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\usepackage{tikz}
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\usepackage{amsmath, amsfonts, amssymb, amsxtra, amsthm}
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\usepackage{amssymb} 
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\newcommand{\Z}{\mathbb Z} 
\renewcommand{\P}{\mathbb P}  
\DeclareMathOperator{\GL}{GL} 
\DeclareMathOperator{\wt}{wt}
\DeclareMathOperator{\QBPD}{QBPD}

\DeclareMathOperator{\BPD}{BPD}
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\usepackage{lipsum}

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\title[Quantum bumpless pipe dreams]{Quantum bumpless pipe dreams}

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%% use \addressmark{1}, \addressmark{2} etc for the institutions, and use \thanks{} for contact details
%% for more than 3 names, use the Oxford comma: One Author, Two Author, Red Author, \and Blue Author
\author[T. Le, S. Ouyang, L. Tao, J. Restivo, A. Zhang]{
\texorpdfstring{Tuong Le\addressmark{1}\thanks{\href{mailto:tuongle@umich.edu}{tuongle@umich.edu}.}, Shuge Ouyang\thanks{\href{mailto:shugeoy@umich.edu}{shugeoy@umich.edu}.}\addressmark{1}, Leo Tao\thanks{\href{mailto:leotao@umich.edu}{leotao@umich.edu}.}\addressmark{1}, Joseph Restivo\thanks{\href{mailto:restivo@umich.edu}{restivo@umich.edu}.}\addressmark{1}, \\ \and Angelina Zhang \thanks{\href{mailto:angzh@umich.edu}{angzh@umich.edu}.}\addressmark{1}}{Tuong Le, Shuge Ouyang, Leo Tao, Joseph Restivo, and Angelina Zhang}}
%% then use \addressmark to match authors to institutions here
\address{\addressmark{1}Department of Mathematics, University of Michigan, Ann Arbor, MI, United States}

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\received{\today}

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%% put your English abstract here, or comment this out if you don't have one yet
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\abstract{Schubert polynomials are polynomial representatives of Schubert classes in the cohomology of the complete flag variety and have a combinatorial formulation in terms of bumpless pipe dreams. Quantum double Schubert polynomials are polynomial representatives of Schubert classes in the torus-equivariant quantum cohomology of the complete flag variety, but no analogous combinatorial formulation had been discovered. We introduce a generalization of the bumpless pipe dreams called quantum bumpless pipe dreams, giving a novel combinatorial formula for quantum double Schubert polynomials as a sum of binomial weights of quantum bumpless pipe dreams. We give a bijective proof for this formula by showing that the sum of binomial weights satisfies a defining transition equation.}

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%\resume{\lipsum[2]}

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\keywords{quantum bumpless pipe dreams, quantum Schubert polynomials, quantum cohomology}

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\begin{document}



%\begin{abstract}
    
    
%\end{abstract}

\maketitle

\section{Introduction}
Schubert polynomials were introduced by Lascoux and Schützenberger \cite{Lascoux82} and they represent cohomology classes called Schubert classes of the complete flag variety. The original definition is algebraic and relies on divided difference operators; however, multiple combinatorial formulas for the monomial expansion of Schubert polynomials have been found \cite{assaf, bergeron, FOMIN1996123,  Lam_2021, magyar}. Two such examples are Schubert polynomials as weight-generating functions of pipe dreams (originally called RC-graphs \cite{bergeron, FOMIN1996123}), or bumpless pipe dreams \cite{Lam_2021}. 
For example, for $\BPD(w)$, the set of (reduced) bumpless pipe dreams of a permutation $w$, one has \begin{equation*}
\g_w(x, y) = \sum_{P\in \BPD(w)}\bwt(P) \,,\end{equation*}
where $\bwt(P)$ is a product of binomials $(x_i-y_j)$ associated to $P\in \BPD(w)$ \cite{Lam_2021}.
These pipe dream and bumpless pipe dream formulations  generalize to some generalizations of Schubert polynomials, such as double Schubert polynomials and double Grothendieck polynomials \cite{KNUTSON2004161,WEIGANDT2021105470}.

Motivated by ideas that stem from string theory, mathematicians defined quantum cohomology rings \cite{vafa, witten}. See, e.g.,~\cite{Givental97} for more on the history of quantum cohomology. In the quantum cohomology of the complete flag variety, the Schubert classes correspond to quantum Schubert polynomials, another generalization of Schubert polynomials \cite{quantum}. Quantum double Schubert polynomials, which generalize both quantum Schubert polynomials and double Schubert polynomials, were defined in \cite{cf, KIRILLOV2000191}. Like Schubert polynomials, there is a quantum double Schubert polynomial for each permutation $w$ of $\{1, 2, \dots, n\}$, denoted $\g_w^q(x,y)$, lying in $\Z[x_1, \dots, x_n, y_1, \dots, y_n, q_1, \dots, q_{n-1}]$. There is no known combinatorial formulation for the monomial expansion of quantum Schubert polynomials or quantum double Schubert polynomials. One major difficulty is the presence of unpredictable signs in the monomial expansion of $\g_w^q(x,y)$. In this paper, we define combinatorial objects called quantum bumpless pipe dreams (QBPDs). They are a generalization of  bumpless pipe dreams, and their weight-generating function gives the quantum double Schubert polynomials, i.e., \begin{equation*}
\g_w^q(x,y) = \sum_{P\in \QBPD(w)}\bwt(P)\,,\end{equation*}
where $\QBPD(w)$ is the set of QBPDs of $w$ and $\bwt(P)$ is a product of $(x_i-y_j)$'s and $q_i$'s. This is stated precisely in \Cref{thm:qdform}. 
Unfortunately, this formula has internal cancellation, but the combinatorics seems quite natural.

%We prove this formula by showing the quantity $\sum\limits_{P\in \QBPD(w)}\bwt(P)$
%satisfies a defining transition equation of the quantum double Schubert polynomials. We do this by constructing four bijections between terms on each side of the recursion, plus a family of additional bijections between terms that cancel out on one side. By showing how these bijections change the binomial weights, we establish the recursion and thus establish the main identity, Theorem ~\ref{thm:qdform}.

We give the necessary background in \Cref{background}. In \Cref{quantumbpd}, we define quantum bumpless pipe dreams, establish fundamental combinatorial properties, and state the main theorem. In \Cref{droopandlift}, we provide a way to generate all QBPDs for a given permutation using droop moves as in \cite{Lam_2021}, as well as new moves called \textit{lift moves}. In \Cref{stability}, we establish that the quantity $\sum\limits_{P\in \QBPD(w)}\bwt(P)$ satisfies the stability condition, i.e., it does not change under the natural inclusion map $i\colon S_n\to S_{n+1}$, which is needed for our proof of \Cref{thm:qdform}. We give an overview of the proof of \Cref{thm:qdform} in \Cref{proof33}. In \Cref{cancel}, we provide some examples of (partial) cancellations of the binomial weight of QBPDs and provide some analysis of cancellations. This is an extended abstract for \cite{le2024qbpd}, and the full proof of \Cref{thm:qdform} and other results stated here can be found there.
%In \Cref{proof33}, we prove \Cref{thm:qdform}.
%In \Cref{cancel}, we provide some examples of (partial) cancellations of the binomial weight of QBPDs and provide tables analyzing the cancellations in small degrees. Finally, we provide some future directions in \Cref{future}.

\section{Background}\label{background}
We use the notation $[n]:=\{1, 2, \dots, n\}.$ Let $S_n$ be the symmetric group on $[n]$, i.e. the set of bijections from $[n]$ to $[n]$. To write a bijection $\sigma\colon [n]\to [n]$, we will use one-line notation, i.e., writing $\sigma(1)\sigma(2)\dots \sigma(n)$. Multiplication of permutations are function composition (in that order), that is, $\sigma\tau = \sigma \circ \tau$ as functions from $[n]$ to $[n]$. We write $t_{ab}$ for the transposition that swaps $a$ and $b$, and write $s_i$ for the adjacent transposition $t_{i, i+1}$. The length of $w\in S_n$, denoted $\ell(w)$, is defined as the minimum number of adjacent transpositions $s_i$ required to express $w$ as their product. Any way to write $w$ as a product of exactly $\ell(w)$ adjacent transpositions is called a \textit{reduced word} for $w$. Let $w_0 := n\: n-1\dots 1$ be the longest permutation in $S_n$.

