%% if you are submitting an initial manuscript then you should have submission as an option here
%% if you are submitting a revised manuscript then you should have revision as an option here
%% otherwise options taken by the article class will be accepted
\documentclass[finalversion]{FPSAC2023}
\articlenumber{89}
%% but DO NOT pass any options (or change anything else anywhere) which alters page size / layout / font size etc

%% note that the class file already loads {amsmath, amsthm, amssymb}

\newtheorem{Theorem}{Theorem}
\newtheorem{Lemma}{Lemma}
\newtheorem{Proposition}{Proposition}
\newtheorem{Definition}{Definition}

\usepackage{array,dashbox}

\newcommand\es{$\emptyset$}


\usepackage{tikz}
\usetikzlibrary{cd}
\usepackage{tikz-cd}
\usetikzlibrary{shapes.geometric}

\definecolor{UF2}{HTML}{FA4616}
\definecolor{UF}{HTML}{0021A5}
\definecolor{lightergray}{rgb}{0.8,0.85,0.85}
\newcommand{\cross}{\begin{tikzpicture}[scale = .4]
		\draw[fill=lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
		\draw[very thick,UF!100!black] (.5,0) -- (.5,1);
		\draw[very thick,UF!100!black] (0,.5) -- (1,.5);
\end{tikzpicture}}
\newcommand{\hwire}{\begin{tikzpicture}[scale = .4]
		\draw[fill=lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
		\draw[very thick,UF!100!black] (0,.5) -- (1,.5);
\end{tikzpicture}}
\newcommand{\vwire}{\begin{tikzpicture}[scale = .4]
		\draw[lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
		\draw[very thick,UF!100!black] (.5,0) -- (.5,1);
\end{tikzpicture}}
\newcommand{\nowire}{\begin{tikzpicture}[scale = .4]
		\draw[lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
\end{tikzpicture}}
\newcommand{\shade}{\begin{tikzpicture}[scale = .4]
		\draw[fill=lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
\end{tikzpicture}}
\newcommand{\notshade}{\begin{tikzpicture}[scale = .4]
		\draw (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
\end{tikzpicture}}
%-----------------------------------------------
\newcommand{\bump}{\begin{tikzpicture}[scale = .4]
		\draw[lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
		\draw[very thick,UF!100!black] (0.5,0) to[out=90,in=180] (1,0.5);
		\draw[very thick,UF!100!black] (0,0.5) to[out=0,in=-90] (0.5,1);
\end{tikzpicture}}
\newcommand{\jay}{\begin{tikzpicture}[scale = .4]
		\draw[lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
		\draw[very thick,UF!100!black] (0,0.5) to[out=0,in=-90] (0.5,1);
\end{tikzpicture}}
\newcommand{\jayshade}{\begin{tikzpicture}[scale = .4]
		\draw[fill=lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
		\draw[very thick,UF!100!black] (0,0.5) to[out=0,in=-90] (0.5,1);
\end{tikzpicture}}
\newcommand{\are}{\begin{tikzpicture}[scale = .4]
		\draw[lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
		\draw[very thick,UF!100!black] (0.5,0) to[out=90,in=180] (1,0.5);
\end{tikzpicture}}
\newcommand{\bcross}{\begin{tikzpicture}[scale = .4]
		\draw[lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
		\draw[very thick,UF!100!black] (.5,0) -- (.5,1);
		\draw[very thick,UF!100!black] (0,.5) -- (1,.5);
\end{tikzpicture}}
\newcommand{\bhwire}{\begin{tikzpicture}[scale = .4]
		\draw[lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
		\draw[very thick,UF!100!black] (0,.5) -- (1,.5);
\end{tikzpicture}}
\newcommand{\bnowire}{\begin{tikzpicture}[scale = .4]
		\draw[fill=lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
\end{tikzpicture}}

\newcommand{\en}{\begin{tikzpicture}[scale = .4]
		\draw[lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
		\draw[very thick,UF!100!black] (0,0.5) to[out=0,in=90] (0.5,0);
\end{tikzpicture}}

\newcommand{\sea}{\begin{tikzpicture}[scale = .4]
		\draw[lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
		\draw[very thick,UF!100!black] (0.5,1) to[out=-90,in=-180] (1,0.5);
\end{tikzpicture}}
%-------------------------------------------------------------------------------
\newcommand{\inlinecross}{\begin{tikzpicture}[scale = .3]
		\draw[lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
		\draw[very thick,UF!100!black] (.5,0) -- (.5,1);
		\draw[very thick,UF!100!black] (0,.5) -- (1,.5);
\end{tikzpicture}}
\newcommand{\inlinenowire}{\begin{tikzpicture}[scale = .3]
		\draw[lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
\end{tikzpicture}}
\newcommand{\inlinebump}{\begin{tikzpicture}[scale = .3]
		\draw[lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
		\draw[very thick,UF!100!black] (0.5,0) to[out=90,in=180] (1,0.5);
		\draw[very thick,UF!100!black] (0,0.5) to[out=0,in=-90] (0.5,1);
\end{tikzpicture}}
\newcommand{\inlineare}{\begin{tikzpicture}[scale = .3]
		\draw[lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
		\draw[very thick,UF!100!black] (0.5,0) to[out=90,in=180] (1,0.5);
\end{tikzpicture}}
\newcommand{\inlinejay}{\begin{tikzpicture}[scale = .3]
		\draw[lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
		\draw[very thick,UF!100!black] (0,0.5) to[out=0,in=-90] (0.5,1);
\end{tikzpicture}}
\newcommand{\inlineshade}{\begin{tikzpicture}[scale = .3]
		\draw[fill=lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
\end{tikzpicture}}
\newcommand{\inlinehwire}{\begin{tikzpicture}[scale = .3]
		\draw[lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
		\draw[very thick,UF!100!black] (0,.5) -- (1,.5);
\end{tikzpicture}}
\newcommand{\inlinevwire}{\begin{tikzpicture}[scale = .3]
		\draw[lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
		\draw[very thick,UF!100!black] (.5,0) -- (.5,1);
\end{tikzpicture}}
\newcommand{\inlineen}{\begin{tikzpicture}[scale = .3]
		\draw[lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
		\draw[very thick,UF!100!black] (0,0.5) to[out=0,in=90] (0.5,0);
\end{tikzpicture}}
\newcommand{\inlinesea}{\begin{tikzpicture}[scale = .4]
    \draw[lightergray] (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);
    \draw[very thick,UF!100!black] (0.5,1) to[out=-90,in=-180] (1,0.5);
  \end{tikzpicture}}

\newcommand\junk[1]{}

%\usepackage{lipsum}

%% define your title in the usual way
\title[Interpolating between classic and bumpless pipe dreams]{Interpolating between \\ classic and bumpless pipe dreams}

%% define your authors in the usual way
%% use \addressmark{1}, \addressmark{2} etc for the institutions, and use \thanks{} for contact details
\author{Allen Knutson\thanks{\href{mailto:allenk@math.cornell.edu}{allenk@math.cornell.edu}. AK was partially supported by NSF Grant DMS--1953948.}\addressmark{1} \and Gabe Udell\thanks{\href{mailto:gru5@cornell.edu}{gru5@cornell.edu}.}\addressmark{1}}

%% then use \addressmark to match authors to institutions here
\address{\addressmark{1}Department of Mathematics, Cornell University}

