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\title{Delta operators at $q=1$ and polyominoes}
\author{Angela
Hicks\thanks{\href{mailto:anh316@lehigh.edu}{anh316@lehigh.edu}}\addressmark{1}\and Marino
Romero\thanks{\href{mailto:mar007@ucsd.edu}{mar007@ucsd.edu}. Partially
supported by NSF Grant 1700233.}\addressmark{2}}
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%\author[Optional shorter names]{Longer names me\thanks{\href{mailto:hello@world.c}{hello@world.c}. Longer names me was partially supported by Grant 2017.11.14.$\partial$\;supp.}\addressmark{1}, \and you\addressmark{2}}

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\address{\addressmark{1}Department of Mathematics, Lehigh University, Bethlehem, PA  \\ \addressmark{2}Department of Mathematics, University of California, San Diego, La Jolla, CA}

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\abstract{For a symmetric function $G$, the Delta operator $\Delta_G$ is defined via its action on modified Macdonald polynomials by setting $\Delta_G \widetilde{H}_\mu = G[B_\mu] \widetilde{H}_\mu$, where $B_\mu$ is a polynomial in $q$ and $t$. Previous work by Haglund, Remmel, Wilson conjectures a combinatorial interpretation for $\Delta_{e_k} e_n$, generalizing the Shuffle Theorem. Here, we prove combinatorial interpretations for $\Delta_{m_\lambda} e_n|_{q=1}$ and $\Delta_{s_{\lambda}} e_n|_{q=1}$, expressing each as weighted sum over (parallelogram) polyominoes in a rectangle, and provide an explicit combinatorial interpretation for their elementary and Schur function expansions.
}
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\keywords{shuffle conjecture, Macdonald Polynomials, delta conjecture, polyominoes}

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\begin{document}
	
	\maketitle
	%% note that you DO NOT have to put your abstract here -- it is generated by \maketitle and the \abstract and \resume commands above
	%\section{Introduction}
	
	\section{Introduction} 
	
	
	Several recent works study the symmetric function expression
	$\Delta_G e_n$ (as defined just below) for various symmetric
	functions $G$.  One is the well known Shuffle Theorem
	(\cite{Shuffle},\cite{CompoShuffle}), recently proven by Carlsson and Mellit (\cite{ShuffleProof}),  which describes $\Delta_{e_n} e_n$ (or $\nabla e_n$ as it is more frequently written) as a weighted sum of parking functions and connects the result (via work of Mark Haiman~(\cite{haiman2002vanishing})) to the Frobenius characteristic of the space of diagonal harmonics.  Another is the Delta Conjecture of Haglund, Remmel, and Wilson ~(\cite{Delta}), which gives a  combinatorial interpretation for the closely related symmetric function $\Delta_{e_k} e_n$. (In particular, the Delta Conjecture has been proved when $q=1$ (\cite{qDelta}) or $q=0$ (\cite{tDelta}).)  In this work, we provide combinatorial interpretations for the more general $\Delta_{m_\lambda} e_n$ and $\Delta_{s_{\lambda}} e_n$ at $q=1$.
    
The operator $\Delta_G$ is defined using plethystic notation. Given a symmetric function $G$ and an expression $E(t_1,t_2,\dots,)$, the plethystic evaluation $G[E]$ is defined as the image of $G$ under the homomorphism mapping the power symmetric function $p_k \mapsto E(t_1^k,t_2^k,\dots)$. (A complete exposition of plethysm is found in \cite{loehr2011computational}, while \cite{Positivity} and \cite{HLOps} contain many well known plethystic identities.)
	We let $\Delta_G$ (\cite{Positivity}) be the Delta operator on symmetric functions (with coefficients in $\mathbb{C}(q,t)$), which acts on the modified Macdonald basis $\{\Ht_\mu\}_\mu$ by
	$$
	\Delta_G \Ht_\mu = G[B_\mu] \Ht_\mu\text{, where }B_\mu = \sum_{(i,j) \in \mu} q^i t^j .$$  
	The sum is over all cells in $\mu$, where we represent $\mu$ by its French diagram, drawn in the first quadrant with lower left corner at $(0,0)$, and $(i,j)$ is the lower left corner of a lattice cell in $\mu$.   The expression $G[B_\mu]$ can be understood as the evaluation of $G$ at all $q^i t^j$ where $(i,j) \in \mu$ (given in any order). Thus for example $B_{3,1}=1+q+q^2+t$ and $e_4[B_{3,1}]=e_4(1,q,q^2,t)=q^3t$.


	Inspired by \cite{StatPoly}, which relates a particular coefficient of $\nabla e_n$ to the $q,t$-Narayana polynomials\textemdash a weighted sum of polyominoes\textemdash along with a conjecture of Adriano Garsia, our main result is that the set of polyominoes $P_{\lambda,\mu}$ (see Section \ref{sec:polyominoes})  whose steps are indexed by partitions $\lambda$ and $\mu$ give
	\begin{theorem}\label{thm:monThm}
		\begin{equation}\label{eq:monThm}
		\Delta_{m_{\lambda}} e_n \Big|_{q=1} = \sum_{\mu \vdash n} e_\mu \sum_{T \in P_{\lambda,\mu}} t^{\area(T)} .
		\end{equation}
	\end{theorem}
	$\Delta_{m_{\lambda}} e_n $ is not generally Schur positive, but
	in \cite{Positivity} it was conjectured that $\Delta_{s_\lambda} e_n$ is Schur positive; thus we end with a similar combinatorial formula for $\Delta_{s_\lambda} e_n|_{q=1}$. (One can also show these polynomials are symmetric in $q$ and $t$, so this of course gives expansions when $t=1$ as well.)   
	
	Section \ref{sec:symfunc} transforms the left hand side of \eqref{eq:monThm} to a weighted sum of ad hoc combinatorial objects, Section \ref{sec:involution} gives a sign reversing involution reducing the number of summands, Section \ref{sec:polyominoes} gives a bijection to polyominoes in the right hand side of \eqref{eq:monThm}, and Section \ref{sec:main} gives a related expression for $\Delta_{s_\lambda}e_n|_{q=1}$.
	
	
	\section{Symmetric function expansions}\label{sec:symfunc}
	
	The modified Macdonald polynomial specializes at $q=1$ (see
	section Integral Forms, \cite[p.\,364]{Macdonald} and \cite{qLagrange}) to 
	$$
	\Ht_{\mu}[X;1,t] = (t:t)_{\mu'} h_{\mu'}\left[\frac{X}{1-t}\right],
	$$
	where $\mu'$ is the conjugate of $\mu$, $(t:t)_a = \prod_{i=1}^a(1-t^i)$ is the $t$-Pochhammer symbol, and $(t:t)_\mu = \prod_{j=1}^{\ell(\mu)}(t:t)_{\mu_j}$. 
	This means that an eigenoperator for the modified Macdonald polynomial becomes at $q=1$ an eigenoperator of $h_{\mu'}[X/(1-t)]$. This gives 
	$$
	\Delta_G h_{\mu}\left[\frac{X}{1-t}\right] \Big|_{q=1}= G\left[  \sum_i [\mu_i]_t \right] h_{\mu}\left[ \frac{X}{1-t} \right].
	$$
	To apply Delta operators to $e_n$ at $q=1$, we can use the Cauchy formula (where $f_{\eta}$ is the forgotten symmetric function) to show both:
	\begin{align*}
	e_n   = \sum_{\eta \vdash n} h_{\eta} \left[\frac{X}{1-t} \right] f_{\eta}[1-t]\text{ and }
	h_{\eta} \left[\frac{X}{1-t} \right]  = \sum_{\nu^i \vdash \eta_i} e_{\nu^1} \cdots e_{\nu^{\ell(\eta)}} [X] f_{\nu^1} \cdots f_{\nu^{\ell(\eta)}} \left[ \frac{1}{1-t} \right].
	\end{align*}
	
	\noindent  In the following, we use $R(\mu)$ for the set of compositions which rearrange to $\mu$ and  $\PR{\mu}$ (respectively $\CR{\mu}$) for the set of ordered lists of possibly empty partitions (respectively compositions), whose concatenation is in $R(\mu)$. For any $G$,
	\begin{align*}
	\Delta_{G } e_n |_{q=1} %& = \sum_{\eta \vdash n} f_{\eta}[1-t] G\left[ \sum [\eta_i]_t \right]  h_{\eta} \left[\frac{X}{1-t} \right]\\
	&= \sum_{\eta \vdash n} f_{\eta}[1-t] G \left[ \sum [\eta_i]_t \right]  \sum_{\nu^i \vdash \eta_i} e_{\nu^1} \cdots e_{\nu^{\ell(\eta)}} [X] f_{\nu^1} \cdots f_{\nu^{\ell(\eta)}} \left[ \frac{1}{1-t} \right]
	%\end{align*}
	%Collecting the elementary symmetric functions, we arrive at
	%\begin{align*}
	%\Delta_{G} e_n |_{q=1} &
	\\& = \sum_{\mu \vdash n} e_{\mu}[X] \sum_{\eta \vdash n} G\left[ \sum [\eta_i]_t \right]
	\sum_{\substack{(\nu^1,\dots, \nu^{\ell(\eta)}) \in \PR{\mu} \\ \nu^i \vdash \eta_i}} f_{\eta}[1-t] f_{\nu^1} \cdots f_{\nu^{\ell(\eta)}} \left[ \frac{1}{1-t} \right].
	\end{align*}
	
	
	\subsection{A forgotten basis expansion}Our first goal will be to describe the product of forgotten symmetric functions. 
	By the Cauchy formula and a plethystic addition formula, we have two expansions:  
	\begin{align}e_n[X(Y+Z)]&  =  \sum_{\nu \vdash n} h_{\nu}[X] f_{\nu}[Y+ Z], \\
	e_n[X(Y+Z)]&  = \sum_{i=0}^n e_{n-i}[XY] e_i [XZ]  = \sum_{i=0}^n \sum_{\alpha \vdash n-i} \sum_{\beta \vdash i} h_{\alpha}[X] f_{\alpha}[Y] h_{\beta}[X] f_{\beta}[ Z].
	\end{align}
	Comparing the coefficients of $h_{\nu}[X]$, \begin{equation}\label{eq:fsum}
	f_{\nu}[Y+Z] = \sum_{(\alpha, \beta) \in \PR{\nu}} f_{\alpha}[Y] f_{\beta}[Z].
	\end{equation}
	
