Representation Theory: a Combinatorial Viewpoint

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REPRESENTATION THEORY: A COMBINATORIAL VIEWPOINT This book discusses the representation theory of symmetric groups, the theory of symmetric functions and the polynomial representation theory of general linear groups. The ﬁrst chapter provides a detailed account of necessary representation-theoretic background. An important highlight of this book is an innovative treatment of the Robinson–Schensted–Knuth correspondence and its dual by extending Viennot’s geometric ideas. Another unique feature is an exposition of the relationship between these correspondences, the representation theory of symmetric groups and alternating groups and the theory of symmetric functions. Schur algebras are introduced very naturally as algebras of distributions on general linear groups. The treatment of Schur–Weyl duality reveals the directness and simplicity of Schur’s original treatment of the subject. This book is suitable for graduate students, advanced undergraduates and non-specialists with a background in mathematics or physics.

Amritanshu Prasad is a mathematician at The Institute of Mathematical Sciences, Chennai. He obtained his PhD from the University of Chicago, where he worked on automorphic forms and representations of p-adic groups. His current research interests include representation theory, combinatorics, harmonic analysis and number theory. Prasad has extensive experience in teaching mathematics to undergraduate and graduate students in the US, Canada and India. He has been an associate of the Indian Academy of Sciences and was awarded the Young Scientist Medal by the Indian National Science Academy.

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Representation Theory

A Combinatorial Viewpoint

Amritanshu Prasad The Institute of Mathematical Sciences, Chennai

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Library of Congress Cataloging-in-Publication Data Prasad, Amritanshu, author. Representation theory : a combinatorial viewpoint / Amritanshu Prasad. pages cm Includes bibliographical references and index. Summary: “Discusses the representation theory of symmetric groups, the theory of symmetric functions and the polynomial representation theory of general linear groups”—Provided by publisher. ISBN 978-1-107-08205-2 (hardback) 1. Combinatorial group theory. 2. Representations of groups. 3. Symmetry groups. 4. Symmetric functions. I. Title. QA182.5.P73 2015 515’.7223—dc23 2014024621

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Contents

List of Tables page vii Preface page ix 1 Basic Concepts of Representation Theory 1 1.1 Representations and Modules 1 1.2 Invariant Subspaces and Simplicity 5 1.3 Complete Reducibility 7 1.4 Maschke’s Theorem 11 1.5 Decomposing the Regular Module 13 1.6 Tensor Products 19 1.7 Characters 22 1.8 Representations over Complex Numbers 29 2 Permutation Representations 32 2.1 Group Actions and Permutation Representations 32 2.2 Permutations 34 2.3 Partition Representations 39 2.4 Intertwining Permutation Representations 41 2.5 Subset Representations 44 2.6 Intertwining Partition Representations 46 3 The RSK Correspondence 51 3.1 Semistandard Young Tableaux 51 3.2 The RSK Correspondence 56 3.3 Classiﬁcation of Simple Representations of S n 68 4 Character Twists 70 4.1 Inversions and the Sign Character 70 4.2 Twisting by a Multiplicative Character 73 4.3 Conjugate of a Partition 75 4.4 Twisting by the Sign Character 79

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vi Contents

4.5 The Dual RSK Correspondence 80 4.6 Representations of Alternating Groups 83 5 Symmetric Functions 96 5.1 The Ring of Symmetric Functions 96 5.2 Other Bases for Homogeneous Symmetric Functions 98 5.3 Specialization to m Variables 107 5.4 Schur Functions and the Frobenius Character Formula 110 5.5 Frobenius’ Characteristic Function 117 5.6 Branching Rules 119 5.7 Littlewood–Richardson Coeﬃcients 120 5.8 The Hook–Length Formula 124 5.9 The Involution sλ 7→ sλ0 127 5.10 The Jacobi–Trudi Identities 129 5.11 The Recursive Murnaghan–Nakayama Formula 132 5.12 Character Values of Alternating Groups 136 6 Representations of General Linear Groups 141 6.1 Polynomial Representations 141 6.2 Schur Algebras 142 6.3 Schur Algebras and Symmetric Groups 148 6.4 Modules of a Commutant 150 6.5 Characters of the Simple Representations 153 6.6 Polynomial Representations of the Torus 155 6.7 Weight Space Decompositions 158

Hints and Solutions to Selected Exercises 160 Suggestions for Further Reading 182 References 185 Index 189

Tables

2.1 Conjugacy class sizes in S 4 39 2.2 Characters of partition representations of S 3 41

2.3 dim HomS n (K[Xk], K[Xl]) 45

2.4 dim HomS n (K[Xk], K[Xl]) 45 2.5 Mλµ for partitions of 3 47 2.6 Characters of K[Xλ] for S 3 48 2.7 The character table of S 3 49 3.1 Number of SSYT of given shape and type 53 5.1 Growth of det X 107 5.2 Partial character table of S 15 137 S.1 Character table of S 4 170 S.2 Inversions for permutations of 4 172

vii

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Preface

This book is based on courses taught to graduate students at The Institute of Mathematical Sciences, Chennai, and undergraduates of Chennai Mathematical Institute. It presents important combinatorial ideas that underpin contemporary research in representation theory in their simplest setting: the representation theory of symmetric groups, the theory of symmetric functions and the polynomial representation theory of general linear groups. Readers who have a knowledge of algebra at the level of Artin’s book [1] (undergraduate honours level) should find this book quite easy to read. However, Artin’s book is not a strict pre-requisite for reading this book. A good understanding of linear algebra and the definitions of groups, rings and modules will suffice.

A Chapterwise Description

The first chapter is a quick introduction to the basic ideas of representation theory leading up to Schur’s theory of characters. This theory is developed using an explicit Wedderburn decomposition of the group algebra. The irreducible characters emerge naturally from this decomposition. Readers should try and get through this chapter as quickly as possible; they can always return to it later when needed. Things get more interesting from Chapter 2 onwards. Chapter 2 focusses on representations that come from group actions on sets. By constructing enough such representations and studying intertwiners between them, the irreducible representations of the first few symmetric groups are classified. A combinatorial criterion for this method to work in general is also deduced. The combinatorial criterion of Chapter 2 is proved using the Robinson– Schensted–Knuth correspondence in Chapter 3. This correspondence is constructed by generalizing Viennot’s light-and-shadows construction of the Robinson–Schensted algorithm. The classification of irreducible representations

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x Preface

of S n by partitions of n along with a proof of Young’s rule are the main results of this chapter. Chapter 4 introduces the sign character of a symmetric group and shows that twisting by the sign character takes the irreducible representation corresponding to a partition to the representation corresponding to its conjugate partition. Young’s construction of the irreducible representations of S n is deduced, and the relationship between these results and the dual RSK correspondence is explained. A light-and-shadows type construction for the dual RSK correspondence is also provided. In the last section of this chapter, the irreducible representations of the alternating groups An are classified, and some ideas involved in computing their character tables are outlined with the help of examples. The complete determination of the character table of An is postponed to Chapter 5. Chapter 5 concerns the algebra of symmetric functions. Bases of this algebra consisting of monomial symmetric functions, elementary symmetric functions, complete symmetric functions, power sum symmetric functions and Schur functions are introduced. Combinatorial interpretations of the transition matrices between these bases are provided. The RSK correspondence and its dual are used to understand and organize these transition matrices. Three different formulae for Schur functions are provided: Kostka’s combinatorial definition using semistandard Young tableaux, Cauchy’s bi-alternant formula and the formulae of Jacobi and Trudi. Frobenius’s beautiful formula for characters of a symmetric group using symmetric functions is a highlight of this chapter. This result moti- vates the definition of Frobenius’s characteristic function, which associates symmetric functions to class functions on S n. Frobenius’s characteristic function is used to deduce branching rules for the restriction of representations of S n to S n−1 and to provide a representation-theoretic interpretation of the Littlewood– Richardson coefficients. Combining the characteristic function with the Jacobi– Trudi identity allows for the deduction of the recursive Murnaghan–Nakayama formula, which is a fast algorithm for computing a character value of a symmetric group. With the help of the recursive Murnaghan–Nakayama formula, the character tables of alternating groups are computed. Chapter 6 treats the polynomial representation theory of general linear groups. Schur algebras are introduced as algebras of homogeneous polynomial distributions on general linear groups. The modules of Schur algebras are shown to correspond to polynomial representations of general linear groups. By interpreting Schur algebras as endomorphism algebras for the actions of symmetric groups on tensor spaces (Schur–Weyl duality), their simple modules are classified. It is shown that polynomial representations of general linear groups are determined by their character values on diagonal matrices or by their restrictions to the subgroup of diagonal matrices (weight spaces). A combinatorial interpretation of the weight

Preface xi

space decomposition of a simple polynomial representation of a general linear group is provided.

About the Exercises

Exercises are interspersed with the text throughout this book. Sometimes important steps in proofs are left as exercises. This gives the reader a chance to think about them carefully. In such cases, unless the exercise is very straightforward, at least a sketch of the solution is always provided. Many other exercises also come with solutions. Readers should make multiple attempts to solve an exercise before looking at the solution. Sometimes reading ahead to the end of the chapter or rereading relevant sections may help in solving them. Exercises are assigned diﬃculty levels from 0 to 5, indicated in square brackets at the beginning. Roughly speaking the diﬃculty levels are decided based on the following key: [0] trivial [1] routine and almost immediate [2] follows from a careful understanding of the material presented [3] a new idea is needed [4] a clever application of a theorem from the text or elsewhere is needed [5] needs sustained work with several new ideas

Acknowledgments

This book would not have been possible had it not been for the support and encouragement that I received from my colleagues K. N. Raghavan and S. Viswanath. Kannappan Sampath read through large parts of this book, discussed them at length with me and suggested many improvements. He also contributed some exercises and many solutions. Important parts of this book have been inﬂuenced by my interactions with my long-time collaborators Uri Onn and Pooja Singla. The idea that the RSK correspondence can be used to classify the simple representations of S n and prove Young’s rule was suggested by their article [23]. Later on, I learned about Schur’s purely algebraic approach to what is now known as Schur–Weyl duality from a series of lectures given by Singla based on Green’s book [10]. She also suggested Exercise 2.4.6 to me. My discussions with T. Geetha were instrumental in shaping many sections of this book, particularly the detailed treatment of the representation theory of alternating groups.

xii Preface

Exercise 1.7.17 was suggested by Jan-Christoph Schlage-Puchta. Ananth Shankar suggested the simple proof of Burnside’s theorem (Theorem 1.5.17) here. Sudipta Kolay contributed the elegant proof to Part 2 of Lemma 3.1.12. The students of IMSc and CMI who took my courses or read early versions of this book were a constant source of inspiration, as where my PhD students C. P. Anilkumar, Kamlakshya Mahatab and Uday Bhaskar Sharma. Steven Spallone went through a preliminary version of this manuscript and made some helpful suggestions for its improvement. The comments of the anonymous referees have also helped improve this book. Manish Chaudhary of Cambridge University Press helped me with my book proposal and through the review process. Suvadip Bhattacharjee and Sana Bhanot did an excellent job editing the manuscript. My wife Anita and my son Kailash were very supportive and understanding throughout the time I was working on this book. It is a great pleasure to thank all these people.

1 Basic Concepts of Representation Theory

This chapter contains a fairly self-contained account of the representation theory of finite groups over a field whose characteristic does not divide the order of the group (the semisimple case). The reader who is already familiar with representations, the group algebra, Schur’s lemma, characters, and Schur’s orthogonality relations could move on to Chapter 2. However, the treatment of these topics in this book may have some new insights for some readers. For instance, the reader will find a careful explanation of why it is that characters (traces of representations) play such an important role in the theory.

1.1 Representations and Modules

Let K be a ﬁeld and G be a ﬁnite group. For a K-vector space V, let GL(V) denote the group of all invertible K-linear maps V → V.

Deﬁnition 1.1.1 (Representation). A representation of G is a pair (ρ, V), where V is a K-vector space and ρ : G → GL(V) is a homomorphism of groups.

Deﬁnition 1.1.2 (Multiplicative character). A multiplicative character of G is a homomorphism χ : G → K∗. Here, K∗ denotes the multiplicative group of non-zero elements of K.

Example 1.1.3. The simplest example of a multiplicative character χ : G → K∗ is given by χ(g) = 1 for all g ∈ G. This is called the trivial character of G.A non-trivial character is any character that is diﬀerent from the trivial character.

Each multiplicative character χ gives rise to a representation as follows: take V to be the one-dimensional vector space K and take ρ to be the homomorphism which takes g ∈ G to the linear automorphism of K, which multiplies each element by χ(g). Conversely, every one-dimensional representation comes from a

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2 Basic Concepts of Representation Theory

multiplicative character. The representation corresponding to the trivial character of G is called the trivial representation of G.

[1] Exercise 1.1.4. Show that each multiplicative character of G contains [G, G] in its kernel (and therefore descends to a multiplicative character G/[G, G] → K∗). Here, [G, G] denotes the subgroup of G generated by elements of the form xyx−1y−1 as x and y run over all elements of G.

[3] Exercise 1.1.5. Let χ : G → K∗ be a non-trivial multiplicative character. Show that X χ(g) = 0. g∈G Representations of groups can be viewed as modules for certain special types of rings called group algebras. It is assumed that the reader is familiar with the definition of rings, ideals and modules. If not, a quick look at the relevant definitions in a standard textbook (for example, Artin [1, Chapter 12, Section 1]) should suffice.

Deﬁnition 1.1.6 (K-algebra). A K-algebra is a ring R whose underlying additive group is a K-vector space and whose multiplication operation R × R → R is K- bilinear. Only unital K-algebras will be considered here, namely those with a multiplicative unit.

Example 1.1.7. The space Mn(K) of n × n matrices with entries in K is a unital K-algebra. If V is an n-dimensional vector space over K, then a choice of basis for V identiﬁes Mn(K) with the algebra EndKV of K-linear maps V → V.

A left ideal of a K-algebra R is a linear subspace of R which is closed under multiplication on the left by elements of R. Similarly, a right ideal is a linear subspace of R which is closed under multiplication on the right by elements of R. A two-sided ideal is a subspace of R which is both a left and a right ideal.

Example 1.1.8. Let W ⊂ Kn be a linear subspace. Then

n {A ∈ Mn(K) | Ax ∈ W for all x ∈ K }

is a right ideal in Mn(K), while

{A ∈ Mn(K) | Ax = 0 for all x ∈ W}

is a left ideal in Mn(K).

[3] Exercise 1.1.9. Show that Mn(K) has no two-sided ideals except for the two obvious ones, namely {0} and Mn(K).

1.1 Representations and Modules 3

[1] Exercise 1.1.10. Show that if R is a K-algebra and I is a two-sided ideal in R, then the product operation of R descends to a bilinear map R/I × R/I → R/I which makes it a K-algebra.

Example 1.1.11. The polynomial algebra K[x1,..., xn] is an infinite dimensional commutative unital K-algebra. Every finitely generated1 commutative K- algebra is a quotient of a polynomial algebra by one of its ideals. The free algebra2 Khx1,..., xni is an infinite dimensional non-commutative algebra. Every finitely generated algebra is a quotient of such an algebra by a two-sided ideal. A K-algebra homomorphism is a homomorphism of rings, which is also K-linear. The usual definition of modules for a ring can be adapted to K-algebras: Definition 1.1.12 (Module). For a K-algebra R, an R-module is a pair (˜ρ, V), where V is a K-vector space andρ ˜ : R → EndKV is a K-algebra homomorphism. We will always assume thatρ ˜ maps the unit of R to the unit of EndKV (such modules are called unital modules). The notion of an R-module in Definition 1.1.12 requires the K-linearity ofρ ˜ and is therefore a little stronger than the general definition of a module for a ring (see, for example, [1, Chapter 12, Section 1]). But the definition above is exactly what is needed to make the correspondence between representations of G and modules of a certain K-algebra K[G] associated to G, as we shall soon see. Example 1.1.13. Every left ideal of R is an R-module. Any subspace of an R-module M, which is closed under the action of R on M, can be viewed as an R-module in its own right and is called a submodule. A quotient of an R-module by a submodule is also an R-module.

n n Example 1.1.14. The vector space K is an Mn(K)-module when vectors in K are written as columns and Mn(K) acts by matrix multiplication on the left. The group algebra K[G] of the group G is a K-algebra whose additive group is the K-vector space with basis

{1g|g ∈ G} and whose product is deﬁned by bilinearly extending

1g1h = 1gh for all g, h ∈ G. (1.1)

1 A subset S of an algebra R is said to be a generating set if R is the smallest algebra containing S . An algebra is said to be ﬁnitely generated if it has a ﬁnite generating subset. 2 The free algebra Khx1,..., xni has as basis words xi1 xi2 ··· xim in the symbols x1, x2,..., xm. Basis elements are multiplied by concatenating the corresponding words.

4 Basic Concepts of Representation Theory

Another useful way of thinking about the group algebra is as the algebra of K-valued functions on G with the product given by convolution: if f1 and f2 are two K-valued functions on G, their convolution f1 ∗ f2 is deﬁned by X f1 ∗ f2(g) = f1(x) f2(y) for all g ∈ G. (1.2) xy=g

[2] Exercise 1.1.15. Identify 1g with the function whose value at g is 1 and which vanishes everywhere else. Under this identiﬁcation, show that the two definitions of the group algebra given above are equivalent.

[3] Exercise 1.1.16. Let n > 1 be an integer. Show that K[Z/nZ] is isomorphic to K[t]/(tn − 1) as an algebra. Here, (tn − 1) denotes the ideal in K[t] generated by tn − 1.

If ρ : G → GL(V) is a representation, and one deﬁnes a K-algebra homomor- phismρ ˜ : K[G] → EndK(V) by X ρ˜ : f 7→ f (g)ρ(g) (1.3) g∈G for each f ∈ K[G], then (˜ρ, V) is a K[G]-module. Conversely, suppose thatρ ˜ : K[G] → EndK(V) is a K[G]-module. Note that if e denotes the identity element of G, then 1e is the multiplicative unit of K[G]. Since we have assumed thatρ ˜(1e) = idV (such a module is called unital), then for any g ∈ G,

ρ˜(1g)˜ρ(1g−1 ) = ρ˜(1e) = idV ,

soρ ˜(1g) ∈ GL(V). Deﬁne a representation ρ of G by

ρ(g) = ρ˜(1g). (1.4) The prescriptions (1.3) and (1.4) deﬁne an equivalence between representations of G and unital K[G]-modules. This correspondence makes it possible to use concepts from ring theory in the study of group representations.

Example 1.1.17 (Regular representation). For each r ∈ R, deﬁne L˜(r) to be the linear endomorphism of R obtained by left multiplication by r. This turns R into an R-module, which is known as the left regular R-module. Let us examine the above construction in the case where R = K[G]. The group ring K[G] becomes a representation of G if we deﬁne L(g): K[G] → K[G] by

L(g)1x = L˜(1g)1x = 1gx. This representation is known as the left regular representation of G. If we deﬁne R : G → GL(K[G]) by

R(g)1x = 1xg−1 ,

1.2 Invariant Subspaces and Simplicity 5

we get another representation of G on K[G], which is known as the right regular representation of G. [1] Exercise 1.1.18. If K[G] is viewed as the space of K-valued functions on G (as in Exercise 1.1.15), then (L(g) f )(x) = f (g−1 x) and (R(g) f )(x) = f (xg).

1.2 Invariant Subspaces and Simplicity

Deﬁnition 1.2.1 (Invariant subspace). A subspace W of V is called an invariant subspace for a representation ρ : G → GL(V) if ρ(g)W ⊂ W for all g ∈ G. Similarly, a subspace W of V is called an invariant subspace for an R-module 3 ρ˜ : R → EndKV ifρ ˜(r)W ⊂ W for all r ∈ R. Example 1.2.2. For the left regular representation (L, K[G]), the subspace of constant functions is a one-dimensional invariant subspace. The subspace n X o K[G]0 = f : G → K | f (g) = 0 g∈G is an invariant subspace of dimension |G| − 1.

[3] Exercise 1.2.3. The subspace K[G]0 has an invariant complement in (L, K[G]) if and only if |G| is not divisible by the characteristic of K (this includes the case where K has characteristic zero). [1] Exercise 1.2.4. Let G = Z/2Z and let K be a field of characteristic two. Show that the subspace of K[G] spanned by 10 + 11 is the only non-trivial proper invariant subspace for the left (or right) regular representation of G. [3] Exercise 1.2.5. Show that if every representation of a group is a sum of one-dimensional invariant subspaces, then the group is abelian. Hint: Use Exercise 1.1.4 and the regular representation. Definition 1.2.6 (Simplicity). A representation or module is said to be simple (or irreducible) if it has no non-trivial proper invariant subspaces. As a convention, the representation or module of dimension zero is not considered to be simple.4 Example 1.2.7. Every one-dimensional representation is simple. [3] Exercise 1.2.8. Every simple module for a finite-dimensional K-algebra is finite dimensional.

3 An invariant subspace of a representation is often called a subrepresentation, and an invariant subspace of a module is usually called a submodule. 4 This is a little bit like 1 not being considered a prime number.

6 Basic Concepts of Representation Theory

[5] Exercise 1.2.9. If K is algebraically closed, and G is abelian, then every simple representation of G is of dimension one. Hint: Show that for any commuting family of matrices in an algebraically closed ﬁeld, there is a basis with respect to which all the matrices in that family are upper triangular.5 Example 1.2.10. The hypothesis that K is algebraically closed is necessary in Exercise 1.2.9. Take for example, G = Z/4Z and ρ : G → GL (R) the represen- ! 2 0 1 tation which takes a generator of Z/4Z to the matrix . Since this matrix −1 0 is a rotation by π/2, no line in R2 is left invariant by it, and so the abelian group Z/4Z admits a simple two-dimensional representation over real numbers.

Deﬁnition 1.2.11 (Intertwiners). Let (ρ1, V1) and (ρ2, V2) be representations of G. A linear transformation T : V1 → V2 is called an intertwiner (or a G-homomorphism) if

T ◦ ρ1(g) = ρ2(g) ◦ T for all g ∈ G. (1.5)

The space of all intertwiners V1 → V2 is denoted HomG(V1, V2). Similarly, for R-modules (˜ρ1, V1) and (˜ρ2, V2), an intertwiner is a linear transformation T : V1 → V2 such that

T ◦ ρ˜1(r) = ρ˜2(r) ◦ T for all r ∈ R.

The space of all such intertwiners is denoted by HomR(V1, V2). The intertwiner condition (1.5) can be visualized as a commutative diagram:

T V1 / V2

ρ1(g) ρ2(g) V / V 1 T 2 If one begins with an element in the top-right corner of this diagram, the images obtained by applying the functions along either of the two paths to the bottom- right corner are the same. [1] Exercise 1.2.12. The kernel of an intertwiner is an invariant subspace of its domain, and the image is an invariant subspace of its codomain. Theorem 1.2.13 (Schur’s lemma I). If K is algebraically closed and V is a ﬁnite-dimensional simple representation of G, then every self-intertwiner T : V → V is a scalar multiple of the identity map. In other words, EndGV = KidV (EndGV denotes HomG(V, V), the self-intertwiners of V, which are also called G-endomorphisms of V).

5 Exercise 1.2.9 becomes much easier if Schur’s lemma (Theorem 1.2.13) is used instead of the hint.

1.3 Complete Reducibility 7

Proof. Since K is algebraically closed, any self-intertwiner T : V → V has an eigenvalue, say λ. Now T − λidV is also an intertwiner. Moreover, it has a non- trivial kernel. Since its kernel is an invariant subspace (Exercise 1.2.12), it must (by the simplicity of V) be all of V. Therefore, T = λidV . A similar statement (with the same proof) holds for simple modules of a K-algebra. [1] Exercise 1.2.14 (Central character). When K is algebraically closed, show that the centre Z(G) of G acts on any simple representation by scalar matrices (if g ∈ Z(G) acts by the scalar matrix λ(g)I, then g 7→ λ(g) is a homomorphism Z(G) → K∗, which is called the central character of the representation). [1] Exercise 1.2.15 (Schur’s lemma for arbitrary ﬁelds). Let K be any ﬁeld (not necessarily algebraically closed). Show that any non-zero self-intertwiner of a simple representation (or module) is invertible.

Deﬁnition 1.2.16 (Isomorphism). We say that representations (or modules) V1 and V2 are isomorphic (and write V1 V2 or ρ1 ρ2) if there exists an invertible intertwiner V1 → V2 (its inverse will be an intertwiner V2 → V1).

Theorem 1.2.17 (Schur’s lemma II). If V1 and V2 are simple, then every non- zero intertwiner T : V1 → V2 is an isomorphism. Consequently, either V1 V2 or there are no non-zero intertwiners V1 → V2. Proof. If T is a non-zero intertwiner, then its kernel is an invariant subspace of V1. Since this kernel cannot be all of V1, it is trivial; hence, T is injective. Its image, being a non-trivial invariant subspace of V2, must be all of V2; therefore, T is an isomorphism. An easy consequence of the two previous results is

Corollary 1.2.18. If K is algebraically closed, V1 and V2 are simple and T : V1 → V2 is any non-trivial intertwiner, then HomG(V1, V2) = KT.

Proof. T is invertible by Schur’s Lemma II. If S : V1 → V2 is another intertwiner, −1 −1 then T ◦ S is a self-intertwiner of V1. By Schur’s Lemma I, T S = λidV1 for some λ ∈ K, whence S = λT.

1.3 Complete Reducibility

Deﬁnition 1.3.1 (Completely reducible module). An R-module is said to be completely reducible if it is a direct sum of simple modules. We have already seen (Exercises 1.2.3 and 1.2.4) that not all modules are completely reducible. From now on, in order to not get distracted by issues that

8 Basic Concepts of Representation Theory

are interesting, but ultimately incidental to the subject matter of this book, we will only consider ﬁnite-dimensional K-algebras and their ﬁnite-dimensional modules until Chapter 6.

[2] Exercise 1.3.2. Assume that every invariant subspace of an R-module V admits an invariant complement. Let W be an invariant subspace of V. Show that every invariant subspace of W admits an invariant complement in W and that every invariant subspace of V/W admits an invariant complement in V/W.

[3] Exercise 1.3.3. Show that an R-module is completely reducible if and only if every invariant subspace has an invariant complement.

[3] Exercise 1.3.4. Show that if the left regular R-module is completely reducible, then every R-module is completely reducible.

If V is a ﬁnite-dimensional completely reducible R-module, then

⊕m1 ⊕m2 ⊕mr V V1 ⊕ V2 ⊕ · · · ⊕ Vr , (1.6) where (by grouping the simple subspaces of V which are isomorphic together) V1, V2,..., Vr is a collection of pairwise non-isomorphic simple R-modules. The number mk is called the multiplicity of Vk in V. We shall refer to (1.6) as the decomposition of V into simple modules with multiplicities. Let W be another ﬁnite-dimensional completely reducible module whose decomposition into simple modules with multiplicities is

⊕n1 ⊕n2 ⊕nr W V1 ⊕ V2 ⊕ · · · Vr (1.7)

(by allowing some of the nk’s and mk’s to be 0, we may assume that the underlying collection V1, V2,..., Vr of simple modules is the same for V and W). Since there are no intertwiners Vi → V j for i , j, any T ∈ HomR(W, V) can be expressed as M T = Tk, k

⊕nk ⊕mk ⊕nk where Tk : Vk → Vk is an intertwiner. Represent an element x ∈ Vk as ∈ ⊕mk ∈ a vector (x1,..., xnk ) and y Vk as y = (y1,..., ymk ), with each xi, yi Vk. Writing these vectors as columns, the intertwiner Tk can itself be expressed as an mk × nk matrix Tk = (Ti j) (where Ti j ∈ EndRVk) using

 T(x)   T T ··· T   x   1   11 12 1nk   1     ···     T(x)2   T21 T22 T2nk   x2    =     .  .   . . .. .   .   .   . . . .   .        T(x)mk Tmk1 Tmk2 ··· Tmknk xnk

1.3 Complete Reducibility 9

Thus, the entries of the matrix, being scalar multiples of the identity idVk , can themselves be thought of as scalars, allowing us to write Mr HomR(W, V) = Mmk×nk (K), k=1

where Mmk×nk denotes the set of mk × nk matrices with entries in K. One easily checks that composition of intertwiners expressed as matrices in the above manner corresponds to multiplication of matrices. Theorem 1.3.5. If K is algebraically closed and V and W have decompositions into sums of simple modules with multiplicities given by (1.6) and (1.7), then X dim HomR(V, W) = dim HomR(W, V) = mini. i In the special case where W = V, we obtain Theorem 1.3.6. Let K be an algebraically closed ﬁeld and R be a K-algebra. If the R-module V is a sum of non-isomorphic simple modules with multiplicities given by (1.6), then EndRV is a sum of matrix algebras (with componentwise multiplication): Mr EndRV Mmi (K), i=1 where the right-hand side should be interpreted as a sum of algebras.

Sum of algebras The notion of a sum of algebras will come up often and therefore deserves a short discussion.

Deﬁnition 1.3.7 (Sum of algebras). If R1, R2,..., Rk are algebras, their sum is the algebra whose underlying vector space is the direct sum R := R1 ⊕ R2 ⊕ · · · ⊕ Rk, with multiplication deﬁned componentwise:

(r1 + r2 + ··· + rk)(s1 + s2 + ··· + sk) = r1 s1 + r2 s2 + ··· + rk sk

Thus, each Ri is a subalgebra of R for each i. If each of the algebras Ri is unital with unit 1i, then the sum

1 := 11 + 12 + ··· + 1k

is the multiplicative unit for R. In particular, R is also unital. If (˜ρi, Mi) is a

unital Ri module (meaning thatρ ˜i(1i) = idMi ), then

M = M1 ⊕ M2 ⊕ · · · ⊕ Mk

10 Basic Concepts of Representation Theory

is also a unital R-module whenρ ˜ : R → EndK M is deﬁned by

ρ˜(r1 + r2 + ··· + rk):= ρ˜1(r1) + ρ˜2(r2) + ··· + ρ˜k(rk).

The Mi’s can be recovered from M by Mi = ρ˜(1i)M. Thus, R-modules correspond precisely to collections of Ri-modules (one for each i).

On a purely combinatorial level

Theorem 1.3.8. Assume that K is algebraically closed. If the R-module V is a sum of non-isomorphic simple modules with multiplicities given by (1.6), then r X 2 dim EndRV = mi . i=1 Recall that the centre of a K-algebra R consists of those elements which commute with every element of R. We all know that the centre of a matrix algebra consists of scalar matrices. The centre of a direct sum of algebras is the direct sum of their centres. It follows that Lr the dimension of the centre of i=1 Mmi (K) is the number of i such that mi > 0. Thus, a consequence of Theorem 1.3.6 is

Theorem 1.3.9. Let R be a K-algebra, with K algebraically closed. If the R-module V is a sum of non-isomorphic simple modules with multiplicities given by (1.6) with all the multiplicities mi > 0, then the dimension of the centre of EndRV is r. The next exercise is a trivial consequence of Theorem 1.3.8

[0] Exercise 1.3.10. Let R be a K-algebra, where K is an algebraically closed ﬁeld. Show that a completely reducible R-module V is simple if and only if

dim EndRV = 1. And similarly, Theorem 1.3.5 can be used to solve the following:

[0] Exercise 1.3.11. Assume that K is algebraically closed, V is simple and W is completely reducible. Then, dim HomR(V, W) is the multiplicity of V in W. For the following exercise, use Theorem 1.3.6

[1] Exercise 1.3.12. Assume that K is algebraically closed. A completely reducible R-module V has a multiplicity-free decomposition (meaning that its decomposition into simple modules with multiplicities is of the form (1.6) with mi = 1 for all i) if and only if its endomorphism algebra EndRV is commutative.

1.4 Maschke’s Theorem 11

[2] Exercise 1.3.13. Assume that K is algebraically closed. If V and W are completely reducible ﬁnite-dimensional R-modules such that

dim EndRV = dim HomR(V, W) = dim EndRW,

then V and W are isomorphic.

[1] Exercise 1.3.14. Two R-modules V and W, satisfying (1.6) and (1.7), where V1, V2,..., Vr are pairwise non-isomorphic simple R-modules, are isomorphic if and only if mi = ni for i = 1, 2,..., r.

[2] Exercise 1.3.15. Assume that K is algebraically closed. Suppose that V1,..., Vr are pairwise non-isomorphic simple R-modules. Show that every

invariant subspace of V = V1 ⊕ · · · ⊕ Vr is of the form Vi1 ⊕ · · · ⊕ Vik for some ⊕n 1 ≤ i1 < ··· < ik ≤ r. In contrast, if n ≥ 2, then V has inﬁnitely many invariant subspaces if K is inﬁnite. Hint: Using Exercise 1.3.11, HomR(Vi, V) is of dimension one.

1.4 Maschke’s Theorem

In Exercise 1.2.3, we saw that when the characteristic of K divides |G| then an invariant subspace of a representation need not contain a complement. On the other hand, we have

Theorem 1.4.1 (Maschke). If (ρ, V) is a representation of G and the characteristic of the ﬁeld K does not divide |G|, then every invariant subspace of V has an invariant complement.

Proof of Maschke’s theorem. Let W be an invariant subspace and let U be any complement (not necessarily invariant) of W. Thus, V = W ⊕ U, meaning that every vector x ∈ V can be uniquely written in the form x = xW + xU . Deﬁne projection operators on V by

PW (x) = xW and PU (x) = xU for all x ∈ V.

Note that PW depends not only on W but also on U. Maschke’s theorem is proved by constructing an invariant complement from the arbitrary complement U by averaging projection operators. The following lemma gives an interpretation of the invariance of U in terms of the projection operators PW and PU :

Lemma 1.4.2. The subspace U is also invariant for ρ if and only if PW ∈ EndGV (equivalently, if and only if PU ∈ EndGV, since PW + PU = idV ).

12 Basic Concepts of Representation Theory

Proof of lemma. Since PW (xW ) = xW , we have

ρ(g)PW (xW ) = ρ(g)xW .

Since W is invariant, ρ(g)xW ∈ W, and therefore,

PW (ρ(g)(xW )) = ρ(g)xW . It follows that

PW ◦ ρ(g)(xW ) = ρ(g) ◦ PW (xW ) for all g ∈ G. (1.8)

On the other hand, since PW (xU ) = 0,

ρ(g)(PW (xU )) = 0.

But PW (ρ(g)xU ) = 0 if and only if ρ(g)xU ∈ U. It follows that the identity

PW (ρ(g)xU ) = ρ(g)(PW (xU )) for all g ∈ G (1.9)

holds for all xU ∈ U if and only if U is an invariant subspace. Therefore, the intertwining property, namely that

ρ(g) ◦ PW = PW ◦ ρ(g) for all g ∈ G being the sum of (1.8) and (1.9), holds if and only if U is an invariant subspace. This completes the proof of the lemma. In order to complete the proof of Maschke’s theorem, deﬁne 1 X P = ρ(g)P ρ(g)−1. W |G| W g∈G Note that the hypothesis that |G| is not divisible by the characteristic is necessary in order to make sense of division by |G|. It is easy to see that PW is a self- intertwiner:

ρ(g)PW = PW ρ(g) for all g ∈ G.

−1 −1 −1 If x ∈ W, then ρ(g) (x) ∈ W. Therefore, PW (ρ(g) (x)) = ρ(g) (x). It follows −1 that ρ(g)PW ρ(g) (x) = x, from which one concludes that for x ∈ W, PW (x) = x. Now, if x ∈ V, then PW (x) ∈ W, and therefore, PW (PW (x)) = PW (x). Thus, 2 PW = PW . 2 Let PU = idV − PW . One easily checks that PU = PU . Let U denote the image of PU . Every vector x ∈ V can be written as

x = PW (x) + PU (x), so V = W + U.

1.5 Decomposing the Regular Module 13

We have seen that if x ∈ W, then PW (x) = x. Therefore,

PU (x) = (1 − PW )(x) = 0 for all x ∈ W. (1.10)

0 0 On the other hand, if x ∈ U, x = PU (x ) for some x ∈ V. Using the fact that 2 PU = PU , 2 0 0 PU (x) = PU (x ) = PU (x ) = x for all x ∈ U. (1.11)

Together, (1.10) and (1.11) imply that W ∩ U = 0. Therefore, V = W ⊕ U (with corresponding projection operators PW and PU ). Thus, by Lemma 1.4.2, U is an invariant complement of W.

Theorem 1.4.3 (Complete reducibility of representations). If (ρ, V) is a ﬁnite- dimensional representation of G and the characteristic of K does not divide |G|, then V is completely reducible.

Proof. Easily follows from Maschke’s theorem (Theorem 1.4.1).

[2] Exercise 1.4.4. Let A be a matrix with entries in an algebraically closed ﬁeld K. Suppose that An = 1 for some n not divisible by the characteristic of K. Use Maschke’s theorem and Exercise 1.2.9 to show that A is diagonalizable.

Definition 1.4.5 (Semisimple algebra). A finite-dimensional K-algebra R is said to be semisimple if every finite-dimensional R-module M is a sum of simple invariant subspaces.

Maschke’s theorem tells us that if the characteristic of K does not divide |G|, then K[G] is a semisimple algebra. The converse also holds if the characteristic of K divides |G|, then K[G] is not semisimple since the regular module admits an invariant subspace that does not have an invariant complement (see Exercise 1.2.3).

1.5 Decomposing the Regular Module

In view of complete reducibility, classifying simple modules up to isomorphism is equivalent to classifying all ﬁnite-dimensional modules up to isomorphism. In order to ﬁnd simple modules, we do not have to look beyond the regular module (namely R itself, thought of as an R-module; see Example 1.1.17):

Theorem 1.5.1. If R is semisimple, then every simple R-module is isomorphic to a submodule of the left regular R-module.

14 Basic Concepts of Representation Theory

Proof. Given a simple R-module V, choose a non-zero vector x ∈ V. Deﬁne φx : R → V by

φx(r) = rx for all r ∈ R.

Then φx is a non-zero intertwiner from the left regular R-module to V. It follows from Theorem 1.3.5 that V is isomorphic to a submodule of R, as claimed. Notice that Theorem 1.5.1 also shows that, up to isomorphism, there are only ﬁnitely many simple R-modules, namely, those which are isomorphic to some simple invariant subspace of the regular R-module. Indeed, if we express the left-regular R-module in the form

⊕m1 ⊕m2 ⊕mr R = V1 ⊕ V2 ⊕ · · · ⊕ Vr (1.12)

with V1, V2,..., Vr pairwise non-isomorphic, then any simple R-module will be isomorphic to one of V1, V2,..., Vr. By Theorem 1.3.6, if (1.12) holds, then Mr EndRR Mmi (K). i=1

Moreover, it is not diﬃcult to relate EndRR to the K-algebra R: [2] Exercise 1.5.2. Let R be a K-algebra with a multiplicative unit. For each r ∈ R, deﬁne ψr : R → R by

ψr(s) = sr for all s ∈ R. (1.13)

Show that r 7→ ψr is a vector-space isomorphism R → EndRR.

Both R and EndRR are K-algebras. How does the isomorphism of Exercise 1.5.2 relate these algebra structures? We calculate:

ψr ◦ ψr0 (s) = ψr0 (s)r = (sr0)r

= ψr0r(s).

Thus, r 7→ ψr is not an algebra homomorphism; rather it reverses the order of multiplication. Reversing the order of multiplication in a K-algebra results in another K-algebra. Deﬁnition 1.5.3 (Opposite algebra). If R is a K-algebra, then its opposite algebra Ropp is the K-algebra whose underlying vector space is R, but whose multiplication operation is given by reversing the order of multiplication in R: rs (product in Ropp) = sr (product in R).

We can now state the exact relationship between the algebras EndRR and R:

1.5 Decomposing the Regular Module 15

Theorem 1.5.4. For any K-algebra R, the map r 7→ ψr deﬁned in (1.13) is an opp isomorphism of algebras R → EndRR. There are many non-commutative algebras which are isomorphic to their opposite algebras:

opp [0] Exercise 1.5.5. An isomorphism of K-algebras Mn(K) → Mn(K) is given by T 7→ T t. [0] Exercise 1.5.6. Let G be a group. An isomorphism of K-algebras K[G] → opp K[G] is defined by 1g 7→ 1g−1 for each g ∈ G. [3] Exercise 1.5.7. Show that the K-algebra of upper triangular n × n matrices is isomorphic to the opposite of the K-algebra of lower triangular n × n matrices. Also, show that the K-algebra of upper triangular n × n matrices is also isomorphic to the K-algebra of lower triangular n × n matrices. Thus, the K-algebra of upper triangular n × n matrices is isomorphic to its own opposite.6 Returning to our semisimple algebra R, we find that R is isomorphic to opp (EndRR) , which in turn is isomorphic to a sum of matrix algebras, which are isomorphic to their own opposites. We get: Theorem 1.5.8 (Wedderburn decomposition). When K is an algebraically closed field, every semisimple K-algebra is isomorphic to a sum of matrix algebras. Thus, matrix algebras are the building blocks of semisimple algebras. If our goal is to understand modules for semisimple algebras, it makes sense to first understand modules for matrix algebras (see the discussion accompanying Definition 1.3.7). To begin with, let us first understand the invariant subspaces in the left regular R-module when R is the matrix algebra Mn(K).

n [2] Exercise 1.5.9. For each linear subspace V of K , deﬁne MV to be the set of all matrices whose rows, when thought of as elements of Kn, lie in V.

1. MV is an invariant subspace of the left regular Mn(K)-module of dimension n dimK V. 2. Every invariant subspace of the left regular Mn(K)-module is of the form MV for some linear subspace V of Kn. 3. MV is simple if and only if V is one dimensional. 4. MV is isomorphic to MW as an Mn(K)-module if and only if the two subspaces V and W have the same dimension.

For example, take Ui to be the line spanned by the ith coordinate vector. Then

MUi consists of all matrices whose non-zero entries all lie in the ith column. By

6 Algebras which are not isomorphic to their opposites do exist. Refer to the discussion at http://mathoverflow.net/q/64370/9672.

16 Basic Concepts of Representation Theory

Exercise 1.5.9, the MUi ’s are simple invariant subspaces which are isomorphic to one another. M (K) = M ⊕ M ⊕ · · · ⊕ M M⊕n n U1 U2 Un U1

is the decomposition of the left regular Mn(K)-module into simple invariant subspaces. n The vector space K is itself an Mn(K)-module when Mn(K) acts by left multi- n n plication on column vectors. Deﬁne a linear map K → MUi by taking x ∈ K to the matrix whose ith column is x and all other columns are 0. This is an isomor- n phism K → MUi of Mn(K)-modules. We obtain the decomposition of the left regular Mn(K)-module:

Theorem 1.5.10. As an Mn(K)-module,

n n n Mn(K) K ⊕ K ⊕ · · · ⊕ K (n times) is a decomposition of the regular representation into a sum of simple representations.

Corollary 1.5.11. The matrix algebra Mn(K) is semisimple. Proof. This is an immediate consequence of Theorem 1.5.10 and Exercise 1.3.4.

By Theorem 1.5.1, every simple module for Mn(K) is isomorphic to the module Kn. From Theorem 1.5.10, we can get the decomposition of the regular module for any sum of matrix algebras: Corollary 1.5.12. Let Mr R = Mmi (K). i=1 m Let (˜ρi, Vi) denote the R-module whose underlying vector space Vi is K i , regarded as column vectors and if

r = r1 + r2 + ··· + rk r j ∈ Mm j (K)),

then ρ˜i(r)(x) = ri x for x ∈ Vi (left multiplication of a column vector by a matrix). The decomposition of the left regular R-module into simples is given by Mr ⊕mi R Vi . i=1 In particular, R is semisimple. Remark 1.5.13. The semisimplicity of R follows from Exercise 1.3.4.

1.5 Decomposing the Regular Module 17

Since every semisimple algebra is a sum of matrix algebras, Corollary 1.5.12 can be rephrased as follows:

Corollary 1.5.14. If K is an algebraically closed ﬁeld and R is a semisimple K-algebra, then there is an isomorphism of algebras M R EndKVi,

(˜ρi,Vi)

where (˜ρi, Vi) runs over the set of isomorphism classes of simple R-modules.

An element ∈ R is said to be idempotent if 2 = . A central idempotent is an idempotent that lies in the centre of R (i.e., it commutes with every r ∈ R). Two central idempotents 1 and 2 are said to be orthogonal if 12 = 0. A central idempotent is said to be primitive if it cannot be written as 1 + 2, where 1 and 2 are non-zero orthogonal central idempotents. The following exercise should convince the reader that the Wedderburn decomposition is unique:

Lk [3] Exercise 1.5.15. In the ring R = i=1 Mmi (K), let i denote the identity matrix of the ith summand. Show that 1, 2, . . . , k are all the primitive central idempotents of R.

The ith matrix algebra can be isolated in R as the two-sided ideal RiR. Clearly, the multiplicative unit 1 of R is the sum of the primitive central idempotents:

1 = 1 + 2 + ··· + k. (1.14)

Corollary 1.5.16. Let G be a ﬁnite group and K be an algebraically closed ﬁeld whose characteristic does not divide |G|. Let (ρi, Vi), i = 1,..., r be a complete set of representatives of the isomorphism classes of simple representations of G. Let di = dim Vi. Then

2 2 1. d1 + ... + dr = |G|. 2. r is the number of conjugacy classes of G.

Proof. By Corollary 1.5.14 and the equivalence between K[G]-modules and representations of G discussed in Section 1.1,

Mr K[G] = EndK(Vi). i=1 Comparing the dimensions of the two sides of the above isomorphism gives (1). Lr Since r is the dimension of the centre of i=1 Mmi (K), which is isomorphic to K[G], it suﬃces to show that the centre of K[G] has dimension equal to the

18 Basic Concepts of Representation Theory P number of conjugacy classes in G. For g ag1g to lie in the centre of K[G], it suﬃces that it commutes with 1x for each x ∈ G, i.e., X X ag1xg = ag1gx for each x ∈ G. g g By re-indexing the sums, we may write X X ax−1g1g = agx−1 1g for each x ∈ G. g g

Comparing the coeﬃcients of 1g on both sides of the above equality gives

ax−1g = agx−1 for all x, g ∈ G.

Replacing g by gx in the above identity gives

ax−1gx = ag for all x, g ∈ G. P Thus, g ag1g is in the centre of K[G] if and only if the function g 7→ ag is constant on the conjugacy classes of G. It follows that the dimension of the centre of K[G] is the number of conjugacy classes of G, completing the proof of (2).

If (˜ρ, V) is an R-module for a ﬁnite-dimensional K-algebra R, andρ ˜(R) = EndKV, then clearly, V cannot admit a non-trivial proper invariant subspace. There- fore, V is simple. The converse is true whenever K is algebraically closed. When R is semisimple, then this is easier to prove and goes by the name of Burnside’s theorem.

Theorem 1.5.17 (Burnside’s theorem). Let K be an algebraically closed ﬁeld and R be a ﬁnite-dimensional semisimple K-algebra. If (˜ρ, V) is a simple R-module, then ρ˜(R) = EndKV.

Proof. Think of EndKV as an R-module using r · T = ρ˜(r) ◦ T. Let x1,..., xn be a basis of V over K. If we write T ∈ EndKV as a matrix with respect to this basis, then each column of this matrix can be thought of as a vector in V, and the action of R on EndKV coincides with that of R on each column under this identiﬁcation. Therefore, as R-modules, ⊕n EndKV V .

Now the image R¯ of R in EndKV, being a subalgebra of EndKV, can be viewed as a submodule of EndKV. Since it is also completely reducible, it must be isomorphic ⊕d to V for some d ≤ n. Therefore, EndR¯ R¯ = EndRR¯ = Md(K). On the other hand, opp EndR¯ R¯ = R¯ . It follows that R¯ is isomorphic to Md(K). But then, since R¯ has an n-dimensional simple module (the module V), d must equal n.

