Part III Representation TheoryLectured by Stuart Martin, Lent Term 2013

Please send any comments and corrections to cn319@cam.ac.uk.

Contents

0 Preliminaries 2

1 Semisimple algebras 4

2 Irreducible modules for the symmetric group 8

3 Standard basis of Specht modules 12

4 Character formula 14

5 The hook length formula 21

6 Multilinear algebra and algebraic geometry 24

Interlude: some reminders about affine varieties 29

7 Schur–Weyl duality 31

8 Tensor decomposition 34

9 Polynomial and rational representations of GL(V ) 37

Interlude: some reminders about characters of GLn(C) 40

10 Weyl character formula 41

11 Introduction to invariant theory and first examples 44

12 First fundamental theorem of invariant theory 50

Example sheet 1 53

Example sheet 2 57

Last updated Thursday 21 March 2013

1

0 Preliminaries

In this course we will study representations of

(a) (finite) symmetric groups

(b) (infinite) general linear groups

all over C, and apply the theory to “classical” invariant theory.

We require no previous knowledge, except for some commutative algebra, ordinary representationtheory and algebraic geometry, as outlined below.

Commutative algebra

References: Part III course, Atiyah–Macdonald.

It is assumed that you will know about rings, modules, homomorphisms, quotients and chain conditions(the ascending chain condition, ACC, and descending chain condition, DCC).

Recall: a chain, or filtration, of submodules of a module M is a sequence (Mj : 0 ≤ j ≤ n) suchthat

M = M0 > M1 > · · · > Mn > 0

The length of the chain is n, the number of links. A composition series is a maximal chain;equivalently, each quotient Mj−1/Mj is irreducible.

Results:

1. Suppose M has a composition series of length n. Then every composition series of M has lengthn, and every chain in M can be extended to a composition series.

2. M has a composition series if and only if it satisfies ACC and DCC.

Modules satisfying both ACC and DCC are called modules of finite length. By (1), all compositionseries of M have the same length, which we denote by `(M) and call the length of M .

3. (Jordan–Holder) If (Mi) and (M ′i) are composition series then there exists a one-to-one cor-respondence between the set of quotients (Mi−1/Mi) and the set of quotients (M ′i−1/M

′i) such

that the corresponding quotients are isomorphic.

Remark. For a vector space V over a field K, the following are equivalent:

(i) finite dimension

(ii) finite length

(iii) ACC

(iv) DCC

and if these conditions are satisfied then `(V ) = dim(V ).

(Ordinary) representation theory (of finite groups)

References: Part II course, James–Liebeck, Curtis–Riener Vol. 1, §1 below.

Work over C or any field K where char(K) - |G|. Let V be a finite-dimensional vector space.

2

GL(V ) is the general linear group of K-linear automorphisms of V .

If dimK(V ) = n, choose a basis e1, . . . , en of V over K to identify it with Kn. Then θ ∈GL(V )↔ Aθ = (aij) ∈ Mn(K), where

θ(ej) =

n∑

i=1

aijei

and, in fact, Aθ ∈ GLn(K). This gives rise to a group isomorphism GL(V ) ∼= GLn(K).

Given G and V , a representation of G on V is a homomorphism

ρ = ρV : G→ GL(V )

G acts linearly on V if there is a linear action G× V → V , given by (g, v) 7→ g · v, where(a) (action) (g1g2) · v = g1 · (g2 · v) and e · v = v

(b) (linearity) g · (v1 + v2) = g · v1 + g · v2, g · (λv) = λ(g · v)

If G acts linearly on V then the map G→ GL(V ) given by g 7→ ρg, where ρg(v) = g · v, definesa representation of G on V . Conversely, given a representation ρ : G→ GL(V ) we have a linearaction of G on V given by g · v = ρ(g)v.

In this case, we say V is a G-space or G-module. By linear extension, V is a KG-module,where KG is the group algebra – more to come later.

For finite groups you should know

If S is an irreducible KG-module then S is a composition factor of KG

(Maschke’s theorem) IfG is finite and char(K) - G, then everKG-module is completely reducible;or equivalently, every submodule is a summand.

The number of inequivalent (ordinary) irreducible representations of G is equal to the numberof conjugacy classes of G.

If S is an irreducible CG-module and M is any CG-module, then the number of compositionfactors of M isomorphic to S is equal to dim HomCG(S,M).

Algebraic geometry

References: Part II course, Reid, §6 below.

You should know about affine varieties, polynomial functions, the Zariski topology (in particular,Zariski denseness) and the Nullstellensatz.

3

1 Semisimple algebras

Conventions All rings are associative and have an identity, which we denote by 1 or 1R. Modulesare always left R-modules.

(1.1) Definition A C-algebra is a ring R which is also a C-vector space, whose notions of additioncoincide, and for which

λ(rr′) = (λr)r′ = r(λr′)

for all r, r′ ∈ R and λ ∈ C. There are obvious notions of subalgebras and algebra homomor-phisms.

Usually, our algebras will be finite-dimensional over C.

(1.2) Examples and remarks

(a) C is a C-algebra, and Mn(C) is a C-algebra.

If G is a finite group then the group algebra CG = ∑g αgg : αg ∈ C is a C-algebra with pointwise addition:

∑g αgg +

∑g βgg =

∑g(αg + βg)g

multiplication:∑g

(∑hh′=g αhβh′

)g

Recall the correspondence between representations ofG over C and finite-dimensional CG-modules.

(b) If R,S are C-algebras then the tensor product R⊗S = R⊗C S is a C-algebra. [More on ⊗ canbe found in Atiyah–MacDonald p. 30.]

(c) If 1R = 0R then R = 0R is the zero ring. Otherwise, for λ, µ ∈ C, we have λ1R = µ1R if andonly if λ = µ, so we can identify λ ∈ C with λ1R ∈ R.

With the identification C → R we can view C as a subalgebra of R.

(d) An R-module M becomes a C-space via λm = (λ1R)m for λ ∈ C,m ∈M .

Given any R-module N , HomR(M,N) is a C-subspace of HomC(M,N). In particular, it is aC-space, with structure

(λϕ)(m) = λϕ(m) = ϕ(λm) ∀λ ∈ C, m ∈M, φ ∈ HomR(M,N)

(e) If M be an R-module. Then EndR(M) = HomR(M,M) is a C-algebra with multiplication givenby composition of maps:

(ϕψ)(m) = ϕ(ψ(m)) ∀m ∈M, φ, ψ ∈ EndR(M)

In particular, if V is a C-space then EndC(V ) is an algebra. Recall that EndC(V ) ∼= Mn(C), wheren = dimC(V ).

(f) Let R be a C-algebra and X ⊆ R be a subset. Define the centraliser of X in R by

cR(X) = r ∈ R : rx = xr ∀x ∈ X

It is a subalgebra of R. The centre of R is cR(R) = Z(R).

(g) Let R be C-algebra and M be a R-module. Then the map αR→ EndC(M) given by α(r)(m) = rmfor r ∈ R, m ∈M is a map of C-algebras.

Now, EndR(M) = cEndC(M)(αR), and so αR ⊆ cEndC(M)(EndR(M)).

(h) An R-module M is itself naturally an EndR(M)-module, where the action is evaluation, andthe action commutes with that of R. The inclusion α as in (g) says that elements of R act asEndR(M)-module endomorphisms of M .

Given an R-module N , HomR(M,N) is also an EndR(M)-module, where the action is compositionof maps.

4

(1.3) Lemma Let R be a finite-dimensional C-algebra. Then there are only finitely many isomor-phism classes of irreducible R-modules. Moreover, all the irreducible R-modules are finite-dimensional.

Proof. Let S be an irreducible R-module and pick 0 6= x ∈ S. Define a map R→ S by r 7→ rx. Thismap has nonzero image, so its image is S by irreducibility. In particular,

dimC(S) ≤ dimC(R) <∞

But S must occur in any composition series of R, so by the Jordan–Holder theorem, there are onlyfinitely many isomorphism classes of irreducible modules.

(1.4) Lemma (Schur’s lemma) Let R be a finite-dimensional C-algebra. Then

(a) If S T are nonisomorphic irreducible R-modules then HomR(S, T ) = 0.

(b) If S is an irreducible R-module then EndR(S) ∼= C.

Proof

(a) If f : R → S is a map of algebras then ker(f) ≤ R and im(f) ≤ S. By irreducibility, ker(f) = 0or R and im(f) = 0 or S. If ker(f) = 0 then R ∼= im(f) = S, contradicting nonisomorphism; sowe must have ker(f) = R and im(f) = 0, i.e. f = 0.

(b) Clearly D = EndR(S) is a division ring. Since S is finite-dimensional, so is D. Thus, if d ∈ D,then the elements 1, d, d2, . . . are linearly dependent. Let p ∈ C[X] be a nonzero polynomial withp(d) = 0.

Since C is algebraically closed, p factors as

p(X) = c(X − a1) · · · (X − an)

for some 0 6= c ∈ C and ai ∈ C. Hence

(d− a11D) · · · (d− an1D) = 0

But, being a division ring, D has no zero divisors, so one of the terms must be zero. Thusd = aj1D ∈ C1D. But d ∈ D was arbitrary, so D = C1D ∼= C.

(1.5) Definition An R-module is semisimple (or completely reducible) if it is a direct sum ofirreducible submodules. A finite-dimensional C-algebra R is semisimple if it is so as an R-module.

(1.6) Proposition

(a) Every submodule of a semisimple module is semisimple, and every such submodule is a directsummand.

(b) Every quotient of a semisimple module is semisimple.

(c) Direct sums of semisimple modules are semisimple.

In fact, M is semisimple if and only if every submodule of M is a direct summand. The proofs of thisand of (1.6) are left as an exercise.

Remarks

1. R a semisimple C-algebra ⇒ any R-module is semisimple.

2. G a finite group ⇒ CG is semisimple. (This is precisely Maschke’s theorem.)

(1.7) Definition If R and S are C-algebras, M is an R-module and N is an S-module, then M ⊗Nhas the structure of an R-module via

r(m⊗ n) = rm⊗ n, r ∈ R, m ∈M, n ∈ N

5

and the structure of an S-module via

s(m⊗ n) = m⊗ sn, s ∈ S, m ∈M, n ∈ N

and these actions commute:

r(s(m⊗ n)) = r(m⊗ sn) = rm⊗ sn = s(rm⊗ n) = s(r(m⊗ n))

hence the images of R,S in EndC(M ⊗N) commute.

If N has basis e1, . . . , ek then the map M⊕k →M ×N given by

(m1, . . . ,mk) 7→ m1 ⊗ e1 + · · ·+mk ⊗ ek

is an isomorphism of R-modules.

Similarly, if M has basis f1, . . . , f` then the map N⊕` →M ⊗N given by

(n1, . . . , n`) 7→ n1 ⊗ f1 + · · ·+ n` ⊗ f`

is an isomorphism of S-modules.

(1.8) Lemma Let M be a semisimple R-module. Then the evaluation map

⊕

S

S ⊗HomR(S,M)→M

is an isomorphism of R-modules and of EndR(M)-modules. Here S runs over a complete set of noni-somorphic irreducible R-modules and we use the action of R on S and of EndR(M) on HomR(S,M).

Proof Check that the map is a map of R-modules and of EndR(M)-modules. It’s an isomorphismof vector spaces, by observing that we can reduce to the case where M is irreducible and apply Schur’slemma (1.4).

(1.9) Lemma Let M be a finite-dimensional semisimple R-module. Then

EndR(M) ∼=∏

S

EndC HomR(S,M)

where S runs over a complete set of nonisomorphic irreducible R-modules.

Proof Observe that the product E =∏S EndC HomR(S,M) acts naturally on S ⊗ HomR(S,M)

and, since this action commutes with that of R, there exists a homomorphism E → EndR(M), whichis injective.

Now count dimensions by (1.8), M is isomorphic to a direct sum of dimC HomR(S,M) copies of eachirreducible module S, so by (1.4),

dimC EndR(S,M) =∑S(dimC HomR(S,M))2 = dimCE

(1.10) Theorem (Artin–Wedderburn Theorem) Any finite-dimensional semisimple C-algebra isisomorphic to a product

R =

k∏

i=1

EndC(Vi)

where Vi are finite-dimensional vector spaces. Conversely, if R has this form then it is semisimple, thenonzero Vis form a complete set of nonisomorphic irreducible R-modules and, as an R-module, R isisomorphic to a direct sum of dimC Vi copies of each Vi.

Proof We’ll take this in steps.

Step 1 (Deal with the product) Without loss of generality, assume each Vi is nonzero. The Vi arenaturally R-modules, with the factors other than EndC(Vi) acting as 0.

6

Now EndC, and hence R, has transitively on Vi − 0, thus the Vi are irreducible R-modules.

If Vi has basis ei1, . . . , eimi then the map

R→ V1 ⊕ · · · ⊕ V1︸ ︷︷ ︸m1 copies

⊕ · · · ⊕ Vk ⊕ · · · ⊕ Vk︸ ︷︷ ︸mk copies

given by(f1, . . . , fk) 7→ (f1(e11), . . . , f1(e1m1), . . . , fk(ek1), . . . , fk(ekmk))

is an injective map of R-modules, and hence is an isomorphism by counting dimensions. Thus R issemisimple.

The Vi form a complete set of irreducible R-modules (by Jordan–Holdeer, as in (1.3)), and they arenonisomorphic: if i 6= j then (0, . . . , 0, 1, 0, . . . , 0) annihilates Vj but not Vi.

Step 2 Let R be any ring. Now the natural map

α : R→ EndEndR(R)(R)

sending r to the map x 7→ rx is an isomorphism. It’s injective, and if θ ∈ EndEndR(R)(R) then itcommutes with the endomorphisms αr ∈ EndR(R), where αr(x) = xr. Now if r ∈ R then

θ(r) = θ(1r) = θ(αr(1)) = αr(θ(1)) = θ(1)r

so θ acts by left-multiplication. Hence θ = α(θ(1)) ∈ α(R), so α is surjective.

Assume R is a semisimple C-algebra. Then EndR(R) is semisimple by (1.9), and so R is a semisimpleEndR(R)-module. By (1.9) and the isomorphism α, R has the required form.

(1.11) Lemma If R is semisimple and M is a finite-dimensional R-module then the natural map

α : R→ EndEndR(M)(M)

sending r to the map m 7→ rm is surjective.

Proof Let I = M = r ∈ R : rM = 0 be the annihilator of M ; then kerα = I. Now, R/I issemisimple and M is an R/I-module, and EndR(M) = EndR/I(M). So we can replace R by R/I andsuppose that M is faithful (and α is injective).

By (1.9), EndR(M) =∏S EndC HomR(S,M) and, since M is faithful, all the spaces HomR(S,M)

are nonzero because they are precisely the simple EndR(M)-modules. By (1.8), M is isomorphic (asan EndR(M)-module) to

⊕S S ⊗ HomR(S,M), so it’s the direct sum of dimC S copies of the simple

module HomR(S,M) for each S.

As in (1.9), this implies that

dimC EndEndR(M)(M) =∑

S

(dimC S)2

But this is equal to dimCR, so α is an isomorphism.

Finally, it’s worth noting:

(1.12) Lemma Let R be a C-algebra and let e ∈ R be ‘almost idempotent’ in the sense that e2 = λefor some 0 6= λ ∈ C. Then for any R-module M , we have an isomorphism of EndR(M)-modules

HomR(Re,M) ∼= eM

Proof eM is an EndR(M)-submodule of M : if ϕ ∈ EndR(M) and em ∈ eM then ϕ(em) = eϕ(m) ∈eM .

Replacing e by eλ , we may suppose that e is idempotent. Now ther exists a map of EndR(M)-modules

Hom(Re,M)→ eM

given by ϕ 7→ ϕ(e), whose inverse sends m to the map r 7→ rm.

7

Chapter I: Representation theory of the symmetric group

2 Irreducible modules for the symmetric group

Recall the correspondence between representations ρ : G→ GL(V ) and CG-modules given by settingg · v = ρ(g)v for g ∈ G and v ∈ V . The trivial representation G→ C× = GL(C) is given by g 7→ 1,and the corresponding CG-module is C, sometimes written CG to emphasise the context.

If X ⊆ 1, . . . , n, write SX for the subgroup of Sn fixing every number outside of X.

Let ε : Sn → ±1 be the sign representation, and write

εσ =∏

1≤i<j≤n

σ(i)− σ(j)

i− j

Throughout this section, write A = CSn; this is a finite-dimensional semisimple C-algebra.

(2.1) Definition For λi ∈ N0, we say λ = (λ1, λ2, λ3, . . . ) is a partition of n, and write λ ` n, ifλ1 ≥ λ2 ≥ λ3 ≥ · · · and

∑i λi = n. Write (λa11 , λa22 , . . . , λakk ) to denote

(λ1, . . . , λ1︸ ︷︷ ︸a1

, λ2, . . . , λ2︸ ︷︷ ︸a2

, . . . , λk, . . . , λk︸ ︷︷ ︸ak

, 0, 0, . . . )

with λ1 > λ2 > · · · > λk.

Partitions of n correspond with conjugacy classes of Sn, e.g.

(5, 22, 1) ↔ (• • • • •)(• •)(•)

If λ, µ ` n, write λ < µ if and only if there is some i ∈ N with λj = µj for all j < i and λi < µi. Thisis the lexicographic (or dictionary) order on partitions. It is a total order on the set of partitionsof n. E.g.

(5) > (4, 1) > (3, 2) > (3, 12) > (22, 1) > (2, 13) > (15)

(2.2) Definition Partitions λ ` n may be geometrically represented by their Young diagram

[λ] = (i, j) : i ≥ 1, 1 ≤ j ≤ λi ⊆ N× N

If (i, j) ∈ [λ] then it is called a node of [λ]. The kth row (resp. column) of a diagram consists ofthose nodes whose first (resp. second) coordinate is k.

Example If λ = (4, 22, 1) then

[λ] =

• • • • ← λ1

• • ← λ2

• •• ← λ3

The height (or length) of λ is the length of the first column of [λ], i.e. maxj : λj 6= 0, and isdenoted by ht(λ).

(2.3) Definition A λ-tableau tλ is one of the n! arrays of integers obtained by replacing each nodein [λ] by one of the integers 1, . . . , n, with no repeats. E.g. the following are (4, 3, 1)-tableaux

t1 =1 2 4 53 6 78

, t2 =4 5 7 32 1 86

Equivalently, a λ-tableau can be regarded as a bijection [λ]→ 1, 2, . . . , n.

