Part III Representation Theory Lectured by Stuart Martin, Lent Term 2013

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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 representation theory 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) such that ≤ ≤ 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 length n, 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 composition series of M have the same length, which we denote by `(M) and call the length of M.

0 3.( Jordan–H¨older) If (Mi) and (Mi ) are composition series then there exists a one-to-one cor- 0 0 respondence between the set of quotients (Mi−1/Mi) and the set of quotients (Mi−1/Mi ) 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 .

n If dimK (V ) = n, choose a basis e1, . . . , en of V over K to identify it with K . Then θ ∈ GL(V ) Aθ = (aij) Mn(K), where ↔ ∈ n θ(ej) = aijei i=1 X 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) g v, where (a)( action)(g1g2) v = g1 (g2 v) and e v = v × → 7→ · · · · · (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 ρg, where ρg(v) = g v, deﬁnes a representation of G on V . Conversely,→ given a representation7→ρ : G GL(V ) we have· a linear action 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) If G is ﬁnite and char(K) - G, then ever KG-module is completely reducible; or equivalently, every submodule is a summand.

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

If S is an irreducible CG-module and M is any CG-module, then the number of composition factors 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. Modules are always left R-modules.

(1.1) Deﬁnition A C-algebra is a ring R which is also a C-vector space, whose notions of addition coincide, and for which λ(rr0) = (λr)r0 = r(λr0) for all r, r0 R and λ C. There are obvious notions of subalgebras and algebra homomorphisms. ∈ ∈

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 = αgg : αg C is a C-algebra with { g ∈ } pointwise addition: g αgg + g βgg = g(αg + βg)g P multiplication: g P hh0=g αhPβh0 g P Recall the correspondenceP P between representations of G 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 can be 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 and { } ∈ only 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 a C-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 given by composition of maps:

(ϕψ)(m) = ϕ(ψ(m)) m M, φ, ψ EndR(M) ∀ ∈ ∈ V V V In particular, if is a C-space then EndC( ) is an algebra. Recall that EndC( ) ∼= Mn(C), where n = 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 End (M) given by α(r)(m) = rm → C for r R, m M is a map of C-algebras. ∈ ∈ Now, EndR(M) = cEnd (M)(αR), and so αR cEnd (M)(EndR(M)). C ⊆ C (h) An R-module M is itself naturally an EndR(M)-module, where the action is evaluation, and the action commutes with that of R. The inclusion α as in (g) says that elements of R act as EndR(M)-module endomorphisms of M.

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

4 (1.3) Lemma Let R be a finite-dimensional C-algebra. Then there are only finitely many isomorphism 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 = x S. Deﬁne a map R S by r rx. This map has nonzero image, so its image is S by irreducibility.6 ∈ In particular, → 7→

dim (S) dim (R) < C ≤ C ∞ But S must occur in any composition series of R, so by the Jordan–H¨oldertheorem, there are only ﬁnitely 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) = 0 → ≤ ≤ or R and im(f) = 0 or S. If ker(f) = 0 then R ∼= im(f) = S, contradicting nonisomorphism; so we 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 with p(d) = 0. ∈ Since C is algebraically closed, p factors as

p(X) = c(X a1) (X an) − ··· − for some 0 = c C and ai C. Hence 6 ∈ ∈

(d a11D) (d an1D) = 0 − ··· − But, being a division ring, D has no zero divisors, so one of the terms must be zero. Thus d = 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 of irreducible 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 direct summand. (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 this and 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 N has 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 End (M N) commute. C ⊗ ⊕k If N has basis e1, . . . , ek then the map M M N given by { } → ×

(m1, . . . , mk) m1 e1 + + mk ek 7→ ⊗ ··· ⊗ is an isomorphism of R-modules.

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

(n1, . . . , n`) n1 f1 + + n` f` 7→ ⊗ ··· ⊗ is an isomorphism of S-modules.

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

S HomR(S, M) M ⊗ → S M is an isomorphism of R-modules and of EndR(M)-modules. Here S runs over a complete set of nonisomorphic 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 isomorphism of vector spaces, by observing that we can reduce to the case where M is irreducible and apply Schur’s lemma (1.4). (1.9) Lemma Let M be a ﬁnite-dimensional semisimple R-module. Then

M S, M EndR( ) ∼= EndC HomR( ) S Y where S runs over a complete set of nonisomorphic irreducible R-modules.

Proof Observe that the product E = End HomR(S, M) acts naturally on S HomR(S, M) S C ⊗ and, since this action commutes with that of R, there exists a homomorphism E EndR(M), which is injective. Q →

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

2 dimC EndR(S, M) = S(dimC HomR(S, M)) = dimC E

(1.10) Theorem (Artin–Wedderburn Theorem)P Any ﬁnite-dimensional semisimple C-algebra is isomorphic to a product k

R = EndC(Vi) i=1 Y where Vi are ﬁnite-dimensional vector spaces. Conversely, if R has this form then it is semisimple, the nonzero Vis form a complete set of nonisomorphic irreducible R-modules and, as an R-module, R is isomorphic 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 are naturally R-modules, with the factors other than EndC(Vi) acting as 0.

6 Now End , and hence R, has transitively on Vi 0 , thus the Vi are irreducible R-modules. C − { }

If Vi has basis ei1, . . . , eim then the map { i } R V1 V1 Vk Vk → ⊕ · · · ⊕ ⊕ · · · ⊕ ⊕ · · · ⊕ m1 copies mk copies given by | {z } | {z } (f1, . . . , fk) (f1(e11), . . . , f1(e1m ), . . . , fk(ek1), . . . , fk(ekm )) 7→ 1 k is an injective map of R-modules, and hence is an isomorphism by counting dimensions. Thus R is semisimple.

The Vi form a complete set of irreducible R-modules (by Jordan–H¨oldeer,as in (1.3)), and they are nonisomorphic: if i = j then (0,..., 0, 1, 0,..., 0) annihilates Vj but not Vi. 6 Step 2 Let R be any ring. Now the natural map α : R End (R) → EndR(R) sending r to the map x rx is an isomorphism. It’s injective, and if θ End (R) then it 7→ ∈ EndR(R) commutes 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 semisimple EndR(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 End (M) → EndR(M) sending r to the map m rm is surjective. 7→ Proof Let I = M ◦ = r R : rM = 0 be the annihilator of M; then ker α = I. Now, R/I is { ∈ } semisimple and M is an R/I-module, and EndR(M) = EndR/I (M). So we can replace R by R/I and suppose 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 (as Q an EndR(M)-module) to S HomR(S, M), so it’s the direct sum of dim S copies of the simple S ⊗ C module HomR(S, M) for each S. L As in (1.9), this implies that

2 dimC EndEndR(M)(M) = (dimC S) S X But this is equal to dimC R, 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 = λe ∈ for some 0 = λ C. Then for any R-module M, we have an isomorphism of EndR(M)-modules 6 ∈ 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. ∈ ∈ ∈

e Replacing e by λ , we may suppose that e is idempotent. Now ther exists a map of EndR(M)-modules Hom(Re, M) eM → given by ϕ ϕ(e), whose inverse sends m to the map r rm. 7→ 7→

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 setting → g v = ρ(g)v for g G and v V . The trivial representation G C× = GL(C) is given by g 1, · ∈ ∈ → 7→ 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 → {± } σ(i) σ(j) ε = − σ i j 1≤i

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, if ∈ a1 a2 ak ` λ1 λ2 λ3 and λi = n. Write (λ , λ , . . . , λ ) to denote ≥ ≥ ≥ · · · i 1 2 k

P (λ1, . . . , λ1, λ2, . . . , λ2, . . . , λk, . . . , λk, 0, 0,... )

a1 a2 ak

with λ1 > λ2 > > λk. | {z } | {z } | {z } ···

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. This is the lexicographic` (or dictionary) order on partitions.∈ It is a total order on the set of partitions of 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 of those nodes∈ whose ﬁrst (resp. second) coordinate is k.

Example If λ = (4, 22, 1) then

λ1 • • • • ← λ2 [λ] = • • ← •• λ3 • ←

The height (or length) of λ is the length of the ﬁrst column of [λ], i.e. max j : λj = 0 , and is denoted by ht(λ). { 6 }

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

1 2 4 5 4 5 7 3 t1 = 3 6 7 , t2 = 2 1 8 8 6

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 and π 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. Let ◦ ` tλ be the standard λ-tableau, i.e. the tableau numbered in order from left to right and from top to bottom, e.g. 1 2 3 4 5 ◦ 6 7 t 2 = (5,2 ,1) 8 9 10

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 σt ∈ ⇔ ∈ { } σ 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

Rt = S S S 1 {1,2,4,5} × {3,6,7} × {8} Ct = S S S S 1 {1,3,8} × {2,6} × {4,7} × {1}

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

h(tλ) = εcrc A ∈ r∈R , c∈C tλX tλ ◦ Write hλ = h(tλ). Examples λ n h h t σ symmetrizer A (a) If = ( ), then = ( (n)) = σ∈Sn , called the in . Since ρ h = h for all ρ Sn, Ah = Ch is the trivial representation. · ∈ P n λ h h t n ε σ alternizer A (b) If = (1 ), then = ( (1 )) = σ∈Sn σ , called the in . Since ρ h = ερh, Ah = Ch is the sign representation. · P

Goal The left-ideals in A of the form Ahλ for λ n (known as Specht modules) form a complete set 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;

−1 −1 (c) Rσ(t) = σRtσ and Cσ(t) = σCtσ ;

−1 (d) h(σtλ) = σh(tλ)σ ;

(e) Ah(tλ) ∼= Ahλ as A-modules.

Proof (a)–(d) are clear.

◦ For (e), tλ = σt for some permutation σ Sn. Postmultiplication by σ then gives an isomorphism λ ∈ Ah(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 is true: ` ≥

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 Rt and c Ct . ∈ λ ∈ λ

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

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

0 0 Continuing in this way we see that λ = µ and there exists c Ctλ such that tλ and c tµ have the 0 ∈ same numbers in each row. Then rtλ = c tµ for some r Rt . ∈ λ Now deﬁne −1 0 −1 −1 0 0 −1 −1 c = r (c ) r r c Ct (c ) r = r Cc0t r = Cc0t = Ct ∈ µ µ µ λ Then tµ = rctλ.

(2.7) Lemma If σ Sn cannot be written as rc for any r Rt and c Ct then there exist ∈ ∈ λ ∈ λ transpositions u Rt and v Ct such that uσ = σv. ∈ λ ∈ λ

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

(2.8) Lemma Given a tableau tλ and a A, the following are equivalent: ∈ (1) rac = εca for all r Rt and c Ct ; ∈ λ ∈ λ (2) a = kh(tλ) for some k C. ∈

Proof

0 0 0 (2) (1) As r runs through Rt so does rr , and as c runs through Ct so does c c with εc0c = εc0 εc. ⇒ λ λ Calculation reveals that rh(tλ)c = εch(tλ).

(1) (2) Let a = aσσ. If σ is not of the form rc then by (2.7) there exist transpositions u Rt ⇒ σ∈Sn ∈ λ and v Ct such that uσv = σ. ∈ Pλ By assumption, uav = εva, and the coeﬃcient of σ gives auσv = εvaσ, so aσ = aσ, and hence − aσ = 0. The coeﬃcient of rc in (1) gives a1 = εcarc C. Thus ∈

a = arcrc = εca1rc = a1h(tλ) r,c rc X X 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 = dimC Ahλ. Then

2 n! (a) h(tλ) = h(tλ) fλ

(b) fλ n! | (c) h(tλ)Ah(tλ) = Ch(tλ)

fλ In particular, h(tλ) is idempotent. n! Proof

10 2 (a) Let h = h(tλ). We already know that h = kh for some k C. Right-multiplication by h induces ∈ a linear map hˆ : A A. For a A,(ah)h = k(ah), so kˆ Ah is multiplication by k. → ∈ | Extend a basis of Ah to a basis of A. With respect to this basis,

kIf hˆ λ ∗ ∼ 0 0 and so tr(hˆ) = kfλ.

