Schur-Weyl duality

These notes are from a graduate student reading group at the University ofUtah happening during the Fall semester 2018.

Contents

1 Introduction 2

2 Modules and representations of groups 32.1 Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Group representations . . . . . . . . . . . . . . . . . . . . . . . . 4

3 Linear algebra 63.1 Tensor products . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3.1.1 Application to representations . . . . . . . . . . . . . . . 83.2 Symmetric products . . . . . . . . . . . . . . . . . . . . . . . . . 103.3 Exterior products . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.4 Determinant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

4 Character theory 134.1 Decomposition of Representations . . . . . . . . . . . . . . . . . 144.2 Character Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4.2.1 Properties of χ . . . . . . . . . . . . . . . . . . . . . . . . 154.2.2 More Properties of χ . . . . . . . . . . . . . . . . . . . . . 16

5 Representations of the symmetric group 16

6 Double Centralizer Theorem 24

7 Schur-Weyl duality 27

8 Determinental rings and applications to commutative algebra 298.1 Determinantal Rings Are Cohen-Macaulay . . . . . . . . . . . . 308.2 Relation to Representation Theory . . . . . . . . . . . . . . . . . 31

9 Decomposing tensor products of Weyl modules 33

1

1 Introduction

Talk by Adam Brown, notes by Sabine Lang

We give here a brief outline of the topics that will be covered during the semester,along with some motivations behind Schur-Weyl duality. The basic idea comesfrom Lie groups: we want to find all the representations of these Lie groups. Inour case, we will focus on only one Lie group: the general linear group.

For V = Cn, we define G = GL(V ) and want to find all the representationsof G, i.e., the continuous homomorphisms π : G → GL(W ) for W a complexvector space. We also say that G acts on W in this case. As a starting point,we know one representation: the canonical or standard representation, definedby Id : G→ GL(V ). Can we use this to construct new representations?

Let us first analyze the operations that we can do with vector spaces. Thedirect sum of V with itself does not give a new representation: G → GL(V ⊕

V ) is given by g →(g 00 g

)which corresponds to two copies of the standard

representation. A more interesting operation is the tensor product: G acts onV ⊗ V by g · (v1 ⊗ v2) = (g · v1)⊗ (g · v2). How can we decompose V ⊗ V intoG-invariant subspaces?

A first subspace which is preserved by the action of G is the space of alternat-ing vectors Λ2V = 〈{v1⊗v2 | v1⊗v2 = −v2⊗v1}〉. Can we decompose V ⊗V asΛ2V ⊕W ? Yes! And W is then equal to Sym2V = 〈{v1⊗v2 | v1⊗v2 = v2⊗v1}〉and is also preserved by the action of G. We obtain V ⊗ V = Λ2V ⊕ Sym2V.

Can we generalize this idea? We can consider

k︷ ︸︸ ︷V ⊗ · · · ⊗ V = V ⊗k. However,

the decomposition that we had for k = 2 is no longer as simple. For k = 3already, V ⊗3 = Λ3V ⊕ Sym3V ⊕W with W 6= 0. Therefore, one of our goals isto decompose V ⊗k into G-invariant subspaces.

Let us consider the symmetric group Sk = { bijections from Fk to Fk}, whereFk is a set with k elements. We have an action of Sk on Fk by σ · x = σ(x) forσ ∈ Sk, x ∈ Fk. However, Fk is a set and not a vector space, so this is not arepresentation. But we can construct a vector space E = {f : Fk → C}, whichhas an action of Sk given by (σ · f)(x) = f(σ−1(x)). This is a representation ofSk.

Going back to V ⊗k, we can use it to construct a representation of Sk: wehave an action of the symmetric group on V ⊗k given by σ · (v1 ⊗ · · · ⊗ vk) =vσ−1(1) ⊗ · · · ⊗ vσ−1(k) on simple tensors, and extended linearly. Now, we can

try to decompose V ⊗k into Sk-invariant subspaces.For example, when k = 2, the group S2 has two elements,

(1 2

)and the

identity. If v1 ⊗ v2 ∈ Λ2V , then(1 2

)· v1 ⊗ v2 = v2 ⊗ v1 = −v1 ⊗ v2.

Therefore, S2 acts on Λ2V by the sign of the permutation, and we can decomposeΛ2V into copies of the sign representation. Now if v1 ⊗ v2 ∈ Sym2V, then(1 2

)· v1 ⊗ v2 = v2 ⊗ v1 = v1 ⊗ v2, and Sym2V decomposes into copies of

the trivial representation. We conclude that V ⊗ V = Λ2V ⊕ Sym2V is a

2

decomposition into invariant subspaces, both for the G-action and for the S2-action.

This turns out to be true in general. This is due to the fact (to be provenlater) that the actions of Sk and G commute: for σ ∈ Sk, g ∈ G we haveσ · (g · (v1 ⊗ · · · ⊗ vk)) = g · (σ · (v1 ⊗ · · · ⊗ vk)). Therefore, the decompositionsinto invariant subspaces should agree. This leads to the main theorem of Schur-Weyl duality:

Theorem (Schur-Weyl duality). V ⊗k decomposes both as ⊕π∈Irr(Sk)Eπ and as⊕φ∈Irr(GL(V ))Uφ.

We are in particular interested in using Sk (it is a finite group, and there is acombinatorial approach to its representations) to understand GL(V ). In conclu-sion, if we have a canonical representation V of G = GL(V ) and an irreduciblerepresentation π of Sk, then we can construct an irreducible representation ofG. More precisely, we can get all polynomial representations of G that way. Wecan realize this using the Schur functor

Sπ : V 7→ HomSk(Eπ, V⊗k),

and HomSk(Eπ, V⊗k) is an irreducible representation of G when dim(V ) = k or

0.This duality has applications to any object with both a GL(V ) and an Sk

action. For example, for a field k, we can define k[

(x yw z

)] (think of k[x], but

we add a matrix instead of a variable x). Then k[

(x yw z

)]/(xz − yw) has a

natural GL(V )-action. We can use Schur-Weyl duality to decompose it, andcompute the syzygies.

2 Modules and representations of groups

Talk by Cameron Zhao, notes by Peter McDonald

Today we are going to introduce some fundamental tools of linear algebra thatwe will be using later in the semester.

2.1 Modules

Definition 2.1.1. Let R be an associative algebra with unity. An R-module isan abelian group M with an action of R on M satisfying

1. The action preserves addition and multiplication in R and addition in M ,i.e., (rs+ t) · (a+ b) = r(s · a) + t · a+ r · (s · b) + t · b for r, s, t ∈ R anda, b ∈M

2. 1R · a = a for all a ∈M

3

Example 2.1.2. Abelian groups are Z-modules

Example 2.1.3. Vector spaces over a field k are k-modules.

Example 2.1.4. Consider the vector space kn for a field k. This is a moduleover the matrix algebra Mn(k).

Example 2.1.5. Take A ∈ Mn(k). Then kn can be viewed as a k[x]-modulewhere x acts as A. This gives us a lot of information about A: rational andJordan canonical forms, minimal polynomial, etc.

Example 2.1.6. If R is an algebra, then R is a left R-module. The submodulesof R are precisely the left ideals of R.

Example 2.1.7. Let R be a commutative ring and M a left R-module. ThenM is also a right R-module where the action is given by a · r = r · a for r ∈ Rand a ∈M . Then M is a bi-module.

Remark 2.1.8. If R is not commutative then the above is not well-definedbecause

a · (rs) = (a · r) · s = (r · a) · s = s · (r · a) = (sr) · a 6= (rs) · a

2.2 Group representations

Definition 2.2.1. Let G be a group. A representation of G is a group homo-morphism p : G→ GL(V ) for some vector space V .

Group representations are precisely modules over a certain algebra, so we canuse the tools to study modules to study representations. What is this algebrathough?

Definition 2.2.2. The group algebra of G over a field k, denoted k[G], is con-structed as the k-vector space spanned by the basis G, i.e., k(G) = spank(G).Then, every element looks like ∑

g∈Gcg · [g].

Multiplication is defined on basis elements by

[g1][g2] = [g1g2],

and extended linearly.

Example 2.2.3. C[Z/3] = {c0[0] + c1[1] + c2[2] : ci ∈ C}.

(2[0] + [1])(i√

2[1] + 3[2]) = 2i√

2[0][1] + 6[0][2] + i√

2[1][1] + 3[1][2]

= 2i√

2[0 + 1] + 6[0 + 2] + i√

2[1 + 1] + 3[1 + 2]

= 3[0] + 2i√

2[1] + (6 + i√

2)[2]

We can show this is an associative algebra with identity because the identity is1k[e] where e ∈ G is the identity element of the group.

4

Proposition 2.2.4. Representations of the group G over the field k are k[G]-modules.

Proof. If we have a representation ρ : G → GL(V ) then it extends linearly toρ̃ : k[G]→ Endk(V ). Then V is a k[G]-module.

