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Tensoring a Verma module by a finite-dimensional module shifts the type

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let V be a finite-dimensional g-module with weight multiset Wt⁡V and let ψ be a weight. Then M(ψ)⊗V is Verma-filtered with Typ⁡(M(ψ)⊗V)=ψ+Wt⁡V={ψ+μ:μ∈Wt⁡V}.

Facts & Assumptions

Given: The Axiom of Choice, a weight ψ, a finite-dimensional g-module V with weight multiset Wt⁡V, and the Verma module M(ψ).

[F1]

Lie's theorem for the solvable algebra b: V has a b-stable flag with one-dimensional quotients; choosing a basis v1,…,vn adapted to the flag, each vk is a weight vector of some weight λk∈Wt⁡V and n+vk⊆span⁡(v1,…,vk−1), because n+=[b,b] acts by zero on the one-dimensional quotients (Finite Lie triangularization and rank-one complete reducibility).

[F2]

The PBW model: u↦uvψ is a vector-space isomorphism U(n−)→∼M(ψ), and U(n−) has a PBW basis with associated graded the polynomial algebra S(n−), a domain (The PBW model of a Verma module, PBW gives an ordered monomial basis for the enveloping algebra).

[F3]

Verma filtrations and their type are as in Type of a module with a Verma filtration.

[F4]

A singular vector of weight η in a g-module determines a unique homomorphism from M(η) sending its highest weight vector to that vector (The universal property of Verma modules).

Proof

1.1givenF2algebra

Put Nk:=U(g) (vψ⊗v1,…,vψ⊗vk) for 0≤k≤n. These are g-submodules forming an increasing filtration 0=N0⊆⋯⊆Nn of M(ψ)⊗V; and Nn=M(ψ)⊗V. To see the latter, induct on the PBW degree of u∈U(n−): the diagonal action satisfies u(vψ⊗vi)=(uvψ)⊗vi plus terms of strictly smaller PBW degree in the first factor. Those terms lie in Nn by induction, and the degree-zero tensors are its generators, so all (uvψ)⊗vi lie in Nn.

2.1F1step 1.1algebra

The vector vψ⊗vk is a weight vector of weight ψ+λk, since h(vψ⊗vk)=(ψ+λk)(h)(vψ⊗vk), and n+(vψ⊗vk)=vψ⊗n+vk∈vψ⊗span⁡(v1,…,vk−1)⊆Nk−1 by [F1]. Hence the class of vψ⊗vk in Nk/Nk−1 is a highest weight vector of weight ψ+λk, and because all the other generators of Nk lie in Nk−1 this class generates Nk/Nk−1 as a U(g)-module.

3.1F2step 2.1algebra

Nk is free over U(n−) on the generators vψ⊗v1,…,vψ⊗vk. Indeed, Nk=U(n−)(vψ⊗v1,…,vψ⊗vk): by [F1] the action of U(b)U(n+) on vψ⊗vi stays in vψ⊗span⁡(v1,…,vi), and U(n−) carries vψ to M(ψ). If ∑iξi(vψ⊗vi)=0 with ξi∈U(n−), choose the largest PBW degree d occurring among the ξi and take the degree-d part of the relation in the associated graded S(n−)⊗V: it reads ∑iσ(ξi)⊗vi=0 with σ(ξi)=0 whenever deg⁡ξi<d and σ(ξi)≠0 for the maximal ones; since S(n−) is a domain and the vi are linearly independent over C, all σ(ξi) vanish, a contradiction.

4.1F2F4step 2.1step 3.1algebra

Consequently Nk/Nk−1 is free of rank one over U(n−), generated by the class ck of vψ⊗vk. By The universal property of Verma modules there is a nonzero (hence surjective) homomorphism M(ψ+λk)→Nk/Nk−1 carrying the highest weight vector to ck; source and target are both free of rank one over U(n−) by [F2] and step 3.1, and the map carries a free generator to a free generator, so it is an isomorphism. Thus Nk/Nk−1≅M(ψ+λk).

5.1F3step 1.1step 4.1∎

The filtration 0=N0⊆⋯⊆Nn=M(ψ)⊗V therefore exhibits M(ψ)⊗V as Verma-filtered with type {ψ+λ1,…,ψ+λn}=ψ+Wt⁡V.

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