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\title[Krystals for rectangular shapes]{K-theoretic crystals for set-valued tableaux of rectangular shapes}  

\author[Oliver Pechenik and Travis Scrimshaw]{Oliver Pechenik\thanks{\href{mailto:pechenik@umich.edu}{pechenik@umich.edu}. Partially supported by the NSF Mathematical Sciences Postdoctoral Research Fellowship \#1703696.}\addressmark{1}
\and
Travis Scrimshaw\thanks{\href{mailto:tcscrims@gmail.com}{tcscrims@gmail.com}. Partially supported by the Australian Research Council DP170102648.}\addressmark{2}}
\address{\addressmark{1}Department of Mathematics, University of Michigan, Ann Arbor, MI, USA \\
\addressmark{2}School of Mathematics and Physics, The University of Queensland, St.\ Lucia, Australia
}

\keywords{Lascoux polynomial, crystal, Kohnert diagram, skyline tableau}
%\subjclass[2010]{05E05, 05A19, 14M15, 17B37}


\abstract{
With C.~Monical (2018), we introduced a notion of K-crystals and conjectured they exist for all rectangular shapes $\lambda$.
Here, we establish this conjecture, yielding the first combinatorial formula (as the sum over flagged set-valued tableaux) for the Lascoux polynomials $L_{w\lambda}$. We then prove corresponding cases of conjectures of Ross--Yong (2015) and Monical (2016).
}

\begin{document}

\maketitle

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction}
\label{sec:introduction}

In classical Schubert calculus, we can study the cohomology ring of the Grassmannian $\Gr(k, n)$, the set of $k$-dimensional subspaces of $\CC^n$, using the basis of Poincar\'e duals of the Schubert varieties $X_{\lambda}$ that decompose $\Gr(k, n)$. The cohomology classes $[X_{\lambda}]$ can be represented by Schur polynomials $s_{\lambda}$, where the partition $\lambda$ sits inside a $k \times (n-k)$ rectangle. A more modern approach is to use connective K-theory, where the Schubert class $[X_{\lambda}]$ is given as the push-forward of the class for any Bott--Samelson resolution of $X_{\lambda}$. Here, representatives are symmetric (or stable) $\beta$-Grothendieck polynomials~\cite{FK94}.

We can describe $s_{\lambda}$ combinatorially as a generating function for semistandard (Young) tableaux of shape $\lambda$ and representation-theoretically as the character of the highest weight representation $V(\lambda)$ of the Lie algebra $\fsl_n$ of traceless $n \times n$ matrices. We can also compute $s_{\lambda}$ by applying a product of Demazure operators $\pi_{w_0}$ for the reverse permutation $w_0$ to the monomial $\xx^{\lambda} := x_1^{\lambda_1} \dotsm x_n^{\lambda_n}$. We can refine $s_{\lambda}$ to the key polynomials $\kappa_{w\lambda} := \pi_w \xx^{\lambda}$ for any permutation $w$, which are characters of Demazure modules $V_w(\lambda)$. 

Combinatorially, A.~Buch~\cite{Buch02} showed the symmetric Grothendieck polynomial $\G_{\lambda}$ is the generating function for semistandard set-valued tableaux of shape $\lambda$. A.~Lascoux~\cite{Lascoux01} deformed the Demazure operators to Demazure--Lascoux operators $\varpi_w$, so that $\G_{\lambda} = \varpi_{w_0} \xx^{\lambda}$. The analogous deformation of key polynomials, the so-called Lascoux polynomials $L_{w\lambda} = \varpi_w \xx^{\lambda}$, remain mysterious as currently there is no known geometric, representation-theoretic, or combinatorial interpretation, despite recent work~\cite{RY15,Kirillov:notes,Monical16,MPS18II}. Yet, combinatorial formulas have been conjectured~\cite{Monical16,RY15,Kirillov:notes}.

One way to connect the combinatorics and representation theory associated to key polynomials is using M.~Kashiwara's crystal bases (see, \textit{e.g.},~\cite{BS17,K93}). Indeed, Kashiwara showed that the Demazure module $V_w(\lambda)$ has a crystal basis and can be described as a subcrystal $B_w(\lambda)$(called a Demazure crystal) of the highest weight crystal $B(\lambda)$~\cite{K93}. For the quantum group $U_q(\fsl_n)$, the crystal $B(\lambda)$ may be realized as the set of semistandard tableaux of shape $\lambda$ and the subcrystal $B_w(\lambda)$ is characterized by key tableaux~\cite{LS90}.

In our previous paper with C.~Monical~\cite{MPS18}, we initiated an analogous approach to Demazure crystals for Lascoux polynomials. We first gave a $U_q(\fsl_n)$-crystal structure to semistandard set-valued tableaux. Then we proposed an enriched crystal structure with the property that the Lascoux polynomials appear as the characters of our K-theoretic analogs of Demazure subcrystals. We coined this enriched structure a K-crystal. We established the existence of K-crystals for single rows and columns, but we discovered that no such structure exists for general shapes.  Nonetheless, we conjectured~\cite[Conjecture~7.12]{MPS18} that K-crystals exist for all rectangular shapes. Our first main result is a proof of this conjecture. Our proof gives rise to a combinatorial formula for the class of Lascoux polynomials indexed by a weight in the Weyl group orbit of a multiple of a fundamental weight (\textit{i.e.}, a rectangular shape partition).  We then use this formula to establish the corresponding cases of Ross--Yong--Kirillov and Monical conjectures.

Let us remark on why our proposed K-crystal structure exists only for rectangular shapes. With C.~Monical~\cite{MPS18}, we proposed a slightly weaker structure for general $\lambda$ that depends on a choice of a reduced expression for $w_0$. The key distinction appears to be that in the rectangular case the minimal-length coset representatives that index Lascoux polynomials are all fully-commutative (\textit{i.e.}, all reduced words differ only by commutations). However, for more general shapes, such as $\lambda = (2,1)$ in~\cite[Figure~6,7]{MPS18}, one needs to apply the braid relations $s_i s_{i+1} s_i = s_{i+1} s_i s_{i+1}$ to get all possible reduced expressions. Subsequently, we believe that, in general, K-crystal structures depend on choosing a commutation class of the reduced words for the appropriate parabolic $w_0$ (see also~\cite[\S7.3]{MPS18}). This fact seems related to a similar dependence for Schubert classes in cohomology theories more general than connective K-theory (see, \textit{e.g.},~\cite{LZ17}). In the rectangular case, we have a flagging condition to characterize the tableaux in the K-Demazure crystal, and we expect a key tableau condition to work for general shapes.

