\section{Bijection between \texorpdfstring{$\BS(Z(d,r))$}{} and \texorpdfstring{$\Red(w^{(d,r)})$}{} via reflection order}\label{sec:trapezoid}

A crucial shape for our analysis is the \emph{trapezoid} 
\[Z(d,r):=(r+2d-1,r+2d-3,\ldots,r+3,r+1)\]
with height $d$ and base lengths $r+2d-1$ and $r+1$. In particular, $Z(d,0)$ is the \textit{double staircase} and every shifted shape is contained in some trapezoid of the same height. 

Set $e_{-j} = -e_j$ for all $j>0$ and $e_0=0$, and consider the labeling $f:Z(d,r) \longrightarrow \Phi^{+}(B_{d+r})$ where
\begin{equation}\label{eqn:defrootlabel}
    \begin{split}
        f(i,j) = 
\begin{cases}
    e_{d+1-i}-e_{j} & \text{if } j\leq 0,\\
    % e_i &  \text{if } j = 0,\\
    e_{d+1-i}+e_{j+d}&\text{if }0<j\leq r,\\
    e_{d+1-i}-e_{j-r}&\text{if }j>r.
\end{cases}
    \end{split}
\end{equation}
Define the permutation $w^{(d,r)}\in W(B_{d+r})$ associated to $Z(d,r)$ by
\begin{equation}\label{eqn:defw}
    \begin{split}
        w^{(d,r)}(i) := 
\begin{cases}
    d+i &\text{if }0<i\leq r,\\
    \overline{i-r} &\text{if }i>r.
\end{cases}
    \end{split}
\end{equation}
\begin{prop}
For all $d>0$ and $r\geq 0$, $f(Z(d,r)) = \Inv(w^{(d,r)})$.
\end{prop}

The labeling $f$ can also be extended to a labeling $\tilde{f}:\tilde{Z}(d,r)\to \Phi^+(B_{d+r})$ where
\[\tilde{Z}(d,r)=Z(d,r)\cup\{(1,\bar d),(2,\overline{d-1}),\ldots(d,\bar1)\}\]
is the extended shape of $Z(d,r)$ with $d$ extra boxes as defined in \cref{sec:prelim}. The extended labeling is given by
\[\tilde{f}(i,j)=\begin{cases}
    2e_{d+1-i} &\text{if }j=\overline{d+1-i},\\
    f(i,j) &\text{otherwise}.
\end{cases}\]

\begin{ex}\label{ex:45123}
For $d = 3$ and $r = 2$, we have $Z(d,r) = (7,5,3)$ and $w^{(3,2)} = 45\bar{1}\bar{2}\bar{3}\in W(B_5)$. See \cref{fig:d=3r=2} for the extended labeling $\tilde f$ in this case.
\begin{figure}[h!]
\centering
\ytableausetup{boxsize=3em}
\begin{ytableau}
\none[2e_3] & e_3{+}e_2 & e_3{+}e_1 & e_3 & e_4{+}e_3 & e_5{+}e_3 & e_3{-}e_1 & e_3{-}e_2 \\
\none & \none[2e_2] & e_2{+}e_1 & e_2 & e_4{+}e_2 & e_5{+}e_2 & e_2{-}e_1\\
\none & \none & \none[2e_1] & e_1 & e_4{+}e_1 & e_5{+}e_1
\end{ytableau}
\caption{The extended labeling $\tilde{f}$ of $\tilde{Z}(3,2)$}
\label{fig:d=3r=2}
\end{figure}
\end{ex}
The filling of a balanced shifted tableau $B\in\BS(Z(d,r))$ can be viewed as a map $B:Z(d,r)\to\mathbb{N}$ by sending a box to its entry. Then the composition $Bf^{-1}:\Inv(w^{(d,r)})\to \mathbb{N}$ encodes an ordering of the roots in $\Inv(w^{(d,r)})$. We will show that this actually gives a reflection order in $\ro(w^{(d,r)})$.
\begin{prop}\label{prop:root-order-balanced}
The map $B\mapsto Bf^{-1}$ is a bijection between $\BS(Z(d,r))$ and $\ro(w^{(d,r)})$, and thus induces a bijection between $\BS(Z(d,r))$ and $\Red(w^{(d,r)})$.
\end{prop}

\begin{comment}
To prove \cref{prop:root-order-balanced}, we need the following technical lemmas.

\begin{lemma}\label{lm:same-col-order}
    Given $B\in \BS(\lambda)$ and $j\geq 0$, if column $j$ and $j+1$ have the same length in $B$, then $B(i,j)<B(i,j+1)$ for all $i$.
\end{lemma}

\begin{cor}\label{cor:same-col-order}
    For any balanced shifted tableaux $B$ of shape $Z(d,r)$ and $1\leq i\leq d$, 
    \[B(i,0)<B(i,1)<B(i,2)<\cdots<B(i,r).\]
\end{cor}

