Multidimensional Lattices in Signal Processing

Riccardo Bernardini [email protected] https://www.linkedin.com/in/riccardobernardini/

March 5, 2019

Abstract

This document collects few notes about the use of lattices as multidimensional sampling/periodicity structures, pointing out the connection with the Uniﬁed Theory.

Contents

1 What is a lattice?2 1.1 Change of basis and sub-lattices...... 4 1.1.1 Change of basis and elementary transformations...... 5 1.2 Rational Lattices...... 5

2 Lattices and uniﬁed signal theory7 2.1 Reciprocal lattice and quantum...... 7 2.2 Integral and convolution...... 8 2.3 Periodic signals...... 8 2.4 Period and fundamental cells...... 9 2.5 Fourier transform...... 11 2.6 Sampling and aliasing...... 11 2.6.1 Sampling resolution...... 11

A Applications 12 A.1 Hexagonal sampling of images...... 12 A.2 Electrons in a crystal...... 15 A.2.1 The case of ﬁnite crystal...... 16 A.3 Spatio-temporal sampling...... 17

B Brief summary about quotient groups 19

1 C Proofs 20

C.1 Density of Z(α,β) ...... 22

1 What is a lattice?

The usual model for a one-dimensional discrete-time signal is a function deﬁned on the set

Z(T) = {nT : n ∈ Z} (1) representing the set of sampling instants. Value T in (1) is the sampling interval, has the dimension of time and it is the reciprocal of the sampling frequency Fc = 1/T. Set (1) formalizes the intuitive idea of “set of regularly spaced points on a line.” In order to extend this idea to a d-dimensional space, a natural approach is to sample regularly along every direction of the d-dimensional space (e.g., along rows and columns for a 2-dimensional signal like an image) obtaining

d Z (T1,...,Td) = {[n1T1,n2T2,...,ndTd] : n1,n2,...nd ∈ Z} (2) = Z(T1) × Z(T2) × ··· × Z(Td) However, while in the 1-dimensional case sets like (1) are the only possibility for the domain of a 1-dimensional signal, in multiple dimensions sets of type (2) are not the only choice. Indeed, when moving to more than one dimension there are many other possibilities for a “regularly spaced collection of points.” Few examples for the two-dimensional case can be seen in Fig.1. In order to ﬁnd a suitable mathematical description for sets like the ones shown in Fig.1, observe that a characteristic shared by the sets shown in Fig.1 is that they can be generated by “walking” along the shown vectors by an integer number of steps. For example, the point marked with a cross in Fig.1b can be reached by moving 2 steps forward along b2 and 1 step backward (that is, −1 steps. . . ) along b1. This observation suggests that the sets shown in Fig.1 can be written as

Λ(b1,b2) := {n1b1 + n2b2 : n1,n2 ∈ Z} (3)

The d-dimensional counterpart of (3) is, quite naturally,

Λ(b1,b2,...,bd) = {n1b1 + n2b2 + ··· + ndbd : n1,n2,...,nd ∈ Z} (4)

Deﬁnition 1. Set Λ(b1,b2,...,bd) in (4) is called the d-dimensional lattice with basis {b1,b2,...,bd}.

Notation (4) is fairly cumbersome. It is more convenient to use a notation that is more compact and also dimension-independent. This can be obtained by considering the vectors bi as column vectors of t matrix M = [b1,b2,...,bd] and collecting n1,...,nd in column vector n = [n1,n2,...,nd] . Equation (4) becomes d Λ(M) = {Mn : n ∈ Z } (5)

We will mostly use notation (5), resorting to the more explicit (4) only when necessary.

2 (a) (b)

Figure 1: Examples of 2-dimensional lattices (a) Orthogonal (b) Quincunx (c) and (d) two generic lattices

3 Figure 2: Three different bases for the quincunx lattice

Remark 1.1 (Lattices vs. vector spaces)

Equations (4) and (5) suggest that a lattice generated by vectors b1,b2,...,bd is similar to the vector space generated by the same set of vectors, but with the constraint that the coefﬁcients must be integer. This seemly small difference has a big impact in the properties of Λ(M). d For example, suppose B = {b1,b2,...,bd} is a set of d linearly independent vectors of R . If one d + + adds to B a vector c ∈ R to obtain B = B ∪{c}, the space generated by B coincides, clearly, with the space generated by B. That is, adding extra vectors to a basis (in the case of vector spaces) does not change anything. With lattices things are much more complex. For example, working for simplicity with the one dimensional case, it is fairly easy to check that

Λ(1/7,1/4) = Λ(1/28) (6)

that is, the set obtained by taking integer linear combinations of 1/7 and 1/4 is the set of multiplies of 1/28. The proof is a (really ) easy exercise. More (much more) tricky is to prove that set

Λ(2,π) ⊂ R (7)

is a dense subset of R, that is, for every x ∈ R and it is possible to ﬁnd u ∈ Λ(2,π) arbitrarily near to x (that is, such that |x − u| ≤ ε with ε > 0 arbitrarily small). See Property6 in AppendixC.

1.1 Change of basis and sub-lattices

In the one-dimensional case the sampling interval T > 0 is uniquely determined by the corresponding

lattice Λ(T) = Z(T). In multiple dimensions this is not true anymore since a lattice can be generated by many different bases. See, for example, Fig.2 that shows three different bases for the quincunx lattice. The set of bases that generate the same lattice has an easy characterization.

d×d Property 1. Two matrices M,N ∈ R are two bases for the same lattice (i.e., Λ(M) = Λ(N)) if and only if d×d M = NH, H ∈ Z ,|detH| = 1 (8)

Property1 is a corollary of the following property.

