modern group theory · representation theory of nite groups. we will begin the course with an...

Modern Group TheoryMAT 4199/5145

Fall 2017

Alistair Savage

Department of Mathematics and Statistics

University of Ottawa

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Contents

Preface 4

1 Representation theory of finite groups 51.1 Basic concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.1.1 Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.1.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.1.3 Intertwining operators . . . . . . . . . . . . . . . . . . . . . . . . . . 91.1.4 Direct sums and Maschke’s Theorem . . . . . . . . . . . . . . . . . . 111.1.5 The adjoint representation . . . . . . . . . . . . . . . . . . . . . . . 121.1.6 Matrix coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.1.7 Tensor products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.1.8 Cyclic and invariant vectors . . . . . . . . . . . . . . . . . . . . . . . 18

1.2 Schur’s lemma and the commutant . . . . . . . . . . . . . . . . . . . . . . . 191.2.1 Schur’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.2.2 Multiplicities and isotypic components . . . . . . . . . . . . . . . . . 211.2.3 Finite-dimensional algebras . . . . . . . . . . . . . . . . . . . . . . . 231.2.4 The commutant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.2.5 Intertwiners as invariant elements . . . . . . . . . . . . . . . . . . . . 28

1.3 Characters and the projection formula . . . . . . . . . . . . . . . . . . . . . 291.3.1 The trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291.3.2 Central functions and characters . . . . . . . . . . . . . . . . . . . . . 301.3.3 Central projection formulas . . . . . . . . . . . . . . . . . . . . . . . 32

1.4 Permutation representations . . . . . . . . . . . . . . . . . . . . . . . . . . . 381.4.1 Wielandt’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391.4.2 Symmetric actions and Gelfand’s lemma . . . . . . . . . . . . . . . . 411.4.3 Frobenius reciprocity for permutation representations . . . . . . . . 421.4.4 The structure of the commutant of a permutation representation . . . 47

1.5 The group algebra and the Fourier transform . . . . . . . . . . . . . . . . . . 501.5.1 The group algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501.5.2 The Fourier transform . . . . . . . . . . . . . . . . . . . . . . . . . . 541.5.3 Algebras of bi-K-invariant functions . . . . . . . . . . . . . . . . . . 57

1.6 Induced representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 611.6.1 Definitions and examples . . . . . . . . . . . . . . . . . . . . . . . . . 611.6.2 First properties of induced representations . . . . . . . . . . . . . . . 631.6.3 Frobenius reciprocity . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

2

Contents 3

1.6.4 Mackey’s lemma and the intertwining number theorem . . . . . . . . 66

2 The theory of Gelfand–Tsetlin bases 692.1 Algebras of conjugacy invariant functions . . . . . . . . . . . . . . . . . . . . 69

2.1.1 Conjugacy invariant functions . . . . . . . . . . . . . . . . . . . . . . 692.1.2 Multiplicity-free subgroups . . . . . . . . . . . . . . . . . . . . . . . . 73

2.2 Gelfand–Tsetlin bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 742.2.1 Branching graphs and Gelfand–Tsetlin bases . . . . . . . . . . . . . . 742.2.2 Gelfand–Tsetlin algebras . . . . . . . . . . . . . . . . . . . . . . . . . 76

3 The Okounkov–Vershik approach 803.1 The Young poset . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

3.1.1 Partitions and conjugacy classes in Sn . . . . . . . . . . . . . . . . . 803.1.2 Young diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 813.1.3 Young tableaux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 823.1.4 Coxeter generators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 833.1.5 The content of a tableau . . . . . . . . . . . . . . . . . . . . . . . . . 863.1.6 The Young poset . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

3.2 The Young–Jucys–Murphy elements and a Gelfand–Tsetlin basis for Sn . . . 893.2.1 The Young–Jucys–Murphy elements . . . . . . . . . . . . . . . . . . . 903.2.2 Marked permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . 913.2.3 Olshanskii’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

3.3 The spectrum of the YJM elements and the branching graph of Sn . . . . . 963.3.1 The weight of a Young basis vector . . . . . . . . . . . . . . . . . . . 973.3.2 The spectrum of the YJM elements . . . . . . . . . . . . . . . . . . 983.3.3 Spec(n) = Cont(n) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

3.4 The irreducible representations of Sn . . . . . . . . . . . . . . . . . . . . . . 1053.4.1 Young’s seminormal form . . . . . . . . . . . . . . . . . . . . . . . . 1053.4.2 Young’s orthogonal form . . . . . . . . . . . . . . . . . . . . . . . . . 1073.4.3 The Young seminormal units . . . . . . . . . . . . . . . . . . . . . . . 1103.4.4 The Theorem of Jucys and Murphy . . . . . . . . . . . . . . . . . . . 113

4 Further directions 1144.1 Schur–Weyl duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1144.2 Categorification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

4.2.1 Symmetric functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 1154.2.2 The Grothendieck group . . . . . . . . . . . . . . . . . . . . . . . . . 1174.2.3 Categorification of the algebra of symmetric functions . . . . . . . . 1184.2.4 The Heisenberg algebra . . . . . . . . . . . . . . . . . . . . . . . . . . 1194.2.5 Categorification of bosonic Fock space . . . . . . . . . . . . . . . . . 1194.2.6 Categorification of the basic representation . . . . . . . . . . . . . . 1214.2.7 Going even further . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

Index 123

Preface

These are notes for the course Modern Group Theory (MAT 4199/5145) at the University ofOttawa. Since the pioneering works of Frobenius, Schur, and Young more than a hundredyears ago, the representation theory of the symmetric group has developed into a huge areaof study, with applications to algebra, combinatorics, category theory, and mathematicalphysics. In this course, we will cover the representation theory of the symmetric groupfollowing modern techniques developed by Vershik, Olshankii, and Okounkov.

Using techniques from algebra, combinatorics, and category theory, we will cover thefollowing topics.

• Representation theory of finite groups. We will begin the course with an introduction tothe representation theory of finite groups. This will include a discussion of irreduciblerepresentations, tensor products, Schur’s lemma, characters, permutation representa-tions, group algebras, and Frobenius reciprocity.

• The theory of Gelfand-Tsetlin bases. We will discuss branching rules for representationsof symmetric groups and see how such branching rules allow one to obtain particularlynice bases for irreducible representations.

• The Okounkov-Vershik approach. We will discuss the combinatorics of Young tableaux,Jucys-Murphy elements, and the Okounkov-Vershik approach to the representationtheory of symmetric groups.

Acknowledgements: These notes closely follow the book [CSST10], which is the recom-mended textbook for the course.

Alistair Savage

Course website: http://alistairsavage.ca/mat5145

4

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Chapter 1

Representation theory of finite groups

In this chapter we discuss some basic facts about the representation theory of finite groups.While we will focus on the symmetric groups later in the course, we work mostly witharbitrary finite groups in this chapter. We closely follow the presentation in [CSST10, Ch. 1].

Throughout this chapter, G will denote a finite group and V,W will denote finite-dimensional complex vector spaces. Unless otherwise specified, we will always work overthe field of complex numbers. So the term vector space means complex vector space.

1.1 Basic concepts

In this section we give the main definitions related to representations of finite groups anddiscuss some examples.

1.1.1 Representations

Recall that the general linear group

GL(V ) := T : V → V : T is an invertible linear map

is a group under composition. Its identity element is the identity map IV . A (linear)representation of G on V is a group homomorphism

σ : G→ GL(V ).

The name arises from the fact that elements g of G are “represented” by linear transformati-ons σ(g) of V . When we wish to make the vector space V explicit, we will sometimes denotethe representation by (σ, V ), or simply by V (with the homomorphism σ understood). Thedimension of the representation σ is the dimension of V .

A subspace W ≤ V is said to be σ-invariant (or G-invariant , when the representation σis understood) if

σ(g)W ⊆ W, for all g ∈ G (i.e. σ(g)w ∈ W for all g ∈ G, w ∈ W ).

5

6 Chapter 1. Representation theory of finite groups

If this is the case, then (σ|W ,W ) is also a representation of G. We say that σ|W is asubrepresentation of G. (Note that we will use the notation ≤ for subspaces and subgroups.We reserve the symbol ⊆ for set inclusion.)

Note that the trivial spaces V and 0 are always invariant. A nonzero representation(σ, V ) is irreducible if V has no nontrivial invariant subspaces; otherwise we say it is reducible.

If (σ, V ) is a representation of G and K ≤ G is a subgroup, the restriction of σ from Gto K, denoted ResGK σ (or ResGK V ) is the representation of K of V defined by the restrictionσ|K : K → GL(V ).

A unitary space is a vector space V endowed with a Hermitian scalar product. Recallthat a Hermitian scalar product (or Hermitian inner product) is a map 〈·, ·〉V : V × V → Csuch that, for all u, v, w ∈ V and α ∈ C, we have

(a) 〈u+ v, w〉 = 〈u,w〉+ 〈v, w〉,(b) 〈u, v + w〉 = 〈u, v〉+ 〈u,w〉,(c) 〈αu, v〉 = α〈u, v〉,(d) 〈v, αv〉 = α〈u, v〉,(e) 〈u, v〉 = 〈v, u〉,(f) 〈u, u〉 ≥ 0, with equality only if u = 0,

where z denotes the complex conjugate of z ∈ C. From now on, the term scalar product willmean Hermitian scalar product.

Suppose V is a unitary space and T : V → V is a linear operator. The adjoint operatorT ∗ is defined by

〈Tu, v〉V = 〈u, T ∗v〉V , for all u, v ∈ V. (1.1)

(See Exercise 1.1.1.)If V is a unitary space, then a linear operator T : V → V is unitary if it preserves the

scalar product, i.e. if

〈Tu, Tv〉V = 〈u, v〉V , for all u, v ∈ V.

(More generally, T : V → W is unitary if 〈Tu, Tv〉W = 〈u, v〉V for all u, v ∈ V .) All unitaryoperators are invertible (Exercise 1.1.2). Furthermore, T ∈ GL(V ) is a unitary operator ifand only if T−1 = T ∗ (Exercise 1.1.3).

Suppose V is a unitary space. A representation (σ, V ) is unitary if σ(g) is a unitaryoperator for all g ∈ G or, in other words, if σ(g−1) = σ(g)∗ for all g ∈ G.

We way a representation (σ, V ) is unitarizable if there exists a scalar product on V withrespect to which σ is unitary.

Lemma 1.1.1. Every finite-dimensional representation of a finite group is unitarizable.

Proof. Let (·, ·) be an arbitrary scalar product on V . (See Exercise 1.1.4.) Then define anew scalar product on V by

〈u, v〉 =∑g∈G

(σ(g)u, σ(g)v), for all u, v ∈ V.

1.1. Basic concepts 7

Then, for all h ∈ G and u, v ∈ V , we have

〈σ(h)u, σ(h)v〉 =∑g∈G

(σ(gh)u, σ(gh)v)

=∑s∈G

(σ(s)u, σ(s)v) (setting s = gh)

= 〈u, v〉.

Hence the representation is unitary with respect to 〈·, ·〉.

In light of Lemma 1.1.1, we can assume representations are unitary, which we will do fromnow on. Note that, for infinite groups, it is not true that all representations are unitarizable.See for example, this Wikipedia entry.

Exercises.

1.1.1. Prove that, given a linear operator T : V → V , there is a unique linear operatorT ∗ : V → V satisfying (1.1).

1.1.2. Prove that all unitary operators are invertible.

1.1.3. Prove that T ∈ GL(V ) is unitary if and only if T−1 = T ∗.

1.1.4. Prove that every finite-dimensional vector space V has a scalar product.

1.1.5. Consider the infinite group Z (under addition). Let V = C2, with elements viewed ascolumn vectors. Prove that

σ : Z→ GL(V ), σ(n)(v) =

(1 n0 1

)v

is a representation of Z on V . Prove that this representation is reducible.

1.1.2 Examples

Example 1.1.2 (Trivial representation). The trivial representation ofG is the one-dimensionalrepresentation (ιG,C) given by ιG(g) = IC, for all g ∈ G.

Example 1.1.3 (Permutation representation (homogeneous space)). Suppose G acts on afinite set X, and let L(X) denote the vector space of all complex-valued functions on X.Then we can define a representation λ of G on L(X) by

(λ(g)f)(x) = f(g−1x), for all g ∈ G, f ∈ L(X), x ∈ X.

https://en.wikipedia.org/wiki/Uniformly_bounded_representation#Examples_of_non-unitarizable_representations


Note that, for all g1, g2 ∈ G, f ∈ L(X), and x ∈ X, we have(λ(g1g2)(f)

)(x) = f

((g1g2)−1x

)= f(g−1

2 g−11 x) =

(λ(g2)(f)

)(g−1x) =

(λ(g1)

(λ(g2)f

))(x).

Hence λ(g1g2) = λ(g1)λ(g2). Also, it is clear that λ(1G) = IL(X). So λ is indeed a represen-tation. We call it the permutation representation of G on L(X).

We can define a scalar product on L(X) by

〈f1, f2〉 =∑x∈X

f1(x)f2(x), for all f1, f2 ∈ L(X). (1.2)

With this scalar product, λ is a unitary representation. (See Exercises 1.1.6 and 1.1.7.)For x ∈ X, the Dirac function δx centered at x is defined by

δx(y) =

1 if y = x,

0 if y 6= x.

Then δX : x ∈ X is an orthonormal basis for L(X). In particular,

f =∑x∈X

f(x)δx, for all f ∈ L(X). (1.3)

Furthermore,λ(g)δx = δgx, for all g ∈ G, x ∈ X. (1.4)

Example 1.1.4. The group G acts on itself by multiplication on the left:

g · h = gh, for all g, h ∈ G. (1.5)

The associated permutation representation is called the left regular representation of G andis typically denoted λ. Explicitly, we have(

λ(g)f)(h) = f(g−1h), for all g, h ∈ G, f ∈ L(G).

Similarly, G acts on itself by multiplication on the right by the inverse:

g · h = hg−1, for all g, h ∈ G. (1.6)

(We must multiply by the inverse in order for this to be a group action. See Exercise 1.1.8.)The associated permutation representation is called the right regular representation of G andis typically denoted ρ. Explicitly, we have(

ρ(g)f)(h) = f(hg), for all g, h ∈ G, f ∈ L(G).

Recall that the symmetric group Sn of degree n is the group of all bijections (called per-mutations) π : 1, 2, . . . , n → 1, 2, . . . , n. Recall that a permutation is even (respectively,odd) if it is a product of an even (respectively, odd) number of transpositions.

Example 1.1.5 (The sign representation). The sign representation (or alternating represen-tation) of Sn is the one-dimensional representation (ε,C) defined by

ε(π) =

IC if π is even,

−IC if π is odd.


Exercises.

1.1.6. Prove that (1.2) defines a scalar product on L(X).

1.1.7. Prove that the permutation representation λ of Example 1.1.3 is unitary.

1.1.8. Prove that (1.5) and (1.6) define actions of G on the set G.

1.1.3 Intertwining operators

We let Hom(V,W ) denote the vector space of all linear maps from V to W . If (σ, V ) and(ρ,W ) are two representations of G, we define

HomG(V,W ) = HomG(σ, ρ) := T ∈ Hom(V,W ) : Tσ(g) = ρ(g)T ∀ g ∈ G.

We call elements of HomG(V,W ) intertwining operators (or simply interwiners), and we saythat they intertwine σ and ρ (or V and W ).

If σ and ρ are unitary, then

HomG(σ, ρ)∼=−→ HomG(ρ, σ), T 7→ T ∗, (1.7)

is an antilinear isomorphism. Here, antilinear means that

(αT1 + βT2)∗ = αT ∗1 + βT ∗2 , for all α, β ∈ C, T1, T2 ∈ HomG(σ, ρ).

To see this, note that (T ∗)∗ = T (so (1.7) is bijective) and

T ∈ HomG(σ, ρ) ⇐⇒ Tσ(g) = ρ(g)T ∀ g ∈ G⇐⇒ σ(g)∗T ∗ = T ∗ρ(g)∗ ∀ g ∈ G (taking the adjoint of both sides)

⇐⇒ σ(g−1)T ∗ = T ∗ρ(g−1) ∀ g ∈ G (since σ and ρ are unitary)

⇐⇒ T ∗ ∈ HomG(ρ, σ).

Two representations (σ, V ) and (ρ,W ) are equivalent if there is a bijective intertwinerT ∈ HomG(V,W ). In this case, we call T an isomorphism of representations and we writeσ ∼ ρ or V ∼= W . If, in addition, σ and ρ are unitary representations and T is a unitaryoperator, then we say that σ and ρ are unitarily equivalent .

Recall that a bijective operator T ∈ Hom(V,W ) has a unique polar decomposition T =U |T |, where |T | ∈ GL(V ) is the (positive definite) square root of the positive operator T ∗T ,and U ∈ Hom(U, V ) is unitary. (See, for example, [Roy, Th. 10.5.5] or [Tri, Th. 3.5].)

Lemma 1.1.6. Two unitary representations of a finite group are equivalent if and only ifthey are unitarily equivalent.


Proof. Since unitarily equivalent representations are equivalent by definition, it suffices toprove the reverse implication. Let (σ, V ) and (ρ,W ) be unitary representations of a finitegroup G and suppose they are equivalent. Then there exists a bijection T ∈ HomG(V,W ),and

T ∗T ∈ HomG(V, V ) ∩GL(V ).

Let T = U |T | be the polar decomposition of T . Then, for all g ∈ G, we have(σ(g−1)|T |σ(g)

)2= σ(g−1)|T |σ(g)σ(g−1)|T |σ(g)

= σ(g−1)|T |2σ(g) (since σ is a group homomorphism)

= σ(g−1)T ∗Tσ(g) (by the definition of |T |)= T ∗ρ(g−1)ρ(g)T (since T and T ∗ are intertwiners)

= T ∗T (since ρ is a group homomorphism).

The uniqueness of positive square root of T ∗T implies that

σ(g−1)|T |σ(g) = |T | =⇒ |T |σ(g) = σ(g)|T |.

Hence |T | ∈ HomG(V, V ). Then, for all g ∈ G, we have

Uσ(g) = T |T |−1σ(g) = Tσ(g)|T |−1 = ρ(g)U.

Thus U is a unitary equivalence from σ to ρ.

Definition 1.1.7. We let Irr(G) denote the set of all (unitary) irreducible representations of

G, and we let G = Irr(G)/ ∼ denote the set of equivalence classes of Irr(G). By a slight abuse

of notation, we will often identity G with a fixed set of representatives of these equivalenceclasses, that is, a set of irreducible (unitary) pairwise inequivalent representations of G.

Remark 1.1.8. For those students who know a bit of category theory, one can define acategory of (finite-dimensional) representations of a fixed group G. The objects are finite-dimensional representations and the morphisms are intertwiners. One can check that theaxioms of a category are satisfied. For example, the composition of intertwiners is again anintertwiners, and this composition is associative (see Exercise 1.1.9).

Exercises.

1.1.9. Suppose that (σi, Vi), i = 1, 2, 3, are representations of G, that S ∈ HomG(σ1, σ2),and that T ∈ HomG(σ2, σ3). Prove that the composition TS is an element of HomG(V1, V3).

1.1.10. Suppose (σ, V ) and (ρ,W ) are representations of G. Prove that if T ∈ HomG(σ, ρ)is invertible, then T−1 ∈ HomG(ρ, σ).


1.1.4 Direct sums and Maschke’s Theorem

Suppose that (σj, Vj), j = 1, 2, . . . ,m, are representations of a group G. Their direct sum is

the representation (σ, V ) =(⊕m

j=1 σj,⊕m

j=1 Vj

)defined by

σ(g)(vj)mj=1 = (σj(g)vj)

mj=1 .

We will often write the element (vj)mj=1 of a direct sum

⊕mj=1 Vj as

∑mj=1 vj.

Conversely, suppose (σ, V ) is a representation of G such that

• V =⊕m

j=1 Vj (as vector spaces), and

• the subspace Vj is σ-invariant for each j = 1, . . . ,m.

Then, for j = 1, . . . ,m, we can define

σj(g) = σ(g)|Vj : Vj → Vj, g ∈ G,

and we have

σ =m⊕j=1

σj.

Theorem 1.1.9 (Maschke’s Theorem). Every finite-dimensional representation of a finitegroup can be decomposed as a direct sum of irreducible representations. Furthermore, thedecomposition can be chosen to be orthogonal (i.e. elements of distinct direct summands areorthogonal to each other).

Proof. Let (σ, V ) be a finite-dimensional representation of G. As noted after Lemma 1.1.1,we may assume that σ is unitary.

We prove the result by induction on the dimension of V . If dimV = 0 or dimV = 1,then V is clearly irreducible. Now suppose dimV > 1, and that the result has been provedfor all representations of dimension less than V .

If V is irreducible, we are done. Thus, we suppose V is reducible. Then there is anontrivial σ-invariant subspace W ≤ V . We claim that its orthogonal complement

W⊥ = v ∈ V : 〈v, w〉V = 0 ∀ w ∈ W

is also σ-invariant. Indeed, for g ∈ G and v ∈ W⊥, we have

〈σ(g)v, w〉V = 〈v, σ(g−1)w〉V = 0, for all w ∈ W,

since σ(g−1)w ∈ W and v ∈ W⊥. Thus σ(g)v ∈ W⊥ and so W⊥ is σ-invariant.By the above, we have the orthogonal decomposition

V = W ⊕W⊥.

Since W is nontrivial, we have dimW, dimW⊥ < dimV . Therefore, by the inductive hypot-hesis, W and W⊥ have orthogonal decompositions into direct sums of irreducible represen-tations. This completes the proof of the inductive step.


Remark 1.1.10. Maschke’s Theorem relies on our assumptions about the group and ourchoice of C as the ground field. More generally, Maschke’s Theorem holds for finite groupsand representations over a field whose characteristic does not divide the order of the group.See Exercise 1.1.12.

Remark 1.1.11. A representation that cannot be decomposed into a direct sum of two nontri-vial representations is said to be indecomposable. Irreducible representations are always in-decomposable (for arbitrary groups and arbitrary ground fields). Maschke’s Theorem saysthat, under additional assumptions, the converse is true: indecomposable representationsare also irreducible. However, this is not true in general. See Exercise 1.1.11.

Exercises.

1.1.11. Show that the representation of Exercise 1.1.5 cannot be decomposed as a direct sumof irreducible representations. Why does this not violate Maschke’s Theorem?

1.1.12. Let p be a prime number. In this exercise, we work over the field Zp with p elements,instead of over C. Suppose G is a finite group whose order is divisible by p. Consider theleft regular representation (λ, L(G)) of G. Let

V =

f ∈ L(G) :

∑g∈G

f(g) = 0

.

(a) Prove that V is G-invariant.

(b) Prove that V ∩W 6= 0 for every nonzero G-invariant subspace W of L(G). Hint :Let f be a nonzero element of W such that f 6∈ V . Define s ∈ L(G) by s(g) = 1 forall g ∈ G. Note that s ∈ V . Prove that s ∈ W by considering

∑g∈G

(λ(g)f

).

(c) Explain how this example shows that Maschke’s Theorem does not hold when thecharacteristic of the field divides the order of the group.

1.1.5 The adjoint representation

Recall that the dual of the vector space V , denoted V ′ is the vector space of all linearfunctions V → C. If V is endowed with scalar product 〈·, ·〉V , then we have the Riesz map

V → V ′, v 7→ θv, where θv(w) := 〈w, v〉V , for all w ∈ V.

This map is antilinear:

θαv+βw = αθv + βθw, for all α, β ∈ C, v, w ∈ V.

Since V is finite-dimensional, this map is also bijective. (This follows from the Riesz repre-sentation theorem. See, for example, [Tri, Th. 2.1]. Alternatively, you can prove it directly.)


We have dual scalar product on V ′ defined by

〈θv, θw〉V ′ = 〈w, v〉V , for all v, w,∈ V. (1.8)

If v1, v2, . . . , vn is an orthonormal basis for V , then the corresponding dual basis of V ′ isθv1 , . . . , θvn. It is characterized by the property that

θvi(vj) = δi,j.

Suppose (σ, V ) is a representation of G. The adjoint (or conjugate, or contragredient)representation is the representation (σ′, V ′) of G defined by

(σ′(g)f) v = f(σ(g−1)v

), for all g ∈ G, f ∈ V ′, v ∈ V. (1.9)

Note that

(σ′(g)θw) v = θw(σ(g−1)v

)= 〈σ(g−1)v, w〉V= 〈v, σ(g)w〉V= θσ(g)w(v).

Thusσ′(g)θw = θσ(g)w, for all g ∈ G, w ∈ V. (1.10)

It follows that σ is irreducible if and only if σ′ is irreducible (Exercise 1.1.13).

Exercises.

1.1.13. Prove that a representation (σ, V ) of G is irreducible if and only if the adjoint repre-sentation (σ′, V ′) is irreducible.

1.1.6 Matrix coefficients

As we know from linear algebra, if we choose a basis for V , then any linear transformationof V can be represented by a matrix. Let (σ, V ) be a unitary representation of G, and letv1, . . . , vn be an orthonormal basis of V . Then the matrix coefficients associated with thisbasis are given by

Ui,j(g) = 〈σ(g)vj, vi〉V , g ∈ G, i, j = 1, . . . , n.

For g ∈ G, let U(g) =(Ui,j(g)

)ni,j=1

∈ Mn,n(C) be the matrix whose entries are the matrix

coefficients of (σ, V ). Recall that if M ∈Mn,n(C), then its conjugate transpose is the matrixM∗ with entries given by

M∗i,j = Mj,i.

The matrix M is unitary if it is invertible and M−1 = M∗.


Lemma 1.1.12. For all g, g1, g2 ∈ G, we have

(a) U(g1g2) = U(g1)U(g2),

(b) U(g−1) = U(g)∗,

(c) U(g) is unitary,

(d) the matrix coefficients of the adjoint representation σ′ with respect to the dual basisθv1 , . . . , θvn are

〈σ′(g)θvj , θvi〉V ′ = Ui,j(g).

Proof. (a) We have

Ui,j(g1g2) = 〈σ(g1g2)vj, vi〉V= 〈σ(g1)σ(g2)vj, vi〉V (σ is a group homomorphism)

=

⟨σ(g1)

(n∑k=1

〈σ(g2)vj, vk〉V vk

), vi

⟩V

(w =

n∑k=1

〈w, vk〉vk for all w ∈ V

)

=n∑k=1

〈σ(g1)vk, vi〉V 〈σ(g2)vj, vk〉V (the scalar product is linear)

=n∑k=1

Ui,k(g1)Uk,j(g2).

(b) We have

Ui,j(g−1) = 〈σ(g−1)vj, vi〉V

= 〈vj, σ(g)vi〉V (σ(g−1) = σ(g)∗)

= 〈σ(g)vi, vj〉V (property of the scalar product)

= Uj,i(g).

(c) It follows from part (a) that U(g−1) = U(g)−1. Then it follows from part (b) thatU(g) is unitary.

(d) We have

〈σ′(g)θvj , θvi〉V ′(1.10)= 〈θσ(g)vj , θvi〉V

(1.8)= 〈vi, σ(g)vj〉V= 〈σ(g)vj, vi〉V (property of the scalar product)

= Ui,j(g)

It follows from Lemma 1.1.12 that we have a group homomorphism

U : G→ U(n), g 7→ U(g),

where U(n) is the group of n × n unitary matrices (under multiplication). This grouphomomorphism is called a unitary matrix realization of σ. Note that it depends on thechoice of orthonormal basis.


Exercises.

1.1.14. Suppose B = v1, . . . , vn and B′ = w1, . . . , wn are two orthonormal bases for V .Then we have a “change of basis matrix” A whose (i, j) entry is given by

Ai,j = 〈vj, wi〉V .

Let U and U ′ be the unitary matrix realizations of a representation σ on V in terms of thebases B and B′, respectively. State and prove an equality relating U(g), U ′(g), and A thatholds for all g ∈ G.

1.1.7 Tensor products

In this section, we introduce the important notion of a tensor product. Rather than give themost general possible definition (that of a tensor product of bimodules over rings), we givea more direct definition that is suitable for our purposes.

Suppose V and W are finite-dimensional unitary spaces over C. The tensor productV ⊗W is the vector space consisting of all maps

B : V ×W → C

that are bi-antilinear:

B(α1v1 + α2v2, w) = α1B(v1, w) + α2B(v2, w),

B(v, α1w1 + α2w2) = α1B(v, w1) + α2B(v, w2),

for all α1, α2 ∈ C, v, v1, v2 ∈ V , and w,w1, w2 ∈ W .For v ∈ V and w ∈ W we define the simple tensor v ⊗ w ∈ V ⊗W by

(v ⊗ w)(v1, w1) = 〈v, v1〉V 〈w,w1〉W .

The mapV ×W → V ⊗W, (v, w) 7→ v ⊗ w,

is bilinear:

(α1v1 + α2v2)⊗ (β1w1 + β2w2) =2∑

i,j=1

αiβjvi ⊗ wj,

for all α1, α2, β1, β2 ∈ C, v1, v2 ∈ V , and w1, w2 ∈ W .

Lemma 1.1.13. If v1, . . . , vn is an orthonormal basis for V and w1, . . . , wm is an ort-honormal basis for W , then

vi ⊗ wj : 1 ≤ i ≤ n, 1 ≤ j ≤ m (1.11)

is a basis for V ⊗W . In particular,

dim(V ⊗W ) = (dimV )(dimW ).


Proof. Suppose B ∈ V ⊗W , v ∈ V , and w ∈ W . Then there exist α1, . . . , αn, β1, . . . , βm ∈ Csuch that

v =n∑i=1

αivi and w =m∑i=1

βiwi.

Then, for i = 1, . . . , n, we have

〈vi, v〉V =n∑k=1

αk〈vi, vk〉V = αk.

Similarly, for j = 1, . . . ,m, we have 〈wj, w〉W = βj. Therefore,

B(v, w) =n∑i=1

m∑j=1

αiβjB(vi, wj) =

(n∑i=1

m∑j=1

B(vi, wj)vi ⊗ wj

)(v, w).

It follows that B =∑n

i=1

∑mj=1B(vi, wj)vi ⊗wj. So every element of V ⊗W can be written

in a unique way as a linear combination of the elements of (1.11). Thus, these elements forma basis for V ⊗W .

Remark 1.1.14. It is important to note that not every element of V ⊗W can be written asa simple tensor. When working with an arbitrary element of a tensor product, one mustconsider finite sums of simple tensors.

In the notation of Lemma 1.1.13, we define a scalar product on V ⊗W by

〈vi ⊗ wk, vj ⊗ w`〉 = 〈vi, vj〉V · 〈wk, w`〉W = δi,jδk,`, (1.12)

and extending by linearity. Thus the basis (1.11) is orthonormal.Suppose V1, V2,W1,W2 are unitary spaces. If T ∈ Hom(V1, V2) and S ∈ Hom(W1,W2),

we define

T ⊗ S ∈ Hom(V1 ⊗W1, V2 ⊗W2), (T ⊗ S)(v ⊗ w) = (Tv)⊗ (Sw), v ∈ V1, w ∈ W1,

and extend by linearity.

Lemma 1.1.15. We have an isomorphism of vector spaces

Hom(V1 ⊗W1, V2 ⊗W2) ∼= Hom(V1, V2)⊗ Hom(W1,W2).

Proof. Since the dimensions of the two spaces are both (dimV1)(dimV2)(dimW1)(dimW2),it suffices to prove that every element of Hom(V1 ⊗W1, V2 ⊗W2) is a linear combination ofelements of the form T ⊗ S for T ∈ Hom(V1, V2) and S ∈ Hom(W1,W2). For i = 1, 2, letvi,1, . . . , vi,ni be a basis of Vi, and let wi,1, . . . , wi,mi be a basis of Wi. Choose

a ∈ 1, . . . , n1, b ∈ 1, . . . , n2, c ∈ 1, . . . ,m1, d ∈ 1, . . . ,m2.

