Download - Part II - Representation Theory - SRCF · 2 Basic de nitions II Representation Theory 2 Basic de nitions We now start doing representation theory. We boringly start by de ning a representation.

Part II — Representation Theory

Based on lectures by S. MartinNotes taken by Dexter Chua

Lent 2016

These notes are not endorsed by the lecturers, and I have modified them (oftensignificantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

Linear Algebra and Groups, Rings and Modules are essential

Representations of finite groupsRepresentations of groups on vector spaces, matrix representations. Equivalence ofrepresentations. Invariant subspaces and submodules. Irreducibility and Schur’sLemma. Complete reducibility for finite groups. Irreducible representations of Abeliangroups.

Character theoryDetermination of a representation by its character. The group algebra, conjugacy classes,and orthogonality relations. Regular representation. Permutation representations andtheir characters. Induced representations and the Frobenius reciprocity theorem.Mackey’s theorem. Frobenius’s Theorem. [12]

Arithmetic properties of charactersDivisibility of the order of the group by the degrees of its irreducible characters.Burnside’s paqb theorem. [2]

Tensor productsTensor products of representations and products of characters. The character ring.Tensor, symmetric and exterior algebras. [3]

Representations of S1 and SU2

The groups S1, SU2 and SO(3), their irreducible representations, complete reducibility.The Clebsch-Gordan formula. *Compact groups.* [4]

Further worked examples

The characters of one of GL2(Fq), Sn or the Heisenberg group. [3]

1

Contents II Representation Theory

Contents

0 Introduction 3

1 Group actions 4

2 Basic definitions 7

3 Complete reducibility and Maschke’s theorem 12

4 Schur’s lemma 16

5 Character theory 21

6 Proof of orthogonality 28

7 Permutation representations 32

8 Normal subgroups and lifting 36

9 Dual spaces and tensor products of representations 399.1 Dual spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399.2 Tensor products . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409.3 Powers of characters . . . . . . . . . . . . . . . . . . . . . . . . . 439.4 Characters of G×H . . . . . . . . . . . . . . . . . . . . . . . . . 459.5 Symmetric and exterior powers . . . . . . . . . . . . . . . . . . . 469.6 Tensor algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479.7 Character ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

10 Induction and restriction 50

11 Frobenius groups 58

12 Mackey theory 61

13 Integrality in the group algebra 65

14 Burnside’s theorem 69

15 Representations of compact groups 7115.1 Representations of SU(2) . . . . . . . . . . . . . . . . . . . . . . 7715.2 Representations of SO(3), SU(2) and U(2) . . . . . . . . . . . . . 85

Index 88

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0 Introduction II Representation Theory

0 Introduction

The course studies how groups act as groups of linear transformations on vectorspaces. Hopefully, you understand all the words in this sentence. If so, this is agood start.

In our case, groups are usually either finite groups or topological compactgroups (to be defined later). Topological compact groups are typically subgroupsof the general linear group over some infinite fields. It turns out the tools wehave for finite groups often work well for these particular kinds of infinite groups.The vector spaces are always finite-dimensional, and usually over C.

Prerequisites of this course include knowledge of group theory (as much as theIB Groups, Rings and Modules course), linear algebra, and, optionally, geometry,Galois theory and metric and topological spaces. There is one lemma wherewe must use something from Galois theory, but if you don’t know about Galoistheory, you can just assume the lemma to be true without bothering yourselftoo much about the proofs.

3

1 Group actions II Representation Theory

1 Group actions

We start by reviewing some basic group theory and linear algebra.

Basic linear algebra

Notation. F always represents a field.

Usually, we take F = C, but sometimes it can also be R or Q. These fieldsall have characteristic zero, and in this case, we call what we’re doing ordinaryrepresentation theory. Sometimes, we will take F = Fp or Fp, the algebraicclosure of Fp. This is called modular representation theory.

Notation. We write V for a vector space over F — this will always be finitedimensional over F. We write GL(V ) for the group of invertible linear mapsθ : V → V . This is a group with the operation given by composition of maps,with the identity as the identity map (and inverse by inverse).

Notation. Let V be a finite-dimensional vector space over F. We write End(V )for the endomorphism algebra, the set of all linear maps V → V .

We recall a couple of facts from linear algebra:If dimF V = n < ∞, we can choose a basis e1, · · · , en of V over F. So we

can identify V with Fn. Then every endomorphism θ ∈ GL(V ) corresponds to amatrix Aθ = (aij) ∈Mn(F ) given by

θ(ej) =∑i

aijei.

In fact, we have Aθ ∈ GLn(F), the general linear group. It is easy to see thefollowing:

Proposition. As groups, GL(V ) ∼= GLn(F), with the isomorphism given byθ 7→ Aθ.

Of course, picking a different basis of V gives a different isomorphism toGLn(F), but we have the following fact:

Proposition. Matrices A1, A2 represent the same element of GL(V ) with respectto different bases if and only if they are conjugate, namely there is some X ∈GLn(F) such that

A2 = XA1X−1.

Recall that tr(A) =∑i aii, where A = (aij) ∈ Mn(F), is the trace of A. A

nice property of the trace is that it doesn’t notice conjugacy:

Proposition.tr(XAX−1) = tr(A).

Hence we can define the trace of an operator tr(θ) = tr(Aθ), which isindependent of our choice of basis. This is an important result. When we studyrepresentations, we will have matrices flying all over the place, which are scary.Instead, we often just look at the traces of these matrices. This reduces ourproblem of studying matrices to plain arithmetic.

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When we have too many matrices, we get confused. So we want to put amatrix into a form as simple as possible. One of the simplest form a matrixcan take is being diagonal. So we want to know something about diagonalizingmatrices.

Proposition. Let α ∈ GL(V ), where V is a finite-dimensional vector space overC and αm = id for some positive integer m. Then α is diagonalizable.

This follows from the following more general fact:

Proposition. Let V be a finite-dimensional vector space over C, and α ∈End(V ), not necessarily invertible. Then α is diagonalizable if and only if thereis a polynomial f with distinct linear factors such that f(α) = 0.

Indeed, we have xm − 1 =∏

(x− ωj), where ω = e2πi/m.Instead of just one endomorphism, we can look at many endomorphisms.

Proposition. A finite family of individually diagonalizable endomorphisms ofa vector space over C can be simultaneously diagonalized if and only if theycommute.

Basic group theory

We will not review the definition of a group. Instead, we look at some of ourfavorite groups, since they will be handy examples later on.

Definition (Symmetric group Sn). The symmetric group Sn is the set of allpermutations of X = {1, · · · , n}, i.e. the set of all bijections X → X. We have|Sn| = n!.

Definition (Alternating group An). The alternating group An is the set ofproducts of an even number of transpositions (i j) in Sn. We know |An| = n!

2 .So this is a subgroup of index 2 and hence normal.

Definition (Cyclic group Cm). The cyclic group of order m, written Cm is

Cm = 〈x : xm = 1〉.

This also occurs naturally, as Z/mZ over addition, and also the group of nth rootsof unity in C. We can view this as a subgroup of GL1(C) ∼= C×. Alternatively,this is the group of rotation symmetries of a regular m-gon in R2, and can beviewed as a subgroup of GL2(R).

Definition (Dihedral group D2m). The dihedral group D2m of order 2m is

D2m = 〈x, y : xm = y2 = 1, yxy−1 = x−1〉.

This is the symmetry group of a regular m-gon. The xi are the rotations andxiy are the reflections. For example, in D8, x is rotation by π

2 and y is anyreflection.

This group can be viewed as a subgroup of GL2(R), but since it also acts onthe vertices, it can be viewed as a subgroup of Sm.

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Definition (Quaternion group). The quaternion group is given by

Q8 = 〈x, y : x4 = 1, y2 = x2, yxy−1 = x−1〉.

This has order 8, and we write i = x, j = y, k = ij, −1 = i2, with

Q8 = {±1,±i,±j,±k}.

We can view this as a subgroup of GL2(C) via

1 =

(1 00 1

), i =

(i 00 −i

), j =

(0 1−1 0

), k =

(0 ii 0

),

−1 =

(−1 00 −1

), −i =

(−i 00 i

), −j =

(0 −11 0

), −k =

(0 −i−i 0

).

Definition (Conjugacy class). The conjugacy class of g ∈ G is

CG(g) = {xgx−1 : x ∈ G}.

Definition (Centralizer). The centralizer of g ∈ G is

CG(g) = {x ∈ G : xg = gx}.

Then by the orbit-stabilizer theorem, we have |CG(g)| = |G : CG(G)|.

Definition (Group action). Let G be a group and X a set. We say G acts onX if there is a map ∗ : G×X → X, written (g, x) 7→ g ∗ x = gx such that

(i) 1x = x

(ii) g(hx) = (gh)x

The group action can also be characterised in terms of a homomorphism.

Lemma. Given an action of G on X, we obtain a homomorphism θ : G →Sym(X), where Sym(X) is the set of all permutations of X.

Proof. For g ∈ G, define θ(g) = θg ∈ Sym(X) as the function X → X by x 7→ gx.This is indeed a permutation of X because θg−1 is an inverse.

Moreover, for any g1, g2 ∈ G, we get θg1g2 = θg1θg2 , since (g1g2)x = g1(g2x).

Definition (Permutation representation). The permutation representation of agroup action G on X is the homomorphism θ : G→ Sym(X) obtained above.

In this course, X is often a finite-dimensional vector space over F (and wewrite it as V ), and we want the action to satisfy some more properties. We willrequire the action to be linear, i.e. for all g ∈ G, v1,v2 ∈ V , and λ ∈ F.

g(v1 + v2) = gv1 + gv2, g(λv1) = λ(gv1).

Alternatively, instead of asking for a map G → Sym(X), we would require amap G→ GL(V ) instead.

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2 Basic definitions II Representation Theory

2 Basic definitions

We now start doing representation theory. We boringly start by defining arepresentation. In fact, we will come up with several equivalent definitions of arepresentation. As always, G will be a finite group and F will be a field, usuallyC.

Definition (Representation). Let V be a finite-dimensional vector space overF. A (linear) representation of G on V is a group homomorphism

ρ = ρV : G→ GL(V ).

We sometimes write ρg for ρV (g), so for each g ∈ G, ρg ∈ GL(V ), and ρgρh = ρghand ρg−1 = (ρg)

−1 for all g, h ∈ G.

Definition (Dimension or degree of representation). The dimension (or degree)of a representation ρ : G→ GL(V ) is dimF(V ).

Recall that ker ρCG and G/ ker ρ ∼= ρ(G) ≤ GL(V ). In the very special casewhere ker ρ is trivial, we give it a name:

Definition (Faithful representation). A faithful representation is a representa-tion ρ such that ker ρ = 1.

These are the representations where the identity is the only element thatdoes nothing.

An alternative (and of course equivalent) definition of a representation is toobserve that a linear representation is “the same” as a linear action of G.

Definition (Linear action). A group G acts linearly on a vector space V if itacts on V such that

g(v1 + v2) = gv1 + gv2, g(λv1) = λ(gv1)

for all g ∈ G, v1,v2 ∈ V and λ ∈ F. We call this a linear action.

Now if g acts linearly on V , the map G → GL(V ) defined by g 7→ ρg,with ρg : v 7→ gv, is a representation in the previous sense. Conversely, given arepresentation ρ : G→ GL(V ), we have a linear action of G on V via gv = ρ(g)v.

In other words, a representation is just a linear action.

Definition (G-space/G-module). If there is a linear action G on V , we say Vis a G-space or G-module.

Alternatively, we can define a G-space as a module over a (not so) cleverlypicked ring.

Definition (Group algebra). The group algebra FG is defined to be the algebra(i.e. a vector space with a bilinear multiplication operation) of formal sums

FG =

∑g∈G

αgg : αg ∈ F

with the obvious addition and multiplication.

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Then we can regard FG as a ring, and a G-space is just an FG-module inthe sense of IB Groups, Rings and Modules.

Definition (Matrix representation). R is a matrix representation of G of degreen if R Is a homomorphism G→ GLn(F).

We can view this as a representation that acts on Fn. Since all finite-dimensional vector spaces are isomorphic to Fn for some n, every representationis equivalent to some matrix representation. In particular, given a linear repre-sentation ρ : G→ GL(V ) with dimV = n, we can get a matrix representationby fixing a basis B, and then define the matrix representation G→ GLn(F) byg 7→ [ρ(g)]B.

Conversely, given a matrix representation R, we get a linear representation ρin the obvious way — ρ : G→ GL(Fn) by g 7→ ρg via ρg(v) = Rgv.

We have defined representations in four ways — as a homomorphism toGL(V ), as linear actions, as FG-modules and as matrix representations. Nowlet’s look at some examples.

Example (Trivial representation). Given any group G, take V = F (the one-dimensional space), and ρ : G→ GL(V ) by g 7→ (id : F→ F) for all g. This isthe trivial representation of G, and has degree 1.

Despite being called trivial, trivial representations are highly non-trivialin representation theory. The way they interact with other representationsgeometrically, topologically etc, and cannot be disregarded. This is a veryimportant representation, despite looking silly.

Example. Let G = C4 = 〈x : x4 = 1〉. Let n = 2, and work over F = C. Thenwe can define a representation by picking a matrix A, and then define R : x 7→ A.Then the action of other elements follows directly by xj 7→ Aj . Of course, wecannot choose A arbitrarily. We need to have A4 = I2, and this is easily seen tobe the only restriction. So we have the following possibilities:

(i) A is diagonal: the diagonal entries can be chosen freely from {±1,±i}.Since there are two diagonal entries, we have 16 choices.

(ii) A is not diagonal: then it will be equivalent to a diagonal matrix sinceA4 = I2. So we don’t really get anything new.

What we would like to say above is that any matrix representation in whichX is not diagonal is “equivalent” to one in which X is. To make this notionprecise, we need to define what it means for representations to be equivalent, or“isomorphic”.

As usual, we will define the notion of a homomorphism of representations,and then an isomorphism is just an invertible homomorphism.

Definition (G-homomorphism/intertwine). Fix a group G and a field F. LetV, V ′ be finite-dimensional vector spaces over F and ρ : G → GL(V ) andρ′ : G → GL(V ′) be representations of G. The linear map ϕ : V → V ′ is aG-homomorphism if

ϕ ◦ ρ(g) = ρ′(g) ◦ ϕ (∗)

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for all g ∈ G. In other words, the following diagram commutes:

V V

V ′ V ′

ρg

ϕ ϕ

ρ′g

i.e. no matter which way we go form V (top left) to V ′ (bottom right), we stillget the same map.

We say ϕ intertwines ρ and ρ′. We write HomG(V, V ′) for the F-space of allthese maps.

Definition (G-isomorphism). A G-homomorphism is a G-isomorphism if ϕ isbijective.

Definition (Equivalent/isomorphic representations). Two representations ρ, ρ′

are equivalent or isomorphic if there is a G-isomorphism between them.

If ϕ is a G-isomorphism, then we can write (∗) as

ρ′ = ϕρϕ−1. (†)

Lemma. The relation of “being isomorphic” is an equivalence relation on theset of all linear representations of G over F.

This is an easy exercise left for the reader.

Lemma. If ρ, ρ′ are isomorphic representations, then they have the same di-mension.

Proof. Trivial since isomorphisms between vector spaces preserve dimension.

The converse is false.

Example. C4 has four non-isomorphic one-dimensional representations: ifω = e2πi/4, then we have the representations

ρj(xi) = ωij ,

for 0 ≤ i ≤ 3, which are not equivalent for different j = 0, 1, 2, 3.

Our other formulations of representations give us other formulations ofisomorphisms.

Given a groupG, field F, a vector space V of dimension n, and a representationρ : G → GL(V ), we fix a basis B of V . Then we get a linear G-isomorphismϕ : V → Fn by v 7→ [v]B, i.e. by writing v as a column vector with respect to B.Then we get a representation ρ′ : G→ GL(Fn) isomorphic to ρ. In other words,every representation is isomorphic to a matrix representation:

V V

Fn Fn

ρ

ϕ ϕ

ρ′

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Thus, in terms of matrix representations, the representations R : G→ GLn(F)and R′ : G→ GLn(F) are G-isomorphic if there exists some non-singular matrixX ∈ GLn(F) such that

R′(g) = XR(g)X−1

for all g.Alternatively, in terms of linear G-actions, the actions of G on V and V ′ are

G-isomorphic if there is some isomorphism ϕ : V → V ′ such that

gϕ(v) = ϕ(gv).

for all g ∈ G,v ∈ V . It is an easy check that this is just a reformulation of ourprevious definition.

Just as we have subgroups and subspaces, we have the notion of sub-representation.

Definition (G-subspace). Let ρ : G → GL(V ) be a representation of G. Wesay W ≤ V is a G-subspace if it is a subspace that is ρ(G)-invariant, i.e.

ρg(W ) ≤W

for all g ∈ G.

Obviously, {0} and V are G-subspaces. These are the trivial G-subspaces.

Definition (Irreducible/simple representation). A representation ρ is irreducibleor simple if there are no proper non-zero G-subspaces.

Example. Any 1-dimensional representation of G is necessarily irreducible, butthe converse does not hold, or else life would be very boring. We will later seethat D8 has a two-dimensional irreducible complex representation.

Definition (Subrepresentation). If W is a G-subspace, then the correspondingmap G → GL(W ) given by g 7→ ρ(g)|W gives us a new representation of W .This is a subrepresentation of ρ.

There is a nice way to characterize this in terms of matrices.

Lemma. Let ρ : G→ GL(V ) be a representation, and W be a G-subspace ofV . If B = {v1, · · · ,vn} is a basis containing a basis B1 = {v1, · · · ,vm} of W(with 0 < m < n), then the matrix of ρ(g) with respect to B has the block uppertriangular form (

∗ ∗0 ∗

)for each g ∈ G.

This follows directly from definition.However, we do not like block triangular matrices. What we really like is

block diagonal matrices, i.e. we want the top-right block to vanish. There isno a priori reason why this has to be true — it is possible that we cannot findanother G-invariant complement to W .

10


Definition ((In)decomposable representation). A representation ρ : G→ GL(V )is decomposable if there are proper G-invariant subspaces U,W ≤ V with

V = U ⊕W.

We say ρ is a direct sum ρu ⊕ ρw.If no such decomposition exists, we say that ρ is indecomposable.

It is clear that irreducibility implies indecomposability. The converse isnot necessarily true. However, over a field of characteristic zero, it turns outirreducibility is the same as indecomposability for finite groups, as we will see inthe next chapter.

Again, we can formulate this in terms of matrices.

Lemma. Let ρ : G→ GL(V ) be a decomposable representation with G-invariantdecomposition V = U ⊕W . Let B1 = {u1, · · · ,uk} and B2 = {w1, · · · ,w`} bebases for U and W , and B = B1 ∪ B2 be the corresponding basis for V . Thenwith respect to B, we have

[ρ(g)]B =

([ρu(g)]B1

00 [ρu(g)]B2

)Example. Let G = D6. Then every irreducible complex representation hasdimension at most 2.

To show this, let ρ : G → GL(V ) be an irreducible G-representation. Letr ∈ G be a (non-identity) rotation and s ∈ G be a reflection. These generate D6.

Take an eigenvector v of ρ(r). So ρ(r)v = λv for some λ 6= 0 (since ρ(r) isinvertible, it cannot have zero eigenvalues). Let

W = 〈v, ρ(s)v〉 ≤ V

be the space spanned by the two vectors. We now check this is fixed by ρ. Firstly,we have

ρ(s)ρ(s)v = ρ(e)v = v ∈W,and

ρ(r)ρ(s)v = ρ(s)ρ(r−1)v = λ−1ρ(s)v ∈W.Also, ρ(r)v = λv ∈ W and ρ(s)v ∈ W . So W is G-invariant. Since V isirreducible, we must have W = V . So V has dimension at most 2.

The reverse operation of decomposition is taking direct sums.

Definition (Direct sum). Let ρ : G → GL(V ) and ρ′ : G → GL(V ′) berepresentations of G. Then the direct sum of ρ, ρ′ is the representation

ρ⊕ ρ′ : G→ GL(V ⊕ V ′)

given by(ρ⊕ ρ′)(g)(v + v′) = ρ(g)v + ρ′(g)v′.

In terms of matrices, for matrix representations R : G → GLn(F) andR′ : G→ GLn′(F), define R⊕R′ : G→ GLn+n′(F) by

(R⊕R′)(g) =

(R(g) 0

0 R′(g)

).

The direct sum was easy to define. It turns out we can also multiply tworepresentations, known as the tensor products. However, to do this, we need toknow what the tensor product of two vector spaces is. We will not do this yet.

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3 Complete reducibility and Maschke’s theorem II Representation Theory

3 Complete reducibility and Maschke’s theorem

In representation theory, we would like to decompose a representation into sumsof irreducible representations. Unfortunately, this is not always possible. Whenwe can, we say the representation is completely reducible.