%% put the date of submission here
\received{\today}

%% leave this blank until submitting a revised version
%\revised{}

%% put your English abstract here, or comment this out if you don't have one yet
%% please don't use custom commands in your abstract / resume, as these will be displayed online
%% likewise for citations -- please don't use \cite, and instead write out your citation as something like (author year)
\abstract{We introduce {\em hybrid pipe dreams,} in which some rows are
  specified to contain classic pipe dream tiles (plus a couple new ones)
  and some to contain bumpless pipe dream tiles (albeit upside down).
  For each of the $2^n$ possibilities of row specification, there is a sum
  over hybrid pipe dreams to compute each double Schubert polynomial.
  We give a bijection to swap the specifications
  on two adjacent rows, demonstrating the equivalence of the many formul\ae.}

%% put your French abstract here, or comment this out if you don't have one
% \resume{\lipsum[2]}

%% put your keywords here, or comment this out if you don't have them yet
\keywords{Schubert polynomials, pipe dreams}

%% you can include your bibliography however you want, but using an external .bib file is STRONGLY RECOMMENDED and will make the editor's life much easier
%% regardless of how you do it, please use numerical citations; i.e., [xx, yy] in the text

%% this sample uses biblatex, which (among other things) takes care of URLs in a more flexible way than bibtex
%% but you can use bibtex if you want
\usepackage[backend=bibtex]{biblatex}
\addbibresource{theBib.bib}
%% note the \printbibliography command at the end of the file which goes with these biblatex commands

\begin{document}

\maketitle
%% note that you DO NOT have to put your abstract here -- it is generated by \maketitle and the \abstract and \resume commands above

\newcommand\defn[1]{{\bf #1}}

\section{Hybrid pipe dreams compute
  Schubert polynomials}

\subsection{Hybrid pipe dreams and the formula}

Define the \defn{classic pipe dream tiles} to be 
\[\cross,\ \hwire,\ \jay,\ \are,\ \bump,\ \nowire \text{ but not } \vwire,\] 
the \defn{(upside down) bumpless pipe dream tiles} to be 
\[\bcross,\ \bhwire,\ \en,\ \sea,\ \vwire,\ \bnowire \text{ but not } \bump,\]
and in each case call the shaded tile(s) \defn{weighty}.

Given $\pi \in S_n$ a permutation, and $\tau \in \{C,B\}^n$ a word in
{\bf C}lassic and {\bf B}umpless, define a \defn{$\tau$-hybrid pipe dream
  for $\pi$} to be an $n\times n$ square made of tiles, such that
\begin{itemize}
    \setlength\itemsep{.1em}
\item if two tiles share an edge, they must agree as to whether it is
  blank or not. Hence the pipes end only on the boundary of the square
\item the $i$th row is called ``a classic row'' or ``a bumpless row''
  depending on $\tau_i$, and uses only tiles from one set or the other
\item each classic row has a number on the left edge, and is blank on
  the right edge, with the opposite being true for bumpless rows;
  this number is the \defn{row label}
\item the top edges are labeled with $\pi^{-1}(1), \pi^{-1}(2), \ldots, \pi^{-1}(n)$ from left to right
\item the bottom edges are blank
\item the labeled boundary edges on the left and right are
  labeled $1\ldots n$ in order, counterclockwise starting from the
  Northwest corner
\item the numbers at the two ends of a pipe agree, and no two pipes cross twice.
\end{itemize}

Some examples are below.
Our first main theorem is the following:

\newcommand\If{&\text{if }}

\begin{Theorem}\label{thm:formula}
  For $\pi \in S_n$ a permutation, and any word $\tau \in \{C,B\}^n$,
  the double Schubert polynomial $S_\pi(x_1,\ldots,x_n,y_1,\ldots,y_n)$
  can be computed as a sum over the $\tau$-hybrid pipe dreams $\delta$
  for $\pi$:
  {\small
 \[ S_\pi = \sum_{\delta} \prod
  \{ x_i - y_j\ :\  i,j \in [n] \text{ and the tile in {\bf physical} column $j$
    and row {\bf labeled} $i$ is weighty}\} 
  \]}

  In the case $\tau = C^n$, this reproduces the pipe dream formula in
  \cite[Prop. 6.2]{FominKirillov} (and \cite{BergeronBilley} in the
  $y_i \equiv 0$ setting).  In the case $\tau = B^n$, this reproduces
  the bumpless pipe dream formula in \cite{LamLeeShimozono} (albeit upside down) and is equivalent to the alternating sign matrix formula from \cite{Weigandt,  Lascoux}.
  The pipes are {\em labeled} differently here (by $\pi^{-1}$) than they
  are in \cite{BergeronBilley,FominKirillov,KM,LamLeeShimozono}, but the
  placements of tiles are exactly the same.
\end{Theorem}

The number of terms in this sum is $S_\pi|_{x_i\equiv 1,y_j\equiv 0}$,
independent of $\tau$. Our other main theorem (in \cref{sec:swapping}),
which implies this one, will be a bijective correspondence.

Here are the hybrid pipe dreams for $\pi=132$ and various choices of $\tau$.