1.6 Tensor Products 19 1.6 Tensor Products

The simplest way to understand tensor products is to work with bases. If V and W are vector spaces with bases x1,..., xn and y1,..., ym, respectively, their tensor product is the vector space whose basis is given by formal symbols xi ⊗ y j and is usually denoted by V ⊗ W. The map on bases

(xi, y j) 7→ xi ⊗ y j

extends uniquely to a bilinear map B : V × W → V ⊗ W, which we shall call the tensor product map. For (x, y) ∈ V × W, the vector B(x, y) in V ⊗ W is usually denoted by x ⊗ y. In general, we can write

x = a1 x1 + ··· + an xn and y = b1y1 + ··· + bmym,

for unique scalars a1,..., an and b1,..., bm. Then

Xn Xm x ⊗ y = aib j (xi ⊗ y j). i=1 j=1

Starting with diﬀerent bases of V and W would have led to an apparently diﬀerent description of the tensor product; let us provisionally denote it by V W. Write B0 for the corresponding tensor product map V × W → V W. However, the tensor product is independent of the choice of basis in the sense that there exists a unique isomorphism φ : V ⊗ W → V W such that B0 = φ ◦ B. This situation is visually described by asserting that the diagram

V × W (1.15) B B0 y % V ⊗ W V W φ /

commutes (in other words, it does not matter along which path you compose the functions; the end result is the same, so φ ◦ B = B0). One may try to work out this isomorphism φ in terms of the change-of-basis matrices. An interesting alternative is to use the following basis-free characterization of the tensor product as MacLane and Birkhoﬀ do [21, Chapter IX, Section 8]:

[2] Exercise 1.6.1 (Universal property of tensor products). Show that the tensor product V ⊗ W has the following property: for every vector space U and every bilinear map D : V × W → U, there exists a unique linear transformation

20 Basic Concepts of Representation Theory

D˜ : V ⊗ W → U such that D = D˜ ◦ B. In other words, there exists a unique linear transformation D˜ such that the diagram commutes.

V × W B D

y # V ⊗ W / U D˜

Applying the result of Exercise 1.6.1 to U = V W and D = B0, we ﬁnd that there exists a unique linear transformation φ : V ⊗ W → V W such that the diagram (1.15) commutes. Exercise 1.6.1 is also valid with V W instead of V ⊗ W. Now taking U = V ⊗ W and D = B, we obtain a linear transformation ψ : V W → V ⊗ W such that the following diagram commutes:

V × W B0 B y % V W V ⊗ W ψ /

By combining the two diagrams, we get

V × W B B B0 y % V ⊗ W V W V ⊗ W φ / ψ /

whence it follows that the composition ψ ◦ φ satisﬁes the commutative diagram:

V × W B B y % V ⊗ W V ⊗ W ψ◦φ /

However, this diagram would still commute if we replace ψ ◦ φ by idV⊗W . The uniqueness assertion in Exercise 1.6.1 with U = V ⊗ W and D = B forces that

ψ◦φ = idV⊗W . Similarly, it follows that φ◦ψ = idVW . Thus, φ is an isomorphism. In the above reasoning, it does not matter how V ⊗W and V W are constructed. The conclusion is that any two models of the tensor product which satisfy the universal property of Exercise 1.6.1 are identified by a unique isomorphism (in this context, a model for the tensor product includes two pieces of information: the vector space V ⊗ W and the bilinear map B : V × W → V ⊗ W). Thus, all the different models of the tensor product are identified with one another via unique isomorphisms.

1.6 Tensor Products 21

[2] Exercise 1.6.2. If X and Y are two ﬁnite sets, construct an isomorphism K[X] ⊗ K[Y] → K[X × Y] (for any set S , K[S ] denotes the K-vector space of K-valued functions on S ).

If S : V1 → V2 and T : W1 → W2, Bi : Vi × Wi → Vi ⊗ Wi are tensor product maps, since (x, y) 7→ B2(S (x), T(y)) is a bilinear map V1 × W1 → V2 ⊗ W2, by Exercise 1.6.1, there exists a unique linear map S ⊗ T : V1 ⊗ W1 → V2 ⊗ W2 such that

(S ⊗ T) ◦ B1 = B2(S (x), T(y)). Classically, S ⊗ T is known as the Kronecker product of S and T. [2] Exercise 1.6.3. Show that (S, T) 7→ S ⊗ T induces an isomorphism Hom(V1, V2) ⊗ Hom(W1, W2) → Hom(V1 ⊗ V2, W1 ⊗ W2). [2] Exercise 1.6.4. Show that trace(S ⊗ T) = (traceS )(traceT). Definition 1.6.5. Suppose (ρ, V) is a representation of G and (σ, W) is a representation of H. Then ρ σ :(g, h) 7→ ρ(g) ⊗ σ(h) is a representation of G × H on V ⊗ W, which is known as the external tensor product of (ρ, V) and (σ, W). Remark 1.6.6. The notion of tensor product defined above for representations of two groups is called the external tensor product. There is also a notion of internal tensor product (ρ ⊗ σ, V ⊗ W), which is defined when both (ρ, V) and (σ, W) are representations of the same group G. This is nothing but the external tensor product (which is a representation of G × G) restricted to {(g, g) | g ∈ G}, the diagonal copy of G inside G × G.

0 [2] Exercise 1.6.7. If V = HomK(V, K) is the dual vector space of a vector space 0 V then for any vector space W, the linear map V ⊗ W → HomK(V, W) induced 0 by the bilinear map V × W → HomK(V, W) deﬁned by (ξ, y) 7→ (x 7→ ξ(x)y) is an isomorphism of vector spaces. [2] Exercise 1.6.8. Let (ρ, V) and (σ, W) be representations of groups G and H, respectively. Then (ρ0 σ, V0 ⊗ W) is a representation of G × H by Exercise 1.6.9. On the other hand, Hom(V, W) (which by Exercise 1.6.7 is canonically isomorphic to the vector space V0 ⊗ W) is also a representation of G × H via τ : G × H → GL(Hom(V, W)) deﬁned by τ(g, h)(T) = σ(h) ◦ T ◦ ρ(g)−1. Show that the isomorphism of Exercise 1.6.7 is in fact an intertwiner of representations of G × H. Thus, V0 ⊗ W and Hom(V, W) are isomorphic representations of G × H.

22 Basic Concepts of Representation Theory

[3] Exercise 1.6.9. Assume that K is algebraically closed and that its characteristic divides neither |G| nor |H|. Show that if ρ and σ are simple, then so is ρ σ. Hint: Use Burnside’s theorem (Theorem 1.5.17) and Exercise 1.6.7. If furthermore (τ, U) and (θ, X) are simple representations of G and H, respectively, such that ρ σ is isomorphic to τ θ, then ρ is isomorphic to τ and σ is isomorphic to θ.

1.7 Characters

In this section, we make the Wedderburn decomposition explicit when R = K[G]. We assume that K is an algebraically closed field whose characteristic does not divide |G|. The simple K[G]-modules are just simple representations of G (under the correspondence discussed in Section 1). Each simple representation (ρ, V) corresponds to a matrix algebra in the Wedderburn decomposition of R (Corollary 1.5.12). The identity matrix in this matrix algebra is a primitive central idempotent of K[G], which we denote by ρ (see Exercise 1.5.15). The element ρ can be viewed as a K-valued function g 7→ ρ(g) on G, as explained in Section 1.1. Our goal shall be the determination of the values ρ(g) in terms of ρ and g. The answer (1.17) highlights the important role that characters play in representation theory. Definition 1.7.1 (Contragredient representation). Let (ρ, V) be a representation of G. The contragredient representation is the representation (ρ0, V0), where V0 is 0 0 −1 the vector space dual to V (V = HomK(V, K)) and ρ (g) is the adjoint of the linear operator ρ(g), namely the linear operator for which (ρ0(g)−1(ξ))(x) = ξ(ρ(g)(x)) for all ξ ∈ V0 and x ∈ V. If (ρ, V) is a representation of G, then (ρ ⊗ ρ0, V ⊗ V0) is a representation of G × G defined by (ρ ⊗ ρ0)(g, g0)(x ⊗ ξ) = [ρ(g)x] ⊗ [ρ0(g0)ξ]. By combining the left and right regular modules, K[G] can also be considered a representation of G × G: 0 T(g, g )1x = 1g0 xg−1 . These two representations of G × G are linked by matrix coefficients: Lemma 1.7.2. Let c : V ⊗ V0 → K[G] be defined by X c(x ⊗ ξ) = ξ(ρ(g)x)1g. g∈G Then c is an intertwiner of representations of G × G.

1.7 Characters 23

Matrix coeﬃcients

0 Suppose x1,..., xn is a basis of V and ξ1, . . . , ξn is the dual basis of V (the linear functional ξi is defined by ξi(x j) = δi j). Then ξi(ρ(g)x j) is the (i, j)th entry of the matrix of ρ(g) with respect to the basis x1,..., xn. Given (x, ξ) ∈ 0 V × V , such that ξ(v) = 1, it is always possible to find a basis x1,..., xn such that x = x1 and ξ = ξ1. Then ξ(ρ(g)v) becomes the (1, 1)th entry of the matrix of ξ with respect to the basis x1,..., xn. For this reason, ξ(ρ(g)x) is called a matrix coefficient of ρ(g).

Theorem 1.7.3. Let (ρ1, V1), (ρ2, V2),..., (ρk, Vk) be a set of representatives for the isomorphism classes of simple representations of G. The linear map k M 0 Φ : Vi ⊗ Vi → K[G] i=1 defined by 0 xi ⊗ ξi 7→ c(xi ⊗ ξi) for xi ∈ Vi, ξi ∈ Vi is an isomorphism of representations of G × G. Proof. Note that Φ is an intertwiner of representations of G×G by Lemma 1.7.2. Since Φ is a linear transformation between two vector spaces which have (by the Wedderburn decomposition) the same dimension, it suffices to show that Φ is injective. 0 By Exercise 1.6.9, the Vi ⊗ Vi are pairwise non-isomorphic simple representations of G × G. By Exercise 1.3.15, the kernel of Φ, being an invariant subspace, 0 must be a sum of some subcollection of the Vi ⊗ Vi. However, none of the sub- 0 spaces Vi ⊗Vi can be contained in the kernel, since for any non-zero vector vi ∈ Vi, 0 there exists ξi ∈ Vi such that Φ(ξi ⊗ vi)(1) = ξi(vi) , 0. [2] Exercise 1.7.4. Work out the isomorphism of Theorem 1.7.3 for G = Z/3Z when K is the field of complex numbers. What about other cyclic groups? If we restrict the representation K[G] of G×G to the second copy of G in G×G, we get the left regular representation of G. On the other hand, the restriction of the representation V ⊗ V0 of G × G to the second copy of G is isomorphic to V0⊕dim V . Theorem 1.7.3, therefore, gives a decomposition of the regular representation: Theorem 1.7.5. In the notation of Theorem 1.7.3, the left regular representation of G has decomposition Mk ⊕ dim Vi K[G] Vi , i=1

24 Basic Concepts of Representation Theory

so each simple representation of G occurs in the regular representation with multiplicity equal to its dimension.

Applying the result of Exercise 1.6.8 to the case where (σ, W) = (ρ, V), we find that V ⊗ V0 and End(V) are isomorphic as representations of G × G. Under this isomorphism of V ⊗ V0 with linear endomorphisms of V, matrix coefficients correspond to trace: Theorem 1.7.6. Let V be a finite dimensional vector space and V0 be its dual. Then for every (ξ, y) ∈ V0 × V, ξ(y) is the trace of the linear operator

Tξ,y : x 7→ ξ(x)y.

Proof. Extend y to a basis of V. The matrix of Tξ,y with respect to any such basis has only one non-zero column, namely the first one. Its trace is, therefore, the element in the first row and first column, which is ξ(y). Theorem 1.7.6 can be restated by saying that the following diagram commutes:

V0 ⊗ V / K (1.16) <

trace

EndKV where the horizontal arrow represents the linear map V0 ⊗ V → K induced by the bilinear map (ξ, x) 7→ ξ(x) and the vertical equality represents the canonical linear 0 isomorphism V ⊗ V → EndKV. 0 Using the fact that EndKV is isomorphic to V ⊗ V as a representation of G ×G, we may rewrite Theorem 1.7.3 as

Theorem 1.7.7. Let (ρ1, V1), (ρ2, V2),..., (ρk, Vk) be a set of representatives for the isomorphism classes of simple representations of G. The linear map

Mk Φ : EndK(Vi) → K[G] i=1 deﬁned by X Ti 7→ trace(ρ(g)Ti)1g for Ti ∈ EndK(Vi) g∈G is an isomorphism of representations of G × G.

[2] Exercise 1.7.8. Show that EndG×G K[G] is the centre of K[G] (you may use Theorem 1.5.4 as a starting point for this).

It follows (using Exercise 1.3.12, or independently from Theorem 1.7.3 and Ex- ercise 1.6.9) that K[G] has a multiplicity-free decomposition into simple

1.7 Characters 25

representations of G × G. Both the Wedderburn decomposition (the version given in Corollary 1.5.14) and Theorem 1.7.7 are decompositions of K[G] into a sum of simple representations of G × G. Therefore, they must coincide on each simple factor up to a scalar. It follows that for each primitive central idempotent i, which is the image of the identity element of EndK(Vi), is given by

i(g) = citrace(ρ(g)),

where ci is some constant depending on i. Note that if V and W are two ﬁnite-dimensional vector spaces over K, T ∈ −1 EndK(V) and S : V → W is an isomorphism, then T and STS have the same trace. It follows that isomorphic representations have the same trace. This allows us to equate the traces on both sides of the isomorphism Mr ⊕ dim Vi K[G] Vi i=1 of Theorem 1.7.5 to obtain the identity 1 Xr X 1 = dim V trace(ρ (g))1 e |G| i i g i=1 g∈G 1 Xr = dim V c−1 . |G| i i i i=1 The right-hand side is a linear combination of linearly independent vectors 1, . . . , r in the vector space K[G]. On the other hand, we already know that Xr 1e = i. i=1

It follows immediately that ci = dim Vi/|G|. In other words, we have Theorem 1.7.9. Let G be a finite group and K be an algebraically closed field whose characteristic does not divide |G|. Let (ρ1, V1),..., (ρr, Vr) be a set of representatives for the isomorphism classes of simple representations of G over the field K. Let dim V (g) = i trace(ρ (g)). (1.17) i |G| i

Then 1, . . . , r are the primitive central idempotents in K[G]. Deﬁnition 1.7.10 (Character). If G is a group and ρ : G → GL(V) is a representation of G on a ﬁnite-dimensional vector space V, then the K-valued function

χρ(g) = trace(ρ(g)) is called the character of ρ.

26 Basic Concepts of Representation Theory

If (ρ1, V1),..., (ρk, Vk) are a set of representatives for the isomorphism classes of simple representations of G, and we write χi(g) for trace(ρi(g); Vi) then the functions χ1, . . . , χr are called the irreducible characters of G. An immediate consequence of Theorem 1.7.9 is that {χ1, . . . , χr} is a linearly independent set in K[G]. Now suppose that a representation (ρ, V) has a decomposition into simples given by

⊕m1 ⊕mr V1 ⊕ · · · ⊕ Vr , then

χρ = m1χ1 + ··· + mrχr.

By the linear independence of irreducible characters, the coefficients m1,..., mr and hence the isomorphism class of (ρ, V) are completely determined by χρ. We have Theorem 1.7.11. Let K be an algebraically closed field whose characteristic does not divide |G|. If two finite-dimensional representations of G have the same character, then they are isomorphic. The most important properties of characters follow from the characterization (1.17) of primitive central idempotents: [2] Exercise 1.7.12. When K is an algebraically closed field whose characteristic does not divide |G|, show that the irreducible characters χ1, . . . , χr form a basis for the centre of K[G] (which is the space of K-valued functions on G which are constant on conjugacy classes, as we have seen in the proof of Corollary 1.5.16) and satisfy the identities 1 X χ (h) χ (g)χ (g−1h) = δ i . (1.18) |G| i j i j χ (1) g∈G i Upon substituting h = 1, (1.18) gives Schur’s orthogonality relations 1 X χ (g)χ (g−1) = δ . (1.19) |G| i j i j g∈G In other words, if we define a bilinear map K[G] × K[G] → K by 1 X h f , f i = f (g) f (g−1), (1.20) 1 2 G |G| 1 2 g∈G then the irreducible characters form an orthonormal set for this pairing. An obvious consequence of Theorem 1.19 is that when the characteristic of an algebraically closed field K does not divide |G|, then the number of isomorphism classes of simple representations is the number of conjugacy classes in G. Fur- ther, each irreducible character of G is a function on the conjugacy classes. The

1.7 Characters 27

character table of G is the square array whose rows are indexed by the isomorphism classes of the simple representations of G and whose columns are indexed by its conjugacy classes. Each entry is the value of the irreducible character of the simple representation indexing its row evaluated at any element of the conjugacy class indexing its column. For example, see Table 2.7, the character table of the group S 3. The importance of the identities (1.19) is that they allow us to calculate the dimensions of intertwiner spaces between representations (and hence also multiplicities of the simple representation in a given representation) using characters.

Theorem 1.7.13. Let K be an algebraically closed ﬁeld whose characteristic does not divide |G|. Let (ρ, V) and (σ, W) be two ﬁnite-dimensional representations of G over K. Then

dim HomG(V, W) = hχρ, χσiG. (1.21)

Proof. Suppose that V1,..., Vr is a set of representatives for the isomorphism classes of simple representations of G. If V and W have decompositions given by Eqs. (1.6) and (1.7), respectively, then

χρ = n1χ1 + ··· + nrχr

χσ = m1χ1 + ··· + mrχr.

Now using the fact that χ1, . . . , χr form an orthonormal set for the pairing (1.20), we have that

hχρ, χσi = m1n1 + ··· + mrnr,

which, by Theorem 1.3.5, is the dimension of HomG(V, W). As a special case, we may compute the multiplicity of a simple representation in any given representation using characters:

Theorem 1.7.14. Let K be an algebraically closed ﬁeld whose characteristic does not divide |G|. Suppose that a simple representation (ρ, V) of G occurs in a representation (σ, W) with multiplicity m. Then

m = hχρ, χσiG. Exercise 1.7.15. Compute the character table for the following groups over the speciﬁed ﬁelds:

1. Finite cyclic groups over complex numbers (difficulty level: 1). 2. The dihedral group of order 8 over any algebraically closed field of characteristic greater than 2 (this is the group of symmetries of a square, see box below; difficulty level: 4).

28 Basic Concepts of Representation Theory

3. The quaternion group over any algebraically closed ﬁeld of characteristic greater than 2. The quaternion group is the subgroup of GL2(C) consisting of the elements {±1, ±i, ±j, ±k}, where ! ! ! ! 1 0 i 0 0 1 0 i 1 = , i = , j = and k = 0 1 0 −i −1 0 i 0 (diﬃculty level: 4).7

Dihedral Groups

The dihedral group D2n is defined to be the group of linear transformations R2 → R2 which fixes the vertices of a regular n-gon centred at the origin. This group consists of rotations (by angles that are multiples of 2π/n) and reflections about the n axes of symmetry of the n-gon and therefore has order 2n. If s denotes any one of the reflections and r a rotation by 2π/n, then D2n has a presentation hs, r | s2 = 1, rn = 1, s−1rs = r−1i.

[2] Exercise 1.7.16. Let g1, g2,..., gr denote representatives of the conjugacy classes of G. Let χ1, χ2, . . . , χr be the characters of the simple representations of G. Let X denote the character table matrix of G, namely the matrix whose (i, j)th element is χi(g j). Let Z denote the diagonal matrix whose ith diagonal entry is the cardinality of the centralizer of gi for each i ∈ {1,..., r}. Let E denote the −1 permutation matrix for which Ei j = 1 if gi lies in the conjugacy class of g j. When K is an algebraically closed ﬁeld whose characteristic does not divide |G|, show that X0X = EZ (here X0 denotes the transpose of X). Use this to deduce the dual orthogonality relations:

r  X |ZG(g j)| if i = j, χ (g−1)χ (g ) =  k i k j  k=1 0 otherwise. Hint: Start oﬀ by computing the matrix XZ−1EX0.

7 Interestingly, the character tables of the dihedral group of order 8 and the quaternion group are isomorphic in the sense that, after rearrangement of rows and columns, they become the same. Yet these groups are not isomorphic (this can be seen, for example, by noting that the dihedral group of order 8 has three elements, namely r2, s and rs, of order 2, while the quaternion group only has only one such element, namely −1). It would, therefore, be fair to say that a group is not determined by its character table. Frobenius’s definition of characters preceded his definition of representations. The original definition of characters was motivated by trying to understand a polynomial known as the group determinant that Dedekind and Frobenius associated to a group. It is now known (Formanek and Sibley [6]) that although the character table of a group does not determine the group up to isomorphism, its group determinant does. We refer the reader to Chapter II of [5] for a nice discussion of this story.

1.8 Representations over Complex Numbers 29

[2] Exercise 1.7.17. Let g1, g2,..., gr denote representatives of the conjugacy classes C1, C2,..., Cr of G. Let χ1, χ2, . . . , χr be the characters of the simple representations of G.

1. Show that the indicator function 1C j (the function that is 1 on C j and 0 outside it) has an expansion r |C j| X 1 = χ (x−1)χ . C j |G| i j i i=1 2. Show that for any z ∈ G,

r −1 −1 |C j||Ck| X χi(x j )χi(xk )χi(z) |{x ∈ C , y ∈ C | xy = z}| = . j k |G| χ (1) i=1 i

Hint: Compute 1C j ∗1Ck (z) using part (1) of this exercise and the identity (1.18).

1.8 Representations over Complex Numbers

In this section, we point out some features of representation theory over complex numbers. Let V be a ﬁnite-dimensional complex vector space. We assume that the reader is familiar with the basic theory of Hermitian inner products on vector spaces. We quickly recall the most important ideas. An Hermitian inner product on V is a function V × V → C denoted by x, y 7→ (x, y) such that

1. the function x 7→ (x, y) is C-linear for all y ∈ V, 2. (y, x) = (x, y), 3. (x, x) ≥ 0 for all x ∈ V, equality holding only if x = 0.

The standard example is the Hermitian dot product Cn × Cn → C given by

(x1,..., xn) · (y1,..., yn) = x1y¯1 + x2y¯2 + ... + xny¯n. (1.22) A finite-dimensional vector space, together with an Hermitian inner product, is called a finite-dimensional Hilbert space. Every finite-dimensional Hilbert space admits an orthonormal basis (namely, a n basis e1,..., en such that (ei, e j) = δi j for all i, j). When V is identified with C by taking coordinates of vectors with respect to an orthonormal basis, then the Hermitian inner is given by (1.22). Thus, every finite-dimensional Hilbert space can be identified with Cn endowed with the Hermitian dot product.

Deﬁnition 1.8.1 (Unitary operator). Let V be a ﬁnite-dimensional Hilbert space and T : V → V be a linear endomorphism of V. Then T is said to be unitary if, for all x, y ∈ V, (T x, Ty) = (x, y).

30 Basic Concepts of Representation Theory

Clearly, the set U(V) of unitary operators on a ﬁnite-dimensional Hilbert space V forms a subgroup of GL(V). An important property of unitary operators is that it is very easy to ﬁnd invariant complements of invariant subspaces:

Lemma 1.8.2. If S is a set of unitary operators on a ﬁnite-dimensional Hilbert space V and W ⊂ V is invariant under each operator in S , then the subspace

W⊥ = {v ∈ V | (v, w) = 0 for all w ∈ W}

is a complement of W that is invariant under every operator in S .

Proof. If v ∈ W⊥, then by the unitarity of ρ(g),

(ρ(g)v, w) = (v, ρ(g−1)w) = 0,

since ρ(g−1)w ∈ W. Therefore, W⊥ is also an invariant subspace. That W⊥ is a complement of W is a standard exercise in linear algebra (see Artin [1, Chapter 7, Proposition 4.20]).

The unitarity of T can be interpreted in terms of matrices as follows: let A denote the matrix of T with respect to any orthonormal basis. Let A∗ denote the adjoint of A, namely the transpose of the entry-wise complex conjugate of A. Then T is unitary if and only if A∗A = I. As a consequence, we have

Lemma 1.8.3. If T is unitary, then traceT −1 = traceT.

The importance of unitarity to complex representations of ﬁnite groups lies in the following fact:

Theorem 1.8.4. Given a representation of a ﬁnite group G on a ﬁnite-dimensional complex vector space V, there exists a Hermitian inner product on V with respect to which ρ(g) is unitary for every g ∈ G.

Proof. Take any Hermitian inner form x, y 7→ (x, y)0 on V and make it G- invariant by summing it over G: X (x, y) = (ρ(g)x, ρ(g)y)0 g∈G

to obtain an Hermitian inner product with respect to which ρ(g) is unitary for all g ∈ G.

[2] Exercise 1.8.5. Use Lemma 1.8.2 and Theorem 1.8.4 to give an alternate (and shorter) proof of Maschke’s theorem for complex representations.

1.8 Representations over Complex Numbers 31

Lemma 1.8.3 and Theorem 1.8.4 imply that for any complex representation (ρ, V) of a finite group G, trace(ρ(g); V) = trace(ρ(g−1); V). (1.23) Moreover, let l2(G) denote the vector space of all complex valued functions on G, endowed with the Hermitian inner product: 1 X ( f, g) = f (x)g(x). G |G| x∈G Then Schur’s orthogonality relations (see Exercise 1.7.12) can be reinterpreted in the following form: Theorem 1.8.6. The characters of the irreducible complex representations of G form an orthonormal basis for the subspace of l2(G) consisting of class functions. We end this section with a result about character values: Theorem 1.8.7 (Integrality of character values). Let (ρ, V) be a representation of a finite group G over a field of characteristic zero. Then, if for any g ∈ G, trace(ρ(g); V) is a rational number, it is an integer. Proof. Since g lies in a finite group, gn = 1 for some non-negative integer n. It n follows that ρ(g) = idV , so the minimal polynomial of ρ(g) divides the polynomial xn − 1. Thus, the roots of the minimal polynomial of ρ(g), which coincide with the roots of its characteristic polynomial, are all roots of unity. Any sum of roots of unity is an algebraic integer. But any algebraic integer, if rational, is an integer (see, for example, [1, Chapter 11, Corollary 6.8]). [2] Exercise 1.8.8. If χ is the character of a representation of a finite group in a complex vector space, then |χ(g)| ≤ χ(1) for all g ∈ G. Equality holds if and only if g lies in the kernel of the corresponding representation. Hint: Show that χ(g) is the trace of a matrix whose eigenvalues are unit complex numbers. [3] Exercise 1.8.9. When K is an algebraically closed field whose characteristic does not divide |G|, show that the centre Z(G) of G consists of those elements for which |χ(g)| = χ(1) for each irreducible character χ over complex numbers.

2 Permutation Representations

2.1 Group Actions and Permutation Representations

Definition 2.1.1 (G-set). Let G be a finite group. A G-set is a finite set X together with an action of G on X, namely a homomorphism a : G → Aut(X), where Aut(X) denotes the group of all bijections X → X. For convenience, we use the notation g · x := a(g)(x). We will use G-sets to construct interesting representations of G (see Definition 2.1.12). Example 2.1.2 (Some G-set constructions). Let X and Y be G-sets. Then if we write g · (x, y) = (g · x, g · y), X × Y becomes a G-set. For a function f : X → Y, if we write (g · f )(x) = g · f (g−1 x), then the set Y X of all functions from X to Y becomes a G-set. Definition 2.1.3 (Orbit). Let X be a G-set and x ∈ X. The G-orbit of x is the set G x := {g · x | g ∈ G}. A subset O of X is said to be a G-orbit in X if it is the G-orbit of some point. [1] Exercise 2.1.4. For two points x and y in a G-set X, write x ∼ y if y is in the G-orbit of x. Show that ‘∼’ is an equivalence relation. The equivalence classes are the G-orbits in X. The set of G-orbits in X is usually denoted by G\X. Each G-orbit of X is a G-set in its own right. Definition 2.1.5 (Transitive G-set). We say that a G-set X is transitive if X has only one G-orbit. [1] Exercise 2.1.6. If X is a G-set and Y is an H-set, then X × Y is a G × H-set under the action (g, h) · (x, y) = (g · x, h · y). Describe the G × H-orbits in X × Y.

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2.1 Group Actions and Permutation Representations 33

Example 2.1.7 (Coset space). Let G be a group and H be a subgroup. Recall that G/H is the set of left cosets of H in G (a left coset is a subset of the form xH for some x ∈ G). G acts on G/H by g · xH = gxH. Clearly, G/H is a transitive G-set. A transitive group action is nothing but the set of cosets with respect to some subgroup. In order to see this, we need the following definition: Definition 2.1.8 (Stabilizer subgroup). If X is a G-set and x ∈ X, then the stabilizer subgroup of x is the subgroup Gx of G consisting of group elements which fix x:

Gx := {g ∈ G | g · x = x}. We also need the following notion of isomorphism of G-sets: Definition 2.1.9 (Isomorphism of G-sets). Two G-sets X and Y are said to be isomorphic if there exists a bijection φ : X → Y, which is compatible with the group actions, i.e., φ(g · x) = g · φ(x) for all g ∈ G, x ∈ X. [2] Exercise 2.1.10. If X is a transitive G-set, then X is isomorphic to the coset space G/Gx for any x ∈ X. [2] Exercise 2.1.11. Let H and K be subgroups of G. Then G/H and G/K are isomorphic as G-sets if and only if H is conjugate to K (i.e., there exists g ∈ G such that H = gKg−1). Definition 2.1.12 (Permutation representation). Let K[X] denote the vector space of all K-valued functions on a G-set X. Define ρX : G → GL(K[X]) by −1 ρX(g) f (x) = f (g · x) for all g ∈ G, f ∈ K[X].

Then (ρX, K[X]) is a representation of G and is called the permutation representa- tion1 associated to the G-set X. 1 A permutation representation for a transitive G-set is a special case of an induced representation. If X is a transitive G-set, then we may view K[X] as functions G → K which are invariant under right translation by elements of H, where H is the stabilizer of some x ∈ X (see Exercise 2.1.10). G In general, let H be a subgroup of G and (ρ, V) be a representation of H. Let IndHV be the space { f : G → V | f (xh) = ρ(h)−1 f (x) for all h ∈ H}. This space can be made into a representation of G by setting

g · f (x) = f (g−1 x). In the language of induced representations, the permutation representation K[G/H] is the induced G representation IndH1. Induced representations play an important role in the theory of groups, in their representations and also in number theory. See, for example, [28, chapters VII and VIII].

34 Permutation Representations

[1] Exercise 2.1.13. For each x ∈ X, let 1x denote the K-valued function on X which takes value 1 at x and 0 everywhere else. Show that ρX(g)1x = 1g·x. [1] Exercise 2.1.14. Show that the character of a permutation representation is given by the number of ﬁxed points:

g trace(ρX(g)) = |X |, where Xg denotes the set {x ∈ X | g · x = x} of points in X which are ﬁxed by g. [0] Exercise 2.1.15. If X and Y are isomorphic G-sets, then K[X] and K[Y] are isomorphic representations of G. [1] Exercise 2.1.16. Let X be a G-set. For each G-orbit O ⊂ X, identify K[O] with the subspace of K[X] consisting of functions supported on O. Then K[O] is an invariant subspace of K[X] and M K[X] = K[O]. O∈G\X Thus, every permutation representation is a direct sum of permutation representations associated to a transitive action. [1] Exercise 2.1.17. The subspace P K[X]0 = f : X → K | x∈X f (x) = 0 is always an invariant subspace of K[X]. It has an invariant complement if |X| is not divisible by the characteristic of K.

2.2 Permutations

For each positive integer n, let n = {1, 2,..., n}. Deﬁnition 2.2.1 (Permutation). A permutation of n objects is a bijection w : n → n. The set of all permutations of n objects forms a group under composition, which is called the symmetric group on n objects and is denoted S n.

A convenient way to describe a permutation w ∈ S n is by writing out its permutation string w(1)w(2) ··· w(n).

For example, the permutation w ∈ S 5 such that w(1) = 4, w(2) = 3, w(3) = 1, w(4) = 5 and w(5) = 2 is denoted by the string 43152

2.2 Permutations 35

[1] Exercise 2.2.2. For w as above, what is the string for the permutation wn, for each integer n? What is the string for w−1?

One systematic way to enumerate all permutations is in lexicographic (or dic- tionary) order of permutation strings:

Deﬁnition 2.2.3 (Lexicographic order). The permutation w(1)w(2) ··· w(n) occurs before v(1)v(2) ··· v(n) in lexicographic order if, for some i ∈ {1,..., n−1},

1. w( j) = v( j) for all j < i, 2. w(i) < w(i).

In this order, the identity permutation with permutation string 12 ··· n is the least element and the order-reversing permutation n(n − 1) ··· 1 is the largest.

[1] Exercise 2.2.4. Show that lexicographic order is a linear ordering on S n. [2] Exercise 2.2.5. What is the permutation that comes after 526431 in lexicographic order? What is the permutation that precedes it?

[5] Exercise 2.2.6. Write out an algorithm (or better still, a computer program) to ﬁnd the successor and predecessor of a permutation in lexicographic order.

[1] Exercise 2.2.7. List the six permutations of S 3 in lexicographic order. Draw the multiplication table for this group.

A permutation w of n objects can be visualized as a directed graph: the graph has vertices numbered 1,..., n. There is a directed edge from i to j whenever w(i) = j. The connected components of this directed graph are all cycles, and we will call them the cycles of w. By arbitrarily picking a starting point for a cycle, we can represent it as a sequence of vertices, which will be enclosed in parentheses; if a cycle is of the form i1 → i2 → · · · → ik → i1, we will represent it by (i1, i2,..., ik). Since there are k choices for the starting point, there are k diﬀerent ways of representing this cycle. The integer k is called the length of the cycle; a cycle of length k is called a k-cycle. The decomposition of a permutation into cycles (which are always disjoint) is called its cycle decomposition. Let λ1, λ2, . . . , λl denote the lengths of the cycles of a permutation w of n objects. Then λ1 +···+λl = n. The order in which the cycles are listed is of no importance, so we may rearrange them so that λ1 ≥ · · · ≥ λl. Deﬁnition 2.2.8 (Partition). A partition of n is a sequence

λ1 ≥ · · · ≥ λl

of positive integers such that λ1 + ··· + λl = n. We write λ ` n to signify that λ is a partition of n.

36 Permutation Representations

More generally, any weakly decreasing sequence of positive integers is called a partition.

The partition whose parts are λ1 ≥ · · · ≥ λl is usually denoted (λ1, . . . , λl), its tuple of parts written in weakly decreasing order. When an entry appears repeat- edly, like the 1 in (4, 3, 1, 1, 1), it is customary to indicate this with a superscript as (4, 3, 13). For example, 3 has three partitions, (3), (2, 1) and (1, 1, 1) = (13). We have just seen how the cycle lengths of a permutation of n objects form a partition of n.

Deﬁnition 2.2.9 (Cycle type). The cycle type of a permutation in S n is the partition of n obtained by listing its cycle lengths in decreasing order.

Example 2.2.10. The permutation 53421 is represented by the graph

The cycle decomposition is (234)(15), and the cycle type is (3, 2). The cycle (234) can also be represented as (342) or (423), but not as (432). On the other hand, the permutation 52314 is represented by the graph

The cycle decomposition is (154)(2)(3), and the cycle type is (3, 1, 1).

2.2 Permutations 37

[3] Exercise 2.2.11. Show that the number of permutations in S n with cycle type (n) is (n − 1)!.

[2] Exercise 2.2.12. Show that the order of an element of S n whose cycle type is λ = (λ1, . . . , λl) is the least common multiple of λ1, λ2, . . . , λl. 2 [2] Exercise 2.2.13. A permutation w ∈ S n is called an involution if w is the identity permutation. The permutation w is said to be ﬁxed-point-free if w(i) , i for every i ∈ n. Show that

1. if n is odd, then S n has no fixed-point-free involutions; 2. if n = 2m, then the number of fixed-point-free involutions in S n is the product of the first m odd positive integers: (2m − 1)(2m − 3) ··· 3 · 1, which is usually denoted by (2m − 1)!!.2

We are now ready to come to the description of conjugacy classes in S n based on the following observation, which is so obvious that it hardly deserves a proof: Lemma 2.2.14. For permutations x and w of n objects, if x(i) = j, then wxw−1(w(i)) = w( j).

It follows that if i1 → i2 → · · · → ik → i1 is a cycle of x, then w(i1) → w(i2) → −1 −1 · · · → w(ik) → w(i1) is a cycle of wxw . In particular, x and wxw have the same cycle type. Conversely, suppose x and y have the same cycle type. Then choose w to be the permutation which takes the cycles of x to the cycles of y. For example, if x = (134)(26)(5) and y = (356)(14)(2), then y = wxw−1 when w is the permutation 315624. We have just proved the following theorem: Theorem 2.2.15. Two permutations are conjugate if and only if they have the same cycle type. The number of conjugacy classes in S n is equal to the number of partitions of n.

[2] Exercise 2.2.16. Show that every element of S n is conjugate to its inverse.

[3] Exercise 2.2.17. Show that if two permutations w1 and w2 of S n generate the same cyclic group, then they are conjugate.3

2 See sequence A001147 in The Online Encyclopedia of Integer Sequences, http:oeis.org. 3 The property of S n proved in Exercise 2.2.17 is known as the cyclic conjugation property. More precisely, a group G is said to have the cyclic conjugation property if any two elements which generate the same cyclic subgroup of G are conjugate in G. A group G is called a Q-group if the character of every complex representation of G takes rational values. It is a well-known result in representation theory that a ﬁnite group G is a Q-group if and only if it has the cyclic conjugation property (see [28, Section 13.1] or [14, Proposition 9]). For symmetric groups, we will see that this is the case quite explicitly; see Exercise 3.3.3. The theory Q-groups is quite well developed. See, for example, Kletzing [14]

38 Permutation Representations

For enumerating partitions, we use reverse lexicographic order:

Deﬁnition 2.2.18. We say that the partition (λ1, . . . , λl) of n comes before the partition (µ1, . . . , µm) of n in reverse lexicographic order if there exists i ∈ {1,..., m} such that λ j = µ j for all j < i and λi > µi. It is routine to see that reverse lexicographic order is a linear order on the set of partitions of n. The ﬁrst partition of n in reverse lexicographic order is (n). The last is the partition (1n).

[1] Exercise 2.2.19. What partition of 11 follows (3, 3, 2, 1, 1, 1) in reverse lexicographic order? What partition precedes it?

[5] Exercise 2.2.20. Write out an algorithm (or better still, a computer program) to ﬁnd the successor and predecessor of any partition in reverse lexicographic order?

[2] Exercise 2.2.21. Let p(n) denote the number of integer partitions of n. Let q(n) denote the number of integer partitions of n with no part equal to 1. Show that for every p(n) = p(n − 1) + q(n). Conclude that the sequence p(n) is strictly increasing for n = 1, 2,... .

For the next exercise, where the cardinality of the centralizer of each element of S n is computed in terms of its cycle type, it is helpful to introduce a new way of writing partitions:

Deﬁnition 2.2.22 (Exponential notation for partitions). For each partition λ, its exponential notation is the sequence (1m1 2m2 ··· kmk ), where k is the largest part of λ, and for each i ∈ {1,..., k}, mi is the number of occurrences of i in λ. Terms of the form i0 will sometimes be omitted from this sequence.

Example 2.2.23. The exponential notation for the partition λ = (6, 5, 3, 3) is (325161).

[4] Exercise 2.2.24. If w ∈ S n is an element with cycle type λ, where λ = (1m1 2m2 ··· ) in exponential notation, show that the cardinality of its centralizer in S n is given by Y mi zλ = mi!i , (2.1) i

where the product runs over those i for which mi > 0. The above exercise also allows us to compute the cardinalities of conjugacy classes in S n, because the class consisting of elements with cycle type λ has n!/zλ elements. The centralizer and conjugacy class cardinalities in S 4 are given in Table 2.1.

2.3 Partition Representations 39

λ Exp. notation zλ Class size (4) (41) 4 6 (3, 1) (1131) 3 8 (2, 2) (22) 8 3 (2, 1, 1) (1221) 4 6 (1, 1, 1, 1) (14) 24 1

Table 2.1 Conjugacy class sizes in S 4

[3] Exercise 2.2.25. For each permutation w ∈ S n and each positive integer i, let mi(w) denote the number of cycles of length i in w. Given a polynomial f ∈ K[x1,..., xn], deﬁne a function f˜ : S n → K by

f˜(w) = f (m1(w),..., mn(w)).

Show that if K has characteristic zero, then every class function S n → K is of the 4 form f˜ for some f ∈ K[x1,..., xn].

2.3 Partition Representations

Let us return to our study of permutation representations through a simple but important example. Recall that

n = {1, 2,..., n}.

By an ordered partition of n with l parts, we mean a decomposition a a a n = S 1 S 2 ··· S l,

into pairwise disjoint subsets. Let λi denote the cardinality of the ith subset S i. The sequence

λ = (λ1, λ2, . . . , λl)

is called the shape of the ordered partition S = (S 1, S 2,..., S l). To combinatori- alists, λ is known as a weak composition of n.

4 Since every class function can be expressed as a polynomial in variables x1, x2,... , this exercise opens up the possibility of investigating the polynomials which represent the irreducible characters of S n. For example, in Exercise 2.5.7, we will see that for every n ≥ 2, there exists a simple representation V1 (which we will later denote by V(n−1,1)) of S n whose character is represented by the polynomial x1 − 1. Thus, we have an inﬁnite collection of simple representations, one for each S n (n suﬃciently large) whose characters are represented by the same polynomial. The theory of FI-modules due to Church, Ellenberg and Farb [4] gives rise to families of representations whose characters are eventually given by the same polynomial.

40 Permutation Representations

Deﬁnition 2.3.1 (Weak composition). A weak composition of n into l parts is a solution λ = (λ1, . . . , λl) to

λ1 + ··· + λl = n in non-negative integers.

[1] Exercise 2.3.2 (A standard exercise in combinatorics). The number of weak n+l−1 compositions of n into l parts is l−1 .

Let X denote the set of all ordered partitions of n. Deﬁne an action of σ ∈ S n on X in the obvious manner:

σ · (S 1, S 2,..., S l) = (σ(S 1), σ(S 2), . . . , σ(S l)).

Clearly, the orbits of the symmetric group S n on the set X are completely determined by the shape function:

Theorem 2.3.3. Two ordered partitions of n are in the same S n-orbit if and only if they have the same shape.

If we denote by Xλ the set of all ordered partitions of n of shape λ, then S n acts on Xλ and we may form the permutation representation K[Xλ] of S n. Deﬁnition 2.3.4 (Partition representation). For each weak composition λ of n, the permutation representation K[Xλ] of S n is called the partition representation of shape λ.

[1] Exercise 2.3.5. Show that K[X(1n)] is isomorphic to the regular representation of S n. Suppose that the weak composition µ is obtained from the weak composition λ by permuting its parts. In other words, there is a permutation σ of the symbols 1,..., l such that if λ = (λ1, . . . , λl), then µ = (µσ(1), . . . , µσ(l)). Then an isomorphism of S n-sets Xλ → Xµ is deﬁned by (S 1,..., S l) 7→ (S σ(1),..., S σ(l)). By Exercise 2.1.15, K[Xλ] and K[Xµ] are isomorphic representations. Rearranging the parts of a weak composition in decreasing order results in a partition. There- fore, every one of the partition representations considered above is isomorphic to a partition representation whose shape is a partition. The determination of the decomposition of K[Xλ] of S n into simple representations (in the semisimple case) for every partition λ of n will result in the classiﬁcation of all the simple representations of S n in Chapter 3. Example 2.3.6 (n = 3). For n = 3, there are three partitions, namely (3), (2, 1) and (1, 1, 1). The group S 3 has three conjugacy classes (recall that the conjugacy classes of S n are determined by the cycle decomposition of elements) indexed by

2.4 Intertwining Permutation Representations 41

these partitions. Since the character of a permutation representation at a group element is just the number of ﬁxed points of the group element (Exercise 2.1.14), it is easy to calculate the characters of the partition representations of S 3. In Table 2.2, the entry in the column indexed by partition λ and row indexed by the partition µ is the value of the character of the partition representation of shape µ at the conjugacy class consisting of elements with cycle decomposition λ.

(3) (2, 1) (1, 1, 1) (3) 1 1 1 (2, 1) 0 1 3 (1, 1, 1) 0 0 6

Table 2.2 Characters of partition representations of S 3

2.4 Intertwining Permutation Representations

In order to understand intertwiners between permutation representations, it is convenient to use the language of integral operators associated to kernels. Let X and Y be two ﬁnite sets. A function k : X × Y → K gives rise to a linear transformation Tk : K[Y] → K[X]: X Tk f (x) = k(x, y) f (y). y∈Y

5 The linear transformation Tk is known as the integral operator with kernel k.

[2] Exercise 2.4.1. Show that the map K[X × Y] → HomK(K[Y], K[X]) given by k 7→ Tk is an isomorphism of vector spaces. The composition of integral kernels is like matrix multiplication: [1] Exercise 2.4.2 (Composition of integral operators). Given three ﬁnite sets X, Y and Z, and kernels k1 ∈ K[X × Y] and k2 ∈ K[Y × Z],

Tk1 ◦ Tk2 = Tk1∗k2 ,

where k1 ∗ k2 : X × Z → K is deﬁned by X (k1 ∗ k2)(x, z) = k1(x, y)k2(y, z). y∈Y

5 Integral operators play an important role in functional analysis. For example, see Zimmer [38, Chapter 3].

42 Permutation Representations

When X and Y are both ﬁnite G-sets, we may ask which kernels give rise to integral operators that intertwine the representations K[Y] and K[X]. In other words, for which k ∈ K[X × Y] do we have

−1 ρX(g) ◦ Tk ◦ ρY (g) = Tk for all g ∈ G? (2.2)

Working out the left-hand side:

−1 (ρX(g) ◦ Tk ◦ ρY (g) f ))(x) = Tk ◦ ρY (g) f (g · x) X = k(g · x, y)(ρY (g) f )(y) y∈Y X = k(g · x, y) f (g−1 · y) y∈Y X = k(g · x, g · y) f (y). y∈Y Therefore, (2.2) holds if and only if we have

k(x, y) = k(g · x, g · y) for all x ∈ X, y ∈ Y, g ∈ G. (2.3)

In this context, a useful notion is that of relative position:

Deﬁnition 2.4.3 (Relative position). If X and Y are G-sets, then X × Y can be viewed as a G-set, with the diagonal action:

g · (x, y) = (g · x, g · y).

We say that two pairs (x, y) and (x0, y0) have the same relative position if they lie in the same G-orbit of X × Y. We write G\(X × Y) for the set of all G-orbits in X × Y.

What we have shown is

Theorem 2.4.4 (Intertwining number theorem). Let X and Y be ﬁnite G-sets. Let (ρX, K[X]) and (ρY , K[Y]) be the corresponding permutation representations. Then

dim HomG(K[Y], K[X]) = |G\(X × Y)|.

[2] Exercise 2.4.5 (Multiplicity of the trivial representation). Show that

dim HomG(K[X], 1) = |G\X|.

Here, 1 denotes the trivial representation of G. When K is algebraically closed and its characteristic does not divide the order of G, conclude that the multiplicity of the trivial representation in K[X] is the same as the number of G-orbits in X.

2.4 Intertwining Permutation Representations 43

[2] Exercise 2.4.6 (Burnside’s lemma). Let X be any ﬁnite G-set. Show that 1 X |G\X| = |Xg|. |G| g∈G Hint: Use Exercise 2.4.5 in conjunction with Theorem 1.7.14.

