II Representation Theory (Full)

Part II — Representation Theory

Based on lectures by S. Martin

Notes taken by Dexter Chua

Lent 2016

These notes are not endorsed by the lecturers, and I have modified them (often

significantly) 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 groups

Representations of groups on vector spaces, matrix representations. Equivalence of

representations. Invariant subspaces and submodules. Irreducibility and Schur’s

Lemma. Complete reducibility for finite groups. Irreducible representations of Abelian

groups.

Character theory

Determination of a representation by its character. The group algebra, conjugacy classes,

and orthogonality relations. Regular representation. Permutation representations and

their characters. Induced representations and the Frobenius reciprocity theorem.

Mackey’s theorem. Frobenius’s Theorem. [12]

Arithmetic properties of characters

Divisibility of the order of the group by the degrees of its irreducible characters.

Burnside’s p

theorem. [2]

Tensor products

Tensor products of representations and products of characters. The character ring.

Tensor, symmetric and exterior algebras. [3]

Representations of S

and SU

The groups

and

(3), their irreducible representations, complete reducibility.

The Clebsch-Gordan formula. *Compact groups.* [4]

Further worked examples

The characters of one of GL

), S

or the Heisenberg group. [3]

Contents

0 Introduction

1 Group actions

2 Basic definitions

3 Complete reducibility and Maschke’s theorem

4 Schur’s lemma

5 Character theory

6 Proof of orthogonality

7 Permutation representations

8 Normal subgroups and lifting

9 Dual spaces and tensor products of representations

9.1 Dual spaces

9.2 Tensor products

9.3 Powers of characters

9.4 Characters of G × H

9.5 Symmetric and exterior powers

9.6 Tensor algebra

9.7 Character ring

10 Induction and restriction

11 Frobenius groups

12 Mackey theory

13 Integrality in the group algebra

14 Burnside’s theorem

15 Representations of compact groups

15.1 Representations of SU(2)

15.2 Representations of SO(3), SU(2) and U(2)

0 Introduction

The course studies how groups act as groups of linear transformations on vector

spaces. Hopefully, you understand all the words in this sentence. If so, this is a

good start.

In our case, groups are usually either finite groups or topological compact

groups (to be defined later). Topological compact groups are typically subgroups

of the general linear group over some infinite fields. It turns out the tools we

have 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 the

IB Groups, Rings and Modules course), linear algebra, and, optionally, geometry,

Galois theory and metric and topological spaces. There is one lemma where

we must use something from Galois theory, but if you don’t know about Galois

theory, you can just assume the lemma to be true without bothering yourself

too much about the proofs.

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

, but sometimes it can also be

. These fields

all have characteristic zero, and in this case, we call what we’re doing ordinary

representation theory. Sometimes, we will take

, the algebraic

closure of F

. This is called modular representation theory.

Notation.

We write

for a vector space over

— this will always be finite

dimensional over

. We write

(

) for the group of invertible linear maps

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

be a finite-dimensional vector space over

. We write

End

(

)

for the endomorphism algebra, the set of all linear maps V → V .

We recall a couple of facts from linear algebra:

dim

n < ∞

, we can choose a basis

, ··· , e

over

. So we

can identify

with

. Then every endomorphism

θ ∈ GL

(

) corresponds to a

matrix A

= (a

) ∈ M

(F ) given by

θ(e

) =

In fact, we have

∈ GL

(

), the general linear group. It is easy to see the

following:

Proposition.

As groups,

(

)

∼

(

), with the isomorphism given by

θ 7→ A

Of course, picking a different basis of

gives a different isomorphism to

(F), but we have the following fact:

Proposition.

Matrices

, A

represent the same element of

(

) with respect

to different bases if and only if they are conjugate, namely there is some

X ∈

(F) such that

= XA

−1

Recall that

(

) =

, where

= (

)

∈ M

(

), is the trace of

. 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

(

) =

(

), which is

independent of our choice of basis. This is an important result. When we study

representations, 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 our

problem of studying matrices to plain arithmetic.

When we have too many matrices, we get confused. So we want to put a

matrix into a form as simple as possible. One of the simplest form a matrix

can take is being diagonal. So we want to know something about diagonalizing

matrices.

Proposition.

Let

α ∈ GL

(

), where

is a finite-dimensional vector space over

C and α

= id for some positive integer m. Then α is diagonalizable.

This follows from the following more general fact:

Proposition.

Let

be a finite-dimensional vector space over

, and

α ∈

End

(

), not necessarily invertible. Then

is diagonalizable if and only if there

is a polynomial f with distinct linear factors such that f (α) = 0.

Indeed, we have x

− 1 =

(x − ω

), where ω = e

2πi/m

Instead of just one endomorphism, we can look at many endomorphisms.

Proposition.

A finite family of individually diagonalizable endomorphisms of

a vector space over

can be simultaneously diagonalized if and only if they

commute.

Basic group theory

We will not review the definition of a group. Instead, we look at some of our

favorite groups, since they will be handy examples later on.

Definition

(Symmetric group

)

The symmetric group

is the set of all

permutations of

{

, ··· , n}

, i.e. the set of all bijections

X → X

. We have

| = n!.

Definition

(Alternating group

)

The alternating group

is the set of

products of an even number of transpositions (

i j

) in

. We know

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

Definition (Cyclic group C

). The cyclic group of order m, written C

= hx : x

= 1i.

This also occurs naturally, as

Z/mZ

over addition, and also the group of

th roots

of unity in

. We can view this as a subgroup of

(

)

∼

. Alternatively,

this is the group of rotation symmetries of a regular

-gon in

, and can be

viewed as a subgroup of GL

(R).

Definition (Dihedral group D

). The dihedral group D

of order 2m is

= hx, y : x

= y

= 1, yxy

−1

= x

−1

This is the symmetry group of a regular

-gon. The

are the rotations and

are the reflections. For example, in

is rotation by

and

is any

reflection.

This group can be viewed as a subgroup of

(

), but since it also acts on

the vertices, it can be viewed as a subgroup of S

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

= hx, y : x

= 1, y

= x

, yxy

−1

= x

−1

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

, with

= {±1, ±i, ±j, ±k}.

We can view this as a subgroup of GL

1 =



1 0

0 1



, i =



i 0

0 −i



, j =



0 1

−1 0



, k =



0 i

i 0



−1 =



−1 0

0 −1



, −i =



−i 0

0 i



, −j =



0 −1

1 0



, −k =



0 −i

−i 0



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

(g) = {xgx

−1

: x ∈ G}.

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

(g) = {x ∈ G : xg = gx}.

Then by the orbit-stabilizer theorem, we have |C

(g)| = |G : C

(G)|.

Definition

(Group action)

Let

be a group and

a set. We say

acts on

X 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

, we obtain a homomorphism

G →

Sym(X), where Sym(X) is the set of all permutations of X.

Proof.

For

g ∈ G

, define

(

) =

∈ Sym

(

) as the function

X → X

x 7→ gx

This is indeed a permutation of X because θ

−1

is an inverse.

Moreover, for any

, g

∈ G

, we get

, since (

)

(

Definition

(Permutation representation)

The permutation representation of a

group action G on X is the homomorphism θ : G → Sym(X) obtained above.

In this course,

is often a finite-dimensional vector space over

(and we

write it as

), and we want the action to satisfy some more properties. We will

require the action to be linear, i.e. for all g ∈ G, v

, v

∈ V , and λ ∈ F.

g(v

+ v

) = gv

+ gv

, g(λv

) = λ(gv

Alternatively, instead of asking for a map

G → Sym

(

), we would require a

map G → GL(V ) instead.

2 Basic definitions

We now start doing representation theory. We boringly start by defining a

representation. In fact, we will come up with several equivalent definitions of a

representation. As always,

will be a finite group and

will be a field, usually

Definition

(Representation)

Let

be a finite-dimensional vector space over

F. A (linear) representation of G on V is a group homomorphism

ρ = ρ

: G → GL(V ).

We sometimes write

for

(

), so for each

g ∈ G

∈ GL

(

), and

and ρ

−1

= (ρ

)

−1

for all g, h ∈ G.

Definition

(Dimension or degree of representation)

The dimension (or degree)

of a representation ρ : G → GL(V ) is dim

(V ).

Recall that

ker ρ C G

and

G/ ker ρ

∼

(

)

≤ GL

(

). In the very special case

where 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 that

does nothing.

An alternative (and of course equivalent) definition of a representation is to

observe that a linear representation is “the same” as a linear action of G.

Definition

(Linear action)

A group

acts linearly on a vector space

if it

acts on V such that

g(v

+ v

) = gv

+ gv

, g(λv

) = λ(gv

)

for all g ∈ G, v

, v

∈ V and λ ∈ F. We call this a linear action.

Now if

acts linearly on

, the map

G → GL

(

) defined by

g 7→ ρ

with

v 7→ gv

, is a representation in the previous sense. Conversely, given a

representation

G → GL

(

), we have a linear action of

via

(

)

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

Definition

(

-space/

-module)

If there is a linear action

, we say

is a G-space or G-module.

Alternatively, we can define a

-space as a module over a (not so) cleverly

picked ring.

Definition

(Group algebra)

The group algebra

is defined to be the algebra

(i.e. a vector space with a bilinear multiplication operation) of formal sums

FG =







g∈G

g : α

∈ F







with the obvious addition and multiplication.

Then we can regard

as a ring, and a

-space is just an

-module in

the sense of IB Groups, Rings and Modules.

Definition

(Matrix representation)

. R

is a matrix representation of

of degree

n if R Is a homomorphism G → GL

(F).

We can view this as a representation that acts on

. Since all finite-

dimensional vector spaces are isomorphic to

for some

, every representation

is equivalent to some matrix representation. In particular, given a linear repre-

sentation

G → GL

(

) with

dim V

, we can get a matrix representation

by fixing a basis

, and then define the matrix representation

G → GL

(

) by

g 7→ [ρ(g)]

Conversely, given a matrix representation

, we get a linear representation

in the obvious way — ρ : G → GL(F

) by g 7→ ρ

via ρ

(v) = R

We have defined representations in four ways — as a homomorphism to

(

), as linear actions, as

-modules and as matrix representations. Now

let’s look at some examples.

Example

(Trivial representation)

Given any group

, take

(the one-

dimensional space), and

G → GL

(

) by

g 7→

(

F → F

) for all

. This is

the trivial representation of G, and has degree 1.

Despite being called trivial, trivial representations are highly non-trivial

in representation theory. The way they interact with other representations

geometrically, topologically etc, and cannot be disregarded. This is a very

important representation, despite looking silly.

Example.

Let

= 1

. Let

= 2, and work over

. Then

we can define a representation by picking a matrix

, and then define

x 7→ A

Then the action of other elements follows directly by

7→ A

. Of course, we

cannot choose

arbitrarily. We need to have

, and this is easily seen to

be the only restriction. So we have the following possibilities:

(i) A

is diagonal: the diagonal entries can be chosen freely from

{±

, ±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 since

= I

. So we don’t really get anything new.

What we would like to say above is that any matrix representation in which

is not diagonal is “equivalent” to one in which

is. To make this notion

precise, 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

(

-homomorphism/intertwine)

Fix a group

and a field

. Let

V, V

be finite-dimensional vector spaces over

and

G → GL

(

) and

G → GL

(

) be representations of

. The linear map

V → V

is a

G-homomorphism if

ϕ ◦ ρ(g) = ρ

(g) ◦ ϕ (∗)

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

V V

ϕ ϕ

i.e. no matter which way we go form

(top left) to

(bottom right), we still

get the same map.

We say

intertwines

and

. We write

Hom

(

V, V

) for the

-space of all

these maps.

Definition

(

-isomorphism)

-homomorphism is a

-isomorphism if

bijective.

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

= ϕρϕ

−1

. (†)

Lemma.

The relation of “being isomorphic” is an equivalence relation on the

set of all linear representations of G over F.

This is an easy exercise left for the reader.

Lemma.

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. C

has four non-isomorphic one-dimensional representations: if

ω = e

2πi/4

, then we have the representations

) = ω

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

Our other formulations of representations give us other formulations of

isomorphisms.

Given a group

, field

, a vector space

of dimension

, and a representation

G → GL

(

), we fix a basis

. Then we get a linear

-isomorphism

V → F

v 7→

[

]

, i.e. by writing

as a column vector with respect to

Then we get a representation

G → GL

(

) isomorphic to

. In other words,

every representation is isomorphic to a matrix representation:

V V

ϕ ϕ

Thus, in terms of matrix representations, the representations

G → GL

(

)

and

G → GL

(

) are

-isomorphic if there exists some non-singular matrix

X ∈ GL

(F) such that

(g) = XR(g)X

−1

for all g.

Alternatively, in terms of linear

-actions, the actions of

and

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 our

previous definition.

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

representation.

Definition

(

-subspace)

Let

G → GL

(

) be a representation of

. We

say W ≤ V is a G-subspace if it is a subspace that is ρ(G)-invariant, i.e.

(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

is irreducible

or simple if there are no proper non-zero G-subspaces.

Example.

Any 1-dimensional representation of

is necessarily irreducible, but

the converse does not hold, or else life would be very boring. We will later see

that D

has a two-dimensional irreducible complex representation.

Definition

(Subrepresentation)

is a

-subspace, then the corresponding

map

G → GL

(

) given by

g 7→ ρ

(

)

gives us a new representation of

This is a subrepresentation of ρ.

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

Lemma.

Let

G → GL

(

) be a representation, and

be a

-subspace of

. If

, ··· , v

}

is a basis containing a basis

, ··· , v

}

(with 0

< m < n

), then the matrix of

(

) with respect to

has the block upper

triangular 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 is

no a priori reason why this has to be true — it is possible that we cannot find

another G-invariant complement to W .

Definition

((In)decomposable representation)

A representation

G → GL

(

)

is decomposable if there are proper G-invariant subspaces U, W ≤ V with

V = U ⊕ W.

We say ρ is a direct sum ρ

⊕ ρ

If no such decomposition exists, we say that ρ is indecomposable.

It is clear that irreducibility implies indecomposability. The converse is

not necessarily true. However, over a field of characteristic zero, it turns out

irreducibility is the same as indecomposability for finite groups, as we will see in

the next chapter.

Again, we can formulate this in terms of matrices.

Lemma.

Let

G → GL

(

) be a decomposable representation with

-invariant

decomposition

U ⊕ W

. Let

, ··· , u

}

and

, ··· , w

}

bases for

and

, and

∪ B

be the corresponding basis for

. Then

with respect to B, we have

[ρ(g)]



[ρ

(g)]

0 [ρ

(g)]



Example.

Let

. Then every irreducible complex representation has

dimension at most 2.

To show this, let

G → GL

(

) be an irreducible

-representation. Let

r ∈ G

be a (non-identity) rotation and

s ∈ G

be a reflection. These generate

Take an eigenvector

(

). So

(

)

λv

for some

λ 6

= 0 (since

(

) is

invertible, it cannot have zero eigenvalues). Let

W = hv, ρ(s)vi ≤ V

be the space spanned by the two vectors. We now check this is fixed by

. 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

−1

ρ(s)v ∈ W.

Also,

(

)

λv ∈ W

and

(

)

v ∈ W

. So

-invariant. Since

irreducible, 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

G → GL

(

) and

G → GL

(

) be

representations 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

G → GL

(

) and

: G → GL

(F), define R ⊕ R

: G → GL

n+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 two

representations, known as the tensor products. However, to do this, we need to

know what the tensor product of two vector spaces is. We will not do this yet.

3 Complete reducibility and Maschke’s theorem

In representation theory, we would like to decompose a representation into sums

of irreducible representations. Unfortunately, this is not always possible. When

we can, we say the representation is completely reducible.

Definition

(Completely reducible/semisimple representation)

A representation

G → GL

(

) is completely reducible or semisimple if it is the direct sum of

irreducible 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, there

are representations of

over

that are not completely reducible, and also a

non-completely reducible representation of C

over F

However, it turns out we have the following theorem:

Theorem.

Every finite-dimensional representation

of a finite group over a

field of characteristic 0 is completely reducible, namely,

∼

⊕ ··· ⊕ V

is a

direct sum of irreducible representations.

By induction, it suffices to prove the following:

Theorem

(Maschke’s theorem)

Let

be a finite group, and

G → GL

(

)

a representation over a finite-dimensional vector space

over a field

with

char F

= 0. If

is a

-subspace of

, then there exists a

-subspace

such that V = W ⊕U.

We will prove this many times.

Proof.

From linear algebra, we know

has a complementary subspace. Let

be any vector subspace complement of W in V , i.e. V = W ⊕ W

as vector

spaces.