8

Sn acts on the set of [λ]-tableaux, e.g. (1 4 7 8 6)(2 5 3) sends t1 to t2. Formally, given a tableau tλ andπ ∈ Sn, the composition of the functions π and tλ gives a new λ-tableau πtλ, where (πtλ)(x) = π(tλ(x)).

Given a partition λ ` n, we want to pick one representative of all the corresponding λ-tableaux. Lettλ be the standard λ-tableau, i.e. the tableau numbered in order from left to right and from top tobottom, e.g.

t(5,22,1) =

1 2 3 4 56 78 910

Define the row stabilizer Rt and column stabilizer Ct of a tableau t by

σ ∈ Rt ⇔ each i ∈ 1, . . . , n is in the same row of t and σtσ ∈ Ct ⇔ each i ∈ 1, . . . , n is in the same column of t and σt

For instance, if t = t1 as in the examples after (2.3) above, then

Rt1 = S1,2,4,5 × S3,6,7 × S8Ct1 = S1,3,8 × S2,6 × S4,7 × S1

(2.4) Definition Let tλ be a λ-tableau. The Young symmetrizer is

h(tλ) =∑

r∈Rtλ , c∈Ctλ

εcrc ∈ A

Write hλ = h(tλ).

Examples

(a) If λ = (n), then h = h(t(n)) =∑σ∈Sn σ, called the symmetrizer in A.

Since ρ · h = h for all ρ ∈ Sn, Ah = Ch is the trivial representation.

(b) If λ = (1n), then h = h(t(1n)) =∑σ∈Sn εσσ, called the alternizer in A.

Since ρ · h = ερh, Ah = Ch is the sign representation.

Goal The left-ideals in A of the form Ahλ for λ ` n (known as Specht modules) form a completeset of nonisomorphic irreducible A-modules: see (2.14) below.

(2.5) Lemma If λ ` n, t = tλ is a tableau, σ ∈ Sn.

(a) Rt ∩ Ct = 1;

(b) The coefficient of 1 in h(tλ) is 1;

(c) Rσ(t) = σRtσ−1 and Cσ(t) = σCtσ

−1;

(d) h(σtλ) = σh(tλ)σ−1;

(e) Ah(tλ) ∼= Ahλ as A-modules.

Proof (a)–(d) are clear.

For (e), tλ = σtλ for some permutation σ ∈ Sn. Postmultiplication by σ then gives an isomorphismAh(tλ) ∼= Ahλ.

The following lemma is sometimes called the basic combinatorial lemma.

(2.6) Lemma Let λ, µ ` n with λ ≥ µ, and let tλ and tµ be tableaux. Then one of the following istrue:

9

(A) There exist distinct integers i and j occurring in the same row of tλ and the same column of tµ;

(B) λ = µ and tµ = rctλ for some r ∈ Rtλ and c ∈ Ctλ .

Proof Suppose (A) is false. If λ1 6= µ1 then λ1 > µ1, so [µ] has fewer columns than [λ]. Hence twoof the numbers in the first row of tλ are in the same column of tµ, so (A) holds, contradicting ourassumption. So λ1 = µ1.

Since (A) fails, some c1 ∈ Ctµ forces c1tµ to have the same members in the first row of tλ. Now ignorethe first rows of tλ and c1tµ. The same argument implies that λ2 = µ2 and there is c2 ∈ Ctµ such thattλ and c2c1tµ have the same numbers in each of their first two rows.

Continuing in this way we see that λ = µ and there exists c′ ∈ Ctλ such that tλ and c′tµ have thesame numbers in each row. Then rtλ = c′tµ for some r ∈ Rtλ .

Now definec = r−1(c′)−1r ∈ r−1c′Ctµ(c′)−1r = r−1Cc′tµr = Cc′tµ = Ctλ

Then tµ = rctλ.

(2.7) Lemma If σ ∈ Sn cannot be written as rc for any r ∈ Rtλ and c ∈ Ctλ then there existtranspositions u ∈ Rtλ and v ∈ Ctλ such that uσ = σv.

Proof Since (2.6)(B) fails for tλ and σtλ, there exist integers i 6= j in the same row of tλ and thesame column of σtλ. Take u = (i j) and v = σ−1uσ.

(2.8) Lemma Given a tableau tλ and a ∈ A, the following are equivalent:

(1) rac = εca for all r ∈ Rtλ and c ∈ Ctλ ;

(2) a = kh(tλ) for some k ∈ C.

Proof

(2)⇒(1) As r′ runs through Rtλ so does rr′, and as c runs through Ctλ so does c′c with εc′c = εc′εc.Calculation reveals that rh(tλ)c = εch(tλ).

(1)⇒(2) Let a =∑σ∈Sn

aσσ. If σ is not of the form rc then by (2.7) there exist transpositions u ∈ Rtλand v ∈ Ctλ such that uσv = σ.

By assumption, uav = εva, and the coefficient of σ gives auσv = εvaσ, so aσ = −aσ, and henceaσ = 0. The coefficient of rc in (1) gives a1 = εcarc ∈ C. Thus

a =∑

r,c

arcrc =∑

rc

εca1rc = a1h(tλ)

as required

(2.9) Lemma If a ∈ A then h(tλ)ah(tλ) = kh(tλ) for some k ∈ C.

Proof x = h(tλ)ah(tλ) satisfies (2.8)(1).

(2.10) Proposition Define fλ = dimCAhλ. Then

(a) h(tλ)2 =n!

fλh(tλ)

(b) fλ | n!

(c) h(tλ)Ah(tλ) = Ch(tλ)

In particular,fλn!h(tλ) is idempotent.

Proof

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(a) Let h = h(tλ). We already know that h2 = kh for some k ∈ C. Right-multiplication by h induces

a linear map h : A→ A. For a ∈ A, (ah)h = k(ah), so k|Ah is multiplication by k.

Extend a basis of Ah to a basis of A. With respect to this basis,

h ∼(kIfλ ∗

0 0

)

and so tr(h) = kfλ.

With respect to the basis Sn of A, h has matrix H whose entries are

Hστ = coefficient of σ in τh

But Hστ = 1, so tr(h) = n!.

Hence k =n!

fλ.

(b) The coefficient of 1 in h2 = kh is∑εc1εc2 ∈ Z where the sum runs over all r1, r2 ∈ Rtλ and

c1, c2 ∈ Ctλ such that r1c1r2c2 = 1.

(c) By (2.9), the only other possibility is hAh = 0; but h2 6= 0.

(2.11) Proposition Ah(tλ) is an irreducible A-module.

Proof Put h = h(tλ). Then Ah 6= 0 and A is semisimple, so it is enough to show that Ah isindecomposable.

Suppose Ah = U ⊕ V . Then Ch = hAh = hU ⊕ hV , so one of hU and hV is nonzero; without loss ofgenerality, hU 6= 0. Then hU = Ch, so Ah = AhU ⊆ U , so U = Ah and V = 0.

(2.12) Lemma If λ > µ are partitions and tλ, tµ are tableaux then h(tµ)Ah(tλ) = 0.

Proof By (2.6) there exist integers in the same row of tλ and the same column of tµ. Let τ ∈Rtλ ∩ Ctµ be the corresponding transposition. Then

h(tµ)h(tλ) = h(tµ)ττh(tλ) = −h(tµ)h(tλ)

so h(tµ)h(tλ) = 0.

Applying this to (σtλ, tµ) for σ ∈ Sn gives

0 = h(tµ)h(tλ) = h(tµ)σh(tλ)σ−1

and so h(tµ)σh(tλ) = 0.

Thus h(tµ)Ah(tλ) = 0.

(2.13) Lemma If λ 6= µ and tλ, tµ are tableaux, then Ah(tλ) Ah(tµ).

Proof Assume without loss of generality that λ > µ. If there exists an A-module isomorphismf : Ah(tλ) ∼= Ah(tµ) then

f(Ch(tλ))(2.10)

= f(h(tλ)Ah(tλ)) = h(tλ)f(Ah(tλ)) = h(tλ)Ah(tµ)(2.12)

= 0

which is a contradiction.

(2.14) Theorem The left-ideals Ahλ : λ ` n form a complete set of nonisomorphic irreducibleA-modules.

Proof They are irreducible by (2.11) and nonisomorphic by (2.13) and, since

|λ : λ ` n| = |ccls of Sn| = |irreducible CSn-modules|they form a complete set.

Further reading Chapter 4 of James’s lecture notes (Springer) – alternative proof using ‘poly-tabloids’ and certain permutation modules.

11

3 Standard basis of Specht modules

We first redefine what we mean by a ‘standard’ tableau.

(3.1) Definition t is a standard tableau if the numbers increase along the rows (left to right) andcolumns (top to bottom) of t.

The standard tableaux of shape λ are ordered such that tλ < t′λ if tλ is smaller than t′λ in the firstplace they differ when you read [λ] ‘like a book’. E.g. the standard (3, 2)-tableaux are

1 2 34 5

<1 2 43 5

<1 2 53 4

<1 3 42 5

<1 3 52 4

Let Fλ be the number of standard λ-tableaux.

Goal Fλ = fλ; recall from (2.10) that fλ = dimCAhλ.

First we show∑

λ`n

F 2λ = n!.

Notation λ/µ means there is some m with λ ` m, µ ` m− 1 and [µ] ⊆ [λ] ⊆ N× N.

This idea will be used in induction arguments.

(3.2) Lemma Let λ ` m. Then Fλ =∑

µ :λ/µ

Fµ.

Proof If tλ is standard then t−1λ (1, 2, . . . ,m − 1) is the diagram of a partition µ ` m − 1; and

tλ|[µ] must be standard. The converse is similar.

(3.3) Lemma If λ 6= π are partitions of m then

|ν : ν/λ and ν/π| = |τ : λ/τ and π/τ| ∈ 0, 1

Proof If ν/λ and ν/π then [ν] ⊇ [λ] ∪ [π]. Since [λ] and [π] differ in at least one place, we must infact have [ν] ⊇ [λ] ∪ [π].

Likewise, if λ/τ and µ/τ then [τ ] ⊇ [λ] ∩ [π] and we must have [τ ] = [λ] ∩ [π].

Now [λ] ∪ [π] and [λ] ∩ [π] are partitions, so

ν exists ⇔ |[λ] ∪ [π]| = m+ 1 ⇔ |[λ] ∩ [π]| = m− 1 ⇔ τ exists

This completes the proof.

(3.4) Lemma If λ ` m then (m+ 1)Fλ =∑

ν : ν/λ

Fν .

Proof by induction on m.

The case m = 1 is clear. Suppose the statement holds true for partitions of m− 1. Then

∑

ν : ν/λ

Fν(3.2)=

∑

ν : ν/λ

∑

π : ν/π

Fπ

= |ν : ν/λ|Fλ +

∑

ν,π : ν/λ,π 6=λ

Fπ

Using the fact that |ν : ν/λ| = |τ : λ/τ| + 1 together with (3.3) gives that our expression isequal to

(|τ : λ/τ|+ 1)Fλ +∑

τ,π :λ/τ,π/τ,π 6=λ

Fπ

= Fλ +∑

τ,π :λ/τ,π/τ

Fπ = Fλ +∑

τ :λ/τ

mFτ = Fλ +mFλ = (m+ 1)Fλ

12

as required.

(3.5) Lemma∑

λ`m

F 2λ = m!

Proof by induction on m.

The case m = 1 is clear. Suppose the statement holds true for partitions of m− 1. Then

∑

λ`m

F 2λ

(3.2)=

∑

λ`m,λ/τ

FλFτ(3.4)=

∑

τ`m−1

mF 2τ = m · (m− 1)! = m!

as required.

(3.6) Lemma If tλ > t′λ are standard then h(tλ)h(t′λ) = 0.

Proof It suffices to show that there are integers i 6= j lying in the same row of t′λ and the samecolumn of tλ. For then the corresponding transposition τ = (i j) ∈ Rt′λ ∩Ctλ gives the result as in theproof of (2.12).

It remains to find such integers i and j. Let x ∈ [λ] be the first place where tλ and t′λ first differ. Lett′λ = i and y = (t′λ)−1(i) ∈ [λ]. Then y must be below and to the left of x since tλ is standard. Inparticular, x cannot lie in the first column or the last row. Let z ∈ [λ] be in the same row as x andthe same column of y, and let j = tλ(z) = t′λ(z) be the common value of tλ and t′λ at z.

It is now elementary to verify that i and j satisfy the above assumption.

(3.7) Theorem CSn =⊕CSnh(tλ), where the direct sum runs over all standard tableaux tλ for

all partitions λ ` n.

Proof The sum is direct: suppose we have a nontrivial relation∑a(tλ)h(tλ) with a(tλ) ∈ CSn

are not all zero. Choose µ maximal (in the ordering on partitions) such that some a(tµ)h(tλ) 6= 0,and for this µ pick t′µ minimal (in the ordering on standard µ-tableaux) such that a(t′µ)h(t′µ) 6= 0.Multiplying the relation on the right by h(t′µ) gives a(t′µ)h(t′µ)2 = 0 by (3.6) and (2.12), and hencea(t′µ)h(t′µ) = 0, contradicting our assumption.

It is clear that the⊕CSnh(tλ) ≤ CSn. By Artin–Wedderburn, CSn ∼=

⊕fλCSnhλ as CSn-modules,

while in our direct sum there are Fλ copies of Chλ. By Jordan–Holder, Fλ ≤ fλ. But fλ ≤ Fλ since,by (3.5),

∑F 2λ = n! =

∑f2λ. Hence Fλ = fλ for each λ ` n, and so CSn ≤

⊕CSnh(tλ).

(3.8) Corollary The number of standard λ-tableaux is equal to dimC(CSnhλ).

Proof Hλ = hλ as in the proof of (3.7).

(3.9) Lemma Let M be a finite-dimensional CSn-module. Then M ∼=⊕h(tλ)M , where the direct

sum runs over all standard tableaux tλ for all partitions λ ` n.

Proof If∑m(tλ) = 0 is a nontrivial relation with m(tλ) ∈ h(tλ)M then choose µ minimal and t′µ

maximal with m(t′µ) 6= 0. Premultiply the relation by h(t′µ) to contradict (3.6) and (2.12). Finally,

⊕h(tλ)M

(1.12)∼=⊕

HomCSn(CSnh(tλ),M)

∼= HomCSn

(⊕CSnh(tλ),M

)∼= HomCSn(CSn,M) ∼= M

13

4 Character formula

Given a finite-dimensional CG-algebra M , its character is defined by

χM (g) = tr

(M →Mm 7→ gm

)

and χM is a class function G→ C, i.e. a function that is constant on the conjugacy classes of G. Ifα is a conjugacy class in G, write χM (α) for the common value of χM on α.

Given λ ` n, denote the character of the irreducible CSn-module CSnhλ by χλ. We will calculateχλ(α) and later use the formula to derive Weyl’s character formula for GLn(C).

Now Sn-class α comprises all permutations with fixed cycle type nαn · · · 2α21α1 , i.e. having αn n-cycles,. . . , α2 2-cycles and α1 1-cycles. Let nα = |α|.

(4.1) Lemma nα =n!

1α12α2 . . . nαn · α1!α2! . . . αn!

Proof Any permutation in α is one of the n! of the form

(∗)(∗) · · · (∗)︸ ︷︷ ︸α1

(∗∗)(∗∗) · · · (∗∗)︸ ︷︷ ︸α2

· · · (n︷ ︸︸ ︷∗ · · · ∗)(

n︷ ︸︸ ︷∗ · · · ∗) · · · (n︷ ︸︸ ︷∗ · · · ∗)︸ ︷︷ ︸

αn

Each such permutation can be represented in 1α1 . . . nαn ·α1! . . . αn! ways: each of the αk k-cycles canbe represented in k ways by cycling their components, and each block of αk k-cycles can be representedin αk! ways by permuting the position of the factors.

(4.2) Theorem (orthogonality relations) Let λ, µ ` n and let α and β be conjugacy classes in Sn.Then

(a)∑

ccls α

nαχλ(α)χµ(β) =

n! if λ = µ

0 if λ 6= µ

(b)∑

λ`n

χλ(α)χλ(β) =

n!/nα if α = β

0 if α 6= β

Proof Every element of Sn is real, i.e. conjugate to its inverse, and so

χλ(σ) = χλ(σ−1) = χλ(σ)

so χλ(σ) ∈ R for all σ ∈ Sn. These relations are then just the usual ones for general finite groups.

Notation Given x = (x1, . . . , xn) ∈ Cm, `1, . . . , `m ∈ Z, define

|x`1 , . . . , x`m | = det(x`ji )

Usually `j ≥ 0, in which case this is a homogeneous polynomial of degree

m∑

j=1

`j in x1, . . . , xm.

Example The Vandermonde determinant is given by

V (x) = |xm−1, . . . , 1|

(4.3) Lemma V = V (x) =∏i<j(xi − xj)

Proof Exercise.

In fact, if `i ≥ 0 for each i then |x`1 , . . . , x`m | is divisible by V . We define the Schur polynomial s`by

s`(x) =|x`1 , . . . , x`m ||xm−1, . . . , 1|

14

This is indeed a polynomial in x1, . . . , xm, provided that `i ≥ 0.

Example s(1,2,...,n)(x) = x1 · · ·xm.

(4.4) Lemma (Cauchy’s lemma) If xi, yj ∈ C for 1 ≤ i, j ≤ m and xiyj 6= 1 for all i, j, then

det

(1

1− xiyj

)=∣∣xm−1, . . . , 1

∣∣ ∣∣ym−1, . . . , 1∣∣

m∏

i,j=1

1

1− xiyj

Proof by induction on m. The case m = 1 is clear. Let m > 1.

Now1

1− xiyj− 1

1− x1yj=

xi − x1

1− x1yj· yj

1− xiyjSo subtracting the first row from each other row in the determinant, one can remove the factor (xi−x1)

from each row i 6= 1 and1

1− x1yjfrom each column. So

det

(1

1− xiyj

)=

(∏

i>1

(xi − x1)

)

m∏

j=1

1

1− x1yj

det

1 1 · · · 1y1

1−x2y1

y21−x2y2

· · · yn1−x2yn

y11−x3y1

y21−x3y2

· · · yn1−x3yn

......

