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Schur-Weyl duality over finite fields

Stephen Doty
September 08, 2014

Schur-Weyl duality over finite fields

Talk at Jena (Algdar2014), 8 September 2014

Stephen Doty

September 08, 2014
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  1. Notation k = any field. V = vector space over

    k (dim k V = n). G = GL(V ) GLn(k), Sr = symmetric group on {1,...,r}. Note: G depends on k. In particular, |G| < ∞ if |k| < ∞. Have natural commuting actions: G V ⊗r Sr . Thus V ⊗r is a (kG,kSr )-bimodule, and the actions induce algebra homomorphisms kG ϕ −→ EndSr (V ⊗r ), kSop r ρ −→ EndG (V ⊗r ).
  2. Defining the actions The left action is by G =

    GL(V ) acting diagonally g ·(x1 ⊗···⊗xr ) = g(x1)⊗···⊗g(xr ), g ∈ GL(V ). The right action of Sr is by place-permutation on the tensor positions (x1 ⊗···⊗xr )·π = x π(1) ⊗···⊗x π(r) , π ∈ Sr . For convenience, let’s change the right action into a left one, by composing with the map π → π−1: so now π ·(x1 ⊗···⊗xr ) = x π−1(1) ⊗···⊗x π−1(r) , π ∈ Sr . This turns the map ρ into a homomorphism kSr ρ −→ EndG (V ⊗r ).
  3. Previous results kG ϕ −→ EndSr (V ⊗r ), kSr

    ρ −→ EndG (V ⊗r ). QUESTION: When are the maps ϕ,ρ surjective? (If so, then each ring of invariants is generated by the image of the other action.) [Schur, 1927]: Both maps are surjective if k = C. [Thrall, 1942]: In positive characteristic p, the map ϕ is surjective if |k| ≥ r and r < 2p. (But this is wrong! Counterexample at the end.) [J.A. Green, 1980]: The map ϕ is surjective if |k| = ∞. [Carter & Lusztig, 1974]: The map ρ is surjective (in fact an isomorphism) if |k| = ∞ and n ≥ r. Their proof relies on [Thrall, 1944]. [De Concini & Procesi, 1976]: The map ρ is surjective if |k| = ∞.
  4. So by 1980 it was known that both maps ϕ,ρ

    are surjective whenever k is an infinite field. (Other proofs, and many generalizations, since 1980.) QUESTION: What about finite fields? Note: We didn’t know of Thrall’s paper when we wrote our first draft. Theorem (Main result) Both maps ϕ,ρ are surjective whenever |k| > r. The bound in the theorem appears to be sharp; i.e., there are counterexamples when |k| = r. The rest of the talk is an outline of the proof.
  5. The Schur algebra Let k be any commutative ring. Let

    A k (n) = k[cij ] be the commutative polynomial algebra in n2 variables cij (1 ≤ i,j ≤ n). This is graded by regarding each generator cij as having degree 1: A k (n) = r≥0 A k (n,r) where A k (n,r) is the sub-coalgebra of homogeneous polynomials of degree r. This is a coalgebra under the coproduct ∆(cij ) = ∑n k=1 cik ⊗ckj . (This defines ∆ on generators; extend to an algebra morphism.) Definition 1 (J.A. Green, 1980) The Schur algebra (in degree r) is the algebra S k (n,r) = A k (n,r)∗. It is well known that the linear dual of any coalgebra is an algebra.
  6. Basis for Schur algebra Put I(n,r) = {1,...,n}r . For

    a fixed basis {v1,...,vn} of V , define vj = vj1 ⊗···⊗vjr , any j = (j1,...,jr ) ∈ I(n,r). Then the set {vj : j ∈ I(n,r)} is a basis of V ⊗r . For i = (i1,...,ir ), j = (j1,...,jr ) in I(n,r) write ci,j = ci1,j1 ···cir ,jr . Then Sr acts by place-permutation on I(n,r) and thus on I(n,r)×I(n,r). We have the equality rule: ci,j = ck,l ⇐⇒ (i,j), (k,l) lie in the same Sr -orbit of I(n,r)×I(n,r). If Ω is a set of Sr -orbit representatives then {ci,j : (i,j) ∈ Ω} is a basis of A k (n,r). Let {ξi,j : (i,j) ∈ Ω} be the dual basis of S k (n,r) = A k (n,r)∗.
  7. Schur–Weyl duality with Schur algebras For any ξ ∈ S