Let

V → W

be the projection of

onto

along

, i.e. if

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→

|G|

g∈G

ρ(g)q(ρ(g

−1

)v).

This is in some sense an averaging operator, averaging over what

(

) does. Here

we need the field to have characteristic zero such that

|G|

is well-defined. In

fact, this theorem holds as long as char F - |G|.

For simplicity of expression, we drop the ρ’s, and simply write

¯q : v 7→

|G|

g∈G

gq(g

−1

v).

We first claim that

¯q

has image in

. This is true since for

v ∈ V

(

−1

)

∈ W

and gW ≤ W . So this is a little bit like a projection.

Next, we claim that for

w ∈ W

, we have

¯q

(

) =

. This follows from the

fact that

itself fixes

. Since

-invariant, we have

−1

w ∈ W

for all

w ∈ W . So we get

¯q(w) =

|G|

g∈G

gq(g

−1

w) =

|G|

g∈G

−1

w =

|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

h¯q

(

) =

¯q

(

), i.e. it is invariant

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

h¯q(v) = h

|G|

g∈G

gq(g

−1

|G|

g∈G

hgq(g

−1

|G|

g∈G

(hg)q((hg)

−1

hv)

We now put

. Since

is invertible, summing over all

is the same as

summing over all g

. So we get

|G|

∈G

q(g

0−1

(hv))

= ¯q(hv).

We are pretty much done. We finally show that

ker ¯q

-invariant. If

v ∈ ker ¯q

and h ∈ G, then ¯q(hv) = h¯q(v) = 0. So hv ∈ ker ¯q.

Thus

V = 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

|G|

to exist, which in particular

means

must be a finite group. There is no obvious way to generalize this to

infinite groups. So let’s try a different proof.

The second proof uses inner products, and hence we must take

. This

can be generalized to infinite compact groups, as we will later see.

Recall the definition of an inner product:

Definition

(Hermitian inner product)

For

a complex space,

h·, ·i

is a

Hermitian inner product if

(i) hv, wi = hw, vi (Hermitian)

(ii) hv, λ

+ λ

i = λ

hv, w

i + λ

hv, w

i (sesquilinear)

(iii) hv, vi > 0 if v 6= 0 (positive definite)

Definition

(

-invariant inner product)

An inner product

h·, ·i

is in addition

G-invariant if

hgv, gwi = hv, wi.

Proposition.

Let

-invariant subspace of

, and

have a

-invariant

inner product. Then W

⊥

is also G-invariant.

Proof.

To prove this, we have to show that for all

v ∈ W

⊥

g ∈ G

, we have

gv ∈ W

⊥

This is not hard. We know

v ∈ W

⊥

if and only if

hv, wi

= 0 for all

w ∈ W

Thus, using the definition of G-invariance, for v ∈ W

⊥

, we know

hgv, gwi = 0

for all g ∈ G, w ∈ W .

Thus for all

∈ W

, pick

−1

∈ W

, and this shows

hgv, w

= 0.

Hence gv ∈ W

⊥

Hence if there is a

-invariant inner product on any complex

-space

then we get another proof of Maschke’s theorem.

Theorem

(Weyl’s unitary trick)

Let

be a complex representation of a finite

group

on the complex vector space

. Then there is a

-invariant Hermitian

inner product on V .

Recall that the unitary group is defined by

U(V ) = {f ∈ GL(V ) : hf(u), f(v)i = hu, vi for all u, v ∈ V }

= {A ∈ GL

†

= I}

= U(n).

Then we have an easy corollary:

Corollary.

Every finite subgroup of

(

) is conjugate to a subgroup of U(

Proof.

We start by defining an arbitrary inner product on

: take a basis

, ··· , e

. Define (

, e

) =

, and extend it sesquilinearly. Define a new inner

product

hv, wi =

|G|

g∈G

(gv, gw).

We now check this is sesquilinear, positive-definite and

-invariant. Sesquilin-

earity and positive-definiteness are easy. So we just check

-invariance: we

have

hhv, hwi =

|G|

g∈G

((gh)v, (gh)w)

|G|

∈G

v, g

= hv, wi.

Note that this trick also works for real representations.

Again, we had to take the inverse

|G|

. To generalize this to compact groups,

we will later replace the sum by an integral, and

|G|

by a volume element. This is

fine since (

v, g

) is a complex number and we know how to integrate complex

numbers. This cannot be easily done in the case of ¯q.

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

FG = he

: g ∈ Gi,

i.e. its basis is in one-to-one correspondence with the elements of

. There is a

linear G-action defined in the obvious way: for h ∈ G, we define

−1

This gives a representation of G.

Definition

(Regular representation and regular module)

The regular represen-

tation of a group

, written

reg

, is the natural action of

is called

the regular module.

It is a nice faithful representation of dimension

|G|

. Far more importantly, it

turns out that every irreducible representation of

is a subrepresentation of

the regular representation.

Proposition.

Let

be an irreducible representation of the finite group

over

a field of characteristic 0. Then ρ is isomorphic to a subrepresentation of ρ

reg

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

Proof.

Take

G → GL

(

) be irreducible, and pick our favorite 0

v ∈ V

Now define θ : FG → V by

7→

(gv).

It is not hard to see this is a

-homomorphism. We are now going to exploit the

fact that

is irreducible. Thus, since

im θ

is a

-subspace of

and non-zero,

we must have

im θ

. Also,

ker θ

is a

-subspace of

. Now let

the

-complement of

ker θ

, which exists by Maschke’s theorem. Then

W ≤ FG is a G-subspace and

FG = ker θ ⊕W.

Then the isomorphism theorem gives

∼

FG/ ker θ

∼

im θ = V.

More generally,

doesn’t have to just act on the vector space generated by

itself. If

acts on any set, we can take that space and create a space acted on

by G.

Definition

(Permutation representation)

Let

be a field, and let

act on a

set X. Let FX = he

: x ∈ Xi with a G-action given by

So we have a

-space on

. The representation

G → GL

(

) is the corre-

sponding permutation representation.

4 Schur’s lemma

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

in representation theory.

Theorem (Schur’s lemma).

(i)

Assume

and

are irreducible

-spaces over a field

. Then any

G-homomorphism θ : V → W is either zero or an isomorphism.

(ii)

is algebraically closed, and

is an irreducible

-space, then any

G-endomorphism V → V is a scalar multiple of the identity map ι

Proof.

(i)

Let

V → W

be a

-homomorphism between irreducibles. Then

ker θ

-subspace of

, and since

is irreducible, either

ker θ

= 0 or

ker θ

Similarly,

im θ

is a

-subspace of

, and as

is irreducible, we must

have

im θ

= 0 or

im θ

. Hence either

ker θ

, in which case

= 0,

or ker θ = 0 and im θ = W , i.e. θ is a bijection.

(ii)

Since

is algebraically closed,

has an eigenvalue

. Then

θ − λι

is a

singular

-endomorphism of

. So by (i), it must be the zero map. So

θ = λι

Recall that the

-space

Hom

(

V, W

) is the space of all

-homomorphisms

V → W . If V = W , we write End

(V ) for the G-endomorphisms of V .

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

dim

Hom

(V, W ) =

(

1 V, W are G-isomorphic

0 otherwise

Proof.

and

are not isomorphic, then the only possible map between

them is the zero map by Schur’s lemma.

Otherwise, suppose

∼

and let

, θ

∈ Hom

(

V, W

) be both non-

zero. By Schur’s lemma, they are isomorphisms, and hence invertible. So

−1

∈ End

(V ). Thus θ

−1

= λι

for some λ ∈ C. Thus θ

= λθ

We have another less obvious corollary.

Corollary.

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-theoretic

knowledge (the existence of a faithful complex irreducible representation) to

group theoretic knowledge (center of group being cyclic). This will become

increasingly common in the future, and is a good thing since representations are

easy and groups are hard.

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

10.

Proof.

Let

G → GL

(

) be a faithful irreducible complex representation. Let

z ∈ Z

(

). So

for all

g ∈ G

. Hence

v 7→ zv

is a

-endomorphism

. Hence by Schur’s lemma, it is multiplication by a scalar

, say. Thus

zv = µ

v for all v ∈ V .

Then the map

σ : Z(G) → C

z 7→ µ

is a representation of

(

). Since

is faithful, so is

. So

(

) =

{µ

z ∈

Z(G)} is isomorphic to a finite subgroup of C

, hence cyclic.

Corollary.

The irreducible complex representations of a finite abelian group

are all 1-dimensional.

Proof.

We can use the fact that commuting diagonalizable matrices are simulta-

neously diagonalizable. Thus for every irreducible

, we can pick some

v ∈ V

that is an eigenvector for each

g ∈ G

. Thus

hvi

is a

-subspace. As

irreducible, we must have V = hvi.

Alternatively, we can prove this in a representation-theoretic way. Let

an irreducible complex representation. For each g ∈ G, the map

: V → V

v 7→ gv

is a

-endomorphism of

, since it commutes with the other group elements.

Since V is irreducible, θ

= λ

for some λ

∈ C. Thus

gv = λ

for any g. As V is irreducible, we must have V = hvi.

Note that this result fails over

. For example,

has a two irreducible real

representations, one of dimension 1 and one of dimension 2.

We can do something else. Recall that every finite abelian group

isomorphic

to a product of abelian groups. In fact, it can be written as a product of

for various primes

and

α ≥

1, and the factors are uniquely determined up to

order.

This you already know from IB Groups Rings and Modules. You might be

born knowing it — it’s such a fundamental fact of nature.

Proposition.

The finite abelian group

× ··· × C

has precisely

|G|

irreducible representations over C.

This is not a coincidence. We will later show that the number of irreducible

representations is the number of conjugacy classes of the group. In abelian

groups, each conjugacy class is just a singleton, and hence this result.

Proof. Write

G = hx

i × ··· × hx

where

. Any irreducible representation

must be one-dimensional. So

we have

ρ : G → C

Let

, ··· , x

, ··· ,

1) =

. Then since

is a homomorphism, we must have

= 1. Therefore λ

is an n

th root of unity.

Now the values (λ

, ··· , λ

) determine ρ completely, namely

ρ(x

, ··· , x

) = λ

···λ

Also, whenever

is an

th root of unity for each

, then the above formula

gives a well-defined representation. So there is a one-to-one correspondence

ρ ↔ (λ

, ··· , λ

), with λ

= 1.

Since for each

, there are

many

th roots of unity, it follows that there

are |G| = n

···n

many choices of the λ

. Thus the proposition.

Example.

(i)

Consider

hxi

. The four 1-dimensional irreducible representations

are given by

1 x x

1 1 1 1

1 i −1 −i

1 −1 1 −1

1 −i −1 i

(ii)

Consider the Klein four group

i × hx

. The irreducible

representations are

1 x

1 1 1 1

1 1 −1 −1

1 −1 1 −1

1 −1 −1 1

These are also known as character tables, and we will spend quite a lot of

time computing these for non-abelian groups.

Note that there is no “natural” one-to-one correspondence between the

elements of

and the representations of

(for

finite-abelian). If we choose

an isomorphism

∼

× ···C

, then we can identify the two sets, but it

depends on the choice of the isomorphism (while the decomposition is unique,

we can pick a different generator of, say,

and get a different isomorphism to

the same decomposition).

Isotypical decompositions

Recall that we proved we can decompose any

-representation into a sum of

irreducible representations. Is this decomposition unique? If it isn’t, can we say

anything about, say, the size of the irreducible representations, or the number of

factors in the decomposition?

We know any diagonalizable endomorphism

V → V

for a vector space

gives us a vector space decomposition

V =

V (λ),

where

V (λ) = {v ∈ V : α(v) = λv}.

This is canonical in that it depends on α alone, and nothing else.

is moreover a

-representation, how does this tie in to the decomposition

of V into the irreducible representations?

Let’s do an example.

Example.

Consider

∼

hr, s

= 1

, rs

−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→ 1

S sign r 7→ 1 s 7→ −1

W 2-dimensional

The last representation is the action of

in the natural way, i.e. the

rotations and reflections of the plane that corresponds to the symmetries of the

triangle. It is helpful to view this as a complex representation in order to make

the matrix look nice. The 2-dimensional representation (

ρ, W

) is defined by

W = C

, where r and s act on W as

ρ(r) =



ω 0

0 ω



, ρ(s) =



0 1

1 0



and

2πi/3

is a third root of unity. We will now show that these are indeed

all the irreducible representations, by decomposing any representation into sum

of these.

So let’s decompose an arbitrary representation. Let (

, V

) be any complex

representation of

. Since

(

) has order 3, it is diagonalizable has eigenvalues

, ω, ω

. We diagonalize

(

) and then

splits as a vector space into the

eigenspaces

V = V (1) ⊕ V (ω) ⊕ V (ω

Since

srs

−1

, we know

(

) preserves

(1) and interchanges

(

) and

V (ω

Now we decompose

(1) into

(

) eigenspaces, with eigenvalues

1. Since

has to act trivially on these eigenspaces,

(1) splits into sums of copies of the

irreducible representations 1 and S.

For the remaining mess, choose a basis

, ··· , v

(

), and let

(

)

. Then

(

) acts on the two-dimensional space

, v



0 1

1 0



, while

(

) acts as



ω 0

0 ω



. This means

(

)

⊕V

(

) decomposes into many copies

of W.

We did this for

by brute force. How do we generalize this? We first have

the following lemma:

Lemma. Let V, V

, V

be G-vector spaces over F. Then

(i) Hom

(V, V

⊕ V

)

∼

Hom

(V, V

) ⊕ Hom

(V, V

)

(ii) Hom

⊕ V

, V )

∼

Hom

, V ) ⊕ Hom

, V ).

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

Define the projection map

: V

⊕ V

→ V

which is the G-linear projection onto V

Then we can define the G-homomorphism

Hom

(V, V

⊕ V

) 7→ Hom

(V, V

) ⊕ Hom

(V, V

)

ϕ 7→ (π

ϕ, π

ϕ).

Then the map (ψ

, ψ

) 7→ ψ

+ ψ

is an inverse.

For the second part, we have the homomorphism

ϕ 7→

(

ϕ|

, ϕ|

) with

inverse (ψ

, ψ

) 7→ ψ

+ ψ

Lemma.

Let

be an algebraically closed field, and

be a representation of

Suppose

i=1

is its decomposition into irreducible components. Then

for each irreducible representation S of G,

|{j : V

∼

S}| = dim Hom

(S, V ).

This tells us we can count the multiplicity of

by looking at the

homomorphisms.

Proof.

We induct on

. If

= 0, then this is a trivial space. If

= 1, then

itself is irreducible, and by Schur’s lemma,

dim Hom

(

S, V

) = 1 if

, 0

otherwise. Otherwise, for n > 1, we have

V =

n−1

i=1

⊕ V

By the previous lemma, we know

dim hom

n−1

i=1

⊕ V

= dim Hom

n−1

i=1

+ dim hom

(S, V

The result then follows by induction.

Definition

(Canonical decomposition/decomposition into isotypical compo-

nents)

A decomposition of

, where each

is (isomorphic to)

copies of the irreducible

(with

∼

for

i 6

) is the canonical decomposition

or decomposition into isotypical components.

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

= dim Hom

, V ),

and hence this decomposition is well-defined.

We’ve finally all the introductory stuff. The course now begins.

5 Character theory

In topology, we want to classify spaces. To do so, we come up with invariants of

spaces, like the number of holes. Then we know that a torus is not the same

as a sphere. Here, we want to attach invariants to a representation

of a finite

group G on V .

One thing we might want to do is to look at the matrix coefficients of

(

However, this is bad, since this is highly highly highly basis dependent. It is not

a true invariant. We need to do something better than that.

Let

, and

be a finite group. Let

G → GL

(

) be a

representation of

. The clue is to look at the characteristic polynomial of the

matrix. The coefficients are functions of the eigenvalues — on one extreme, the

determinant is the product of all eigenvalues; on the other extreme, and the

trace is the sum of all of them. Surprisingly, it is the trace that works. We don’t

have to bother ourselves with the other coefficients.

Definition

(Character)

The character of a representation

G → GL

(

written χ

= χ

= χ, is defined as

χ(g) = tr ρ(g).

We say ρ affords the character χ.

Alternatively, the character is

tr R

(

), where

(

) is any matrix representing

ρ(g) with respect to any basis.

Definition (Degree of character). The degree of χ

is dim V .

Thus, χ is a function G → C.

Definition

(Linear character)

We say

is linear if

dim V

= 1, in which case

χ is a homomorphism G → C

= GL

(C).

Various properties of representations are inherited by characters.

Definition

(Irreducible character)

A character

is irreducible if

is irreducible.