	\noindent From the Jacobi--Trudi determinant (see details in \cite{qDelta}), we get 
	$\displaystyle
	f_{\nu}[1] = (-1)^{|\nu| - \ell(\nu)} |R(\nu)|.$
	Using \eqref{eq:fsum} repeatedly, treating each term $x_i$ as a new alphabet, we get a formula for the forgotten symmetric functions (see \cite{Brick} for an alternate combinatorial description):
	\begin{equation}\label{eq:fexp}
	f_{\nu}[X] = (-1)^{|\nu| - \ell(\nu)} \sum_{(\alpha^1,\alpha^2,\dots) \in \CR{\nu}} x_1^{|\alpha^1|} x_2^{|\alpha^2|} \cdots.
	\end{equation}
	Substituting $x_i = t^i$ for all $i$, we get a formula for $f_\nu[1/(1-t)]$. For a composition $\gamma$ in $R(\nu)$ define $T\in CC_\gamma$, the Column Composition Tableau of horizontal composition (\hct) $\gamma$: 
	\begin{enumerate}
		\item Begin with a ``bottom'' row of size $|\nu|$ and draw a horizontal line segment above.
		\item  Draw vertical line segments between the columns to denote $\gamma$, adding lines after the first $\gamma_1$ columns, after the next $\gamma_2$, and so on.
		\item   To each set of columns corresponding to some $\gamma_i$, add a weakly increasing (as $i$ increases) number of squares.  All the columns with $j-1$ boxes will together form a subsequence of $\gamma$, which we call $\alpha^j$.  
	\end{enumerate}
	The cells above columns corresponding to $\alpha^j$ thus correspond to $x_j^{|\alpha^j|}$ in some summand in \eqref{eq:fexp}, so we let $w(T) = |\alpha^2| + 2 \cdot |\alpha^3 | + 3 \cdot |\alpha^4|+ \cdots,$ the number of squares above the bottom row.  We use $\fr(T)$ to denote the number of cells in the first column of $T$.  ($\fr(T)\neq 0$ if and only if $\alpha^1=()$, the empty composition.) 
	
	For example, when $\nu=(3,2,2,2,1,1,1)$, we may start with  $\gamma= (2,1,2,1,2,1,3)\in R(v)$ and add boxes to indicate the compositions $\alpha^1 = (2,1)$, $\alpha^2 = (2,1,2)$, $\alpha^3=()$, and $\alpha^4= (1,3)$. Then $T$ is given by
	$$
	\scalebox{.6}{
		\begin{tikzpicture}[xscale=-1,scale=.6]\path[use as bounding box] (0,0) rectangle (12,4); % T
		\partitionfr{12,9,4,4}
		\draw[line width = .95mm] (-.25,1) -- (12.25,1);
		\draw[line width = .75mm] (10,-.25) -- (10,1.25);
		\draw[line width = .75mm] (9,-.25) -- (9,2.25);
		\draw[line width = .75mm] (7,-.25) -- (7,2.25);
		\draw[line width = .75mm] (6,-.25) -- (6,2.25);
		\draw[line width = .75mm] (4,-.25) -- (4,4.25);
		\draw[line width = .75mm] (3,-.25) -- (3,4.25);
		\end{tikzpicture}}
	$$  with $w(T) = 17$ and $\fr(T)=0$.  Let $\overline{CC_\gamma}$ be the subset of $CC_\gamma$ where $\fr(T)=0$. Then
	\begin{align*}
	F_{\nu} &= (-1)^{|\nu|-\ell(\nu)} f_{\nu}\left[ \frac{1}{1-t} \right] = \sum_{\gamma \in R(\nu)}\sum_{T \in CC_\gamma} t^{w(T)}\text{, and }\\
	\overline{F_{\nu}}& =  (1-t^{|\nu|})(-1)^{|\nu|-\ell(\nu)} f_{\nu}\left[ \frac{1}{1-t} \right] = \sum_{\gamma \in R(\nu)}\sum_{T \in \overline{CC_\gamma}} t^{w(T)}.
	\end{align*}
	
	Additionally, we will need the fact that
	$$
	f_{\eta}[1-t] = (-1)^{|\eta| - \ell(\eta)} \sum_{\beta \in R(\eta)} (1-t^{\beta_1}). 
	$$
	The derivation, which we omit for this extended abstract, is similar to \eqref{eq:fexp} and follows from work in \cite{qDelta}.
	
	We can therefore interpret the product of forgotten functions as a generating series for sequences  of column composition tableaux:
	\begin{align*}
	\sum_{\substack{(\nu^1,\dots, \nu^{\ell(\eta)}) \in \PR{\mu} \\ \nu^i \vdash \eta_i}} & f_{\eta}[1-t] f_{\nu^1} \cdots f_{\nu^{\ell(\eta)}} \left[ \frac{1}{1-t} \right] %\\&=  (-1)^{n - \ell(\eta)}\sum_{\beta \in R(\eta)}\sum_{\substack{ \gamma^i \models \beta_i \\ (\gamma^1,\dots,\gamma^{\ell(\eta)}) \in \CR{\mu} }} (-1)^{n - \ell(\mu)} (1-t^{|\gamma_1|}) F_{\gamma^1} \cdots F_{\gamma^{\ell(\gamma)}}
	\\ &= (-1)^{\ell(\eta) - \ell(\mu)} \sum_{\beta \in R(\eta)}\sum_{\substack{ \gamma^i \models \beta_i \\ (\gamma^1,\dots,\gamma^{\ell(\eta)}) \in \CR{\mu} }} \overline{F_{\gamma^1}}F_{\gamma^{2}} \cdots F_{\gamma^{\ell(\gamma)}}.
	\end{align*}
	
	Next, setting $G=m_{\lambda}$, we interpret $$\left(\Delta_{m_\lambda } e_n |_{q=1}\right)\Big|_{e_\mu}=\sum_{\eta \vdash n} m_\lambda\left[ \sum [\eta_i]_t \right]
	\sum_{\substack{(\nu^1,\dots, \nu^{\ell(\eta)}) \in \PR{\mu} \\ \nu^i \vdash \eta_i}} f_{\eta}[1-t] f_{\nu^1} \cdots f_{\nu^{\ell(\eta)}} \left[ \frac{1}{1-t} \right].$$  
	
	\noindent We construct the set of $( \lambda, \mu)$-lists of labeled column composition tableaux $CC(\lambda, \mu)$:
	\begin{enumerate}
		\item Select a partition $\eta \vdash n$ and $\beta \in R(\eta)$.
		\item For $i =1 ,\dots, \ell(\eta)$, select $\gamma^i \models \beta_i$ such that $(\gamma^1,\dots \gamma^{\ell(\gamma)})\in \CR{(\mu)}$.
		\item Select $T_1 \in \overline{CC_{\gamma^1}}$; and for $i =2 ,\dots, \ell(\eta)$, select $T_i \in CC_{\gamma^i}$. 
		\item Distribute a rearrangement of the list $\lambda_1,\dots, \lambda_{\ell(\lambda)}, 0^{n-\ell(\lambda)}$
		along the cells in the bottom rows of $T_1,\dots, T_{\ell(\eta)}$ to obtain a sequence of labeled tableaux $(\text{lT}_1,\dots, \text{lT}_{\ell(\eta)}),$ which we denote by $S$. (Here and elsewhere, we use variable names beginning with `l' for labeled objects, so $\text{lT}$ is formed by labeling column composition tableau $T$.) 
		
	\end{enumerate}
	These new labels correspond to the weight $m_\lambda\left[ \sum [\eta_i]_t \right]$, so define
	$$
	w(\text{lT}_i) = w(T_i) + \sum_{\text{label $j$ in $\text{lT}_i$}} j \cdot (\text{ the number of cells to the right of $j$ in $\text{lT}_i$ }),
	$$
	and $w(S) = w(\text{lT}_1) + \cdots + w(\text{lT}_{\ell(\eta)})$. The sign of $S$ is given by $\sign(S) = (-1)^{\ell(\eta) - \ell(\mu)}$.
	
	
	For example, suppose $\eta= (5,3,1)$, $\mu =(3,2,1,1,1,1)$, and $\lambda = (4,3,2,1,1)$. Suppose we select: $\beta=(3,1,5)$, then select $\gamma^1 \models 3, \gamma^2 \models 1, \gamma^3 \models 5$: $(1,2), (1),$ and $(1,3,1)$ respectively.  Finally, we select $T_1 \in \overline{CC_{(1,2)}}, T_2 \in CC_{(1)}$ and $T_3 \in CC_{(1,3,1)}$ and  a rearrangement of $\lambda,0^{n-\ell(\lambda)} =( 4,3,2,1,1,0,0,0,0)$ to place in the bottom rows. 
	%\begin{align*}
	%\scalebox{.6}{
	%\begin{tikzpicture}[ xscale=-1,scale=.8]\path[use as bounding box, draw] (0,0) rectangle (3,2); % T
	%\partitionfr{3,2}
	%\draw[line width = .75mm] (-.25,1) -- (3.25,1);
	%\draw[line width = .75mm] (2,-.25) -- (2,2.25);
	%\end{tikzpicture}} &&
	%\scalebox{.6}{
	%\begin{tikzpicture}[ xscale=-1,scale=.8]\path[use as bounding box, draw] (0,0) rectangle (1,2); % T
	%\partitionfr{1,1}
	%\draw[line width = .75mm] (-.25,1) -- (1.25,1);
	%\end{tikzpicture}} &&
	%\scalebox{.6}{
	%\begin{tikzpicture}[ xscale=-1,scale=.8]\path[use as bounding box] (0,0) rectangle (5,2); % T
	%\partitionfr{5,5,4}
	%\draw[line width = .75mm] (-.25,1) -- (5.25,1);
	%\draw[line width = .75mm] (1,-.25) -- (1,3.25);
	%\draw[line width = .75mm] (4,-.25) -- (4,3.25);
	%\end{tikzpicture}}
	%\end{align*}
	%For example, selecting $(0,2,0,1,3,0,4,0,1)$ would give $S =$
	For example, let $S=$
	\begin{align}\label{eq:runningex}
	\scalebox{.6}{
		\begin{tikzpicture}[ xscale=-1,scale=.8]
		\partitionfr{3,2}
		\draw[line width = .75mm] (-.25,1) -- (3.25,1);
		\draw[line width = .75mm] (2,-.25) -- (2,2.25);
		\draw (.5,.5) node {$0$};
		\draw (1.5,.5) node {$2 $};
		\draw (2.5,.5) node {$0 $};
		\end{tikzpicture}} &&
	\scalebox{.6}{
		\begin{tikzpicture}[ xscale=-1,scale=.8]
		\partitionfr{1,1}
		\draw[line width = .75mm] (-.25,1) -- (1.25,1);
		\draw (.5,.5) node {$1 $};
		\end{tikzpicture}} &&
	\scalebox{.6}{
		\begin{tikzpicture}[ xscale=-1,scale=.8]
		\path[use as bounding box] (0,0) rectangle (5,2); % T
		\partitionfr{5,5,4}
		\partitionfr{5,5,4}
		\draw[line width = .75mm] (-.25,1) -- (5.25,1);
		\draw[line width = .75mm] (1,-.25) -- (1,3.25);
		\draw[line width = .75mm] (4,-.25) -- (4,3.25);
		\draw (.5,.5) node {$1$};
		\draw (1.5,.5) node {$0 $};
		\draw (2.5,.5) node {$4 $};
		\draw (3.5,.5) node {$0 $};
		\draw (4.5,.5) node {$3 $};
		\end{tikzpicture}}
	\end{align}
	We can calculate the weight of this list by first counting the number of cells above the bottom rows. In this case we would have $2, 1, $ and $5+4=9$ respectively. The weight corresponding to the new labels would be:
	\begin{align*}
	0(2) + 2(1) + 0(0) =  2, && 1(0)=0, &&\text{and}   && 3(4) + 0(3) + 4(2) + 0(1) + 1(0) = 20.
	\end{align*}
	Thus $w(S) = (2+1+9) + (2+0+20) = 34$ and $\sign(S) = (-1)^{(3-6)} = -1$.
	