This extended abstract of~\cite{PS19} (where we refer the reader to for more details) is organized as follows.
In \cref{sec:background}, we recall the necessary background.
In \cref{sec:K_rectangles}, we construct a K-crystal structure on set-valued tableaux of rectangular shapes.
In \cref{sec:Kohnert} (resp.\ \cref{sec:skyline_tableaux}), we prove the conjectural combinatorial interpretation of Lascoux polynomials for rectangular shapes due to Ross--Yong--Kirillov (resp.\ Monical).
In \cref{sec:Key_tableaux}, we describe our conjecture for key tableaux of set-valued tableaux and their relationship with Lascoux polynomials.







%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Background}
\label{sec:background}

Let $\sym_n$ be the symmetric group with simple transpositions $\{ s_i \mid 1 \leq i < n \}$ and longest element $w_0 = [n, \dotsc, 2, 1]$.
Let $v \leq w$ be the (strong) Bruhat order, which means there is a reduced word for $v$ that is a subword of a reduced word for $w$.
Let $\xx = (x_1, x_2, x_3, \ldots)$ be a countable vector of indeterminants.
For a tuple $\alpha = (\alpha_1, \alpha_2, \ldots)$, define $\xx^{\alpha} = x_1^{\alpha_1} x_2^{\alpha_2} \cdots$.

We use the English convention for both partitions and tableaux.
Consider a partition $\lambda$ as a word of length $n$ by appending $0$'s as necessary, whence it carries a natural $\sym_n$-action.
Let $\Stab_n(\lambda) = \{w \in \sym_n \mid w \lambda = \lambda\}$ denote the stabilizer of $\lambda$.
Let $\sym_n^{\lambda}$ denote the set of minimal length coset representatives $\{ \lfloor w \rfloor \mid w \in \sym_n \}$ of $\sym_n / \Stab_n(\lambda)$.

A \defn{(semistandard) set-valued tableau of shape $\lambda$} is a filling $T$ of the boxes of $\lambda$ by finite nonempty sets of positive integers so that for every set $A$ to the left of a set $B$ in the same row, we have $\max A \leq \min B$, and for $C$ below $A$ in the same column, we have $\max A < \min C$.
We say an integer $a \in T$ if there exists a box of $T$ containing a set $A$ with $a \in A$.
A set-valued tableau is a \defn{semistandard Young tableau} if all sets have size $1$.
Let $\svt^n(\lambda)$ denote the set of all set-valued tableaux of shape $\lambda$ with entries at most $n$.

We recall the crystal structure on $\svt^n(\lambda)$ from~\cite{MPS18}. We refer to~\cite{BS17} for more details on crystals. First, we recall the \defn{crystal operators} $e_i, f_i \colon \svt^n(\lambda) \to \svt^n(\lambda) \sqcup \{0\}$, where $i \in I := \{1, \dotsc, n-1\}$.
Begin by constructing a sequence by writing $\bplus$ (resp.~$\bminus$) above each column of $T$ containing $i$ but not $i+1$ (resp.\ $i+1$ but not $i$) and canceling ordered pairs $\bminus \bplus$.
If every $\bplus$ (resp.\ $\bminus$) thereby cancels, then $f_i T = 0$ (resp.\ $e_i T = 0$). Otherwise,
\begin{itemize}
\item if there exists a box $\bbb'$ immediately to the right (resp.\ left) of $\bbb$ that contains an $i$ (resp.\ $i+1$), then remove the $i$ (resp.\ $i+1$) from $\bbb'$ and add an $i+1$ (resp.\ $i$) to $\bbb$;
\item otherwise replace the $i$ in $\bbb$ with an $i+1$ (resp.\ $i+1$ in $\bbb$ with an $i$);
\end{itemize}
where $\bbb$ is the box of the rightmost uncanceled $\bplus$ (resp.\ leftmost uncanceled $\bminus$), and the result is $f_i T$ (resp.\ $e_i T$).
See \cref{fig:Krystal_ex_22} for an example.

Identifying $\ZZ^n$ with the multiplicative group generated by $(x_1, \dotsc, x_n)$, we define the weight function $\wt \colon \svt^n(\lambda) \to \ZZ^n$ by $\wt(T) = \prod_{i=1}^n x_i^{c_i}$, where $c_i$ is the number of $A \in T$ such that $i \in A$.
Define $|\wt(T)| = \sum_{i=1}^n c_i$.


\begin{thm}[{\cite[Theorem~3.9]{MPS18}}]
$\svt^n(\lambda)$ is isomorphic to a direct sum of highest weight crystals.
\end{thm}


For $1 \leq i < n$, the \defn{Demazure--Lascoux operator} $\varpi_i$  acts on $\ZZ[\beta][x_1, \dotsc, x_n]$ by
\[
\varpi_i f
= \pi_i\bigl( (1  + \beta x_{i+1}) \cdot f \bigr) = \pi_i f + \beta \cdot \pi_i (x_{i+1} \cdot f),  \text{ where } \pi_i f = \dfrac{x_i \cdot f - x_{i+1} \cdot s_i f}{x_i - x_{i+1}},
\]
is the \defn{Demazure operator}.
The Demazure--Lascoux operators (and Demazure operators) satisfy the braid relations. Thus for any $w \in \sym_n$, one may unambiguously define $\varpi_w := \varpi_{i_1} \dotsm \varpi_{i_{\ell}}$, where $s_{i_1} \dotsm s_{i_{\ell}}$ is some reduced expression for $w$ (and similarly for $\pi_w$).


Since $\varpi_w$ does not depend on the choice of reduced expression, we can define the \defn{Lascoux polynomials}~\cite{Lascoux01} for any $a \in \ZZ_{\geq 0}^n$ as
\[
L_a(\xx; \beta) := \varpi_w \xx^{\lambda},
\]
where $\lambda$ is the sorting of $a$ to a partition and $w \in \sym_n^{\lambda}$ is the unique element such that $a = w \lambda$.
The \defn{symmetric Grothendieck polynomial} can be defined as the $n$ variable truncation of $L_{w_0\lambda}(\xx; \beta)$ and is known~\cite[Theorem~3.1]{Buch02} to be given combinatorially by 
\begin{equation}
\label{eq:Grothendieck_defn}
L_{w_0\lambda}(\xx; \beta) = \sum_{T \in \svt^n(\lambda)} \wt_{\beta}(T), \quad \text{ where } \wt_{\beta}(T) := \beta^{|\wt(T)| - |\lambda|} \wt(T) \text{ is the \defn{$\beta$-weight}}.
\end{equation}
We now recall two conjectural combinatorial descriptions of Lascoux polynomials.

