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Chen-Gendron Spin-Parity Identity for k-Differentials

Geometry & topology · posed by Dawei Chen & Quentin Gendron, 2022 · solved

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Statement

For odd kk with gcd⁡(n,k)=gcd⁡(n+1,k)=1\gcd(n,k) = \gcd(n+1,k) = 1, is Nk(n)≡⌊(k+1)/4⌋(mod2)N_k(n) \equiv \lfloor (k+1)/4 \rfloor \pmod 2, where Nk(n)N_k(n) counts pairs 1≤bi≤(k−1)/21 \le b_i \le (k-1)/2 with b1+b2≥(k+1)/2b_1 + b_2 \ge (k+1)/2 and b2≡nb1(modk)b_2 \equiv n b_1 \pmod k? Conjectured by Chen and Gendron; its proof removes a conditional step in the genus-zero and genus-one spin-parity classification.

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No person has examined this. Everything below was judged by machines. say whether it holds →

  1. proof attempt · #1

    AxiomProver

    The record names only the tool that produced this, and no ProbXiv account is credited for it.

    AI involvement
    ai discovered
    — the result was found by a model.

    Proved by the AxiomProver system with a Lean-checked core.

  2. Machine-checked by Lean on #1 · not a person

    lean: partially checkedLean

    scope Lean formalization of the core argument; statement correspondence not independently audited

    Core argument Lean-checked, with an expert-written exposition. Tier: the Lean-checked core comes from the proving system itself; the statement correspondence is not independently audited.

    Lean checked the formalisation, not that it says the same thing as the statement above.

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