Representation Theory of Symmetric Groups

Home , Character theory, Equivariant map, Real representation

MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS

LECTURES BY PROF. ANDREW SNOWDEN; NOTES BY ALEKSANDER HORAWA

These are notes from Math 711: Representation Theory of Symmetric Groups taught by Profes- sor Andrew Snowden in Fall 2017, LATEX’ed by Aleksander Horawa (who is the only person responsible for any mistakes that may be found in them). This version is from December 18, 2017. Check for the latest version of these notes at

http://www-personal.umich.edu/~ahorawa/index.html If you find any typos or mistakes, please let me know at [email protected]. The first part of the course will be devoted to the representation theory of symmetric groups and the main reference for this part of the course is [Jam78]. Another standard reference is [FH91], although it focuses on the characteristic zero theory. There is not general reference for the later part of the course, but specific citations have been provided where possible.

Contents

1. Preliminaries2 2. Representations of symmetric groups9 3. Young’s Rule, Littlewood–Richardson Rule, Pieri Rule 28

4. Representations of GLn(C) 42 5. Representation theory in positive characteristic 54 6. Deligne interpolation categories and recent work 75 References 105

Introduction. Why study the representation theory of symmetric groups? (1) We have a natural isomorphism V ⊗ W ∼ W ⊗ V v ⊗ w w ⊗ v

We then get a representation of S2 on V ⊗V for any vector space V given by τ(v⊗w) = ⊗n w ⊗ v. More generally, we get a representation of Sn on V = V ⊗ · · · ⊗ V . If V is | {z } n times Date: December 18, 2017. 1 2 ANDREW SNOWDEN

⊗n a representation of G, then V is a representation of G ⊗ Sn. Hence representations of Sn arise naturally when studying any representations, especially when considering tensor products. (2) Suppose X is a topological space and let Confn(X) be the conﬁguration space of n points in X, deﬁned by Xn \{locus where two coordinates are equal}.

The symmetric group Sn acts natrually on this conﬁguration space, giving a repre- i sentation of Sn on H (Confn(X), C). (3) We actually do not need a real reason, the subject is fundamental in itself.

1. Preliminaries

Let G be a group, k be a field, and V be a vector space over k. Definition 1.1. A representation of G on V is any (and all) of the following • A homomorphism G → GL(V ), the group of automorphisms of V , • A linear action of G on V , i.e. a map G × V → V such that – (gh)v = g(hv), – 1 · v = v, – g(αv + βw) = α(gv) + β(gw). • A k[G]-module structure on V , where k[G] is the group algebra: ( ) X k[G] = ag[g]: ag ∈ k, ag = 0 for all but finitely many g g∈G ! ! ! X X X ag[g] bh[h] = agbh[gh] . g h g,h

If V,W are representations of G, we can form the following constructions • V ⊕ W is a representation of G by g(v ⊕ w) = gv ⊕ gw, • V ⊗ W is a representation of G by g(v ⊗ w) = gv ⊗ gw, • V ∗ is a representation of G by (gλ)(v) = λ(g−1v), • Hom(V,W ) = {all linear maps V → W } is a representation of G by (gf)(v) = gf(g−1v),

• HomG(V,W ) = {all linear maps f : V → W s.t. f(gv) = gf(v) for all v ∈ V }, • invariant subspace: V G = {v ∈ V | gv = v for all g ∈ G}, • coinvariant subspace: V V = . G span(gv − v | g ∈ G, v ∈ V ) Fact 1.2. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 3

G • HomG(V,W ) = Hom(V,W ) . ∗ G ∗ • (V ) = (VG) . G ∗ ∗ • Not true that (V ) = (V )G in general, but true if dim(V ) < ∞. Deﬁnition 1.3. • The trivial representation of G is V = k, gv = v for all g ∈ G, v ∈ V . • The left regular representation of G is V = k[G] and G acts by left multiplication. For the right regular representation, ! ! X X −1 g · ahh = ahh g . h h • If G acts on a set X, the permutation representation associated to X is V = k[X] (vector space with basis X), and G acts by left multiplication X X g · ax[x] = ax[gx]. x∈X x∈X Deﬁnition 1.4. • A representation V of G is irreducible if V 6= 0 and the only subrepresentations of V are 0 and V (or, equivalently, V is a simple k[G]-module). • A representation V is semi-simple (or completely reducible) if it is a direct sum of simple representations. Fact 1.5. Any subrepresentation or quotient representation of a semi-simple representation is semi-simple. Proposition 1.6. The following are equivalent: • Every representation of G is semi-simple. • If V is a representation of G, W ⊆ V subrepresentation, then there exists a subrepresentation W 0 ⊆ V such that V = W ⊕ W 0. • Every extension of representations splits, i.e. if i p 0 V1 V2 V3 0

is an exact sequence of representations then there exists a map s: V3 → V2 such that

ps = idV3 (called a splitting). • Every representation of G is projective (injective). Deﬁnition 1.7. An R-module M is projective if every exact sequence

0 N N 0 M 0 splits. Proposition 1.8. The following are equivalent: (1) Every representation of G is semi-simple. (2) The trivial representation is projective. 4 ANDREW SNOWDEN

Proof. By Proposition 1.6, (1) implies (2). Conversely, assume (2) and let V be a representation of G with subrepresentation W ⊆ V . We then have a surjective map Hom(V,W ) Hom(W, W ). Taking G-invariants, we get a map

HomG(V,W ) → HomG(W, W ). As the trivial representation is projective, we get a map

k · idW

Hom(V,W ) Hom(W, W )

Thus the map HomG(V,W ) → HomG(W, W ) is surjective and the identity idW comes from a map s ∈ HomG(V,W ), which provides a splitting for W ⊆ V . Theorem 1.9. Suppose G is ﬁnite and |G| 6= 0 in k. Then every representation of G is semi-simple.

Proof. By Proposition 1.8, we need to show the trivial representation is projective, which is equivalent to showing that if f : V → W is a surjection of representations, then V G → W G is surjective.

G Given w ∈ W , pick v0 ∈ V such that f(v0) = w. Now, deﬁne 1 X v = gv . |G| 0 g∈G For all h ∈ G we then have 1 X hv = (hg)v |G| 0 g∈G 1 X = gv |G| 0 g∈G = v and so v ∈ V G. Finally, 1 X f(v) = gf(v ) |G| 0 g∈G 1 X = w |G| g∈G = w, completing the proof.

Remark 1.10. Theorem 1.9 is not true if |G| = 0 in k. For example, let G = Z/p, k = Fp, 1 a V = 2 = e ⊕ e with action of a ∈ G given by the matrix . Then there exists Fq Fp 1 Fp 2 0 1 a short exact sequence MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 5

0 triv V triv 0

but V 6∼= triv ⊕ triv.

Remark 1.11. In general, if p = char(k) |G|, then k[G] is not semi-simple. We show that G ∼ P k[G] = k. If x = ahh, then h∈G X X gx = ag−1hh = ahh, h∈G h∈H P so x is G-invariant if and only if ag−1h = ah for all g. Thus x is a scalar multiple of y = g. g∈G Hence indeed k[G]G = ky. We deﬁne the augmentation map by : k[G] → k, (g) = 1. Then (y) = |G| = 0 in this case, and hence does not split.

Suppose for now k and G are arbitrary. Let {Li}i∈I be the representatives of isomorphism classes of irreducible representations of G. Suppose V is a semi-simple representation. Then there exists an isomorphism M ⊕ni f : Li → V i∈I for some ni’s. This isomorphism is not canonical.

⊕ni Fact 1.12. The embedding f(Li ) ⊆ V is canonical.

It is the sum of images of all maps Li → V , called the Li-isotypic piece of V . Thus there exists a canonical decomposition M V = (Li-isotypic piece of V ), i∈I

⊕ni and each Li-isotypic piece of V admits a non-canonical decomposition as Li for some ni. Proposition 1.13 (Schur’s Lemma). Suppose V and W are irreducible representations. (1) Any G-map V → W is either 0 or an isomorphism. (2) EndG(V ) is a division algebra. (3) If V is ﬁnite dimensional and k is algebraically closed then EndG(V ) = k.

Proof. For (1), say f 6= 0. Then ker(f) ( V , so ker(f) = 0 as V is irreducible, and 0 6= im(f) ⊆ W so im(f) = W . Then (2) follows from (1). For (3), let f : V → V be a G-endomorphism. Let λ be an eigenvalue of f. Then ker(f − λid) 6= 0, so f = λid. Example 1.14 ((3) does not hold in general). Let G = C∗ and consider C as a 2-dimensional real representation. Then V is irreducible. But EndG(V ) = C. The same example works for ∼ G = {1, i, −1, −i} = Z/4. 6 ANDREW SNOWDEN

Remark 1.15. There are examples where EndG(V ) is genuinely not a ﬁeld. For instance, try to think of a representation whose endomorphism ring is the quaternions.

Assume k = k¯ and let V be a ﬁnite-dimensional isotypic representation of G (so there is only 1 non-zero isotypic piece). Say Li is the irreducible in V .

Deﬁnition 1.16. The Li multiplicity space in V is HomG(Li,V ). Fact 1.17. If V isotypic, then the canonical map

HomG(Li,V ) ⊗ Li → V is an isomorphism.

To show this, reduce to the case where V is a direct sum of Li’s and then to V being just Li, in which case, HomG(Li,V ) = k by Schur’s Lemma 1.13 (3), so the isomorphism is k ⊗Li → V .

Let V be a ﬁnite-dimensional semi-simple representation and Mi be the multiplicity space of Li, i.e. Mi = HomG(Li,V ). Then the canonical map M Mi ⊗ Li → V i∈I is an isomorphism. Remark 1.18. If V is not semisimple, the map is still injective and its image is the socle of V , the maximal semi-simple subrepresentation of V . Proposition 1.19. Let k be algebraically closed and G, H be groups. Any ﬁnite-dimensional irreducible representation of G×H has the form V ⊗W with V an irreducible representation of G and W an irreducible representation of H. Conversely, any representation of this form is irreducible.

Proof. We ﬁrst show that V ⊗ W is irreducible. Let U ⊆ V ⊗ W be a subrepresentation. As a representation of H, V ⊗ W is isotypic and its W -multiplicity space is V , so ∼ V = HomH (W, V ) ⊗ W.

Now, HomH (W, V ) is a G-subrepresentation of V , so it is 0 or V , and hence V is 0 or V ⊗W . Suppose U is a ﬁnite-dimensional irreducible representation of G×H. Let W be an irreducible representation of H contained in U. We then have an injective map

HomH (W, V ) ⊗W → U | {z } a rep. of G and it is actually surjective as U is irreducible and the map is non-zero. We now assume G is finite, k = k¯, |G| 6= 0 in k. The goal is to understand k[G], the left regular representation of G. We first think of k[G] as a representation of G × G, with action defined by (g, h) · x = gxh−1 (so both the left and the right regular representations). We know that k[G] is a semi-simple representation of G × G. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 7

Lemma 1.20. Let V be a representation of G × G. We have that G HomG×G(k[G],V ) = V , where G ⊆ G × G is the diagonal copy of G.

G Proof. The image of 1 under any map in HomG×G(k[G],V ) gives an element of V . The map deﬁned this way is the isomorphism.

Therefore, we have that k if L ∼= L∗, Hom (k[G],L ⊗ L ) = (L ⊗ L )G = Hom (L∗,L ) = j i G×G i j i j G i j 0 otherwise, where the last equality follows from Schur’s Lemma 1.13. Therefore, as G×G-representations: ∼ M ∗ k[G] = Li ⊗ Li . i∈I In fact, we get a canonical isomorphism M k[G] → End(Li). i∈I

By considering these vector spaces as representations of G, we get the following theorem.

L ⊕ dim Li Theorem 1.21. The left regular representation k[G] of G decomposes as Li . i∈I Corollary 1.22. We have that X 2 dim(Li) = |G|. i∈I Remark 1.23. These results are sometimes proven using character theory.

Induction. Let R → S be a ring homomorphism. By restriction of modules, we get a functor: ModS → ModR . It has a left and a right adjoint:

• The left adjoint is extension of scalars S ⊗R −:

HomS(S ⊗R M,N) = HomR(M,N).

• The right adjoint is co-extension of scalars HomR(S, −):

HomS(N, HomR(S,M)) = HomR(N,M). Now, suppose H ⊆ G is a subgroup, and left R = k[H], S = k[G]. Then we get a functor G ResH : Rep(G) → Rep(H). Its left adjoint is G IndH (V ) = k[G] ⊗k[H] V, i.e. we have G G HomG(IndH (V ),W ) = HomH (V, ResH (W )). 8 ANDREW SNOWDEN

Its right adjoint is G CoIndH (V ) = Homk[H](k[G],V ), i.e. we have G G HomG(W, CoIndH (V )) = HomH (ResH (W ),V ). The two adjunction statements are called Frobenius reciprocity. Fact 1.24.

G ∼ G • If [H : G] < ∞ then IndH = CoIndH . G G • Both IndH and CoIndH are exact. • Induction is transitive: if K ⊆ H ⊆ G, then G H G IndH IndK = IndK .

The proof is left as an exercise. Examples 1.25.

G • Ind{1}(k) = k[G], where k is the trivial representation of H = {1} G • IndH (k) = k[G/H], the permutation representation of G acting on G/H G • IndH (k[H]) = k[G]

Character theory. Let V be a ﬁnite-dimensional representation of G. The character of V is the function χV : G → k given by χV (g) = tr(g|V ). This is a class function, i.e. −1 ξV (hgh ) = ξV (g). Remark 1.26. These characters are only interesting in characteristic zero. In positive characteristic, one has to use Brauer characters.

Now assume k = C and G is ﬁnite. Fact 1.27.

• χV ⊕W = χV + χW

• χV ⊗W = χV · χW

• χV ∗ = χV

• χHom(V,W ) = χV χW

Proof. Over k = C, the trace is the sum of the eigenvalues, and the ﬁrst three formulas follow immediately. The ﬁnal formula follows from the previous ones by noting Hom(V,W ) ∼= ∗ V ⊗ W .

For ϕ, ψ : G → C, put 1 X hϕ, ψi = ϕ(g)ψ(g). |G| g∈G MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 9

Proposition 1.28. Let V , W be ﬁnite-dimensional representations of G. Then

hχV , χW i = dim HomG(V,W ).

HomG(V,W ) = HomG(triv, Hom(V,W )) and hence using the previous part, we obtain

dim HomG(V,W ) = hχtriv, χHom(V,W )i = hχtriv, χV χW i = hχV , χW i, completing the proof. Corollary 1.29. If V , W are irreducible, then 1 if V ∼= W hχ , χ i = V W 0 otherwise

Proof. This follows from Proposition 1.28 and Schur’s Lemma 1.13.

Corollary 1.30. The set {χV | V irreducible} is a basis for the space of class functions on G.

Thus, the data about representations of a group can be arranged into a character table of G: — conjugacy classes of G — | irreducible representations of G |

2. Representations of symmetric groups

Every element σ of Sn admist a decomposition σ = C1 ...Cr, where the Ci’s are disjoint cycles and every element of {1, . . . , n} appears once. A partition of n is an unordered collection of positive integers summing to n:

n = m1 + ··· + mr, where we typically take m1 ≥ m2 ≥ · · · ≥ mr. 10 ANDREW SNOWDEN

We associate to σ the partition

p(σ) = #C1 + ··· + #Cr of n. (The notation p(σ) is not standard but we assume it temporarily.) Note that −1 τ τ τστ = C1 ··· Cr and hence p(τστ −1) = p(σ). We therefore get a map

p: {conjugacy classes in Sn} → {partitions of n}. Fact 2.1. This is a bijection.

Corollary 2.2. The number of conjugacy classes in Sn is the number of partitions of n.

Corollary 2.3. The number of complex irreducible representations of Sn is the number of partitions of n.

Proof. This follows from Corollary 2.2 and Corollary 1.30. Examples 2.4. • The trivial representation. • We have the sign homomorphism

sgn: Sn → {±1}, deﬁning a 1-dimensional sign (or alternating) representation.

• Since Sn naturally permutes the set {1, . . . , n}, we get a permutation representation n n n of Sn on C . If e1, . . . , en is the standard basis of C , then σ(ei) = eσ(i). Then C is the permutation representation of Sn. n X n • en is Sn-invariant, and hence C is not irreducible unless n = 1. i=1 n • Deﬁne : C → C by (ei) = 1 (the augmentation map). Then ker() is an Sn- subrepresentation. Hence n C = ker() ⊕ C. Proposition 2.5. For n > 1, ker() is irreducible.

Proof. Let V ⊆ ker() be a non-zero subrepresentation. Let r X v = aiei i=1

be a nonzero element of V with minimal r (so no ai is 0 after permuting the ei’s).

Trivially, r 6= 1 or otherwise V = 0. If r = 2, v is a scalar multiple of e1 −e2, which generates ker(), so V = ker().

Suppose r > 2. There exist 1 ≤ i, j ≤ r such that ai 6= aj. We may assume that i = r − 1, j = r. We then note that ar−1v − ar(r r − 1)v ∈ V MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 11

but it is equal to r−2 X 2 2 (ar−1 − ar) aiei +(ar−1 − ar)er−1. i=1 | {z } 6=0 This contradicts the minimality of r.

Deﬁnition 2.6. The representation ker() is called the standard representation of Sn.

We list the irreducible representation of Sn for small n.

n number list notes 1 1 triv 2 2 triv, sgn = std 3 3 triv, sgn, std In particular, we can conclude that sgn ⊗ std ∼= std, because our list is complete. 4 5 triv, sgn, std, To see that sgn ⊗ std 6∼= std, we compute the characters: sgn ⊗ std, ? χstd((12)) = 1, χstd ⊗ sgn((12)) = −1. We note that there is one irreducible representation still missing. Remark 2.7. Note that if G acts on X, the character of the permutation representation at g ∈ G is the number of ﬁxed points of g on X. We can use that to compute the character of the standard representation.

We write down the character table of S4 for the representations that we know. partition 14 (2, 1, 1) (2, 2) (3, 1) (4) permutation 1 (12) (12)(34) (123) (1234) triv 1 1 1 1 1 sgn 1 −1 1 1 −1 std 3 1 −1 0 −1 std ⊗ sgn 3 −1 −1 0 1 ? Deﬁnition 2.8. A perfect matching on a set S is a graph (undirected with no loops) in which every vertex belongs to exactly 1 edge.

Let Mn = {matchings on {1, 2, . . . , n}}. Clearly, Sn acts on Mn, so we get the permutation representation C[Mn].

Let us think about the n = 4 case. The set M4 has 3 elements:

1 2 1 2 1 2

3 4 3 4 3 4 12 ANDREW SNOWDEN

Let : C[M4] → C be the augmentation map. Is V = ker irreducible? To check, we compute the character by looking at how many matchings are ﬁxed by the consecutive permutations and subtracting 1:

• 1 fixes all 3 matchings, • (12) fixes only the second matching, • (12)(34) fixes all 3 matchings, • (123) does not fix any matchings, • (1234) fixes only the first matching.

Hence the character χV of V is

partition 14 (2, 1, 1) (2, 2) (3, 1) (4) permutation 1 (12) (12)(34) (123) (1234)

χV 2 0 2 −1 0

Looking at the value on (123), we see that χV is not a positive linear combination of triv and sgn, and it is 2-dimensional, so it is irreducible.

Thus the complete character table of S4 is

partition 14 (2, 1, 1) (2, 2) (3, 1) (4) permutation 1 (12) (12)(34) (123) (1234) triv 1 1 1 1 1 sgn 1 −1 1 1 −1 std 3 1 −1 0 −1 std ⊗ sgn 3 −1 −1 0 1 χV 2 0 2 −1 0

We will ﬁnd more irreducible representations of Sn inside of permutation representations. For example, for a subgroup H ⊆ Sn, we can look at the permutation representation associated to the action of Sn on the cosets Sn/H. Hence we want a good source of subgroups of Sn.

Deﬁnition 2.9. Given a partition λ = (λ1, λ2, . . . , λr) of n, the Young subgroup Sλ of Sn is

Sλ1 × Sλ2 × · · · × Sλr ⊆ Sn.

We can now consider the permutation representation C[Sn/Sλ]. Recall that the number of irreducible representations of Sn is the number of partitions of Sn, so we get the correct number of representations; however, they will not be irreducible. A λ-partition of {1, 2, . . . , n} is a decomposition

r a {1, 2, . . . , n} = Ai with #Ai = λi. i=1 MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 13

The set of λ-partitions has a natural Sn-action, and if r ( i i ) a X X P = λj + 1,..., λj + λi+1 i=1 j=1 j=1

then Sλ is the stabilizer of P . Thus

C[Sn/Sλ] = C[set of all λ-partitions]. (This is sometimes called the generalized orbit-stablizer theorem.) Examples 2.10.

• For λ = (n), Sλ = Sn, and C[Sn/Sλ] = triv. n • For λ = (1 ), Sλ = {1}, and C[Sn/Sλ] is the regular representation. n • For λ = (n − 1, 1), Sλ = Sn−1, and C[Sn/Sλ] = C = triv ⊕ std. • For λ = (2, 2), n = 4, Sλ = S2 × S2, and C[Sn/Sλ] = triv ⊕ std ⊕V , where V = ker for the augmentation map : C[M4] → C. Note that C[Sn/Sλ] has a representation of S4 × S2. The decomposition as a representation of S4 × S2, [(triv ⊕V ) triv] ⊕ [(std sgn)], where by V W we mean the representation of S4 × S2 obtained from tensoring V as a representation of S4 with W as a representation of S2. (We use to distinguish this from tensoring two representations of S4 × S2.)

Deﬁnition 2.11. The Young diagram of a partition λ is a diagram with λ1 boxes in the ﬁrst row, λ2 in the second and so on. Examples 2.12. The Young diagram of λ = (3, 1) is

The Young diagram of λ = (2, 1, 1) is

Note that Young diagrams have a symmetry: we can ﬂip along the diagonal line:

This gives an involution on the set of partitions , called the conjugation or transpose, λ 7→ λ†. Explicitly, † (λ )i = max j such that λj ≥ i. 14 ANDREW SNOWDEN

Deﬁnition 2.13. If λ, µ are two partitions, we say that λ dominates µ and write λ µ if

λ1 + ··· + λk ≥ µ1 + ··· + µk for all k. This deﬁned a partial ordering on the set of partitions. Example 2.14. The partitions of 6 form the following graph where λ is above µ if and only if λ µ: (6)

(5, 1)

(4, 2)

(3, 3) (4, 12)

(3, 2, 1)

(3, 13) (23)

(22, 12)

(2, 14)

(16)

We make the following observation:  sum of irreducibles,   each of which appears  C[Sn/Sλ] = (some irreducible) ⊕    in some C[Sn/Sµ]  with µ λ The evidence for this is given by Examples 2.10. Examples 2.15.

• C[Sn/Sn−1] = std ⊕ triv, where triv appears in C[Sn/Sn], and std is new. • C[S4/S(2,2)] = V ⊕ std ⊕ triv . |{z} |{z} |{z} new C[S4/S(3,1)] C[S4/S(4)] MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 15

We will see later that this observation is actually true. Deﬁnition 2.16. A λ-tableau is a way of ﬁlling the Young diagram of λ with numbers 1, . . . , n such that each number appears once. Examples 2.17. The following are two λ-tableux obtained from the same Young diagram

1 2 3 4 1 2 4 3 Deﬁnition 2.18. A λ-tabloid is the same, but the order within the rows is irrelevant.

A λ-tabloid is drawn with only horizontal lines. Example 2.19. The ﬁrst two λ-tabloids are the same, while the last one is diﬀerent:

1 2 3 1 3 2 4 1 2 4 4 3

Note that Sn permutes the λ-tableaux and λ-tabloids, and the associated representations are: C[λ-tableaux] = regular representation, λ M := C[λ-tabloids] = C[Sn/Sλ]. Recall that we expect M λ to contain some simple Lλ, and all others appear in M µ for some µ λ, µ 6= λ. Example 2.20 (n = 4). For the diﬀerent partitions λ of 4, we get the following simple modules Lλ: • M (4) = triv, so L(4) = triv, • M (3,1) = triv ⊕ std, so L(3,1) = std, • M (2,2) = triv ⊕ std ⊕V , so L(2,2) = V , • M (2,1,1) = triv ⊕ std⊕2 ⊕V ⊕ (std ⊗ sgn), so L(2,1,1) = std ⊗ sgn, (14) (14) • M = C[S4], so L = sgn.