4 Property 2. Lattice Λ(M) is a sub-lattice of Λ(N) (that is, Λ(M) ⊆ Λ(N) if and only if

d×d M = NH, H ∈ Z (9)

Remark 1.2 Note the similarity of Property2 with the one-dimensional case where Z(U) is a subset of Z(T) if and only if U is a multiple of T, that is, U = NT, N ∈ Z.

The proofs of Property1 and Property2 are in AppendixC, Proof C.1 and Proof C.2, respectively.

1.1.1 Change of basis and elementary transformations

There are three simple elementary transformations that can be done on a basis M in order to obtain an equivalent one; they are

1. Multiply a column by −1

2. Swap two columns

3. Add to column i column j multiplied by an integer.

It is (really ) easy to verify that the three operations above correspond to right-multiplying M by a matrix H with |detH| = 1. Therefore, by doing the operations above we do not change the generated lattice. Moreover, it is possible to prove that every matrix H with |detH| = 1 can be obtained by using the elementary operations above.

Remark 1.3 The proof of the latter result is deﬁnitively not trivial. . . A possibility is to use the Hermite normal forma) of an integer matrix to prove that every matrix H with |detH| = 1 can be reduced to the identity matrix by elementary transformations. It is less trivial than it seems since we can multiply only by integers and cannot do divisions. ahttps://en.wikipedia.org/wiki/Hermite_normal_form

1.2 Rational Lattices

It is often convenient to suppose that the lattice of interest is a rational lattice.

Deﬁnition 2. Lattice Λ(M) is said to be rational if for every r = 1,...,d of M, the elements of row r are in rational ratio. More precisely, Λ(M) is rational if

[M]r,c2 ∀r,c1,c2 ∈ {1,...,d} [M]r,c1 = 0 ∨ ∈ Q (10) [M]r,c1 where ∨ denotes, as usual, logical OR.

5 Remark 1.4 Definition2 seems to depend on the specific basis of the lattice. However, it is easy to show that if d×d H ∈ Z , |detH| = 1, then M satisfies (10) if and only if MH does. In other words, if a basis of Λ(M) satisfies (10), then all the bases of Λ(M) satisfy it; if a basis of Λ(M) does not satisfies (10), then no basis of Λ(M) satisfy it.

The class of rational lattices is very large, but not every lattice is a rational lattice; for example, the lattice of basis  √  1 2  √  (11) π 3 √ is clearly non rational since 1/ 2 6∈ Q. However, it is also clear that in practical applications it is quite unlikely to ﬁnd lattices with non rational bases. Therefore, the hypothesis of rational lattice is quite cheap, since we can expect that no case of practical interest will be left out. Rational lattices enjoy the following property

Property 3. Every basis M of a rational lattice Λ(M) can be written as

M = DK (12)

d×d d×d with D = diag(D1,...,Dd) ∈ R a diagonal matrix and K ∈ Z . Moreover, there is another basis d×d N of the same lattice Λ(M) (that is, N = MH, H ∈ Z , |detH| = 1) such that

N = DKb (13) with Kb in lower triangular (actually Kb can be made in Hermite normal form1).

The proof is in AppendixC, Proof C.3. Informally, Property3 can be rephrased that if Λ(M) is rational, one can choose the units of measure along the d axis in order to have a lattice with integer coordinates. The new units of measure are, of course, the entries of matrix D. An easy (and often useful) corollary of Property3 is that a rational lattice has always an orthogonal sub-lattice and an orthogonal super-lattice.

Corollary 1. For every rational lattice Λ(M) it is possible to ﬁnd two orthogonal lattices Λ(A) and Λ(B) such that Λ(A) ⊆ Λ(M) ⊆ Λ(B) (14)

Proof. Let D and K as in (12). A suitable choice for A is A = D; in order to ﬁnd B, let γ ∈ Z such − d×d that γK 1 ∈ Z and B = M(γK−1) = DKγK−1 = γD (15)

1https://en.wikipedia.org/wiki/Hermite_normal_form

6 2 Lattices and uniﬁed signal theory

We saw in the previous section that lattices are a very general way to construct “regular distribution of points” in d-dimensional space. In this section we are going to show that they can be used as signal domain in Uniﬁed Signal Theory (UST). This will allow us to “import” at once all the results of UST. It is easy to verify that every lattice Λ(M) satisﬁes the following properties

0 ∈ Λ(M) (16a)

x ∈ Λ(M) ⇒ −x ∈ Λ(M) (16b)

x,y ∈ Λ(M) ⇒ x + y ∈ Λ(M) (16c)

Equations (16) (together with the commutative property of the sum) show that every lattice is a (locally compact) Abelian group. Therefore, lattices are suitable signal domains in the context of UST and every result derived in UST holds also for signals deﬁned on multidimensional lattices. In the following we will show how the concepts of UST specialize in the case of lattices.

2.1 Reciprocal lattice and quantum

One key concept of UST is the concept of reciprocal group that in the case of lattices specializes as follows.

Deﬁnition 3. The reciprocal lattice of a d-dimensional lattice Λ(M) is deﬁned as the set of d-dimensional ∗ d ∗ vectors Λ ⊂ R such that every product between an element of Λ(M) and an element of Λ is an integer, that is ∗ t ∀k ∈ Λ ,q ∈ Λ(M) k q ∈ Z (17)

It turns out that the reciprocal Λ∗ of lattice Λ(M) is a lattice too.