Define

Ta,c ∈ Hom(V1,W1), T (v1,i) =

w1,c if i = a,

0 if i 6= a,


Sb,d ∈ Hom(V2,W2), T (v2,j) =

w2,d if j = b,

0 if j 6= b,

Then

(Ta,c ⊗ Sb,d)(v1,i ⊗ v2,j) =

w1,c ⊗ w2,d if i = a, j = b,

0 otherwise.

Since v1,i ⊗ v2,j : 1 ≤ i ≤ n1, 1 ≤ j ≤ n2 is a basis of V1 ⊗ V2 and w1,i ⊗ w2,j : 1 ≤i ≤ m1, 1 ≤ j ≤ m2 is a basis of W1 ⊗W2, very element of Hom(V1 ⊗ V2,W1 ⊗W2) canbe written as a linear combination of the above maps (for various choices of a, b, c, d). Thiscompletes the proof.

Now suppose G1 and G2 are finite groups. Let (σi, Vi) be a representation of Gi fori = 1, 2. The outer tensor product of σ1 and σ2 is the representation σ1 σ2 of G1 × G2

given by

σ1 σ2 : G1 ×G2 → GL(V1 ⊗ V2),

(σ1 σ2)(g1, g2) = σ1(g1)⊗ σ2(g2), for all g1 ∈ G1, g2 ∈ G2.

Recall that we have the diagonal embedding (a group homomorphism)

G→ G×G, g 7→ (g, g).

Suppose (σ1, V1) and (σ2, V2) are two representations of the same group G. Then the inter-nal tensor product of σ1 and σ2, denoted σ1 ⊗ σ2, is the representation of G given by thecomposition

G→ G×G σ1σ2−−−→ GL(V1 ⊗ V2).

In other words, the σ1 ⊗ σ2 is given by

(σ1 ⊗ σ2)(g) = σ1(g)⊗ σ2(g).

Exercises.

1.1.15. Suppose (σi, Vi) is a representation of Gi for i = 1, 2. Let Wi be a σi-invariantsubspace of Vi for i = 1, 2. Prove that W1⊗W2 is a σ1σ2-invariant subspace of V1⊗V2. (Herewe identify W1⊗W2 with the subspace of V1⊗V2 spanned by w1⊗w2 : w1 ∈ W1, w2 ∈ W2.)

1.1.16. Prove that the scalar product defined in (1.12) is indeed a Hermitian scalar producton V ⊗W .


1.1.8 Cyclic and invariant vectors

Let (σ, V ) be a unitary representation of G. For v ∈ V ,

〈σ(g)v : g ∈ G〉

is a σ-invariant subspace of V , called the subspace generated by v. (We use the notation 〈〉to denote the C-span.) If this space is all of V , then we say that v is a cyclic vector.

We say that a vector v ∈ V is σ-invariant or fixed if

σ(g)v = v, for all g ∈ G.

We let

V G = v ∈ V : σ(g)v = v ∀ g ∈ G

denote the subspace of all σ-invariant vectors. More generally, if K ≤ G is a subgroup, welet

V K := v ∈ V : σ(k)v = v ∀ k ∈ K (1.13)

be the subspace of K-invariant vectors.

Lemma 1.1.16. Suppose that u ∈ V G, u 6= 0. If v ∈ V is orthogonal to u (i.e. 〈u, v〉V = 0),then v is not cyclic.

Proof. For all g ∈ G, we have

〈u, σ(g)v〉V = 〈σ(g−1)u, v〉V= 〈u, v〉V (since u ∈ V G)

= 0.

Hence σ(g)v ∈ 〈u〉⊥. So

〈σ(g)v : g ∈ G〉 ⊆ 〈u〉⊥ V,

and hence v is not cyclic.

The following corollary will be useful in our study of the representation theory of thesymmetric group.

Corollary 1.1.17. Suppose that there exists a cyclic vector v ∈ V , and g ∈ G, λ ∈ C,λ 6= 1, such that σ(g)v = λv. Then V G = 0.

Proof. Suppose u ∈ V G. Thus σ(g)u = u. So u and v are eigenvectors for the unitaryoperator with distinct eigenvalues. Hence there are orthogonal. By Lemma 1.1.16, u = 0.

1.2. Schur’s lemma and the commutant 19

Exercises.

1.1.17. Show that if dimV G ≥ 2, then V has no cyclic vectors. Hint : Suppose u,w ∈ V G

are nonzero and orthogonal. Then, for v ∈ V \ V G, we have dim〈u,w, v〉 = 3, and so thereexists a nonzero u0 ∈ 〈u,w〉 ⊆ V G such that 〈u0, v〉 = 0.

1.1.18. Suppose G acts transitively on a set X.

(a) Show that dimL(X)G = 1.

(b) Show that the vectors δx are cyclic.

1.2 Schur’s lemma and the commutant

In this section we prove Schur’s lemma, which is a very useful result in representation theory,even though its proof is quite simple. We also discuss decompositions of representations intoisotypic components, commutants, and how these relate to spaces of intertwiners.

1.2.1 Schur’s lemma

Lemma 1.2.1 (Schur’s lemma). Suppose (σ, V ) and (ρ,W ) are irreducible representationsof G.

(a) Every nonzero element of HomG(σ, ρ) is an isomorphism.

(b) We have HomG(σ, σ) = CIV .

Proof. (a) Suppose T ∈ HomG(σ, ρ). Then KerT ≤ V and ImT ≤ W are G-invariant(Exercise 1.2.1). Since V is irreducible, this implies that KerT = V or KerT = 0. IfKerT = V , then T = 0. Otherwise, T is injective. Also, if T 6= 0, then ImT 6= 0, and soImT = W since W is irreducible. So T is also surjective.

(b) Suppose T ∈ HomG(σ, σ). Since C is algebraically closed, T has at least one eigen-value. So there exists λ ∈ C such that Ker(T − λIV ) 6= 0. This implies that T − λIV is notinjective. So, by part (a), T − λIV = 0. Hence T = λIV .

Corollary 1.2.2. If σ and ρ are irreducible representations of G, then

dim HomG(σ, ρ) =

1 if σ ∼ ρ,

0 if σ 6∼ ρ.

Corollary 1.2.3. Every irreducible representation of an abelian group is one-dimensional.


Proof. Suppose (σ, V ) is an irreducible representation of an abelian group G. For everyh ∈ G, we have

σ(h)σ(g) = σ(hg) = σ(gh) = σ(g)σ(h), for all g ∈ G.

Thus σ(h) ∈ HomG(σ, σ). By Lemma 1.2.1(b), we have σ(h) = χ(h)IV for some χ(h) ∈ C.Since this holds for all h ∈ G, it follows that every subspace of V is invariant. Since V isirreducible, we must have dimV = 1.

Example 1.2.4 (Representations of cyclic groups). Suppose

Cn = 〈a : an = 1〉

is a cyclic group of order n. By Corollary 1.2.3, every irreducible representation correspondsto a group homomorphism

χ : Cn → GL(C) = C× := C \ 0.

Such a map is uniquely determined by χ(a). Since the only relation in the group is an = 1,we can choose any value for χ(a) such that χ(a)n = 1, that is, such that χ(a) is an n-th rootof unity. So every representation is of the form

χj : Cn → C, χj(ak)

= χj(a)k = exp

(k

2πij

n

), k ∈ Z,

for some j ∈ 0, 1, . . . , n − 1. Since one-dimensional representations are equivalent if andonly if they corresponds to the same χ (Exercise 1.2.2), we have

Cn = χ0, . . . , χn−1.

In particular, the only irreducible representations of C2, up to isomorphism are the trivialrepresentation χ0 = ιC2 (Example 1.1.2) and the sign representation χ1 = ε (Example 1.1.5).

Exercises.

1.2.1. Suppose (σ, V ) and (ρ,W ) are representations of G and T ∈ HomG(V,W ). Prove thatKerT is a σ-invariant subspace of V and that ImT is a ρ-invariant subspace of W .

1.2.2. Prove that two one-dimensional representations (σ,C) and (ρ,C) are equivalent if andonly if they are equal.


1.2.2 Multiplicities and isotypic components

A linear transformation E ∈ Hom(V, V ) is a projection if it is idempotent : E2 = E. If, inaddition, ImE is orthogonal to KerE, we say that E is an orthogonal projection of V ontoImE. It is not hard to verify that a projection E is orthogonal if and only if it is self-adjoint ,that is, E = E∗. (See Exercise 1.2.3.)

Now suppose (σ, V ) is a representation of G. By Maschke’s Theorem (Theorem 1.1.9),we have an orthogonal decomposition

V =⊕ρ∈G

Vρ,

whereVρ ∼= W⊕mρ

ρ := Wρ ⊕ · · · ⊕Wρ︸︷︷︸mρ summands

is an orthogonal direct sum ofmρ copies of an irreducible representationWρ in the equivalence

class ρ ∈ G, for some mρ ∈ N = Z≥0. We adopt the convention that U⊕0 = 0 for a vectorspace U . The summand Vρ is called the ρ-isotypic component of V , and mρ is called themultiplicity of ρ in σ (or of Wρ in V ).

LetGσ = ρ ∈ G : mρ ≥ 1

be the set of isomorphism classes of irreducible representations of G that appear with nonzeromultiplicity in σ. The inclusions of the summands yield interwiners

Iρ,j ∈ HomG(Wρ, V ), ρ ∈ Gσ, 1 ≤ j ≤ mρ.

So we have

V =⊕ρ∈Gσ

mρ⊕j=1

Iρ,jWρ. (1.14)

Thus, every v ∈ V can be written uniquely in the form

v =∑ρ∈Gσ

mρ∑j=1

vρ,j, vρ,j ∈ Iρ,jWρ.

For each ρ ∈ Gσ and 1 ≤ j ≤ mρ, we have the orthogonal projection

Eρ,j ∈ HomG(V, V ), Eρ,j

∑η∈Gσ

mη∑j=1

vη,j

= vρ,j.

It follows that

IV =∑ρ∈Gσ

mρ∑j=1

Eρ,j. (1.15)


Lemma 1.2.5. For ρ ∈ Gσ, the intertwiners Iρ,1, . . . , Iρ,mρ form a basis of HomG(Wρ, V ).In particular, mρ = dim HomG(Wρ, V ).

Proof. Suppose T ∈ HomG(Wρ, V ). Then

T = IV T =∑η∈Gσ

mη∑j=1

Eη,jT.

The domain of Eη,jT is Wρ, while its image is Iη,jWη∼= Wη. Thus, by Schur’s lemma

(Lemma 1.2.1), we have

Eη,jT =

0, if η 6= ρ,

αjIρ,j for some αj ∈ C, if η = ρ.

Hence,

T =

mρ∑j=1

αjIρ,j.

Since this decomposition is unique, the lemma follows.

Corollary 1.2.6. With notation as in Lemma 1.2.5, we have mρ = dim HomG(V,Wρ).

Proof. This follows from Lemma 1.2.5 and (1.7). Alternatively, it can be proved directly,using an argument analogous to that used to prove Lemma 1.2.5 (Exercise 1.2.4).

The isotypic summands Vρ of V are unique. However, the decomposition of Vρ into asum of irreducible representations is not unique when mρ > 1; it corresponds to a choice ofbasis for HomG(Wρ, V ).

For ρ ∈ Gσ and 1 ≤ j, k ≤ mρ, consider the intertwiner T ρk,j ∈ HomG(V, V ) defined to bethe composition

T ρk,j : V Iρ,jWρ

Iρ,kI−1ρ,j−−−−→ Iρ,kWρ → V, (1.16)

where and → denote the projection onto and inclusion of the given summand of V , andI−1ρ,j denotes the inverse of Iρ,j when its codomain is replaced by its image (so that it becomes

invertible, since it is injective). It follows that

T ρk,jTηs,t = δρ,ηδj,sT

ρk,t. (1.17)

Note also that T ρj,j = Eρ,j. Furthermore, it follows from Corollary 1.2.2 that

HomG(Iρ,jWρ, Iρ,kWρ) = CT ρk,j.


Exercises.

1.2.3. Prove that a projection is orthogonal if and only if it is self-adjoint.

1.2.4. Prove Corollary 1.2.6 directly, using an argument analogous to that used to proveLemma 1.2.5.

1.2.5. Prove that a representation (σ, V ) is reducible if and only if HomG(σ, σ) has nontrivialidempotents. (The trivial idempotents are 0 and IV .)

1.2.3 Finite-dimensional algebras

An (associative) algebra over C is a vector space A with bilinear product A×A → A suchthat A is a ring (possibly without unit) with respect to this product and the vector addition.

A subalgebra of A is a subspace B ≤ A that is closed under multiplication. An involutionof A is a bijective map A 7→ A∗ such that

• (A∗)∗ = A,

• (αA+ βB)∗ = αA∗ + βB∗, and

• (AB)∗ = B∗A∗,

for all α, β ∈ C and A,B ∈ A. An algebra with involution is called an involutive algebra or∗-algebra. An element A in an involutive algebra is self-adjoint if A∗ = A.

The algebra A is commutative if it is commutative as a ring:

AB = BA, for all A,B ∈ A.

The center of A is the commutative algebra

Z(A) = B ∈ A : AB = BA for all A ∈ A.

The algebra A is unital if there exists I ∈ A such that

AI = IA = A, for all A ∈ A.

The unit is unique: If I ′ is another unit, then

I = II ′ = I ′.

Furthermore,I∗ = I∗I = ((I∗I)∗)∗ = (I∗(I∗)∗)∗ = (I∗I)∗ = (I∗)∗ = I.

So the unit is self-adjoint.Suppose A1 and A2 are algebras. A map φ : A1 → A2 is an algebra homomorphism if

• φ is linear,


• φ is multiplicative: φ(AB) = φ(A)φ(B) for all A,B ∈ A1.

If A1 and A2 are involutive, then φ is a ∗-homomorphism if it is an algebra homomorphismand it preserves the involution:

φ(A∗) = φ(A)∗, for all A ∈ A1.

If, in addition, φ is bijective, then we call it a ∗-isomorphism, and we say that A1 and A2

are ∗-isomorphic.A map φ : A1 → A2 is a ∗-anti-homomorphism if it satisfies the conditions of a ∗-

homomorphism, except that we replace the multiplicative property by the anti-multiplicativeproperty

φ(AB) = φ(B)φ(A).

If φ is also bijective, it is called a ∗-anti-isomorphism, and we say that A1 and A2 are∗-anti-isomorphic..

Suppose A1,A2, . . . ,Ak, k ≥ 2, are algebras. Their direct sum, denoted A1 ⊕ · · · ⊕ Ak,is equal to A1 ⊕ · · · ⊕ Ak as a vector space, with componentwise product:

(A1, . . . , Ak)(B1, . . . , Bk) = (A1B1, . . . , AkBk), for all Ai, Bi ∈ Ai, 1 ≤ i ≤ k.

The algebra generated by a subset B ⊆ A, denoted 〈B〉, is the smallest subalgebra of Acontaining B. Explicitly, 〈B〉 is the set of all linear combinations of products of elements ofB.

The dimension of A is its dimension as a vector space. Suppose A is finite dimensional,and let e1, . . . , ed be a basis of A. Then the structure coefficients ci,j,k ∈ C, 1 ≤ i, j, k ≤ ddefined by

eiej =d∑

k=1

ci,j,kek

uniquely determine the product in A.

Example 1.2.7 (Endomorphism algebra). The endomorphism algebra End(V ) := Hom(V, V )is a unital ∗-algebra with the usual vector space structure and multiplication (compositionof operators). The involution is the map T 7→ T ∗, where T ∗ is the adjoint of T . Moreover, if(σ, V ) is a unitary representation of G, then the subalgebra HomG(V, V ) of End(V ) is alsoa unital ∗-algebra.

Example 1.2.8 (Matrix algebra). The matrix algebra Mm,m(C) of complex m×m matrices isa unital ∗-algebra under matrix multiplication, where the involution is givin by the conjugatetranspose. If V = Cm, then Mm,m(C) ∼= End(V ) (∗-isomorphism). The center of Mm,m(C)is

Z(Mm,m(C)) = λI : λ ∈ C ∼= C,

where I is the identity matrix (Exercise 1.2.6).


Exercises.

1.2.6. Prove that Z(Mm,m(C)) = λI : λ ∈ C.

1.2.7. Prove that if φ : A1 → A2 is a homomorphism of algebras, then Imφ is a subalgebraof A2. If, in addition, φ is a ∗-homomorphism, prove that Imφ is an involutive algebra, withinvolution induced by the involution on A2.

1.2.8. A subspace J of an algebra A is a (two-sided) ideal of A if

AB ∈ J and BA ∈ J, for all A ∈ A, B ∈ J.

(a) Prove that if φ : A1 → A2 is a homomorphism of algebras, then Kerφ is an ideal of A1.

(b) If J is an ideal of A, define a natural algebra structure on the quotient vector spaceA/J .

(c) If J is an ideal of A that is ∗-invariant (i.e. A∗ ∈ J for all A ∈ J), define a naturalinvolution on the quotient algebra A/J and prove that this gives the quotient thestructure of an involutive algebra.

(d) If φ : A1 → A2 is a ∗-homomorphism, prove that Imφ ∼= A1/Kerφ as ∗-algebras.

1.2.9. Let m ≥ 2 and consider the algebra C[x]/(xm), where (xm) = C[x]xm is the idealof C[x] generated by xm. Prove that C[x]/(xm) is not isomorphic as an algebra to Cm =⊕m

k=1C, even though these two algebras are both commutative and both have dimension m.Hint : The image of x in the quotient C[x]/(xm) has a property that no element of Cm has.

1.2.10. An element E of an algebra A is idempotent if E2 = E. Of course, every unitalalgebra has the trivial idempotents 0 and I. Suppose a unital algebra A has a nontrivialcentral idempotent E, that is, E is a nontrivial idempotent in the center of A. Show that

EAE := EAE : A ∈ A and (I − E)A(I − E) := (I − E)A(I − E) : A ∈ A

are subalgebras of A with units. (Here we do not require the units of the subalgebras to bethe same as the unit of A.) Furthermore, prove that A = EAE ⊕ (I − E)A(I − E) (directsum of algebras). Thus, central idempotents allow one to decompose algebras (or rings) asdirect sums of smaller algebras (or rings).

1.2.4 The commutant

Suppose (σ, V ) is a representation of G.

Definition 1.2.9 (Commutant). The algebra EndG(V ) := HomG(V, V ) is called the com-mutant of (σ, V ).

Recall the elements T ρk,j ∈ EndG(V ) defined in (1.16).


Theorem 1.2.10. The set

T ρk,j : ρ ∈ Gσ, 1 ≤ k, j ≤ mρ (1.18)

is a basis for EndG(V ). Furthermore, the map

EndG(V )→⊕ρ∈Gσ

Mmρ,mρ(C),∑ρ∈Gσ

mρ∑k,j=1

αρk,jTρk,j 7→

⊕ρ∈Gσ

(αρk,j)mρk,j=1

, (1.19)

is an isomorphism of algebras.

Proof. Suppose T ∈ EndG(V ). Then

T = IV TIV(1.15)=

∑ρ∈G

mρ∑k=1

Eρ,k

T

∑η∈G

mη∑j=1

Eη,j

=∑ρ,η∈G

mρ∑k=1

mη∑j=1

Eρ,kTEη,j.

Note that Im(Eρ,kTEη,j) ≤ Iρ,kWρ. Thus, the restriction of Eρ,kTEη,j to Iη,jWη is an intert-wining operator from Iη,jWη to Iρ,kWρ. Therefore, by Corollary 1.2.2, we have

Eρ,kTEη,j =

0, if η 6∼ ρ,

αρk,jTρk,j for some αρk,j ∈ C, if ρ ∼ η.

This proves that the set (1.18) spans EndG(V ).It remains to prove that the set (1.18) is linearly independent. Suppose that

∑ρ∈Gσ

mρ∑k,j=1

αρk,jTρk,j = 0.

For ρ ∈ Gσ, choose a nonzero v ∈ Iρ,jWρ. Since T ηk,`v = 0 for η 6∼ ρ or ` 6= j, we have

mρ∑k=1

αρk,jTρk,jv = 0

Since the T ρk,jv, 1 ≤ k ≤ mρ, belong to different summands in a decomposition of V intoirreducible subrepresentations, they are linearly independent. Hence αρk,j = 0 for all 1 ≤ k ≤mρ.

The fact that (1.19) is an isomorphism of algebras follows from (1.17).

Corollary 1.2.11. We have dim EndG(V ) =∑

ρ∈Gσ m2ρ.

Definition 1.2.12. We say that the representation (σ, V ) is multiplicity free if mρ ≤ 1 for

all ρ ∈ G, or equivalently, if mρ = 1 for all ρ ∈ Gσ.

Corollary 1.2.13. The representation (σ, V ) is multiplicity free if and only if its commutantEndG(V ) is commutative.


Proof. This follows from Theorem 1.2.10 and the fact that the matrix algebra Mm,m(C) iscommutative if and only if m = 1.

One of the nice properties of multiplicity free representations is that their decompositioninto irreducible subrepresentations is unique, since it is the same as their decomposition intoisotypic components.

Note that, for ρ ∈ Gσ,

Eρ :=

mρ∑j=1

Eρ,j =

mρ∑j=1

T ρj,j

is the projection from V onto the ρ-isotypic component Vρ =⊕mρ

j=1 Iρ,jWρ. The projectionEρ is called the minimal central projection or minimal central idempotent corresponding toρ. The Eρ,j are called minimal projections or minimal idempotents .

Corollary 1.2.14. The center Z(EndG(V )) is isomorphic, as an algebra, to CGσ . Further-

more, the minimal central projections Eρ : ρ ∈ Gσ are a basis for the center.

Proof. This follows from Theorem 1.2.10 and Exercise 1.2.6.

Exercises.

1.2.11. Let (σ, V ) be a representation of G and ρ ∈ Gσ. Suppose Eρ = A + B for centralidempotents A,B ∈ EndG(V ) (i.e. A2 = A, B2 = B, and A,B ∈ Z(EndG(V ))). Prove thateither Eρ = A (hence B = 0) or Eρ = B (hence A = 0). This justifies the term minimalcentral idempotent.

1.2.12. Let (σ, V ) be a representation of G. Prove that Eρ,j, ρ ∈ Gσ, 1 ≤ j ≤ mρ, cannotbe written as a sum of two nontrivial orthogonal idempotents. In other words, prove that ifEρ,j = A + B for A,B ∈ EndG(V ) with A2 = A, B2 = B, and AB = BA = 0 (we say theidempotents A and B are orthogonal if AB = BA = 0), then Eρ,j = A (hence B = 0) orEρ,j = B (hence A = 0). This justifies the term minimal idempotent (sometimes also calleda primitive idempotent .)

1.2.13. Suppose (σ, V ) and (η, U) are two representations of G with decompositions

V =⊕ρ∈Gσ

W⊕mρρ and U =

⊕ρ∈Gη

W⊕nρρ

into irreducible subrepresentations. Prove that we have an isomorphism of vector spaces

HomG(V, U) ∼=⊕

ρ∈Gσ∩Gη

Mnρ,mρ(C).


1.2.5 Intertwiners as invariant elements

We have a canonical (i.e. basis independent) isomorphism of vector spaces

W ′ ⊗ V ∼= Hom(W,V ), ϕ⊗ v 7→ Tϕ,v, where (1.20)

Tϕ,vw = ϕ(w)v, for all w ∈ W.

(See Exercise 1.2.14.) It is important to remember here that not all elements of W ′ ⊗ V aresimple tensors of the form ϕ ⊗ v. However, it is enough to define a linear map on simpletensors, since we then extend it to all of the tensor product by linearity.

Suppose that (σ, V ) and (ρ,W ) are two representations of G. We define a representationof G on Hom(V,W ) by

η(g)T = σ(g)Tρ(g−1), for all g ∈ G, T ∈ Hom(W,V ). (1.21)

Lemma 1.2.15. The map (1.20) is an isomorphism from ρ′ ⊗ σ to η.

Proof. For g ∈ G, ϕ ∈ W ′, v ∈ V , and w ∈ W , we have

(η(g)Tϕ,v)w = σ(g)Tϕ,vρ(g−1)w

= σ(g)ϕ(ρ(g−1)w)v

= ϕ(ρ(g−1)w)σ(g)v

= (ρ′(g)ϕ)(w)σ(g)v

= Tρ′(g)ϕ,σ(g)vw.

Since(ρ′ ⊗ σ)(g)(ϕ⊗ v) = (ρ′(g)ϕ)⊗ (σ(g)v),

this proves that the map (1.20) is an intertwiner. Since it is an isomorphism of vector spaces,it follows that it is an isomorphism of representations.

Corollary 1.2.16. We have

HomG(W,V ) = Hom(W,V )G ∼= HomG(ιG, ρ′ ⊗ σ),

where the isomorphism is one of vector spaces and ιG is the trivial representation of G.

Proof. For T ∈ Hom(W,V ), we have

T ∈ HomG(W,V ) ⇐⇒ σ(g)T = Tρ(g), for all g ∈ G⇐⇒ σ(g)Tρ

(g−1)

= T, for all g ∈ G⇐⇒ η(g)T = T, for all g ∈ G⇐⇒ T ∈ Hom(W,V )G.

Thus HomG(W,V ) = Hom(W,V )G.Note that Hom(W,V )G is precisely the ιG-isotypic component of Hom(W,V ) (Exer-

cise 1.2.15). Thus,

dim Hom(W,V )G = dim Hom(W,V )ιG = dim HomG(ιG, η) = dim HomG(ιG, ρ′ ⊗ σ)

by Lemmas 1.2.5 and 1.2.15. Therefore, we have an isomorphism of vector spaces

Hom(W,V )G ∼= HomG(ιG, ρ′ ⊗ σ).

1.3. Characters and the projection formula 29

Exercises.

1.2.14. Prove that (1.20) is an isomorphism of vector spaces.

1.2.15. Suppose (σ, V ) is a representation of G. Prove that V G is the ιG-isotypic componentof V . (Recall that ιG is the trivial representation of G.)

1.3 Characters and the projection formula

1.3.1 The trace

Suppose v1, . . . , vn is an orthonormal basis of V . The trace is the linear map

tr : End(V )→ C, tr(T ) =n∑j=1

〈Tvj, vj〉. (1.22)

The trace is independent of the chosen basis. In fact, it is uniquely determined by theproperties

(a) tr(TS) = tr(ST ) for all S, T ∈ End(V ), and

(b) tr(IV ) = dimV .

(See Exercise 1.3.1.)

If T ∈ End(W ) and S ∈ End(V ), then T ⊗S ∈ End(W ⊗V ). Choose orthonormal basesv1, . . . , n and w1, . . . , wm for V and W , respectively. Then

wi ⊗ vj : 1 ≤ i ≤ m, 1 ≤ j ≤ n

is an orthonormal basis for W ⊗ V , and so

tr(T ⊗ S) =m∑i=1

n∑j=1

〈(T ⊗ S)(wi ⊗ vj), wi ⊗ vj〉W⊗V

=m∑i=1

n∑j=1

〈Twi, wi〉W 〈Svj, vj〉V

= tr(T ) tr(S).


Exercises.

1.3.1. Prove that the trace map is uniquely characterized by the fact that it is linear andsatisfies the properties

(a) tr(TS) = tr(ST ) for all S, T ∈ End(V ), and

(b) tr(IV ) = dimV .

More precisely, show that any map with these properties is given by (1.22) for any choice oforthonormal basis.

1.3.2 Central functions and characters

Definition 1.3.1 (Class function). A function f ∈ L(G) is central (or is a class function)if f(gh) = f(hg) for all g, h ∈ G.

Lemma 1.3.2. A function f ∈ L(G) is a central if and only if it is conjugacy invariant:

f(g−1hg

)= f(h), for all g, h ∈ G.

In other words, a function is a central if and only if it is constant on each conjugacy class.

Proof. First suppose f is central. Then, for all g, h ∈ G, we have

f(g−1hg

)= f

(hgg−1

)= f(h),

and so f is conjugacy invariant. On the other hand, if f is conjugacy invariant, then, for allg, h ∈ G, we have

f(gh) = f(ghgg−1

)= f(hg),

and so f is central.

Lemma 1.3.2 justifies the terminology class function since one can think of such a functionas a function from the set of conjugacy classes to C.

Definition 1.3.3 (Character of a representation). The character of a representation (σ, V )of G is the function

χσ : G→ C, χσ(g) = tr(σ(g)).

In other words, χσ = tr σ. A character of an irreducible representation is called an irredu-cible character .

The basic properties of characters are given in the following proposition.

Proposition 1.3.4. Suppose (σ, V ) and (ρ,W ) are two representations of G.

(a) χσ(1G) = dimV


(b) χσ (g−1) = χσ(g) for all g ∈ G.

(c) χσ is a central function.

(d) χσ⊕ρ = χσ + χρ.

(e) If σi is a representation of Gi for i = 1, 2, then

χσ1σ2(g1, g2) = χσ1(g1)χσ2(g2), for all (g1, g2) ∈ G1 ×G2.

(f) χσ′(g) = χσ(g) for all g ∈ G. (Recall that σ′ is the adjoint of σ.)

(g) χσ⊗ρ = χσχρ.

(h) If η is the representation of G on Hom(V,W ) given by (1.21), then χη = χρχσ.

Proof. Choose an orthonormal basis v1, . . . , vd for V .

(a) We have χσ(1G) = tr(σ(1G)) = tr(IV ) = dimV .

(b) For g ∈ G, we have

χσ(g−1)

=d∑i=1

⟨σ(g−1)vi, vi

⟩=

d∑i=1

〈vi, σ(g)vi〉 =d∑i=1

〈σ(g)vi, vi〉 = χσ(g).

(c) For g, h ∈ G, we have

χσ(gh) = tr(σ(gh)) = tr(σ(g)σ(h)) = tr(σ(h)σ(g)) = tr(σ(hg)) = χσ(hg).

(d) For g ∈ G, we have

χσ⊕ρ(g) = tr(σ(g)⊕ ρ(g)) = tr(σ(g)) + tr(ρ(g)) = χσ(g) + χρ(g).

(e) For (g1, g2) ∈ G1 ×G2, we have

χσ1σ2(g1, g2) = tr(σ1(g1)⊗ σ2(g2)) = tr(σ1(g1)) tr(σ2(g2)) = χσ1(g1)χσ2(g2).

(f) Let θv1 , . . . , θvd be the basis of V ′ dual to the chosen basis of V . Then

χσ′(g) = tr(σ′(g))

=d∑i=1

〈σ′(g)θvi , θvi〉V ′

(1.10)=

d∑i=1

〈θσ(g)vi , θvi〉V ′

(1.8)=

d∑i=1

〈vi, σ(g)vi〉V

=d∑i=1

〈σ(g)vi, vi〉V

= χσ(g).


(g) This follows from (e).

(h) By Lemma 1.2.15 and (f) and (g), we have

χη = χρ′⊗σ = χρ

′χσ = χρχσ.

Exercises.

1.3.2. Compute the character of the sign representation ε of Sn (Example 1.1.5).

1.3.3 Central projection formulas

If E : V → V is a projection, then

dim(ImE) = tr(E). (1.23)

(Exercise 1.3.3.) Recall that we may (and will) assume all representations are unitary byLemma 1.1.1.

Lemma 1.3.5 (Basic projection formula). Suppose (σ, V ) is a representation of G andK ≤ G. Then

EKσ :=

1

|K|∑k∈K

σ(k)

is the orthogonal projection of V onto V K.

Proof. Let E = EKσ . For v ∈ V and g ∈ K, we have

σ(g)Ev =1

|K|∑k∈K

σ(gk)v = Ev.

Thus Ev ∈ V K . On the other hand, if v ∈ V K , then

Ev =1

|K|∑k∈K

σ(k)v =1

|K|∑k∈K

v = v.

Thus E is a projection from V onto V K . To see that E is orthogonal, we compute

E∗ =1

|K|∑k∈K

σ(k)∗ =1

|K|∑k∈K

σ(k−1)

= E.