Consider the following example of how k[G] acts on V . Consider how c1g1 +c2g2 ∈ k[G] acts on v ∈ V :

(c1g1 + c2g2) · v = ρ̃(c1g1 + c2g2) · v= c1ρ(g1) · v + c2ρ(g2) · v= ρ̃(c1g2) · v + ρ̃(c2g2) · v= (c2g2) · v + (c2g2) · v

Now, given a k[G]-module V , we get a ring homomorphism ρ̃ : k[G] →Endk(V ) where ρ̃ maps groups elements of G to invertible transformations:

ρ̃([g])ρ̃([g−1]) = ρ̃([g][g−1]) = ρ̃(e) = Id.

Because ρ̃ preserves multiplication, ρ = ρ̃|G is a group homomorphism. Thenρ : G→ GL(V ) is a representation.

Essentially, instead of thinking of G acting on a set, we are thinking aboutG as giving us functions on our set. Now we need to make sure that k[G]-homomorphisms are the same as homomorphisms on representations.

Definition 2.2.5. A homomorphism of representations ϕ : V → W is a k-linear map that commutes with the group action, i.e., the following diagramcommutes:

Vϕ−−−−→ W

g

y g

yV

ϕ−−−−→ W

This is also called a G-equivariant map.

Definition 2.2.6. An R-module homomorphism ϕ : M → N is a Z-linear mapthat commutes with the ring action, i.e., r · ϕ(m) = ϕ(r · m) for r ∈ R andm ∈M . This is also called an R-linear map.

So a k[G]-linear map ϕ : V →W is one such that for all∑g∈G

cg[g] ∈ k[G] the

following diagram commutes:

Vϕ−−−−→ W∑

g∈Gcg[g]

y ∑g∈G

cg[g]y

Vϕ−−−−→ W

5

Take 1[e] = [e], then considering the following diagram we can see that ϕ isk-linear:

Vϕ−−−−→ W

[e]

y [e]

yV

ϕ−−−−→ W

Furthermore, if you take [g], then ϕ becomes a homomorphism betweenrepresentations, so a k[G]-module homomorphism is also a homomorphism ofrepresentations:

Vϕ−−−−→ W

g

y g

yV

ϕ−−−−→ W

3 Linear algebra

Talk by Cameron Zhao, notes by Cameron Zhao

(Some of the contents in this section were skipped in the talk).

3.1 Tensor products

Recall from last time: we showed that k[G]-modules are the same as representa-tions of G, and that k[G]-module maps are the same as homomorphisms betweenrepresentations. In other words, the category of k[G]-modules is isomorphic tothe category of representations of G.

The direct product M×N of two modules is again a module, where r(m,n) =(rm, rn). It is also called the direct sum, denoted M ⊕N (note: direct sumsand direct products of modules only differ when infinitely many modules areconsidered). The idea is that when we take the direct sum of two vector spaces,the resulting dimension is the sum of the dimension of the two spaces. Soone can try to construct a vector space whose dimension is the product of twosmaller spaces. A natural choice of basis on such a space is {(ei, fj)}. This isthe tensor product.

This can be done for general modules. It should be an abelian group satis-fying some conditions. So to do this, we take elements in the free abelian groupgenerated by M ×N and impose the relations we want:

Definition 3.1.1. Let M be a right R-module and N a left R-module. Thetensor product is defined to be M ⊗R N := F (M × N)/I, where F (M × N)is the free abelian group generated by M × N , and I is the ideal generated byelements of the form

(m1 +m2, n)− (m1, n)− (m2, n),

6

(m,n1 + n2)− (m,n1)− (m,n2),

(mr, n) = (m, rn).

The image of (m,n) in the quotient is denoted by m⊗ n. So M ⊗R N consistsof elements of the form

∑imi ⊗ ni subject to the relations

(m1 +m2)⊗ n = m1 ⊗ n+m2 ⊗ n,

m⊗ (n1 + n2) = m⊗ n1 +m⊗ n2,

mr ⊗ n = m⊗ rn.

An element of the form m⊗n is called a simple tensor or a rank one tensor.So M ⊗R N consists of R-linear combinations of simple tensors.

The most important property of the tensor product is its universal property.Many characterizations of the tensor product can be deduced from it.

Definition 3.1.2. A map ϕ : M ×N → L is called R-balanced if it is Z-linearin both arguments and ϕ(mr, n) = ϕ(m, rn).

Theorem 3.1.3 (Universal Property of Tensor Product). Let R be an algebrawith 1, M a right module, N a left module, and L an abelian group. Then wehave a 1-1 correspondence correspondence

{R-balanced maps M ×N ϕ−→ L} ←→ {group homomorphisms M ⊗R Nφ−→ L}

ϕ 7→ φ,where φ(m⊗ n) = ϕ(m,n)

such that the following diagram commute:

M ×N M ⊗R N

L

ϕ

ι

φ

where ι(m,n) = m⊗ n.

Proof. Given φ, let ϕ = φ◦ ι, then ϕ is R-balanced because ι(mr, n) = mr⊗n =m ⊗ rn = ι(m, rn). Given ϕ, it extends to a group homomorphism ϕ̃ : F (M ×N)→ L from the free abelian group F (M ×N). We want it to factor throughM ⊗R N , so that it gives a map φ : M ⊗R N → L. Indeed, the generators ofthe ideal defining the tensor product vanish under ϕ̃.

The universal property can be understood in the following way: we wantto study bilinear maps, but bilinear maps are not module maps. The universalproperty states that using the tensor product we can encodes all information inthe bilinear map in a module map.

A commutative version of this is: bilinear maps M × N → L are the sameas linear maps M ⊗R N → L.

Using this we can prove many nice properties of the tensor product:

7

Proposition 3.1.4.

1. (Extension of scalars) If M is a left R-module and S is an R-algebra, thenS ⊗RM is the “smallest” S-module containing M .

2. (Tensor products are associative) (M ⊗R N)⊗R L ∼= M ⊗R (N ⊗R L) forany left module M , bimodule N and right module L.

3. (Tensor products are commutative) If R is commutative, then M ⊗RN ∼=N ⊗RM .

4. (Tensor product distributes with direct sums) (M ⊕ N) ⊗R L ∼= (M ⊗RL)⊕ (N ⊗R L).

5. (Vector spaces) If k is a field and V,W are vector spaces with bases{ei}, {fj} respectively, then V ⊗kW is a vector space with basis {ei⊗ fj},so dim(V ⊗W ) = (dimV ) · (dimW ).

Intuitively, when tensoring over fields, the product is not “shrunk” by rela-tions. But over general algebras the product will “shrink”. For example:

Example 3.1.5. Let R be a commutative ring, I ⊂ R an ideal, M a module.Then R/I ⊗R M ∼= M/IM . So if J is another ideal, then R/I ⊗R R/J ∼=(R/J)/(I/J) ∼= R/(I + J).

If we take a bimodule M we can tensor it with itself over and over again.Since tensor products are associative, we have formed the tensor powers M⊗n.We can sum up all the tensor powers and form the tensor algebra

T (M) = R⊕M ⊕M⊗2 ⊕ · · ·

where the multiplication is given by concatenation of tensors. This is an asso-ciative algebra with 1, and is the “largest” associative algebra containing M .Note: Countable direct sum is defined to be

∞⊕i=1

Mi := {(m1,m2, . . .) | mi ∈Mi, only finitely many entries are nonzero},

whereas countable direct product is

∞∏i=1

Mi := {(m1,m2, . . .) | mi ∈Mi},

without the finiteness assumption.

3.1.1 Application to representations

If A : M1 →M2, B : N1 → N2 are linear maps, then we can construct the tensorproduct A⊗B : M1 ⊗N1 →M2 ⊗N2,m⊗ n 7→ A(m)⊗B(n). So if A′, B′ arealso linear maps, then (A ⊗ B) ◦ (A′ ⊗ B) = (A ◦ A′) ⊗ (B ◦ B′). Over fields

8

and finite dimensional vector spaces, A,B can be represented as matrices. Thematrix of A⊗B is a11B a12B · · · a1lB

......

...as1B as2B · · · aslB

,assuming A = (aij)i=1,...,s,j=1,...,l. In particular if A is s × s, B is l × l, thentr(A⊗B) = trA trB, det(A⊗B) = (detA)s(detB)l.

If ρ : G → GL(V ), σ : G → GL(W ) are two representations, then for anyg ∈ G, ρ(g)⊗ σ(g) is an invertible transformation on V ⊗kW . Let (ρ⊗ σ)(g) =ρ(g)⊗ σ(g), then ρ⊗ σ : G→ GL(V ⊗k W ) is another representation. We willsimply write V ⊗W .

(Note: V ⊗k W is different from V ⊗k[G] W ! For example if V = Vtrv isthe trivial representation, then Vtrv ⊗k W ∼= W , but Vtrv ⊗k[G] W ∼= WG :=W/〈gw − w | g ∈ G〉.)

