Representation Theory - homepages.ucl.ac.ukucahato/RepresentationTheory.pdf · Representation Theory Alice Pozzi, Alex Torzewski MATH0073 Contents 1 Introduction 1 2 Subrepresentations

Representation Theory

Alice Pozzi, Alex Torzewski

MATH0073

Contents

1 Introduction1 Introduction 1

2 Subrepresentations and irreducible representations2 Subrepresentations and irreducible representations 3

3 Modules3 Modules 6

4 Homs and tensor products4 Homs and tensor products 10

5 Semisimplicity5 Semisimplicity 14

6 Characters6 Characters 19

7 Character tables7 Character tables 22

8 Inner product of characters8 Inner product of characters 248.1 More examples8.1 More examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1 Introduction

A representation of a group consists of a group acting on a vector space. Representations areubiquitous within mathematics. Understanding representations has applications to all other areasof mathematics.

Representations have very different behaviour for infinite groups to finite groups and the theoryis also very sensitive to the field over which the vector spaces are defined. In this course, the focuswill be on representations of finite groups on finite dimensional C-vector spaces. In this setting weshall be able to say a lot about the representations.

NOTATION 1.1 Let k be a field and consider the vector space V = kn. The group of invertible linearmaps kn ∼−→ kn is GLn(k) (this is the automorphism group of the vector space kn). We shall alsointerchangeably write this as GL(V ). Going forward, for a vector space V , we shall write GL(V )when we don’t have a fixed choice of basis of V in mind. A choice of basis e1, ..., en for V is thesame as fixing an isomorphism V ∼= kn and so an isomorphism GL(V ) ∼= GLn(k).

DEFINITION 1.2 A representation of a group G is a group homomorphism ρ : G→ GL(V ) for somevector space V .

1

1 Introduction

So a representation assigns a linear map V → V for every element g ∈ G. This is what wemean by the action of g on V .

If we have chosen a basis, a representation is precisely a group homomorphism ρ : G→ GLn(k)for some n.

EXAMPLE 1.3 i) Let k = R and G = C2. We can let C2 = 〈σ〉 act on the plane R2 = {(x, y)} byletting σ act as reflection in the line y = x (a linear map). This interchanges (1, 0) and (0, 1)so in this basis

ρ(σ) =

(0 11 0

).

As a group homomorphism, we must have ρ(e) = id. To check that this is a representation wemust check that ρ(g)ρ(h) = ρ(gh) for all g, h ∈ C2. This is easy to check. For example,

ρ(σ)ρ(σ) =

(0 11 0

)(0 11 0

)=

(1 00 1

)= ρ(e)

= ρ(σ2)

ii) Let k = C and G = Cn = 〈σ〉. Let V = C and ζn be a primitive nth root of unity. There is arepresentation χ : Cn → GL1(C) = C× given by sending σk 7→ ζkn.

REMARK 1.4 Given that representations are group homomorphisms it suffices to define ρ only ongenerators ofG. Moreover, if we specify ρ only on generators, to check ρ is a group homomorphismit suffices to only check the relations between generators.

EXAMPLE 1.5 Let G = S3 = D6 = 〈σ, τ |σ3, τ2, στ = τσ−1〉. The group S3 is the symmetry group ofan equilateral triangle (σ acts by rotation, τ be reflection). We can use this to define a representa-tion of S3 on R2. This is easiest if we chose our basis so that e1 and e2 are corners of the triangle.The final corner is then −e1 − e2. We then have

ρ(σ) =

(0 −11 −1

)ρ(τ) =

(1 −10 −1

)By the above remark this completely determines ρ. We must still check the relations in order for ρto be a group homomorphism. Namely that

ρ(σ)3 = id

ρ(τ)2 = id

ρ(σ)ρ(τ) = ρ(τ)ρ(σ)−1.

As an example, in the last case(0 −11 −1

)(1 −10 −1

)=

(0 11 0

)whilst (

1 −10 −1

)(0 −11 −1

)2

=

(0 11 0

)as desired.

2

2 Subrepresentations and irreducible representations

DEFINITION 1.6 Given a representation ρ : G → GL(V ) with n = dimk V , we call n the dimensionof V .

DEFINITION 1.7 For any group G and field k, we call the one-dimensional representation ρ : G →GL1(k) given by ρ(g) 7→ 1 the trivial representation. We usually denote it by 1.


Recall that in Example 1.31.3 we defined the representation of C2 = 〈σ〉 given by σ acting as areflection in the line x = y. Specifically,

ρ(σ) =

(0 11 0

).

Lets still consider the σ acting as a reflection in the line x = y but instead the basis (1, 1), (1,−1).Then (1, 1) 7→ (1, 1) whilst (1,−1) 7→ −(1,−1). So with respect to this basis we obtain a represen-tation ρ′ with

ρ′(σ) =

(1 00 −1

).

In studying representations we are interested in actions of groups on vector spaces. For all intentsand purposes, we should consider ρ and ρ′ to be given isomorphic representations (they have thesame action on V ). This motivates the following definition.

DEFINITION 2.1 An isomorphism between two representations ρ, ρ′ : G → GLn(k) is given by aninvertible matrix P such that

ρ′(g) = P−1ρ(g)P ∀g ∈ G.

When such a P exists, we say that ρ and ρ′ are isomorphic.

REMARK 2.2 One should think of P as being a change of basis matrix. For ρ and ρ′ to be isomorphicwe are saying that there is some change of basis under which ρ(g) is the same as ρ′(g) and thisworks for all g simultaneously.

EXAMPLE 2.3 The change of basis matrix from (1, 1), (1,−1) to (1, 0), (0, 1) is given by:(1 11 −1

).

Accordingly

P−1ρ(σ)P =1

−2

(−1 −1−1 1

)(0 11 0

)(1 11 −1

)=

(1 00 −1

)= ρ(σ)

and we also have P−1ρ(e)P = P−1P = id = ρ′(e).

DEFINITION 2.4 i) Let ρ : G → GL(V ) be a representation. A subrepresentation is a subspaceW ⊂ V that is stable under G, i.e. for which ρ(g) ·W ⊆W for all g ∈ G.

ii) Such representation is called irreducible or simple if its only subrepresentations are 0 and V .

3


LEMMA 2.5 A representation ρ : G→ GL(V ) has a subrepresentation W ⊆ V if and only if, for somechoice of basis {e1, ..., en} of W and {f1, ..., fm} such that {e1, ..., en, f1, ..., fm} is a basis of V , eachelement of G acts as

A(g) ∗

0 C(g)

,

{ei} {fj}

for some matrices A(g), C(g). Moreover, the maps g 7→ A(g), g 7→ C(g) are themselves representa-tions.

Proof. In fact any such choice will work. By assumption ρ(g) · ei ∈ W and so when written in thebasis {e1, ..., en, f1, ..., fm} the coefficients of all fj are zero.

The last statement is just the observation that for block upper-triangular matrices(A B0 C

)(A′ B′

0 C ′

)=

(AA′ AB′ +BC ′

0 CC ′

).

REMARK 2.6 So a subrepresentation W is a representation in its own right. Similarly, we call arepresentation arising as C(g) above a quotient representation. We can quotient a representationby an arbitrary subrepresentation by following this procedure, i.e. choosing an extension of a basisof the subrepresentation.

EXAMPLE 2.7 i) Consider again the representation ρ : C2 → GL2(R) of Example 1.31.3 i) given by

σ 7→ ρ(σ) =

(0 11 0

).

Then as we saw this is isomorphic to the representation given by

σ 7→ ρ(σ) =

(1 00 −1

)In this basis it the representation is of the form required in the lemma and so ρ has a non-trivialsubrepresentation.

ii) Consider the representation ρ : S3 → GL2(R) of Example 1.51.5. We claim that ρ is irreducible.Suppose it had a non-trivial subrepresentation, i.e. a one dimensional subspace W ⊂ R2 forwhich ρ(g) ·W = W for all g ∈ S3, and let w be a generator. In order for ρ(τ) to stablise W ,w must be an eigenvector of ρ(τ). The eigenvectors of(

1 −10 −1

)are (

10

),

(12

).

So the only candidates for W are the subspaces generated by these elements. Since ρ(σ) doesnot preserve them, neither can be a subrepresentation.

Later we shall find much easier tools for checking irreducibility.

4


EXERCISE 2.8 Show, for any representation ρ : G → GL(V ), that the fixed points V G := {v ∈ V |ρ(g)v = v ∀g ∈ G} is a sub-vector space of V . Moreover, show that V G is a subrepresentationisomorphic to some number of copies of 1.

DEFINITION 2.9 Given two representations ρ1 : G → GLn1(k), ρ2 : G → GLn2

(k) we can define anew representation called the direct sum ρ1 ⊕ ρ2 : G→ GLn1+n2

(k). It is given by

(ρ1 ⊕ ρ2)(g) =

(ρ1(g) 0

0 ρ2(g)

)

In other words, a representation is a direct sum if it is built out of two smaller representationsacting on complementary sub-vectorspaces.