Recall that by Corollary 2.3 there exists a bijection between irreducibles and partitions—this example shows a speciﬁc bijection between them. Note that M λ = IndSn (triv). Sλ We also deﬁne † N λ = IndSn (sgn) = M λ ⊗ sgn . Sλ† Example 2.21 (n = 4). We have:

• N (14) = sgn = L(14) • N (2,1,1) = sgn ⊕ std ⊕ sgn = L(14) ⊕ L(2,1,1) 16 ANDREW SNOWDEN

It appears that N λ = Lλ ⊕ (L Lµ) with µ λ, whereas M λ = Lλ ⊕ (L Lµ) with µ λ. Based on this observation, we expect there to be a unique map (up to scaling) N λ → M λ, whose image will be Lλ. Think of elements of N λ as tableaux of shape λ such that permuting a column changes by sign of that permutation (“signed tabloids”): 1 3 5 2 3 5 = − 2 4 1 4

We will deﬁne an averaging map to get the following diagram

averaging map N λ C[tableaux] M λ

Let t be a tableau. Deﬁne

•{ t} ∈ M λ the tabloid defined by t, •{ t}0 ∈ N λ the signed tabloid defined by t, • Ct the column stabilizer of t, i.e. the subgroup of Sn fixing columns of t, • Rt the row stablizer of t, X • κt = (sgn σ)σ ∈ C[Sn]. σ∈Ct Note that λ M = C[λ-tabloids] = C[λ-tableaux]/hσt − σ | σ ∈ Rt, t λ-tableaui, λ N = C[λ-tableuax]/hσt − sgn(σ)t | σ ∈ Ct, t λ-tableaui. We define the averaging map by 0 {t} 7→ κtt λ for any t. In other words, a tableau t goes et = κt{t} ∈ M , which is a polytabloid. λ λ λ The image of N → M is the span of the et’s. It is called the Specht module and denoted S . Example 2.22. Let t be

2 4 1 3 5

Then

Ct = h(23), (54)i ⊆ S5 and so

κt = 1 − (23) − (54) + (23)(54).

Thus et is 2 4 1 3 4 1 2 5 1 3 5 1 − − + 3 5 2 5 3 4 2 4 MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 17

Lemma 2.23. Let λ, µ be partitions of n, t a λ-tableau, t0 a µ-tableau. Suppose that, for all i, all numbers in the ith row of t0 appear in diﬀerent columns in t. Then λ µ Proof. We rearrange the numbers step by step:

0 (1) We can use Ct to move all numbers in the first row of t into the first row of t. Thus λ1 ≥ µ1. 0 (2) We can use Ct to move all numbers in the second row of t to the first or second row of t. Thus λ1 + λ2 ≥ µ1 + µ2.

Continuing this way, we get the result. Lemma 2.24. Let λ and µ be partitions of n, t a λ-tableau, t0 a µ-tableau. Suppose 0 κt{t }= 6 0. 0 Then λ µ, and if λ = µ then κt{t } = ±et. Proof. Let a and b be two numbers in the same row of t0. Then (a b){t0} = {t0}. Hence a, b are in diﬀerent columns of t, or otherwise (a b) ∈ Ct and we have 0 0 0 0 κt{t } = (a b)κt{t } = sgn((a b))κt{t } = −κt{t }, 0 contradicting κt{t }= 6 0. By Lemma 2.23, λ µ. 0 0 If λ = µ, then t ∈ Ctt by the above argument, and so κt{t } = ±κt{t} = ±et. λ Corollary 2.25. Given m ∈ M and a λ-tableau t, κtm ∈ Cet.

Proof. By definition, m is a linear combination of {t0} for λ-tableaux t0. Moreover, we know 0 that κt{t } = 0 or ±et by Lemma 2.24, which completes the proof. Definition 2.26. Define a symmetric bilinear inner product h−, −i on M λ by 0 0 1 if {t} = {t } h{t}, {t }i = δ 0 = {t},{t } 0 otherwise.

Note that this product is invariant under Sn. Proposition 2.27. Suppose V ⊆ M λ is a subrepresentation. Then V ⊇ Sλ or V ⊆ (Sλ)⊥.

0 Proof. Let v ∈ V . If κtv 6= 0 then et ∈ V by Corollary 2.25, and so et0 ∈ V for all t , since et’s λ form a single orbit under Sn. Thus S ⊆ V .

Otherwise, κtv = 0 for any v ∈ V and t, and we obtain

0 = hκtv, {t}i = hv, κt{t}i = hv, eti. | {z } et λ ⊥ λ ⊥ Thus v ∈ (S ) , showing that V ⊇ (S ) . Corollary 2.28. The Specht module Sλ is irreducible. 18 ANDREW SNOWDEN

Proof. Suppose V ( Sλ is a subrepresentation. By Proposition 2.27, V ⊆ (Sλ)⊥. Then V ⊆ Sλ ∩ (Sλ)⊥ = 0 because everything is defined over R and h−, −i is positive definite over R. (Alternatively, we can define the inner product above to be Hermitian instead and use this.)

We therefore have M λ = Sλ ⊕ (Sλ)⊥ and the ﬁrst summand is irreducible. Lemma 2.29. Suppose f : M λ → M µ is a map of representations and Sλ 6⊂ ker(f). Then

λ µ and if λ = µ, then fSλ is a scalar. Proof. Let t be a λ-tableau. If et 6∈ ker(f), then

0 6= f(et) = f(κt{t}) = κtf({t}). Since f{t} is a linear combination of µ-tabloids, we can use Lemma 2.24 to conclude that λ µ. If λ = µ then κtf({t}) ∈ Cet by Lemma 2.24. Thus f(et) is a scalar multiple of et for any t, and since Sn acts transitively on et, it must be the same scalar α, so f(v) = αv for λ any v ∈ S . Corollary 2.30. We have that M λ = Sλ ⊕ (Sλ)⊥ and (Sλ)⊥ = sum of (Sµ)0s for µ λ, µ 6= λ. λ Furthermore, every irreducible of Sn is isomorphic to some S .

Proof. We claim that if Sλ ∼= Sµ, then λ = µ. Indeed, choose f : M λ → M µ by extending this isomorphism (this is possible because the characteristic is 0):

f M λ N λ

∼ Sλ = Sµ

By Lemma 2.29, λ µ. By symmetry, µ λ, and hence λ = µ. λ Thus every irreducible of Sn is S , since there are the correct number of them (Corollary 2.3). Finally, suppose Sµ appears in (Sλ)⊥. Then choose an f extending the inclusion, i.e. so that

f M µ M λ

Sµ (Sλ)⊥

λ λ ⊥ commutes. By Lemma 2.29, µ λ and we cannot have µ = λ since S ∩ (S ) = 0. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 19

Remark 2.31. We will use the following fact in the next propositon. Recall that M λ = [S /S ] = IndSn (triv), C n λ Sλ and so by Frobenius reciprocity

λ Sλ HomSn (M ,V ) = HomSλ (triv,V ) = V via the map (f : M λ → V ) 7→ f({t}) ∈ V Sλ

(noting that Sλ = Rt, the row stabilizer). Proposition 2.32. We have that Sλ ⊗ sgn = Sλ† .

† Proof. Fix a λ-tableau and let t be the transpose tableau. Then there is a unique Sn- equivariant map f : M λ† → Sλ ⊗ sgn † f({t }) = et ⊗ 1. because et ⊗ 1 is invariant under Rt† = Ct (note: et is skew-invariant under Ct, so et ⊗ 1 is invariant). Then † f(et† ) = f(κt† {t }) = κt† (et ⊗ 1) = (ρtet) ⊗ 1.

We show that ρtet 6= 0:

hρtet, {t}i = het, ρt{t}i = |Rt|het, {t}i = |Rt|= 6 0. λ† λ Hence we have shown the map fSλ† : S → S ⊗ sgn is non-zero, and since both sides are irreducible, it gives an isomorphism by Schur’s Lemma 1.13. Example 2.33.

• S(n) = triv, since M (n) = triv. • S(1n) = sgn, since M (1n) is the regular representation and its elements have the shape

with n boxes, so Ct = Sn, and et projects {t} ∈ C[Sn] to sgn component. • We claim that S(n−1,1) = std. First, M (n−1,1) is naturally the permutation representation Cn via

········· n 7→ vi, the ith basis vector of C . i If t is the tableau i ··· j Then 20 ANDREW SNOWDEN

i ··· j ··· et = − j i

and hence et corresponds to vj − vi ∈ ker .

We know σet = sgn(σ)et for σ ∈ Ct. But note that σet 6= et for σ ∈ Rt, in general. We now want to ﬁnd more linear relations between the et’s. Example 2.34. For the regular representation, we have the relation

(vk − vj) + (vj − vi) = (vk − vi). and we get the following linear relation in S(n−1,1):

j i ··· i i ··· i j ··· + − = 0. k j k

after applying e.

Let us generalize this to arbitrary Sλ. Let t be a tableau of shape λ. Pick sets X and Y as follows:

ith (i + 1)st column of t

† λi+1 X ⊆ {numbers in ith column}

† λi X Y ⊆ {numbers in (i + 1)st column}

Deﬁne X GX,Y = (sgn σ)σ,

σ∈SX∪Y called a Garnir element. † Proposition 2.35. Assume |X ∪ Y | ≥ λi . Then GX,Y et = 0.

This is called a Garnir relation.

Proof. Let σ ∈ Ct. There exist a, b ∈ X ∪ Y such that a and b are in the same row of σt (by the pigeon hole principle). Then (a b){σt} = {σt} MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 21 and hence GX,Y {σt} = 0 because (a b) ∈ SX∪Y . Since et is a linear combination of {σt}’s with σ ∈ Ct, we get GX,Y et = 0.

Remark 2.36. Note that SX × SY ⊆ Ct, so it acts on et through sgn. Thus GX,Y acts by X |SX × SY | · (sgn σ)σ,

σ∈SX∪Y /SX ×SY | {z } Garnir element where the sum is over a set of coset representatives. Example 2.37. • Let t be the tableau 1 2 3 4 and X = {1, 2}, Y = {3}. Then

SX∪Y /SX × SY = {1, (23), (13)} and the Garnir relation is 1 3 4 1 2 4 3 1 4 − − = 0 2 3 2 (after applying e). • Let t be the tableau 1 4 2 5 3 6 and X = {1, 2, 3}, Y = {4}. Then

SX∪Y /SX × SY = {1, (14), (24), (34)} and the Garnir relation is 1 4 4 1 1 2 1 3 2 5 = 2 5 + 4 5 + 2 5 3 6 3 6 3 6 4 6 (after applying e). • For X = {2, 3}, Y = {4, 5}, we get a relation with 6 terms. Deﬁnition 2.38. A tableau t is standard if rows and columns of t are increasing. Example 2.39. The tableau

1 2 6 3 4 5 22 ANDREW SNOWDEN

is standard, while

1 4 6 2 3 5

is not standard.

λ Theorem 2.40. The set {et}t standard is a basis of S .

Proof. We deﬁne a total order < on tabloids. Let ri({t}) be the row in which i appears in t. Deﬁne {t} < {t0} if 0 0 (rn({t}), rn−1({t}),...) < (rn({t }), rn−1({t }),...) in the lexicographical order, i.e., explicitly, {t} < {t0} if

0 • rn({t}) < rn({t }), 0 0 • or rn({t}) = rn({t }) and rn−1({t}) < rn−1({t }), • or ...

We observe that if t is standard, then {t} is the biggest tabloid (with respect to <) appearing λ in Ct{t}. Therefore, {et}t standard are linearly independent in S , since

et = {t} + (smaller things under < ). 0 We need to show that for any tableau t, et is a linear combination of et0 with t standard. We do this by induction on t, using the transpose of the total order <. Fix t. We may assume that the columns of t are increasing and that t is not standard. We can ﬁnd adjact columns where a row decreases: suppose ak > bk in the following adjacent columns

a1 b1

a2 b2 . . . .

ak bk ...... bs . . . .

Let X be the blue set above and Y be the red set above. We apply the Garnir relation on 0 X, Y . This writes et as a linear combination of et0 s, where t is bigger than t in our order (because as get moved into b slots). This completes the proof by induction. Corollary 2.41. For any partition λ, dim Sλ = #{standard λ-tableaux}. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 23

Moreover, by the proof of Theorem 2.40 we also get the following result.

0 Corollary 2.42. Every et is a Z-linear combination of et0 s with t standard. Thus the matrix λ of any σ ∈ Sn in S with respect to the standard basis has integer entries. Example 2.43. Let λ = (3, 2). The standard tableaux are

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

Thus dim S(3,2) = 5. By taking transpositions, we note that S(2,2,1) = S(3,2) ⊗ sgn, so dim(2, 2, 1) = 5. Example 2.44. Let λ = (3, 1, 1). The picture is

1 x1 x2

(3,1,1) 4 and choosing any x1 and x2 determines y1 and y2. Thus dim S = 2 = 6.

We now know what the irreducible representations of Sn are and how to compute their dimensions. However, we have no relation between the representations of Sn and Sk ⊆ Sn for k < n.

λ Important question. Let λ be a partition of n. How does S |Sn−1 decompose? λ L µ Answer. We will show S |Sn−1 = S , where µ ⊂ λ is containment of Young diagrams. µ⊂λ

Let r1 < ··· < rk be the rows of a tableau obtained from removing a box from λ. Deﬁne a ﬁltration i λ F S = span of all et’s where the number n appears in row ≤ ri in t. i λ Clearly, F S is Sn−1-stable. i λ Lemma 2.45. The space F S has a basis {et} where t is standard and n appears in row ≤ i.

Proof. Linear independence follows from Theorem 2.40. Suppose t is not standard and n is in row ≤ ri. Applying a Garnir relation, similarly as in the proof of Theorem 2.40, writes et 0 as a linear combination of et0 s where t still has the property that n is in row ≤ ri.

Lemma 2.46. As Sn−1-representations, we have that F iSλ/F i−1Sλ ∼= Sλ(i),

where λ(i) is λ with a box in row ri removed.

Proof. Deﬁne a linear map f : Sλ(i) → F iSλ/F i−1Sλ by

f(et) = et0 24 ANDREW SNOWDEN

where t is a standard λ(i)-tableau and t0 is t with n placed in missing box. This is clearly an isomorphism of vector spaces. We have to show it is Sn−1-equivariant. We only present a sketch of this proof.

We want to show f(eσt) = σf(et) = σet0 = eσt0 for σ ∈ Sn−1 and t standard. The map does the following

−→ n

To see where it maps eσt, we have to use Garnir relations to express it in terms of standard λ(i)-tableau. The point is that the Garnir relations are the same on either side when working i−1 modulo F Sλ, which gives the desired equivariance.

Since we are in characteristic 0, we have a splitting F iSλ ∼= (F i−1Sλ) ⊕ (F iSλ/F i−1Sλ), and hence λ M µ S |Sn−1 = S . µ⊂λ |µ|=n−1

Example 2.47. The Specht module associated to the tableau

restricted to S6 splits as

⊕ ⊕

Remark 2.48. We could have attempted to prove the above using the map f mapping et i λ to et0 ∈ F S .

Sλ(i) ∼ F iSλ/F i−1Sλ

f F iSλ MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 25

However, the map f is not Sn−1-equivariant. (2) (2,1) Indeed, consider f : S → S mapping et to et0 . In the discussion below, to simplify λ notation, we will identify t with its image et in S . The map f is deﬁned by

1 2 1 2 −→ 3

1 2 2 1 1 2 1 3 But (12) applied to gives which splits as − in F 1Sλ. 3 3 3 4 Proposition 2.49 (Pieri’s rule). Let µ be a partition of n − 1. Then M IndSn Sµ = Sλ. Sn−1 µ⊂λ |λ|=n

Proof. By Frobenius reciprocity, the multiplicity of Sλ in IndSn (Sµ) is the multiplicity of Sn−1 µ λ S in S |Sn−1 , which is 1 if µ ⊂ λ, 0 otherwise, completing the proof.

2.1. The Grothendieck group. Let G be a ﬁnite group and k be a ﬁeld.

Deﬁnition 2.50. The representation ring of G over k, denoted Rk(G), is

free Z-module on symbols [V ] for V ﬁnite-dimensional representation of G

[V2] = [V1] + [V3] whenever we have a short exact sequence

0 V1 V2 V3 0

We deﬁne multiplication on this ring by [V ][W ] = [V ⊗k W ].

Exercise. Verify this works: Rk(G) is a well-deﬁned, commutative, associative, unital ring. Remark 2.51. To motivate the deﬁnition, we make the following observations.

• If char(k) = 0, then V2 = V1 ⊕ V3 if we have a short exact sequence above, so + is simply ⊕.

• A generalized Euler characteristic for representations of G is a way to assign a number χ(V ) to a representation V of G such that χ(V2) = χ(V1) + χ(V3) if we have a short exact sequence above. Hence giving an Euler characteristic χ is the same as giving a homomorphism χ: Rk(G) → C.

More generally, we can make the following construction for any (essentially small) abelian category A. Deﬁnition 2.52. The Grothendieck group of A, denoted K(A) is 26 ANDREW SNOWDEN

free Z-module on symbols [M] for M ∈ A

[M2] = [M1] + [M3] whenever we have a short exact sequence

0 M1 M2 M3 0

In general, K(A) is just an abelian group. If A has a symmetric ⊗-product that is exact, then ⊗ induces a ring structure on K(A) via [M][N] = [M ⊗ N]. Remark 2.53. More generally, if ⊗ is not exact, but it is right exact, and derived functors Tor∗ exist, and Torn = 0 for n 0, then we can given K(A) a ring structure by X p [M][N] = (−1) [Torp(M,N)]. p≥0

fd Remark 2.54. When A = Repk (G), we recover the representation ring of G: fd Rk(G) = K(Repk (G)).

Let G be a ﬁnite group and let L1,...,Lr be the irreducible complex representations. Then r M RC(G) = Z[Li] i=1

and hence RC(G) is free of rank r. It is clear that [Li] generate Rk(G) under +. To show that there are no linear relations, we present two similar arguments (1) We have an exact functor fd Hom(Li, −): RepC (G) → Vec and this induces

RC(G) → RC(triv) = Z since Li 7→ 1 and Lj 7→ 0 if i 6= j. Thus there are no linear relations. (2) Taking characters gives a map

RC(G) → {class functions G → C}.

Since χLi are linearly independent, there cannot be a linear relation between [Li]. Remark 2.55. If A is an abelian category and all its objects have ﬁnite length, then M K(A) = Z[Li] i∈I

where {Li}i∈I are the simple objects of A. This follows from the Jordan-H¨older Theorem. ∼ 3 Example 2.56. We note that RC(S3) = Z and it has a basis s(3), s(13), s(2,1), where we λ write sλ = [S ]. We have that

s(3) = [triv] = identity in the ring, 2 s(13) = 1, s(13)s(2,1) = s(2,1), 2 s(2,1) = s(3) + s(13) + s(2,1). Exercise. Try to realize the last decomposition explicitly. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 27

Hence [a, s] R (S ) = Z . C 3 (a2 = 1, as = s, s2 = 1 + a + s) ⊗2 Remark 2.57. More generally, if V is self-dual, irreducible representation, then (V )G 6= 0.

In general, M RC(Sn) = Zsλ |λ|=n and X sλsµ = gλµνsν, |ν|=n where gλµν are the Kronecker coeﬃcients and

λ µ ν Sn gλµν = dim(S ⊗ S ⊗ S ) because representations are self-dual. Remark 2.58. The Kronecker coeﬃcients are poorly understood in general, so it is hard to understand multiplication in a representation ring of Sn.

A different ring is better understood and more important. Define M M Λ = RC(Sn) = Zsλ. λ For |λ| = n, |µ| = m, we define h i s s = IndSn+m (Sλ ⊗ Sµ) . λ µ Sn×Sm

Exercise. This makes Λ a commutative, associative, unital ring with s(0) = [trivial representation of S0] as the unit.

Remark 2.59. The ring Λ is graded: sµ has degree |µ|.

Proposition 2.60. The ring Λ is a polynomial ring in the elements {s(n)}n≥1, where s(n) = [trivial representation of Sn].

Proof. Suppose λ = (λ1, . . . , λr), |λ| = n. We have that n s s . . . s = IndS (triv) = M λ = s + (sum of s ’s with µ λ). (λ1) (λ2) λr Sλ λ µ Thus we have upper triangular relations between the sλ’s and monomials in s(n)’s. X Proposition 2.61. We have that s(1)sλ = sµ. λ⊂µ |µ|=|λ|+1

M Proof. This follows from IndSn Sλ = Sλ, Pieri’s rule 2.49. Sn−1 λ⊂µ |µ|=|λ|+1 Remark 2.62. 28 ANDREW SNOWDEN

• General Pieri’s rule describes s(n)sλ. • Littlewood-Richardson rule describes sλsµ.

Question. Is Λ naturally the Grothendieck group of something?

Deﬁnition 2.63. A representation of S∗ is a sequence (Mn)n≥0, where Mn is a representation of Sn.

The category RepC(S∗) is an abelian category. Every ﬁnite-dimensional representation of S∗ λ is a direct sum of S ’s, a representations of S∗ concentrated in degree |λ|. We then have that fd K(RepC ) = Λ, which answers the above questions.

For M∗,N∗ ∈ Rep(S∗), deﬁne M (M ⊗ N ) = IndSn (M ⊗ N ). ∗ ∗ n Si×Sj i j i+j=n Fact 2.64. This deﬁnes a ⊗-product on Rep(S ), which gives K(Repfd) a ring structure, ∗ C with which K(Repfd) = Λ is a ring isomorphism. C fd Theorem 2.65. The category Rep∗ is the universal C-linear ⊗-category, i.e. given a C- linear ⊗-category A, the functor {⊗-functors Repfd(S ) → A} → A C ∗ F 7→ F(S1) is an equivalence of categories.

This is analogous to the statement ∼ {ring homomorphisms Z[x] → R} = R.

3. Young’s Rule, Littlewood–Richardson Rule, Pieri Rule

We start with considerations that are purely combinatorial and we later apply them to representation theory of symmetric groups.

Let µ = (µ1, µ2 ...) be a sequence of non-negative integers. A sequence of positive integers has type µ if it has exactly µi i’s.

Deﬁnition 3.1. Let a = (a1, a2 ...) be a sequence of positive integers. We deﬁne each term of a to be good as follows:

• All 1s in a are good. • An (i + 1) is good if and only if #{previous good i’s} ≥ #{previous good (i + 1)’s}.

If a term of a is bad if it is not good. Example 3.2. The sequence a = (1, 2, 1, 1, 2) has type (3, 2) and all terms are good. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 29

One way to think of good terms which will be useful later is to replace all the good i’s be left parenthesis and all the i + 1’s as right parenthesis. Then an i + 1 is bad if it does not complete a correct expression. Example 3.3. Consider the sequence a = (1, 2, 3, 1, 1, 3, 2) of type (3,2,2). To check which 2s are good, we consider the parentheses ()3((3). Thus all 2s are good in a. To check which 3s are good, we consider the parentheses 1()11)(. Thus the ﬁrst 3 is good but the second 3 is bad.

0 Deﬁnition 3.4. A pair of partitions is a pair (µ, µ ) where µ = (µ1, µ2,...) is a sequence of non-negative integers and µ0 is a non-increasing sequence of non-negative integers such that 0 0 0 µi ≤ µi. We write s(µ, µ ) for the set of type µ sequences such that #(good i’s) ≥ µi.

Note that:

• s(µ, 0) is the set of all type µ sequences, 0 0 0 • s(µ, µ ) = s(µ, (µ1, µ2, µ3,...)) because all 1’s are good, so we may (and will) always 0 assume µ1 = µ1. We represent (µ, µ0) graphically as follows: we draw µ like a tabloid with vertical lines 0 indicating µis, and add black circles to represent the empty spaces. Example 3.5. The pair (µ, µ0) with µ = (3, 2) and µ0 = (2, 1) can be represented as follows1

• • •

• •

0 0 Deﬁnition 3.6. Let (µ, µ ) be a pair of partitions. Let c > 1 be such that µc−1 = µc−1 and 0 µc < µc.

