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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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Serre elements vanish before Serre generation

Statement

For every finite GCM and i≠j, the maximal-ideal quotient satisfies (ad⁡ei)1−aijej=0 and (ad⁡fi)1−aijfj=0. This asserts vanishing, without yet asserting generation of the defining ideal.

Facts & Assumptions

Given: The standard simple generators in g(A), and i≠j.

[F1]

A negative vector killed by all opposite simple generators is zero. (The opposite simple centralizer in a Kac Moody half vanishes).

[F2]

The algebra g(A) is the quotient of g~(A) by the largest Cartan-disjoint ideal and retains the generator names h,ei,fi. (Kac moody algebra associated to a gcm).

[F3]

Before the quotient, the generators satisfy [h,h′]=0, [h,ei]=αi(h)ei, [h,fi]=−αi(h)fi, and [ei,fj]=δijhi. (Contragredient lie algebra before the maximal ideal quotient).

Proof

1.1F2F3given

The relations of F3 descend through the quotient of F2. For operators E,F,H with [E,F]=H and [H,F]=−2F, the identity [E,Fm]=mFm−1(H−m+1) follows from [E,Fm+1]=[E,Fm]F+FmH and HF=F(H−2), starting at m=1. Take the adjoint operators of ei,fi,hi. On v=fj, Ev=0 and Hv=−aijv. For m=1−aij, the identity gives EFmv=m(−aij−m+1)Fm−1v=0.

2.1F2F3givenstep 1.1

For k∉{i,j}, ad⁡ek commutes with ad⁡fi and kills fj, so kills Fmv. For k=j, it commutes with F and sends fj to hj, so sends Fmv to Fmhj. Now Fhj=ajifi and F2hj=0. If m≥2 this vanishes; if m=1, aij=0 and the symmetric-zero axiom gives aji=0. Thus all ek kill the negative Serre vector.

3.1F1F2step 1.1step 2.1∎

F1 kills this vector in n−. The sign-changing involution descending in F2 proves the positive relation with the same exponent. At no point has an ideal been asserted to be generated by these vectors.

Sources

Source comparison: Kleshchev, §1.4 Serre vanishing and Lemma 3.1.1; Perrin Propositions 4.2.6–4.2.7, pp.35–37.

Additional source: Perrin, section 4.2, at the numbered locators above.

Depends on

Used by

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Sources