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\title[Plane partitions and rowmotion]{Plane partitions and rowmotion on rectangular and trapezoidal posets}

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\author{Joseph Johnson\thanks{\href{mailto:josjohn@kth.se}{josjohn@kth.se}}\addressmark{1} \and Ricky Ini Liu\thanks{\href{mailto:riliu@uw.edu}{riliu@uw.edu}. The second author was partially supported by grants from the National Science Foundation (DMS 1700302/2204415 and CCF-1900460).}\addressmark{2}}

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\address{\addressmark{1}Department of Mathematics, KTH Royal Institute of Technology, Stockholm, Sweden \\ \addressmark{2}Department of Mathematics, University of Washington, Seattle, WA}

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\abstract{We define a birational map between labelings of a rectangular poset and its associated trapezoidal poset. This map tropicalizes to a bijection between the plane partitions of these posets of fixed height, giving a new bijective proof of a result by Proctor. We also show that this map is equivariant with respect to birational rowmotion, resolving a conjecture of Williams and implying that birational rowmotion on trapezoidal posets has finite order. }

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\keywords{birational rowmotion, plane partitions, trapezoidal posets}

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\begin{document}

\maketitle
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\section{Introduction}


For a finite poset $P$, a \emph{plane partition} of $P$ (also known as a \emph{$P$-partition}) is an order-preserving labeling of $P$ with nonnegative integers. %Plane partitions and their variations arise in a number of important contexts in combinatorics, representation theory, and related areas; see, for instance, \cite{stanley4} for an overview. 
When $P$ is the \emph{rectangular poset} $R_{r,s}$, the Cartesian product of two chains of $r$ and $s$ elements, an elegant product formula for the number of plane partitions of $P$ with maximum label at most $\ell$ was given by MacMahon \cite{macmahon}. Surprisingly, Proctor \cite{proctor} showed that there is another poset, namely the \emph{trapezoidal poset} $T_{r,s}$, that has the same number of plane partitions with maximum label at most $\ell$ for all $\ell$. (See Figure~\ref{fig:rectangleandtrapezoid} for a depiction of $R_{4,3}$ and $T_{4,3}$.)

Proctor's proof relies on a branching rule for Lie algebra representations and is not bijective.
%---he describes the question of giving a combinatorial correspondence between these two sets of plane partitions as ``a complete mystery.'' 
Partial bijections were later constructed by Stembridge \cite{stembridge} and Reiner \cite{reiner} for $\ell=1$, and Elizalde \cite{elizalde} for $\ell=2$, but a full bijection for all $\ell$ was not given until work of Hamaker, Patrias, Pechenik, and Williams \cite{hamakerpatriaspechenikwilliams} using $K$-theoretic jeu de taquin. 

Although the bijection given in \cite{hamakerpatriaspechenikwilliams} has many nice properties, it also has some shortcomings. First, it cannot be extended in a natural way to a continuous piecewise-linear map on real-valued labelings of the rectangle and trapezoid. As a result, it cannot be written using expressions in the tropical semiring (that is, using the operations addition, subtraction, and maximum). Second, it does not appear to be generally well-behaved with respect to a certain map on posets called \emph{rowmotion}.

(Combinatorial) rowmotion is a term coined by Striker and Williams \cite{strikerwilliams} to describe a map first studied by Brouwer and Schrijver \cite{brouwerschrijver} that permutes the set of order ideals of a poset, sending an order ideal $I \subseteq P$ to the order ideal generated by the minimal elements of $P \setminus I$. It was shown in \cite{brouwerschrijver} that the action of rowmotion on order ideals of the rectangle $R_{r,s}$ has order exactly $r+s$. 
%Since then, rowmotion has received much study; see, for instance, \cite{cameronfonderflaass, fonderflaass, joseph, propproby, striker, strikerwilliams, thomaswilliams}. 
Einstein and Propp \cite{einsteinpropp1} observed that one can generalize combinatorial rowmotion to a piecewise-linear map or birational map. Results about birational rowmotion then descend to results for piecewise-linear rowmotion (via tropicalization) and further to combinatorial rowmotion. Birational rowmotion on rectangular posets is closely related to the \emph{birational Robinson--Schensted--Knuth (RSK) correspondence}, also known as \emph{tropical} or \emph{geometric RSK}---see \cite{noumiyamada} for some discussion. 

For rectangular posets, birational rowmotion maintains many of the important dynamical properties of combinatorial rowmotion. For example, Grinberg and Roby \cite{grinbergroby2} showed that birational rowmotion on $R_{r,s}$ still has finite order $r+s$, which was observed by Glick and Grinberg \cite{grinbergroby3, musikerroby} to be equivalent to a phenomenon from discrete dynamics known as \emph{type AA Zamolodchikov periodicity} (see Volkov \cite{volkov}). However, the class of posets for which birational rowmotion is known to have finite order is very small \cite{grinbergroby2, grinbergroby1}. Grinberg and Roby conjecture that birational rowmotion on $T_{r,s}$ likewise has finite order $r+s$. (See also \cite{galashinpylyavskyy} for more on conjectured good behavior of the related \emph{$R$-systems}.) %(For more on the conjectural good behavior of birational rowmotion and the related \emph{$R$-systems}, see Galashin--Pylyavskyy~\cite{galashinpylyavskyy}.)

Given the apparent close relationship between the rectangular and trapezoidal posets, Williams conjectures (as noted in \cite{grinbergroby2}, based on work in \cite{williams}) that there should exist a birational map between labelings of $R_{r,s}$ and $T_{r,s}$ that intertwines with the action of rowmotion (see also Hopkins \cite{hopkins2} for further discussion). In particular, such a map would prove that birational rowmotion on $T_{r,s}$ has finite order $r+s$.  In work of Dao, Wellman, Yost-Wolff, and Zhang \cite{daowellmanyostwolffzhang}, it was shown that the bijection given in \cite{hamakerpatriaspechenikwilliams} does intertwine with combinatorial rowmotion on plane partitions of height 1, thereby showing that combinatorial rowmotion on $T_{r,s}$ has the correct order. However, they also note that it does not respect piecewise-linear or birational rowmotion, so it cannot be used to prove periodicity on $T_{r,s}$ in these cases.

Our main result is to settle these questions. We construct a birational map between labelings of $R_{r,s}$ and $T_{r,s}$ using birational toggles \cite{cameronfonderflaass,einsteinpropp1}. We then show that this map:
\begin{itemize}
	\item tropicalizes to a continuous, piecewise-linear map that restricts to a bijection between plane partitions of $R_{r,s}$ and $T_{r,s}$ of height at most $\ell$, and
	\item is equivariant with respect to rowmotion on $R_{r,s}$ and $T_{r,s}$, implying that birational (as well as piecewise-linear and combinatorial) rowmotion on $T_{r,s}$ has order $r+s$.
\end{itemize}
%Our map is primarily built from \emph{birational toggles} \cite{cameronfonderflaass,einsteinpropp1}. 
We also generalize the \emph{chain shifting lemma} proved by the current authors in \cite{johnsonliu} (see also the noncommutative conversion lemma by Grinberg--Roby \cite{grinbergroby3}), which is closely related to Sch\"utzenberger promotion on semistandard Young tableaux \cite{johnsonliu2}.
%, and also appears in a different form in the study of the hidden invariance of last passage percolation by Dauvergne \cite{dauvergne}. 
In particular, we derive a new, simple proof of this lemma based on the duality of plane trees.

The full version of this paper, which includes proofs, can be found at \cite{johnsonliu3}.

%To prove that our birational map respects rowmotion, we utilize an alternate construction of rowmotion known as \emph{birational antichain rowmotion}, introduced by Joseph and Roby \cite{josephroby2}. Our construction deforms labelings of the trapezoid into labelings of the rectangle via certain intermediate posets, and this construction allows us to define a version of rowmotion on these intermediate posets that respects these deformations.

%\medskip

%The outline of this paper is as follows. In Section~\ref{sec:paper3Background}, we introduce some background and preliminary results about posets and rowmotion. In Section~\ref{section:skewChainShifting}, we show how the action of rowmotion on chains in the poset can be described in terms of weighted arborescences. We then define a key bijection $\aleph$ on arborescences and show how it can be used to give a straightforward combinatorial proof of the chain shifting lemma from \cite{johnsonliu} as well as a number of important generalizations to other skew shapes, including trapezoids. In Section~\ref{section:equivariantMap}, we use rowmotion to construct a birational map between labelings of the rectangle and trapezoid and use the chain shifting lemma to prove that, when tropicalized, it can be used to give a bijection between plane partitions of the same height. (A detailed example of the bijection on plane partitions is given in Section~\ref{sec:polytopes}.) In Section~\ref{sec:rowmotionequivariance}, we prove that the map intertwines with rowmotion on the rectangle and trapezoid, which in particular implies that birational rowmotion on the trapezoid has finite order. We conclude in Section~\ref{sec:conclusion} with some possible directions for future study.



\section{Background}
\label{sec:paper3Background}

%\subsection{Posets and labelings}

%We will assume all posets to be finite. 
%Given a poset $P$ and a set $K$, a \emph{$K$-labeling} (or just a \emph{labeling} if $K$ is clear) of $P$ is an element $x = (x_p)_{p \in P} \in K^P$, that is, an assignment of an element of $K$ to each element of $P$.
%
%\begin{Rem} \label{rem:semiring}
%	We usually take $K$ to be the positive real numbers $(\RR_+,+,\cdot)$, except when discussing plane partitions or polytopes when $K$ will be the \emph{tropical semiring} $(\RR, \max, +)$.
%	Typically $K$ can be a field or a semifield (commutative semiring with multiplicative inverses) such as the positive real numbers $(\RR_+, +, \cdot)$. For simplicity, we will state most of our results using $K = \RR_+$, except when discussing plane partitions or polytopes in which case we will let $K$ be the \emph{tropical semiring} $(\RR, \max, +)$. Regardless, most of our results will hold in general for fields (provided one takes the necessary precautions such as using rational maps) and semirings. We refer the reader to Grinberg-Roby \cite{grinbergroby1} for further discussion, as well as Noumi-Yamada \cite{noumiyamada} for discussion of tropicalization of rational expressions.
%	
%	In the case of semirings, all of our formulas will be subtraction-free until Section 5 so the same proofs will apply. However, even in Section 5, as noted in \cite[Section 3]{grinbergroby1}, any equality between subtraction-free rational expressions that holds over every field  also holds over semirings, so in particular Theorem~\ref{thm:equivariance} will hold for semirings as well. 
%\end{Rem}

%A poset $P$ is called \emph{bounded} if it has unique minimum and maximum elements. 
%We can augment a poset $P$ to a poset $\widehat P$ by adding a minimum element $\hat 0$ and a maximum element $\hat 1$ such that $\hat 0 < p < \hat 1$ for all $p \in P$. 
%We will naturally identify $P$ with the induced subposet $\widehat P \setminus \{\hat 0, \hat 1\}$. %Thus any labeling of $\widehat P$ can be restricted to a labeling of $P$, and any labeling of $P$ yields a labeling of $\widehat P$ if we specify the labels at $\hat 0$ and $\hat 1$.

%We will call an induced subposet $I$ of $P$ \emph{saturated} if all of its cover relations $p \lessdot q$ are cover relations of $P$. In other words, $I$ is saturated if the Hasse diagram of $I$ is a subgraph of the Hasse diagram of $P$. %Note that any maximal chain of a saturated subposet $I$ is a saturated chain in $P$. By default, we will assume that all subposets of interest are saturated, and we will use the word \emph{chain} to refer to a saturated chain.


%\subsection{Rectangles and Trapezoids}

%Throughout this paper, fix positive integers $r \geq s$. 
%We will use $[n]$ to denote the set $\{1, 2, \dots, n\}$ with the usual total order.

%Given two posets $P$ and $Q$, the \emph{Cartesian product} $P \times Q$ is the poset of pairs $(p,q)$ for $p \in P$ and $q \in Q$ with $(p,q) \leq (p', q')$ if $p \leq p'$ in $P$ and $q \leq q'$ in $Q$.

We begin with some background on posets and rowmotion. 
Fix positive integers $r \geq s$.

\begin{Def}
	The \emph{rectangle poset} $R_{r,s}$ is the Cartesian product of chains $[r] \times [s]$.
	
	The \emph{right trapezoid poset} $RT_{r,s}$ is the induced subposet
	$\{(i,j) \mid i-j < r\} \subset R_{r+s-1,s}$.
	
	The \emph{trapezoid poset} $T_{r,s}$ is the induced subposet
	$\{(i,j) \mid i+j > s\} \subset RT_{r,s}$.
\end{Def}



%Note that $R_{r,s}$ is also an induced subposet of $RT_{r,s}$, and $|R_{r,s}| = |T_{r,s}| = rs$.

See Figure~\ref{fig:rectangleandtrapezoid}. We draw our posets oriented in the plane so that the first coordinate increases to the northwest and the second coordinate increases to the northeast.

\begin{figure}
	\centering
	\begin{tikzpicture}[scale = 0.6]
		%VERTICAL
		\foreach \x in {0,...,2}
		\draw (\x,\x)--(-3,3+2*\x);
		
		%HORIZONTAL
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		\draw (-\x,\x)--(2-\x,2+\x);
		
		\foreach \x in {0,...,1}
		\draw (-3,5+2*\x)--(-2-\x,6+\x);
		
		\foreach [count=\i,
		evaluate=\i as \xmin using int(2-\y)] \y in {0,...,2} {
			\foreach \x in {\xmin,...,3} {
				\draw [fill=black] (- \x + \y, \x + \y) circle [radius=0.15cm];
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		\foreach \x in {0,...,1}
		\foreach \y in {\x,...,1}
		\draw [fill=black] (-\x + \y-3,\x + \y+5) circle [radius=0.15cm];
		
		\draw (0,0) [fill=black] circle [radius=0.15cm];
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		\node at (-3.9,3) {(4,1)};
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		\node at (-3.9,7) {(6,3)};
		
		
		
		\begin{scope}[shift = {(7,0)}]
			%VERTICAL
			\foreach \x in {0,...,2}{
				\draw (\x,\x)--(-3+\x,3+\x);
				\draw [dashed] (-3+\x,3+\x)--(-3,3+2*\x);
			};
			
			%HORIZONTAL
			\foreach \x in {0,...,3}
			\draw (-\x,\x)--(2-\x,2+\x);
			
			\foreach \x in {0,...,1}
			\draw [dashed] (-3,5+2*\x)--(-2-\x,6+\x);
			
			%CIRCLES
			\foreach \x in {0,...,3}
			\foreach \y in {0,...,2}
			\draw [fill=black] (-\x + \y,\x + \y) circle [radius=0.15cm];
			
			\foreach \x in {0,...,1}
			\foreach \y in {\x,...,1}
			\draw [fill=white] (-\x + \y-3,\x + \y+5) circle [radius=0.15cm];
		\end{scope}
		
		
		
		
		
		
		\begin{scope}[shift = {(14,0)}]
			%VERTICAL
			\foreach \x in {0,...,2}{
				\draw (-2+2*\x,2)--(-3,3+2*\x);
				\draw [dashed] (\x,\x)--(-2+2*\x,2);
			};
			
			%HORIZONTAL
			\foreach \x in {0,...,2}{
				\draw (2-2*\x,2)--(2-\x,2+\x);
				\draw [dashed] (-\x,\x)--(2-2*\x,2);
			};
			
			\foreach \x in {-1,...,1}
			\draw (-3,5+2*\x)--(-2-\x,6+\x);
			
			%COLUMN 1
			\foreach \x in {0,...,1}
			\draw [fill=white] (-\x ,\x) circle [radius=0.15cm];
			
			\foreach \x in {2,...,3}
			\draw [fill=black] (-\x ,\x) circle [radius=0.15cm];
			
			%COLUMN 2
			\draw [fill=white] (1,1) circle [radius=0.15cm];
			
			\foreach \x in {1,...,4}
			\draw [fill=black] (-\x+1 ,\x+1) circle [radius=0.15cm];
			
			%COLUMN 3
			\foreach \x in {0,...,5}
			\draw [fill=black] (-\x+2 ,\x+2) circle [radius=0.15cm];
		\end{scope}
		
	\end{tikzpicture}
	\caption{The right trapezoid $RT_{4,3}$, the rectangle $R_{4,3}$, and the trapezoid $T_{4,3}$. %The labeled points $(i,j)$ on the left border of the right trapezoid satisfy $i-j = r-1$.
	}
	\label{fig:rectangleandtrapezoid}
\end{figure}


%\subsubsection{Chain and order polytopes}

%Following Stanley \cite{stanley2}, we can associate with a poset $P$ two polytopes called the chain and order polytopes.
%
%\begin{Def}
%	The \emph{chain polytope} $\cp \subset \RR^P$ is the set of all $\RR$-labelings $x = (x_p)_{p \in P}$ such that $x_p \geq 0$ for all $p \in P$, and $\sum_{p \in C} x_p \leq 1$ for all (maximal) chains $C \subset P$.
%	
%	The \emph{order polytope} $\op \subset \RR^P$ is the set of all $\RR$-labelings $y = (y_p)_{p \in P}$ such that $0 \leq y_p \leq 1$ for all $p \in P$, and $y_p \leq y_q$ if $p \leq q$ in $P$.
%\end{Def}
%
%\begin{Def}
%	\label{Def:planePartition}
%	A \emph{plane partition} of $P$ (or \emph{$P$-partition}) is a $\mathbb Z_{\geq 0}$-labeling $x \in \mathbb{Z}_{\geq 0}^{P}$ such that $x_p \leq x_q$ whenever $p \leq q$ in $P$.
%	
%	We say a plane partition $x$ has \emph{height $\ell$} if the maximum label appearing in $x$ is $\ell$.
%\end{Def}
%
%Note that plane partitions of height at most $\ell$ are exactly the lattice points in $\ell \mathcal O(P)$. As shown by Stanley \cite{stanley2}, there exists a continuous, piecewise-linear, volume-preserving map between the chain and order polytopes called the \emph{transfer map}, which we will define in Section~\ref{sec:rowmotion}.


% Stanley \cite{stanley2} defines a continuous, piecewise-linear, volume-preserving map $\phi \colon \RR^P \to \RR^P$ called the \emph{transfer map} by
% \[\phi(x)_p = x_p - \max\limits_{q \lessdot p} x_q,\]
% where we interpret an empty $\max$ as $0$.
% The inverse of this map is given by
% \[\phi^{-1}(x)_p = \max\limits_{c_1 < c_2 < \dots < c_k = p} \left( \sum\limits_{i=1}^k x_{c_i} \right).\]
% As shown in \cite{stanley2}, $\phi$ maps $\op$ bijectively onto $\cp$, so these two polytopes have the same volume. Even more, $\phi$ preserves the lattice $\frac 1k \ZZ^{P}$ for any positive integer $k$, so $k \cp$ and $k \op$ have the same number of lattice points for all positive integers $k$. In other words, $\cp$ and $\op$ have the same Ehrhart polynomial.





%For the rectangle poset $R_{r,s}$, the number of plane partitions is given by the following formula.
%\begin{Th}[MacMahon \cite{macmahon}]
%	The number of plane partitions of $R_{r,s}$ of height at most $\ell$ is
%	\[N(r,s,\ell) = \prod_{i=1}^r \prod_{j=1}^s \prod_{k=1}^\ell \frac{i+j+k-1}{i+j+k-2}.\]
%\end{Th}

%In 1983, Proctor used a branching rule for the inclusion of Lie algebras $\mathfrak{sp}_{2n} (\mathbb C) \hookrightarrow \mathfrak{sl}_{2n}(\mathbb C)$ to prove the following result.
%
%\begin{Th}[Proctor \cite{proctor}]
%	\label{Th:ppsamenumber}
%	The posets $R_{r,s}$ and $T_{r,s}$ have the same number of plane partitions of height $\ell$ for all $\ell \in \mathbb{Z}_{\geq 0}$.
%\end{Th}
%
%Since Proctor's proof is not bijective, this raises the following natural question.
%
%\begin{Prob}
%	\label{Prob:pps}
%	Find a bijective proof that $R_{r,s}$ and $T_{r,s}$ have the same number of plane partitions of height $\ell$ for all $\ell \in \mathbb Z_{\geq 0}$.
%\end{Prob}
%
%Understanding the relationship between the plane partitions of the rectangle and trapezoid has been the subject of a number of papers over the past 40 years. Partial results giving bijections for low heights were given by Stembridge \cite{stembridge}, Reiner \cite{reiner}, and Elizalde \cite{elizalde} before a bijective proof for all heights was found in 2016 by Hamaker, Patrias, Pechenik, and Williams \cite{hamakerpatriaspechenikwilliams} in terms of $K$-theoretic jeu de taquin. 



%\subsection{Rowmotion} \label{sec:rowmotion}
%In this section we define birational toggles and rowmotion. 
%Note that we define these as maps on $\RR_+$-labelings of $P$, but one can also define them as rational maps on $\RR$-labelings of $P$.  (See Remark~\ref{rem:semiring}.) 
We can augment a poset $P$ to a poset $\widehat P$ by adding minimum and maximum elements $\hat 0$ and $\hat 1$. 
Throughout, we will identify the labelings in $\RR_+^P$ with the corresponding labelings in $\RR_+^{\widehat P}$, where the labels at $\hat 0$ and $\hat 1$ are both $1$. 
%(Although rowmotion and related notions are sometimes defined with different labels at $\hat 0$ and $\hat 1$, this does not have any significant effect on the generality of results when $P$ is graded since one can renormalize as discussed in \cite{einsteinpropp1}.) 


\begin{Def}
	For any $p \in P$, the \emph{(birational) toggle} $t_p \colon \RR_+^{P} \to \RR_+^{P}$ is the map that acts on $y \in \RR_+^P$ by fixing all coordinates except $y_p$ and sending
	$y_p \mapsto \Bigl(\sum\limits_{\scriptscriptstyle{q \gtrdot p}} \frac{1}{y_q} \Bigr)^{-1} \Bigl( \sum\limits_{\scriptscriptstyle q \lessdot p} y_q \Bigr) \frac{1}{y_p}.$
	
	The \emph{rowmotion map} $\rho \colon \RR_+^P \to \RR_+^P$ is the composition
	$\rho = t_{L^{-1}(1)} \circ t_{L^{-1}(2)} \circ \cdots \circ t_{L^{-1}(n)}$ for any linear extension $L \colon P \to [n]$.
\end{Def}

%It is easy to verify that toggles $t_p$ and $t_q$ commute if and only if $p$ and $q$ are not adjacent in the Hasse diagram of $P$. From this, one can deduce that $\rho$ does not depend on the choice of linear extension.

%There is an alternate form of rowmotion obtained using the following transfer map.

\begin{Def}
	The \emph{transfer map} $\psi^{-1} \colon \RR_+^P \to \RR_+^P$ is defined coordinatewise by
	$\psi^{-1}(y)_p= \frac{y_p}{\sum\limits_{q \lessdot p} y_q}.$
	Its inverse $\phibar$ acts by
	$\displaystyle \psi(x)_p = \sum\limits_{\scriptscriptstyle\hat 0 \lessdot q_1 \lessdot \dots \lessdot q_n = p} \prod\limits_{i=1}^n x_{q_i}.$
	%Thus $\phibar(x)_p$ is the total weight of all maximal chains in the interval $[\hat 0, p]$ in $x$.
\end{Def}

%We similarly define dual maps ${\psi^*}^{-1}$ and $\psi^*$ as the conjugates of $\psi^{-1}$ and $\psi$ under poset duality (i.e., by reversing the cover relations in the definition above).

%If one works over the tropical semiring (and sets $x_{\hat 0} = 0$ and $x_{\hat 1}=1$), then the transfer map defines a piecewise-linear, continuous bijection from the order polytope of $P$ to the chain polytope of $P$, as shown by Stanley \cite{stanley2}.

It will be convenient for us to work with the conjugate of rowmotion under the transfer map, $\tilde\rho = \psi^{-1} \circ \rho \circ \phibar$. (This map was called \emph{birational antichain rowmotion} or \emph{barmotion} by Joseph and Roby \cite{josephroby2}.) An important property of $\tilde\rho$ is the following identity.

%\begin{Def}
%	The \emph{(antichain) rowmotion map} is defined by $\tilde\rho = \psi^{-1} \circ \rho \circ \phibar$.
%\end{Def}

%\begin{Def}
%	The \emph{dual transfer map} ${\psi^*}^{-1} \colon \RR_+^P \to \RR_+^P$ is defined coordinatewise by, for $p \in P$,
%	\[x_p \mapsto \frac{x_p}{\sum_{q \gtrdot p} x_q}.\]
%	Its inverse $\phibar^*$ acts by
%	\[x_p \mapsto \sum_{\hat 1 \gtrdot q_1 \gtrdot \dots \gtrdot q_n = p} \prod_{i=1}^n x_{q_i}.\]
%	Thus $\phibar^*(x)_p$ is the total weight of all maximal chains in the interval $[p,\hat 1]$ in $x$.
%\end{Def}

%The following lemma is originally due to Einstein and Propp in a slightly different form (see \cite{einsteinpropp2,josephroby2}).

