Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)
How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

BGG reciprocity

Statement

Assume the Axiom of Choice (The Axiom of Choice). For all weights λ and μ, (P(λ):Δ(μ))=[Δ(μ):L(λ)]=[M(μ):L(λ)], where P(λ) is the projective cover of L(λ), the left-hand multiplicity is the Verma-flag multiplicity of Projectives in category O have finite Verma flags, and the right-hand multiplicity is the simple composition multiplicity of the Verma module M(μ) (Standard and costandard objects).

Facts & Assumptions

Given: The Axiom of Choice, weights λ,μ, the projective cover P(λ) of L(λ), and the costandard object ∇(μ)=D(M(μ)).

[F1]

P(λ) is Verma-filtered, so dim⁡CHom⁡O(P(λ),∇(μ))=(P(λ):Δ(μ)) and Ext⁡O1(P(λ),∇(μ))=0 (Projectives in category O have finite Verma flags, Hom to costandards counts Verma-flag factors).

[F2]

For every finite-length object X one has dim⁡CHom⁡O(P(λ),X)=[X:L(λ)] (Hom from a projective counts simple composition factors).

[F3]

Restricted duality D is an exact contravariant involution preserving composition multiplicities and D(L(λ))≅L(λ); hence [∇(μ):L(λ)]=[D(M(μ)):L(λ)]=[M(μ):L(λ)] (Restricted duality is exact and involutive on O, Restricted self-duality of simple highest-weight modules).

Proof

technique · direct: convert the flag multiplicity into a Hom dimension, then count with the projective-cover Hom formula and duality
1.1F1given

By [F1], (P(λ):Δ(μ))=dim⁡CHom⁡O(P(λ),∇(μ)).

1.2F2F3given

Since ∇(μ)=D(M(μ)) is an object of O of finite length, [F2] gives dim⁡CHom⁡O(P(λ),∇(μ))=[∇(μ):L(λ)], and by [F3] this equals [M(μ):L(λ)].

2.1step 1.1step 1.2∎

Combining steps 1.1 and 1.2 gives (P(λ):Δ(μ))=[M(μ):L(λ)]; since Δ(μ)=M(μ), the middle and right multiplicities agree, so all three quantities are equal.

Depends on

Used by

Dependency tree · two levels

29 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