The first conjectural combinatorial rule was introduced in~\cite{RY15}. To state it, we begin by recalling the notion of a \defn{K-Kohnert diagram} to be a subset $D$ of $\ZZ^n_{>0}$, which we realize as boxes, and a subset $M \subseteq D$ of boxes that are marked. Now start with some $a = (a_1, \dotsc, a_n) \in \ZZ^n_{\geq 0}$ and draw the initial K-Kohnert diagram as a \defn{skyline diagram} by putting a box at each position $\{(i,y) \mid i \in [n], 1 \leq y \leq a_i\}$ (in Cartesian coordinates), marking no boxes. Then we successively apply any sequence of the following operations.
\begin{description}
\item[\defn{Kohnert move}:] Move any unmarked box at the top of a column into the rightmost open position to its left and in the same row so that it does not pass over a marked box.\footnote{In~\cite{RY15}, it is misstated that a Kohnert move could move the unmarked box over a marked box.}
\item[\defn{K-Kohnert move}:] Perform a Kohnert move but leave a marked box behind.
\end{description}
Let $\mcD_a$ denote the resulting set of K-Kohnert diagrams obtainable from the original skyline diagram for $a$.
Define the $\beta$-weight of $D \in \mcD_a$ by $\wt_{\beta}(D) := \beta^e \prod_{i=1}^n x_i^{c_i}$, where $e$ (resp.~$c_i$) is the number of marked boxes (resp.\ boxes in column $i$) in $D$.

\begin{conj}[{\cite[Conjecture~1.4]{RY15},\cite[Fn.~14]{Kirillov:notes}}]
\label{conj:K_skyline}
We have
$
L_a(\xx; \beta) = \sum_{D \in \mcD_a}  \wt_{\beta}(D).
$
\end{conj}

The second conjectural combinatorial rule is from~\cite{Monical16}. We fill a skyline diagram with finite nonempty sets of positive integers that satisfy the following conditions. Call the largest entry in a box the \defn{anchor} and the other entries \defn{free}.
(S.1): Entries do not repeat in a row.
(S.2): If $B$ is below $A$, then $\min B \geq \max A$ (\textit{i.e.}, the columns are weakly increasing top-to-bottom in the set-valued sense).
(S.3): For every triple of boxes of the form
\[
\begin{array}{c@{\hspace{40pt}}c}
\begin{array}{|c|c|c|}
\cline{1-1} \cline{3-3}
C & \cdots & A
\\ \cline{1-1} \cline{3-3}
\multicolumn{1}{c}{} & & B
\\ \cline{3-3}
\end{array}
&
\begin{array}{|c|c|c|}
\cline{1-1} \cline{3-3}
A & \cdots & C
\\ \cline{1-1} \cline{3-3}
B & \multicolumn{1}{c}{}
\\ \cline{1-1}
\end{array}
\\[10pt]
\text{right column weakly taller}
&
\text{left column strictly taller}
\end{array}
\]
the anchors $a,b,c$ of $A,B,C$, respectively, must satisfy either $c < a$ or $b < c$.
(S.4): Every free entry is in the leftmost cell of its row such that~(S.2) is not violated.
(S.5): Anchors in the bottom row equal their column index.
We call such a tableau a \defn{(semistandard) set-valued skyline tableau}, and let $\skyline_a$ denote those of shape $a$.
We define the weight, excess, and $\beta$-weight for a set-valued skyline tableau the same way as for a set-valued tableau.

Let $\overline{\varpi}_i = \varpi_i - 1$.
Define the \defn{Lascoux atom} to be
$
\overline{L}_{w\lambda}(\xx; \beta) := \overline{\varpi}_w x^{\lambda}.
$

\begin{conj}[{\cite[Conjecture~5.2]{Monical16}}]
\label{conj:skyline_tableaux}
We have
$
\overline{L}_{w\lambda} = \sum_{S \in \skyline_{w\lambda}} \wt_{\beta}(S).
$
\end{conj}

Note \cref{conj:skyline_tableaux} is equivalent to~\cite[Conjecture~5.3]{Monical16} by~\cite{MPS18II}.
Also from~\cite[Theorem~5.1]{Monical16},
\begin{equation}
\label{eq:lascoux_to_atoms}
L_{w\lambda}(\xx; \beta) = \sum_{v \leq w} \overline{L}_{v \lambda}(\xx; \beta),
\end{equation}
where the inequality is (strong) Bruhat order on permutations.








%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{K-crystals for rectangular shapes}
\label{sec:K_rectangles}


We aim to prove the proposed K-theory analog of crystals from~\cite{MPS18} exists on $\svt^n(\lambda)$ when $\lambda$ is a rectangle.
Recall that a $U_q(\fsl_n)$-crystal $B$ is called a \defn{K-crystal} if it is enhanced with \defn{K-crystal operators}, $e_i^K, f_i^K \colon B \to B \sqcup \{0\}$ that satisfy the following properties:
\begin{enumerate}
\item[(K.1)] \label{Krystal:connected} The set $B$ is generated by a unique element $u \in B$ that satisfies $e_i u = 0$ and $e_i^K u = 0$ for all $i \in I$. The element $u$ is called the \defn{minimal highest weight element}.
\item[(K.2)] \label{Krystal:demazure} The \defn{K-Demazure crystal}
  $
  B_w := \{ b \in B \mid (e_{i_{\ell}}^K)^{\max} e_{i_{\ell}}^{\max} \cdots (e_{i_1}^K)^{\max} e_{i_1}^{\max} b = u \}
  $
  does not depend on the choice of reduced expression $w = s_{i_1} \cdots s_{i_{\ell}}$. Moreover, $B_{w_0} = B$.
\item[(K.3)] \label{Krystal:character} Let $\lambda = \wt(u)$. The \defn{$\beta$-character}
$
\ch_{\beta}(B_w) := \sum_{b \in B_w} \beta^{|\wt(b)| - |\lambda|} \xx^{\wt(b)} = L_{w\lambda}(\xx; \beta).
$
\end{enumerate}