% \begin{Lemma} \label{lemma:dualtransfer}
	% For any $x \in \mathbb{R}_{+}^{P}$ and $p \in P$,
	% \[\rho \circ \phibar(x)_p = \frac{1}{\phibar^*(x)_p}.\]
	% \end{Lemma}

%\begin{Lemma} \label{lemma:dualtransfer}
%	Let $x \in \RR_+^P$, $z=\tilde \rho^{-1}(x)$, and $p \in P$. Then $\psi(x)_p\psi^*(z)_p = 1$.
%\end{Lemma}
%\begin{proof}
%	Assume by induction that the claim holds for all elements of $P$ above $p$. Since $\psi(x) = \rho(\psi(z))$, the value of $\psi(x)_p$ is determined when toggling $p$ while performing rowmotion on $\psi(z)$. At this point, the labels above $p$ agree with $\psi(x)$, while the labels below $p$ agree with $\psi(z)$. Therefore we find that
%	\begin{align*}
%		\psi(x)_p &= \left(\sum_{q \gtrdot p} \frac{1}{\psi(x)_q}\right)^{-1} \cdot \left(\sum_{q \lessdot p} \psi(z)_q\right)\cdot \psi(z)_p^{-1}\\
%		&= \left(\sum_{q \gtrdot p} \psi^*(z)_q\right)^{-1} \cdot z_p^{-1}\\
%		&= \psi^*(z)_p^{-1},
%	\end{align*}
%	where we used the fact that $\psi(z)_p = z_p \cdot \sum_{q \lessdot p} \psi(z)_q$, and similarly for $\psi^*(z)_p$.
%\end{proof}

%\begin{Ex}
%	Let $P = RT_{3,2}$, and let $x \in \RR_+^P$ be labeled as in the leftmost diagram of Figure~\ref{fig:r32}. Then we can compute $y=\psi(x)$, $\rho^{-1}(y)$, and $z = \psi^{-1}(\rho^{-1}(y))$ as shown. Note that the values of $\psi^*(z)_p$ agree with the values of $y_p^{-1} = \psi(x)_p^{-1}$. For instance, when $p = (2,2)$, \[\psi^*(z)_{22} = \frac{1}{a(bd+be+ce)fg} \cdot g \cdot \frac{(bd+be+ce)f}{(b+c)e} = \frac{1}{a(b+c)e} = y_{22}^{-1}.\]
%\end{Ex}
	\begin{figure}
		\centering
		\begin{tikzpicture}[scale=.7, transform shape]
			\begin{scope}[shift={(-.5,0)}]
				\draw (0,0)--(-2,2) (1,1)--(-2,4) (0,0)--(1,1) (-1,1)--(0,2) (-2,2)--(-1,3);
				\node[circle, fill=black, inner sep=2pt, label=left:$a$] at (0,0){};
				\node[circle, fill=black, inner sep=2pt, label=left:$b$] at (-1,1){};
				\node[circle, fill=black, inner sep=2pt, label=left:$d$] at (-2,2){};
				\node[circle, fill=black, inner sep=2pt, label=right:$c$] at (1,1){};
				\node[circle, fill=black, inner sep=2pt, label=right:$e$] at (0,2){};
				\node[circle, fill=black, inner sep=2pt, label=right:$f$] at (-1,3){};
				\node[circle, fill=black, inner sep=2pt, label=right:$g$] at (-2,4){};
			\end{scope}
			\begin{scope}[shift={(4.5,0)}]
				\draw (0,0)--(-2,2) (1,1)--(-2,4) (0,0)--(1,1) (-1,1)--(0,2) (-2,2)--(-1,3);
				\node[circle, fill=black, inner sep=2pt, label=left:$a$] at (0,0){};
				\node[circle, fill=black, inner sep=2pt, label=left:$ab$] at (-1,1){};
				\node[circle, fill=black, inner sep=2pt, label=left:$abd$] at (-2,2){};
				\node[circle, fill=black, inner sep=2pt, label=right:$ac$] at (1,1){};
				\node[circle, fill=black, inner sep=2pt, label=right:$a(b+c)e$] at (0,2){};
				\node[circle, fill=black, inner sep=2pt, label=right:$a(bd+be+ce)f$] at (-1,3){};
				\node[circle, fill=black, inner sep=2pt, label=right:$a(bd+be+ce)fg$] at (-2,4){};
			\end{scope}
			\begin{scope}[shift={(10,0)}]
				\draw (0,0)--(-2,2) (1,1)--(-2,4) (0,0)--(1,1) (-1,1)--(0,2) (-2,2)--(-1,3);
				\node[circle, fill=black, inner sep=2pt, label=left:$\frac{bc}{b+c}$] at (0,0){};
				\node[circle, fill=black, inner sep=2pt, label=left:$\frac{bcde}{bd+be+ce}$] at (-1,1){};
				\node[circle, fill=black, inner sep=2pt, label=left:$cef$] at (-2,2){};
				\node[circle, fill=black, inner sep=2pt, label=right:$be$] at (1,1){};
				\node[circle, fill=black, inner sep=2pt, label=right:$b(d+e)f$] at (0,2){};
				\node[circle, fill=black, inner sep=2pt, label=right:$(bd+be+ce)fg$] at (-1,3){};
				\node[circle, fill=black, inner sep=2pt, label=right:$\frac{1}{a}$] at (-2,4){};
			\end{scope}
			\begin{scope}[shift={(16.5,0)}]
				\draw (0,0)--(-2,2) (1,1)--(-2,4) (0,0)--(1,1) (-1,1)--(0,2) (-2,2)--(-1,3);
				\node[circle, fill=black, inner sep=2pt, label=left:$\frac{bc}{b+c}$] at (0,0){};
				\node[circle, fill=black, inner sep=2pt, label=left:$\frac{(b+c)de}{bd+be+ce}$] at (-1,1){};
				\node[circle, fill=black, inner sep=2pt, label=left:$\frac{(bd+be+ce)f}{bd}$] at (-2,2){};
				\node[circle, fill=black, inner sep=2pt, label=right:$\frac{(b+c)e}{c}$] at (1,1){};
				\node[circle, fill=black, inner sep=2pt, label=right:$\frac{(bd+be+ce)f}{(b+c)e}$] at (0,2){};
				\node[circle, fill=black, inner sep=2pt, label=right:$g$] at (-1,3){};
				\node[circle, fill=black, inner sep=2pt, label=right:$\frac{1}{a(bd+be+ce)fg}$] at (-2,4){};
			\end{scope}
		\end{tikzpicture}
		\caption{Labelings of $RT_{3,2}$: $x$, $y=\psi(x)$, $\rho^{-1}(y)$, and $z =\tilde\rho^{-1}(x) = \psi^{-1}(\rho^{-1}(y))$.}
		\label{fig:r32}
	\end{figure}




%\subsubsection{Antichain rowmotion}
%
%Lemma~\ref{lemma:dualtransfer} allows us to give an alternate formulation of rowmotion via its conjugation under the transfer map. (This map was called \emph{birational antichain rowmotion} or \emph{barmotion} by Joseph and Roby \cite{josephroby2}.)
%
%\begin{Def}
%	For any $p \in P$, the \emph{(antichain) toggle} $\tau_p \colon \RR_+^{P} \to \RR_+^{P}$ is the map that changes the $p$-coordinate of $x \in \RR_+^{P}$ by
%	\[x_p \mapsto \left(\sum_{C} \prod_{q \in C} x_q\right)^{-1},\]
%	where $C$ ranges over all maximal chains of $P$ containing $p$, while keeping all other coordinates fixed.
%\end{Def}
%
%\begin{Def}
%	The \emph{(antichain) rowmotion map} $\tilde\rho \colon \RR_+^P \to \RR_+^P$ is the composition
%	\[\tilde\rho = \tau_{L^{-1}(n)} \circ \cdots \circ \tau_{L^{-1}(2)}\circ \tau_{L^{-1}(1)}\] for any linear extension (order-preserving bijection) $L \colon P \to [n]$.
%\end{Def}
%%It is easy to verify that toggles $\tau_p$ and $\tau_q$ commute if and only if $p$ and $q$ are incomparable in $P$, which implies that $\tilde \rho$ does not depend on the choice of linear extension. Note that the $\tau_p$ here are applied from bottom to top as opposed to the $t_p$ in the definition of $\rho$, which were applied from top to bottom.
%
%In fact, $\tilde\rho$ and $\rho$ are conjugate to one another, as shown in the following proposition due to Joseph and Roby \cite{josephroby2}.
%\begin{Prop} \label{prop:rhotilde}
%	The equality $\tilde \rho = \psi^{-1} \circ \rho \circ \phibar$ holds.
%\end{Prop}
%%\begin{proof}
%%	Suppose $\tilde \rho(z) = x$. During the application of $\tilde \rho$ to $z$, when applying $\tau_p$ to find $x_p$ for some $p \in P$, the labels below $p$ agree with $x$ while the labels above $p$ agree with $z$. Each maximal chain in $\widehat P$ through $p$ can be written as a union of a maximal chain from $[\hat 0, p]$ and one from $[p, \hat 1]$ (that overlap at $p$), so before applying $\tau_p$, the weight of the maximal chains through $p$ is
%%	$\frac{\psi(x)_p \psi^*(z)_p}{x_p}$. But this equals $x_p^{-1}$ by the definition of $\tau_p$, so we find that $\psi(x)_p \psi^*(z)_p = 1$. Comparing with Lemma~\ref{lemma:dualtransfer} then gives the result. 
%%\end{proof}



%One important property of $\tilde\rho$ is the following identity.

\begin{Prop}\label{cor:edgeweight}
	Let $x \in \RR_+^P$, $y = \phibar(x)$, and $z = \tilde \rho^{-1}(x)$. Then for $p \in P$,
	\[
		x_p^{-1} = \sum_{q \lessdot p} \frac{y_q}{y_p} \qquad \text{and} \qquad
		z_p^{-1} = \sum_{q \gtrdot p} \frac{y_p}{y_q}.
	\]
\end{Prop}
See Figure~\ref{fig:r32} for some example calculations involving these maps on $RT_{3,2}$.
%\begin{proof}
%	The first equation follows by applying the definition of $\psi^{-1}$ to $x = \psi^{-1}(y)$. For the second, by Proposition~\ref{prop:rhotilde} and Lemma~\ref{lemma:dualtransfer} we have
%	$\phibar^*(z)_p = \frac{1}{y_p}$.
%	Therefore the equation $z = {\psi^*}^{-1}(\phibar^*(z))$ gives
%	\[z_p = \frac{\frac{1}{y_p}}{\sum_{q \gtrdot p} \frac{1}{y_q}} = \frac{1}{\sum_{q \gtrdot p} \frac{y_p}{y_q}}.\]
%	Inverting gives the desired result.
%\end{proof}

%\begin{Ex}
%	Figure~\ref{fig:r32} shows a labeling $x$ of $RT_{3,2}$, together with $y=\psi(x)$, $\rho^{-1}(y)$, and $z = \tilde\rho^{-1}(x) = \psi^{-1}(\rho^{-1}(y))$. 
%	One can compute $z$ from $y$ using Corollary~\ref{cor:edgeweight}. For example, 
%	\[z_{21}^{-1} = \frac{y_{21}}{y_{31}}+\frac{y_{21}}{y_{22}} = \frac{ab}{abd} + \frac{ab}{a(b+c)e} = \frac{bd+be+ce}{(b+c)de}.\]
%\end{Ex}

%We will also need the following easy fact about $\tilde\rho^{-1}$ for later.
%\begin{Prop} \label{prop:arbitrarya}
%	Suppose $P$ has a unique minimum element $a$. Let $x \in \RR_+^P$ and $z = \tilde \rho^{-1}(x)$. Then for any nonmaximal $p \in P$, $z_p$ does not depend on $x_a$.
%	
%	Similarly, if $P$ has a unique maximum element $b$, $z \in \RR_+^P$, and $x = \tilde \rho(z)$, then for any nonminimal $p \in P$, $x_p$ does not depend on $z_b$.
%\end{Prop}
%\begin{proof}
%	By the definition of $\phibar$, since $a$ is a minimum of $P$, $y_p=\phibar(x)_p$ is a multiple of $x_a$ for all $p \in P$, but $\frac{y_p}{x_a}$ is independent of $x_a$. This factor of $x_a$ then cancels in every term in the expression for $z_p^{-1}$ in Corollary~\ref{cor:edgeweight}. 
%	
%	The second statement is proved similarly, using the fact that $\frac{1}{y_p} = \phibar^*(z)_p$ for $p \in P$.
%\end{proof}







\section{Chain Shifting in Skew Shapes}
\label{section:skewChainShifting}

%In this section we prove a chain shifting lemma for rowmotion on skew shapes, generalizing a lemma for rectangles proved by the current authors in \cite{johnsonliu}, as well as in the noncommutative setting by Grinberg and Roby \cite{grinbergroby3}. As we will see, the proof given here is simpler than those earlier proofs and relies mainly on a duality between plane trees. (Notably, this proof does not require the use of subtraction.)

%\subsection{Arborescences}


%For any subset $S \subset P$ and $x \in \RR_+^P$, define the \emph{weight} $w_S(x) = \prod_{p \in S} x_p$.

Let $P$ be a poset and $x \in \RR^P_+$. For a subset $S \subset P$, define the \emph{weight} of $S$ to be $w_S(x) = \prod_{p \in S} x_p$. We first show how to relate weights of certain subsets of $P$ to weights of certain arborescences with respect to $y=\phibar(x)$. (Recall that we set $y_{\hat 0} = y_{\hat 1} = 1$.)

%We first prove two propositions expressing the weights of chain statistics in terms of weights of trees. Fix a finite poset $P$. Throughout this section, all subposets that we consider are induced and have cover relations that also appear in the Hasse diagram of $P$. That is, the Hasse diagrams of all subposets we consider are induced subgraphs of the Hasse diagram of $P$. Throughout this section all empty sums evaluate to 1.

%\begin{Def}
%	\label{Def:upanddowndegree}
%	Let $G$ be a (not necessarily induced) subgraph of the Hasse diagram of $P$. (The edges of $G$ are therefore cover relations of $P$).
%	The \emph{up degree} $\updeg_G(p)$ of a vertex $p \in G$ is the number of edges in $G$ that connect $p$ to an element that covers $p$ in $P$. Similarly, the \emph{down degree} $\downdeg_G(p)$ is the number of edges connecting $p$ to an element covered by $p$.
%\end{Def}

\begin{Def}
	\label{Def:spanningTree}
	An \emph{upward arborescence of $P$} is a subgraph of $\widehat P \setminus \{\hat 1\}$ such that every element of $P$ has down degree $1$. Similarly, a \emph{downward arborescence of $P$} is a subgraph of $\widehat P \setminus \{\hat 0\}$ such that every element of $P$ has up degree $1$.	
	We denote the set of upward and downward arborescences of $P$ by $U_P$ and $D_P$, respectively.
\end{Def}

% \begin{Not}
	% Let $I$ be an induced subposet of $P$ with minimum $m$ and maximum $M$. 
	% \begin{itemize}
		%     \item We let $U_P(I)$ denote the set of all upward spanning trees of $P$ with a distinguished chain from $m = p_1 \lessdot p_2 \lessdot \dots \lessdot p_k = M$ contained in $I$.
		%     \item We let $D_P(I)$ denote the set of all downward spanning trees of $P$ with a distinguished chain from $m = p_1 \lessdot p_2 \lessdot \dots \lessdot p_k = M$ contained in $I$.
		%     \item Let $U_P$ and $D_P$ denote the sets of all upward and downward spanning trees of $P$ with no particular distinguished chain.
		%     \item Let $MC(I)$ denote the set of maximal chains in $I$.
		% \end{itemize}  
	% \end{Not}

Define the \emph{weight} (with respect to $y$) of the edge $e$ corresponding to the cover relation $p \lessdot q$ to be $\omega_e(y) = \frac{y_p}{y_q}$ and the \emph{weight} of an arborescence $T$ to be $\omega_T(y) = \prod_{e \in E(T)} \omega_e(y)$. (Note: The weight $\omega_T$ of an arborescence is different from the weight $w_S$ of a subset.)
%Using Corollary~\ref{cor:edgeweight}, one can show that the total weight of all upward arborescences is $w_P(x)^{-1}$, while the total weight of all downward arborescences is $w_P(z)^{-1}$ where $z = \tilde \rho^{-1}(x)$.
\begin{Ex}
	The upward and downward arborescences of $RT_{3,2}$ are shown in Figure~\ref{fig:upwardSpanningTree} with their weights. The weight of the first upward arborescence can be computed as
	\[\frac{1}{y_{11}} \cdot \frac{y_{11}}{y_{21}} \cdot \frac{y_{11}}{y_{12}} \cdot \frac{y_{21}}{y_{31}} \cdot \frac{y_{21}}{y_{22}} \cdot \frac{y_{31}}{y_{32}} \cdot \frac{y_{32}}{y_{42}} = \frac{y_{11}y_{21}}{y_{12}y_{22}y_{42}}.\]
%	
%	Let us take $y=\psi(x)$ as in Figure~\ref{fig:r32}. Then the total weight of all four upward arborescences is\[\frac{y_{11}(y_{21}+y_{12})(y_{31}+y_{22})}{y_{12}y_{22}y_{31}y_{42}}=\frac{a(ab+ac)(abd+a(b+c)e)}{ac \cdot a(b+c)e \cdot abd \cdot a(bd+be+ce)f} = \frac{1}{abcdef},\] which is $w_P(x)^{-1}$.
\end{Ex}
\begin{figure}
	\centering
	\begin{tikzpicture}[scale = 0.55]
		%TREE #1
		\begin{scope}[shift={(0,0)}]
			%UP-DIAGONALS
			\draw[gray] (0,0)--(1,1);
			\draw[gray] (-1,1)--(0,2);
			\draw[gray] (-2,2)--(-1,3);
			
			%DOWN-DIAGONALS
			\draw[gray] (0,0)--(-2,2);
			\draw[gray] (1,1)--(-2,4);
			
			\draw [red, ultra thick] (0,0)--(1,1);
			\draw [red, ultra thick] (0,0)--(-2,2);
			\draw [red, ultra thick] (-1,3)--(-2,4);
			
			\draw [red, ultra thick] (-2,2)--(-1,3);
			\draw [red, ultra thick] (-1,1)--(0,2);
			
			\draw [red, ultra thick] (0,0)--(0,-1);
			
			\node[wB] at (0,0) {};
			\node[wB] at (1,1) {};
			\node[wB] at (-1,1) {};
			\node[wB] at (0,2) {};
			\node[wB] at (-2,2) {};
			\node[wB] at (-1,3) {};
			\node[wB] at (-2,4) {};
			\node[wB] at (0,-1) {};
			
			\node[scale=1.5,transform shape] at (0,-2) {\large$\frac{y_{11}y_{21}}{y_{12}y_{22}y_{42}}$};
		\end{scope}
		
		%TREE #2
		\begin{scope}[shift={(5,0)}]
			%UP-DIAGONALS
			\draw[gray] (0,0)--(1,1);
			\draw[gray] (-1,1)--(0,2);
			\draw[gray] (-2,2)--(-1,3);
			
			%DOWN-DIAGONALS
			\draw[gray] (0,0)--(-2,2);
			\draw[gray] (1,1)--(-2,4);
			
			\draw [red, ultra thick] (0,0)--(1,1);
			\draw [red, ultra thick] (0,0)--(-2,2);
			\draw [red, ultra thick] (-1,3)--(-2,4);
			
			\draw [red, ultra thick] (-2,2)--(-1,3);
			\draw [red, ultra thick] (1,1)--(0,2);
			
			\draw [red, ultra thick] (0,0)--(0,-1);
			
			\node[wB] at (0,0) {};
			\node[wB] at (1,1) {};
			\node[wB] at (-1,1) {};
			\node[wB] at (0,2) {};
			\node[wB] at (-2,2) {};
			\node[wB] at (-1,3) {};
			\node[wB] at (-2,4) {};
			\node[wB] at (0,-1) {};
			
			\node[scale=1.5,transform shape] at (0,-2) {\large$\frac{y_{11}}{y_{22}y_{42}}$};
		\end{scope}
		
		%TREE #3
		\begin{scope}[shift={(10,0)}]
			%UP-DIAGONALS
			\draw[gray] (0,0)--(1,1);
			\draw[gray] (-1,1)--(0,2);
			\draw[gray] (-2,2)--(-1,3);
			
			%DOWN-DIAGONALS
			\draw[gray] (0,0)--(-2,2);
			\draw[gray] (1,1)--(-2,4);
			
			\draw [red, ultra thick] (0,0)--(1,1);
			\draw [red, ultra thick] (0,0)--(-2,2);
			\draw [red, ultra thick] (-1,3)--(-2,4);
			
			\draw [red, ultra thick] (0,2)--(-1,3);
			\draw [red, ultra thick] (-1,1)--(0,2);
			
			\draw [red, ultra thick] (0,0)--(0,-1);
			
			\node[wB] at (0,0) {};
			\node[wB] at (1,1) {};
			\node[wB] at (-1,1) {};
			\node[wB] at (0,2) {};
			\node[wB] at (-2,2) {};
			\node[wB] at (-1,3) {};
			\node[wB] at (-2,4) {};
			\node[wB] at (0,-1) {};
			
			\node[scale=1.5,transform shape] at (0,-2) {\large$\frac{y_{11}y_{21}}{y_{12}y_{31}y_{42}}$};
		\end{scope}
		
		%TREE #4
		\begin{scope}[shift={(15,0)}]
			%UP-DIAGONALS
			\draw[gray] (0,0)--(1,1);
			\draw[gray] (-1,1)--(0,2);
			\draw[gray] (-2,2)--(-1,3);
			
			%DOWN-DIAGONALS
			\draw[gray] (0,0)--(-2,2);
			\draw[gray] (1,1)--(-2,4);
			
			\draw [red, ultra thick] (0,0)--(1,1);
			\draw [red, ultra thick] (0,0)--(-2,2);
			\draw [red, ultra thick] (-1,3)--(-2,4);
			\draw [red, ultra thick] (0,2)--(-1,3);
			\draw [red, ultra thick] (1,1)--(0,2);
			
			\draw [red, ultra thick] (0,0)--(0,-1);
			
			\node[wB] at (0,0) {};
			\node[wB] at (1,1) {};
			\node[wB] at (-1,1) {};
			\node[wB] at (0,2) {};
			\node[wB] at (-2,2) {};
			\node[wB] at (-1,3) {};
			\node[wB] at (-2,4) {};
			\node[wB] at (0,-1) {};
			
			\node[scale=1.5,transform shape] at (0,-2) {\large$\frac{y_{11}}{y_{31}y_{42}}$};
		\end{scope}
		
		%TREE #1
		\begin{scope}[shift={(0,-8)}]
			%UP-DIAGONALS
			\draw[gray] (0,0)--(1,1);
			\draw[gray] (-1,1)--(0,2);
			\draw[gray] (-2,2)--(-1,3);
			
			%DOWN-DIAGONALS
			\draw[gray] (0,0)--(-2,2);
			\draw[gray] (1,1)--(-2,4);
			
			\draw [blue, ultra thick] (0,0)--(1,1);
			\draw [blue, ultra thick] (1,1)--(-1,3);
			\draw [blue, ultra thick] (-1,3)--(-2,4);
			\draw [blue, ultra thick] (-2,2)--(-1,3);
			\draw [blue, ultra thick] (-1,1)--(0,2);
			\draw [blue, ultra thick] (-2,4)--(-2,5);
			
			\node[wB] at (0,0) {};
			\node[wB] at (1,1) {};
			\node[wB] at (-1,1) {};
			\node[wB] at (0,2) {};
			\node[wB] at (-2,2) {};
			\node[wB] at (-1,3) {};
			\node[wB] at (-2,4) {};
			\node[wB] at (-2,5) {};
			
			\node[scale=1.5,transform shape] at (0,-1) {\large$\frac{y_{11}y_{21}y_{31}}{y_{22}y_{32}}$};
		\end{scope}
		
		%TREE #2
		\begin{scope}[shift={(5,-8)}]
			%UP-DIAGONALS
			\draw[gray] (0,0)--(1,1);
			\draw[gray] (-1,1)--(0,2);
			\draw[gray] (-2,2)--(-1,3);
			
			%DOWN-DIAGONALS
			\draw[gray] (0,0)--(-2,2);
			\draw[gray] (1,1)--(-2,4);
			
			\draw [blue, ultra thick] (0,0)--(-1,1);
			\draw [blue, ultra thick] (1,1)--(-1,3);
			\draw [blue, ultra thick] (-1,3)--(-2,4);
			\draw [blue, ultra thick] (-2,2)--(-1,3);
			\draw [blue, ultra thick] (-1,1)--(0,2);
			
			\draw [blue, ultra thick] (-2,4)--(-2,5);
			
			\node[wB] at (0,0) {};
			\node[wB] at (1,1) {};
			\node[wB] at (-1,1) {};
			\node[wB] at (0,2) {};
			\node[wB] at (-2,2) {};
			\node[wB] at (-1,3) {};
			\node[wB] at (-2,4) {};
			\node[wB] at (-2,5) {};
			
			\node[scale=1.5,transform shape] at (0,-1) {\large$\frac{y_{11}y_{12}y_{31}}{y_{22}y_{32}}$};
		\end{scope}
		
		%TREE #3
		\begin{scope}[shift={(10,-8)}]
			%UP-DIAGONALS
			\draw[gray] (0,0)--(1,1);
			\draw[gray] (-1,1)--(0,2);
			\draw[gray] (-2,2)--(-1,3);
			
			%DOWN-DIAGONALS
			\draw[gray] (0,0)--(-2,2);
			\draw[gray] (1,1)--(-2,4);
			
