MA3413: Group representations Lecturer: Prof. Vladimir Dotsenko Michaelmas term 2012

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MA3413: Group representations Lecturer: Prof. Vladimir Dotsenko

Michaelmas term 2012

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1Transcribed by Stiofáin Fordham. Last updated 16:19 Monday 21^st October, 2013

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The question is then, if we repeatedly do this, will the values somehow stabilise after a long time? This can be solved using basic analysis but it is dicult. Instead we consider the generalised case of the n-polygon with rotational symmetry. It turns out that applying this procedure adds additional symmetry that, when investigated, gives an easy solution.

a

b c

b+c 2

a+c 2

a+b 2

● Take a nite group Gand write out its Cayley table. For exampleZ/3Z

¯0 ¯1 ¯2

¯0 ¯0 ¯1 ¯2

¯1 ¯1 ¯2 ¯0

¯2 ¯2 ¯0 ¯1

or for Z/2Z

¯0 ¯1

¯0 ¯0 ¯1

¯1 ¯1 ¯0

Now for each g ∈G, introduce the variable xg. For example in the case Z/2Z

(x¯0 x¯1

x¯1 x¯0)

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and investigate the determinant. Dedekind observed that the determinant seemed to factor very nicely in general. In the above case

det(x¯0 x¯1

x¯1 x¯0) =x²¯0−x²¯1= (x¯0+x¯1)(x¯0−x¯1)

He sent a letter to Frobenius describing this. The latter then tried to explain this and in doing so essentially founded the area of representation theory.

● We denote the symmetric group onnobjects by S_n. Consider the cross- product, it satises

(x1×x2) ×x3+ (x2×x3) ×x1+ (x3×x1) ×x2=0

This identity is invariant under permutation of indices (up to, perhaps, multiplication by−1). Objects like this, that do not change much under permutation of objects are studied extensively - they have signicance in physics also. We would like to be able to classify these types of symmetries. Say we have a function of two arguments f(x1, x2). We call functions that satisfy f(x1, x2) =f(x2, x1) symmetric. Functions that satisfy f(x1, x2) = −f(x2, x1) are called skew-symmetric. It is fairly easy to proove that every function of two arguments can be represented as the sum of a skew-symmetric function and a symmetric function. But for more than two variables the problem becomes much more dicult.

The prerequisites for this course

(1) Linear algebra: trace, determinant, change of basis, diagonalisation of matrices, vector spaces over arbitrary elds.

(2) Algebra: groups (for the most part up to order 8)¹, conjugacy classes &

the class formula

∣G∣ = ∑

conj. classc

∣G∣

∣Z(c)∣

where Z(c)is the centraliser of any element of the conjugacy class c. So in the case ofg0∈c

Z(g₀) = {g∈G∣g₀g=gg₀} Lecture 2

24/9 4pm

2.1. Examples of representations. As a rst approximation, a representation tries to understand all about a group by looking at it as a vector space.

Denition 1. Let G be a group, and V a vector space (over a eld k), then a representation ofGonV is a group homomorphism

ρ∶G→GL(V)

where GL(V)is the group of invertible linear transformations of the spaceV. Examples

(1) Let G be arbitrary, and let V = k. We want a group homomorphism G→k^×=k/{0}. The trivial representation isρ(g) =1 for everyg∈G.

1Up to isomorphism: order 2 isZ/2Z, odd primes areZ/nZby the corollary to Lagrange's theorem, order 4 are the Kleinsche-vierergruppeZ/2Z×Z/2ZandZ/4Z, order 6 isZ/6ZandS3≅D3

(the dihedral group: the symmetries of a regularn-gon) and for order 8 are, in the Abelian case, Z/8Z,Z/4Z×Z/2ZandZ/2Z×Z/2Z×Z/2Z, and in the non-Abelian caseD4and the quaternions Q8.

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(2) (a) Let G be arbitrary, and let V =kG:= the vector space with a basis {eg}g∈G. It is enough to understand how the element acts on each basis element ρl(g)eh := egh. We must check that ρl(g1g2) = ρl(g1)ρl(g2); it is enough to determine the eect of these elements on basis elements. We have

ρl(g1g2)eh=e_(g₁_g₂_)h=e_g₁_(g₂_h)

=ρ_l(g₁)e_g₂_h=ρ_l(g₁)ρ_l(g₂)e_h

So ρl(g1g2) and ρl(g1)ρl(g2) map every basis vector to the same vector, hence are equal. This representation is called the left regular representation ofG.

(b) G is arbitrary, V = kG. Let ρr(g)eh = e_hg−1 (Exercise: show that dening this the other way around does not work.). We have

ρr(g1g2)eh=e_h(g₁_g₂)⁻¹=e_hg⁻1

2 g⁻₁¹=ρr(g1)e_hg⁻1 2

=ρr(g1)ρr(g2)eh

(3) Let G = Z/2Z and V = k. The only constraint is ρ(¯1)² = 1, so two dierent one-dimensional representations for chark≠2 (there is only one one-dimensional representation for chark=2). We have

ρ(¯0) =1 ρ(¯1) =1 (4) LetG=Z/2ZandV =k². We have

ρ(¯0) = (1 0

0 1) ρ(¯1) = (1 0 0 −1) or

ρ(¯0) = (1 0

0 1) ρ(¯1) = (0 1 1 0)

(5) Take G = S₃ and V = R². Take an arbitrary triangle in the plane, let the origin be the center of the triangle. Rotations of the triangle are represented by

(1 0

0 1), (cos(^2π₃) −sin(^2π₃ )

sin(^2π₃) cos(^2π₃ )), (cos(^4π₃) −sin(^4π₃ ) sin(^4π₃) cos(^4π₃ ))

Reections are also linear transformations, so represented by matrices accordingly.

