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\begin{document}

\newbox\Adr
\setbox\Adr\vbox{
\centerline{\sc Jeffrey Remmel$^1$} 
\vskip18pt
\centerline{\small Department of Mathematics}
\centerline{\small University of California, San Diego}
\centerline{\small La Jolla, CA 92093-0112. USA}
\centerline{\small \texttt{jremmel@ucsd.edu}}
}

\title{The Combinatorics of Macdonald's $D_n^1$ operator} 

\author[Jeffrey Remmel]{\box\Adr}

\subjclass[2000]{05E05 05E10} 

\thanks{$^1$ Partially supported by NSF Grants DMS 0400507
and DMS 0502858}

\dedicatory{Dedicated to Adriano Garsia who taught me the beauty of 
combinatorics and symmetric functions.}

\begin{abstract}
To prove the existence of the Macdonald polynomials $\{P_{\la}(x;q,t)\}_{\la \vdash n}$, Macdonald 
[S\'{e}minaire Lotharingien Combin.\ {\bf20} (1988), Article~B20a;
{\it``Symmetric functions and Hall polynomials"}, 
2nd ed., Clarendon Press, New York, 1995] 
introduced an operator $D^1_n$ and proved that for any  
Schur function $s_\la(x_1, \ldots, x_n)$, $D^1_n(s_\la(x_1, \ldots, x_n)) = 
\sum_{\mu} d_{\la,\mu}(q,t) s_\mu(x_1,\ldots, x_n)$ where the sum runs over all partitions $\mu$ of $n$ which are less than or equal to $\la$ in the dominance order and the $d_{\la,\mu}(q,t)$ are polynomials in $q$ and $t$ with integer coefficients.  We give an explicit combinatorial formula for the $d_{\la,\mu}(q,t)$'s.
\end{abstract}

\maketitle

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\markright{\its S\'eminaire Lotharingien de
Combinatoire \bfs 54A \rms (2007), Article~B54As\hfill}
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\section{Introduction}

In 1988, I.G. Macdonald \cite{Mac} 
introduced a remarkable new basis for the space of  symmetric functions.
The elements of this basis are denoted $P_{\lambda}(\overline{x};q,t)$ 
where  $\la$ is a partition, $\ox = (x_1, \ldots, x_N)$,  and $p,q$ are two free parameters.
The $P_{\lambda}(\overline{x};q,t)$'s,  which are now called ``{\it Macdonald polynomials}'', specialize
to many of the well-known bases for the symmetric functions  
by suitable choices of the parameters $q$ and $t$. In fact, we can obtain in this manner
the Schur functions, the Hall-Littlewood symmetric functions, the Jack symmetric functions, the zonal symmetric functions, 
the zonal spherical functions, and the elementary and 
monomial symmetric functions. 


We will use standard notation from the theory of symmetric functions. 
We let $\Lambda_n(\overline{x}) = \Lambda_n(x_1, x_2, \ldots, x_N)$ 
denote the space of 
homogeneous symmetric functions of degree $n$ where $N \geq n$. Then the  
dimension 
of $\Lambda_n(\overline{x})$ is the number of partitions of $n$ and 
$\Lambda_n(\overline{x})$ has many well known bases including the monomial  symmetric functions  
$\{m_\lambda(\overline{x})\}_\lambda$, the power symmetric functions  
$\{p_\lambda(\overline{x})\}_\lambda$, the elementary symmetric functions 
$\{e_\lambda(\overline{x})\}_\lambda$, and the homogeneous symmetric functions, $\{h_\lambda(\overline{x})\}_\lambda$, where in each case 
$\lambda$ ranges over the set $Ptn(n)$ of partitions of $n$.  Here 
$h_n(\ox) = \sum_{1\leq i_1 \leq \cdots \leq i_n \leq N} 
x_{i_1} \cdots x_{i_n}$, 
$e_n(\ox) = \sum_{1\leq i_1 < \cdots < i_n  \leq N} x_{i_1} \cdots x_{i_n}$, 
and 
$p_n(\ox) = \sum_{1=1}^N x_i^n$. Then for a partition $\lambda = 
(\lambda_1, \ldots, \lambda_k)$, $m_\lambda(\ox)$ is the sum over all monomials
$x_1^{a_1}\cdots x_N^{a_N}$ so that the non-zero terms in $a_1, \ldots, a_N$ can be rearranged to $\lambda$,  
$h_\lambda (\ox) = \prod_{i=1}^k h_{\lambda_i}(\ox)$, 
$e_\lambda (\ox) = \prod_{i=1}^k e_{\lambda_i}(\ox)$, and 
$p_\lambda (\ox) = \prod_{i=1}^k p_{\lambda_i}(\ox)$.


Given two partitions $\la, \mu$, we write $\la \subseteq \mu$ provided the 
Ferrers diagram of $\la$ fits inside the Ferrers diagram of $\mu$.  If 
$\la \subseteq \mu$, we let $|\mu/\la|=|\mu| - |\la|$ and we associate 
$\mu / \la$ with the cells in the Ferrers diagram of $\mu$ that are not in the
Ferrers diagram of $\la$.  The resultant cells are known as the skew shape 
$\mu/\la$.  

A column strict tableau $T$ of shape $\mu / \la$ is a filling of the skew 
shape $\mu / \la$ with positive integers such that the integers weakly  
increase when read from left to right and strictly increase when read from 
bottom to top.  Let $CS(\mu / \la)$ be the set of all column strict tableaux 
of shape $\mu / \la$.  Given $T \in CS(\mu/ \la)$, let $w_i(T)$ be the number 
of occurrences of $i$ in $T$ and let $w(T) = \prod_i x_i^{w_i(T)}$.  
Figure \ref{Gar1} gives an example of a column strict tableau $T$ of shape
$(1,2,3,4,4)/(1,3)$ with $w(T) 
= x_1^3 x_2^2 x_3^3 x_4^2$.

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\markboth{\SMALL JEFFREY REMMEL}{\SMALL THE COMBINATORICS OF MACDONALD'S $D_n^1$ OPERATOR}
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\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.75in]{Gar1.pdf}
%\epsfysize=1.75in \epsffile{Gar1.eps}
$$
\caption{A column strict tableau}
\label{Gar1}
\end{figure}


Then the skew Schur function \label{sschur} $s_{\mu / \la}$ can be defined by
\begin{equation*}
s_{\mu/\la}(x_1,x_2,\dots)=\sum_{T\in CS(\mu/\la)}w(T).
\end{equation*}
When $\la = \emp$, this coincides with the definition of 
$s_\mu(x_1,x_2,\dots)$.  


Macdonald \cite{Mac,Macb} 
introduced a new scalar product $\langle , \rangle_{q,t}$ on 
$\Lambda_n(\ox)$ by declaring that  
\begin{equation}\label{eqn:Mac1}
\langle p_\lambda(\ox), p_\mu(\ox) \rangle = z_{\lambda} \delta_{\lambda,\mu}
\prod_{i=1}^{\ell(\lambda)} \frac{1-q^{\lambda_i}}{1-t^{\lambda_i}}.
\end{equation}
Here we let $\ell(\la)$ denote the number of parts of $\la$ so that 
$\la =(\la_1, \ldots, \la_{\ell(\la)})$. 
Macdonald then proved the existence of a unique family of polynomials 
$\{P_\la(\ox;q,t)\}_{\la \vdash n}$ which is a basis of 
$\Lambda(x_1, \ldots, x_n)$
such that for all partitions $\lambda$ and $\mu$ of $n$, there exists 
rational functions  $\xi_{\lambda,\mu}(q,t)$ so that 
\begin{equation}\label{eqn:Mac2}
P_\lambda(\overline{x};q,t) = s_{\lambda} + \sum_{\mu \vdash n,\mu <_D \lambda} s_{\mu}(\overline{x})
\xi_{\lambda,\mu}(q,t)
\end{equation}
and 
\begin{equation}\label{eqn:Mac3} 
\langle P_\lambda(\overline{x};q,t),P_\mu(\overline{x};q,t) \rangle_{q,t} = 0 \ \text{if } \lambda \neq \mu.  \end{equation} 
Here $<_D$ is the dominance order. That is, given two partitions of 
$n$, $\la = (\la_1 \geq \la_2 \geq \cdots \geq \la_n)$ and 
$\mu = (\mu_1 \geq \mu_2 \geq \cdots \geq \mu_n)$, we define 
$\la \geq_D \mu$ if and only if for $i =1, \ldots, n$, 
$\sum_{j=1}^i \la_j \geq \sum_{j=1}^i \mu_i$. 

In general, conditions (\ref{eqn:Mac2}) and (\ref{eqn:Mac3}) overdetermine the family $\{P_\la(\ox;q,t)\}_{\la \vdash n}$ so the proof of existence is non-trivial. 
To prove the existence of the family of polynomials $\{P_\la(\ox;q,t)\}_{\la \vdash n}$,
Macdonald introduced a very useful operator 
$D^1_n$ which is defined as follows. 
Let $[n]= \{1,\ldots,n\}$. Then given a polynomial $P(x_1, \ldots, x_n)$, let 
\begin{equation}\label{Tqs}
T^{(s)}_q P(x_1, \ldots, x_n) = 
P(x_1, \ldots,x_{s-1},qx_s,x_{s+1}, \ldots, x_n)
\end{equation}
and
\begin{align}\notag
D^1_n P(x_1,\ldots,x_n) &= \sum_{s=1}^n \prod_{i \in [n]-\{s\}}
\frac{tx_s -x_i}{x_s-x_i} T^{(s)}_q P(x_1, \ldots,x_n)\\
&= \sum_{s=1}^n 
\frac{T_t^{(s)} \Delta(x_1,\ldots,x_n)} {\Delta(x_1,\ldots,x_n)} 
T^{(s)}_q P(x_1, \ldots,x_n) \label{D1}
\end{align}
where $\Delta(x_1, \ldots, x_n) = \prod_{1 \leq i < j \leq n}(x_i-x_j)$ is the 
Vandermonde determinant. 
Macdonald proved the following facts about the operator $D^1_n$. Given 
any sequence of integers $p= (p_1,\ldots, p_n)$, let $\omega_p(q,t) = 
\sum_{i=1}^n t^{n-i}q^{p_i}$. For any partition $\la = (\la_1 \geq \la_2 \geq \cdots \geq \la_n)$ we shall write $\la(p) = \la$ if the decreasing rearrangement of $p$ equals $\la$. 
Then Macdonald proved that for all partitions $\la$ of $n$, 
\begin{equation}\label{eqn:D2}
D^1_n (m_\la(x_1, \ldots, x_n))= \omega_\la(q,t)m_\la(x_1, \ldots, x_n) + 
\sum_{\mu <_D \la} m_\mu \sum_{p =(p_1, \ldots,p_n), \la(p) = \la} \epsilon_\mu(p) \omega_p(q,t)
\end{equation}
where $\epsilon_\mu(p) =sgn(\sigma)$ if there is a permutation $\sigma$ such 
that $p_{\sigma_i}+n-i =\mu_i +n-i$ for $i =1, \ldots, n$ and 
$\epsilon_\mu(p) = 0$ if there is no such $\sigma$. It immediately 
follows that for all partitions $\la$ of $n$
\begin{equation}\label{fund}
D^1_n(s_\la(x_1, \ldots, x_n)) = \omega_\la(q,t) s_\la(x_1, \ldots, x_n) + 
\sum_{\mu <_D \la} d_{\la,\mu}(q,t) s_\mu (x_1, \ldots, x_n)
\end{equation} for some  polynomials $d_{\la,\mu}(q,t)$.
In fact, Macdonald proved that if  
\begin{equation}\label{zla}
z_\la(x) = \prod_{\mu \in Ptn(n) - \{\la\}} \frac{x-\omega_{\mu}(q,t)}{\omega_\la(q,t) - \omega_\mu(q,t)},
\end{equation}
then
\begin{equation}\label{eqn:D3}
P_\la(x_1, \ldots, x_n;q,t) = z_\la(D^1_n)s_\la(x_1,\ldots,x_n).
\end{equation}

The main goal of this paper is to find a combinatorial interpretation 
for the polynomials $d_{\la,\mu}(q,t)$. Our combinatorial interpretation of  
$d_{\la,\mu}(q,t)$ will be defined as 
a product of weights of certain rim hooks 
associated to $\mu$ which depend on $\lambda$, see 
Theorem~\ref{main}. Given a Ferrers diagram 
$\la$, a {\em rim hook} $h$ of $\la$ is a consecutive sequence of cells along the northeast boundary of $\la$ such that any two consecutive cells of $h$ share an edge and the removal of cells of $h$ from $\lambda$ results in a Ferrers diagram corresponding to another partition. A {\em broken rim hook} (BRH) of $\la$ is a union of rim hooks of $\la$. See Figure \ref{brim}. 

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=2.75in]{brim.pdf}
$$
\caption{A broken rim hook}
\label{brim}
\end{figure}




This given, the first step in finding our combinatorial interpretation for 
the $d_{\la,\mu}(q,t)$ is to use a $\Lambda$-ring expression for 
$D_n^1$ due to 
Garsia and Haiman \cite{GH} to prove that  
\begin{multline}\label{comdef1}
D_n^1(s_\la(x_1, \ldots, x_n)) = \frac{s_\la(x_1, \ldots, x_n)}{1-t} \\
+\frac{t^{|\la|}}{t-1}  
\underset{\gamma/\mu \ \text{is a BRH of}\  \gamma}
{\sum_{ \la/\mu \ \text{is a BRH of}\  \la }} w_r(\la/\mu) w_b(\gamma/\mu) s_\gamma(x_1, \ldots, x_n) 
\end{multline}
where the sum runs over all $\mu$ and $\gamma$ such that 
$|\la/\mu| = |\gamma/\mu|$ and, for $u \in \{r,b\}$,
\begin{equation}\label{comdef2}
w_u(\la/\mu) = \prod_{s \in \la/\mu} w_{u,\la/\mu}(s)
\end{equation}
where 
\begin{equation}\label{r-weight}
w_{r,\la/\mu}(s) = \begin{cases} \frac{q}{t} & \text{if $s$ has a square in $\la/\mu$ to its right},\\
\frac{-1}{t} & \text{if $s$ has a square in $\la/\mu$ below it},\\
\frac{(q-1)}{t} & \text{otherwise}
\end{cases}
\end{equation}
and
\begin{equation}\label{a-weight}
w_{b,\la/\mu}(s) = \begin{cases} t & \text{if $s$ has a square in $\la/\mu$ to its right},\\
-1  & \text{if $s$ has a square in $\la/\mu$ below it},\\
(t-1) & \text{otherwise}.
\end{cases}
\end{equation}  
The next step in the evaluation of the right-hand side of (\ref{comdef1}) is to give combinatorial 
proofs of the following two facts:
\begin{enumerate}
\item $d_{\la,\la}(q,t) = \sum_{i=1}^n t^{n-i}q^{\la_i}$ and 
\item if $\la \neq \gamma$ and $d_{\la,\gamma}(q,t) \neq 0$, then $\gamma$ satisfies the following condition:

\bigskip
\noindent
(*) $\gamma$ arises from $\la$ by first removing a broken rim hook $R$ 
of $\la$ from $\lambda$ to get a partition $\mu$ and then 
adding a broken rim hook 
$B$ of $\gamma$ on the outside of $\mu$ to obtain obtain $\gamma$ so that 
rows occupied by $B$ lie strictly above the rows occupied by $R$ and 
$|B| =|R|$.
\end{enumerate}

Note that any $\gamma$ that satisfies condition (*) is automatically dominated 
by $\lambda$. If $\gamma$ satisfies condition (*), then our combinatorial 
interpretation for $d_{\la,\gamma}(q,t)$ is best described by example. 
Suppose 
$$
\la = (1,3,4,5^2,6,7,8^2,11^2,13^4,16^3,17,18) $$
and
$$
\gamma = (1,5^4,8^4,11^3,12^2,13,15^2,16,17,18),
$$
 see Figure \ref{gamma}. 
We shall think of the cells of the broken rim hook $R$ as being colored red 
so we will put an $r$ in the cells corresponding to $R$. We shall think of 
the cells of the broken rim hook $B$ as being colored blue so that we will put a $b$ in the cells corresponding to $B$. Thus in 
Figure \ref{gamma}, $\la$ corresponds to all the cells except the blue 
cells and $\gamma$ corresponds to all the cells except the red cells. As required by condition (*), $|R| =|B|$ and all the blue cells lie strictly above any red cells. If we consider the 
right-hand side of (\ref{comdef1}), it is clear 
that the red and blue cells completely determine $\la$ and $\gamma$, but there are many $\mu$'s in the sum that could give rise to the same 
$\la$ and $\gamma$. That is, if we consider the rim hooks which form the connecting pieces between colored cells, it could be that some of these cells were originally removed from $\la$ to form $\mu$ and then added back 
to form $\gamma$. Thus in Figure \ref{gamma}, we know what weight to assign any blue square which has a blue square to its right or below it and, similarly, 
we know what weight to assign any red square which has a red square to its right or below it. However, we do not know how to assign a weight to a blue or 
red square that is at the top of one of the rim hooks which connect colored cells because the next square could have been part of a $\mu$ that appears on the right-hand side of 
(\ref{comdef1}) or not. Thus for each such 
connecting rim hook, we have to consider the sum of the weights 
over all possible choices of which cells in that rim hook are part of the corresponding $\mu$ on the right-hand side of (\ref{comdef1}). 
In Figure \ref{gamma}, we have pictured the possible connecting rim hooks 
to the right of the diagram and labeled these connecting rim hooks from 
top to bottom with the numbers 1 through 5.  We shall make the convention that the top colored square, if any, is part of the connecting rim hook while the bottom colored square, if it exists, is not part of the connecting rim hook. Nevertheless, we shall always draw a connecting rim hook with its colored square at the bottom, if it exists, because the ultimate contribution of that connecting rim hook to the coefficient of $s_\gamma(x_1,\ldots,x_n)$ on the right-hand side of (\ref{comdef1}) depends on the color of the bottom square. Thus in Figure \ref{gamma}, 
we shall think of the connecting rim hook 1 as blank-blue connecting rim hook, 
the connecting rim hook 2 as blue-blue connecting rim hook, the connecting rim hook 3 as blue-red connecting rim hook, the connecting rim hook 4 as red-red  connecting rim hook, and the connecting rim hook 5 as red-blank connecting rim hook. These are all the possible connecting rim hooks that can arise from a $\gamma$ that satisfies condition (*). 