Our construction of the K-crystal operators are based off the heuristics given in~\cite{MPS18}, which come from the following K-theory analog of the decomposition of a crystal into $i$-strings (\textit{i.e.}, restricting to the action of $e_i$ and $f_i$ for a fixed $i \in I$) based on the definition of the Demazure--Lusztig operators. Indeed, by considering only the action of a fixed $i \in I$, the K-crystal is expected to decompose into (maximal) subcrystals of the form
\[
\begin{tikzpicture}[xscale=2,yscale=-1.0]
\node (top) at (0,0)  {$b$};
\node (f) at (1,0)  {$\bullet$};
\node (ff) at (2,0)  {$\bullet$};
\node (dots) at (3,0) {$\cdots$};
\node (fend) at (4,0)  {$\bullet$};
\node (ffend) at (5,0)  {$\bullet$};
\node (K) at (0,1)  {$\bullet$};
\node (fK) at (1,1)  {$\bullet$};
\node (ffK) at (2,1)  {$\bullet$};
\node (dotsK) at (3,1) {$\cdots$};
\node (fendK) at (4,1)  {$\bullet$};
\draw[->,blue] (top) -- node[midway,above] {$i$} (f);
\draw[->,blue] (f) -- node[midway,above] {$i$} (ff);
\draw[->,blue] (ff) -- node[midway,above] {$i$} (dots);
\draw[->,blue] (dots) -- node[midway,above] {$i$} (fend);
\draw[->,blue] (fend) -- node[midway,above] {$i$} (ffend);
\draw[->,blue] (K) -- node[midway,above] {$i$} (fK);
\draw[->,blue] (fK) -- node[midway,above] {$i$} (ffK);
\draw[->,blue] (ffK) -- node[midway,above] {$i$} (dotsK);
\draw[->,blue] (dotsK) -- node[midway,above] {$i$} (fendK);
\draw[->,blue,dashed] (top) -- node[midway,left] {$i$} (K);
\end{tikzpicture}
\]
where the solid (resp.\ dashed) arrow represents the $f_i$ (resp.~$f_i^K$) action.
Such a subcrystal was coined an \defn{$i$-K-string} in~\cite{MPS18}. We say an $i$-K-string has \defn{length}
$
\ell := \max \{k \mid f_i^k b \neq 0\}.
$
Note that $f_i^{\ell-1} f_i^K b \neq 0$ and $f_i^{\ell} f_i^K b = 0$. It is easy to see that for $b$ such that $e_i b = 0$, the $\beta$-character of the $i$-K-string starting at $b$ equals $\varpi_i \wt(b)$.

\begin{lemma}
\label{lemma:coset_repr}
Let $\lambda$ be an $r \times s$ rectangle.
For any $w \in \sym_n^{\lambda}$, there exists a reduced word of $w$ equal to
$
(s_{i_k} \dotsm s_{r-k+1} s_{r-k}) \dotsm (s_{i_1} \dotsm s_{r} s_{r-1}) (s_{i_0} \dotsm s_{r+1} s_r)
$
for some $k < r$ and $i_k < \cdots < i_0 \leq n$.
\end{lemma}

\begin{dfn}
\label{def:K_crystal_ops}
Let $T \in \svt^n(\lambda)$ and fix some $i \in I$.
\begin{description}
\item[\defn{$f_i^K$}:] If $i \notin T$ or $e_i T \neq 0$, then $f_i^K T = 0$. Otherwise, let $\bbb$ be the rightmost box that contains an $i$ corresponding to an unpaired $\bplus$.
If $i$ and $i+1$ are both in a box weakly to the right of $\bbb$, then $f_i^K T = 0$.
Otherwise, define $f_i^K T$ by adding an $i+1$ to $\bbb$.
\item[\defn{$e_i^K$}:] If there does not exist a box with both an $i$ and $i+1$ or $e_i T \neq 0$, then $e_i^K T = 0$. Otherwise, let $\bbb$ be the rightmost box that has both an $i$ and $i+1$. If there exists an $i$ corresponding to an unpaired $\bplus$ strictly to the right of $\bbb$, then $e_i^K T = 0$. Otherwise, define $e_i^K T$ by removing the $i+1$.
\end{description}
\end{dfn}

\begin{figure}[t]
\[
\ytableausetup{boxsize=2.0em}
\begin{tikzpicture}[>=latex,scale=2,every node/.style={scale=0.7},baseline=-50]
\node (m) at (-1,-1) {$\ytableaushort{11,22}$};
\node (f2m) at (0,-1) {$\ytableaushort{11,23}$};
\node (f22m) at (1,-2) {$\ytableaushort{11,33}$};
\node (f12m) at (-1,-2) {$\ytableaushort{12,23}$};
\node (f122m) at (0,-3) {$\ytableaushort{12,33}$};
\node (f1122m) at (-1,-3) {$\ytableaushort{22,33}$};
\draw[->,red] (m) -- node[midway,above] {\small $2$} (f2m);
\draw[->,red] (f2m) -- node[midway,above right] {\small $2$} (f22m);
\draw[->,blue] (f2m) -- node[midway, above left] {\small $1$} (f12m);
\draw[->,red] (f12m) -- node[midway,below left] {\small $2$} (f122m);
\draw[->,blue] (f22m) -- node[midway, below right] {\small $1$} (f122m);
\draw[->,blue] (f122m) -- node[midway, below] {\small $1$} (f1122m);
%
\node (t) at (-2.5,-1) {$\ytableaushort{11,2{2,\!3}}$};
\node (f2t) at (-3.5,-1) {$\ytableaushort{11,{2,\!3}3}$};
\node (f12t) at (-2.5,-2) {$\ytableaushort{12,{2,\!3}3}$};
\draw[->,red] (t) -- node[midway,above] {\small $2$} (f2t);
\draw[->,blue] (f2t) -- node[midway, below left] {\small $1$} (f12t);
%
\node (b) at (2.5,-1) {$\ytableaushort{1{1,\!2},23}$};
\node (f2b) at (2.5,-2) {$\ytableaushort{1{1,\!2},33}$};
\node (f12b) at (2.5,-3) {$\ytableaushort{{1,\!2}2,33}$};
\draw[->,red] (b) -- node[midway,right] {\small $2$} (f2b);
\draw[->,blue] (f2b) -- node[midway, right] {\small $1$} (f12b);
%
\node (all) at (4,-1) {$\ytableaushort{1{1,\!2},{2,\!3}3}$};
% K-arrows
\draw[->,dashed,red] (m) -- node[midway,above] {\small $2$} (t);
\draw[->,dashed,blue] (f2m) -- node[midway,above] {\small $1$} (b);
\draw[->,dashed,blue] (f22m) -- node[midway,above] {\small $1$} (f2b);
\draw[->,dashed,red] (f12m) -- node[midway,above] {\small $2$} (f12t);
\draw[->,dashed,red] (b) -- node[midway,above] {\small $2$} (all);
\draw[->,dashed,blue] (f2t) .. controls (-2.2,-0.2) and (2.2,-0.2) .. node[midway,below] {\small $1$} (all);
\end{tikzpicture}
\]
\ytableausetup{boxsize=0.8em}
\caption{The K-crystal for $\svt^3\left( \raisebox{3pt}{$\ydiagram{2,2}$} \right)$ with the K-crystal operators as dashed lines.}
\label{fig:Krystal_ex_22}
\end{figure}

For an example, see \cref{fig:Krystal_ex_22}; additional examples may be found in~\cite{MPS18}. Note that it is clear that if $f_i^K T \neq 0$ (resp.\ $e_i^K T \neq 0$), then $f_i^K T \in \svt^n(\lambda)$ (resp.\ $e_i^K T \in \svt^n(\lambda)$).

\begin{lemma}
Let $\lambda$ be an $r \times s$ rectangle.
Let $T, T' \in \svt^n(\lambda)$.
We have $e_i^K T' = T$ if and only if $T' = f_i^K T$.
Moreover, for any $i \in I$, we have that $\svt^n(\lambda)$ is a union of $i$-K-strings.
\end{lemma}