			\draw [blue, ultra thick] (0,0)--(1,1);
			\draw [blue, ultra thick] (1,1)--(-1,3);
			\draw [blue, ultra thick] (-1,3)--(-2,4);
			\draw [blue, ultra thick] (-2,2)--(-1,3);
			\draw [blue, ultra thick] (-1,1)--(-2,2);
			
			\draw [blue, ultra thick] (-2,4)--(-2,5);
			
			\node[wB] at (0,0) {};
			\node[wB] at (1,1) {};
			\node[wB] at (-1,1) {};
			\node[wB] at (0,2) {};
			\node[wB] at (-2,2) {};
			\node[wB] at (-1,3) {};
			\node[wB] at (-2,4) {};
			\node[wB] at (-2,5) {};
			
			\node[scale=1.5,transform shape] at (0,-1) {\large$\frac{y_{11}y_{21}}{y_{32}}$};
		\end{scope}
		
		%TREE #4
		\begin{scope}[shift={(15,-8)}]
			%UP-DIAGONALS
			\draw[gray] (0,0)--(1,1);
			\draw[gray] (-1,1)--(0,2);
			\draw[gray] (-2,2)--(-1,3);
			
			%DOWN-DIAGONALS
			\draw[gray] (0,0)--(-2,2);
			\draw[gray] (1,1)--(-2,4);
			
			\draw [blue, ultra thick] (-2,2)--(-1,3);
			\draw [blue, ultra thick] (0,0)--(-2,2);
			\draw [blue, ultra thick] (-1,3)--(-2,4);
			\draw [blue, ultra thick] (0,2)--(-1,3);
			\draw [blue, ultra thick] (1,1)--(0,2);
			
			\draw [blue, ultra thick] (-2,4)--(-2,5);
			
			\node[wB] at (0,0) {};
			\node[wB] at (1,1) {};
			\node[wB] at (-1,1) {};
			\node[wB] at (0,2) {};
			\node[wB] at (-2,2) {};
			\node[wB] at (-1,3) {};
			\node[wB] at (-2,4) {};
			\node[wB] at (-2,5) {};
			
			\node[scale=1.5,transform shape] at (0,-1) {\large$\frac{y_{11}y_{12}}{y_{32}}$};
		\end{scope}
		
	\end{tikzpicture}
	\caption{The four upward arborescences in $U_{RT_{3,2}}$ and the four downward arborescences in $D_{RT_{3,2}}$, together with their weights. %The bijection $\aleph$ maps each upward arborescence to the downward arborescence below it. This multiplies each weight by $\frac{y_{12}y_{31}y_{42}}{y_{32}}$ in accordance with Lemma~\ref{lemma:weightChange}.
	}
	\label{fig:upwardSpanningTree}
\end{figure}

%The following proposition is a simple consequence of Corollary~\ref{cor:edgeweight}.
%
%\begin{Prop} \label{prop:weightofP}
%	Let $P$ be a poset and $x \in \RR_+^P$, and let $y=\psi(x)$ and $z = \tilde \rho^{-1}(x)$. Then
%	\begin{align*}
%		\frac{1}{w_P(x)} &= \sum_{T \in U_P} \omega_T(y),\\
%		\frac{1}{w_P(z)} &= \sum_{T \in D_P} \omega_T(y).
%	\end{align*}
%\end{Prop}
%\begin{proof}
%	For the first equation, by Corollary~\ref{cor:edgeweight},
%	\begin{align*}
%		\frac{1}{w_P(x)} = \prod_{p \in P} x_p^{-1} = \prod_{p \in P} \sum_{q \lessdot p} \frac{y_q}{y_p}. 
%	\end{align*}
%	Each upward arborescence contains exactly one edge downward from each element of $P$, so the right hand side is the total weight of all upward arborescences in $U_P$, which gives the first equation. The proof of the second equation is similar.
%\end{proof}

For any $y \in \RR_+^P$, we can use $\omega_T(y)$ to define probability measures on $U_P$ and $D_P$: for subsets $U \subset U_P$ and $D \subset D_P$, define
\[
	\mu_y(U) = \frac{\sum_{T \in U} \omega_T(y)}{\sum_{T \in U_P} \omega_T(y)}, \qquad
	\mu_y(D) = \frac{\sum_{T \in D} \omega_T(y)}{\sum_{T \in D_P} \omega_T(y)}.
\]
%We will show that these are closely related to the weights of chains in $P$.

%\subsubsection{Weights of chains}
%Define a \emph{$k$-chain} $C = (C_1, \dots, C_k)$ to be a $k$-tuple of disjoint saturated chains $C_i$, and define $\min(C) = (a_1, \dots, a_k)$ and $\max(C) = (b_1, \dots, b_k)$, where $a_i$ and $b_i$ are the minimum and maximum elements of $C_i$, respectively.

%We will abusively treat $C$ as either the subset $C_1 \sqcup \cdots \sqcup C_k \subset P$ or as the subgraph of the Hasse diagram of $P$ on this subset whose edges are the cover relations of the $C_i$. (This subgraph is therefore a union of $k$ vertex-disjoint paths). Let $U_P(C)$ and $D_P(C)$ denote the sets of all upward and downward arborescences that contain $C$ as a subgraph.


%For any saturated chain $C \subset P$, we say that an arborescence \emph{contains} $C$ if it contains all of the edges representing cover relations in $C$. 
Given a collection of saturated chains $\mathscr C$, let $U_P(\mathscr C)$ and $D_P(\mathscr C)$ denote the sets of all upward and downward arborescences that contain (the edges of) some chain in $\mathscr C$, and let $w_\mathscr C(x)$ denote the total weight of all chains in $\mathscr C$ (as subsets of $P$).
%The following proposition relates the weight of $C$ (as a subset of $P$) to the weights of the arborescences containing $C$.
%\begin{Prop} \label{prop:chainarborescence}
%	Let $P$ be a poset, $x \in \RR_+^P$, $y = \phibar(x)$, and $z = \tilde\rho^{-1}(x)$. Let $C$ be a saturated chain in $P$ with minimum $a$ and maximum $b$. Then
%	\begin{align*}
%		w_C(x) &= \left(\sum_{a' \lessdot a} \frac{y_{a'}}{y_b}\right)^{-1} \cdot \mu_y(U_P(C)),\tag{$*$}\label{eq:wcx}\\
%		w_C(z) &= \left(\sum_{b' \gtrdot b} \frac{y_{a}}{y_{b'}}\right)^{-1} \cdot \mu_y(D_P(C)).\tag{$**$}\label{eq:wcz}
%	\end{align*}
%\end{Prop}
% \begin{Prop} \label{prop:chainarborescence}
	%     Let $P$ be a poset, $x \in \RR_+^P$, $y = \phibar(x)$, and $z = \tilde\rho^{-1}(x)$. Let $C = (C_1, \dots, C_k)$ be a $k$-chain, where $C_i$ has minimum $a_i$ and maximum $b_i$. Then
	%     \begin{align*}
		%     w_C(x) &= \prod_{i=1}^k\left(\sum_{a' \lessdot a_i} \frac{y_{a'}}{y_{b_i}}\right)^{-1} \cdot \mu_y(U_P(C)),\tag{$*$}\label{eq:wcx}\\
		%     w_C(z) &= \prod_{i=1}^k\left(\sum_{b' \gtrdot b_i} \frac{y_{a_i}}{y_{b'}}\right)^{-1} \cdot \mu_y(D_P(C)).\tag{$**$}\label{eq:wcz}
		%     \end{align*}
	% \end{Prop}
%\begin{proof}
%	The product of the weights of the edges in $C$ is $\frac{y_{a}}{y_{b}}$. The elements of $U_P(C)$ are obtained from $C$ by adding one downward edge from each element of $(P \setminus C) \cup \{a\}$. It follows from Corollary~\ref{cor:edgeweight} that
%	\begin{align*}
%		\sum_{T \in U_P(C)} \omega_T(y) &= \left(\frac{y_{a}}{y_{b}} \cdot \sum_{a' \lessdot a} \frac{y_{a'}}{y_{a}} \right) \cdot \prod_{p \in P \setminus C} \sum_{q \lessdot p} \frac{y_q}{y_p}\\
%		&=\sum_{a' \lessdot a} \frac{y_{a'}}{y_{b}} \cdot \prod_{p \in P \setminus C} x_p^{-1}\\
%		&=\sum_{a' \lessdot a} \frac{y_{a'}}{y_{b}} \cdot \frac{w_{C}(x)}{w_P(x)}.
%		%&= \prod_{i=1}^k\sum_{a' \lessdot a_i} \frac{y_{a'}}{y_{b_i}} \cdot w_C(x) \cdot \sum_{T \in U_P} \omega(T;y).
%	\end{align*}
%	Subsitituting for $w_P(x)$ using Proposition~\ref{prop:weightofP} and
%	rearranging gives the result.
%	The other equation is proved similarly.
%\end{proof}

%\begin{Ex}
%	Consider again the four upward arborescences in $P=RT_{3,2}$ as depicted in Figure~\ref{fig:upwardSpanningTree}, and let $x$ and $y = \psi(x)$ be as shown in Figure~\ref{fig:r32}. 
%	
%	Suppose $C$ is the chain $(2,1) \lessdot (3,1) \lessdot (3,2)$. Only the first two arborescences shown contain $C$, and the fraction of the total weight covered by these two is
%	\[\mu_y(U_P(C)) = \frac{y_{31}}{y_{31}+y_{22}}.\]
%	Plugging in the coordinates from Figure~\ref{fig:r32} gives
%	\[\frac{y_{31}}{y_{31}+y_{22}} = \frac{abd}{abd+a(b+c)e} = bdf \cdot \frac{a}{a(bd+be+ce)f} = w_C(x) \cdot \frac{y_{11}}{y_{32}}\]
%	as required by Proposition~\ref{prop:chainarborescence}.
%	%= \frac{a(ac+ab)}{ac \cdot a(b+c)e \cdot a(bd+be+ce)f} = \frac{1}{ace(bd+be+ce)f} = \frac{bd}{bd+be+ce} \cdot \frac{1}{abcdef}.\]
%\end{Ex}



\iffalse
% In fact, arborescences can be used to give more refined information about weights in subposets of $P$. We first introduce some terminology for our subposets of interest.

% \begin{Def}
	% A subposet $I \subset P$ is \emph{saturated} if it is an induced subposet of $P$, and every cover relation in $I$ is a cover relation of $P$.
	% \end{Def}

% Put another way, an induced subposet $I$ is saturated if the Hasse diagram of $I$ is a subgraph of the Hasse diagram of $P$. Note that if $I$ is saturated, then every maximal chain of $I$ is a saturated chain in $P$.

% If $\mathscr C$ is a collection of saturated chains in $P$ such that each chain in $\mathscr C$ has the same minimum $a$ and the same maximum $b$, we will say that $\mathscr C$ is \emph{bounded} by $(a,b)$. In particular, this holds if $\mathscr C$ is the set of all maximal chains of a saturated subposet $I \subset P$ with minimum $a$ and maximum $b$, in which case we write $\mathscr C = \mathscr C(I)$.


% Define
% \[w_{\mathscr C}(x) = \sum_{C \in \mathscr C} w_C(x).\]
% Let $U_P(\mathscr C)$ be the set of all upward arboresences of $P$ that contain (i.e., contain the edges of) some element of $\mathscr C$, and similarly define $D_P(\mathscr C)$.



% %For any saturated subposet $I \subset P$, let $C(I)$ denote the set of all maximal chains in the interval $I$, and let \[H_I(x) = \sum_{C \in C(I)} w_C(x).\]
% %In addition, let $U_P(I)$ be the set of all upward arborescences of $P$ that contain the edges of some maximal chain in $I$, and similarly define $D_P(I)$.

% \begin{Prop} \label{prop:chainarborescence}
	% Let $P$ be a poset, $x \in \RR_+^P$, $y=\psi(x)$, and $z=\tilde\rho^{-1}(x)$. Suppose $\mathscr C$ is a collection of saturated chains in $P$ bounded by $(a,b)$. Then
	% \begin{align*}
		%     \frac{w_{\mathscr C}(x)}{w_P(x)} &= \left(\sum_{a' \lessdot a} \frac{y_{a'}}{y_b}\right)^{-1} \cdot  \sum_{T\in U_P(\mathscr C)} w(T;y),\\
		%     \frac{w_{\mathscr C}(z)}{w_P(z)} &= \left(\sum_{b' \gtrdot b} \frac{y_{a}}{y_{b'}}\right)^{-1} \cdot  \sum_{T\in D_P(\mathscr C)} w(T;y).
		% \end{align*}
	% \end{Prop}
% \begin{proof}
	% The product of the weights of the edges in any $C \in \mathscr C$ is $\frac{y_a}{y_b}$ (which does not depend on $C$). The elements of $U_P(\mathscr C)$ that contain the edges in $C$ are then obtained by adding one downward edge from each element of $(P \setminus C) \cup \{a\}$. It follows that
	% \begin{align*}\sum_{T \in U_P(\mathscr C)} w(T;y) &= \sum_{C \in \mathscr C} \left(\frac{y_a}{y_b} \cdot \sum_{a' \lessdot a} \frac{y_{a'}}{y_a} \cdot \prod_{p \in P \setminus C} \sum_{q \lessdot p} \frac{y_q}{y_p}\right)\\
		% &=\sum_{a' \lessdot a} \frac{y_{a'}}{y_b} \cdot \sum_{C \in \mathscr C} \prod_{p \in P \setminus C} x_p^{-1}\\
		% &=\sum_{a' \lessdot a} \frac{y_{a'}}{y_b} \cdot \frac{w_{\mathscr C}(x)}{w_P(x)}.
		% \end{align*}
	% The other equation is proved similarly.
	% \end{proof}

% In light of Proposition~\ref{prop:weightofP}, we can rephrase Proposition~\ref{prop:chainarborescence} as follows. For a subset $U \subset U_P$ and any $y \in \RR_+^P$, define
% \begin{align*}
	%     \mu_y(U) &= \frac{\sum_{T \in U} w(T;y)}{\sum_{T \in U_P} w(T;y)},
	% \end{align*}
% the proportion of the weight of $U_P$ contained in $U$ or $D$. (Hence $\mu_y$ is a probability measure on $U_P$.) Similarly define $\mu_y(D)$ for any subset $D \subset D_P$. Then we have the following corollary.

% \begin{Cor}
	%     Let $P$ be a poset, $x \in \RR_+^P$, $y = \psi(x)$, and $z = \tilde\rho^{-1}(x)$. Then
	%     \[\mu_y(U_P(\mathscr C)) = w_{\mathscr C}(x) \cdot \sum_{a' \lessdot a} \frac{y_{a'}}{y_b}.\]
	% \end{Cor}
\fi

%Given a collection $\mathscr C$ of (saturated) chains, define
%\[U_P(\mathscr C) = \bigsqcup_{C \in \mathscr C} U_P(C) \quad \text{and} \quad D_P(\mathscr C) = \bigsqcup_{C \in \mathscr C} D_P(C).\]
%We will typically be interested in finding the total weight \[w_{\mathscr C}(x) = \sum_{C \in \mathscr C} w_C(x).\] 
%Note that if $a = \min(C)$ and $b=\max(C)$ are fixed for all $C \in \mathscr C$, then it is easy to sum \eqref{eq:wcx} (and similarly \eqref{eq:wcz}) over all $C \in \mathscr C$ since the first factor on the right hand side will be fixed. Thus the result will be the same equation \eqref{eq:wcx} but with $C$ replaced by $\mathscr C$. In addition, each arborescence will contain at most one $C \in \mathscr C$, so there would be no need to worry about multiplicities in $U_P(\mathscr C)$.


One can exploit the symmetry of Proposition~\ref{cor:edgeweight} to compute the following result.
\begin{Cor} \label{cor:bijectionimpliesweight}
	Let $P$ be a poset, $x \in \RR_+^P$, and $z = \tilde \rho^{-1}(x)$. Let $m,m',M,M' \in P$ such that $m'$ is the unique element covered by $m$ and $M$ is the unique element covering $M'$. Let $\mathscr C$ be any collection of saturated chains from $m$ to $M$, and similarly define $\mathscr C'$. Then 
	$\frac{\mu_y(U_P(\mathscr C))}{w_{\mathscr C}(x)} %= \frac{y_{m'}}{y_M} 
	= \frac{\mu_y(D_P(\mathscr C'))}{w_{\mathscr C'}(z)}$.
%	the equality $\mu_y(U_P(\mathscr C)) = \mu_y(D_P(\mathscr C'))$ implies $w_{\mathscr C}(x) = w_{\mathscr C'}(z)$.
	
\end{Cor}
%\begin{proof}
%	Summing \eqref{eq:wcx} and \eqref{eq:wcz} over all elements of $\mathscr C$ and $\mathscr C'$, respectively, gives
%	\[w_{\mathscr C}(x) = \left(\frac{y_{m'}}{y_{M}}\right)^{-1} \cdot \mu_y(U_P(\mathscr C)) \quad \text{and} \quad w_{\mathscr C'}(z) = \left(\frac{y_{m'}}{y_{M}}\right)^{-1} \cdot \mu_y(D_P(\mathscr C')).\qedhere\]
%\end{proof}

% \begin{Ex}
	%      Consider again $P = RT_{3,2}$ with arborescences shown in Figure~\ref{fig:upwardSpanningTree}. Let $\mathscr C$ be the set of two chains from $(2,1)$ to $(3,2)$, and let $\mathscr C'$ be the set of two chains from $(1,1)$ to $(2,2)$. Then $\mathscr C$ and $\mathscr C'$ satisfy the conditions of Corollary~\ref{cor:bijectionimpliesweight} with $m=(2,1)$, $m'=(1,1)$, $M = (3,2)$, and $M'=(2,2)$.
	
	%      In addition, $U_P(\mathscr C)$ consists of the first three upward arborescences shown, while $D_P(\mathscr C')$ consists of the first three downward arborescences shown.
	% \end{Ex}

%It is not difficult to generalize Proposition~\ref{prop:chainarborescence} to unions of nonintersecting chains, but we will leave this until Section~\ref{sec:nonintersecting}.

%An important case occurs when $\mathcal C$ is the set of all maximal chains in a saturated subposet $I \subset P$. In this case, we will write $\mathcal C=\mathcal C(I)$.

% \begin{Prop}
	% \label{Prop:upwardSpanningTrees}
	% Let $P$ be a poset with minimum element $\,\widehat{0}$. Let $I$ be an induced subposet of $P$ with minimum $m$ and maximum $M$ and suppose that $m$ covers at most 1 element of $P$. Let $m' \lessdot m$ if such $m'$ exists and let $y_{m'} = 1$ otherwise. Then
	% \[H_I(x;1) = w_P(x) \frac{y_M}{y_{m'}y_{\;\widehat{0}}} \sum\limits_{T \in U_P(I)} w(T;x).\]
	% \end{Prop}

% \begin{proof}
	% We have
	% \[H_I(x;1) = \sum\limits_{C \in MC(I)} \prod\limits_{p \in C} x_p = w_P(x) \sum\limits_{C \in MC(I)} \prod\limits_{p \in P \setminus C} \frac{1}{x_p}.\]
	% Since $x = \phi(y)$, $x_p = \frac{y_p}{\sum\limits_{q \lessdot p} y_q}$ for all $p \in P$ (where we interpret the empty sum as $1$). The reciprocal of this expression is a Laurent polynomial, which we substitute into the previous expression. This yields
	% \begin{align*}
		% w_P(x) \sum\limits_{C \in MC(I)} \prod\limits_{p \in P \setminus C} \frac{1}{x_p} &= w_P(x) \sum\limits_{C \in MC(I)} \prod\limits_{p \in P \setminus C} \sum\limits_{q \lessdot p} \frac{y_q}{y_p} \\
		% &= w_P(x) \frac{y_M}{y_{m'}} \sum\limits_{C \in MC(I)} \frac{y_{m'}}{y_M} \prod\limits_{p \in P \setminus C} \sum\limits_{q \lessdot p} \frac{y_q}{y_p}
		% \end{align*}
	% It remains to translate this expression into tree weights. That is, we will show
	% \[\frac{1}{y_{\;\widehat{0}}} \sum\limits_{T \in U_P(I)} w(T;x) = \sum\limits_{C \in MC(I)} \frac{y_{m'}}{y_M} \prod\limits_{p \in P \setminus C} \sum\limits_{q \lessdot p} \frac{y_q}{y_p}.\]
	% Given a tree $T \in U_P(I)$, each non-minimal vertex $p \in T$ has exactly one edge $q \lessdot p$ in $T$. If $C$ is the distinguished chain of $T$, then we multiply the edge weights of the edges $q \lessdot p$ with $p \in C$. This product telescopes. Since there is at most 1 vertex covered by $m$, we know that this product is $\frac{y_{m'}}{y_M}$. For each non-minimal $p \in P \setminus C$, $T$ contains one edge $q \lessdot p$ which yields a choice of monomial in the sum $\sum\limits_{q \lessdot p} \frac{y_q}{y_p}$. For the minimal vertex, $\frac{1}{y_{\;\widehat{0}}}$ appears as an additional variable in every Laurent monomial of the right hand side that is not accounted for by the tree weight. Hence $\frac{1}{y_{\widehat{0}}} w(T;x)$ is a monomial appearing in the right hand side. 
	
	% Additionally, any monomial in the right hand side gives rise to a tree in $U_P(I)$. To choose a monomial, we must fix a chain $C \in MC(I)$ and an edge $q \lessdot p$ for each $p \in P \setminus C$. We form a tree $T$ with the chosen edges, the edges of the fixed chain, and the unique edge $m' \lessdot m$ downward in $P$ from $m$. Then $\frac{1}{y_{\;\widehat{0}}} w(T;x)$ is the monomial we started with.
	% \end{proof}

% Similarly, we can express chain statistics after the application of rowmotion in terms of weights of downward spanning trees.

% \begin{Prop}
	% \label{Prop:downwardSpanningTrees}
	% Let $P$ be a poset with maximum element $\widehat{1}$. Let $I$ an induced subposet with minimum $m$ and maximum $M$ and suppose that $M$ is covered by at most 1 element of $P$. Let $M \lessdot M'$ if such $M'$ exists and let $y_{M'} = 1$ otherwise. Then
	% \[H_I(\phi \circ \rho^{-1} \circ \phi^{-1}(x);1) = w_P(x) \frac{y_{M'}y_{\;\widehat{1}}}{y_m} \sum\limits_{T \in D_P(I)} w(T;x).\]
	% \end{Prop}

% \begin{proof}
	% Let $\widetilde{x} = \phi \circ \rho^{-1} \circ \phi^{-1}(x)$. Then
	% \[H_I(\widetilde{x};1) = \sum\limits_{C \in MC(I)} \prod\limits_{p \in C} \widetilde{x}_p = w_P(\widetilde{x}) \sum\limits_{C \in MC(I)} \prod\limits_{p \in P \setminus C} \frac{1}{\widetilde{x}_p}\]
	% But $\widetilde{x} = \phi \circ \rho^{-1}(y)$. For all $p \in P$
	% \[\rho^{-1}(y)_p = \frac{1}{\sum\limits_{q \gtrdot p} \frac{1}{y_q}} \left(\sum\limits_{q \lessdot p} \rho^{-1}(y)_q \right) \frac{1}{y_p}.\]
	% After applying $\phi$, we obtain
	% \[\widetilde{x}_p = \frac{1}{\sum\limits_{q \gtrdot p} \frac{1}{y_q}}  \left(\frac{1}{y_p}\right) = \frac{1}{\sum\limits_{q \gtrdot p} \frac{y_p}{y_q}}.\]
	% The reciprocal of $\widetilde{x}_p$ is a Laurent polynomial in $y_q$. Similar to the proof of Proposition~\ref{Prop:upwardSpanningTrees}, we obtain
	% \begin{align*}
		% w_P(\widetilde{x}) \sum\limits_{C \in MC(I)} \prod\limits_{p \in P \setminus C} \frac{1}{\widetilde{x}_p} &= w_P(\widetilde{x}) \sum\limits_{C \in MC(I)} \prod\limits_{p \in P \setminus C} \sum\limits_{q \gtrdot p} \frac{y_p}{y_q} \\
		% &= w_P(\widetilde{x}) \frac{y_{M'}}{y_m} \sum\limits_{C \in MC(I)} \frac{y_m}{y_{M'}} \prod\limits_{p \in P \setminus C} \sum\limits_{q \gtrdot p} \frac{y_p}{y_q}
		% \end{align*}
	% A bijection of Laurent monomials analogous to the proof of Proposition~\ref{Prop:upwardSpanningTrees} yields the result.
	% \end{proof}

% Furthermore $w_P(x)$, the vertex weight of the entire poset, can be described in terms of tree weights.

% \begin{Cor}
	%     \label{Cor:weightofp}
	%     Let $P$ be a poset with maximum and minimum element. Then
	%     \[w_P(x)^{-1} = \frac{1}{y_{\;\widehat{0}}} \sum\limits_{T \in U_P} w(T;x) = y_{\;\widehat{1}} \sum\limits_{T \in D_P} w(T;x).\]
	% \end{Cor}

% \begin{proof}
	% We first prove
	% \[w_P(x)^{-1} = \frac{1}{y_{\;\widehat{0}}} \sum\limits_{T \in U_P} w(T;x).\]
	% We use the notation of Proposition~\ref{Prop:upwardSpanningTrees} and apply Proposition~\ref{Prop:upwardSpanningTrees} in the case of $I=P$. Since the minimum element of $P$ does not cover any elements, $y_{m'} = 1$. By definition of $\phi^{-1}$ we know that $y_M = H_P(x;1)$. By Proposition~\ref{Prop:upwardSpanningTrees} we obtain
	% \[y_M = w_P(x) \frac{y_M}{y_{\;\widehat{0}}} \sum\limits_{T \in U_P} w(T;x).\]
	% Canceling the instances of $y_M$ and solving for $w_P(x)$ yields the result. The other equation involving downward spanning trees follows via a similar argument.
	% \end{proof}

%Proposition~\ref{Prop:upwardSpanningTrees}, Proposition~\ref{Prop:downwardSpanningTrees}, and Corollary~\ref{Cor:weightofp} suggest a strategy for proving equations involving chain statistics. If one can find a bijection between upward and downward spanning trees, one can keep track of how statistics $H_I(x;1)$ change as we apply rowmotion iteratively. For many posets, such a bijection will not exist because either there will not be the same number of Laurent monomials associated to different intervals or it will be difficult to keep track of changes of weights under a bijection. However, for skew shapes there exists a simple bijection which modifies weight in a predictable way.





%\subsection{Skew shapes}

 %We identify certain elements of a skew shape poset as \emph{inward} or \emph{outward corner points} as in Figure~\ref{fig:cornerPoints}. 


%\begin{figure}
%	\centering
%	\begin{tikzpicture}[scale = 0.5]
%		%DOWN-DIAGONALS
%		\draw (0,0)--(-2,2);
%		\draw (1,1)--(-3,5);
%		\draw (2,2)--(-2,6);
%		\draw (3,3)--(-2,8);
%		\draw (4,4)--(-1,9);
%		\draw (3,7)--(0,10);
%		
%		%UP-DIAGONALS
%		\draw (0,0)--(4,4);
%		\draw (-1,1)--(3,5);
%		\draw (-2,2)--(3,7);
%		\draw (-2,4)--(2,8);
%		\draw (-3,5)--(1,9);
%		\draw (-2,8)--(0,10);
%		
%		\node[bigB] at (4,4) {};
%		\node[bigB] at (3,7) {};
%		\node[bigB] at (-2,2) {};
%		\node[bigB] at (-3,5) {};
%		\node[bigB] at (-2,8) {};
%		
%		\node[bigW] at (2,6) {};
%		\node[bigW] at (-1,3) {};
%		\node[bigW] at (-1,7) {};
%	\end{tikzpicture}
%	\caption{A skew shape and its corner points. Outward corner points are marked in black and inward corner points are marked in white.}
%	\label{fig:cornerPoints}
%\end{figure}


%\subsection{The main bijection}

Let $S$ be a \emph{skew shape poset} as in Figure~\ref{fig:treeBijection1}.
%In this section, we will exhibit a bijection between $U_P$ and $D_P$ in the case that $P$ is a skew shape and show that this bijection multiplies weight by a constant factor. (Such a bijection does not exist for most posets $P$, so this is a special feature of skew shapes.) This will allow us to use Propositions~\ref{prop:weightofP} and \ref{prop:chainarborescence} to relate the chain statistics $w_{\mathscr C}(x)$ before and after rowmotion.
%Fix a skew shape $S \subset R_{r,s}$. 
Note that if $q \in S$ only covers a single element $p$, then any element of $U_S$ must contain the edge $p \lessdot q$, so we call this edge \emph{forced} for $U_S$. Likewise, the edge $p \lessdot q$ is \emph{forced} for $D_S$ if $q$ is the only element covering $p$.

We define a bijection $\aleph \colon U_S \to D_S$ as follows. Translate $T \in U_S$ in the plane by the vector $(-\frac12, -\frac12)$ (i.e., downward) to $\overline T$. Then form $\aleph(T)$ by taking all edges of $S$ that do not intersect $\overline{T}$, together with all forced edges for $D_S$. See Figure~\ref{fig:treeBijection1} for an example. %(Aside from some special behavior along the boundary of $S$, $\aleph$ is essentially the standard bijection between spanning trees of a planar graph and its dual graph.) 

We show that $\aleph$ affects the weight of each arborescence in a uniform way.

\begin{figure}
	\centering
	\begin{tikzpicture}[scale=0.45]
		\begin{scope}
			\draw [gray] 
			%down diagonals
			(0,0)--(-2,2)
			(1,1)--(-3,5)
			(2,2)--(-2,6)
			(3,3)--(-2,8)
			(4,4)--(-1,9)
			(3,7)--(0,10)
			%up diagonals
			(0,0)--(4,4)
			(-1,1)--(3,5)
			(-2,2)--(3,7)
			(-2,4)--(2,8)
			(-3,5)--(1,9)
			(-2,8)--(0,10);
			\draw [ultra thick, red] 
			(0,-1)--(0,0)
			(0,0)--(4,4) (-1,1)--(1,3) (-1,3)--(1,5) (2,6)--(3,7) (-2,6)--(1,9) (-2,8)--(0,10)
			(0,0)--(-2,2) (0,2)--(-3,5) (0,4)--(-2,6) (3,3)--(2,4) (1,5)--(0,6) (-1,7)--(-2,8) (4,4)--(1,7) (3,7)--(2,8);
			\node[scale=1.5, transform shape] at (-1.5,.5){$m$};
			\node[scale=1.5, transform shape] at (1.5,9.5){$M$};
		\end{scope}
		\begin{scope}[shift={(9,-1)}]
			\draw [ultra thick, red!50] 
			(0,-1)--(0,0)
			(0,0)--(4,4) (-1,1)--(1,3) (-1,3)--(1,5) (2,6)--(3,7) (-2,6)--(1,9) (-2,8)--(0,10)
			(0,0)--(-2,2) (0,2)--(-3,5) (0,4)--(-2,6) (3,3)--(2,4) (1,5)--(0,6) (-1,7)--(-2,8) (4,4)--(1,7) (3,7)--(2,8);
		\end{scope}
		\begin{scope}[shift={(9,0)}]
			\draw [gray, dashed] 
			%down diagonals
			(0,0)--(-2,2)
			(1,1)--(-3,5)
			(2,2)--(-2,6)
			(3,3)--(-2,8)
			(4,4)--(-1,9)
			(3,7)--(0,10)
			%up diagonals
			(0,0)--(4,4)
			(-1,1)--(3,5)
			(-2,2)--(3,7)
			(-2,4)--(2,8)
			(-3,5)--(1,9)
			(-2,8)--(0,10);
			\draw [ultra thick, blue]
			(0,0)--(2,2) (0,2)--(2,4) (-2,2)--(-1,3) (-1,5)--(2,8) (-3,5)--(1,9) (-2,8)--(0,10)--(0,11)
			(-1,1)--(-2,2) (-1,3)--(-3,5) (2,2)--(1,3) (0,4)--(-1,5) (3,3)--(0,6) (4,4)--(1,7) (3,7)--(0,10);
		\end{scope}
		\begin{scope}[shift={(18,0)}]
			\draw [gray] 
			%down diagonals
			(0,0)--(-2,2)
			(1,1)--(-3,5)
			(2,2)--(-2,6)
			(3,3)--(-2,8)
			(4,4)--(-1,9)
			(3,7)--(0,10)
			%up diagonals
			(0,0)--(4,4)
			(-1,1)--(3,5)
			(-2,2)--(3,7)
			(-2,4)--(2,8)
			(-3,5)--(1,9)
			(-2,8)--(0,10);
			\draw [ultra thick, blue]
			(0,0)--(2,2) (0,2)--(2,4) (-2,2)--(-1,3) (-1,5)--(2,8) (-3,5)--(1,9) (-2,8)--(0,10)--(0,11)
			(-1,1)--(-2,2) (-1,3)--(-3,5) (2,2)--(1,3) (0,4)--(-1,5) (3,3)--(0,6) (4,4)--(1,7) (3,7)--(0,10);
			\node[scale=1.5, transform shape] at (-0.5,-.5){$m'$};
			\node[scale=1.5, transform shape] at (2.5,8.5){$M'$};
		\end{scope}
	\end{tikzpicture}
	\caption{An upward arborescence $T$ (in red) and its image $\aleph(T)$ (in blue).}
	\label{fig:treeBijection1}
\end{figure}

% \begin{figure}
	%     \centering
	% \begin{tikzpicture}[scale = 0.6]
		% %DOWN-DIAGONALS
		% \draw [gray] (0,0)--(-2,2);
		% \draw [gray] (1,1)--(-2,4);
		% \draw [gray] (2,2)--(-1,5);
		% \draw [gray] (3,3)--(0,6);
		% \draw [gray] (3,5)--(1,7);
		
		% %UP-DIAGONALS
		% \draw [gray] (0,0)--(3,3);
		% \draw [gray] (-1,1)--(3,5);
		% \draw [gray] (-2,2)--(2,6);
		% \draw [gray] (-2,4)--(1,7);
		
		% %RED
		% \draw [ultra thick, red] (0,0)--(-2,2);
		% \draw [ultra thick, red] (0,2)--(-2,4);
		% \draw [ultra thick, red] (0,4)--(-1,5);
		% \draw [ultra thick, red] (3,3)--(2,4);
		
		% \draw [ultra thick, red] (0,0)--(3,3);
		% \draw [ultra thick, red] (-1,1)--(1,3);
		% \draw [ultra thick, red] (2,4)--(3,5);
		% \draw [ultra thick, red] (-1,3)--(2,6);
		% \draw [ultra thick, red] (-1,5)--(1,7);
		
		% \draw[ultra thick, red] (0,-1)--(0,0);
		
		% %\node at (-1.5,0.9) {$m$};
		% %\node at (2.5,6.1) {$M$};
		
		% \begin{scope}[shift={(7,-1)}]
			% %RED
			% \draw [ultra thick, red] (0,0)--(-2,2);
			% \draw [ultra thick, red] (0,2)--(-2,4);
			% \draw [ultra thick, red] (0,4)--(-1,5);
			% \draw [ultra thick, red] (3,3)--(2,4);
			
			% \draw [ultra thick, red] (0,0)--(3,3);
			% \draw [ultra thick, red] (-1,1)--(1,3);
			% \draw [ultra thick, red] (2,4)--(3,5);
			% \draw [ultra thick, red] (-1,3)--(2,6);
			% \draw [ultra thick, red] (-1,5)--(1,7);
			
			% \draw[ultra thick, red] (0,-1)--(0,0);
			
			% %BLUE
			% \begin{scope}[shift={(0,1)}]
				% \draw [ultra thick, blue] (-1,1)--(-2,2);
				% \draw [ultra thick, blue] (-1,3)--(-2,4);
				% \draw [ultra thick, blue] (2,2)--(1,3);
				% \draw [ultra thick, blue] (3,3)--(2,4);
				% \draw [ultra thick, blue] (3,5)--(1,7);
				
				% \draw [ultra thick, blue] (0,0)--(2,2);
				% \draw [ultra thick, blue] (0,2)--(3,5);
				% \draw [ultra thick, blue] (-2,2)--(-1,3);
				% \draw [ultra thick, blue] (0,4)--(2,6);
				% \draw [ultra thick, blue] (-2,4)--(1,7);
				
				% \draw[ultra thick, blue] (1,7)--(1,8);
				% \end{scope}
			% \end{scope}
		
		% \begin{scope}[shift={(14,0)}]
			% %DOWN-DIAGONALS
			% \draw [gray] (0,0)--(-2,2);
			% \draw [gray] (1,1)--(-2,4);
			% \draw [gray] (2,2)--(-1,5);
			% \draw [gray] (3,3)--(0,6);
			% \draw [gray] (3,5)--(1,7);
			
			% %UP-DIAGONALS
			% \draw [gray] (0,0)--(3,3);
			% \draw [gray] (-1,1)--(3,5);
			% \draw [gray] (-2,2)--(2,6);
			% \draw [gray] (-2,4)--(1,7);
			
			% \draw [ultra thick, blue] (-1,1)--(-2,2);
			% \draw [ultra thick, blue] (-1,3)--(-2,4);
			% \draw [ultra thick, blue] (2,2)--(1,3);
			% \draw [ultra thick, blue] (3,3)--(2,4);
			% \draw [ultra thick, blue] (3,5)--(1,7);
			
			% \draw [ultra thick, blue] (0,0)--(2,2);
			% \draw [ultra thick, blue] (0,2)--(3,5);
			% \draw [ultra thick, blue] (-2,2)--(-1,3);
			% \draw [ultra thick, blue] (0,4)--(2,6);
			% \draw [ultra thick, blue] (-2,4)--(1,7);
			
			% \draw[ultra thick, blue] (1,7)--(1,8);
			
			% %\node at (0.6,0) {$m'$};
			% %\node at (3.5,5) {$M'$};
			% \end{scope}
		% \end{tikzpicture}
	% \caption{An upward arborescence $T$ (in red) and its image $\aleph(T)$ (in blue). After shifting $T$ down, we form $\aleph(T)$ by taking all edges that do not intersect $T$, together with all forced edges. Note that the only intersecting edges occur near the inner corners of $S$.}
	% \label{fig:treeBijection1}
	% \end{figure}

%As an abuse of language, we say edges $e \in T$ and $e' \in \aleph(T)$ \emph{intersect} if the shifted edge $\overline e \in \overline T$ intersects $e'$. If $e$ is, say, the covering relation $(i,j) \lessdot (i,j+1)$, then it only has the potential to intersect $e'$ if $e'$ is the edge $(i-1,j) \lessdot (i,j)$. Moreover, these edges are either both forced (if $(i,j)$ is a right inward corner) or neither is forced. Therefore, $e \in T$ and $e' \in \aleph(T)$ can only intersect if they are both forced edges incident to an inward corner of $S$.

%\begin{Prop} \label{prop:alephbijection}
%	The map $\aleph \colon U_S \to D_S$ is a well-defined bijection.
%\end{Prop}
%\begin{proof}
%	We need to check that for any $T \in U_S$, we have $\aleph(T) \in D_S$, that is, for any $a \in S$, there is exactly one edge $a \lessdot b$ in $\aleph(T)$.
%	
%	If $a$ is only covered by one element, then this is clear since all forced edges are in $\aleph(T)$, so suppose $a$ is covered by two elements $b$ and $c$. Then $b$ and $c$ are both covered by a unique element $d\in S$. By the nonintersecting condition,
%	$\aleph(T)$ contains the edge $a \lessdot b$ if and only if $T$ does not contain the edge $b \lessdot d$, and similarly with the edges $a \lessdot c$ and $c \lessdot d$. Since $T \in U_S$, it contains exactly one of $b \lessdot d$ and $c \lessdot d$, so $\aleph(T)$ will contain exactly one of $a \lessdot b$ and $a \lessdot c$, as desired.
%	
%	To see that $\aleph$ is a bijection, one can define $\aleph^{-1}$ in an analogous manner to $\aleph$ by rotating the plane by a half turn.
%\end{proof}





 %Recall that for any (not necessarily induced) subgraph $G$ of $S$, $\updeg_G(i,j)$ and $\downdeg_G(i,j)$ denote the up degree and down degree of $(i,j)$ in $G$.

\begin{Lemma}
	\label{lemma:weightChange}
	Let $T \in U_S$ and $y \in \mathbb{R}^{S}_+$. Then there exists a Laurent monomial $y^{\alpha(S)}$ depending only on $S$ such that $\omega_{\aleph(T)}(y) = \omega_T(y)\cdot  y^{\alpha(S)}$ for all $T \in U_S$.
%	Specifically,
%	\[\frac{ \omega_{\aleph(T)}(y) }{ \omega_T(y) }  = y_{11}^{2-\updeg_S(1,1)}y_{rs}^{2-\downdeg_S(r,s)} \cdot 
%	\frac{\prod_{p \in \outcorner(S)} y_p }{  \prod_{p \in \incorner(S)} y_p }.\]
%	%where $u = \updeg_S(1,1)$ and $d =\downdeg_S(r,s)$. 
\end{Lemma}
%
%\begin{proof}
%	Note that $\omega_T(y)$ contains a factor of $y_p$ for each edge $p \lessdot q$ in $T$ and a factor of $y_p^{-1}$ for each edge $q \lessdot p$ in $T$. Therefore we can write the exponent of $y_p$ in $\omega_{\aleph(T)}(y)/\omega_T(y)$ as
%	%\[\frac{ w(\aleph(T);y) }{ w(T;y) } = \prod_{p \in S} y_{p}^{\alpha_{p}},\]
%	%where
%	\begin{align*}
%		\alpha_{p} &= (\updeg_{\aleph(T)}(p) - \downdeg_{\aleph(T)}(p))-(\updeg_T(p) - \downdeg_T(p))\\
%		&= 2-\downdeg_{\aleph(T)}(p)-\updeg_T(p).
%	\end{align*}
%	
%	If $p = (1,1)$, then $\downdeg_{\aleph(T)}(1,1)=0$ while $\updeg_T(1,1)=\updeg_S(1,1)$ since the upward edges from $(1,1)$ are forced in $T$. Thus $\alpha_{11} = 2-\updeg_S(1,1)$. Similarly for the maximum element $(r,s)$, we have $\alpha_{rs} = 2-\downdeg_S(r,s)$.
%	
%	For any other element $p = (i,j)$, we can consider the contributions to $2-\alpha_p= \downdeg_{\aleph(T)}(p)+\updeg_{T}(p)$ from the possible edges to the right, namely $(i,j) \lessdot (i,j+1)$ and $(i-1,j) \lessdot (i,j)$.
%	\begin{itemize}
%		\item If $p$ is a right outward corner point, then clearly there is no contribution.
%		\item If $p$ is a right inward corner point, then both edges are forced in their respective arborescences, so the contribution is $2$.
%		\item If only one of the two edges, say $(i,j) \lessdot (i,j+1)$, exists in $S$, then it is a forced edge for $T$, so the contribution is $1$ (and similarly if the other edge is the only one present).
%		\item If both edges exist in $S$ but $p$ is not a right outward corner point, then neither edge is forced. Then by the construction of $\aleph$, $(i,j) \lessdot (i,j+1)$ is an edge of $T$ if and only if $(i-1,j) \lessdot (i,j)$ is not an edge of $\aleph(T)$. It follows that the contribution is $1$ in this case as well.
%	\end{itemize}
%	Since the contributions from the possible edges to the left of $p$ can be obtained symmetrically, we find that $2-\alpha_p$ is usually $2$, but we must subtract $1$ if $p$ is an outward corner point and add $1$ if $p$ is an inward corner point (these adjustments cancel out if $p$ is both an inward and an outward corner point). The result follows easily.
%\end{proof}

%In other words, $\aleph$ scales the weight of every arborescence by the same amount.

\begin{Cor} \label{cor:measurepreserving}
	The bijection $\aleph \colon U_S \to D_S$ is measure-preserving: $\mu_y(U) = \mu_y(\aleph(U))$ for all $y \in \RR_+^S$ and subsets $U \subset U_S$.
\end{Cor}
%\begin{proof}
%	By Lemma~\ref{lemma:weightChange}, both the numerator and denominator of $\mu_y(\aleph(U))$ are obtained from those of $\mu_y(U)$ by scaling by $y^{\alpha(S)}$, so these factors cancel out.
%\end{proof}

\begin{Ex}
	Consider again the arborescences for $RT_{3,2}$ in Figure~\ref{fig:upwardSpanningTree}. The bijection $\aleph$ sends each upward arborescence to the downward arborescence directly below it. In each case, $\aleph$ multiplies the weight by $y_{42} \cdot \frac{y_{12}y_{31}}{y_{32}}$, as predicted by Lemma~\ref{lemma:weightChange}. %As a consequence, each upward arborescence occupies the same fraction of weight as the corresponding downward arborescence.
\end{Ex}

%\subsection{Chain shifting} \label{sec:chainshifting}

%We are now ready to prove a chain shifting lemma for skew shapes by combining Corollaries~\ref{cor:bijectionimpliesweight} and \ref{cor:measurepreserving}.

%The following lemma summarizes the properties we need of $\aleph$. Given a saturated chain $C \subset S$, we say that $p \in S$ lies \emph{southeast} (resp. \emph{southwest}) of $C$ if it lies to the east (resp. west) of an element of the downward shift $\overline C$. We say that $p$ \emph{lies beyond} $C$ if it does not lie southeast or southwest of $C$, that is, when $\overline C$ does not contain an element at the same height as $p$. 
%
%\begin{Lemma} \label{lemma:chaincross}
%	Let $T \in U_S$ and $C$ a saturated chain contained in $T$. Let $p \in S$, and let $C'$ be the unique chain upwards from $p$ to $\hat 1$ in $\aleph(T) \in D_S$.
%	
%	If $p$ lies southeast of $C$, and $C$ contains no (forced) edge northeast from a right inward corner point, then every element of $C'$ lies to the southeast of or beyond $C$. In particular, if $C$ contains the rightmost element at any rank above $p$, then $C'$ must contain its southeast neighbor.
%	
%	The analogous statement obtained by reflecting across the vertical axis also holds.
%\end{Lemma}
%\begin{proof}
%	The only way that $C'$ can go from an element southeast of $C$ to an element southwest of $C$ is along an edge of $\aleph(T)$ to the northwest that intersects an edge of $C \subset T$ to the northeast. But this can only occur if the edge in $C$ is a forced edge of the given type.
%\end{proof}


%If $p = (i,j)$, denote by $se(p)$ the set of elements of $S$ of the form $(i-1-k,j+k)$ for $k \geq 0$. (These are the elements that lie east of the downward shift $\overline p$.) For instance, $se(p) \neq 0$ if and only if $p$ has a southeast neighbor.
%
%Similarly denote by $sw(p)$ the set of elements $(i-1-k,j+k)$ with $k < 0$, and define $nw(p) = \{q \mid p \in se(q)\}$ and $ne(p) = \{q \mid p \in sw(q)\}$. 

%\subsubsection{Simple chain shifting} \label{sec:simplechainshifting}

We are now ready to prove a chain shifting lemma for skew shapes.
% by combining Corollaries~\ref{cor:bijectionimpliesweight} and \ref{cor:measurepreserving}.
Our first form is a generalization of the chain shifting lemma for rectangles proven by the current authors in \cite{johnsonliu} (and in the noncommutative setting by Grinberg and Roby \cite{grinbergroby3}) to skew shapes $S$.


For $p \in S$, write $se(p) \neq \varnothing$ if $p$ has a southeast neighbor and $se(p) = \varnothing$ otherwise. Given elements $p < q$ in $S$, let $\mathscr C_{p,q}$ denote the set of all saturated chains from $p$ to $q$, and let $\mathscr C^{se}_{p,q} \subset \mathscr C_{p,q}$ denote the subset consisting of those chains $C$ for which $se(r) \neq \varnothing$ for all $r \in C$. (Also define the analogous notation for the directions $sw$, $ne$, and $nw$.)

% Fix a skew shape $S \subset R_{r,s}$. For integers $1 < i_1 \leq i_2 \leq r$, define
% \begin{align*}
	%     I &= \{(a,b) \in S \mid i_1 \leq a \leq i_2 \text{ and } (a-1,b) \in S\},\\
	%     I' &= \{(a,b) \in S \mid (a+1,b) \in I\},
	% \end{align*}
% so that $I$ and $I'$ are translates of one another, and assume that $I$ and $I'$ are saturated subposets of $S$. See Figure~\ref{fig:skewChainShifting} for an example. We will let $m$ and $M$ (resp.\ $m'$ and $M'$) denote the minimum and maximum elements of $I$ (resp.\ $I'$). Note that $m'$ is the unique element covered by $m$, and $M$ is the unique element covering $M'$. 
%By construction, $I$ does not contain any edges that go northeast from a right inward corner, which are forced for upward arborescences. Similarly, $I'$ does not contain any edges that go southwest from a left inward corner, which are forced for downward arborescences.