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

Hστ = coeﬃcient of σ in τh

But Hστ = 1, so tr(hˆ) = n!. n! Hence k = . fλ 2 (b) The coeﬃcient of 1 in h = kh is εc1 εc2 Z where the sum runs over all r1, r2 Rtλ and ∈ ∈ c1, c2 Ct such that r1c1r2c2 = 1. ∈ λ P (c) By (2.9), the only other possibility is hAh = 0; but h2 = 0. 6

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

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

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

(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 ∈ Rt Ct be the corresponding transposition. Then λ ∩ µ 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 ∈ −1 0 = h(tµ)h(tλ) = h(tµ)σh(tλ)σ

and so h(tµ)σh(tλ) = 0.

Thus h(tµ)Ah(tλ) = 0.

(2.13) Lemma If λ = µ and tλ, tµ are tableaux, then Ah(tλ) Ah(tµ). 6 Proof Assume without loss of generality that λ > µ. If there exists an A-module isomorphism f : Ah(tλ) ∼= Ah(tµ) then (2.10) (2.12) f(Ch(tλ)) = f(h(tλ)Ah(tλ)) = h(tλ)f(Ah(tλ)) = h(tλ)Ah(tµ) = 0 which is a contradiction.

(2.14) Theorem The left-ideals Ahλ : λ n form a complete set of nonisomorphic irreducible A-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) and columns (top to bottom) of t.

0 0 The standard tableaux of shape λ are ordered such that tλ < tλ if tλ is smaller than tλ in the ﬁrst place they diﬀer when you read [λ] ‘like a book’. E.g. the standard (3, 2)-tableaux are

1 2 3 1 2 4 1 2 5 1 3 4 1 3 5 < < < < 4 5 3 5 3 4 2 5 2 4

Let Fλ be the number of standard λ-tableaux.

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

2 First we show Fλ = n!. λ`n X 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 = Fµ. ` µX: λ/µ −1 Proof If tλ is standard then t ( 1, 2, . . . , m 1 ) is the diagram of a partition µ m 1; and λ { − } ` − tλ must be standard. The converse is similar. |[µ] (3.3) Lemma If λ = π are partitions of m then 6 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 in fact 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 = Fν . ` νX: ν/λ Proof by induction on m.

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

(3.2) Fν = Fπ = ν : ν/λ Fλ + Fπ   |{ }| νX: ν/λ νX: ν/λ πX: ν/π ν,π : ν/λ,πX 6=λ   Using the fact that ν : ν/λ = τ : λ/τ + 1 together with (3.3) gives that our expression is equal to |{ }| |{ }| ( τ : λ/τ + 1)Fλ + Fπ |{ }| τ,π : λ/τ,π/τ,πX 6=λ

= Fλ + Fπ = Fλ + mFτ = Fλ + mFλ = (m + 1)Fλ τ,π :Xλ/τ,π/τ τX: λ/τ

12 as required.

2 (3.5) Lemma Fλ = m! λ`m X Proof by induction on m.

The case m = 1 is clear. Suppose the statement holds true for partitions of m 1. Then − 2 (3.2) (3.4) 2 F = FλFτ = mF = m (m 1)! = m! λ τ · − λ`m τ`m−1 X λ`Xm,λ/τ X as required.

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

Proof It suﬃces to show that there are integers i = j lying in the same row of t0 and the same 6 λ column of t . For then the corresponding transposition τ = (i j) R 0 C gives the result as in the λ tλ tλ proof of (2.12). ∈ ∩

0 It remains to find such integers i and j. Let x [λ] be the first place where tλ and tλ first differ. Let 0 0 −1 ∈ tλ = i and y = (tλ) (i) [λ]. Then y must be below and to the left of x since tλ is standard. In particular, x cannot lie in∈ the first column or the last row. Let z [λ] be in the same row as x and 0 ∈ 0 the same column of y, and let j = tλ(z) = tλ(z) be the common value of tλ and tλ 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. ` L 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) = 0, 0 P 0 0 6 and for this µ pick tµ minimal (in the ordering on standard µ-tableaux) such that a(tµ)h(tµ) = 0. 0 0 0 2 6 Multiplying the relation on the right by h(tµ) gives a(tµ)h(tµ) = 0 by (3.6) and (2.12), and hence 0 0 a(tµ)h(tµ) = 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–H¨older, Fλ fλ. But fλ Fλ since, 2 L 2 L ≤ ≤ by (3.5), Fλ = n! = fλ. Hence Fλ = fλ for each λ n, and so CSn CSnh(tλ). ` ≤ P P L (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. ` L 0 Proof If m(tλ) = 0 is a nontrivial relation with m(tλ) h(tλ)M then choose µ minimal and tµ maximal with m(t0 ) = 0. Premultiply the relation by h(t0 )∈ to contradict (3.6) and (2.12). Finally, P µ 6 µ (1.12) h(t )M Hom ( S h(t ),M) λ ∼= CSn C n λ M M Hom S h(t ),M Hom ( S ,M) M ∼= CSn C n λ ∼= CSn C n ∼= M

13 4 Character formula

Given a ﬁnite-dimensional CG-algebra M, its character is deﬁned by M M χ (g) = tr M m →gm 7→ and χM is a class function G C, i.e. a function that is constant on the conjugacy classes of G. If → α is a conjugacy class in G, write χM (α) for the common value of χM on α.

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

αn α2 α1 Now Sn-class α comprises all permutations with ﬁxed cycle type n 2 1 , i.e. having αn n-cycles, ··· ... , α2 2-cycles and α1 1-cycles. Let nα = α . | | n! (4.1) Lemma nα = α α α 1 1 2 2 . . . n n α1!α2! . . . αn! · Proof Any permutation in α is one of the n! of the form n n n ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ∗ ∗ ··· ∗ ∗∗ ∗∗ ··· ∗∗ ··· ∗ · · · ∗ ∗ · · · ∗ ··· ∗ · · · ∗ α1 α2 z }| { z }|αn{ z }| { α1 αn Each such permutation| can{z be represented} | {z in 1 }. . . n| α1! . . . α{zn! ways: each} of the αk k-cycles can · be represented in k ways by cycling their components, and each block of αk k-cycles can be represented in 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 `

λ µ n! if λ = µ (a) nαχ (α)χ (β) = λ µ ccls α (0 if = X 6 n!/n if α = β (b) χλ(α)χλ(β) = α α β λ`n (0 if = X 6

Proof Every element of Sn is real, i.e. conjugate to its inverse, and so χλ(σ) = χλ(σ−1) = χλ(σ)

λ so χ (σ) R for all σ Sn. These relations are then just the usual ones for general ﬁnite groups. ∈ ∈ m Notation Given x = (x1, . . . , xn) C , `1, . . . , `m Z, deﬁne ∈ ∈ ` x`1 , . . . , x`m = det(x j ) | | i m

Usually `j 0, in which case this is a homogeneous polynomial of degree `j in x1, . . . , xm. ≥ j=1 X Example The Vandermonde determinant is given by V (x) = xm−1,..., 1 | |

(4.3) Lemma V = V (x) = (xi xj) i

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

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

Example s (x) = x1 xm. (1,2,...,n) ··· (4.4) Lemma (Cauchy’s lemma) If xi, yj C for 1 i, j m and xiyj = 1 for all i, j, then ∈ ≤ ≤ 6 m 1 1 det = xm−1,..., 1 ym−1,..., 1 1 x y 1 x y i j i,j=1 i j − Y −

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

Now 1 1 xi x1 yj = − 1 xiyj − 1 x1yj 1 x1yj · 1 xiyj − − − − So subtracting the ﬁrst row from each other row in the determinant, one can remove the factor (xi x1) 1 − from each row i = 1 and from each column. So 6 1 x1yj − 1 1 1 y1 y2 ··· yn m 1−x2y1 1−x2y2 1−x2yn 1 1  y1 y2 ··· yn  det = (xi x1) det 1−x3y1 1−x3y2 ··· 1−x3yn 1 xiyj − !  1 x1yj   . . . .  − i>1 j=1 −  . . .. .  Y Y      y1 y2 yn   1−xny1 1−xny2 ··· 1−xnyn    =∆

Now subtract the ﬁrst column in the matrix above from the| other columns{z and use the fact that}

yj y1 yj y1 1 = − 1 xiyj − 1 xiy1 1 xiy1 · 1 xiyj − − − − This gives

1 0 0 0 1 1 ··· 1 m 1−x2y2 1−x2y3 1−x2yn 1 ∗ 1 1 ··· 1  ∆ = (yj y1) det ∗ 1−x3y2 1−x3y3 ··· 1−x3yn  −  1 xiy1 ! . . . . .  j>1 i=2 − ......  Y Y      1 1 1   1−xny2 1−xny3 1−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 ∈ ≤ ≤ 1 det = x`1 , . . . , x`m y`1 , . . . , y`m 1 xiyj `1>`2>···>`m≥0 − X

Proof We use the expansion 1 ` = (xiyj) 1 xiyj `≥0 − X The determinant is thus m 2 εσ 1 + xiy + (xiy ) + σ(i) σ(i) ··· σ∈S i=1 Xm Y `1 `m and the monomial x1 xm (ì N) occurs with coefficient ··· ∈ m ì `1 `m yσ(i) = y , . . . , y σ∈S i=1 Xm Y

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

1 `1 `m `1 `m det = x1 . . . xm y , . . . , y 1 xiyj `1,...,`m distinct − X

` ` π(1) π(m) `π(1) `π(m) = x1 . . . xm y , . . . , y ` >···>` ≥0 π∈S ! 1 Xm Xm

` ` π(1) π(m) `1 `m = x1 . . . xm επ y , . . . , y ` >···>` ≥0 π∈S ! 1 Xm Xm = x`1 , . . . , x`m y`1 , . . . , y` m ` >···>` ≥0 1 Xm

which is what we needed.

m Notation If λ = (λ1, . . . , λm) Z , 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 ≥ · · · ≥ ··· x`1 , . . . , x`m (2) If λi = n and λi 0 then is a polynomial of degree m in the xis. i ≥ xm−1,..., 1 | | P 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 ∈ ∈ i i i i si = x + + x , ti = y + + t 1 ··· m 1 ··· m These are called the power sums (sometimes referred to in the literature as Newton sums).

x`1 , . . . , x`m y`1 , . . . , y`m 1 α1 αn α1 αn (4.7) Lemma = nαs . . . s t . . . t xm−1,..., 1 ym−1,..., 1 n! 1 n 1 n + ccls α λ∈ΛX(m,n) | | | | X 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 have modulus < 1. Then n n 1 x y (x y )2 (x y )3 log = i j + i j + i j + 1 x y 1 2 3 ··· i,j=1 i j i,j=1 Y − X s t s t s t = 1 1 + 2 2 + 3 3 + 1 2 3 ··· and hence n 1 s t s t s t = exp 1 1 + 2 2 + 3 3 + 1 xiyj 1 2 3 ··· i,j=1 − Y ∞ 1 s t s t s t n = 1 1 + 2 2 + 3 3 + n! 1 2 3 n=0 ··· X 1 n! s t α1 s t α2 = 1 1 2 2 n! · α1!α2! 1 2 ··· X(∗) ···

where ( ) denotes the sum over all n and all sequences α1, α2,... of nonnegative integers with only ∗ ﬁnitely many nonzero terms and α1 + α2 + = n. But then ··· n 1 sα1 sα2 tα1 tα2 = 1 2 ··· 1 2 ··· 1 x y 1α1 2α2 α !α ! i,j=1 i j 1 2 Y − X(∗) ··· ···

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

`1 `m `1 `m m α1 α2 α1 α2 x , . . . , x y , . . . , y 1 s1 s2 t1 t2 m−1 m−1 = = α α ··· ··· x ,..., 1 y ,..., 1 1 xiyj 1 1 2 2 α1!α2! ` >` >···>` ≥0 i,j=1 1 2 X m | | | | Y − X(∗) ··· ··· so we can equate terms which are of degree n in the xi to obtain the desired result.

m (4.8) Deﬁnition For λ = (λ1, . . . , λm) Z and α a conjugacy class in Sn, let ψλ(α) be the λ1 λm α∈1 αn coeﬃcient of the monomial x1 . . . xm in s1 . . . sn , i.e.

α1 αn λ1 λm s1 . . . sn = ψλ(α)x1 . . . xm λ∈ m XZ

Remarks

(1) ψλ is a class function for Sn.

m (2) ψλ(α) = 0 if any λi is negative or if λi = n. 6 i=1 X (3) ψλ is a symmetric function of the λis.