Denition 2. Let (V1, ρ1)and(V2, ρ2)be two representations of the same group G. Then their direct sum(V1⊕V2, ρ1⊕ρ2)is dened as follows: the vectors of the representation areV1⊕V2, the homomorphismsG→GL(V)(V1⊕V2)are dened as

(ρ1⊕ρ2)(g) = (ρ₁(g) 0 0 ρ₂(g)) We check

(ρ₁⊕ρ₂)(g₁g₂) = (ρ1(g1g2) 0 0 ρ2(g1g2))

= (ρ₁(g₁)ρ₁(g₂) 0 0 ρ2(g1)ρ2(g2))

= (ρ₁(g₁) 0 0 ρ₂(g₁)) (

ρ₁(g₂) 0 0 ρ₂(g₂))

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So it is a homomorphism.

Denition 3. Suppose that (V, ρ) is a representation of G. It is said to be re- ducible if there exists a subspaceU⊂V withU ≠ {0}orV, which is invariant with respect to all ρ(g)i.e. ρ(g)U⊂U forg∈G. Otherwise it is said to be irreducible.

The representation given above forS3 is irreducible.

Lecture 3 27/9 1pm

When we stopped on Monday, we had given the denition of an irreducible representation and one example. We will now discuss some more examples.

Recall that the sum of two representations is just the direct sum of the vectors spaces - the block matrices, and an irreducible representation is a representation without a non-trivial invariant subspace. Aside: note that we will only discuss representations of positive dimension. V1⊕V2 has invariant subspaces V1, V2 so is not irreducible.

Last lecture I stated that the representation that I gave forS3constructed last time is irreducible. To see this: if it were not, all operators representing elements ofS3would have a common eigenvector.

Over real numbers, each reection has just two possible directions for eigenvectors so altogether no common eigenvectors. Over C: real eigenvectors of reections remain the only eigenvectors, so the same argument works.

3.1. Tensor products. Recall the tensor product: letV, W be vectors spaces, then V ⊗W is the span of vectors (v, w)for v∈ V and w∈W quotiented out by the relations

v⊗ (w1+w2) =v⊗w1+v⊗w2

(v₁+v₂) ⊗w=v₁⊗w+v₂⊗w λ(v⊗w) = (λv) ⊗w=v⊗ (λw) to obtain the tensor product.

Denition 4. If (V, ρ) and(W, ϕ)are two representations of the same group G, then the tensor product (V ⊗W, ρ⊗ϕ)is dened as follows

(ρ⊗ϕ)(g)(v⊗w):= (ρ(g)(v)) ⊗ (ϕ(g)(w)) extended to allV ⊗W by linearity.

ρ⊗ϕis a group homomorphism ifρandϕare (it is extremely tedious to show that this is a homomorphism so we will not do it here).

Exercise: Let us x a basis{ei}ofV, a basis{fj}ofW, and order a basis of V ⊗W as follows

e1⊗f1, e1⊗f2, . . . , e1⊗fm, e2⊗f1, . . . , e2⊗fm, . . . , . . . , en⊗f1, . . . , en⊗fm

Suppose thatρ(g)has the matrixArelative to{ei},ϕ(g)has the matrixBrelative to {fj}

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(1) Write down the matrix of(ρ⊗ϕ)(g)relative to the basis above.

(2) tr((ρ⊗ϕ)(g)) =tr(ρ(g)) ⋅tr(ϕ(g)).

Denition 5. If we have two representations (V₁, ρ₁) and (V₂, ρ₂) of the same group G, then they are said to be isomorphic or equivalent if there exists a vector space isomorphism ϕ∶V1≃V2such that for eachg∈G

ρ2(g) ○ϕ=ϕ○ρ1(g) V1

ρ₁(g)//

ϕ

V1 ϕ

V₂

ρ2(g)

//V₂

In other words, if we work with matrices, two representations are equivalent if they can be identied by change of coordinates.

To illustrate this denition, let us go back to one of the representations that we discussed on Monday.

Proposition 1. For any G,(kG, ρl) ≃ (kG, ρr).

Proof. Let us dene ϕ∶kG→kG as follows: ϕ(eg) =e_g⁻1. This is invertible because if you apply it twice you get the identity. Let us check that it agrees with what we have here: we must show thatϕ○ρl(g) =ρr(g) ○ϕ. It is enough to check this for the basis

ϕ○ρl(g)(eh) =ϕ(egh) =e_(gh)−1

ρr(g) ○ϕ(eh) =ρr(g)(eh⁻¹) =eh⁻¹g⁻¹

because (gh)⁻¹=h⁻¹g⁻¹, the result follows.

We will now deal with the question: is (kG, ρl)irreducible? The answer is no.

For the case where Gis nite, consider v= ∑

g∈G

e_g

Then for eachh∈G,

ρ_l(h)(v) =v because

ρl(h)⎛

⎝∑

g∈G

eg

⎞

⎠= ∑

g∈G

ehg= ∑

g∈G

eg=v

Now regarding textbooks: there is one by Curtis and Reiner Methods of Rep- resentation Theory: With Applications to Finite Groups and Orders; very compre- hensive - ten times more than we could cover. Another is Linear Representations of Finite Groups by J.P. Serre - very concise, and clear but covers more than we can cover. And of course the textbook by Fulton and Harris.

Lecture 4 1/10 12pm

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4.1. Decomposing the left-regular representation. The last thing we discussed on Thursday was the equivalence of representations which is just the same sort of equivalence that you know from linear algebra, it is just equivalence under change of coordinates essentially. There was an example over 1-dimensional subspaces of the left regular representation which looked exactly like the trivial representation. Let us look at that again because we will use this again often.

LetGbe a nite group, and consider againkGequipped with the left regular representationρland letv∈kG. Consider

v= ∑

g∈G

eg

For each g∈G we haveρ_l(g)v=v. The linear span of kv ⊂kGis a 1-dimensional subspace, invariant under all ρ_l(g) and as a representation is isomorphic to the trivial representation.