	In conclusion, we have: 
	\begin{equation}
	\Delta_{m_{\lambda} } e_n \Big|_{q=1} = \sum_{\mu \vdash n} e_\mu ~ \left(  \sum_{S \in CC(\lambda, \mu)} \sign(S) t^{w(S)}\right).\label{eq:preinv}
	\end{equation}
	
	\section{A weight-preserving, sign-reversing involution}\label{sec:involution}
	We will find a sign reversing involution $\phi: CC(\lambda, \mu) \rightarrow CC(\lambda, \mu)$ which simplifies the right hand side of \eqref{eq:preinv}.  The goal will be to only sum over a subset of the $CC(\lambda, \mu)$ where each $\text{lT}_i$ has no vertical bars.  In addition to continuing to use $\fr(\text{lT}_i)$ for the number of cells in the first column of each labeled column composition tableau, we will use $|\text{lT}_i|$ for the sum of its labels, so that in \eqref{eq:runningex}, we have $\fr(\text{lT}_1)=0$ and $|\text{lT}_1|=0+2+0=2$.    %with the following properties: 
	
	% Let $S \in CC(\lambda, \mu)$.
	%\begin{enumerate}
	%\item If $\sign(S) = -1$, then $\sign(\phi(S)) = 1$. 
	%\item If $\sign(S) = \sign(\phi(S)) = 1$, then $\phi(S) =S$.
	%\item And $w(S) = w(\phi(S))$.
	%\end{enumerate}
	
	First, we will define the ``separation'' operation. Let $\gamma$ be a composition of length larger than $1$, and let $\text{lT}$ be a labeled column composition tableau of horizontal composition $\gamma$.  Then split $\text{lT}$ at the first vertical line segment, dividing it into two labeled column composition tableaux $\text{lS}_1$ and $\text{lS}_2$.  Modify $\text{lS}_2$ by adding $|\text{lS}_1|$ cells to each of its columns.  Here is an example:
	
	\begin{align}\label{eq:separate}
	\scalebox{.6}{
		\begin{tikzpicture}[ xscale=-1,scale=.8]
		\partitionfr{5,5,3,1}
		\draw[line width = .75mm] (-.25,1) -- (5.25,1);
		\draw[line width = .75mm] (1,-.25) -- (1,4.25);
		\draw[line width = .75mm] (3,-.25) -- (3,3.25);
		\draw (.5,.5) node {$3$};
		\draw (1.5,.5) node {$0 $};
		\draw (2.5,.5) node {$4 $};
		\draw (3.5,.5) node {$0 $};
		\draw (4.5,.5) node {$1 $};
		\end{tikzpicture}}
	&&  \scalebox{2}{
		\begin{tikzpicture}
		\draw (0,0) node { };
		\draw (0,.75) node {$\rightarrow $};
		\end{tikzpicture}
	}
	&&
	\scalebox{.6}{
		\begin{tikzpicture}[ xscale=-1,scale=.8]
		\partitionfr{2,2}
		\draw[line width = .75mm] (-.25,1) -- (2.25,1);
		\draw (.5,.5) node {$0$};
		\draw (1.5,.5) node {$1$};
		\end{tikzpicture}}
	& \scalebox{2}{
		\begin{tikzpicture}
		\draw (0,0) node { };
		\draw (0,.25) node {,};
		\end{tikzpicture}
	}
	&\scalebox{.6}{
		\begin{tikzpicture}[ xscale=-1,scale=.8] % T
		\path[use as bounding box] (0,0) rectangle (3,4); % T
		\partitionfr{3,3,3,3,1}
		\draw[line width = .75mm] (-.25,1) -- (3.25,1);
		\draw[line width = .75mm] (1,-.25) -- (1,5.25);
		\draw (.5,.5) node {$3$};
		\draw (1.5,.5) node {$0 $};
		\draw (2.5,.5) node {$4 $};
		\end{tikzpicture}}&
	&&
	\end{align}
	\\
	Note that $w(\text{lT})=w(\text{lS}_1)+w(\text{lS}_2)$; the additional cells in $w(\text{lS}_2)$ exactly make up for the loss of weight by the labels in $\text{lS}_1$.  Moreover, since $\text{lT}$ must have weakly increasing numbers of squares from one column to the next, before we add the new cells we have $\fr(\text{lS}_1)\leq \fr(\text{lS}_2)$ and thus after we add the cells, $$\fr(\text{lS}_1)+|\text{lS}_1|\leq \fr(\text{lS}_2).$$ %Note that the weight contributed by the labels on the right of the first vertical line of $\text{lT}$ are not affected by the map. But the labels on the left lost weight. In particular, let us suppose that a label $i$ on the left has $a+b$ cells to its right, where $a$ cells occur to the left of the vertical line segment, and $b$ occur to the right. Then $i \cdot (a+b) = i \cdot a + i \cdot b.$ In the image, the weight of the label is $i \cdot a$. So to preserve weight, we have added $i$ rows to the shape of $\text{lS}_2$ ( that is, we have added $i \cdot b$ cells in total). This shows that the map is weight preserving. Also note that the number of full rows in $\text{lT}$ equals the number of full rows of $\text{lS}_1$. 
	
	To form the inverse operation, let $\text{lS}_1$ and $\text{lS}_2$ be given, where $\text{lS}_1$ consists of a single part (so there are no vertical line segments) and $ \fr(\text{lS}_1)+|\text{lS}_1|\leq \fr(\text{lS}_2)$. Then we say that we can ``join'' $\text{lS}_1$ and $\text{lS}_2$, producing a single labeled column composition tableau $\text{lT}$, by first removing $|\text{lS}_1|$ cells from each column of $\text{lS}_2$, then concatenating $\text{lS}_1$ on the left with a vertical line segment. For example, note that this describes the inverse of \eqref{eq:separate}. %For example,\\
	%\begin{align*}
	%\scalebox{.6}{
	%\begin{tikzpicture}[ xscale=-1,scale=.8]
	%\partitionfr{2,2}
	%\draw[line width = .75mm] (-.25,1) -- (2.25,1);
	%\draw (.5,.5) node {$1$};
	%\draw (1.5,.5) node {$1$};
	%\end{tikzpicture}}
	%& \scalebox{2}{
	%\begin{tikzpicture}
	%\draw (0,0) node { };
	%\draw (0,.25) node {,};
	%\end{tikzpicture}
	%}
	%&\scalebox{.6}{
	%\begin{tikzpicture}[ xscale=-1,scale=.8]
	%\partitionfr{3,3,3,3,1}
	%\draw[line width = .75mm] (-.25,1) -- (3.25,1);
	%\draw[line width = .75mm] (1,-.25) -- (1,5.25);
	%\draw (.5,.5) node {$3$};
	%\draw (1.5,.5) node {$0 $};
	%\draw (2.5,.5) node {$4 $};
	%\end{tikzpicture}}
	%&&
	%&&  \scalebox{2}{
	%\begin{tikzpicture}
	%\draw (0,0) node { };
	%\draw (0,.75) node {$\rightarrow $};
	%\end{tikzpicture}
	%}
	%&&
	%\scalebox{.6}{
	%\begin{tikzpicture}[ xscale=-1,scale=.8]
	%\partitionfr{5,5,1}
	%\draw[line width = .75mm] (-.25,1) -- (5.25,1);
	%\draw[line width = .75mm] (1,-.25) -- (1,3.25);
	%\draw[line width = .75mm] (3,-.25) -- (3,2.25);
	%\draw (.5,.5) node {$3$};
	%\draw (1.5,.5) node {$0 $};
	%\draw (2.5,.5) node {$4 $};
	%\draw (3.5,.5) node {$1 $};
	%\draw (4.5,.5) node {$1 $};
	%\end{tikzpicture}}
	%&&
	%\end{align*}
	
	Define $\phi:CC(\lambda, \mu) \rightarrow CC(\lambda, \mu)$ by scanning $(\text{lT}_1,\dots, \text{lT}_{\ell})$, from left to right, for the first $i$ such that either 
	$\text{lT}_i$ can be separated or $\text{lT}_i$ and $\text{lT}_{i+1}$ can be joined; separate or join accordingly. If no such $i$ is found, leave the sequence fixed. We have
\begin{proposition}
The map $\phi$ is a weight-preserving, sign-reversing involution.
\end{proposition}
\begin{proof} We have seen that the map is weight-preserving. Since we are changing the number of tableaux in the list by $1$, we are reversing the sign. 
	We need only check that it is an involution, which means  we want to make sure we encounter the same $i$ when we apply $\phi$ again. 
Recall we can separate $\text{lT}_i$ if and only if $\text{lT}_i$ is of horizontal composition $\nu$ where $l(\nu)>1$ and we can join $\text{lS}_i$ and $\text{lS}_{i+1}$ if and only if $\text{lS}_i$ is of horizontal composition $(k)$ for some integer $k$ and  $\fr(\text{lS}_i)+|\text{lS}_i| \leq \fr(\text{lS}_{i+1})$.
			
			First, if  $(\text{lT}_1,\text{lT}_2, \text{lT}_3)$ produces $(\text{lT}_1, \text{lS})$ under $\phi$, then $\text{lT}_1$ and $\text{lT}_2$ both have a single part, and $\text{lT}_1$ cannot join with $\text{lT}_2$, so $\fr(\text{lT}_2)  < \fr(\text{lT}_1) + |\text{lT}_1|$. Thus $\fr(\text{lS}) = \fr(\text{lT}_2) < \fr(\text{lT}_1) +|\text{lT}_1|$, so $\text{lT}_1$ cannot join with $\text{lS}$ as desired. %Thus if $\text{lT}_1$ and $\text{lT}_2$ cannot join, then $\text{lT}_1$ and $\text{lS}$ cannot join. 
			If $(\text{lT}_1, \text{lS})$ produces $(\text{lT}_1, \text{lT}_2, \text{lT}_3)$, then $\text{lT}_1$ must again have horizontal composition $(k)$ and since $\text{lT}_1$ cannot join with $\text{lS}$, $\fr(\text{lT}_2) = \fr(\text{lS}) < \fr(\text{lT}_1) +|\text{lT}_1|$. Thus $\text{lT}_1$ and $\text{lT}_2$ cannot join. 
			
		\end{proof}

	
	Let $M_{\lambda, \mu}$ give the set of fixed points $S = (\text{lT}_1, \text{lT}_2,\dots, \text{lT}_\ell)$ of $\phi$, that is $S$ such that:
	\begin{enumerate}
		\item For all $i$, $\hc(\text{lT}_i)$ has a single part. Thus $S$ is an element of $CC(\lambda, \mu)$, where in the first step of the construction, we select $\eta = \mu$ and $\sign(S) = (-1)^{\ell(\mu) - \ell(\mu)} = 1.$ 
		\item For $i =1,\dots, \ell(\lambda)-1$, $\fr(\text{lT}_i) + |\text{lT}_i| > \fr(\text{lT}_{i+1})$. Also, $\fr(\text{lT}_1) = 0$.
	\end{enumerate}
	
	Let $M_{\lambda, \mu}$ denote the collection of fixed points. For example, let $\mu = (4,3,1,1)$ and $\lambda = (3,2,1,1,1)$. Then $(\text{lT}_1,\text{lT}_2,\text{lT}_3,\text{lT}_4) =$
	
	\begin{align*}
	\scalebox{.6}{
		\begin{tikzpicture}[ xscale=-1,scale=.8]
		\partitionfr{3}
		\draw[line width = .75mm] (-.25,1) -- (3.25,1);
		\draw (.5,.5) node {$3$};
		\draw (1.5,.5) node {$0 $};
		\draw (2.5,.5) node {$1 $};
		\end{tikzpicture}} &&
	\scalebox{.6}{
		\begin{tikzpicture}[ xscale=-1,scale=.8]
		\partitionfr{1,1,1,1}
		\draw[line width = .75mm] (-.25,1) -- (1.25,1);
		\draw (.5,.5) node {$1$};
		\end{tikzpicture}} &&
	\scalebox{.6}{
		\begin{tikzpicture}[ xscale=-1,scale=.8]
		\partitionfr{4,4,4,4}
		\draw[line width = .75mm] (-.25,1) -- (4.25,1);
		\draw (.5,.5) node {$0$};
		\draw (1.5,.5) node {$1 $};
		\draw (2.5,.5) node {$2 $};
		\draw (3.5,.5) node {$0 $};
		\end{tikzpicture}}&&
	\scalebox{.6}{
		\begin{tikzpicture}[ xscale=-1,scale=.8]
		\partitionfr{1,1,1}
		\draw[line width = .75mm] (-.25,1) -- (1.25,1);
		\draw (.5,.5) node {$0$};
		\end{tikzpicture}} 
	\end{align*}
	is an element of $M_{\lambda,\mu}$. To check this, we see first that every tableau corresponds to a part of $\mu$. To check the second condition, we start with  $\fr(\text{lT}_1) = 0$,  $\fr(\text{lT}_2)=  3$, $\fr(\text{lT}_3) = 3$, $\fr(\text{lT}_4) = 2,$
	$|\text{lT}_1| =4$, $|\text{lT}_2| =  1$, $|\text{lT}_3| = 3$, and   $|\text{lT}_4| = 0$.  In summary, we have shown:
	\begin{theorem}\label{thm:comb}
		Let $\lambda$ be a partition. Then 
		$$
		\Delta_{m_{\lambda}} e_n \Big|_{q=1}= \sum_{\mu \vdash n} e_{\mu} \sum_{S \in M_{\lambda,\mu}} t^{w(S)}. 
		$$
	\end{theorem}
	
	\section{A bijection}\label{sec:polyominoes} Next, we biject from $M_{\lambda,\mu}$ to a subset of the polyominoes.
	Let $P_{m,n}$ be the set of (parallelogram) polyominoes in the $m\times n$ rectangle, i.e.\ the collection of pairs $(P,Q)$ of lattice paths consisting of North and East steps from $(0,0)$ to $(m,n)$ such that 
	\begin{enumerate}
		\item $P$ and $Q$ only touch at $(0,0)$ and $(m,n)$.
		\item $P$ is above $Q$. 
	\end{enumerate}
	