0 0 0 0 0 • Suppose µc < µc−1. Deﬁne Acµ to be µ but with cth entry changed to µc + 1. (We 0 think of Ac as Adding a box to µ in the cth row.) 0 0 0 0 This gives a new pair (µ, Acµ ). By convention (µ, Acµ ) = (0, 0) if µc = µc−1.

0 0 • Similarly, Rcµ is obtained from µ by changing µc to µc and µc−1 to µc−1 + (µc − µc). 0 (We think of Rc as Raising boxes in row c to the right of µ to row c − 1.) 0 This gives a new pair (Rcµ, µ ).

Example 3.7. The following diagram shows what the operations Ac and Rc do to some pairs (µ, µ0).

1The author apologizes for the quality of the pictures of pairs (µ, µ0). Making these proved more diﬃcult than excepted. If you a better way to produce these, please let me know! 30 ANDREW SNOWDEN

µ=(3,2) • • • R2 µ=(5,0) µ0=(3,0) • • • • • µ0=(5,0) • •

µ=(3,2) • • • R2 • • • • µ=(4,1) µ0=(3,1) µ0=(4,1) • • •

µ=(3,2) • • • µ=(3,2) • •

0 0 A priori, the images of applying R2 would have µ = (3, 0) and µ = (3, 1), but we always 0 normalize to assume that µ1 = µ1 Remark 3.8.

• Starting with any (µ, µ0) and applying a suﬃciently long sequence of A’s and R’s, we eventually get a pair (λ, λ). This is because applying both of the operators gets you closer to having all the boxes in the µ’ part.

• Given any (µ, µ0), there exists a sequence ν and a sequence of A’s and R’s that takes 0 (ν, (ν1, 0,..., 0)) to (µ, µ ). 0 0 0 0 Take ν = (µ1, . . . , µr, µ1 − µ1, . . . , µn − µn). First apply all the As to get the prime piece correct and the R’s to get extra boxes in the correct positions.

0 Theorem 3.9. Let (µ, µ ) be a set of partitions such that Ac and Rc are deﬁned. We have a bijection 0 0 0 s(µ, µ ) \ s(µ, Acµ ) → s(Rcµ, µ ) given by changing all the bad c’s to (c − 1)’s.

0 0 0 Proof. Let a ∈ s(µ, µ ) \ s(µ, Acµ ). Suppose Ac is non-trivial, so µc−1 = µc−1. Then the 0 number of good c − 1’s in a is at least µc−1 = µc−1, so it is exactly µc−1. Moreover, note 0 0 0 0 that a ∈ s(µ, Acµ ) if the number of good i’s in a is at least µi for i 6= 0, and the number 0 0 0 0 of good c’s in a is at least µc + 1. Hence an element a ∈ s(µ, µ ) \ s(µ, Acµ ) has to contain 0 exactly µc good c’s. To sum up, we have that:

• a contains µc−1 (c − 1)’s, all of which are good, 0 • a contains µc good c’s, 0 • a contains µc − µc bad c’s.

Let b be obtained from a by changing the bad c’s to (c − 1)’s. Then b has type Rcµ, since a 0 0 contained µc − µc bad c’s, and hence the number of (c − 1)’s in b is µc−1 + µc − µc, and the 0 number of c’s in b is µc. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 31

0 To prove that the map is well-deﬁned, we still have to show b has time (Rc, µ ). We will ﬁrst prove the following statement

(∗) For all j, #(good (c − 1)’s before jth term of b) ≥ #(good (c − 1)’s before jth term of a). by induction on j. It is clear for j = 1. Assume (∗) holds for j = i. The inequality is obviously true for j + 1 unless jth term is (c − 1) which is good in a and bad in b. But in this case, the inequality at j is strict because the number of (c − 2)’s before j is the same in a and b. Hence (∗) holds for j + 1, completing the induction.

0 Applying (∗), b has at least µc−1 good (c − 1)s, because a does, and all c’s in b are good. If 0 i 6= c, c−1, then an i in b is good if and only if it is good in a. This shows that b ∈ s(Rcµ, µ ). Hence the map is well-deﬁned.

0 0 Now, we will show that it is a bijection. Let a ∈ s(µ, µ ) \ s(µ, Acµ ) and b be its image. We think about replacing the (c − 1)’s in b with left parenthesis and c’s with right parenthesis. Since all c’s in b are good, all the right parentheses are matched. The sequence b now looks as follows

b0(b1(... (br where each bi is a correctly parenthesized expression, and “(” are the unmatched left parentheses.

0 We claim that the first µc − µc unmatched left parenthesis were flipped from a. Otherwise, we would create a new good c or destroy an existing good c. Indeed, in a we know that all the left parenthesis are good (since all (c − 1)’s are good). If a looked like a0(a1)a2, then we would not flip the right parenthesis around b1, because it corresponded to a good c. This shows that the map is injective, because it determines b from a.

0 If b ∈ s(Rcµ, µ ) is given, we say that a (c−1) is red if it is an unmatched and green otherwise. 0 0 Let a be the sequence obtained by changing the ﬁrst µc − µc red( c − 1)’s to c’s. This is the 0 0 0 candidate for the inverse image of b. We need to check that a ∈ s(µ, µ ) \ s(µ, Acµ ). We must show

(∗∗) Every c − 1 in a0 is good.

To do that, we proceed in the following steps.

(1) For any j, #(green( c − 1)’s before j) ≤ #(good (c − 1)’s before j). Suppose the jth term is red( c − 1). Then #(green( c − 1)’s before j) = #(c’s before j) ≤ #(good (c − 1)’s before j). The same argument applies to the entire sequence, so the inequality holds if j is maximal. Now, we complete the proof by argueing by descending induction on j, using the step above. (2) (Assume c > 2.) For every c − 1 in a0, #(previous good (c − 2)’s) > #(previous good (c − 1)’s). 0 0 As b ∈ s(Rcµ, µ ), b contains at most µc − µc bad (c − 1)’s. For any c − 1 in b, 32 ANDREW SNOWDEN

0 #(previous good (c − 2)’s) > #(previous (c − 1)’s) −(µc − µc). 0 Therefore, the inequality in the claim holds holds after the ﬁrst µc − µc red( c − 1)’s. 0 0 Suppose the jth term of a is c − 1 that is before the (µc − µc)th red c − 1. Then #((c − 1)’s before j in a0) = #(green( c − 1)’s before j in b) ≤ #(good (c − 1)’s before j in b) ≤ #(good (c − 2)’s before j in b) and the last two inequalities are strict if j is a good (c − 1) in b.

Then (2) proves (∗∗), and hence completes the proof.

λ Recall Λ is the free Z-module on {sλ}λ partition, where sλ is the class of S in the Grothendieck 0 ring. Given a pair (µ, µ ), deﬁne a linear endomorphism Eµ,µ0 of Λ by X Eµ,µ0 (sλ) = aνsν, λ⊂ν where aν is the number of ways to ﬁll the boxes of ν \ λ with numbers such that (a) rows are non-decreasing, (b) columns are increasing, (c) by reading rows top to bottom and right to left, we get a sequence in s(µ, µ0).

In particular, (c) shows that |ν| = |λ| + |µ|.

Lemma 3.10. Let µ be a partition. Then Eµ,µ(s0) = sµ.

Proof. We have that X Eµ,µ(s0) = aνsν. ν

We will show that aν = 0 for ν 6= µ and aµ = 1. Consider a filling of ν that satisfies (a)–(c). We show that the only possible filling has to have 1’s in the first row, 2’s in the second row, etc. Suppose some i appears in row j of ν and j < i, and take a minimal such i. Then no (i−1)’s appear above i by minimality, and none appear to to the right by (a). Therefore, this i is bad. Hence the sequence is not in s(µ, µ0), contradicting (c). We also know that i does not appear in row j with j > i by (b). Therefore, every i appears in the ith row. This implies that µ = ν and aν = 1.

Remark 3.11. Suppose µ = (µ1, . . . , µr). Then

Eµ,0(s0) = sµ1 . . . sµr . | {z } [M µ]

µ P This gives us the decomposition of M into irreducibles. We have that Eµ,0 = aνsν, and 0 aν is the number of ﬁllings of diagram ν such that (a) and (b) holds, and the number of i s is µi. This is the number of semi-standard (i.e. satisfying (a) and (b)) tableaux of shape ν and type µ. Thus this result is equivalent to the following statement. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 33

Theorem 3.12 (Young’s Rule). The representation M µ decomposes into irreducibles as M M µ = (Sν)⊕aν ,

where aν is the number of semistandard tableaux of shape ν and type µ.

We will only prove this theorem towards the end of this chapter. We ﬁrst use Theorem 3.9 to

0 0 0 prove that Eµ,µ = Eµ,Acµ + ERcµ,µ . This will allow us to prove the Littlewood-Richardson rule 3.14 using Young’s rule 3.12.

0 Proposition 3.13. Assume Ac and Rc are deﬁned on (µ, µ ). Then

0 0 0 Eµ,µ = Eµ,Acµ + ERcµ,µ .

Proof. Let |µ| = r and let λ, ν be partitions of n − r and n with λ ⊂ ν. Fill ν \ λ such that 0 0 we get a sequence in s(µ, µ ) \ s(µ, Acµ ). We claim that changing all the bad c’s to (c − 1)’s gives a filling satisfying (a)–(c) if and only if the original filling satisfies (a)–(c). This will prove the proposition by Theorem 3.9. Throughout the proof, we mark the bad c’s with red and the good c’s with green. Suppose first that (a)–(c) holds for the initial filling. There could be two possible problems for the modified filling.

• There could be a bad c to the right of a good c in the same row. c c This is impossible because a c to the left of a bad c cannot be good. • There could be a bad c in box (i, j) and c − 1 in box (i − 1, j). Let m − 1 be number of c’s following the c at (i, j).

c−1 c c c c c x Above each of these c’s is a c − 1 by (a) and (b). All of the (c − 1)’s are good by assumption.

≤c−1 c−1 c−1 c−1 c−1 c−1

c c c c c x When we hit box (i, j + m − 1), we have built up m good (c − 1)’s, so the next m c’s are good. Hence at (i, j) the c is actually good.

Now, suppose after changing bad c’s to (c − 1)’s, properties (a)–(c) hold. Let us show the original filling satisfied (a)–(c). Again, there could be two possible problems with the original filling.

• There could be a bad c to the left of a c − 1 in the same row. c c−1 But this cannot happen: the c − 1 is good, so the c is also good. • There could be a bad c directly above a good c, say (i − 1, j) and (i, j): 34 ANDREW SNOWDEN

c c We have that #((c − 1)’s to the left of (i − 1, j) in row i − 1) ≥ #(good c’s in ith row). Below every c − 1 in (i − 1)st row, we must have a c by (a) and (b), and they must be good because of the conditions (a) and (b) after the change. enough (c − 1)’s here c c’s (must all be good) c

This completes the proof. Theorem 3.14 (Littlewood-Richardson rule). We have that X sλsµ = Eµ,µ(sλ) = aνsν, λ⊂ν where aν is the number of semi-standard ﬁllings of ν \ λ such that the resulting sequence has type µ and all terms in it are good.

ν The coeﬃcients aν are sometimes written cλ,µ and called the Littlewood-Richardson coeﬃ- cients. Before giving the proof, we present a few special cases, one of which we will have to prove before we prove the theorem. Corollary 3.15 (Pieri rule). We have that X s(n)sλ = sν λ⊂ν where |ν| = |λ| + n and ν \ λ has no two boxes in the same column.

Note that Pieri rule is multiplicity free.

Example 3.16. If ν = and λ = , then

ν \ λ =

This is what we mean by ν \ λ having no two boxes in the same column. Example 3.17. We have that

· = + + +

Remark 3.18. Pieri rule 3.15 is much easier to use in practice than the Littlewood-Richardson rule 3.14, because the statement does not involve as much counting. It is easy to make mistakes when using the Littlewood-Richardson rule 3.14. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 35

Remark 3.19 (Alternative Pieri rule). We have that X s(1n)sλ = sν

λ⊂sν for |ν| = |λ| + n and ν \ λ has no two boxes in the same row.

Corollary 3.20 (Pieri rule II). We have that

µ X [M ]sλ = b(ν \ λ, µ)sν, λ\µ

where b(ν \ λ, µ) is the number of semi-standard ﬁllings of ν \ λ of type µ.

Proof. This follows from iterating the normal Pieri rule 3.15.

We now restate Pieri’s rule II as a lemma to Theorem 3.14. We will prove it using Young’s rule 3.12: µ X [M ] = b(ν, µ)sν, which we assume for now and prove later.

Lemma 3.21 (Pieri rule II). We have that

µ X [M ]sλ = b(ν \ λ, µ)sν, λ\µ

where b(ν \ λ, µ) is the number of semi-standard ﬁllings of ν \ λ of type µ.

Proof. We prove this by induction on λ (using the dominance order). We have that

X λ µ λ µ∪λ X b(ν, µ)[M ]sν = [M ][M ] = [M ] = b(ω, λ ∪ µ)sω. ν ω To show that this is equal to X b(ν, µ)b(ω, ν, λ)sω, ν,ω we just have to show that X (1) b(ω, µ ∪ λ) = b(ν, µ)b(ω \ ν, λ). ν Filling ω with type µ ∪ λ is the same as choosing a subset ν ⊆ ω and ﬁrst ﬁlling ω with type µ and then ω \ ν with type λ. If µ = (µ1, . . . , µr), λ = (λ1, . . . , λs), this is summarized by the following diagram 36 ANDREW SNOWDEN

numbers 1 to r numbers ν r + 1 to r + s

ω \ ν

This gives a bijection between the element on the left hand size and right hand side of equation (1), which shows that it holds.

Proof of Littlewood-Richardson rule 3.14. Let ν be a partition. We apply a long sequence of A’s and R’s to (ν, 0) to end up with (ω, ω)’s. By Proposition 3.13, X Eν,0 = aωEω,ω. ω ν Therefore, we have an upper-triangular change of basis. Write X Eµ,µ = bαEα,0 for bα ∈ Z. α We have that X Eµ,µ(sλ) = bαEα,0(sλ) X = bαEα,0(s0) sλ

= Eµ,µ(s0)sλ

= sµsλ by Lemma 3.10

This gives the result.

λ µ λ µ We ﬁnally prove Young’s rule 3.12, i.e. the multiplicity of S in M , which is dim HomSn (S ,M ), is the number of semistandard tableau of shape λ, type µ. Let T (λ, µ) = {tableaux of shape λ and type µ}. n Fix a tableau t of shape λ and type 1 (which labels the boxes of λ). This makes Sn permute the boxes of λ, and hence we get an Sn action on T (λ, µ).

The action is transitive and the stabilizer of an element on Sµ is µ ∼ M = C[T (λ, µ)]. We say that T,T 0 ∈ T (λ, µ) are row equivalent if T = σT 0 for some σ ∈ Rt (and similarly for column equivalent). MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 37

λ µ Deﬁnition 3.22. Let T ∈ T (λ, µ) and deﬁne θT : M → M to be the unique Sn-equivariant map taking X {t} 7→ T 0, T 0 row equiv. to T

and θbT = (θT )|Sλ . Theorem 3.23. The set n o θbT | T is semistandard forms a basis for λ µ HomSn (S ,M ). Remark 3.24. This theorem implies Young’s rule 3.12.

Write [T ] for the column equivalence class of T . Deﬁne an order on these equivalence classes generated by letting [T1] [T2] if [T2] is obtained from [T1] by switching a and b where a < b and the columns of a is to the right of the column of b. Remark 3.25. If T is semistandard and T 0 is row equivalent to T , then [T 0] [T ]. Lemma 3.26. The set n o θbT | T is semistandard is linearly independent.

X 2 0 Proof. Suppose aT θbT = 0 is a nontrivial linear relation. Let [T ] be maximal with T sstd aT 0 6= 0. We have that ! X 0 00 0 00 aT θbT ({t}) = aT 0 T + (linear combination of T s.t. [T ] 6 [T ]), T sstd and hence applying κt ! X 0 00 0 00 aT θbT (et) = aT 0 κtT +(linear combination of κt(T ) s.t. [T ] 6 [T ]) 6= 0, |{z} T sstd 6=0 which is a contradiction.

Now, we have an injective map

M λ µ S ⊗ (span of θbT with T semistandard) ,→ M . λ

Remark 3.27. To show that the θbT ’s with T semistandard span, it suﬃces to show that the dimensions above are equal. We have that

2We write sstd for semistandard in the summation. 38 ANDREW SNOWDEN

[ standard tableau of semistandard tableau of LHS has a basis shape λ shape λ type µ λ

RHS has a basis tabloids of shape µ

The RSK (Robinson–Schensted–Knuth) correspondence gives an explicit bijection. This argument will not be presented here.

λ λ Lemma 3.28. Let θb ∈ HomSn (S ,M ) 6= 0. Put X θb(et) = cT T. T ∈T (λ,µ) Then

0 (1) cT 0 = 0 if T has a repeat in some column, 0 (2) there exists a semistandard T with cT 0 6= 0.

Proof. For (1), suppose that T 0 has a repeat in the boxes labeled by i and j in t in the same column. Then X X cT (i j)T = (i j)θb(et) = θb((i j)et) = −θb(et) = − cT T.

0 0 Since (i j)T = T , cT 0 = −cT 0 .

To show (2), we note that, more generally, if σ ∈ Ct, then cσT = ±cT for any T . Let [T1] be maximal with cT1 6= 0. By the above, we can assume that the columns of T1 are increasing. We want to show that the rows of T1 are semistandard. Assume not, so say there exists a decrease in row q, going from column j to column j + 1. We have the following situation

a1 b1

a2 b2 . . . .

aq bq ...... br . . . .

with aq > bq and a1 < . . . , < as and b1 < . . . < br. Let X be the blue set and Y be the red set of T1. Consider the Garnir relation 2.35 for X ∪ Y : X (sgn σ)σet = 0.

σ∈SX∪Y /SX ×SY MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 39

Apply θb to get X X CT (sgn σ)σT = 0.

T ∈T (λ,µ) σ∈SX∪Y /SX ×SY

Consider the coeﬃcient of T1 in this sum. We know it is 0. Hence there exists a nonzero term canceling c1T1, and hence there is a σ ∈ SX∪Y /SX × SY such that cσT1 6= 0. But [T1] [σT1], a contradiction. Lemma 3.29. The set n o θbT | T is semistandard spans Hom(Sλ,M µ).

Proof. Consider a nonzero θb ∈ Hom(Sλ,M µ), and write X θb(et) = cT T. T ∈T (λ,µ)

Let T1 be semistandard and [T1] be maximal with cT1 6= 0. Then

θbT1 (et) = κtT1 +(smaller things), |{z} coeﬀ. of T1 is 1 and hence in

(θb− cT1 θbT1 )(et)

the coeﬃcient of T1 is 0, and we have modiﬁed by elements T1. We can now continue inductively to prove the assertion.

Altogether, this completes the proof of Young’s rule 3.12.

3.1. Hook length formula. Deﬁnition 3.30. The (i, j) hook of a partition λ consists of the box at (i, j) and all boxes directly below or directly right of it. The hook length at (i, j) is the number of boxes in the hook at (i, j). Example 3.31. For λ = (5, 3), (i, j) = (1, 2) the hook is marked in blue below

and the hook length is 5. Deﬁnition 3.32. A boder strip (or skew hook) of λ is a shape of the form λ \ µ such that it is connected and contains no 2 × 2 box.

Not that is not connected.

Example 3.33. A border strip of λ = (5, 3) is marked in red below 40 ANDREW SNOWDEN

Note that there is a bijection between the hooks and the border strips, and this preserves the number of boxes. We know that λ 0 [M ] = sλ + (sum of sµs with µ λ), where the sum is determined by Young’s rule 3.12. Hence there is an upper-triangular change λ of basis between [M ] = s(λ1) . . . s(λr) and sλ if λ = (λ1, . . . , λr). λ Now we want to invert this and express sλ in terms of the [M ]’s. Theorem 3.34. We have that

sλ = det ((sλi−i+j)1≤i,j≤r)

where s(k) = 0 if k < 0, s(0) = 1.

Proof. Let A(λ) be the matrix (sλi−i+j)1≤i,j≤r. We will compute det(A(λ)) by Laplace expression in ﬁnal column. Consider the matrix obtained from A(λ) by deleting the ﬁnal column and kth row. The (i, j) entry of this matrix is

s(λi−i+jif i < k

s(λi+1−1−i+j)if i ≥ k and hence this is the matrix

A(λ1, . . . , λk−1, λk+1 − 1, . . . , λr − 1). This partition is obtained from λ by deleting the kth row and subtracting 1 from all the following rows. Thus this corresponds to deleting the hook of λ at (k, 1), and shifting the boxes up to get a Young diagram.

This is the same as removing the corresponding border strip βk.

Hence this matrix is det(λ \ βk). We have that

det(A(λ)) = det(A(λ \ βr))s(λr) − det(A(λ \ βr−1))sλr−1+1 + · · · ± det(A(λ \ β1))sλ1+r−1

= sλ\βr shr,1 − sλ\βr−1 shr−1,1 + · · · ± sλ\β1 shr,1 ,

by induction on the size of the partition, and writing |βk| = hk,1. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 41

We compute the products using Pieri rule 3.15. We show that sλ appears once, everything else cancels.

First, sλ appears in sλ\βr shr,1 with multiplicity 1 and does not appear in any other term because βk with k < r has boxes in the same column.

Let Xk be the diagrams in sλ\βk s(hk,1) that contain ﬁnal box in row k of λ, and let Yk be the complement of Xk. Then Xr = {λ} and Y1 = ∅.

⊗

One can now show that Xk−1 = Yk. This shows that everything but Xr cancels, we we get sλ.

Corollary 3.35. For a partition λ = (λ1, . . . , λn), 1 dim Sλ = n! det , (λi − i + j)! 1≤i,j≤r 1 where m! = 0 if m < 0.

Proof. Consider the additive function

ϕ:Λ → Q dim V V 7→ n!

for any representation V of Sn.

We show that this is a ring homomorphism. If V is a representation of Sn and W is a representation of Sm, then h i [V ][W ] = IndSn+m (V ⊗ W ) , Sn×Sm and the dimension is (n + m)! (dim V )(dim W ) . n!m! Hence (dim V )(dim W ) (n+m)! ϕ([V ][W ]) = n!m! = ϕ([V ])ϕ([W ]). (n + m)!

To get the result, apply ϕ to sλ = det(sλi−i+j) and note that s(m) is the trivial representation 1 of Sm, so ϕ(s(m)) = m! . 42 ANDREW SNOWDEN

Example 3.36. Let λ = (3, 2). Then s(3) s(4) s(3,2) = det = s(2)s(2) − s(4)s(1) = s(3,2) + s(4,1) + s(5) − s(5) − s(4,1), s(1) s(2) which agree with Theorem 3.34. By Corollary 3.35, the dimension is 1 1 1 1 5! (3,2) 3! 4! dim S = 5! det 1 = 5! − = = 5. 1 2! 12 24 24 Theorem 3.37 (Hook length formula). For a partition λ, n! dim Sλ = . Q(all hook lengths) Example 3.38. We have the following hook lengths at every box of λ = (3, 2)

4 3 1 2 1

and hence the dimension is 5! dim Sλ = = 5. 4 · 3 · 1 · 2 · 1 This agrees with Example 3.36.

Proof of Theorem 3.37. We take r = 3 for exposition, the general case is treated the same way. By Corollary 3.35, it is enough to show that 1 1 det = Q . (λi − i + j)! (hook lengths)

Note that λi − i + j = hi,1 + j − r. We have that

 1 1 1  h (h − 1) h 1 (h11−2)! (h11−1)! (h11)! 1 11 11 11  1 1 1  h (h − 1) h 1 det (h21−2)! (h21−1)! (h21)! = det  21 21 21   1 1 1  h11!h21!h31! h31(h31 − 1) h31 1 (h31−2)! (h31−1)! (h31)! (h − h )(h − h )(h − h ) = 11 21 11 31 21 31 h11!h21!h31! which shows the result.

