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Commutator nilpotency for somewhere-to-below shuffles

Published 11 Sep 2023 in math.CO and math.RA | (2309.05340v2)

Abstract: Given a positive integer $n$, we consider the group algebra of the symmetric group $S_{n}$. In this algebra, we define $n$ elements $t_{1},t_{2},\ldots,t_{n}$ by the formula [ t_{\ell}:=\operatorname*{cyc}\nolimits_{\ell}+\operatorname*{cyc}\nolimits_{\ell,\ell+1}+\operatorname*{cyc}\nolimits_{\ell,\ell+1,\ell+2}+\cdots+\operatorname*{cyc}\nolimits_{\ell,\ell+1,\ldots,n}, ] where $\operatorname*{cyc}\nolimits_{\ell,\ell+1,\ldots,k}$ denotes the cycle that sends $\ell\mapsto\ell+1\mapsto\ell+2\mapsto\cdots\mapsto k\mapsto\ell$. These $n$ elements are called the somewhere-to-below shuffles due to an interpretation as card-shuffling operators. In this paper, we show that their commutators $\left[ t_{i},t_{j}\right] =t_{i}t_{j}-t_{j}t_{i}$ are nilpotent, and specifically that [ \left[ t_{i},t_{j}\right] {\left\lceil \left( n-j\right) /2\right\rceil +1}=0\ \ \ \ \ \ \ \ \ \ \text{for any }i,j\in\left{ 1,2,\ldots,n\right} ] and [ \left[ t_{i},t_{j}\right] {j-i+1}=0\ \ \ \ \ \ \ \ \ \ \text{for any }1\leq i\leq j\leq n. ] We discuss some further identities and open questions.

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