EXERCISE 2.10 Show that for any ρ1, ρ2, then ρ1 ⊕ ρ2 is isomorphic to ρ2 ⊕ ρ1.

DEFINITION 2.11 A decomposition of a representation ρ is an isomorphism of ρ with a direct sumρ1 ⊕ ...⊕ ρk.

We will be particularly interested in decomposing representations into irreducibles (but is thisalways possible? is it unique?).

EXAMPLE 2.12 Consider again the representation ρ : C2 → GL2(R) of Example 1.31.3 i). As we haveseen it is isomorphic to the representation

ρ(e) =

(1 00 1

)ρ′(σ) =

(1 00 −1

).

As such it is a sum of two one dimensional representations ρ1, ρ2, where ρ1(σ) = 1 and ρ2(σ) = −1.So ρ is reducible as a sum of two one dimensional irreducible representations (all one dimensionalrepresentations must be irreducible).

EXAMPLE 2.13 In this example, we classify all complex representations of Cn = 〈σ〉 up to isomor-phism. Let ρ : Cn → GLk(C) be a representation. As ρ(σ)n = 1, ρ(σ) satisfies the polynomialXn − 1 and then the minimal polynomial must divide this, the minimal polynomial must havedistinct roots. A matrix is diagonalisable if and only if its minimal polynomial has no repeatedroots, so ρ(σ) is diagonalisable.

Therefore, there is some choice of basis such that ρ(σ)

P−1ρ(σ)P =

λ1 0 0 . . . 00 λ2 0 . . . 00 0 λ3 . . . 0...

. . ....

0 . . . . . . . . . λk

.

Since P−1ρ(σt)P = P−1ρ(σ)tP = (P−1ρ(σ)P )t and the latter is diagonal, the same choice ofP also diagonalises every other matrix in the image of the representation. So by definition ρ isisomorphic to a direct sum of one dimensional representations.

We claim that there is a unique one dimensional irreducible, up to isomorphism, for everychoice of (not necessarily primitive) nth root of unity. Indeed, for any ζ there is a representationχ : Cn → C× given by setting ρ(σ) = ζ. This determines the entire representation as χ(σk) = χ(σ)k

5

3 Modules

for all k. Conversely, as χ(σ)n = 1 always, χ(σ) is always an nth root of unity. To see that none ofthese are isomorphic, simply note that for any change of basis Q we have that Q−1χ(σ)Q = χ(σ).

In conclusion, any representation ofCn is isomorphic to some direct sum of the n one-dimensionalirreducibles.

LEMMA 2.14 Let ρ : G→ GLn(C) be any complex representation of a finite group. Then

i) for all g, ρ(g) is diagonalisable,

ii) the eigenvalues of ρ(g) are nth -roots of unity, where n = |g|.

Proof. For any g, we can consider the representation of Cn = 〈g〉 ⊆ G defined by forgetting that ρis defined on all of G. This is certainly a representation. Now use the previous example.

Diagonalisability can fail for infinite groups. Note it is certainly not always possible to diago-nalise all ρ(g) simultaneously. If this were true, then ρ would have to have abelian image.

3 Modules

Rings act on modules.

DEFINITION 3.1 Let R be a (not necessarily commutative) ring. A (left) R-module is an abeliangroup M equipped with an action · : R×M →M such that for all r, s ∈ R and m,n ∈M ,

i) r · (m+ n) = r ·m+ r · n,

ii) (r + s) ·m = r ·m+ s ·m,

iii) (rs) ·m = r · (m · s),

iv) 1R ·m = m.

A submodule N ⊆M is just a subgroup of M for which r ·N ⊆ N for all r ∈ R.

EXAMPLE 3.2 i) If R = k is a field, then a k-module is precisely the same as a k-vector space.

ii) For any ring R, R is itself a left R-module acting by left multiplication. When thought of amodule, submodules of R are precisely left ideals of R (exercise).

DEFINITION 3.3 For any group G and field k, we write k[G] for the |G|-dimensional vector space⊕g∈G eg · k. We give additionally this the structure of a (not necessarily commutative) ring by

specifying that the product of basis elements is given by eg · eh = egh and extending linearly (aswe must). This means that∑

g∈Gλg · eg

∑g∈G

µg · eg

=∑g∈G

(∑h∈G

λgh−1µh

)eg.

The multiplicative identity is then e1 where 1 is the identity of the group. Since the multiplicationof basis elements is associative the same is also true of general elements and it is automatic thatwe have the required distributivity.

We shall often drop the eh notation and simply use h to denote the element of the group andthe corresponding basis element.

We shall be only interested in k[G]-modules that are finite dimensional as a k-vector space.From now on we omit the words finite dimensional when speaking of k[G]-modules.

6

3 Modules

LEMMA 3.4 There is an equivalence

{k[G]-modules} ←→ {representations of G over k} ,

by which we mean that to a k[G]-module we can associate a representation and vice versa and theseconstructions are inverse.

Proof. Given a k[G]-module M , we define a representation ρ : G → GL(M) (here we also use M

to denote the underlying vector space) by setting ρ(g) to be the linear map Meg·(−)−→ M . This is

a group homomorphism as ρ(1) is the identity by condition iv) of being a module whilst ρ(g)ρ(h)

represents the linear map Meh·(−)−→ M

eg·(−)−→ M , i.e. the map given by eg · eh · (−), which bycondition iii) of a module is egh · (−) and so is the map ρ(gh).

Conversely, given a representation ρ : G→ GL(V ), we can define a module action by letting egact as ρ(g) and then extending linearly so that∑

g∈Gλg · eg acts as

∑g∈G

λgρ(g).

Checking that this defines a module is left as an exercise.

That the constructions are inverse is almost tautological. For example, start with a k[G]-moduleM . Then the associated representation has ρ(g) given by the multiplication by eg linear map. Thenits associated k[G]-module is defined by letting multiplication by eg coincide with ρ(g), which wasmultiplication by eg in the first place!

EXAMPLE 3.5 Consider the representation ρ : C2 → GL2(R). The corresponding R[C2]-module hasthe action map given by

R[C2]× R⊕2 −→R⊕2

(ae1 + beσ, v) 7−→(a

(1 00 1

)+ b

(0 11 0

))· v

=

(a bb a

)· v,

i.e. the action of multiplication by a general element (ae1 +beσ) ∈ R[C2] is given by the linear mapR⊕2 → R⊕2 with matrix (

a bb a

).

This enables us to freely change between the two concepts. For some tasks it is easier to thinkin terms of representations and others in terms of k[G]-modules.

Note that both the k[G]-module and the associated representation have the same underlyingvector space.

DEFINITION 3.6 Any remarked previously, any ring is a module over itself, acting by left multiplica-tion. The representation we obtain by considering k[G] as a module over itself is called the regularrepresentation. It will be very important going forward and understanding it will help understandall other representations.

EXAMPLE 3.7 The regular representation of C3 = {1, σ, σ2} over C is given by ρ(e) = id,

7

3 Modules

0 0 1

1 0 0

0 1 0

,

e1 eσ eσ2

ρ(σ) =

0 1 0

0 0 1

1 0 0

.

e1 eσ eσ2

ρ(σ2) =

For C2, this is actually the previous example, but we went in reverse, so the associated module wasR[C2] acting on itself by left multiplication.

LEMMA 3.8 Let V be a k[G]-module and W ⊆ V a sub-vector space. Then W is a k[G]-submodule ifand only W is a subrepresentation of the representation defined by V as a k[G]-module.

Proof. Suppose that W is a k[G]-submodule, i.e. W is stable under multiplication by all elementsof k[G]. Then in particular it is stable under multiplication by eg for g ∈ G and so under ρ(g). Butthis is the definition of a a subrepresentation. Conversely, suppose W is stable under the action ofρ(g). Then eg ·w ∈W for all w ∈W so

(∑g∈G λgeg

)·w ∈W as well as it is a linear combination

of elements of W .

A module with no non-zero proper submodules is called simple. From the above, we see that ak[G]-module is simple if and only if its associated representation is irreducible.

Similarly, all notions of representations should have an incarnation in terms of modules andvice versa. One important concept for modules are homomorphisms.

DEFINITION 3.9 Let M,N be R-modules. An R-module homomorphism is a group homomorphismf : M −→ N for which f(rm) = rf(m) for all r ∈ R and m ∈M .

EXERCISE 3.10 If f, f ′ : M → N are two R-module homomorphisms, then f + f ′ : m 7→ f(m) +f ′(m) is also an R-module homomorphism.

NOTATION 3.11 Let HomR(M,N) denote the set of all R-module homomorphisms from M → N .By the last exercise this is an abelian group. When R = k[G] we shall write HomG(M,N) :=Homk[G](M,N).

EXERCISE 3.12 For k[G]-modules, we can also scale homomorphisms f : M → N by setting (λf)(m) :=λf(m). This gives HomG(M,N) the structure of a k-vector space.