Consider some $w \in \sym_n^{\lambda}$, and let $i_k < \cdots < i_0$ be from the reduced expression of $w$ given by \cref{lemma:coset_repr}. For all $k < j < r$, define $i_j = r - j$.
Define $F(\lambda; w)$ to be the subset of $\svt^n(\lambda)$ such that row $r-j$ has all entries at most $i_j + 1$. We call such a set-valued tableau a \defn{flagged set-valued tableau}. This flagging characterizes the K-Demazure crystal.

\begin{lemma}
\label{lemma:flag_condition}
Let $\lambda$ be an $r \times s$ rectangle.
For $w \in \sym_n$, then
$
\svt^n_w(\lambda) = \svt^n_{\lfloor w \rfloor}(\lambda) = F(\lambda; \lfloor w \rfloor).
$
\end{lemma}


\begin{thm}
\label{thm:Krystal_rectangles}
Let $\lambda$ be an $r \times s$ rectangle.
Then $\svt^n(\lambda)$ is a K-crystal.
\end{thm}

We also have the following K-theoretic analog of~\cite[Proposition~3.3.4]{K93}. 

\begin{cor}
\label{cor:demazure_strings}
Let $\lambda$ be an $r \times s$ rectangle.
Consider an $i$-K-string $S$ of $\svt^n(\lambda)$, and let $b$ be the highest weight element of $S$. Then, the set $\svt^n_w(\lambda) \cap S$ is either empty, $S$, or $\{b\}$.
\end{cor}

We also have the following interpretation of certain Lascoux polynomials as instances of \defn{($\beta$-)Grothendieck polynomials}, which recall from~\cite{Lascoux90,LS82,LS83,FK94} are defined by
\[
\G_{w_0 s_{i_1} \cdots s_{i_{\ell}}} := \partial^{\beta}_{i_1} \cdots \partial^{\beta}_{i_{\ell}} x_1^{n-1} \dotsm x_{n-1}^1 x_n^0,
\quad \text{where }
\partial_i^{\beta} f = \frac{(1 + \beta x_i) \cdot f - (1 + \beta x_{i+1}) \cdot s_i f}{x_i - x_{i+1}}.
\]

\begin{cor}
\label{cor:vexillary_Lascoux}
Let $\lambda$ be an $r \times s$ rectangle. Let
$
w = (s_k \dotsm s_2 s_1) \dotsm (s_{k+r-1} \dotsm s_{r+1} s_r) $
for some $k \geq 1$, and let
$
\widetilde{w} = s_{m-1} (s_{m-2} s_{m-1}) \dotsm (s_{r+1} \dotsm s_{m-1}) (s_r \dotsm s_{k-1}) \dotsm (s_1 \dotsm s_{k-1})
\in S_m
$
where $m = s + k + 1$.
Then, we have
$
L_{w\lambda}(\xx; \beta) = \G_{w_0\widetilde{w}^{-1}}(\xx; \beta).
$
\end{cor}

The permutations $w_0\widetilde{w}^{-1}$ appearing in \cref{cor:vexillary_Lascoux} are vexillary (\textit{i.e.}, $2143$-avoiding). Since the greatest term of $L_{w\lambda}(\xx; 0) $ in reverse lexicographic order is $\xx^{w \lambda}$ and the greatest term of $\G_{w_0\widetilde{w}^{-1}}(\xx; 0)$ in the same order is the Lehmer code of $w_0\widetilde{w}^{-1}$, we see $w \lambda$ is the Lehmer code of $w_0\widetilde{w}^{-1}$. Hence, $w_0\widetilde{w}^{-1}$ are Grassmannian, and so the Grothendieck polynomials from \cref{cor:vexillary_Lascoux} are actually \emph{symmetric} Grothendieck polynomials, but symmetric only in some initial segment of the variables $\xx$.

\begin{ex}
For $\lambda$ be a $2 \times 2$ rectangle,
$
L_{s_1 s_2 \lambda}(\xx; \beta) = \G_{w_0 (s_2 s_1) s_2 (s_4 s_3) s_4}(x_1, \dotsc, x_5; \beta),
$
and
$
L_{s_2 s_1 s_3 s_2 \lambda}(\xx; \beta) = \G_{w_0 (s_3 s_2 s_1) (s_3 s_2) (s_5 s_4 s_3) (s_5 s_4) s_5}(x_1, \dotsc, x_6; \beta).
$
\end{ex}

The Lascoux polynomials from \cref{cor:vexillary_Lascoux} are are not the only ones equal to a Grothendieck polynomial; \textit{e.g.},
\ytableausetup{boxsize=0.8em}
$
L_{s_2 \lambda}(\xx; \beta) = \G_{w_0 (s_2 s_4 s_3 s_4)}(x_1, \dotsc, x_5; \beta)
$
for $\lambda = \raisebox{3pt}{$\ydiagram{4,2}$}$.
Yet, this is the only such coincidence when $\lambda$ is a rectangle.
%
T.~Matsumura and S.~Su\-gi\-mo\-to have informed the authors~\cite[Theorem~3.3]{Matsumura18} can be extended to show every flagged Grothendieck polynomial is a Lascoux polynomial and will appear in their future work. 



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Bijection with K-Kohnert diagrams}
\label{sec:Kohnert}

Recall that there is a natural bijection between the set of semistandard Young tableaux of shape $1^r$ with entries at most $n$ and the collection of subsets of $\{1, \dotsc, n\}$ of size $r$.
For row $i$ (starting from the bottom row and going up) of a K-Kohnert diagram $D$, consider  the subset of $\{1, \dotsc, n\}$ given by the horizontal coordinates of the unmarked boxes. Construct column $i$ (from right to left) of a tableau $T$ by applying the natural bijection given above to this subset. Now, for every marked box in position $(x,i)$ of $D$, there is a rightmost unmarked box $(x',i)$ to the left of $(x,i)$. Insert $x$ into the cell of column $i$ containing $x'$.
In other words, we insert $x$ into the topmost possible cell of column $i$ such that the column is semistandard. Write $\phi(D)$ for the resulting tableau $T$.

It is straightforward to see that the map $\phi$ is invertible and $\beta$-weight preserving. We will show below that $\phi(D)$ is in fact always a semistandard set-valued tableau.

\begin{prop}
\label{prop:K_skyline_bijection}
Let $\lambda$ be an $r \times s$ rectangle.
For any $w \in \sym_n^{\lambda}$, $\phi$ restricts to a $\beta$-weight preserving bijection
$
\phi \colon \mcD_{w\lambda} \to \svt^n_w(\lambda).
$
\end{prop}

\begin{ex}
\label{ex:phi_bijection}
Consider $\lambda$ be a $2 \times 2$ square and $w = s_2$.
Under $\phi$ described above,
\[
\newcommand{\kohnert}[1]{\begin{tikzpicture}[scale=0.5] #1 \draw[very thin] (0,0) grid (3,2); \end{tikzpicture}}
\ytableausetup{boxsize=1.5em}
\begin{array}{c@{\qquad}c@{\qquad}c@{\qquad}c@{\qquad}c}
\kohnert{ \fill[darkred!60] (0,0) rectangle (1,2); \fill[darkred!60] (2,0) rectangle (3,2); } &
\kohnert{ \fill[darkred!60] (0,0) rectangle (1,2); \fill[darkred!60] (2,0) rectangle (3,1); \fill[darkred!60] (1,1) rectangle (2,2); } &
\kohnert{ \fill[darkred!60] (0,0) rectangle (1,2); \fill[darkred!60] (1,1) rectangle (2,2); \fill[darkred!60] (2,0) rectangle (3,2); \fill[black] (2.5,1.5) circle (0.2); } &
\kohnert{ \fill[darkred!60] (0,0) rectangle (1,2); \fill[darkred!60] (1,0) rectangle (2,2); } &
\kohnert{ \fill[darkred!60] (0,0) rectangle (2,2); \fill[darkred!60] (2,0) rectangle (3,1); \fill[black] (2.5,0.5) circle (0.2); }
\\
\ytableaushort{11,33} &
\ytableaushort{11,23} &
\ytableaushort{11,{2,\!3}3} &
\ytableaushort{11,22} &
\ytableaushort{11,2{2,\!3}}
\end{array}
\]
where we have shaded in the selected boxes and put a $\bullet$ in the marked boxes. 

We continue to $w' = s_1 s_2$ to obtain all of $\svt^3_{w'}(\lambda) =\svt^3(\lambda)$ under $\phi$: 
\[
\newcommand{\kohnert}[1]{\begin{tikzpicture}[scale=0.5] #1 \draw[very thin] (0,0) grid (3,2); \end{tikzpicture}}
\begin{array}{c@{\quad}c@{\quad\;}c@{\quad\;}c@{\quad\;}c@{\quad\;}c@{\quad\;}c@{\quad\;}c}
\kohnert{ \fill[darkred!60] (1,0) rectangle (3,2); }
&
\kohnert{ \fill[darkred!60] (2,0) rectangle (3,2); \fill[darkred!60] (1,0) rectangle (2,1); \fill[darkred!60] (0,1) rectangle (1,2); }
&
\kohnert{ \fill[darkred!60] (1,0) rectangle (3,2); \fill[darkred!60] (0,1) rectangle (1,2); \fill[black] (1.5,1.5) circle (0.2);}
&
\kohnert{ \fill[darkred!60] (0,1) rectangle (1,2); \fill[darkred!60] (1,0) rectangle (2,2); \fill[darkred!60] (2,0) rectangle (3,1); }
&
\kohnert{ \fill[darkred!60] (0,1) rectangle (3,2); \fill[darkred!60] (1,0) rectangle (3,1); \fill[black] (2.5,1.5) circle (0.2); }
&
\kohnert{ \fill[darkred!60] (0,0) rectangle (1,2); \fill[darkred!60] (2,0) rectangle (3,2); \fill[darkred!60] (1,0) rectangle (2,1); \fill[black] (1.5,0.5) circle (0.2); } &
\kohnert{ \fill[darkred!60] (0,0) rectangle (2,2); \fill[darkred!60] (2,0) rectangle (3,1); \fill[black] (1.5,0.5) circle (0.2); } &
\kohnert{ \fill[darkred!60] (0,0) rectangle (3,2); \fill[black] (1.5,0.5) circle (0.2); \fill[black] (2.5,1.5) circle (0.2);}
\\
\ytableaushort{22,33} &
\ytableaushort{12,33} &
\ytableaushort{{1,\!2}2,33} &
\ytableaushort{12,23} &
\ytableaushort{12,{2,\!3}3}
&
\ytableaushort{1{1,\!2},33} &
\ytableaushort{1{1,\!2},23} &
\ytableaushort{1{1,\!2},{2,\!3}3}
\end{array}
\]
\end{ex}

To prove \cref{prop:K_skyline_bijection}, we construct (K-)Kohnert moves on set-valued tableaux.

\begin{dfn}
\label{def:Kohnert_moves_tableaux}
Let $T \in \svt^n(\lambda)$. Consider $x \in \ZZ$ such that $x \in T$. Let $\mcC$ be the leftmost column of $T$ containing an $x$ in box $\bbb$. Let $x'$ be minimal such that $x'+1, x'+2, \dotsc, x \in \mcC$, and let $\bbb'$ be the box in $\mcC$ containing $x'+1$. If $\{x'+1\}, \dotsc, \{x-1\}$ are not in $\mcC$ (\textit{i.e.}, the corresponding boxes only have 1 entry), $x' = 0$, or $x \neq \min \bbb$, then we do not have a (K-)Kohnert move. Otherwise, define the \defn{Kohnert move} on $T$ to remove $x$ from $\bbb$, moving all entries $x'+1, \dotsc, x-1$ down one row (which inserts $x-1$ into $\bbb$), and inserting $x'$ into $\bbb'$.
A \defn{K-Kohnert move} is the same as before except we leave $x \in \bbb$.
\end{dfn}

\begin{lemma}
\label{lem:Kohnert_svt}
Let $T \in \svt^n(\lambda)$. Applying any (K-)Kohnert move to $T$ results in $T' \in \svt^n(\lambda)$.
\end{lemma}

Now we prove \cref{prop:K_skyline_bijection} by using our flagging characterization of K-Demazure crystals from \cref{lemma:flag_condition} and showing that $\phi$ intertwines the (K-)Kohnert moves on K-Kohnert diagrams with the (K-)Kohnert moves on set-valued tableaux.

\begin{ex}
\ytableausetup{boxsize=1.5em}
Let $\lambda = 3^2$ be a $2 \times 3$ rectangle and consider $n = 4$. We exhibit the sequence of (K-)Kohnert moves described in the proof of \cref{prop:K_skyline_bijection} to obtain the element $\scalebox{0.7}{$\ytableaushort{1{1,\!2}2,{2,\!3}3{3,\!4}}$} \in \svt^4_{s_1 s_3 s_2} (\lambda)$ from the initial tableau $\scalebox{0.7}{$\ytableaushort{222,444}$}$ (below is the diagram under $\phi^{-1}$):
\[
\newcommand{\kohnert}[1]{\begin{tikzpicture}[scale=0.48,baseline=13] #1 \draw[very thin] (0,0) grid (4,3); \end{tikzpicture}}
\scalebox{0.83}{$
\begin{array}{c@{\;\longleftarrow\;}c@{\;\longleftarrow\;}c@{\;\longleftarrow\;}c@{\;\longleftarrow\;}c@{\;\longleftarrow\;}c@{\;\longleftarrow\;}c}
\ytableaushort{1{1,\!2}2,{2,\!3}3{3,\!4}} &
\ytableaushort{1{1,\!2}2,{2,\!3}34} &
\ytableaushort{1{1,\!2}2,{2,\!3}44} &
\ytableaushort{1{1,\!2}2,344} &
\ytableaushort{1{1,\!2}2,444} &
\ytableaushort{122,444} &
\ytableaushort{222,444}
\\[18pt]
\kohnert{ \fill[darkred!60] (0,1) rectangle (3,3); \fill[darkred!60] (1,0) rectangle (4,1); \fill[black] (1.5,1.5) circle (0.2); \fill[black] (2.5,2.5) circle (0.2); \fill[black] (3.5,0.5) circle (0.2);} &
\kohnert{ \fill[darkred!60] (0,1) rectangle (3,3); \fill[darkred!60] (0,1) rectangle (1,3); \fill[black] (1.5,1.5) circle (0.2); \fill[darkred!60] (1,0) rectangle (2,1); \fill[darkred!60] (3,0) rectangle (4,1); \fill[black] (2.5,2.5) circle (0.2); } &
\kohnert{ \fill[darkred!60] (1,0) rectangle (2,2); \fill[darkred!60] (0,1) rectangle (1,3); \fill[black] (1.5,1.5) circle (0.2); \fill[darkred!60] (3,0) rectangle (4,2); \fill[darkred!60] (1,2) rectangle (3,3); \fill[black] (2.5,2.5) circle (0.2); } &
\kohnert{ \fill[darkred!60] (1,0) rectangle (2,2); \fill[darkred!60] (0,1) rectangle (1,3); \fill[black] (1.5,1.5) circle (0.2); \fill[darkred!60] (3,0) rectangle (4,2); \fill[darkred!60] (2,2) rectangle (3,3); } &
\kohnert{ \fill[darkred!60] (1,0) rectangle (2,2); \fill[darkred!60] (0,1) rectangle (1,3); \fill[black] (1.5,1.5) circle (0.2); \fill[darkred!60] (3,0) rectangle (4,3); } &
\kohnert{ \fill[darkred!60] (1,0) rectangle (2,2); \fill[darkred!60] (0,2) rectangle (1,3); \fill[darkred!60] (3,0) rectangle (4,3); } &
\kohnert{ \fill[darkred!60] (1,0) rectangle (2,3); \fill[darkred!60] (3,0) rectangle (4,3); }
\end{array}
$}
\]
\end{ex}

\begin{remark}
\label{rem:Key_property}
The proof of the intertwining of (K-)Kohnert moves did not require $\lambda$ to be a rectangle, but $\phi$ does require it as otherwise the image might not be a partition.
\end{remark}

\begin{thm}
The Ross--Yong--Kirillov Conjecture (\cref{conj:K_skyline}) holds for $L_{a}$ when $a$ is any weak composition with a unique nonzero part size; \textit{i.e.},
$
L_a(\xx; \beta) = \sum_{D \in \mcD_a}  \wt_{\beta}(D).
$
\end{thm}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Bijection with set-valued skyline tableaux}
\label{sec:skyline_tableaux}

\cref{conj:skyline_tableaux} is equivalent to showing that
\begin{equation}
\label{eq:equivalent_conjecture}
\overline{L}_{w\lambda} = \ch_{\beta} \left( \overline{\svt}^n_w(\lambda) \right), \quad \text{ where } \overline{\svt}^n_w(\lambda) := \svt^n_w(\lambda) \setminus \bigcup_{v < w} \svt^n_v(\lambda),
\end{equation}
with the union taken over all $v$ strictly less that $w$ in Bruhat order, by inclusion-exclusion, applying M\"obius inversion on (strong) Bruhat order, and Equation~\eqref{eq:lascoux_to_atoms}

\begin{prop}
\label{prop:skyline_tableaux_bijection}
Let $\lambda$ be an $r \times s$ rectangle.
For any $w \in \sym_n^{\lambda}$, there exists a $\beta$-weight preserving bijection
$
\psi \colon \skyline_{w\lambda} \to \overline{\svt}^n_w(\lambda).
$
\end{prop}

We prove \cref{prop:skyline_tableaux_bijection} by explicitly defining the bijection as follows. Consider some $S \in \skyline_{w\lambda}$ and define $T := \psi(S)$ by
(1) sorting the anchor entries in each row in increasing order left to right;
(2) placing each free entry $f$ in the leftmost box of its row such that $f$ is less than the anchor entry; 
(3) constructing the $i$-th column of $T$ from the $(r+1-i)$-th row of the result from the previous step, as in \cref{sec:Kohnert}.

\begin{ex}
Let $\lambda$ be a $2 \times 2$ rectangle and $n = 3$. Then the set-valued skyline tableaux $\skyline_{s_2\lambda}$ and their corresponding elements in $\overline{\svt}^3_{s_2}(\lambda)$ under $\psi$ are given by
\[
\ytableausetup{boxsize=1.5em}
\scalebox{0.85}{$
\begin{ytableau}
1 & \cdot & 3 \\
1 & \cdot & 3
\end{ytableau}
\longmapsto
\ytableaushort{11,33}\,,
\quad 
\begin{ytableau}
1 & \cdot & 2 \\
1 & \cdot & 3
\end{ytableau}
\longmapsto
\ytableaushort{11,23}\,,
\quad
\begin{ytableau}
1 & \cdot & 2,\!3 \\
1 & \cdot & 3
\end{ytableau}
\longmapsto
\ytableaushort{11,{2,\!3}3}\,,
\quad
\begin{ytableau}
1 & \cdot & 2 \\
1 & \cdot & 2,\!3
\end{ytableau}
\longmapsto
\ytableaushort{11,2{2,\!3}}\,.
$}
\]
\end{ex}


\begin{thm}
Monical's Skyline Conjecture (\cref{conj:skyline_tableaux}) holds for $L_{a}$ when $a$ is any weak composition with a unique nonzero part size; \textit{i.e.},
$
\overline{L}_{a} = \sum_{S \in \skyline_{a}} \wt_{\beta}(S).
$
\end{thm}




               


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{K-key tableaux}
\label{sec:Key_tableaux}

A \defn{key tableau} $K$ is a semistandard tableau such that the entries in the $j$-th column of $K$ are a subset of those in the $(j-1)$-st column of $K$.
Every semistandard tableau $T$ has a unique \defn{(right) key tableau} $k(T)$ associated with it, and a Demazure atom can be computed as a generating function for all semistandard tableaux $T$ with $k(T) = K_{w\lambda}$~\cite{LS90}.
Let $\prec$ denote the partial order on semistandard tableaux of shape $\lambda$ such that $T \preceq T'$ if and only if every entry of $T$ is at most the corresponding entry in $T'$.
A Demazure character $\kappa_{w\lambda}$ can be given by summing over all semistandard Young tableaux $T$ of shape $\lambda$ such that $k(T) \preceq K_{w\lambda}$, where $K_{w\lambda}$ is the unique key tableau of shape $\lambda$ and weight $w\lambda$~\cite{LS90}.

Based on the bijection from \cref{prop:skyline_tableaux_bijection} and the (K-)Kohnert moves on set-valued tableaux (\cref{def:Kohnert_moves_tableaux}), the following is a natural possible extension of key tableaux to the K-theory setting.
For $T \in \svt^n(\lambda)$, define $\mcK(T) := k\bigl( \max(T) \bigr)$, where $\max(T)$ is semistandard tableau obtained by taking the greatest entry in each box of $T$.
Thus \cref{thm:Krystal_rectangles} and \cref{lemma:flag_condition} imply that for $\lambda$ and $r \times s$ rectangle
\begin{equation}
\label{eq:rectangle_lascoux}
L_{w\lambda}(\xx; \beta) = \sum_{\substack{T \in \svt^n(\lambda) \\ \mcK(T) \preceq K_{w\lambda}}} \wt_{\beta}(T),
\qquad\qquad
\overline{L}_{w\lambda}(\xx; \beta) = \sum_{\substack{T \in \svt^n(\lambda) \\ \mcK(T) = K_{w\lambda}}} \wt_{\beta}(T),
\end{equation}
or equivalently summed over $\svt^n_w(\lambda)$ and $\overline{\svt}^n_w(\lambda)$ respectively.
However, these formulas do not work for general $\lambda$ as, for example,
$
\ytableausetup{boxsize=2.0em}
\mcK\left(
\scalebox{0.7}{$\ytableaushort{{1}{1,\!2,\!3},{2,\!3}}$}
\right) = \scalebox{0.7}{$\ytableaushort{13,3}$}\,, \ytableausetup{boxsize=1.5em}
$
but it can only contribute to the Lascoux polynomial/atom corresponding to $w_0 \lambda$, where $\lambda = 21$, as it has an excess of $3$.
Moreover, the weak K-crystal in~\cite[Figure~7]{MPS18} does \emph{not} decompose the K-crystal into atoms, as \scalebox{0.7}{$\ytableaushort{1{2,\!3},3}$} should not be in the atom for to $w_0$.

Instead, we conjecture that~\eqref{eq:rectangle_lascoux} modifies as follows.
Recall that the Lusztig involution on the highest weight crystal $B(\mu)$ is defined by sending the highest weight element $U$ to the lowest weight element $U^*$ and extending to $B(\mu)$ by
\begin{equation}
\label{eq:lusztig_involution}
e_i(T^*) = (f_{n-i} T)^*,
\qquad\quad
f_i(T^*) = (e_{n-i} T)^*,
\qquad\quad
\wt(T^*) = w_0 \wt(T).
\end{equation}
We can extend this naively to $\svt^n(\lambda)$ by acting on each irreducible component $B(\mu)$.
Define the \defn{(right) K-key tableau} of a set-valued tableau $T \in \svt^n(\lambda)$ by
\[
K(T) := k(\min(T^*)^*),
\]
where $\min(T)$ is obtained from $T$ by taking the least entry in each box of $T$.

\begin{conj}
\label{conj:Key}
Let $\lambda$ be a partition.
Define the sets
\begin{align*}
\svt^n_w(\lambda) & := \{ T \in \svt^n(\lambda) \mid K(T) \preceq K_{w\lambda} \},
& \quad \overline{\svt}^n_w(\lambda) & := \{ T \in \svt^n(\lambda) \mid K(T) = K_{w\lambda} \}.
\end{align*}
Then we have
$
L_{w\lambda}(\xx; \beta) = \sum_{T \in \svt^n_w(\lambda)} \wt_{\beta}(T),
$
and
$
\overline{L}_{w\lambda}(\xx; \beta) = \sum_{T \in \overline{\svt}^n_w(\lambda)} \wt_{\beta}(T).
$
\end{conj}

We show~\eqref{eq:rectangle_lascoux} establishes \cref{conj:Key} when $\lambda$ is a rectangle by constructing a \defn{K-Lusztig involution}
$\star \colon \svt^n(\lambda) \to \svt^n(\lambda)$
that also satisfies~\eqref{eq:lusztig_involution}.
However, it is a twist of the Lusztig involution by permuting the connected $U_q(\fsl_n)$-components of $\svt^n(\lambda)$.
Let $\lambda$ be a rectangle and $T \in \svt^n(\lambda)$. Define $T^{\star}$ to be the set-valued tableau obtained by rotating the tableau $180^{\circ}$ and then replacing each $i \mapsto n+1-i$.
We note this is a well-known description of the Lusztig involution on semistandard tableaux of shape $\lambda$.

\begin{prop}
\label{prop:Lusztig_reversal}
Let $\lambda$ be a rectangle.
The K-Lusztig involution $\star$ satisfies~\eqref{eq:lusztig_involution}.
For $T \in \svt^n(\lambda)$ as a tensor product of rows $T = R_1 \otimes \cdots \otimes R_k$, we have
$
T^{\star} = R_k^* \otimes \cdots \otimes R_1^*.
$
\end{prop}


This also suggests that \cref{conj:Key} holds for a definition of a (right) K-key tableau
$
K'(T) := k(\min(T^{\dagger})^*),
$
where $T^{\dagger}$ is constructed from $T$ according to \emph{any} automorphism of $\svt^n(\lambda)$ such that $\wt(T^{\dagger}) = w_0 \wt(T)$.
However, given a (weak) K-crystal structure on $\svt^n(\lambda)$, it would be preferable to have a $T^{\dagger}$ construction that matches the labeling of tableaux $T$ by K-keys $K'(T)$ with the decomposition of the K-crystal by K-Demazure subcrystals, as is the case with our K-Lusztig involution $T^\star$.
Furthermore, it is likely that in general we want $T^{\star} = R_k^* \otimes \cdots \otimes R_1^*$ as in \cref{prop:Lusztig_reversal}, but this would require an appropriate K-rectification or insertion scheme in order to obtain a result back in $\svt^n(\lambda)$.


%%%%%%%%%%
\acknowledgements{
OP is grateful for interesting conversations with B.~Proctor.  TS thanks T.~Ikeda, T.~Ma\-tsu\-mu\-ra, and S.~Sugimoto for stimulating discussions. The authors thank C.~Monical for useful discussions. This work benefited from computations using \textsc{SageMath}~\cite{sage}.
}


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