    k (n,r) and any j ∈ I(n,r) define an action of S k (n,r) on V ⊗r by ξ ·vj = ∑i∈I(n,r) ξ(ci,j )vi . This action commutes with the place-permutation action of Sr , so we have induced maps S k (n,r) → EndSr (V ⊗r ), kSr → EndS k (n,r) (V ⊗r ). Lemma 1 (Green, Bryant) (a) For any commutative ring k, the first map is an isomorphism; i.e., we have S k (n,r) ∼ = EndSr (V ⊗r ). (b) For an arbitrary field k, the second map kSr → EndS k (n,r) (V ⊗r ) is surjective. Part (a) goes back to Schur. The argument given in J.A. Green, 1980 [SLN 830, (2.6c)] works in this generality. Part (b) appears in a paper of R.M. Bryant, 2009 [Beitr¨ age Algebra Geom. 50].
  8. The main issue Recall the map kG ϕ −→ EndSr

    (V ⊗r ). The central problem is that the image ϕ(kG) does not have to agree with the Schur algebra, if k is a finite field. Because in that case we cannot necessarily identify A k (n) = k[cij ] with the ring of polynomial functions on G = GL(V ). Goal: We will show that ϕ(kG) is isomorphic to the Schur algebra, provided that the field k has at least r +1 elements. The argument is based on the Chevalley group construction. It is a modification of an argument in Section 3.1 of [Carter-Lusztig, 1974].
  9. Enveloping algebras We need to pay attention to change of

    scalars, so from now on we write V k instead of V , to emphasize the ground ring k. Consider the complex Lie group GLn(C) and its Lie algebra g := gln (C). Let U = U C (g) be its universal enveloping algebra over C. This is the associative algebra generated by eij (1 ≤ i,j ≤ n) subject to the defining relations [eij ,eab] = δaj eib −δibeaj where δ is Kronecker’s delta function. Let U be the subalgebra of U generated by all eij with i = j. Then we can identify U with U C (sln).
  10. Algebras of distributions Let U Z be the algebra of

    distributions (the Kostant Z-form) on the algebraic group (scheme) GLn defined by the Hopf algebra A Z (n)det. This is the subring of U generated by all divided powers (1) em ij m! (1 ≤ i = j ≤ n, m ≥ 0) along with the binomial coefficients (2) eii m := eii (eii −1)···(eii −m +1) m! (1 ≤ i ≤ n, m ≥ 0). Let U Z be the algebra of distributions on the algebraic group (scheme) SLn defined by the Hopf algebra A Z (n)/(det−1). This is the subring of U Z generated by only the divided powers in (1) above.
  11. Z-forms and k-forms Let V (I am now changing the

    definition of V ) be an n-dimensional complex vector space, and identify V with Cn via a fixed basis {v1,...,vn}. Make g = gln (C) act on V by eij ·va = δaj vi . This is the “natural” action if we identify gln (C) with gl(V ) via the basis. Then V , and hence also V ⊗r , become U-modules. They are also U -modules by restriction. Put V Z := ∑n j=1 Zvj . This is stable under the action of U Z ⊂ U. Hence V ⊗r Z is also stable. Hence for any commutative ring k, V k := k⊗ Z V Z and V ⊗r k ∼ = k⊗ Z V ⊗r Z are U k -modules, where U k = k⊗ Z U Z is the algebra of distributions on the algebraic k-group (scheme) GLn,k . They become also U k -modules by restriction, where U k = k⊗ Z U Z is the algebra of distributions on the algebraic k-group (scheme) SLn,k .
  12. Second Lemma The action of U k on V ⊗r

    k commutes with the place-permutation action of the symmetric group Sr . Hence we have induced algebra homomorphisms U k ψ −→ EndSr (V ⊗r k ), U k ψ −→ EndSr (V ⊗r k ) where the second is the restriction of the first. By part (a) of the preceding lemma, these maps are going into the Schur algebra S k (n,r). Lemma 2 For any m > r the elements em ij /(m!), eii m ∈ U act as zero on V ⊗r . Hence the corresponding elements 1⊗em ij /(m!), 1⊗ eii m ∈ U k act as zero on V ⊗r k . This is proved by some easy computations with weight vectors in tensor space.
  13. The Chevalley groups GL(V k ), SL(V k ) Henceforth