Notation Set

ωλ(α) = επψ(`π(1)+1−m, `π(2)+2−m, ... , `π(m))(α) π∈S Xm

λ The idea now is to show that ωλ = χ whenever λ n. ` (4.9) Lemma (Character formula)

α1 αn m−1 `1 `m s1 . . . sn x ,..., 1 = ωλ(α) x , . . . , x λ∈Λ+(m,n) X

Proof The left-hand side is equal to

λ1 λm m−1 0 ψλ(α)x1 . . . xm ετ xτ(1) . . . xτ(m) λ∈ m τ∈S XZ Xm which, by combining terms, is equal to

λτ(1)+m−1 λτ(m) ετ ψλ(α)xτ(1) . . . xτ(m) λ∈ m τ∈S XZ Xm

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

`1 `m ψ(`1+1−m, ... , `m)(α)xτ(1) . . . xτ(m) λ∈ m τ∈S XZ Xm

`1 `m = ψ(`1+1−m, ... , `m)(α) x , . . . , x λ∈ m XZ Whenever the `i are not distinct, the corresponding term gives zero. So setting λi = `i + i m and − using the fact that `1 > > `m if and only if λ1 λm, our expression is equal to ··· ≥ · · · ≥

`1 `m ψ(`π(1)+1−m, ... , `π(m))επ x , . . . , x λ ≥···≥λ π∈S 1 X m Xm

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

17 (4.10) Lemma (Orthogonality of ω) Let λ, λ0 n be partitions of m parts. Then ` ≤ n! if λ = λ0 nαωλ(α)ωλ0 (α) = λ λ0 ccls α (0 if = X 6

Proof By (4.7) we have

`1 `m `1 `m 1 α1 αn α1 αn m−1 m−1 x , . . . , x y , . . . , y = nαs . . . s t . . . t x ,..., 1 y ,..., 1 n! 1 n 1 n λ∈Λ+(m,n) ccls α X X

By (4.9) this is equal to

0 0 1 `1 `m ` ` nαωλ(α)ωλ0 (α) x , . . . , x y 1 , . . . , y m n!   ccls α λ,λ0∈Λ+(m,n) X X   0 where `i come from λi and `i come from λi in the usual way.

0 ` ` `0 `0 As λ, λ vary, the polynomials x 1 , . . . , x m and y 1 , . . . , y m are linearly independent in the poly-

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

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

σ∈Rt◦ Xλ

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

Sn = giR i=1 [ for the transversal 1 = g1, g2, . . . , gN . (Recall: a transversal is a minimal set of coset representa- { } tives.) Then CSnrλ has basis girλ 1≤i≤N , which we now use to calculate the traces. The CSn-action { } is given by σgirλ = gjrλ for σ Sn, where σgi gjR. Thus ∈ ∈ −1 χ(α) = 1 i N : g σgi R { ≤ ≤ i ∈ }

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

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

1 −1 Q χ(α) = m g Sn : g σg R 1 λj! { ∈ ∈ }

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

1 α1 α2 αn χ(α) = 1 2 . . . n α1!α2! . . . αn! α R m λ ! | ∩ | 1 j since CSn (σ) = n!/nα. Now a permutationQ τ α R restricts to a permutation of the numbers in th| | ◦ ∈ ∩ the i 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! λ2! λm! 1α11 . . . nα1n α ! . . . α ! 1α21 . . . nα2n α ! . . . α ! ··· 1αm1 . . . nαmn α ! . . . α ! 11 1n 21 2n m1 mn and so

α1! αm! χ(α) = α α α α ( ) 1 11 . . . n 1n α11! . . . α1n! ··· 1 m1 . . . n mn αm1! . . . αmn! † X where the sum runs over all αij satisfying ( ). By the multinomial theorem, ∗ α xα11 . . . xα1m α xαn1 . . . xαnm α1 αn 1! 1 m m! 1 m s1 . . . sn = α α α α 1 11 . . . n 1n α11! . . . α1n! ··· 1 m1 . . . n mn αm1! . . . αmn! X where the sum runs over all αij N satisfying ∈

α1j + α2j + + αmj = αj ···

λ1 λm for all 1 j n. So ψλ(α), which is the coeﬃcient of x1 . . . xm , 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 a ` ≤ submodule 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 same −1 ◦ ∈ −1 column of σ t , so if τ is their transposition then στσ Ct◦ . And µ ∈ µ −1 hµσrλ = hµστσ στrλ = hµσrλ hµσrλ = 0 − ⇒ Thus 0 = h S r Hom ( S h , S r ) by (1.12). µC n λ ∼= CSn C n µ C n λ Conversely suppose µ = λ. Then 2 hλrλ εσσ = h = 0 λ 6 σ∈Ct◦ Xλ so 0 = hλCSnrλ = Hom Sn (CSnhλ, CSnrλ). 6 ∼ C Note that, in general, CSnhλ is not a submodule of CSnrλ.

+ λ (4.13) Lemma If λ Λ (n, m) then ωλ = χ . ∈ 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 ∈ ` + 1 m, . . . , ` π(1) − π(m) −1 i.e. with parts λi + π (i) 1. Since ψλ(α) is symmetric in the λi, we can write −

ωλ = επψµπ π∈S Xn If π = 1 then µπ = λ, so we’re done.

−1 −1 −1 If π = 1, then λ1+π (1) 1 λ (with equality if and only if π (1) = 1; and then λ2+π (2) 2 λ2 6 − ≥−1 − ≥ with equality if and only if π (2) = 2; and so on. Thus µπ > λ. /

ν Step 2 ωλ = kλν χ with kλν Z, kλλ > 0 and kλν = 0 if ν < λ. ∈ ν`n X µ By (4.11) and (4.12), ψλ is an N-linear combination χ s with µ λ and with nonzero coeﬃcient of λ ν ≥ λ χ . Thus ωλ is a Z-linear combination of χ s with ν λ and with positive coeﬃcient of χ . / ≥

19 Step 3 The result follows.

We know by (4.10) that n! if λ = µ nαωλ(α)ωµ(α) = λ µ ccls α (0 if = X 6 λ 2 In the case λ = µ, orthogonality of χ implies k = 1, and thus kλν = 0 if λ = µ and kλλ = 1. / λν 6 ν`n X So we’re done.

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

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

α1 αn m−1 λ `1 `n (4.14) Theorem s1 . . . sn x ,..., 1 = χ (α) x , . . . , x λ∈Λ+(m,n) X

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 partitions of n. ≥

+ λ `1 `m (2) If λ Λ (m, n) then χ (α) is the coeﬃcient of the monomial x1 . . . xm in the expansion of α1 ∈ αn s1 . . . sn .

20 5 The hook length formula

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

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

(`i `j) f = deg χλ = n! 1≤i

n m−1 n τ(1)−1 τ(m)−m (x1 + + xm) x ,..., 1 = (x1 + + xm) ετ x . . . x ··· ··· 1 m τ∈S Xn

By the multinomial theorem, this coeﬃcient equals n! (`1 + 1 τ(1))! ... (`m + 1 τ(m))! τ∈S Xn − − 1 1 1 = n! ,..., , (` m + 1)! (` 1)! `!

− − 1 = n! m . . . , `(` 1), `, 1 · j=1 `j! | − |

1 m−1 2 = n! Qm ` , . . . , ` , `, 1 · j=1 `j!

Q 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 corner with [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 0 − 0 0 (the arm) and the (µj i) nodes below it (the leg). [Here, [µ] has column lengths µ1, µ2,... .] − 0 The length of the (i, j)-hook is hij = µi + µ + 1 i j. j − − (c) Replacing the (i, j)-node of [µ] by hij for each node gives the hook graph. For example, the hook graph of (42, 3) is 6 5 4 2 5 4 3 1 3 2 1

(d)A skew hook is a connected part of the rim of [µ] which can be removed to leave a proper diagram. 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]

n! fλ = (i,j)∈[λ] hij Q Proof Suppose λ has m parts. By (5.1) we have to show that

m λi (`i `k) hij = `i! for each i − k=i+1 j=1 Y Y

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

ì `m > ì `m−1 > > ì ì+1 − − ··· −

hi1 > hi2 > > hiλ ··· i 0 0 Since λ has m parts and λ1 is the length of the first column, λ1 = m, and hi1 = λi. So each term is ì. Thus it is enough to show that hij = ì `k for any j, k. ≤ 6 − 0 However if r = λ then λr j and λr+1 < j, so j ≥

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

hij ì + `r+1 = (λi + r 1 j + 1) (λi + m i) + (λr+1 + m r 1) = λr+1 j < 0 − − − − − − − − So ì `r < hij < ì `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 1 4 3 1 2 1

and so 11! fλ = = 990 8.7.5.4.32.22.13

Remarks

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

1 fλ = n! (λ i + j)! i − + where λ Λ (n, m) and the determinant on the right is of an m m matrix and we adopt the ∈ 1 × convention = 0 if m < 0. For a proof see e.g. Sagan or James §19. m!

2 (2) The equation fλ = n! can be proved as a purely combinatorial statement without reference to λ`n representations,X 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 with ρ an r-cycle and π a permutation of the remaining n r numbers then ∈ − χλ(πρ) = ( )iχν (π) ν − X where ν runs over partitions for which [λ] [ν] is a skew r-hook of leg length i. (Note: leg length refers to the leg of the hook with which the− skew hook corresponds.)

22 For example,

conj. class 2 2 χ(5,4 )((5, 4, 3, 1)) = χ(5,3) + χ(3 ,2) on (4, 3, 1) − 2 2 z }| { = χ(2 ) (χ(3,1) + χ(2,1 )) on (3, 1) − from (5,3) from (32,2) = |χ{z(1)}+ 0 + 0| {z } 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 the following: Tensor products: V W is a CG-module with the diagonal action g(v w) = gv gw. ⊗C ⊗ ⊗ Basic properties: – V C = C ⊗ ∼ – V W = W V ⊗ ∼ ⊗ – (V W ) Z = V (W Z) ⊗ ⊗ ∼ ⊗ ⊗ – If θ : V V 0 and ϕ : W W 0 are CG-maps then there is a CG-map defined by → → θ ϕ : V W V 0 W 0 ⊗ v ⊗ w → θ(v) ⊗ ϕ(w) ⊗ 7→ ⊗ Hom spaces: HomC(V,W ) is a CG-module via

−1 αigi (f)(v) = αigi(f(gi v)) i ! i X X ∗ 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 map ∗ V W HomC(V,W ) f ⊗w → (v f(v)w) ⊗ 7→ 7→ is a CG-isomorphism. Tensor powers: The nth tensor power of V is deﬁned by

n ⊗n 0 T V = V = V V ,T V = C ⊗ · · · ⊗ n copies Basic properties: | {z } ⊗n – If V has basis ei : 1 i m then V has basis { ≤ ≤ } ei ei : 1 i1, . . . , in m { 1 ⊗ · · · ⊗ n ≤ ≤ } and hence dim(V ⊗n) = mn. ⊗n – V is naturally a left CSn-module via

σ(v1 vn) = v −1 v −1 ⊗ · · · ⊗ σ (1) ⊗ · · · σ (n) – The actions of CSn and CG commute, i.e. ⊗n σgx = gσx for all g G, σ Sn, x V ∈ ∈ ∈

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

⊗n X = span x εσσx : x V , σ Sn C{ − ∈ ∈ } n The image of v1 vn in Λ V under the quotient map will be denoted by v1 vn. ⊗ · · · ⊗ ∧ · · · ∧ Deﬁne the antisymmetric tensors by

n ⊗n T Vanti = x V : σx = εσx for all σ Sn { ∈ ∈ }

(6.2) Lemma

24 (i) v1 vn = εσv −1 v −1 for all σ Sn. ∧ · · · ∧ σ (1) ∧ · · · ∧ σ (n) ∈ This tells us, by considering transpositions, that v1 vn = 0 whenever vi = vj for some i = j. ∧ · · · ∧ 6 n n (ii) T Vanti and Λ V are CG-modules.

n n m (iii)Λ V has basis ei ei : 1 i1 < < in n , and so dim Λ V = . { 1 ∧ · · · ∧ n ≤ ··· ≤ } 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 = span v1 vn : vi = vj for some i = j C{ ⊗ · · · ⊗ 6 } If char k = 2 then this deﬁnition reduces to (6.2). 6 n ⊗n (6.3) Lemma T Vanti = aV , where a = εσσ is the alternizer [cf. (2.4)], and the natural

σ∈Sn n n X map T Vanti Λ V is an isomorphism of CG-modules. → Proof If x is antisymmetric then

ax = εσσx = εσ(εσx) = x = n! x σ∈S σ∈S σ∈S Xn Xn Xn n ⊗n and so T Vanti aV . ⊆ ⊗n Conversely, since σa = εσa for all σ Sn, any element of aV must be antisymmetric. ∈ We have a map aV ⊗n , V ⊗n ΛnV → → 1 which is a CG-homomorphism and whose kernel is X aV ⊗n aX, since x = ax for x antisym- ∩ ≤ n! 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 ∈ 1 ∈ of some y V ⊗n, but then x is the image of ay aV ⊗n. So the map is an isomorphism. ∈ n! ∈ n ∗ n ∗ (6.4) Lemma Λ (V ) ∼= (Λ V ) . ⊗n n Proof The quotient map V Λ V is surjective, so induces an inclusion (ΛnV )∗ , (T nV )∗ = T n(V ∗) → Now by the universal property of ΛnV (i.e. any multilinear map V V C factors through × · · · × → n copies n n ∗ n ∗ Λ V ), the image of the inclusion is T (V )anti, which is isomorphic to Λ (V ). | {z } Now prepare for some d´ej`avu in (6.5)–(6.8)...

(6.5) Deﬁnition The nth symmetric power of V is SnV = V ⊗n/Y , where ⊗n Y = span x σx : x V , σ Sn C{ − ∈ ∈ } n The image of v1 vn in S V under the quotient map will be denoted by v1 vn. ⊗ · · · ⊗ ∨ · · · ∨ Deﬁne the symmetric tensors by n ⊗n T Vsym = x V : σx = x for all σ Sn { ∈ ∈ }

(6.6) Lemma

25 (i) v1 vn = v −1 v −1 for all σ Sn. ∨ · · · ∨ σ (1) ∨ · · · ∨ σ (n) ∈ n n (ii) T Vsym and S V are CG-modules.

n n m + n 1 (iii)S V has basis ei ei : 1 i1 in n , and so dim S V = − . { 1 ∨ · · · ∨ n ≤ ≤ · · · ≤ ≤ } n

Proof An easy exercise.

n ⊗n (6.7) Lemma T Vsym = sV , where s = σ is the symmetrizer [cf. (2.4)], and the natural

σ∈Sn n n X map T Vsym S V is an isomorphism of CG-modules. → Proof If x is symmetric then sx = σx = x = n! x σ∈S σ∈S Xn Xn n ⊗n and so T Vsym sV . ⊆ ⊗n Conversely, since σs = s for all σ Sn, any element of sV must be symmetric. ∈ We have a map sV ⊗n , V ⊗n SnV → → 1 which is a CG-homomorphism and whose kernel is Y aV ⊗n sY , since x = sx for x symmetric. ∩ ≤ n! 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 ∈ 1 ∈ of some y V ⊗n, but then x is the image of sy sV ⊗n. So the map is an isomorphism. ∈ n! ∈ n ∗ n ∗ (6.8) Lemma S (V ) ∼= (S V ) . ⊗n n Proof The quotient map V S V is surjective, so induces an inclusion (SnV )∗ , (T nV )∗ = T n(V ∗) → n Now by the universal property of S V (i.e. any multilinear map V V C factors through × · · · × → n copies n n ∗ n ∗ S V ), the image of the inclusion is T (V )sym, which is isomorphic to S (V ). | {z }

Polynomial maps between vector spaces

(6.9) Deﬁnition Let V and W be ﬁnite-dimensional C-spaces with bases ei : 1 i m and { ≤ ≤ } fr : 1 r h , respectively. A function ϕ : V W is a polynomial map (resp. n-homogeneous { ≤ ≤ } → map) 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~ abbreviates the m-tuple (X1,...,Xm).]

(6.10) Lemma The deﬁnitions in (6.9) do not depend on the choice of bases. That is, if there exist such functions ϕi with respect to some bases ei , fr , then there exist such functions with respect to any bases e0 , f 0 . { } { } { i} { r} Proof Write m h 0 0 ei = pjiej and fr = qsrfs j=1 s=1 X X

26 Then

m m 0 ϕ Xie = ϕ Xipjiej i   i=1 ! i,j=1 X X h m  m

= ϕr Xi1 p1i1 ,..., Xim pmim fr r=1 i =1 i =1 ! X X1 Xm h m m 0 = ϕr Xi1 p1i1 ,..., Xim pmim qsrfs r,s=1 i =1 i =1 ! X X1 Xm h and the ϕr qsr are polynomials (resp. n-homogeneous polynomials) like the ϕr. r=1 ··· X

Let PC(V,W ) and HC,n(V,W ) denote the spaces of polynomial maps and n-homogeneous maps, respectively, 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→n0-homogeneous map W Z→is an nn0-homogeneous map X Z. → → → Proof An easy exercise. Basic examples V,W W (1)H C,0( ) ∼=

(2)H C,1(V,W ) = HomC(V,W ) n (3)∆ H ,n(V, S V ), where ∆ : v v v. For ∈ C 7→ ∨ · · · ∨

∆ Xiei = Xi ...Xi ei ei = λ~Xi ...Xi ei ei 1 n 1 ∨ · · · ∨ n i 1 n 1 ∨ · · · ∨ n i ! ~ i ≤···≤i X Xi 1 X n ~ for suitable λ~ C. [Here i abbreviates the n-tuple (i1, . . . , in).] i ∈

(6.12) Theorem For C-spaces V and W , the map ψ ψ ∆ induces an isomorphism 7→ ◦ n ∼= Hom (S V,W ) H ,n(V,W ) C −→ C 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’s true for i + 1, then for any α C we have ∈ 0 = ψ((x + αy) (x + αy) vi+2 vn) ∨ · · · ∨ ∨ ∨ · · · ∨ i+1 i+1 | j i + 1 {z } = α ψ(x x y y vi+2 vn) j ∨ · · · ∨ ∨ ∨ · · · ∨ ∨ ∨ · · · ∨ j=0 X i+1−j j Since this holds true for any| α{z C,} each| term{z must} be zero. In particular ∈ i + 1 ψ(x x y vi+2 vn) = 0 1 ∨ · · · ∨ ∨ ∨ ∨ · · · ∨ i If dim V = m and dim W = h, say, then| {z } m + n 1 dim Hom (SnV,W ) = h − = dim H (V,W ) C n C,n

27 So the map is an isomorphism. (6.13) Corollary SnV = span v v : v V C{ ∨ · · · ∨ ∈ } Proof Take W = C in (6.12). If the elements v v do not span then there exists a nonzero n ∨ · · · ∨ map S V C whose composition with ∆ is zero, contradicting (6.12). →

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

ϕ(X1e1 + + Xmem) = ϕ1(X1,...,Xm)f1 + + ϕk(X1,...,Xm)fh ··· ··· n with ϕi homogeneous. The total polarisation P ϕ Hom (S V,W ) of ϕ is ∈ C h n ∂ ϕj P ϕ(ei ei ) = fj 1 ∨ · · · ∨ n ∂X . . . ∂X j=1 i1 in X ∈C | {z } (The partial derivative is formal, is symmetric in the Xik , and lies in C since ϕj has degree N.) Now

(P ϕ)(∆(X1e1 + + Xmem)) = Xi ...Xi (P ϕ)(ei ei ) ··· 1 n 1 ∨ · · · ∨ n i ,...,i 1X n ∂nϕ = X ...X j f i1 in ∂X . . . ∂X j j i ,...,i i1 in X 1X n = n! ϕj(X1,...,Xm)fj j X = n!ϕ(X1e1 + + Xmem) ··· So P ϕ ∆ = n! ϕ. The penultimate equality follows from Euler’s theorem, which states that if F ◦ ∂F is r-homogeneous then Xi = rF . i ∂Xi

Example Let ϕ : V PC be a quadratic form: → m

ϕ Xiei = aijXiXj, aij = aji i=1 ! i,j X X

Then (P ϕ) ( Xiei) YjeJ = 2 aijXiYj is twice the corresponding symmetric bilinear i ∨ j i,j form. h P P i P

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.

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

Define a map ideals I of R subsets of An { I } ←→ { V (I) } 7−→ 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, say ⊆ I = (f1, . . . , fm), so n V (I) = p A : fj(p) = 0 for each 1 j m { ∈ ≤ ≤ } is 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. The Zariski topology on An induces a topology on any algebraic set X An: the closed subsets of X are the algebraic subsets of X. ⊆

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

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

ideals I of R subsets of An { (X) } ←→ { X } I ←− where (X) = f R : f(p) = 0 for all p X . I { ∈ ∈ } 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 (X) is a prime ideal ⇔ I

If I E R then the radical √I of I is given by √I = f R : f m 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) − − n for some p = (a1, . . . , an) A . ∈ (b) If (1) = I R then V (I) = ∅. 6 E 6 (c) (V (I)) = √I for all ideals I. I

Proof Omitted. Let W a ﬁnite-dimensional K-space. A map f : W K is regular (or polynomial) if it is deﬁned by a polynomial in the coordinates with respect to→ a given basis of W .(Check: it doesn’t matter which 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 corresponding dual basis of W ∗ then K[W ] = K[x1, . . . , xm]

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

K[W ] = K[W ]d d≥0 M is a graded K-algebra, i.e. if f, g are homogeneous of degree df , dg, respectively, then fg is homogeneous of degree df + dg.

d1 dm ∗ The monomials x1 xm with d1 + +dm = d form a basis of K[W ]d. In particular, K[W ]1 = W . ··· ··· d ∗ This extends to a canonical identiﬁcation of K[W ]d with S (W ), so that

d ∗ ∗ K = K[W ]d = S (W ) = S(W ) d≥0 d≥0 M M 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 ] vanishing on X is the⊆ zero function. So every polynomial function f K[W ] is completely∈ determined by its ∈ restriction f X to a Zariski-dense subset X W . | ⊆ Now let X be an arbitrary subset of W . As above, the ideal of X

(X) = f K[W ]: f(a) = 0 for all a X = ma I { ∈ ∈ } a∈X \ where ma = ( a ) is the maximal ideal of functions vanishing at a, i.e. I { } ev : K[W ] m = ker a C a f → f(a) 7→

We say X Y is Zariski-dense in Y if (X) = (Y ). ⊆ I I 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. ∈ ⊗n 1 V is a right CSn-module via the action

(v1 vn)σ = v v ⊗ · · · ⊗ σ(1) ⊗ · · · ⊗ σ(n) So there is a map

S End V ⊗n C n C —(1) σ → (v vσ) 7→ 7→ Regarding V as a representation of GL(V ) in the natural way, V ⊗n is a (left) CGL(V )-module via the action g(v1 vn) = gv1 gvn ⊗ · · · ⊗ ⊗ · · · ⊗ So there is a map

GL(V ) End V ⊗n C C —(2) g → (v gv) 7→ 7→ The actions (1) and (2) commute, so by abusing notation slightly, (1) induces a map

⊗n ϕ : CSn End GL(V )(V ) → C and (2) induces a map ⊗n ψ : CGL(V ) End Sn (V ) → C ⊗n Schur–Weyl duality asserts that the images of CSn and CGL(V ) in EndC(V ) are each other’s centralizers.

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

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

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

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

⊗n SC(m, n) = EndCSn (V ) It is called the Schur algebra. [For a deﬁnition over arbitrary ﬁelds, see J.A. Green’s book Polynomial representations of GLn, SLN 830.]

