = dim(Πn(R )). { ∈ + | | }

We view such functions c : ∆n R as polynomials of degree n in d–variables by identifying c with the unique polynomial→ of degree n on the d–dimensional afﬁne d+1 subspace x R : x0 + x1 + + xd = n which takes the value cα at α ∆n. { ∈ } ∈ For example, by the multinomial theorem f = 1 = ∑ α =n Bα , and so 1 corresponds to the constant polynomial c : α 1. More generally,| we| have: → Π d Proposition 15.1. Suppose that f = ∑ α =n cα Bα n(R ) and 0 s n. Then f | | ∈ ≤ ≤ has degree s if and only if c : α cα is a polynomial of degree s. → Π d Proof. The polynomials (Bβ ) β =s are a basis for s(R ), and can be expressed | | 1, α = β; Bβ = ∑ bα Bα , bα := α =s 0, otherwise. | |

Let j := n s. Then by (15.4), the Bernstein coefﬁcients of Bβ = ∑ α =n cα Bα are − | | α α j α j j ( )γ j! ( )α β j!( 1) ( )β cα =(R b)α = ∑ − bα γ = − − = − − . γ α − α β β β γ = j ( ) j ( )! ( n) j ( n) j ( ) | | −| | − − − −

Since ∆n R : α ( α)β , β = s, is a polynomial of degree s, we obtain the correspondence.→ → − | | ⊓⊔ We define an inner product on the space of polynomials ∆n R of degree n by → ν ( )α α α f ,g ν,n := ∑ α f ( )g( ). (15.5) α =n ! | | The orthogonal polynomials of degree s corresponding to the discrete measure ν,n above are called the Hahn polynomials, and we denote them by Ps , 0 s n. ≤ ≤ The adjoint R∗ν of the degree raising operator R with respect to (15.5) is defined by Rc,b ν,n = c,R∗ν b ν,n 1, c : ∆n 1 R, b : ∆n R. (15.6) − − → → A simple calculation (Exer. 15.1) shows the powers of R∗ν are given by β ν j ( + )γ j ∆ ((Rν∗ ) b)β = ∑ β γ bβ+γ , b : n R, 0 j n. (15.7) γ = j ( + 1) j → ≤ ≤ | | | | 15.1 Jacobi polynomials and their Bernstein coefficients 433

We can now show that Jacobi polynomials are characterised by the fact their Bernstein coefﬁcients are a Hahn polynomial (and vice versa). This is a relatively new result (see [Cie87] for the univariate case, and [Wal06]). ν Π d Theorem 15.1. Fix > 0. Let f = ∑ α =n cα Bα n(R ), c =(cα ), and 0 s n. Then the following are equivalent | | ∈ ≤ ≤ Pν 1. f s (Jacobi polynomials). ∈ ν,n 2. c Ps (Hahn polynomials). ∈ n s+1 3. (R∗ν ) − c = 0. Pν Proof. (1 2) We have f s if and only if it is orthogonal to the spanning set ξ β ⇐⇒Π d ∈ β

~~With pβ : ∆n R : α (ν + α)β , this orthogonality condition can be written as → →~~

~~c, pβ ν n = 0, β < s. , | | ν,n Since pβ β~~

ν α (ν)β γ ( ) α + β γ c,qβ ν,n = ∑ α cα ( )β = ∑ β γ cβ+γ ( )β α =n ! − γ =k ( + )! − − |α | β | | ≥ β ( β γ)β ν ν β ν ν β ( 1)| | =( )β ∑ ( + )γ cβ+γ −β −γ =( )β ∑ ( + )γ cβ+γ −γ γ =k ( + )! γ =k ! | | | | s 1 ν β ν ( 1) − ( + )γ k! =( )β − (s)k ∑ β cβ+γ γ k! γ =k ( + 1)k ! | | | | s 1 n k =(ν)β ( 1) − ((R∗ν ) c)β . − s 1 − ν,n Since qβ β =s 1 is a basis for the polynomials of degree < s, we have c Ps { }| | − k ∈ if and only if c,qβ ν,n = 0, β = s 1, which is equivalent to (Rν∗ ) c = 0, by the calculation above. | | − ⊓⊔ This association between Jacobi and Hahn polynomials preserves inner products:

Theorem 15.2. ([Wal06]) Suppose that f = ∑ α =n cα ( f )Bα and g = ∑ α =n cα (g)Bα . If f or g belongs to Pν , 0 s n, then | | | | s ≤ ≤ (n!)2 (ν)α (n!)2 f ,g ν = ν ∑ α cα ( f )cα (g)= ν c( f ),c(g) ν,n. (n s)!( )n+s α =n ! (n s)!( )n+s − | | | | − | | 434 15 Tight frames of orthogonal polynomials on the simplex 15.2 The Bernstein–Durrmeyer operator

To motivate the tight frame of Jacobi polynomials on a simplex in 15.3, we consider ν § the Bernstein–Durrmeyer operator Mn (see [Der85], [BX91]). This is deﬁned on the continuous functions on the simplex T with the Jacobi inner product (15.2) by ξ α ν ν f , ν ξ α ( )n n! ξ α Mn f := ∑ α Bα = ∑ f , ν | | . ξ ν ν α α α =n 1, α =n ( ) ! | | | | This self adjoint operator is a natural generalisation of the Bernstein operator (see 4.10). It is positive, i.e., § ν f 0 = M f 0, ≥ ⇒ n ≥ but does not reproduce the nonconstant linear polynomials. It is degree reducing, and satisﬁes an analogue of Theorem 4.6. It can be viewed as a de la Vallee–Poussin´ Pν mean of the orthogonal projections onto s 0 s n. From this, or by a simple ≤ ≤ ν calculation (see [Der85]), it follows that the eigenvalues of Mn are

λ ν n! 1 s(Mn )= , 0 s n, (n s)! (n + ν )s ≤ ≤ − | | with corresponding eigenspace the Jacobi polynomials Pν , i.e., for 0 s n s ≤ ≤ ξ α ν 1 ξ α Pν f =(n s)!( )n+s ∑ f , ν , f s . (15.8) α ν α − | | α =n ! ( ) ∀ ∈ | | ν ν Let Qs be the orthogonal projection onto P . Then for f P , s ∈ s α α α f ,ξ ν = Qs f ,ξ ν = f ,Qs(ξ ) ν , and so from (15.8), we obtain α 1 α ξ f =(n s)!( ν )n+s ∑ f ,Qs(ξ ) ν α ν α − | | α =n ! ( ) | | ν ξ α ξ α ν ( )α Qs( ) Qs( ) Pν =(n s)!( )n+s ∑ f , ν , f s . (15.9) α ν α ν α − | | α =n ! ( ) ( ) ∀ ∈ | | Pν In other words, (15.9) gives a tight frame expansion for s . In the next section, we α obtain an explicit formula for Qs(ξ ) when n = s (the case with the fewest vectors). This result was found independently by [Ros99] (not presented in terms of tight frames) and by [XW01], [PW02]. The presentation in terms of the Bernstein– ν Durrmeyer operator Mn given here is adapted from [Wal06]. 15.3 Tight frames of Jacobi polynomials with symmetries 435 15.3 Tight frames of Jacobi polynomials with symmetries

We now give the main result. Pν Theorem 15.3. A tight frame expansion for the Jacobi polynomials n is given by ν ν ( )α φ ν φ ν Pν f =( )2n ∑ α f , α ν α , f n , (15.10) | | α =n ! ∀ ∈ | | where n β ν ( 1) (n + ν 1) β ( α)β ξ φ | |− | | − α := −ν ∑ ν β (15.11) (n + 1)n β α ( )β ! | |− ≤ ξ α ν Pν is the orthogonal projection of /( )α onto s . Pν Proof. In view of the tight frame expansion (15.9) for n (take s = n), it sufﬁces φ ν ξ α ν Pν to prove that α is the orthogonal projection of /( )α onto n . Since ξ α ν d φα + Πn 1(R ), ∈ (ν)α −

φ ν ξ γ Π d this reduces to showing that α is orthogonal to the basis ( ) γ =n 1 for n 1(R ). Suppose that γ = n 1. Then | | − − | | − ν ν ν ν ν ν γ ( ) β + γ =( )n 1( + n 1) β , ( )β+γ =( )γ ( + )β , | | | | | | | | − | | − | | and so, by (15.3), we have

n ν α ν ν γ ( 1) (n + 1) β ( )β ( )β+γ φ ξ ν | |− | | − α , = −ν ∑ ν β ν (n + 1)n β α ( )β ! ( ) β + γ | |− ≤ | | | | | | n ν γ ( α)β ( 1) ( ) ν γ = −ν ν ∑ ν− β ( + )β . (n + 1)n ( )n 1 β α ( )β ! | |− | | − ≤ By the Chu–Vandermonde identity, the last sum above simpliﬁes to

~~( α)β (ν + γ)β ( α)β (ν + γ)β ν ν γ α γ α ν γ ( ( + )) ( ) ∑ − ν β ( + )β = ∑ − ν β = − ν = −ν . β α ( )β ! β ( )β ! ( )α ( )α ≤~~

Since γ < α = n, we must have γ j < α j for some j, i.e., ( γ j)α = 0, and so | | | | − J ( γ)α = 0, which completes the proof. − ⊓⊔ φ ν Pν The Jacobi polynomials α in the tight frame for n can be written

n ν ( 1) φα = − FA(n + ν 1, α,ν;ξ), (n + ν 1)n | |− − | |− where FA is the Lauricella function of type A, which is given by 436 15 Tight frames of orthogonal polynomials on the simplex α (b)α x d+1 FA(a,b;c;x) := ∑ (a) α , a R, b,c,x R . | | (c)α α! ∈ ∈ α Zd+1 ∈ +

This is a multivariate generalisation of the hypergeometric function 2F1(a,b;c;x). We observe FA(a, β,c;x) is a polynomial of degree β in x when β is a multiindex. − | | Example 15.3. (Univariate Jacobi polynomials). The univariate Jacobi polynomials (α,β) Pn for the weight ν =(α + 1,β + 1) on [ 1,1], and barycentric coordinates ξ given by (15.1), can be written − α α,β ( + 1)n 1 P (x)= 2F1 n,1 + α + β + n;α + 1; (1 x) . (15.12) n n! − 2 − The Jacobi polynomial for the multiindex (n,0) given by (15.11) is

n n ν ξ j φ ν ( 1) (n + 1) j( n) j 0(x) (n,0)(x)= − ∑ | |− − (n + ν 1)n (ν0) j j! | |− j=0 n n α β 1 j ( 1) ( n) j(n + + + 1) j ( 2 (1 x)) = α− β ∑ − α − (n + + + 1)n j=0 ( + 1) j j! ( 1)n 1 = − 2F1( n,1 + α + β + n;α + 1; (1 x)) (n + α + β + 1)n − 2 − n ( 1) n! (α,β) = − Pn (x) (n + α + β + 1)n (α + 1)n

Pν φ ν α,β Since n is one–dimensional, each (n k,k), 0 k n, is a scalar multiple of Pn . − ≤ ≤ Pν The tight frame expansion for the Jacobi polynomials s given by (15.9) for α 0 s < n has more vectors than that of (15.10). By calculating Qs(ξ ) explicitly (see≤ [Wal06]), one can generalise Theorem 15.3, as follows.

Theorem 15.4. A tight frame for the Jacobi polynomials Pν ,s n, is given by s ≤ ν ν ( )α φ ν,s φ ν,s Pν f =(n s)!( )n+s ∑ α f , α ν α , f s , (15.13) − | | α =n ! ∀ ∈ | | where s n ν α ξ β ν,s ( 1) s (s + 1) β ( )β ( s) β φα := − ∑ | |− | | − − | | (15.14) s ν 1 ν 2s ν β n β β! ( + )s ( + )n s β α ( ) ( ) | |− | | − β≤ s − | | | |≤ ξ α ν Pν is the orthogonal projection of /( )α onto s . We also have

(n s)! ν φ ν,s Pν f = − ( )n+s ∑ f , α ν Bα , f s . (15.15) n! | | α =n ∀ ∈ | | 15.4 The orthogonal polynomials of Appell and Proriol 437 15.4 The orthogonal polynomials of Appell and Proriol

Pν We now consider the symmetries of tight frames for n given by Theorem 15.4, and compare them with the previously known expansions. The afﬁne maps g Aff(Rd) of the simplex T onto itself are uniquely determined ∈ by their action on the vertices v0,v1,...,vd or the barycentric coordinates, i.e., { }

~~gv j = vσ j, g ξ j = ξσ j, where σ Sd+1 is a permutation (of the vertices). Under this identiﬁcation, the symmetry∈ group of the Jacobi measure is~~

G := σ Sd 1 : νσ j = ν j, j , { ∈ + ∀ } Pν and the Jacobi polynomials n are a G–invariant space, where the unitary action of G is given by 1 d σ f := f g− , f Π(R ). ◦ ∈ With σα :=(σα0,...,σαd), applying σ ξ j = ξσ j to (15.14) gives ν,s σν,s ν,s σ φα = φσα = φσα , σ G, ∀ ∈ i.e., the tight frames of Theorem given by Theorem 15.4 are G–invariant, i.e., the G–orbit of some smaller number of vectors.

Example 15.4. (Legendre polynomials) Suppose that all the ν j are equal e.g., the Legendre polynomials given by ν j = 1 (giving Lebesgue measure on T). Then φ ν the symmetry group is G = Sd+1, and the tight frame ( α ) α =n of Theorem 15.3 is the orbit of p(n) polynomials, where p(n) is the partition| function| (the number of partitions of n). For example, 3,2 + 1,1 + 1 + 1 are the three partitions of n = 3, Pν and so the tight frame for the cubic Jacobi polynomials 3 is the orbit of three polynomials for any dimension d. Example 15.5. (Quadratic Jacobi polynomials) From (15.11), we have

ξ 2 ξ φ ν 0 2 0 1 (2,0,...,0) = + , ν0(ν0 + 1) − (2 + ν ) ν0 (1 + ν )(2 + ν ) ξ ξ |ξ| ξ | | | | φ ν 0 1 1 0 1 1 (1,1,,0,...,0) = + + . ν0ν1 − (2 + ν ) ν0 ν1 (1 + ν )(2 + ν ) | | | | | | Φ φ ν The remaining quadratic Jacobi polynomials in the tight frame =( α ) α =2 for Pν | | 2 are obtained from these by making the substitution

ν0 ν j, ν1 νk, k = j. → → ν ν ν ν Pν When 0 = j, 1 = k, then this corresponds to a symmetry of 2 . In any case, formulas for all the polynomials in Φ can be obtained from those two above, by substitutions, independent of the dimension d. 438 15 Tight frames of orthogonal polynomials on the simplex

The Appell polynomials (introduced in [AKdF26]) are the nonorthogonal basis Pν Φ φ ν for n given by the subset of the tight frame =( α ) α =n for which the k–th component of α is zero, i.e., | |

ν Φk := φα : α = n,αk = 0 . { | | } There is an explicit formula for the polynomials in the dual basis (see [AKdF26], [FL74], [KMT91]). The Appell basis and its dual basis are invariant under the action of subgroup H of G (the symmetry group of the measure) given by

H := σ G : νσk = νk . { ∈ } The Proriol polynomials (introduced in [Pro57]) are an orthonormal basis for Pν n given explicitly, but by complicated formulas (see [DX01]). These polynomials are not invariant under any of the (nonidentity) symmetries of the Jacobi weight. Φ φ ν Pν The basis =( α ) α =n for n has the following desirable properties: | | It is a tight frame. • Its polynomials are given explicitly in the Bernstein form by using a multivariate • version of the 1F2 hypergeometric function (the Lauricella function FA). These formulas do not become more complicated as d becomes large. It is invariant under all of the symmetries of the Jacobi weight. • As discussed, the bases of Appell and Proriol do not share all of these properties. Pν Example 15.6. Consider the three–dimensional space 2 of quadratic Legendre polynomials on a triangle. The symmetry group of the weight is S3 (order 6), i.e., all permutations of the vertices give symmetries. The tight frame Φ is given by the orthogonal projections of ξ 2 ξ 2 ξ 2 ξ ξ ξ ξ ξ ξ 0 , 1 , 2 , 1 2, 0 2, 0 1 Pν onto 2 . This is the S3–orbit of two polynomials. The Appell basis is the orthogonal projection of ξ 2 ξ 2 ξ ξ 0 , 1 , 0 1, which is invariant under subgroup of order 2 generated by the permutation (0 1). The Proriol basis is given by the orthogonal projection of

~~2 2 2 ξ , ξ0(ξ0 + 2ξ1), 4ξ + 2ξ0ξ1 ξ . 0 1 − 0~~

~~Notes~~

The idea of using ﬁnite tight frame expansions for spaces of multivariate orthogonal polynomials (not to be confused with multiple orthogonal polynomials [MFVA16]) appeared independently in [Ros99], and [XW01], [PW02]. A detailed account of 15.4 The orthogonal polynomials of Appell and Proriol 439 the multivariate orthogonal polynomials, which includes the systems of Appell and Prorial, is given in [DX01]. The presentation in terms of the Bernstein–Durrmeyer ν operator Mn that is given here is adapted from [RW04], [Wal06]. There are similar expansions for the multivariate Hahn and continuous Hahn polynomials [RW04], and for the multivariate orthogonal polynomials for a radially symmetric weight (see Chapter 16). Tight frames allow for optimal expansions for spaces of multivariate orthogonal polynomials for speciﬁc weights, e.g., see [Dun87] and 10.10, 10.14. § §

~~Exercises~~

15.1. Let R be the degree raising operator given by (15.4), and Rν∗ be its adjoint as given by (15.6). Show that the j–th power of Rν∗ is given by (15.7), i.e., β ν j ( + )γ j ∆ ((R∗ν ) b)β = ∑ β γ bβ+γ , b : n R, 0 j n. γ = j ( + 1) j → ≤ ≤ | | | |

~~Chapter 16 Continuous tight frames for ﬁnite dimensional spaces~~

~~The tight frame expansion for n equally spaced unit vectors in R2 is~~

n 2π j 2 2 cos n f = ∑ f ,u j u j, f R , u j := π . (16.1) n ∀ ∈ sin 2 j j=1 n We may take the limit of this, as n ∞, to obtain → 2π θ 2 θ 2 cos f = f ,uθ uθ d , f R , uθ := θ , (16.2) 2π 0 ∀ ∈ sin which is the prototypical example of a continuous tight frame expansion. A key feature of this expansion is that the vectors (uθ )0 θ 2π are invariant under the symmetries O(2) of R2 (see the comments at the start≤ of≤ 9). Indeed, one can argue that this is the natural representation for the space H =§ R2, which follows directly from its symmetries (apply O(2) to any unit vector), and that expansions like (16.1) then follow by a process of discretisation (called sampling).

~~16.1 Continuous and discrete frames~~

The sum ∑ j J in the deﬁnition of a frame can be a more general integral. ∈ Deﬁnition 16.1. Let H be a Hilbert space, and (J,S , µ) be a measure space. A (generalised) frame for H with respect to µ is a family ( f j) j J for which ∈ 1. For each f H , J F : j f , f j is S –measurable on J. 2. There exist∈ (frame bounds)→ A→,B > 0 such that

~~2 2 2 A f f , f j dµ( j) B f , f H . ≤ J | | ≤ ∀ ∈~~

~~441 442 16 Continuous tight frames for ﬁnite dimensional spaces~~

It is tight if one can choose A = B. Further, we will refer to the frame as being discrete if µ( j ) > 0, j J, and continuous if µ( j )= 0, j J. { } ∀ ∈ { } ∀ ∈ Example 16.1. If µ is the counting measure on a set J, then a generalised frame is precisely a frame.

~~Example 16.2. The vectors (uθ )0 θ 2π of (16.2) form a continuous tight frame for R2 (with respect to the Lebesgue≤ measure),≤ since~~

2π 2π 2 2 2 2 x,uθ dθ = (x1 cosθ + x2 sinθ) dθ,= π x , x R . 0 | | 0 ∀ ∈ The basic results on frames extend in the obvious fashion, i.e., replace the sum over J by an integral. In particular (see Exercises 16.1 and 16.2), we have:

Proposition 16.1. Let Φ =( f j) j J be a generalised frame with respect to µ for H . Then ∈ Sf := f , f j f j dµ( j) J deﬁnes a bounded invertible self adjoint operator S = SΦ µ : H H , for which , → 1 1 f = f ,S− f j f j dµ( j)= f , f j S− f j dµ( j) J J 1 1 = S− 2 f , f j S− 2 f j dµ( j), f H . J ∀ ∈ 1 1 We call ( f˜j)=(S− f j) the dual frame, and (S− 2 f j) the canonical tight frame.

~~Proposition 16.2. (Variational characterisation) Let ( f j) j J be a generalised frame with respect to µ for a d–dimensional space H . Then ∈~~

2 2 1 2 f j, fk dµ( j)dµ(k) f j dµ( j) , J J | | ≥ d J with equality if and only if ( f j) is tight. Π d Example 16.3. Let n◦(R ) be the space of homogeneous polynomials of degree n on Rd, i.e., those polynomials f satisfying x f (x)= x n f , x = 0. x In view of this, these polynomials are determined by their values on the unit sphere S := x Rd : x = 1 , and so we can deﬁne an inner product on them by { ∈ } Φ n f ,g = S f g (Lebesgue integration on S). The ridge polynomials =( ,v )v S give a continuous frame for Π (Rd) (it is well known that they span). A calculation ∈ n◦ shows that Φ is tight for n = 0,1 (but not for n 2). In 16.5, we will consider continuous tight frames for the this space (and the≥ space of§ complex homogeneous polynomials). 16.2 The analysis and synthesis operators 443 16.2 The analysis and synthesis operators

~~Let Φ =( fJ) j J be a generalised tight frame for H with respect to µ. Then the synthesis operator∈ can be deﬁned in the natural way (see Exer. 16.1)~~

~~V = VΦ : L2(µ) H : a a j f j dµ( j). → → J From this, one can deﬁne the Gramian and the canonical Gramian~~

1 Gram(Φ)= V ∗V : L2(µ) L2(µ), PΦ = V ∗S− V : L2(µ) L2(µ), → → where S = VV ∗ : H H is the frame operator. The canonical Gramian PΦ is an → 1 orthogonal projection. It can be represented by the “matrix” [PΦ ]=[ fk,S− f j ] j,k J (see Exer. 16.3), where ∈

1 PΦ a =[PΦ ] µ a := [PΦ ] jkak dµ(k)= fk,S− f j ak dµ(k). J J d d Example 16.4. (Short-time Fourier transform) Let H = L2(R ), g L2(R ) be 2πiω t ∈ nonzero, and Mω Txg(t) := e g(t x). Then − 2 2 2 d f ,Mω Txg dxdω = g f , f L2(R ), d d | | ∀ ∈ R R so that (Mω Txg)(x,ω) Rd Rd is a tight continuous frame with respect to Lebesgue d d ∈ × d d d measure on R R . The analysis operator V : L2(R ) L2(R R ) given by × ∗ → × 2πiω t d (V ∗ f )(x,ω)= f ,Mω Txg f (t)g(t x)e− dt, x,ω R d − ∈ R is called the short-time Fourier transform (STFT) of f with respect to g (the window function). See [Gro01]¨ for details.

In principle, the deﬁnition of a continuous tight frame could be extended, so as to include many important integral transforms which behave in a similar way. For example, the Fourier transform fˆ of f L2(R), ∈ ∞ ∞ 2πiω 2πiω fˆ(ω) := f (t)e− dt = f ,e , f ,g := f (t)g(t)dt ∞ ∞ − − gives the reconstruction formula

∞ ∞ 1 2πiωt 1 2πiω 2πiωt f (t)= fˆ(ω)e dω = f ,e e dω, f L2(R), 2π ∞ 2π ∞ ∀ ∈ − − 2πiω which would be a continuous tight frame expansion if e L2(R). ∈ 444 16 Continuous tight frames for ﬁnite dimensional spaces 16.3 Reproducing kernels

~~Many interesting continuous tight frames come from reproducing kernels. Some, such as the family of zonal harmonics predate the theory of frames.~~

Definition 16.2. A Hilbert space H , of functions defined on some set X, is called a reproducing kernel Hilbert space if each of the point evaluations x f (x) , x X → ∈ is continuous, and hence has a Riesz representer Kx H . For such a space, the function K : X X F defined by ∈ × →

~~K(x,y) := Ky,Kx = Ky(x), x,y X ∀ ∈ is called the reproducing kernel.~~

~~The reproducing kernel is Hermitian, i.e.,~~

~~Kx(y)= Kx,Ky = Ky,Kx = Ky(x). Example 16.5. Let H be a reproducing kernel Hilbert space of functions X F, with the inner product given by a measure µ, i.e., →~~

~~f ,g := f (y)g(y)dµ, (16.3) X If K is the reproducing kernel, then~~

~~f (x)= f ,Kx = f (y)Kx(y)dµ(y)= f ,Ky Ky(x)dµ(y), X X i.e.,~~

~~f = f ,Ky Ky dµ(y), f H , (16.4) X ∀ ∈ and so (Ky)y X is a normalised generalised tight frame with respect to the measure µ, which we∈ will refer to as the reproducing kernel tight frame.~~

~~Clearly, every ﬁnite dimensional space with an inner product of the form (16.3) has a reproducing kernel, and hence a natural generalised tight frame.~~

Example 16.6. Let H = Fd, a space of functions on J = 1,2,...,n , with the inner product given by the counting measure. The Riesz represente{ r of the} point evaluation j x j = x,e j is the standard basis vector e j, and so the reproducing kernel is →

K( j,k)= ek( j)= δ jk, Kj = e j, and the corresponding generalised tight frame is the standard orthonormal basis n H n (e j) j=1. More generally, if is a subspace of F , then the reproducing kernel tight n H frame is (Pe j) j=1, where P is the orthogonal projection onto (see Exer. 16.5). A formula for the reproducing kernel can be computed from any frame. 16.4 Zonal harmonics 445

~~Proposition 16.3. Suppose that K is the reproducing kernel for H , and ( f j) is a ﬁnite frame for H , with alternate dual frame (g j), e.g., g j = f˜j, then~~

~~K(x,y)= Ky,Kx = ∑g j(x) f j(y). j~~

~~Proof. Expand Kx H in terms of this frame ∈~~

~~Kx = ∑ Kx, f j g j = ∑ f j,Kx g j = ∑ f j(x)g j. j j j~~

~~Using this, and the frame expansion, we obtain~~

~~Ky,Kx = ∑ f j(y)g j,∑ fk(x)gk = ∑ f j(y)∑ g j,gk fk(x)= ∑ f j(y)g j(x), j k j k j as claimed. ⊓⊔~~

~~16.4 Zonal harmonics~~

~~The extension of (16.2) to Rd is most easily obtained by using zonal harmonics. We denote the (unit) sphere in Rd by~~

d 1 d S = S − := x R : x = 1 . { ∈ } A function f : X R, X Rd is harmonic if it satisﬁes Laplace’s equation, i.e., → ⊂ ∆ f = 0, ∆ := D2 + D2. 1 d d Let Hk = Hk(R ) be the space of homogeneous harmonic polynomials of degree k. The map f f S, of restriction of a function to the sphere, applied to Hk has trivial kernel, so→ that|

dim(Hk)= dim(Hk(S)), Hk(S) := f : f Hk . { |S ∈ } d The spaces Hk(R ) and Hk(S) are the (solid and surface) spherical harmonics of degree k. They are invariant under the action of O(d) and SO(d) (which are absolutely irreducible), and have dimension

k + d 1 k + d 3 dim(Hk)= − − . (16.5) d 1 − d 1 − − Spherical harmonics of different degrees are orthogonal to each other with respect to the inner product 446 16 Continuous tight frames for ﬁnite dimensional spaces 1 f ,g := f gdσ = f (ξ)g(ξ)dξ, (16.6) S area(S) S S where σ is normalised surface-area measure on the sphere S, dξ denotes Lebesgue measure on the sphere, and the area of the sphere is

~~d 2π 2 area(S) := 1dξ = . (16.7) Γ d S ( 2 ) The spherical harmonics give the orthogonal decomposition~~

∞

L2(S)= Hk, (16.8) k=0 of L2(S) into absolutely irreducible SO(d)–invariant subspaces. The spherical harmonics of a degree k with the inner product of (16.6) form a reproducing kernel Hilbert space, and so have a natural continuous tight frame. Deﬁnition 16.3. The zonal harmonic1 of degree k with pole ξ S is the Riesz (k) ∈ representer of point evaluation at ξ, i.e., the unique Z = Z Hk with ξ ∈

~~(k) (k) f (ξ)= f ,Z = f Z dσ(ξ), f Hk. (16.9) ξ S ξ ∀ ∈ S Thus (16.4) gives the following reproducing kernel tight frame:~~

~~(k) Example 16.7. The zonal harmonics (Zξ )ξ S are a continuous tight frame for the ∈ space Hk of spherical harmonics of degree k, i.e.,~~

~~(k) (k) f = f ,Z Z dσ(ξ), f Hk. (16.10) ξ S ξ ∀ ∈ S~~

We recall some basic facts about zonal harmonics (cf [SW71] and[ABR01]). Definition 16.4. A function f defined on a O(d)–invariant subspace of Rd (such as S or Rd) is zonal with pole ξ S if it can be written in the form ∈ f (x)= g( x,ξ , x ). This definition is equivalent to: f is invariant under the action of the subgroup of O(d) which fixes ξ. • f constant on parallels of the sphere, i.e., f is constant on H S, where H is any • hyperplane in Rd which is orthogonal to the vector ξ, and S is∩ a sphere (on which f is defined).

~~1 (k) The zonal harmonic Zξ is is also commonly deﬁned for unnormalised surface-area measure, which adds a scaling factor to the formulas for it presented here. 16.4 Zonal harmonics 447~~

~~Zonal functions generalise ridge functions and radial functions, which have the form~~

~~f (x)= g( x,ξ ) (ridge function), f (x)= g( x ) (radial function). (k) The zonal harmonic Z Hk is a zonal function, as can be seen from the following ξ ∈ explicit formula (see Exer. 16.8)~~

d ξ d ξ (k) k ( 2 ) x, k ( 2 ) x, Zξ (x)= x Ck x Ck 2 , (16.11) x − − x (λ) λ where Ck are the ultraspherical (Gegenbauer) polynomials, with Ck := 0, k < 0. If fη is zonal with pole η, then we can move the pole of fη to be ξ by applying any g O(d) with ξ = gη. We use the notation fξ to denote the corresponding (well ∈ 1 deﬁned) zonal function fξ := gfη := fη g , g O(d), ξ = gη. In particular, ◦ − ∈ (k) (k) (k) 1 Z = gZ := Z g− , g O(d). (16.12) gξ ξ ξ ◦ ∀ ∈ (k) ξ The zonal function Zξ is peaked at (see Exer. 16.13), i.e.,

(k) 2 (k) (k) (k) Z = Z ,Z = Z (ξ)= dim(Hk), (16.13) ξ S ξ ξ S ξ (k) Z (η) dim(Hk), η S. (16.14) | ξ |≤ ∀ ∈ Example 16.8. (Homogeneous linear polynomials). For k = 1, Hk is the space of homogeneous linear polynomials in d real variables, and (16.11) gives

~~Z(1)(x)= d x,ξ , ξ and (16.10) becomes~~

(1) (1) d f = f ,Z Z dσ = f (ξ) ,ξ dξ, f H1. ξ S ξ area(S) ∀ ∈ S S Let f be homogeneous linear polynomial ,x , to obtain d d ,x = x,ξ ,ξ dξ = , x,ξ ξ dξ , area(S) area(S) S S and we deduce the following generalisation of (16.2) to Rd. ξ Proposition 16.4. The vectors ( )ξ S of the unit sphere are a continuous tight frame ∈ for Rd, i.e., d x = x,ξ ξ dξ, x Rd. (16.15) area(S) ∀ ∈ S By linearity, the expansion (16.15) also holds for x Cd, and an analogue exists ∈ for S replaced by the complex unit sphere SC (see Exer. 16.12). 448 16 Continuous tight frames for ﬁnite dimensional spaces 16.5 Homogeneous polynomials

Π d d Let n◦(R ) be the space of homogeneous polynomials on R of degree n, with the inner product f ,g S deﬁned by (16.6). We will ﬁnd the reproducing kernel tight Π d frame for this space. Every homogeneous polynomial p n◦(R ) can be written uniquely ∈ n [ 2 ] 2 j 2 j p(x)= ∑ x pn 2 j(x)= ∑ x pn 2 j(x), (16.16) − − 0 j n j=0 ≤ ≤ 2 d where pn 2 j Hn 2 j(R ). This and the density of the polynomials in L2(S) gives the orthogonal− ∈ decomposition− of (16.8). Since the restriction map

~~d Π ◦(R ) L2(S) : f f n → → |S is injective (see Example 16.3), from (16.16) we obtain the natural identiﬁcation~~

Π d Π H n◦(R ) n◦(S)= n 2 j(S). (16.17) ≈ − 0 j n ≤≤ 2 We say that a space is rotationally invariant if it is SO(d)–invariant. The desired Π d tight frame for the rotationally invariant subspace k◦(R ) of L2(S) is special case of the following general result.

~~Theorem 16.1. Let H be a rotationally invariant subspace of L2(S). Then~~

~~H = H j, (16.18) j J ∈ for some subset J of N, and we have the generalised tight frame expansion~~

~~( j) ( j) ξ ( j) ( j) ξ f = ∑ f ,Zξ SZξ d = ∑ f ,Zξ SZξ d , f H . (16.19) j J S S j J ∀ ∈ ∈ ∈ Moreover, for H ﬁnite-dimensional, i.e., J ﬁnite, let~~

~~J ( j) Zξ = Zξ := ∑ Zξ H . (16.20) j J ∈ ∈~~

~~Then (Zξ )ξ S is the reproducing kernel tight frame for H , i.e., ∈~~

~~f = f (ξ)Zξ dξ = f ,Zξ Zξ dξ, f H . (16.21) S ∀ ∈ S S~~

Proof. The action of SO(d) on H j is absolutely irreducible, and so the orthogonal projection of H onto H j is either 0 or H j, and (16.18) holds (see Exer. 16.7). ( j) Recall (see Example 16.7) that (Zξ )ξ S is a continuous tight frame for Hk, i.e., ∈ 16.5 Homogeneous polynomials 449

~~( j) 2 2 f ,Z dσ(ξ)= f , f H j. | ξ | ∀ ∈ S Let f H , and Pj be the orthogonal projection onto H j. Then ∈ ( j) ( j) ( j) f ,Z = f ,PjZ = Pj f ,Z , ξ S ξ S ξ S so that~~

( j) 2 ξ ( j) 2 ξ 2 2 ∑ f ,Zξ d = ∑ Pj f ,Zξ d = ∑ Pj f = f . j J S | | j J S | | j J ∈ ∈ ∈ By Fubini, we can exchange the sum and integral above, and so we have (16.19). ( j) Now suppose that J is ﬁnite. Since f ,Z = f ,Zξ , from (16.19) we obtain ξ ( j) ( j) σ ξ ( j) σ ξ σ ξ f = ∑ f ,Zξ SZξ d ( )= ∑ f ,Zξ SZξ d ( )= f ,Zξ SZξ d ( ), S j J S j J S ∈ ∈ which is (16.21). By Exer. 16.6 the reproducing kernel for H is the sum of the reproducing kernels for H j, j J, and so we recognise (Zξ )ξ S as the reproducing kernel tight frame for H . ∈ ∈ ⊓⊔ This generalises Example 16.7, which is the special case J = k , H = Hk. In addition to (16.11), the solid zonal harmonics are given by the formulas{ } x Z(k)(x)= x kZ(k) ξ ξ x [k/2] j d(d + 2) (d + 2k 2 j 4) k 2 j 2 j =(d + 2k 2) ∑ ( 1) − − x,ξ − x − − 2 j j!(k 2 j)! j=0 − 2k + d 2 ( d 2 ) x,ξ = − x kC −2 (Funk–Hecke formula). (16.22) d 2 k x − We now consider Π (Rd), i.e., the case J = n 2 j : 0 j n . n◦ { − ≤ ≤ 2 } Corollary 16.1. (Homogeneous polynomials) The reproducing kernel tight frame Π d expansion for n◦(R ) is given by

~~d f = f ,Zξ Zξ dσ(ξ), f Π ◦(R ), (16.23) S ∀ ∈ n S where~~

~~d ξ 2 j (n 2 j) n ( 2 ) x, d Zξ (x) := ∑ x Zξ − (x)= x Cn , x R . (16.24) 0 j n x ∈ ≤ ≤ 2~~

~~Proof. The reproducing kernel tight frame (pξ )ξ S for Hn 2 j(S) is given by ∈ − 0 j n ≤≤ 2 450 16 Continuous tight frames for ﬁnite dimensional spaces~~

~~(n 2 j) pξ = ∑ Zξ − S. 0 j n | ≤ ≤ 2 A polynomial in the form (16.16) is mapped by the identiﬁcation as follows~~

~~Π d H 2 j n◦(R ) n 2 j(S) : p = ∑ pn 2 j ∑ pn 2 j S. → − n − → n − | 0 j n 0 j 0 j ≤≤ 2 ≤ ≤ 2 ≤ ≤ 2~~

~~The inverse of pξ under this identiﬁcation map is Zξ , which establishes (16.23). The second formula in (16.24) follows from (16.11), which gives the telescoping sum~~

d ξ d ξ 2 j n 2 j ( 2 ) x, n 2 j ( 2 ) x, Zξ (x)= ∑ x x − Cn 2 j x − Cn 2 j 2 , 0 j n − x − − − x ≤ ≤ 2 λ which simpliﬁes to (16.24), since C( ) := 0, k < 0. k ⊓⊔ Example 16.9. (Spherical polynomials) In view of (16.16), the space polynomials of degree n on the sphere is given by the choice J = 0,1,...,n , i.e., { } Π H H H Π Π n(S)= 0 1 n = n◦(S) n◦ 1(S). ⊕ ⊕⊕ ⊕ − From (16.24), we obtain

~~d d ( 2 ) ξ ( 2 ) ξ Zξ (x)= Cn ( x, )+Cn 1( x, ), x S. − ∈ Example 16.10. (Polynomial wavelets) For J = n + 1,n + 2,...,n + s , we have { }~~

~~H = Πn s(S) Πn(S)= Hn 1 Hn 2 Hn s, + ⊖ + ⊕ + ⊕⊕ + and from (16.11) and (16.22) we obtain~~

~~d d d d ( 2 ) ξ ( 2 ) ξ ( 2 ) ξ ( 2 ) ξ Zξ (x)= Cn+s( x, )+Cn+s 1( x, ) Cn ( x, ) Cn 1( x, ) − − − − n+s 2k + d 2 ( d 2 ) = ∑ − C −2 ( x,ξ ). d 2 k k=n+1 −~~

This zonal function Zξ is localised in space near ξ. The coefﬁcients in the second formula can be modiﬁed to obtain a zonal polynomial wavelet for Πn+s(S) Πn(S) which has good space–frequency localisation [Fer07]. ⊖

Example 16.11. (Poisson kernel) It is natural to take J = N to obtain a reproducing kernel for H = L2(S). In this case, the series for Zξ does not converge in L2(S). Nevertheless, it sums to give the Poisson kernel (see Exer. 16.8)

~~∞ 2 j 1 x P(x,ξ)= ∑ Z( )(x)= − . ξ x ξ d j=0 − 16.6 Orthogonal polynomials for a radially symmetric weight 451 16.6 Orthogonal polynomials for a radially symmetric weight~~

Here we consider the multivariate orthogonal polynomials Pn (see 10.10) for a radially symmetric measure µ on Rd, i.e., one for which the symmetry§ group of the measure is O(d), and so Pn is O(d)–invariant. For simplicity, we suppose µ is Lebesgue integration with a nonnegative radial d weight function w : [0,R) R on the ball BR := x R : x < R , 0 < R ∞, → { ∈ } ≤ (which deﬁnes an inner product on Πn), i.e.,

f ,g = f ,g w := f (x)g(x)w( x )dx. (16.25) B R By using the characterisation of G–frames (Theorem 10.9) extended to the inﬁnite compact group G = SO(d), we ﬁnd a single polynomial p Pn of unit norm, for ∈ which (gp)g G is a continuous tight frame for Pn, i.e., ∈

~~f = dim(Pn) f ,gp gp dνd(g), f Pn, (16.26) g SO(d) ∀ ∈ ∈ where νd is the normalised Haar measure on SO(d). By choosing p = pξ to be zonal, we obtain~~

~~f = dim(Pn) f , pξ pξ dσ(ξ), f Pn, ∀ ∈ S with one choice for pξ giving the reproducing kernel tight frame. Both of these expansions can be discretised to obtain ﬁnite tight frames.~~

~~16.6.1 The O(d)–invariant subspaces of Pn~~

To ﬁnd a p Pn giving (16.26), we need the SO(d)–invariant subspaces Vj of Pn, ∈ i.e., the O(d)–invariant subspaces (we use the subgroup SO(d) for convenience). We will repeatedly use the fact that if a function f deﬁned on BR can be factored into a radial and angular part x f (x)= R( x )Θ , x then (by Fubini’s theorem) it can be integrated

R R d 1 d 1 f (x)dx = R(r)Θ(ξ)r − dr dξ = R(r)r − dr Θ(ξ)dξ . BR S 0 0 S In particular, for j = k, polynomials in H j and Hk multiplied by radial polynomials are orthogonal with respect to the inner product (16.25) for Pn. This leads to the decomposition of Pn into its O(d)–invariant subspaces. 452 16 Continuous tight frames for ﬁnite dimensional spaces

~~n (n) Lemma 16.1. For 0 j , let Pj = P be an orthogonal polynomial of degree j ≤ ≤ 2 j for the univariate weight on [0,R2) given by~~

~~n 2 j+ d 2 t t − −2 w(√t). (16.27) →~~

Then Pn is the orthogonal direct sum of the absolutely irreducible O(d)–invariant subspaces P (n) (n) (n) 2 H n = Vj , Vj := Pj ( ) n 2 j, (16.28) − 0 j n ≤≤ 2 (n) where the inner product on Vj = Vj is given by

2 2 2 H h1Pj( ),h2Pj( ) = area(S) h1,h2 S Pj w,n 2 j, h1,h2 n 2 j, − ∀ ∈ (16.29)− where 2 1 R d 2 2 2 k+ −2 p w,k := p(t) t w(√t)dt. (16.30) 2 0 | | (n) Proof. We have already observed that the Vj = Vj are orthogonal to each other. Since Hn 2 j is an absolutely irreducible O(d)–invariant subspace, it follows that Vj is also. Moreover,− by (16.17), we have

~~H Π d P ∑ dim(Vj)= ∑ dim( n 2 j)= dim( n◦(R )= dim( n). − 0 j n 0 j n ≤ ≤ 2 ≤ ≤ 2~~

Hence to prove (16.28), it sufﬁces to show that Vj Pn, which we now do. ⊂ Let hα Hα , hβ Hβ and p and q be polynomials, then ∈ ∈ 2 2 hα p( ),hβ q( ) α x 2 β x 2 = x hα p( x ) x hβ q( x )w( x )dx B x x R R 2 2 α+β+d 2 = p(r )q(r )r − w(r)rdr hα (ξ)hβ (ξ)dξ 0 S 2 R α+β d 2 dt + − = area(S) p(t)q(t)t 2 2 w(√t) hα ,hβ S. (16.31) 0 2 Let hn 2 j Hn 2 j and p j be a univariate polynomial of degree j. In view of (16.16), − ∈ − 2 the polynomial p j( )hn 2 j is in Pn if and only if −

2 2ℓ k p j( )hn 2 j, hk 2ℓ = 0, hk 2ℓ Hk 2ℓ, 0 k < n, 0 ℓ . − − − ∈ − ≤ ≤ ≤ 2 By (16.31) and orthogonality of spherical harmonics of different degrees, this holds except for when n 2 j = k 2ℓ, in which case ℓ = j 1 (n k) < j, and we require − − − 2 − 16.6 Orthogonal polynomials for a radially symmetric weight 453

R2 ℓ n 2 j+ d 2 p j(t)t t − −2 w(√t)dt = 0, 0 which is satisﬁed by the choice p j = Pj. The formulas (16.29) and (16.30) are special case of (16.31). ⊓⊔ Example 16.12. (Gegenbauer polynomials). The orthogonal polynomials on the unit ball B1 corresponding to the weight

α β d w(r) :=(1 r2) r2 , α > 1, β > , (16.32) − − − 2 are the generalised Gegenbauer polynomials, Gegenbauer polynomials (when β = 0), and Legendre polynomials (when α = β = 0). For this weight

~~n 2 j+ d 2 n 2 j+ d 2 α β α n 2 j+ d 2 +β t − −2 w(√t)= t − −2 (1 t) t =(1 t) t − −2 , − − so that~~

α d 2 β ( ,n 2 j+ −2 + ) Pj = P − (2( ) 1), j − Γ α Γ d β 2 1 ( j + + 1) (n j + 2 + ) Pj w,n 2 j = d − d , (16.33) − 2 (α + n + + β) j!Γ (α + n j + + β) 2 − 2 (α,β) where Pj are the univariate Jacobi polynomials given by (15.12).