. . ....

y11−xny1

y21−xny2 · · · yn

1−xnyn

︸ ︷︷ ︸=∆

Now subtract the first column in the matrix above from the other columns and use the fact that

yj1− xiyj

− y1

1− xiy1=

yj − y1

1− xiy1· 1

1− xiyjThis gives

∆ =

∏

j>1

(yj − y1)

(

m∏

i=2

1

1− xiy1

)det

1 0 0 · · · 0∗ 1

1−x2y21

1−x2y3· · · 1

1−x2yn

∗ 11−x3y2

11−x3y3

· · · 11−x3yn

......

.... . .

...∗ 1

1−xny21

1−xny3 · · · 11−xnyn

The result now follows by applying the induction hypothesis to this last determinant.

(4.5) Lemma Choose xi, yj ∈ C for 1 ≤ i, j ≤ n with modulus < 1. Then

det

(1

1− xiyj

)=

∑

`1>`2>···>`m≥0

∣∣x`1 , . . . , x`m∣∣ ∣∣y`1 , . . . , y`m

∣∣

Proof We use the expansion1

1− xiyj=∑

`≥0

(xiyj)`

The determinant is thus∑

σ∈Sm

εσ

m∏

i=1

(1 + xiyσ(i) + (xiyσ(i))

2 + · · ·)

and the monomial x`11 · · ·x`mm (`i ∈ N) occurs with coefficient

∑

σ∈Sm

m∏

i=1

y`iσ(i) =∣∣y`1 , . . . , y`m

∣∣

15

In particular, it this term is zero unless the `i are all distinct. So

det

(1

1− xiyj

)=

∑

`1,...,`m distinct

x`11 . . . x`mm∣∣y`1 , . . . , y`m

∣∣

=∑

`1>···>`m≥0

( ∑

π∈Sm

x`π(1)

1 . . . x`π(m)m

∣∣y`π(1) , . . . , y`π(m)∣∣)

=∑

`1>···>`m≥0

( ∑

π∈Sm

x`π(1)

1 . . . x`π(m)m επ

∣∣y`1 , . . . , y`m∣∣)

=∑

`1>···>`m≥0

∣∣x`1 , . . . , x`m∣∣ ∣∣y`1 , . . . , y`m

∣∣

which is what we needed.

Notation If λ = (λ1, . . . , λm) ∈ Zm, let `i = λi + m − i, i.e. `1 = λ1 + m − 1, . . . , `m = λm.Sometimes in the literature, the `i are denoted by βi and are thus called ‘β-numbers’. Note that

(1) λ1 ≥ · · · ≥ λm if and only if `1 > · · · > `m

(2) If∑i λi = n and λi ≥ 0 then

∣∣x`1 , . . . , x`m∣∣

|xm−1, . . . , 1| is a polynomial of degree m in the xis.

Write Λ+(m,n) for the set of all partitions λ of n into ≤ m parts.

(4.6) Definition Given x1, . . . , xm, y1, . . . , ym ∈ C, then for any i ∈ N, set

si = xi1 + · · ·+ xim, ti = yi1 + · · ·+ tim

These are called the power sums (sometimes referred to in the literature as Newton sums).

(4.7) Lemma∑

λ∈Λ+(m,n)

∣∣x`1 , . . . , x`m∣∣

|xm−1, . . . , 1|

∣∣y`1 , . . . , y`m∣∣

|ym−1, . . . , 1| =1

n!

∑

ccls α

nαsα11 . . . sαnn tα1

1 . . . tαnn

Note that both quotients on the left-hand side genuinely are polynomials (in fact, Schur polynomials),so they make sense even if the xi, yi are not distinct.

Proof Both sides are polynomials, so without loss of generality we may take all the xi, yi to havemodulus < 1. Then

log

n∏

i,j=1

1

1− xiyj=

n∑

i,j=1

(xiyj

1+

(xiyj)2

2+

(xiyj)3

3+ · · ·

)

=s1t1

1+s2t2

2+s3t3

3+ · · ·

and hence

n∏

i,j=1

1

1− xiyj= exp

(s1t1

1+s2t2

2+s3t3

3+ · · ·

)

=

∞∑

n=0

1

n!

(s1t1

1+s2t2

2+s3t3

3+ · · ·

)n

=∑

(∗)

(1

n!· n!

α1!α2! · · ·

(s1t1

1

)α1(s2t2

2

)α2

· · ·)

where (∗) denotes the sum over all n and all sequences α1, α2, . . . of nonnegative integers with onlyfinitely many nonzero terms and α1 + α2 + · · · = n. But then

n∏

i,j=1

1

1− xiyj=∑

(∗)

sα11 sα2

2 · · · tα11 tα2

2 · · ·1α12α2 · · ·α1!α2! · · ·

16

By (4.4)(Cauchy) and (4.5), we have

∑

`1>`2>···>`m≥0

∣∣x`1 , . . . , x`m∣∣

|xm−1, . . . , 1|

∣∣y`1 , . . . , y`m∣∣

|ym−1, . . . , 1| =

m∏

i,j=1

1

1− xiyj=∑

(∗)

sα11 sα2

2 · · · tα11 tα2

2 · · ·1α12α2 · · ·α1!α2! · · ·

so we can equate terms which are of degree n in the xi to obtain the desired result.

(4.8) Definition For λ = (λ1, . . . , λm) ∈ Zm and α a conjugacy class in Sn, let ψλ(α) be thecoefficient of the monomial xλ1

1 . . . xλmm in sα11 . . . sαnn , i.e.

sα11 . . . sαnn =

∑

λ∈Zmψλ(α)xλ1

1 . . . xλmm

Remarks

(1) ψλ is a class function for Sn.

(2) ψλ(α) = 0 if any λi is negative or if

m∑

i=1

λi 6= n.

(3) ψλ is a symmetric function of the λis.

Notation Setωλ(α) =

∑

π∈Sm

επψ(`π(1)+1−m, `π(2)+2−m, ... , `π(m))(α)

The idea now is to show that ωλ = χλ whenever λ ` n.

(4.9) Lemma (Character formula)

sα11 . . . sαnn

∣∣xm−1, . . . , 1∣∣ =

∑

λ∈Λ+(m,n)

ωλ(α)∣∣x`1 , . . . , x`m

∣∣

Proof The left-hand side is equal to

∑

λ∈Zmψλ(α)xλ1

1 . . . xλmm∑

τ∈Sm

ετxm−1τ(1) . . . x

0τ(m)

which, by combining terms, is equal to

∑

λ∈Zm

∑

τ∈Sm

ετψλ(α)xλτ(1)+m−1

τ(1) . . . xλτ(m)

τ(m)

Let `i = λτ(i) +m−i—note that this is not our usual convention—then since the ψλ(α) are symmetric,our expression is equal to

∑

λ∈Zm

∑

τ∈Sm

ψ(`1+1−m, ... , `m)(α)x`1τ(1) . . . x`mτ(m)

=∑

λ∈Zmψ(`1+1−m, ... , `m)(α)

∣∣x`1 , . . . , x`m∣∣

Whenever the `i are not distinct, the corresponding term gives zero. So setting λi = `i + i −m andusing the fact that `1 > · · · > `m if and only if λ1 ≥ · · · ≥ λm, our expression is equal to

∑

λ1≥···≥λm

∑

π∈Sm

ψ(`π(1)+1−m, ... , `π(m))επ∣∣x`1 , . . . , x`m

∣∣

The terms for which λ is not a partition of n are zero, for if λm < 0 then `m +π−1(m)−m < 0. Thuswe have the desired equality.

17

(4.10) Lemma (Orthogonality of ω) Let λ, λ′ ` n be partitions of ≤ m parts. Then

∑

ccls α

nαωλ(α)ωλ′(α) =

n! if λ = λ′

0 if λ 6= λ′

Proof By (4.7) we have

∑

λ∈Λ+(m,n)

∣∣x`1 , . . . , x`m∣∣ ∣∣y`1 , . . . , y`m

∣∣ =1

n!

∑

ccls α

nαsα11 . . . sαnn tα1

1 . . . tαnn∣∣xm−1, . . . , 1

∣∣ ∣∣ym−1, . . . , 1∣∣

By (4.9) this is equal to

1

n!

∑

ccls α

∑

λ,λ′∈Λ+(m,n)

nαωλ(α)ωλ′(α)∣∣x`1 , . . . , x`m

∣∣∣∣∣y`′1 , . . . , y`′m

∣∣∣

where `i come from λi and `′i come from λi in the usual way.

As λ, λ′ vary, the polynomials∣∣x`1 , . . . , x`m

∣∣ and∣∣∣y`′1 , . . . , y`′m

∣∣∣ are linearly independent in the poly-

nomial ring C[x1, . . . , xm, y1, . . . , ym], so the result follows.

Now that we have trudged through lots of combinatorics we will reintroduce the symmetric group.

(4.11) Lemma If λ ∈ Λ+(n,m) then ψλ(α) is the character of the CSn-module CSnrλ, where

rλ =∑

σ∈Rtλ

σ

Proof (X-rated) Take the character χ and of CSnrλ and suppose σ ∈ α, where α is a conjugacyclass in Sn. Write R = Rtλ , and find a coset decomposition

Sn =

N⋃

i=1

giR

for the transversal 1 = g1, g2, . . . , gN. (Recall: a transversal is a minimal set of coset representa-tives.) Then CSnrλ has basis girλ1≤i≤N , which we now use to calculate the traces. The CSn-actionis given by σgirλ = gjrλ for σ ∈ Sn, where σgi ∈ gjR. Thus

χ(α) =∣∣1 ≤ i ≤ N : g−1

i σgi ∈ R∣∣

We will now take this and bash it very hard lots of times until enough juice comes out to give us whatwe want.

Now g−1σg ∈ R if and only if g ∈ giR, where g−1i σgi ∈ R, and |R| = ∏m

1 λj !, so

χ(α) =1∏m

1 λj !

∣∣g ∈ Sn : g−1σg ∈ R∣∣

Now since g−1σg = h−1σh if and only if gh−1 ∈ CSn(σ), and each value taken by g−1σg for g ∈ Sn istaken by |CSn | elements. By (4.1) we have

χ(α) =

(1∏m

1 λj !1α12α2 . . . nαnα1!α2! . . . αn!

)|α ∩R|

since |CSn(σ)| = n!/nα. Now a permutation τ ∈ α ∩ R restricts to a permutation of the numbers inthe ith row of tλ. If this restriction involves, say, αij j-cycles, then the αij satisfy

αi1 + 2αi2 + · · ·+ nαin = λi (1 ≤ i ≤ m)α1j + α2j + · · ·+ αmj = αj (1 ≤ j ≤ n)

(∗)

18

The number of permutations in R of this type is(

λ1!

1α11 . . . nα1nα11! . . . α1n!

)(λ2!

1α21 . . . nα2nα21! . . . α2n!

)· · ·(

λm!

1αm1 . . . nαmnαm1! . . . αmn!

)

and so

χ(α) =∑[(

α1!

1α11 . . . nα1nα11! . . . α1n!

)· · ·(

αm!

1αm1 . . . nαmnαm1! . . . αmn!

)](†)

where the sum runs over all αij satisfying (∗). By the multinomial theorem,

sα11 . . . sαnn =

∑[(α1!xα11

1 . . . xα1mm

1α11 . . . nα1nα11! . . . α1n!

)· · ·(

αm!xαn11 . . . xαnmm

1αm1 . . . nαmnαm1! . . . αmn!

)]

where the sum runs over all αij ∈ N satisfying

α1j + α2j + · · ·+ αmj = αj

for all 1 ≤ j ≤ n. So ψλ(α), which is the coefficient of xλ11 . . . xλmm , is equal to the right-hand side of

(†), and hence it equals χ(α).

(4.12) Lemma Let λ, µ ` n with µ ≤ λ. The irreducible module CSnhµ is isomorphic to asubmodule of CSnrλ if and only if λ = µ.

Proof With µ < λ and σ ∈ Sn, by (2.6) there exist two integers in the same row of tλ and the samecolumn of σ−1tµ, so if τ is their transposition then στσ−1 ∈ Ctµ . And

hµσrλ = hµστσ−1στrλ = −hµσrλ ⇒ hµσrλ = 0

Thus 0 = hµCSnrλ ∼= HomCSn(CSnhµ,CSnrλ) by (1.12).

Conversely suppose µ = λ. Then

hλrλ∑

σ∈Ctλ

εσσ = h2λ 6= 0

so 0 6= hλCSnrλ ∼= HomCSn(CSnhλ,CSnrλ).

Note that, in general, CSnhλ is not a submodule of CSnrλ.

(4.13) Lemma If λ ∈ Λ+(n,m) then ωλ = χλ.

Proof We’ll take this in steps.

Step 1 ωλ is a Z-linear combination of ψν with ν ≥ λ and with coefficient of ψλ equal to 1.

If π ∈ Sn and µπ is the partition with parts

`π(1) + 1−m, . . . , `π(m)

i.e. with parts λi + π−1(i)− 1. Since ψλ(α) is symmetric in the λi, we can write

ωλ =∑

π∈Sn

επψµπ

If π = 1 then µπ = λ, so we’re done.

If π 6= 1, then λ1+π−1(1)−1 ≥ λ (with equality if and only if π−1(1) = 1; and then λ2+π−1(2)−2 ≥ λ2

with equality if and only if π−1(2) = 2; and so on. Thus µπ > λ. /

Step 2 ωλ =∑

ν`n

kλνχν with kλν ∈ Z, kλλ > 0 and kλν = 0 if ν < λ.

By (4.11) and (4.12), ψλ is an N-linear combination χµs with µ ≥ λ and with nonzero coefficient ofχλ. Thus ωλ is a Z-linear combination of χνs with ν ≥ λ and with positive coefficient of χλ. /

19

Step 3 The result follows.

We know by (4.10) that∑

ccls α

nαωλ(α)ωµ(α) =

n! if λ = µ

0 if λ 6= µ

In the case λ = µ, orthogonality of χλ implies∑

ν`n

k2λν = 1, and thus kλν = 0 if λ 6= µ and kλλ = 1. /

So we’re done.

The kλνs in the above proof are sometimes referred to as Kostka numbers in the literature.

At long last, take m ∈ N, x1, . . . , xm ∈ C arbitrary, `i = λi +m− i for λ ` n, and si = xi1 + · · ·+ xim.

(4.14) Theorem sα11 . . . sαnn

∣∣xm−1, . . . , 1∣∣ =

∑

λ∈Λ+(m,n)

χλ(α)∣∣x`1 , . . . , x`n

∣∣

Proof (4.9) and (4.13).

Remarks

(1) With m ≥ n we can ensure that the right-hand side of the equation in (4.14) involves all partitionsof n.

(2) If λ ∈ Λ+(m,n) then χλ(α) is the coefficient of the monomial x`11 . . . x`mm in the expansion ofsα1

1 . . . sαnn .

20

5 The hook length formula

We already know by (3.8) that the dimension of an irreducible CSn-module is the number of standardtableaux. We give some results which allow for easy calculation of this dimension.

(5.1) Theorem (Young–Frobenius formula) If λ ∈ Λ+(n,m) then

fλ = degχλ = n! ·∏

1≤i<j≤n(`i − `j)∏ni=1 `i!

Proof degχλ = χλ(1) is the coefficient of x`11 . . . x`mm in the expansion of

(x1 + · · ·+ xm)n∣∣xm−1, . . . , 1

∣∣ =∑

τ∈Sn

(x1 + · · ·+ xm)nετxτ(1)−11 . . . xτ(m)−m

m

By the multinomial theorem, this coefficient equals

∑

τ∈Sn

n!

(`1 + 1− τ(1))! . . . (`m + 1− τ(m))!

= n!

∣∣∣∣1

(`−m+ 1)!, . . . ,

1

(`− 1)!,

1

`!

∣∣∣∣

= n! · 1∏mj=1 `j !

|. . . , `(`− 1), `, 1|

= n! · 1∏mj=1 `j !

∣∣`m−1, . . . , `2, `, 1∣∣

This determinant is a Vandermonde determinant, so applying (4.3) gives the result.

(5.2) Definitions

(a) Let λ ` n. The (i, j)-hook of λ is the intersection of an infinite Γ–shape having (i, j) as its cornerwith [λ]. For example, the (2, 2)–hook of (42, 3) is shown as stars (?) in the following tableau:

• • • •• ? ? ?• ? •

(b) The (i, j)-hook of [µ] comprises the (i, j)-node along with the (µi − j) nodes to the right of it(the arm) and the (µ′j − i) nodes below it (the leg). [Here, [µ] has column lengths µ′1, µ

′2, . . . .]

The length of the (i, j)-hook is hij = µi + µ′j + 1− i− j.

(c) Replacing the (i, j)-node of [µ] by hij for each node gives the hook graph. For example, thehook graph of (42, 3) is

6 5 4 25 4 3 13 2 1

(d) A skew hook is a connected part of the rim of [µ] which can be removed to leave a properdiagram. For example, the skew 4-hooks of (42, 3) are

• • • •• • ? ?• ? ?

and• • • ?• • ? ?• • ?

There is a natural one-to-one correspondence between hooks and skew hooks in a given tableau.For each node (i, j) there is a unique skew hook starting in the ith row and ending in the jth

column, and it corresponds with the (i, j)-hook.

21

Theorem (5.3) (hook length formula) [Frame, Robinson & Thrall, 1954]

fλ =n!∏

(i,j)∈[λ] hij

Proof Suppose λ has m parts. By (5.1) we have to show that

m∏

k=i+1

(`i − `k)

λi∏

j=1

hij = `i! for each i

The (combined) product on the left-hand side is a product of λi +m− 1 = `i terms, so it suffices toshow that they are precisely 1, 2, . . . , `i in some order. Now

`i − `m > `i − `m−1 > · · · > `i − `i+1

hi1 > hi2 > · · · > hiλi

Since λ has m parts and λ′1 is the length of the first column, λ′1 = m, and hi1 = λi. So each term is≤ `i. Thus it is enough to show that hij 6= `i − `k for any j, k.

However if r = λ′j then λr ≥ j and λr+1 < j, so

hij − `i + `r+1 = (λi + r − 1− j + 1)− (λi +m− i) + (λr +m− r) = λr + 1− j > 0

hij − `i + `r+1 = (λi + r − 1− j + 1)− (λi +m− i) + (λr+1 +m− r − 1) = λr+1 − j < 0

So `i − `r < hij < `i − `r+1.

This theorem is often referred to as the FRT theorem in the literature.