$$
\begin{matrix}
  \tau=CCC:
  &
  \begin{tikzpicture}[scale=0.4]
    \node at (0,3) {1}; \node at (1,3) {3}; \node at (2,3) {2}; 
    \node at (0,2) {\bump}; \node at (1,2) {\bump}; \node at (2,2) {\jay}; 
    \node at (0,1) {\cross}; \node at (1,1) {\jay}; \node at (2,1) {\nowire}; 
    \node at (0,0) {\jay}; \node at (1,0) {\nowire}; \node at (2,0) {\nowire}; 
    \node at (-1,0) {3}; \node at (-1,1) {2}; \node at (-1,2) {1}; 
  \end{tikzpicture}
  &
  \begin{tikzpicture}[scale=0.4]
    \node at (0,3) {1}; \node at (1,3) {3}; \node at (2,3) {2}; 
    \node at (0,2) {\bump}; \node at (1,2) {\cross}; \node at (2,2) {\jay}; 
    \node at (0,1) {\bump}; \node at (1,1) {\jay}; \node at (2,1) {\nowire}; 
    \node at (0,0) {\jay}; \node at (1,0) {\nowire}; \node at (2,0) {\nowire}; 
    \node at (-1,0) {3}; \node at (-1,1) {2}; \node at (-1,2) {1}; 
  \end{tikzpicture}  
&\qquad\qquad&
               \tau=CCB: &
\begin{tikzpicture}[scale=0.4]
  \node at (0,3) {1}; \node at (1,3) {3}; \node at (2,3) {2}; 
  \node at (0,2) {\bump}; \node at (1,2) {\bump}; \node at (2,2) {\jay}; 
  \node at (0,1) {\cross}; \node at (1,1) {\jay}; \node at (2,1) {\nowire}; 
  \node at (0,0) {\sea}; \node at (1,0) {\bhwire}; \node at (2,0) {\bhwire}; 
  \node at (3,0) {3}; \node at (-1,1) {2}; \node at (-1,2) {1}; 
\end{tikzpicture}
&
\begin{tikzpicture}[scale=0.4]
  \node at (0,3) {1}; \node at (1,3) {3}; \node at (2,3) {2}; 
  \node at (0,2) {\bump}; \node at (1,2) {\cross}; \node at (2,2) {\jay}; 
  \node at (0,1) {\bump}; \node at (1,1) {\jay}; \node at (2,1) {\nowire}; 
  \node at (0,0) {\sea}; \node at (1,0) {\bhwire}; \node at (2,0) {\bhwire}; 
  \node at (3,0) {3}; \node at (-1,1) {2}; \node at (-1,2) {1}; 
\end{tikzpicture} 
\\
\tau=CBC:
&
\begin{tikzpicture}[scale=0.4]
  \node at (0,3) {1}; \node at (1,3) {3}; \node at (2,3) {2}; 
  \node at (0,2) {\bump}; \node at (1,2) {\bump}; \node at (2,2) {\jay}; 
  \node at (0,1) {\sea}; \node at (1,1) {\bcross}; \node at (2,1) {\bhwire}; 
  \node at (0,0) {\hwire}; \node at (1,0) {\jay}; \node at (2,0) {\nowire}; 
  \node at (3,1) {3}; \node at (-1,0) {2}; \node at (-1,2) {1}; 
\end{tikzpicture}
&
\begin{tikzpicture}[scale=0.4]
  \node at (0,3) {1}; \node at (1,3) {3}; \node at (2,3) {2}; 
  \node at (0,2) {\bump}; \node at (1,2) {\cross}; \node at (2,2) {\jay}; 
  \node at (0,1) {\vwire}; \node at (1,1) {\sea}; \node at (2,1) {\bhwire}; 
  \node at (0,0) {\jay}; \node at (1,0) {\nowire}; \node at (2,0) {\nowire}; 
  \node at (3,1) {3}; \node at (-1,0) {2}; \node at (-1,2) {1}; 
\end{tikzpicture} 
&\hspace{1 in}&
\tau=CBB:
&
\begin{tikzpicture}[scale=0.4]
  \node at (0,3) {1}; \node at (1,3) {3}; \node at (2,3) {2}; 
  \node at (0,2) {\bump}; \node at (1,2) {\bump}; \node at (2,2) {\jay}; 
  \node at (0,1) {\sea}; \node at (1,1) {\bcross}; \node at (2,1) {\bhwire}; 
  \node at (0,0) {\bnowire}; \node at (1,0) {\sea}; \node at (2,0) {\bhwire}; 
  \node at (3,1) {3}; \node at (3,0) {2}; \node at (-1,2) {1}; 
\end{tikzpicture}
&
\begin{tikzpicture}[scale=0.4]
  \node at (0,3) {1}; \node at (1,3) {3}; \node at (2,3) {2}; 
  \node at (0,2) {\bump}; \node at (1,2) {\cross}; \node at (2,2) {\jay}; 
  \node at (0,1) {\vwire}; \node at (1,1) {\sea}; \node at (2,1) {\bhwire}; 
  \node at (0,0) {\sea}; \node at (1,0) {\bhwire}; \node at (2,0) {\bhwire}; 
  \node at (3,1) {3}; \node at (3,0) {2}; \node at (-1,2) {1}; 
\end{tikzpicture}
\\
\tau=BCB:&
\begin{tikzpicture}[scale=0.4]
  \node at (0,3) {1}; \node at (1,3) {3}; \node at (2,3) {2}; 
  \node at (0,2) {\vwire}; \node at (1,2) {\sea}; \node at (2,2) {\bcross}; 
  \node at (0,1) {\jay}; \node at (1,1) {\are}; \node at (2,1) {\jay}; 
  \node at (0,0) {\bnowire}; \node at (1,0) {\sea}; \node at (2,0) {\bhwire}; 
  \node at (3,2) {3}; \node at (3,0) {2}; \node at (-1,1) {1}; 
\end{tikzpicture} &
\begin{tikzpicture}[scale=0.4]
  \node at (0,3) {1}; \node at (1,3) {3}; \node at (2,3) {2}; 
  \node at (0,2) {\vwire}; \node at (1,2) {\sea}; \node at (2,2) {\bcross}; 
  \node at (0,1) {\bump}; \node at (1,1) {\hwire}; \node at (2,1) {\jay}; 
  \node at (0,0) {\sea}; \node at (1,0) {\bhwire}; \node at (2,0) {\bhwire}; 
  \node at (3,2) {3}; \node at (3,0) {2}; \node at (-1,1) {1}; 
\end{tikzpicture} & \hspace{1in}&
\tau=BBC: &
\begin{tikzpicture}[scale=0.4]
  \node at (0,3) {1}; \node at (1,3) {3}; \node at (2,3) {2}; 
  \node at (0,2) {\vwire}; \node at (1,2) {\sea}; \node at (2,2) {\bcross}; 
  \node at (0,1) {\vwire}; \node at (1,1) {\bnowire}; \node at (2,1) {\sea}; 
  \node at (0,0) {\jay}; \node at (1,0) {\nowire}; \node at (2,0) {\nowire}; 
  \node at (3,2) {3}; \node at (3,1) {2}; \node at (-1,0) {1}; 
\end{tikzpicture} &
\begin{tikzpicture}[scale=0.4]
  \node at (0,3) {1}; \node at (1,3) {3}; \node at (2,3) {2}; 
  \node at (0,2) {\vwire}; \node at (1,2) {\sea}; \node at (2,2) {\bcross}; 
  \node at (0,1) {\sea}; \node at (1,1) {\en}; \node at (2,1) {\sea}; 
  \node at (0,0) {\hwire}; \node at (1,0) {\jay}; \node at (2,0) {\nowire}; 
  \node at (3,2) {3}; \node at (3,1) {2}; \node at (-1,0) {1}; 
\end{tikzpicture} 
\end{matrix}
$$

\subsection{The all-classic case $\tau=C^n$}

%Ordinary pipe dreams were introduced as RC-graphs in \cite{BergeronBilley}. 



The boundary conditions mean that every edge on North and West is
crossed by a pipe, whereas every edge on South and East is blank.

\begin{Proposition}\label{prop:ordinary}
  A $\tau$-hybrid pipe dream for $\tau=C^n$ has
  \begin{itemize}
    \setlength\itemsep{.1em}
  \item only {\tt j}-tiles $\jay$ on the antidiagonal,
    and only blank tiles $\nowire$ below the antidiagonal;
  \item only crosses $\cross$ and elbows $\bump$ above the antidiagonal.
  \end{itemize}
  In particular the $\hwire$ and {\tt r}-tile $\are$,
  despite being allowed, are never used.
\end{Proposition}

\begin{proof}
  Consider the rightmost column, from top to bottom. The Northeast tile
  has a number on its North edge and blank on its East edge, and cannot
  be a $\vwire$-tile (as those are forbidden in classic rows), so must be
  a $\jay$-tile. The next tile below it is blank on North and East, so must
  be $\nowire$, and the same argument holds all the way down
  the rightmost column.

  Now consider the first row, from right to left, starting just West of
  the Northeast tile. The next tile has pipes crossing its North and East
  side, so must be a $\cross$ or an $\bump$. The same argument holds
  all the way to the left in the first row.

  Remove the last column and first row and do induction on the
  remaining $(n-1)\times (n-1)$ square.
\end{proof}

This picture is slightly different {\em mathematically} from the
classic pipe dreams in, say, \cite{BergeronBilley} which fill the
fourth quadrant and have only crossing and elbow tiles. That formulation
is based on the forward-stability of Schubert polynomials,
$S_\pi = S_{\pi \oplus I_m}$ (where the $\oplus$ is on the permutation
matrices). However, even in \cite{BergeronBilley} they omit ``the `sea'
of wavy strands'' and obtain as pictures {\em exactly} the
$C^n$-hybrid pipe dreams here.

\subsection{The all-bumpless case $\tau=B^n$}

There is less to say here -- our definition of $B^n$-hybrid pipe dreams
exactly matches that of \cite{LamLeeShimozono}, except for being upside down
and for the shift in labeling (they have $\pi(1)\ldots \pi(n)$ along
the East side, where we have $n\ldots 1$).