Recall that our ultimate goal is to understand all representations of a nite group by looking at the irreducible representations. We now want to do this with this representation. We make some assumptions: eitherchark=0or charkis not a divisor of the number of elements in our group. Then there exists a subspace W ⊂kGwhich is invariant under allρl(g)such thatkG≃kv⊕W - an isomorphism of representations, where we have

ρl≃trivial⊕ρl∣_W Let us proove this

Proof. DeneW := {∑^g∈Gcgeg∶ ∑^g∈Gcg=0}which is invariant under allρl(g) ρl(h) (∑cgeg) = ∑cgehg

= ∑

g∈G

c_h⁻1hgehg

= ∑c_h−1g^′e_g^′

with ∑g^′∈Gc_h−1g^′ =0. There are now many ways to proceed. We havedimkv =1 and dimW = ∣G∣ −1 so it is enough to show that their intersection is equal to zero i.e. kv∩W = {0}. For ∑g∈Gc_ge_g ∈kv, allc_g are equal to each other. Take g₀∈G then

∑

g∈G

cg= ∣G∣ ⋅cg₀

If ∑cgeg ∈W we conclude that∣G∣ ⋅cg0 =0 since in W all coecients are equal to zero. Ifc_g₀=0, then allc_g=0, and we proved thatkv∩W = {0}. Otherwise∣G∣ =0

in kwhich is a contradiction.

There is an alternative argument going in the other direction. We used the fact that if you have two subspaces of complementary dimensions, then their direct sum is the entire space i they have zero intersection. Take some typical vector in your vector spacekG

∑

g∈G

ageg∈kG then their sum can be written

∑

g∈G

⎛

⎝ag− 1

∣G∣ ∑h∈G

ah

⎞

⎠eg+ ∑

g∈G

⎛

⎝ 1

∣G∣ ∑h∈G

ah

⎞

⎠eg

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where the average element given here forces: the rst element belongs toW and the second one belongs to k⋅v so we have kG=W+kv²and since the dimensions are complementary, the sum must be direct.

One example that shows that our assumptions about characteristics are important: takeG=Z/2Zandk=F2. SoGconsists of the elements{¯0,¯1}andF2consists of{0,1}. kGis a 2-dimensional representation ofGoverk. Thenv=e¯0+e¯1 which is an invariant subspace. What happens now is that there is no way to choose a complementary subspace. In fact, the only invariant subspace is the one just given.

How to see this? Any invariant subspace must be one-dimensional, supposeae¯0+be¯1

spans it. Let us take this element under the action ρ_l(¯1)(ae¯0+be¯1) =ae¯1+be¯0

So we must check under what conditionsae¯1+be¯0 is proportional toae¯0+be¯1. The coecient of proportionality is an element of k so therefore it is either 0 or 1. If it is 0 then a =b = 0 and it is not a 1-dimensional subspace, and if it is 1 then a=bso it'sk⋅v. So we see in positive characteristic we get these undesirable things happening.

Let us formulate a more general result. From now on (except for rare explicit cases) we assume that 1/∣G∣ make sense in in our eld k - note this also assumes our group is nite.

Theorem 2. Let (V, ρ)be a representation ofGand letU⊂V be invariant under all ρ(g). Then there exists a subspaceW ⊂V that is also invariant under all ρ(g) and such that V ≃U⊕W with

ρ≃ρ∣_U⊕ρ∣_W Lecture 5

1/10 4pm

5.1. Complements of invariant subspaces. Continuing from last time the proof of the theorem. We will give two proofs. One will only work overRand with a modication over C, whereas the other will work in generality.

Proof #1: Take k = R. Recall the proof in linear algebra that over real numbers all symmetric matrices are diagonalisable. Our proof will mirror this almost exactly.

Suppose that there exists a scalar product on our vector space which is invariant with respect to the group action

(ρ(g)(v), ρ(g)(w)) = (v, w)

i.e. they are all representable as orthogonal matrices. Then if U ⊂V is invariant, we shall show that U^⊥ is invariant. Indeed, letv∈U^⊥ andu∈U. We have

(ρ(g)(v), u) = (ρ(g)(v), ρ(g)ρ(g)⁻¹(u))

= (v, ρ(g)⁻¹(u))

= (v, ρ(g⁻¹)(u))

=0

2If we set cg = ag − ¹

∣G∣∑_h∈Ga_h then ∑_g∈Gcg = ∑_g∈G(ag − ¹

∣G∣∑_h∈Ga_h) = ∑_g∈Gag −

∣G∣

∣G∣∑_h∈Gah=0and the coecients _∣G∣¹ ∑_h∈Gahclearly don't depend ong∈G.

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because ofρ(g⁻¹)(u) ∈U sinceU is invariant. It remains to show that an invariant scalar product always exists. How to show this? Take some scalar product(v, w) ↦ f(v, w)and dene a new one

(v, w) = 1

∣G∣ ∑g∈G

f(ρ(g)(v), ρ(g)(w)) If we now apply someρ(h)

(ρ(h)(v), ρ(h)(w)) = 1

∣G∣ ∑g∈G

f(ρ(g)ρ(h)(v), ρ(g)ρ(h)(w))

= (v, w)

This works on R because we have positive deniteness - a total ordering on R. Otherwise we can have some non-trivial intersection of a subspace and its orthogonal complement: if we take C² and (z1, w1),(z2, w2) ↦z1z2+w1w2 then (1, i)is orthogonal to itself.

Proof #2: To dene a direct sum decomposition onV ≃U⊕W is equivalent to nding a linear operator P such that

● P²=P

● kerP=W

● ImP =U

Remark. So how does it work? SupposeP²=Pthen we want to show that V =kerP⊕Im P. We want to check whetherkerP∪ImP = ∅.

Letv∈kerP. ThenP v=0 and letv=P w then

0=P v=P P w=P²w=P w=v

Then we writev= (Id−P)v+P vand the rst one is in the kernel and the second is in the image.

Moreover, ifP ρ(g) =ρ(g)P for allg, then the subspaceW =kerP is invariant.

Indeed, if w∈W then we have

P ρ(g)(w) =ρ(g)P(w) =ρ(g)(0) =0

which implies thatρ(g)(w) ∈kerP=W. Therefore, it is enough to show that there exists an operator P such that

(1) P²=P (2) U =ImP

(3) ρ(g)P=P ρ(g)for allg∈G

The idea is to take some P₀ satisfying 1,2 - this is not dicult because this just means nding a decomposition on the level of the vector spaces. Now put

P := 1

∣G∣ ∑g∈G

ρ(g)P₀ρ(g)⁻¹

Condition three is satised (amounts to checking the same thing as in last lecture).

