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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)
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Positive root strings sum the Freudenthal correction

Statement

Assume the Axiom of Choice. Keep the notation of The Casimir comparison on a weight space: g is a finite-dimensional complex semisimple Lie algebra with Cartan subalgebra h and positive system Φ+, the vectors eα∈gα and fα∈g−α satisfy B(eα,fα)=1, so that [eα,fα]=Hα is the Killing-dual vector of α, and mλ(μ)=dim⁡L(λ)μ for λ∈Λ+. Then for every λ∈Λ+, μ∈h∗ and α∈Φ+, tr⁡L(λ)μ(eαfα+fαeα)=mλ(μ)(μ,α)+2∑j≥1mλ(μ+jα)(μ+jα,α), the sum being finite because L(λ) has only finitely many weights.

Facts & Assumptions

Given: The Axiom of Choice, such g,h,Φ+, vectors eα,fα with B(eα,fα)=1 for a fixed α∈Φ+, a dominant integral weight λ, and an element μ∈h∗.

[A1]

The Axiom of Choice is assumed; it enters only through the published classification and Casimir suppliers used in [F4] (The Axiom of Choice).

[F1]

On the weight space L(λ)ν the Cartan element Hα acts by the scalar (ν,α), and [eα,fα]=Hα (The Casimir comparison on a weight space, Opposite root spaces bracket to the Killing-dual line).

[F2]

eα maps L(λ)ν into L(λ)ν+α and fα maps it into L(λ)ν−α (Root vectors shift weights, Weight and weight space).

[F3]

The weight set of L(λ) is finite, since its distinct nonzero weight spaces are independent in a finite-dimensional vector space (Weight and weight space, Highest-weight classification).

[F4]

L(λ) is a finite-dimensional irreducible highest weight module of highest weight λ, so all its weight spaces are finite-dimensional and the traces below are finite sums of matrix traces (Highest-weight classification, The Casimir comparison on a weight space).

Proof

technique · direct
1.1F2F3F4A1

Fix α∈Φ+ and set Vi=L(λ)μ+iα for every i∈Z, allowing Vi=0. Let Ai:Vi→Vi+1 and Bi+1:Vi+1→Vi be the actions of eα,fα, and put Ti=tr⁡Vi(Bi+1Ai). By [F3] there is an integer N≥0 such that Vi=0 for all i>N, even if the whole line contains no weights; in particular TN=0.

2.1F1F4step 1.1algebra

For maps A:U→Z and B:Z→U between finite-dimensional spaces, tr⁡U(BA)=tr⁡Z(AB): in bases both traces equal ∑p,qBpqAqp, including zero-dimensional spaces. Applying this to Ai,Bi+1 and using [eα,fα]=Hα on Vi+1 gives Ti=Ti+1+mλ(μ+(i+1)α)(μ+(i+1)α,α). Telescoping from i=0 to N therefore gives T0=∑j≥1mλ(μ+jα)(μ+jα,α), with only finitely many nonzero terms.

3.1F1step 1.1step 2.1algebra∎

On V0, the commutator identity gives tr⁡(eαfα)=tr⁡(fαeα)+mλ(μ)(μ,α)=T0+mλ(μ)(μ,α). Adding the trace of fαeα and substituting step 2.1 proves the asserted formula, including absent weights and an empty line.

Depends on

Used by

Dependency tree · two levels

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Sources