Definition (Completely reducible/semisimple representation). A representationρ : G→ GL(V ) is completely reducible or semisimple if it is the direct sum ofirreducible representations.

Clearly, irreducible implies completely reducible.Not all representations are completely reducible. An example is to be found

on example sheet 1. These are in fact not too hard to find. For example, thereare representations of Z over C that are not completely reducible, and also anon-completely reducible representation of Cp over Fp.

However, it turns out we have the following theorem:

Theorem. Every finite-dimensional representation V of a finite group over afield of characteristic 0 is completely reducible, namely, V ∼= V1 ⊕ · · · ⊕ Vr is adirect sum of irreducible representations.

By induction, it suffices to prove the following:

Theorem (Maschke’s theorem). Let G be a finite group, and ρ : G→ GL(V )a representation over a finite-dimensional vector space V over a field F withcharF = 0. If W is a G-subspace of V , then there exists a G-subspace U of Vsuch that V = W ⊕ U .

We will prove this many times.

Proof. From linear algebra, we know W has a complementary subspace. LetW ′ be any vector subspace complement of W in V , i.e. V = W ⊕W ′ as vectorspaces.

Let q : V →W be the projection of V onto W along W ′, i.e. if v = w + w′

with w ∈W,w′ ∈W ′, then q(v) = w.The clever bit is to take this q and tweak it a little bit. Define

q : v 7→ 1

|G|∑g∈G

ρ(g)q(ρ(g−1)v).

This is in some sense an averaging operator, averaging over what ρ(g) does. Herewe need the field to have characteristic zero such that 1

|G| is well-defined. In fact,

this theorem holds as long as charF - |G|.For simplicity of expression, we drop the ρ’s, and simply write

q : v 7→ 1

|G|∑g∈G

gq(g−1v).

We first claim that q has image in W . This is true since for v ∈ V , q(g−1v) ∈W ,and gW ≤W . So this is a little bit like a projection.

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Next, we claim that for w ∈ W , we have q(w) = w. This follows from thefact that q itself fixes W . Since W is G-invariant, we have g−1w ∈ W for allw ∈W . So we get

q(w) =1

|G|∑g∈G

gq(g−1w) =1

|G|∑g∈G

gg−1w =1

|G|∑g∈G

w = w.

Putting these together, this tells us q is a projection onto W .Finally, we claim that for h ∈ G, we have hq(v) = q(hv), i.e. it is invariant

under the G-action. This follows easily from definition:

hq(v) = h1

|G|∑g∈G

gq(g−1v)

=1

|G|∑g∈G

hgq(g−1v)

=1

|G|∑g∈G

(hg)q((hg)−1hv)

We now put g′ = hg. Since h is invertible, summing over all g is the same assumming over all g′. So we get

=1

|G|∑g′∈G

g′q(g′−1(hv))

= q(hv).

We are pretty much done. We finally show that ker q is G-invariant. If v ∈ ker qand h ∈ G, then q(hv) = hq(v) = 0. So hv ∈ ker q.

ThusV = im q ⊕ ker q = W ⊕ ker q

is a G-subspace decomposition.

The crux of the whole proof is the definition of q. Once we have that,everything else follows easily.

Yet, for the whole proof to work, we need 1|G| to exist, which in particular

means G must be a finite group. There is no obvious way to generalize this toinfinite groups. So let’s try a different proof.

The second proof uses inner products, and hence we must take F = C. Thiscan be generalized to infinite compact groups, as we will later see.

Recall the definition of an inner product:

Definition (Hermitian inner product). For V a complex space, 〈 · , · 〉 is aHermitian inner product if

(i) 〈v,w〉 = 〈w,v〉 (Hermitian)

(ii) 〈v, λ1w1 + λ2w2〉 = λ1〈v,w1〉+ λ2〈v,w2〉 (sesquilinear)

(iii) 〈v,v〉 > 0 if v 6= 0 (positive definite)

Definition (G-invariant inner product). An inner product 〈 · , · 〉 is in additionG-invariant if

〈gv, gw〉 = 〈v,w〉.

13


Proposition. Let W be G-invariant subspace of V , and V have a G-invariantinner product. Then W⊥ is also G-invariant.

Proof. To prove this, we have to show that for all v ∈ W⊥, g ∈ G, we havegv ∈W⊥.

This is not hard. We know v ∈W⊥ if and only if 〈v,w〉 = 0 for all w ∈W .Thus, using the definition of G-invariance, for v ∈W⊥, we know

〈gv, gw〉 = 0

for all g ∈ G,w ∈W .Thus for all w′ ∈ W , pick w = g−1w′ ∈ W , and this shows 〈gv,w′〉 = 0.

Hence gv ∈W⊥.

Hence if there is a G-invariant inner product on any complex G-space V ,then we get another proof of Maschke’s theorem.

Theorem (Weyl’s unitary trick). Let ρ be a complex representation of a finitegroup G on the complex vector space V . Then there is a G-invariant Hermitianinner product on V .

Recall that the unitary group is defined by

U(V ) = {f ∈ GL(V ) : 〈f(u), f(v)〉 = 〈u,v〉 for all u,v ∈ V }= {A ∈ GLn(C) : AA† = I}= U(n).

Then we have an easy corollary:

Corollary. Every finite subgroup of GLn(C) is conjugate to a subgroup of U(n).

Proof. We start by defining an arbitrary inner product on V : take a basise1, · · · , en. Define (ei, ej) = δij , and extend it sesquilinearly. Define a new innerproduct

〈v,w〉 =1

|G|∑g∈G

(gv, gw).

We now check this is sesquilinear, positive-definite and G-invariant. Sesquilin-earity and positive-definiteness are easy. So we just check G-invariance: wehave

〈hv, hw〉 =1

|G|∑g∈G

((gh)v, (gh)w)

=1

|G|∑g′∈G

(g′v, g′w)

= 〈v,w〉.

Note that this trick also works for real representations.Again, we had to take the inverse 1

|G| . To generalize this to compact groups,

we will later replace the sum by an integral, and 1|G| by a volume element. This is

fine since (g′v, g′w) is a complex number and we know how to integrate complexnumbers. This cannot be easily done in the case of q.

14


Recall we defined the group algebra of G to be the F-vector space

FG = 〈eg : g ∈ G〉,

i.e. its basis is in one-to-one correspondence with the elements of G. There is alinear G-action defined in the obvious way: for h ∈ G, we define

h∑g

ageg =∑g

agehg =∑g′

ah−1g′eg′ .

This gives a representation of G.

Definition (Regular representation and regular module). The regular represen-tation of a group G, written ρreg, is the natural action of G on FG. FG is calledthe regular module.

It is a nice faithful representation of dimension |G|. Far more importantly, itturns out that every irreducible representation of G is a subrepresentation ofthe regular representation.

Proposition. Let ρ be an irreducible representation of the finite group G overa field of characteristic 0. Then ρ is isomorphic to a subrepresentation of ρreg.

This will also follow from a more general result using character theory later.

Proof. Take ρ : G → GL(V ) be irreducible, and pick our favorite 0 6= v ∈ V .Now define θ : FG→ V by ∑

g

ageg 7→∑

ag(gv).

It is not hard to see this is a G-homomorphism. We are now going to exploit thefact that V is irreducible. Thus, since im θ is a G-subspace of V and non-zero,we must have im θ = V . Also, ker θ is a G-subspace of FG. Now let W bethe G-complement of ker θ in FG, which exists by Maschke’s theorem. ThenW ≤ FG is a G-subspace and

FG = ker θ ⊕W.

Then the isomorphism theorem gives

W ∼= FG/ ker θ ∼= im θ = V.

More generally, G doesn’t have to just act on the vector space generated byitself. If G acts on any set, we can take that space and create a space acted onby G.

Definition (Permutation representation). Let F be a field, and let G act on aset X. Let FX = 〈ex : x ∈ X〉 with a G-action given by

g∑x

axex =∑x

axegx.

So we have a G-space on FX. The representation G → GL(FX) is the corre-sponding permutation representation.

15

4 Schur’s lemma II Representation Theory

4 Schur’s lemma

The topic of this chapter is Schur’s lemma, an easy yet extremely useful lemmain representation theory.

Theorem (Schur’s lemma).

(i) Assume V and W are irreducible G-spaces over a field F. Then anyG-homomorphism θ : V →W is either zero or an isomorphism.

(ii) If F is algebraically closed, and V is an irreducible G-space, then anyG-endomorphism V → V is a scalar multiple of the identity map ιV .

Proof.

(i) Let θ : V →W be a G-homomorphism between irreducibles. Then ker θ isa G-subspace of V , and since V is irreducible, either ker θ = 0 or ker θ = V .Similarly, im θ is a G-subspace of W , and as W is irreducible, we musthave im θ = 0 or im θ = W . Hence either ker θ = V , in which case θ = 0,or ker θ = 0 and im θ = W , i.e. θ is a bijection.

(ii) Since F is algebraically closed, θ has an eigenvalue λ. Then θ − λιV is asingular G-endomorphism of V . So by (i), it must be the zero map. Soθ = λιV .

Recall that the F-space HomG(V,W ) is the space of all G-homomorphismsV →W . If V = W , we write EndG(V ) for the G-endomorphisms of V .

Corollary. If V,W are irreducible complex G-spaces, then

dimC HomG(V,W ) =

{1 V,W are G-isomorphic

0 otherwise

Proof. If V and W are not isomorphic, then the only possible map betweenthem is the zero map by Schur’s lemma.

Otherwise, suppose V ∼= W and let θ1, θ2 ∈ HomG(V,W ) be both non-zero. By Schur’s lemma, they are isomorphisms, and hence invertible. Soθ−1

2 θ1 ∈ EndG(V ). Thus θ−12 θ1 = λιV for some λ ∈ C. Thus θ1 = λθ2.

We have another less obvious corollary.

Corollary. If G is a finite group and has a faithful complex irreducible repre-sentation, then its center Z(G) is cyclic.

This is a useful result — it allows us transfer our representation-theoreticknowledge (the existence of a faithful complex irreducible representation) togroup theoretic knowledge (center of group being cyclic). This will becomeincreasingly common in the future, and is a good thing since representations areeasy and groups are hard.

The converse, however, is not true. For this, see example sheet 1, question10.

16


Proof. Let ρ : G→ GL(V ) be a faithful irreducible complex representation. Letz ∈ Z(G). So zg = gz for all g ∈ G. Hence φz : v 7→ zv is a G-endomorphismon V . Hence by Schur’s lemma, it is multiplication by a scalar µz, say. Thuszv = µzv for all v ∈ V .

Then the map

σ : Z(G)→ C×

z 7→ µg

is a representation of Z(G). Since ρ is faithful, so is σ. So Z(G) = {µz : z ∈Z(G)} is isomorphic to a finite subgroup of C×, hence cyclic.

Corollary. The irreducible complex representations of a finite abelian group Gare all 1-dimensional.

Proof. We can use the fact that commuting diagonalizable matrices are simulta-neously diagonalizable. Thus for every irreducible V , we can pick some v ∈ Vthat is an eigenvector for each g ∈ G. Thus 〈v〉 is a G-subspace. As V isirreducible, we must have V = 〈v〉.

Alternatively, we can prove this in a representation-theoretic way. Let V bean irreducible complex representation. For each g ∈ G, the map

θg : V → V

v 7→ gv

is a G-endomorphism of V , since it commutes with the other group elements.Since V is irreducible, θg = λgιV for some λg ∈ C. Thus

gv = λgv

for any g. As V is irreducible, we must have V = 〈v〉.

Note that this result fails over R. For example, C3 has a two irreducible realrepresentations, one of dimension 1 and one of dimension 2.

We can do something else. Recall that every finite abelian group G isomorphicto a product of abelian groups. In fact, it can be written as a product of Cpα

for various primes p and α ≥ 1, and the factors are uniquely determined up toorder.

This you already know from IB Groups Rings and Modules. You might beborn knowing it — it’s such a fundamental fact of nature.

Proposition. The finite abelian group G = Cn1 × · · · × Cnr has precisely |G|irreducible representations over C.

This is not a coincidence. We will later show that the number of irreduciblerepresentations is the number of conjugacy classes of the group. In abeliangroups, each conjugacy class is just a singleton, and hence this result.

Proof. WriteG = 〈x1〉 × · · · × 〈xr〉,

where |xj | = nj . Any irreducible representation ρ must be one-dimensional. Sowe have

ρ : G→ C×.

17


Let ρ(1, · · · , xj , · · · , 1) = λj . Then since ρ is a homomorphism, we must haveλnjj = 1. Therefore λj is an njth root of unity.

Now the values (λ1, · · · , λr) determine ρ completely, namely

ρ(xj11 , · · · , xjrr ) = λj11 · · ·λjrr .

Also, whenever λi is an nith root of unity for each i, then the above formulagives a well-defined representation. So there is a one-to-one correspondenceρ↔ (λ1, · · · , λr), with λ

njj = 1.

Since for each j, there are nj many njth roots of unity, it follows that thereare |G| = n1 · · ·nr many choices of the λi. Thus the proposition.

Example.

(i) Consider G = C4 = 〈x〉. The four 1-dimensional irreducible representationsare given by

1 x x2 x3

ρ1 1 1 1 1ρ2 1 i −1 −iρ2 1 −1 1 −1ρ2 1 −i −1 i

(ii) Consider the Klein four group G = VR = 〈x1〉 × 〈x2〉. The irreduciblerepresentations are

1 x1 x2 x1x2

ρ1 1 1 1 1ρ2 1 1 −1 −1ρ2 1 −1 1 −1ρ2 1 −1 −1 1

These are also known as character tables, and we will spend quite a lot oftime computing these for non-abelian groups.

Note that there is no “natural” one-to-one correspondence between theelements of G and the representations of G (for G finite-abelian). If we choosean isomorphism G ∼= Cn1

× · · ·Cnr , then we can identify the two sets, but itdepends on the choice of the isomorphism (while the decomposition is unique,we can pick a different generator of, say, Cn1

and get a different isomorphism tothe same decomposition).

Isotypical decompositions

Recall that we proved we can decompose any G-representation into a sum ofirreducible representations. Is this decomposition unique? If it isn’t, can we sayanything about, say, the size of the irreducible representations, or the number offactors in the decomposition?

We know any diagonalizable endomorphism α : V → V for a vector space Vgives us a vector space decomposition

V =⊕λ

V (λ),

18


whereV (λ) = {v ∈ V : α(v) = λv}.

This is canonical in that it depends on α alone, and nothing else.If V is moreover a G-representation, how does this tie in to the decomposition

of V into the irreducible representations?Let’s do an example.

Example. Consider G = D6∼= S3 = 〈r, s : r3 = s2 = 1, rs = sr−1〉. We have

previously seen that each irreducible representation has dimension at most 2.We spot at least three irreducible representations:

1 triad r 7→ 1 s 7→ 1S sign r 7→ 1 s 7→ −1W 2-dimensional

The last representation is the action of D6 on R2 in the natural way, i.e. therotations and reflections of the plane that corresponds to the symmetries of thetriangle. It is helpful to view this as a complex representation in order to makethe matrix look nice. The 2-dimensional representation (ρ,W ) is defined byW = C2, where r and s act on W as

ρ(r) =

(ω 00 ω2

), ρ(s) =

(0 11 0

),

and ω = e2πi/3 is a third root of unity. We will now show that these are indeedall the irreducible representations, by decomposing any representation into sumof these.

So let’s decompose an arbitrary representation. Let (ρ′, V ) be any complexrepresentation of G. Since ρ′(r) has order 3, it is diagonalizable has eigenvalues1, ω, ω2. We diagonalize ρ′(r) and then V splits as a vector space into theeigenspaces

V = V (1)⊕ V (ω)⊕ V (ω2).

Since srs−1 = r−1, we know ρ′(s) preserves V (1) and interchanges V (ω) andV (ω2).

Now we decompose V (1) into ρ′(s) eigenspaces, with eigenvalues ±1. Since rhas to act trivially on these eigenspaces, V (1) splits into sums of copies of theirreducible representations 1 and S.

For the remaining mess, choose a basis v1, · · · ,vn of V (ω), and let v′j =

ρ′(s)vj . Then ρ′(s) acts on the two-dimensional space 〈vj ,v′j〉 as

(0 11 0

), while

ρ′(r) acts as

(ω 00 ω2

). This means V (ω)⊕V (ω2) decomposes into many copies

of W.

We did this for D6 by brute force. How do we generalize this? We first havethe following lemma:

Lemma. Let V, V1, V2 be G-vector spaces over F. Then

(i) HomG(V, V1 ⊕ V2) ∼= HomG(V, V1)⊕HomG(V, V2)

(ii) HomG(V1 ⊕ V2, V ) ∼= HomG(V1, V )⊕HomG(V2, V ).

19


Proof. The proof is to write down the obvious homomorphisms and inverses.Define the projection map

πi : V1 ⊕ V2 → Vi,

which is the G-linear projection onto Vi.Then we can define the G-homomorphism

HomG(V, V1 ⊕ V2) 7→ HomG(V, V1)⊕HomG(V, V2)

ϕ 7→ (π1ϕ, π2ϕ).

Then the map (ψ1, ψ2) 7→ ψ1 + ψ2 is an inverse.For the second part, we have the homomorphism ϕ 7→ (ϕ|V1 , ϕ|V2) with

inverse (ψ1, ψ2) 7→ ψ1π1 + ψ2π2.

Lemma. Let F be an algebraically closed field, and V be a representation of G.Suppose V =

⊕ni=1 Vi is its decomposition into irreducible components. Then

for each irreducible representation S of G,

|{j : Vj ∼= S}| = dim HomG(S, V ).

This tells us we can count the multiplicity of S in V by looking at thehomomorphisms.

Proof. We induct on n. If n = 0, then this is a trivial space. If n = 1, then Vitself is irreducible, and by Schur’s lemma, dim HomG(S, V ) = 1 if V = S, 0otherwise. Otherwise, for n > 1, we have

V =

(n−1⊕i=1

Vi

)⊕ Vn.

By the previous lemma, we know

dim homG

(S,

(n−1⊕i=1

Vi

)⊕ Vn

)= dim HomG

(S,

n−1⊕i=1

Vi

)+ dim homG(S, Vn).

The result then follows by induction.

Definition (Canonical decomposition/decomposition into isotypical compo-nents). A decomposition of V as

⊕Wj , where each Wj is (isomorphic to) nj

copies of the irreducible Sj (with Sj 6∼= Si for i 6= j) is the canonical decompositionor decomposition into isotypical components.

For an algebraically closed field F, we know we must have

nj = dim HomG(Sj , V ),

and hence this decomposition is well-defined.We’ve finally all the introductory stuff. The course now begins.

20

5 Character theory II Representation Theory

5 Character theory

In topology, we want to classify spaces. To do so, we come up with invariants ofspaces, like the number of holes. Then we know that a torus is not the sameas a sphere. Here, we want to attach invariants to a representation ρ of a finitegroup G on V .

One thing we might want to do is to look at the matrix coefficients of ρ(g).However, this is bad, since this is highly highly highly basis dependent. It is nota true invariant. We need to do something better than that.

Let F = C, and G be a finite group. Let ρ = ρV : G → GL(V ) be arepresentation of G. The clue is to look at the characteristic polynomial of thematrix. The coefficients are functions of the eigenvalues — on one extreme, thedeterminant is the product of all eigenvalues; on the other extreme, and thetrace is the sum of all of them. Surprisingly, it is the trace that works. We don’thave to bother ourselves with the other coefficients.

Definition (Character). The character of a representation ρ : G → GL(V ),written χρ = χv = χ, is defined as

χ(g) = tr ρ(g).

We say ρ affords the character χ.Alternatively, the character is trR(g), where R(g) is any matrix representing

ρ(g) with respect to any basis.

Definition (Degree of character). The degree of χv is dimV .

Thus, χ is a function G→ C.

Definition (Linear character). We say χ is linear if dimV = 1, in which caseχ is a homomorphism G→ C× = GL1(C).

Various properties of representations are inherited by characters.

Definition (Irreducible character). A character χ is irreducible if ρ is irreducible.

Definition (Faithful character). A character χ is faithful if ρ is faithful.

Definition (Trivial/principal character). A character χ is trivial or principal ifρ is the trivial representation. We write χ = 1G.

χ is a complete invariant in the sense that it determines ρ up to isomorphism.This is staggering. We reduce the whole matrix into a single number — thetrace, and yet we have not lost any information at all! We will prove this later,after we have developed some useful properties of characters.

Theorem.

(i) χV (1) = dimV .

(ii) χV is a class function, namely it is conjugation invariant, i.e.

χV (hgh−1) = χV (g)

for all g, h ∈ G. Thus χV is constant on conjugacy classes.

21


(iii) χV (g−1) = χV (g).

(iv) For two representations V,W , we have

χV⊕W = χV + χW .

These results, despite being rather easy to prove, are very useful, since theysave us a lot of work when computing the characters of representations.

Proof.

(i) Obvious since ρV (1) = idV .

(ii) Let Rg be the matrix representing g. Then

χ(hgh−1) = tr(RhRgR−1h ) = tr(Rg) = χ(g),

as we know from linear algebra.

(iii) Since g ∈ G has finite order, we know ρ(g) is represented by a diagonalmatrix

Rg =

λ1

. . .

λn

,

and χ(g) =∑λi. Now g−1 is represented by

Rg−1 =

λ−11

. . .

λ−1n

,

Noting that each λi is an nth root of unity, hence |λi| = 1, we know

χ(g−1) =∑

λ−1i =

∑λi =

∑λi = χ(g).

(iv) Suppose V = V1 ⊕ V2, with ρ : G→ GL(V ) splitting into ρi : G→ GL(Vi).Pick a basis Bi for Vi, and let B = B1 ∪ B2. Then with respect to B, wehave

[ρ(g)]B =

([ρ1(g)]B1 0

0 [ρ2(g)]B2

).

So χ(g) = tr(ρ(g)) = tr(ρ1(g)) + tr(ρ2(g)) = χ1(g) + χ2(g).

We will see later that if we take characters χ1, χ2 of G, then χ1χ2 is also acharacter of G. This uses the notion of tensor products, which we will do later.

Lemma. Let ρ : G→ GL(V ) be a complex representation affording the characterχ. Then

|χ(g)| ≤ χ(1),

with equality if and only if ρ(g) = λI for some λ ∈ C, a root of unity. Moreover,χ(g) = χ(1) if and only if g ∈ ker ρ.

22


Proof. Fix g, and pick a basis of eigenvectors of ρ(g). Then the matrix of ρ(g)is diagonal, say

ρ(g) =

λ1

. . .

λn

,

Hence|χ(g)| =

∣∣∣∑λi

∣∣∣ ≤∑ |λi| =∑

1 = dimV = χ(1).

In the triangle inequality, we have equality if and only if all the λi’s are equal,to λ, say. So ρ(g) = λI. Since all the λi’s are roots of unity, so is λ.

And, if χ(g) = χ(1), then since ρ(g) = λI, taking the trace gives χ(g) = λχ(1).So λ = 1, i.e. ρ(g) = I. So g ∈ ker ρ.

The following lemma allows us to generate new characters from old.

Lemma.

(i) If χ is a complex (irreducible) character of G, then so is χ.

(ii) If χ is a complex (irreducible) character of G, then so is εχ for any linear(1-dimensional) character ε.

Proof.

(i) If R : G→ GLn(C) is a complex matrix representation, then so is R : G→GLn(C), where g 7→ R(g). Then the character of R is χ

(ii) Similarly, R′ : g 7→ ε(g)R(g) for g ∈ G is a representation with characterεχ.

It is left as an exercise for the reader to check the details.

We now have to justify why we care about characters. We said it was acomplete invariant, as in two representations are isomorphic if and only if theyhave the same character. Before we prove this, we first need some definitions

Definition (Space of class functions). Define the complex space of class functionsof G to be

C(G) = {f : G→ C : f(hgh−1) = f(g) for all h, g ∈ G}.

This is a vector space by f1 + f2 : g 7→ f1(g) + f2(g) and λf : g 7→ λf(g).

Definition (Class number). The class number k = k(G) is the number ofconjugacy classes of G.

We can list the conjugacy classes as C1, · · · , Ck. wlog, we let C1 = {1}. Wechoose g1, g2, · · · , gk to be representatives of the conjugacy classes. Note alsothat dim C(G) = k, since the characteristic function δj of the classes form a basiswhere

δj(g) =

{1 g ∈ Cj0 g 6∈ Cj .

23


We define a Hermitian inner product on C(G) by

〈f, f ′〉 =1

|G|∑g∈G

f(g)f ′(g)

=1

|G|

k∑j=1

|Cj |f(gj)f′(gj)

By the orbit-stabilizer theorem, we can write this as

=

k∑j=1

1

|CG(gj)|f(gj)f

′(gj).

In particular, for characters,

〈χ, χ′〉 =

k∑j=1

1

|CG(gj)|χ(g−1

j )χ′(gj).

It should be clear (especially using the original formula) that 〈χ, χ′〉 = 〈χ′, χ〉.So when restricted to characters, this is a real symmetric form.

The main result is the following theorem:

Theorem (Completeness of characters). The complex irreducible characters ofG form an orthonormal basis of C(G), namely

(i) If ρ : G → GL(V ) and ρ′ : G → GL(V ′) are two complex irreduciblerepresentations affording characters χ, χ′ respectively, then

〈χ, χ′〉 =

{1 if ρ and ρ′ are isomorphic representations

0 otherwise

This is the (row) orthogonality of characters.

(ii) Each class function of G can be expressed as a linear combination ofirreducible characters of G.

Proof is deferred. We first look at some corollaries.

Corollary. Complex representations of finite groups are characterised by theircharacters.

Proof. Let ρ : G → GL(V ) afford the character χ. We know we can writeρ = m1ρ1 ⊕ · · · ⊕mkρk, where ρ1, · · · , ρk are (distinct) irreducible and mj ≥ 0are the multiplicities. Then we have

χ = m1χ1 + · · ·+mkχk,

where χj is afforded by ρj . Then by orthogonality, we know

mj = 〈χ, χj〉.

So we can obtain the multiplicity of each ρj in ρ just by looking at the innerproducts of the characters.

24


This is not true for infinite groups. For example, if G = Z, then therepresentations

1 7→(

1 00 1

), 1 7→

(1 10 1

)are non-isomorphic, but have the same character 2.

Corollary (Irreducibility criterion). If ρ : G → GL(V ) is a complex repre-sentation of G affording the character χ, then ρ is irreducible if and only if〈χ, χ〉 = 1.

Proof. If ρ is irreducible, then orthogonality says 〈χ, χ〉 = 1. For the otherdirection, suppose 〈χ, χ〉 = 1. We use complete reducibility to get

χ =∑

mjχj ,

with χj irreducible, and mj ≥ 0 the multiplicities. Then by orthogonality, weget

〈χ, χ〉 =∑

m2j .

But 〈χ, χ〉 = 1. So exactly one of mj is 1, while the others are all zero, andχ = χj . So χ is irreducible.

Theorem. Let ρ1, · · · , ρk be the irreducible complex representations of G, andlet their dimensions be n1, · · · , nk. Then

|G| =∑

n2i .

Recall that for abelian groups, each irreducible character has dimension 1,and there are |G| representations. So this is trivially satisfied.

Proof. Recall that ρreg : G→ GL(CG), given by G acting on itself by multipli-cation, is the regular representation of G of dimension |G|. Let its character beπreg, the regular character of G.

First note that we have πreg(1) = |G|, and πreg(h) = 0 if h 6= 1. The firstpart is obvious, and the second is easy to show, since we have only 0s along thediagonal.

Next, we decompose πreg as

πreg =∑

ajχj ,

We now want to find aj . We have

aj = 〈πreg, χj〉 =1

|G|∑g∈G

πreg(g)χj(g) =1

|G|· |G|χj(1) = χj(1).

Then we get

|G| = πreg(1) =∑

ajχj(1) =∑

χj(1)2 =∑

n2j .

From this proof, we also see that each irreducible representation is a subrep-resentation of the regular representation.

25


Corollary. The number of irreducible characters of G (up to equivalence) is k,the number of conjugacy classes.

Proof. The irreducible characters and the characteristic functions of the conju-gacy classes are both bases of C(G).

Corollary. Two elements g1, g2 are conjugate if and only if χ(g1) = χ(g2) forall irreducible characters χ of G.

Proof. If g1, g2 are conjugate, since characters are class functions, we must haveχ(g1) = χ(g2).

For the other direction, let δ be the characteristic function of the class of g1.Then since δ is a class function, we can write

δ =∑

mjχj ,

where χj are the irreducible characters of G. Then

δ(g2) =∑

mjχj(g2) =∑

mjχj(g1) = δ(g1) = 1.

So g2 is in the same conjugacy class as g1.

Before we move on to the proof of the completeness of characters, we firstmake a few more definitions:

Definition (Character table). The character table of G is the k × k matrixX = (χi(gj)), where 1 = χ1, χ2, · · · , χk are the irreducible characters of G, andC1 = {1}, C2, · · · , Ck are the conjugacy classes with gj ∈ Cj .

We seldom think of this as a matrix, but as an actual table, as we will see inthe table below.

We will spend the next few lectures coming up with techniques to computethese character tables.

Example. Consider G = D6∼= S3 = 〈r, s | r3 = s2 = 1, srs−1 = r−1〉. As in all

computations, the first thing to do is to compute the conjugacy classes. This isnot too hard. We have

C1 = {1}, C2 = {s, sr, sr2}, C2 = {r, r−1}.

Alternatively, we can view this as S3 and use the fact that two elements areconjugate if and only if they have the same cycle types. We have found thethree representations: 1, the trivial representation; S, the sign; and W, thetwo-dimensional representation. In W , the reflections srj acts by matrix witheigenvalues ±1. So the sum of eigenvalues is 0, hence χw(sr2) = 0. It also alsonot hard to see

rm 7→(

cos 2mπ3 − sin 2mπ

3sin 2mπ

3 cos 2mπ3

).

So χw(rm) = −1.Fortunately, after developing some theory, we will not need to find all the

irreducible representations in order to compute the character table.

26


1 C2 C31 1 1 1S 1 −1 1χw 2 0 −1

We see that the sum of the squares of the first column is 12 + 12 + 22 = 6 = |D6|,as expected. We can also check that W is genuinely an irreducible representation.Noting that the centralizers of C1, C2 and C3 are of sizes 6, 2, 3. So the innerproduct is

〈χW , χW 〉 =22

6+

02

2+

(−1)2

3= 1,

as expected.

So we now need to prove orthogonality.

27

6 Proof of orthogonality II Representation Theory

6 Proof of orthogonality

We will do the proof in parts.

Theorem (Row orthogonality relations). If ρ : G → GL(V ) and ρ′ : G →GL(V ′) are two complex irreducible representations affording characters χ, χ′

respectively, then

〈χ, χ′〉 =

{1 if ρ and ρ′ are isomorphic representations

0 otherwise.

Proof. We fix a basis of V and of V ′. Write R(g), R′(g) for the matrices of ρ(g)and ρ′(g) with respect to these bases respectively. Then by definition, we have

〈χ′, χ〉 =1

|G|∑g∈G

χ′(g−1)χ(g)

=1

|G|∑g∈G

∑1≤i≤n′1≤j≤n

R′(g−1)iiR(g)jj .

For any linear map ϕ : V → V ′, we define a new map by averaging by ρ′ and ρ.

ϕ : V → V ′

v 7→ 1

|G|∑

ρ′(g−1)ϕρ(g)v

We first check ϕ is a G-homomorphism — if h ∈ G, we need to show

ρ′(h−1)ϕρ(h)(v) = ϕ(v).

We have

ρ′(h−1)ϕρ(h)(v) =1

|G|∑g∈G

ρ′((gh)−1)ϕρ(gh)v

=1

|G|∑g′∈G

ρ′(g′−1)ϕρ(g′)v

= ϕ(v).

(i) Now we first consider the case where ρ, ρ′ is not isomorphic. Then bySchur’s lemma, we must have ϕ = 0 for any linear ϕ : V → V ′.

We now pick a very nice ϕ, where everything disappears. We let ϕ = εαβ ,the operator having matrix Eαβ with entries 0 everywhere except 1 in the(α, β) position.

Then εαβ = 0. So for each i, j, we have

1

|G|∑g∈G

(R′(g−1)EαβR(g))ij = 0.

Using our choice of εαβ , we get

1

|G|∑g∈G

R′(g−1)iαR(g)βj = 0

28


for all i, j. We now pick α = i and β = j. Then

1

|G|∑g∈G

R′(g−1)iiR(g)jj = 0.

We can sum this thing over all i and j to get that 〈χ′, χ〉 = 0.

(ii) Now suppose ρ, ρ′ are isomorphic. So we might as well take χ = χ′, V = V ′

and ρ = ρ′. If ϕ : V → V is linear, then ϕ ∈ EndG(V ).

We first claim that tr ϕ = trϕ. To see this, we have

tr ϕ =1

|G|∑g∈G

tr(ρ(g−1)ϕρ(g)) =1

|G|∑g∈G

trϕ = trϕ,

using the fact that traces don’t see conjugacy (and ρ(g−1) = ρ(g)−1 sinceρ is a group homomorphism).

By Schur’s lemma, we know ϕ = λιv for some λ ∈ C (which depends onϕ). Then if n = dimV , then

λ =1

ntrϕ.

Let ϕ = εαβ . Then trϕ = δαβ . Hence

εαβ =1

|G|∑g

ρ(g−1)εαβρ(g) =1

nδαβι.

In terms of matrices, we take the (i, j)th entry to get

1

|G|∑

R(g−1)iαR(g)βj =1

nδαβδij .

We now put α = i and β = j. Then we are left with

1

|G|∑g

R(g−1)iiR(g)jj =1

nδij .

Summing over all i and j, we get 〈χ, χ〉 = 1.

After learning about tensor products and duals later on, one can provide ashorter proof of this result as follows:

Alternative proof. Consider two representation spaces V and W . Then

〈χW , χV 〉 =1

|G|∑

χW (g)χV (g) =1

|G|∑

χV⊗W∗(g).

We notice that there is a natural isomorphism V ⊗ W ∗ ∼= Hom(W,V ), andthe action of g on this space is by conjugation. Thus, a G-invariant elementis just a G-homomorphism W → V . Thus, we can decompose Hom(V,W ) =HomG(V,W )⊕U for some G-space U , and U has no G-invariant element. Hencein the decomposition of Hom(V,W ) into irreducibles, we know there are exactly

29


dim HomG(V,W ) copies of the trivial representation. By Schur’s lemma, thisnumber is 1 if V ∼= W , and 0 if V 6∼= W .

So it suffices to show that if χ is a non-trivial irreducible character, then∑g∈G

χ(g) = 0.

But if ρ affords χ, then any element in the image of∑g∈G ρ(g) is fixed by G.

By irreducibility, the image must be trivial. So∑g∈G ρ(g) = 0.

What we have done is show the orthogonality of the rows. There is a similarone for the columns:

Theorem (Column orthogonality relations). We have

k∑i=1

χi(gj)χi(g`) = δj`|CG(g`)|.

This is analogous to the fact that if a matrix has orthonormal rows, then italso has orthonormal columns. We get the extra factor of |CG(g`)| because ofthe way we count.

This has an easy corollary, which we have already shown previously usingthe regular representation:

Corollary.

|G| =k∑i=1

χ2i (1).

Proof of column orthogonality. Consider the character table X = (χi(gj)). Weknow

δij = 〈χi, χj〉 =∑`

1

|CG(g`)|χi(g`)χk(g`).

ThenXD−1XT = Ik×k,

where

D =

|CG(g1)| · · · 0...

. . ....

0 · · · |CG(gk)|

.

Since X is square, it follows that D−1XT is the inverse of X. So XTX = D,which is exactly the theorem.

The proof requires that X is square, i.e. there are k many irreducible repre-sentations. So we need to prove the last part of the completeness of characters.

Theorem. Each class function of G can be expressed as a linear combinationof irreducible characters of G.

30


Proof. We list all the irreducible characters χ1, · · · , χ` of G. Note that we don’tknow the number of irreducibles is k. This is essentially what we have to provehere. We now claim these generate C(G), the ring of class functions.

Now recall that C(G) has an inner product. So it suffices to show that theorthogonal complement to the span 〈χ1, · · · , χ`〉 in C(G) is trivial. To see this,assume f ∈ C(G) satisfies

〈f, χj〉 = 0

for all χj irreducible. We let ρ : G → GL(V ) be an irreducible representationaffording χ ∈ {χ1, · · · , χ`}. Then 〈f, χ〉 = 0.

Consider the function

ϕ =1

|G|∑g

f(g)ρ(g) : V → V.

For any h ∈ G, we can compute

ρ(h)−1ϕρ(h) =1

|G|∑g

f(g)ρ(h−1gh) =1

|G|∑g

f(h−1gh)ρ(h−1gh) = ϕ,

using the fact that f is a class function. So this is a G-homomorphism. So as ρis irreducible, Schur’s lemma says it must be of the form λιV for some λ ∈ C.

Now we take the trace of this thing. So we have

nλ = tr

(1

|G|∑g

f(g)ρ(g)

)=

1

|G|∑g

f(g)χ(g) = 〈f, χ〉 = 0.

So λ = 0, i.e.∑g f(g)ρ(g) = 0, the zero endomorphism on V . This is valid for

any irreducible representation, and hence for every representation, by completereducibility.

In particular, take ρ = ρreg, where ρreg(g) : e1 7→ eg for each g ∈ G. Hence∑f(g)ρreg(g) : e1 7→

∑g

f(g)eg.

Since this is zero, it follows that we must have∑f(g)eg = 0. Since the eg’s are

linearly independent, we must have f(g) = 0 for all g ∈ G, i.e. f = 0.

31

7 Permutation representations II Representation Theory

7 Permutation representations

In this chapter, we are going to study permutation representations in a bit moredetail, since we can find some explicit formulas for the character of them. Thisis particularly useful if we are dealing with the symmetric group.

Let G be a finite group acting on a finite set X = {x1, · · · , xn} (sometimesknown as a G-set). We define CX to be the complex vector space with basis{ex1

, · · · , exn}. More explicitly,

CX =

∑j

ajexj : aj ∈ C

.

There is a corresponding representation permutation

ρX : G→ GL(CX)

g 7→ ρ(g)

where ρ(g) : exj 7→ egxj , extended linearly. We say ρX is the representationcorresponding to the action of G on X.

This is a generalization of the group algebra — if we let G act on itself byleft multiplication, then this is in fact the group algebra.

The corresponding matrix representations ρX(g) with respect to the basis{ex}x∈X are permutation matrices — it is 0 everywhere except for one 1 in eachrow and column. In particular, (ρ(g))ij = 1 precisely when gxj = xi. So theonly non-zero diagonal entries in ρ(g) are those i such that xi is fixed by g. Thuswe have

Definition (Permutation character). The permutation character πX is

πX(g) = |fix(g)| = |{x ∈ X : gx = x}|.

Lemma. πX always contains the trivial character 1G (when decomposed in thebasis of irreducible characters). In particular, span{ex1 + · · ·+ exn} is a trivialG-subspace of CX, with G-invariant complement {

∑x axex :

∑ax = 0}.

Lemma. 〈πX , 1〉, which is the multiplicity of 1 in πX , is the number of orbitsof G on X.

Proof. We write X as the disjoint union of orbits, X = X1 ∪ · · · ∪ X`. Thenit is clear that the permutation representation on X is just the sum of thepermutation representations on the Xi, i.e.

πX = πX1 + · · ·+ πx` ,

where πXj is the permutation character of G on Xj . So to prove the lemma, itis enough to consider the case where the action is transitive, i.e. there is justone orbit.

So suppose G acts transitively on X. We want to show 〈πX , 1〉 = 1. By

32


definition, we have

〈πX , 1〉 =1

|G|∑g

πX(g)

=1

|G||{(g, x) ∈ G×X : gx = x}|

=1

|G|∑x∈X|Gx|,

where Gx is the stabilizer of x. By the orbit-stabilizer theorem, we have |Gx||X| =|G|. So we can write this as

=1

|G|∑x∈X

|G||X|

=1

|G|· |X| · |G|

|X|= 1.

So done.

Lemma. Let G act on the sets X1, X2. Then G acts on X1 ×X2 by

g(x1, x2) = (gx1, gx2).

Then the characterπX1×X2

= πX1πX2

,

and so〈πX1 , πX2〉 = number of orbits of G on X1 ×X2.

Proof. We know πX1×X2(g) is the number of pairs (x1, x2) ∈ X1 ×X2 fixed by

g. This is exactly the number of things in X1 fixed by g times the number ofthings in X2 fixed by g. So we have

πX1×X2(g) = πX1

(g)πX2(g).

Then using the fact that π1, π2 are real, we get

〈πX1, πX2

〉 =1

|G|∑g

πX1(g)πX2

(g)

=1

|G|∑g

πX1(g)πX2

(g)1G(g)

= 〈πX1πX2

, 1〉= 〈πX1×X2

, 1〉.

So the previous lemma gives the desired result.

Recall that a 2-transitive action is an action that is transitive on pairs, i.e.it can send any pair to any other pair, as long as the elements in the pair aredistinct (it can never send (x, x) to (x′, x′′) if x′ 6= x′′, since g(x, x) = (gx, gx)).More precisely,

33


Definition (2-transitive). Let G act on X, |X| > 2. Then G is 2-transitive onX if G has two orbits on X ×X, namely {(x, x) : x ∈ X} and {(x1, x2) : xi ∈X,x1 6= x2}.