~~Example 16.13. (Hermite polynomials). The orthogonal polynomials on Rd = B∞ corresponding to the weight~~

~~2β r2 w(r)= r e− , β 0, (16.34) ≥ are called generalised Hermite polynomials, and Hermite polynomials (β = 0). For these, we can take~~

~~d 2 β Γ n j d β (n 2 j+ −2 + ) 2 1 ( + 2 + ) Pj = L j − , Pj w,n 2 j = − , (16.35) − 2 j!~~

~~(α) where Ln are the generalised Laguerre polynomials given by~~

n ∞ α 1 α d α α α Γ (α + n + 1) L( )(x) := x ex (xn+ e x), (L( )(x))2 x e x dx = . n n! − dxn − n − n! 0 (n) P Example 16.14. In view of (16.28), an orthonormal basis for Vj (and hence n) can be obtained from an orthonormal basis (Yβ )=(Sn 2 j,β ) for Hn 2 j (see [DX01] − for details). This basis is not O(d)–invariant (none can− be).

~~We now ﬁnd a continuous tight frame for Pn which is O(d)–invariant. 454 16 Continuous tight frames for ﬁnite dimensional spaces~~

~~16.6.2 Continuous tight frames for Pn~~

We now give a continuous tight frame (pξ )ξ S for the orthogonal polynomials Pn, ∈ which is O(d)–invariant. This is obtained by considering all group frames (gp)g G ∈ for Pn, where G is the continuous group G = SO(d). We assume the decomposition of Pn = Vj given by Lemma 16.1, and suppose that P is normalised so that its leading term has a positive coefﬁcient. Let ν = ν j d be the normalised Haar measure on SO(d).

~~Theorem 16.2. Let p Pn be any unit norm polynomial of the form ∈~~

~~dim(Hn 2 j) (n) p ∑ − p p V p 1 (16.36) = P j, j j , j = . 0 j n dim( n) ∈ ≤ ≤ 2~~

~~Then gp g SO(d) is an equal–norm continuous tight frame for Pn, i.e., { } ∈~~

~~f = dim(Pn) f ,gp gpdν(g), f Pn, (16.37) SO d ∀ ∈ ( )~~

~~and these are all such p Pn. Moreover, p can be chosen to be zonal, in which case ∈~~

~~f = dim(Pn) f , pξ pξ dσ(ξ), f Pn. (16.38) ∀ ∈ S There are a ﬁnite number of such p = pξ with a given pole ξ. Of these, we call~~

~~(n 2 j) − 2 1 Zξ Pj( ) pξ := ∑ . (16.39) P dim( n) 0 j n area(S) Pj w,n 2 j ≤ ≤ 2 − the canonical choice of p (the leading term of Pj has a positive coefﬁcient).~~

Proof. Let G = SO(d). We observe that Theorem 10.9 extends to this G (or any 1 ∑ inﬁnite compact group), with the ﬁnite sum G g G replaced by integration with respect to the normalised Haar measure on G.| | ∈ The action of G on Pn is unitary, and is absolutely irreducible on the subspaces (n) Vj = Vj . Moreover, none of the Vj are CG–isomorphic to each other (see [FH87]). This is easily seen for d 3 where (16.5) implies the Vj have different dimensions, ≥ and for d = 2 from following the explicit description of Vj

~~n 2 j 2 2 Vj = span (x,y) ℜ((a + ib)(x + iy) − )Pj(x + y ) : a,b R . { → ∈ }~~

Hence, by Theorem 10.9, the choice (16.36) gives (16.37). Here we write dim(Vj) as dim(Hn 2 j), and normalise p so that it is a unit vector. A simple calculation shows − that p is zonal with a pole at ξ if and only if each p j is. The space of zonal functions ξ (n 2 j) in Hn 2 j with pole is one dimensional, and spanned by Zξ − . Thus, by (16.29) − 16.6 Orthogonal polynomials for a radially symmetric weight 455 and (16.13)

(n 2 j) 2 (n 2 j) 2 Zξ − Pj( ) Zξ − Pj( ) p j = = , (n 2 j) 2 H ± Z − Pj( ) ± area(S) dim( n 2 j) Pj w ξ − where the ‘+’ choice gives the canonical choice of pξ (which is most peaked at ξ). Finally, suppose p is zonal, so that pξ is zonal with pole ξ S. Let Gξ be the subgroup of G = SO(d) which ﬁxes the point ξ, i.e., ∈

~~Gξ := g SO(d) : gξ = ξ = SO(d 1). { ∈ } ∼ − Then (see [SD80] for details) the integral of (16.37) can be computed~~

f ,gp gpdνd(g)= f ,gp gpdνd 1(g)dσ(ξ) SO(d) S SO(d 1) − − = f , pξ pξ dνd 1(g)dσ(ξ)= f , pξ pξ dσ(ξ), G − S ξ S which gives (16.38). ⊓⊔ The ﬁrst few zonal harmonics on Rd are given by

d + 2 Z(0) = 1, Z(1) = d x,ξ , Z(2) = d x,ξ 2 x 2 , ξ ξ ξ 2 − d(d + 4) Z(3) = x,ξ (d + 2) x,ξ 2 3 x 2 , (16.40) ξ 6 − d(d + 6) Z(4) = (d2 + 6d + 8) x,ξ 4 (6d + 12) x,ξ 2 x 2 + 3 x 4 . ξ 24 − Example 16.15. Let P4 be the quartic Legendre polynomials on the unit disc. Then by (16.33) and (16.40), the summands of the canonical choice pξ are

1 4 2 2 4 p0(x)= 16 x,ξ 16 x x,ξ + 2 x , √π − 1 2 2 2 p1(x)= 4 x,ξ 2 x 4 x 3 , √π − − 1 4 2 p2(x)= 6 x 6 x + 1 , √π − and so the canonical choice is a ridge polynomial

1 4 2 pξ (x)= p0(x)+ p1(x)+ p2(x)= 16 x,ξ 12 x,ξ + 1 . √π − The choice qξ := p0 + p1 p2 is is not a ridge polynomial (see Fig. 16.1). − 456 16 Continuous tight frames for ﬁnite dimensional spaces

~~Fig. 16.1: Contour plots of the quartic Legendre polynomials pξ and qξ from Example 16.15 for ξ =(1,0). Clearly the canonical choice pξ is a ridge function.~~

For the Legendre polynomials (constant weight) the canonical choice for pξ is always a ridge polynomial. This is not the case for the Gegenbauer polynomials for a nonconstant weight (see [Wal09]). The corresponding continuous tight frame of ridge polynomials for the Legendre polynomials was used by Petrushev [Pet99] to study approximation by ridge functions and neural networks on the unit ball. Corollary 16.2. (Legendre polynomials). For the constant weight 1 on the unit ball, the canonical choice for p in Theorem 16.2 is the ridge polynomial given by

~~√ d 2n + d ( 2 ) pξ (x)= Cn ( x,ξ ). (16.41) area(S) dim(Pn)~~

~~Example 16.16. For the Legendre polynomials on the disc in R2, (16.41) gives~~

√2n + 2 (1) 1 pξ (x)= Cn ( x,ξ )= Un( x,ξ ), (16.42) √2π√n + 1 √π where Un are the Chebyshev polynomials of the second kind (see Example 16.15). In this case, both (16.37) and (16.38) reduce to

~~n + 1 2π f = f ,Rθ p Rθ pdθ, f Pn, (16.43) 2π 0 ∀ ∈ where Rθ is rotation by θ and p = pξ (for any ξ). For the Legendre polynomials on the unit ball in R3, (16.41) gives~~

√ 3 2n + 3 ( 2 ) pξ (x)= Cn ( x,ξ ). √ π n+2 4 2 In this case, the integral in (16.37) is over the manifold SO(3) of dimension 3, and the integral in (16.38) is over S which has dimension 2. 16.6 Orthogonal polynomials for a radially symmetric weight 457

~~16.6.3 The reproducing kernel for Pn~~

We now consider the reproducing kernel tight frame for Pn (and associated spaces). Lemma 16.2. Let p = 0 be a univariate polynomial. Then the reproducing kernel 2 for the absolutely irreducible SO(d)–invariant subspace p( )Hk of L2(µ) is p( x 2)p( y 2) K(x,y)= Z(k)(x,y), area(S) p 2 w,k (k) where p w k is given by (16.30) and Z is the polynomial given by . (k) k k (k) y k k (k) x Z (x,y) := x y Z x = x y Z y . (16.44) x y y x Proof. Let Ky(x) := K(x,y), so that

~~2 2 p( )p( y ) k (k) Ky = 2 y Z y , y = 0. area(S) p y w,k~~

Then, for f Hk, using (16.31) and (16.9), we calculate ∈ 2 2 p( y ) k 2 2 (k) p( ) f ,Ky = 2 y p( ) f , p( )Z y area(S) p y w,k 2 k (k) = p( y ) y f ,Z y S y y = p( y 2) y k f = p( y )2 f (y), y 2 so that K(x,y) is the reproducing kernel for p( )Hk. ⊓⊔ An explicit formula for Z(k)(x,y) is given by (16.22), i.e.,

(k) j d(d + 2) (d + 2k 2 j 4) k 2 j 2 j 2 j Z (x,y)=(d + 2k 2) ∑ ( 1) − − x,y − x y . − − 2 j j!(k 2 j)! 0 j k ≤ ≤ 2 − (n) (n) 2 H Example 16.17. The reproducing kernel for the subspace Vj = Pj ( ) n 2 j − of Pn given in Lemma 16.1 is

~~P(n) x 2 P(n) y 2 (n) j ( ) j ( ) (n 2 j) K (x,y)= Z − (x,y), j (n) 2 area(S) Pj w,n 2 j − where Z(k) is the polynomial given by (16.44).~~

~~We now give two SO(d)–invariant formulas for the reproducing kernel for Pn, which follow from Theorem 16.2. 458 16 Continuous tight frames for ﬁnite dimensional spaces~~

~~Theorem 16.3. The reproducing kernel for Pn is given by the formulas~~

Kn(x,y)= dim(Pn) pξ (x)pξ (y)dσ(ξ) S P(n)( x 2)P(n)( y 2) ∑ j j Z(n 2 j) x y (16.45) = n − ( , ), n ( ) 2 0 j 2 area(S) Pj w,n 2 j ≤ ≤ − (k) where pξ is the canonical choice and Z is the polynomial given by (16.44).

~~Proof. Expanding (16.38) gives~~

f (y)= dim(Pn) f (x)pξ (x)dµ(x) pξ (y)dσ(ξ) S BR = f (x) dim(Pn) pξ (x)pξ (y)dσ(ξ) dµ(x), BR S so that the reproducing kernel for Pn is given by the ﬁrst formula. Since Pn is the orthogonal direct sum (16.28), i.e.,

~~P (n) n = Vj , 0 j n ≤≤ 2~~

(n) its reproducing kernel is the sum of the reproducing kernels for the Vj given in Example 16.17 (see Exer. 16.6), which gives the second formula. ⊓⊔ The reproducing kernel for a ﬁnite dimensional rotationally invariant subspace of L2(µ) can be calculated by using the above techniques. This is a little more involved than for L2(S) (see Theorem 16.1), since the homogeneous components of the CSO(d)–module L2(µ) contain more than a single copy of each irreducible. Indeed, the homogeneous components, as deﬁned by (10.14), are

~~∞ 2 j ∑ Hk, k = 0,1,.... j=0~~

~~Example 16.18. Since the polynomials of degree k are the orthogonal direct sum~~

k d (n) Πk(R )= P0 P1 Pk = V , (16.46) ⊕ ⊕⊕ j n=0 0 j n ≤≤ 2 d the reproducing kernel of Πk(R ) L2(µ) is ⊂ k P(n)( x 2)P(n)( y 2) K x y ∑ ∑ j j Z(n 2 j) x y (16.47) ( , )= n − ( , ). n ( ) 2 n=0 0 j 2 area(S) Pj w,n 2 j ≤ ≤ − We now consider the reproducing kernel for the homogeneous polynomials. 16.6 Orthogonal polynomials for a radially symmetric weight 459

~~Example 16.19. By (16.16) and (16.31), the homogeneous polynomials of degree n are the orthogonal direct sum~~

~~Π d 2 jH n◦(R )= n 2 j. − 0 j n ≤≤ 2 Π d Hence, by Lemma 16.2, the reproducing kernel of n◦(R ) is~~

~~2 j 2 j x y (n 2 j) K(x,y)= ∑ j Z − (x,y). 0 j n area(S) ( ) w,n 2 j ≤ ≤ 2 −~~

~~16.6.4 Finite tight frames for Pn~~

~~We now outline how the continuous tight frame expansions for Pn of 16.6.2 can be discretised (sampled) to obtain a ﬁnite tight frame expansion. § d For a ﬁxed x R and f Pn, (16.38) gives ∈ ∈~~

f (x)= dim(Pn) f , pξ pξ (x)dξ. (16.48) S The above integral of the polynomial ξ f , pξ pξ (x) of degree 2n can be replaced by a quadrature rule (spherical design)→ toobtain a discrete form of (16.38).

~~Deﬁnition 16.5. A ﬁnite subset Θ of S together with weights cξ R, ξ Θ is called a quadrature (or cubature) rule of degree k for the sphere if ∈ ∈~~

~~d f dσ(ξ)= ∑ cξ f (ξ), f Πk(R ). S ∀ ∈ ξ Θ ∈~~

An equal weight quadrature rule, i.e., one with cξ = 1/ Θ , ξ Θ is known as a spherical k–design (see 6.4). There is an extensive literature| | ∀ on∈ quadrature rules for the sphere (see [Str71],§ [CR93], [Coo99]). Theorem 16.4. (Finite tight frame). Let Θ S be a cubature rule of degree 2n for ⊂ the sphere S with weights (cξ )ξ Θ , and pξ be the canonical choice (16.39). Then ∈

~~f = dim(Pn) ∑ cξ f , pξ pξ , f Pn. (16.49) ξ Θ ∀ ∈ ∈ Proof. Apply the quadrature rule of degree 2n to (16.48) to obtain~~

~~dim(Pn) f (x)= f , pξ pξ (x)dξ = dim(Pn) ∑ cξ f , pξ pξ (x), area(S) S ξ Θ ∈ which is (16.49). ⊓⊔ 460 16 Continuous tight frames for ﬁnite dimensional spaces~~

Example 16.20. For equal weight quadrature rules, i.e., spherical designs, (16.49) reduces to P dim( n) P f = Θ ∑ f , pξ pξ , f n. (16.50) ξ Θ ∀ ∈ | | ∈ We now consider the bivariate polynomials, i.e., d = 2. Here

~~cosθ sinθ SO(2)= Rθ : 0 θ < 2π , Rθ := − , { ≤ } sinθ cosθ where Rθ is rotation through θ.~~

Example 16.21. (The circle) Let Θk be any set of k equally spaced points on the unit circle S (d = 2). These give an equal weight quadrature rule of degree k 1 for S (see − Exer. 6.10). Hence (16.50) holds for Θ = Θk, k 2n + 1. This also extends to when ≥ n k n + 1 and k is odd by by writing Θ2k = Θk RπΘk, and using Rπ pξ =( 1) pξ . ≥ ∪ − Example 16.22. (Logan–Shepp) Taking k = 2(n + 1) in Example 16.21 gives two n P copies (up to a scalar 1) of the orthonormal basis (R jπ p0) j=0 for n. ± n+1 In particular, for the Legendre polynomials on the unit disc (constant weight), the continuous tight frame expansion (16.43) can be discretised to the orthogonal expansion

n + 1 2π 1 f = f ,R jπ p R jπ p, f Pn, p(x,y) := Un(x), 2π 0 n+1 n+1 ∀ ∈ √π of Logan and Shepp [LS75]. In a similar vein, one can obtain discrete versions of (16.37). Definition 16.6. A finite subgroup G of SO(d) generates a spherical t–design if the set Θ = gη g G is a spherical t–design for some (and hence every) η S. { } ∈ ∈ Such groups are said to be t–homogeneous (see [Ban84], [dlHP04]). Corollary 16.3. Let G be a finite subgroup of SO(d) which generates a spherical 2n–design, and p = pξ the canonical choice (16.39). Then

dim(Pn) f = ∑ f ,gp gp,, f Pn. (16.51) G g G ∀ ∈ | | ∈ Θ ξ Proof. Let = g g G in (16.50), and use gpξ = pgξ . { } ∈ ⊓⊔ Example 16.23. (The circle) Similarly to Example 16.21, (16.51) holds for

G = R 2π SO(2), k ⊂ the cyclic group of rotations through multiples of 2π/k (of order k), where k n+1 and k is odd, or k 2n + 1 and k is even. ≥ ≥ 16.7 Functions on the complex sphere 461 16.7 Functions on the complex sphere

~~We now outline how the orthogonal decomposition (16.8) of L2(S) into spherical harmonics can be extended to complex valued functions on the complex sphere~~

~~S = S˜ := x Cd : x = 1 . C { ∈ } For more detail see [Rud80]. The inner product (16.6) is replaced by~~

~~f ,g := f gdσ, (16.52) SC SC where σ is normalised surface-area measure on the sphere in R2d. From (16.8), we have the orthogonal decomposition into absolutely irreducible O(2d)–subspaces ∞~~

L2(SC)= Hk(SC), (16.53) k=0 d where Hk(SC)= Hk(C ) is the complex vector space of all harmonic (as functions on R2d) homogeneous (with respect to real scalars) polynomials of degree k on R2d (which is identiﬁed with Cd). The monomials z zα zβ , α + β = k, are an d → | | | | orthogonal basis for Hk(SC)= Hk(C ), and a calculation shows

α (d 1)!α! z 2 dσ(z)= − . (16.54) S | | (d 1 + α )! − | | Let U = U (d) be the group of all unitary operators on Cd, which is a compact subgroup of O(2d). The U –invariant subspaces Hk(S ), k = 0, are not irreducible. C Let H(p,q) be the subspace of Hk(SC), k = p + q, consisting of all polynomials on Cd that have bidegree (p,q), i.e.,

α β H(p,q) := span z z z : α = p, β = q . { → | | | | } Π d H This is the space p◦,q(C ) considered in Exer. 6.17. Then k(SC) is the orthogonal direct sum Hk(SC)= H(p,q), p+q=k of absolutely irreducible U –invariant subspaces, so that

~~∞ ∞~~

L2(SC)= Hk(SC)= H(p,q), (16.55) k=0 k=0 p+q=k where none of the H(p,q) in this orthogonal direct sum are CU –isomorphic. We now consider the reproducing kernel tight frames for various U –invariant α β α β subspaces of L2(S ). Here we write z z for the monomial z z z . C → 462 16 Continuous tight frames for ﬁnite dimensional spaces

~~Lemma 16.3. The reproducing kernel for the irreducible U –invariant subspace H(p,q) of L2(SC) is~~

(d 1 + p + q)! zα zβ wα wβ Kpq(z,w)= − ∑ ∑ α β . (16.56) (d 1)! α =p β =q ( + )! − | | | | Proof. We ﬁrst observe that by (16.54) α β α β α β α β α β σ α+β 2 σ (d 1)!( + )! z z ,z z SC = z z z z d (z)= z d (z)= − . S S | | (d 1 + α + β )! C C − | | α β Since (z z ) α =p, β =q is an orthogonal basis for H(p,q), Proposition 16.3 gives | | | |

α β α β (d 1 + p + q)! Kpq(z,w)= ∑ ∑ z z w w − α β , α =p β =q (d 1)!( + )! | | | | − which is (16.56). ⊓⊔ Example 16.24. (Holomorphic polynomials). The space H(p,0) is the holomorphic homogeneous polynomials of degree p. Its reproducing kernel is

d 1 + p α α p d 1 + p p Kp0(z,w)= − ∑ z w α = − z,w . (16.57) d 1 α =p d 1 − | | − In view of (16.55), the reproducing kernel for the space of holomorphic polynomials of degree n is ≤ n d 1 + p K(z,w)= ∑ − z,w p. d 1 p=0 − Summing over all p gives the Szego¨ kernel for the holomorphic functions in L2(SC) ∞ ∞ d 1 + p p 1 S(z,w)= ∑ Kp0(z,w)= ∑ − z,w = . d 1 (1 z,w )d p=0 p=0 − − d Example 16.25. (Spherical harmonics) The reproducing kernel for Hk(SC)= Hk(C ) is given by

~~k (d) z,w + w,z k (d) z,w + w,z ∑ Kpq(z,w)= z Ck z Ck 2 . p q k 2 z − − 2 z + =~~

~~Summing over all p and q gives the Poisson kernel for SC~~

~~1 z 2 P(z,w)= − , z B, w S. w z 2d ∈ ∈ − The details of these examples are given in Exer. 16.10. 16.8 G–frames for inﬁnite groups 463 16.8 G–frames for inﬁnite groups~~

The previous constructions of tight G–frames for the continuous group G = SO(d) relied on the extension of Theorem 10.8 to infinite groups G. Motivated by these, we now briefly discuss such extensions in generality. We first suppose that:

G is a locally compact (topological) group, and µ is the associated (left) Haar • measure, which is normalised so that µ(G)= 1 when G is compact. There is a unitary continuous representation of G on a Hilbert space H , i.e., a • unitary representation ρ for which the map (g,v) gv := ρ(g)v is continuous. → It is natural the role played by (algebraically) irreducible subspaces is replaced by topologically irreducible subspaces. A continuous representation of G on V is said to be topologically irreducible if V does not contain a proper closed G–invariant subspace (for V ﬁnite dimensional there is no distinction between the two). The closure of a topologically irreducible subspace is a topologically irreducible subspace. For a continuous unitary action on H , the orthogonal complement of a G–invariant subspace is G–invariant.

If G is compact, then all topologically irreducible unitary representations are ﬁnite dimensional. This leads to an orthogonal decomposition of H into ﬁnite dimensional irreducible G–invariant subspaces (Peter–Weyl theorem).

If G is not compact, e.g., G = Z, then the situation is far more complicated: Irreducible representations may not exist, and they can be inﬁnite dimensional. • The frame operator for a G–orbit does not obviously converge. • Here is an example of G noncompact (see Exer. 16.11 for details).

j Example 16.26. Let G = Z act on H = ℓ2(Z) via the bilateral shift j v = S v, where Sek := ek+1. This gives a unitary continuous action. Elementary calculations show j that (S v) j J is a tight frame if and only if it is an orthogonal basis, and the only ∈ v with ﬁnite support giving a tight frame are the standard basis vectors v = ek (and multiples of them). There exist vectors with inﬁnite support giving tight frames, e.g., 1 1 1 1 1 v =(..., ,0, ,0, 1,0,1,0, ,0, ,0, ,...). (16.58) −5 −3 − 3 5 7

j The vectors v for which (S v) j Z is a tight frame (orthogonal basis) can be better understood by identifying them∈ as the Fourier coefﬁcients of a 2π–periodic function j f L2(T), where v j = f (z),z (T =[0,2π] has the normalised Haar measure). ∈ L2(T) Here the shift S corresponds to multiplication of f (z) by z = eit . The condition for v to give a tight frame is that f have constant modulus. The examples v = ek correspond to f (z)= zk (the only trigonometric polynomials with modulus 1), and the v of (16.58) to the function f = 1 on [0,π] and f = 1 on [π,2π] (up to a scalar). − 464 16 Continuous tight frames for ﬁnite dimensional spaces

We now consider the case when G is compact, e.g., G = SO(d). The Peter–Weyl theorem ensures: A continuous action of G (compact) on H can be taken to be unitary. • For a continuous unitary action of G on H , there is an orthogonal decomposition • of H into ﬁnite dimensional absolutely irreducible subspaces.

Theorem 16.5. Let G be a compact group with a continuous unitary action on H . Then the orbit (gv)g G is a (generalised) tight frame for H if and only if H is ﬁnite dimensional, and the∈ conditions of Theorem 10.8 hold, where (10.16) is replaced by

v j,gv j gvk dµ(g)= 0. (16.59) G Proof. We ﬁrst suppose that H is ﬁnite dimensional, and so can be written as an orthogonal direct sum H = V1 Vm of irreducible G–invariant spaces. Since ⊕⊕ f ,gv 2 dµ(g) f 2 gv 2 dµ(g)= µ(G) f 2 v 2, G | | ≤ G

~~the frame operator for Φ =(gv)g G is well deﬁned (see Exer. 16.1) by ∈~~

~~SΦ ( f )= f ,gv gvdµ(g), f H . G ∀ ∈~~

The argument of Theorem 10.8 then follows with ∑g G replaced by integration with ∈ respect to the Haar measure. In particular, we observe trace(SΦ )= µ(G) v 2 and Schur’s lemma (Lemma 10.4) also holds for G infinite (and Vj finite dimensional). We now suppose that H is infinite dimensional. Since the orthogonal projection of a tight frame onto a closed subspace is a tight frame (for the subspace), we can assume without loss of generality that

H = V1 V2 V3 , ⊕ ⊕ ⊕ a countable direct sum of (ﬁnite dimensional) irreducible G–invariant subspaces. Suppose that (gv)g G is tight frame for H , and write v = v1 + v2 + , v j Vj. ∈ ∈ Then (gw)g G, w := v1 + + vm, is a tight frame for the ﬁnite dimensional space ∈ V1 Vm, and so the conditions of Theorem 10.8 hold. In particular, ⊕⊕ 2 v j dim(Vj) v j = 0, j, 2 = . ∀ vk dim(Vk) This implies 2 ∞ 2 2 v1 w = ∑ v j = ∑ dim(Vj)= ∞, j dim(V1) j=1

~~which is not possible. Thus there can be a tight G–frame (gw)g G for H only when H has ﬁnite dimension. ∈ ⊓⊔ 16.8 G–frames for inﬁnite groups 465~~

~~The considerations of 10.13 on when a G–frame exists extend, e.g., §~~

~~For G compact and H ﬁnite dimensional, there is a tight G–frame for H if and only if each irreducible CG–module W in H has multiplicity dim(W). ≤~~

This result is closely connected with the result of [GM71] (for G compact) on the existence of cyclic vectors, i.e., v for which the closed span of (gv)g G is H . The condition for V = H to have cyclic vectors is that it has countable dimension,∈ and

~~mult(W,V) dim(W), for every irreducble CG–module W. ≤~~

By considering the projection vW of a cyclic vector v V onto its homogeneous ∈ components HV (W), it is easy enough to see that this is a necessary condition. Conversely, by choosing vW so that (gvW )g G is a tight frame for HV (W), and 2 ∞ ∈ ∑W vW < , one can construct a cyclic vector v = ∑W vW . For Ga countable group, G–frames (for the counting measure) are studied up to unitary equivalence in [HL00] (using Von Neumann algebras), and for G compact they are studied in [Ive15] (using the Zak transform). Both approaches make use of the the left regular representation of G, which plays the same role as the group algebra CG in the case when G is ﬁnite–dimensional.

~~d Example 16.27. Let G = SO(d) act on the space of polynomials Πk(R ), with a radially symmetric measure. From the orthogonal direct sum (16.46), we have~~

d (2) 2 Π2(R )= H0 H1 P ( )H0 H2. ⊕ ⊕ 1 ⊕ d Since dim(H0)= 1 and the multiplicity of the irreducible H0 in Π2(R ) is two, d d there is no G–frame for the quadratic polynomials Π2(R ) (or Πk(R ), k 2). Since d ≥ Πk(R ) is ﬁnite dimensional, it is possible to construct a reproducing kernel tight frame (see Example 16.18).

~~Notes~~

Continuous tight frames were introduced and studied in detail in [AAG93]. A good account of zonal harmonics is given in [ABR01]. The section on the orthogonal polynomials for a radially symmetric weight was adapted from [Wal09]. I thank Tom Ter Elst and Joey Iverson for many useful discussions about this chapter. 466 16 Continuous tight frames for ﬁnite dimensional spaces Exercises

~~16.1. Let ( f j) j J be a generalised frame with respect to µ for a Hilbert space H (possibly inﬁnite∈ dimensional). (a) Show that~~

~~g f , f j g, f j dµ( j)=: g,Sf → J deﬁnes a bounded linear functional on H , and denote its Riesz representer by~~

~~Sf = f , f j f j dµ( j). J (b) Show this deﬁnes a linear map S : H H , with → A f 2 Sf , f B f 2, f H . ≤ ≤ ∀ ∈ (c) Show that S has a bounded inverse, and~~

1 1 f = f ,S− f j f j dµ( j)= f , f j S− f j dµ( j) J J 1/2 1/2 = f ,S− f j S− f j dµ( j), f H . J ∀ ∈ (d) Show that the synthesis and analysis operators (see 2.4) can be generalised to §

~~V : ℓ2(µ) H : a a j f j dµ( j), → → J~~

~~V ∗ : H ℓ2(µ) : f ( f , f j ) j J, → → ∈ where Va is deﬁned as the Riesz representer of~~

~~f f ,a j f j dµ( j)=: f ,Va . → J~~

Remark: Here S = VV , and one can deﬁne the Gramian as V V : ℓ2(µ) ℓ2(µ). ∗ ∗ → 16.2. Let ( f j) j J be a generalised frame with respect to µ for a d–dimensional Hilbert space H∈ . Show that the variational characterisation (Theorem 6.1) extends in the obvious way (Proposition 16.2), i.e.,

~~2 2 1 2 f j, fk dµ( j)dµ(k) f j dµ( j) , J J | | ≥ d J with equality if and only if ( f j) j J is tight. ∈~~

16.3. The canonical Gramian of a generalised frame Φ =( f j) j J with respect to µ is given by ∈ 1 PΦ = V ∗S− V : L2(µ) L2(µ), → 16.8 G–frames for inﬁnite groups 467 where S = VV ∗ is the frame operator and V is the synthesis operator of Φ. (a) Show that PΦ is an orthogonal projection. 1 (b) Show that the “matrix” [PΦ ] :=[ fk,S− f j ] j,k J represents PΦ in the sense ∈ 1 Φ Φ µ Φ µ µ P a =[P ] a := [P ] jkak d (k) j J = fk,S− f j ak d (k) j J. J J ∈ ∈ 1 (c) Let vk :=( fk,S− f j ) j J be the k–th “column” of [PΦ ]. Show that vk L2(µ), ∈ 1/2 ∈ and (v j) j J gives a copy of the canonical tight frame (S− f j) j J, i.e., ∈ ∈

~~vr,vs µ = fr, fs , r,s J. L2( ) ∀ ∈~~

16.4. Suppose ( f j) j J is a unit–norm generalised tight frame for a d–dimensional ∈ space H , i.e., f j = 1, j J. Show that µ is ﬁnite, and ∀ ∈ d f = f , f j f j dµ( j), f H . µ(J) J ∀ ∈ 16.5. Let H be a subspace of Fn, and P be the orthogonal projection onto H . n (a) Show that the tight frame (Kj) j=1 corresponding to the reproducing kernel for H is given by Kj = Pe j. (b) Find this normalised tight frame explicitly for

n H = x F : x1 + + xn = 0 . { ∈ } 16.6. Suppose that H is a reproducing kernel Hilbert space. (a) Show that any subspace of H is again a reproducing kernel Hilbert space. (b) Suppose H = j H j, an orthogonal direct sum of subspaces. Show that the reproducing kernel of H is K = ∑ K , where K is the reproducing kernel of H . j j j j 16.7. Use the orthogonal decomposition L2(S)= j H j of L2(S) into absolutely irreducible rotationally invariant subspaces, to show that the rotationally invariant subspaces of L2(S) have the form

~~H = H j, for some J N. j J ⊂ ∈~~

16.8. The Poisson kernel for the unit ball B = x Rd : x < 1 is given by { ∈ } 1 x 2 1 x 2 P(x,ξ)= = , x B, ξ S. − ξd − d x (1 2 x,ξ + x 2) 2 ∈ ∈ − − It has the property that for every u which is harmonic on the closed unit ball

u(x)= u(ξ)P(x,ξ)dξ. S From this, and (16.8), (16.9), (16.7), it follows that 468 16 Continuous tight frames for ﬁnite dimensional spaces ∞ ξ (k) ξ P(x, )= ∑ Zξ (x), x B, S, k=0 ∈ ∈ where the series converges absolutely and locally uniformally. Use the generating function for the Gegenbauer polynomials

~~∞ 1 λ = ∑ C( )(y)tk (1 2yt +t2)λ k − k=0 to expand the Poisson kernel in terms of the zonal harmonics, to obtain the formula~~

d ξ d ξ (k) k 2 x, k 2 x, Zξ (x)= x Ck x Ck 2 . x − − x d 16.9. (Linear polynomials on the sphere). The spaces Π1(S) and Π1(R ) of linear polynomials on the sphere and on Rd have dimension d + 1. (a) Find the reproducing kernel tight frame (Zξ ) for these spaces, with the norms S and w, respectively. 3 (b) Let ξ1,...,ξ4 be four points on the sphere in R . Show that the zonal functions 4 (Zξ ) are a basis for Π1(S) if and only if the points ξ j do not lie on a circle. j j=1 { } 4 Π ξ (c) Show that (Zξ j ) j=1 is an orthogonal basis for 1(S) if and only if the points j are the vertices of a regular tetrahedron. { }

16.10. The U –invariant subspaces of L2(SC) are given by the subsums of (16.55). Thus the reproducing kernel of such a space is the sum of the reproducing kernels of its summands (cf Theorem 16.1). (a) Show that the reproducing kernel of H(p,0) is zonal. (b) Find the reproducing kernel for the holomorphic polynomials of degree n. (c) Show that the sum of the reproducing kernels of the homogeneous holomorphic≤ functions of all degrees (these are orthogonal) is the the Szego¨ kernel 1 S(z,w) := , z B, w S. (1 z,w )d ∈ ∈ − It has the property that for every f which is holomorphic on the closed unit ball

~~f (z)= f (w)S(z,w)dσ(w). SC (d) Expand the Poisson kernel~~

~~∞ ∞ 1 z 2 P(z,w)= ∑ ∑ K (z,w)= − , pq w z 2d p=0 q=0 − d to ﬁnd a formula for the reproducing kernel for Hk(SC)= Hk(C ). 16.8 G–frames for inﬁnite groups 469~~

j 16.11. Let S be the bilateral shift on ℓ2(Z), given by Se j := e j 1. Then j v = S v + deﬁnes unitary continuous action of the noncompact group G = Z on H = ℓ2(Z). j (a) Let v ℓ2(Z) be nonzero. Show that if (S v) j Z is a tight frame for ℓ2(Z), then the frame∈ bound is A = v 2, i.e., (S jv) is orthogonal∈ basis. (b) Let v = vaea + vbeb = 0, a = b. Show j 2 2 2 a b ∑ x,S v = v x + 2ℜ vavb S − x,x , j | | j and conclude that (S v) j Z is a tight frame for ℓ2(Z) if and only if va = 0 or vb = 0. ∈ j (c) Find all ﬁnitely supported vectors v for which (S v) j Z is a tight frame for ℓ2(Z). 1 1 1 1 1 ∈ j (d) Let v =(..., 5 ,0, 3 ,0, 1,0,1,0, 3 ,0, 5 ,0, 7 ,...). Show that (S v) j Z is a − − − ∈ tight frame for ℓ2(Z). j (e) Determine all vectors v for which (S v) j Z is a tight frame for ℓ2(Z). ∈ d d 2d 16.12. Let SC := z C : z = 1 be the complex unit sphere in C R , and { ∈ } ≈2d σ be Lebesgue surface area measure on SC viewed as a unit sphere in R . Deduce the analogue of (16.15), i.e.,

d z = z,ξ ξ dσ(ξ), z Cd. area(S ) ∀ ∈ C SC 16.13. Show that the zonal harmonic Z(k) is localised at ξ S in the following sense. ξ ∈ (a) Z(k)(x) < Z(k)(ξ), x = ξ, x S. | ξ | ξ ∀ ∈ (k) (k) 2 (b) Z (ξ)= Z = dim(Hk). ξ ξ S (c) The maximum

~~d 1 d max p(ξ) : p = 1, p Hk(S − ) = dim(Hk)= O(k 2 ), k ∞ { S ∈ } → 1 (k) is attained if and only if p = Zξ . √dim(Hk)~~

~~d 16.14. The variational condition (6.5) for ( f j) to be a ﬁnite tight frame for F is~~

~~2 2 1 ∑∑ f j, fk = ∑ f j, f j . j k | | d j Show that the following analogous condition holds~~

1 2 x,y 2 dσ(x)dσ(y)= x,x dσ(x) . S S | | d S Remark: This can be interpreted as saying that although a generic frame of n unit vectors for Fd is not tight, for n d it is close to being tight (also see Exer. 6.3). ≫ 16.15. Let , ,k denote the apolar inner product of (6.19) on H(k,0) (the space of holomorphic homogeneous◦ polynomials of degree k). 470 16 Continuous tight frames for ﬁnite dimensional spaces

~~(a) Use (6.21), to show that~~

~~k + d 1 f ,g ,k = − f ,g S . ◦ d 1 C −~~

~~(b) Express the reproducing kernel tight frame given by (16.57) in terms of , ,k. (c) Use the formula ◦~~

~~2π 1 d 1 2 d 2 iθ f ( ξ,η )dσ(ξ)= − (1 r ) − f (re )rdrdθ, η SC, π 0 0 − ∈ SC to calculate ,η k . SC Chapter 17 Solutions~~

~~Exercises of Chapter 1~~

1.1 Since ∑ f ,u j u j = ∑ (u f )u j = V(V f )=(VV ) f , we obtain the matrix j j ∗j ∗ ∗ form. Verify that (1.1) holds for a particular choice of u1,u2,u3. If R is a rotation, 3 3 then [Ru1,Ru2,Ru3]= RV, and (RV)(RV)∗ = R(VV ∗)R∗ = 2 RR∗ = 2 I. Thus (1.1) holds for Ru1,Ru2,Ru3. 2 1.2 (a) Suppose f = f ,u1 v1 + f ,u2 v2, f R . Since the RHS is linear in f , ∀ ∈ 1 1 it suffices that this hold for the basis u1,u2 i.e., u1 = v1 2 v2, u2 = 2 v1 + v2, 2 {2 } − − which leads to v1 = 3 (2u1 + u2), v2 = 3 (u1 + 2u2). (b) Yes. Since u1 + u2 + u3 = 0, the coefficients satisfy ∑ j f ,u j = 0. Hence if one is changed, then this sum will no longer be zero. (c) No. Two coefficients can be changed while preserving ∑ f ,u j = 0. j 1.3 Equiangularity/equispacing. (a) Verify these are a tight frame by using the matrix form of Exer. 1.1. (b), (c) Direct computation. (d) Though equally spaced from each other, these vectors do not “fill up” the space 2 C , e.g., if w =( 1,0), then v j w = 2 + √2 > √3, j. − − ∀ 1.4 Gabor and Wavelet systems. 2πib x 2πib a 2πia b (a) TaMb f (x)=(Mb f )(x a)= e e f (x a)= e MbTa f (x). − − − − (b) Since TaTb = Ta+b and MaMb = Ma+b, a,b, the set G is closed under multiplication and inversion, and so forms a group∀ (generated by any set of generators for the subgroups Ta a A and Mb b B). D D{j T} ∈ T k { } ∈ (c) Since 2 j = 2 , k = 1 , and

j j k D T 2 j 2 j T D 2 j k f = 2 f (2 k)= 2 f (2 ( j )) = k j f , − − 2 2 j the group generated by D2 and T1 contains the translates of the diadic integers T D T k j,k Z, which are not contained in the set 2 j k j,k Z. { 2 j } ∈ { } ∈