⊗n CSn is semisimple and V is a ﬁnite-dimensional CSn-module, so by (1.9),SC(m, n) is a semisimple C-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 End (V ⊗n) given by → C

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

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

1 Note 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)∗ ∼ ⊗ V ⊗n ∼= EndC( )

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

n (7.4) Lemma SC(m, n) = α(T Wsym)

⊗n n Proof SC(m, n) is the set of all x EndC(V ) fixed under the action of Sn. By (6.5), T Wsym is ⊗n ∈ ⊗n the set of all y W fixed under the action of Sn (which is induced by α, i.e. y σ W corresponds with ϕ(σ)α(y)ϕ∈(σ−1) End (V ⊗n). · ∈ ∈ C 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. ∪ n Proof The ring of regular functions on A is R = C[X1,...,Xn]. If X and Y are zero-sets of ideals I,J E R then, by assumption, any maximal ideal contains either I or J. If I,J are both nonzero then pick 0 = f I and 0 = g J. Any maximal ideal contains fg, so 6 ∈ 6 ∈ fg(a1, . . . , an) = 0 for all a1, . . . , an C ∈ By Nullstellensatz, fg 0 = 0 , contradicting the fact that R is an integral domain. ∈ { } { } p d d d (7.6) Lemma Let Y C be a subspace. Then Y is Zariski-closed, where C is identiﬁed with A in the canonical way. ≤

d d Proof Chose a basis ϕ1, . . . , ϕh of Hom (C /Y, C) and view them as maps C C. Then Y is the C → zero set of the ϕi. n n (7.7) Lemma T WSym = span ϕ ϕ : ϕ GL(V ) C { ⊗ · · · ⊗ ∈ } n z }| { Proof Deﬁne a subspace X of T Wsym by n X = span ϕ ϕ : ϕ GL(V ) C { ⊗ · · · ⊗ ∈ } z }| { n There is a map α : W W ⊗n given by α(ϕ) = ϕ ϕ, which is a regular map of the aﬃne 2 →2n ⊗ · · · ⊗ m W m W ⊗n spaces A ∼= and A ∼= . z }| { 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 determinant ∪ { } is a union of Zariski-closed subsets.

2 Since Am is irreducible (by (7.5)), α−1(X) = W . Thus X contains all maps of the form ϕ ϕ n n ⊗ · · · ⊗ for ϕ W . But they span T Wsym since the images ϕ ϕ span S W by (6.13). ∈ ∨ · · · ∨ In other words. . . n (7.8) Theorem S (m, n) = span ϕ ϕ : ϕ GL(V ) C C { ⊗ · · · ⊗ ∈ } z }| { Proof of (7.1) The assertion that the image of CGL(V ) is the centralizer of CSn is a reformulation n of the assertion that S (m, n) = span ϕ ϕ : ϕ GL(V ) , which is (7.8). C C { ⊗ · · · ⊗ ∈ } z }| { 32 ⊗n Conversely, S (m, n) is a semisimple -algebra. By (1.11), Sn maps onto EndS (m,n)(V ), and C C C C since the image of GL(V ) spans SC(m, n), we must have

⊗n ⊗n EndS (m,n)(V ) = End GL(V )(V ) C C

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

⊗n ⊗n V = h(tλ)V —(1) where the direct sum runs over standard tableauxL tλ with λ n. ` n ⊗n n n Examples If λ = (1 ) then hλV ∼= T Vanti ∼= Λ V . ⊗n n n If λ = (n) then hλV ∼= T VSsym ∼= S V . In particular,when n = 2, this tells us that

2 2 2 n V V = T Vanti T Vsym = Λ V S V ⊗ ∼ ⊕ ∼ ⊕ ⊗n ⊗n Now the actions of Sn and GL(V ) on V commute. So if λ n and tλ is a λ-tableau, then h(tλ)V ` is a CGL(V )-submodule of V ⊗n. Observe that ⊗n ⊗n h(tλ)V ∼= hλV as CGL(V )-modules −1 since σh(tλ)σ = hλ for some σ Sn. So premultiplication by σ induces an isomorphism ∈ ⊗n ∼= ⊗n h(tλ)V hλV −→

⊗n (8.1) Lemma The nonzero hλV 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 the ⊗n natural map CGL(V ) S (m, n) then M is isomorphic to a direct sum of copies of the hλV . → C ⊗n Proof Recall that SC(m, n) = EndCSn (V ) and

(1.12) h V ⊗n Hom ( S h ,V ⊗n) λ ∼= CSn C n λ ⊗n By (1.9) and (1.10)(Artin–Wedderburn), the nonzero spaces hλV 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 from the fact (at the end of §7) that the map CGL(V ) S (m, n) is surjective. → C

(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)-submodule of M.

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

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

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

0 if λm+1 = 0 ⊗n 6 dim hλV = 1 if λm+1 = 0 andλ1 = = λm C  ···  2 otherwise ≥  First we have a slightly technical result. Take a partition λ n such that [λ] is partitioned into two nonempty parts, say of sizes i and j = n i, by a vertical wall.` For example, if λ = (5, 3, 2, 1) 11 we 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 the ` { } ` corresponding partition, so that tµ = tλ . Let ν j correspond to the right-hand part, with ↓[µ] ` tableau tν . This is a map

0 0 0 tν :[ν] 1 , . . . , j where k = k + i for 1 k j → { } ≤ ≤

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

⊗i ⊗j ⊗n ((h(tµ)V ) (h(tν )V ) h(tλ)V —( ) ⊗ → ∗ 0 0 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 which commute. Also Ct = Ct Ct and H = Rt Rt Rt λ µ × ν µ × ν ≤ λ

and H permutes each side of the wall. Take a transversal Rtλ = i riH. Then

(2.4) S h(tλ) = εcrc r∈R c∈C Xtλ Xtλ 0 00 0 00 = εc0 εc00 rir r c c i r0∈R r00∈R c0∈C c00∈C X Xtµ Xtν Xtµ Xtν = rih(tµ)h(tν ) i X

So 2 2 (2.10) h(tλ)h(tµ)h(tλ) = rih(tµ) h(tν ) = kh(tλ) i X i!j! where k = . We thus have a CGL(V )-map fµfν

⊗i ⊗j ⊗n V V h(tλ)V ⊗ →

given by premultiplication by h(tλ). Restriction of this map to the left-hand side of ( ) is surjective, since ∗ 1 h(tλ)(x y) = h(tλ) (h(tµ)x h(tν )y) ⊗ · k ⊗ (8.4) Lemma

⊗n (1) If λm+1 = 0 then hλV = 0 6 ⊗n (2) If n > 0 and λm = 0 then dim hλV 2 C ≥

Proof

◦ (1) Let ij be the row in which j occurs in t and let x = ei ei . For σ Sn λ 1 ⊗ · · · ⊗ n ∈ −1 σx = x ij = iσ−1(j) j j, σ (j) are in the same row σ Rt◦ ⇔ ∀ ⇔ ⇔ ∈ λ ⊗n So the coeﬃcient of x in the decomposition of hλx with respect to the standard basis of V is Rt◦ = 0. So hλx = 0. λ 6 6 (2) Let y = e1+i e1+i and adapt the argument above, to see that hλx and hλy are linearly 1 ⊗ · · · ⊗ n independent.

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

⊗n m ⊗(n−m) hλV = Λ V h V ∼ ⊗ (λ1−1,...,λm−1)

35 Proof Put a wall in [λ] between the first column and the rest. Let tλ be a tableau whose first column consists of 1, . . . , m . By (8.3) there exists a surjection { } m ⊗(n−m) ⊗n Λ V h(tν )V h(tλ)V ⊗ where ν = (λ1 1, . . . , λm 1). Hence, with the usual identifications, there exists a map − − m ⊗(n−m) ⊗n Λ V hν V hλV ⊗ ⊗(n−m) ⊗n Both hν V and hλV are nonzero, hence they are irreducible CGL(V )-modules by (8.1). Since m Λ V is one-dimensional, both sides are irreducible and the map is an isomorphism.

Proof of (8.2) If λm+1 = 0 then [λ] has i > m rows and as in (8.5) there exists a surjection 6 i ⊗(n−i) ⊗n Λ V hν V hλV ⊗ i ⊗n but Λ V = 0, so hλV = 0.

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

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

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) functions 2 fij(Xrs) for 1 i, j h in m variables Xrs for 1 r, s m such that the map ≤ ≤ ≤ ≤

basis ei representation basis wj GLm(C) GL(V ) End (M) Mh(C) −−−−−→ −−−−−−−−−→ C −−−−−→

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

Exercise Check that the deﬁnitions in (9.1) are independent of the choice of bases ei and wj of V and 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) modules are 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 n0-homogeneous) then M N is rational (resp. polynomial, (n + n0)-homogeneous). ⊗ (d) If n = n0 and M is both n- and n0-homogeneous then M = 0. 6 (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 λi = n, set ≥ · · · ≥ ≥ i P ⊗n Dλ1,...,λm (V ) = h(λ1,...,λm)(V )

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-homogeneous CGL(V )-modules.

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

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

ρ0 : GL(V ) End (M) → C Consider the following diagram

37 , n ∼= n ∼= GL(V ) → EndC(V ) S (EndC V ) T (EndC V )sym S(m, n)

ρ0 ρ1 ρ2 ρ3 ρ4

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

In the top row all the maps are natural, and the composite γ : GL(V ) S(m, n) is the natural map → we 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-homogeneous map ρ1. For ρ2, use the universal property of symmetric powers. Then ρ3, ρ4 are transported by the isomorphisms.

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

ρ4(1) = ρ4(γ(1)) = ρ0(1) = 1 and 0 0 0 0 ρ4(γ(gg )) = ρ0(gg ) = ρ0(g)ρ0(g ) = ρ4(γ(g))ρ4(γ(g )) for g, g0 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 submodules on 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 polynomial → ∼ in g. Choose N N such that each ρ(g)ij has degree < N. Then M is a CG-module by restriction, where ∈ 2π ij × G = e N : 0 j N C { ≤ ≤ } ⊆ is a cyclic group of order N. Now M = M1 Mh ⊕ · · · ⊕ nj as a CG-module with each Mj one-dimensional, and g G acts as multiplication by g on Mj for ∈ 0 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) with respect to this basis then gni if i = j ρ(g)ij = 0 if i = j ( 6 2π ij × for g e N (0 j < N), and hence for all g C since the ρ(g)ij are polynomials of degree < N ∈ ≤ ∈ in 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. We → have an inclusion C× , GL(V ), so that one can regard U as a polynomial representation of C×, so by (9.5) U will split as→ U = U0 UN ⊕ · · · ⊕ n with α1 GL(V ) acting as multiplication by α 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 ∈ ≤ ≤ ∈ N n n u0 + αu1 + + α uN = α u0 + + α uN ··· ··· and hence ui = 0 for i = n. 6 n Thus gu Un and the Un are CGL(V )-submodules of U. Since α1 acts as multiplication by α on ∈ Un, 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. The modules Dλ ,...,λ (V ) with λ1 λm 0 are a complete set of nonisomorphic irreducible 1 m ≥ · · · ≥ ≥ polynomial CGL(V )-modules.

(9.8) Deﬁnition If n Z then the one-dimensional CGL(V )-module corresponding to the representation ∈ GL(V ) C⊗ g → (det g)n 7→ 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 ≥ · · · ≥ λm Dλ ,...,λ = Dλ −λ ,...,λ −λ ,0(V ) det 1 m 1 m m−1 m ⊗

Remark If λ1 λm > 0 then by iterating (8.5) (as in the proof of (8.2)) we know that ≥ · · · ≥ λm Dλ ,...,λ (V ) = Dλ −λ ,...,λ −λ ,0(V ) det 1 m ∼ 1 m m−1 m ⊗

(9.10) Theorem Every rational CGL(V )-module is a direct sum of irreducible submodules. The modules Dλ ,...,λ (V ) for λ1 λm form a complete set of nonisomorphic irreducible (rational) 1 m ≥ · · · ≥ CGL(V )-modules. p Proof The rational functions f : GL(V ) C are of the form f = with p a polynomial function. → deti N Thus if M is a rational CGL(V )-module then M1 = M det is a polynomial CGL(V )-module for ⊗ some N. Since M1 decomposes as a direct sum of irreducibles, so does M. If M is irreducible then so N is M1, since det is one-dimensional, and thus M1 is of the form

M1 = Dµ ,...,µ (V ) for some µ1 µm ∼ 1 m ≥ · · · ≥

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

n (9.11) Theorem The one-dimensional rational CGL(V )-modules are precisely the det 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 λi = n (9.4) ≥ · · · ≥ i polynomial λ1 λm 0 (9.7) ≥ · · · ≥ ≥ P 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 diagonal matrices.≤

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

λ1 . χρ :(x1, . . . , xn) tr ρ . 7→  .  λ  n   sometimes denoted by χW .

Basic properties

0 0 Let ρ, ρ be rational representations of GLn(C) associated with W, W . −1 −1 (a) χρ Z[x1, x1 , . . . , xn, xn ], 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 of the diagonal entries.

0 (c) ρ = ρ χρ = χρ0 ; will be (10.6). ∼ ⇒ ⇐ (d) χW ⊕W 0 = χW + χW 0 and χW ⊗W 0 = χW χW 0 .