Recall that ιG is the trivial representation of a group G, with ιG(g) = IC for all g ∈ G.Under the natural identification of End(C) with C, ιG corresponds to the element of L(g)which is the constant function 1.


Corollary 1.3.6. If (σ, V ) is a representation of G and K ≤ G, then

dimV K =1

|K|

⟨χResGK σ, ιK

⟩L(K)

.

Proof. We have

dimV K = tr

(1

|K|∑k∈K

σ(k)

)((1.23) and Lemma 1.3.5)

=1

|K|∑k∈K

χσ(k) (linearity of tr and definition of χσ)

=1

|K|

⟨χResGK σ, ιK

⟩L(K)

(definition (1.2) of scalar product on L(K)).

Corollary 1.3.7 (Orthogonality of irreducible characters). Suppose (σ, V ) and (ρ,W ) areirreducible representations of G. Then

1

|G|〈χσ, χρ〉L(G) =

1 if σ ∼ ρ,

0 if σ 6∼ ρ.

Proof. Let η be the representation of G on Hom(V,W ) given by (1.21). Then

1

|G|〈χσ, χρ〉L(G) =

1

|G|〈χρχσ, ιG〉L(G)

=1

|G|〈χη, ιG〉L(G) (Proposition 1.3.4(h))

= dim Hom(ρ, σ)G (Corollary 1.3.6)

= dim HomG(ρ, σ) (Corollary 1.2.16)

=

1 if σ ∼ ρ,

0 if σ 6∼ ρ(Schur’s lemma).

Corollary 1.3.8. Let (σ, V ) be a representation of G with multiplicities mρ, ρ ∈ G. Then

(a) mρ = 1|G|〈χ

ρ, χσ〉L(G) for all ρ ∈ G,

(b) 1|G|〈χ

σ, χσ〉L(G) =∑

ρ∈Gm2ρ, and

(c) 1|G|〈χ

σ, χσ〉L(G) = 1 if and only if σ is irreducible.

Proof. We leave the proof of this corollary as an exercise (Exercise 1.3.4).

Note that Corollary 1.3.8 implies that σ is determined uniquely, up to equivalence, by itscharacter. Since characters are readily computable, this is a very useful fact.


Lemma 1.3.9 (Fixed points character formula). Suppose G acts on a finite set X, and let(λ, L(X)) be the corresponding permutation representation (Example 1.1.3). Then

χλ(g) = |x ∈ X : gx = x|.

Proof. Recall that the Dirac functions, δx, x ∈ X, form an orthonormal basis for L(X). Sowe compute

χλ(g) =∑x∈X

〈λ(g)δx, δx〉L(X)

=∑x∈X

〈δgx, δx〉L(X) (by (1.4))

= |x ∈ X : gx = x|.

Corollary 1.3.10. The multiplicity of an irreducible representation (ρ, Vρ) in the left regularrepresentation (λ, L(G)) is equal to dρ := dimVρ. In other words,

L(G) ∼=⊕ρ∈G

V ⊕dρρ (as representations of G).

In particular |G| =∑

ρ∈G(dimVρ)2.

Proof. By Lemma 1.3.9, we have

χλ(g) = |h ∈ G : gh = h| =

|G| if g = 1G,

0 otherwise,= |G|δ1G(g).

Thus, by Corollary 1.3.8(a), we have

mρ =1

|G|〈χρ, χλ〉L(G) = χρ(1G) = dimVρ,

where the last equality follows from Proposition 1.3.4(a).

Corollary 1.3.11. Suppose G acts transitively on a finite set X. Choose x0 ∈ X, and let

K = g ∈ G : gx0 = x0

be the stabilizer of x0. Then the character of the permutation representation λ of G on Xis given by

χλ(h) =|X||C||C ∩K|, for all h ∈ G,

where C is the conjugacy class of G containing h.

Proof. Suppose x ∈ X. Since the action of G on X is transitive, there exists s ∈ G suchthat sx0 = x. Then we have a bijection

g ∈ C : gx = x → g ∈ C : gx0 = x0, g 7→ s−1gs.


Thus,|g ∈ C : gx = x| = |g ∈ C : gx0 = x0| = |C ∩K|. (1.24)

Then, for h ∈ G, we have

χλ(h) = |x ∈ X : hx = x (Lemma 1.3.9)

=1

|C||(x, g) ∈ X × C : gx = x

=1

|C|∑x∈X

|g ∈ C : gx = x|

=|X||C||C ∩K|.

Definition 1.3.12 (Fourier transform). Suppose (σ, V ) is a representation of G and f ∈L(G). The operator

σ(f) :=∑g∈G

f(g)σ(g) ∈ End(V )

is called the Fourier transform of f at σ.

Lemma 1.3.13 (Fourier transform of central functions). If f ∈ L(G) is central and (ρ,W )is an irreducible representation, then

ρ(f) =1

dρ

⟨χρ, f

⟩L(G)

IW .

Proof. First we show that ρ(f) is an intertwiner. For h ∈ G, we have

ρ(f)ρ(h) =∑g∈G

f(g)ρ(gh)

=∑s∈G

f(sh−1)ρ(s) (s = gh)

=∑s∈G

f(h−1s)ρ(s) (f is central)

=∑g∈G

f(g)ρ(hg) (g = h−1s)

= ρ(h)ρ(f).

Thus ρ(f) ∈ EndG(W ). By Schur’s lemma, there exists c ∈ C such that

ρ(f) = cIW .

It remains to compute c. We have

cdρ = tr(cIW ) = tr(ρ(f)) = tr

(∑g∈G

f(g)ρ(g)

)=∑g∈G

f(g)χρ(g) =⟨χρ, f

⟩L(G)

.


Thus

c =1

dρ

⟨χρ, f

⟩L(G)

,

as desired.

Corollary 1.3.14. If (σ, V ) and (ρ,W ) are irreducible representations, then

σ (χρ) =

|G|dσIV if σ ∼ ρ,

0 otherwise.

Proof. By Lemma 1.3.13 and Corollary 1.3.7, we have

σ (χρ) =1

dσ〈χσ, χρ〉 IV =

|G|dσIV if σ ∼ ρ,

0 otherwise.

Recall from Corollary 1.2.14 that, if (σ, V ) is a representation of G, then the minimal

central projections Eρ ∈ EndG(V ), ρ ∈ Gσ, form a basis for the center Z(EndG(V )). We cannow give an explicit expression for these minimal central projections.

Corollary 1.3.15 (Projection onto an isotypic component). Suppose (σ, V ) is a represen-tation of G. Then, for (ρ,W ) an irreducible representation of G, the orthogonal projectiononto the ρ-isotypic component of V is given by

Eρ =dρ|G|

σ (χρ) .

Proof. Let σ =⊕

η∈Gσ

⊕mηj=1 ηj be a decomposition of σ into irreducible subrepresentations,

where (ηj,Wη,j) is a representation in the equivalence class η for each 1 ≤ j ≤ mη (see(1.14)). Then

dρ|G|

σ (χρ) =dρ|G|

∑η∈Gσ

mη∑j=1

ηj (χρ)

=

mρ∑j=1

Eρ,j (Corollary 1.3.14)

= Eρ.

We would now like to determine the cardinality of G, that is, the number of irreduciblenonequivalent representations of G. Note that, for f ∈ L(G), we have

f(1.3)=∑g∈G

f(g)δg(1.4)=∑g∈G

f(g)λ(g)δ1G = λ(f)δ1G . (1.25)

Proposition 1.3.16. The characters χρ, ρ ∈ G, form an orthonormal basis for the space ofcentral functions of G. In particular, |G| is equal to the number of conjugacy classes of G.


Proof. The elementsχρ, ρ ∈ G,

are orthogonal by Corollary 1.3.7. Since orthogonal vectors are linearly independent, itsuffices to show that the characters span the space of central functions in L(G). For this, itsuffices to prove that the orthogonal complement to their span is zero.

Suppose f ∈ L(G) is central and

〈f, χρ〉L(G) = 0, for all ρ ∈ G.

Then

λ(f) =∑ρ∈G

dρρ(f) (Corollary 1.3.10)

=∑ρ∈G

〈χρ, f〉L(G) (Lemma 1.3.13)

=∑ρ∈G

〈f, χρ〉L(G)

=∑ρ∈G

〈f, χρ′〉L(G) (Proposition 1.3.4(f))

= 0.

Therefore, f = 0 by (1.25).If C denotes the set of all conjugacy classes of G, then the characteristic functions 1C ,

C ∈ C, defined by

1C(g) =

1 if g ∈ C,0 if g /∈ C,

form another basis for the space of central functions by Lemma 1.3.2. Thus, the dimensionof this space is |C|, and so |G| = |C|.

Example 1.3.17. If G is abelian, then every conjugacy class is a singleton, and all elementsof L(G) are central. It follows that |G| = |G|. For the cyclic group with n elements, wefound the n inequivalent irreducible representations in Example 1.2.4.

The next theorem gives a precise relation between the representations of two groups andtheir product.

Theorem 1.3.18. Suppose G1 and G2 are finite groups. Then we have a bijection of sets

G1 × G2 → G1 ×G2, (ρ1, ρ2) 7→ ρ1 ρ2.

Proof. For ρ1, σ1 ∈ G1 and ρ2, σ2 ∈ G2, we have

1

|G1 ×G2|⟨χρ1ρ2 , χσ1σ2

⟩L(G1×G2)


=1

|G1 ×G2|〈χρ1χρ2 , χσ1χσ2〉L(G1×G2) (Proposition 1.3.4(e))

=1

|G1|〈χρ1 , χσ1〉L(G1)

1

|G2|〈χρ2 , χσ2〉L(G2) (Exercise 1.3.5)

= δρ1,σ1δρ2,σ2 (Corollary 1.3.7).

It then follows from Corollary 1.3.7 and Corollary 1.3.8(c) that

ρ1 ρ2, ρ1 ∈ G1, ρ2 ∈ G2,

are pairwise inequivalent irreducible representations of G1 ×G2.

Now, by Proposition 1.3.16, |G1 ×G2| is equal to the number of conjugacy classes ofG1 ×G2, which is the product of the number of conjugacy classes of G1 and the number ofconjugacy classes of G2 (Exercise 1.3.6.) By Proposition 1.3.16, this is equal to |G1| · |G1|.So we have found all of the irreducible representations, completing the proof.

Example 1.3.19. By the fundamental theorem of finite abelian groups (see, for example,[Jud, Th. 13.4]), every finite abelian group is a product of cyclic groups. Thus, Exam-ple 1.2.4 and Theorem 1.3.18 completely classify the irreducible representations of finiteabelian groups, up to isomorphism.

Exercises.

1.3.3. Prove (1.23).

1.3.4. Prove Corollary 1.3.8.

1.3.5. Suppose X1 and X2 are finite sets. Any element f ∈ L(X1) can be viewed naturallyas an element of L(X1 ×X2) by setting f(x1, x2) = f(x1) for (x1, x2) ∈ X1 ×X2. Similarly,elements of L(X2) can also be viewed as elements of L(X1 ×X2). Prove that

〈f1f2, g1g2〉L(X1×X2) = 〈f1, g1〉L(X1)〈f2, g2〉L(X2), for all f1, g1 ∈ L(X1), f2, g2 ∈ L(X2).

1.3.6. Suppose G1 and G2 are finite groups. Show that the conjugacy classes of G1×G2 areprecisely the sets C1 × C2, where Ci is a conjugacy class of Gi for i = 1, 2.

1.3.7. Consider the natural action of the symmetric group Sn on the set X = 1, . . . , n.Using the theory of characters discussed in this section, compute the multiplicity of thetrivial representation in the corresponding permutation representation (λ, L(X)) of Sn.

1.4 Permutation representations

In this section, we study permutation representations (Example 1.1.3) in more detail.

1.4. Permutation representations 39

1.4.1 Wielandt’s lemma

Suppose G acts on a finite set X and let (λ, L(X)) denote the corresponding permutationrepresentation (Example 1.1.3).

Define a product on L(X,X) by:

(F1F2)(x, y) =∑z∈X

F1(x, z)F2(z, y), for all F1, F2 ∈ L(X ×X), x, y ∈ X. (1.26)

Under this product and pointwise addition and scalar multiplication, L(X×X) is an algebra(Exercise 1.4.1). It may be viewed as the algebra of X ×X matrices with coefficients in C.

Lemma 1.4.1. We have an isomorphism of algebras

L(X ×X)→ End(L(X)), F 7→ TF , where

(TFf)(x) =∑y∈X

F (x, y)f(y). (1.27)

Proof. The proof of this lemma is left as an exercise (Exercise 1.4.1).

The group G acts diagonally on X ×X:

g(x, y) = (gx, gy), for all g ∈ G, x, y ∈ X.

Then we have the associated permutation representation on L(X × X). We can also viewEnd(L(X)) = Hom(L(X), L(X)) as a representation of G as in (1.21).

Lemma 1.4.2. The isomorphism of Lemma 1.4.1 is an intertwiner and hence an isomor-phism of representations of G.

Proof. Let λX denote the permutation representation on L(X) and let λX×X denote thepermutation representation on L(X ×X). Furthermore, let η denote the representation ofG on End(L(X)) defined in (1.21). Then, for F ∈ L(X ×X), f ∈ L(X), g ∈ G, and x ∈ X,we have (

(η(g)TF )f)(x) =

(λX(g)TF

(λX(g−1)f

))(x)

= λX(g)∑y∈X

F (x, y)(λX(g−1)f)(y)

= λX(g)∑y∈X

F (x, y)f(gy)

=∑y∈X

F (g−1x, y)f(gy)

=∑z∈X

F (g−1x, g−1z)f(z) (z = gy)

=∑z∈X

(λX×X(g)F

)(x, z)f(z)

=(TλX×X(g)Ff

)(x).

Thus, the isomorphism of Lemma 1.4.1 intertwines the G-action.


Corollary 1.4.3. We have EndG(L(X)) ∼= L(X ×X)G as algebras.

Proof. By Lemma 1.4.2 and Corollary 1.2.16, we have isomorphisms of algebras

L(X ×X)G ∼= End(L(X))G = EndG(L(X)).

Corollary 1.4.4 (Wielandt’s lemma). Suppose G acts on a finite set X, and let λ =⊕ρ∈Gλ ρ

⊕mρ be the decomposition of the associated permutation representation into irre-ducibles. Then ∑

ρ∈Gλ

m2ρ = number of orbits of G on X ×X.

Proof. Since the characteristic functions of the orbits of G on X ×X form a basis of L(X ×X)G, we have

number of orbits of G on X ×X = dimL(X ×X)G

= dim EndG(L(X)) (Corollary 1.4.3)

=∑ρ∈Gλ

m2ρ (Corollary 1.2.11).

Example 1.4.5. Consider the natural action of the symmetric group Sn on X = 1, 2, . . . , n.We have an orthogonal direct sum decomposition (as representations)

L(X) = V0 ⊕ V1, where (1.28)

V0 = f ∈ L(X) : f(i) = f(j) for all i, j ∈ X,V1 = f ∈ L(X) :

∑nj=1 f(j) = 0.

On the other hand Sn has precisely two orbits on X ×X, given by

Ω0 = (i, i) : i ∈ X and Ω1 = (i, j) : i, j ∈ X, i 6= j.

Thus, by Wiedlandt’s lemma (Corollary 1.4.4), (1.28) is a decomposition into irreduciblesubrepresentations.

Exercises.

1.4.1. Show that L(X,X) is an algebra under the product (1.26) and pointwise addition andscalar multiplication. Then prove Lemma 1.4.1.

1.4.2 ([CSST10, Ex. 1.4.2]). Suppose G acts on a finite set X. Show that the permutationreputation of G on L(X × X) is equivalent to the tensor product of the permutation re-presentation L(X) with itself. In other words, show that L(X × X) ∼= L(X) ⊗ L(X) asrepresentations of G.


1.4.3 ([CSST10, Ex. 1.4.3]). Suppose G acts on finite sets X and Y . Show that

HomG(L(X), L(Y )) ∼= L(X × Y )G

as vector spaces.

1.4.4. Verify the details of Example 1.4.5. More precisely, do the following:

(a) Prove that V0 and V1 are Sn-invariant subspaces of L(X).

(b) Prove that one has an orthogonal direct sum decomposition L(X) = V0 ⊕ V1.

(c) Prove that Ω0 and Ω1 are Sn-orbits for the diagonal action of Sn on X ×X.

1.4.5 ([CSST10, Ex. 1.4.6]). Suppose G acts transitively on a finite set X with at least twoelements. As in Example 1.4.5, define

V0 = f ∈ L(X) : f(x) = f(y) for all x, y ∈ X and V1 =

f ∈ L(X) :

∑x∈X

f(x) = 0

.

We say that the action of G on X is doubly transitive if

∀ x, y, z, u ∈ X such that x 6= y and z 6= u,∃ g ∈ G such that g(x, y) = (z, u).

Prove that L(X) = V0⊕V1 is the decomposition of L(X) into irreducible subrepresentationsif and only if the action of G on X is doubly transitive.

1.4.6 ([CSST10, Ex. 1.4.7]). Suppose G acts on finite sets X and Y . For ρ ∈ G, let mρ andm′ρ denote the multiplicities of ρ in L(X) and L(Y ), respectively. Show that∑

ρ∈G

mρm′ρ = number of orbits of G on X × Y .

1.4.2 Symmetric actions and Gelfand’s lemma

A subset A ⊆ X ×X is symmetric if

(x, y) ∈ A =⇒ (y, x) ∈ A.

The action of G on a set X is symmetric if all orbits of G on X × X are symmetric. Afunction F ∈ L(X ×X) is symmetric if

F (x, y) = F (y, x), for all x, y ∈ X.

Proposition 1.4.6 (Gelfand’s lemma; symmetric case). Suppose that the action of G on afinite set X is symmetric. Then EndG(L(X)) is commutative and the permutation represen-tation of G on L(X) is multiplicity free.


Proof. Since the action of G is symmetric, any F ∈ L(X × X)G is symmetric since it isconstant on G-orbits. Therefore, for all F1, F2 ∈ L(X ×X)G and x, y ∈ X, we have

(F1F2)(x, y) =∑z∈X

F1(x, z)F2(z, y)

=∑z∈X

F1(z, x)F2(y, z)

= (F2F1)(y, x)

= (F2F1)(x, y) (F2F1 ∈ L(X ×X)G by Corollary 1.4.3).

Hence L(X ×X)G is commutative. Thus, by Corollary 1.4.3, EndG(L(X)) is commutative.Then the result follows from Corollary 1.2.13.

Proposition 1.4.6 corresponds to the following fact for matrices: If A is a subalgebra ofMn,n(C) consisting of symmetric matrices, then A is commutative, since

AB = ATBT = (BA)T = BA, for all A,B ∈ A.

1.4.3 Frobenius reciprocity for permutation representations

In this section we will assume that G acts transitively on a finite set X. We fix x0 ∈ X andlet

K = g ∈ G : gx0 = x0

denote its stabilizer. Then the map

G/K → X, gK 7→ gx0,

is a bijection of G-sets (i.e. it is a bijection of sets that commutes with the G-actions on Xand on the set of right cosets G/K). So we may identify X with the space G/K of rightcosets in G. We have ∑

x∈X

f(x) =1

|K|∑g∈G

f(gx0), for all f ∈ L(X). (1.29)

Definition 1.4.7 (Gelfand pair). Suppose G acts transitively on X. We say that (G,K) isa Gelfand pair if the permutation representation L(X) is multiplicity free. If the action ofG on X is symmetric, we say that (G,K) is a symmetric Gelfand pair . (Note that, in thiscase, L(X) is multiplicity free by Gelfand’s lemma (Proposition 1.4.6).)

Example 1.4.8. Consider the natural action of the symmetric group Sn on 1, 2, . . . , n. Fixk ∈ Z, 0 ≤ k ≤ n/2, and define

Ωn−k,k = A ⊆ 1, 2, . . . , n : |A| = k.

For A ∈ Ωn−k,k and π ∈ Sn, we have

πA = π(j) : j ∈ A ∈ Ωn−k,k,


so Sn acts on the set Ωn−k,k.Fix A0 ∈ Ωn−k,k and let K denote its stabilizer. Then

K ∼= Sn−k ×Sk (as groups),

where the first factor is the symmetric group on A0 := 1, 2, . . . , n \ A0 and the secondfactor is the symmetric group on A0. Since the action of Sn on Ωn−k,k is transitive, we mayidentify

Ωn−k,k = Sn/ (Sn−k ×Sk) .

Claim: Two elements (A,B) and (A′, B′) of Ωn−k,k × Ωn−k,k are in the same Sn-orbit(under the diagonal action) if and only if |A ∩B| = |A′ ∩B′|.

Proof of claim: The “only if” part is clear. Suppose |A ∩ B| = |A′ ∩ B′|. Consider thedecomposition

1, 2, . . . , n = (A ∪B) t(A \ (A ∩B)

)t(B \ (A ∩B)

)t (A ∩B)

= (A′ ∪B′) t(A′ \ (A′ ∩B′)

)t(B′ \ (A′ ∩B′)

)t (A′ ∩B′).

We can choose π ∈ Sn such that

• π(A ∩B) = A′ ∩B′,• π(A \ (A ∩B)) = A′ \ (A′ ∩B′), and

• π(B \ (A ∩B)) = B′ \ (A′ ∩B′),

since we know that each pair of sets involved have the same cardinality. It follows thatπ(A,B) = (A′, B′). This proves the claim.

Now define

Θj = (A,B) ∈ Ωn−k,k × Ωn−k,k : |A ∩B| = j, 0 ≤ j ≤ k,

so that the decomposition of Ωn−k,k × Ωn−k,k into Sn-orbits is given by

Ωn−k,k × Ωn−k,k =k⊔j=0

Θj.

(We use the fact that k ≤ n/2 here to conclude that the Θj are all nonempty.) Since|A ∩ B| = |B ∩ A|, every orbit is symmetric. So (Sn,Sn−k × Sk) is a symmetric Gelfandpair.

Since there are precisely k+ 1 orbits of Sn on Ωn−k,k×Ωn−k,k, Wielandt’s lemma (Corol-lary 1.4.4) implies that L(Ωn−k,k) decomposes into k + 1 pairwise inequivalent irreducibleSn-representations. When k = 1, this recovers the result of Example 1.4.5.

Suppose (ρ,W ) is an irreducible representation of G. Define dρ = dimW and supposethat WK is nontrivial. For every v ∈ WK , define a linear map

Tv : W → L(X), (Tvu)(gx0) =

√dρ|X|〈u, ρ(g)v〉W , for all g ∈ G, u ∈ W. (1.30)


This is defined on all of X since the action of G on X is transitive. Furthermore, if g, h ∈ Gand gx0 = hx0, then g−1h ∈ K, and thus

(Tvu)(hx0) =

√dρ|X|〈u, ρ(h)v〉W

=

√dρ|X|〈u, ρ(g)ρ(g−1h)v〉W

=

√dρ|X|〈u, ρ(g)v〉W (since v ∈ V K)

= (Tvu)(gx0).

Hence Tvu is well defined.

Theorem 1.4.9 (Frobenius reciprocity for permutation representations). With notation asabove, we have the following.

(a) Tv ∈ HomG(W,L(X)) for all v ∈ WK.

(b) (Orthogonality relations) For all v, u ∈ WK and w, z ∈ W , we have

〈Tuw, Tvz〉L(X) = 〈w, z〉W 〈v, u〉W .

(c) The mapWK → HomG(W,L(X)), v 7→ Tv, (1.31)

is an antilinear isomorphism. In particular, the multiplicity of ρ in the permutationrepresentation L(X) is equal to dimWK.

Proof. (a) For g, h ∈ G and w ∈ W , we have

(λ(g)Tvw) (hx0) = (Tvw)(g−1hx0

)=

√dρ|X|〈w, ρ(g−1)ρ(h)v〉W

=

√dρ|X|〈ρ(g)w, ρ(h)v〉W

= (Tvρ(g)w) (hx0).

Hence λ(g)Tv = Tvρ(g) for all g ∈ G, and so Tv ∈ HomG(W,L(X)).

(b) For u, v ∈ WK , define a linear map

Ru,v : W → W, Ru,vw = 〈w, u〉Wv, for all w ∈ W.

Choosing an orthonormal basis w1, . . . , wdρ for W , we compute

tr(Ru,v) =

dρ∑j=1

〈Ru,vwj, wj〉W =

dρ∑j=1

〈wj, u〉W 〈v, wj〉W =

⟨v,

dρ∑j=1

〈u,wj〉wj

⟩W

= 〈v, u〉W .


While Ru,v ∈ End(W ), in general Ru,v is not an element of EndG(W ). However, we canproject it onto EndG(W ) = End(W )G (see Corollary 1.2.16) to get

R :=1

|G|∑g∈G

ρ(g)Ru,vρ(g−1) ∈ EndG(W ). (1.32)

(Here we use Lemma 1.3.5 and the action on End(W ) given by (1.21).) SinceW is irreducible,by Schur’s lemma (Lemma 1.2.1) we have R = cIW for some c ∈ C. Taking the trace of bothsides of (1.32), we have

cdρ = tr(cIW )

=1

|G|∑g∈G

tr(ρ(g)Ru,vρ(g−1)

)=

1

|G|∑g∈G

tr(Ru,v)

= 〈v, u〉W .

Thus c = 1dρ〈v, u〉W , and so

R =1

dρ〈v, u〉W IW . (1.33)

Then, for w, z ∈ W , we have

〈Tuw, Tvz〉L(X) =∑x∈X

(Tuw)(x)(Tvz)(x)

=1

|K|∑g∈G

(Tuw)(gx0)(Tvz)(gx0) (by (1.29))

=dρ|G|

∑g∈G

〈w, ρ(g)u〉W 〈z, ρ(g)v〉W (since |K| · |X| = |G|)

=dρ|G|

∑g∈G

〈〈w, ρ(g)u〉Wρ(g)v, z〉W

=dρ|G|

∑g∈G

⟨ρ(g)〈ρ(g−1)w, u〉Wv, z

⟩W

=dρ|G|

∑g∈G

〈ρ(g)Ru,vρ(g−1)w, z〉W

= dρ〈Rw, z〉W (by (1.32))

= 〈w, z〉W 〈v, u〉W (by (1.33)).

(c) The map (1.31) is antilinear since the bilinear form is antilinear in the second ar-gument. We now show that it is bijective. Fix T ∈ HomG(W,L(X)) and consider thecomposition of linear maps

WT−→ L(X)

f 7→f(x0)−−−−−→ C, u 7→ (Tu)(x0),


As discussed in Section 1.1.5, this implies that there exists a unique v ∈ W such that

(Tu)(x0) = 〈u, v〉W , for all u ∈ W.

Then

(Tu)(gx0) =(λ(g−1)Tu

)(x0)

=(Tρ(g−1)u

)(x0) (T ∈ HomG(W,L(X)))

= 〈ρ(g−1)u, v〉W= 〈u, ρ(g)v〉W ,

which implies that

T =

√|X|dρTv.

We also have v ∈ WK since, for k ∈ K,

〈u, ρ(k)v〉W = (Tu)(kx0) = (Tu)(x0) = 〈u, v〉W , for all u ∈ W,

and so ρ(k)v = v by the nondegeneracy of the scalar product. Since the vector v was uniquelydetermined by T , we see that (1.31) is bijective.

Finally, by Lemma 1.2.5, the multiplicity of ρ in the permutation representation on L(X)is equal to

dim HomG(W,L(X)) = dimWK .

Corollary 1.4.10. The pair (G,K) is a Gelfand pair if and only if dimWK ≤ 1 for everyirreducible G-representation W . In particular, (G,K) is a Gelfand pair if and only if

dimWK = 1 ⇐⇒ W is a subrepresentation of L(X).

Exercises.

1.4.7 ([CSST10, Ex. 1.4.11]). (a) Show that, if 0 ≤ h ≤ k ≤ n/2, then Sn has preciselyh+ 1 orbits on Ωn−k,k × Ωn−h,h.

(b) Suppose that L(Ωn−k,k) =⊕k

j=0 Vk,j is the decomposition of L(Ωn−k,k) into irreduciblesubrepresentations (see Example 1.4.8). Use part (a) and Exercise 1.4.6 to show thatit is possible to number the representations Vk,0, . . . , Vk,j in such a way that Vh,j ∼= Vk,j(as representations) for all j = 0, 1, . . . , h and 0 ≤ h ≤ k ≤ n/2. Hint : Everysubrepresentation of L(Ωn−h,h) is also a subrepresentation of L(Ωn−k,k).


1.4.4 The structure of the commutant of a permutation represen-tation

The goal of this subsection is to give an explicit form for the operators T ρk,j defined in (1.16).Recall that, by Theorem 1.2.10, these operators give a basis for the commutant EndG(V ).

As in Section 1.4.3, we assume that G acts transitively on a finite set X. We fix x0 ∈ X,and let K be the stabilizer of x0, so that we can identify X with G/K.

For ρ ∈ G, let mρ be the multiplicity of ρ in the permutation representation (λ, L(X)).

For each ρ ∈ Gλ, by Theorem 1.4.9, we have dimWKρ = mρ, so we can choose an orthonormal

basis

vρ1 , . . . , vρmρ

for WKρ . Let

T ρj := Tvρj ∈ HomG(Wρ, L(X)), ρ ∈ Gλ, 1 ≤ j ≤ mρ,

be the intertwiners defined in (1.30) (which are intertwiners by Theorem 1.4.9(a)), so that

(T ρj u

)(gx0) =

√dρ|X|

⟨u, ρ(g)vρj

⟩Wρ, for all g ∈ G, u ∈ Wρ. (1.34)

Recall that if U and V are unitary spaces, an isometric immersion of U into V is a linearmap T : U → V such that

〈Tu1, Tu2〉V = 〈u1, u2〉U , for all u1, u2 ∈ U.

The term immersion comes from the fact that such a map is necessarily injective (Exer-cise 1.4.8).

Lemma 1.4.11. We have that

L(X) =⊕ρ∈Gλ

mρ⊕j=1

T ρj Wρ (1.35)

is an orthogonal decomposition of L(X) into irreducible subrepresentations. Furthermore,every T ρj is an isometric immersion of Wρ into L(X).

Proof. It follows from Theorem 1.4.9(b) that⊕

ρ∈Gλ

⊕mρj=1 T

ρj Wρ is an orthogonal decompo-

sition (i.e. the summands are orthogonal) and that the T ρj are isometric immersions. Then,since

dim

⊕ρ∈Gλ

mρ⊕j=1

T ρj Wρ

=∑ρ∈Gλ

mρ∑j=1

dim T ρj Wρ =∑ρ∈Gλ

mρ∑j=1

dimWρ = dimL(X),

we see that we have the equality (1.35).


Now, for ρ ∈ Gλ and 1 ≤ i, j ≤ mρ, define

φρi,j ∈ L(X ×X), φρi,j(gx0, hx0) =dρ|X|

⟨ρ(h)vρj , ρ(g)vρi

⟩Wρ, g, h ∈ G. (1.36)

Note that φρi,j is well defined since vρi and vρj are K-invariant. Moreover, φρi,j ∈ L(X ×X)G,since, for all s, g, h ∈ G, we have

φρi,j(sgx0, shx0) =dρ|X|

⟨ρ(s)ρ(h)vρj , ρ(s)ρ(g)vρi

⟩Wρ

= φρi,j(gx0, hx0),

since ρ is unitary.Therefore, as in (1.27) and Corollary 1.4.3, we can define

Φρi,j ∈ EndG(L(X)),(

Φρi,jf)

(x) =∑y∈X

φρi,j(x, y)f(y), for all f ∈ L(X), x ∈ X.

Lemma 1.4.12. For all g ∈ G and f ∈ L(X), we have

Φρi,jf =

√dρ√

|X| |K|T ρi

(∑h∈G

ρ(h)f(hx0)vρj

).