\subsection{Repeating letters}

We inverted the labeling principally to simplify the description of the
labels on the boundary (as $1\ldots n$ counterclockwise). There is another
benefit (though not one we will pursue here);
in the case that $\pi$ has descents in at most the positions
$k_1, k_1+k_2, \ldots, k_1+k_2+k_3+\ldots+k_d$ it is natural to write $\pi^{-1}$
as a \defn{string} with content $1^{k_1}2^{k_2}\cdots d^{k_d} (d+1)^{n-\sum k_i}$.
Put another way, $\pi$ is the shortest representative inside a coset
$S_n / \prod_i S_{k_i}$ (where $k_{d+1} := n-\sum_{i=1}^d k_i$), which is
an equivalence class within which the {\em positions} are ambiguous,
and its inverse has ambiguous {\em values}.
 If one allows in this way multiple pipes to have the same label, one must forbid
 those pipes from crossing even {\em once}.
% Pipes with equal labels may not cross.

Placing the ambiguity on values is more natural in these scattering-diagram-inspired
contexts, in that physics is comfortable with indistinguishable particles
but not with ambiguous positions.

\section{Equivalence of the $2^n$ formul\ae}
\label{sec:swapping}








In this section we give combinatorial proofs of the equivalence of the
$2^n$ formul\ae. For this purpose, we introduce
\textbf{decorated hybrid pipe dreams}, as hybrid pipe dreams with
decorations of $x$ or $-y$ on the weighty tiles. Following the conventions
in \cite{Huang} (which introduced this notion in the bumpless case), 
we write the set of decorated $\tau$-hybrid pipe dreams for $\pi$ as
\[\widetilde{\mathrm{HPD}}(\tau,\pi) := \left\{(D,f)\ \mid\ D\in
    \mathrm{HPD}(\tau,\pi),\ f:\mathrm{weighty}(D)\to \{x,-y\} \right\}\]
where $\mathrm{HPD}(\tau,\pi)$ is the set of $\tau$-hybrid pipe dreams for
$\pi$, and $\mathrm{weighty}(D)$ is the set of weighty tiles in $D$.
Define the \defn{monomial of a decorated hybrid pipe dream} to be
\[\mathrm{mon}(D,f) := \prod_{\substack{(i,j)\in \mathrm{weighty}(D)\\f(i,j)=x}}x_i 
  \prod_{\substack{(i,j)\in \mathrm{weighty}(D)\\f(i,j)=-y}}(-y_j),
  \qquad\text{so}\
S_\pi =\sum_{(D,f)\in \mathrm{HPD}(\tau,\pi)}\mathrm{mon}(D,f)\]
as follows from \cref{thm:formula}.

We prove the equivalence of these $2^n$ formul\ae\ by constructing a
bijection
\[\phi:\widetilde{\mathrm{ HPD}}(\tau_1,\pi)\to \widetilde{\mathrm
  HPD}(\tau_2,\pi)\] for any $\tau_1,\tau_2\in \{C,B\}^n$ while making sure that
$\mathrm{mon}(\phi(D,f))=\phi(\mathrm{mon}(D,f))$ for all
$(D,f)\in \widetilde{\mathrm{HPD}}(\tau_1,\pi)$. We will do this in
two steps (each then iterated many times).
First, we %create a bijection between decorated
% $\tau$-hybrid pipe dreams and decorated $\tau'$-hybrid pipe dreams
handle the case 
where $\tau,\tau'$ differ only in the last letter.
That is, we describe how to change the last row from classic to
bumpless, and in fact, we do this in such a way that $\phi(D,f)$
differs from $(D,f)$ only in its final row.
Second, we %create bijections between decorated $\tau$-hybrid pipe dreams and
%decorated $\tau'$-hybrid pipe dreams
handle the case
where $\tau'$ is the same as
$\tau$ but with some (consecutive) substring $CB$ replaced by $BC$.
That is, we describe how to switch the order of two adjacent
rows, one bumpless and one classic. We do this in such a way that
$\phi(D,f)$ differs from $(D,f)$ only in the rows that were swapped.
The composite bijections $\phi$ are automatically
coherent in the sense that any triangle 
$
\widetilde{\mathrm{ HPD}}(\tau_1,\pi)
\to \widetilde{\mathrm  {HPD}}(\tau_2,\pi)
\to \widetilde{\mathrm  {HPD}}(\tau_3,\pi) $
commutes.





\subsection{Monk's rule and the Gao--Huang bijection}

%As far as the authors are aware, the construction in this paper is the first explicit bijection between decorated ordinary and decorated bumpless pipe dreams evidencing the equivalence of formulas for double Schubert polynomials. However, it is not the first explicit bijection 

Hybrid pipe dreams were created to help understand the connection between pipe dreams and bumpless pipe dreams, but they are not the first tool for doing so. Another approach to this connection involves Monk's rule.



\begin{Theorem}[Monk's rule for double Schubert polynomials]
	For $\pi$ a permutation, $\alpha$ a positive integer, $t_{a,b}$ the transposition of $a$ and $b$, and $\gtrdot$ the Bruhat covering relation, 
	\[(x_\alpha-y_{\pi(\alpha)}) S_\pi + \sum_{\substack{s<\alpha \\ \pi t_{s,\alpha}\gtrdot \pi}}S_{\pi t_{s,\alpha}} = \sum_{\substack{l>\alpha\\ \pi t_{\alpha,l}\gtrdot \pi }}S_{\pi t_{\alpha, l}}.\]
\end{Theorem}
Monk's rule for single Schubert polynomials can be proven bijectively
using either pipe dreams \cite{BergeronBilley,BilleyHolroydYoung,Huang}
or bumpless pipe dreams \cite{Huang} by using, for either $\tau=C^n$ or $\tau=B^n$, maps
$x_\alpha\leadsto:\mathrm{HPD}(\tau,\pi)\to
\bigcup_{\substack{\alpha<l\\ \pi t_{\alpha,l}\gtrdot \pi
  }}\mathrm{HPD}(\tau,\pi t_{\alpha,l})$ and
$m_{s,\beta}:\mathrm{HPD}(\tau,\pi t_{s,\beta})\to
\bigcup_{\substack{l>\beta \\ \pi t_{\beta,l}\gtrdot \pi
  }}\mathrm{HPD}(\tau,\pi t_{\beta,l})$ when
$s<\beta, \pi t_{s,\beta}\gtrdot \pi$.
% In \cite{Huang}, Huang noted that the bijections (in the single Schubert polynomial context) could be used to inductively construct a bijection between ordinary and bumpless pipe dreams preserving $\mathrm{mon}(D,f)$ when all decorations are $x$. Building on this 
In \cite{GaoHuang}, Gao and Huang constructed a bijection $\phi_{GH}$
between bumpless and classic pipe dreams preserving
$\mathrm{mon}(D,f)$ when all decorations are $x$. Furthermore, they proved
their bijection to be canonical in the sense that it commutes with
$x_\alpha\leadsto$ and $m_{s,\beta}.$ When $\tau_1=B^n$, $\tau_2=C^n$
and either all decorations being $x$ or all decorations being $-y$, we
get specializations $\phi_x, \phi_y$ of the hybrid pipe dream
bijection. We conjecture that
$\phi_{GH}(D)=\phi_x(D)=(\phi_y(D^{\mathsf{T}}))^{\mathsf{T}}$ for any
bumpless pipe dream $D$, where $D^{\mathsf{T}}$ denotes the transpose
of a pipe dream or bumpless pipe dream.

