\documentclass[ALCO, Unicode]{cedram} \usepackage{algorithm} \usepackage{algpseudocode} \usepackage{pgf} \usepackage{tikz} \usepackage{tikz-cd} \usetikzlibrary{positioning,shapes,shadows,arrows} \usepackage{bbm,bm} \usepackage{enumitem} \usepackage{mathabx} \usepackage{ytableau} \usepackage{thmtools} \usepackage[matrix,arrow,tips,curve]{xy} \equalenv{cor}{coro} \equalenv{exa}{exam} \equalenv{ourthm}{theo} \equalenv{defn}{defi} \equalenv{lem}{lemm} \equalenv{rem}{rema} \equalenv{prob}{ques} \newcommand{\NN}{\mathbb{N}} \newcommand{\PP}{\mathbb{P}} \newcommand{\BPD}{\mathsf{BPD}} \newcommand{\LTBPD}{\mathsf{LTBPD}} \newcommand{\df}{:=} \newcommand{\fG}{\mathfrak{G}} \newcommand{\fGb}{\fG} \newcommand{\fL}{\mathfrak{L}} \newcommand{\fLb}{\fL} \newcommand{\topLas}{\widehat{\fL}} \newcommand{\topGro}{\widehat{\fG}} \newcommand{\fS}{\mathfrak{S}} \newcommand{\id}{\mathsf{id}} \newcommand{\PT}{\mathsf{PerfectTab}_\downarrow} \newcommand{\myref}[1]{\href{#1}{#1}} \newcommand{\pab}{\partial^{(\beta)}} \newcommand{\pib}{\pi^{(\beta)}} \newcommand{\rev}{\mathsf{rev}} \newcommand{\RSVT}{\mathsf{RSVT}} \newcommand{\RSSYT}{\mathsf{RSSYT}} \newcommand{\supp}{\mathsf{supp}} \newcommand{\word}{\mathsf{word}} \newcommand{\wt}{\mathsf{wt}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\C}{\mathcal{C}} \newcommand{\Tt}{\tilde{\mathcal{T}}} \newcommand{\ex}{\mathsf{ex}} \newcommand{\cloud}{\mathsf{cloud}} \newcommand{\dark}{\mathsf{dark}} \newcommand{\snow}{\mathsf{snow}} \newcommand{\rajcode}{\mathsf{rajcode}} \newcommand{\raj}{\mathsf{raj}} \newcommand{\Inv}{\mathsf{Inv}} \newcommand{\graph}{\mathsf{graph}} \newcommand{\snowflake}{\Asterisk} \newcommand{\leqe}{\leqslant_e} \newcommand{\geqe}{\geqslant_e} \newcommand{\topV}{\widehat{V}} \newcommand{\invcode}{\mathsf{invcode}} \newcommand{\inv}{\mathsf{inv}} \newcommand{\std}{\mathsf{std}} \newcommand{\Rook}{\mathsf{Rook}} \newcommand{\Stair}{\mathsf{Stair}} \newcommand{\GR}{\mathsf{GR}} \newcommand{\NW}{\mathsf{NW}} \newcommand{\conv}{\mathsf{conv}} \newcommand{\SYT}{\mathsf{SYT}} \newcommand{\shift}{\mathsf{shift}} \newcommand{\btile}{ \begin{tikzpicture}[x=1em,y=1em,thick,color = blue] \draw[step=1,gray,thin] (0,0) grid (1,1); \draw[color=black, thick, sharp corners] (0,0) rectangle (1,1); \end{tikzpicture}} \newcommand{\htile}{ \begin{tikzpicture}[x=1em,y=1em,thick,color = blue] \draw[step=1,gray,thin] (0,0) grid (1,1); \draw[color=black, thick, sharp corners] (0,0) rectangle (1,1); \draw(1.0,0.5)--(0.0,0.5); \end{tikzpicture} } \newcommand{\vtile}{ \begin{tikzpicture}[x=1em,y=1em,thick,color = blue] \draw[step=1,gray,thin] (0,0) grid (1,1); \draw[color=black, thick, sharp corners] (0,0) rectangle (1,1); \draw(0.5,1.0)--(0.5,0.0); \end{tikzpicture}} \newcommand{\ptile}{ \begin{tikzpicture}[x=1em,y=1em,thick,color = blue] \draw[step=1,gray,thin] (0,0) grid (1,1); \draw[color=black, thick, sharp corners] (0,0) rectangle (1,1); \draw(0.5,1.0)--(0.5,0.0); \draw(1.0,0.5)--(0.0,0.5); \end{tikzpicture}} \newcommand{\rtile}{ \begin{tikzpicture}[x=1em,y=1em,thick,color = blue] \draw[step=1,gray,thin] (0,0) grid (1,1); \draw[color=black, thick, sharp corners] (0,0) rectangle (1,1); \draw(1.0,0.5)--(0.5,0.5)--(0.5,0.0); \end{tikzpicture}} \newcommand{\jtile}{ \begin{tikzpicture}[x=1em,y=1em,thick,color = blue] \draw[step=1,gray,thin] (0,0) grid (1,1); \draw[color=black, thick, sharp corners] (0,0) rectangle (1,1); \draw(0.5,1.0)--(0.5,0.5)--(0.0,0.5); \end{tikzpicture}} \definecolor{darkred}{rgb}{0.7,0,0} \newcommand{\definitionx}[1]{{\color{darkred}\emph{#1}}} \title [Schubert and top Lascoux] {Connection between Schubert polynomials and top Lascoux polynomials} \author{\firstname{Tianyi} \lastname{Yu}} \address{Laboratoire d'Alg\`{e}bre, de Combinatoire et d'Informatique Math\'{e}matique (LACIM)\\ 201 avenue du Pr\'{e}sident Kennedy\\ Montr\'{e}al, Qu\'{e}bec, H2X 3Y7, Canada} \email{yu.tianyi@uqam.ca} \keywords{Schubert polynomials, Lascoux polynomials, Key polynomials} \subjclass{05E05} \begin{document} \begin{abstract} Schubert polynomials form a basis of the polynomial ring $\mathbb{Q}[x_1, x_2, \dots]$. This basis and its structure constants have received extensive study. Recently, Pan and Yu initiated the study of top Lascoux polynomials. These polynomials form a basis of the vector space $\widehat{V}$, a sub-algebra of $\mathbb{Q}[x_1, x_2, \dots]$ where each graded piece has finite dimension. This paper connects Schubert polynomials and top Lascoux polynomials via a simple operator. We use this connection to show these two bases share the same structure constants. We also translate several results on Schubert polynomials to top Lascoux polynomials, including combinatorial formulas for their monomial expansions and supports. \end{abstract} \maketitle \section{Introduction} For a permutation $w$, Lascoux and Sch\"utzenberger~\cite{LS:Schubert} recursively defined the \definitionx{Schubert polynomial} $\fS_w$ using \definitionx{divided difference operators}. These polynomials represent Schubert cycles in flag varieties and have been extensively investigated from various perspectives. We summarize some significant results on Schubert polynomials relevant to this paper. \begin{enumerate} \item The set of all Schubert polynomials forms a basis of the polynomial ring $\mathbb{Q}[x_1, x_2, \dots]$. Products of Schubert polynomials can be expanded positively into Schubert polynomials (i.e. the expansion only involves positive integer coefficients): $$ \fS_u \fS_v = \sum_{w} c_{u,v}^w \fS_w, $$ The coefficient $c^{w}_{u,v}$ is known as the \definitionx{Schubert structure constant}. A major open problem in algebraic combinatorics is to compute $c^{w}_{u,v}$ combinatorially. \item Lam, Lee and Shimozono~\cite{LLS} introduced the \definitionx{(reduced) bumpless pipedreams (BPD)} to compute the monomial expansion of Schubert polynomials. \item The Schubert polynomial can be expanded positively into \definitionx{key polynomials}~\cite{RS}. \item The dual character of the flagged Weyl module of a diagram $D$ is denoted as $\chi_D$. The Schubert polynomial $\fS_w$ is $\chi_{RD(w)}$ where $RD(w)$ is the Rothe diagram of $w$~\cite{KP}. \item Adve, Robichaux, and Yong~\cite{ARY} introduced \definitionx{perfect tableaux} to compute the support of Schubert polynomials. \item The Schubert polynomials have the \definitionx{saturated Newton polytope (SNP)} property~\cite{FMS}. \end{enumerate} The key polynomials mentioned above are denoted as $\kappa_\alpha$, where $\alpha$ is a \definitionx{weak composition}. They are the characters of Demazure modules~\cite{Dem}. Lascoux~\cite{Las} introduced an inhomogeneous analogue of $\kappa_\alpha$ known as the \definitionx{Lascoux polynomial} $\fLb_\alpha$. The lowest-degree terms of $\fLb_\alpha$ form $\kappa_\alpha$. Recently, Pan and Yu~\cite{PY2} introduced the \definitionx{top Lascoux polynomial} $\topLas_\alpha$ which consists of the highest-degree terms of $\fLb_\alpha$. Let $\widehat{V}$ be the $\Q$-span of all top Lascoux polynomials. Unlike the Schubert polynomials, the set of all top Lascoux polynomials is not linearly independent. To resolve this, Pan and Yu called a weak composition \definitionx{snowy} if its positive entries are distinct. Then $\{\topLas_\alpha: \alpha \textrm{ is snowy}\}$ forms a basis of $\widehat{V}$. By~\cite[Theorem 1.2]{PY2}, every top Lascoux polynomial is a top Lascoux indexed by snowy weak composition multiplied by an integer. In the rest of this paper, we only focus on $\topLas_\alpha$ when $\alpha$ is snowy. Pan and Yu showed that $\widehat{V}$ is closed under multiplication. Thus, $\widehat{V}$ can be viewed as a graded algebra where the grading is given by degrees of polynomials. Each graded piece of $\widehat{V}$ has finite dimension. Pan and Yu computed the Hilbert series of $\widehat{V}$. Intuitively, $\widehat{V}$ is much smaller than the polynomial ring $\Q[x_1, x_2, \dots]$ which has no Hilbert series. Just like the Schubert polynomials, $\topLas_\alpha$ can be defined recursively using divided difference operators (see~(\ref{EQ: Define top Las})). This resemblance leads to a strong connection between Schubert polynomials and top Lascoux polynomials, which is the main focus of this paper. \begin{defn} Define the following involution on polynomials in $\Q[x_1, \dots, x_n]$ where each variable has degree at most $m$: $$r_{m,n}(f) := (x_1\cdots x_n)^m f(x_n^{-1}, \dots, x_1^{-1}).