We can use the notion of a homomorphism of k[G]-modules and Lemma 3.43.4 to define the corre-sponding notion of a homomorphism of representations. Given representations ρ : G→ GL(V ) andρ′ : G→ GL(W ), it is simply a linear map f : V →W for which ρ′(g)f(v) = f(ρ(g)v) for all v ∈ V ,g ∈ G. This property is sometimes called G-equivariance In other words, the homomorphisms ofrepresentations is precisely HomG(V,W ) where V,W are the corresponding k[G]-modules.

EXAMPLE 3.13 We already have a notion of an isomorphism of representations. We must check thetwo notions coincide. Suppose we have V = Cn and W = Cn and with representations ρ, ρ′. Theabove definition says that a homomorphism is a linear map f : Cn → Cn satisfying the formula

ρ′(g)f(v) = f(ρ(g)v) ∀g ∈ G.

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3 Modules

Suppose that the matrix representing f is T . Then the formula for all g ∈ G is equivalent to thestatement that

ρ′(g)T = Tρ(g).

Now suppose f is an isomorphism, i.e. has an inverse homomorphism. Then T must be invertible,so we have ρ(g) = T−1ρ′(g)T just as in Definition 2.12.1.

Conversely, an invertible T for which ρ(g) = T−1ρ′(g)T defines an invertible linear mapf : Cn ∼−→ Cn which is a homomorphism of representations. Note that if f is a linear map whichis invertible and f is a homomorphism of representations then so is f−1!

The direct sum ofR-modulesM,N is defined by taking the direct sum of the underlying abeliangroupsM⊕N and lettingR-act on both coordinates “diagonally”. Again, under the correspondenceof Lemma 3.43.4, this coincides with the definition for representations. One can build up a dictionaryto go between the two sides. For any given definition it is fairly easy to see how to translate to theother language, so we will leave it to the reader to make sure they are comfortable using such adictionary.

representations modulessubrepresentation submodule

quotient representation quotient moduleirreducible simple

trivial representation trivial module...

...

EXERCISE 3.14 Given a homomorphism of R-modules f : M → N ,

i) ker f is a submodule of M ,

ii) im f is a submodule of N .

LEMMA 3.15 Any irreducible representation is a quotient of the regular representation.

Proof. It suffices to instead show any simple k[G]-module M is a quotient of k[G] as a module. Letm ∈M be non-zero and consider the map

k[G](−)·m−→ M

x 7−→ x ·m.

This is a k[G]-module homomorphism (for example, (rx) · m = r · (x · m)). So the image is asubmodule. This is called the submodule generated by m. It is the smallest submodule containingm (exercise).

Since M is simple, and k[G] ·m is a non-zero submodule, it must be all of M . In other words,the map is surjective and M is a quotient of k[G].

LEMMA 3.16 (Schur’s Lemma) If M,N are simple k[G]-modules, then every homomorphism M →N is either zero or an isomorphism. Moreover, if k is an algebraically closed field, then

HomG(M,N) ∼=

{k M ∼= N

0 M 6∼= N.

Here, if M = N , the k denotes the scalar maps m 7→ λm for λ ∈ k.

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4 Homs and tensor products

Proof. Suppose f : M → N is a homomorphism. If f isn’t an isomorphism, it is either not surjectiveor not injective. If it is not surjective, then the image is a proper submodule of N and so is zeroand f = 0. If f is not injective, then the kernel of f is a non-zero submodule of M and so all of Mand again f = 0.

This deals with all but the isomorphic case over an algebraically closed field. For this, firstsuppose M = N . Since any non-zero f : M →M is an isomorphism, its matrix T must have someeigenvalue λ which is non-zero. Since k is algebraically closed, this eigenvalue is defined over kand since scalars are certainly endomorphisms T −λ · idM is another element of HomG(M,M). Bydefinition this is not of full rank, its kernel consists of eigenvectors with eigenvalue λ, so is not anisomorphism. Therefore it must be zero and T = λ · idM is a scalar.

Lastly, we can reduce from the case of M ∼= N to M = N by fixing a single isomorphismϕ : N

∼→M and noting that this defines a (non-canonical) isomorphism

HomG(M,N)∼−→ HomG(M,M)

f 7−→ ϕ ◦ f.

REMARK 3.17 For any module M , EndG(M) := HomG(M,M) has the structure of a ring given byf · g = f ◦ g. If M = 1

⊕n is some number of copies of the trivial representation, then EndG(M) =Mn(k) as a ring. In fact, if k is an algebraically closed field this also holds with 1 replaced by anysimple module. (Exercise: try and prove this using the above lemma.)

WARNING 3.18 If k is not algebraically closed, it is not the case that EndG(M) = k for simplek[G]-modules. It must always contain k (scalars), but may be much larger. What is true is that it isalways a “division algebra”. This will be revisited later in the course .


Before moving on we give three ways of creating new modules (and so representations) fromold ones.

CONSTRUCTION 4.1 Let V1, V2 be k[G]-modules with associated representations ρ1, ρ2. Let Homk(V1, V2)denote the space of all k-linear maps ψ : V1 → V2. This is a vector space of dimension (dimV1) ×(dimV2). We let G act on this by

(v 7→ ψ(v)) 7→ (v 7→ ρ2(g)ψ(ρ1(g−1)v)).

This is a composite of three linear maps and so is itself a linear map and element of Homk(V1, V2).Moreover, the action ofG is linear with respect to addition and scaling of elements of Homk(V1, V2).Finally, for g, h ∈ G

g · (h · ψ) = g · (v 7→ ρ2(h)ψ(ρ1(h−1)v))

= (v 7→ ρ2(g)ρ2(h)ψ(ρ1(h−1)ρ1(g−1)v))

= (v 7→ ρ2(gh)ψ(ρ1(h−1g−1)v))

= (v 7→ ρ2(gh)ψ(ρ1((gh)−1)v))

= (gh) · ψ.

So this defines a representation of G on Homk(V1, V2).

In the special case when V2 = 1 is the trivial representation, we call Homk(V,1) the dualrepresentation of V and denote it by V ∨.

10


WARNING 4.2 The two spaces HomG(V1, V2) and Homk(V1, V2) are not the same. The first con-sists of linear maps that are additionally homomorphisms of k[G]-modules and does not have any(interesting) G-action. The latter consists of just linear maps but is itself a representation.

LEMMA 4.3 HomG(ρ1, ρ2) = Homk(ρ1, ρ2)G

Proof. Write Vi for the underlying module of ρi. Let ψ : V1 → V2 be a homomorphism of modules,i.e. an element of the left hand side. Then ψ(gv) = gψ(v) for all g, v. So gψ(g−1v) = ψ(v) andψ ∈ HomC(ρ1, ρ2)G. Conversely, if we have a linear map ψ lying in the fixed points of HomC(ρ1, ρ2),then the fact that gψ(g−1v) = ψ(v) ensures that ψ is a k[G]-module homomorphism.

EXERCISE 4.4 Show that Homk(ρ1, ρ2⊕ ρ3) = Homk(ρ1, ρ2)⊕Homk(ρ1, ρ3). Deduce that the sameis true for HomG(−,−).

CONSTRUCTION 4.5 Let V1, V2 be k[G]-modules. Let {e1, ..., en} be a basis of V1 and {f1, ..., fm} bea basis of V2. Let V1 ⊗ V2 denote the (dimk V1) × (dimk V2)-dimensional vector space whose basisis the symbols ei ⊗ fj for 1 ≤ i ≤ n, 1 ≤ j ≤ m. We call V1 ⊗ V2 the tensor product of V1 and V2.

We also let ourselves write elements of the form v1 ⊗ v2 for arbitrary v1 =∑i λiei ∈ V1 and

v2 =∑j µjfj ∈ V2 by using the formula

v1 ⊗ v2 = (∑i

λiei)⊗ (∑j

µjfj) :=∑(i,j)

(λiµj · ei ⊗ fj).

This is just notation. The element v1 ⊗ v2 is a label for the vector on the right hand side.

Important warning:

We can give V1 ⊗ V2 an action of G by acting “diagonally”. Namely, if g maps er to∑i λiei and

fs to∑j µjfj , we stipulate that

ei ⊗ fj 7→ (g · ei)⊗ (g · fj) = (∑i

λiei)⊗ (∑j

µjfj) :=∑(i,j)

(λiµj · ei ⊗ fj) (1)

and extend linearly to all elements of V1 ⊗ V2.

WARNING 4.6 Elements of V1⊗V2 are not the same as the pairs (v1, v2) making up V1⊕V2. Indeed,dimk(V1⊗ V2) = (dimk V1)× (dimk V2) whereas dimk(V1⊕ V2) = (dimk V1) + (dimk V2). Explicitly,

(v1 + v′1, v2 + v′2) = (v1, v2) + (v′1, v′2)

whereas

(v1 + v′1)⊗ (v2 + v′2) := v1 ⊗ v2 + v1 ⊗ v′2 + v′1 ⊗ v2 + v′1 ⊗ v′26= v1 ⊗ v2 + v′1 ⊗ v′2.