But we must check that replacingP0 byP does not break conditions one or two.

So: why is P²=P? First we show thatImP ⊂U: if v∈V thenρ(g)⁻¹(v) ∈V so P0ρ(g)⁻¹(v) ∈Ubecause forP0, property two holds. Thereforeρ(g)P0ρ(g)⁻¹(v) ∈U (due to invariance ofU). SoP(v) ∈U. Next, supposeu∈U thenρ(g)⁻¹(u) ∈U due to invariance of U. Therefore P0ρ(g)⁻¹(u) =ρ(g⁻¹)(u) because of properties one

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and two for P₀³. Finally,ρ(g)P₀ρ(g)⁻¹(u) =ρ(g)ρ(g⁻¹)u=uso as a consequence P u=u. Finally, nally,P²(v) =P(P(v)) =P(v)sinceP(v) ∈U.

Finally, nally, nally,ImP =U because P(u) =uonU. Lecture 6

4/10

6.1. Intertwining operators and Schur's lemma. Last time we had: let V be a representation ofGandU⊂V an invariant subspace such thatV ≃U⊕W. Our strategy was to take an operator P0 such thatP0=P₀² withImP0=U. Then we average overG

P= 1

∣G∣ ∑g∈G

ρ(g)P0ρ(g)⁻¹ We showed that

● P²=P

● ImP =U

● ρ(g)P=P ρ(g)forg∈G

Denition 6. Suppose that(V, ρ)is a representation of a groupG, then operators T satisfying

ρ(g)T=T ρ(g) for allg∈G

are called intertwining operators for V. More generally, if(V, ρ)and(W, ϕ)are two representations of the groupGthenT∶V →W satisfying

T ρ(g) =ϕ(g)T are called intertwining operators fromV toW. Theorem 3 (Schur's lemma).

(1) Suppose that (V1, ρ1) and (V2, ρ2) are irreducible representations of a group G. Then any intertwining operator T∶V1 → V2 is either zero or an isomorphism.

(2) Over the complex numbers (or any algebraically closed eld), any intertwining operatorT∶V1→V2 is scalar.

Proof. Consider the kernel ofT, it satiseskerT⊂V1. Our claim is that it is invariant. Let v∈kerT

T ρ1(g)(v) =ρ2(g)T(v) =ρ2(g)(0) =0

Therefore v∈ kerT ⇒ρ₁(g)(v) ∈kerT. But it is irreducible. Therefore there are two cases: kerT =V₁ or kerT = {0}. If kerT =V₁ then T = 0. If kerT = {0}, T is injective. Why is T surjective? Well, let's consider the image, I claim that the image is an invariant subspace of V2. Let's suppose thatw∈Im T and w=T(v) then we have

ρ₂(g)(w) =ρ₂(g)T(v)

=T ρ1(g)(v) ∈ImT

because T is injective. So either ImT =0 or Im T =V2. ButIm T = {0} ⇒T = 0. And Im T = v2 ⇒ T is surjective. So it is injective and surjective, thus an isomorphism as required.

For the second, part we will use the fact that every operator on a nite dimensional vector space has an eigenvalue. Assume thatT ≠0 so it is an isomorphism.

Without loss of generality, V1 =V2 i.e. T∶V1 →V1. Let λbe an eigenvalue of T.

3If P² =P andU=ImP thenP =IdonU: letv∈V andv≠0andP v=w≠0say, then P²v=P v=wandP(P v) =P wthereforeP w=w.

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Then consider the operator T−λ⋅Idwhich is an intertwining operator. It has a non-zero kernel

ker(T−λ⋅Id) ≠ {0}

and thus cannot be an isomorphism, so by the rst part is zero, i.e. T=λ⋅Id. We will see on Monday that this result allows us to build all the representation theory over complex numbers in a nice and simple way. For the next week and a half we will only look at representations where the ground eld is the complex numbers, k=C.

Informally, our goal is the following: sinceChas characteristic zero, our theorem from Monday about invariant complements holds. Our strategy is, if we have a complex representation then use this theorem, we will get two complementary subspaces which are invariant, then we apply it again and again until everything is irreducible. This will work since we are working over nite dimensional vector spaces.

Recall the left regular representationkG. Let us view this in the following way:

this vector space has the structure of an algebra over k. The product is given on basis elements in the most straight-forward way as egeh:=egh. Associativity is a consequence of the associativity of the group operation. Then representations of a groupGare in a one-to-one correspondence with left modules over this algebrakG -kG⊗M →M orkG⊗kG⊗M ⇉kG⊗M. kGhas a bilinear form

⎛

⎝∑

g∈G

a_ge_g,∑

g∈G

b_ge_g⎞

⎠:= 1

∣G∣ ∑g∈G

a_gb_g−1

Theorem 4 (Orthogonality relations for matrix elements). Suppose that (V1, ρ1) and(V₂, ρ₂)are irreducible representations of a groupG. Fix a basis{e_i}i=1,...,n of V₁ and{f_i}i=1,...,m of V₂. We have matrices(ρ₁(g))i,j=,1...,n and (ρ₂(g))i,j=,1...,m

representing group elements. Fix i, j between 1 and n and k, l between 1 and m. We have E_ij⁽¹⁾, E_kl⁽²⁾∈kGwith (whereE_ij⁽¹⁾(g)are the matrix elements ofρ₁(g)with respect to the basis {e} and vice versaE_kl⁽²⁾ andρ2(g)and{f})

E⁽¹⁾_ij (g):= ∑

g∈G

ρ1(g)ijeg

E⁽²⁾_kl (g):= ∑

g∈G

ρ₂(g)kle_g Then

(1) (E_ij⁽¹⁾, E_kl⁽²⁾) =0 if V₁≠V₂ (2) (E_ij⁽¹⁾, E_kl⁽¹⁾) = _dim^δ^jk^δ_V^il₁

Lecture 7 8/10 12pm

We will proove the theorem stated at the end of the last class.