$$ \end{defn} In~\S\ref{S: Relations}, we show that each top Lascoux polynomial can be realized as $r_{m,n}(\fS_w)$ for some $m, n, w$ and vice versa. Following this connection, we translate the results on $\fS_w$ summarized above to $\topLas_\alpha$. \begin{enumerate} \item Products of top Lascoux polynomials can be expanded positively into top Lascoux polynomials: $$ \topLas_\alpha \topLas_\gamma = \sum_{\sigma} d^\sigma_{\alpha, \gamma} \topLas_\sigma. $$ We call the coefficient \definitionx{top Lascoux structure constants}. Every $d^\sigma_{\alpha, \gamma}$ is $c^{w}_{u,v}$ for some explicit permutations $u, v, w$ and vice versa (see~\S\ref{S: structure constants}). \item We give a monomial expansion of top Lascoux polynomials using (modified) bumpless-pipedreams (see~\S\ref{S: BPD}). \item The top Lascoux polynomials can be expanded positively into key polynomials (see~\S\ref{S: key expansion}). \item The top Lascoux polynomial $\topLas_\alpha = \chi_{\snow(D(\alpha))}$ where $\snow(D(\alpha))$ is some diagram defined in~\cite{PY2}(See~\S\ref{S: character}). \item The support of top Lascoux polynomials can be computed using perfect tableaux (see~\S\ref{S: character}). \item The top Lascoux polynomials have the SNP property (See~\S\ref{S: character}). \end{enumerate} These properties of the top Lascoux basis precisely mirror those of the Schubert basis. It may be interesting to do Schubert calculus in the graded ring $\widehat{V}$ which as a defined and understood Hilbert series. By (1), computing the Schubert structure constants is the same as computing the top Lascoux structure constants. Another potential application of our results is to understand the Grothendieck polynomial $\fGb_w$, the $K$-theoretic analogue of $\fS_w$. By Shimozono and Yu~\cite{SY}, $\fGb_w$ positively into $\fLb_\alpha$. Consequently, the top degree component of $\fGb_w$, denoted as $\topGro_w$, expands positively into $\topLas_\alpha$. There has been a recent surge in the study of $\topGro_w$~\cite{CY, CY2, DMS, Haf, HMSS, PSW, PY, RRRSW, RRW}. Together with the expansion of $\topGro_w$ into $\topLas_\alpha$, one would translate our results on $\topLas_\alpha$ to $\topGro_w$. In particular, when $w$ is vexillary ($2143$-avoiding), Pechenik and Scrimshaw~\cite{PS} showed that $\fGb_w$ is just a Lascoux polynomial. We may then translate our results to $\topGro_w$ for vexillary $w$. See Corollary~\ref{C: HMSS} for one such application. The rest of the paper is structured as follows. In~\S\ref{S: Background}, we provide an overview of the necessary background information. In~\S\ref{S: Relations}, we use $r_{m,n}$ to relate the Schubert polynomials and top Lascoux polynomials. The subsequent sections explore various applications of this relationship. In~\S\ref{S: structure constants}, we examine the connection between the structure coefficients of top Lascoux polynomials and Schubert polynomials. In~\S\ref{S: BPD}, we derive a combinatorial formula for top Lascoux polynomials from the BPD formula of Schubert polynomials. In~\S\ref{S: key expansion}, we translate the key expansion of Schubert polynomials to obtain a key expansion of top Lascoux polynomials. In~\S\ref{S: character}, we show the top Lascoux polynomials are certain dual characters of the flagged Weyl modules and characterize the support of top Lascoux polynomials. \section{Background} \label{S: Background} \subsection{Schubert polynomials} Let $S_+$ be the group of permutations of $\{1, 2, \dots\}$ where only finitely many elements are permuted. The simple transpositions $s_1, s_2, \dots$ where $s_i = (i, i+1)$ generate $S_+$. For any positive number $n$, $S_n$ is a subgroup of $S_+$ consisting of $w$ that only permutes $[n] = \{1, 2, \dots, n\}$. We represent $w \in S_+$ by its \definitionx{one-line notation} $[w(1), \dots, w(n)]$ for some $n$ large enough such that $w \in S_n$. A \definitionx{weak composition} $\alpha = (\alpha_1, \alpha_2, \dots )$ is an infinite sequence of non-negative numbers with finitely many positive entries. The \definitionx{support} of $\alpha$ is $\supp(\alpha) := \{i: \alpha_i > 0\}$. We represent $\alpha$ as $(\alpha_1, \alpha_2, \dots, \alpha_n)$ where $\supp(\alpha) \subseteq [n]$. Let $x^\alpha := x_1^{\alpha_1}x_2^{\alpha_2}\cdots$ and~$|\alpha| := \sum_{i\geqslant 1}\alpha_i$. We say $(i,j)$ is an inversion of $w \in S_+$ if $i < j$ and $w(i) > w(j)$. The \definitionx{inversion code} of $w$, denoted as $\invcode(w)$ is a weak composition defined as $$\invcode(w)_i := |\{j: (i,j) \textrm{ is an inversion of }w\}|.$$ The \definitionx{Schubert polynomials} $\fS_w$ are indexed by permutations from $S_+$. When a weak composition is weakly decreasing, we say it is a \definitionx{partition}. When $\invcode(w)$ is a partition, we say $w$ is a \definitionx{dominant permutation}. Define the \definitionx{Newton divided difference operator}: $$\partial_i(f) := \dfrac{f - s_i f}{x_i-x_{i+1}},$$ where $s_i$ is the operator that swaps $x_i$ and $x_{i+1}$. Now we can define the \definitionx{Schubert polynomial} of $w \in S_+$ recursively~\cite{LS:Schubert}. \begin{align*} \fS_w = \begin{cases} x^{\invcode(w)} & \text{if $w$ is dominant} \\ \partial_i (\fS_{w s_i} ) &\text{if $w(i) < w(i+1)$.} \end{cases} \end{align*} The set of Schubert polynomials forms a $\mathbb{Q}$-basis of the polynomial ring $\mathbb{Q}[x_1, x_2, \dots]$. For $u, v \in S_+$, the product $\fS_u \fS_v$ can be expanded into Schubert polynomials. Let $c^{w}_{u,v}$ be the coefficient of $\fS_w$ in this expansion. By geometric results, $c^w_{u,v}$ is a non-negative integer known as the \definitionx{Schubert structure constants}. \subsection{Key polynomials and top Lascoux polynomials} The \definitionx{key polynomials} $\kappa_\alpha$ are indexed by weak compositions. Lascoux and Sch{\"u}tzenberger~\cite{LS:Groth} define the key polynomials recursively. using the operator $\pi_i(f) := \partial_i(x_i f)$: \begin{align*} \kappa_\alpha := \begin{cases} x^\alpha & \text{if $\alpha$ is a partition,} \\ \pi_i (\kappa_{s_i\alpha} ) &\text{if $\alpha_i<\alpha_{i+1}$,} \end{cases} \end{align*} where $s_i$ swaps the $i^\textsuperscript{th}$ and $(i+1)^\textsuperscript{th}$ entries of $\alpha$. The \definitionx{top Lascoux polynomials} $\topLas_\alpha$ are homogeneous polynomials indexed by \definitionx{snowy} weak compositions: weak compositions whose positive entries are distinct. Following~\cite[Lemma 4.23]{PY2}, we may define these polynomials recursively. Define the operator~$\widehat{\pi}_i$ as $$\widehat{\pi}_i(f) := \pi_i(x_{i+1}f)= x_i x_{i+1} \partial_i(f).$$ Then define \begin{equation} \label{EQ: Define top Las} \topLas_\alpha := \begin{cases} x^\alpha & \text{if $\alpha$ is a partition,} \\ \widehat{\pi}_i (\topLas_{s_i\alpha}) &\text{if $\alpha_i<\alpha_{i+1}$.