an introduction to representation theory of finite...

An Introduction to Representation Theory of FiniteGroups

Pooja Singla

Ben-Gurion University of the NegevBe’er Sheva

Israel

February 28, 2011

Pooja Singla (BGU) Representation Theory February 28, 2011 1 / 37

History

The groups discussed were mainly

Symmteric groups

Z/nZ and (Z/nZ)?

GLn(C)

Cayley (1894) gave definition of abstract group

(Group) A set G is called group if there exists an operation? : G × G → G such that(1) ? is associative.(2)There exists an element 1 ∈ G such that 1a = a1 = a for all a ∈ G .(3) For every a ∈ G there exists a unique b ∈ G such thatab = ba = 1.

Simple Groups - Group having no proper normal subgroups.

Question

What are all the finite simple groups?

To answer this question various tools were discussed andRepresentation theory of finite groups is one of these.

Other motivation of representation theory comes from the study of groupactions.

Basic Definitions

G - Always finite group.

Definition

A representation of G is a homomorphism from G to the set ofautomorphisms of a finite dimensional complex vector space V , i.e.

φ : G 7→ GL(V )

φ(g1g2) = φ(g1)φ(g2).

V is called a representation space for φ.

For a fixed choice of basis GL(V ) ∼= GLn(C).

dimension of φ := dimension of V .

Notation: (φ,V ) or φ or V .

Note: For all g ∈ G , φ(g)V = V .

Basic Definitions

Definition

φ : G 7→ GL(V )

φ(g1g2) = φ(g1)φ(g2).

Basic Definitions

Definition

φ : G 7→ GL(V )

φ(g1g2) = φ(g1)φ(g2).

Basic Definitions

Definition

φ : G 7→ GL(V )

φ(g1g2) = φ(g1)φ(g2).

Basic Definitions

Definition

φ : G 7→ GL(V )

φ(g1g2) = φ(g1)φ(g2).

Basic Definitions

Definition

φ : G 7→ GL(V )

φ(g1g2) = φ(g1)φ(g2).

Examples:

For any group G , the map φ : G → C?, given by φ(g) = 1 for all g .dim(φ) = 1.

Let Cn - Cyclic group with n elements. Let φ : Cn → C? ahomomorphism, then

φ(x)n = 1 for all x ∈ Cn.

Let ω be a generator of Cn and ζn be n− th primitive root of 1. Thenφ is completely determined by φ(ω).

Let φ(ω) = ζ in then we denote φ by φi .

Examples:

Examples

Let S3 - permutations of {1, 2, 3}. Define φ : S3 → GL3(C) by

(1)→

1 0 00 1 00 0 1

, (12)→

0 1 01 0 00 0 1

(13)→

0 0 10 1 01 0 0

, (23)→

1 0 00 0 10 1 0

(123)→

0 0 11 0 00 1 0

, (132)→

0 1 00 0 11 0 0

.dim(φ) = 3.

Examples

(Permutation Representation) Suppose G acts on finite set X , that isfor each s ∈ G , there is given a permutation x 7→ sx of X satisfying

1x = x , s(tx) = (st)x , s, t ∈ G , x ∈ X .

Let V be complex vector space with basis (ex)x∈X . For s ∈ G , let

ρ : G → GL(V );

ρ(s) : ex 7→ esx .

dim(ρ) = |X |.(Regular Representation) If V is space with basis (eg )g∈G , then aboveaction is called regular representation of G .

Examples

1x = x , s(tx) = (st)x , s, t ∈ G , x ∈ X .

ρ : G → GL(V );

ρ(s) : ex 7→ esx .

dim(ρ) = |X |.

(Regular Representation) If V is space with basis (eg )g∈G , then aboveaction is called regular representation of G .

Examples

1x = x , s(tx) = (st)x , s, t ∈ G , x ∈ X .

ρ : G → GL(V );

ρ(s) : ex 7→ esx .

dim(ρ) = |X |.(Regular Representation) If V is space with basis (eg )g∈G , then aboveaction is called regular representation of G .

Question

What are all the complex representations of a finite group G?

Definition

(G-invariant Space) A space V is called G -invariant if there existsφ : G → Aut(V ) such that

φ(g)V = V for all g ∈ G .

