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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)
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Verma-flag multiplicities are independent of the flag

Statement

Assume the Axiom of Choice (The Axiom of Choice). If X∈O admits two finite Verma flags with corresponding multiplicities mμ and mμ′ (Finite Verma flags and their multiplicities), then mμ=mμ′ for every weight μ. Hence the multiplicity (X:Δ(μ)) of Finite Verma flags and their multiplicities is well defined.

Facts & Assumptions

Given: The Axiom of Choice, an object X with two finite Verma flags and their multiplicity functions m,m′.

[F1]

If 0=X0⊆X1⊆⋯⊆Xn=X is a Verma flag with factors Xi/Xi−1≅Δ(μi), then [X]=∑i=1n[Δ(μi)] in the Grothendieck group, where [Δ(μ)]=[M(μ)], and mμ=#{i:μi=μ} is the multiplicity; all but finitely many mμ vanish (Finite Verma flags and their multiplicities, The Grothendieck group and character of O).

[F2]

The classes [M(λ)], equivalently the classes [Δ(λ)], form a Z-basis of K0(O) (Simple and standard bases of K0(O)).

Proof

technique · direct comparison of two basis expansions in the Grothendieck group
1.1F1given

The two flags give two finite expansions of the same class, [X]=∑μmμ[Δ(μ)] and [X]=∑μmμ′[Δ(μ)], in K0(O).

2.1F2step 1.1∎

Since the standard classes [Δ(μ)]=[M(μ)] form a Z-basis of K0(O), the coefficient of each basis element in a class is uniquely determined. Comparing the two expansions of [X] from step 1.1 therefore gives mμ=mμ′ for every weight μ, so the multiplicity (X:Δ(μ)) is independent of the chosen flag.

Depends on

Used by

Cited to discharge well-definedness by Finite Verma flags and their multiplicities.

Dependency tree · two levels

17 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