} \end{cases} \end{equation} By~\cite{PY2}, the vector space \begin{equation} \label{EQ: widehat V} \widehat{V} := \mathbb{Q}\textrm{-}\textsf{span} \{ \topLas_\alpha: \alpha \textrm{ is a snowy weak composition}\} \end{equation} is a sub-algebra of $\Q[x_1, x_2, \dots]$. Its basis is given by the spanning set in (\ref{EQ: widehat V}) and its Hilbert series is $\prod_{m > 0} \left( 1 + \frac{q^m}{1 - q} \right)$. In particular, each graded piece of $\widehat{V}$ has finite dimension. For weak compositions $\alpha, \gamma,$ and $\delta$, let $d^\delta_{\alpha, \gamma}$ be the coefficient of $\topLas_\delta$ in the expansion of $\topLas_\alpha \times \topLas_\gamma$. We call them the \definitionx{top Lascoux structure constants}. Later in~\S\ref{S: structure constants}, we show each $d^\delta_{\alpha, \gamma}$ is the Schubert structure constant $c^w_{u,v}$ for some permutations $u, v, w$ and vice versa. \subsection{Bumpless pipedreams} The \definitionx{(reduced) bumpless pipedreams (BPD)}, introduced by Lam, Lee and Shimozono~\cite{LLS}, are combinatorial objects that give a monomial expansion of a Schubert polynomial. For a permutation $w \in S_n$, a BPD is an $n \times n$ grid built by the following six tiles: $$ \btile, \quad \rtile, \quad \jtile, \quad \ptile, \quad \htile, \quad \vtile. $$ We adopt the convention that row $1$ is the topmost row and column $1$ is the leftmost column. For each $i \in [n]$, we require a pipe to enter from the bottom of column $i$ and end at the rightmost edge of row $w(i)$. Moreover, two pipes cannot cross more than once. \begin{exa} \label{E: BPD} There are three BPDs for the permutation $w \in S_4$ with one-line notation $[2,1,4,3]$. \[ \begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue] \draw[step=1,gray,thin] (0,0) grid (4,4); \draw[color=black, thick, sharp corners] (0,0) rectangle (4,4); \draw(.5, 0)--(.5,2.5)--(4,2.5); \draw(1.5, 0)--(1.5,3.5)--(4,3.5); \draw(2.5, 0)--(2.5,0.5)--(4,0.5); \draw(3.5, 0)--(3.5,1.5)--(4,1.5); \node[color=black] at (0.5,-.5) {$1$}; \node[color=black] at (1.5,-.5) {$2$}; \node[color=black] at (2.5,-.5) {$3$}; \node[color=black] at (3.5,-.5) {$4$}; \node[color=black] at (5.5,3.5) {$w(1) = 2$}; \node[color=black] at (5.5,2.5) {$w(2) = 1$}; \node[color=black] at (5.5,1.5) {$w(3) = 4$}; \node[color=black] at (5.5,0.5) {$w(4) = 3$}; \end{tikzpicture} \quad\quad \begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue] \draw[step=1,gray,thin] (0,0) grid (4,4); \draw[color=black, thick, sharp corners] (0,0) rectangle (4,4); \draw(.5, 0)--(.5,1.5)--(2.5,1.5)--(2.5,2.5)--(4,2.5); \draw(1.5, 0)--(1.5,3.5)--(4,3.5); \draw(2.5, 0)--(2.5,0.5)--(4,0.5); \draw(3.5, 0)--(3.5,1.5)--(4,1.5); \node[color=black] at (0.5,-.5) {$1$}; \node[color=black] at (1.5,-.5) {$2$}; \node[color=black] at (2.5,-.5) {$3$}; \node[color=black] at (3.5,-.5) {$4$}; \node[color=black] at (5.5,3.5) {$w(1) = 2$}; \node[color=black] at (5.5,2.5) {$w(2) = 1$}; \node[color=black] at (5.5,1.5) {$w(3) = 4$}; \node[color=black] at (5.5,0.5) {$w(4) = 3$}; \end{tikzpicture} \quad\quad \begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue] \draw[step=1,gray,thin] (0,0) grid (4,4); \draw[color=black, thick, sharp corners] (0,0) rectangle (4,4); \draw(.5, 0)--(.5,2.5)--(4,2.5); \draw(1.5, 0)--(1.5,1.5)--(2.5,1.5)--(2.5,3.5)--(4,3.5); \draw(2.5, 0)--(2.5,0.5)--(4,0.5); \draw(3.5, 0)--(3.5,1.5)--(4,1.5); \node[color=black] at (0.5,-.5) {$1$}; \node[color=black] at (1.5,-.5) {$2$}; \node[color=black] at (2.5,-.5) {$3$}; \node[color=black] at (3.5,-.5) {$4$}; \node[color=black] at (5.5,3.5) {$w(1) = 2$}; \node[color=black] at (5.5,2.5) {$w(2) = 1$}; \node[color=black] at (5.5,1.5) {$w(3) = 4$}; \node[color=black] at (5.5,0.5) {$w(4) = 3$}; \end{tikzpicture} \] \end{exa} We let $\BPD(w)$ be the set of BPDs of a permutation $w$. We call $\btile$ a \definitionx{blank}. The \definitionx{blank-weight} of a BPD $D$ is a weak composition where the $i^\textsuperscript{th}$ entry counts the number of $\btile$ in row $i$. We denote it as $\wt_{\Box}(D)$ to emphasize that the weight comes from the blanks. Then BPD gives a combinatorial formula for Schubert polynomials. \begin{theo}[\cite{LLS}] For a permutation $w \in S_n$, $$ \fS_w = \sum_{D \in \BPD(w)} x^{\wt_\square(D)}. $$ \end{theo} Continuing Example~\ref{E: BPD}, $\fS_{[2,1,4,3]} = x_1x_3 + x_1 x_2 + x_1^2$. \subsection{Diagrams} A \definitionx{diagram} is a finite subset of $\NN \times \NN$. We may represent a diagram by putting a cell at row $r$ and column $c$ for each $(r,c)$ in the diagram. The \definitionx{weight} of a diagram $D$, denoted as $\wt(D)$, is a weak composition whose $i$\textsuperscript{th} entry is the number of boxes in its $i$\textsuperscript{th} row. Each weak composition $\alpha$ is associated with a diagram $D(\alpha)$, the unique left-justified diagram with weight $\alpha$. Each permutation $w \in S_n$ or $S_+$ is associated with a diagram called the \definitionx{Rothe diagram} $$RD(w) := \{(r,c): w(r) > c, w(i) \neq c \textrm{ for any } i \in [r]\}.$$ \begin{exa} We provide examples of two diagrams. For clarity, we put an ``$i$'' on the left of the $i$\textsuperscript{th} row and put a small dot in each cell. \begin{align*} D((0,2,4,0,1)) = \raisebox{1cm}{ \begin{ytableau} \none[1]\cr \none[2] & \cdot & \cdot \cr \none[3] & \cdot & \cdot & \cdot & \cdot \cr \none[4] \cr \none[3] & \cdot \end{ytableau}} \quad,\quad\quad RD([4,1,5,3,2]) = \raisebox{1cm}{ \begin{ytableau} \none[1] & \cdot & \cdot & \cdot \cr \none[2] & \none & \none & \none \cr \none[3] & \none & \cdot & \cdot \cr \none[4] & \none & \cdot & \none \cr \none[5] & \none & \none & \none \end{ytableau}}\quad. \end{align*} \end{exa} The Rothe diagram can characterize one special term in a Schubert polynomial. We consider the \definitionx{tail-lexicographical order} on weak compositions: For two weak compositions $\alpha, \gamma$, we say $\alpha$ is larger than $\gamma$ if there exists $i$ such that $\alpha_j = \gamma_j$ for all $j > i$ and $\alpha_i > \gamma_i$. For a polynomial $f$, the \definitionx{support} of $f$, denoted as $\supp(f)$, is the set of weak compositions $\alpha$ such that $x^\alpha$ has non-zero coefficient in $f$. The \definitionx{leading monomial} of $f$ is $x^\alpha$ such that $\alpha$ is the largest in $\supp(\alpha)$. By Lascoux and Sch{\"u}tzenberger~\cite{LS:Schubert}, the leading monomial of $\fS_w$ is $x^{\wt(RD(w))}$ with coefficient $1$. To describe the leading monomial of a top Lascoux polynomial, Pan and Yu~\cite{PY2} introduce the \definitionx{snow diagram}. For each diagram $D$, its snow diagram $\snow(D)$ is a diagram together with some labels in its cells. Each cell can be unlabeled, or labeled by $\bullet$ or $\snowflake$. We only consider the snow diagram of $D(\alpha)$ where $\alpha$ is a snowy weak composition. In this case, $\snow(D(\alpha))$ can be defined as follows. In $D(\alpha)$, label the rightmost cell on each row with $\bullet$. Then put a cell labeled by $\snowflake$ in empty spaces above each $\bullet$. \begin{exa} Let $\alpha = (2,0,4,0,1)$. Then $D(\alpha)$ and $\snow(D(\alpha))$ are depicted as follows. \begin{align*} D(\alpha) = \raisebox{0.95cm}{ \begin{ytableau} \none[1] & \cdot& \cdot \cr \none[2] \cr \none[3] & \cdot& \cdot& \cdot& \cdot \cr \none[4] \cr \none[5] & \cdot\cr \end{ytableau}}\quad, \quad \quad \snow(D(\alpha)) = \raisebox{0.95cm}{ \begin{ytableau} \none[1] & \cdot & \bullet & \none & \snowflake \cr \none[2] & \snowflake & \none & \none & \snowflake\cr \none[3] & \cdot& \cdot& \cdot& \bullet \cr \none[4] & \snowflake \cr \none[5] & \bullet \cr \end{ytableau}}\quad. \end{align*} \end{exa} For a snowy $\alpha$, define $$\rajcode(\alpha) := \wt(\snow(D(\alpha))). \textrm{ Equivalently, } \rajcode(\alpha)_i := \alpha_i + |\{j > i: \alpha_j > \alpha_i \}|.