Schur-Weyl duality and the free Lie algebra


Stephen Doty

July 14, 2016


  1. 1.

    Schur–Weyl duality and the free Lie algebra S.R. Doty (Joint

    with J. Matthew Douglass) – Preprint on arXiv – Representations of Algebraic Groups York University 14 July 2016
  2. 2.

    Schur–Weyl duality Let k be a field and V a

    finite dimensional vector space over k. Put E = End k (V ). For any subset X of E we have the centralizer algebra ZE (X) = {ϕ ∈ E : ϕx = xϕ, all x ∈ X}. Suppose further that AVB is an (A, B)-bimodule, where A and B are algebras over k. The bimodule structure on V induces representations (algebra morphisms) A Φ −→ E Ψ ←− Bop. Write A = Φ(A), B = Ψ(B) for the images of A, B in E, noting that B = Bop as sets. The commuting actions of A and B induce inclusions (∗) A ⊆ EndB(V ), B ⊆ EndA(V ) where we write EndA(V ) = ZE (A), EndB(V ) = ZE (B). Whenever the inclusions in (∗) are equalities, we say that the (A, B)-bimodule AVB satisfies Schur–Weyl duality (SWD). This means that the image of each action is the full centralizer of the other. It implies that ZE (ZE (A)) = A, ZE (ZE (B)) = B.
  3. 3.

    The free Lie algebra Suppose that k is a field

    and let V be a fixed n-dimensional vector space over k. Let T(V ) = r≥0 Tr (V ) = r≥0 V ⊗r be the tensor algebra of V , where Tr (V ) = V ⊗r = V ⊗ · · · ⊗ V (r times) is the rth tensor power of V . Regard T(V ) as a Lie algebra by means of the Lie bracket [a, b] := ab − ba. By definition, the free Lie algebra Lie(V ) on V is the Lie subalgebra of T(V ) generated by V . The grading on T(V ) naturally induces a corresponding grading Lie(V ) = r≥0 Lier (V ) on the free Lie algebra, where Lier (V ) := Lie(V ) ∩ Tr (V ).
  4. 4.

    Lie polynomials If we fix a basis X = {x1,

    . . . , xn} of V then we identify V = kn via the basis. Then the tensor algebra T(V ) may be identified with T(V ) = k X where k X is the free associative algebra on the set X. In this viewpoint, elements of T(V ) = k X are regarded as noncommutative polynomials in the variables X = {x1, . . . , xn}. Then elements of Lie(V ) are known as Lie polynomials and the subspace Lier (V ) is the subspace of homogeneous Lie polynomials of degree r. The general linear group GL(V ) acts naturally on V and hence on T(V ). The action preserves the Lie bracket; i.e., g · [a, b] = [ga, gb] (g ∈ GL(V ), a, b ∈ T(V )) so Lie(V ) is a (left) k GL(V )-module. Since the action also preserves Tr (V ), we have an induced action of GL(V ) on Lier (V ), making Lier (V ) a k GL(V )-module.
  5. 5.

    Classical Schur–Weyl duality The symmetric group Sr on r symbols

    acts naturally (on the right) on the rth tensor power Tr (V ), by place-permutation. This action commutes with the left action of GL(V ), so Tr (V ) is a (k GL(V ), kSr )-bimodule. Classical Schur–Weyl duality is the statement that for the bimodule Tr (V ), the image of each action in End k (Tr (V )) is equal to the full centralizer of the other. In other words, we have representations (k-algebra morphisms) k GL(V ) Φ −→ End k (Tr (V )) Ψ ←− (kSr )op such that Φ(k GL(V )) = EndSr (Tr (V )), Ψ(kSr ) = EndGL(V ) (Tr (V )). This holds for any field k in arbitrary characteristic, provided only that |k| > r [Benson–D, 2009]. (Over C this is due to Schur; for an infinite field see [Green 1980, DeConcini–Procesi 1976].)
  6. 6.

    Questions Since GL(V ) acts on the rth homogeneous component

    Lier (V ) of the free Lie algebra, it is natural to ask what happens when the classical bimodule k GL(V ) Φ −→ End k (Tr (V )) Ψ ←− kSr is replaced by some (yet to be described) bimodule structure k GL(V ) Φ −→ End k (Lier (V )) Ψ ←− B. In other words: 1 What is an algebra B to put on the right hand side in the restricted setting, so that its image computes the invariants EndGL(V ) (Tr (V ))? 2 If such a set of operators B can be described, does it produce a new instance of Schur–Weyl duality in which the rth tensor power Tr (V ) is replaced by the rth Lie power Lier (V )?
  7. 7.

    Lie idempotents A Lie idempotent is an idempotent e ∈

    kSr such that Tr (V )e = Lier (V ). Lemma (see e.g., Garsia, Reutenauer) If the characteristic of k does not divide r then the element e = er defined by the product e = er = 1 r (1 − γ2)(1 − γ3) · · · (1 − γr ) is a Lie idempotent, where γk = (k, k − 1, · · · , 2, 1) is a descending k-cycle, in the cycle notation for permutations. This idempotent is known as the Dynkin–Specht–Wever idempotent; it has a long history. Remark. If the field k is big enough and its characteristic does not divide r then there is another Lie idempotent, due to Klyachko. But for now we stick to the idempotent e in the Lemma.
  8. 8.