Then we have the following lemma:

Lemma. Let G act on X, with |X| > 2. Then

πX = 1G + χ,

with χ irreducible if and only if G is 2-transitive on X.

Proof. We knowπX = m11G +m2χ2 + · · ·+m`χ`,

with 1G, χ2, · · · , χ` distinct irreducible characters and mi ∈ N are non-zero.Then by orthogonality,

〈πX , πX〉 =

j∑i=1

m2i .

Since 〈πX , πX〉 is the number of orbits of X ×X, we know G is 2-transitive onX if and only if ` = 2 and m1 = m2 = 1.

This is useful if we want to understand the symmetric group, since it isalways 2-transitive (modulo the silly case of S1).

Example. Sn acting on X = {1, · · · , n} is 2-transitive for n ≥ 2. So πX = 1+χ,with χ irreducible of degree n− 1 (since πX has degree n). This works similarlyfor An, whenever n ≥ 3.

This tells us we can find an irreducible character of Sn by finding πX , andthen subtracting 1 off.

Example. Consider G = S4. The conjugacy classes correspond to differentcycle types. We already know three characters — the trivial one, the sign, andπX − 1G.

1 3 8 6 61 (1 2)(3 4) (1 2 3) (1 2 3 4) (1 2)

trivial χ1 1 1 1 1 1sign χ2 1 1 1 −1 −1

πX − 1G χ3 3 −1 0 −1 1χ4

χ5

We know the product of a one-dimensional irreducible character and anothercharacter is another irreducible character, as shown on the first example sheet.So we get

1 3 8 6 61 (1 2)(3 4) (1 2 3) (1 2 3 4) (1 2)

trivial χ1 1 1 1 1 1sign χ2 1 1 1 −1 −1

πX − 1G χ3 3 −1 0 −1 1χ3χ2 χ4 3 −1 0 1 −1

χ5

34


For the last representation we can find the dimension by computing 24− (12 +12 + 32 + 32) = 22. So it has dimension 2. To obtain the whole of χ5, we canuse column orthogonality — for example, we let the entry in the second columnbe x. Then column orthogonality says 1 + 1− 3− 3 + 2x = 0 . So x = 2. In theend, we find

1 3 8 6 61 (1 2)(3 4) (1 2 3) (1 2 3 4) (1 2)

trivial χ1 1 1 1 1 1sign χ2 1 1 1 −1 −1

πX − 1G χ3 3 −1 0 −1 1χ3χ2 χ4 3 −1 0 1 −1

χ5 2 2 −1 0 0

Alternatively, we have previously shown that χreg = χ1 + χ2 + 3χ3 + 3χ4 + 2χ5.By computing χreg directly, we can find χ5.

Finally, we can obtain χ5 by observing S4/V4∼= S3, and “lifting” characters.

We will do this in the next chapter. In fact, this χ5 is the degree-2 irreduciblerepresentation of S3

∼= D6 lifted up.

We can also do similar things with the alternating group. The issue of courseis that the conjugacy classes of the symmetric group may split when we move tothe alternating group.

Suppose g ∈ An. Then

|CSn(g)| = |Sn : CSn(g)||CAn(g)| = |An : CAn(g)|.

We know CSn(g) ⊇ CAn(g), but we need not have equality, since elements neededto conjugate g to h ∈ CSn(g) might not be in An. For example, considerσ = (1 2 3) ∈ A3. We have CA3(σ) = {σ} and CS3(σ) = {σ, σ−1}.

We know An is an index 2 subgroup of Sn. So CAn(g) ⊆ CSn(g) either hasindex 2 or 1. If the index is 1, then |CAn | = 1

2 |CSn |. Otherwise, the sizes areequal.

A useful criterion for determining which case happens is the following:

Lemma. Let g ∈ An, n > 1. If g commutes with some odd permutation in Sn,then CSn(g) = CAn(g). Otherwise, CSn splits into two conjugacy classes in An ofequal size.

This is quite useful for doing the examples on the example sheet, but we willnot go through any more examples here.

35

8 Normal subgroups and lifting II Representation Theory

8 Normal subgroups and lifting

The idea of this section is that there is some correspondence between theirreducible characters of G/N (with N C G) and those of G itself. In mostcases, G/N is a simpler group, and hence we can use this correspondence to findirreducible characters of G.

The main result is the following lemma:

Lemma. Let N CG. Let ρ : G/N → GL(V ) be a representation of G/N . Thenthe composition

ρ : G G/N GL(V )natural ρ

is a representation of G, where ρ(g) = ρ(gN). Moreover,

(i) ρ is irreducible if and only if ρ is irreducible.

(ii) The corresponding characters satisfy χ(g) = χ(gN).

(iii) degχ = deg χ.

(iv) The lifting operation χ 7→ χ is a bijection

{irreducibles of G/N} ←→ {irreducibles of G with N in their kernel}.

We say χ lifts to χ.

Proof. Since a representation of G is just a homomorphism G → GL(V ), andthe composition of homomorphisms is a homomorphisms, it follows immediatelythat ρ as defined in the lemma is a representation.

(i) We can compute

〈χ, χ〉 =1

|G|∑g∈G

χ(g)χ(g)

=1

|G|∑

gN∈G/N

∑k∈N

χ(gk)χ(gk)

=1

|G|∑

gN∈G/N

∑k∈N

χ(gN)χ(gN)

=1

|G|∑

gN∈G/N

|N |χ(gN)χ(gN)

=1

|G/N |∑

gN∈G/N

χ(gN)χ(gN)

= 〈χ, χ〉.

So 〈χ, χ〉 = 1 if and only if 〈χ, χ〉 = 1. So ρ is irreducible if and only if ρis irreducible.

(ii) We can directly compute

χ(g) = tr ρ(g) = tr(ρ(gN)) = χ(gN)

for all g ∈ G.

36


(iii) To see that χ and χ have the same degree, we just notice that

degχ = χ(1) = χ(N) = deg χ.

Alternatively, to show they have the same dimension, just note that ρ andρ map to the general linear group of the same vector space.

(iv) To show this is a bijection, suppose χ is a character of G/N and χ is itslift to G. We need to show the kernel contains N . By definition, we knowχ(N) = χ(1). Also, if k ∈ N , then χ(k) = χ(kN) = χ(N) = χ(1). SoN ≤ kerχ.

Now let χ be a character of G with N ≤ kerχ. Suppose ρ : G → GL(V )affords χ. Define

ρ : G/N → GL(V )

gN 7→ ρ(g)

Of course, we need to check this is well-defined. If gN = g′N , theng−1g′ ∈ N . So ρ(g) = ρ(g′) since N ≤ ker ρ. So this is indeed well-defined.It is also easy to see that ρ is a homomorphism, hence a representation ofG/N .

Finally, if χ is a character of ρ, then χ(gN) = χ(g) for all g ∈ G bydefinition. So χ lifts to χ. It is clear that these two operations are inversesto each other.

There is one particular normal subgroup lying around that proves to beuseful.

Lemma. Given a group G, the derived subgroup or commutator subgroup

G′ = 〈[a, b] : a, b ∈ G〉,

where [a, b] = aba−1b−1, is the unique minimal normal subgroup of G such thatG/G′ is abelian. So if G/N is abelian, then G′ ≤ N .

Moreover, G has precisely ` = |G : G′| representations of dimension 1, allwith kernel containing G′, and are obtained by lifting from G/G′.

In particular, by Lagrange’s theorem, ` | G.

Proof. Consider [a, b] = aba−1b−1 ∈ G′. Then for any h ∈ G, we have

h(aba−1b−1)h−1 =(

(ha)b(ha)−1b−1)(bhb−1h−1

)= [ha, b][b, h] ∈ G′

So in general, let [a1, b1][a2, b2] · · · [an, bn] ∈ G′. Then

h[a1, b1][a2, b2] · · · [an, bn]h−1 = (h[a1, b1]h−1)(h[a2, b2]h−1) · · · (h[an, bn]h−1),

which is in G′. So G′ is a normal subgroup.Let N C G. Let g, h ∈ G. Then [g, h] ∈ N if and only if ghg−1h−1 ∈ N if

and only if ghN = hgN , if and only if (gN)(hN) = (hN)(gN) by normality.Since G′ is generated by all [g, h], we know G′ ≤ N if and only if G/N is

abelian.

37


Since G/G′, is abelian, we know it has exactly ` irreducible characters,χ1, · · · , χ`, all of degree 1. The lifts of these to G also have degree 1, and by theprevious lemma, these are precisely the irreducible characters χi of G such thatG′ ≤ kerχi.

But any degree 1 character of G is a homomorphism χ : G → C×, henceχ(ghg−1h−1) = 1. So for any 1-dimensional character, χ, we must have G′ ≤kerχ. So the lifts χ1, · · · , χ` are all 1-dimensional characters of G.

Example. Let G be Sn, with n ≥ 3. We claim that S′n = An. Since sgn(a) =sgn(a−1), we know that [a, b] ∈ An for any a, b ∈ Sn. So S′n ≤ An (alternatively,since Sn/An ∼= C2 is abelian, we must have S′n ≤ An). We now notice that

(1 2 3) = (1 3 2)(1 2)(1 3 2)−1(1 2)−1 ∈ S′n.

So by symmetry, all 3-cycles are in S′n. Since the 3-cycles generate An, we knowwe must have S′n = An.

Since S′n/An∼= C2, we get ` = 2. So Sn must have exactly two linear

characters, namely the trivial character and the sign.

Lemma. G is not simple if and only if χ(g) = χ(1) for some irreducible characterχ 6= 1G and some 1 6= g ∈ G. Any normal subgroup of G is the intersection ofthe kernels of some of the irreducible characters of G, i.e. N =

⋂kerχi.

In other words, G is simple if and only if all non-trivial irreducible charactershave trivial kernel.

Proof. Suppose χ(g) = χ(1) for some non-trivial irreducible character χ, and χis afforded by ρ. Then g ∈ ker ρ. So if g 6= 1, then 1 6= ker ρCG, and ker ρ 6= G.So G cannot be simple.

If 1 6= N CG is a non-trivial proper subgroup, take an irreducible characterχ of G/N , and suppose χ 6= 1G/N . Lift this to get an irreducible character χ,afforded by the representation ρ of G. Then N ≤ ker ρCG. So χ(g) = χ(1) forg ∈ N .

Finally, let 1 6= N CG. We claim that N is the intersection of the kernels ofthe lifts χ1, · · · , χ` of all the irreducibles of G/N . Clearly, we have N ≤

⋂i kerχi.

If g ∈ G \N , then gN 6= N . So χ(gN) 6= χ(N) for some irreducible χ of G/N .Lifting χ to χ, we have χ(g) 6= χ(1). So g is not in the intersection of thekernels.

38

9 Dual spaces and tensor products II Representation Theory

9 Dual spaces and tensor products of represen-tations

In this chapter, we will consider several linear algebra constructions, and seehow we can use them to construct new representations and characters.

9.1 Dual spaces

Our objective of this section is to try to come up with “dual representations”.Given a representation ρ : G→ GL(V ), we would like to turn V ∗ = Hom(V,F)into a representation. So for each g, we want to produce a ρ∗(g) : V ∗ → V ∗. Sogiven ϕ ∈ V ∗, we shall get a ρ∗(g)ϕ : V → F. Given v ∈ V , a natural guess forthe value of (ρ∗(g)ϕ)(v) might be

(ρ∗(g)ϕ)(v) = ϕ(ρ(g)v),

since this is how we can patch together ϕ, ρ, g and v. However, if we do this, wewill not get ρ∗(g)ρ∗(h) = ρ∗(gh). Instead, we will obtain ρ∗(g)ρ∗(h) = ρ∗(hg),which is the wrong way round. The right definition to use is actually thefollowing:

Lemma. Let ρ : G → GL(V ) be a representation over F, and let V ∗ =HomF(V,F) be the dual space of V . Then V ∗ is a G-space under

(ρ∗(g)ϕ)(v) = ϕ(ρ(g−1)v).

This is the dual representation to ρ. Its character is χ(ρ∗)(g) = χρ(g−1).

Proof. We have to check ρ∗ is a homomorphism. We check

ρ∗(g1)(ρ∗(g2)ϕ)(v) = (ρ∗(g2)ϕ)(ρ(g−11 )(v))

= ϕ(ρ(g−12 )ρ(g−2

1 )v)

= ϕ(ρ((g1g2)−1)(v))

= (ρ∗(g1g2)ϕ)(v).

To compute the character, fix a g ∈ G, and let e1, · · · , en be a basis of eigenvectorsof V of ρ(g), say

ρ(g)ej = λjej .

If we have a dual space of V , then we have a dual basis. We let ε1, · · · , εn bethe dual basis. Then

(ρ∗(g)εj)(ei) = εj(ρ(g−1)ei) = εj(λ−1i ei) = λ−1

i δij = λ−1j δij = λ−1

j εj(ei).

Thus we getρ∗(g)εj = λ−1

j εj .

Soχρ∗(g) =

∑λ−1j = χρ(g

−1).

Definition (Self-dual representation). A representation ρ : G → GL(V ) isself-dual if there is isomorphism of G-spaces V ∼= V ∗.

39


Note that as (finite-dimensional) vector spaces, we always have V ∼= V ∗.However, what we require here is an isomorphism that preserves the G-action,which does not necessarily exist.

Over F = C, this holds if and only if

χρ(g) = χρ∗(g) = χρ(g−1) = χρ(g).

So a character is self-dual if and only if it is real.

Example. All irreducible representations of Sn are self-dual, since every elementis conjugate to its own inverse. This is not always true for An.

9.2 Tensor products

The next idea we will tackle is the concept of tensor products. We first introduceit as a linear algebra concept, and then later see how representations fit into thisframework.

There are many ways we can define the tensor product. The definition wewill take here is a rather hands-on construction of the space, which involvespicking a basis. We will later describe some other ways to define the tensorproduct.

Definition (Tensor product). Let V,W be vector spaces over F. SupposedimV = m and dimW = n. We fix a basis v1, · · · ,vm and w1, · · · ,wn of Vand W respectively.

The tensor product space V ⊗W = V ⊗F W is an nm-dimensional vectorspace (over F) with basis given by formal symbols

{vi ⊗wj : 1 ≤ i ≤ m, 1 ≤ j ≤ n}.

ThusV ⊗W =

{∑λijvi ⊗wj : λij ∈ F

},

with the “obvious” addition and scalar multiplication.If

v =∑

αivi ∈ V, w =∑

βjwj ,

we definev ⊗w =

∑i,j

αiβj(vi ⊗wj).

Note that note all elements of V ⊗W are of this form. Some are genuinecombinations. For example, v1 ⊗w1 + v2 ⊗w2 cannot be written as a tensorproduct of an element in V and another in W .

We can imagine our formula of the tensor of two elements as writing(∑αivi

)⊗(∑

βjwj

),

and then expand this by letting ⊗ distribute over +.

Lemma.

(i) For v ∈ V , w ∈W and λ ∈ F, we have

(λv)⊗w = λ(v ⊗w) = v ⊗ (λw).

40


(ii) If x,x1,x2 ∈ V and y,y1,y2 ∈W , then

(x1 + x2)⊗ y = (x1 ⊗ y) + (x2 ⊗ y)

x⊗ (y1 + y2) = (x⊗ y1) + (x⊗ y2).

Proof.

(i) Let v =∑αivi and w =

∑βjwj . Then

(λv)⊗w =∑ij

(λαi)βjvi ⊗wj

λ(v ⊗w) = λ∑ij

αiβjvi ⊗wj

v ⊗ (λw) =∑

αi(λβj)vi ⊗wj ,

and these three things are obviously all equal.

(ii) Similar nonsense.

We can define the tensor product using these properties. We consider thespace of formal linear combinations v ⊗ w for all v ∈ V,w ∈ W , and thenquotient out by the relations re had above. It can be shown that this producesthe same space as the one constructed above.

Alternatively, we can define the tensor product using its universal property,which says for any U , a bilinear map from V ×W to U is “naturally” equivalentto a linear map V ⊗W → U . This intuitively makes sense, since the distributivityand linearity properties of ⊗ we showed above are exactly the properties requiredfor a bilinear map if we replace the ⊗ with a “,”.

It turns out this uniquely defines V ⊗W , as long as we provide a sufficientlyprecise definition of “naturally”. We can do it concretely as follows — in ourexplicit construction, we have a canonical bilinear map given by

φ : V ×W → V ⊗W(v,w) 7→ v ⊗w.

Given a linear map V ⊗W → U , we can compose it with φ : V ×W → V ⊗Win order to get a bilinear map V ×W → U . Then universality says every bilinearmap V ×W → U arises this way, uniquely.

In general, we say an equivalence between bilinear maps V ×W → U andlinear maps V ⊗W → U is “natural” if it is be mediated by such a universalmap. Then we say a vector space X is the tensor product of V and W if thereis a natural equivalence between bilinear maps V ×W → U and linear mapsX → U .

We will stick with our definition for concreteness, but we prove basis-independence here so that we feel happier:

Lemma. Let {e1, · · · , em} be any other basis of V , and {f1, · · · , fm} be anotherbasis of W . Then

{ei ⊗ fj : 1 ≤ i ≤ m, 1 ≤ j ≤ n}

is a basis of V ⊗W .

41


Proof. Writing

vk =∑

αikei, w` =∑

βj`f`,

we havevk ⊗w` =

∑αikβjlei ⊗ fj .

Therefore {ei ⊗ fj} spans V ⊗W . Moreover, there are nm of these. Thereforethey form a basis of V ⊗W .

That’s enough of linear algebra. We shall start doing some representationtheory.

Proposition. Let ρ : G→ GL(V ) and ρ′ : G→ GL(V ′). We define

ρ⊗ ρ′ : G→ GL(V ⊗ V ′)

by

(ρ⊗ ρ′)(g) :∑

λijvi ⊗wj 7→∑

λij(ρ(g)vi)⊗ (ρ′(g)wj).

Then ρ⊗ ρ′ is a representation of g, with character

χρ⊗ρ′(g) = χρ(g)χρ′(g)

for all g ∈ G.

As promised, the product of two characters of G is also a character of G.Just before we prove this, recall we showed in sheet 1 that if ρ is irreducible

and ρ′ is irreducible of degree 1, then ρ′ρ = ρ ⊗ ρ′ is irreducible. However, ifρ′ is not of degree 1, then this is almost always false (or else we can producearbitrarily many irreducible representations).

Proof. It is clear that (ρ ⊗ ρ′)(g) ∈ GL(V ⊗ V ′) for all g ∈ G. So ρ ⊗ ρ′ is ahomomorphism G→ GL(V ⊗ V ′).

To check the character is indeed as stated, let g ∈ G. Let v1, · · · ,vm bea basis of V of eigenvectors of ρ(g), and let w1, · · · ,wn be a basis of V ′ ofeigenvectors of ρ′(g), say

ρ(g)vi = λivi, ρ′(g)wj = µjwj .

Then

(ρ⊗ ρ′)(g)(vi ⊗wj) = ρ(g)vi ⊗ ρ′(g)wj

= λivi ⊗ µjwj

= (λiµj)(vi ⊗wj).

Soχρ⊗ρ′(g) =

∑i,j

λiµj =(∑

λi

)(∑µj

)= χρ(g)χρ′(g).

42


9.3 Powers of characters

We work over C. Take V = V ′ to be G-spaces, and define

Notation.V ⊗2 = V ⊗ V.

We define τ : V ⊗2 → V ⊗2 by

τ :∑

λijvi ⊗ vj 7→∑

λijvj ⊗ vi.

This is a linear endomorphism of V ⊗2 such that τ2 = 1. So the eigenvalues are±1.

Definition (Symmetric and exterior square). We define the symmetric squareand exterior square of V to be, respectively,

S2V = {x ∈ V ⊗2 : τ(x) = x}Λ2V = {x ∈ V ⊗2 : τ(x) = −x}.

The exterior square is also known as the anti-symmetric square and wedge power.

Lemma. For any G-space V , S2V and Λ2V are G-subspaces of V ⊗2, and

V ⊗2 = S2V ⊕ Λ2V.

The space S2V has basis

{vivj = vi ⊗ vj + vj ⊗ vi : 1 ≤ i ≤ j ≤ n},

while Λ2V has basis

{vi ∧ vj = vi ⊗ vj − vj ⊗ vi : 1 ≤ i < j ≤ n}.

Note that we have a strict inequality for i < j, since vi ⊗ vj − vj ⊗ vi = 0 ifi = j. Hence

dimS2V =1

2n(n+ 1), dim Λ2V =

1

2n(n− 1).

Sometimes, a factor of 12 appears in front of the definition of the basis elements

vivj and vi ∧ vj , i.e.

vivj =1

2(vi ⊗ vj + vj ⊗ vi), vi ∧ vj =

1

2(vi ⊗ vj − vj ⊗ vi).

However, this is not important.