We want 1

∣G∣ ∑g∈G

E_ij⁽¹⁾(g)E_kl⁽²⁾(g⁻¹) =0 (7.1)

1

∣G∣ ∑g∈G

E_ij⁽¹⁾(g)E_kl⁽¹⁾(g⁻¹) = δjkδil

dimV₁ =⎧⎪⎪

⎨⎪⎪⎩

1/dimV₁ i=l, j=k

0 otherwise

(7.2)

Proof. We want to proove the above identities for all i, j <n and k, l <m. Let us take the operator T0∶V1 → V2 with T0(ei) =fl and T0(ep) = 0 for p≠ i.

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We want to average over the elements so that the group action commutes, like an intertwining operator.

Dene an operator T by T= 1

∣G∣ ∑g∈G

ρ₂(g)⁻¹T₀ρ₁(g)

Recall, that this averaging operation has the following eect: the operator T now satises

ρ2(g)T=T ρ1(g) for allg∈G

So T is an intertwining operator. SinceV1 andV2 are non-isomorphic, by Schur's lemma, the operator must be zero. So now, let us write down this condition and try to extract some information from it. If this operator is written as a sum of basis elements then we get the zero matrix so we will have a lot of conditions. So we should apply T to a basis vector. Let's consider

ρ1(g)(ej) = ∑ⁿ

k=1

ρ1(g)kjek

⇒T₀ρ₁(g)(e_j) =ρ₁(g)ijf_l

because it is only non-zero on e_i from above. Now we apply the inverse ρ2(g)⁻¹T0ρ1(g)(ej) =ρ2(g⁻¹)(ρ1(g)ijfl)

=ρ₁(g)ij m

∑

k=1

ρ₂(g⁻¹)klf_k

We considered the term ρ2(g)⁻¹T0ρ1(g)(ej). Now we write, using the above T(ej) = 1

∣G∣ ∑g∈G m

∑

k=1

E_ij⁽¹⁾(g)E_kl⁽²⁾(g⁻¹)fk

This is a linear combination offk, so if it is equal to zero, then every coecient must be zero. This gives us precisely the relations (7.1). We can get the second identity by adapting the above argument. First of all take T0(ei) =el and T0(ep) =0 for p≠iand

T= 1

∣G∣ ∑g∈G

ρ1(g)⁻¹T0ρ1(g)

Our conclusion using Schur's lemma is not true now - we cannot conclude that it is zero. Instead it acts like a scalar. We need to see what kind of scalar we get, consider⁴

tr(T) = 1

∣G∣ ∑g∈G

tr(ρ₁(g⁻¹)T₀ρ₁(g))

= 1

∣G∣∣G∣ ⋅tr(T0) =δil

Which is clear from the denition of T₀ given above. If T is a scalar λ then tr(T) =λ⋅dimV₂ so we have

λ= δil

dimV1

Similar to the above we have

ρ1(g)⁻¹T0ρ1(g)(ej) =ρ1(g⁻¹)(ρ1(g)ijfl)

=ρ1(g)ij m

∑

k=1

ρ1(g⁻¹)klfk 4Use the cycle property of traces heretr(abc) =tr(bca).

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and eventually

T(ej) = 1

∣G∣ ∑g∈G m

∑

k=1

E_ij⁽¹⁾(g)E_kl⁽¹⁾(g⁻¹)ek

=λe_j= δ_il dimV1

e_j

which is actually equivalent to (7.2).

Now we have one of the main denitions needed for the classication of irreducible representations of nite groups. In the next class we will discuss applications.

7.1. Characters.

Denition 7. For a representation(V, ρ)ofG, we dene the characterχ_V as the function fromGto the complex numbersχ_V∶G→Cgiven by

χV(g) =trV(ρ(g))

It is clear this does not depend on choice of basis, and only depends on the isomorphism classes of representations. Two basic properties that we will state before the class ends

(1) χV1=χV2 ifV1≃V2 which is obvious.

(2) χ(g⁻¹) =χ(g)for allg∈G.

Previously, we had the alternate proof of the decomposition into irreducibles which only worked for real or complex numbers, and we showed that there always exists an invariant Hermitian scalar product. The idea is that: ifvis an eigenvector ofρ(g), then

λ(v, v) = (ρ(g)(v), v) = (ρ(g)(v), ρ(g)ρ(g⁻¹)(v)) = (v, ρ(g⁻¹)(v))

= (v, λ⁻¹v)

=λ⁻¹(v, v) Soλ= (λ⁻¹). Soχ(g⁻¹) =χ(g).

Lecture 8 8/10 4pm

8.1. Orthogonality relations. The last thing was about characters: for every representation we have χV(g⁻¹) = χV(g). There is another reason why this should hold. Basically if you take the matrixρ(g), then because of one of the basic results of group theory, that is if you raise it to the power of the order of the group ρ(g)^∣G∣, then

ρ(g)^∣G∣=ρ(g^∣G∣) =ρ(1) =Id

The conclusion is that for every element of a nite group, some power of the representative matrix is the identity, thus every representative matrix is diagonalisable.