4. Representations of GLn(C)

Everything in this chapter is over C.

Definition 4.1. An m-dimensional representation of GLn(C) is algebraic if the homomorphism GLn(C) → GLm(C) induced by a map of varieties GLn → GLm. An infinite dimensional representation is algebraic if it is a union of finite-dimensional representations. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 43

Remark 4.2. Equivalently, an algebraic representation of GLn is a comodule over C[GLn], the coordinate ring of GLn. To define a comodule, first note that an algebra A comes with a map A ⊗ A → A, and a coalgebra will be equipped with a map A → A ⊗ A satisfying similar axioms. A module over an algebra comes with a map A ⊗ M → M, whereas a comodule comes with a map M → A ⊗ M. The reason the arrows are flipped is that O(−) is a contravariant functor. Example 4.3. The following representations are algebraic:

n • C , the standard representation of GLn, • (Cn)∗, det, n n ∗ Vk n • any ⊗-construction on algebraic representations; e.g. C ⊗ (C ) , (C ), ...

∗ 1 A non-algebraic representation of GL1(C) = C = S ⊗ R is 1 y (x, y) 7→ . 0 1

Theorem 4.4. Algebraic representation of GLn are semisimple.

Proof. Weyl’s unitary trick. Let Un ⊆ GLn(C) be the unitary group. Then Un is compact, so there is a Haar measure dg on Un.

Let V be an algebraic representation of GLn and let W ⊆ V be a subrepresentation. Let h, i0 be a hermitian form on V . We average it out over Un: deﬁne Z hv, wi = hgv, gwi0dg, Un a non-degenerate hermitian form on V which is Un-invariant. If W ⊥ is the orthogonal space to W with respect to h , i. Then V = W ⊕ W ⊥ ⊥ because h i is Un-invariant, W is Un-stable. ⊥ Since Un is Zariski-dense in GLn, any Un-stable subgroup of V is GLn-stable, so W is a GLn-subrepresentation. Remark 4.5. Weyl’s trick can be summarized as follows: semisimplicity can be transferred from a Zariski-dense subgroup to the whole group.

Now, we just need to classify the simple algebraic representations of GLn. The ﬁrst step is the following proposition.

Proposition 4.6. Suppose V is a simple algebraic representation of GLn. Then there exists n ⊗k ∼ m m ∈ Z and k ∈ N and a simple subrepresentation of (C ) such that V = W ⊗ det .

Proof. Let v ∈ V . Then we get an algebraic function GLn → V mapping g 7→ gv. Then ∗ V → C[GLn] λ 7→ (g 7→ λ(gv)) 44 ANDREW SNOWDEN

is GLn-equivariant. Hence algebraic irreducible is contained in C[GLn] with left regular action. Since

GLn = {m ∈ Mn | det(m) 6= 0}, we C[GLn] is the localization at det: 1 [GL ] = [M ] . C n C n det

Thus if V is an irreducible algebraic representation of GLn, it is ﬁnite-dimensional, and hence m det ⊗V ⊆ C[Mn] for m 0. Since n n ∗ Mn = C ⊗ (C ) , we have that n n ∗ M k n n ∗ C[Mn] = Sym(C ⊗ (C ) ) = Sym (C ⊗ (C ) ). k≥0 Finally, k n n ∗ n n ⊗k M n ⊗k Sym (C ⊗ (C ) ) ⊆ (C ⊗ (C )) = (C ) , m n ⊗k and hence det ⊗V is contained in one of these piecies, (C ) for some k.

n ⊗k It is hence enough to understand the structure of (C ) . This is a representation of GLn×Sk, where Sk permutes the ⊗-factors.

Deﬁnition 4.7. Given a partition λ of k, then we deﬁne an algebraic GLn-representation n λ n ⊗k Sλ(C ) to be the multiplicity space of S in (C ) , i.e. n λ n ⊗k Sλ(C ) = HomSk (S , (C ) ).

Let `(λ) be the number of parts of λ, i.e. the number of rows in the Young diagram. Theorem 4.8.

n • The representation Sλ(C ) is 0 if `(λ) > n, and irreducible if `(λ) ≤ n n ∼ n • If λ 6= µ and both have ` ≤ n, then Sλ(C ) 6= Sµ(C ). n m Corollary 4.9. Every irreducible algberaic representation of GLn has the form Sλ(C )⊗det for some λ, m ∈ Z.

n ⊗k M λ n n Proof. This is because (C ) = S ⊗Sλ(C ) (because Sλ(C ) is deﬁned as a multiplicity λ space), and it follows from Theorem 4.8. n Remark 4.10. Note that λ and m are no quite unique: for example, S1n (C ) = det = n S0(C ) ⊗ det. n Lemma 4.11. For any λ, Sλ(C ) is 0 or irreducible.

Proof. We have that n ⊗k M λ n (C ) = S ⊗ Sλ(C ). λ MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 45

Hence n ⊗k M n EndSk ((C ) ) = End(Sλ(C )), λ and n ⊗k n ⊗ Sk n ⊗k Sk EndSk ((C ) ) = End((C ) ) = (End(C ) ) . This is spanned as a C-vector space by elements of the form a ⊗ · · · ⊗ a for a ∈ End(Cn).

For 1 ≤ i ≤ k, let xi(a) = 1 ⊗ · · · ⊗ a ⊗ · · · ⊗ 1 with a in the ith position. Then note that

a ⊗ · · · ⊗ a = x1 ⊗ · · · ⊗ xk.

Sk The ring of symmetric polynomials C[x1, . . . , xk] is generated as a C-algebra by elements r r n ⊗k Sk of the form x1 + ··· + xk for all r ≥ 0. Hence power sums in xi(a)’s generate (End(C ) ) r r as a C-algebra. Since xi(a) = xi(a ), we only need the ﬁrst powers. Hence the elements n x1(a) + x2(a) + ··· + xn(a) for a ∈ End(C ) = gln (Lie algebra) generated the invariant n ⊗k algebra. Note that the element x1(a) + ··· + xk(a) is the action of a ∈ gln on (C ) . Therefore, we have shown that the map n ⊗k U(gln) → EndSk ((C ) ) is surjective, where U(g) is the universal enveloping algebra: T (g) U(g) = , (X ⊗ Y − Y ⊗ X) where T (g) is the tensor algebra. Thus n U(gln) End(Sλ(C )). n n Hence Sλ(C ) is either 0 or irreducible as a representation of gln. Thus Sλ(C ) is either 0 or an irreducible GLn-representation. n Lemma 4.12. If n ≤ `(λ) then Sλ(C ) = 0.

Proof. We need to show that λ n ⊗k HomSk (S , (C ) ) = 0, λ n ⊗k so consider a map f : S → (C ) . It suﬃces to show that f(et) = 0 for any tableau t. Recall that et = κt{t} and κtet = |Ct|et. Hence

1 κt f(et) = f(κtet) = f(et). |Ct| |ct| n ⊗k n Thus it suffices to show that κt(C ) = 0. Let e1, . . . , en be a basis of C . Then ei ⊗ ej is a n n ⊗k basis for C , so ei1 ⊗ · · · ⊗ eik for i1, . . . , ik ∈ {1, . . . , n} is a basis for (C ) . Consider some basis vector. Since `(λ) > n, there exist distinct numbers j and j0 in the first column of t 0 such that ij = ij0 . Then the basis vector is fixed by (j j ), and hence it is killed by κt.

n The algebraic irreducibles of Gm = GL1 are 1-dimensional and given by z 7→ z for n ∈ Z. n Similarly, the irreducibles of Gm are 1-dimensional and given by

a1 an (z1, . . . , zn) 7→ z1 . . . zn n for a = (a1, . . . , an) ∈ Z . 46 ANDREW SNOWDEN

n Deﬁnition 4.13. The group Gm is called a torus and its 1-dimensional algebraic representations are called weights.

n Let V be an algebraic representation of GLn ⊇ Gm (diagonal matrices). Then M V = Va, a∈Zn

n where Va is the isotypic piece for V|Gm corresponding to the weight a. Then Va is called the weight space for a.

k n Lemma 4.14. Let λ be a partition of k. Suppose `(λ) ≤ n. The 1 -weight space of Sλ(C ) λ is naturally isomorphic to S as a representation of Sk.

n Remark 4.15. Recall that Sn ⊆ GLn as permutation matrices and Sn normalizes Gm, so k Sn acts on the weights. In this case, 1 is ﬁxes by Sk ⊆ Sn, so Sk acts on the weight spaces.

k n Proof of Lemma 4.14. The 1 -weight space of Sλ(C ) is λ k n ⊗k HomSk (S , (1 -weight space of (C ) )). n ⊗k A basis element ei1 ⊗ · · · ⊗ eik of (C ) has weight a = (a1, . . . , an) where aj is the number k n ⊗k of is equal to j. Thus the 1 -weights space of (C ) has a basis ei1 ⊗ · · · ⊗ eik where

{i1, . . . , ik} = {1, . . . , k}.

This has an action of Sk × Sk, where the ﬁrst Sk is the one we Hom over and it acts by permuting the ⊗-factors, and the second Sk is the subgroup of GLn and it acts by relabeling the indices. As an Sk × Sk-set, the set of basis vectors looks like Sk with two Sk’s acts by k n ⊗k right and left multiplication. Hence the 1 -weight space of (C ) is isomorphic to C[Sk] with the usual action of Sk × Sk. Thus λ k n ⊗k ∼ λ HomSk (S , (1 -weight space of (C ) )) = S . This completes the proof. n Corollary 4.16. If `(λ) ≤ n, then Sλ(C ) is irreducible.

n Proof. By Lemma 4.14, Sλ(C ) 6= 0, and hence it is irreducible by Lemma 4.11. n ∼ n Corollary 4.17. If λ 6= µ and ` ≤ n, then Sλ(C ) 6= Sµ(C ).

This completes the proof of Theorem 4.8. Proposition 4.18. For |λ| = k, |µ| = `,

M ν n n ∼ n cλ,µ Sλ(C ) ⊗ Sµ(C ) = Sν(C ) , ν

ν where cλ,µ are the Littlewood-Richardson coeﬃcients (see the Littlewood-Richardson rule 3.14). MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 47

Proof. We have that λ n ⊗k µ n ` ∼ λ µ n ⊗(k+`) HomSk (S , (C ) ) ⊗ HomSk (S , (C ) ) = HomSk×S` (S ⊗ S , (C ) ) ∼= Hom (IndSk+` (Sλ ⊗ Sµ), ( n)⊗(k+`)) Sk+` Sk×S` C (by Frobenius reciprocity) M ν ∼ ν ⊕cλ,µ n ⊗k+` = HomSk+` ( (S ) , (C ) ) ν (by Littlewood-Richardson rule 3.14)

M n ⊕cν = Sν(C ) λ,µ ν which completes the proof. n n Corollary 4.19. We have that Sλ(C ) ⊗ det = Sλ+1n (C ).

n n n Proof. Recall that S1n (C ) = det and Sλ(C ) ⊗ S1n (C ) decomposes according to the Pieri n rule 3.15. Since Sν(C ) = 0 if `(ν) > n, we must put the n boxes in the ﬁrst n rows. n Corollary 4.20. Every algebraic irreducible representation of GLn has the form Sλ(C ) ⊗ m det where λn = 0 and m ∈ Z and λ, m are unique.

Therefore, we have established a bijection

   λ   irreducibles of GLn   irreducibles S  appearing in (Cn)⊗k of Sk  for some k   with `(λ) ≤ n 

This motivates the following deﬁnition.

Deﬁnition 4.21. A representation of GLn is polynomial if it can be realized as a subquotient of a direct product of tensor powers of the standard representation.

The condition “appearing in (Cn)⊗k for some k” in the bijection above reduces to saying they are polynomial. Examples 4.22.

n • The representations Sλ(C ) are polynomial. • The representation (Cn)∗ is not polynomial.

Remark 4.23. The following are equivalent for m-dimensional representation of GLn: (1) V is polynomial, (2) V is algebraic and all weights in V are non-negative, (3) in the homomorphism GLn → GLm the entries of GLm are polynomials in the entries of GLm L Deﬁnition 4.24. A polynomial representation of GL∞ = n≥1 GLn is one appearing as a ∞ S n subquotient of ⊕ of ⊗-powers of C = n≥1 C . pol We write Rep (GL∞) for the category of polynomial representations. 48 ANDREW SNOWDEN

pol ∞ Theorem 4.25. The category Rep (GL∞) is semi-simple and its simple objects are Sλ(C ) = λ ∞ ⊗k HomSk (S , (C ) ) for all partitions λ.

Proof. We have that ∞ [ n Sλ(C ) = Sλ(C ). n≥1 ∞ n Suppose V ⊆ Sλ(C ) is non-zero and GL∞-stable. For n 0, V ∩ Sλ(C ) is non-zero and n ∞ GLn-stable, so Sλ(C ) ⊆ V , whence V = Sλ(C ). Hence Sλ(C∞) is simple. Moreover, ∞ ⊗k M λ ∞ (C ) = S ⊗ Sλ(C ), λ ∞ ⊗k so (C ) is a semisimple representation of GL∞. Any direct sum, subrepresentation or pol quotient of semisimple objects is semisimple, hence every object of Rep (GL∞) is semisimple. Remark 4.26. We have that

λ λ ∗ Sk λ Sk HomSk (S , −) − ((S ) ⊗ −) = (S ⊗ −) , and sometimes it is preferable to use the latter form rather than the former, because it is covariant. Theorem 4.27 (Schur–Weyl duality). We have an equivalence of categories

pol Rep(S∗) Rep (GL∞)

M ∞ ⊗k Sk (Mk)k≥0 (Mk ⊗ (C ) ) k≥0

n (1 weight space of V )n≥0 V

which is compatible with ⊗-products (using induction ⊗-product on the left hand side).

Proof. Both sides are semisimple abelian categories, so we just need to check that the bijective functor between the two sides sends simple objects to simple objects, which we showed above.

We deﬁne a third category, which will also be equivalent to the above two. For any vector ⊗n λ Sn space V , put Sλ = (V ⊗ S ) . This deﬁnes a functor

f Sλ : Vec → Vec, called a Schur functor. We will deﬁne a class of functors from Vec to Vec which are the most natural consider in this context, called polynomial functors. Note ﬁrst that the set of functors Vec → Vec forms MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 49

an abelian category. Let f Tn : Vec → Vec V 7→ V ⊗n. Deﬁnition 4.28. A functor F : Vecf → Vec is polynomial if it ccurs as a subquotient (in the functor category) of direct sum of Tn’s. Let P be the abelian category of polynomial functors.

Exercise. The functors TnSλ are polynomial. Remark 4.29. Consider a functor F : Vecf → Vecf . Then the following are equivalent:

(1) F is a subquotient of ⊕ of (Tn)|Vecf , (2) for any V,W ∈ Vecf , F : Hom(V,W ) → Hom(FV, FW ) is a polynomial function.

Theorem 4.30. The category P is semisimple and the Schur functors Sλ are simple.

Proof. We ﬁrst prove that Sλ is simple. Suppose F ⊆ Sλ is a nonzero subobject. There n n n n exists an n such that F (C ) 6= 0. Then F (C ) = Sλ(C ), because Sλ(C ) is irreducible as a n m GLn-representation. Suppose m > n. We have the standard inclusion i: C → C and the standard projection p: Cm → Cn, with p ◦ i = id. Then the diagram

n = n F (C ) Sλ(C )

F (i) Sλ(i)

m m F (C ) Sλ(C )

commutes. Clearly, F (i) is injective, because F (p) ◦ F (i) = F (p ◦ i) = id. Hence F (Cm) 6= 0 m m and so F (C ) = Sλ(C ). m m m Now suppose m < n. If m < `(λ), then Sλ(C ) = 0, so F (C ) = Sλ(C ). So suppose `(λ) ≤ m < n. Then the diagram

n = n F (C ) Sλ(C )

F (p) Sλ(p)

m m F (C ) Sλ(C )

m commutes. Since Sλ(p) ◦ Sλ(i) = id, Sλ(p) 6= 0. Thus F (p) 6= 0, and hence F (C ) 6= 0, and m n hence F (C ) = Sλ(C ). This proves Sλ is simple.

Note that Sn acts on Tn in the category P. Then M λ Tn = S ⊗ Sλ |λ|=n 50 ANDREW SNOWDEN holds functorially, and hence Tn is a direct sum of simples, and hence it is semisimple. Any ⊕ or subquotient of semisimple objects is semisimple, so P is semisimple.

To summarize, we have the following equivalences of categories

n (Mn)n≥0 Rep(S∗) (1 weight space in V )n≥n

M ⊗n Sn pol (V 7→ (Mn ⊗ V ) ) P Rep (GL∞) V n≥0

F F ( ∞) = lim F ( n) C −→ C

f We have an analogy between Vec and A1, summarized by the following table

f Vec is an abelian ⊗-category A1 is a ring variety P are its automorphisms C[t] and its automorphisms (1) Given V ∈ Vec, evv : F 7→ F (V ) (1) Given a ∈ C, eva : C[t] → C (2) F,G ∈ P, F ◦ G ∈ P (2) f, g ∈ C[t], f ◦ g ∈ C[t] (3) P is a symmetric abelian ⊗-category with (3) C[t] is a ring (F ⊗ G)(V ) = F (V ) ⊗ G(V ) (4) have coaddition P → P2 given by F 7→ (4) C[t] has coaddition t 7→ t ⊗ 1 + 1 ⊗ t and ((V,W ) 7→ (V ⊕ W )) and comultiplication comultiplication t 7→ t ⊗ t P → P2 given by F 7→ ((V,W ) 7→ (V ⊗ W ))

In (4) above, we deﬁne Pn as the category of polynomial functors (Vecf )n → Vec similarly to P, with

⊗k1 ⊗kn Tk1,...,kn (V1,...,Vn) = V1 ⊗ · · · ⊗ Vn .

n Fact 4.31. The category P is semisimple category which objects are Sλ1 ⊗ · · · ⊗ Sλn .

Two interesting questions to ask are

(1) Describe how the extra structure described above works on Sλ’s. pol (2) Describe how the extra structure works on Rep(S∗) or Rep (GL∞). MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 51

Let us start with describing coaddition on Sλ with |λ| = n. In other words, consider the functor (V,W ) 7→ Sλ(V ⊕ W ). We have that

λ ⊗n Sλ(V ⊕ W ) = HomSn (S , (V ⊕ W ) ) ! λ M = HomSn S , (⊗-strings with i V ’s and j W ’s) i+j=n ! M = Hom Sλ, IndSn (V ⊗i ⊗ W ⊗j) Sn Si×Sj i+j=n M = Hom (ResSn (Sλ),V ⊗i ⊗ W ⊗j) Si×Sj Si×Sj i+j=n M M λ µ ν ⊗i ⊗j = Cµ,ν ⊗ HomSi×Sj (S ⊗ S ,V ⊗ W ) i+j=n µ,ν and hence

M ⊕gλ,µ,ν Sλ(V ⊗ W ) = (Sµ(V ) ⊗ Sν(W )) , µ,ν |µ|=|ν|=|λ| where gλ,µ,ν are the Kronecker coeﬃcients.

n M Exercise. Show that Sym (V ⊗ W ) = Sλ(V ) ⊗ Sλ(W ), which is known as the Cauchy |λ|=n rule.

Decomposing Sλ ◦ Sν is called the plethysm problem, and it is more or less impossible to do in general.

n 2 M Exercise. Sym ◦ Sym = Sλ |λ|=2n all points of λ even If A is an abelian symmetric ⊗-category, we can do commutative algebra within A. For example, a commutative algebra in A is an object A of A with multiplication m: A⊗A → A satisfying the usual rules. For example, it is commutative

A ⊗ A

symmetry m of ⊗ A ⊗ A m A

Similarly, we can talk about A-modules in A. Of course, general abelian symmetric ⊗-categories are diﬃcult to study, so we specialize to ∼ pol A = Rep(S∗) = Rep (GL∞).

Deﬁnition 4.32. A twisted commutative algebra (tca) is a commutative algebra in Rep(S∗). 52 ANDREW SNOWDEN L A tca A is a graded vector space A = n≥0 An with an action of Sn on An. We have multiplication A ⊗ A → A given by M IndSn (A ⊗ A ) → A , Si×Sj i j n i+j=n which is the same as giving Si × Sj-equivariant maps

Ai ⊗ Aj → Ai+j for any i and j. These maps make A into an associative, unital graded C-algebra. The commutativity law is

m x ⊗ y Ai ⊗ Aj Ai+j

σ m y ⊗ x Aj ⊗ Ai Ai+j where σ is a twist given by σ(1) = j + 1, . . . , σ(i) = i + j, σ(i + 1) = 1, . . . , σ(i + j) = j. This justiﬁes the name twisted commutative algebra. ∼ pol Under the equivalence Rep(S∗) = Rep (GL∞) tca’s correspond to commutative associative unital C-algebras with polynomial action of GL∞. ⊗n Example 4.33. Let U be a vector space and A = T (U) be the tensor algebra, i.e. An = U with the permutation action of Sn. Then A is a tca. In fact, A = Sym(U ⊗ S(1)). Under the Schur-Weyl duality 4.27, A corresponds to Sym(U ⊗ C∞).

Deﬁnition 4.34. A tca A is noetherian if ModA is locally noetherian, i.e. every object is lim of noetherian objects. −→ Proposition 4.35. If dim(U) < ∞, then A = T (U) is noetherian as a tca.

Sketch of a proof. Let M be a ﬁnitely generated A-module. Then M is a quotient of A ⊗ V for some ﬁnite length object V in Rep(S∗). (The objects A ⊗ V are the projective objects in this category.) We have that M ∞ A = Sλ(U) ⊗ Sλ(C ) `(λ)≤d by Cauchy rule. Hence all partitions in A have ≤ d rows. Hence all partitions in M have ≤ N rows for some N. Now, think of Schur functors. Note that if L ⊆ L0 ⊆ M and L(CN ) = L0(CN ), then L = L0. This is because we can decompose these as multiplicity spaces and they all have to agree. Therefore, there exists an order-preserving injection N N {submodules of M} → {A(C )-submodules of M(C )}. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 53

Since A(CN ) is a finitely-generated C-algebra, M(CN ) is a finitely-generated module over M(CN ), and hence the right hand side satisfies the ascending chain condition. Hence the left hand side satisfies the ascending chain condition.

Let us specialize to dim U = 1. Then A = C[t] and Sn acts trivially on An. An A-module 3 is a sequence (Mn) in Rep(S∗) with transition maps Mn → Mn+1 which are Sn-equivariant. These sequences have to satisfy a few extra conditions which we omit here. We present an alternative way to view A-modules in this case. Definition 4.36. Define FI to be the category whose objects are finite sets and maps are injections between the sets. An FI-module is a functor FI → Vec.

(The notion of FI-modules was introduced by Church–Ellenberg–Farb around 2012.) If M is an FI-module, then

Mn = M([n]) if an Sn-representation, where [n] = {1, . . . , n} and the inclusion [n] → [n+1] induces a transition map Mn → Mn+1. ∼ Fact 4.37. We have an equivalence of categories ModC[t] = ModFI .

Every A-module is a quotient of direct sums of A ⊗ Sλ. We can decompose this using the Pieri rule 3.15. The following shapes appear in Pieri rule 3.15 for λ:

Let Lλ be the sum of all of these representations in A ⊗ S6λ. λ Fact 4.38. The representation Lλ is an A-quotient of A ⊗ S .

Fact 4.39. If M is a finitely-generated A-module, then there exists a filtration 0 = F0 ⊂ ∼ · · · ⊂ Fn = M such that Fi/Fi−1 = Lλ (up to finite error, i.e. truncating to ≥ n on either side).

Deﬁnition 4.40. Let M, N be A-modules. Let M N be deﬁned by (M N)n = Mn ⊗Nn.

Note that A A = A, so if M, N are A-modules, so is M N. Fact 4.41. If M,N are ﬁnitely-generated, so is M N.

3 These correspond to multiplication by t ∈ C[t], and it is enough to say what the generator does. 54 ANDREW SNOWDEN

The idea is to reduce to the case M = A ⊗ C[Sm] and N = A ⊗ C[Sn]. For these modules, HomA(M, −) = (−)m, so these are analogs of free modules. In this case, we explicitly compute.

