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The triangular restriction on projective Verma flags

Statement

Assume the Axiom of Choice (The Axiom of Choice). If (P(λ):Δ(μ)) is nonzero then μ≥λ, that is, μ−λ∈Q+. Moreover (P(λ):Δ(λ))=1, so exactly one factor of every Verma flag of P(λ) has label λ.

Facts & Assumptions

Given: The Axiom of Choice, weights λ,μ, and the Verma-filtered projective cover P(λ) of L(λ).

[F1]

(P(λ):Δ(μ))=[Δ(μ):L(λ)]=[M(μ):L(λ)], where the right-hand side is the composition multiplicity of the simple module L(λ) in the Verma module M(μ)=Δ(μ) (BGG reciprocity, Projectives in category O have finite Verma flags).

[F2]

The weights of M(μ) are exactly μ−Q+ and M(μ)μ=Cvμ; L(μ) is the unique simple quotient of M(μ), with highest weight μ (Weights of a Verma module lie below lambda, A Verma module has a unique simple quotient).

[F3]

μ≥λ means μ−λ∈Q+, and the order is a partial order (Root order on weights).

Proof

technique · direct: identify the flag multiplicity with a Verma composition multiplicity and compare weights
1.1F1F2F3given

If (P(λ):Δ(μ))≠0, then by [F1] the simple module L(λ) is a composition factor of M(μ), hence its highest weight λ is a weight of M(μ); by [F2] every weight of M(μ) lies in μ−Q+, so λ≤μ, that is, μ−λ∈Q+.

1.2F1F2

(P(λ):Δ(λ))=[M(λ):L(λ)]=1: the kernel J(λ) of the quotient map is the sum of all proper submodules, so J(λ) contains no highest-weight vector of weight λ and J(λ)λ=0; since M(λ)=n−M(λ)⊕Cvλ by [F2], this gives J(λ)⊆n−M(λ). The highest weight of any composition factor of J(λ) is a weight of J(λ) and is therefore different from λ, so no factor is isomorphic to L(λ); from 0→J(λ)→M(λ)→L(λ)→0 the multiplicity [M(λ):L(λ)] is exactly one.

2.1F3step 1.1step 1.2∎

Thus a nonzero multiplicity (P(λ):Δ(μ)) forces μ≥λ, and the label λ occurs exactly once in every Verma flag of P(λ).

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