% \begin{figure}
	%     \centering
	% \begin{tikzpicture}[scale = 0.5]
		% %DOWN-DIAGONALS
		% \draw (0,0)--(-2,2);
		% \draw (1,1)--(-2,4);
		% \draw (2,2)--(-1,5);
		% \draw (3,3)--(0,6);
		% \draw (3,5)--(1,7);
		
		% %UP-DIAGONALS
		% \draw (0,0)--(3,3);
		% \draw (-1,1)--(3,5);
		% \draw (-2,2)--(2,6);
		% \draw (-2,4)--(1,7);
		
		% \node at (-1,-0.1) {\footnotesize$(1,1)$};
		% \node at (-2,0.9) {\footnotesize$(2,1)$};
		% \node at (2,0.9) {\footnotesize$(1,2)$};
		% \node at (3.8,5) {\footnotesize$(2,5)$};
		
		% \draw [ultra thick] (-1,1)--(-2,2);
		% \draw [ultra thick] (0,2)--(-1,3);
		% \draw [ultra thick] (1,3)--(0,4);
		% \draw [ultra thick] (2,4)--(1,5);
		
		% \draw [ultra thick] (-1,1)--(2,4);
		% \draw [ultra thick] (-2,2)--(2,6);
		
		% \begin{scope}[shift={(9,0)}]
			% %DOWN-DIAGONALS
			% \draw (0,0)--(-2,2);
			% \draw (1,1)--(-2,4);
			% \draw (2,2)--(-1,5);
			% \draw (3,3)--(0,6);
			% \draw (3,5)--(1,7);
			
			% %UP-DIAGONALS
			% \draw (0,0)--(3,3);
			% \draw (-1,1)--(3,5);
			% \draw (-2,2)--(2,6);
			% \draw (-2,4)--(1,7);
			
			% \draw [ultra thick] (0,0)--(-1,1);
			% \draw [ultra thick] (1,1)--(0,2);
			% \draw [ultra thick] (2,2)--(1,3);
			% \draw [ultra thick] (3,3)--(2,4);
			
			% \draw [ultra thick] (0,0)--(3,3);
			% \draw [ultra thick] (-1,1)--(3,5);
			% \end{scope}
		% \end{tikzpicture}
	% \caption{The subposets $I$ (left) and $I'$ (right) in Proposition~\ref{prop:alephbijection2}. The element $(2,5) \in S$ is omitted from $I$ since it has no shifted counterpart in $S$.}
	% \label{fig:skewChainShifting}
	% \end{figure}

%Recall that if $\mathscr C$ is a collection of chains, then $U_S(\mathscr C) = \bigsqcup_{C \in \mathscr C} U_S(C)$ is the set of all arborescences containing some chain in $\mathscr C$, and $D_S(\mathscr C)$ is defined similarly.

\begin{Lemma} \label{lemma:skewchainshifting}
	Let $m' \lessdot m$ and $M' \lessdot M$ be elements of $S$ such that $sw(m) = ne(M') = \varnothing$.
	\begin{enumerate}[label={(\alph*)},itemsep=0pt,parsep=0pt,topsep=4pt]
		\item The bijection $\aleph$ restricts to a bijection from $U_S(\mathscr C^{se}_{m,M})$ to $D_S(\mathscr C^{nw}_{m',M'})$.
		\item Let $x \in \RR_+^S$ and $z = \tilde\rho^{-1}(x)$. Then $w_{\mathscr C^{se}_{m,M}}(x) = w_{\mathscr C^{nw}_{m',M'}}(z)$.
	\end{enumerate}
\end{Lemma}
\begin{proof}[Proof sketch]
	Let $T \in U_S(\mathscr C^{se}_{m,M})$, and let $C$ be the chain in $T$ from $m$ to $M$. By the construction of $\aleph$, $\aleph(T) \in D_S$ contains a saturated chain upward from $m'$ that does not cross $C$, so it must pass through $M'$. It follows that $\aleph(T) \in D_S(\mathscr C^{nw}_{m',M'})$. The reverse argument shows that $\aleph$ is a bijection, and (b) then follows from Corollaries~\ref{cor:measurepreserving} and \ref{cor:bijectionimpliesweight}.
\end{proof}
%\begin{proof}
%	Note that the conditions imply that $m'$ is the only element covered by $m$ (with $m'$ to the southeast) and $M$ is the only element covering $M'$ (with $M'$ to the southeast). Choose any $T \in U_S(\mathscr C^{se}_{m,M})$, and let $C$ be the chain in $T$ from $m$ to $M$. By definition of $\mathscr C^{se}_{m,M}$, $C$ does not contain any (forced) edge northeast from a right inward corner point. Hence by Lemma~\ref{lemma:chaincross}, $\aleph(T) \in D_S$ contains a saturated chain $C'$ from $m'$ to $M'$ that lies entirely southeast of $C$. Thus $C' \in \mathscr C^{nw}_{m',M'}$ and hence $\aleph(T) \in D_S(\mathscr C^{nw}_{m',M'})$. 
%	A similar argument shows that $\aleph^{-1}$ sends $D_S(\mathscr C^{nw}_{m',M'})$ to $U_S(\mathscr C^{se}_{m,M})$, completing part (a). Part (b) then follows from part (a), Corollary~\ref{cor:measurepreserving}, and Corollary~\ref{cor:bijectionimpliesweight}.
%\end{proof}

%Choose any $T \in U_S(I)$, and consider its shift $\overline T$ in the overlay of $\overline S$ and $S$. Then $\overline T$ contains a chain $\overline C \in C(\overline I)$ from $\overline m$ to $\overline M$. We also have that $\aleph(T) \in D_S$ contains some saturated chain $C'$ from $m'$ to $\hat 1$. Note that the minima of $C$ and $C'$, namely $\overline m$ and $m'$, are at the same height with $m'$ to the right.

%Let $m$ and $m'$ be the minimum elements of $I$ and $I'$, and let $M$ and $M'$ be their maximum elements. For any $T \in U_S(I)$, $T$ contains a maximal chain $C \in C(I)$ from $m$ to $M$, while $\aleph(T) \in D_S(I')$ likewise contains a chain $C' \in C(I')$ from $m'$ to $M'$. Let $\overline C$ be the shift of $C$ in $\overline T$.

%We claim that $\overline C$ and $C'$ do not intersect. Indeed, if they did, then $C$ would have an edge $(i,j) \lessdot (i,j+1)$ that intersects an edge $(i-1,j) \lessdot (i,j)$ in $C'$, where $(i,j)$ is a right inward corner point, and so $(i-1,j+1) \notin S$. But $(i,j+1) \in C \subset I$, so this contradicts the definition of $I$. It follows that $\overline C$ and $C'$ do not intersect and so $C'$ is always to the right of $C$. In particular, $C'$ must contain $M'$ since this is the only point of $S$ to the right of $\overline M$ at the same height. In addition, $C' \cap [m',M']$ lies entirely to the right of $C$, so it is contained within $I'$. Thus $\aleph(T)$ contains a maximal chain in $C(I')$, so $\aleph(T) \in D_S(I')$.



% The following chain shifting lemma follows immediately.

% \begin{Lemma} \label{Lemma:skewchainshifting}
	% Let $x \in \RR^S_+$ and $z = \tilde\rho^{-1}(x)$, and let $\mathscr C$ and $\mathscr C'$ be the sets of maximal chains in $I$ and $I'$, respectively. Then $w_{\mathscr C}(x)=w_{\mathscr C'}(z)$.
	% \end{Lemma}
% \begin{proof}
	%     By Proposition~\ref{prop:alephbijection2} and Corollary~\ref{cor:measurepreserving}, $\mu_y(U_S(\mathscr C)) = \mu_y(D_S(\mathscr C'))$. The result then follows from Corollary~\ref{cor:bijectionimpliesweight}.
	% \end{proof}

% \begin{proof}
	% By Proposition~\ref{prop:chainarborescence} (and the ensuing discussion),
	% \begin{align*}
		%     w_{\mathscr C}(x) &= \left(\frac{y_{m'}}{y_{M}}\right)^{-1} \cdot \mu_y(U_S(\mathscr C)),\\
		%     w_{\mathscr C'}(z) &= \left(\frac{y_{m'}}{y_{M}}\right)^{-1} \cdot \mu_y(D_S(\mathscr C')).
		% \end{align*}
	% where $y = \psi(x)$. By Proposition~\ref{prop:alephbijection2}, $\aleph(U_S(\mathscr C)) = D_S(\mathscr C')$, so the result follows from Corollary~\ref{cor:measurepreserving}.
	% \end{proof}
% \begin{figure}
	%     \centering
	% \begin{tikzpicture}[scale = 0.6]
		% %DOWN-DIAGONALS
		% \draw [gray] (0,0)--(-2,2);
		% \draw [gray] (1,1)--(-2,4);
		% \draw [gray] (2,2)--(-1,5);
		% \draw [gray] (3,3)--(0,6);
		% \draw [gray] (3,5)--(1,7);
		
		% %UP-DIAGONALS
		% \draw [gray] (0,0)--(3,3);
		% \draw [gray] (-1,1)--(3,5);
		% \draw [gray] (-2,2)--(2,6);
		% \draw [gray] (-2,4)--(1,7);
		
		% %RED
		% \draw [ultra thick, red] (0,0)--(-2,2);
		% \draw [ultra thick, red] (0,2)--(-2,4);
		% \draw [ultra thick, red] (0,4)--(-1,5);
		% \draw [ultra thick, red] (3,3)--(2,4);
		
		% \draw [ultra thick, red] (0,0)--(3,3);
		% \draw [ultra thick, red] (-1,1)--(1,3);
		% \draw [ultra thick, red] (2,4)--(3,5);
		% \draw [ultra thick, red] (-1,3)--(2,6);
		% \draw [ultra thick, red] (-1,5)--(1,7);
		
		% \node at (-1.5,0.9) {$m$};
		% \node at (2.5,6.1) {$M$};
		
		% \begin{scope}[shift={(7,-1)}]
			% %RED
			% \draw [ultra thick, red] (0,0)--(-2,2);
			% \draw [ultra thick, red] (0,2)--(-2,4);
			% \draw [ultra thick, red] (0,4)--(-1,5);
			% \draw [ultra thick, red] (3,3)--(2,4);
			
			% \draw [ultra thick, red] (0,0)--(3,3);
			% \draw [ultra thick, red] (-1,1)--(1,3);
			% \draw [ultra thick, red] (2,4)--(3,5);
			% \draw [ultra thick, red] (-1,3)--(2,6);
			% \draw [ultra thick, red] (-1,5)--(1,7);
			
			% %BLUE
			% \begin{scope}[shift={(0,1)}]
				% \draw [ultra thick, blue] (-1,1)--(-2,2);
				% \draw [ultra thick, blue] (-1,3)--(-2,4);
				% \draw [ultra thick, blue] (2,2)--(1,3);
				% \draw [ultra thick, blue] (3,3)--(2,4);
				% \draw [ultra thick, blue] (3,5)--(1,7);
				
				% \draw [ultra thick, blue] (0,0)--(2,2);
				% \draw [ultra thick, blue] (0,2)--(3,5);
				% \draw [ultra thick, blue] (-2,2)--(-1,3);
				% \draw [ultra thick, blue] (0,4)--(2,6);
				% \draw [ultra thick, blue] (-2,4)--(1,7);
				% \end{scope}
			% \end{scope}
		
		% \begin{scope}[shift={(14,0)}]
			% %DOWN-DIAGONALS
			% \draw [gray] (0,0)--(-2,2);
			% \draw [gray] (1,1)--(-2,4);
			% \draw [gray] (2,2)--(-1,5);
			% \draw [gray] (3,3)--(0,6);
			% \draw [gray] (3,5)--(1,7);
			
			% %UP-DIAGONALS
			% \draw [gray] (0,0)--(3,3);
			% \draw [gray] (-1,1)--(3,5);
			% \draw [gray] (-2,2)--(2,6);
			% \draw [gray] (-2,4)--(1,7);
			
			% \draw [ultra thick, blue] (-1,1)--(-2,2);
			% \draw [ultra thick, blue] (-1,3)--(-2,4);
			% \draw [ultra thick, blue] (2,2)--(1,3);
			% \draw [ultra thick, blue] (3,3)--(2,4);
			% \draw [ultra thick, blue] (3,5)--(1,7);
			
			% \draw [ultra thick, blue] (0,0)--(2,2);
			% \draw [ultra thick, blue] (0,2)--(3,5);
			% \draw [ultra thick, blue] (-2,2)--(-1,3);
			% \draw [ultra thick, blue] (0,4)--(2,6);
			% \draw [ultra thick, blue] (-2,4)--(1,7);
			
			% \node at (0.6,0) {$m'$};
			% \node at (3.5,5) {$M'$};
			% \end{scope}
		% \end{tikzpicture}
	% \caption{An upward spanning tree (in red) and its dual (in blue). Under the map $\aleph$ an upward spanning tree with chain in $\mathscr C^{se}_{m,M}$ maps to a downward spanning tree with a chain in $\mathscr C^{nw}_{m',M'}$. The blue chain lies visually to the right of the red chain in the overlayed diagram.}
	% \label{fig:treeBijection}
	% \end{figure}

%\begin{Ex}
%	As an illustration of Lemma~\ref{lemma:skewchainshifting}, consider Figure~\ref{fig:treeBijection1} with the elements $m$, $M$, $m'$, and $M'$ indicated. The upward arborescence $T$ depicted contains a chain $C \in \mathscr C^{se}_{m,M}$. When applying $\aleph$ as in the center picture, we find that the chain in $\aleph(T)$ upward from $m'$ must remain to the right of $\overline{C}$ and therefore pass through $M'$. Thus $\aleph(T)$ contains a chain $C' \in \mathscr C^{nw}_{m',M'}$ as in Lemma~\ref{lemma:skewchainshifting}.
%\end{Ex}

%\begin{Ex}
%	Consider again $S=RT_{32}$ with arborescences shown in Figure~\ref{fig:upwardSpanningTree}. If we let $m = (2,1)$, $M = (3,2)$, $m'=(1,1)$, and $M' = (2,2)$, then the conditions of Lemma~\ref{lemma:skewchainshifting} are satisfied. Only the first three upward arborescences contain a chain from $m$ to $M$ (which are all in $\mathscr C^{se}_{m,M}$), and $\aleph$ sends these to the first three downward arborescences, which are the only ones that contain a chain in $\mathscr C^{nw}_{m',M'}$.
%	
%	We can verify that Lemma~\ref{lemma:skewchainshifting}(b) holds in this case using the labels in Figure~\ref{fig:r32}:
%	\begin{align*}
%		z_{11}z_{21}z_{22}+z_{11}z_{12}z_{22} &= \frac{bc}{b+c} \left( \frac{(b+c)de}{bd+be+ce}+\frac{(b+c)e}{c}\right) \frac{(bd+be+ce)f}{(b+c)e}\\
%		&= \frac{bcdf}{b+c} + \frac{b(bd+be+ce)f}{b+c}\\
%		&= b(d+e)f = x_{21}x_{31}x_{32}+x_{21}x_{22}x_{32}.
%	\end{align*}
%	This equality holds even though neither term on the left is equal to a term on the right. (This is because $\aleph$ does not restrict to a bijection between $U_S(C)$ and $D_S(C')$ for any single chains $C \in \mathscr C^{se}_{m,M}$ and $C'\in \mathscr C^{nw}_{m',M'}$).
%\end{Ex}

\begin{Ex}
	Let $S = RT_{32}$ and take $m = (2,1)$, $M = (3,2)$, $m'=(1,1)$, and $M' = (2,2)$. We can verify that Lemma~\ref{lemma:skewchainshifting}(b) holds in this case using the labels in Figure~\ref{fig:r32}:
		\begin{align*}
				z_{11}z_{21}z_{22}+z_{11}z_{12}z_{22} %&= \frac{bc}{b+c} \left( \frac{(b+c)de}{bd+be+ce}+\frac{(b+c)e}{c}\right) \frac{(bd+be+ce)f}{(b+c)e}\\
				= \tfrac{bcdf}{b+c} + \tfrac{b(bd+be+ce)f}{b+c}
				= bdf+bef = x_{21}x_{31}x_{32}+x_{21}x_{22}x_{32}.
		\end{align*}
	We can also verify Lemma~\ref{lemma:skewchainshifting}(a) using Figure~\ref{fig:upwardSpanningTree} by noting that $U_S(\mathscr C_{m,M}^{se})$ and $D_S(\mathscr C_{m',M'}^{nw})$ are the arborescences in the leftmost three columns, which are in bijection via $\aleph$.
\end{Ex}


The bijection $\aleph$ is a powerful tool for relating weights of subsets of $P$ with respect to $x$ and $z = \tilde\rho(x)$. The general strategy is simple: relate the quantities of interest to the weights of certain subsets of $U_P$ and $D_P$, then show that these subsets are in bijection via $\aleph$. In this way, one can easily prove many previously established results about rowmotion on rectangles as well as further generalizations. %We give a variety of examples of this in the rest of this section.
%For example, one can similarly prove the following result for the right trapezoid $S=RT_{r,s}$.


%\subsubsection{Chains in trapezoids} \label{sec:chainsintrapezoids}

%A particular case that will be important for us concerns chains in the right trapezoid $S=RT_{r,s}$.

For another example of this, define the \emph{left border} of the right trapezoid $RT_{r,s}$ to be the set of elements of the form $L = \{(\ell+r-1,\ell) \mid 1 \leq \ell \leq s\}$. For $p,q \in RT_{r,s}$, let $\mathscr C^L_{p,q}$ be the subset of $\mathscr C_{p,q}$ consisting of chains that intersect $L$.

%In other words, $L$ is the set of all left outer corners of $RT_{r,s}$ together with the maximum element.



\begin{Lemma} \label{Lemma:rtchainshifting}
	Let $S=RT_{r,s}$, and let $m' \lessdot m$ and $M' \lessdot M$ such that $se(m) = ne(M') = \varnothing$.
	\begin{enumerate}[label={(\alph*)},itemsep=0pt,parsep=0pt,topsep=4pt]
		\item The bijection $\aleph$ restricts to a bijection from $U_S(\mathscr C^L_{m,M})$ to $D_S(\mathscr C^L_{m',M'})$.
		\item Let $x \in \RR_+^S$ and $z = \tilde \rho^{-1}(x)$. Then $w_{\mathscr C^L_{m,M}}(x) = w_{\mathscr C^L_{m',M'}}(z)$.
	\end{enumerate}
	% let  $x \in \RR_+^{S}$ and $z = \tilde \rho^{-1}(x)$.  Let $\mathscr C$ be the set of all saturated chains from $m$ to $M$ that intersect $L$, and let $\mathscr C'$ be the set of all saturated chains from $m'$ to $M'$ that intersect $L$. Then $w_{\mathscr C}(x)=w_{\mathscr C'}(z)$.
\end{Lemma}
%\begin{proof}
%	%    We apply Proposition~\ref{prop:chainarborescence} as in the proof of Lemma~\ref{Lemma:skewchainshifting}.
%	%     \begin{align*}
%		%     w_{\mathscr C}(x) &= \left(\frac{y_{m'}}{y_{M}}\right)^{-1} \cdot \frac{\sum_{T \in U_S(\mathscr C)} w(T;y)}{\sum_{T \in U_S}w(T;y)},\\
%		%     w_{\mathscr C'}(z) &= \left(\frac{y_{m'}}{y_{M}}\right)^{-1} \cdot \frac{\sum_{T \in D_S(\mathscr C')} w(T;y)}{\sum_{T \in D_S}w(T;y)}.
%		% \end{align*}
%	Choose any chain $C \in \mathscr C^L_{m,M}$, and let $c$ be the lowest element of $C \cap L$. Let $c'$ be the left inner corner covered by $c$, and let $c''$ be the element of $L$ covered by $c'$.
%	By Lemma~\ref{lemma:chaincross}, $\aleph(T)$ contains a chain $C'$ that starts at $m'$, remains southwest of $C$ until it reaches $c''$, at which point it crosses over (using a forced edge) to $c'$, and then remains southeast of $C$ until reaching $M'$. It follows that $\aleph(T) \in D_S(\mathscr C^L_{m',M'})$, and the highest element of $C' \cap L$ is $c''$.
%	%    , so that $(\ell+r-2,\ell) \in C$ is an inner corner of $S$. By Lemma~\ref{lemma:chaincross}, $\aleph(T)$ contains a chain $C'$ that starts at $m'$, remains southwest of $C$ until it reaches $(\ell+r-2,\ell-1) \in L$, at which point it crosses over (using a forced edge) to $(\ell+r-2,\ell)$, and then remains southeast of $C$ until reaching $M'$. It follows that $\aleph(T) \in D_S(\mathscr C')$. 
%	A similar argument applied to $\aleph^{-1}$ shows that $\aleph$ gives a bijection from $U_S(\mathscr C^L_{m,M})$ to $D_S(\mathscr C^L_{m',M'})$, as desired.
%	
%	Part (b) then follows from part (a) by using Corollaries~\ref{cor:bijectionimpliesweight} and \ref{cor:measurepreserving}.
%\end{proof}

%See Figure~\ref{fig:rtchainshifting} for an illustration. We can visualize this by breaking the chains in $\mathscr C$ immediately before they first intersect $L$ and then shifting the lower portions to the southwest and the upper portions to the southeast using Lemma~\ref{lemma:skewchainshifting}.

%In particular, this gives an alternate proof of Lemma~\ref{Lemma:rtchainshifting}(b): for $c \in L$, define $c'$ and $c''$ as in the proof of Lemma~\ref{Lemma:rtchainshifting}. Then using Lemma~\ref{lemma:skewchainshifting},
%\[w_{\mathscr C^L_{m,M}}(x) = \sum_c w_{\mathscr C^{sw}_{m,c'}}(x)w_{\mathscr C^{se}_{c,M}}(x) = \sum_c w_{\mathscr C^{ne}_{m',c''}}(z) w_{\mathscr C^{nw}_{c',M'}}(z) = w_{\mathscr C^L_{m',M'}}(z).\]
%This also makes it clear that $\mathscr C^L_{m,M}$ and $\mathscr C^L_{m',M'}$ contain the same number of chains.

%\begin{Ex}
%	A simple example of Lemma~\ref{Lemma:rtchainshifting} can be seen in $S=RT_{3,2}$ using the arborescences in Figure~\ref{fig:upwardSpanningTree}. If $m = (1,2)$ and $M=(4,2)$, then there is a unique chain in $\mathscr C^L_{m,M}$, and the only upward arborescence containing this chain is the fourth one. Since $m' = (1,1)$ and $M' = (3,2)$, there is also a unique chain in $\mathscr C^L_{m',M'}$ (it must pass through $(3,1) \in L$), and the only downward arborescence containing this chain is again the fourth one, the image under $\aleph$ of the upward arborescence found above. Algebraically, we can check using the labelings given in Figure~\ref{fig:r32} that Lemma~\ref{Lemma:rtchainshifting}(b) holds here:
%	\[z_{11}z_{21}z_{31}z_{32} = \frac{bc}{b+c} \cdot \frac{(b+c)de}{bd+be+ce} \cdot \frac{(bd+be+ce)f}{bd} \cdot g = cefg = x_{12}x_{22}x_{32}x_{42}.\]
%\end{Ex}

%\begin{Rem}Though we will not need it for our purposes, it is straightforward to apply this technique more generally to write down similar chain shifting lemmas for chains that touch the boundary of a general skew shape $S$, possibly in several places.
%\end{Rem}

% \subsubsection{Nonintersecting chains} \label{sec:nonintersecting}
% We can generalize Proposition~\ref{prop:chainarborescence} to apply not just to chains but also to disjoint unions of chains. %Suppose $C = C_1 \sqcup C_2 \sqcup \dots \sqcup C_k$ is a disjoint union of saturated chains $C_i \subset P$. Let $a_i$ and $b_i$ be the minimum and maximum of $C_i$.

% \begin{Prop}
	%     Let $P$ be a poset, $x \in \RR_+^P$, $y = \psi(x)$, and $z = \tilde\rho^{-1}(x)$. Let $C = C_1 \sqcup C_2 \sqcup \dots \sqcup C_k$ be a disjoint union of saturated chains $C_i$ with minima $a_i$ and maxima $b_i$. Then
	%         \begin{align*}
		%     w_C(x) &= \prod_{i=1}^k\left(\sum_{a' \lessdot a_i} \frac{y_{a'}}{y_{b_i}}\right)^{-1} \cdot \mu_y(U_P(C)),\\
		%     w_C(z) &= \prod_{i=1}^k\left(\sum_{b' \gtrdot b_i} \frac{y_{a_i}}{y_{b'}}\right)^{-1} \cdot \mu_y(D_P(C)).
		%     \end{align*}
	% \end{Prop}

% Suppose $I \subset P$ is a saturated subposet with distinguished elements $a_1, \dots, a_k \in I$ and $b_1, \dots, b_k \in I$. Let $C(I;k)$ denote the set of all collections of $k$ disjoint saturated chains, each from $a_i$ to $b_i$ for some $i$. Then define
% \[H_I(x;k) = \sum_{\mathcal C \in C(I;k)} \prod_{C \in \mathcal C} w_C(x).\]
% We also let $U_P(I;k)$ be the set of all upward arborescences of $P$ that contain all the edges of the chains in some element of $C(I;k)$, and similarly for $D_P(I;k)$.

% \begin{Prop}
	% Let $P$ be a poset and $I \subset P$ a saturated subposet with distinguished elements $a_1, \dots, a_k$ and $b_1, \dots, b_k$ as described above. Let $x \in \RR_+^P$, $y = \phibar(x)$, and $z = \tilde\rho^{-1}(x)$. Then
	% \begin{align*}
		%     \frac{H_I(x;k)}{w_P(x)} &= \prod_{i=1}^k\left(\sum_{a' \lessdot a_i} \frac{y_{a'}}{y_{b_i}}\right)^{-1} \cdot  \sum_{T\in U_P(I;k)} w(T;y),\\
		%     \frac{H_I(z;k)}{w_P(z)} &= \prod_{i=1}^k\left(\sum_{b' \gtrdot b_i} \frac{y_{a_i}}{y_{b'}}\right)^{-1} \cdot  \sum_{T\in D_P(I;k)} w(T;y).
		% \end{align*}
	% \end{Prop}
% \begin{proof}
	% Similar to the proof of Proposition~\ref{prop:chainarborescence}.
	% \end{proof}

% One particular case of interest is when $P$ is a skew shape $S$.