Property 4. The reciprocal of lattice Λ(M) is the lattice with basis M−t, that is,

Λ∗ = Λ(M−t) (18)

The proof is an easy exercise and it can be found in AppendixC, Proof C.4.

Remark 2.1 (Link with the 1-dimensional case) Compare Property4 with the one-dimensional case where the reciprocal of Z(T) is Z(1/T).

In unified signal theory the integral on a discrete domain is defined as the sum on the domain multiplied by the quantum of the domain. In the one dimensional case Z(T) the quantum is T, in the d-dimensional case the quantum is defined as |detM|.

7 Remark 2.2 (Link with the 1-dimensional case) Value |detM| can be interpreted as the volume of the parallelotopea identiﬁed by the vectors of the basis M. Compare this with the 1-dimensional case where T is the measure of the interval between two consecutive sampling instants.

aA parallelotope is the d-dimensional generalization of a parallelogram/parallelepiped.

Remark 2.3 (Anonymous notation) Often, when it will not be necessary to write explicitly a basis, it is convenient to resort to a “anonymous” and handier notation. More precisely, the generic lattice will be denoted with a Greek capital letter Λ, Γ, . . . The reciprocal lattice will be denoted as Λ∗, Γ∗, . . . . Finally, the modulus will be denoted as |detΛ| to be understood as |detΛ| := |detM| (19)

where M is any basis of Λ. Deﬁnition (19) is well-posed since Property1 grants that |detM| does not depend on the chosen basis. Note also that

1 |detΛ| = (20) |detΛ∗|

2.2 Integral and convolution

The deﬁnition of the integral on Λ = Λ(M) is similar to the deﬁnition for Z(T), with the role of the quantum T played by |detΛ| = |detM|, that is,

Z x = ∑ |detΛ| x(q) = ∑ |detM| x(Mn) (21) Λ(M) q∈Λ n∈Zd

It follows that the convolution of two signals x : Λ → C, y : Λ → C is

x ∗ y(q) = ∑ |detΛ|x(r)y(q − r) q ∈ Λ (22) r∈Λ

2.3 Periodic signals

The deﬁnition of periodic signal is exactly the same deﬁnition given in UST.

d Deﬁnition 4. Let D be a d-dimensional domain (that is, D = R or D = Λ(N)) and let Λ = Λ(M) be a lattice such that Λ ⊆ D (23)

A signal x : D → R is said to be periodic with period Λ if

x(t) = x(t + q) ∀t ∈ D,q ∈ Λ (24)

8 Figure 3: Examples of periodic patterns

As in the one-dimensional case, a signal x deﬁned on D and periodic on Λ can be interpreted as a signal deﬁned on the quotient group2 D/Λ. Therefore, if x is periodic with period Λ one can write x : D/Λ → C.

Remark 2.4 d d×d Condition (23) is automatically satisﬁed if D = R while it requires M = NH, H ∈ Z if D = Λ(N), because of Property2.

Remark 2.5 Multidimensional signals periodic on a lattice are deﬁnitively more common than signals deﬁned on lattices. See, for example, Fig.3. Another example of signal periodic on a lattice is the potential of an ideal crystal that has the periodicity of the crystal itself. This has consequences on the description of the behavior of an electron in a crystal, see Appendix A.2.

2.4 Period and fundamental cells

In the one-dimensional case, a signal x : R/Z(T) → C, deﬁned on R and periodic with period T is uniquely determined by the values assumed in a period that is usually deﬁned as the interval [0,T) or [−T/2,T/2), depending on what is more convenient.

2A brief summary of the idea of quotient group can be found in AppendixB.

9 (a) (b) (c)

Figure 4: Three elementary cells for an hexagonal lattice. (c) shows the Voronoi cell

The key properties of a one-dimensional period are that when a period is repeated over Z(T) (that is, translated of multiples of T)

• It fully covers R without leaving holes, • There is no overlapping between different translations.

The two properties above define what is a called in UST an elementary cell of R/Z(T). It is easy to check that an equivalent definition is that an elementary cell is a set that contains one and only one representative for every class3 of R/Z(T). The concept of elementary cell gives us a wide choice for the set representing the period of a periodic signal. For example, it is easy to verify (exercise!) that the set J = [0,T/2) ∪ [(3/2)T,2T) is an elementary cell of R/Z(T). However, this kind of freedom is not usually exploited since it is difficult to imagine a one-dimensional case where it would be more convenient to use J as period rather than the more usual [0,T) or [−T/2,T/2). The freedom in the definition of elementary cell has, however, an in impact in multiple dimensions. See Fig.4 for some examples of elementary cells for an hexagonal lattice.

• The elementary cell shown in Fig.4a is the parallelogram identiﬁed by the basis vectors. This is maybe the natural extension of the 1-dimensional cell [0,T).

• The elementary cell shown in Fig.4b is interesting because — despite the non-orthogonal lattice — it is a rectangle and this makes it easier to iterate over the points in the cell.

• The elementary cell shown in Fig.4c is special because it consists of all points that are closer to the origin than to any other lattice point. It is called with different names, depending on the context. In UST is called the Voronoi cell, in crystal theory is called Wigner-Seitz cell or Brillouin zone, depending on the considered space (crystal lattice or reciprocal space). This also explain why the lattice in Fig.4 is called the hexagonal lattice.

Finally, it is worth giving the following result

3See AppendixB.