G -invariant ↔ G -representation

(Subrepresentation) Any G invariant subspace of V is calledsubrepresentation.

(Irreducible Representation) A representation is called irreducible if ithas no proper subrepresentations.

One dimensional representations are irreducible.

Definition

Two representations (φ1,V1) and (φ2,V2) are said to be equivalent ifthere exists an isomorphism T : V1 7→ V2 such that φ2(g)T = Tφ1(g) forall g ∈ G .

T��

φ1(g) // V1

T��

V2φ2(g) // V2

In matrix notations: If dim(V1) = dim(V2) = n. Then(φ1,V1) ∼= (φ2,V2), if there exists X ∈ GLn(C) such that

Xφ1(g)X−1 = φ2(g).

Definition

T��

φ1(g) // V1

T��

V2φ2(g) // V2

Xφ1(g)X−1 = φ2(g).

Definition

T��

φ1(g) // V1

T��

V2φ2(g) // V2

Xφ1(g)X−1 = φ2(g).

Examples

Consider φ1 : S2 → GL2(C) by

(1) 7→[

1 00 1

], (12) 7→

[0 11 0

and φ2 : S2 → GL2(C) by

(1) 7→[

1 00 1

], (12) 7→

[1 00 −1

Claim: φ1∼= φ2.

Take X =

[1 11 −1

], then

[0 11 0

]X−1 =

[1 00 −1

Examples

(1) 7→[

1 00 1

], (12) 7→

[0 11 0

(1) 7→[

1 00 1

], (12) 7→

[1 00 −1

Claim: φ1∼= φ2.

Take X =

[1 11 −1

], then

[0 11 0

]X−1 =

[1 00 −1

Examples

(1) 7→[

1 00 1

], (12) 7→

[0 11 0

(1) 7→[

1 00 1

], (12) 7→

[1 00 −1

Claim: φ1∼= φ2.

Take X =

[1 11 −1

], then

[0 11 0

]X−1 =

[1 00 −1

Examples

(1) 7→[

1 00 1

], (12) 7→

[0 11 0

(1) 7→[

1 00 1

], (12) 7→

[1 00 −1

Claim: φ1∼= φ2.

Take X =

[1 11 −1

], then

[0 11 0

]X−1 =

[1 00 −1

Examples

Consider one dimensional representations φi and φj of cyclic group Cn

given byφi (ω) = ζ i

n , φj(ω) = ζ jn

where ω is generator of Cn.

Claim: For i 6= j , φi � φj .

Proof:If f : C→ C is an isomorphism then f (x) = λx for someλ ∈ C?.

��

ζ in // C

λ��

Cζ jn // C

φi∼= φj implies λζ i

n = ζ jnλ, which is not true.

Examples

n , φj(ω) = ζ jn

��

ζ in // C

λ��

Cζ jn // C

Examples

n , φj(ω) = ζ jn

��

ζ in // C

λ��

Cζ jn // C

Definition

(G -linear map) Let (φ1,V1) and (φ2,V2) be two representations of finitegroup G . Then a map T : V1 → V2 is called G-linear if

1 T is C-linear.

2 T ◦ φ1(g) = φ2(g) ◦ T .

The kernel and image of G -linear map are G -invariant subspaces.

Proof:

Let W1 ⊂ V1 be the kernel of T .

Take v ∈W1, we prove φ1(g)v ∈W1 for all g ∈ G .

T (φ1(g)v) = φ2(g)(T (v)) = 0.

φ1(g)v ∈W1.

Similar argument for the image.

Definition

1 T is C-linear.

2 T ◦ φ1(g) = φ2(g) ◦ T .

Proof:

T (φ1(g)v) = φ2(g)(T (v)) = 0.

φ1(g)v ∈W1.

Definition

1 T is C-linear.

2 T ◦ φ1(g) = φ2(g) ◦ T .

Proof:

T (φ1(g)v) = φ2(g)(T (v)) = 0.

φ1(g)v ∈W1.

Definition

1 T is C-linear.

2 T ◦ φ1(g) = φ2(g) ◦ T .

Proof:

T (φ1(g)v) = φ2(g)(T (v)) = 0.