%Setting $\tau_1=B^n$ and $\tau_2=O^n,$ our construction in terms of hybrid pipe dreams gives a bijection from decorated bumpless pipe dreams to decorated ordinary pipe dreams. 
%Denote by $\phi_x$ and $\phi_y$ the induced bijections (via hybrid p) from bumpless pipe dreams to ordinary ones corresponding to all decorations being $x$ or all decorations being $-y$ respectively. Denote by $D^\mathsf{T}$ either the transpose of a pipe dream about its diagonal or the transpose of an (upside down) bumpless pipe dream about its antidiagonal. We conjecture that $\phi_{GH}(D)=\phi_x(D)=(\phi_y(D^{\mathsf{T}}))^{\mathsf{T}}$ for any bumpless pipe dream $D$.


%In \cite{Huang}, a bijective proof of Monk's rule was given using bumpless pipe dreams. In particular, there  

%Monk's rule for single Schubert polynomials is obtained by setting the $y$ variables equal to 0. 


\subsection{Changing the bottom row}

A single picture shows the full gamut of possibilities for the
bottom rows, in the $C$ and $B$ cases:
$$
\begin{tikzpicture}[scale=0.4]
  \node at (-1,0) {5};
  \node at (0,0) {\hwire};
  \node at (1,0) {\hwire};
  \node at (2,0) {\jay};
  \node at (3,0) {\nowire};
  \node at (4,0) {\nowire};
  \node at (5,0) {\nowire};
  \node at (6,0) {\nowire};
\end{tikzpicture}
\qquad\Leftrightarrow\qquad
\begin{tikzpicture}[scale=0.4]
  \node at (0,0) {\bnowire};
  \node at (1,0) {\bnowire};
  \node at (2,0) {\sea};
  \node at (3,0) {\bhwire};
  \node at (4,0) {\bhwire};
  \node at (5,0) {\bhwire};
  \node at (6,0) {\bhwire};
  \node at (7,0) {5};
\end{tikzpicture}
$$

\junk{
  {\bf not sure what to keep of the following}
  
  Given all but the last row of a hybrid pipe dream, there is a unique
  classic last row and a unique bumpless last row. Say the label of the
  last row is $i$, and the $i$-pipe enters the last row in the $j$th
  column. If the last row is classic then it must consist of $j-1$
  $\inlinehwire$ tiles followed by a $\inlinejay$ in the $j$th column
  and $n-j$ blank tiles in the last row. If the last row is bumpless
  then it must consist of $j-1$ $\inlinenowire$ tiles followed by a
  $\inlinesea$ tile in the $j$th column and $\inlinehwire$ tiles in the
  last $n-j$ columns.
}

\noindent
\begin{minipage}{.65\linewidth}
Note that this correspondence preserves the positions of the weighty tiles,
and hence the associated monomials (assuming we keep the same decoration).

We give an in-context example of the correspondence:
\end{minipage}
\hfill
\begin{minipage}{.325\linewidth}
$
  \begin{tikzpicture}[scale=0.4]
    \node at (0,4) {2}; \node at (1,4) {1}; \node at (2,4) {3}; \node at (3,4){4}; 
    \node at (0,3) {\cross}; \node at (1,3) {\bump}; \node at (2,3) {\cross}; \node at (3,3){\jay}; 
    \node at (0,2) {\vwire}; \node at (1,2) {\sea}; \node at (2,2){\bcross}; \node at (3,2){\bhwire}; 
    \node at (0,1) {\jay}; \node at (1,1) {\are}; \node at (2,1) {\jay}; \node at (3,1){\nowire}; 
    \node at (0,0) {\hwire}; \node at (1,0) {\jay}; \node at (2,0) {\nowire}; \node at (3,0) {\nowire}; 
    \node at (-1,3) {1}; \node at (4,2) {4}; \node at (-1,1) {2}; \node at (-1,0) {3};
  \end{tikzpicture}
  \begin{tikzpicture}[scale=0.4]
    \node at (0,4) {2}; \node at (1,4) {1}; \node at (2,4) {3}; \node at (3,4){4}; 
    \node at (0,3) {\cross}; \node at (1,3) {\bump}; \node at (2,3) {\cross}; \node at (3,3){\jay}; 
    \node at (0,2) {\vwire}; \node at (1,2) {\sea}; \node at (2,2){\bcross}; \node at (3,2){\bhwire}; 
    \node at (0,1) {\jay}; \node at (1,1) {\are}; \node at (2,1) {\jay}; \node at (3,1){\nowire}; 
    \node at (0,0) {\bnowire}; \node at (1,0) {\sea}; \node at (2,0) {\bhwire}; \node at (3,0) {\bhwire};
    \node at (-1,3) {1}; \node at (4,2) {4}; \node at (-1,1) {2}; \node at (4,0) {3};
  \end{tikzpicture}
$
\end{minipage}


\junk{
Since there is a unique final classic row and a unique final bumpless row, we have no choice but to map the former to the latter. It remains to describe how to map a decoration of the former to a decoration of the latter in a weight preserving manner.
When the last row is classic, the weighty tiles are the $\inlinehwire$ tiles in the first $j-1$ columns. When the last row is bumpless, the weighty tiles are the $\inlinenowire$ tiles, also occurring in the first $j-1$ columns. Since the locations of the weighty tiles are the same, the decoration of $\phi(D,f)$ can be exactly the same as the decoration $f$ of $D$ and we will have $\mathrm{mon}(\phi(D,f))=\mathrm{mon}(D,f)$. In fact we can see that this was the only choice we could have made since the coefficients in the expansion of $\prod_{k=1}^{j-1}(x-y_i)$ are all 1.
}


 %\begin{tikzpicture}[scale=0.3] 
%\node at (-1,0) {i}
%\node at (0,0) {\inlinehwire};
%\node at (0,1) {\inlinehwire};
%\node at (0,2) {\inlinenowire};
%\node at (0,2) {\dots};
%\node at (0,3) {\inlinejay};
% \end{tikzpicture}} where ther







\subsection{Swapping a classic row with an adjacent
  bumpless row}

\newcommand\mapstoo[2]{ \begin{array}{ccc}
  #1 & \raisebox{.8cm}{$\!\!\!\mapsto\!\!\!$} & #2
                        \end{array}}

\newcommand\tilecol[2]        
{	\begin{tikzpicture}[scale=0.4]
    \node at (0,0) {#1}; \node at (0,1) {#2};
  \end{tikzpicture} }

\newcommand\tilecool[8]
{\begin{tikzpicture}[scale=0.4]
    \node at ( 0,0) {#1}; \node at (0, 1) {#2};
    \node at (-1,1) {#3}; \node at (0,-1) {#5}; \node at ( 1,0) {#6};
    \node at ( 1,1) {#7}; \node at (0, 2) {#8}; \node at (-1,0) {#4};
  \end{tikzpicture}}

    %To discuss the row swapping bijection we will focus on the possibilities for tiled $2\times k$ rectangles that could
    %could appear in the $i$th and $(i+1)st$ rows of some $\tau$-hybrid pipe dream. 
	
	%Define a (decorated) \textbf{full $OB$-dream} to be a tiled $2\times n$ rectangle that could appear as the $i$th an $(i+1)$st rows of some (decorated) $\tau$-hybrid pipe dream with $\tau_i=O$ and $\tau_{i+1}=B$. 
%	Define (decorated) \textbf{full $B\atop O$ sections} similarly but with $\tau_i=B$ and $\tau_{i+1}=O.$
%	Define an $OB$-dream ($B\atop O$ section) to be any contiguous $k$ columns of an $OB$-dream satisfying the technical requirement that some pipe connects to one of the two vertical edges on the left (right) of the 
	
	
	
%	With these definitions in mind, the goal of this section is to produce a weight preserving bijection between decorated full $OB$-dreams and decorated full $B\atop O$ sections which have the same labels in the same locations along their top and bottom edges.