    Main Result If e ∈ kSr is a Lie idempotent

    then GL(V ) acts (on the left) on Lier (V ) = Tr (V )e and B = e(kSr )e acts (on the right). These actions commute, so Lier (V ) is a (k GL(V ), e(kSr )e)-bimodule. In other words, we have representations k GL(V ) Φ −→ End k (Lier (V )) Ψ ←− e(kSr )e. Theorem (Schur–Weyl duality for Lier (V )) Suppose that |k| > r and characteristic(k) does not divide r, and e ∈ kSr is a Lie idempotent. 1 Then the image of Ψ is equal to EndGL(V ) (Lier (V )). 2 If Tr (V ) is completely reducible as k GL(V )-module, then the image of Φ is equal to Ende(kSr )e (Lier (V )).
  9. 9.

    Proof sketch First, we prove a slightly more general result

    that will imply Part 1. Lemma Let AVB be a bimodule such that SWD holds. Suppose that e ∈ B an idempotent such that Ve = 0. Let Ψe : eBe → EndA(Ve) be the representation corresponding to the right eBe-module structure on Ve. Then Ψe(eBe) = EndA(Ve). The proof boils down to a commutative diagram A Φ // Φe && Endk(V ) Π  B Ψ oo Πe  Endk(Ve) eBe Ψe oo in which the vertical maps are given by appropriate restrictions.
  10. 10.

    Proof sketch, continued Take V = Lier (V ), A

    = k GL(V ), B = kSr . Let e ∈ B be a Lie idempotent. Then by the Lemma we get the surjectivity Ψe(eBe) = EndA(Ve) = EndGL(V ) (Lier (V )). This proves Part 1. Part 2 follows from Part 1 and the semisimplicity assumption, by standard arguments.
  11. 11.

    Conjecture QUESTION: Does Part 2 of the theorem hold without

    the semisimplicity hypothesis? In other words, does the second half of SWD hold for Lie powers Lier (V ) in non-semisimple cases? The conjecture is that it does, although there has to be some restriction on the characteristic (e.g., characteristic(k) cannot divide r). The rest of the talk is about this question/conjecture. We can make some reductions on this question using a different Lie idempotent, namely, the Klyachko idempotent, κ.
  12. 12.

    The Klyachko idempotent If σ ∈ Sr is a permutation,

    let D(σ) = {k : σ(k) > σ(k + 1)} (the descent set of σ). The major index maj(σ) is defined by maj(σ) := k∈D(σ) k. Assume that k contains a primitive rth root of unity ζ (and that its characteristic does not divide r, as before). Then the Klyachko idempotent κ is κ = 1 r σ∈Sr ζmaj(σ) σ ∈ kSr . In 1974 Klyachko showed that this is a Lie idempotent. Moreover, if we fix any r-cycle γ ∈ Sr then f = 1 r r k=1 ζ−kγk is an idempotent satisfying κf = f and f κ = κ [Reutenauer, 8.4].
  13. 13.

    Description of the corner algebra For any Lie idempotent e,

    it is easy to check that e(kSr )e ∼ = κ(kSr )κ ∼ = f (kSr )f ∼ = End kSr (IndSr Γ ζ) where Γ = γ is the cyclic subgroup generated by the r-cycle γ.
  14. 14.

    Reformulation of the conjecture Whenever e, e are two Lie

    idempotents, then the (k GL(V ), e(kSr )e)-bimodule Lier (V ) satisfies SWD if and only if the (k GL(V ), e (kSr )e )-bimodule Lier (V ) does. So in the theorem we can replace the Dynkin–Specht–Wever idempotent e by the Klyachko idempotent κ. Theorem The (k GL(V ), κ(kSr )κ)-bimodule Lier (V ) satisfies SWD if and only if the restriction map Θf : EndSr (Tr (V )) → Endf (kSr )f (Tr (V )f ) is surjective. NOTE: The algebra EndSr (Tr (V )) is the Schur algebra, studied in [J.A. Green, SLN 830] and many subsequent papers.
  15. 15.

    A question on symmetric groups Further analysis of the question

    of the surjectivity of Θf leads us to the following result. Let Λ(n, r) = {n-part compositions of r} and let Mλ = Tr (V )λ ∼ = indSr Sλ k be the transitive permutation module determined by λ ∈ Λ(n, r). Theorem Assume that e is a Lie idempotent. Then the (k GL(V ), e(kSr )e)-bimodule Lier (V ) satisfies SWD if and only if the restriction maps HomSr (Mλ, Mµ) → Homf (kSr )f (Mλf , Mµf ) are surjective for all λ, µ ∈ Λ(n, r). So the question is reduced to a question entirely about representations of symmetric groups.