~~471 472 17 Solutions Exercises of Chapter 2~~

2.1 Polarisation identity.1 Expand, and use g, f = f ,g , to get f + g 2 f g 2 = 2( f ,g + g, f )= 2( f ,g + f ,g )= 4ℜ f ,g . − − If the inner product is complex, then this gives

f + ig 2 f ig 2 = 4ℜ f ,ig = 4ℜ( i f ,g )= 4ℑ f ,g . − − − 2.2 Take the inner product of Parseval with g to obtain Plancherel. The reverse implication follows by the uniqueness of the Riesz representation and 1 1 f ,g = ∑ f , f j f j,g = ∑ f , f j f j,g , g H . A j J A j J ∀ ∈ ∈ ∈ Taking g = f in Plancherel gives the tight frame condition. Conversely, given a tight frame, the polarisation identity gives

2 2 1 2 2 4ℜ f ,g = f + g f g = ∑ f + g, f j f g, f j − − A j J | | −| − | ∈ 1 1 = ∑ 4ℜ( f , f j f j,g )= 4ℜ ∑ f , f j f j,g , A j J A j J ∈ ∈ ℑ ℑ 1 ∑ H and 4 f ,g = 4 ( A j J f , f j f j,g ), when is complex. ∈ Remark. This result extends to countably (or even uncountably) inﬁnite tight frames, with the interchange of sums and inner products being justiﬁed by considering the appropriate limits of partial sums. 1 2.3 Clearly, P = AVV ∗ is self adjoint, i.e., P∗ = P. Since ( f j) is a tight frame for K , V is onto K and Parseval gives VV K = AIK , so that ∗|

~~2 1 1 1 P =( VV ∗ K )( VV ∗)= IK VV ∗ = P. A | A A 2.4 (a) Given the uniqueness of the coefﬁcients in the orthogonal expansion~~

f , f j f = ∑ f j, f H , j J f j, f j ∀ ∈ ∈ 2 by Parseval, ( f j) j J is a tight frame if and only if f j = f j, f j = A, j, and this ∈ ∀ is normalised if and only if f j = √A = 1, j. ∀ (b) Taking f = f j/ f j in the normalised tight frame condition gives 1 A real or complex normed linear space (X, ) is an inner product space, with the inner product given by the polarisation identity, if and only if it satisﬁes the parallelogram identity f + g 2 + f g 2 = 2 f 2 + 2 g 2, f ,g X. − ∀ ∈ 17 Solutions 473

2 f j 2 2 1 2 1 = f = ∑ , fk = f j + 2 ∑ f j, fk , k J | f j | f j k= j | | ∈ so we must have f j 1, with equality if and only if f j, fk = 0, k = j. ≤ | | ∀ 2.5 (a) Since unitary maps preserve norms: U f = f , and we obtain ∗ 2 2 2 2 A f = A U∗ f = ∑ U∗ f , f j = ∑ f ,Uf j , f . j | | j | | ∀

(b) Since (Tf j) is a tight frame for the ﬁnite dimensional space H , the map T is onto, and hence invertible. Using the normalised tight frame property of (Tf j), followed by that of ( f j), we obtain

~~1 1 T − f = ∑ f ,Tf j T − (Tf j)= ∑ T ∗ f , f j f j = T ∗ f , f H , j j ∀ ∈~~

~~1 so that T − = T ∗, and T is unitary. 2.6 Apply P to the tight frame expansion, and use f = Pf , f H , to obtain ∈~~

~~f = P ∑ f , f j f j = ∑ Pf , f j Pf j = ∑ f ,Pf j Pf j, f H . j j j ∈ 474 17 Solutions~~

2.7 (a)= (b) Suppose that Q is a partial isometry. Clearly, (QQ∗)∗ = QQ∗. Let Q = Q ⇒ , and x ker(Q )= ran(Q) , y ran(Q). Then | |ran(Q∗) ∈ ∗ ⊥ ∈ 2 (QQ∗) (x + y)= QQ∗QQ∗y = Q(Q ∗Q )Q∗y = QQ∗y, | | so that QQ∗ is the orthogonal projection onto ran(Q). (b)= (c) Suppose that QQ∗ is an orthogonal projection. Clearly, Q∗Q is Hermitian. Let x⇒ ker(Q) and y = Q z ker(Q) = ran(Q ). Then ∈ ∗ ∈ ⊥ ∗ 2 2 (Q∗Q) (x + y)= Q∗QQ∗QQ∗z = Q∗(QQ∗) z = Q∗QQ∗z = Q∗Qy, so that Q∗Q is the orthogonal projection onto ran(Q∗). (c)= (a) Suppose that Q∗Q is an orthogonal projection, so that (Q∗Q)∗ = Q∗Q, (Q Q⇒)2 = Q Q. Let y = Q z ker(Q) = ran(Q ). Then ∗ ∗ ∗ ∈ ⊥ ∗ 2 2 2 Qy = QQ∗z,QQ∗z = (QQ∗) z,z = QQ∗z,z = Q∗z,Q∗z = y , so that Q is a partial isometry. (a)= (d) If Q is a partial isometry, then it can be factored Q = UP, where P is the ⇒ orthogonal projection onto (kerQ)⊥ and U : (kerQ)⊥ Q(H ) is unitary, so by Exer. 2.5 and 2.6, it follows QΦ is a normalised tight frame→ (for its span). (d)= (b) Suppose Φ =( f j) and QΦ =(Qf j) are normalised tight frames. We will ⇒ show that QQ∗ is the orthogonal projection onto ran(Q). For y ran(Q)⊥ = ker(Q∗), we have QQ y = Q0 = 0, and so it sufﬁces to show QQ y =∈y, y = Qf ran(Q). ∗ ∗ ∀ ∈ This follows by ﬁrst expanding in the normalised tight frame (Qf j), and then in ( f j) (and using linearity):

~~y = Qf = ∑ Qf ,Qf j Qf j = Q∑ Q∗Qf , f j f j = Q(Q∗Qf )= QQ∗y. j j~~

2.8 (a) For a given U of size n, a random equal–norm tight frame for Cd is given by the matlab code: J=randperm(n), J=J(1:d), V=U(J,:). (b) The Fourier matrix F is given the matlab code: F=fft(eye(n))’/sqrt(n) or w=exp(2*pi*i/n), F=w.ˆ([0:n-1]’*[0:n-1])/sqrt(n). Since ωn = 1 and 1 + ω + ω2 + + ωn 1 = 0 (for any n–root of unity ω = 1), we have − 1 jℓ kℓ 1 k j ℓ 1 (F∗F) jk = ∑(F∗) jℓFℓk = ∑ω− ω = ∑ (ω − ) = dδ jk. ℓ d ℓ d 0 ℓ

~~4 2 2 (F ) jk =(F F ) jk = ∑δ j+ℓ,0δk+ℓ,0 = δ jk. ℓ~~

~~2 2.9 (a) Let z j := x j + iy j, so w j = z j , and the frame operator is 17 Solutions 475~~

∑ z z 2 1 ∑ z2 z 2 z1+z1 zn+zn 1 j( j + j) i j( j j ) 2 2 S = VV ∗ = 1 2 2 − 2 , V = z1 z1 zn zn . 4 ∑ (z z j ) ∑ (z j z j) − − i j j − − j − 2i 2i This is diagonal, with equal diagonal entries if and only if

~~2 2 2 2 2 ∑z j ∑z j = 0, ∑z j + ∑z j = 0 ∑w j = ∑z j = 0, j − j j j ⇐⇒ j j~~

1 ∑ 2 in which case S = 2 j z j I. | | 2 (b) Tight frames ( f j) and (g j) for R are projectively unitarily equivalent if and only if g j = cα jUf j, j, where U is unitary, c > 0, α j = 1. Assume wlog that ∀ iθ ± U is rotation by θ, so this becomes w j = ce α jz j, j, where z j := ℜ f j + iℑ f j, 2 ∀ 2 2 2iθ w j := ℜg j + iℑg j, i.e., the diagram vectors satisfy w = ζz , j, ζ = c e . j j ∀ (c) Three unit vectors are a tight frame for R2 if and only if the sum of their diagram 2 2 2 vectors is zero: z1 + z2 + z3 = 0. By (b), we may assume that these diagram vectors are the third roots of unity, so z1,z2,z3 are distinct sixth roots of unity, none of which is the negatives of another, and so the vectors in R2 they represent are projectively equivalent to three equally spaced vectors. (d) Four unit vectors (v j) are a tight frame if and only if the sum of their diagram 2 vectors (z j ) is zero in C, i.e., the Argand diagram of their sum is a parallelogram. 2 2 2 2 2 2 Suppose wlog that z1 = z2 =(iz2) and z3 = z4 =(iz4) . Then z1 = iz2, z3 = − − ± 2 iz4, i.e., v1 v2, v3 v4, and so (v j) is a union of two orthonormal bases for R . (e)± By construction,⊥ ⊥ the sum of the diagram vectors is zero, and so they give a tight frame (see Figure 17.1). Moreover, none contains an orthonormal basis (these correspond to diagram vectors which are negatives of each other). The inner product between the ﬁrst and last has modulus cosθ, which gives the projective unitary inequivalence.

~~Fig. 17.1: The diagram vectors, and their sum (above) for the tight frames of Exer. 2.9.~~

2.10 (a) The diagram vectors of a unit–norm tight frame for R2 can be ordered 2 iθ z = e j , 0 θ1 θ2 θn < 2π, so that so the Argand diagram for their sum j ≤ ≤ ≤≤ z2 + +z2 = 0 is a convex n–gon with unit length sides. This polygon has the same 1 n 476 17 Solutions

2 σ shape as that obtained from (czσ j), where c = 1 and is a permutation of the indices, and so, by Exer. 2.9 (b), depends only on± the equivalence class. Conversely, such a convex polygon is uniquely determined by the sequence of its exterior angles θ ψ ψ 2 i j θ ψ 1,..., n (taken in either angular direction), and z j := e , j : ∑1 ℓ j j is a sequence of diagram vectors (since the exterior angles add to 2π) which≤ ≤ correspond to this polygon. (b) Since the angle between vectors is half the angle between their diagram vectors, two vectors are orthogonal to each other if and only if the angle between their diagram vectors is π, i.e., they are negatives of each other, and hence correspondtoa pair of parallel sides of the diagram vector polygon. 2π (c) Suppose wlog the angles of n equally spaced vectors are n j, 1 j n, so the 4π ≤ ≤ angles of the diagram vectors are n j, 1 j n. For n odd this is original angles (reordered) so the polygon the regular n–gon.≤ For≤ n = 4 even this is two copies of the angles 2π k, 1 k n , so that polygon a the regular n –gon of side length 2 (where n/2 ≤ ≤ 2 2 each ‘side’ is two collinear unit length edges). For n = 4 the polygon is degenerate, a ﬂat parallelogram. (d) Yes. Suppose all vectors are nonzero (to avoid edges of zero length), then, 2 2 iθ j as in (a), we can order the diagram vectors: z1,...,zn, where z j = r je , and 0 θ1 θ2 θn < 2π. As before, the corresponding convex polygon (with ≤ ≤ ≤≤2 sides of length z j ) depends only on the equivalence class. | | 2.11 (a) C(αv + βw)= αv + βw = α v + β w = αC(v)+ βC(w). (b) SΦ (Cf )= ∑ j f , f j f j = ∑ j f , f j f j = C(SΦ f ), and f j, fk = f j, fk . (c) With this deﬁnition of v, we need only check that the operations of (2.18) hold: v + w = (v j + w j)=(v j + w j)=(v j + w j)=(v j)+(w j)= v + w, etc.

2.12 (a) trace(S)= trace(VV ∗)= trace(V ∗V)= ∑ j f j, f j . 2 2 (b) Since S is self adjoint, ( S F ) = trace(SS )= trace(S ), and we get ∗ 2 2 2 trace(S )= trace(V ∗V(V ∗V)∗)= V ∗V F = ∑∑ f j, fk . j k | |

~~2.13 Let V =[ f j]. By Parseval: VV ∗ = IH , so L = LVV ∗, and we have~~

~~trace(L)= trace(V ∗LV)= ∑e∗j (V ∗LV)e j = ∑e∗j (V ∗Lf j)= ∑ Lf j, f j . j j j~~

2.14 Let(ek) be an orthonormal basis. 1 1 2 1 3 (a) e1,( ) e1,( ) e1,... e2,...,ed is a normalised tight frame for H . { √2 √2 √2 } ∪ { } (b) The trace formula (2.9) holds for inﬁnite frames, since (by Plancherel) trace(S)= ∑ Sek,ek = ∑ ∑ ek, f j f j,ek = ∑ ∑ f j,ek ek, f j = ∑ f j, f j . k k j j k j 2 Therefore ∑ f j = dA < ∞, and in particular f j 0. j → (c) Since f j 0 for an inﬁnite frame, it can not have equal (nonzero) norms. → 17 Solutions 477

~~2.15 If f j = c, j, then the trace formula (2.9) gives ∀ dA ∑ f 2 = nc2 = dA = c = . j n j ⇒~~

2 dA 2.16 Since ( f j) is equiangular, Exer. 2.15 gives f j, f j = f j = , j. Suppose n ∀ that f j, fk = c, j = k, then the variational formula (2.10) gives | | ∀ dA 2 1 dA 2 A d(n d) (n2 n)c2 + n = n = c = − . − n d n ⇒ n n 1 − 2.17 (a) By Theorem 2.1, Φ is a normalised tight frame for H if and only if the Hermitian matrix P = Gram(Φ) is an orthogonal projection of rank d, i.e., its eigenvalues are 0 and 1, with exactly d being nonzero. (b) If all eigenvalues of SΦ are 1, it is the identity, giving the Parseval identity. (c),(d) The singular values of V and V ∗ are precisely the (nonzero) eigenvalues of the frame operator S = VV ∗, or, equivalently the Gramian P = V ∗V.

2.18 (a)= (b) Since VV ∗ = I, V ∗ f ,V ∗g = VV ∗ f ,g = f ,g , f ,g. ⇒ 2 ∀ 2 (b)= (c) Take f = g, so that V ∗ f = V ∗ f ,V ∗ f = f , f = f , f . ⇒ 2 2 2 2 ∀ (c)= (a) Expand V f =( ( f , f j ) 2) = ∑ f , f j = f , f . ⇒ ∗ j | | ∀ 2.19 If there is such a unitary U, then clearly g j,gk = Uf j,Ufk = f j, fk , j,k. ∀ Now suppose g j,gk = f j, fk , j,k holds. Assume, wlog, that f1,..., fd is a basis for H , and deﬁne a linear map∀ U by { }

~~U : H K , Uf j := g j, 1 j d. → ≤ ≤ ∑d α ∑d β Then U is unitary, since for h1 = j=1 j f j, h2 = k=1 k fk,~~

~~Uh1,Uh2 = ∑α jβk g j,gk = ∑α jβk f j, fk = h1,h2 . j,k j,k~~

~~For k > d and 1 j d, we have ≤ ≤~~

Ufk gk,g j = Ufk,g j gk,g j = Ufk,Uf j gk,g j − − − = fk, f j gk,g j = 0, − which implies Ufk = gk since g1,...,gd is a basis for K . { } 2.20 (a) Φ =( f j) j J and Ψ =(gk)k K are unitarily equivalent up to reordering if ∈ ∈ and only if there is a bijection σ : J K for which Φ and (gσ j) j J are unitarily equivalent, i.e., by Corollary 2.1, → ∈

~~Gram(Φ)= Gram((gσ j) j J)=([gk]Q)∗[gk]Q = Q∗ Gram(Ψ)Q, ∈ 478 17 Solutions~~

K J where Q = Qσ F × is the permutation matrix Q :=[eσ j] j J. ∈ ∈ (b) They are projectively unitarily equivalent up to reordering if and only if gσ j = α jUf j, j, with σ : J K a bijection, α j = 1, j, and U unitary, i.e., (α j f j) j J ∀ → | | ∀ ∈ and (gσ j) j J are unitarily equivalent, i.e., with Λ := diag(α j) j J and Q as above ∈ ∈

~~Λ ∗ Gram(Φ)Λ = ([ f j]Λ)∗[ f j]Λ = ([gk]Q)∗[gk]Q = Q∗ Gram(Ψ)Q.~~

~~(c) If these are projectively unitarily equivalent as above, then~~

~~gσ j,gσk = α jUf j,αkUfk = α jαk f j, fk = gσ j,gσk = f j, fk . ⇒ | | | | The equiangular harmonic frames with the Gramians of (2.12) and (2.17), i.e.,~~

2 1 1 2 ω2 ω 3 3 3 3 −3 3 − − ω −ω2 2π 1 2 1 , 2 , ω := e 3 3 3 3  −3 3 3  − 1 1 −2  ω2 ω −2 3 3 3 − − −  3 − 3 3      satisfy this condition, but are not projectively unitarily equivalent. Remark: Projective unitary equivalence is determined by the m–products (see 8). § 2.21 (a) By Exer. 2.20, normalised tight frames Φ and Ψ are projectively unitarily equivalent (up to reordering) if and only if their Gramians PΦ and PΨ satisfy

Λ ∗PΦΛ = Q∗PΨ Q Λ ∗(I PΦ )Λ = Q∗(I PΨ )Q, ⇐⇒ − − i.e., the complementary normalised tight frames are projectively unitarily equivalent (up to reordering). (b) The complementary normalised tight frame of an equal–norm tight frame of n = d + 1 vectors for Fd consists of d + 1 scalars with modulus 1/√d + 1. Since all such frames are projectively unitarily equivalent, the result follows by (a). (c) We observe that up to a scalar multiple (α j) is given by

~~α jαk f j, fk = α jUf j,αkUfk = g j,gk .~~

Taking α1 = 1, we have ω ω2 g2,g1 1 + 2 g3,g1 1 + α2 = = = ω , α3 = = = ω, f2, f1 1 f3, f1 1 − − 1 with U given by U(α j f j)= g j, and so U =[α1 f1,α2 f2]− [g1,g2]= I. 2.22 Let a,b,c C be the complementary normalised tight frame. Since unitary maps preserve{ the}∈ Gramiam, we can assume that c 0, and since it is normalised a 2 + b 2 + c 2 = 1, thus the complementary frame≥ has frame operator | | | | | | V =[a,b, 1 a 2 b 2], a 2 + b 2 1. −| | −| | | | | | ≤ 17 Solutions 479

~~Thus all Gramians for normalised tight frames of three vectors for C2 are given by~~

1 a 2 ab a a 2 + b 2 −| | − 2 − | | | | I V ∗V = ba 1 b b a 2 + b 2 . −  − −| | − | | | |  a a 2 + b 2 b a 2 + b 2 a 2 + b 2 − | | | | − | | | | | | | |  2.23 Let P be the Gramian of a normalised tight frame. Then I P is the Gramian of the complementary tight frame, and these frames are unitarily− equivalent if and 1 only if their Gramians are equal (Corollary 2.1), i.e., P = I P. This implies P = 2 I, which is not an orthogonal projection matrix. − The projectively unitarily equivalent normalised tight frames of two vectors for F1 given by V =[ 1 , 1 b] and W =[ 1 , 1 b], b = 1, are complements of each √2 √2 √2 − √2 | | other. Further examples might be given by equiangular tight frames of 2d vectors for Fd (a calculation excludes the case d = 2).

2.24 The complementary frame to V =[ f j] is null(V)’ d n 2.25 (a) In terms of the synthesis matrix V =[ f j] j=1n C × , we seek a d n matrix V with columns of equal (nonzero) length, say d,∈ and rows of equal length× which are orthogonal. For n d = 1, we can take V to be any 1 n matrix with entries of unit–modulus. Suppose≥ n d 2, and write n = kd + d×+ r, 0 r < d. ≥ ≥ ≤ We seek a V of the form V = d[V1,V2,...,Vk,W], where each Vj is a unitary matrix (say I). This V has the desired properties provided the d (d + r) matrix W has × columns of length 1 (the same as those of Vj) and rows of equal length, i.e., there exists an equal–norm tight frame of d +r vectors in Fd. Such a frame is given by an orthonormal basis when r = 0, or by the complement of an equal–norm tight frame of d + r vectors in Fr, which we can construct by (strong) induction since r < d. (b) The following matlab code gives such a function function V = ENTF(n,d) if d==1, V=ones(1,n); end; if n==d, V=d*eye(d); end; if d=2*d & d>1, V=[d*eye(d) ENTF(n-d,d)]; end; Here eye(d) can be replaced by any d d unitary matrix and ones(1,n) any 1 n matrix with unit–modulus entries . × (c)× Now an example

~~take add complement √3 √3 0 columns 2 0 √3 √3 0 [1,1,1] 1− 1 2 021− 1 2 −→ − −→ −~~

take 3 0 √3 √3 0 add 30030 √3 √3 0 complement 0 3 − 1 1 2 columns V = 03003 − 1 1 2 ,  − −   − −  −→ 0 0 √5 √5 √5 −→ 00300 √5 √5 √5     2.26 (a) Each of the maps A j : f f , f j f j is a (bounded) positive operator since 2 → A j f , f = f , f j 0, and hence so are the differences Ft Fs, s < t (which are | | ≥ − 480 17 Solutions

ﬁnite sums of these). Clearly, F0 = OH and Fn = IH . ∑n (b) We have the telescoping sum j=1(Fj Fj 1)= Fn F0 = IH , and Q j := E j − − − − E j 1 (which satisﬁes Q∗j = Q j) is an orthogonal projection, since − 2 2 2 (E j E j 1) = E j E jE j 1 E j 1E j + E j 1 = E j E j 1 E j 1 + E j 1 = Q j. − − − − − − − − − − − −

~~Similarly, if j < k, then Q jQk = QkQ j = 0. (c) Let K˜ K be the orthogonal complement of the kernel of the linear map ⊂ n n L : K C : f ( f , f j ) . → → j=1~~

~~Since f ker(L) H = f = ∑ f , f j f j = 0, we have H K˜. Further, ∈ ∩ ⇒ j ⊂ ran(Qk) H = 0 , since otherwise, ran(Qk) H , so that PQk = 0, and ∩ { } ⊥~~

fk, fk fk =(Fk Fk 1) fk = PQk fk = 0, − − which contradicts fk = 0. Thus we may choose f ran(Qk), with Pf = 0, so that ∈ n Lf = LPf is a nonzero scalar multiple of the standard basis vector ek for C , and hence L maps onto Cn. A dimension count gives dim(K˜)= rank(L)= n, and so we may replace K by K˜. d (d) Suppose that ( f j) is a normalised tight frame for F , i.e., the rows of V = d n [ f1,..., fn] F × are orthonormal. Extend V by adding a further n d orthonormal rows, to obtain∈ a unitary matrix U. The columns of U are an orthonormal− basis for n F , and their orthogonal projection onto the ﬁrst d components gives ( f j).

~~2.27 Since F :=[ f j]= U + iV, U :=[u j], V :=[v j], the normalised tight frame condition for ( f j) is~~

FF∗ =(U + iV)(U + iV)∗ = UU∗ +VV ∗ + i(VU∗ UV ∗)= I, − i.e., UU∗ +VV ∗ = I, and VU∗ = UV ∗. The sequence of 2n real vectors has synthesis operator S =[U,V], which satisﬁes SS∗ = UU∗ +VV ∗ = I, and so is a tight frame. d d 2.28 Let ( f j) be a normalised tight frame for R . Write f C as f = u + iv, u,v Rd. Then ∈ ∈ 2 2 2 2 2 2 2 ∑ f , f j = ∑ u + iv, f j = ∑ u, f j + v, f j = u + v = f . j | | j | | j | | | |

~~2.29 (a) Let V =[ f j], W =[gk]. As in Exer. 2.13, we calculate~~

trace(M∗L)= trace(M∗LVV ∗)= trace((MV)∗LV)= ∑ Lf j,Mf j , j trace(M∗L)= trace(M∗WW ∗L)= trace((L∗W)∗M∗W)= ∑ M gk,L gk . k ∗ ∗ (b) Let L L (H ,K ). We verify the Parseval identity for L (applied to f ). Observe ∈

~~L,gk f ∗ HS = trace((gk f ∗)∗L)= trace( f jg∗L)= trace(g∗Lf j)= Lf j,gk , j j k k 17 Solutions 481 and so, by linearity, we have~~

~~∑ L,gk f j∗ HS(gk f j∗) f = ∑ Lf j,gk gk f , f j = ∑ f , f j ∑ Lf j,gk gk j k j k j k , , = ∑ f , f j Lf j = L ∑ f , f j f j = Lf . j j~~

2.30 (a) VV ∗ = I gives Vx = V(V ∗ f )= f , WAV ∗ = W(W ∗LV)V ∗ = L. (b) We have [αL + βM]= W ∗(αL + βM)V = αW ∗LV + βW ∗MV = α[L]+ β[M], and [L∗]= V ∗L∗W =(W ∗LV)∗ =[L]∗. (c) Suppose that M : K L , (hℓ)ℓ L is a normalised tight frame for L , and let → ∈ X =[hℓ]ℓ L. Then [ML]= X∗LMV =(X∗MW)(W ∗LV)=[M][L]. (d) With x∈=[ f ], part (a) gives Ax =[L][ f ]=[Lf ] and λx = λ[ f ]=[λ f ]. But f [ f ] is 1–1, so that Ax = λx if and only if Lf = λ f . → (e) The singular values of A are the positive square roots of the nonzero eigenvalues of A∗A =(W ∗LV)∗W ∗LV = V ∗L∗WW ∗LV = V ∗L∗LV, with the eigenspaces giving the corresponding singular vectors. If x is an eigenvector of A∗A for λ = 0, then f = Vx = 0, and we may apply (d). 482 17 Solutions Exercises of Chapter 3

3.1 Let g denote f f ,g j , so that A j = f jg . ∗j → ∗j (a) trace(A j)= trace( f jg )= trace(g f j)= f j,g j . ∗j ∗j 2 f ,g j (b) Since A f = f j(g f j)g f = f j,g j Af , take c = f j,g j , P : f f j. ∗j ∗j f j,g j → f , f j (c) Since A = g j f , we have P f = g j, so P = P if and only if f ,g j f j = ∗j j∗ ∗ g j, f j ∗ f , f j g j, f , i.e., f j and g j are scalar multiples of each other. ∀ (d) trace(A jA )= trace( f j(g gk) f )= gk,g j trace( f f j)= gk,g j f j, fk . k∗ ∗j k∗ k∗ 3.2 Least squares solution. (a) V(λa +(1 λ)b)= λVa +(1 λ)Vb = λ f +(1 λ) f = f . 1 − 1 − − (b) V(V ∗S− f )= VV ∗(VV ∗)− f = f . 1 1 (c) Since V ∗S− f is a solution, A = V ∗S− f +ker(V). Since ran(V ∗) is orthogonal to ker(V), for any a = V S 1 f + b A , b ker(V), Pythagorus gives ∗ − ∈ ∈ 2 2 1 2 2 ∑ c j = c = V ∗S− f + b , j | | so the unique solution of minimal ℓ2–norm is obtained by choosing b = 0. † 1 † 1 † 1 3.3 (a) With A := A∗(AA∗)− , we have AA = AA∗(AA∗)− = I, A A = A∗(AA∗)− A are Hermitian, and AA†A = IA = A, A†AA† = A†I = A†. (b) With P† := P, PP† = P†P = P is Hermitian, PP†P = P, P†PP† = P†. † 1 1 2 (c) Let A = Gram(Φ)= V ∗V, and A := Gram(Φ˜ )=(S− V)∗(S− V)= V ∗S− V, † † 2 1 S = VV ∗. We verify A is the pseudoinverse of A: AA = V ∗VV ∗S− V = V ∗S− V † 2 1 and A A = V ∗S− VV ∗V = V ∗S− V are Hermitian, and

~~† † 1 AA A =(AA )A =(V ∗S− V)V ∗V = V ∗V = A,~~

~~† † 1 2 2 † A AA =(V ∗S− V)V ∗S− V = V ∗S− V = A .~~

~~can 1 (d) The synthesis operator of Φ is S− 2 V, S = VV ∗, so by (a) and (c), we have~~

~~can 1 1 1 1 † Gram(Φ )=(S− 2 V)∗S− 2 V = V ∗S− V = V ∗(VV ∗)− V = V V 1 2 † = V ∗S− V = V ∗V(V ∗S− )= Gram(Φ)Gram(Φ) .~~

~~3.4 Expanding Tf in the normalised tight frame ( f j) gives~~

~~1 1 1 f = T − (Tf )= ∑ Tf , f j T − f j = ∑ f ,T ∗ f j T − f j. j j~~

~~1 Replace T ∗ by T − to obtain the other equality. Let V =[ f j]. Then frame operator of (T ∗ f j) is T ∗V(T ∗V)∗ = T ∗(VV ∗)T = T ∗T, and so the dual frame is~~

~~1 1 (T ∗T)− T ∗ f j = T − f j. 17 Solutions 483~~

2 2 2 2 2 3.5 Since ∑ j g,Qf j = ∑ j Q∗g, f j BΦ Q∗g BΦ Q g , we have 2| | | 2| ≤ ≤ Ψ BΦ BΦ Q . Similarly, ∑ j g,Qf j AΦ Q∗g . If g span , then g = Qf . ≤ † † | † | ≥ ∈ † † But Q = QQ Q =(QQ )∗Q =(Q )∗Q∗Q, and so we obtain g = (Q )∗Q∗Qf (Q )∗ Q∗g 2 † 2 2 † 2 ≤ Thus ∑ g,Qf j AΦ Q g , giving AΨ AΦ Q . j | | ≥ − ≥ − 2 n f , f j 3.6 (a)= (b) ∑ | | A, gives the lower frame bound. For the upper bound: ⇒ j=1 f 2 ≥ n 2 n 2 2 2 n 2 ∑ f , f j ∑ f f j = B f , B := ∑ f j < ∞. j=1 | | ≤ j=1 j=1 (b)= (e) (f) V ∗ is 1–1 if and only if its kernel is zero. The lower frame bound ⇒ ⇐⇒2 2 2 gives: V f = ∑ f , f j A f > 0, f = 0, i.e., ker(V )= 0. ∗ j | | ≥ ∀ ∗ (c) (d) (e) ( f j) spans H if and only if V is onto iff V ∗ is 1–1. ⇐⇒ ⇐⇒ 2 ∑n f , f j 2 (f)= (a) Since inf f =0 j=1 | 2| = inf g =1 g, f j , and g : g = 1 is com- ⇒ f | | { } pact (H is ﬁnite dimensional), the inﬁmum is attained, and so is nonzero.

~~3.7 Since S is positive deﬁnite, it is unitarily diagonalisable S = UΛU∗, where U =[u1,...,ud] and Λ = diag(λ1,...,λd), with λ j > 0, so that~~

~~2 Sf , f = UΛU∗ f , f = ΛU∗ f ,U∗ f = ∑λ j f ,u j . j | |~~

~~Let λmin = minλ j and λmax = maxλ j. Then we have~~

~~2 2 2 2 λmin f = λmin ∑ f ,u j Sf , f λmax ∑ f ,u j = λmax f , j | | ≤ ≤ j | |~~

~~with equality if (and only if) f is an eigenvector for λmin, λmax, respectively.~~

~~3.8 Let f = f j in the frame bound inequality~~

2 4 2 2 A f j f j + ∑ f j, fk B f j . ≤ k= j | | ≤ (a) Immediate. 2 (b) Equality in (a) if and only if ∑k= j f j, fk = 0, i.e., f j fk, j = k. 2 4 | 2 | ⊥ ∀ (c) Since f j < A gives f j < A f j , we must have 2 ∑ f j, fk > 0 = f j is not orthogonal to spank= j fk = k= j | | ⇒ ⇒

~~3.9 Let S = VV ∗. The commutativity relations of Exer. 3.11 give~~

~~1 1 ˜ 1 Φ † can 2 Φ † 2 [ f j]= S− V = V Gram( ) , [ f j ]= S− V = V(Gram( ) ) .~~

~~Expanding gives 484 17 Solutions~~

˜ 1 Σ Σ 1 ΣΣ 1 Σ [ f j]= S− V =(U1 U2∗U2 ∗U1∗)− V = U1( ∗)− U1∗(U1 U2∗) σ 2 σ σ = U1 diag(1/ 1 ,...)diag( 1,...)U2∗ = U1 diag(1/ 1,...)U2∗, 1 1 can 2 ΣΣ 1 2 Σ [ f j ]= S− V =(U1( ∗)− U1∗) (U1 U2∗) σ σ Σ = U1 diag(1/ 1,...)diag( 1,...) U2∗ = U1 diag(1,...)U2∗.

1 3.10 Let S = SΦ = VV ∗, where V =[ f j] j J, so that SΦ˜ = S− , and we have ∈ ˜can 1 1 1 1 can (a) [ f j ]=(S− )− 2 S− V = S− 2 V =[ f j ]. 1 1 2 (b) GΦ = Gram(Φ) = V ∗V, GΦ˜ = Gram(Φ˜ )=(S− V)∗S− V = V ∗S− V, and can 1 1 1 GΦcan = Gram(Φ )=(S− 2 V)∗S− 2 V = V ∗S− V, so we obtain

~~2 1 2 GΦ GΦ˜ = V ∗VV ∗S− V = V ∗S− V = GΦcan = V ∗S− VV ∗V = GΦ˜ GΦ .~~

~~1 1 (c) Use f˜j = SΦ− f j and SΦ− = SΦ˜ .~~

3.11 (a) Observe SV =(VV ∗)V = V(V ∗V)= VG, and use induction. (b) By (a), p(S)V = V p(G) for all polynomials p. Now use Weierstrass density. (c) Similarly, approximate A† via the Tikhonov regularisation formula

† 1 1 A = lim(A∗A + δI)− A∗ = lim A∗(AA∗ + δI)− . (17.1) δ 0 δ 0 → → 2 3.12 P := VW = WV = P , and P K = I, gives P = I K (VW )= P. ∗ ∗ ∗ | | ∗ 1 3.13 Let S be the frame operator of ( f j). Since S− is a positive deﬁnite operator, 1 1 1 can can can 2 can f j, f˜j = f j,S f j = S 2 f j,S 2 f j = f , f = f 0. Since ( f ) − − − j j j ≥ j is a normalised tight frame, it satiﬁes (2.9), i.e., ∑ f can 2 = d. j j 3.14 Let T be the frame operator for ( f j), i.e., T = S K . The assertions amount to 1 † † | the claim T − = S K . Now ran(S )= ran(S∗)= ran(S)= K , so for y = Sx K , † | † † ∈ we have (TS K )y = SS Sx = Sx = y, i.e., TS K = IK . Also see Corollary 3.5. | | Φ 3.15 The ( j,k)–entry of Gram( ) is fk, f j = v∗j vk = vk,v j (so the columns of L give a copy of Φ). Now apply Exer. 3.14 with V = L.

3.16 Let S = VV ∗. (a) Ψ is a frame since its synthesis operator QV is onto H . 1 1 1 1 1 1 1 (b)g ˜ j =(QVV ∗Q∗)− Qf j =(Q∗)− S− Q− Qf j =(Q∗)− S− f j =(Q∗)− f˜j. can 1 1 1 1 can (c) g j =(QVV ∗Q∗)− 2 Qf j =(QSQ∗)− 2 QS 2 (S− 2 f j)= Uf j , where the matrix 1 1 1 1 1 1 U =(QSQ∗)− 2 QS 2 is unitary, since UU∗ =(QSQ∗)− 2 QS 2 S 2 Q∗(QSQ∗)− 2 = I. 1 1 2 1 1 1 1 (d) Let Q = cU, then (QSQ∗)− 2 QS 2 =(c USU∗)− 2 cUS 2 = US− 2 U∗US 2 = U, 1 1 j j with US 2 U∗ =(USU∗) 2 following from US U∗ =(USU∗) , j = 1,2,... and the Weierstrass density theorem. d n 3.17 Deﬁne W˜ F by W = U1WU˜ . Then ∈ × 2∗ ˜ ˜ WW ∗ = U1∗WU2U2∗W ∗U1 = U1∗(WW ∗)U1 = AU1∗U1 = AI, 17 Solutions 485

~~so that~~

~~Σ ˜ Σ ˜ min V W F = min U1 U2∗ U1WU2∗ F = min W F . d n ˜ d n ˜ d n W F × ,A>0 − W F × ,A>0 − W F × ,A>0 − ∈WW AI ∈˜ ˜ ∈˜ ˜ ∗= WW∗=AI WW∗=AI~~

~~Let Σ = diag(σ1,σ2,...) and W˜ =[w jk]. Then~~

Σ W˜ F = trace((Σ W˜ )(Σ W˜ )∗)= trace(ΣΣ ∗ W˜ Σ ∗ ΣW˜ +WW˜ ) − − − − − ∗ = trace(VV ∗ W˜ Σ ∗ ΣW˜ ∗ + AI) − − = trace(VV ∗)+ dA trace(W˜ Σ ∗ + ΣW˜ ∗), − where d trace(W˜ Σ ∗ + ΣW˜ ∗)= 2 ∑ σ jℜ(w˜ jj). j=1

The condition W˜ W˜ ∗ = AI, says that W˜ has orthogonal rows of length √A, so the unique choice maximising ℜ(w˜ jj) isw ˜ jj = √A,w ˜ jk = 0, j = k. Thus V W F is minimised, when − dA 2∑σ j√A = ∑(A 2σ j√A) − j j − √ 1 ∑ σ is maximised, which occurs when A = d j j. Thus, with A deﬁned as above, the unique minimiser W is given by

~~W = U1[√AI,0]U2∗ = √AU2∗.~~

This gives Corollary 3.3, since the singular values σ j of V are the square roots of the eigenvalues λ j of the frame operator S = VV ∗, and the canonical tight frame for V =[ f j] is given by (3.7). ℜ∑ ∑ ∑ λ µ 3.18 It sufﬁces to show that j f j,g j k1 k2 k1 √ k2 uk1 ,vk2 . By Cauchy–Schwarz, ≤ | | 1 1 2 2 2 2 ∑ f j,uk1 vk2 ,g j ∑ f j,uk1 ∑ vk2 ,g j j | | ≤ j | | j | | = SΦ uk ,uk SΨ vk ,vk = λk √µk , 1 1 2 2 1 2 which gives

~~ℜ ∑ f j,g j ∑ f j,g j = ∑∑∑ f j,uk1 uk1 ,vk2 vk2 ,g j j ≤ j j k1 k2 ∑∑ uk1 ,vk2 ∑ f j,uk1 vk2 ,g j ≤ | | | | k1 k2 j~~

~~∑∑ uk ,vk λk √µk . ≤ | 1 2 | 1 2 k1 k2 486 17 Solutions~~

3.19 (a) Let e =(1,...,1). It is easy to check that Me =(na a + 1)e, and so − e is an eigenvector for na a + 1. If v e, i.e., v,e = v1 + + vd = 0, then − ⊥ (Mv) j = av1 + +avd av j +v j =(1 a)v j, so that v is an eigenvector for 1 a. (b) The matrix M is positive− semideﬁnite− if and only if its eigenvalues are nonneg-− 1 ative: na a + 1,1 a 0, i.e., n− 1 a 1. Its rank is the number of nonzero − − ≥ − ≤ ≤ 1 eigenvalues which is either 1 (a = 1), n 1(a = n− 1 ), or n (otherwise). 1 − − (c) If rank(M)= n 1, then a = − , and so M is the Gramian of the n = d + 1 − n 1 vertices of a simplex in Rd, see (2.14).− † 1 (d) If rank(M)= n, then M is invertible, and so the Gramian of (u˜ j) is M = M− . Let B be the n n matrix with diagonal entries b, and off diagonal entries c. Then × b +(n 1)ac, j = k; (MB) jk = − c + ab +(n 2)ac, j = k −

~~It is easy to check that MB = I, i.e., B = Gram((v˜j)), when~~

na 2a + 1 a b = − , c = − . (1 a)(na a + 1) (1 a)(na a + 1) − − − − Thus (v˜j) is (a scalar multiple) of an isogonal conﬁguration of vectors. 3.20 Recall a frame Φ is real if and only if Gram(Φ) has real entries. (a) Use (3.4), and observe that the Tikhonov regularisation formula (17.1) implies the pseudoinverse of a matrix is real if and only the matrix itself is. (b) Use (3.9), i.e., Gram(Φcan)= Gram(Φ)Gram(Φ˜ ). (c) No. Suppose that Φ is a basis for which the Gramian has complex entry (easily constructed), then Φcan is an orthonormal basis, which is real.

3.21 Let V =[ f j]. The dual frame of (Pf j) and (Pg j) are equal if and only their 1 1 synthesis operators are, i.e., (PV(PV)∗)− PV = P(V ∗)− , which gives (a) PV(V )=(PVV P)P PS = PSP. ∗ ∗ ⇐⇒ (b) Taking the adjoint of (a) gives SP =(PS)∗ =(PSP)∗ = PSP. (c) By (a) and (b), we have PS = SP, and conversely, given this, we may obtain (a) and (b) by right and left multiplication by P. Thus (a),(b),(c) are equivalent. (d) Since ran(P)= H , (d) follows from any of (a),(b),(c). Conversely, suppose that SH H . Then SPf H , so that SPf = PSPf , f , which is (b). ⊂ ∈ ∀ 3.22 Let L = UΣV ∗ be a singular value decomposition, where the diagonal entries of Σ are σ1,...,σm, and V =[v1,...,vm]. Since U is unitary, we have

~~m 1 Σ Σ σ m σ 2 2 2 Lx = U V ∗x = V ∗x = ( j x,v j ) j=1 = ∑ j x,v j . j=1 | | Hence (since V is unitary), we obtain 17 Solutions 487~~

~~m 1 2 2 Lx M ∑ x,v j = M V ∗x = M x , M := maxσ j, ≤ j=1 | |~~

with equality if and only if x,v j = 0 for every singular value σ j less than the maximum, i.e., x is right–singular vector for the maximum singular value. Similarly, since (kerL)⊥ is the orthogonal direct sum of the right–singular vector subspaces for the nonzero singular values, for a nonzero x (kerL) ∈ ⊥ m 1 2 2 Lx m ∑ x,v j = m V ∗x = m x , m := min σ j, σ 0 ≥ j=1 | | j=

with equality if and only if x,v j = 0 for every singular value σ j greater than m, the minimum nonzero singular value, i.e., x is right–singular vector for m. 3.23 Since the norm is unitarily invariant, it sufﬁces to consider the ﬁrst inequality. Let m and M be the smallest and largest singular values of L, so, by Exer. 3.22,

m x Lx M x , x. ≤ ≤ ∀ Then (L U)x Lx Ux = Lx x , so that − ≥ − − Lx L U max 1 max m 1 , M 1 = L I . − ≥ x=0 x − ≥ {| − | | − |} − 3.24 (a) If L : H K is a linear map between ﬁnite–dimensional Hilbert spaces, → with singular values σ1 σ2 σm, then there are the sharp inequalities ≥ ≥≥

min σ j x Lx maxσ j x , (kerL)⊥ = ran(L∗). σ j=0 ≤ ≤ j ∀ Σ (see Exer. 3.22 for details). Suppose that V = U1 U2∗ is a singular value decomposi- Σ λ λ tion of V, where = diag(√λ1,..., λd) 0 and 1,..., d > 0 are the eigenvalues of S = VV . The inequalities (and their sharpness) then follow by taking L to be ∗ V, V ∗, S = VV ∗ and G = Gram(Φ)= V ∗V, and observing that these have singular value decompositions Σ Σ ΣΣ Σ Σ V = U1 U2∗, V ∗ = U2 ∗U1∗, S = U1 ∗U1∗, G = U2 ∗ U2∗.