$$ By~\cite{PY2}, $x^{\rajcode(\alpha)}$ is the leading monomial of $\topLas_\alpha$. Moreover, any two distinct snowy weak compositions have different $\rajcode$. We end this subsection with a simple property of $\rajcode$ that will be useful in~\S\ref{S: structure constants}. \begin{lemm} \label{L: rajcode tail-lex} Let $\alpha, \gamma$ be two snowy weak compositions. If $\alpha$ is larger than $\gamma$ in tail-lexicographical order, then $\rajcode(\alpha)$ is also larger than $\rajcode(\gamma)$. \end{lemm} \begin{proof} Find $i$ such that $\alpha_j = \gamma_j$ for all $j > i$ and $\alpha_i > \gamma_i$. Clearly, $\rajcode(\alpha)_j = \rajcode(\gamma)_j$ for all $j > i$ and $\rajcode(\alpha)_i > \rajcode(\gamma)_i$. \end{proof} \subsection{Dual character of the flagged Weyl module} Let $B$ be the group of $n \times n$ upper triangular matrices over $\mathbb{C}$. Each diagram $D$ that lies in $[n] \times [n]$ is associated with a representation of $B$ known as the \definitionx{flagged Weyl module}. See~\cite{Mac, RS_Percent} for a detailed construction of this module. Let $D$ be a diagram. We use $\chi_D$ to denote the dual character of the flagged Weyl module associated with $D$. This is a family of polynomials in $\mathbb{Q}[x_1, x_2, \dots]$ containing the Schubert polynomials and key polynomials. \begin{theo}[{\cite{Dem, KP, Mag}}] \label{T: Schuberts are chars} For $w \in S_n$, $\chi_{RD(w)} = \fS_w$. For a weak composition $\alpha$, $\chi_{D(\alpha)} = \kappa_\alpha$. \end{theo} The polynomials $\chi_D$ can be computed recursively for certain diagrams $D$. To characterize them, we define the following moves on diagrams. \begin{defn}[\cite{Mag}] Let $D$ be a diagram. We say $D'$ is obtained from $D$ via an \definitionx{orthodontic move} if one of the following holds. \begin{itemize} \item Suppose there is $r$ such that $(1,r') \in D$ if and only if $r' \in [r]$. Then one may remove all cells in column $1$ of $D$ and shift all remaining cells leftward by $1$ to obtain $D'$. \item Assume there exists $r$ such that $(r, c) \in D$ implies $(r+1,c) \in D$ for any $c \in \mathbb{Z}_{>0}$. Then one may swap row $r$ and row $r+1$ to obtain $D'$. \end{itemize} \end{defn} The effect of each orthodontic move on $\chi_D$ can be characterized as follows. \begin{theo}[{\cite{Mag}}] \label{T: ort} Let $D$ be a diagram and $D'$ is obtained from $D$ via an orthodontic move. \begin{itemize} \item If $D'$ is obtained from $D$ by removing cells $(1, 1), \dots, (r,c)$ and shifting all cells left, then $\chi_{D} = \chi_{D'} x_1 \cdots x_r$. \item If $D'$ is obtained from $D$ by swapping row $r$ and $r+1$, then $\chi_{D} = \pi_i(\chi_{D'})$. \end{itemize} \end{theo} Together with the base case $\chi_{\emptyset} = 1$, one may compute $\chi_D$ using Theorem~\ref{T: ort} if one can obtain the empty diagram from $D$ using the orthodontic moves. Such diagrams can be described as follows. \begin{defn}[\cite{RS_Percent}] A diagram $D$ is \definitionx{percent-avoiding} if $(i_1,j_1), (i_2, j_2) \in D$ with $i_1 < i_2$ and $j_2 < j_1$ implies either $(i_1, j_2) \in D$ or $(i_2, j_1) \in D$. \end{defn} \begin{theo}[\cite{RS_Percent}] \label{T: PA and OM} Let $D$ be a diagram. One may apply orthodontic moves on~$D$ repetitively and eventually obtain the empty diagram if and only if $D$ is percent-avoiding. In particular, Rothe diagrams are such diagrams. \end{theo} We also need a well-known property of $\chi_D$ that follows immediately from its construction. \begin{theo} \label{T: Permuting columns} If $D$ and $D'$ differ by permuting the columns, then $\chi_D = \chi_{D'}$. \end{theo} \section{Relations between top Lascoux polynomials and Schubert polynomials} \label{S: Relations} This section describes the relationship between top Lascoux and Schubert polynomials. \subsection{The reverse complement involution on polynomials} We introduce a linear operator on polynomials that later will be used to transform a top Lascoux polynomial into a Schubert polynomial. We begin with an involution on certain weak compositions. \begin{defn} Let $m, n$ be positive integers. Define the \definitionx{reverse complement} operator~$r_{m,n}$ on the set of weak compositions $\alpha$ such that $\supp(\alpha) \subseteq [n]$ and $\alpha_i \leq m$ for all $i$. We define \begin{align*} r_{m,n}(\alpha) := (m - \alpha_n, \dots, m - \alpha_1). \end{align*} \end{defn} Next, we analogously define $r_{m,n}$ on certain polynomials. \begin{defn} Let $m, n$ be positive integers. We extend $r_{m, n}$ to the set of polynomials in $x_1, \dots, x_n$ where the power of any $x_i$ is at most $m$. We define it as the linear operator such that $r_{m,n}(x^\alpha) := x^{r_{m,n}(\alpha)}$. Equivalently, we can define $r_{m,n}$ as \begin{align*} r_{m,n}(f) & := x_1^m \cdots x_n^m f(x_n^{-1}, \dots, x_1^{-1}). \end{align*} \end{defn} \begin{rem} The operator $r_{m,n}$ on polynomials is similar to the operator $$f \mapsto x_1^n \cdots x_n^n f(x_1^{-1}, \dots, x_n^{-1})$$ considered by Huh, Matherne, M{\'e}sz{\'a}ros and St. Dizier~\cite{HMMS}. In \cite[Theorem 6]{HMMS}, the authors apply this operator on a Schubert polynomial $\fS_w$ with $w \in S_n$ and show the resulting polynomial is Lorentzian after normalization. Our $r_{m,n}$ is also similar to the operator $$f \mapsto f(x_1^{-1}, \dots, x_n^{-1})$$ which sends the character of a $\textrm{GL}_n$ module to the character of its dual. A more generalized operator $f \mapsto x^\alpha f(x_1^{-1}, \dots, x_n^{-1})$ and its action on Schubert polynomials was studied by Fan, Guo and Liu~\cite{FGL}. \end{rem} Next, we investigate how to swap $r_{m,n}$ with the operators: $\partial_i$, $\pi_i$, and $\widehat{\pi}_i$. Even though the proofs are straightforward, these identities are crucial in the rest of the paper. \begin{lemm} \label{L: commute operators} Suppose $r_{m,n}$ is defined on a polynomial $f$. Take $i \in [n-1]$. Then $r_{m,n}$ is also defined on $\partial_i(f), \pi_i(f)$, and $\widehat{\pi}_i(f)$, with: \begin{align*} r_{m,n} (\partial_i(f)) &= \widehat{\pi}_{n-i} (r_{m,n}(f)),\\ r_{m,n} (\pi_i(f)) &= \pi_{n-i} (r_{m,n}(f)),\\ r_{m,n} (\widehat{\pi}_i(f)) &= \partial_{n-i} (r_{m,n}(f)). \end{align*} \end{lemm} \begin{proof} It is enough to assume $f = x^\alpha$. Clearly, $\partial_i(f), \pi_i(f)$, and $\widehat{\pi}_i(f)$ will involve monomials in $x_1, \dots, x_n$ where the power of any $x_i$ is at most $m$. Thus, $r_{m,n}$ is defined on these polynomials. The three equations follow from a routine check. \end{proof} \subsection{Relating Schubert polynomials to top Lascoux polynomials} In this subsection, we establish that each Schubert polynomial is the reverse complement of a top Lascoux polynomial and vice versa. We start by describing a variation of the map introduced by Fulton~\cite[(3.4)]{ful}. \begin{defn} Let $\alpha$ be a snowy weak composition. Take any $m, n$ such that $\supp(\alpha) \subseteq [n]$ and $m \geq \max(\alpha)$. The \definitionx{$(m,n)$-standardization} of $\alpha$, denoted as $\std_{m,n}(\alpha)$ is the unique permutation $w$ satisfying $w(n+1) < w(n+2) < \cdots$ and \begin{align*} w(i) := \begin{cases} r_{m+1, n}(\alpha)_i & \text{if $r_{m+1, n}(\alpha)_i \leq m$} \\ m + |\{j \in [i]: r_{m+1, n}(\alpha)_j = m+1\}| &\text{if $r_{m+1, n}(\alpha)_i = m+1$,} \end{cases} \end{align*} for any $i \in [n]$. \end{defn} \begin{exa} \label{E: std} Let $\alpha = (2,4,0,6,0,0,1)$, $m = 6$ and $n = 7$. To compute $w = std_{m,n}(\alpha)$, we start with $r_{m+1, n}(\alpha) = (6,7,7,1,7,3,5)$. Based on the definition, the first $n$ entries in the one-line notation of $w$ can be obtained by changing those $m+1$ in $r_{m+1, n}(\alpha)$ into $m+1, m+2, \dots, $. Thus, we get $(6,7,8,1,9,3,5)$. Finally, since $w(n+1) < w(n+2) < \cdots$, we know $w$ has one-line notation $[6,7,8,1,9,3,5,2,4]$. \end{exa} Then we can describe the relation between top Lascoux polynomials and Schubert polynomials. \begin{theo} \label{T: topLas as reverse Schubert} Let $\alpha$ be a snowy weak composition. Take any $m, n$ such that \mbox{$\supp(\alpha) \subseteq [n]$} and $m \geq \max(\alpha)$. Let $w$ be the permutation $\std_{m,n}(\alpha)$. Then $$ r_{m, n}(\topLas_{\alpha}) = \fS_w. $$ \end{theo} For instance, let $\alpha = (2,4,0,6,0,0,1)$, $m = 6$ and $n = 7$. The theorem concludes that $r_{6,7}(\topLas_\alpha) = \fS_w$ where $w = \std_{m,n}(\alpha) = [6,7,8,1,9,3,5,2,4]$ from the computation in Example~\ref{E: std}. \begin{proof} Prove by induction on $\alpha$. For the base case, assume $\alpha$ is a partition with $\supp(\alpha) = [k]$. Then $r_{m+1, n}(\alpha) = ( m+1, \dots, m+1, m+1 - \alpha_k, \dots, m+1 - \alpha_1)$. The first $n$ numbers in the one-line notation of $w$ are $$m+1, m+2, \dots, m + n - k, m+1 - \alpha_k, \dots, m+1 - \alpha_1.