Moreover, not all elements of V1⊗V2 are “simple tensors”, i.e. of the form v1⊗v2, though necessarilythey are all linear combinations of such elements. This is in contrast to the case of V1 ⊗ V2.

EXAMPLE 4.7 Let ρ be the representation C2 → GL1(C) = C× given by σ 7→ −1. We now describeρ ⊗ ρ. Fix a basis vector v0 of C. Then a basis for C ⊗ C (which is one dimensional) is given byv0 ⊗ v0. The action of σ is then

v0 ⊗ v0 7→ (−v0)⊗ (−v0) = ((−1)× (−1)) · (v0 ⊗ v0) = v0 ⊗ v0.

So σ acts as the identity and ρ⊗ ρ is the trivial representation. More generally, any tensor productof one dimensional representations is just multiplication of matrices.

11


EXAMPLE 4.8 Let ρ : S3 → GL2(R) be the two-dimensional representation that with respect to abasis {e1, e2} is

ρ(σ) =

(0 −11 −1

)ρ(τ) =

(1 −10 −1

),

as given in Example 1.51.5. We now calculate ρ⊗ρ. A basis of V1⊗V2 is given by {e1⊗e1, e1⊗e2, e2⊗e1, e2 ⊗ e2}. Since ρ(σ) acts as e1 7→ e2, e1 7→ −e1 − e2 applying (11) we have that under σ:

e1 ⊗ e1 7−→ e2 ⊗ e2e1 ⊗ e2 7−→ e2 ⊗ (−e1 − e2) = −e2 ⊗ e1 − e2 ⊗ e2e2 ⊗ e1 7−→ (−e1 − e2)⊗ e2 = −e1 ⊗ e2 − e2 ⊗ e1e2 ⊗ e2 7−→ (−e1 − e2)⊗ (−e1 − e2) = e1 ⊗ e1 + e1 ⊗ e2 + e2 ⊗ e1 + e2 ⊗ e2.

So the matrix (ρ⊗ ρ)(σ) is given by

0 0 0 1

0 0 −1 1

0 −1 0 1

1 −1 −1 1

.

e1 ⊗ e1 e1 ⊗ e2 e2 ⊗ e1 e2 ⊗ e2

(ρ⊗ ρ)(σ) =

.

Note that this is a block matrix with 2x2 blocks for which the (i, j)th block is ρ(σ) scaled by the(i, j)-entry of ρ(σ). In fact this is true more generally, the matrices of the tensor product are theKronecker product of the matrices of the constituents (exercise* convince yourself of this). Usingthis, the matrix for τ is given by

1 −1 −1 1

0 −1 0 1

0 0 −1 1

0 0 0 1

.

e1 ⊗ e1 e1 ⊗ e2 e2 ⊗ e1 e2 ⊗ e2

(ρ⊗ ρ)(τ) =

.

EXAMPLE 4.9 Let ρ1 : G → C× be a one-dimensional representation and ρ2 : G → GLn(C) be anyrepresentation. After choosing bases, then the matrix (ρ1 ⊗ ρ2)(g) is the matrix of ρ2(g) scaled byρ1(g).

EXERCISE 4.10 If ρ is an irreducible representation and χ is a one dimensional representation, thenχ⊗ ρ is also an irreducible representation.

LEMMA 4.11 Homk(V1, V2) ∼= V ∨1 ⊗ V2.

12


Proof. We want to show that Homk(V1, V2) ∼= Homk(V1,1) ⊗ V2 as k[G]-modules. First we definean isomorphism as vector spaces. Let f1, ..., fm be a basis of V2. Then elements of the right handside are of the form ∑

j

ψj ⊗ fj

(this is taking a general element of a tensor product but then collecting all terms of the form(−)⊗ fj into a single ψj ∈ V ∨1 ). In other words, they are collections of m maps V1 → 1.

Given such a choice of maps ψj : V1 → k for each 1 ≤ j ≤ m. Then ψ : V1 → V2 defined byψ(v) =

∑j ψj(v)·fj is a well-defined linear map (in other words ψ(v) = (ψ1(v), ψ2(v), ..., ψm(v)) in

the fj-basis). Moreover, by extracting the coefficient of the (fj)th basis element of any ψ : V1 → V2

we can reverse this process to write an arbitrary ψ as a sum of m maps V1 → k. The isomorphismis then given by

Homk(V1, V2) −→ Homk(V1,1)⊗ V2 (2)

ψ 7−→∑j

ψj ⊗ fj .

It remains to show that this is G-equivariant. More precisely, we must show that the image of(v 7→ gψ(g−1v)) under (22) is g ·

(∑j ψj(v)⊗ fj

)=∑j ψj(g

−1v)⊗ g · fj . Firstly,

gψ(g−1v) = g ·∑j

ψj(g−1v) · fj

=∑j

ψj(g−1v) · (g · fj).

Now suppose that g · fj = ρ2(g)⊗ fj =∑k µk,jfk. Then continuing the above

=∑j

(ψj(g

−1v) · (∑k

µk,jfk)

)=∑k

∑j

(ψj(g

−1v) · (µk,jfk))

=∑k

∑j

ψj(g−1v)µk,j

· fk

which by construction maps to

∑k

∑j

ψj(g−1v)µk,j

⊗ fk

(taking everything inside the inner bracket to be our jth coefficient function in (22)). On the otherhand ∑

j

ψj(g−1v)⊗ g · fj =

∑j

(ψj(g

−1v)⊗ (∑k

µk,jfk)

)

=∑k

∑j

ψj(g−1v)µk,j

⊗ fk

13

5 Semisimplicity

as desired.

REMARK 4.12 We did not check that the tensor product is independent of the choice of basis. Ourdefinition is very ad hoc and only makes sense for our narrow purposes. There is a much moregeneral construction which is obviously independent of the choices. In fact, we can actually deduceindependence of the basis from Lemma 4.114.11, but this is a bit of an ugly route.

5 Semisimplicity

For our purposes the following are pathological examples:

EXAMPLE 5.1 • Consider the representation ρ : Z→ GL2(C) given by

1 7→(

1 10 1

).

Since this is of the form specified in Lemma 2.52.5, there is a one-dimensional subrepresentation

generated by(

10

). This is the only one-dimensional representation as if

(ab

)is a basis

element of a one-dimensional subspace, then(1 10 1

)(ab

)=

(a+ bb

)

and this lies in⟨(

ab

)⟩only when b = 0.

In particular, ρ is not irreducible, but there is no complementary subrepresentation for⟨(10

)⟩and there does not exist a decomposition of ρ into irreducibles.

• There following example is very similar. For p prime, there is a representation of Cp over Fp(the field with p elements) given by

ρ : Cp −→ GL2(Fp)

1 7→(

1 10 1

).

This is a representation as

1 + 1 + ...+ 1︸︷︷︸p

7→(

1 p0 1

)≡(

1 00 1

)

(check!). This suffers from the same problem as above, i.e. is not irreducible but does notdecompose.

DEFINITION 5.2 If a representation can be decomposed as a direct sum of irreducible representa-tions it is said to be semisimple.

The representations of Example 5.15.1 is not semisimple. The main result of this section is thatany representation of a finite group over a field of characteristic not dividing |G| is semisimple.

14

5 Semisimplicity

THEOREM 5.3 (Maschke’s Theorem) Let M be any k[G]-module. If the characteristic of k does notdivide |G|, then any submodule V ⊆M has a complementary submodule W ⊆M with M ∼= V ⊕W .

Proof. As a vector spaces we can choose some projection map π : M → V ⊆ M , i.e. a linear mapwith image V and π(v) = v for v ∈ V . This will not in general be a module homomorphism, butwe may modify it by “averaging” over G. Let π′ : M → V ⊆M be defined by

π′(s) =1

|G|∑g∈G

g · π(g−1 · s),

(where we write g · (−) and eg · (−) interchangeably). Then π′ remains a linear map and

π′(h · s) =1

|G|∑g∈G

g · π(g−1 · h · s)

=1

|G|∑g′∈G

hg′ · π(g′−1 · s)

= h · π′(s).

So π′ is a module homomorphism. We claim it is still a projection, i.e. imπ′ = V and π′(v) = vfor v ∈ V . The first statement follows as all terms of the sum lie in V (why?). For the second, asg−1 · v ∈ V we have π(g−1 · v) = g−1 · v so g−1 · v ∈ V and π(g−1 · v) = g−1 · v. So

π′(v) =1

|G|∑g∈G

g · g−1 · v = v.

Now consider (idM −π′) : M → M , this is also a module homomorphism. We claim its kernelis precisely V . As a projection V ⊆ ker(id−π′). On the other hand,

dimkM = rk idM ≤ rk(idM −π′) + rkπ′ = rk(idM −π′) + dimk V,

so the dimension of the kernel is at most dimk V . The image of (idM −π′) is then the desiredcomplementary submodule W for V and the isomorphism M → V ⊕W is given by (π′, idM −π′).

