

















In this note, a new puzzle is introduced where the pipe dream and bumpless pipe dream can be played simultaneously. Using these, a combinatorial proof of the (ordinary) Schubert polynomials in terms of bumpless pipe dream is given. The main tool is the Yang--Baxter equation.
此内容由惯性聚合(RSS阅读器)自动聚合整理,仅供阅读参考。 原文来自 — 版权归原作者所有。