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=2.25in]{gamma.pdf}
$$
\caption{Connecting rim hooks for $\gamma$}
\label{gamma}
\end{figure}



This given, we shall show that for $\gamma \neq \la$, 
$d_{\la,\gamma}(q,t)$ is equal to 
$\frac{t^{|\la|}}{t-1}$ times the product of the weights assigned to the connecting rim hooks  times the product of the weights of all the blue squares which are not part of any connecting rim hook according their weights relative to the weight function 
$w_b$ times  
the product of the weights of all the red squares which are not part of any connecting rim hook according their weights relative to the weight 
function $w_r$. The weight 
$W(H)$ of a connecting rim hook $H$ is obtained by summing the weights of its squares over all possible $\mu$ that give rise to $\la$ and $\gamma$ in equation (\ref{comdef1}). We shall show that we can find a closed expression for such a sum for each type of connecting rim hook. These closed expressions are listed below. Given any connecting rim hook 
$h$, we let $r(h)$ and $c(h)$ denote the  number of rows and columns of $h$ respectively. Again we emphasize that the bottom colored square, if any, is not considered part of the connecting rim hook. Thus for example, the first rim 
hook $h_1$ in Figure \ref{gamma} has $r(h_1) =2$ and $c(h_1) =3$. Similarly, 
if $h_2$ is the second connecting rim hook in Figure \ref{gamma}, then 
$r(h_2)=4$ and $c(h_2)=2$. For each non-negative integer $n$, we let $[n]_t = \frac{1-t^n}{1-t} = 1+t + \cdots + t^{n-1}$.

\bigskip
\noindent
{\bf Weights of connecting rim hooks}
\begin{enumerate}
\item If $H$ is a blank-blue connecting rim hook, then $W(H)=q^{c(H)}$. 

\item If $H$ is a blue-blue connecting rim hook, then $W(H)= (t-q)q^{c(H)-1}$.

\item If $H$ is a blue-red connecting rim hook, then 
\begin{equation*}
W(H) = \frac{(t-1)(q-t)}{t(q-1)}\left( 
\frac{[r(H) -1]_t -q[r(H)-2]_t}{t^{r(H) -2}} -q^{c(H)} - (q-t)
\sum_{i=1}^{\ell(\beta)} \frac{q^{\beta_i} -1}{t^i}\right),
\end{equation*}
where $\beta$ is the partition whose diagram is obtained by removing the first 
row and column of the smallest shape $\alpha$ which contains the connecting rim hook $H$, see Figure \ref{gamma}. Moreover, we assume that the sum 
$\sum_{i=1}^{\ell(\beta)} \frac{q^{\beta_i} -1}{t^i}$ is equal to 0 if $\beta$ is empty.

\item If $H$ is a red-red connecting rim hook, then $W(H)= \frac{(q-t)}{t^{r(H)}}$.

\item If $H$ is a red-blank connecting rim hook, then $W(H)= \frac{(q-1)}{t^{r(H)}}$.
\end{enumerate}

For example, if we let $H_i$ denote the connecting rim hook numbered by $i$ in 
Figure \ref{gamma}, then
\begin{align*}
 W(H_1) &= q^3, \\
 W(H_2) &= (t-q)q, \\
 W(H_3) &= \frac{(t-1)(q-t)}{t(q-1)}\left(\frac{[5]_t-q[4]_t}{t^4} - q^4 -
(q-t)(\frac{q^3-1}{t} + \frac{q^3-1}{t^2})\right),\\
W(H_4) &= \frac{q-t}{t^3}, \quad  \text{and} \\
 W(H_5) &= \frac{q-1}{t^4}.
\end{align*}
Now the weights of the blue squares not in any connecting rim hook reading 
from top to bottom in Figure \ref{gamma} are $t$, $-1$, $t$, and $-1$ respectively, and the weights of the red squares not in any connecting rim hook reading 
from top to bottom in Figure \ref{gamma} are $q/t$, $1/t$, $1/t$, and $1/t$, respectively. Thus for our example, 
$d_{\la,\gamma}(q,t) = 
\frac{t^{|\la|}}{t-1} \frac{q}{t^2} W(H_1) \cdots W(H_5)$.
We should note that if $\gamma \neq \la$, then there must always  
be a blue-red connecting rim hook $H$ for $\gamma$. Thus the factor of $(t-1)$ 
which occurs in $W(H)$ will always cancel the factor of $t-1$ which appears 
in $\frac{t^{|\la|}}{t-1}$. 
In addition, there will either be a red-blank connecting rim hook or a bottom red square which will contribute a factor of $q-1$ to 
cancel the $q-1$ that appears in the denominator of $W(H)$. 
Then it is not difficult to see that the factor $t^{|\la|}$ is enough to cancel all the powers of $t$ that occur in the denominators of 
any of the weights of the connecting rim hooks or isolated blue and red squares. Thus our result will show that $d_{\la,\gamma}(q,t)$ is always just a polynomial in $t$ and $q$. 




\section{$\la$-Ring Notation}
\label{sec:lambda ring}

Let $A$ be a set of formal commuting variables and $A^*$ denote 
the set of all words over  
$A$.  The empty word will be identified with ``1''.  Let $c \in \mathbb{C}$, 
$\gm = (\gm_1, \dots, \gm_\ell) \vdash n$, $x=a_1 a_2 \dots a_i$ be any word 
in $A^*$, and $X,X_1,X_2, \dots$ be any sequence of formal sums of the words 
in $A^*$ with complex coefficients. We define 
 the $\la$-ring notation for the power 
symmetric functions by 
\begin{align*}
p_r[0] & = 0, & 
p_r[1] & = 1, \\ 
p_r[x] & = x^r = a_1^r a_2^r \dots a_i^r, & 
p_r[c \, X] & = c \, p_r[X], \\
p_r \left[ \sum_i X_i \right] & = \sum_i p_r[X_i], & 
p_\gm [X] & = p_{\gm_1}[X] \cdots p_{\gm_{\ell}} [X],
\end{align*}
where $r$ is a nonnegative integer.  These definitions imply that $p_r[X X_1] 
= p_r[X] p_r[X_1]$ and therefore $p_\gm [X X_1] = p_\gm[X] p_\gm[X_1]$.  These
definitions also imply that for any complex number $c$  and $\gm \vdash n$, 
$p_\gm[c X] = c^{\ell(\gm)} p_\gm[X]$.  

When $X=x_1+\dots+x_N$, then our definitions ensure that 
\begin{equation*}
p_k[X] = \sum_{i=1}^N x_i^k,
\end{equation*} 
which is the usual power symmetric function $p_k(x_1, \ldots, x_N)$.  
Furthermore, for any partition 
$\lambda = (\lambda_1, \ldots, \lambda_{\ell})$,  
\begin{equation*}
p_\lambda[X] = p_\lambda(x_1, \ldots, x_N).
\end{equation*}

The power symmetric functions are a basis for the ring of symmetric functions,
so if $Q$ is a symmetric function, then there are unique coefficients $a_\la$ 
such that $Q = \sum_\la a_\la p_\la$.  We then define 
 $Q[X]= \sum_\la a_\la p_\la[X]$.
It follows that in the special case where $X=x_1+\dots+x_N$ is a sum of 
letters in $A$, $Q[X]$ is simply the symmetric function $Q(x_1,\dots,x_N)$.  
We note that if $X=x_1+x_2+\cdots$ as an infinite sum of letters, the same 
reasoning will show that for any symmetric function $Q$, $Q[X] = Q(x_1,x_2, \ldots )$. 

Define  
\begin{equation}
\Omega[A] = e^{\sum_{k\geq 1} \frac{p_k[A]}{k}}.
\end{equation}
Next we state two well known theorems from the theory of $\la$-rings, 
see \cite{Kn} or \cite{MRW}.

\begin{thm} 
\label{thm:lambda ring} 
For $X,Y$ formal sums of words in $A^*$ with complex coefficients,
\begin{align}
s_{\mu/\la}[X+Y] & = 
\sum_{\la \subseteq \dl \subseteq \mu} s_{\mu/\dl}[X] s_{\dl/\la}[Y],
\label{eqn:X+Y} \\ 
s_{\mu/\la}[-X] & = (-1)^{|\mu'/\la'|} s_{\mu'/\la'}[X], \label{eqn:-X}\\
s_{\mu/\la}[X-Y] & = 
\sum_{\la \subseteq \dl \subseteq \mu} s_{\mu/\dl}[X] (-1)^{|\dl/\la|}
s_{\dl'/\la'}[Y].\label{eqn:X-Y}
\end{align}
where $\la'$ denotes the conjugate partition of  $\la$. 
\end{thm}


\begin{thm}\label{thm:omega}
 If $X = x_1 + x_2 + \cdots$ and $Y = y_1 + y_2+ \cdots$, then  
\begin{align}
\Omega[X+Y ] &= \Omega[X]\Omega[Y],\\
\Omega[X] &= \prod_i \frac{1}{1-x_i} = \sum_{n\geq 0} h_n[X],\\
\Omega[-X] &= \prod_i (1-x_i) = \sum_{n\geq 0} (-1)^n e_n[X].
\end{align}
\end{thm}

Garsia and Haiman \cite{GH} proved the following $\lambda$-ring formulation 
of the operator $D_1^n$. 
\begin{thm}\label{thm:GH}
If $X = x_1 + x_2 + \cdots + x_n$, then for any symmetric polynomial $Q[X]$,
\begin{equation}\label{eq:GH1}
D_n^1(Q[X]) = \frac{Q[X]}{1-t} + \frac{t^n}{t-1}Q\left[X-\frac{1-q}{tz}\right]\Omega[(t-1)Xz]\bigg|_{z^0}.
\end{equation}
\end{thm}

For the rest of this paper, we shall let $X = x_1 + x_2 + \cdots + x_n$. Then 
for any Schur function $s_\la(x_1,\ldots,x_n) = s_\la[X]$ and  we have 
by Theorems~(\ref{thm:lambda ring}) and (\ref{thm:omega}) that 
\begin{align}\nonumber
D_n^1(s_\la[X]) &= \frac{s_\la[X]}{1-t} + \frac{t^n}{t-1}s_\la\left[X-\frac{1-q}{tz}\right]\Omega[(t-1)Xz]\bigg|_{z^0} \\
&= \frac{s_\la[X]}{1-t} + \frac{t^n}{t-1}\left(\sum_{\mu \subseteq \la}s_\mu[X] 
s_{\la/\mu}\left[\frac{q-1}{tz}\right]\right)
\Omega[tXz]\Omega[ -Xz]\Bigg|_{z^0}  \nonumber \\
&= \frac{s_\la[X]}{1-t} + \frac{t^n}{t-1}\left(\sum_{\mu \subseteq \la}s_\mu[X] 
\left(\frac{1}{tz}\right)^{|\la/\mu|}s_{\la/\mu}[q-1]\right) \nonumber \\
&\kern2cm \times\left(\sum_{n\geq 0} h_n[X](tz)^n\right) \left(\sum_{n\geq 0} e_n[X](-z)^n\right)\Bigg|_{z^0} \nonumber \\
&= \frac{s_\la[X]}{1-t} + \frac{t^n}{t-1}\Bigg(\sum_{\mu \subseteq \la}s_\mu[X] 
\left(\frac{1}{t}\right)^{|\la/\mu|}s_{\la/\mu}[q-1] \nonumber \\
&\kern4cm \times\sum_{k=0}^{|\la/\mu|} t^k h_k[X] (-1)^{|\la/\mu| -k}e_{|\la/\mu| -k}[X]\Bigg). 
\label{eq:GH2}
\end{align}

By (\ref{eqn:X-Y}),
\begin{equation}
s_{\la/\mu}[q-1] = \sum_{\mu \subseteq \beta \subseteq \la} 
s_{\beta/\mu}[q](-1)^{|\la/\beta|}s_{\la'/\beta'}[1].
\end{equation}
A skew diagram is a {\em horizontal strip} (respectively {\em vertical   strip}) 
if it contains no two cells in the same column (row). It is 
easy to see that $s_{\beta/\mu}[q] = 0$ unless $\beta/\mu$ is a horizontal 
strip. If $\beta/\mu$
is a horizontal strip, then $s_{\beta/\mu}[q] = q^{|\beta/\mu|}$. Similarly, 
$s_{\la'/\beta'}[1] = 0$ unless $\la/\beta$ is a vertical strip. 
If $\la/\beta$ is a vertical strip, then $s_{\la/\beta}[1] = 1$. 
This means that if $\mu$ arises from $\la$ by first removing a vertical 
strip to obtain $\beta$ and then removing a horizontal strip to obtain 
$\mu$, then in the last summand of (\ref{eq:GH2}), $s_\mu$ would 
come with a weight $\frac{1}{t^{|\la/\mu|}}(-1)^{|\la/\beta|}q^{|\beta/\mu|}$. 
We can picture this situation as in Figure \ref{GV1} by placing a $-1$ in each 
cell of $\la/\beta$ and $q$ in each cell of $\beta/\mu$. It easy to see 
that in such situation, $\la/\mu$ is just a broken rim hook (BRH) of $\la$ 
and that 
\begin{description}
\item[(i)] if $s$ is a cell which has a cell in $\la/\mu$ to its right, 
then $s$ must have a $q$ in it, 
\item[(ii)] if $s$ is a cell which has a cell in $\la/\mu$ to below it, 
then $s$ must have a $-1$ in it, and 
\item[(iii)] if $s$ is a cell which has neither a cell in $\la/\mu$ to its right or below it, i.e. if $s$ is the lowest cell in one of the rim hooks of 
$\la/\mu$, then $s$ could have either a $q$ or a $-1$ in it. 
\end{description}



\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=2.25in]{GV1.pdf}
$$
\caption{A term from $s_{\la/\mu}[q-1]$}
\label{GV1}
\end{figure}



It follows that 
\begin{multline}\label{eq:GH3}
D_n^1(s_\la[X]) = \frac{s_\la[X]}{1-t} \\+ 
\frac{t^n}{t-1}
\left(\sum_{\mu \subseteq \la, \la/\mu \ \text{is a BRH of $\la$}} 
w_r(\la/\mu) s_\mu[X]
\sum_{k=0}^{|\la/\mu|} t^k h_k[X] (-1)^{|\la/\mu| -k} e_{|\la/\mu| -k}[X]\right). 
\end{multline} 
Here 
\begin{equation}\label{r-wieght}
w_r(\la/\mu) = \prod_{s \in \la/\mu} w_{r,\la/\mu}(s)
\end{equation}
where for each cell $s \in \la/\mu$, 
\begin{equation}\label{eq:GH4}
w_{r,\la/\mu}(s)= \begin{cases} \frac{q}{t} & \text{if $s$ has a cell of 
$\la/\mu$ to its right,} \\
\frac{-1}{t} & \text{if $s$ has a cell of 
$\la/\mu$ below it, and} \\
\frac{q-1}{t} & \text{if $s$ is lowest cell of a rim hook in $\la/\mu$}.
\end{cases}
\end{equation}

Next we consider a term of the form 
\begin{equation}\label{eq:GH5}
s_\mu[X]\left(\sum_{k=0}^{|\la/\mu|} t^k h_k[X] (-1)^{|\la/\mu| -k}
e_{|\la/\mu| -k}[X]\right).
\end{equation}
The Pieri rules state that 
\begin{equation}
s_\mu[X]h_k[X] = \sum_{\mu \subseteq \beta, \beta/\mu \ \text{is a horizontal strip of size $k$}} s_\beta \label{eq:GH6} 
\end{equation}
and
\begin{equation}
s_\mu[X]e_k[X] = \sum_{\mu \subseteq \beta, \beta/\mu \ \text{is a vertical strip of size $k$}} s_\beta. \label{eq:GH7}
\end{equation}
It follows that if $\gamma$ arises from $\mu$ by first adding 
a horizontal strip of size $k$ to the outside of $\mu$ to get a shape 
$\alpha$ and then adding of vertical strip of size $|\la/\mu| -k$ on 
the outside of $\alpha$ to get $\gamma$, then $s_\gamma$ will appear in 
the product $s_\mu[X] t^k h_k[X] (-1)^{|\la/\mu| -k}
e_{|\la/\mu| -k}[X])$ with weight $t^k(-1)^{|\la/\mu| -k}$. We can 
picture this situation in Figure \ref{GV2} where we place a $t$ in 
each square of the horizontal strip and $-1$ is each cell of the vertical strip. It easy to see 
that in such situation that $\gamma/\mu$ is just a broken rim hook 
(BRH) of $\gamma$ and that  
\begin{description}
\item[(a)] if $s$ is a cell which has a cell in $\gamma/\mu$ to its right, 
then $s$ must have a $t$ in it, 
\item[(b)] if $s$ is a cell which has a cell in $\gamma/\mu$ below it, 
then $s$ must have a $-1$ in it, and 
\item[(c)] if $s$ is a cell which has neither a cell in $\gamma/\mu$ to its right or below it, i.e. if $s$ is the lowest cell in one of the rim hooks of 
$\gamma/\mu$, then $s$ could have either a $t$ or a $-1$ in it. 
\end{description}

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=2.25in]{GV2.pdf}
$$
\caption{A term from $s_\mu[X] t^k h_k[X] (-1)^{|\la/\mu| -k}
e_{|\la/\mu| -k}[X])$}
\label{GV2}
\end{figure}


It follows that 
\begin{equation}\label{eq:GH8}
s_\mu[X]\left(\sum_{k=0}^{|\la/\mu|} t^k h_k[X] (-1)^{|\la/\mu| -k}
e_{|\la/\mu| -k}[X]\right) =
\sum_{\mu \subseteq \gamma, \gamma/\mu \ \text{is a BRH of $\gamma$ of size 
$|\la/\mu|$}} 
 w_b(\gamma/\mu) s_\gamma[X] 
\end{equation}
where 
\begin{equation}\label{b-wieght}
w_b(\gamma/\mu) = \prod_{s \in \la/\mu} w_{b,\la/\mu}(s)
\end{equation}
and for each cell $s \in \gamma/\mu$, 
\begin{equation}\label{eq:GH9}
w_{b,\la/\mu}(s)= \begin{cases} t  & \text{if $s$ has a cell of 
$\gamma/\mu$ to its right,} \\
-1  & \text{if $s$ has a cell of 
$\gamma/\mu$ below it, and} \\
t-1 & \text{if $s$ is lowest cell of a rim hook in $\gamma/\mu$}.
\end{cases}
\end{equation}
Thus combining (\ref{eq:GH3}) and (\ref{eq:GH8}), we see that 
\begin{equation}
D_n^1(s_\la[X]) =  \frac{s_\la[X]}{1-t}  +
\frac{t^n}{t-1}
\left(
\underset{\mu \subseteq \gamma, \gamma/\mu \ \text{is a BRH of $\gamma$ of 
size $|\la/\mu|$}}
{\sum_{\mu \subseteq \la, \la/\mu \ \text{is a BRH of $\la$}}} 
w_r(\la/\mu) w_b(\gamma/\mu)s_\gamma[X]\right).
\label{eq:GH10}
\end{equation} 

Each $\gamma$ that appears on the right-hand side of (\ref{eq:GH10}) can be 
constructed  by 
first removing a broken rim hook $\la/\mu$ from $\la$ and then 
adding back a broken rim hook $\gamma/\mu$ of size $|\la/\mu|$ 
on the outside of $\mu$. We shall think of the cells of $\la/\mu$ as 
being colored red and the cells of $\gamma/\mu$ as being colored blue. 
This is pictured in Figure \ref{GV3} by putting an $r$ in the cells 
of $\la/\mu$ and putting a $b$ in the cells of $\gamma/\mu$. In 
this case we see that $\la = (2^2, 6^3,9^2,10^2,14^3 ,16,17)$, 
$\mu = (1,4,5^3,9,10^2,11, 14^2, 15,17)$, and 
$\gamma = (1,6^3,10^4, 14^3,17^2)$. 


\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=2.25in]{GV3.pdf}
$$
\caption{A term from the sum in (\ref{eq:GH10})}
\label{GV3}
\end{figure}




We should note however that we cannot determine $\mu$ from 
$\la$ and $\gamma$. That is, in Figure \ref{GV3}, we can determine 
the cells that are colored with just red and the cells that are colored 
with just blue from $\la$ and $\gamma$.  However $\la$ and $\gamma$ do not 
give us enough information to distinguish a cell which 
is colored both red and blue from a cell that is blank. That is, a cell 
that is colored both red and blue represents a cell that we first removed 
in the process of forming $\mu$ but then added back in the process of 
forming $\gamma$. Such a cell must lie on the outer boundary of $\gamma$. 
However, there may be other cells that are on the outer boundary of 
$\gamma$ that have no color and hence are not a part of $\mu$.   For example, 
suppose $\la = (2^3,4)$ and $\gamma = (1,2^2,5)$, then we know that 
cell $(2,4)$ must be red and cell $(5,1)$ must be blue. However there 
are various possibilities of 
for $\mu$ which are pictured in Figure \ref{GV4}. For the rim 
hook $H$ of $(2^3,5)$ which includes the  cell (2,4) plus all cells on the boundary 
between cell (2,4) and cell (5,1), there are 10 possible ways to color the 
remaining squares of $H$ with either red-blue or 
blank so that the shape $\mu$ that results by removing all the squares which are colored red, blue, or red-blue from $\la$ corresponds to a partition.


\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=2.75in]{GV4.pdf}
$$
\caption{Possible $\mu$ for $\la = (2^3,4)$ and $\gamma = (1,2^2,5)$}
\label{GV4}
\end{figure}





Thus the final weight associated with $s_\gamma$ in (\ref{eq:GH10}) in 
the example of Figure \ref{GV4} is  
the sum $w_r(\la/\mu)w_b(\gamma/\mu)$ over all such $\mu$. 
In our case, if we number the ten diagrams with the numbers 1 through 10 reading from left to right and 
then from top to bottom, then table 1 gives the 
corresponding weights of $w_r(\la/\mu)$, $w_b(\gamma/\mu)$ and 
$w_r(\la/\mu)w_b(\gamma/\mu)$ for each of the 10 diagrams. 




\begin{center}
\begin{tabular}{| l | l | l | l |}
\hline {\bf diagram}  & $w_r(\la/\mu)$ & $w_b(\gamma/\mu)$ &  $w_r(\la/\mu)w_b(\gamma/\mu)$ \\
\hline (1) & $\frac{(q-1)}{t}$ & $(t-1)$ & $\frac{(q-1)(t-1)}{t}$ \\
\hline (2) & $\frac{-(q-1)}{t^2}$ & $(t-1)^2$ & $\frac{-(q-1)(t-1)^2}{t^2}$ \\
\hline (3) & $\frac{(q-1)}{t^3}$ & $-(t-1)^2$ & $\frac{(q-1)(t-1)^2}{t^3}$ \\
\hline (4) & $\frac{(q-1)^2}{t^2}$ & $t(t-1)$ & $\frac{(q-1)^2t(t-1)}{t^2}$ \\
\hline (5) & $\frac{-(q-1)^2}{t^3}$ & $t(t-1)^2$ & $\frac{-(q-1)^2t(t-1)^2}{t^3}$ \\
\hline (6) & $\frac{(q-1)^2}{t^4}$ & $-t(t-1)^2$ & $\frac{-(q-1)^2t(t-1)^2}{t^4}$ \\
\hline (7) & $\frac{q(q-1)^2}{t^3}$ & $t^2(t-1)$ & $\frac{q(q-1)^2t^2(t-1)}{t^3}$ \\
\hline (8) & $\frac{-q(q-1)^2}{t^4}$ & $t^2(t-1)^2$ & $\frac{-q(q-1)^2t^2(t-1)^2}{t^4}$ \\
\hline (9) & $\frac{q(q-1)^2}{t^5}$ & $-t^(t-1)^2$ & $\frac{-q(q-1)^2t^2(t-1)^2}{t^5}$ \\
\hline (10) & $-\frac{q^2(q-1)}{t^6}$ & $t^3(t-1)$ & $\frac{-q^2(q-1)t^3(t-1)}{t^6}$ \\
\hline
\end{tabular}\\[18pt]
Table 1 \end{center}


Now the weight associated to $s_\gamma[X]$ in (\ref{eq:GH10}) would be the sum of all 
the polynomials in the last column. It is also easy to see that the bottom blue 
cell contributes a factor of $(t-1)$ to $w_b(\gamma/\mu)$, regardless of 
what $\mu$ is, since it will never have 
cell to its right or below it in $\gamma/\mu$. 
In this case, there is only one connecting rim hook, namely $H$.
Since $H$ does not include cell (5,1), it follows that we can associate a weight $W(H)$ by summing over 
all the terms in column four of the table and dividing by $(t-1)$.
In this way, we can associate a final weight $W(H)$ with each connecting 
rim hook $H$ of $\gamma$. In fact, in general there will 9 types of connecting rim 
hooks associated an $s_\gamma[X]$ that appears on the right-hand side of (\ref{eq:GH10}) 
depending on the color of the top and bottom squares of the connecting 
rim hooks. These are pictured in Figure \ref{nine}.

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=2.75in]{nine.pdf}
$$
\caption{Possible connecting rim hooks}
\label{nine}
\end{figure}



Now suppose that  relative to $\la$, the connecting rim hooks of $\gamma$ are 
$H_1, \ldots, H_a$, the red cells which are not part of any 
connecting rim hooks are 
$r_1, \ldots, r_b$ and blue cells which are not part of any connecting rim hooks are 
$b_1,\ldots,b_c$. Then it is easy to see 
that since the color of the cells which lie either immediately to the right of directly below each $r_i$ and $b_j$ is 
completely determined by $\la$ and $\gamma$, 
 the weights of $w_{r,\la/\mu}(r_i)$ and $w_{b,\gamma/\mu}(b_j)$ will be 
always be the same for all $\mu$. Thus the total weight associated with $s_\gamma[X]$ in 
(\ref{eq:GH10}) can be factored as  
$$\left(\prod_{s=1}^a W(H_i)\right) \left(\prod_{i=1}^b w_{r,\la/\mu}(r_i)\right) 
\left(\prod_{j=1}^c w_{b,\gamma/\mu}(b_j)\right).$$




\section{The leading coefficient}
In this section, we shall directly compute the coefficient of $s_\la$ in 
$D^1_n(s_\la)$.  To do this, we shall need to compute $W(H)$ when $H$ is 
a blank-blank connecting rim hook. Indeed the only $s_\gamma$ that 
appears in (\ref{eq:GH10}) that involves 
a blank-blank connecting rim hook is when $\gamma = \la$. We then have the following result. 

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{border.pdf}
$$
\caption{The connecting rim hook $B_{\la}$}
\label{border}
\end{figure}



\begin{thm}\label{blank-blank}
Let $\la = (\la_1 \geq \cdots \geq \la_{\ell(\al)})$ be a partition of $n$ and let 
$B_{\la}$ be the outside border of $\la$ as pictured in Figure \ref{border}. Then 
\begin{equation}\label{eq:bl-bl}
W(B_\la) = 1+(t-1) \sum_{i=1}^{\ell(\la)} \frac{q^{\la_i} -1}{t^i}.
\end{equation}
\end{thm}
\begin{proof}

We proceed by induction on the number of non-zero parts of $\la$. 
Let $BRH(B_\la)$ denote the set of broken rim hooks in $B_\la$. 
Note that 
\begin{align}\label{eq:bl-bl1}
W(B_\la) &= \sum_{\mu \subseteq \la, \la/\mu \ \text{is a BRH}}
w_r(\la/\mu) w_b(\la/\mu) \nonumber \\
&= 1 + \sum_{h \in BRH(B_\la), h \neq \emptyset} 
w_r(h) w_b(h)
\end{align}
That is, the $\mu = \la$ term above corresponds to the empty broken rim 
hook which  
contributes 1 to the sum since all the cells would be blank in this case.  
Thus to prove (\ref{eq:bl-bl}), we must 
show that     
\begin{equation}\label{eq:bl-bl2}
\sum_{h \in BRH(B_\la),h \neq \emptyset} w_r(h) w_b(h) =
(t-1)\sum_{i=1}^{\ell(\la)} \frac{q^{\la_i}-1}{t_i}.
\end{equation}
Now if $\la =(n)$, then $B_\la = \la$  and the only non-empty 
$h \in BRH(B_\la)$ are of form $h= (n)/(n-k)$ some $k >  0$. 
For each cell $c$ that has a cell to its right in 
$h$, 
$$w_{r,(n)/(n-k)}(c) w_{b,(n)/(n-k)}(c) = \frac{q}{t} t = q.$$
 For the 
last red-blue cell in $(n)/(n-k)$, 
$w_{r,(n)/(n-k)}(c) w_{b,(n)/(n-k)}(c) = \frac{q-1}{t} (t-1)$. 
Thus in this case,
$$w_r((n)/(n-k))w_b((n)/(n-k)) =(q-1)(t-1)q^{k-1}/t.
$$
Hence 
\begin{align*}
\sum_{h \in BRH(B_\la), h \neq \emptyset} 
w_r(h) w_b(h) &= \sum_{k=1}^n (q-1)(t-1)q^{k-1}/t \\
&= \frac{(t-1)}{t} (q-1)(1 + q + \cdots +q^{n-1}) \\
&= \frac{(t-1)}{t} (q^n-1) = (t-1) \frac{q^n-1}{t}.
\end{align*}

Now assume $\ell(\la) \geq 2$ and that 
for all $k$ and all partitions $\gamma=(\gamma_1 \geq \cdots \geq \gamma_{\ell(\gamma)})$ of $k$ such that $\ell(\gamma) < \ell(\la)$, we have 
\begin{equation}\label{eq:bl-bl3}
\sum_{h \in BRH(B_\gamma),h \neq \emptyset} w_r(h) w_b(h) =
(t-1) \sum_{i=1}^{\ell(\gamma)} \frac{q^{\gamma_i} -1}{t^i}.
\end{equation}
It follows that if $\ell(\gamma) < \ell(\la)$, then  
\begin{equation}\label{eq:bl-bl4}
\sum_{h \in BRH(B_\gamma),\,h \ \text{starts in row 1}} w_r(h) w_b(h) =
(t-1) \sum_{i=1}^{\ell(\gamma)} \frac{q^{\gamma_i} -1}{t^i} - 
(t-1) \sum_{i=1}^{\ell(\gamma)-1} \frac{q^{\gamma_{i+1}} -1}{t^i}.
\end{equation}

We shall classify the non-empty $h \in BRH(B_\la)$ into two types depending 
on whether $h$ contains the left most cell $s$ in the first row or not.
We let $\gamma$ be the partition that results by removing $\la_1$ from 
$\la$. Thus $\gamma = (\la_2 \geq \cdots \geq \la_{\ell(\la)})$. 
First suppose $\la_1 \neq \la_2$. Then we have two types of configurations 
as pictured in Figure \ref{border2} where we indicate the red-blue squares 
by shading the square. 

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{border2.pdf}
$$
\caption{Two types of configuration for $B_{\la}$}
\label{border2}
\end{figure}


If $h$ does not contain $s$ then 
\begin{align}\label{eq:bl-bl5}
\sum_{h \in BRH(B_\la),h \neq \emptyset, s \notin h}&w_r(h) w_b(h) \nonumber \\&= 
\left(\sum_{h_1 \in BRH(B_\gamma)}w_r(h_1) w_b(h_1)\right)
\left(\sum_{h_2 \in BRH(B_{(\la_1-\la_2)})}w_r(h_2) w_b(h_2)\right) -1 
\nonumber \\
&= \left(1+(t-1)\sum_{i=1}^{{\ell(\la)}-1}\frac{q^{\la_{i+1}}-1}{t^i}\right)
\left(1 +(t-1) \frac{q^{\la_1-\la_2}-1}{t}\right) -1 \nonumber \\
&= \left(1+t(t-1)\sum_{i=2}^{n}\frac{q^{\la_{i}}-1}{t^i}\right)
\left(1 +(t-1) \frac{q^{\la_1-\la_2}-1}{t}\right) -1 \nonumber \\
&= (t-1)\frac{q^{\la_1-\la_2}-1}{t} + t(t-1) \sum_{i=2}^{{\ell(\la)}} 
\frac{q^{\la_i}-1}{t^i} \nonumber \\
&\kern4cm
+ (q^{\la_1-\la_2}-1)(t-1)^2 \sum_{i=2}^{{\ell(\la)}} \frac{q^{\la_i}-1}{t^i}  \nonumber \\
&= (t-1)\frac{q^{\la_1-\la_2}-1}{t} + (t-1) \sum_{i=2}^{{\ell(\la)}} 
\frac{q^{\la_i}-1}{t^i} \nonumber \\
&\kern4cm
+ q^{\la_1-\la_2}(t-1)^2 \sum_{i=2}^{{\ell(\la)}} \frac{q^{\la_i}-1}{t^i}.
\end{align}
Here the $-1$ in the first equality comes from subtracting away the 
configuration corresponding to the empty BRH which has weight 1. 
Now if $\la_1 =\la_2$, then it is easy to see that 
\begin{align*}
\sum_{h \in BRH(B_\la),h \neq \emptyset, s \notin h}w_r(h) w_b(h) &=
\sum_{h \in BRH(B_\gamma),h \neq \emptyset}w_r(h) w_b(h) \\
&=(t-1) \sum_{i=1}^{n-\la_1} \frac{q^{\gamma_i}-1}{t_i}\\
&=t(t-1) \sum_{i=2}^{n} \frac{q^{\la_i}-1}{t^i}.
\end{align*}
which is equal to the expression in line 4 of our string of equalities in 
(\ref{eq:bl-bl5}). Thus (\ref{eq:bl-bl5}) is also true when 
$\la_1=\la_2$. 