EXERCISE 5.4 Where does the proof use that G is finite and the order is not divisible by the char-acteristic?

Maschke’s Theorem also commonly refers to the following corollaries:

COROLLARY 5.5 If the characteristic of k does not divide |G|, then

i) all representations of G are semisimple

ii) every representation of G has a decomposition⊕k

i=1 ρ⊕nii with the ρi pairwise non-isomorphic

irreducibles and this is unique up to replacing the ρi with isomorphic representations and reorder-ing them.

Proof. For i), repeatedly apply Maschke’s theorem until only simple summands remain.

For ii), suppose ρ ∼=⊕k

i=1 ρ⊕nii and ρ ∼=

⊕sj=1 ρ

′⊕mj

j . Inverting the isomorphism, we obtain aninclusion

ρ′⊕mj

j ↪→k⊕i=1

ρ⊕nii . (3)

15

5 Semisimplicity

By Schur’s lemma 3.163.16, there are no non-zero homomorphisms ρ′j → ρi unless ρ′j ∼= ρi. So (33)forces some ρ′j ∼= ρi and the image of (33) to lie only in the ith factor. By dimension reasons, wemust have ni ≥ nj and the converse must also be true by symmetry. So all factors ρi appear onboth sides with the same exponents. This completes the proof.

REMARK 5.6 So to classify representations of finite groups over fields of characteristic zero it suf-fices to classify the irreducible representations. Moreover, by Lemma 3.153.15 all simple k[G]-modulesare quotients of the regular representation, so all irreducible representations must appear as sum-mands of the regular representation (by Schur’s lemma 3.163.16). In particular, there are finitely manyirreducible representations of a finite group over a field of characteristic not dividing |G|.

REMARK 5.7 Aside: The uniqueness of decompositions in iii) is called the Krull–Schmidt property.In much more pathological settings we can have that

M ⊕N ∼= M ′ ⊕N ′

with M,N,M ′, N ′ not decomposing further but for which M is not isomorphic to either M ′ or N ′.

Semisimple algebras

In Maschke’s theorem we considered k[G] as a module. We can say something much strongerif we attempt to understand how k[G] decomposes as a ring.

The ring k[G] is in fact a k-algebra, i.e. a ring R with an inclusion k ↪→ Z(R) of k into its.

DEFINITION 5.8 A k-algebra R is called semisimple if R is semisimple when considered as a moduleover itself acting by left multiplication.

FACT 5.9 (Artin–Wedderburn Theorem) A finite dimensional k-algebra is semisimple if and only ifit is a product of matrix rings over division rings over k. Moreover, this decomposition is unique up toreordering.

Here a division ring is a ring for which every non-zero element has a multiplicative inverse. Acommutative division ring is the same as a field. It is a fact that the only division ring which is aC-algebra is C itself.

COROLLARY 5.10 If G is any finite group, then there is a decomposition of algebras

C[G] ∼=k∏i=1

Mni(C) (4)

and this is unique up to reordering the summands.

Proof. By Maschke’s theorem, C[G] is semisimple so we may apply Artin–Wedderburn to see thatC[G] is a product of matrix rings over division rings. As C[G] is a C-algebra we have

C ↪→Z

(k∏i=1

Mni(Di)

)

=

k∏i=1

Z(Mni(Di))

=

k∏i=1

Z(Di)

16

5 Semisimplicity

as the centre of a matrix ring consists of scalar matrices. So each of the Di is a C-algebra which isa division ring and so C itself.

EXAMPLE 5.11 • Since C3 is abelian, C[C3] is abelian and the only possible k-algebra decom-position of the form (44) is

C[C3] ∼= C× C× C.

Of course, this works for any abelian group.

• Since S3 is not abelian, C[S3] cannot be abelian and some ni appearing in (44) must be > 1.Counting dimensions we see that

C[S3] ∼= C× C×M2(C).

LEMMA 5.12 Let M be a module over a ring of the form R × S. Then (1R, 0) ·M is a submodule ofM and there is an isomorphism of R× S-modules:

ϕ : M −→ (1R, 0) ·M ⊕ (0, 1S) ·Mm 7−→ ((1R, 0) ·m, (0, 1S) ·m)

Proof. For the first statement, (r, s)·(1R, 0)·m = (1R, 0)·(r, 0)·m ∈ (1R, 0)·M . It is easy to see that ϕis a homomorphism of R×S-modules. To see the map is injective note that 1R×S = (1R, 0)+(0, 1S)so

m = (1R, 1S) ·m = (1R, 0) ·m+ (0, 1S) ·m,

and form to lie in the kernel the RHS is zero and som itself is 0. Conversely, let ((1R, 0)u, (0, 1S)w) ∈(1R, 0) ·M ⊕ (0, 1S) ·M , then

(1R, 0)u+ (0, 1S)w 7−→((1R, 0)(1R, 0)u+ (1R, 0)(0, 1S)w, (0, 1S)(1R, 0)u+ (0, 1S)(0, 1S)w)

= ((1R, 0)u, (0, 1S)w).

So ϕ is surjective.

Note that in the above lemma, S acts trivially on (1R, 0) ·M .

CONSTRUCTION 5.13 Let Vn denote the n-dimensional vector space of column vectors for Mn(C).It has a canonical structure of an Mn(C)-module given by left multiplication (Vn is sometimescalled the standard or tautological module of Mn(C)). This is a simple Mn(C)-module as for anynon-zero v ∈ Vn, there is some matrix taking v to any other w ∈ V . As a module acting on itselfMn(C) decomposes as V ⊕nn (it is n sets of column vectors).

LEMMA 5.14 Isomorphism classes of irreducible representations of C[G] are in bijection with sum-mands of the decomposition C[G] ∼=

∏ki=1 Mni

(C).

Proof. By Lemma 5.125.12, any simple module over a product of rings can only be acted on non-trivially by a single factor. So simple

∏ki=1 Mni(C)-modules are given by a choice of simple Mni(C)-

module for some choice of i. There is an isomorphism

Mni(C) ∼= V ⊕ni

ni.

Simple Mni(C)-modules are quotients of Mni

(C) = V ⊕nini

(the argument of Lemma 3.153.15 worksverbatim!), there is exactly one simple Mni

(C)-module, Vni.

For distinct i, the Vniare all non-isomorphic as Mnj

(C) acts trivially on Vniif and only if i = j.

So the simple k[G] =∏ki=1 Mni

(C)-modules are {Vn1, ..., Vnk

}.

17

5 Semisimplicity

This argument also demonstrates the uniqueness part of Corollary 5.105.10.

This result has a number of easy but important consequences:

COROLLARY 5.15 i) As a representation, the regular representation decomposes as

C[G] ∼=k⊕i=1

V⊕ dimVnini .

ii) If n1, ..., nk denotes the dimensions of all the non-isomorphic irreducible representations of G overC, then

|G| =k∑i=1

n2i . (5)

iii) If G is abelian, then there are |G| non-isomorphic C-representations and all are one dimensional.

We will repeatedly make use of (55).

EXAMPLE 5.16 From Example 5.115.11 we can now classify all irreducible representations of S3 over C.We know to expect two one-dimensional representations and one two-dimensional irreducible. Thelatter was described in Example 1.51.5 and shown to be irreducible in Example 2.72.7. There is alwaysa one dimensional representation given by setting ρ(g) = id for all g. The last representation isgiven by S3 � C2 ↪→ C× where the first map is the quotient by the C3-subgroup and the secondsends the non-trivial element to −1.

THEOREM 5.17 The number of isomorphisms classes of irreducible representations over C of G isequal to the number of conjugacy classes of G.

Proof. We have already seen that the number of irreducible representations is the same as thenumber of factors in the decomposition C[G] ∼=

∏ki=1 Mni

(C). Since the centre of Mni(C) consists

of the scalar matrices, we thus have

|{C-irreducible representations}| = dimC Z(C[G]).

We claim that dimC Z(C[G]) is also equal to the number of conjugacy classes of G. WriteG =

⋃tj=1 Cj with Ct a conjugacy class. For any element h ∈ G

hCjh−1 = Cj . (6)

The containment ⊆ is obvious. The converse follows as both sides have the same size as if hgh−1 =hg′h−1 then g = g′. As a result, the element tj :=

∑g∈Cj

g lies in the centre of C[G]. Indeed,

h

∑g∈Cj

g

=∑g∈Cj

hgh−1h

=∑g′∈Cj

g′h

=

∑g′∈Cj

g′

h

18

6 Characters

where we rewrite the sum using (66).

Clearly, the elements tj are all linearly independent so we have shown that

dimC Z(C[G]) ≥ |{conjugacy classes}|.

