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Chen-Gendron Spin-Parity Identity for k-Differentials

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chen-gendron-spin-parityGeometry & topologyposed by Dawei Chen & Quentin Gendron, 2022recorded: solved

1 attempt · 1 machine check · no person has looked

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 1bi(k1)/21 \le b_i \le (k-1)/2 with b1+b2(k+1)/2b_1 + b_2 \ge (k+1)/2 and b2nb1(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.

Context

A concrete identity conjectured in the strata-of-differentials literature.

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1 attempt

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  • #1

    Attempt 1

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

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