% \begin{Lemma}
	% \label{Lemma:skewchainshifting}
	% Let $1 < i_1 \leq i_2 \leq r$,
	% \[I = \{(a,b) \in S: a \in [i_1,i_2]~\mbox{and}~ (a-1,b) \in S\}, \quad~\mbox{and}~\quad I' = \{(a,b) \in S: (a+1,b) \in I\}.\]
	% Then for any labeling $x \in \mathbb{R}_{>0}^S$
	% \[H_{I'}(\phi \circ \rho^{-1} \circ \phi^{-1}(x);1) = H_I(x;1).\]
	% \end{Lemma}

\iffalse
\begin{Ex}
	For this example we consider the skew shape in Figure~\ref{fig:skewChainShifting} and fix $i_1 = 2$ and $i_2 = 3$. Figure~\ref{fig:skewChainShifting} depicts the subposets $I$ and $I'$ as defined in Lemma~\ref{Lemma:skewchainshifting}. If $x \in \mathbb{R}_{>0}^{S}$ and $\widetilde{x} = \phi \circ \rho^{-1} \circ \phi^{-1}(x)$, then Lemma~\ref{Lemma:skewchainshifting} states that
	\begin{align*}
		&\widetilde{x}_{11} \widetilde{x}_{21} \widetilde{x}_{22} \widetilde{x}_{23} \widetilde{x}_{24} \widetilde{x}_{25} + \widetilde{x}_{11} \widetilde{x}_{12} \widetilde{x}_{22} \widetilde{x}_{23} \widetilde{x}_{24} \widetilde{x}_{25} + \widetilde{x}_{11} \widetilde{x}_{12} \widetilde{x}_{13} \widetilde{x}_{23} \widetilde{x}_{24} \widetilde{x}_{25} + \widetilde{x}_{11} \widetilde{x}_{12} \widetilde{x}_{13} \widetilde{x}_{14} \widetilde{x}_{24} \widetilde{x}_{25} \\
		= &x_{21} x_{31} x_{32} x_{33} x_{34} x_{35} + x_{21} x_{22} x_{32} x_{33} x_{34} x_{35} + x_{21} x_{22} x_{23} x_{33} x_{34} x_{35} + x_{21} x_{22} x_{23} x_{24} x_{34} x_{35}.
	\end{align*}
	Here the variables within a monomial are ordered from the bottom of their respective chain to the top.
	
	
\end{Ex}
\fi

% Let $S$ be a skew shape and let $T \in U_S$. We say an edge $p \lessdot q$ is \emph{forced} if there is no other $p' \in S$ such that $p' \lessdot q$. Similarly we have forced edges in $D_S$.

% Let $I$ and $I'$ be defined as in Lemma~\ref{Lemma:skewchainshifting}. We define a bijection $\aleph:U_S(I) \to D_S(I')$ as follows. Let $T \in U_S(I)$. Form a tree $\overline{T}$ by shifting $T$ in the plane by the vector $(-0.5,-0.5)$ (and let $\overline{S}$ be the shifted version of $S$). We form a graph $\aleph(T)$ on $V(T)$ whose edges are those that do not intersect $\overline{T}$, together with all forced edges (see Figure~\ref{fig:treeBijection1} for an example). Let $\aleph^*$ denote $\aleph$ on $S^*$ and let $T^*$ denote the tree analogous to $T$ in $S^*$. Our next goal is to show that $\aleph(T)$ is indeed a downward spanning tree and that $T$ and $\aleph(T)$ are dual in the sense that $\aleph^*( \aleph(T)^* ) = T^*$. As an abuse of language, we say edges $a \lessdot b$ in $T$ and $c \lessdot d$ in $\aleph(T)$ \emph{intersect} if they intersect after shifting $T$ by $(-0.5,-0.5)$. Note that parallel edges cannot intersect.

% \begin{Obs}
	% \label{Obs:edgeIntersection}
	%     Edges $(i,j) \lessdot (i+1,j)$ in $\overline{S}$ and $(a,b) \lessdot (a,b+1)$ in $S$ intersect if and only if $a = i+1$ and $b = j$. Edges $(i,j) \lessdot (i,j+1)$ in $\overline{S}$ and $(a,b) \lessdot (a+1,b)$ in $S$ intersect if and only if $a = i$ and $b = j+1$.
	% \end{Obs}

% A tree $T$ and its image $\aleph(T)$ often intersect. See Figure~\ref{fig:treeBijection1}. From the definition of $\aleph(T)$, it is clear that the only edges of $\aleph(T)$ that can intersect edges of $T$ are the forced edges. However, the forced edges of $T$ are also the only edges that can intersect $\aleph(T)$. Indeed, if $(i,j) \lessdot (i+1,j)$ in $T$ intersects $(i,j-1) \lessdot (i,j)$ in $\aleph(T)$, then $(i,j-1) \lessdot (i,j)$ must be a forced edge of $\aleph(T)$ and so $(i+1,j-1) \not\in S$. It follows that $(i,j)$ is the unique element covered by $(i+1,j)$ and the edge $(i,j) \lessdot (i+1,j)$ is forced. The intersections of $T$ and $\aleph(T)$ are described in terms of \emph{corner points} of the skew shape.





% \begin{Prop}
	% \label{Prop:intersectionofdualtrees}
	%     Edges $(i,j) \lessdot (i,j+1)$ in $T$ and $(i-1,j) \lessdot (i,j)$ in $\aleph(T)$ intersect if and only if $(i,j)$ is a right inward corner point of $S$.
	% \end{Prop}

% A similar statement holds for left inward corner points.

% \begin{proof}
	%     Suppose that $(i,j) \lessdot (i,j+1)$ in $T$ and $(i-1,j) \lessdot (i,j)$ in $\aleph(T)$ intersect. This implies that both edges are forced. Consequently $(i,j)$ is the unique element covered by $(i,j+1)$ in $S$ and $(i,j)$ is the unique element covering $(i-1,j)$ in $S$. It follows that $(i-1,j+1) \not\in S$, and so $(i,j)$ is a right inward corner point.
	
	%     Conversely, let $(i,j)$ be a right inward corner point, so that $(i,j) \lessdot (i,j+1)$ in $T$ and $(i-1,j) \lessdot (i,j)$ in $\aleph(T)$ are forced. After applying a $(-0.5,-0.5)$ shift to $T$, these edges intersect at $(i-0.5,j)$.
	%\end{proof}
	
	% \begin{Ex}
		% Consider the pair of trees in Figure~\ref{fig:treeBijection}. If $T$ is the red tree, we saw previously that $\aleph(T)$ is the blue tree. As we showed in Proposition~\ref{Prop:intersectionofdualtrees}, the intersections of blue and red edges occur southwest of left inward corner points in $S$ and southeast of right inward corner points.
		
		% Consider the subposets $I$ and $I'$ in Figure~\ref{fig:skewChainShifting}, which have minima and maxima as labeled in Figure~\ref{fig:treeBijection}. In the overlaid middle diagram the chain from $m$ to $M$ is always to the left of the chain from $m'$ to $M'$.
		
		% \end{Ex}
	
	% We now prove that $\aleph$ maps upward spanning trees to downward spanning trees.
	
	% \begin{Prop}
		% \label{Prop:alephBijection}
		%     $\aleph$ maps $U_S(I)$ to $D_S(I')$ bijectively.
		% \end{Prop}
	
	% \begin{proof}
		% Let $(i,j) \in S$ be non-maximal. Then there exists at least one edge upward from $(i,j)$. If this edge is unique, then the edge is forced and appears in $\aleph(T)$. Otherwise $(i+1,j)$ and $(i,j+1)$ cover $(i,j)$. We know that $(i+1,j+1) \in S$ because $S$ is a skew shape (with maximum and minimum). Since $T \in U_S$, there exists a unique edge downward from $(i+1,j+1)$ in $T$. Without loss of generality, assume $(i+1,j) \lessdot (i+1,j+1)$ is not in $T$. By Observation~\ref{Obs:edgeIntersection}, this is the only edge intersecting $(i,j) \lessdot (i+1,j)$, and so $(i,j) \lessdot (i+1,j)$ is an edge of $\aleph(T)$. The edge $(i,j) \lessdot (i,j+1)$ is not an edge of $\aleph(T)$ since it intersects $(i,j+1) \lessdot (i+1,j+1)$, which is an edge of $T$. Since each non-maximal vertex has a unique edge upward in $S$ from it, $\aleph(T) \in D_S$.
		
		% To see that $\aleph(T) \in D_S(I')$ we have the following set up:
		% \begin{enumerate}
			%     \item Let $m$ and $M$ be the minimum and maximum elements of $I$.
			%     \item Let $m'$ and $M'$ be the minimum and maximum elements of $I'$.
			%     \item Let $C$ be the distinguished maximal chain of $I$ containing $m$.
			%     \item Let $C'$ be the maximal chain in $\aleph(T)$ containing $m'$.
			% \end{enumerate}
		
		% Since $m' \lessdot m$, $m'$ and $m$ are horizontally aligned in the shifted overlay of $\overline{S}$ and $S$ (and similarly for $M' \lessdot M$). Suppose that $C$ and $C'$ intersect. Since $m'$ is right of $m$ in the shifted overlay, we must have intersecting edges $(i,j) \lessdot (i,j+1)$ in $T$ and $(i-1,j) \lessdot (i,j)$ in $\aleph(T)$. Since these edge intersect, $(i,j)$ is a right inward corner point, and so $(i-1,j+1) \not\in S$. But $(i,j+1) \in C \subset I$, and so by the definition of $I$ we must have $(i-1,j+1) \in S$, which is a contradiction. Hence it must be the case that $C$ and $C'$ do not intersect. 
		
		% Consequently $C'$ is right of $C$ at each height in the shifted overlay of $\overline{S}$ and $S$. In particular, $C'$ must contain $M'$ since $M$ is the rightmost point in $S$ at the height of $M'$. So $C'$ contains $m'$ and $M'$. Since $C'$ and $C$ do not intersect, we conclude that $C' \cap [m',M']$ is a chain contained in $I'$. Hence $\aleph(T) \in D_S(I')$.
		
		% By Proposition~\ref{Prop:intersectionofdualtrees}, we know where intersections of $T$ and $\aleph(T)$ occur, and that all edges of $S$ not intersecting $T$ are contained in $\aleph(T)$. Thus $T$ can be reconstructed from $\aleph(T)$ via the analogous map on the dual of $S$.
		% \end{proof}
	
	
	% We are now ready to prove the chain shifting lemma for skew shapes.
	
	% \begin{proof}[Proof of Lemma~\ref{Lemma:skewchainshifting}]
		% Let the notation be as in the statement of Lemma~\ref{Lemma:skewchainshifting}. By Proposition~\ref{Prop:upwardSpanningTrees} and Corollary~\ref{Cor:weightofp},
		% \[H_I(x;1) = w_S(x) \frac{y_M}{y_{m'} y_{\;\widehat{0}}} \sum\limits_{T \in U_S(I)} w(T;x) = \frac{ \frac{y_M}{y_{m'} y_{\;\widehat{0}}} \sum\limits_{T \in U_S(I)} w(T;x) }{ \frac{1}{y_{\;\widehat{0}}} \sum\limits_{T \in U_S} w(T;x)}.\]
		% By Lemma~\ref{Lemma:weightChange}, we know that for each $T$, $w(\aleph(T);x) = x^{\alpha(S)} w(T;x)$ for some Laurent monomial $x^{\alpha(S)}$ which does not depend on $T$. 
		% \[H_I(x;1) = \frac{ \frac{y_M}{y_{m'}} \sum\limits_{T \in U_S(I)} x^{-\alpha(S)} w(\aleph(T);x) }{\sum\limits_{T \in U_S} x^{-\alpha(S)} w(\aleph(T);x)}.\]
		% Since $\alpha(S)$ is independent of tree, we factor $x^{-\alpha(S)}$ out of the sum and cancel. By Proposition~\ref{Prop:alephBijection}, we know that $\aleph:U_S(I) \to D_S(I')$ is a bijection and so we may reindex this sum:
		% \[H_I(x;1) = \frac{ \frac{y_M}{y_{m'}} \sum\limits_{T \in D_S(I')} w(T;x) }{\sum\limits_{T \in D_S}  w(T;x)} = \frac{ \frac{y_M y_{\widehat{1}}}{y_{m'}} \sum\limits_{T \in D_S(I')} w(T;x) }{ y_{\widehat{1}} \sum\limits_{T \in D_S}  w(T;x)}.\]
		% By Proposition~\ref{Prop:downwardSpanningTrees} and Corollary~\ref{Cor:weightofp} this expression is just $H_{I'}(\phi \circ \rho^{-1} \circ \phi^{-1}(x);1)$.
		% \end{proof}
	
	% By gluing intervals together, we can further derive chain shifting lemmas for skew shapes. At certain locations in a skew shape, chains may switch from running southeast to northwest to running southwest to northeast (and vice versa). See Figure~\ref{fig:rtchainshifting}. We prove a special case of this phenomenon for the right trapezoid.
	
	% \begin{Def}
		% Let $r,s > 0$. The \emph{left border} of $RT_{r,s}$ is the set of element $\{(r+i,1+i) \in RT_{r,s}: i \in \mathbb{Z}_{\geq 0}\}$.
		% \end{Def}
	
	% \begin{Lemma}[Right trapezoid chain shifting]
		% \label{Lemma:rtchainshifting}
		% Let $(1,j),(i,s) \in RT_{r,s}$ for some $i,j > 1$, $J = [(1,j),(i,s)]$, and $J' = [(1,j-1),(i-1,s)]$. Let $L$ denote the left border of $RT_{r,s}$. Then
		% \[\sum\limits_{\substack{C \in MC(J) \\ C \cap L \neq \emptyset}} \prod\limits_{p \in C} x_p = \sum\limits_{\substack{C \in MC(J') \\ C \cap L \neq \emptyset}} \prod\limits_{p \in C} \phi \circ \rho^{-1} \circ \phi^{-1}(x)_p.\]
		% \end{Lemma}
	
	% \begin{proof}
		% For each $\ell \in [1,s]$, let 
		% \begin{align*}
			% &A_\ell = [(1,j),(r+\ell,1+\ell)] \setminus L & B_\ell = [(r+\ell,1+\ell),(i,s)] \\
			% &A'_\ell = \{(a,b) \in RT_{r,s}: (a,b+1) \in A_\ell \} & B'_\ell = \{(a,b) \in RT_{r,s}: (a+1,b) \in B_\ell\}
			% \end{align*}
\iffalse
		\begin{figure}
			\centering
			\begin{tikzpicture}[scale = 0.35]
				%FIGURE 1
				\begin{scope}[shift = {(0,0)}]
					\draw (-3,3)--(-6,6)--(-5,7)--(-6,8)--(-5,9)--(-6,10)--(-5,11)--(-6,12)--(-5,13)--(-6,14)--(-5,15)--(-6,16)--(-5,17)--(-6,18)--(-5,19)--(-6,20)--(-5,21)--(-6,22);
					\draw (-3,3)--(5,11)--(-6,22);
					\draw[ultra thick] (-6,14)--(-5,13);
					\node[label=left:$c$] at (-6,14) {};
					\node[label={[label distance=-8pt]above right:$c'$}] at (-5,13) {};
					\node[label={[label distance=-4pt]left:$c''$}] at (-6,12) {};
					
					\filldraw [ultra thick, red, fill=red!40] (-1,5)--(-5,9)--(-4,10)--(-5,11)--(-4,12)--(-5,13)--(1,7)--cycle;
					\node[label={[label distance=-6pt]below right:$m$}] at (-1,5) {};
					
					\filldraw [ultra thick, blue, fill=blue!40] (-6,14)--(-5,15)--(-6,16)--(-5,17)--(-6,18)--(-4,20)--(-2,18)--cycle;
					\node[label={[label distance=-6pt]above right:$M$}] at (-4,20) {};
				\end{scope}
				
				
				\begin{scope}[shift = {(17,0)}]
					\draw (-3,3)--(-6,6)--(-5,7)--(-6,8)--(-5,9)--(-6,10)--(-5,11)--(-6,12)--(-5,13)--(-6,14)--(-5,15)--(-6,16)--(-5,17)--(-6,18)--(-5,19)--(-6,20)--(-5,21)--(-6,22);
					\draw (-3,3)--(5,11)--(-6,22);
					\draw[ultra thick] (-6,12)--(-5,13);
					\node[label=left:$c$] at (-6,14) {};
					\node[label={[label distance=-4pt]left:$c''$}] at (-6,12) {};
					\node[label={[label distance=-10pt]below right:$c'$}] at (-5,13) {};
					
					\begin{scope}[shift={(-1,-1)}]
						\filldraw [ultra thick, red, fill=red!40] (-1,5)--(-5,9)--(-4,10)--(-5,11)--(-4,12)--(-5,13)--(1,7)--cycle;
						\node[label={[label distance=-10pt]below right:$m'$}] at (-1,5) {};
					\end{scope}
					
					\begin{scope}[shift={(1,-1)}]
						\filldraw [ultra thick, blue, fill=blue!40] (-6,14)--(-5,15)--(-6,16)--(-5,17)--(-6,18)--(-4,20)--(-2,18)--cycle;
						\node[label={[label distance=-6pt]above right:$M'$}] at (-4,20) {};
					\end{scope}
				\end{scope}
			\end{tikzpicture}
			\caption{Chain shifting in the right trapezoid. The total weight of all chains in the highlighted regions on the left (passing through $c'$ and $c$) with respect to $x$ equals the total weight of all chains in the highlighted regions on the right (passing through $c''$ and $c'$) with respect to $z = \tilde\rho^{-1}(x)$.}
			\label{fig:rtchainshifting}
		\end{figure}
\fi
	
	
	


\section{A map between the rectangle and trapezoid}
\label{section:equivariantMap}

In this section, we use the chain shifting lemmas (Lemmas~\ref{lemma:skewchainshifting} and \ref{Lemma:rtchainshifting}) to define a birational map $\themap$ between labelings of the rectangle $R_{r,s}$ and trapezoid $T_{r,s}$. In the tropical setting, this map will become a continuous, piecewise-linear, volume-preserving map between the chain polytopes of these two posets, which can be used to give a bijection between the plane partitions of $R_{r,s}$ and $T_{r,s}$ of height $\ell$. %We will express this map as a composition of rowmotion-like maps between labelings of certain intermediate posets.
%The statements and proofs in this section are given in the piecewise-linear realm. However, the map we define also detropicalizes to a birational map $\mathbb{R}_{>0}^T \to \mathbb{R}_{>0}^R$.
%
%\subsection{Intermediate posets and maps}
To construct $\themap$, we need to construct certain intermediate posets as induced subposets of the right trapezoid $RT_{r,s}$.


\begin{figure}
	\centering
	\begin{tikzpicture}[scale = 0.47]
		%FIRST INTERMEDIATE POSET
		%UP-DIAGONALS 
		\draw [dashed] (0,0)--(3,3);
		\draw [dashed] (-1,1)--(1,3);
		\draw (1,3)--(2,4);
		\draw [dashed] (-2,2)--(-1,3);
		\draw (-1,3)--(1,5);
		\draw (-3,3)--(0,6);
		\draw (-3,5)--(-1,7);
		\draw (-3,7)--(-2,8);
		
		%DOWN-DIAGONALS
		\draw [dashed] (0,0)--(-3,3);
		\draw [dashed] (1,1)--(-1,3);
		\draw (-1,3)--(-3,5);
		\draw [dashed] (2,2)--(1,3);
		\draw (1,3)--(-3,7);
		\draw (3,3)--(-3,9);
		
		%WHITE
		\node[wW] at (0,0) {};
		\node[wW] at (1,1) {};
		\node[wW] at (2,2) {};
		\node[wW] at (-1,1) {};
		\node[wW] at (0,2) {};
		\node[wW] at (-2,2) {};
		
		%BLACK
		\node[wB] at (3,3) {};
		\node[wB] at (1,3) {};
		\node[wB] at (2,4) {};
		\node[wB] at (-1,3) {};
		\node[wB] at (0,4) {};
		\node[wB] at (1,5) {};
		\node[wB] at (-3,3) {};
		\node[wB] at (-2,4) {};
		\node[wB] at (-1,5) {};
		\node[wB] at (0,6) {};
		\node[wB] at (-3,5) {};
		\node[wB] at (-2,6) {};
		\node[wB] at (-1,7) {};
		\node[wB] at (-3,7) {};
		\node[wB] at (-2,8) {};
		\node[wB] at (-3,9) {};
		
		%SECOND INTERMEDIATE POSET
		\begin{scope}[shift={(7.5,0)}]
			
			\tikzstyle{s}=[line width = .5cm, color=red!30]
			\filldraw[s](-1,1)--(-3,3)--(-3,7)--(1,3)--cycle;
			
			%UP-DIAGONALS 
			\draw [dashed] (0,0)--(3,3);
			\draw [dashed] (-1,1)--(0,2);
			\draw (0,2)--(1,3);
			\draw [dashed] (1,3)--(2,4);
			\draw (-2,2)--(1,5);
			\draw (-3,3)--(0,6);
			\draw (-3,5)--(-1,7);
			\draw (-3,7)--(-2,8);
			
			%DOWN-DIAGONALS
			\draw [dashed] (0,0)--(-2,2);
			\draw (-2,2)--(-3,3);
			\draw [dashed] (1,1)--(0,2);
			\draw (0,2)--(-3,5);
			\draw (2,2)--(-3,7);
			\draw [dashed] (3,3)--(1,5);
			\draw (1,5)--(-2,8);
			\draw [dashed] (-2,8)--(-3,9);
			
			%WHITE
			\node[wW] at (0,0) {};
			\node[wW] at (1,1) {};
			\node[wW] at (3,3) {};
			\node[wW] at (-1,1) {};
			\node[wW] at (2,4) {};
			\node[wW] at (-3,9) {};
			
			%BLACK
			\node[wB] at (2,2) {};
			\node[wB] at (0,2) {};
			\node[wB] at (1,3) {};
			\node[wB] at (-2,2) {};
			\node[wB] at (-1,3) {};
			\node[wB] at (0,4) {};
			\node[wB] at (1,5) {};
			\node[wB] at (-3,3) {};
			\node[wB] at (-2,4) {};
			\node[wB] at (-1,5) {};
			\node[wB] at (0,6) {};
			\node[wB] at (-3,5) {};
			\node[wB] at (-2,6) {};
			\node[wB] at (-1,7) {};
			\node[wB] at (-3,7) {};
			\node[wB] at (-2,8) {};
			
			
		\end{scope}
		
		%THIRD INTERMEDIATE POSET
		\begin{scope}[shift={(15,0)}]
			
			\tikzstyle{s}=[line width = .5cm, color=red!30]
			\filldraw[s](-1,1)--(-3,3)--(-3,7)--(1,3)--cycle;
			
			%UP-DIAGONALS 
			\draw [dashed] (0,0)--(3,3);
			\draw (-1,1)--(2,4);
			\draw (-2,2)--(1,5);
			\draw (-3,3)--(0,6);
			\draw (-3,5)--(-1,7);
			\draw [dashed] (-3,7)--(-2,8);
			
			%DOWN-DIAGONALS
			\draw [dashed] (0,0)--(-1,1);
			\draw (-1,1)--(-3,3);
			\draw (1,1)--(-3,5);
			\draw [dashed] (2,2)--(1,3);
			\draw (1,3)--(-2,6);
			\draw [dashed] (-2,6)--(-3,7);
			\draw [dashed] (3,3)--(2,4);
			\draw (2,4)--(-1,7);
			\draw [dashed] (-1,7)--(-3,9);
			
			%WHITE
			\node[wW] at (0,0) {};
			\node[wW] at (2,2) {};
			\node[wW] at (3,3) {};
			\node[wW] at (-3,7) {};
			\node[wW] at (-2,8) {};
			\node[wW] at (-3,9) {};
			
			%BLACK
			\node[wB] at (1,1) {};
			\node[wB] at (-1,1) {};
			\node[wB] at (0,2) {};
			\node[wB] at (1,3) {};
			\node[wB] at (2,4) {};
			\node[wB] at (-2,2) {};
			\node[wB] at (-1,3) {};
			\node[wB] at (0,4) {};
			\node[wB] at (1,5) {};
			\node[wB] at (-3,3) {};
			\node[wB] at (-2,4) {};
			\node[wB] at (-1,5) {};
			\node[wB] at (0,6) {};
			\node[wB] at (-3,5) {};
			\node[wB] at (-2,6) {};
			\node[wB] at (-1,7) {};
		\end{scope}
		
		%FOURTH INTERMEDIATE POSET
		\begin{scope}[shift={(22.5,0)}]
			%UP-DIAGONALS 
			\draw (0,0)--(3,3);
			\draw (-1,1)--(2,4);
			\draw (-2,2)--(1,5);
			\draw (-3,3)--(0,6);
			\draw [dashed] (-3,5)--(-1,7);
			\draw [dashed] (-3,7)--(-2,8);
			
			%DOWN-DIAGONALS
			\draw (0,0)--(-3,3);
			\draw (1,1)--(-2,4);
			\draw [dashed] (-2,4)--(-3,5);
			\draw (2,2)--(-1,5);
			\draw [dashed] (-1,5)--(-3,7);
			\draw (3,3)--(0,6);
			\draw [dashed] (0,6)--(-3,9);
			
			%WHITE
			\node[wW] at (-3,5) {};
			\node[wW] at (-2,6) {};
			\node[wW] at (-1,7) {};
			\node[wW] at (-3,7) {};
			\node[wW] at (-2,8) {};
			\node[wW] at (-3,9) {};
			
			%BLACK
			\node[wB] at (0,0) {};
			\node[wB] at (1,1) {};
			\node[wB] at (2,2) {};
			\node[wB] at (3,3) {};
			\node[wB] at (-1,1) {};
			\node[wB] at (0,2) {};
			\node[wB] at (1,3) {};
			\node[wB] at (2,4) {};
			\node[wB] at (-2,2) {};
			\node[wB] at (-1,3) {};
			\node[wB] at (0,4) {};
			\node[wB] at (1,5) {};
			\node[wB] at (-3,3) {};
			\node[wB] at (-2,4) {};
			\node[wB] at (-1,5) {};
			\node[wB] at (0,6) {};
		\end{scope}
	\end{tikzpicture}
	\caption{The four intermediate posets $I_4, \dots, I_1$ lying inside of $RT_{4,4}$ with $M_2$ highlighted in red. The map $\themap_2$ acts on labelings of $I_3$ by applying $\tilde\rho_2^{-1}$ inside $M_2$ and shifting the other entries parallel to the sides.}
	\label{fig:intermediatePosets}
\end{figure}


\begin{Def}
	\label{Def:intermediatePosets}
	Let $k \leq s \leq r$ be positive integers. The $k$th \emph{intermediate poset} $I_k = I_{r,s,k}$ is the induced subposet of $RT_{r,s}$ on $T_{r,k} \cup [(k,k+1), (r+k-1, s)]$.
	% elements $(i,j)$ satisfying
	% \begin{itemize}
		%     \item $i \leq r+s-k$,
		%     \item $i+j > s+1-k$,
		%     \item and either $i \geq s+1-k$ or $j \leq s+1-k$.
		% \end{itemize}
\end{Def}

See Figure~\ref{fig:intermediatePosets}. %Note that when $k=1$, we always have $i \leq r+s-1$ and $j \leq s$, so the only restriction is $i+j > s$, which implies that $I_1 = T_{r,s}$. Similarly, if $k=s$, then we always have $i+j > 1$ and $i \geq 1$, so the only restriction is $i \leq r$, which implies that $I_s = R_{r,s}$. It is straightforward to verify that $I_k$ always contains exactly $rs$ elements.
Note that the leftmost minimal element of $I_k$ is $(k,1)$. 
One can easily verify that $I_1 = R_{r,s}$, $I_s = T_{r,s}$, and $|I_k| = rs$ for all $k$.