The equivalence of these inequalities with the frame bounds follows from Exer. 3.7. Σ 3.25 Let V = U1 U2∗ be a singular value decomposition of V =[ f j]. Then Σ Σ ΣΣ SΦ = VV ∗ = U1 U2∗U2 ∗U1∗ = U1 ∗U1∗,

~~which is a unitary diagonalisation of SΦ . Let Λ := ΣΣ ∗ = diag(λ1,...,λd). Since 1 Λ 1 Λ 1 SΦ˜ = SΦ− =(U1 U1∗)− = U1 − U1∗, we have 488 17 Solutions~~

1 αSΦ + βSΦ˜ + γI = αU1ΛU∗ + βU1Λ − U∗ + γU1U∗ 1 1 1 1 = αΛ + βΛ − + γI (U1 is unitary) αλ βλ 1 γ = max j + j− + ( is the spectral norm). 1 j d ≤ ≤ Φ Σ Σ Φ˜ Σ Σ † Σ Σ † We have Gram( )= U2 ∗ U2∗, Gram( )=(U2 ∗ U2∗) = U2( ∗ ) U2∗, and Φcan Φ Φ˜ Σ Σ Σ Σ † Gram( )= Gram( )Gram( )= U2 ∗ ( ∗ ) U2∗.

1/2 † 1 Since Σ has the block form Σ = Λ 0 , Σ ∗Σ = diag(Λ,0), (Σ ∗Σ) = diag(Λ − ,0), and we obtain α Gram(Φ)+ β Gram(Φ˜ )+ γ Gram(Φcan) Λ 1 Λ − I = U2 α + β + γ U∗ 0 0 0 2 Λ 1 Λ − I 1 = α + β + γ = αΛ + βΛ − + γI . 0 0 0 3.26 Since the function λ λ 1 , λ > 0, is increasing, and changes sign at λ = 1, → − λ the maximum is either 1 A, where A 1, and we need to prove A − ≤ 1 B A A − (√AB 1)(√AB + A2) 0, A − ≥ √AB ⇐⇒ − ≤ or it is B 1 , where B 1, and we need to prove − B ≥ 1 B A B − (√AB 1)(√AB + B2) 0. − B ≥ √AB ⇐⇒ − ≥ The ﬁrst clearly holds for B 1, and the second for A 1 (with equality iff AB = 1). It therefore sufﬁces to consider≤ the case A < 1, B > 1.≥ Here either 1 1 1 1 1 A = B A √AB 1, A − A − 1/A ≥ − B ⇐⇒ ≤ B ⇐⇒ ≤ 1 1 1 1 1 A = B A √AB 1, A − A − 1/A ≤ − B ⇐⇒ B ≤ ⇐⇒ ≥ and again the corresponding inequality holds via

1 1 1 1 1 1 B A A √AB, B B A − , A − ≥ A − B ≥ A − B − B ≥ − ≥ √AB (with equality iff AB = 1). Alternatively, suppose that the inequality is false, i.e., 17 Solutions 489 1 1 B A A , B < − , − A − B √AB then one obtains the contradiction 1 2 1 2 (B A)2 (A2 + B2)(AB 1)2 A + B < 2 − − < 0. − A − B AB ⇐⇒ A2B2 3.27 (a) Consider the first minimisation. Assume t > 0. From B A 2 tB 1 2 tA 1 2 = t − t , | − | −| − | A + B − A + B we conclude that 2 2 tB 1 tA 1 , t , tB 1 tA 1 , t . | − |≥| − | ∀ ≥ A + B | − |≤| − | ∀ ≤ A + B Hence B A min max tA 1 , tB 1 = min tB 1 = min (tB 1)= − , t 2 {| − | | − |} t 2 | − | t 2 − A + B ≥ A+B ≥ A+B ≥ A+B B A min max tA 1 , tB 1 = min tA 1 = min (1 tA)= − , t 2 {| − | | − |} t 2 | − | t 2 − A + B ≤ A+B ≤ A+B ≤ A+B 2 each of which is attained if and only if t = A+B , which gives the first minimum. For the second one, observe that 0 < 1 1 , and apply the first to obtain B ≤ A 1 1 1 1 B A minmax 1 , 1 = A − B = − , t>0 { tA − tB − } 1 + 1 A + B A B 1 2 A+B which is attained if and only if t = 1 1 , i.e., t = 2AB . A + B λ 2λ 1 1 λ 1 1 λ 1 (b) Since [A,B] R : c 1 and [ B , A ] R : − c2 − 1 attain their maxima at→ an endpoint,→ | from− (3.31)| and (3.32)→ we obtain→ | − |

dist(cΦ,Φcan)= max c2A 1 , c2B 1 , {| − | | − |} 1 1 1 dist( Φ˜ ,Φcan)= max 1 , 1 . c { c2A − c2B − } By part (a), with t = √c, the minima of these distances over c> 0 is B A , and this A−+B is attained as claimed. (c) The four quantities in the min–max are continuous functions of t, which are monotone at every point except the value of t which makes them zero. Hence, for the minima to occur two of these functions must equal the maximum (if not, then the one nonzero value equal to the maximum could be reduced slightly, so as to reduce the maximum of all four values). Thus it sufﬁces to consider only the values 490 17 Solutions of t which make (at least) two of these functions equal, and the common value M (a potential min–max). Calculations (or a sketch) shows the minimum M is given by

1 B 1 tB 1 = 1 , M = 1 at t = . | − | tA − A − √AB (d) By the formulas in (b) above, we have 1 max dist(cΦ,Φcan),dist( Φ˜ ,Φcan) c 1 1 = max c2A 1 , c2B 1 , 1 , 1 . {| − | | − | c2A − c2B − } By part (c), with t = c2, the minimum of this over c > 0 is B 1, and this is A − attained precisely when c2 = 1 . √AB (e) We observe that Gram(cUΦcan)= c2 Gram(Φcan), and the eigenvalues of SΦ 1 satisfy B = = λ1 λ2 λd = A, so Exer. 3.25 gives A ≥ ≥ can 2 can dist(Φ,cUΦ )= SΦ cI = Gram(Φ) c Gram(Φ ) − − 2 2 1 2 = max λ j c = max A c , c , j | − | | − | A − Φ˜ Φcan λ 1 2 2 1 2 dist( ,cU )= max j− c = max A c , c . j | − | | − | A − By the reasoning of (c), the minimum occurs when 1 A2 + 1 A c2 = c2 c2 = . | − | A − ⇐⇒ 2A 3.28 Since 1 1 , we may assume that B 1. Suppose that B ≤ A ≥ 1 1 A 1 B 1 max A 1 , B 1 = max 1 , 1 = max | − |, | − | . | − | | − | A − B − A B Consider cases. If A 1 B 1, then A 1, which gives | − |≥| − | ≤ A 1 A 1 = | − | = A = 1, B = 1. | − | A ⇒ If A 1 B 1 , then there are two possibilities | − |≤| − | B 1 B 1 = | − | = B = 1, A = 1, | − | B ⇒ A 1 B 1 = | − | = AB A = A 1 = AB 1,2A 1 . | − | A ⇒ | − | | − | ⇒ ∈ { − } 17 Solutions 491

When AB = 2A 1, we have − 2A 1 A B = − = (A 1)2 = A2 2A + 1 0 = A = 1, B = 1, ≤ A ⇒ − − ≤ ⇒ and so AB = 1 in all cases. can 3.29 Recall, from (3.13), that ran(V ∗)= ran(Gram(Φ )). (a)= (c) W = QV gives ran(W ∗)= ran((QV)∗)= ran(V ∗Q∗) ran(V ∗). ⇒ † ⊂ † (c)= (a) Suppose that ran(W ∗) ran(V ∗). Let Q = WV . Since V maps ran(V) ⇒ ⊂ † onto ker(V)⊥ = ran(V ∗), we can decompose x as x = V a+b, b ker(V) ker(W). Then QVx = W(V †VV †)a +WV †Vb = WV †a + 0 = Wx, x, so∈ that QΦ ⊂= Ψ. (b) (c) The Gramians are orthogonal projections (determined∀ by their ranges), and⇐⇒ so their product PQ equals Q if and only if ran(Q) ran(P). (c)= (d) W is 1–1 and ran(W ) ran(V )= ker(V)⊂, so that VW is 1–1. ⇒ ∗ ∗ ⊂ ∗ ⊥ ∗ (d) (e) Immediate, since VW g = ∑ g,g j f j. ⇐⇒ ∗ j 3.30 Let V =[ f j], W =[g j]. Then the closeness condition is

(W V)c λ Vc , c. − ≤ ∀ (a) If Vc = 0, then this implies (W V)c = 0, and so Wc =(W V)c +Vc = 0. Thus ker(V) ker(W), and Exer. 3.29− gives Ψ = QΦ. − (b) In view of⊂ (a), W = QV and the closeness condition is equivalent to

~~(Q I)Vc λ Vc , c ker(V)⊥ = ran(V ∗). − ≤ ∀ ∈~~

Since Φ is a frame, S = VV ∗ is onto, and so Vc above can be an arbitrary element of H . Thus the closeness condition is equivalent to λ being an upper bound for Q I , and cl(Ψ,Φ)= Q I . (c) Since− Vc = (V W)c−+Wc (V W)c + Wc , we have − ≤ − λ (W V)c λ (V W)c + Wc = (V W)c Wc , − ≤ − ⇒ − ≤ 1 λ − Φ Ψ Φ Ψ λ i.e., is close to , with cl( , ) 1 λ . ≤ − 3.31 (a) We have M Sx MSx x (I MS)x (1 I MS ) x , i.e., ≥ ≥ − − ≥ − − 1 I MS Sx − − x , x X. ≥ M ∀ ∈ (b) F(g1) F(g2) = (I MS)(g1 g2) I MS I MS . (c) By (a), MS− is bounded below:− MSx− ( ≤1 −I MS)−x , and so ≥ − −

~~1 1 (MS)− . ≤ 1 I MS − − 3.32 (a) The eigenvalues of S = SΦ satisfy A λ j B, so that ≤ ≤ 492 17 Solutions~~

2 2λ j 2λ I S = max 1 max 1 . − A + B j − A + B ≤ A λ B − A + B ≤ ≤ The maximum of λ 1 2λ above occurs at the endpoints, and so it is → − A+B 2A 2B B A max 1 , 1 = − . − A + B − A + B A + B (b) The error in the fixed point iteration gn+1 := F(gn) for a Banach contraction F with constant κ and fixed point g can be estimated by either of κn n gn g κ g g0 , gn g g1 g0 . − ≤ − − ≤ 1 κ − − 2 B A Using the estimate κ = I S − < 1, from the second we obtain − A+B ≤ A+B n B A n A−+B 2 h B A gn g B A h = − . − ≤ 1 − A + B A A + B − A+B Remark. The choice h = Sf , i.e., g = f , gives the frame algorithm (3.37), for which the first estimate gives B A n gn f f − . − ≤ A + B 1 3.33 Let S = UΛU∗ be a unitary diagonalisation of S, and f (x) := x− 2 , x > 0. Since

~~j j ( 1) (2 j)! 2 j+1 f ( )(x)= − x 2 , 22 j j! −~~

~~A+B the Taylor series expansion of f about c = 2 is~~

∞ j 2 j+1 1 ( 1) (2 j)! A + B 2 A + B j 2 − x− = ∑ −2 j 2 x j=0 2 ( j!) 2 − 2 ∞ 2 ( 1) j(2 j)! 2 j = ∑ −2 j 2 1 x , A + B j=0 2 ( j!) − A + B which (by the ratio test) is absolutely convergent for 0 < x < A + B. By Exer. 3.32, 1 B A 2 2 I S − = 1 A = 1 B − A + B ≤ A + B − A + B − A + B 1 and so the series for S− 2 is absolutely convergent. Its partial sums satisfy

~~n j n j 2 (2 j)! 2 2 (2 j)! 2 Λ ∑ 2 j 2 I S =U ∑ 2 j 2 I U∗. A + B j=0 2 ( j!) − A + B A + B j=0 2 ( j!) − A + B 17 Solutions 493~~

Since the eigenvalues of S (diagonal entries of Λ) satisfy 0 < A λ j B < A + B, 1 ≤ ≤ the partial sums in the converge to Λ − 2 (by considering entries), and so the 1 { } 1 series converges to S− 2 = UΛ − 2 U∗.

~~3.34 Let V =[ f j], and use equivalence (f) of Proposition 3.4. We calculate~~

~~1 1 1 1 1 1 I PΦ = I V ∗S− V = 1 1 1 = vv∗, v := (1,1,1), − − 3 1 1 1 √3~~

~~Thus all possible 2 3 matrices L(I PΦ ) are determined by z = Lv, i.e., × − 1 L(I PΦ )= L(vv∗)=(Lv)v∗ = zv∗ = [z,z,z], − √3~~

~~1 2 and so all possible duals are given by W =[ f˜1 + w, f˜2 + w, f˜3 + w], w = z R . √3 ∈ (b) Let W =[g j]. Then (g j) is a pseudo–dual if and only if~~

~~(2/√3)det(VW ∗)= w11w22 w21w12 + w13(w21 w22)+ w23(w12 w11) = 0. − − −~~

The left hand side above is zero if g1 = g2, i.e., w11 = w12 and w21 = w22, otherwise the coefﬁcient of w13 or of w23 is nonzero, and g3 =(w13,w23) can be chosen to make the left hand side nonzero. 1 3.35 (a) This follows immediately from WV ∗ =(WQ− )(VQ∗)∗. (b) The synthesis operator of the canonical dual of the frame given by VQ∗ is

~~1 U =(VQ∗QV ∗)− VQ∗.~~

By (3.40), this dual frame uniquely minimises U over all dual frames to VQ∗. 1 Thus, by (a), W = UQ =(VQ∗QV ∗)− VQ∗Q is the unique dual frame of Φ which 1 minimises W Q = WQ = U . − 3.36 (a) To ( f , fˆj )=( f , f j ) ( f , f j fˆj ) apply the triangle inequality − − 1 1 1 2 2 2 2 2 2 ∑ f , fˆj ∑ f , f j + ∑ f , f j fˆj √B f + √R f , j | | ≤ j | | j | − | ≤ and the reverse triangle inequality

1 1 1 2 2 2 2 2 2 √A f √R f ∑ f , f j ∑ f , f j fˆj ∑ f , fˆj . − ≤ j | | − j | − | ≤ j | | 2 2 Thus (√A √R) and (√B + √R) are lower and upper frame bounds for ( fˆj). − (b) Let V =[ f j], W =[ fˆj]. Then the synthesis operator of the canonical dual of Φˆ is U =(WW ) 1W, and so UV I = UV UW U V W . Now ∗ − ∗ − ∗ − ∗ ≤ ∗ − ∗ 494 17 Solutions

~~1 2 2 V ∗ W ∗ = sup (V ∗ W ∗) f = sup ∑ f j fˆj, f √R. − f =1 − f =1 j | − | ≤~~

Similarly, U = U∗ is bounded by the square root of an upper frame bound for the canonical dual of Φˆ (see Exer. 3.24), which by (a) is 1/(√A √R). Thus ≤ − UV ∗ I < 1, so that the canonical dual of Φˆ and Φ are approximate duals, if √R/(√−A √R) < 1, which holds if and only if R < A . − 4 3.37 (a) Let V and W = cV be the synthesis operators of Φ and cΦ. Then

~~WV ∗ I = cVV ∗ I = max cλ j 1 = max cA 1 , cB 1 − − j | − | {| − | | − |}~~

(see Exer. 3.27 for details). We consider cases: c 0 gives cB 1 = 1 cB 1, 0 < c < 2 gives 1 < cA 1 cB 1 < 1, and≤ c 2 gives| −cB| 1− 1.≥ The B − − ≤ − ≥ B − ≥ minimum occurs when cA 1 = cB 1 , i.e., c = 2 . | − | | − | A+B 3.38 By replacing H by V + W , we may assume without loss of generality that H is ﬁnite dimensional. Let d1 = dim(V ), d2 = dim(W ) and n = dim(H ). (a) Suppose that V W = W V = 0. Then the algebraic direct sums V a W ∩ ⊥ ∩ ⊥ ⊕ ⊥ and W a V are subspaces H , and so their dimensions satisfy ⊕ ⊥

d1 +(n d2) n, d2 +(n d1) n = d1 = d2. − ≤ − ≤ ⇒ Thus a dimension count gives (ii). Conversely, if (ii) holds, then (i) is immediate. (b) The ﬁrst part was proved in (a). Let Q = P∗, P = PV W , and w W , v⊥ V ⊥. , ⊥ Then ∈ ∈ Qv⊥,z = v⊥,Pz = 0, z H = Qv⊥ = 0, ∀ ∈ ⇒ Qz,w⊥ = z,Pw⊥ = z,0 = 0, w⊥ W ⊥ = Qz (W ⊥)⊥ = W . ∀ ∈ ⇒ ∈ Since P and Q have the same rank, we conclude that Q W is a bijection W W . 2 2 | → Since P = P implies Q = Q, we conclude that Q W = IW . Thus Q = PW V . | , ⊥ 3.39 Let V =[ f j] and W be the synthesis operator of a ﬁnite frame for W . Then † [g j]= W(V ∗W) , and so the canonical oblique dual of (g j) has synthesis operator

~~† † † † † U = V((W(V ∗W) )∗V) = V((W ∗V) W ∗V) = V(W ∗V) W ∗V.~~

We observe that ker(W ∗)= W ⊥, so that W ∗ is 1–1 on ran(V) (since W ⊥ V = 0), ∩ † and ker(W ∗V)= ker(V). Now right multiplication of the above by (W ∗V) , using † † † † † A AA = A , gives U(W ∗V) = V(W ∗V) . Thus it sufﬁces to show that U = V on

† † (ran((W ∗V) )⊥ = ker((V ∗W) )= ker(W ∗V)= ker(V), which follows immediately from the formula for U (which has right factor of V). 3.40 (a) A†AA† = A† gives P2 = V(ΛV)†ΛV(ΛV)†Λ = V(ΛV)†Λ = P. (b) AA†A = A gives ΛPV = ΛV(ΛV)†ΛV = ΛV, i.e., Λ(Pf )= Λ( f ), f V . (c) By (b), ΛPV = ΛV so that rank(P) rank(ΛV), while P = V(Λ∀V)†∈V gives ≥ 17 Solutions 495 rank(P) rank((ΛV)†)= rank(ΛV), so rank(P)= rank(ΛV). ≤ (d) If dim(L V ) = dim(Λ V ) dim(V ), then Λ V is 1–1, so that ΛV maps | | ≥ † | ker(V)⊥ bijectively onto ΛV , hence (ΛV) maps ΛV onto ker(V)⊥, and so P maps onto V = V(ker(V)⊥). (e) If dim(L ) dim(L V ), then ran(Λ)= ran(Λ V ), so for x X there is v V with Λx = Λv,≤ which gives| ΛP(x v)= ΛV(ΛV)†|Λ(x v)= 0∈= Λ(x v), while∈ (b) gives ΛPv = Λv. Adding these− gives ΛPx = ΛP (x −v)+ v = Λx.− { − } † (f) If λk( f )= f ,gk , then Λ = W , where W =[gk], so P = V(W V) W . ∗ ∗ ∗ 3.41 If Φ =( f j) is a basis and V =[ f j], then the orthonormal basis given by the 1 1 canonical tight frame has synthesis operator S− 2 V = VG− 2 , where G = Gram(Φ) (see Exer. 3.11). For the Legendre weight w = 1 on [ 1,1], applying Gram–Schmidt and then Lowdin¨ orthogonalisation to the ﬁrst four normalised− monomials gives following sequences of polynomials As one would expect, for Lowdin¨ orthogonalisa-

Fig. 17.2: Gram–Schmidt and Lowdin¨ orthogonalisation of the ﬁrst ﬁve monomials tion, different sequences of polynomials of degree 0,1,...,n give different orthonormal bases (in contrast to Gram–Schmidt). Starting instead with the Bernstein basis, one obtains Notice that here, Gram–Schmidt does not preserve the symmetries of the Bern- stein basis, whilst Lowdin¨ does. These latter polynomials presumably maintain good conditioning and have the advantage of being orthogonal. They also seem to maintain the partition of unity property 496 17 Solutions

~~Fig. 17.3: Gram–Schmidt and Lowdin¨ orthogonalisation of the Bernstein basis 17 Solutions 497 Exercises of Chapter 4~~

~~4.1 (a) (b) (c) Since f X is determined by (λ( f ))λ X , ⇐⇒ ⇐⇒ ∈ ∈ ′~~

VΛ = IX f = IX f = VΛ f = ∑ λ j( f ) f j, f ⇐⇒ j ∀ λ( f )= ∑ λ j( f )λ( f j)=(∑ λ( f j)λ j)( f ), f X, λ ⇐⇒ j j ∀ ∈ ∀ λ = ∑ λ( f j)λ j, λ. ⇐⇒ j ∀ (a) (d) (e) (f) Since V is onto X, we can multiply and cancel by V on the right,⇐⇒ and since⇐⇒ Λ⇐⇒is 1–1, we can multiply and cancel by Λ on the left. Thus

2 VΛ = IX G = Λ(VΛ)V = Λ(IX )V = G (multiply by Λ and V) ⇐⇒ V = V(ΛV)= VG (cancel Λ) ⇐⇒ Λ =(ΛV)Λ = GΛ (multiply by Λ, cancel V). ⇐⇒ Φ 1 1 1 1 4.2 Since c j ( f )= f ,S− f j = S− 2 f ,S− 2 f j , where (S− 2 f j) is the canonical tight frame, we compute

~~1 1 1 1 f ,g Φ = ( S− 2 f ,S− 2 f j ),( S− 2 g,S− 2 f j ) 1 1 1 1 1 1 = ∑ S− 2 f ,S− 2 f j S− 2 f j,S− 2 g = S− 2 f ,S− 2 g (Plancherel). j~~

4.3 By Proposition 4.9, ξ j = ξk if and only if ψ j = ψk = v. In this case, write ξ j = ξk as fv, to obtain x = ∑ξ j(x)ψ j = ∑ mv fv(x)v. j v Θ ∈ Since the barycentric coordinates for Θ are unique, we have that mv fv = ℓv. 498 17 Solutions Exercises of Chapter 5

~~5.1 Suppose that ( f j) is a disjoint union m f j : j Jm of tight frames, with frame ∪ { ∈ } bounds Am. Then for f j, j Jm, we have ∈~~

~~Sf j = ∑ f j, fk fk = ∑ f j, fk fk = Am f j. k k Jm ∈~~

Conversely, suppose that each f j is an eigenvector of S. Let λm be the eigenvalues { } of S, and Φm := j : Sf j = λm f j . Since S is Hermitian, its eigenspaces span(Φm) { } are orthogonal, so that for f j Φm, we have ∈

~~SΦ f j = ∑ f j, fk fk = ∑ f j, fk fk = Sf j = λm f j, m f Φm k k∈ and so (by linearity) each Φm is a tight frame. Thus ( f j) is the disjoint union of the tight frames Φm.~~

~~5.2 (a) Since f + g,φ j = f ,φ j and f + g,ψk = g,ψk , we obtain~~

~~S( f + g)= ∑ f + g,φ j φ j + ∑ f + g,ψk ψk = SΦ ( f )+ SΨ (g). j k~~

(b) The dual vector to φ j + 0 Φ Ψ is ∈ ∪ 1 1 1 S− (φ j + 0)= SΦ− (φ j)+ SΨ− (0)= φ˜j + 0, and similarly for 0 + ψk Φ Ψ. ∈ ∪ (c) Since φ j +0,0+ψk = 0, the entries of the Gramian are zero if they correspond to a vector in Φ and one in Ψ.

~~5.3 Let P1 and P2 be the orthogonal projections onto H1 and H2. Then~~

~~( f + g,φ j + ψ j )=( f ,φ j )+( g,ψ j )= V ∗ f +W ∗g, so that~~

(φ j + ψ j) is a frame for H1 H2 ⊕ J U =[φ j + ψ j] : F H1 H2 is onto ⇐⇒ → ⊕ U = V P1 +W P2 : f + g ( f + g,φ j + ψ j ) is 1–1 ⇐⇒ ∗ ∗ ∗ → J J ran(V ∗) ran(W ∗)= 0 (since V : H1 F and W : H2 F are 1–1). ⇐⇒ ∩ ∗ → ∗ → 2 2 2 5.4 (a) Since f + g = f + g , f + g H1 H2, we may add the frame bounds ∀ ∈ ⊕

~~2 2 2 2 AΦ f SΦ f , f BΦ f , AΨ g SΨ g,g BΨ g , ≤ ≤ ≤ ≤ to obtain the sharp inequalities 17 Solutions 499~~

~~2 2 min AΦ ,AΨ f + g S( f + g), f + g max BΦ ,BΨ f + g . { } ≤ ≤ { }~~

~~Thus the sequence is frame for H1 H2, with the above frame bounds. ⊕ 1 (b) Suppose S( f + g)= SΦ ( f )+ SΨ (g). Then S (and hence S− ) maps H j H j, so that →~~

1 1 1 f + g = S− (SΦ ( f )+ SΨ (g)) = S− SΦ ( f )+ S− SΨ (g) 1 1 = f = S− SΦ ( f ), g = S− SΨ (g) ⇒ 1 1 1 1 = S− ( f )= SΦ− ( f ), S− (g)= SΨ− (g). ⇒ 1 1 1 Similarly, S− ( f +g)= SΦ− ( f )+SΨ− (g) implies that S( f )= SΦ ( f ), S(g)= SΨ (g).

~~5.5 Suppose that ( f j) is a simple lift, say f j = φ j +αψ H K , ∑ φ j = 0. Then ∈ ⊕ j 2 ∑ f j = α J ψ = 0, ∑ f j, fk = α J ψ,φk + αψ = α J ψ,ψ . j | | j | | | | | |~~

~~Conversely, suppose that ψ := ∑ f j = 0 and ∑ f j, fk = C, k. Let P be the j j ∀ orthogonal projection onto ψ⊥. Then ( f j) is the direct sum~~

~~f j,ψ C f j = Pf j +(I P) f j = Pf j + ψ = Pf j + ψ, − ψ,ψ ψ,ψ which is a simple lift of (Pf j). 500 17 Solutions Exercises of Chapter 6~~

6.1 Since A is Hermitian, it is unitarily diagonalisable, with real eigenvalues λ1,...,λn, and an orthonormal basis of eigenvectors u1,...,un . The inequality { } follows from the Cauchy-Schwartz inequality applied to (1,...,1) and (λ1,...,λn), with equality if and only if λ1 = = λr = c. Thus d r λ A = ∑ ju ju∗j = ∑ cu ju∗j = c[u1,...,ur][u1,...,ur]∗ = cUU∗. j=1 j=1

2 6.2 With K := ∑ f j, f j = ∑ f j > 0, we have j j 2 2 2 2 2 K (n n)max f j, fk + ∑ f j, f j ∑∑ f j, fk . − j=k | | | | ≥ | | ≥ d j j k 6.3 In matlab, we take a quotient, with value d (with equality for a tight frame) c=[]; m=100; d=7; ≤ for n=d:m, V=rand(d,n)-1/2; c(n)=trace(V’*V)ˆ2/norm(V’*V,’fro’)ˆ2; end; c(1)=0, c(m+1)=d, plot(c) Large numbers of vectors do give frames which are close to tight. The fact that tight frames correspond to cubature rules gives a heuristic explanation for this. A more precise treatment of random frames is given in [Ehl12]. 6.4 (a) For F1 = F, the condition for perfect tightness becomes

4 4 v j 1 v j = , n 1 d n 1 − − which can only hold (for v j = 0) if d = 1. (c) An equal–norm frame is perfect if and only if it is an equiangular tight frame. (d) This is an open question.

6.5 If ( f j) is orthogonal, i.e., f j, fk = 0, j = k, then FP( f1,..., fn)= n, while any other choice gives a larger value.| | 6.6 The lower bound, and equality, follow directly from Theorem 6.1. For the upper bound, Cauchy–Schwarz gives 17 Solutions 501

n n 2 n n 2 2 ∑ j 1 ∑ f j, fk ∑ j 1 ∑ f j fk = k=1 | | = k=1 = 1, n 2 2 n 2 2 ∑ f j ≤ ∑ f j j=1 j=1 with equality if and only if all f j and fk are scalar multiples of each other. 6.7 The integral over S is invariant under a unitary change of variables U. (a) Choose U with Ue1 = y/ y , so that y,Ux = y e1,x , which gives 2 2 2 py dσ = (py U)dσ = y pe dσ = y pe dσ = c y . ◦ 1 1 S S S S Thus 1 2 1 2 ∑ y,φ j = ∑ py(φ j)= py dσ = c y , n | | n j j S i.e., Φ is a tight frame. d d (b) Since Π (R )= span py : y R , it sufﬁces to consider p = py 2◦ { ∈ } n n 2 1 2 c py dσ = c y = c ∑ y,φ j = ∑ py(φ j). A | | A S j=1 j=1 6.8 By Proposition 6.1, we seek a balanced equal–norm tight frame of n (distinct) vectors for Rd. We must have n > d, since the sum of the vectors in an orthonormal d basis is nonzero. Let V =[v1,...,vn] be a balanced normalised tight frame for R . The condition V be balanced is that it is orthogonal to the normalised tight frame U =[1/√n,...,1/√n] (see Lemma 5.1). Let W =[w1,...,wn] be the complement n 1 d of the direct sum of U and V, which is a balanced tight frame for R − − . Since there is no balanced tight frame of n vectors for R1 when n is odd, we cannot have n 1 d = 1 and n odd. In all the other cases, 2–designs can be obtained by taking an−unlifted− real cyclic harmonic frame of n vectors for Rd (see Corollary 11.3).

~~6.9 (a) For t = 1 or d = 1, we have ct (d,R)= ct (d,C). For t > 1, we have~~

t ct (d,C) ct 1(d,C) d+t 1 ct 1(d,R) t(d + 2t 2) = − − = − − , c (d,R) 2t 1 c (d,C) t(d + 2t 2)+(t 1)(d 1) t ct 1(d,R) d+2(−t 1) t 1 − − − − − − and so, by induction, there is strict inequality for t > 1, d > 1. 1 3 5 (t 1) t t t 2t (b) Using c t (d,R)= − , x⊗ 2 x⊗ 2 = x⊗ , ∑k fk = n, we get the 2 d(d+2) (d+t 2) ⊗ following conditions, which are equivalent− to being an order t spherical half–design,

n n t 2 1 3 5 (t 1) ∑ ∑ f j, fk = n − , d(d + 2) (d +t 2) j=1 k=1 − n t t 1 d(d + 2) (d +t 2) t d x = x,x 2 = − ∑ x, f j , x R , n 1 3 5 (t 1) | | ∀ ∈ − j=1 502 17 Solutions

1 n x t dσ(x)= ∑ f t . ⊗ n j⊗ S j=1 α α α α (c) For a monomial x , ( x) =( 1)| |x , so the space of odd polynomials is Π Π Π , and by− (6.26),− we have 1◦ ⊕ 3◦ ⊕ 5◦ ⊕ n/2 α 1 α α x dσ(x)= 0 = ∑ ( f j) +( f j) , α = 1,3,5,.... n − | | S j=1 Thus (6.33) integrates the odd monomials, and hence all odd polynomials. 2 6.10 (a) The claim is that for all bivariate polynomials p Πn 1(R ) ∈ − 1 2π 1 n 2π 2π p(x,y)dσ(x,y)= p(cosθ,sinθ)dθ = ∑ p cos j,sin j . 2 2π n n n S(R ) 0 j=1 By identiﬁng (x,y) R2 with z = x + iy C, this can be written as ∈ ∈ n 2π 1 j j 2 q(z,z)dσ(z)= ∑ q ω ,ω , ω := e n , q Πn 1(R ). n ∀ ∈ − S(C) j=1 Let q(z,z) := zα zβ , α + β < n. Then α = β or α β 0 (mod n), so that − ≡ n n n 1 1 α β 1 α β 0, α = β; ∑ q(z,z)= ∑(ω j) (ω j) = ∑ ω( ) j = − − α β n j=1 n j=1 n j=1 1, = .

α β σ This equals the integral S(C) z z d (z), which is given by taking d = 1 in (6.25). 2 (b) It sufﬁces to show (6.9), i.e., that (w j) is a spherical (t,t)–design for R . Using 2 θ cos2θ+1 the trigonometric identity cos = 2 , and the cubature rule of (a), we have

π π π π 2t 2t ∑∑ w j,wk = ∑∑ cos j cosk + sin j sink j k | | j k n n n n π π 2t cos 2 ( j k)+ 1 t = ∑∑ cos( j k) = ∑∑ n − j k − n j k 2 π cos 2 j + 1 t 2π θ t 2 1 n 2 1 cos + 1 θ = n ∑ = n π d . n j 2 2 0 2 By (6.25), with α =(t,0) and d = 2, the integral simpliﬁes to

1 2π cosθ + 1 t 1 2π θ 2t dθ 1 π dθ = cos = (cosx)2t dx 2π 2 π 2 2 π 0 0 0 2π 1 1 2t ( 2 )t 1 3 2t 3 2t 1 = (cosx) dx = 2 = − − = ct (d,R). 2π 0 ( ) 2 4 2t 2 2t 2 t − 17 Solutions 503

2 6 6.11 We verify (6.39), noting that c2(d,C)= d(d+1) , c3(d,C)= d(d+1)(d+2) . 4 3 (a) d2 (d2 1) 1 + 14 = 2d = 2 (d2)2. − √d+1 d+1 d(d+1) 4 (b) md (md d) 1 + 14 +(d 1)04 = m(m + d 1)= 2 (md)2 if and − √d − − d(d+1) only ifm = d + 1(d =1). 6 2 2 (c) d(d + 1) (d(d + 1) d) 1 + 16 +(d 1)06 =(d + 1)2 = 6d (d+1) if − √d − d(d+1)(d+2) and only if d= 2. 6.12 Verify that 40(1 + 27( 1 )2t )= c (4,C)402 holds for t = 1,2,3. √3 t

~~1 d(n d) 1 6.13 The common angle is v ,v = − = , j = k (see Exer. 2.16). j k d n 1 √d+2 | | − Thus~~

4 4 2 1 d(n d) ∑∑ v j,vk =(n n) − + n, j k | | − d n 1 − 2 4 1 3 2 c2(d,R) ∑ v = n , ℓ d(d + 2) ℓ 1 and a simple calculation shows that these are equal if and only if n = 2 d(d + 1).

6.14 (a) It is easy to see the vectors in Φ0 =(v j) (64 of type 1 and 56 of type 2) satisfy v j,vk 0,1,2 . For v j ﬁxed, a simple calculation shows that the number | |∈{ } of times v j,vk takes the values 0,1,2 is 63,56,1. Thus we verify (6.39) as follows | | 6 6 6 6 ∑∑ v j,vk = 120(63 0 + 56 1 + 1 2 ) j k | | 2 1 3 5 3 2 6 = (120 2 ) = c3(8,R) ∑ v . 8 10 12 ℓ ℓ (b) By (a), Φ0 and hence Φ integrates all homogeneous polynomials of degree 6, and therefore of degrees 0,2,4 also. Since Φ is centrally symmetric it integrates all homogeneous polynomials of odd order, and in particular those of orders 1,3,5,7. t t 6.15 Take f = x⊗ , g = y⊗ in the Plancherel identity (2.4) for the tight frame for Symt (H ), using (2.9) and (6.16) to determine A

t t 1 t t t t t + d 1 t 2 x⊗ ,y⊗ = ∑ x⊗ , f ⊗ f ⊗ ,y⊗ , − A = ∑ f ⊗ . A j j t ℓ j ℓ After simpliﬁcation by (6.17) this gives (a). Setting y = x gives (b). 6.16 Recall (Exer. 2.11) that H has inner product given by x,y := x,y , x,y H , and so the apolar inner product on Symt (H ) Symt (H ) satisﬁes ∈ ⊗ t t t t t t t x⊗ y⊗ ,v⊗ w⊗ = x,v y,w =( x,v y,w ) , x,y,v,w H . ⊗ ⊗ ◦ ∈ 504 17 Solutions

Write ξ = a b. Using (6.29), we obtain − t t t t 2t a,a = x⊗ x⊗ ,y⊗ y⊗ dσ(x)dσ(y)= x,y dσ(x)dσ(y) ⊗ ⊗ ◦ | | S S S S 2t = y ct (d,F)dσ(y)= ct (d,F), S 1 t t t t 1 2t ⊗ ⊗ b,b = 2 ∑∑ f j⊗ f j , fk⊗ fk = 2 ∑∑ f j, fk , C j k ⊗ ⊗ ◦ C j k | | 1 t t t t σ 1 2t σ a,b = ∑ x⊗ x⊗ , f j⊗ f j⊗ d (x)= ∑ x, f j d (x) C ⊗ ⊗ ◦ C | | j S j S 1 2t = ∑ f j ct (d,F)= ct (d,F)= b,a , C j

~~t t which combine to give the result for ξ. Let Px := ,x x . The Frobeneous inner ⊗ ⊗ product between Px and Py is (cf. Exer. 3.1)~~

~~t t t t 2t Px,Py F = trace(PxP∗)= x⊗ ,y⊗ y⊗ ,x⊗ = x,y . y | | Using this, a simple modiﬁcation of the above argument gives the result for~~

~~1 n Q = Px dσ(x) ∑ Pf . − C j S j=1~~

β 6.17 (a) The linear functionals f ∂ α ∂ f (0), α = t, β = r are dual to these monomials. Thus counting gives the→ dimension. | | | | (b) Recall (Exer. 2.11) that v,w := w,v . Thus the inner product induced by taking t r the apolar inner product on Sym (H ∗) and Sym (H ∗) is given by

t r t r t t t r ,v x, , ,w y, = ,v ⊗ ,x ⊗ , ,w ⊗ ,y ⊗ ◦ ⊗ ⊗ = ,v , ,w t ,x , ,y r = w,v t y,x r. = w,v t x,y r. (c) Let p = ,v t x, r and y = w in (6.71), to obtain p, ,w t w, r = w,v t x,w r = p(w). ◦ (d) If p Π (Cd) is orthogonal to P, then by (c) we have ∈ t◦,r t r d p(w)= p, ,w w, = 0, w = P⊥ = 0 = Π ◦ (C )= P. ∀ ⇒ ⇒ t,r (e) It sufﬁces to consider p = ,v t x, r, q = ,w t y, r. We have

~~∂,v ( z,w )=(∑ v j∂ j)(∑ wkzk)= v1w1 + + vdwd = w,v , j k~~

~~∂,x ( z,y )=(∑ x j∂ j)(∑ y zk)= x1y + + xdy = x,y . j k k 1 d 17 Solutions 505~~

Sinceq ˜(z)= z,w t y,z r = z,w t z,y r, and p(∂)q˜ is a constant, we obtain p(∂)q˜(z)= ∂,v t ∂,x r( z,w t z,y r)= t! w,v t r! x,y r = t!r! p,q . ◦ 6.18 Making the substitution (6.41) in Theorem 6.7 gives: n 2t 2t d (a) ct (d,F) x = ∑ w j x,φ j , x F . j=1 | | ∈ n t t t d (b) ct (d,F) x,y = ∑ w j x,φ j φ j,y , x F . j=1 ∈ n (c) p(x)dσ(x)= ∑ w j p(φ j), p Π ◦ (S). ∀ ∈ t,t S j=1 n t t t t (d) x⊗ x⊗ dσ(x)= ∑ w j (φ ⊗ φ j⊗ ). ⊗ j ⊗ S j=1 n t t t t (e) ,x⊗ x⊗ dσ(x)= ∑ w j ,φ ⊗ φ ⊗ . j j S j=1 n n 2 2 (f) g x,y dσ(y)dσ(x)= ∑ ∑ w jwk g φ j,φk , g Πt (R). | | | | ∀ ∈ S S j=1 k=1 2t t 2t 2t 6.19 (a) Since x =(x1 + + xd) , x,v =(v1x1 + vdxd) , we have ∂ ∂ 2 2t t 1 2t t 1 t 2 x = 2tx j x − , x = 2t x − + 2tx j(2(t 1)x j x − ), ∂x j ∂x j − ∂ ∂ 2 2t 2t 1 2t 2t 2 x,v = 2tv j x,v − , x,v = 2tv j(2(t 1)v j) x,v − . ∂x j ∂x j − 2 2 (b) Since z = z1z1 + + zdzd, z,v =(z1v1 + + zdvd)(z1v1 + + zdvd) | | 2t 2(t 1) 2t 2t 2 2(t 2) ∂ j( z )= tz j z − , ∂ j∂ j( z )= t z − +tz j(t 1)z j z − , − 2t 2(t 1) ∂ j( z,v )= t z,v − v j z,v , | | | | 2t 2(t 1) 2(t 2) ∂ j∂ j( z,v )= t z,v − v jv j +t(t 1) z,v − z,v v jv j z,v . | | | | − | | (c) Applying the Laplacian in each case gives

n 2(t 1) 1 2(t 1) 2 ct (d,R)2t(d + 2t 2) x − = ∑ 2t(2t 1) x, f j − f j , − ∑ f 2t − | | ℓ ℓ j=1 n 2(t 1) 1 2 2(t 1) 2 ct (d,C)4t(t + d 1) z − = ∑ 4t z, f j − f j , − ∑ f 2t | | ℓ ℓ j=1 which is the Bessel identity for t replaced by t 1, since − d + 2t 2 t + d 1 ct (d,R) − = ct 1(d,R), ct (d,C) − = ct 1(d,C). 2t 1 − t − − 506 17 Solutions

~~Remark. This (with induction) provides an alternative proof of Proposition 6.2. 2t 2t 6.20 We verify the generalised Bessel identity (6.31). Let C = ∑ j v j = ∑k wk . Then~~

2t 1 1 2t 1 2t ct (d,F) x = ∑ x,v j + ∑ x,wk 2 C j | | C k | | 1 2t 2t d = ∑ x,v j + ∑ x,wk , x F , 2C j | | k | | ∀ ∈ 2t 2t where 2C = ∑ v j + ∑ wk . j k n 6.21 Suppose that ( f j) j=1 is a weighted (t,t)–design with the minimal number 2t Π d of vectors. We claim that the polynomials x x, f j in t◦,t (F ) are linearly independent, and hence obtain the result by→ a dimension | | count using (6.16) and (6.70). First scale ( f j) so that the generalised Bessel identity (6.31) becomes

~~n 2t 2t d x = ∑ x, f j , x F . j=1 | | ∀ ∈~~

2t Now, by way of contradiction, suppose that ∑ c j x, f j = 0, where c j R, and, j | | ∈ without loss of generality, c j cn = 1, j. Subtracting this from the above gives ≤ ∀ n 1 n 1 2t − 2t − 2t d x = ∑ (1 c j) x, f j = ∑ x,v j , x F . j=1 − | | j=1 | | ∀ ∈

1 n 1 where v j :=(1 c j) 2t f j. Thus, by the generalised Bessel identity (6.31), (v j) − is − j=1 a (t,t)–design of n 1 vectors for Fd, which contradicts the minimality of n. − 6.22 The generalised Plancherel identity (6.32) implies the holomorphic polynomial t t x x,y is in the span of the polynomials x x, f j , 1 j n. By Exer. 6.17, the→ polynomials x x,y : y Fd span Π (→Cd )= Π (C≤d), with≤ a similar result { → ∈ } t◦ t◦,0 for Rd, and so we have

d t + d 1 n dim(Π ◦(F )) = − . ≥ t d 1 − 6.23 Suppose that ( f j) is the tight frame corresponding to a (t,t)–design (g j), i.e., t 1 t f j = g j − g j (see Proposition 6.2). Taking norms gives f j = g j , and hence t 1 g j = f j/ f j −t . Substituting this into (6.39) gives the condition n n 2t n 2 f j fk fℓ 2t ∑ ∑ t 1 , t 1 = ct (d,F) ∑ t 1 . j=1 k 1 f −t f −t 1 f −t = j k ℓ= ℓ 17 Solutions 507

2 6.24 (a) Px,Py = trace(xx∗yy∗)= trace(y∗xx∗y)= x,y y,x = x,y . 2 | | (b) By (a), we have ρ(Px,Py) = Px Py,Px Py = 2 2 Px,Py . ρ −2 − − x y 2 (c) By (a) and (b), we have (Px,Py) = 2 2 Px,Py = 2 2 x , y . 1 1 − − | | (d) Clearly trace(Px d I)= 1 d d = 0, and we calculate 1 1− 2 − 2 (Px I) (Py I) = Px Py,Px Py = 2 2 Px,Py = ρ(Px,Py) . − d − − d − − − 6.25 For F = R,C, each p = 0 can be expressed as a ﬁnite sum of monomials: j α β p(x)= ∑c jx (c j = 0), p(z)= ∑ cα,β z z (cα,gb = 0). j (α,β)

~~Since these monomials are linearly independent, we have~~

~~j j j p(ax)= ∑(a c j)x = p(x), a = c ja = c j, a = p is even, j ∀ ⇒ ∀ ⇒~~

y j+a y j = b = yk+a yk = (y j yk)(ya 1)= 0, − − ⇒ − − so that either j = k or a = 0. Thus for (a,b) =(0,0) there is at most one solution to f (x + a) f (a)= b, and so f is 1–uniform. − 6.27 The equations have d + 2d(d 1)= d(2d 1) solutions of the form − − (w,x,y,z)=(w,w,w,w), (w,x,y,z)=(w,x,w,x) or (w,x,x,w).

