发表机构
Peking University; Lanzhou University（北京大学; 兰州大学）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
本文证明了偶数阶万花筒 Yang-Baxter 代数中链词消失的精确阈值，刻画了非零词的秩，并揭示了其单调路径代数结构及根的性质。
AI 中文摘要
Qiu、Guan 和 Yu (2026) 提出的万花筒 Yang-Baxter 方程是 Gaudin 万花筒模型中多重散射的一致性条件。在偶数阶 $N$ 时，它涉及两个矩阵：一个移位矩阵和一个带一个复参数的平方零矩阵。他们猜想：当幂次达到 $N/2$ 时，每个链词（即平方零矩阵与移位矩阵整数幂交替的乘积）都会消失，并已通过阶数至十的验证。我们证明该猜想对每个偶数阶和每个矩阵存在的参数都成立。该界限是精确的。在比该界限低一步时，一个词非零当且仅当没有任何因子的幂次乘以其位置能被阶数的一半整除，且此时其秩为二。证明依赖于一个旗，每个因子使旗下降一步。这两个矩阵生成的代数是一个单调路径代数：其基路径仅在水平保持方向时相乘。它与参数无关，维数为 $N^2-N+2$。其根（由平方零矩阵生成）的幂零指数为阶数的一半加一，其表示类型为无限型。在任意域上的四个条件重现了该阈值和根。每个条件都不能省略，而单位根模型是其中一个实例。
英文摘要
The Kaleidoscope Yang-Baxter equation of Qiu, Guan, and Yu (2026) is the consistency condition of multiple scattering in Gaudin's kaleidoscope models. At an even order $N$ it involves two matrices: a shift, and a square-zero matrix with one complex parameter. They conjectured that every chained word, the square-zero matrix alternating with integer powers of the shift, vanishes once the number of powers reaches $N/2$, verified through order ten. We prove it for every even order and every parameter where the matrices exist. The bound is sharp. One step below it, a word is nonzero exactly when no factor's power times its position is divisible by half the order, and then has rank two. The proof rests on one flag that every factor lowers by a step. The algebra the two matrices generate is a monotone-path algebra: its basis paths multiply only while their level keeps direction. It is independent of the parameter and has dimension $N^2-N+2$. Its radical, generated by the square-zero matrix, is nilpotent of index one more than half the order, and its representation type is infinite. Four conditions over an arbitrary field reproduce the threshold and the radical. None can be dropped, and the root-of-unity model is one instance.
Comments17 pages, 0 figures