\begin{lemma}[Strongly balanced conditions]\label{lm:balance-equiv}
    A shifted tableaux $B$ of shape $Z(d,r)$ is balanced if and only if the following holds at $f^{-1}(\alpha)$:
    %the map $Bf^{-1}$ satisfies the following conditions (called strongly balanced conditions):
    \begin{enumerate}
        \item(Right Staircase) For any $\alpha = e_i-e_j$ where $1\leq j<i\leq d$, %$\alpha=e_i-e_j$,
        %satisfies the following condition
        %\begin{enumerate}
            %\item 
            $Bf^{-1}(\alpha)$ lies between $Bf^{-1}(e_i-e_k)$ and $Bf^{-1}(e_k-e_j)$ for all $j<k<i$;
        %\end{enumerate}
        \item(Rectangle) For any $\alpha=e_i+e_p$ where $1\leq i\leq d<p\leq d+r$, %$\alpha=e_i+e_p$ satisfies the following conditions
        \begin{enumerate}
            \item $Bf^{-1}(\alpha)<Bf^{-1}(e_i+e_q)$ for all $p<q\leq d+r$;
            \item $Bf^{-1}(\alpha)$ lies between $Bf^{-1}(e_i-e_k)$ and $Bf^{-1}(e_k+e_p)$ for all $1\leq k<i$;
        \end{enumerate}
        \item(Column $0$) For any $\alpha=e_i$ where $1\leq i\leq d$, %$\alpha=e_i$ satisfies the following conditions
        \begin{enumerate}
            \item $Bf^{-1}(\alpha)<Bf^{-1}(e_i+e_q)$ for all $d< q\leq d+r$;
            \item $Bf^{-1}(\alpha)$ lies between $Bf^{-1}(e_i-e_k)$ and $Bf^{-1}(e_k)$ for all $1\leq k<i$;
        \end{enumerate}
        \item(Left Staircase) For any $\alpha=e_i+e_j$ where $1\leq j<i\leq d$, % $\alpha=e_i+e_j$ satisfies the following conditions
        \begin{enumerate}
            \item $Bf^{-1}(\alpha)<Bf^{-1}(e_i+e_q)$ for all $d<q\leq d+r$;
            \item $Bf^{-1}(\alpha)<Bf^{-1}(e_j+e_q)$ for all $d<q\leq d+r$;
            \item $Bf^{-1}(\alpha)$ lies between $B\Tilde{f}^{-1}(e_i-e_k)$ and $B\Tilde{f}^{-1}(e_j+e_k)$ for all $-j<k<i$.
        \end{enumerate}
    \end{enumerate}
\end{lemma}

\begin{ex}
Let $d=3$ and $r=2$. \cref{fig:balance-equiv-ex} illustrates an example of the strongly balanced conditions of \cref{lm:balance-equiv} at the box $(1,-1)$ (or at root $\alpha=e_3+e_1$), labeled by $\ast$ in the diagram. Its extended hook is colored yellow. The strongly balanced conditions for box $\ast$ are
\begin{itemize}
    \item \underline{Condition (4a) and (4b):} Box $\ast$ is smaller than the boxes labeled by ${+}$;
    \item \underline{Condition (4c):} Box $\ast$ lies between the pair of $a$'s, pair of $b$'s and pair of $c$'s.
\end{itemize}
\begin{figure}[h!]
\centering
\ytableausetup{boxsize=1.5em}
\begin{ytableau}
{} & *(yellow)\ast & *(yellow)c & *(yellow){+} & *(yellow){+} & *(yellow)b & *(yellow)a\\
\none & *(yellow)a & {} & {} & {} & {}\\
\none & \none[\circled{b}] & *(yellow)c & *(yellow){+} & *(yellow){+}
\end{ytableau}
\caption{An example of strongly balanced condition for box $\ast$}
\label{fig:balance-equiv-ex}
\end{figure}
\end{ex}
\end{comment}

Since there is a natural bijection between $\ro(w)$ and $\Red(w)$, \cref{prop:root-order-balanced} implies:
\begin{cor}\label{cor:reduced-word-balanced}
The map $B\mapsto \ro^{-1}(Bf^{-1})$ is a bijection between $\BS(Z(d,r))$ and $\Red(w^{(d,r)})$. 
\end{cor}

\begin{ex}
Assume we started with the following balanced tableau of shape $Z(3,2)$.
\begin{center}
\ytableausetup{boxsize=1.5em}
$B=$
\begin{ytableau}
4 & 8 & 7 & 10 & 13 & 5 & 15\\
\none & 3 & 2 & 6 & 9 & 1\\
\none & \none & 11 & 12 & 14
\end{ytableau}
\end{center}
\end{ex}
The corresponding reflection order $Bf^{-1}$ is given as follows:
\[
\begin{tikzcd}[row sep = small]
12345 \arrow[r, "e_2-e_1"]& 21345 \arrow[r, "e_2"] & \bar21345 \arrow[r, "e_2+e_1"] & 1\bar2345 \arrow[r, "e_3+e_2"] & 13\bar245 \arrow[r, "e_3-e_1"] & 31\bar245\\
\arrow[r, "e_4+e_2"]& 314\bar25 \arrow[r, "e_3"] & \bar314\bar25 \arrow[r, "e_3+e_1"] & 1\bar34\bar25 \arrow[r, "e_5+e_2"] & 1\bar345\bar2 \arrow[r, "e_4+e_3"] & 14\bar35\bar2\\
\arrow[r, "e_1"]& \bar14\bar35\bar2 \arrow[r, "e_4+e_1"] & 4\bar1\bar35\bar2 \arrow[r, "e_5+e_3"] & 4\bar15\bar3\bar2 \arrow[r, "e_5+e_1"] & 45\bar1\bar3\bar2 \arrow[r, "e_3-e_2"] & 45\bar1\bar2\bar3.\\
\end{tikzcd}
\]
Therefore, we can read off a reduced word $\mathbf{a}$ of $w^{(3,2)}=45\bar1\bar2\bar3$ as
\[\mathbf{a}=\ro^{-1}(Bf^{-1})=101213014201324\in\Red(w^{(3,2)}).\]
.