Proof. This is elementary linear algebra. For the decomposition V ⊗2, givenx ∈ V ⊗2, we can write it as

x =1

2(x + τ(x))︸︷︷︸∈S2V

+1

2(x− τ(x))︸︷︷︸∈Λ2V

.

How is this useful for character calculations?

43


Lemma. Let ρ : G → GL(V ) be a representation affording the character χ.Then χ2 = χS+χΛ where χS = S2χ is the character of G in the subrepresentationon S2V , and χΛ = Λ2χ the character of G in the subrepresentation on Λ2V .Moreover, for g ∈ G,

χS(g) =1

2(χ2(g) + χ(g2)), χΛ(g) =

1

2(χ2(g)− χ(g2)).

The real content of the lemma is, of course, the formula for χS and χΛ.

Proof. The fact that χ2 = χS + χΛ is immediate from the decomposition ofG-spaces.

We now compute the characters χS and χΛ. For g ∈ G, we let v1, · · · ,vn bea basis of V of eigenvectors of ρ(g), say

ρ(g)vi = λivi.

We’ll be lazy and just write gvi instead of ρ(g)vi. Then, acting on Λ2V , we get

g(vi ∧ vj) = λiλjvi ∧ vj .

ThusχΛ(g) =

∑1≤i<j≤n

λiλj .

Since the answer involves the square of the character, let’s write that down:

(χ(g))2 =(∑

λi

)2

=∑

λ2i + 2

∑i<j

λiλj

= χ(g2) + 2∑i<j

λiλj

= χ(g2) + 2χΛ(g).

Then we can solve to obtain

χΛ(g) =1

2(χ2(g)− χ(g2)).

Then we can get

χS = χ2 − χΛ =1

2(χ2(g) + χ(g2)).

Example. Consider again G = S4. As before, we have the following table:

1 3 8 6 61 (1 2)(3 4) (1 2 3) (1 2 3 4) (1 2)

trivial χ1 1 1 1 1 1sign χ2 1 1 1 −1 −1

πX − 1G χ3 3 −1 0 −1 1χ3χ2 χ4 3 −1 0 1 −1

χ5

44


We now compute χ5 in a different way, by decomposing χ23. We have

1 3 8 6 61 (1 2)(3 4) (1 2 3) (1 2 3 4) (1 2)

χ23 9 1 0 1 1

χ3(g2) 3 3 0 −1 3S2χ3 6 2 0 0 2Λ2χ3 3 −1 0 1 −1

We notice Λ2χ3 is just χ4. Now S2χ3 is not irreducible, which we can easilyshow by computing the inner product. By taking the inner product with theother irreducible characters, we find S2χ3 = 1 + χ3 + χ5, where χ5 is our newirreducible character. So we obtain

1 3 8 6 61 (1 2)(3 4) (1 2 3) (1 2 3 4) (1 2)

trivial χ1 1 1 1 1 1sign χ2 1 1 1 −1 −1

πX − 1G χ3 3 −1 0 −1 1χ3χ2 χ4 3 −1 0 1 −1

S3χ3 − 1− χ3 χ5 2 2 −1 0 0

9.4 Characters of G×H

We have looked at characters of direct products a bit before, when we decomposedan abelian group into a product of cyclic groups. We will now consider this inthe general case.

Proposition. Let G and H be two finite groups with irreducible charactersχ1, · · · , χk and ψ1, · · · , ψr respectively. Then the irreducible characters of thedirect product G×H are precisely

{χiψj : 1 ≤ i ≤ k, 1 ≤ j ≤ r},

where(χiψj)(g, h) = χi(g)ψj(h).

Proof. Take ρ : G → GL(V ) affording χ, and ρ′ : H → GL(W ) affording ψ.Then define

ρ⊗ ρ′ : G×H → GL(V ⊗W )

(g, h) 7→ ρ(g)⊗ ρ′(h),

where(ρ(g)⊗ ρ′(h))(vi ⊗wj) 7→ ρ(g)vi ⊗ ρ′(h)wj .

This is a representation of G × H on V ⊗W , and χρ⊗ρ′ = χψ. The proof issimilar to the case where ρ, ρ′ are both representations of G, and we will notrepeat it here.

45


Now we need to show χiψj are distinct and irreducible. It suffices to showthey are orthonormal. We have

〈χiψj , χrψs〉G×H =1

|G×H|∑

(g,h)∈G×H

χiψj(g, h)χrψs(g, h)

=

1

|G|∑g∈G

χi(g)χr(g)

( 1

|H|∑h∈H

ψj(h)ψs(h)

)= δirδjs.

So it follows that {χiψj} are distinct and irreducible. We need to show this iscomplete. We can consider∑i,j

χiψj(1)2 =∑

χ2i (1)ψ2

j (1) =(∑

χ2i (1)

)(∑ψ2j (1)

)= |G||H| = |G×H|.

So done.

9.5 Symmetric and exterior powers

We have introduced the objects S2V and Λ2V . We can do this in a more generalsetting, for higher powers. Consider a vector space V over F, and dimF V = d.We let {v1, · · · ,vd} be a basis, and let

TnV = V ⊗n = V ⊗ · · · ⊗ V︸︷︷︸n times

.

The basis for this space is just

{vi1 ⊗ · · · ⊗ vin : i1, · · · , in ∈ {1, · · · , d}}.

First of all, there is an obvious Sn-action on this space, permuting the multipli-cands of V ⊗n. For any σ ∈ Sn, we can define an action σ : V ⊗n → V ⊗n givenby

v1 ⊗ · · ·vn 7→ vσ(1) ⊗ · · · ⊗ vσ(n).

Note that this is a left action only if we compose permutations right-to-left. Ifwe compose left-to-right instead, we have to use σ−1 instead of σ (or use σ andobtain a right action). In fact, this defines a representation of Sn on V ⊗n bylinear representation.

If V itself is a G-space, then we also get a G-action on V ⊗n. Let ρ : G →GL(V ) be a representation. Then we obtain the action of G on V ⊗n by

ρ⊗n : v1 ⊗ · · · ⊗ vn 7→ ρ(g)v1 ⊗ · · · ⊗ ρ(g)vn.

Staring at this carefully, it is clear that these two actions commute with each other.This rather simple innocent-looking observation is the basis of an importanttheorem by Schur, and has many many applications. However, that would befor another course.

Getting back on track, since the two actions commute, we can decomposeV ⊗n as an Sn-module, and each isotypical component is a G-invariant subspaceof V ⊗n. We don’t really need this, but it is a good generalization of what we’vehad before.

46


Definition (Symmetric and exterior power). For a G-space V , define

(i) The nth symmetric power of V is

SnV = {x ∈ V ⊗n : σ(x) = x for all σ ∈ Sn}.

(ii) The nth exterior power of V is

ΛnV = {x ∈ V ⊗n : σ(x) = (sgnσ)x for all σ ∈ Sn}.

Both of these are G-subspaces of V ⊗n.

Recall that in the n = 2 case, we have V ⊗2 = S2V ⊕ Λ2V . However, forn > 2, we get SnV ⊕ ΛnV � V ⊗n. In general, lots of others are obtained fromthe Sn-action.

Example. The bases for SnV and ΛnV are obtained as follows: take e1, · · · , eda basis of V . Then{

1

n!

∑σ∈Sn

eiσ(1) ⊗ · · · ⊗ eiσ(n): 1 ≤ i1 ≤ · · · ≤ in ≤ d

}

is a basis for SnV . So the dimension is

dimSnV =

(d+ n− 1

n

).

For ΛnV , we get a basis of{1

n!

∑σ∈Sn

(sgnσ)eiσ(1) ⊗ · · · ⊗ eiσ(n): 1 ≤ i1 < · · · < in ≤ d

}.

The dimension is

dim ΛnV =

{0 n > d(dn

)n ≤ d

Details are left for the third example sheet.

9.6 Tensor algebra

This part is for some unknown reason included in the schedules. We do not needit here, but it has many applications elsewhere.

Consider a field F with characteristic 0.

Definition (Tensor algebra). Let TnV = V ⊗n. Then the tensor algebra of V is

T ·(V ) = T (V ) =⊕n≥0

TnV,

with T 0V = F by convention.This is a vector space over F with the obvious addition and multiplication by

scalars. T (V ) is also a (non-commutative) (graded) ring with product x·y = x⊗y.This is graded in the sense that if x ∈ TnV and y ∈ TmV , x ·y = x⊗y ∈ Tn+mV .

47


Definition (Symmetric and exterior algebra). We define the symmetric algebraof V to be

S(V ) = T (V )/(ideal of T (V ) generated by all u⊗ v − v ⊗ u).

The exterior algebra of V is

Λ(V ) = T (V )/(ideal of T (V ) generated by all v ⊗ v).

Note that v and u are not elements of V , but arbitrary elements of TnV forsome n.

The symmetric algebra is commutative, while the exterior algebra is graded-commutative, i.e. if x ∈ Λn(V ) and y ∈ Λm(V ), then x · y = (−1)n+m+1y · x.

9.7 Character ring

Recall that C(G) is a commutative ring, and contains all the characters. Also,the sum and products of characters are characters. However, they don’t quiteform a subring, since we are not allowed to subtract.

Definition (Character ring). The character ring of G is the Z-submodule ofC(G) spanned by the irreducible characters and is denoted R(G).

Recall we can obtain all characters by adding irreducible characters. However,subtraction is not allowed. So the charactering ring includes something that isnot quite a character.

Definition (Generalized/virtual characters). The elements of R(G) are calledgeneralized or virtual characters. These are class functions of the form

ψ =∑χ

nχχ,

summing over all irreducibles χ, and nχ ∈ Z.

We know R(G) is a ring, and any generalized character is a difference of twocharacters (we let

α =∑nχ≥0

nχχ, β =∑nχ<0

(−nχ)χ.

Then ψ = α− β, and α and β are characters). Then the irreducible characters{χi} forms a Z-basis for R(G) as a free Z-module, since we have shown thatthey are independent, and it generates R(G) by definition.

Lemma. Suppose α is a generalized character and 〈α, α〉 = 1 and α(1) > 0.Then α is actually a character of an irreducible representation of G.

Proof. We list the irreducible characters as χ1, · · · , χk. We then write

α =∑

niχi.

Since the χi’s are orthonormal, we get

〈α, α〉 =∑

n2i = 1.

So exactly one of ni is ±1, while the others are all zero. So α = ±χi for some i.Finally, since α(1) > 0 and also χ(1) > 0, we must have ni = +1. So α = χi.

48


Henceforth we don’t distinguish between a character and its negative, andoften study generalized characters of inner product 1 rather than irreduciblecharacters.

49

10 Induction and restriction II Representation Theory

10 Induction and restriction

This is the last chapter on methods of calculating characters. Afterwards, weshall start applying these to do something useful.

Throughout the chapter, we will take F = C. Suppose H ≤ G. How docharacters of G and H relate? It is easy to see that every representation ofG can be restricted to give a representation of H. More surprisingly, given arepresentation of H, we can induce a representation of G.

We first deal with the easy case of restriction.

Definition (Restriction). Let ρ : G→ GL(V ) be a representation affording χ.We can think of V as an H-space by restricting ρ’s attention to h ∈ H. We geta representation ResGH ρ = ρH = ρ ↓H , the restriction of ρ to H. It affords thecharacter ResGH χ = χH = χ ↓H .

It is a fact of life that the restriction of an irreducible character is in generalnot irreducible. For example, restricting any non-linear irreducible character tothe trivial subgroup will clearly produce a reducible character.

Lemma. Let H ≤ G. If ψ is any non-zero irreducible character of H, then thereexists an irreducible character χ of G such that ψ is a constituent of ResGH χ, i.e.

〈ResGH χ, ψ〉 6= 0.

Proof. We list the irreducible characters of G as χ1, · · · , χk. Recall the regularcharacter πreg. Consider

〈ResGH πreg, ψ〉 =|G||H|

ψ(1) 6= 0.

On the other hand, we also have

〈ResGH πreg, ψ〉H =

k∑1

degχi〈ResGH χi, ψ〉.

If this sum has to be non-zero, then there must be some i such that 〈ResGH χi, ψ〉 6=0.

Lemma. Let χ be an irreducible character of G, and let

ResGH χ =∑i

ciχi,

with χi irreducible characters of H, and ci non-negative integers. Then∑c2i ≤ |G : H|,

with equality iff χ(g) = 0 for all g ∈ G \H.

This is useful only if the index is small, e.g. 2 or 3.

50


Proof. We have

〈ResGH χ,ResGH χ〉H =∑

c2i .

However, by definition, we also have

〈ResGH χ,ResGH χ〉H =1

|H|∑h∈H

|χ(h)|2.

On the other hand, since χ is irreducible, we have

1 = 〈χ, χ〉G

=1

|G|∑g∈G|χ(g)|2

=1

|G|

∑h∈H

|χ(h)|2 +∑

g∈G\H

|χ(g)|2

=|H||G|

∑c2i +

1

|G|∑

g∈G\H

|χ(g)|2

≥ |H||G|

∑c2i .

So the result follows.

Now we shall go in the other direction — given a character of H, we want toinduce a character of the larger group G. Perhaps rather counter-intuitively, it iseasier for us to first construct the character, and then later find a representationthat affords this character.

We start with a slightly more general notion of an induced class function.

Definition (Induced class function). Let ψ ∈ C(H). We define the induced classfunction IndGH ψ = ψ ↑G= ψG by

IndGH ψ(g) =1

|H|∑x∈G

ψ(x−1gx),

where

ψ(y) =

{ψ(y) y ∈ H0 y 6∈ H

.

The first thing to check is, of course, that IndGH is indeed a class function.

Lemma. Let ψ ∈ CH . Then IndGH ψ ∈ C(G), and IndGH ψ(1) = |G : H|ψ(1).

Proof. The fact that IndGH ψ is a class function follows from direct inspection ofthe formula. Then we have

IndGH ψ(1) =1

|H|∑x∈G

ψ(1) =|G||H|

ψ(1) = |G : H|ψ(1).

51


As it stands, the formula is not terribly useful. So we need to find analternative formula for it.

If we have a subset, a sensible thing to do is to look at the cosets of H.We let n = |G : H|, and let 1 = t1, · · · , tn be a left transversal of H in G, i.e.t1H = H, · · · , tnH are precisely the n left-cosets of H in G.

Lemma. Given a (left) transversal t1, · · · , tn of H, we have

IndGH ψ(g) =

n∑i=1

ψ(t−1i gti).

Proof. We can express every x ∈ G as x = tih for some h ∈ H and i. We thenhave

ψ((tih)−1g(tih)) = ψ(h−1(t−1i gti)h) = ψ(t−1

i gti),

since ψ is a class function of H, and h−1(t−1i gti)h ∈ H if and only if t−1

i gti ∈ H,as h ∈ H. So the result follows.

This is not too useful as well. But we are building up to something useful.

Theorem (Frobenius reciprocity). Let ψ ∈ C(H) and ϕ ∈ C(G). Then

〈ResGH ϕ,ψ〉H = 〈ϕ, IndGH ψ〉G.

Alternatively, we can write this as

〈ϕH , ψ〉 = 〈ϕ,ψG〉.

If you are a category theorist, and view restriction and induction as functors,then this says restriction is a right adjoint to induction, and thsi is just a specialcase of the tensor-Hom adjunction. If you neither understand nor care aboutthis, then this is a good sign [sic].

Proof. We have

〈ϕ,ψG〉 =1

|G|∑g∈G

ϕ(g)ψG(g)

=1

|G||H|∑x,g∈G

ϕ(g)ψ(x−1gx)

We now write y = x−1gx. Then summing over g is the same as summing over y.Since ϕ is a G-class function, this becomes

=1

|G||H|∑x,y∈G

ϕ(y)ψ(y)

Now note that the sum is independent of x. So this becomes

=1

|H|∑y∈G

ϕ(y)ψ(y)

Now this only has contributions when y ∈ H, by definition of ψ. So

=1

|H|∑y∈H

ϕ(y)ψ(y)

= 〈ϕH , ψ〉H .

52


Corollary. Let ψ be a character of H. Then IndGH ψ is a character of G.

Proof. Let χ be an irreducible character of G. Then

〈IndGH ψ, χ〉 = 〈ψ,ResGH χ〉.

Since ψ and ResGH χ are characters, the thing on the right is in Z≥0. Hence IndGHis a linear combination of irreducible characters with non-negative coefficients,and is hence a character.

Recall we denote the conjugacy class of g ∈ G as CG(g), while the centralizeris CG(g). If we take a conjugacy class CG(g) in G and restrict it to H, thenthe result CG(g) ∩H need not be a conjugacy class, since the element neededto conjugate x, y ∈ CG(g) ∩H need not be present in H, as is familiar from thecase of An ≤ Sn. However, we know that CG(g) ∩ H is a union of conjugacyclasses in H, since elements conjugate in H are also conjugate in G.

Proposition. Let ψ be a character of H ≤ G, and let g ∈ G. Let

CG(g) ∩H =

m⋃i=1

CH(xi),

where the xi are the representatives of the H conjugacy classes of elements of Hconjugate to g. If m = 0, then IndGH ψ(g) = 0. Otherwise,

IndGH ψ(g) = |CG(g)|m∑i=1

ψ(xi)

|CH(xi)|.

This is all just group theory. Note that some people will think this proofis excessive — everything shown is “obvious”. In some sense it is. Some stepsseem very obvious to certain people, but we are spelling out all the details sothat everyone is happy.

Proof. If m = 0, then {x ∈ G : x−1gx ∈ H} = ∅. So ψ(x−1gx) = 0 for all x. SoIndGH ψ(g) = 0 by definition.

Now assume m > 0. We let

Xi = {x ∈ G : x−1gx ∈ H and is conjugate in H to xi}.

By definition of xi, we know the Xi’s are pairwise disjoint, and their union is{x ∈ G : x−1gx ∈ H}. Hence by definition,

IndGH ψ(g) =1

|H|∑x∈G

ψ(x−1gx)

=1

|H|

m∑i=1

∑x∈Xi

ψ(x−1gx)

=1

|H|

m∑i=1

∑x∈Xi

ψ(xi)

=

m∑i=1

|Xi||H|

ψ(xi).

53


So we now have to show that in fact

|Xi||H|

=|CG(g)||CH(xi)|

.

We fix some 1 ≤ i ≤ m. Choose some gi ∈ G such that g−1i ggi = xi. This exists

by definition of xi. So for every c ∈ CG(g) and h ∈ H, we have

(cgih)−1g(cgih) = h−1g−1i c−1gcgih

We now use the fact that c commutes with g, since c ∈ CG(g), to get

= h−1g−1i c−1cggih

= h−1g−1i ggih

= h−1xih.

Hence by definition of Xi, we know cgih ∈ Xi. Hence

CG(g)giH ⊆ Xi.

Conversely, if x ∈ Xi, then x−1gx = h−1xih = h−1(g−1i ggi)h for some h. Thus

xh−1g−1i ∈ CG(g), and so x ∈ CG(g)gih. So

x ∈ CG(g)giH.

So we concludeXi = CG(g)giH.

Thus, using some group theory magic, which we shall not prove, we get

|Xi| = |CG(g)giH| =|CG(g)||H|

|H ∩ g−1i CG(g)gi|

Finally, we noteg−1i CG(g)gi = CG(g−1

i ggi) = CG(xi).

Thus

|Xi| =|H||CG(g)||H ∩ CG(xi)|

=|H||CG(g)||CH(xi)|

.

Dividing, we get|Xi||H|

=|CG(g)||CH(xi)|

.

So done.

To clarify matters, if H,K ≤ G, then a double coset of H,K in G is a set

HxK = {hxk : h ∈ H, k ∈ K}

for some x ∈ G. Facts about them include

(i) Two double cosets are either disjoint or equal.

(ii) The sizes are given by

|H||K||H ∩ xKx−1|

=|H||K|

|x−1Hx ∩K|.

54


Lemma. Let ψ = 1H , the trivial character of H. Then IndGH 1H = πX , thepermutation character of G on the set X, where X = G/H is the set of leftcosets of H.

Proof. We let n = |G : H|, and t1, · · · , tn be representatives of the cosets. Bydefinition, we know

IndGH 1H(g) =

n∑i=1

1H(t−1i gti)

= |{i : t−1i gti ∈ H}|

= |{i : g ∈ tiHt−1i }|

But tiHt−1i is the stabilizer in G of the coset tiH ∈ X. So this is equal to

= |fixX(g)|= πX(g).

By Frobenius reciprocity, we know

〈πX , 1G〉G = 〈IndGH 1H , 1G〉G = 〈1H , 1H〉H = 1.

So the multiplicity of 1G in πX is indeed 1, as we have shown before.

Example. Let H = C4 = 〈(1 2 3 4)〉 ≤ G = S4. The index is |S4 : C4| = 6. Weconsider the character of the induced representation IndGH(α), where α is thefaithful 1-dimensional representation of C4 given by

α((1 2 3 4)) = i.