φ1(g)v ∈W1.

Definition

1 T is C-linear.

2 T ◦ φ1(g) = φ2(g) ◦ T .

Proof:

T (φ1(g)v) = φ2(g)(T (v)) = 0.

φ1(g)v ∈W1.

Definition

1 T is C-linear.

2 T ◦ φ1(g) = φ2(g) ◦ T .

Proof:

T (φ1(g)v) = φ2(g)(T (v)) = 0.

φ1(g)v ∈W1.

Definition

1 T is C-linear.

2 T ◦ φ1(g) = φ2(g) ◦ T .

Proof:

T (φ1(g)v) = φ2(g)(T (v)) = 0.

φ1(g)v ∈W1.

More Tools

Definition

(Direct Sum of Representations) If (φ,V ) and (ψ,W ) be tworepresentations of group G , then (φ⊕ ψ,V ⊕W ) given by

[(φ⊕ ψ)(g)](v ,w) = (φ(g)v , ψ(g)w)

is a representation of G .

Observe dim(φ⊕ ψ) = dim(φ) + dim(ψ).

In terms of matrices

(φ⊕ ψ)(g) =

[φ(g) 0

0 ψ(g)

More Tools

Definition

[(φ⊕ ψ)(g)](v ,w) = (φ(g)v , ψ(g)w)

(φ⊕ ψ)(g) =

[φ(g) 0

0 ψ(g)

More Tools

Definition

[(φ⊕ ψ)(g)](v ,w) = (φ(g)v , ψ(g)w)

(φ⊕ ψ)(g) =

[φ(g) 0

0 ψ(g)

More Tools

Definition

[(φ⊕ ψ)(g)](v ,w) = (φ(g)v , ψ(g)w)

(φ⊕ ψ)(g) =

[φ(g) 0

0 ψ(g)

Proposition

Let (φ,V ) be a complex representation of finite group G . The followingare equivalent :

1 (φ,V ) is irreducible.

2 (φ,V ) can not be written as direct sum of two propersubrepresentations.

Proof: Let W be a G -invariant subspace of V . For proof we show thatthere is a complimentary invariant subspace W ′ such that

V = W ⊕W ′.

Let U be an arbitrary complement of W in V , let

π0 : V →W

be the projection given by the direct sum decomposition V = W ⊕ U.

Proposition

Let (φ,V ) be a complex representation of finite group G . The followingare equivalent :

1 (φ,V ) is irreducible.

2 (φ,V ) can not be written as direct sum of two propersubrepresentations.

Proof: Let W be a G -invariant subspace of V . For proof we show thatthere is a complimentary invariant subspace W ′ such that

V = W ⊕W ′.

Let U be an arbitrary complement of W in V , let

π0 : V →W

be the projection given by the direct sum decomposition V = W ⊕ U.

Average the map π0 over G , that is an onto map π : V →W by,

π(v) =1

|G |∑g∈G

φ(g)(π0(φ(g)−1v)).

Then π is a G -linear. Therefore its kernel is the required G -invariantcomplement of W .

Theorem

Every complex representation of finite group G is direct sum of irreduciblerepresentations.

Proof:(For cyclic group case) Let φ : Cn → GLm(C)- homomorphism.

Step 1. Every finite order complex matrix is diagonalizable.

Step 2. The matrices φ(x) are pairwise commuting and diagonalizable,hence are simultaneously diagonalizable. Therefore φ is easily seen to bedirect sum of one dimensional representations. That is

φ = ⊕iφ⊕mii

where mi is the multiplicity.

General Proof Decompose V into irreducible representation by usinglast proposition.

Theorem

φ = ⊕iφ⊕mii

Theorem

φ = ⊕iφ⊕mii

Theorem

φ = ⊕iφ⊕mii

Theorem

φ = ⊕iφ⊕mii

So the question is...

Question

What are all the finite dimensional inequivalent irreducible complexrepresentations of a given finite group G ?

Observe we have answered it already for cyclic groups.

So the question is...

Question

What are all the finite dimensional inequivalent irreducible complexrepresentations of a given finite group G ?

Observe we have answered it already for cyclic groups.