ν λ µ Recall the Kronecker coefficient gλ,µ,ν is the multiplicity of S in S ⊗ S , where |λ| = |µ| = |ν|. We define λ[n] = (λ1 + n, λ2 + n, . . .). We mentioned before that these coefficents are really hard to understand. However, there are partial results towards understanding them, the first of which is the following theorem.

Theorem 4.42 (Murnaghan). Given any λ, µ, ν, gλ[n],µ[n],ν[n] is constant for n 0.

Remark 4.43. The eventual value is called the stable Kronecker coeﬃcient Gλ,µ,ν.

Instead of presenting the original proof by Murnaghan, we show an easier proof by Church– Ellenberg–Farb which uses the introduced theory.

Proof. Since Lλ Lµ is a ﬁnitely-generated A-module, we have that

[Lλ Lµ] = [Lν1 ] + ··· + [Lνk ] in the Grothendieck group K(A) (up to ﬁnite error). For n large,

gλ[n],µ[n],ν[n] = #{i | νi = ν}, completing the proof.

5. Representation theory in positive characteristic

We finally move on to representation theory in characteristic p > 0. We first review a few basic notions. Let k be an algebraically closed field of characteristic p > 0 and G be a finite group. p 6 | |G| p | |G| Representations of G over k are Representations of G over k are not semisimple. semisimple in general. The number of simples is the number The number of simples is the number of conjugacy classes of G. of p-regular4 conjugacy classes. Example 5.1. If G is a p-group, then the number of p-regular conjugacy classes is 1, so there is only one simple, the trivial representations.

Example 5.2. Let G = S3 If p = 2, then the 2-regular conjugacy classes are 1, (123). Then 2 simples are triv and std. Indeed, std = ker , the kernel of the augmentation map : k3 → k , and the augmentation 3 S3 map splits as e1 + e2 + e3 ∈ (k ) has non-zero image under as 3 6= 0 in characterstic 2. Thus k3 = k ⊕ std, and it is not hard to show that std is simple. If p = 3, then the 3-regular conjugacy classes are 1, (12), so there are 2 simples: triv and sgn. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 55

Therefore, std is an extension of the two. We have std = ker but

(e1 + e2 + e3) = 3 = 0 in characteristic 3, so triv = k(e1 + e2 + e3) ⊆ std .

The quotient is 1-dimensional, spanned by the image of e1 − e2, so it must be either triv or sgn. Since (12)(e1 − e2) = −(e1 − e2), the quotient must be sgn.

First goal. Understand the simples of Sn in characteristic p > 0.

The number of p-regular conjugacy classes in Sn is the number of partitions of n into parts prime to p. However, we will actually parametrize the irreducibles by a diﬀerent class of partitions.

Deﬁnition 5.3. A partition is p-singular if some part occurs at least p times (i.e. λi = λi+1 = ··· = λi+p−1) and p-regular otherwise.

We will index the irreducibles of Sn in characteristic p by p-regular partitions. Let us ﬁrst see that their number is correct. Note that Y 1 Y = (1 + xn + x2n + ··· ), 1 − xn n≥1 n≥1 and hence the coeﬃcient of xn is the number of partitions of n. In other words, this is the generating function for the number of partitions of n. Lemma 5.4. The number of p-regular partitions of n is the number of p-regular conjugacy classes in Sn.

Proof. Consider the product Y 1 − xnp P = 1 − xn n≥1 in two ways to get the result. (1) By canceling the fractions, we obtain Y 1 Y P = = (1 + xn + x2n + ··· ), 1 − xn p - n p - n n so the coeﬃcient of x is the number of p-regular conjugacy classes in Sn. (2) By expanding 1 − xnp, we obtain Y P = (1 + xn + x2n + ··· + x(p−1)n) n≥1 so the coeﬃcient of xn is the number of p-regular partitions of n.

This shows the desired equality. We deﬁne the space of tabloids M λ and the Specht module Sλ ⊂ M λ as in characteristic 0. Remark 5.5. Some of the results/proofs in characteristic 0 are still valid in characteristic p > 0: 56 ANDREW SNOWDEN

• Garnir relations 2.35, • Standard basis 2.40 (so dim Sλ is the same in all characteristics), λ • First version of the Pieri rule 3.15: if |λ| = n, then S |Sn−1 has a ﬁltration where graded pieces are Sµ where µ are what appears in the Pieri rule.

We have an invariant form h , i on M λ given by 1 if {t} = {t0} h{t}, {t0}i = 0 otherwise which induces a form on Sλ. Proposition 5.6. If V ⊆ M λ is a subrepresentation, then V ⊇ Sλ or V ⊆ (Sλ)⊥.

Proof. (Same as in characteristic 0.) Pick v ∈ V , tableau t. Then κtv is a scalar multiple λ of et. If κtv 6= 0 for some v, t, then et ∈ V , so S ⊂ V . If κtv = 0 for all v, t, then λ ⊥ 0 = hκtv, {t}i = hv, eti = 0 for all v, t, so V ⊆ (S ) . Proposition 5.7. The quotient Sλ/(Sλ ∩ (Sλ)⊥) is either 0 or irreducible.

Proof. Let V ⊆ Sλ/(Sλ ∩ (Sλ)⊥) be subrepresentations, and Sλ ∩ (Sλ) ⊆ Ve ⊆ Sλ be the inverse image under the quotient map. By Proposition 5.6, either Ve ⊇ Sλ, and then λ λ ⊥ λ λ ⊥ Ve = S , and V is the whole space, or Ve ⊆ (S ) , and then Ve = S ∩ (S ) , so V = 0.

The diﬀerence in positive characteristic is that the quotient could actually be 0. We will try to understand when this happens. Proposition 5.8. The inner product h , i is 0 on Sλ if and only if λ is p-singular.

Proof. Suppose λ is p-singular. We consider het, et0 i. If a part of λ appears with multiplicity k, we get an action of Z/kZ on the set of tabloids in both et and et0 by cycling the relevant rows. For example, in

we have an action of Z/3Z on the three middle rows.

Then het, et0 i is a multiple of k. Taking k = p (this is possible, since λ is p-singular), we get het, et0 i = 0. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 57

Suppose λ is p-regular. Let t be a tableau and t0 the tableau obtained from t by reversing each row:

1 2 3 4 4 3 2 1 t = 5 6 7 t0 = 7 6 5 8 9 10 10 9 8

Consider het, et0 i. Suppose σ ∈ Ct and satisﬁes i and σ(i) below to the rows of same length for all i. Then σ{t} appears both in et and in et0 . In the example above, σ = (58) is allowed but σ = (15) is not allowed. One can see that these σ{t} account for all tabloids in common between t and t0. Then het, et0 i is the number of such σ. This is equal to Y (#(rows of length n)!)n, n≥1

and #(rows of length n) < p, because λ is p-regular, which shows that het, et0 i= 6 0. Deﬁnition 5.9. For λ p-regular, put Dλ = Sλ/(Sλ ∩ (Sλ)⊥).

λ By Propositions 5.7 and 5.8, D is an irreducible representation of Sn (in particular, it is nonzero). The next step will be to show that all Dλ’s are distinct. Note that there are the correct number of Dλ’s, so if we show they are all distinct, this will show that they are all the irreducibles. Lemma 5.10. Let λ, µ be partitions of n with λ p-regular. Consider a nonzero homomorphism θ : Sλ → M µ/U µ Sµ+U for some U ⊂ M . Then λ µ, and if λ = µ then im θ = U . Recall the following Lemma we proved in Chapter 2. Lemma (Lemma 2.24). Let λ and µ be partitions of n, t a λ-tableau, t0 a µ-tableau. Suppose 0 κt{t }= 6 0. 0 Then λ µ, and if λ = µ then κt{t } = ±et. The proof still works in any characteristic, so we may use it here.

Proof of Lemma 5.10. Let t be a λ-tableau, t0 be the row revesal of t, as in th proof of Proposition 5.7. Then het, et0 i= 6 0. We have that 0 0 6= het0 , κt{t}i = hκtet, {t}i. 0 Hence κtet 6= 0, so κtet0 = het for some nonzero h ∈ k. We then have that

hθ(et) = θ(κtet0 ) = κtθ(et0 ) 6= 0, 58 ANDREW SNOWDEN

λ WIndeed, we have θ(et) 6= 0 because θ 6= 0 and et0 generates S , and hence h 6= 0. Thus µ µ θ(et0 ) ∈ M /U is not killed by κt. Hence κtM 6= 0, and by Lemma 2.24, λ µ. −1 If λ = µ, then θ(et) = h κtθ(et0 ) and by Lemma 2.24, this is a scalar multiple of et (mod U), Sµ+U so im θ = U . Corollary 5.11. Let λ, µ, U be as in Lemma 5.10. Suppose θ : Dλ → M µ/U is nonzero. Then λ µ and if U ⊃ Sµ then µ 6= λ. λ λ θ µ Proof. Consider S D → M /U. By Lemma 5.10, we get µ µ, and the other assertion is clear. Theorem 5.12.

λ (1) Every irreducible of Sn is isomorphic to D for some λ p-regular. (2) No two distinct Dλ’s are isomorphic. λ (3) Each D is self-dual, absolutely irreducible, deﬁned over Fp.

Proof. For (2), suppose Dλ ∼= Dµ. This gives a nonzero map M µ Dλ → Dµ ⊂ , Sµ ∩ (Sµ)⊥ so Corollary 5.11 gives λ µ, and we also get µ λ by symmetry so λ = µ. Then (1) follows by counting the number of irreducibles. λ Finally for (3), it is obvious that D is deﬁned over Fp, and it is absolutely irreducible (because we never used properties of k, just that is has characteristic p). Finally, h i is a perfect pairing on M λ, which induces a perfect pairing on Sλ/(Sλ ∩ (Sλ)⊥), so Dλ is self-dual. Theorem 5.13. All the Jordan-H¨older constituents of M µ have the form Dλ with λ µ. µ The module D occurs if and only if µ is p-regular, in which case it has multiplicity 1.

Proof. What we just proved shows that all constituents of M µ/Sµ are Dλ with λ µ. Indeed, λ λ µ µ if D is a constituent, we get an injection D → M /U for some U ⊃ S , and λ µ follows from Corollary 5.11. We have an isomorphism (Sµ)⊥ ∼= (M µ/Sµ)∗ coming from the inner product h , i. Hence all the constituents of (Sµ)⊥ have the form (constituent of M µ/Sµ)∗ = (Dλ)∗ = Dλ with λ µ. The only left over part of M µ is 0 µ is p-singular, Sµ/(Sµ ∩ (Sµ)⊥) = Dµ otherwise. Indeed, we have the short exact sequences MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 59

0 Sµ M M/Sµ 0

0 Sµ ∩ (Sµ)⊥ Sµ Sµ/(Sµ ∩ (Sµ)⊥) 0

This completes the proof. Remark 5.14. It is in general very hard to know the multiplicity of Dµ in Sλ, as we did in characteristic 0. Remark 5.15. For any ﬁnite group G, we have a well-deﬁned map

K(Rep (G)) K(Rep (G)) Q Fp

[V ] [M/pM]

where M ⊆ V is a G-stable lattice. (We can always get such a lattice by taking any lattice in V and averaging over the group G.)

In K(RepQ(Sn)), we have Young’s rule [M µ] = [Sµ] + (sum of [Sµ]’s with λ µ) . Therefore, the same holds in K(Rep )(S ), because M µ andSµ are deﬁned integrally, so Fp n they are preserved by the functor. We can use this to prove that the Dλ’s are all the irreducibles without counting. We will prove by induction on µ that all constituents of M µ are Dλ’s. We have that [M µ] = [Sµ] + (sum of [Sµ]’s with λ µ), | {z } sum of [Dν ]’s by induction and [Sµ] = [Sµ ∩ (Sµ)⊥] + [Sµ/Sµ ∩ (Sµ)⊥], | {z } =0 or Dµ and Sµ ∩ (Sµ)⊥ ⊆ (Sµ)⊥ ∼= (M µ/Sµ)∗, which is a sum of [Dν]’s by induction. 1n λ λ Note that M = k[Sn] and all the constituents of D ’s, so D ’s are all irreducible.

Question. When is Sλ irreducible? Let us give an example where we can answer this question.

(n−1,1) Lemma 5.16. The standard representation S of Sn is irreducible if and only if p - n.

Proof. Suppose n std X z }|n { 0 6= x = aiei ∈ ker(: k → k) . i=1 Because p - n, there exist i 6= j such that ai 6= aj. Then

(i j)x − x = (ai − aj)(ej − ei). 60 ANDREW SNOWDEN

Hence if V ⊆ std is a nonzero subrepresentation, then V contains ei − ej for some i 6= j. Then V contains ei − ej for all i, j by using Sn, so V = std.

If p|n, then k(e1 + ··· + en) ⊂ std is Sn-stable.

Next, we will give a necessary and suﬃcient condition for a general Sλ for p-regular λ to be irreducible.

µ 5 µ Lemma 5.17. Suppose EndSn (S ) = k. Then S is irreducible if and only if it is self dual.

Proof. Any irreducible is self-dual by Theorem 5.12. Suppose Sµ self-dual. Let U ⊆ Sµ be an irreducible submodule. Then

Sµ ∼= (Sµ)∗ → U µ ∼= U ⊆ Sµ. | {z } | {z } assumption U irreducible

Hence the composition Sµ → Sµ must be a scalar multiple of id because End(Sµ) = k. This µ shows U = S .

µ 0 µ µ Let g = gcd(het, et0 i | t, t any µ-tableaux), computed in S over Q. We showed that p|g if and only if µ is p-singular. In fact, our proof showed more: if zj = #(parts of µ = j), then

Y µ Y j zj! g (zj!)

where we got the ﬁrst one by counting the common tabloids in et and et0 , and the second one 0 by looking at het, et0 i where t was the row reversal of t (see the proof of Proposition 5.8). Recall that X κt = (sgn σ)σ,

σ∈Ct

X ρt = σ.

σ∈Rt Lemma 5.18.

µ† (1) The gcd of the coeﬃcients of the tabloids in ρtκt{t} is g . Y (2) We have that κtρtκt = (hook lengths of µ) κt{t}.

Proof. Suppose t is a µ-tableau.

5This assumption holds always in characteristic p 6= 2 and often in characteristic 2. This follows from our results on semistandard homomorphisms. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 61

µ† For (1), note that g = gcd het† , πet† i. Then π∈Sn † † sgn(π)het† , πet† i = sgn(π)h{t }, κt† πκt† {t }i X = sgn(σπτ)

σ,τ∈Ct† σπτ∈Rt† X = sgn(ω)

ω∈Rt† τ∈Ct† −1 −1 ωτ π ∈Ct† X = sgn(ω)

ω∈Ct τ∈Rt ωπ−1τ −1 −1 = h{t}, π ρtκt{t}i

= h{π{t}, ρtκt{t}}

µ† Hence g = gcdπ∈Sn hπ{t}, ρtκt{t}i = gcd of coeﬃcients in ρtκt{t}.

For (2), we know that κtρtκt{t} = cκt{t} for some c. We have a map µ M → Q[Sn]

σ{t} 7→ σρt µ 2 and hence S = Q[Sn]κtρt. We know that (κtρt) = cκtρt. Let U be the complement in Q[Sn] to Q[Sn]κtρt. Consider right multiplication by κtρt on Q[Sn], and compute its trace in two ways.

First, Q[Sn] = Q[Sn]κtρt ⊕ U. Since κtρt is a projector onto its image, it acts by 0 on U. The matrix for this operator is hence

c 0 0 0

µ Hence the trace of right multiplication by κtρt is c · dim S .

We compute the trace in another way. The coeﬃcient of 1 in κtρt is 1 because Rt ∩Ct = {1}.

Hence the coeﬃcient of σ in σκtρt is 1 for all σ ∈ Sn. Hence in the basis {σ}σ∈Sn of Q[Sn], the matrix for κtρt has 1s on the diagonals, which shows that its trace is n!. This shows that c · dim Sµ = n! which, by the Hook Length Formula 3.37, implies the result Q c = (hook lengths in µ). µ µ Proposition 5.19. Let θ : Mk → Sk be the map 1 {t} 7→ ρ κ {t} mod p. gµ† t t | {z } computed in M µ Z 62 ANDREW SNOWDEN

µ ⊥ µ ⊥ µ Theorem 5.20. We have that ker θ ⊇ (Sk ) and if ker θ = (Sk ) and End(S ) = k then im θ = Sµ and Sµ is irreducible.

Proof. Let θ : M µ → Sµ be given by the same formula. Then Q Q Q

θQ(κt{t}) = (nonzero rational number) · et, so it is nonzero. Hence θQ is a nonzero multiplication of the standard projection, which shows that ker(θ ) = (Sµ)⊥. Q Q Consider (Sµ)⊥ ∩ M µ, a free -module of rank (dim Sµ)⊥. Consider a -basis. Its reduction Q Z Z Q Z µ ⊥ µ ⊥ mod p is linearly independent in (Sk ) , and hence it spans a subspace of (Sk ) of dimension equal to dim(Sµ)⊥ = (Sµ)⊥, since they both have basis given by semistandard tableaux. Q k Hence element of (Sµ)⊥ lifts to an element of (Sµ)⊥. Thus ker(θ) ⊇ (Sµ)⊥. k Q k µ ⊥ If ker θ = (Sk ) , then θ induces an isomorphism M µ/(Sµ)⊥ → Sµ, | {z } =∼(Sµ)∗ µ which shows that S is self-dual, and hence it is irreducible by Lemma 5.17. Theorem 5.21. Suppose µ is p-regular. Then Sµ is reducible if and only if p divides Q(hook lengths in µ) . gµ† | {z } this is an integer by Lemma 5.18

µ µ ⊥ Sµ+(Sµ)⊥ Proof. By Theorem 5.20, S is reducible if and only if ker θ ) (S ) . Note that (Sµ)⊥ is the unique minimal submodule of M µ/(Sµ)⊥. Hence Sµ is irreducible if and only if ker θ ⊃ Sµ. Finally, we have that 1 Q(hook lengths) θ(κ {t}) = κ ρ κ {t} = {t} = 0 t gµ† t t t gµ† Q if and only if p divides (hook lengths) . gµ† Theorem 5.22 (James and Murphy, conjectured by Carter). Consider the diagram obtained from λ by placing in each box the p-adic valuation of the hook length. Then the following are equivalent:

(1) the numbers are constant along the columns, (2) λ is p-regular and Sλ is irreducible.

5.1. Polynomial representations of GLm. Recall that in characteristic 0 we had an equiv- pol alence between Rep (GL∞) and Rep(S∗). We will see what happens in characteristic p > 0. Fix a ﬁeld k of characteristic p > 0. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 63

Definition 5.23. An algebraic representation of GLm is a comodule of the coordinate ring k[GLm]. If k is infinite, a finite-dimensional algebraic representation is the same as a representation o fGLn(k) such that the matrix entries are rational functions. (This is not true is k is finite.)

Example 5.24. Let m = 1, i.e. consider the algebraic representations of GL1 = Gm, k[Gm] = k[t, t−1]. The comultiplication is

∆: k[Gm] → k[Gm] ⊗ k[Gm] t 7→ t ⊗ t and the counit is

η : k[Gm] → k t 7→ 1.

Suppose M is a comodule for k[Gm] with the comodule structure given by ∆: M → k[Gm] ⊗ M satisfying the approporiate axioms. Let

n Mn = {m ∈ M | ∆(m) = t ⊗ m}. M We claim that M = Mn. For m ∈ M, n∈Z

X n ∆(m) = t ⊗ mn for some mn ∈ M. n∈Z By one of the comodule axioms (∆ ⊗ 1)∆ = (1 ⊗ ∆)∆, which gives

X n n X n t ⊗ t ⊗ mn = t ⊗ ∆mn, n∈Z n∈Z n which shows that ∆mn = t ⊗ mn, i.e. mn ∈ Mn. By the counit axiom, we obtain X m = (η ⊗ 1)∆m = mn, n∈Z which proves the claim. This shows that alg ∼ Rep (Gm) = (graded vector spaces) so it is semisimple, and for each n ∈ Z there is a simple which has a 1-dimensional graded vector space concentrated in degree n.

We present two applications of this example.

(1) If we consider Gm ⊆ GLm as scalar matrices. Every algebraic GLm-representation M n of V breaks up as V = Vn, where Gm acts by t 7→ t in Vn. n∈Z (2) The same analysis applies to:

m m Rep(Gm) = (Z -graded vector spaces) 64 ANDREW SNOWDEN

m and (Gm) ⊆ GLm as diagonal matrices. Hence if V is an algebraic representation of GLm, we get a decomposition of vector spaces M V = Vλ, λ∈Zm m n where Gm acts on Vλ via (t1, . . . , tn) 7→ t . This is the weight decomposition of V .

p−1 Remark 5.25. Note that t 7→ 1 and t 7→ t are two non-isomorphic representation of Gm, but they are isomorphic representations of Gm(Fp). Examples 5.26.

• The usual symmetric and exterior powers are representations of GLm. • Symmetric powers are not irreducible in general. Let V = km be the standard representation of GLm and consider

p Sym (V ) = span of degree p polynomials in variables x1, . . . , xm. p Note that the span of the xi is GLm-stable when k is inﬁnite. The reason is that if g ∈ GLm(k) and

a11 ··· a21 ··· g =  .   .  an1 ··· then

p p p p p p p p p gx1 = (gx1) = (a11x1 + a21x2 + ··· + am1xm) = a11x1 + a21x2 + ··· + am1xm. Thus we have an exact sequence 0 (span of pth powers) Symp(V ) (quotient) 0. One can see that this sequence is not split. Indeed, it is enough to show that any p p non-zero subrepresentation of Sym (V ) contains xi . This can be done by considering d the operators Ei,j = xj . dxi alg Hence Rep (GLm) is not semisimple for m > 1. • We deﬁne n ⊗n Sym (V ) = (V )Sn (the coninvariants)

Dn(V ) = (V ⊗n)Sn (the invariants) where the latter is called the nth divided power. Then we get a diagram

Dn(V ) V ⊗n Symn(V ) where the composition is an isomorphism in characteristic 0 but not in characteristic p > 0 (in general). We also get the short exact sequence MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 65

0 (kernal) Dp(V ) (span of pth powers) 0

p xi ⊗ · · · ⊗ xi xi kills other basis vectors This shows that Dp(V ) is not isomorphic to Symp(V ) even by a diﬀerent isomorphism. • If V is any algebraic representation of GLm, we can precompose with the Frobenius p map F : GLm → GLm, taking g = (gij) = (gij), to get a new representation F (V ). For example, when m = 2, F (std) = ke1 ⊕ ke2, and a b ap bp · e = · e = ape + cpe . c d 1 cp dp 1 1 2 | {z } | {z } in F (std) in std Hence F (std) = (sum of pth powers) ⊂ Symp. This is what appeared in two of the exact sequences above. • For any algebraic representation V , V ∗ is naturally an algebraic representation. Note that if V = ke1 ⊕ · · · ⊕ kem is the standard representation then

a1   a2   .  ei = aiei,  ..  am and the weight is λ = (0,..., 1,..., 0) with 1 in the ith spot. Hence the weights in V have one entry 1 and the rest 0s. A similar argument shows that the weights in V ∗ have one entry −1 and the rest 0s. Therefore, dualizing can take representations with positive weights to ones that have negative weights. When we restrict our attention to polynomial representations later, we will want the representations which have positive weights, and hence we define a different dual. Define V ∨ to be V ∗ precomposed with the automorphism t −1 (−) : GLm → GLm. Then V ∨ is V ∗ as a vector space with the action (g · λ)(v) = λ(tgv). We then have that Symn(V )∨ = Dn(V ∨)

for all vector spaces V , canonically. If V is the standard representation of GLm, then ∼ ∨ V = V as GLm-representations.

Deﬁnition 5.27. A representation of GLm is polynomial if it occurs as a subquotient of direct sums of ⊗-powers of std. Equivalently, a polynomial representation of GLm is a comodule for k[Mm] (where the comomdule structure comes from matrix multiplication).