$$ Thus, we have $\invcode(w) = (m, \dots, m, m - \alpha_k, \dots, m - \alpha_1) = r_{m,n}(\alpha)$, so $w$ is a dominant permutation. By the definition of Schubert polynomials, $$\fS_w = x^{r_{m,n}(\alpha)} = r_{m,n}(x^\alpha) = r_{m,n}(\topLas_\alpha).$$ Now suppose $\alpha_i < \alpha_{i+1}$. It is routine to check $w(n-i) < w(n-i+1)$ and $\std_{m,n}(s_i \alpha) = ws_{n-i}$. Then the proof is finished by Lemma~\ref{L: commute operators}: \begin{align*} \fS_w & = \partial_{n-i} (\fS_{w s_{n-i}}) = \partial_{n-i} (\fS_{\std_{m,n}(s_i \alpha)})\\ & = \partial_{n-i}(r_{m,n}(\topLas_{s_i \alpha})) = r_{m,n}(\widehat{\pi}_i(\topLas_{s_i \alpha})) = r_{m,n}(\topLas_\alpha). \qedhere \end{align*} \end{proof} \begin{exa} We can understand Theorem~\ref{T: topLas as reverse Schubert} via a commutative diagram. For instance, the equation $r_{4,5}(\topLas_{(2,0,4,0,1}) = \topLas_{[4,5,1,6,3,2]}$ is implied by the following commutative diagram. {\small \[ \begin{tikzcd} \fS_{[5,6,4,3,1,2]} \arrow[r,"\partial_2"] \arrow[d,"r_{4,5}"] & \fS_{[5,4,6,3,1,2]} \arrow[r,"\partial_1"] \arrow[d,"r_{4,5}"] & \fS_{[4,5,6,3,1,2]} \arrow[r,"\partial_4"] \arrow[d,"r_{4,5}"] & \fS_{[4,5,6,1,3,2]} \arrow[r,"\partial_3"] \arrow[d,"r_{4,5}"] & \fS_{[4,5,1,6,3,2]} \arrow[d,"r_{4,5}"] \\ \topLas_{(4,2,1,0,0)} \arrow[r,"\widehat{\pi}_3"] \arrow[u] & \topLas_{(4,2,0,1,0)} \arrow[r,"\widehat{\pi}_4"] \arrow[u] & \topLas_{(4,2,0,0,1)} \arrow[r,"\widehat{\pi}_1"] \arrow[u] & \topLas_{(2,4,0,0,1)} \arrow[r,"\widehat{\pi}_2"] \arrow[u] & \topLas_{(2,0,4,0,1)} \arrow[u] \end{tikzcd} \]} \end{exa} Consequently, every Schubert polynomial is the reverse complement of a top Lascoux polynomial. \begin{coro} Consider $w \in S_n$. Let $\alpha = (n+1 - w(n), \dots, n+1 - w(2), n+1-w(1))$. Then $\fS_w = r_{n,n}(\topLas_\alpha)$. \end{coro} \begin{proof} Notice that $\std_{n,n}(\alpha) = w$. Then the proof is finished by Theorem~\ref{T: topLas as reverse Schubert}. \end{proof} \subsection{A diagrammatic perspective of standardization} We can interpret the standardization map $\std_{m,n}$ in terms of diagrams. Recall that each snowy weak composition $\alpha$ is associated with a labeled diagram $\snow(D(\alpha))$. Each permutation $w$ is associated with the Rothe diagram $RD(w)$. We describe the relationship between $\snow(D(\alpha))$ and $RD(\std_{m,n}(\alpha))$. \begin{exa} \label{E: compare diagrams} Consider the snowy weak composition $\alpha = (0,4,2)$. Let $m = 4$ and $n = 3$. Let $w = \std_{m,n}(\alpha) = [3,1,5,2,4]$. We depict $\snow(D(\alpha))$ and $RD(w)$ as follows: $$ \snow(D(\alpha)) = \raisebox{0.45cm}{ \begin{ytableau} \none[1] & \none & \snowflake & \none & \snowflake \cr \none[2] & \cdot & \cdot & \cdot & \bullet\cr \none[3] & \cdot & \bullet \end{ytableau}} \quad\quad RD(w) = \raisebox{0.45cm}{ \begin{ytableau} \none[1] & \cdot & \cdot \cr \none[2] \cr \none[3] & \none& \cdot & \none & \cdot \end{ytableau}} $$ We observe that both diagrams live in the first $3$ rows and $4$ columns. Imagine that we ignore the labels of $\snow(D(\alpha))$ and put it in an $3 \times 4$ box. Then we take its complement within the box and rotate the box by $180^\circ$. What we get is exactly~$RD(w)$. $$ \begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue] \draw[step=1,gray,ultra thin,dashed] (0,0) grid (4,3); \draw[color=red, very thick, sharp corners] (0,0) rectangle (4,3); \draw[color=black, thick, sharp corners] (0,1) rectangle (1,2); \filldraw[black] (0.5,1.5) circle (0.2pt); \draw[color=black, thick, sharp corners] (0,0) rectangle (1,1); \filldraw[black] (0.5,0.5) circle (0.2pt); \draw[color=black, thick, sharp corners] (1,2) rectangle (2,3); \filldraw[black] (1.5,2.5) circle (0.2pt); \draw[color=black, thick, sharp corners] (1,1) rectangle (2,2); \filldraw[black] (1.5,1.5) circle (0.2pt); \draw[color=black, thick, sharp corners] (1,0) rectangle (2,1); \filldraw[black] (1.5,0.5) circle (0.2pt); \draw[color=black, thick, sharp corners] (2,1) rectangle (3,2); \filldraw[black] (2.5,1.5) circle (0.2pt); \draw[color=black, thick, sharp corners] (3,2) rectangle (4,3); \filldraw[black] (3.5,2.5) circle (0.2pt); \draw[color=black, thick, sharp corners] (3,1) rectangle (4,2); \filldraw[black] (3.5,1.5) circle (0.2pt); \end{tikzpicture} \quad \raisebox{0.8cm}{$\xrightarrow{\textrm{complement}}$} \quad \begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue] \draw[step=1,gray,ultra thin,dashed] (0,0) grid (4,3); \draw[color=red, very thick, sharp corners] (0,0) rectangle (4,3); \draw[color=black, thick, sharp corners] (0,2) rectangle (1,3); \filldraw[black] (0.5,2.5) circle (0.2pt); \draw[color=black, thick, sharp corners] (2,2) rectangle (3,3); \filldraw[black] (2.5,2.5) circle (0.2pt); \draw[color=black, thick, sharp corners] (2,0) rectangle (3,1); \filldraw[black] (2.5,0.5) circle (0.2pt); \draw[color=black, thick, sharp corners] (3,0) rectangle (4,1); \filldraw[black] (3.5,0.5) circle (0.2pt); \end{tikzpicture} \quad \raisebox{0.8cm}{$\xrightarrow{\textrm{rotate}}$} \quad \begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue] \draw[step=1,gray,ultra thin,dashed] (0,0) grid (4,3); \draw[color=red, very thick, sharp corners] (0,0) rectangle (4,3); \draw[color=black, thick, sharp corners] (0,2) rectangle (1,3); \filldraw[black] (0.5,2.5) circle (0.2pt); \draw[color=black, thick, sharp corners] (1,2) rectangle (2,3); \filldraw[black] (1.5,2.5) circle (0.2pt); \draw[color=black, thick, sharp corners] (1,0) rectangle (2,1); \filldraw[black] (1.5,0.5) circle (0.2pt); \draw[color=black, thick, sharp corners] (3,0) rectangle (4,1); \filldraw[black] (3.5,0.5) circle (0.2pt); \end{tikzpicture} $$ \end{exa} To characterize this relation in general, we define the operator $r_{m,n}$ on certain diagrams. \begin{defn} Let $D$ be a diagram that lives in the first $n$ rows and first $m$ columns. Let $r_{m,n}(D)$ be the diagram obtained from the following process: Place $D$ in a $n \times m$ box, take its complement within the box, and rotate the box by $180^\circ$. \end{defn} \begin{lemm} \label{L: compare diagram} Let $\alpha$ be a snowy weak composition. Take $n$ with $\supp(\alpha) \subseteq [n]$ and $m$ with $\max(\alpha) \leq m$. Let $w = \std_{m,n}(\alpha)$. Then $RD(w) = r_{m,n}(\snow(D(\alpha)))$. \end{lemm} \begin{proof} If we ignore the labels, $\snow(D(\alpha))$ consists of cells $(r,c)$ such that $\alpha_r \geq c$ or~$\alpha_i = c$ for some $i > r$. Clearly, $\snow(D(\alpha))$ lies in the first $n$ rows and $m$ columns. If we complement $\snow(D(\alpha))$ within the $n \times m$ box, the resulting diagram consists of $(r,c) \in [m] \times [n]$ such that $\alpha_r < c$ and $\alpha_i \neq c$ for any $i > r$. Let $D' = r_{m,n}(\snow(D(\alpha)))$. We check $D'$ and $RD(w)$ agree row by row. They clearly agree under row $n$. Consider $r \in [n]$. If~$\alpha_{n+1-r} = 0$, then $D'$ has no cells in row~$r$. Also, $w(r) \geq m+1$. By the definition of standardization map, there are no inversions of $w$ of the form $(r, r')$. Thus, $RD(w)$ also has no cells in row $r$. If $\alpha_{n+1-r} > 0$, $D'$ has cells in column $c$ for $c$ such that $c < m+1 - \alpha_{n+1-r}$ and $c \neq m+1 - \alpha_{n+1-i}$ for all $i \in [r]$. In other words, that is all $c$ such that $c < w(r)$ and~$c$ is not in $w(1), \dots, w(r)$. Therefore, $D'$ and $RD(w)$ agree on row $r$. \end{proof} \section{Relations between top Lascoux structure constants and the Schubert structure constants} \label{S: structure constants} Recall that $\{\topLas_\alpha: \alpha \textrm{ is snowy}\}$ forms a basis of the algebra $\widehat{V}$. The top Lascoux structure constant $d^{\delta}_{\alpha, \gamma}$ is the coefficient of $\topLas_\delta$ in the expansion of $\topLas_\alpha \topLas_\gamma$. At this point, we do not have any reason to believe that they are positive integers. Moreover, it is well-known that key