    assume that k is a field. Define elements Eij (t) ∈ End k (V k ) by Eij (t) = 1+tψ(1⊗eij ), t ∈ k. Then for i = j we have Eij (t)Eij (t ) = Eij (t +t ). In particular, Eij (t)Eij (−t) = 1 = Eij (−t)Eij (t), so Eij (t) ∈ G = GL(V k ) for any i = j. In fact, Eij (t) ∈ SL(V k ) and SL(V k ) is generated by these elements. Furthermore, Eii (t) ∈ G = GL(V k ) provided t = −1 in k. Easy to check that G = GL(V k ) is generated by SL(Vk) and the Eii (t) for t = −1. (By Gaussian elimination.)
  14. Relating the maps ψ and ϕ Recall the map kG

    = kGL(V k ) ϕ −→ EndSr (V ⊗r k ). Let kSL(V k ) ϕ −→ EndSr (V ⊗r k ) be its restriction. We also have maps U k ψ −→ EndSr (V ⊗r k ) and U k ψ −→ EndSr (V ⊗r k ). Lemma 3 Suppose that k is a field. (a) If the order of k is strictly bigger than r +1 then ϕ is surjective if and only if ψ is surjective. (b) If the order of k is strictly bigger than r then ϕ is surjective if and only if ψ is surjective.
  15. Proof sketch For the proof, we use the equation (∗)

    ϕ(Eij (t)) = r ∑ m=0 tm ψ 1⊗ em ij m! , i = j. This can be checked by direct computation; e.g., see formula (25) of [Carter–Lusztig, 1974]. We also use (∗∗) ϕ(Eii (t)) = r ∑ m=0 tm ψ 1⊗ eii m , t = −1 which is also easily checked directly. Since the elements ψ 1⊗ em ij m! for i = j, 0 ≤ m ≤ r generate the image of ψ and the elements ϕ(Eij (t)) for i = j, t ∈ k generate the image of ϕ , it follows from (∗) that imϕ ⊂ imψ .
  16. Proof sketch, continued The image of ψ is generated by

    the image of ψ along with the elements ψ 1⊗ eii m and the image of ϕ is generated by the image of ϕ along with the elements ϕ(Eii (t)), for t = −1. Hence (∗∗) implies that imϕ ⊂ imψ. Pick r +1 distinct elements t0, t1,...,tr of k (possible by hypothesis). Then (∗) gives the system of linear equations ϕ(Eij (t0)) = 1+t0ψ(1⊗eij )+t2 0 ψ(1⊗ e2 ij 2! )+···+tr 0ψ(1⊗ er ij r! ) ϕ(Eij (t1)) = 1+t1ψ(1⊗eij )+t2 1 ψ(1⊗ e2 ij 2! )+···+tr 1ψ(1⊗ er ij r! ) . . . . . . . . . ϕ(Eij (tr )) = 1+tr ψ(1⊗eij )+t2 r ψ(1⊗ e2 ij 2! )+···+tr r ψ(1⊗ er ij r! ).
  17. Proof sketch, continued The coefficient matrix of the linear system

    is a Vandermonde matrix of the form      1 t0 t2 0 ··· tr 0 1 t1 t2 1 ··· tr 1 . . . . . . . . . ... . . . 1 tr t2 r ··· tr r      . Since the values t0,t1,...,tr are distinct, the matrix is invertible over k, so there are scalars bml ∈ k such that ψ(1⊗ em ij m! ) = r ∑ l=0 bml ϕ(Eij (tl )) for each m. This shows that the generators ψ(1⊗ em ij m! ) of the image of ψ all lie in the image of ϕ on SL(Vk), proving that imψ ⊂ imϕ (typo in the paper!). This completes the proof of part (b) of the lemma. To complete the proof of part (a) one applies the same technique to (∗∗), avoiding t = −1, in order to get the opposite inclusion imψ ⊂ imϕ.
  18. Surjectivity over Z Lemma 4 (Carter–Lusztig, 1974) The maps U