	The above conditions force $Q$ to begin with an East step. From here on, when we say the path $Q$, we mean the portion of $Q$ which begins at $(1,0)$ and ends at $(m,n)$.  
	For $(P,Q) \in P_{m,n}$, let 
	$$
	\area((P,Q)) = (\text{the number of lattice cells between $P$ and $Q$}) ~ - ~ (m+n-1).
	$$
	%For example, $T=$
	%\begin{align*}
	%\begin{tikzpicture}[scale=.75]
	%\dyckpathmn{0,0}{6}{3}{1,1,0,0,0,1,0,0,0};
	%\draw[line width=2pt,black] (0,0) -- (2,0) -- (2,1) -- (5,1) -- (5,2)--(6,2) -- (6,3) ;
	%\draw[dashed,black] (.5,.5) -- (.5,1.5) -- (4.5,1.5) -- (4.5,2.5) -- (5.5,2.5) ;
	%\end{tikzpicture}
	%\end{align*} 
	%For example, see $T$, the first diagram in Figure \ref{fig:Psi}, in $P_{9,9}$, which has
	%, which is an element of $P_{9,9}$. The dashed line represents the ``middle'' path of $T$. The middle path can be constructed from $T$ by:
	%\begin{enumerate}
	%	\item Shade any cell directly beside each North step, highlighting a number of disjoint columns.  
	%	\item Shift each column except the first right, until the column hits $Q$.   
	%	\item Trace the middle path by drawing a vertical line through the shifted columns, and connecting each shifted column to the next by a horizontal line. 
	%\end{enumerate}
	%	\begin{center}\begin{tabular}{ccc}\scalebox{.4}{
	%			\begin{tikzpicture}	
	%			\dyckpathmn{0,0}{9}{9}{1,1,1,0,1,0,1,1,1,1,0,0,0,0,1,0,0,0};
	%			\draw[line width=4pt,black] (0,0) -- (2,0) -- (2,2) -- (5,2) -- (5,3)--(6,3) -- (6,5) -- (8,5) -- (8,6) -- (9,6) -- (9,9) ;
	%			\draw[dashed,black] (.5,.5) -- (.5,2.5) -- (4.5,2.5) -- (4.5,3.5) -- (5.5,3.5)-- (5.5,7.5) -- (8.5,7.5) -- (8.5,8.5);
	%			\end{tikzpicture}}&
	%		\scalebox{.4}{
	%			\begin{tikzpicture}
	%		
	%			\dyckpathmn{0,0}{9}{9}{1,1,1,0,1,0,1,1,1,1,0,0,0,0,1,0,0,0};
	%			\draw[line width=4pt,black] (0,0) -- (2,0) -- (2,2) -- (5,2) -- (5,3)--(6,3) -- (6,5) -- (8,5) -- (8,6) -- (9,6) -- (9,9) ;
	%			\draw (.5,.5) node {$1 $};
	%			\draw (.5,1.5) node {$0 $};
	%			\draw (.5,2.5) node {$3 $};
	%			\draw (4.5,3.5) node {$1 $};
	%			\draw (5.5,4.5) node {$0 $};
	%			\draw (5.5,5.5) node {$2 $};
	%			\draw (5.5,6.5) node {$1 $};
	%			\draw (5.5,7.5) node {$0 $};
	%			\draw (8.5,8.5) node {$0 $};
	%			\end{tikzpicture}}&
	%		\scalebox{.4}{
	%			\begin{tikzpicture}[rotate=-90]
	%			\dyckpathmn{0,0}{9}{9}{1,1,1,0,1,0,1,1,1,1,0,0,0,0,1,0,0,0};
	%			\draw[line width=4pt,black] (0,0) -- (2,0) -- (2,2) -- (5,2) -- (5,3)--(6,3) -- (6,5) -- (8,5) -- (8,6) -- (9,6) -- (9,9) ;
	%			\draw (.5,.5) node {$1 $};
	%			\draw (.5,1.5) node {$0 $};
	%			\draw (.5,2.5) node {$3 $};
	%			\draw (4.5,3.5) node {$1 $};
	%			\draw (5.5,4.5) node {$0 $};
	%			\draw (5.5,5.5) node {$2 $};
	%			\draw (5.5,6.5) node {$1 $};
	%			\draw (5.5,7.5) node {$0 $};
	%			\draw (8.5,8.5) node {$0 $};
	%			\draw[line width=4pt,black] (0,0) -- (1,0) -- (1,3) -- (0,3) -- (0,0) ;
	%			\draw[line width=4pt,black] (1,3) -- (5,3) -- (5,4) -- (1,4) -- (1,3);
	%			\draw[line width=4pt,black] (2,4) -- (6,4 ) -- (6,8) -- (2,8) -- (2,4) ;
	%			\draw[line width=4pt,black](6,8) -- (9,8) -- (9,9) -- (6,9) -- (6,8);
	%			\end{tikzpicture}}\\1.&2.&3.\\
	%	\end{tabular}
	%\end{center}
	\begin{figure}
		\begin{center}\begin{tabular}{cccc}\scalebox{.4}{
					\begin{tikzpicture}
					\dyckpathmn{0,0}{9}{9}{1,1,1,0,1,0,1,1,1,1,0,0,0,0,1,0,0,0};
					\draw[line width=4pt,black] (0,0) -- (2,0) -- (2,2) -- (5,2) -- (5,3)--(6,3) -- (6,5) -- (8,5) -- (8,6) -- (9,6) -- (9,9) ;
					%\draw[dashed,black] (.5,.5) -- (.5,2.5) -- (4.5,2.5) -- (4.5,3.5) -- (5.5,3.5)-- (5.5,7.5) -- (8.5,7.5) -- (8.5,8.5);
					\end{tikzpicture}}&
				\scalebox{.4}{
					\begin{tikzpicture}
					\dyckpathmn{0,0}{9}{9}{1,1,1,0,1,0,1,1,1,1,0,0,0,0,1,0,0,0};
					\draw[line width=4pt,black] (0,0) -- (2,0) -- (2,2) -- (5,2) -- (5,3)--(6,3) -- (6,5) -- (8,5) -- (8,6) -- (9,6) -- (9,9) ;
					\draw (.5,.5) node {$1 $};
					\draw (.5,1.5) node {$0 $};
					\draw (.5,2.5) node {$3 $};
					\draw (1.5,3.5) node {$1 $};
					\draw (2.5,4.5) node {$0 $};
					\draw (2.5,5.5) node {$2 $};
					\draw (2.5,6.5) node {$1 $};
					\draw (2.5,7.5) node {$0 $};
					\draw (6.5,8.5) node {$0 $};
					\end{tikzpicture}}&\scalebox{.4}{
					\begin{tikzpicture}
					\dyckpathmn{0,0}{9}{9}{1,1,1,0,1,0,1,1,1,1,0,0,0,0,1,0,0,0};
					\draw[dashed,black] (.7,.5) -- (.7,2.5) -- (4.7,2.5) -- (4.7,3.5) -- (5.7,3.5)-- (5.7,7.5) -- (8.7,7.5) -- (8.7,8.5);
					\draw[line width=4pt,black] (0,0) -- (2,0) -- (2,2) -- (5,2) -- (5,3)--(6,3) -- (6,5) -- (8,5) -- (8,6) -- (9,6) -- (9,9) ;
					\draw (.5,.5) node {$1 $};
					\draw (.5,1.5) node {$0 $};
					\draw (.5,2.5) node {$3 $};
					\draw (4.5,3.5) node {$1 $};
					\draw (5.5,4.5) node {$0 $};
					\draw (5.5,5.5) node {$2 $};
					\draw (5.5,6.5) node {$1 $};
					\draw (5.5,7.5) node {$0 $};
					\draw (8.5,8.5) node {$0 $};
					\end{tikzpicture}}&
				\scalebox{.4}{
					\begin{tikzpicture}[rotate=-90]
					\dyckpathmn{0,0}{9}{9}{1,1,1,0,1,0,1,1,1,1,0,0,0,0,1,0,0,0};
					\draw[line width=4pt,black] (0,0) -- (2,0) -- (2,2) -- (5,2) -- (5,3)--(6,3) -- (6,5) -- (8,5) -- (8,6) -- (9,6) -- (9,9) ;
					\draw (.5,.5) node {$1 $};
					\draw (.5,1.5) node {$0 $};
					\draw (.5,2.5) node {$3 $};
					\draw (4.5,3.5) node {$1 $};
					\draw (5.5,4.5) node {$0 $};
					\draw (5.5,5.5) node {$2 $};
					\draw (5.5,6.5) node {$1 $};
					\draw (5.5,7.5) node {$0 $};
					\draw (8.5,8.5) node {$0 $};
					\draw[line width=4pt,black] (0,0) -- (1,0) -- (1,3) -- (0,3) -- (0,0) ;
					\draw[line width=4pt,black] (1,3) -- (5,3) -- (5,4) -- (1,4) -- (1,3);
					\draw[line width=4pt,black] (2,4) -- (6,4 ) -- (6,8) -- (2,8) -- (2,4) ;
					\draw[line width=4pt,black](6,8) -- (9,8) -- (9,9) -- (6,9) -- (6,8);
					\end{tikzpicture}}\\1.&2.&3.&4.\\
			\end{tabular}
			\begin{align*}
			\scalebox{.6}{
				\begin{tikzpicture}[ xscale=-1,scale=.8]
				\partitionfr{3}
				\draw[line width = .75mm] (-.25,1) -- (3.25,1);
				\draw (.5,.5) node {$3$};
				\draw (1.5,.5) node {$0 $};
				\draw (2.5,.5) node {$1 $};
				\end{tikzpicture}} &&
			\scalebox{.6}{
				\begin{tikzpicture}[ xscale=-1,scale=.8]
				\partitionfr{1,1,1,1}
				\draw[line width = .75mm] (-.25,1) -- (1.25,1);
				\draw (.5,.5) node {$1$};
				\end{tikzpicture}} &&
			\scalebox{.6}{
				\begin{tikzpicture}[ xscale=-1,,scale=.8]
				\partitionfr{4,4,4,4}
				\draw[line width = .75mm] (-.25,1) -- (4.25,1);
				\draw (.5,.5) node {$0$};
				\draw (1.5,.5) node {$1 $};
				\draw (2.5,.5) node {$2 $};
				\draw (3.5,.5) node {$0 $};
				\end{tikzpicture}}&&
			\scalebox{.6}{
				\begin{tikzpicture}[ xscale=-1,scale=.8]
				\partitionfr{1,1,1}
				\draw[line width = .75mm] (-.25,1) -- (1.25,1);
				\draw (.5,.5) node {$0$};
				\end{tikzpicture}} 
			\end{align*}\\5.\end{center}
		\caption{Constructing $\Psi$. The numbers correspond to steps.}\label{fig:Psi}
	\end{figure}
	Define $\colcomp(P)$ (respectively $\rowcomp(Q)$) to be the composition whose $i$th part is the number of North (respectively East) steps in $P$ (in $Q$) occurring along $x=i-1$ ($y=i-1$), where as usual we do not count the first step of $Q$. We can now define $P_{\lambda,\mu}$ of Theorem 1 to be the set of all $(P,Q) \in P_{|\lambda|+1,|\mu|}$ such that $\colcomp(P)$ rearranges to $\mu$ and $\rowcomp(Q)$ rearranges to $\lambda$.  
	
	For example, if $(P,Q)$ is the first diagram in Figure \ref{fig:Psi}, then $(P,Q)$ is in $P_{9,9}$, $\area(P,Q) \\= 24$, $\colcomp(P)=(3,1,4,0,0,0,1,0,0)$, and $\rowcomp(Q)=(1,0,3,1,0,2,1,0,0)$. 
Thus, $(P,Q) \in P_{(3,2,1,1,1),(4,3,1,1)}$. 
	