polynomials do not have positive structure constants. Surprisingly, the connection between top Lascoux polynomials and Schubert polynomials establishes a bridge between $d^{\delta}_{\alpha, \gamma}$ and the Schubert structure constants $c^w_{u,v}$. First, we describe a necessary condition for $d^{\delta}_{\alpha, \gamma}$ to be non-zero. \begin{lemm} \label{L: condition for d nonzero} Let $\alpha$, $\gamma$ and $\delta$ be snowy weak compositions. Find $m_1, m_2$ and $n$ such that $m_1 \geq \max(\alpha)$, $m_2 \geq \max(\gamma)$, $\supp(\alpha) \subseteq [n]$ and $\supp(\gamma) \subseteq [n]$. If $d^{\delta}_{\alpha, \gamma} \neq 0$, we must have $\supp(\delta) \subseteq [n]$ and $\max(\delta) \leq m_1 + m_2$. \end{lemm} The proof relies heavily on the statistic $\rajcode$. \begin{proof} First, expand \begin{equation} \label{EQ: product of top L} \topLas_\alpha \times \topLas_\gamma = \sum_{\sigma} d^{\sigma}_{\alpha, \gamma} \topLas_\sigma. \end{equation} We know the left hand side uses only variables $x_1, \dots, x_n$. Moreover, in any monomial on the left hand side, each variable has power at most $m_1 + m_2$. Let $S$ be the set of all $\sigma$ with $d^{\sigma}_{\alpha, \gamma} \neq 0$. Among $S$, find the largest $\sigma$ in tail-lexicographical order. By Lemma~\ref{L: rajcode tail-lex}, $\rajcode(\sigma)$ is also larger than $\rajcode(\sigma')$ for any $\sigma' \in S$. Thus, $x^{\rajcode(\sigma)}$ has non-zero coefficient on the right hand side of (\ref{EQ: product of top L}), so $\supp(\sigma) \subseteq [n]$. It follows that $\supp(\sigma') \subseteq [n]$ for any $\sigma' \in S$. Now find $\sigma \in S$ with the largest $\max(\sigma)$, break ties by picking the largest in tail-lexicographical order. Say $\max(\sigma) = m$. In $\rajcode(\sigma)$, one entry is $m$. We can see $\rajcode(\sigma)$ cannot appear in $\topLas_{\sigma'}$ for any other $\sigma' \in S$: If so, then $\sigma'$ has an entry at least $m$ and $\rajcode(\sigma')$ is larger than $\rajcode(\sigma)$, contradicting to the maximality of $\sigma$. Thus, $x^\rajcode(\sigma)$ has non-zero coefficient on the right hand side of (\ref{EQ: product of top L}), so $m \leq m_1 + m_2$. It follows that $\max(\sigma') \leq m_1 + m_2$ for any $\sigma' \in S$. \end{proof} Now we describe the main theorem of this section. \begin{theo} \label{T: structure constants} Let $\alpha$, $\gamma$ be snowy weak compositions. Find $m_1, m_2$ and $n$ such that $m_1 \geq \max(\alpha)$, $m_2 \geq \max(\gamma)$, $\supp(\alpha) \subseteq [n]$ and $\supp(\gamma) \subseteq [n]$. Let $u = \std_{m_1, n}(\alpha)$ and $v = \std_{m_2, n}(\gamma)$. For any snowy weak composition $\delta$ with $\supp(\delta) \subseteq [n]$ and $\max(\delta) \leq m_1 + m_2$, we let $w = \std_{m_1 + m_2, n}(\delta)$. Then $d^{\delta}_{\alpha, \gamma} = c^w_{u,v}$. \end{theo} \begin{proof} First, we have $\topLas_\alpha \times \topLas_\gamma = \sum_{\sigma} d^{\sigma}_{\alpha, \gamma} \topLas_\sigma$. By Lemma~\ref{L: condition for d nonzero}, the sum is over all $\sigma$ with $\supp(\sigma) \subseteq [n]$ and $\max(\sigma) \leq m_1 + m_2$. Apply $r_{m_1 + m_2, n}$ on both sides. Using Theorem~\ref{T: topLas as reverse Schubert}, the left hand side becomes \begin{align*} r_{m_1 + m_2,n}(\topLas_\alpha \times \topLas_\gamma) =r_{m_1,n}(\topLas_\alpha) r_{m_2,n}(\topLas_\gamma) = \fS_u \fS_v. \end{align*} The right hand becomes \begin{align*} r_{m_1 + m_2,n} \left( \sum_{\sigma} d^{\sigma}_{\alpha, \gamma} \topLas_\sigma \right) = \sum_{\sigma} d^{\sigma}_{\alpha, \gamma} r_{m_1 + m_2,n}(\topLas_\sigma) = \sum_{\sigma} d^{\sigma}_{\alpha, \gamma} \fS_{\std_{m_1 + m_2, n}(\sigma)}. \qquad \qedhere \end{align*} \end{proof} \begin{exa} \label{E: structure constants permutation} Let $\alpha = (2,3,1,4)$ and $\gamma = (2,1,4,3)$. We can let $m_1 = m_2 = n = 4$. Then $u = \std_{4,4}(\alpha) = [1,4,2,3]$ and $v = \std_{4,4}(\gamma) = [2,1,4,3]$. We compute \begin{align*} \topLas_{\alpha} \times \topLas_{\gamma} = \topLas_{(8,6,5,7)} + \topLas_{(6,8,4,7)} + \topLas_{(7,8,5,6)} + \topLas_{(7,6,8,5)} + \topLas_{(6,7,8,4)}.\\ \fS_{u} \times \fS_{v} = \fS_{[2,4,3,1]} + \fS_{[2,5,1,3,4]} + \fS_{[3,4,1,2]} + \fS_{[4,1,3,2]} + \fS_{[5,1,2,3,4]}. \end{align*} We check Theorem~\ref{T: structure constants} when $\delta = (8,6,5,7)$. We have $d^\delta_{\alpha, \gamma} = 1$. Now we compute $w = \std_{4 + 4, 4}(\delta) = [2,4,3,1]$. Indeed, $c^w_{u,v} = 1$. \end{exa} Theorem~\ref{T: structure constants} can express each Schubert structure constant as a top Lascoux structure constant. \begin{coro} Take $u, v \in S_n$ and $w \in S_{2n}$. If $w(n+1) < \cdots < w(2n)$. Then $c^w_{u,v} = d^{\delta}_{\alpha, \gamma}$, where \begin{align*} \alpha & = (n+1 - u(n), \dots, n+1-u(1)),\\ \gamma & = (n+1 - v(n), \dots, n+1-v(1)), \textrm{ and}\\ \delta & = (n+1 - w(n), \dots, n+1-w(1)). \end{align*} \end{coro} \section{Bumpless pipedream formula} \label{S: BPD} We discuss another application of the relationship between top Lascoux polynomials and Schubert polynomials. Bumpless pipedreams (BPD) give a formula to compute the monomial expansion of Schubert polynomials. After ``reversing the BPDs'', we get a formula for top Lascoux polynomials. \begin{defn} Let $\alpha$ be a snowy weak composition. Find the smallest $n$ such that $\supp(\alpha) \subseteq [n]$ and let $m = \max(\alpha)$. A \definitionx{left-to-top BPD} of $\alpha$ is a grid with $n$ rows and $m$ columns built by tiles $\btile, \rtile, \jtile, \ptile, \htile$ and $\vtile$. For each $i \in [n]$ with $\alpha_i > 0$, there is a pipe entering from the left in row $i$ and goes to the top of column $\alpha_i$. Moreover, no two pipes can cross more than once. Let $\LTBPD(\alpha)$ be the set of all left-to-top BPDs of $\alpha$. \end{defn} \begin{exa} \label{E: BPD topL} Consider $\alpha = (0,3,0,2)$. Then $n = 4$ and $m = 3$. The set $\LTBPD(\alpha)$ has the following five elements. We write $\alpha$ vertically on the left. The $3$ (resp. $2$) indicates that the pipe from that row should end in column $3$ (resp. $2$). The $0$ indicates that there is no pipe entering from that row. {\tiny \[ \begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue] \draw[step=1,gray,thin] (0,0) grid (3,4); \draw[color=black, thick, sharp corners] (0,0) rectangle (3,4); \draw(0, 2.5)--(2.5,2.5)--(2.5,4); \draw(0, 0.5)--(1.5,0.5)--(1.5,4); \node[color=black] at (-.5, 3.5) {$0$}; \node[color=black] at (-.5, 2.5) {$3$}; \node[color=black] at (-.5, 1.5) {$0$}; \node[color=black] at (-.5, .5) {$2$}; \end{tikzpicture} \quad\quad\quad \begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue] \draw[step=1,gray,thin] (0,0) grid (3,4); \draw[color=black, thick, sharp corners] (0,0) rectangle (3,4); \draw(0, 2.5)--(2.5,2.5)--(2.5,4); \draw(0, 0.5)--(0.5,0.5)--(0.5,1.5)--(1.5,1.5)--(1.5,4); \node[color=black] at (-.5, 3.5) {$0$}; \node[color=black] at (-.5, 2.5) {$3$}; \node[color=black] at (-.5, 1.5) {$0$}; \node[color=black] at (-.5, .5) {$2$}; \end{tikzpicture} \quad\quad\quad \begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue] \draw[step=1,gray,thin] (0,0) grid (3,4); \draw[color=black, thick, sharp corners] (0,0) rectangle (3,4); \draw(0, 2.5)--(2.5,2.5)--(2.5,4); \draw(0, 0.5)--(0.5,0.5)--(0.5,3.5)--(1.5,3.5)--(1.5,4); \node[color=black] at (-.5, 3.5) {$0$}; \node[color=black] at (-.5, 2.5) {$3$}; \node[color=black] at (-.5, 1.5) {$0$}; \node[color=black] at (-.5, .5) {$2$}; \end{tikzpicture} \quad\quad\quad \begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue] \draw[step=1,gray,thin] (0,0) grid (3,4); \draw[color=black, thick, sharp corners] (0,0) rectangle (3,4); \draw(0, 2.5)--(0.5,2.5)--(0.5,3.5)--(2.5,3.5)--(2.5,4); \draw(0, 0.5)--(1.5,0.5)--(1.5,4); \node[color=black] at (-.5, 3.5) {$0$}; \node[color=black] at (-.5, 2.5) {$3$}; \node[color=black] at (-.5, 1.5) {$0$}; \node[color=black] at (-.5, .5) {$2$}; \end{tikzpicture} \quad\quad\quad \begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue] \draw[step=1,gray,thin] (0,0) grid (3,4); \draw[color=black, thick, sharp corners] (0,0) rectangle (3,4); \draw(0, 2.5)--(0.5,2.5)--(0.5,3.5)--(2.5,3.5)--(2.5,4); \draw(0, 0.5)--(0.5,0.5)--(0.5,1.5)--(1.5,1.5)--(1.5,4); \node[color=black] at (-.5, 3.5) {$0$}; \node[color=black] at (-.5, 2.5) {$3$}; \node[color=black] at (-.5, 1.5) {$0$}; \node[color=black] at (-.5, .5) {$2$}; \end{tikzpicture} \]} \end{exa} \begin{rem} Readers might notice that left-to-top BPDs look like the top left part of a BPD after rotation. Keep $\alpha$, $m$ and $n$ from Example~\ref{E: BPD topL}. Consider the classical BPDs of the permutation $\std_{m,n}(\alpha) = [2,4,1,5,3]$. There are five of them: {\tiny \[ \begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue] \draw[step=1,gray,thin] (0,0) grid (5,5); \draw[color=black, thick, sharp corners] (0,0) rectangle (5,5); \draw[color=red, ultra thick, sharp corners] (0,1) rectangle (3,5); \draw(1.5, 0)--(1.5,4.5)--(5,4.5); \draw(3.5, 0)--(3.5,3.5)--(5,3.5); \draw(0.5, 0)--(0.5,2.5)--(5,2.5); \draw(4.5, 0)--(4.5,1.5)--(5,1.5); \draw(2.5, 0)--(2.5,0.5)--(5,0.5); \end{tikzpicture} \quad \begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue] \draw[step=1,gray,thin] (0,0) grid (5,5); \draw[color=black, thick, sharp corners] (0,0) rectangle (5,5); \draw[color=red, ultra thick, sharp corners] (0,1) rectangle (3,5); \draw(1.5, 0)--(1.5,3.5)--(2.5,3.5)--(2.5,4.5)--(5,4.5); \draw(3.5, 0)--(3.5,3.5)--(5,3.5); \draw(0.5, 0)--(0.5,2.5)--(5,2.5); \draw(4.5, 0)--(4.5,1.5)--(5,1.5); \draw(2.5, 0)--(2.5,0.5)--(5,0.5); \end{tikzpicture} \quad \begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue] \draw[step=1,gray,thin] (0,0) grid (5,5); \draw[color=black, thick, sharp corners] (0,0) rectangle (5,5); \draw[color=red, ultra thick, sharp corners] (0,1) rectangle (3,5); \draw(1.5, 0)--(1.5,1.5)--(2.5,1.5)--(2.5,4.5)--(5,4.5); \draw(3.5, 0)--(3.5,3.5)--(5,3.5); \draw(0.5, 0)--(0.5,2.5)--(5,2.5); \draw(4.5, 0)--(4.5,1.5)--(5,1.5); \draw(2.5, 0)--(2.5,0.5)--(5,0.5); \end{tikzpicture} \quad \begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue] \draw[step=1,gray,thin] (0,0) grid (5,5); \draw[color=black, thick, sharp corners] (0,0) rectangle (5,5); \draw[color=red, ultra thick, sharp corners] (0,1) rectangle (3,5); \draw(1.5, 0)--(1.5,4.5)--(5,4.5); \draw(3.5, 0)--(3.5,3.5)--(5,3.5); \draw(0.5, 0)--(0.5,1.5)--(2.5,1.5)--(2.5,2.5)--(5,2.5); \draw(4.5, 0)--(4.5,1.5)--(5,1.5); \draw(2.5, 0)--(2.5,0.5)--(5,0.5); \end{tikzpicture} \quad \begin{tikzpicture}[x=1.5em,y=1.5em,thick,color = blue] \draw[step=1,gray,thin] (0,0) grid (5,5); \draw[color=black, thick, sharp corners] (0,0) rectangle (5,5); \draw[color=red, ultra thick, sharp corners] (0,1) rectangle (3,5); \draw(1.5, 0)--(1.5,3.5)--(2.5,3.5)--(2.5,4.5)--(5,4.5); \draw(3.5, 0)--(3.5,3.5)--(5,3.5); \draw(0.5, 0)--(0.5,1.5)--(2.5,1.5)--(2.5,2.5)--(5,2.5); \draw(4.5, 0)--(4.5,1.5)--(5,1.5); \draw(2.5, 0)--(2.5,0.5)--(5,0.5); \end{tikzpicture} \]} The top-to-left BPDs in Example~\ref{E: BPD topL} are obtained by rotating the red part of these BPDs. \end{rem} This pattern holds in general. \begin{prop} \label{P: rotate BPD} Let $\alpha$ be a snowy weak composition. Find smallest $n$ such that $\supp(\alpha) \in [n]$ and let $m = \max(\alpha)$. Let $w = \std_{m,n}(\alpha)$. Then $\LTBPD(\alpha)$ is formed by rotating the first $n$ rows, $m$ columns of BPDs in $\BPD(w)$. \end{prop} \begin{proof} Immediate from Lemma~\ref{L: compare diagram}. \end{proof} Now we are ready to introduce a combinatorial formula for the top Lascoux polynomials. We just need to specify the ``weight'' of a left-to-top BPD. \begin{defn} Let $D$ be a left-to-top BPD for some snowy weak composition. The \definitionx{non-blank weight} of $D$, denoted as $\wt_{\boxdot}(D)$, is a weak composition where the $i^\textsuperscript{th}$ entry counts the number of \textbf{non-blank tiles} in row $i$ of $D$. \end{defn} \begin{theo} \label{T: BPD topL} Let $\alpha$ be a snowy weak composition. Then $$ \topLas_\alpha = \sum_{D \in \LTBPD(\alpha)} \wt_{\boxdot}(D). $$ \end{theo} For instance, Example~\ref{E: BPD topL} yields $$ \topLas_{(0,3,0,2)} = x_1^2x_2^3x_3x_4^2 + x_1^2 x_2^3x_3^2x_4 + x_1^3 x_2^3 x_3x_4 + x_1^3x_2^2x_3x_4^2 + x_1^3 x_2^2 x_3^2 x_4. $$ \begin{proof} Follows from Theorem~\ref{T: topLas as reverse Schubert} and Proposition~\ref{P: rotate BPD}. \end{proof} \section{Key expansion of top Lascoux polynomials} \label{S: key expansion} We expand top Lascoux polynomials positively into key polynomials. Our main tool is the Schubert-to-key expansion: \begin{theo}[\cite{RS}] For $w \in S_n$, the Schubert polynomial $\fS_w$ can be expanded as $\sum_{\alpha} c^w_\alpha \kappa_\alpha$, where the coefficients $c^w_\alpha$ are non-negative integers counting certain tableaux. \end{theo} We will translate this expansion to top Lascoux polynomials. The first step is to understand how the $r_{m,n}$ operator affects a key polynomial. \begin{prop} \label{P: reverse key} Let $\alpha$ be any weak composition. Let $m, n$ be positive integers. Then~$r_{m,n}$ is defined on $\alpha$ if and only if $r_{m,n}$ is defined on $\kappa_\alpha$. If this is the case, $r_{m,n}(\kappa_\alpha) = \kappa_{r_{m,n}(\alpha)}$. \end{prop} The proof is essentially the same as the proof of Theorem~\ref{T: topLas as reverse Schubert}. \begin{proof} The first claim is immediate. We prove the equation by induction on $\alpha$. For the base case, assume $\alpha$ is a partition. So is $r_{m,n} (\alpha)$. We have $r_{m,n}(\kappa_\alpha) = r_{m,n}(x^\alpha) = x^{r_{m,n}(\alpha)} = \kappa_{r_{m,n}(\alpha)}$. Now assume $\alpha_i < \alpha_{i+1}$ for some $i$. For our inductive hypothesis, assume $\kappa_{s_i \alpha} = r_{m,n}(\kappa_{r_{m,n}(s_i \alpha)})$. By Lemma~\ref{L: commute operators}, \begin{align*} \kappa_\alpha & = \pi_i(\kappa_{s_i \alpha}) = \pi_i (r_{m,n} (\kappa_{r_{m,n}(s_i \alpha)}))\\ & = r_{m,n} (\pi_{n-i} (\kappa_{s_{n-i} (r_{m,n}(\alpha)}))) = r_{m,n} (\kappa_{r_{m,n}(\alpha)}). \qedhere \end{align*} \end{proof} \begin{coro} Let $\alpha$ be a snowy weak composition. Find $m, n$ such that $\alpha \subseteq [n]$ and $\max(\alpha) \leq m$. Let $w$ be the permutation $\std_{m,n}(\alpha)$. The top Lascoux polynomial~$\topLas_\alpha$ can be expanded as $\sum_{\gamma} c^w_\gamma \kappa_{r_{m,n}(\gamma)}$ where the coefficient $c^w_\gamma$ is the coefficient of $\kappa_\gamma$ in the expansion of $\fS_w$. \end{coro} \begin{proof} By Theorem~\ref{T: topLas as reverse Schubert}, we know $\topLas_{\alpha} = r_{m,n}(\fS_w)$. By the Schubert-to-key expansion and Proposition~\ref{P: reverse key}, \begin{align*} \topLas_{\alpha} = r_{m,n}\left(\sum_{\gamma} c^w_\gamma \kappa_{\gamma}\right) = \sum_{\gamma} c^w_\gamma r_{m,n}(\kappa_{\gamma}) = \sum_{\gamma} c^w_\gamma \kappa_{r_{m,n}(\gamma)}. \qquad \qedhere \end{align*} \end{proof} \section{Top Lascoux polynomials as dual-characters of flagged Weyl modules} \label{S: character} We show $\topLas_\alpha$ agrees with $\chi_{\snow(D(\alpha))}$. Our proof relies on the following relation between the $r_{m,n}$ on polynomials and the $r_{m,n}$ on diagrams. \begin{prop} \label{P: chi rotate} Let $D$ be a diagram in the first $n$ rows and $m$ columns. Assume one can obtain the empty diagram by applying orthodontic moves from $D$. Then \begin{align} \label{EQ: chi reverse} \chi_{r_{m,n}(D)} = r_{m,n}(\chi_D). \end{align} \end{prop} \begin{proof} We prove by induction on $m$. The case when $m = 0$ is trivial. Suppose $m > 0$. If $D$ is the empty diagram, then $r_{m,n}(D)$ is the $n \times m$ box. We have $\chi_{r_{m,n}(D)} = x_1^m \cdots x_n^m$, which agrees with $r_{m,n}(\chi_D) = r_{m,n}(1)$. Now suppose $D$ is not empty and let $D'$ be a diagram obtained from $D$ after one orthodontic move. It remains to check~(\ref{EQ: chi reverse}) while assuming $\chi_{r_{m,n}(D')} = r_{m,n}(\chi_{D'})$. We consider the two types of orthodontic