~~We can rewrite the equations as~~

w z = y x = a, f (w) f (z)= f (y) f (x)= b, − − − − for some a G, b H. We now treat cases. For a = 0, we have z = w, y = x. For a = 0, we have∈ w =∈z + a, y = x + a, so that f (z + a) f (z)= f (x + a) f (x)= b − − and the 1–uniformity of f implies z = x, and so w = y. Thus there are only the d(2d 1) solutions ﬁrst mentioned. − 6.28 (a) Subtracting the component of the force in the direction a gives

~~EFF(a,b)= v v,a a = a,b ( b,a a b). − −~~

~~(b) Since a j is a λ–eigenvector of the frame operator S, and EFF(a j,a j)= 0, 508 17 Solutions~~

∑ EFF(a j,ak)= ∑ a j,ak ( ak,a j a j ak)= Sa j,a j a j Sa j = λa j λa j = 0. k= j k − − − ∂ ∂ 6.29 A simple calculation, using ∂z z = 0, ∂z z = 1, gives ∂ ∂ δ ∂ v j,vk = ∂ ∑vsjvsk = kβ vα j. vαβ vαβ s Therefore (using the chain rule), we calculate ∂ ∂ p 2(t 1) ∂ (V)= ∑∑t v j,vk − ∂ v j,vk vk,v j vαβ j k | | vαβ 2(t 1) = ∑∑t v j,vk − δkβ vα j vk,v j + v j,vk δ jβ vαk j k | | 2(t 1) 2(t 1) = ∑t v j,vβ − vα j vβ ,v j + ∑t vβ ,vk − vβ ,vk vαk j | | k | | 2(t 1) = 2t ∑ v j,vβ − vβ ,v j vα j, j | | ∂ ∂ g 2(t 1) 2(t 1)δ 2(t 1) ∂ (V)= ∑t vℓ − ∂ vℓ,vℓ = ∑t vℓ − ℓβ vαℓ = t vβ − vαβ . vαβ ℓ vαβ ℓ 6.30 To ﬁnd the entries of the Hessian, we use the Wirtinger calculus:

~~∂ 1 ∂ ∂ = + i , ∂z 2 ∂x ∂x which gives~~

∂ 2h ∂ 2h ∂ 2h ∂ 2h = 2ℜ , = 2ℑ , ∂xab∂xαβ ∂vab∂xαβ ∂yab∂xαβ ∂vab∂xαβ ∂ 2h ∂ 2h ∂ 2h ∂ 2h = 2ℜ , = 2ℑ . ∂xab∂yαβ ∂vab∂yαβ ∂yab∂yαβ ∂vab∂yαβ We now take h = p and h = g. By Exer. 6.29, we have ∂ ∂ p ℜ p 2(t 1) ∂ (V)= 2 ∂ (V) = 2t ∑ v j,vβ − vα j vβ ,v j + vα j v j,vβ , xαβ vαβ j | | ∂ ∂ g g 2(t 1) (V)= 2ℜ (V) = t vβ − (vαβ + vαβ ). ∂xαβ ∂vαβ Differentiating these gives 17 Solutions 509 ∂ 2 ∂ p 2(t 1) ∂ ∂ (V)= 2t ∑ v j,vβ − ∂ vα j vβ ,v j + vα j v j,vβ vab xαβ j | | vab ∂ ℜ 2(t 2) + 2t ∑2 vα j vβ ,v j (t 1) v j,vβ − ∂ v j,vβ vβ ,v j j − | | vab 2(t 1) = 2t ∑ v j,vβ − vα jδbjvaβ + δaα δbj v j,vβ + vα jδbβ vaj j | | 2(t 2) + 4t ∑ℜ vα j vβ ,v j (t 1) v j,vβ − v j,vβ δbjvaβ + δbβ vaj vβ ,v j j − | | 2(t 1) 2(t 1) = 2t vb,vβ − vαbvaβ + δaα vb,vβ + 2tδbβ ∑ v j,vβ − vα jvaj | | j | | 2(t 2) + 4t(t 1) vb,vβ − ℜ vαb vβ ,vb v β vb,vβ − | | a δ 2(t 2)ℜ + 4t(t 1) bβ ∑ v j,vβ − vα j vβ ,v j vaj vβ,v j , − j | | ∂ 2 ∂ g 2(t 1) (V)= t vβ − (vαβ + vαβ ) ∂vab∂xαβ ∂vab ∂ 2(t 2) +t(t 1) vβ − (vαβ + vαβ ) vβ ,vβ − ∂vab 2(t 1) 2(t 2) = t vβ − δaα δ β +t(t 1) vβ − (vαβ + vαβ )δ β v β . b − b a Similarly, since 2ℑ(z)= i(z z), we have − ∂ ∂ p ℑ p 2(t 1) ∂ (V)= 2 ∂ (V) = 2t ∑ v j,vβ − i vα j v j,vβ vα j vβ ,v j , yαβ vαβ j | | − ∂ ∂ g g 2(t 1) (V)= 2ℑ (V) = t vβ − i(vαβ vαβ ). ∂yαβ ∂vαβ − Differentiating these gives

∂ 2 p 2(t 1) δ δ δ δ ∂ ∂ (V)= 2t ∑ v j,vβ − i aα bj v j,vβ + vα j bβ vaj vα j bjvaβ vab yαβ j | | − 2(t 2) + 4t ∑ℑ vα j vβ ,v j (t 1) v j,vβ − v j,vβ δbjvaβ + δbβ vaj vβ ,v j j − | | 2(t 1) 2(t 1) = 2ti vb,vβ − δaα vb,vβ vαbvaβ + 2tiδbβ ∑ v j,vβ − vα jvaj | | − j | | 2(t 2) + 4t(t 1) vb,vβ − ℑ vαb vβ ,vb v β vb,vβ − | | a δ 2(t 2)ℑ + 4t(t 1) bβ ∑ v j,vβ − vα j vβ ,v j vaj vβ,v j , − j | | ∂ 2 g 2(t 1) 2(t 2) (V)= tδaα δbβ i vβ − + 2tℑ(vαβ )(t 1) vβ − δbβ vaβ . ∂vab∂yαβ − 510 17 Solutions

From these formulas one can calculate the Hessian matrix of p and g. To ﬁnd the d 2 Hessian matrix of f := p ct (d,F )g , use the fact − ∂ 2 ∂ ∂g ∂ 2g ∂g ∂g (g2)= 2g = 2g(V) + 2 , r,s X. ∂r∂s ∂r ∂s ∂r∂s ∂r ∂s ∈ 6.31 (a) Let f1 =(1,0), f2 =(1,1). Then

~~2 2x2 + 2xy + y2 4 2 g(x,y)= , g(1,0)= , g(0,1)= . 3 x2 + y2 3 3~~

~~2 (b) Expand ∑ c j f j = ∑ c j f j,∑ ck fk = ∑ ∑ c jck f j, fk , so that j j k j k~~

~~2 ∑c j f j ∑∑ c j ck f j, fk ∑∑ c j f j, fk ck f j, fk . ≤ | || || | ≤ | | | | | | j j k j k By Cauchy–Schwarz, using the fact that both norms are equal, we have~~

~~2 2 2 ∑c j f j ∑∑ c j f j, fk = ∑ c j ∑ f j, fk . j ≤ j k | | | | j | | k | |~~

1 1 2 2 2 2 2 2 ∑c j f j ∑∑ c j ck f j, fk ∑∑ c j ck f j, fk j ≤ j k | || || | ≤ j k | | | | | | 1 2 2 2 = ∑ c j f j, fk . j | | | | 2 2 2 2 1 Now take c j = x, f j in ∑ c j x, f j x (∑ c j )( f j, fk ) 2 . | j | ≤ j | | | | 17 Solutions 511 Exercises of Chapter 7

~~7.1 (a) A block matrix calculation gives~~

U11 U12 I 0 (I 0) =(U11 U12)=(I 0) = U = . U21 U22 ⇒ 0 U22 (b) Since dim(U(Fn)) dim(U(Fn d)), we calculate − − 1 1 1 dim(N d )= n(n 1) (n d)(n d 1)= d(2n d 1), n,R 2 − − 2 − − − 2 − −

~~2 2 dim(N d )= n (n d) = d(2n d)= 2dim(N d )+ d. n,C − − − n,R 7.2 (a) The derivative~~

~~2 2 2 2 2 f ′(t)= t 1 t ( v w )+(1 2t )α , t = 1, √1 t2 − − − − can be zero only when~~

t2(1 t2)β =(1 2t2)2α2, β :=( v 2 w 2)2. − − − For 4α2 + β = 0, this gives 1 r β t2 = ± , r := . 2 β + 4α2 For this t, we obtain 1 r f (t)= ( v 2 + w 2) ( v 2 w 2)+ 1 r2α. 2 ± 2 − − This is clearly maximised by the choice for which ( v 2 w 2)= β (and is minimised by the other choice). Thus possible local maximum± − and minimum values are 1 r 1 1 β + 4 α α f (t)= ( v 2 + w 2) β + 1 r2α = ( v 2 + w 2)+ ± | | . 2 ± 2 − 2 2 β + 4α2 (b) The choices of α that give the maximum and minimum value of the possible local maximum value and possible local minimum value from (a) are

~~α = σ v,w = v,w , α = σ v,w = v,w , | | −| | which give the values 1 1 f (t)= ( v 2 + w 2) β + 4α2. 2 ± 2 512 17 Solutions~~

(c) We have f (0)= w 2, f (1)= v 2. Since β + 4α2 β = v 2 w 2 , the maximum and minimum values of (b) are the global maximum≥ and| minimum. − | 7.3 (a),(b) Since the vectors of V (k) satify (7.7), the condition (i) gives

~~n n (k) 2 ∑ v j = ∑ a j. j=k+1 j=k+1~~

~~(k) 2 (k) Suppose that v < ak 1 and (a) does not hold, i.e., v = 0, j k + 2. Then k+1 + j ≥ n n (k) 2 (k) 2 ∑ v j = vk+1 < ak+1 ∑ a j, j=k+1 ≤ j=k+1~~

(k) 2 which is a contradiction. Now suppose that v > ak 1 and (b) does not hold, k+1 + (k) 2 i.e., v > ak 1, j k + 2. Then we again obtain a contradiction j + ∀ ≥ n n n (k) 2 ∑ v j > ∑ ak+1 ∑ a j. j=k+1 j=k+1 ≥ j=k+1 17 Solutions 513 Exercises of Chapter 8

8.1 (a) The forward implication follows by taking the inner product with f , and the reverse follows from the polarisation identity (as in Exer. 2.2). (b) It follows from (8.30) that set of c =(c j) satisfying (8.29) is an afﬁne subspace n of F , and so has an element of minimial ℓ2–norm. If c =(c j) satisﬁes (8.30), then so does (ℜ(c j)). Since ℜ(c j) c j , the c with minimal ℓ2–norm must have real entries. Taking the trace| of the|≤| linear| operators in (8.29) gives

~~2 trace(IH )= dim(H )= ∑c j trace(φ jφ ∗)= ∑c j φ j . j j~~

8.2 (a) S is Hermitian, and (8.31) gives AIH S BIH . Thus S has eigenvalues ≤ ≤ 1 0 < A λ B, and so is invertible, with (1/B)IH S (1/A)IH . ≤ ≤ ≤ − ≤ (b) The signed frame operator of ( f j), σ can be written S = VΛσV ∗, where V =[ f j] and Λσ is diagonal with diagonal entries σ. Thus the signed frame operator of ( f˜j), σ is 1 1 1 1 1 1 1 (S− V)Λσ (S− V)∗ = S− (VΛσV ∗)S− = S− SS− = S− . 1 1 Expanding IH = SS− = S− S gives the signed frame expansion, e.g,

~~1 1 1 f = S(S− f )= ∑σ j S− f , f j f j = ∑σ j f ,S− f j f j = ∑σ j f , f˜j f j. j j j~~

~~can 1/2 Since S is positive deﬁnite, the canonical tight signed frame f j := S− f j is well can deﬁned. The signed frame operator of ( f j ) is~~

1/2 1/2 1/2 1/2 1/2 1/2 (S− V)Λσ (S− V)∗ = S− (VΛσV ∗)S− = S− SS− = IH , i.e., ( f jcan, σ is a tight signed frame. − n d 8.3 Suppose that ( f j) j=1 is a tight fusion frame for F with signature

~~σ1 = = σm = 1, σm 1 = = σn = 1. + − Then, by Cauchy–Schwarz, we have~~

n m m 2 2 2 2 2 d f = ∑ σ j f , f j ∑ f , f j ∑ f j f , f F , j=1 | | ≤ j=1 | | ≤ j=1 ∀ ∈ d so that ( f1,..., fm) is frame for F , and hence we must have m d. ≥ d Now let ( f1,..., fm), m d, be any normalised tight frame for F , and σ be the signature above. Using the≥ deﬁnitions and results of Exercise 8.2, we observe that d ( f1,..., fn) will be a signed frame for F with signature σ provided that

n n n 2 2 2 2 2 2 ∑ σ j f , f j = f ∑ f , f j 1 ∑ f j f A f , f , j=1 | | − j=m+1 | | ≥ − j=m+1 ≥ ∀ 514 17 Solutions for some A > 0. This is easy achieved by choosing ( fm+1,..., fn) sufﬁciently small. The canonical tight frame for this signed frame is then a tight signed frame with the desired signature. 8.4 We consider the ( j,k)–entries

(aa∗) (bb∗) =(aa∗) jk(bb∗) jk =(a j1ak1)(b j1bk1)=(a j1b j1)(ak1bk1) ◦ jk =(a b) j1 (a b)∗ = (a b)(a b)∗ . ◦ ◦ 1k ◦ ◦ jk Π d 8.5 We recall (see Exer. 6.17) that r◦,s(C ) has dimension n, and p j is the Riesz representer of the point evaluation δ j (for the apolar inner product). Thus (a) and (b) are equivalent. Suppose that (a), and hence (b), holds. Then each p Π (Cd) has a ∈ r◦,s unique expansion p = ∑ j ck pk, where the coefﬁents are uniquely determined by the linear system ∑ck pk(v j)= p(v j), 1 j n, k ≤ ≤ r s r s i.e., Ac =[p(v j)], where a jk = δ j(vk)= v j,vk vk,v j = v j,vk v j,vk . Since n [p(v j)]=[δ j(p)] can be any vector in C , we conclude A is invertible. Conversely, suppose that A =[δ j(vk)] is invertible, then its rows are linearly independent. Since the rows of A are the linear functionals δ j restricted to (vk), we conclude that the δ j Π d are linearly independent, and hence are basis for r◦,s(C )′. 17 Solutions 515 Exercises of Chapter 9

9.1 (a) Let Φ and Ψ be orthonormal bases for Cd1 and Cd2 , then Φ Ψ is an or- d1+d2 ∪ thonormal basis for C , and we have Sd1 Sd2 Sd1+d2 . (b) If Ψ = Φ, then there are additional symmetries× which swap the two copies of Φ. (c) Let Φ and Ψ be orthonormal bases for Cd1 and Cd2 , then Φ Ψ is an orthonor- d1d2 ⊗ mal basis for C , and we have Sd1 Sd2 Sd1d2 . (d) Let Φ and Ψ be complements, so that× Φ Ψ is a basis, and so every permutation gives a symmetry of Φ Ψ. ⊕ ⊕ 9.2 (a) This is immediate, since Sym(Φ) is a subgroup of Sn, which has order n!. (b) If πΦ (σ)= Lσ = I, then fσ j = f j, j (since the vectors are distinct), so that ∀ σ must be the identity. Each Lσ = πΦ (σ) is determined by its action on a basis of vectors for H taken from ( f j), which gives another basis of vectors in ( f j). If there are only m vectors in ( f j), then number of possible choices for the image of the ﬁrst basis vector is m, for the second m 1, and so on (for the d elements of the basis). − 9.3 (a) Let Φ =( f j) be a ﬁnite frame for H . Then for σ,τ Sym(Φ), we have ∈

~~(Lσ Lτ ) f j = Lσ (Lτ f j)= Lσ fτ j = fστ j = Lστ f j, which implies πΦ (στ)= πΦ (σ)πΦ (τ), since ( f j) spans H . (b) We calculate~~

~~1 1 Φ ∑ ∑ Φ H S (gf )= gf , f j f j =(g∗)− f ,g∗ f j g∗ f j =(g∗)− Sg∗ ( f ), f . j j ∀ ∈~~

~~1 (c) If Φ is tight, then g = Lσ is unitary, i.e., g∗ = g− , and from (b) we have~~

~~SΦ (σ f )= SΦ (gf )= gS 1Φ ( f )= gSσ 1Φ ( f )= gSΦ ( f )= σSΦ ( f ). g− ) −~~

9.4 In view of Theorem 4.3 and Example 9.1, we may assume that Φ =(v j) is tight frame which is not a basis. Since each permutation is a symmetry v j,vk = R = 0, j = k, i.e., Φ is equiangular, and its frame graph Γ is complete.| Let|T be the spanning tree for Γ with root v1, and edges v1,v j , j = 1. By Theorem 8.2, we can { } assume that v1,v j = r, j = 1, where r > 0. Since symmetries preserve the triple products for distinct vectors,− these must all be equal to some real C. Hence C v1,v j v j,vk vk,v1 = C = v j,vk = , j = k, j,k = 1. ⇒ r2 By Exer. 3.19, we have that if C = r3, then Gram(Φ) has rank n, which is not 2 possible (Φ is not a basis), and so we must have C/r = r, so that v j,vk = r, j = k, and therefore Φ is the vertices of the simplex. − − ∀ 9.5 Let Φ be an α–partition frame. Since a frame and it complement have the same symmetry group, it sufﬁces to calculate the symmetry group of the complementary tight frame (2.15). Let M be the set values α j , and m# := j : α j = m , m M. { } |{ }| ∈ 516 17 Solutions

Since the action of the symmetry group is unitary, it must map the vectors k e j/√α j with α j = m to themselves. The subspaces Hm := span e j : α j = m R are orthogonal to each other, and so the symmetry group (and p{rojective symmetry}⊂ group) is the product of the symmetry groups for the equal norm tight frames of the m m# vectors contained in Hm. There are m#! unitary maps which map these vectors to themselves. For each of these maps, the image of the m copies of each of e j the m# vectors Hm (which are equal) can be reordered in m! ways, giving √m ∈

~~m# Sym(Φ) = SymP(Φ) = ∏ m#!(m!) . | | | | m M ∈~~

In particular, if the α j are distinct, then the symmetry group is Sα Sα Sα , 1 × 2 ×× k and for a proper α–partition frame, i.e., α j 2, j, we have the lower estimate ≥ ∀ m# n Sym(Φ) = SymP(Φ) ∏ 2 = 2 > n. | | | |≥ m M ∈

9.6 From the Gramians (see Example 2.8), it is easy to see that Sym(Φ)= S3, and Sym(Ψ) has order 3 and is generated by the cycle a =(123). The faithful actions of these symmetry groups are given by the generators

0 1 ω 1 πΦ 12 πΦ 123 πΨ 123 ( ) = 1 0 , ( ) = ω2 , ( ) = ω . These tight frames are projectively equivalent (see Exer. 2.21, or observe that all the triple products are equal to 1). Thus SymP(Φ)= SymP(Ψ)= S3. − 9.7 Since the action of G = a,b is unitary, 1 g1w1,g2w2 = w1,g− g2w2 , g1,g2 G, w1,w2 v,e1,e2 . | | | 1 | ∈ ∈ { } Using 1 + ω + ω2 + ω3 + ω4 = 0 to simplify, these angles can be computed as

~~5 √5 5 + √5 v,av = v,a4v = − , v,a2v = v,a3v = , | | | | √10 | | | | √10 ω2 j + ζ 2 4 14 v,a jbv = | |, 0 j 4, ∑ v,a jbv 8 = , | | √5 ≤ ≤ j=0 | | 25~~

~~j j 2 e1,a bv = e2,a v = , 0 j 4, | | | | 10 + 2√5 ≤ ≤~~

~~j j 1 + √5 e2,a bv = e1,a v = , 0 j 4. | | | | 10 + 2√5 ≤ ≤ From these we verify the condition for being a (4,4)–design, i.e., 17 Solutions 517~~

8 5 √5 8 5 + √5 8 14 ∑∑ u j,uk = 12 1 + 10 2 − + 2 + + 2 0 j k | | √10 √10 25 8 8 2 1 + √5 144 2 + 20 + = = c4(2,C)(12) . 10 + 2√5 10 + 2√5 5 518 17 Solutions Exercises of Chapter 10

~~1 10.1 Since the action is unitary, hf ,gv j = f ,h∗gv j = f ,h− gv j , and so we can repeat the argument of Lemma 10.1, which gives~~

~~1 1 S(hf )= ∑ hf ,gw gv = h∑ f ,h− gw h− gv = h∑ f ,kw kv = h(Sf ). g g k~~

~~10.2 Consider the FG–homomorphism σkS : Vj W given by →~~

~~σkSf j = σk ∑ f j,gv j gvk . g G ∈ By Lemma 10.4,~~

~~2 σkS = cσ j, c = σkSv j,σ jv j / σ jv j . Using Theorem 10.5, we calculate~~

σkSv j,σ jv j = ∑ v j,gv j gσkvk,σ jv j = ∑ v j,gv j gσkvk,σ jv j g G g G ∈ ∈ 1 1 1 1 = ∑ g− v j,v j σkvk,σ jg− v j = σkvk,σ j ∑ v j,g− v j g− v j g G g G ∈ ∈ 2 2 G v j G v j = σkvk,σ j | | v j = | | σkvk,σ jv j . dim(Vj) dim(Vj)

~~Therefore, the condition (10.16), i.e., Sv j = 0, is~~

~~2 σ σ G v j kvk, jv j σ 1σ σ σ | | 2 k− jv j = 0 jv j, kvk = 0. dim(Vj) σ jv j ⇐⇒ 10.3 (a) We observe ρ(σ) : R8 R8 maps H to H . This gives an action since →~~

ρ(σ)ρ(τ)x = ρ(σ)(xτ j)=(xστ j)= ρ(στ)x = ρ(σ)ρ(τ)= ρ(στ), ⇒ ρ σ ρ σ which is unitary since ( )x, ( )y = ∑ j xσ jyσ j = x jy j = x,y . (b) The stabiliser of v consists of all 2!6! permutations which map 1,2 to itself. Thus, by the orbit size theorem, the orbit has 8!/(2!6!)= 28 vectors.{ } 1 1 10.4 (a) We have that x,x ρ = ∑ ρ(g)x,ρ(g)x = ∑ ρ(g)x 2 0, with G g G g ≥ equality possible if and only if each| | term is zero, i.e., x = 0| (since| each ρ(g) is 1–1). (b) Since ρ(g)ρ(h)= ρ(gh), we obtain 1 ρ(h)x,ρ(h)y ρ = ∑ ρ(g)ρ(h)x,ρ(g)ρ(h)y = x,y ρ . G | | g (c) Clearly ρ˜ is a group homomorphism, and each ρ˜(g) is unitary, since by (b) 17 Solutions 519

1 1 1 1 ρ˜(g)x,ρ˜(g)y = Aρ(g)B− x,ρ(g)B− y = ρ(g)B− x,ρ(g)B− y ρ 1 1 1 1 = B− x,B− y ρ = A− x,y ρ = A(A− x),y = x,y . 1 1 1 (d) We have B− Ψ = B− (Bρ(g)B− Bv)g G =(ρ(g)v)g G = Φ, so that Φ is similar to Ψ. Since Ψ is a G–frame, Theorem 10.1∈ implies that∈Ψ can is a G–frame. 10.5 , is G–invariant if and only if for each g G ∈ ρ(g)x,ρ(g)y = x,y Mρ(g)x,ρ(g)y ρ = Mx,y ρ , x,y ⇐⇒ ∀ 1 ρ(g)− Mρ(g)x,y ρ = Mx,y ρ , x,y ⇐⇒ ∀ 1 ρ(g)− Mρ(g)= M. ⇐⇒ By Schur’s lemma, the last condition implies that M is nonzero scalar multiple of the identity matrix when ρ is absolutely irreducible, i.e., in this case all G–invariant inner products are scalar multiples of each other. 10.6 (a) Since the action of G on H is unitary, we have

2 1 2 2 ∑∑ gv,hv = ∑∑ v,g− hv = G ∑ v,gv g h | | g h | | | | g | | 1 2 1 1 ∑ gv,gv = ( G v 2)2 = G 2 v 4. ≥ d g d | | d | | (b) As above, the (t,t)–design condition simpliﬁes, since

2 2 ∑∑ gv,hv 2t = G ∑ v,gv 2t , ∑ gv 2t = G v 2t = G 2 v 4t . g h | | | | g | | g G | | | | ∈ 10.7 Use the notation v w and v w for elements of H1 H2 and H1 H2. ⊕ ⊗ ⊕ ⊗ (a) If Φ =(gv)g G and Ψ =(gw)g G are disjoint G–frames, then their direct sum is ∈ ∈ Φ Ψ =(g(v w))g G, where g(x y) := gx gy is a unitary action. ⊕ Φ ⊕ ∈ ⊕ ⊕ (b),(c) If j =(g jv j)g j G j are G–frames, then ∈ 1 1 Φ1+ˆ Φ2 = (g1,g2)( v1 v2) (g1,g2) G1 G2 G2 ⊕ G1 ∈ × | | | | Φ Φ 1 2 = (g1,g2)(v1 v2) (g ,g) G G ⊗ ⊗ 1 2 ∈ 1× 2 where (g1,g2)(x y) := g1x g2y and (g1,g2)(x y) := g1x g2y are unitary. ⊕ ⊕ ⊗ ⊗ (d) Let Ψ be the complement of a normalised tight G–frame Φ =(φg)g G. Since Gram(Φ)+ Gram(Ψ) is the identity matrix, we have ∈

φ φ 1 φ φ ψ ψ h, g = g− h 1, 1 , h = g; h, g = − − 1 1 φh,φg = 1 g hφ1,φ1 , h = g − − − so that Gram(Ψ) is a G–matrix, and, by Theorem 10.3, Ψ is G–frame. 520 17 Solutions

10.8 If the action was not irreducible, then there would be a plane P R3 that was G–invariant (otherwise R3 would be a sum of 1–dimensional G–invariant⊂ subspaces, and so G would be abelian). The restriction of the action of G to this plane P would be a faithful representation of G, and so G would be isomorphic to a finite subgroup of U (R2). This is impossible, since the finite subgroups of U (R2) are cyclic or dihedral. 10.9 (a) Suppose V is an FG–module. Since v gv is linear map, ρ(g)v := gv defines a map ρ : G GL(V), and since (g,v) →gv is an action → → ρ(g)ρ(h)v = g(hv)=(gh)v = ρ(gh)v = ρ is a homomorphism. ⇒ Conversely, suppose that ρ is representation, and let gv := ρ(g)v. Then

g(hv)= ρ(g)ρ(h)v = ρ(gh)v =(gh)v, 1v = ρ(1)v = v so (g,v) gv is an action, and ρ(g) : v gv is a linear map (by deﬁnition). (b) By deﬁnition:→ W is a FG–submodule→ of V if it is closed under vector addition and scalar multiplication, i.e., is a vector subspace, and it is closed under the action of G, i.e., is G–invariant. 1 10.10 First observe that τ− is an FG–isomorphism, since

1 1 1 1 1 τg = gτ = g = τ− gτ = gτ− = τ− gττ− = τ− g. ⇒ ⇒ 1 The map στ− :Vj Vj is an FG–isomorphism (check taking the inverse commutes with the action of →G). Let λ C be an eigenvector (for a nonzero eigenvalue) of 1 ∈ στ− . Then (gv)g G spans Vj, and we commute ∈ 1 1 στ− gv = gστ− v = g(λv)= λ(gv),

~~στ 1 λ so that − = IVj , as claimed. 10.11 We have the equivalences~~

1 1 g1H = g2H g− g1 H g− g1v = v g1v = g2v. ⇐⇒ 2 ∈ ⇐⇒ 2 ⇐⇒ (a) The forward direction implies C H : gH gv is a well defined map onto the vectors in Φ, and the reverse direction→ that this is→ 1–1. (b) Let N be the kernel of ρ, so that N H. If G/H U (H ) : gH ρ(g) defines a group homomorphism, then by the⊂ first isomorphism→ theorem the→ composition G/N G/H U (H ) : gN gH ρ(g) is a well defined injective map, and so we must→ have →H = N. → → (c) If G is abelian, then h(gv)=(hg)v =(gh)v = g(hv)= gv, h H,g G, and so ∀ ∈ ∈ H stabilises H = span gv g G. { } ∈ 10.12 (a) By Exercise 10.6, the condition for (gv)g G to be a (t,t)–design is that ∈ 17 Solutions 521

1 2t 4t ∑ v,gv = ct (d,F) v . G g G | | | | ∈ This will hold for every v if the above polynomials of degree 4t in the entries of v and v are equal. This can be veriﬁed in Magma for t = 5, but not t = 6. An easy way to do this is to check equality at a set of points v on which a polynomial in Π d 2◦t,2t (C ) is determined by its values. (b) We observe that a has order 5 and b2 = I. Hence if v is an eigenvector of a, − then (gv)g G consists of 120/10 = 12 lines (from each of which we can select a vector to obtain∈ (5,5)–design of 12 vectors). d 10.13 Let G = Cn = a act on a θ–isogonal conﬁguration (x j) via a cyclic shift: j=1 ax j := x j+1, xd+1 := x1. This action is unitary, since

~~a(∑α jx j),a(∑βkxk) = ∑α jx j+1,a∑βkxk+1 j k j k~~

~~= ∑ ∑ α jβk cosθ + ∑α jβ j = ∑α jx j,∑βkxk , j k= j j j k and so (x j) is a Cn–frame. 522 17 Solutions Exercises of Chapter 11~~

~~11.1 (a) Let~~

~~1 ω ω2 1 1 1 π u v ω : e2 i/3 ( j)= 1 , ω2 , ω , ( j)= 1 , ω , ω2 , = . Then we calculate~~

~~u j uk = √6, j = k, v j vk = √3, j = k, − −~~

and u j,uk = v j,vk = 1, j = k. | | | | √ √5 1 (b) A calculation, see (c), gives w j wk = 5, j = k. Here w j,wk = 2± . j (d 1)j − 2πi/(2d 1) | | (c) Since v j =(1,ω ,...,ω − ), with ω := e − , we calculate

~~d 1 2d 1 2 − aj ak 2 − m( j k) v j vk = ∑ ω ω = 2d 1 ∑ ω − = 2d 1, j = k. − a=0 | − | − − m=1 −~~

~~j 2 j dj 2πi/(2d+1) Since w j =(ω ,ω ,...,ω ), with ω := e , we calculate~~

d 2d 2 aj ak 2 m( j k) w j wk = ∑ ω ω = 2d ∑ ω − = 2d ( 1)= 2d + 1, j = k. − a=1 | − | − m=1 − − Remark: Notice that the separation between points in Ψ is larger than that for those in Φ (which has fewer points), and so equispacing is perhaps not the most useful notion of “equal spacing”. The difﬁculty here is that though an equal distance from each other, the points are clumped in a particular cone. This phenomenon can be seen for the standard basis vectors in C3, which are equally spaced, and all lie in the ﬁrst octant. 11.2 (a) The symmetry group has order 384 (it is the group <384,5557>), which is the largest for the class of cyclic harmonic frames of this size. This frame consists of scalar multiples of an orthonormal basis. (b) The symmetry groups are both <200,31>, which is the largest for its class. The projective symmetry groups are also both <200,31>. 11.3 Since U = U 1, Uf = ∑ Uf ,U jv U j = U ∑ f ,U j 1v U j 1, which gives ∗ − j j − − n 1 n 1 − j 1 j 1 − j j f = ∑ f ,U − v U − v = ∑ f ,U v U v, f . j=0 j=0 ∀

1 1 n 1 n 1 Cancelling terms in the above gives f ,U− v U− v = f ,U − v U − v, so that Unv = cv, for some c F. For (U jv) to be a frame, v must be nonzero, and so n ∈ k n 1 v = U v = cv = c v , implies that c = 1. On the spanning set U v k=−0, we have | | | | { } 17 Solutions 523

Un(Ukv)= Un ∑ Uk,U jv U jv = ∑ Uk,U jv U j(cv)= c(Ukv), j j so that Un = cI. If 1/α = √n c is an n–th root of c, then (αU)n = I, and so n 1 α j − α α2 2 αn 1 n 1 ( U) v j=0 =(v, Uv, U v,..., − U − v) is a cyclic harmonic frame. 11.4 LetΦ be the harmonic frame of n vectors given by a G. For the vectors { }⊂ of Φ to be distinct, a must generate G, which implies that G = Zn with a a unit. The ωaj ω 2πi/n corresponding harmonic frames are ( ) j Zn , where = e , i.e., the n–th roots of unity. ∈ 11.5 Let G be an abelian group of order n. If J = a,b G gives a real harmonic frame of n distinct vectors, then b = a (to be real){ }⊂ and a generates G (to have − distinct vectors). Thus we must have G = Zn, with a a unit. For such a given subset a, a Zn, the corresponding cyclic harmonic frame is { − }⊂ a 2a (d 1)a 1 ω ω ω − , ω a , 2a , (d 1)a , 1 − ω− ω − − which is unique up to reordering. This harmonic frame must be the n equally spaced unit vectors in R2, since the equally spaced unit vectors are the orbit of a vector 2π under the unitary action of the cyclic group of order n given by rotations through n (and therefore are a real cyclic harmonic frame). 11.6 Representatives of the six multiplicatively inequivalent 2–element subsets of Z6 (the unique abelian group of order 6) together with the distances v j vk and − angles v j,vk , j = k are | | 0,1 1,√3,2, 0,1,√3, 1,2 2,√6, 0,1,√3, { } { } 1,3 √3,√5,2√2, 1,2, 1,4 2,√6, 0,2, { } { } 1,5 √2,√6,2√2, 1,2, 3,4 √3,2,√7, 0,1,√3. { } { }

~~Since the distances and angles are ﬁxed under v j v j, we conclude that none of these harmonic frames are conjugates of each other.→~~

~~11.7 Since matrix multiplication in G commutes, for g1,g2 G, we have ∈~~

g1φ = g2φ = g1hv = g2hv, h G = g1 = g2. ⇒ ∀ ∈ ⇒ If G is nonabelian, then this does not hold. For example, let G be the dihedral group of the square acting on R2 in the usual way, with a reﬂection ﬁxing the x–axis. Then the orbit of v = e1 consists of four equally spaced vectors, each repeated twice. 11.8 Suppose that G = n, J = d. Then the normalised tight frame given by J 1 | | | | is (ξ J)ξ ˆ , and similarly for the one given by G J. The (ξ,η)–entry of the √n | G \ Gramians for∈ these normalised tight frames are 524 17 Solutions

η J,ξ J = ∑ η( j)ξ( j), η G J,ξ G J = ∑ η( j)ξ( j). | | j J | \ | \ j G J ∈ ∈ \ These sum to 1 η,ξ , which is the identity matrix (orthogonality of the characters). n Φ a 11.9 Let n = and v j := v j, 1 a d. Then (up to a scalar multiple) the vectors of Ψ are | | ≤ ≤ 1 2 d (v ,v ,...,v ), 1 j1, j2,..., jd n. j1 j2 jd ≤ ≤ We think of these vectors as being indexed by the pairs

~~(a, ja) . 1 a d,1 ja n ≤ ≤ ≤ ≤ There are symmetries given by~~

d (a, ja) (τa, ja), τ Sd, (a, ja) (a,σa ja), (σ1,...,σd) Sym(Φ) . → ∈ → ∈ The subgroups of the symmetries of these two types have trivial intersection, and so d d Sym(Ψ) Sd Sym(Φ) = d!m . | |≥| || | Ψ ψ ψ ξ d 11.10 Let ξ ,...,ξ =( g)g G, g :=( j(g)) j=1 denote the harmonic frame given { 1 d } ∈ by a choice of characters ξ1,...,ξd Gˆ. Ψ { }⊂ (a) For (vg)g G = ξ ,...,ξ , we calculate ∈ { 1 d }

va j,va k = ξ1(a + j)ξ1(a + k)+ + ξd(a + j)ξd(a + k) + + = ξ1(a)ξ1( j)ξ1(a)ξ1(k)+ + ξd(a)ξd( j)ξd(a)ξd(k) = ξ1( j)ξ1(k)+ + ξd( j)ξd(k)= v j,vk . Ψ Ψ (b) Let (vg)g G = ξ ,...,ξ and (wg)g G = η1,...,η . If there is unitary U and a ∈ { 1 d } ∈ { d } permutation σ : G G with vg = Uwσg, g G, then, by (a), we calculate → ∀ ∈

v j,vk = Uwσ j,Uwσk = wσ j,wσk = wσ j σk+b,wb , − and so we may take a = σ j σk + b. − (c) Take j = 1, k = 0, b = 0 in (b), to obtain v1,v0 = wa,w0 , i.e., d d ∑ ξ j(1)= ∑ η j(a). j=1 j=1

~~Moreover, if G = Zn and σ is an automorphism, then a = σ1 is a unit.~~

11.11 (a) The frame given by J Zp is real if and only if J is closed under taking inverses, i.e., J = J. Since p is an⊂ odd prime, 2 is a unit, and so j = j if and only − − if j = 0. Thus J has the stated form. Since Z = Zp 0 , K generates Zp for d > 1. ∗p \{ } (b) The unit group Z∗p is cyclic of even order p 1. Let a Z∗p have order j. The p 1 − ∈ action of a on Zp 0 gives − orbits of size j. We now count the number of \ { } j 17 Solutions 525 orbits of the d–element subsets of Zp which give real harmonic frames (as above) under the action of Z∗p, which is the number real harmonic frames. First consider the case when d is even. Suppose that a ﬁxes some J = K K, j d ∪− J = d. If j is even, then 1 = a 2 , so J consists of orbits under the action of a | | − j (half the orbit in K and the other half in K), and we must have j d. If j is odd, d/2 − | d/2 then K consists of j orbits of size j (their negatives give the remaining j orbits which make up J), and we must have j (d/2). Thus Burnside orbit counting gives | p 1 p 1 1 − − hR = ∑ j ϕ( j)+ ∑ 2 j ϕ( j) . p,d p 1 d d j gcd(p 1,d) j j gcd(p 1, d ) 2 j − | j even− | − 2 j odd

When d is odd, the subsets J giving real frames have the form J = 0 K K. These are multiplicatively equivalent if and only if the sets K K are,{ }∪ and∪− so we may apply the previous count (with d replaced by d 1). The∪− formula so obtained also holds for d = 1. − 526 17 Solutions Exercises of Chapter 12

~~d d 12.1 If A =[a jk] F is Hermitian, i.e., a jk = akj, then it is determined by ∈ ×~~

~~a jj R, 1 j d, a jk F, 1 j < k d. ∈ ≤ ≤ ∈ ≤ ≤ Thus the real vector space of Hermitian matrices has dimension 1 1 1 d + d(d 1)= d(d + 1) (F = R), d + 2 d(d 1)= d2 (F = C). 2 − 2 2 −~~

~~12.2 (a) Take the trace of Pk = ∑ j c jPjPk to obtain~~

2 2 2 1 = ∑ c jα + ck = ∑c jα + ck(1 α ), j=k j − d ∑ so that the ck are constant, hence equal to n (by taking the trace of I = j c jPj). d (b) Substituting c j = n above, and multiplying by n gives

n = ndα2 + d(1 α2). − Thus n(1 dα2)= d(1 α2) > 0, so α2 < 1 , and we may solve for n. − − d 2 2 12.3 By (2.9), we have f j + ∑k= j fk = d, and so 2 2 4 2 2 2 4 2 2 2 f j = ∑ f j, fk = f j + ∑ C f j fk = f j +C f j (d f j ). k | | k= j − 2 2 2 Thus 1 C d = f j (1 C ). If C = 1, then the vectors are all collinear (which is − − 2 not possible), and so we conclude that f j is independent of j. 12.4 By construction, two of the x j are 3 and the other six are 1. If two such vectors are not equal, then the number of times they can have a 3 in− same position is 0 or 1. The corresponding inner products for these cases are 1 1 1 1 (4 3( 1)+ 4 ( 1)2)= , (1 32 + 2 3( 1)+ 5 ( 1)2)= . 24 − − −3 24 − − 3

12.5 Consider the n lines given by the f j (some might be coincident). The acute 1 1 angle between the lines given by f j and fk is θ jk = cos− ( f j, fk ). Since cos− is strictly decreasing on [0,1], the asserted inequality is equivalent| to| π π 1 1 cos− max f j, fk cos− cos minθ jk . j=k | | ≤ n ⇐⇒ j=k ≤ n Rotate the ( f j) so that one line is the x–axis. Then the lines partition the upper half plane into n sectors. The angle of the smallest sector gives min j=k θ jk. This must be π , since otherwise the sum of the angles of the n sectors would exceed π. There ≤ n 17 Solutions 527

~~π is equality if and only if all the n sectors have angle n . The strict inequality follows by calculus.~~

Φ Ψ d n d 12.6 The normalisations of and are ( n f j) and ( −n g j). Thus n d n d d d d − g j, − gk = f j, fk = g j,gk = f j, fk . n n − n n ⇒ | | n d | | − Thus if Φ minimises M∞(Φ) over all unit–norm tight frames of n vectors for Fd, n d then Ψ minimises M∞(Ψ) over all unit–norm tight frames of n vectors for F − . 12.7 (a) Let Σ be the Seidel matrix of the 5–cycle (1,2,3,4,5), i.e.