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

Definition

(Trivial/principal character)

A character

is trivial or principal if

ρ is the trivial representation. We write χ = 1

is a complete invariant in the sense that it determines

up to isomorphism.

This is staggering. We reduce the whole matrix into a single number — the

trace, 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) χ

(1) = dim V .

(ii) χ

is a class function, namely it is conjugation invariant, i.e.

(hgh

−1

) = χ

(g)

for all g, h ∈ G. Thus χ

is constant on conjugacy classes.

(iii) χ

−1

) = χ

(g).

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

V ⊕W

= χ

+ χ

These results, despite being rather easy to prove, are very useful, since they

save us a lot of work when computing the characters of representations.

Proof.

(i) Obvious since ρ

(1) = id

(ii) Let R

be the matrix representing g. Then

χ(hgh

−1

) = tr(R

−1

) = tr(R

) = χ(g),

as we know from linear algebra.

(iii)

Since

g ∈ G

has finite order, we know

(

) is represented by a diagonal

matrix













and χ(g) =

. Now g

−1

is represented by

−1







−1







Noting that each λ

is an nth root of unity, hence |λ

| = 1, we know

χ(g

−1

) =

−1

= χ(g).

(iv)

Suppose

⊕V

, with

G → GL

(

) splitting into

G → GL

(

Pick a basis

for

, and let

∪ B

. Then with respect to

, we

have

[ρ(g)]



[ρ

(g)]

0 [ρ

(g)]



So χ(g) = tr(ρ(g)) = tr(ρ

(g)) + tr(ρ

(g)) = χ

(g) + χ

(g).

We will see later that if we take characters

, χ

, then

is also a

character of

. This uses the notion of tensor products, which we will do later.

Lemma.

Let

G → GL

(

) be a complex representation affording the character

χ. Then

|χ(g)| ≤ χ(1),

with equality if and only if

(

) =

λI

for some

λ ∈ C

, a root of unity. Moreover,

χ(g) = χ(1) if and only if g ∈ ker ρ.

Proof.

Fix

, and pick a basis of eigenvectors of

(

). Then the matrix of

(

)

is diagonal, say

ρ(g) =













Hence

|χ(g)| =



≤

|λ

| =

1 = dim V = χ(1).

In the triangle inequality, we have equality if and only if all the

’s are equal,

to λ, say. So ρ(g) = λI. Since all the λ

’s are roots of unity, so is λ.

And, if

(

) =

(1), then since

(

) =

λI

, taking the trace gives

(

) =

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)

is a complex (irreducible) character of

, then so is

εχ

for any linear

(1-dimensional) character ε.

Proof.

(i)

G → GL

(

) is a complex matrix representation, then so is

G →

(C), where g 7→ R(g). Then the character of

R is ¯χ

(ii)

Similarly,

g 7→ ε

(

)

(

) for

g ∈ G

is a representation with character

εχ.

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 a

complete invariant, as in two representations are isomorphic if and only if they

have the same character. Before we prove this, we first need some definitions

Definition

(Space of class functions)

Define the complex space of class functions

of G to be

C(G) = {f : G → C : f(hgh

−1

) = f(g) for all h, g ∈ G}.

This is a vector space by f

+ f

: g 7→ f

(g) + f

(g) and λf : g 7→ λf(g).

Definition

(Class number)

The class number

(

) is the number of

conjugacy classes of G.

We can list the conjugacy classes as

, ··· , C

. wlog, we let

{

}

. We

choose

, g

, ··· , g

to be representatives of the conjugacy classes. Note also

that

dim C

(

) =

, since the characteristic function

of the classes form a basis

where

(g) =

(

1 g ∈ C

0 g 6∈ C

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

hf, f

i =

|G|

g∈G

f(g)f

(g)

|G|

j=1

|f(g

)

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

j=1

f(g

In particular, for characters,

hχ, χ

i =

j=1

χ(g

−1

)χ

It should be clear (especially using the original formula) that

hχ, χ

hχ

, χi

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 of

G form an orthonormal basis of C(G), namely

(i)

G → GL

(

) and

G → GL

(

) are two complex irreducible

representations affording characters χ, χ

respectively, then

hχ, χ

i =

(

1 if ρ and ρ

are isomorphic representations

0 otherwise

This is the (row) orthogonality of characters.

(ii)

Each class function of

can be expressed as a linear combination of

irreducible characters of G.

Proof is deferred. We first look at some corollaries.

Corollary.

Complex representations of finite groups are characterised by their

characters.

Proof.

Let

G → GL

(

) afford the character

. We know we can write

⊕ ··· ⊕ m

, where

, ··· , ρ

are (distinct) irreducible and

≥

are the multiplicities. Then we have

χ = m

+ ··· + m

where χ

is afforded by ρ

. Then by orthogonality, we know

= hχ, χ

So we can obtain the multiplicity of each

just by looking at the inner

products of the characters.

This is not true for infinite groups. For example, if

, then the

representations

1 7→



1 0

0 1



, 1 7→



1 1

0 1



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

Corollary

(Irreducibility criterion)

G → GL

(

) is a complex repre-

sentation of

affording the character

, then

is irreducible if and only if

hχ, χi = 1.

Proof.

is irreducible, then orthogonality says

hχ, χi

= 1. For the other

direction, suppose hχ, χi = 1. We use complete reducibility to get

χ =

with

irreducible, and

≥

0 the multiplicities. Then by orthogonality, we

get

hχ, χi =

But

hχ, χi

= 1. So exactly one of

is 1, while the others are all zero, and

χ = χ

. So χ is irreducible.

Theorem.

Let

, ··· , ρ

be the irreducible complex representations of

, and

let their dimensions be n

, ··· , n

. Then

|G| =

Recall that for abelian groups, each irreducible character has dimension 1,

and there are |G| representations. So this is trivially satisfied.

Proof.

Recall that

reg

G → GL

(

), given by

acting on itself by multipli-

cation, is the regular representation of

of dimension

|G|

. Let its character be

reg

, the regular character of G.

First note that we have

reg

(1) =

|G|

, and

reg

(

) = 0 if

h 6

= 1. The first

part is obvious, and the second is easy to show, since we have only 0s along the

diagonal.

Next, we decompose π

reg

We now want to find a

. We have

= hπ

reg

, χ

i =

|G|

g∈G

reg

(g)χ

(g) =

|G|

· |G|χ

(1) = χ

(1).

Then we get

|G| = π

reg

(1) =

(1)

From this proof, we also see that each irreducible representation is a subrep-

resentation of the regular representation.

Corollary.

The number of irreducible characters of

(up to equivalence) is

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

, g

are conjugate if and only if

(

) =

(

) for

all irreducible characters χ of G.

Proof.

, g

are conjugate, since characters are class functions, we must have

χ(g

) = χ(g

For the other direction, let

be the characteristic function of the class of

Then since δ is a class function, we can write

δ =

where χ

are the irreducible characters of G. Then

δ(g

) =

) = δ(g

) = 1.

So g

is in the same conjugacy class as g

Before we move on to the proof of the completeness of characters, we first

make a few more definitions:

Definition

(Character table)

The character table of

is the

k × k

matrix

= (

(

)), where 1 =

, χ

, ··· , χ

are the irreducible characters of

, and

= {1}, C

, ··· , C

are the conjugacy classes with g

∈ C

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

the table below.

We will spend the next few lectures coming up with techniques to compute

these character tables.

Example.

Consider

∼

hr, s | r

= 1

, srs

−1

. As in all

computations, the first thing to do is to compute the conjugacy classes. This is

not too hard. We have

= {1}, C

= {s, sr, sr

}, C

= {r, r

−1

Alternatively, we can view this as

and use the fact that two elements are

conjugate if and only if they have the same cycle types. We have found the

three representations: 1, the trivial representation; S, the sign; and W, the

two-dimensional representation. In

, the reflections

acts by matrix with

eigenvalues

1. So the sum of eigenvalues is 0, hence

(

) = 0. It also also

not hard to see

7→



cos

2mπ

−sin

2mπ

sin

2mπ

cos

2mπ



So χ

) = −1.

Fortunately, after developing some theory, we will not need to find all the

irreducible representations in order to compute the character table.

1 C

1 1 1 1

S 1 −1 1

2 0 −1

We see that the sum of the squares of the first column is 1

+ 1

+ 2

= 6 =

as expected. We can also check that W is genuinely an irreducible representation.

Noting that the centralizers of

, C

and

are of sizes 6

3. So the inner

product is

hχ

, χ

i =

(−1)

= 1,

as expected.

So we now need to prove orthogonality.

6 Proof of orthogonality

We will do the proof in parts.

Theorem

(Row orthogonality relations)

G → GL

(

) and

G →

(

) are two complex irreducible representations affording characters

χ, χ

respectively, then

hχ, χ

i =

(

1 if ρ and ρ

are isomorphic representations

0 otherwise

Proof.

We fix a basis of

and of

. Write

(

)

, R

(

) for the matrices of

(

)

and ρ

(g) with respect to these bases respectively. Then by definition, we have

hχ

, χi =

|G|

g∈G

−1

)χ(g)

|G|

g∈G

1≤i≤n

1≤j≤n

−1

)

R(g)

For any linear map

V → V

, we define a new map by averaging by

and

˜ϕ : V → V

v 7→

|G|

−1

)ϕρ(g)v

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

−1

) ˜ϕρ(h)(v) = ˜ϕ(v).

We have

−1

) ˜ϕρ(h)(v) =

|G|

g∈G

((gh)

−1

)ϕρ(gh)v

|G|

∈G

0−1

)ϕρ(g

= ˜ϕ(v).

(i)

Now we first consider the case where

ρ, ρ

is not isomorphic. Then by

Schur’s lemma, we must have ˜ϕ = 0 for any linear ϕ : V → V

We now pick a very nice

, where everything disappears. We let

αβ

the operator having matrix

αβ

with entries 0 everywhere except 1 in the

(α, β) position.

Then ˜ε

αβ

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

|G|

g∈G

−1

αβ

R(g))

= 0.

Using our choice of ε

αβ

, we get

|G|

g∈G

−1

)

iα

R(g)

βj

= 0

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

|G|

g∈G

−1

)

R(g)

= 0.

We can sum this thing over all i and j to get that hχ

, χi = 0.

(ii)

Now suppose

ρ, ρ

are isomorphic. So we might as well take

and ρ = ρ

. If ϕ : V → V is linear, then ˜ϕ ∈ End

(V ).

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

tr ˜ϕ =

|G|

g∈G

tr(ρ(g

−1

)ϕρ(g)) =

|G|

g∈G

tr ϕ = tr ϕ,

using the fact that traces don’t see conjugacy (and

(

−1

) =

(

)

−1

since

ρ is a group homomorphism).

By Schur’s lemma, we know

˜ϕ

λι

for some

λ ∈ C

(which depends on

ϕ). Then if n = dim V , then

λ =

tr ϕ.

Let ϕ = ε

αβ

. Then tr ϕ = δ

αβ

. Hence

˜ε

αβ

|G|

ρ(g

−1

)ε

αβ

ρ(g) =

αβ

ι.

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

|G|

R(g

−1

)

iα

R(g)

βj

αβ

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

|G|

R(g

−1

)

R(g)

Summing over all i and j, we get hχ, χi = 1.

After learning about tensor products and duals later on, one can provide a

shorter proof of this result as follows:

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

hχ

, χ

i =

|G|

(g)χ

(g) =

|G|

V ⊗W

∗

(g).

We notice that there is a natural isomorphism

V ⊗ W

∗

∼

Hom

(

W, V

), and

the action of

on this space is by conjugation. Thus, a

-invariant element

is just a

-homomorphism

W → V

. Thus, we can decompose

Hom

(

V, W

) =

Hom

(

V, W

)

⊕U

for some

-space

, and

has no

-invariant element. Hence

in the decomposition of

Hom

(

V, W

) into irreducibles, we know there are exactly

dim Hom

(

V, W

) copies of the trivial representation. By Schur’s lemma, this

number 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

affords

, then any element in the image of

g∈G

(

) is fixed by

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 similar

one for the columns:

Theorem (Column orthogonality relations). We have

i=1

)χ

) = δ

)|.

This is analogous to the fact that if a matrix has orthonormal rows, then it

also has orthonormal columns. We get the extra factor of

(

)

because of

the way we count.

This has an easy corollary, which we have already shown previously using

the regular representation:

Corollary.

|G| =

i=1

(1).

Proof of column orthogonality.

Consider the character table

= (

(

)). We

know

= hχ

, χ

i =

)χ

Then

−1

= I

k×k

where

D =







)| ··· 0

0 ··· |C







Since

is square, it follows that

−1

is the inverse of

. So

which is exactly the theorem.

The proof requires that

is square, i.e. there are

many irreducible repre-

sentations. So we need to prove the last part of the completeness of characters.

Theorem.

Each class function of

can be expressed as a linear combination

of irreducible characters of G.

Proof.

We list all the irreducible characters

, ··· , χ

. Note that we don’t

know the number of irreducibles is

. This is essentially what we have to prove

here. We now claim these generate C(G), the ring of class functions.

Now recall that

(

) has an inner product. So it suffices to show that the

orthogonal complement to the span

hχ

, ··· , χ

(

) is trivial. To see this,

assume f ∈ C(G) satisfies

hf, χ

i = 0

for all

irreducible. We let

G → GL

(

) be an irreducible representation

affording χ ∈ {χ

, ··· , χ

}. Then hf, χi = 0.

Consider the function

ϕ =

|G|

f(g)ρ(g) : V → V.

For any h ∈ G, we can compute

ρ(h)

−1

ϕρ(h) =

|G|

f(g)ρ(h

−1

gh) =

|G|

f(h

−1

gh)ρ(h

−1

gh) = ϕ,

using the fact that

is a class function. So this is a

-homomorphism. So as

is irreducible, Schur’s lemma says it must be of the form λι

for some λ ∈ C.

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

nλ = tr

|G|

f(g)ρ(g)

|G|

f(g)χ(g) = hf, χi = 0.

= 0, i.e.

f(g)ρ

(

) = 0, the zero endomorphism on

. This is valid for

any irreducible representation, and hence for every representation, by complete

reducibility.

In particular, take ρ = ρ

reg

, where ρ

reg

(g) : e

7→ e

for each g ∈ G. Hence

f(g)ρ

reg

(g) : e

7→

f(g)e

Since this is zero, it follows that we must have

f(g)e

= 0. Since the

’s are

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

7 Permutation representations

In this chapter, we are going to study permutation representations in a bit more

detail, since we can find some explicit formulas for the character of them. This

is particularly useful if we are dealing with the symmetric group.

Let

be a finite group acting on a finite set

, ··· , x

}

(sometimes

known as a

-set). We define

to be the complex vector space with basis

, ··· , e

}. More explicitly,

CX =







: a

∈ C







There is a corresponding representation permutation

: G → GL(CX)

g 7→ ρ(g)

where

(

) :

7→ e

, extended linearly. We say

is the representation

corresponding to the action of G on X.

This is a generalization of the group algebra — if we let

act on itself by

left multiplication, then this is in fact the group algebra.

The corresponding matrix representations

(

) with respect to the basis

}

x∈X

are permutation matrices — it is 0 everywhere except for one 1 in each

row and column. In particular, (

(

))

= 1 precisely when

. So the

only non-zero diagonal entries in

(

) are those

such that

is fixed by

. Thus

we have

Definition (Permutation character). The permutation character π

(g) = |fix(g)| = |{x ∈ X : gx = x}|.

Lemma. π

always contains the trivial character 1

(when decomposed in the

basis of irreducible characters). In particular,

span{e

···

}

is a trivial

G-subspace of CX, with G-invariant complement {

= 0}.

Lemma. hπ

, which is the multiplicity of 1 in

, is the number of orbits

of G on X.

Proof.

We write

as the disjoint union of orbits,

∪ ··· ∪ X

. Then

it is clear that the permutation representation on

is just the sum of the

permutation representations on the X

, i.e.

= π

+ ··· + π

where

is the permutation character of

. So to prove the lemma, it

is enough to consider the case where the action is transitive, i.e. there is just

one orbit.

So suppose

acts transitively on

. We want to show

hπ

= 1. By

definition, we have

hπ

, 1i =

|G|

(g)

|G|

|{(g, x) ∈ G × X : gx = x}|

|G|

x∈X

where

is the stabilizer of

. By the orbit-stabilizer theorem, we have

||X|

|G|. So we can write this as

|G|

x∈X

|G|

|X|

|G|

· |X| ·

|G|

|X|

= 1.

So done.

Lemma. Let G act on the sets X

, X

. Then G acts on X

× X

g(x

, x

) = (gx

, gx

Then the character

×X

= π

and so

hπ

, π

i = number of orbits of G on X

× X

Proof.

We know

×X

(

) is the number of pairs (

, x

)

∈ X

× X

fixed by

. This is exactly the number of things in

fixed by

times the number of

things in X

fixed by g. So we have

×X

(g) = π

(g)π

(g).

Then using the fact that π

, π

are real, we get

hπ

, π

i =

|G|

(g)π

(g)

|G|

(g)π

(g)1

(g)

= hπ

, 1i

= hπ

×X

, 1i.

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 are

distinct (it can never send (

x, x

) to (

, x

) if

, since

(

x, x

) = (

gx, gx

)).

More precisely,

Definition

(2-transitive)

Let

act on

|X| >

2. Then

is 2-transitive on

has two orbits on

X × X

, namely

{

(

x, x

) :

x ∈ X}

and

{

(

, x

) :

∈

X, x

6= x

Then we have the following lemma:

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

= 1

+ χ,

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

Proof. We know

= m

+ m

+ ··· + m

with 1

, χ

, ··· , χ

distinct irreducible characters and

∈ N

are non-zero.

Then by orthogonality,

hπ

, π

i =

i=1

Since

hπ

, π

is the number of orbits of

X × X

, we know

is 2-transitive on

X if and only if ` = 2 and m

= m

= 1.

This is useful if we want to understand the symmetric group, since it is

always 2-transitive (modulo the silly case of S

Example. S

acting on

{

, ··· , n}

is 2-transitive for

n ≥

2. So

= 1+

with

irreducible of degree

n −

1 (since

has degree

). This works similarly

for A

, whenever n ≥ 3.

This tells us we can find an irreducible character of

by finding

, and

then subtracting 1 off.

Example.

Consider

. The conjugacy classes correspond to different

cycle types. We already know three characters — the trivial one, the sign, and

− 1

1 3 8 6 6

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

trivial χ

1 1 1 1 1

sign χ

1 1 1 −1 −1

− 1

3 −1 0 −1 1

We know the product of a one-dimensional irreducible character and another

character is another irreducible character, as shown on the first example sheet.

So we get

1 3 8 6 6

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

trivial χ

1 1 1 1 1

sign χ

1 1 1 −1 −1

− 1

3 −1 0 −1 1

3 −1 0 1 −1

For the last representation we can find the dimension by computing 24

−

+ 3

) = 2

. So it has dimension 2. To obtain the whole of

, we can

use column orthogonality — for example, we let the entry in the second column

. Then column orthogonality says 1 + 1

−

3 + 2

= 0 . So

= 2. In the

end, we find

1 3 8 6 6

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

trivial χ

1 1 1 1 1

sign χ

1 1 1 −1 −1

− 1

3 −1 0 −1 1

3 −1 0 1 −1

2 2 −1 0 0

Alternatively, we have previously shown that

reg

+ 3

+ 2

By computing χ

reg

directly, we can find χ

Finally, we can obtain

by observing

∼

, and “lifting” characters.

We will do this in the next chapter. In fact, this

is the degree-2 irreducible

representation of S

∼

lifted up.

We can also do similar things with the alternating group. The issue of course

is that the conjugacy classes of the symmetric group may split when we move to

the alternating group.

Suppose g ∈ A

. Then

(g)| = |S

: C

(g)|

(g)| = |A

: C

(g)|.

We know

(

)

⊇ C

(

), but we need not have equality, since elements needed

to conjugate

h ∈ C

(

) might not be in

. For example, consider

σ = (1 2 3) ∈ A

. We have C

(σ) = {σ} and C

(σ) = {σ, σ

−1

We know

is an index 2 subgroup of

. So

(

)

⊆ C

(

) either has

index 2 or 1. If the index is 1, then

. Otherwise, the sizes are

equal.

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

Lemma.

Let

g ∈ A

n >

1. If

commutes with some odd permutation in

then

(

) =

(

). Otherwise,

splits into two conjugacy classes in

equal size.

This is quite useful for doing the examples on the example sheet, but we will

not go through any more examples here.

8 Normal subgroups and lifting

The idea of this section is that there is some correspondence between the

irreducible characters of

G/N

(with

N C G

) and those of

itself. In most

cases,

G/N

is a simpler group, and hence we can use this correspondence to find

irreducible characters of G.

The main result is the following lemma:

Lemma.

Let

N C G

. Let

˜ρ

G/N → GL

(

) be a representation of

G/N

. Then

the 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) deg χ = 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

is just a homomorphism

G → GL

(

), and

the composition of homomorphisms is a homomorphisms, it follows immediately

that ρ as defined in the lemma is a representation.

(i) We can compute

hχ, χi =

|G|

g∈G

χ(g)χ(g)

|G|

gN∈G/N

k∈N

χ(gk)χ(gk)

|G|

gN∈G/N

k∈N

˜χ(gN) ˜χ(gN)

|G|

gN∈G/N

|N|

˜χ(gN) ˜χ(gN)

|G/N|

gN∈G/N

˜χ(gN) ˜χ(gN)

= h˜χ, ˜χi.

hχ, χi

= 1 if and only if

h˜χ, ˜χi

= 1. So

is irreducible if and only if

˜ρ

is irreducible.

(ii) We can directly compute

χ(g) = tr ρ(g) = tr(˜ρ(gN)) = ˜χ(gN)

for all g ∈ G.

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

deg χ = χ(1) = ˜χ(N) = deg ˜χ.

Alternatively, to show they have the same dimension, just note that

and

˜ρ 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

is its

lift to

. We need to show the kernel contains

. By definition, we know

˜χ

(

) =

(1). Also, if

k ∈ N

, then

(

) =

˜χ

(

) =

˜χ

(

) =

(1). So

N ≤ ker χ.

Now let

be a character of

with

N ≤ ker χ

. Suppose

G → GL

(

)

affords χ. Define

˜ρ : G/N → GL(V )

gN 7→ ρ(g)

Of course, we need to check this is well-defined. If

, then

−1

∈ N

. So

(

) =

(

) since

N ≤ ker ρ

. So this is indeed well-defined.

It is also easy to see that

˜ρ

is a homomorphism, hence a representation of

G/N .

Finally, if

˜χ

is a character of

˜ρ

, then

˜χ

(

) =

(

) for all

g ∈ G

definition. So

˜χ

lifts to

. It is clear that these two operations are inverses

to each other.

There is one particular normal subgroup lying around that proves to be

useful.

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

= h[a, b] : a, b ∈ Gi,

where [

a, b

] =

aba

−1

, is the unique minimal normal subgroup of

such that

G/G

is abelian. So if G/N is abelian, then G

≤ N.

Moreover,

has precisely

representations of dimension 1, all

with kernel containing G

, and are obtained by lifting from G/G

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

Proof. Consider [a, b] = aba

−1

∈ G

. Then for any h ∈ G, we have

h(aba

−1



(ha)b(ha)

−1



bhb

−1



= [ha, b][b, h] ∈ G

So in general, let [a

, b

][a

, b

] ···[a

, b

] ∈ G

. Then

h[a

, b

][a

, b

] ···[a

, b

−1

= (h[a

, b

−1

)(h[a

, b

−1

) ···(h[a

, b

−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

−1

∈ N

and only if ghN = hgN , if and only if (gN)(hN) = (hN)(gN) by normality.

Since

is generated by all [

g, h

], we know

≤ N

if and only if

G/N

abelian.

Since

G/G

, is abelian, we know it has exactly

irreducible characters,

˜χ

, ··· , ˜χ

, all of degree 1. The lifts of these to

also have degree 1, and by the

previous lemma, these are precisely the irreducible characters

such that

≤ ker χ

But any degree 1 character of

is a homomorphism

G → C

, hence

(

ghg

−1

) = 1. So for any 1-dimensional character,

, we must have

≤

ker χ. So the lifts χ

, ··· , χ

are all 1-dimensional characters of G.

Example.

Let

, with

n ≥

3. We claim that

. Since

sgn

(

) =

sgn

(

−1

), we know that [

a, b

]

∈ A

for any

a, b ∈ S

. So

≤ A

(alternatively,

since S

∼

is abelian, we must have S

≤ A

). We now notice that

(1 2 3) = (1 3 2)(1 2)(1 3 2)

−1

(1 2)

−1

∈ S

So by symmetry, all 3-cycles are in

. Since the 3-cycles generate

, we know

we must have S

= A

Since

∼

, we get

= 2. So

must have exactly two linear

characters, namely the trivial character and the sign.

Lemma. G

is not simple if and only if

(

) =

(1) for some irreducible character

χ 6

= 1

and some 1

g ∈ G

. Any normal subgroup of

is the intersection of

the kernels of some of the irreducible characters of G, i.e. N =

ker χ

In other words,

is simple if and only if all non-trivial irreducible characters

have trivial kernel.

Proof.

Suppose

(

) =

(1) for some non-trivial irreducible character

, and

is afforded by

. Then

g ∈ ker ρ

. So if

g 6

= 1, then 1

ker ρ C G

, and

ker ρ 6

So G cannot be simple.

If 1

N C G

is a non-trivial proper subgroup, take an irreducible character

˜χ

G/N

, and suppose

˜χ 6

= 1

G/N

. Lift this to get an irreducible character

afforded by the representation

. Then

N ≤ ker ρ C G

. So

(

) =

(1) for

g ∈ N.

Finally, let 1

N C G

. We claim that

is the intersection of the kernels of

the lifts

, ··· , χ

of all the irreducibles of

G/N

. Clearly, we have

N ≤

ker χ

g ∈ G \ N

, then

gN 6

. So

˜χ

(

)

˜χ

(

) for some irreducible

˜χ

G/N

Lifting

˜χ

, we have

(

)

(1). So

is not in the intersection of the

kernels.

9 Dual spaces and tensor products of represen-

tations

In this chapter, we will consider several linear algebra constructions, and see

how 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

G → GL

(

), we would like to turn

∗

Hom

(

V, F

)

into a representation. So for each

, we want to produce a

∗

(

) :

∗

→ V

∗

. So

given

ϕ ∈ V

∗

, we shall get a

∗

(

)

V → F

. Given

v ∈ V

, a natural guess for

the 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

. However, if we do this, we

will not get

∗

(

)

∗

(

) =

∗

(

). Instead, we will obtain

∗

(

)

∗

(

) =

∗

(

which is the wrong way round. The right definition to use is actually the

following:

Lemma.

Let

G → GL

(

) be a representation over

, and let

∗

Hom

(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) = χ

−1

Proof. We have to check ρ

∗

is a homomorphism. We check

∗

)(ρ

∗

)ϕ)(v) = (ρ

∗

)ϕ)(ρ(g

−1

)(v))

= ϕ(ρ(g

−1

)ρ(g

−2

)v)

= ϕ(ρ((g

)

−1

)(v))

= (ρ

∗

)ϕ)(v).

To compute the character, fix a

g ∈ G

, and let

, ··· , e

be a basis of eigenvectors

of V of ρ(g), say

ρ(g)e

= λ

If we have a dual space of

, then we have a dual basis. We let

, ··· , ε

the dual basis. Then

(ρ

∗

(g)ε

)(e

) = ε

(ρ(g

−1

) = ε

(λ

−1

) = λ

−1

= λ

−1

= λ

−1

Thus we get

∗

(g)ε

= λ

−1

∗

(g) =

−1

= χ

−1

Definition

(Self-dual representation)

A representation

G → GL

(

) is

self-dual if there is isomorphism of G-spaces V

∼

∗

Note that as (finite-dimensional) vector spaces, we always have

∼

∗

However, what we require here is an isomorphism that preserves the

-action,

which does not necessarily exist.

Over F = C, this holds if and only if

(g) = χ

∗

(g) = χ

−1

) = χ

(g).

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

Example.

All irreducible representations of

are self-dual, since every element

is conjugate to its own inverse. This is not always true for A

9.2 Tensor products

The next idea we will tackle is the concept of tensor products. We first introduce

it as a linear algebra concept, and then later see how representations fit into this

framework.

There are many ways we can define the tensor product. The definition we

will take here is a rather hands-on construction of the space, which involves

picking a basis. We will later describe some other ways to define the tensor

product.

Definition

(Tensor product)

Let

V, W

be vector spaces over

. Suppose

dim V

and

dim W

. We fix a basis

, ··· , v

and

, ··· , w

and W respectively.

The tensor product space

V ⊗ W

V ⊗

is an

-dimensional vector

space (over F) with basis given by formal symbols

⊗ w

: 1 ≤ i ≤ m, 1 ≤ j ≤ n}.

Thus

V ⊗ W =

⊗ w

: λ

∈ F

with the “obvious” addition and scalar multiplication.

v =

∈ V, w =

we define

v ⊗ w =

i,j

⊗ w

Note that note all elements of

V ⊗ W

are of this form. Some are genuine

combinations. For example,

⊗ w

cannot be written as a tensor

product of an element in V and another in W .

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





⊗





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).

(ii) If x, x

, x

∈ V and y, y

, y

∈ W , then

+ x

) ⊗ y = (x

⊗ y) + (x

⊗ y)

x ⊗ (y

+ y

) = (x ⊗ y

) + (x ⊗ y

Proof.

(i) Let v =

and w =

. Then

(λv) ⊗ w =

(λα

)β

⊗ w

λ(v ⊗ w) = λ

⊗ w

v ⊗ (λw) =

(λβ

⊗ w

and these three things are obviously all equal.

(ii) Similar nonsense.

We can define the tensor product using these properties. We consider the

space of formal linear combinations

v ⊗ w

for all

v ∈ V, w ∈ W

, and then

quotient out by the relations re had above. It can be shown that this produces

the same space as the one constructed above.

Alternatively, we can define the tensor product using its universal property,

which says for any

, a bilinear map from

V ×W

is “naturally” equivalent

to a linear map

V ⊗W → U

. This intuitively makes sense, since the distributivity

and linearity properties of

⊗

we showed above are exactly the properties required

for a bilinear map if we replace the ⊗ with a “,”.

It turns out this uniquely defines

V ⊗W

, as long as we provide a sufficiently

precise definition of “naturally”. We can do it concretely as follows — in our

explicit 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

V × W → V ⊗ W

in order to get a bilinear map

V ×W → U

. Then universality says every bilinear

map V × W → U arises this way, uniquely.

In general, we say an equivalence between bilinear maps

V × W → U

and

linear maps

V ⊗ W → U

is “natural” if it is be mediated by such a universal

map. Then we say a vector space

is the tensor product of

and

if there

is a natural equivalence between bilinear maps

V × W → U

and linear maps

X → U.

We will stick with our definition for concreteness, but we prove basis-

independence here so that we feel happier:

Lemma.

Let

, ··· , e

}

be any other basis of

, and

, ··· , f

}

be another

basis of W . Then

⊗ f

: 1 ≤ i ≤ m, 1 ≤ j ≤ n}

is a basis of V ⊗ W .

Proof. Writing

, w

we have

⊗ w

⊗ f

Therefore

⊗ f

}

spans

V ⊗ W

. Moreover, there are

of these. Therefore

they form a basis of V ⊗ W .

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

theory.

Proposition. Let ρ : G → GL(V ) and ρ

: G → GL(V

). We define

ρ ⊗ ρ

: G → GL(V ⊗ V

)

(ρ ⊗ ρ

)(g) :

⊗ w

7→

(ρ(g)v

) ⊗ (ρ

(g)w

Then ρ ⊗ ρ

is a representation of g, with character

ρ⊗ρ

(g) = χ

(g)χ

(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

is irreducible

and

is irreducible of degree 1, then

ρ ⊗ ρ

is irreducible. However, if

is not of degree 1, then this is almost always false (or else we can produce

arbitrarily many irreducible representations).

Proof.

It is clear that (

ρ ⊗ ρ

)(

)

∈ GL

(

V ⊗ V

) for all

g ∈ G

. So

ρ ⊗ ρ

is a

homomorphism G → GL(V ⊗ V

To check the character is indeed as stated, let

g ∈ G

. Let

, ··· , v

a basis of

of eigenvectors of

(

), and let

, ··· , w

be a basis of

eigenvectors of ρ

(g), say

ρ(g)v

= λ

, ρ

(g)w

= µ

Then

(ρ ⊗ ρ

)(g)(v

⊗ w

) = ρ(g)v

⊗ ρ

(g)w

= λ

⊗ µ

= (λ

)(v

⊗ w

ρ⊗ρ

(g) =

i,j







= χ

(g)χ

(g).

9.3 Powers of characters

We work over C. Take V = V

to be G-spaces, and define

Notation.

⊗2

= V ⊗ V.

We define τ : V

⊗2

→ V

⊗2

τ :

⊗ v

7→

⊗ v

This is a linear endomorphism of

⊗2

such that

= 1. So the eigenvalues are

±1.

Definition

(Symmetric and exterior square)

We define the symmetric square

and exterior square of V to be, respectively,

V = {x ∈ V

⊗2

: τ(x) = x}

V = {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 , S

V and Λ

V are G-subspaces of V

⊗2

, and

⊗2

= S

V ⊕ Λ

The space S

V has basis

= v

⊗ v

+ v

⊗ v

: 1 ≤ i ≤ j ≤ n},

while Λ

V has basis

∧ v

= v

⊗ v

− v

⊗ v

: 1 ≤ i < j ≤ n}.

Note that we have a strict inequality for

i < j

, since

⊗ v

− v

⊗ v

= 0 if

i = j. Hence

dim S

V =

n(n + 1), dim Λ

V =

n(n − 1).

Sometimes, a factor of

appears in front of the definition of the basis elements

and v

∧ v

, i.e.

⊗ v

+ v

⊗ v

), v

∧ v

⊗ v

− v

⊗ v

However, this is not important.

Proof.

This is elementary linear algebra. For the decomposition

⊗2

, given

x ∈ V

⊗2

, we can write it as

x =

(x + τ(x))

| {z }

∈S

(x − τ(x))

| {z }

∈Λ

How is this useful for character calculations?

Lemma.

Let

G → GL

(

) be a representation affording the character

Then

where

is the character of

in the subrepresentation

, and

= Λ

the character of

in the subrepresentation on Λ

Moreover, for g ∈ G,

(g) =

(χ

(g) + χ(g

)), χ

(g) =

(χ

(g) − χ(g

)).

The real content of the lemma is, of course, the formula for χ

and χ

Proof.

The fact that

is immediate from the decomposition of

G-spaces.

We now compute the characters

and

. For

g ∈ G

, we let

, ··· , v

a basis of V of eigenvectors of ρ(g), say

ρ(g)v

= λ

We’ll be lazy and just write

instead of

(

)

. Then, acting on Λ

, we get

g(v

∧ v

) = λ

∧ v

Thus

(g) =

1≤i<j≤n

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

(χ(g))





+ 2

i<j

= χ(g

) + 2

i<j

= χ(g

) + 2χ

(g).

Then we can solve to obtain

(g) =

(χ

(g) − χ(g

)).

Then we can get

= χ

− χ

(χ

(g) + χ(g

)).

Example. Consider again G = S

. As before, we have the following table:

1 3 8 6 6

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

trivial χ

1 1 1 1 1

sign χ

1 1 1 −1 −1

− 1

3 −1 0 −1 1

3 −1 0 1 −1

We now compute χ

in a different way, by decomposing χ

. We have

1 3 8 6 6

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

9 1 0 1 1

) 3 3 0 −1 3

6 2 0 0 2

3 −1 0 1 −1

We notice Λ

is just

. Now

is not irreducible, which we can easily

show by computing the inner product. By taking the inner product with the

other irreducible characters, we find

= 1 +

, where

is our new

irreducible character. So we obtain

1 3 8 6 6

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

trivial χ

1 1 1 1 1

sign χ

1 1 1 −1 −1

− 1

3 −1 0 −1 1

3 −1 0 1 −1

− 1 − χ

2 2 −1 0 0

9.4 Characters of G × H

We have looked at characters of direct products a bit before, when we decomposed

an abelian group into a product of cyclic groups. We will now consider this in

the general case.

Proposition.

Let

and

be two finite groups with irreducible characters

, ··· , χ

and

, ··· , ψ

respectively. Then the irreducible characters of the

direct product G × H are precisely

{χ

: 1 ≤ i ≤ k, 1 ≤ j ≤ r},

where

(χ

)(g, h) = χ

(g)ψ

(h).

Proof.

Take

G → GL

(

) affording

, and

H → GL

(

) affording

Then define

ρ ⊗ ρ

: G × H → GL(V ⊗ W )

(g, h) 7→ ρ(g) ⊗ρ

(h),

where

(ρ(g) ⊗ ρ

(h))(v

⊗ w

) 7→ ρ(g)v

⊗ ρ

(h)w

This is a representation of

G × H

V ⊗ W

, and

ρ⊗ρ

χψ

. The proof is

similar to the case where

ρ, ρ

are both representations of

, and we will not

repeat it here.

Now we need to show

are distinct and irreducible. It suffices to show

they are orthonormal. We have

hχ

, χ

G×H

|G × H|

(g,h)∈G×H

(g, h)χ

(g, h)





|G|

g∈G

(g)χ

(g)





|H|

h∈H

(h)ψ

(h)

= δ

So it follows that

{χ

}

are distinct and irreducible. We need to show this is

complete. We can consider

i,j

(1)

(1)ψ

(1) =



(1)



(1)



= |G||H| = |G × H|.

So done.

9.5 Symmetric and exterior powers

We have introduced the objects

and Λ

. We can do this in a more general

setting, for higher powers. Consider a vector space

over

, and

dim

We let {v

, ··· , v

} be a basis, and let

V = V

⊗n

= V ⊗ ··· ⊗ V

| {z }

n times

The basis for this space is just

⊗ ··· ⊗ v

: i

, ··· , i

∈ {1, ··· , d}}.

First of all, there is an obvious

-action on this space, permuting the multipli-

cands of

⊗n

. For any

σ ∈ S

, we can define an action

⊗n

→ V

⊗n

given

⊗ ···v

7→ v

σ(1)

⊗ ··· ⊗ v

σ(n)

Note that this is a left action only if we compose permutations right-to-left. If

we compose left-to-right instead, we have to use

−1

instead of

(or use

and

obtain a right action). In fact, this defines a representation of

⊗n

linear representation.

itself is a

-space, then we also get a

-action on

⊗n

. Let

G →

GL(V ) be a representation. Then we obtain the action of G on V

⊗n

: v

⊗ ··· ⊗ v

7→ ρ(g)v

⊗ ··· ⊗ ρ(g)v

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 important

theorem by Schur, and has many many applications. However, that would be

for another course.

Getting back on track, since the two actions commute, we can decompose

⊗n

as an

-module, and each isotypical component is a

-invariant subspace

⊗n

. We don’t really need this, but it is a good generalization of what we’ve

had before.

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

(i) The nth symmetric power of V is

V = {x ∈ V

⊗n

: σ(x) = x for all σ ∈ S

(ii) The nth exterior power of V is

V = {x ∈ V

⊗n

: σ(x) = (sgn σ)x for all σ ∈ S

Both of these are G-subspaces of V

⊗n

Recall that in the

= 2 case, we have

⊗2

V ⊕

. However, for

n >

2, we get

V ⊕

V  V

⊗n

. In general, lots of others are obtained from

the S

-action.

Example.

The bases for

and Λ

are obtained as follows: take

, ··· , e

a basis of V . Then

(

σ∈S

σ(1)

⊗ ··· ⊗ e

σ(n)

: 1 ≤ i

≤ ··· ≤ i

≤ d

)

is a basis for S

V . So the dimension is

dim S

V =



d + n − 1



For Λ

V , we get a basis of

(

σ∈S

(sgn σ)e

σ(1)

⊗ ··· ⊗ e

σ(n)

: 1 ≤ i

< ··· < i

≤ d

)

The dimension is

dim Λ

V =

(

0 n > d





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 need

it here, but it has many applications elsewhere.

Consider a field F with characteristic 0.

Definition

(Tensor algebra)

Let

⊗n

. Then the tensor algebra of

(V ) = T (V ) =

n≥0

with T

V = F by convention.

This is a vector space over

with the obvious addition and multiplication by

scalars.

(

) is also a (non-commutative) (graded) ring with product

x·y

x⊗y

This is graded in the sense that if

x ∈ T

and

y ∈ T

x·y

x⊗y ∈ T

n+m

Definition

(Symmetric and exterior algebra)

We define the symmetric algebra

of 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

and

are not elements of

, but arbitrary elements of

for

some n.

The symmetric algebra is commutative, while the exterior algebra is graded-

commutative, i.e. if x ∈ Λ

(V ) and y ∈ Λ

(V ), then x · y = (−1)

n+m+1

y · x.

9.7 Character ring

Recall that

(

) is a commutative ring, and contains all the characters. Also,

the sum and products of characters are characters. However, they don’t quite

form a subring, since we are not allowed to subtract.

Definition

(Character ring)

The character ring of

is the

-submodule of

C(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 is

not quite a character.

Definition

(Generalized/virtual characters)

The elements of

(

) are called

generalized or virtual characters. These are class functions of the form

ψ =

χ,

summing over all irreducibles χ, and n

∈ Z.

We know

(

) is a ring, and any generalized character is a difference of two

characters (we let

α =

≥0

χ, β =

(−n

)χ.

Then

α − β

, and

and

are characters). Then the irreducible characters

{χ

}

forms a

-basis for

(

) as a free

-module, since we have shown that

they are independent, and it generates R(G) by definition.

Lemma.

Suppose

is a generalized character and

hα, αi

= 1 and

(1)

Then α is actually a character of an irreducible representation of G.

Proof. We list the irreducible characters as χ

, ··· , χ

. We then write

α =

Since the χ

’s are orthonormal, we get

hα, αi =

= 1.

So exactly one of

1, while the others are all zero. So

±χ

for some

Finally, since

(1)

0 and also

(1)

0, we must have

= +1. So

Henceforth we don’t distinguish between a character and its negative, and

often study generalized characters of inner product 1 rather than irreducible

characters.

10 Induction and restriction

This is the last chapter on methods of calculating characters. Afterwards, we

shall start applying these to do something useful.

Throughout the chapter, we will take

. Suppose

H ≤ G

. How do

characters of

and

relate? It is easy to see that every representation of

can be restricted to give a representation of

. More surprisingly, given a

representation of H, we can induce a representation of G.

We first deal with the easy case of restriction.

Definition

(Restriction)

Let

G → GL

(

) be a representation affording

We can think of

as an

-space by restricting

’s attention to

h ∈ H

. We get

a representation

Res

ρ ↓

, the restriction of

. It affords the

character Res

χ = χ

= χ ↓

It is a fact of life that the restriction of an irreducible character is in general

not irreducible. For example, restricting any non-linear irreducible character to

the trivial subgroup will clearly produce a reducible character.

Lemma.

Let

H ≤ G

. If

is any non-zero irreducible character of

, then there

exists an irreducible character

such that

is a constituent of

Res

, i.e.

hRes

χ, ψi 6= 0.

Proof.

We list the irreducible characters of

, ··· , χ

. Recall the regular

character π

reg

. Consider

hRes

reg

, ψi =

|G|

|H|

ψ(1) 6= 0.

On the other hand, we also have

hRes

reg

, ψi

deg χ

hRes

, ψi.

If this sum has to be non-zero, then there must be some

such that

hRes

, ψi 6

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

Res

χ =

with χ

irreducible characters of H, and c

non-negative integers. Then

≤ |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.

Proof. We have

hRes

χ, Res

χi

However, by definition, we also have

hRes

χ, Res

χi

|H|

h∈H

|χ(h)|

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

1 = hχ, χi

|G|

g∈G

|χ(g)|

|G|





h∈H

|χ(h)|

g∈G\H

|χ(g)|





|H|

|G|

g∈G\H

|χ(g)|

≥

|H|

|G|

So the result follows.

Now we shall go in the other direction — given a character of

, we want to

induce a character of the larger group

. Perhaps rather counter-intuitively, it is

easier for us to first construct the character, and then later find a representation

that affords this character.

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

Definition

(Induced class function)

Let

ψ ∈ C

(

). We define the induced

class function Ind

ψ = ψ ↑

= ψ

Ind

ψ(g) =

|H|

x∈G

ψ(x

−1

gx),

where

ψ(y) =

(

ψ(y) y ∈ H

0 y 6∈ H

The first thing to check is, of course, that Ind

is indeed a class function.

Lemma. Let ψ ∈ C

. Then Ind

ψ ∈ C(G), and Ind

ψ(1) = |G : H|ψ(1).

Proof.

The fact that

Ind

is a class function follows from direct inspection of

the formula. Then we have

Ind

ψ(1) =

|H|

x∈G

ψ(1) =

|G|

|H|

ψ(1) = |G : H|ψ(1).

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

alternative formula for it.

If we have a subset, a sensible thing to do is to look at the cosets of

We let

, and let 1 =

, ··· , t

be a left transversal of

, i.e.

H = H, ··· , t

H are precisely the n left-cosets of H in G.

Lemma. Given a (left) transversal t

, ··· , t

of H, we have

Ind

ψ(g) =

i=1

ψ(t

−1

Proof.

We can express every

x ∈ G

for some

h ∈ H

and

. We then

have

ψ((t

−1

g(t

h)) =

ψ(h

−1

)h) =

ψ(t

−1

since

is a class function of

, and

−1

(

−1

)

h ∈ H

if and only if

−1

∈ 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

hRes

ϕ, ψi

= hϕ, Ind

ψi

Alternatively, we can write this as

hϕ

, ψi = hϕ, ψ

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 special

case of the tensor-Hom adjunction. If you neither understand nor care about

this, then this is a good sign [sic].

Proof. We have

hϕ, ψ

i =

|G|

g∈G

ϕ(g)ψ

(g)

|G||H|

x,g∈G

ϕ(g)

ψ(x

−1

gx)

We now write

−1

. Then summing over

is the same as summing over

Since ϕ is a G-class function, this becomes

|G||H|

x,y∈G

ϕ(y)

ψ(y)

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

|H|

y∈G

ϕ(y)

ψ(y)

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

ψ. So

|H|

y∈H

ϕ(y)ψ(y)

= hϕ

, ψi

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

ψ is a character of G.

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

hInd

ψ, χi = hψ, Res

χi.

Since

and

Res

are characters, the thing on the right is in

≥0

. Hence

Ind

is a linear combination of irreducible characters with non-negative coefficients,

and is hence a character.

Recall we denote the conjugacy class of

g ∈ G

(

), while the centralizer

(

). If we take a conjugacy class

(

) in

and restrict it to

, then

the result

(

)

∩ H

need not be a conjugacy class, since the element needed

to conjugate

x, y ∈ C

(

)

∩ H

need not be present in

, as is familiar from the

case of

≤ S

. However, we know that

(

)

∩ H

is a union of conjugacy

classes 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

(g) ∩ H =

[

i=1

where the

are the representatives of the

conjugacy classes of elements of

conjugate to g. If m = 0, then Ind

ψ(g) = 0. Otherwise,

Ind

ψ(g) = |C

(g)|

i=1

ψ(x

)

This is all just group theory. Note that some people will think this proof

is excessive — everything shown is “obvious”. In some sense it is. Some steps

seem very obvious to certain people, but we are spelling out all the details so

that everyone is happy.

Proof.

= 0, then

{x ∈ G

−1

gx ∈ H}

∅

. So

(

−1

) = 0 for all

. So

Ind

ψ(g) = 0 by definition.

Now assume m > 0. We let

= {x ∈ G : x

−1

gx ∈ H and is conjugate in H to x

By definition of

, we know the

’s are pairwise disjoint, and their union is

{x ∈ G : x

−1

gx ∈ H}. Hence by definition,

Ind

ψ(g) =

|H|

x∈G

ψ(x

−1

gx)

|H|

i=1

x∈X

ψ(x

−1

gx)

|H|

i=1

x∈X

ψ(x

)

i=1

|H|

ψ(x

So we now have to show that in fact

|H|

(g)|

We fix some 1

≤ i ≤ m

. Choose some

∈ G

such that

−1

. This exists

by definition of x

. So for every c ∈ C

(g) and h ∈ H, we have

(cg

−1

g(cg

h) = h

−1

gcg

We now use the fact that c commutes with g, since c ∈ C

(g), to get

= h

−1

cgg

= h

−1

= h

−1

Hence by definition of X

, we know cg

h ∈ X

. Hence

(g)g

H ⊆ X

Conversely, if

x ∈ X

, then

−1

(

−1

)

for some

. Thus

−1

∈ C

(g), and so x ∈ C

(g)g

h. So

x ∈ C

(g)g

So we conclude

= C

(g)g

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

| = |C

(g)g

H| =

(g)||H|

|H ∩ g

−1

(g)g

Finally, we note

−1

(g)g

= C

−1

) = C

Thus

| =

|H||C

(g)|

|H ∩ C

|H||C

(g)|

Dividing, we get

|H|

(g)|

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|

−1

Hx ∩ K|

Lemma.

Let

= 1

, the trivial character of

. Then

Ind

, the

permutation character of

on the set

, where

G/H

is the set of left

cosets of H.

Proof.