More tools and Interesting Results

Theorem

(Schur’s Lemma) Let φ1 : G → GL(V1) and φ2 : G → GL(V2) be twoirreducible representations of G , and let T be a linear mapping of V1 intoV2 such that φ2(g) ◦ T = T ◦ φ1(g) for all g ∈ G . Then :

1 If φ1 and φ2 are not isomorphic, we have T = 0.

2 If V1 = V2 and φ1 = φ2, then T (x) = λx for x ∈ V and for somescalar λ ∈ C.

Proof:

Suppose T 6= 0. The G -linearity of T implies that both kernel andimage of T are G -invariant subspaces.

Theorem

Proof:

Theorem

Proof:

More interesting tools...

The irreduciblity of φ1 and φ2 implies T is an isomorphism.

If V1 = V2, φ1 = φ2. Let λ be an eigenvalue of T (recall field is C).

Put T ′ = T − λ. Then Ker(T ′) 6= 0.

Also φ2(g) ◦ T ′ = T ′ ◦ φ1(g).

By the first part, we must have T ′ = 0. That is T = λ.

Theorem

The number of inequivalent irreducible representations of finite group isequal to the number of its conjugacy classes.

Put T ′ = T − λ. Then Ker(T ′) 6= 0.

Also φ2(g) ◦ T ′ = T ′ ◦ φ1(g).

Theorem

Put T ′ = T − λ. Then Ker(T ′) 6= 0.

Also φ2(g) ◦ T ′ = T ′ ◦ φ1(g).

Theorem

Put T ′ = T − λ. Then Ker(T ′) 6= 0.

Also φ2(g) ◦ T ′ = T ′ ◦ φ1(g).

Theorem

Put T ′ = T − λ. Then Ker(T ′) 6= 0.

Also φ2(g) ◦ T ′ = T ′ ◦ φ1(g).

Theorem

Put T ′ = T − λ. Then Ker(T ′) 6= 0.

Also φ2(g) ◦ T ′ = T ′ ◦ φ1(g).

Theorem

If ρ1, ρ2, . . . , ρt are all the inequivalent irreducible representations of groupG then

|G | =t∑

dim(ρi )2.

Consider the group S3.

|S3| = 6.

S3 has three conjugacy classes given by (1), (12), (123).

Define σ : S3 → C? by

σ(1) = 1, σ(123) = σ(132) = 1,

σ(12) = σ(23) = σ(13) = −1.

Theorem

If ρ1, ρ2, . . . , ρt are all the inequivalent irreducible representations of groupG then

|G | =t∑

dim(ρi )2.

Consider the group S3.

|S3| = 6.

S3 has three conjugacy classes given by (1), (12), (123).

Define σ : S3 → C? by

σ(1) = 1, σ(123) = σ(132) = 1,

σ(12) = σ(23) = σ(13) = −1.

σ and trivial representations are two one dimensional inequivalentirreducible representations of S3 of dimension one.

The only way to write 6 as sum of three squares is 6 = 1 + 1 + 22.

Recall that the permutation representation of S3 which maps eachpermutation to corresponding permutation matrices.

This is not direct sum of one dimensional representations.

The only decomposition possible is 3 = 1 + 2.

So its decomposition with above observations will give all theirreducible representations of S3.

Permutation Representation

Let S3 - permutations of {1, 2, 3}. Define φ : S3 → GL3(C) by

(1)→

1 0 00 1 00 0 1

, (12)→

0 1 01 0 00 0 1

(13)→

0 0 10 1 01 0 0

, (23)→

1 0 00 0 10 1 0

(123)→

0 0 11 0 00 1 0

, (132)→

0 1 00 0 11 0 0

.dim(φ) = 3.

Groups GL2(K )

Set G = GL2(K ), |K | = q where q = pr for an odd prime p.

Question

What are all the irreducible complex representations of G ?

We shall construct a very special class of representations of these groups.

Remark

A representation of dimension one of G is simply a homomorphism φ of Gto C?.

Let G0 = {xyx−1y−1 | x , y ∈ G} be the derived subgroup of G .

Let π : G → G/G0 be the natural epimorphism.

Proposition

(One dimensional representations of G) Let φ : G → C? be ahomomorphism.