0 1 1 1 1 1− 0 1 1− 1 Σ = −1 1− 0 1 1 . 1− 1 1− 0 1    1 1− 1 1− 0  − −    This has eigenvalues √5, √5,0,√5,√5. The Gramian of the corresponding nontight equiangular frame− Φ−is 1 Gram(Φ)= Q = I + Σ. √5 From this, we calculate (see Exer. 3.3)

2 5 abba 2 a 5 a b b †   3 √5 3 + √5 Gram(Φ˜ )= Q = b a 2 a b , a := − , b := , 5 20 20 b b a 2 a  5  abba 2   5    3 5 a b b a −3 − a 5 a b b can † − −  √5 1 √5 + 1 Gram(Φ )= QQ = b a 3 a b , a := − , b := . − 5 − 10 10  b b a 3 a  − 5 −   a b b a 3   5  − −  Thus the minimal angle between the lines given by Φ, Φ˜ and Φcan are

~~1 1 1 3 + √5 1 1 + √5 cos− 63.4349◦, cos− 49.1176◦, cos− 57.3610◦. √5 ≈ 8 ≈ 6 ≈~~

(b) It is easy to verify that Φ =(v j) is given by vectors that lie in ﬁve of the six diagonals of the regular icosahedron, e.g., (using the hint) we may take 528 17 Solutions 0 0 c 1 c − 1 + √5 V =[v j]= 1 1 0 c 0 , c := . c c 1 0 1 2 − − can (c) It is easy to verify that Φ =(v j) is the harmonic frame given by 0,2,3 Z5, i.e., { }⊂

11 1 1 1 π 2 4 3 2 i V =[v j]= 1 ω ω ω ω , ω := e 5 , 1 ω3 ω ω4 ω2 which is the lifted fifth roots of unity (see Example 5.9). 12.8 This does not hold in general. For the four isogonal vectors in R3 the dual frame is equiangular (see Example 3.9), but for the five equiangular lines in R3 given by five of the six diagonals of the regular icosahedron the dual frame has two 1 angles v j,vk = (3 √5) (see Exer. 12.7). | | 8 ± 1 12.9 (a) Adding the last d rows of A gives (1,1,...,1). Subtracting the multiple d of this from the first n rows, gives the row equivalent matrix

(α2 1 )J +(1 α2)I 0 B = d n . − 1 J− I d d (b) Since B is block upper triangular, its eigenvalues are those of the leading blocks: α2 α 1 α2 (1 ) with multiplicity n 1, n( d )+(1 ) with multiplicity 1, and 1 with multiplicity− d. Thus, the rank− of A is− either n +−d 1 if − 1 n(α )+(1 α2)= 0, − d − which is equivalent to Φ being tight, or it has rank n + d. In any case

~~n + d 1 rank(A) d2. − ≤ ≤ 12.10 The vectors have equal norms, since~~

~~2 2 1 3 1 v˜a j = va j,va j + 2 = r + 2, w˜ = ν + = (ν + 3)= r + 2. , , , ℓ 2 2 2 By (12.15) and (12.16), we have~~

~~(a) (a) (a) (a) (a) (a) v˜a j,v˜a k = h h + 2h h = h h , j = k, , , − 0, j 0,k 0, j 0,k 0, j 0,k~~

(a) (b) v˜a j,v˜b k = h h , a = b. , , β, j β,k 2 Since w ,wm = α αm, ℓ = m, w = ν, we have ℓ − ℓ ℓ 1 3 w˜ ,w˜m = ( α αm)+ α αm = α αm, ℓ = m. ℓ 2 − ℓ 2 ℓ ℓ 17 Solutions 529

Finally, (a) 1 (a) v˜a, j,w˜ℓ = √2h ea, w j = h (w j)a, 0, j √2 0, j and so the vectors are equiangular (Hadamard matrices have entries of modulus 1). The space CB CV C has dimension ⊕ ⊕ 1 1 d := B + V + 1 = ν(ν 1)+ ν + 1 = (ν + 2)(ν + 3), | | | | 6 − 6 and the frame ( f j) :=(v˜a j) (w˜ ) has , ∪ ℓ 1 n := ν(r + 1)+(ν + 1)= (ν + 1)(ν + 2) 2

~~2 1 vectors with f j = r +2 = (ν +3), and f j, fk = 1, j = k. This frame is a tight 2 | | frame for Cd, by the variational characterisation (Theorem 6.1), i.e.,~~

2 1 2 2 ∑∑ f j, fk = n ( (ν + 3)) +(n n) 1 j k | | 2 − 2 3 2 1 1 2 2 = (ν + 1) (ν + 2)(ν + 3)= n (ν + 3) = ∑ f j . 8 d 2 j 2 1 The common angle is f j, fk / f j = , j = k. | | r+2 T T 12.11 (a) Since Σ is Hermitian, we have iC = Σ ∗ = Σ = iC, so that C = C. (b) The condition of Theorem 12.7 for Σ−to give an equiangular tight− frame of n vectors for Cd is

CTC = C2 = Σ 2 =(n 1)I + µΣ =(n 1)I + µiC, − − − which that implies CTC =(n 1)I (since µ is real), i.e., C is a conference matrix, where µ = 0, i.e., n = 2d. − T T 12.12 (a) Expanding out C C = nI, using C = C and 11∗ = J, we have − 0 1 0 1 n 1 A n 0 ∗ ∗ = ∗ = . 1 − A 1 A A1 J− A2 0 nI − − − Equating the blocks gives J A2 = nI and A1 = 0. From A1 = 0, we have −

AJ =(A1)1∗ = 0, JA =( AJ)∗ = 0. − (b) By construction, Σ is a signature matrix, and so it sufﬁces to verify the second condition of Theorem 12.7. Using J2 = nJ, AJ = JA = 0 and A2 = J nI, we have − 530 17 Solutions 1 Σ 2 = (J I)2 nA2 i√n(J I)A i√nA(J I) n + 1 − − ± − ± − 1 1 = nJ 2J + I n(J nI) 2i√nA = (n2 1)I + 2(I J i√nA) n + 1 − − − ∓ n + 1 − − ∓ 2 2 =(n 1)I (J I)+ i√nA =(n 1)I Σ. − ∓ n + 1 ± − − ∓ √n + 1 Thus the second condition holds, with 2 n nµ n 1 µ = , d = = . ∓√n + 1 2 − 2 4(n 1)+ µ2 2 ± 2 − (c) Since µ 2, we can choose ζ = 1 ( µ + 4 µ 2i) in the third condition. | |≤ 2 − −| | 12.13 (a) Since 1 has no square root in GF(q), it follows that N = S. Hence, if we add 0 to the difference− set S, then the additional differences 0 s−,s 0, s S, give exactly one extra copy of each nonzero element of G. − − ∈ (b) Since G = N 0 S (disjoint union), N and N 0 are the complementary difference sets to∪S { }∪0 and S (respectively). Since∪x { } x is an automorphism of G, it follows that∪ the { harmonic} frames given by S,N and→ − by S 0 ,N 0 are unitarily equivalent up to reordering by an automorphism (Theorem∪ { 11.4).} ∪ { } (c) Let z = x + iy = ξ S,η S be the inner product between two distinct vectors of Φ | |ξ the harmonic frame S =( S)ξ Gˆ . The inner product between the corresponding | ∈ vectors of ΦS 0 is z + 1. Since each of these frames is equiangular (they are given ∪{ } 2 2 n 1 by difference sets), z + 1 z = 2x + 1 is constant, i.e., with d = − , we have | | −| | 2 (d + 1)(n d 1) d(n d) 1 2x + 1 = z + 1 2 z 2 = − − − = 0 = x = . | | −| | n 1 − n 1 ⇒ −2 − − Thus the real parts of the signature matrices for the equiangular harmonic frames given by S and S 0 are harmonic frames given by S and S 0 are ∪ { } ∪ { } 1 d(n 1) 1 1 1 (d + 1)(n 1) 1 − 2 − = , − 2 − = . d n d −√n + 1 d + 1 n d 1 √n + 1 − − − These signature matrices can be calculated explicitly by using the the formula for quadratic Gauss sums (see [Kal06], [Ren07]), or by (12.63). ξ (d) Let W =[ S]ξ Gˆ be the synthesis map for the equiangular harmonic frame Φ ξ | ∈ S =( S)ξ Gˆ , and 1 =(1,...,1). By (c), the of diagonal entries of W ∗W have the form | ∈ q 1 q 1 1 1 − (q − ) q z = + iy, + y2 = 2 − 2 = y2 = . −2 4 q 1 ⇒ 4 − Since ΦS is unlifted, we have W1∗ = 0. Hence the frame of 4m = q+1 vectors with synthesis operator √q 1∗ V = 0 √2W 17 Solutions 531 is a tight frame for C2m (it has orthogonal rows of equal norm). Its Gramian is

~~√q 0 √q 1∗ q √q1∗ V ∗V = = . 1 √2W 0 √2W √q1 11∗ + 2W ∗W ∗~~

~~The off diagonal entries of 11∗ + 2W ∗W have the form~~

1 q 1 + 2 + iy = 2iy, 2iy = 2 = √q, −2 | | 4 and so it is equiangular. This example is due to Zauner [Zau99]. For an equiangular tight frame of n = 2d vectors for Cd whose signature matrix has off diagonal entries i, it can be shown that the canonical tight frame of the frame obtained by removing ±a vector is an equiangular tight frame of n 1 vectors for Cd (see [Ren07], [Str08]). − 12.14 By Theorem 12.6, the Gramian of the normalised equiangular tight frame is

d n d P = I +C Σˆ , C := − . n n,d n,d d(n 1) − d Since n d Cn,d = Cn,n d, the complementary orthogonal projection is − − d d n d I P = I I Cn,dΣˆ = − I +Cn,n d( Σˆ ) , − − n − n n − − so the complementary equiangular tight frame has reduced signature matrix Σˆ . The complement of a srg(ν,k,λ,ν) is a srg(ν,ν k 1,ν 2 2k+µ,ν 2k+−λ), which gives the desired parameters. Alternatively,− a− count−shows− the vertex− degree of the complementary graph is k′ =(n 1) k 1, and the other parameters follow from (12.35) − − − 3(n k 2) n n k 2 λ = − − − , µ = − − . ′ 2 ′ 2

12.15 Without loss of generality, we may assume that Qn r is the leading principal submatrix, so that − Qn r V n r r Q = − , V C − × . V ∗ B ∈ Thus (by block multiplication), the leading block of of dQ2 nQ gives − 2 dQn r nQn r = dVV ∗. − − − −

Since rank(VV ∗)= rank(V ∗V) min r,n r , the result follows, and the equations 1 ≤ { − } are nontrivial only when r < 2 n. 12.16 (a) Since G is Hermitian with constant diagonal 1 and constant modulus off diagonal entries, we have all but the unitary property. Since the frame is tight, its d 2 d canonical Gramian is an orthogonal projection, i.e. ( n G) = n G, and we calculate 532 17 Solutions

2 d d U∗U = U = I 4 G + 4 G = I. − n n (b) By construction G is Hermitian, with constant diagonal 1 and constant modulus d off diagonal entries. Thus it sufﬁces to show that n G is an orthogonal projection, 2 n 2 i.e., G = d G. Since U = U∗U = I, we calculate n 2 n 2 n G2 = (I 2U +U2)= 2(I U)= G. 2d − 2d − d 12.17 By the variational characterisation of tight frames, we have

2 2 2 n ∑∑ v j,vk = ∑ v j,vk + n = . j k | | j,k | | d j=k We recall that in Holder’s¨ inequality with one vector being (1,...,1) there is equality if and only if the other vector has entries of constant modulus. 2 2 (a) In Holder’s¨ inequality take the conjugate exponents 2 p and p , to obtain − p p 2 1 2 2 2 p p 2 p − p p ∑∑ v j,vk = ∑ v j,vk + n ∑(1) − ∑( v j,vk ) + n j k | | j,k | | ≤ j,k j,k | | j=k j=k j=k p n2 p p ( n) 2 2 1 2 2 2 d = n n − ∑ v j,vk + n = − p + n, 2 1 − j,k | | (n n) 2 − j=k − with equality if and only if (v j) is equiangular. p p (b) Take the conjugate exponents p 2 and 2 in Holder’s¨ inequality, to obtain − 2 2 2 n p 1 p p p 2 p 2 − 2 2 = ∑ v j,vk + n ∑(1) − ∑( v j,vk ) + n d j,k | | ≤ j,k j,k | | j=k j=k j=k 2 2 2 1 p p p = n n − ∑ v j,vk + n, − j,k | | j=k which gives the desired inequality, with equality if and only if (v j) is equiangular. (c) For p = 2, equality in (a) and (b) is the variational characterisation of being tight. 12.18 (a) Since two frames are unitarily equivalent if and only if their Gramians are equal, the condition (12.65) is equivalent to Λ Λ Pσ 1 ∗ Pσ 1 (U[V1,W1])∗U[V1,W1]= [V2,W2] [V2,W2] . PτΛ2 PτΛ2 Block multiplying out, using Vj∗Vj = I, gives 17 Solutions 533

I V W I (Pσ Λ ) V W (PτΛ ) 1∗ 1 = 1 ∗ 2∗ 2 2 , W ∗V1 I (PτΛ2)∗W ∗V1(Pσ Λ1) I 1 1 Λ 1 1 Λ which is equivalent to H1 = 1− Pσ− H2Pτ 2 (Hadamard matrix equivalence). (b) The nonzero 4–products for the vectors (V1e j) and (W1e j) of the ﬁrst pair of MUBs are

V1ek,W1e j W1e j,V1em V1em,W1e W1e ,V1ek =(H1) jk(H1)mj(H1) m(H1)k . ℓ ℓ ℓ ℓ Since the vectors in the pairs of MUBs are projectively unitarily equivalent via UV1e j = α jV2eσ j, UW1e j = β jW2eτ j, the condition for equivalence of Hadamard matrices is

~~(H1) jk(H1)kℓ(H1)ℓm(H1)mj =(H2)σ j,τk(H2)σk,τℓ(H2)σℓ,τm(H2)σm,τ j.~~

It is enough to check this condition for j,k,l,m corresponding to a basis of 4–cycles for the cycle space of the frame graph (the complete bipartite graph Kd,d), which has dimension d2 2d + 1 =(d 1)2. − − 12.19 (a) Multiplying out, using µ2 = ω, gives

~~d 1 2d 1 1 a 1 − js as(s+d) sk 1 − as(s+d)+2s(k j) (F− R F) jk = ∑ ω− µ ω = ∑ µ − . d s=0 2d s=0~~

~~(b) Suppose d is odd. Since the s = β and s = β + d terms in the second sum above are equal, this implies that the sum above is twice the sum over the even terms, i.e.,~~

d 1 d 1 − a(2r)2+(ad+2(k j))(2r) − 2ar2+2(k j)r G(a,ad + 2(k j),2d)= 2 ∑ µ − = 2 ∑ ω − − r=0 r=0 = 2G(2a,2(k j),d). − 1 d Since 2 is a unit, with 2 = −2 , completing the square gives the desired formula

~~d 1 − 2a (r+2a(k j))2 (2a(k j))2 2a(k j)2 G(2a,2(k j),d)= ∑ ω { − − − } = ω− − G(2a,0,d). − r=0 Suppose d is even. Then we can complete the square as follows.~~

~~2d 1 − a (s+ d +a(k j))2 ( d +a(k j))2 G(a,ad + 2(k j),2d)= ∑ µ { 2 − − 2 − } − s=0 a( d +a(k j))2 = µ− 2 − G(a,0,2d).~~

~~1 a (c) By (a) and (b), the modulus of the entries of F− R F for d odd and even are 1 1 1 1 2 √d = , √2√2d = . 2d √d 2d √d 534 17 Solutions Exercises of Chapter 14~~

1 14.1 (a) Expanding gives Pj c1I,Pk c2I = d+1 c1 c2 + dc1c2 = 0. − − 1 − − 1 (b) For c1 = 0, we solve the above to get c2 = . Thus (Pj I) is the dual d+1 − d+1 basis to (Pj). (c) Since the vectors giving (Pj) are a tight frame, we have ∑ j Pj = dI, i.e., the linear 1 dependency ∑ (Pj I)= 0. Suppose that ∑ a j(Pj cI)= 0 for some scalars a j, j − d j − i.e., ∑ j a jPj = c∑ j a jI. Since (Pj) is a basis and ∑ j Pj = dI, the a j must be constant, α α α 2 α say a j = , and the condition becomes ∑ j Pj = dI = d c I. Thus (a j) can be 1 nonzero if and only if c = d . 1 2 1 (d) By (a) and (c), (Pj cI) is orthogonal if and only if 2c + dc = 0, c = , − d+1 − d i.e., c = 1 (1 1 ). d ± √d+1 (e) Since the traceless matrices are the orthogonal complement of I, the projection is Pj cI,I 1 1 Pj cI − I = Pj cI (1 cd)I = Pj I. − − I,I − − d − − d Remark: By (d), c can be chosen so (Pj cI) is an orthogonal basis, and so this 1 − shows that (Pj ) is an (equiangular) tight frame for the traceless matrices. − d 14.2 (a) Expanding, and using the trigonometric identities sin2x = 2sinxcosx, 2cos2 x = 1 + cos2x, we have

~~2 θ θ θ iφ cos 2 sin 2 cos 2 e− 1 1 + c a ib vv∗ = θ θ φ θ = − . sin cos ei sin2 2 a + ib 1 c 2 2 2 −~~

~~(b) Using (a), the image of vv∗ is~~

√2 c a ib A = − , 2 a + ib c − 2 1 2 2 2 2 2 2 where A F = 2 (c +(a + b )+(a + b )+ c )= 1. The form of A clearly shows that the map is onto. (c) Since A is Hermitian, it sufﬁces to calculate its trace and Frobenius norm:

d 1 d d trace(A)= vv∗ I = 1 = 0, d 1 − d d 1 − d − − 2 2 d 2 1 d 2 d A = trace(A )= trace vv∗ vv∗ + I = 1 + = 1. F d 1 − d d2 d 1 − d d2 − − d 1 1 d If the map is onto, then P = −d A+ d I is a rank one orthogonal projection on C for any traceless Hermitian matrix A with unit Frobenius norm. If A is zero except for the principal 2 2 submatrix B, then × 17 Solutions 535

d 1 B 0 1 I 0 d 1 B + 1 I 0 P = − + 2 = −d d 2 , d 0 0 d 0 Id 2 1 − 0 d Id 2 − which clearly has rank larger than one for d > 3. See [SoS16] for more detail about the Bloch sphere and its relationship with SICs. 14.3 (a) The synthesis map of the Bloch vectors for Φ =(v,Sv,Ωv,SΩv) is

1 1 1 1 1 1 1 −1− 1 , √3 1 −1− 1 1 − − which clearly gives the vertices of a regular tetrahedron. 1 + c a ib (b) Taking the ﬁrst column of 2vv = gives the SIC ∗ a + ib 1− c − 1 3 √3 1 3 + √3 − , . 18 6√3 √6 18 + 6√3 i√6 − ± ± j k j k kb kb 14.4 (a) Since (S Ω )ab = ∑α (S )aα (Ω )αb = ∑α δa,α+ jω δαb = ω δa,b+ j,

~~k j k j ka k(b+ j) jk j k (Ω S )ab = ∑(Ω )aα (S )αb = ∑ω δaα δα,b+ j = ω δa,b+ j =(ω S Ω )ab. α α~~

~~(b) Use (a) and induction on r (it clearly holds for r = 0,1):~~

j r 1 (r 1)(r 2) jk (r 1) j (r 1)k j k (S Ω) = ω 2 − − S − Ω − S Ω 1 (r 1)(r 2) jk (r 1) j (r 1) jk j (r 1)k k 1 r(r 1) jk rj rk = ω 2 − − S − (ω − S Ω − )Ω = ω 2 − S Ω .

~~In particular, for d even, we have (SΩ)d = I, and so SΩ has order 2d. (c) Since cc = 1, repeated application of (a)− gives~~

j k 1 a b j k b a a b j a(k b) a k b h(S Ω )h− = S Ω S (Ω Ω − S− )= S Ω S (ω− − S− Ω − ) a(k b) a b j a k b a(k b)+b( j a) j k bj ak j k = ω− − S (Ω S − )Ω − = ω− − − S Ω = ω − S Ω .

~~jΩ k ωβkδ 14.5 Using, (S )αβ = α,β+ j, we have 536 17 Solutions~~

~~ωaα ω bβ jΩ k 1 jΩ k ωβkδ − (F(S )F− )ab = ∑∑(F)aα (S )αβ (F∗)βb = ∑∑ α,β+ j α β α β √d √d~~

~~1 aα (α j)k b(α j) jk bj 1 (a+k b)α = ∑ω ω − ω− − = ω− ω ∑ω − d α d α jk bj jk k j = ω− ω δa,b k =(ω− S− Ω )ab, −~~

j k 1 j k (R(S Ω )R− )ab = ∑∑(R)aα (S Ω )αβ (R∗)βb α β µa(a+d)δ ωβkδ µ b(b+d)δ = ∑∑ aα α,β+ j − βb α β (b+ j)(b+ j+d) bk b(b+d) j( j+d)+2bj bk = µ ω µ− δa,b+ j = µ ω δa,b+ j j( j+d) b( j+k) j( j+d) j j+k = µ ω δa,b+ j =(µ S Ω )ab,

~~j k 1 1 j k 1 jk k j 1 M(S Ω )M− = R(F− S Ω F)R− = R(ω− S− Ω )R− jk k( k+d) k j k k(k 2 j+d) k j k = ω− µ− − S− Ω − = µ − S− Ω − ,~~

~~j k 1 j k (Pσ (S Ω )Pσ− )ab = ∑∑(Pσ )aα (S Ω )αβ (Pσ∗)βb α β~~

~~βk σ 1kb = ∑∑δa σα ω δα β δ σβ = ω − δσ 1 σ 1 , , + j b, − a, − b+ j α β σ 1 σ σ 1 − bk j − k = ω δa,b+σ j =(S Ω )ab.~~

~~d 14.6 The matrices Rd and Ω 2 are diagonal, with~~

~~d j( j+d)d j( j+d) j2 j d (R ) jj = µ =( 1) =( 1) =( 1) =(Ω 2 ) jj. − − − d 1 d From (14.13), we have F(S 2 )F− = Ω 2 . 14.7 We have associativity~~

~~(A,zA)(B,zB) (C,zC)= AB,(zA B)zB (C,zC)= ABC,(zA BC)(zB C)zC , ◦ ◦ ◦~~

~~(A,zA) (B,zB)(C,zC) =(A,zA) BC,(zB C)zC = ABC,(zA BC)(zB C)zC, ◦ ◦ ◦ and identity (I,1)~~

~~(I,1)(A,zA)=(IA,(1 A)zA)=(A,zA), (A,zA)(I,1)=(AI,(zA I)1)=(A,zA), ◦ ◦ 1 1 1 1 1 1 1 with inverse (A,ZA) =(A ,z− A )=(A ,(ZA A ) ) − − A ◦ − − ◦ − − 17 Solutions 537~~

1 1 1 1 1 1 1 (A,ZA)− (A,ZA)=(A− ,Z− A− )(A,ZA)=(A− A,(Z− A− A)ZA)=(I,1), A ◦ A ◦ 1 1 1 1 1 1 1 1 (A,ZA)(A,ZA)− =(A,ZA)(A− ,Z− A− )=(AA− ,(ZA A− )(Z− A− ))=(I,1). A ◦ ◦ A ◦ 14.8 (a) Expanding, and using det(A)= αδ βγ = 1, gives −

~~cA(p,q)=(γ p1 + δ p2)(αq1 + βq2) p2q1 − = αγ p1q1 + βγ p1q2 +(αδ 1)p2q1 + βδ p2q2 − T = αγ p1q1 + βγ p1q2 + βγ p2q1 + βδ p2q2 = p σAq.~~

~~(b) Since cA(p + q, p + q)= cA(p, p)+ 2cA(p,q)+ cA(q,q), we have~~

zˆ (p + q) z (p + q) ( µ)cA(p,p)( µ)cA(q,q) ωcA(p,q) a a 1 = − c p−q p q = 2c p q = . zˆa(p)zˆa(q) za(p)za(q) ( µ) A( + , + ) ( µ) A( , ) − − Let B := ψb. Substituting into zab(p)= za(Bp)zb(p) gives

( µ)(ABp)1(ABp)2 ( µ)(ABp)1(ABp)2 ( µ)(Bp)1(Bp)2 zâb(p) − = zâ(Bp) − zˆb(p) − , ( µ)p1 p2 ( µ)(Bp)1(Bp)2 ( µ)p1 p2 − − − and cancellation givesz âb(p)= zâ(Bp)zˆb(p). From (14.19), (14.20), we get

jk jk ( k) j jk jk zˆF ( j,k)= ω− ( µ) − − = ω− ω = 1, − j( j+d) jk j( j+k) j( j+d) j2 j2 jd j zˆR( j,k)= µ ( µ) − = µ µ− ( 1)− = µ ( 1) = 1. − − − j k Since M = RF, this givesz ˆM =(zˆR ψF )zˆF = 1. For a = S Ω , ψa = I, so that ◦ zˆS jΩ k = zS jΩ k . A simple calculation (see Exer. 14.4) gives zS jΩ k , i.e.,

j k p1 p2 j k 1 kp1 jp2 p1 p2 kp1 jp2 (S Ω )S Ω (S Ω )− = ω − S Ω = z jΩ k (p)= ω − . ⇒ S α β T (c) For any B = SL2(Z2d), expanding (p + dj) σB(p + dj) gives γ δ ∈ αγ 2 2 2 βγ 2 (p1 + 2dj1 p1 + d j1)+ 2 (p1 p2 + dj1 p2 + dj2 p1 + d j1 j2) 2 2 2 2 2 T + βδ (p + 2dj2 p2 + d j ) αγ p + 2βγ p1 p2 + βδ p p σB p mod 2d. 2 2 ≡ 1 2 ≡ Hence the argument of (b) goes through.

~~14.9 (a) From zab(p)= zb(p)za(ψb(p)) and ψab(p)= ψa(ψb(p)), we obtain~~

~~ψ ψ p1 p2 p1 p2 b(p)1 b(p)2 c1 c2 c1 c2 c1 c2 zab(p)= zb(p) za(ψb(p)), ψab(p)1 ψab(p)2 ψb(p)1 ψb(p)2 ψa(ψb(p))1 ψa(ψb(p))2 c1 c2 c1 c2 c1 c2~~

~~a b j k a b 1 bj ak j k which isz ˆab = zˆb(zˆa ψb). Since S Ω (S Ω )(S Ω ) = ω S Ω , we have ◦ − − ωbj ak zˆSaΩ b ( j,k)= zSaΩ b ( j,k)= − . 538 17 Solutions~~

(b) Since det(S)= det(Ω)=( 1)d 1, and ( µ)d =( 1)d µd =( 1)d+1, we have − − − − − j k j k d( j+k) (d 1)( j+k) 2d( j+k) det(Uˆ p)= det(c c S Ω )=( µ) ( 1) − =( 1) = 1. 1 2 − − − d ˆ ˆ 1 Hence za(p) = det(za(p)Uψa(p))= det(aUpa− )= 1, i.e,.z ˆa(p) is a d–th root of unity. From Lemma 14.1, we calculate

~~µ(d+1)( j+k) zˆ ( j,k)= ω jk = ω j jk. F µ(d+1)(k j) − − − µ(d+1)( j+k) zˆ ( j,k)= µ j( j+d) = µ j( j 1). R µ(d+1)( j+ j+k) −~~

~~µ(d+1)( j+k) zˆ ( j,k)= µk(k 2 j+d) = µk(k 2 j+3). M µ(d+1)( k+ j k) − − − − (d+1)( j+k) µ σ 1 σ µ(d+1)((1 − ) j+(1 )k) zˆM( j,k)= σ 1 σ = − − . µ(d+1)( − j+ k)~~

1 These are indeed powers of ω (for the last, if d is even then σ,σ − are odd). α β T 14.10 (a) For M = γ δ , the condition M AM = A gives T α β 0 1 α β 0 αδ βγ 0 1 = = , γ δ 1 0 γ δ (αδ βγ) −0 1 0 − − − − which is equivalent to det(M)= αδ βγ = 1. − (b) The inverse of M SL2(Z) is given by the formula ∈ α β 1 δ β δ β γ δ = αδ βγ γ− α = γ− α . − − − (c) As in (b), the inverse in SL2(Zd) is given by α β δ β γ δ = γ− α . − a 14.11 Suppose that b Z∗ has odd order, say 2a + 1. Since b is a unit, we have ∈ d′

d 1 d 1 d 1 π bj( j+d) − bba j(ba j+d) − j2+ba+1dj − i ( j2+ba+1dj) √dcb,d = ∑ µ = ∑ µ = ∑ µ = ∑ e d . j Z j=0 j=0 j=0 ∈ d Evaluating the Gauss sum using the quadratic reciprocity law (14.108) gives

~~π i (d (ba+1d)2) 1 b2a+2d √dcb,d = √de 4d − = √d(√i) − . 17 Solutions 539~~

For d even, b must be odd, and for d odd, we may assume that b is odd (by replacing 2 1 d it by b + d if need be). Thus b 1 mod 8, so that cb d =(√i) . ≡ , − 14.12 (a) The coefﬁcient matrix of the k k linear system ×

1 1 1 P1 I λ λ λ 1 2 k P2 A  λ 2 λ 2 λ 2    2  1 2 k P3 = A . . . .  . .  .   .   k 1 k 1 k 1   k 1 λ λ λ Pk A   1− 2− k −    −       is the transpose of a Vandermonde matrix, and so is invertible. Thus the system can 2 k 1 be solved to give each Pj as a unique linear combination of I,A,A ,...,A − . (b) For the eigenvalues 1,ω,ω2,...,ωk 1 the coefﬁcient matrix is the k k Fourier − × matrix Fk (multiplied by √k). Since Fk is unitary, we can compute the inverse of the coefﬁcient matrix by taking its Hermitian transpose, to solve for Pω j

P 1 1 1 I 1 k 1 Pω 1 ω ω − A 1  2 2(k 1) 2  Pω2  1 ω ω −  A  = . . k . .  .  .  . .  .     2  k 1 Pωk 1  1 ωk 1 ω(k 1) A −   −   − −        (c) If A is d d, then by (b) the multiplicity of the eigenvalue λ is × 1 2 2 k 1 k 1 trace(Pλ )= d + λ trace(A)+ λ trace(A )+ + λ − trace(A − ) . k If A is unitary, i.e., A 1 = A , and k <ℓ k 1, then Aℓ = Aℓ k =(A )k ℓ gives − ∗ 2 ≤ − − ∗ −

~~ℓ k ℓ k ℓ k trace(A )= trace((A − )∗)= trace(A ), 1 k ℓ< . − ≤ − 2~~

~~2 3 1 (d) It follows from F = P 1 that F has order 4. Since F = F− = F, we have −~~

1 2 2 3 mλ = d + λ trace(F)+ λ trace(F )++λ trace(F) , 4 where, by the quadratic reciprocity law (14.108), 1 1 trace(F)= G(1,0,d)= (1 + i) 1 +( i)d , √d 2 − and 1 α 1 α 1 trace(F2)= ∑ ∑(ω2 j) = ∑ ∑1 = 3 +( 1)d . j d α d j:2 j=0 α 2 − 540 17 Solutions

~~Thus mλ depends on d modulo 4, as given by Table 14.6. 3 2 1 (e) Since Z = I, Z = Z− = Z∗, we have~~

~~1 2 mλ = d + λ trace(Z)+ λ trace(Z) . 3~~

~~2πi With τ = e 3 , the quadratic reciprocity law (14.108), for a = 3, b = c = d, gives~~

2 d 1 1 j( j+d)+2 j2 d 1 1 d 3 d 2dk(k+1) trace(Z)= ζ − ∑ µ = ζ − ζ − ∑ τ− √ √ 3 d j d k=0 2 + τ, d 0 mod 3; ζ 2 d 1 2 d ≡ = (2 + τ− )= (1 τ )(2 + τ− )= 1, d 1 mod 3; √3 3 −  ≡ 1 + τ, d 2 mod 3. ≡ Thus mλ depends on d modulo 3, as given by Table 14.7.

j2 14.13 (a) It sufﬁces to show that R has order d . Since R jj =( µ) and µ is a ′ − − primitive d′–th root of unity, the diagonal matrix R has order d′ (consider j = 1). 2πi (b),(c) Let τ = e ℓ , a primitive ℓ–th root of unity. Take a = ℓ, b = ℓd + 2(k j), c = d in the quadratic reciprocity law (14.108), to get −

~~d 1 1 ℓ 1 ℓ 1 − 2α j ℓα(α+d) 2αk (F− R F) jk = ∑∑(F− )α j(R )αβ (F)βk = ∑ µ− µ µ α β d α=0~~

π ℓ 1 π 1 d i ℓd (ℓd+2(k j))2 − i (dβ 2+(ℓd+2(k j))β) = e 4ℓd − − ∑ e− ℓ − d ℓ β=0 ℓ 1 1 d 1 ℓd j k 1 (k j)2 − d β 2+dβ ( j k)β = (√i) − ( 1) − µ− ℓ − ∑ ( 1) ℓ τ − . d ℓ − β=0 −

~~For d odd, and for d even and ℓ odd, we have~~

d β 2+dβ d β+dβ β( d +d) ( 1) ℓ =( 1) ℓ =( 1) ℓ = 1, − − − j k β j k so that the last sum becomes ∑β (τ − ) , which is ℓ if τ − = 1, i.e., j k 0 mod ℓ, and is zero otherwise, which gives (b). For d even and ℓ even, we have− ≡

π d β 2+dβ d β 2 i d β d β ( 1) ℓ =( 1) ℓ = e− ℓ 2 = τ− 2 , − − d β ∑ τ j k 2 d so the sum becomes β ( − − ) , which is ℓ if j k 2 0 mod ℓ, and is zero otherwise, which gives (c). − − ≡ 14.14 (a) From Exer. 14.13 (b), we have 17 Solutions 541

µ 1 ( j k)2 1 3 3 1 3d ( )− 3 − , j k 0 mod 3; (F− R F) jk = (√i) − − − ≡ d 0, j k 0 mod 3, − ≡ j k 1 ( j k) 1 ( j k)2 since ( 1) − =( 1) 3 − =( 1)− 3 − for j k 0 (mod 3). Multiplication −d 1 −1 3d − − ≡ by ( 1) − =( 1) − , followed by left and right multiplication by the diagonal − 2d a − matrices R 3 and R gives the desired formula for (Wa) jk. 2d a 1 3 3 (b) The symplectic operation [M], M := R 3 F− R FR, has order 3, i.e., (cM) = I, for some scalar c T, since it has the following symplectic index of order 3 ∈ 2d a 1 3 2 3 1 0 3 0 1 − 1 0 0 1 1 0 w = − − = . a 1 + 2d a 1 1 1 1− 0 1 1 1− 0 1 1 3 2 1 1 Since cM is unitary, we can determine c from (cM) =(cM)− = c− M∗ (at some nonzero entry). By the formula of (a) divided by ( 1)d 1 =( 1)1 3d, we calculate − − − − d 1 2 2 − 3 1 3d 1 α2+α2 1 α2+ 2d aα2 (M ) = ∑ M α Mα = (√i) ∑ ( µ) 3 3 3 . 00 0 0 d − − − α=0 α 0 mod 3 − ≡ 2 Since ( 1)3β =( 1)β = µdβ and 3 α, the sum over α 0 (mod 3) evaluates to − − | ≡ d 1 d 1 d 1 3 3 3 πi 2 d 1 α2 − 3β 2 − 3β 2 dβ − (β + β) d 1 d ∑( µ) 3 = ∑ ( µ) = ∑ µ + = ∑ e d/3 3 = (√i) 3 , 3 − α − β=0 − β=0 β=0

3 (1 3d) by the quadratic reciprocity law (14.108) Since (M∗)00 = d (√i)− − , we have (1 3d) (M∗)00 (√i)− − d d 3 √ 4( 3 1) 3 1 d 1 c = 2 = d =( i) − =( 1) − =( 1) − , (M )00 2(1 3d) 1 3 − − (√i) − (√i) −

d 1 so that c =( 1) , and Wa has order 3. − − 1 3 3 1 3d (c) Since F− R F is circulant, with constant diagonal d (√i) − , we have d 1 2d a+1 1 3 d 1 3 1 3d 2d a+1 trace(Wa)=( 1) − trace(R 3 F− R F)=( 1) − (√i) − trace(R 3 ). − − d 2d Since τ3 = 1, µ 3 = τ, and 3 d, we calculate | d 1 d 1 d 1 2d − 2d − 2d − 2 2 trace(R 3 a+1)= ∑ µ( 3 a+1) j( j+d) = ∑ (τaµ)( 3 a+1) j( j+d) = ∑ τaj µ j +dj j=0 j=0 j=0 d d 2 3 1 2 3 1 − 2 2 2 2 − 2 = ∑ ∑ τa(3k+c) µ(3k+c) +d(3k+c) = ∑ τac µcd+c ∑ (µ3)3k +(d+2c)k. c=0 k=0 c=0 k=0 542 17 Solutions

~~By the quadratic reciprocity law, the sum over k can be written as~~

d 1 3 πi 2 π 2 π − (3k +(d+2c)k) d i d d 2c 2 i d β 2 d 2c β ∑ e d/3 = e 4d ( ( + ) ) ∑ e 3 ( 3 +( + ) ) 9 − − k=0 β=0 2 d cd c2 1 d d β 2 d 2c β = µ (√i) ∑ ( τ) 3 +( + ) , 9 − − − β=0 −

~~d c d +c where the sum over β simpliﬁes to 1 + τ 3 − + τ 3 . Thus~~

2 d 1 3 1 3d d 1 d ac2 d c d c trace(W )=( 1) (√i) (√i) ∑ τ 1 + τ 3 + τ 3 + a − d − 9 − − − c=0 i d a d = 1 + 2τ 3 + 2τ (1 τ 3 ) . −√3 − b b 2b b 2 2 2 (d) Since τ = τ− = τ =(τ ) , the eigenvalues λ = 1,τ,τ of Wa satisfy λ = λ , and so, by Exer. 14.12, the multiplicity of the eigenvalue λ of Wa is

~~1 2 mλ = d + λ trace(W )+ λtrace(W ) . 3 a a~~

d which depends on a and 3 (mod 3). d d (i) For d 3 mod 9, d = 3, 1 (mod 3), so that τ 3 = τ, trace(W1)= 3, and ≡ 3 ≡ d+6 λ 1 2 3 , = 1; mλ = d + 3(λ + λ) = d 3 λ τ τ2 3 −3 , = , . d (ii) For d 6 mod 9, 2 (mod 3), so that trace(W2)= 3, and ≡ 3 ≡ − d 6 λ 1 2 −3 , = 1; mλ = d 3(λ + λ) = d+3 λ τ τ2 3 − 3 , = , . 1 1 1 (e) The fact cW0 =(R− F)− Z(R− F), with c a scalar, follows from the conjugacy

~~1 0 1 1 0 − 0 1 w g 1zg g : 0 = − , = 1− 1 = 1 1 1− 0 . 1 d 1 1 1 By matrix multiplication, τ − W0 =(R− F)− Z(R− F) is equivalent to~~