%Let $\phi_y$ denote the restriction of a bijection $\phi:\widetilde{HPD}(\tau_1 OB \tau_2,\pi)\to \widetilde{HPD}(\tau_1 BO \tau_2,\pi)$ to decorated hybrid pipe dreams where all decorations are $-y$. 
%$\phi_y$ is entirely local in the sense that the $i$th 
%Not only is there a unique w $\phi_y$ is entirely local in the sense that the tiles in column $i$ of $\phi()

%this bijection is local in the sense that the $i$th column of $\phi(D)$ depends only on the $i$th column of $D$. %The catch is that it is important  

%the rows being changed depend only on the two tiles of $D$ in the $i$th column of the rows being changed. The catch is that these tiles do depend on knowing 

Define a $C\atop B$ \textbf{section} to be a tiled $2\times k$ rectangle 
that could appear in rows $i$ and $i+1$ (and $k$ contiguous columns)
of some $\tau$-hybrid pipe dream, where $\tau_i=C$ and $\tau_{i+1}=B$.
Define \textbf{$B\atop C$ sections} similarly. We say the pipe
connecting $i$ on the top of a hybrid pipe dream to $i$ on a side is
\defn{labeled $i$} or is \defn{the $i$ pipe}. Where a pipe enters/exits an
$C\atop B$ section ($B \atop C$ section) we will place a number
indicating the label of the pipe. These numbers are just to
distinguish which pipes connect to each other outside the
$C\atop B$ section ($B \atop C$ section). That is, if we take an
$C\atop B$ section ($B\atop C$ section) and apply a permutation to
each number on its boundary, we will not distinguish this from the
original.  
%We will place labels around an $O\atop B$ section ($B\atop O$ section) to indicate labels of pipes passing entering/exciting at the . 

 %the labels of pipes entering or exiting the $O\atop B$ section (BO-dream) through its boundary edges. This is to indicate when two pipes both passing through left/right boundaries are the same pipe. The specific numerical labels on pipes won't matter.


\begin{Theorem}\label{thm:justy}
  Assuming all the decorations are $-y$s, there exists {\em uniquely}
  a monomial-preserving
  bijection $\phi_y: \widetilde{HPD}(\tau_1 CB \tau_2,\pi)\to \widetilde{HPD}(\tau_1 BC \tau_2,\pi)$ changing only those
  two rows, using the replacements below:

  \newcommand\Dashbox[1]{\dashbox{$#1$}}

% \begin{tabular}{m{3em} m{1em} m{3em} m{3em} m{3em} m{1em} m{3em} m{3em} m{3em} m{1em} m{3em}}
$$
  \arraycolsep=3pt
\begin{array}{ccccccc} % \def\arraystretch{2.2}
  \mapstoo{ \tilecool\bcross\cross{1}{3}{2}{3}{1}{2}  }
  {\tilecool\cross\bcross{3}{1}{2}{1}{3}{2} }
  & \qquad &
      \mapstoo{  \tilecool\sea\cross{1}{\es}{\es}{2}{1}{2} }
             { \tilecool\hwire\sea{\es}{1}{\es}{1}{2}{2} }
    & \qquad &
{      \mapstoo{ \tilecool\vwire\cross{1}{\es}{2}{\es}{1}{2}  }
               { \tilecool\cross\vwire{\es}{1}{2}{1}{\es}{2}}
               }
    & \qquad &
    \mapstoo{ \tilecool\bhwire\hwire{1}{2}{\es}{2}{1}{\es}}
    {\tilecool\hwire\bhwire{2}{1}{\es}{1}{2}{\es}}
    \\
\boxed{      \mapstoo{ \tilecool\bhwire\hwire{1}{1}{\es}{1}{1}{\es}}
      {\tilecool\nowire\bnowire{\es}{\es}{\es}{\es}{\es}{\es}}  }
    & &
     \mapstoo{ \tilecool\en\hwire{1}{2}{2}{\es}{1}{\es}}
      {\tilecool\cross\en{2}{1}{2}{1}{\es}{\es}} 
    & &
\boxed{    \mapstoo{ \tilecool\en\hwire{1}{1}{1}{\es}{1}{\es}}
    {\tilecool\are\bnowire{\es}{\es}{1}{1}{\es}{\es}} }
    & &
      \mapstoo{ \tilecool\bnowire\hwire{1}{\es}{\es}{\es}{1}{\es}}
      {\tilecool\hwire\bnowire{\es}{1}{\es}{1}{\es}{\es}}
    \\
      \mapstoo{  \tilecool\bhwire\jay{1}{2}{\es}{2}{\es}{1}}
      {\tilecool\jay\bcross{2}{1}{\es}{\es}{2}{1}}
    & &
    \mapstoo{ \tilecool\bhwire\jay{1}{1}{\es}{1}{\es}{1} }
    {\tilecool\nowire\sea{\es}{\es}{\es}{\es}{1}{1}}
    & &
      \mapstoo{ \tilecool\en\jay{1}{2}{2}{\es}{\es}{1}}
      {\tilecool\bump\bcross{2}{1}{2}{2}{2}{1}}
    & &
      \mapstoo{ \tilecool\en\jay{1}{1}{1}{\es}{\es}{1}}
      {\tilecool\are\sea{\es}{\es}{1}{1}{1}{1}}
    \\
\boxed{    \mapstoo{ \tilecool\bnowire\jay{1}{\es}{\es}{\es}{\es}{1}}
    {\tilecool\hwire\sea{\es}{1}{\es}{1}{1}{1}} }
    & &
      \mapstoo{ \tilecool\bcross\are{\es}{2}{1}{2}{1}{\es}}
      {\tilecool\are\bhwire{2}{\es}{1}{1}{2}{\es}}
    & &
\Dashbox{      \mapstoo{ \tilecool\sea\are{\es}{\es}{\es}{1}{1}{\es}}
      {\tilecool\jay\en{0}{0}{\es}{\es}{\es}{\es}} }
    & &
\Dashbox{    \mapstoo{ \tilecool\vwire\are{\es}{\es}{1}{\es}{1}{\es}}
    {\tilecool\bump\en{0}{0}{1}{1}{\es}{\es}} }
    \\
      \mapstoo{ \tilecool\bcross\bump{1}{3}{2}{3}{2}{1}}
      {\tilecool\bump\bcross{3}{1}{2}{2}{3}{1}}
    & &
      \mapstoo{ \tilecool\bcross\bump{2}{2}{1}{2}{1}{2}}
      {\tilecool\are\sea{\es}{\es}{1}{1}{2}{2}}
    & &
\Dashbox{    \mapstoo{ \tilecool\sea\bump{1}{\es}{\es}{2}{2}{1}}
    {\tilecool\jay\vwire{\es}{1}{\es}{\es}{\es}{1}} }
    & &
      \mapstoo{ \tilecool\vwire\bump{1}{\es}{2}{\es}{2}{1}}
      {\tilecool\bump\vwire{\es}{1}{2}{2}{\es}{1}}
    \\
      \mapstoo{ \tilecool\bhwire\nowire{\es}{1}{\es}{1}{\es}{\es}}
      {\tilecool\nowire\bhwire{1}{\es}{\es}{\es}{1}{\es}}
    & &
    \mapstoo{ \tilecool\en\nowire{\es}{1}{1}{\es}{\es}{\es}}
    {\tilecool\are\bhwire{1}{\es}{1}{1}{1}{\es}}
  & &
      \boxed{
      \mapstoo{ \tilecool\bnowire\nowire{\es}{\es}{\es}{\es}{\es}{\es}}
      {\tilecool\hwire\bhwire{1}{1}{\es}{1}{1}{\es}} }
  \end{array}
  $$
\end{Theorem}

The proof is a not-too-laborious case check, determining all the ways to 
glue a classic tile atop a bumpless tile (or vice versa), then to
specify the possible connectivities; finally to match the 23 of each,
preserving the net connectivity (up to potential pipes entering and leaving
on the same side!) and the number of weighty tiles.
It takes some additional analysis to confirm that the replacement tiles
glue together. Uniqueness follows (slightly non-trivially) from the rest of the lemmas and theorems of this section and the observation that each of the 23 pairs of tiles has a distinct connectivity (i.e. no two have the same labels around the outside). 
The four \fbox{boxed} replacements are the ones that would
not preserve the monomial if the weighty tiles were labeled $x$
rather than $-y$. We spend the rest of the abstract dealing with them
(and bringing the \dashbox{dash-boxed} replacements along for the ride).