Now if the broken rim hook $h$ does contain $s$, then all the squares directly above and to the right of $s$ must be colored red-blue. For the cell $s_a$ directly 
above $s$, $w_{r,h}(s_a)w_{b,h}(s_a)= \frac{-1}{t}(-1)= \frac{1}{t}$. 
Then we can argue as above that the cells in the first row contribute a 
factor of $q^{\la_1-\la_2}\frac{(q-1)(t-1)}{t}$. Now let $h'$ be the BRH which 
consists of all the cells of $h$ except those in the first row. In that case 
it follows that  $w_{r,h'}(s_a)w_{b,h'}(s_a)= 
\frac{(q-1)(t-1)}{t}$. Thus $w_r(h)w_b(h) = \frac{q^{\la_1 -\la_2}}{t}
w_r(h')w_b(h')$. Again this holds for $\la_1 = \la_2$. 
Hence 
\begin{align}
\sum_{h \in BRH(B_\la),h \neq \emptyset, s \in h}w_r(h) w_b(h) \nonumber 
&= \frac{q^{\la_1 -\la_2}}{t}\sum_{h' \in BRH(B_\gamma),\,h \ \text{starts in row 1}}w_r(h') w_b(h')\nonumber \\
&=\frac{q^{\la_1 -\la_2}}{t} (t-1)\left( \sum_{i=1}^{{\ell(\la)}-1}
\frac{q^{\la_{i+1}}-1}{t^i} - \sum_{i=1}^{{\ell(\la)}-2}\frac{q^{\la_{i+2}}-1}{t^i}\right) 
\nonumber \\
&=q^{\la_1-\la_2}(t-1)\left (\sum_{i=2}^{\ell(\la)} \frac{q^{\la_i}-1}{t^i}-t 
\sum_{i=3}^{\ell(\la)} \frac{q^{\la_i}-1}{t^i}\right).
\label{eq:bl-bl6}
\end{align}
Thus combining (\ref{eq:bl-bl5}) and (\ref{eq:bl-bl6}), we see that 
\begin{align*}
\sum_{h \in BRH(B_\la), h \neq \emptyset} &w_r(h)w_b(h) \\
&=\frac{t-1}{t} (q^{\la_1-\la_2}-1)+(t-1)\sum_{i=2}^{\ell(\la)} \frac{q^{\la_i}-1}{t^i}
+q^{\la_1-\la_2}(t-1)^2\sum_{i=2}^{\ell(\la)} \frac{q^{\la_i}-1}{t^i}  \\
&\kern3cm
+q^{\la_1-\la_2}(t-1)\sum_{i=2}^{\ell(\la)} \frac{q^{\la_i}-1}{t^i} 
-q^{\la_1-\la_2}(t^2-t)\sum_{i=3}^{\ell(\la)} \frac{q^{\la_i}-1}{t^i} 
\\
&=\frac{t-1}{t} (q^{\la_1-\la_2}-1)+t(t-1)\sum_{i=2}^{\ell(\la)} \frac{q^{\la_i}-1}{t^i} \\
&\kern3cm
+q^{\la_1-\la_2}(t^2-t)\sum_{i=2}^{\ell(\la)} \frac{q^{\la_i}-1}{t^i} 
-q^{\la_1-\la_2}(t^2-t)\sum_{i=3}^{\ell(\la)} \frac{q^{\la_i}-1}{t^i} 
\\
&=\frac{t-1}{t} (q^{\la_1-\la_2}-1)+(t-1)\sum_{i=2}^{\ell(\la)} \frac{q^{\la_i}-1}{t^i}
+(t^2-t)\frac{q^{\la_2}-1}{t^2}\\
&=\frac{t-1}{t} (q^{\la_1-\la_2}-1)+(t-1)\sum_{i=2}^{\ell(\la)} \frac{q^{\la_i}-1}{t^i}
+q^{\la_1-\la_2}\frac{(t-1)}{t}(q^{\la_2}-1)\\
&=\frac{t-1}{t} (q^{\la_1-\la_2}-1)+(t-1)\sum_{i=2}^{\ell(\la)} \frac{q^{\la_i}-1}{t^i}
+\frac{(t-1)}{t}(q^{\la_1} - \frac{(t-1)}{t}q^{\la_1-\la_2})\\
&=(t-1)\frac{q^{\la_1}-1}{t} +(t-1) \sum_{i=2}^{\ell(\la)} \frac{q^{\la_i}-1}{t^i} \\
&=(t-1) \sum_{i=1}^{\ell(\la)} \frac{q^{\la_i}-1}{t^i}.
\end{align*}

\end{proof}


It follows that the coefficient of $s_\la$ in $D^1_n(s_\la)$ is 
\begin{align*}
\frac{1}{1-t} + \frac{t^n}{(t-1)}
\left(1+ (t-1)\sum_{i=1}^n \frac{q^{\la_i}-1}{t^i}\right) 
&=\frac{t^n-1}{t-1} + \sum_{i=1}^n q^{\la_i}t^{n-i} -t^{n-i}  \\
&= \sum_{i=1}^n q^{\la_i}t^{n-i}.
\end{align*}

Thus we have proved the following. 

\begin{thm}\label{thm:leading}
The coefficient of $s_\la$ in $D^1_n(s_\la)$ equals $\sum_{i=1}^n q^{\la_i}t^{n-i}$.
\end{thm}


\section{Computing weights of simple connecting rim hooks}

The goal of the next 4 sections is compute explicit formulas for 
the weights over all such connecting rim hooks as described in section 2. We shall start by 
computing the weights of certain simple connecting rim hooks.



We start by considering the connecting rim hooks which are just a column.  
That is, in Figure \ref{vert}, we consider connecting rim hooks which 
consists of a column 
of $L$ squares that can be colored either red-blue or blank and 
then we add either a 
square colored red, a square colored blue, a square colored red-blue, or nothing on top. 
We shall let $rV(L)$, $bV(L)$, $rbV(L)$, and $V(L)$ denote 
the connecting rim hooks pictured 
in Figure \ref{vert} reading from left to right. We then have the following lemma.

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=2.25in]{vert.pdf}
$$
\caption{Weights of vertical connecting rim hooks}
\label{vert}
\end{figure}



\begin{lem}\label{lem:vert} 
\begin{align}  
W(rV(L)) &= \frac{(q-1)}{t^{L+1}}, \label{eq:r-vert_L} \\
W(bV(L)) &= (t-q) + \frac{(q-1)}{t^{L}}, \label{eq:b-vert_L} \\
W(rbV(L)) &= (q-1)\left(1-\frac{1}{t^{L+1}}\right), \label{eq:rb-vert_L} \\
W(V(L)) &= 1+(q-1)\left(1-\frac{1}{t^{L}}\right). \label{eq:vert_L} 
\end{align}
\end{lem}
\begin{proof}
We will do the calculation for $W(rV(L))$ and leave the rest of the 
calculations 
to the reader as they are all done in the same way.  Let the cells of 
$rV(L)$ be labeled as $c_0,c_1, \ldots, c_L$ reading from top to bottom. 
Relative to $rV(L)$, the possible $\mu$ are determined by how many 
$(r,b)$ cells there are immediately 
below the cell $c_0$ which contains the red cell at the top of $rV(L)$. 
If there are $k$ such cells, then the factor of $w_r(\la/\mu)$ contributed by 
these cells is 
$ \prod_{i=0}^k w_{r,\la/\mu}(c_i) = \frac{(-1)^k(q-1)}{t^{k+1}}$. 
The factor of $w_b(\gamma/\mu)$ contributed by 
these cells is $1$ if $k=0$ and $ \prod_{i=1}^k w_{b,\gamma/\mu}(c_i) = 
(-1)^{k-1}(t-1)$. Thus if $k =0$, $w_{r,\la/\mu}(c_0) = \frac{q-1}{t}$ and 
if $0 < k \leq L$, 
$$ \left(\prod_{i=0}^k w_{r,\la/\mu}(c_i)\right) 
\left(\prod_{i=1}^k w_{b,\gamma/\mu}(c_i)\right) = 
\frac{(-1)(q-1)(t-1)}{t^{k+1}}.$$ 
Thus 
\begin{align*}
W(rV(L)) &= \frac{(q-1)}{t} + \sum_{k=1}^L \frac{(-1)(q-1)(t-1)}{t^{k+1}}\\
&= (q-1)\left( \frac{1}{t} - \sum_{k=1}^L (\frac{1}{t^k} - \frac{1}{t^{k+1}})\right) \\
&= \frac{(q-1)}{t^{L+1}}.
\end{align*}
\end{proof}
  
In Figure \ref{horz}, we consider the connecting rim hooks which consist of 
a row of $L$ squares 
which can be colored with either red-blue or blank and then we add either a 
square colored red, a square colored blue, a square colored red-blue, or nothing to the right. 
We shall let $Hr(L)$, $Hb(L)$, $Hrb(L)$ and $H(L)$ denote the connecting rim hooks pictured 
in Figure \ref{vert} reading from top to bottom. Recall that by our conventions, these rim hooks 
do not include the colored square on the right if it exists. 
We then have the following lemma.

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=2.25in]{horz.pdf}
$$
\caption{Possible connecting horizontal rim hooks}
\label{horz}
\end{figure}



\begin{lem}\label{lem:horz} 
\begin{align}  
W(Hr(L)) &= \frac{(q-t) +(t-1)q^{L+1}}{t(q-1)}, \label{eq:L-horz-r} \\
W(Hb(L)) &= q^L, \label{eq:L-horz-b} \\
W(Hrb(L)) &= \frac{q^{L+1}-1}{q-1}, \label{eq:L-horx-rb} \\
W(H(L)) &= 1+\frac{(t-1)(q^L-1)}{t}. \label{eq:L-horz} 
\end{align}
\end{lem}
\begin{proof}
We will do the calculation for $W(Hr(L))$ and leave the rest 
of the calculations 
to the reader.  Let the cells of 
$Hr(L)$ be labeled as $c_0,c_1, \ldots, c_L$ reading from right to left. 
Relative to $Hr(L)$, the possible $\mu$ are determined by how many 
$(r,b)$ cells there are immediately 
to the left of the cell $c_0$ which contains the red cell at the right of 
$Hr(L)$. 
If there are $k$ such cells, then the factor of $w_r(\la/\mu)$ contributed by 
these cells is $ \prod_{i=1}^k w_{r,\la/\mu}(c_i) = (q/t)^k$. 
The factor of $w_b(\gamma/\mu)$ contributed by 
these cells is $1$ if $k=0$ and $ \prod_{i=1}^k w_{b,\gamma/\mu}(c_i) = 
t^{k-1}(t-1)$ if $k >0$. Thus if $k =0$, the contribution is 1 and 
if $0 < k \leq L$, the contribution is 
$$ \left(\prod_{i=1}^k w_{b,\gamma/\mu}(c_i) w_{r,\la/\mu}(c_i)\right) = 
\frac{q^k(t-1)}{t}.$$ 
Thus 
\begin{align*}
W(Hr(L)) &= 1 + \frac{q(t-1)}{t} \sum_{k=0}^{L-1}q^k\\
&= 1 +  \frac{q(t-1)(q^L-1)}{t(q-1)}\\
&= \frac{tq -t + tq^{L+1} -qt - q^{L+1}+q}{t(q-1)} \\
&= \frac{q-t + (t-1)q^L}{t(q-1)}.
\end{align*}
\end{proof}


There is one more simple rim hook 
whose weight we want to compute for future computations. This rim hook 
pictured in Figure \ref{invertL} which we call an {\em inverted-L of type} 
$(a,b)$ and denote by $INVL(a,b)$. 

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.75in]{invertL.pdf}
$$
\caption{An inverted $L$ of type $(a,b)$}
\label{invertL}
\end{figure}


\begin{lem}\label{lem:inverted-L}
\begin{equation}\label{eq:invertL}
W(INVL(a,b)) = q^{a+1} + \frac{1}{t^{b+1}} - \frac{q^{a+1}}{t^{b+1}}.
\end{equation}
\end{lem}
\begin{proof}
There are two cases. Namely if the top corner square is blank, then 
it must be case that all the squares are blank. That is, the red squares 
represent those squares which we first removed from $\la$ to form 
$\la/\mu$ so that a square which is colored red or red-blue cannot lie 
to the left or below a blank square. Thus if the top corner square is blank, 
then the  corresponding  configuration would have weight 1. 

Next if the top corner square is colored red-blue, then the configuration 
breaks up into two connecting rim hooks, namely, $Hrb(a)$ and 
$rbV(b)$. Thus by our previous lemma, 
the sum of $w_r(\la/\mu) w_r(\gamma/\mu)$ over all possible $\mu$'s in this 
case is just 
$$\frac{q^{a+1}-1}{q-1} \cdot (q-1)\left(1 - \frac{1}{t^{b+1}}\right) = 
(q^{a+1}-1)\left(1 - \frac{1}{t^{b+1}}\right).$$
Thus 
\begin{align*}
W(H) &= 1 + (q^{a+1}-1)(1 - \frac{1}{t^{b+1}}) \\
&= 1 + (q^{a+1}-1) - \frac{q^{a+1}}{t^{b+1}} + \frac{1}{t^{b+1}}\\
&= q^{a+1} + \frac{1}{t^{b+1}} - \frac{q^{a+1}}{t^{b+1}}.
\end{align*}

\end{proof}


\section{$W(H)$ for red-blue, blue-blue, and blank-blue connecting rim hooks.}

In this section, we shall compute $W(H)$ 
for red-blue, blue-blue, and blank-blue 
connecting rim hooks. Recall that by our conventions, 
the blue square at the end 
is not part of the connecting rim hook. We also note that in such a case, 
we cannot have the last square of the connecting rim hook be immediately above the blue square for this would mean 
that there is blank square immediately above the blue square. But then 
this blank square would be part of both $\la$ and $\mu$ which would violate 
the condition 
that the blue squares are added on the outside of $\mu$. 


Our first step is to compute a reduction lemma which will allows 
to reduce the number of lower corner squares that are part of 
the connecting rim hook. To this end, we consider the 
connecting rim hooks $H_1$ and 
$H_2$ pictured in Figures \ref{endb1} and \ref{endb9} respectively. 

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.75in]{endb1.pdf}
$$
\caption{The connecting rim hook $H_1$}
\label{endb1}
\end{figure}

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.75in]{endb9.pdf}
$$
\caption{The connecting rim hook $H_2$}
\label{endb9}
\end{figure}

In Figure \ref{endb1}, we shall label the corner squares on the lower 
side of $H_1$ by $s$, $t$, and $w$ reading from top to bottom. 
In Figure \ref{endb9}, we shall label the corner squares on the lower 
side of $H_2$ by $u$ and $v$ reading from top to bottom. 
We shall refer to the portion of the connecting rim hook that lies above 
above the corner labeled $s$ in 
Figure \ref{endb1} as $Up(H_1,s)$ and the portion of the connecting rim hook 
that lies above the corner labeled $u$ in Figure \ref{endb9} 
as $Up(H_2,u)$. Then we shall assume that $Up(H_1,s) = Up(H_2,u)$ where  
we allow the possibility that both $Up(H_1,s)$ and $Up(H_2,u)$ are empty.
While we have drawn our pictures so that 
$E \neq 0$, we also allow the possibility that $E=0$. 

\begin{lem}\label{blue-reduction}
If $H_1$ and $H_2$ are the connecting rim hooks pictured in Figures \ref{endb1} and \ref{endb9}, respectively, and $Up(H_1,s) = Up(H_2,u)$, then 
$W(H_1) = W(H_2)$.
\end{lem}

We note that import of Lemma~\ref{blue-reduction} is that it says 
that to compute $W(H)$ for any blue-blue, red-blue, or blank-blue connecting 
rim hook, we need only consider the cases where $H$ has at most 
2 lower corner squares. For example, to compute a formula 
for $W(H)$ for red-blue connecting rim hooks, we need only consider 
the two types of connecting rim hooks pictured in Figure \ref{2-typesrb}.

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{2-typesrb.pdf}
$$
\caption{Limiting cases of red-blue connecting rim hooks}
\label{2-typesrb}
\end{figure}



\begin{proof}
Let ${\bf A}= q^{A+1}$, ${\bf B}= t^{B+1}$, ${\bf C}=q^{C+1}$, ${\bf D}= 
t^{D+1}$ and ${\bf E}=q^{E}$.  To  compute $W(H_i)$, 
we need to compute the sum of $\prod_{c \in H_1} w_{r,\la/\mu}(c)w_{b,\gamma/\mu}(c)$ over all possible choices of $\mu$ in (\ref{comdef1}).  This is equivalent 
to considering all possible colorings of $H_i$, using either the colors 
blank or red-blue for all the cells of $H_i$, except possibly the top cell of $H_i$, that could correspond to such $\mu$. Thus we shall think of the weights of colorings of the cells of $H_i$ where the weight of any blank cell is 1 and 
the weight of any red-blue cell $c$ is $w_{r,\la/\mu}(c)w_{b,\gamma/\mu}(c)$.