	
	The goal of this section is to prove the following theorem:
	\begin{theorem}\label{thm:toparal}
		$$\sum_{S \in M_{\lambda,\mu}} t^{w(S)} = \sum_{P \in P_{\lambda,\mu}} t^{\area(P)}$$
	\end{theorem}
	We will do this by producing a weight-preserving bijection $\psi:P_{\lambda,\mu} \rightarrow M_{\lambda,\mu}$. 
	See the example in Figure \ref{fig:Psi} as we work through the bijection.  Let $(P,Q) \in P_{\lambda,\mu}$. 
	%\begin{figure}
	%	\begin{center}\begin{tabular}{ccc}\scalebox{.4}{
	%			\begin{tikzpicture}
	%			\dyckpathmn{0,0}{9}{9}{1,1,1,0,1,0,1,1,1,1,0,0,0,0,1,0,0,0};
	%			\draw[line width=4pt,black] (0,0) -- (2,0) -- (2,2) -- (5,2) -- (5,3)--(6,3) -- (6,5) -- (8,5) -- (8,6) -- (9,6) -- (9,9) ;
	%			\draw[dashed,black] (.5,.5) -- (.5,2.5) -- (4.5,2.5) -- (4.5,3.5) -- (5.5,3.5)-- (5.5,7.5) -- (8.5,7.5) -- (8.5,8.5);
	%			\end{tikzpicture}}&
	%		\scalebox{.4}{
	%			\begin{tikzpicture}
	%			\dyckpathmn{0,0}{9}{9}{1,1,1,0,1,0,1,1,1,1,0,0,0,0,1,0,0,0};
	%			\draw[line width=4pt,black] (0,0) -- (2,0) -- (2,2) -- (5,2) -- (5,3)--(6,3) -- (6,5) -- (8,5) -- (8,6) -- (9,6) -- (9,9) ;
	%			\draw (.5,.5) node {$1 $};
	%			\draw (.5,1.5) node {$0 $};
	%			\draw (.5,2.5) node {$3 $};
	%			\draw (4.5,3.5) node {$1 $};
	%			\draw (5.5,4.5) node {$0 $};
	%			\draw (5.5,5.5) node {$2 $};
	%			\draw (5.5,6.5) node {$1 $};
	%			\draw (5.5,7.5) node {$0 $};
	%			\draw (8.5,8.5) node {$0 $};
	%			\end{tikzpicture}}&
	%		\scalebox{.4}{
	%			\begin{tikzpicture}[rotate=-90]
	%			\dyckpathmn{0,0}{9}{9}{1,1,1,0,1,0,1,1,1,1,0,0,0,0,1,0,0,0};
	%			\draw[line width=4pt,black] (0,0) -- (2,0) -- (2,2) -- (5,2) -- (5,3)--(6,3) -- (6,5) -- (8,5) -- (8,6) -- (9,6) -- (9,9) ;
	%			\draw (.5,.5) node {$1 $};
	%			\draw (.5,1.5) node {$0 $};
	%			\draw (.5,2.5) node {$3 $};
	%			\draw (4.5,3.5) node {$1 $};
	%			\draw (5.5,4.5) node {$0 $};
	%			\draw (5.5,5.5) node {$2 $};
	%			\draw (5.5,6.5) node {$1 $};
	%			\draw (5.5,7.5) node {$0 $};
	%			\draw (8.5,8.5) node {$0 $};
	%			\draw[line width=4pt,black] (0,0) -- (1,0) -- (1,3) -- (0,3) -- (0,0) ;
	%			\draw[line width=4pt,black] (1,3) -- (5,3) -- (5,4) -- (1,4) -- (1,3);
	%			\draw[line width=4pt,black] (2,4) -- (6,4 ) -- (6,8) -- (2,8) -- (2,4) ;
	%			\draw[line width=4pt,black](6,8) -- (9,8) -- (9,9) -- (6,9) -- (6,8);
	%			\end{tikzpicture}}\\1.&2.&3.\\
	%		\end{tabular}
	%	\begin{align*}
	%	\scalebox{.6}{
	%		\begin{tikzpicture}[ xscale=-1,scale=.8]
	%		\partitionfr{3}
	%		\draw[line width = .75mm] (-.25,1) -- (3.25,1);
	%		\draw (.5,.5) node {$3$};
	%		\draw (1.5,.5) node {$0 $};
	%		\draw (2.5,.5) node {$1 $};
	%		\end{tikzpicture}} &&
	%	\scalebox{.6}{
	%		\begin{tikzpicture}[ xscale=-1,scale=.8]
	%		\partitionfr{1,1,1,1}
	%		\draw[line width = .75mm] (-.25,1) -- (1.25,1);
	%		\draw (.5,.5) node {$1$};
	%		\end{tikzpicture}} &&
	%	\scalebox{.6}{
	%		\begin{tikzpicture}[ xscale=-1,,scale=.8]
	%		\partitionfr{4,4,4,4}
	%		\draw[line width = .75mm] (-.25,1) -- (4.25,1);
	%		\draw (.5,.5) node {$0$};
	%		\draw (1.5,.5) node {$1 $};
	%		\draw (2.5,.5) node {$2 $};
	%		\draw (3.5,.5) node {$0 $};
	%		\end{tikzpicture}}&&
	%	\scalebox{.6}{
	%		\begin{tikzpicture}[ xscale=-1,scale=.8]
	%		\partitionfr{1,1,1}
	%		\draw[line width = .75mm] (-.25,1) -- (1.25,1);
	%		\draw (.5,.5) node {$0$};
	%		\end{tikzpicture}} 
	%	\end{align*}\\4.\end{center}
	%	\caption{Constructing $\Psi$. The numbers correspond to steps.}\label{fig:Psi}
	%\end{figure}
	\begin{enumerate}
		\item Draw $(P,Q)$, as in Figure \ref{fig:Psi}, step 1.
		%\begin{align*}
		%\scalebox{.4}{
		%\begin{tikzpicture}
		%\dyckpathmn{0,0}{9}{9}{1,1,1,0,1,0,1,1,1,1,0,0,0,0,1,0,0,0};
		%\draw[line width=4pt,black] (0,0) -- (2,0) -- (2,2) -- (5,2) -- (5,3)--(6,3) -- (6,5) -- (8,5) -- (8,6) -- (9,6) -- (9,9) ;
		%\draw[dashed,black] (.5,.5) -- (.5,2.5) -- (4.5,2.5) -- (4.5,3.5) -- (5.5,3.5)-- (5.5,7.5) -- (8.5,7.5) -- (8.5,8.5);
		%\end{tikzpicture}}
		%\end{align*}
		
		\item Place the sequence $\rowcomp(Q)$ along the North steps of $P$. Then  for $T$ in Figure \ref{fig:Psi}, $ \rowcomp(Q) = (1,0,3,1,0,2,1,0,0)$.  Thus  we get Figure \ref{fig:Psi}, step 2.
		%\begin{align*}
		%\scalebox{.4}{
		%\begin{tikzpicture}
		%\dyckpathmn{0,0}{9}{9}{1,1,1,0,1,0,1,1,1,1,0,0,0,0,1,0,0,0};
		%\draw[line width=2pt,black] (0,0) -- (2,0) -- (2,2) -- (5,2) -- (5,3)--(6,3) -- (6,5) -- (8,5) -- (8,6) -- (9,6) -- (9,9) ;
		%\draw (.5,.5) node {$1 $};
		%\draw (.5,1.5) node {$0 $};
		%\draw (.5,2.5) node {$3 $};
		%\draw (4.5,3.5) node {$1 $};
		%\draw (5.5,4.5) node {$0 $};
		%\draw (5.5,5.5) node {$2 $};
		%\draw (5.5,6.5) node {$1 $};
		%\draw (5.5,7.5) node {$0 $};
		%\draw (8.5,8.5) node {$0 $};
		%\end{tikzpicture}}
		%\end{align*}
		\item Next, shift the integers by columns, keeping integers in the first column fixed, but otherwise moving the integers in a given column right until the bottom integer's cell touches the bottom path $Q$.  See Figure \ref{fig:Psi}, step 3.  Call the cells containing the integers, and the sequences of horizontal squares which connect them, the ``middle path.''  See the dashed line in our running example.
		\item Turn the diagram $90$ degrees clockwise.
		Rotating $(P,Q)$ gives Figure \ref{fig:Psi}, step 4.
		%\begin{align*}
		%\scalebox{.4}{
		%\begin{tikzpicture}[rotate=-90]
		%\dyckpathmn{0,0}{9}{9}{1,1,1,0,1,0,1,1,1,1,0,0,0,0,1,0,0,0};
		%\draw[line width=2pt,black] (0,0) -- (2,0) -- (2,2) -- (5,2) -- (5,3)--(6,3) -- (6,5) -- (8,5) -- (8,6) -- (9,6) -- (9,9) ;
		%\draw (.5,.5) node {$1 $};
		%\draw (.5,1.5) node {$0 $};
		%\draw (.5,2.5) node {$3 $};
		%\draw (4.5,3.5) node {$1 $};
		%\draw (5.5,4.5) node {$0 $};
		%\draw (5.5,5.5) node {$2 $};
		%\draw (5.5,6.5) node {$1 $};
		%\draw (5.5,7.5) node {$0 $};
		%\draw (8.5,8.5) node {$0 $};
		%\draw[line width=2pt,black] (0,0) -- (1,0) -- (1,3) -- (0,3) -- (0,0) ;
		%\draw[line width=2pt,black] (1,3) -- (5,3) -- (5,4) -- (1,4) -- (1,3);
		%\draw[line width=2pt,black] (2,4) -- (6,4 ) -- (6,8) -- (2,8) -- (2,4) ;
		%\draw[line width=2pt,black](6,8) -- (9,8) -- (9,9) -- (6,9) -- (6,8);
		%\end{tikzpicture}}
		%\end{align*}
		
		\item To get $\psi((P,Q)) = (\text{lT}_1,\dots, \text{lT}_{\ell(\lambda)})$ we look at the cells bounded from above by the vertical segments of $P$ (which will now be horizontal since we rotated the picture), and bounded from below by the middle path. The cells along the middle path are the labeled bottom rows of $\text{lT}_1,\dots, \text{lT}_{\ell(\lambda)}$. Adding the cells between $P$ and the middle path produces the element in Figure \ref{fig:Psi}, step 5.
		%\begin{align*}
		%\scalebox{.6}{
		%\begin{tikzpicture}[ xscale=-1,scale=.8]
		%\partitionfr{3}
		%\draw[line width = .75mm] (-.25,1) -- (3.25,1);
		%\draw (.5,.5) node {$3$};
		%\draw (1.5,.5) node {$0 $};
		%\draw (2.5,.5) node {$1 $};
		%\end{tikzpicture}} &&
		%\scalebox{.6}{
		%\begin{tikzpicture}[ xscale=-1,scale=.8]
		%\partitionfr{1,1,1,1}
		%\draw[line width = .75mm] (-.25,1) -- (1.25,1);
		%\draw (.5,.5) node {$1$};
		%\end{tikzpicture}} &&
		%\scalebox{.6}{
		%\begin{tikzpicture}[ xscale=-1,,scale=.8]
		%\partitionfr{4,4,4,4}
		%\draw[line width = .75mm] (-.25,1) -- (4.25,1);
		%\draw (.5,.5) node {$0$};
		%\draw (1.5,.5) node {$1 $};
		%\draw (2.5,.5) node {$2 $};
		%\draw (3.5,.5) node {$0 $};
		%\end{tikzpicture}}&&
		%\scalebox{.6}{
		%\begin{tikzpicture}[ xscale=-1,scale=.8]
		%\partitionfr{1,1,1}
		%\draw[line width = .75mm] (-.25,1) -- (1.25,1);
		%\draw (.5,.5) node {$0$};
		%\end{tikzpicture}} 
		%\end{align*}
	\end{enumerate}
	
	To show that this process is the required bijection we must show that the resulting element satisfies the conditions of $M_{\lambda,\mu}$ and that the process is invertible and weight preserving. 
	\begin{proof} % Let $(P,Q) \in P_{\lambda,\mu}$. Then when read from left to right,  the vertical segments of $P$ have lengths $a_1,\dots, a_{\ell(\lambda)}$ which rearrange to $\lambda$. To create $\psi((P,Q)) = (\text{lT}_1,\dots, \text{lT}_{\ell(\lambda)})$, begin by drawing the bottom rows of the sequence. The bottom rows will have lengths (from left to right) given by $a_1,\dots, a_{\ell(\lambda)}$. Label the cells of these bottom rows (again from left to right) by the sequence $s(Q) = (s_1,\dots, s_n)$. So $\text{lT}_1$ will be labeled by $s_1,\dots, s_{a_1}$, then $\text{lT}_2$ will be labeled by $s_{a_1+1}, \dots, s_{a_1+a_2}$, and so on. 
		%The number of full rows above $\text{lT}_i$ will then equal to the number of lattice cells between $P$ and the middle path of $(P,Q)$ along the vertical segment of length $a_i$.  
		