[2] Exercise 2.4.7. The dihedral group D2n of symmetries of the regular n- gon (see Section 1.7) acts on the set V of n vertices. Compute the cardinality of |D2n\(V × V)|. What about |D2n\(V × V × V)|? Example 2.4.8 (Relative position in a group). Let G be a group, which is a G- set in its own right under left translation (g · x = gx for g, x ∈ G). Then two pairs (x, y) and (x0, y0) in G × G have the same relative position if and only if x−1y = x0−1y0. More speciﬁcally, if G = R3 (three-dimensional Euclidean space), then the relative position of two vectors ~x and ~y is determined by the diﬀerence ~y − ~x, which is consistent with the colloquial use of the term ‘relative position’. Example 2.4.9 (Klein’s Erlanger program). Take for X the three-dimensional Euclidean space R3 and for G the groups of all rigid motions (translations com- posed with rotations). Then two pairs of vectors (~x,~y) and (x~0, y~0) have the same relative position if and only if the Euclidean distance between ~x and ~y is the same as the Euclidean distance between x~0 and y~0. Klein [13, Part III, I.1] postulated that a geometry is characterized by its group of symmetries. Here, we see that the fundamental invariant of Euclidean geometry, namely Euclidean distance, can be recovered from the symmetry group of Euclidean geometry by using the notion of relative position. Example 2.4.10 (Doubly transitive actions). For any G-set X which has at least two elements, |G\(X × X)| is always at least two, because (x, x) and (x, y), where y , x, cannot have the same relative position. If K is algebraically closed, then it follows from Schur’s lemma (Theorem 1.2.13) that a permutation representation over K is simple if and only if X is singleton. The subspace of constant functions and K[X]0 (see Exercise 2.1.17) are always proper invariant subspaces (they sometimes coincide; see Exercise 1.2.4). However, there are times when the next- best thing happens: K[X]0 is simple. If K[X] is completely reducible and K is algebraically closed, then K[X]0 is simple if and only if |G\(X × Y)| = 2, in other words, if and only if, whenever (x, y) and (x0, y0) are two pairs of distinct elements of X then there exists g ∈ G such that g · x = x0 and g · y = y0. Such an action is called doubly transitive.

[2] Exercise 2.4.11. Show that the action of S n on n is doubly transitive. Con- clude that if K is an algebraically closed ﬁeld whose characteristic is greater than n, then S n has a simple representation of dimension n − 1 for each n.

44 Permutation Representations 2.5 Subset Representations

We shall now use the intertwining number theorem (Theorem 2.4.4) to ﬁnd the decomposition into simple representations of a subclass of the set of partition representations of S n. This serves to illustrate one of the key steps towards the decomposition of all partition representations into simple representations in a much simpler context. For each 0 ≤ k ≤ n, let Xk denote the set of all subsets of n of cardinality k. ` If S ∈ Xk, then n = S (n − S ) is a partition of n of shape (k, n − k). Thus, as an S n-set, Xk is isomorphic to X(k,n−k) (and also to X(n−k,k) and therefore, to Xn−k), introduced in Section 2.3.

Theorem 2.5.1. For non-negative integers k, l ≤ n, two pairs (S, T) and (S 0, T 0) 0 0 in Xk × Xl have the same relative position if and only if S ∩ T and S ∩ T have the same cardinality.

Proof. If (S, T) and (S 0, T 0) have the same relative position, then there exists 0 0 0 0 g ∈ S n such that g(S ) = S and g(T) = T . Therefore, g(S ∩ T) = S ∩ T . Thus, g restricts to a bijection S ∩ T → S 0 ∩ T 0. Conversely, if S ∩T has the same cardinality as S 0 ∩T 0, then any bijection from S ∩ T to S 0 ∩ T 0 extends to a permutation which takes S to S 0 and T to T 0. Corollary 2.5.2. If k, l ≤ n/2, then

dimK HomS n (K[Xk], K[Xl]) = min{k, l} + 1. Proof. This follows from Theorem 2.4.4 and Theorem 2.5.1, because, for (S, T) ∈ Xk × Xl, the cardinality of S ∩ T can take any integer value between 0 and min{k, l}. The information provided by Corollary 2.5.2 is enough to establish the following:

Theorem 2.5.3 (Decomposition of subset representations). Assume that representations K[Xk], 0 ≤ k ≤ n/2, are all completely reducible. There exist simple representations V0,..., Vbn/2c of S n such that for each integer 0 ≤ k ≤ n/2.

K[Xk] = V0 ⊕ · · · ⊕ Vk.

Proof. Table 2.3 collects the dimensions of HomS n (K[Xk], K[Xl]). Since

dim EndS n K[X0] = 1, K[X0] is a simple representation (by Exercise 1.3.10). An easier way to see this would be to note that K[X0] is the trivial representation, but later in the proof, we will need to rely only on the dimension of the space of

endomorphisms. Set V0 = K[X0]. Since dim HomS n (K[X0], K[Xl]) = 1, V0 occurs with multiplicity one in K[Xl] for each l ≥ 1 (Exercise 1.3.11). Therefore,

2.5 Subset Representations 45

K[X0] K[X1] K[X2] K[X3] ···

K[X0] 1 1 1 1 ··· K[X1] 1 2 2 2 ··· K[X2] 1 2 3 3 ··· K[X3] 1 2 3 4 ··· ......

Table 2.3 dim HomS n (K[Xk], K[Xl])

there exist representations K[Xl]0 such that K[Xl] = V0 ⊕ K[Xl]0 for all l ≥ 1. Since V0 had multiplicity one in K[Xl], it does not occur in K[Xl]0. Since the

contribution of V0 to dim HomS n (K[Xk], K[Xl]) is 1 (in the expression for such a dimension in Theorem 1.3.5, each simple module contributes the term mini).

Therefore, dim HomS n (K[Xk]0, K[Xl]0) = dim HomS n (K[Xk], K[Xl]) − 1. These are shown in Table 2.4 (note the shift in the indices associated to the rows and columns). Proceeding as before, we ﬁnd that V1 := K[X1]0 is simple and occurs

K[X1]0 K[X2]0 K[X3]0 K[X4]0 ···

K[X1]0 1 1 1 1 ··· K[X2]0 1 2 2 2 ··· K[X3]0 1 2 3 3 ··· K[X4]0 1 2 3 4 ··· ......

Table 2.4 dim HomS n (K[Xk], K[Xl])

with multiplicity one in K[Xl]0 for all l ≥ 2. Continuing in this manner, we construct the representations V2, V3 and so on, proving Theorem 2.5.3.

The characters of the representation K[Xk] at an element of S n can be computed at the number of ﬁxed points of this element in Xk (Exercise 2.1.14). Using this, the characters of the simple representations can be computed eﬀectively.

[1] Exercise 2.5.4. Show, for 0 ≤ k ≤ bn/2c, that Vk has dimension n!(n − 2k + 1) . k!(n − k + 1)!

[1] Exercise 2.5.5. Show that x ∈ Xk is ﬁxed by an element g ∈ S n if and only if x is a union of cycles of g. Use this to show that, for each partition λ of n, the

46 Permutation Representations

number of elements of Xk ﬁxed by an element of S n with cycle decomposition λ is the number of ways of adding some parts of λ to get k. [1] Exercise 2.5.6. Let K be an algebraically closed ﬁeld of characteristic greater than 3. Compute the character of the simple representation V1 of S 3.

[2] Exercise 2.5.7. Show that the character value of the representation V1 of S n at a permutation with cycle type λ is m1(λ) − 1, where m1(λ) is the number of times that 1 occurs in λ.

2.6 Intertwining Partition Representations

Let λ = (λ1, . . . , λl) and µ = (µ1, . . . , µm) be weak compositions of a positive integer n. Consider the S n-sets Xλ and Xµ deﬁned in Section 2.3. In Theorem 2.4.4,

we saw that dim HomS n (K[Xλ], K[Xµ]) is the number of orbits for the diagonal action of S n on Xλ × Xµ. Our goal now is to obtain a combinatorial interpretation of S n\(Xλ × Xµ). If S = (S 1,..., S l) and T = (T1,..., Tm) are elements of Xλ and Xµ, respectively, consider the cardinalities of intersections:

ri j(S, T) = |S i ∩ T j|. Observe that

ri1(S, T) + ··· + rim(S, T) = |S i ∩ T1| + ··· + |S i ∩ Tl|

= |S i|

= λi

and similarly, r1 j(S, T) + ··· + rl j(S, T) = µ j. Thus, the l × m matrix r(S, T) = (ri j(S, T)) is a matrix with non-negative integer entries, whose row sums are given by the weak composition λ and column sums are given by the weak composition µ. We call r(S, T) the matrix of intersection numbers of (S, T). Deﬁnition 2.6.1 (λ×µ matrix). Given weak compositions λ and µ of n as above, a λ × µ matrix is an l × m matrix whose entries are all non-negative integers, such that the sum of the ith row is λi for 1 ≤ i ≤ l and the sum of the jth column is µ j for 1 ≤ j ≤ m. We denote the set of λ × µ matrices by Mλµ and the cardinality of Mλµ by Mλµ. Theorem 2.6.2. The map (S, T) 7→ r(S, T) deﬁned above induces a bijection S n\(Xλ × Xµ) → Mλµ.

2.6 Intertwining Partition Representations 47

check that r is injective, we need to show that if (S, T) and (S 0, T 0) are two pairs 0 0 of partitions in Xλ × Xµ such that |S i ∩ T j| = |S i ∩ T j| for each 1 ≤ i ≤ l and 0 0 1 ≤ j ≤ m, then there exists g ∈ S n such that g(S i) = S i and g(T j) = T j for ` ` 0 0 all i and j. To see this, observe that both n = i, j S i ∩ T j and n = i, j S i ∩ T j are partitions of n with parts of the same cardinality (with possibly some empty 0 0 parts). Any function which maps each S i ∩ T j bijectively onto S i ∩ T j for all i and 0 0 j is in fact a permutation g ∈ S n which takes (S, T) to (S , T ). Now suppose that r ∈ Mλµ is any λ × µ matrix. In order to show that (S, T) 7→ r(S, T) is onto Mλµ, we need to exhibit a pair (S, T) ∈ Xλ×Xµ such that r(S, T) = r. P ` Note that n = i, j ri j. Let n = i, j Ai j be any partition (with possibly some empty parts) such that |Ai j| = ri j for each i and j. Now deﬁne a S i = Ai j j a T j = Ai j i

for each 1 ≤ i ≤ l and 1 ≤ j ≤ m. Then, since S i ∩ T j is obviously Ai j,(S, T) is the required pair of partitions. Now using Theorem 2.6.2 together with Theorem 2.4.4, we have Corollary 2.6.3. For any two weak compositions λ and µ of n,

dim HomS n (K[Xλ], K[Xµ]) = Mλµ.

Let us examine the partition representations for n = 3. We tabulate Mλµ as µ and λ range over partitions of 3 in Table 2.5. Assuming that the representations K[Xλ]

(3) (2, 1) (1, 1, 1) (3) 1 1 1 (2, 1) 1 2 3 (1, 1, 1) 1 3 6

Table 2.5 Mλµ for partitions of 3

are all completely reducible, we shall now imitate the reasoning in the proof of Theorem 2.5.3. Since M(3)(3) = dim EndG(K[X(3)]) = 1, K[X(3)] must be simple. Set V(3) = K[X(3)]. We know from the table that V(3) has multiplicity one in K[X(2,1)] and K[X(1,1,1)]. Thus, there exist representations K[Xλ](3) such that K[Xλ] K[Xλ](3) ⊕ V(3) for λ = (2, 1) or (1, 1, 1). Moreover, V0 does not occur K[Xλ](3) for either λ. The

following table gives the values of dim HomS n (K[Xλ](3), K[Xµ](3)):

48 Permutation Representations

(2, 1) (1, 1, 1) (2, 1) 1 2 (1, 1, 1) 2 5

It follows that V(2,1) := K[X(2,1)](3) is simple and occurs with multiplicity two in K[X(1,1,1)](3). Thus, there exists a representation K[X(1,1,1)](2,1) in which neither ⊕2 V(3) nor V(2,1) occurs and such that K[X(1,1,1)](3) = V(2,1) ⊕ K[X(1,1,1)](2,1). Using Theorem 1.3.5, we get

dim EndS n K[X(1,1,1)](3) = dim EndS n K[X(1,1,1)](2,1) + 2 × 2,

2 whence dim EndS n K[X(1,1,1)](2,1) = 5 − 2 = 1, so that V(1,1,1) := K[X(1,1,1)](2,1) is also simple. Thus, we have proved

Theorem 2.6.4. There exist irreducible representations V(3),V(2,1) and V(1,1,1) of S 3 such that

K[X(3)] = V(3)

K[X(2,1)] = V(3) ⊕ V(2,1) ⊕2 K[X(1,1,1)] = V(3) ⊕ V(2,1) ⊕ V(1,1,1).

Since K[X(1,1,1)] is nothing but the left regular representation of S 3 (Exer- cise 2.3.5), V(3), V(2,1) and V(1,1,1) form a complete set of representatives for the isomorphism classes of simple representations if S 3 and the multiplicity of each simple in K[X(1,1,1)] coincides with its dimension (see Theorem 1.7.5). By computing the number of ﬁxed points, we may easily calculate the characters of the representations K[Xλ] for n = 3. These character values are listed in Table 2.6, where the columns represent the cycle types of the conjugacy classes of permutations. Now, using the data from Theorem 2.6.4, we can easily calculate

(3) (2, 1) (1, 1, 1)

K[X(3)] 1 1 1 K[X(2,1)] 0 1 3 K[X(1,1,1)] 0 0 6

Table 2.6 Characters of K[Xλ] for S 3

the character table of S 3 (Table 2.7). In this table, the last column corresponds to the identity permutation and gives the dimensions of these representations. The representation V(2,1) is the same as the representation V1 studied in Exercise 2.5.6. The representation V(1,1,1) is a one-dimensional representation, which the reader

2.6 Intertwining Partition Representations 49

(3) (2, 1) (1, 1, 1)

V(3) 1 1 1 V(2,1) −1 0 2 V(1,1,1) 1 −1 1

Table 2.7 The character table of S 3

may recognize as the sign representation (this representation will be discussed at length in Chapter 4).

[3] Exercise 2.6.5. Perform the analogous analysis for partitions of 4. The order in which the partitions are considered will play an important role in the analysis. They should be considered in reverse lexicographic order.

In each of the three examples that we have considered so far (Theorem 2.5.3, Theorem 2.6.4 and Exercise 2.6.5), we had a collection {Uλ}λ∈P of completely reducible representations of a group G. The indexing set P came with an ordering ‘≤’. Simply by knowing dim HomG(Uλ, Uµ), we were able to deduce the existence of a family of simple representations {Vµ}µ∈P such that

M ⊕Kµλ Uλ = Vµ with Kλλ = 1 (2.4) µ≤λ

and at the same time determine the multiplicities Kµλ. This technique works under a simple combinatorial condition:

Theorem 2.6.6 (Combinatorial resolution theorem). Suppose (P, ≤) is a partially ordered set, and {Uλ}λ∈P is a family of completely reducible representations of a group G. Let Mλµ be the dimension of HomG(Uλ, Uµ). If there exist non-negative integers Kµλ for all µ ≤ λ in P such that Kλλ = 1 for each λ ∈ P and X Mµλ = KνµKνλ for all µ, λ ∈ P, (2.5) ν≤µ, ν≤λ

then, for every µ ∈ P, there exists a simple representation Vµ such that

M ⊕Kµλ Uλ = Vµ for all λ ∈ P. µ≤λ Proof. Let λ be a minimal element of P. Eq. (2.5) implies that M = K2 = 0 λ0λ0 λ0λ0

1. Therefore, Vλ0 := Uλ0 is a simple representation. From (2.5) we also have that

Mλ0λ = Kλ0λ for each λ ∈ P, from which we deduce that Vλ0 occurs in Uλ exactly 0 Kλ0λ times. Therefore, there exist representations Uλ in which Vλ0 does not occur ⊕K and such that U = U0 ⊕ V λ0λ . λ λ λ0λ

50 Permutation Representations

0 0 0 0 0 Let P = P − {λ0}. Let Mλµ = dim HomG(Uλ, Uµ) for all λ, µ ∈ P . Then 0 Mλµ = Mλµ − Kλ0λKλ0µ X = KνλKνµ.

λ0<ν≤λ, λ0<ν≤µ { 0} Therefore, Uλ λ∈P0 is a smaller collection of representations of G satisfying the hypotheses of the theorem. The theorem, therefore, follows by induction on |P|.

Thus, in order to resolve the family of partition representations of S n (and hence also to construct all the simple representations of S n, since K[X(1n)] is the left regular representation), we only need to find a partial order on the set of partitions of n and a collection of non-negative integers Kµλ for µ ≤ λ which satisfy (2.5). This is exactly what it provided by the first of two purely combinatorial Robinson– Schensted–Knuth (RSK) correspondences, which will be the subject of discussion of the next chapter. Thus, the results of the next chapter will enable us to classify the simple representations of S n over any field of characteristic greater than n (see Theorem 3.3.2).

3 The RSK Correspondence

The results of the ﬁrst two sections of this chapter are purely combinatorial and do not use any results from preceding chapters. In the third section, we return to the representation theory of symmetric groups, the primary application of these results. Recall (Deﬁnition 2.2.8) that a partition of n is just a weakly decreasing sequence of non-negative integers which add up to n. We will denote partitions by Greek letters λ, µ, ν, etc., and follow the conventions

λ = (λ1, . . . , λl), (µ1, . . . , µm), and ν = (ν1, . . . , νn).

3.1 Semistandard Young Tableaux

Partitions can be represented by pictures called Young diagrams:

Definition 3.1.1 (Young Diagram). A finite collection of boxes or cells arranged in an array of left justified rows in such a way that each row has at most as many cells as the row above it is called a Young diagram. The shape of a Young diagram is the partition (λ1, λ2,...), where λi is the number of boxes in the ith row.

For example, the Young diagram of shape (6, 5, 3, 3) is

Downloaded from https:/www.cambridge.org/core. Columbia University Libraries, on 16 Jun 2017 at 04:44:52, subject to the Cambridge Core terms of use, available at https:/www.cambridge.org/core/terms. https://doi.org/10.1017/CBO9781139976824.004 Cambridge7A Chapter 03 2014/8/22 15:20 Page 52 #2

52 The RSK Correspondence

A Young tableau1 is a Young diagram in which the boxes are filled in by integers. Of special interest to us will be the class of semistandard Young tableaux: Definition 3.1.2 (Semistandard Young Tableaux). A semistandard Young tableau (SSYT) is a Young diagram in which the boxes are filled in with positive integers in such a way that

1. the entries increase strictly down each column; 2. the entries increase weakly (from left to right) along each row.

A Young tableau has two basic invariants, its shape and its type. The shape of the Young tableau is deﬁned to be the shape of the underlying Young diagram. The type of a Young tableau is deﬁned to be the sequence (λ1, λ2,... ), where λi is the number of times that the integer i occurs in the tableau. We will only consider those tableaux for which the type is a partition (in other words, λ1 ≥ λ2 ≥ · · · ). For example, 1 1 1 1 2 3 (3.1) 2 2 3 3 4 4 5 6 5 6 7 is an SSYT of shape (6, 5, 3, 3) and type (4, 3, 3, 2, 2, 2, 1). Another somewhat silly example of an SSYT is 1 1 1 1 1 1 (3.2) 2 2 2 2 2 3 3 3 4 4 4 which has both shape and type (6, 5, 3, 3). We shall see in Lemma 3.1.12 that, for each partition λ, there is only one SSYT whose shape and type are λ. Example 3.1.3. Consider SSYT of shape µ and type λ, where λ and µ are partitions of 3. For shape µ = (3), there is exactly one SSYT of each type λ: 1 1 1 of type (3), 1 1 2 of type (2, 1) and 1 2 3 of type (1, 1, 1). For shape µ = (2, 1), there is no SSYT of type (3), because putting a 1 in the second row would violate the condition that columns are strictly increasing.

1 In common English usage (see http://en.wiktionary.org/wiki/tableau), the word tableau is a noun which stands for: 1. A striking and vivid representation, a picture. 2. A vivid graphic scene of a group of people arranged as in a painting. It derives from the French word, and in mathematical usage, it is pluralized as tableaux, following the French pluralization.

3.1 Semistandard Young Tableaux 53

(3) (2, 1) (1, 1, 1) (3) 1 1 1 (2, 1) 0 1 2 (1, 1, 1) 0 0 1

Table 3.1 Number of SSYT of given shape and type

However, there is one tableau 1 1 of type (2, 1) and two tableaux 1 2 and 1 3 of 2 3 2 type (1, 1, 1). For shape µ = (1, 1, 1), there is only one SSYT, 1 , which is of type (1, 1, 1). 2 3 The number of SSYT of shape µ and type λ is given in Table 3.1, whose rows are indexed by shapes and columns by types. [1] Exercise 3.1.4. Suppose that λ = (n − l, l) and µ = (n − m, m) for 0 ≤ l, m ≤ bn/2c. Show that there exists a unique SSYT of shape µ and type λ if and only if m ≤ l and if m > l, then there is no SSYT of shape µ and type λ. Given an SSYT of type λ and shape µ, one can construct a sequence of partitions: ∅ ⊂ λ(0) ⊂ λ(1) ⊂ · · · ⊂ λ(m) = λ (3.3) where λ(i) is the partition corresponding to the Young diagram obtained by taking the boxes of the tableau whose entries are at most i for each i ∈ {1,..., m}. For the tableau (3.1), we have the following sequence of Young diagrams:

∅ ⊂ ⊂ ⊂ ⊂ ⊂

⊂ ⊂ ⊂ .

The fact that the columns of an SSYT are strictly increasing implies that each column of λ(i) contains at most one more box than the corresponding column of λ(i−1). It is easy to recover the original SSYT from the sequence of Young diagrams. The i’s are ﬁlled into the boxes which are in λ(i) but not in λ(i−1). [1] Exercise 3.1.5. Write down the SSYT corresponding to the following sequence of Young diagrams: ∅ ⊂ ⊂ ⊂

We may summarize the above discussion as the following theorem, which is easy to prove:

54 The RSK Correspondence

Theorem 3.1.6. There is a bijective correspondence between the set of SSYT of shape λ and type µ and sequences of Young diagrams of the form (3.3), where λ(i) (i) is a Young diagram with µ1 + ··· + µi boxes and each column of λ has at most one cell more than the corresponding column of λ(i−1) for each i = 1,..., m. Let

Kµλ := The number of SSYT of type λ and shape µ. Deﬁne a partial order on the set of partitions of a given integer n as follows: Deﬁnition 3.1.7 (Reverse dominance order). Let λ and µ be partitions of n. Say that µ ≤ λ if µ1 + ··· + µi ≥ λ1 + ··· + λi for each 1 ≤ i ≤ min{l, m}. Then ‘≤’ is a partial order on the set of partitions of n known as the reverse dominance order. [3] Exercise 3.1.8. Show that the reverse dominance order is indeed a partial order on the set of all partitions of n. Find the least value of n for which it is not a total order. [1] Exercise 3.1.9. Show that if µ ≤ (n − k, k) (for k ≤ n − k) in the reverse dominance order, then µ = (n − l, l) for some 0 ≤ l ≤ k. [1] Exercise 3.1.10. Let λ be a hook, i.e., λ = (m, 1k) for some positive integers m and k. Which are the partitions µ ≤ λ? [1] Exercise 3.1.11. Show that if µ ≤ λ in the reverse dominance order, then the number of parts in µ is at most the number of parts in λ.

The numbers Kµλ will satisfy the hypothesis of the combinatorial resolution theorem (Theorem 2.6.6): Lemma 3.1.12 (Triangularity lemma). 1. For every partition λ, the only SSYT of shape λ and type λ is the one whose ith row has all entries equal to i for all i [for example, as in the SSYT of (3.2)]. In particular, Kλλ = 1 for every partition λ. 2 2. For any partitions λ and µ,Kµλ > 0 if and only if µ ≤ λ. Proof. Suppose Y is an SSYT of shape λ and type µ. The main observation for both statements is that, because the columns are strictly increasing, an integer i must go into the first i rows of an SSYT. Thus if λ = µ, then the 1’s, which all go into the first row, exhaust the first row. The 2’s, which, a priori, go into the first two rows, are forced into the second row and exhaust it. By induction, the first i − 1 rows are exhausted by the integers

2 The ‘if’ part of this theorem is generally considered the non-trivial part, but it is not needed for any of the main results in this book. It was conjectured by Snapper [29] in 1971 and proved by Liebler and Vitale [18] in 1973. The simple argument given here was suggested by Sudipta Kolay. For other proofs, see Lam [17] and Hazewinkel and Vorst [11].

3.1 Semistandard Young Tableaux 55

1,..., i − 1 and the λi i’s are forced into the ith row and they fill it up completely. This proves that Y must be, as claimed, in the first part of Lemma 3.1.12. Since the integers 1,..., i must be squeezed into the first i rows, we have

λ1 + ··· + λi ≤ µ1 + ··· + µi (3.4) for every i. Thus, if there exists an SSYT of shape λ and type µ, then (3.4) holds for all i. Conversely, suppose that (3.4) holds for all i. We begin with a Young diagram of shape µ. We put λl l’s into squares that are at the bottom of a column of the Young diagram, starting at the bottommost row and going up, ﬁlling each row from right to left until there are no more bottom squares in that row. For example, if λ = (4, 4, 4) and µ = (7, 3, 2), we get 3 3 3 3 Observe that this procedure ensures that each l is entered at the bottom of a column and that if the boxes containing l are removed from the original Young diagram, then we are left with a smaller Young diagram, which we denote byµ ˜. In the example,µ ˜ has Young diagram

In general,µ ˜ is determined as follows: we know that λl ≤ µ1. Therefore, there exists a unique i ≥ 2 such that µi−1 ≥ λl > µi (if µm ≥ λl, set i = m). Then the l’s occupy all the bottommost squares in the rows i to m and occupy the rightmost λl − µi squares in the (i − 1)th row. Therefore,µ ˜ is given by  µ j if j < i − 1  µ˜ = µ + µ − λ if j = i − 1 j  i−1 i l  µ j+1 if j ≥ i. ˜ Let λ = (λ1, . . . , λl−1). It now suﬃces to show that Kµ˜λ˜ > 0. This will follow by induction on the size of the partitions, provided we can show that

µ˜1 + ··· + µ˜ j ≥ λ1 + ··· + λ j for all j. This is clearly true for j < i − 1. For j = i − 1, we have

µ˜1 + ··· + µ˜i−1 = µ1 + ··· + µi−1 + µi − λl

≥ λ1 + ··· + λi − λl

≥ λ1 + ··· + λi−1 [since λi ≥ λl].

56 The RSK Correspondence

For j ≥ i, the veriﬁcation is similar:

µ˜ 1 + ··· + µ˜ j = µ1 + ··· + (µi−1 + µi − λl) + µi+1 + ··· + µ j+1

≥ λ1 + ··· + λ j+1 − λl

≥ λ1 + ··· + λ j [since λ j ≥ λl]. In the running example, this algorithm would have resulted in an SSYT of shape (7, 3, 2) and type (4, 4, 4).

1 1 1 1 2 2 3 2 2 3 3 3

3.2 The RSK Correspondence

We now come to the main topic of this chapter, namely, the Robinson–Schensted– Knuth (RSK) correspondence. This is a construction which associates each matrix A with non-negative integer entries a pair (P; Q) of SSYT of the same shape. The number of times that an integer i occurs in Q is the sum of the entries of the ith row of A, and the number of times that an integer j occurs in P is the sum of the entries in the jth column of A. Furthermore, this construction is reversible; given a pair (P, Q) of SSYT of the same shape, it is possible to work backwards to uniquely determine the matrix A to which would yield P and Q under the RSK correspondence. Recall that Mλµ denotes the number of λ × µ matrices, namely matrices with non-negative integer entries such that the entries in the ith row sum to µi for i = 1,..., m and the entries in the jth column sum to λ j for j = 1,..., l (see Deﬁnition 2.6.1). The RSK correspondence gives the identity X Mλµ = KνλKνµ, ν≤λ, ν≤µ which is precisely the hypothesis for the combinatorial resolution theorem 3 (Theorem 2.6.6) for the family Uλ = K[Xλ] of S n.

3 The RSK correspondence is a generalization due to Knuth [15] of the Robinson–Schensted correspondence. The Robinson–Schensted correspondence is a correspondence between permutations of n and pairs of standard Young tableaux (see Definition 3.2.10) of the same shape. A permutation matrix is nothing but a λ × µ matrix, where λ = µ = (1n). A standard Young tableau is an SSYT of shape (1n). Thus, the RSK correspondence reduces to the Robinson–Schensted correspondence when λ = µ = (1n). In his paper, Knuth actually introduced two different generalizations of the Robinson–Schensted correspondence, which are now known as the RSK correspondence and the dual RSK correspondence. We will consider the dual RSK correspondence in Chapter 4. Knuth’s definitions

3.2 The RSK Correspondence 57 3.2.1 The Shadow Path Let A be a matrix with non-negative integer entries (henceforth referred to simply as an integer matrix). Fix an entry (i, j) of A (the entry in the ith row and jth column). Following Viennot [37], we call the set of entries (i0, j0) for which i ≤ i0 and j ≤ j0 the shadow of the entry (i, j). One should think of the shadow as deﬁning a partial order on the set of entries of A:(i, j) ≥ (i0, j0) if (i0, j0) lies in the shadow of (i, j), or equivalently,

(i, j) ≥ (i0, j0) if i0 ≥ i and j0 ≥ j. (3.5)

The shadow of the entry (3, 3) is the set of entries in the lower-right rectangular region bounded by (and including) the grey lines in the matrix below:    0 0 0 0 0 0 0     0 0 0 0 0 2 0     0 0 1 1 0 1 0     0 0 1 0 1 0 0     2 1 0 1 0 0 0     0 0 0 0 0 0 1  (3.6)

We say that an entry is maximal if it is non-zero and does not lie in the shadow of any other non-zero entry. This is nothing but an entry which is maximal among the non-zero entries of A with respect to the partial order (3.5). The matrix above has three maximal entries: (5, 1), (3, 3) and (2, 6). Now, suppose that (i1, j1) and (i2, j2) are distinct maximal entries of A, then they cannot lie in the same row or column. For example, suppose j1 = j2. Then if, for example, i1 < i2,(i2, j2) would lie in the shadow of (i1, j1) contradicting the maximality of (i2, j2). So suppose that j1 < j2. Then the maximality of (i2, j2) would force i1 > i2. Suppose that (i1, j1),..., (ir, jr) are the maximal entries of an integer matrix, then we may rearrange them in such a way that

j1 < j2 < ··· < jr.

of these correspondences use Schensted’s insertion algorithm and a variant and are quite different from the ones given here. Our definitions are generalizations of Viennot’s elegant ‘light-and-shadows’ descriptions of the Robinson–Schensted algorithm [37]. A similar generalization of Viennot’s algorithm for the RSK correspondence and its dual already exists; it is Fulton’s ‘matrix-ball construction’ [8]. Fulton’s constructions involve replacing an integer k in the integer matrix by k ‘balls’ in a certain configuration and then applying Viennot’s construction. The method described in this book involves working directly with the matrices themselves. There is yet another construction of the Robinson–Schensted correspondence due to Fomin using ‘growth diagrams’ and ‘local rules’ (see Stanley [34]).

58 The RSK Correspondence

It follows that we would have

i1 > i2 > ··· > ir. The shadow path of A is deﬁned to be the zigzag path obtained by joining the vertices:

(i1, j1), (i1, j2), (i2, j2), (i2, j3), (i3, j3),..., (ir−1, jr), (ir, jr). It bounds the union of the shadows of all the positive entries of A. In the matrix of (3.6), the shadow path is depicted by a grey line.    0 0 0 0 0 0 0     0 0 0 0 0 2 0     0 0 1 1 0 1 0    (3.7)  0 0 1 0 1 0 0     2 1 0 1 0 0 0     0 0 0 0 0 0 1 

The vertices of this path include the original r maximal vertices, as well as r − 1 new vertices

(i1, j2), (i2, j3),..., (ir−1, jr),

which we will call the shadow points of the shadow path. The column j1 is called the terminal column and the row ir is called the terminal row of the shadow path. In (3.7), the shadow points are the entries (5, 3) and (3, 6). The ﬁrst column is the terminal column, while the second row is the terminal row. The shadow path was completely determined by the maximal entries of the matrix. It turns out that the shadow path (and hence the set of maximal entries) can be recovered from the set of shadow points, together with the terminal row and terminal column. To see this, deﬁne the reverse shadow of an entry (i, j) to be the set of all entries (i0, j0) such that i0 ≤ i and j0 ≤ j. Thus, the reverse shadow of the (3, 6)th entry of the matrix A of (3.7) is bounded by the grey lines below:    0 0 0 0 0 0 0     0 0 0 0 0 2 0     0 0 1 1 0 1 0    (3.8)  0 0 1 0 1 0 0     2 1 0 1 0 0 0     0 0 0 0 0 0 1 

Now take the matrix A and append to it an extra row below and an extra column to the right. In the extra row, set all entries zero, except for a ∗ in the terminal column. In the extra column, set all entries zero, except for a ∗ in the terminal

3.2 The RSK Correspondence 59

row. The original shadow path of (3.7) coincides with the path that bounds the reverse shadows of the shadow points (shown in boldface) and the two new entries marked with ∗ (the reverse shadow path):    0 0 0 0 0 0 0 0     0 0 0 0 0 2 0 *     0 0 1 1 0 1 0 0     0 0 1 0 1 0 0 0  (3.9)    2 1 0 1 0 0 0 0     0 0 0 0 0 0 1 0     * 0 0 0 0 0 0 0 

To summarize the preceding discussion, we introduced the shadow of an entry of A and used it to define the shadow partial order (3.5) on the entries of a matrix. The maximal entries of A were defined to be maximal non-zero entries with respect to the shadow partial order. The shadow path was defined to be the zigzag path bounding the shadows of the r maximal entries. The shadow points were the r −1 vertices of the shadow path apart from the maximal entries. We also defined the terminal row and terminal column of A as the first non-zero row and first non-zero column. We found that the set of maximal entries could be recovered from the set of shadow points and the terminal row and column using reverse shadows.

[2] Exercise 3.2.1. For any two elements x and y in a partially ordered set, x ∧ y (the greatest lower bound) of x and y is deﬁned to be the maximal element of the set {z | z ≤ x and z ≤ y},

provided that such an element exists.

1. What is (i, j) ∧ (i0, j0) in the shadow partial order? 2. Show that the shadow points of A are the maximal elements of the set

{(i, j) ∧ (i0, j0) | (i, j) and (i0, j0) are non-zero entries of A}.

3.2.2 Algorithm for Obtaining P and Q In the Viennot version of the RSK correspondence, in order to obtain each row of the SSYT’s P and Q, we keep track of two matrices, A and S (S stands for shadow), and two sequences of integers, p and q (these sequences will be the ﬁrst row of P and the ﬁrst row of Q).

60 The RSK Correspondence

To begin with, A is set to be the original integer matrix, S is set to be the zero matrix and p and q are set to be empty sequences. For example, take 0 0 1 0 0 0 0 0 0 0 0 0 0 0     0 0 0 0 0 2 0 0 0 0 0 0 0 0     1 1 1 1 0 1 0 0 0 0 0 0 0 0 A =   , S =   , 0 0 1 0 0 0 0 0 0 0 0 0 0 0     2 1 0 1 0 0 0 0 0 0 0 0 0 0     0 0 0 0 0 0 1 0 0 0 0 0 0 0

p = ∅ q = ∅ in order to compute the SSYT corresponding to the matrix A above. The shadow path of A is given by    0 0 1 0 0 0 0     0 0 0 0 0 2 0     1 1 1 1 0 1 0     0 0 1 0 0 0 0     2 1 0 1 0 0 0     0 0 0 0 0 0 1 

and gives maximal entries at (3, 1) and (1, 3). Subtract 1 from these entries of A. Because there is a shadow point at (3, 3), add 1 to the (3, 3) entry of S . The terminal column is 1 so append 1 to p. The terminal row is 1 so append 1 to q. At the end of the ﬁrst step, we have 0 0 0 0 0 0 0 0 0 0 0 0 0 0     0 0 0 0 0 2 0 0 0 0 0 0 0 0     0 1 1 1 0 1 0 0 0 1 0 0 0 0 A =   , S =   , 0 0 1 0 0 0 0 0 0 0 0 0 0 0     2 1 0 1 0 0 0 0 0 0 0 0 0 0     0 0 0 0 0 0 1 0 0 0 0 0 0 0

p = 1 q = 1. Now the maximal path of A is given by    0 0 0 0 0 0 0     0 0 0 0 0 2 0     0 1 1 1 0 1 0     0 0 1 0 0 0 0     2 1 0 1 0 0 0     0 0 0 0 0 0 1 

3.2 The RSK Correspondence 61

Subtract 1 from the maximal entries (5, 1), (3, 2) and (2, 6) of A, add 1 to the shadow entries (5, 2) and (3, 6) and append 1, the number of the terminal column to p, and 2, the number of the terminal row to Q. At the end of the second step, this gives

0 0 0 0 0 0 0 0 0 0 0 0 0 0     0 0 0 0 0 1 0 0 0 0 0 0 1 0     0 0 1 1 0 1 0 0 0 1 0 0 0 0 A =   , S =   , 0 0 1 0 0 0 0 0 0 0 0 0 0 0     1 1 0 1 0 0 0 0 1 0 0 0 0 0     0 0 0 0 0 0 1 0 0 0 0 0 0 0

p = 11 q = 12.

This process is continued until A is reduced to 0. The final values of p and q at this stage are the first row of P and the first row of Q. The reader is urged to continue this process and tally the results with Figure 3.1. In order to obtain the second rows of P and Q, the same process is repeated, but this time, the shadow matrix S obtained in the computation of the first row is used as the matrix A, and S is initialized to 0, p and q to ∅ as before:

0 0 0 0 0 0 0 0 0 0 0 0 0 0     0 0 0 0 0 0 0 0 0 0 0 0 0 0     0 0 1 0 0 2 0 0 0 0 0 0 0 0 A =   , S =   , 0 0 0 1 0 0 0 0 0 0 0 0 0 0     0 1 2 0 0 1 0 0 0 0 0 0 0 0     0 0 0 0 0 0 0 0 0 0 0 0 0 0

p = ∅ q = ∅.

Applying the same algorithm that was used to obtain the ﬁrst rows of P and Q to this initial data results in the second rows of P and Q, which (the reader is urged to verify) are p = 2335 and q = 3335

and a shadow matrix 0 0 0 0 0 0 0   0 0 0 0 0 0 0   0 0 0 0 0 0 0   0 0 0 0 0 0 0   0 0 2 0 0 1 0   0 0 0 0 0 0 0

62 The RSK Correspondence

A S p q

0 0 1 0 0 0 0 0 0 0 0 0 0 0     0 0 0 0 0 2 0 0 0 0 0 0 0 0 1 1 1 1 0 1 0 0 0 0 0 0 0 0     ∅ ∅ 0 0 1 0 0 0 0 0 0 0 0 0 0 0     2 1 0 1 0 0 0 0 0 0 0 0 0 0     0 0 0 0 0 0 1 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0     0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 1 1 1 0 1 0 0 0 1 0 0 0 0     1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0     2 1 0 1 0 0 0 0 0 0 0 0 0 0     0 0 0 0 0 0 1 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0     0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 0 1 0 0 1 0     11 12 0 0 1 0 0 0 0 0 0 0 0 0 0 0     1 1 0 1 0 0 0 0 1 0 0 0 0 0     0 0 0 0 0 0 1 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0     0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 0 2 0     111 122 0 0 1 0 0 0 0 0 0 0 0 0 0 0     0 1 0 1 0 0 0 0 1 1 0 0 0 0     0 0 0 0 0 0 1 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0     0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 2 0     1112 1223 0 0 0 0 0 0 0 0 0 0 1 0 0 0     0 0 0 1 0 0 0 0 1 2 0 0 0 0     0 0 0 0 0 0 1 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0     0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 2 0     11124 12233 0 0 0 0 0 0 0 0 0 0 1 0 0 0     0 0 0 0 0 0 0 0 1 2 0 0 1 0     0 0 0 0 0 0 1 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0     0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 2 0     111247 122336 0 0 0 0 0 0 0 0 0 0 1 0 0 0     0 0 0 0 0 0 0 0 1 2 0 0 1 0     0 0 0 0 0 0 0 0 0 0 0 0 0 0

Figure 3.1 Generating the ﬁrst rows of P and Q

3.2 The RSK Correspondence 63

This matrix is in turn used to obtain the third rows of P and Q as

p = 36 and q = 55.

This time the resulting shadow matrix is 0, and the process stops. Thus, we have obtained two semistandard Young tableaux from the matrix

0 0 1 0 0 0 0   0 0 0 0 0 2 0   1 1 1 1 0 1 0 A =   , 0 0 1 0 0 0 0   2 1 0 1 0 0 0   0 0 0 0 0 0 1 namely

1 1 1 2 4 7 1 2 2 3 3 6 P = and Q = 2 3 3 6 3 3 3 5 3 6 5 5

The algorithm that was explained with an example above is summarized below.

Algorithm for generating a row (AROW) Initialization. Initialize S = 0, p = ∅, q = ∅.

While A , 0 keep repeating the following step:

Main Step. Compute the shadow path of A. Reduce by 1 the maximal entries of A. Add 1 to the entries of S corresponding to the shadow points of A. Append the terminal column number of the shadow path to p and the terminal row number to the shadow path of q.

When A = 0, output S , p and q.

Viennot-RSK algorithm (VRSK)

Initialization. A = input matrix, S = 0, P = ∅, Q = ∅ While A , 0 keep repeating the following step:

Main step. Apply the algorithm AROW to A and obtain outputs S, p, q. Replace A by S , append p as a new row to P and q as a new row to Q.

When A = 0, output P and Q.

Theorem 3.2.2 (RSK correspondence). The VRSK algorithm gives a bijection A → (P, Q) from the set of all λ × µ matrices (Deﬁnition 2.6.1) to pairs of SSYT P and Q of the same shape, with P of type µ and Q of type λ.

64 The RSK Correspondence

Before starting the proof, it is necessary to introduce some notation: First, given two integer matrices A and B, say that A ≥ B if each entry of A is at least as large as the corresponding entry of B (in other words, the matrix A − B has non- negative entries). Starting with the matrix A, an application of the main step of AROW modiﬁes it. Denote the modiﬁed matrix by A◦. Thus, AROW generates a sequence of matrices:

A ≥ A◦ ≥ A◦◦ ≥ · · · ≥ A(i−1) ≥ A(i) ≥ · · ·

ending in the zero matrix. Also, given the matrix A, let S (A) denote the matrix S output by the algorithm AROW (S (A) is the shadow matrix of A).

P and Q have weakly increasing rows

For this, it suffices to show that the sequences p = p1 p2 ... and q = q1q2 ... generated by AROW are always weakly increasing. Now pi is the number of the (i) (i) first non-zero column of A , and qi is the number of the first non-zero row of A . (i) (i+1) Since A ≥ A , it follows that pi ≤ pi+1 and qi ≤ qi+1.

P and Q have strictly increasing columns For each integer matrix A, let L(A) denote the matrix whose (i, j)th entry is 1 if (i, j) is a maximal entry, and zero otherwise. Thus, A◦ = A − L(A).

Lemma 3.2.3. For every integer matrix A, L(S (A)) ≥ S (L(A)).

Proof. Suppose that (i, j) is a non-zero entry of S (L(A)). Then, from the deﬁni- tion of shadow points, it follows that there exist maximal entries (i, j0) and (i0, j) of A such that i0 < i and j0 < j, and moreover, A has no maximal entries in the rectangle [i0, i] × [ j0, j] (the grey rectangle in Figure 3.2). In order to prove the lemma, we must show that (i, j) is a maximal entry of S (A). If (i, j) is not a maximal entry of S (A), then there exists a non-zero entry (k, l) of S (A) in whose shadow (i, j) lies. Since (k, l) is a shadow point of A, there exist non-zero entries (k0, l) and (k, l0) of A with k0 < k and l0 < l. Now these two must all lie in the upper-left quadrant from (i, j) (since (i, j) lies in the shadow of (k, l) and, therefore, also in the shadow of these two points). But they cannot lie in the rectangle [i0, i] × [ j0, j]. This means that each of them contains either (i0, j) or (i, j0) in its shadow, contradicting the maximality of (i0, j) and (i, j0).

[3] Exercise 3.2.4. Can it happen that L(S (A)) > S (L(A))? If yes, construct an example where this happens.

3.2 The RSK Correspondence 65    ···             (i0, j0) (i0, j)   ···                   i j0 i, j   ( , ) ( )     .   .   .               ···       

Figure 3.2 The shadow point (i, j) comes from maximal entries (i, j0) and (i0, j)

Lemma 3.2.5. For any integer matrix A, S (A◦) ≥ S (A)◦.

Proof. We have

S (A)◦ = S (A) − L(S (A)) ≤ S (A) − S (L(A)) [by Lemma 3.2.3] = S (A◦),

thereby proving the lemma.

Corollary 3.2.6. For each integer matrix A and integer i > 0,S (A(i)) ≥ S (A)(i).

Proof. This is proved by induction on i. The base case i = 1 is exactly Lemma 3.2.5. Inductively, we have

S (A(i)) = S (A(i−1)◦) ≥ S (A(i−1))◦ [by Lemma 3.2.5] ≥ S (A)(i−1)◦ [by induction hypothesis] = S (A)(i),

and the corollary is proved.

66 The RSK Correspondence

The ith entry of the first row of P is the first non-zero column of A(i), while the ith entry of the second row of P is the first non-zero column of S (A)(i). Since S (A)(i) ≤ S (A(i)), the first non-zero column of S (A)(i) cannot come before the first non-zero column of S (A(i)), but that comes strictly after the first non-zero column of A(i) (the way in which shadow points are defined means that each non-zero entry of S (A) is strictly to the right of some non-zero entry of A). Since the ith row of the tableau P associated to A is the first row of the tableau P associated to S i(A), the argument above, applied to S i(A) for i > 0, proves that the columns of P are strictly increasing. A similar argument works for Q as well. This proves that P and Q are semistandard Young tableaux.

Conserved Quantities We now check that the type of P is µ and the type of Q is λ. Given a quadruple (A, S, p, q), where A and S are integer matrices and p and q are rows of positive integers, deﬁne

Ri(A, S, p, q) = ri(A) + ri(S ) + ni(q)

C j(A, S, p, q) = c j(A) + c j(S ) + n j(p),

where the function ri takes a matrix to the sum of the ith row, the function c j takes a matrix to the sum of its jth column and ni takes an SSYT to the number of occurrences of i in it. If the main step of AROW takes (A, S, p, q) to (A◦, S ◦, p◦, q◦), then ◦ ◦ ◦ ◦ Ri(A , S , p , q ) = Ri(A, S, p, q) ◦ ◦ ◦ ◦ (3.10) C j(A , S , p , q ) = C j(A, S, p, q) for all i, j. Indeed, the row sum goes down by 1 for the terminal row, but the row number of the terminal row gets added to q. For the other rows with maximal ◦ ◦ entries, ri(A) + ri(S ) = ri(A ) + ri(S ) because, in the notation of Section 3.2.1, while the (is, js)th entry of A is reduced by 1, the (is, js+1)th entry of S is increased by 1. It is now easy to see why P is of type µ and Q of type λ. When we begin the VRSK algorithm, we start with (A, S, P, Q) = (A, 0, ∅, ∅), with A the input matrix. So Ri is the sum of the ith row of A. When VRSK ends, we have (0, 0, P, Q), so Ri is the number of i’s in Q. Therefore, (3.10) implies that Q is of type λ. Similarly, P must be of type µ. At the end of Section 3.2.1, it was pointed out that the shadow points of the shadow path, together with the numbers of terminal row and column, can be used to recover the shadow path. This enables us to reverse the main step of AROW. In this manner, it is possible to reverse the algorithm AROW itself: the original matrix A can be recovered from the outputs S , p and q of AROW. Since the

3.2 The RSK Correspondence 67

outputs P and Q of VRSK are obtained by repeated application of AROW, it is also possible to recover A uniquely from P and Q. The reader is invited to do this in an example: [1] Exercise 3.2.7. Find the 5 × 4 matrix A to which VRSK would associate the SSYT’s 1 1 2 3 1 1 3 4 P = , Q = 2 4 2 5

[4] Exercise 3.2.8. Find the matrix A for which the types of P and Q are the same as their (common) shape. Since the VRSK algorithm is reversible, we may start with any pair (P, Q) of the SSYT of the same shape and recover uniquely the matrix A to which VRSK would associate P and Q. This shows that the correspondence

VRSK A / (P, Q) (3.11) is injective as well as surjective (onto the set of pairs of all SSYT of the same shape). This completes the proof of Theorem 3.2.2 Write RSK(A) = (P, Q) to denote the situation of (3.11). [1] Exercise 3.2.9. Show that if RSK(A) = (P, Q), then RSK(A0) = (Q, P) (as usual, A0 denotes the transpose of A). Deﬁnition 3.2.10 (Standard Young tableau). Let λ be a partition of n. A standard Young tableau (SYT) of shape λ is an SSYT of shape λ and type (1n). Each of the integers 1, 2,..., n occurs exactly once in an SYT. Therefore, the rows of an SYT are also forced to strictly increasing. Denote by fλ the number of SYT of shape λ. We have:

fλ = Kλ,(1n). A notable special case of the RSK correspondence, when λ = µ = (1n), is the Robinson-Schensted correspondence: Theorem 3.2.11. The VRSK algorithm gives a bijection from the set of permutation matrices to the set of pairs of SYT of the same shape.