Then there exists a homomorphism φ : G/G0 → C? such thatφ.π = φ.

Conversely, any homomorphism from G/G0 to C? produces ahomomorphism from G to C? by composition with π.

Remark

Proposition

Remark

Proposition

Definition

(Induced Representations) Let H be a subgroup of a finite group G, andlet (ψ,U) be a representation of H. Let

V = {f : G → U|f (hg) = ψ(h)f (g), h ∈ H, g ∈ G}.

Then G acts on V by right translations; φ : G → Aut(V ) by

[φ(g)(f )](g ′) = f (g ′g) g , g ′ ∈ G , f ∈ V .

It follows that (φ,V ) is a representation of G . This is called inducedrepresentation of (ψ,U) from H to G , denoted by IndG

H(ψ).

dim(IndGH(ψ)) = |G/H|.dim(ψ).

By restricting the action of φ on a subgroup H of G , we obtain arepresentation of H, denoted by ResGH(φ).

Definition

[φ(g)(f )](g ′) = f (g ′g) g , g ′ ∈ G , f ∈ V .

H(ψ).

Definition

[φ(g)(f )](g ′) = f (g ′g) g , g ′ ∈ G , f ∈ V .

H(ψ).

Set G = GL2(K ), |K | = q where q = pr for an odd prime p.

Question

What are all the irreducible complex representations of G ?

Let K ∗ be invertible elements of K .

Recall K ? is a cyclic group of order q − 1.

Let K ? is the set of one dimensional representations of K ?.

B = {(

a b0 c

): a, c ∈ K ∗, b ∈ K}.

Choose µ1, µ2 ∈ K ∗.Define

µ : B 7→ C?;

(a b0 c

)7→ µ1 (a)µ2 (c) .

Writeµ = (µ1, µ2).

SetVµ = {f : G → C | f (bg) = µ(b)f (g)}.

Now define an action of G on Vµ by

(µ(g)f )(x) = f (xg) , for all x , g ∈ G .

dimension of µ = |G ||B| = (q2−1)(q2−q)

(q−1)2q= q + 1.

Question

Is µ irreducible ?

(q−1)2q= q + 1.

Question

Is µ irreducible ?

(q−1)2q= q + 1.

Question

Is µ irreducible ?

(q−1)2q= q + 1.

Question

Is µ irreducible ?

Case 1. For µ1 6= µ2, the G -representation (µ,Vµ) is irreducible.Furthermore,

(µ1, µ2) ∼= (µ′1, µ′2)

if and only if either

µ1 = µ′1 and µ2 = µ

orµ1 = µ

′2 and µ2 = µ

This gives (q−1)(q−2)2 inequivalent (q + 1) dimensional irreducible

representations.

Case 2. For µ1 = µ2, the G -representation (µ,Vµ) is not irreducible.In this case

µ = φ1 ⊕ φ2,

where φ1 is one dimensional and φ2 is q dimensional.

Suppose µ = φ1 ⊕ φ2 and ν = ψ1 ⊕ ψ2.If µ � ν then

φ1 � ψ1 and φ2 � ψ2.

This gives (q − 1) one dimensional representations and (q − 1)inequivalent q dimensional representations.

Table of Conjugacy classes

conjugacy class representative No. of classes

central semisimple

(a 00 a

)q − 1

unitary

(a 10 a

)q − 1

non central semisimple

(a 00 b

), a 6= b (q−1)(q−2)

anisotropic

(0 −αα1 α + α

)q2−q

Mackey’s intertwining Theorem

Let H and K be subgroups of G . If g ∈ G then the setHgK = {hgk | h ∈ H, k ∈ K} is called a double coset with respect tosubgroup H and K . The element g is called its representative.

A complete set of representatives of all (H,K )-double cosets isdenoted by H\G/K .

For s ∈ H\G/K we set Hs = sHs−1 ∩ K .

Consider the representation (τ,W ) of H.

The subgroup Hs has a natural representation (τ s ,W ) defined by

τ s(x) = τ(s−1xs), x ∈ Hs

If (φ,V ) and (ψ,W ) are two G -representations then

HomG (φ, ψ) = {T : V →W | T is G − linear}.

Theorem

(Mackey’s Intertwining Theorem) Let H and K be subgroups of G and(U, σ) a representation of K and (W , τ) a representation of H. Then:

HomG (IndGK (σ), IndG

H(τ)) ∼= ⊕s∈H\G/K

HomsHs−1∩K (σ, τ s).

Continuation for Cases I and II

For our case H = K = B, where B = {(

a b0 c

)| a, c ,∈ K ?, b ∈ K}

and σ = τ = (µ1, µ2).

For c 6= 0:(a bc d

(b − ac−1d a

)(0 11 0

)(1 c−1d0 1

)The double coset representatives B\GL2(K )/B correspond to

1 00 1

(0 11 0

Let s =

(0 11 0

). Then

B ∩ sBs−1 = T = {(

a 00 b

)| a.b ∈ K ?}.

a b0 c

)| a, c ,∈ K ?, b ∈ K}

and σ = τ = (µ1, µ2).For c 6= 0:(

a bc d

(b − ac−1d a

)(0 11 0

)(1 c−1d0 1

The double coset representatives B\GL2(K )/B correspond to

1 00 1

(0 11 0

Let s =

(0 11 0

). Then

B ∩ sBs−1 = T = {(

a 00 b

)| a.b ∈ K ?}.

a b0 c

)| a, c ,∈ K ?, b ∈ K}

and σ = τ = (µ1, µ2).For c 6= 0:(

a bc d

(b − ac−1d a

)(0 11 0

)(1 c−1d0 1

1 00 1

(0 11 0

Let s =

(0 11 0

). Then

B ∩ sBs−1 = T = {(

a 00 b

)| a.b ∈ K ?}.

a b0 c

)| a, c ,∈ K ?, b ∈ K}

and σ = τ = (µ1, µ2).For c 6= 0:(

a bc d

(b − ac−1d a

)(0 11 0

)(1 c−1d0 1

1 00 1

(0 11 0

Let s =

(0 11 0

). Then

B ∩ sBs−1 = T = {(

a 00 b

)| a.b ∈ K ?}.

Observe that

(µ1, µ2)s

(x 00 y

)= µ2(x)µ1(y).

Hence by Mackey’s intertwining theorem we obtain that

HomG ( (µ1, µ2), (µ1, µ2)) = [HomB((µ1, µ2), (µ1, µ2))]

⊕[HomT ((µ1, µ2), (µ2, µ1))]

The T -representations (µ1, µ2) and (µ2, µ1) are equivalent if and only ifµ1 = µ2.

This implies dimC(HomG ( (µ1, µ2), (µ1, µ2))) is equal to one for µ1 6= µ2

and is equal to two for µ1 = µ2.

This combined with Schur’s Lemma gives the result.

Observe that

(µ1, µ2)s

(x 00 y

)= µ2(x)µ1(y).

Hence by Mackey’s intertwining theorem we obtain that

HomG ( (µ1, µ2), (µ1, µ2)) = [HomB((µ1, µ2), (µ1, µ2))]

⊕[HomT ((µ1, µ2), (µ2, µ1))]

The T -representations (µ1, µ2) and (µ2, µ1) are equivalent if and only ifµ1 = µ2.

This implies dimC(HomG ( (µ1, µ2), (µ1, µ2))) is equal to one for µ1 6= µ2

and is equal to two for µ1 = µ2.

This combined with Schur’s Lemma gives the result.

Summary

dimension No of irreducible namerepresentations

1 q − 1 one dimensional

q q − 1 special

q + 1 (q−1)(q−2)2 regular principal series

q − 1 q2−q2 cuspidal

References

Fulton, William and Harris, Joe,Representation theory,Graduate Texts in Mathematics, Springer-Verlag, 1991.

Serre, Jean-PierreLinear representations of finite groupsSpringer-Verlag 1977.

Ilya Piatetski-Shapirocomplex representations of GL(2,K ) for finite fields KContemporary Mathematics, 1983.

Aburto, Luisa and Johnson, Roberto and Pantoja, JoseeThe complex linear representations of GL(2, k), k a finite field,Proyecciones. Journal of Mathematics,Vol. 25, 2006, 307–329.

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