Conversely, let x =∑g∈G λg · g ∈ Z(C[G]). We claim that the coefficients λg are constant on

conjugacy classes. Indeed, by assumption hxh−1 = x and hxh−1 is given by∑g∈G

λg · hgh−1 =∑g′∈G

λh−1g′h · g′

using the same rewriting as in (66). This forces

dimC Z(C[G]) ≤ |{conjugacy classes}|.

6 Characters

To decide if representations are isomorphic we should look for isomorphism invariants.

LEMMA 6.1 If A,P ∈Mn(k) thentr(A) = tr(P−1AP ).

Proof. The trace of a matrix is the sum of its eigenvalues, but eigenvalues are preserved underchanges of basis.

DEFINITION 6.2 Let ρ : G → GLn(k) be a representation over k. The function χ : G → k given byg 7→ tr(ρ(g)) is called the character of ρ.

LEMMA 6.3 If ρ1 : G → GL(V ) and ρ2 : G → GL(V ) are isomorphic representations, then the char-acters χ1 and χ2 are equal.

Proof. Direct from Definition 2.12.1 using the previous lemma.

In other words, characters are an isomorphism invariant. This also shows that the character isindependent of the choice of basis.

EXAMPLE 6.4 For any group G, the character χ of the regular representation k[G] is

χ(g) =

{|G| g = e

0 g 6= e.

For g = e, ρ(g) = id|G| and this is clear. Using the basis of k[G] given by eh for h ∈ G, for any g 6= emultiplication by g takes eh to egh 6= eh the matrix ρ(g) has only one 1 in any column and it doesnot lie on the diagonal. In particular, the trace is zero.

EXAMPLE 6.5 For any one dimensional representation ρ, the character is precisely the same as ρ.So in this case the character is a group homomorphism, but this is not true in general.

REMARK 6.6 In general, χ(e) is the dimension of the representation giving rise to χ. As such werefer to value of χ(e) as the dimension or degree of χ.

DEFINITION 6.7 A class function over a field k is a map of setsG→ k which is constant on conjugacyclasses.

19

6 Characters

REMARK 6.8 We already saw a class function in the proof of Theorem 5.175.17. The coefficients of anelement of Z(C[G]) define a class function.

It is possible to add and scale class functions. As such, the space of all possible class functionsitself forms a vector space of dimension equal to the number of conjugacy classes of G.

Whilst the trace of a matrix is not a strong invariant of matrices. The character will be a stronginvariant of representations. The following exercise motivates this.

EXERCISE 6.9 It is a fact that the characteristic polynomial of an (n × n)-matrix A is determinedby the set of {tr(A), tr(A2), tr(A3), ...} for k > 0. Verify this for (2× 2)-matrices. Explain why thismeans that the character of a representation ρ also encodes the characteristic polynomials of ρ(g),and so the diagonalisation of ρ(g), for all g.

LEMMA 6.10 Characters are class functions.

Proof. We must show for a representation ρ that ρ(g) and ρ(h−1gh) have the same trace. But

tr(ρ(h−1gh)) = tr(ρ(h−1)ρ(g)ρ(h))

= tr(ρ(h)−1ρ(g)ρ(h))

= tr(ρ(g))

using Lemma 6.16.1.

So we can really think of characters as functions on the conjugacy classes of G.

EXAMPLE 6.11 The conjugacy classes of S3 = 〈σ, τ |σ3, τ2, στ = τσ−1〉 are {e}, {σ, σ2}, {τ, στ, σ2τ}.The representation ρ of Example 1.51.5 then has character

e τ σχ 2 0 −1

(It is common to denote a conjugacy class by a representative.)

LEMMA 6.12 Let χ be a character and g ∈ G have order n. Then

i) χ is a sum of χ(1) nth roots of unity

ii) χ(g−1) = χ(g)

Proof. Let χ be the character of ρ. We saw in Lemma 2.142.14 that the eigenvalues of ρ(g) are nth rootsof unity. Since the trace is the sum of the eigenvalues and ρ is χ(1)-dimensional we have i). Forii) note that the eigenvalues of ρ(g−1) are the inverses of the eigenvalues of ρ(g) (diagonalisingmakes this obvious!). Finally, ζ−1 = ζ̄ for any root of unity.

NOTATION 6.13 Given a k[G]-module V , we write χV for its character.

LEMMA 6.14 For a finite group G we have

i) χV⊕W = χV + χW

ii) χV⊗W = χV · χW ,

iii) χV ∨ = χV ,

iv) χHomk(V,W ) = χV χW .

20

6 Characters

Proof. The first part is obvious. For the second, suppose that V,W are two k[G]-modules withchosen basis {e1, ..., en}, {f1, ..., fm} respectively. Fix g ∈ G and suppose that g sends er ⊗ fs 7→∑i,j λ

(r,s)(i,j)ei ⊗ fj with λ(r,s)(i,j) ∈ k. So χV⊗W (g), i.e. the trace of g acting on V ⊗W , is

∑r,s λ

(r,s)(r,s).

On the other hand,

(er ⊗ fs) 7−→(g · er)⊗ (g · fs)

=

(∑i

µ(r)(i) ei

)⊗

∑j

t(s)(j)fj

,

where the µ(r)(i) ∈ k are the matrix coefficients of the action of g on V and the t(s)(j) are those for the

action of g on W . So the coefficient λ(r,s)(r,s) = µ(r)(r)t

(s)(s). In conclusion,

χV (g)χW (g) =

(∑r

µ(r)(r)

)(∑s

t(s)(s)

)=∑r,s

µ(r)(r)t

(s)(s) = χV⊗W (g).

The proof of iii) will be an exercise (using the previous lemma).

For iv) we can use that Homk(V,W ) ∼= V ∨ ⊗W (Lemma 4.114.11).

Since all representations are direct sums of irreducible representations (Maschke’s Theorem5.35.3), the class functions which are characters are precisely the sums of the characters of irre-ducibles. Note also that the product of characters is a character (a priori it was only a classfunction). We shall later find a criterion for telling if a class function comes from a representation.

The following is one of the most important theorems in the course:

THEOREM 6.15 Let ρ1 : G→ GL(W ) and ρ2 : G→ GL(V ) two representations over C, then ρ1 ∼= ρ2if and only if χ1 = χ2.

Proof. We have already seen isomorphic representations have the same characters (Lemma 6.36.3).So we show the converse.

Since the eg for g ∈ G are a basis of C[G] and traces are additive for linear maps, knowing

the traces of Vg·(−)−→ V for a representation also fixes the traces of V

x·(−)−→ V for any x ∈ C[G].Consequently, if the characters of ρ1 and ρ2 maps are equal, then the traces of multiplication by xfor any x ∈ C[G] also agree.

By Corollary 5.105.10, it suffices to show that modules over A :=∏ki=1 Mni

(C) are determinedup to isomorphism by knowing all traces of multiplication by x for x ∈ A. We have already seenthat the irreducible modules over A are just the vector spaces of column vectors Vni

for eachfactor 1 ≤ i ≤ k (Lemma 5.145.14). Arbitrary A-modules are of the form

⊕ki=1 V

⊕aini

. The trace ofmultiplication by

t1 :=

1 0 . . . 00 0 . . . 0...

.... . .

...0 0 . . . 0

, 0, ..., 0

is 1 for Vn1 and 0 for all other Vni . As a result, the trace of t1 acting on an A-module M countshow many summands of M are isomorphic to Vn1

. Repeating this for the other Mni(C) appearing

in A we can completely decompose M using traces.

21

7 Character tables

DEFINITION 6.16 A character is called irreducible if it is the character of an irreducible representa-tion.

So by the theorem to the decomposition of a C-representations of a group G can be read off bywriting its character as a sum of irreducible characters.

The following is a slight strengthening of the theorem.

COROLLARY 6.17 Irreducible characters form a basis of the space of class functions over C.

Proof. The space of class functions has the same dimension as the number of conjugacy classes andso, by Theorems 5.175.17 and 6.156.15, the number of irreducible characters. So we just need to show thatthe irreducible characters are linearly independent. This is a consequence of the proof of Theorem6.156.15. Specifically, if the character of some Vnj was a linear combination of the characters of theVni

for i 6= j, then the trace of tj acting on Vnjwould be a combination of the trace of tj acting on

Vnifor i 6= j. This is impossible as the first is 1 whilst the others are all 0.

7 Character tables

CONSTRUCTION 7.1 For a finite group G we construct its character table as follows. It is the tableformed with one row for each of the irreducible characters of G over C and one column for eachconjugacy class. The entries are then the values of the character on the conjugacy class.

C1 C2 ... Ckχ1 χ1(C1) χ1(C2) ... χ1(Ck)χ2 χ2(C1) ......

.... . .

...χk χk(C1) . . . χk(Ck)

Note that this is always square as the number of conjugacy classes equals the number of irreduciblerepresentations.

We do not specify an ordering, but is normal to put the conjugacy class {e} at the start and toorder the characters starting with the trivial character and then by ascending dimensions. It is alsocommon to write a representative of the conjugacy class instead of Ci and to label the rows usingrepresentations rather than characters (which we may do by Theorem 6.156.15).