10 Property 5. Let Γ ⊂ Λ. Any elementary cell of Λ/Γ has |detΓ|/|detΛ| elements.

Remark 2.6 Let Γ = Λ(N) and Λ = Λ(M). Note that Γ ⊂ Λ implies that N = MH for some integer matrix H and this implies that |detΓ| = |detH| ∈ (25) |detΛ| N+

2.5 Fourier transform

d ∗ The Fourier transform of a signal x : Λ → C is the function X : R /Λ → C deﬁned as

t d ∗ X(f) := ∑ |detΛ|x(q)exp(− j2π f q) f ∈ R /Λ (26) q∈Λ The periodicity of X over the reciprocal of the domain of x is the counterpart of the fact that the Fourier transform of a signal deﬁned on Z(T) is periodic with period 1/T.

2.6 Sampling and aliasing

d d Let xc : R → C be a signal deﬁned on continuous domain R and let x : Λ → C its sampled version deﬁned as

x(q) = xc(q) q ∈ Λ (27)

The sampling theory of uniﬁed theory specializes in this case to

d X(f) = ∑ Xc(f + k) f ∈ R (28) k∈Λ∗ d d ∗ where Xc : R → C and X : R /Λ → C are, clearly, the Fourier transforms of, respectively, xc and x. Equation (28) is clearly the equivalent of Shannon’s theorem in multiple dimensions. It follows that xc is exactly recoverable from its sampled version as long as the support of Xc is contained inside any d ∗ elementary cell of R /Λ .

2.6.1 Sampling resolution

Observe that if we sample with lattice Λ, with a sensor for every lattice point (e.g., like pixels in a camera), the maximum area available for the sensor is the area of an elementary cell (different sensors d cannot overlap) of R /Λ, that is, |detΛ|. The resolution, measured as number of sensors per unit of area is, therefore, 1/|detΛ|. By remembering that 1/|detΛ| = |detΛ∗| (see (20)) it is immediate to recognize the following fact.

Fact 1. The resolution (measured in sensors per unit of d-dimensional volume) associated with a lattice d ∗ Λ is equal to the area of an elementary cell of the frequency domain R /Λ . Therefore, the minimum resolution required for sampling a given signal is equal to the area of the frequency support of the signal.

11 Remark 2.7 Fact1 above can be seen as the generalization of the Shannon theorem. In the one dimensional case

the volume of an elementary cell of R/Z(Fc) is clearly equal to the sampling frequency Fc that can be interpreted as the number of samples taken in a unit of time. Moreover, if a signal has frequency support

[−Fmax,Fmax] the area of the support is 2Fmax and Fact1 gives the well known Shannon theorem. However, while in the one-dimensional case it is always possible to work at the optimal resolution (practical considerations aside), we will see that in the multidimensional case, depending on the shape of the support, it can be impossible to reach the maximum efﬁciency (see Appendix A.1).

A Applications

A.1 Hexagonal sampling of images

The increased ﬂexibility of multiple dimensions can be exploited to do a more efﬁcient sampling. 2 Suppose d = 2 and let xc : R → R be an image whose Fourier transform is contained inside a circle of radius Fmax, that is,

kfk > Fmax ⇒ Xc(f) = 0 (29)

Using a separable lattice Suppose we want to sample xc on a separable lattice Λ(diag(∆x,∆y)), where

∆x and ∆y are the sampling steps along the two dimensions. The Fourier transform of the sampled version 2 x is clearly deﬁned on R /Λ(diag(1/∆x,1/∆y)), that is, X is periodic on a separable lattice.

How do we choose ∆x and ∆y? We want to choose them as large as possible (since this means a larger sensor and less sensors per unit of area), but not so large to have aliasing. It is easy to see that the optimal choice is such that the Fourier transform of x is like in Fig.5c.1. The Fourier transform in

Fig.5c.1 is periodic on Λ(diag(2Fmax,2Fmax)). It follows that the optimal choice is ∆x = ∆y = 1/(2Fmax), similarly to the 1-dimensional case where the sampling frequency needs to be at least twice the maximum frequency of the signal. The area of the sensor can be at most 1 1 1 × = 2 (30) 2Fmax 2Fmax 4Fmax 2 and the required resolution is 4Fmax. Since according to Fact1 the minimum required resolution for xc 2 is πFmax, the overhead required by an orthogonal sampling 2 4Fmax 2 ≈ 1.273 (31) πFmax that is, the resolution required by an orthogonal sampling is 27% larger than the minimum.

Using an hexagonal lattice Suppose now to sample the signal using an hexagonal lattice with basis   ∆x ∆x/2 M =   (32) 0 ∆y

12 Figure 5: (a.1) Orthogonal sampling lattice (b.1) Aliasing corresponding to sampling lattice (a.1); (c.1) Optimal aliasing for lattice (a.1); (a.2) Quincunx sampling lattice ; (b.2) Aliasing corresponding to sampling lattice (a.2); (c.2) Optimal aliasing for lattice (a.2)

13 See also Fig.5a.2. By calling Fx = 2/∆x and Fy = 1/(2∆y) the basis of the reciprocal lattice is

  −t −t ∆x 1 1/2 Mb := M =    ∆y 1    1/∆x 1 =    1/∆y −1/2 1   (33) 1/∆x 0 =   −1/(2∆y) 1/∆y   Fx/2 0 =   −Fy 2Fy that is equivalent (in the sense that it generates the same lattice) to      Fx Fx/2 Fx/2 0 2 1  =   (34) Fy −Fy 2Fy 1 1 | {z } | {z }| {z } Nb Mb H with |detH| = 1 and comparing Nb in (34) with M in (32) it is possible to see that Λ∗ is still of quincunx type, see Fig.5a.2 and Fig.5b.2.