We shall organize our computation of $W(H_1)$ and $W(H_2)$ by first 
considering those colorings for which the square $s$ and the square $u$ 
are blank. In such a situation, we let $U$ denote the sum of 
the weights over all possible colorings of $Up(H_1,s) = UP(H_2,u)$ given that 
the cells $s$ and $u$ are blank. This given, we can then classify 
such colorings for $H_1$ according to whether cells $t$ and $w$ are 
blank or red-blue. In the pictures below, we shall picture a 
red-blue cell as a filled cell. There are clearly four such cases.


In Case 1, we are assuming that $t$ and $w$ are blank, see Figure 
\ref{1endb1}. Thus 
the sum $S_1$ of the weights of the colorings in this case 
is 
\begin{equation}\label{eq:1-S1}
S_1 = W(INVL(A,B)) W(INVL(C,D)) W(Hb(E))U.
\end{equation}
Thus using Lemmas~\ref{lem:vert}, \ref{lem:horz}, and \ref{lem:inverted-L}, 
we see that 
\begin{align}
S_1 &= W(INVL(A,B)) W(INVL(C,D)) W(Hb(E))U \nonumber \\
&= 
\left(q^{A+1}+\frac{1}{t^{B+1}} - \frac{q^{A+1}}{t^{B+1}}\right)
\left(q^{C+1}+\frac{1}{t^{D+1}} - \frac{q^{C+1}}{t^{D+1}}\right)q^E U \nonumber \\
&= \left({\bf A}+ \frac{1}{{\bf B}} -\frac{{\bf A}}{{\bf B}}\right)
\left({\bf C}+\frac{1}{{\bf D}} -\frac{{\bf C}}{{\bf D}}\right){\bf E}U .
\label{eq:2-S1}
\end{align}

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{endb1.pdf}
$$
\caption{Case 1 for $H_1$, $s$ is blank}
\label{1endb1}
\end{figure}



In Case 2, we are assuming that cell $t$ is colored red-blue 
and that cell $w$ is  blank, see Figure \ref{endb3}. 
It follows that the cells directly above cell $t$ and 
the cells directly to the right of cell $t$ must be colored red-blue. Among those 
cells it is easy to see from our definitions of $w_{r,\la/\mu}(c)$ and 
$w_{b,\la/\mu}(c)$ that the cells directly above cell $t$ contribute a factor 
of $\frac{1}{{\bf B}}= \frac{1}{t^{B+1}}$ and cell $t$ plus all the cells to the right of 
cell $t$ except the last cell in its row contribute a factor of 
${\bf C} = q^{C+1}$. Note we consider the last cell in the row of 
cell $t$ to be part of the $rbV(D)$ connecting rim hook. 
 Thus 
the sum $S_2$ of the weights of the colorings in this case 
is 
\begin{equation}\label{eq:1-S2}
S_2 = W(Hrb(A)) W(rbV(D)) W(Hb(E))U\frac{{\bf C}}{{\bf B}}.
\end{equation}
Again using Lemmas~\ref{lem:vert}, \ref{lem:horz}, and \ref{lem:inverted-L}, 
we see that 
\begin{align}\label{eq:2-S2}
S_2 &= W(Hrb(A)) W(rbV(D)) W(Hb(E))U\frac{{\bf C}}{{\bf B}} \nonumber \\
&= 
\frac{(q^{A+1}-1)}{(q-1)} (q-1)\left(1 - \frac{1}{t^{D+1}}\right)q^e U \frac{{\bf C}}{{\bf B}}\nonumber 
\\
&= \left({\bf A}-1\right)\left(1 - \frac{1}{{\bf D}}\right){\bf E}U\frac{{\bf C}}{{\bf B}}.
\end{align}

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{endb3.pdf}
$$
\caption{Case 2 for $H_1$, $s$ is blank}
\label{endb3}
\end{figure}

In Case 3, we are assuming that cell $w$ is colored red-blue 
and that cell $t$ is  blank, see Figure \ref{endb4}. 
It follows that the cells directly above cell $w$ and 
the cells directly to the right of $w$ must be colored red-blue. Among those 
cells it is easy to see from our definitions of $w_{r,\la/\mu}(c)$ and 
$w_{b,\la/\mu}(c)$ that the cells directly above cell $w$ contribute a factor 
of $\frac{1}{{\bf D}}= \frac{1}{t^{D+1}}$ and cell $w$ plus all the cells to the right of 
cell $w$ except the last cell in its row contribute a factor of 
${\bf E}(q-1) = q^E(q-1)$. Here the $(q-1)$ corresponds to the weight of 
the cell directly to left of the blue cell at the bottom of the diagram.
Thus the sum $S_3$ of the weights of the colorings in this case 
is 
\begin{equation}\label{eq:1-S3}
S_3 = W(INVL(A,B)) W(Hrb(C)) U\frac{{\bf E}}{{\bf D}}(q-1).
\end{equation}
Again using Lemmas~\ref{lem:vert}, \ref{lem:horz}, and \ref{lem:inverted-L}, 
we see that 
\begin{align}
S_3 &= W(INVL(A,B)) W(Hrb(C)) U\frac{{\bf E}}{{\bf D}}(q-1)
 \nonumber \\
&= 
\left(q^{A+1}+\frac{1}{t^{B+1}} - \frac{q^{A+1}}{t^{B+1}}\right) 
\frac{(q^{C+1} -1)}{(q-1)}U\frac{{\bf E}}{{\bf D}}(q-1) \nonumber 
\\
&= \left({\bf A}+\frac{1}{{\bf B}} -\frac{{\bf A}}{{\bf B}}\right)
({\bf C}-1)U\frac{{\bf E}}{{\bf D}}.
\label{eq:2-S3}
\end{align}

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{endb4.pdf}
$$
\caption{Case 3 for $H_1$, $s$ is blank}
\label{endb4}
\end{figure}


In Case 4, we are assuming that both cell $t$ and cell $w$ are 
colored red-blue, see Figure \ref{endb7}. 
Thus all the cells of $H_1$ that lie  
either directly above or to the right of cells $t$ and $w$ must 
be colored red-blue. The product of the weights of such cells is 
$\frac{{\bf C}{\bf E}}{{\bf B}{\bf D}}(q-1) = \frac{q^{A+1} q^{D}}{t^{B+1}t^{D+1}}(q-1)$. 
Thus the sum $S_4$ of the weights of the colorings in this case 
is 
\begin{equation}\label{eq:1-S4}
S_4 = W(Hrb(A))U \frac{{\bf C}{\bf E}}{{\bf B}{\bf D}}(q-1).
\end{equation}
Using Lemma~\ref{lem:horz}, 
we see that 
\begin{align}
S_4 &= W(Hrb(A))U \frac{{\bf C}{\bf E}}{{\bf B}{\bf D}}(q-1) \nonumber \\
&= \frac{(q^{A+1} -1)}{(q-1)}U\frac{{\bf C}{\bf E}}{{\bf B}{\bf D}}(q-1) \nonumber
\\
&= ({\bf A}-1)U\frac{{\bf C}{\bf E}}{{\bf B}{\bf D}}.
\label{eq:2-S4}
\end{align}

A little algebra will show that 
\begin{equation}\label{S1+S4}
S_1 + S_2 +S_3 +S_4 = {\bf A}{\bf C}{\bf E}U.
\end{equation}

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{endb7.pdf}
$$
\caption{Case 4 for $H_1$, $s$ is blank}
\label{endb7}
\end{figure}


Next we consider the cases for $H_1$ where cell $s$ is colored red-blue. 
This means that all the cells directly above or directly to the right 
of $s$ must be colored red-blue. In particular, all the cells in same 
row as $s$ except for the last cell contributes a factor of 
${\bf A}= q^{A+1}$ to the weight.  We let $V$ denote the sum of 
weights over all possible colorings of $Up(H_1,s) = UP(H_2,u)$ given that 
the cells $s$ and $u$ are colored red-blue. Then we have four more cases 
to consider, depending on the colors of cells $t$ and $w$. 




In Case 5, we are assuming that $t$ and $w$ are blank, see Figure 
\ref{endb2}. Thus 
the sum $S_5$ of the weights of the colorings in this case 
is 
\begin{equation}\label{eq:1-S5}
S_5 = W(rbV(B)) W(INVL(C,D)) W(Hb(E)){\bf A}V.
\end{equation}
Thus using Lemmas~\ref{lem:vert}, \ref{lem:horz}, and \ref{lem:inverted-L}, 
we see that 
\begin{align}
S_5 &= W(rbV(B)) W(INVL(C,D)) W(Hb(E)){\bf A}V \nonumber \\
&= 
\left(q-1)(1 - \frac{1}{t^{B+1}}\right)
\left(q^{C+1}+\frac{1}{t^{D+1}} - \frac{q^{D+1}}{t^{D+1}}\right)q^E {\bf A}V \nonumber \\
&= (q-1)\left(1 - \frac{1}{{\bf B}}\right)\left({\bf C}+\frac{1}{{\bf D}} -\frac{{\bf C}}{{\bf D}}\right){\bf E}{\bf A}V. 
\label{eq:2-S5}
\end{align}





\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{endb2.pdf}
$$
\caption{Case 5 for $H_1$, $s$ is colored red-blue}
\label{endb2}
\end{figure}

In Case 6, we are assuming that cell $t$ is colored red-blue 
and that cell $w$ is  blank. 
It follows that the cells directly above cell $t$ and 
the cells directly to the right of cell $t$ must be colored red-blue. It is easy to see from our definitions of $w_{r,\la/\mu}(c)$ and 
$w_{b,\la/\mu}(c)$ that the cells directly above cell $t$ contribute a factor 
of $\frac{1}{{\bf B}}= \frac{1}{t^{B+1}}$ and cell $t$ plus all the cells to 
the right of 
cell $t$ except the last cell in its row contribute a factor of 
${\bf C} = q^{C+1}$. Again, we shall consider the last cell in the row of 
cell $t$ to be part of the $rbV(D)$ connecting rim hook. 
 Thus 
the sum $S_6$ of the weights of the colorings in this case 
is 
\begin{equation}\label{eq:1-S6}
S_6 = W(rbV(D))W(Hb(E)){\bf A}V\frac{{\bf C}}{{\bf B}}.
\end{equation}
Using Lemmas~\ref{lem:vert} and \ref{lem:horz}, 
we see that 
\begin{align}
S_6 &= W(rbV(D)) W(Hb(E)){\bf A}V\frac{{\bf C}}{{\bf B}} \nonumber \\
&= (q-1)\left(1 - \frac{1}{t^{D+1}}\right)q^E {\bf A}V \frac{{\bf C}}{{\bf B}}\nonumber 
\\
&= (q-1)\left(1-\frac{1}{{\bf D}}\right){\bf E}{\bf A}V\frac{{\bf C}}{{\bf B}}.
\label{eq:2-S6}
\end{align}

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{endb5.pdf}
$$
\caption{Case 6 for $H_1$, $s$ is colored red-blue}
\label{endb5}
\end{figure}


In Case 7, we are assuming that cell $w$ is colored red-blue 
and that cell $t$ is  blank, see Figure \ref{endb6}. 
It follows that the cells directly above cell $w$ and 
the cells directly to the right of $w$ must be colored red-blue. Among those 
cells it is easy to see from our definitions of $w_{r,\la/\mu}(c)$ and 
$w_{b,\la/\mu}(c)$ that the cells directly above cell $w$ contribute a factor 
of $\frac{1}{{\bf D}}= \frac{1}{t^{D+1}}$ and cell $w$ plus all the cells to 
the right of 
cell $w$ except the last cell in its row contribute a factor of 
${\bf E}(q-1) = q^E(q-1)$. Here the $(q-1)$ corresponds to the weight of 
the cell directly to left of the blue cell at the bottom of the diagram.
Thus the sum $S_7$ of the weights of the colorings in this case 
is 
\begin{equation}\label{eq:1-S7}
S_7 = W(rbV(B))W(Hrb(C)) {\bf A}V\frac{{\bf E}}{{\bf D}}(q-1).
\end{equation}
Again using Lemmas~\ref{lem:vert}, \ref{lem:horz}, and \ref{lem:inverted-L}, 
we see that 
\begin{align}
S_7 &= W(rbV(B)) W(Hrb(C)) {\bf A}V\frac{{\bf E}}{{\bf D}}(q-1)
 \nonumber \\\nonumber
&= 
(q-1)\left(1 - \frac{1}{t^{B+1}}\right) 
\frac{(q^{c+1} -1)}{(q-1)}{\bf A}V\frac{{\bf E}}{{\bf D}}(q-1)
\\
&= (q-1)\left(1 -\frac{1}{{\bf B}}\right)({\bf C}-1){\bf A}V\frac{{\bf E}}{{\bf D}}.
\label{eq:2-S7}
\end{align}



\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{endb6.pdf}
$$
\caption{Case 7 for $H_1$, $s$ is colored red-blue}
\label{endb6}
\end{figure}

In Case 8, we are assuming that both cell $t$ and cell $w$ are 
colored red-blue, see Figure \ref{endb8}. Thus all the cells of $H_1$ that 
lie 
either directly above or to the right of cells $t$ and $w$ must 
be colored red-blue. The product of the weights of such cells is 
$\frac{{\bf C}{\bf E}}{{\bf B}{\bf D}}(q-1) = \frac{q^{A+1} q^{E}}{t^{B+1}t^{D+1}}(q-1)$. 
Thus the sum $S_8$ of the weights of the colorings in this case 
is 
\begin{equation}\label{eq:1-S8}
S_8 = (q-1){\bf A}V \frac{{\bf C}{\bf E}}{{\bf B}{\bf D}}.
\end{equation}

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{endb8.pdf}
$$
\caption{Case 8 for $H_1$, $s$ is red-blue}
\label{endb8}
\end{figure}


Then it is easy to check that 
\begin{equation}
S_5+S_6+S_7+S_8 = (q-1){\bf A}{\bf C}{\bf E}V
\end{equation}
and, hence,
\begin{equation}\label{wH1}
W(H_1) = {\bf A}{\bf C}{\bf E}U + (q-1){\bf A}{\bf C}{\bf E}V.
\end{equation}


We organize the computation of $W(H_2)$ in the same way. 
That is, we first consider those colorings where cell $u$ is 
blank. In this situation, there are two cases. 

In Case I, we assume that cell $v$ is blank, see Figure \ref{1endb9}. Thus the sum $T_1$ of 
weights of the colorings in this case is 
\begin{align}\nonumber 
T_1 &= W(INVL(A+C+1,B+D+1)) W(Hb(E))U \\
&= \left(q^{A+C+2}+ \frac{1}{t^{B+D+2}} -  \frac{q^{A+C+2}}{t^{B+D+2}}\right)q^E U 
\nonumber \\
&= \left({\bf AC} + \frac{1}{{\bf BD}} - \frac{{\bf AC}}{{\bf BD}}\right){\bf E}U.
\label{eq:1-T1}
\end{align}

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{endb9.pdf}
$$
\caption{Case I for $H_2$, $u$ is blank}
\label{1endb9}
\end{figure}






In Case II, we assume that $v$ is colored red-blue, see Figure \ref{endb11}.
Thus all the cells in the same row and column as $v$ must be colored red-blue.
These cells always contribute a factor of 
$\frac{1}{t^{B+D+2}}q^E(q-1)=\frac{{\bf E}}{{\bf BD}} $ to the 
weight of any coloring. Thus the sum $T_2$ of the weights of the possible 
colorings in this case is 
\begin{align}\nonumber
T_2 &= W(Hrb(A+C+1))U (q-1)\frac{{\bf  E}}{{\bf BD}}\\
&= \frac{(q^{A+C+2}-1)}{(q-1)}U (q-1)\frac{{\bf E}}{{\bf BD}}\nonumber\\
&= ({\bf AC}-1)U \frac{{\bf E}}{{\bf BD}}
\label{eq:1-T2}
\end{align}

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{endb11.pdf}
$$
\caption{Case II for $H_2$, $u$ is blank}
\label{endb11}
\end{figure}


Then it is easy to check that 
$T_1 +T_2 =   {\bf ACE}U$.

Next we consider the case where $u$ is colored red-blue. Then all 
the cells in the same row or column as $u$ must be colored red-blue. 
In this case, the cells in the same row as $u$ except for 
the last cell contribute a factor of  
$q^{A+C+2} = {\bf AC}$ to the weight of any coloring. Again there are two 
cases depending on the color of 
$v$. 