We let

, and

, ··· , t

be representatives of the cosets. By

definition, we know

Ind

(g) =

i=1

−1

)

= |{i : t

−1

∈ H}|

= |{i : g ∈ t

−1

But t

−1

is the stabilizer in G of the coset t

H ∈ X. So this is equal to

= |fix

(g)|

= π

(g).

By Frobenius reciprocity, we know

hπ

, 1

= hInd

, 1

= h1

, 1

= 1.

So the multiplicity of 1

in π

is indeed 1, as we have shown before.

Example.

Let

(1 2 3 4)

i ≤ G

. The index is

= 6. We

consider the character of the induced representation

Ind

(

), where

is the

faithful 1-dimensional representation of C

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)

1 i −1 −i

What is the induced character in S

? We know that

Ind

(α)(1) = |G : H|α(1) = |G : H| = 6.

So the degree of

Ind

(

) is 6. Also, the elements (1 2) and (3 4) are not

conjugate to anything in C

. So the character vanishes.

For (1 2)(3 4), only one of the three conjugates in

lie in

(namely

(1 3)(2 4)). So, using the sizes of the centralizers, we obtain

Ind

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



−1



= −2.

For (1 2 3 4), it is conjugate to 6 elements of

, two of which are in

, namely

(1 2 3 4) and (1 4 3 2). So

Ind

(1 2 3 4) = 4



−i



= 0.

So we get the following character table:

1 6 8 3 6

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

Ind

(α) 6 0 0 −2 0

Induced representations

We have constructed a class function, and showed it is indeed the character of

some representation. However, we currently have no idea what this corresponding

representation 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 also

need this explicit construction when proving Mackey’s theorem later.

Let

H ≤ G

have index

, and 1 =

, ··· , t

be a left transversal. Let

a H-space. Define the vector space

V = Ind

W = W ⊕ (t

⊗ W ) ⊕ ··· ⊕ (t

⊗ W ),

where

⊗ W = {t

⊗ w : w ∈ W }.

So dim V = n dim W .

To define the

-action on

, let

g ∈ G

. Then for every

, there is a unique

such that t

−1

∈ H, i.e. gt

H = t

H. We define

g(t

⊗ w) = t

⊗ ((t

−1

)w).

We tend to drop the tensor signs, and just write

g(t

w) = t

−1

w).

We claim this is a G-action. We have

w) = g

−1

)w),

where j is the unique index such that g

H = t

H. This is then equal to

((t

−1

)(t

−1

)w) = t

((t

−1

)w) = (g

)(t

w),

where

is the unique index such that

(and hence

H = t

H).

Definition

(Induced representation)

Let

H ≤ G

have index

, and 1 =

, ··· , t

be a left transversal. Let

be a

-space. Define the induced

representation to be the vector space

Ind

W = W ⊕ t

⊗ W ⊕ ··· ⊕ t

⊗ W,

with the G-action

g : t

w 7→ t

−1

)W,

where t

is the unique element (among t

, ··· , t

) such that t

−1

∈ H.

This has the “right” character. Suppose

has character

. Since

w 7→

(

−1

), the contribution to the character is 0 unless

, i.e. if

−1

∈ H

Then it contributes

ψ(t

−1

Thus we get

Ind

ψ(g) =

i=1

ψ(t

−1

Note that this construction is rather ugly. It could be made much nicer if we

knew a bit more algebra — we can write the induced module simply as

Ind

W = FG ⊗

where we view

as a left-

module, and

as a (

FG, FH

)-bimodule. Alter-

natively, this is the extension of scalars from FH to FG.

11 Frobenius groups

We will now use character theory to prove some major results in finite group

theory. We first do Frobenius’ theorem here. Later, will prove Burnside’s

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 unavoidable

ingredient of it.

Theorem

(Frobenius’ theorem (1891))

Let

be a transitive permutation group

on a finite set

, with

|X|

. Assume that each non-identity element of

fixes at most one element of

. 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

is even a subgroup at all. It turns out

normality isn’t really hard to prove — the hard part is indeed showing it is a

subgroup.

Note that we did not explicitly say

is finite. But these conditions imply

G ≤ S

, which forces G to be finite.

Proof.

The idea of the proof is to construct a character Θ whose kernel is

First note that by definition of K, we have

G = K ∪

[

α∈X

where

is, as usual, the stabilizer of

. Also, we know that

∩ G

{

}

α 6= β by assumption, and by definition of K, we have K ∩ G

= {1} as well.

Next note that all the

are conjugate. Indeed, we know

is transitive,

and

−1

gα

. We set

for some arbitrary choice of

. Then the

above tells us that

|G| = |K| − |X|(|H| − 1).

On the other hand, by the orbit-stabilizer theorem, we know

|G|

|X||H|

. So it

follows that we have

|K| = |X| = n.

We first compute what induced characters look like.

Claim. Let ψ be a character of H. Then

Ind

ψ(g) =











nψ(1) g = 1

ψ(g) g ∈ H \ {1}

0 g ∈ K \ {1}

Since every element in

is either in

or conjugate to an element in

, this

uniquely specifies what the induced character is.

This is a matter of computation. Since [

] =

, the case

= 1

immediately follows. Using the definition of the induced character, since any

non-identity in

is not conjugate to any element in

, we know the induced

character vanishes on K \ {1}.

Finally, suppose

g ∈ H \{

}

. Note that if

x ∈ G

, then

xgx

−1

∈ G

xα

. So this

lies in H if and only if x ∈ H. So we can write the induced character as

Ind

ψ(g) =

|H|

g∈G

ψ(xgx

−1

) =

|H|

h∈H

ψ(hgh

−1

) = ψ(g).

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

θ = ψ

− ψ(1)(1

)

+ ψ(1)1

Then θ is a character, and

θ(g) =

(

ψ(h) h ∈ H

ψ(1) k ∈ K

Note that we chose the coefficients exactly so that the final property of

holds. This is a matter of computation:

1 h ∈ H \ {1} K \ {1}

nψ(1) ψ(h) 0

ψ(1)(1

)

nψ(1) ψ(1) 0

ψ(1)1

ψ(1) ψ(1) ψ(1)

ψ(1) ψ(h) ψ(1)

The less obvious part is that

is a character. From the way we wrote it, we

already know it is a virtual character. We then compute the inner product

hθ, θi

|G|

g∈G

|θ(g)|

|G|





g∈K

|θ(g)|

g∈G\K

|θ(g)|





|G|





n|ψ(1)|

+ n

h6=1∈H

|ψ(h)|





|G|

h∈H

|ψ(h)|

|G|

(n|H|hψ, ψi

)

= 1.

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

Finally, we have

Claim.

Let

, ··· , ψ

be the irreducible representations of

, and

be the

corresponding representations of G constructed above. Set

Θ =

i=1

(1)θ

Then we have

θ(g) =

(

|H| g ∈ K

0 g 6∈ K

From this, it follows that the kernel of the representation affording

, and

in particular K is a normal subgroup of G.

This is again a computation using column orthogonality. For 1

h ∈ H

, we

have

Θ(h) =

i=1

(1)ψ

(h) = 0,

and for any y ∈ K, we have

Θ(y) =

i=1

(1)

= |H|.

Definition

(Frobenius group and Frobenius complement)

A Frobenius group

is a group

having a subgroup

such that

H ∩ gHg

−1

= 1 for all

g 6∈ H

. We

say 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 the

Frobenius complement satisfies the hypothesis of Frobenius’ theorem.

Definition

(Frobenius kernel)

The Frobenius kernel of a Frobenius group

the K obtained from Frobenius’ theorem.

Proof.

Let

be a Frobenius group, having a complement

. Then the action of

on the cosets

G/H

is transitive. Furthermore, if 1

g ∈ G

fixes

and

then we have

g ∈ xHx

−1

∩ yHy

−1

. This implies

H ∩

(

−1

)

(

−1

)

−1

= 1.

Hence xH = yH.

Note that J. Thompson (in his 1959 thesis) proved any finite group having

a fixed-point free automorphism of prime order is nilpotent. This implies

nilpotent, which means K is a direct product of its Sylow subgroups.

12 Mackey theory

We work over

. We wish to describe the restriction to a subgroup

K ≤ G

of an

induced representation

Ind

, i.e. the character

Res

Ind

. In general,

and

can be unrelated, but in many applications, we have

, in which

case we can characterize when Ind

W is irreducible.

It is quite helpful to first look at a special case, where

= 1

is the trivial

representation. Thus

Ind

is the permutation representation of

G/H

(coset action on the left cosets of H in G).

Recall that by the orbit-stabilizer theorem, if

is transitive on the set

and

for some

α ∈ X

, then the action of

is isomorphic the action

on G/H, namely the correspondence

gα ↔ gH

is a well-defined bijection, and commutes with with the

-action (i.e.

(

gα

) =

(xg)α ↔ x(gH) = (xg)H).

We now consider the action of

G/H

and let

K ≤ G

. Then

also acts

G/H

, and

G/H

splits into

-orbits. The

-orbit of

contains precisely

kgH for k ∈ K. So it is the double coset

KgH = {kgh : k ∈ K, h ∈ H}.

The set of double cosets

K\G/H

partition

, and the number of double cosets

|K\G/H| = hπ

G/K

, π

G/H

We don’t need this, but this is true.

If it happens that

, and

is normal, then we just have

K\G/H

G/H.

What about stabilizers? Clearly,

gHg

−1

. Thus, restricting to the

action of K, we have K

= gHg

−1

∩ K. We call H

= K

So by our correspondence above, the action of

on the orbit containing

is isomorphic to the action of

(

gHg

−1

∩ K

) =

K/H

. From this,

and using the fact that

Ind

(

G/H

), we get the special case of Mackey’s

theorem:

Proposition.

Let

be a finite group and

H, K ≤ G

. Let

, ··· , g

be the

representatives of the double cosets K\G/H. Then

Res

Ind

∼

i=1

Ind

−1

∩K

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

Theorem

(Mackey’s restriction formula)

In general, for

K, H ≤ G

, we let

S = {1, g

, ··· , g

} be a set of double coset representatives, so that

G =

[

We write

gHg

−1

∩ K ≤ G

. We let (

ρ, W

) be a representation of

. For

each

g ∈ G

, we define (

, W

) to be a representation of

, with the same

underlying vector space W , but now the action of H

(x) = ρ(g

−1

xg),

where h = g

−1

xg ∈ H by construction.

This is clearly well-defined. Since

≤ K

, we obtain an induced representa-

tion Ind

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

Res

Ind

W =

g∈S

Ind

We will defer the proof of this for a minute or two. We will first derive some

corollaries of this, starting with the character version of the theorem.

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

Res

Ind

ψ =

g∈S

Ind

where ψ

is the class function (and a character) on H

given by

(x) = ψ(g

−1

xg).

These characters

are sometimes known as the conjugate characters. This

obviously follows from Mackey’s restriction formula.

Corollary

(Mackey’s irreducibility criterion)

Let

H ≤ G

and

be a

-space.

Then V = Ind

W is irreducible if and only if

(i) W is irreducible; and

(ii)

For each

g ∈ S\H

, the two

spaces

and

Res

have no irreducible

constituents in common, where H

= gHg

−1

∩ H.

Note that the set

of representatives was chosen arbitrarily. So we may

require (ii) to hold for all

g ∈ G \ H

, instead of

g ∈ S \ H

. While there are

more things to check in

, it is sometimes convenient not have to pick out an

explicitly.

Note that the only

g ∈ S ∩ H

is the identity, since for any

h ∈ H

KhH

KH = K1H.

Proof.

We use characters, and let

afford the character

. We take

Mackey’s restriction formula. Then we have H

= gHg

−1

∩ H.

Using Frobenius reciprocity, we can compute the inner product as

hInd

ψ, Ind

ψi

= hψ, Res

Ind

ψi

g∈S

hψ, Ind

g∈S

hRes

ψ, ψ

= hψ, ψi +

g∈S\H

hRes

ψ, ψ

We can write this because if g = 1, then H

= H, and ψ

= ψ.

This is a sum of non-negative integers, since the inner products of characters

always are. So

Ind

is irreducible if and only if

hψ, ψi

= 1, and all the other

terms in the sum are 0. In other words,

is an irreducible representation of

and for all g 6∈ H, W and W

are disjoint representations of H

Corollary.

Let

H C G

, and suppose

is an irreducible character of

. Then

Ind

is irreducible if and only if

is distinct from all its conjugates

for

g ∈ G \ H (where ψ

(h) = ψ(g

−1

hg) as before).

Proof.

We take

H C G

. So the double cosets are just left cosets. Also,

= H for all g. Moreover, W

is irreducible since W is irreducible.

So, by Mackey’s irreducible criterion,

Ind

is irreducible precisely if

W 6

∼

for all g ∈ G \ H. This is equivalent to ψ 6= ψ

Note that again we could check the conditions on a set of representatives of

(double) cosets. In fact, the isomorphism class of

(for

g ∈ G

) depends only

on the coset gH.

We now prove Mackey’s theorem.

Theorem

(Mackey’s restriction formula)

In general, for

K, H ≤ G

, we let

S = {1, g

, ··· , g

} be a set of double coset representatives, so that

G =

[

We write

gHg

−1

∩ K ≤ G

. We let (

ρ, W

) be a representation of

. For

each

g ∈ G

, we define (

, W

) to be a representation of

, with the same

underlying vector space W , but now the action of H

(x) = ρ(g

−1

xg),

where h = g

−1

xg ∈ H by construction.

This is clearly well-defined. Since

≤ K

, we obtain an induced representa-

tion Ind

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

Res

Ind

W =

g∈S

Ind

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

Proof.

Write

Ind

. Pick

g ∈ G

, so that

KgH ∈ K\G/H

. Given a left

transversal 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 coset

representatives, by collecting together the t ⊗ W ’s with t ∈ KgH. We define

V (g) =

t∈KgH∩T

t ⊗ W.

Now each

(

) is a

-space — given

k ∈ K

and

t ⊗w ∈ t ⊗W

, since

t ∈ KgH

we have

kt ∈ KgH

. So there is some

∈ T

such that

ktH

. Then

∈ ktH ⊆ KgH. So we can define

k · (t ⊗ w) = t

⊗ (ρ(t

0−1

kt)w),

where t

kt ∈ H.

Viewing V as a K-space (forgetting its whole G-structure), we have

Res

V =

g∈S

V (g).

The left hand side is what we want, but the right hand side looks absolutely

nothing like Ind

. So we need to show

V (g) =

t∈KgH∩T

t ⊗ W

∼

Ind

as K representations, for each g ∈ S.

Now for

fixed, each

t ∈ KgH

can be represented by some

kgh

, and by

restricting to elements in the traversal

, we are really summing over cosets

kgH

. Now cosets

kgH

are in bijection with cosets

(

gHg

−1

) in the obvious way.

So we are actually summing over elements in

(

gHg

−1

∩ K

) =

K/H

. So we

write

V (g) =

k∈K/H

(kg) ⊗ W.

We claim that there is a isomorphism that sends

k ⊗W

∼

(

)

⊗W

. We define

k ⊗W

→

(

)

⊗W

k ⊗w 7→ kg ⊗w

. This is an isomorphism of vector spaces

almost by definition, so we only check that it is compatible with the action. The

action of x ∈ K on the left is given by

(x)(k ⊗ w) = k

⊗ (ρ

0−1

xk)w) = k

⊗ (ρ(g

−1

0−1

xkg)w),

where

∈ K

is such that

0−1

xk ∈ H

, i.e.

−1

0−1

xkg ∈ H

. On the other

hand,

ρ(x)(kg ⊗ w) = k

⊗ (ρ(k

−1

(kg))w),

where

∈ K

is such that

00−1

xkg ∈ H

. Since there is a unique choice of

(after picking a particular transversal), and

works, we know this is equal to

g ⊗ (ρ(g

−1

0−1

xkg)w).

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

Then

V (g) =

k∈K/H

k ⊗ W

= Ind

as required.

Example.

Let

and

. Consider a

σ ∈ S

. Then its conjugacy

class in S

is determined by its cycle type.

If the class of

splits into two classes in

, and

is an irreducible character

that takes different values on the two classes, then by the irreducibility

criterion, Ind

χ is irreducible.

13 Integrality in the group algebra

The next big result we are going to have is that the degree of an irreducible

character always divides the group order. This is not at all an obvious fact. To

prove this, we make the cunning observation that character degrees and friends

are not just regular numbers, but algebraic integers.