Then the character of α is

1 (1 2 3 4) (1 3)(2 4) (1 4 3 2)

χα 1 i −1 −i

What is the induced character in S4? We know that

IndGH(α)(1) = |G : H|α(1) = |G : H| = 6.

So the degree of IndGH(α) is 6. Also, the elements (1 2) and (3 4) are notconjugate to anything in C4. So the character vanishes.

For (1 2)(3 4), only one of the three conjugates in S4 lie in H (namely(1 3)(2 4)). So, using the sizes of the centralizers, we obtain

IndGH((1 3)(2 4)) = 8

(−1

4

)= −2.

For (1 2 3 4), it is conjugate to 6 elements of S4, two of which are in C4, namely(1 2 3 4) and (1 4 3 2). So

IndGH(1 2 3 4) = 4

(i

4+−i4

)= 0.

So we get the following character table:

1 6 8 3 61 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)

IndGH(α) 6 0 0 −2 0

55


Induced representations

We have constructed a class function, and showed it is indeed the character ofsome representation. However, we currently have no idea what this correspondingrepresentation is. Here we will try to construct such a representation explicitly.This is not too enlightening, but its nice to know it can be done. We will alsoneed this explicit construction when proving Mackey’s theorem later.

Let H ≤ G have index n, and 1 = t1, · · · , tn be a left transversal. Let W bea H-space. Define the vector space

V = IndGHW = W ⊕ (t2 ⊗W )⊕ · · · ⊕ (tn ⊗W ),

whereti ⊗W = {ti ⊗w : w ∈W}.

So dimV = ndimW .To define the G-action on V , let g ∈ G. Then for every i, there is a unique j

such that t−1j gti ∈ H, i.e. gtiH = tjH. We define

g(ti ⊗w) = tj ⊗ ((t−1j gti)w).

We tend to drop the tensor signs, and just write

g(tiw) = tj(t−1j gtiw).

We claim this is a G-action. We have

g1(g2tiw) = g1(tj(t−1j g2ti)w),

where j is the unique index such that g2tiH = tjH. This is then equal to

t`((t−1` gtj)(t

−1j g2ti)w) = t`((t

−1` (g1g2)ti)w) = (g1g2)(tiw),

where ` is the unique index such that g1tjH = t`H (and hence g1g2tiH =g1tjH = t`H).

Definition (Induced representation). Let H ≤ G have index n, and 1 =t1, · · · , tn be a left transversal. Let W be a H-space. Define the inducedrepresentation to be the vector space

IndGHW = W ⊕ t2 ⊗W ⊕ · · · ⊕ tn ⊗W,

with the G-actiong : tiw 7→ tj(t

−1j gti)W,

where tj is the unique element (among t1, · · · , tn) such that t−1j gti ∈ H.

This has the “right” character. Suppose W has character ψ. Since g : tiw 7→tj(t−1j gti), the contribution to the character is 0 unless j = i, i.e. if t−1

i gti ∈ H.Then it contributes

ψ(t−1i gti).

Thus we get

IndGH ψ(g) =

n∑i=1

ψ(t−1i gti).

56


Note that this construction is rather ugly. It could be made much nicer if weknew a bit more algebra — we can write the induced module simply as

IndGHW = FG⊗FH W,

where we view W as a left-FH module, and FG as a (FG,FH)-bimodule. Alter-natively, this is the extension of scalars from FH to FG.

57

11 Frobenius groups II Representation Theory

11 Frobenius groups

We will now use character theory to prove some major results in finite grouptheory. We first do Frobenius’ theorem here. Later, will prove Burnside’s paqb

theorem.This is a theorem with lots and lots of proofs, but all of these proofs involves

some sort of representation theory — it seems like representation is an unavoidableingredient of it.

Theorem (Frobenius’ theorem (1891)). Let G be a transitive permutationgroup on a finite set X, with |X| = n. Assume that each non-identity elementof G fixes at most one element of X. Then the set of fixed point-free elements(“derangements”)

K = {1} ∪ {g ∈ G : gα 6= α for all α ∈ X}

is a normal subgroup of G with order n.

On the face of it, it is not clear K is even a subgroup at all. It turns outnormality isn’t really hard to prove — the hard part is indeed showing it is asubgroup.

Note that we did not explicitly say G is finite. But these conditions implyG ≤ Sn, which forces G to be finite.

Proof. The idea of the proof is to construct a character Θ whose kernel is K.First note that by definition of K, we have

G = K ∪⋃α∈X

Gα,

where Gα is, as usual, the stabilizer of α. Also, we know that Gα ∩Gβ = {1} ifα 6= β by assumption, and by definition of K, we have K ∩Gα = {1} as well.

Next note that all the Gα are conjugate. Indeed, we know G is transitive,and gGαg

−1 = Ggα. We set H = Gα for some arbitrary choice of α. Then theabove tells us that

|G| = |K| − |X|(|H| − 1).

On the other hand, by the orbit-stabilizer theorem, we know |G| = |X||H|. So itfollows that we have

|K| = |X| = n.

We first compute what induced characters look like.

Claim. Let ψ be a character of H. Then

IndGH ψ(g) =

nψ(1) g = 1

ψ(g) g ∈ H \ {1}0 g ∈ K \ {1}

.

Since every element in G is either in K or conjugate to an element in H, thisuniquely specifies what the induced character is.

58


This is a matter of computation. Since [G : H] = n, the case g = 1immediately follows. Using the definition of the induced character, since anynon-identity in K is not conjugate to any element in H, we know the inducedcharacter vanishes on K \ {1}.

Finally, suppose g ∈ H \ {1}. Note that if x ∈ G, then xgx−1 ∈ Gxα. So thislies in H if and only if x ∈ H. So we can write the induced character as

IndGH ψ(g) =1

|H|∑g∈G

ψ(xgx−1) =1

|H|∑h∈H

ψ(hgh−1) = ψ(g).

Claim. Let ψ be an irreducible character of H, and define

θ = ψG − ψ(1)(1H)G + ψ(1)1G.

Then θ is a character, and

θ(g) =

{ψ(h) h ∈ Hψ(1) k ∈ K

.

Note that we chose the coefficients exactly so that the final property of θholds. This is a matter of computation:

1 h ∈ H \ {1} K \ {1}ψG nψ(1) ψ(h) 0

ψ(1)(1H)G nψ(1) ψ(1) 0ψ(1)1G ψ(1) ψ(1) ψ(1)θi ψ(1) ψ(h) ψ(1)

The less obvious part is that θ is a character. From the way we wrote it, wealready know it is a virtual character. We then compute the inner product

〈θ, θ〉G =1

|G|∑g∈G|θ(g)|2

=1

|G|

∑g∈K|θ(g)|2 +

∑g∈G\K

|θ(g)|2

=1

|G|

n|ψ(1)|2 + n∑

h6=1∈H

|ψ(h)|2

=1

|G|

(n∑h∈H

|ψ(h)|2)

=1

|G|(n|H|〈ψ,ψ〉H)

= 1.

So either θ or −θ is a character. But θ(1) = ψ(1) > 0. So θ is a character.Finally, we have

59


Claim. Let ψ1, · · · , ψt be the irreducible representations of H, and θi be thecorresponding representations of G constructed above. Set

Θ =

t∑i=1

θi(1)θi.

Then we have

θ(g) =

{|H| g ∈ K0 g 6∈ K

.

From this, it follows that the kernel of the representation affording θ is K, andin particular K is a normal subgroup of G.

This is again a computation using column orthogonality. For 1 6= h ∈ H, wehave

Θ(h) =

t∑i=1

ψi(1)ψi(h) = 0,

and for any y ∈ K, we have

Θ(y) =

t∑i=1

ψi(1)2 = |H|.

Definition (Frobenius group and Frobenius complement). A Frobenius groupis a group G having a subgroup H such that H ∩ gHg−1 = 1 for all g 6∈ H. Wesay H is a Frobenius complement of G.

How does this relate to the previous theorem?

Proposition. The left action of any finite Frobenius group on the cosets of theFrobenius complement satisfies the hypothesis of Frobenius’ theorem.

Definition (Frobenius kernel). The Frobenius kernel of a Frobenius group G isthe K obtained from Frobenius’ theorem.

Proof. Let G be a Frobenius group, having a complement H. Then the action ofG on the cosets G/H is transitive. Furthermore, if 1 6= g ∈ G fixes xH and yH,then we have g ∈ xHx−1 ∩ yHy−1. This implies H ∩ (y−1x)H(y−1x)−1 6= 1.Hence xH = yH.

Note that J. Thompson (in his 1959 thesis) proved any finite group havinga fixed-point free automorphism of prime order is nilpotent. This implies K isnilpotent, which means K is a direct product of its Sylow subgroups.

60

12 Mackey theory II Representation Theory

12 Mackey theory

We work over C. We wish to describe the restriction to a subgroup K ≤ G of aninduced representation IndGHW , i.e. the character ResGK IndGHW . In general, Kand H can be unrelated, but in many applications, we have K = H, in whichcase we can characterize when IndGHW is irreducible.

It is quite helpful to first look at a special case, where W = 1H is the trivialrepresentation. Thus IndGH 1H is the permutation representation of G on G/H(coset action on the left cosets of H in G).

Recall that by the orbit-stabilizer theorem, if G is transitive on the set X,and H = Gα for some α ∈ X, then the action of G on X is isomorphic the actionon G/H, namely the correspondence

gα↔ gH

is a well-defined bijection, and commutes with with the g-action (i.e. x(gα) =(xg)α↔ x(gH) = (xg)H).

We now consider the action of G on G/H and let K ≤ G. Then K also actson G/H, and G/H splits into K-orbits. The K-orbit of gH contains preciselykgH for k ∈ K. So it is the double coset

KgH = {kgh : k ∈ K,h ∈ H}.

The set of double cosets K\G/H partition G, and the number of double cosetsis

|K\G/H| = 〈πG/K , πG/H〉.We don’t need this, but this is true.

If it happens that H = K, and H is normal, then we just have K\G/H =G/H.

What about stabilizers? Clearly, GgH = gHg−1. Thus, restricting to theaction of K, we have KgH = gHg−1 ∩K. We call Hg = KgH .

So by our correspondence above, the action of K on the orbit containinggH is isomorphic to the action of K on K/(gHg−1 ∩K) = K/Hg. From this,

and using the fact that IndGH 1H = C(G/H), we get the special case of Mackey’stheorem:

Proposition. Let G be a finite group and H,K ≤ G. Let g1, · · · , gk be therepresentatives of the double cosets K\G/H. Then

ResGK IndGH 1H ∼=k⊕i=1

IndKgiHg

−1i ∩K

1.

The general form of Mackey’s theorem is the following:

Theorem (Mackey’s restriction formula). In general, for K,H ≤ G, we letS = {1, g1, · · · , gr} be a set of double coset representatives, so that

G =⋃KgiH.

We write Hg = gHg−1 ∩K ≤ G. We let (ρ,W ) be a representation of H. Foreach g ∈ G, we define (ρg,Wg) to be a representation of Hg, with the sameunderlying vector space W , but now the action of Hg is

ρg(x) = ρ(g−1xg),

61


where h = g−1xg ∈ H by construction.This is clearly well-defined. Since Hg ≤ K, we obtain an induced representa-

tion IndKHg Wg.Let G be finite, H,K ≤ G, and W be a H-space. Then

ResGK IndGHW =⊕g∈S

IndKHg Wg.

We will defer the proof of this for a minute or two. We will first derive somecorollaries of this, starting with the character version of the theorem.

Corollary. Let ψ be a character of a representation of H. Then

ResGK IndGH ψ =∑g∈S

IndKHg ψg,

where ψg is the class function (and a character) on Hg given by

ψg(x) = ψ(g−1xg).

These characters ψg are sometimes known as the conjugate characters. Thisobviously follows from Mackey’s restriction formula.

Corollary (Mackey’s irreducibility criterion). Let H ≤ G and W be a H-space.Then V = IndGHW is irreducible if and only if

(i) W is irreducible; and

(ii) For each g ∈ S\H, the two Hg spaces Wg and ResHHg W have no irreducible

constituents in common, where Hg = gHg−1 ∩H.

Note that the set S of representatives was chosen arbitrarily. So we mayrequire (ii) to hold for all g ∈ G \ H, instead of g ∈ S \ H. While there aremore things to check in G, it is sometimes convenient not have to pick out an Sexplicitly.

Note that the only g ∈ S ∩H is the identity, since for any h ∈ H, KhH =KH = K1H.

Proof. We use characters, and let W afford the character ψ. We take K = H inMackey’s restriction formula. Then we have Hg = gHg−1 ∩H.

Using Frobenius reciprocity, we can compute the inner product as

〈IndGH ψ, IndGH ψ〉G = 〈ψ,ResGH IndGH ψ〉H=∑g∈S〈ψ, IndHHg ψg〉H

=∑g∈S〈ResHHg ψ,ψg〉Hg

= 〈ψ,ψ〉+∑

g∈S\H

〈ResHHg ψ,ψg〉Hg

We can write this because if g = 1, then Hg = H, and ψg = ψ.This is a sum of non-negative integers, since the inner products of characters

always are. So IndGH ψ is irreducible if and only if 〈ψ,ψ〉 = 1, and all the otherterms in the sum are 0. In other words, W is an irreducible representation of H,and for all g 6∈ H, W and Wg are disjoint representations of Hg.

62


Corollary. Let H CG, and suppose ψ is an irreducible character of H. ThenIndGH ψ is irreducible if and only if ψ is distinct from all its conjugates ψg forg ∈ G \H (where ψg(h) = ψ(g−1hg) as before).

Proof. We take K = H C G. So the double cosets are just left cosets. Also,Hg = H for all g. Moreover, Wg is irreducible since W is irreducible.

So, by Mackey’s irreducible criterion, IndGHW is irreducible precisely ifW 6∼= Wg for all g ∈ G \H. This is equivalent to ψ 6= ψg.

Note that again we could check the conditions on a set of representatives of(double) cosets. In fact, the isomorphism class of Wg (for g ∈ G) depends onlyon the coset gH.

We now prove Mackey’s theorem.

Theorem (Mackey’s restriction formula). In general, for K,H ≤ G, we letS = {1, g1, · · · , gr} be a set of double coset representatives, so that

G =⋃KgiH.

We write Hg = gHg−1 ∩K ≤ G. We let (ρ,W ) be a representation of H. Foreach g ∈ G, we define (ρg,Wg) to be a representation of Hg, with the sameunderlying vector space W , but now the action of Hg is

ρg(x) = ρ(g−1xg),

where h = g−1xg ∈ H by construction.This is clearly well-defined. Since Hg ≤ K, we obtain an induced representa-

tion IndKHg Wg.Let G be finite, H,K ≤ G, and W be a H-space. Then

ResGK IndGHW =⊕g∈S

IndKHg Wg.

This is possibly the hardest and most sophisticated proof in the course.

Proof. Write V = IndGHW . Pick g ∈ G, so that KgH ∈ K\G/H. Given a lefttransversal T of H in G, we can obtain V explicitly as a direct sum

V =⊕t∈T

t⊗W.

The idea is to “coarsen” this direct sum decomposition using double cosetrepresentatives, by collecting together the t⊗W ’s with t ∈ KgH. We define

V (g) =⊕

t∈KgH∩Tt⊗W.

Now each V (g) is a K-space — given k ∈ K and t⊗w ∈ t⊗W , since t ∈ KgH,we have kt ∈ KgH. So there is some t′ ∈ T such that ktH = t′H. Thent′ ∈ ktH ⊆ KgH. So we can define

k · (t⊗w) = t′ ⊗ (ρ(t′−1kt)w),

63


where t′kt ∈ H.Viewing V as a K-space (forgetting its whole G-structure), we have

ResGK V =⊕g∈S

V (g).

The left hand side is what we want, but the right hand side looks absolutelynothing like IndKHg Wg. So we need to show

V (g) =⊕

t∈KgH∩Tt⊗W ∼= IndKHg Wg,

as K representations, for each g ∈ S.Now for g fixed, each t ∈ KgH can be represented by some kgh, and by

restricting to elements in the traversal T of H, we are really summing over cosetskgH. Now cosets kgH are in bijection with cosets k(gHg−1) in the obvious way.So we are actually summing over elements in K/(gHg−1 ∩K) = K/Hg. So wewrite

V (g) =⊕

k∈K/Hg

(kg)⊗W.

We claim that there is a isomorphism that sends k⊗Wg∼= (kg)⊗W . We define

k⊗Wg → (kg)⊗W by k⊗w 7→ kg⊗w. This is an isomorphism of vector spacesalmost by definition, so we only check that it is compatible with the action. Theaction of x ∈ K on the left is given by

ρg(x)(k ⊗w) = k′ ⊗ (ρg(k′−1xk)w) = k′ ⊗ (ρ(g−1k′−1xkg)w),

where k′ ∈ K is such that k′−1xk ∈ Hg, i.e. g−1k′−1xkg ∈ H. On the otherhand,

ρ(x)(kg ⊗w) = k′′ ⊗ (ρ(k′′x−1(kg))w),

where k′′ ∈ K is such that k′′−1xkg ∈ H. Since there is a unique choice of k′′

(after picking a particular transversal), and k′g works, we know this is equal to

k′g ⊗ (ρ(g−1k′−1xkg)w).

So the actions are the same. So we have an isomorphism.Then

V (g) =⊕

k∈K/Hg

k ⊗Wg = IndKHg Wg,

as required.

Example. Let g = Sn and H = An. Consider a σ ∈ Sn. Then its conjugacyclass in Sn is determined by its cycle type.

If the class of σ splits into two classes in An, and χ is an irreducible characterof An that takes different values on the two classes, then by the irreducibilitycriterion, IndSnAn χ is irreducible.

64

13 Integrality in the group algebra II Representation Theory

13 Integrality in the group algebra

The next big result we are going to have is that the degree of an irreduciblecharacter always divides the group order. This is not at all an obvious fact. Toprove this, we make the cunning observation that character degrees and friendsare not just regular numbers, but algebraic integers.

Definition (Algebraic integers). A complex number a ∈ C is an algebraic integerif a is a root of a monic polynomial with integer coefficients. Equivalently, a issuch that the subring of C given by

Z[a] = {f(a) : f ∈ Z[X]}

is finitely generated. Equivalently, a is the eigenvalue of a matrix, all of whoseentries lie in Z.

We will quote some results about algebraic integers which we will not prove.

Proposition.

(i) The algebraic integers form a subring of C.

(ii) If a ∈ C is both an algebraic integer and rational, then a is in fact aninteger.

(iii) Any subring of C which is a finitely generated Z-module consists of algebraicintegers.

The strategy is to show that given a group G and an irreducible character χ,

the value |G|χ(1) is an algebraic integer. Since it is also a rational number, it must

be an integer. So |G| is a multiple of χ(1).We make the first easy step.

Proposition. If χ is a character of G and g ∈ G, then χ(g) is an algebraicinteger.

Proof. We know χ(g) is the sum of roots nth roots of unity (where n is the orderof g). Each root of unity is an algebraic integer, since it is by definition a root ofxn − 1. Since algebraic integers are closed under addition, the result follows.

We now have a look at the center of the group algebra CG. Recall that thegroup algebra CG of a finite group G is a complex vector space generated byelements of G. More precisely,

CG =

∑g∈G

αgg : αg ∈ C

.

Borrowing the multiplication of G, this forms a ring, hence an algebra. We nowlist the conjugacy classes of G as

{1} = C1, · · · , Ck.

Definition (Class sum). The class sum of a conjugacy class Cj of a group G is

Cj =∑g∈Cj

g ∈ CG.

65


The claim is now Cj lives in the center of CG (note that the center of thegroup algebra is different from the group algebra of the center). Moreover, theyform a basis:

Proposition. The class sums C1, · · · , Ck form a basis of Z(CG). There existsnon-negative integers aij` (with 1 ≤ i, j, ` ≤ k) with

CiCj =

k∑`=1

aij`C`.

The content of this proposition is not that CiCj can be written as a sum ofC`’s. This is true because the C`’s form a basis. The content is that these areactually integers, not arbitrary complex numbers.

Definition (Class algebra/structure constants). The constants aij` as definedabove are the class algebra constants or structure constants.

Proof. It is clear from definition that gCjg−1 = Cj . So we have Cj ∈ Z(CG).

Also, since the Cj ’s are produced from disjoint conjugacy classes, they are linearlyindependent.

Now suppose z ∈ Z(CG). So we can write

z =∑g∈G

αgg.

By definition, this commutes with all elements of CG. So for all h ∈ G, we musthave

αh−1gh = αg.

So the function g 7→ αg is constant on conjugacy classes of G. So we can writeαj = αg for g ∈ Cj . Then

g =

k∑j=1

αjCj .