We write $\Z[x]$ for $\Z[x_1, \dots, x_n]$,  $\Z[x, y]$ for $\Z[x][y_1, \dots, y_n]$, $\Z[x, q]$ for \\$\Z[x][q_1, \dots, q_{n-1}]$ and $\Z[x, y, q]$ for $\Z[x, y][q_1, \dots, q_{n-1}]$.
\subsection{Rothe Diagrams}
Throughout this paper, we use matrix coordinate notation: $(i, j)$ means the box on row $i$ column $j$. For a permutation $\sigma : [n]\to [n]$, the \textit{Rothe diagram} of $\sigma$ is defined as follows.  The set of boxes $\{(i,\sigma(i)):i\in [n]\}$ are first marked with a dot in the grid. Then, starting from each dot and ending on edges of the grid, vertical lines are drawn downward and horizontal lines are drawn rightward. The resulting figure is the Rothe diagram for $\sigma$. 

Note that the Rothe diagram for a permutation $\sigma$ can be turned into a bumpless pipe dream by ``smoothing'' the corners into \begin{tikzpicture}[scale=\inlinescale]
\draw (0, 0) grid (1,1);
\draw[blue](1,0.5) arc (90:180:0.5);
\end{tikzpicture} tiles, as in \Cref{fig:rothe}. (For the definition of bumpless pipe dream, see \cite{Lam_2021}, or \Cref{def:qbpd} below.)

\begin{figure}[!ht]
        \centering
        \begin{tikzpicture}[scale=\figscale]
            \node at (0.5,-0.3){1};
\node at (1.5,-0.3){2};
\node at (2.5,-0.3){3};
\node at (3.5,-0.3){4};
\node at (4.3,0.5){4};
\node at (4.3,1.5){3};
\node at (4.3,2.5){2};
\node at (4.3,3.5){1};
            \draw (0, 0) grid (4,4);
            
            \draw[blue] (3.5, 3.5)--(4, 3.5); 
            \draw[blue] (3.5, 3.5)--(3.5, 0); 

            \draw[blue] (2.5, 0.5)--(4, 0.5);
            \draw[blue] (2.5, 0.5)--(2.5, 0); 

            \draw[blue] (1.5, 2.5)--(4, 2.5); 
            \draw[blue] (1.5, 2.5)--(1.5, 0); 

            \draw[blue] (0.5, 1.5)--(4, 1.5); 
            \draw[blue] (0.5, 1.5)--(0.5, 0); 
            
            \node[blue] at (3.5,3.5)[circle,fill,inner sep=1.5pt]{};
            \node[blue] at (2.5,0.5)[circle,fill,inner sep=1.5pt]{};
            \node[blue] at (1.5,2.5)[circle,fill,inner sep=1.5pt]{};
            \node[blue] at (0.5,1.5)[circle,fill,inner sep=1.5pt]{};
        \end{tikzpicture}\:\:\:\:
        \begin{tikzpicture}[scale=\figscale]
\node at (0.5,-0.3){1};
\node at (1.5,-0.3){2};
\node at (2.5,-0.3){3};
\node at (3.5,-0.3){4};
\node at (4.3,0.5){4};
\node at (4.3,1.5){3};
\node at (4.3,2.5){2};
\node at (4.3,3.5){1};
\draw (0, 0) grid (4,4);
\draw[blue](1,1.5) arc (90:180:0.5);
\draw[blue](0.5,0)--(0.5,1);
\draw[blue](2,2.5) arc (90:180:0.5);
\draw[blue](1.5,1)--(1.5,2);
\draw[blue](1,1.5)--(2,1.5);
\draw[blue](1.5,0)--(1.5,1);
\draw[blue](2,2.5)--(3,2.5);
\draw[blue](2,1.5)--(3,1.5);
\draw[blue](3,0.5) arc (90:180:0.5);
\draw[blue](4,3.5) arc (90:180:0.5);
\draw[blue](3.5,2)--(3.5,3);
\draw[blue](3,2.5)--(4,2.5);
\draw[blue](3.5,1)--(3.5,2);
\draw[blue](3,1.5)--(4,1.5);
\draw[blue](3.5,0)--(3.5,1);
\draw[blue](3,0.5)--(4,0.5);
\end{tikzpicture}
        \caption{Rothe diagram for $4213$}
        \label{fig:rothe}
    \end{figure}
\subsection{Double Schubert polynomials}

Consider the action of $S_n$ on $\Z[x, y]$ by permuting the $y$ variables; in particular, $s_i$ swaps $y_i$ and $y_{i+1}$:\begin{equation*}
s_if(x, y_{1}, \dots, y_{i}, y_{i+1}, \dots, y_n) = f(x, y_{1}, \dots, y_{i+1}, y_{i}, \dots, y_n).\end{equation*} 
We define divided difference operators $\partial_i^y$ as follows \begin{equation*}
    \de_i^y(f) := \frac{f-s_if}{y_i-y_{i+1}}. 
\end{equation*}
The divided difference operators $\partial_w^y$ for $w\in S_n$ are defined as follows. Let $ s_{a_1}\cdots s_{a_k}$ be any reduced word for $w$. Then, \begin{equation*}\de_{w}^y = \de_{a_1}^y\cdots \de_{a_k}^y.\end{equation*} The double Schubert polynomials are then defined as follows:
\begin{equation}\g_w(x,y) = \begin{cases}
     \prod\limits_{i+j\leq n}(x_i-y_j), & \text{if }w = w_0,\\
    (-1)^{\ell(w_{0})-\ell(w)}\de_{ww_0}^y\g_{w_0}(x,y), & \text{otherwise}.
\end{cases}\end{equation}
Specializing the $y$ variables to $0$ recovers the Schubert polynomials. 
%\begin{theorem}[Monk's rule for double Schubert polynomials]
%     For any $k$ and any permutation $w$,
% \begin{equation}
% \g_{s_k}(x,y)\g_{w}(x,y) = \sum_{\substack{a \leq k < b,\\ \ell(w  t_{ab}) = \ell(w)+1}}\g_{w\cdot t_{ab}}(x,y)+\sum_{i=1}^k(y_{w(i)}-y_i)\g_w(x, y).\end{equation}
% \end{theorem}

\subsection{Quantum double Schubert polynomials}
As in \cite{quantum}, we define $E_i^k(x_1, \dots, x_k)$ to be the coefficient of $\lambda^i$ in the characteristic polynomial $\det(1+\lambda G_k)$ where \begin{equation*}G_k = \begin{bmatrix}
    x_1 & q_1 & 0 &\dots & 0\\
    -1 & x_2 & q_2 &\dots & 0\\
    0 & -1 & x_3 &\dots & 0\\
    \vdots & \vdots &\vdots&\ddots & \vdots \\
    0 & 0 & 0 &\dots & x_k
\end{bmatrix}.\end{equation*}
These $E_i^k$ are quantum analogues of the elementary symmetric polynomials $e_i^k$'s which are defined by
\[e_i^k := \sum\limits_{1\le a_1 < \dots < a_i\le k}x_{a_1}\cdots x_{a_i}.\]
Indeed, specializing all the $q$ variables to $0$ turns $E_i^k$ into $e_i^k$. Fomin, Gelfand and Postnikov \cite{quantum} originally defined quantum Schubert polynomials by passing the Schubert polynomials through a quantization map that sends $e_i^k$ to $E_i^k$. There is, however, a way to define the quantum double Schubert polynomials via divided difference operators \cite{cf}. For the longest permutation $w_0$, we have \begin{equation}\g_{w_0}^q(x, y):=\prod_{k = 1}^{n-1} E_k^k(x_1-y_{n-k}, \dots, x_k-y_{n-k}),\end{equation}
and, for any permutation $w$, we have \begin{equation}
\g_{w}^q(x, y) = (-1)^{\ell(w_0)-\ell(w)}\partial_{ww_0}^y\g_{w_0}^q(x,y).    
\end{equation}
Specializing the $q$ variables to $0$ recovers the double Schubert polynomials while specializing the $y$ variables to $0$ recovers the quantum Schubert polynomials. Setting both $y$ and $q$ variables to $0$ recovers the Schubert polynomials. 
\begin{theorem}[{Monk's rule for quantum double Schubert polynomials \cite{qbsm}}] \label{thm:monk}
    For any $k$ and any permutation $w$,
\begin{align*}
    \g_{s_k}^q(x, y)\g_{w}^q(x,y) &= \sum_{\substack{a \leq k < b,\\ \ell(w  t_{ab}) = \ell(w)+1}}\g_{w t_{ab}}^q(x,y)+\sum_{\substack{c \leq k < d,\\ \ell(w t_{cd}) = \ell(w)-\ell(t_{cd})}}q_{cd}\g_{w t_{cd}}^q(x,y)\\
    &+\sum_{i=1}^k(y_{w(i)}-y_i)\g_w^q(x, y),
\end{align*}
where $q_{cd}:= q_{c}q_{c+1}\dots q_{d-1}$.
\end{theorem}