Note that if W is a polynomial representation, so is W ∨, since std∨ ∼= std and ∨ ∼ ∨ ∨ (W1 ⊗ W2) = W1 ⊗ W2 . 66 ANDREW SNOWDEN

pol Proposition 5.28. The category Rep (GLm)n (the degree n piece) is equivalent to the category of modules over Dn(End(V )) = (End(V )⊗n)Sn , V = km.

Proof. The objects in Reppol(GLm) are comodules over

M n k[Sm] = Sym(End V ) = Sym (End(V )), n≥0

pol n a sum of coalgebras. Hence objects in Rep (GLm)n are comodules over Sym (End(V )), which are modules over Symn(End(V ))∗ = Dn((End V )∗) = Dn(End V ).

This completes the proof.

For a partition λ = (λ1, . . . , λr), put Dλ(V ) = Dλ1 (V ) ⊗ · · · ⊗ Dλr (V ). For λ = (n), we see that Dλ(V ) = Dn(V ); for λ = (1,..., 1), Dλ = V ⊗n.

λ pol Proposition 5.29. Let |λ| ≤ m = dim(V ). Then D (V ) is projective in Rep (GLm). Hence every degree n object is the quotient of direct sums of Dλ’s with |λ| = n.

Proof. Trivially, Dn(End(V )) is projective as a Dn(End(V ))-module. The multiplication on Dn(End(V )) is Dn(End(V )) ⊗ Dn(End(V )) → Dn(End(V )) | {z } | {z } Dn(V ⊗V ∗) Dn(V ⊗V ∗) is induced by pairing the V with the V ∗. As a module, M Dn(End(V )) = Dn(V ⊕m) = Dλ(V ). |λ|=n//`(λ)≤m

λ This shows that D is projective if `(λ) ≤ m as direct summand of a projective module. Corollary 5.30. The modules Symλ(V ) = Symλ1 (V ) ⊗ · · · ⊗ Symλr (V ) are injective for |λ| ≤ m = dim(V ). Every degree n representation with n ≤ m injects into a direct sum of these.

Warning. The modules Dλ and Symλ are not indecomposable in general. Remark 5.31. For m0 ≥ m, we have an exact functor pol pol Rep (GLm0 )n → Rep (GLm)n sum of weight spaces V 7→ Wλ where λ0 = 0 for i > m One can show that this is an equivalence for m0, m 0.

We have functors for m ≥ n MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 67

pol Φ Rep (GLm)n Rep(Sn)

W (1n weight space in W )

Ψ pol Rep(Sn) Rep (GLm)n

M ((km)⊗n ⊗ M)Sn

We note that Φ is exact and Φ is left exact. Note that if V is any polynomial representation the weights in F (V ) are divisible by p, where F is the Frobenius twist. Hence Φ(F (V )) = 0, so Φ is not an equivalence. Proposition 5.32.

• The functors (Φ, Ψ) are an adjoint pair. • The counit ΦΨ → id is an isomorphism.

Proof. We have that ΦΨM = (1n weight space in (V ⊗n ⊗ M)Sn ) = ((1n weight space in V ⊗n) ⊗M)Sn ( it is important that m ≥ n here) | {z } k[Sn] ∼= M ∼ Given a GLm-equivariant map W → ΨM, apply Φ to get ΦW → ΦΨM = M. This gives a map

(∗) HomGLm (W, ΨM) → HomSn (ΦW, M). We want to show (∗) is an isomorphism. Given W , pick a presentation A → B → W → 0, where A and B are sums of Dλ’s (cf. Proposition 5.29). To given an exact sequence ΦA → ΦB → ΦW → 0 by exactness of Φ. We then obtain

0 0

Hom(W, ΨM) Hom(ΨW, M)

Hom(B, ΨM) Hom(ΨB,M)

Hom(A, ΨM) Hom(ΨA, M) 68 ANDREW SNOWDEN

with exact columns, and hence, by the Five Lemma, it is enough to show that the two bottom horizontal maps are isomorphisms. Therefore, it is enough to show that (∗) is an isomorphism when W = Dλ(V ). Note that ΦDn = triv .

Note. If V and W are polynomial representations of GLm of degrees n1, n2 summing to at most m, then Φ(V ⊗ W ) = Φ(V ) ⊗ Φ(W ) . | {z } in Rep(S∗) This is because M (1n weight space in V ⊗ W ) = (1A weight space in V ) ⊗ (1B weight space in W ) [n]=AqB S = Ind n1+n2 ((1n1 weight space in V ) ⊗ (1n2 weight space in W )) Sn1 ×Sn2 S = Ind n1+n2 Φ(V ) ⊗ Φ(W ). Sn1 ×Sn2 Therefore, ΦDλ = M λ = IndSn (triv), Sλ which shows that ΨΦDλ = (V ⊗n ⊗ IndSn (triv))Sn = (V ⊗n ⊗ triv)Sλ = Dλ(V ). Sλ Exercise. This gives the inverse to (∗).

5.2. Serre quotient categories. We will now see that if we are in this situation, we can recover the target category from the original category via Serre quotients. In other words, pol our next goal is to prove that Rep(Sn) is a quotient of Rep (GLm)n for m ≥ n. The reference for this section is [Gab62]. We omit a lot of the proofs here and the reader is referred to this paper for details. Suppose that Φ: A → B is an exact functor of abelian categories. Deﬁne ker Φ = {M ∈ A | Φ(M) = 0}. Then ker Φ is closed under sub/quotients and extensions:

(1) If M ∈ ker(Φ) and N is a sub or quotient of M, then N ∈ ker Φ. (2) Suppose

0 M1 M2 M3 0

is exact with M1,M3 ∈ ker Φ. Then M2 ∈ ker Φ. Both of these properties follow from exactness. Deﬁnition 5.33. Let A be an abelian category. A Serre subcategory of A is a full subcategory K satisfying (1) and (2).

Serre quotient construction. Let K be a Serre subcategory of A. We deﬁne a new category A/K, called the Serre quotient, by MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 69

Objects: same as objects of A, Morphisms: Hom (M,N) = lim Hom (M 0,N 0), where the colimit is over all A/K −→ A subobjects M 0 of M with M/M 0 ∈ K and all quotients N 0 of N with

0 ker(N N ) ∈ K, where one can think of the following picture N 0

M N

M 0 Composition: induced by the following pull back L00 M 0 N 0

L M N

L0 M 00

L00

Fact. The Serre quotient A/K is an abelian category. We have a functor T : A → A/K which is the identity on objects and the natural map on morphisms. Proposition 5.34. The functor T is exact.

Proof. Let us show T (ker(f)) = ker T (f), where f : M → N is a morphism. We have the commutative diagram

f 0 ker(f 0) M 0 N 0

f ker(f) M N

X0 X

Using this, we get a map X → ker(f) in A/K. Proposition 5.35. We have that ker(T ) = K and if f : M → N is a morphism, then T (f) is an isomorphism if and only if ker(f) and coker(f) are in K. 70 ANDREW SNOWDEN

Proposition 5.36 (Mapping property). Let K ⊆ A and Φ: A → B such that Φ is exact and ker Φ ⊇ K. Then there is a unique functor Φ0 : A/K → B such that Φ = Φ0 ◦ T .

Proof. We have the commutative diagram

Φ HomA(M,N) HomB(ΦM, ΦN)

0 0 Φ 0 0 HomA(M ,N ) HomB(ΦM , ΦN )

and the right vertical arrow is a bijection because the maps ΦN → ΦN 0 and ΦM 0 → ΦM are isomorphisms. Then Φ induces a map lim Hom (M 0,N 0) → Hom (ΦM, ΦN), −→ A A/K 0 which gives the desired Φ . Deﬁnition 5.37. We say that K is a localizing subcategory if it is a Serre subcategory and T has a right adjoint S : A/K → A, called the section functor. Remark 5.38. The section functor S be left exact and takes injectives to injectives. Example 5.39. Let R be an integral domain and K = Frac(R). Let

A = ModR, K = torsion modules. We have an exact functor

Modr → VecK

M 7→ K ⊗R M with ker Φ ⊇ K (actually, they are equal). Therefore, we get a functor 0 Φ : A/K → VecK . Exercise. The functor Φ0 is actually an equivalence of categories. (More generally, we can treat localization with respect to the multiplicative subset R.)

Finally, the restriction functor S : VecK → ModR is the section functor. Example 5.40. If X is a topological space and i: U,→ X is the inclusion of an open subset, then we have the adjoint functors

i∗ Sh(X) Sh(X) i∗ It is (typically) true that Sh(X) Sh(U) ∼= . ker(i∗) ∗ We recall that ker(i ) consists of sheaves F on X such that Fx = 0 for all x ∈ U. The analogous statement is true for quasi-coherent sheaves on a scheme. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 71

Proposition 5.41. Let M ∈ A. The following are equivalent:

(1) the unit M → STM is an isomorphism. i (2) ExtA(K,M) = 0 for all K ∈ K and i = 0, 1, (3) for any f : A → B in A such that T (f) is an isomorphism, the induced map

=∼ HomA(B,M) → HomA(A, M), ∼ (4) HomA(N,M) = HomA/K(TN,TM) for all N ∈ A. Deﬁnition 5.42. An object M ∈ A is saturated (or K-saturated) if it satisﬁes conditions (1)–(4). Proposition 5.43. For any M ∈ A/K, SM is saturated. Proposition 5.44.

(1) The counit TS → idA/K is an isomorphism. (2) The kernel and cokernel of a unit idA → ST always belongs to K. Proposition 5.45. Suppose Φ: A → B is exact with right adjoint Ψ: B → A such that the counit ΦΨ → id is an isomorphism. Then the canonical functor A/ ker Φ → B is an equivalence and ker Φ is a localizing subcategory.

Back to our case. We had functors

pol Φ: Rep (GL∞) → Rep(S∗),

pol Ψ: Rep(S∗) → Rep (GL∞), where Φ is exact, (Φ, Ψ) is an adjoint pair, and the counit ΦΨ → id is an isomorphism.

pol By Proposition 5.45, Rep(S∗) is the Serre quotient of Rep (GL∞) by ker Φ = {V | (1n-weight space in V ) = 0 for all n}.

5.3. Highest weight structure. Motivation. Let g be a complex semisimple Lie algebra, Borel subalgebra b with radical u, and Cartan subalgebra h.

Example. Let g = sln. Then b consists of the upper triangular matrices, u consists of the strictly upper triangular matrice, and h are the traceless diagonal matrices. Deﬁnition 5.46. The category O is the category of g-modules satisfying:

(a) ﬁnitely generated over U(g), the universal enveloping algebra, (b) have weight decomposition, i.e. semisimple as h-modules, (c) action of u is locally nilpotent, i.e. every vector is killed by a power of u. Deﬁnition 5.47. An element M ∈ O is a highest weight module of weight λ if M u is one- dimensional of weight λ and generates M. 72 ANDREW SNOWDEN

− Let u be the opposite of u (for example, for sln take the strictly lower triangular matrices). Then U(g) = U(u−)U(h)U(u).

Suppose M is a highest weight module of weight λ if N and N 0 are submodule of M without λ as a weight, then N +N 0 satisﬁes the same condition, so there exists a maximal submodule N of M without λ as a weight. If K is any submodule of M then either K ⊂ N or K = M, hence M/N is simple with λ as the highest weight. If λ is a weight, we can regard λ as a representation of b via the homomorphism b → h ∼= b/u.

Deﬁnition 5.48. The Verma module V (λ) is U(g) ⊗U(b) Cλ.

This is a highest weight module of weight λ. Moreover, for any module M. we have

u Homg(V (λ),M) = {λ-weight vectors in M } We get a simple quotient L(λ) of V (λ) with highest weight λ, and the L(λ) are the simple modules. The weights appearing in ker(V (λ) → L(λ)) are smaller than λ, so the simple constituents of V (λ) other than L(λ) have the form L(µ) with µ < λ. Fact. The category O has enough projectives. However, V (λ) are typically not projective (or injective). Let P (λ) be a projective cover of L(λ) of ﬁnite length. This gives a surjection P (λ) → V (λ). We actually have a ﬁltration · · · ⊇ F 1 ⊇ F 0 = P (λ) such that F 0/F 1 = V (λ) and F i/F i+1 = V (µ) with µ > λ for i ≥ 1.

Fact. The category O has a duality. Therefore, we it has enough injectives, and co-Verma modules, giving a similar picture to the above. This finishes the motivation and we return to the general case. The following definition, given by Cline–Parshall-Scott, allows to generalize the notion of highest weights. Definition 5.49. A highest weight category is a k-linear (where k is a field) locally artinian abelian category A satisfying (AB5), i.e. the flitered colimits are exact and it has enough injectives, with a partially ordered set Λ (of weights) such that:

(1) The simples are indexed by Λ: {L(λ)}λ∈Λ, (2) For each λ ∈ Λ, there exists an object A(λ) with an embedding L(λ) ⊂ A(λ) such that all composition factors of A(λ)/L(λ) have the form L(µ) with µ < λ. We also require that [A(λ): L(µ)] < ∞ and dim Hom(A(λ),A(µ)) < ∞. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 73

(3) There is an injective envelope of I(λ) with a ﬁltration

0 = F0 ⊂ F1 ⊂ · · · such that (a) F1 = A(λ), (b) Fi/Fi−1 = A(µ) with µ > λ for i ≥ 2, (c)[ I(λ): A(µ)] = (the number of times Fi/Fi−1 = A(µ)) < ∞, [ (d) I(λ) = Fi. i≥1

Motivating example. The category O is highest weight category. pol Our motivation. The category Rep (GL∞) is a highest weight category. Proposition 5.50. Suppose A is a highest weight category with #Λ < ∞. Then gldimA < ∞.

Proof. We ﬁrst show that A(λ) has ﬁnite injective dimension by descending induction on λ. We have an exact sequence

0 A(λ) I(λ) Q 0,

where Q is the quotient, and we know that Q has pieces A(µ) with µ > λ. Hence Q has ﬁnite injective dimension by inductive hypothesis, and so injdimA(λ) ≤ injdimQ + 1 < ∞.

We next show that L(λ) has ﬁnite injective dimension by ascending induction on λ:

0 L(λ) A(λ) Q 0,

where Q has pieces L(µ) with µ < λ. By inductive hypothesis Q has ﬁnite injective dimension, and hence injdimL(λ) ≤ max(injdimA(λ), 1 + injdimQ(λ)) < ∞. This completes the proof. Deﬁnition 5.51. A ring R is (left) hereditary if any submodule of a (left) projective module is again projective. Equivalently, lgldimR ≤ 1.

Suppose R is a hereditary ﬁnite-dimensional k-algebra. We claim that ModR is a highest weight category. Let J be the socle of R, i.e. the maximal semisimple subobject of R. Fact 5.52. • The socle J is a 2-sided ideal. • The ring R/J is hereditary. • The modules J and R/J have no common composition factors. 74 ANDREW SNOWDEN

Deﬁne 0 = J0 ⊂ J1 = J ⊂ · · · ⊂ Jr = R by Jn/Jn−1 = soc(R/Jn−1).

For a simple L, let let n(L) be the n such that L is a summand of Jn/Jn−1. Partially order the simples by L < L0 if n(L) > n(L0).

If n(L) = n, then L is a quotient of Jn, so P (L) surjects onto L, ker ⊆ Jn−1, so it only has larger simples. Take linear duals to get a highest weight category, with A(L) = I(L). Deﬁnition 5.53. A 2-sided ideal J in a ring R is hereditary if (1) J is projective as a left R-module, (2) HomR(J, R/J) = 0, (3) Jrad(R)J = 0. Deﬁnition 5.54. A ring R is quasi-hereditary if there is a chain of 2-sided ideals in R

0 = J0 ⊂ J1 ⊂ · · · ⊂ Jr = R

such that Jn/Jn−1 ⊂ R/Jn−1 is a hereditary ideal.

R Example 5.55. If R is hereditary, we can take Jn to be the inverse image of soc . Jn−1 This gives a chain as above, so R is quasi-hereditary.

Theorem 5.56. Let R be a ﬁnite-dimensional k-algebra. Then ModR is a highest weight category if and only if R is quasi-hereditary.

pol 5.4. Highest weight structure on Rep (GLm)n. We omit most of the proofs in this section. Parts of it follow [Wey03], but precise references are not provided.

m Let λ be a partition of n with r rows and s columns, and V = k . The Weyl module Kλ is the image of

Dλ1 (V ) ⊗ · · · ⊗ Dλr (V )

V ⊗n

λ† λ† V 1 ⊗ · · · ⊗ V s (V )

The Schur module Lλ is the image of

† λ λ† ^1 ^s ⊗ · · · ⊗ (V ) → Symλ1 (V ) ⊗ · · · ⊗ Symλr (V ).

It is easy to see that Lλ and Kλ both have λ as a weight with multiplicity 1, and all other weights ae smaller.

Fact 5.57. The λ-weight vector generates Kλ and cogenerates Lλ. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 75

Hence Kλ has a simple quotient with highest weight λ and Lλ has a simple submodule with highest weight λ.

Remark 5.58. The Schur module Lλ has a basis indexed by semistandard tableaux of shape λ ﬁlled with numbers 1, . . . , m = dim V . In particular, the weight decomposition of Lλ, Kλ is independent of characteristic. n Vn Examples 5.59. The Weyl module K(n) is D (V ) and similarly K(1n) = (V ). For Schur n n modules: L(n) = Sym (V ) and L(1n) = Λ (V ). ∼ ∼ In characteristic 0, Lλ = Kλ = Sλ(V ).

We have shown that Dλ = Dλ1 ⊗ · · · ⊗ Dλr (V ) is a projective object of Reppol.

n Version of Pieri rule. The module Kλ ⊗ D has a ﬁltration where the graded pieces are Kµ’s where µ’s are as in the usual Pieri rule 3.15, the top piece (quotient) is least dominant (boxes go as low as possible).

λ Corollary 5.60. The module D has a filtration with graded pieces equal to Kλ. We can deduce a similar filtration for summands of Dλ. pol Theorem 5.61. The category Rep (GLm)n is a highest weight category. Its standard objects are Kλ’s (filter projectives, analogous to Verma). Its costandard objects are Lλ’s (filter injectives, A(λ)’s in definition of highest weight category). pol Corollary 5.62. The category Rep (GLm)n has finite global dimension.

Proof. This following by combining Proposition 5.50 with Theorem 5.61.

Remark 5.63. Totaro computed the global dimension in the 90s. Let αp(n) be the sum of the digits in base p expansion of n. Then

gldim = 2(n − αp(n)).

pol Note that if n < p then αp(n) ≥ n, so the global dimension is 0. Hence Rep (GLm)n is semi-simple. Corollary 5.64 (to Theorem 5.61). The Schur algebra Dn(End V ) is quasi-hereditary. Remark 5.65. Let λ be a partition of n and m ≥ n. Then 0 H (ﬂag variety for GLm, L(λ)) = Kλ and higher cohomology groups vanish, where L(λ) is a G-equivariant line bundle.

6. Deligne interpolation categories and recent work

Let C be a category. We will deﬁne a monoidal structure on C. Consider a functor ⊗: C × C → C. We want associativity, but equality X ⊗ (Y ⊗ Z) = (X ⊗ Y ) ⊗ Z is too much to ask for. Instead, we want an isomorphism α: X ⊗ (Y ⊗ Z) → (X ⊗ Y ) ⊗ Z 76 ANDREW SNOWDEN

of functors (considering both sides are functors C3 → C). We also require the following diagram to commute

W ⊗ (X ⊗ (Y ⊗ Z))

1⊗α α

W ⊗ ((X ⊗ Y ) ⊗ Z) (W ⊗ X) ⊗ (Y ⊗ Z)

α α

(W ⊗ (X ⊗ Z)) ⊗ Z α⊗1 ((W ⊗ X) ⊗ Y ) ⊗ Z

This is called them pentagon axiom (for obvious resons). We also want commutativity, which corresponds to a functorial isomorphism

β : X ⊗ Y → Y ⊗ X

such that

β β X ⊗ Y X,Y Y ⊗ X Y,X X ⊗ Y

and the following diagram commutes

X ⊗ (Y ⊗ Z) α (X ⊗ Y ) ⊗ Z

1⊗β β

X ⊗ (Z ⊗ Y ) Z ⊗ (X ⊗ Y )

α α β⊗1 (X ⊗ Z) ⊗ Y (Z ⊗ X) ⊗ Y

This is called the hexagon axiom. Given C with ⊗, α, β, a unit object of C is an object 1 with an isomorphism 1 → 1 ⊗ 1 such that X 7→ X ⊗ 1 is an equivalence of categories C → C.

Deﬁnition 6.1. A symmetric monoidal category is a category C with functor ⊗ and α, β, 1 as above. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 77

Remark 6.2. If C is a symmetric monoidal category, I is a ﬁnite set, then there exists a functor CI → C O (Xi)i∈I 7→ Xi. i∈I Deﬁnition 6.3. The internal Hom, denoted Hom(X,Y ) is the right adjoint to − ⊗ X: Hom(T, Hom(X,Y )) ∼= Hom(T ⊗ X,Y ).

Taking T = Hom(X,Y ), idT corresponds to a map ev: Hom(X,Y ) ⊗ X → Y called the evaluation map. Remark 6.4. We have that:

• Hom(1,X) ∼= X, • Hom(1, Hom(X,Y )) ∼= Hom(X,Y ). Definition 6.5. The dual of X, denoted X∨, is Hom(X, 1). We have an evaluation map ev: X ⊗ X∨ → 1. There is a canonical map X → X∨∨ and we say that X is reflective if this is an isomorphism. Definition 6.6. A symmetrical monoidal category is called rigid if (a) Hom(X,Y ) exists for all X,Y , (b) all objects are reflective, (c) Hom(X,Y ) ⊗ Hom(X0,Y 0) → Hom(X ⊗ X0,Y ⊗ Y 0) is an isomorphism

In particular, (c) for Y = 1, X0 = 1 implies that X∨ ⊗ Y 0 ∼= Hom(X,Y 0).

Assume C is a rigid symmetric monoidal category. Then we have the evaluation map ev Hom(X,X) ∼= X ⊗ X∨ → 1. Apply Hom(1, −) to get a map

trX : End(X) → End(1) called the trace. The rank of X is

rank(X) = trX (idX ) ∈ End(1). Deﬁnition 6.7. A tensor category (or ⊗-category) is an additive symmetric monoidal category such that ⊗ is bi-additive. Remark 6.8. There is no consistent deﬁnition of a tensor category in the literature. Some papers require it to be abelian, some require it to be rigid. Examples 6.9.

fd • For a ﬁeld k, Veck is a rigid abelian ⊗-category. 78 ANDREW SNOWDEN

fg • For a commutative ring R, ModR is an ⊗-category and it is abelian if and only if R is Noetherian. • For a commutative ring R, the category of projective, finitely-generated R-modules, fg ProjR , is a rigid ⊗-category (although it is typically not abelian). fd • If k is a field and G is a group scheme over k, Repk (G) is a rigid abelian ⊗-category. fd • The category SVeck of finite-dimensional super vector sapces over k.A super vector space is a Z/2-graded vector space:

V = V0 ⊕ V1. fd Then SVeck is a symmetric monoidal category, using usual ⊗ product of graded vector spaces with usual α and 1 but with modiﬁed β: β : X ⊗ Y → Y ⊗ X x ⊗ y 7→ (−1)deg x deg yy ⊗ x.

Then rank(V ) = dim(V0) − dim(V1) ∈ Z, but it can be negative. Remark 6.10. Suppose C is a monoidal category, graded by a group G: every object has a degree in G, and if we tensor two objects their degree multiply in G. We can try to modify α to get a new monoidal category: α0(X,Y,Z) = φ(deg X, deg Y, deg Z)α(X,Y,Z), 3 where φ: G → Aut(idC). This satisfies the hexagon axiom if φ is a 3-cocycle, and ones that differ by a coboundary give the same monoidal structure. Therefore, get new monoidal structures parameterized by 3 H (G, Aut(idC)). For symmetric monoidal categories graded by abelian groups G, we can try to modify both α and β: β0(X,Y ) = ψ(deg X, deg Y )β(X,Y ). The hexagon axiom translates to an identity on ψ. The new structures are parameterized 3 by the abelian cohomology Hab(G, Aut(idC)), which is defined appropriately for the axioms to hold. Definition 6.11. A fiber functor on a k-linear ⊗-category is an exact faithful k-linear ⊗- functor fd ω : C → Veck . Theorem 6.12 (Main Theorem of Tanaka Duality). If C is a rigid, abelian k-linear ⊗- 1 ∼ category such that End( ) = k with fiber functor ω, then C = Repk(G) for some group scheme G (where G = Aut⊗(ω), defined in an appropriate way).