In Case III, we assume that cell $v$ is blank, see Figure \ref{endb10}. Thus the sum $T_3$ of 
weights of the colorings in this case is 
\begin{align}\nonumber 
T_3 &= W(rbV(B+D+1) W(Hb(E))V {\bf AC}\\
&= (q-1) \left(1- \frac{1}{t^{B+D+2}}\right)q^E V {\bf AC}\nonumber \\
&= (q-1)\left(1-\frac{1}{{\bf BD}}\right){\bf E}V{\bf AC}.
\label{eq:1-T3}
\end{align}

 \begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{endb10.pdf}
$$
\caption{Case III for $H_1$, $u$ is red-blue}
\label{endb10}
\end{figure}
 
In Case IV, we assume that $v$ is colored red-blue, see Figure \ref{endb12}.
In this case, the sum $T_4$ of the weights of the coloring is 
\begin{equation}\label{eq:1-T4}
T_4 = q^{A+C+2}\frac{1}{t^{B+D+2}}q^E(q-1)V= (q-1)\frac{{\bf ACE}}{{\bf BD}}V.
\end{equation}


\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{endb12.pdf}
$$
\caption{Case 4 for $H_1$, $u$ is red-blue}
\label{endb12}
\end{figure}

Then it is easy to check that 
$$
T_3+T_4 = (q-1){\bf ACE}V$$
so that 
\begin{equation}\label{wH2}
W(H_2) ={\bf ACE}U + (q-1){\bf ACE}V.
\end{equation}

Thus comparing (\ref{wH1}) and (\ref{wH2}), we see that 
$W(H_1) =W(H_2)$ as claimed.
\end{proof}


We now are in position to prove the main result of this section.
\begin{thm}\label{thm:rbcon}
\begin{enumerate}

\item If $H$ is red-blue connecting rim hook, then $W(H) = 0$.
\item If $H$ is a blank-blue connecting rim hook, then $W(H) =q^{c(H)}$.

\item If $H$ is a blue-blue connecting rim hook, then 
$W(H) = (t-q)q^{c(H) -1}$.
\end{enumerate}
\end{thm}
\begin{proof}
We shall prove only (1) since the exact same ideas can be used to prove 
(2) and (3) and, hence, we will leave those details to the reader.
As noted above, we need only consider connecting rim hooks $H$ where there 
are either one  or two  lower corner squares as pictured Figure 
\ref{2-typesrb}. 

Let us first consider the case where the red-blue connecting rim hook 
$H$ has one lower corner square. 
There are two basic cases, 
namely, the case where the corner square is blank and 
the corner square is colored red-blue, see Figure \ref{rbhook}. 

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{rbhook.pdf}
$$
\caption{Red-blue 1 lower corner shapes}
\label{rbhook}
\end{figure}

If the corner square is blank, then it is easy to see that sum of the weights 
of the possible colorings is 
$$
W(rV(A))W(Hb(B)) = \frac{(q-1)}{t^{A+1}}q^B.
$$
If the corner cell is colored red-blue, then the top red cell has 
weight $\frac{-1}{t}$ and the remaining $A$ cells is the vertical 
part each contribute a weight $\frac{-1}{t}(-1) = \frac{1}{t}$. 
Each cell in the bottom row except for the cell that immediately 
precedes the blue cell, contributes a weight of $\frac{q}{t}t = q$. The 
cell that immediately precedes the blue cell contributes a weight of 
$\frac{(q-1)}{t} t = (q-1)$. The weight corresponding to this configuration 
is $-\frac{(q-1)q^B}{t^{A+1}}$. Thus $W(H) = 0$ in this case. 


Next we consider the case where the red-blue connecting rim hook 
has two corner squares. Thus there are four cases depending 
on whether the two lower corner squares are colored red-blue or blank as 
pictured in Figure \ref{rb2corner}.

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=2.25in]{rb2corner.pdf}
$$
\caption{Red-blue 2 lower corner shapes}
\label{rb2corner}
\end{figure}


In Case 1 of Figure \ref{rb2corner}, we have the sum of 
the weights of the colorings in this case is 
\begin{align}\nonumber
W(rV(A))W(INVTL(B,C))W(Hb(D)) &= \frac{(q-1)}{t^{A+1}}
\left(q^{B+1} +\frac{1}{t^{C+1}}-\frac{q^{B+1}}{t^{C+1}}\right)q^D  \nonumber \\
&=\frac{(q-1)q^D}{t^{A+1}}\left(q^{B+1} +\frac{1}{t^{C+1}}-\frac{q^{B+1}}{t^{C+1}}\right).
\label{rb2corner1}
\end{align}

In Case 2 of Figure \ref{rb2corner}, we have the sum of 
the weights of the colorings in this case is 
\begin{align}
\frac{-1}{t^{A+1}} q^{B+1} W(rbV(C))W(Hb(D)) &=
 \frac{-q^{B+1}}{t^{A+1}}
 (q-1)\left( 1 - \frac{1}{t^{C+1}}\right)q^D \nonumber \\
&=\frac{-(q-1)q^D}{t^{A+1}}\left(q^{B+1} -\frac{q^{B+1}}{t^{C+1}}\right).
\label{rb2corner2}
\end{align} 
It is then easy to see that the sum of the weights of the colorings 
in Cases~1 and 2 is simply
\begin{equation}\label{rb2corner3}
\frac{(q-1)q^D}{t^{A+1}}\frac{1}{t^{C+1}} = \frac{(q-1)q^D}{t^{A+C+2}}.
\end{equation}

In Case 3 of Figure \ref{rb2corner}, we have the sum of 
the weights of the colorings in this case is 
\begin{align}
W(rbV(A))W(Hrb(B)) \frac{1}{t^{C+1}}q^D(q-1)&= 
 \frac{(q-1)}{t^{A+1}}\frac{(q^{B+1}-1)}{(q-1)} 
(q-1)\frac{q^D}{t^{C+1}}  \nonumber \\
&=\frac{(q-1)q^D}{t^{A+C+2}}(q^{B+1} -1).
\label{rb2corner4}
\end{align} 

Finally in  case 4 of Figure \ref{rb2corner}, we have the sum of 
the weights of the colorings in this case is 
\begin{equation}\label{rb2corner5}
\frac{-1}{t}\frac{1}{t^A} q^{B+1}\frac{1}{t^{C+1}}q^D(q-1) = 
\frac{-(q-1)q^D}{t^{A+C+2}}q^{B+1}.
\end{equation} 
It is then easy to see that the sum of the weights of the colorings 
in Cases~3 and 4 is simply
\begin{equation}\label{rb2corner6}
\frac{-(q-1)q^D}{t^{A+C+2}}.
\end{equation}

Thus comparing equations (\ref{rb2corner3}) and 
(\ref{rb2corner6}), we see that $W(H)=0$ in this case as well.
\end{proof}

We note that Theorem~\ref{thm:rbcon} is important because 
it says that the coefficient of any $s_\gamma$ in (\ref{comdef1}) such 
that $\gamma$ has a red-blue connecting rim hook 
is 0. First, if $\gamma$ is such that there is a red square which 
lies above a blue square, then the $\gamma$ must have a red-blue connecting 
rim hook. Thus the the coefficient of $s_\gamma$ in (\ref{comdef1}) is automatically 0. Hence there is no loss in restricting ourselves to $\gamma$ that 
satisfies condition $(*)$. Moreover, it follows that there is no 
use in computing $W(H)$ when $H$ is either a blank-red or blue-blank connecting rim hook since any $s_\gamma$ in (\ref{comdef1}) for which $\gamma$ has either a blank-red or blue-blank connecting rim hook must also have red-blue connecting rim hook and, hence, the coefficient of $s_\gamma$ in (\ref{comdef1}) is automatically 0. Thus, of the nine types of connecting rim hooks, we are left 
with computing $W(H)$ when $H$ is red-red, red-blank, or blue-red connecting rim hooks. In the next section, we shall compute $W(H)$ for red-red and red-blank connecting rim hooks. 

\section{$W(H)$ for red-red and red-blank connecting rim hooks.}

The main goal of this section is to compute $W(H)$ for red-red and red-blank connecting rim hooks. Our first step to prove a reduction lemma which is 
similar to Lemma~\ref{blue-reduction} for those 
connecting rim hooks which start with a red square.  
We note that in such a case, 
we cannot have the next square of the connecting rim hook be immediately to the right of  the red square for this would mean 
that there is blank square immediately to the right of the red square which 
is impossible. 
We want to consider the connecting rim hooks $G_1$ and 
$G_2$ pictured in Figures \ref{rst1} and \ref{rst9} respectively. 

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.75in]{rst1.pdf}
$$
\caption{The connecting rim hook $G_1$}
\label{rst1}
\end{figure}

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.75in]{rst9.pdf}
$$
\caption{The connecting rim hook $G_2$}
\label{rst9}
\end{figure}

In Figure \ref{rst1}, we shall label the corner squares on the lower 
side of $G_1$ by $s$, $t$, and $w$ reading from bottom to top. 
In Figure \ref{rst9}, we shall label the corner squares on the lower 
side of $G_2$ by $u$ and $v$ reading from bottom to top. 

We shall refer to the portion of the connecting rim hook that lies to the  
right of the corner labeled $s$ in 
Figure \ref{rst1} as $Rt(G_1,s)$ and the portion of the connecting rim hook 
that lies to the right of the corner labeled $u$ in Figure \ref{rst9} 
as $Rt(G_2,u)$. Then shall assume that $Rt(G_1,s) = Rt(G_2,u)$ and 
we allow the possibility that both $Rt(G_1,s)$ and $Rt(G_2,u)$ are empty.
While we have drawn our pictures so that 
$A \neq 0$, we also allow the possibility that $A=0$. 

\begin{lem}\label{red-reduction}
If $G_1$ and $G_2$ are the connecting rim hooks pictured in Figures 
\ref{rst1} and \ref{rst9}, respectively, and $Rt(G_1,s) = Rt(G_2,u)$, then 
$W(G_1) = W(G_2)$.
\end{lem}
\begin{proof}

The proof of Lemma~\ref{red-reduction} is very similar to 
the proof of Lemma~\ref{blue-reduction} so we shall suppress some of 
the details. We let 
 ${\bf A}= q^{A+1}$, ${\bf B}= t^{B+1}$, ${\bf C}=q^{C+1}$, ${\bf D}= 
t^{D+1}$ and ${\bf E}=q^{E+1}$. 


We shall organize our computation of $W(G_1)$ and $W(G_2)$ by first 
considering those colorings for which the square $s$ and the square $u$ 
are blank. In such a situation, we let $U$ denote the sum of the  
weights over all possible colorings of $Rt(G_1,s) = Rt(G_2,u)$ given that 
the cells $s$ and $u$ are blank. This given, we can then classify 
such colorings for $H_1$ according to whether cells $t$ and $w$ are 
blank or red-blue. In the pictures, below we shall picture a 
red-blue cell as a filled cell. There are clearly four such cases.


In Case 1, we are assuming that $t$ and $w$ are blank, see Figure 
\ref{1rst1}. Thus 
the sum $P_1$ of the weights of the colorings in this case 
is 
\begin{align}
P_1 &= W(rV(A))W(INVL(B,C)) W(INVL(D,E))U \nonumber \\
&=
\frac{(q-1)}{t^{A+1}}\left(q^{B+1}+\frac{1}{t^{C+1}} - \frac{q^{B+1}}{t^{C+1}}\right)
\left(q^{D+1}+\frac{1}{t^{E+1}} - \frac{q^{D+1}}{t^{E+1}}\right)U \nonumber \\
&= \frac{(q-1)}{{\bf A}}\left({\bf B}+ \frac{1}{{\bf C}} -\frac{{\bf B}}{{\bf C}}\right)\left({\bf D}+\frac{1}{{\bf E}} -\frac{{\bf D}}{{\bf E}}\right)U. 
\label{eq:2-P1}
\end{align}

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{rst1.pdf}
$$
\caption{Case 1 for $G_1$, $s$ is blank}
\label{1rst1}
\end{figure}



In Case 2, we are assuming that cell $t$ is colored red-blue 
and that cell $w$ is  blank, see Figure \ref{rst3}. 
It follows that the cells directly above cell $t$ and 
the cells directly to the right of cell $t$ must be colored red-blue. It is easy to see from our definitions of $w_{r,\la/\mu}(c)$ and 
$w_{b,\la/\mu}(c)$ that the cells directly above cell $t$ contribute a factor 
of $\frac{1}{{\bf C}}= \frac{1}{t^{C+1}}$ and cell $t$ plus all the cells to the right of 
cell $t$ except the last cell in its row contribute a factor of 
${\bf D} = q^{D+1}$. Note we consider the last cell in the row of 
cell $t$ to be part of the $rbV(E)$ connecting rim hook. 
 Thus 
the sum $P_2$ of the weights of the colorings in this case 
is 
 \begin{align}
P_2 &= W(rV(A)) W(Hrb(B)) W(rbV(E))U\frac{{\bf D}}{{\bf C}} \nonumber \\
&= \frac{(q-1)}{t^{A+1}}
(q-1)\frac{(q^{B+1}-1)}{(q-1)} \left(1 - \frac{1}{t^{E+1}}\right)U \frac{{\bf D}}{{\bf C}}\nonumber 
\\
&= (q-1)({\bf B}-1)\left(1 - \frac{1}{{\bf E}}\right)U\frac{{\bf D}}{{\bf AC}}.
\label{eq:2-P2}
\end{align}

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{rst3.pdf}
$$
\caption{Case 2 for $G_1$, $s$ is blank}
\label{rst3}
\end{figure}

In Case 3, we are assuming that cell $w$ is colored red-blue 
and that cell $t$ is  blank, see Figure \ref{rst2}. 
It follows that the cells directly above cell $w$ and 
the cells directly to the right of $w$ must be colored red-blue. It is easy to see from our definitions of $w_{r,\la/\mu}(c)$ and 
$w_{b,\la/\mu}(c)$ that the cells directly above cell $w$ contribute a factor 
of $-\frac{1}{t^{A+1}} = -\frac{1}{{\bf A}}$ and cell $t$ plus all the cells to the right of 
cell $t$ except the last cell in its row contribute a factor of 
$ q^{B+1}= {\bf B}$. 
Thus the sum $P_3$ of the weights of the colorings in this case 
is 
\begin{align}
P_3 &= W(rbV(C)) W(INVL(D,E)) U(-1)\frac{{\bf B}}{{\bf A}}
 \nonumber \\
&= 
(q-1)\left(1 - \frac{1}{t^{C+1}}\right)\left(q^{D+1}+\frac{1}{t^{E+1}} - \frac{q^{D+1}}{t^{E+1}}\right) U(-1)\frac{{\bf B}}{{\bf A}} \nonumber 
\\
&= -(q-1)\left(1 - \frac{1}{{\bf C}}\right)\left({\bf D}+\frac{1}{{\bf E}} -
\frac{{\bf D}}{{\bf E}}\right)U\frac{{\bf B}}{{\bf A}}.
\label{eq:2-P3}
\end{align}

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{rst2.pdf}
$$
\caption{Case 3 for $G_1$, $s$ is blank}
\label{rst2}
\end{figure}


In Case 4, we are assuming that both cell $t$ and cell $w$ are 
colored red-blue, see Figure \ref{rst5}. 
Thus all the cells of $G_1$ that lie 
either directly above or to the right of cells $t$ and $w$ must 
be colored red-blue. The product of the weights of such cells is 
$-\frac{q^{B+1} q^{D+1}}{t^{A+1}t^{C+1}}=-\frac{{\bf B}{\bf D}}{{\bf A}{\bf C}}(q-1)$. 
Thus the sum $P_4$ of the weights of the colorings in this case 
\begin{align}
P_4 &= W(rbV(E))U (-1)\frac{{\bf B}{\bf D}}{{\bf A}{\bf C}} \nonumber \\
&= (q-1)\left(1-\frac{1}{t^{E+1}}\right)U(-1)\frac{{\bf B}{\bf D}}{{\bf A}{\bf C}} \nonumber
\\
&= -(q-1)\left(1 - \frac{1}{{\bf E}}\right)U\frac{{\bf B}{\bf D}}{{\bf A}{\bf C}}.
\label{eq:2-P4}
\end{align}

A little algebra will show that 
\begin{equation}\label{P1+P4}
P_1 + P_2 +P_3 +P_4 = (q-1)\frac{1}{{\bf ACE}}U.
\end{equation}

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{rst5.pdf}
$$
\caption{Case 4 for $G_1$, $s$ is blank}
\label{rst5}
\end{figure}


Next we consider the cases for $G_1$ where cell $s$ is colored red-blue. 
This means that all the cells directly above or directly to the right 
of $s$ must be colored red-blue. In particular, all the cells in the same 
column as $s$ except for $s$ itself contribute a factor of 
$\frac{1}{{\bf E}}= \frac{1}{t^{E+1}}$ to the weight.  We let $V$ denote the sum of the 
weights over all possible colorings of $Rt(G_1,s) = Rt(G_2,u)$ given that 
the cells $s$ and $u$ are colored red-blue. Then we have four more cases 
to consider for $G_1$, depending on the colors of cells $t$ and $w$. 