Similarly, if σ : H → GL(W ) is a representation of another group, thenρ ⊗ σ : G ×H → GL(V ⊗k W ) is a representation of G ×H. This is denotedby V �W .

Now we look at Hom sets. Let N∗ := HomR(N,R) be the dual space of N .

Proposition 3.1.6. Let M be a free module, i.e., M ∼= R⊕m for some m. Thereis a Z-linear isomorphism N ⊗RM∗ → HomR(M,N), (n⊗f) 7→ (m 7→ nf(m)).If R is commutative, then it is also R-linear.

Proof. For simplicity, we prove it for the case when R = k is a field. Any linearmap A : V → W is uniquely determined by its value on basis elements. Ifspan{ei} = V , span{fj} = W , then A is uniquely determined by the coefficientcij ’s where Aei =

∑j cijfj . So A is the image of

∑ij cijfj ⊗ e∗i .

This is very useful. Recall that any linear map A : V → W induces a mapA∗ : W ∗ → V ∗, f 7→ f ◦ A. If V,W are finite dimensional, then the matrix ofA∗ is the transpose of the matrix of A. If ρ : G → GL(V ), σ : G → GL(W )are representations, then ρ∗ : G→ GL(V ∗), g 7→ −◦ρ(g−1) is a representation.Therefore Homk(V,W ) = W ⊗k V ∗ is also a representation. The link of thisto characters is that dimk(Homk(V,W )G) = 〈χV , χW 〉, where Homk(V,W )G isthe invariant subspace of Homk(V,W ) consisting of the elements that are fixedunder all g ∈ G.

We mentioned that extension of scalars can be done using tensor products.If H ≤ G is a subgroup, then k[H] ⊂ k[G] is a subalgebra. So if V is anH-module, then we can extend the scalars to k[G] by taking the tensor product.

Definition 3.1.7. Let H ≤ G be a subgroup, and V be an H-module. The in-duced representation from H to G obtained from V is defined to be IndGH V :=k[G] ⊗k[H] V . The coinduced representation from H to G obtained from V

is CoindGH V := HomH(k[G], V ).

This definition is much cleaner than the one that does not use tensor product.An important fact about induction is the Frobenius reciprocity, which is a directcollorary of the following general fact:

9

Proposition 3.1.8 (Adjoint pairs). Given a right R-module M , an (R,S)-bimodule N and a right S-module L, there is a unique abelian group isomorphism

η : HomS(M ⊗R N,L) ∼−−→ HomR(M,HomS(N,L)),

given by

η(f)(m)(n) = f(m⊗ n), for f ∈ HomS(M ⊗R N,L),m ∈M,n ∈ N.

Pictorially

M M ⊗R N

HomS(N,L) L

η(f) f

As a result,

Corollary 3.1.9. If H ≤ G is a subgroup, W is a G-module, V is an H-module,then

HomH(V,ResGHW ) ∼= HomG(IndGH V,W ).

Proof. We only need to show that ResGHW = HomG(k[G],W ). In general, if

M is an R-module, then HomR(R,M) ∼−→M,f 7→ f(1). So HomG(k[G],W ) ∼=W as vector spaces. The action of H on k[G] makes HomG(k[G],W ) an H-module.

3.2 Symmetric products

In general m ⊗ m′ 6= m′ ⊗ m in M ⊗R M . But we can make them equal bytaking a quotient.

Definition 3.2.1. Let R be commutative, and let M be an R-module. Thesymmetric power SymmM is the quotient of the tensor power M⊗m by thesubmodule generated by elements of the form

m1 ⊗ · · · ⊗mi ⊗mi+1 ⊗ · · · ⊗mm −m1 ⊗ · · · ⊗mi+1 ⊗mi ⊗ · · · ⊗mm.

So in SymmM the tensor factors commute. We write m1 · · ·mm instead ofm1⊗ · · · ⊗mm. There is also a universal property for symmetric powers. Manyproperties of the symmetric powers can be proved using this.

Theorem 3.2.2 (Universal Property). Let M,N be R-modules. Then sym-metric multilinear maps Mm → N are the same as R-module homomorphismsSymmM → N .

Proposition 3.2.3. Let V be a vector space over k with basis {e1, . . . , en},then {ei1 · · · eim | 1 ≤ ij ≤ ij+1 ≤ n} is a basis of Symm V . As a result,dimk Symm V =

(n+m−1

m

). We also have

Symm V ∼−−→ k[x1, . . . , xn]m, ei 7→ xi

where k[x1, . . . , xn]m is the degree m part of the polynomial ring.

10

There is another way to construct symmetric tensors. For any m1⊗· · ·⊗mm,simply sum up all of its permutations. Then we get

∑σ∈Sm mσ(1)⊗· · ·⊗mσ(m),

which is clearly invariant under permutations. As one can expect,

Proposition 3.2.4. If M ∼= Rl is a free module, then we have an injection

SymnM ↪→M⊗n, m1 · · ·mn 7→∑σ∈Sn

mσ(1) ⊗ · · · ⊗mσ(n).

In particular this is true for vector spaces. If n! is invertible in R, then thefollowing map is also injective:

SymnM ↪→M⊗n, m1 · · ·mn 7→1

n!

∑σ∈Sn

mσ(1) ⊗ · · · ⊗mσ(n).

Moreover, the composition of this map with the projection M⊗n � SymnM isthe identity on SymmM .

We can also form the symmetric algebra SymM := R⊕M⊕Sym2M⊕· · · .This is the “largest” commutative algebra containing M . An insightful fact isthat Sym(V ) ∼= k[V ∗] canonically for a vector space V .

Recall that if A ∈ Endk(V ), then A⊗m acts on V ⊗m by v1 ⊗ · · · ⊗ vm 7→Av1 ⊗ · · · ⊗ Avm. So this action passes down to the quotient on Symm V . Forthe same reason, if V is a representation of G, then so is Symm V .

Example 3.2.5. GL2(C)

� C2 naturally. So GL2(C)

�

Sym2(C2) ∼= k[x, y]2.Explicitly,

ρ(g)(ax2 + bxy + cy2) = a(gx)2 + b(gx)(gy) + c(gy)2.

Under the basis {x2, xy, y2}, we can write down the matrix for ρ(g):

ρ

[g11 g12g21 g22

]=

g211 g11g12 g2122g11g21 g11g22 + g12g21 2g12g21g221 g21g22 g222

which is a polynomial representation.

Lastly, we have the following decomposition.

Proposition 3.2.6. If V,W are finite dimensional vector spaces, then we havea canonical isomorphism

Symn(V ⊕W ) ∼−−→n⊕a=0

Syma V ⊗ Symn−aW,

v1 · · · vaw1 · · ·wn−a ← [ (v1 · · · va)⊗ (w1 · · ·wn−a).

11

Proof. The map is defined in a coordinate-free way, so it is canonical. To seethat it is an isomorphism, note that if {ei}, {fj} are basis of V,W respectively,then we have a correspondence of basis

ei1 · · · eiafj1 · · · fjn−a ←[ (ei1 · · · eia)⊗ (fj1 · · · fjn−a)

for all 0 ≤ a ≤ n, 1 ≤ i1 ≤ · · · ≤ ia ≤ a, 1 ≤ j1 ≤ · · · ≤ n− a.

Corollary 3.2.7. If V,W are also representations of G, then the above decom-position of vector spaces is also a decomposition of representations.

Proof. We only need that the isomorphism is G-equivariant. Indeed, the follow-ing diagram commute:

Symn(V ⊕W )⊕n

a=0 Syma V ⊗ Symn−aW

Symn(V ⊕W )⊕n

a=0 Syma V ⊗ Symn−aW

g g

ei1 · · · eiafj1 · · · fjn−a (ei1 · · · eia)⊗ (fj1 · · · fjn−a)

(gei1) · · · (geia)(gfj1) · · · (gfjn−a) (gei1 · · · geia)⊗ (gfj1 · · · gfjn−a)

g g

3.3 Exterior products

In the exterior power we require instead that tensors anti-commute, i.e., wewant m⊗m′ = −m′ ⊗m.

Definition 3.3.1. Let R be commutative, M an R-module. The exteriorpower

∧mM is the quotient of the tensor power M⊗m by the ideal generated

by elements of the form

m1 ⊗ · · · ⊗mi ⊗mi+1 ⊗ · · · ⊗mm +m1 ⊗ · · · ⊗mi+1 ⊗mi ⊗ · · · ⊗mm.

We shall write m1 ∧ · · · ∧mm instead of m1 ⊗ · · · ⊗mm.

There are parallel results for exterior products.

Theorem 3.3.2 (Universal Property). Let M,N be R-modules. Then alter-nating multilinear maps Mm → N are the same as R-module homomorphisms∧m

M → N .

Proposition 3.3.3. Let V be a vector space over k with basis {e1, . . . , en}, then{ei1 · · · eim | 1 < ij < ij+1 ≤ n} is a basis of

∧mV . As a result, dimk

∧mV =(

nm

)for m ≤ n and 0 for m > n.

