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Preprint

Rogers--Ramanujan identities from the geometry of $X^a=Y^b$

Sep 2026 · 0 citations · 25 references
Mathematics

Abstract

We prove the conjecture of Huang, Jiang, and Oblomkov (HJO) giving a geometric extension of the Rogers--Ramanujan and Andrews--Gordon identities for every torus-knot singularity $X^a=Y^b$ with coprime $1<a<b.$ For a prime power $q$, let $\mathcal{NC}_n^{a,b}(\mathbb F_q)$ denote the set of pairs of commuting nilpotent $n\times n$ matrices $(A,B)$ over $\mathbb F_q$ satisfying $A^a=B^b$. We establish the threefold equality between their normalized counts, the HJO $q$-series $Z_{a,b}$, and the explicit infinite product $P_{a,b}$: \[ \underbrace{\vphantom{\Bigg|} \prod_{m\geq1}(1-q^{-m}) \Biggl(\sum_{n=0}^{\infty} \frac{\lvert\mathcal{NC}_n^{a,b}(\mathbb F_q)\rvert} {\lvert\operatorname{GL}_n(\mathbb F_q)\rvert}\Biggr) }_{\text{point count}} = \underbrace{\vphantom{\Bigg|}Z_{a,b}(q^{-1}) }_{\text{\(q\)-series}} = \underbrace{\vphantom{\Bigg|}P_{a,b}(q^{-1}) }_{\text{infinite product}}. \] Our main result is a stronger finite identity: the rank $N$ HJO sum equals $(q;q)_N$ times the generating function for balanced cylindric partitions with entries bounded by $N$. Taking $N\to\infty$ yields the HJO conjecture. The proof combines the compositional rational shuffle theorem of Bergeron--Garsia--Leven--Xin and Mellit with a multiplicativity theorem for slope operators and a determinantal model for bounded cylindric partitions, linked by a common $q$-difference equation. The finite identity and the HJO conjecture have been formalized in Lean by AxiomProver, conditional on two stated literature inputs.

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