moves. \begin{itemize} \item Suppose $D'$ is obtained from $D$ by removing cells $(1,1), \dots, (1,r)$ and shift all remaining cells leftward. Thus, $D'$ lives in the $n \times (m-1)$ box. The diagram $r_{m,n}(D)$ is obtained by adding cells $(1,m), \dots, (n-r, m)$ to $r_{m-1, n}(D')$. Let~$D''$ be the diagram obtained by moving column $m$ of $r_{m,n}(D)$ to column $1$ and shift all other column rightward by $1$. By Theorem~\ref{T: Permuting columns}, $\chi_{r_{m,n}(D)} = \chi_{D''}$. Notice that $r_{m-1,n}(D')$ is obtained from $D''$ via an orthodontic move, so $\chi_{D''} = x_1\cdots x_{n-r}\chi_{r_{m-1,n}(D')}$. We have \begin{align*} r_{m,n}(\chi_D) & = r_{m,n}(x_1 \cdots x_r \chi_{D'}) = x_1\cdots x_{n-r} r_{m-1,n}(\chi_{D'})\\ & = x_1\cdots x_{n-r} \chi_{r_{m-1,n}(D')} = \chi_{r_{m,n}(D)}. \end{align*} \item Suppose $D'$ is obtained from $D$ by swapping row $r$ and row $r+1$. Then~$r_{m,n}(D')$ is obtained from $r_{m,n}(D)$ via an orthodontic move that swaps row $n-r$ and row $n-r+1$. By Lemma~\ref{L: commute operators}, we have \begin{align*} r_{m,n}(\chi_D) = r_{m,n}(\pi_r(\chi_{D'})) = \pi_{n-r}(r_{m,n}(\chi_{D'})) = \chi_{r_{m,n}(D)}. \qquad \qedhere \end{align*} \end{itemize} \end{proof} \begin{rem} Fix an $n \in \mathbb{Z}_{>0}$. The symmetrization operator $\pi_{w_0}$ can be defined as $$ \pi_{w_0} := (\pi_1 \cdots \pi_{n-1}) (\pi_1 \cdots \pi_{n-2}) \cdots (\pi_1 \pi_2) (\pi_1). $$ It turns a polynomial in $x_1, \dots, x_n$ into a polynomial symmetric in these variables. Magyar~\cite{Mag} showed that for any diagram $D$ living in the first $n$ rows and $m$ columns, $$ \pi_{w_0}(\chi_{r_{m,n}(D)}) = r_{m,n}(\pi_{w_0}(\chi_D)). $$ Our statement in Proposition~\ref{P: chi rotate} can be viewed as the ``non-symmetric refinement'' of Magyar's result. \end{rem} \begin{coro} \label{C: top Las char} Let $\alpha$ be a snowy weak composition. We have $\topLas_\alpha = \chi_{\snow(D(\alpha))}$. \end{coro} \begin{proof} Take $n$ with $\supp(\alpha) \subseteq [n]$ and $m$ with $\max(\alpha) \leq m$. Let $w = \std_{m,n}(\alpha)$. By Lemma~\ref{L: compare diagram}, $RD(w) = r_{m,n}(\snow(D(\alpha))$. By Theorem~\ref{T: PA and OM} and Proposition~\ref{P: chi rotate}, $r_{m,n}(\chi_{RD(w)}) = \chi_{\snow(D(\alpha))}$. On the other hand, by Theorem~\ref{T: Schuberts are chars}, $\chi_{RD(w)} = \fS_w$. Thus, \begin{align*} \topLas_\alpha = r_{m,n}(\fS_w) = r_{m,n}(\chi_{RD(w)}) = \chi_{\snow(D(\alpha))}. \qquad \qedhere \end{align*} \end{proof} In~\cite{HMSS}, the authors focused on Grothendieck polynomials indexed by vexillary (i.e. $2143$-avoiding) permutations. For $w$ vexillary, they constructed a diagram $D^{top}(w)$. Then they showed the polynomials $\topGro_w$ and $\chi_{D^{top}(w)}$ have the same support in Theorem 1.2. Finally, they conjectured $\topGro_w$ is a scalar multiple of $\chi_{D^{top}(w)}$ in Conjecture 1.5. We can resolve this conjecture using Corollary~\ref{C: top Las char}. \begin{coro} \label{C: HMSS} Conjecture 1.5 of~\cite{HMSS} is correct. \end{coro} \begin{proof} By Pechenik and Scrimshaw~\cite{PS}, in this case, $\fG_w$ is a Lascoux polynomial. Then by~\cite[Theorem 1.2]{PY2}, its top degree component $\topGro_w$ is a scalar multiple of $\topLas_\alpha$ for some snowy~$\alpha$. By Corollary~\ref{C: top Las char}, $\topGro_w = \chi_{\snow(D(\alpha))}$. By~\cite[Theorem 1.2]{HMSS}, $\supp(\chi_{D^{top}(w)}) = \supp(\topGro_w) = \supp(\chi_{\snow(D(\alpha))})$. Finally,~\cite[Theorem 1.4]{HMSS} says that if the dual characters of the two diagrams have the same support, then the two dual characters must agree. We have $\chi_{D^{top}(w)} = \chi_{\snow(D(\alpha))}$, so $\topGro_w$ is a multiple of~$\chi_{D^{top}(w)}$. \end{proof} We conclude this section by deriving a few properties related to the support of $\topLas_\alpha$, showing that its support has similar properties as the Schubert polynomials. Adve, Robichaux, and Yong~\cite{ARY} characterized the support of Schubert polynomials using certain fillings: \begin{defn}\cite{ARY} Let $D$ be a diagram and $\alpha$ be a weak composition. Then $\PT(D,\alpha)$ is the set of fillings of $D$ satisfying all of the following: \begin{itemize} \item For each $k$, the number of cells in $D$ filled by $k$ is $\alpha_k$. \item In each column, numbers are increasing from top to bottom. \item Any entry in row $i$ is at most $i$. \end{itemize} \end{defn} \begin{exa}\cite[Example 1.4]{ARY} \label{E: PT Rothe} Consider $D = RD((3,1,5,2,4))$. We enumerate the six elements in $\bigcup_\alpha \PT(D, \alpha)$: $$ \begin{ytableau} \none[1] & 1 & 1 \cr \none[2] \cr \none[3] & \none& 2 & \none & 1 \end{ytableau} \quad\quad\quad\quad \begin{ytableau} \none[1] & 1 & 1 \cr \none[2] \cr \none[3] & \none& 2 & \none & 2 \end{ytableau} \quad\quad\quad\quad \begin{ytableau} \none[1] & 1 & 1 \cr \none[2] \cr \none[3] & \none& 2 & \none & 3\cr \none \end{ytableau} $$ $$ \begin{ytableau} \none[1] & 1 & 1 \cr \none[2] \cr \none[3] & \none& 3 & \none & 1 \end{ytableau} \quad\quad\quad\quad \begin{ytableau} \none[1] & 1 & 1 \cr \none[2] \cr \none[3] & \none& 3 & \none & 2 \end{ytableau} \quad\quad\quad\quad \begin{ytableau} \none[1] & 1 & 1 \cr \none[2] \cr \none[3] & \none& 3 & \none & 3 \end{ytableau} $$ \end{exa} \begin{theo}\cite[Theorem 1.3]{ARY} For a permutation $w$, the support of $\fS_w$ is the set of~$\alpha$ such that $\PT(RD(w), \alpha) \neq \emptyset$. \end{theo} For instance, Example~\ref{E: PT Rothe} says that $\supp(\fS_{(3,1,5,2,4)})$ consists of $$(3,1), (2,2), (2,1,1), (3,0,1) \textrm{ and }(2,0,2).$$ Fink, M{\'e}sz{\'a}ros, and St. Dizier~\cite{FMS} extended this characterization to $\chi_D$. \begin{theo}\cite[Theorem 7]{FMS} For a diagram $D$, $\chi_D$ has SNP and its support $\chi_D$ is the set of $\alpha$ such that $\PT(D, \alpha) \neq \emptyset$. \end{theo} Consequently, the support of a top Lascoux polynomial can be characterized in the same manner. \begin{coro} \label{C: support of topL} For a snowy weak composition $\alpha$, $\topLas_\alpha$ has saturated Newton polytope. View $\snow(D(\alpha))$ as a diagram with labels erased. The support of $\topLas_\alpha$ is the set of $\gamma$ such that $\PT(\snow(D(\alpha)), \gamma) \neq \emptyset$. \end{coro} \begin{exa} \label{E: PT snow} Consider the snowy weak composition $(0,4,2)$. We enumerate the six elements in $\bigcup_\alpha \PT(\snow(D(\alpha)), \alpha)$: $$ \begin{ytableau} \none[1] & \none & 1 & \none & 1 \cr \none[2] & 1 & 2 & 1 & 2\cr \none[3] & 2 & 3 \end{ytableau} \quad\quad\quad\quad \begin{ytableau} \none[1] & \none & 1 & \none & 1 \cr \none[2] & 1 & 2 & 1 & 2\cr \none[3] & 3 & 3 \end{ytableau} \quad\quad\quad\quad \begin{ytableau} \none[1] & \none & 1 & \none & 1 \cr \none[2] & 2 & 2 & 1 & 2\cr \none[3] & 3 & 3 \end{ytableau} $$ $$ \begin{ytableau} \none[1] & \none & 1 & \none & 1 \cr \none[2] & 1 & 2 & 2 & 2\cr \none[3] & 2 & 3 \end{ytableau} \quad\quad\quad\quad \begin{ytableau} \none[1] & \none & 1 & \none & 1 \cr \none[2] & 1 & 2 & 2 & 2\cr \none[3] & 3 & 3 \end{ytableau} \quad\quad\quad\quad \begin{ytableau} \none[1] & \none & 1 & \none & 1 \cr \none[2] & 2 & 2 & 2 & 2\cr \none[3] & 3 & 3 \end{ytableau} $$ By Proposition~\ref{C: support of topL}, the support of $\topLas_{(0,4,2)}$ consists of $(4,3,1), (4,2,2), (3,3,2)$, $(3,4,1)$ and $(2,4,2)$. \end{exa} \longthanks{We are grateful to Elena Hafner, Jianping Pan, Brendon Rhoades, Travis Scrimshaw and Avery St. Dizier for carefully reading an earlier version of this paper and giving many useful comments. We also thank Sara Billey, Zachary Hamaker, Daoji Huang, Oliver Pechenik, Mark Shimozono, and Alexander Yong for helpful conversations. We thank the two referees for providing many valuable suggestions. A special thanks to Linus Setiabrata who pointed out that our results can imply Conjecture 1.5 of~\cite{HMSS} (see Corollary~\ref{C: HMSS}).} \bibliographystyle{mersenne-plain-nobysame} \bibliography{ALCO_Yu_1195} \end{document}