Finally, the center Z(CG) is an algebra. So

CiCj =

k∑`=1

aij`C`

for some complex numbers aij`, since the Cj span. The claim is that aij` ∈ Z≥0

for all i, j`. To see this, we fix g` ∈ C`. Then by definition of multiplication, weknow

aij` = |{(x, y) ∈ Ci × Cj : xy = g`}|,

which is clearly a non-negative integer.

Let ρ : G→ GL(V ) be an irreducible representation of G over C affordingthe character χ. Extending linearly, we get a map ρ : CG→ EndV , an algebrahomomorphism. In general, we have the following definition:

Definition (Representation of algebra). Let A be an algebra. A representationof A is a homomorphism of algebras ρ : A→ EndV .

66


Let z ∈ Z(CG). Then ρ(z) commutes with all ρ(g) for g ∈ G. Hence, bySchur’s lemma, we can write ρ(z) = λzI for some λz ∈ C. We then obtain thealgebra homomorphism

ωρ = ωχ = ω : Z(CG)→ Cz 7→ λz.

By definition, we have ρ(Ci) = ω(Ci)I. Taking traces of both sides, we know∑g∈Ci

χ(g) = χ(1)ω(Ci).

We also know the character is a class function. So we in fact get

χ(1)ω(Ci) = |Ci|χ(gi),

where gi ∈ Ci is a representative of Ci. So we get

ω(Ci) =χ(gi)

χ(1)|Ci|.

Why should we care about this? The thing we are looking at is in fact analgebraic integer.

Lemma. The values of

ωχ(Ci) =χ(g)

χ(1)|Ci|

are algebraic integers.

Note that all the time, we are requiring χ to be an irreducible character.

Proof. Using the definition of aij` ∈ Z≥0, and the fact that ωχ is an algebrahomomorphism, we get

ωχ(Ci)ωχ(Cj) =

k∑`=1

aij`ωχ(C`).

Thus the span of {ω(Cj) : 1 ≤ j ≤ k} is a subring of C and is finitely generatedas a Z-module (by definition). So we know this consists of algebraic integers.

That’s magic.In fact, we can compute the structure constants aij` from the character table,

namely for all i, j, `, we have

aij` =|G|

|CG(gi)||CG(gj)|

k∑s=1

χs(gi)χs(gj)χs(g−1` )

χs(1),

where we sum over the irreducible characters of G. We will neither prove it noruse it. But if you want to try, the key idea is to use column orthogonality.

Finally, we get to the main result:

67


Theorem. The degree of any irreducible character of G divides |G|, i.e.

χj(1) | |G|

for each irreducible χj .

It is highly non-obvious why this should be true.

Proof. Let χ be an irreducible character. By orthogonality, we have

|G|χ(1)

=1

χ(1)

∑g∈G

χ(g)χ(g−1)

=1

χ(1)

k∑i=1

|Ci|χ(gi)χ(g−1i )

=

k∑i=1

|Ci|χ(gi)

χ(1)χ(gi)

−1.

Now we notice|Ci|χ(gi)

χ(1)

is an algebraic integer, by the previous lemma. Also, χ(g−1i ) is an algebraic

integer. So the whole mess is an algebraic integer since algebraic integers areclosed under addition and multiplication.

But we also know |G|χ(1) is rational. So it must be an integer!

Example. Let G be a p-group. Then χ(1) is a p-power for each irreducible χ.If in particular |G| = p2, then χ(1) = 1, p or p2. But the sum of squares of thedegrees is |G| = p2. So it cannot be that χ(1) = p or p2. So χ(1) = 1 for allirreducible characters. So G is abelian.

Example. No simple group has an irreducible character of degree 2. Proof isleft as an exercise for the reader.

68

14 Burnside’s theorem II Representation Theory

14 Burnside’s theorem

Finally, we get to Burnside’s theorem.

Theorem (Burside’s paqb theorem). Let p, q be primes, and let |G| = paqb,where a, b ∈ Z≥0, with a+ b ≥ 2. Then G is not simple.

Note that if a+ b = 1 or 0, then the group is trivially simple.In fact even more is true, namely that G is soluble, but this follows easily

from above by induction.This result is the best possible in the sense that |A5| = 60 = 22 · 3 · 5, and

A5 is simple. So we cannot allow for three different primes factors (in fact, thereare exactly 8 simple groups with three prime factors). Also, if a = 0 or b = 0,then G is a p-group, which have non-trivial center. So these cases trivially work.

Later, in 1963, Feit and Thompson proved that every group of odd order issoluble. The proof was 255 pages long. We will not prove this.

In 1972, H. Bender found the first proof of Burnside’s theorem without theuse of representation theory, but the proof is much more complicated.

This theorem follows from two lemmas, and one of them involves some Galoistheory, and is hence non-examinable.

Lemma. Suppose

α =1

m

m∑j=1

λj ,

is an algebraic integer, where λnj = 1 for all j and some n. Then either α = 0 or|α| = 1.

Proof (non-examinable). Observe α ∈ F = Q(ε), where ε = e2πi/n (since λj ∈ Ffor all j). We let G = Gal(F/Q). Then

{β ∈ F : σ(β) = β for all σ ∈ G} = Q.

We define the “norm”N(α) =

∏σ∈G

σ(α).

Then N(α) is fixed by every element σ ∈ G. So N(α) is rational.Now N(α) is an algebraic integer, since Galois conjugates σ(α) of algebraic

integers are algebraic integers. So in fact N(α) is an integer. But for α ∈ G, weknow

|σ(α)| =∣∣∣∣ 1

m

∑σ(λj)

∣∣∣∣ ≤ 1.

So if α 6= 0, then N(α) = ±1. So |α| = 1.

Lemma. Suppose χ is an irreducible character of G, and C is a conjugacy classin G such that χ(1) and |C| are coprime. Then for g ∈ C, we have

|χ(g)| = χ(1) or 0.

Proof. Of course, we want to consider the quantity

α =χ(g)

χ(1).

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14 Burnside’s theorem II Representation Theory

Since χ(g) is the sum of degχ = χ(1) many roots of unity, it suffices to showthat α is an algebraic integer.

By Bezout’s theorem, there exists a, b ∈ Z such that

aχ(1) + b|C| = 1.

So we can write

α =χ(g)

χ(1)= aχ(g) + b

χ(g)

χ(1)|C|.

Since χ(g) and χ(g)χ(1) |C| are both algebraic integers, we know α is.

Proposition. If in a finite group, the number of elements in a conjugacy classC is of (non-trivial) prime power order, then G is not non-abelian simple.

Proof. Suppose G is a non-abelian simple group, and let 1 6= g ∈ G be living inthe conjugacy class C of order pr. If χ 6= 1G is a non-trivial irreducible characterof G, then either χ(1) and |C| = pr are not coprime, in which case p | χ(1), orthey are coprime, in which case |χ(g)| = χ(1) or χ(g) = 0.

However, it cannot be that |χ(g)| = χ(1). If so, then we must have ρ(g) = λIfor some λ. So it commutes with everything, i.e. for all h, we have

ρ(gh) = ρ(g)ρ(h) = ρ(h)ρ(g) = ρ(hg).

Moreover, since G is simple, ρ must be faithful. So we must have gh = hg for allh. So Z(G) is non-trivial. This is a contradiction. So either p | χ(1) or χ(g) = 0.

By column orthogonality applied to C and 1, we get

0 = 1 +∑

16=χ irreducible, p|χ(1)

χ(1)χ(g),

where we have deleted the 0 terms. So we get

−1

p=∑χ 6=1

χ(1)

pχ(g).

But this is both an algebraic integer and a rational number, but not integer.This is a contradiction.

We can now prove Burnside’s theorem.

Theorem (Burside’s paqb theorem). Let p, q be primes, and let |G| = paqb,where a, b ∈ Z≥0, with a+ b ≥ 2. Then G is not simple.

Proof. Let |G| = paqb. If a = 0 or b = 0, then the result is trivial. Supposea, b > 0. We let Q ∈ Sylq(G). Since Q is a p-group, we know Z(Q) is non-trivial.Hence there is some 1 6= g ∈ Z(Q). By definition of center, we know Q ≤ CG(g).Also, CG(g) is not the whole of G, since the center of G is trivial. So

|CG(g)| = |G : CG(g)| = pr

for some 0 < r ≤ a. So done.

This is all we’ve got to say about finite groups.

70

15 Representations of compact groups II Representation Theory

15 Representations of compact groups

We now consider the case of infinite groups. It turns out infinite groups don’tbehave well if we don’t impose additional structures. When talking aboutrepresentations of infinite groups, things work well only if we impose a topologicalstructure on the group, and require the representations to be continuous. Thenwe get nice results if we only concentrate on the groups that are compact. Thisis a generalization of our previous results, since we can give any finite group thediscrete topology, and then all representations are continuous, and the group iscompact due to finiteness.

We start with the definition of a topological group.

Definition (Topological group). A topological group is a group G which is alsoa topological space, and for which multiplication G×G→ G ((h, g) 7→ hg) andinverse G→ G (g 7→ g−1) are continuous maps.

Example. Any finite group G with the discrete topology is a topological group.

Example. The matrix groups GLn(R) and GLn(C) are topological groups

(inheriting the topology of Rn2

and Cn2

).

However, topological groups are often too huge. We would like to concentrateon “small” topological groups.

Definition (Compact group). A topological group is a compact group if it iscompact as a topological space.

There are some fairly nice examples of these.

Example. All finite groups with discrete topology are compact.

Example. The group S1 = {z ∈ C : |z| = 1} under multiplication is a compactgroup. This is known as the circle group, for mysterious reasons. Thus the torusS1 × S1 × · · · × S1 is also a compact group.

Example. The orthogonal group O(n) ≤ GLn(R) is compact, and so is SO(n) ={A ∈ O(n) : detA = 1}. These are compact since they are closed and bounded

as a subspace of Rn2

, as the entries can be at most 1 in magnitude. Note thatSO(2) ∼= S1, mapping a rotation by θ to eiθ.

Example. The unitary group U(n) = {A ∈ GLn(C) : AA† = I} is compact,and so is SU(n) = {A ∈ U(n) : detA = 1}.

Note that we also have U(1) ∼= SO(2) ∼= S1. These are not only isomorphismsas group, but these isomorphisms are also homeomorphisms of topological spaces.

The one we are going to spend most of our time on is

SU(2) = {(z1, z2) ∈ C2 : |z1|2 + |z2|2 = 1} ≤ R4 = C2.

We can see this is homeomorphic to S3.It turns out that the only spheres on which we can define group operations

on are S1 and S3, but we will not prove this, since this is a course on algebra,not topology.

We now want to develop a representation theory of these compact groups,similar to what we’ve done for finite groups.

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Definition (Representation of topological group). A representation of a topolog-ical group on a finite-dimensional space V is a continuous group homomorphismG→ GL(V ).

It is important that the group homomorphism is continuous. Note that for ageneral topological space X, a map ρ : X → GL(V ) ∼= GLn(C) is continuous ifand only if the component maps x 7→ ρ(x)ij are continuous for all i, j.

The key that makes this work is the compactness of the group and thecontinuity of the representations. If we throw them away, then things will govery bad. To see this, we consider the simple example

S1 = U(1) = {g ∈ C× : |g| = 1} ∼= R/Z,

where the last isomorphism is an abelian group isomorphism via the mapx 7→ e2πix : R/Z→ S1.

What happens if we just view S1 as an abstract group, and not a topologicalgroup? We can view R as a vector space over Q, and this has a basis (by Hamel’sbasis theorem), say, A ⊆ R. Moreover, we can assume the basis is ordered, andwe can assume 1 ∈ A.

As abelian groups, we then have

R ∼= Q⊕⊕

α∈A\{1}

Qα,

Then this induces an isomorphism

R/Z ∼= Q/Z⊕⊕

α∈A\{1}

Qα.

Thus, as abstract groups, S1 has uncountably many irreducible representations,namely for each λ ∈ A \ {1}, there exists a one-dimensional representation givenby

ρλ(e2πiµ) =

{1 µ 6∈ Qλe2πiµ µ ∈ Qλ

We see ρλ = ρλ′ if and only if Qλ = Qλ′. Hence there are indeed uncountablymany of these. There are in fact even more irreducible representation we haven’tlisted. This is bad.

The idea is then to topologize S1 as a subset of C, and study how it actsnaturally on complex spaces in a continuous way. Then we can ensure that wehave at most countably many representations. In fact, we can characterize allirreducible representations in a rather nice way:

Theorem. Every one-dimensional (continuous) representation S1 is of the form

ρ : z 7→ zn

for some n ∈ Z.

This is a nice countable family of representations. To prove this, we needtwo lemmas from real analysis.

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Lemma. If ψ : (R,+) → (R,+) is a continuous group homomorphism, thenthere exists a c ∈ R such that

ψ(x) = cx

for all x ∈ R.

Proof. Given ψ : (R,+) → (R,+) continuous, we let c = ψ(1). We now claimthat ψ(x) = cx.

Since ψ is a homomorphism, for every n ∈ Z≥0 and x ∈ R, we know

ψ(nx) = ψ(x+ · · ·+ x) = ψ(x) + · · ·+ ψ(x) = nψ(x).

In particular, when x = 1, we know ψ(n) = cn. Also, we have

ψ(−n) = −ψ(n) = −cn.

Thus ψ(n) = cn for all n ∈ Z.We now put x = m

n ∈ Q. Then we have

mψ(x) = ψ(nx) = ψ(m) = cm.

So we must haveψ(mn

)= c

m

n.

So we get ψ(q) = cq for all q ∈ Q. But Q is dense in R, and ψ is continuous. Sowe must have ψ(x) = cx for all x ∈ R.

Lemma. Continuous homomorphisms ϕ : (R,+)→ S1 are of the form

ϕ(x) = eicx

for some c ∈ R.

Proof. Let ε : (R,+)→ S1 be defined by x 7→ eix. This homomorphism wrapsthe real line around S1 with period 2π.

We now claim that given any continuous function ϕ : R → S1 such thatϕ(0) = 1, there exists a unique continuous lifting homomorphism ψ : R → Rsuch that

ε ◦ ψ = ϕ, ψ(0) = 0.

(R,+) 0

(R,+) S1

ε

ϕ∃!ψ

The lifting is constructed by starting with ψ(0) = 0, and then extending a smallinterval at a time to get a continuous map R → R. We will not go into thedetails. Alternatively, this follows from the lifting criterion from IID AlgebraicTopology.

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We now claim that if in addition ϕ is a homomorphism, then so is itscontinuous lifting ψ. If this is true, then we can conclude that ψ(x) = cx forsome c ∈ R. Hence

ϕ(x) = eicx.

To show that ψ is indeed a homomorphism, we have to show that ψ(x+ y) =ψ(x) + ψ(y).

By definition, we know

ϕ(x+ y) = ϕ(x)ϕ(y).

By definition of ψ, this means

ε(ψ(x+ y)− ψ(x)− ψ(y)) = 1.

We now look at our definition of ε to get

ψ(x+ y)− ψ(x)− ψ(y) = 2kπ

for some integer k ∈ Z, depending continuously on x and y. But k can only bean integer. So it must be constant. Now we pick our favorite x and y, namelyx = y = 0. Then we find k = 0. So we get

ψ(x+ y) = ψ(x) + ψ(y).

So ψ is a group homomorphism.

With these two lemmas, we can prove our characterization of the (one-dimensional) representations of S1.

Theorem. Every one-dimensional (continuous) representation S1 is of the form

ρ : z 7→ zn

for some n ∈ Z.

Proof. Let ρ : S1 → C× be a continuous representation. We now claim that ρactually maps S1 to S1. Since S1 is compact, we know ρ(S1) has closed andbounded image. Also,

ρ(zn) = (ρ(z))n

for all n ∈ Z. Thus for each z ∈ S1, if |ρ(z)| > 1, then the image of ρ(zn) isunbounded. Similarly, if it is less than 1, then ρ(z−n) is unbounded. So we musthave ρ(S1) ⊆ S1. So we get a continuous homomorphism

R→ S1

x 7→ ρ(eix).

So we know there is some c ∈ R such that

ρ(eix) = eicx,

Now in particular,1 = ρ(e2πi) = e2πic.

This forces c ∈ Z. Putting n = c, we get

ρ(z) = zn.

74


That has taken nearly an hour, and we’ve used two not-so-trivial facts fromanalysis. So we actually had to work quite hard to get this. But the result isgood. We have a complete list of representations, and we don’t have to fiddlewith things like characters.

Our next objective is to repeat this for SU(2), and this time we cannot justget away with doing some analysis. We have to do representation theory properly.So we study the general theory of representations of complex spaces.

In studying finite groups, we often took the “average” over the group viathe operation 1

|G|∑g∈G something. Of course, we can’t do this now, since the

group is infinite. As we all know, the continuous analogue of summing is calledintegration. Informally, we want to be able to write something like

∫G

dg.This is actually a measure, called the “Haar measure”, and always exists as

long as you are compact and Hausdorff. However, we will not need or use thisresult, since we can construct it by hand for S1 and SU(2).

Definition (Haar measure). Let G be a topological group, and let

C(G) = {f : G→ C : f is continuous, f(gxg−1) = f(x) for all g, x ∈ G}.

A non-trivial linear functional∫G

: C(G)→ C, written as∫G

f =

∫G

f(g) dg

is called a Haar measure if

(i) It satisfies the normalization condition∫G

1 dg = 1

so that the “total volume” is 1.

(ii) It is left and right translation invariant, i.e.∫G

f(xg) dg =

∫G

f(g) dg =

∫G

f(gx) dg

for all x ∈ G.

Example. For a finite group G, we can define∫G

f(g) dg =1

|G|∑g∈G

f(g).

The division by the group order is the thing we need to make the normalizationhold.

Example. Let G = S1. Then we can have∫G

f(g) dg =1

2π

∫ 2π

0

f(eiθ) dθ.

Again, division by 2π is the normalization required.

75


We have explicitly constructed them here, but there is a theorem thatguarantees the existence of such a measure for sufficiently nice topological group.

Theorem. Let G be a compact Hausdorff topological group. Then there existsa unique Haar measure on G.

From now on, the word “compact” will mean “compact and Hausdorff”, sothat the conditions of the above theorem holds.

We will not prove theorem, and just assume it to be true. If you don’t likethis, whenever you see “compact group”, replace it with “compact group withHaar measure”. We will explicitly construct it for SU(2) later, which is all wereally care about.

Once we know the existence of such a thing, most of the things we know forfinite groups hold for compact groups, as long as we replace the sums with theappropriate integrals.

Corollary (Weyl’s unitary trick). Let G be a compact group. Then everyrepresentation (ρ, V ) has a G-invariant Hermitian inner product.

Proof. As for the finite case, take any inner product ( · , · ) on V , then define anew inner product by

〈v,w〉 =

∫G

(ρ(g)v, ρ(g)w) dg.

Then this is a G-invariant inner product.

Theorem (Maschke’s theoerm). Let G be compact group. Then every repre-sentation of G is completely reducible.

Proof. Given a representation (ρ, V ). Choose a G-invariant inner product. IfV is not irreducible, let W ≤ V be a subrepresentation. Then W⊥ is alsoG-invariant, and

V = W ⊕W⊥.Then the result follows by induction.

We can use the Haar measure to endow C(G) with an inner product given by

〈f, f ′〉 =

∫G

f(g)f ′(g) dg.

Definition (Character). If ρ : G→ GL(V ) is a representation, then the char-acter χρ = tr ρ is a continuous class function, since each component ρ(g)ij iscontinuous.

So characters make sense, and we can ask if all the things about finite groupshold here. For example, we can ask about orthogonality.

Schur’s lemma also hold, with the same proof, since we didn’t really needfiniteness there. Using that, we can prove the following:

Theorem (Orthogonality). Let G be a compact group, and V and W beirreducible representations of G. Then

〈χV , χW 〉 =

{1 V ∼= W

0 V 6∼= W.

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Now do irreducible characters form a basis for C(G)?

Example. We take the only (infinite) compact group we know about — G = S1.We have found that the one-dimensional representations are

ρn : z 7→ zn

for n ∈ Z. As S1 is abelian, these are all the (continuous) irreducible representa-tions — given any representation ρ, we can find a simultaneous eigenvector foreach ρ(g).

The “character table of S1” has rows χn indexed by Z, with χn(eiθ) = einθ.Now given any representation of S1, say V , we can break V as a direct sum

of 1-dimensional subrepresentations. So the character χV of V is of the form

χV (z) =∑n∈Z

anzn,

where an ∈ Z≥0, and only finitely many an are non-zero (since we are assumingfinite-dimensionality).

Actually, an is the number of copies of ρn in the decomposition of V . Wecan find out the value of an by computing

an = 〈χn, χV 〉 =1

2π

∫ 2π

0

e−inθχV (eiθ) dθ.

Hence we know

χV (eiθ) =∑n∈Z

(1

2π

∫ 2π

0

χV (eiθ′)e−inθ

′dθ′)einθ.