Example When n = 11 and λ = (6, 3, 2), the hook graph is given by

8 7 6 5 3 2 14 3 12 1

and so

fλ =11!

8.7.5.4.32.22.13= 990

Remarks

(1) The determinantal form of the Young–Frobenius formula is given by

fλ = n!

∣∣∣∣1

(λi − i+ j)!

∣∣∣∣

where λ ∈ Λ+(n,m) and the determinant on the right is of an m ×m matrix and we adopt the

convention1

m!= 0 if m < 0. For a proof see e.g. Sagan or James §19.

(2) The equation∑

λ`n

f2λ = n! can be proved as a purely combinatorial statement without reference to

representations, using the so-called ‘RSK algorithm’ (for Robinson–Schensted–Knuth).

(3) The Murnaghan–Nakayama rule (e.g. example sheet 1, question 9) states that if πρ ∈ Sn withρ an r-cycle and π a permutation of the remaining n− r numbers then

χλ(πρ) =∑

ν

(−)iχν(π)

where ν runs over partitions for which [λ]− [ν] is a skew r-hook of leg length i. (Note: leg lengthrefers to the leg of the hook with which the skew hook corresponds.)

22

For example,

χ(5,42)(

conj. class︷ ︸︸ ︷(5, 4, 3, 1)) = −χ(5,3) + χ(32,2) on (4, 3, 1)

= χ(22)

︸ ︷︷ ︸from (5,3)

− (χ(3,1) + χ(2,12)

︸ ︷︷ ︸from (32,2)

) on (3, 1)

= −χ(1) + 0 + 0 on (1)

= −1

23

Chapter II: Representation theory of general linear groups

6 Multilinear algebra and algebraic geometry

Let V and W be finite-dimensional CG-modules, where G is any (possibly infinite) group. Recall thefollowing:

Tensor products: V ⊗C W is a CG-module with the diagonal action g(v ⊗ w) = gv ⊗ gw.

Basic properties:– V ⊗ C ∼= C– V ⊗W ∼= W ⊗ V– (V ⊗W )⊗ Z ∼= V ⊗ (W ⊗ Z)

– If θ : V → V ′ and ϕ : W →W ′ are CG-maps then there is a CG-map defined by

θ ⊗ ϕ : V ⊗W → V ′ ⊗W ′v ⊗ w 7→ θ(v)⊗ ϕ(w)

Hom spaces: HomC(V,W ) is a CG-module via(∑

i

αigi

)(f)(v) =

∑

i

αigi(f(g−1i v))

Dual space: V ∗ = HomC(V,C).

Basic properties:– If V is 1-dimensional then V ∗ ⊗ V ∼= C.

– If θ : V →W is a CG-map then the dual θ∗ : W ∗ → V ∗ is too.

– The mapV ∗ ⊗W → HomC(V,W )f ⊗ w 7→ (v 7→ f(v)w)

is a CG-isomorphism.

Tensor powers: The nth tensor power of V is defined by

TnV = V ⊗n = V ⊗ · · · ⊗ V︸ ︷︷ ︸n copies

, T 0V = C

Basic properties:– If V has basis ei : 1 ≤ i ≤ m then V ⊗n has basis

ei1 ⊗ · · · ⊗ ein : 1 ≤ i1, . . . , in ≤ m

and hence dim(V ⊗n) = mn.

– V ⊗n is naturally a left CSn-module via

σ(v1 ⊗ · · · ⊗ vn) = vσ−1(1) ⊗ · · · vσ−1(n)

– The actions of CSn and CG commute, i.e.

σgx = gσx for all g ∈ G, σ ∈ Sn, x ∈ V ⊗n

(6.1) Definition The nth exterior power of V is ΛnV = V ⊗n/X, where

X = spanCx− εσσx : x ∈ V ⊗n, σ ∈ Sn

The image of v1 ⊗ · · · ⊗ vn in ΛnV under the quotient map will be denoted by v1 ∧ · · · ∧ vn.

Define the antisymmetric tensors by

TnVanti = x ∈ V ⊗n : σx = εσx for all σ ∈ Sn

(6.2) Lemma

24

(i) v1 ∧ · · · ∧ vn = εσvσ−1(1) ∧ · · · ∧ vσ−1(n) for all σ ∈ Sn.

This tells us, by considering transpositions, that v1 ∧ · · · ∧ vn = 0 whenever vi = vj for somei 6= j.

(ii) TnVanti and ΛnV are CG-modules.

(iii) ΛnV has basis ei1 ∧ · · · ∧ ein : 1 ≤ i1 < · · · < in ≤ n, and so dim ΛnV =

(m

n

).

In particular, ΛmV is one-dimensional and Λm+1V = Λm+2V = · · · = 0.

Proof An easy exercise.

Remark If V is m-dimensional over a field k then usually we define ΛnV = V ⊗n/X, where

X = spanCv1 ⊗ · · · ⊗ vn : vi = vj for some i 6= jIf char k 6= 2 then this definition reduces to (6.2).

(6.3) Lemma TnVanti = aV ⊗n, where a =∑

σ∈Sn

εσσ is the alternizer [cf. (2.4)], and the natural

map TnVanti → ΛnV is an isomorphism of CG-modules.

Proof If x is antisymmetric then

ax =∑

σ∈Sn

εσσx =∑

σ∈Sn

εσ(εσx) =∑

σ∈Sn

x = n!x

and so TnVanti ⊆ aV ⊗n.

Conversely, since σa = εσa for all σ ∈ Sn, any element of aV ⊗n must be antisymmetric.

We have a mapaV ⊗n → V ⊗n → ΛnV

which is a CG-homomorphism and whose kernel is X ∩ aV ⊗n ≤ aX, since x =1

n!ax for x antisym-

metric. But for y ∈ V ⊗n,

a(y − εσσy) = ay − εσ(εσa)y = ay − ay = 0

so ax = 0 (since x ∈ X). Hence the map is injective. And it is surjective: any x ∈ ΛnV is the image

of some y ∈ V ⊗n, but then x is the image of1

n!ay ∈ aV ⊗n. So the map is an isomorphism.

(6.4) Lemma Λn(V ∗) ∼= (ΛnV )∗.

Proof The quotient map V ⊗n ΛnV is surjective, so induces an inclusion

(ΛnV )∗ → (TnV )∗ = Tn(V ∗)

Now by the universal property of ΛnV (i.e. any multilinear map V × · · · × V︸ ︷︷ ︸n copies

→ C factors through

ΛnV ), the image of the inclusion is Tn(V ∗)anti, which is isomorphic to Λn(V ∗).

Now prepare for some deja vu in (6.5)–(6.8). . .

(6.5) Definition The nth symmetric power of V is SnV = V ⊗n/Y , where

Y = spanCx− σx : x ∈ V ⊗n, σ ∈ SnThe image of v1 ⊗ · · · ⊗ vn in SnV under the quotient map will be denoted by v1 ∨ · · · ∨ vn.

Define the symmetric tensors by

TnVsym = x ∈ V ⊗n : σx = x for all σ ∈ Sn

(6.6) Lemma

25

(i) v1 ∨ · · · ∨ vn = vσ−1(1) ∨ · · · ∨ vσ−1(n) for all σ ∈ Sn.

(ii) TnVsym and SnV are CG-modules.

(iii) SnV has basis ei1 ∨ · · · ∨ ein : 1 ≤ i1 ≤ · · · ≤ in ≤ n, and so dim SnV =

(m+ n− 1

n

).

Proof An easy exercise.

(6.7) Lemma TnVsym = sV ⊗n, where s =∑

σ∈Sn

σ is the symmetrizer [cf. (2.4)], and the natural

map TnVsym → SnV is an isomorphism of CG-modules.

Proof If x is symmetric then

sx =∑

σ∈Sn

σx =∑

σ∈Sn

x = n!x

and so TnVsym ⊆ sV ⊗n.

Conversely, since σs = s for all σ ∈ Sn, any element of sV ⊗n must be symmetric.

We have a mapsV ⊗n → V ⊗n → SnV

which is a CG-homomorphism and whose kernel is Y ∩ aV ⊗n ≤ sY , since x =1

n!sx for x symmetric.

But for y ∈ V ⊗n,s(y − σy) = sy − sy = 0

so sx = 0 (since x ∈ Y ). Hence the map is injective. And it is surjective: any x ∈ SnV is the image

of some y ∈ V ⊗n, but then x is the image of1

n!sy ∈ sV ⊗n. So the map is an isomorphism.

(6.8) Lemma Sn(V ∗) ∼= (SnV )∗.

Proof The quotient map V ⊗n SnV is surjective, so induces an inclusion

(SnV )∗ → (TnV )∗ = Tn(V ∗)

Now by the universal property of SnV (i.e. any multilinear map V × · · · × V︸ ︷︷ ︸n copies

→ C factors through

SnV ), the image of the inclusion is Tn(V ∗)sym, which is isomorphic to Sn(V ∗).

Polynomial maps between vector spaces

(6.9) Definition Let V and W be finite-dimensional C-spaces with bases ei : 1 ≤ i ≤ m andfr : 1 ≤ r ≤ h, respectively. A function ϕ : V →W is a polynomial map (resp. n-homogeneousmap) if, for all X1, . . . , Xm ∈ C,

ϕ(X1e1 + · · ·+Xmem) = ϕ1( ~X)f1 + · · ·+ ϕh( ~X)fh

for some polynomials (resp. homogeneous polynomials of degree n) ϕ1, . . . , ϕh. [Here ~X abbreviatesthe m-tuple (X1, . . . , Xm).]

(6.10) Lemma The definitions in (6.9) do not depend on the choice of bases. That is, if there existsuch functions ϕi with respect to some bases ei, fr, then there exist such functions with respectto any bases e′i, f ′r.

Proof Write

e′i =

m∑

j=1

pjiej and fr =

h∑

s=1

qsrf′s

26

Then

ϕ

(m∑

i=1

Xie′i

)= ϕ

m∑

i,j=1

Xipjiej

=

h∑

r=1

ϕr

(m∑

i1=1

Xi1p1i1 , . . . ,

m∑

im=1

Ximpmim

)fr

=

h∑

r,s=1

ϕr

(m∑

i1=1

Xi1p1i1 , . . . ,

m∑

im=1

Ximpmim

)qsrf

′s

and the

h∑

r=1

ϕr

(· · ·)qsr are polynomials (resp. n-homogeneous polynomials) like the ϕr.

Let PC(V,W ) and HC,n(V,W ) denote the spaces of polynomial maps and n-homogeneous maps, re-spectively, from V to W . They are C-spaces, as can be easily verified.

(6.11) Lemma The composite of polynomial maps X → W → Z is a polynomial map X → Z.Likewise, the composite of an n-homogeneous map X →W and a n′-homogeneous map W → Z is annn′-homogeneous map X → Z.

Proof An easy exercise.

Basic examples

(1) HC,0(V,W ) ∼= W

(2) HC,1(V,W ) = HomC(V,W )

(3) ∆ ∈ HC,n(V,SnV ), where ∆ : v 7→ v ∨ · · · ∨ v. For

∆

(∑

i

Xiei

)=∑

~i

Xi1 . . . Xinei1 ∨ · · · ∨ ein =∑

i1≤···≤in

λ~iXi1 . . . Xinei1 ∨ · · · ∨ ein

for suitable λ~i ∈ C. [Here ~i abbreviates the n-tuple (i1, . . . , in).]

(6.12) Theorem For C-spaces V and W , the map ψ 7→ ψ ∆ induces an isomorphism

HomC(SnV,W )∼=−→ HC,n(V,W )

called restitution.

Proof It’s injective: suppose ψ ∆ = 0. We use descending induction on i to show that ψ(v1 ∨ · · · ∨vn) = 0 whenever i of the terms are equal. When i = n this is precisely our assumption. Suppose it’strue for i+ 1, then for any α ∈ C we have

0 = ψ((x+ αy) ∨ · · · ∨ (x+ αy)︸ ︷︷ ︸i+1

∨vi+2 ∨ · · · ∨ vn)

=

i+1∑

j=0

αj(i+ 1

j

)ψ(x ∨ · · · ∨ x︸ ︷︷ ︸

i+1−j

∨ y ∨ · · · ∨ y︸ ︷︷ ︸j

∨vi+2 ∨ · · · ∨ vn)

Since this holds true for any α ∈ C, each term must be zero. In particular(i+ 1

1

)ψ(x ∨ · · · ∨ x︸ ︷︷ ︸

i

∨y ∨ vi+2 ∨ · · · ∨ vn) = 0

If dimV = m and dimW = h, say, then

dim HomC(SnV,W ) = h

(m+ n− 1

n

)= dim HC,n(V,W )

27

So the map is an isomorphism.

(6.13) Corollary SnV = spanCv ∨ · · · ∨ v : v ∈ V

Proof Take W = C in (6.12). If the elements v ∨ · · · ∨ v do not span then there exists a nonzeromap SnV → C whose composition with ∆ is zero, contradicting (6.12).

Remark We can construct the inverse explicitly. Let ϕ ∈ HC,n(V,W ), so

ϕ(X1e1 + · · ·+Xmem) = ϕ1(X1, . . . , Xm)f1 + · · ·+ ϕk(X1, . . . , Xm)fh

with ϕi homogeneous. The total polarisation Pϕ ∈ HomC(SnV,W ) of ϕ is

Pϕ(ei1 ∨ · · · ∨ ein) =

h∑

j=1

∂nϕj∂Xi1 . . . ∂Xin︸ ︷︷ ︸

∈C

fj

(The partial derivative is formal, is symmetric in the Xik , and lies in C since ϕj has degree N .) Now

(Pϕ)(∆(X1e1 + · · ·+Xmem)) =∑

i1,...,in

Xi1 . . . Xin(Pϕ)(ei1 ∨ · · · ∨ ein)

=∑

j

∑

i1,...,in

Xi1 . . . Xin

∂nϕj∂Xi1 . . . ∂Xin

fj

=∑

j

n!ϕj(X1, . . . , Xm)fj

= n!ϕ(X1e1 + · · ·+Xmem)

So Pϕ ∆ = n!ϕ. The penultimate equality follows from Euler’s theorem, which states that if Fis r-homogeneous then

∑i∂F∂Xi

Xi = rF .

Example Let ϕ : V → C be a quadratic form:

ϕ

(m∑

i=1

Xiei

)=∑

i,j

aijXiXj , aij = aji

Then (Pϕ)[(∑iXiei) ∨

(∑j YjeJ

)]= 2

∑i,j aijXiYj is twice the corresponding symmetric bilinear

form.

28

Interlude: some reminders about affine varieties

Let K be an infinite field. Then R = K[X1, . . . , Xn] a Noetherian ring by Hilbert’s basis theorem.

Define AnK = Kn, the n-dimensional affine space over K.

If f ∈ R and p = (a1, . . . , an) ∈ An then the element f(p) = f(a1, . . . , an) ∈ K is the evaluation off at p.

Define a mapideals I of R ←→ subsets of An

I 7−→ V (I)

where V (I) = p ∈ An : f(p) = 0 for all f ∈ I.

A subset X ⊆ An is an algebraic set if X = V (I) for some ideal I. But I is finitely generated, sayI = (f1, . . . , fm), so

V (I) = p ∈ An : fj(p) = 0 for each 1 ≤ j ≤ mis a locus of points satisfying a finite number of polynomial equations.

Algebraic subsets of An form the closed sets of a topology on An, called the Zariski topology. TheZariski topology on An induces a topology on any algebraic set X ⊆ An: the closed subsets of X arethe algebraic subsets of X.

Example Zariski-closed subsets of A1 are A1 and finite subsets of A1; that is, the Zariski topologyon A1 is the cofinite topology.

Returning to our correspondence, we have a map in the other direction

ideals I of R ←→ subsets of AnI(X) ←− X

where I(X) = f ∈ R : f(p) = 0 for all p ∈ X.

An algebraic set X is irreducible if there does not exist a decomposition X = X1∪X2 with X1, X2 (X proper algebraic subsets. Note

X is irreducible ⇔ I(X) is a prime ideal

If I E R then the radical√I of I is given by

√I = f ∈ R : fm ∈ I for somem

and I is a radical ideal if I =√I. Note that prime ideals are automatically radical.

Theorem (Nullstellensatz) Let K be algebraically closed. Then

(a) Maximal ideals of R are of the form

mj = (X1 − a1, . . . , Xn − an)

for some p = (a1, . . . , an) ∈ An.

(b) If (1) 6= I E R then V (I) 6= ∅.

(c) I(V (I)) =√I for all ideals I.

Proof Omitted.

Let W a finite-dimensional K-space. A map f : W → K is regular (or polynomial) if it is definedby a polynomial in the coordinates with respect to a given basis of W . (Check: it doesn’t matterwhich basis we pick.)

29

Define the coordinate ring (or ring of regular functions) of W , denoted K[W ], to be the K-algebra of polynomial functions on W . If e1, . . . , em is a basis of W and x1, . . . , xm is the correspondingdual basis of W ∗ then

K[W ] = K[x1, . . . , xm]

We say f ∈ K[W ] is d-homogeneous if f(tw) = tdf(w) for all d ∈ C and w ∈ W . Write K[W ]d forthe subspace of K[W ] of all d-homogeneous polynomials. Then

K[W ] =⊕

d≥0

K[W ]d

is a gradedK-algebra, i.e. if f, g are homogeneous of degree df , dg, respectively, then fg is homogeneousof degree df + dg.

The monomials xd11 · · ·xdmm with d1 + · · ·+dm = d form a basis of K[W ]d. In particular, K[W ]1 = W ∗.This extends to a canonical identification of K[W ]d with Sd(W ∗), so that

K =⊕

d≥0

K[W ]d =⊕

d≥0

Sd(W ∗) = S(W ∗)

We call S(W ∗) the symmetric algebra of W ∗.

Zariski-dense subsets

A subset X ⊆W , with W finite-dimensional, is Zariski-dense if every function f ∈ K[W ] vanishingon X is the zero function. So every polynomial function f ∈ K[W ] is completely determined by itsrestriction f |X to a Zariski-dense subset X ⊆W .

Now let X be an arbitrary subset of W . As above, the ideal of X

I(X) = f ∈ K[W ] : f(a) = 0 for all a ∈ X =⋂

a∈Xma

where ma = I(a) is the maximal ideal of functions vanishing at a, i.e.

ma = ker

(eva : K[W ] → C

f 7→ f(a)

)

We say X ⊆ Y is Zariski-dense in Y if I(X) = I(Y ).