The optimal choice of ∆x and ∆y gives rise to the situation in Fig.5c.2. Intuitively we expect Fig.5c.2 to be more efﬁcient than Fig.5c.1, since the amount of “unused area” is smaller. The lattice in Fig.5c.2 has basis   2Fmax Fmax Mb =  √  (35) 0 3Fmax According to Fact1, the resolution (number of sensors per unit of area) required is equal to the determi- nant of (35), that is, √ 2 2 |detMb | = 2 3Fmax ≈ 3.46Fmax (36) that is 13% smaller than the resolution required by a separable lattice. Equivalently, with an hexagonal sampling we can use a sensor that is 15% larger than the sensor used with an orthogonal sampling. Finally, the overhead associated with the hexagonal lattice is √ 2 2 3Fmax 2 ≈ 1.10 (37) πFmax that is 10%, to be compared with the 27% overhead of the orthogonal lattice. Finally, note that the quincunx packing is optimal. This means that if the signal has frequency 2 support as in (29), it is not possible to sample it with the optimal resolution of πFmax pixel per unit of area.

14 Problem 1. Processing signals deﬁned on an orthogonal lattice is somehow simpler than processing signals on a generic lattice. Suppose that after acquiring an image on an hexagonal lattice we want to interpolate it to an orthogonal one for convenience of processing.

• Find the smallest orthogonal lattice Γ that includes the hexagonal lattice Λ used for sampling

• The interpolation can be done, as usual, by doing a “pure interpolator” from Λ to Γ and then by applying an anti-aliasing ﬁlter. Fill the missing details, that is: how is the “pure interpolation” done? What is the ideal frequency response of the anti-aliasing ﬁlter?

A.2 Electrons in a crystal

Consider a crystal whose atoms are regularly placed in space over the points of a lattice Λ. We want to determine the behavior of an electron in the crystal. The Schrodinger¨ equation in this case can be written as d h¯ 2 ih¯ ψ(t) = − ∇2 +V ψ(t) (38) dt 2m | {z } H where V : R3 → R is the potential due to the atoms in the crystal and H is the Hamiltonian of the system. Since the physical system is stationary, the Hamiltonian H in (38) does not depend on time and this allows to separate the solution of (38) in a stationary part and a part that depends only on time. More precisely, it is possible to show that the solution ψ to equation (38) can be written as Ent ψ(t) = cn exp −i φn (39) ∑ h¯ n∈N where cn are constants to be determined according to the boundary conditions, while φn and En are, respectively, the eigenvectors and eigenvalues of the Hamiltonian H, that is,

Hφn = Enφn (40)

It is possible to show that every eigenvector φn can be interpreted as a stationary state of the system and that the corresponding eigenvalue En is its energy. Note that since H does not depend on time, time dependency of ψ(t) in (39) is due to the exponentials in the sum. Since the crystal is periodic with period Λ, V will be periodic too with the same period, that is,

τqV = V ∀q ∈ Λ (41)

Since ∇2 is translation invariant (in the sense that commutes with translations, that is, τq∇2 = ∇2τq), it is immediate to show that periodicity (41) is inherited by the Hamiltonian in the sense that

τqH = Hτq (42)

15 Property (42) imposes some strong constraints on the eigenvectors φn of H. These constraints are de- 4 scribed in the well known Bloch’s theorem that claims that if H satisﬁes (42) then every eigenvector φn of H can be written as t φn(x) = exp( j 2π k x)un,k(x) (43)

3 3 for some k ∈ R and some un,k : R /Λ → C periodic with the same periodicity of the crystal, that is,

q τ un,k = un,k q ∈ Λ (44)

Note that the real “physical” object that determine the behavior of the electron (and that can be poten- tially experimentally measured) is φn in (43). One could wonder if φn uniquely determines k and un,k. It is easy to verify this is not the case; indeed, by multiplying and dividing (43) by exp( j 2π rt x) where r ∈ Λ∗ one obtains

t φn(x) = exp( j 2π k x)un,k(x)

t t t = exp( j 2π k x)exp( j 2π r x) exp(− j 2π r x)un,k(x) | {z } (45) vr(x) = exp( j 2π (k + r)t x) v(x)

It is immediate to verify that vr has still the same periodicity of the crystal, indeed given q ∈ Λ

t vr(x + q) = exp(− j 2π r (x + q))un,k(x + q)

t t = exp(− j 2π r x) exp(− j 2π r q)un,k(x + q) | {z }| {z } 1 un,k(x) (46) t = exp(− j 2π r x)un,k(x)

= vr(x)

∗ Note how — in the second line — r ∈ Λ is exploited to prove that vr is still periodic with period Λ.

It follows that the same φn can be decomposed as in (45) both using k or k + r where r is any vector ∗ of Λ . In other words, any eigenfunction φn of H does not uniquely identify a speciﬁc k, but the coset ∗ k + Λ that contains k and that any element of the coset can be used to decompose φn in (43). In other d ∗ words, we can think k in (43) as belonging to the quotient group R /Λ . The Voronoi elementary cell d ∗ of R /Λ is called in this context the Brillouin zone; a cell of R3/Λ is called Wigner-Seitz cell.