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

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

(g) When V = Cn we have m χ ⊗n = (x1 + + xn) , V ··· χS2V = xixj, χΛ2V = xixj, i≤j i

n−1 j χΛ V = x1 ... xi . . . xn, χS V = hj(x1, . . . , xn) = xi1 . . . xij i=1 i ≤···≤i X 1 X j The hj are called complete symmetricb functions. We could (but won’t) study the theory of symmetric functions and apply it here; see R. Stanley’s book for more.

40 10 Weyl character formula

m Let V be an m-dimensional C-space. If λ = (λ1, . . . , λm) Z and λ1 λm then the character ⊗n ∈ ≥ · · · ≥ of the CGL(V )-module Dλ1,...,λm (V ) (= hλV if λm 0 and n = λi) will be denoted by φλ. ≥ i

(10.1) Lemma If ξ End (V ) then φλ(ξ) is a symmetric rationalP function of the eigenvalues of ξ. ∈ C Proof The function P (x1, . . . , xm) = φλ(diag(x1, . . . , xm))

is a rational function of the x1, . . . , xm, and it is symmetric since diag(x1, . . . , xm) is conjugate to diag(x , . . . , x ) for any τ Sm. τ(1) τ(m) ∈ Now choose a basis for V such that the matrix A1 of ξ is in Jordan normal form, and for t C let At ∈ be the matrix obtained from A1 by replacing the upperdiagonal 1s by ts. Let ξt be the endomorphism corresponding to At. For t = 0, At is conjugate to A1, and so φλ(ξt) = φλ(ξ). Since φλ is a rational function it is continuous (wherever6 it is deﬁned), and so

φλ(ξ) = lim φλ(ξt) = φλ(lim ξt) = φλ(ξ0) = P (x1, . . . , xm) t→0 t→0 which is a symmetric function of the eigenvalues of ξ.

αn α1 (10.2) Lemma Let α be an Sn-conjugacy class with cycle type n ,..., 1 , and let ξ be an i i endomorphism of V with eigenvalues x1, . . . , xm. If si = x + + x then 1 ··· m

α1 αn λ s1 . . . sn = χ (α)φλ(ξ) + λ∈ΛX(m,n) (Compare this with (4.14).)

Proof Pick g α. Suppose ξ has matrix diag(x1, . . . , xm) with respect to the standard basis ∈ ⊗n e1, . . . , em of V . We compute the trace of the endomorphism of V sending x (gξ)x = ξgx in two ways. 7→

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

⊗n ⊗n V = CSnhλ Hom Sn (CSnhλ,V ) ∼ ⊗ C λ`n M which, by (1.12), becomes ⊗n ⊗n V = (CSnhλ hλV ) ∼ ⊗ λ`n M ⊗n ⊗n This is an isomorphism both of CSn- and EndCSn (V )-modules. Since the action of GL(V ) on V commutes with that of Sn, the corresponding action of gξ on the right-hand side is given by the action ⊗n of g on CSnhλ and of ξ on hλV . So the trace of the action is

λ λ∈Λ+(n,m) ξ (α)φλ(ξ) —(1)

Way 2 (direct computation) P

gξ(ei ei ) = xi . . . xi ei ei 1 ⊗ · · · ⊗ n 1 n g−1(1) ⊗ · · · ⊗ g−1(n) and so

tr(gξ) = xi1 . . . xin D X where D = (i1, . . . , in) : 1 i1, . . . , in m and i −1 = ij j . { ≤ ≤ g (j) ∀ }

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

α1 αn tr(gξ) = s1 . . . sn —(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 = i i ∈ ∈ x1 + + xm, then ··· λ α α χ (α) s1 1 sn n φλ(ξ) = α1! . . . αn! 1 ··· n ccls α X

µ Proof Take the formula of (10.2), multiply it by nαχ (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 ∈ x`1 , . . . , x`m φλ(ξ) = m−1 x ,..., 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 the symmetric group≥ (4.14) we know that∈

x`1 , . . . , x`m α1 αn λ λ s . . . s = χ (α) = χ (α)φλ(α) 1 n xm−1,..., 1 + + λ∈ΛX(n,m) | | λ∈ΛX(n,m) Then the orthogonality of χλs allows us to equate coeﬃcients.

(b) For general λ, since (9.9) λm Dλ = Dλ −λ ,...,λ −λ (V ) det ∼ 1 m m m ⊗ it follows that

x`1−λm , . . . , x`m−λm x`1 , . . . , x`m λm φλ(ξ) = m−1 (x1 . . . xm) = m−1 x ,..., 1 x ,..., 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 in GLm(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)

1≤i

42 (10.6) Theorem If two rational CGL(V )-modules have the same characters then they are isomorphic.

Proof As λ varies, the rational functions in Weyl’s character formula are linearly independent elements of C(x1, . . . , xm). (10.7) Lemma D (V )∗ D (V ) λ1,...,λm ∼= −λm,...,−λ1

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

x−`1 , . . . , x−`m x−`m , . . . , x−`1 ψ ξ φ ξ−1 ( ) = λ( ) = −(m−1) = −(m−1) x ,..., 1 1, . . . , x m−1−` m−1 −` x m , . . . , x 1 = m−1 = φµ(ξ) x ,..., 1 | | 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 ∈ min{p,q} Dp,0(V ) Dq,0(V ) ∼= Dp+q−r,r(V ) ⊗ r=0 M Proof Exercise; write down the characters and use (10.6).

(10.9) Examples of rational CGL(V )-modules C ∼= D0,0,...,0,0(V ) i det ∼= Di,i,...,i,i(V ) V ∼= D1,0,...,0,0(V ) n S V ∼= Dn,0,...,0,0(V ) n Λ V = D1,...,1,0,...,0(V ) where n ‘1’s and m n ‘0’s appear, and n m ∼ − ≤ ∗ V ∼= D0,0,...,0,−1(V ) n ∗ (S V ) ∼= D0,0,...,0,−n(V ) n ∗ (Λ V ) = 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 = 0, with 6 (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 . ∈ ∈ ∈ dim W C[W ] is the ring of regular functions of the aﬃne variety A ∼= W . (6.12) W W, C n W ∗ W ∗ C[ ] ∼= HC,n( ) ∼= S ( ) = S( ) n≥0 n≥0 M M This latter isomorphism is known as polarisation. ∗ ∗ ∗ If W has basis w1, . . . , wh then the map

C[X1,...,Xh] C[W ] → ∗ Xi w 7→ 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 . ∈ ∈ ∈ C[W ] C[W ] For g G, the map → is a C-algebra automorphism, since ∈ f gf 7→ g(ff 0)(w) = (ff 0)(g−1w) = f(g−1w)f 0(g−1w) = (gf)(w)(gf 0)(w) = ((gf)(gf 0))(w)

and −1 g1C[W ](w) = 1C[W ](g w) = 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 ≡ V SnW (3) The map ∆ : is an n-homogeneous concomitant. w → w w 7→ ∨ · · · ∨ (4) C[W ]G = f C[W ]: gf = f for all g G is a C-subalgebra of C[W ], called the polynomial { ∈ ∈ } invariant ring. It is a graded C-algebra:

G G C[W ] = C[W ]d d≥0 M 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 char K = 0 then for any representation W of the finite group G, the invariant ring K[W ]G is generated by the invariants of degree G . ≤ | | [Hilbert] If W is a rational CGL(V )-module then C[W ]GL(V ) and C[W ]SL(V ) are finitely-generated C-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 injective dimension).

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

Examples

n (11.3) Symmetric polynomials W = C with basis e1, . . . , en, is naturally a CSn-module. If ∗ ξ1, . . . , ξn is the corresponding dual basis for W , then the isomorphism

∼= C[X1,...,Xn] C[W ] −→ Xi ξi 7→ allows an identiﬁcation C[W ] C[X1,...,Xn]. The polynomial invariant ring of the CSn-module W is ↔ Sn C[X1,...,Xn] = p C[X1,...,Xn]: p symmetric in the Xis { ∈ }

Sn We have elementary symmetric functions ej(X1,...,Xn) C[X1,...,Xn] such tha ∈ n n−1 (t + X1) (t + Xn) = t + e1(X~ )t + + en(X~ ) ··· ··· where X~ = (X1,...,Xn). That is,

e1(X~ ) = X1 + + Xn, ei(X~ ) = Xj Xj ...Xj , en(X~ ) = X1 ...Xn ··· 1 2 i j

Sn C[Y1,...,Yn] C[X1,...,Xn] → Yi ei(X~ ) 7→

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

Recall for p(t) C[t] monic of degree n, say ∈ n n n−1 p(t) = t + a1t + + an = (t + λi) ··· i=1 Y its discriminant is 2 disc(p) = (λi λj) − i

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 of An ≥ invariants is C[X1,...,Xn] . Consider the Vandermonde determinant

n−1 X ,..., 1 = (Xi Xj) − i

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

Claim The C-algebra map

An θ : C[Y1,...,Yn,Z] C[X1,...,Xn] → Yi ei 7→ Z Xn−1,..., 1 7→ 2 is surjective, and ker(θ) = (Z D(Y1,...,Yn)). − Proof We prove the claim in three steps.

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

2 τfsym = τf + τ f = τf + f = fsym

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

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

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

falt (Xi Xj) = (Xi Xj) ∈ − − q and so (Xi Xj) falt for all i = j. In particular, − | 6 n−1 falt = X ,..., 1 g

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

1 1 n−1 f = (fsym + falt) = (fsym + X ,..., 1 g) 2 2

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

Step 2 (Z2 D(Y~ )) ker(θ). This is clear since − ⊆ 2 n−1 2 θ(Z D(Y~ )) = X ,..., 1 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~ )(Z2 D(Y~ )) + A(Y~ ) + B(Y~ )Z − quotient remainder

We show that if P ker(θ) has the form| {zA(}Y~ ) + B(Y~ )Z then| P ={z 0. } ∈ Suppose a1, . . . , an C, and let ∈ n n−1 f(t) = t + a1t + + an = (t + λ1) (t + λn) ··· ···

46 Then n−1 P (a1, . . . , an, λ ,..., 1 ) = θ(P )(λ1, . . . , λn) = 0

since P ker(θ). Exchanging λ1 and λ2 changes the sign of the Vandermonde determinant, so ∈ n n−1 1 P (a1, . . . , an, δ) = 0, where δ = disc(t + a1t + + an) 2 . 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

n n n−1 (a1, . . . , an) C : disc(t + a1t + + an) = 0 { ∈ ··· 6 } n Then A = B = 0 everywhere on C , 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). · m−n Suppose χθ(t) = det(t1V + θ) is the characteristic polynomial of θ and cn(θ) is the coeﬃcient of t in χθ(t). Then −1 −1 χg·θ(t) = det(t1V + gθg ) = det(g(t1V + θ)g ) = 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) = (t + λi) i=1 Y so that cn(θ) = en(λ1, . . . , λm); in particular

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

GL(V ) C[Y1,...,Ym] C[U] Claim The C-algebra map → is an isomorphism. Yi ci 7→ Sketch proof If f is a polynomial invariant for U then f(θ) is a symmetric polynomial function of the eigenvalues of θ. Mimic the proof that the characters of rational CGL(V )-modules are symmetric rational 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 = H ,2(V, C) = quadratic forms on V C { } By polarisation (see end of §6) identify U with the symmetric m m matrices via f A, where × 7→ f(x) = xtAx for x Cm a column vector. ∈ Deﬁne disc(f) = det(A).

GLm(C) acts on U by

−1 t −1t −1 m (gf)(x) = f(g x) = x (g Ag )x, f U, g GLm(C), x C ∈ ∈ ∈ so t disc(gf) = det(g−1 Ag−1) = (det g)−2 disc(f)

47 and therefore disc : U C is an SLm(C)-invariant (but not a GLm(C)-invariant). → [X] [U]SLm(C) Claim The -algebra map C C is an isomorphism. C X → disc 7→ Proof It is injective: if there is a polynomial P such that P (disc f) = 0 for all f U then P (λ) = 0 ∈ for all λ C, since the quadratic form ∈ 2 2 2 fλ(x1, . . . , xm) = x + + x + λx 1 ··· m−1 m has discriminant λ, so P = 0.