Proof. Indeed, every M(1n)×(1n) matrix is a permutation matrix.

This gives the identity X 2 n! = fλ , (3.12) λ where λ runs over the set of partitions of n.

68 The RSK Correspondence

[2] Exercise 3.2.12. Show that the number of involutions in S n (an involution is an element whose square is the identity; see Exercise 2.2.13) is equal to the number of SYT of size n (of any shape).

3.3 Classiﬁcation of Simple Representations of S n

Consider the family K[Xλ] of representations of S n deﬁned in Section 2.3 as λ runs over the set of partitions of n. We found that given two partitions λ and µ of

n, the dimension of HomS n (K[Xλ], K[Xµ]) is the number Mλµ of λ × µ matrices (Section 2.6.2). In the ﬁrst two sections of this chapter, we have shown that the set of λ × µ matrices is in bijective correspondence with pairs (P, Q) of SSYT of the same shape ν, where P is of type µ and Q is of type λ. By Lemma 3.1.12, we must have that ν ≤ µ and ν ≤ λ in the reverse dominance order. Thus, we have proved that X Mλµ = KνλKνµ, ν≤λ, ν≤µ

with Kλλ = 1 for each λ. These are precisely the hypotheses required in the combinatorial resolution theorem (Theorem 2.6.6). Therefore, we have proved

Theorem 3.3.1 (Young’s rule). Let K be a ﬁeld of characteristic > n. For every partition λ of n, there exists a unique simple representation Vλ of S n such that L M Kνλ K[Xλ] = Vν ν≤λ

The representations Vν, as ν runs over the set of partitions of n, form a complete system of representatives for the isomorphism classes of simple representations of S n.

Recall (Exercise 2.3.5) that K[X(1n)] is the regular representation of S n. Thus, as a special case of Young’s rule, we get the decomposition of the regular representation of S n: M ⊕ fλ K[S n] = Vλ λ Comparing this result with Theorem 1.7.5, we get

Theorem 3.3.2. The collection {Vλ} as λ runs over all the partitions of n is a complete collection of irreducible representations of S n over any ﬁeld K of characteristic > n. The dimension of Vλ is fλ.

3.3 Classiﬁcation of Simple Representations of S n 69

[2] Exercise 3.3.3. Show that every irreducible character of S n (when K is of characteristic greater than n) takes integer values. Hint: Use Young’s rule and induction.

[1] Exercise 3.3.4. The only partitions λ of n for which fλ = 1 are λ = (n) and n λ = (1 ). Therefore, S n has exactly two one-dimensional representations, V(n) and V(1n).

[2] Exercise 3.3.5. Show that the simple representation Vk of Theorem 2.5.3 is the representation V(n−k,k) for k ≤ n − k. Hint: Use Exercise 3.1.9.

We know that V(n) = K[X(n)] is the trivial representations of S n. The representation V(1n) comes from a non-trivial multiplicative character of S n. This character, known as the sign character, is constructed in Section 4.1.

4 Character Twists

4.1 Inversions and the Sign Character

1 Deﬁnition 4.1.1 (Inversion). An inversion for a permutation w ∈ S n is a pair of integers (i, j) such that 1 ≤ i < j ≤ n and w(i) > w( j).

[1] Exercise 4.1.2. If a permutation has no inversions, then it is the identity.

The following table lists the inversions for the permutations in S 3:

Permutation Inversions 123 None 132 (2, 3) 213 (1, 2) 231 (1, 3), (2, 3) 312 (1, 2), (1, 3) 321 (1, 2), (1, 3), (2, 3)

[1] Exercise 4.1.3. Generate a similar table for S 4.

Deﬁnition 4.1.4 (Inversion number). The total number of inversions for a permutation w is called its inversion number and denoted i(w).

We see that in S 3 there is one permutation with no inversions, two each with one and two inversions and one with three inversions.

Deﬁnition 4.1.5 (Transposition). For 1 ≤ i ≤ n−1, let si denote the permutation of n which interchanges i with i + 1 and maps each of the remaining elements of n to itself. The permutations s1,..., sn−1 are called the transpositions in S n.

1 For an enjoyable discussion of some basic combinatorial properties of inversions, see Knuth [16, Section 5.1]

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4.1 Inversions and the Sign Character 71

Example 4.1.6. The four transpositions in S 5 in terms of permutation strings are given by

s1 = 21345, s2 = 13245, s3 = 12435 and s4 = 12354.

[2] Exercise 4.1.7. The transpositions s1,..., sn−1 are the only elements of S n with exactly one inversion.

Visualizing Inversions

A permutation w ∈ S n can be represented by a bipartite graph with 2n vertices. Draw the vertices in two rows, each of length n. Join the ith vertex of the ﬁrst row to the w(i)th vertex in the second row. For example, the permutation 32154 is represented by the graph

1 2 3 4 5

Ð 1w 2 ' 3 4 5 The number of pairs of crossing arrows in this graph is the inversion number of w. Thus, 32154 has four inversions.

[2] Exercise 4.1.8. For any permutation w,

1. the permutation string for wsi is obtained by interchanging the ith and (i + 1)st terms in the permutation string for w; 2. if w(i) > w(i + 1), then i(wsi) = i(w) − 1; otherwise i(wsi) = i(w) + 1. Theorem 4.1.9. Every permutation w is a composition of i(w) transpositions. It cannot be written as a composition of fewer than i(w) transpositions.

Proof. The second assertion follows from part 2 of Exercise 4.1.8, which implies that a product of l transpositions cannot have more than l inversions. For the ﬁrst assertion, proceed by induction on the number of inversions. The base case (one inversion) is taken care of by Exercise 4.1.7. Now suppose w is a permutation which has l inversions with l > 1. Then, the permutation string for w cannot be strictly increasing. Therefore, there must exist i such that w(i) > w(i+1). By part 1 of Exercise 4.1.8, i(wsi) = l − 1. By induction hypothesis, wsi can be written as the product of l − 1 transpositions. Therefore, w = wsi si can be written as the product of l transpositions.

Deﬁnition 4.1.10 (Reduced word). An expression for w ∈ S n as a product of i(w) transpositions is called a reduced word for w.

72 Character Twists

Example 4.1.11. Exercise 4.1.8 gives rise to an algorithm for writing down a reduced word for a permutation. For example, let’s write down a reduced word for the permutation w = 32154

with four inversions. Exercise 4.1.8 tells us that ws1 = 23154 and has only three inversions. Next, ws1 s2 = 21354 has two inversions, ws1 s2 s1 = 12354 has one inversion and ﬁnally ws1 s2 s1 s4 = 12345 is the identity element. So

w = s4 s1 s2 s1 is a reduced word for w. We could have made a diﬀerent choice at each stage for the i such that w(i) > w(i + 1) to get a diﬀerent reduced word.

Example 4.1.12. S 3 has two transpositions s1 and s2. The following table gives all the reduced words for each element of S 3. Permutation Reduced word 123 1 132 s2 213 s1 231 s1 s2 312 s2 s1 321 s1 s2 s1, s2 s1 s2 Note the non-uniqueness of reduced words for 321.

[3] Exercise 4.1.13. Find all the reduced words for the permutation 4321.2

[3] Exercise 4.1.14. Find an algorithm (or write a computer program) to enumerate all the reduced words for a given permutation.

Consider the function : S n → {±1} given by (w) = (−1)i(w). (4.1)

A permutation w is said to be even if (w) = 1 and odd if (w) = −1.

Theorem 4.1.15. For w1, w2 ∈ S n,

(w1w2) = (w1)(w2).

In other words, is a multiplicative character of S n (see Deﬁnition 1.1.2).

The multiplicative character is called the sign character of S n.

2 For every positive integer n, the permutation n(n − 1) ··· 21 is known to have n n−1 n−2 1 2 !/1 3 ··· (2n − 3) reduced words; see Stanley [33].

4.2 Twisting by a Multiplicative Character 73

Proof. By part 2 of Exercise 4.1.8, if w1 = si1 si2 ··· sin (not necessarily a reduced

word), then n ≡ i(w1) mod 2. Similarly, if w2 = s j1 s j2 ··· s jm , then m ≡ i(w2) mod 2. Since

w1w2 = si1 si2 ··· sin s j1 s j2 ··· s jm ,

m + n ≡ i(w1w2) mod 2. Therefore, i(w1) + i(w2) ≡ i(w1w2) mod 2, from which the lemma follows. When the characteristic of K is greater than n, the irreducible representations of S n are parameterized by partitions of n (Theorem 3.3.2). By Exercise 3.3.4, the one-dimensional representation corresponding to is the representation V(1n). The set of even permutations, being the kernel of , is clearly a normal subgroup of S n. This group is called the alternating group and is denoted by An.

[3] Exercise 4.1.16. Show that [S n, S n] = An (see Exercise 1.1.4).

4.2 Twisting by a Multiplicative Character

Any representation can be twisted by a multiplicative character: Definition 4.2.1 (Twist of a representation by a character). Let (ρ, V) be a representation of a group G over a field K. Let χ : G → K∗ be a multiplicative character. Define ρ ⊗ χ : G → GL(V) by ρ ⊗ χ(g) = χ(g)ρ(g). Then ρ ⊗ χ is a representation of G called the twist of ρ by χ. If the symbol ρ is suppressed in the notation, then we write V ⊗ χ to signify that the action of G on V has been twisted by χ. [0] Exercise 4.2.2. In the notation of the above definition, if ρ is simple, then so is ρ ⊗ χ. In Section 2.4, we were able to understand intertwiners between permutation representations in terms of relative position. We shall now carry out a similar analysis of twists of permutation representations to obtain Theorem 4.2.3. Let X and Y be finite G-sets, and let χ be a character of G. We saw in Section 2.4 that all linear maps K[Y] → K[X] are of the form Tk for some k ∈ K[X × Y]. Now Tk ∈ HomG(K[Y], K[X] ⊗ χ) if and only if

−1 χ(g)ρX(g) Tk ◦ ρY (g) = Tk. In this case, the condition (2.3) is slightly modiﬁed: k(g · x, g · y) = χ(g)k(x, y) for all g ∈ G, (x, y) ∈ X × Y. (4.2)

74 Character Twists

Thus, it is still true that k is completely determined on a G-orbit in X × Y from its value at any one point in that orbit, whence

dim HomG(K[Y], K[X] ⊗ χ) ≤ |G\(X × Y)|. (4.3)

However, there is an obstruction to assigning a non-zero value to k(x, y): suppose we take g ∈ G such that g · x = x and g · y = y. Combining this property of g with (4.2) gives k(x, y) = k(g · x, g · y) = χ(g)k(x, y).

If χ(g) , 1, then k(x, y) is forced to be 0. We will soon see that this is the only problem: for an element x in a G-set X, let StabG(x) denote the stabilizer of x, namely the subgroup

StabG(x) = {g ∈ G|g · x = x}.

We have

Theorem 4.2.3. For G sets X and Y, let

X t Y := {(x, y)|StabG(x) ∩ StabG(y) ⊂ ker χ}.

Then X t Y is a G-stable subset of X × Y (meaning that it is a union of G-orbits), and

dim HomG(K[Y], K[X] ⊗ χ) = |G\(X t Y)|. (4.4)

Proof. For (x, y) ∈ X t Y, deﬁne k ∈ K[X × Y] by

k(g · x, g · y) = χ(g)k(x, y) (4.5)

on the G-orbit of (x, y) and to be 0 on the rest of X × Y. If g, g0 ∈ G are such 0 0 −1 0 that g · x = g · x and g · y = g · y, then g g ∈ StabG(x) ∩ StabG(y). Since (x, y) ∈ X t Y, χ(g−1g0) = 1. Therefore, χ(g) = χ(g0). Thus, the value assigned to k(g · x, g · y) depends only on the point (g · x, g · y) and not on the choice of g, making it well deﬁned. As (x, y) ranges over a set of representatives for the G-orbits in X t Y, the functions k that we have just constructed form a basis of the space of functions satisfying (4.2).

Let us now examine the case which is of greatest interest to us, namely where ∗ G = S n, X = Xλ, Y = Xµ and χ is the sign character : S n → K .

Theorem 4.2.4. Let λ and µ be partitions of n. For (S, T) ∈ Xλ × Xµ with S = ` ` ` ` S 1 ··· S l and T = T1 ··· Tm, (S, T) ∈ Xλ t Xµ if and only if |S i ∩T j| ≤ 1 for all 1 ≤ i ≤ l, 1 ≤ j ≤ m.

4.3 Conjugate of a Partition 75

Proof. Suppose there exist i, j such that S i ∩ T j has more than one element, say

α and β. Then the two cycle (αβ) lies in StabS n (S )∩StabS n (T). On the other hand, ((αβ)) = −1 (every two cycle is conjugate to s1 and conjugate elements in S n have the same sign, since An is a normal subgroup of S n).

Conversely, suppose |S i ∩ T j| ≤ 1 for each i, j. If g ∈ StabS n (S ) ∩ StabS n (T), then g must ﬁx each of the subsets S i ∩ T j. But every individual element of n is the sole element of some S i ∩ T j. Therefore, g must be the identity element of S n, which is certainly in ker .

Deﬁnition 4.2.5 (Transverse pairs). A pair (S, T) ∈ Xλ × Xµ is called a transverse pair if it lies in Xλ t Xµ.

Deﬁnition 4.2.6 (0-1 λ × µ matrix). A 0-1 λ × µ matrix is a λ × µ matrix (see Deﬁnition 2.6.1) all of whose entries are either 0 or 1.

Let Nλµ denote the number of 0-1 λ × µ matrices.

Corollary 4.2.7. For all partitions λ, µ of n,

dim HomS n (K[Xµ], K[Xλ] ⊗ ) = Nλµ.

Proof. By Theorem 4.2.3, we need to show that

|S n\(Xλ t Xµ)| = Nλµ.

In Theorem 2.6.2, we saw that the S n-orbits in Xλ × Xµ are parametrized by λ × µ matrices. By Theorem 4.2.4, the orbits of transverse pairs correspond to the 0-1 matrices among these.

4.3 Conjugate of a Partition

Deﬁnition 4.3.1 (Conjugate of a partition). If λ = (λ1, . . . , λl) is a partition, its 0 0 0 0 conjugate is deﬁned as the partition λ = (λ1, . . . , λs), where λ j is the number of parts of λ which are greater than or equal to j:

0 λ j = |{1 ≤ i ≤ l : λi ≥ j}|.

0 [0] Exercise 4.3.2. Show that λ1 is the number of non-zero parts in λ.

The conjugate of a partition is best understood (and most easily computed) if visualized in terms of Young diagrams (Deﬁnition 3.1.1). The Young diagram of

76 Character Twists

shape λ0 is obtained from the Young diagram of shape λ by reﬂecting it about the principal diagonal. For example, the Young diagram of shape (2, 2, 1) is

Flipping it about its principal diagonal gives

which is the Young diagram of shape (3, 2), the partition conjugate to (2, 2, 1).

[1] Exercise 4.3.3. Let m and n be arbitrary positive integers. Show that the number of partitions on n with parts bounded by m is equal to the number of partitions of n with at most m parts.

[1] Exercise 4.3.4. For each partition λ, show that fλ = fλ0 (recall that fλ is the number of standard Young tableaux of shape λ; see Deﬁnition 3.2.10).

[4] Exercise 4.3.5. Show that λ ≤ µ if and only if µ0 ≤ λ0.

Lemma 4.3.6. For all partitions λ and µ of n, Nλµ > 0 if and only if Kµ0λ > 0.

Proof. Suppose that Nλµ > 0. Then there exist ordered partitions S = (S 1,..., S l) and T = (T1,..., Tm) of n of shape λ and µ, respectively, such that S i ∩ T j has at most one element for each (i, j). Consider a Young tableau YS whose ith row is 0 ﬁlled by the elements of S i and a Young Tableau YT whose jth column is ﬁlled by the elements of T j. For example, if a a S = {1, 2, 3} {4, 6, 7} {5, 8} a a a T = {2, 5, 7} {4, 1} {3, 8} {6}

then

1 2 3 0 2 4 3 6 YS = and YT = . 4 6 7 5 1 8 5 8 7

0 0 The tableau YS is of shape λ, while the tableau YT is of shape µ . Since each element of S 1 lies in a different part of T, each entry of the first row of YS comes 0 0 from a different column of YT . By permuting the entries of each column of YT

4.3 Conjugate of a Partition 77

(which has no eﬀect on T), we can move all the elements of S 1 into the ﬁrst row 0 of YT . In the running example, we get

2 1 3 6 5 4 8 7

Since each element of S 2 lies in a different part of T, each entry of the second row 0 of YS lies in a different column of YT . By permuting the entries of each column 0 of YT , while leaving the entries which come from S 1 undisturbed, we may bring each element of S 2 either to the top row or directly below an element of S 1. 0 Continuing in this manner, we may permute each column of YT in such a way that each element of S i lies either in the first row or directly below an element of ` ` S 1 ··· S i−1. In the example, we get

2 1 3 6 7 4 8 5

` ` 0 Thus, all the elements of S 1 ··· S i are in the ﬁrst i rows of YT for each i, 0 0 0 0 whence λ1 + ··· + λi ≤ µ1 + ··· + µi for each i. In other words, µ ≤ λ, so Kµ λ > 0 (Lemma 3.1.12, part 2). For the converse, suppose there exists an SSYT of shape µ0 and type λ. We shall use this SSYT to algorithmically construct a transverse pair of ordered partitions (S, T) in Xλ × Xµ. Give distinct labels from n to each of the boxes in the SSYT. For example, given the SSYT

1 1 1 2 2 2 3 3

we use the most prosaic labelling (labels are indicated as superscripts)

11 12 13 24 25 26 37 38

Let S i consist of the labels corresponding to all entries that are equal to i. Since the SSYT has type λ, |S i| = λi, so S = (S 1, S 2,... ) ∈ Xλ. Let T j consist of the

78 Character Twists

labels from the jth column of the labelled tableau. Since the SSYT has shape µ0, |T j| = µ j, so T = (T1, T2,... ) ∈ Xµ. In the example, we get a a S = {1, 2, 3} {4, 5, 6} {7, 8} a a a T = {1, 5, 8} {2, 6} {3, 7} {4}. Since each occurrence of a number i in the SSYT is the only one in its column, just one element of S i can occur in any T j, whence S i ∩T j has at most one element for all (i, j), so S is transverse to T. In conjunction with Lemma 3.1.12, Lemma 4.3.6 gives rise to the Gale–Ryser theorem

Theorem 4.3.7 (Gale–Ryser). For any partitions λ and µ of n, Nλµ > 0 if and only if µ0 ≤ λ.

Theorem 4.3.8. For any two partitions µ and λ of n, Kµ0λ > 0 if and only if Kλ0µ > 0.

Proof. By Lemma 4.3.6, Kµ0λ > 0 if and only if there exists a transverse pair (S, T) in Xλ × Xµ. Since transversality is a symmetric relationship, (T, S ) is a transverse pair in Xµ × Xλ, which by Lemma 4.3.6 is equivalent to Kλ0µ > 0. In view of Theorem 3.1.12, Theorem 4.3.8 gives an elegant solution to Exercise 4.3.5.

Lemma 4.3.9. For every partition λ of n, Nλλ0 = 1. An example is far more illuminating than the proof. If λ = (6, 5, 3, 3), then λ0 = (4, 4, 4, 2, 2, 1). The unique 0-1 λ × λ0 matrix is 1 1 1 1 1 1   1 1 1 1 1 0   1 1 1 0 0 0   1 1 1 0 0 0 0 Proof. We prove this by induction on n. If λ = (λ1, . . . , λl) and A is a 0-1 λ × λ 0 matrix, then A is an l × λ1 matrix (since λ has λ1 parts). Moreover, the first row 0 sums to λ1 and the first column to l = λ1. Therefore, these rows are forced to be all 1. Now consider the (l − 1) × (λ1 − 1) submatrix obtained by leaving out the first row and first column. In this submatrix, the entries in the last few rows (the 0 rows i for which λi = 1) and last few columns (the columns j for which λ j = 1) are all forced to be 0 since the first row/column (which we have removed) has already contributed 1 to the sum. Leave out these rows and columns as well. Then 0 the matrix that remains is a 0-1 λ˜ × λ˜ matrix, where λ˜ i = λi − 1 for each i such that λi > 1. But λ˜ is a partition of some integer strictly less than n, so the lemma follows by induction.

4.4 Twisting by the Sign Character 79 4.4 Twisting by the Sign Character

We saw in Theorem 3.3.2 that when K is an algebraically closed ﬁeld of characteristic greater than n, then simple representations of S n are indexed by partitions of n. We denoted the representation corresponding to the partition λ by Vλ. On the other hand, by Corollary 4.2.7 and Lemma 4.3.9,

0 dim HomS n (K[Xλ], K[Xλ ] ⊗ ) = 1,

so there exists a unique simple representation Uλ which occurs in both K[Xλ] and K[Xλ0 ] ⊗ , and it occurs in both these representations with multiplicity one (Theorem 1.3.5).

Theorem 4.4.1. The representations Uλ and Vλ are isomorphic. ˜ Proof. Let λ be the partition for which Uλ Vλ˜ . Since Uλ occurs in K[Xλ], we know from Theorem 3.3.2 that λ˜ ≤ λ. By the same theorem, we also know that

Uλ Vλ˜ does occur in K[Xλ˜ ]. By deﬁnition it occurs in K[Xλ0 ] ⊗ . Therefore,

0 dim HomS n (K[Xλ˜ ], K[Xλ ] ⊗ ) > 0

which implies (by Corollary 4.2.7) that Nλλ˜ 0 > 0. By Lemma 4.3.6, we get ˜ ˜ Kλλ˜ > 0, which implies that λ ≤ λ. Thus λ = λ.

Theorem 4.4.2. For every partition λ of n, the simple representation Vλ ⊗ of S n is isomorphic to Vλ0 .

Proof. Since Vλ Uλ, both Vλ ⊗ and Vλ0 would be the unique simple representation occurring in K[Xλ0 ] and K[Xλ] ⊗ . A direct consequence of Theorem 4.4.2 is that

M ⊕Kν0µ K[Xµ] ⊗ = Vν . ν0≤µ Therefore, Theorem 1.3.5 gives X 0 dim HomS n (K[Xλ], K[Xµ] ⊗ ) = KνλKν µ ν≤λ, ν0≤µ which, by Corollary 4.2.7 and Exercise 4.3.5, is X Nλµ = KνλKν0µ, (4.6) µ0≤ν≤λ which is precisely the identity that will be obtained from the dual RSK correspondence in Section 4.5. [1] Exercise 4.4.3. Deduce Lemmas 4.3.6 and 4.3.9 directly from (4.6). This is circular logic, but the dual RSK correspondence of the next section will give an independent, purely combinatorial proof of (4.6).

80 Character Twists

[1] Exercise 4.4.4. Use (4.6) to strengthen the Gale–Ryser theorem (Theorem 0 4.3.7): show that Nλµ ≥ 2 for µ < λ.

4.5 The Dual RSK Correspondence

The dual RSK correspondence, a variant of the RSK correspondence of Section 3.2, is the second generalization of the RSK correspondence due given by Knuth in [15]. While the RSK correspondence applies to matrices whose entries are non-negative integers, the dual RSK correspondence applies to matrices whose entries are all either 0 or 1. The dual RSK correspondence associates to each 0-1 λ × µ matrix, a pair (P, Q) of SSYT such that P and Q have mutually conjugate shapes, Q has type λ and P has type µ. Thus, the dual RSK correspondence gives a bijective proof of the identity (4.6). We now proceed directly to the definition of the dual RSK correspondence, which is quite similar to that of the RSK correspondence, except that it involves modifying the partial order (3.5) used to define the RSK correspondence. We now set (i, j) > (i0, j0) if i0 ≥ i and j0 > j. Define the dual shadow of an entry (i, j) to be the set of entries (i0, j0) such that (i0, j0) ≤ (i, j). The dual shadow of (i, j) consists of all the points (i0, j0) in the usual shadow of (i, j) except for the points with i0 > i and j0 = j, the entries in the same column as (i, j) which lie strictly below it. Consider, for example, the matrix 0 0 1 0 0   1 1 0 1 0   A = 1 0 0 0 0   0 0 0 0 1   0 0 1 0 0 The dual shadow of the (2, 1)th entry is represented in boldface below: 0 0 1 0 0   1 1 0 1 0   A = 1 0 0 0 0 (4.7)   0 0 0 0 1   0 0 1 0 0 Thus, there are three non-zero entries in A which do not lie in the dual shadow of any other non-zero entry: (1, 3), (2, 1) and (3, 1). We refer to these entries as the dual maximal entries.

4.5 The Dual RSK Correspondence 81

Because of the way in which the dual shadows have been deﬁned, it is quite possible that there is more than one maximal entry in any given column. There cannot, however, be two maximal entries in the same row. Suppose that (i1, j1),..., (ir, jr) are the dual maximal entries of a 0-1 matrix. We may rearrange them in such a way that

j1 ≤ j2 ≤ · · · ≤ jr and

i1 > i2 > ··· > ir. As before, the dual shadow path is deﬁned to be the zigzag path obtained by joining the vertices

(i1, j1), (i1, j2), (i2, j2), (i2, j3), (i3, j3),..., (ir−1, jr), (ir, jr), (but now, some of these vertices may lie in a vertical line) and its dual shadow points are deﬁned to be the entries

(i1, j2), (i2, j3),..., (ir−1, jr). For the matrix A in (4.7), the dual shadow path (with dual shadow points under- lined) is (3, 1), (3, 1), (2, 1), (2, 3), (1, 3). Modify the algorithm AROW of Section 3.2 to DAROW by replacing the shadow path to dual shadow path: Algorithm for generating a row (DAROW) Initialization. Initialize S = 0, p = ∅, q = ∅.

While A , 0 keep repeating the following step:

Main Step. Compute the dual shadow path of A. Reduce by 1 the maximal entries of A. Add 1 to the entries of S corresponding to the dual shadow points of A. Append the terminal column number of the shadow path to p and the terminal row number to the shadow path of q.

When A = 0, output S , p and q.

And modify the algorithm VRSK to VDRSK by using DAROW instead of AROW: Viennot-dual RSK algorithm (VDRSK) Initialization. A = input matrix, S = 0, P0 = ∅, Q = ∅ While A , 0 keep repeating the following step:

Main step. Apply the algorithm DAROW to A and obtain outputs S, p, q. Replace A by S , append p as a new row to P0 and q as a new row to Q.

82 Character Twists

When A = 0, output P (the reﬂection of P0 about its diagonal) and Q.

[1] Exercise 4.5.1. Verify that when VDRSK is applied to the matrix A of (4.7), the resulting Young tableaux are

1 1 2 1 2 2 3 P = Q = 2 4 2 5 3 3 5

Theorem 4.5.2 (Dual RSK correspondence). The VDRSK algorithm gives a bijection A → (P, Q) from the set of all 0-1 λ × µ matrices to pairs (P, Q) of SSYT of mutually conjugate shapes, with P of type µ and Q of type λ.

Proof. The proof is quite analogous to that of Theorem 3.2.2 of (the RSK correspondence). One shows that for the tableaux P0 and Q, Q is semistandard, while P0 has strictly increasing rows and weakly increasing columns. Starting with a 0-1 matrix A, let (exactly as in the proof of Theorem 3.2.2) A◦ denote the matrix obtained by applying DAROW to it. Iterating this process, the repeated application of DAROW results in a sequence of matrices

A ≥ A◦ ≥ A◦◦ ≥ · · · ≥ A(i−1) ≥ A(i) ≥ · · ·

ending in the zero matrix. Also, let S (A) denote the matrix S output by the algorithm DAROW when the input is A [we now call S (A) the dual shadow matrix].

P0 has strictly increasing rows and Q has weakly increasing rows

For this it suffices to show that the sequence p1 p2 ··· generated by DAROW is strictly increasing, while the sequence q1, q2 ··· is weakly increasing. The latter assertion follows exactly as in the corresponding part of the proof of (i) Theorem 3.2.2. Now pi is the first non-zero column of A , and all the non-zero entries of this column are maximal. These are therefore reduced to 0 by the main step of DAROW. It follows that the first non-zero column of A(i+1) is strictly to the right of pi, i.e., pi+1 > pi.

P0 has weakly increasing rows and Q has strictly increasing columns Lemmas 3.2.3 and 3.2.5, and Corollary 3.2.6, go through verbatim for DAROW as they did for AROW. However, it does not follow that the columns of P are strictly increasing. This is because, in the case where A has several maximal entries in one column, it is possible that the ﬁrst non-zero row of S (A) lies in the same row as

4.6 Representations of Alternating Groups 83

the ﬁrst non-zero row of A [this happens when S (A) is calculated for A in (4.7)]. However, one can still conclude that the columns of P are weakly increasing. The proof that the columns of Q are strictly increasing goes through exactly as in the proof of Theorem 3.2.2.

Types of P0 and Q The arguments concerning the types of P0 and Q are exactly the same as in the proof of Theorem 3.2.2.

Reversibility The reversibility of the VDRSK algorithm is also similar to that of the VRSK algorithm. It is only necessary to specify what notion of reverse shadow allows us to recover the dual path from its dual shadow points and its extremal rays. We say that (i0, j0) lies in the reverse dual shadow of (i, j) if

j0 < j and i0 ≤ i.

This completes the proof of Theorem 4.5.2. [2] Exercise 4.5.3. Find the matrix A on which VDRSK outputs

1 1 2 3 1 1 P = Q = 2 4 2 4 3 5

[2] Exercise 4.5.4. Show that the RSK correspondence and its dual coincide on permutation matrices. This is to be interpreted in a sensible way: if A is a permutation matrix such that (P1, Q1) is the output when VRSK is applied to 0 A and (P2, Q2) is the output when VDRSK is applied to A, then P2 = P1 and Q2 = Q1.

4.6 Representations of Alternating Groups

In this section, we discuss representations of alternating groups An. The group An is the subgroups of S n consisting of even permutations (see Section 4.1). The method involves studying the restrictions of representations of S n to An. An important tool is the twisting of representations of An by conjugation by odd permutations in S n. Throughout this section, we will assume that n > 2.

84 Character Twists 4.6.1 Outer Twists Given a representation (ρ, V) of a group G and a group automorphism α : G → G, one may deﬁne a new representation αρ of G on V by

αρ(g) = ρ(α(g)).

[0] Exercise 4.6.1. Let (ρ, V) be a simple representation of G and α : G → G be any automorphism. Show that (αρ, V) is also a simple representation of G.

Lemma 4.6.2. If α is an inner automorphism, meaning that α(g) = x−1gx for some x ∈ G, then (αρ, V) and (ρ, V) are isomorphic representations of G.

Proof. We have

αρ(g) = ρ(x−1gx) = ρ(x)−1ρ(g)ρ(x),

which may be written as

ρ(x)αρ(g) = ρ(g)ρ(x),

which says that ρ(x) ∈ GL(V) intertwines (αρ, V) with (ρ, V) (Deﬁnition 1.2.11).

It follows that twists by automorphisms of G which diﬀer by an inner automorphism are isomorphic:

Lemma 4.6.3. If α and β are automorphisms of G such that αβ−1 is an inner automorphism, then (αρ, V) is isomorphic to (βρ, V).

− Proof. This follows from Lemma 4.6.2 since αρ = αβ 1 (βρ), so αρ is obtained by twisting βρ by an inner automorphism.

We will now apply twists to representations of An. Let x be any odd permutation −1 (an element of S n that is not in An). The map α : w 7→ xwx is an automorphism of An which is not inner (for n > 2). Then for any representation (ρ, V) of An, α twisting by α gives rise to a representation ( ρ, V) of An. Since for any two odd permutations x and y, xy−1 is even, the isomorphism class of (αρ, V) does not depend on the choice of the odd permutation x. Moreover, since the square of any permutation is even, α(αρ) is always isomorphic to ρ. Thus, ρ 7→ αρ is an involution on the set of isomorphism classes of simple representations of An. In terms of characters, we have

trace(αρ(w); V) = trace(ρ(xwx−1); V). (4.8)

4.6 Representations of Alternating Groups 85

4.6.2 Restriction to An

Let x ∈ S n be an odd permutation. Let α : An → An be the automorphism α(w) = x−1wx. Lemma 4.6.4. Let K be an algebraically closed ﬁeld of characteristic greater than n. Let (σ, W) be a simple representation of S n. If a simple representation (ρ, V) of S n occurs in the restriction of σ to An, then the representation of An on the image ρ(x)(V) of V under ρ(x) is isomorphic to (αρ, V). Proof. This is proved by observing that if ρ is realized on a subspace V of W, then ρ(x)(V) is also an An-invariant subspace of W. Indeed, if v ∈ V, then for any w ∈ An, ρ(w)ρ(x)v = ρ(x)ρ(x−1wx)(v). (4.9)

−1 −1 Since x wx ∈ An, ρ(x wx)v ∈ V. Therefore, ρ(w)ρ(x)v ∈ ρ(x)(V), showing that ρ(x)(V) is an invariant subspace for An. The identity (4.9) shows that ρ(x) intertwines (αρ, V) with (ρ, ρ(x)(V)). Theorem 4.6.5. For each partition λ, the restriction of the irreducible representation Vλ of S n to An is

+ − 1. a sum of two non-isomorphic irreducible representations Vλ and Vλ of An if 0 − + λ = λ . The representation Vλ is the twist of Vλ by conjugation by an odd permutation in S n; 0 2. an irreducible representation of An if λ , λ .

0 Moreover, the restrictions of Vλ and Vµ to An are disjoint unless µ = λ or µ = λ . Deﬁnition 4.6.6 (Self-conjugate partition). A partition λ for which λ = λ0 is called a self-conjugate partition. Proof of Theorem 4.6.5. By Theorem 4.4.2, for each odd partition3 µ,

χλ(wµ) = −χλ0 (wµ). (4.10)

Suppose that λ is self-conjugate. Then (4.10) implies that χλ(wµ) = 0 for each odd partition µ. Thus,

X 2 X 2 |χλ(w)| = |χλ(w)| = n!,

w∈An w∈S n

since χλ is irreducible. Therefore,

1 X 2 |χλ(w)| = 2, |An| w∈An

3 A partition µ is said to be an odd partition if any element wµ of cycle type µ is odd.

86 Character Twists

from which it follows that the restriction of Vλ to An is the sum of two non- + − isomorphic simple representations Vλ and Vλ of An. Lemma 4.6.4 implies that − + Vλ is the twist of Vλ by α. If λ is not self-conjugate, then the characters χλ and χλ0 of S n cannot be supported on An (for if they were, they would be equal, because is trivial on An, and we know, from Theorem 3.3.2, that Vλ and Vλ0 are non-isomorphic). Therefore, we have

1 X 2 1 X 2 |χλ(w)| < |χλ(w)| = 1. |S n| |S n| w∈An w∈S n Consequently,

1 X 2 |χλ(w)| < 2. |An| w∈An Since this sum is the sum of squares of multiplicities of the irreducible representations of An into which Vλ decomposes, it has to be a positive integer and hence must be equal to one. Thus, the restriction of Vλ to An is simple, proving (2). In general, if λ , µ, then

X −1 0 = χλ(w)χµ(w )

w∈S n X −1 X −1 = χλ(w)χµ(w ) + χλ(w)χµ(w ). (4.11)

w∈An w

X −1 X −1 0 = χλ0 (w)χµ(w ) + χλ0 (w)χµ(w ). (4.12)

w∈An w

Since χλ0 (w) = (w)χλ(w), (4.12) becomes

X −1 X −1 0 = χλ(w)χµ(w ) − χλ(w)χµ(w ). (4.13)

w∈An w

X −1 χλ(w)χµ(w ) = 0,

w∈An proving the last assertion of the theorem.

4.6.3 Classiﬁcation of Simple Representations

Since every simple representation of An is a subrepresentation of its regular representation, which in turn is a subrepresentation of the regular representation of S n, it is a subrepresentation of some representation of S n and hence of some simple

4.6 Representations of Alternating Groups 87

representation of S n. It follows that Theorem 4.6.5 gives a description of all the simple representations of An in terms of the simple representations of S n. Theorem 4.6.7. Let K be an algebraically closed ﬁeld of characteristic greater + − than n. For each self-conjugate partition λ of n, let Vλ and Vλ denote the two irreducible representations of An which occur in the restriction of the irreducible ± 0 representation Vλ of S n to An. Then, the representations Vλ for λ = λ and the restrictions of the simple representations Vλ of S n to An (which we will also denote 0 by Vλ) for those λ for which λ precede λ in the reverse lexicographic order form a complete set of representatives for the set of isomorphism classes of simple representations of An. Example 4.6.8. The partitions of 3 are (3), (2, 1), (1, 1, 1). We have (3)0 = (1, 1, 1), (2, 1)0 = (2, 1). Therefore, by Theorem 4.6.7, the simple representations of A3 are + − V(3), V(2,1) and V(2,1). The partitions of 4 are (4), (3, 1), (2, 2), (2, 1, 1), (1, 1, 1, 1). We have (4)0 = (1, 1, 1, 1), (3, 1)0 = (2, 1, 1) and (2, 2)0 = (2, 2). Therefore, the simple representations of A4 are + − V(4), V(3,1), V(2,2) and V(2,2). The partitions of 5 are (5), (4, 1), (3, 2), (3, 1, 1), (2, 2, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1). We have (5)0 = (1, 1, 1, 1, 1), (4, 1)0 = (2, 1, 1, 1), (3, 2)0 = (2, 2, 1), (3, 1, 1)0 = (3, 1, 1). Therefore, the simple representations of A5 are + − V(5), V(4,1), V(3,2), V(3,1,1) and V(3,1,1).

4.6.4 Conjugacy Classes in An

If two elements w1 and w2 are two elements of An which are conjugate in An, then they are also conjugate in S n. However, the converse is not true.

Example 4.6.9. The permutations (written as permutation strings) w1 = 231 and w2 = 312 are both 3-cycles, namely (123) and (321), respectively, and −1 hence are conjugate in S 3. The elements x such that xw1 x = w2 are 321 and 213, which are both odd, so 231 and 312 are not conjugate in An. The conjugacy

88 Character Twists

class in S 3 consisting of 231 and 312 splits into two singleton conjugacy classes in A3.

[1] Exercise 4.6.10. 1. Given an even permutation w ∈ An, show that the following are equivalent:

1. No odd permutation in S n commutes with w.

2. The conjugacy class of w in S n splits into a union of two conjugacy classes −1 −1 in An, namely the class {xwx | x ∈ S n is odd} and {xwx | x ∈ S n is even}, which are of equal cardinality.

2. Given an even permutation w ∈ An, show that the following are equivalent:

1. There exists an odd permutation in S n which commutes with w.

2. The conjugacy class of w in S n is also its conjugacy class in An.

Since An is a normal subgroup of S n, conjugate elements of S n have the same sign. Thus, permutations with the same cycle type have the same sign. A partition λ of n is said to be even if any permutation w ∈ S n with cycle type λ is even.

Example 4.6.11. The even partitions of 3 are (1, 1, 1) and (3). The only element of cycle type (1, 1, 1) is the identity, which commutes with all the odd permutations in S 3. An element of cycle type (3) commutes only with the identity, itself and its square, all of which are even permutations. Therefore, the class in S n consisting of 3-cycles (which consists of the two elements 231 and 312) splits into two singleton conjugacy classes in A3. Thus, all the classes in A3 are singletons (we already knew this, because A3, being a group of order 3 is cyclic). The even partitions of 4 are (1, 1, 1, 1), (2, 2) and (3, 1). Again the element of type (1, 1, 1, 1) is the identity and commutes with all odd partitions. An element of type (2, 2) commutes with either of the 2-cycles in its cycle decomposition, so its class in S n is a class in An. However, an element of type (3, 1) commutes only with the identity, itself and its square, which are all even permutations. Therefore, the class of (3, 1), which has eight elements, splits into two conjugacy classes in A4, each consisting of four elements. Therefore, A4 has four conjugacy classes: the class of the identity element (with one element), the class of permutations with cycle type (2, 2) (with three elements) and the two classes of permutations with cycle type (3, 1) (consisting of four elements each).

By Exercise 4.6.10, there are two types of conjugacy classes of even elements in S n: those that remain conjugacy classes in An and those that split into the union of two conjugacy classes in An. In order to classify the conjugacy classes of An, we only need to determine the even partitions λ of n for which an element with cycle type λ does not commute with any odd permutation (these are the classes in S n that split into two classes of An of equal cardinality).

4.6 Representations of Alternating Groups 89

Theorem 4.6.12. Let λ be any partition. Then an element with cycle type λ does not commute with any odd permutation if and only if λ has distinct odd parts. Proof. Any permutation commutes with each of its cycles (see Lemma 2.2.14 and the discussion that follows). Therefore, if a permutation has an even cycle, it commutes with an odd permutation (an even cycle is an odd permutation). Now suppose that a permutation w has two odd cycles of the same size, say (a1,..., a2l+1) and (b1,..., b2l+1), then (once again using Lemma 2.2.14) w commutes with the odd permutation (a1b1)(a2b2) ··· (a2l+1b2l+1). Therefore, the only possibilities for permutations that do not commute with odd permutations are the permutations with distinct odd parts. Any permutation that commutes with such a permutation will have to be a composition of powers of its cycles. Since these cycles are of odd length, they are all even permutations, and therefore so are their powers and compositions of their powers. It follows that a permutation with distinct odd parts only commutes with even partitions.

Therefore, among the conjugacy classes in S n of even partitions, the classes corresponding to the partitions with distinct odd parts split. We have Theorem 4.6.13. For each partition of n with distinct odd parts, the set of per- + − mutations with cycle type λ splits into two conjugacy classes in An,Cλ and Cλ of equal cardinality. In fact, for any odd permutation x ∈ S n, − + −1 Cλ = xCλ x . For any other even partition λ, the set of permutations with cycle type λ is a conjugacy class in An.

Corollary 4.6.14. The number of conjugacy classes in An is the sum of the number of even partitions and the number of partitions with distinct odd parts. Comparing the above corollary and Theorem 4.6.7 and using the fact that the number of simple representations of a ﬁnite group is equal to the number of conjugacy classes (Theorem 1.5.16), we have the following purely combinatorial result: Theorem 4.6.15. For every positive integer n, the following identity holds:

p(n) − pself-conjugate(n) + 2p (n) = p (n) + p (n). 2 self-conjugate even distinct odd parts As usual, p(n) denotes the number of partitions of n, and (as most readers would have guessed) pself-conjugate(n) is the number of self-conjugate partitions of n, peven(n) is the number of even partitions of n and pdistinct odd parts(n) is the number of partitions of n with distinct odd parts. The following combinatorial result allows us to put this strange identity into a more elegant form:

90 Character Twists

Lemma 4.6.16. The number of partitions of n with distinct odd parts is equal to the number of self-conjugate partitions of n.

Proof. This result has a very elegant proof in terms of Young diagrams. A strip consisting of an odd number of squares can be transformed by ‘folding’ it into a self-conjugate hook:

→

In the same way, the rows of a Young diagram consisting of partitions with distinct odd parts can be folded to give a general self-conjugate partition as in the example below:

1 1 1 1 1 1 1 1 1 → 1 1 1 1 1 2 2 2 2 2 1 2 2 2 3 3 3 1 2 3 3 1 2 3 1

We have put numbers in the boxes only to indicate how the three rows of the Young diagram on the left get arranged as three nested hooks in the Young diagram on the right. For the nesting of hooks to result in the Young diagram of a partition, the parts of the original partition should decrease by at least two. Thus, this construction takes Young diagrams of partitions with distinct odd parts to Young diagrams of self-conjugate partitions. The inverse of this process is the ‘unfolding’ of hooks, which allows one to recover the Young diagram of the original partition with distinct odd parts, establishing the desired bijection.

In view of Lemma 4.6.16, the identity of Theorem 4.6.15 becomes

p(n) + 3pself-conjugate(n) = 2peven(n) + 2pself-conjugate(n),

which, after noting that p(n) = peven(n) + podd(n) and simplifying, gives

peven(n) − podd(n) = pself-conjugate(n). (4.14)

In Exercise 5.9.7, we will see an alternative proof of this identity using symmetric functions. We have seen that every element of S n is conjugate to its inverse (see Exercise 2.2.16). This is not true for An. Consider, for example, the cycle (123)

4.6 Representations of Alternating Groups 91

−1 in A3. Its inverse is (321), which is (13)(123)(13) . In the notation of Theorem 4.6.13, − −1 + C(3) = {w | w ∈ C(3)}.

[3] Exercise 4.6.17. 1. Show that for an odd integer k > 2, a k-cycle in An is conjugate to its inverse if and only if bk/2c is even. 2. An element of An with cycle type λ whose parts are distinct and odd is conju- P gate to its inverse if and only if ibλi/2c is even.

4.6.5 Character Values (First Steps) We will now try to write out the character tables for the alternating groups when K is the ﬁeld C of complex numbers. First, if λ is not self-conjugate, then the simple representation Vλ of S n, when restricted to An, remains simple, so the problem of computing these characters reduces to the problem of computing irreducible + − characters of the symmetric group. Now if λ is self-conjugate, then Vλ = Vλ ⊕Vλ . The character of χλ of Vλ therefore decomposes into a sum + − χλ = χλ + χλ (4.15)

of two simple characters of An, such that for every w ∈ An, − + −1 χλ = χλ (xwx ) (4.16) for any odd permutation x. If w has cycle type µ which does not have distinct odd −1 parts, then w and xwx are conjugate in An, so it follows that 1 χ+(w) = χ−(w) = χ (w) λ λ 2 λ ± for such w. Thus, our problem is only to determine the character values of Vλ at elements whose cycle types have distinct odd parts. Let us begin by determining the character values of A3 (pretending not to have noticed that A3 is a cyclic group of order 3). Most of the character table is already in place from the observations listed above. In the table below, the columns represent the conjugacy classes, while the rows represent the irreducible characters. + − The identity (4.16) relating χ(2,1) and χ(2,1) has already been taken into considera- tion in the 2 × 2 lower-right corner. 123 231 312

χ(3) 1 1 1 + χ(2,1) 1 u v + χ(2,1) 1 v u The identity (4.15) tells us that

u + v = χ(2,1)(231) = −1. (4.17)

92 Character Twists

On the other hand, the orthogonality relations (1.19) tell us that

|u|2 + |v|2 = 2 andvu ¯ + uv¯ = −1. (4.18)

If we write u = x + iy, then (4.17) implies that v = −(1 + x) − iy. In terms of x and y, both the identities (4.18) reduce to the quadratic equation

2(x2 + y2) + 2x = 1. (4.19)

The last piece of information that allows us to compute the values of u and v is

that the permutations 231 and 312 are mutual inverses, so by Lemma 1.8.3,√ we 1 3 have v = u¯. It follows that x = −1 − x, or x = − 2 , and subsequently that y = ± 2 . Thus, the character table of A3 is

123 231 312

χ(3) 1 1 1 + 2 χ(2,1) 1 ω ω + 2 χ(2,1) 1 ω ω

√ with ω = (−1 + i 3)/2. Coming to A4, we have once again that 4 has only one self-conjugate partition, namely the partition (2, 2), and only one partition with distinct odd parts, namely ± the partition (3, 1). Our goal is to compute the values of the characters χ(3,3) at the elements 2314 and 3124, which represent the two classes in A4 which make up the elements of cycle type (3, 1). Let us denote these classes by (3, 1)+ and (3, 1)−, respectively. Using the information from the earlier parts of this section, we have the incomplete character table:

Class (14) (2, 2) (3, 1)+ (3, 1)− cardinality 1 3 4 4

χ(4) 1 1 1 4 χ(3,1) 3 −1 0 0 + χ(2,2) 1 1 u v + χ(2,2) 1 1 v u

[2] Exercise 4.6.18. Determine u and v to complete the character table of A4. − Hint: Note once again that the elements of C(3,1) are the inverses of the elements + of C(3,1), so v = u¯.