Definition

(Algebraic integers)

A complex number

a ∈ C

is an algebraic integer

is a root of a monic polynomial with integer coefficients. Equivalently,

such that the subring of C given by

Z[a] = {f(a) : f ∈ Z[X]}

is finitely generated. Equivalently,

is the eigenvalue of a matrix, all of whose

entries 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)

a ∈ C

is both an algebraic integer and rational, then

is in fact an

integer.

(iii)

Any subring of

which is a finitely generated

-module consists of algebraic

integers.

The strategy is to show that given a group

and an irreducible character

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.

is a character of

and

g ∈ G

, then

(

) is an algebraic

integer.

Proof.

We know

(

) is the sum of roots

th roots of unity (where

is the order

). Each root of unity is an algebraic integer, since it is by definition a root of

−

1. Since algebraic integers are closed under addition, the result follows.

We now have a look at the center of the group algebra

. Recall that the

group algebra

of a finite group

is a complex vector space generated by

elements of G. More precisely,

CG =







g∈G

g : α

∈ C







Borrowing the multiplication of

, this forms a ring, hence an algebra. We now

list the conjugacy classes of G as

{1} = C

, ··· , C

Definition

(Class sum)

The class sum of a conjugacy class

of a group

g∈C

g ∈ CG.

The claim is now

lives in the center of

(note that the center of the

group algebra is different from the group algebra of the center). Moreover, they

form a basis:

Proposition.

The class sums

, ··· , C

form a basis of

(

). There exists

non-negative integers a

ij`

(with 1 ≤ i, j, ` ≤ k) with

`=1

ij`

The content of this proposition is not that

can be written as a sum of

’s. This is true because the

’s form a basis. The content is that these are

actually integers, not arbitrary complex numbers.

Definition

(Class algebra/structure constants)

The constants

ij`

as defined

above are the class algebra constants or structure constants.

Proof.

It is clear from definition that

−1

. So we have

∈ Z

(

Also, since the

’s are produced from disjoint conjugacy classes, they are linearly

independent.

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

z =

g∈G

By definition, this commutes with all elements of

. So for all

h ∈ G

, we must

have

−1

= α

So the function

g 7→ α

is constant on conjugacy classes of

. So we can write

= α

for g ∈ C

. Then

g =

j=1

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

`=1

ij`

for some complex numbers

ij`

, since the

span. The claim is that

ij`

∈ Z

≥0

for all

i, j`

. To see this, we fix

∈ C

. Then by definition of multiplication, we

know

ij`

= |{(x, y) ∈ C

× C

: xy = g

}|,

which is clearly a non-negative integer.

Let

G → GL

(

) be an irreducible representation of

over

affording

the character

. Extending linearly, we get a map

CG → End V

, an algebra

homomorphism. In general, we have the following definition:

Definition

(Representation of algebra)

Let

be an algebra. A representation

of A is a homomorphism of algebras ρ : A → End V .

Let

z ∈ Z

(

). Then

(

) commutes with all

(

) for

g ∈ G

. Hence, by

Schur’s lemma, we can write

(

) =

for some

∈ C

. We then obtain the

algebra homomorphism

= ω

= ω : Z(CG) → C

z 7→ λ

By definition, we have ρ(C

) = ω(C

)I. Taking traces of both sides, we know

g∈C

χ(g) = χ(1)ω(C

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

χ(1)ω(C

) = |C

|χ(g

where g

∈ C

is a representative of C

. So we get

ω(C

) =

χ(g

)

χ(1)

Why should we care about this? The thing we are looking at is in fact an

algebraic integer.

Lemma. The values of

) =

χ(g)

χ(1)

are algebraic integers.

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

Proof.

Using the definition of

ij`

∈ Z

≥0

, and the fact that

is an algebra

homomorphism, we get

)ω

) =

`=1

ij`

Thus the span of

{ω

(

) : 1

≤ j ≤ k}

is a subring of

and is finitely generated

as a

-module (by definition). So we know this consists of algebraic integers.

That’s magic.

In fact, we can compute the structure constants

ij`

from the character table,

namely for all i, j, `, we have

ij`

|G|

)||C

s=1

)χ

−1

)

(1)

where we sum over the irreducible characters of

. We will neither prove it nor

use it. But if you want to try, the key idea is to use column orthogonality.

Finally, we get to the main result:

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

(1) | |G|

for each irreducible χ

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

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

|G|

χ(1)

g∈G

χ(g)χ(g

−1

)

χ(1)

i=1

|χ(g

)χ(g

−1

)

i=1

|χ(g

)

χ(1)

χ(g

)

−1

Now we notice

|χ(g

)

χ(1)

is an algebraic integer, by the previous lemma. Also,

(

−1

) is an algebraic

integer. So the whole mess is an algebraic integer since algebraic integers are

closed under addition and multiplication.

But we also know

|G|

χ(1)

is rational. So it must be an integer!

Example.

Let

be a

-group. Then

(1) is a

-power for each irreducible

If in particular

|G|

, then

(1) = 1

, p

. But the sum of squares of the

degrees is

|G|

. So it cannot be that

(1) =

. So

(1) = 1 for all

irreducible characters. So G is abelian.

Example.

No simple group has an irreducible character of degree 2. Proof is

left as an exercise for the reader.

14 Burnside’s theorem

Finally, we get to Burnside’s theorem.

Theorem

(Burside’s

theorem)

Let

p, q

be primes, and let

|G|

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

is soluble, but this follows easily

from above by induction.

This result is the best possible in the sense that

= 60 = 2

5, and

is simple. So we cannot allow for three different primes factors (in fact, there

are exactly 8 simple groups with three prime factors). Also, if

= 0 or

= 0,

then

is a

-group, which have non-trivial center. So these cases trivially work.

Later, in 1963, Feit and Thompson proved that every group of odd order is

soluble. The proof was 255 pages long. We will not prove this.

In 1972, H. Bender found the first proof of Burnside’s theorem without the

use of representation theory, but the proof is much more complicated.

This theorem follows from two lemmas, and one of them involves some Galois

theory, and is hence non-examinable.

Lemma. Suppose

α =

j=1

is an algebraic integer, where

= 1 for all

and some

. Then either

= 0 or

|α| = 1.

Proof (non-examinable).

Observe

α ∈ F

(

), where

2πi/n

(since

∈ F

for 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

(

) is an algebraic integer, since Galois conjugates

(

) of algebraic

integers are algebraic integers. So in fact

(

) is an integer. But for

α ∈ G

, we

know

|σ(α)| =



σ(λ

)



≤ 1.

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

Lemma.

Suppose

is an irreducible character of

, and

is a conjugacy class

in 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)

Since

(

) is the sum of

deg χ

(1) many roots of unity, it suffices to show

that α is an algebraic integer.

By B´ezout’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 class

C is of (non-trivial) prime power order, then G is not non-abelian simple.

Proof.

Suppose

is a non-abelian simple group, and let 1

g ∈ G

be living in

the conjugacy class

of order

. If

χ 6

= 1

is a non-trivial irreducible character

, then either

(1) and

|C|

are not coprime, in which case

p | χ

(1), or

they are coprime, in which case |χ(g)| = χ(1) or χ(g) = 0.

However, it cannot be that

|χ

(

)

(1). If so, then we must have

(

) =

λI

for 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

is simple,

must be faithful. So we must have

for all

. So

(

) is non-trivial. This is a contradiction. So either

p | χ

(1) or

(

) = 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

−

χ6=1

χ(1)

χ(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

theorem)

Let

p, q

be primes, and let

|G|

where a, b ∈ Z

≥0

, with a + b ≥ 2. Then G is not simple.

Proof.

Let

|G|

. If

= 0 or

= 0, then the result is trivial. Suppose

a, b >

0. We let

Q ∈ Syl

(

). Since

is a

-group, we know

(

) is non-trivial.

Hence there is some 1

g ∈ Z

(

). By definition of center, we know

Q ≤ C

(

Also, C

(g) is not the whole of G, since the center of G is trivial. So

(g)| = |G : C

(g)| = p

for some 0 < r ≤ a. So done.

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

15 Representations of compact groups

We now consider the case of infinite groups. It turns out infinite groups don’t

behave well if we don’t impose additional structures. When talking about

representations of infinite groups, things work well only if we impose a topological

structure on the group, and require the representations to be continuous. Then

we get nice results if we only concentrate on the groups that are compact. This

is a generalization of our previous results, since we can give any finite group the

discrete topology, and then all representations are continuous, and the group is

compact due to finiteness.

We start with the definition of a topological group.

Definition

(Topological group)

A topological group is a group

which is also

a topological space, and for which multiplication

G × G → G

((

h, g

)

7→ hg

) and

inverse G → G (g 7→ g

−1

) are continuous maps.

Example.

Any finite group

with the discrete topology is a topological group.

Example.

The matrix groups

(

) and

(

) are topological groups

(inheriting the topology of R

and C

However, topological groups are often too huge. We would like to concentrate

on “small” topological groups.

Definition

(Compact group)

A topological group is a compact group if it is

compact as a topological space.

There are some fairly nice examples of these.

Example. All finite groups with discrete topology are compact.

Example.

The group

{z ∈ C

|z|

= 1

}

under multiplication is a compact

group. This is known as the circle group, for mysterious reasons. Thus the torus

× S

× ··· × S

is also a compact group.

Example.

The orthogonal group O(

)

≤ GL

(

) is compact, and so is

(

) =

{A ∈

) :

det A

= 1

}

. These are compact since they are closed and bounded

as a subspace of

, as the entries can be at most 1 in magnitude. Note that

SO(2)

∼

, mapping a rotation by θ to e

iθ

Example.

The unitary group U(

) =

{A ∈ GL

(

) :

†

is compact,

and so is SU(n) = {A ∈ U (n) : det A = 1}.

Note that we also have

(1)

∼

(2)

∼

. These are not only isomorphisms

as 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) = {(z

, z

) ∈ C

: |z

+ |z

= 1} ≤ R

= C

We can see this is homeomorphic to S

It turns out that the only spheres on which we can define group operations

on are

and

, 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.

Definition

(Representation of topological group)

A representation of a topolog-

ical group on a finite-dimensional space

is a continuous group homomorphism

G → GL(V ).

It is important that the group homomorphism is continuous. Note that for a

general topological space

, a map

X → GL

(

)

∼

(

) is continuous if

and only if the component maps x 7→ ρ(x)

are continuous for all i, j.

The key that makes this work is the compactness of the group and the

continuity of the representations. If we throw them away, then things will go

very bad. To see this, we consider the simple example

= U(1) = {g ∈ C

: |g| = 1}

∼

R/Z,

where the last isomorphism is an abelian group isomorphism via the map

x 7→ e

2πix

: R/Z → S

What happens if we just view

as an abstract group, and not a topological

group? We can view

as a vector space over

, and this has a basis (by Hamel’s

basis theorem), say,

A ⊆ R

. Moreover, we can assume the basis is ordered, and

we can assume 1 ∈ A.

As abelian groups, we then have

∼

Q ⊕

α∈A\{1}

Qα,

Then this induces an isomorphism

R/Z

∼

Q/Z ⊕

α∈A\{1}

Qα.

Thus, as abstract groups,

has uncountably many irreducible representations,

namely for each

λ ∈ A \{

}

, there exists a one-dimensional representation given

2πiµ

) =

(

1 µ 6∈ Qλ

2πiµ

µ ∈ Qλ

We see

if and only if

Qλ

. Hence there are indeed uncountably

many of these. There are in fact even more irreducible representation we haven’t

listed. This is bad.

The idea is then to topologize

as a subset of

, and study how it acts

naturally on complex spaces in a continuous way. Then we can ensure that we

have at most countably many representations. In fact, we can characterize all

irreducible representations in a rather nice way:

Theorem.

Every one-dimensional (continuous) representation

is of the form

ρ : z 7→ z

for some n ∈ Z.

This is a nice countable family of representations. To prove this, we need

two lemmas from real analysis.

Lemma.

: (

→

(

+) is a continuous group homomorphism, then

there exists a c ∈ R such that

ψ(x) = cx

for all x ∈ R.

Proof.

Given

: (

→

(

+) continuous, we let

(1). We now claim

that 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 =

∈ Q. Then we have

mψ(x) = ψ(nx) = ψ(m) = cm.

So we must have





= c

So we get

(

) =

for all

q ∈ Q

. But

is dense in

, and

is continuous. So

we must have ψ(x) = cx for all x ∈ R.

Lemma. Continuous homomorphisms ϕ : (R, +) → S

are of the form

ϕ(x) = e

icx

for some c ∈ R.

Proof.

Let

: (

→ S

be defined by

x 7→ e

. This homomorphism wraps

the real line around S

with period 2π.

We now claim that given any continuous function

R → S

such that

(0) = 1, there exists a unique continuous lifting homomorphism

R → R

such that

ε ◦ ψ = ϕ, ψ(0) = 0.

(R, +)

∃!ψ

The lifting is constructed by starting with

(0) = 0, and then extending a small

interval at a time to get a continuous map

R → R

. We will not go into the

details. Alternatively, this follows from the lifting criterion from IID Algebraic

Topology.

We now claim that if in addition

is a homomorphism, then so is its

continuous lifting

. If this is true, then we can conclude that

(

) =

for

some c ∈ R. Hence

ϕ(x) = e

icx

To show that

is indeed a homomorphism, we have to show that

(

) =

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

and

. But

can only be

an integer. So it must be constant. Now we pick our favorite

and

, namely

x = 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 S

Theorem.

Every one-dimensional (continuous) representation

is of the form

ρ : z 7→ z

for some n ∈ Z.

Proof.

Let

→ C

be a continuous representation. We now claim that

actually maps

. Since

is compact, we know

(

) has closed and

bounded image. Also,

ρ(z

) = (ρ(z))

for all

n ∈ Z

. Thus for each

z ∈ S

, if

|ρ

(

)

| >

1, then the image of

(

) is

unbounded. Similarly, if it is less than 1, then

(

−n

) is unbounded. So we must

have ρ(S

) ⊆ S

. So we get a continuous homomorphism

R → S

x 7→ ρ(e

So we know there is some c ∈ R such that

ρ(e

) = e

icx

Now in particular,

1 = ρ(e

2πi

) = e

2πic

This forces c ∈ Z. Putting n = c, we get

ρ(z) = z

That has taken nearly an hour, and we’ve used two not-so-trivial facts from

analysis. So we actually had to work quite hard to get this. But the result is

good. We have a complete list of representations, and we don’t have to fiddle

with things like characters.

Our next objective is to repeat this for

(2), and this time we cannot just

get 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 via

the operation

|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 called

integration. Informally, we want to be able to write something like

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 this

result, since we can construct it by hand for S

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

: C(G) → C, written as

f =

f(g) dg

is called a Haar measure if

(i) It satisfies the normalization condition

1 dg = 1

so that the “total volume” is 1.

(ii) It is left and right translation invariant, i.e.

f(xg) dg =

f(g) dg =

f(gx) dg

for all x ∈ G.

Example. For a finite group G, we can define

f(g) dg =

|G|

g∈G

f(g).

The division by the group order is the thing we need to make the normalization

hold.

Example. Let G = S

. Then we can have

f(g) dg =

2π

f(e

iθ

) dθ.

Again, division by 2π is the normalization required.

We have explicitly constructed them here, but there is a theorem that

guarantees the existence of such a measure for sufficiently nice topological group.

Theorem.

Let

be a compact Hausdorff topological group. Then there exists

a unique Haar measure on G.

From now on, the word “compact” will mean “compact and Hausdorff”, so

that the conditions of the above theorem holds.

We will not prove theorem, and just assume it to be true. If you don’t like

this, whenever you see “compact group”, replace it with “compact group with

Haar measure”. We will explicitly construct it for

(2) later, which is all we

really care about.

Once we know the existence of such a thing, most of the things we know for

finite groups hold for compact groups, as long as we replace the sums with the

appropriate integrals.

Corollary

(Weyl’s unitary trick)

Let

be a compact group. Then every

representation (ρ, V ) has a G-invariant Hermitian inner product.

Proof.

As for the finite case, take any inner product (

·, ·

) on

, then define a

new inner product by

hv, wi =

(ρ(g)v, ρ(g)w) dg.

Then this is a G-invariant inner product.

Theorem

(Maschke’s theoerm)

Let

be compact group. Then every repre-

sentation of G is completely reducible.

Proof.

Given a representation (

ρ, V

). Choose a

-invariant inner product. If

is not irreducible, let

W ≤ V

be a subrepresentation. Then

⊥

is also

G-invariant, and

V = W ⊕W

⊥

Then the result follows by induction.

We can use the Haar measure to endow

(

) with an inner product given by

hf, f

i =

f(g)f

(g) dg.

Definition

(Character)

G → GL

(

) is a representation, then the char-

acter

tr ρ

is a continuous class function, since each component

(

)

continuous.