A.2.1 The case of ﬁnite crystal

The theory above worked implicitly with an infinite crystal. Indeed, if the crystal was finite it would not be possible to have periodicity. The classical approach used in the literature to handle a finite crystal is to consider a cube with dimensions Lx, Ly and Lz and “repeat it by periodicity” (that is, over the orthogonal

4Called Foquet’s theorem by mathematicians

16 lattice Γ = Λ(diag(Lx,Ly,Lz))) in order to ﬁll the whole space. This is done by imposing boundary conditions that make eigenvectors φn periodic. The idea is that in a practical case even small crystals (e.g., 1 mm size) has dimensions that are several order of magnitudes larger than the size of a cell of

R3/Λ. Therefore, the impact of the speciﬁc boundary conditions on an electron in the core of the crystal are considered negligible.

Remark A.1 (A technical point) The approach usually found in the literature neglects a technical point: the periodic repetition must be done over a lattice Γ that is a super-lattice of Λ, that is, Γ ⊇ Λ, otherwise the periodicity over the crystal

lattice would be spoiled. However, this is just a minor point. Indeed, even if Lx, Ly and Lz are such that

Γ = Λ(diag(Lx,Ly,Lz)) is not a super-lattice of Λ, one can twiddle the crystal sizes Lx, Ly and Lz a tiny bit (of the order of a cell size) in order to satisfy the super-lattice constraint.

If periodic boundary conditions are imposed, then eigenvalues φn must be periodic with period Γ, that is, q τ φn = φn, ∀q ∈ Γ (47)

By using this constraint in Bloch’s theorem and observing that un,k is periodic on Γ ⊃ Λ, one obtains that also the exponential in decomposition (43) must be periodic on Γ. In other words, k must satisfy

∀q ∈ Γ exp( j 2π kt x) = exp( j 2π kt (x − q)) (48) = exp( j 2π kt x)exp(− j 2π kt q) that implies exp(− j 2π kt q) = 1, that is, k ∈ Γ∗. In other words, in the case of ﬁnite crystal with periodicity conditions imposed, we obtain that k must belong to the ﬁnite quotient group Γ∗/Λ∗ that is the dual of Λ/Γ. The cardinality of Γ∗/Λ∗ is clearly equal to the cardinality of Λ/Γ that is equal to

This explains why the density of states (in reciprocal space) is often given as a “number of states per unit of volume” (e.g., in units of cm−3) which at ﬁrst seems strange since the reciprocal space where k lives has units that are reciprocal to the units of space.

A.3 Spatio-temporal sampling

Suppose we are interesting in measuring a physical variable (e.g., pressure, temperature, . . . ) alongs a line at different times. This could be, for example, the monitoring of the pressure along a gas pipeline. 2 Let xc(s,t) be value of the variable at time t ∈ R and position s ∈ R. Let Xc : R → C be the Fourier transform of xc.

17 Suppose we place sensors along the line with distance ∆ between sensors and suppose every sensor acquire a meausure at a sampling frequency Fc = 1/T. For uniformity, we deﬁne also the “spatial sampling frequency” Fs = 1/∆. The measures taken at different locations and times make a two-dimensional signal x : Λ(∆,T) → R deﬁned on the orthogonal lattice Λ(∆,T). Suppose we have N sensors and that the available bandwidth is half the bandwidth that is required to send the data of the N sensors every T seconds. We want to analyze two possible solutions

1. Sample at half rate, that is with sampling frequency Fc/2. This corresponds to sampling with lattice Λ(∆,2T).

2. Have the sensors send their data in an alternate fashion, that is, at “even” times 2nT, n ∈ Z the even-numbered sensors transmit, at “odd” times (2n + 1)T the odd-numbered ones transmit. This corresponds to sampling over the lattice with basis      2∆ ∆ ∆ 2 1 M =   =    (50) T T 1

The basis of the reciprocal lattice is        −1 −t ∆ 1/2 Fs/2 Fs Fs/2 Mb = M =    =   ≡   (51) −1 T −1/2 1 −Fc/2 Fc Fc/2

where the equivalence sign ≡ is to be interpreted that they generate the same lattice. Clearly, the lattice generated by basis (51) is a quincunx lattice.

Suppose now, for the sake of simplicity, that we choose ∆ e T as unit of measures for length and time.

Correspondly, we will have Fs and Fc as measure of spatial and time frequency. Suppose that with this 2 2 choice of units, the Fourier transform Xc : R → C of the continuous-domain signal xc : R → C is “band limited” in the sense that Xc(f) = 0 for every f such that kfk > Fmax.

Clearly, it must be Fmax ≤ 1/2 in order not to have aliasing when sampling on the original lattice 2 Λ(∆,T) (that with the chosen units becomes Z ). How does it change the te maximum Fmax when we work With the two alternative lattices?

1. In the ﬁrst case, with the chosen units, we have aliasing on lattice Λ(1,1/2). This means that Fmax must be less than 1/4 = 0.25, otherwise we would have aliasing in the temporal direction.

2. In the second case, basis (51) becomes   1 1/2   (52) 0 1/2

It is easy to recognize that a point of the reciprocal lattice closest to the origin is [1/2,1/2] whose √ √ length is 2/2. Therefore, it must be Fmax < 2/4 ≈ 0.35 which is larger than the maximum Fmax for the ﬁrst solution.

18 Figure 6: Example of a two-dimensional lattice Λ (small, light dots) and a corresponding coset k + Λ ∈ R2/Λ (large, dark dots)

It follows that there is a range of Fmax that can be sampled with the quincunx lattice, but not the orthogonal one.