Further reading Chapter 4 of Reid’s book.

30

7 Schur–Weyl duality

Let V be an m-dimensional C-space and let n ∈ N.

V ⊗n is a right CSn-module via the action1

(v1 ⊗ · · · ⊗ vn)σ = vσ(1) ⊗ · · · ⊗ vσ(n)

So there is a map

CSn → EndC V⊗n

σ 7→ (v 7→ vσ)—(1)

Regarding V as a representation of GL(V ) in the natural way, V ⊗n is a (left) CGL(V )-module viathe action

g(v1 ⊗ · · · ⊗ vn) = gv1 ⊗ · · · ⊗ gvnSo there is a map

CGL(V ) → EndC V⊗n

g 7→ (v 7→ gv)—(2)

The actions (1) and (2) commute, so by abusing notation slightly, (1) induces a map

ϕ : CSn → EndCGL(V )(V⊗n)

and (2) induces a mapψ : CGL(V )→ EndCSn(V ⊗n)

Schur–Weyl duality asserts that the images of CSn and CGL(V ) in EndC(V ⊗n) are each other’scentralizers.

(7.1) Theorem (Schur–Weyl duality) [Schur, 1927]

ψ(CGL(V )) = EndCSn(V ⊗n) and ϕ(CSn) = EndCGL(V )(V⊗n)

Such an assertion looks innocuous, but it is fundamental in explaining why the symmetric group andthe general linear group are so closely related.

(7.2) Definition The C-algebra SC(m,n) is the subalgebra of EndC(V ⊗n) consisting of endomor-phisms which commute with the image of CSn, i.e.

SC(m,n) = EndCSn(V ⊗n)

It is called the Schur algebra. [For a definition over arbitrary fields, see J.A. Green’s book Polynomialrepresentations of GLn, SLN 830.]

CSn is semisimple and V ⊗n is a finite-dimensional CSn-module, so by (1.9), SC(m,n) is a semisimpleC-algebra.

Put W = EndC(V ). This is a CGL(V )-module under the conjugation action (check this).

(7.3) Lemma There is a map α : W⊗n → EndC(V ⊗n) given by

f1 ⊗ · · · ⊗ fn 7→(v1 ⊗ · · · ⊗ vn 7→ f1(v1)⊗ · · · ⊗ fn(vn)

)

which is an isomorphism both of CSn-modules and of CGL(V )-modules.

1Note that, because this is a right-action, σ1σ2 denotes the effect of first applying σ1 and then applying σ2, which is‘backwards’ compared to normal.

31

Proof

W⊗n = W ⊗ · · · ⊗W ∼= (V ⊗ V ∗)⊗ · · · ⊗ (V ⊗ V ∗)∼= (V ⊗ · · · ⊗ V )⊗ (V ∗ ⊗ · · · ⊗ V ∗)∼= V ⊗n ⊗ (V ⊗n)∗

∼= EndC(V ⊗n)

It is left as an exercise to check that this isomorphism is the map we think it is, that W⊗n hasthe natural structure of a CSn-module and EndC(V ⊗n) inherits structure from V ⊗n via θ · σ =ϕ(σ)θϕ(σ−1), and finally that α is a CSn-map.

(7.4) Lemma SC(m,n) = α(TnWsym)

Proof SC(m,n) is the set of all x ∈ EndC(V ⊗n) fixed under the action of Sn. By (6.5), TnWsym isthe set of all y ∈W⊗n fixed under the action of Sn (which is induced by α, i.e. y ·σ ∈W⊗n correspondswith ϕ(σ)α(y)ϕ(σ−1) ∈ EndC(V ⊗n).

Recall affine n-space An from §6.

(7.5) Lemma An is irreducible; i.e. if An = X∪Y with X,Y Zariski-closed then X = An or Y = An.

Proof The ring of regular functions on An is R = C[X1, . . . , Xn]. If X and Y are zero-sets of idealsI, J E R then, by assumption, any maximal ideal contains either I or J . If I, J are both nonzero thenpick 0 6= f ∈ I and 0 6= g ∈ J . Any maximal ideal contains fg, so

fg(a1, . . . , an) = 0 for all a1, . . . , an ∈ CBy Nullstellensatz, fg ∈

√0 = 0, contradicting the fact that R is an integral domain.

(7.6) Lemma Let Y ≤ Cd be a subspace. Then Y is Zariski-closed, where Cd is identified with Adin the canonical way.

Proof Chose a basis ϕ1, . . . , ϕh of HomC(Cd/Y,C) and view them as maps Cd → C. Then Y is thezero set of the ϕi.

(7.7) Lemma TnWSym = spanC(

n︷ ︸︸ ︷ϕ⊗ · · · ⊗ ϕ : ϕ ∈ GL(V )

)

Proof Define a subspace X of TnWsym by

X = spanC(

n︷ ︸︸ ︷ϕ⊗ · · · ⊗ ϕ : ϕ ∈ GL(V )

)

There is a map α : W → W⊗n given by α(ϕ) =

n︷ ︸︸ ︷ϕ⊗ · · · ⊗ ϕ, which is a regular map of the affine

spaces Am2 ∼= W and Am2n ∼= W⊗n.

Now X is a subspace, so is Zariski-closed by (7.6), and hence α−1(X) is Zariski-closed. Thus,

W = α−1(X) ∪ endomorphisms with zero determinantis a union of Zariski-closed subsets.

Since Am2

is irreducible (by (7.5)), α−1(X) = W . Thus X contains all maps of the form ϕ⊗ · · · ⊗ ϕfor ϕ ∈W . But they span TnWsym since the images ϕ ∨ · · · ∨ ϕ span SnW by (6.13).

In other words. . .

(7.8) Theorem SC(m,n) = spanC(

n︷ ︸︸ ︷ϕ⊗ · · · ⊗ ϕ : ϕ ∈ GL(V )

)

Proof of (7.1) The assertion that the image of CGL(V ) is the centralizer of CSn is a reformulation

of the assertion that SC(m,n) = spanC(

n︷ ︸︸ ︷ϕ⊗ · · · ⊗ ϕ : ϕ ∈ GL(V )

), which is (7.8).

32

Conversely, SC(m,n) is a semisimple C-algebra. By (1.11), CSn maps onto EndSC(m,n)(V⊗n), and

since the image of GL(V ) spans SC(m,n), we must have

EndSC(m,n)(V⊗n) = EndCGL(V )(V

⊗n)

Thus CSn maps onto EndCGL(V )(V⊗n), i.e. the image of CSn in EndC(V ⊗n) is the centralizer of the

image of CGL(V ).

33

8 Tensor decomposition

Let V be an m-dimensional C-space. We know from (3.9) that

V ⊗n =⊕h(tλ)V ⊗n —(1)

where the direct sum runs over standard tableaux tλ with λ ` n.

Examples If λ = (1n) then hλV⊗n ∼= TnVanti

∼= ΛnV .

If λ = (n) then hλV⊗n ∼= TnVSsym

∼= SnV .

In particular,when n = 2, this tells us that

V ⊗ V ∼= T 2Vanti ⊕ T 2Vsym∼= Λ2V ⊕ SnV

Now the actions of Sn and GL(V ) on V ⊗n commute. So if λ ` n and tλ is a λ-tableau, then h(tλ)V ⊗n

is a CGL(V )-submodule of V ⊗n. Observe that

h(tλ)V ⊗n ∼= hλV⊗n as CGL(V )-modules

since σh(tλ)σ−1 = hλ for some σ ∈ Sn. So premultiplication by σ induces an isomorphism

h(tλ)V ⊗n∼=−→ hλV

⊗n

(8.1) Lemma The nonzero hλV⊗n with λ ` n are nonisomorphic irreducible CGL(V )-modules.

Moreover, if M is an SC(m,n)-module and M is viewed as a CGL(V )-module by restriction via thenatural map CGL(V )→ SC(m,n) then M is isomorphic to a direct sum of copies of the hλV

⊗n.

Proof Recall that SC(m,n) = EndCSn(V ⊗n) and

hλV⊗n

(1.12)∼= HomCSn(CSnhλ, V⊗n)

By (1.9) and (1.10)(Artin–Wedderburn), the nonzero spaces hλV⊗n are a complete set of noniso-

morphic irreducible SC(m,n)-modules.

Now SC(m,n) is semisimple, so the result follows from (2) and (3) below, both of which derive fromthe fact (at the end of §7) that the map CGL(V )→ SC(m,n) is surjective.

(2) If M is an SC(m,n)-module and N is a CGL(V )-module of M then N is an SC(m,n)-submoduleof M .

(3) If M and N are SC(m,n)-modules and θ : M → N is a CGL(V )-module map then it is anSC(m,n)-map.

This implies that (1) is a decomposition of V ⊗n into CGL(V )-submodules which are either zero orirreducible. Which of these submodules are nonzero? We’ll prove:

(8.2) Theorem Let λ ` n and m = dimC V . Then

dimC hλV⊗n =

0 if λm+1 6= 0

1 if λm+1 = 0 andλ1 = · · · = λm

≥ 2 otherwise

First we have a slightly technical result. Take a partition λ ` n such that [λ] is partitioned into twononempty parts, say of sizes i and j = n − i, by a vertical wall. For example, if λ = (5, 3, 2, 1) ` 11we could have

i = 7 −→• • • • •• • •• ••

←− j = 4

34

For λ ` n, let tλ be a tableau whose entries in the left-hand part are 1, . . . , i. Let µ ` i be thecorresponding partition, so that tµ = tλ ↓[µ]. Let ν ` j correspond to the right-hand part, withtableau tν . This is a map

tν : [ν]→ 1′, . . . , j′ where k′ = k + i for 1 ≤ k ≤ j

(8.3) Lemma There exists a surjective CGL(V )-module map

((h(tµ)V ⊗i)⊗ (h(tν)V ⊗j)→ h(tλ)V ⊗n —(∗)

Proof Let Si = Aut(1, . . . , i) and Sj = Aut(1′, . . . , j′).

Si and Sj are embedded in Sn and are disjoint, so we regard CSi and CSj as subsets of CSn whichcommute. Also

Ctλ = Ctµ × Ctν and H = Rtµ ×Rtν ≤ Rtλand H permutes each side of the wall. Take a transversal Rtλ =

⋃i riH. Then

h(tλ)(2.4)=

∑

r∈Rtλ

∑

c∈Ctλ

εcrc

=∑

i

∑

r′∈Rtµ

∑

r′′∈Rtν

∑

c′∈Ctµ

∑

c′′∈Ctν

εc′εc′′rir′r′′c′c′′

=∑

i

rih(tµ)h(tν)

So

h(tλ)h(tµ)h(tλ) =∑

i

rih(tµ)2h(tν)2 (2.10)= kh(tλ)

where k =i!j!

fµfν. We thus have a CGL(V )-map

V ⊗i ⊗ V ⊗j → h(tλ)V ⊗n

given by premultiplication by h(tλ). Restriction of this map to the left-hand side of (∗) is surjective,since

h(tλ)(x⊗ y) = h(tλ) · 1

k(h(tµ)x⊗ h(tν)y)

(8.4) Lemma

(1) If λm+1 = 0 then hλV⊗n 6= 0

(2) If n > 0 and λm = 0 then dimC hλV⊗n ≥ 2

Proof

(1) Let ij be the row in which j occurs in tλ and let x = ei1 ⊗ · · · ⊗ ein . For σ ∈ Sn

σx = x ⇔ ij = iσ−1(j) ∀j ⇔ j, σ−1(j) are in the same row ⇔ σ ∈ RtλSo the coefficient of x in the decomposition of hλx with respect to the standard basis of V ⊗n is∣∣Rtλ

∣∣ 6= 0. So hλx 6= 0.

(2) Let y = e1+i1 ⊗ · · · ⊗ e1+in and adapt the argument above, to see that hλx and hλy are linearlyindependent.

(8.5) Lemma If λm+1 = 0 and λm > 0 then

hλV⊗n ∼= ΛmV ⊗ h(λ1−1,...,λm−1)V

⊗(n−m)

35

Proof Put a wall in [λ] between the first column and the rest. Let tλ be a tableau whose firstcolumn consists of 1, . . . ,m. By (8.3) there exists a surjection

ΛmV ⊗ h(tν)V ⊗(n−m) h(tλ)V ⊗n

where ν = (λ1 − 1, . . . , λm − 1). Hence, with the usual identifications, there exists a map

ΛmV ⊗ hνV ⊗(n−m) hλV⊗n

Both hνV⊗(n−m) and hλV

⊗n are nonzero, hence they are irreducible CGL(V )-modules by (8.1). SinceΛmV is one-dimensional, both sides are irreducible and the map is an isomorphism.

Proof of (8.2) If λm+1 6= 0 then [λ] has i > m rows and as in (8.5) there exists a surjection

ΛiV ⊗ hνV ⊗(n−i) hλV⊗n

but ΛiV = 0, so hλV⊗n = 0.

If λm+1 = 0 then iterating (8.5) we have

dimC hλV⊗n = dimC h(λ1−λm,...,λm−1−λm)V

⊗(n−mλm)

=

1 if λ1 = · · · = λm

≥ 2 if not, by (8.4)

36

9 Polynomial and rational representations of GL(V )

Let V be an m-dimensional C-space with basis e1, . . . , em.

(9.1) Definition A finite-dimensional CGL(V )-module M , with basis w1, . . . , wn, is rational (resp.polynomial, n-homogeneous) if there exist rational (resp. polynomial, n-homogeneous) functionsfij(Xrs) for 1 ≤ i, j ≤ h in m2 variables Xrs for 1 ≤ r, s ≤ m such that the map

GLm(C)basis ei−−−−−→ GL(V )

representation−−−−−−−−−→ EndC(M)basis wj−−−−−→ Mh(C)

sends (Ars)1≤r,s≤m 7→ (fij(Ars))1≤i,j≤h.

Exercise Check that the definitions in (9.1) are independent of the choice of bases ei and wj of Vand M .

Examples

(0) The natural representation of GL(V ) on V ⊗n is polynomial. In fact. . .

(1) V ⊗n is n-homogeneous; but. . .

(2) C⊕ V is polynomial but not homogeneous; and. . .

(3) V ∗ is rational but not polynomial (or homogeneous).

(9.2) Lemma

(a) Submodules, quotients and direct sums of rational (resp. polynomial, n-homogeneous) modulesare rational (resp. polynomial, n-homogeneous).

(b) If M is rational then so is M∗.

(c) If M and N are rational (resp. polynomial, n- and n′-homogeneous) then M⊗N is rational (resp.polynomial, (n+ n′)-homogeneous).

(d) If n 6= n′ and M is both n- and n′-homogeneous then M = 0.

(e) if M is rational (resp. polynomial) then ΛnM and SnM are rational (resp. polynomial).

Proof Easy exercise.

(9.3) Definition For λ1 ≥ · · · ≥ λm ≥ 0 with∑i λi = n, set

Dλ1,...,λm(V ) = h(λ1,...,λm)(V⊗n)

This is sometimes referred to in the literature as the Weyl module or Schur module.

(9.4) Theorem Every n-homogeneous CGL(V )-module is a direct sum of irreducible submodules.Moreover, the modules of (9.3) form a complete set of nonisomorphic irreducible n-homogeneousCGL(V )-modules.

Proof By (8.1) it suffices to show that any n-homogeneous CGL(V )-module is obtained from someS(m,n)-module (by restriction).

Let M be an n-homogeneous CGL(V )-module. M corresponds to a map

ρ0 : GL(V )→ EndC(M)

Consider the following diagram

37

GL(V ) EndC(V ) Sn(EndC V ) Tn(EndC V )sym S(m,n)

EndC(M) EndC(M) EndC(M) EndC(M) EndC(M)

→ ∼= ∼=

ρ0 ρ1 ρ2 ρ3 ρ4

In the top row all the maps are natural, and the composite γ : GL(V )→ S(m,n) is the natural mapwe use for restricting S(m,n)-modules to CGL(V )-modules.

Claim There exist ρ1, ρ2, ρ3, ρ4 as above making the diagram commute.

Since M is n-homogeneous, we can extend the domain of definition of ρ0 to obtain an n-homogeneousmap ρ1. For ρ2, use the universal property of symmetric powers. Then ρ3, ρ4 are transported by theisomorphisms.

We check that ρ4 is a map of C-algebras. Well

ρ4(1) = ρ4(γ(1)) = ρ0(1) = 1

andρ4(γ(gg′)) = ρ0(gg′) = ρ0(g)ρ0(g′) = ρ4(γ(g))ρ4(γ(g′))

for g, g′ ∈ GL(V ). But γ(GL(V )) spans S(m,n) by (7.8).

So M is a CS(m,n)-module and the restriction to GL(V ) is the module we started with.

(9.5) Lemma Every polynomial module for GL(C) = C× decomposes as a direct sum of submoduleson whih g ∈ C× acts as multiplication by gr for some r.

Proof If ρ : C× → GL(M) ∼= GLh(C) is a polynomial representation then each ρ(g)ij is polynomialin g. Choose N ∈ N such that each ρ(g)ij has degree < N . Then M is a CG-module by restriction,where

G = e2π ijN : 0 ≤ j ≤ N ⊆ C×

is a cyclic group of order N . NowM = M1 ⊕ · · · ⊕Mh

as a CG-module with each Mj one-dimensional, and g ∈ G acts as multiplication by gnj on Mj for0 ≤ j < N . (These are the possible irreducible CG-modules.)

Choosing nonzero elements of the Mjs gives a basis for M , and if (ρ(g)ij) is the matrix of ρ(g) withrespect to this basis then

ρ(g)ij =

gni if i = j

0 if i 6= j

for g ∈ e2π ijN (0 ≤ j < N), and hence for all g ∈ C× since the ρ(g)ij are polynomials of degree < Nin g.

(9.6) Lemma Every polynomial decomposes as a direct sum of n-homogeneous modules.

Proof Tke a representation ρ : GL(V )→ GL(U) be a representation of GL(V ) on a C-space U . Wehave an inclusion C× → GL(V ), so that one can regard U as a polynomial representation of C×, soby (9.5) U will split as

U = U0 ⊕ · · · ⊕ UNwith α1 ∈ GL(V ) acting as multiplication by αn on Un.

Let u ∈ Un and let g ∈ GL(V ). Write

gu = u0 + · · ·+ uN

38

where ui ∈ Ui for 0 ≤ i ≤ N . Now (α1)gu = g(α1)u for α ∈ C×, so

u0 + αu1 + · · ·+ αNuN = αnu0 + · · ·+ αnuN

and hence ui = 0 for i 6= n.

