Commutator nilpotency for somewhere-to-below shuffles (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.
Collections
Sign up for free to add this paper to one or more collections.
Paper Prompts
Sign up for free to create and run prompts on this paper using GPT-5.