EXAMPLE 7.2 • For C2 we havee σ

1 1 1ε 1 −1

where 1 denotes the character of the trivial representation and ε the character of the repre-sentation σ 7→ −1.

• For S3 we havee τ σ

1 1 1 1ε 1 −1 1ρ 2 0 −1

where ρ denotes the character of the two dimensional irreducible.

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7 Character tables

EXAMPLE 7.3 LetQ8 be the quaternion group. It has eight elements {1, i, j, ij,−1,−i,−j,−ij}withmultiplication defined via the relations 〈i, j | i2 = j2 = −1, ij = −ji〉. One can then check thereare five conjugacy classes {{1}, {−1}, {i,−i}, {j,−j}, {ij,−ij}} and so there are five irreduciblerepresentations. Using the formula (55) we know there must be four one-dimensional irreduciblerepresentations and one two-dimensional.

There is a map Q8 � C2 × C2. We may lift the four one dimensional characters of C2 × C2 toobtain the four one-dimensional irreducibles. Inputting this into the character table we have

1 −1 i j ij1 1 1 1 1 1χ1 1 1 −1 1 −1χ2 1 1 1 −1 −1χ1χ2 1 1 −1 −1 1ρ 2

It remains to fill in the final line. One option is to note that you found an irreducible 2-dimensionalrepresentation of Q8 in the exercises (which we now know must be ρ as it is unique) and to takethe trace (do this yourself!). In fact we can find the character of ρ without finding ρ. As we shallnow see.

We know thatC[Q8] = 1⊕ χ1 ⊕ χ2 ⊕ χ1χ2 ⊕ ρ⊕2.

So by linearity,χC[Q8] = 1 + χ1 + χ2 + χ1χ2 + 2χρ. (7)

We also saw that the character of the regular representation C[Q8] is given by

{8 g = 1

0 g 6= 1(this was

Example 6.46.4). Plugging in the values we have already found we can now solve the equation (77) tofind that ρ(g). For example, for ρ(−1),

0 = 1 + 1 + 1 + 1 + 2ρ(−1)

so ρ(−1) = −2 (this method is also how we really found that ρ(1) = 2).

Alternatively, we can also use the following trick for find the remaining values of ρ(g). If χis any of the four one-dimensional characters then χρ is also the character of an irreducible (useExercise 4.104.10 and Proposition 6.146.14). Since we also know that there is only one two dimensionalirreducible we must have that χρ = ρ for any χ. Since there is some χ with χ(g) 6= 1, this forcesρ(g) = 0 unless g = ±1.

In conclusion, the character table of Q8 is

1 −1 i j ij1 1 1 1 1 1χ1 1 1 −1 1 −1χ2 1 1 1 −1 −1χ1χ2 1 1 −1 −1 1ρ 2 −2 0 0 0

Note that whilst we now know from the character table that Q8 has a two-dimensional irre-ducible and its character, it does not find the associated representation. This requires a bit morework, but for many purposes just knowing its character suffices.

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8 Inner product of characters


CONSTRUCTION 8.1 For class functions (and in particular characters) we define an inner product〈−,−〉 by setting

〈χ1, χ2〉 =1

|G|∑g∈G

χ1(g)χ2(g).

When we want to specify the group we may write 〈−,−〉G for the above.

This definition makes sense for any field k whose characteristic does not divide |G| (why?). Weshall always assume this unless explicitly stated otherwise. We are most concerned with k = C.

LEMMA 8.2 The function 〈−,−〉 is a well-defined non-degenerate hermitian inner product.

Proof. We must check

i) 〈χ1 + χ2, χ3〉 = 〈χ1, χ3〉+ 〈χ2, χ3〉,

ii) 〈λ · χ1, χ2〉 = λ · 〈χ1, χ2〉,

iii) 〈χ, χ〉 > 0 for χ 6= 0,

iv) 〈χ1, χ2〉 = 〈χ2, χ1〉.

These are all easy. For example in iii) use that

〈χ, χ〉 =1

|G|∑g∈G

χ(g)χ(g) > 0

as χ(g)χ(g) is > 0 if χ(g) 6= 0 and some χ(g) must not be zero.

THEOREM 8.3 (Row orthogonality) If χ1, χ2 are irreducible characters over C, then

〈χ1, χ2〉 =

{1 χ1 = χ2

0 χ1 6= χ2

.

We first prove two lemmas. Firstly, we consider the case when one of the characters is thetrivial representation:

LEMMA 8.4 For any C[G]-module V we have that 〈χV ,1〉 = dimk VG

Proof. Let a = 1|G|∑g∈G g. Consider the matrix ρ(a) = 1

|G|∑g∈G ρ(g) defining its action on V .

Thentr(ρ(a)) =

1

|G|∑g∈G

tr(ρ(g)) =1

|G|χV (g) = 〈χV ,1〉. (8)

We now calculate tr(ρ(a)) directly.

Note that a = 1|G|∑g∈G g has the property that ag = g for all g ∈ G (by reordering the sum). As

a result it is an idempotent, i.e. a2 = a. This also means that matrix, ρ(a) =∑g ρ(g), representing

the action of a on V is also an idempotent. This means that the matrix ρ(g) satisfies the polynomialX2 −X.

The eigenvalues of a matrix are the roots of its characteristic polynomial and all these rootsappear in its minimal polynomial. The matrix itself satisfies the minimal polynomial and any

24


polynomial satisfied by the matrix is divisible by the minimal polynomial. Putting this together wehave that the only possible eigenvalues of ρ(a) are roots of X2 −X i.e. 0 and 1.

We claim that V G is precisely the 1-eigenspace for ρ(a) (i.e. the elements fixed by ρ(a)). If vlies in the 1-eigenspace, then

ρ(g)v = ρ(ga)v = ρ(a)v = v.

Conversely, if v ∈ V G,

ρ(a)v =1

|G|∑g∈G

ρ(g)v =1

|G|∑g∈G

v = v

Finally, as tr(ρ(a)) is the sum of the eigenvalues of ρ(a), tr(ρ(a)) = dimk VG and combining this

with (88) gives the result.

LEMMA 8.5 If V,W are representations of G over C, then

〈χV , χW 〉 = dimC HomG(W,V ).

Proof. By Lemma 6.146.14 we have that 〈χV , χW 〉 = 1|G|χV (g)χW (g) = 1

|G|∑g∈G χHomk(W,V ) =

〈χHomk(V,W ),1〉. By the previous lemma, this equals dimk Homk(V,W )G. Now use that HomG(V,W ) =Homk(V,W )G (Lemma 4.34.3).

From the lemma it seems as if we should have defined the inner product the other wayround. The definition we gave is consistent with most texts. In any case, we could have useddimC HomG(−,−) as the definition of the inner product, which would maybe be more conceptual.

Proof of Theorem 8.38.3. We have now seen that 〈χ1, χ2〉 = dimC HomG(ρ1, ρ2) where ρ1, ρ2 are therepresentations corresponding to those characters. Now use the ρi are irreducible so HomG(ρ1, ρ2) =0 unless they are isomorphic in which case Schur’s lemma 3.163.16 gives that it is isomorphic to C.

In particular, the inner products of characters of any representation is always an integer. Thiswas not obvious given the terms appearing lie in C!

COROLLARY 8.6 The characters of irreducible representations form an orthonormal basis of the spaceof class functions with respect to 〈−,−〉G.

COROLLARY 8.7 If χ is an irreducible character and ψ any character, then 〈χ, ψ〉 is equal to thenumber of summands of χ in the decomposition of ψ into irreducibles.

Proof. Suppose that the module V corresponding to ψ decomposes into irreducibles as V ⊕n11 ⊕ ...⊕

V ⊕nk

k , so ψ =∑i niχVi

. Now

〈χ, ψ〉 = 〈χ,∑i

niχVi

=∑i

ni〈χ, χVi〉

= nj

where j is the unique choice such that χ = χVj(using Theorem 8.38.3).

EXAMPLE 8.8 There is a representation ρ of S4 on a four dimensional vector space given letting acycle interchange the basis elements e1, e2, e3, e4 just as it acts on {1, 2, 3, 4}. For example,

ρ((1, 2, 3, 4)) =

0 0 0 11 0 0 00 1 0 00 0 1 0

.

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It is easy to see that its character is given χ(g) = |{fixed points of g}| as a cycle on {1, 2, 3, 4}, e.g.χ((1, 2, 3)(4)) = 1. As a result,

〈χ, χ〉 =1

|G|∑g∈G

χ(g)χ(g)

=1

24

∑g∈S4

g has 4 fixed points

42 +∑g∈S4


32 +∑g∈S4


22 +∑g∈S4

g has 1 fixed point

12

=

1

24

(42 + 0 + 6× 22 + 8× 1

)= 2

Now suppose that χ =∑ki=1 ni · χi when written as a sum of irreducible characters. Then

〈χ, χ〉 = 〈k∑i=1

ni · χi,k∑i=1

ni · χi〉

=∑i,j

(ninj〈χi, χj〉)

=∑i

n2i

using row orthogonality. So since 〈χ, χ〉 = 2, we must have that ρ is a sum of exactly two irreduciblerepresentations.

Now, if we instead calculate 〈χ,1〉 then we obtain

〈χ,1〉 =1

|G|∑g∈G

χ(g)

=1

24

∑g∈S4


4 +∑g∈S4


3 +∑g∈S4


2 +∑g∈S4

g has 1 fixed point

1

=

1

24(4 + 0 + 6× 2 + 8× 1)

= 1

So one of the summands of χ is 1 and ρ ∼= 1⊕ ρ′ where ρ′ is a three dimensional irreducible withcharacter χ′(g) = χ(g)− 1.

PROPOSITION 8.9 (Column orthogonality) Let χ1, ..., χk be the complex irreducible characters of afinite group G. Let g1, ..., gk be representatives of each conjugacy class and mi the orders of each class.Then

1

|G|

k∑i=1

χi(gr)χi(gs) =

{1mr

r = s

0 r 6= s. (9)

Proof. This is actually a corollary of row orthogonality. To see this define a new matrix T whoseentries are

Ti,j =

(mj

|G|

) 12

χi(gj).

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Then ∑j

Tk,jTl,j =∑j

(mj

|G|χk(gj)χl(gj)

)= 〈χk, χl〉= δkl

In other words TT t = Ik. Therefore we also have that T tT = Ik and moreover T tT = Ik, i.e.

δrs =∑j

(mr

|G|

) 12(ms

|G|

) 12

χj(gr)χj(gs)

=

√mrms

|G|∑j

χj(gr)χj(gs)

So the left hand side of (99) is zero unless r = s in which case it is 1/mr.

EXAMPLE 8.10 We shall now calculate the character table of the alternating group A4. This hasfour conjugacy classes with representatives given by 1, (12)(34), (123), (132) (exercise). Using thesums of squares formula (55), we see that there must be three one-dimensional representations andone three-dimensional representation.

Let ζ3 be a choice of third root of unity. In addition to 1, there is a non-trivial one dimensionalrepresentation defined by setting χ((12)(34)) = 1 and χ((123)) = ζ3 (exercise). Since products ofcharacters are also characters we must have that χ2 is the remaining one-dimensional representa-tion. So far we have

1 (12)(34) (123) (132)1 1 1 1 1χ 1 1 ζ3 ζ23χ2 1 1 ζ23 ζ3ψ 3

We can now use column orthogonality to fill in the remaining entries. For example, pairing thefirst and second column gives

0 =1

|G|

k∑i=1

χi(1)χi((12)(34))

=1

|G|(1 + 1 + 1 + 3ψ((12)(34)))

So ψ((12)(34)) = −1. In conclusion, we find that

1 (12)(34) (123) (132)1 1 1 1 1χ 1 1 ζ3 ζ23χ2 1 1 ζ23 ζ3ψ 3 −1 0 0

8.1 More examples

EXAMPLE 8.11 Let H3 ⊂ GL3(F3) be the subgroup of strictly upper triangular matrices as definedin the exercises. In this example we calculate its character table. This is harder than is expected inthis course, but will show lots of techniques put together.

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In the exercises it is checked that Hab3∼= Z/3Z × Z/3Z so there are nine one-dimensional

C-irreducibles. Applying (55) we find that the there are two three dimensional irreducibles withcharacters ρ1, ρ2 and these are all the irreducibles.

Using this, we can also calculate the conjugacy classes of H3. We now know there are 11conjugacy classes as this is the number of irreducibles. Consider π : H3 � Z/3Z × Z/3Z. Sinceelements (a, b) ∈ Z/3Z×Z/3Z are their own conjugacy class, every conjugacy class of H3 must liecompletely in π−1((a, b)) for some (a, b). Additionally, all elements of Z(H3) ∼= Z/3Z individuallyform their own conjugacy class. Together these three conjugacy classes form π−1((0, 0)). It remainsto decide if π−1((a, b)) for (a, b) 6= (0, 0) decomposes as union of conjugacy classes or is a singleconjugacy class. Since there are only eleven, in fact none decomposes. Fix representatives:

{e, σ, σ2, g(0,1), g(0,2), g(1,0), g(1,1), g(1,2), g(2,0), g(2,1), g(2,2)}

where σ is a generator of Z(H3) and g(a,b) is a choice of representative of π−1((a, b)) for (a, b) 6=(0, 0).

We now calculate the character table as follows. Firstly, there are nine one-dimensional char-acters inflated from Z/3Z× Z/3Z:

e σ σ2 g(0,1) g(0,2) g(1,0) g(1,1) g(1,2) g(2,0) g(2,1) g(2,2)1 1 1 1 1 1 1 1 1 1 1 1χ1 1 1 1 1 1 ζ3 ζ3 ζ3 ζ23 ζ23 ζ23χ2 1 1 1 ζ3 ζ23 1 ζ3 ζ23 1 ζ3 ζ23...

......

......

......

......

......

...χ21χ

22 1 1 1 ζ23 ζ3 ζ23 ζ3 1 ζ3 1 ζ23

ρ1 3ρ2 3

Here ζ3 is a fixed choice of primitive 3rd root of unity and χ1 is the character coming from the firstfactor and χ2 the second. All the nine other one-dimensional characters are of the form χi1χ

j2 for

0 ≤ i, j ≤ 2 (all this follows from the exercises).

It remains to find the last two rows. Unfortunately, column orthogonality with the first columnwill not find the remaining entries as the resulting equations have too many unknowns. Firstly,we can use the “tensor product trick” to show that ρi(g(a,b)) = 0 for (a, b) 6= (0, 0). Specifically, asρ1 is irreducible and χ1 one-dimensional, ρ1 ⊗ χ1 is also irreducible and so equal to either ρ1 orρ2. The same is true of ρ1 ⊗ χ2

1. Suppose that ρ1(g(1,0)) 6= 0, then (ρ1 ⊗ χ1)((1, 0)) = ζ3ρ((g(1,0)))whilst (ρ1 ⊗ χ1)((1, 0)) = ζ23ρ((g(1,0))). But then we have three distinct values of characters ofthree-dimensional irreducibles on g(1,0) but only two three-dimensionals. This is a contradictionand ρ1(g(1,0)) = 0. The same argument shows ρ2(g(1,0)) = 0 and, possibly choosing a differentone-dimensional character, more generally shows that ρ1 and ρ2 vanish on all g(a,b) with (a, b) 6= 0.


......

......

......

......

......

...χ21χ

22 1 1 1 ζ23 ζ3 ζ23 ζ3 1 ζ3 1 ζ23

ρ1 3 ? ? 0 0 0 0 0 0 0 0ρ2 3 ? ? 0 0 0 0 0 0 0 0

Finally, we have to find ρi(σ), ρi(σ2). The above trick will not work for these and there are still too

many unknowns for orthogonality. Fortunately, we have one more advantage: σ2 = σ−1. We have

28


already seen that for any character χ(g−1) = χ(g) (Lemma 6.126.12 ii)). This allows us to reduce theunknowns by one.

Row orthogonality now gives

1 = 〈ρ1, ρ1〉 =1

27

(32 + ρ1(σ)ρ1(σ) + ρ1(σ)ρ1(σ)

)so

ρ1(σ)ρ1(σ) = 9,

and the same is true for ρ2. So |ρ1(σ)| = 3. On the other hand, ρ1(σ) is a sum of three 3rd rootsof unity (Lemma 6.126.12 i)). This means that ρ1(σ) = 3, 3 · ζ3 or 3 · ζ23 . We know that ρ1(σ) cannotequal 3, for example as then the ρ1-row would not be orthogonal to 1 (or more fundamentally, thecharacters would not form a basis of the space of class functions). So ρ1(σ) = 3 · ζ3 or 3 · ζ23 . Thesame is also true of ρ2. If ρ1(σ) = ρ2(σ), then ρ1 = ρ2 which would contradict Theorem 6.156.15. SoWLOG ρ1(σ) = 3 · ζ3 and ρ2(σ) = 3 · ζ23 . In conclusion, we have:


......

......

......

......

......

...χ21χ

22 1 1 1 ζ23 ζ3 ζ23 ζ3 1 ζ3 1 ζ23

ρ1 3 3 · ζ3 3 · ζ23 0 0 0 0 0 0 0 0ρ2 3 3 · ζ23 3 · ζ3 0 0 0 0 0 0 0 0

DEPARTMENT OF MATHEMATICS, UNIVERSITY COLLEGE LONDON, GOWER STREET, LONDON,WC1E 6BT, UKE-mail address: [email protected], [email protected]

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Representation Theory - homepages.ucl.ac.ukucahato/RepresentationTheory.pdf · Representation Theory Alice Pozzi, Alex Torzewski MATH0073 Contents 1 Introduction 1 2 Subrepresentations