Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Homomorphism spaces between Verma modules have dimension at most one

Statement

For all weights λ,μ, dim⁡Hom⁡g(M(μ),M(λ))≤1.

Facts & Assumptions

Given: A simple Verma submodule exists by Every Verma module contains a simple Verma submodule, and maps from one are unique up to scalar by Homomorphisms from a simple Verma module have dimension at most one.

[L1]

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

Proof

technique · direct
1.1

Choose a simple Verma submodule S⊂M(μ). The restrictions of any two maps f,g:M(μ)→M(λ) are proportional, say f∣S=cg∣S.

givenchoose
2.1

Then (f−cg)∣S=0. If f−cg were nonzero, [L1] would make it injective, so it could not vanish on nonzero S. Hence f=cg, proving the bound.

L1step 1.1contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources