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CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedverified 2026-09-23 (gpt-6-sol)
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Equal central character does not give every Verma embedding direction

Statement refuted

Equal central character of λ and μ implies both M(μ)→M(λ) and M(λ)→M(μ) are nonzero.

Counterexample

Given: A regular dominant integral weight λ in type A2 and a simple root α1. Verma modules use their supplied triangular decomposition.

Proof technique: direct.

1.1given

Put m=⟨λ+ρ,α1∨⟩∈Z>0 and μ=s1⋅λ=λ−mα1. The singular-vector lemma The simple-root singular vector in a Verma module gives a nonzero highest vector f1mvλ of weight μ in M(λ). The Verma universal property The universal property of Verma modules therefore gives a nonzero map M(μ)→M(λ).

2.1step 1.1

Every central element acts by a scalar on each cyclic highest-weight module by The Harish-Chandra projection computes the highest-weight scalar. The nonzero map in step 1.1 intertwines these scalar actions, so χμ=χλ.

3.1step 1.1step 2.1∎

The PBW weight description The PBW model of a Verma module puts every weight of M(μ) in μ−Q+. Since λ=μ+mα1 lies strictly above μ, M(μ) has no vector of weight λ. A map M(λ)→M(μ) must therefore kill its highest vector and is zero. Thus equal central characters do not give both embedding directions.

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