So characters make sense, and we can ask if all the things about finite groups

hold here. For example, we can ask about orthogonality.

Schur’s lemma also hold, with the same proof, since we didn’t really need

finiteness there. Using that, we can prove the following:

Theorem

(Orthogonality)

Let

be a compact group, and

and

irreducible representations of G. Then

hχ

, χ

i =

(

1 V

∼

0 V 6

∼

Now do irreducible characters form a basis for C(G)?

Example.

We take the only (infinite) compact group we know about —

We have found that the one-dimensional representations are

: z 7→ z

for

n ∈ Z

. As

is abelian, these are all the (continuous) irreducible representa-

tions — given any representation

, we can find a simultaneous eigenvector for

each ρ(g).

The “character table of

” has rows

indexed by

, with

(

iθ

) =

inθ

Now given any representation of

, say

, we can break

as a direct sum

of 1-dimensional subrepresentations. So the character χ

of V is of the form

(z) =

n∈Z

where

∈ Z

≥0

, and only finitely many

are non-zero (since we are assuming

finite-dimensionality).

Actually,

is the number of copies of

in the decomposition of

. We

can find out the value of a

by computing

= hχ

, χ

i =

2π

−inθ

iθ

) dθ.

Hence we know

iθ

) =

n∈Z



2π

iθ

−inθ

dθ



inθ

This is really just a Fourier decomposition of

. This gives a decomposition

into irreducible characters, and the “Fourier mode”

is the number of

each irreducible character occurring in this decomposition.

It is possible to show that

form a complete orthonormal set in the Hilbert

space

(

), i.e. square-integrable functions on

. This is the Peter-Weyl

theorem, 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 ∈ GL

†

A = I, det A = I}.

Note that if

A =



a b

c d



∈ SU(2),

then since det A = 1, we have

−1



d −b

−c a



So we need d = ¯a and c = −

b. Moreover, we have

a¯a + b

b = 1.

Hence we can write SU(2) explicitly as

G =



a b

−

b ¯a



: a, b ∈ C, |a|

+ |b|

= 1



Topologically, we know G

∼

⊆ C

∼

Instead of thinking about

in the usual vector space way, we can think of

it as a subgroup of M

H =



z w

−¯w ¯z



: w, z ∈ C



≤ M

(C).

This is known as Hamilton’s quaternion algebra. Then

is a 4-dimensional

Euclidean space (two components from z and two components from w), with a

norm on H given by

kAk

= det A.

We now see that

(2)

≤ H

is exactly the unit sphere

kAk

= 1 in

. If

A ∈ SU

(2) and

x ∈ H

, then

kAxk

kxk

since

kAk

= 1. So elements of

acts

as isometries on the space.

After normalization (by

2π

), the usual integration of functions on

defines

a Haar measure on

. It is an exercise on the last example sheet to write this

out explicitly.

We now look at conjugacy in G. We let

T =



a 0

0 ¯a



: a ∈ C, |a|

= 1



∼

This is a maximal torus in

, and it plays a fundamental role, since we happen

to know about S

. 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) s

= −I ∈ Z(SU(2)).

(iii) The normalizer

(T ) = T ∪sT =



a 0

0 ¯a





0 a

−¯a 0



: a ∈ C, |a| = 1



(iv)

Every conjugacy class

(2) contains an element of

, 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}.

(vi) There is a bijection

{conjugacy classes in SU(2)} ↔ [−1, 1],

given by

A 7→

tr A.

We can see that if

A =



λ 0



then

tr A =

(λ +

λ) = Re(λ).

Note that what (iv) and (v) really say is that matrices in

(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

has an

orthonormal basis of eigenvectors, and hence is conjugate in

(2) to one

in T , say

QXQ

†

∈ T.

We now want to force Q into SU(2), i.e. make Q have determinant 1.

We put

det Q

. Since

is unitary, i.e.

†

, we know

|δ|

= 1. So

we let ε be a square root of δ, and define

= ε

−1

Then we have

†

∈ T.

(v)

We let

g ∈ G

, and suppose

g ∈ C

. If

±I

, then

C ∩T

{g}

. Otherwise,

has two distinct eigenvalues

λ, λ

−1

. Note that the two eigenvlaues must

be inverses of each other, since it is in SU(2). Then we know

C =





λ 0

0 λ

−1



−1

: h ∈ G



Thus we find

C ∩ T =



λ 0

0 λ

−1





−1

0 λ



This is true since eigenvalues are preserved by conjugation, so if any



µ 0

0 µ

−1



then

{µ, µ

−1

}

{λ, λ

−1

}

. Also, we can get the second matrix from the

first by conjugating with s.

(vi) Consider the map

tr : {conjugacy classes} → [−1, 1].

By (v), matrices are conjugate in

iff they have the same set of eigenvalues.

Now



λ 0

0 λ

−1



(λ +

λ) = Re(λ) = cos θ,

where λ = e

iθ

. Hence the map is a surjection onto [−1, 1].

Now we have to show it is injective. This is also easy. If

and

have the

same image, i.e.

tr g =

tr g

then g and g

have the same characteristic polynomial, namely

− (tr g)x + 1.

Hence they have the same eigenvalues, and hence they are similar.

We now write, for t ∈ (−1, 1),



g ∈ G :

tr g = t



In particular, we have

= {I}, C

−1

= {−I}.

Proposition. For t ∈ (−1, 1), the class C

∼

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

be the complex space of all homogeneous polynomials of degree

in variables x, y, i.e.

= {r

+ r

n−1

y + ··· + r

: r

, ··· , r

∈ C}.

This is an

+ 1-dimensional complex vector space with basis

, x

n−1

y, ··· , y

We want to get an action of

(2) on

. It is easier to think about the

action of

(

) in general, and then restrict to

(2). We define the action of

: GL

)

given by the rule



a b

c d



f(x, y) = f(ax + cy, bx + dy).

In other words, we have



a b

c d





x y



= f





x y





a b

c d



where the multiplication in f is matrix multiplication.

Example. When n = 0, then V

∼

C, and ρ

is the trivial representation.

When n = 1, then this is the natural two-dimensional representation, and



a b

c d



has matrix



a b

c d



with respect to the standard basis {x, y} of V

= C

More interestingly, when n = 2, we know



a b

c d



has matrix





ab b

2ac ad + bc 2bd

cd d





with respect to the basis

, xy, y

. We obtain the matrix by

computing, e.g.

(g)(x

) = (ax + cy)

= a

+ 2acxy + c

Now we know

(2)

≤ GL

(

). So we can view

as a representation of

SU(2) by restriction.

Now we’ve got some representations. The claim is that these are all the

irreducibles. Before we prove that, we look at the character of these things.

Lemma.

A continuous class function

G → C

is determined by its restriction

to T , and F |

is even, i.e.



λ 0

0 λ

−1



= f



−1

0 λ



Proof.

Each conjugacy class in

(2) meets

. So a class function is determined

by its restriction to

. Evenness follows from the fact that the two elements are

conjugate.

In particular, a character of a representation (ρ, V ) is also an even function

: S

→ C.

Lemma.

is a character of a representation of

(2), then its restriction

χ|

is a Laurent polynomial, i.e. a finite N-linear combination of functions



λ 0

0 λ

−1



7→ λ

for n ∈ Z.

Proof.

is a representation of

(2), then

Res

SU(2)

is a representation

, and its character

Res

SU(2)

is the restriction of

. But every

representation of T has its character of the given form. So done.

Notation. We write N[z, z

−1

] for the set of all Laurent polynomials, i.e.

N[z, z

−1

] =

: a

∈ N : only finitely many a

non-zero

We further write

N[z, z

−1

]

= {f ∈ N[z, z

−1

] : f(z) = f(z

−1

)}.

Then by our lemma, for every continuous representation

(2), the

character χ

∈ N[z, z

−1

]

(by identifying it with its restriction to T ).

We now actually calculate the character

of (

, V

) as a representation of

SU(2). We have

(g) = tr ρ

(g),

where

g ∼



z 0

0 z

−1



∈ T.

Then we have



z 0

0 z

−1



) = (zx

)(z

−1

) = z

i−j

So each

is an eigenvector of

(

), with eigenvalue

i−j

. So we know

(

)

has a matrix







n−2

2−n

−n







with respect to the standard basis. Hence the character is just



z 0

0 z

−1



= z

+ z

n−2

+ ··· + z

−n

n+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

(2)

→ GL

(

) of dimension

+ 1 are

irreducible for n ∈ Z

≥0

Again, we get a complete set (completeness proven later). A complete list of

all irreducible representations, given in a really nice form. This is spectacular.

Proof.

Let 0

W ≤ V

be a

-invariant subspace, i.e. a subrepresentation of

. We will show that W = V

All we know about

is that it is non-zero. So we take some non-zero vector

of W .

Claim. Let

0 6= w =

j=0

n−j

∈ W.

Since this is non-zero, there is some

such that

= 0. The claim is that

n−i

∈ W .

We argue by induction on the number of non-zero coefficients

. If there

is only one non-zero coefficient, then we are already done, as

is a non-zero

scalar multiple of x

n−i

So assume there is more than one, and choose one

such that

= 0. We

pick z ∈ S

with z

, z

n−2

, ··· , z

2−n

, z

−n

all distinct in C. Now



−1



w =

n−2j

n−j

∈ W.

Subtracting a copy of w, we find



−1



w − z

n−2i

w =

n−2j

− z

n−2i

n−j

∈ W.

We now look at the coefficient

n−2j

− z

n−2i

This is non-zero if and only if

is non-zero and

j 6

. So we can use this to

remove any non-zero coefficient. Thus by induction, we get

n−j

∈ W

for all j such that r

6= 0.

This gives us one basis vector inside W , and we need to get the rest.

Claim. W = V

We now know that x

n−i

∈ W for some i. We consider



√



1 −1

1 1



n−i

√

(x + y)

n−i

(−x + y)

∈ W.

It is clear that the coefficient of

is non-zero. So we can use the claim to

deduce x

∈ W .

Finally, for general a, b 6= 0, we apply



a −

b ¯a



= (ax + by)

∈ W,

and the coefficient of everything is non-zero. So basis vectors are in

. So

W = V

This proof is surprisingly elementary. It does not use any technology at all.

Alternatively, to prove this, we can identify

cos θ



A ∈ G :

tr A = cos θ



with the 2-sphere

{|Im a|

+ |b|

= sin

θ}

of radius

sin θ

. So if

is a class function on

is constant on each class

cos θ

It turns out we get

f(g) dg =

2π



iθ

−iθ



4π sin

θ dθ.

This is the Weyl integration formula, which is the Haar measure for

(2). Then

is the character of

, we can use this to show that

hχ

, χ

= 1. Hence

is irreducible. We will not go into the details of this construction.

Theorem.

Every finite-dimensional continuous irreducible representation of

is one of the ρ

: G → GL(V

) as defined above.

Proof.

Assume

G → GL

(

) is an irreducible representation affording a

character

∈ N

[

z, z

−1

]

. We will show that

for some

. Now we see

= 1

= z + z

−1

= z

+ 1 + z

−2

form a basis of

[

z, z

−1

]

, which is a non-finite dimensional vector space over

Q. Hence we can write

a finite sum with finitely many

= 0. Note that it is possible that

∈ Q

. So

we clear denominators, and move the summands with negative coefficients to

the left hand side. So we get

mχ

i∈I

j∈J

with I, J disjoint finite subsets of N, and m, m

, n

∈ N.

We know the left and right-hand side are characters of representations of

So we get

mV ⊕

Since

is irreducible and factorization is unique, we must have

∼

for some

n ∈ J.

We’ve got a complete list of irreducible representations of

(2). So we can

look at what happens when we take products.

We know that for V, W representations of SU(2), if

Res

SU(2)

∼

Res

SU(2)

then in fact

∼

This gives us the following result:

Proposition.

Let

(2) or

, and

V, W

are representations of

Then

V ⊗W

= χ

Proof.

By the previous remark, it is enough to consider the case

. Suppose

V and W have eigenbases e

, ··· , e

and f

, ··· , f

respectively such that

ρ(z)e

= z

, ρ(z)f

= z

for each i, j. Then

ρ(z)(e

⊗ f

) = z

⊗ f

Thus the character is

V ⊗W

(z) =

i,j









= χ

(z)χ

(z).

Example. We have

⊗V

(z) = (z + z

−1

)

= z

+ 2 + z

−2

= χ

+ χ

So we have

⊗ V

∼

⊕ V

Similarly, we can compute

⊗V

(z) = (z

+ 1 + z

−2

)(z + z

−1

) = z

+ 2z + 2z

−1

+ z

−3

= χ

+ χ

So we get

⊗ V

∼

⊕ V

Proposition (Clebsch-Gordon rule). For n, m ∈ N, we have

⊗ V

∼

n+m

⊕ V

n+m−2

⊕ ··· ⊕ V

|n−m|+2

⊕ V

|n−m|

Proof.

We just check this works for characters. Without loss of generality, we

assume n ≥ m. We can compute

(χ

)(z) =

n+1

− z

−n−1

z − z

−1

+ z

m−2

+ ··· + z

−m

)

j=0

n+m+1−2j

− z

2j−n−m−1

z − z

−1

j=0

n+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

and

(2). We see if

we can make some deductions about some related groups. We will not give too

many details about these groups, and the proofs are just sketches. Moreover, we

will rely on the following facts which we shall not prove:

Proposition. There are isomorphisms of topological groups:

(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 show

this, we can use the fact that a continuous bijection from a Hausdorff space to a

compact 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:

: SO(3) → GL(V

for some m ≥ 0, where V

are the irreducible representations of SU(2).

Proof.

Irreducible representations of

(3) correspond to irreducible representa-

tions of

(2) such that

−I

acts trivially by lifting. But

−I

acts on

−

when n is odd, and as 1 when n is even, since

ρ(−I) =







(−1)

n−2

(−1)

−n







= (−1)

For the sake of completeness, we provide a (sketch) proof of the isomorphism

SO(3)

∼

SU(2)/{±I}.

Proposition. SO(3)

∼

SU(2)/{±I}.

Proof sketch.

Recall that

(2) can be viewed as the sphere of unit norm

quaternions H

∼

Let

= {A ∈ H : tr A = 0}.

These are the “pure” quaternions. This is a three-dimensional subspace of

. It

is not hard to see this is

= R



i 0

0 −i





0 1

−1 0





0 i

i 0



= R hi, j, ki,

where Rh···i is the R-span of the things.

This is equipped with the norm

kAk

= det A.

This gives a nice 3-dimensional Euclidean space, and

(2) acts as isometries

on H

by conjugation, i.e.

X · A = XAX

−1

giving a group homomorphism

ϕ : SU(2) → O(3),

and the kernel of this map is

(

(2)) =

{±I}

. We also know that

(2) is

compact, and O(3) is Hausdorff. Hence the continuous group isomorphism

¯ϕ : SU(2)/{±I} → im ϕ

is a homeomorphism. It remains to show that im ϕ = SO(3).

But we know

(2) is connected, and

det

(

)) is a continuous function

that can only take values 1 or

−

1. So

det

(

)) is either always 1 or always

−

But det(ϕ(I)) = 1. So we know det(ϕ(X)) = 1 for all X. Hence im ϕ ≤ SO(3).

To show that equality indeed holds, we have to show that all possible rotations

are possible. We first show all rotations in the

i, j

-plane are implemented

by elements of the form

, and similarly for any permutation of

i, j, k

. Since

all such rotations generate SO(3), we are then done. Now consider



iθ

0 e

−iθ



ai b

−

b −ai



iθ

0 e

iθ





ai e

2iθ

−

−2iθ

−ai





iθ

0 e

−iθ



acts on

Rhi, j, ki

by a rotation in the (

j, k

)-plane through an angle 2

. We can

check that



cos θ sin θ

−sin θ cos θ





cos θ i sin θ

i sin θ cos θ



act by rotation of 2

in the (

i, k

)-plane and (

i, j

)-plane respectively. So done.

We can adapt this proof to prove the other isomorphisms. However, it is

slightly more difficult to get the irreducible representations, since it involves

taking some direct products. We need a result about products

G × H

of two

compact groups. Similar to the finite case, we get the complete list of irreducible

representations by taking the tensor products

V ⊗ W

, where

and

range

over the irreducibles of G and H independently.

We will just assert the results.

Proposition.

The complete list of irreducible representations of

(4) is

×ρ

where m, n > 0 and m ≡ n (mod 2).

Proposition. The complete list of irreducible representations of U(2) is

det

⊗m

⊗ ρ

where

m, n ∈ Z

and

n ≥

0, and

det

is the obvious one-dimensional representation.