Remark 6.13. We can recover G from Repk(G), the tensor structure, and ω.

If k = C, we do not need ω—just the symmetric monoidal structure on Repk(G). However, we really do need the symmetric monoidal structure. There are examples of ﬁnite groups G, H such that ∼ RepC(G) = RepC(H) as monoidal categories but G 6∼= H. (The ﬁrst example is due to Etingof and Gelaki). MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 79

It is important to know when C has a ﬁber functor. A necessary condition is that rank(any object) = non-negative integer ∈ End(1). Theorem 6.14 (Deligne, [Del90]). If C is a rigid abelian k-linear, ⊗-category, char(k) = 0, End(1) = k, k is algebraically closed, then the following are equivalent:

(1) C has a fiber functor, (2) for all objects M, rank(M) is a non-negative integer, (3) for all objects M, there exists n 0 such that Vn(M) = 0. Remark 6.15 (Rough idea of the proof). In the proof, Deligne mimics algebraic geometry. If you have an object, you can pass to an affine variety, and then your fiber functor maps points to fibers over that points.

In the setting of Deligne’s Theorem 6.14, is condition (2) automatic? No, because of the example of super vector spaces (cf. Examples 6.9).

If V is a super vector space, write V [1] for the super vector space with V [1]i = Vi+1. We can show

Sλ(V [1])) = Sλ† (V )[|λ|], so for example n ^ (V [1]) = Symn(V )[n].

If V is a ﬁnite-dimensional super vector space such that V1 6= 0 and V0 6= 0, then n ^ (V ) 6= 0, Symn(V ) 6= 0 for all n. This shows that condition (3) in Deligne’s Theorem 6.14 does not hold. From this, one easily sees that there is no ﬁber functor on super vector spaces. Question. Is the rank always an integers in the setting of Deligne’s Theorem 6.14? Answer. No. Deligne’s interpolation categories give a counterexample. From now on, we work over k = C.

Question. How can we construct the category Rep(Sn) without talking about Sn? n Answer. We will start with representations C of Sn and its ⊗-powers. Lemma 6.16. Let G be a ﬁnite group (or, more generally, a reductive algebraic group) and let V be a faithful representation of G. Then every irreducible representation of G is a quotient of a Symn(V ) for some n.

Proof. Pick v ∈ V with trivial stabilizer in G. We get a map G → V given by g 7→ gv. This is a closed embedding (thinking of these as varieties), so we get a surjection of coordinate rings Sym(V ∗)) → Fun(G). Since any representation appears in Fun(G), this completes the proof. 80 ANDREW SNOWDEN

r n ⊗r Hence all irreducible representations of Sn appear in some T = (C ) . We now want to understand maps between these representations:

r r r • Sr acts on T gives maps T → T , 1 0 • T → T = C, the augmentation map ei 7→ 1, n 0 1 P • T → T , 1 7→ ei, i=1 r 0 r P • αr : T → T , 1 7→ ei ⊗ · · · ⊗ ei, i=1 1 if i = ··· = i , • β : T r → T 0, e ⊗ · · · ⊗ e 7→ 1 r r i1 ir 0 otherwise, 1 2 • T → T , ei 7→ ei ⊗ ei, 2 1 • T → T , ei ⊗ ej 7→ δi,jei, • if r, s > 0, deﬁne r s γr,s : T → T

ei1 ⊗ · · · ⊗ eir 7→ δi1,...,ir ei1 ⊗ · · · ⊗ eir We write [n] = {1, . . . , n}. Given a partition of the set [r] q [s], we deﬁne a map T r → T s as follows:

• for a part of size k contained in [r], use βk, • for a part of size k contained in [s], use αk, • for a part with k > 0 vertices in [r], and ` > 0 vertices in [s], use γk,l. Example 6.17. Consider the partition of [4] q [3] given by

T 4

T 3

Then the associated map T 3 → T 4 is given by n X ei ⊗ ej ⊗ ek 7→ δj,k ei ⊗ ej ⊗ ei `=1 Example 6.18. Note that

T 2

∅ T 0 gives the map n n X X 1 7→ e` ⊗ ek `=1 k=1 MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 81 whereas

T 2

∅ T 0 gives the map n X 1 7→ e` ⊗ e`. `=1

Let Hr,s be the vector space with basis given by these diagrams. We have now constructed a map r s Hr,s → HomSn (T ,T ). Proposition 6.19. This map is always surjective, and an isomorphism if n ≥ r + s.

s Proof. Moving T over with a dual, we can reduce to r = 0. Note that Sn acts on basis s s Sn s vectors of T , so (T ) has a basis consisting of orbits of Sn acting on [n] . These orbits correspond to partitions of the set [s]. The associated invariants are related to the partitions in H0,s by an upper triangular change of variables.

Question. How does composition Hr,s × Hs,t → Hr,t work in terms of diagrams? Example 6.20. We have the following example:

T 1 T 1

T 2 T 2

T 4 T 4 and the composition is T 4 → T 2 → T 1

ei ⊗ ej ⊗ ek ⊗ e` 7→ δi,jδk,l which does correspond to the diagram on the right.

Generalization: merge two partitions together, delete middle row vertices. If only one partition remains, and touches top or bottom row, that is the answer. Example 6.21. Following this rule, we have the diagrams 82 ANDREW SNOWDEN

T 0 ∅ ∅ T 0

T 2 T 2

T 4 T 4 and the composition is indeed T 4 → T 2 → T 0 ei ⊗ ej ⊗ ek ⊗ e` 7→ δijδk`eie` 7→ δijδklδil. Examples 6.22. We give a few more examples to understand the general rule of composition. We have that

T 0 ∅

T 0 T 1 ∅ n times T 1 ∅ T 0

T 0 ∅ since this is given by T 0 → T 1 → T 0 n P 1 7→ ei 7→ n. i=1

We have that

T 0 ∅

T 0 T 2 ∅ n times T 2 ∅ T 0

T 0 ∅ since this is given by T 0 → T 2 → T 0 n P 1 7→ ei ⊗ ei 7→ n. i=1

We have that MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 83

T 0 ∅

T 0 T 2 ∅ n times T 2 ∅ T 0

T 0 ∅ since this is given by T 0 → T 2 → T 0 n P 1 7→ ei ⊗ ei 7→ n. i=1

We have that

T 0 ∅

T 0 T 2 ∅ n times T 2 ∅ T 0

T 0 ∅ since this is given by T 0 → T 2 → T 0 n P 2 1 7→ ei ⊗ ej 7→ n . i,j=1

General rule of composition: Overlay the two diagrams and (merge partitions). For each partition entirely in the middle row contributes to a factor of n. Discard these and what is left is the answer.

Example 6.23. The two diagrams

T 4

T 5 T 5

T 6 give 84 ANDREW SNOWDEN

n times

We work over C and ﬁx t ∈ C.

Deﬁne the category Rep(St)0 as follows • objects: T r for r ≥ 0, r s • morphisms: Hom(T ,T ) = Hr,s, • composition: the rule above with n changed to t.

This category is C-linear but not additive. We have a ⊗-product on Rep(St)0: • T r ⊗ T s = T rs, •⊗ : Hr,s × Hr0,s0 → Hr+r0,s+s0 is deﬁned by putting the two diagrams next to each other.

This satisﬁes the conditions to be rigid. Moreover, (T r)∨ = T r and ∨ (−) : Hr,s → Hs,r ﬂips the diagram upside down. If t ∈ N, we have a functor

Φ: Rep(St)0 → Rep(St) r t ⊗r T 7→ (C ) This functor is full by Proposition 6.19 (but not faithful).

We will now try to make an abelian category from Rep(St)0. Suppose A is an Ab-category (enriched over Ab). The additive envelope of A is a new category:

• objects: formal direct sums M1 ⊕ · · · ⊕ Mr, Mi ∈ A, • morphisms:

Hom(M1 ⊕ · · · ⊕ Mr,N1 ⊕ · · · ⊕ Ns) = s × r matrices with entries in HomA(Mi,Nj), • composition: deﬁned by matrix multiplication.

This is an additive category. Say A is an additive category. We deﬁne the Karoubian envelope of A as follows:

• objects: pairs (M, e) for M ∈ A, e ∈ End(M) is an idempotent [think of these as representing eM], 0 0 • morphisms: Hom((M, e), (N, e )) = e HomA(M,N)e . MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 85

The Karoubian envelope is additive and all idempotent endomorphisms have images.

Deﬁnition 6.24. We deﬁne the Deligne category Rep(St) as the Karoubian envelope of the additive envelope of Rep(St)0.

This category has a structure of a rigid ⊗-category.

6.1. Relation to Rep(Sn). Remark 6.25. For t ∈ N, we still have an ⊗-functor

Φ: Rep(St) → Rep(St) which is still full. But now, it is also surjective on objects (which is why we took the Karoubian envelope).

r Let Pr(t) = End(T ) in Rep(St). This is called the partition algebra. For t ∈ N, we have a surjection t ⊗r Pr(T ) → EndSt ((C ) ). An elementary argument to justify this is that any idempotent of the target lifts to and idempotent of the source.

Corollary 6.26. The functor Rep(St) → Rep(St) is essentially surjective.

λ λ t ⊗r Proof. Consider simple object S of Rep(St). We know that S is a summand of some (C ) , λ t ⊗r t ⊗r so S = (C ) for some idempotent ∈ EndSt ((C ) ). Let ˜ ∈ Pr(t) be a lift of . Then Φ((T r, ˜)) ∼= Sλ, so this functor is essentially surjective on objects. We already saw that it is surjective on morphisms in Remark 6.25. Deﬁnition 6.27. A morphism f : X → Y (in some rigid ⊗-category) is negligible if tr(gf) = 0 for all g : Y → X.

1 Example 6.28. In Rep(St), consider End(T ). It has a basis α

β = idT 1 and α2 = tα. We can compute tr(α) = tr(β) = t. We have

α,β⊗id T 0 T 1 ⊗ (T 1)∨ T 1 ⊗ (T 1)∨ T 0

corresponding to the diagram

∅ ∅ 86 ANDREW SNOWDEN

whose composition is t. Consider f = α − β. Then f is negligible if and only if tr(αf) = tr(βf) = 0 and tr(βf) = tr(f) = tr(α) − tr(β) = 0 tr(αf) = tr(α2 − α) tr((t − 1)α) = (t − 1)t. Hence f is negligible if and only if t = 0 or t = 1. Remark 6.29. Note that tr(β) = t says rank(T 1) = t.

Proposition 6.30. Let A and B be rigid ⊗-categories with EndA(1) = End(B)(1), and Φ: A → B is a full ⊗-functor. Then for f ∈ Mor(A), f is negligible if and only if Φ(f) is negligible.

Proof. Because Φ is an ⊗-functor, it preserves traces. Then tr(gf) = tr(Φ(g)Φ(f)) for f : X → Y , so tr(gf) = 0 for all g : Y → X if and only if tr(Φ(g)Φ(f)) = 0 for all Φ(g): Φ(Y ) → Φ(X). Since Φ is full, Φ(g) are all morphisms Φ(Y ) → Φ(X).

Let t ∈ N and N be the class of negligible morphisms in Rep(St).

Theorem 6.31. There is an equivalence of categories Rep(St)/N → Rep(St).

Proof. Let Φ: Rep(St) → Rep(St) be the usual functor. It suﬃces to show Φ(f) = 0 if and only if f is negligible. If Φ(f) = 0 then Φ(f) is negligible, and so f is negligible by the previous proposition. If f is negligible then Φ(f) is negligible, and so Φ(f) = 0. (Note: if C is a semisimple C-linear abelian rigid ⊗-category, then all negligible morphisms are 0. Indeed, if f : X → Y is non-zero, there exists g : Y → X such that gf is a nonzero projector so it has nonzero trace, so f is not negligible).

Note that this theorem is not true in positive characteristic.

6.2. Universal property of Deligne Categories. Let C be a rigid ⊗-category. Suppose X is a commutative unital associative algebra in C (so we have µ: X ⊗X → X and i: 1 → X satisfying the necessary properties). Let tr: X → 1 be the composite

µ⊗id 1 id⊗coev X ⊗ X ⊗ X∨ X ⊗ X∨ ev 1

This gives a pairing

X ⊗ X 1

µ tr X MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 87

which gives a map X → X∨. Deﬁnition 6.32. The algebra X is a Frobenius algebra if X → X∨ is an isomorphism. We denote by Ft(C) the category of Frobenius algebras in C of rank t. Remark 6.33. If X is a Frobenius algebra, X ∼= X∨, so we get a coalgebra structure on X by dualizing the algebra structure. (Frobenius algebras are often deﬁned in this way.) Examples 6.34.

n n n Q (1) The permutation representation C ∈ Rep(Sn) is a Frobenius algebra and C = C i=1 as algebras. 1 1 1 1 (2) The object T ∈ Rep(St) is a Frobenius algebra with µ: T ⊗ T → T given by the diagram

It is clear that if Φ: A → B is an ⊗-functor of rigid ⊗-categories then Φ carries a Frobenius algebra to a Frobenius algebra.

Hence if A = Rep(St), we get a functor ⊗ Fun (Rep(St), C) → Ft(C)(∗) Φ 7→ Φ(T 1). Theorem 6.35. If C is an additive Karoubian category then (∗) is an equivalence.

Remark 6.36. This is the universal property of Rep(St). In words, giving a ⊗-functor Rep(St) → C is the same as giving a Frobenius algebra in C of rank t.

Proof of Theorem 6.35 (sketch). Let X ∈ Ft(C). We get a map ⊗r ⊗s Hr,s → HomC(X ,X ). Idea: basic diagrams correspond to basic operations on X (multiplication, comultiplication, unit, counit). For example, when r = 2, s = 3, the diagram

is the composition of the diagrams

iterated comultiplication X → X ⊗ X → X ⊗ X ⊗ X

multiplication X ⊗ X → X 88 ANDREW SNOWDEN

This deﬁnes a functor Rep(St)0 → C and it is easy to see it is a ⊗-functor. This extends to Rep(St) because C is additive and Karoubian. Therefore, we have now constructed a functor ⊗ Ft(C) → Fun (Rep(St), C) and now one shows that it is equivalent to (∗).

6.3. Representations of S∞. We give an overview of the topic, without proofs or details. See [SS15] for a more in depth treatment of the subject. r ∞ ⊗r ∞ S n S Let T = (C ) with the natural action of S∞. Recall that C = C , S∞ = Sn. n≥1 n≥1

Deﬁnition 6.37. A representation of S∞ is algebraic if it is a subquotient of a direct sum r of T ’s. Let Rep(S∞) be the category of all algebraic representations of S∞.

This is an abelian ⊗-category, but it is not rigid.

Remark 6.38. The category Rep(S∞) is not semisimple. For example, the augmentation map : T 1 → T 0 has no section, because (T 1)S∞ = 0.

Can describe maps between T r’s using the partition diagrams:

T 1 δijei

T 2 ei ⊗ ej

2 T ei ⊗ ei

T 1 ei

T 0 ∅ 1

T 1 ei

However, we do not have the opposite map

T 1

T 0 ∅

Deﬁnition 6.39. The downwards partition category (dp) is the following

• objects: ﬁnite sets, • a map S → T is a partition of S q T such that all parts meet S, • composition is deﬁned as before using the diagrams. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 89

Remark 6.40. Because all partitions meet the bottom row, no parameter appears.

We have a functor

κ: (dp) → Rep(S∞) {1, . . . , r} = [r] 7→ T r

Deﬁnition 6.41. A (dp)-module is a functor (dp) → Vec. Let Mod(dp) be the category of (dp)-modules.

We then get a functor op Mod(dp) → Rep(S∞)

M 7→ Hom(dp)(M, κ). Theorem 6.42 (Sam–Snowden). This functor is an equivalence of categories of ﬁnite length objects.

Deﬁne (up) like (dp) but with the opposite condition, i.e. (up) ∼= (dp)op. Then f ∼ op f Mod(up) = (Moddp) M 7→ (S 7→ M(S)∗), where by f we mean the ﬁnite length objects. Therefore: f ∼ f Rep(S∞) = Mod(up) (with no op). Remark 6.43. This equivalence is of ⊗-categories, once one describes an ⊗-structure on the right hand side.

General construction. If F : C → D is a functor of (essentially) small categories, we get a pullback functor ∗ F : ModD → ModC which has a left adjoint F! and a right adjoint F∗ (for example, by theory of Grothendieck abelian categories). Therefore, if C is a monoidal category with monoidal structure q, then we get two ⊗- structures on ModC by using q∗ or q!. If M,N are C-modules,

M ⊗ N = q∗(M N) or q! (M N) where M N : C × C → Vec is given by (x, y) 7→ M(x) ⊗ N(y).

We can apply this to (up) with q given by disjoint union to get an ⊗-structure on Mod(up) (use q∗). This corresponds to the usual ⊗ on Rep(S∗) under the equivalence.

We have the following universal property of Rep(S∞).

Theorem 6.44. Giving a left-exact ⊗-functor Rep(S∞) → C, where C is a C-linear symmetric ⊗-category is the same as giving an object A ∈ C equipped with maps m: A ⊗ A → A, η : A → 1, ∆: A → A ⊗ A such that ... (the obvious properties hold: m is associative, ∆ is co-associative, etc). 90 ANDREW SNOWDEN

This can be proved easily using the modules over (up) perspective, but it is much harder when working with Rep(S∞) directly. Note that A is like a Frobenius algebra, but without a unit.

n We can apply the universal property with C = Rep(Sn) and A = C . This gives a left exact ⊗-functor

Γn : Rep(S∞) → Rep(Sn), r n ⊗r T 7→ (C ) called the specialization functor.

One can give an elementary description of this functor. Let Hn ⊂ S∞ be the subgroup ﬁxing ∼ each of 1, . . . , n (so Hn = S∞) and Gn ⊂ S∞ be the subgroup ﬁxing each of n + 1, n + 2,... ∼ (so Gn = Sn). We can then consider Gn × Hn ⊆ S∞, which is the correct analog of a Young subgroup.

Hn Hence if M is an S∞-representation, M is a Gn-representation, and hence an Sn-representation.

Hn Theorem 6.45. For any S∞-representation, Γn(M) = M .

Hn ∼ Hn Hn Remark 6.46. This implies (M ⊗ N) = M ⊗ N for M,N ∈ Rep(S∞).

The simple objects of Rep(S∞) are parameterized by partitions, called them Lλ. i Theorem 6.47. One can compute R Γn(Lλ) using a Borel–Weil–Bott-style formula.

f ∼ f We have seen that Rep(S∞) = Mod(up), which is useful for understanding the category f Rep(S∞) and its monoidal structure by analogy to Deligne interpolation categories. How- ever, it is not useful in understanding the abelian structure. f Note that ModC is easy to understand if all morphisms in C go in one direction. If x, y ∈ C, define x ≤ y if HomC(x, y) 6= ∅. Assume that this is a partial order. In this case, we call C directed. For fixed x, if M is a C-module, we can define a submodule M 0 by

My if y ≥ x, 0 otherwise. Therefore, if M is simple, it is concentrated in one degree. Under this assumption, simple C-modules correspond to pairs (x, V ), where x is an object of C and V is an irreducible representation of Aut(x). However, (up) is not directed. For example, we have the maps

T 1 δijei

T 2 ei ⊗ ej

2 T ei ⊗ ei

T 1 ei MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 91 so [1] ≤ [2] ≤ [1], but [2] 6= [1] in (up). Recall that FI is the category whose objects and ﬁnite sets and who morphisms are injections. Note that FI is directed. Deﬁne a functor

Φ: ModFI → Rep(S∞) M 7→ lim M . −→ n n→∞ This functor is exact.

Deﬁne Pn ∈ ModFI by (Pn)k = C[HomFI ([n], [k])]. Then Hom(Pn,M) = Mn for all FI- modules M. Therefore, these objects are projective.

Recall that x ∈ Mn is torsion if x maps to 0 ∈ Mn for some m ≥ n, and M is torsion if all its elements are. This functor is exact and kills the torsion subcategory. By the mapping property of Serre quotients, we have the induced functor

ModFI Φ: tors → Rep(S∞). ModFI | {zgen } ModFI Theorem 6.48. The functor Φ is an equivalent.

Idea of proof. We deﬁne

Ψ: Rep(S∞) → ModFI

V 7→ (Γn(V )), ∼ where Γn is the specialization functor deﬁned before. It is easy to see that ΦΨ = id and Φ and Ψ are adjoint to each other. The theorem now following from general properties of Serre quotients.

Recall for a partition λ, we have an FI-module Lλ given by setting (Lλ)n to be

n − |λ|

for n ≥ |λ| + λ1, and 0 otherwise. Recall that this is the sort of picture that appeared in the Pieri rule. We showed that every finitely generated FI-module has a finite filtration with graded pieces Lλ, up to ignoring finitely many degrees. Therefore, the Lλ’s are the simple objects of gen ModFI . 92 ANDREW SNOWDEN

∼ ∞ Recall that ModFI = ModA, where A = Sym(C ) = C[x1, x2,...] and ModA is the category of A-modules with compatible polynomial representations of GL∞. Under this identiﬁcation: torsion FI-modules ↔ torsion A-modules,

where the torsion A-modules are ones where every element is annihilated by a power of m = (x1, x2,...). Roughly: ∞ ModA ↔ GL∞-equivariant quasicoherent sheaves on A (≈ Spec(A)) torsion ↔ supported at 0. gen Therefore, we expend ModA to correspond to GL∞-equivariant quasicoherent sheaves on A∞ \{0}. gen This suggests that ModA is equivalent to representations of stabilizers of GL∞ on some ∞ point in A \{0}. Consider the ﬁrst basis vector. Then Stab(e1) in GL∞ is a generally aﬃne group GA  1 ∗   0   .   . ∗  0 ∼ ∞ and GA = GL∞ n C . The actual theorem is the following. gen ∼ pol Theorem 6.49. We have that ModA = Rep (GA).

∞ Giving a representation of GA is the same as giving a representation of C without a GL∞- equivariance. Therefore, representations of C∞ correspond to Sym(C∞)-modules. pol ∼ tors Theorem 6.50. We have that Rep (GA) = ModA . gen ∼ tors Corollary 6.51. We have that ModA = ModA .

It is easy to see that every ﬁnite length A-module has ﬁnite injective dimension.

Recall we have the specialization functor Γn : Rep(S∞) → Rep(Sn), which is left exact.

Problem: compute the derived functors of Γn. Recall that we have

=∼ gen =∼ gen V Rep(S∞) ModFI ModA

S S

(Γn(V )) ModFI ModA

i i Therefore, to compute R Γn, it is enough to compute R S. We have

S=j ∗ gen ModA ModA T =j∗ MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 93

where j : A∞ \{0} ,→ A. i i i ∞ Therefore, R S corresponds to R j∗, which is just H on A \{0}. We have     M  ∞ ∞ Spec  O(n) = A \{0} → P   n∈Z  | {z } =B gen and one can show that the simples Lλ in ModA correspond to

B ⊗ Sλ(R). Then short exact sequence

0 R C∞ O(1) 0

then shows that M i ∼ i ∞ R ΓnLλ = H (P ,B ⊗ Sλ(R)), n≥0 which is computed by BWB.

Theorem 6.52 (Deligne). If t 6∈ N then Rep(St) is semisimple and abelian.

n Basic idea: inductively decompose T into direct sums of objects Lλ and show Hom(Lλ,Lµ) = 0.

If t ∈ N, then Rep(St) is not abelian. For example,

f T 1 T 0 g

and fg = t. If t = 0, ker(f) does not exist. For other t, there are other examples that show it is not abelian. Recall that for an Ab-category, we constructed an additive envelope and then a Karoubian envelope, but it is more diﬃcult to construct an “abelian envelope” in general.

ab In this case, Deligne constructed a rigid abelian ⊗-category Rep(St) with a fully faithful ⊗-functor ab Rep(St) → Rep(St) .