1 1 d 1 1 d RFRF− RF− =( τζ) − I =(√i) − I. − 1 d From the calculation (14.62), we have RFRFRF =(√i) − I, and so it sufﬁces to 1 1 prove that F− RF− = FRF. By the quadratic reciprocity law, we have 17 Solutions 543

d 1 d 1 π 1 1 1 − 2 jα α(α+d) 2kα 1 − i (α2+α(d 2 j 2k)) (F− RF− ) jk = ∑ µ− µ µ− = ∑ e d − − d α=0 d α=0 π 1 i (d (d 2 j 2k)2) 1 1 d j+k ( j+k)2 = √de 4d − − − = (√i) − ( 1) µ− . d √d −

~~1 1 1 1 From this, we have (F− RF− ) jk =(F− RF− ) j, k =(FRF) jk. − − 14.15 By the commutativity relation (14.5), we have~~

p1 p2+q1q2 p1 p2 q1 q2 p1 p2+q1q2 p2q1 p1+q1 p2+q2 Dˆ pDˆ q =( µ) S Ω S S =( µ) ω S Ω − − p2q1 p1q2 (p1+p2)(q1+q2) p1+q1 p2+q2 p,q =( µ) − ( µ) S Ω =( µ) Dˆ p q. − − − + From this we obtain

~~ˆ ˆ µ p, p ˆ ˆ 1 ˆ DpD p =( ) − Dp p = I = D−p = D p, − − − ⇒ − p,q q,p 2 p,q p,q Dˆ pDˆ q =( µ) − Dˆ qDˆ p =( µ) Dˆ qDˆ p = ω Dˆ qDˆ p. − − Now~~

(p1+dq1)(p2+dq2) p1 p2 d(p1q2+p2q1+dp1q1) Dˆ p dq =( µ) S Ω =( µ) Dˆ p. + − − Since ( µ)d =( 1)d+1, for d odd the scalar on the left hand side is 1, and for d even it is− − p q +p q p q p q p,q ( 1) 1 2 2 1 =( 1) 2 1− 1 2 =( 1) . − − − p p p p 14.16 (a) Since Dˆ p =( µ) 1 2Uˆ p, where Uˆ p := Up mod d = S 1 Ω 2 , we have −

~~p1 p2 1 χ,Bp (Bp)1(Bp)2 2 a( µ) Uˆ pa− = ω ( µ) UˆBp, p Z . − − ∀ ∈~~

Since Bp Ap mod d, we have ψa = A, and ≡ χ,Ap (Bp)1(Bp)2 p1 p2 2 za(p)= ω ( µ) − , p Z , − ∈ d where cB(p, p) :=(Bp)1(Bp)2 p1 p2, − is calculated modulo d′ (but depends only on p mod d). As in Exer. 14.8 (a), we T have cB(p, p)= p σB p. (b) An element [B, χ] ker f provided θ([B, χ])=(I,1). For d odd, this implies ∈ χ,p B = ψa = I, while za(p)= ω = 1 gives χ = 0, and so f is 1–1. For d even, we obtain (14.45) since

~~1 + rd sd ψa = B mod d = I, det(B)= 1 = B = , ⇒ td 1 + rd~~

~~td dj2 d j and using σB = and ( µ) = ω 2 gives sd − 544 17 Solutions~~

~~d χ 2 2 χ χ d d ,p dtp +dsp 2 p1 1 p2 tp1+ sp2 s 2 za(p)= ω ( µ) 1 2 = 1 = ω − ω 2 2 = χ = . − ⇒ t d 2 (c) Each Appleby index has the form [B,0], so that~~

m (d+1)m za(p)=( µ) = µ , m :=(Bp)1(Bp)2 p1 p2. − − (i) Since 2d,d2 0 mod d , dj2 dj mod d , we calculate ≡ ′ ≡ ′ 2 m =(d 1)p1((d + 1)p1 +(d 1)p2) p1 p2 p 2p1 p2 mod d′, − − − ≡ 2 − 2 2 2 (d + 1)m p 2p1 p2 + dp 2dp1 p2 p 2p1 p2 + dp2 mod d′, ≡ 2 − 2 − ≡ 2 − so that 0 1 ψ z j k µk(k 2 j+d) a = 1 −1 , a( , )= − . − Thus a = Z (up to a scalar). (ii) We have 4 m =(p1 +(d + 3)p2)(( d 1)p1 +(d 2)p2) p1 p2 3 − − − 4 2 2 ( d 1)p 6p1 p2 +(d 6)p mod d′, ≡ 3 − 1 − − 2

~~1 2 2 which gives (d + 1)m ( d 1)p 6p1 p2 +(d 6)p mod d , so that ≡ 3 − 1 − − 2 ′~~

~~1 3 1 d 1 j2 6 jk d 6 k2 ψ z j k µ( 3 ) +( ) a = 3α 2 , a( , )= − − − . − We have the factorisation~~

d 1 1 3 d d + 1 3 1 0 − 0 1 − 1 0 − 0 1 1 0 3 (d + 1)Fa = d 3 = − − , − d 2 1 1 1 0 1 1 1 0 1 1 3 − d 1 1 3 d and so, by Lemma 14.3, we have a = R − F− R− FR 3 (up to a scalar multiple). (iii) Let β = √d + 1. We have

m =( √d + 1p1 + dp2)(dp1 +(d √d + 1)p2) − − 2 2 d√d + 1(p + p1 p2 + p )+ dp1 p2 mod d′, ≡− 1 2 2 For d odd, m(d + 1) (d + d)p1 p2 0 mod 2d, so that za(p)= 1. For d even, ≡ ≡ 2 2 (d+1)m m √d+1(p +p1 p2+p )+p1 p2 za(p)= µ = µ =( 1)− 1 2 − p2 p p p2 p p p p d p p =( 1) 1+ 1 2+ 2+ 1 2 =( 1) 1+ 2 = ω 2 ( 1+ 2). − − Thus 17 Solutions 545

~~β d 1, d odd; ψ ω 2 (d+1)( j+k) a = − β , za( j,k)= = ( 1) j+k, d even, − − and we can take~~

1 d(d+1) 1 d(d+1) P β , d odd; a = S 2 Ω 2 P β = −d d − S 2 Ω 2 P β , d even. − (iv) We observe that if d is even, then k is even and κ is odd, so that κd d mod d . ≡ ′ Since det(Fc)= 1, the operation is antiunitary. We have the factorisation − d 2κ κ 0 1 1 0 F = , c d− κ d + 2κ 1− 0 0 1 − − where

d+2 1 d 2 d 2κ κ 1 0 κ 0 0 1 − 1 0 − = . d− κ d + 2κ 1 1 0 d 3κ 1− 0 1 1 − − Since κ(d 3κ) 1 mod d , κ is a unit in Z , and so we may take a = Rd+2Pκ F 1Rd 2FC. − ≡ ′ d′ − − 14.17 (a) The commutativity follows from (14.5), i.e.,

~~(SΩ ℓ)(SΩ m)= S(ωℓSΩ ℓ)Ω m = ωℓS2Ω ℓ+m.~~

(b) Since A,B = trace(AB ), we have ∗ ℓ m ℓ m 1 ℓ m SΩ ,SΩ = trace(SΩ Ω − S− )= trace(Ω − )= 0, ℓ m Z∗. − ∈ d ℓ 1 1 ℓ (c) A symplectic index for a =(R F)− = F− R− is given by

~~0 1 1 0 ℓ 1 B = = , 1 0 ℓ 1 −1 0 − − − so that ( ℓ j+k)( j) jk ℓ j2 2 jk za( j,k)=( µ) − − − =( µ) − . − − Thus~~

ℓ 1 ℓ ℓ ℓ 1 ℓ1+ℓ 1 ℓ 1 (R F)− SΩ (R F)= aSΩ a− = za(1,ℓ)S− Ω − =( µ)− Ω − , − ℓ ℓ j (d 1)ℓ j i.e., SΩ has distinct eigenvalues ( µ)− ω− = µ − ω− with corresponding ℓ − eigenvectors R Fe j. (d) The matrices in the ﬁrst set are diagonal, and so have eigenvectors e j . Up to ℓ { } multiplication by a scalar, the matrices in Mℓ are powers of SΩ , and so have the same eigenvectors. We can compute the diagonalisation by

ℓ r 1 r(r 1)ℓ r rℓ ℓ 1 r rℓ r (SΩ ) = ω 2 − S Ω = ( µ)− Ω − =( µ)− Ω − , − − 546 17 Solutions or ℓ 1 r rℓ ℓ ℓr+rℓ r r2ℓ r (R F)− S Ω (R F)= za(r,rℓ)S− Ω − =( µ)− Ω − . − The eigenvalues will not all be distinct if r is not a unit. 14.18 (a) Recall the determinant is the product of the eigenvalues. We have

~~d 1 1 − j ∑ j 1 d(d 1) d(d 1) d 1 det(S)= det(FSF− )= det(Ω)= ∏ ω = ω j = ω 2 − = µ − =( 1) − . j=0 −~~

~~From the formula for the sum of consecutive squares, we have~~

~~d 1 1 1 ∑ j( j + d)= d(d + 1)(2d + 1)+ d2(d + 1)= d(d + 1)(5d + 1). j=1 6 2 6~~

~~2πi 3 Since ζ = e 24 = τ2√i , we have that det(R) depends on d mod 12, i.e.,~~

~~d j( j+d) 1 d(d+1)(5d+1) 2(d+1)(5d+1) 1 d2 3(d+1)2 det(R)= ∑ µ = µ 6 = ζ = τ − i . j=1~~

~~The determinant of F and Z can be calculated from the Tables 14.6 and 14.7. Since det(F) depends on d mod 8, we ﬁrst consider det(Z), for which the table gives~~

~~(d 1)2 1, d 1 mod 3 det(Z)= τ − = ≡ τ otherwise.~~

~~d 1 Since Z = ζ − RF, we have~~

~~d 1 1 1 d(d 1) 2(d+1)(5d+1) 8(d 1)2 det(F)= det(ζ − I)− det(R)− det(Z)= ζ − − ζ − ζ − 2 3d2 3d+6 3 3d2 3d+6 d d+2 1 1 d(d+1) = ζ − − =(√i )− − = √i− − = i − 2 .~~

~~1 d (b) Since det(c− M)= c− det(M), c is a d–th root of det(M). Since µ is a d–th root of 1, we can take − S, d odd; Ω, d odd; Sˆ := Ωˆ := µS, d even; µΩ, d even.~~

For F, we can take c = ia to be a fourth root of unity when d 0 mod 4, by solving ≡ 2ad d2 d+2 cd = √i = √i− − 2a d2 d + 2 mod 8, ⇐⇒ ≡− − 1 e.g, for d 3 mod 8, 6a 6 gives a = 1, Fˆ = c− F = iF. For d 0,4 mod 8, we must take≡c to be a 4d–th≡ root of unity. By similar calculations− for≡Z, we obtain 17 Solutions 547

1 µ 2 F, d 0 mod 8; − ≡ F, d 1,6 mod 8; 1  ≡ ω 3 , d 0 mod 3;  − ˆ iF, d 2,7 mod 8; ˆ ≡ F =  ≡ Z = Z, d 1 mod 3;  iF, d 3 mod 8; ≡ −1 ≡ τZ, d 2 mod 3; µ 2 F, d 4 mod 8; ≡  ≡   F, d 5 mod 8;  − ≡   1 and one can take Rˆ = ZˆFˆ − . (c) The subgroups of Sˆ,Ωˆ ,Fˆ,Rˆ containing Sˆ,Ωˆ are canonical abstract error groups. For d = 2, the canonical abstract error group, index group pairs are

~~8,4 , 4,2 16,9 , 8,3 , 24,3 , 12,3 , 48,28 , 24,12 . These include the two which are not dicyclic groups (see Example 13.21). For d = 3,~~

27,3 , 9,2 , 54,8 , 18,4 , 81,9 , 27,3 , 108,15 , 36,9 , 162,14 , 54,5 , 216,88 , 72,41 , 648,532 , 216,153 . For d = 4, there are 19 canonical abstract error groups. The ﬁrst few are

64,19 , 16,2 , 128,749 , 32,34 , 128,782 , 32,31 , 128,545 , 32,24 , 192,4 , 48,3 , 256,24064 , 64,242 , 256,21237 , 64,236 , 256,17275 , 64,216 , 256,217 , 64,18 , 256,395 , 64,34 . The remaining ones have orders 384, 512, 768, 1024, 1536, 3072.

14.19 (a) Since F and R have ﬁnite orders (4 and d′) their eigenvalues, and hence determinants, are roots of unity. Thus F and R can be multiplied by appropriate roots of unity to obtain matrices Fˆ and Rˆ with determinant 1 (see Exer. 14.18). The group Fˆ,Rˆ contains all elements of F,R up to multiplication by an n–th root of unity (where n is ﬁxed). Since Fˆ,Rˆ SLd(C), the only scalar matrices that can belong to it are those given by d–th roots⊂ of unity. Since

CSp(d) F,R Fˆ,Rˆ,ωI = = , [I] ∼ [I] F,R ∼ ωI ∩ and CSp(d)/[I] is finite, we conclude that F,R must be finite. 1 2 (b) Since F commutes with P 1 =(P 1)− = F , it suffices to show that R does also − − δ µα(α+d)δ δ µ j( j+d)δ µ j( j+d)δ (P 1RP 1) jk = ∑∑ j, α α,β β, k = − − jk = jk =(R) jk. − − α β − −

~~14.20 (a) Since A = A∗, we have g(A)∗ = g(A)= g(A∗) if and only if~~

g(akj)= g(akj), j,k, ∀ 548 17 Solutions which is equivalent to g commuting with conjugation on Q( a jk ) (since g fixes Q). { } T T (b) If g commutes with conjugation, then g(v∗)= g(v) = g(v) = g(v)∗, so that g(vv∗)= g(v)g(v∗)= g(v)g(v)∗, g(v)∗g(v)= g(v∗)g(v)= g(v∗v)= g(1)= 1, i.e., g(vv∗) is a rank one orthogonal projection, with the necessity that g commute with conjugation from (a). (c) Since g(Π) is an orthogonal projection, by (b), g must commute with complex conjugation on Q(Π). Since g is an automorphism of E fixing 1 Q, it maps µ to another 2d–th root of unity, and so g commutes with complex conjugation∈ on Q(µ), and hence on the SIC field E = Q(Π, µ). 0 1 14.21 We observe that b F 1C has the (extended) symplectic index B . = − = 1− 0 (a) It suffices to show the symplectic indices of b and Z do not commute:−

0 1 0 1 1 1 1 0 0 1 0 1 Bz = − − = − = = − − = zB. 1 0 1 1 0 1 1 1 1 1 1 0 − − − − − From (a), it follows that the subgroup generated by b and Z is nonabelian (for d 2). (b) Since b and Z are symmetries of all the d = 2,3 SICs, their symmetry groups≥ are nonabelian. (c) If b and Z are symmetries of a SIC, then the symmetry group is nonabelian. Sim- ilarly, if b and M! are symmetries of a SIC, then the symmetry group is nonabelian, since their symplectic indices do not commute:

~~3 d d + 1 3 − d + 2 3 d 1 d + 1 3 B d 3 = 3 = − 3−d = d 3 B. − d 2 d 1 3 d + 2 − − d 2 3 − − − 3 3 −~~

~~14.22 Since cc = 1, c T, [a] Π =[a] (vv∗) is well defined, and we observed that it maps (Weyl–Heisenberg)∈ SIC fiducials to SIC fiducials. For a unitary, we have~~

~~1 [a] Π =(av)(av)∗ = a(vv∗)a∗ = aΠa− . 1 1 1 1 For aC antiunitary, we observe (aC)− = C− a− and Π = CΠC− , so that~~

1 1 [aC] Π =(aCv)(aCv)∗ =(av)(av)∗ = avv a∗ = aΠa− =(aC)Π(aC)− . ∗ Thus, for [a] PEC(d), [a] Π = aΠa 1, and so is an action on the SIC fiducials. ∈ − 14.23 (a) It suffices to show that the conjugates of the generators of E are in E. 1 T T Clearly µ = µ− E. Since Π ∗ =(vv∗)∗ = vv∗ = Π, we have Π =(Π ∗) = Π , i.e., the conjugate∈ of an entry of Π is an entry of Π. (b) Since [a] (vv∗) :=(av)(av)∗, [a] PEC(d), defines an action of the extended Clifford group on the Weyl–Heisenberg∈ SIC fiducial projectors (see Exer. 14.22), it sufficies to show that the entries of Π ′ =[a] Π belong to E = Q(Π, µ) for the 1 nonscalar generators a = S,Ω,F,R,C (Theorem 14.1). For a unitary, Π ′ = aΠa− , 17 Solutions 549 so the entries of Π ′ belong to E provided that the entries of a do. This works for a = S,Ω,R (since µ E by definition). For a = F, we modify this argument. We ∈ 1 observe that Π ′ =(√dF)Π(√dF)− , where √dF has entries in E (formerly the SIC field was defined to contain √d, so that F itself would have entries in the SIC field [AYAZ13]). Finally, for a = C, Π ′ = Π, which has entries in E by (a). 1 Π [a]− 1 1 14.24 (a) We have χp = trace(a− ΠaDˆ p)= trace(ΠaDˆ pa− ), so (14.43) gives

~~1 Π χ[a]− Πω b,Bp ˆ ω b,Bp χΠ p = trace( DBp)= Bp.~~

~~Replacing Π by [a] Π, gives the stated formula. T T (b) Since Π ∗ = Π, we have Π = Π . Using trace(A)= trace(A ), we calculate~~

~~χΠ χΠ T Π T ˆ ˆ T Π Π ˆ T p = p = trace( Dp)= trace(Dp )= trace( Dp ).~~

p p p p T 1 T Since Dˆ p =( µ) 1 2 S 1 Ω 2 and S = S , Ω = Ω, (14.5) gives − − ˆ T µ p1 p2 Ω p2 p1 µ p1 p2 ω p1 p2 p1 Ω p2 µ p1 p2 p1 Ω p2 ˆ Dp =( ) S− =( ) − S− =( )− S− = D Jp, − − − − so that χΠ Π ˆ T Π ˆ χΠ p = trace( Dp )= trace( D Jp)= Jp. − − (c) By part (a), the formula holds for an Appleby index with det(B)= 1. Hence it sufﬁces to consider an antiunitary [aC], with Appleby index [BJ,b]=[B,b][J,0], 1 1 where det(BJ)= 1. Now [aC] Π = aCΠCa− = aΠa− =[a] Π. Since, a C(d) has Appleby index−[B,b], by (a) and (b), we have ∈

χΠ χΠ ω b,Bp χ[a] Π ω b,Bp χ[aC] Π Jp = p = Bp = Bp . − Replacing p by Jp above gives − Π Π b, BJp [aC] Π b,det(BJ)BJp [aC] Π χ χ ω χ ω χ p = ( J)2 p = − BJp = det(BJ)BJp. − − 0 1 α β 14.25 First suppose that B = Fz = − . Then a matrix A = GL2(Zd ) 1 1 γ δ ∈ ′ − is in C(Fz) if and only if AFz = FzA, i.e., γ = β, α + β = δ, which gives − α β 1 0 0 1 A = = α + β = αI βFz. β α + β 0 1 1 1 − − −

~~Therefore C(Fz)= αI + βFz : α,β Zd GL2(Zd ). By the Cayley–Hamilton {2 ∈ ′ }∩ ′ theorem, we have Fz + Fz + I = 0, so that~~

(α1I + β1Fz)(α2I + β2Fz)= α1α2I +(α1β2 + α2β1)Fz + β1β2( Fz I) − − =(α2I + β2Fz)(α1I + β1Fz), 550 17 Solutions and C(Fz) is abelian. Thus C(Fz) is the unique maximal abelian subgroup containing Fz, and hence containing S0(Π) and S(Π) (which are abelian and contain Fz). 1 Now suppose that B = AFzA , for some A ESL2(Z ). Then − ∈ d′ 1 1 C(B)= AC(Fz)A− = αI + βAFzA− : α,β Z GL2(Z ), { ∈ d′ }∩ d′ which is abelian, since it is the conjugate of an abelian group. It is the maximal abelian subgroup containing B, and hence containing S0(Π) and S(Π), as before. 14.26 (a) It sufﬁces to show this for the generators a = S,Ω,F,R (Theorem 14.1). d 1 d d d d For these, we have S,Ω,R,√i − F Q(µ) × E × . (b) It sufﬁces to consider the generators∈ of part⊂ (a). By (14.86), we have g(S)= S, 2 2 kg j kg j kg g(Ω)= Ω , and, similarly g(( µ) δ jk)=( µ) δ jk gives g(R)= R . Since g commutes with conjugation, we− have −

~~d 1 d 1 (g(√i − F))∗g(√i − F)= g(F∗F)= g(I)= I,~~

~~d 1 so that g(√i − F) is unitary, and it is in the Clifford group, since~~

~~d 1 j k d 1 1 j k 1k 1 k 1 jk k 1k j g(√i − F)S Ω g(√i − F)− = g(FS Ω g− F− )= g(ω− g− S g− Ω − ) 1 jk k k kg j = ω− S g− Ω − .~~

(c) We observe g([a]) is well deﬁned, since if b [a] with b = za, z E, z = 1, then g(z) 2 = g(z)g(z)= g(z)g(z)= g(zz)= g(1)=∈1, so that g(b)= g∈(z)g(|a|) [g(a)]. | | ∈ This deﬁnes an action of Gc, since 1([a])=[1(a)]=[a] and

~~g(h([a])) = g([h(a)])=[g(h(a))] = [(gh)(a)]=(gh)([a]).~~

~~(d) The Appleby index multiplication [B1,b1][B2,b2]=[B1B2,b1 + B1b2] satisﬁes~~

~~1 1 1 [HgB1Hg− ,Hgb1][HgB2Hg− ,Hgb2]=[Hg(B1B2)Hg− ,Hg(b1 + B1b2)], and so it sufﬁces to show the formula for the generators. From (b), we have~~

~~g( f ([0, p])) = g([Dp])=[DHg p]= f ([0,Hg p]), and with Ba denoting the symplectic index of a symplectic operation [a],~~

kg 1 0 1 0 1 0 1 0 1 Bg(R) = B kg =(BR) = = 1 = HgBRHg− , R kg 1 0 kg 1 1 0 k− g 1 0 kg− 1 0 0 1 1 0 1 Bg(F) = = 1 = HgBF Hg− . kg 0 0 kg 1 0 0 k− − − g 14.27 (a) Since [a] Π = aΠa 1, we have − 17 Solutions 551

1 1 1 1 1 1 [b] SΠ bΠb− = Π aba− (aΠa− )(aba− )− = aΠa− ∈ ⇐⇒ ⇐⇒ 1 1 [a][b][a]− =[aba− ] S[a] Π . ⇐⇒ ∈ (b) By Exer. 14.26, for any [a] PEC(d), we may assume that Q(a) E. Thus ∈ ⊂ 1 [a] SΠ [a] Π = Π aΠa− = Π, where Q(a) E ∈ ⇐⇒ ⇐⇒ ⊂ 1 g(a)g(Π)g(a)− = g(Π) [g(a)] g(Π)= g(Π) ⇐⇒ ⇐⇒ g([a])=[g(a)] S Π . ⇐⇒ ∈ g( )

~~(c) The ﬁducial g(Π j) is on some orbit Ok, and so there is [a] PEC(d) with ∈~~

~~g(Π j)=[a] Πk. By (a) and (b), we have~~

~~1 Π Π Π Sg(Π j) = S[a] Π = g(S j )=[a] S k =[a]S k [a]− . k ⇒~~

~~Let [aL] SΠ be canonical order 3, with symplectic index L. Then by the above, ∈ j there is some [aM] SΠ , with symplectic index M, and ∈ k 1 g([aL])=[g(aL)]=[a][aM][a]− .~~

~~1 Let [B,q] be an Appleby index of a. Since HgLHg− is a symplectic index of g([aL]), it follows from the above that the Appleby indices satisfy~~

1 1 1 [HgLH− ,0] [B,q][HgMH− ,0][B,q]− mod K, g ≡ g where K is the kernel of the Appleby indexing homomorphism fE of (14.57), which 1 is the kernel of f (see Theorem 14.2), and a b (mod K) means that a− b K. 1 1≡ ∈ Expanding [HgLH ,0][B,q] [B,q][HgMH ,0], gives g− ≡ g− 1 1 1 [HgLH− B,HgLH− q] [BHgMH− ,q] mod K. g g ≡ g 1 For d odd, K is trivial, so that HgLHg− q = q. For d even, multiplication by an element of K adds (s d ,t d ), s,t 0,1 , to the second index, so that 2 2 ∈ { }

~~1 0 mod d, d is odd; (A I)q =(HgLHg− I)q = d − − 0 mod 2 , d is even,~~

1 where A := HgLH mod d has trace 1 (since aL is canonical order 3). By (14.71), g− − we have A2 = A I, so that ( A 2I)(A I)= 3I, and we have − − − − − 0 mod d, d is odd; 3q =( A 2I)(A I)q = d − − − 0 mod 2 , d is even. 552 17 Solutions

~~(d) If d is odd, then 3 is a unit, so that 3q = 0 gives q = 0. If d is even, we have~~

s d q = 2 , s,t 0,1 . t d ∈ { } 2 1 + rd sd s d Multiplication of [B,q] by [ , 2 ] K gives a symplectic index. td 1 + rd t d ∈ 2 Hence, if Π = Π j is a centred fiducial, then g(Π)=[a] Πk, with [a] symplectic, so that 1 Sg(Π) = S[a] Π =[a]SΠk [a]− , k which consists of symplectic operations, i.e., g(Π) is a centred fiducial. d Remark: If d 0 mod 3, then it can be shown that q is unique, with q 0 mod 3 . (e) We show that≡ if Π is strongly centred, then all centred fiducials on≡ its extended Clifford orbit are strongly centred. By (c), every centred fiducial on the extended Clifford orbit of Π has the form [a] Π, where [a] is symplectic, with symplectic index B. By (14.89), we have

[a] Π Π χ = χ E1, det(B)Bp p ∈ so that [a] Π is strongly centred. (f) For the ﬁrst part, we need only consider the case d 0 mod 3. Here, we have Π Π ≡ d g( j)=[a] k, where [a] has extended Appleby index [B,q] with q 0 mod 3 , d ≡ i.e., q j = α j , α j 0,1,2 . By (14.89), 3 ∈ { } Π χΠ ω q,det(B)Bp χ[a] k 2 p = , p Z , det(B)Bp ∀ ∈ d′ π q 2 i α which implies that the third roots of unity ω j =(e 3 ) j are in E1. Now

2πi 2d 2d 2πi g1(e 3 )= g1(( µ) 3 )=( µ)− 3 = e− 3 , − − q so ω j is ﬁxed by g1 (is in E1) if and only if a j = 0, i.e., q = 0 and [a] is symplectic. Now consider all d. Since g1 G0, we have g1(Π j)=[a] Π j, where [a] is sympletic, ∈ with sympletic index B. Let Π = Π j. Then by (14.90) and (14.89), we have

g (Π) Π Π [a] Π g (Π) χ 1 = χ , χ = χ = χ 1 , p, Jp p p det(B)Bp det(B)Bp ∀ so that J = det(B)B, which gives B = J. The symplectic operation with sympletic − index B = J is P 1C, so that − − Π Π Π 1 Π g1( )=[P 1C] = P 1gc( )P−1 = P 1 P 1. − − − − − 14.28 (a) The composition is given by the multiplication formula, since

~~(a1g1) (a2g2)(v)=(a1g1)(a2g2(v)) = a1g1(a2)g1(g2(v)) = a1g1(a2)g1g2(v), ◦ 17 Solutions 553~~

~~and the product is in X (see Exer. 14.26). Since composition is associative, it sufﬁces to verify the formula given for the inverse, i.e.,~~

~~1 1 1 1 1 1 1 1 1 1 (g− (a− )g− )(ag)= g− (a− )g− (a)g− g = g− (a− a)1 = g− (I)= I,~~

1 1 1 1 1 1 1 (ag)(g− (a− )g− )= ag(g− (a− ))gg− = aa− 1 = I. (b) The multiplication is well deﬁned, since for [a] PC(d) one can always choose d d ∈ a E (see Exer. 14.26), and in its deﬁnition a j is unique up to multiplication by ∈ × a unit scalar c j E, and therefore ∈

~~([c1a1]g1)([c2a2]g2)=[c1a1g1(c2a2)]g1g2 =[c1c2a1g1(a2)]g1g2 =[a1g1(a2)]g1g2.~~

d d The group properties can veriﬁed by taking representatives in E × , and using (a). It is clear PC(d) and Gc are subgroups. PEC(d) appears as the subgroup EC(d) gc (it is easily checked the multiplication is consistent). × (c) The formula for [a]g Π is well deﬁned (multiplying a by a unit scalar in E doesn’t change its value). It gives a group action, since [I]1 Π = Π, and 1 1 1 [a1]g1 ([a2]g2 Π)=[a1]g1 (a2g2(Π)a− )= a1g1(a2)g1(g2(Π))g1(a− )a− 2 2 1 1 =(a1g1(a2))(g1g2)(Π)(a1g1(a2))− = ([a1]g1)([a2]g2) Π.

If Π = vv∗ is stabilised by ag, i.e., (agv)∗(agv)= vv∗, then unit vectors agv and v gives the same rank one orthogonal projector, and so must be unit scalar multiples of each other. Since [a]g has ﬁnite order (Gc is assumed to be a ﬁnite abelian group), the scalar is a root of unity. (d) It is easy to verify that this multiplication gives a group, with identity [I,0,1], and inverse 1 1 1 1 1 1 [B,b,g]− =[H− B− Hg, H− B− b,g− ]. g − g Further, we have

~~1 fG B b g B b g f B H B H b B H b g g c ([ 1, 1, 1][ 2, 2, 2]) = ([ 1 g1 2 g−1 , 1 + 1 g1 2]) 1 2, and since f is a homomorphism~~

fGc ([B1,b1,g1]) fGc ([B2,b2,g2]) = f ([B1,b1])g1 f ([B2,b2])g2 f B b g f B b g g f B b f H B H 1 H b g g = ([ 1, 1]) 1( ([ 2, 2])) 1 2 = ([ 1, 1]) ([ g1 2 g−1 , g1 2]) 1 2 f B b H B H 1 H b g g f B H B H 1 b B H b g g = ([ 1, 1][ g1 2 g−1 , g1 2]) 1 2 = ([ 1 g1 2 g−1 , 1 + 1 g1 2]) 1 2.

~~Thus fGc is a homomorphism, with kernel ker( fGc )= [B,b,1] : [B,b] ker( f ) . (e) The map Θ is a homomorphism, since { ∈ }~~

Θ j1 j2 Θ j1 j1 j1 j1+ j2 ([B1,b1,gc ][B2,b2,gc ]) = ([B1J B2J− ,b1 + B1J b2,gc ]) j1 j1 j1+ j2 j1 =[B1J B2J− J ,b1 + B1J b2] j1 j2 Θ j1 Θ j2 =[B1J ,b1][B2J ,b2]= ([B1,b1,gc ]) ([B2,b2,gc ]). 554 17 Solutions

~~j j An index [B,b,gc] is in its kernel if and only if [BJ ,b]=[I,0], i.e., b = 0, and~~

~~BJ j = I = ( 1) j = det(BJ j)= det(I)= 1 = j = 0, B = I, ⇒ − ⇒ j so that [B,b,gc]=[I,0,1], and therefore Θ is an isomorphism. 17 Solutions 555 Exercises of Chapter 15~~

15.1 Suppose 0 j n. Then for c : ∆n j R and b : ∆n R, we calculate ≤ ≤ − → → ν ν α j ( )α j ( )α j ( )γ R c,b ν,n = ∑ (R c)α bα = ∑ ∑ − cα γ bα α α γ α − α =n ! α =n ! γ = j ( ) j | | | | | | −| | ν ( )β+γ j ( β γ)γ = ∑ ∑ β γ γ − − cβ bβ+γ β =n j γ = j ( + )! ( n) j | | − | | − (ν)β ν β γ β γ γ β ( + ) j ( ) = ∑ β cβ ∑ ! β γ γ − − bβ+γ , β =n j ! γ = j ( + )! ( n) j | | − | | − j and so (Rν∗ ) is given by ν β β γ j β ( + )γ j ( )γ ((R∗ν ) b)β = ∑ ! β γ γ − − bβ+γ γ = j ( + )! ( n) j | | − j β ν ν β j ( 1) ( + )γ j = ∑ ( + )γ γ − bβ+γ = ∑ β γ bβ+γ . γ = j ( n) j γ = j ( + 1) j | | − | | | | 556 17 Solutions Exercises of Chapter 16

~~16.1 (a) The integral converges, and deﬁnes a bounded linear map, since~~

~~2 2 2 2 2 f , f j g, f j dµ( j) f , f j dµ( j) g, f j dµ( j) B f g . J ≤ J | | J | | ≤ (b) The linearity of S, and inequality follow from~~

~~2 Sf ,g = g,Sf = f , f j g, f j dµ( j), Sf , f = f , f j dµ( j). J J | | 1 1 (c) Expand f = SS− f = S− Sf , and use~~

~~1 1 1 1 S− f , f j = f ,S− f j , S− f , f j f j dµ( j)= f , f j S− f j dµ( j). J J (d) The linear functional deﬁning Va is bounded, since (by Cauchy–Schwarz)~~

~~2 µ 2 µ 2 2 2 f , f j a j d ( j) f , f j d ( j) a ℓ (µ) B f a ℓ (µ). J ≤ J | | 2 ≤ 2 The formula for V ∗ f follows from~~

~~µ V ∗ f ,a = f ,Va ℓ2(µ) = f , f j a j d ( j)= ( f , f j ) j J,a ℓ2(µ). J ∈ 16.2 With S the frame operator (see Exer. 16.1), the proof of Theorem 6.1 gives~~

~~trace(S)2 d trace(S2), ≤ with equality if and only if ( f j) j J is tight. Let (eℓ) be an orthonormal basis, then the trace formula (see Exer. 2.13)∈ and Plancherel, gives~~

~~2 2 trace(S)= ∑ Seℓ,eℓ = ∑ f j,eℓ dµ( j)= ∑ f j,eℓ dµ( j) J | | J | | ℓ ℓ ℓ 2 = f j dµ( j), J~~

~~2 trace(S )= ∑ S(Seℓ),eℓ = ∑ eℓ, fk fk, f j dµ(k) f j,eℓ dµ( j) J J ℓ ℓ~~

~~= ∑ eℓ, fk f j,eℓ fk, f j dµ(k)dµ( j) J J ℓ 2 = fk, f j dµ(k)dµ( j). J J | | 17 Solutions 557~~

Substituting these into the inequality above gives the result. 1 16.3 We recall that S (and hence S− ) is Hermitian. 1 1 2 1 1 1 (a) PΦ∗ =(V ∗S− V)∗ = V ∗S− V = PΦ , PΦ = V ∗S− VV ∗S− V = V ∗S− V = PΦ . (b) We have

~~1 1 1 (PΦ a) j = S− Va, f j = Va,S− f j = ak fk,S− f j dµ(k) J 1 = fk,S− f j ak dµ(k)=([PΦ ] µ a) j. J (c) From the frame deﬁnition, we have~~

2 1 2 µ 1 2 µ 1 2 vk L µ) = fk,S− f j d ( j)= S− fk, f j d ( j) B S− fk , ( J | | J | | ≤ so that vk L2(µ). We calculate ∈ 1 1 µ vr,vs L2(µ) = fr,S− f j fr,S− f j d ( j) J 1 1 1 1 1 1 = S− 2 fr,S− 2 f j S− 2 f j,S− 2 fr dµ( j)= S− 2 fr,S− 2 fs , J 1 by the Plancherel identity for the normalised tight frame (S− 2 f j). 16.4 The frame operator S = AI, and so has ﬁnite trace:

~~2 traceS = trace , f j f j dµ( j)= f j dµ( j)= µ(J)= Ad, J J 1 d which gives A = µ(J) . 16.5 (a) Let x H , so that x = Px, and point evaluation of x at j is given by ∈~~

~~j x j = x,e j = Px,e j = x,Pe j , → which implies Kj = Pe j. (b) Since H is the orthogonal complement of the vector (1,1,...,1), one computes~~

n 1 −n , k = j; Kj(k)= 1 , otherwise. − n This frame is the vertices of the simplex. 16.6 (a) Let P be the orthogonal projection onto a subspace K of H , and K be the reproducing kernel for H . For f K , we have ∈

f (x)= f ,Kx = Pf ,Kx = f ,PKx , 558 17 Solutions so that K is reproducing kernel Hilbert space, with kernel KK (x,)= PKy,PKx . (b) Let Pj be the orthogonal projection onto H j, and write f = ∑ f j, f j H j. Then j ∈ f (x)= ∑ f j(x)= ∑ f j,(Kj)x = ∑ Pj f ,(Kj)x = ∑ f ,Pj(Kj)x = ∑ f ,(Kj)x j j j j j

~~= f ,(∑Kj)x , f H , j ∀ ∈ so that ∑ j Kj is the reproducing kernel of H .~~

16.7 Every subspace H of L2(S) is the orthogonal direct sum of its projections onto the H j. If H is rotationally invariant, then so is its projection onto the rotationally invariant subspace Hk, which is either Hk or 0 (by the irreducibility of Hk). Thus H has the asserted form. λ d x,y 16.8 Substitute = 2 , t = x , y = x , to obtain ∞ ∞ d ξ (k) 2 ( 2 ) x, k ∑ Zξ (x)=(1 x ) ∑ Ck x . k 0 − k 0 x = = 16.9 (a) For Π1(S)= H0 H1, (16.20) and (16.40) give ⊕ 0,1 (0) (1) Zξ (x)= Z{ }(x)= Z (x)+ Z (x)= 1 + d x.ξ . ξ ξ ξ

d (0) (1) For Π1(R )= P0 P1 = V V , (16.47) gives the reproducing kernel ⊕ 0 ⊕ 0 Z(0)(x,y) Z(1)(x,y) 1 1 d x,ξ K(x,y)= 2 + 2 = 2 + 2 , area(S) 1 w 0 area(S) 1 w 1 area(S) 1 w 0 1 w 1 , , , , with the corresponding tight frame given by Zξ = K( ,ξ). ξ 4 (b) Let j =(x j,y j,z j), then the Gramian of (Zξ j ) j=1 is

1 + 3 ξ1,ξ1 1 + 3 ξ1,ξ2 1 + 3 ξ1,ξ3 1 + 3 ξ1,ξ4 1 + 3 ξ2,ξ1 1 + 3 ξ2,ξ2 1 + 3 ξ2,ξ3 1 + 3 ξ2,ξ4 . 1 + 3 ξ3,ξ1 1 + 3 ξ3,ξ2 1 + 3 ξ3,ξ3 1 + 3 ξ3,ξ4  1 + 3ξ ,ξ 1 + 3ξ ,ξ 1 + 3ξ ,ξ 1 + 3ξ ,ξ  4 1 4 2 4 3 4 4    This factors as A∗A, with A column equivalent to B, where 1 1 1 1 1 0 0 0 √3x1 √3x2 √3x3 √3x4 √3x1 x2 x1 x3 x1 x4 x1 A =  , B =  − − − . √3y1 √3y2 √3y3 √3y4 √3y1 y2 y1 y3 y1 y4 y1 − − − √3z1 √3z2 √3z3 √3z4 √3z1 z2 z1 z3 z1 z4 z1     − − −      4 The polynomials (Zξ j ) j=1 are a basis if and only if their Gramian is invertible, i.e., B is invertible. By taking a cofactor expansion along the ﬁrst row, it follows that B is invertible if and only if the vectors ξ2 ξ1, ξ3 ξ1, ξ4 ξ1 are not coplanar. − − − 17 Solutions 559

4 (c) The basis (Zξ j ) j=1 is orthogonal if the off diagonal entries of its Gramian are 0, i.e., 1 1 + 3 ξ j,ξk = 0, j = k = ξ j,ξk = , j = k, ⇒ −3 ξ 4 so that ( j) j=1 are the vertices of a regular simplex. 16.10 (a) By (16.56) and the multinomial identity, we have

d 1 + p α α p d 1 + p p Kp0(z,w)= − ∑ z w α = − z,w . d 1 α =p d 1 − | | − (b) Since the holomorphic polynomials of degrees 0,1,...,n are orthogonal, we add their reproducing kernels, to get

n d 1 + p K(z,w)= ∑ − z,w p. d 1 p=0 − (c) Differentiating the (absolutely convergent) geometric series 1 1 + x + x2 + = , x < 1, 1 x | | − d 1 times gives − d! (d + 1)! (d 1)! (d 1)! + x + x2 + = − , x < 1, − 1! 2! (1 x)d | | − i.e., ∞ d 1 + p 1 ∑ − xp = , x < 1. p (1 x)d | | p=0 − Thus, we have ∞ d 1 + p p 1 ∑ Kp0(z,w)= − z,w = . d 1 (1 z,w )d p=0 − − (d) Expanding the Poisson kernel, as in Exer. 16.8, gives

1 z 2 1 z 2 P(z,w)= − = − w z 2d (1 z,w w,z + z 2)d − − − 1 z 2 z,w + w,z = − , y = , t = z , λ = d, (1 2yt +t2)λ 2 z − ∞ 2 (d) z,w + w,z k =(1 z ) ∑ Ck z . − k 0 2 z = Equating the homogeneous polynomials of degree k in z gives 560 17 Solutions

~~k (d) z,w + w,z k (d) z,w + w,z ∑ Kpq(z,w)= z Ck z Ck 2 . p q k 2 z − − 2 z + =~~