In Case 5, we are assuming that $t$ and $w$ are blank, see Figure 
\ref{rst4}. Thus 
the sum $P_5$ of the weights of the colorings in this case 
is  
\begin{align}
P_5 &= W(rV(A)) W(INVL(B,C)) W(Hrb(D))V \frac{1}{{\bf E}}\nonumber \\
&= 
\frac{(q-1)}{t^{A+1}}
\left(q^{B+1}+\frac{1}{t^{C+1}} - \frac{q^{B+1}}{t^{C+1}}\right)
\frac{q^{D+1}-1}{(q-1)} V \frac{1}{{\bf E}}\nonumber \\
&= \left({\bf B}+\frac{1}{{\bf C}} -\frac{{\bf B}}{{\bf C}}\right)({\bf D} -1)V 
\frac{1}{{\bf AE}}.
\label{eq:2-P5}
\end{align}





\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{rst4.pdf}
$$
\caption{Case 5 for $G_1$, $s$ is colored red-blue}
\label{rst4}
\end{figure}

In Case 6, we are assuming that cell $t$ is colored red-blue 
and that cell $w$ is  blank, see Figure \ref{rst7}. 
It follows that the cells directly above cell $t$ and 
the cells directly to the right of cell $t$ must be colored red-blue. It is easy to see from our definitions of $w_{r,\la/\mu}(c)$ and 
$w_{b,\la/\mu}(c)$ that the cells directly above cell $t$ contribute a factor 
of $\frac{1}{t^{C+1}}=\frac{1}{{\bf C}}$ and cell $t$ plus all the cells to the right of 
cell $t$ except the last cell in its row contribute a factor of 
$ q^{D+1}= {\bf D}$. 
 Thus 
the sum $P_6$ of the weights of the colorings in this case 
is 
\
\begin{align}
P_6 &= W(rV(A)) W(Hrb(B))V\frac{{\bf D}}{{\bf CE}} \nonumber \\
&= \frac{(q-1)}{t^{A+1}} \frac{(q^{B+1}-1)}{(q-1)}V \frac{{\bf D}}{{\bf CE}}\nonumber 
\\
&= ({\bf B}-1)\frac{{\bf D}}{{\bf ACE}}V.
\label{eq:2-P6}
\end{align}

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{rst7.pdf}
$$
\caption{Case 6 for $G_1$, $s$ is colored red-blue}
\label{rst7}
\end{figure}


In Case 7, we are assuming that cell $w$ is colored red-blue 
and that cell $t$ is  blank, see Figure \ref{rst6}. 
It follows that the cells directly above cell $w$ and 
the cells directly to the right of $w$ must be colored red-blue. It is easy to see from our definitions of $w_{r,\la/\mu}(c)$ and 
$w_{b,\la/\mu}(c)$ that the cells directly above cell $w$ contribute a factor 
of $= -\frac{1}{t^{A+1}}= -\frac{1}{{\bf A}}$ and cell $w$ plus all the cells to the right of 
cell $w$ except the last cell in its row contribute a factor of 
$q^{B+1}= {\bf B}$. 
Thus the sum $P_7$ of the weights of the colorings in this case 
is 
\begin{align}
P_7 &= W(rbV(C)) W(Hrb(D)) V(-1)\frac{{\bf B}}{{\bf AE}}
 \nonumber \\\nonumber
&= 
-(q-1)\left(1 - \frac{1}{t^{C+1}}\right) 
\frac{(q^{D+1} -1)}{(q-1)}{\bf A}V\frac{{\bf B}}{{\bf AE}}
\\
&= -\left(1 -\frac{1}{{\bf C}}\right)({\bf D}-1)V\frac{{\bf B}}{{\bf AE}}.
\label{eq:2-P7}
\end{align}



\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{rst6.pdf}
$$
\caption{Case 7 for $G_1$, $s$ is colored red-blue}
\label{rst6}
\end{figure}

In Case 8, we are assuming that both cell $t$ and cell $w$ are 
colored red-blue, see Figure \ref{rst8}. Thus all the cells of $G_1$ that 
lie 
either directly above or to the right of cells $t$ and $w$ must 
be colored red-blue. The product of the weights of such cells is 
$-\frac{q^{B+1} q^{D+1}}{t^{A+1}t^{C+1}}=-\frac{{\bf B}{\bf D}}{{\bf A}{\bf C}}$. 
Thus the sum $P_8$ of the weights of the colorings in this case 
is 
\begin{equation}\label{eq:1-P8}
P_8 = -V \frac{{\bf B}{\bf D}}{{\bf AC}{\bf E}}.
\end{equation}

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{rst8.pdf}
$$
\caption{Case 8 for $H_1$, $s$ is red-blue}
\label{rst8}
\end{figure}

\noindent
Then it is easy to check that 
\begin{equation}
P_5+P_6+P_7+P_8 = \frac{-1}{{\bf A}{\bf C}{\bf E}}V
\end{equation}
and, hence,
\begin{equation}\label{wG1}
W(G_1) = \frac{(q-1)}{{\bf A}{\bf C}{\bf E}}U - 
\frac{1}{{\bf A}{\bf C}{\bf E}}V.
\end{equation}


We organize the computation of $W(G_2)$ in the same way. 
That is, we first consider those colorings where cell $u$ is 
blank. In this situation, there are two cases. 

In Case I, we assume that cell $v$ is blank, see Figure \ref{1rst9}. Thus the sum $Q_1$ of 
weights of the colorings in this case is 
\begin{align}\nonumber 
Q_1 &= W(rV(A+C+1))W(INVL(B+D+1,E))U \\
&= \frac{(q-1)}{t^{A+C+2}}\left(q^{B+D+2}+ \frac{1}{t^{E+1}} -  
\frac{q^{B+D+2}}{t^{E+1}}\right)U 
\nonumber \\
&= (q-1) \frac{1}{{\bf AC}}\left( {\bf BD} + \frac{1}{{\bf E}} - \frac{{\bf BD}}{{\bf E}}\right)U.
\label{eq:1-Q1}
\end{align}

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{rst9.pdf}
$$
\caption{Case I for $G_2$, $u$ is blank}
\label{1rst9}
\end{figure}






In Case II, we assume that $v$ is colored red-blue, see Figure \ref{rst10}.
Thus all the cells in the same row and column as $u$ must be colored red-blue.
These cells always contribute a factor of 
$-\frac{q^{B+D+2}}{t^{A+C+2}} = -\frac{{\bf BD}}{{\bf AC}} $ to the 
weight of any coloring. Thus the sum $Q_2$ of the weights of the possible 
colorings in this case is 
\begin{align}\nonumber
Q_2 &= W(rbV(E))U (-1)\frac{{\bf BD}}{{\bf AC}}\\
&= -(q-1)\left(1 - \frac{1}{t^{E+1}}\right)U \frac{{\bf BD}}{{\bf AC}}\nonumber\\
&= -(q-1)\left(1 - \frac{1}{{\bf E}}\right)U\frac{{\bf BD}}{{\bf AC}}.
\label{eq:1-Q2}
\end{align}

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{rst10.pdf}
$$
\caption{Case II for $G_2$, $u$ is blank}
\label{rst10}
\end{figure}

\noindent
Then it is easy to check that 
$Q_1 +Q_2 =   (q-1)\frac{1}{{\bf ACE}}U$.

Next we consider the case where $u$ is colored red-blue. Then all 
the cells in the same row or column as $u$ must be colored red-blue. 
The factor that all the cells in the same column as $u$ 
contributes to the weights of the colorings in this situation is 
$\frac{1}{t^{E+1}} = \frac{1}{{\bf E}}$. 
Again there are two cases depending on the color of 
$v$. 


In Case III, we assume that cell $v$ is blank, see Figure \ref{rst11}. Thus the sum $Q_3$ of the 
weights of the colorings in this case is 
\begin{align}\nonumber 
Q_3 &= W(rV(A+C+1)) W(Hrb(B+D+1))V\frac{1}{{\bf E}}\\
&= \frac{(q-1)}{t^{A+C+2}} \frac{(q^{B+D+2} -1)}{(q-1)}V\frac{1}{{\bf E}}\nonumber \\
&= ({\bf BD}-1)V\frac{1}{{\bf ACE}}.
\label{eq:1-Q3}
\end{align}

 \begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{rst11.pdf}
$$
\caption{Case III for $G_2$, $u$ is red-blue}
\label{rst11}
\end{figure}
 
In Case IV, we assume that $v$ is colored red-blue, see Figure \ref{rst12}.
In this case, the sum $Q_4$ of the weights of the coloring is 
\begin{equation}\label{eq:1-Q4}
Q_4 = q^{B+D+2}\frac{1}{t^{A+C+2}}V\frac{1}{{\bf E}}= -\frac{{\bf BD}}{{\bf ACE}} V.
\end{equation}


\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{endb12.pdf}
$$
\caption{Case IV for $G_2$, $u$ is red-blue}
\label{rst12}
\end{figure}

\noindent
Then it is easy to check that 
$$
Q_3+Q_4 = \frac{-1}{{\bf ACE}}V.$$
so that 
\begin{equation}\label{wG2}
W(H_2) =\frac{(q-1)}{{\bf ACE}}U - \frac{1}{{\bf ACE}}V.
\end{equation}

Thus comparing (\ref{wG1}) and (\ref{wG2}), we see that 
$W(G_1) =W(G_2)$ as claimed.
\end{proof}






We now are in position to prove the main result of this section.
\begin{thm}\label{thm:rrcon2}
\begin{enumerate}

\item If $H$ is a red-red connecting rim hook, then $W(H) = \frac{(q-t)}
{t^{r(h)}}$.
\item If $H$ is a red-blank connecting rim hook, then $W(H) =\frac{(q-1)}
{t^{r(h)}}$.
\end{enumerate}
\end{thm}
\begin{proof}
We shall prove only (1) since the proof of (2) is similar.
It follows from Lemma~\ref{red-reduction} that we need only consider 
the connecting rim hooks $H$ where there 
are either one or two lower corner squares.

Let us first consider the case where the red-red connecting rim hook 
$H$ has one lower corner square. 
There are two basic cases, 
namely, the case where the corner square is blank and 
the corner square is colored red-blue, see Figure \ref{rrhook}. 

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{rrhook.pdf}
$$
\caption{Red-red 1 lower corner shapes}
\label{rrhook}
\end{figure}

If the corner square is blank, then it is easy to see that sum of the weights 
of the possible colorings is 
$$
W(rV(A))W(Hr(B)) = \frac{(q-1)}{t^{A+1}}\frac{(q-t) + (t-1)q^{B+1}}{t(q-1)} = 
\frac{(q-t) + (t-1)q^{B+1}}{t^{A+2}}.
$$
If the corner cell is colored red-blue, then top red cell has 
weight $\frac{-1}{t}$ and the remaining $A$ cells in the vertical 
part each contribute a weight $\frac{-1}{t}(-1) = \frac{1}{t}$. 
Each cell in the bottom row except for the cell that immediately 
precedes the red cell, contributes a weight of $\frac{q}{t}t = q$. The 
cell that immediately precedes the red cell contributes a weight of 
$\frac{q}{t} (t-1)= \frac{(t-1)q}{t}$. Thus the weight corresponding to this configuration 
is $- \frac{(t-1)q^{B+1}}{t^{A+2}}$. Hence 
$W(H)=\frac{q-t}{t^{A+2}} = \frac{(q-t)}{t^{r(H)}}$ in this case. 


Next we consider the case where the red-red connecting rim hook 
has two corner squares. Thus there are four cases depending 
on whether the two lower corner squares are colored red-blue or blank as 
pictured in Figure \ref{rr2corner}.

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=2.25in]{rr2corner.pdf}
$$
\caption{Red-red 2 lower corner shapes}
\label{rr2corner}
\end{figure}


In Case 1 of Figure \ref{rr2corner}, we have the sum of 
the weights of the colorings in this case is 
\begin{align}
 W(rV(A))&W(INVTL(B,C))W(Hr(D))\\
 &=
\frac{(q-1)}{t^{A+1}}
\left(q^{B+1} +\frac{1}{t^{C+1}}-\frac{q^{B+1}}{t^{C+1}}\right)
\frac{(q-t) + (t-1)q^{D+1}}{t(q-1)}  \nonumber \\
&=\frac{(q^{B+1}t^{C+1} +1-q^{B+1})((q-t)+(t-1)q^{D+1})}{t^{A+C+3}}.
\label{rr2corner1}
\end{align}

In Case 2 of Figure \ref{rr2corner}, we have the sum of 
the weights of the colorings in this case is 
\begin{align}
\frac{-1}{t^{A+1}} q^{B+1} W(rbV(C))W(Hr(D)) &=
 \frac{-q^{B+1}}{t^{A+1}}
 (q-1)\left( 1 - \frac{1}{t^{C+1}}\right)\frac{(q-t) + (t-1)q^{D+1}}{t(q-1)}\nonumber \\
&=\frac{-q^{B+1}(t^{C+1}-1)((q-t) + (t-1)q^{D+1})}{t^{A+C+3}}.
\label{rrcorner2}
\end{align} 
It is then easy to see that the sum of the weights of the colorings 
in Cases~1 and 2 is simply
\begin{equation}\label{rr2corner3}
\frac{(q-t) + (t-1)q^{D+1}}{t^{A+C+3}}.
\end{equation}

In Case 3 of Figure \ref{rr2corner}, we have the sum of 
the weights of the colorings in this case is 
\begin{align}
W(rV(A))W(Hrb(B)) \frac{1}{t^{C+1}}q^D\frac{q(t-1)}{t}&=
\frac{(q-1)}{t^{A+1}}\frac{(q^{B+1}-1)}{(q-1)} 
 \frac{1}{t^{C+1}}q^D\frac{q(t-1)}{t} \nonumber \\
&= \frac{(t-1)(q^{B+1}-1)q^{D+1}}{t^{A+C+3}}.
\label{rr2corner4}
\end{align} 

Finally in  case 4 of Figure \ref{rr2corner}, we have the sum of 
the weights of the colorings in this case is 
\begin{equation}\label{rr2corner5}
\frac{-1}{t}\frac{1}{t^A} q^{B+1}\frac{1}{t^{C+1}}q^D\frac{q(t-1)}{t}= 
\frac{-(t-1)q^{D+1}q^{B+1}}{t^{A+C+3}}.
\end{equation} 
It is then easy to see that the sum of the weights of the colorings 
in Cases~3 and 4 is simply
\begin{equation}\label{rr2corner6}
\frac{-(t-1)q^{D+1}}{t^{A+C+2}}.
\end{equation}

Thus comparing equations (\ref{rr2corner3}) and 
(\ref{rr2corner6}), we see that 
$W(H) = \frac{(q-t)}{t^{A+C+3}} = \frac{(q-t)}{t^{r(H)}}$ in this case as well.
\end{proof}

At this point, to complete our calculation, we need only compute 
the weights of blue-red connecting rim hooks.

\section{$W(H)$ for blue-red connecting rim hooks.}

In this section, we shall compute $W(H)$ when $H$ is a blue-red 
connecting rim hook. As one might expect from the formula given 
in section 1, this is the most involved computation for the 
weights of connecting 
rim hooks. 

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{brhook.pdf}
$$
\caption{Blue-red 1 lower corner shape}
\label{br1corner}
\end{figure}


We start with the case when $H$ has one lower corner, see Figure 
\ref{br1corner}. 
\begin{thm}\label{thm:br1corner}
If $H$ is a blue-red connecting rim hook with one lower corner, then 
\begin{equation}\label{eq:br1corner}
W(H)= \frac{(q-t)(t-1)}{(q-1)t}
\left(\frac{[r(H)-1]_t -q[r(H)-2]}{t^{r(H)-2}} -
q^{c(H)}\right).
\end{equation}
\end{thm} 
\begin{proof}
We shall consider two cases. First suppose that the corner cell is blank.
Then the sum of the weights of the colorings in this case is 
\begin{align}
\nonumber
W(&bV(A))W(Hr(B))\\ 
&= \left((t-q) +\frac{(q-1)}{t^A}\right)\frac{((q-t) + (t-1)q^{B+1})}{t(q-1)} 
\nonumber \\
&= \frac{((q-1) -(q-t)t^A)((q-t) + (t-1)q^{B+1})}{(q-1)t^{A+1}} \nonumber \\
&= \frac{((q-1)(q-t) + (q-1)(t-1)q^{B+1}t^A - (q-t)^2t^A -(q-t)(t-1)q^{B+1}t^A)}{(q-1)t^{A+1}}.
\label{eq:br1}
\end{align}

If the corner square is colored red-blue, then all the squares except the top 
blue square and the bottom red square must be colored red-blue so that 
the weight of this configuration is 
\begin{equation}\label{eq:br2}
(-1)\frac{1}{t^A} q^B \frac{q(t-1)}{t} = \frac{-((q-1)(t-1)q^{B+1}t^A)}{(q-1)t^{A+1}}.
\end{equation}

Combining (\ref{eq:br1}) and (\ref{eq:br2}), we see that 
\begin{align*}
W(H) &= \frac{1}{(q-1)t^{A+1}}((q-1)(q-t) - 
(q-t)^2t^A -(q-t)(t-1)q^{B+1}t^A)\\
&= \frac{(q-t)}{(q-1)t^{A+1}}((q-1) - (q-t)t^A -(t-1)q^{B+1}t^A)\\
&= \frac{(q-t)(t-1)}{(q-1)t}\left(\frac{(q-1) - (q-t)t^A}{t^A(t-1)} -q^{B+1}\right)\\
&= \frac{(q-t)(t-1)}{(q-1)t}
\left(\frac{(t^{A+1}-1) - q(t^A-1)}{(t-1)t^A} - q^{B+1}\right)\\
&= \frac{(q-t)(t-1)}{(q-1)t}\left(\frac{[r(H)-1]_t -q[r(H)-2]_t}{t^{r(H)-2}} - q^{c(H)}\right).
\end{align*}
\end{proof}


Now we consider the general case. 
\begin{thm}\label{thm:brgeneral}
Let $H$ be a blue-red rim hook with at least two lower corners and suppose 
that $\alpha$ is the smallest partition which contains $H$ and 
$\beta$ is the partition that results by removing the first row and column 
from $\alpha$. 
Then 
\begin{equation}\label{eq;genbr1}
W(H)= \frac{(q-t)(t-1)}{(q-1)t}\left(\frac{[r(H)-1]_t -q[r(H)-2]}{t^{r(H)-2}} -
q^{c(H)} -(q-t) \sum_{i=1}^{\ell(\beta)} \frac{q^{\beta_i}-1}{t^i}\right).
\end{equation}
\end{thm}
\begin{proof}


We first consider the case where the blue-red connecting rim hook 
has exactly two corner squares. Thus there are four cases depending 
on whether the two lower corner squares are colored red-blue or blank as 
pictured in Figure \ref{br2corner}. We let ${\bf A} = t^A$, 
${\bf B} = q^{B+1}$, ${\bf C}=t^{C+1}$, and ${\bf D} = q^{D+1}$. 