%	For any $\pi\in S_n$ and any $\tau_1,\tau_2\in \{O,B\}^n,$ decorated $\tau_1$-hybrid pipe dreams for $\pi$, and decorated $\tau_2$-hybrid pipe dreams compute the same thing so there must exist some weight preserving bijection between them. Moreover, for $\pi$ such that $\mathcal{S}_\pi$ has coefficients greater than 1 in absolute value, there will necessarily exist multiple such bijections. In fact, there are often multiple row swapping bijections available that only modify the two rows being swapped. 
	
	
	
	
	
	
%	We introduce an additional desirable property such that our row-swapping bijection is the unique weight-preserving row-swapping bijection which has this property. In order to introduce this property we first define $OB$-\textbf{dreams} to be $2\times k$ rectangles of tiles that could appear as a contiguous subrectangle of some $\tau$-hybrid pipe dream contained in the $i$th and $i+1$th row for some $i$ such that $\tau_i=O$ and $\tau_{i+1}=B$. Similarly, we define $B\atop O$ sections as $2\times k$ rectangles of tiles that could appear as a contiguous subrectangle of some $\tau$-hybrid pipe dream contained in the $i$th and $i+1$th row for some $i$ such that $\tau_i=O$ and $\tau_{i+1}=B$. Given the desire for the row swapping bijection to affect only the rows being swapped, it makes sense for us to focus attention on pairs of adjacent rows, one of type $O$ and one of type $B$, occurring within some hybrid pipe dream; these constitute a proper subset of $O\atop B$ sections and $B\atop O$ sections which we will call \textbf{full} $O\atop B$ sections and \textbf{full} $B\atop O$ sections. The row-swapping bijection we seek only needs to be defined on full $O\atop B$ sections; however, we define our bijection by defining it for a subset of small $O\atop B$ sections and describing how to break a full $O\atop B$ section into these pieces. We show that our bijection between full $O\atop B$ sections and full $B\atop O$ sections is the only possible bijection which can be computed by breaking up an $O\atop B$ section into these smaller pieces and computing the bijection on these pieces. \textcolor{red}{Some care needs to be taken and I need to think carefully about what exactly is true and how to say it.}


%A bijection that only affects two rows at a time is appealing because it is somewhat local in nature. %In a completely local bijection f $i$th and $(i+1)$st row would give the tiles in positions $(i,j)$ and $(i+1,j)$ after swapping the rows i

%the $i$'th column of the two rows being swapped only in terms of the tiles in the $i$th column of the 

% Ideally we would be able to find a completely local bijection, i.e. one that woul



\junk{Define a $CB$-\textbf{dream} to be a tiled $2\times k$ rectangle 
that could appear in the $i$th and $(i+1)$st row (and $k$ contiguous columns)
of some $\tau$-hybrid pipe dream, where $\tau_i=C$ and $\tau_{i+1}=B$ .
 Define $B\atop C$ sections similarly.}
 
%as $2\times k$ rectangles of tiles that could appear as a contiguous
%subrectangle of some $\tau$-hybrid pipe dream contained in the $i$th
%and $i+1$th row for some $i$ such that $\tau_i=O$ and
%$\tau_{i+1}=B$.
%Given the desire for the row swapping bijection to
%affect only the rows being swapped, it makes sense for us to focus
%attention on pairs of adjacent rows, one of type $O$ and one of type
%$B$, occurring within some hybrid pipe dream; these constitute a
%proper subset of $O\atop B$ sections and $B\atop O$ sections which we will call
%\textbf{full} $O\atop B$ sections and \textbf{full} $B\atop O$ sections.

When a $C\atop B$ section or $B\atop C$ section is of (full) width $n$, call it \defn{full}. The major property that full $C\atop B$ sections ($B\atop C$ sections) have,
that non-full ones need not, 
is that a full
$C\atop B$ section ($B\atop C$ section) has blank labels on the right edge of its top
row and the left edge of its bottom row (left edge of its top row and
right edge of its bottom row) and blank labels on its other vertical
edges, whereas an arbitrary $C\atop B$ section ($B\atop C$ section) may have any
combination of blank and non-blank labels on its vertical edges. 

% {\bf picture}

Two $C\atop B$ sections ($B\atop C$ sections) are said to be the same if they have the
same tiles and the same connectivity. As in \cref{thm:justy}, we put labels on $C\atop B$ sections
($B\atop C$ sections) in order to show their connectivity and not to indicate
the labels the pipes may have had within some larger hybrid pipe dream.


%We (re)label all $O\atop B$ sections ($B\atop O$ sections) in a consistent manner so its easy to compare them.
%The labels tell us where pipes are and how they connect, but they
%are not a completely essential part of the data encoding an $O\atop B$ section
%($B\atop O$ section). That is, the data of a full $OB$-deam ($B\atop O$ section) is
%the $2\times k$ grid of tiles making it up, and we will say that two
%$O\atop B$ sections are the same if they have the same tiles, even if we think
%about them as subrectangles of different hybrid pipe dreams (that may
%have given different labels to the pipes). \textcolor{red}{this is
%  blatantly false but I'm leaving it here for now so I can hopefully
 % get to other stuff.} Although the precise labels on an $O\atop B$ section
%aren't important, we still want to be able to describe the
%connectivity of them. To do this, we will apply labels in a consistent
%manner. In particular, we label the pipes from $0,1,\dots, p,p+1$
%based on where they exit the $O\atop B$ section ($B\atop O$ section) so that a pipe
%that exits on the left edge of the ordinary row is labeled $1$ (unless
%it is in a $B\atop O$ section and originally entered through the left edge of
%the bumpless row, in which case we label it $0$), labels of pipes are
%increasing left to right across the bottom edge of the $O\atop B$ section
%($B\atop O$ section), a pipe exiting the $O\atop B$ section ($B\atop O$ section) through the
%right edge of the bumpless row is labeled $p$, and pipes have the same
%label iff they are the same pipe, and all labels between $1$ and $p$
%are used. The final condition of the previous sentence is equivalent
%to saying that for both $O\atop B$ sections and $B\atop O$ sections, $p$ will be the
%number of pipes or 1 less than the number of pipes if it is a $BO$
%dream where a pipe enters through the left edge of the bumpless row
%and exits through the right left edge of the ordinary row.
%\textcolor{red}{This is kind of a lot of explaining to do for
%  something that is just notational and not all that important. Soooo
%  it turns out this labeling scheme needs to change. A pipe needs to
%  be labeled by its final exit and not its first exit}
% \textcolor{red}{Here are some examples (TODO):}




\junk{If $D$ is a $2\times k$ $C\atop B$ section and $D'$ is a $2\times k$ $B\atop C$ section then we will say \defn{the boundaries of $D$ and $D'$ are the same} and write $\partial D = \partial D'$ if the labels along the top and bottom edges of $D$ and $D'$ are the same, $C_0(D)=C_0(D')$,  and $C_k(D)=C_k(D')$. \textcolor{red}{Once again I need to return to this definition once proofs are finished so that it gets defined in whichever way turns out to be best.}}%end of junk

