II Representation Theory

4Schur's lemma

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.