This can be seen by taking powers of the Jordan normal form - we can't have

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elements outside of the main diagonal otherwise. We have

ρ(g)is represented by

⎛⎜⎜

⎜⎜⎝

χ1 0 . . . 0

0 ⋱ ⋱ ⋮

⋮ ⋱ ⋱ 0

0 . . . 0 χ_n

⎞⎟⎟

⎟⎟⎠

relative to some basis

ρ(g⁻¹)is represented by

⎛⎜⎜

⎜⎜⎝

χ⁻¹₁ 0 . . . 0

0 ⋱ ⋱ ⋮

⋮ ⋱ ⋱ 0

0 . . . 0 χ⁻¹_n

⎞⎟⎟

⎟⎟⎠

=

⎛⎜⎜

⎜⎜⎝

χ1 0 . . . 0

0 ⋱ ⋱ ⋮

⋮ ⋱ ⋱ 0

0 . . . 0 χ_n

⎞⎟⎟

⎟⎟⎠ and the roots of unity all belong to the unit circle⁵∣z∣ =1⇔zz¯=1⇔z¯=z⁻¹. In particular we have

1

∣G∣ ∑g∈G

χV₁(g)χV₂(g⁻¹) = 1

∣G∣ ∑g∈G

χV₁(g)χV₂(g) Since we have

χ_V₁(g) =^dimV∑¹

i=1

E⁽¹⁾_ii (g)

χV2(g) =^dimV∑²

j=1

E⁽²⁾_jj (g) so we conclude that

1

∣G∣ ∑g∈G

χV₁(g)χV₂(g⁻¹) =0 ifV1 is not isomorphic toV2 (V1, V2 irreducible) 1

∣G∣ ∑g∈G

χV₁(g)χV₁(g⁻¹) =1 for an irreducible representationV1

The rst statement is easy because of orthogonality for matrix elements. The second:

1

∣G∣ ∑g∈G

χ_V₁(g)χ_V₁(g⁻¹) = 1

∣G∣ ∑g∈G

∑

j,k

E_jj(g)E_kk(g⁻¹) (j=1. . .dimV₁, k=1. . .dimV₁)

= 1

dimV1

⋅dimV₁=1 Dene for ϕ, ψ∶G→Cthe following

(ϕ, ψ) = 1

∣G∣ ∑g∈G

ϕ(g)ψ(g) Then ifV1, V2are irreducible representations, we have

(χV₁, χV₂) =⎧⎪⎪

⎨⎪⎪⎩

1, V1≃V2

0, V₁/≃V₂ Clearly, if V ≃V1⊕V1⊕ ⋅ ⋅ ⋅ ⊕V1

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

a₁

⊕V2⊕V2⊕ ⋅ ⋅ ⋅ ⊕V2

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

a₂

⊕ ⋅ ⋅ ⋅ ⊕Vn⊕Vn⊕ ⋅ ⋅ ⋅ ⊕Vn

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

a_n

where we assume that theViare pairwise non-isomorphic irreducible representations, then we have

χV =a1χ1+ ⋅ ⋅ ⋅ +anχn

5Since the group has nite order, the eigenvalues must have nite order, so they must be roots of unity etc.

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and (χ_V, χ_V_i) =a_i. Furthermore, let us consider the representation (V =CG, ρ_l) (the left regular representation) and also W some irreducible representation. We will compute

(χ_CG, χW)

The character of the left-regular representation is very easy to compute:

ρ_l(g)e_h=e_gh

If g ≠ 1, all diagonal entries of ρ_r(g) are zero⁶, so χ_C_G(g) =0. The case of the identity element is easy since it goes to the identity matrix, so the value is just χ_C_G(1) = ∣G∣.

(χ_CG, χW) = 1

∣G∣ ∑g∈G

χ_CGχW(g)

= 1

∣G∣χ_CG(1)χW(1) =dimW So we have proven something rather important here.

Theorem 5. The decomposition of the left regular representation into a sum of irreducibles contains a copy of every irreducible representation with multiplicity equal to the dimension.

This is a very powerful result, because up to now we had no reason to assume there were even nitely many non-isomorphic representations. Let us look at an example: takeG=S3, we have∣G∣ =6. Our theorem tells us there are no irreducible representations of dimension three - if there were then we would have three copies of a three dimensional space inside a six dimensional space. If G has two non- isomorphic two-dimensional irreducible representation sayU1, U2, thenCGcontains U1⊕U1⊕U2⊕U2 so dim≥8a contradiction.

We know one two-dimensional representationU, soCG≃U⊕U⊕V1⊕ ⋅ ⋅ ⋅ ⊕Vm

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

1d. irreducible

and 6=2+2+mso m=2, that is, there are two non-isomorphic irreducibles. The trivial representation is an example of a one-dimensional representation. Another is the sign representation - even permutations are mapped to 1 and odd permutations are mapped to -1 i.e. ρ(σ) = sgn(σ). Then our claim is that these are all the irreducible representations ofS3 and hence we have classied all representations of S3.

In conclusion,S3has three irreducible representations (up to isomorphism): the trivial representation, the sign representation and the representation by symmetries of a triangle.

In principle, the above theorem is enough to determine all irreducible representations, but we would prefer to have some more results.

One simple observation is that characters are invariant under conjugation χ_V(ghg⁻¹) =χ_V(h) g, h∈G

This is simple because

trV(ρ(ghg⁻¹)) =trV(ρ(g)ρ(h)ρ(g)⁻¹)

=tr_V(ρ(h)) =χ_V

In fact this gives us an upper bound on the number of irreducible representations:

it is bounded by the number of conjugacy classes. We will proove this next time and show that this bound is sharp.

6It is clear from the way thatρl(g)acts that the diagonal entries are either 1 or 0. If it was 1 then there would be someek that would get mapped toek, which is only possible ifg=1.

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Lecture 9 11/10

9.1. The number of irreducible representations. Last time we discussed some things about characters of representations.

Theorem 6. The number of non-isomorphic irreducible representations of a nite group GoverCis equal to the number of conjugacy classes ofG.

Proof. For any character χ_V of a representationV we have χV(ghg⁻¹) =χV(h)

(9.3)

Functions that satisfy equation (9.3) are called class functions. We shall proove that characters of irreducibles form a basis in the space of class functions. Since the dimension of the space of class functions is clearly equal to the number of conjugacy classes, this will complete the proof. We have

χ_V(ghg⁻¹) =tr_V(ρ(ghg⁻¹)) =tr_V(ρ(g)ρ(h)ρ(g)⁻¹) =tr_V(ρ(h)) =χ_V(h) It is harder to proove the part about bases. We will need to use the following lemma Lemma 7. Let (V, ρ)be a representation ofG,ϕ∶G→Cbe a class function, then

Tϕ,ρ= ∑

g∈G

ϕ(g)ρ(g) is an intertwining operator.

Proof.