~~16.11 (a) For x = ek, we have ∞ j 2 2 2 2 ∑ x,S v = ∑ v j = v x . j= ∞ | | j | | − Scaling so that v = 1, gives the orthogonality (see Exer. 2.4). 1 (b) Since S∗ = S− , we calculate~~

j 2 2 ∑ x,S v = ∑ xa+ jva + xb+ jvb j | | j | | 2 2 2 2 = ∑ xa+ j va + ∑ xb+ j vb + ∑ xa+ jxb+ jvavb + xa+ jxb+ jvavb j | | | | j | | | | j 2 2 b a a b = v x + vavb S− x,S− x + vavb S− x,S− x 2 2 a b = v x + 2ℜ vavb S − x,x . Here the absolute convergence of the terms in the bracketed sum follows by the Cauchy–Schwarz inequality. For x = zea + eb, z C, we have ∈ a b S − x,x = ze2a b + ea,zea + eb = ea,zea = z. − j 2 Thus for (S v) to be a tight frame (with frame bound v ), we must have vavb = 0. (c) The previous argument can be modiﬁed to show that a v with ﬁnite support gives j a tight frame (S v) for ℓ2(Z) if and only if v is a nonzero scalar multiple of a standard basis vector. Here, a calculation gives

j 2 2 2 a b ∑ x,S v = v x + 2 ∑ ℜ vavb S − x,x . j | | a,b supp(v) { }⊂a=b Since a b could be equal for different pairs, we have to be careful with our choices for x. Let−a and b be the minimum and the maximum of supp(v) (the support of v). If supp(v) has at least two elements, i.e., a < b, then choosing x = zea +eb above leads to the necessary condition vavb = 0 for a tight frame, which cannot be satisfied. j k ℓ k ℓ (d) By (a), it suffices to show (S v) is orthogonal. Clearly, S v,S = S − v,v = 0 for k ℓ odd, and so it remains to consider the case k ℓ = 2m, m= 0. Here − − 2m 1 1 1 1 1 S v,v = ∑vkvk+2m = ∑ = ∑ , k n Z 2n + 1 2n + 2m + 1 2m n 2n + 1 − 2n + 2m + 1 ∈ which is zero, since the series converges absolutely. (e) To find all such v ℓ2(Z), we identify each v ℓ2(Z) with a f L2(T) via the ∈ j ∈j ∈ Fourier transform, i.e., f (z)= ∑ v jz , v j = f (z),z , where j L2(T) 17 Solutions 561

2π 1 it it f (z),g(z) L2(T) := f (e )g(e )dt, 2π 0 and v = f L2(T). Here the shift S on ℓ2(Z) corresponds to multiplication by z on j j L2(T), so that (S v) gives a tight frame for ℓ2(Z) if and only if (z f (z)) j Z gives a ∈ tight frame for L2(T), i.e.,

~~∑ g(z),z j f (z) 2 = f 2 g 2 , g L (T). L2(T) L2(T) L2(T) 2 j | | ∀ ∈~~

~~Now~~

~~∑ g(z),z j f (z) 2 = ∑ g(z) f (z),z j 2 = g(z) f (z) 2 , L2(T) L2(T) L2(T) j | | j | | so that f must satisfy~~

f g = f g , g L2(T), L2(T) L2(T) L2(T) ∀ ∈ i.e., f must have constant modulus. Thus the v for which (S jv) is a tight frame for ℓ2(Z) are precisely those v which are the Fourier coefﬁcients of a nonzero function with constant modulus. The example of (d) is (up to a scalar) given by f = 1 on [0,π] and f = 1 on [π,2π]. Since the only trigonometric polynomials with unit modulus on T are− the monomials, we also recover (b) and (c). d 16.12 Let z =(z1,...,zd) C , and ﬁx j. We need to show that ∈ d d z j = z,ξ ξ j dσ(ξ)= (z1ξ1 + + zdξd)ξ j dσ(ξ). ˜ ˜ ˜ ˜ area(S) S area(S) S This follows from the simple calculations

~~ξkξ j dσ(ξ)= 0, k = j, ˜ S ˜ ξ 2 σ ξ 1 σ ξ area(S) ξ 2 j d ( )= 1d ( )= (since ∑ j j = 1). ˜ | | d ˜ d | | S S~~

~~16.13 For p Hk, Cauchy–Schwarz gives ∈ p(ξ) = p,Z(k) p Z(k) , | | | ξ S| ≤ S ξ S (k) (k) with equality if and only if p is a scalar multiple of Zξ . Taking p = Zx gives~~

(k) (k) (k) (k) (k) 2 (k) Z (x) = Zx (ξ) Zx Z = Z = Z (ξ), | ξ | | ||≤ S ξ S ξ S ξ with equality if and only if x = ξ, which gives (a). Let (u j) be an orthonormal basis for Hk. Then taking the trace of linear operators in (16.10) gives (b), i.e., 562 17 Solutions

(k) (k) (k) (k) dim(Hk)= ∑ u j, u j,Z Z dσ(ξ) = ∑ u j, u j,Z Z dσ(ξ) ξ S ξ S ξ S ξ S j S j S (k) 2 (k) 2 (k) 2 (k) = ∑ u j,Z dσ(ξ)= Z dσ(ξ)= Z = Z (ξ). ξ S ξ S ξ S ξ S j S (k) By the argument for (a) and (b), we have that the maximum is Z = dim(Hk), ξ S (k) (k) (k) which is attained if and only if p = Z / Z S (the unique scalar multiple of Z ξ ξ ξ which is positive at ξ and has unit norm). ξ 16.14 This the variational characterisation (see Proposition 16.2) of ( )ξ S being ∈ a continuous tight frame for Fd (Proposition 16.4, Exer. 16.12). It can be proved directly by using (6.29), i.e.,

~~2 2 2 1 1 x,y dσ(x)dσ(y)= y c1(d,F)dσ(y)= = x,x dσ(x) . S S | | S d d S α 16.15 The monomials (z ) α =k are orthogonal with respect to , ,k and , SC , | | ◦~~

α α α! α α (d 1)!α! (d 1)! α ! α α z ,z ,k = , z ,z S = − = − | | z ,z ,k. ◦ α ! C (d 1 + α )! (d 1 + α )! ◦ | | − | | − | | (a) This follows from the above by linearity. (b) By the reproducing property, (a) and (6.20), we have

k + d 1 k k + d 1 k σ f = f , − ,w SC − ,w d (w) S d 1 d 1 C − − k + d 1 k k = − f , ,w ,k ,w dσ(w) d 1 S ◦ − C k + d 1 = − f (w) ,w k dσ(w), f H(k,0). d 1 S ∀ ∈ − C (c) From the formula, we obtain a Beta integral

2π 1 η k 2 ξ η 2k σ ξ d 1 2 d 2 2k θ , S = , d ( )= − (1 r ) − r rdrd C | | π 0 0 − SC 1 1 2 d 2 2k d 2 k =(d 1) (1 r ) − r 2rdr =(d 1) (1 t) − t dt − 0 − − 0 − 1 (d 2)!k! k + d 1 − =(d 1) − − . − (k + d 1)! d 1 − − References 563 References

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~~Index~~

(1,1)–design, 126 G–orbit, 209 ( f˜j), 31, 32 G(m, p,d), 345 can ( f ), 35 Gg = Gg j GL2(Z ), 406 j , ∈ d′ (n,d,λ)–difference set, 269 H, Hˆ Heisenberg group, 366 1–uniform, 132 H(p,q), 461 2–optimal frame, 268 HV (W), 225 24–cell, 22 Hg GL2(Z ), 403 ∈ d′ A = AΦ , B = BΦ , 32 Lσ f j = fσ j, 191 A†, 61 M, 369 Θ B Bα , 92 α = M1 = aFa , 389 Θ ν Bn, = Bn, 94 Mn , 434 C : Cd Cd ;z z , 382 N(Π)= N(S (Π)) = N(S(Π)), 407 → → 0 C : v v, 190 PΦ , 72 → D square–free part of (d 3)(d + 1), 412 Pσ , 369 − D3, 213, 219, 331, 335 R, 369 D4, 231 S(Π) ESL2(Z ), 407 ⊂ d′ Dv f , 93 S = S(v) symmetries of a fiducial v, 385 F, 369 S0(Π) ESL2(Zd ), 405 F a b c;x , 435 ⊂ ′ A( , , ) SJ, Sn, 190 Ff , 413 SΠ symmetries of a SIC fiducial, 404 Fg = Fg j ESL2(Z ), 406 , ∈ d′ Tj Chebyshev polynomials of the first kind, Fz, Fa, Fb, Fc, Fd , Fe, 389 413 G–equivariant map, 225 V, V ∗, S = VV ∗, 12 G–frame, 210 Z, W1, W2, 395 G–frame of multivariate orthogonal Z(k), 446 polynomials, 231 ξ G–frame, characterisation, 214 Adj(Γ ), 291 G–frame, complement, 215 Ang(Φ), 253 G–frames for infinite groups, 463 Aut(Γ ), 224 G–invariant frame, 209, 234 C(d) Clifford group, 368 G–invariant polynomials, 239 CG, 332 G–invariant set, 210 CG–module, 227 d 1 G–invariant subspace, 210 CP − , 129 G–map, 225 EC(d) extended Clifford group, 368 G–matrix, 332 EC(d) group of unitary and antiunitary maps, G–module, 210 368 Π G G–morphism, 225 E0 = E0 fixed field of 0, 404

~~575 576 Index~~

E1 fixed field of G1, 406 G = Gal(E/Q) Galois group of the SIC field, Emax maximal SIC field, 410 402 G G Π Emin minimal SIC field, 410 0 = 0 , 404 EGL(H ), 190 G1 = g1 , 406 EU(H ), 190 H , 7 FG–homomorphism, 225 H C = H iH , 227 ∼ d⊕ FG–isomorphism, 225 Hk, Hk(R ), Hk(S), 445 FG–module, 210 IG, 240 FG–submodule, 210 Lt (H ,F), 120 F C, 72 M (d), 22 F ⊂the finite field of order q, 297 N q n,Fd , 152 FF(a,b) frame force, 135, 138 O1,...,Oℓ, 406 FP(Φ) frame potential, 115 Pν , 431 Φ k Gram( ), 13 Pn, 451 d H d ν,n Harmk(R )= k(R ), 117 Ps , 432 Hom (Rd ), 116 χΠ Π ˆ j p := trace( Dp) overlap, 402 Hom V W , 228 FG( , ) trC :C(d) Zd , 392 Ind(d), 374 dep(Φ), 72→ K := Q( (d 3)(d + 1)), 401 ∆(v ,v ,...,v ), 167 − j1 j2 jm O(d)–invariant subspaces of Pn, 451 ∆ (v ,v ,...,v ), 176 C j1 j2 jm PC(d) := C(d)/[I] projective Clifford group, Θ : GL (Z ) GL (Z ), 407 2 d′ 2 d′ 368 α : SL (Z ) → (d)/[I], 378 2 d′ CSp PEC(d) := EC(d)/[I] extended projective α–partition frame,→ 20 Clifford group, 368 ω primitive d–th root of unity, 366 d PU(C ), 365 ∇ f , 136 Φ =(φg)g G, 213 σ σ σ σ σ σ ∈ 1 = x, 2 = y, 3 = z, 353 Φ Ψ, 99 σ ∪ A, 370 Φ +ˆ Ψ, 106 2 θ : SL2(Zd ) ⋉Z Ind(d), 376 Φ Ψ, 100 ′ d → Gˆ, 133, 246 Φ ⊕Ψ, 108 Dˆ displacement operator, 374 Φcan⊗ , 35 p ψˆ : CSp(d)/[I] SL2(Zd ), 378 Π G, 240 → Π G FPˆ (Φ) normalised frame potential, 115 + , 240 Π Π I , Φ , 80 G = / G, 240 Π d , , 120 t◦(R ), 117 ◦ Π d f dσ, 116 t◦r(C ), 145 S , E, 400 Π (Rd)= Π (Rd) , 122 t◦,t 2◦t µ primitive 2d–th root of unity, 366 Ψ , 370 a µ(n) Mobius¨ function, 81 Q(ω), 81 mult(W,H ), 238 SL2(Zd ), 370 d πΦ : Sym(Φ) GL(H ) : σ Lσ , 191 S = S , 445 ψ → → Γ :C(d) SL2(Zd ), 372 Seid( ), 291 2πi→ √ 8 CSp(d) := F,R,[I] , 372 i = e , 85 Sym(Φ), 191 srg(ν,k,λ, µ), 291 Sym(µ), 231 θ–isogonal configuration, 212 ϕ SymE(Φ), 191 (n) Euler phi function, 81 SymEP(Φ), 195 vold (T), 430 ξΘ ξ SymP(Φ), 195 (x)=( j(x)), 87 α Symt (H ), 120 ξ , 92 T, 366 a b, 83 ∗ Z4–Kerdock codes, 325 a b, 183 Φ◦ area(S), 446 c ( f ), 74 F ( f1,..., fn), 113 ct = ct (d,F), 122 F n,Fd , 153 d–polytope–configuration, 344 Index 577 d–polytope–configuration, flag of, 344 analytic SIC, 361 d–polytope–configuration, regular, 344 angle between subspaces/vectors, 266 d–polytope–configuration, symmetry group of, angle multiset, 253 344 anti equivalent, 11 f : SL (Z ) ⋉Z2 C(d)/[I] Appleby antilinear map, 11, 190 2 d′ d indexing homomorphism,→ 375 antisymmetry, 191 2 fE : ESL2(Zd ) ⋉Zd EC(d)/[I], 382 antiunitary map, 190 1′ → g1 := ggcg− , 405 antiunitary symplectic, 382 gc complex conjugation, 400, 402 apolar inner product, 120 u l hn,d , hp,d , hp,d , 258 Appell polynomials, 438 j–faces, 344 Appell’s biorthogonal system, 429 k–flag, 201 Appleby index, 375 kg Z∗ , 403 Appleby index, extended, 382 d′ m–distance∈ set, 309 Appleby indexing of the Clifford operations, m–distance tight frame, 310 374 m–products, 165, 167 approximate inverse, 53, 66 m–th root signature matrix, 312 approximate left inverse, 66 m : SL (Z ) SL (Z ), 378 approximately dual frames, 56 d 2 d′ 2 d n–th roots of unity,→ 263 association schemes, 22 p–groups, 256 automorphism group of a graph, 224 p(∂), 146 automorphism of a graph, 224 q–Paley graph, 297 s–angular frame, 318 balanced frame, 104 zψZ , m1, m2, 394 Bargmann invariants, 167 za, 370 barycentric coordinates, 87, 430 2F1, 429 basis, 73 Bernstein coefficients, 433 POVM, 364 Bernstein frame, 93 Bernstein frame, properties, 93 abelian difference set, 269 Bernstein operator, 94 abelian field extension, 412 Bernstein polynomials, 431 absolute bound (on number of equiangular Bernstein–Bezier´ coefficients, 431 lines), 268 Bernstein–Durrmeyer operator, 429, 434 absolutely irreducible, FG–module, Bessel identity, 8 representation, 228 Bessel identity, generalised, 121, 123 ADE classification of the finite subgroups of bilateral shift, 469 SL2(C), 357 binary icosahedral group, 243 adjacency matrix of a graph, 224, 291 biorthogonal system, 33 adjacent vertices of a graph, 291 Bloch sphere, 363 affine generalised barycentric coordinates, 87 block designs, 275 affine group, 261 block matrix calculation, 316 affine independence, 98 blocks, 275 affine map, 91 Bombieri inner product, 120 affinely equivalent subsets of a finite abelian Bombieri’s inequality, 149 group, 261 boring strongly regular graphs, 293 affinely independent points, 98 bra w , 364 | algebra of group matrices, 332 bra–ket notation, 364 algebraic equations for tight complex Bruck–Ryser–Chowla theorem, 269 equiangular lines, 316 algebraic variety, 151 canonical m–products, 176 algebraic variety of SIC fiducials, 414 canonical coordinates, 71, 74, 215 alternate dual, 54 canonical coordinates and linear maps, 79 analysis operator, 12, 27 canonical coordinates for cyclotomic fields, 81 analysis operator of a generalised frame, 443 canonical coordinates, characterisation, 76 578 Index canonical coordinates, properties, 78 Clifford trace, 392 canonical copy of a tight frame, 15, 17 close frames, 51 canonical dual frame, 32, 54, 56 closeness bound, 51, 66 canonical dual functionals, 74 code division multiple access systems, 113 canonical expansion, 74 coinvariant space, 240 canonical factorisation, 74 coinvariants of a finite reflection group, 240 canonical Gramian, 54, 72, 74 coisometry, 13, 25 canonical Gramian of a generalised frame, 443 column orthogonality (of irreducible canonical inner product, 80, 98 characters), 246 canonical inner product, properties, 83 column, of a linear map on ℓ2(J), 12 canonical isomorphism between a vector space commutativity relation for S and Ω, 366 and its bidual, 76 commutativity relations for V,S,G, 62 canonical matrix, 75 complement of a G–frame, 215 canonical oblique dual, 59, 68 complement of a difference set, 269 canonical order 3 Clifford operation, 394 complement of a frames, 105 canonical order 3 symplectic unitaries, complement of a harmonic frame, 250 conjugates of, 398 complement of a tight frame, 18 canonical tight frame, 31, 35, 62, 66, 186 complementary tight frame, 18 canonical tight frame of a G–frame, 211 complete frame graph, 168 canonical tight frame of a generalised frame, complex (t,t)–design, 126 442 complex t–design, 126 Cauchy–Schwarz inequality, 38, 114, 143, 149 complex conjugate equivalence, 11 Cayley graph, 309 complex conjugate of a Hilbert space, 26 Cayley transform, 155 complex conjugation map, 11, 190 Cayley transform, truncated, 155 complex equiangular tight frames, 310 CDMA systems, 113 complex frame, 41, 63 central G–frame, 230, 336 complex Hadamard matrix, 281 central force, 135 complex polytope, 343 central group frame, 230 complex projective sphere, 129 central tight G–frames, classification, 337 complex reflection, 240, 343 centrally symmetric, 116 complex reflection group, 240 centred SIC fiducial, 405 complex tight frame, 21 character group, 133, 246 complex unit sphere, 122 character of a representation, 334 complexification, 227 character table of a finite abelian group, 246 compressed sensing, 4 characters of a finite abelian group, 246 condition number of a frame, 46 Chebyshev method, 53 condition number of the frame operator, 47 Chebyshev polynomials, 430 conductor, of a ray class field, 412 Chebyshev polynomials of the first kind, 413 conference graph, 296 Chinese remainder theorem, 398 conference matrix, 296, 314, 327 chordal graph, 174 conjugate gradient method, 53 Chu–Vandermonde identity, 435 conjugates of the canonical order 3 symplectic circulant graph, 303 unitaries, 398 circulant matrix, 213 conjugation map, 26 class function, 336 connected components of a graph, 168 Clifford action, 386 connection set of a graph, 309 Clifford action of the extended Clifford group, constant diagonal Hadamard matrix, 282 385 continuous frame, 442 Clifford action on SICs, 386 continuous generalised frame, 441 Clifford group, 361, 368 continuous tight frame, 2 Clifford group, generators for, 372 continuous tight frame expansion, 441 Clifford group, monomial representation of, continuous tight frame of zonal harmonics, 446 391 continuous tight frames for Pn, 454 Clifford operation, 368 contraction map, 66 Index 579 coordinate functionals, 33 direct sum, 102 coordinate star, 15 direct sum of frames, 100 coordinates, 12 direct sum of harmonic frames, 250 coordinates with respect to a tight frame, 29 directional derivative, 93 Coulomb force, 135 discrete Fourier transform, 25 cross in Rn, 16, 346 discrete Fourier transform matrix, 369 cross–correlation, 3, 22, 113 discrete frame, 442 cubature formula, 119, 122 discrete generalised frame, 441 cubature formula, existence of, 119 discrete Heisenberg group, 366 cubature rule, 119, 122, 123 discrete inner product, 251 cubature rule for the sphere, 459 disjoint union of frames, 99 cube, 222, 252, 346 disjointness, 101 cube roots, 283 displacement operation, 372 cube, vertices of, 107, 248 displacement operator, 374 cuboctahedron, 346 distance between frames, 47 cycle space of a finite graph, 171 division ring, 228 cyclic convolution, 83 dodecahedron, 222, 252 cyclic difference set, 269 dual basis, 33, 75 cyclic frame, 247 dual frame, 31, 32 cyclic group, 210, 213 dual frame of a G–frame, 211 cyclic harmonic frame, 247 dual frame of a generalised frame, 442 cyclic harmonic frames for C1, 255 dual frames, 54 cyclic harmonic frames for C2, 255 dual Riesz basis, 45 cyclic shift, 81 dual sequences, 98 cyclic shift matrix, 366 dual signed frame, 186 cyclic shift operator S, 85 cyclic vector, 465 effective frame force, 138, 148 cyclotomic field, 80, 81, 374, 401 eigensteps, 154 cyclotomic field Q(ω), 81 elementary abelian p–group, 250 elliptic curve, 401 de la Vallee–Poussin´ mean, 434 empty frame graph, 168 decomposition of frames, 103, 110 ENPTF, 23 degree elevation of the Bernstein operator, 429 equal–norm frame, 9 degree of a character, 334 equal–norm tight frame, 27 degree raising operator, 431 equally spaced unit vectors in R2, 199 degree reducing, 95, 96 equally spaced vectors, 251 degree reducing operator, 434 equations for SIC fiducials, 414 demicube, 347 equiangular lines, 174, 265, 266 density matrix, 364 equiangular tight frame, 27, 73 density of rational tight frames, 156 equiangular tight frames from block designs, dependencies, 72 275 dependencies dep(Φ), 72 equiangular vectors, 266 determining set of m–products, 167 equiangularity, 3, 4 DFT matrix, 369 equispaced equal–norm frame, 263 diagram vectors, 9, 25 equispacing, 4 dicyclic group, 357 equivalence of tight frames, 23 difference set, 269 equivalence, unitary, 10 difference sets, infinite families of, 273 equivalent difference sets, 271 differentially 1–uniform, 132 equivalent group representations, 210 dihedral group, 193, 213 equivalent Hadamard matrices, 321 dilation, 2 erasures, 262 dimension towers, 413 error operator basis, 353 Diophantine equation, 269 Eulerian function, 254 Dirac notation, 364 Eulerian subgraph, 171 580 Index eutactic star, 16 frame with s angles, 318 existence of equal–norm tight frames, 157 frame with maximal projective symmetry, 198 expressible by radicals, 400, 401 frame, continuous, 442 extended Appleby index, 382 frame, discrete, 442 extended Clifford group, 368 frame, fusion, 9 extended projective Clifford group, 368 frame, generalised, 442 extended projective symmetry group of a frame, semicritical, 110 frame, 195 frames as orthogonal projections, 43 extended projective symmetry group of a frames invariant under the unitary action of an sequence of vectors, 195 abelian group, 236 extended symmetry group of a frame, 191 full octahedral group, 346 extended symmetry group of a sequence of functions on the complex sphere, 461 vectors, 191 fundamental cycle basis, 171 fundamental cycle of an edge, 170 facts, about SICs, 401 fusion frame, 9, 179 faithful representation, 210 fusion frame operator, 180 Fano plane, 273 fusion frame system, 180 FF–critical, 138 fiducial vector, 361, 365 Gabor system, 2, 5, 209, 366 fiducial vector, real, 417 Galois automorphism, 400 field trace, 325 Galois correspondence of SIC subfields, 410 finite abelian extensions of Q, 401 Galois field, 325 finite field of order q, 297 Galois group of the SIC field, 402 finite Fourier methods, 414 Galois multiplet, 403 finite reflection group, 343 Galois symmetries of a SIC, 361 finite reflection groups, classfication of, 345 Galois theory, 400 finite tight frame, 7 Gamma function, 430 finite tight frame, early examples, 2 Gauss sum, 381 finite tight frame, prototypical example, 1 Gauss sums, 330 finite tight frame, second prototypical example, Gegenbauer polynomials, 447, 453 3 generalised barycentric coordinates, 87 finite tight frames for Pn, 459 generalised barycentric coordinates, properties, Fisher inner product, 120 88 five vectors in C3, 199 generalised Bernstein operator, 94 flag of a d–polytope–configuration, 344 generalised cross, 346 forward cyclic shift, 81 generalised frame, 441, 442 Fourier matrix, 9, 245, 246, 324, 369 generalised Gauss sums, 330 Fourier transform, 443 generalised Gegenbauer polynomials, 453 Fourier transform matrix, 25 generalised Hermite polynomials, 453 frame, 32 generalised Pauli group, 366 frame algorithm, 53 generalised quaternian group, 357 frame bound, 7 generalised resolution of the identity, 28 frame bounds, 32, 46, 64 generating set of m–products, 167 frame force, 115, 135 generic SIC, 401, 424 frame force critcal sequence, 138 geometrically uniform frame, 247 frame graph, 100, 165 Gerzon bound, 268 frame homotopy problem, 153 Givens rotation, 159 frame operator, 12, 27, 32, 62 global field, 412 frame potential, 115 golden ratio, 206, 243, 327 frame potential, global minimisers, 136 Gram matrix, 13 frame potential, local minimisers, 136 Gram–Schmidt algorithm, 69, 72 Frame Research Centre, 186 Gramian, 13, 27, 62, 102 frame theory, history, 2 Gramian of a generalised frame, 443 frame transform operator, 12 Grassmann manifold of d–planes in Fn, 153 Index 581

Grassmannian, 42 index group of the Clifford operations, 374 Grassmannian frame, 268 index group, of a nice error frame, 353 Grassmannian packing problem, 268 index map [a] (ψa,za) of the Clifford Grassmannian space, 268 operations,→ 374 group algebra, 210, 215 inner direct sum of frames, 100 group algebra over C, 332 inner product preserving map, 27 group covariance, 361, 365 inner product, apolar, 120 group frame, 209, 210 inner product, Bombieri, 120 group matrix, G–matrix, 213 inner product, Fisher, 120 group of rotations, 2 intermediate SIC field, 410 growing a flag set, 201 invariant polynomials, 239 invariant subspace, 44 Haar measure, 463 irreducibility, 85 Haar measure on SO(d), 451 irreducible G–frame, 218 Hadamard conjecture, 25 irreducible character, 334 Hadamard difference set, 273 irreducible characters of a finite abelian group, Hadamard matrix, 9, 25, 278 246 Hadamard matrix of (v j) and (w j), 320 irreducible group representation/action, 218 Hadamard product of matrices, 183 irreducible tensors, 108 Hadamard type difference set, 273 isogonal configuration, 41, 63, 212, 265, 309 Hahn polynomials, 429, 433 isolated point (of a graph), 294 Hall’s Eulerian function, 254 isometric embedding, 131 Hamiltonian circulant graph, 297 isometry, 8, 13, 25, 27 harmonic frame, 9, 229, 245, 247 harmonic frame with distinct vectors, 249 Jacobi inner product, 430 harmonic frame with real vectors, 249 Jacobi measure, 431 harmonic function, 445 Jacobi polynomials, 429, 433, 437 harmonic polynomials, 117 Jordan–Holder¨ theorem, 226 harmonic upper bound, 304 Heisenberg frame, 366 ket v , 364 Heisenberg group, 366 Kirkman| frame, 312 Heisenberg group, normaliser of, 368 Kronecker product, 252 Heisenberg operation, 372 Kronecker–Weber theorem, 401 Hermite polynomials, 453 Hermitian complex Hadamard matrix, 281 Lowdin¨ orthogonalisation, 35, 60, 68 Hessian matrix, 141, 149 La Jolla Difference Set Repository, 273 hexadecachoron, 346 Laplace’s equation, 445 highly symmetric tight frame, 223, 349 Laplacian, 147 highly symmetric tight frames, construction, Lauricella function of type A, 429, 435 342 least squares solution, 60 Hilbert class field, 412 Legendre polynomials, 429, 430, 437, 453 Hilbert’s 12–th problem, 401 Legendre polynomials on a square, 231 Hilbert–Schmidt inner product, 29 Legendre symbol, 399 Hoggar lines, 365, 367, 399 Lie group, 152, 163 homogeneous G–frame, 226 lift, 110 homogeneous components, 225 lift of a frame, 104 homogeneous polynomials, 117 lifted equally spaced vectors, 105 homogeneous polynomials on a Hilbert space, lifted roots of unity, 105 120 lifting a frame, 104 Householder transformation, 155, 159 linear action, 210 hyperplane, 343 linear dependencies, 72 linear map, 79 icosahedron, 116, 222, 252, 288 linear reproduction, 97 imprimitive finite reflection group, 345 LLL (Lenstra–Lenstra–Lovasz)´ algorithm, 404 582 Index local compact (topological) group, 463 nested roots, 400 local frame, 180 nested SIC fields, 413 Lucas–Fibonnacci SIC, 413 nice (unitary) error basis, 353 nice (unitary) error frame, 353 Mobius¨ function, 81 nice error bases, 221, 353 Magma, 342, 347 nice error basis, 222 majorisation, 157 nice error frames, 353 Maschke’s theorem, 225, 227 non-adjacent vertices of a graph, 291 matrices, 75 noncanonical dual frame, 54 matrices with respect to a normalised tight noncyclic harmonic frame, 248, 256 frame, 29 normal equations, 183 matrix with respect to a tight frame, 29 normal field extension, 401 maximal abelian –algebra, 322 normalised eutactic star, 16 maximal number∗ of MUBs, 174 normalised frame, 48 maximal SIC field, 410 normalised frame potential, 115 maximally symmetric frame, 198 normalised surface area measure, 122 maximally symmetric tight frame, 198 normalised tight frame, 7, 8 maximum cross–correlation, 266 normalised tight frames and linear mappings, maximum size g(d) of a spherical two–distance 29 set, 304 normalised tight frames, algebraic variety of, Mercedes–Benz frame, 3, 189, 288 151, 152 minimal angle, 266 normalising a frame, 47 minimal SIC field, 410 number of angles, 318 minor of a graph, 316 number of generators of a G–invariant frame, modulation, 2 234 modulation matrix, 366 number of harmonic frames, 258 monomial representations of the Clifford numerical SIC, 361 group, 391 monomials, 121 oblique dual frames, 57 mother wavelet, 366 oblique projection, 58 MUB, 22, 127, 174, 209, 318 octahedron, 222, 252, 346 MUB problem, 22, 325 octaplex, 346 MUBs, 22 optimal frame bounds, 32, 49, 50, 52, 61, 64 multi-indices, 92 optimal Grassmannian frame, 268 multinomial identity, 96 order of a difference set, 269 multiple orthogonal polynomials, 438 orthogonal basis, 100 multiplet, 403 orthogonal compression, 101 multiplicative equivalence, 254 orthogonal decomposition into homogeneous multiplicatively equivalent subsets of a group, components, 226 254 orthogonal dilation, 101 multiresolution analysis, 2 orthogonal frames, 42 multivariate continuous Hahn polynomials, orthogonal polynomials, 69, 231 439 orthogonal polynomials of Appell, 437 multivariate Hahn polynomials, 439 orthogonal polynomials of Proriol, 437 multivariate Jacobi polynomials, 431 orthogonal polynomials on the simplex, tight multivariate orthogonal polynomials, 231, 438 frames for, 429 multivariate shifted factorial, 95 orthogonal polynomials on the triangle, 429 mutually unbiased bases, 22, 174, 318 orthogonal projection, 25 MWBE codebook, 271 orthogonal projection formula, 24, 62 orthogonal projection, one–dimensional, 267 Na˘ımark’s theorem, 15 orthogonal resolution of the identity, 28 near frames, 51 orthogonality (of irreducible characters), 246 neighbourhood of a switching class of graphs, orthogonality of frames, 101 294 orthonormal bases, 248, 267, 271 Index 583 orthonormal basis, 9, 24, 189 primitive strongly regular graphs, 293 outer direct sum of frames, 100 principal submatrix of a matrix, 316 overlaps, 402 probability distribution, 364 overlaps phases, 402 projections of normalised tight frames, 25 projective, 9 Paley graph, 297, 303 projective (t,t)–designs, 130 Paley–type difference set, 273 projective Clifford group, 368 parabolic subgroup of a finite reflection group, projective equivalence, 23 344 projective invariants, 165 Parseval frame, 8, 23 projective linear group, 351 Parseval identity, 8, 13, 24 projective plane, 273 partial difference set, 308 projective representation, 351 partial isometry, 8, 25, 36, 61 projective similarity, 176 partition frame, 20, 105 projective similarity of vectors, 165 partition of unity, 92 projective symmetry group of a frame, 189, path graph, 290 190, 195 path–connected, 153 projective symmetry group of a harmonic path–connectivity, 153 frame, 262 Pauli matrices, 353 projective unitary equivalence, 11, 26, 165, Pell’s equation, 412 166 perfectly tight frame, 143 projective unitary equivalence of harmonic permutation matrices, 193 frames, 260 permutation matrices in the Clifford group, projective unitary group, 351, 365 380 projectively equivalent, 11 permutations, 190 projectively repeated, 262 perturbation of a frame, 68 projectively similar, 176 perturbation of a normalised tight frame, 61 projectively unitarily equivalent, 11 phase-permutation representation, 391 Proriol polynomials, 438 planar difference set, 273 Proriol’s orthogonal basis, 429 Plancherel identity, 8, 24 prototypical example, first, 1 Plancherel identity, generalised, 121, 123 prototypical example, second, 3 plane wave, 145, 146 pruning a set of flags, 201 Platonic solid, 209, 222, 252 pseudo–dual frames, 56 Platonic solid, symmetry group, 222 pseudodual, 57 Pochhammer symbol, 122, 431 pseudoinverse, 31, 60 Pohlke normal star, 15 pseudoreflection, 343 point evaluation, 78, 187 PSLQ algorithm, 404 pointwise product of matrices, 183 Poisson kernel P(x,ξ), 450, 467 quadratic distance (between frames), 51 polar decomposition, 36 quadratic reciprocity law, 418 polarisation identity, 7, 24 quadratic residue, 297 pole, 446 quadrature rule for the sphere, 459 pole of a zonal function, 446 quantum measurements, 364 polynomials on projective spaces, 148 quantum state determination, 22 Pontryagin duality, 246, 270 quantum system, 364 Pontryagin duality map, 133, 246 positive operator, 434 radial function, 447 positive operator valued measure, 364 radially symmetric weights, 439 potential, 135 ramification, of a ray class field, 412 pre-frame operator, 12 rank one orthogonal projection, 31 precision bumping algorithm, 404, 410 rank one projection, 60 Prime Power Conjecture, 273 rational tight frames, density of, 155 primitive finite reflection group, 345 ray class conjecture, 412 primitive roots of unity, 81 ray class field, 401, 412 584 Index ray class group, 412 Scott–Grassl numerical SIC, 385 real algebraic variety, 152 seeding, 15 real cyclic harmonic frame, 251 Seidel matrix, 289 real fiducial vectors, 417 Seidel spectrum, 290 real frame, 41, 63 semi–regular complex polytopes, 345 real harmonic frame, 251 semicritical frame, 110 real MUBs, 22 semidirect product, 375 real orthogonal group, 189 sensor networks, 180 real projective space, 129 Shephard–Todd classification, 345, 347 real projective sphere, 129 shift invariant tight frame, 86 real spherical t–design, 116 shift matrix, 366 real spherical half–design, 116 shifted factorial, 95 real tight frame, 21 short-time Fourier transform, 443 real unit sphere, 122 SIC, 3, 22, 127, 174, 209, 363, 364 reconstruction operator, 12 SIC facts, 401 rectified tesseract, 346 SIC fiducials, algebraic variety of, 414 reduced Hadamard matrix, 322 SIC field E = Q(Π, µ), 400 reduced signature matrix of an equiangular SIC field of a multiplet, 410 frame, 284 SIC field, minimal, intermediate, maximal, 410 redundancy, 7, 47 SIC problem, 22, 361 reflection group, 343 SIC problem, addictiveness, 361 regular, 308 SIC-POVM, 363, 364 regular d–polytope–configuration, 344 signal transmission with erasures, 248 regular character (of an abelian group), 270 signal transmission with quantization, 248 regular polygon, 90 signature of a signed frame, 181, 186 regular two–graph, 294 signed frame, 181, 186 reindexing a frame, 254 signed frame operator, 186 relative bound (on number of equiangular similar frames, 41, 42 lines), 287 similarity, 42, 73, 101, 191 reordering, 10 simple lift, 110 reordering a frame, 254 simple lift of a frame, 104 representation of a group, 210 simplex, 73, 286, 290 representation theory, 209, 210 simplex, vertices of, 271 reproducing kernel, 118, 444 Singer difference sets, 273 reproducing kernel for Pn, 457 singular value decomposition, 62, 65 reproducing kernel Hilbert space, 444 Sixteen equiangular lines in R6, 205 reproducing kernel tight frame, 444 size of a Galois multiplet, 403 ridge function, 145, 146, 447 skew Hermitian matrix, 155 ridge polynomial, 146, 184, 442 Sobolev dual, 68 Riesz basis, 45, 51 solid spherical harmonics, 445 Riesz representer, 31, 120 spanning set, 31, 32, 61, 73 ring of coinvariants, 240 spanning tree for a graph, 170 robust signal transmission, 248 sparsness, 3 robustness to erasures, 1, 3, 4 special unitary group, 152 roots of unity, 107 sphere in Rd , 445 rotation, 2, 10 spherical (t,t)–designs, numerical construc- row construction of a tight frame, 17 tion, 139 spherical 2–design, 117 sampling, 441 spherical m–distance set, 309 scalable frame, 182 spherical t–design, 116, 119 scaling to a tight frame, 182 spherical t–designs, characterisations, 117 Schur’s lemma, 226, 228 spherical design of harmonic index t, 118, 119 Schur–Horn majorisation, 157 spherical half–design, 116 Schweinler–Wigner orthogonalisation, 60 spherical half–design of order t, 119 Index 585 spherical harmonics, 445 symmetry group of a measure µ , 231 spherical two–distance set, 304 symmetry group of a Platonic solid, 222 spherical two–distance tight frame, 304 symmetry group of the square, 233 sporadic SIC, 401, 416, 424 symmetry group of the triangle, 233 square free integer, 257 symplectic index, 377 stability, 3 symplectic index, table of, 381 stable isogon, 359 symplectic operation, 372 stabliser, 342 symplectic spreads, 325 standard m–distance tight frame, 310, 346 symplectic unitaries, 372, 378 standard terminology for frames, 23 symplectic unitaries of order 3, 392 state, 364 synthesis operator, 12, 27, 62 state, of a quantum system, 364 synthesis operator of a generalised frame, 443 Steinberg’s fixed point theorem, 345 Szego¨ kernel, 468 Steiner equiangular tight frame, 277 Steiner system, 275 Tammes’ problem, 200, 269 STFT, 443 Teichmuller¨ set, 325 strongly centred SIC fiducial, 406 tensor product integration formula, 123 strongly disjoint frames, 42 tensor product of frames, 108 strongly regular graph, 291 tensor product of harmonic frames, 250 structured frames, 209 tensor product of Hilbert spaces, 108 subsimplicial equiangular tight frame, 275 tensor product of nice error frames, 355 subsimplicial frame, 275 tensor product of vector spaces, 108 sum of frames, 106 tesseract, 346 sum of frames, direct, 100 tetrahedron, 222, 252 sum of harmonic frames, 250 tight G–frames as idempotents of the group surface area measure, 122 algebra, 333 surface area measure, invariance of, 123 tight equiangular lines, 265 surface spherical harmonics, 445 tight frame, 7 switching, 290 tight frame, canonical, 35 switching class (two–graph), 290 tight frame, canonical copy, 15 switching equivalent graphs, 290 tight frame, continuous, 2 symmetric t–linear maps, 120 tight frame, copy of, 15 symmetric block design, 269 tight frame, early examples, 2 symmetric Gram–Schmidt algorithm, 35 tight frame, equal–norm, 9 symmetric group on a set, 190 tight frame, repeated vectors, 8 symmetric informationally complete positive tight frame, unit–norm, 9 operator valued measure, 363 tight frames of orthogonal polynomials on the symmetric orthogonalisation, 35, 60 simplex, 429 symmetric tensors, 120 tight frames, construction of, 158 symmetries of a basis, 192 tight frames, existence of, 157 symmetries of a SIC fiducial, 404 tight fusion frame, 179 symmetries of a SIC fiducial vector, 385, 386 tight generalised frame, 441 symmetries of the vertices of a simplex, 192 tight signed frame, 181 symmetries of the Weyl–Heisenberg SICs, 389 tightness, 47 symmetries, of a fiducial, 385 Tikhonov regularization (of the pseudoinverse), symmetry, 191 484 symmetry group of n equally spaced vectors, time–frequency localisation, 1 193 topologically irreducible, 463 symmetry group of a d–polytope– torsion point, 401 configuration, 344 total potential, 135 symmetry group of a complementary frame, trace formula, 13, 26 196 translate, 260 symmetry group of a frame, 189–191 translation, 2, 260 symmetry group of a harmonic frame, 250 Tremain equiangular tight frame, 278 586 Index

Tremain frame, 278 unitary group, projective, 365 triangular graph, 301, 310 unitary images of tight frames, 24 triangulated graph, 174 unitary map, 211 trigonal bipyramid, 199, 343 unitary matrix, 152 trigonometric polynomials, 251, 463 univariate Jacobi polynomials, 430 triple products, 165, 167, 174, 284, 365 univariate orthogonal polynomials, 68, 69 trivial difference sets, 271 two–distance set, 304 variational characterisation of tight frames, two–graph (switching class), 290 114 variational formula, 13 ultraspherical polynomials, 447 vertex–transitive graphs, 224 unfaithful action, 190 vertices of the cube, 107, 248, 264 unimodular simplex, 278 vertices of the simplex, 19, 20, 248 union of frames, 99 Von Neumann algebra, 465 unit sphere, 122 unit sphere in Rd , 445 unit sphere, complex, 122 Waring formula, 116 unit sphere, real, 122 wavelet system, 2, 5, 209 unit–norm frame, 9 wavelets, 1 unital equiangular tight frame, 312 WBE sequences, 113 unitarily equivalent frames, 40 Wedderburn’s Theorem, 238 unitarily equivalent tight frames, 10 weighted (t,t)–design, 128 unitarily equivalent tight frames up to Welch bound, 113 reordering, 10 Welch bound equality sequences, 113 unitarily equivalent via an automorphism, 254 Weyl displacement operation, 372 Weyl–Heisenberg SIC, 366 unitarily inequivalent tight D3–frames, 331 unitary action, 211 Weyl–Heisenberg SIC, symmetries of, 389 unitary continuous representation, 463 Wirtinger calculus, 146 unitary equivalence, 23 Wirtinger complex differential operators, 146 unitary equivalence of harmonic frames, 253 unitary equivalence of tight frames, 10 Zak transform, 465 unitary equivalence of tight frames up to Zauner matrix Z, 384 reordering, 10 Zauner’s conjecture, 22, 361 unitary equivalence up to reordering, 10 Zauner’s conjecture, stronger form, 385 unitary group, 152 zonal function, 446 unitary group, parametrisations of, 155 zonal harmonic, 446