		Briefly, we begin by inverting $\Psi$.  Start with
		$(\text{lT}_1,\dots \text{lT}_n)$ in $M_{\lambda,\mu}$.
		By construction, its labels give $\rowcomp(Q)$ which
		immediately determines $Q$ which (as desired) rearranges
		to $\lambda$. Next, rotating each $\text{lT}_i$
		counterclockwise and stacking them from bottom to top
		will give our path $P$, if we can determine where to
		align the ``right column'' (formerly ``bottom row'') of each $\text{lT}_i$ in our diagram to create the middle path.  $\text{lT}_1$ must be left justified.  For $i>1$, each $\text{lT}_i$ should be placed as far West as possible among all positions where its Southeast corner still touches the path $Q$. In particular, for all $j$, say $\text{lT}_j$'s bottom right (respectively bottom left) corner is placed along the line $x=c_j$ (respectively $x=d_j$). Then $c_j=|\text{lT}_1|+|\text{lT}_2|+\cdots |\text{lT}_{j-1}|+1$.  The resulting path $P$ will fail to have $\colcomp(P)$ rearrange to $\mu$ (or worse, will fail to be a connected lattice path) only if the left column (formerly top row) of some $\text{lT}_i$ is weakly East of the left of column of $\text{lT}_{i+1}$ so that $d_i\geq d_{i+1}$. Since the bottom left corner of $\text{lT}_j$ is by construction $|\fr(\text{lT}_j)|$ left of its right corner, $d_j=c_j-\fr(\text{lT}_j)$.  Thus the left column of $\text{lT}_i$ is weakly East of the left of column of $\text{lT}_{i+1}$ if and only if \begin{align*}|\text{lT}_1|+|\text{lT}_2|+\cdots |\text{lT}_{i-1}|+1-\fr(\text{lT}_i)&\geq|\text{lT}_1|+|\text{lT}_2|+\cdots |\text{lT}_{i}|+1-\fr(\text{lT}_{i+1})\Leftrightarrow\\\fr(\text{lT}_{i+1})&\geq \fr(T_i)+|\text{lT}_i| \end{align*}
		\noindent This, of course, is exactly when $(\text{lT}_1,\dots \text{lT}_n)$ was not a fixed point of our involution and thus was not in $M_{\lambda,\mu}$.
		
		%from $(\text{lT}_1,\dots \text{lT}_n)$ immediately). Since $(P,Q) \in P_{\lambda,\mu}$, the labels must be a rearrangement of $\mu$ with $0$'s added. Similarly, the bottom rows of $(\text{lT}_1,\dots \text{lT}_n)$ have lengths which form a rearrangement of $\lambda$. Therefore, to show that $\psi((P,Q)) \in M_{\lambda,\mu}$, we need only check the inequality conditions between $\text{lT}_i$ and $\text{lT}_{i+1}$. That is, we must check that
		%$$
		%\fr(\text{lT}_i) + |\text{lT}_i| > \fr(\text{lT}_{i+1}).
		%$$
		%For all $i$, let $a_i$ be the $i$th vertical segment (from left to right) in $P$ and $b_i$ the corresponding $i$th vertical segment in the middle path, such that number of complete cells between the two in any given row corresponds by definition to $\fr(\text{lT}_i)$.  Moreover $b_{i+1}$ is East of $b_i$ by $|\text{lT}_i|$ many cells by construction, but $a_{i+1}$ is at least one cell further East than $a_i$.  Thus $\fr(\text{lT}_{i+1})$, as the number of complete cells between $a_{i+1}$ and $b_{i+1}$ and is at most $|\text{lT}_i|-1$ larger than $\fr(\text{lT}_i)$.%  Then the inequality is equivalent to the fact that the vertical segment $a_i$ occurs to the left of $a_{i+1}$ in $P$.   In particular, first note that the columns of $b_{i+1}$ can only be shifted to the right of $b_i$ by $|\text{lT}_i|$ spaces, and second, that  the number of complete cells in each row between $a_i$ and $b_i$ is $|T_i|$ and the number of cells  if $a_i$ occurs along the line $x=c$ and $b_i$ occurs along the line $x=d$, More precisely, the portion of the middle path which runs to the East of the segment of length $a_i$ occurs in column $s_1+\cdots + s_{a_1+\cdots +a_{i-1}}$. And the portion of the middle path East of the segment of length $a_{i+1}$ is in column $s_1+\cdots + s_{a_1+\cdots +a_{i}}$. The difference between the two is $ s_{a_1+\cdots +a_{i-1} +1} + \cdots + s_{a_1+\cdots +a_{i-1} + a_i} = |\text{lT}_i|$.  Thus the column for the segment of length $a_i$ is $\fr(\text{lT}_i) + |\text{lT}_i|$ columns West of the portion of the middle path which is East to the segment of length $a_{i+1}$, and the column for the segment of length $a_{i+1}$ is $\fr(\text{lT}_{i+1})$ columns West. Since the portion of length $a_{i+1}$ is East of the portion of length $a_i$, we necessarily have
		%$$
		%\fr(\text{lT}_i) + |\text{lT}_i| > \fr(\text{lT}_{i+1}).
		%$$
		
		To see that $\psi$ is weight preserving, first note that the middle path goes through $m+n-1$ cells; thus area counts the number of cells between (P,Q) but not on the middle path.   The cells West of the middle path clearly correspond to the cells above the bottom rows in $(\text{lT}_1,\dots \text{lT}_n)$, so we need to show the cells South/East of the middle path correspond to the weight contributed by the labels in $(\text{lT}_1,\dots \text{lT}_n)$.  This is exactly the case by construction.  %the  we look at the portion of $(P,Q)$ corresponding to $\text{lT}_i$. This portion of $(P,Q)$ occurs between the lines $y=s_1+\cdots +s_{a_1+\cdots +a_{i-1} }$ and $y=s_1+\cdots +s_{a_1+\cdots +a_{i} }$. Note that the middle path crosses through the cells in the top row of this section between $P$ and $Q$. To the left of the middle path, there are \fr\text{lT}_i) \cdot a_i$ lattice cells. This is the part of the weight $w(\text{lT}_i)$ which is attained by counting the number of cells above the bottom row of $\text{lT}_i$. To the right of the middle path, we note that every East segment of $Q$ along the line $y=s_1+\cdots +s_{a_1+\cdots +a_{i-1} } + r$ ($r=0,\dots, a_i-1$) has precisely $a_i - r -1$ lattice cells above which are strictly below the middle path. Therefore, if $j$ occurs along the middle path so that there are $c$ other labels directly above, then there are $j$ East steps on the line $y= s_1+\cdots +s_{a_1+\cdots +a_{i-1} } + (a_i-b -1) $; and the number of cells above these $j$ East steps is 
		%$$
		%j \cdot( a_i - (a_i -b -1) -1) = j \cdot b.
		%$$
		%This is exactly the contribution of $j$ in $w(\text{lT}_i)$:
		%$$
		%j \cdot (\text{ the number of cells to the right of $j$ in $\text{lT}_i$ }).
		%$$
		%The last part we must check is the invertibility of $\psi$.  Given $(\text{lT}_1,\dots, \text{lT}_{\ell(\lambda)}) \in M_{\lambda,\mu}$, the labels of $\text{lT}_1,\dots, \text{lT}_{\ell(\lambda)}$ uniquely determines the path $Q$. If $\text{lT}_i$ has bottom row of length $a_i$, then the vertical segments in $P$ and the middle path are of length $a_1,\dots a_{\ell(\lambda)}$. We can reconstruct the exact East-West placement of the vertical segments in the middle path once we know $Q$.  Finally, the distance (number of full squares) between those vertical segments and the vertical segments in $P$, as previously observed is $\fr(T_i)$.   %determines the vertical segments of $P$. This is enough to determine the middle path of $(P,Q)$, and thus \fr\text{lT}_i)$ determines the number of lattice cells between the middle path and the portion of length $a_i$ in $P$. This process is valid so long as the segment of length $a_{i+1}$ is placed East of the segment of length $a_i$. This is, as we observed, given precisely by the inequality \fr\text{lT}_i) + |\text{lT}_i| >\fr\text{lT}_{i+1}).$
	\end{proof}
	\noindent This proves Theorem \ref{thm:toparal}. Theorems \ref{thm:comb} and \ref{thm:toparal} imply Theorem \ref{thm:monThm}.
	\section{The main result}\label{sec:main}
	
	The previous theorem may now be rewritten as 
	\begin{corollary}
		Let $\nu$ be a partition. Then 
		$$
		\Delta_{m_{\nu}} e_n \Big|_{q=1}= \sum_{\eta \vdash n} e_{\eta} \sum_{S \in P_{\nu,\eta}} t^{w(S)}. 
		$$
	\end{corollary}
	To get a Schur function expansion of $\Delta_{s_\lambda} e_n|_{q=1}$, recall that for any $\zeta,\kappa\vdash n$, $\alpha\in R(\kappa)$, the Kostka number $K_{\zeta,\kappa}=K_{\zeta,\alpha}$ counts the set of semi-standard tableaux of shape $\zeta$ and content $\alpha$, $T_{\zeta,\alpha}$.  Also recall %. or equivalently the number of ways of filling the rows of $\eta$ by a Yamanouchi word of shape $\nu$, 
	%occurs in the expansions
	\begin{align*}
	e_{\eta} = \sum_{\mu' \vdash n} K_{\mu,\eta} s_{\mu'} && \text{ and } && s_{\lambda} = \sum_{\nu \vdash n} K_{\lambda, \nu} m_{\nu}
	\end{align*}
	
	\noindent We have 
	\begin{align*}
	\Delta_{s_{\lambda}} e_n \Big|_{q=1}  & = \sum_{\nu \vdash n} K_{\lambda,\nu} \Delta_{ m_\nu} e_n \Big |_{q=1} 
	\\& =  \sum_{\nu \vdash n} K_{\lambda,\nu} \sum_{\eta \vdash n} e_{\eta} \sum_{S \in P_{\nu,\eta}} t^{w(S)}  \\
	& =  \sum_{\nu \vdash n} K_{\lambda,\nu} \sum_{\eta \vdash n} \sum_{\mu \vdash n} K_{\mu',\eta} s_{\mu} \sum_{S \in P_{\nu,\eta}} t^{w(S)} 
	\\& =   \sum_{\mu \vdash n} s_{\mu}  \sum_{\nu \vdash n}  \sum_{\eta \vdash n}  \sum_{S \in P_{\nu,\eta}} K_{\lambda,\nu} K_{\mu',\eta} t^{w(S)}.
	\end{align*}
	We can use lattice words to give a nice combinatorial interpretation of $$\sum_{\nu \vdash n} \sum_{\eta \vdash n}  \sum_{S \in P_{\nu,\eta}} K_{\lambda',\nu} K_{\mu',\eta} t^{w(S)},$$ using labeled polynomials  (note that here we work with $\lambda'$ for convenience).
	