ρ(h)T_ϕ,ρρ(h)⁻¹= ∑

g∈G

ρ(h)ϕ(g)ρ(g)ρ(h)⁻¹

= ∑

g∈G

ϕ(g)ρ(hgh⁻¹)

= ∑

g∈G

ϕ(hgh⁻¹)ρ(hgh⁻¹)

= ∑

g^′∈G

ϕ(g^′)ρ(g^′) =Tϕ,ρ In particular, if V is irreducible, then by Schur's lemma, this operatorTϕ,ρis a scalarλso

dimV ⋅λ=tr(Tϕ,ρ)

= ∑

g∈G

ϕ(g)trρ(g) = ∑

g∈G

ϕ(g)χ_V(g)

= ∣G∣ ⋅ (ϕ,χ¯V)

⇒λ= ∣G∣

dimV(ϕ,χ¯_V)

Now, suppose that characters of irreducibles do not span the space of class functions. Then their complex conjugates do not span the space either. Therefore there exists a non-zero class function orthogonal to all χ¯V for irreducible V with respect to (,). Denote it by ψ. We haveTψ,ρ =0 for irreducible representations (V, ρ). Because of our formula forTϕ,ρit follows thatTψ,ρ=0for all representations (V, ρ).

Our objective is to show Tψ,ρ is zero to get a contradiction. Let us consider (CG, ρ_l), then

0=Tψ,ρ_l= ∑

g∈G

ψ(g)ρl(g)

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In particular we have

0=Tψ,ρ(e1) = ∑

g∈G

ψ(g)ρl(g)(e1) = ∑

g∈G

ψ(g)eg

Therefore ψ = 0, because the eg are linearly independent. Since characters of irreducibles are orthogonal they are linearly independent, therefore they form a

basis of the space of class functions.

Let us now summarise our knowledge for the specic case of S3 which we discussed on Monday. We know that it has three irreducible representations. So it has three conjugacy classes: the three cycles, the two cycles (transpositions) and the identity. So let us write down the character table forS3.

0 Id (12),(13),(23) (123),(132)

Trivial 1 1 1

Sign 1 -1 1

2d. U 2 0⁷ 2cos(^2π₃ ) = −1

(χ_triv, χ_sign) = 1

6(1⋅1+3⋅1(−1) +2⋅1⋅1) The middle term above is

χ_triv(12)χ_sign(12) +χ_triv(13)χ_sign(13) +χ_triv(23)χ_sign(12) We also have

(χ_triv, χ_U) = 1

6(1⋅2+3⋅1⋅0+2⋅1⋅ (−1))

(these are weighted sums: the middle term is multiplied by three because there are three terms in that conjugacy class etc.) and

(χ_U, χ_U) = 1

6(2⋅2+3⋅0⋅0+2⋅ (−1) ⋅ (−1))

Lecture 10 15/10 12pm

The goal today is to show how characters can also be used to decompose a given representation into a direct sum of irreducible representations. We know this is true in principle, but we do not have a direct method for doing this.

Aside: suppose thatV, W are two representations of a nite groupG. As usual all our representations are overC, unless stated otherwise. Then of course we know that they can be decomposed into direct sums of irreducibles

V ≃ ⊕iV_i^⊕aⁱ W ≃ ⊕iV_i^⊕bⁱ

This is sort of like decomposing integers into primes. {Vi} is a list of irreducible representations of G. But we have not shown that these numbers ai, bi are well- dened - that is, we have not shown that this decomposition is unique. The analogy is like the fundamental theorem of arithmetic, that decomposition into a prime factorisation is unique. To proove this, we use the orthogonality of characters

(χ_V, χ_V_i) =a_i

(χV_i, χV_j) =δij (orthogonality)

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More generally, if we compute the scalar product ofV, W (χV, χW) = ∑

i

aibi

Our criterion for irreducibility is

V is irreducible⇔ (χV, χV) =1

Proof. We haveV ≃ ⊕V_i^⊕aⁱ then(χV, χV) = ∑ⁱa²_i. Now, given some arbitrary representation, it is dicult to determine whether a representation is irreducible by hand. But the above criterion makes it very easy.

This alone might be enough to highlight the importance of characters.

It is possible to view the above formulas in a more conceptual way. We note that the characters are integers. There is a philosophy in mathematics called cate- gorication: when we have integers and there's no particular reason to assume that they should all be integers, then we can interpret these integers as dimensions.

Theorem 8. If V ≃ ⊕V_i^⊕aⁱ andW ≃ ⊕V_i^⊕bⁱ then (χV, χW) = ∑

i

aibi

=dim HomG(V, W)

where HomG(V, W)is the space of intertwining operators betweenV andW. Remark. This can be interpreted as an alternative way to view Schur's lemma.

Proof. Linear operators between V andW are blockN×N matrices

A=

⎛⎜⎜

⎜⎜⎝ Aij

⎞⎟⎟

⎟⎟⎠ and A_ij is itself a block matrix

A_ij=

⎛⎜⎜

⎜⎜⎝ T_pq^ij

⎞⎟⎟

⎟⎟⎠

(where T_pq^ij is a matrix of a linear operator from thep^th copy ofVi to theq^th copy ofVj) and of course group elementsg∈Gacting onV, W are represented by block diagonal matrices. The intertwining condition is

ρ_W(g)A=Aρ_V(g) for allg∈G

Recall that the multiplication rules for block matrices imply that all Tpqare intertwining operators iArepresents an intertwining operator.

(Here is an example to illustrate this (r1 0

0 r2) (a b

c d) = (r1a r2b r1c r2d) (a b

c d) (s1 0

0 s2) = (as1 bs2

cs1 ds2) and we equate these.)

ThereforeArepresents an intertwining operator(s) iT_pq^ij=0fori≠jandT_pq^ijis a scalar wheneveri=jby Schur's lemma. Aiidepends onai×biparameters (because p, q range overai, bi values respectively), sodim HomG(V, W) = ∑ⁱaibi.

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Let us now describe how to decompose a representation into invariant subspaces using characters in a canonical fashion.