12

Proposition 3.3.4. If M ∼= Rl is a free module, then we have an injection∧nM ↪→M⊗n, m1 ∧ · · · ∧mn 7→

∑σ∈Sn

ε(σ)mσ(1) ⊗ · · · ⊗mσ(n).

In particular this is true for vector spaces. If n! is invertible in R, then thefollowing map is also injective:∧n

M ↪→M⊗n, m1 ∧ · · · ∧mn 7→1

n!

∑σ∈Sn

ε(σ)mσ(1) ⊗ · · · ⊗mσ(n).

Moreover, the composition of the latter map with the projection M⊗n �∧n

Mis the identity on

∧mM .

Proposition 3.3.5. If V,W are finite dimensional vector spaces, then we havea canonical isomorphism

∧n(V ⊕W ) ∼−−→

n⊕a=0

∧aV ⊗

∧n−aW,

v1 ∧ · · · ∧ va ∧ w1 ∧ · · · ∧ wn−a ←[ (v1 ∧ · · · ∧ va)⊗ (w1 ∧ · · · ∧ wn−a).

If V,W are representations of G, then the above is a decomposition of represen-tations.

3.4 Determinant

It is worth mentioning that the exterior powers can be used to develop acoordinate-free version of linear algebra. The determinant is one of the in-stances.

Let V be an n-dimensional vector space over k. Notice that dimk

∧nV =(

nn

)= 1, so for any linear map A ∈ Endk V , its exterior product

∧nA :

∧nV →∧n

V is canonically identified with a number. If we choose a basis and writedown the formula for

∧nA, we will see that this number is exactly detA.

4 Character theory

Talk by Sam Swain, notes by Peter McDonald

Recall, a representation of a finite group G is a homomorphism ρ : G→ GL(V )where V is a vector space. We often say that V is a representation.

Definition 4.0.1. A subrepresentation of a group G is a subspace W of arepresentation V such that ρ(g)(W ) ⊂W for all g ∈ G.

Definition 4.0.2. A representation is called irreducible if its only subrepre-sentations are 0 and itself.

13

4.1 Decomposition of Representations

Let W ⊂ V be a subrepresentation of a representation V of a group G. Since Gis finite, then W has a complement W ′ such that V = W ⊕W ′. Let π : V →Wbe the projection map. Then we can define

π′ :=1

|G|∑g∈G

ρ(g)πρ(g−1)

If we take x ∈W then ρ(g−1)(x) ∈W because W is a subrepresentation. So

πρ(g−1)x = ρ(g−1)x

because π is a projection. Then

ρ(g)πρ(g−1)x = ρ(g)ρ(g−1)x = x

Then

π′(x) =1

|G|∑g∈G

x = x

so π′ is a projection.We now show that ρ(g)π′ = π′ρ(g) for all g ∈ G. Note that

ρ(h)π′ρ(h−1) =1

|G|∑g∈G

ρ(h)ρ(g)π′ρ(g−1)ρ(h−1)

=1

|G|∑g∈G

ρ(hg)π′ρ((hg)−1)

= π′

Take W⊥ = kerπ′. Take x ∈ W⊥. Then π′ρ(g)(x) = ρ(g)π′(x) = 0 ∈ W⊥.Then W⊥ is a subrepresentation. So V = W ⊕W⊥. We should note that thisdecomposition does not depend on our choice of π.

If we keep breaking up V into these direct summands, we can eventuallywrite V = V ⊕m1

1 ⊕ · · · ⊕ V ⊕mkk . Note, this is not always the case when ourgroup is infinite. Consider the following example:

Example 4.1.1. Consider ρ : R→ GL(C2) given by ρ(x) =

(1 x0 1

). Consider

the subspace spanned by (10

).

This is invariant under ρ and so is a subrepresentation. However, its comple-ment would be the subspace spanned by(

01

),

but R acts on this subgroup nontrivially.

14

Proposition 4.1.2 (Schur’s Lemma). Let V and W be irreducible representa-tions. Let ϕ : V → W be an intertwining operator, i.e., the following diagramcommutes

Vϕ−−−−→ W

ρ1(g)

y yρ2(g)V

ϕ−−−−→ W

Then

1. ϕ is either the 0 map or an isomorphism

2. If ϕ is an isomorphism, then ϕ = λId.

Proof. 1. We claim that kerϕ and Imϕ are subrepresentations of V and Wrespectively. Take x ∈ kerϕ. Then ρ2(g)ϕ(x) = 0 = ϕρ1(g)(x) whichmeans that ρ1(g) ∈ kerϕ and therefore kerϕ is a subrepresentation.Similarly, w ∈ Imϕ implies there is some v ∈ V such that ϕ(v) = w.Then ρ2(g)(w) = ϕ(ρ1(g)(v)) which means that ρ2(g)(w) ∈ Imϕ andImϕ is a subrepresentation. By irreducibility, the only possibilities arekerϕ = 0, Imϕ = W or kerϕ = V, Imϕ = 0.

2. Assume V = kn where k is an algebraically closed field. Then ϕ has aneigenvalue λ in k. Then det(ϕ−λId) = 0 means ker(ϕ−λId) is nontrivial.Then by (1) ker(ϕ− λId) = V and ϕ = λId.

If we have an isomorphism between representations ϕ : V →W with

V = V ⊕m11 ⊕ · · · ⊕ V ⊕mkk

W = W⊕n11 ⊕ · · · ⊕W⊕njj

then we can restrict ϕ to each irreducible component.

4.2 Character Theory

Another way of extracting irreducible representations from other representationsis through the use of characters.

Definition 4.2.1. If G is a finite group and ρ : G→ GL(V ) is a representation,the character of V , denoted χV , is a map χV : G → k given by χV (g) =tr(ρ(g)).

4.2.1 Properties of χ

1. χV (e) = dim(V ).

2. χV (g−1) = χV (g).

3. χV (xgx−1) = χV (g)

15

Proof. 1. Since ρ is a representation, we have ρ(e) = Id. Hence, χ(e) =tr(Id) = dim(V )

2.χV (g) = tr(ρ(g)) =

∑λi

We claim this is equal to∑λ−1i . To see this, ρ(g)|G| = ρ(g|G|) = ρ(e) has

eigenvalues 1. We also know that the eigenvalues of Ak are the eigenvaluesof A raised to the k. Then |λi| = 1 for all ρ(g) and so λi = λ−1i . Hence

χV (g) = tr(ρ(g)) =∑

λi =∑

λ−1i = tr(ρ(g−1)) = χV (g−1)

3. tr(AB) = tr(BA) so tr(ABA−1) = tr(B).

4.2.2 More Properties of χ

If V and W are two representations with respective characters χV and χW , then

1. χV⊕W = χV + χW

2. χV⊗W = χV · χWProof. 1. χV⊕W (g) is just the trace of the following matrix:(

ρV (g) 00 ρW (g)

)Then the trace of this matrix is just tr(ρV ) + tr(ρW ) = χV (g) + χW (g).

2. Look at the Kronecker product

5 Representations of the symmetric group

Talk by Faith Pearson, notes by Faith Pearson

(Additional proofs and examples are covered in these notes that were not coveredin the talk.)

Our goal in this section is to classify all of the irreducible representationsof the symmetric group. Given any finite group G, the number of irreduciblerepresentations of G is equal to the number of conjugacy classes of G. Thoughthis is true for every finite group G, it is not always possible to create anexplicit bijection between its irreducible representations and conjugacy classes.However, we will see that each irreducible representation of the symmetric grouphas a one-to-one correspondence with a combinatorial object called a Youngdiagram that corresponds to a given conjugacy class.

In order to study the irreducible representations of the symmetric group, wemust first establish its conjugacy classes.

16

Definition 5.0.1. Suppose σ ∈ Sn is a product of cycles σ1σ2 · · ·σk and let λibe the length of σi. We may assume that λi ≥ λi+1 for all i since disjoint cyclescommute. Then we define λ(σ) = {λ1, λ2, . . . , λk} to be the cycle type of σ.

Definition 5.0.2. For a given σ ∈ Sn, the conjugacy class Cσ is the set ofall elements σ′ = τστ−1 for some τ ∈ Sn.

We can now determine the conjugacy classes of the symmetric group Sn. Westart by noticing that any conjugate of a k-cycle is also a k-cycle.

Lemma 5.0.3. Let α, τ ∈ Sn, where α is a k-cycle (a1, a2, . . . , ak). Then

τατ−1 = (τ(a1), τ(a2), . . . , τ(ak)).

Proof. Consider τ(ai) such that 1 ≤ i ≤ k. Then we have τ−1τ(ai) = ai, andα(ai) = ai+1 mod k. Then τατ−1(τ(ai)) = τ(ai+1) mod k. Now take any jwhere j ∈ {1, 2, . . . , n}, and where j 6= ai for any i. Then α(j) = j because j isnot in the k-cycle defining α, and so τατ−1(τ(j)) = τ(j). Hence, τατ−1 fixesany number which is not of the form τ(ai) for some i, and thus

τατ−1 = (τ(a1), τ(a2), . . . τ(ak)).

Lemma 5.0.4. The conjugate of a product of k-cycles is equivalent to the prod-uct of the conjugates of k-cycles. That is, for αi disjoint, we have that

τα1α2 . . . αnτ−1 = (τα1τ

−1)(τα2τ−1) . . . (ταnτ

−1).

Finally, we can describe the conjugacy classes of the symmetric group.

Proposition 5.0.5. The conjugacy classes of Sn are determined by cycle type.That is, if σ has cycle type (λ1, λ2, . . . , λ`), and if ρ is any other element of Snwith cycle type (λ1, λ2, . . . , λ`), then σ is conjugate to ρ.

Proof. Suppose that σ is a product of disjoint cycles σ = α1α2 . . . α`, with cycletype (λ1, λ2, . . . , λ`), where αi is a λi-cycle. By Lemma 5.0.4,

τστ−1 = (τα1τ−1)(τα2τ

−1) . . . (ταnτ−1),

and by Lemma 5.0.3, ταiτ−1 is a λi-cycle. For any i, j ∈ {1, 2, . . . , n} where

i 6= j, αi and αj are disjoint. Then since τ is a bijection, this implies ταiτ−1

and ταjτ−1 must also be disjoint. Thus the conjugate τστ−1 is a product of

disjoint λi-cycles, and has cycle type (λ1, λ2, . . . , λ`).

Conversely, suppose σ and ρ both have cycle type (λ1, λ2, . . . , λ`). Let σ =α1α2 . . . α` and let ρ = β1β2 . . . β`, where αi and βi are λi cycles. Then αi =(ai,1, ai,2, . . . , ai,λi) and bi = (bi,1, bi,2, . . . , bi,λi). We can chose τ to be the mapsuch that τ(ai,k) = bi,k. Because αi are mutually disjoint and similarly βi aremutually disjoint, and σ and ρ are permutations on {1, 2, . . . , n}, τ is well definedand also a permutation. Thus by Lemma 5.0.3, τστ−1 = ρ.

17

Thus, we see that any two permutations in Sn are conjugate if and only ifthey have the same cycle type.

Recall the definition of the group algebra, so we can define a few new repre-sentations.

Definition 5.0.6. The group algebra CSn is the set of all finite formal sumsof the form ∑

σ∈Sn

zσeσ for zσ ∈ C

where eσ are basis elements indexed by σ ∈ Sn. Multiplication is defined onbasis elements by (∑

zieσi

)(∑yjeσj

)=∑

ziyjeσieσj

which we can expand linearly.

Definition 5.0.7. We may also think of CSn as a complex vector space withbasis elements indexed by each eσ. Then we can define a representation

ρ : Sn → GL(CSn) ∼= GL(Cn!),

which is called the regular representation.

Let us define two more representations that will come up again in a laterexample.

Definition 5.0.8. For Sn, the alternating representation (or sign represen-tation) is C equipped with the action

σ · v =

{v, if σ is an even permutation

−v, if σ is an odd permutation

or equivalently, ρ(σ) = sgn(σ)I for every σ ∈ Sn. Remark that any Sn wheren ≥ 2 has the alternating representation, and since this representation is onedimensional, it is irreducible.

Definition 5.0.9. For any n, let {e1, e2, . . . , en} be the standard basis for Cn,and define the action of Sn on Cn to be

σ(a1e1 + a2e2 + ·+ anen) = a1eσ(1) + a2eσ(2) + ·+ aneσ(n).

This is a permutation representation of Sn. Remark that the one-dimensionalsubspace of C spanned by e1 + e2 + · · ·+ en is invariant under the action of Sn,and so its orthogonal complement V = {(x1, x2, . . . , xn)|x1 + x2 + · · ·+ xn = 0}is also invariant, and therefore a subrepresentation. We call V the standardrepresentation of Sn.

Definition 5.0.10. A partition of a positive integer n = λ1 + λ2 + · · ·+ λk isan ordered set λ = (λ1, λ2, . . . , λk) of positive integers such that λi ≥ λi+1 forevery 1 ≤ i ≤ k.

18

Recall that in Sn, the number of its irreducible representations is equal tothe number of its conjugacy classes, and each conjugacy class is a cycle-typeequivalence class. Now, there is also a bijective correspondence between the setof cycle-types and the ways n can be written as the sum of positive integers.For example, S4 has the following five cycle-types:

Cycle Notation Alternate Form Corresponding Sumε (1)(2)(3)(4) 1+1+1+1

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

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

A fundamental tool for studying the representations of Sn is the Youngdiagram.

Definition 5.0.11. A Young diagram is a graphical representation of a par-tition λ = (λ1, λ2, . . . , λk), as an array of boxes. This array is constructed bydrawing a row of λ1 boxes, then beneath it drawing a row of λ2 boxes, and so onuntil the last row contains λk boxes, and each row is as long as or shorter thanthe one above it.

Example 5.0.12. The Young Diagrams corresponding to the partitions of n = 4are:

.

So we see each conjugacy class of Sn corresponds to a Young diagram, andnow we can find a method that will generate all of the irreducible representationsof Sn. (For a proof of why this method works, see Section 4.2 of Fulton andHarris.) To continue, we must define a Young tableau.

Definition 5.0.13. A Young tableau is a Young diagram whose boxes arelabeled in any way with each of the numbers 1, . . . , n.

For our purposes, we will fill in our Young diagrams in the natural way,starting with 1 in the upper left and increasing by 1 as we move from left toright and top to bottom.

Example 5.0.14. For n = 9, the partition λ = (3, 2, 2, 1, 1) induces the Youngdiagram,

19

and its associated Young tableau would be

1 2 34 56 789 .

Given any Young tableau, we can define

Pλ = {σ ∈ Sn | σ preserves each row}, and

Qλ = {σ ∈ Sn | σ preserves each column}.

Example 5.0.15. If we were working in S6, and wanted to find the irreduciblerepresentations corresponding to the partition λ = (3, 2, 1), our Young tableauwould be

1 2 34 56 .

Then we would have

Pλ = {e, (12), (23), (13), (123), (132), (45), (12)(45), (23)(45), (13)(45),

(123)(45), (132)(45)},

Qλ = {e, (14), (16), (46), (146), (164), (25), (14)(25), (16)(25), (46)(25),

(146)(25), (164)(25)}.

We can use Pλ and Qλ to define the elements aλ, bλ in the group algebraCSn to be

aλ :=∑σ∈Pλ

eσ

bλ :=∑σ∈Qλ

sgn(σ)eσ

Finally, we can define a key tool in finding the irreducible representations of Sn.

Definition 5.0.16. The Young symmetrizer is

cλ := aλ · bλ.

Theorem 5.0.17. Given Sn, let λ be a partition of n. Let Vλ = CSn · cλ be thesubspace of CSn spanned by the Young symmetrizer cλ. Then

1. Vλ is an irreducible representation of Sn.

20

2. If λ and µ are distinct partitions of n, then Vλ and Vµ are not isomorphic.

3. The Vλ account for all of the irreducible representations of Sn.

We will follow Sean McAfee’s proof. The full proof can be found in Sections4.1 and 4.2 in Fulton and Harris.

Lemma 5.0.18. For all x ∈ CSn, cλ · x · cλ is a scalar multiple of cλ.

Lemma 5.0.19. If λ 6= µ, then cλ · CSn · cµ = 0.

Assuming these two lemmas to be true, we can now prove Theorem 2.8.

Proof. 1. Take Vλ = CSn · cλ for a given Young symmetrizer cλ. Then byLemma 2.9, we have that

cλVλ ⊆ Ccλ.

Let W be a nonzero subrepresentation of Vλ. We want to show W =Vλ. First, we claim that cλVλ and cλW are both nonzero. Suppose thatcλVλ = 0. Then

VλVλ = CSn(cλVλ) = 0.

Considering CSn and CSn · cλ as subspaces, there exists a projection π :CSn → CSn · cλ that commutes with the action of Sn. This projectioncan be described as right multiplication on the group algebra CSn by anelement x ∈ CSn by letting x := π(1). Since x = 1 · x, this x must be inVλ. Then by the definition of projection,

x = x2 ∈ VλVλ = 0,

and thus x must equal zero, which is a contradiction since the nonzero cλitself is in CSn · cλ. Therefore we must have cλVλ 6= 0. Next we will showcλW 6= 0.

Since we have that W is a subspace of Vλ, cλVλ ⊆ Ccλ, and that cλW 6= 0,we must have cλW = Ccλ. Therefore,

Vλ = CSn · cλ = CSn(CCcλ) = CSn(cλW ) ⊆W

where the inclusion on the right follows from the fact that W is a subrep-resentation of Vλ, that is, W is invariant under the action of CSn. Thus,we have Vλ = W , which completes the proof that Vλ is irreducible in Sn.

2. Let λ and µ be distinct partitions of n, and let Vλ and Vµ be their corre-sponding representations. By (a), we have that cλVλ = Ccλ 6= 0, and byLemma 2.10, we have that

cλVµ = cλCSncµ = 0.

Thus, Vλ and Vµ cannot be isomorphic, and therefore, if λ and µ aredistinct partitions of n, then Vλ and Vµ are not isomorphic.

21

3. We know that each partition λ of n is in one-to-one correspondence with adistinct conjugacy class of Sn. We know from (b) that the Vλ determine bysuch partitions are all inequivalent. Then since the number of conjugacyclasses of a finite group is equal to the number of irreducible representa-tions, we have therefore accounted for all of the irreducible representationsof Sn.

Therefore we have proved that the subspace CSn · cλ is an irreducible rep-resentation of Sn, and distinct partitions of λ correspond to distinct irreduciblerepresentations. Furthermore, the CSn · cλ account for all irreducible represen-tations of Sn.

Example 5.0.20. Let us use what we have learned to find all of the irreduciblerepresentations of S3. There are three Young diagrams corresponding to thethree partitions λ = (3), µ = (2, 1), ω = (1, 1, 1), which are respectively

1 2 31 23

123 .

In the first Young tableau, since all of the numbers are in the same row, anypermutation will preserve the row. However, the only permutation that willpreserve the columns is the identity. Thus, Pλ = S3, and Qλ = ε. So we have

aλ = eε + e(12) + e(23) + e(13) + e(123) + e(132),

bλ = eε,

cλ = (eε + e(12) + e(23) + e(13) + e(123) + e(132))eε

= eε + e(12) + e(23) + e(13) + e(123) + e(132).

(1)

Therefore CS3 ·cλ = C ·cλ = 〈cλ〉 is the associated irreducible representationsince multiplying by any element in the basis of CS3 will simply rearrange theaddends of cλ, however it will not change the sum. Remark that the subspacegenerated by cλ is one dimensional, and because σ · rcλ = rcλ for any σ ∈ S3

and any r ∈ C, the action of every σ leaves every vector in 〈cλ〉 fixed. Therefore〈cλ〉 is the trivial representation.

In the second Young diagram, we have Pµ = {ε, (12)}, and Qµ = {ε, (13)}.Then we obtain

aµ = eε + e(12),

bµ = eε − e(13),cµ = (eε + e(12))(eε − e(13))

= eε − e(13) + e(12) − e(132).

(2)

The associated irreducible representation is CS3 · cµ. To find out what thissubspace is, we multiply cµ by the basis elements of CS3, and obtain

22

eε(eε − e(13) + e(12) − e(132)) = eε − e(13) + e(12) − e(132),e(12)(eε − e(13) + e(12) − e(132)) = e(12) − e(132) + eε − e(13),e(13)(eε − e(13) + e(12) − e(132)) = e(13) − eε + e(123) − e(23)e(23)(eε − e(13) + e(12) − e(132)) = e(23) − e(123) + e(132) − e(12)e(123)(eε − e(13) + e(12) − e(132)) = e(123) − e(23) + e(13) − eεe(132)(eε − e(13) + e(12) − e(132)) = e(132) − e(12) + e(23) − e(123).

(3)

The matrix corresponding to the above set of equations is1 1 −1 0 −1 01 1 0 −1 0 −1−1 −1 1 0 1 00 0 −1 1 −1 10 0 1 −1 1 −1−1 −1 0 1 0 1

.

Since the set is spanned by the first and third vectors, CS3 ·cµ is the subspace

〈eε − e(13) + e(12) − e(132), e(13) − eε + e(123) − e(23)〉.

This must be the standard representation, since it is the only two-dimensionalrepresentation of S3.

In the third Young diagram, any permutation in S3 will preserve the column,however only the identity will fix the rows. So we have Pω = {ε}, and Qω = S3.Then we obtain

aω = eε,

bω = eε − e(12) − e(23) − e(13) + e(123) + e(132),

cω = eε(eε − e(12) − e(23) − e(13) + e(123) + e(132))

= eε − e(12) − e(23) − e(13) + e(123) + e(132).

(4)

Once again, CS3 · cω = C · cω = 〈cω〉 is the associated irreducible represen-tation since multiplying by any element in the basis of CS3 will rearrange theaddends of cω and negate their signs. This subspace is also one dimensional,and for any σ ∈ S3, and any r ∈ C, we have σ · rcω = rcω if σ is even, andσ · rcω = −rcω if σ is odd. Thus, 〈cω〉 is the alternating representation.

Thus, the three representations of S3 are the trivial representation, the stan-dard representation, and the alternating representation. This method will stillhold for any symmetric group, however the Young symmetrizers size becomelarge very quickly.

23

6 Double Centralizer Theorem

Talk by Sabine Lang, notes by Peter McDonald

In this section, let k be an algebraically closed field, A a finite dimensionalk-algebra, and ρ : A→ End(V ) is a representation of A.

Definition 6.0.1. The radical of A is the set

Rad(A) := {x ∈ A : xM = 0 ∀M irreducible left A-modules}

If Rad(A) = 0 we say that A is semisimple.

Theorem 6.0.2. Up to isomorphism A has finitely many irreducible represen-tations Vi with dim(Vi) <∞ for all i and A/Rad(A) ∼=

⊕ni=1 End(Vi)

Proof. Let Vi be an irreducible representation of A. If 0 6= v ∈ Vi, then 0 6=Av ⊆ Vi, hence Av = Vi. So dim(Vi) = dim(Av) < ∞ because A is finitedimensional. By linear algebra⊕

i∈Iρi : A→

⊕i∈I

End(Vi)

is surjective. Then

|I| = # of irreducible representations ≤∑i∈I

dim(End(Vi)) < dim(A) <∞

Finally,

Rad(A) = ker

(⊕i∈I

ρi

),

so by the first isomorphism theorem

A/Rad(A) ∼=n⊕i=1

End(Vi).

Corollary 6.0.3.

dim(A)− dim(Rad(A)) =∑i∈I

dim2(Vi) =∑i∈I

dim End(Vi).

Theorem 6.0.4. If A is finite dimensional, the following are equivalent.

1. A is semisimple

2.∑ni=1 dim2(Vi) = dim(A)

3. A ∼=⊕

i Matdi(k)

24

4. Any finite dimensional representation of A is completely reducible.

5. A is completely reducible as a representation of A.

Definition 6.0.5. Let E be a finite dimensional k-vector space. ConsiderEnd(E) and A ⊆ End(E) a subalgebra. Then the centralizer of A in End(E) isthe set

EndA(E) := {f : E → E : f is linear and a morphism of A-representations}= the centralizer of A

Theorem 6.0.6 (Double Centralizer Theorem). Let E be a finite dimensionalk-vector space and A ⊆ End(E) a subalgebra such that A is semisimple. LetB = EndA(E). Then

1. EndB(E) = A

2. B is semisimple

3. E =⊕n

i=1(Vi ⊗Wi) as a representation of A ⊗ B where the Vi are irre-ducible representations of A and the Wi are irreducible representations ofB. This gives a bijection between the irreducible representations of A andthe irreducible representations of B.

Proof. Let V1, . . . , Vn be irreducbile representations of A. Then, because A issemisimple

A ∼=n⊕i=1

End(Vi).

We can also decompose E using these representations of A:

E ∼=n⊕i=1

(Vi ⊗HomA(Vi, E))

LetWi = HomA(Vi, E).

25

Then

B = EndA(E)

= HomA(E,E)

∼= HomA

(n⊕i=1

(Vi ⊗Wi), E

)

=

n⊕i=1

HomA(Vi ⊗Wi, E)

=

n⊕i=1

HomA(Wi ⊗ Vi, E)

∼=n⊕i=1

HomA(Wi,HomA(Vi, E))

=

n⊕i=1

HomA(Wi,Wi)

=

n⊕i=1

EndA(Wi)

We now show that the Wi are irreducible B-modules. Let f, f ′ ∈ Wi =HomA(Vi, E). We know Vi is irreducible as an A-module, so given 0 6= v ∈ Vi,Vi = Av, which means that f and f ′ are determined by f(v) and f ′(v) respec-tively. Then Af(v) ⊆ E is an A-invariant subspace and there is an invariantcomplement W . Then E = Af(v)⊕W .

Let T : E → E be defined by

af(v) 7−→ af ′(v)

W 7−→W

Then T ◦ f = f ′. T is an A-homomorphism, so T ∈ EndA(E) = B. Then Wi isan irreducible B-module. Because B =

⊕ni=1 EndA(Wi), B is semisimple.

We are now ready to show EndB(E) = A. As B-modules

E ∼=n⊕i=1

(Wi ⊗HomB(Wi, E))

We know Wi = HomA(Vi, E) are irreducible B-modules. Comparing the twodecomposition of E as

E ∼=n⊕i=1

(Wi ⊗HomB(Wi, E)) ∼=n⊕i=1

(Vi ⊗Wi) ,

26

we can deduce that Vi ∼= HomB(Wi, E) (Note: this can be seen in severaldifferent ways. One of them is to use the decomposition as an A-module, withthe fact that each irreducible for A is one of V1, . . . , Vn.). Hence,

E ∼=n⊕i=1

(Vi ⊗Wi)

is a decomposition of E as an A⊗B-module. Then

EndB(E) = HomB(E,E)

= HomB

(n⊕i=1

(V⊗Wi), E

)∼=⊕

Hom(Vi,HomB(Wi, E))

=

n⊕i=1

End(Vi)

7 Schur-Weyl duality

Talk by Chengyu Du, notes by Sam Swain

Recall the Double Centralizer Theorem from last week:

Theorem 7.0.1. Given a finite-dimensional vector space E and A ⊆ End(E)semisimple, then defining B := EndA(E), we have that

1. A = EndB(E)

2. B is semisimple

3. E ∼=⊕r

i=1(Vi ⊗C Wi) where the Vi are irreducible A-modules and the Wi

are irreducible B-modules and Wi = Hom(Vi, E).

We now consider the special case where E = V ⊗d with dim(V ) = n and

A = CSd =⊕|λ|=d

CSdcλ

Theorem 7.0.2 (Schur-Weyl Duality). Given A = CSd =⊕|λ|=d CSdcλ and

E = V ⊗d. ThenE = V ⊗d ∼=

⊕|λ|=d

(SλV )

27

Proof. A is semi-simple as it is the direct sum of irreducible A-modules. Wenow define an A-action on V ⊗d. Consider σ ∈ Sd. Then define

σ(v1 ⊗ v2 ⊗ · · · ⊗ vd) := vσ−1(1) ⊗ · · · ⊗ vσ−1(d)

which we can expand linearly to V ⊗d. Then dim(D) = d · n <∞ which meanswe can apply the double centralizer theorem.

In this case B = EndA(V ⊗d). Then

EndC(V ⊗d) ∼= (V ⊗d)∗ ⊗C V⊗d

∼= (V ∗)⊗d ⊗C V⊗d

∼= (V ∗ ⊗C V )⊗d

∼= (End(V ))⊗d

Then B consists of the elements in End(V )⊗d that commute with the Sd action.In general, these elements look like

φ = φ1 ⊗ φ2 ⊗ · · · ⊗ φd, φi ∈ End(V )

In fact, it turns out that B is spanned by elements like φ⊗ φ⊗ · · · ⊗ φ (this isnot trivial, but was not covered in the talk) so B ' End(V ).

Then, letting Vλ = CSλ · cλ we have

V ⊗d ∼=⊕

(Vλ ⊗C HomA(Vλ, V⊗d)) ∼=

⊕|λ|=d

(V ⊗C SλV )

due to the fact that

HomA(Vλ, V⊗d) ∼= V ⊗d ⊗A V ∗λ∼= V ⊗d ⊗A Vλ∼= V ⊗d ⊗A A · cλ∼= V ⊗d · cλ∼= Imcλ

Note that the transition from the third-to-last to the second-to-last step is nottrivial.

We should also note that Fulton-Harris proves this using the fact that wecan think of

⊕|λ|=d(SλV )⊕mλ as an End(V )-module where mλ = dimVλ =

dimSλV. However, this loses the A-module structure.

Example 7.0.3. V = span{e1, e2}.

Proof. We will use the irreducible representation of S3:

CS3 · c(3) ' Sym3V

CS3 · c(2,1) '?

CS3 · c(1,1,1) ' Λ3V

28

Note that Λ3V = 0 because dim(V ) = 2 < 3.Now

Imc(3) = Sym3V = span

e1 ⊗ e1 ⊗ e1,e2 ⊗ e2 ⊗ e2,

e1 ⊗ e1 ⊗ e2 + e1 ⊗ e2 ⊗ e1 + e2 ⊗ e1 ⊗ e1,e2 ⊗ e2 ⊗ e1 + e2 ⊗ e1 ⊗ e2 + e1 ⊗ e2 ⊗ e2

so dim Sym3V = 4. Now dimV ⊗3 = 8 and

V ⊗3 ∼=⊕|λ|=d

(SλV )mλ =

(Sym3V

)1⊕?⊕ 0

From Faith’s talk, we know dim(V(2,1)) = 2, and after lots of computation wecan get

Imc(2,1) = span{e1 ⊗ e1 ⊗ e2 − e2 ⊗ e1 ⊗ e1, e2 ⊗ e2 ⊗ e1 − e1 ⊗ e2 ⊗ e2}

8 Determinental rings and applications to com-mutative algebra

Talk by Jenny Kenkel, notes by Jenny Kenkel

My objects of interest are the ring of a field adjoin nm many variables, and theideal generated by minors:

R = F

x11 . . . x1n...

...xm1 . . . xmn

, Ir+1 = ( size r + 1 minors )

Note: R/Ir+1 is setting all size r+1 minors to zero, so it corresponds to matricesof rank r or less.The example I consider the most is:

R =

[u v wx y z

], I2 = (∆1 = vz − wy,∆2 = wx− uz,∆3 = uy − vx)

Properties of R/I: in general, this ring is nice but not too nice.

• I is prime (nice)

• u∆1 + v∆2 + w∆3 = 0 and x∆1 + y∆2 + z∆3 = 0 (not too nice)

• dim(R/I) = 4 as a ring, that is, we can find a chain of prime ideals withfour containments (this ring is infinite dimensional as a vector space)

(∆1,∆2,∆3) ( (∆1,∆2,∆3, v − x) ( (∆1,∆2,∆3, v − x,w − y)

( (∆1,∆2,∆3, v − x,w − y, u− z) ( (u, v, w, x, y, z)

29

8.1 Determinantal Rings Are Cohen-Macaulay

The following section is a description of why determinantal rings are interestingto algebraists, and an example of the way commutative algebraists often thinkof rings.Local Ring Fact: Define the radical of J , denoted

√J to be

√J = {r|rn ∈ J}

Then I can always find some ideal that is generated by dim(S) many elementsthat’s radical is the maximal ideal. Those dim(S) elements are referred to as asystem of parameters.In the case of

R =

[u v wx y z

], I2 = (∆1 = vz − wy,∆2 = wx− uz,∆3 = uy − vx)

I claim√

(u, v − x,w − y, z) = (u, v, w, x, y, z), the maximal (homogenous) idealof R/I.Note: the ring R has infinitely many maximal ideals, but only one maximalhomogenous ideal. For many purposes, then, it can be considered a local ring.

Proof. Note that√

(u, v − x,w − y, z) ⊆ (u, v, w, x, y, z), so it suffices to showthat

(u, v, w, x, y, z) ⊆√

(u, v − x,w − y, z)

For purposes of this proof, let n = (u, v − x,w − y, z).Certainly, u, z ∈ n. As u ∈ n, we have that uy ∈ n. Recall that uy−vx = 0 ∈ n,so vx ∈ n. Now,

v(v − x) = v2 − vx ∈ n so v2 ∈ n and similarly

x(v − x) = vx− x2 ∈ n so x2 ∈ n

In a symmetric argument, since z ∈ n, vz ∈ n and so wy ∈ n. Thus,

w(w − y) = w2 − wy ∈ n so w2 ∈ n

y(w − y) = wy − y2 ∈ n so y2 ∈ n

Thus, we have shown that u, v2, w2, x2, y2 and z are in n, and so u, v, w, x, y andz are in

√n.

Definition 8.1.1. Regular Sequence

A regular sequence in a ring S is a sequence of elements, x1, . . . , xn suchthat

• x1 is not a zero divisor in S and x1S 6= S

• x2 is not a zero divisor in S/(x1) and (x1, x2) 6= S

30

• xi is not a zero divisor in S/(x1, . . . , xi−1 and (x1, . . . , xi) 6= S

A very neat property about the ring R/I is that the system of parameters wediscussed above is in fact a regular sequence!Sketch of Proof that the system of parameters is a regular sequence: Since I isa prime ideal, R/I is a domain, so there are no zero divisors, and in particular,u is not a zero divisor in R/I.Now consider (R/I)/u. We are setting u = 0, so

R/(I, u) ∼= F[

0 v wx y z

]/(vz − wy,wx, vx)

Certainly w, v and x are zero divisors in the above ring. But the element v−x isnot. One can convince oneself of this fact by multiplying v− x by any variable,v, w, x, y or z and getting something that is not a zero divisor.Now consider (R/I)/(u, v − x). We are setting v − x = 0, or in other words,v = x. So

R/(I, u, v − x) = F[0 v wv y z

]/(vz − wy, v2, wv)

Certainly, w and v are zero divisors in the above ring, but again, we can convinceourselves that w − y is not a zero divisor.