B Brief summary about quotient groups

As known, quotient group D/Λ is a group built on the set of the equivalence classes of D modulo Λ. Formally, we will say that two elements a,b ∈ D are equivalent modulo Λ and write

a ≡ b (mod Λ) (53) if they differ by an element of Λ a − b ∈ Λ (54)

Clearly, the following properties hold a ≡ a (mod Λ) (55)

a ≡ b (mod Λ) ⇒ b ≡ a (mod Λ) (56)

 a ≡ b (mod Λ) ⇒ a ≡ c (mod Λ) (57) b ≡ c (mod Λ) that is, the relation of equivalence modulo Λ is an equivalence relation. It follows that D is partitioned into equivalence classes.

19 The set of equivalence classes will be denoted as D/Λ; that is, an element of D/Λ is a subset of D that includes elements of D that differ by vectors belonging to Λ. An element of D/Λ is also called coset and it can be depicted as a translated version of Λ, see Fig.6. Usually no special notation is used to distinguish an element a ∈ D from the class it belongs to, but in this comment, for the sake of clarity, we will denote the equivalence class that includes a as [a]M.

Note that if x is a signal periodic on Λ, x(a) depends only on the class [a]M that includes a, not on the specific element of [a]M. This is why we can think x as defined on D/Λ. In order to use the result of the unified theory is necessary to give a group structure to D/Λ, that is, to define a sum between the elements of D/Λ (that, remember, are subsets of D). We do this operating by “representatives,” that is,

[a]M + [b]M = [a + b]M (58)

In other words, in order to compute the sum of two elements of D/Λ ([a]M and [b]M) we take one element of each class (that is, a and b), sum them (with the usual sum on D) and take the class of the result. See Fig.6b. Note that the two “+” in (58) represent two different operations: on the right hand side it represents the usual sum of d-dimensional vector, while on the left hand side represents the sum between cosets, that is, elements of D/Λ. A problem that could arise with definition (58) is the possibility that the result on the right hand side depends on the chosen representative of [a]M or [b]M. Indeed, it is quite obvious that [a]M has d infinite representatives since [a]M = [a + Mn]M for every n ∈ Z and that there is no “natural,” uniquely identified representative. However, it is (really ) easy to prove that even if a and b in (58) are replaced with different representatives of the same classes (that is, aˆ and bˆ with [a]M = [aˆ]M and [b]M = [bˆ]M) the class on the right hand side of (58) does not change.

C Proofs

Proof C.1. Proof of Property2 Step 1: Λ(M) ⊆ Λ(N) ⇒ M = NH

Let mi, i = 1,...,d be the columns of M, that is, M = [m1,...,md]. Every mi clearly belongs to Λ(M); d since Λ(M) ⊆ Λ(N) it follows that mi belongs to Λ(N) too, that is, there is an integer vector ni ∈ Z such that

mi = Nni, i = 1,...,d (59)

By collecting equations (59) in matrix form we obtain

[m1,...,md] = M = N [n1,...,nd] (60) | {z } H

20 d×d where, clearly, H ∈ Z . Therefore, we showed that if Λ(M) is a sub-lattice of Λ(N) then there is an integer matrix H such that M = NH. Now we need to prove the other implication. Step 2: M = NH ⇒ Λ(M) ⊆ Λ(N) Suppose x ∈ Λ(M) we want to show that it belongs to Λ(N). Since x ∈ Λ(M), there is an integer vector d n ∈ Z such that x = Mn. It follows that

x = Mn = N Hn = Nk (61) |{z} k d Since k = Hn ∈ Z is an integer vector it follows that x ∈ Λ(N), showing that Λ(M) ⊆ Λ(N).

Proof C.2. Proof of Property1 Clearly, Λ(M) = Λ(N) if and only if

Λ(M) ⊆ Λ(N) (62a)

Λ(N) ⊆ Λ(M) (62b)

Because of Property2,(62) are true if and only if

M = NH (63a)

N = MK (63b)

d×d for some integer matrices H,K ∈ Z . From (63) we deduce that it must be

K = H−1 (64)

Equation (64) shows that both H and H−1 (or, equivalently, K and K−1) must be integer matrices, but this is true if and only if |detH| = 1.

Proof C.3. Proof of Property3 For every r = 1,...,d let Cr be such that [M]r,Cr 6= 0. There is at least one Cr since otherwise the r-th row of M would be zero and detM = 0. Deﬁne matrix U as

[M]r,c [U]r,c = (65) [M]r,Cr d×d Since Λ(M) is rational, every entry of U is rational, therefore U ∈ Q . Moreover,   [M]1,C1    [M]   2,C2  M =  U (66)  ..   .    [M]d,Cd | {z } Db

Let Lr be, for every r = 1,...,d, the least common multiple of the denominators of the elements of row d×d r of U. Let L = diag(L1,...,Ld) and observe that L U ∈ Z and that

M =DbL −1 L U= DK (67) | {z } |{z} D K

21 Proving (12). In order to prove (13) apply the procedure to bring K in Hermite normal form by elementary operations on columns. This corresponds to right multiplying K by a unimodular matrix, preserving the generated lattice.

Proof C.4. Proof of Property4 We need to prove two facts:

− t − 1. Every vector k ∈ Λ(M t) is such that k q ∈ Z for every q ∈ Λ(M). This will imply that Λ(M t) is contained in Λ∗

− t 2. For every vector kˆ 6∈ Λ(M t) one can ﬁnd qˆ ∈ Λ(M) such that kˆ qˆ 6∈ Z. This will imply that no vector outside Λ(M−t) belongs to Λ∗

− d The ﬁrst implication is quite easy: if k ∈ Λ(M t) and q ∈ Λ(M) then there are n,m ∈ Z such that k = M−tm and q = Mn. It follows

t −t t t k q = (M m) (Mn) = m n ∈ Z (68)

Therefore, every k ∈ Λ(M−t) belongs to Λ∗. In order to prove the second implication, suppose kˆ 6∈ Λ(M−t) and deﬁne

u = (M−t)−1kˆ (69)

− d − Note that kˆ = M tu and that u 6∈ Z since otherwise kˆ would belong to Λ(M t). This implies that at least one component of u (say, the i-th one) is not integer. Let ei be the vector with all entries zero, but the i-th one that is equal to 1 and let qˆ = Mei ∈ Λ(M). It follows

t −t t t kˆ qˆ = (M u) (Mei) = u ei = [u]i 6∈ Z (70)

Therefore, since qˆ ∈ Λ(M), kˆ cannot belong to Λ∗.

