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CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07
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Antidominant regular Verma modules are simple

Statement

If ⟨λ+ρ,α∨⟩<0 for every α∈Φ+, then M(λ) is simple. Thus every regular antidominant weight, in this explicit sense, has simple Verma module.

Facts & Assumptions

Given: Strong linkage The strong linkage order on weights, singular vectors Every nonzero Verma submodule contains a singular vector, the universal property The universal property of Verma modules, and embedding necessity A Verma embedding implies strong linkage.

[L1]

Every nonzero homomorphism between Verma modules is injective (A nonzero homomorphism between Verma modules is injective).

Proof

technique · contradiction
1.1

If a proper nonzero submodule existed, it would contain a singular vector of some weight μ≠λ; the universal property gives a nonzero map M(μ)→M(λ).

givenassume-contra
2.1

By [L1] the map from step 1.1 is an embedding, so μ↑λ. A nonempty linkage chain begins with a positive-integral pairing for λ, contradicting the strictly negative antidominant inequalities. Thus no proper nonzero submodule exists.

L1step 1.1contradictiondischarge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources