\section{Bijection between \texorpdfstring{$\SYT(Z(d,r))$}{} and \texorpdfstring{$\Red(w^{(d,r)})$}{} via the Kra\'skiewicz's insertion}\label{sec:kraskiewicz}

We will follow the notations as recorded in Section~1.3 of \cite{Lam1995thesis}. For a shifted tableau $T$ of shape $\lambda = (\lambda_1,\ldots,\lambda_d)$, define $\pi(T) = T_d T_{d-1},\ldots,T_1$ to be the reading word of $T$ obtained by reading left to right along rows and from bottom to top, where $T_i$ represents the $i$-th row. For a unimodal sequence of integers
\[\mathbf{R} = (r_1>r_2>\ldots>r_k<r_{k+1}<\ldots<r_m),\]
we define the decreasing part of $\mathbf{R}$ to be $\mathbf{R}^{\downarrow} = (r_1>r_2>\ldots>r_k)$, and the increasing part of $\mathbf{R}$ to be $\mathbf{R}^{\uparrow} = (r_{k+1}<r_{k+2}<\ldots<r_m)$.
Note that we include the minimal integer of the sequence in $\mathbf{R}^{\downarrow}$.

Let $w\in W(B_n)$ and $\mathbf{a} = a_1a_2\ldots a_{\ell(w)}\in \Red(w)$, we define the \emph{Kra\'skiewicz's insertion algorithm} recursively.
Set $(P^{(0)},Q^{(0)}) := (\emptyset,\emptyset)$, for any $i\in [\ell(w)]$, define the insertion
\[(P^{(i-1)},Q^{(i-1)}) \leftarrow a_i =: (P^{(i)},Q^{(i)})\]
%of $a_i$ into $(P^{(i-1)},Q^{(i-1)})$
as follows:

Step 1: Set $\mathbf{R}$ to be the first row of $P^{(i-1)}$ and $a = a_i$.

Step 2: Insert $a$ into $\mathbf{R}$ as follows:
\begin{itemize}
    \item Case 1 ($\mathbf{R}a$ is unimodal): Append $a$ to the right of $\mathbf{R}$ to obtain $P^{(i)}$. Then define $Q^{(i)}$ from $Q^{(i-1)}$ by adding $i$ to the unique box in $P^{(i)}/P^{(i-1)}$. Stop.
    \item Case 2 ($\mathbf{R}a$ is not unimodal): Let $b$ be the smallest number in $\bf{R}^{\uparrow}$ such that $b\geq a$.
        \begin{itemize}
            \item Case 2.1 ($a=0$ and $\mathbf{R}$ contains $101$ as a subsequence): We leave $\mathbf{R}$ unchanged and return to start of Step 2 with $a=0$ and $\mathbf{R}$ equals the next row.
            \item Case 2.2 (otherwise): Replace $b$ with $a$. Set $c=b$ if $a\neq b$ or $c=b+1$ if $a=b$.
        \end{itemize}
    We now insert $c$ into $\mathbf{R}^{\downarrow}$. Let $d$ be the largest integer such that $d\leq c$. This number always exists since $\mathbf{R}^{\downarrow}$ contains the smallest number in the row. Replace $d$ with $c$. Set $a'=d$ if $c\neq d$ or $a'=d+1$ if $c=d$.
\end{itemize}

Step 3: Repeat Step 2 with $a = a'$ and $\bf{R}$ the next row.

Define $P(\mathbf{a}) = P^{(\ell(w))}$ to be the \emph{insertion tableau} and $Q(\mathbf{a}) = Q^{(\ell(w))}$ to be the \emph{recording tableau}. 

\begin{ex}
Let $w = w^{(3,2)} = 45\bar{1}\bar{2}\bar{3}$ as in \cref{ex:45123}. Consider the reduced word $\mathbf{a} = 010121012342312 \in \Red(w)$. Following the above insertion algorithm, we obtain
\begin{center}
\ytableausetup{boxsize=1em}
$P^{(0)} = \emptyset \xrightarrow{0}$
\begin{ytableau}
0
\end{ytableau}
$\xrightarrow{1}$ 
\begin{ytableau}
0 & 1
\end{ytableau}
$\xrightarrow{0}$
\begin{ytableau}
1 & 0\\
\none & 0
\end{ytableau}
$\xrightarrow{1}$
\begin{ytableau}
1 & 0 & 1\\
\none & 0
\end{ytableau}
$\xrightarrow{2}$
\begin{ytableau}
1 & 0 & 1 & 2\\
\none & 0
\end{ytableau}
$\xrightarrow{1}$
\begin{ytableau}
2 & 0 & 1 & 2\\
\none & 0 & 1
\end{ytableau}
$\xrightarrow{0}$
\begin{ytableau}
2 & 1 & 0 & 2\\
\none & 1 & 0\\
\none & \none &0
\end{ytableau}
\vspace{0.2cm}
$\xrightarrow{1}$
\begin{ytableau}
2 & 1 & 0 & 1\\
\none & 1 & 0 & 1\\
\none & \none &0
\end{ytableau}
$\xrightarrow{2}$
\begin{ytableau}
2 & 1 & 0 & 1 & 2\\
\none & 1 & 0 & 1\\
\none & \none &0
\end{ytableau}
\vspace{0.2cm}
$\xrightarrow{3}$
\begin{ytableau}
2 & 1 & 0 & 1 & 2 & 3\\
\none & 1 & 0 & 1\\
\none & \none &0
\end{ytableau}
$\xrightarrow{4}$
\begin{ytableau}
2 & 1 & 0 & 1 & 2 & 3 & 4\\
\none & 1 & 0 & 1\\
\none & \none &0
\end{ytableau}
$\xrightarrow{2}$
\begin{ytableau}
3 & 1 & 0 & 1 & 2 & 3 & 4\\
\none & 1 & 0 & 1 & 2\\
\none & \none &0
\end{ytableau}
$\xrightarrow{3}$
\begin{ytableau}
4 & 1 & 0 & 1 & 2 & 3 & 4\\
\none & 1 & 0 & 1 & 2 & 3\\
\none & \none &0
\end{ytableau}
$\xrightarrow{1}$
\begin{ytableau}
4 & 2 & 0 & 1 & 2 & 3 & 4\\
\none & 2 & 0 & 1 & 2 & 3\\
\none & \none &0 & 1
\end{ytableau}
$\xrightarrow{2}$
\begin{ytableau}
4 & 3 & 0 & 1 & 2 & 3 & 4\\
\none & 3 & 0 & 1 & 2 & 3\\
\none & \none &0 & 1 & 2
\end{ytableau}.
\end{center}
We then have
\[\ytableausetup{boxsize=1.1em}
P(\mathbf{a}) = \begin{ytableau}
4 & 3 & 0 & 1 & 2 & 3 & 4\\
\none & 3 & 0 & 1 & 2 & 3\\
\none & \none &0 & 1 & 2
\end{ytableau},\ 
Q(\mathbf{a}) = \begin{ytableau}
1 & 2 & 4 & 5 & 9 & 10 & 11\\
\none & 3 & 6 & 8 & 12 & 13\\
\none & \none &7 & 14 & 15
\end{ytableau},
\]
and the reading word $\pi(P(\mathbf{a})) = 012301234301234$.
\end{ex}


\begin{defin}\label{def:SDT}
    For a shifted tableau $T$ with $m$ rows, we say $T$ is a \emph{standard decomposition tableau} of $w\in W(B_n)$ if
    \begin{enumerate}
        \item $\pi(T) = T_m T_{m-1},\ldots,T_1$ is a reduced word of $w$,
        \item $T_{i}$ is a unimodal subsequence of maximal length in $T_{m}T_{m-1}\ldots T_{i}$.
    \end{enumerate}
Define the set of all such tableaux to be $\SDT(w)$.
\end{defin}

\begin{theorem}[Theorem~5.2, \cite{K89}]\label{thm:insertion}
    The Kra\'skiewicz's insertion gives a bijection between $\{\mathbf{a}\in \Red(w)\}$ and the pairs of tableaux $(P(\mathbf{a}),Q(\mathbf{a}))$ where $P(\mathbf{a})\in \SDT(w)$ and $Q(\mathbf{a})$ is a standard tableau of the same shape.
\end{theorem}

\begin{comment}
For $w\in W(B_n)$ and any strict partition $\lambda$, let $F^{C}(w)$ and $Q_{\lambda}$ be the corresponding \emph{type C Stanley symmetric function} and \emph{Schur-Q function} respectively. See \cite{BLvexillary} for the exact definitions. 
\begin{theorem}[Theorem~3.12, \cite{Lam96}]
    $F^{C}(w) = \sum_{T\in \SDT(w)}Q_{\sh(T)}$.
\end{theorem}

\begin{defin}
    A signed permutation $w\in W(B_n)$ is said to be \emph{vexillary} if $\SDT(w)$ consists of exactly one shifted tableau. We denote this tableau as $P(w)$.
\end{defin}

\begin{defin}
    For any $w\in W(B_n)$ and any $v\in W(B_m)$ such that $m\leq n$, we say $w$ \emph{pattern embeds} $v$ if the following is true for some $1\leq i_1<i_2<\ldots<i_m\leq n$:
    \begin{enumerate}
        \item $w(i_j)$ has the same sign as $v(j)$,
        \item For all $j, k$, $|w(i_j)|<|w(i_k)|$ if and only if $|v(j)|<|v(k)|$.
    \end{enumerate}
We say $w$ \emph{pattern avoids} $v$ if $w$ does not pattern embed $v$.
\end{defin}

\begin{theorem}[Theorem~7, \cite{BLvexillary}]\label{thm:Bvexillary}
    A signed permutation $w\in W(B_n)$ is vexillary if and only if $w$ pattern avoids the following permutations:
    \[\begin{matrix}
        \bar{3}2\bar{1} & \bar{3}21 & 32\bar{1} & 321 & 3\bar{1}2 &
        \bar{2}31 & \bar{1}32 & \bar{4}\bar{1}\bar{2}3 & \bar{4}1\bar{2}3 \\ \bar{3}\bar{4}\bar{1}\bar{2} &
        \bar{3}\bar{4}1\bar{2} & 3\bar{4}\bar{1}\bar{2} & 3\bar{4}1\bar{2} & 3142 & \bar{2}\bar{3}4\bar{1} &
        2413 & 2\bar{3}4\bar{1} & 2143 
    \end{matrix}\,.\]
\end{theorem}
By comparing \eqref{eqn:defw} with the above list of patterns, we directly see that $w^{(d,r)}$ is vexillary. In particular, this means that any $\mathbf{a}\in \Red(w^{(d,r)})$ has the same $P$ tableau.

\begin{prop}\label{prop:shape-of-P(w)}
For $d>0$ and $r\geq0$, $\sh(P(w^{(d,r)})) = Z(d,r)$. Moreover, \begin{equation*}
        \begin{split}
            P(w^{(d,r)})(i,j) = 
            \begin{cases}
            r-j &\text{ if }j<0,\\
            j &\text{ if }j\geq 0.
            \end{cases}
        \end{split}
    \end{equation*}
\end{prop}
See \cref{fig:trapezoidT} for this tableaux. A concrete example is also shown in \cref{ex:w-lam}.

\begin{figure}[h!]
\centering
\ytableausetup{boxsize=2.5em}
\begin{ytableau}
\scriptstyle{r{+}d{-}1} & \none[\cdots] & r{+}1 & 0 & \none[\cdots] & r & r{+}1& \none[\cdots] & \scriptstyle{r{+}d{-}1}\\
\none & \none[\ddots] & \none[\vdots] & \none[\vdots] & \none & \none[\vdots] &\none[\vdots] & \none[\iddots]\\
\none & \none & r{+}1 & 0 & \none[\cdots] & r & r{+}1\\
\none & \none & \none & 0 & \none[\cdots] & r
\end{ytableau}
\caption{The insertion tableaux $P(w^{(d,r)})$ of shape $Z(d,r)$}
\label{fig:trapezoidT}
\end{figure}
\end{comment}

\begin{lemma}\label{lm:vexillary}
$\SDT(w^{(d,r)})$ consists of exactly one shifted tableau. In other words, any $\mathbf{a}\in \Red(w^{(d,r)})$ has the same $P$ tableau.
\end{lemma}

\begin{cor}\label{cor:uniquePtableau}
By restricting to the recording tableau, Kra\'skiewicz's insertion gives a bijection $\mathbf{a}\mapsto Q(\mathbf{a})$ between $\Red(w^{(d,r)})$ and $\SYT(Z(d,r))$.
\end{cor}

Combining \cref{cor:reduced-word-balanced} and \cref{cor:uniquePtableau}, we derive \cref{thm:main} for the trapezoid shape $Z(d,r)$.
.
