LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-11-04 (gpt-6.1-sol)

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.

Simultaneous fusion-matrix diagonalization implies the Verlinde formula

Statement

Let $I$ be a finite set with a distinguished label $0$ and an involution $i \mapsto i^{*}$ . Define the duality matrix $C$ by $C_{i j} = δ_{i, j^{*}}$ . Let $S$ be a symmetric invertible complex matrix indexed by $I$ such that $S^{2} = C$ and $S_{0 x} \neq 0$ for every $x \in I$ . For coefficients $N_{i j}^{k}$ , define the fusion matrices by $(N_{i})_{j, k} = N_{i j}^{k}$ . If

$N_{i} S = S diag ⁣ {(\frac{S_{i x}}{S_{0 x}})}_{x \in I} for every i \in I,$

then

$N_{i j}^{k} = \sum_{x \in I} \frac{S_{i x} S_{j x} S_{k^{*} x}}{S_{0 x}} .$

This is an algebraic implication for the stated matrix data; it does not assert that a VOA supplies those data.

Facts & Assumptions

Given: The finite label set, distinguished unit label, duality involution, matrix $S$ , and fusion coefficients are the ones in the Statement.

[A1]

The duality matrix is a permutation matrix for an involution, so $C^{2} = I$ .

[A2]

$S$ is symmetric, $S^{2} = C$ , $S_{0 x} \neq 0$ , and every $N_{i}$ satisfies the displayed simultaneous diagonalization identity.

Proof

1.1givenA1A2algebra

By [A1] and [A2], $S^{4} = C^{2} = I$ , so $S^{- 1} = S^{3}$ . Since $S^{2} = C$ , $S$ commutes with $C$ ; hence $S^{- 1} = C S$ . For any $x, k \in I$ , $(S^{- 1})_{x k} = S_{x^{*}, k} = S_{k, x^{*}} = S_{k^{*}, x}$ , using the symmetry of $S$ and the equality $C S = S C$ .

2.1givenA2step 1.1algebra∎

From $N_{i} S = S D_{i}$ , where $(D_{i})_{x x} = S_{i x} / S_{0 x}$ , multiply on the right by $S^{- 1}$ to get $N_{i} = S D_{i} S^{- 1}$ . Its $j, k$ entry is therefore $\sum_{x} S_{j x} (S_{i x} / S_{0 x}) (S^{- 1})_{x k}$ . Substituting step 1.1 gives the formula in the Statement.

Proof status

This proves the finite-dimensional algebraic reduction used by R3-14. It does not prove simultaneous diagonalization, symmetry, $S^{2} = C$ , or nonvanishing of $S_{0 x}$ for VOA characters.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Yi-Zhi Huang, Vertex operator algebras and the Verlinde conjecture