\section{Quantum bumpless pipe dreams}\label{quantumbpd}
We will now define \textit{quantum bumpless pipe dreams}, the central objects of this paper.
\begin{definition}\label{def:qbpd}
A \textit{quantum bumpless pipe dream} (QBPD) is a tiling of an $n\times n$ grid filled with tiles
\begin{center}
    \begin{tikzpicture}[scale=\figscale]
\draw (0, 0) grid (1,1);
\end{tikzpicture}\:\:
\begin{tikzpicture}[scale=\figscale]
\draw (0, 0) grid (1,1);
\draw[blue](1,0.5) arc (90:180:0.5);
\end{tikzpicture}\:\:
\begin{tikzpicture}[scale=\figscale]
\draw (0, 0) grid (1,1);
\draw[blue](0.5,0)--(0.5,1);
\draw[blue](0,0.5)--(1,0.5);
\end{tikzpicture}\:\:
\begin{tikzpicture}[scale=\figscale]
\draw (0, 0) grid (1,1);
\draw[blue](0,0.5) arc (-90:0:0.5);
\end{tikzpicture}\:\:
\begin{tikzpicture}[scale=\figscale]
\draw (0, 0) grid (1,1);
\draw[blue](0.5,0)--(0.5,1);
\end{tikzpicture}\:\:
\begin{tikzpicture}[scale=\figscale]
\draw (0, 0) grid (1,1);
\draw[blue](0,0.5)--(1,0.5);
\end{tikzpicture}\:\:
\begin{tikzpicture}[scale=\figscale]
\draw (0, 0) grid (1,1);
\draw[blue](0.5,0) arc (0:90:0.5);
\end{tikzpicture}\:\:
\begin{tikzpicture}[scale=\figscale]
\draw (0, 0) grid (1,1);
\draw[blue](0.5,1) arc (180:270:0.5);
\end{tikzpicture}\:\:
\begin{tikzpicture}[scale=\figscale]
\draw (0, 0) grid (1,1);
\draw[fill=white,draw=black](0,0) rectangle (1,2);
\end{tikzpicture}
\end{center}

so that
\begin{itemize}
    \item the tiling forms $n$ pipes;
    \item each pipe starts horizontally at the right edge of the grid and ends vertically at the bottom edge of the grid;    \item the pipes only move upward, downward, or leftward (but not rightward) when moving from the right edge to the bottom edge;
    \item no two pipes cross more than once.
\end{itemize}

 The last tile is a $2\times 1$ domino tile, which occupies two vertically adjacent empty squares in the grid. A (non-quantum) \textit{bumpless pipe dream} (BPD) (as defined in \cite{Lam_2021}), is a QBPD in which the last three tiles above (\begin{tikzpicture}[scale=\inlinescale]
\draw (0, 0) grid (1,1);
\draw[blue](0.5,0) arc (0:90:0.5);
\end{tikzpicture}, 
\begin{tikzpicture}[scale=\inlinescale]
\draw (0, 0) grid (1,1);
\draw[blue](0.5,1) arc (180:270:0.5);
\end{tikzpicture}, and the domino tile) do not appear, so in a BPD pipes only move downward and leftward. %The last condition that no two pipes cross more than once is the \textit{reducedness} condition. Throughout this paper, all quantum bumpless pipe dreams are reduced %, though in some proofs we may use the term ``reduced quantum bumpless pipe dream'' to stress that the reducedness condition holds when it is not immediately clear why.
\end{definition}
\begin{figure}[ht!]
    \centering
    \begin{tikzpicture}[scale=\figscale]
\draw (0, 0) grid (5,5);
\draw[blue](1,4.5) arc (90:180:0.5);
\draw[blue](0.5,3)--(0.5,4);
\draw[blue](0.5,2)--(0.5,3);
\draw[blue](0.5,1)--(0.5,2);
\draw[blue](0.5,0)--(0.5,1);
\draw[blue](1,4.5)--(2,4.5);
\draw[blue](2,3.5) arc (90:180:0.5);
\draw[blue](1.5,3) arc (180:270:0.5);
\draw[blue](2,1.5) arc (90:180:0.5);
\draw[blue](1.5,0)--(1.5,1);
\draw[blue](2.5,4) arc (0:90:0.5);
\draw[blue](2,3.5) arc (-90:0:0.5);
\draw[blue](2,2.5)--(3,2.5);
\draw[blue](2,1.5)--(3,1.5);
\draw[blue](3,0.5) arc (90:180:0.5);
\draw[blue](4,4.5) arc (90:180:0.5);
\draw[blue](3.5,3)--(3.5,4);
\draw[blue](3.5,2)--(3.5,3);
\draw[blue](3,2.5)--(4,2.5);
\draw[blue](3.5,1)--(3.5,2);
\draw[blue](3,1.5)--(4,1.5);
\draw[blue](3.5,0)--(3.5,1);
\draw[blue](3,0.5)--(4,0.5);
\draw[blue](4,4.5)--(5,4.5);
\draw[blue](5,3.5) arc (90:180:0.5);
\draw[blue](4.5,2)--(4.5,3);
\draw[blue](4,2.5)--(5,2.5);
\draw[blue](4.5,1)--(4.5,2);
\draw[blue](4,1.5)--(5,1.5);
\draw[blue](4.5,0)--(4.5,1);
\draw[blue](4,0.5)--(5,0.5);
\end{tikzpicture}
    \caption{A non-example of a QBPD}
    \label{fig:nonexqbpd}
\end{figure}
\begin{example}
    \Cref{fig:nonexqbpd} shows a non-example of a QBPD. The pipe starting on row $3$ moves rightward in the tiles $(2,2)$ and $(2,3)$, which violates \Cref{def:qbpd}. See \Cref{fig:qbpd} for examples of valid QBPD.
\end{example}
\begin{definition}\label{def:bwt}
    The \textit{binomial weight} for a QBPD $P$, denoted $\bwt(P)$, is the product of weights contributed by the following tiles:
\end{definition}

\begin{itemize}
    \item An empty tile \begin{tikzpicture}[scale=\inlinescale]
\draw (0, 0) grid (1,1);
\end{tikzpicture} on row $i$ and column $j$ contributes $x_i-y_j$;
    \item A domino tile
    \begin{tikzpicture}[scale=\inlinescale]
\draw (0, 0) grid (1,1);
\draw[fill=white,draw=black](0,0) rectangle (1,2);
\end{tikzpicture}
    whose upper cell is on row $i$ contributes $q_i$;
    \item A cross tile \begin{tikzpicture}[scale=\inlinescale]
\draw (0, 0) grid (1,1);
\draw[blue](0.5,0)--(0.5,1);
\draw[blue](0,0.5)--(1,0.5);
\end{tikzpicture} on row $i$ where the vertical strand moves upwards contributes $q_i$;
    \item A southwest elbow \begin{tikzpicture}[scale=\inlinescale]
\draw (0, 0) grid (1,1);
\draw[blue](0.5,0) arc (0:90:0.5);
\end{tikzpicture} on row $i$ contributes $-q_i$;
    \item A vertical tile \begin{tikzpicture}[scale=\inlinescale]
\draw (0, 0) grid (1,1);
\draw[blue](0.5,0)--(0.5,1);
\end{tikzpicture} on row $i$ where the strand moves upward contributes $-q_i$.
\end{itemize}
In other words, let $P(i, j)$ denote the cell on row $i$ and column $j$, and let \begin{align*}E(P) &:= \{(i, j):P(i,j)\text{ is a single empty cell}\},\\
Q(P) &:= \{(i, j):P(i, j)\text{ is the upper cell of a domino or a \begin{tikzpicture}[scale=\inlinescale]
\draw (0, 0) grid (1,1);
\draw[blue](0.5,0)--(0.5,1);
\draw[blue](0,0.5)--(1,0.5);
\end{tikzpicture} tile }\\
&\text{in which the vertical strand moves upward}\}, \\
NQ(P) &:= \{(i, j):P(i, j)\text{ is a \begin{tikzpicture}[scale=\inlinescale]
\draw (0, 0) grid (1,1);
\draw[blue](0.5,0) arc (0:90:0.5);
\end{tikzpicture} tile or a \begin{tikzpicture}[scale=\inlinescale]
\draw (0, 0) grid (1,1);
\draw[blue](0.5,0)--(0.5,1);
\end{tikzpicture} tile }\\
&\text{in which the strand moves upward}\}.\end{align*}
Then, \begin{align*}\bwt(P)&:= \prod_{(i, j)\in E(P)}(x_i-y_j)\prod_{(i, j)\in Q(P)}q_i\prod_{(i, j)\in NQ(P)}(-q_i)\notag\\
&= (-1)^{|NQ(P)|}\prod_{(i, j)\in E(P)}(x_i-y_j)\prod_{(i, j)\in Q(P)\cup NQ(P)}q_i \,.\end{align*}

A QBPD $P$ is said to be associated with a permutation $w$ if the pipe starting on the right on row $i$ ends up on column $w(i)$. 

Let $\QBPD(w)$ denote the set of QBPDs associated to $w$. Our main result is the following:
\begin{theorem}\label{thm:qdform}
The quantum double Schubert polynomial indexed by $w \in S_n$ is the sum of binomial weights of all QBPDs associated to $w$:
\begin{equation*}
    \g^q_w(x,y) = \sum_{P\in \QBPD(w)}\bwt(P)\,.
\end{equation*}
\end{theorem}
\begin{figure}[ht]
    \centering
\begin{tikzpicture}[scale=\figscale]
\node at (0.5,-0.3){1};
\node at (1.5,-0.3){2};
\node at (2.5,-0.3){3};
\node at (3.5,-0.3){4};
\node at (4.3,0.5){4};
\node at (4.3,1.5){3};
\node at (4.3,2.5){2};
\node at (4.3,3.5){1};
\draw (0, 0) grid (4,4);
\draw[blue](1,1.5) arc (90:180:0.5);
\draw[blue](0.5,0)--(0.5,1);
\draw[blue](2,2.5) arc (90:180:0.5);
\draw[blue](1.5,1)--(1.5,2);
\draw[blue](1,1.5)--(2,1.5);
\draw[blue](1.5,0)--(1.5,1);
\draw[blue](2,2.5)--(3,2.5);
\draw[blue](2,1.5)--(3,1.5);
\draw[blue](3,0.5) arc (90:180:0.5);
\draw[blue](4,3.5) arc (90:180:0.5);
\draw[blue](3.5,2)--(3.5,3);
\draw[blue](3,2.5)--(4,2.5);
\draw[blue](3.5,1)--(3.5,2);
\draw[blue](3,1.5)--(4,1.5);
\draw[blue](3.5,0)--(3.5,1);
\draw[blue](3,0.5)--(4,0.5);
\end{tikzpicture}
\begin{tikzpicture}[scale=\figscale]
\node at (0.5,-0.3){1};
\node at (1.5,-0.3){2};
\node at (2.5,-0.3){3};
\node at (3.5,-0.3){4};
\node at (4.3,0.5){4};
\node at (4.3,1.5){3};
\node at (4.3,2.5){2};
\node at (4.3,3.5){1};
\draw (0, 0) grid (4,4);
\draw[fill=white,draw=black](0,2) rectangle (1,4);
\draw[blue](1,1.5) arc (90:180:0.5);
\draw[blue](0.5,0)--(0.5,1);
\draw[blue](2,2.5) arc (90:180:0.5);
\draw[blue](1.5,1)--(1.5,2);
\draw[blue](1,1.5)--(2,1.5);
\draw[blue](1.5,0)--(1.5,1);
\draw[blue](2,2.5)--(3,2.5);
\draw[blue](2,1.5)--(3,1.5);
\draw[blue](3,0.5) arc (90:180:0.5);
\draw[blue](4,3.5) arc (90:180:0.5);
\draw[blue](3.5,2)--(3.5,3);
\draw[blue](3,2.5)--(4,2.5);
\draw[blue](3.5,1)--(3.5,2);
\draw[blue](3,1.5)--(4,1.5);
\draw[blue](3.5,0)--(3.5,1);
\draw[blue](3,0.5)--(4,0.5);
\end{tikzpicture}
\begin{tikzpicture}[scale=\figscale]
\node at (0.5,-0.3){1};
\node at (1.5,-0.3){2};
\node at (2.5,-0.3){3};
\node at (3.5,-0.3){4};
\node at (4.3,0.5){4};
\node at (4.3,1.5){3};
\node at (4.3,2.5){2};
\node at (4.3,3.5){1};
\draw (0, 0) grid (4,4);
\draw[blue](1,1.5) arc (90:180:0.5);
\draw[blue](0.5,0)--(0.5,1);
\draw[blue](2,3.5) arc (90:180:0.5);
\draw[blue](1.5,2)--(1.5,3);
\draw[blue](1.5,1)--(1.5,2);
\draw[blue](1,1.5)--(2,1.5);
\draw[blue](1.5,0)--(1.5,1);
\draw[blue](2.5,3) arc (0:90:0.5);
\draw[blue](2.5,3) arc (180:270:0.5);
\draw[blue](2,1.5)--(3,1.5);
\draw[blue](3,0.5) arc (90:180:0.5);
\draw[blue](4,3.5) arc (90:180:0.5);
\draw[blue](3.5,2)--(3.5,3);
\draw[blue](3,2.5)--(4,2.5);
\draw[blue](3.5,1)--(3.5,2);
\draw[blue](3,1.5)--(4,1.5);
\draw[blue](3.5,0)--(3.5,1);
\draw[blue](3,0.5)--(4,0.5);
\end{tikzpicture}
\begin{tikzpicture}[scale=\figscale]
\node at (0.5,-0.3){1};
\node at (1.5,-0.3){2};
\node at (2.5,-0.3){3};
\node at (3.5,-0.3){4};
\node at (4.3,0.5){4};
\node at (4.3,1.5){3};
\node at (4.3,2.5){2};
\node at (4.3,3.5){1};
\draw (0, 0) grid (4,4);
\draw[fill=white,draw=black](0,2) rectangle (1,4);
\draw[blue](1,1.5) arc (90:180:0.5);
\draw[blue](0.5,0)--(0.5,1);
\draw[blue](2,3.5) arc (90:180:0.5);
\draw[blue](1.5,2)--(1.5,3);
\draw[blue](1.5,1)--(1.5,2);
\draw[blue](1,1.5)--(2,1.5);
\draw[blue](1.5,0)--(1.5,1);
\draw[blue](2.5,3) arc (0:90:0.5);
\draw[blue](2.5,3) arc (180:270:0.5);
\draw[blue](2,1.5)--(3,1.5);
\draw[blue](3,0.5) arc (90:180:0.5);
\draw[blue](4,3.5) arc (90:180:0.5);
\draw[blue](3.5,2)--(3.5,3);
\draw[blue](3,2.5)--(4,2.5);
\draw[blue](3.5,1)--(3.5,2);
\draw[blue](3,1.5)--(4,1.5);
\draw[blue](3.5,0)--(3.5,1);
\draw[blue](3,0.5)--(4,0.5);
\end{tikzpicture}
\begin{tikzpicture}[scale=\figscale]
\node at (0.5,-0.3){1};
\node at (1.5,-0.3){2};
\node at (2.5,-0.3){3};
\node at (3.5,-0.3){4};
\node at (4.3,0.5){4};
\node at (4.3,1.5){3};
\node at (4.3,2.5){2};
\node at (4.3,3.5){1};
\draw (0, 0) grid (4,4);
\draw[blue](1,3.5) arc (90:180:0.5);
\draw[blue](0.5,2)--(0.5,3);
\draw[blue](0.5,1)--(0.5,2);
\draw[blue](0.5,0)--(0.5,1);
\draw[blue](1,3.5)--(2,3.5);
\draw[blue](2,2.5) arc (90:180:0.5);
\draw[blue](1.5,1)--(1.5,2);
\draw[blue](1.5,0)--(1.5,1);
\draw[blue](2.5,3) arc (0:90:0.5);
\draw[blue](2.5,2)--(2.5,3);
\draw[blue](2,2.5)--(3,2.5);
\draw[blue](2.5,2) arc (180:270:0.5);
\draw[blue](3,0.5) arc (90:180:0.5);
\draw[blue](4,3.5) arc (90:180:0.5);
\draw[blue](3.5,2)--(3.5,3);
\draw[blue](3,2.5)--(4,2.5);
\draw[blue](3.5,1)--(3.5,2);
\draw[blue](3,1.5)--(4,1.5);
\draw[blue](3.5,0)--(3.5,1);
\draw[blue](3,0.5)--(4,0.5);
\end{tikzpicture}
    \caption{QBPDs for $4213$}
    \label{fig:qbpd}
\end{figure}
\begin{example}From \Cref{fig:qbpd}, we have\begin{align*}
\g^q_{4213}(x,y) &= (x_1-y_1)(x_1-y_2)(x_1-y_3)(x_2-y_1)+ q_1(x_1-y_2)(x_1-y_3)\\&+(x_1-y_1)(x_2-y_1)(-q_1) + q_1(-q_1)+(-q_1)q_2\,.\end{align*}
\end{example}
As a corollary, we have the following formula.

\begin{corollary}\label{ds}
The quantum Schubert polynomial indexed by $w \in S_n$ is the sum of monomial weights of all QBPDs associated to $w$:
\begin{equation*}\g^q_w(x) = \sum_{P\in \QBPD(w)}\wt(P)\end{equation*}
where \begin{equation*}\wt(P) := \prod_{(i, j)\in E(P)}x_i\prod_{(i, j)\in Q(P)}q_i\prod_{(i, j)\in NQ(P)}(-q_i)\,.\end{equation*}
\end{corollary}
 \subsection{Droop moves and lift moves}\label{droopandlift}

Droop moves on bumpless pipe dreams are defined in  \cite{Lam_2021}. They are moves of the form illustrated in \Cref{fig:droop}. Droop moves can be extended to the QBPD setting. We allow a droop move only if the result is a valid and reduced QBPD.
\begin{figure}[ht!]
    \centering
    \begin{tikzpicture}[scale=\figscale]
    \draw (0, 0) grid (3,4);
    \draw[blue, ultra thick] (0.5, 1)--(0.5, 3); 
    \draw[blue] (0.5, 0)--(0.5, 1);
    \draw[blue](2,3.5) arc (-90:0:0.5);
    \draw[blue, ultra thick] (1,3.5) arc (90:180:0.5);
    \draw[blue, ultra thick] (1, 3.5)--(2, 3.5);
    \draw[blue] (2, 3.5)--(3, 3.5);
    \draw[blue](0,0.5) arc (-90:0:0.5);
    \draw[->] (3.5,2)--(4.5,2);
    \draw (5, 0) grid (8,4);
    \draw[blue, ultra thick] (7.5, 1)--(7.5, 3); 
    \draw[blue] (7.5, 3)--(7.5, 4);
    \draw[blue](6,0.5) arc (90:180:0.5);
    \draw[blue, ultra thick] (7,0.5) arc (-90:0:0.5);
    \draw[blue, ultra thick] (6, 0.5)--(7, 0.5);
    \draw[blue] (5, 0.5)--(6, 0.5);
    \draw[blue](8,3.5) arc (90:180:0.5);
\end{tikzpicture}
\caption{A droop move (light color indicates possibilities)}
    \label{fig:droop}
\end{figure}

In \cite{Lam_2021}, Lam, Lee, and Shimozono proved that any bumpless pipe dream of a given permutation can be obtained from the Rothe diagram by a sequence of droops. To generate all QBPDs, we introduce other moves called \textit{lift moves}.

The lift moves are moves in which a horizontal segment of a strand is ``lifted up'' into a detour that goes up, moves left, and then moves back down. \Cref{fig:lift} shows an example of a lift move. There might be other unpictured pipes in the picture as long as the result is a valid and reduced QBPD.
\begin{figure}[ht]
    \centering
    \begin{tikzpicture}[scale=\figscale]
    \draw (0, 0) grid (4,3);
    \draw[blue, ultra thick] (6.5, 1)--(6.5, 2); 
    \draw[blue] (0, 0.5)--(1, 0.5);
    \draw[blue](3.5,0) arc (0:90:0.5);
    \draw[blue, ultra thick] (7,2.5) arc (90:180:0.5);
    \draw[blue, ultra thick] (1, 0.5)--(3, 0.5);
    \draw[blue] (3, 0.5)--(4, 0.5);
    \draw[blue](0.5,0) arc (180:90:0.5);
    \draw[->] (4.5,1.5)--(5.5,1.5);
    \draw (6, 0) grid (10,3);
    \draw[blue, ultra thick] (9.5, 1)--(9.5, 2); 
    \draw[blue] (6.5, 0)--(6.5, 1);
    \draw[blue](6,0.5) arc (-90:0:0.5);
    \draw[blue, ultra thick] (9,2.5) arc (90:0:0.5);
    \draw[blue, ultra thick] (7, 2.5)--(9, 2.5);
    \draw[blue] (9.5, 0)--(9.5, 1);
    \draw[blue](10, 0.5) arc (-90:-180:0.5);
\end{tikzpicture}
\caption{A lift move (light color indicates possibilities)}
    \label{fig:lift}
\end{figure}

A QBPD is said to be \textit{unpaired} if it has no domino tile. To generate all QBPDs, we can generate all unpaired QBPDs and find all ways to pair empty boxes into dominos. Note that droops moves and lift moves preserve the permutation associated to the QBPD. %It turns out %(\Cref{lem:droopliftgen} below) 
%that all unpaired QBPD for a permutation can be generated from the Rothe diagram using only droop and lift moves. %To prove this, we first need some technical definitions and lemmas.

% \begin{definition}\label{def:simple}
%     A \textit{leftmost region} is a region given by all columns to the left of a given column. A \textit{simple region} is a region in which the tiles containing pipes are all \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](0,0.5)--(1,0.5);
% \end{tikzpicture}, \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](0.5,0)--(0.5,1);
% \end{tikzpicture}, \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](0.5,0)--(0.5,1);
% \draw[blue](0,0.5)--(1,0.5);
% \end{tikzpicture}, or \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](1,0.5) arc (90:180:0.5);
% \end{tikzpicture} tiles. A strand is said to have the \textit{simple form} if it goes straight left, takes a \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](1,0.5) arc (90:180:0.5);
% \end{tikzpicture} turn, then goes straight down.
% \end{definition}
% \begin{lemma}\label{simple}
%     If \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](0,0.5) arc (-90:0:0.5);
% \end{tikzpicture} and \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](0.5,0) arc (0:90:0.5);
% \end{tikzpicture} tiles do not exist in a leftmost region of a QBPD, then that region of the QBPD must be simple. Furthermore, every strand has the simple form in that leftmost region.

%  \begin{proof}
% If \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](0,0.5) arc (-90:0:0.5);
% \end{tikzpicture} and \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](0.5,0) arc (0:90:0.5);
% \end{tikzpicture} do not exist in some leftmost region of the QBPD, then there is also no \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](0.5,1) arc (180:270:0.5);
% \end{tikzpicture} tile, since if a \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](0.5,1) arc (180:270:0.5);
% \end{tikzpicture} exists, then the pipe of that tile must move upward and eventually take a left to produce a \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](0.5,0) arc (0:90:0.5);
% \end{tikzpicture} tile within the region. Thus, the only possible elbow tile is the \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](1,0.5) arc (90:180:0.5);
% \end{tikzpicture} tile, and all strands must have the ``simple'' form in which it only has a single turn. In particular, if the region is the entire QBPD, then the entire QBPD has the ``simple'' form and it is exactly the Rothe diagram.
%  \end{proof}
% \end{lemma}
\begin{lemma}\label{lem:droopliftgen}
    All unpaired QBPDs can be generated from the Rothe diagram using a sequence of droop and lift moves.
%     \begin{proof}
%         It suffices to show that, given any QBPD, we can use a sequence of valid inverse droop and inverse lift moves to get back to the Rothe diagram. Given a QBPD that is not the Rothe diagram, we find the leftmost \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](0,0.5) arc (-90:0:0.5);
% \end{tikzpicture} or \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](0.5,0) arc (0:90:0.5);
% \end{tikzpicture}
%  tile. If no such tile exists, then the QBPD is just the Rothe diagram by \Cref{simple}. Otherwise, if there are ties, take the topmost one.
 
% Now look at the region strictly to the left of that leftmost chosen tile, this region is free of \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](0,0.5) arc (-90:0:0.5);
% \end{tikzpicture} and \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](0.5,0) arc (0:90:0.5);
% \end{tikzpicture}
%  tiles by definition, and thus is simple by \Cref{simple}. Let $R$ denotes this simple region. By \Cref{simple}, every strand in $R$ has the simple form. 
%  %\begin{itemize}
 
%      We first consider the case where the leftmost chosen tile is \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](0,0.5) arc (-90:0:0.5);
% \end{tikzpicture}. Call this tile $e$. Then, tracing this pipe to the left, it eventually takes a \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](1,0.5) arc (90:180:0.5);
% \end{tikzpicture} turn---call this elbow $a$---and go straight down since this is $R$. Tracing this pipe up from tile $e$, it eventually takes a \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](1,0.5) arc (90:180:0.5);
% \end{tikzpicture} turn---call this elbow $b$. Consider the rectangle bounded by $a,b, e$. We claim we can undroop $e$ into the topleft corner of this rectangle. Call this topleft corner $c$. Note that every other pipe in $R$ has the simple form, so its behavior in $R$ is determined by the location of its \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](1,0.5) arc (90:180:0.5);
% \end{tikzpicture} tile, which cannot be on the row of $b$ or column of $a$ (since the pipe would run into either $a$ or $b$ otherwise). This means there are no pipes moving horizontally in the $b\to c$ horizontal segment, there are no pipes moving vertically in the $c\to a$ horizontal segment, and $c$ is empty. Thus, the undroop does not conflict with other pipes. After undrooping, the changed strand is simple in $R$ and all other strands are still simple, so in $R$, they cross at most once. There could be at most one other crossing since crossings out of $R$ exist before the undroop. Then if any two pipes cross twice after the undroop, they must cross once before the undroop, which is a contradiction since the parity of the number of crossings cannot change.

% Now consider the case where the leftmost chosen tile is \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](0.5,0) arc (0:90:0.5);
% \end{tikzpicture}. Call this tile $e$.
% In this case, trace the pipe down and find the \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](0.5,1) arc (180:270:0.5);
% \end{tikzpicture} tile and call this tile $a$. Trace the pipe to the left and find the \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](1,0.5) arc (90:180:0.5);
% \end{tikzpicture} turn. Call this tile $b$. After $b$, it goes straight down since this is in $R$. Make the inverse lift by unlifting the $e\to b$ horizontal segment down to the row of $a$, i.e., making the pipe at $a$ move to the left until it hits the column of $b$ instead of moving up to $e$. As in the previous case, other pipes in $R$ have the simple form and the \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](1,0.5) arc (90:180:0.5);
% \end{tikzpicture} tile of another strand in $R$ cannot be on the row of $a$. So, the inverse lift does not conflict with other pipes. After inverse lifting, the strand is simple in $R$, so a similar argument to the previous case shows it is reduced.
%  %\end{itemize}
 
%  After each inverse droop or inverse lift above, the number of \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](0.5,0) arc (0:90:0.5);
% \end{tikzpicture} or \begin{tikzpicture}[scale=\inlinescale]
% \draw (0, 0) grid (1,1);
% \draw[blue](0,0.5) arc (-90:0:0.5);
% \end{tikzpicture} tiles decreases, so we eventually arrive at the Rothe diagram.
%     \end{proof}
\end{lemma}
\subsection{Stability}\label{stability}
Given a QBPD of $w\in S_n$, we can think of $w$ as being in $S_{n+1}$. In terms of QBPDs, we can extend an $n\times n$ QBPD $P$ to a $(n+1)\times (n+1)$ QBPD as in \Cref{fig:sta}. 
\begin{figure}[ht!]
    \centering
    \begin{tikzpicture}[scale=\figscale]
        \draw (0,0) rectangle (3,3);
        \node at (1.5,1.5) {$P$};
        \draw[->] (3.5,1.25)--(4.5,1.25);
        \draw (5,-1) grid (9, 3);
        \draw[fill=white] (5,0) rectangle (8,3);
        \node at (6.5,1.5) {$P$};
        \draw[blue, ultra thick] (7.5,0)--(7.5,-1); 
        \draw[blue, ultra thick] (5.5,0)--(5.5,-1); 
        \draw[blue, ultra thick] (6.5,0)--(6.5,-1); 
        \draw[blue, ultra thick] (8,0.5)--(9,0.5); 
        \draw[blue, ultra thick] (8,1.5)--(9,1.5);  
        \draw[blue, ultra thick] (8,2.5)--(9,2.5); 
        \draw[blue, ultra thick] (8.5,-1) arc (180:90:0.5);
    \end{tikzpicture}
    \caption{Extending a $n\times n$ QBPD to a $(n+1)\times (n+1)$ QBPD}
    \label{fig:sta}
\end{figure}

The reverse is also possible:
\begin{lemma}\label{lem:stab}
    Given $w\in S_{n+1}$ such that $w(n+1)= n+1$, for any $(n+1)\times (n+1)$ QBPD of $w$, we can restrict it to an $n\times n$ QBPD of $w$ restricted to $S_n$.
\end{lemma}
\subsection{Overview of the proof of \texorpdfstring{\Cref{thm:qdform}}{Theorem 3.4}}\label{proof33}
%From \Cref{thm:monk}, one can obtain a defining transition equation by specializing at appropriate special case. 
We prove \Cref{thm:qdform} by showing the quantity $\sum\limits_{P\in \QBPD(w)}\bwt(P)$
satisfies a defining transition equation of the quantum double Schubert polynomials obtained by taking appropriate special cases of \Cref{thm:monk}. 
For example, the transition equation for $3241$ is as follows:$
%\begin{equation*}
    \g_{3241} = (x_3-y_1)\g_{3214} + q_{2}\g_{3124}+q_{1}q_2\g_{1234}\,.
%\end{equation*}
$

In general, the transition equation has four families of terms on the right hand side. For details see \cite[Proposition 3.14]{le2024qbpd}. We show the quantity satisfies the transition equation by splitting the QBPDs of the permutations on the left hand side into four sets, and constructing four bijections between those four sets with the four families on the right hand side. However, portions of the last the family on the right hand side will remain, but they cancel out using another family of bijections.
%In general, the transition equation has four terms on the right hand side, for details see Proposition 3.11 of \cite{le2024qbpd}.
%We do this by spliting the QBPDs of the permutations on the left hand side into four sets, and constructing four bijections between those fours with the four terms on the right hand side. However, portions of the last the term in the right hand side will remains, and they will be shown to cancel out using another family of bijections.
These bijections are constructed for permutations $\pi$ with $\pi(n)\ne n$ which we can assume without loss of generality by \Cref{lem:stab}. 
\begin{figure}[ht!]
    \centering
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\draw[blue](2.5,2)--(2.5,3);
\draw[blue](2,2.5)--(3,2.5);
\draw[blue](2.5,1)--(2.5,2);
\draw[blue](2.5,0)--(2.5,1);
\draw[blue](2,0.5)--(3,0.5);
\draw[blue](3,3.5)--(4,3.5);
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\draw[blue](2,2.5)--(3,2.5);
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\draw[blue](2,1.5)--(3,1.5);
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\draw[blue](3,3.5)--(4,3.5);
\draw[blue](3,2.5)--(4,2.5);
\draw[blue](3,1.5)--(4,1.5);
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\end{scope}
\end{tikzpicture}\quad\quad
\begin{tikzpicture}[scale=\figscale]
\node at (0.5,-0.3){1};
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\draw[blue](3,3.5) arc (90:180:0.5);
\draw[blue](2.5,2)--(2.5,3);
\draw[blue](2,2.5)--(3,2.5);
\draw[blue](2.5,1)--(2.5,2);
\draw[blue](2.5,0)--(2.5,1);
\draw[blue](2,0.5)--(3,0.5);
\draw[blue](3,3.5)--(4,3.5);
\draw[blue](3,2.5)--(4,2.5);
\draw[blue](4,1.5) arc (90:180:0.5);
\draw[blue](3.5,0)--(3.5,1);
\draw[blue](3,0.5)--(4,0.5);
\draw[->] (4.7,2)--(5.7,2);
\begin{scope}[shift={(6.2,0)}]
\node at (0.5,-0.3){1};
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\draw[blue](1,1.5)--(2,1.5);
\draw[blue](1.5,0)--(1.5,1);
\draw[blue](3,3.5) arc (90:180:0.5);
\draw[blue](2.5,2)--(2.5,3);
\draw[blue](2,2.5)--(3,2.5);
\draw[blue](2.5,1)--(2.5,2);
\draw[blue](2,1.5)--(3,1.5);
\draw[blue](2.5,0)--(2.5,1);
\draw[blue](3,3.5)--(4,3.5);
\draw[blue](3,2.5)--(4,2.5);
\draw[blue](3,1.5)--(4,1.5);
\draw[blue](4,0.5) arc (90:180:0.5);
\end{scope}
\end{tikzpicture}

\begin{tikzpicture}[scale=\figscale]
\node at (0.5,-0.3){1};
\node at (1.5,-0.3){2};
\node at (2.5,-0.3){3};
\node at (3.5,-0.3){4};
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\node at (4.3,3.5){1};
\draw (0, 0) grid (4,4);
\draw[fill=white,draw=black](0,1) rectangle (1,3);
\draw[blue](1,0.5) arc (90:180:0.5);
\draw[blue](2,2.5) arc (90:180:0.5);
\draw[blue](1.5,1)--(1.5,2);
\draw[blue](1.5,0)--(1.5,1);
\draw[blue](1,0.5)--(2,0.5);
\draw[blue](3,3.5) arc (90:180:0.5);
\draw[blue](2.5,2)--(2.5,3);
\draw[blue](2,2.5)--(3,2.5);
\draw[blue](2.5,1)--(2.5,2);
\draw[blue](2.5,0)--(2.5,1);
\draw[blue](2,0.5)--(3,0.5);
\draw[blue](3,3.5)--(4,3.5);
\draw[blue](3,2.5)--(4,2.5);
\draw[blue](4,1.5) arc (90:180:0.5);
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\begin{scope}[shift={(6.2,0)}]
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\draw[blue](2.5,0)--(2.5,1);
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\draw[blue](3,1.5)--(4,1.5);
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\end{scope}
\end{tikzpicture}\quad\quad
\begin{tikzpicture}[scale=\figscale]
\node at (0.5,-0.3){1};
\node at (1.5,-0.3){2};
\node at (2.5,-0.3){3};
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\draw[blue](0.5,2)--(0.5,3);
\draw[blue](0.5,1)--(0.5,2);
\draw[blue](0.5,0)--(0.5,1);
\draw[blue](1.5,3) arc (0:90:0.5);
\draw[blue](1.5,3) arc (180:270:0.5);
\draw[blue](2,1.5) arc (90:180:0.5);
\draw[blue](1.5,0)--(1.5,1);
\draw[blue](3,3.5) arc (90:180:0.5);
\draw[blue](2.5,2)--(2.5,3);
\draw[blue](2,2.5)--(3,2.5);
\draw[blue](2.5,1)--(2.5,2);
\draw[blue](2,1.5)--(3,1.5);
\draw[blue](2.5,0)--(2.5,1);
\draw[blue](3,3.5)--(4,3.5);
\draw[blue](3,2.5)--(4,2.5);
\draw[blue](3,1.5)--(4,1.5);
\draw[blue](4,0.5) arc (90:180:0.5);
\draw[->] (4.7,2)--(5.7,2);
\begin{scope}[shift={(6.2,0)}]
\node at (0.5,-0.3){1};
\node at (1.5,-0.3){2};
\node at (2.5,-0.3){3};
\node at (3.5,-0.3){4};
\node at (4.3,0.5){4};
\node at (4.3,1.5){3};
\node at (4.3,2.5){2};
\node at (4.3,3.5){1};
    \draw (0, 0) grid (4,4);
\draw[blue](1,3.5) arc (90:180:0.5);
\draw[blue](0.5,2)--(0.5,3);
\draw[blue](0.5,1)--(0.5,2);
\draw[blue](0.5,0)--(0.5,1);
\draw[blue](1,3.5)--(2,3.5);
\draw[blue](2,2.5) arc (90:180:0.5);
\draw[blue](1.5,1)--(1.5,2);
\draw[blue](1.5,0)--(1.5,1);
\draw[blue](2,3.5)--(3,3.5);
\draw[blue](2,2.5)--(3,2.5);
\draw[blue](3,1.5) arc (90:180:0.5);
\draw[blue](2.5,0)--(2.5,1);
\draw[blue](3,3.5)--(4,3.5);
\draw[blue](3,2.5)--(4,2.5);
\draw[blue](3,1.5)--(4,1.5);
\draw[blue](4,0.5) arc (90:180:0.5);
\end{scope}
\end{tikzpicture}
    \caption{Bijections for the transition equations of $3241$.}
    \label{fig:3241bij}
\end{figure}

For example, the bijections for $3241$ are illustrated in \Cref{fig:3241bij}. The top row (top left and top right correspondence) is one bijection that sends two QBPDs of $3241$ to the two QBPDs of $3214$ which corresponds to the $(x_3-y_1)\g_{3214}$ term on the right hand side. The bottom left correspondence is one bijection that sends the one remaining QBPD of $3241$ to one QBPD of $3124$, so the remaining part of the left hand side corresponds to part of the $q_2\g_{3124}$ term on the right hand side. Finally, the bottom right correspondence sends the one remaining QBPD of $3124$ to the one QBPD of $1234$, which means the remaining parts of the $q_2\g_{3124}$ term on the right hand side cancel outs with the $q_1q_2\g_{1234}$ term.


In general, these bijections are more complicated, as some cases do not show up in the example above. The full details of these bijections can be found in \cite[Section 3.3]{le2024qbpd}.
%By showing how these bijection changes the binomial weight, we establish the transition equation and thus establish Theorem 
By showing how these bijections change the binomial weight, we establish that the quantum bumpless pipe dreams formula satisfies the transition equation and thus establish \Cref{thm:qdform}.
\section{Cancellation Analysis}\label{cancel}
There is no cancellation in the bumpless pipe dream formula for the double Schubert polynomials. Quantum bumpless pipe dreams provide a combinatorial formula for the monomial expansion of the quantum double Schubert polynomial, but this formula is not cancellation-free. In this section, we give some analysis of how much and what kind of cancellation occurs.

\Cref{tab:s4cancel} lists all the permutations in $S_4$ for which the QBPD formula gives cancellation when computing the quantum double Schubert polynomials. 
\begin{table}[ht!]
    \centering
    \begin{tabular}{ c c c c c }
     Permutation &   Monomials & QBPD Monomials &  Cancellations & Number of QBPDs\\
     $[4,1,3,2]$ & $50$ & $54$ & $2$ & $9$\\
     $[3,1,4,2]$ & $18$ & $20$ & $1$ & $4$\\
     $[1,4,3,2]$ & $46$ & $48$ & $1$ & $9$\\
     $[2,1,4,3]$ & $12$ & $14$ & $1$ & $5$\\
\end{tabular}
    \caption{Nonzero cancellations for QBPDs in $S_4$, when considering the generated quantum double Schubert polynomial.}
    \label{tab:s4cancel}
\end{table}
\begin{example}
    Permutation $615432$ has $97032$ monomials in its quantum double Schubert polynomial, while the total number of monomials generated from QBPDs is $140052$. $21510$ pairs of monomials cancel out, and the number of QBPDs is $1038$. This is the permutation with the most cancellation in $S_6$.
\end{example}
\begin{table}[ht!]
    \centering
    \begin{tabular}{ c c c c c c }
         & Total & Average per Permutation & Permutation of Max & Max\\
         $S_3$ & $0$ & $-$ & $-$ & $-$\\
         $S_4$ & $5$ & $0.208$ & $[4,1,3,2]$ & $2$\\ 
         $S_5$ & $1350$ & $11.25$ & $[5,1,4,3,2]$ & $153$\\ 
         $S_6$ & $570549$ & $792.43$ & $[6,1,5,4,3,2]$ & $21510$
    \end{tabular}
    \caption{Cancellations in $S_n$ for $n=3,4,5,6$}
    \label{tab:cancelstats}
\end{table}
As shown in \Cref{tab:cancelstats}, the number of cancellations per permutation grows larger with greater $n$.  In particular, we can observe that there are several ways that cancellations occur. Two QBPDs could completely cancel each other out in both the single and double quantum Schubert polynomial case, as in \Cref{fig:completecancel}, or they could partially cancel. In the case of partial cancellation, \Cref{fig:binomial-cancel} illustrates the cancellation of binomial terms and no cancellation in monomial terms, while \Cref{fig:12543} illustrates the cancellation of monomial terms but not binomial terms. 
\begin{figure}[ht!]
    \centering
    \begin{tikzpicture}[scale=\figscale]
\draw (0, 0) grid (4,4);
\draw[blue](1,3.5) arc (90:180:0.5);
\draw[blue](0.5,2)--(0.5,3);
\draw[blue](0.5,1)--(0.5,2);
\draw[blue](0.5,0)--(0.5,1);
\draw[blue](1.5,3) arc (0:90:0.5);
\draw[blue](1.5,3) arc (180:270:0.5);
\draw[blue](2,1.5) arc (90:180:0.5);
\draw[blue](1.5,0)--(1.5,1);
\draw[blue](3,3.5) arc (90:180:0.5);
\draw[blue](2.5,2)--(2.5,3);
\draw[blue](2,2.5)--(3,2.5);
\draw[blue](2,1.5) arc (-90:0:0.5);
\draw[blue](3,0.5) arc (90:180:0.5);
\draw[blue](3,3.5)--(4,3.5);
\draw[blue](3,2.5)--(4,2.5);
\draw[blue](4,1.5) arc (90:180:0.5);
\draw[blue](3.5,0)--(3.5,1);
\draw[blue](3,0.5)--(4,0.5);
\end{tikzpicture}
\begin{tikzpicture}[scale=\figscale]
\draw (0, 0) grid (4,4);
\draw[fill=white,draw=black](0,2) rectangle (1,4);
\draw[blue](1,1.5) arc (90:180:0.5);
\draw[blue](0.5,0)--(0.5,1);
\draw[blue](2,3.5) arc (90:180:0.5);
\draw[blue](1.5,2)--(1.5,3);
\draw[blue](1.5,1)--(1.5,2);
\draw[blue](1,1.5)--(2,1.5);
\draw[blue](1.5,0)--(1.5,1);
\draw[blue](2,3.5)--(3,3.5);
\draw[blue](3,2.5) arc (90:180:0.5);
\draw[blue](2,1.5) arc (-90:0:0.5);
\draw[blue](3,0.5) arc (90:180:0.5);
\draw[blue](3,3.5)--(4,3.5);
\draw[blue](3,2.5)--(4,2.5);
\draw[blue](4,1.5) arc (90:180:0.5);
\draw[blue](3.5,0)--(3.5,1);
\draw[blue](3,0.5)--(4,0.5);
\end{tikzpicture}
    \caption{Two QBPDs for $2143$ whose binomial weights cancel each other out completely. The left contributes $-q_1$, and the right contributes $q_1$.}
    \label{fig:completecancel}
\end{figure}
\begin{figure}[ht!]
    \centering
    \begin{tikzpicture}[scale=\figscale]
\draw (0, 0) grid (4,4);
\draw[fill=white,draw=black](0,2) rectangle (1,4);
\draw[blue](1,1.5) arc (90:180:0.5);
\draw[blue](0.5,0)--(0.5,1);
\draw[blue](2,2.5) arc (90:180:0.5);
\draw[blue](1,1.5) arc (-90:0:0.5);
\draw[blue](2,0.5) arc (90:180:0.5);
\draw[blue](3,3.5) arc (90:180:0.5);
\draw[blue](2,2.5) arc (-90:0:0.5);
\draw[blue](3,1.5) arc (90:180:0.5);
\draw[blue](2.5,0)--(2.5,1);
\draw[blue](2,0.5)--(3,0.5);
\draw[blue](3,3.5)--(4,3.5);
\draw[blue](4,2.5) arc (90:180:0.5);
\draw[blue](3.5,1)--(3.5,2);
\draw[blue](3,1.5)--(4,1.5);
\draw[blue](3.5,0)--(3.5,1);
\draw[blue](3,0.5)--(4,0.5);
\end{tikzpicture}
\begin{tikzpicture}[scale=\figscale]
\draw (0, 0) grid (4,4);
\draw[blue](1,3.5) arc (90:180:0.5);
\draw[blue](0.5,2)--(0.5,3);
\draw[blue](0.5,1)--(0.5,2);
\draw[blue](0.5,0)--(0.5,1);
\draw[blue](1.5,3) arc (0:90:0.5);
\draw[blue](1.5,3) arc (180:270:0.5);
\draw[blue](2,0.5) arc (90:180:0.5);
\draw[blue](3,3.5) arc (90:180:0.5);
\draw[blue](2,2.5) arc (-90:0:0.5);
\draw[blue](3,1.5) arc (90:180:0.5);
\draw[blue](2.5,0)--(2.5,1);
\draw[blue](2,0.5)--(3,0.5);
\draw[blue](3,3.5)--(4,3.5);
\draw[blue](4,2.5) arc (90:180:0.5);
\draw[blue](3.5,1)--(3.5,2);
\draw[blue](3,1.5)--(4,1.5);
\draw[blue](3.5,0)--(3.5,1);
\draw[blue](3,0.5)--(4,0.5);
\end{tikzpicture}
    \caption{Two QBPDs for $1432$ whose monomial weights do not cancel out, but binomial weights partially cancel out. The left QBPD contributes $x_1q_1-y_2q_1$, and the right QBPD contributes $-x_3q_1+y_2q_1$.}
    \label{fig:binomial-cancel}
\end{figure}
\begin{figure}[ht!]
    \centering
    \begin{tikzpicture}[scale=\figscale]
\draw (0, 0) grid (5,5);
\draw[fill=white,draw=black](0,3) rectangle (1,5);
\draw[blue](1,2.5) arc (90:180:0.5);
\draw[blue](0.5,1)--(0.5,2);
\draw[blue](0.5,0)--(0.5,1);
\draw[blue](2,4.5) arc (90:180:0.5);
\draw[blue](1.5,3)--(1.5,4);
\draw[blue](1,2.5) arc (-90:0:0.5);
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\draw[blue](1.5,0)--(1.5,1);
\draw[blue](2,4.5)--(3,4.5);
\draw[blue](3,3.5) arc (90:180:0.5);
\draw[blue](2.5,2)--(2.5,3);
\draw[blue](2,1.5) arc (-90:0:0.5);
\draw[blue](3,0.5) arc (90:180:0.5);
\draw[blue](3,4.5)--(4,4.5);
\draw[blue](3,3.5)--(4,3.5);
\draw[blue](4,1.5) arc (90:180:0.5);
\draw[blue](3.5,0)--(3.5,1);
\draw[blue](3,0.5)--(4,0.5);
\draw[blue](4,4.5)--(5,4.5);
\draw[blue](4,3.5)--(5,3.5);
\draw[blue](5,2.5) arc (90:180:0.5);
\draw[blue](4.5,1)--(4.5,2);
\draw[blue](4,1.5)--(5,1.5);
\draw[blue](4.5,0)--(4.5,1);
\draw[blue](4,0.5)--(5,0.5);
\end{tikzpicture}
\begin{tikzpicture}[scale=\figscale]
\draw (0, 0) grid (5,5);
\draw[blue](1,4.5) arc (90:180:0.5);
\draw[blue](0.5,3)--(0.5,4);
\draw[blue](0.5,2)--(0.5,3);
\draw[blue](0.5,1)--(0.5,2);
\draw[blue](0.5,0)--(0.5,1);
\draw[blue](1.5,4) arc (0:90:0.5);
\draw[blue](1.5,4) arc (180:270:0.5);
\draw[blue](2,1.5) arc (90:180:0.5);
\draw[blue](1.5,0)--(1.5,1);
\draw[blue](3,4.5) arc (90:180:0.5);
\draw[blue](2,3.5) arc (-90:0:0.5);
\draw[blue](3,2.5) arc (90:180:0.5);
\draw[blue](2,1.5) arc (-90:0:0.5);
\draw[blue](3,0.5) arc (90:180:0.5);
\draw[blue](3,4.5)--(4,4.5);
\draw[blue](4,3.5) arc (90:180:0.5);
\draw[blue](3,2.5) arc (-90:0:0.5);
\draw[blue](4,1.5) arc (90:180:0.5);
\draw[blue](3.5,0)--(3.5,1);
\draw[blue](3,0.5)--(4,0.5);
\draw[blue](4,4.5)--(5,4.5);
\draw[blue](4,3.5)--(5,3.5);
\draw[blue](5,2.5) arc (90:180:0.5);
\draw[blue](4.5,1)--(4.5,2);
\draw[blue](4,1.5)--(5,1.5);
\draw[blue](4.5,0)--(4.5,1);
\draw[blue](4,0.5)--(5,0.5);
\end{tikzpicture}
    \caption{Two QBPDs for $12543$ whose monomial weights cancel each other out, but binomial weights do not cancel completely. The left QBPD contributes $x_3q_1-y_4q_1$, and the right contributes $-x_3q_1+y_2q_1$.}
    \label{fig:12543}
\end{figure}
\FloatBarrier

\acknowledgements{
This project started in the Fall 2023 Lab of Geometry at the University of Michigan, and the initial problem was suggested by George H. Seelinger. We are thankful to him for his mentorship during the project and for helpful comments, as well as suggestions on the writing of this paper. We would also like to thank Sergey Fomin for helpful thoughts and comments. %Finally, we thank the anonymous referees for their detailed and useful suggestions.
}

\printbibliography
\end{document}