4.6 Representations of Alternating Groups 93

[2] Exercise 4.6.19. Complete the character table of A5 given below by determining the values u and v:

Class (15) (22, 1) (3, 12) (5)+ (5)− cardinality 1 15 20 12 12

χ(5) 1 1 1 1 1 χ(4,1) 4 0 1 −1 −1 χ(3,2) 5 1 −1 0 0 + χ(3,1,1) 3 −1 0 u v − χ(3,1,1) 3 −1 0 v u

+ − Hint: In this case, note that the elements of the classes C(5) and C(5) are conjugate to their inverses, so u and v are real.

All the integers up to eight have only one self-conjugate partition and therefore also only one partition with distinct odd parts: indeed, (3, 2, 1) is the only self- conjugate partition of 6, and (4, 13) is the only self-conjugate partition of 7. The character tables of these alternating groups can be determined in the same way as those of A3, A4 and A5. For n = 8, there are two self-conjugate partitions, namely (4, 2, 12) and (32, 2). Under the correspondence of Lemma 4.6.16, they correspond to the partitions (7, 1) and (5, 3) with distinct odd parts. This, being the ﬁrst non-trivial example, was in fact used by Frobenius [7] to illustrate his method. The unknown part of the character table is the following:

Class (7, 1)+ (7, 1)− (5, 3)+ (5, 3)− cardinality 2880 2880 1344 1344 + ¯ χ(4,2,1,1) a a¯ b b − ¯ χ(4,2,1,1) a¯ a b b + ¯ χ(3,3,2) c c¯ d d − ¯ χ(3,3,2) c¯ c d d

Here, we have already taken into account the fact that an element with cycle decomposition µ = (7, 1) or µ = (5, 3) is not conjugate to its inverse, so the + + + − value of χλ on wµ is the conjugate of χλ on wµ . Also, from the character theory of symmetric groups (in principle, we know how to compute characters of S n, but a faster algorithm is given in Section 5.12), we ﬁnd that

χ(4,2,1,1)(w(7,1)) = χ(3,3,2)(w(5,3)) = −1, (4.20)

χ(3,3,2)(w(7,1)) = χ(4,2,1,1)(w(5,3)) = 0. (4.21)

Thus, a + a¯ = −1, b + b¯ = 0, c + c¯ = −1 and d + d¯ = 0.

94 Character Twists

In order to compute a, b, c and d, it is helpful to consider (for λ ∈ {(7, 1), (5, 3)}) the absolute diﬀerence character

+ − + − + − 2 ∆λ = (χλ − χλ )(χλ − χλ ) = |χλ − χλ | . Since it is the product of an integer linear combination of simple characters of + An, it is also an integer linear character of An. Moreover, it is clear that ∆λ(wµ ) = − α − α + ∆λ(wµ ) for µ ∈ {(7, 1), (5, 3)}. Therefore, ∆λ = ∆λ. Since χη = χη , we have

+ − h∆λ, χη iAn = h∆λ, χη iAn . (4.22) for η ∈ {(4, 2, 1, 1), (3, 3, 2)}. Viewing ∆λ as a class function on S n which is supported on An, we have

h∆λ, χηiS n = h∆λ, χη|An iAn /2 + − = (h∆λ, χη iAn + h∆λ, χη iAn )/2 + = h∆λ, χη iAn .

+ Since ∆λ is an integer linear combination of characters of An, h∆λ, χη iAn is an

integer. Therefore, h∆λ, χηiS n is also an integer. In particular, we have 5760 h∆ , χ i = − |a − a¯|2 ∈ Z, (4,2,1,1) (4,2,1,1) S 8 8! 2688 h∆ , χ i = − |b − b¯|2 ∈ Z, (4,2,1,1) (3,2,2) S 8 8! 5760 h∆ , χ i = − |c − c¯|2 ∈ Z, (3,2,2) (4,2,1,1) S 8 8! 2688 h∆ , χ i = − |d − d¯|2 ∈ Z. (3,2,2) (3,2,2) S 8 8! h − − − −i Also, since χλ χλ , χλ χλ A8 = 2, 2 × 5760 2 × 2688 |a − a¯|2 + |b − b¯|2 = 2, 8! 8! 2 × 5760 2 × 2688 |c − c¯|2 + |d − d¯|2 = 2. 8! 8! 5760 2 2688 ¯ 2 It follows that one of α = 8! |a − a¯| and β = 8! |b − b| is 1, while the other is 5760 2 2688 ¯ 2 0, and one of γ = 8! |c − c¯| and δ = 8! |d − d| is 1, while the other is 0. Suppose that β , 1. Then since α = 0,

+ + χ(4,2,1,1)(w(3,3,2)) = χ(4,2,1,1)(w(3,3,2))/2 = −1/2, which contradicts Theorem 1.8.7. Therefore, we have α = δ = 1 and β = γ = 0. 2 ¯ 2 We get |a − a¯| = 7 and |d − d| = 15. Combining√ this with the fact√ that a + a¯ = d + d¯ = −1/2, we ﬁnd that a = (−1 ± i 7)/2 and d = (−1 ± i 15)/2.

4.6 Representations of Alternating Groups 95

Therefore, the 4 × 4 part of the character table that we needed to resolve comes out as Class (7, 1)+ (7, 1)− (5, 3)+ (5, 3)− cardinality 2880 2880 1344 1344 √ √ χ+ −1+i 7 −1−i 7 0 0 (4,2,1,1) 2 √ 2 √ χ− −1−i 7 −1+i 7 0 0 (4,2,1,1) 2 2 √ √ χ+ 0 0 −1+i 15 −1−i 15 (3,3,2) 2√ 2√ − −1−i 15 −1+i 15 χ(3,3,2) 0 0 2 2

In writing down the character table of A8, we have developed almost all the ideas needed to deal with the general case. What is missing, however, is a technique to easily compute the character values χλ(wµ), when λ is a self-conjugate partition and µ is a partition with distinct odd parts. This technique is provided by the recursive Murnaghan–Nakayama formula (Section 5.11). Therefore, a general solution of the problem of writing down the character table of An is postponed to Section 5.12.

5 Symmetric Functions

5.1 The Ring of Symmetric Functions

We will consider symmetric polynomials in ‘infinitely many variables’. These are formal expressions, almost always with infinitely many terms. Let x = (x1, x2,... ) be an infinite sequence of commuting variables (or indeter- minates). A multi-index is an infinite sequence α = (α1, α2,... ) of non-negative integers, of which only finitely many are positive. For each multi-index α, let xα α1 α2 denote the formal product x1 x2 ··· , which is called a monomial. Its degree is, by definition, the sum α1 +α2 +... , denoted by |α|. Given a multi-index α, its positive terms can be arranged in weakly decreasing order to obtain a partition, which is called the shape of α. For example, if α = (0, 2, 0, 0, 3, 1, 5, 0, 0,... ), then

α 2 3 5 x = x2 x5 x6 x7, a monomial of degree 11 and shape (5, 3, 2, 1). Note that the shape of the multi- index (0, 0,... ) with all entries 0 is the unique partition of 0 (called the empty partition and denoted by ∅). The corresponding monomial x∅ is the constant monomial 1 of degree 0.

Definition 5.1.1 (Homogeneous symmetric function of degree n). A homogeneous symmetric function of degree n with coefficients in any field (or ring) K is a formal sum X α f (x) = cα x , |α|=n

where α ranges over the set of all multi-indices with sum n and each cα ∈ K depends only on the shape of α. The space of all homogeneous symmetric func- n tions of degree n is denoted by ΛK. It should be understood from the above deﬁnition that homogeneous symmetric 0 functions of degree 0 are all constant, in other words, ΛK = K. Since there are

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5.1 The Ring of Symmetric Functions 97

infinitely many multi-indices of shape λ for any non-empty partition λ, a homogeneous symmetric function of positive degree is an infinite sum. Yet it always m makes sense to multiply two homogeneous symmetric functions f ∈ ΛK and n P α P α α g ∈ ΛK: if f = cα x and = dα x , the coefficient of x in f g is X cβdγ, β+γ=α

a ﬁnite sum. Clearly, f g is a homogeneous symmetric function of degree m + n. Let ∞ M n ΛK = ΛK. (5.1) n=0

Then ΛK becomes a graded K-algebra, meaning that the multiplication map ΛK × m n m+n ΛK → ΛK takes ΛK × ΛK into ΛK .

Graded Algebra A graded algebra A is an algebra that can be decomposed into a direct sum: M∞ A = An, n=0

with the property that the product of an element of An with an element of Am lies in Am+n.

Deﬁnition 5.1.2 (Monomial symmetric function). For every partition λ of n, let

X α mλ = x . shape(α)=λ

Then mλ is a homogeneous symmetric function of degree n, called the monomial symmetric function of shape λ.

For example,

m∅ = 1 X m(1) = xi i X 2 m(2) = xi i

98 Symmetric Functions X m(1,1) = xi x j i< j X 2 m(2,1) = xi x j i, j

[2] Exercise 5.1.3. Express the product m(2,1)m(1,1) as a linear combination of monomial symmetric functions. [3] Exercise 5.1.4. For positive integers m and n, show that min(m,n) ! X m + n − 2k m m m n = m k m+n−2k . (1 ) (1 ) m − k (2 ,1 ) k=0

[2] Exercise 5.1.5. Given partitions λ and µ, show that the coeﬃcient of mν in the expansion of mλmµ, in terms of monomial symmetric functions, is the number of ways of expressing the multi-index ν = (ν1, . . . , νn, 0,... ) as a sum α + β of multi-indices, where α has shape λ and β has shape µ. Theorem 5.1.6. For every n ≥ 0, the set

{mλ | λ is a partition of n},

n is a basis of ΛK. Proof. Indeed, if X α f (x) = cα x |α|=n is a homogeneous symmetric function of degree n, then X f (x) = cλmλ λ`n is the unique expression of f (x) as a linear combination of monomial symmetric functions. Corollary 5.1.7. For every n ≥ 0,

n dim ΛK = p(n), the number of partitions of n.

5.2 Other Bases for Homogeneous Symmetric Functions

Deﬁnition 5.2.1 (Elementary symmetric functions). For each positive integer n, deﬁne

en = m(1n) for each n ≥ 0.

5.2 Other Bases for Homogeneous Symmetric Functions 99

For each partition λ = (λ1, λ2,... ), deﬁne

eλ = eλ1 eλ2 ··· .

Then eλ, which is a homogeneous symmetric function of degree |λ|, is called the elementary symmetric function of shape λ.

In other words, the elementary symmetric function eλ is the product X X X x ··· x x ··· x ··· x ··· x . i1 iλ1 i1 iλ2 i1 iλl ··· ··· ··· i1<

where Nλµ denotes the number of 0-1 λ × µ matrices (Section 4.2).

µ µ1 µ2 Proof. We wish to show that the coeﬃcient of x = x1 x2 ··· in eλ is Nλµ. Consider the array of variables

eλ1 : x1 x2 ··· xm

eλ2 : x1 x2 ··· xm ......

eλl : x1 x2 ··· xm

whose rows are indexed by the n factors eλ1 , ··· , eλl in eλ. Each monomial in the expansion of eλ involving the variables x1,..., xm is obtained by choosing λ1 entries from the ﬁrst row, λ2 entries from the second row and so on, and multiply- µ ing them together. The resulting monomial is x precisely when x1 is chosen µ1 times, x2 is chosen µ2 times and so on. Each such choice can be recorded in a 0-1 λ × µ matrix, whence the result follows. It is convenient to think of (5.2) in a matrix form. In order to do this, we will need to enumerate the partitions of n in such a way that λ appears before µ in this enumeration if λ ≤ µ. This can be achieved in many ways, but a nice concrete method is to use the reverse lexicographic order (Deﬁnition 2.2.18). [2] Exercise 5.2.3. Show that if µ ≤ λ (in reverse dominance order), then λ precedes µ in the reverse lexicographic order. Now enumerate the partitions of n in the reverse lexicographic order. The corresponding enumeration of monomial symmetric functions (mλ) forms a row vector, which we denote by m.

100 Symmetric Functions

Similarly, we denote by e the row vector (eλ), with the indices λ enumerated in reverse lexicographic order. Let N denote the matrix with entries (Nλµ) (since Nµλ = Nλµ, this is a symmetric matrix). The identity (5.2) can now be expressed in a matrix form e = mN. (5.3) The dual RSK correspondence gives rise to the identity X Nλµ = KνλKν0µ. (5.4) ν≤λ ν0≤µ

Let K denote the matrix whose (λ, µ)th entry is Kλµ. We know from Lemma 3.1.12 that K is an upper triangular matrix with 1’s along the diagonal and therefore has determinant 1. Let J denote the permutation matrix whose (λ, µ)th entry is δλ0µ. Left multiplication by J has the effect of interchanging the νth row of a matrix with the ν0th row for each partition ν. The identity (5.4) can be rewritten as N = K0 JK, (5.5) where K0 denotes the transpose of the matrix K. Theorem 5.2.4 (Fundamental theorem of symmetric functions). For any field n K, the elementary symmetric functions of degree n form a basis of ΛK. If we take n n K = Q, then they generate the free abelian subgroup ΛZ of ΛQ consisting of symmetric functions with integer coefficients.1 Proof. By (5.5), the determinant of N is ±1, whence its inverse is also a matrix with integer entries (this fact is also evident from Lemmas 4.3.6 and 4.3.9, which directly show that N is a lower triangular matrix with 1’s along the diagonal). The invertibility of N implies the first assertion. For the second assertion, note that symmetric functions with integer coefficients form the subgroup spanned by the monomial symmetric functions. The invertibility of N in integer coefficients therefore implies the second assertion.

Since each eλ is a monomial in the en’s, Theorem 5.2.4 can be viewed as a statement about algebras:

Corollary 5.2.5. For any field K, the algebra ΛK of symmetric functions is the polynomial algebra in the elementary symmetric functions e1, e2,... with coefficients in K. Also, ΛZ is the polynomial algebra in the elementary symmetric functions e1, e2,... with integer coefficients. In the above statement, a polynomial is to be interpreted as a linear combination of monomials in the variables e1, e2,... .

1 See Artin [1], Chapter 14, Section 3 for a very diﬀerent approach to this result and Corollary 5.2.5.

5.2 Other Bases for Homogeneous Symmetric Functions 101

Historically, the importance of the fundamental theorem of symmetric functions lies in the following fact: let K be an algebraically closed field, and suppose n n−1 that ξ1, . . . , ξn are the roots of the polynomial x + a1 x + ··· + an. We have n Y n n−1 (x − ξi) = x + a1 x + ··· + an, i=1 i so that ai = (−1) ei(ξ1, . . . , ξn, 0 ··· ). Since any elementary symmetric function (Definition 5.2.1) is a monomial in the en’s, the fundamental theorem of symmetric functions just says that every polynomial in the coefficients of a polynomial can be expressed as a symmetric polynomial in its roots. Definition 5.2.6 (Complete symmetric functions). For each positive integer n, define X hn = mλ for each n ≥ 0. λ`n

For each partition λ = (λ1, λ2, ··· ), deﬁne

hλ = hλ1 hλ2 ··· .

Then hλ, which is a homogeneous symmetric function of degree |λ|, is called the complete symmetric function of shape λ.

In other words, the complete symmetric function hλ is the product X X X x ··· x x ··· x ··· x ··· x i1 iλ1 i1 iλ2 i1 iλl ≤···≤ ≤···≤ ≤···≤ i1 iλ1 i1 iλ2 i1 iλl Theorem 5.2.7. For every n ≥ 0 and every partition λ of n, X hλ = Mλµmµ, (5.6) µ

where Mλµ denotes the number of λ × µ matrices (Deﬁnition 2.6.1). Proof. The proof is similar to that of Theorem 5.2.2. We wish to show that the µ1 µ2 coeﬃcient of x1 x2 ··· in hλ is Mλµ. Consider the array of variables

hλ1 : x1 x2 ··· xm

hλ2 : x1 x2 ··· xm ......

hλl : x1 x2 ··· xm

where each row is indexed by the n factors hλ1 ,..., hλl in hλ. Each monomial in the expansion of hλ is obtained by choosing λ1 entries from the ﬁrst row, λ2 entries from the second row and so on, but this time with repetition, and multiplying them µ together. The resulting monomial is x precisely when x1 is chosen µ1 times, x2 is

102 Symmetric Functions

chosen µ2 times and so on. Each such choice can be recorded in a λ×µ matrix (its entries record the number of times a variable from the array was chosen), whence the result follows.

Let h denote the row vector (hλ). The identity (5.6) can be written in a matrix form as h = mM, (5.7)

where M is the matrix whose (λ, µ)th entry is Mλµ. On the other hand, the RSK correspondence (Theorem 3.2.2) gives the identity M = K0K. (5.8) It follows immediately that det M = 1. We get the analog of the fundamental theorem of symmetric functions for complete symmetric functions Theorem 5.2.8. For any ﬁeld K, the complete symmetric functions of degree n n form a basis of ΛK. If we take K = Q, then they span the same free abelian n n subgroup ΛZ of ΛQ consisting of symmetric functions with integer coeﬃcients. The analog of Corollary 5.2.5 holds for complete symmetric functions:

Corollary 5.2.9. For any field K, the algebra ΛK of symmetric functions is the polynomial algebra in the complete symmetric functions h1, h2,... with coefficients in K. Also, ΛZ is the polynomial algebra in the complete symmetric functions h1, h2,... with integer coefficients. Definition 5.2.10 (Power sum symmetric functions). For each positive integer n, define n n pn = m(n) = x1 + x2 + ··· .

For each partition λ = (λ1, λ2,... ), deﬁne

pλ = pλ1 pλ2 ··· .

Then pλ, which is a homogeneous symmetric function of degree |λ|, is called the power sum symmetric function of shape λ. It is in the study of power sum symmetric functions that we will see the direct connection of the theory of symmetric function with characters of symmetric groups. For each partition µ of n, let wµ denote an element of S n with cycle type µ. If λ is another partition of n, let

Pλµ = trace(wµ; K[Xλ])

the trace of the permutation wµ when it acts on the representation K[Xλ] (recall that this is the number of points in Xλ that are ﬁxed by wµ; see Exercise 2.1.14).

5.2 Other Bases for Homogeneous Symmetric Functions 103

Theorem 5.2.11. For every n ≥ 0 and every partition λ of n, X pµ = Pλµmλ. (5.9) λ

Proof. Take wµ to be the product of disjoint cycles

wµ = (1 ··· µ1)(µ1 + 1 ··· µ1 + µ2)(µ1 + µ2 + 1 ··· µ1 + µ2 + µ3) ··· Consider the following array of variables whose rows correspond to the factors

pµ1 , pµ2 ,... of pµ, along with a subset of n associated to each row:  1  µ µ µ . 1 1 1 ··· . pµ1 : x x x . 1 2 3   µ1  µ + 1  1 µ µ µ . 2 2 2 ··· . pµ2 : x x x . 1 2 3   µ1 + µ2  µ + µ + 1  1 2 µ µ µ . 3 3 3 ··· . pµ3 : x x x . 1 2 3   µ1 + µ2 + µ3 . .

A monomial in pµ is obtained by picking one entry from each row and multiplying λ1 λ2 them out. The monomial so obtained is x1 x2 ··· if and only if, for each j, the µi’s for those i’s, where x j is picked, add up to λ j. Construct a partition of n of shape λ as follows: the jth part consists of the subsets associated to those rows where x j is picked. Since the parts of this partition are unions of the cycles in the cycle decomposition of wµ, this partition is fixed by wµ. One easily verifies (for example, by reversing it) that this construction gives rise to a bijection between the set of partitions of n with shape λ that are fixed by wµ and the number of ways λ1 λ2 of getting x1 x2 ··· in the expansion of pµ. In matrix notation, we may rewrite (5.9) as p = mP, (5.10)

where p = (pλ) and P = (Pλµ). [2] Exercise 5.2.12 (Polya´ enumeration theorem2). Let G be a finite group and X be a finite G-set with n elements. Let C = {c1,..., cm} be a finite set of 2 See Polya´ [24] and also an expository article by Read [25].

104 Symmetric Functions

‘colours’. A colouring of X is deﬁned to be a function c : X → C. Two colourings c and c0 are said to be equivalent if there exists g ∈ G such that, for all x ∈ X, 0 c (g · x) = c(x). Let (λ1, . . . , λm) be a weak composition of n with m parts. Let N(λ) be the number of equivalence classes of colourings of X, where λi elements of X have colour ci for each i = 1,..., m. For each g ∈ G, let µ(g) denote the cycle of g (viewed as a permutation of the n-element set X). Show that N(λ) is the coeﬃcient of the monomial xλ in the polynomial 1 X p . |G| µ(g) g∈G Hint: Use Burnside’s lemma (Exercise 2.4.6) together with Theorem 5.2.11.

Let

Xλµ = trace(wµ; Vλ). (5.11)

The matrix X = (Xλµ) is the character table of S n. By Young’s rule (Theorem 3.3.1),

Pλµ = trace(wµ; K[Xλ]) X = trace(wµ; Vν)Kνλ ν X = XνµKνλ. ν

Here, we have followed the convention of setting Kνλ = 0 if ν λ. In matrix notation, this becomes P = K0X. (5.12)

When the characteristic of K does not divide G, the characters of a ﬁnite group G form a basis of the space of class functions. It follows that the character table (which is determined by the group up to permutation of rows and columns) is non-singular. Since K is upper triangular with diagonal entries 1, we ﬁnd that P is non-singular if the characteristic of K exceeds n. Moreover, det P = det X. We have

Theorem 5.2.13. If the characteristic of K is greater than n, then the power sum n symmetric functions form a basis of ΛK. It is not so hard to calculate det P (and hence det X) explicitly. The ﬁrst thing to note is

Lemma 5.2.14. If Pλµ > 0, then λ ≤ µ.

5.2 Other Bases for Homogeneous Symmetric Functions 105

Proof. Recall that trace(wµ; K[Xλ]) is just the number of ordered partitions of n that are ﬁxed by wµ. An ordered partition a a n = S 1 S 2 ···

is ﬁxed by wµ if and only if each of its parts S i is a union of the cycles in the disjoint cycle decomposition of wµ. Write Ti = {µ1 + ··· + µi−1 + 1, . . . , µ1 + ··· + µi} for the elements in the ith cycle of wµ. We shall construct a tableau Y of shape λ and type µ (which is not necessarily semistandard) as follows: If ` ` S 1 = Ti ··· Ti , put in µi 1’s, µi 2’s and so on in the ﬁrst row. Similarly, if 1 ` ` k 1 2 S = T 0 ··· T 0 , then put in µ 0 (k + 1)’s, µ 0 (k + 2)’s and so on in the second 2 i1 ik0 i1 i2 row. Continuing in this manner, we obtain the Young tableau Y of shape λ and type µ. For example, if λ = (6, 5, 3, 3) and µ = (4, 4, 3, 2, 2, 1, 1), then

wµ = (1 2 3 4)(5 6 7 8)(9 10 11)(12 13)(14 15)(16)(17). Suppose that a a a S = {5, 6, 7, 8, 14, 15} {1, 2, 3, 4, 17} {12, 13, 16} {9, 10, 11},

then 2 2 2 2 5 5 Y = 1 1 1 1 7 4 4 6 3 3 3

Now, rearrange the entries of each column of Y so that they are in increasing order. In the running example, Y becomes

1 1 1 1 5 5 2 2 2 2 7 3 3 3 4 4 6

Since each column has each integer appearing at most once, after sorting, all the 1’s will be in the first column of Y, all the 2’s will be in the first two columns of Y and more generally, all the i’s will be in the first i columns of Y. It follows that µ1 ≤ λ1, µ1 + µ2 ≤ λ1 + λ2 and so on, showing that λ ≤ µ.

Lemma 5.2.15. For each partition λ of n, let mi(λ) denote the number of times that i occurs in λ. Y∞ Pλλ = mi(λ)! i=1

106 Symmetric Functions

Proof. We have already observed that the only ordered partitions that are fixed by wλ are those whose parts are unions of the cycles of λ. If, furthermore, such a partition must have shape λ, it must be obtained from the cycle decomposition of wλ by permuting the parts of the same size. The number of ways to do this is Q i mi(λ)!. Corollary 5.2.16. The matrix X of (5.11) has the property Y Y∞ det X = mi(λ)!. λ`n i=1 Proof. This follows from the fact that P is triangular (Lemma 5.2.14) together with Lemma 5.2.15 and (5.12). The following sequence of four exercises will derive an alternative formula for det X from first principles. The first exercise is a special case of the Frobenius reciprocity theorem (see Bump [2, Section 4.1]). [2] Exercise 5.2.17 (A special case of Frobenius reciprocity). Let G be a finite group and H be any subgroup. Let X denote the G-set G/H. Let (ρ, V) be any representation of G. For any H-invariant vector ξ in the contragredient representation 0 V , define φξ ∈ HomK(V, K[G/H]) by −1 φξ(v)(xH) = ξ(ρ(x )v).

0 H Show that ξ 7→ φξ is an isomorphism of vector spaces (V ) → HomG(V, K[G/H]). Here, (V0)H denotes the space of H-invariant vectors in V0. [2] Exercise 5.2.18. Now assume that K is an algebraically closed ﬁeld whose characteristic does not divide |G| and V is a simple representation of G. Use Exer- cise 5.2.17 to show that the multiplicity of the trivial representation in the restriction of (ρ, V) to H is equal to the multiplicity of V in K[G/H].

[2] Exercise 5.2.19. Let η be a partition of n. Take any element S ∈ Xη and

define S η = StabS n S (a different choice of S would give rise to a subgroup that is conjugate to S η; the subgroups S η are called Young subgroups). Show that the multiplicity of the trivial representation in the restriction of Vλ to S η is equal to Kλη. [3] Exercise 5.2.20. For partitions λ and µ of n, define

Aλµ = |{w ∈ S λ | w has cycle type µ}| and Bλµ = |S µ|Kλµ and form the matrices A and B in the usual manner. Show that XA0 = B. Deduce that (see Sagan [26, Section 2.12, Problems 12 & 13]) Y Y det X = λi, λ`n i

5.3 Specialization to m Variables 107

the product of all parts of all partitions of n. The ﬁrst few values of det X are given in Table 5.1.3

n det X 1 1 2 2 3 6 4 96 5 2880 6 9953280 7 100329062400 8 10651768002183168000 9 150283391703941024789299200000 10 9263795272057860957392207640004657152000000000

Table 5.1 Growth of det X

5.3 Specialization to m Variables

It is not always necessary to work with inﬁnitely many variables. Denote the ring of symmetric polynomials in m variables and coeﬃcients in K by ΛK,m. These are polynomials of the form

X α f (x1,..., xm) = cα x , |α|=n

where α runs over the set of all weak compositions (α1, . . . , αm) of n into m parts (see Definition 2.3.1) such that cα = cw·α for all w ∈ S m. Here, for α = (α1, . . . , αm), w · α denotes (αw(1),...,w(m)). Analogous to (5.1), ∞ M n ΛK,m = ΛK,m, n=0 n where ΛK,m denotes the subspace of homogeneous polynomials of degree n in ΛK,m. Definition 5.3.1 (Specialization). Given a polynomial in infinitely many variables, its specialization to m variables is obtained by setting the variables xk = 0 for k ≥ m. In other words, monomials which involve variables other than

3 See sequence A007870 in The Online Encyclopedia of Integer Sequences, http://oeis.org. There is an interesting discussion on the relationship between the formula of Corollary 5.2.16 and the formula of Exercise 5.2.20 at http://mathoverflow.net/q/99271.

108 Symmetric Functions

x1,..., xm are deleted, leaving only a ﬁnite formal sum. We will denote the specialization of f ∈ ΛK to m variables by f (x1,..., xm).

[1] Exercise 5.3.2. 1. Compute the specialized symmetric functions mλ(x1, x2, x3), eλ(x1, x2, x3), hλ(x1, x2, x3) and pλ(x1, x2, x3) for all partitions λ of 3.

2. Compute the specialized symmetric functions mλ(x1, x2), eλ(x1, x2), hλ(x1, x2) and pλ(x1, x2) for all partitions λ of 3.

The specialization of mλ to m variables is

X α mλ(x1,..., xm) = x , α

where the sum is over all multi-indices (α1, . . . , αm) of shape λ. Since no weak composition of n with m parts can have a shape λ with more than m parts, mλ(x1,..., xm) = 0 whenever λ has more than m parts. Similar to symmetric functions in inﬁnitely many variables, we have

Theorem 5.3.3. For any partition λ and any positive integer m, mλ(x1,..., xm) = 0 if λ has more than m parts. For any positive integer n, the set

{mλ(x1,..., xm) | λ is a partition of n with at most m parts}

n is a basis of ΛK,m.

Since specialization maps mλ to mλ(x1,..., xm), we have

Theorem 5.3.4. Specialization deﬁnes a surjective linear map

n n ΛK → ΛK,m.

This map is a linear isomorphism if and only if m ≥ n.

n To get a basis of ΛK,m from specializations of elementary symmetric functions, we observe ﬁrst that en(x1,..., xm) = 0 if n > m . Therefore, eλ(x1,..., xm) = 0 if λ1 > m. Since specialization is surjective and {eλ | λ is a partition of n} is a basis n for ΛK,

S = {eλ(x1,..., xm) | λ is a partition of n with λ1 ≤ m}

n 0 spans ΛK,m. The conjugation operation λ → λ deﬁnes a bijection from the set of partitions λ with largest part less than or equal to m onto the set of partitions with at most m parts (see Exercise 4.3.3). Therefore, the set S deﬁned above has n the same cardinality as the dimension of ΛK,m. It follows that S is in fact a basis. Thus, we have

5.3 Specialization to m Variables 109

Theorem 5.3.5. For any partition λ and any positive integer m, eλ(x1,..., xm) = 0 if λ1 > m. For every non-negative integer n,

{eλ(x1,..., xm) | λ is a partition of n with λ1 ≤ m} n is a basis of ΛK,m.

A similar argument would fail for the complete symmetric functions hλ, because hλ(x1,..., xm) , 0 for every partition λ of n. Instead, we resort to an analysis of the transition matrix (5.7). The method is best illustrated by considering a more general situation. Let V be a vector space with basis v1,..., vr. For some s ≤ r, let V¯ denote the quotient space of V by the subspace spanned by vs+1,..., vr, with quotient map q : V → V¯ . Then q(v1),..., q(vs) is a basis for V¯ . Let w ,..., w be another set of r vectors in V. Consider the row vectors v = 1 r v1 ... vr , w = w1 ... wr . Each of the vectors in w can be expressed in a unique manner as a linear combination of vectors in v. These expansions can be expressed using an r × r transition matrix A: w = vA. (5.13) The vectors in w form a basis of V if and only if A is non-singular. By applying q to both sides of (5.13), we get

(q(w1),..., q(ws)) = (q(v1),..., q(vs))A¯, where A¯ is the s × s upper-left submatrix of A. Thus, we have

Lemma 5.3.6. The vectors q(w1),..., q(ws) form a basis of V¯ if and only if the s × s upper-left submatrix of A is non-singular. In our situation of specialization of symmetric functions, we take r to be the number of partitions of n and s to be the number of partitions of n with at most m parts. We need to show that the r × r upper-left submatrix of the matrix M in Eq. (5.7) is non-singular. By the RSK correspondence, M has a factorization M = K0K, where K is upper triangular. We now use the following exercise: Exercise 5.3.7. Suppose A = LU is a factorization of an r × r matrix into the product of a lower triangular matrix L and an upper triangular matrix U, with both L and U non-singular. Then for every s ≤ r, the r × r upper-left submatrix of A is non-singular. As a consequence, we have the following theorem: Theorem 5.3.8. For every non-negative integer n,

{hλ(x1,..., xm) | λ is a partition of n with at most m parts} n is a basis of ΛK,m.

110 Symmetric Functions

From the upper triangularity of P (Lemma 5.2.14), it is quite easy to deduce the following result for specializations of power sum symmetric functions: Theorem 5.3.9. If the characteristic of K is greater than n, then

{pλ(x1,..., xm) | λ is a partition of n with at most m parts} n is a basis for ΛK,m.

5.4 Schur Functions and the Frobenius Character Formula

In Section 5.2, we computed transition matrices [Eqs. (5.3), (5.7) and (5.10)] and their factorizations [Eqs. (5.5), (5.8) and (5.12)], whence we obtained the following identities between various bases of the space of symmetric functions: e = mK0 JK (5.14) h = mK0K (5.15) p = mK0X (5.16) Thus, all the transitions from monomial symmetric functions to the other families of symmetric functions proceed through a common ﬁrst stage, namely mK0. If we deﬁne s = mK0, (5.17) we get the simpler identities e = sJK (5.18) h = sK (5.19) p = sX. (5.20)

Here, s is a row vector whose entries sλ, λ ` n are symmetric functions known as Schur functions. The deﬁnition (5.17) is equivalent to X sλ = Kλµmµ, (5.21) µ≥λ

and sλ is called the Schur function of shape λ. Eq. (5.20), which is equivalent to X pµ = trace(wµ, Vλ)sλ, (5.22) λ

says that when the power sum symmetric function pµ is expanded in Schur functions, the coeﬃcient of sλ is the character value of Vλ at an element of cycle type µ and is called the Murnaghan–Nakayama rule.

5.4 Schur Functions and the Frobenius Character Formula 111

Example 5.4.1. For each positive integer n, we have X s(n) = mµ = hn and s(1n) = m(1n) = en. µ`n

[1] Exercise 5.4.2. Compute the Schur function s(2,1) in terms of monomial symmetric functions.

An immediate consequence of (5.17) is

Theorem 5.4.3. For any ﬁeld K, the Schur polynomials of degree n form a basis n n n of ΛK. If we take K = Q, then they span the same free abelian subgroup ΛZ of ΛQ consisting of symmetric functions with integer coeﬃcients.

By specializing the identity (5.17) to m variables, we get X sλ(x1,..., xm) = Kλµmµ(x1,..., xm) µ≥λ X = Kλµmµ(x1,..., xm). (5.23) µ0≤λ0 0 0 0 0 0 Now suppose that λ has more than m parts, i.e., λ1 > m. If µ ≤ λ , then µ1 ≥ λ1, which means that µ has more than m parts, and therefore by Theorem 5.3.3, mµ(x1,..., xm) = 0. It follows from (5.23) that sλ(x1,..., xm) = 0. Similar to Theorem 5.3.3, we get the following result:

Theorem 5.4.4. For any partition λ and any positive integer m, sλ(x1,..., xm) = 0 if λ has more than m parts. For every positive integer n,

{sλ(x1,..., xm) | λ is a partition of n with at most m parts}

n is a basis of ΛK,m. The description of Schur polynomials given above is a combinatorial interpretation (due to Kostka) of Cauchy’s original definition, which we now recall. For the Cauchy version, we will work with finitely many variables x1,..., xm, where m ≥ n. m For a α = (α1, . . . , αm) ∈ N , define α α1 α2 ··· m x1 x1 x1 α α α x 1 x 2 ··· x m 2 2 2 α j aα = . . . . = det(x ) . . .. . i . . . α1 α2 αm xm xm ··· xm

Observe that aα is an alternating polynomial in the variables x1,..., xm:

aα(xw(1),..., xw(m)) = (w)aα(x1,..., xm) for any w ∈ S m. (5.24)

112 Symmetric Functions

For example, if we deﬁne

δ = (m − 1, m − 2,..., 1, 0) (5.25)

then we have m−1 m−2 ··· x1 x1 x 1 m−1 m−2 x x ··· x2 1 Y 2 2 aδ = . . . . . = (xi − x j), ...... i< j m−1 m−2 xm xm ··· xm 1 the Vandermonde determinant. Schur functions were ﬁrst introduced by Cauchy in 1815. In this exposition, his deﬁnition becomes the following theorem:

Theorem 5.4.5 (Cauchy’s deﬁnition of Schur functions). Suppose that λ is a partition of n with at most m parts. Then

sλ(x1,..., xm) = aλ+δ/aδ. (5.26)

In the above theorem, it should be understood that δ is as deﬁned in (5.25) and that we think of λ = (λ1, . . . , λl) with l ≤ m as an m-tuple (λ1, . . . , λm) by appending (m − l) 0’s at the end, and that

λ + δ = (λ1 + m − 1, λ2 + m − 2, . . . , λm−1 + 1, λm).

Proof of Theorem 5.4.5. The identity e = sJK characterizes Schur functions. n n Since m ≥ n, specialization to m variables is an isomorphism ΛK → ΛK,m (Theorem 5.3.4), the specializations of Schur functions to m variables are also characterized by the identities X eµ(x1,..., xm) = Kλ0µ sλ(x1,..., xm) for all µ ` n. λ`n

It therefore suffices to prove that the above identities hold with sλ(x1,..., xm) replaced by aλ+δ/aδ. In other words, it suffices to show that X aδeµ = Kλ0µaλ+δ for all µ ` n. (5.27) λ`n In the above equation, both sides are alternating polynomials. In alternating polynomials, only monomials with distinct powers can have non-zero coefficients (except when the characteristic of K is 2):

[1] Exercise 5.4.6. Assume that the characteristic of K is not 2. Then if f (x1,..., xm) is an alternating polynomial, the coeﬃcient of a monomial α1 α2 αm x1 x2 ··· xm is zero whenever αi = α j for any i , j in {1,..., m}.

5.4 Schur Functions and the Frobenius Character Formula 113

Our proof will compare the coefficients of certain special monomials, which we call decreasing monomials, on both sides of (5.27): Definition 5.4.7 (Decreasing monomial). A decreasing monomial is a monomial 2 α α1 α αm of the form x = x1 x2 ··· xm , where α1 > α2 > ··· > αm. Alternating polynomials are determined by the coefficients of their decreasing α monomials: given any monomial x , where the powers α1, . . . , αm are all distinct, it is possible to find a permutation w ∈ S m which sorts them, i.e., αw(1) > αw(2) > . . . , αw(m). The identity (5.24) now allows us to recover the coefficient of α1 α2 αm αw(1) αw(2) αw(m) x1 x2 ··· xm from the coefficient of the decreasing monomial x1 x2 ··· xm . Therefore, it suffices to show that the coefficients of the decreasing monomials on both sides of (5.27) are the same. We will also need the following observation about polynomials of the form aλ+δ:

[1] Exercise 5.4.8. Show that the only decreasing monomial that appears in aλ+δ with non-zero coefficients is xλ+δ, and it appears with coefficient one. Hint: Use the expansion of the determinant as an alternating sum over S m. It follows that the coefficient of the decreasing monomial xλ+δ on the right- hand side of (5.27) is Kλ0µ. In order to prove (5.27), it suffices to show that the λ+δ coefficient of x on the left-hand side is also Kλ0µ. The left-hand side of (5.27) is a product

aδeµ1 (x1,..., xm)eµ2 (x1,..., xm) ··· eµm (x1,..., xm).

Let f0 = aδ, f1 = aδeµ1 (x1,..., xm) and more generally let

fi = fi−1(x1,..., xm)eµi (x1,..., xm) for i = 1,..., m. These are all alternating polynomials with

fm(x1,..., xm) = aδeµ(x1,..., xm).

λ+δ For any partition λ, the decreasing monomial x in fm always comes from a β monomial x in fm−1(x1,..., xm) by incrementing the powers of µi of the variables x1,..., xm by one. It follows that β is either an increasing monomial or has two variables raised to the same power. But since fm−1(x1,..., xm) is alternating, the second possibility is ruled out. Therefore, β is decreasing, and so it can be written λ(m−1)+δ (m−1) as x for some partition λ of n − µm. Working backwards in this way gives rise to a sequence of partitions ∅ = λ(0) ⊂ λ(1) ⊂ · · · ⊂ λ(m−1) ⊂ λ(m) = λ (5.28)

(i) (i−1) such that λ is obtained by adding 1 to µi parts of λ . In other words, µi of the parts of λ(i−1) are increased by one to get λ(i). The coeﬃcient of xλ+δ in

114 Symmetric Functions

aδeµ(x1,..., xm) is precisely the number of such sequences (5.28) (we will illustrate this in Example 5.4.9). In the sequence ∅ = λ(0)0 ⊂ λ(1)0 ⊂ · · · ⊂ λ(m−1)0 ⊂ λ(m)0 = λ0 obtained by transposing the ones in (5.28), the length of each column increases by at most one in each step. By Theorem 3.1.6, the number of such sequences is nothing but Kλ0µ. Example 5.4.9. Take m = 4, λ = (3, 2, 2, 1) and µ = (3, 3, 2). We have

δ 3 2 x eµ(x1, x2, x3, x4) = x1 x2 x3(x1 x2 x3 + x1 x2 x4 + x1 x3 x4 + x2 x3 x4)×

(x1 x2 x3 + x1 x2 x4 + x1 x3 x4 + x2 x3 x4)(x1 x2 + x1 x3 + x1 x4 + x2 x3 + x2 x4 + x3 x4)

λ+δ 6 4 3 δ In order to get the monomial x = x1 x2 x3 x4 from x upon multiplication by eµ(x1,..., xm), one could choose the boxed monomials from the factors

3 2 x1 x2 x3 ( x1 x2 x3 + x1 x2 x4 + x1 x3 x4 + x2 x3 x4)×

(x1 x2 x3 + x1 x2 x4 + x1 x3 x4 + x2 x3 x4)(x1 x2 + x1 x3 + x1 x4 + x2 x3 + x2 x4 + x3 x4), so that the sequence of monomials after each multiplication would be

3 2 4 3 2 5 4 2 6 4 3 x1 x2 x3 → x1 x2 x3 → x1 x2 x3 x4 → x1 x2 x3 x4 which is the same as xδ → xδ+(1,1,1,0) → xδ+(2,2,1,1) → xδ+(3,2,2,1) corresponding to the semistandard Young tableau

1 1 1 2 2 2 3 3 of shape (3, 2, 2, 1) and type (3, 3, 2). Alternatively one could choose

3 2 x1 x2 x3 ( x1 x2 x3 + x1 x2 x4 + x1 x3 x4 + x2 x3 x4)×

( x1 x2 x3 + x1 x2 x4 + x1 x3 x4 + x2 x3 x4)(x1 x2 + x1 x3 + x1 x4 + x2 x3 + x2 x4 + x3 x4) to obtain the sequence of monomials

3 2 4 3 2 5 4 3 6 4 3 x1 x2 x3 → x1 x2 x3 → x1 x2 x3 → x1 x2 x3 x4 which is xδ → xδ+(1,1,1,0) → xδ+(2,2,2,0) → xδ+(3,2,2,1)

5.4 Schur Functions and the Frobenius Character Formula 115

corresponding to the Young tableau

1 1 1 3 2 2 2 3

Indeed, there are exactly two semistandard Young tableaux of λ0 = (4, 3, 1) and λ+δ type µ = (3, 3, 2), and the coeﬃcient of a in aδeµ(x1,..., xm) is 2.

The following theorem is another consequence of the proof of Theorem 5.4.5:

n Theorem 5.4.10. If f ∈ ΛK has an expansion X f = cλ sλ for cλ ∈ K λ`n

λ+δ then cλ is the coeﬃcient of x in f (x1,..., xm)aδ.

Proof. By Cauchy’s definition of Schur functions (Theorem 5.4.5), the symmetric function f satisfies X f (x1,..., xm)aδ = cλ f (x1,..., xm)aλ+δ. λ The λth summand on the right-hand side has only one decreasing monomial, λ+δ λ+δ namely x . Therefore, the coefficient of x on the right-hand side is cλ.

Since the character values of symmetric groups describe the expansion of power sum symmetric functions in terms of Schur functions (5.20) X pµ = trace(wµ; Vλ)sλ, λ a consequence of Theorem 5.4.10 is the following beautiful formula of Frobenius for the characters of symmetric groups:

Theorem 5.4.11 (Frobenius character formula). For any two partitions λ and µ of n, the character of the simple representation Vλ of S n evaluated at an element wµ ∈ S n with cycle decomposition µ is given by

λ+δ trace(wµ; Vλ) = coeﬃcient of x in pµaδ. (5.29)

Example 5.4.12. We illustrate this rule with the trivial example of S 2

2 p(1,1)(x1, x2)aδ = (x1 + x2) (x1 − x2) 3 2 2 3 = x1 + x1 x2 − x1 x2 − x2.

116 Symmetric Functions

3 (2)+δ 2 (1,1)+δ Both decreasing monomials x1 = x and x1 x2 = x have coeﬃcient 1, reﬂecting the fact that all the irreducible characters of S 2 are one dimensional. On the other hand,

2 2 p(2)(x1, x2)aδ = (x1 + x2)(x1 − x2) 3 2 2 3 = x1 − x1 x2 + x1 x2 − x2 3 (2)+δ 2 (1,1)+δ with decreasing monomials x1 = x and x1 x2 = x having coeﬃcients +1 and −1, respectively, which give the values of the irreducible characters of S 2 on the non-trivial element.

Its conceptual utility is illustrated by the following exercise:

[3] Exercise 5.4.13. Show that

 k k (−1) if λ = (n − k, 1 ) for some 0 ≤ k ≤ n − 1, trace(w(n); Vλ) =  0 otherwise. Another beautiful application of Theorem 5.4.10 is a simple proof of Pieri’s rule for Schur functions:

Theorem 5.4.14 (Pieri’s rule). For every partition µ of n, X sµ s(1) = sλ. λ∈µ+ Here, µ+ denotes the set of partitions of n+1 whose Young diagrams are obtained by adding a box to the Young diagram of µ.

Before we prove Pieri’s rule, we clarify the deﬁnition of µ+ with an example and an exercise:

Example 5.4.15. Take µ = (6, 5, 3, 3) with Young diagram

If a new box is added to any row of this diagram which is strictly shorter than the row above it, the result is again a Young diagram. Also a new box can be added just below the leftmost box of the bottom row. Thus, the partitions in µ+ are those with Young diagrams

× , , and , × ×

× the new boxes being identiﬁed by an ‘×’.

5.5 Frobenius’ Characteristic Function 117

[2] Exercise 5.4.16. Show that the cardinality of µ+ is one more than the number of distinct parts in µ. Proof of Pieri’s rule. We may work with specializations to m = n + 1 variables. Note that

s(1)(x1,..., xm) = x1 + ··· + xm.

λ+δ By Theorem 5.4.10, the coeﬃcient of sλ in sµ s(1) is the coeﬃcient of x in µ+δ aµ+δ(x1 + ... + xm). The decreasing monomials in aµ+δ(x1 + ... + xm) are x xi, where i is such that if a box is added to the ith row of the Young diagram of µ, it remains a Young diagram. But these are just the decreasing monomials xλ+δ, where λ ∈ µ+.

5.5 Frobenius’ Characteristic Function

The Murnaghan–Nakayama rule (5.22) tells us that the coeﬃcient of sλ when pµ is expanded in Schur functions is the character value of Vλ at an element wµ of cycle type µ. For convenience, introduce a bilinear form on ΛK,n for which the Schur functions {sλ | λ ` n} form an orthonormal basis  1 if λ = µ hsλ, sµi =  0 otherwise.