\junk{Clearly, if $D$ and $D'$ are both $C\atop B$ sections ($B\atop C$ sections) and
$\partial D'=\partial D,$ then $i$ is separating for $D$ iff it is
separating for $D'$. Less obviously, if $D$ is a $C\atop B$ section and $D'$
is a $B\atop C$ section such that $\partial D= \partial D',$ then $i$ will be
separating for $D$ iff $i$ is separating for $D'$.  Further, if
$C_i(D)=(x,y)$ where $x\neq y$, then not only is $i$ separating for
both $D$ and any $C\atop B$ section ($B\atop C$ section) $D'$ with
$\partial D'=\partial D$, but also $C_i(D')=C_i(D).$ When $i$ is
separating for a full $C\atop B$ section ($B\atop C$ section) $D$, $D'$ is some full
$B\atop C$ section ($C\atop B$ section) with $\partial D'=\partial D$, then
$C_i(D)=(\emptyset, \emptyset)$ iff $C_i(D')=(y,y)$ for some $y$ and
$C_i(D)=(x,x)$ for some $x$ iff $C_i(D')=(\emptyset,\emptyset)$. These
statements are not particularly obvious and we will delay their proofs
until later. For now, they motivate the following definition.
}%Junk closed here


\junk{\begin{Definition}
  If $\phi$ is a map from (some subset of) $C\atop B$ sections to $B\atop C$ sections such that
  \[
    C_i(\phi_y(D))
    = \begin{cases}
        C_i(D) & C_i(D)=(x,y) \text{ with } x\neq y \\
        (\emptyset,\emptyset) & C_i(D)=(x,x) \text{ for some } x\\
        (y,y) & C_i(D)=(\emptyset, \emptyset)
      \end{cases}
    \]
  then we say $\phi$ \defn{separates} $D$ at $i$. 
  Note that the preceding definition is an abuse of notation because
  $y$ is an unbound variable; what is meant is simply that when
  $C_i(D)=(\emptyset,\emptyset),$ $C_i(\phi_y(D))=(y,y)$ for some $y$.
\end{Definition}} %Junked ending
	
%The weighty tiles in a $C\atop B$ section ($B\atop C$ section) are
%$\inlinehwire$ and $\inlinecross$ in the ordinary row and
%$\inlinenowire$ in the bumpless row.
We attach decorations $f$ to
$C\atop B$ sections ($B\atop C$ sections) in the same way that we did with hybrid pipe
dreams. To assign weights to $C\atop B$ sections ($B\atop C$ sections) in the context
of a complete hybrid pipe dream we need to know the labels on the two
rows. However, for the purpose of verifying that our bijection
preserves weights, we will say the classic row is labeled $c$ and the
bumpless row is labeled $b$ so that we attach the weight $x_c$ to
weighty tiles decorated with $x$ in the classic row and $x_b$ to
weighty tiles decorated with $x$ in the bump
less row.

%\input{former.tex}



\input{later.tex}

%\input{diag2.tex}


\acknowledgements{We thank Daoji Huang and Paul Zinn-Justin for many
  pertinent conversations. We are grateful to Zachary Hamaker and Adam
  Gregory for writing some of the \LaTeX{} macros used here.
  Finally, we thank the reviewers for improvement of the exposition.}

%% if you use biblatex then this generates the bibliography
%% if you use some other method then remove this and do it your own way
\printbibliography



\end{document}



Notes: 
-Replace `good' with `separating'
-use \defn in Definitions
-x_o and x_b
-can you correspond HPDs to polytopes so bijection is piecewise linear continuous maps between things....relate to gelfand cetlin










\section{Membrane dreams ($3$-d pipe dreams)}


\section{A first proof of the formula}

We will prove a stronger inductive statement.

\begin{Theorem}\label{thm:inductive}
  Let $\lambda$ be a string of length $n$ in the alphabet $1,2,\ldots,d$
  and
  ``\ '' (blank). Order the alphabet as $1<2<\ldots<d<$``\ '',
  and break ties in $\lambda$ without creating inversions, e.g.
  $$ \lambda = 1\ 2\ 1 - 1 - 2 \qquad \mapsto \qquad
  \lambda_+ = 1.0\ \ 2.0\ \ 1.1\ \ 3.0\ \ 1.2\ \ 3.1\ \ 2.1
  \qquad \sim 1426375 \text{ as an order} $$
  then let $\pi(\lambda)$ be the inverse permutation to this order ($1352746$
  in this example). In one-line notation, $\pi(\lambda)$ lists the positions of
  the $1$s (from left to right), then the $2$s and so on, and finally
  the blanks.

  Let $\tau \in \{C,B\}^{n-\#\text{blanks in $\lambda$}}$, and consider a
  rectangular variant on hybrid pipe dreams:
  \begin{itemize}
    \setlength\itemsep{.1em}
  \item the rows are classic vs.\ bumpless according to $\tau$, the
    bottom edges are again all blank, and no two pipes cross twice
  \item the top side is labeled with $\lambda$, from left to right
  \item the labels on the left and right sides are the non-blank letters
    in $\lambda$, but sorted to be weakly increasing
  \item no two pipes with the same label may cross even once.
  \end{itemize}

  Then $S_{\pi(\lambda)}$ is given by the same formula as in \cref{thm:formula}, except as a sum over these
  $(n-\#\text{blanks in $\lambda$})\times n$ rectangular $\tau$-hybrid
  pipe dreams, where each weighty tile contributes a factor of
  $x_{i+\#\text{blanks in $\lambda$}} - y_j$
  where $i$ is determined by the row label {\em as tie-broken in $\lambda_+$}.
\end{Theorem}

{\em Example: $\lambda = 1\ 2\ 1 - 1 - 2$, $\tau = CBCBC$.}

\begin{proof}
  The proof is by induction on the number of non-blank labels in
  $\lambda$.  In the base case, $\lambda$ is all blanks, $\pi$ is the
  identity, and $S_\pi=1$ arises as the sum over the unique pipe dream
  (a $0\times n$ rectangle).

  When $\lambda$ is not all blanks, we group the rectangular pipe
  dreams based on their first row $R$ of tiles. Each such $R$ is easily
  seen to be determined by its top edge labels $top(R) = \lambda$ and
  its bottom edge labels $bot(R)$. Let $\ell$ be the end label
  ($1$, on the left, if $\tau$ starts with $C$;
  $\#$(non-blanks in $\lambda$), on the right, if $\tau$ starts with $B$).
  Then by induction the theorem reduces to proving
  \begin{equation}     \tag{$\ast$}     \label{eq:onerow} 
    S_\pi = \sum_R S_{\pi(bot(R))}   \prod_{\text{weighty tiles in $R$}}
    (x_{\ell + \#\text{blanks in $\lambda$}} - y_{\text{column}}) 
  \end{equation}

  Note that the right-hand side only depends on the first letter in $\tau$;
  in particular it is enough to prove ($\ast$) when 
  {\em $\tau$ is all $C$s or all $B$s.} 
  We now split into those two cases.

  {\em $\tau$ begins with $C$.} 

  {\em $\tau$ begins with $B$.} 
\end{proof}

One can also handle the two cases in the proof directly from the
cotransition and transition formul\ae, respectively, but this is not an
especially different approach since the classic and bumpless pipe dream
formul\ae\ amount to unwindings of cotransition and transition
(\cite{cotransition} and \cite{Weigandt} respectively).