    Z ψZ −→ EndSr (V ⊗r Z ), U Z ψZ −→ EndSr (V ⊗r Z ) are surjective. The proof relies on the following fact: If A f −→ B is any homomorphism of free abelian groups of finite rank, then f is surjective iff k⊗ Z A 1⊗f −→ k⊗ Z B is surjective for every algebraically closed field k. Suppose that k is an algebraically closed field. Since ϕ is surjective it follows from Lemma 3 that ψ = 1⊗ψZ is, too. Since k is algebraically closed, the surjectivity of ϕ follows from the surjectivity of ϕ, as GL(V k ) is generated by SL(V k ) and the scalar matrices. Thus, Lemma 3 implies that ψ = 1⊗ψZ is surjective.
  19. Proof of the main result For any commutative ring k

    it follows from Lemma 4 that the maps (†) U k ψ −→ EndSr (V ⊗r k ), U k ψ −→ EndSr (V ⊗r k ) are surjective, by right exactness of tensor product (since ψ = 1 k ⊗ψZ , ψ = 1 k ⊗ψZ ). Suppose now that k is a field with at least r +1 elements. Then the maps in (†) are surjective. By Lemma 3, kSL(V k ) ϕ −→ EndSr (V ⊗r k ) is surjective. Since ϕ is the restriction of kGL(V k ) ϕ −→ EndSr (V ⊗r k ), we get surjectivity of ϕ, too. Put G = GL(V k ). Then EndG (V ⊗r k ) = End kG (V ⊗r k ) = End ϕ(kG) (V ⊗r k ) = EndS k (n,r) (V ⊗r k ). Thus kSr ρ −→ EndG (V ⊗r k ) is surjective, by Lemma 1.
  20. Summary Let’s summarize what we proved. Theorem If |k| >

    r then ϕ(kG) = EndSr (V ⊗r k ) and ρ(kSr ) = EndG (V ⊗r k ). A similar statement holds if G = GL(V k ) is replaced by SL(V k ). Note that we got a bit more than what we started out to show, namely, that this holds for SL(V k ) as well.
  21. Schur–Weyl duality for U Z Replacing the operators GL(V k

    ) by U k leads to a formulation of Schur–Weyl duality that works for any commutative ring k. Corollary The natural map ZSr σ −→ EndU Z (V ⊗r Z ) is surjective. The same holds with U Z in place of U Z . It is enough to prove that the corresponding map kSr 1 k ⊗σ −→ EndU k (V ⊗r k ) is surjective for every algebraically closed field k. But if k = k then EndU k (V ⊗r k ) = End ψ(U k ) (V ⊗r k ) ∼ = EndS k (n,r) (V ⊗r k ) ∼ = EndG (V ⊗r k ) so the result follows for U Z . The argument for U Z is similar.
  22. Schur–Weyl duality for Uk If the group G = GL(V

    k ) is replaced by the algebra U k of distributions then we have a formulation of Schur–Weyl duality that works over any commutative ring k. There is no restriction on k. Corollary Let k be any commutative ring. Then the images of the commuting actions of U k and kSr on V ⊗r k are EndSr (V ⊗r k ) and EndU k (V ⊗r k ) respectively. The same holds if U k is replaced by U k . This follows from the preceding corollary by right exactness of tensor product. Note that the first surjectivity over Z is Lemma 4.
  23. Counterexample for Thrall’s Theorem Return to the group GL(V k

    ) acting. The following result is Theorem III of [Thrall, Annals of Math. 45 (1942)], translated into modern language. Theorem (Thrall) Suppose |k| ≥ r and r < 2p, where k has characteristic p > 0. Then the map kGL(V k ) → EndSr (V ⊗r k ) is surjective. Take k = F2, the field of 2 elements. Let dim k V k = 2 (so n = 2). Let r = 2 as well. Then dim k kGL(V k ) = 6 and dim k S k (2,2) = 10. So the map kGL(V k ) → EndS2 (V k ⊗V k ) ∼ = S k (2,2) induced by the diagonal action of G = GL(V k ) on V k ⊗V k cannot be surjective. So the theorem cannot be correct as stated. Note: Thrall’s methods are very different from ours.