AI 中文总结
该研究围绕Rogers–Ramanujan恒等式，证明了Huang等人提出的关于Z_{a,b}(q)的猜想的部分情形，相关恒等式已被形式化验证。
AI 中文摘要
Rogers–Ramanujan恒等式将指数由二次型控制的q级数与支撑在模5的两个剩余类上的无穷乘积相等。这类恒等式十分稀少，核心问题是识别能成族产生它们的结构。Huang、Jiang和Oblomkov提出了一种新来源：对每对互素整数a,b>1，他们构造了无穷秩q级数Z_{a,b}(q)，其由有限域上满足A^a=B^b的交换幂零矩阵对(A,B)的计数组装而成，并猜想其等于(a-1)(b-1)/2个水平为a+b的模单位的显式乘积。a=2的情形是Andrews–Gordon恒等式，此前a>2的情形均未知。我们证明了(a,b)=(3,4)、(3,5)、(3,7)和(3,8)时的猜想。我们的证明依赖一个更精细的级数间恒等式，我们猜想该恒等式对所有与3互素的b成立，并在q=1时对所有b证明了它。此后Lau和Ono证明了该恒等式的一般情形，进而证明了完整的a=3情形。这些恒等式已由AxiomProver在Lean中形式化并验证。
英文摘要
The Rogers--Ramanujan identities equate a $q$-series whose exponents are governed by a quadratic form with an infinite product supported on two residue classes modulo~$5$. Identities of this shape are scarce, and a central problem is to identify the structures that produce them in families. Huang, Jiang, and Oblomkov have proposed a source of a new kind: to each pair of coprime integers $a,b>1$ they attach an infinite-rank $q$-series $Z_{a,b}(q)$, assembled from counts of commuting nilpotent matrix pairs $(A,B)$ with $A^a=B^b$ over finite fields, and they conjecture that it equals an explicit product of $(a-1)(b-1)/2$ modular units of level $a+b$. The $a=2$ cases are the Andrews--Gordon identities; no case with $a>2$ was known. We prove the conjecture for $(a,b)=(3,4)$, $(3,5)$, $(3,7)$, and $(3,8)$. Our proofs pass through a finer sum-to-sum identity, which we conjecture for all $b$ coprime to $3$ and establish for all $b$ when $q=1$. Lau and Ono have since proved that identity in general, and with it the full $a=3$ case. These identities have been formalized and verified in Lean by AxiomProver.
Comments21 pages