10.1. Canonical decomposition of a representation. SupposeV ≃U⊕U where U is irreducible. We can say that the rst factor contains all vectors of the form(u,0)and the second, all vectors(0, u). Alternatively we can takeU1= {(u,0)}

andU2= {(u, u)}. These obviously have zero intersection, soU1, U2are invariant, so V =U1⊕U2. So in general the direct sum decomposition cannot be unique. However, what is true is that if we x some irreducible representation and take the direct sum of all invariant subspaces isomorphic to that space, then the decomposition will be well-dened. That is, if V ≃ ⊕iV_i^⊕aⁱ then for each i,V_i^⊕aⁱ is a well-dened subspace, i.e. it does not depend on the choice of decomposition of V. What we shall proove in the next class is the following

Theorem 9. Let us dene (for V a representation ofG) p_i=dimVi

∣G∣ ⋅ ∑

g∈G

χ_V_i(g)ρ_V(g) then p²_i =p_i,p_i∈Hom_G(V, V)andImp_i=V_i^⊕aⁱ

Lecture 11 15/10 4pm We will proove the theorem from the last class.

Proof. First of all, let us proove that pi ∈ HomG(V, V) i.e. that pi is an intertwining operator. Forχ_i (χi∶=χV_i) then

T_χ

i,ρV = ∑

g∈G

χ_i(g)ρ_V(g) (11.4)

is an intertwining operator, which we prooved last week. If V is an irreducible representation, then

T_χ

i,ρV = ∣G∣

dimV(χ_V_i, χ_V)

So, it is equal to zero if V_i /≃ V and equal to ∣G∣/dimV if V ≃ V_i, under the assumption of irreducibility, but of course in general then (11.4) tells us thatT_χ

i,ρV

is equal to∣G∣/dimV_i on each constituentV_i and equal to zero otherwise. Then pi= dimVi

∣G∣ Tχ_i,ρ_V

is equal to 1 on each irreducible constituent isomorphic to Vi, and to zero on all constituentV_j forj≠i. Now parts (1) to (3) follow.

In particular, from this formula, we can extract a sum of all the copies of the trivial representation. If we consider the operator

p= 1

∣G∣ ∑g∈G

ρV(g)

is the projector on the subspace of invariant vectors inV (ρV(g)v=v).

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11.1. Further decomposition into irreducibles. Our set-up is that V =

⊕iV⁽ⁱ⁾ with V⁽ⁱ⁾ ≃ V_i^⊕aⁱ. So V⁽ⁱ⁾ = Im p_i constructed in the previous theorem.

Now let us assume that we know irreducibles Vi such that we know the matrices ρ_i(g) = (E_kl⁽ⁱ⁾(g)) wherek, l=1, . . . , n_iwheren_i=dimV_i. Now x one of theiand dene the operators

P_kl⁽ⁱ⁾=dimVi

∣G∣ ∑g∈G

E_lk⁽ⁱ⁾(g⁻¹)ρV(g)

(note the form of the matrix element in the sum - it is the transpose inverse) Theorem 10. (1) P_kk⁽ⁱ⁾ are projectors, (P_kk⁽ⁱ⁾)² = P_kk⁽ⁱ⁾ (but not intertwining

operators). The subspaces V^(i,k) := Im P_kk⁽ⁱ⁾ ⊂ V⁽ⁱ⁾ for k =1, . . . , n_i and V⁽ⁱ⁾= ⊕^m_k=1^jV^(i,k) (but not invariant).

(2) P_kl⁽ⁱ⁾ is zero onV^(j), forj≠iandImP_kl⁽ⁱ⁾∣V^(i,l)⊂V^(i,k) andP_kl⁽ⁱ⁾∶V^(i,l)→ V^(i,k) is an isomorphism of vector spaces.

(3) Take x1 ≠ 0 ∈ V^(i,1) and let xk = P_k1⁽ⁱ⁾(x1) ∈ V^(q,k). Then V(x1) = span(x1, . . . , xn)is an invariant subspace where the representation V1 is realised. Finally, if x11, x12, . . . , x1m is a basis ofV^(i,1), then

V⁽ⁱ⁾=V(x11) ⊕V(x12) ⊕ ⋅ ⋅ ⋅ ⊕V(x1m)

is a decomposition of V⁽¹⁾ into a direct sum of irreducibles.

Remark. The nal statement says that nding irreducible subspaces is as easy as nding bases.

Proof. First of all, P_kl⁽ⁱ⁾ is dened for anyV, and to compute them, we shall rst consider an irreducible V, and then we use linearity in V. Suppose V is irreducible, and let f₁, . . . , f_r be a basis ofV and suppose that ρ_V(g) = (E_pq(g)) with respect to {f_j}, then

P_kl⁽ⁱ⁾(fs) = dimVi

∣G∣ ∑g∈G

E_lk⁽ⁱ⁾(g⁻¹)ρV(g)(fs)

= dimV_i

∣G∣ ∑g∈G

∑

t

E_lk⁽ⁱ⁾(g⁻¹)E_ts(g)f_t

=⎧⎪⎪⎪⎪⎪

⎨⎪⎪⎪⎪⎪⎩

0 ifV /≃V_i

fk ifl=sandV ≃Vi

0 ifl≠sandV ≃V_i

Lecture 12 18/10

First, we will nish that disastrously technical result we started last time. Recall we were considering the operators

P_kl⁽ⁱ⁾= dimV₁

∣G∣ ∑g∈G

E_lk⁽ⁱ⁾(g⁻¹)ρ(g)

Now, assume that V is irreducible and letf1, . . . , fk be a basis then

P_kl⁽ⁱ⁾(fr) =⎧⎪⎪⎪⎪⎪

⎨⎪⎪⎪⎪⎪⎩

0 V /≃V_i

fk V ≃Vi andl=r 0 V ≃Vi andl≠r

MA3413: Group representations Lecturer: Prof. Vladimir Dotsenko Michaelmas term 2012

MA3413: Group representations Lecturer: Prof. Vladimir Dotsenko

Michaelmas term 2012

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