Let θ : U C be a polynomial SLn(C)-invariant and deﬁne → C C F : → λ θ(fλ) 7→ This is a polynomial function since θ is a polynomial map. We prove that θ(f) = F (disc f). It is enough to show this for f with disc f = 0. For, if this is case is known, then 6 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 1 ..  .  1 diag(1,..., 1, 0,..., 0) =    0     .   ..     0     Thus if f U corresponds to the matrix A and λ = disc f = det A = 0 then there exists B GLm(C) ∈ t −1 6 ∈ such that B AB = I. Now if C = diag(1,..., 1, (det B) ), then BC SLm(C), and ∈ (BC)tA(BC) = CtBtABC = diag(1,..., 1, (det B)−2) = diag(1,..., 1, λ)

and so θ(f) = θ(fλ) = F (λ), as required. (11.7) Discriminant of binary forms

Deﬁne 2 homogeneous polynomials of degree n U = HC,n(C , C) = in two variables X1,X2 Take f U. Then f has the form ∈ n n−1 n f(X1,X2) = a0X + a1X X2 + + anX 1 1 ··· 2 = b(λ1X1 + µ1X2)(λ2X1 + µ2X2) (λnX1 + µnX2) ··· and deﬁne 2n−2 2 disc(f) = b (λiµj λjµi) − i

2 2 3 3 2 2 Example (n = 3) disc(f) = 27a a + 18a0a1a2a3 4a0a 4a a3 + a a − 0 3 − 2 − 1 1 2 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 form U V C. ⊕ → Examples

48 U V Every invariant θ for U gives rise to a covariant C (u,⊕ v) → θ(u) 7→ U V If U = H (V, ) then there is an ‘evaluation covariant’ ev : C . C,n C (f,⊕ v) → f(v) 7→

(11.9) The Hessian

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

∂2f H(f) = det ∂X ∂X i j

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

m Let U = HC,n(C , C), the n-homogeneous polynomials in X1,...,Xm. The Hessian deﬁnes a polynomial map U Cm C (f,⊕ v) → H(f)(v) 7→ Now U is naturally a CGLm(C)-module via the action

−1 m (gf)(v) = f(g v), f U, g GLm(C), v C ∈ ∈ ∈ 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 invariants when the module is the direct sum of copies of V and V ∗. The hope was then in principle to compute invariants of an arbtirary rational module, but in practice this proved to be impossible. We will state and 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, ∈ n ∗ r (i) Hom GL(V )(T (V ) T (V ), C) = 0 if n = r; C ⊗ 6 n ∗ n (ii) Hom GL(V )(T (V ) T (V ), C) = span µσ : σ Sn , where C ⊗ C{ ∈ }

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

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 ∗ ∗ ∗ ∗∗ ∗ ∗ Hom (X Y, C) = (X Y ) = X Y = X Y = Hom (Y,X) C ⊗ ⊗ ∼ ⊗ ∼ ⊗ ∼ C Taking G-ﬁxed points, we obtain

∗ Hom G(X Y, C) = Hom G(Y,X) C ⊗ ∼ C n ∗ n ∗ With this and the isomorphism T (V ) ∼= (T V ) we have an isomorphism ∼ r n = n ∗ r HomCGL(V )(T V,T V ) HomCGL(V )(T (V ) T (V ), C) π : −→ ⊗ f ϕ1 ϕn ~v (ϕ1 ϕn)(f(v)) 7−→ ⊗ · · · ⊗ ⊗ 7→ ⊗ · · · ⊗ Now (i) If n = r then Hom (T rV,T nV ) = 0 because Hom (S0,S) = 0 if S0 (resp. S) is an 6 CGL(V ) CGL(V ) irreducible r-homogeneous (resp. n-homogeneous) CGL(V )-module. n (ii) is just a restatement of Schur–Weyl duality. To see this, recall that EndCGL(V )(T V ) is spanned by λσ for σ Sn, where ∈

λσ(v1 vn) = v −1 v −1 ⊗ · · · ⊗ σ (1) ⊗ · · · ⊗ σ (n) and now π(λσ) = µσ.

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

Note that HomCG,n(U, W ) is a CG-module with conjugation action −1 (gf)(u) = gf(g u), g G, u U, f Hom G,n(U, W ) ∈ ∈ ∈ C G Note also that HomCG,n(U, W ) = HC,n(U, W ) .

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

U G U, C[ ] ∼= HomCG( C) n≥0 M Proof We know that U U, C[ ] ∼= HC,n( C) n≥0 M

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

Hom (SnU, W ) Hom (U, W ) Hom (U, W ) Hom (SnU, W ) CG CG,n and CG,n CG ψ → ψ ∆ f → 1 P f 7→ ◦ 7→ n! U SnU where ∆ : and P f is the total polarisation of f (see after (6.13)). u 7→ u u 7→ ∨ · · · ∨ n Proof Clearly the maps induce isomorphisms between HomC(S U, W ) and HC,n(U, W ), so we’ll prove: (i) ψ concomitant ψ ∆ concomitant; ⇒ ◦ (ii) f concomitant 1 P f concomitant. ⇒ n!

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 P f (not recommended). 1 Approach 2 If f Hom G,n(U, W ) then f = P f ∆, so ∈ C n! ◦ 1 1 P f (g(u u)) = P f (g∆(u)) n! ∨ · · · ∨ n! 1 = P f ∆(g(u)) n! = f(g(u)) = gf(u) 1 = g P f (u u) n! ∨ · · · ∨ n Approach 3 HomC(S U, W ) and HC,n(U, W ) are CG-modules, with G acting by conjugation, 1 and ψ ψ ∆ is a CG-map and an isomorphism. So its inverse f P f is also a CG-map. 7→ ◦ 7→ n! These maps restrict to isomorphisms between sets of G-ﬁxed points.

δ natural Remark ∆ factors as U T nU SnU, where −→ −−−−→ δ(u) = u u ⊗ · · · ⊗ n so composition with δ induces a surjection | {z }

n Hom G(T U, W ) Hom G(U, W ) C −→ C We’ll use this reformulation of (12.3) to prove

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

∗ ∗ GL(V ) Let U = V V V V . Then C[U] is generated as a C-algebra by elements ρij ⊕ · · · ⊕ ⊕ ⊕ · · · ⊕ p q for 1 i p, 1 j q, where ≤ ≤| {z≤ ≤} | {z } ρij(v1, . . . , vp, ϕ1, . . . , ϕq) = ϕj(vi)

Proof Clearly the ρij are polynomial invariants, since

−1 gϕj(gvi) = ϕj(g gvi) = ϕj(vi)

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

n Hom GL(V )(T U, C) Hom GL(V ),n(U, C) C −→ C ∗ p+q Put V1 = = Vp = V and Vp+1 = = Vp+q = V , so that U = Vi. ··· ··· i=1 This induces a decomposition of T nU as L

p+q p+q p+q n T U = Vi Vi Vi ··· 1 ⊗ 2 ⊗ · · · ⊗ n i =1 i =1 i =1 M1 M2 Mn Thus n Hom GL(V )(T U, C) = Hom GL(V )(Vi1 Vin , C) C C ⊗ · · · ⊗ i ,...,i 1M n

Focus on one of the summands. By (12.1), either Hom GL(V )(Vi1 Vin , C) = 0, or n = 2k, k of C ⊗ · · · ⊗ the 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, ≤ Hom GL(V )(Vi1 Vin , C) is spanned by the k! maps τσ, for σ Sk, given by C ⊗ · · · ⊗ ∈ τσ(v1 vn) = ϕβ (vα ) ϕβ (vα ) ⊗ · · · ⊗ σ(1) 1 ··· σ(k) k The n-homogeneous invariant of U corresponding to τσ is just

ρi ,i ρi ,i α1 βσ(1)−p ··· αk βσ(k)−p and we’re done.

(12.5) Theorem If r N and U = End V End V then C[U]GL(V ) is generated as a ∈ C ⊕ · · · ⊕ C r C-algebra by invariants | U {z C} ti1,...,ik : → (θ1, . . . , θr) tr(θi θi 7→ 1 ··· k where k 1 and 1 i1 < < ik r. ≥ ≤ ··· ≤ Remarks

GL(V ) m (1) [Procesi, 1976] In fact, C[U] is generated by the ti1,...,ik with 1 k 2 1. ≤ ≤ − (2) Procesi also shows that (12.5) is equivalent to the ﬁrst fundamental theorem. This links invariant theory to the representation theory of bouquets, a certain type of quiver.

Proof By (12.2) and (12.3) we must compute Hom (T n(End V, ) which, since End V CGL(V ) C C C ∼= V ∗ V , is isomorphic to Hom (T nV ∗ T nV, C). By the multilinear ﬁrst fundamental theorem (12.1), ⊗ C ⊗ this is spanned by the µσ for σ Sn, where ∈ µσ(ϕ1 ϕn v1 vn) = ϕ (v1) ϕ (vn) ⊗ ⊗ ⊗ · · · ⊗ σ(1) ··· σ(n) We now compute the corresponding map

n νσ : T (End V ) C C −→ ∗ Let V have basis e1, . . . , em and V have dual basis η1, . . . , ηm. Let (Aij) be the matrix of θ EndC V with respect to these bases, i.e. ∈ θ(v) = Aijηj(v)ei i,j X ∗ so that the corresponding element of V V is Aijηj ei. ⊗ i,j ⊗ k P If θk has matrix A then θ1 θn corresponds with ij ⊗ · · · ⊗ 1 n n ∗ n A A ηb ηb ea ea T V T V a1b1 ··· anbn 1 ⊗ · · · ⊗ n ⊗ 1 ⊗ · · · ⊗ n ∈ ⊗ a ,...,a b ,...,b 1X n 1X n

52 so we have

1 n νσ(θ1 θn) = A A ηb (ea ) ηb (ea ) ⊗ · · · ⊗ a1b1 ··· anbn σ(1) 1 ··· σ(n) n ~ X~a,b = A1 An bσ(1)b1 ··· bσ(n)bn ~ Xb

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

i1 i2 ik j1 j` νσ(θ1 θn) = Ab b Ab b Ab b Ab b Ab b ⊗ · · · ⊗ i2 i1 i3 i2 ··· i1 ik j2 j1 ··· j1 j` ··· =X tr(θi θi ) tr(θj θj ) k ··· 1 ` ··· 1 ··· 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 V ∗ V ∗ ⊕ · · · ⊕ ⊕ ⊕ · · · ⊕ p q

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

(v1, , vp, ϕ1, , ϕq) ϕj(vi) for 1 i p, 1 j q; ··· ··· 7→ ≤ ≤ ≤ ≤ (v1, , vp, ϕ1, , ϕq) vi , . . . , vi for 1 i1 < < im p ··· ··· 7→ | 1 m | ≤ ··· ≤ (v1, , vp, ϕ1, , ϕq) ϕj , . . . , ϕj for 1 j1 < < jm q ··· ··· 7→ | 1 m | ≤ ··· ≤

th where v is the determinant of the matrix whose n column is the coordinate of the vi with respect | | n to e1, . . . , em and ϕ likewise with respect to ηj , . . . , ηj . | | 1 m

End of course.

