Complements of Schubert polynomials

Neil J. Y. Fan, Peter L. Guo, Nicolas Y. Liu

Published 2020-01-12Version 1

Let $\mathfrak{S}_w(x)=\mathfrak{S}_w(x_1,\ldots,x_n)$ be the Schubert polynomial for a permutation $w$ of $\{1,2,\ldots,n\}$. For a composition $\mu=(\mu_1,\ldots,\mu_n)\in \mathbb{Z}_{\geq 0}^n$, write $x^\mu=x_1^{\mu_1}\cdots x_n^{\mu_n}$. We say that $x^\mu\mathfrak{S}_w(x^{-1})=x^\mu \mathfrak{S}_w(x_1^{-1},\ldots,x_n^{-1})$ is the complement of $\mathfrak{S}_w(x)$ with respect to $\mu$. Huh, Matherne, M\'esz\'aros and St.Dizier proved that $\mathrm{N}(x_1^{n-1}\cdots x_n^{n-1} \mathfrak{S}_w(x^{-1}))$ is a Lorentzian polynomial, where $\mathrm{N}$ is a linear operator sending a monomial $x_1^{\mu_1}\cdots x_n^{\mu_n}$ to $\frac{x_1^{\mu_1}\cdots x_n^{\mu_n}}{\mu_1!\cdots \mu_n!}$. They further conjectured that $\mathrm{N}(\mathfrak{S}_w(x))$ is Lorentzian. Motivated by this conjecture, we investigate the problem when $x^\mu \mathfrak{S}_w(x^{-1})$ is still a Schubert polynomial. If $x^\mu \mathfrak{S}_w(x^{-1})$ is a Schubert polynomial, then $\mathrm{N}(\mathfrak{S}_w(x))$ will be Lorentzian. In this paper, we pay attention to the typical case that $\mu=\delta_n=(n-1,\ldots, 1,0)$ is the staircase partition. Our result shows that $x^{\delta_n} \mathfrak{S}_w(x^{-1})$ is a Schubert polynomial if and only if $w$ avoids the two patterns 132 and 312.