Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The opposite simple centralizer in a Kac Moody half vanishes

Statement

If xn+ satisfies [fi,x]=0 for every i, then x=0. Likewise xn with [ei,x]=0 for every i is zero.

Facts & Assumptions

Given: A finite GCM and its maximal Cartan-disjoint quotient.

[F1]

Every nonzero ideal meets the Cartan. (Kac moody algebra associated to a gcm).

[F2]

Weights have one sign and distinct root coordinates. (Kac moody root spaces are finite dimensional).

Proof

1.1

Decompose x into finitely many weights. For fixed i, the brackets of its distinct components with fi have distinct weights, so each bracket is zero. It suffices to treat a homogeneous x of positive degree β. Let J be the span of x and all iterated adei applied to x. Every such vector has degree β+γ with γQ+, so Jn+.

F2given
2.1

The space J is stable under all ei by construction and under h by its homogeneous spanning vectors. Induct on the number of positive adjoints to prove stability under fi. The base is [fi,x]=0. For y already treated, [fi,[ej,y]]=δij[hi,y]+[ej,[fi,y]] lies in J. Thus J is an ideal, and Jh=0 by its positive degrees. F1 forces J=0, hence x=0. The sign-changing involution interchanges the two conclusions.

F1F2step 1.1

Sources

Source comparison: Kleshchev, Perrin Lemma 4.2.8, p.36; maximal-ideal argument as in Kleshchev §1.4.

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

Depends on

Used by

Dependency tree · two levels

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Sources