Thus gu ∈ Un and the Un are CGL(V )-submodules of U . Since α1 acts as multiplication by αn onUn, it follows that Un is an n-homogeneous CGL(V )-module.

(9.7) Theorem Every polynomial CGL(V )-module is a direct sum of irreducible submodules. Themodules Dλ1,...,λm(V ) with λ1 ≥ · · · ≥ λm ≥ 0 are a complete set of nonisomorphic irreduciblepolynomial CGL(V )-modules.

(9.8) Definition If n ∈ Z then the one-dimensional CGL(V )-module corresponding to the repre-sentation

GL(V ) → C⊗g 7→ (det g)n

is denoted by detn. Note that

det1 ∼= ΛmV, detn ∼= (det1)⊗n if n ≥ 0, detn ∼= (det−n)∗ if n ≤ 0

(9.9) Definition If λ1 ≥ · · · ≥ λm but λm < 0 then define

Dλ1,...,λm = Dλ1−λm,...,λm−1−λm,0(V )⊗ detλm

Remark If λ1 ≥ · · · ≥ λm > 0 then by iterating (8.5) (as in the proof of (8.2)) we know that

Dλ1,...,λm(V ) ∼= Dλ1−λm,...,λm−1−λm,0(V )⊗ detλm

(9.10) Theorem Every rational CGL(V )-module is a direct sum of irreducible submodules. Themodules Dλ1,...,λm(V ) for λ1 ≥ · · · ≥ λm form a complete set of nonisomorphic irreducible (rational)CGL(V )-modules.

Proof The rational functions f : GL(V )→ C are of the form f =p

detiwith p a polynomial function.

Thus if M is a rational CGL(V )-module then M1 = M ⊗ detN is a polynomial CGL(V )-module forsome N . Since M1 decomposes as a direct sum of irreducibles, so does M . If M is irreducible then sois M1, since detN is one-dimensional, and thus M1 is of the form

M1∼= Dµ1,...,µm(V ) for some µ1 ≥ · · · ≥ µm

Finally, M = Dµ1−N,...,µm−N (V ) by the above remark.

(9.11) Theorem The one-dimensional rational CGL(V )-modules are precisely the detn for n ∈ Z.

Proof As above, pass to polynomial modules and then apply (8.2).

Remark The characterisation of irreducible CGL(V )-modules can be summarised by:

n-homogeneous λ1 ≥ · · · ≥ λm > 0 with∑i λi = n (9.4)

polynomial λ1 ≥ · · · ≥ λm ≥ 0 (9.7)rational λ1 ≥ · · · ≥ λm (9.10)

39

Interlude: some reminders about characters of GLn(C)

Let Tn ≤ GLn(C) be the n-dimensional torus, i.e. the subgroup of GLn(C) consisting of diagonalmatrices.

If ρ : GLn(C)→ GL(W ) is a rational representation then the character of ρ is the function

χρ : (x1, . . . , xn) 7→ tr ρ

λ1

. . .

λn

sometimes denoted by χW .

Basic properties

Let ρ, ρ′ be rational representations of GLn(C) associated with W,W ′.

(a) χρ ∈ Z[x1, x−11 , . . . , xn, x

−1n ], and if ρ is polynomial then χρ ∈ Z[x1, . . . , xn].

(b) χρ is a symmetric function under the conjugation action of Sn on Tn, i.e. under permutations ofthe diagonal entries.

(c) ρ ∼= ρ′ ⇒ χρ = χρ′ ; ⇐ will be (10.6).

(d) χW⊕W ′ = χW + χW ′ and χW⊗W ′ = χWχW ′ .

(e) If W is an irreducible polynomial representation of degree m then χW is a homogeneous polynomialof degree m.

(f) χW∗(x1, . . . , xn) = χW (x−11 , . . . x−1

n ).

(g) When V = Cn we haveχV ⊗n = (x1 + · · ·+ xn)m,

χS2V =∑

i≤j

xixj , χΛ2V =∑

i<j

xixj ,

χdet = x1 . . . xn, χV ∗ = x−11 + · · ·+ x−1

n ,

χΛn−1V =

n∑

i=1

x1 . . . xi . . . xn, χSjV = hj(x1, . . . , xn) =∑

i1≤···≤ij

xi1 . . . xij

The hj are called complete symmetric functions. We could (but won’t) study the theory ofsymmetric functions and apply it here; see R. Stanley’s book for more.

40

10 Weyl character formula

Let V be an m-dimensional C-space. If λ = (λ1, . . . , λm) ∈ Zm and λ1 ≥ · · · ≥ λm then the characterof the CGL(V )-module Dλ1,...,λm(V ) (= hλV

⊗n if λm ≥ 0 and n =∑i λi) will be denoted by φλ.

(10.1) Lemma If ξ ∈ EndC(V ) then φλ(ξ) is a symmetric rational function of the eigenvalues of ξ.

Proof The functionP (x1, . . . , xm) = φλ(diag(x1, . . . , xm))

is a rational function of the x1, . . . , xm, and it is symmetric since diag(x1, . . . , xm) is conjugate todiag(xτ(1), . . . , xτ(m)) for any τ ∈ Sm.

Now choose a basis for V such that the matrix A1 of ξ is in Jordan normal form, and for t ∈ C let Atbe the matrix obtained from A1 by replacing the upperdiagonal 1s by ts. Let ξt be the endomorphismcorresponding to At. For t 6= 0, At is conjugate to A1, and so φλ(ξt) = φλ(ξ). Since φλ is a rationalfunction it is continuous (wherever it is defined), and so

φλ(ξ) = limt→0

φλ(ξt) = φλ(limt→0

ξt) = φλ(ξ0) = P (x1, . . . , xm)

which is a symmetric function of the eigenvalues of ξ.

(10.2) Lemma Let α be an Sn-conjugacy class with cycle type nαn , . . . , 1α1 , and let ξ be anendomorphism of V with eigenvalues x1, . . . , xm. If si = xi1 + · · ·+ xim then

sα11 . . . sαnn =

∑

λ∈Λ+(m,n)

χλ(α)φλ(ξ)

(Compare this with (4.14).)

Proof Pick g ∈ α. Suppose ξ has matrix diag(x1, . . . , xm) with respect to the standard basise1, . . . , em of V . We compute the trace of the endomorphism of V ⊗n sending x 7→ (gξ)x = ξgx in twoways.

Way 1 V ⊗n is a CSn-module, so by (1.8),

V ⊗n ∼=⊕

λ`n

CSnhλ ⊗HomCSn(CSnhλ, V⊗n)

which, by (1.12), becomes

V ⊗n ∼=⊕

λ`n

(CSnhλ ⊗ hλV ⊗n)

This is an isomorphism both of CSn- and EndCSn(V ⊗n)-modules. Since the action of GL(V ) on V ⊗n

commutes with that of Sn, the corresponding action of gξ on the right-hand side is given by the actionof g on CSnhλ and of ξ on hλV

⊗n. So the trace of the action is

∑λ∈Λ+(n,m) ξ

λ(α)φλ(ξ) —(1)

Way 2 (direct computation)

gξ(ei1 ⊗ · · · ⊗ ein) = xi1 . . . xineig−1(1)⊗ · · · ⊗ eig−1(n)

and sotr(gξ) =

∑

D

xi1 . . . xin

where D = (i1, . . . , in) : 1 ≤ i1, . . . , in ≤ m and ig−1(j) = ij ∀j.

The condition ig−1(j) = ij for all j is equivalent to requiring that the function j 7→ ij be constant onthe cycles involved in g. Hence

tr(gξ) = sα11 . . . sαnn —(2)

41

Equating (1) and (2) gives the result.

(10.3) Theorem Let λ ∈ Λ+(m,n) and let ξ ∈ EndC(V ) with eigenvalues x1, . . . , xm. If si =xi1 + · · ·+ xim, then

φλ(ξ) =∑

ccls α

χλ(α)

α1! . . . αn!

(s1

1

)α1

· · ·(snn

)αn

Proof Take the formula of (10.2), multiply it by nαχµ(α) and sum over α; then use orthogonality

of the χs.

(10.4) Theorem (Weyl’s character formula for GLn) Let ξ ∈ GL(V ) with eigenvalues x1, . . . , xm.Then

φλ(ξ) =

∣∣x`1 , . . . , x`m∣∣

|xm−1, . . . , 1|where `i = λi +m− i.

Proof

(a) Suppose λi ≥ 0 for all i, so that λ ∈ Λ+(n,m). By (10.2) and the character formula for thesymmetric group (4.14) we know that

sα11 . . . sαnn =

∑

λ∈Λ+(n,m)

χλ(α)

∣∣x`1 , . . . , x`m∣∣

|xm−1, . . . , 1| =∑

λ∈Λ+(n,m)

χλ(α)φλ(α)

Then the orthogonality of χλs allows us to equate coefficients.

(b) For general λ, since

Dλ

(9.9)∼= Dλ1−λm,...,λm−λm(V )⊗ detλm

it follows that

φλ(ξ) =

∣∣x`1−λm , . . . , x`m−λm∣∣

|xm−1, . . . , 1| (x1 . . . xm)λm =

∣∣x`1 , . . . , x`m∣∣

|xm−1, . . . , 1|

Remark This short proof uses the character formula of Sn from §4. One can do this in reverse,using integration to compute the Weyl formula for the compact subgroup Un of unitary matrices inGLm(C), and then translate that to GLm(C). [See also Telemann’s notes for the 2005 Part II course.]You then use (10.2) to pass to Sn. See also Weyl’s book.

(10.5) Theorem (Weyl’s dimension formula)

deg φλ = dimDλ1,...,λm(V ) =

∏1≤i<j≤m(`i − `j)∏1≤i<j≤m(j − i)

Proof For t ∈ C set xm = 1, xm−1 = et, . . . , x1 = e(m−1)t. Then

∣∣x`1 , . . . , x`m∣∣ =

∏

1≤i<j≤m

(e`it − e`jt)

since it’s the transpose of a Vandermonde matrix. The term of lowest degree is∏i<j((`i − `j)t).

Also∣∣xm−1, . . . , 1

∣∣ =∏i<j(e

(m−i)t − e(m−j)t) and the term of lowest degree in t is∏i<j((j − i)t).

If ξt = diag(x1, . . . , xm) then by (10.4),

deg φλ = φλ(1) = limt→0 φλ(ξt) =∏i<j(`i−`j)∏i<j(j−i)

42

(10.6) Theorem If two rational CGL(V )-modules have the same characters then they are isomor-phic.

Proof As λ varies, the rational functions in Weyl’s character formula are linearly independentelements of C(x1, . . . , xm).

(10.7) Lemma Dλ1,...,λm(V )∗ ∼= D−λm,...,−λ1(V )

Proof Let ψ be the character of Dλ1,...,λm(V ). Then

ψ(ξ) = φλ(ξ−1) =

∣∣x−`1 , . . . , x−`m∣∣

∣∣x−(m−1), . . . , 1∣∣ =

∣∣x−`m , . . . , x−`1∣∣

∣∣1, . . . , x−(m−1)∣∣

=

∣∣xm−1−`m , . . . , xm−1−`1∣∣

|xm−1, . . . , 1| = φµ(ξ)

where µ = (−λm, . . . ,−λ1). Hence we have the required isomorphism by (10.6).

Finally we have the famous Clebsch–Gordan formula.

(10.8) Theorem (Clebsch–Gordan formula) Let V = C2 and p, q ∈ N. Then

Dp,0(V )⊗Dq,0(V ) ∼=minp,q⊕

r=0

Dp+q−r,r(V )

Proof Exercise; write down the characters and use (10.6).

(10.9) Examples of rational CGL(V )-modules

C ∼= D0,0,...,0,0(V )

deti ∼= Di,i,...,i,i(V )

V ∼= D1,0,...,0,0(V )

SnV ∼= Dn,0,...,0,0(V )

ΛnV ∼= D1,...,1,0,...,0(V ) where n ‘1’s and m− n ‘0’s appear, and n ≤ m

V ∗ ∼= D0,0,...,0,−1(V )

(SnV )∗ ∼= D0,0,...,0,−n(V )

(ΛnV )∗ ∼= D0,...,0,−1,...,−1(V ) where m− n ‘0’s and n ‘−1’s appear, and n ≤ m

43

Chapter III: Invariant theory

11 Introduction to invariant theory and first examples

Let W be a finite-dimensional C-space and C[W ] be its coordinate ring (cf. §6).

Facts about C[W ]

C[W ] is a commutative C-algebra, which is infinite-dimensional if W 6= 0, with

(λf)(w) = λf(w), (f + g)(w) = f(w) + g(w),

(fg)(w) = f(w)g(w), 1C[W ](w) = 1

for all λ ∈ C, f, g ∈ C[W ] and w ∈W .

C[W ] is the ring of regular functions of the affine variety AdimW ∼= W .

C[W ] ∼=⊕

n≥0

HC,n(W,C)(6.12)∼=

⊕

n≥0

Sn(W ∗) = S(W ∗)

This latter isomorphism is known as polarisation.

If W ∗ has basis w∗1 , . . . , w∗h then the map

C[X1, . . . , Xh] → C[W ]Xi 7→ w∗i

is a C-algebra homomorphism.

If W is a CG-module then C[W ] is a CG-module via the action

gf(w) = f(g−1w)

for g ∈ G, f ∈ C[W ] and w ∈W .

For g ∈ G, the mapC[W ] → C[W ]f 7→ gf

is a C-algebra automorphism, since

g(ff ′)(w) = (ff ′)(g−1w) = f(g−1w)f ′(g−1w) = (gf)(w)(gf ′)(w) = ((gf)(gf ′))(w)

andg1C[W ](w) = 1C[W ](g

−1w) = 1 = 1C[W ](w)

(11.1) Definition If V,W are CG-modules then f : W → V is concomitant if

f(gw) = gf(w) for all w ∈W, g ∈ G

A G-invariant for the CG-module W is a concomitant map f : W → C.

Remarks

(1) f is invariant if and only if it is a fixed point under the G-action, i.e. gf = f for all g ∈ G.

(2) linear concomitant ≡ CG-homomorphism

(3) The map ∆ :V → SnWw 7→ w ∨ · · · ∨ w is an n-homogeneous concomitant.

(4) C[W ]G = f ∈ C[W ] : gf = f for all g ∈ G is a C-subalgebra of C[W ], called the polynomialinvariant ring. It is a graded C-algebra:

C[W ]G =⊕

d≥0

C[W ]Gd

with grading inherited by that of C[W ].

44

(11.2) Problem Compute C[W ]G, describing its generators and relations.

What follows are some classical results which we will state but not prove.

[Noether] If G is finite then C[W ]G is finitely generated as a C-algebra.

[Noether, 1916] If charK = 0 then for any representation W of the finite group G, the invariantring K[W ]G is generated by the invariants of degree ≤ |G|.

[Hilbert] IfW is a rational CGL(V )-module then C[W ]GL(V ) and C[W ]SL(V ) are finitely-generatedC-algebras. In fact [Hochster–Roberts, 1974] they are also Cohen–Macaulay rings. Also, C[W ]SL(V )

is a unique factorisation domain, and hence it is a Gorenstein ring (i.e. it has finite injectivedimension).

[Popov, Asterisque, 1988] There is an explicit bound on the number of generators for C[W ]SL(V ).

Examples

(11.3) Symmetric polynomials W = Cn with basis e1, . . . , en, is naturally a CSn-module. Ifξ1, . . . , ξn is the corresponding dual basis for W ∗, then the isomorphism

C[X1, . . . , Xn]∼=−→ C[W ]

Xi 7→ ξi

allows an identification C[W ]↔ C[X1, . . . , Xn]. The polynomial invariant ring of the CSn-module Wis

C[X1, . . . , Xn]Sn = p ∈ C[X1, . . . , Xn] : p symmetric in the Xis

We have elementary symmetric functions ej(X1, . . . , Xn) ∈ C[X1, . . . , Xn]Sn such tha

(t+X1) · · · (t+Xn) = tn + e1( ~X)tn−1 + · · ·+ en( ~X)

where ~X = (X1, . . . , Xn). That is,

e1( ~X) = X1 + · · ·+Xn, ei( ~X) =∑

j1<j2<···<ji

Xj1Xj2 . . . Xji , en( ~X) = X1 . . . Xn

The fundamental theorem of symmetric functions computes the polynomial invariants for W ;namely, it shows that the map

C[Y1, . . . , Yn] → C[X1, . . . , Xn]Sn

Yi 7→ ei( ~X)

is a C-algebra isomorphism. The standard way to prove this is by showing that C[X1, . . . , Xn]Sn =C[e1, . . . , en] by induction on n.

Recall for p(t) ∈ C[t] monic of degree n, say

p(t) = tn + a1tn−1 + · · ·+ an =

n∏

i=1

(t+ λi)

its discriminant isdisc(p) =

∏

i<j

(λi − λj)2

Now, since the polynomial∣∣Xn−1, . . . , 1

∣∣2 =∏i<j(Xi − Xj)

2 is symmetric in the Xis, it may berepresented as a polynomial in the eis, say

∏

i<j

(Xi −Xj)2 = D(e1( ~X), . . . , en( ~X))

45

Also, disc(p) is a polynomial in the coefficients of p, for example

disc(t2 + bt+ c) = b2 − 4c, disc(t3 + bt+ c) = −4b3 − 27b2

(11.4) Alternating groups

Restriction in (11.3) enables consideration of W as a representation of An (n ≥ 2). The ring ofinvariants is C[X1, . . . , Xn]An . Consider the Vandermonde determinant

∣∣Xn−1, . . . , 1∣∣ =

∏

i<j

(Xi −Xj)

∣∣Xn−1, . . . , 1∣∣2 is Sn-invariant since it is equal to D(e1, . . . , en) as above.

Claim The C-algebra map

θ : C[Y1, . . . , Yn, Z] → C[X1, . . . , Xn]An

Yi 7→ eiZ 7→

∣∣Xn−1, . . . , 1∣∣

is surjective, and ker(θ) = (Z2 −D(Y1, . . . , Yn)).

Proof We prove the claim in three steps.