This is really just a Fourier decomposition of χV . This gives a decompositionof χV into irreducible characters, and the “Fourier mode” an is the number ofeach irreducible character occurring in this decomposition.

It is possible to show that χn form a complete orthonormal set in the Hilbertspace L2(S1), i.e. square-integrable functions on S1. This is the Peter-Weyltheorem, and is highly non-trivial.

15.1 Representations of SU(2)

In the rest of the time, we would like to talk about G = SU(2).Recall that

SU(2) = {A ∈ GL2(C) : A†A = I, detA = I}.

Note that if

A =

(a bc d

)∈ SU(2),

then since detA = 1, we have

A−1 =

(d −b−c a

).

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So we need d = a and c = −b. Moreover, we have

aa+ bb = 1.

Hence we can write SU(2) explicitly as

G =

{(a b−b a

): a, b ∈ C, |a|2 + |b|2 = 1

}.

Topologically, we know G ∼= S3 ⊆ C2 ∼= R4.Instead of thinking about C2 in the usual vector space way, we can think of

it as a subgroup of M2(C) via

H =

{(z w−w z

): w, z ∈ C

}≤M2(C).

This is known as Hamilton’s quaternion algebra. Then H is a 4-dimensionalEuclidean space (two components from z and two components from w), with anorm on H given by

‖A‖2 = detA.

We now see that SU(2) ≤ H is exactly the unit sphere ‖A‖2 = 1 in H. IfA ∈ SU(2) and x ∈ H, then ‖Ax‖ = ‖x‖ since ‖A‖ = 1. So elements of G actsas isometries on the space.

After normalization (by 12π2 ), the usual integration of functions on S3 defines

a Haar measure on G. It is an exercise on the last example sheet to write thisout explicitly.

We now look at conjugacy in G. We let

T =

{(a 00 a

): a ∈ C, |a|2 = 1

}∼= S1.

This is a maximal torus in G, and it plays a fundamental role, since we happento know about S1. We also have a favorite element

s =

(0 1−1 0

)∈ SU(2).

We now prove some easy linear algebra results about SU(2).

Lemma (SU(2)-conjugacy classes).

(i) Let t ∈ T . Then sts−1 = t−1.

(ii) s2 = −I ∈ Z(SU(2)).

(iii) The normalizer

NG(T ) = T ∪ sT =

{(a 00 a

),

(0 a−a 0

): a ∈ C, |a| = 1

}.

(iv) Every conjugacy class C of SU(2) contains an element of T , i.e. C ∩ T 6= ∅.

(v) In fact,C ∩ T = {t, t−1}

for some t ∈ T , and t = t−1 if and only if t = ±I, in which case C = {t}.

78


(vi) There is a bijection

{conjugacy classes in SU(2)} ↔ [−1, 1],

given by

A 7→ 1

2trA.

We can see that if

A =

(λ 00 λ

),

then1

2trA =

1

2(λ+ λ) = Re(λ).

Note that what (iv) and (v) really say is that matrices in SU(2) are diago-nalizable (over SU(2)).

Proof.

(i) Write it out.

(ii) Write it out.

(iii) Direct verification.

(iv) It is well-known from linear algebra that every unitary matrix X has anorthonormal basis of eigenvectors, and hence is conjugate in U(2) to onein T , say

QXQ† ∈ T.We now want to force Q into SU(2), i.e. make Q have determinant 1.

We put δ = detQ. Since Q is unitary, i.e. QQ† = I, we know |δ| = 1. Sowe let ε be a square root of δ, and define

Q1 = ε−1Q.

Then we haveQ1XQ

†1 ∈ T.

(v) We let g ∈ G, and suppose g ∈ C. If g = ±I, then C ∩ T = {g}. Otherwise,g has two distinct eigenvalues λ, λ−1. Note that the two eigenvlaues mustbe inverses of each other, since it is in SU(2). Then we know

C =

{h

(λ 00 λ−1

)h−1 : h ∈ G

}.

Thus we find

C ∩ T =

{(λ 00 λ−1

),

(λ−1 0

0 λ

)}.

This is true since eigenvalues are preserved by conjugation, so if any(µ 00 µ−1

),

then {µ, µ−1} = {λ, λ−1}. Also, we can get the second matrix from thefirst by conjugating with s.

79


(vi) Consider the map

1

2tr : {conjugacy classes} → [−1, 1].

By (v), matrices are conjugate in G iff they have the same set of eigenvalues.Now

1

2tr

(λ 00 λ−1

)=

1

2(λ+ λ) = Re(λ) = cos θ,

where λ = eiθ. Hence the map is a surjection onto [−1, 1].

Now we have to show it is injective. This is also easy. If g and g′ have thesame image, i.e.

1

2tr g =

1

2tr g′,

then g and g′ have the same characteristic polynomial, namely

x2 − (tr g)x+ 1.

Hence they have the same eigenvalues, and hence they are similar.

We now write, for t ∈ (−1, 1),

Ct =

{g ∈ G :

1

2tr g = t

}.

In particular, we haveC1 = {I}, C−1 = {−I}.

Proposition. For t ∈ (−1, 1), the class Ct ∼= S2 as topological spaces.

This is not really needed, but is a nice thing to know.

Proof. Exercise!

We now move on to classify the irreducible representations of SU(2).We let Vn be the complex space of all homogeneous polynomials of degree n

in variables x, y, i.e.

Vn = {r0xn + r1x

n−1y + · · ·+ rnyn : r0, · · · , rn ∈ C}.

This is an n+ 1-dimensional complex vector space with basis xn, xn−1y, · · · , yn.We want to get an action of SU(2) on Vn. It is easier to think about the

action of GL2(C) in general, and then restrict to SU(2). We define the action ofGL2(C) on Vn by

ρn : GL2(C)→ GL(Vn)

given by the rule

ρn

((a bc d

))f(x, y) = f(ax+ cy, bx+ dy).

In other words, we have

ρn

((a bc d

))f((x y

))= f

((x y

)(a bc d

)),

where the multiplication in f is matrix multiplication.

80


Example. When n = 0, then V0∼= C, and ρ0 is the trivial representation.

When n = 1, then this is the natural two-dimensional representation, and

ρ1

((a bc d

))has matrix (

a bc d

)with respect to the standard basis {x, y} of V1 = C2.

More interestingly, when n = 2, we know

ρ2

((a bc d

))has matrix a2 ab b2

2ac ad+ bc 2bdc2 cd d2

,

with respect to the basis x2, xy, y2 of V2 = C3. We obtain the matrix bycomputing, e.g.

ρ2(g)(x2) = (ax+ cy)2 = a2x2 + 2acxy + c2y2.

Now we know SU(2) ≤ GL2(C). So we can view Vn as a representation ofSU(2) by restriction.

Now we’ve got some representations. The claim is that these are all theirreducibles. Before we prove that, we look at the character of these things.

Lemma. A continuous class function f : G→ C is determined by its restrictionto T , and F |T is even, i.e.

f

((λ 00 λ−1

))= f

((λ−1 0

0 λ

)).

Proof. Each conjugacy class in SU(2) meets T . So a class function is determinedby its restriction to T . Evenness follows from the fact that the two elements areconjugate.

In particular, a character of a representation (ρ, V ) is also an even functionχρ : S1 → C.

Lemma. If χ is a character of a representation of SU(2), then its restrictionχ|T is a Laurent polynomial, i.e. a finite N-linear combination of functions(

λ 00 λ−1

)7→ λn

for n ∈ Z.

Proof. If V is a representation of SU(2), then ResSU(2)T V is a representation

of T , and its character ResSU(2)T χ is the restriction of χV to T . But every

representation of T has its character of the given form. So done.

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Notation. We write N[z, z−1] for the set of all Laurent polynomials, i.e.

N[z, z−1] ={∑

anzn : an ∈ N : only finitely many an non-zero

}.

We further write

N[z, z−1]ev = {f ∈ N[z, z−1] : f(z) = f(z−1)}.

Then by our lemma, for every continuous representation V of SU(2), thecharacter χV ∈ N[z, z−1]ev (by identifying it with its restriction to T ).

We now actually calculate the character χn of (ρn, Vn) as a representation ofSU(2). We have

χVn(g) = tr ρn(g),

where

g ∼(z 00 z−1

)∈ T.

Then we have

ρn

((z 00 z−1

))(xiyj) = (zxi)(z−1yj) = zi−jxiyj .

So each xiyj is an eigenvector of ρn(g), with eigenvalue zi−j . So we know ρn(g)has a matrix

zn

zn−2

. . .

z2−n

z−n

,

with respect to the standard basis. Hence the character is just

χn

((z 00 z−1

))= zn + zn−2 + · · ·+ z−n =

zn+1 − z−(n+1)

z − z−1,

where the last expression is valid unless z = ±1.We can now state the result we are aiming for:

Theorem. The representations ρn : SU(2) → GL(Vn) of dimension n + 1 areirreducible for n ∈ Z≥0.

Again, we get a complete set (completeness proven later). A complete list ofall irreducible representations, given in a really nice form. This is spectacular.

Proof. Let 0 6= W ≤ Vn be a G-invariant subspace, i.e. a subrepresentation ofVn. We will show that W = Vn.

All we know about W is that it is non-zero. So we take some non-zero vectorof W .

Claim. Let

0 6= w =

n∑j=0

rjxn−jyj ∈W.

Since this is non-zero, there is some i such that ri 6= 0. The claim is thatxn−iyi ∈W .

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We argue by induction on the number of non-zero coefficients rj . If there isonly one non-zero coefficient, then we are already done, as w is a non-zero scalarmultiple of xn−iyi.

So assume there is more than one, and choose one i such that ri 6= 0. Wepick z ∈ S1 with zn, zn−2, · · · , z2−n, z−n all distinct in C. Now

ρn

((z

z−1

))w =

∑rjz

n−2jxn−jyj ∈W.

Subtracting a copy of w, we find

ρn

((z

z−1

))w − zn−2iw =

∑rj(z

n−2j − zn−2i)xn−jyj ∈W.

We now look at the coefficient

rj(zn−2j − zn−2i).

This is non-zero if and only if rj is non-zero and j 6= i. So we can use this toremove any non-zero coefficient. Thus by induction, we get

xn−jyj ∈W

for all j such that rj 6= 0.This gives us one basis vector inside W , and we need to get the rest.

Claim. W = Vn.

We now know that xn−iyi ∈W for some i. We consider

ρn

(1√2

(1 −11 1

))xn−iyi =

1√2

(x+ y)n−i(−x+ y)i ∈W.

It is clear that the coefficient of xn is non-zero. So we can use the claim todeduce xn ∈W .

Finally, for general a, b 6= 0, we apply

ρn

((a −bb a

))xn = (ax+ by)n ∈W,

and the coefficient of everything is non-zero. So basis vectors are in W . SoW = Vn.

This proof is surprisingly elementary. It does not use any technology at all.Alternatively, to prove this, we can identify

Ccos θ =

{A ∈ G :

1

2trA = cos θ

}with the 2-sphere

{| Im a|2 + |b|2 = sin2 θ}of radius sin θ. So if f is a class function on G, f is constant on each class Ccos θ.It turns out we get∫

G

f(g) dg =1

2π2

∫ 2π

0

1

2f

((eiθ

e−iθ

))4π sin2 θ dθ.

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This is the Weyl integration formula, which is the Haar measure for SU(2). Thenif χn is the character of Vn, we can use this to show that 〈χn, χn〉 = 1. Henceχn is irreducible. We will not go into the details of this construction.

Theorem. Every finite-dimensional continuous irreducible representation of Gis one of the ρn : G→ GL(Vn) as defined above.

Proof. Assume ρV : G → GL(V ) is an irreducible representation affording acharacter χV ∈ N[z, z−1]ev. We will show that χV = χn for some n. Now we see

χ0 = 1

χ1 = z + z−1

χ2 = z2 + 1 + z−2

...

form a basis of Q[z, z−1]ev, which is a non-finite dimensional vector space overQ. Hence we can write

χV =∑n

anχn,

a finite sum with finitely many an 6= 0. Note that it is possible that an ∈ Q. Sowe clear denominators, and move the summands with negative coefficients tothe left hand side. So we get

mχV +∑i∈I

miχi =∑j∈J

njχj ,

with I, J disjoint finite subsets of N, and m,mi, nj ∈ N.We know the left and right-hand side are characters of representations of G.

So we get

mV ⊕⊕I

miVi =⊕J

njVj .

Since V is irreducible and factorization is unique, we must have V ∼= Vn for somen ∈ J .

We’ve got a complete list of irreducible representations of SU(2). So we canlook at what happens when we take products.

We know that for V,W representations of SU(2), if

ResSU(2)T V ∼= Res

SU(2)T W,

then in factV ∼= W.

This gives us the following result:

Proposition. Let G = SU(2) or G = S1, and V,W are representations of G.Then

χV⊗W = χV χW .

84


Proof. By the previous remark, it is enough to consider the case G = S1. SupposeV and W have eigenbases e1, · · · , en and f1, · · · , fm respectively such that

ρ(z)ei = zniei, ρ(z)fj = zmj fj

for each i, j. Thenρ(z)(ei ⊗ fj) = zni+mjei ⊗ fj .

Thus the character is

χV⊗W (z) =∑i,j

zni+mj =

(∑i

zni

)∑j

zmj

= χV (z)χW (z).

Example. We have

χV1⊗V1(z) = (z + z−1)2 = z2 + 2 + z−2 = χV2 + χV0 .

So we haveV1 ⊗ V1

∼= V2 ⊕ V0.

Similarly, we can compute

χV1⊗V2(z) = (z2 + 1 + z−2)(z + z−1) = z3 + 2z + 2z−1 + z−3 = χV3

+ χV1.

So we getV1 ⊗ V2

∼= V3 ⊕ V1.

Proposition (Clebsch-Gordon rule). For n,m ∈ N, we have

Vn ⊗ Vm ∼= Vn+m ⊕ Vn+m−2 ⊕ · · · ⊕ V|n−m|+2 ⊕ V|n−m|.

Proof. We just check this works for characters. Without loss of generality, weassume n ≥ m. We can compute

(χnχm)(z) =zn+1 − z−n−1

z − z−1(zm + zm−2 + · · ·+ z−m)

=

m∑j=0

zn+m+1−2j − z2j−n−m−1

z − z−1

=

m∑j=0

χn+m−2j(z).

Note that the condition n ≥ m ensures there are no cancellations in the sum.

15.2 Representations of SO(3), SU(2) and U(2)

We’ve now got a complete list of representations of S1 and SU(2). We see ifwe can make some deductions about some related groups. We will not give toomany details about these groups, and the proofs are just sketches. Moreover, wewill rely on the following facts which we shall not prove:

Proposition. There are isomorphisms of topological groups:

85


(i) SO(3) ∼= SU(2)/{±I} = PSU(2)

(ii) SO(4) ∼= SU(2)× SU(2)/{±(I, I)}

(iii) U(2) ∼= U(1)× SU(2)/{±(I, I)}

All maps are group isomorphisms, but in fact also homeomorphisms. To showthis, we can use the fact that a continuous bijection from a Hausdorff space to acompact space is automatically a homeomorphism.

Assuming this is true, we obtain the following corollary;

Corollary. Every irreducible representation of SO(3) has the following form:

ρ2m : SO(3)→ GL(V2m),

for some m ≥ 0, where Vn are the irreducible representations of SU(2).

Proof. Irreducible representations of SO(3) correspond to irreducible representa-tions of SU(2) such that −I acts trivially by lifting. But −I acts on Vn as −1when n is odd, and as 1 when n is even, since

ρ(−I) =

(−1)n

(−1)n−2

. . .

(−1)−n

= (−1)nI.

For the sake of completeness, we provide a (sketch) proof of the isomorphismSO(3) ∼= SU(2)/{±I}.

Proposition. SO(3) ∼= SU(2)/{±I}.

Proof sketch. Recall that SU(2) can be viewed as the sphere of unit normquaternions H ∼= R4.

LetH0 = {A ∈ H : trA = 0}.

These are the “pure” quaternions. This is a three-dimensional subspace of H. Itis not hard to see this is

H0 = R⟨(

i 00 −i

),

(0 1−1 0

),

(0 ii 0

)⟩= R 〈i, j,k〉 ,

where R〈· · ·〉 is the R-span of the things.This is equipped with the norm

‖A‖2 = detA.

This gives a nice 3-dimensional Euclidean space, and SU(2) acts as isometrieson H0 by conjugation, i.e.

X ·A = XAX−1,

giving a group homomorphism

ϕ : SU(2)→ O(3),

86


and the kernel of this map is Z(SU(2)) = {±I}. We also know that SU(2) iscompact, and O(3) is Hausdorff. Hence the continuous group isomorphism

ϕ : SU(2)/{±I} → imϕ

is a homeomorphism. It remains to show that imϕ = SO(3).But we know SU(2) is connected, and det(ϕ(X)) is a continuous function

that can only take values 1 or −1. So det(ϕ(X)) is either always 1 or always −1.But det(ϕ(I)) = 1. So we know det(ϕ(X)) = 1 for all X. Hence imϕ ≤ SO(3).

To show that equality indeed holds, we have to show that all possible rotationsin H0 are possible. We first show all rotations in the i, j-plane are implementedby elements of the form a+ bk, and similarly for any permutation of i, j,k. Sinceall such rotations generate SO(3), we are then done. Now consider(

eiθ 00 e−iθ

)(ai b−b −ai

)(eiθ 00 eiθ

)=

(ai e2iθb

−be−2iθ −ai

).

So (eiθ 00 e−iθ

)acts on R〈i, j,k〉 by a rotation in the (j,k)-plane through an angle 2θ. We cancheck that (

cos θ sin θ− sin θ cos θ

),

(cos θ i sin θi sin θ cos θ

)act by rotation of 2θ in the (i,k)-plane and (i, j)-plane respectively. So done.

We can adapt this proof to prove the other isomorphisms. However, it isslightly more difficult to get the irreducible representations, since it involvestaking some direct products. We need a result about products G ×H of twocompact groups. Similar to the finite case, we get the complete list of irreduciblerepresentations by taking the tensor products V ⊗W , where V and W rangeover the irreducibles of G and H independently.

We will just assert the results.

Proposition. The complete list of irreducible representations of SO(4) is ρm×ρn,where m,n > 0 and m ≡ n (mod 2).

Proposition. The complete list of irreducible representations of U(2) is

det⊗m ⊗ ρn,

where m,n ∈ Z and n ≥ 0, and det is the obvious one-dimensional representation.

87

Index II Representation Theory

Index

2-transitive action, 34Cg, 6G-homomorphism, 8G-invariant inner product, 13G-isomorphism, 9G-module, 7G-space, 7G-subspace, 10R(G), 48S2V , 43S2χ, 44SnV , 47Tnv, 46V ⊗2, 43V ⊗n, 46FG, 7Λ2V , 43Λ2χ, 44ΛnV , 47χS2 , 44χΛ2 , 44C(G), 23CG, 6k(G), 23paqb theorem, 70

afford, 21algebraic integer, 65alternating group, 5

Burside’s theorem, 70

canonical decomposition, 20centralizer, 6character, 21character ring, 48character table, 26circle group, 71class algebra constant, 66class function, 21, 23class number, 23class sum, 65Clebsch-Gordon rule, 85column orthogonality, 30commutator subgroup, 37compact group, 71completely reducible representation,

12

conjugacy class, 6conjugate character, 62cyclic group, 5

decomposable representation, 11degree, 7degree of character, 21derived subgroup, 37dihedral group, 5dimension, 7direct sum, 11double coset, 54dual representation, 39

equivalent representation, 9exterior algebra, 48exterior power, 47exterior square, 43

faithful character, 21faithful representation, 7Frobenius complement, 60Frobenius group, 60Frobenius kernel, 60Frobenius reciprocity, 52Frobenius’ theorem, 58

generalized character, 48group action, 6group algebra, 7

Haar measure, 75Hamilton’s quaternion algebra, 78Hermitian inner product, 13

indecomposable representation, 11induced class function, 51induced representation, 56inner product of class functions, 24intertwine, 8irreducible character, 21irreducible representation, 10isomorphic representation, 9isotypical components, 20

left transversal, 52lifting, 36linear action, 7linear character, 21linear representation, 7

88

Index II Representation Theory

Mackey’s restriction formula, 61Maschke’s theorem, 12matrix representation, 8maximal torus, 78

permutation character, 32permutation representation, 6, 15,

32principal character, 21

quaternion algebra, 78quaternion group, 6

regular module, 15regular representation, 15representation, 7, 72representation of algebra, 66restriction, 50row orthogonality, 28

Schur’s lemma, 16semisimple representation, 12simple representation, 10structure constant, 66subrepresentation, 10symmetric algebra, 48symmetric group, 5symmetric power, 47symmetric square, 43

tensor algebra, 47tensor product, 40topological group, 71trivial character, 21trivial representation, 8

virtual character, 48

Weyl’s unitary trick, 14

89