C.1 Density of Z(α,β)

Property 6. Let α,β ∈ R+. Set Z(α,β) is dense in R if and only if ρ = β/α is irrational.

Proof. Without loss of generality, we can consider only the case Z(1,ρ) since, obviously,

Z(α,β) = α Z(1,ρ) (71)

(that is, we can obtain Z(α,β) by multiplying every element of Z(1,ρ) by α) and this does not change the property of being dense. We need to prove two implications.

If ρ ∈ Q, then S is not dense: This implication is fairly easy. Since ρ ∈ Q, ρ > 0, we can ﬁnd

mutually prime positive integers p,q ∈ N+ such that ρ = p/q. The generic element of Z(1,ρ) can then be written as p nq + mp n + mρ = n + m = n,m ∈ (72) q q Z

22 and by the Chinese Remainder Theorem every multiple of 1/q can be written as (72) for suitable n,m. Therefore, Z(1,ρ) = Z(1/q) (73) that is not dense.

If ρ ∈/ Q then Z(1,ρ) is dense: Our strategy will be to consider the subset of positive elements of Z(1,ρ) R := {x ∈ Z(1,ρ) : x > 0} (74) and proving that its inf r := infR (75) is equal to 0. Our ﬁrst step will be to prove that this strategy is correct.

Step 1: If r = 0 then Z(1,ρ) is dense We will prove density by showing that given any ε > 0 as small as desired and any x ∈ R, we can find xˆ ∈ Z(1,ρ) that approximates x with error at most ε, that is |xˆ− x| ≤ ε. Choose any a ∈ R such that a < ε. A suitable a exists since, by hypothesis, infR = 0 and infR is a limit point of R. Let n = bx/ac ∈ Z. Clearly, since a ∈ R ⊂ Z(1,ρ), na ∈ Z(1,ρ). Moreover, |x − na| = x−na < a ≤ ε. Therefore, na is the desired element of Z(1,ρ) that approximates x, that is, xˆ = na, showing that Z(1,ρ) is dense in R. The first step in proving r = 0 will be to prove that R has not a minimum. (Note that if R had a minimum, r would be equal to said minimum and, therefore, could not be equal to 0, since every element of R is positive). Toward this objective we first prove that if r ∈ R, then Z(1,ρ) = Z(r). Step 2: If r ∈ R, that is, r = minR, then Z(1,ρ) = Z(r)

Suppose R has a minimum, that is, r = infR = minR = n + mρ for some n,m ∈ Z, we want to prove that every element of Z(1,ρ) is multiple of r. Suppose it exists u ∈ Z(1,ρ) such that u is not multiple of r, we want to show that we can ﬁnd q ∈ R smaller than r, so that r could not be the inf of R. Note that we can suppose without loss of generality that u > 0.

Let s = bu/rc ∈ Z be the result of the integer division of u by r and let q = u − sr be the remainder. Clearly q ∈ Z(1,ρ) since both u and r belong to Z(1,ρ); moreover, q is positive, therefore q ∈ R. Finally, q < r, showing that r cannot be the minimum of R. Therefore, if r = minR, then every element of Z(1,ρ) must be a multiple of r. Now, by exploiting this result, we can show that if ρ is irrational than R has no minimum.

Step 3: If ρ 6∈ Q, then r 6∈ R We will prove the contrappositive, that is, if r ∈ R then ρ is rational.

If r ∈ R, then Z(1,ρ) = Z(r) as proved above. Let n,m ∈ Z such that r = n + mρ. Since obviously ρ ∈ Z(1,ρ) = Z(r), ρ must be a multiple of r, that is, there is k ∈ Z such that

ρ = kr = k(n + mρ) (76) that implies kn ρ = ∈ (77) 1 − km Q

23 Therefore, by contrappositive, if ρ is irrational, R has no minimum. The ﬁnal step will be to show that if R has no minimum the inf of R cannot be positive. Step 4: If R has no minimum, then infR = 0

Supposer ˆ > 0 is the inf of R. We will show how to ﬁnd q ∈ R such that q < rˆ. The idea is to exploit the fact that infR is a limit point of R to take two elements of R close enough tor ˆ so that their difference (that still belongs Z(1,ρ) since Z(1,ρ) is a group) is smaller thanr ˆ. Choose any a ∈ R such thatr ˆ < a < 2ˆr and any b ∈ R such thatr ˆ < b < a. Note that

0 < a − b < (2ˆr) − rˆ = rˆ (78)

and that a − b ∈ Z(1,ρ) (79)

Equations (78) and (79) show that a − b is an element of R (since it belongs to Z(1,ρ) and it is positive) smaller thanr ˆ. Therefore,r ˆ cannot be the inf of R. Since this is true for any positiver ˆ, it follows that it must be infR ≤ 0, but infR cannot be negative since every element of R is positive, therefore infR = 0.