1 0 ⊕(t+1) Basic idea. Work in Rep(S−1), which is abelian. Consider A = T ⊕ (T ) . This has rank t because T 1 has rank −1. We can give this the structure of a Frobenius algebra, which gives a functor

Rep(St) → ModA, ab where ModA are the A-modules in Rep(S−1). The category ModA is close to Rep(St) , but one still has to modify it slightly to make it rigid. Deligne also conjectured a universal property, which was later proved by Comes-Ostrik. 94 ANDREW SNOWDEN

Theorem 6.53. Let C be a rigid abelian ⊗-category and A ∈ C be a Frobenius algebra of rank t. Let F : Rep(St) → C be the ⊗-functor corresponding to A. Then F factors uniquely ab through either Rep(St) or Rep(St).

6.4. Some generalities on embedding into abelian categories. Suppose A is an additive category. Let us try to realize A as a full subcategory of a Grothendieck abelian category B such that every object of B is a quotient of a directs sum of objects of A. Given such an embedding, we can consider the class E of morphisms in A that are epimorphisms in B.

Remark 6.54. Note that E ⊆ {class of all epimorphisms in A}, but it can be a proper subclass. For example, if A is the category of free Z-modules, B is the category of all Z- modules. The map Z → Z given by multiplication by 2 is an epimorphism in A, but not in B.

The class E is obtained from an embedding endow A is a subcanonical6 Grothendieck topol- ogy7 We take the morphisms in E to be the covering maps. Now suppose A, E are given and E gives a subcanonical Grothendieck topology on A. We can take B to be the category of additive sheaves on abelian groups, i.e. its objects are F : Aop → Ab, where F is an additive functor and a sheaf, and the morphisms are morphisms of sheaves. For example, we can always take E to be the split epimorphisms, whence B = Fun(Aop, Ab).

We cannot always take E to be all epimorphisms. However, for A = Rep(St), you can, and f ab then B = Rep(St) .

Additional remarks on Rep(St):

• Indecomposable objects are Xλ, parameterized by partitions λ. λ λ[t] • If t ∈ N, the image of X in Rep(St) is S , where

λ[t] = (t − |λ1|, λ1, . . . , λr), if this is a partition, and it is 0 otherwise. • The dimension dim Sλ[t] is given by the Hook length formula 3.37. It is clearly a polynomial in t, say Pλ(t). For example, if λ = (2, 1), we have

6This means that the representable presheaves are sheaves. 7The notion of a Grothendieck topology can be translated into a few simple axioms. For example, if X → Y is a morphism and U → Y is in E, there exists V such that the square V U

X Y commutes. Note that this is not quite a ﬁber product, since we only assert the existence of a such a square. The other axioms also translate to similar properties. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 95

t−1 t−3 t−5 ... 2 1 2 1 1 and the Hook Length Formula shows that t! t(t − 2)(t − 4) dim Sλ[t] = = . 2 · (t − 5)! · (t − 3) · (t − 1) 2 λ • The rank rank(X ) is Pλ(t) for any t ∈ C.

There is a connection between Rep(S∞) and Deligne categories for t ∈ N. We have an ⊗-functor ab F : Rep(S∞) → Rep(St) . Theorem 6.55 (Barter, Entoru-Eizenbud, Heidersdorf). The functor F is exact.

Variant: orthogonal case. Note that O(n), the orthogonal group, acts on Cn and let T r = (Cn)⊗r. Question: what is Hom(T r,T s)? An example of such a map is s = 6 an r = 4, we have

and more generally these maps should be matchings. Similarly as before, we can deﬁne compositions using these diagrams.

≤r 6.5. Harman’s Theorem. Deﬁne RepC(Sn) to be the category of representations that are direct sums of Sλ[n] where |λ| ≤ r. Theorem 6.56. We have an equivalence of categories ≤r ≤r RepC(Sn) → RepC(Sn+1) Sλ[n] 7→ Sλ[n+1] for n ≥ 2r.

Goal. Work towards a theorem of Nate Harman [Har15] that extends this to positive characteristic. In this case, we do not get stability but instead periodicity.

Application. If M is a finitely generated FI-module in characteristic 0, Mn+1 is obtained k from Mn for n 0 via this functor. In characteristic p > 0, for n ≡ n mod p , get Mm from Mn via Nate’s equivalence. We start by treating permutation modules. Fix a field k of characteristic p > 0. Define ≤r λ[n] Permn to be the full subcategory of Repk(Sn) on objects that are direct sums of M ’s with |λ| ≤ r.

Theorem 6.57. For 2r < m < n, suppose pdlogp(r)e|n − m. Then there is an equivalence ≤r ≤r Permm → Permn M λ[m] 7→ M λ[n]. 96 ANDREW SNOWDEN

λ We will take M to be a vector space with basis hA1,...,Ari where A1 q · · · q Ar = [n] and |Ai| = λi. (So Ai is the ith row of the tabloid). Let τ be a tabloid of shape µ type λ. Deﬁne hτ : M µ → M λ X hA1,...,Asi 7→ hB1,...,Bri.

|Ai∩Bj |=τij τ µ λ Theorem 6.58. The maps h form a basis of HomSn (M ,M ).

The proof of this theorem is deferred for now, but it is similar to what we have done before. The set of tabloids of shape µ[n] and type λ[n] stabilizes once n ≥ 2 max(|µ|, |λ|). We call these the stable tabloids of shape µ, type λ. We can think of these as tabloids of shape µ[∞], type λ[∞]:

∞ long

ﬁlled in with numbers so that:

• number of 1s is λ[∞]1 = ∞, • number of 2s is λ[∞]2 = λ1, • and so on. For a stable tabloid τ, we write τ[n] for the truncation of τ to size n. Lemma 6.59. The composition

σ τ M ν h M µ h M λ

with `(ν) = t, `(µ) = s, `(λ) = r is equal to X ρ ρ cσ,τ h ρ where       s   ρ Y  X Y ρi,j  cσ,τ =    αi,j  k=1  α=(αi,j ) i,j  P   αi,j =τk,j  i P  αi,j =σi,k j MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 97

Proof. We have the following ! X 1 if A ∩ B = σ hτ (hσ(hA ,...,A i)) = hτ i j i,j hB ,...,B i 1 t 0 otherwise 1 s B X 1 if A ∩ B = σ 1 if B ∩ C = τ = i j i,j i j i,j hC ,...,C , i 0 otherwise 0 otherwise 1 r B,C X Ai ∩ Bj = σi,j = # B hC1,...,Cri. Bi ∩ Cj = τi,j C | {z } need to count this

Fix A, C and put ρi,j = |Ai ∩ Cj|. Possible choices for B1:

B1 = (A1 ∩ C1 ∩B1) q · · · q (A1 ∩ Cr ∩B1) | {z } | {z } ρ1,1 ρ1,r . .

q (At ∩ C1 ∩ B1) q · · · q (At ∩ Cr ∩ B1)

Let αi,j = |Ai ∩ Cj ∩ B1|. The result then follows.

An integer-valued polynomial is a polynomial F ∈ Q[t] such that F (n) ∈ Z for all n ∈ Z. Let R be the ring of integer-valued polynomials.

ρ[n] Lemma 6.60. Let ρ, σ, τ be stable tabloids. Then cσ[n],τ[n] is an integer valued polynomial.

Proof. By Lemma 6.59:         s  ρ[n]  ρ[n] Y  X Y i,j  Cσ[n],τ[n] = .  αi,j  k=1  α=(αi,j ) i,j  P   αi,j =τ[n]k,j  i P  αi,j =σ[n]i,k j It is clearly integer valued, and we just need to show it is a polynomial. Suppose k > 1. Then X αi,j = τ[n]k,j i is independent of n. Hence αi,j is bounded and independent of n. Moreover, constant if i 6= 1 or j 6= 1 ρ[n] = i,j constant + n if i = j = 1 Therefore ( constant ρ[n] if i 6= 1 or j 6= 1 i,j = constant constant+n αi,j constant if i = j = 1 This is clearly a polynomial in n. 98 ANDREW SNOWDEN P Now suppose k = 1. For j > 1, αi,j = τ[n]i,j is constant, so αi,j is bounded, independent P i of n. For i > 1, j σ[n]i,k is constant, so αi,j is bounded independent of n.

Hence all αi,j’s except α1,1 are bounded independent of n. Then X (α11 − n) + αi,1 = τ[n]1,1 − n = constant. i6=1

Hence α1,1 = constant + n. One then proceeds with a similar argument to the above to conclude the result. x Theorem 6.61 (Polya). The ring R has a Z-basis of x 7→ k for k ≥ 0. Proposition 6.62. For f ∈ R of degree d, f(n) mod p are periodic with period dividing pdlogp(d)e.

Proof. Lucas congruence: if r n = n0 + n1p + ··· + nrp , r m = m0 + m1p + ··· + mrp , then n n n n ≡ 0 1 ··· r mod p, m m0 m1 mr

n dlogp(d)e so m mod p only depends on the class on n modulo p .

Theorem 6.57 now follows for permutation modules. We will now show how to go from the permutation modules to the general case. Let C be a highest weight category with poset Λ, ∆(λ) be the standard object and ∇(λ) be the costandard object. Then let C∆ = category of objects with filtration having ∆(λ) as graded pieces, C∆ = category of objects with filtration having ∇(λ) as graded pieces. Proposition 6.63. For any λ ∈ Λ, there is a unique T (λ) ∈ C (up to isomorphism) such that (1) T (λ) ∈ C∆ ∩ C∇, (2) simple constituents of T (λ) are L(λ) with multiplicity 1 and L(µ) with µ < λ, (3) T (λ) is indecomposable. Definition 6.64. Modules of the form T (λ), or direct sums of T (λ)’s, are called tilting modules.A full tilting module is one that contains all of T (λ)’s as summands.

Theorem 6.65 (Ringel duality). Let T be a full tilting module, A = EndC(T ). Then

op (1) ModA is a highest weight category with poset Λ , whose standard objects are HomC(∆(λ),T ), (2) ModA is independent of the choice of T , up to equivalence; this category is denoted C∨ and called the Ringel dual, (3) we have an equivalence C ∼= (C∨)∨.

Let Tilt(C) be the full subcategory on tilting modules. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 99

Corollary 6.66. Let C and C0 be highest weight categories with poset Λ. Then any equivalence Tilt(C) ∼= Tilt(C0) taking T (λ) to T 0(λ) extends to an equivalence C ∼= C0 respecting the highest weight structure.

Suppose C is a highest weight category with poset Λ. Let Λ0 ⊂ Λ be downwards closed (i.e. 0 0 0 if λ ∈ Λ and µ ≤ λ then µ ∈ Λ ). Let CΛ0 be a Serre subcategory of C on L(λ) with λ ∈ Λ . 0 Proposition 6.67. The category CΛ0 is a highest weight category with poset Λ .

Schur-Weyl duality. pol Let Cn = Rep (GL∞)n. We have

T : Cn → Rep(Sn) V 7→ 1n-weight space of V

S : Rep(Sn) → Cn V 7→ ((k∞)⊗n ⊗ V )Sn

Then Cn is a highest weight category with Λ consisting of partitions of n where ∆(λ) is the λ† λ Weyl module, S(S ) and M is the permutation module in Rep(Sn). Fact 6.68. There are unique indecomposable summands Y λ (the Young submodules) of M λ containing Sλ.

Let T 0 = T ⊗ sgn, S0 = S(− ⊗ sgn). Proposition 6.69. In characteristic at least 5, we have

λ† † † D if λ isp-regular (1) T 0(∆(λ)) = Sλ , T 0(P (λ)) = Y (λ) ⊗ sgn, T 0(L(λ)) = 0 otherwise, T 0(T (λ)) = Y (λ†), (2) S0(Sλ† ) = ∆(λ), S0(Y (λ†)) = T , 0 0 ∆ (3) S , T are mutually inverse equivalences on the subcategory Cn and the subcategory of Rep(Sn) on modules admitting a ﬁltration by Specht modules.

We previously saw that ≤r ∼ ≤r Permn = Permm M λ[n] 7→ M λ[m]

if pdlogp(r)e|n − m.

≤r λ[n] Let Youngn be the subcategory of Rep(Sn) on all direct sums of Y with |λ| ≤ r.

≤r ∼ ≤r dlogp(r)e Corollary 6.70. We have that Youngn = Youngm if p |n − m.

≤r ≤r Proof. This follows from the above since Youngn is the Karoubian envelope of Permn .

≤r † Let Cn be the Serre subcategory of Cn on L(λ[n] ) with |λ| ≤ r. 100 ANDREW SNOWDEN

Theorem 6.71. In characteristic at least 5, we have ≤r ∼ ≤r Cn = Cm as highest weight categories, if pdlogp(r)e|n − m.

≤r † Proof. Note that Cn is a highest weight category with poset {λ[n] | |λ| ≤ r}, independent ≤r † of n for n 0. Tilting modules for Cn are T (λ[n]) with |λ| ≤ r. We have that ≤ ∼ ≤r ∼ ≤r ∼ ≤r Tilt(Cn ) = Youngn = Youngm = Tilt(Cm ) and the equivalence sends T (λ[n]†) 7→ T (λ[m]†). ≤r ∼ ≤r By Corollary 6.66, we obtain Cn = Cm

≤r ≤r Let Rep(Sn)im be the essential image of Cn . ≤r ∼ ≤r Theorem 6.72. We have Rep(Sn)im = Rep(Sm)im extending the equivalence on permutation categories if pdlogp(r)e|n − m and characteristic is at least 5.

≤r sing ≤r † Proof. Deﬁne (Cn ) to be the Serre subcategory of Cn on L(λ[n] ) with λ p-singular. (These simples are in ker(T 0).) We have an equivalence ≤r ∼ ≤r ≤r sing Rep(Sn)im = Cn /(Cn ) . 0 ≤r ≤r ≤r sing The reason is that T : Cn → Rep(Sn)im is an exact functor which kills (Cn ) , and 0 ≤r ≤r 0 S : Rep(Sn)m → Cn is adjoint to T . The equivalence then follows from general facts about Serre quotients. ≤r ∼ ≤r Under the equivalence Cn = Cm , the singular subcategories coincide, because the highest weight structure is respected. The result now follows.

≤r ≤r Let Rep(Sn) be the smallest abelian subcategory of Rep(Sn) containing Permn . ≤r ∼ ≤r Theorem 6.73. We have an equivalence Rep(Sn) = Rep(Sn) extending the equivalence on the permutation categories if pdlogp(r)e|n − m and the characteristic is at least 5.

Proof. We know from the previous theorem that

≤r =∼ ≤r Rep(Sn)im Rep(Sm)im

≤r =∼ ≤r Permn Permm and this proves the result.

One can then deduce the following theorem. Theorem 6.74. The following sequences are periodic with period equal to a power of p: (1) Sλ[n] : Dλ[n], (2) M λ[n] : Y µ[n], MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 101

(3) modular kronecker coeﬃcients Dλ[n] ⊗ Dµ[n] : Dν[n], (4) modular versions of Littlewood-Richardson coeﬃcients: h i IndSn (Dν ⊗ Dµ[n−k]): Dλ[n] . Sk×Sn−k

Deﬁne Xi : Sn → Z by Xi(σ) = #(i-cycles in σ). This is a class function.

Fact 6.75. In characteristic 0, there is a polynomial Pλ ∈ Q[X1,X2,...] such that

χSλ[n] (σ) = Pλ(X1(σ),X2(σ),...) for n 0. Theorem 6.76 (Nate). Given λ, there exists ` ∈ N and P ∈ [X ,X ,...] for 0 ≤ j ≤ p` λij Q 1 2 such that

χ λ (σ) = Pλ (X1(σ),X2(σ),...) bD [n] ij ` for n 0, j ≡ n mod p , where by χbDλ[n] is a Brauer character.

λ[n] µ[n] Proof. We can express [D ] in K(Sn) in terms of [S ]’s, and this expression is periodic in p. We get a similar relation betweenχ ˆDλ[n] andχ ˆSµ[n] = χSµ[n] , where the latter is a characteristic 0 character. Then Theorem 6.76 follows from Fact 6.75.

Applications to FI-modules. If M is a ﬁnitely-generated FI-module in characteristic 0, there exist λ1, . . . , λr such that

∼ λ1[n] λr[n] Mn = S ⊕ · · · ⊕ S for n 0. It is not obvious how to carry this over to characteristic p > 0.

n Example 6.77. Let Mn = k , a ﬁnitely-generated FI-module (where char(k) = p > 0).

(1) Decomposition into irreducibles: ( 2[Dn] + [D(n−1,1)] if p|n [Mn] = [Dn] + [D(n−1,1)] otherwise which is not stable but it is periodic mod p. (2) Decomposition into Specht modules:

n (n−1,1) [Mn] = [S ] + [S ] independently of n.

Both of these properties hold for general ﬁnitely-generated FI-modules. We start by explaining (1). We actually give a more general result of which (1) is a corollary. Theorem 6.78. Suppose M is a ﬁnitely-generated FI-module. Then there is an r such ≤r ∼ ≤r that Mn and Mm correspond under Rep(Sn) = Rep(Sm) for all n, m 0 with n ≡ m mod pdlogp re. 102 ANDREW SNOWDEN

Proof. Let I(q) be a free FI-module in degree q with ≤q I(q)n = k[HomFI ([q], [n])] ∈ Permn .

Since Repk(FI) is locally noetherian, we have a presentation

P1 → P0 → M → 0 where each Pi is a ﬁnite direct sum of I(q)’s. The result now follows. Corollary 6.79. Let M be a ﬁnitely generated FI-modules.

µ[n] (1) The expression for [Mn] in terms of [D ]’s is periodic. (This gives the analog of (1) above.) (2) Decomposing Mn into indecomposables is also periodic. i (3) For ﬁxed i, dim H (Sn,Mn) is periodic in n.

It is not clear how to deduce (3) from the above, but Harman claims it does follows. It is however deﬁnitely true and it was proved independently by Nagpal [Nag15].

For (2), we need a Theorem of Nagpal [Nag15]. If V is a representation of Sq, deﬁne an FI-module I(V ), an induced FI-module, by I(V ) = IndSn (V ⊗ triv). q Sq×Sn−q

For example, if V = k[Sq], I(V ) = I(q). We say that an FI-module M is semi-induced if it has a finite filtration such that the graded pieces are induced. Theorem 6.80 (Nagpal). Let M be a finitely-generated FI-module. Then there exists a complex

0 → M → P0 → · · · → Pt → 0 with Pi semi-induced and a complex is exact in degree n 0.

In characteristic 0, this is a theorem by Snowden and Sam and one can even take Pi to be induced. Theorem 6.81 (Harman). Let M be a ﬁnitely-generated FI-module. Then there exist partitions λ1, . . . , λr and integers c1, . . . , cr such that

λ[n] λr[n] [Mn] = c1[S ] + ··· + cr[S ] for all n 0. In fact, we may assume that λi are p-regular (and hence the c’s are unique).

Proof. By Nagpal’s Theorem 6.80, we can reduce to the can M = I(V ) with V = Sλ. Then M = IndSn (Sλ ⊗ triv) n Sq×Sn−q and the result follows from Pieri’s rule.

One can prove the periodicity of cohomology claimed in Corollary 6.79 (3) in another way. We present the sketch of an argument following [NS17]. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 103

Suppose M is an FI-module. Then

M i i H (Sn,Mn) = Γ (M) n≥0 by deﬁnition. Question. What structure does Γi(M) have that implies periodicity of dimensions?

0 N Recall that A = k[t] is a twisted commutative algebra. Note that Γ : Rep(S∗) → Vec is a tensor functor. Hence Γi(M) is a Γ0(A)-module.

Note that D = Γ0(A) is a divided power algebra. Let k be a commutative ring. A divided power algebra D has a k-basis x[n] with n + m x[n] · x[m] = x[n+m]. m If k is a field of characteristic p, p ∼ k[y0,...]/(y0,...) = D [pi] yi 7→ x . Theorem 6.82. The divided power algebra D is coherent (i.e. finitely generated ideals are finitely presented). Theorem 6.83. If M is a finitely generated FI-module over a noetherian ring k, then Γi(M) is (almost) a finitely-presented D-module.

Main problems for twisted commutative algebras.

(1) Noetherian property. (2) Structure of modules. Conjecture 6.84. All ﬁnitely generated twisted commutative algebras are noetherian.

Status.

• Any ﬁnitely generated bounded twisted commutative algebra is noetherian. For example, Sym(C∞), Sym(U ⊗ C∞) for dim U < ∞ are noetherian. • We have that 2 ∞ L ∞ Sym(Sym C ) = λ Sλ(C ) all parts even Sym(Λ2C∞) are both noetherian. This is a theorem due to Nagpal–Sam–Snowden [NSS15]. • In Summer 2017, Draisma proved [Dra17] that ﬁnitely-generated twisted commutative algebras are topologically noetherian.

∞ Structure of ModA for A = Sym(C ). 104 ANDREW SNOWDEN

b Theorem 6.85 (Structure theorem). Let M ∈ Dfg(A), the bounded, ﬁnitely generated derived category of A. Then there exists an exact triangle

T M F T [−1]

b with T,F ∈ Dfg(A) and T a complex of torsion modules, F a complex of projective modules. fg Corollary 6.86. The Λ-module K0(ModA ) is free of rank 2 with basis [C] and [A].

As an application, for an A-module M, deﬁne the Hilbert series as X tn H (t) = dim M M n n! n≥0

and note that HM⊗N = HM · HN . t Theorem 6.87. If M is ﬁnitely-generated, then HM (t) = p(t)e + q(t) for p, q ∈ Q[t].

Deﬁne the generalized Hilbert series as X tλ HeM (t1, t2,...) = tr(Cλ|M ) λ! λ partition where Cλ is a conjugacy class in S|λ| corresponding to λ, λ m1(λ) m2(λ) t = t1 t2 ..., λ! = m1(λ)!m2(λ)! ...,

mi = #(is in λ). Theorem 6.88. If M is ﬁnitely-generated, then

T0 HeM (t) = p(t)e q(t) P where p, q ∈ Q[t1, t2,...] and T0 = i≥1 ti. Moreover, p(t) is essentially a characteristic polynomial (up to conjugation of variables).

0 i Local cohomology. Let Hm(M) be a torsion submodule of M and Hm be the right derived 0 functor of Hm. i Theorem 6.89. If M is ﬁnitely-generated, Hm(M) is ﬁnite length, vanishes for i 0. Remark 6.90. We can recover q’s in the Hilbert series from local cohomology.

General remarks. Let A be an abelian ⊗-category with some extra properties. Let A be a unit object. Then we can think of A as ModA. An ideal of A is a subobject of A. For ideals a, b ⊆ A, we get a map a ⊗ b → A ⊗ A = A and we can deﬁne ab to be the image of this map. Deﬁnition 6.91. • The object A is a domain if ab = 0 implies a = 0 or b = 0. MATH 711: REPRESENTATION THEORY OF SYMMETRIC GROUPS 105

• An ideal p ⊂ A is prime if A/p is a domain. • We let Spec(A) be the set of all prime ideals with the usual Zariski topology. • For a domain A, we can deﬁne Frac(A) by

ModA ModFrac(A) = tors . ModA • For a prime p, we get a residue “ﬁeld” given by Frac(A/p).

Let A = Sym(U ⊗ C∞) for U ﬁnite dimensional. Then dim U [ Spec(A) = Gr(U) = Grr(U), r=0

the total Grassmanian. We can deﬁne the Serre subquotient ModA,r corresponding to Grr. Proposition 6.92. We have that ∼ 0 ModA,r = ModB ∞ where B = Sym(Q ⊗ C ) on Grr(U) and the 0 means supported at 0.

Corollary 6.93. We have that K0(ModA,r) = Λ ⊗ K0(Grr(U)), where Λ is the ring of symmetric functions. dim U L dim U Corollary 6.94. We have that K0(ModA) = is free of rank 2 as a Λ-module. r=0

For details of this, see [SS17]. Consider A = Sym(Sym2). Then 2 Spec(A) = N ∪ {∞}. The “residue ﬁeld” at (r, s) is closely related to Rep(OSp(r|s)), the ortho-symplectic group. alg The residue ﬁeld at ∞ is Rep (O∞).

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