Proof. We compute(Φρi,jf)

(gx0) =∑y∈X

φρi,j(gx0, y)f(y)

=1

|K|∑h∈G

φρi,j(gx0, hx0)f(hx0)

=dρ

|K| |X|∑h∈G

⟨ρ(h)vρj , ρ(g)vρi

⟩Wρf(hx0) (by (1.36))

=

√dρ√

|X| |K|

∑h∈G

(T ρi ρ(h)vρj

)(gx0) · f(hx0) (by (1.34))

=

√dρ√

|X| |K|T ρi

(∑h∈G

ρ(h)f(hx0)vρj

)(gx0).

Using Lemma 1.4.12, we can now show that the Φρi,j are the operators T ρi,j of Theo-

rem 1.2.10.

Theorem 1.4.13. For all σ, ρ ∈ Gλ, 1 ≤ i, j ≤ mρ, and 1 ≤ s, r,≤ mσ, we have

(a) Im Φρi,j = T ρi Wρ,

(b) Φρi,jΦ

σs,r = δj,sδρ,σΦρ

i,r.

(c) Ker Φρi,j = L(X) T ρj Wρ (here means we remove the summand T ρj Wρ from L(X)),

and


Proof. (a) Since Wρ is irreducible, the G-invariant subspace generated by vρj ∈ Wρ is allof Wρ. Thus, it follows from Lemma 1.4.12 that Im Φρ

i,j = T ρi Wρ.

(b) For all g, h ∈ G, we compute the product:(φρi,jφ

σs,r

)(gx0, hx0)

=1

|K|∑t∈G

φρi,j(gx0, tx0)φσs,r(tx0, hx0)

=dρdσ|X|2|K|

∑t∈G

〈ρ(t)vρj , ρ(g)vρi 〉Wρ〈σ(h)vσr , σ(t)vσs 〉Wσ (by (1.36))

=

√dρdσ

|X| |K|∑t∈G

(T ρj ρ(g)vρi

)(tx0) (T σs ρ(h)vσr ) (tx0) (by (1.34))

=

√dρdσ

|X|⟨T σs ρ(h)vσr , T

ρj ρ(g)vρi

⟩L(X)

= δσ,ρdρ|X|

⟨vρj , v

ρs

⟩Wρ〈ρ(h)vρr , ρ(g)vρi 〉Wρ

(Th. 1.4.9(b), Lem. 1.4.11)

= δσ,ρδj,sφρi,r(gx0, hx0) (by (1.36)).

Thus φρi,jφσs,r = δσ,ρδj,sφ

ρi,rφ

ρi,r. So the result follows from Lemma 1.4.1.

(c) Suppose σ ∈ Gλ and 1 ≤ k ≤ mσ such that σ 6= ρ or that σ = ρ, j 6= k. Then, byparts (a) and (b), we have

Φρi,jT σk Wσ = Im Φρ

i,jΦσk,k = 0.

Since Φρi,j is not the zero map by part (a), it must be nonzero on the remaining irreducible

summand T ρj Wρ.

Corollary 1.4.14. (a) The map Φρi,i is the orthogonal projection onto T ρi Wρ.

(b) The map∑mρ

i=1 Φρi,i is the orthogonal projection onto the ρ-isotypic component.

Exercises.

1.4.8. Prove that every isometric immersion is injective.

1.4.9. Prove that

(Φρi,jf)

(gx0) =

√dρ|X|

⟨f, T ρj ρ(g)vρi

⟩L(X)

, for all g ∈ G, f ∈ L(X).

Hint : Follow the method of Lemma 1.4.12.


1.5 The group algebra and the Fourier transform

In this section we consider the special case of Section 1.4 where X = G. That is, we studythe left regular representation (λ, L(G)) of a the finite group G.

1.5.1 The group algebra

Recall the left regular representation λ and right regular representation ρ of G on L(G) fromExample 1.1.4. In the language of Section 1.4, we take x0 = 1G, so that K = 1G.

We define a convolution product ∗ on L(G) by

(f1 ∗ f2)(g) =∑h∈G

f1(gh−1)f2(h), for all f1, f2 ∈ L(G), g ∈ G. (1.37)

One can show that L(G) is a unital algebra with this product (Exercise 1.5.1). It is calledthe group algebra (or the convolution algebra) of G. The unit of this algebra is δ1G .

Remark 1.5.1. The group algebra of G is sometimes defined to be set of formal linear com-binations

CG =

∑g∈G

αgg : αg ∈ C for all g ∈ G

,

with multiplication given by(∑g∈G

αgg

)(∑h∈G

βhh

)=∑g,h∈G

(αgβh)(gh).

We have an obvious isomorphism

CG∼=−→ L(G),

∑g∈G

αgg 7→∑g∈G

αgδg,

so this is essentially the same construction.

Note that we can also express the convolution product as

(f1 ∗ f2)(g) =∑

s,t∈G:st=g

f1(s)f2(t) =∑h∈G

f1(h)f2(h−1g), for all f1, f2 ∈ L(G), g ∈ G.

It follows that(f1 ∗ f2) =

∑h∈G

f1(h)λ(h)f2.

For g ∈ G and f ∈ L(G), we have

δg ∗ f =∑h∈G

δg(h)λ(h)f2 = λ(g)f. (1.38)

Similarly,f ∗ δg = ρ(g−1)f.

1.5. The group algebra and the Fourier transform 51

In particular,δg ∗ δh = δgh, for all g, h ∈ G. (1.39)

It follows thatρ(g)λ(h) = λ(h)ρ(g), for all g, h ∈ G.

In other words, the left and right regular actions commute.

Lemma 1.5.2. The map

L(G)→ L(G), ψ 7→ ψ, where

ψ(g) = ψ(g−1), for all g ∈ G,

is an involution on the algebra L(G).

Proof. For ψ1, ψ2 ∈ L(G) and g ∈ G, we have(ψ1 ∗ ψ2

)(g) =

∑s∈G

ψ1(gs)ψ2(s−1)

=∑s∈G

ψ1(s−1g−1)ψ2(s)

= (ψ2 ∗ ψ1)(g−1)

= (ψ2 ∗ ψ1)ˇ(g).

We leave it as an exercise to verify the remaining properties of an involution (Exercise 1.5.2).

Remark 1.5.3. Note that

f ∈ Z(L(G)) ⇐⇒ δg ∗ f = f ∗ δg, for all g ∈ G⇐⇒ λ(g)f = ρ(g−1)f, for all g ∈ G⇐⇒ f(g−1h) = f(hg−1), for all g, h ∈ G⇐⇒ f is central.

Proposition 1.5.4. The map

L(G)→ EndG(L(G)), ψ 7→ Tψ, where

Tψf = f ∗ ψ, for all f ∈ L(G),

is a ∗-anti-isomorphism of algebras.

Proof. We leave it as an exercise (Exercise 1.5.4) to show that the linear map

EndG(L(G))→ L(G), T 7→ T (δ1G). (1.40)

is inverse to the linear map ψ 7→ Tψ.For ψ1, ψ2, f ∈ L(G), we have

Tψ1 (Tψ2f) = (f ∗ ψ2) ∗ ψ1 = f ∗ (ψ2 ∗ ψ1) = Tψ2∗ψ1f.


Thus Tψ1Tψ2 = Tψ2∗ψ1 , and so the map ψ 7→ Tψ is an anti-multiplicative. Furthermore, forf1, f2, ψ ∈ L(G), we have

〈Tψf1, f2〉 = 〈f1 ∗ ψ, f2〉

=∑g∈G

(f1 ∗ ψ)(g) f2(g)

=∑g∈G

∑s∈G

f1(gs)ψ(s−1) f2(g)

=∑t∈G

∑s∈G

f1(t)ψ(s−1) f2(ts−1) (t = gs)

=∑t∈G

∑s∈G

f1(t) ψ(s)f2(ts−1)

=∑t∈G

f1(t) (f2 ∗ ψ)(t)

=⟨f1, Tψf2

⟩L(G)

.

Hence (Tψ)∗ = Tψ.

For every ρ ∈ G, fix an orthonormal basis

vρ1 , vρ2 , . . . , v

ρdρ

for the representation space Wρ. Then define the corresponding unitary matrix coefficients

ϕρi,j(g) =⟨ρ(g)vρj , v

ρi

⟩Wρ. (1.41)

We deduced some basic properties of matrix coefficients in Lemma 1.1.12. In the casethat the representations are irreducible, we have the following additional properties.

Lemma 1.5.5. Let ρ and σ be two irreducible representations of G.

(a) (Orthogonality relations)⟨ϕρi,j, ϕ

σs,t

⟩L(G)

=|G|dρδρ,σδi,sδj,t.

(b) ϕρi,j ∗ ϕσs,t =|G|dρδρ,σδj,sϕ

ρi,t.

Proof. (a) We have

〈ϕρi,j, ϕσs,t〉L(G) =∑g∈G

ϕρi,j(g)ϕσs,t(g)

=∑g∈G

⟨ρ(g)vρj , v

ρi

⟩Wρ〈ρ(g)vσt , v

σs 〉Wρ

=

√|G|dρ

√|G|dσ

∑g∈G

(T ρvρjvρi )(g)(T σvσt v

σs )(g) (see (1.30))


=

√|G|dρ

√|G|dσ

⟨T σvσt v

σs , T

ρvρjvρi

⟩L(G)

=|G|dρδρ,σδi,sδj,t (Theorem 1.4.9(b)).

(b) We have(ϕρi,j ∗ ϕσs,t

)(g) =

∑h∈G

ϕρi,j(gh)ϕσs,t(h−1)

=∑h∈G

dρ∑k=1

ϕρi,k(g)ϕρk,j(h)ϕσt,s(h) (Lemma 1.1.12)

=|G|dρ

dρ∑k=1

ϕρi,kδρ,σδk,tδj,s (part (a))

=|G|dρδρ,σδj,sϕ

ρi,t(g).

Corollary 1.5.6. The set ϕρi,j : ρ ∈ G, 1 ≤ i, j ≤ dρ

is an orthogonal basis for L(G).

Proof. The given elements are orthogonal, hence linearly independent, by Lemma 1.5.5. ByCorollary 1.3.10, dimL(G) =

∑ρ∈G d

2ρ, which is equal to the cardinality of the given set.

Exercises.

1.5.1. Prove that L(G) is an algebra under the product ∗ defined in (1.37) and componentwiseaddition and scalar multiplication. Prove that it is commutative if and only if G is abelian.

1.5.2. Complete the proof of Lemma 1.5.2.

1.5.3. Suppose A is a unital algebra and let Aop denote the opposite algebra. Precisely, Aop

is equal to A as a vector space, but the multiplication in Aop is given by

a · b = ba, a, b ∈ A,

where · denotes the multiplication in Aop and juxtaposition denotes the multiplication in A.Note that an algebra anti-homomorphism A → B is the same as an algebra homomorphismA → Bop.

Let EndA denote the algebra of linear maps from A to A, and define

EndAA = f ∈ EndA : f(ab) = af(b) ∀ a, b ∈ A.


This is a subalgebra of EndA. (In fact, EndAA is the algebra of A-module endormorphismsof A, considered as a module over itself.)

(a) Show that EndAA ∼= Aop as algebras.

(b) In the case that A is the group algebra L(G), prove that EndL(G) L(G) = EndG(L(G)).Here, EndG(L(G)) refers to the commutant of the left regular representation (λ, L(G)).

(c) Using the above facts, give another proof that the map of Proposition 1.5.4 is an anti-isomorphism of algebras. (You are not asked to give a new proof that the map respectsthe involutions.)

1.5.4. Show that (1.40) is inverse to the map ψ 7→ Tψ of Proposition 1.5.4.

1.5.2 The Fourier transform

Suppose (σ, V ) is a representation of G and recall the definition (Definition 1.3.12) of theFourier transform of elements of L(G) at σ. For f1, f2 ∈ L(G), we have

σ(f1 ∗ f2) =∑g∈G

(f1 ∗ f2)(g)σ(g)

=∑g,h∈G

f1(gh)f2(h−1)σ(gh)σ(h−1) (1.42)

= σ(f1)σ(f2).

The Fourier transform is the algebra homomorphism

F : L(G)→ A(G) :=⊕ρ∈G

End(Wρ), f 7→⊕ρ∈G

ρ(f).

We define a scalar product on A(G) by⟨⊕ρ∈G

Tρ,⊕σ∈G

Sσ

⟩=

1

|G|∑ρ∈G

dρ tr(TρS

∗ρ

). (1.43)

(See Exercise 1.5.5.)

Recall that vρ1 , . . . , vρdρ is an orthonormal basis for Wρ. For ρ ∈ G and 1 ≤ i, j ≤ dρ,

defineT ρi,j ∈ A(G), T ρi,jw = δρ,σ

⟨w, vρj

⟩Wρvρi , w ∈ Wσ, σ ∈ G.

Thus, the map T ρi,j sends vρj to vρi and vσk to zero for all σ 6= ρ, 1 ≤ k ≤ dσ or σ = ρ, k 6= ij.

Theorem 1.5.7. The Fourier transform F is an isometric ∗-isomorphism between the al-gebras L(G) and A(G). Furthermore,

F ϕρi,j =|G|dρT ρi,j, for all ρ ∈ G, 1 ≤ i, j ≤ dρ. (1.44)


Proof. A proof of this statement can be found in [CSST10, Th. 1.5.11].

Let B′ ⊆ A(G) be the subalgebra consisting of elements that are diagonal in the basesvρ1 , . . . , v

ρdρ. Then B′ is a maximal commutative subalgebra of A(G). (See Exercise 1.5.6.)

DefineB = F−1(B′),

so that B is a maximal commutative subalgebra of L(G). Note that B depends on our choice

of bases for the Wρ, ρ ∈ G.The primitive idempotent associated with the vector vρj is the group algebra element

eρj :=dρ|G|

ϕρj,j ∈ L(G). (1.45)

Thus

eρj (g) =dρ|G|

ϕρj,j(g) =dρ|G|⟨ρ(g)wρj , w

ρj

⟩Wρ, for all g ∈ G. (1.46)

Proposition 1.5.8. (a) The seteρj : ρ ∈ G, j = 1, 2, . . . , dρ

is a vector space basis for B.

(b) For all ρ, σ ∈ G, 1 ≤ j ≤ dρ, 1 ≤ i ≤ dσ, we have

eρj ∗ eσi = δρ,σδj,ieρj .

(So the eρj are orthogonal idempotents.)

(c) For all ρ, σ ∈ G, 1 ≤ j ≤ dρ, 1 ≤ i ≤ dσ, we have

σ(eρj )vσi = δρ,σδj,iv

ρj .

In particular, ρ(eρj ) : Wρ → Wρ is the orthogonal projection onto Cvρj .

(d) If f ∈ B satisfies

ρ(f)vρj = λρjvρj , for all ρ ∈ G, 1 ≤ j ≤ dρ,

for some λρj ∈ C, then

f =∑ρ∈G

dρ∑j=1

λρjeρj (Fourier inversion formula in B)

andf ∗ eρj = λρje

ρj .

Proof. (a) By (1.44) and (1.45), we have

F(eρj ) = T ρj,j,

which is the diagonal matrix acting as 1 on vρj and as 0 on vσi for σ 6= ρ or i 6= j. Since suchmatrices form a basis for B′, it follows from Theorem 1.5.7 that the eρj form a basis for B.


(b) We have

eρj ∗ eσi =dρdσ|G|2

ϕρj,j ∗ ϕσi,i

=dρ

|G|δρ,σδj,iϕ

ρj,j (Lemma 1.5.5(b))

= δρ,σδj,ieρj .

(c) We have⊕σ∈G

σ(eρj )vσi =

(Feρj

)vσi = T ρj,jv

σi = δρ,σ〈vρi , v

ρj 〉Wρv

ρj = δρ,σδi,jv

ρj ∈ Wρ.

Comparing Wσ components, we see that

σ(eρj )vσi = δρ,σδj,iv

ρj .

(d) Suppose f ∈ B satisfies

ρ(f)vρj = λρjvρj , for all ρ ∈ G, 1 ≤ j ≤ dρ,

for some λρj ∈ C. By part (a), we have

f =∑ρ∈G

dρ∑j=1

µρjeρj

for some µρj ∈ C. Then, for σ ∈ G and 1 ≤ i ≤ dσ, we have

λσi vσi = σ(f)vσi

=∑ρ∈G

dρ∑j=1

µρjσ(eρj )vσi

=∑ρ∈G

dρ∑j=1

µρjδρ,σδj,ivρj (part (c))

= µσi vσi .

Thus µσi = λσi , as desired.Finally, we have

f ∗ eρj =

∑σ∈G

dσ∑i=1

λσi eσi

∗ eρj = λρjeρj ,

by part (b).


Exercises.

1.5.5. Prove that (1.43) defines a scalar product on A(G).

1.5.6. Let A be the subalgebra of Mn,n(C) consisting of diagonal matrices. Prove that Ais a maximal commutative subalgebra of Mn,n(C). In other words, prove that if B is acommutative subalgebra of Mn,n(C) containing A, then B = A.

1.5.7 ([CSST10, Ex. 1.5.18]). (a) Consider the decomposition (1.35) in the case of thegroup algebra (so X = G with the left regular action). Show that ψ ∈ L(G) be-

longs to T ρj Wρ if and only if ψ ∗ eσi = δσ,ρδi,jψ for all σ ∈ G and 1 ≤ i ≤ dσ.

(b) Show that f ∈ L(G) belongs to B if and only if each subspace T ρj Wρ is an eigenspace for

the associated convolution operator Tf : ψ 7→ ψ ∗ f ; moreover, if f ∈∑

ρ∈G∑dρ

j=1 λρjeρj ,

then the eigenvalue of Tf corresponding to T ρj Wρ is λρj .

1.5.3 Algebras of bi-K-invariant functions

Suppose G acts transitively on a finite set X. Fix x0 ∈ X and let K be the stabilizer of x0.Then, as before, we can identify X with G/K.

Let S be a set of representatives of the double cosets K\G/K of K in G. In other words,KsK : s ∈ S are the equivalences classes for the relation on G defined by

g ∼ h ⇐⇒ ∃ k1, k2 ∈ K such that g = k1hk2

So we haveG =

⊔s∈S

KsK.

For s ∈ S, define

Ωs := Ksx0 = KsKx0 = ksx0 : k ∈ K,Θs := G(x0, sx0) = (gx0, gsx0) : g ∈ G ⊆ X ×X.

Lemma 1.5.9. (a) X =⊔s∈S Ωs is the decomposition of X into K-orbits.

(b) X × X =⊔s∈S Θs is the decomposition of X × X into G-orbits (under the diagonal

action).

Proof. (a) For g ∈ G, Kgx0 is the K-orbit of gx0. Furthermore, for h ∈ G, we have

Kgx0 = Khx0 ⇐⇒ ∃ k1 ∈ K such that gx0 = k1hx0

⇐⇒ ∃ k1 ∈ K such that (k1h)−1gx0 = x0

⇐⇒ ∃ k1, k2 ∈ K such that (k1h)−1g = k2

⇐⇒ ∃ k1, k2 ∈ K such that g = k1hk2

⇐⇒ g ∈ KhK.


Since S is a set of representatives of K\G/K, it follows that Ksx0, s ∈ S, are the K-orbitson X.

(b) Suppose x, y ∈ X. Then x = g0x0 for some g0 ∈ G. Since the action of G on X istransitive, we can choose g ∈ G such that gx0 = g−1

0 y. Then

G(x, y) = G(g0x0, g0gx0) = G(x0, gx0).

Thus, every G-orbit on X ×X is of the form G(x0, gx0) for some g ∈ G.Furthermore, for g, h ∈ G, we have

G(x0, gx0) = G(x0, hx0) ⇐⇒ ∃ k1 ∈ K such that (x0, gx0) = (x0, k1hx0)

⇐⇒ ∃ k1, k2 ∈ K such that g = k1hk2.

Definition 1.5.10 (Left, right and bi-K-invariant functions). Let K be a subgroup of afinite group G.

• f ∈ L(G) is right K-invariant if f(gk) = f(g) for all g ∈ G, k ∈ K.

• f ∈ L(G) is left K-invariant if f(kg) = f(g) for all g ∈ G, k ∈ K.

• f ∈ L(G) is bi-K-invariant if it is both left and right K-invariant.

We let L(G/K) denote the subspace of rightK-invariant functions and let L(K\G/K) denotethe subspace of bi-K-invariant functions. (Note that this notation corresponds to the factthat right K-invariant functions can be viewed as functions on G/K and vice versa, andsimilarly for bi-K-invariant functions.)

It is straightforward to verify (Exercise 1.5.9) that the space L(G/K) is a left ideal inL(G) and hence, by (1.38), it is invariant under the left regular action. It follows that wecan restrict the permutation representation of G on L(G) to obtain a representation of Gon L(G/K). Similarly, L(K\G) is a right ideal in L(G). It follows that L(K\G/K) is asubalgebra of L(G).

Theorem 1.5.11. (a) The map

L(X)→ L(G/K), f 7→ f , where

f(g) = f(gx0), for all g ∈ G,

is an isomorphism of G-representations.

(b) The map

L(X ×X)G → L(K\G/K), F 7→ F , where

F (g) =1

|K|F (x0, gx0), for all g ∈ G,

is an isomorphism of algebras.

Proof. (a) This follows immediately from the isomorphism of G-sets X ∼= G/K.


(b) The map F 7→ F is clearly linear. It then follows from Exercise 1.5.8 that it is anisomorphism of vector spaces. Now suppose F1, F2 ∈ L(X ×X)G and let

F (x, y) =∑z∈X

F1(x, z)F2(z, y), for all x, y ∈ X,

so that F is the product of F1 and F2 in L(X ×X)G. Then, for all g ∈ G,

F (g) =1

|K|F (x0, gx0)

=1

|K|2∑h∈G

F1(x0, hx0)F2(hx0, gx0) (note we sum over G)

=1

|K|2∑h∈G

F1(x0, hx0)F2(x0, h−1gx0)

(since F2 ∈ L(X ×X)G

)=∑h∈G

F1(h)F2(h−1g)

=(F1 ∗ F2

)(g).

By Theorem 1.5.11 and Lemma 1.5.9 we have

L(X ×X)G ∼= L(K\G/K) ∼= L(X)K ,

where the first isomorphism is one of algebras, and the second is one of vector spaces. Usingthe second isomorphism, one can endow L(X)K with the structure of an algebra.

Corollary 1.5.12. We have that (G,K) is a Gelfand pair if and only if the algebra L(K\G/K)is commutative.

Proof. We have

(G,K) is a Gelfand pair ⇐⇒ L(X) is multiplicity free (definition of Gelfand pair)

⇐⇒ EndG(L(X)) is commutative (Corollary 1.2.13)

⇐⇒ L(X ×X)G is commutative (Corollary 1.4.3)

⇐⇒ L(K\G/K) is commutative (Theorem 1.5.11(b)).

We are now able to prove the general form of Gelfand’s lemma (see Proposition 1.4.6).

Proposition 1.5.13 (Gelfand’s lemma). Suppose G is a finite group and K ≤ G is a sub-group. Furthermore, suppose there exists an automorphism τ of G such that g−1 ∈ Kτ(g)Kfor all g ∈ G. Then (G,K) is a Gelfand pair.

Proof. If f ∈ L(K\G/K), then f(τ(g)) = f(g−1) for all g ∈ G. Thus, for all f1, f2 ∈L(K\G/K) and g ∈ G, we have

(f1 ∗ f2)(τ(g)) =∑s∈G

f1(τ(g)s)f2

(s−1)


=∑h∈G

f1(τ(gh))f2(τ(h−1)) (h = τ−1(s))

=∑h∈G

f1

((gh)−1

)f2(h)

=∑h∈G

f2(h)f1

(h−1g−1

)= (f2 ∗ f1)

(g−1)

= (f2 ∗ f1)(τ(g)),

and so L(K\G/K) is commutative. Then the result follows from Corollary 1.5.12.

We say that (G,K) is a weakly symmetric Gelfand pair if it satisfies the hypotheses ofProposition 1.5.13.

Example 1.5.14. The group G×G acts on G by

(g1, g2) · g = g1gg−12 , for all g1, g2, g ∈ G.

The stabilizer of 1G is the diagonal subgroup

G = (g, g) : g ∈ G ≤ G×G,

and thus G ∼= (G×G)/G as G-sets. Now consider the flip automorphism

τ : G×G→ G×G, τ(g1, g2) = (g2, g1), for all (g1, g2) ∈ G×G.

Then, for all g1, g2 ∈ G, we have

(g1, g2)−1 =(g−1

1 , g−12

)=(g−1

1 , g−11

)(g2, g1)

(g−1

2 , g−12

)∈ Gτ(g1, g2)G.

Therefore, (G×G, G) is a weakly symmetric Gelfand pair.

Exercises.

1.5.8 ([CSST10, Ex. 1.5.20]). For every orbit Ω of K on X, set

ΘΩ = (gx0, gx) : g ∈ G, x ∈ Ω.

Show that the map Ω 7→ ΘΩ is a bijection from the set of K-orbits on X to the set of G-orbitson X ×X.

1.5.9. Verify that L(G/K) is a left ideal in L(G) and that L(K\G) is a right ideal in L(G).Then prove that L(K\G/K) is a subalgebra of L(G).

1.6. Induced representations 61

1.5.10 ([CSST10, Ex. 1.5.25]). Show that (G,K) is a symmetric Gelfand pair if and only ifg−1 ∈ KgK for all g ∈ G. Note that this corresponds to the case τ = IG in Proposition 1.5.13.

1.5.11 ([CSST10, Ex. 1.5.27]). A group G is ambivalent if g−1 is conjugate to g for everyg ∈ G. We adopt the notation of Example 1.5.14. Show that the Gelfand pair (G×G, G) issymmetric if and only if G is ambivalent.

1.6 Induced representations

1.6.1 Definitions and examples

Let K be a subgroup of G and let (ρ, V ) be a representation of K. Consider the action ofK on G given by

k · g = gk−1, k ∈ K, g ∈ G.

Definition 1.6.1 (Induced representation). The representation induced by (ρ, V ) is theG-representation (σ, Z) defined by

Z = FunK(G, V ) :=f : G→ V : f(gk) = ρ(k−1)f(g), for all g ∈ G, k ∈ K

(1.47)

and(σ(g1)f) (g2) = f

(g−1

1 g2

), for all g1, g2 ∈ G, f ∈ Z. (1.48)

(Note that Z is a vector space under pointwise addition and scalar multiplication, and thatσ(g)f ∈ Z for all g ∈ G and f ∈ Z.) We introduce the notation

IndGK ρ = σ and IndGK V = Z.

One can also give another description of induced representations. Fix a set S of repre-sentatives for the set G/K of right cosets of K in G, so that

G =⊔s∈S

sK. (1.49)

For every v ∈ V , define

fv : G→ V, fv(g) =

ρ (g−1) v if g ∈ K,0 if g /∈ K.

(1.50)

Then fv ∈ Z. Furthermore, the space

V = fv : v ∈ V (1.51)

is a K-invariant subspace of Z and the map

V → V , v 7→ fv,

is an isomorphism of K-representations. (See Exercise 1.6.1.)


We claim that

Z =⊕s∈S

σ(s)V (1.52)

as vector spaces. Indeed, for every f ∈ Z and s ∈ S, let vs = f(s). Then we have

f =∑s∈S

σ(s)fvs (1.53)

and this is the unique way to write f as a sum of elements of σ(s)V , s ∈ S. (See Exer-cise 1.6.1.) Furthermore, for all g ∈ G and s ∈ S, by (1.49) there exist unique elementsts ∈ S and ks ∈ K such that gs = tsks. Then we have

σ(g)f =∑s∈S

σ(gs)fvs =∑s∈S

σ(ts)σ(ks)fvs =∑s∈S

σ(ts)fρ(ks)vs ,

where in the last equality we used the fact that the map v 7→ fv is an isomorphism ofK-representations.

The following lemma is a converse to the above.

Lemma 1.6.2. Let K ≤ G, and let S be a set of representatives of G/K. Let (τ,W ) be arepresentation of G. Suppose that V ≤ W is a K-invariant subspace and that

W =⊕s∈S

τ(s)V

as vector spaces. Then W ∼= IndGK V as representations of G.

Proof. Define V as in (1.51). For each s ∈ S, let vs = f(s). Then we have the isomorphismof vector spaces.

W =⊕s∈S

τ(s)V →⊕s∈S

σ(s)V = Z,∑s∈S

τ(s)vs 7→∑s∈S

σ(s)fvs , vs ∈ V ∀ s ∈ S. (1.54)

For g ∈ G, we have

τ(g)

(∑s∈S

τ(s)vs

)=∑s∈S

τ(gs)vs =∑s∈S

τ(ts)τ(ks)vs

7→∑s∈S

σ(ts)fτ(ks)vs =∑s∈S

σ(ts)σ(ks)fvs = σ(g)∑s∈S

σ(s)fvs .

Hence (1.54) is an isomorphism of G-representations.

Recall that the index of the subgroup K ≤ G, is defined to be [G : K] = |G/K|. Since|S| = |G/K|, it follows immediately from Lemma 1.6.2 that

dim(IndGK V

)= [G : K] dimV. (1.55)


Proposition 1.6.3. Suppose K ≤ G and let X = G/K. Then the permutation represen-tation of G on L(X) is isomorphic to IndGK ι, where (ι,C) is the trivial representation ofK.

Proof. By definition,

IndGK C = f : G→ C : f(gk) = f(g), for all g ∈ G, k ∈ K

is the space of right K-invariant functions on G. Thus the proposition follows from Theo-rem 1.5.11(a).

Exercises.

1.6.1. (a) Prove that fv, as defined in (1.50), is an element of Z.

(b) Prove that V , as defined in (1.51), is a K-invariant subspace of Z.

(c) Prove that the map V → V , v 7→ fv, is an isomorphism of K-representations.

(d) Prove that equality (1.53) holds and that this is the unique way to write f ∈ Z as asum of elements of σ(s)V , s ∈ S. Hint : f is uniquely determined by its values on S.

1.6.2 ([CSST10, Ex. 1.6.3]). Suppose K is a subgroup of G and let S be a set of representa-tives of G/K. Let (π,W ) be a representation of G, suppose that V ≤ W is K-invariant, anddenote by (ρ, V ) the corresponding K-representation. Prove that if W = 〈π(s)V : s ∈ S〉,then there exists a surjective intertwiner from IndGK ρ to π.

Hint : If we let σ = IndGK ρ, then the required surjective intertwiner is the map

σ(s)fv 7→ π(s)v, s ∈ S, v ∈ V,

extended by linearity.

1.6.2 First properties of induced representations

Induction is transitive in the following sense.

Proposition 1.6.4 (Transitivity of induction). Suppose K ≤ H ≤ G are subgroups and let(ρ, V ) be a representation of K. Then

IndGH(IndHK V

) ∼= IndGK V

as representations of G.


Proof. Let σ = IndHK ρ. Consider the linear map

IndGH(IndHK V

)= FunH (G,FunK(H, V ))→ Fun(G, V ) := f : G→ V ,f 7→ f , where f(g) = f(g)(1G), g ∈ G.

For k ∈ K, g ∈ G, and f ∈ HomH (G,HomK(H,V )), we have

f(gk) = f(gk)(1G)

=(σ(k−1)f(g)

)(1G) (since f ∈ HomH(G,HomK(H, V )))

= f(g)(k) (by (1.48))

= ρ(k−1)(f(g)(1G)

)(since f(g) ∈ HomK(H,V ))

= ρ(k−1)f(g)

Thus, f ∈ FunK(G, V ).Let η = IndGH σ and π = IndGK ρ. Then, for g1, g2 ∈ G and f ∈ HomH (G,HomK(H, V )),

we have (π(g1)f

)(g2) = f

(g−1

1 g2

)= f

(g−1

1 g2

)(1G)

=(η(g1)f

)(g2)(1G)

= ˜(η(g1)f)(g2).