%See Figure~\ref{fig:intermediatePosets} for examples of more intermediate posets. We note that in Figure~\ref{fig:intermediatePosets} all intermediate posets contain the same number of nodes and at least as many maximal chains as the trapezoid and rectangle (which we later show always holds). 


We now define maps $\themap_k \colon \RR_+^{I_{k+1}} \to \RR_+^{I_{k}}$ as follows. Consider the interval %\[M_k = \{(i,j) \in RT_{r,s} \mid  (i,j) \succeq (s-k,1) \text{ and } j \leq s+1-k\},\]
$M_k = [(k,1),(r+k,k+1)] \subset RT_{r,s}$.
%Note that $M_k \setminus \{(k,1)\} \subset I_{k+1}$, $M_k \setminus \{(r+k,k+1)\} \subset I_k$, and 
%which is isomorphic to the smaller right trapezoid $RT_{r-k+1,k+1}$.
For any $x \in \RR_+^{I_{k+1}}$, let $\bar x \in \RR_+^{M_k}$ be obtained by restricting $x$ to $M_k \setminus \{(k,1)\} \subset I_{k+1}$ and setting $\bar x_{k,1}$ to be an arbitrary $a \in \RR_+$ (say, $1$). Let $\tilde\rho_k \colon \RR_+^{M_k} \to \RR_+^{M_k}$ be the antichain rowmotion map on $M_k$. %(Recall that one can compute $\tilde\rho_k^{-1}$ with relative ease using Corollary~\ref{cor:edgeweight}.) 
Then we define $\themap_k(x) \in \RR_+^{I_{k}}$ by
\[\themap_k(x)_{ij} = \begin{cases} \tilde\rho_k^{-1}(\bar x)_{ij}&\text{if }(i,j) \in I_k \cap M_k = M_k \setminus \{(r+k,k+1)\}, \\ 
	x_{i+1,j}&\text{if }(i,j) \in I_{k} \setminus M_k\text{ and }j > k+1, \\
	x_{i,j+1}&\text{if }(i,j) \in I_{k} \setminus M_k\text{ and }i < k.\end{cases}\]
%We say that the \emph{labels below $M_k$} (with $i < k$) are shifted southwest and the \emph{labels above $M_k$} (with $j > k+1$) are shifted southeast. 
%See Figure~\ref{fig:psiMapExample} for an illustration. 
See Figure~\ref{fig:intermediatePosets}. One can show that $\themap_k$ does not depend on the choice of $a$. 

%By Proposition~\ref{prop:arbitrarya}, $\themap_k(x)$ does not depend on the choice of $a$. Similarly the inverse map $\themap_k^{-1}$ is also well-defined (apply $\tilde\rho_k$ on $M_k$ with an arbitrary label at $(r+k,k+1)$ and shift the labels outside $M_k$ appropriately). 

%\begin{Ex}
%	Let $r=s=4$ and $k=2$, and consider $\themap_2\colon \mathbb{R}^{I_3} \to \mathbb{R}^{I_2}$ as shown in Figure~\ref{fig:psiMapExample}. Here $M_2 = [(2,1),(6,3)]$ is shown in red. Outside of $M_2$, the coordinates of the labeling shift parallel to the sides. Inside of $M_2$ we apply antichain rowmotion $\tilde \rho_2^{-1}$ (ignoring the labels outside of $M_2$).
%	%
%	%See also Figure~\ref{fig:schematic} below for a schematic version of this diagram.
%\end{Ex}

\iffalse
\begin{figure}
	\centering
	\begin{tikzpicture}[scale = 0.52]
		
		%SECOND INTERMEDIATE POSET
		\begin{scope}[shift={(0,0)}]
			%UP-DIAGONALS 
			\draw [dashed] (0,0)--(3,3);
			\draw [dashed, red, very thick] (-1,1)--(0,2);
			\draw [red, very thick] (0,2)--(1,3);
			\draw [dashed] (1,3)--(2,4);
			\draw [red, very thick] (-2,2)--(0,4);
			\draw (0,4)--(1,5);
			\draw [red, very thick] (-3,3)--(-1,5);
			\draw (-1,5)--(0,6);
			\draw [red, very thick] (-3,5)--(-2,6);
			\draw (-2,6)--(-1,7);
			\draw (-3,7)--(-2,8);
			
			%DOWN-DIAGONALS
			\draw [dashed] (0,0)--(-1,1);
			\draw [dashed, red, very thick] (-1,1)--(-2,2);
			\draw [red, very thick] (-2,2)--(-3,3);
			\draw [dashed] (1,1)--(0,2);
			\draw [red, very thick] (0,2)--(-3,5);
			\draw (2,2)--(1,3);
			\draw [red, very thick] (1,3)--(-3,7);
			\draw [dashed] (3,3)--(1,5);
			\draw (1,5)--(-2,8);
			\draw [dashed] (-2,8)--(-3,9);
			
			%WHITE
			\node[wW] at (0,0) {};
			\node[wW] at (1,1) {};
			\node[wW] at (3,3) {};
			\node[wW] at (2,4) {};
			\node[wW] at (-3,9) {};
			
			%BLACK
			\node[wB] at (2,2) {};
			\node[wB] at (1,5) {};
			\node[wB] at (0,6) {};
			\node[wB] at (-1,7) {};
			\node[wB] at (-2,8) {};
			
			%RED
			\node[wR] at (-1,1) {};
			\node[wR] at (0,2) {};
			\node[wR] at (1,3) {};
			\node[wR] at (-2,2) {};
			\node[wR] at (-1,3) {};
			\node[wR] at (0,4) {};
			\node[wR] at (-3,3) {};
			\node[wR] at (-2,4) {};
			\node[wR] at (-1,5) {};
			\node[wR] at (-3,5) {};
			\node[wR] at (-2,6) {};
			\node[wR] at (-3,7) {};
			
			%LABELS
			\node at (2.8,2) {$x_{13}$};
			\node at (1.8,5) {$x_{34}$};
			\node at (0.8,6) {$x_{44}$};
			\node at (-0.2,7) {$x_{54}$};
			\node at (-1.2,8) {$x_{64}$};
		\end{scope}
		
		%ARROW
		\draw [-Stealth, ultra thick] (4.5,4.5)--(6.5,4.5);
		
		%THIRD INTERMEDIATE POSET
		\begin{scope}[shift={(11,0)}]
			%UP-DIAGONALS 
			\draw [dashed] (0,0)--(3,3);
			\draw [red, very thick] (-1,1)--(1,3);
			\draw (1,3)--(2,4);
			\draw [red, very thick] (-2,2)--(0,4);
			\draw (0,4)--(1,5);
			\draw [red, very thick] (-3,3)--(-1,5);
			\draw (-1,5)--(0,6);
			\draw [red, very thick] (-3,5)--(-2,6);
			\draw (-2,6)--(-1,7);
			\draw [dashed] (-3,7)--(-2,8);
			
			%DOWN-DIAGONALS
			\draw [dashed] (0,0)--(-1,1);
			\draw [red, very thick] (-1,1)--(-3,3);
			\draw (1,1)--(0,2);
			\draw [red, very thick] (0,2)--(-3,5);
			\draw [dashed] (2,2)--(1,3);
			\draw [red, very thick] (1,3)--(-2,6);
			\draw [dashed, red, very thick] (-2,6)--(-3,7);
			\draw [dashed] (3,3)--(2,4);
			\draw (2,4)--(-1,7);
			\draw [dashed] (-1,7)--(-3,9);
			
			%WHITE
			\node[wW] at (0,0) {};
			\node[wW] at (2,2) {};
			\node[wW] at (3,3) {};
			\node[wW] at (-3,7) {};
			\node[wW] at (-2,8) {};
			\node[wW] at (-3,9) {};
			
			%BLACK
			\node[wB] at (1,1) {};
			\node[wB] at (2,4) {};
			\node[wB] at (1,5) {};
			\node[wB] at (0,6) {};
			\node[wB] at (-1,7) {};
			
			%RED
			\node[wR] at (-1,1) {};
			\node[wR] at (0,2) {};
			\node[wR] at (1,3) {};
			\node[wR] at (-2,2) {};
			\node[wR] at (-1,3) {};
			\node[wR] at (0,4) {};
			\node[wR] at (-3,3) {};
			\node[wR] at (-2,4) {};
			\node[wR] at (-1,5) {};
			\node[wR] at (-3,5) {};
			\node[wR] at (-2,6) {};
			\node[wR] at (-3,7) {};
			
			%LABELS
			\node at (1.8,1) {$x_{13}$};
			\node at (2.8,4) {$x_{34}$};
			\node at (1.8,5) {$x_{44}$};
			\node at (0.8,6) {$x_{54}$};
			\node at (-0.2,7) {$x_{64}$};
		\end{scope}
	\end{tikzpicture}
	\caption{A right trapezoid poset $M_2$ (red) inside of $RT_{4,4}$. When applying $\themap_2$, $\tilde\rho^{-1}$ is applied inside $M_2$ (ignoring all other coordinates) and then coordinates outside of $M_2$ shift parallel to the sides of $M_2$.}
	\label{fig:psiMapExample}
\end{figure}
\fi

We then define the birational map $\themap = \themap_{1} \circ \themap_{2} \circ \dots \circ \themap_{s-1}$
from $\RR_+^{T_{r,s}}$ to $\RR_+^{R_{r,s}}$.

\begin{Ex}
	Figure~\ref{fig:zeta3} shows the result of applying $\themap = \themap_1 \circ \themap_2$ to a labeling $x \in \RR_+^T$ when $T = T_{3,3}$. Note that $x_{13} = c = \themap_2(x)_{12}$ as this label lies below $M_2$. Similarly $\themap_2(x)_{j+1,3} = \themap(x)_{j3}$ for $j=1,2,3$ as these labels lie above $M_1$.
\end{Ex}

\begin{figure}
	\centering
	\begin{tikzpicture}[scale=.75, transform shape]
		\begin{scope}
			\draw (2,0)--(3,1) (0,0)--(2,2) (0,2)--(1,3) (2,0)--(0,2) (4,0)--(0,4);
			\node[v,label=below:$a$] at (0,0){};
			\node[v,label=below:$b$] at (2,0){};
			\node[v,label=below:$c$] at (4,0){};
			\node[v,label=left:$d$] at (1,1){};
			\node[v,label=right:$e$] at (3,1){};
			\node[v,label=left:$f$] at (0,2){};
			\node[v,label=right:$g$] at (2,2){};
			\node[v,label=right:$h$] at (1,3){};
			\node[v,label=above:$i$] at (0,4){};
		\end{scope}
		\begin{scope}[shift={(7,1)}]
			\draw (1,-1)--(3,1) (0,0)--(2,2) (0,2)--(1,3) (1,-1)--(0,0) (3,-1)--(0,2) (3,1)--(1,3);
			\node[v,label=below:$\frac{ab}{a+b}$] at (1,-1){};
			\node[v,label=below:$c$] at (3,-1){};
			\node[v,label=left:$\frac{(a+b)d}{a}$] at (0,0){};
			\node[v,label=right:$\frac{(a+b)de}{ad+bd+be}$] at (2,0){};
			\node[v,label=left:$\frac{(ad+bd+be)fg}{adf+adg+bdf+bdg+beg}$] at (1,1){};
			\node[v,label=right:$\frac{(ad+bd+be)g}{be}$] at (3,1){};
			\node[v,label=left:$\frac{(adf+adg+bdf+bdg+beg)h}{(a+b)df}$] at (0,2){};
			\node[v,label=right:$\frac{(adf+adg+bdf+bdg+beg)h}{(ad+bd+be)g}$] at (2,2){};
			\node[v,label=above:$i$] at (1,3){};
		\end{scope}
		\begin{scope}[shift={(3.5,-3.2)}]
			\draw (2,-2)--(4,0) (1,-1)--(3,1) (0,0)--(2,2) (2,-2)--(0,0) (3,-1)--(1,1) (4,0)--(2,2);
			\node[v,label=below:$\frac{abc}{ab+ac+bc}$] at (2,-2){};
			\node[v,label=left:$\frac{(ab+ac+bc)de}{(bd+be+ce)a}$] at (1,-1){};
			\node[v,label=right:$\frac{(ab+ac+bc)de}{(ad+bd+be)c}$] at (3,-1){};
			\node[v,label=left:$\frac{(a+b)(bd+be+ce)fg}{(adf+adg+bdf+bdg+beg)b}$] at (0,0){};
			\node[v,label=left:$\frac{(adf+adg+bdf+bdg+beg)h}{(a+b)df}$] at (1,1){};
			\node[v,label=right:$\frac{(ad+bd+be)g}{be}$] at (4,0){};
			\node[v] (x) at (2,0){};
			\node[v,label=right:$\frac{(adf+adg+bdf+bdg+beg)h}{(ad+bd+be)g}$] at (3,1){};
			\node[v,label=above:$i$] at (2,2){};
			\node (l) at (6.5,-2.5){$\frac{(a+b)(ad+bd+be)(bd+be+ce)fg}{(ab+ac+bc)(adf+adg+bdf+bdg+beg)e}$};
			\draw (x) to [out = -60, in=180] (l);
		\end{scope}
	\end{tikzpicture}
	\caption{Applying $\themap_2$ and then $\themap_1$ to a labeling $x$ of $T_{3,3}$, resulting in $\themap(x)$.}
	\label{fig:zeta3}
\end{figure}

The key property of $\themap$ that we will need to prove is that $\themap$ preserves the total weight of all maximal chains. However, this does not hold for the maps $\themap_k$ unless we restrict to a certain special class of polygonal chains.

%\subsection{Polygonal chains}
%We define a special collection of maximal chains within each intermediate poset which are in bijection with chains in the trapezoid and rectangle.

\begin{Def}
	A maximal chain $C \subset I_k \subset RT_{r,s}$ is \emph{polygonal} if $C$ intersects $L$ (the left border of $RT_{r,s}$) or if $(k,1) \in C$.
\end{Def}

%See Figure~\ref{fig:polygonalChains} for the case $k=2$, $r=s=3$.
Note that all chains in the trapezoid and rectangle are polygonal (when $k=s$ or $1$). %Indeed, in the trapezoid $I_s$, all maximal chains contain the maximum element, which lies in $L$. Similarly, in the rectangle $I_1$, all maximal chains contain the minimum element $(1,1)$.

%\begin{Ex}
%	The polygonal chains in $I_2 \subseteq RT_{3,3}$ are shown in Figure~\ref{fig:polygonalChains}. %Note that $I_2$ has two other maximal chains that are not polygonal, as they start at $(1,2)$ and do not intersect $L$.
%\end{Ex}

\iffalse
\begin{figure}
	\centering
	\begin{tikzpicture}[scale = 0.6]
		
		\foreach \x in {0,4,8,12,16,20}{
			%VERTICAL
			\draw (-1+\x,1)--(-2+\x,2);
			\draw (1+\x,1)--(-2+\x,4);
			\draw (1+\x,3)--(-1+\x,5);
			
			%HORIZONTAL
			\draw (-1+\x,1)--(1+\x,3);
			\draw (-2+\x,2)--(0+\x,4);
			\draw (-2+\x,4)--(-1+\x,5);
			
			%CIRCLES
			\draw [fill=black] (\x+1,1) circle [radius=0.13cm];
			
			\draw [fill=black] (\x-1,1) circle [radius=0.13cm];
			\draw [fill=black] (\x,2) circle [radius=0.13cm];
			\draw [fill=black] (\x+1,3) circle [radius=0.13cm];
			
			\draw [fill=black] (\x-2,2) circle [radius=0.13cm];
			\draw [fill=black] (\x-1,3) circle [radius=0.13cm];
			\draw [fill=black] (\x,4) circle [radius=0.13cm];
			
			\draw [fill=black] (\x-2,4) circle [radius=0.13cm];
			\draw [fill=black] (\x-1,5) circle [radius=0.13cm];
		}
		
		%FIRST CHAIN
		\draw [red, very thick] (-1,1)--(-2,2)--(-1,3)--(-2,4)--(-1,5);
		
		\draw [fill=red] (-1,1) circle [radius=0.13cm];
		\draw [fill=red] (-2,2) circle [radius=0.13cm];
		\draw [fill=red] (-1,3) circle [radius=0.13cm];
		\draw [fill=red] (-2,4) circle [radius=0.13cm];
		\draw [fill=red] (-1,5) circle [radius=0.13cm];
		
		%SECOND CHAIN
		\begin{scope}[shift={(4,0)}]
			\draw [red, very thick] (-1,1)--(-2,2)--(-1,3)--(0,4)--(-1,5);
			
			\draw [fill=red] (-1,1) circle [radius=0.13cm];
			\draw [fill=red] (-2,2) circle [radius=0.13cm];
			\draw [fill=red] (-1,3) circle [radius=0.13cm];
			\draw [fill=red] (0,4) circle [radius=0.13cm];
			\draw [fill=red] (-1,5) circle [radius=0.13cm];
		\end{scope}
		
		%THIRD CHAIN
		\begin{scope}[shift={(8,0)}]
			\draw [red, very thick] (-1,1)--(0,2)--(-1,3)--(-2,4)--(-1,5);
			
			\draw [fill=red] (-1,1) circle [radius=0.13cm];
			\draw [fill=red] (0,2) circle [radius=0.13cm];
			\draw [fill=red] (-1,3) circle [radius=0.13cm];
			\draw [fill=red] (-2,4) circle [radius=0.13cm];
			\draw [fill=red] (-1,5) circle [radius=0.13cm];
		\end{scope}
		
		%FOURTH CHAIN
		\begin{scope}[shift={(12,0)}]
			\draw [red, very thick] (-1,1)--(0,2)--(-1,3)--(-0,4)--(-1,5);
			
			\draw [fill=red] (-1,1) circle [radius=0.13cm];
			\draw [fill=red] (0,2) circle [radius=0.13cm];
			\draw [fill=red] (-1,3) circle [radius=0.13cm];
			\draw [fill=red] (0,4) circle [radius=0.13cm];
			\draw [fill=red] (-1,5) circle [radius=0.13cm];
		\end{scope}
		
		%FIFTH CHAIN
		\begin{scope}[shift={(16,0)}]
			\draw [red, very thick] (-1,1)--(0,2)--(1,3)--(-0,4)--(-1,5);
			
			\draw [fill=red] (-1,1) circle [radius=0.13cm];
			\draw [fill=red] (0,2) circle [radius=0.13cm];
			\draw [fill=red] (1,3) circle [radius=0.13cm];
			\draw [fill=red] (0,4) circle [radius=0.13cm];
			\draw [fill=red] (-1,5) circle [radius=0.13cm];
		\end{scope}
		
		%SIXTH CHAIN
		\begin{scope}[shift={(20,0)}]
			\draw [red, very thick] (1,1)--(0,2)--(-1,3)--(-2,4)--(-1,5);
			
			\draw [fill=red] (1,1) circle [radius=0.13cm];
			\draw [fill=red] (0,2) circle [radius=0.13cm];
			\draw [fill=red] (-1,3) circle [radius=0.13cm];
			\draw [fill=red] (-2,4) circle [radius=0.13cm];
			\draw [fill=red] (-1,5) circle [radius=0.13cm];
		\end{scope}
		
	\end{tikzpicture}
	\caption{The six polygonal chains in $I_2 \subseteq RT_{3,3}$.}
	\label{fig:polygonalChains}
\end{figure}
\fi

The following result relates the weights of polygonal chains under $\themap_k$. Since the only complicated part of $\themap_k$ occurs inside $M_k$, it follows readily from Lemmas~\ref{lemma:skewchainshifting} and \ref{Lemma:rtchainshifting}.

%\begin{figure}
%	\centering
%	\begin{tikzpicture}[scale=.3]
%		\begin{scope}
%			\draw[red] (-1,1)--(-3,3)--(-3,15)--(6,6)--(1,1)--(-1,1);
%			\draw[red,dashed] (-1,1)--(0,0)--(1,1);
%			\node[red] at (0,7) {$M_k$};
%			\draw (3,1)--(7,5)--(11,1)--(3,1); %lower part
%			\draw (6,8)--(-2,16)--(0,18)--(8,10)--(6,8); %upper part
%			\draw (3,1)--(2,2) (4,2)--(3,3) (5,3)--(4,4) (6,4)--(5,5) (7,5)--(6,6);
%			\foreach \x in {1, ..., 9}
%			\draw (6-\x,6+\x)--(7-\x,7+\x);
%			\node[label={[label distance=-10pt]above left:$\scriptstyle m$}] at (3,3){};
%			\node[label={[label distance=-10pt]below right:$\scriptstyle{\widetilde m}$}] at (4,2){};
%			\node[label={[label distance=-10pt]below left:$\scriptstyle M$}] at (-1,13){};
%			\node[label={[label distance=-10pt]above right:$\scriptstyle {\widetilde M}$}] at (0,14){};
%			\node at (4,-2){$I_{k+1}$};
%		\end{scope}
%		\draw[->] (11.5,7)--(14.5,7);
%		\node at (13,8){$\zeta_k$};
%		\begin{scope}[shift={(19,0)}]
%			\draw[red] (0,0)--(-3,3)--(-3,13)--(-2,14)--(6,6)--(0,0);
%			\draw[red,dashed] (-3,13)--(-3,15)--(-2,14);
%			\node[red] at (0,7) {$M_k$};
%			\draw (2,0)--(6,4)--(10,0)--(2,0); %lower part
%			\draw (7,7)--(-1,15)--(1,17)--(9,9)--(7,7); %upper part
%			\draw (2,0)--(1,1) (3,1)--(2,2) (4,2)--(3,3) (5,3)--(4,4) (6,4)--(5,5);
%			\foreach \x in {0, ..., 8}
%			\draw (6-\x,6+\x)--(7-\x,7+\x);
%			\node[label={[label distance=-10pt]above left:$\scriptstyle {m'}$}] at (2,2){};
%			\node[label={[label distance=-10pt]below right:$\scriptstyle{\widetilde m'}$}] at (3,1){};
%			\node[label={[label distance=-10pt]below left:$\scriptstyle {M'}$}] at (0,12){};
%			\node[label={[label distance=-10pt]above right:$\scriptstyle {\widetilde M'}$}] at (1,13){};
%			\node at (4,-2){$I_{k}$};
%		\end{scope}
%	\end{tikzpicture}
%	\caption{Schematic of the map $\themap_k$. Polygonal chains in $I_{k+1}$ that intersect $M_k$ in a chain from $m$ to $M$ get shifted to polygonal chains in $I_k$ that intersect $M_k$ in a chain from $m'$ to $M'$.}
%	\label{fig:schematic}
%\end{figure}

