ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

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.

An evaluation module for affine sl2

Example

On $V = C^{2}$ , let $e = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ , $f = (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix})$ and $h = diag (1, - 1)$ . For $a \neq 0$ set $x_{m} v = a^{m} x v$ and $c v = 0$ . This is a level-zero representation of the derived affine algebra and has no compatible action of $d$ .

Facts & Assumptions

Given: The displayed matrices, $a \in C^{\times}$ and $V = C^{2}$ .

[F1]

Evaluation representations are Evaluation module at a nonzero loop parameter.

[F2]

The exact obstruction to adjoining $d$ is Evaluation modules have level zero and do not extend canonically over d.

Verification

1.1

Direct multiplication gives $[e, f] = h$ , $[h, e] = 2 e$ , $[h, f] = - 2 f$ . Thus these matrices represent ${s l}_{2}$ , and $[a^{m} x, a^{n} y] = a^{m + n} [x, y]$ . This is the image of the affine mode bracket because the added central term acts zero. F1 consequently gives the claimed representation for all positive, zero and negative modes. The central operator is zero, so its level is zero.

F1givenalgebra

2.1

The matrix $e$ is nonzero, so F2 excludes an extension. Explicitly, mode zero would give $[D, e] = 0$ , while mode one would give $[D, a e] = a e$ . Bilinearity makes the latter left side zero, but $a e \neq 0$ , a contradiction. This holds also at $a = 1$ , and negative powers are defined precisely because $a \neq 0$ . All vectors and operators are explicit; no AC is used.

F2step 1.1algebra∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Kleshchev, Lectures on Infinite Dimensional Lie Algebras, Sections 7.1-7.2 (standard reference, not scraped)
Perrin, Introduction to Kac-Moody Groups and Lie Algebras, Definition 12.2.3 (standard reference, not scraped)