Thus the map f 7→ f is an intertwiner

IndGH(IndHK V

)= FunH (G,HomK(H,V ))→ FunK(G, V ) = IndGK V.

Since

dim IndGH(IndHK V

)= [G : H] dim IndHK V = [G : H][H : K] dimV

= [G : K] dimV = IndGK V,

to show that f 7→ f is an isomorphism, it suffices to prove that it is injective. Supposef = 0. Then, for all g ∈ G and h ∈ H, we have

f(g)(h) =(σ(h−1)f(g)

)(1G) = f(gh)(1G) = f(gh).

Thus f = 0, as desired.

Theorem 1.6.5 (Frobenius character formula for induced representations). Let (ρ,W ) be arepresentation of K, where K ≤ G. Then

χIndGK ρ(g) =∑

s∈S:s−1gs∈K

χρ(s−1gs

), (1.56)

where S is any system of representatives for G/K.

Proof. A proof of this result can be found in [CSST10, Th. 1.6.7].


1.6.3 Frobenius reciprocity

The following fundamental result gives a precise relationship between the operations of in-duction and restriction.

Theorem 1.6.6 (Frobenius reciprocity). Let G be a finite group, K ≤ G a subgroup, (σ,W )a representation of G, and (ρ, V ) a representation of K. Then we have an isomorphism ofvector spaces

HomG

(W, IndGK V

)→ HomK

(ResGKW,V

), T 7→ T , where

T : W → V, Tw = (Tw)(1G).

Proof. We first check that T ∈ HomK

(ResGKW,V

). Let τ = IndGK ρ. For k ∈ K and w ∈ W ,

we have

T σ(k)w =(T (σ(k)w)

)(1G)

=(τ(k)(Tw)

)(1G)

(T ∈ HomG

(IndGK V

))= (Tw)(k−1) (by (1.48))

= ρ(k)(Tw(1G)

)(by (1.47))

= ρ(k)Tw,

as desired.Now consider the map

HomK

(ResGKW,V

)→ HomG

(W, IndGK V

), U 7→ U , where(

Uw)

(g) = Uσ(g−1)w, for all g ∈ G, w ∈ W.

It is straightforward to verify that U ∈ HomG

(W, IndGK V

)(Exercise 1.6.3). Let T ∈

HomG

(W, IndGK V

), and set U = T . Then(

Uw)

(g) = T σ(g−1)w =(Tσ(g−1)w

)(1G) =

(τ(g−1)(Tw)

)(1G) = (Tw)(g).

Thus U = T . Similarly, let U ∈ HomK

(ResGKW,V

)and set T = U . One can verify that

T = U (Exercise 1.6.4).

Remark 1.6.7. In the language of category theory, Theorem 1.6.6 (together with a “na-turality” statement) says that induction is right adjoint to restriction. Induction is alsoleft adjoint to restriction, but this is an easier result that holds in a much greater level ofgenerality. The fact that induction is right adjoint is an important property of what arecalled Frobenius extensions. The group algebras of nested finite groups are special cases ofa Frobenius extensions.

Corollary 1.6.8. Suppose, in the setting of Theorem 1.6.6 that W and V are irreducible.Then the multiplicity of W in IndGK V is equal to the multiplicity of V in ResGKW .

Proof. This follows immediately from Theorem 1.6.6 and Lemma 1.2.5.


Exercises.

1.6.3. Prove that U , as defined in the proof of Theorem 1.6.6 is an element of HomG

(W, IndGK V

).

1.6.4. In the notation of the proof of Theorem 1.6.6, let U ∈ HomK

(ResGKW,V

)and set

T = U . Prove that T = U .

1.6.4 Mackey’s lemma and the intertwining number theorem

We now discuss a sort of “commutation” property for induction and restriction. Suppose Hand K are two subgroups of G and let (ρ,W ) be a representation of K. Let S be a systemof representatives for the double cosets in H\G/K, so that

G =⊔s∈S

HsK.

For each s ∈ S, letGs = sKs−1 ∩H ≤ G,

and define a representation (ρs,Ws) of Gs by setting Ws = W and

ρs(t)w = ρ(s−1ts)w, for all t ∈ Gs, w ∈ Ws.

Theorem 1.6.9 (Mackey’s lemma). With notation as above, we have an isomorphism ofH-representations

ResGH IndGK ρ∼=⊕s∈S

IndHGs ρs.

Proof. Let

Zs =F : G→ W : F (hs′k) = δs,s′ρ(k−1)F (hs) ∀h ∈ H, k ∈ K, s′ ∈ S

.

Comparing to (1.47), we see that Zs is the subspace of Z = IndGKW consisting of thosefunctions that vanish outside HsK. Thus

Z =⊕s∈S

Zs as vector spaces.

For each s ∈ S, the space Zs is H-invariant (Exercise 1.6.6(a)). Therefore, it suffices toprove that, for all s ∈ S,

Zs ∼= IndHGsWs as representations of H.

Consider the linear map

Zs → Fun(H,W ) = f : H → W, F 7→ fs, where fs(h) = F (hs), ∀ h ∈ H. (1.57)


For t ∈ Gs, we have s−1ts ∈ K, and so

fs(ht) = F (hts) = F(hss−1ts

)= ρ

(s−1t−1s

)F (hs) = ρs

(t−1)fs(h).

Thus fs ∈ FunGs(H,Ws) = IndHGsWs. Hence (1.57) is a linear map

Zs → IndHGsWs.

We will now construct the inverse to (1.57). First consider the map

IndHGsWs → Fun(G,W ), f 7→ Fs ∈ Fun(G,W ), where (1.58)

Fs(hs′k) = δs,s′ρ(k−1)f(h), for all k ∈ K, h ∈ H, s′ ∈ S.

We must verify that Fs is well defined. If hsk = h1sk1 for some h, h1 ∈ H and k, k1 ∈ K,then t := skk−1

1 s−1 = h−1h1 ∈ Gs, and so

ρ(k−1

1

)f(h1) = ρ

(k−1) (ρ(s−1h−1h1s

)f(h1)

)(since k−1

1 = k−1s−1h−1h1s)

= ρ(k−1)

(ρs(t)f(h1))

= ρ(k−1)f(h1t−1)

= ρ(k−1)f(h).

Thus Fs is well defined. For h ∈ H, k ∈ K, and s′ ∈ S, we have

Fs(hs′k) = δs,s′ρ(k−1)f(h) = δs,s′ρ(k−1)Fs(hs),

so Fs ∈ Zs. Hence (1.58) is a linear map

IndHGsWs → Zs.

It is straightforward to verify that the maps (1.57) and (1.58) are mutually inverse andthat (1.57) intertwines the H-action (Exercise 1.6.6(b)). This completes the proof.

Let us now consider the special case where (ρ,W ) is the permutation representation onL(X), with X = G/K. Let x0 = K ∈ G/K, so that K is the stabilizer of x0. As above, welet S be a set of representatives for the double cosets H\G/K. By Exercise 1.6.5, Gs is thestabilizer of sx0 for each s ∈ S. Then

X =⊔s∈S

Ωs, Ωs = Hsx0 = hsx0 : h ∈ H,

is the decomposition of X into H-orbits.Now let ρ = ιK be the trivial representation of K. Then ρs is the trivial representation

of Gs, and IndHGs ρs is the permutation representation of H on L(Ωs) by Proposition 1.6.3.Thus, Mackey’s lemma gives the decomposition

ResGH L(X) =⊕s∈S

L (Ωs) . (1.59)

We finish this chapter with an important application of Mackey’s lemma.


Theorem 1.6.10 (Intertwining number theorem). With the same hypotheses as in Mackey’slemma (Theorem 1.6.9), assume that σ is a representation of H. Then

dim HomG

(IndGH σ, IndGK ρ

)=∑s∈S

dim HomGs

(ResHGs σ, ρs

).

Proof. We have

dim HomG

(IndGH σ, IndGK ρ

)= dim HomH

(σ,ResGH IndGK ρ

)(Frobenius reciprocity (Theorem 1.6.6))

=∑s∈S

dim HomH

(σ, IndHGs ρs

)(Mackey’s lemma (Theorem 1.6.9))

=∑s∈S

dim HomGs

(ResHGs σ, ρs

)(Frobenius reciprocity (Theorem 1.6.6)).

Exercises.

1.6.5 ([CSST10, Ex. 1.6.13]). Identify H\G/H with the set of H-orbits on X = G/K. Provethat Gs is the stabilizer in H of the point xs = sK. (Compare with Lemma 1.5.9.)

1.6.6. (a) Prove that Zs is H-invariant.

(b) Verify that (1.57) and (1.58) are mutually inverse, and that (1.57) intertwines theH-action.

Chapter 2

The theory of Gelfand–Tsetlin bases

Our goal in this chapter is to develop the theory of Gelfand–Tsetlin bases for group algebrasand permutation representations. These are bases that are well suited to the restriction of arepresentation to certain chains of subgroups. We closely follow the presentation in [CSST10,Ch. 2]. Throughout this chapter we again suppose that G is a finite group.

2.1 Algebras of conjugacy invariant functions

2.1.1 Conjugacy invariant functions

Suppose H is a subgroup of G. A function f ∈ L(G) is H-conjugacy invariant if

f(h−1gh

)= f(g), for all h ∈ H, g ∈ G.

We let C(G,H) denote the set of all H-conjugacy invariant functions on G. This is asubalgebra of L(G) (under convolution). Indeed, for f1, f2 ∈ C(G,H), we have

(f1 ∗ f2)(h−1gh

)=∑s∈G

f1

(h−1ghs

)f2

(s−1)

=∑s∈G

f1

(ghsh−1

)f2

(hs−1h−1

)(since f1, f2 ∈ C(G,H))

=∑t∈G

f1(gt)f2

(t−1) (

t = hsh−1)

= (f1 ∗ f2)(g).

Hence f1 ∗ f2 ∈ C(G,H). Note that C(G,G) is the algebra of central functions on G (Defi-nition 1.3.1).

Consider the action of G×H on G defined by

(g, h) · g0 = gg0h−1, for all g, g0 ∈ G, h ∈ H. (2.1)

We denote the associated permutation representation by η, so that(η(g, h)f

)(g0) = f

(g−1g0h

), for all f ∈ L(G), g, g0 ∈ G, h ∈ H. (2.2)

69

70 Chapter 2. The theory of Gelfand–Tsetlin bases

Lemma 2.1.1. (a) The stabilizer of 1G under the action (2.1) is

H = (h, h) : h ∈ H ≤ G×H.

(b) Let L(H\(G×H)/H

)denote the algebra of bi-H-invariant functions on G×H. Then

we have an isomorphism of algebras

Φ: L(H\(G×H)/H

)→ C(G,H), Φ(F )(g) = |H|F (g, 1G).

Proof. Part (a) is clear since

1G = (g, h) · 1G = gh−1 ⇐⇒ g = h.

We now prove part (b).Suppose F ∈ L

(H\(G×H)/H

)and let f = Φ(F ). Then

f(h−1gh

)= |H|F

(h−1gh, h−1h

)= |H|F (g, 1G) (since F is bi-H-invariant)

= f(g).

Thus f ∈ C(G,H). It is clear that Φ is linear. To see that Φ is injective, note that

F (g, h) = F(gh−1, 1G

)(since F is right H-invariant)

=1

|H|f(gh−1

)(where f = Φ(F )).

Thus F is uniquely determined by f . To prove Φ is surjective, let f ∈ C(G,H). Then, if wedefine F (g, h) = 1

|H|f (gh−1), we have that F is bi-H-invariant, and f = Φ(F ).

It remains to prove that Φ is multiplicative. Let F1, F2 ∈ L(H\(G×H)/H

). Then, for

all g ∈ G, we have(Φ(F1 ∗ F2)

)(g) = |H| (F1 ∗ F2)(g, 1G)

= |H|∑s∈G

∑h∈H

F1(gs, h)F2

(s−1, h−1

)= |H|

∑s∈G

∑h∈H

F1(gsh−1, 1G)F2

(hs−1, 1G

)(F1, F2 are bi-H-invariant)

= |H|2∑t∈G

F1(gt, 1G)F2

(t−1, 1G

) (sh−1 = t

)=∑t∈G

(Φ(F1)(gt))(Φ(F2)(t−1)

)=(Φ(F1) ∗ Φ(F2)

)(g).

We now wish to decompose the permutation representation η of (2.2) into irreducibleG×H-subrepresentations. Recall that, by Theorem 1.3.18, every irreducible representationof G × H is of the form σ ρ for some σ ∈ G and ρ ∈ H. Note also that the adjoint ofResGH σ is ResGH σ

′, where σ′ is the adjoint of σ (Exercise 2.1.1).

2.1. Algebras of conjugacy invariant functions 71

Theorem 2.1.2. Suppose (σ, V ) ∈ G and (ρ,W ) ∈ H. Let (ρ′,W ′) ∈ H denote the adjointof (ρ,W ). Then we have an isomorphism of vector spaces

HomG×H(σ ρ, η)→ HomH

(ρ,ResGH σ

′) , T 7→ T , where(Tw)(v) =

(T (v ⊗ w)

)(1G), for all v ∈ V, w ∈ W.

Proof. Note that a linear map T : V ⊗W → L(G) lies in HomG×H(σ ρ, η) if and only if(T (σ(g)v⊗ ρ(h)w)

)(g0) =

(T (v⊗w)

) (g−1g0h

), ∀ g, g0 ∈ G, h ∈ H, v ∈ V, w ∈ W. (2.3)

Let T ∈ HomG×H(σ ρ, η). For h ∈ H, v ∈ V , and w ∈ W , we have(T ρ(h)w

)(v) =

(T (v ⊗ ρ(h)w)

)(1G)

=(T (v ⊗ w)

)(h) (by (2.3))

=(T (σ(h−1)v ⊗ w)

)(1G) (by (2.3))

=(Tw) (σ(h−1)v)

=(σ′(h)Tw

)(v) (by (1.9)).

Thusσ′(h)T = T ρ(h), for all h ∈ H,

and so T ∈ HomH

(ρ,ResGH σ

′), as desired.

To see that the map T 7→ T is injective, note that, by (2.3), we have(T (v ⊗ w)

)(g) =

(T(σ(g−1)v ⊗ w

))(1G) =

(Tw) (σ(g−1)v). (2.4)

Thus T is uniquely determined by T .Since the map T 7→ T is clearly linear, it remains to show it is surjective. Suppose

S ∈ HomH

(ρ,ResGH σ

′). Define T ∈ Hom(V ⊗W,L(G)) by(T (v ⊗ w)

)(g0) = (Sw)

(σ(g−1

0

)v), for all g0 ∈ G, v ∈ V, w ∈ W. (2.5)

Then, for all g ∈ G, h ∈ H, v ∈ V , and w ∈ W , we have(T (σ(g)v ⊗ ρ(h)w)

)(g0) = (Sρ(h)w)

(σ(g−1

0 g)v)

= (σ′(h)Sw)(σ(g−1

0 g)v) (

S ∈ HomH

(ρ,ResGH σ

′))= (Sw)

(σ(h−1g−1

0 g)v)

(by (1.9))

= (T (v ⊗ w))(g−1g0h

).

Hence T satisfies (2.3), and so T ∈ HomG×H(σ ρ, η). Comparing (2.4) and (2.5), we seethat T = S.

Corollary 2.1.3. The multiplicity of σ ρ in η is equal to the multiplicity of ρ in ResGH σ′.

Proof. This follows from Lemma 1.2.5 and Theorem 2.1.2.


For σ ∈ G and ρ ∈ H, let

mρ,σ = dim HomH

(ρ,ResGH σ

′)denote the multiplicity of ρ in ResGH σ

′.

Corollary 2.1.4. The decomposition of η into irreducible subrepresentations of G × H isgiven by

η ∼=⊕σ∈G

⊕ρ∈H

(σ ρ)⊕mρ,σ .

We now consider the case where H = G, so that (η, L(G)) is a representation of G×G.We showed in Example 1.5.14 that (G×G, G) is a Gelfand pair. This also follows from thefollowing result.

Corollary 2.1.5. Suppose that H = G. We have a decomposition into G × G-invariantsubspaces

L(G) =⊕ρ∈G

Mρ,

where Mρ is the subspace of L(G) spanned by all matrix coefficients

ϕ(g) = 〈ρ(g)v, w〉Wρ , v, w ∈ Wρ.

Furthermore, the action of G × G on the summand Mρ is isomorphic to ρ′ ρ. Hence wehave an isomorphism of G×G-representations

η ∼=⊕ρ∈G

ρ′ ρ.

Proof. We have

mρ,σ = dim HomG(ρ, σ′) =

1 if ρ ∼ σ′,

0 otherwise.

Define T ∈ HomG×G(ρ′ ρ, η) as in (2.5) with σ = ρ′ and S = IWρ . Then, for all v, w ∈ Wρ

and g ∈ G, we have (T (θv ⊗ w)

)(g) =

(ρ′(g−1)θv)

(w)

= θv(ρ(g)w)

= 〈ρ(g)w, v〉,

where, in the first equality, we use the natural identification of (ρ′)′ with ρ (i.e. of the doubledual of a vector space with the vector space itself.) Hence T (θv ⊗ w) ∈Mρ.

Since ρ′ ρ is irreducible and dimMρ = d2ρ (a basis for Mρ is given by the matrix

coefficients ϕρi,j, 1 ≤ i, j ≤ dρ, by Corollary 1.5.6), it follows that T (W ′ρ ⊗ Wρ) = Mρ,

completing the proof.

2.1. Algebras of conjugacy invariant functions 73

Exercises.

2.1.1. Prove that if σ is a representation of G with adjoint σ′, then the adjoint of ResGH σ isResGH σ

′.

2.1.2. (a) Using the properties of the matrix coefficients, prove directly that Mρ∼= W ′

ρ ⊗Wρ, in the language of the proof of Corollary 2.1.5. Then deduce Corollary 2.1.5.

(b) Use part (a) to prove Corollary 2.1.4.

2.1.2 Multiplicity-free subgroups

Definition 2.1.6 (Multiplicity-free subgroup). A subgroup H of G is said to be multiplicity

free if, for every σ ∈ G, the restriction ResGH σ is multiplicity free or, equivalently, if

dim HomH

(ρ,ResGH σ

)≤ 1, for all ρ ∈ H, σ ∈ G.

Theorem 2.1.7. The following conditions are equivalent:

(a) The algebra C(G,H) is commutative.

(b) (G×H, H) is a Gelfand pair. (Recall that H = (h, h) : h ∈ H.)

(c) H is a multiplicity-free subgroup of G.

Proof. The equivalence of (b) and (c) follows from Corollary 2.1.4, Lemma 2.1.1(a), and thedefinition of a Gelfand pair (Definition 1.4.7). The equivalence of (a) and (b) follows fromLemma 2.1.1(b) and Corollary 1.5.12.

Remark 2.1.8. (a) When H = G, the conditions of Theorem 2.1.7 are always satisfied:

• C(G,G) is the space of central functions, which is the centre of the group algebra L(G)(see Remark 1.5.3). Thus it is commutative.

• (G×G, G) is a Gelfand pair by Example 1.5.14.

• Clearly G is a multiplicity-free subgroup of itself.

(b) When H = 1G, the conditions in Theorem 2.1.7 are equivalent to G being abeliansince C(G, 1G) is commutative if and only if G is commutative. (This follows, for example,from (1.39).)

Proposition 2.1.9. We have that (G×H, H) is a symmetric Gelfand pair if and only if

∀ g ∈ G ∃h ∈ H such that hgh−1 = g−1

(that is, every element of G is H-conjugate to its inverse). Moreover, if this is the case, thenH is a multiplicity free subgroup of G.


Proof. By Exercise 1.5.10, the pair (G × H, H) is symmetric if and only if for all (g, h) ∈G×H, there exist h1, h2 ∈ H such that(

g−1, h−1)

= (g, h)−1 = (h1, h1)(g, h)(h2, h2) = (h1gh2, h1hh2). (2.6)

Taking h = 1G, we obtain h2 = h−11 and so g−1 = h1gh

−11 .

To prove the converse implication, suppose that every element of G is H-conjugate to itsinverse. Then, for (g, h) ∈ G×H, we can choose t ∈ H such that(

gh−1)−1

= t(gh−1

)t−1.

Thus, taking h1 = h−1t and h2 = h−1t−1, we see that

(h1gh2, h1hh2) =(h−1tgh−1t−1, h−1thh−1t−1

)=(h−1hg−1, h−1

)=(g−1, h−1

).

Hence (2.6) is satisfied.

Exercises.

2.1.3. Suppose G is abelian. Show that (G×H, H) is a symmetric Gelfand pair if and onlyif every g ∈ G satisfies g2 = 1G.

2.2 Gelfand–Tsetlin bases

2.2.1 Branching graphs and Gelfand–Tsetlin bases

A chain

1G = G1 ≤ G2 ≤ · · · ≤ Gn−1 ≤ Gn = G (2.7)

of subgroups is said to be multiplicity free if Gk−1 is a multiplicity-free subgroup of Gk for1 < k ≤ n. Note that, by Remark 2.1.8((b)), if (2.7) is multiplicity free, then G2 is abelian.

From now on, we fix a multiplicity-free chain (2.7). The branching graph of this chain isthe directed graph with vertex set

n⊔k=1

Gk

and edge set(ρ, σ) ∈ Gk × Gk−1 : σ is a subrepresentation of ResGkGk−1

ρ, 2 ≤ k ≤ n.

We will write ρ→ σ if (ρ, σ) is an edge of the branching graph.

2.2. Gelfand–Tsetlin bases 75

Suppose (ρ, Vρ) ∈ Gn. Then

ResGnGn−1Vρ =

⊕σ∈Gn−1:ρ→σ

Vσ

is an orthogonal decomposition. Then, for each σ ∈ Gn−1 we have an orthogonal decompo-sition

ResGn−1

Gn−2Vσ =

⊕θ∈Gn−2:σ→θ

Vθ.

We continue in this way until, after the restriction from G2 to G1, we are left with sums ofone-dimensional trivial representations.

To keep track of the restrictions, let

T (ρ) = T : T = (ρ = ρn → ρn−1 → ρn−2 → · · · → ρ2 → ρ1), ρk ∈ Gk, 1 ≤ k ≤ n− 1.

denote the set of all paths T in the branching graph. (We will always be interested in directedpaths.) Then we have

Vρ =⊕ρn−1:ρ→ρn−1

Vρn−1 =⊕ρn−1:ρ→ρn−1

⊕ρn−2:

ρn−1→ρn−2

Vρn−2 = · · · =⊕

T∈T (ρ)

Vρ1 . (2.8)

Since ρ1 is the trivial representation of G1 = 1G, it is one-dimensional. Therefore, for eachT ∈ T (ρ), we can choose vT ∈ Vρ1 with ‖vT‖ :=

√〈vT , vT 〉 = 1. (Thus vT is defined up to a

scalar factor of norm one.) Then (2.8) becomes

Vρ =⊕

T∈T (ρ)

CvT . (2.9)

In other words,vT : T ∈ T (ρ)

is an orthonormal basis of V , called a Gelfand–Tsetlin basis for Vρ with respect to themultiplicity-free chain (2.7).

The branching graph and Gelfand–Tsetlin bases are well adapted to restriction to thesubgroups appearing in the corresponding multiplicity free chain. In particular, if θ ∈ Gk

for some 1 ≤ k ≤ n− 1, then the multiplicity of θ in ResGnGk ρ is equal to the number of pathsfrom ρ to θ in the branching graph. Furthermore, we obtain an orthogonal decomposition ofthe θ-isotypic component of ResGnGk ρ into irreducible Gk-subrepresentations (each isomorphicto Vθ). Namely, we have a unique component Vθ for Vρ for each path from ρ to θ, and thecomponents corresponding to distinct paths are orthogonal.

For j = 1, 2, . . . , n and ρ ∈ G, we let

Tj(ρ) = S : S = (ρ = ρn → ρn−1 → · · · → ρj+1 → ρj), ρk ∈ Gk, j ≤ k ≤ n− 1.

In particular, T1(ρ) = T (ρ). For

T = (ρ = ρn → ρn−1 → · · · → ρ2 → ρ1) ∈ T (ρ),


we define the j-th truncation of T to be

Tj = (ρ = ρn → ρn−1 → · · · → ρj+1 → ρj) ∈ Tj(ρ).

For 1 ≤ j ≤ n and S ∈ Tj(ρ), we define

VS =⊕

T∈T (ρ):Tj=S

Vρ1 , ρS =(

ResGGj ρ)∣∣∣

VS. (2.10)

Then (ρS, VS) is an irreducible Gj-representation, and ρS ∼ ρj.

Exercises.

2.2.1. Fix a positive integer n and let Gn be the cyclic group of order 2n with generator a.For 0 ≤ k ≤ n, let Gk be the subgroup of Gn generated by a2n−k .

(a) Prove that 1G = G0 ≤ G1 ≤ · · · ≤ Gn is a multiplicity-free chain of subgroups.

(b) Describe the branching graph for this chain.

(c) Draw the branching graph for n = 3.

2.2.2 Gelfand–Tsetlin algebras

Let H be a subgroup of G. We can extend any f ∈ L(H) by zero to a function fGH ∈ L(G)defined by

fGH (g) =

f(g) if g ∈ H,0 otherwise.

If H ≤ K ≤ G, it is straightforward to verify that, for all f, f1, f2 ∈ L(H) and α1, α2 ∈ C,we have

•(fKH)GK

= fGH ,

• (f1 ∗ f2)GH = (f1)GH ∗ (f2)GH (Exercise 2.2.2), and

• (α1f1 + α2f2)GH = α1(f1)GH + α2(f2)GH .

Thus we can view L(H) as a subalgebra of L(G). By an abuse of notation, we will oftendenote the extension fGH again by f .

Recalling Definition 1.3.12, for all ρ ∈ G and f ∈ L(H), we have

ρ(fGH)

=∑g∈G

fGH (g)ρ(g) =∑h∈H

f(h)ρ(h) =∑h∈H

f(h)(ResGH ρ

)(h) =

(ResGH ρ

)(f).


Definition 2.2.1 (Gelfand–Tsetlin algebra). For 1 ≤ k ≤ n, we let Z(k) denote the centerof the group algebra L(Gk), that is, Z(k) is the subalgebra of central functions on Gk (seeRemark 1.5.3). The Gelfand–Tsetlin algebra GZ(n) associated with the multiplicity-freechain (2.7) is the subalgebra of L(Gn) generated by the subalgebras

Z(1), Z(2), . . . , Z(n).

Theorem 2.2.2. The Gelfand–Tsetlin algebra GZ(n) is a maximal commutative subalgebraof L(Gn). Furthermore

GZ(n) = f ∈ L(G) : ρ(f)vT ∈ CvT , for all ρ ∈ Gn and T ∈ T (ρ). (2.11)

In other words, GZ(n) is the subalgebra of L(Gn) consisting of those f ∈ L(Gn) whose

Fourier transforms ρ(f), ρ ∈ Gn act diagonally on the Gelfand–Tsetlin basis of Vρ.

Proof. For fi ∈ Z(i) and fj ∈ Z(j) with i ≤ j, we have fi ∈ L(Gi) ≤ L(Gj), and so fi ∗ fj,since fj ∈ Z(j). Thus GZ(n) is commutative and spanned by the products

f1 ∗ f2 ∗ · · · ∗ fn, fk ∈ Z(k), 1 ≤ k ≤ n.

Let A denote the right-hand side of (2.11). Then A is an algebra by the multiplicativeproperty of the Fourier transform (see (1.42)).

Suppose 1 ≤ j ≤ n. Let fj ∈ Z(j), ρ ∈ Gn, and T ∈ T (ρ). Let S = Tj. Then

ρ(fj)vT =∑g∈G

fj(g)ρ(g)vT

=∑g∈Gj

fj(g)(

ResGGj ρ)

(g)vT (since fj ∈ L(Gj))

=∑g∈Gj

fj(g)ρS(g)vT (by (2.10), since vT ∈ VS) (2.12)

= ρS(fj)vT

∈ CvT (Lemma 1.3.13).

Thus Z(j) ⊆ A for each 1 ≤ j ≤ n. Hence GZ(n) ⊆ A.

We now prove that A ⊆ GZ(n). Let ρ ∈ Gn and

T = (ρ = ρn → ρn−1 → · · · → ρ1) ∈ T (ρ).

By Theorem 1.5.7, we can choose fj ∈ L(Gj), 1 ≤ j ≤ n, such that, for all σ ∈ Gj, we have

σ(fj) =

IVρj if σ = ρj,

0 otherwise.

LetFT = f1 ∗ f2 ∗ · · · ∗ fn.


As in (2.12), for S ∈ T (ρ), we have

ρ(FT )vS =

vT if S = T,

0 otherwise.(2.13)

Therefore, FT : T ∈ T (ρ), ρ ∈ Gn is a basis for A, and so A ⊆ GZ(n).By Theorem 1.5.7, we have

L(Gn) ∼=⊕ρ∈Gn

Hom(Vρ, Vρ) ∼=⊕ρ∈Gn

Mdρ,dρ(C).

Thus A is a maximal commutative subalgebra of L(Gn) by Exercise 1.5.6.

Corollary 2.2.3. Every element of the Gelfand–Tsetlin basis of Vρ is a common eigenvectorfor all ρ(f), f ∈ GZ(n). In particular, it is uniquely determined, up to a scalar factor, bythe corresponding eigenvalues.

Proof. The last statement follows from (2.13).

Suppose f1, f2, . . . , fn ∈ GZ(n). By Corollary 2.2.3, for all ρ ∈ Gn, T ∈ T (ρ), and1 ≤ i ≤ n, we have

ρ(fi)vρT = αρ,T,iv

ρT for some αρ,T,i ∈ C, (2.14)

where vρT is the GZ-vector associated with the path T .When the map

Gn × T (ρ)→ Cn, (ρ, T ) 7→ (αρ,T,1, αρ,T,2, . . . , αρ,T,n) ,

is injective, we say that f1, . . . , fn separate the vectors of the GZ-bases vρT : ρ ∈ Gn, T ∈T (ρ).

Proposition 2.2.4. Let G1 ≤ G2 ≤ · · · ≤ Gn−1 ≤ Gn be a multiplicity-free chain of groups.Then C(Gn, Gn−1) ⊆ GZ(n).

Proof. By Theorem 2.2.2, it suffices to prove that, for all f ∈ C(Gn, Gn−1), the elements vT ,

T ∈ T (ρ), of the Gelfand-Tsetlin basis are eigenvectors for ρ(f), for all ρ ∈ Gn.Note that

f ∈ C(Gn, Gn−1) ⇐⇒ f(h−1gh) = f(g), for all h ∈ Gn−1, g ∈ Gn

⇐⇒ f(gh) = f(hg), for all h ∈ Gn−1, g ∈ Gn

⇐⇒ (f ∗ δh)(g) = (δh ∗ f)(g), for all h ∈ Gn−1, g ∈ Gn

⇐⇒ f ∗ δh = δh ∗ f, for all h ∈ Gn−1.

Let f ∈ C(Gn, Gn−1). Then, for all ρ ∈ Gn and h ∈ Gn−1, we have

ρ(h)ρ(f) = ρ(δh ∗ f)

= ρ(f ∗ δh)


= ρ(f)ρ(h).

Thusρ(f) ∈ EndGn−1

(ResGnGn−1

ρ).

Since ResGnGn−1is multiplicity free, it then follows from Schur’s lemma that

ρ(f)Vσ ⊆ Vσ, for all f ∈ C(Gn, Gn−1), σ ∈ Gn−1, Vσ ≤ Vρ.

Now, note that

C(Gn, Gn−1) ⊆ C(Gn, Gk), for all 1 ≤ k ≤ n− 1.

Therefore, iterating the argument above, we see that every vector vT of the Gelfand-Tsetlinbasis is an eigenvector for ρ(f), as desired.

Exercises.

2.2.2. Suppose H ≤ G. Verify that, for f1, f2 ∈ L(H), we have (f1 ∗ f2)GH = (f1)GH ∗ (f2)GH .

2.2.3 ([CSST10, Ex. 2.2.5]). Prove that if the functions f1, . . . , fn ∈ GZ(n) separate thevectors of the GZ-bases, then the set δ1G , f1, f2, . . . , fn generates GZ(n) as an algebra.

Hint : For ρ ∈ Gn and T ∈ T (ρ), define Fρ,T to be the convolution of all

fi − ασ,S,iδ1G

αρ,T,i − ασ,S,i, σ ∈ Gn, S ∈ T (σ), 1 ≤ i ≤ n, such that(σ, S) 6= (ρ, T ) and ασ,S,i 6= αρ,T,i,

using the notation of (2.14). (Note that the order of convolution is irrelevant since GZ(n) iscommutative.) Show that Fρ,T is given by (2.13).

Chapter 3

The Okounkov–Vershik approach

In this chapter, we study the representation theory of the symmetric groups following theapproach of Okounkov and Vershik [OV96, VO04, Ver06]. We closely follow the presentationin [CSST10, Ch. 3]. The reference [Py03] may also be helpful to the reader.

3.1 The Young poset

In this section we introduce some algebraic and combinatorial concepts that will be used inour study of the representation theory of the symmetric group.

3.1.1 Partitions and conjugacy classes in Sn

Definition 3.1.1. Suppose n is a positive integer. A partition of n is a sequence λ =(λ1, λ2, . . . , λh) of positive integers such that λ1 ≥ λ2 ≥ · · · ≥ λh such that λ1 +λ2 + · · ·λh =n. We call h the length of the partition λ, and denote it `(λ). We adopt the convention thatλr = 0 for r > h. We write λ ` n to indicate that λ is a partition of n. We also call |λ| := nthe size of λ.

Recall that Sn is the symmetric group of all permutations of the set 1, 2, . . . , n. Apermutation γ ∈ Sn is called a cycle of length t if there exist pairwise distinct elementsa1, a2, . . . , at ∈ 1, 2, . . . , n such that

γ(ai) = ai+1, 1 ≤ i ≤ t− 1, γ(at) = a1,

γ(b) = b, b ∈ 1, 2, . . . , n \ a1, a2, . . . , at.

We denote this cycle by γ = (a1, a2, . . . , at), or sometimes by

(a1 → a2 → · · · at → a1). (3.1)

A transposition is a cycle of length two.Two cycles γ = (a1, . . . , at) and θ = (b1, . . . , bs) are disjoint if

a1, . . . , ar ∩ b1, . . . , bs = ∅.

80

3.1. The Young poset 81

If γ and θ are disjoint cycles, then they commute: γθ = θγ.Every π ∈ Sn can be written as a product of disjoint cycles

π = (a1,1, a1,2, . . . , a1,µ1)(a2,1, a2,2, . . . , a2,µ2) · · · (ak,1, ak,2, . . . , ak,µk), (3.2)

where the list

a1,1, . . . , a1,µ1 , . . . , ak,1, . . . , ak,µk

is a permutation of 1, 2, . . . , n. In particular∑k

i=1 µi = n. Rearranging the cycles if necessary,we may assume that µ1 ≥ µ2 ≥ · · · ≥ µk > 0. The partition µ = (µ1, . . . , µk) is called thecycle type of π. The expression (3.2) is called a cycle decomposition of π. It is unique up tocyclic permutation of the elements of the cycles and permutation of cycles of equal size.

We say that i ∈ 1, 2, . . . , n is a fixed point of π ∈ Sn if π(i) = i. Thus i is a fixed pointof π if and only if i appears in a cycle of length 1 in the cycle decomposition of π.

If σ ∈ Sn, then

σπσ−1 =(σ(a1,1), . . . , σ(a1,µ1)

)(σ(a2,1), . . . , σ(a2,µ2)

)(σ(ak,1), . . . , σ(ak,µk)

).

It follows that two elements π, π′ ∈ Sn are conjugate if and only if they have the same cycletype.

Proposition 3.1.2. The conjugacy classes of Sn are parameterized by partitions of n. Theconjugacy class associated to λ ` n consists of all permutations of cycle type λ.

Proof. This follows from the above discussion.

Exercises.

3.1.1. Let λ ` n. Deduce an explicit formula for the number of permutations of cycle typeλ.

3.1.2 Young diagrams

Let λ = (λ1, . . . , λh) ` n. The Young diagram associated to λ, also called the Young diagramof shape λ, is the array consisting of n boxes, with h left-justified rows, the i-th row containingλi boxes for 1 ≤ i ≤ h. It follows that the Young diagram has λ1 columns. We will oftenabuse terminology and refer to λ itself as a Young diagram.

For example, the Young diagram associated to the partition λ = (5, 4, 1, 1) ` 11 is

.

82 Chapter 3. The Okounkov–Vershik approach

The box in row i (with row 1 at the top of the diagram) and column j (with column 1 onthe left of the diagram) will be said to be in position (i, j).

We say that a box in position (i, j) is removable if removing this box results in a Youngdiagram. In other words, it is removable if there are no boxes in position (i + 1, j) and(i, j + 1). Similarly, the position (i, j) is addable if we can add a box in position in position(i, j) and end up with a Young diagram. In other words, it is addable if

λi = j − 1 < λi−1 or (i = h+ 1 and j = 1).

Exercises.

3.1.2. For a Young diagram λ, let a(λ) be the number of addable positions of λ and let r(λ)be the number of removable boxes of λ. Show that a(λ)− r(λ) = 1.

3.1.3 Young tableaux

Suppose λ ` n. A (bijective) Young tableau of shaped λ is a bijection between the boxes ofthe Young diagram of shape λ and the set 1, 2, . . . , n. It is depicted by filling the boxes ofthe Young diagram with the numbers 1, 2, . . . , n, with exactly one number in each box. Forexample,

4 5 1 6 72 3 89

(3.3)

is a Young tableau of shape (5, 3, 1). The plural of Young tableau is Young tableaux.A Young tableau is standard if the numbers in the boxes are increasing along the rows

(from left to right) and down the columns (from top to bottom). For instance, the Youngtableau of (3.3) is not standard, while

1 3 5 8 92 4 76

is.Note that in a standard Young tableau, the number 1 is always in position (1, 1), while

n is always in a removable box.We denote by Tab(λ) the set of all standard tableaux of shape λ, and we let

Tab(n) =⋃λ`n

Tab(λ).


Exercises.

3.1.3. A Young diagram is called a hook if every row below the first has one box. Forexample,

, , , and

are hooks. Let λ be a hook. Deduce an explicit expression for the number of standardtableaux of shape λ.

3.1.4 Coxeter generators

The elements

si = (i, i+ 1) ∈ Sn, i = 1, 2, . . . , n− 1,

are called simple transpositions (or adjacent transpositions). They are also called the Coxetergenerators of Sn. (This is because of the connection to the more general theory of Coxetergroups.) The term “generator” will be justified in Proposition 3.1.3.

If T is a Young tableaux with n boxes and π ∈ Sn, then πT will denote the tableau obtai-ned from T by replacing i with π(i) for i = 1, 2, . . . , n. For example, if π = (156)(78)(234)(9)and

T =4 5 1 6 72 3 89

, then πT =2 6 5 1 83 4 79

.

If T is standard, then an admissible transposition for T is a simple transposition si suchthat siT is standard. Thus, si is admissible for T if and only if i and i+ 1 belong neither tothe same row, nor the same column of T .

An inversion for π ∈ Sn is a pair (i, j) with 1 ≤ i, j ≤ n such that

i < j and π(i) > π(j).

Let I(π) denote set of all inversions for π and let

`(π) = |I(π)|

denote the number of inversions of π.

The length of π ∈ Sn is the smallest integer k such that π can be written as a productof k simple transpositions: π = si1si2 · · · sik .

Proposition 3.1.3. The length of π ∈ Sn is equal to `(π).


Proof. We first claim that, for any π ∈ Sn, we have

`(πsi) =

`(π)− 1 if (i, i+ 1) ∈ I(π),

`(π) + 1 if (i, i+ 1) /∈ I(π).(3.4)

First consider k satisfying 1 ≤ k < i. There are three possibilities:

• π(k) < minπ(i), π(i+ 1),• π(k) > maxπ(i), π(i+ 1),• minπ(i), π(i+ 1) < π(k) < maxπ(i), π(i+ 1).

In the first case, (k, i) and (k, i + 1) are neither inversions for π nor for πsi. In the secondcase, (k, i) and (k, i+ 1) are inversions for both. In the third case, exactly one of (k, i) and(k, i + 1) is an inversion for π, while only the other one is an inversion for πsi. Thus, in allthree cases, the number of inversions in the set (k, i), (k, i+1) is the same for π and πsi. Asimilar argument gives the same result for the case that i+1 < k ≤ n. Since (i, i+1) ∈ I(π)if and only if (i, i+ 1) /∈ I(πsi), the claim follows.

Now suppose π = si1si2 · · · sik is a minimal representation of π as a product of simpletranspositions. Then, by (3.4), we have

`(π) = `(si1si2 · · · sik−1)± 1 ≤ `(si1si2 · · · sik−1

) + 1 ≤ · · · ≤ k.

Therefore, the length of π is greater than or equal to `(π).It remains to prove the reverse inequality, which we do by induction on n. It is clear for

n = 2 that π ∈ S2 can be written as a product of `(π) transpositions. (Simply consider thetwo cases π = (1)(2) and π = (12).) Now suppose n ≥ 3 and that any π ∈ Sn−1 can bewritten as a product of `(π) transpositions.

Fix π ∈ Sn. Let jn = π−1(n) and let

πn = πsjnsjn+1 · · · sn−1.

Thus πn(n) = n and so, by (3.4), we have

`(πnsn−1) = `(πn) + 1.

Now, since πnsn−1(n− 1) = n, we also have, by (3.4),

`(πnsn−1sn−2) = `(πnsn−1) + 1 = `(πn) + 2.

Continuing in this way, we see that

`(πn) = `(π)− (n− jn).

Now, since πn(n) = n, it can naturally be viewed as an element of Sn−1. By our inductionhypothesis, it can be written as a product of `(πn) simple transpositions. But then, sinceπ = πnsn−1sn−2 · · · sjn , we have that π can be written as a product of `(πn) + (n− jn) = `(π)transpositions. This completes the proof of the induction step.


For λ ` n, let T λ denote the standard tableau of shape λ, where we number the boxes1, 2, . . . , n from left-to-right starting in the top row, then continuing in the second row, etc.For example,

if λ = (5, 3, 1) then T λ =1 2 3 4 56 7 89

.

For T ∈ Tab(λ), we let πT ∈ Sn denote the unique permutation such that πTT = T λ.

Theorem 3.1.4. Suppose T ∈ Tab(λ). Then there exists a sequence of `(πT ) admissibletranspositions transforming T into T λ.

Proof. We prove the result by induction on |λ|. If |λ| = 1, then T λ is the only tableau ofshape λ and we are done. Now suppose |λ| > 1 and that the result holds for all tableau withfewer than |λ| boxes.

Suppose λ = (λ1, λ2, . . . , λk) ` n and T ∈ Tab(λ). Let j denote the entry in the rightmostbox of the bottom row of T . If j = n then, since the box is removable, we can consider thestandard tableau T ′ of shape λ′ = (λ1, λ2, . . . , λk−1) obtained by removing that box. By ourinduction hypothesis, there exists a sequence of `(πT ′) admissible transpositions transformingT ′ into T λ

′. This same sequence transforms T into T λ and `(πT ′) = `(πT ).

Now suppose j 6= n. Then sj is admissible for T . Similarly, sj+1 is admissible for sjT , . . . ,sn−1 is admissible for sn−2 . . . sj1sjT . Now, T ′′ = sn−1sn−2 · · · sjT contains n in the rightmostbox of the bottom row. Therefore, by the previous case, there exists a sequence of `πT ′′ simpletranspositions transforming T ′′ into T λ. It follows from (3.4) that `(πT ) = `πT ′′ + n − j,completing the proof of the induction step.

Corollary 3.1.5. For any T, S ∈ Tab(λ), there is a sequence of admissible transpositionstransforming S into T .

Remark 3.1.6. Note that the proof of Theorem 3.1.4 gives a standard procedure to write πTas a product of `(πT ) admissible transpositions. We shall use this procedure in what follows.

Exercises.

3.1.4. Show that the simple transpositions in Sn satisfy the relations

s2i = 1, 1 ≤ i ≤ n− 1,

sisj = sjsi, 1 ≤ i, j ≤ n− 1, |i− j| > 1,

sisi+1si = si+1sisi+1, 1 ≤ i ≤ n− 2.

(In fact, the above form a complete set of relations for the simple transpositions. That is,Sn is the group with generators si, 1 ≤ i ≤ n− 1, and relations as given above.)


3.1.5 The content of a tableau

Suppose T ∈ Tab(λ) for some λ ` n. For 1 ≤ t ≤ n, we let row(t) and col(t) denote the rowand column of the box of T containing t. For example,

if T =1 2 4 5 73 6 89

then row(8) = 2, col(8) = 3.

Given a box in the Young diagram λ with coordinates (i, j) we define the content of the boxto be

c(i, j) := j − i.

For example, the contents of the boxes in λ = (5, 3, 1) are as indicated:

0 1 2 3 4-1 0 1-2

So, essentially, the content of a box corresponds to the diagonal on which it lies.We define the content of the tableau T to be

C(T ) :=(c(row(1), col(1)), c(row(2), col(2)), . . . , c(row(n), col(n))

)∈ Zn.

For example,

if T =4 5 1 6 72 3 89

then c(T ) = (2,−1, 0, 0, 1, 3, 4, 1,−2).

Note that, for a fixed partition λ, the entries of the contents of the tableau of shape λ arethe same, but potentially in different orders (i.e. they are permutations of one another).

Definition 3.1.7 (Cont(n)). Let Cont(n) be the set of all (a1, a2, . . . , an) ∈ Cn such that

(a) a1 = 0,

(b) aj + 1, aj − 1 ∩ a1, a2, . . . , aj−1 6= ∅ for all j > 1,

(c) if ai = aj for some i < j then aj − 1, aj + 1 ⊆ ai+1, ai+2, . . . , aj−1.

It follows from the definition that, in fact, Cont(n) ⊆ Zn. For example,

Cont(1) = 0 and Cont(2) = (0, 1), (0,−1). (3.5)

For α, β ∈ Cont(n), we will write α ≈ β if πβ = α for some π ∈ Sn (here Sn acts on Znby permuting the entries). Note that ≈ is an equivalence relation on Cont(n), but that Sn

does not act on Cont(n) since there are α ∈ Cont(n) and π ∈ Sn such that πα /∈ Cont(n).For example, if π = (1, 2) ∈ S2, then

α = (0, 1) ∈ Cont(2) but π(0, 1) = (1, 0) /∈ Cont(2).


Theorem 3.1.8. For any T ∈ Tab(n), we have C(T ) ∈ Cont(n). Furthermore, the map

Tab(n)→ Cont(n), T 7→ C(T ),

is a bijection. In addition, for T, S ∈ Tab(n), we have

C(T ) ≈ C(S) ⇐⇒ T and S are tableaux of the same shape.

Proof. A proof of this theorem can be found in [CSST10, Th. 3.1.10].

For α ∈ Cont(n), we say that a simple transposition si is admissible for α if it is admissiblefor the unique T ∈ Tab(n) such that α = C(T ).

Corollary 3.1.9. For α, β ∈ C(T ), we have α ≈ β if and only if there exists a sequence ofadmissible transpositions transforming α into β.

Proof. This follows from Corollary 3.1.5 and Theorem 3.1.8.

Corollary 3.1.10. The cardinality of the quotient set Cont(n)/ ≈ is equal to the number ofpartitions of n.

Proof. This follows from Theorem 3.1.8.

Exercises.

3.1.5. Suppose that α = (α1, α2, . . . , αn) ∈ Cont(n) and 1 ≤ i ≤ n − 1. Prove that si isadmissible for α if and only if ai+1 6= ai ± 1.

3.1.6 The Young poset

LetY = λ : λ ` n, n ∈ Z>0

be the set of all partitions. Equivalently, Y is the set of all Young diagrams. We define apartial ordering on Y by saying that, for µ, λ ∈ Y, µ λ if and only if the Young diagramof µ is contained in the Young diagram of λ. In other words, if µ = (µ1, . . . , µk) ` n andλ = (λ1, . . . , λh) ` m, then

µ λ ⇐⇒ m ≥ n and λj ≥ µj for all j = 1, 2, . . . , k.

For example,(4, 3, 1) (5, 3, 1, 1).

For µ, λ ∈ Y, we say that λ covers µ if µ is obtained from λ by removing a single box.In particular, this implies that µ λ.


The Hasse diagram of Y, also called the Young (branching) graphs , is the oriented graphwith vertex set Y and an arrow from λ to µ if and only if λ covers µ. The bottom of theYoung graph is as follows:

$$ ||

(One sometimes includes the empty Young diagram ∅ at the bottom of the graph.)A path in the Young graph is a sequence

p =(λ(n) → λ(n−1) → · · · → λ(1)

)of partitions λ(k) ` k such that λ(k) covers λ(k−1) for k = 2, 3, . . . , n. We call `(p) := n thelength of the path p. We let Πn(Y) denote the set of all paths in the Young graph of lengthn and let

Π(Y) =∞⋃n=1

Πn(Y).

To any standard Young tableau T of shape λ ` n we can associate a path

λ = λ(n) → λ(n−1) → · · · → λ(1)

by letting λ(k), 1 ≤ k ≤ n, be the Young diagram formed by the boxes of T with labels lessthan or equal to k. For example, to the standard tableau

1 2 5 63 4 78

3.2. The Young–Jucys–Murphy elements and a Gelfand–Tsetlin basis for Sn 89

we associate the path

(4, 3, 1)→ (4, 3)→ (4, 2)→ (3, 2)→ (2, 2)→ (2, 1)→ (2)→ (1).

In this way we have a natural bijection

Πn(Y)↔ Tab(n) (3.6)

which gives a bijection

Π(Y)↔∞⋃n=1

Tab(n). (3.7)

Combining (3.6) with the bijection in Theorem 3.1.8 yields a bijection

Πn(Y)↔ Cont(n). (3.8)

Proposition 3.1.11. Suppose α, β ∈ Cont(n) correspond to the paths

λ(n) → λ(n−1) → · · · → λ(1) and µ(n) → µ(n−1) → · · · → µ(1),

respectively. Then α ≈ β if and only if λ(n) = µ(n).

Proof. This follows immediately from Theorem 3.1.8.

Exercises.

3.1.6. Suppose λ is a hook (see Exercise 3.1.3). How many paths in the Young graph arethere that start at λ?

3.2 The Young–Jucys–Murphy elements and a Gelfand–

Tsetlin basis for Sn

Our goal in this section is to prove that the chain

S1 ≤ S2 ≤ · · · ≤ Sn ≤ Sn+1 ≤ · · ·

is multiplicity free. This allows use to use the techniques of Chapter 2. In particular, wewill study the Gelfand–Tsetlin algebra associated to this chain.


3.2.1 The Young–Jucys–Murphy elements

In the remainder of these notes, we will identity the Dirac function δπ, π ∈ Sn, with π.Thus, elements f ∈ L(Sn) will be written as formal sums

f =∑π∈Sn

f(π)π.

Similarly, the characteristic function of A ⊆ Sn will be denoted simply by A, so that

A =∑π∈A

π.

In addition, the convolution of f1, f2 ∈ L(Sn) will be denoted f1 ·f2 and written as a productof formal sums:

f1 · f2 =∑π∈Sn

∑σ,θ∈Sn:σθ=π

f1(σ)f2(θ)

π.

The Young–Jucys–Murphy (YJM) elements of L(Sn) are defined by

X1 = 0, Xk = (1, k) + (2, k) + · · ·+ (k − 1, k), k = 2, . . . , n.

In some places in the literature, these elements are simply called Jucys–Murphy elements .

Exercises.

3.2.1. Show that the following relations hold in L(Sn):

Xi+1si = siXi + 1, 1 ≤ i ≤ n− 1,

Xjsi = siXj, 1 ≤ i ≤ n− 1, 1 ≤ j ≤ n, j 6= i, i+ 1.

3.2.2. Show that XiXj = XjXi for all 1 ≤ i, j ≤ n.

3.2.3. Fix n ≥ 2. The degenerate affine Hecke algebra Hn is the C-algebra generated byelements

ti, 1 ≤ i ≤ n− 1, and xj, 1 ≤ j ≤ n,

subject to the relations

t2i = 1, 1 ≤ i ≤ n− 1,

titj = tjti, 1 ≤ i, j ≤ n− 1, |i− j| > 1,

titi+1ti = ti+1titi+1, 1 ≤ i ≤ n− 2.

xi+1ti = tixi + 1, 1 ≤ i ≤ n− 1,

xjti = tixj, 1 ≤ i ≤ n− 1, 1 ≤ j ≤ n, j 6= i, i+ 1,


xixj = xjxi, 1 ≤ i, j ≤ n.

(This means that for any C-algebra A with elements t′i, x′j satisfying the above relations,

there is a unique algebra homomorphism Hn → A such that ti 7→ t′i and xj 7→ x′j.) There isan injective algebra homomorphism L(Sn) → Hn given by si 7→ ti and we use this to viewL(Sn) as a subalgebra of Hn.

Let I be the ideal ofHn generated by x1. (See Exercise 1.2.8.) Prove thatHn/I ∼= L(Sn),as algebras (or as rings). You may use the fact that Hn has a basis given by the elements

xk11 xk22 . . . xknn π, k1, k2, . . . , kn ∈ Z≥0, π ∈ Sn, (3.9)

where we view the element π ∈ Sn (identified with δπ ∈ L(Sn)) as an element of Hn asexplained above.

3.2.2 Marked permutations

Fix `, k ≤ 1. In this section, we let S`+k, S`, and Sk denote the symmetric groups on thesets 1, 2, . . . , `+ k, 1, 2, . . . , ` and `+ 1, `+ 2, . . . , `+ k, respectively. Thus

S`,Sk ≤ S`+k and S` ∩Sk = 1.

We letZ(`, k) = C(S`+k,S`)

denote the algebra of all S`-conjugacy invariant functions in L(S`+k).We wish to analyze the algebra Z(`, k). The first step is to parameterize the orbits

of the S`-conjugacy action on S`+k. Recall that, for π ∈ S` and θ = S`+k, the cycledecomposition of πθπ−1 is obtained from the cycle decomposition of θ by replacing 1, 2, . . . , `with π(1), π(2), . . . , π(`). Thus, the S`-orbit of θ is obtained from its cycle decompositionby permuting in all possible ways the elements 1, 2, . . . , `, leaving the remaining elements`+ 1, `+ 2, . . . , `+ k unchanged.

Consider the cycle decomposition of a permutation using our notation (3.1) for cycles:

(a1,1 → a1,2 → · · · → a1,µ1 → a1,1)(a2,1 → · · · → a2,µ2 → a2,1) · · · (ar,1 → · · · → ar,µr → ar,1).(3.10)

A marked permutation is a permutation as in (3.10), together with a labeling of all thearrows by nonnegative integers, called tags , potentially will some additional empty cyclesalso labeled with tags. For example, the permutation (3.10) may be marked as follows:

(a1,1u1,1−−→ a2

u1,2−−→ · · ·u1,µ1−1−−−−→ a1,µ1

u1,µ1−−−→ a1,1)(a2,1u2,1−−→ · · ·

u2,µ2−1−−−−→ a2,µ2

u2,µ2−−−→ a2,1)

· · · (ar,1ur,1−−→ · · · ur,µr−1−−−−→ ar,µr

ur,µr−−−→ ar,1)(v1−→)(

v2−→) · · · ( vs−→).(3.11)

The orbits of the conjugacy action of S` on S`+k are in natural one-to-one correspondencewith the set of all marked permutations of `+ 1, . . . , `+ k such that the sum of the tags isequal to `. The orbit corresponding to a given marked permutation is obtained by inserting,in all possible ways, the elements 1, . . . , ` into the marked permutation, with the numberof elements added at any particular arrow equal to the label of that arrow.


Example 3.2.1. If ` = 12 and k = 8, then the marked permutation

(192−→ 15

1−→ 130−→ 14

0−→ 19)(162−→ 20

1−→ 171−→ 16)(18

2−→ 18)(2−→))(

1−→)

corresponds to the orbit consisting of all permutations of the form

(19→ y1 → y2 → 15→ y3 → 13→ 14→ 19)(16→ y4 → y5 → 20→ y6 → 17→ y7 → 16)

· (18→ y8 → y9 → 18)(y10 → y11 → y10)(y12 → y12),

where y1, y2, . . . , y12 = 1, 2, . . . , 12.

We will typically omit trivial cycles of the form (a0−→ a) and (

1−→). Note that, whenomitting such trivial cycles, the sum of the tags is ≤ `, with strict inequality a possibility.

Theorem 3.2.2. Let Sn−1 be the symmetric group on 1, 2, . . . , n − 1. Then (Sn ×Sn−1, Sn−1) is a symmetric Gelfand pair.

Proof. In light of Proposition 2.1.9, it suffices to prove that every π ∈ Sn is Sn−1-conjugateto π−1. Suppose π has cycle decomposition

π = (n = a1,1 → a1,2 → · · · → a1,µ1 → n)(a2,1 → · · · → a2,µ2 → a2,1)

· · · (ar,1 → · · · → ar,µr → ar,1).

Then π belongs to the Sn−1-conjugacy class corresponding to the marked permutation

(nµ1−1−−−→ n)(

µ2−→) · · · ( µr−→).

Then

π−1 = (n→ a1,µ1 → · · · → a1,1 = n)(a2,1 → a2,µ2 → · · · → a2,1) · · · (ar,1 → ar,µr · · · → ar,1)

clearly belongs to the same conjugacy class.

Corollary 3.2.3. The algebra C(Sn,Sn−1) is commutative, and Sn−1 is a multiplicity-freesubgroup of Sn. Thus

S1 ≤ S2 ≤ · · · ≤ Sn−1 ≤ Sn ≤ · · ·is a multiplicity-free chain.

Proof. This follows from Theorems 2.1.7 and 3.2.2.

Example 3.2.4. (a) For j = 1, . . . , k, the YJM element X`+j can be written as

X`+j = (`+ j1−→ `+ j) +

`+j−1∑h=`+1

(`+ j0−→ h

0−→ `+ j). (3.12)

(Recall our convention that, for A ⊆ S`+k, we denote the characteristic function of Asimply by A.) In particular, X`+1, X`+2, . . . , X`+k ∈ Z(`, k) since these elements areall sums of characteristic functions of orbits under the conjugacy action of S`.


(b) Any σ ∈ Sk forms a one-element orbit of S`. Thus, viewing σ as an element of L(Sk),we have Sk ⊆ Z(`, k).

(c) It is clear that Z(`) := Z(L(S`)) ⊆ Z(`, k).

It follows from Example 3.2.4 that

〈X`+1, X`+2, . . . , X`+k,Sk, Z(`)〉 ⊆ Z(`, k), (3.13)

where, on the left side, the angled brackets denote the subgroup of L(S`+k) generated bythe elements/sets inside the brackets.

Exercises.

3.2.4. For σ ∈ Sk, describe the marked permutation corresponding to the S`-orbit σ (seeExample 3.2.4(b)).

3.2.5. Suppose that Cλ is the conjugacy class of S` corresponding to λ ` ` (see Proposi-tion 3.1.2). Describe the marked permutation corresponding to the S`-orbit formed by Cλ(see Example 3.2.4(c)).

3.2.3 Olshanskii’s Theorem

The goal of this subsection is to prove the reverse of the inclusion (3.13). Precisely, we wantto prove that Z(`, k) is generated by X`+1, X`+2, . . . , X`+k, Sk, and Z(`).

Let Zh(`, k) be the subspace of Z(`, k) spanned by the S`-conjugacy classes consisting ofpermutations with at least `+ k− h fixed points (equivalently, moving at most h elements).Then we have a filtration

C1 = Z0(`, k) ⊆ Z1(`, k) ⊆ Z`+k−1(`, k) ⊆ Z`+k(`, k) = Z(`, k). (3.14)

We will essentially be interested in “leading terms” with respect to this filtration. Moreprecisely, for f1, f2, f3 ∈ Z(`, k), we will write

f1 · f2 = f3 + lower terms

if there exists h such that

f3 ∈ Zh(`, k)− Zh−1(`, k) and f1 · f2 − f3 ∈ Zh−1(`, k).

Lemma 3.2.5. Let i, j ≥ 1. Suppose a1, a2, . . . , ai and b1, b2, . . . , bj are each sequences ofdistinct elements in `+1, . . . , `+k. (Note that we do not require the ak to be distinct fromthe bk.) Suppose that, in Sk, we have

(a1 → a2 → · · · → ai → a1)(b1 → b2 → · · · → bj → b1) = (c1 → c2 → · · · → ch → c1),


withh = |a1, a2, . . . , ai ∪ b1, b2, . . . , bj| ≤ i+ j.

Let u1, u2, . . . , ui, v1, v2, . . . , vj ≥ 0 such that

u1 + u2 + · · ·+ ui + v1 + v2 + · · ·+ vj ≤ `.

Then, in Z(`, k), we have(a1

u1−→ a2u2−→ · · · ui−1−−→ ai

ui−→ a1

)(b1

v1−→ b2v2−→ · · · vj−1−−→ bj

vj−→ v1

)=(c1

w1−→ c2w2−→ · · · wh−1−−−→ ch

wh−→ c1

)+ lower terms,

where

ws =

vt if cs = bt and bt+1 /∈ a1, a2, . . . , ai,vt + um if cs = bt and bt+1 = am,

um if cs = am /∈ b1, b2, . . . , bj.(3.15)

for 1 ≤ s ≤ h.

Proof. Consider a product of the form

(a1, x1,1, x1,2, . . . , x1,u1 , a2, x2,1, . . . , x2,u2 , . . . , ai, xi,1, . . . , xi,ui)

· (b1, y1,1, . . . , y1,v1 , . . . , bj, yj,1, . . . , yj,vj). (3.16)

If the numbers x1,1, . . . , xi,ui , y1,1, . . . , yj,vj are all distinct, which is possible since u1 + · · ·+ui + v1 + · · ·+ vj ≤ `, then (3.16) is equal to a permutation of the form

(c1, z1,1, z1,2, . . . , z1,w1 , c2, . . . , ch, zh,1, . . . , zh,wh),

where w1, w2, . . . , wh are given by (3.15).Otherwise, the product (3.16) moves fewer than h+ w1 + w2 + · · ·+ wh elements.

Example 3.2.6. (a) For a ∈ `+ 1, `+ 2, . . . , `+ k and 0 ≤ u ≤ `, we have(a

1−→ a)u

=(a

u−→ a)

+ lower terms.

(b) For a, b ∈ `+ 1, `+ 2, . . . , `+ k, a 6= b, and 0 ≤ u ≤ `, we have(b

u−→ b)(

a0−→ b

0−→ a)

=(a

u−→ b0−→ a).

Note that there are no lower order terms in this case.

(c) For a1, a2, . . . , ai ∈ ` + 1, ` + 2, . . . , ` + k, pairwise distinct, and u1, u2, . . . , ui ≥ 0with u1 + u2 + · · ·ui ≤ `, we have(

a1ui−→ a1

)(a1

ui−1−−→ ai0−→ a1

)· · ·(a1

u2−→ a30−→ a1

)(a1

u1−→ a20−→ a1

)=(a1

u1−→ a2u2−→ a3

u3−→ · · · ui−1−−→ aiui−→ a1

)+ lower terms.


Theorem 3.2.7 (Olshanskii’s Theorem). The centralizer algebra Z(`, k) is generated by theYoung–Jucys–Murphy elements X`+1, X`+2, . . . , X`+k, the subgroup Sk, and the center Z(`)of S`. In other words, we have

Z(`, k) = 〈X`+1, X`+2, . . . , X`+k,Sk, Z(`)〉.

Proof. LetA = 〈X`+1, X`+2, . . . , X`+k,Sk, Z(`)〉.

We have A ⊆ Z(`, k) from (3.13).Keeping in mind the filtration (3.14), we will prove by induction on h that Zh(`, k) ⊆ A

for all h = 0, 1, . . . , `+ k. Since Z0(`, k) = C1 ⊆ A, our base case is proved.Now suppose that 1 ≤ h ≤ ` + k, and that the result holds for h − 1. For a, j ∈

`+ 1, `+ 2, . . . , `+ k with a 6= j, we have(a

0−→ j0−→ a)∈ Sk.

Thus, by (3.12), we have

(a

1−→ a)

= Xa −a−1∑j=`+1

(a

0−→ j0−→ a)∈ A.

Now, the typical orbit in Zh(`, k)− Zh−1(`, k) is of the form

(a1,1u1,1−−→ a1,2

u1,2−−→ · · ·u1,µ1−1−−−−→ a1,µ1

u1,µ1−−−→ a1,1)(a2,1u2,1−−→ · · ·

u2,µ2−1−−−−→ a2,µ2

u2,µ2−−−→ a2,1)

· · · (ar,1ur,1−−→ · · · ur,µr−1−−−−→ ar,µr

ur,µr−−−→ ar,1)(v1−→)(

v2−→) · · · ( vs−→) (3.17)

for pairwise distinct elements

a1,1, a1,2, . . . , a1,µ1 , . . . , ar,1, . . . , ar,µr ∈ `+ 1, `+ 2, . . . , `+ k

and

u1,1, u1,2, . . . , u1,µ1 , . . . , ur,1, . . . , ur,µr , v1, v2, . . . , vs ∈ Z≥0

such thatr∑p=1

µp∑q=1

up,q +s∑

p=1

vp = h ≤ `.

Repeated application of Lemma 3.2.5 (see also Example 3.2.6) gives that(a1,1

1−→ a1,1

)u1,µ1 (a1,µ1

1−→ a1,µ1

)u1,µ1−1(a1,1

0−→ a1,µ10−→ a1,1

)· · ·

· · ·(a1,3

1−→ a1,3

)u1,2 (a1,1

0−→ a1,30−→ a1,1

)(a1,2

1−→ a1,2

)u1,1 (a1,1

0−→ a1,20−→ a1,1

)· · ·

· · ·

· · ·(ar,1

1−→ ar,1

)ur,µr (ar,µr

1−→ ar,µr

)ur,µr−1(ar,1

0−→ ar,µr0−→ ar,1

)· · ·


· · ·(ar,3

1−→ ar,3

)ur,2 (ar,1

0−→ ar,30−→ ar,1

)(ar,2

1−→ ar,2

)ur,1 (ar,1

0−→ ar,20−→ ar,1

)· · ·

· · ·(v1−→)(

v2−→)· · ·(vs−→)

is equal to (3.17) modulo terms. Thus, by the inductive hypothesis, we have that (3.17) isan element of A, completing the proof.

Corollary 3.2.8. The Gelfand–Tsetlin algebra GZ(n) of the multiplicity free chain

S1 ≤ S2 ≤ · · · ≤ Sn

is generated by the Young–Jucys–Murphy elements X1, X2, . . . , Xn.

Proof. For 2 ≤ k ≤ n, let Tk denote the set of all transpositions (not necessarily simple) inSk. Then Tk ∈ Z(k) (recall that we identify a subset of Sk with its characteristic function).Thus

Xk = Tk − Tk−1 ∈ GZ(n).

Hence 〈X1, X2, . . . , Xn〉 ∈ GZ(n).We now prove by induction on n that GZ(n) = 〈X1, X2, . . . , Xn〉. The result for n = 1 is

trivial. Assume n > 1 and that the result holds for n− 1. Then

Z(n) = C(Sn,Sn)

⊆ C(Sn,Sn−1)

= Z(n− 1, 1)

= 〈Z(n− 1), Xn〉, (by Theorem 3.2.7).

Thus, using the induction hypothesis, we have

GZ(n) = 〈GZ(n− 1), Z(n)〉 = 〈GZ(n− 1), Xn〉 = 〈X1, X2, . . . , Xn−1, Xn〉,completing the proof of the induction step.

Exercises.

3.2.6. Use Corollary 3.2.8 to give an alternate proof that Z(n − 1, 1) = C(Sn,Sn−1) iscommutative (see Corollary 3.2.3).

3.3 The spectrum of the Young–Jucys–Murphy ele-

ments and the branching graph of Sn

In this section, our goal is to show that the spectrum (i.e. set of possible eigenvalues) of theYJM elements is given by Cont(n) (see Definition 3.1.7) and prove that the branching graphof the multiplicity-free chain

S1 ≤ S2 ≤ · · · ≤ Sn ≤ · · ·is precisely the Young graph.

3.3. The spectrum of the YJM elements and the branching graph of Sn 97

3.3.1 The weight of a Young basis vector

For ρ ∈ Sn, we have the Gelfand–Tsetlin basis vT : T ∈ T (ρ) associated with the mul-tiplicity free chain S1 ≤ S2 ≤ · · · ≤ Sn. In this setting, we also call this basis the Youngbasis for Vρ.

By Theorem 2.2.2 and Corollary 3.2.8, every vT is an eigenvector for ρ(Xj) for all 1 ≤j ≤ n. We define the weight of vT to be

α(T ) = (a1, a2, . . . , an), where ρ(Xj) = ajvT , 1 ≤ j ≤ n. (3.18)

Since X1, . . . , Xn generate GZ(n) by Corollary 3.2.8, it follows from Corollary 2.2.3 that vTis determined, up to a scalar factor, by α(T ).

It follows from Theorem 1.5.7 that ρ(Xj) is self-adjoint for all Xj and all representationsρ of Sn. (See Exercise 3.3.1.)

Proposition 3.3.1. Suppose ρ ∈ Sn and

T = (ρ = ρn → ρn−1 → · · · → ρ2 → ρ1) ∈ T (ρ).

Then ρ(sk)vT is a linear combination of vectors vT ′ with T ′ of the form

T ′ = (σ = σn → σn−1 → · · · → σ2 → σ1) ∈ T (ρ), such that σi = ρi for all i 6= k.

Proof. For 1 ≤ j ≤ n, let Vj denote the representation space of ρj. Then

Vj = ρj(f)vT : f ∈ L(Sj),

since the right side is a Sj-invariant subspace generated by a nonzero element of Vj.If j > k, then sk ∈ Sj, and so ρj(sk) ∈ Vj. Thus

σj = ρj for all j = k + 1, k + 2, . . . , n.

Now suppose j < k. Then sk and Sj commute. If we define

Wj = ρj(f)ρ(sk)vT : f ∈ L(Sj) = ρ(sk)Vj,

then we have an isomorphism of Sj-representations

Vj → Wj ρj(f)vT 7→ ρj(f)ρ(sk)vT .

Therefore ρ(sk)vT belongs to the ρj-isotypic component of ResSnSjρ, and so σj = ρj.

Exercises.

3.3.1. Use Theorem 1.5.7 to prove that ρ(Xj) is self-adjoint for all 1 ≤ j ≤ n and allrepresentations ρ of Sn.


3.3.2 The spectrum of the YJM elements

We define the spectrum of the YJM elements to be

Spec(n) = α(T ) : T ∈ T (ρ), ρ ∈ Sn,

where α(T ) is the weight of vT , as in (3.18). Since the elements of the Young basis areuniquely determined by their weight, we have

| Spec(n)| =∑ρ∈Sn

dimVρ.

In particular, Spec(n) is in natural bijection with the set of all paths in the branching graphof the chain S1 ≤ S2 ≤ · · · ≤ Sn. We let Tα denote the path corresponding to α ∈ Spec(n),and we let vα denote the Young basis vector corresponding to Tα.

We define an equivalence relation on Spec(n) by declaring that α ∼ β if vα and vβ belongto the same irreducible Sn-representation. Equivalently, α ∼ β if the corresponding pathsin the branching graph start at the same vertex. It follows that

| Spec(n)/ ∼ | =∣∣∣Sn

∣∣∣ . (3.19)

We would now like to deduce an explicit description of Spec(n) and ∼ using the relations(see Exercise 3.2.1)

Xi+1si = siXi + 1, 1 ≤ i ≤ n− 1, (3.20)

Xjsi = siXj, 1 ≤ i ≤ n− 1, 1 ≤ j ≤ n, j 6= i, i+ 1. (3.21)

Note that (3.20) is equivalent to

siXi+1 − 1 = Xisi, 1 ≤ i ≤ n− 1, 1 ≤ j ≤ n, j 6= i, i+ 1. (3.22)

In what follows, if vα is a vector of the Young basis of an irreducible representation ρ ofSn, we denote ρ(si)vα and ρ(Xi)vα by sivα and Xivα, respectively. (In other words, we willuse the notation of modules, which is equivalent to that of representations.)

Proposition 3.3.2. Suppose α = (a1, a2, . . . , an) ∈ Spec(n) and 1 ≤ i ≤ n− 1.

(a) ai 6= ai+1.

(b) ai+1 = ai ± 1 if and only if sivα = ±vα.

(c) If ai+1 6= ai ± 1, then

α′ := siα = (a1, a2, . . . , ai−1, ai+1, ai, ai+2, . . . , an) ∈ Spec(n),

α ∼ α′, and we have

vα′ = sivα −1

ai+1 − aivα (3.23)


(up to a scalar factor). Moreover, the space 〈vα, vα′〉 is invariant under the action ofXi, Xi+1, and si, and in the basis vα, vα′, these operators are given by the matrices(

ai 00 ai+1

),

(ai+1 0

0 ai

), and

(1

ai+1−ai 1− 1(ai+1−ai)2

1 1ai−ai+1

),

respectively.

Proof. It follows immediately from the definitions of α and vα that

Xivα = aivα and Xi+1vα = ai+1vα.

It also follows from (3.20) and (3.22) that 〈vα, sivα〉 is invariant under the action of Xi andXi+1. It is also clearly invariant under the action of si.

Suppose that sivα and vα are linearly independent and ai+1 = ai ± 1. Then a directcomputation shows that the only line stable under the action of the subalgebra A of L(Sn)generated by si, Xi, and Xi+1 is the line spanned by sivα ∓ vα. This contradicts the factthat representations of A are completely reducible (since si, Xi, and Xi+1 act by unitaryoperators). Thus, if ai+1 = ai ± 1, then the vectors sivα and α are proportional, in whichcase (3.20) implies that

aisivα + vα = ai+1sivα, (3.24)

and so sivα = ±vα. Conversely, if sivα = λvα, then the fact that s2i = 1 implies that λ2 = 1,

and so λ = ±1. Then (3.24) implies that ai+1 = ai ± 1. This proves (b).Now suppose that ai+1 6= ai ± 1. Then, by the above, we have

dim〈vα, sivα〉 = 2.

By (3.20) and (3.22), we have

Xisivα = −vα + ai+1sivα,

Xi+1sivα = vα + aisivα.

Thus the actions of si, Xi, and Xi+1 on 〈vα, sivα〉 are represented, with respect to the basisvα, sivα by the matrices(

0 11 0

),

(ai −10 ai+1

), and

(ai+1 1

0 ai

),

respectively.Now, we know that the actions of Xi and Xi+1 are diagonalizable. This implies that

ai 6= ai+1, proving (a). Then we check directly that

v′ := sivα −1

ai+1 − aivα

is an eigenvector of Xi and Xi+1 with eigenvalues ai+1 and ai, respectively. In addition (3.21)implies that

Xjv′ = ajv

′, for all j 6= i, i+ 1.


Thusα′ := (a1, a2, . . . , ai−1, ai+1, ai, ai+2, . . . , an) ∈ Spec(n),

and v′ = vα′ is a vector of the Young basis. We leave it as an exercise (Exercise 3.3.2) toverify the formula for the action of si in the basis vα, vα′.

Exercises.

3.3.2. In the notation of the proof of Proposition 3.3.2, prove that the action of si on thespace 〈vα, vα′〉, with respect to the basis vα, vα′, is represented by the matrix(

1ai+1−ai 1− 1

(ai+1−ai)2

1 1ai−ai+1

).

3.3.3 Spec(n) = Cont(n)

Letα = (a1, a2, . . . , an) ∈ Spec(n).

We say that si is an admissible transposition for α if ai+1 6= ai ± 1. In this case, byProposition 3.3.2(c), we have

siα = (a1, . . . , ai−1, ai+1, ai, ai+2, . . . , an) ∈ Spec(n).

The Coxeter generators satisfy the relations (see Exercise 3.1.4)

sisj = sjsi, 1 ≤ i, j ≤ n− 1, |i− j| > 1, (3.25)

sisi+1si = si+1sisi+1, 1 ≤ i ≤ n− 2. (3.26)

Lemma 3.3.3. Let α = (a1, a2, . . . , an) ∈ Cn. If ai = ai+2 = ai+1 ± 1 for some i ∈1, 2, . . . , n− 2, then α /∈ Spec(n).

Proof. Suppose, towards a contradiction, that ai = ai+2 = ai+1 − 1 and α ∈ Spec(n). Then,by Proposition 3.3.2(b), we have

sivα = vα and si+1vα = −vα.

Then, by (3.26), we have

vα = si+1sisi+1vα = sisi+1sivα = −vα,

which contradicts the fact that vα 6= 0. The proof of the case that ai = ai+2 = ai+1 + 1 issimilar.

Lemma 3.3.4. (a) For all (a1, a2, . . . , an) ∈ Spec(n), we have a1 = 0.


(b) If (a1, a2, . . . , an) ∈ Spec(n), then (a1, a2, . . . , an−1) ∈ Spec(n− 1).

(c) We have Spec(2) = (0, 1), (0,−1).

Proof. (a) This follows immediately from the fact that X1 = 0.

(b) This follows from the fact that X1, X2, . . . , Xn−1 ∈ L(Sn−1) and Xjvα = ajvα for all1 ≤ j ≤ n− 1.

(c) The group S2 has two irreducible representations: the trivial representation ι and thesign representation ε (see Example 1.1.5). The branching graph of S1 ≤ S2 is

ι

ε

ι0

where ι0 is the trivial representation of S1. Since X2 = (1, 2), we have

X2v =

v if v ∈ Vι,−v if v ∈ Vε.

Lemma 3.3.5. (a) For all n ≥ 1, we have Spec(n) ⊆ Cont(n).

(b) If α ∈ Spec(n), β ∈ Cont(n), and α ≈ β, then β ∈ Spec(n) and α ∼ β.

Proof. We prove (a) by induction on n. The case n = 1 is trivial, while the case n = 2follows from Lemma 3.3.4(c) and (3.5).

Suppose that Spec(n− 1) ⊆ Cont(n− 1), and let

α = (a1, a2, . . . , an) ∈ Spec(n).

By Lemma 3.3.4(a), we have a1 = 0, which corresponds to (a) of Definition 3.1.7. ByLemma 3.3.4(b) and our induction hypothesis, to prove that α ∈ Spec(n), it suffices to provethat conditions (b) and (c) of Definition 3.1.7 are satisfied for j = n.

First suppose, towards a contradiction, that α does not satisfy Definition 3.1.7(b). Inother words, we suppose that

an − 1, an + 1 ∩ a1, a2, . . . , an−1 = ∅. (3.27)

By Proposition 3.3.2(c), the transposition (n− 1, n) is admissible for α, that is

(a1, a2, . . . , an−2, an, an−1) ∈ Spec(n).

Then, by Lemma 3.3.4(b) and our induction hypothesis, we have

(a1, a2, . . . , an−2, an) ∈ Spec(n− 1) ⊆ Cont(n− 1).

By (3.27), we havean − 1, an + 1 ∩ a1, a2, . . . , an−2 = ∅.


But this contradicts Definition 3.1.7(b) for Cont(n − 1). This completes the proof that (b)of Definition 3.1.7 is satisfied for Cont(n).

Now suppose, towards a contradiction, that α does not satisfy Definition 3.1.7(c) forj = n. In other words, we suppose that ai = an = a for some i < n and, for instance,

a− 1 /∈ ai+1, ai+2, . . . , an−1.

(The case a + 1 /∈ ai+1, ai+2, . . . , an−1 is similar and will be omitted.) We choose i to bemaximal with the above properties, so that we also have

a /∈ ai+1, ai+2, . . . , an−1. (3.28)

By the inductive hypothesis, we have (a1, a2, . . . , an−1) ∈ Cont(n − 1). Thus a + 1 mayappear at most once in ai+1, ai+2, . . . , an−1 since, if it appeared more than once, Defini-tion 3.1.7(c) for Cont(n− 1) would imply that a appears between two occurrences of a + 1in ai+1, ai+2, . . . , an−1, contradicting (3.28). Suppose

a+ 1 /∈ ai+1, ai+2, . . . , an−1.

Then(ai, ai+1, . . . , an) = (a, ∗, . . . , ∗, a)

where all the entries ∗ are different from a, a+ 1, and a−1. Then, by a sequence of n− i−1admissible transpositions, we get

α ∼ (. . . , a, a, . . . ) ∈ Spec(n),

which contradicts Proposition 3.3.2(a).Similarly, if

a+ 1 ∈ ai+1, ai+2, . . . , an−1,

then(ai, ai+1, . . . , an) = (a, ∗, . . . , ∗, a+ 1, ∗, . . . , ∗, a),

where, as before, each ∗ represents a number not equal to a, a + 1, or a − 1. Then, by asequence of admissible transpositions, we get

α ∼ (. . . , a, a+ 1, a, . . . ) ∈ Spec(n),

which contradicts Lemma 3.3.3. This completes the proof that (c) of Definition 3.1.7 issatisfied for Cont(n), completing the proof of part (a) of the current lemma.

Part (b) of the current lemma is an immediate consequence of part (a), Corollary 3.1.9,and Proposition 3.3.2(c).

Theorem 3.3.6. (a) We have Spec(n) = Cont(n).

(b) The equivalence relations ∼ and ≈ are the same.

(c) The Young graph Y is isomorphic to the branching graph of the multiplicity-free chainS1 ≤ S2 ≤ · · · ≤ Sn ≤ Sn+1 ≤ · · · .


Proof. First note that

|Cont(n)/ ≈ | = number of partitions of n (Corollary 3.1.10)

= number of conjugacy classes of n (Proposition 3.1.2)

=∣∣∣Sn

∣∣∣ (Proposition 1.3.16)

= | Spec(n)/ ∼ | (by (3.19)).

By Lemma 3.3.5, each equivalence class in Cont(n)/ ≈ is either disjoint from Spec(n)or is contained in a single equivalence class in Spec(n) ∼. In particular, the partition ofSpec(n) induced by ≈ is finer than the partition of Spec(n) induced by ∼. Therefore

| Spec(n)/ ∼ | ≤ | Spec(n)/ ≈ | ≤ |Cont(n)/ ≈ | = | Spec(n)/ ∼ |.

It follows that the two inequalities above must be equality, proving (a) and (b).

As explained at the beginning of Section 3.3.2, Spec(n) parameterizes the paths in thebranching graph. By (3.8), Cont(n) parameterizes the paths in Y. Thus, by part (a), we havea bijective correspondence between the paths in Y and the paths in the branching graph.By Proposition 3.1.11 and the definition of ∼, this yields a bijective correspondence betweenthe vertices of these graphs. This correspondence is clearly a graph isomorphism.

It follows from Theorem 3.3.6 that we have a natural correspondence between Sn andthe n-th level of the branching graph Y, that is, the set of all partitions of n.

Definition 3.3.7 (The irreducible representations Sλ). Given a partition λ ` n, we defineSλ to be the irreducible representation of Sn spanned by the vectors vα, with α ∈ Spec(n) =Cont(n) corresponding to a standard tableau of shape λ.

Proposition 3.3.8. We have dimSλ = |Tab(λ)|. In other words, the dimension of Sλ isequal to the number of standard tableaux of shape λ.

Proof. This follows immediately from Definition 3.3.7.

Proposition 3.3.2 will allow us to give explicit formulas for the action of the Coxetergenerators (and hence any element of Sn) on the irreducible representations Sλ. See Theo-rems 3.4.2 and 3.4.4.

Corollary 3.3.9. Suppose 0 ≤ k < n, λ ` n, and µ ` k. The multiplicity of Sµ in ResSnSkSλ

is equal to the number of paths in Y from λ to µ.

Proof. We have

ResSnSkSλ = Res

Sk+1

SkRes

Sk+2

Sk+1· · ·ResSnSn−1

Sλ.

At each step of the right side, the decomposition is multiplicity free and according to thebranching graph Y. Thus, the multiplicity of Sµ in ResSnSk

Sλ is equal to the number of pathsin Y that start at λ and end at µ.


Corollary 3.3.10 (Branching rule). For λ ` n,

ResSnSn−1Sλ =

⊕µ`n−1:λ→µ

Sµ. (3.29)

The sum above runs over all partitions µ ` n− 1 obtained from λ by removing a single box.Moreover, for all µ ` n− 1, we have

IndSnSn−1

Sµ =⊕λ`n:λ→µ

Sλ. (3.30)

Proof. Corollary 3.3.9 immediately implies (3.29). Now suppose µ ` n− 1 and λ ` n. Thenwe have

dim HomSn

(Sλ, IndSn

Sn−1Sµ)

= dim HomSn−1

(ResSnSn−1

Sλ, Sµ)

(Frobenius reciprocity (Theorem 1.6.6))

= dim HomSn−1

⊕ν`n−1:λ→ν

Sν , Sµ

(by (3.29))

=

1, if λ→ µ,

0, otherwise.(Schur’s lemma (Corollary 1.2.2)).

Thus (3.30) follows from Corollary 1.2.6.

We have defined two notions of admissible transposition, one for Spec(n), and one forCont(n) (coming from the notion of admissible transposition for tableaux). The followingresult states that these coincide.

Lemma 3.3.11. Suppose T ∈ Tab(n) is a standard tableau with content α = C(T ) =(a1, a2, . . . , an) ∈ Spec(n). For 1 ≤ i ≤ n− 1, the simple transposition si is admissible for Tif and only if ai+1 6= ai ± 1.

Proof. We have

ai+1 = ai ± 1 ⇐⇒ the box labelled i+ 1 is immediately to the right of

or immediately below the box labelled i in T

⇐⇒ i and i+ 1 belong to the same row or column of T

⇐⇒ si is not admissible for T .

3.4. The irreducible representations of Sn 105

Exercises.

3.3.3. Fix n ≥ 2. Determine the partitions λ, µ ` n such that Sλ is the trivial representationand Sµ is the sign representation (see Example 1.1.5). Hint : Use Proposition 3.3.2.

3.3.4. Fix n ≥ 2. Prove that Sn has exactly two inequivalent one-dimensional irreduciblerepresentations.

3.4 The irreducible representations of Sn

In this section we will deduce explicit descriptions of the irreducible representations. Inparticular, we will compute matrix coefficients for the simple transpositions. We will alsoderive a formula for the primitive idempotents in terms of the YJM elements. Finally, westate a theorem of Jucys and Murphy relating the centre of the group algebra L(Sn) andthe YJM elements.

3.4.1 Young’s seminormal form

For a partition λ ` n, recall the tableau T λ defined in Section 3.1.4. Furthermore, recallthat for any T ∈ Tab(λ), there exists a unique permutation πT ∈ Sn such that πTT = T λ.

Recall that the Young vector vT associated to a tableau T is defined up to a scalar factor(of norm one if the Young vectors are normalized).

Proposition 3.4.1. It is possible to choose the vectors vT , T ∈ Tab(n), such that, for everyT ∈ Tab(λ), λ ` n, we have

π−1T vTλ = vT +

∑R∈Tab(λ):`(πR)<`(πT )

γRvR,

for some γR ∈ C.

Proof. We prove the result by induction on `(πT ). At each step in the induction, we willchoose the vectors vT for all T with `(vT ) = `.

If `(πT ) = 1, then πT is an admissible transposition for T λ. In this case, the resultfollows from Proposition 3.3.2 (and Lemma 3.3.11). In particular, we choose vT to be thevα′ appearing in (3.23).

Now suppose πT = si1si2 · · · si`−1sj is the standard decomposition of πT into a product

of admissible transpositions (see Remark 3.1.6). Then πT = πT1sj, where T1 = sjT is astandard tableau and `(πT1) = `(πT )− 1.

By the inductive hypothesis, we have

π−1T1vTλ = vT1 +

∑R∈Tab(λ):

`(πR)<`(πT1 )

γ(1)R vR, (3.31)


for some γ(1)R ∈ C. Since T = sjT1, then as in (3.23) we can choose vT such that

sjvT1 = vT +1

aj+1 − ajvT1 , (3.32)

where (a1, a2, . . . , an) = C(T1) is the content of T1. Then we have

π−1T vTλ = sjπ

−1T1vTλ

(3.31)= sjvT1 +

∑R∈Tab(λ):

`(πR)<`(πT1 )

γ(1)R sjvR

(3.32)= vT +

1

aj+1 − ajvT1 +

∑R∈Tab(λ):

`(πR)<`(πT1 )

γ(1)R sjvR

Then the result follows by using Proposition 3.3.2 to compute the terms sjvR.

Theorem 3.4.2 (Young’s seminormal form). Choose the vectors of the Young basis accordingto Proposition 3.4.1. If T ∈ Tab(λ) has content C(T ) = (a1, a2, . . . , an), then the simpletransposition sj acts on vT as follows:

(a) If aj+1 = aj ± 1, then sjvT = ±vT .

(b) If aj+1 6= aj ± 1, then, setting T ′ = sjT , we have

sjvT =

1

aj+1−aj vT + vT ′ if `(πT ′) > `(πT ),

1aj+1−aj vT +

(1− 1

(aj+1−aj)2

)vT ′ if `(πT ′) < `(πT ).

(3.33)

Proof. Part (a) follows immediately from Proposition 3.3.2(b).Suppose that `(πT ′) > `(πT ). By Proposition 3.3.2(c), we have

sjvT = cvT ′ +1

ai+1 − aivT

for some nonzero c ∈ C (since (3.23) holds up for vα′ up to a scalar factor).Since πT ′ = πT sj, we have, by Proposition 3.4.1

vT ′ +∑

R′∈Tab(λ):`(πR′ )<`(πT ′ )

γ′R′vR′ = π−1T ′ vTλ = sjπ

−1T vTλ = sjvT +

∑R∈Tab(λ):`(πR)<`(πT )

γRsjvR

Thus c = 1.The case `(πT ′) < `(πT ) is similar, starting from πT = πT ′sj and applying Proposi-

tion 3.3.2 with α = C(T ′). (See Exercise 3.4.1.)

Corollary 3.4.3. In the orthogonal bases of Proposition 3.4.1 and Theorem 3.4.2 the matrixcoefficients of the irreducible representations of Sn are rational numbers. In particular, thecoefficients γR in Proposition 3.4.1 are rational numbers.


Exercises.

3.4.1. Complete the proof of Theorem 3.4.2(b) by treating the case `(πT ′) < `(πT ).

3.4.2 Young’s orthogonal form

The orthogonal bases of Proposition 3.4.1 and Theorem 3.4.2 do not consist of unit vectorsin general. Given an arbitrary invariant scalar product ‖ ‖ on Sλ that makes it a unitaryrepresentation of Sn (see Lemma 1.1.1) we can, of course, normalize the basis. If λ ` n andvT : T ∈ Tab(λ) is the basis as in Proposition 3.4.1 and Theorem 3.4.2, we define

wT =vT‖vT‖

, T ∈ Tab(λ).

Let T be a standard tableau and let C(T ) = (a1, a2, . . . , an) be its content. For i, j ∈1, 2, . . . , n, the axial distance from j to i in T is the integer aj − ai. Geometrically, wemove from j to i in the tableau T , counting each step left or downwards as +1 and eachstep right or upwards as −1. The resulting integer is aj − ai (and is independent of the pathchosen). For example, if we have

j

i,

then the axial distance from j to i is aj − ai = 5. On the other hand, if we have

ji ,

then the axial distance from j to i is aj − ai = −2.

Theorem 3.4.4 (Young’s orthogonal form). In the orthonormal basis wT : T ∈ Tab(n)we have

sjwT =1

rwT +

√1− 1

r2wsjT , (3.34)

where r is the axial distance from j + 1 to j in T . In particular, if aj+1 = aj ± 1, we haver = ±1 and sjwT = ±wT .

Proof. Let T ′ = sjT and suppose `(πT ′) > `(πT ). By Theorem 3.4.2, we have

‖vT ′‖2 =

∥∥∥∥sjvT − 1

rvT

∥∥∥∥2

= ‖sjvT‖2 − 1

r〈sjvT , vT 〉 −

1

r〈vT , sjvT 〉+

1

r2‖vT‖2

= ‖vT‖2 − 1

r

⟨1

rvT + vT ′ , vT

⟩− 1

r

⟨vT ,

1

rvT + vT ′

⟩+

1

r2‖vT‖2


=

(1− 1

r2

)‖vT‖2,

where we have used the fact that vT ⊥ vT ′ . Thus we have

wT =vT‖vT‖

and wT ′ =vT ′

‖vT ′‖=

vT ′√1− 1

r2‖vT‖

.

In this basis, the first case in (3.33) becomes (3.34).

The proof in the case `(πT ′) < `(πT ) is similar (Exercise 3.4.2).

It follows from Theorem 3.4.4 that

sjwT =1

rwT +

√1− 1

r2wsjT , sjwsjT = −1

rwsjT +

√1− 1

r2wT , (3.35)

where r is the axial distance from j + 1 to j. Thus, in the bases wT , wsjT, the action of sjis given by the orthogonal matrix 1

r

√1− 1

r2√1− 1

r2−1r

.

Example 3.4.5. The only standard tableau of shape (n) is

T = T (n) = 1 2 n .

Its content is C(T ) = (0, 1, 2, . . . , n−1). Thus aj+1 = aj +1 for all 1 ≤ j ≤ n−1. By (3.34),

sjwT = wT , for all 1 ≤ j ≤ n− 1.

Hence S(n) is the trivial representation of Sn.

Example 3.4.6. The only standard tableau of shape (1, 1, . . . , 1) is

T = T (1,1,...,1) =

12

n

.

Its content is C(T ) = (0,−1,−2, . . . ,−n+ 1). Thus aj+1 = aj − 1 for all 1 ≤ j ≤ n− 1. By(3.34),

sjwT = −wT , for all 1 ≤ j ≤ n− 1.

Hence S(1,1,...,1) is the sign representation of Sn.


Example 3.4.7. Consider the representation S(n−1,1). The standard tableaux of shape (n −1, 1) are

Tj = 1 2 j −1 j +1 nj

, 2 ≤ j ≤ n.

For 2 ≤ j ≤ n, we have

C(Tj) = (0, 1, . . . , j − 2,−1, j − 1, j, . . . , n− 2), (3.36)

where the entry −1 is in the j-th position. Let wj = wTj for 2 ≤ j ≤ n. Then the Youngorthogonal form becomes

sjwj =1

jwj +

√1− 1

j2wj+1, (3.37)

sj−1wj = − 1

j − 1wj +

√1− 1

(j − 1)2wj−1, (3.38)

skwj = wj, k 6= j − 1, j. (3.39)

The branching rule gives

ResSnSn−1S(n−1,1) = S(n−1) ⊕ S(n−2,1).

We claim that S(n−1,1) is isomorphic to the representation V1 of Example 1.4.5. We provethis claim by giving an explicit isomorphism. Let X = 1, 2, . . . , n and recall that

V1 =

f ∈ L(X) :

n∑j=1

f(j) = 0

.

For 2 ≤ j ≤ n, define

wj =1√

j(j − 1)1j−1 −

√j − 1

jδj, (3.40)

where δj is the Dirac function at j, and 1j = δ1 + δ2 + · · ·+ δj.One can check (Exercise 3.4.3) that wj : 2 ≤ j ≤ n is an orthonormal basis for V1.