R/(I, u, v − x,w − y) ∼= F[0 v wv w z

]/(vz − w2, v2, wv)

Finally,

R/(I, u, v − x,w − y, z) ∼= F[0 v wv w 0

]/(w2, v2, wv)

Since every non-unit is a zerodivisor in the above ring, not only do we have aregular sequence, but we have a maximal regular sequence.

Definition 8.1.2. Cohen-Macaulay

If there is some (equivalently, every) system of parameters that is a regularsequence, than the ring is called Cohen-Macaulay.

Thus, the ring R/I is Cohen-Macaulay (nice!). It is not, however, Goren-stein, a definition beyond the scope of this talk (not too nice!).

8.2 Relation to Representation Theory

Notice that

2× 3 matrices of elements of C = Hom(C3,C2

)=(C3)∗ ⊗ C2

where a matrix acts on an element of C3 by multiplication on the left.Define the group GL(3)×GL(2) action on

(C3)∗ ⊗ C2 in the following way.

Let φ ∈ Hom(C3,C2). Then

(g1, g2)φ 7→ g−12 φ(g−11 ).

31

After fixing a basis, let Sym(V ) denote the symmetric algebra on the basis ofV .

Sym(Hom(C3,C2)) = Sym(2× 3 matrices with elements in C

= C[u v wx y z

]= R

Then we can think of R/I as a subrepresentation of R, that is, all matrices ofrank 1 or less, or as a quotient representation of R. If M is some matrix thathas rank 1 or less, then action by GL(3)×GL(2) will take this matrix to someother element of rank 1 or less.Let E be the vector space C3 and F be the vector space C2, and let e1, e2, e3be a basis for E and f1, f2 be a basis for F . Then Sym1(E∗ ⊗ F ) ∼= E∗ ⊗ Fhas basis: [

e∗1 ⊗ f1 e∗2 ⊗ f1 e∗3 ⊗ f1e∗1 ⊗ f2 e∗2 ⊗ f2 e∗3 ⊗ f2

]and degree 1 polynomials in R = C

[u v wx y z

]have, as a vector space basis,[

u v wx y z

].

We might expect symmetric algebras to play nicely with tensor products, andthat Sym2(E∗⊗F ) might be the same as Sym2(E∗)⊗Sym2(F ). However, inSym2(E∗ ⊗ F ),

(e∗1 ⊗ f1)(e∗2 ⊗ f2) 6= (e∗1 ⊗ f2)(e∗2 ⊗ f1)

but in Sym2(E∗) ⊗ Sym2(F ) the analagous element for both of the aboveelements is:

e∗1e∗2 ⊗ f1f2

In other words, in Sym2(E∗ ⊗ F ), uy 6= vx, but Sym2(E∗) ⊗ Sym2(F ) actsjust like R/I!

Suppose we want to understand polynomials in degree m in C[u v wx y z

]∼=

Symm(E∗ ⊗ F ). One way to do so is to understand its decomposition intosymmetric products of E∗ and symmetric products of F . The Cauchy Formulatells us how to do that.Cauchy Formula If F a field of characteristic 0, E,F vector spaces, then

Symm(E ⊗ F ) =⊕|λ|=m

SλE ⊗ SλF

where Sλ(V ) is the Schur functor acting on the vector space V , that is, Sλ(V ) =Im(cλ|V ⊗d), where cλ is the Young symmetrizer.

32

9 Decomposing tensor products of Weyl mod-ules

Talk by Peter McDonald, notes by Peter McDonald

Throughout this section, let λ = (λ1, . . . , λk) be a partition of d, let µ =(µ1, . . . , µk) be a partition of d and let ν = (ν1, . . . , νk) be a partition of d+m.

Recall that given λ, we can construct a Young tableau and define the fol-lowing sets

Pλ = {σ ∈ Sd : σ preserves the rows of the Young tableau}Qλ = {σ ∈ Sd : σ preserves the columns of the Young tableau}

Letting

aλ =∑σ∈Pλ

eσ

bλ =∑σ∈Qλ

sgn(σ)eσ

define the Young symmetrizer

cλ = aλ · bλ

Considering the action of cλ on V ⊗d, we get a Weyl module which we denote

SλV = cλ(V ⊗d)

which we can use as a building block of our representations. Then Schur-Weylduality gives us that

V ⊗d ∼=⊕|λ|=d

SλV mλ

where mλ is the dimension of Vλ, the irreducible representation of Sd corre-sponding to λ.

Now that we have these Weyl modules, we would like to understand how thetensor product of two Weyl modules behave. Intuitively, the tensor product oftwo Weyl modules Sλ and Sµ can be decomposed into components of V ⊗d+m,but we want to know exactly which components their product corresponds to.In fact, given λ a partition of d and µ a partition of m, we have the followingisomorphism

SλV ⊗ SµV ∼=⊕

|ν|=d+m

NλµνSνV

where Nλµν are numbers determined by the Littlewood-Richardson rule. Whilewe will not prove this formula, we will investigate the Littlewood-Richardsonrule and look at examples in the case where µ = (m) and µ = (1, . . . , 1).

33

Definition 9.0.1. Given an endomorphism g of V , we have an induced endo-morphism g′ of SλV . Let χSλV (g) denote the trace of g′. This will be a sym-metric polynomial of degree d in k-variables, each representing an eigenvalueof x. This polynomial with indeterminates x1, . . . , xk is known as the Schurpolynomial and is denoted Sλ.

Remark 9.0.2. {Sλ}|λ|=d is a basis for the symmetric polynomials of degree d.

Given that Schur polynomials form a basis for the symmetric polynomials,we would like a systematic way to express the product of two Schur polynomialsin terms of the basis for the corresponding degree of the product. It turns outthat

Sλ · Sµ =∑ν

NλµνSν

so we will need to understand how to calculate these Nλµν .

Proposition 9.0.3 (Pieri’s Formula). Sλ · S(m) =∑ν

Sν where ν ranges over

the partitions of d+m whose Young diagrams are obtained from λ’s by addingm boxes, no two in the same column, i.e., all ν = (ν1, . . . , νk) with

ν1 ≥ λ1 ≥ ν2 ≥ λ2 ≥ · · · ≥ νk ≥ λk ≥ 0

Example 9.0.4. S(2,1) · S(2) = S(4,1) + S(3,2) + S(3,1,1) + S(2,2,1)

Proof. The corresponding Young diagrams are, where the x’s denote the twoboxes added:

− − x x− ,

− − x− x ,

− − x−x ,

− −− xx .

While this can be combined with the determinental formula to find theproduct of any two Schur polynomials, there is an easier way.

Definition 9.0.5. Given λ = (λ1, . . . , λk) a partition of d and µ = (µ1, . . . , µk)a partition of m, a µ-extension of the Young diagram for λ is obtained by thefollowing:

1. Add µ1 boxes to λ’s Young diagram according to Pieri’s formula, markingthe boxes with a 1

2. Add µ2 boxes to the above Young diagram according to Pieri’s formula,marking the boxes with a 2

3. Continue this process for all remaining µi

We say that a µ-expansion is strict if, when reading the diagram from right toleft and top to bottom, at any point in the reading the integer p appears at leastas any times as the integer p+ 1 for 1 ≤ p ≤ k − 1.

34

Proposition 9.0.6 (Littlewood-Richardson Rule). Given λ a partition of d, µa partition of m, and ν a partition of d + m, Nλνµ is the number of ways theYoung diagram for λ can be expanded to a Young diagram for ν by a µ-strictexpansion.

Recall our formula for the product of two Schur polynomials

Sλ · Sµ =∑ν

NλµνSν

We compute an example

Example 9.0.7. S(2,1) ·S(2,1) = S(4,2) +S(4,1,1) +S(3,3) + 2S(3,2,1) +S(3,1,1,1) +S(2,2,2) + S(2,2,1,1)

Proof. The (2, 1)-extensions of (2, 1) are listed below:

− − 1 1− 2 ,

− − 1 1−2 ,

− − 1− 1 2 ,

− − 1− 12 ,

− − 1− 21 ,

− − 1−12 ,

− −− 11 2 ,

− −− 112

The fact that these Littlewood-Richardson coefficients come from the mul-tiplication of Schur polynomials is no mistake. Because Schur polynomials cor-respond to the characters of the Weyl modules, the character of the product oftwo Weyl modules is the product of their characters. Now that we understandwhere the coefficients Nλµν come from, we can compute a few decompositionsof the tensor products of Weyl-modules.

Example 9.0.8.

Sλ ⊗ S(m) = Sλ ⊗ SymmV ∼=⊕ν

NλµνSν

where ν is all partitions of d+m whose Young diagrams are obtained from λ’sby adding m boxes, no two in the same column.

Example 9.0.9.

Sλ ⊗ S(1,...,1) = Sλ ⊗ ΛmV ∼=⊕ν

NλµνSν

where ν is all partitions of d+m whose Young diagrams are obtained from λ’sby adding m boxes, no two in the same row.

35