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Hom to costandards counts Verma-flag factors

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let X∈O be Verma-filtered (Finite Verma flags and their multiplicities). Then for every weight ν dim⁡CHom⁡O(X,∇(ν))=(X:Δ(ν)),Ext⁡O1(X,∇(ν))=0. The result is stated for all weights, including equal and incomparable labels.

Facts & Assumptions

Given: The Axiom of Choice, a Verma-filtered object X, and a weight ν.

[F1]

A Verma flag of length n has a top step 0→K→X→Δ(μ)→0 in which K is Verma-filtered of length n−1, and the multiplicities are additive along the step: (X:Δ(ν))=(K:Δ(ν))+δμν; the zero object has the empty flag and all multiplicities zero (Finite Verma flags and their multiplicities).

[F2]

For all weights μ,ν one has dim⁡Hom⁡O(Δ(μ),∇(ν))=δμν and Ext⁡O1(Δ(μ),∇(ν))=0; and Hom⁡O(0,−)=0=Ext⁡O1(0,−) (Standard-costandard Hom and Ext-one orthogonality).

[F3]

Category O is abelian and has enough projectives (Category O is abelian and extension closed among weight modules, Category O has enough projectives). The exact contravariant equivalence D exchanges projectives and injectives: Hom⁡(−,D(P))≅Hom⁡(P,D(−)) is exact for projective P. Dualizing a projective epimorphism P↠D(Y) therefore embeds Y into the injective D(P), proving enough injectives (Restricted duality is exact and involutive on O). Finitely generated U(g)-modules have a set of representatives, since they are quotients of U(g)n for finite n. Work on a set-sized skeleton of O; under AC choose a projective epimorphism onto and an injective embedding of each object, then recursively cover kernels and embed cokernels to supply resolutions. AC implies DC by selecting successors in any serial relation. Fix these resolution systems and use the canonical comparison identifications of The balanced Ext bifunctor. For a short exact sequence 0→M′→M→M′′→0 in O and every N there is a natural exact sequence 0→Hom⁡(M′′,N)→Hom⁡(M,N)→Hom⁡(M′,N)→Ext⁡1(M′′,N)→Ext⁡1(M,N)→Ext⁡1(M′,N) (The long exact Ext sequence in the first variable).

Proof

technique · induction on the length of a Verma flag, using the long exact sequence and the $\Delta$-$\nabla$ orthogonality
1.1F1F2base

If X=0 has the empty flag, then Hom⁡(X,∇(ν))=0 and Ext⁡1(X,∇(ν))=0 while all multiplicities (X:Δ(ν)) vanish, so both formulas hold.

1.2F1givenih

Let X have a Verma flag of length n≥1 with top step 0→K→X→Δ(μ)→0; then K has a Verma flag of length n−1 and (X:Δ(ν))=(K:Δ(ν))+δμν. Assume as induction hypothesis that the two formulas hold for K.

2.1F2F3step 1.2algebra

The long exact sequence of [F3] for the top step begins 0→Hom⁡(Δ(μ),∇(ν))→Hom⁡(X,∇(ν))→Hom⁡(K,∇(ν))→Ext⁡1(Δ(μ),∇(ν))→Ext⁡1(X,∇(ν))→Ext⁡1(K,∇(ν)). By [F2] and the induction hypothesis of step 1.2 the fourth and sixth terms vanish, so the sequence gives the short exact sequence 0→Hom⁡(Δ(μ),∇(ν))→Hom⁡(X,∇(ν))→Hom⁡(K,∇(ν))→0 and the vanishing of Ext⁡1(X,∇(ν)). Taking dimensions and using [F2] and step 1.2, dim⁡Hom⁡(X,∇(ν))=δμν+(K:Δ(ν))=(X:Δ(ν)).

3.1step 1.1step 2.1discharge-induction: step 2.1∎

By induction on the flag length, steps 1.1 and 2.1 prove dim⁡Hom⁡O(X,∇(ν))=(X:Δ(ν)) and Ext⁡O1(X,∇(ν))=0 for every Verma-filtered X and every weight ν.

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