Now,

1

jwj +

√1− 1

j2wj+1 =

1

j√j(j − 1)

1j−1 −1

j

√j − 1

jδj +

1

j

√j − 1

j1j −

√j − 1

jδj+1

=1√

j(j − 1)1j−1 −

√j − 1

jδj+1 = sjwj.

Thus, the vectors wj satisfy (3.37). The proof that they satisfy (3.38) is similar. That factthat they satisfy (3.39) is straightforward. Therefore we have an isomorphism of represen-tations

V1 → S(n−1,1), wj 7→ wj, 2 ≤ j ≤ n.


Exercises.

3.4.2. Complete the proof of Theorem 3.4.4 in the case `(πT ′) < `(πT ).

3.4.3. We adopt the notation of Example 3.4.7.

(a) Verify that wj : 2 ≤ j ≤ n is an orthonormal basis for V1.

(b) Verify that the vectors wj satisfy (3.38).

3.4.4. Prove that wj, as defined in (3.40), corresponds to the path

(n− 1, 1)→ (n− 2, 1)→ · · · → (j, 1)→ (j − 1, 1)→ (j − 1)→ (j − 2)→ · · · → (2)→ (1)

in the Young graph by examining the action of Sn ≥ Sn−1 ≥ · · · ≥ S1. This gives anotherway of identifying wj with wj.

3.4.5. Define wj as in (3.40). Show, by direct computation, that

Xjwj =

(i− 1)wj for i < j,

−wj for i = j,

(i− 2)wj for i > j.

In other words, prove that α(Tj) = (0, 1, 2, . . . , j − 2,−1, j − 1, j, . . . , n − 2) ∈ Spec(n).Compare with (3.36).

3.4.3 The Young seminormal units

In this section we will deduce an expression, in terms of YJM elements, for the primitiveidempotents of L(Sn) corresponding to the Gelfand–Tsetlin bases for the irreducible repre-sentations. (See Proposition 1.5.8.)

For λ ` n, letdλ = dimSλ.

For each T = Tab(n) of shape λ, the primitive idempotent in L(Sn) corresponding to theGelfand–Tsetlin vector wT is given by (see (1.46))

eT (π) =dλn!〈πwT , wT 〉Sλ , π ∈ Sn. (3.41)

Following the notation of this chapter, we will identify eT with the formal sum∑π∈Sn

eT (π)π.

For S ∈ Tab(n), we let

eTwS =∑π∈Sn

eT (π)πws


denote the action of eT (more precisely, of its Fourier transform) on wS. Furthermore, eT eSdenotes the convolution of eT and eS.

By Proposition 1.5.8, for all S, T ∈ Tab(n) we have

eT eS = δT,SeT and (3.42)

eTwS = δT,SwT . (3.43)

Note that (3.43) uniquely characterizes eT among elements of L(Sn). Also, it follows from(3.43) and Theorem 2.2.2 that

eT : T ∈ Tab(n)is a basis of the Gelfand–Tsetlin algebra GZ(n). The elements eT , T ∈ Tab(n), are calledYoung seminormal units .

For T ∈ Tab(n), let T ∈ Tab(n − 1) denote the tableau obtained from T by removingthe box labelled n. We also denote by aT (j) the j-th component of C(T ); in other words,

C(T ) = (aT (1), aT (2), . . . , aT (n)) .

It follows thatXkwT = aT (k)wT , for all T ∈ Tab(n), 1 ≤ k ≤ n.

The following theorem gives a recursive formula for the Young seminormal units in termsof the YJM elements.

Theorem 3.4.8. We have eT = 1 for the unique T ∈ Tab(1). For n ≥ 2 and T ∈ Tab(n),we have

eT = eT∏

S∈Tab(n):

S=T , S 6=T

Xn − aS(n)

aT (n)− aS(n). (3.44)

Proof. Let eT denote the element of L(Sn) recursively defined by the right side of (3.44).We will show that eTwS = δT,SwS for all S ∈ Tab(n). By the characterizing property (3.43),this will imply that eT = eT .

We proceed by induction on n. The result is clearly true for n = 1. Thus, we assumen ≥ 2 and that the result is true for n− 1.

Suppose S 6= T . TheneTwS = eTwS = 0.

The first equality above follows from the fact that eT ∈ L(Sn−1), and so, to compute theaction of eT on wS, we first restrict the irreducible representation containing wS to Sn−1,obtaining the vector wS. The second equality above follows from the induction hypothesis.

Now suppose S = T , but S 6= T . Then

XnwS = aS(n)wS,

and so eTwS = 0 since the factor Xn − aS(n) in the right side of (3.44) acts as zero on wS.Finally, note that XnwT = aT (n)wT , and so

Xn − aS(n)

aT (n)− aS(n)wT = wT


for all S ∈ Tab(n) such that S = T and S 6= T . Hence

eTwT = eT ·∏

S∈Tab(n):

S=T , S 6=T

Xn − aS(n)

aT (n)− aS(n)wT

= eTwT

= eTwT= wT ,

where the final two equalities follow from restriction to Sn−1 and the induction hypothesis.

Corollary 3.4.9. For 1 ≤ k ≤ n, we have

Xk =∑

T∈Tab(n)

aT (k)eT .

Proof. This follows immediately from Theorem 3.4.8 and Proposition 1.5.8(d).

Exercises.

3.4.6 ([CSST10, Ex. 3.4.13]). (a) Let T be the unique standard tableau of shape n (seeExample 3.4.5). Show that

eT =1

n!

n∏j=1

(1 +Xj).

Prove also that eT = 1n!

∑π∈Sn π in two ways: (i) by means of the representation

theory of Sn, and (ii) as an algebraic identity in Sn. Hint : If π ∈ Sn, then thereexists a unique σ ∈ Sn−1 such that σ ∈ Sn−1 and j ∈ 1, 2, . . . , n − 1 such thatπ = σ(j → n→ j).

(b) Let T be the unique standard of shape (1n) = (1, . . . , 1) ` n (see Example 3.4.6). Showthat

eT =1

n!

n∏j=1

(1−Xj).

As in (a), give two proofs of the fact that eT = 1n!

∑π∈Sn(−1)`(π)π.

(c) Let Tj be the standard tableau of Example 3.4.7. Show that

eTj = − (j − 2)!

n!(n− 2)!

(j−1∏i=1

(Xi + 1)

)· (Xj − j + 1) ·

(n∏

i=j+1

Xi(Xi + 2)

).


3.4.4 The Theorem of Jucys and Murphy

Consider the polynomial algebra C[y1, . . . , ym]. The symmetric group Sm acts on C[y1, . . . , ym]by permuting the indeterminates y1, . . . , ym. In other words, for π ∈ Sm, we have a uniquealgebra isomorphism

C[y1, . . . , ym]→ C[y1, . . . , ym], f 7→ π · f,

determined by π · yi = yπ(i).The subalgebra

C[y1, . . . , ym]Sm = f ∈ C[y1, . . . , ym] : π · f = f for all π ∈ Sm ⊆ C[y1, . . . , ym]

is called the algebra of symmetric polynomials . So an element f ∈ C[y1, . . . , ym] is a symme-tric polynomial if and only if it is left unchanged by any permutation of the indeterminatesy1, . . . , ym.

Recall that Z(n) = Z(L(Sn)) is the center of the group algebra L(Sn). We say anelement f ∈ Z(n) is a symmetric polynomial in the YJM elements if

f = p(X2, . . . , Xn) for some p ∈ C[y1, . . . , yn−1]Sn−1 .

(In fact, recalling that X1 = 0, one can show that this is equivalent to requiring thatf = p(X1, X2, . . . , Xn) for some p ∈ C[y1, . . . , yn]Sn .)

Theorem 3.4.10 (Theorem of Jucys and Murphy). The center Z(n) of the group algebraof the symmetric group Sn is precisely the algebra of all symmetric polynomials in the YJMelements X2, X3, . . . , Xn.

Proof. Due to lack of time, we will not prove this theorem in this course. A proof can befound in [CSST10, Th. 4.4.5] or in [Mur83, Th. 1.9].

Chapter 4

Further directions

In this final chapter, we briefly touch on some more advanced topics related to the represen-tation theory of the symmetric group and related algebras. We will omit the proofs of mostresults.

4.1 Schur–Weyl duality

Schur–Weyl duality is a result that gives a precise relationship between the representationtheory of the symmetric group and the representation theory of the general linear group.

Fix n ≥ 1. The general linear group GLn(C) is the group of invertible n × n complexmatrices, under multiplication. It is an important example of a Lie group. The groupGLn(C) acts naturally on the space Cn (thought of as consisting of column vectors) viamatrix multiplication. It then acts on the space

V := Cn ⊗ Cn ⊗ · · · ⊗ Cn︸︷︷︸k factors

(4.1)

by simultaneous matrix multiplication:

g(v1 ⊗ v2 ⊗ · · · ⊗ vk) = gv1 ⊗ gv2 ⊗ · · · gvk, v1, . . . , vk ∈ Cn, g ∈ G,

extended by linearity.On the other hand, the symmetric group Sk also acts naturally on V by permuting the

factors

π(v1 ⊗ v2 ⊗ · · · ⊗ vk) = vπ−1(1) ⊗ vπ−1(2) ⊗ · · · ⊗ vπ−1(k), v1, . . . , vk ∈ Cn, π ∈ Sk,

extended by linearity.It is straightforward to verify that the actions of GLn(C) and Sk on V commute:

gπv = πgv, for all g ∈ GLn(C), π ∈ Sk, v ∈ V.

Thus we have maps to the commutants:

GLn(C)→ EndSk V and Sk → EndGLn(C) V.

114

4.2. Categorification 115

Schur–Weyl duality asserts that the images of each of these maps generate the codomain (asan algebra). Furthermore, V decomposes as

V =⊕λ

Sλ ⊗ Lλ, (4.2)

where the sum is over all Young diagrams (equivalently, partitions) with k boxes and at mostn rows. In this decomposition, Sk acts on the first factor, while GLn(C) acts on the secondfactor. The Lλ are pairwise inequivalent irreducible representations of GLn(C) (just as theSλ are pairwise inequivalent irreducible representations of Sk). It follows, for instance, thatthe multiplicity of Sλ in V is dimLλ, and that the multiplicity of Lλ in V is dimSλ.

Example 4.1.1. Suppose that k = 2 and n ≥ 2. We know that S2 has exactly two irreduciblerepresentations: the trivial representation and the sign representation. Thus we have

Cn ⊗ Cn = S2Cn ⊕ Λ2Cn,

where S2Cn is the space of symmetric tensors (the subspace of Cn ⊗ Cn on which S2 actstrivially) and Λ2Cn is the space of antisymmetric tensors (the subspace of Cn⊗Cn on whichS2 acts via the sign representation). Each of these summands is an irreducible representationof GLn(C).

4.2 Categorification

Categorification is a powerful tool for relating mathematical structures that may appear onthe surface to be completely unrelated. It also reveals hidden mathematical structure andprovides a method to study and organize the representation theory of important algebras. Inthis section we will give a very brief overview of some examples of categorification, includingcategorification of symmetric functions, bosonic Fock space, and representations of certainLie algebras. For further details we refer the reader to the expository references [Kle05,LS12, Sav17] and to the original research papers [Gei77, LS13, Kho14, RS17]

4.2.1 Symmetric functions

For n ≥ 0, let Symn denote the space of all degree n elements of

CJx1, x2, . . .K

that remain invariant under any permutation of the indeterminates x1, x2, . . . . For example,we have the n-th power sum

pn =∞∑i=1

xni ∈ Symn, n ∈ Z>0.

We set p0 = 1. The algebra of symmetric functions is

Sym :=∞⊕n=0

Symn,

116 Chapter 4. Further directions

together with the natural sum and product of formal power series.Let P denote the set of all partitions, including the empty partition ∅ of 0. For λ =

(λ1, λ2, . . . , λk) ∈ P , we define the power sum symmetric function

pλ := pλ1pλ2 · · · pλk .

One can show that pλ : λ ∈ P is a basis for Sym, so that

Sym =⊕λ∈P

Cpλ,

It follows that Sym is isomorphic as an algebra to the polynomial algebra in the pn:

Sym ∼= C[p1, p2, . . . ].

We define an inner product on Sym by declaring

〈pλ, pµ〉 = δλ,µzλ, zλ =

λ1∏i=1

imi(λ)mi(λ)!,

where mi(λ) denotes the number of parts of the partition λ = (λ1, λ2, . . . , λk) equal to i.(Note that |λ|!/zλ is the number of partitions of |λ| that have cycle type λ. See Exercise 3.1.1.)

For λ ∈ P , a semistandard tableau of shape of λ is a filling T of the boxes of λ (consideredas a Young diagram) with elements of Z>0 such that the entries are weakly increasing fromleft to right along rows and strictly increasing down columns. For example

1 1 2 2 3 52 3 3 3 63 46

is a semistandard tableau of shape (6, 5, 2, 1).For a semistandard tableau T , define

xT :=∞∏i=1

xtii ∈ CJx1, x2, . . .K,

where ti is the number of occurrences of i in the tableau T . (Note that the above product isactually finite since ti = 0 for all but finitely many values of i.)

The Schur function corresponding to λ ∈ P is

sλ :=∑T

xT ,

where the sum is over all semistandard tableaux of shape λ. It can be shown that the Schurfunctions form an orthonormal basis for Sym.

For n ∈ Z>0, let

hn := s(n) =∑

1≤i1≤i2≤···≤in

xi1xi2 · · ·xin .


We define h0 = 1. For every partition λ = (λ1, λ2, . . . , λk) we have the correspondingcomplete symmetric function

hλ = hλ1hλ2 · · ·hλk .One can show that hλ : λ ∈ P is a basis for Sym. It follows that

Sym = C[h1, h2, . . . ].

(In fact, the complete symmetric functions have the slightly better property that they gene-rate Sym over Z, which makes them more naturally suited to categorification.)

4.2.2 The Grothendieck group

For the remainder of these notes, we let S0 denote the trivial group, so that S0 = S1.Clearly, S0 has one irreducible representation, which we denote by S∅, where ∅ is theempty partition of 0.

The (finite-dimensional) representations of Sn, together with homomorphisms of repre-sentations, form a category RepSn. For a representation V of Sn, let [V ] denote its iso-morphism class. Consider the free vector space Fn on the set of isomorphism classes ofrepresentations of Sn:

Fn :=⊕[V ]

C[V ], (4.3)

where the sum is over all isomorphism classes [V ].Now define

Fn = 〈[V1 ⊕ V2]− [V1]− [V2] : V1, V2 reps of Sn〉 ⊆ Fn. (4.4)

The Grothendieck group of RepSn is defined to be

K(Sn) := Fn/Fn.

Equivalently, K(Sn) is the quotient of Fn by the relation

[V1 ⊕ V2] = [V1] + [V2], for all representations V1 and V2 of Sn.

We will denote the image in K(Sn) of an isomorphism class [V ] again by [V ]. The groupoperation on K(Sn) is the vector space addition.

One can show that K(Sn) has a basis given by the classes of the irreducible representa-tions. Thus

K(Sn) =⊕λ`n

C[Sλ].

Define

K(S) :=∞⊕n=0

K(Sn),

so thatK(S) =

⊕λ∈P

C[Sλ].


Taking the Grothendieck group of the category RepSn is an example of decategorification.It takes a category and produces a vector space. One can take the Grothendieck group ofother categories, provided they have enough structure, but that is beyond the scope of ourcurrent discussion.

4.2.3 Categorification of the algebra of symmetric functions

Recall that for `, k ∈ Z≥0, we can view S`×Sk as a subgroup of S`+k, where S` permutes theelements 1, 2, . . . , ` and Sk permutes the elements `+1, `+2, . . . , `+k. Therefore, givena representation U of S` and a representation V of Sk, we have the outer tensor productrepresentation U V of S` ×Sk (see Section 1.1.7), and hence the induced representation

IndS`+kS`×Sk(U V )

of S`+k. Since the operations of outer tensor product and induction preserve isomorphismand direct sums, we have an induced bilinear map[

IndS`+kS`×Sk

]: K(S`)⊗K(Sk)→ K(S`+k).

Taking these maps for all `, k gives a binary operation on K(S). One can show that thisoperation is associative and unital. The unit element is class of the trivial representation ofK(S0).

The following theorem was first proved by Geissinger in [Gei77].

Theorem 4.2.1 (Categorification of the algebra of symmetric functions). The linear map

Φ: K(S)→ Sym

determined by Φ([Sλ]) = sλ, λ ∈ P, is an isomorphism of algebras.

Theorem 4.2.1 is an example of a categorification. We have a category (namely, thecategory of representations of symmetric groups) with some extra structure coming frominduction. Decategorifying, i.e. passing to the Grothendieck group, recovers the algebra ofsymmetric functions. Thus, the category of representations of symmetric groups categorifiesthe algebra of symmetric functions. In fact, if we also consider the restriction functorsRes

S`+kS`×Sk , we obtain the structure of a coproduct on K(Sym). Then Theorem 4.2.1 can be

strengthened to state that we have an isomorphism of Hopf algebras.We can also categorify the inner product on Sym as follows. For U, V representations of

Sn, we define〈[U ], [V ]〉 = dim HomSn(U, V ). (4.5)

One can show that this does indeed define an inner product on K(S) (we declare elementsof K(Sn) to be orthogonal to elements of K(Sm) for n 6= m). For λ, µ ∈ P , it follows fromSchur’s lemma that

〈[Sλ], [Sµ]〉 = δλ,µ.

Thus, the [Sλ], λ ∈ P , form an orthonormal basis for K(S). It follows that the isomorphismΦ of Theorem 4.2.1 is an isometry (i.e. it respects the inner products).


4.2.4 The Heisenberg algebra

Heisenberg algebras play a fundamental role in mathematics and mathematical physics. The(infinite rank) Heisenberg algebra H is the unital associative C-algebra with generators pn,p∗n, n ∈ Z>0, and relations

pnp∗m = p∗mpn + δn,m1, pnpm = pmpn, p∗np

∗m = p∗mp

∗n, n,m ∈ Z>0. (4.6)

The first relation in (4.6) is often called the canonical commutation relation in the physicsliterature, where the generators p∗n and pn correspond to position and momentum operatorsin a single particle system with a countable infinite number of degrees of freedom. TheHeisenberg algebra is also crucial in the study of the quantum harmonic oscillator.

There is another, more presentation independent, way to describe the Heisenberg algebraH. Any f ∈ Sym acts on Sym via multiplication. Let f ∗ denote the operator on Sym adjointto multiplication by f :

〈f ∗(g), h〉 = 〈g, fh〉 for all f, g, h ∈ Sym.

Then H is the subalgebra of EndC Sym generated by the operators f and f ∗, for f ∈ Sym.The tautological action of H on Sym is called the bosonic Fock space representation. Anychoice of generating set for Sym yields a presentation of H. In particular, if we choosepower sums, we recover the presentation (4.6). If we instead choose the complete symmetricfunctions, we see that H is the unital associative C-algebra generated by hn, h∗n, n ∈ Z>0,are relations

hnh∗m =

min(m,n)∑r=0

h∗m−rhn−r, hnhm = hmhn, h∗nh∗m = h∗mh

,n n,m ∈ Z>0. (4.7)

Note, in particular, that h1 and h∗1 satisfy the canonical commutation relation:

h1h∗1 = h∗1h1 + 1 (4.8)

4.2.5 Categorification of bosonic Fock space

By Corollary 3.3.10, for λ ` n, we have

ResSnSn−1Sλ =

⊕µ`n−1:λ→µ

Sµ and IndSn+1

SnSλ =

⊕µ`n+1:µ→λ

Sµ.

Restriction and induction respect isomorphism, in the sense that, if V1∼= V2 are isomorphic

representations of Sn, then

ResSnSn−1V1∼= ResSnSn−1

V2 and IndSn+1

SnV1∼= Ind

Sn+1

SnV2.

Thus, restriction and induction induce linear maps (see (4.3)):

[IndSn+1

Sn] : Fn → Fn+1, [ResSnSn−1

] : Fn → Fn−1.


Restriction and induction also respect direct sums, in the sense that

ResSnSn−1(V1 ⊕ V2) ∼=

(ResSnSn−1

V1

)⊕(

ResSnSn−1V2

), and

IndSn+1

Sn(V1 ⊕ V2) ∼=

(Ind

Sn+1

SnV1

)⊕(

IndSn+1

SnV2

).

It follows that (see (4.4))

[IndSn+1

Sn](Fn) ⊆ Fn+1, [ResSnSn−1

](Fn) ⊆ Fn−1.

Therefore, we have induced maps on the quotients:

[IndSn+1

Sn] : K(Sn)→ K(Sn+1), [ResSnSn−1

] : K(Sn)→ K(Sn−1).

Then we have linear maps

[Ind] : K(S)→ K(S), [Ind] =∞∑n=0

[Ind

Sn+1

Sn

]and

[Res] : K(S)→ K(S), [Res] =∞∑n=1

[ResSnSn−1

]where we adopt the convention that [Res] acts as zero on K(S0).

Using the combinatorics of the Young graph, one can show that, for all λ ` n, n ≥ 1, wehave

IndSnSn−1

ResSnSn−1Sλ ∼=

(Res

Sn+1

SnInd

Sn+1

SnSλ)⊕ Sλ.

(See Exercise 4.2.1.) It follows that

[Ind][Res] = [Res][Ind] + 1 (4.9)

as linear operators on K(S). Note that these are precisely the canonical commutationrelations satisfied by the generators e1 and e∗1 of the Heisenberg algebra (see (4.8))!

What about other generators? Recall from Section 4.2.3 that we have defined a producton K(S). Thus, for n ∈ Z>0 we have the operator given by multiplication by [S(n)]:

an : K(S)→ K(S), an([V ]) = [V ] · [S(n)] =[Ind

Sk+nSk×Sn

(V S(n)

)], V a rep of Sk.

Since we also have an inner product on K(S) as in (4.5), we can consider operators a∗nadjoint to an. These can also be described directly using restriction. One can then showthat these operators satisfy the relations (4.7):

ana∗m =

min(m,n)∑r=0

a∗m−ran−r, anam = aman, a∗na∗m = a∗ma

,n n,m ∈ Z>0. (4.10)

It follows that we have an action of the Heisenberg algebra on K(Sym). In other words, wecan view K(Sym) as a representation of H

Theorem 4.2.2 (Categorification of bosonic Fock space). The map Φ of Theorem 4.2.1 isan isomorphism of representations of the Heisenberg algebra H from K(S) to the bosonicFock space representation of H on Sym.


Exercises.

4.2.1. Prove that for all λ ` n, n ≥ 1, we have

IndSn+1

SnResSnSn−1

Sλ ∼=(

ResSnSn−1Ind

Sn+1

SnSλ)⊕ Sλ.

4.2.6 Categorification of the basic representation

The categorification of bosonic Fock space described in Section 4.2.5 can be refined somewhat.Suppose V is a representation of Sn for some n ∈ Z>0. Since the YJM element Xn commuteswith Sn−1 (see Exercise 3.2.1), the action of Xn on ResSnSn−1

V gives an Sn−1-intertwiner.We know that Xn acts diagonally, with integral eigenvalues. For i ∈ Z, let proji denote theprojection onto the eigenspace corresponding to eigenvalue i. If we define

Resi V := proji ResSnSn−1V,

it follows that

ResSnSn−1V =

⊕i∈Z

Resi V.

(Note that Resi V = 0 for all but finitely many values of i.) We call Resi i-restriction. Thisis a functor, with an adjoint functor i-induction, denoted Indi. In terms of the combinatoricsof standard tableaux, Resi acts on irreducible representations by removing a box of contenti (if such a box exists). That is,

Resi Sλ =

Sµ if µ is obtained from λ by removing a box of content i,

0 if µ has no removable boxes of content i.

Similarly, Indi adds a box of content i (if possible).As in Section 4.2.5, we have induced maps

[Indi], [Resi] : K(S)→ K(S).

Let sl∞ denote the space of trace zero Z × Z matrices with a finite number of nonzeroentries. In other words, elements of sl∞ are matrices X = (Xi,j)i,j∈Z such that Xi,j = 0 forall but finitely many pairs (i, j) ∈ Z×Z, and such that

∑i∈ZXi,i = 0. This is a Lie algebra

with Lie bracket

[X, Y ] = XY − Y X,

where juxtaposition denotes matrix multiplication.One can prove that the action of [Indi] and [Resi] on K(S) define an action of the

Lie algebra sl∞ on K(S). The particular representation one obtains is called the basicrepresentation. Thus, the representation theory of symmetric groups yields a categorificationof the basic representation of sl∞.


4.2.7 Going even further

The constructions discussed briefly above are just the start of the deep and extremely activearea of categorification. We mention here some related constructions. For details on the firsttwo, we refer the reader to the book [Kle05].

Positive characteristic. If we work over a field of positive characteristic p instead ofover the complex numbers, we can repeat the categorification of the basic representationsketched in Section 4.2.6. Then the eigenvalues of the YJM elements lie in Z/pZ. We

obtain a categorification of the affine Lie algebra slp. Note that, in positive characteristic,Maschke’s Theorem fails (see Exercise 1.1.12). It is no longer true that every representationof Sn decomposes as a sum of irreducible ones. So here the representation theory is muchmore complicated. But categorification provides some very useful tools for studying theserepresentations.

Higher level cyclotomic quotients. The group algebra L(Sn) of the symmetric groupis a quotient of the degenerate affine Hecke algebra (see Exercise 3.2.3) by the ideal generatedby x1. More generally one can take the cyclotomic quotient by the ideal generated by∏

i∈Z

(x1 − i)µi ,

for some µi ∈ Z≥0, i ∈ Z, where all but finitely many of the µi are equal to zero. Such aquotient is called a degenerate cyclotomic Hecke algebra. One can repeat the constructions ofSections 4.2.5 and 4.2.6 in this setting and obtain categorifications of other representationsof the Heisenberg algebra and of sl∞ (or of slp if we work in positive characteristic). Thestudy of these categorification is currently an active area research.

Wreath product algebras. One can replace the group algebra L(Sn) by wreath productalgebras . These are algebras of the form

F⊗n ⊗ L(Sn),

where the multiplication involves the action of Sn on F⊗n by permuting the factors (similarto a semidirect product of groups). We have a chain

C ⊆ F ⊆ F⊗2 ⊗ L(S2) ⊆ F⊗3 ⊗ L(S3) ⊆ · · · .

Then one can examine the functors of induction and restriction. There are deep connectionsto representation, geometry, and algebraic combinatorics. We refer the reader to [RS17] forfurther details.

Index

≈, 86〈〉, 18≤, 6⊥, 11∼, 98⊆, 6

addable, 82adjacent transposition, 83adjoint, 6adjoint representation, 13admissible, 87admissible transposition, 83, 100, 104algebra, 23∗-algebra, 23algebra homomorphism, 23alternating representation, 8ambivalent, 61∗-anti-homomorphism, 24∗-anti-isomorphic, 24∗-anti-isomorphism, 24antilinear, 9antisymmetric tensors, 115associative algebra, 23α(T ), 97axial distance, 107

basic representation, 121bi-K-invariant, 58branching graph, 74branching rule for Sn, 104

C×, 20categorification, 118

symmetric functions, 118category, 117center, 23central function, 30, 69

C(G,H), 69character, 30class function, 30col, 86commutant, 25commutative algebra, 23complete symmetric function, 117conjugacy invariant, 69conjugate representation, 13conjugate transpose, 13Cont(n), 86content, 86contragredient representation, 13convolution, 50convolution algebra, 50covers, 87Coxeter generator, 83cycle, 80cycle decomposition, 81cycle type, 81cyclic, 18cyclotomic quotient, 122

dρ, 34decategorification, 118degenerate cyclotomic Hecke algebra, 122δx, 8dimension, 5Dirac function, 8direct sum, 11

of algebras, 24disjoint cycles, 80doubly transitive, 41dual, 12dual basis, 13

EndG(V ), 25endomorphism algebra, 24

123

124 Index

End(V ), 24equivalent, 9even permutation, 8

fixed point, 81fixed points character formula, 34fixed vector, 18Fourier inversion formula in B, 55Fourier transform, 35, 54Frobenius character formula for induced re-

presentations, 64Frobenius extension, 65Frobenius reciprocity, 44, 65

G, 10Gσ, 21G-invariant, 5Gelfand pair, 42

weakly symmetric, 60Gelfand’s lemma, 59

symmetric case, 41Gelfand–Tsetlin algebra, 77Gelfand–Tsetlin basis, 75general linear group, 5, 114GL(V ), 5Grothendieck group, 117group algebra, 50GZ(n), 77

Hasse diagram, 88Heisenberg algebra, 119Hermitian inner product, 6Hermitian scalar product, 6Hom(V,W ), 9homomorphism

of algebras, 23∗-homomorphism, 24hook, 83

IV , 5i-induction, 121i-restriction, 121ideal, 25idempotent, 21, 25indecomposable, 12index of a subgroup, 62

induced representation, 61induction

transitivity, 63inner product

Hermitian, 6internal tensor product, 17intertwine, 9intertwining number theorem, 68intertwining operator, 9interwiner, 9G-invariant, 5σ-invariant, 5invariant vector, 18inversion, 83involution, 23involutive algebra, 23I(π), 83Irr(G), 10irreducible character, 30irreducible representation, 6isometric immersion, 47isomorphism, 9∗-isomorphism, 24isotypic component, 21

Jucys–Murphy elements, 90

left K-invariant, 58left regular representation, 8length, 88length of a partition, 80length of a permutation, 83Lie algebra, 121Lie group, 114L(X), 7

Mackey’s lemma, 66marked permutation, 91Maschke’s Theorem, 11matrix algebra, 24matrix coefficients, 13minimal central idempotent, 27minimal central projection, 27minimal idempotent, 27minimal projection, 27multiplicity, 21

Index 125

multiplicity free, 26, 74multiplicity-free subgroup, 73

N, 21

odd permutation, 8Olshanskii’s Theorem, 95operator

unitary, 6opposite algebra, 53orthogonal projection, 21outer tensor product, 17

P , 116partition, 80path, 88permutation, 8

marked, 91permutation representation, 7polar decomposition, 9power sum, 115power sum symmetric function, 116primitive idempotent, 27, 55projection, 21

onto an isotypic component, 36

reducible representation, 6removable, 82representation, 5

unitary, 6ResGK , 6restriction, 6Riesz map, 12Riesz representation theorem, 12right K-invariant, 58right regular representation, 8row, 86

Sλ, 103scalar product, 6

Hermitian, 6Schur function, 116Schur’s lemma, 19self-adjoint, 21, 23semistandard tableau, 116separate, 78

sign representation, 8simple tensor, 15simple transposition, 83size of a partition, 80Sn, 8, 80Spec(n), 98spectrum, 98stabilizer, 34standard Young tableau, 82structure coefficients, 24subalgebra, 23subrepresentation, 6Sym, 115symmetric, 41symmetric functions, 115symmetric Gelfand pair, 42symmetric group, 8symmetric polynomials, 113symmetric tensors, 115

T λ, 85Tα, 98Tab(λ), 82Tab(n), 82tag, 91tensor product, 15

internal, 17outer, 17

Theorem of Jucys and Murphy, 113trace, 29transitivity of induction, 63transposition, 80trivial idempotent, 23trivial representation, 7truncation, 76

unital algebra, 23unitarily equivalent, 9unitarizable, 6unitary

matrix, 13unitary matrix realization of σ, 14unitary operator, 6unitary representation, 6unitary space, 6

126 Index

vα, 98

weakly symmetric Gelfand pair, 60weight, 97Wielandt’s lemma, 40wreath product algebras, 122

Young (branching) graphs, 88Young basis, 97Young diagram, 81Young seminormal units, 111Young tableau, 82

standard, 82Young’s orthogonal form, 107Young’s seminormal form, 106Young–Jucys–Murphy (YJM) elements, 90

Z(`, k), 91Z(n), 77, 113

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