\begin{Prop} \label{prop:intermediate}
	Let $\mathscr P_{k}$ be the collection of polygonal chains in $I_k$. For $x \in \RR_+^{I_{k+1}}$, let $z = \themap_k(x)$. Then $w_{\mathscr P_{k+1}}(x) = w_{\mathscr P_{k}}(z)$.
\end{Prop}
%\begin{proof}
%	Let $\mathscr P_{k+1} = \mathscr P_{k+1}^{(1)} \sqcup \mathscr P_{k+1}^{(2)}$, where $\mathscr P_{k+1}^{(1)}$ consists of the chains that contain $(k+1,1)$ and $\mathscr P_{k+1}^{(2)}$ contains the ones that do not. Similarly, write $\mathscr P_{k} = \mathscr P_{k}^{(1)} \sqcup \mathscr P_{k}^{(2)}$, where $\mathscr P_{k}^{(1)}$ consists of the chains that do not intersect $L$ and $\mathscr P_{k}^{(2)}$ contains the ones that do. (Note that all chains in $\mathscr P_{k+1}^{(2)}$ intersect $L$, while all chains in $\mathscr P_{k}^{(1)}$ contain $(k,1)$.)
%	We will show that $w_{\mathscr P_{k+1}^{(i)}}(x) = w_{\mathscr P_{k}^{(i)}}(z)$ for $i=1,2$.
%	
%	For any chain $C \in \mathscr P_{k+1}$, $C \cap M_k$ is a chain with minimum $m$ and maximum $M$ such that either $m=(k+1,1)$ or $m = (k,j+1)$ for some $1\leq j \leq k$, and $M = (i+1,k+1)$ for some $k\leq i< r+k$. 
%	%(See Figure~\ref{fig:schematic} for the case $m = (k,j+1)$.) Let $m'$ be the unique element of $M_k$ covered by $m$, and let $M'$ be the southeast neighbor of $M$. Let $\widetilde M$ and $\widetilde M'$ be the northeast neighbors of $M$ and $M'$ (if they exist), and let $m$ and $m'$ have southeast neighbors $\widetilde m$ and $\widetilde m'$ (if they exist). 
%	
%	Suppose first that $m = (k,j+1)$ and $M=(i+1,k+1)$. Let $m' = (k,j)$ and $M' = (i,k+1)$. Also let $\widetilde m$ and $\widetilde m'$ be the southeast neighbors of $m$ and $m'$, and let $\widetilde M$ and $\widetilde M'$ be the northeast neighbors of $M$ and $M'$ (if they exist). See Figure~\ref{fig:schematic}.
%	
%	Every chain in $\mathscr P_{k+1}^{(2)}$ is the union of a chain in some $\mathscr C^L_{m,M}$, a chain down from $\widetilde m$, and a chain up from $\widetilde M$, while every chain in $\mathscr P_{k}^{(2)}$ is the union of a chain in some $\mathscr C^L_{m',M'}$, a chain down from $\widetilde m'$, and a chain up from $\widetilde M'$. By Lemma~\ref{Lemma:rtchainshifting} and the fact that $\themap_k$ shifts labels outside $M_k$,
%	\begin{align*}
%		w_{\mathscr P_{k+1}^{(2)}}(x) &= \sum_{m,M} w_{\mathscr C^L_{m,M}}(x) \cdot \psi_{I_{k+1}}(x)_{\widetilde m}\psi_{I_{k+1}}^*(x)_{\widetilde M}\\
%		&= \sum_{m,M} w_{\mathscr C^{L}_{m',M'}}(z)  \cdot \psi_{I_k}(z)_{\widetilde m'}\psi_{I_k}^*(z)_{\widetilde M'}= w_{\mathscr P_k^{(2)}}(z).
%	\end{align*}    
%	(By convention we set the value of $\psi$ or $\psi^*$ to be $1$ if the label does not exist.)
%	
%	A similar argument using Lemma~\ref{lemma:skewchainshifting} shows that $w_{\mathscr P_{k+1}^{(1)}}(x) = w_{\mathscr P_k^{(1)}}(z)$.
%	% Suppose first that $m = (k+1,1)$ (so that $m'=(k,1)$). Every chain in $\mathscr P_{k+1}^{(1)}$ is the union of a chain in some $\mathscr C_{m,M}$ and a chain up from $\widetilde M$ (empty if $k=s-1$). Similarly, every chain in $\mathscr P_k^{(1)}$ is the union of a chain in some $\mathscr C^{\overline L}_{m',M'}$ and a chain up from $\widetilde M'$. Hence by Lemma~\ref{lemma:skewchainshifting} and the fact that $\themap_k$ shifts labels above $M_k$,
%	% \begin{align*}
%		%     w_{\mathscr P_{k+1}^{(1)}}(x) &= \sum_M w_{\mathscr C_{m,M}}(x) \cdot\psi_{I_{k+1}}^*(x)_{\widetilde M}\\
%		%     &= \sum_M w_{\mathscr C^{\overline L}_{m',M'}}(z) \cdot\psi_{I_k}^*(z)_{\widetilde M'}\\
%		%     &= w_{\mathscr P_k^{(1)}}(z).
%		% \end{align*}
%	% By Lemma~\ref{Lemma:rtchainshifting}, $w_{\mathscr C^L_{m,M}}(x) = w_{\mathscr C^L_{m',M'}}(\themap_k(x))$. Since $\themap_k$ again shifts the coordinates of $x$ above $M_k$ to the southeast as well as the coordinates below $M_k$ to the southwest (by the same shift that takes $m$ to $m'$), summing over all possible $C \setminus M_k$ and all possible $m$ and $M$ gives $w_{\mathscr P_{k+1}^{(2)}}(x) = w_{\mathscr P_{k}^{(2)}}(\themap_k(x))$.
%\end{proof}



%As a remark, note that since in Lemmas~\ref{lemma:skewchainshifting} and \ref{Lemma:rtchainshifting}, the two collections of chains always have the same size, the technique in the above proof also shows that the intermediate posets $I_k$ all contain the same number of polygonal chains.

It is now simple to deduce the following theorem.

\begin{Th}\label{thm:main}
	Let $\mathscr C$ and $\mathscr C'$ be the sets of all maximal chains in $T_{r,s}$ and $R_{r,s}$, respectively. Then for all $x \in \RR_+^{T_{r,s}}$, $w_{\mathscr C}(x) = w_{\mathscr C'}(\themap(x))$.
\end{Th}
%\begin{proof}
%	Apply Proposition~\ref{prop:intermediate} to $\themap = \themap_{1} \circ \themap_{2} \circ \dots \circ \themap_{s-1}$.
%\end{proof}

\subsection{Polytopes and plane partitions} \label{sec:polytopes}
We now examine the consequences of Proposition~\ref{prop:intermediate} and Theorem~\ref{thm:main} in the piecewise-linear case. In this section, we will take all maps to be their piecewise-linear counterparts. In particular, the definition of $\themap_k$ inside $M_k$ utilizes the map $\tilde \rho^{-1}$ on $M_k$. By tropicalizing Proposition~\ref{cor:edgeweight}, we can compute $z = \tilde\rho^{-1}(x)$ via
\begin{align}
z_p &= - \max_{q \gtrdot p} \{y_p-y_q\} = \min_{q \gtrdot p} \{y_q\}-y_p, \quad \text{where}\label{eq:yz} \tag{$\dagger$}\\
y_p &= \phibar(x)_p = \max_{\hat 0 \lessdot q_1 \lessdot \dots \lessdot q_n = p} \sum_i x_{q_i}. \label{eq:xy} \tag{$\ddagger$}
\end{align}

%Note that if $x$ has nonnegative coordinates, then so does $z$ (except at a maximal element), from which it follows that $\zeta_k$ also preserves nonnegativity. 
%(This nonnegativity property also follows from the fact that piecewise-linear toggles map the order polytope to itself.) 
%It is also clear that $\themap_k$ is \emph{lattice-preserving} in that it sends lattice points to lattice points.

\begin{Ex}
	An example calculation of $\themap$ when $r=5$ and $s=4$ is given in Figure~\ref{fig:plexample}. Each $\zeta_k$ can be computed by applying \eqref{eq:xy} and then \eqref{eq:yz} on $M_k$. (The minimum element of $M_k$ is arbitrarily given the label $0$, and the label of the maximum element of $M_k$ is discarded after $\themap_k$ is applied.) The entries outside of $M_k$ are shifted downward appropriately.
\end{Ex}

\begin{figure}
	\centering
	\begin{tikzpicture}[scale=.3, every node/.style={scale=2.5, transform shape}]
		\tikzstyle{s}=[line width = .4cm, color=red!30]
		\begin{scope}
			\filldraw[s](-2,4)--(-4,6)--(-4,12)--(1,7)--cycle;
			\foreach \i/\j/\l in {4/1/0,5/1/1,3/2/1,4/2/1,5/2/0,6/2/1,2/3/2,3/3/0,4/3/0,5/3/1,6/3/0,7/3/1,1/4/0,2/4/1,3/4/0,4/4/0,5/4/1,6/4/0,7/4/0,8/4/0}
			\node at (\j-\i,\i+\j){$\l$};
		\end{scope}
		\draw[->] (-2,2.5)--(-1,1);
		%\node at (0,2){$\psi_{M_3}$};
		\draw[->] (3.5,1)--(4.5,2.5);
		\begin{scope}[shift={(4,-12)}]
			\foreach \i/\j/\l in {3/1/0,4/1/0,5/1/1,3/2/1,4/2/2,5/2/2,6/2/3,3/3/1,4/3/2,5/3/3,6/3/3,7/3/4,3/4/1,4/4/2,5/4/4,6/4/4,7/4/4,8/4/4}
			\node at (\j-\i,\i+\j){$\l$};
		\end{scope}
		\draw[->](3,8)--(4.5,8);
		\node at (3.75,9){$\themap_3$};
		\begin{scope}[shift={(10,1)}]
			\filldraw[s](-1,3)--(-4,6)--(-4,10)--(1,5)--cycle;
			\foreach \i/\j/\l in {3/1/0,4/1/1,5/1/1,2/2/2,3/2/0,4/2/0,5/2/1,6/2/0,1/3/0,2/3/1,3/3/0,4/3/0,5/3/0,6/3/1,7/3/0,3/4/1,4/4/2,5/4/0,6/4/0,7/4/0}
			\node at (\j-\i,\i+\j){$\l$};
		\end{scope}
		\draw[->] (9,2.5)--(10,1);
		%\node at (11,2){$\psi_{M_2}$};
		\draw[->] (14.5,1)--(15.5,2.5);
		\begin{scope}[shift={(15,-10)}]
			\foreach \i/\j/\l in {2/1/0,3/1/0,4/1/1,5/1/2,2/2/2,3/2/2,4/2/2,5/2/3,6/2/3,2/3/3,3/3/3,4/3/3,5/3/3,6/3/4,7/3/4}
			\node at (\j-\i,\i+\j){$\l$};
		\end{scope}
		\draw[->](12.5,8)--(14,8);
		\node at (13.25,9){$\themap_2$};
		\begin{scope}[shift={(19.5,2)}]
			\filldraw[s](0,2)--(-4,6)--(-4,8)--(1,3)--cycle;
			\foreach \i/\j/\l in {2/1/0,3/1/1,4/1/1,5/1/1,1/2/0,2/2/0,3/2/0,4/2/1,5/2/0,6/2/1,2/3/0,3/3/0,4/3/0,5/3/1,6/3/0,2/4/1,3/4/2,4/4/0,5/4/0,6/4/0}
			\node at (\j-\i,\i+\j){$\l$};
		\end{scope}
		\draw[->] (19.5,2.5)--(20.5,1);
		%\node at (21.5,2){$\psi_{M_1}$};
		\draw[->] (25,1)--(26,2.5);
		\begin{scope}[shift={(25.5,-8)}]
			\foreach \i/\j/\l in {1/1/0,2/1/0,3/1/1,4/1/2,5/1/3,1/2/0,2/2/0,3/2/1,4/2/3,5/2/3,6/2/4}
			\node at (\j-\i,\i+\j){$\l$};
		\end{scope}
		\draw[->](22.5,8)--(24,8);
		\node at (23.25,9){$\themap_1$};
		\begin{scope}[shift={(29,3)}]
			\foreach \i/\j/\l in {1/1/0,2/1/0,3/1/0,4/1/1,5/1/0,1/2/0,2/2/1,3/2/2,4/2/0,5/2/1,1/3/0,2/3/0,3/3/0,4/3/1,5/3/0,1/4/1,2/4/2,3/4/0,4/4/0,5/4/0}
			\node at (\j-\i,\i+\j){$\l$};
		\end{scope}
	\end{tikzpicture}
	\caption{Example calculation of $\themap$ for $r=5$ and $s=4$. The top row shows the labelings of the intermediate posets obtained from $x$ when applying $\themap_3$, $\themap_2$, and $\themap_1$ with entries in $M_k$ highlighted. The vertical maps show applications of \eqref{eq:xy} and \eqref{eq:yz} on $M_k$.}
	\label{fig:plexample}
\end{figure}

%Just as we defined the chain polytopes of the rectangle and trapezoid, 
To each intermediate poset $I_k \subseteq RT_{r,s}$, we can associate a \emph{polygonal chain polytope}. This coincides with the chain polytope (as defined by Stanley \cite{stanley2}) when $k=1$ or $k=s$.

\begin{Def}
	The \emph{polygonal chain polytope} $\widetilde{\mathcal C}(I_k) \subseteq \mathbb{R}^{I_k}$ is the set of all $\RR$-labelings $x = (x_p)_{p \in I_k}$ such that $x_p \geq 0$ for all $p \in I_k$, and $\sum_{p \in C} x_p \leq 1$ for all polygonal chains $C \subset I_k$.
\end{Def}

%When $I_k$ is either $R_{r,s}$ or $T_{r,s}$, the polygonal chain polytope coincides with the chain polytope, but in general $\widetilde{\mathcal C}(I_k)$ will be larger than the chain polytope of $I_k$.

%\begin{Ex}
%	Consider the poset $I_2 \subseteq RT_{3,3}$ pictured in Figure~\ref{fig:polygonalChains}. The polygonal chain polytope $\widetilde {\mathcal C}(I_2)$ is defined by the inequalities $x_p \geq 0$ for all $p$, and
%	\begin{align*}
%		&x_{21}+x_{31}+x_{32}+x_{42}+x_{43} \leq 1, &x_{21}+x_{31}+x_{32}+x_{33}+x_{43} \leq 1, \\
%		&x_{21}+x_{22}+x_{32}+x_{42}+x_{43} \leq 1, &x_{21}+x_{22}+x_{32}+x_{33}+x_{43} \leq 1, \\
%		&x_{21}+x_{22}+x_{23}+x_{33}+x_{43} \leq 1, &x_{12}+x_{22}+x_{32}+x_{42}+x_{43} \leq 1.
%	\end{align*}
%\end{Ex}

Although $\widetilde{\mathcal C}(I_k)$ is a lattice polytope when $k = 1$ or $k = s$ (when it is an ordinary chain polytope), this is not true in general. %(for instance, when $r=s=4$ and $k=3$). 
Nevertheless, for fixed $r$ and $s$, these polytopes all have the same volume and Ehrhart polynomial. % by Theorem~\ref{thm:polytope} below. 
In particular, if $\widetilde{\mathcal C}(I_k)$ is not a lattice polytope, then it exhibits \emph{period collapse} of its Ehrhart quasi-polynomial.
%---see, for instance, \cite{haasemcallister} for some discussion of this phenomenon.

\begin{Th} \label{thm:polytope}
	The map $\themap_k \colon \RR^{I_{k+1}} \to \RR^{I_{k}}$ defines a continuous, piecewise-linear, and lattice-preserving bijection from $\ell \cdot \widetilde {\mathcal C}(I_{k+1})$ to $\ell \cdot \widetilde {\mathcal C}(I_{k})$ for all $\ell \in \ZZ_{\geq 0}$.
	Hence, for fixed $r$ and $s$, the rational polytopes $\widetilde{\mathcal C}(I_k)$ 
	%exhibit Ehrhart quasi-polynomial period collapse and 
	share the same Ehrhart polynomial for all $k$.
\end{Th}
%\begin{proof}
%	As seen above, $\themap_k$ is a continuous, piecewise-linear, lattice-preserving map that preserves nonnegativity. The tropicalized version of Proposition~\ref{prop:intermediate} states that for any $x \in \RR^{I_{k+1}}$ and $z = \themap_k(x)$,
%	\[\max_{C \in \mathscr P_{k+1}} \sum_{p \in C} x_p = \max_{C \in \mathscr P_{k}} \sum_{p \in C} z_p.\]
%	In particular, this quantity is at most $\ell$ if and only if $x \in \ell \cdot \widetilde {\mathcal C}(I_{k+1})$ by the left hand side, but also if and only if $z \in \ell \cdot \widetilde {\mathcal C}(I_{k})$ by the right hand side. The result follows.
%\end{proof}

The following corollary then gives a bijective proof of the result of Proctor~\cite{proctor}.

\begin{Cor}
	The continuous, piecewise-linear map $\psi \circ \themap \circ \psi^{-1} \colon \RR^{T_{r,s}} \to \RR^{R_{r,s}}$ defines a bijection between plane partitions of $T_{r,s}$ and $R_{r,s}$ of height $\ell$ for all $\ell \in \ZZ_{\geq 0}$.
\end{Cor}
%\begin{proof}
%	By Theorem~\ref{thm:polytope}, the composition $\themap = \themap_1 \circ \cdots \circ \themap_k$ defines a continuous, piecewise-linear bijection from the lattice points in $\ell \cdot \mathcal C(T_{r,s})$ to the lattice points in $\ell \cdot \mathcal C(R_{r,s})$. But the inverse transfer map $\psi$ is a continuous, piecewise-linear bijection from the lattice points in $\ell \cdot \mathcal C(P)$ to the lattice points in $\ell \cdot \mathcal O(P)$ (which are plane partitions of $P$ of height at most $\ell$) for any poset $P$, as shown by Stanley \cite{stanley2}.
%\end{proof}

\begin{Ex}
	An example application of $\psi \circ \themap \circ \psi^{-1}$ is given in Figure~\ref{fig:planepartition} (obtained by applying $\psi$ to Figure~\ref{fig:plexample}). As required, both plane partitions have the same height.
\end{Ex}

\begin{figure}
	\centering
	\begin{tikzpicture}[scale=.35, every node/.style={scale=2.5, transform shape}]
		\begin{scope}
			\foreach \i/\j/\l in {4/1/0,5/1/1,3/2/1,4/2/2,5/2/2,6/2/3,2/3/2,3/3/2,4/3/2,5/3/3,6/3/3,7/3/4,1/4/0,2/4/3,3/4/3,4/4/3,5/4/4,6/4/4,7/4/4,8/4/4}
			\node at (\j-\i,\i+\j){$\l$};
		\end{scope}
		\draw[->] (3.5,8)--(8.5,8);
		\node at (6,8.8){$\psi \circ \themap \circ \psi^{-1}$};
		\begin{scope}[shift={(14,3)}]
			\foreach \i/\j/\l in {1/1/0,2/1/0,3/1/0,4/1/1,5/1/1,1/2/0,2/2/1,3/2/3,4/2/3,5/2/4,1/3/0,2/3/1,3/3/3,4/3/4,5/3/4,1/4/1,2/4/3,3/4/3,4/4/4,5/4/4}
			\node at (\j-\i,\i+\j){$\l$};
		\end{scope}
	\end{tikzpicture}
	\caption{Example of the bijection $\psi \circ \themap \circ \psi^{-1}$ obtained by applying $\psi$ to the labelings in Figure~\ref{fig:plexample}. Note that both plane partitions have the same height.}
	\label{fig:planepartition}
\end{figure}





\section{Rowmotion equivariance} \label{sec:rowmotionequivariance}

To prove that the map $\themap$ defined in the previous section is equivariant with respect to the action of rowmotion $\tilde \rho$ (or, equivalently, that $\psi \circ \themap \circ \psi^{-1}$ is equivariant with respect to $\rho$), we define a modified version of rowmotion on the intermediate posets $I_k$ that is respected by the maps $\themap_k$.

%\subsection{Polygonal chain rowmotion}

As above, let $k \leq s \leq r$ and consider $I_k \subset RT_{r,s}$. Let $\mathscr P_k$ denote the set of polygonal chains in $I_k$, and let $\mathscr P_k(p)$ denote the subset of those chains that contain $p$.

\begin{Def}
	\label{Def:polygonalChainToggle}
	For any $p \in I_k$, the \emph{(birational) polygonal toggle} $\tau'_p\colon \mathbb{R}_+^{I_k} \to \mathbb{R}_+^{I_k}$ is the map that changes the $p$-coordinate of $x \in \mathbb{R}_+^{I_k}$ by
	$x_p \mapsto %\Bigl(\sum_{C \in \mathscr P_k(p)} \prod_{q \in C} x_q\Bigr)^{-1} 
	w_{\mathscr P_k(p)}(x)^{-1}$ while keeping all other coordinates fixed.	
	The \emph{(birational) polygonal rowmotion} map $\spicyrho_k:\mathbb{R}_+^{I_k} \to \mathbb{R}_+^{I_k}$ is the composition
	$\spicyrho_k = \tau'_{L^{-1}(|I_k|)} \circ \dots \circ \tau'_{L^{-1}(1)}$
	for any linear extension $L$ of $I_k$.
\end{Def}

%As with ordinary toggles, it is easy to verify that the polygonal toggles $\tau'_p$ and $\tau'_q$ commute if $p$ and $q$ are incomparable (or more generally, if no polygonal chain contains both $p$ and $q$). It follows that $\spicyrho_k$ is well-defined.

%Note that Definition~\ref{Def:polygonalChainToggle} differs from the definition of the usual antichain toggle only in that the sum is taken only over polygonal chains rather than all maximal chains.
When $k=1$ or $s$ (that is, on the rectangle or trapezoid), the set of polygonal chains is just the set of all maximal chains, and so $\spicyrho_k=\tilde\rho_k$ as shown by Joseph and Roby~\cite{josephroby2}.
%Therefore, in the case when $k=1$ (when $I_k = R_{r,s}$) or $k=s$ (when $I_k = T_{r,s}$), we have $\spicyrho_k = \tilde \rho_k$ since $\mathscr P_k$ is just the set of all maximal chains.

\iffalse
\begin{Ex}
	Let $r = s = 3$ and let $x \in \mathbb{R}^{I_2}$. There are $6$ polygonal chains in $I_2$ (see Figure~\ref{fig:polygonalChains}). Applying the polygonal toggle at the maximum element $(4,3)$ gives
	\[
	t_{43}(x)_{43} = \frac{1}{ \splitfrac{x_{21}x_{31}x_{32}x_{42}x_{43} + x_{21}x_{31}x_{32}x_{33}x_{43} + x_{21}x_{22}x_{32}x_{42}x_{43}}{ +x_{21}x_{22}x_{32}x_{33}x_{43} + x_{21}x_{22}x_{23}x_{33}x_{43} + x_{12}x_{22}x_{32}x_{42}x_{43} } } 
	\]
	Note that the terms $x_{12}x_{22}x_{32}x_{33}x_{43}$ and $x_{12}x_{22}x_{23}x_{33}x_{43}$ that would appear in the usual notion of antichain toggles are absent here. Applying polygonal chains in the order of a linear extension from bottom to top gives piecewise-linear polygonal antichain rowmotion.
\end{Ex}
\fi

%It is important to note that unlike ordinary rowmotion, there does not appear to be a nice ``order'' version of $\spicyrho$ (as $\rho$ is to $\tilde\rho$). The reason for this is that the inverse transfer and dual transfer maps $\phibar$ and $\phibar^*$ are particularly well suited for working with the set of all maximal chains. Indeed, these two maps enumerate the ``bottom'' and ``top'' parts of maximal chains through $p$, and any bottom part can be combined with any top part to yield a maximal chain. However, the same is not true for polygonal chains: the bottom part of one polygonal chain may not be compatible with the top part of another. Nevertheless, we will see that not all hope is lost because interchangeable parts of chains will satisfy an appropriate chain shifting lemma when considered as a group.

%Our main result for this section will be to prove the following theorem. % hopefully!
We then prove the following theorem.

\begin{Th} \label{thm:equivariance}
	The maps $\themap_k \colon \RR_+^{I_{k+1}} \to \RR_+^{I_{k}}$ are equivariant with respect to polygonal rowmotion:
	\[\themap_k \circ \spicyrho_{k+1} = \spicyrho_{k} \circ \themap_k.\]
\end{Th}

The proof involves categorizing the possible bottom and top parts of polygonal chains and expressing their weights using ``partial transfer maps''. We then use the bijection $\aleph$ to formulate and apply chain shifting results for these maps. (This proof requires the use of subtraction.) The following corollary follows immediately.
\begin{Cor}
	Let $T = T_{r,s}$ and $R=R_{r,s}$ be the rectangle and trapezoid poset. Then the map $\themap \colon \RR_+^{T} \to \RR_+^{R}$ is equivariant with respect to birational (antichain) rowmotion:
	\[\themap \circ \tilde\rho_T = \tilde\rho_R \circ \themap.\]
	In particular, birational rowmotion on the trapezoid ($\tilde \rho_T$ or $\rho_T$) has order $r+s$.
\end{Cor}
%\begin{proof}
%	Since the maps $\themap_k$ are equivariant with respect to polygonal rowmotion by Theorem~\ref{thm:equivariance}, so is their composition $\themap$. The first part follows since $\spicyrho = \tilde\rho$ on both $T$ and $R$. The second claim follows since birational rowmotion on the rectangle has order $r+s$ as shown by Grinberg and Roby \cite{grinbergroby2}.
%\end{proof}

%\begin{Cor}
%	Polygonal rowmotion $\spicyrho_k$ has order $r+s$ on $I_k = I_{r,s,k}$.
%\end{Cor}
%\begin{proof}
%	Similarly follows from Theorem~\ref{thm:equivariance} since birational rowmotion on the rectangle $R_{r,s}$ has order $r+s$.
%\end{proof}




\acknowledgements{The authors would like to thank Darij Grinberg, Sam Hopkins, Tom Roby, and Sylvester Zhang for interesting conversations. }

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