	
	
	We will call $(\text{lP},\text{lQ})$ a labeled polyomino, if $P$ (respectively $Q$) has positive integer labels beside each North step in $P$ (respectively each East step except the first in $Q$) such that the integers in $P$ (respectively $Q$) are increasing from South to North (respectively East to West).  We use $\ctype{(P)}$ (respectively $\ctype(Q)$) for the composition whose $i$th part gives the number of labels $i$ along $P$ (respectively $Q$).  
	
	If we start with $(P,Q)\in P_{\nu,\eta}$, let $\colcomp(P)=\alpha$ and $\rowcomp(Q)=\beta$ so that $\alpha\in R(\eta)$ and $\beta\in R(\nu)$. Then we can construct a labeled polyomino $(\text{lP},\text{lQ})$, by the following:   Select $T_1 \in T_{\mu',\alpha}$, and $T_2\in  K_{\lambda',\beta}$.  Then working through the columns of $T_1$ from left to right, if $i$ is in the $c$th column of $T_i$, place a $c$ by a north step in the $i$th column of $P$ (within a column, fill cells from South to North to force an increasing column condition).  Repeat the process with $Q$ and $T_2$, this time filling the cells above East steps in an analogous manner.  Call the result $(\text{lP},\text{lQ})$. 
	For instance,\begin{align}\label{PQTab}\begin{tikzpicture}[scale=.75]
	\dyckpathmn{0,0}{6}{3}{1,1,0,0,0,1,0,0,0};
	\draw[line width=3pt,black] (0,0) -- (2,0) -- (2,1) -- (5,1) -- (5,2)--(6,2) -- (6,3) ;
	\end{tikzpicture},&\scalebox{.65}{
		\begin{tikzpicture}%[xscale=-1]
		\partitionfr{2,1}
		\draw (.5,.5) node {$1 $};
		\draw (1.5,.5) node {$1 $};
		\draw (.5,1.5) node {$4 $};
		\end{tikzpicture}
	}, &
	\scalebox{.65}{
		\begin{tikzpicture}
		\partitionfr{3,2}
		\draw (.5,.5) node {$1 $};
		\draw (.5,1.5) node {$2 $};
		\draw (1.5,.5) node {$2 $};
		\draw (1.5,1.5) node {$3 $};
		\draw (2.5,.5) node {$2 $};
		\end{tikzpicture}
	}
	\scalebox{2}{
		\begin{tikzpicture}
		\draw (0,0) node { };
		\draw (0,.45) node {$\rightarrow $};
		\end{tikzpicture}
	}&\begin{tikzpicture}[scale=.75]
	\dyckpathmn{0,0}{6}{3}{1,1,0,0,0,1,0,0,0};
	\draw[line width=3pt,black] (0,0) -- (2,0) -- (2,1) -- (5,1) -- (5,2)--(6,2) -- (6,3) ;
	\draw (.35,.5) node {$1 $};
	\draw (.35,1.5) node {$2$};
	\draw (3.35,2.5) node {$1 $};
	\draw (1.5,.35) node {$1$};
	\draw (2.5,1.35) node {$1 $};
	\draw (3.5,1.35) node {$2 $};
	\draw (4.5,1.35) node {$3$};
	\draw (5.5,2.35) node {$2 $};
	\end{tikzpicture}
	\end{align}
	
	
	
	Note that we have exactly one label for each North step since $\colcomp(\text{lP})= \\\type(T_1)$. Similarly we have $\rowcomp(\text{lQ})=\type(T_2)$.   Note that the conjugate of the shape of $T_1$, $\mu$, is  equal to $\ctype(\text{lP})$ and similarly $\ctype(\text{lQ})$ is equal to $\lambda$.
	Finally, note that if we read the labels from bottom to top along $\text{lP}$ and right to left along $\text{lQ}$, each word must be a lattice word, since the columns of $T_1$ and $T_2$ are strictly increasing.  Thus we call a labeled polyomino $(\text{lP},\text{lQ})$ with $\ctype(\text{lP})=\mu$ and $\ctype(\text{lQ})=\lambda$ a lattice labeled polyomino in $\llp_{\lambda,\mu}$  if the word for each path is a lattice word.
	%For instance,
	%
	%is a labeled polyomino. The word of $P$ is the word formed by reading the labels of $P$ bottom to top. Similarly, the word of $Q$ is formed by reading the labels of $Q$ left to right. So in the case above, we would have words
	%$
	%1~2~2 \text{ and } 1~1~2~3~2
	%$
	%for $P$ and $Q$ respectively. 
	%\begin{align*}
	%	\begin{tikzpicture}[scale=.75]
	%	\dyckpathmn{0,0}{6}{3}{1,1,0,0,0,1,0,0,0};
	%	\draw[line width=3pt,black] (0,0) -- (2,0) -- (2,1) -- (5,1) -- (5,2)--(6,2) -- (6,3) ;
	%	\draw (.35,.5) node {$1 $};
	%	\draw (.35,1.5) node {$2$};
	%	\draw (3.35,2.5) node {$2 $};
	%	\draw (1.5,.35) node {$1$};
	%	\draw (2.5,1.35) node {$1 $};
	%	\draw (3.5,1.35) node {$2 $};
	%	\draw (4.5,1.35) node {$3$};
	%	\draw (5.5,2.35) node {$2 $};
	%	\end{tikzpicture}
	%\end{align*}
	% For a word $w = w_1~\dots ~ w_r$ of $P$ (respectively of $Q$), we can interpret each $w_i$ as an instruction for building a tableau. In particular, if $w_i$ is in column $j$ (respectively row $j$), then in a tableau $T$ place a $j$ in row $w_i$.  Note that the result will only be a semistandard Young tableau if $w$ is a lattice word (with the type of $w$ giving the shape, the lattice condition giving the weakly increasing columns, and the strict increasing columns/rows condition in $P$ or $Q$ being required for injectivity).   The words for $P$ and $Q$ above respectively would produce, with the word of $P$, as it is not a lattice word, failing to create a tableau:
	%\begin{align*}
	%\scalebox{.65}{
	%	\begin{tikzpicture}[xscale=-1]
	%	\partitionfr{2,1}
	%	\draw (.5,.5) node {$1 $};
	%	\draw (1.5,.5) node {$1 $};
	%	\draw (.5,1.5) node {$4 $};
	%	\end{tikzpicture}
	%} &&
	%\scalebox{.65}{
	%	\begin{tikzpicture}
	%	\partitionfr{3,2}
	%	\draw (.5,.5) node {$1 $};
	%	\draw (.5,1.5) node {$2 $};
	%	\draw (1.5,.5) node {$2 $};
	%	\draw (1.5,1.5) node {$3 $};
	%	\draw (2.5,.5) node {$2 $};
	%	\end{tikzpicture}
	%}
	%\end{align*}
	%Moreover by construction the shape of the resulting tableau to the type of the word of $P$ (respectively $Q$) and the type of the tableaux reflects the lengths of the columns in $P$ (respectively rows in $Q$)  Reading these of as a composition,  
	%
	%We can use the A word $w = w_1,\dots,w_r$ is Yamanouchi of shape $\lambda$ and content $\mu$ if it produces a semistandard tableau of shape $\lambda$ with content $\mu$ (i.e. the number of $i$'s equals $\mu_i$ ). For partitions $\nu,\eta$ we will call a labeled polyomino Yamanouchi of type $(\nu,\eta)$ if the word of $P$ is Yamanouchi of shape $\eta$, and the word of $Q$ is Yamanouchi of shape $\nu$.  The above example is not a Yamanouchi polyomino since even though its word for $Q$ is Yamanouchi of type $(2,2,1)$, its word for $P$ is not Yamanouchi. 
	In particular we get
	$$
	\sum_{\nu \vdash n}  \sum_{\eta \vdash n} K_{\lambda,\nu} K_{\mu',\eta} \sum_{S \in P_{\nu,\eta}} t^{w(S)}
	=\sum_{P \in \llp_{\lambda',\mu} }t^{\area(P)}.
	$$
	
	\begin{theorem} Let $\lambda$ be a partition. Then
		$$\Delta_{s_{\lambda} }e_n \Big|_{q=1} = \sum_{\mu \vdash n} s_{\mu} \sum_{P \in \llp_{\lambda',\mu} } t^{\area(P)}.$$
	\end{theorem}
	
	% To prove this, we will first expand $ \Delta_{ m_\nu} e_n |_{q=1}$ by constructing a certain set of combinatorial objects. We will then use a weight-preserving, sign-reversing involution to get
	% a set of fixed points which expand this expression in terms of elementary symmetric functions.
	% These fixed points will then be in bijection with a set of polyominoes. The main result will follow from observing that 
	% $$
	% \Delta_{s_\lambda} e_n \Big |_{q=1} =\Delta_{ \sum_{\nu \vdash n} K_{\lambda,\nu} m_\nu} e_n \Big |_{q=1}
	% =\sum_{\nu \vdash n} K_{\lambda,\nu} \Delta_{ m_\nu} e_n \Big |_{q=1}.
	% $$
	It is an interesting open question as to what second statistic on the labeled parallelogram polyominoes gives an interpretation for $\Delta_{s_{\lambda}} e_n$  (i.e.\ not just when $q=1$).  Several of the previous conjectures/theorems in \cite{Delta} or \cite{StatPoly} about $\Delta_{e_k} e_n$  can be seen to be consistent with this conjecture at $q=1$, and thus provide clues in specific cases; we omit the details for this extended abstract. 

	\acknowledgements{The authors wish to express their thanks to Adriano Garsia, who first conjectured that the $\Delta_G$ operator could be explored using polyominoes.}
	
	
	
	
	
	
	\printbibliography
	
	
	
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