Step 1 θ is surjective. Take a transposition τ ∈ Sn. Let f ∈ C[X1, . . . , Xn]An . Then fsym = f + τfis a symmetric polynomial, since Sn is generated by τ and An, and

τfsym = τf + τ2f = τf + f = fsym

and if σ ∈ An thenσfsym = σf + τ(τ−1στ)f = f + τf = fsym

Similarly, falt = f − τf is an alternating polynomial.

Observe that falt(X1, . . . Xn) = 0 whenever Xi = Xj for i 6= j. Hence by Nullstellensatz

falt ∈√

(Xi −Xj) = (Xi −Xj)

and so (Xi −Xj) | falt for all i 6= j. In particular,

falt =∣∣Xn−1, . . . , 1

∣∣ g

for some polynomial g. Now g is symmetric, and hence

f =1

2(fsym + falt) =

1

2(fsym +

∣∣Xn−1, . . . , 1∣∣ g)

Since fsym and g lie in im(θ), so does f .

Step 2 (Z2 −D(~Y )) ⊆ ker(θ). This is clear since

θ(Z2 −D(~Y )) =∣∣Xn−1, . . . , 1

∣∣2 −D(e1, . . . , en) = D(e1, . . . , en)−D(e1, . . . , en) = 0

Step 3 ker(θ) ⊆ (Z2 −D(~Y )). Any polynomial P (~Y , Z) is of the form

P (~Y , Z) = Q(~Y , Z)︸ ︷︷ ︸quotient

(Z2 −D(~Y )) +A(~Y ) +B(~Y )Z︸ ︷︷ ︸remainder

We show that if P ∈ ker(θ) has the form A(~Y ) +B(~Y )Z then P = 0.

Suppose a1, . . . , an ∈ C, and let

f(t) = tn + a1tn−1 + · · ·+ an = (t+ λ1) · · · (t+ λn)

46

ThenP (a1, . . . , an,

∣∣λn−1, . . . , 1∣∣) = θ(P )(λ1, . . . , λn) = 0

since P ∈ ker(θ). Exchanging λ1 and λ2 changes the sign of the Vandermonde determinant, so

P (a1, . . . , an,±δ) = 0, where δ = disc(tn + a1tn−1 + · · ·+ an)

12 . Using the given form for P ,

A(a1, . . . , an)±B(a1, . . . , an)δ = 0

and hence A = B = 0 on the Zariski-dense open subset

(a1, . . . , an) ∈ Cn : disc(tn + a1tn−1 + · · ·+ an) 6= 0

Then A = B = 0 everywhere on Cn, so P = 0.

(11.5) Characteristic polynomial

Let V be an m-dimensional C-space. GL(V ) acts naturally on U = EndC(V ) by conjugation, i.e.

(g · θ)(v) = g · θ(g−1 · v)

(or, in shorthand, g · θ = gθg−1).

Suppose χθ(t) = det(t1V + θ) is the characteristic polynomial of θ and cn(θ) is the coefficient of tm−n

in χθ(t). Thenχg·θ(t) = det(t1V + gθg−1) = det(g(t1V + θ)g−1) = χθ(t)

so cn : U → C is an (n-homogeneous) invariant.

If θ has eigenvalues lambda1, . . . , λm, then putting θ in Jordan normal form gives

χθ(t) =

m∏

i=1

(t+ λi)

so that cn(θ) = en(λ1, . . . , λm); in particular

c1(θ) = λ1 + · · ·+ λm = tr(θ) and cm(θ) = λ1 . . . λm = det(θ)

Claim The C-algebra mapC[Y1, . . . , Ym] → C[U ]GL(V )

Yi 7→ ciis an isomorphism.

Sketch proof If f is a polynomial invariant for U then f(θ) is a symmetric polynomial function ofthe eigenvalues of θ. Mimic the proof that the characters of rational CGL(V )-modules are symmetricrational functions of the eigenvalues of g ∈ GL(V ) as done in (10.1).

(11.6) Discriminant of quadratic forms

Let V be an m-dimensional C-space and let

U = HC,2(V,C) = quadratic forms on V

By polarisation (see end of §6) identify U with the symmetric m×m matrices via f 7→ A, where

f(x) = xtAx

for x ∈ Cm a column vector.

Define disc(f) = det(A).

GLm(C) acts on U by

(gf)(x) = f(g−1x) = xt(g−1tAg−1)x, f ∈ U, g ∈ GLm(C), x ∈ Cm

sodisc(gf) = det(g−1tAg−1) = (det g)−2 disc(f)

47

and therefore disc : U → C is an SLm(C)-invariant (but not a GLm(C)-invariant).

Claim The C-algebra mapC[X] → C[U ]SLm(C)

X 7→ discis an isomorphism.

Proof It is injective: if there is a polynomial P such that P (disc f) = 0 for all f ∈ U then P (λ) = 0for all λ ∈ C, since the quadratic form

fλ(x1, . . . , xm) = x21 + · · ·+ x2

m−1 + λx2m

has discriminant λ, so P = 0.

Let θ : U → C be a polynomial SLn(C)-invariant and define

F :C → Cλ 7→ θ(fλ)

This is a polynomial function since θ is a polynomial map. We prove that θ(f) = F (disc f). It isenough to show this for f with disc f 6= 0. For, if this is case is known, then

U = f : disc f = 0 ∪ f : θ(f) = F (disc f)is a union of Zariski-closed subsets, but U is irreducible, so that U = f : θ(f) = F (disc f).

Recall that any symmetric matrix is congruent (over C) to a matrix of the form

diag(1, . . . , 1, 0, . . . , 0) =

1. . .

10

. . .

0

Thus if f ∈ U corresponds to the matrix A and λ = disc f = detA 6= 0 then there exists B ∈ GLm(C)such that BtAB = I. Now if C = diag(1, . . . , 1, (detB)−1), then BC ∈ SLm(C), and

(BC)tA(BC) = CtBtABC = diag(1, . . . , 1, (detB)−2) = diag(1, . . . , 1, λ)

and so θ(f) = θ(fλ) = F (λ), as required.

(11.7) Discriminant of binary forms

Define

U = HC,n(C2,C) =

homogeneous polynomials of degree n

in two variables X1, X2

Take f ∈ U . Then f has the form

f(X1, X2) = a0Xn1 + a1X

n−11 X2 + · · ·+ anX

n2

= b(λ1X1 + µ1X2)(λ2X1 + µ2X2) · · · (λnX1 + µnX2)

and definedisc(f) = b2n−2

∏

i<j

(λiµj − λjµi)2

This is well-defined since it is unchanged if two terms are exchanged or if one term is enlarged andanother is reduced by the same factor.

Example (n = 3) disc(f) = −27a20a

23 + 18a0a1a2a3 − 4a0a

32 − 4a3

1a3 + a21a

22

Claim disc : U → C is a (2n− 2)-homogeneous SL2(C)-invariant.

Proof Exercise for the daring.

(11.8) Definition) A covariant for a CSL(V )-module U is a polynomial invariant of the formU ⊕ V → C.

Examples

48

Every invariant θ for U gives rise to a covariantU ⊕ V → C(u, v) 7→ θ(u)

If U = HC,n(V,C) then there is an ‘evaluation covariant’ ev :U ⊕ V → C(f, v) 7→ f(v)

.

(11.9) The Hessian

If f is a function of X1, . . . , Xm then the Hessian of f is

H(f) = det

(∂2f

∂Xi∂Xj

)

It is itself a function of X1, . . . , Xm.

Let U = HC,n(Cm,C), the n-homogeneous polynomials in X1, . . . , Xm. The Hessian defines a polyno-mial map

U ⊕ Cm → C(f, v) 7→ H(f)(v)

Now U is naturally a CGLm(C)-module via the action

(gf)(v) = f(g−1v), f ∈ U, g ∈ GLm(C), v ∈ Cm

By the chain rule,H(gf)(gv) = (det g)−2H(f)(v)

and so H is an SL2(C)-covariant.

49

12 First fundamental theorem of invariant theory

The first fundamental theorem of invariant theory gives generators for the set of polynomial invariantswhen the module is the direct sum of copies of V and V ∗. The hope was then in principle to computeinvariants of an arbtirary rational module, but in practice this proved to be impossible. We will stateand prove the first fundamental theorem for GL(V ) and state it for SL(V ).

As usual, V is an m-dimensional C-space with basis e1, . . . , em.

(12.1) Theorem (multilinear first fundamental theorem for the general linear group)

For n, r ∈ N,

(i) HomCGL(V )(Tn(V ∗)⊗ T r(V ),C) = 0 if n 6= r;

(ii) HomCGL(V )(Tn(V ∗)⊗ Tn(V ),C) = spanCµσ : σ ∈ Sn, where

µσ(ϕ1 ⊗ · · · ⊗ ϕn ⊗ v1 ⊗ · · · ⊗ vn) = ϕσ(1)(v1) · · ·ϕσ(n)(vn)

Remark A proof in arbitrary characteristic can be found in de Cocini and Procesi (1976).

Proof Let X,Y be CG-modules, where G is some group. We know that

HomC(X∗ ⊗ Y,C) = (X∗ ⊗ Y )∗ ∼= X∗∗ ⊗ Y ∗ ∼= X ⊗ Y ∗ ∼= HomC(Y,X)

Taking G-fixed points, we obtain

HomCG(X∗ ⊗ Y,C) ∼= HomCG(Y,X)

With this and the isomorphism Tn(V ∗) ∼= (TnV )∗ we have an isomorphism

π :HomCGL(V )(T

rV, TnV )∼=−→ HomCGL(V )(T

n(V ∗)⊗ T r(V ),C)

f 7−→(ϕ1 ⊗ · · · ⊗ ϕn ⊗ ~v 7→ (ϕ1 ⊗ · · · ⊗ ϕn)(f(v))

)

Now

(i) If n 6= r then HomCGL(V )(TrV, TnV ) = 0 because HomCGL(V )(S

′, S) = 0 if S′ (resp. S) is anirreducible r-homogeneous (resp. n-homogeneous) CGL(V )-module.

(ii) is just a restatement of Schur–Weyl duality. To see this, recall that EndCGL(V )(TnV ) is spanned

by λσ for σ ∈ Sn, where

λσ(v1 ⊗ · · · ⊗ vn) = vσ−1(1) ⊗ · · · ⊗ vσ−1(n)

and now π(λσ) = µσ.

Notation For CG-modules U andW , let HomCG,n(U,W ) denote the vector space of all n-homogeneousconcomitants U →W (cf. (11.1)).

Note that HomCG,n(U,W ) is a CG-module with conjugation action

(gf)(u) = gf(g−1u), g ∈ G, u ∈ U, f ∈ HomCG,n(U,W )

Note also that HomCG,n(U,W ) = HC,n(U,W )G.

(12.2) Lemma If U is a CG-module then

C[U ]G ∼=⊕

n≥0

HomCG(U,C)

Proof We know thatC[U ] ∼=

⊕

n≥0

HC,n(U,C)

50

is an isomorphism of CG-modules; now take G-fixed points.

(12.3) Lemma If U and W are CCG-modules then there exist invrse isomorphisms of vector spaces

HomCG(SnU,W ) → HomCG,n(U,W )ψ 7→ ψ ∆

andHomCG,n(U,W ) → HomCG(SnU,W )

f 7→ 1n!Pf

where ∆ :U 7→ SnUu 7→ u ∨ · · · ∨ u and Pf is the total polarisation of f (see after (6.13)).

Proof Clearly the maps induce isomorphisms between HomC(SnU,W ) and HC,n(U,W ), so we’llprove:

(i) ψ concomitant ⇒ ψ ∆ concomitant;

(ii) f concomitant ⇒ 1n!Pf concomitant.

So. . .

(i) is clear since ∆ is a concomitant.

(ii) To prove this, there are (at least) three possible approaches.

Approach 1 Use the formula for Pf (not recommended).

Approach 2 If f ∈ HomCG,n(U,W ) then f =(

1n!Pf

)∆, so

(1

n!Pf

)(g(u ∨ · · · ∨ u)) =

(1

n!Pf

)(g∆(u))

=

(1

n!Pf

)∆(g(u))

= f(g(u))

= gf(u)

= g

(1

n!Pf

)(u ∨ · · · ∨ u)

Approach 3 HomC(SnU,W ) and HC,n(U,W ) are CG-modules, with G acting by conjugation,and ψ 7→ ψ ∆ is a CG-map and an isomorphism. So its inverse f 7→ 1

n!Pf is also a CG-map.These maps restrict to isomorphisms between sets of G-fixed points.

Remark ∆ factors as Uδ−→ TnU

natural−−−−→ SnU , where

δ(u) = u⊗ · · · ⊗ u︸ ︷︷ ︸n

so composition with δ induces a surjection

HomCG(TnU,W ) −→ HomCG(U,W )

We’ll use this reformulation of (12.3) to prove

(12.4) Theorem (first fundamental theorem for the general linear group)

Let U = V ⊕ · · · ⊕ V︸ ︷︷ ︸p

⊕V ∗ ⊕ · · · ⊕ V ∗︸ ︷︷ ︸q

. Then C[U ]GL(V ) is generated as a C-algebra by elements ρij

for 1 ≤ i ≤ p, 1 ≤ j ≤ q, where

ρij(v1, . . . , vp, ϕ1, . . . , ϕq) = ϕj(vi)

Proof Clearly the ρij are polynomial invariants, since

gϕj(gvi) = ϕj(g−1gvi) = ϕj(vi)

51

By (12.2) we must show that any n-homogeneous invariant f is a linear combination of products ofthe ρij . By (12.3) there exists a surjection

HomCGL(V )(TnU,C) −→ HomCGL(V ),n(U,C)

Put V1 = · · · = Vp = V and Vp+1 = · · · = Vp+q = V ∗, so that U =⊕p+q

i=1 Vi.

This induces a decomposition of TnU as

TnU =

p+q⊕

i1=1

p+q⊕

i2=1

· · ·p+q⊕

in=1

Vi1 ⊗ Vi2 ⊗ · · · ⊗ Vin

ThusHomCGL(V )(T

nU,C) =⊕

i1,...,in

HomCGL(V )(Vi1 ⊗ · · · ⊗ Vin ,C)

Focus on one of the summands. By (12.1), either HomCGL(V )(Vi1 ⊗ · · · ⊗ Vin ,C) = 0, or n = 2k, k ofthe ij are ≤ p and k of the ij are > p.

In the latter case, say ij ≤ p for j = α1, . . . , αk and ij > p for j = β1, . . . , βk. In this case,HomCGL(V )(Vi1 ⊗ · · · ⊗ Vin ,C) is spanned by the k! maps τσ, for σ ∈ Sk, given by

τσ(v1 ⊗ · · · ⊗ vn) = ϕβσ(1)(vα1) · · ·ϕβσ(k)(vαk)

The n-homogeneous invariant of U corresponding to τσ is just

ρiα1 ,iβσ(1)−p· · · ρiαk ,iβσ(k)−p

and we’re done.

(12.5) Theorem If r ∈ N and U = EndC V ⊕ · · · ⊕ EndC V︸ ︷︷ ︸r

then C[U ]GL(V ) is generated as a

C-algebra by invariants

ti1,...,ik :U → C

(θ1, . . . , θr) 7→ tr(θi1 · · · θikwhere k ≥ 1 and 1 ≤ i1 < · · · < ik ≤ r.

Remarks

(1) [Procesi, 1976] In fact, C[U ]GL(V ) is generated by the ti1,...,ik with 1 ≤ k ≤ 2m − 1.

(2) Procesi also shows that (12.5) is equivalent to the first fundamental theorem. This links invarianttheory to the representation theory of bouquets, a certain type of quiver.

Proof By (12.2) and (12.3) we must compute HomCGL(V )(Tn(EndC V,C) which, since EndC V ∼=

V ∗⊗V , is isomorphic to HomC(TnV ∗⊗TnV,C). By the multilinear first fundamental theorem (12.1),this is spanned by the µσ for σ ∈ Sn, where

µσ(ϕ1 ⊗ ϕn ⊗ v1 ⊗ · · · ⊗ vn) = ϕσ(1)(v1) · · ·ϕσ(n)(vn)

We now compute the corresponding map

νσ : Tn(EndC V ) −→ C

Let V have basis e1, . . . , em and V ∗ have dual basis η1, . . . , ηm. Let (Aij) be the matrix of θ ∈ EndC Vwith respect to these bases, i.e.

θ(v) =∑

i,j

Aijηj(v)ei

so that the corresponding element of V ∗ ⊗ V is∑i,j Aijηj ⊗ ei.

If θk has matrix Akij then θ1 ⊗ · · · ⊗ θn corresponds with

∑

a1,...,an

∑

b1,...,bn

A1a1b1 · · ·Ananbnηb1 ⊗ · · · ⊗ ηbn ⊗ ea1 ⊗ · · · ⊗ ean ∈ TnV ∗ ⊗ TnV

52

so we have

νσ(θ1 ⊗ · · · ⊗ θn) =∑

~a,~b

A1a1b1 · · ·Ananbnηbσ(1)(ea1) · · · ηbσ(n)

(ean)

=∑

~b

A1bσ(1)b1

· · ·Anbσ(n)bn

[The latter equality comes from the definition of dual basis.] Writing σ = (i1 · · · ik)(j1 · · · j`) · · · as aproduct of disjoint cycles, we reorder this to get

νσ(θ1 ⊗ · · · ⊗ θn) =∑

Ai1bi2bi1Ai2bi3bi2

· · ·Aikbi1bikAj1bj2bj1

· · ·Aj`bj1bj` · · ·= tr(θik · · · θi1) tr(θj` · · · θj1) · · ·

which is what we required.

Theorem (first fundamental theorem for the special linear group)

Let V be an m-dimensional C-space with basis e1, . . . , em and let V ∗ have dual basis η1, . . . , ηm. If

U = V ⊕ · · · ⊕ V︸ ︷︷ ︸p

⊕V ∗ ⊕ · · · ⊕ V ∗︸ ︷︷ ︸q

then C[U ]SL(V ) is generated as a C-algebra by the polynomial invariants which send

(v1, · · · , vp, ϕ1, · · · , ϕq) 7→ ϕj(vi) for 1 ≤ i ≤ p, 1 ≤ j ≤ q;

(v1, · · · , vp, ϕ1, · · · , ϕq) 7→ |vi1 , . . . , vim | for 1 ≤ i1 < · · · < im ≤ p

(v1, · · · , vp, ϕ1, · · · , ϕq) 7→ |ϕj1 , . . . , ϕjm | for 1 ≤ j1 < · · · < jm ≤ q

where |v| is the determinant of the matrix whose nth column is the coordinate of the vin with respectto e1, . . . , em and |ϕ| likewise with respect to ηj1 , . . . , ηjm .

End of course.

53

MATHEMATICAL TRIPOS PART III, 2013REPRESENTATION THEORY

This sheet deals with the representation theory and combinatorics of Sn and of GLn.Work over C unless told otherwise.

1 Let H be a subgroup of Sn; define the element H− =∑

h∈H εhh which lies in the groupalgebra CSn. Prove that

πH− = H−π = επH− for any π ∈ H,

andπH−π−1 = (πHπ−1)− for any π ∈ Sn.

2 Let r be a natural number. Let λ = (λ1, . . . λ`) be a sequence of positive integers. We

call λ a composition of r into ` parts if∑`

i=1 λi = r. The elements λi are called the parts ofλ.

(a) Find a formula for the number of compositions of r.(b) Find a formula for the number of compositions of r with at most k parts.

3 Let r, s be non-negative integers. A partition of the form λ = (r, 1s) is called a hookpartition; we will write H(n) for the set of all hook partitions of n. Let fλ be the number ofstandard tableaux of shape λ.

(a) Find a formula for the number of standard tableaux of shape λ = (r, 1s) where λ isa partition of n. Justify your answer.

(b) Prove that∑

λ∈H(n) fλ = 2n−1.

(c) Prove that∑

λ∈H(n)(fλ)2 =

(2(n−1)n−1

).

4 Let n ∈ N, and let λ = (λi)i∈N and µ = (µj)j∈N be partitions of n. The conjugatepartition λ′ is defined to be the partition λ′ = (λ′1, λ

′2, . . .) where λ′i = |j|λj > i|. The

dominance order on the set of partitions of n is defined as follows:λ µ if and only if for all i ∈ N:

∑

j6iλj 6

∑

j6iµj.

(a) Prove that λ µ if and only if µ′ λ′.(b) We say that a partition λ is self-conjugate if λ = λ′. Show that there are as many

self-conjugate partitions of n as there are partitions of n with distinct odd parts (i.e. all λiare odd and distinct).

5 (a) Assuming k > 3, prove that then f (3k) > 3k.

(b) Let λ = (l, . . . , l) = (lk) be a partition of n = kl and λ = (l + 1, . . . , l + 1) be apartition of n+ k. Prove (by using the Young-Frobenius formula (5.1)) that

f λ

fλ=

(n+ k)!

n!· l!

(l + k)!.

(c) Let k, l > 3 with kl = n. Prove that then f (lk) > n.

1

2 MATHEMATICAL TRIPOS PART III, 2013 REPRESENTATION THEORY

6 Let Ck be the set of all standard tableaux of shape (k, k) whose entry in box (2,1) issome odd number.

(a) Find |Ck|.(b) For a partition λ = (r, s) with two parts, write down f (r,s) (e.g. from the Young-

Frobenius formula). Can you find limk→∞ |Ck|/f (k,k)?

7 In this question λ = (λ1λ2, . . . , λk) is a partition of n, and Mλ is the permutation modulewith basis the λ-tabloids (a λ-tabloid is an equivalence class of numberings of a Young diagram(with distinct numbers 1,...,n), two being equivalent if the corresponding rows containing thesame entries.)

(a) Show that Mλ is isomorphic to the permutation module C∆ where ∆ = cos(Sn : Sλ),(the coset space with Sλ the Young subgroup).

(b) Show that

dimMλ =n!∏ki=1 λi!

.

(c) In the case λ = (n− 2, 2) has two parts, show that Mλ ∼= CΩ(2) where Ω(2) is the setof 2-element subsets of Ω = 1, . . . , n (with the natural action).

8 For a λ-tableau T with n boxes, define the signed column sum κT =∑

σ∈CTεσσ . Let

eT = κTT be the λ-polytabloid. Let 1 denote the identity element of Sn.(a) List the columns of T as C1, C2, . . . , Cs. Prove that κT = κC1 · · ·κCs .(b) Let H 6 Sn and let (a b) ∈ H. Define H− =

∑h∈H εhh, as in Q.1. Prove that

H− = s · (1− (a b)) for some s ∈ CSn.(c) Let π ∈ Sn and σ ∈ CT . Show that

κπT = πκTπ−1, κTσ = σκT = εσκT

andeπT = πeT ; ; σeT = εσeT .

9 *[The Murnaghan-Nakayama Rule.]Part (a) below is an optional task for enthusiasts only. You will need to read about

James’ method (lecture notes pp 79–83) using the Littlewood-Richardson Rule. A goodaccount appears in Sagan’s Springer GTM, p.182. This beautiful result has generalisationsto the irreducible characters of exotic structures called Iwahori-Hecke algebras of type An−1due to Ram (who appealed to Schur-Weyl duality), and it appears to have been known toYoung in his work on hyperoctahedral groups (the so-called Weyl groups of type Bn).

(a) Consider the character χλ(w) =: χλµ where w ∈ Sn has cycle type (µ1, µ2, . . . , µ`).Suppose there are precisely t partitions λ(1), . . . , λ(t) which are obtained by removing a rim

hook of length µ1 from [λ]. Find an expression for χλµ in terms of χλ(i)(µ2,µ3...,µ`)

.

(b) Let n > 6. Prove that

χ(n−3,3)(2,1n−2) = (n− 3)(n− 4)(n− 5)/6.

MATHEMATICAL TRIPOS PART III, 2013 REPRESENTATION THEORY 3

10 Given some CSn-module M , with action written as π · m (π ∈ Sn,m ∈ M). Define anew CSn-module M ε as the module whose underlying vector space is M and with new actionπ ∗m := εππ ·m (π ∈ Sn,m ∈M).

(a) Verify that M ε is a CSn-module.(b) Prove that M is irreducible if and only if M ε is irreducible.(c) Let χ be the character of M , and let χ(1n) be the character of the Specht module

S(1n). Prove that the character χε of M ε is given by χε(π) = χ(1n)(π) for π ∈ Sn.[Why do you now know that the trivial module S(n) and the sign module S(1n) are the only1-dimensional CSn-modules.]

11 Let ρ :GL(V ) →GLn(C) be an irreducible polynomial representation. Show that thematrix entries ρij ∈ C[GL(V )] are homogeneous polynomials of the same degree. Moregenerall, show that every polynomial representation ρ is a direct sum of representations ρ(i)

whose matrix entries are homogeneous polynomials of degree i.

12 Give a proof of the result proved in lectures that affine n-space An is irreducible withoutappealing to the Nullstellensatz.

13 Prove that the 1-dimensional rational representations of C∗ =GL1(C) are power mapsof the form u 7→ ur where r ∈ Z.

14 (a) Show that the set of diagonalisable matrices is Zariski dense in Mn(C) (hence everyinvariant function on Mn(C) is completely determined by its restriction to Tn, the diagonalmatrices. [Actually this is nothing more than Jordan decomposition; if you work a bit harderyou can show this is true over any field, not necessarily algebraically closed].

(b) Let ρ :GLn →GL(W ) be a rational representation. Show that ρ is polynomial if andonly if ρ|Tn is polynomial. Show also that ρ is polynomial if and only if its character χρ is apolynomial.

(c) Deduce that every 1-dimensional rational representation of GL(V ) is of the formdetr :GL(V )→GL1, r ∈ Z.

15 Consider the linear action of GL(V ) on End(V ) by conjugation.(a) What are the orbits of this action? Any element α ∈End(V ) has a characteristic

polynomial of the form pα(t) = tn +∑n

i=1(−1)iei(α)tn−i, where n = dimV , showing that theei are invariant polynomial functions on End(V ).

(b) [HARD] Find a proof of the fact that the ring of invariants for the conjugation actionis generated by the si without using properties of symmetric functions.

(c) Show that the functions Tr1 . . . ,Trn generate the invariant ring C[End(V )]GL(V ). HereTrk :End(V )→ C is the map A 7→TrAk, (k = 1, 2, . . .).

SM, Lent Term 2013Comments on and corrections to this sheet may be emailed to sm@dpmms.cam.ac.uk

MATHEMATICAL TRIPOS PART III, 2013REPRESENTATION THEORY

This sheet deals mostly with polynomial invariant theory. Work over C.

1 Let W be finite-dimensional over C. Check that the coordinate functions x1, . . . , xm ∈C[W ] are algebraically independent (equivalently, if f(a1, . . . , am) = 0 for a polynomial f andall a = (a1, . . . , am) ∈ Cm then f is the zero polynomial). Hint: use induction on the numberof variables and the fact that a non-zero polynomial in one variable has only finitely manyzeroes.

2 (a) Show that the natural representation of SL2(C) on C2 is self-dual (i.e. equivalent toits dual representation). Hint: find a non-singular matrix P such that (A−1)t = PAP−1.

(b) Show that this representation has two orbits.

3 Consider GL2 acting on C[X, Y ] (induced by the natural representation of GL2 on C2).(a) What is the image of X and Y under the action of any g ∈GL2?

(b) Calculate C[X, Y ]GL2 and C[X, Y ]SL2 .(c) Calculate C[X, Y ]U where U is the subgroup of upper triangular unipotent matrices.

4 Consider the representation of SLn(C) in its action on Mn(C), as discussed in lectures.Prove that the invariant ring is generated by the determinant. You may need the fact thatGLn is Zariski-dense in Mn.

5 Determine the invariant rings of C[M2(C)]U and C[M2(C)]T under left multiplication bythe subgroup U of upper triangular unipotent matrices and the 2-torus T =diag(t, t−1) ofdiagonal matrices of SL2.

6 Let Tn be the subgroup of GLn of non-singular diagonal matrices. Choose the standardbasis in W = Cn and the dual basis in W ∗ and thus identify the coordinate ring C[W ⊕W ∗]with C[x1, . . . xn, ζ1, . . . , ζn]. Show that

C[W ⊕W ∗]Tn = C[x1ζ1, . . . , xnζn].

What happens if you replace Tn by the subgroup T ′n of diagonal matrices with determinant1?

1

2 MATHEMATICAL TRIPOS PART III, 2013 REPRESENTATION THEORY

7 Recall that for each j > 1 the symmetric functions nj(x) = xj1 +xj2 + · · · xjn are known aspower sums (or Newton functions). As before the ej are the elementary symmetric functions.

(a) Show that

(−1)j+1jej = nj − e1nj−1 + e2nj−2 − · · ·+ (−1)j−1ej−1n1

for all j = 1 . . . , n. This is known as the Newton-Girard identity. [Hint: let j = n, considerf(t) =

∏i(t− xi) and calculate

∑i f(xi). For j < n what can you say about the right-hand

side?](b) Here is Weyl’s original proof1 of the identity in (a). Define the polynomial

ψ(t) =n∏

i=1

(1− txi) = 1− e1t+ e2t2 − · · ·+ (−1)nent

n,

where the ei are the elementary symmetric functions. Determine its logarithmic derivative

−ψ′(t)ψ(t)

as a formal power series. Deduce (a).

(c) Show that (in characteristic 0) the power sums generate the symmetric functions.

8 (a) Let A be a commutative algebra and let G be a group algebra of automorphisms ofA. Assume the representation of G on A is completely reducible. Show that the subalgebraAG of invariants has a canonical G-stable complement and the corresponding G-equivariantprojection π : A → AG satisfies π(hf) = hπ(f) for h ∈ AG, f ∈ A. This projection issometimes called the Reynolds operator.

(b) Let A = ⊕j>0Aj be a graded K-algbbra (meaning AiAj ⊂ Ai+j). Assume theidea A+ = ⊕j>0Aj is finitely-generated. Show that A is finitely-generated as an algebraover A0. (In fact if the ideal A+ is generated by the homogeneous elements a1, . . . , an, thenA = A0[a1, . . . , an].)

(c) Deduce Hilbert’s theorem: if W is a G-module and the representation of G on K[W ]is completely reducible then the invariant ring K[W ]G is finitely-generated. [Hint: you willneed to use Hilbert’s Basis Theorem.] This shows also that K[W ]G is noetherian.

(d) Hilbert’s Theorem is applicable to finite groups provided Maschke’s Theorem holds(i.e we require that |G| is invertible in K). In the special case of characteristic 0, show thatfor any representation W of G, the ring of invariants K[W ]G is generated by invariants ofdegree less than or equal to |G| (Noether’s Theorem).

Schmidt introduced a numerical invariant β(G) for every finite group G. It is definedto be the minimal number m such that for every representation W of G, the invariant ringK[W ]G is generated by the invariants of degree 6 m. With this definition, Noether’s Theoremasserts that β(G) 6 |G|. In fact Schmidt showed that β(G) = |G| if and only if G is cyclic2.It is also a fact that if G is actually abelain then β(G) coincides with the so-called Davenportconstant.

In the next few questions we’ll write V = C2, G =SL(V ) =SL2(C), and

Cn = HC,n(V,C) ∼= (SnV )∗ ∼= Sn(V ∗),

which can be identified with the set of homogeneous polynomials of degree n in variablesX1, X2. Note that Cn : n ∈ N are the non-isomorphic simple CG-modules. For a fixed n,recall that a covariant for Cn is a polynomial CSL(V )-invariant Cn ⊕ V → C. The followingquestions sketch a classification of the generators for C[Cn ⊗ V )]G.

1Weyl, The Classical Groups, II.A.32see Finite groups and invariant theory in Seminaire d’Algebre Paul Dubreil et M.-P. Malliavin (Springer

Lecture Notes 1478, Springer 1991)

MATHEMATICAL TRIPOS PART III, 2013 REPRESENTATION THEORY 3

9 Suppose f, g are functions of X1, X2 and r ∈ N. Define the rth transvectant of f, g as

τr(f, g) =r∑

j=0

(−1)j

j!(r − j)!∂rf

∂Xr−j1 ∂Xj

2

∂rg

∂Xj1∂X

r−j2

.

=1

r!

(∂

∂X1

∂

∂Y2− ∂

∂X2

∂

∂Y1

)r(f(X1, X2)g(Y1, Y2) |Y1=X1,Y2=X2 .

You should observe τ0(f, g) = fg, τ1(f, g) is the Jacobian of f, g and τ2(f, f) is the Hessianof f . Show that

(a) if f, g are homogeneous polynomials of degrees p, q then τr(f, g) = 0 unless r 6minp, q, in which case it is a homogeneous polynomial of degree p+ q − 2r;

(b) if r 6 minp, q then the map τr : Cp ⊗ Cq → Cp+q−2r sending f ⊗ g 7→ τr(f, g) is anon-zero map of CG-modules;

(c) any CG-module map Cp ⊗ Cq → Cp+q−2r is a multiple of τr. [Hint: use Clebsch-Gordan];

(d) the τr give an isomorphism of CG-modules

Cp ⊗ Cq →minp,q⊕

r=0

Cp+q−2r.

10 Let p, q, n, r ∈ N and r 6 minq, n. Set N = max0, r−p. Show that there are scalarsλN , . . . , λr ∈ C, with λr 6= 0, such that

fτr(g, h) =r∑

j=N

λjτj(τr−j(f, g), h),

for all f ∈ Cp, g ∈ Cq, h ∈ Cn. [HINT: Schur’s Lemma.] Thus the τj are trying hard to beassociative.

11 Fix n ∈ N. Let R = C[Cn ⊕ V ], so that RG is the set of covariants of Cn. If ϕ ∈ R andf ∈ Cn, define ϕ(f) ∈ C[V ] by ϕ(f)(x) = ϕ(f, x). If r ∈ N and φ, ψ ∈ R define τr(φ, ψ) asthe map Cn ⊕ V → C by (f, x) 7→ τr(φ(f), ψ(f))(x). Finally if d, i ∈ N, let Rdi be the set ofall φ ∈ R which are homogeneous, of degree d in Cn and degree i in V . Show that

(a) Rdi are CG-submodules of R, and R is graded, R = ⊕∞d,i=0Rdi, hence so too is RG;(b) RdiRej ⊆ Rd+e,i+j and τr(Rdi, Rej) ⊆ Rd+e,i+j−2r for r 6 mini, j;(c) the assignment φ 7→ (f 7→ φ(f)) induces an isomorphism

RGdi∼= homCG,d(Cn, Ci),

and deduce that(d) any covariant θ ∈ RG

di (d > 1) can be expressed as a linear combination

θ =

minn,i∑

r=0

τn−r(φr, E)

with φr ∈ RGd−1,n+i−2r and E is the evaluation map E(f, v) = f(v) [you should note that (c)

says that RG1,i = C.E when i = n and 0 if i 6= n.]

4 MATHEMATICAL TRIPOS PART III, 2013 REPRESENTATION THEORY

12 [Gordan’s Theorem3 - a weak form] If S is a C-subalgebra of RG with the property thatτr(φ,E) ∈ S whenever r ∈ N and φ ∈ S, then S = RG. [Remark: after the last question theproof should reduce to a two- line argument.]

13 [REALLY HARD] This uses Gordan’s Theorem to find a set of generators of RG in thecase n = 3. Given n = 3, show that RG is generated by the covariants

E ∈ RG13;

H = τ2(E,E), the Hessian, ∈ RG22;

t = τ1(H,E) ∈ RG33;

D = τ3(t, E), the discriminant (×48), ∈ RG40

14 [REALLY REALLY HARD4 ] Take n = 4 and repeat the last question to generate RG

using quartic forms.Classically, the (still unsolved) problem of computing all covariants of binary forms of

degree n was tackled by something called the symbolic method (using polarisation to reduceit to the FFT). Associated to the symbolic method is the symbolic notation, designed tomake the calculations easier, but still non-trivial. You can consult any old text on InvariantTheory, such as Grace and Young if interested.

SM, Lent Term 2013Comments on and corrections to this sheet may be emailed to sm@dpmms.cam.ac.uk

3Paul Gordan, dubbed the ‘King of Invariant Theory’ is perhaps better known for being Emmy Noether’sthesis advisor. The result appeared in 1868 in Crelle’s Journal as Beweis, dass jede Covariante und Invari-ante einer binaren Form eine ganze Funktion mit numerischen Coeffizienten einer endlichen Anzahl solcherFormen ist.

4J.J. Sylvester’s collected works (four volumes, edited by H.F. Baker) include tables with details abouthigher degree forms. However his published tables for forms of degree larger than 6 appear to be totally wrong.Corrections appear in von Gall (1888) and Dixmier/Lazard (1986), Shioda (1967) and Brouwer/Popoviciu(2010).