For any f ∈ ΛK,n and partition λ of n, h f, sλi is the coeﬃcient of f when it is expanded in terms of Schur functions. The Murnaghan–Nakayama rule can be rewritten as

hpµ, sλi = trace(wµ; Vλ). (5.30)

Let 1µ ∈ K[S n] denote the function which is 1 on elements of S n with cycle class type µ and 0 on all other elements of S n. Then (5.30) can be rewritten as n! h1µ, χλiS n = hpµ, sλi, cλ

where cλ denotes the number of elements of S n with cycle type λ. Note that n!/cλ = zλ, where zλ is the cardinality of the centralizer of any element with cycle type λ in S n (see Exercise 2.2.24 for the calculation of zλ). This suggests the following device for using symmetric functions to compute the multiplicity of Vλ in a representation: given a class function f ∈ K[S n], deﬁne a symmetric function X 1 ch ( f ) = f (w )p . (5.31) n z µ µ µ`n µ

118 Symmetric Functions

Then X 1 hch ( f ), s i = f (w )hp , s i n λ z µ µ λ µ`n µ 1 X = c f (w )χ (w ) n! µ µ λ µ µ`n 1 X = f (w)χ (w) n! λ w∈S n

= h f, χλiS n .

In the last step, we used the fact that any element of S n is conjugate to its inverse −1 (Exercise 2.2.16), so χλ(w) = χλ(w ).

Theorem 5.5.1. For any class function f ∈ K[S n],

h f, χλiS n = hchn( f ), sλi. (5.32)

[1] Exercise 5.5.2. Show that chn(χλ) = sλ for every partition λ of n. Hint: Calculate hchn(χλ), sµi.

[1] Exercise 5.5.3. Show that for all class functions f, g ∈ K[S n],

hchn( f ), chn(g)i = h f, giS n .

Theorem 5.5.4. Let RK(S n) denote the space of class functions on S n. The linear map chn : RK(S n) → ΛK,n deﬁned by (5.31) is a linear isomorphism such that, for λ+δ every function f ∈ RK(S n), the coeﬃcient of x in chn( f )aδ is h f, χλiS n .

It is convenient to think of chn as an operator on representations: for any representation (ρ, V) of S n, deﬁne

chn(ρ, V) = chn(w 7→ trace(ρ(w); V)).

We will abbreviate chn(ρ, V) to ρ(V) when the action ρ is clear from the context. Thus, for example, the fact that chn(χλ) = sλ can be represented as

chn(Vλ) = sλ.

[2] Exercise 5.5.5. For every partition λ of n, show that

1. chn(K[Xλ]) = hλ.

2. chn(K[Xλ] ⊗ ) = eλ.

[3] Exercise 5.5.6. For which partitions λ does there exist a representation Uλ such that ch(Uλ) = pλ?

5.6 Branching Rules 119 5.6 Branching Rules

In this section, we will use the Murnaghan–Nakayama rule (5.22) and Theorem 5.5.1 on Frobenius’s characteristic function to see how a representation of S n decomposes into irreducibles when restricted to S n−1. Given a partition λ of n, let χλ ↓ denote the restriction of its character χλ to S n−1. This is nothing but the character of the representation obtained by viewing Vλ as a representation of S n−1. Here, we think of S n−1 as the subgroup of S n which ﬁxes n. For any partition µ = (µ1, . . . , µm) of n−1, an element wµ ∈ S n−1 with cycle type µ, when viewed as an element of S n has cycle type (µ1, . . . , µm, 1) [for brevity, we will denote this partition by (µ, 1)]. Therefore, X 1 ch (χ ↓) = χ (w )p . n−1 λ z λ (µ,1) µ µ`(n−1) µ

By the Murnaghan–Nakayama rule, χλ(w(µ,1)) is the coefficient of sλ in the expansion of p(µ,1) in terms of Schur functions. In terms of the bilinear form defined in Section 5.5, this coefficient is hp(µ,1), sλi. Note that p(µ,1) = pµ p(1) = pµ s(1). Therefore, for any partition η of n − 1, * + X 1 hch (χ ↓), s i = hp s , s ip , s n−1 λ η z µ (1) λ µ η µ`(n−1) µ X 1 = hp s , s ihp , s i z µ (1) λ µ η µ`(n−1) µ * + X 1 = hp , s ip s , s z µ η µ (1) λ µ`(n−1) µ * + X 1 = χ (w )p s , s z η µ µ (1) λ µ`(n−1) µ

= hchn(χη)s(1), sλi

= hsη s(1), sλi.

By Pieri’s rule,

 + 1 if λ ∈ η hsη s(1), sλi =  0 otherwise.

Let λ− denote the set of partitions whose Young diagrams can be obtained from the Young diagram of λ by removing one box. Then clearly,

λ ∈ η+ if and only if η ∈ λ−.

120 Symmetric Functions

We therefore have, for every partition η of n − 1,

 − 1 if η ∈ λ hch(χλ ↓), sηi =  0 otherwise. We have proved the following theorem: Theorem 5.6.1 (Branching rules). The restriction of the simple representation − Vλ of S n to S n−1 is a direct sum over all η ∈ λ of the simple representations Vη of S n−1. In particular, the restriction of Vλ to S n−1 is multiplicity free. An immediate consequence of this theorem is a recursive method for computing the dimension of Vλ: Theorem 5.6.2. For every partition λ of n, X dim Vλ = Vµ. µ∈λ− Example 5.6.3. We have

dim V(2,2,1) = dim V(2,2) + dim V(2,1,1)

= 2 dim V(2,1) + dim V(1,1,1)

= 2 dim V(2) + 3 dim V(1,1)

= 5 dim V(1) = 5.

[2] Exercise 5.6.4. Let fλ denote the number of standard Young tableaux of shape λ (see Deﬁnition 3.2.10). Show that, for every partition λ, X fλ = fµ. µ∈λ−

As a consequence, recover the result that dim Vλ = fλ (see Theorem 3.3.2).

5.7 Littlewood–Richardson Coeﬃcients

Definition 5.7.1 (Littlewood–Richardson coefficient). For partitions µ of m, ν of λ n and λ of m + n, the Littlewood–Richardson coefficient cµ,ν is defined by λ cµν = hsλ, sµ sνi. λ Example 5.7.2. The Littlewood–Richardson coefficient cµ,(1) is determined by Pieri’s rule:   ∈ + λ 1 if λ µ cµ,(1) =  0 otherwise.

5.7 Littlewood–Richardson Coeﬃcients 121

In this section, we will use Frobenius’s characteristic function to give a λ representation-theoretic interpretation of cµν. The argument is a generalization of the trick used in Section 5.6 to convert hch(χλ) ↓, sηi into hsλ, sη s(1)i. It will follow that this coefficient is a non-negative integer. Yet another interpretation of λ cµν will be given in Chapter 6 (in terms of polynomial representations of general linear groups; see Theorem 6.5.8).4 Think of S m × S n as a subgroup of S m+n, where S m permutes the first m elements of {1,..., m + n} and S n permutes the last n elements. We will denote this inclusion map S m × S n → S m+n by φ. The external tensor product Vµ Vν (see Definition 1.6.5) is an irreducible representation of S m × S n. Theorem 5.7.3. For non-negative integers m and n, let λ be a partition of m + n, µ be a partition of m and ν be a partition of n. The multiplicity of Vµ Vν in the λ restriction of Vλ to S m × S n is the Littlewood–Richardson coefficient cµν.

For the proof, we need to extend Frobenius’s characteristic function to R(S m × S n), the space of class functions on S m × S n. [1] Exercise 5.7.4. Let G and H be ﬁnite groups with spaces of class functions R(G) and R(H), respectively. Show that R(G × H) = R(G) ⊗ R(H), and for f1, f2 ∈ R(G), g1, g2 ∈ R(H),

h f1 ⊗ g1, f2 ⊗ g1iG×H = h f1, f2iGhg1, g2iH, where f ⊗ g is the function deﬁned by ( f ⊗ g)(x, y) = f (x)g(y).

4 Theorem 5.7.3 tells us that the Littlewood–Richardson coefficients are non-negative integers. The Littlewood–Richardson rule is the answer to the question: What do Littlewood–Richardson coefficients count? Before we can state the rule, we need some definitions. For two partitions λ and µ, we say that µ ⊂ λ if µi ≤ λi for all i. Let µ ⊂ λ.A skew diagram of shape λ/µ is the set of boxes left behind when the boxes in the Young diagram of µ are removed from the Young diagram of λ. For instance, the skew diagram of shape (4, 4, 2, 1)/(2, 1) is given by .

A semistandard Young tableau of shape λ/µ is a Young diagram of shape λ/µ with all the boxes ﬁlled in by integers in such a way that the integers increase weakly along rows and strictly along columns. The type of such a Young tableau is the sequence ν = (ν1, ν2,... ), where νi is the number of times that i occurs in the tableau truncated after the last non-zero term. Thus, for example, T = 1 1 is a semistandard Young tableau of shape (4, 4, 2, 1)/(2, 1). 1 2 2 1 3 2 The reverse reading word of a Young tableau is the string formed by reading its rows right to left, starting at the top row and moving down to the bottom row. For example, the reverse reading word of the tableau T from the previous paragraph is 11221312. Finally, a lattice permutation is deﬁned to be a sequence a1a2 ··· an such that in any initial part a1a2 ··· a j, the number of i’s is at least as large as the number of (i + 1)’s. For example, 11221312 is a lattice permutation. λ We can now state the Littlewood–Richardson rule: cµν is equal to the number of semistandard Young tableaux of shape λ/µ and type ν whose reverse reading word is a lattice permutation. For a simple and elegant proof of this result, we refer the reader to Loehr [19], where it is proved along with many other interesting Schur function identities.

122 Symmetric Functions

Deﬁne the characteristic function chm,n : R(S m × S n) → ΛK,m ⊗ ΛK,n as the tensor chm ⊗ chn of the characteristic maps chm : R(S m) → ΛK,m and chn : R(S n) → ΛK,n. If f ∈ R(S m × S n), then X X 1 1 ch ( f ) = f (φ(w , w ))p ⊗ p . m,n z z θ η θ η θ η θ η

On ΛK,m ⊗ ΛK,n, deﬁne an bilinear form by

h f1 ⊗ f2, g1 ⊗ g2i = h f1, g1ih f2, g2i for f1, f2 ∈ ΛK,m, g1, g2 ∈ ΛK,n.

Lemma 5.7.5. For any f ∈ R(S m × S n), and for any partitions µ and ν of m and n, respectively,

h f, χµ ⊗ χνiS m+n = hchm,n( f ), sµ ⊗ sνi.

Proof. It is easy to see that the lemma holds for f of the form f1 ⊗ f2, where f1 ∈ R(S m) and F2 ∈ R(S n), and therefore, by linearity, it holds for all f ∈ R(S m × S n).

Proof of Theorem 5.7.3. The multiplicity of Vµ Vν in the restriction of Vλ to S m × S n is given by

S m+n hRes χ , χ ⊗ χ i × . S m×S n λ µ ν S m S n For partitions θ and η of m and n, respectively, by the Murnaghan–Nakayama rule,

χλ(φ(wθ, wη)) = hpθ pη, sλi.

Therefore, the characteristic function of the restriction of χλ to S m ×S n is given by X X 1 1 ch (χ ) = hp p , s ip ⊗ p . m,n λ z z θ η λ θ η θ η θ η By Lemma 5.7.5, X X S m+n 1 1 hRes χλ, χµ ⊗ χνiS ×S = hpθ pη, sλihpθ, sµihpη, sνi S m×S n m n z z θ η θ η X 1 X 1 = hp , s i hp , s ihp p , s i z θ µ z η ν θ η λ θ θ η η X 1 = hp , s ihp ch (χ ), s i z θ µ θ n ν λ θ θ

= hchm(χµ)chn(χν), sλi

= hsµ sν, sλi

as required.

5.7 Littlewood–Richardson Coeﬃcients 123

Consider the vector space M∞ RK = R(S n). n=0

Putting together the characteristic functions ch : R(S n) → ΛK,n for all n, we get a linear isomorphism

ch : RK → ΛK

The graded algebra structure on ΛK can be transferred back to RK to deﬁne an associative product: given f ∈ R(S m), g ∈ R(S n), deﬁne f ∗ g ∈ R(S m+n) by requiring

chm+n( f ∗ g) = chm( f )chn(g) (5.33)

turning RK into a graded algebra. For partitions µ and ν of m and n, respectively, we have

λ hχµ ∗ χν, χλiS m+n = cµν for every partition λ of m + n.

It follows that χµ ∗ χν is the character of a representation of S m+n, namely the representation λ M ⊕cµν Vµ ∗ Vν := Vλ . λ`(m+n)

Theorem 5.7.6. Suppose U and V are representations of S m and S n, respectively. Then there exists a representation U ∗ V of S m+n with chm+n(U ∗ V) = chm(U)chn(V).

L ⊕mλ L ⊕nλ Proof. Suppose that U = λ Vλ and V = λ Vλ are decompositions of U and V into simple representations with multiplicity. Then we may take M M ⊕mµnν U ∗ V = (Vµ ∗ Vν) , µ ν because X X chm+n(U ∗ V) = mµnνchm+n(Vµ ∗ Vν) µ ν X X = mµnνchm(Vµ)chn(Vν) µ ν X X = mµchm(Vµ) mνchn(Vν) µ ν

= chm(U)chn(V)

as required.

124 Symmetric Functions

There are some classes of representations for which the ∗ representation is easy to compute. Given partitions µ = (µ1, . . . , µr) and ν = (ν1, . . . , νs), let (µ, ν) denote the partition obtained by reordering the sequence (µ1, . . . , µr, ν1, . . . , νs) in weakly decreasing order. With this notation, we have hµhν = h(µ,ν) and eµeν = e(µ,ν). Therefore, by Exercise 5.5.5, we have

K[Xµ] ∗ K[Xν] = K[X(µ,ν)] and (5.34)

(K[Xµ] ⊗ ) ∗ (K[Xν] ⊗ ) = K[X(µ,ν)] ⊗ . (5.35)

5.8 The Hook–Length Formula

The Frobenius character formula (Theorem 5.4.11) gives, as a special case, the dimension of the simple representation Vλ of S n:

Lemma 5.8.1. For every partition λ of n, the dimension of Vλ is the coeﬃcient λ+δ n of x in (x1 + ... + xn) aδ(x1,..., xn).

We know this coeﬃcient to be the number (usually denoted fλ) of standard Young tableaux of shape λ. We may write

λ+δ n fλ = coeﬃcient of x in (x1 + ... + xn) aδ(x1,..., xn). (5.36) Theorem 5.8.2 (Frobenius dimension formula). For each partition λ of n, n! fλ = Qn aδ(λ1 + n − 1, λ2 + n − 2, . . . , λn). (5.37) i=1(λi + n − i)!

Proof. Expanding out the determinant aδ in the polynomial in (5.36) and using the notation δi = n − i from (5.25) gives X n δw(1) δw(2) δw(n) (x1 + ··· + xn) (w)x1 x2 ··· xn . w∈S n n Expanding out (x1 + ... + xn) using the multinomial theorem gives X X n! α1+δw(1) α2+δw(2) αn+δw(n) (w)x1 x2 ··· xn , α1! ··· αn! α w∈S n

the sum being over all vectors (α1, . . . , αn) with non-negative integer coeﬃcients adding up to n. The coeﬃcient of xλ+δ in this polynomial is therefore X n! f = (w), (5.38) λ α ! ··· α ! α,w 1 n where the sum is over pairs (α, w) such that

αi + δw(i) = λi + δi. (5.39)

5.8 The Hook–Length Formula 125

Qn Multiplying and dividing by the desired denominator i=1(λi + δi)! of the Frobenius dimension formula gives n! X Yn (λ + δ )! f = (w) i i λ Q (λ + δ )! α ! i i i α,w i=1 i n! X Yn = Q (w) (λi + δi)(λi + δi − 1) ··· (λi + δi − δw(i) + 1). i(λi + δi)! w∈S n i=1 In the second step, we used the identity (5.39), which determines α given w. The condition that αi ≥ 0 for each i is automatically taken care of, since one of the factors in the product

(λi + δi)(λi + δi − 1) ··· (λi + δi − δw(i) + 1)

is zero whenever αi = λi + δi − δw(i) < 0. If we deﬁne polynomials

f j(x) = x(x − 1) ··· (x − j + 1), then we have shown that n! X Yn fλ = Qn (w) fδw(i) (λi + δi). (5.40) (λi + n − i)! i=1 w∈S n i=1 The sum in the above expression is clearly the determinant

f − (x ) f − (x ) ··· f (x ) n 1 1 n 2 1 0 1 f (x ) f (x ) ··· f (x ) n−1 2 n−2 2 0 2 . . . . (5.41) . . .. .

fn−1(xn) fn−2(xn) ··· f0(xn)

evaluated at xi = λi + δi for each i. Now, f j is a monic polynomial of degree i for each j (in particular, f0 is the constant polynomial 1). Therefore, by subtracting from each column an appropriate linear combination of columns to the right of it, the determinant (5.41) can be reduced to the Vandermonde determinant. Thus, (5.40) reduces (5.37). Given a box in a Young diagram (see Deﬁnition 3.1.1), its hook length is the number of boxes of the Young diagram that lie either below it or to the right of it plus one (to account for the original box itself). For the Young diagram of shape λ = (6, 5, 3, 3), each box is ﬁlled in with its own hook length as follows: 9 8 7 4 3 1 7 6 5 2 1 4 3 2 3 2 1

126 Symmetric Functions

If the Young diagram has shape (λ1, . . . , λl), all the numbers λ1 + l − 1, λ1 + l − 2, all the way down to 1 appear as hook lengths of boxes in the first row, except for jumps each time that number of boxes in a column decreases. Going from the ( j − 1)st box in the first row to the jth box in the first row, we skip as many numbers as there are i such that λi = j. In effect, there is one missing number for each row of λ; the missing number corresponding to the ith row is λl + l − λi − i. Thus, the missing hook lengths in the first row are

(λ1 + l) − (λl−1 − l − 1), (λ1 + l) − (λl−2 − l − 2),..., (λl + l) − (λ2 − 1).

In a similar manner, hook lengths in the second row are all the numbers from λ2 + l − 1 down to 1, except for the missing numbers

(λ2 + l − 1) − (λl−1 − l − 1), (λ2 + l − 1) − (λl−2 − l − 2),..., (λ2 + l − 1) − (λ3 − 2).

Continuing in this manner, we ﬁnd that the hook lengths in the ith row are the numbers

(λi + l − i) − (λ j + l − j), where j > i.

Adding 0 parts to the end of λ [for example, writing (6, 5, 3, 3) as (6, 5, 3, 3, 0, 0, .., 0)] does not really change the validity of these arguments (the numbers from λ1 +n−1 down to λ1 +l are skipped), so we may assume that l = n. We get

Qn i=1(λi + n − i)! product of hook lengths = Q . i< j[(λi + n − i) − (λ j + n − j)] The denominator is the Vandermonde determinant

aδ(λ1 + n − 1, λ2 + n − 2, . . . , λn).

Thus, the Frobenius dimension formula (5.37) becomes the hook-length formula of Frame, Robinson and Thrall:

Theorem 5.8.3 (Hook-length formula). For each partition λ of n, n! fλ = Q , i, j hi j

where hi j denotes the hook length of the box in the ith row and jth column of the Young diagram for λ, and the product is over all the boxes in the Young diagram of λ.

[2] Exercise 5.8.4. Suppose that λ is a hook, i.e., λ is of the form (m, 1k) for some positive integers m and k. What is the dimension of Vλ?

5.9 The Involution sλ 7→ sλ0 127

[2] Exercise 5.8.5. Suppose that λ is of the form (n − k, k), with k ≤ n − k. We already know (from Exercise 2.5.4 and Exercise 3.3.5) that n!(2n − k + 1) dim V = . λ k!(n − k + 1)! Deduce this from the hook-length formula.

[2] Exercise 5.8.6. Show that, when λ = (m, m−1,..., 1), then Vλ has dimension m+1 ! 2 , (2m − 1)!!(2m − 3)!! ··· 3!!1!! where the double factorial (2k − 1)!! denotes the product of the ﬁrst k odd positive integers (we encountered these in Exercise 2.2.13).

5.9 The Involution sλ 7→ sλ0 We now study the analog of twisting by the sign character (Section 4.4) on sym- n metric functions. Since Schur functions of degree n form a basis of ΛK, there n n exists a unique linear involution ω : ΛK → ΛK such that ω(sλ) = sλ0 for each partition λ of n. We may also view ω as an involution of ΛZ (the space of symmetric functions with integer coeﬃcients), and all the results of this section remain true when ΛK is replaced by ΛZ. In the language of transition matrices, ω(s) = sJ. (5.42)

Lemma 5.9.1. For every partition λ,

ω(hλ) = eλ. Proof. From (5.15) and the deﬁnition of Schur functions (5.17), we have

h = sK. (5.43)

Applying ω to both sides of (5.43) and using its linearity gives

ω(h) = ω(s)K,

which by (5.42) gives ω(h) = sJK,

which by (5.14) and (5.17) gives

ω(h) = e

as claimed.

128 Symmetric Functions

[2] Exercise 5.9.2. Show (without using Lemma 5.9.1) that for any class function f ∈ K[S n],

ω(chn( f )) = chn( f ).

Recall that is the sign function on S n [see (4.1)]. Use this to deduce Lemma 5.9.1 from the deﬁnition of ω. Hint: Use Exercise 5.5.5. Corollary 5.9.3. The involution ω (a priori, just a linear map) is an algebra isomorphism ΛK → ΛK. Proof. The algebra of symmetric functions is the polynomial algebra in the elementary symmetric functions e1, e2,... as well as the complete symmetric functions h1, h2,... (see Corollaries 5.2.5 and 5.2.9). Since ω simply interchanges these sets of free generators of the algebra, it is an isomorphism of algebras.

Deﬁnition 5.9.4 (Sign of a partition). For each partition λ of n, its sign λ is deﬁned to be the value (w) of the sign character of S n evaluated at any permutation whose conjugacy class corresponds to the partition λ. In the parlance of Section 4.6.4, even partitions have sign +1 and odd partitions have sign −1.

[1] Exercise 5.9.5. For a partition λ = (1m1 , 2m2 ,... ),

m2+m4+... λ = (−1) Theorem 5.9.6. For every partition λ,

ω(pλ) = λ pλ. Proof. Let E denote the diagonal matrix whose rows and columns are indexed by partitions of n with diagonal entries

Eλλ = λ. It follows from Theorem 4.4.2, that JX = XE. We have ω(p) = ω(sX) = ω(s)X = sJX = sXE = pE, which is nothing but the identity in the theorem.

5.10 The Jacobi–Trudi Identities 129

[2] Exercise 5.9.7. Let peven(n) and podd(n) denote the number of even and odd permutations of n. Let pself-conjugate(n) denote the number of partitions λ of n which are self-conjugate (see Deﬁnition 4.6.6). Show that

peven(n) − podd(n) = pself-conjugate(n).

Recall that we have already encountered this identity in the context of representations of alternating groups [see (4.14)]. Hint: Compare the trace of ω on Λn when it is computed with respect to the basis of Schur functions with the same trace when it is computed with respect to the basis of power sum symmetric functions.5

5.10 The Jacobi–Trudi Identities

The formulas (5.14) and (5.15) allow us to express elementary symmetric functions and complete symmetric functions in terms of Schur functions. The Jacobi– Trudi identities allow us to invert these relations; to express Schur functions in terms of the elementary symmetric functions and complete symmetric functions

s = det(h ) = det(e 0 ). (5.44) λ λi−i+ j λi −i+ j The determinants are those of m × m matrices, where m is any integer greater than or equal to the number of parts of λ (the partition λ is extended to an m-tuple by appending 0’s on the right if necessary). In these formulas, hr and er are taken to be 0 for negative values of r. Each monomial in the expansion of the determinant is a product of functions of the form hr and is therefore a complete symmetric function of the form hµ for some partition µ. One easily checks that the degree of each of these monomials is n. By the results of the previous section, applying ω to the ﬁrst identity sλ = det(h ) gives the second identity s = det(e 0 ). Therefore, we shall focus λi−i+ j λ λi −i+ j only on the ﬁrst identity.

Example 5.10.1. For n = 3, we get

h3 h4 h5

s(3) = 0 h0 h1 = h(3),

0 0 h0

h2 h3 h4

s(2,1) = h0 h1 h2 = h(2,1) − h(3),

0 0 h0

5 See also Stanley [35, Exercise 1.22(b)] for a diﬀerent approach. For a bijective proof due to van Leeuwen, visit http://math.stackexchange.com/a/102293/10126.

130 Symmetric Functions

h1 h2 h3

s(1,1,1) = h0 h1 h2 = h(1,1,1) − 2h(2,1) + h(3).

0 h0 h1 The most obvious way to try to prove the Jacobi–Trudi identity (following Trudi’s method adapted from Stanley [32]) is to start with the determinant aλ+δ in the numerator of Cauchy’s formula for Schur functions and try to divide out the factors xi − x j of the denominator aδ from it. Two other proofs can be found in Stanley’s book [34, Section 7.16]. In what follows, hn(x1,..., xk) represents the specialization of the nth complete symmetric polynomial to k variables (Deﬁnition 5.3.1). For example, hn(x1) is n just x1. Subtracting the ﬁrst row from the ith row of the determinant

λ m− λ m− λ x 1+ 1 x 2+ 2 ··· x m 1 1 1 xλ1+m−1 xλ2+m−2 ··· xλm 2 2 2 ...... λ1+m−1 λ2+m−2 λm xm xm ... xm

and dividing the ith row of the resulting matrix by (xi − x1) for i > 1 gives (in the agreed-upon notation)

h − (x ) h − (x ) ··· h λ1+m 1 1 λ2+m 2 1 λm n ··· Y hλ1+m−2(x1, x2) hλ2+m−3(x1, x2) hλm−1(x1, x2) a = (x − x ) λ+δ i 1 . . .. . i=2 . . . .

hλ1+m−2(x1, xm) hλ2+m−3(x1, x2) ··· hλm−1(x1, x2)

Here, we have used the fact that hn(xi)−hn(x1) = (xi −x1)hn−1(x1, xi). This identity actually generalizes to

hn(x1,..., xk−1, xi) − hn(x1,..., xk−1, xk) = (xi − xk)hn−1(x1,..., xk, xi) which will be used in later steps. Now subtracting the second row from the ith row for each i > 2 (and using the above identity) gives

Ym Ym aλ+δ = (xi − x1)(x j − x2)× i=2 j=3

hλ +m−1(x1) hλ +m−2(x1) ··· hλ (x1) 1 2 m h − (x , x ) h − (x , x ) ··· h − (x , x ) λ1+m 2 1 2 λ2+m 3 1 2 λm 1 1 2 h (x , x , x ) h (x , x , x ) ··· h (x , x , x ) λ1+m−3 1 2 3 λ2+m−4 1 2 3 λm−2 1 2 3 ......

hλ1+m−3(x1, x2, x3) hλ2+m−4(x1, x2, x3) ··· hλm−2(x1, x2, x3)

5.10 The Jacobi–Trudi Identities 131

Continuing in this manner results in the identity Y aλ+δ = (xi − x j) det(hλ j+m−i− j+1(x1,..., xi)). i> j Reversing the order of the rows and interchanging the rows with the columns gives Y bm/2c aλ+δ = (xi − x j)(−1) det(hλi−i+ j(x1,..., xm− j+1)). i> j Q Dividing by the Vandermonde determinant i< j(xi − x j) [which is the same as m m (2) Q (−1) i> j(x1 − x j)] and using the fact that 2 and bm/2c always have the same parity gives

sλ(x1,..., xm) = det(hλi−i+ j(x1,..., xm− j+1))

by Cauchy’s formula for Schur functions (Theorem 5.4.5). Here, sλ(x1,..., xm) denotes the Schur function specialized to m variables (Deﬁnition 5.3.1). In expanded form, the determinant above is

h (x ,..., x ) ··· h − (x , x ) h − (x ) λ1 1 m λ1+m 2 1 2 λ1+m 1 1 h (x ,..., x ) ··· h (x , x ) h (x ) λ2−1 1 m λ2+m−3 1 2 λ2+m−2 1 ......

hλm−m+1(x1,..., xm) ··· hλm−1(x1, x2) hλm (x1)

Adding x1 times the (m − 1)st column to the last column gives

h (x ,..., x ) ··· h − (x , x ) h − (x , x ) λ1 1 m λ1+m 2 1 2 λ1+m 1 1 2 h (x ,..., x ) ··· h (x , x ) h (x , x ) λ2−1 1 m λ2+m−3 1 2 λ2+m−2 1 2 ......

hλm−m+1(x1,..., xm) ··· hλm−1(x1, x2) hλm (x1, x2)

Here, we have used the fact that x2hn−1(x1, x2) + hn(x1) = hn(x1, x2). This generalizes to the identity

xkhn−1(x1,..., xk) + hn(x1,..., xk−1) = hn(x1,..., xk),

which will be used in later steps. Now adding x3 times the (m − 2)nd column to each of the last two columns (and using the above identity) allows us to write power sum symmetric functions in the three variables x1, x2 and x3 in the last three columns. Continuing in this manner (adding xi times the (m − i + 1)st column to each column to its right as i decreases from m − 1 to 1) allows us to write each of the terms as a complete symmetric function in x1,..., xm. From this, the Jacobi–Trudi identity follows for specializations to m variables. Because of the triangularity of the transition between Schur functions and complete symmetric functions, the expansion of (the unspecialized) sλ, when λ has at most n parts, in

132 Symmetric Functions

terms of complete symmetric functions, involves only those hµ for which µ has at most m parts. By the linear independence of such complete symmetric functions (see Theorem 5.3.8), the Jacobi–Trudi identity (5.44) in inﬁnitely many variables also holds. An application of Theorem (5.44) is an expression of a simple representation of S n as a Z-linear combination of the permutation representations K[Xλ] (which can be computed by counting ﬁxed points). Let σλ denote the character of K[Xλ] for each partition λ. We have:

Theorem 5.10.2. For every partition λ of n, the simple character χλ of S n is given by

χλ = det(σ(λi−i+ j)). (5.45)

In the determinant on the right-hand side in the identity above, product should be interpreted as the ∗-product deﬁned by the identity (5.33). By Exercise (5.34), σµ ∗ σν = σ(µ,ν), so the right-hand side is really an alternating sum of σµ’s. For example, similar to Example 5.10.1, we have

χ(3) = σ(3),

χ(2,1) = σ(2,1) − σ(3),

χ(1,1,1) = σ(1,1,1) − 2σ(2,1) + σ(3).

Proof of Theorem 5.10.2. We know that two class functions on S n are equal if and only if their characteristics are equal (Theorem 5.5.4). Taking characteristics of both sides of (5.45) gives the Jacobi–Trudi identity, so the identity (5.45) also holds.

5.11 The Recursive Murnaghan–Nakayama Formula

Theorem 5.10.2 expresses the values of the irreducible characters of the symmetric groups in terms of conceptually simpler quantities, the characters of the permutation representations K[Xλ] (which just count fixed points in Xλ). In prac- tice, this is a cumbersome process. However, a slight modification of this formula leads to an extremely efficient rule for computing the irreducible characters of symmetric groups. This is the recursive Murnaghan–Nakayama rule. Recall that for each box in a Young diagram, the hook length is the number of boxes in the Young diagram that lie directly below or directly to its right plus one. The hook length of the box in the ith row and jth column is denoted hi j. Let rimi j denote the set of boxes in positions (k, l) such that k ≥ i, l ≥ j and the Young diagram does not have a box in position (i + 1, j + 1).

5.11 The Recursive Murnaghan–Nakayama Formula 133

For example, for λ = (6, 5, 3, 3), h2,2 = 6 and rim2,2 consists of the six boxes marked with an ‘×’:

• × × × × × ×

It is easy to see that the number of boxes in rimi j is hi j. Also, if the boxes in rimi j are removed from the Young diagram of a partition λ of n, what remains is the Young diagram of a partition of n − hi j, which we denote by λ − rimi j. Finally, deﬁne the leg-length li j of the box in the ith row and jth column to be the number 0 of boxes that lie directly below it plus one (so li j = λ j − i + 1). In our example with λ = (6, 5, 3, 3), λ − rim2,2 has the diagram

so (6, 5, 3, 3) − rim2,2 = (6, 2, 2, 2, 1). Also, l2,2 = 3. With these deﬁnitions out of the way, we are ready to state the recursive Murnaghan–Nakayama rule.

Theorem 5.11.1. Let λ = (λ1, . . . , λl) and µ = (µ1, . . . , µm) be partitions of n. For any i ∈ {1,..., m}, let µˆi denote the partition obtained from µ by removing its ith part. Then X li j−1 χλ(wµ) = (−1) χλ−rimi j (wµˆi ). hi j=µi

As usual, wµ is the element with cycle decomposition

(1 ··· µ1)(µ1 + 1 ··· µ1 + µ2) ··· (µ1 + ··· + µm−1 + 1 ··· n). Example 5.11.2. We have

χ(3,3,3)(w(4,3,2)) = χ(2,2,1)(w(3,2)) − χ(3,2)(w(3,2))

= −χ(2)(w(2)) + χ(1,1)(w(2)) = −1 − 1 = −2.

[1] Exercise 5.11.3. Use the recursive Murnaghan–Nakayama rule to compute

1. χ(8,17)(w(7,5,3)). 2. χ(2,2,1,1)(w(2,2,2)).

134 Symmetric Functions

The ﬁrst step towards proving the recursive Murnaghan–Nakayama formula is to obtain recursive rule for the character σλ of the permutation representation K[Xλ] of Section 2.3:

Lemma 5.11.4. For any weak compositions λ = (λ1, . . . , λl) and µ = (µ1, . . . , µm) of n, X σλ(wµ) = trace(wµˆ j ; K[Xλ−µ jei ]). λi>µ j

Proof. The character value σλ(wµ) = trace(wµ; K[Xλ]) is the number of points in Xλ that are fixed by wµ. An element (S 1,..., S l) ∈ Xλ (which is an ordered partition of n) is fixed by wµ if and only if each of the subsets S 1,..., S l is a union of cycles of wµ. The number of such partitions such that the jth cycle of wµ (which has length µ j) occurs in S i is the same as the number of fixed points for a permutation with cycle decompositionµ ˆ j acting on ordered partitions of shape

(λ1, . . . , λi−1, λi − µ j, λi+1, . . . , λl), which is trace(wµˆ j ; K[Xλ−µ jei ]) (here, ei denotes the ith coordinate vector). Summing over all i gives the formula in the lemma. Another tool in the proof of the recursive Murnaghan–Nakayama formula is a generalization of the character χλ of Vλ to the case where λ is not a partition. For any tuple (α1, . . . , αl) of integers, deﬁne

χα = det(σαi−i+ j),

interpreting (as usual) σk as 0 when k is negative. When α is a partition, Theorem 5.10.2 tells us that this notation is consistent with our convention of denoting the character of Vλ by χλ. When α is not a partition, we can try to transform the determinant representing χα using row transformations. The ﬁrst basic observation is

χ(α1,...,αl) = −χ(α1,...,αi−1,αi+1−1,αi+1,αi+1,...,αl), (5.46) which holds because the determinant representing the right-hand side is obtained from the determinant representing the left-hand side by interchanging the ith row and the (i + 1)st row.

Example 5.11.5. In many cases, we can use (5.46) to transform χα to χλ, where λ is a partition

χ(−1,3,3) = −χ(2,0,3) = χ(2,2,1). The second observation is

If αi+1 = αi + 1 for any i, then χα = 0, (5.47)

because the ith and (i + 1)st rows of the determinant representing χα are the same.

5.11 The Recursive Murnaghan–Nakayama Formula 135

Example 5.11.6. In many cases, we can use (5.47) together with (5.46) to show that χα = 0:

χ(1,3,3) = χ(2,2,3) = 0.

Proof of the recursive Murnaghan–Nakayama Formula. Suppose that λ and µ are partitions of n. By Lemma 5.11.4, we have

Xn χλ(wµ) = χλ−µ jei (wµˆ j ). i=1

Now consider the entries of the ﬁrst column of the determinant representing

χλ−µ jei . They are of the form σk, where k takes the values

λ1, λ2 − 1, . . . , λi−1 − (i − 1) + 1, λi − i + 1 − µ j, λi+1 − (i + 1) + 1, . . . , λl − l + 1.

If two of these indices coincide, then the corresponding rows are the same, and

χλ−µ jei = 0. If, on the other hand, these indices are distinct, they can be rearranged into strictly decreasing order by moving the ith index (the only one that is out of order) down until it ﬁts between two other entries so that

λk − k + 1 > λi − i + 1 − µ j > λk+1 − (k + 1) + 1,

or to the last place

λl − l + 1 > λi − i + 1 − µ j. (5.48)

By moving the ith row past the rows i + 1, i + 2,..., k (where k is taken to be l in case (5.48) holds), the integer vector λ − µ jei becomes the partition

λ1, . . . , λi − 1, λi+1 − 1, λi+2 − 1, λk − 1, λi − µ j + k − i, λk+1, . . . , λl.

At the level of Young diagrams, this means that the boxes in each row [starting with the ith, up to the (k − 1)st] are replaced by the number of boxes in the row below it less one. Totally, µ1 boxes are removed. Therefore, the eﬀect of this operation is the removal of the rim of a hook of length µ j, which has boxes in rows i to k. The number of times rows are swapped is k − i, which is one less than the leg-length of the hook.

Example 5.11.7. Taking λ = (6, 5, 3, 3), i = 2 and µ j = 6, we start with α = λ − µ je2 = (6, −1, 3, 3). Applying a sequence of moves of type (5.46) leads to the sequence

(6, −1, 3, 3) → (6, 2, 0, 3) → (6, 2, 2, 1).

136 Symmetric Functions

The boxes in the rim of the hook of length six that was removed by these opera- tions are marked by ‘×’ in the Young diagram below:

. × × × × × ×

For any hook in λ of length µ j, the partition obtained from λ by removing a hook of length µ j whose top-most boxes are in the ith row is obtained by such a sequence of moves from λ − µ jei. This establishes a correspondence between

the terms χλ−µ jei which do not cancel out and the hooks of length µ j, thereby establishing the recursive Murnaghan–Nakayama formula.

m [1] Exercise 5.11.8. Show that χ(m+1,1m)(w(2m+1)) = (−1) .

[2] Exercise 5.11.9. For each positive integer n, show that

bn/2c χ(nn)(w(2n−1,2n−3,...,3,1)) = (−1) .

Here, (nn) denotes the partition of n2 all of whose parts are equal to n.

5.12 Character Values of Alternating Groups

The recursive Murnaghan–Nakayama rule gives enough information to be able to complete a project that was begun in Section 4.6. In that section, we had found that when K is an algebraically closed ﬁeld of characteristic greater than n, the irreducible characters of the alternating group An consist of

• Restrictions of the irreducible characters χλ of S n to An for partitions λ of n which are not self-conjugate + − • Two irreducible characters χλ and χλ for each self-conjugate partition λ of n. − −1 + Moreover, for any odd permutation v, χλ (vwv ) = χλ (w).

As for the conjugacy classes of An, we found

• For every even partition with distinct odd parts, the set of permutations of cycle + type λ splits into two conjugacy classes in An of equal cardinality. If wλ lies in − + −1 one of these classes, then wλ = vwλ v lies in the other for any odd permutation v. • For all other even partitions, the set of permutations of cycle type λ form a single conjugacy class.

5.12 Character Values of Alternating Groups 137

In this section, we will assume that K is the ﬁeld C of complex numbers. The approach here is based on James and Kerber [12, Section 2.5]. For brevity, let SP(n) denote the set of partitions of n which are self-conjugate. Let DOP(n) denote the set of partitions of n which have distinct odd parts. The ‘folding’ algorithm (see Lemma 4.6.16) deﬁnes a bijection φ : DOP(n)→˜ SP(n). (5.49)

[1] Exercise 5.12.1. Given a partition µ with distinct odd parts (2m1 + 1, 2m2 + 1,... ), show that P i mi χφ(µ)(wµ) = (−1) . Hint: Use the method of Exercise 5.11.8 recursively. [2] Exercise 5.12.2. Suppose that θ and µ are partitions with distinct odd parts and θ comes before µ in reverse lexicographic order. Show that χφ(µ)(wθ) = 0.

Remark 5.12.3. For n < 15, it happens to be true that χφ(µ)(wθ) = 0 unless µ = θ for all µ, θ ∈ DOP(n). However, for n = 15, the values of χφ(µ)(wθ) are as given in Table 5.2 (the µ’s are the row indices and the θ’s are the column indices).

(15) (11, 3, 1) (9, 5, 1) (7, 5, 3) (15) −1 0 0 2 (11, 3, 1) 0 1 0 0 (9, 5, 1) 0 0 1 2 (7, 5, 3) 0 0 0 1

Table 5.2 Partial character table of S 15

Exercises 5.12.1 and 5.12.2 are the last two ingredients that are needed to complete a description of the character tables of all the alternating groups. As in Section 4.6, we consider the absolute diﬀerence characters

+ − + − + − 2 ∆λ = (χλ − χλ )(χλ − χλ ) = |χλ − χλ |

for each λ ∈ SP(n). Then ∆λ is a real-valued character of An supported on permu- + − − + + + − − tations with cycle type in DOP(n). Since χλ (wµ ) = χλ (wµ ) and χλ (wµ ) = χλ (wµ ) for each w ∈ DOP(n), we have + − ∆λ(wµ ) = ∆λ(wµ ) (5.50) for each µ ∈ DOP(n). Thus, ∆λ can be viewed as a class function on S n which is supported on An. As in Section 4.6, one may deduce that

h∆λ, χηiS n ∈ Z (5.51) for all η ∈ SP(n).

138 Symmetric Functions

Lemma 5.12.4. For all λ ∈ SP(n) and µ ∈ DOP(n),

cµ ∆ (w ) is a non-negative integer. n! λ µ

Here cµ denotes the number of permutations of cycle type µ. Proof. Let µ be the maximal element of DOP(n) in reverse lexicographic order. By Exercise 5.12.2, χφ(µ)(wθ) = 0 for all θ , µ. Therefore, 1 X h∆ , χ i = c ∆ (w )χ (w )| λ φ(µ) n! θ λ θ φ(µ) θ θ∈DOP(n) cµ = ± ∆ (w ), n! λ µ

since χφ(µ)(wµ) = ±1 by Exercise 5.12.1. Therefore, by (5.51), cµ∆λ(wµ)/n! is an integer for every λ ∈ SP(n). Now proceed by induction. Suppose that for every θ which comes before µ in reverse lexicographic order. We have shown that cθ∆λ(wθ)/n! is an integer for all λ ∈ SP(n). Then by (5.51), 1 X h∆ , χ i = c ∆ (w )χ (w ) λ φ(µ) n! θ λ θ φ(µ) θ θ∈DOP(n) cµ X c = ± ∆ (w ) + θ ∆ (w )χ (w ) n! λ µ n! λ θ φ(µ) θ θ,µ is an integer. By Exercise 5.12.2, the terms in the last sum above are non-zero only for those θ which come after µ in reverse lexicographic order. The induction hypothesis tells us that the corresponding values of cθ∆λ(wθ)/n! are all integers, and of course, we know that the character values χφ(µ)(wθ) are all integers. It follows that cµ∆λ(wµ)/n! is also an integer. Its non-negativity follows directly from the deﬁnition of ∆λ as a complex absolute value. We also know from Schur’s orthogonality relation that

+ − + − hχλ − χλ , χλ − χλ iAn = 2. Expanding out the left-hand side from deﬁnition gives 1 X c ∆ (w ) = 2, |A | µ λ µ n µ`n or equivalently X cµ ∆ (w ) = 1. n! λ µ µ`n If a sum of non-negative integers is 1, then exactly one of them can be equal to 1 and the others must all be 0. Therefore, for each λ ∈ SP(n), there exists a unique

5.12 Character Values of Alternating Groups 139

cµ µ ∈ DOP(n) such that n! ∆λ(wµ) = 1 and ∆λ(wθ) = 0 for all other θ ∈ DOP(n) which are different from µ. We claim that this µ is the unique partition distinct odd parts for which φ(µ) = λ. For if it were not, then we would have ∆φ(µ)(wµ) = 0, and + − + − so χφ(µ)(wµ) − χφ(µ)(wµ) = 0. Since χφ(µ)(wµ) + χφ(µ)(wµ) = ±1 (Exercise 5.12.1), + 1 it would follow that χφ(µ)(wµ) = ± 2 , which is not possible, since character values which are rational must be integers (see Theorem 1.8.7). We may therefore conclude that for every λ ∈ SP(n),  cµ 1 if µ ∈ DOP(n) and satisfies φ(µ) = λ, ∆λ(wµ) =  n! 0 otherwise. As a consequence, for any λ ∈ SP(n), 1 χ+(w ) = χ−(w ) = χ (w ) λ µ λ µ 2 λ µ unless µ ∈ DOP(n) and φ(µ) = λ. For each µ ∈ DOP(n), say µ = (2m1 + 1, 2m2 + 1,... ), define P mi µ = (−1) . (5.52)

We know from Exercise 5.12.1 that µ is the character value χφ(µ)(wµ). So + − χφ(µ)(wµ) + χφ(µ)(wµ) = µ. (5.53)

We also know from Exercise 4.6.17 that wµ is conjugate in An to its own inverse if and only if µ = +1. So

 + χ (wµ) if µ = +1, χ+ (w ) = χ+ (w−1) =  φ(µ) (5.54) φ(µ) µ φ(µ) µ  − χφ(µ)(wµ) if µ = −1.

The fact that ∆φ(µ)(wµ) = cµ/n! leads to s + − n! |χφ(µ)(wµ) − χφ(µ)(wµ)| = . (5.55) cµ The conditions (5.53), (5.54) and (5.55) uniquely (up to interchange) determine ± χφ(µ)(wµ) as  s  1  n! ±  ± µ  χφ(µ)(wµ) = µ  . (5.56) 2  cµ 

Theorem 5.12.5. For every partition λ of n such that λ , λ0, the irreducible character χλ of S n restricts to an irreducible character of An. For every partition 0 + − λ of n such that λ = λ , there exists a pair of irreducible characters χλ and χλ which satisfy + − χλ (w) = χλ (w) = χλ(w)/2

140 Symmetric Functions

unless the cycle type of w is a partition µ with distinct odd parts whose folding φ(µ) [see (5.49)] is λ. If the cycle type of w is a partition µ with distinct odd parts such that φ(µ) = λ, then  s  1  n! ± +  µ  χλ (wµ ) = µ ±  , 2  cµ 

± − ∓ + and χλ (wµ ) = χλ (wµ ). Here, µ is as deﬁned by (5.52).

[2] Exercise 5.12.6. Compute the following values of characters of A13: ± ± χ(4,4,3,2)(w(7,5,1)).

6 Representations of General Linear Groups

6.1 Polynomial Representations

We assume that K is an infinite field, so that there is no difference between formal polynomials and the functions that they give rise to.

Deﬁnition 6.1.1 (Polynomial representation). A polynomial representation of GLm(K) is a representation (ρ, V) of GLm(K), where V is a ﬁnite-dimensional vector space over K such that for each v ∈ V and ξ ∈ V0, the function

g 7→ hξ, ρ(g)vi

is a polynomial function in the entries of the matrix g. If, for all v ∈ V and ξ ∈ V0, this polynomial is homogeneous of degree n, then (ρ, V) is said to be a homogeneous polynomial representation of degree n.

Here, as usual, V0 denotes the space of K-linear maps V → K.

