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geometry-topology / Flat surfaces & moduli

Chen-Gendron Spin-Parity Identity for k-Differentials

For odd $k$ with $\gcd(n,k) = \gcd(n+1,k) = 1$, is $N_k(n) \equiv \lfloor (k+1)/4 \rfloor \pmod 2$, where $N_k(n)$ counts pairs $1 \le b_i \le (k-1)/2$ with $b_1 + b_2 \ge (k+1)/2$ and $b_2 \equiv n b_1 \pmod k$? 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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geometry-topologyFeb 3, 2026Significance 10/100Registry: lean checked

Chen-Gendron Spin-Parity Identity for k-Differentials

Prior state unknown→proved

For odd $k$ with $\gcd(n,k) = \gcd(n+1,k) = 1$, is $N_k(n) \equiv \lfloor (k+1)/4 \rfloor \pmod 2$, where $N_k(n)$ counts pairs $1 \le b_i \le (k-1)/2$ with $b_1 + b_2 \ge (k+1)/2$ and $b_2 \equiv n b_1 \pmod k$? Conjectured by Chen and Gendron; its proof removes a conditional step in the genus-zero and genus-one spin-parity classification.

Source authenticatedChen-Gendron Spin-Parity Identity for k-Differentials →
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For odd $k$ with $\gcd(n,k) = \gcd(n+1,k) = 1$, is $N_k(n) \equiv \lfloor (k+1)/4 \rfloor \pmod 2$, where $N_k(n)$ counts pairs $1 \le b_i \le (k-1)/2$ with $b_1 + b_2 \ge (k+1)/2$ and $b_2 \equiv n b_1 \pmod k$? 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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