53 MATHEMATICAL TRIPOS PART III, 2013 REPRESENTATION 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 group ∈ algebra CSn. Prove that P πH− = H−π = ε H− for any π H, π ∈ and 1 1 πH−π− = (πHπ− )− for any π S . ∈ n

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 of λ. (a) Find a formula for the numberP 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 hook partition; we will write H(n) for the set of all hook partitions of n. Let f λ be the number of standard tableaux of shape λ. (a) Find a formula for the number of standard tableaux of shape λ = (r, 1s) where λ is a partition of n. Justify your answer. λ n 1 (b) Prove that λ H(n) f = 2 − . ∈ λ 2 2(n 1) (c) Prove that λ H(n)(f ) = n −1 . P ∈ − P 4 Let n N, and let λ = (λi)i N and µ = (µj)j N be partitions of n. The conjugate ∈ ∈ ∈ partition λ0 is deﬁned to be the partition λ0 = (λ10 , λ20 ,...) where λi0 = 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: ∈ λj 6 µj. j i j i X6 X6 (a) Prove that λ ¤ µ if and only if µ0 ¤ λ0. (b) We say that a partition λ is self-conjugate if λ = λ0. Show that there are as many self-conjugate partitions of n as there are partitions of n with distinct odd parts (i.e. all λi are odd and distinct).

k 5 (a) Assuming k > 3, prove that then f (3 ) > 3k. (b) Let λ = (l, . . . , l) = (lk) be a partition of n = kl and λ˜ = (l + 1, . . . , l + 1) be a partition of n + k. Prove (by using the Young-Frobenius formula (5.1)) that f λ˜ (n + k)! l! = . f λ n! · (l + k)! k (c) Let k, l > 3 with kl = n. Prove that then f (l ) > 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) is some 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- (k,k) Frobenius formula). Can you ﬁnd limk Ck /f ? →∞ | | λ 7 In this question λ = (λ1λ2, . . . , λk) is a partition of n, and M is the permutation module with 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 the same entries.) λ (a) Show that M is isomorphic to the permutation module C∆ where ∆ = cos(Sn : Sλ), (the coset space with Sλ the Young subgroup). (b) Show that λ n! dim M = k . i=1 λi! (c) In the case λ = (n 2, 2) has two parts, show that M λ Ω(2) where Ω(2) is the set Q ∼= C of 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 C εσσ . Let ∈ T eT = κT T be the λ-polytabloid. Let 1 denote the identity element of Sn. (a){ List} the columns of T as C ,C ,...,C . Prove that κ = κ κ P. 1 2 s T C1 ··· Cs (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 S and σ C ∈. Show that P ∈ n ∈ T 1 κπT = πκT π− , κT σ = σκT = εσκT and eπT = 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 good account appears in Sagan’s Springer GTM, p.182. This beautiful result has generalisations to the irreducible characters of exotic structures called Iwahori-Hecke algebras of type An 1 due to Ram (who appealed to Schur-Weyl duality), and it appears to have been known− to Young in his work on hyperoctahedral groups (the so-called Weyl groups of type Bn). (a) Consider the character χλ(w) =: χλ where w S has cycle type (µ , µ , . . . , µ ). µ ∈ n 1 2 ` Suppose there are precisely t partitions λ(1), . . . , λ(t) which are obtained by removing a rim hook of length µ from [λ]. Find an expression for χλ in terms of χλ(i) . 1 µ (µ2,µ3...,µ`) (b) Let n > 6. Prove that (n 3,3) χ −n 2 = (n 3)(n 4)(n 5)/6. (2,1 − ) − − − 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 a ε · ∈ ∈ new CSn-module M as the module whose underlying vector space is M and with new action π m := εππ m (π Sn, m M). ∗ · ∈ ε ∈ (a) Verify that M 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 (1n) ε ε ε (1n) S . Prove that the character χ of M is given by χ (π) = χ (π) for π Sn. [Why do you now know that the trivial module S(n) and the sign module S∈(1n) are the only 1-dimensional CSn-modules.]

11 Let ρ :GL(V ) GLn(C) be an irreducible polynomial representation. Show that the → matrix entries ρij C[GL(V )] are homogeneous polynomials of the same degree. More ∈ generall, 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 without appealing to the Nullstellensatz.

13 Prove that the 1-dimensional rational representations of C∗ =GL1(C) are power maps of the form u ur where r Z. 7→ ∈

14 (a) Show that the set of diagonalisable matrices is Zariski dense in Mn(C) (hence every invariant function on Mn(C) is completely determined by its restriction to Tn, the diagonal matrices. [Actually this is nothing more than Jordan decomposition; if you work a bit harder you can show this is true over any ﬁeld, not necessarily algebraically closed]. (b) Let ρ :GL GL(W ) be a rational representation. Show that ρ is polynomial if and n → only if ρ Tn is polynomial. Show also that ρ is polynomial if and only if its character χρ is a polynomial.| (c) Deduce that every 1-dimensional rational representation of GL(V ) is of the form r det :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 n n i n i ∈ polynomial of the form p (t) = t + ( 1) e (α)t − , where n = dim V , showing that the α i=1 − i ei are invariant polynomial functions on End(V ). (b) [HARD] Find a proof of theP fact that the ring of invariants for the conjugation action is generated by the si without using properties of symmetric functions. GL(V ) (c) Show that the functions Tr1 ...,Trn generate the invariant ring C[End(V )] . Here k Trk :End(V ) C is the map A TrA ,(k = 1, 2,...). → 7→

SM, Lent Term 2013 Comments on and corrections to this sheet may be emailed to [email protected] MATHEMATICAL TRIPOS PART III, 2013 REPRESENTATION 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 and m all a = (a1, . . . , am) C then f is the zero polynomial). Hint: use induction on the number of variables and the∈ fact that a non-zero polynomial in one variable has only ﬁnitely many zeroes.

2 2 (a) Show that the natural representation of SL2(C) on C is self-dual (i.e. equivalent to 1 t 1 its dual representation). Hint: ﬁnd a non-singular matrix P such that (A− ) = P AP − . (b) Show that this representation has two orbits.

2 3 Consider GL2 acting on C[X,Y ] (induced by the natural representation of GL2 on C ). (a) What is the image of X and Y under the action of any g GL ? ∈ 2 GL2 SL2 (b) Calculate C[X,Y ] and C[X,Y ] . (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 that GLn is Zariski-dense in Mn.

U T 5 Determine the invariant rings of C[M2(C)] and C[M2(C)] under left multiplication by 1 the subgroup U of upper triangular unipotent matrices and the 2-torus T =diag(t, t− ) of diagonal matrices of SL2.

6 Let Tn be the subgroup of GLn of non-singular diagonal matrices. Choose the standard n basis in W = C and the dual basis in W ∗ and thus identify the coordinate ring C[W W ∗] ⊕ with C[x1, . . . xn, ζ1, . . . , ζn]. Show that Tn C[W W ∗] = C[x1ζ1, . . . , xnζn]. ⊕ What happens if you replace Tn by the subgroup Tn0 of diagonal matrices with determinant 1?

1 2 MATHEMATICAL TRIPOS PART III, 2013 REPRESENTATION THEORY

7 Recall that for each j 1 the symmetric functions n (x) = xj +xj + xj are known as > j 1 2 ··· n power sums (or Newton functions). As before the ej are the elementary symmetric functions. (a) Show that j+1 j 1 ( 1) jej = nj e1nj 1 + e2nj 2 + ( 1) − ej 1n1 − − − − − · · · − − for all j = 1 . . . , n. This is known as the Newton-Girard identity. [Hint: let j = n, consider f(t) = i(t xi) and calculate i f(xi). For j < n what can you say about the right-hand side?] − (b)Q Here is Weyl’s original proofP 1 of the identity in (a). Deﬁne the polynomial n ψ(t) = (1 tx ) = 1 e t + e t2 + ( 1)ne tn, − i − 1 2 − · · · − n i=1 Y where the ei are the elementary symmetric functions. Determine its logarithmic derivative ψ0(t) as a formal power series. Deduce (a). − ψ(t) (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 of A. Assume the representation of G on A is completely reducible. Show that the subalgebra AG of invariants has a canonical G-stable complement and the corresponding G-equivariant projection π : A AG satisﬁes π(hf) = hπ(f) for h AG, f A. This projection is sometimes called the→ Reynolds operator. ∈ ∈

(b) Let A = j>0Aj be a graded K-algbbra (meaning AiAj Ai+j). Assume the + ⊕ ⊂ idea A = j>0Aj is finitely-generated. Show that A is finitely-generated as an algebra ⊕ + over A0. (In fact if the ideal A is generated by the homogeneous elements a1, . . . , an, then A = 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 will need 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 that for any representation| |W of G, the ring of invariants K[W ]G is generated by invariants of degree less than or equal to G (Noether’s Theorem). Schmidt introduced a numerical| | invariant β(G) for every finite group G. It is defined to be the minimal number m such that for every representation W of G, the invariant ring K[W ]G is generated by the invariants of degree 6 m. With this definition, Noether’s Theorem 2 asserts that β(G) 6 G . In fact Schmidt showed that β(G) = G if and only if G is cyclic . It is also a fact that if| G| is actually abelain then β(G) coincides| with| the so-called Davenport constant.

2 In the next few questions we’ll write V = C , G =SL(V ) =SL2(C), and n n Cn = HC,n(V, C) ∼= (S V )∗ ∼= S (V ∗), which can be identified with the set of homogeneous polynomials of degree n in variables X1,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 following G ⊕ → questions sketch a classification of the generators for C[Cn V )] . ⊗

1Weyl, The Classical Groups, II.A.3 2see Finite groups and invariant theory in S´eminaired’Alg`ebrePaul 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 ∈ r ( 1)j ∂rf ∂rg τr(f, g) = − r j j j r j . j!(r j)! − − j=0 ∂X1 ∂X2 ∂X1 ∂X2 X − 1 ∂ ∂ ∂ ∂ r = (f(X ,X )g(Y ,Y ) . r! ∂X ∂Y − ∂X ∂Y 1 2 1 2 |Y1=X1,Y2=X2 1 2 2 1 You should observe τ0(f, g) = fg, τ1(f, g) is the Jacobian of f, g and τ2(f, f) is the Hessian of f. Show that (a) if f, g are homogeneous polynomials of degrees p, q then τr(f, g) = 0 unless r 6 min p, q , in which case it is a homogeneous polynomial of degree p + q 2r; { } − (b) if r 6 min p, q then the map τr : Cp Cq Cp+q 2r sending f g τr(f, g) is a { } ⊗ → − ⊗ 7→ non-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 min p,q { } Cp Cq Cp+q 2r. ⊗ → − r=0 M

10 Let p, q, n, r N and r min q, n . Set N = max 0, r p . Show that there are scalars ∈ 6 { } { − } λN , . . . , λr C, with λr = 0, such that ∈ 6 r

fτr(g, h) = λjτj(τr j(f, g), h), − j=N X for all f Cp, g Cq, h Cn. [HINT: Schur’s Lemma.] Thus the τj are trying hard to be associative.∈ ∈ ∈

G 11 Fix n N. Let R = C[Cn V ], so that R is the set of covariants of Cn. If ϕ R and ∈ ⊕ ∈ f Cn, deﬁne ϕ(f) C[V ] by ϕ(f)(x) = ϕ(f, x). If r N and φ, ψ R deﬁne τr(φ, ψ) as ∈ ∈ ∈ ∈ the map Cn V C by (f, x) τr(φ(f), ψ(f))(x). Finally if d, i N, let Rdi be the set of ⊕ → 7→ ∈ all φ R which are homogeneous, of degree d in Cn and degree i in V . Show that ∈ G (a) Rdi are CG-submodules of R, and R is graded, R = ∞ Rdi, hence so too is R ; ⊕d,i=0 (b) RdiRej Rd+e,i+j and τr(Rdi,Rej) Rd+e,i+j 2r for r 6 min i, j ; (c) the assignment⊆ φ (f φ(f)) induces⊆ an isomorphism− { } 7→ 7→ G Rdi ∼= homCG,d(Cn,Ci), and deduce that (d) any covariant θ RG (d 1) can be expressed as a linear combination ∈ di > min n,i { } θ = τn r(φr,E) − r=0 X G with φr Rd 1,n+i 2r and E is the evaluation map E(f, v) = f(v) [you should note that (c) ∈ G− − says that R1,i = C.E when i = n and 0 if i = n.] 6 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 that G τr(φ, E) S whenever r N and φ S, then S = R . [Remark: after the last question the proof should∈ reduce to a∈ two- line argument.]∈

13 [REALLY HARD] This uses Gordan’s Theorem to ﬁnd a set of generators of RG in the case n = 3. Given n = 3, show that RG is generated by the covariants G E R13; ∈ G H = τ2(E,E), the Hessian, R22; G ∈ t = τ1(H,E) R33; D = τ (t, E),∈ the discriminant ( 48), RG 3 × ∈ 40 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 reduce it to the FFT). Associated to the symbolic method is the symbolic notation, designed to make the calculations easier, but still non-trivial. You can consult any old text on Invariant Theory, such as Grace and Young if interested.

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3Paul Gordan, dubbed the ‘King of Invariant Theory’ is perhaps better known for being Emmy Noether’s thesis advisor. The result appeared in 1868 in Crelle’s Journal as Beweis, dass jede Covariante und Invari- ante einer bin¨aren Form eine ganze Funktion mit numerischen Coeﬃzienten einer endlichen Anzahl solcher Formen ist. 4J.J. Sylvester’s collected works (four volumes, edited by H.F. Baker) include tables with details about higher 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).