[1] Exercise 6.1.2. A representation (ρ, V) is a polynomial representation if and 0 only if, for any basis e1,..., em of V, taking ξ1, . . . , ξm to be the dual basis of V , the function

g 7→ hξi, ρ(g)e ji (6.1)

is a polynomial function of the entries of g for all i, j ∈ {1,..., m}. Furthermore, ρ(g) is homogeneous of degree d if and only if the polynomial in (6.1) is homogeneous of degree d for all i, j ∈ {1,..., m}.

m Example 6.1.3 (The deﬁning representation of GLm(K)). Let V = K . View the elements of V as column vectors and let g ∈ GLm(K) act on v ∈ V by ρ1(g)v = gv n (matrix multiplication). Taking ei to be the basis of K given by the coordinate vectors,

hξi, ρ1(g)e ji = gi j,

141

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142 Representations of General Linear Groups

n where gi j is the (i, j)th entry of the matrix g ∈ GLm(K). Therefore, (ρ1, K ) is a homogeneous polynomial representation of GLm(K) of degree 1. It is often called the deﬁning representation of GLm(K).

∗ Example 6.1.4 (The determinant). The determinant function GLm(K) → K is a multiplicative character which is also a polynomial of degree m in the entries. Therefore, it deﬁnes a one-dimensional homogeneous polynomial representation of degree m. More generally, g 7→ det(g)k is a homogeneous polynomial representation of degree mk for each non-negative integer k. We will see (Exercise 6.5.5) that these are the only one-dimensional polynomial representations of GLm(K). [1] Exercise 6.1.5 (Subrepresentations and quotient representations of polynomial representations). Show that every invariant subspace of a polynomial representation is a polynomial representation. Also, show that the quotient of a polynomial representation modulo an invariant subspace is a polynomial representation. If the ambient representation is homogeneous, then the invariant subspace and the quotient modulo the invariant subspace are homogeneous of the same degree.

[1] Exercise 6.1.6 (Direct sums of polynomial representations). Show that the direct sum of a ﬁnite number of polynomial representations of GLm(K) is a polynomial representation. Furthermore, if all these representations are homogeneous of degree n, then the direct sum is also homogeneous of degree n.

[1] Exercise 6.1.7 (Tensor product of polynomial representations). Show that a ﬁnite tensor product of polynomial representations is a polynomial representation. Furthermore, if each of these polynomial representations is homogeneous, then their tensor product is homogeneous with degree equal to the sum of their degrees.

[1] Exercise 6.1.8 (Contragredient of a polynomial representation). Show that the contragredient (Deﬁnition 1.7.1) of a homogeneous polynomial representation is not a polynomial representation unless the original representation has degree 0.

6.2 Schur Algebras

Schur algebras S K(m, n) are algebras whose finite-dimensional modules, considered collectively for all n ≥ 0, are precisely the polynomial representations of GLm(K) (see Theorem 6.2.9). Thus, in the polynomial representation theory of GLm(K), Schur algebras play the role that group algebras play in the representation theory of finite groups. 2 Let AK(m) denote the algebra of polynomial functions in m variables, which should be thought of as polynomials in the entries of m × m matrices. The space S K(m) is defined to be the set of all linear functionals α : AK(m) → K. Taking a

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6.2 Schur Algebras 143

1 leaf out of Schwartz’s theory of distributions, for α ∈ S K(m) and f ∈ AK(m), we will use the notation Z α( f ) = f (g)dα(g). G The notation signiﬁes that elements of the Schur algebra should be thought of as formal measures on G = GLm(K) with respect to which polynomials can be integrated.

Example 6.2.1 (Dirac delta function). For any g ∈ GLm(K), δg ∈ S K(m) deﬁned by Z f (x)δg(x) = f (g) G

is an element of S K(m). It is called the Dirac delta function at x. One reason for introducing this notation is that when the product on the Schur algebra is defined as Z Z Z f (g)d(αβ)(g) = f (xy)dα(x)dβ(y), (6.2) G G G it looks just like the formula for convolving measures on a topological group. The double integral on the right-hand side of the formula should be interpreted as follows: the function f (xy) may be viewed as a polynomial in y whose coefficients are polynomials in x. The integral Z f (xy)dα(x) G should be interpreted as the polynomial in y obtained by replacing each of those coefficients by their integrals with respect to x. This in turn can be integrated with respect to the variable y to obtain the double integral in question.

[1] Exercise 6.2.2. Show that for any g, h ∈ G, δgδh = δgh.

[1] Exercise 6.2.3. Show that the product (6.2) turns S K(m) into an unital associative (inﬁnite dimensional) K-algebra, where the unit is given by δI (here, I denotes the identity matrix in GLm(K)).

1 The theory of distributions (or generalized functions) allows us to make things that Physicists have formally manipulated mathematically rigorous, for instance, the Dirac delta function and its derivatives. Its systematic study was begun by Schwartz [27]. It turns out to be a powerful tool in the theory of diﬀerential equations and Fourier transforms. By deﬁnition, distributions are linear functionals on a space of test functions. Usually, the space of test functions consists of smooth and compactly supported functions on a domain in Rn or on a smooth manifold. One may think of the value of a linear functional on a test function as the integral of the product of the distribution and the test function. In this book, the space of test functions is the space of polynomial functions in the entries of a matrix.

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144 Representations of General Linear Groups

Let (ρ, V) be a polynomial representation of GLm(K). For each α ∈ S K(m), the formula Z hξ, ρ˜(α)vi = hξ, ρ(g)vidα(g) for all v ∈ V and ξ ∈ V0 (6.3) G defines a linear endomorphismρ ˜(g): V → V. Indeed, the right-hand side makes perfect sense, because the hypothesis that (ρ, V) is a polynomial representation ensures that the integrand hξ, ρ(g)vi is a polynomial function of g. The vector ρ˜(α)v is completely determined by the values hξ, ρ˜(α)vi as ξ ranges over V0 by the relatively trivial finite-dimensional version of the celebrated Riesz representation theorem. To see that (˜ρ, V) is indeed an S K(m)-module in the sense of Definition 1.1.12, it is necessary to verify the identity

hξ, ρ˜(αβ)vi = hξ, ρ˜(α)˜ρ(β)vi.

The proof has an air of inevitability about it: Z hξ, ρ˜(α)˜ρ(β)vi = hξ, ρ(x)˜ρ(β)vidα(g) G Z = hρ0(x)−1ξ, ρ˜(β)vidα(x) G Z Z = hρ0(x)−1ξ, ρ(y)vidα(x)dβ(y) G G Z Z = hξ, ρ(xy)vidα(x)dβ(y) G G Z = hξ, ρ(g)vid(αβ)(g) G = hξ, ρ˜(αβ)vi.

Conversely, given an S K(m)-module (˜ρ, V), we know that hξ, ρ˜(α)vi should give us the value of ρ(g)v in the following manner: the polynomial hξ, ρ(g)vi must be such that, for each α ∈ S K(m), the identity (6.3) holds. But such a polynomial need not exist. For indeed, to begin with, α 7→ hξ, ρ˜(α)vi is just some linear functional on S K(m). But not all linear functionals on S K(m) are polynomials (the double dual of an inﬁnite dimensional vector space is always larger than the original vector space). Thus, only a restricted class of S K(m)-modules come from polynomial representations of GLm(K). In order to describe this class, it is necessary to break down AK(m) into ﬁnite-dimensional pieces and study their duals individually. 2 Let AK(m, n) denote the subspace of homogeneous polynomials in m variables of degree n. Let S K(m, n) denote the subspace of S K(m) consisting of linear functionals α : AK(m) → K which vanish outside AK(m, n). Thus, S K(m, n) is the vector space dual to AK(m, n).

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6.2 Schur Algebras 145

Lemma 6.2.4. For each n ≥ 0,S K(m, n) is a subalgebra of S K(m).

0 Proof. Suppose α, β ∈ S K(m, n). We wish to show that αβ ∈ S K(m, n). If n , n, 0 then for any f ∈ AK(m, n ), Z Z f (xy)dα(x)dβ(y) = 0 G G because x 7→ xy being linear, its composition with f , which is x 7→ f (xy), is also R of degree n0, whence the inner integral f (xy)dα(x) vanishes identically.

2 Example 6.2.5. The algebra S K(m, 1) is just the linear dual of the m -dimensional vector space spanned by the variables xi j, 1 ≤ i, j ≤ m (each of the variables xi j is a polynomial of degree one). Given a matrix ξ = (ξi j) in Mm(K), use the symbol ξ to also represent the element of S K(m, 1) which takes xi j to ξi j for each 1 ≤ i, j ≤ m. In integral notation, Z xi jdξ(x) = ξi j. G

Similarly, let ζ = (ζi j) denote the linear functional for which xi j 7→ ζi j. Then Z Z Z xi jdξdζ(x) = (xy)i jdξ(x)dζ(y) G G G Z Z Xm = xikyk jdξ(x)dζ(y) G G k=1 Xm = ξikζk j. k=1

It follows that S K(m, 1) is isomorphic to the algebra Mm(K) of m × m matrices m with entries in K. In other words, S K(m, 1) is isomorphic to the algebra End(K ) of linear endomorphisms of the vector space Km. In Theorem 6.3.6, we will generalize this from S K(m, 1) to S K(m, n) for all n ≥ 1.

Every polynomial (not necessarily homogeneous) f ∈ AK(m) of degree d can be written as a sum Xd f = fn, n=0

for homogeneous polynomials fn ∈ AK(m, n). For each α ∈ S K(m), deﬁne αn ∈ S K(m) by Z Z f (g)dαn(g) = fn(g)dα(g). G G

Clearly, αn ∈ S K(m, n).

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146 Representations of General Linear Groups

[1] Exercise 6.2.6. For any element g = (gi j) ∈ GLm(K), show that Z  gi j if n = 1, xi jdδg,n(x) =  G 0 otherwise.

The map α 7→ αn is a K-algebra homomorphism. The unit δI of S K(m) maps onto δI,n, the unit of S K(m, n). If (ρ, V) is homogeneous of degree n0, then Z hξ, ρ˜(δI,n)vi = hξ, ρ(g)vidδI,n(g) G  0 hξ, vi if n = n =  0 otherwise.

Thus, if (ρ, V) is a homogeneous polynomial representation of degree n of GLm(K), then

ρ˜(δI,n)v = v for all v ∈ V (6.4)

and (˜ρ, V) is a unital module for S K(m, n). The δI,n’s constitute a family of central idempotents of S K(m) (see Exercise 1.5.15). These idempotents are also pairwise orthogonal, in the sense that 0 δI,nδI,n0 = 0 whenever n , n . For any unital ﬁnite-dimensional S K(m)-module (˜ρ, V), let Vn = ρ˜(δI,n)V. The fact that δI,n’s are central implies that the Vn’s are S K(m)-invariant subspaces of V. The fact that they are pairwise orthogonal idempotents implies that these Vn’s are mutually disjoint. Write Vpoly for the sum of the subspaces Vn over all non- negative integers n: M∞ Vpoly = Vn. n=0

The space Vpoly is called the space of polynomial vectors in V. We say that the S K(m)-module V is polynomial if

V = Vpoly.

Since V is ﬁnite dimensional, only ﬁnitely many of the spaces Vn are non-zero.

Lemma 6.2.7. If (ρ, V) is a polynomial representation of GLm(K). Then the associated S K(m)-module (˜ρ, V) deﬁned by (6.3) is polynomial.

Proof. Let (ρ, V) be a polynomial representation of GLm(K). Let v1,..., vr be a basis of V, and ξ1, . . . , ξr be the dual basis. Let d be the maximum of the degrees of the polynomials hξi, ρ(g)v ji for all 1 ≤ i, j ≤ m. Then, by the bilinearity of the map (ξ, v) 7→ hξ, ρ(g)vi, hξ, ρ(g)vi is a polynomial of degree at most d for all v ∈ V and ξ ∈ V0.

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6.2 Schur Algebras 147

We have Z hξ, ρ˜(α)vi = hξ, ρ(g)vidα(g) G Z Xd = hξ, ρ(g)vindα(g) G n=0 Xd Z = hξ, ρ(g)vidαn(g) n=0 Xd = hξ, ρ˜(αn)vi. n=0

0 for all ξ ∈ V and v ∈ V. In particular, taking α = δI and writing vn forρ ˜(δI,n)v ∈ Vn gives Xd hξ, vi = hξ, vni n=0

0 Ld for all ξ ∈ V and v ∈ V, whence it follows that V = n=0 Vd.

Lemma 6.2.8. If (˜ρ, V) is a polynomial representation of S K(m) and ρ(g) = ρ˜(δg), then (ρ, v) is the polynomial representation (ρ, V) of GLm(K) to which (˜ρ, V) is associated via (6.3).

Proof. Since V is a finite-dimensional polynomial representation of S K(m), V = Ld n=0 Vn for some positive integer d. Since a direct sum of polynomial representations is polynomial, it suffices to show that each Vn comes from a polynomial representation of GLm(K). We therefore assume that V = Vn. It follows that the linear functional α 7→ hξ, ρ˜(α)vi on S K(m) vanishes outside S K(m, n) and may therefore be viewed as a linear functional on S K(m, n), and hence an element of AK(m, n); in other words, there exists a homogeneous polynomial cξ,v(g) of degree n in the variables xi j which satisfies Z cξ,v(g)dα(g) = hξ, ρ˜(α)vi for every α ∈ S K(m, n). (6.5) G

When ρ(g) is deﬁned asρ ˜(δg), Exercise 6.2.2 and (6.4) imply that (ρ, V) is a representation of G. The identity (6.5) with α = δg says that Z hξ, ρ(g)vi = cξ,v(x)dδg(x) = cξ,v(g). G

Therefore, (ρ, V) is the polynomial representation of GLm(K) for which (˜ρ, V) is the S K(m)-module associated by (6.3).

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148 Representations of General Linear Groups

Lemmas 6.2.7 and 6.2.8 complete the proof of the main theorem of this section, which reduces the study of polynomial representations of GLm(K) to the study of S K(m, n)-modules for all n ≥ 0:

Theorem 6.2.9. 1. Every polynomial representation (ρ, V) of GLm(K) can be written as a direct sum M∞ V = Vn, n=0

of invariant subspaces, where the restriction of ρ to Vn is homogeneous of degree n. 2. The correspondence which takes a homogeneous polynomial representation (ρ, V) of degree n to the S K(m, n) module deﬁned by Z hξ, ρ˜(α)vi = hξ, ρ(g)vidα G

is the inverse of the correspondence which deﬁnes for each S K(m, n)-module (˜ρ, V) the homogeneous polynomial representation (ρ, V) of GLm(K) by ρ(g) = ρ˜(δg,n). Thus, there is an equivalence between the class of homogeneous polynomial representations of GLm(K) of degree n and ﬁnite-dimensional S K(m, n)- modules.

6.3 Schur Algebras and Symmetric Groups

We shall see in this section that the representation theory of the Schur algebras is closely related to that of symmetric groups. The vector space AK(m, n) is spanned by monomials of the form

xij = xi1 j1 ... xin jn

n for some i = (i1,..., in) and j = ( j1,..., jn) elements of I(m, n):= {1,..., m} . Clearly, xij = xi0j0 if and only if there exists a permutation w ∈ S n such that 0 0 iw(k) = ik and jw(k) = jk for each k = 1,..., n. Another way of viewing the situation is the following: let S n act on I(m, n) by permuting the Cartesian factors. 2 Thus, S n also acts on I(m, n) by the diagonal action: w · (i, j) = (w · i, w · j). Then 0 0 2 xij = xi0j0 if and only if (i, j) and (i , j ) lie in the same S n-orbits of I(m, n) . In other words, if and only if (i, j) and (i0, j0) have the same relative position (see Deﬁnition 2.4.3). If this happens, we write (i, j) ∼ (i0, j0). 0 0 Let ij denote the element of S K(m, n) such that ij(xi0j0 ) is zero unless (i , j ) has the same relative position as (i, j). Thus, as (i, j) ranges over a set of representatives 2 for S n-orbits in I(m, n) , the elements ij enumerate a basis of S K(m, n).

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6.3 Schur Algebras and Symmetric Groups 149

Lemma 6.3.1. For any i, j, k, l ∈ I(m, n), X ijkl = cr,srs, r,s where

cr,s = |{t ∈ I(m, n)|(r, t) ∼ (i, j) and (t, s) ∼ (k, l)}|.

Proof. The coeﬃcient cr,s is just i,jk,l(xrs), which is, by deﬁnition, Z Z X (xy)rsdij(x)dkl(y) = xrtytsdij(x)dkl(y) G G t∈I(m,n) = |{t ∈ I(m, n)|(r, t) ∼ (i, j) and (t, s) ∼ (k, l)}|

as claimed. Comparing the above result with the description of the endomorphism algebra of a permutation representation (Exercise 2.4.2) gives

Corollary 6.3.2. The Schur algebra S K(m, n) is the endomorphism algebra of the permutation representation K[I(m, n)] of S n:

S K(m, n) = EndS n K[I(m, n)]. An immediate corollary is the semisimplicity of the Schur algebras:

Theorem 6.3.3. Assume that K is an algebraically closed ﬁeld of characteristic greater than n. Then the Schur algebra S K(m, n) is a sum of matrix algebras over K. In particular, it is semisimple.

Proof. If K is algebraically closed of characteristic greater than n, then every irreducible representation of S n over K is completely reducible by Maschke’s theorem (Theorem 1.4.1). Therefore, by Theorem 1.3.6, S K(m, n), being the endomorphism algebra of a completely reducible module over the algebraically closed ﬁeld K, is a sum of matrix algebras over K.

Given i ∈ I(m, n), let λk be the number of times that k appears in i. Then (λ1, . . . , λm) is a weak composition (see Deﬁnition 2.3.1) of n into m parts, which we call the shape of i. Clearly, i and j are in the same S n-orbit in I(m, n) if and only if they have the same shape. Thus, if we write I(m, n)λ for the subset of I(m, n) consisting of tuples of shape λ, then the decomposition of I(m, n) into S n orbits is given by a I(m, n) = I(m, n)λ, λ where λ runs over all weak compositions of n into m parts.

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150 Representations of General Linear Groups

For each weak composition λ of n into m parts, let Xλ denote the set of all ordered partitions of shape λ (this extends the deﬁnition of Section 2.3 to weak compositions). Now given an ordered partition a a n = S 1 ··· S m (6.6)

with |S i| = λi, deﬁne an element i = (i1,..., im) ∈ I(m, n) by setting

ik = r if k ∈ S r. (6.7)

The partition (6.6) of n can be recovered from i = (i1,..., in) by setting

S i = {k | ik = i}. (6.8) We have Lemma 6.3.4. For every weak composition λ of n with m parts, the bijective correspondence I(m, n)λ → Xλ deﬁned by (6.8) is an isomorphism of S n-sets (see Deﬁnition 2.1.9). Therefore, we have Theorem 6.3.5. For any positive integer m and any non-negative integer n, M S K(m, n) = EndS n K[Xλ] , λ where λ runs over the set of weak compositions of n into m parts. Another interpretation of K[I(m, n)] is as the n-fold tensor power of the vector m space V = K . Indeed, K[I(m, n)] has a basis consisting of functions δi, i ∈

I(m, n). Mapping δi to ei1 ⊗ · · · ⊗ ein (here, e1,..., em are the coordinate vectors m of K ) gives such an isomorphism. The permutation representation of S n just corresponds to the permutation of factors in a tensor:

w · (v1 ⊗ · · · ⊗ vn) 7→ vw(1) ⊗ · · · ⊗ vw(n). (6.9) We may restate Corollary 6.3.2 in a more standard form:

m ⊗n Theorem 6.3.6. Consider the action (6.9) of S n on (K ) . Then

m ⊗n S K(m, n) = EndS n (K ) .

6.4 Modules of a Commutant

m ⊗n In the previous section, we had a completely reducible K[S n] module (K ) whose intertwiner algebra was S K(m, n). This is a special case of a more general situation.

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6.4 Modules of a Commutant 151

Let R and S be K-algebras. The vector space R ⊗ S becomes a K-algebra when the product is deﬁned by linearly extending

(r ⊗ s)(r0 ⊗ s0) = (rr0) ⊗ (ss0) for all r, r0 ∈ R, s, s0 ∈ S.

[1] Exercise 6.4.1. Let G and H be ﬁnite groups. Then the algebras K[G]⊗K[H] and K[G × H] are isomorphic.

Now suppose that (˜ρ, V) is an R-module and (σ, ˜ V) is an S -module (both have the same underlying vector space V) such thatρ ˜(r) commutes withσ ˜ (s) for all r ∈ R and all s ∈ S . We may think of V as an R ⊗ S -module under the actionρ ˜ ⊗ σ˜ deﬁned by (˜ρ ⊗ σ˜ )(r ⊗ s)v = ρ˜(r)σ ˜ (s)v.

Theorem 6.4.2. Let K be an algebraically closed ﬁeld and R be a K-algebra. Suppose that (σ, ˜ V) is a completely reducible R-module and that S = EndRV. Then S is a semisimple K-algebra. We may think of V as a left S -module and write ρ˜ : S → EndKV for the action of S on V. Let {(ρλ, Vλ)}λ∈Λ be a set of representatives for the isomorphism classes of simple R-modules which occur in V. Then, for each λ ∈ Λ, there exists a simple S -module (σλ, Wλ) of S such that, as an R ⊗ S -module, M (˜ρ ⊗ σ,˜ V) = (˜ρλ ⊗ σ˜ λ, Vλ ⊗ Wλ). λ∈Λ

Furthermore, {(σ ˜ λ, Wλ)}λ∈Λ is a complete set of representatives for the set of isomorphism classes of simple S -modules.

Proof. Suppose that V has a decomposition into simple modules with multiplicities given by M ⊕mλ V = Vλ

then by Theorem 1.3.6, S = EndRV is a sum of matrix algebras, and therefore, it is semisimple (see Corollary 1.5.12). ⊕m Think of V λ as V ⊗ Kmλ (by mapping the v in the jth copy of V to v ⊗ e ). Lλ i λ j Then S = M (K) acts on Kmλ via its λth matrix summand, so W := Kmλ λ mλ L λ are the simple modules for S , and V = Vλ ⊗ Wλ.

m ⊗n Now consider the case where ρ is the representation of S n on (K ) as in the m ⊗n previous section. Then Theorem 6.3.6 says that EndS n (K ) is the Schur algebra S (m, n). If the characteristic of K is greater than n, then the representation L K λ K[Xλ] is completely reducible with one simple representation for each partition λ of n with at most m parts. Applying Theorem 6.4.2 to this situation gives,

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152 Representations of General Linear Groups

for every partition λ with at most m parts, a simple S K(m, n)-module (σ ˜ λ, Wλ) such that, as a K[S n] ⊗ S K(m, n)-modules,

m ⊗n X (K ) = Vλ ⊗ Wλ (6.10) λ is a decomposition into simple modules. We have

Corollary 6.4.3. If the characteristic of K is greater than n, then (σ ˜ λ, Wλ), as λ runs over the set of partitions of n with at most m parts, forms a complete set of representatives of the set of simple S K(m, n)-modules.

Combining this with Theorem 6.2.9 gives

Theorem 6.4.4. Let (σλ, Wλ) be the polynomial representation of degree n of GLm(K) corresponding to the S K(m, n)-module (σ ˜ λ, Wλ). Then (σλ, Wλ), as λ runs over the set of partitions of n with at most m parts, is a complete set of representatives for the isomorphism classes of simple polynomial representations of GLm(K) of degree n.

m ⊗n The representation Vλ ⊗ Wλ is realized in (K ) as the Vλ-isotypic part of m ⊗n m ⊗n (K ) , or in other words, as ρ(λ)(K ) , where λ denotes the primitive central idempotent in K[S n] corresponding to the representation (ρλ, Vλ) (see Theo- rem 1.7.9). Since V(n) and V(1n) are one dimensional, W(n) and W(1n) are realized onρ ˜((n)) m ⊗n m ⊗n (K ) andρ ˜((1n))(K ) , respectively. By Theorem 1.7.9, 1 1 (w) = and n (w) = (w) (n) n! (1 ) n! (on the right-hand side of the second equation above, denotes the sign character of S n, see Section 4.1). We have 1 X ρ( )(e ⊗ · · · ⊗ e ) = e ⊗ · · · ⊗ e , (n) i1 in n! iw(1) iw(n) w∈S n while 1 X ρ( n )(e ⊗ · · · ⊗ e ) = (w) e ⊗ · · · ⊗ e . (1 ) i1 in n! iw(1) iw(n) w∈S n Deﬁnition 6.4.5 (Symmetric tensor). An element of (Km)⊗n is said to be a symmetric n-tensor if it is invariant under ρ(w) for every w ∈ S n.

[2] Exercise 6.4.6. Show that the space of symmetric tensors in (Km)⊗n has dimension equal to the number of weak compositions of n with at most m parts.

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6.5 Characters of the Simple Representations 153

m ⊗n [2] Exercise 6.4.7. Show that the image of ρ((n)) in (K ) is the space of symmetric tensors. Conclude that dim(W(n)) is the number of weak compositions of n with m parts (see Exercise 2.3.2). It follows that the space of symmetric m ⊗n tensors in (K ) has no invariant subspaces for the action of GLm(K).

Deﬁnition 6.4.8 (Alternating tensor). An element x of (Km)⊗n is said to be an alternating tensor if it transforms under the action of S n by

ρ(w)(x) = (w)x for each w ∈ S n.

[3] Exercise 6.4.9. Assume that K has characteristic diﬀerent from 2. Show that m ⊗n m the space of alternating tensors in (K ) has dimension equal to n .

m ⊗n [2] Exercise 6.4.10. Show that the image of ρ((1n)) in (K ) is the space of m alternating tensors. Conclude that dim(W(1n)) is n . It follows that the space of m ⊗n alternating tensors in (K ) has no invariant subspaces for the action of GLm(K).

6.5 Characters of the Simple Representations

Lemma 6.5.1. Let∆(x1,..., xm)denotethediagonalmatrixwithentries x1,..., xm along the diagonal. Then

m ⊗n X trace(ρ(w)σ(∆(x1,..., xm)); (K ) ) = trace(w; K[Xλ])mλ(x1,..., xm). λ

Proof. By Lemma 6.3.4, we have an isomorphism of S n-sets

m ⊗n M (K ) = K[I(m, n)] K[Xλ]. λ

m ⊗n The basis element ei1 ⊗ · · · ⊗ ein of (K ) is an eigenvector with eigenvalue λ x for σ(∆(x1,..., xm)). For any w ∈ S n, ρ(w) transforms this eigenvector to

eiw(1) ⊗ · · · ⊗ eiw(n) . Thus, the trace of ρ(wµ)σ(∆(x1,..., xm)) computed with respect to this basis vector is X xi1 ··· xin . (i1,...,in)=(iw(1),...,iw(n))

λ Given (i1,..., in) ∈ I(m, n), the monomial xi1 ··· xim is x , where λ is the weak composition of n with m parts corresponding to (i1,..., in) in the bijection of Lemma 6.3.4. Therefore, the coeﬃcient of the monomial xλ in the above sum is the number of elements of Xλ that are ﬁxed by w, which is nothing but the trace of the action of w on K[Xλ] (see Exercise 2.1.14).

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154 Representations of General Linear Groups

Corollary 6.5.2.

m ⊗n X trace(ρ(wµ)σ(∆(x1,..., xm)); (K ) ) = sλ(x1,..., xm)χλ(w). (6.11) λ Proof. By Young’s rule (Theorem 3.3.1), X X X trace(w; K[Xλ])mλ = Kµλχµ(w)mλ λ λ µ≤λ X X = χµ(w) Kµλmλ µ λ≥µ X = χµ(w)sµ [by (5.21)] µ whence Corollary 6.5.2 follows from Lemma 6.5.1. The notion of character of a polynomial representation is slightly different from the corresponding notion for a representation of a finite group which was introduced in Section 1.7. Definition 6.5.3 (Character of a polynomial representation). The character of the polynomial representation (ρ, V) of GLm(K) is defined as the polynomial function charρ ∈ K[x1,..., xm] defined by

charρ(x1,..., xm) = trace(ρ(∆(x1,..., xm)); V). By (6.10), m ⊗n X trace(ρ(wµ)σ(∆(x1,..., xm)); (K ) ) = charσλ(x1,..., xn)χλ(w). λ Therefore, X X sλ(x1,..., xm)χλ(w) = charσλ(x1,..., xn)χλ(w), (6.12) λ λ

and identity in ΛK,m[S n], the space of all ΛK,m-valued functions on S n. Since the characters of S n form a basis of this free ΛK,m-module, the identities (6.11) and (6.12) imply the following theorem: Theorem 6.5.4 (Schur). For every partition λ of n with no more than m parts,

charσλ(x1,..., xm) = sλ(x1,..., xm).

The identity (6.12) also implies that sλ(x1,..., xm) = 0 for all partitions λ with more than m parts.

[3] Exercise 6.5.5. Show that the representation Wλ of GLm(K) has dimension 1 if and only if λ = (km) for some non-negative integer k. Conclude that the only one-dimensional polynomial representations of GLm(K) are powers of the determinant.

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6.6 Polynomial Representations of the Torus 155

[2] Exercise 6.5.6. Compute the dimension of the representation W(2,1) of GL2(K) and the representation W(2,1) of GL3(K).

Theorem 6.5.7 (Schur). Two polynomial representations of GLm(K) are isomorphic if and only if their characters are equal.

Proof. This is a consequence of the linear independence part of Theorem 5.4.4; for if M M 0 ⊕nλ ⊕nλ V = Wλ and W = Wλ λ λ have the same characteristics, then

X X 0 nλ sλ(x1,..., xm) = nλ sλ(x1,..., xm), λ λ 0 from which it follows that nλ = nλ for every λ, whence the representations are isomorphic. Theorem 6.5.8. For partitions µ of m and ν of n,

λ M ⊕cµν Wµ ⊗ Wν Wλ , λ`m+n λ where cµν is the Littlewood–Richardson coeﬃcient of Section 5.7.

Proof. By Exercise 1.6.4 and Theorem 6.5.4, char(σµ ⊗ σν) = sµ sν, which is the λ L ⊕cµν same as the character of the representation λ`m+n Wλ .

6.6 Polynomial Representations of the Torus

Let Tm(K) denote the subgroup of GLm(K) consisting of diagonal matrices. The letter T here stands for ‘torus’. In this section, we will see that every polynomial representation of Tm(K) is a sum of simple representations and we will classify all the simple polynomial representations of Tm(K). Since the theory is analogous to (and easier than) what we did in Sections 6.1 and 6.2 for GLm(K), many details are skipped. For convenience, we will denote the matrix in Tm(K) with diagonal entries t1, t2,..., tm as a vector (t1, t2,..., tm).

Deﬁnition 6.6.1 (Polynomial representation of Tm(K)). A polynomial representation of Tm(K) is a pair (ρ, V), where V is a ﬁnite-dimensional vector space over K, ρ : Tm(K) → GLK(V) is a homomorphism, and for every vector v ∈ V and linear functional ξ : V → K, the function

t 7→ hξ, ρ(t)vi

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156 Representations of General Linear Groups

is a polynomial in the entries of t ∈ Tm(K). If these polynomials are homogeneous of degree n for all v ∈ V and ξ : V → K, then ρ is said to be a homogeneous representation of degree n.

Example 6.6.2. Let µ = (µ1, . . . , µm) be a weak composition of n with m parts. ∗ Deﬁne rµ : Tm(K) → K by

µ1 µm rµ(t1,..., tm) = t1 ··· tm . ∗ Since K = GL1(K), rµ is a one-dimensional polynomial representation of Tm(K), which is homogeneous of degree n.

In order to understand the polynomial representations of Tm(K), it is helpful to deﬁne the analogue of the Schur algebra for this group. Consider the vector space XK(m) of all linear functionals α : K[t1,..., tm] → K. For f ∈ K[t1,..., tm] and α ∈ XK(m), we use the notation Z f (t)dα(t) T

to denote the evaluation α( f ) of α at f . We may think of XK(T) as an algebra in the usual manner Z Z Z f (t)d(αβ)(t) = f (uv)dα(u)dβ(v), (6.13) T T T the double integral being interpreted like (6.2). Given a polynomial representation (ρ, V) of T, deﬁne a XK(m)-module (˜ρ, V) by Z hξ, ρ˜(α)vi = hξ, ρ(t)vidα(t). T

Let XK(m, n) denote the subalgebra of X(m) consisting of linear functionals which vanish on homogeneous polynomials in K[t1,..., tm] of degree diﬀerent from n. For each α ∈ XK(m), deﬁne αn ∈ XK(m, n) by Z Z f (t)dαn(t) = fn(t)dα(t), T T

where fn is the degree n homogeneous part of f . For any s ∈ Tm(K), let δs ∈ XK(m) be the ‘evaluation at s’ functional: Z f (t)dδt(s) = f (s). T

In particular, δI denotes evaluation at the identity. Here, I = (1,..., 1) is the identity matrix. For an XK(m)-module (˜ρ, V), let

Vn = ρ˜(δI,n)V.

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6.6 Polynomial Representations of the Torus 157 L∞ Say that (˜ρ, V) is a polynomial representation if V = n=0 Vn. Also say that (˜ρ, V) is homogeneous of degree n if V = Vn. Analogous to Theorem 6.2.9, we have

Theorem 6.6.3. 1. Every polynomial representation (ρ, V) of Tm(K) can be written as a direct sum M∞ V = Vn n=0

of invariant subspaces, where the restriction of ρ to Vn is homogeneous of degree n. 2. The correspondence which takes a homogeneous polynomial representation (ρ, V) of Tm(K) of degree n to the XK(m, n) module deﬁned by Z hξ, ρ˜(α)vi = hξ, ρ(t)vidα(t) T

is the inverse of the correspondence which deﬁnes for each XK(m, n)-module (˜ρ, V) the homogeneous polynomial representation (ρ, V) of Tm(K) deﬁned by ρ(s) = ρ˜(δs,n) for all s ∈ Tm(K).

The structure of the algebra XK(m, n) is relatively simple. The set

{tµ | µ is a weak composition of n with m parts}

is a basis for the space K[t1,..., tm]n of homogeneous polynomials of degree n. Let

{τµ | µ is a weak composition of n with m parts} (6.14)

denote the dual basis of XK(m, n). In other words,  Z  ν 1 if µ = ν, t dτµ(t) =  T 0 otherwise.

Exercise 6.6.4. Let (rµ, K) be the one-dimensional polynomial representation of Tm(K) from Example 6.6.2. Show that  idK if µ = ν, r˜µ(τν) =  0 otherwise.

One easily calculates the product τµτν using (6.13): Z Z Z θ θ θ x d(τµτν) = u v dτµdτν T T T  1 if µ = ν = θ, =  0 otherwise.

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158 Representations of General Linear Groups

In other words,  τµ if µ = ν, τµτν =  0 otherwise.

Therefore, the basis (6.14) is a basis of central idempotents of XK(m, n). From the general representation theory of semisimple algebras (see the box ‘Sum of algebras’ in Section 1.3), there is one simple module for XK(m, n) for each τµ. This module has dimension 1, τµ acts on it as the identity and τν acts on it as 0 for all ν , µ. This of course is none other than the module (˜rµ, K) from Exercise 6.6.4. We therefore have the following classiﬁcation of polynomial representations of T: Theorem 6.6.5. Every polynomial representation (ρ, V) of T is isomorphic to a sum

M ⊕nµ ρ rµ , (6.15) µ∈W(m,n) where the sum is over a ﬁnite collection of weak compositions µ with m parts, and for each of these weak compositions, nµ is a positive integer.

6.7 Weight Space Decompositions

Suppose (ρ, V) is any polynomial representation of GLm(K). The restriction of ρ to Tm(K) is a polynomial representation of Tm(K). By Theorem 6.6.5,

M ⊕nµ ρ|Tm(K) = rµ , µ the sum being over a ﬁnite collection of weak compositions with m parts for each µ of which mµ is a positive integer. Since rµ(x1,..., xm) is the monomial x , we have

X µ trace(ρ(∆(x1,..., xm)); V) = charρ(x1,..., xm) = nµ x . µ

Since characters determine polynomial representations of GLm(K) up to isomorphism (Theorem 6.5.7), we obtain

Theorem 6.7.1. Two polynomial representations of GLm(K) are isomorphic if and only if their restrictions to Tm(K) are isomorphic.

The decomposition of the restriction to Tm(K) of the simple polynomial representation (σλ, Wλ) of GLm(K), where λ is a partition of n with at most m parts, can be deduced from its character. If, as a representation of Tm(K), M nθ(λ) Wλ = rµ , θ

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6.7 Weight Space Decompositions 159

then X θ1 θm charσλ(x1,..., xm) = nθ x1 ... xm . (6.16) θ On the other hand, we have already seen [see Eq. (5.21)]

charσλ(x1,..., xm) = sλ(x1,..., xm) X = Kλµmµ(x1,..., xm). µ≥λ Comparing the coefficient of xθ in the above identity with its coefficient in (6.16), we find Theorem 6.7.2. For any weak composition θ of n with m parts, the multiplicity of rθ in the restriction of (ρλ, Wλ) to Tm(K) is the Kostka number Kλµ, where µ is the partition of n obtained from θ by rearranging its parts in weakly decreasing order. In particular, there is, up to scaling, a unique vector wλ ∈ Wλ such that each t ∈ Tm(K) acts on wλ by multiplication by rλ(t).

Deﬁnition 6.7.3 (Weights). Let (ρ, V) be a polynomial representation of GLm(K). The weak compositions θ for which rθ occurs in the restriction of (ρ, V) to Tm(K) are known as the weights of V. The multiplicity of a weight θ in V is the number of times that rθ occurs in V.

Example 6.7.4 (Weights for representations of GL2(K)). The partitions of n with at most two parts are (n), (n − 1, 1), (n − 2, 2),..., (n − bn/2c, bn/2c). We have, for 0 ≤ k ≤ bn/2c,

s(n−k,k) = m(n−k,k) + m(n−(k+1),k+1) + ··· + m(n−bn/2c,bn/2c) n−k k n−(k+1) k+1 k+1 n−(k+1) k n−k = x1 x2 + x1 x2 + ··· + x1 x2 + x1 x2 .

Thus, the weights of W(n−k,k) are the n − 2k + 1 weak compositions (n − k, k), (n − (k + 1), k + 1),..., (k + 1, n − (k + 1)), (k, n − k).

In particular, W(n−k,k) has dimension n − 2k + 1.

[1] Exercise 6.7.5. Determine the weights of the representations W(n) and W(1n) of GLm(K) along with their multiplicities.

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Hints and Solutions to Selected Exercises (With Contributions from Kannappan Sampath)

Chapter 1

Solution 1.1.4. Let χ : G → K∗ be a multiplicative character. Since χ(x) and χ(y) are scalars, they commute, so

χ(xyx−1y−1) = χ(x)χ(y)χ(x)−1χ(y)−1 = χ(x)χ(x)−1χ(y)χ(y)−1 = 1.

Thus, xyx−1y−1 ∈ ker(χ) for any x, y ∈ G; hence, the subgroup generated by these elements [G, G] is contained in ker(χ).

Solution 1.1.5. Since χ is non-trivial, there exists g0 ∈ G such that χ(g0) , 1. We have X X χ(g) = χ(gg0) g∈G g∈G X = χ(g0) χ(g). g∈G P P Therefore, (1 − χ(g0)) χ(g) = 0, whence χ(g) = 0.

Solution 1.1.9. Let J be a non-zero two-sided ideal in Mn(K). It suﬃces to show that the identity matrix lies in J. Since J is non-zero, there is a matrix A in J with at least one non-zero entry a = alm. Write Ei j for the matrix whose (i, j)th entry is 1, and every other entry is 0. Let M be any matrix. Observe that, Ei j M is the matrix whose only possibly non-zero row is the ith row which is the jth row of the matrix M and MEi j is the matrix

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Hints and Solutions to Selected Exercises 161

whose only possibly non-zero column is the jth column, which is the ith column of the matrix M. Therefore, n −1 X I = a EklAEmk, k=1 whence I ∈ J. Solution 1.1.16. Construct an isomorphism from K[Z/nZ] → K[t]/(tn − 1) by mapping 11 to t.

Solution 1.2.3. Since K[G]0 has codimension one in K[G], the complement must be one dimensional. Suppose that it is spanned by some f ∈ K[G], f not identically 0. Since the line spanned by f is G-invariant, for every g ∈ G, there exists a scalar λ(g) ∈ K such that g · f = λ(g) f . One easily checks that λ(g1g2) = λ(g1)λ(g2) for all g1, g2 ∈ G, and that λ(e) = 1. It follows that λ is a multi- ∗ plicative character G → K . Since f is not identically zero, a := f (g0) , 0 for −1 some g0 ∈ G. Now, for every g ∈ G, f (g g0) = g · f (g0) = λ(g) f for some function λ : G → K. Therefore,

X X −1 f (g) = f (g g0) g∈G g∈G X = f (g0) λ(g). g∈G

The latter sum is 0 for all non-trivial characters of G and is |G| f (g0) if χ is trivial. Thus, if λ is non-trivial, then the subspace spanned by f is always contained in K[G]0 and so cannot be a complement of K[G]0. If λ is trivial, then the subspace spanned by f consists of constant functions. If the characteristic of K does not divide |G| (or K has characteristic 0), then this subspace is the unique invariant complement of K[G]0. On the other hand, if the characteristic of K does divide |G|, even this subspace is contained in K[G]0, and therefore, K[G]0 has no complement. Solution 1.2.4. See solution to Exercise 1.2.3. The subspace described in this exercise is not complemented (consistent with Exercise 1.2.3). Solution 1.2.5. First observe that the regular representation is faithful (a representation (ρ, V) of G is faithful if ρ : G → GL(V) is injective). From the hypothesis, the regular representation is a sum of one-dimensional representations. Since [G, G] lies in the kernel of each of these one-dimensional representations, it follows that, [G, G] lies in the kernel of the regular representation, which is trivial. So [G, G] = {1} and G is abelian. This problem can also be solved without using Exercise 1.1.4. Since L is a sum of one-dimensional representations, K[G] has a basis of eigenvectors for L. It follows that L(g) is a diagonal matrix for each g ∈ G. Thus, L : G → GL(K[G])

162 Hints and Solutions to Selected Exercises

is an injective homomorphism from G into a group of diagonal matrices, showing that G is abelian. Solution 1.2.8. Let v ∈ V be a non-zero vector in the simple R-module V. Since R is a finite-dimensional K-algebra, it is spanned by a finite set {e1, e2,..., en} ⊂ R. The K-span of the vectors S = {ρ˜(ei)(v) | 1 ≤ i ≤ n} is easily seen to be an invariant subspace in V. This is a non-zero subspace in V, sinceρ ˜(1R)(v) = v. By the simplicity of V, one concludes that S spans V, so V is finite dimensional. Solution 1.2.9. We give two proofs, one which is simpler, but requires Schur’s Lemma (Theorem 1.2.13). We’ll first see a proof based on a result from linear algebra which is interesting in its own right and then the proof that uses Schur’s lemma. Using Linear Algebra We first prove the following proposition (which is a bit more than what is needed for this exercise, but is instructive): Theorem. Let S be a family of pairwise commuting linear operators on a finite- dimensional vector space V over an algebraically closed field K. Then, there is a common basis with respect to which they are all triangular, and the diagonalizable operators are diagonal.

Proof. If every element of S was a scalar multiple of the identity map, then every element of S would be diagonal with respect to any basis, and there would be nothing to prove. So assume that there exists T0 ∈ S such that T0 is not a scalar multiple of the identity map. Since we are working over an algebraically closed ﬁeld, T0 has an eigenvalue; call it λ. Since T0 is not a scalar multiple of the identity, the λ-eigenspace of T0, namely Vλ := ker(T0 − λid), is a proper subspace of V. Since any T ∈ S commutes with T0, for any v ∈ Vλ,

T0Tv = TT0v = λTv.

Thus, Tv ∈ Vλ. So Vλ is a proper invariant subspace for the operators in S . Arguing inductively, assume that Vλ has a basis {e1,..., ek} with respect to which the restrictions of all operators T ∈ S are upper triangular. Similarly, assume that the quotient space V/Vλ has a basis {e¯k+1,..., e¯n} with respect to which the operators induced by all operators T ∈ S are upper triangular. Lifting the elements e¯i ∈ V/Vλ to arbitrary preimages ei ∈ V for i = k + 1,..., n gives rise to a basis {e1,..., en} with respect to which all the operators in S are upper triangular. The same argument can be carried out with diagonalizability in mind. If all the diagonalizable operators in S are scalar multiples of the identity, proceed exactly as above. Otherwise, choose T0 to be diagonalizable but not scalar. Then Vλ has 0 a T0-invariant complement Vλ (namely the sum of all the other eigenspaces of T0), which is also invariant under all the operators in S . The induction hypothe- 0 sis implies that Vλ and Vλ have bases with respect to which all operators in S are

Hints and Solutions to Selected Exercises 163

upper triangular and all diagonalizable operators are diagonal. Concatenating these bases gives a basis of V with respect to which all the operators in S are upper triangular and all the diagonalizable operators in S are diagonal.

Now, applying this theorem to the family of operators representing an abelian group we have that with respect to some basis, every matrix in the representation can be upper triangularized. In other words, the subspace spanned by any initial subset of basis vectors is invariant, contradicting simplicity unless the representation has dimension one. Using Schur’s Lemma Let ρ be a simple representation. Then ρ(g) is an intertwiner, because, for any h ∈ G,

ρ(g)ρ(h) = ρ(gh) = ρ(hg) = ρ(h)ρ(g).

By Schur’s Lemma I (Theorem 1.2.13), ρ(g) is a scalar matrix for all g ∈ G. Simplicity forces the dimension of ρ to be one. Solution 1.2.14. Show that the centre of G acts by intertwiners and apply Schur’s Lemma I (Theorem 1.2.13). Solution 1.2.15. Let V be a simple representation. Exercise 1.2.12 says that for a self-intertwiner, T : V → V, ker(T) and Im(T) are invariant subspaces of V. Since T is non-zero and V is simple, we have ker(T) = {0V } and Im(T) = V so T is invertible. Solution 1.3.2. Let X be any invariant subspace of W. Then X admits a complement Y in V. It is easy to check that Y ∩ W is a complement of X in W. If X is an invariant subspace of V/W, let X˜ be the preimage of X in V. Let Y˜ be a complement of X˜ in V. It is easy to see that Y = (Y˜ + W)/W is a complement of X in V/W. Solution 1.3.3. Suppose that V is completely reducible. Let W be an invariant subspace of V. Let U be a maximal invariant subspace of V such that U ∩V = {0}. We claim that W + U = V, so U is a complement of W. Let

V = V1 ⊕ V2 ⊕ · · ·

be the decomposition of V into a sum of simple modules. If W + U , V, then Vi ) Vi ∩ (W + U) for at least one i. Since Vi is simple, this would mean that Vi ∩ (W + U) = {0}. It follows that (Vi + U) ∩ W = {0}, contradicting the maximality of U. For the converse, choose a maximal collection {V1,..., Vn} of mutually disjoint n simple R-invariant subspaces of V. Let W = ⊕i=1Vi. We claim that W = V. If not, let U be an invariant complement of W in V. Let Vn+1 be a simple invariant subspace of U (if U is simple, take Vn+1 = U). Then {V1,..., Vn+1} is again a

164 Hints and Solutions to Selected Exercises

mutually disjoint collection of simple R-invariant subspaces of V, contradicting the maximality of {V1,..., Vn}.

Solution 1.3.4. Let e1,..., en be a basis of an R-module V. Since R is a completely ⊕n reducible R-module, so is U = R . Now T : U → V deﬁned by (r1,..., rn) Pn 7→ i=1 riei is a surjective R-module homomorphism, so V, being a quotient a completely reducible R-modules, is again completely reducible (by Exercise 1.3.2).

Solution 1.3.12. Use Theorem 1.3.6 along with the fact that Mn(K) is commutative if and only if n = 1. Solution 1.3.13. Using Theorem 1.3.5, we have that

X 2 X X 2 mi = mini = ni . P Taking diﬀerences of successive terms in the above identity gives mi(mi − ni) = P P 2 ni(mi −ni) = 0. Taking the diﬀerence of these two identities gives (mi −ni) = 0, whence mi = ni for all i, so V and W are isomorphic. Solution 1.3.14. The ‘if’ part is trivial. For the ‘only if’ part, observe that if V is isomorphic to W, then by Exercise 1.3.11,

mi = dim Hom(Vi, V) = dim Hom(Vi, W) = ni.

Solution 1.3.15. Let W be an invariant subspace of V. If Vi occurs in W, then since dim Hom(Vi, V) = 1, the image of Vi in V is fixed. Thus, W is the sum of these fixed images of the simple modules which occur in it and therefore is completely determined by which of the simple constituents of V occur in W. Solution 1.4.4. Setting ρ(1) = A defines a representation of the cyclic group Z/nZ over K. By Maschke’s theorem, this representation is a sum of simples, which by Exercise 1.2.9 are all one dimensional. The matrix A is diagonal with respect to any basis formed by taking one vector from each simple summand of this decomposition into one-dimensional simples. Solution 1.5.2. The inverse of this linear map is given by ψ 7→ ψ(1) for each ψ ∈ EndRR. Solution 1.5.7. The transpose map gives an isomorphism from the algebra of upper triangular matrices to the algebra of lower triangular matrices. Conjugation by the antidiagonal matrix (the n × n matrix (ai j) where ai j = 1 if j = n − i + 1 and ai j = 0 otherwise) gives an isomorphism from the algebra of upper triangular matrices to the algebra of lower triangular matrices.

Solution 1.5.15. The only non-trivial thing to show is that i cannot be written as a sum i1 + i2, where i1 and i2 are non-zero orthogonal central idempotents. We have i1i = i1(i1 + i2) = i1. Therefore, i1 j = i1i j = 0 if j , i. It follows

Hints and Solutions to Selected Exercises 165

that i1 (and similarly i2) lies in the ith matrix summand in the Wedderburn decomposition of R. But the only central idempotents in a matrix algebra are 0 and the identity matrix. Solution 1.6.2. Use the bilinear map K[X] × K[Y] → K[X × Y] deﬁned by

( f (x), g(y)) 7→ f (x)g(y)

and the universal property of tensor products. Solution 1.6.4. Work with bases.

Solution 1.6.9. By Burnside’s theorem (Theorem 1.5.17),ρ ˜(K[G]) = EndKV andσ ˜ (K[H]) = EndKW. It follows that ρ] σ(K[G × H]) = EndK(V ⊗ W), so ρ σ is simple. For the second part, note that if ρ σ are isomorphic τ θ, then so are ⊕ dim W their restrictions to G × {eh} and {eG} × H. But ρ σ|G×{eH } σ , while ⊕ dim W τ θ|G×{eh} τ . It follows that ρ and τ are isomorphic. Similarly, σ and θ are isomorphic.

Solution 1.7.4. The spaces V1, V2 and V3 are all one-dimensional equal to K, 2πi jk/3 ⊗ ∈ ⊗ ⊗ 0 ρ j(k) = e idV j for j = 1, 2, 3. Writing e j for 1 1 K K = V j V j, we have

2 X 2πi jk/3 c(e j) = e 1k. k=0 For Z/nZ in general, the answer is similar:

n−1 X eπi jk/n c(e j) = e 1k, for j = 1,..., n. k=0

Solution 1.7.12. Substitute the explicit form of i from Theorem 1.7.9 in the orthogonality relations for primitive central idempotents: i j = δi ji. Solution 1.7.15. 1. The classes are just the elements, indexed 0, 1,..., n−1, the 2πi jk/n characters are χ0, χ1, . . . , χn−1 with χ j(k) = e . 2. This group G has eight elements. Think of G as the group of motions in the plane which ﬁx a unit square with vertices (±1, ±1). Let r denote clockwise rotation by π/2 and s denote reﬂection about the Y-axis. Then

2 3 2 3 G = {eG, r, r , r , sr, sr , sr }. This group has ﬁve conjugacy classes:

1. {eG} 2. {sr, sr3} 3. {s, sr2} 4. {r, r3} 5. {r2}

166 Hints and Solutions to Selected Exercises

The subgroup generated by r is normal of order 4. Therefore, : G → {±1} deﬁned by (r) = 1 and (s) = −1 is a multiplicative character of G. On the other hand, there is an obvious two-dimensional representation (ρ, K2) of G coming from the way we deﬁned it. In this representation, ! ! 0 1 −1 0 ρ(r) = and ρ(s) = . −1 0 0 1

If χρ is the character of this representation, we find hχρ, χρiG = 1, so this representation is simple. Since G has five conjugacy classes, it must have five simple representations, and by using Corollary 1.5.16, the remaining two must have dimension one. One easily sees that setting χ(r) = −1 and χ(s) = 1 gives rise to a multiplicative character of G. Another one is the product χ. Therefore, the character table is given by

2 eG sr s r r 1 1 1 1 1 1 1 −1 −1 1 1 ρ 2 0 0 0 −2 χ 1 −1 1 −1 1 χ 1 1 −1 −1 1

3. The conjugacy classes in Q = {±1, ±i, ±j, ±k} are easily seen to be 1. {1} 2. {−1} 3. {±i} 4. {±j} 5. {±k} Besides the trivial representation, there is another obvious representation ρ, which comes from the way in which we have deﬁned Q. This is the two- dimensional complex representation where each element of Q is represented by the matrix that was used to deﬁne it. Corollary 1.5.16 can be used to conclude that there are three more one-dimensional representations. A few minutes of trial and error will reveal these to be χi, χj and χk given in the table below: 1 −1 ±i ±j ±k 1 1 1 1 1 1 ρ 2 −2 0 0 0 χi 1 1 1 −1 −1 χj 1 1 −1 1 −1 χk 1 1 −1 −1 1

Hints and Solutions to Selected Exercises 167

Solution 1.7.16. Schur’s orthogonality relations imply that XZ−1EX0 = I, whence X0X = EZ (note that E2 = I).

Solution 1.7.17. 1. Since the irreducible characters χ1, . . . , χr form an orthonormal basis with respect to the non-degenerate bilinear for h·, ·iG, we have Xr 1C j = h1C j , χiiGχi, i=1 which upon expanding using (1.20) gives the required formula. 2. Follows easily by using the hint. Solution 1.8.5. By Theorem 1.8.4, there exists a Hermitian inner product with respect to which any complex representation of a finite group is unitary. Every invariant subspace for a unitary representation has an invariant complement (namely its orthogonal complement). Therefore, by Exercise 1.3.3, the representation is completely reducible. Solution 1.8.8. Since G is finite, each element g ∈ G has finite order, say m. It follows that χ(g) is the trace of a matrix which satisfies the equation xn = 1. It follows that every characteristic root of this matrix is a root of unity and therefore a unit complex number. Noting that the number of roots is χ(1), the dimension of the representation, the result follows from the triangle inequality. Solution 1.8.9. If g lies in the centre of G, then g acts by a scalar matrix on any simple representation (Exercise 1.2.14). Since g has finite order, this scalar is a root of unity and therefore has absolute value 1. It follows that |χ(g)| = χ(1). The dual orthogonality relations of Exercise 1.7.16 imply that r X 2 |χk(g)| = |ZG(g)|, k=1

where χ1, . . . , χr are the simple characters of G. If |χk(g)| = χk(1) for each k, then P 2 the left-hand side of the above equation would be k |χk(1)| , which by Corol- lary 1.5.16 is |G|. Therefore, |G| = |ZG(g)|, whence g lies in the centre of G.

Chapter 2

Solution 2.1.10. Use the map gGx 7→ g · x. Solution 2.1.11. Suppose K = xHx−1, then in the transitive G-space G/H, the point xH has stabilizer K. It follows that G/H is isomorphic to G/K by Exer- cise 2.1.10. Conversely, if G/H is isomorphic to G/K as a G-set, then since G/K has a point with stabilizer K, G/H must also have such a point. If xH is this point, then K = xHx−1.

168 Hints and Solutions to Selected Exercises

Solution 2.1.14. Compute the trace with respect to the basis {1x | x ∈ X}. Solution 2.2.2. We have  12345 if n ≡ 0 mod 5,  43152 if n ≡ 1 mod 5,  wn = 51423 if n ≡ 2 mod 5,   24531 if n ≡ 3 mod 5,  35214 if n ≡ 4 mod 5.

In particular, w−1 = 35214. Solution 2.2.5. The permutation that comes after 526431 in lexicographic order is 531246. The permutation preceding it is 526413. Solution 2.2.7. The products vw, where rows are indexed by v and columns are indexed by w, are given as follows: 123 132 213 231 312 321 123 : 123 132 213 231 312 321 132 : 132 123 231 213 321 312 213 : 213 312 123 321 132 231 231 : 231 321 132 312 123 213 312 : 312 213 321 123 231 132 321 : 321 231 312 132 213 123

Solution 2.2.11. Each such cycle has a unique representative (1, x2, x3,..., xn), where x2,..., xn is a permutation of {2, 3,..., n}. Solution 2.2.12. Suppose a permutation w has cycle type λ. The order of w is the least k such that if w is applied k times, each element of n comes back to itself. An element in a cycle of length l comes back to itself upon the application of w k times if and only if k is a multiple of l. Therefore, every element of n comes back to itself after w is applied k times if and only if k is a multiple of the order of each cycle of w. Therefore, the order of w is the LCM of the cycle lengths of w. Solution 2.2.13. 1. The cycle type of a ﬁxed-point-free involution is always of the form (2k), so we must have n = 2k. 2. This is easily proved by induction on m. Solution 2.2.16. The cycles of w−1 are the cycles of w taken in reverse order. Therefore, w and w−1 have the same cycle type. k Solution 2.2.17. If v and w generate the same cyclic subgroup of S n, then v = w , where k is coprime to the order of w. By Exercise 2.2.12, k is coprime to the order of each cycle of w. Now v is the composition of the kth powers of the cycles of w. But a cycle, when raised to a power that is coprime to its order, remains a cycle

Hints and Solutions to Selected Exercises 169

of the same length. Therefore, the cycle type of v is the same as the cycle type of w, whence v and w are conjugate. Solution 2.2.19. The partition succeeding it is (3, 3, 1, 1, 1, 1, 1), and the partition preceding it is (3, 3, 2, 2, 1). Solution 2.2.24. Suppose that x commutes with w. If w(i) = j, then w(x(i)) = x(w(i)) = x( j). So if (i1,..., ik) is a cycle for w, then (x(i1),..., x(ik)) is also a cycle for w. Thus, x commutes with w if and only if it takes cycles of w to cycles of w (not just as sets, but as sets with a cyclic order). Thus, all the elements commuting with w can be constructed by taking a set of representatives of the cycles of w and choosing for each such representative an element of a cycle of w of the same size. The underlying permutation on cycles Q allows for i mi! possibilities, and for each representative of a cycle of length i, there are i possibilities for the element of the target cycle to which it is mapped, Q mi giving i mi! i as the total number of possibilities. Solution 2.2.25. It suffices to show that for every partition λ of n, there exists a polynomial fλ ∈ K[x1,..., xn] such that f˜(w) is 1 if w has cycle type λ and 0 otherwise. Since K has characteristic zero, it contains Q as its prime field, so it suffices to exhibit such a polynomial with rational coefficients. By appending zeroes at the end, think of each partition of n as an element of Qn. The polynomial n Y X 2 gλ(x1,..., xn) = (xi − mi(µ)) µ`n, µ,λ i=1 vanishes at the points in Qn represented by partitions of n that are different from λ, but is non-zero at the point represented by λ (here, mi(µ) denote the multiplicity of i in µ). Therefore, we may take fλ to be

gλ(x1,..., xn) fλ(x1,..., xn) = . gλ(λ1, . . . , λn)

Solution 2.4.1. One way to solve this problem would be to show that k 7→ Tk is injective and use a dimension count. Another would be to construct an inverse. Solution 2.4.5. Think of the trivial representation as K[Y], where Y is the singleton G-set and apply Theorem 2.4.4. Solution 2.4.11. Show that K[n] = 1 ⊕ V, where V is a simple representation.

Solution 2.5.4. Note that dim Vk = dim K[Xk] − dim Vk−1 − · · · − dim V0 = n n dim K[Xk] − dim K[Xk−1] = k − k−1 .

Solution 2.5.6. The three conjugacy classes in S 3 are represented by 123, 213 and 231. The character of V1, which is χ(123) = 2, χ(213) = 0 and χ(231) = −1, is computed by computing the character of K[X1] and subtracting the character of K[X0] (the trivial representation).

170 Hints and Solutions to Selected Exercises

Solution 2.5.7. By Exercise 2.5.5, the number of ﬁxed points in X1 for a permutation with cycle type λ is m1(λ).

Solution 2.6.5. The character table of S 4 is given in Table S.1. The partitions indexing the columns are the cycle types of permutations. The row indexed by λ is the irreducible representation Vλ in K[Xλ], which does not occur in K[Xµ] for any µ which precedes λ in reverse lexicographic order.

(14) (2, 13) (2, 2) (3, 1) (4) (14) 1 −1 1 1 −1 (2, 13) 3 −1 −1 0 1 (2, 2) 2 0 2 −1 0 (3, 1) 3 1 −1 0 −1 (4) 1 1 1 1 1

Table S.1 Character table of S 4

Chapter 3

Solution 3.1.5. 1 1 1 2 . 2 3 3 Solution 3.1.8. The least n for which incomparable partitions exists is n = 6; (4, 1, 1) and (3, 3) are incomparable. If n has incomparable partitions, so does n + 1; just add trailing 1’s to both partitions. Thus, incomparable partitions exist for all n ≥ 6. Solution 3.1.10. The partitions which have largest part at least m. Solution 3.1.11. Suppose that λ and µ are partitions of n and that λ has l parts. Then µ1 + ··· + µl ≥ λ1 + ··· + λl = n, so µ has at most l parts. Solution 3.2.4. Consider   0 0 1   A = 1 1 0   1 0 0 Then L(S (A)) > S (L(A)). Solution 3.2.7. 1 1 0 0   1 0 0 0   A = 0 1 0 0   0 0 0 1   0 0 1 0

Hints and Solutions to Selected Exercises 171

Solution 3.2.8. If λ = (λ1, λ2,... ) is the common shape and type of P and Q, the entries of A are given by ai j = λi+ j−1 − λi+ j. Solution 3.2.12. A permutation matrix A is an involution if and only if A = A0. By Exercise 3.2.9, the RSK correspondence gives a bijection between symmetric permutation matrices and pairs of SYT of the form (P, P).

Solution 3.3.3. Suppose we have shown that for all µ < λ in reverse dominance order that the character of Vµ takes integer values. Then Young’s rule: K[Xλ] = Vλ ⊕ (⊕µ<λVµ), together with the fact that the character of K[Xλ] takes integers values (Exercise 2.1.14), implies that the character of Vλ also takes integer values.

Solution 3.3.4. If λ is not of the form (n) or (1n), then the top-left corner of its Young diagram contains . This can be filled in as 1 2 or 1 3 and then 3 2 extended by filling in the remaining integers in the remaining boxes in increasing order going from top row to bottom row, filling each row from left to right.

Solution 3.3.5. Use the fact that Xk and X(n−k,k) are isomorphic S n-sets.

Chapter 4

Solution 4.1.3. See Table S.2.

Solution 4.1.7. Suppose the inversion is (i, j). Then for any i < k < j, if w(k) < w(i), (i, k) is an inversion. If w(k) ≥ w(i), then since w(i) > w( j), w(k) > w( j), so (k, j) is an inversion. Thus, if w has only one inversion, that inversion must be (i, i + 1) for some i ∈ {1,..., n − 1}, whence w = si.

Solution 4.1.8. 1. Under wsi, i 7→ w(i + 1), and i + 1 7→ w(i). All other elements of n have the same image under w and wsi. 2. Follows easily from 1.

Solution 4.1.13. s1 s2 s3 s1 s2 s1, s1 s2 s1 s3 s2 s1, s2 s1 s2 s3 s2 s1, s2 s3 s1 s2 s3 s1, s2 s1 s3 s2 s3 s1, s1 s2 s3 s2 s1 s2, s3 s1 s2 s3 s1 s2, s1 s3 s2 s3 s1 s2, s3 s1 s2 s1 s3 s2, s1 s3 s2 s1 s3 s2, s3 s2 s1 s2 s3 s2, s2 s3 s1 s2 s1 s3, s2 s1 s3 s2 s1 s3, s2 s3 s2 s1 s2 s3, s3 s2 s3 s1 s2 s3, s3 s2 s1 s3 s2 s3.

−1 −1 Solution 4.1.16. Since (xyx y ) = 1 for all x, y ∈ S n,[S n, S n] ⊂ An. The commutator of the 2-cycles (i, j) and ( j, k) is the 3-cycle (i, k, j). Thus, every 3-cycle belongs to [S n, S n]. Now the following lemma completes the proof:

Lemma. An is generated by 3-cycles.

172 Hints and Solutions to Selected Exercises

Permutation Inversions 1234 1243 (3, 4) 1324 (2, 3) 1342 (2, 4), (3, 4) 1423 (2, 3), (2, 4) 1432 (2, 3), (2, 4), (3, 4) 2134 (1, 2) 2143 (1, 2), (3, 4) 2314 (1, 3), (2, 3) 2341 (1, 4), (2, 4), (3, 4) 2413 (1, 3), (2, 3), (2, 4) 2431 (1, 4), (2, 3), (2, 4), (3, 4) 3124 (1, 2), (1, 3) 3142 (1, 2), (1, 4), (3, 4) 3214 (1, 2), (1, 3), (2, 3) 3241 (1, 2), (1, 4), (2, 4), (3, 4) 3412 (1, 3), (1, 4), (2, 3), (2, 4) 3421 (1, 3), (1, 4), (2, 3), (2, 4), (3, 4) 4123 (1, 2), (1, 3), (1, 4) 4132 (1, 2), (1, 3), (1, 4), (3, 4) 4213 (1, 2), (1, 3), (1, 4), (2, 3) 4231 (1, 2), (1, 3), (1, 4), (2, 4), (3, 4) 4312 (1, 2), (1, 3), (1, 4), (2, 3), (2, 4) 4321 (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)

Table S.2 Inversions for permutations of 4

Proof. By deﬁnition, every element in An is a product of an even number of transpositions. Now

si si+1 = (i, i + 2, i + 1)

is a 3-cycle, and when j > i + 1, we have,

z}|{ z }| { z}|{ s j si = si s j = si si+1 si+1 si+2 ··· s j−1 s j

so any product of an even number of transpositions is a product of 3-cycles.

Therefore, An is generated by 3-cycles.

Solution 4.3.3. By Exercise 4.3.2, λ 7→ λ0 gives a bijection from the set of partitions with parts bounded by m to the set of partitions with at most m parts. Solution 4.3.5. Following MacDonald [20, Chapter I, 1.11], one can prove the contrapositive: suppose that λ0 µ0. Then there exists a least integer k such that

0 0 0 0 λ1 + ··· + λk > µ1 + ··· + µk.

Hints and Solutions to Selected Exercises 173

0 0 By the minimality of k, we have λk > µk. Since λ and µ have the same size, it follows that 0 0 0 0 λk+1 + λk+2 + ··· < µk+1 + µk+2 + ··· . (S.1) The two sides of the above inequality are the number of nodes which lie to the 0 right of the kth column in λ and in µ. Since λk is the number of rows in λ with at least k columns, 0 λk 0 0 X λk+1 + λk+2 + ··· = (λi − k). i=1 Similarly, 0 µk 0 0 X µk+1 + µk+2 + ··· = (µi − k). i=1 Therefore, λ0 µ0 Xk Xk (λi − k) < (µi − k). i=1 i=1 0 0 Since λk > µk, the above inequality implies that

µ0 µ0 Xk Xk (λi − k) < (µi − k). i=1 i=1 or equivalently, µ0 µ0 Xk Xk λi < µi, i=1 i=1 whence λ µ. Solution 4.5.3. 0 0 1 1   0 1 0 0   A = 1 0 0 0   0 1 0 0   1 0 0 0 Solution 4.5.4. If a 0-1-matrix has at most one 1 in each row and column, then the shadow paths and dual shadow paths coincide. This shows that the ﬁrst rows of 0 P1 and P2 coincide, and the ﬁrst rows of Q1 and Q2 coincide. Also, from (3.10), it follows that if A has at most one 1 in each row and column, then so does its 0 shadow. It follows that the subsequent rows of P1 and P2 as well as Q1 and Q2 coincide.

174 Hints and Solutions to Selected Exercises

+ −1 − −1 Solution 4.6.10. Consider the sets C1 = {xwx | x is even} and C1 = {xwx | x is odd}. We have ywy−1 = (yx−1)xwx−1(yx−1)−1, so if yx−1 is even, ywy−1 and −1 + xwx are conjugate in An. From this, it follows that all the elements of C are − conjugate in An, as are all the elements of C . Suppose that some odd permutation x commutes with w. Then w ∈ C1 and −1 w = xwx ∈ C2, so C1 = C2. This shows that 1 =⇒ 2 in the second part of the exercise and by contradiction that 2 =⇒ 1 in the first part. − + If an element of C is conjugate to an element of C in An, then it is conju- + gate to w in An since all the elements of C are conjugate to w in An. In other words, yxwx−1y−1 = w for some even y and odd x. Then yx is an odd permutation that commutes with w, so 2 =⇒ 1 in the second part of the exercise and, by contradiction 1 =⇒ 2 in the first part. −1 Solution 4.6.17. 1. Let w = (i1, i2,..., ik). Then w = (ik, ik−1,..., i1). If x = −1 −1 (i1, ik)(i2, ik−1) ··· (ibk/wc, in−bk/2c), then xwx = w . Thus, if bk/2c is even then −1 w is conjugate to w in An. −1 Conversely, suppose that there exists y ∈ An such that ywy = w. Taking x as before, we find that x−1y commutes with w. Since the centralizer of w consists of its powers, x−1y, being a power of w is even. Therefore, x is also even, so bk/2c is an even integer.

2. For each cycle C j = (i1,..., iλ j ) of w, deﬁne x j = (i1, ik)(i2, ik−1) ··· (ibλ j/2c, −1 −1 P iλ j−bλ j/2c). Let x = x1 x2 ··· . Then xwx = w . It follows that if bλi/2c is even, then w is conjugate to its inverse in An. −1 Conversely, suppose that there exists y ∈ An such that ywy = w. Taking x exactly as before, we ﬁnd that x−1y commutes with w. By Theorem 4.6.12, −1 P x y and hence also x must be even permutations. It follows that bλi/2c is even. √ √ Solution 4.6.18. We have u = (−1 + i 3)/2, v = (−1 − i 3)/2. √ √ Solution 4.6.19. We have u = (1 + 5)/2, v = (1 − 5)/2.

Chapter 5

Solution 5.1.3. 3m(2,1,1,1) + 2m(2,2,1) + 2m(3,1,1) + m(3,2). Solution 5.1.4. Any monomial in the product involves at most m + n variables, has total degree m + n, has each variable appearing with power at most two and has at most min(m, n) variable appearing with power equal to two. Therefore, the

monomials functions in the product must be of the form m(2k,1m+n−2k) for 0 ≤ k ≤ min(m, n). In order to find the coefficient of m(2k,1m+n−2k) in the product, it suffices 2 2 to find the coefficient of the monomial x1 ··· xk xk+1 ··· xm+n−k in m(1m)m(1n). Such a monomial corresponds to a choice of subsets A and B of positive integers such

Hints and Solutions to Selected Exercises 175

that |A| = m, |B| = n, and such that A∩B = {1,..., k} and A∪B = {1,..., m+n−k}. Obviously {1,..., k} ⊂ A. The number of ways of choosing the remaining m − k m+n−2k elements of A is m−k . The stated formula follows.

Solution 5.1.5. The coefficient of mν in mλmµ is the same as the coefficient of the monomial xν in the product. The coefficient of this monomial is the number of pairs of monomials (xα, xβ) such that α has shape λ, β has shape µ and xα xβ = xν, as desired.

Solution 5.2.17. To see that ξ 7→ φξ is a well-deﬁned linear map is routine. That it is an isomorphism can be seen by constructing an inverse: given φ ∈ 0 HomG(V, K[G/H]) deﬁne ξφ ∈ V by

ξφ(v) = φ(v)(H). Solution 5.2.18. This follows by noting that dim(V0)H is the multiplicity of the 0 trivial representation in V (and hence also in V), and dim HomG(V, K[G/H]) is the multiplicity of V in K[G/H].

Solution 5.2.19. Take G = S n, H = S η and V = Vλ in Exercise 5.2.18. Solution 5.2.18. This is the equality of the dimensions of the two sides of Exercise 5.2.19. Solution 5.2.20. We have

0 X (XA )λµ = trace(wη; Vλ)Aµη η X = trace(w; Vλ)

w∈S µ

= |S µ| × multiplicity of the trivial representation in Vλ

= |S µ|Kλµ. The last step used Exercise 5.2.19, so XA0 = B. It follows that det X = det B/ det A. Since K is triangular, so is B, and it has diagonal entries |S λ| as λ runs over parti- Q Q tions of n. Therefore, det B = λ`n i λi!. We claim that Aλµ = 0 if λ comes after µ in reverse lexicographic order. For indeed, if λi = µi for i < j and λ j < µ j, no element with cycle type µ can lie in S λ. Now Aλλ is the number of elements of S λ with cycle type λ. The cycles of such elements must be

(1, . . . , λ1), (λ1 + 1, . . . , λ1 + λ2), (λ1 + λ2 + 1, . . . , λ1 + λ2 + λ3),....

The number of cycles of length λi in S λ is (λi − 1)! (see Exercise 2.2.11). There- Q i fore, Aλλ = i(λi − 1)! and Y Y det A = (λi − 1)! λ`n i

176 Hints and Solutions to Selected Exercises

We have det B Y Y λ ! Y Y det X = = i = λ . det A (λ − 1)! i λ`n i i λ`n i Solution 5.3.2. 1. The specializations to three variables of monomial, elementary, complete and power sum symmetric functions of degree 3 are as follows: • Monomial symmetric functions:

λ mλ(x1, x2, x3) 3 3 3 (3) x1 + x2 + x3 2 2 2 2 2 2 (2, 1) x1 x2 + x1 x2 + x1 x3 + x2 x3 + x1 x3 + x2 x3 (1, 1, 1) x1 x2 x3 • Elementary symmetric functions:

λ eλ(x1, x2, x3) (3) x1 x2 x3 2 2 2 2 2 2 (2, 1) x1 x2 + x1 x2 + x1 x3 + 3x1 x2 x3 + x2 x3 + x1 x3 + x2 x3 3 2 2 3 2 2 2 2 3 (1, 1, 1) x1 + 3x1 x2 + 3x1 x2 + x2 + 3x1 x3 + 6x1 x2 x3 + 3x2 x3 + 3x1 x3 + 3x2 x3 + x3 • Complete symmetric functions:

λ hλ(x1, x2, x3) 3 2 2 3 2 2 2 2 3 (3) x1 + x1 x2 + x1 x2 + x2 + x1 x3 + x1 x2 x3 + x2 x3 + x1 x3 + x2 x3 + x3 3 2 2 3 2 2 2 2 3 (2, 1) x1 + 2x1 x2 + 2x1 x2 + x2 + 2x1 x3 + 3x1 x2 x3 + 2x2 x3 + 2x1 x3 + 2x2 x3 + x3 3 2 2 3 2 2 2 2 3 (1, 1, 1) x1 + 3x1 x2 + 3x1 x2 + x2 + 3x1 x3 + 6x1 x2 x3 + 3x2 x3 + 3x1 x3 + 3x2 x3 + x3 • Power sum symmetric functions:

λ pλ(x1, x2, x3) 3 3 3 (3) x1 + x2 + x3 3 2 2 3 2 2 2 2 3 (2, 1) x1 + x1 x2 + x1 x2 + x2 + x1 x3 + x2 x3 + x1 x3 + x2 x3 + x3 3 2 2 3 2 2 2 2 3 (1, 1, 1) x1 + 3x1 x2 + 3x1 x2 + x2 + 3x1 x3 + 6x1 x2 x3 + 3x2 x3 + 3x1 x3 + 3x2 x3 + x3 2. The specializations to two variables of monomial, elementary, complete and power sum symmetric functions of degree 3 are as follows: • Monomial symmetric functions:

λ mλ(x1, x2) 3 3 (3) x1 + x2 2 2 (2, 1) x1 x2 + x1 x2 (1, 1, 1) 0 • Elementary symmetric functions:

λ eλ(x1, x2) (3) 0 2 2 (2, 1) x1 x2 + x1 x2 3 2 2 3 (1, 1, 1) x1 + 3x1 x2 + 3x1 x2 + x2

Hints and Solutions to Selected Exercises 177

• Complete symmetric functions:

λ hλ(x1, x2) 3 2 2 3 (3) x1 + x1 x2 + x1 x2 + x2 3 2 2 3 (2, 1) x1 + 2x1 x2 + 2x1 x2 + x2 3 2 2 3 (1, 1, 1) x1 + 3x1 x2 + 3x1 x2 + x2

• Power sum symmetric functions:

λ pλ(x1, x2) 3 3 (3) x1 + x2 3 2 2 3 (2, 1) x1 + x1 x2 + x1 x2 + x2 3 2 2 3 (1, 1, 1) x1 + 3x1 x2 + 3x1 x2 + x2

Solution 5.4.2. We have s(2,1) = m(1,1,1) + m(2,1). Solution 5.4.13. Each monomial in

n n (x1 + ··· + xn)aδ

n n−1 n−1 is of the form xi xw(1) xw(2) ··· xw(n−1). The only variable in this monomial having a power greater than n − 1 is xi. Therefore, for the monomial to be decreasing, we must have i = 1, and if w( j) = 1, we must have

w(2) > w(3) > ··· > w( j − 1) > w( j + 1) > ··· > w(n).

It follows that w can only be a cycle of the form (1, 2, ··· , j). The corresponding monomial is

j−1 n+(n− j) n n−1 n− j+1 n− j n− j−1 (−1) x1 x2 x3 ··· x j x j+1 x j+1 ··· xn−1,

which is (−1) j−1 xλ+δ with λ = (n − j + 1, 1 j−1). Taking k = j − 1 and applying Theorem 5.4.11 gives the character values in the exercise. Solution 5.5.2. By Theorem 5.5.1,

hchn(χλ), sµi = hχλ, χµiS n = δλµ,

for all µ ` n, whence chn(χλ) = sλ.

Solution 5.5.3. Since the identity holds when elements of the basis {χλ}λ`n of the space of class functions is substituted for f and g, it holds for all class functions f and g by bilinearity.

178 Hints and Solutions to Selected Exercises

Solution 5.5.5. 1. By Young’s rule (Theorem 3.3.1), Exercise 5.5.2 and (5.19), X ch(K[Xλ]) = Kµλch(Vµ) µ≤λ X = Kµλ sµ µ≤λ

= hλ.

2. Twisting Young’s rule by the sign character, using Theorem 4.4.2 and (5.18), X ch(K[Xλ] ⊗ ) = Kµλch(Vµ ⊗ ) µ≤λ X = Kµλ sµ0 µ≤λ

= eλ.

Solution 5.5.6. The problem is equivalent to asking for which partitions λ pλ is a non-negative linear combination of Schur functions. The only such partition is λ = (1n). To see this, suppose that λ is a partition diﬀerent from (1n). By the dual orthogonality relations (Exercise 1.7.16), X χµ(1)χµ(wλ) = 0. µ`n

Since χµ(1) > 0 (it is the dimension of Vµ) for each µ and χ(n)(wλ) = 1 (V(n) is the trivial representation), there must exist µ such that χµ(wλ) is negative in order that the sum can be 0. Thus, there exists µ such that Xλµ < 0. By (5.22) pλ cannot be a non-negative integer combination of Schur functions. Solution 5.6.4. Given a standard Young tableau with shape λ, remove the box numbered n. This gives rise to a standard Young tableau of shape µ with µ ∈ λ−. This gives rise to a bijection from the set of standard Young tableaux of shape λ onto the disjoint union over all µ ∈ λ− of the sets of standard Young tableaux of shape µ.

n m Solution 5.8.4. k m+k .

Solution 5.9.2. We have ω(chn(χλ)) = ω(sλ) = sλ0 = chn(χλ0 ) = chn(χλ) by Theorem 4.4.2. Thus, the identity holds for χλ for each partition λ of n. Since the χλ’s form a basis for the space of class functions on S n, it holds for all class functions.

Hints and Solutions to Selected Exercises 179

By Exercise 5.5.5,

ω(hλ) = ω(chn(K[Xλ]))

= chn(K[Xλ] ⊗ )

= eλ.

Solution 5.11.3. 1. χ(8,17)(w(7,5,3)) = 2. 2. χ(2,2,1,1)(w(2,2,2)) = −3.

Solution 5.12.1. Removing the rim of the unique (2m1 + 1)-hook of φ(µ) gives rise to φ(˜µ), whereµ ˜ = (2m2+1, 2m3+1,... ). Therefore, the recursive Murnaghan– Nakayama rule (Theorem 5.11.1) gives

m1 χφ(µ)(wµ) = (−1) χφ(˜µ)(wµ˜ ). The formula in the exercise now follows by induction on k.

Solution 5.12.2. If θ = (2t1 + 1, 2t2 + 1,... ) comes before µ = (2m1 + 1, 2m2 + 1,... ) in reverse lexicographic order, then there exists i such that t j = m j for j < i and ti > mi. The recursive Murnaghan–Nakayama rule (Theorem 5.11.1) gives P j

The largest hook of φ(2mi + 1, 2mi+1 + 1,... ) is of length 2mi + 1, so it has no hooks of length 2ti + 1, and hence, by the recursive Murnaghan–Nakayama rule (Theorem 5.11.1), χφ(µ)(wθ) = 0. √ Solution 5.12.6. We have χ+ (w± ) = 1 (−1 ± i 13!/35), χ− (w±) = √ (4,4,3,2) (7,5,1) 2 (4,4,3,2) λ 1 2 (−1 ∓ i 13!/35).

Chapter 6

Solution 6.1.5. Consider a polynomial representation (ρ, V) and an invariant subspace W of V. Let v ∈ W be any vector and ξ ∈ W0 be any linear functional. Let ξ˜ be any extension of ξ to V. Then hξ, ρ(g)vi = hξ, ρ(g)v. Thus, the matrix coefficients of W are also matrix coefficients of V, and the result follows. A slight variant of this argument works of the quotient V/W. Solution 6.1.6. Show that a matrix coefficient of a direct sum is a sum of matrix coefficients of the summands. Solution 6.1.7. Show that a matrix coefficient of a tensor product is a linear combination of products of matrix coefficients of the factors. Solution 6.1.8. Show that the matrix coefficients of the contragredient are rational functions of degree equal to the negative of the degree of the original representation.

180 Hints and Solutions to Selected Exercises

Solution 6.2.2. We have Z Z Z f (x)d(δgδh)(x) = f (xy)dδg(x)dδh(y) G G G Z = f (gy)dδh(y) G = f (gh) Z = f (x)dδgh(x). G

Therefore, δgh = δgδh.

Solution 6.4.6. Writing i for (i1,..., in) and ei for ei1 ⊗ · · · ⊗ ein , every element of (Km)⊗n can be uniquely written as X x = xiei. (S.2) i∈I(m,n)

m ⊗n The vector x is invariant under the action of S n on (K ) if and only if xi = xj whenever i and j lie in the same orbit for the action of S n ∈ I(m, n). We have already seen in Section 6.3 that the number of such orbits is the number of weak compositions of n with at most m parts.

Solution 6.4.7. The operator ρ((n)) replaces the coefficient of ei in (S.2) by the average of the coefficients of ej for all j in the S n-orbit of i. Therefore, its image consists of vectors where the coefficient of ei depends only on the orbit of i under the action of S n on I(m, n). From the solution to Exercise 6.4.6, it follows that its image is precisely the subspace of symmetric tensors. Solution 6.4.9. If the vector in (Km)⊗n with coordinates X x = xiei i∈I(m,n)

is an alternating tensor, then xi determines xj whenever i and j lie in the S n orbit. If j = w · i, then xj = (w)xi. Thus, the dimension of the space of alternating tensors cannot be more than the dimension of the space of symmetric tensors. However, suppose that there exists an odd permutations w such that w · i = i. Then xi = −xi, so unless K has characteristic 2, xi = 0. Now fix a representative for each S n-orbit in I(m, n) whose elements are not fixed by any odd permutations. If i is such a representative, and if j = w1i and −1 j = w2i for two permutations w1 and w2, then w2 w1 fixes i and is therefore even. It follows that (w1) = (w2). Therefore, having assigned a value to xi, setting xj = (w)xi for any w such that j = w · i gives rise to a well-defined vector. It follows that the dimension of the space of alternating tensors in (Km)⊗n is equal to the number of orbits in I(m, n) whose elements are not fixed by odd permutations.

Hints and Solutions to Selected Exercises 181

One easily shows that i = (i1,..., in) is not ﬁxed by an odd permutation if and only if i1,..., in are distinct elements of {1,..., m}. The S n-orbits of such elements of I(m, n) are completely determined by the underlying set {i1,..., in}, which has cardinality m. Thus, the dimension of the space of alternating tensors in (Km)⊗n m is n .

Solution 6.4.10. Show that ρ((1n)) is an idempotent operator which ﬁxes every alternating tensor. Solution 6.5.5. By Schur’s theorem on characters of simple polynomial representations (Theorem 6.5.4),

dim Wλ = sλ(1,..., 1) X = Kλµmµ(1,..., 1) µ≥λ

≥ mλ(1,..., 1). The polynomials above are specializations of symmetric functions to m variables. Therefore, if dim Wλ = 1, then mλ(1, 1,..., 1) = 1. But mλ(1,..., 1) = 1 only if there is exactly one monomial in m variables of shape λ. This happens only if m m λ = (k ) for some non-negative integer k. Thus if Wλ is one dimensional, λ = (k ) for some non-negative integer k. In particular, the degree of a one-dimensional representation is always a multiple of m, and at most one such representation exists in each degree. But g 7→ det(g)k is a one-dimensional homogeneous polynomial representation of GLm(K) of degree mk and therefore must be the representation W(km). Thus, W(km) is one-dimensional and is the kth power of the determinant.

Solution 6.5.6. The dimension of the representation W(2,1) of GL2(K) is s(2,1)(1, 1) = 2, while the dimension of the representation W(2,1) of GL3(K) is s(2,1)(1, 1, 1) = 8. P µ Solution 6.7.5. We have (see Example 5.4.1) char(W(n)) = s(n) = hn = µ x , where µ runs over all weak compositions of n with m parts. Therefore, the weights of W(n) are the weak compositions of n with at most m parts, and moreover, each weight occurs with multiplicity one. For example, the weights of the representation W(2) of GL3(K) are (2, 0, 0), (0, 2, 0), (0, 0, 2), (1, 1, 0), (1, 0, 1) and (0, 1, 1). P µ Also by Example 5.4.1, char(W(1n)) = s(1n) = en = µ x , where µ runs over all vectors in Zm which have n entries equal to 1 and the remaining entries equal to 0. Thus, these vectors form the weights of W(1n), and moreover each weight occurs with multiplicity one. For example, the weights of the representation W(1,1) of GL3(K) are (1, 1, 0), (1, 0, 1) and (0, 1, 1).

Suggestions for Further Reading

We list a few books and major articles that the reader who wishes to go deeper into the subject may enjoy reading. Needless to say, this is a very subjective list, based on the author’s tastes and interests.

1. The Symmetric Group: Representations, Combinatorial Algorithms, and Symmetric Functions, by Bruce E. Sagan [26]. Sagan’s book has a large overlap with this one. It has a chapter on the general theory of group representations where character theory and the decomposition of group algebras is discussed. It also discusses the classification of simple representations of S n, its connection to the combinatorics of Young tableaux and the theory of symmetric functions. However, the overall arrange- ment and emphasis is quite different and nicely complements the material here. Notably, it contains a very nice treatment of the Littlewood–Richardson rule using Schutzenberger’s¨ jeu de taquin. 2. Enumerative Combinatorics, 2, by Richard P. Stanley [34]. The last chapter of this book is titled Symmetric Functions and has an extremely systematic treatment of the theory of symmetric functions, the RSK correspondence and its dual and the applications to the representation theory of symmetric groups. The notes at the end of this chapter provide a detailed history of the subject. There are more than a hundred exercises which bring out many interesting applications. An appendix by Sergey Fomin on Knuth equivalence, jeu de taquin and the Littlewood–Richardson rule is a highlight of this book. Here, Fomin explains the construction of the Robinson–Schensted correspondence using growth diagrams and local rules. These ideas are also used to prove the Littlewood–Richardson rule. 3. Symmetric Functions and Hall Polynomials by Ian G. Macdonald [20]. This classic book has been the standard reference for the theory of symmetric functions ever since its first edition was published in 1979. It contains a discussion of the Hall algebra with a view towards giving a more conceptual

182

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Suggestions for Further Reading 183

treatment of Green’s computation of the irreducible characters of general linear groups over finite fields and the representation theory of general linear groups over local fields. It takes considerable mathematical maturity to be able to appreciate Macdonald’s book. We hope that the present volume will have prepared readers somewhat for this task. 4. Representation Theory: A First Course by William Fulton and Joe Harris [9]. This book emphasizes the analytic theory of representations of Lie groups via Lie algebras. The classical groups are treated individually in great detail. The authors beautifully synthesize algebraic and combinatorial methods with analytic ones. 5. A New Approach to the Representation Theory of Symmetric Groups by Anatoly Vershik and Andrei Okounkov [36]. In this article, the authors use the branching rules (Theorem 5.6.1) for the restriction of a representation of S n to S n−1 as a starting point for develop- ing the representation theory of symmetric groups. In this approach, standard Young tableaux appear quite naturally into the theory. Essential use is made of the fact that the symmetric groups are not individual groups but form an infinite nested sequence of groups. Other expositions of their method can be found in Murali Srinivasan’s concise online notes [31] and the full-length book by Tullio Ceccherini-Silberstein, Fabio Scarabotti and Filippo Tolli [3].

6. Polynomial Representations of GLn by James A. Green [10]. Green’s book was a strong influence on our approach to Schur–Weyl duality and the polynomial representation theory of GLn. Green begins by following Schur’s original papers quite closely, but works carefully with infinite fields of arbitrary characteristic with a view towards modular (positive characteristic, non-semisimple) representation theory. 7. Modular Representations, Old and New by Bhama Srinivasan [30]. Bhama Srinivasan’s article tells the story of modular representations of symmetric groups, Hecke algebras and related objects. It has historical information as well as a discussion of recent developments and open problems. 8. Pioneers of Representation Theory: Frobenius, Burnside, Schur and Brauer, by Charles W. Curtis [5]. Curtis describes the history of representation theory starting with the work of Frobenius in the 1890s and ending with Brauer’s work on modular representation theory in the 1960s, placing it in the context of contemporary mathematical, scientific and political events, thereby bringing to life the people who contributed to this area. At the same time, it is a book full of mathematics, complete with definitions, theorems and many a beautiful proof.

184 Suggestions for Further Reading

9. Iwahori-Hecke Algebras and Schur Algebras of the Symmetric Group, by Andrew Mathas [22]. Mathas uses a cellular-algebras approach due to Graham and Lehrer to study the modular representation theory of Iwahori–Hecke algebras and q-Schur algebras. These algebras are deformations of the group algebra of S n and the Schur algebra, respectively. The appendix on the basic representation theory of ﬁnite-dimensional algebras is one of the best introductions to the subject.

References

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Index

0-1 λ × µ matrix, 75 cycle decomposition, 35 algebra, 2 cycle type, 36 homomorphism, 3 cyclic conjugation property, 37 unital, 2 decreasing monomial, 113 alternating group, 73, 136 deﬁning representation, 141, 142 character table of, 91 determinant, 142 AROW, 63 diagonal action, 42 branching rules, 119 dihedral group, 27, 43 Burnside’s lemma, 43 Dirac delta function, 143 Burnside’s theorem, 18 dominance order, 54 and conjugation, 76 central character, 7 doubly transitive action, 43 central idempotent, 17 centre of an algebra, 10 e, 100 eλ, 99 chn, 117 char, 154 endomorphism, 6 character, 25 Erlanger program, 43 integrality of, 31 even permutation, 72 multiplicative, 1 fλ, 67 of polynomial representation, 154 Frobenius character formula, 115 character table, 27 Frobenius dimension formula, 124 of A3, 92 G-set, 32 of A4, 92 Gale–Ryser theorem, 78 of A5, 93 group action, 32 of A8, 93 group algebra, 3 of alternating groups, 91, 136 h, 102 of D8, 166 of quaternion group, 166 hλ, 101 Hermitian inner product, 29 of S 3, 48 homogeneous symmetric function, 96 of S 4, 170 characteristic function of Frobenius, 117 homomorphism, 6 combinatorial resolution theorem, 49 hook-length, 125 complete reducibility, 7, 13 hook-length formula, 126 conjugate partition, 75 ideal contragredient representation, 22 left, 2 convolution, 4, 143 right, 2 coset space, 33 two-sided, 2

189

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190 Index

idempotent, 17 permutation, 34 integral kernel, 41 centralizer of, 38 integral operator, 41 string notation, 35 intersection numbers, 46 permutation representation, 32 intertwiner, 6 permutation string, 34 invariant subspace, 5 Pieri’s rule, 116 inversion, 70 Polya enumeration theorem, 103 inversion number, 70 polynomial distributions, 142 involution, 37 polynomial representation, 141 involution of symmetric functions, 127 polynomial vectors, 146 isomorphism, 7 primitive central idempotent, 17 Jacobi–Trudi identity, 129 Q-group, 37 K , 54 λµ recursive Murnaghan–Nakayama rule, 132 Klein, Felix, 43 regular representation, 4 ΛK , 97 decomposition of, 13 λ × µ matrix, 46 relative position, 42 left coset, 33 representation, 1 lexicographic order complex, 29 of partitions, 38 defining, 141 of permutations, 35 faithful, 161 Littlewood–Richardson coefficient, 120 polynomial, 141 Mλµ, 46, 56, 101 homogeneous, 141 m, 99 trivial, 2 mλ, 97 reverse dominance order, 54 Maschke’s theorem, 11 and conjugation, 76 matrix algebra, 2, 9, 15 reverse lexicographic order, 38 matrix coefficient, 22 Robinson–Schensted correspondence, 67 module, 3 Robinson–Schensted–Knuth correspondence, 56 monomial symmetric function, 97 RSK correspondence, 56 multiplicity, 8 dual, 80 multiplicity-free decomposition, 10 symmetry property of, 67 Murnaghan–Nakayama rule, 110 s, 110 N , 75, 78, 99 λµ sλ, 110 n, 34 Schur algebras, 142 odd permutation, 72 Schur function, 110 opposite algebra, 14 Schur’s lemma, 6, 7 orbit, 32 Schur’s orthogonality relations, 26 ordered partition, 39 self-conjugate partition, 85 semisimple algebra, 13 Pλµ, 102 p, 103 shadow pλ, 102 matrix, 64 partition, 35, 39 path, 57 even, 88, 129 reverse, 58 exponential notation, 38 shape, 39 odd, 129 sign character, 72 self-conjugate, 85 simple algebra, 15 sign of a, 128 simple module, 5 with distinct odd parts, 89 simple representation, 5 partition function, 98 simple transposition, 70 partition representation, 40 specialization, 107

Index 191

stabilizer subgroup, 33 tensor space, 150 standard Young tableau, 67 terminal column, 58 submodule, 5 terminal row, 58 subrepresentation, 5 transitive action, 32 subset representation, 44 transverse pair, 75 sum of algebras, 9 twist of a representation, 73 symmetric function, 96 unital module, 3, 4 complete, 101 unitary operator, 30 elementary, 98 monomial, 97 VRSK, 63 power sum, 102 weak composition, 40 Schur, 110 Wedderburn decomposition, 9, 15 symmetric functions fundamental theorem of, 100 Xλµ, 104 tensor product, 19 Young diagram, 51 external, 21 Young subgroups, 106 of representations, 21 Young tableau, 51 universal property of, 19 zλ, 38