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=2.25in]{br2corner.pdf}
$$
\caption{Blue-red 2 lower corners shape}
\label{br2corner}
\end{figure}


In Case 1 of Figure \ref{br2corner}, we have the sum of 
the weights of the colorings in this case is 
\begin{align}
\nonumber
W(bV(A))&W(INVTL(B,C))W(Hr(D)) \\
&=
 \left((t-q) + \frac{q-1}{t^A}\right)\left(q^{B+1} + 
\frac{1}{t^{C+1}} -\frac{q^{B+1}}{t^{C+1}}\right) 
\frac{(q-t) + (t-1)q^{D+1}}{t(q-1)} \nonumber \\
&= \frac{((q-1) - (q-t)t^A)}{t^A} \frac{(q^{B+1}t^{C+1}+1-q^{B+1})}{t^{C+1}} 
\frac{((q-t) + (t-1)q^{D+1})}{t(q-1)} \nonumber \\
&= \frac{(q-t) + (t-1){\bf D}}{t(q-1){\bf AC}}((q-1) -(q-t){\bf A})
({\bf BC} +1 - {\bf B}).
\label{br2corner1}
\end{align}

In Case 2 of Figure \ref{br2corner}, we have the sum of 
the weights of the colorings in this case is 
\begin{align}
\nonumber
(-1)\frac{1}{t^{A}} q^{B+1} &W(rbV(C))W(Hr(D))\\ 
&= \frac{-q^{B+1}}{t^{A}}
 (q-1)\left( 1 - \frac{1}{t^{C+1}}\right)\frac{(q-t) 
+ (t-1)q^{D+1}}{t(q-1)} \nonumber \\
&= \frac{-q^{B+1}}{t^{A}}
\frac{((q-1)t^{C+1}-(q-1))}{t^{C+1}}\frac{((q-t) + (t-1)q^{D+1})}{t(q-1)} 
 \nonumber \\
&= -\frac{(q-t) + (t-1){\bf D}}{t(q-1){\bf AC}} ((q-1){\bf BC} -(q-1){\bf B}).
\label{brcorner2}
\end{align} 


Since 
\begin{multline*}
((q-1) -(q-t){\bf A})({\bf BC} +1 - {\bf B})) \\=
(q-1){\bf BC} +(q-1) -(q-1){\bf B}-(q-t){\bf ABC} -(q-t){\bf A} +
(q-t){\bf AB},
\end{multline*}
it is easy to see that sum of the weights of the coloring in Cases~1 and 2 
is 
\begin{equation}\label{br2corner3}
\frac{(q-t) + (t-1){\bf D}}{(t(q-1){\bf AC}}
((q-1) -(q-t){\bf ABC} -(q-t){\bf A} + (q-t){\bf AB}).
\end{equation}







In Case 3 of Figure \ref{br2corner}, we have the sum of 
the weights of the colorings in this case is 
\begin{align}
\nonumber 
W&(bV(A))W(Hrb(B)) \frac{1}{t^{C+1}}q^D\frac{q(t-1)}{t} \\ 
&=\left((t-q) + \frac{q-1}{t^A}\right)\frac{(q^{B+1}-1)}{(q-1)} 
\frac{q^{D+1}(t-1)}{t^{C+2}} \nonumber \\
&=\frac{((q-1) -(q-t)t^A)}{t^A}\frac{(q^{B+1}-1)}{(q-1)} 
\frac{q^{D+1}(t-1)}{t^{C+2}} \nonumber \\
&= \frac{(t-1){\bf D}}{t(q-1){\bf AC}} ((q-1) -(q-t){\bf A})({\bf B}-1) 
 \nonumber \\
&= \frac{((q-1)(t-1){\bf BD} -(q-1)(t-1){\bf D} - 
(q-t)(t-1){\bf ABD} + (q-t)(t-1){\bf AD})}{t(q-1){\bf AC}}.
\label{br2corner4}
\end{align} 

Finally in  Case 4 of Figure \ref{br2corner}, we have the sum of 
the weights of the colorings in this case is 
\begin{equation}\label{br2corner5}
(-1)\frac{1}{t^A} q^{B+1}\frac{1}{t^{C+1}}q^D\frac{q(t-1)}{t}= 
\frac{-1}{t(q-1){\bf AC}}((q-1)(t-1){\bf BD}).
\end{equation} 

It is then easy to see that the sum of the weights of the colorings 
in Cases~3 and 4 is simply
\begin{equation}\label{br2corner6}
\frac{1}{t(q-1){\bf AC}}((q-t)(t-1){\bf AD}  -(q-1)(t-1){\bf D} - 
(q-t)(t-1){\bf ABD}).
\end{equation}

If we factor out the common factor of $\frac{1}{t(q-1){\bf AC}}$ in 
(\ref{br2corner3}) and (\ref{br2corner6}), we are left with 
\begin{multline*}
 (q-t)(q-1) -(q-t)^2{\bf ABC} -(q-t)^2{\bf A} + (q-t)^2{\bf AB})\\
 +((q-1)(t-1){\bf D} -(q-t)(t-1){\bf ABCD} -(q-t)(t-1){\bf AD} + 
(q-t)(t-1){\bf ABD})\\
+(q-t)(t-1){\bf AD}  -(q-1)(t-1){\bf D} - (q-t)(t-1){\bf ABD} \\
= (q-t)(q-1) -(q-t)^2{\bf ABC} -(q-t)^2{\bf A} + (q-t)^2{\bf AB})-
(q-t)(t-1){\bf ABCD}.
\end{multline*} 
It follows that 
\begin{align*}
W(H) &= \frac{(q-t)}{t(q-1)}\left( \frac{(q-1)}{{\bf AC}} -(q-t)
\frac{({\bf ABC} +{\bf A} -{\bf AB})}{{\bf AC}} -(t-1){\bf BD}\right)\\
&=\frac{(q-t)}{t(q-1)}\left( \frac{(q-1)}{{\bf AC}} -(q-t)
\left({\bf B} +\frac{1}{{\bf C}} -\frac{{\bf B}}{{\bf C}}\right)-(t-1){\bf BD}\right).
\end{align*}

In this case $\beta = (B+1)^{C+1}$ and the border of $\beta$, $B_\beta$ is 
equal to $INVL(B,C)$. Note that by Theorem~\ref{blank-blank} and 
Lemma~\ref{lem:inverted-L}, we have that 
\begin{align}
W(B_\beta) &= 1+(t-1) \sum_{i=1}^{\ell(\beta)} \frac{q^{\beta_i} -1}{t^i} \nonumber \\
&= 1 + (t-1) \sum_{i=1}^{C+1} \frac{q^{B+1}-1}{t^i} \nonumber \\
&= q^{B+1} + \frac{1}{t^{C+1}} - \frac{q^{B+1}-1}{t^{C+1}} \nonumber \\
&= 1+(q^{B+1}-1)(t-1)\sum_{i=1}^{C+1}\frac{1}{t^i} \nonumber \\
&= 1+ (q^{B+1}-1)(1 -\frac{1}{t^{C+1}}) \nonumber \\
&= \left({\bf B} + \frac{1}{{\bf C}} - \frac{{\bf B}}{{\bf C}}\right).
\label{br2corner7}
\end{align}
It follows that 
\begin{align*}
W(H) &=\frac{(q-t)}{t(q-1)}\left( \frac{(q-1)}{{\bf AC}} -(q-t)
\left({\bf B} +\frac{1}{{\bf C}} -\frac{{\bf B}}{{\bf C}}\right)-(t-1){\bf BD}\right) \\ 
&= \frac{(q-t)}{t(q-1)}\left( \frac{(q-1)}{{\bf AC}} -(q-t)\frac{{\bf AC}}{{\bf AC}}-(q-t)(t-1)\left(\sum_{i=1}^{\ell(\beta)} \frac{q^{\beta_i} -1}{t^i}\right)-(t-1){\bf BD}\right) \\
&= \frac{(q-t)(t-1)}{t(q-1)}\left( \frac{({\bf tAC} -1) -q({\bf AC} -1)}{(t-1){\bf AC}} -(q-t)\left(\sum_{i=1}^{\ell(\beta)} \frac{q^{\beta_i} -1}{t^i}\right)-
{\bf BD}\right). 
\end{align*}
Note that  $t{\bf AC} = t^{A+C +2} =t^{r(H)-1}$, ${\bf AC} = t^{r(H)-2}$, and 
${\bf BC} = q^{B+C+2} = q^{c(H)}$. Thus it follows that 
\begin{align*}
W(H) &= \frac{(q-t)(t-1)}{t(q-1)}\left( \frac{t^{r(H)-1} -1}{(t-1)t^{r(H)-1}} - 
\frac{q(t^{r(H)-2} -1)}{(t-1)t^{r(H) -1}} -q^{c(H)} -(q-t)\sum_{i=1}^{\ell(\beta)} \frac{q^{\beta_i} -1}{t^i}\right) \\
&= \frac{(q-t)(t-1)}{t(q-1)}\left(\frac{[r(H)-1]_t -q[r(H)-2]_t}{t^{r(H)-1}} - 
q^{c(H)} -(q-t)\sum_{i=1}^{\ell(\beta)} \frac{q^{\beta_i} -1}{t^i}\right)
\end{align*}
which is what we wanted to prove. 


Finally we consider the case where $H$ has 3 or more lower corner squares. 
Again we shall let $\alpha$ be the smallest partition that contains $H$ and 
$\beta$ be the partition whose diagram is the result of removing the first row and column of $\alpha$.  We let 
$u$ denote the top lower corner of $H$ and $v$ denote the bottom lower 
corner of $H$. We let $B_\beta$ denote the border of $\beta$, $a$ denote the first cell of $B_\beta$, and $b$ denote the last cell of $B_\beta$. See Figure \ref{br3corner}. We let $B_\beta^{a}$ denote the connecting rim hook which 
results from $B_\beta$ by coloring cell $a$ red-blue. 
We let $B_\beta^{b}$ denote the connecting rim hook which 
results from $B_\beta$ by first removing the vertical portion corresponding to $C$ at the bottom of diagram and then coloring the cell at the top of column that contains $b$ red-blue. Note, in this case, 
we do not consider the top cell in the column above $b$ to be part of  
$B_\beta^{b}$. Finally 
we let $B_\beta^{ab}$ denote the connecting rim hook which 
results from $B^{b}_\beta$ by coloring the cell $a$ red-blue. Again, in 
this case, we do consider the cell that is at the top of the column that contains 
$b$ to be part of $B_\beta^{ab}$.


\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=2.25in]{br3corner.pdf}
$$
\caption{Blue-red $\geq 3$ lower corners shape}
\label{br3corner}
\end{figure}





We consider 4 cases depending on whether cells $u$ and $v$ are colored 
red-blue. In fact, we shall also need to consider similar cases for two 
other connecting rim hooks derived from $H$, namely, $bbH$ which is the same as $H$ except that the bottom cell is colored blue  instead of red and $rH/row1$ which is same as $H$ except that the top cell is red instead of blue and the first row is removed. See Figure \ref{derived}.

\bigskip


\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=2.25in]{derived.pdf}
$$
\caption{The shapes $bbH$ and $rH/row1$}
\label{derived}
\end{figure}






\noindent 
{\bf Case 1.} Both $u$ and $v$ are blank. See Figure \ref{br3corner1}.

\bigskip

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{br3corner1.pdf}
$$
\caption{Blue-red $\geq$ 3 lower corners shape: Case 1}
\label{br3corner1}
\end{figure}
 

We shall refer to the sum of the weights of all the colorings in this case as 
$W(H(1))$. Similarly, we shall refer to the similar sum of the weights 
of coloring for $bbH$ and $rH/row1$ as $W(bbH(1))$ and $W(rH/row1(1))$ respectively. 
Then it is easy to see that 
\begin{align}
W(H(1)) &= W(bV(A)) W(B_\beta)W(Hr(D)) \nonumber \\
&=\left((t-q) + \frac{(q-1)}{t^A}\right)\left(\frac{(q-t)+(t-1)q^{D+1}}{t(q-1)}\right)W(B_\beta).
\label{genH1}
\end{align}
and 
\begin{align}
W(bbH(1)) &= W(bV(A)) W(B_\beta)W(Hb(D)) \nonumber \\
&=\left((t-q) + \frac{(q-1)}{t^A}\right)q^D W(B_\beta).
\label{genH1bb}
\end{align}
Similarly,
\begin{align}
W(rH/row(1)) &= W(rV(A)) W(B_\beta) \nonumber \\
&=\frac{(q-1)}{t^{A+1}} W(B_\beta).
\label{genrH/row11}
\end{align}
Note that 
\begin{align*}
\left((t-q) + \frac{(q-1)}{t^A}\right)&
\left(\frac{(q-t)+(t-1)q^{D+1}}{t(q-1)}\right) \\
&\kern-2cm
=\left((t-q) + \frac{(q-1)}{t^A}\right)
\left(\frac{t(q-1) -q(t-1) +(t-1)q^{D+1}}{t(q-1)}\right)\\
&\kern-2cm
=\left((t-q) + \frac{(q-1)}{t^A}\right)
\left(1+ \frac{q(t-1)(q^{D}-1)}{t(q-1)}\right)\\
&\kern-2cm
=\left((t-q) + \frac{(q-1)}{t^A}\right)
\left(\frac{q(t-1)}{t(q-1)}q^D +1 - 
\frac{q(t-1)}{t(q-1)}\right)\\
&\kern-2cm
=\frac{q(t-1)}{t(q-1)} \left((t-q) + \frac{(q-1)}{t^A}\right)q^D +
\left(1 -\frac{q(t-1)}{t(q-1)}\right)
\left((t-q) + \frac{(q-1)}{t^A}\right)\\
&\kern-2cm
= \frac{q(t-1)}{t(q-1)}\left((t-q) + \frac{(q-1)}{t^A}\right)q^D  \\
&\kern-2cm
\kern2cm
+ t\left(1 -\frac{q(t-1)}{t(q-1)}\right) \frac{(q-1)}{t^{A+1}}+ 
(t-q)\left(1-\frac{q(t-1)}{t(q-1)}\right).
\end{align*}
It then follows that 
\begin{multline}\label{genH12}
W(H(1)) =
\frac{q(t-1)}{t(q-1)}W(bbH(1))\\
 + t\left(1- \frac{q(t-1)}{t(q-1)}\right)W(rH/row1(1)) + (t-q)\left(1-\frac{q(t-1)}{t(q-1)}\right)W(B_\beta).
\end{multline}

\bigskip
\noindent 
{\bf Case 2.} Cell $u$ is colored red-blue and $v$ are blank. See Figure \ref{br3corner2}.

\bigskip

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{br3corner2.pdf}
$$
\caption{Blue-red $\geq$ 3 lower corners shape: Case 2}
\label{br3corner2}
\end{figure}
 

We shall refer the sum of the weights of all the colorings in this case as 
$W(H(2))$. Similarly, we shall refer to the similar sum of the weights 
of coloring for $bbH$ and $rH/row1$ as $W(bbH(2))$ and $W(rH/row1(2))$ respectively. 
Then it is easy to see that 
\begin{align}
W(H(2)) &= (-1)\frac{1}{t^A} q^{B+1}W(B^{a}_\beta)W(Hr(D)) \nonumber \\
&=\frac{-q^{B+1}}{t^A}\left(\frac{(q-t)+(t-1)q^{D+1}}{t(q-1)}\right)W(B^{a}_\beta)
\label{genH2}
\end{align}
and 
\begin{align}
W(bbH(1)) &= (-1)\frac{1}{t^A} q^{B+1}W(B^{a}_\beta)W(Hb(D)) \nonumber \\
&=\frac{-q^{B+1}}{t^A}\left((t-q) + \frac{(q-1)}{t^A}\right)q^D W(B^{a}_\beta).
\label{genH2bb}
\end{align}
Similarly,
\begin{align}
W(rH/row(1)) &= \frac{-1}{t}\frac{1}{t^A} q^{B+1}W(B^a_\beta) \nonumber \\
&= \frac{-q^{B+1}}{t^{A+1}}W(B^a_\beta).
\label{genrH/row12}
\end{align}
Next observe that 
\begin{align*}
\left(\frac{(q-t)+(t-1)q^{D+1}}{t(q-1)}\right) &=
\left(\frac{t(q-1) -q(t-1) +(t-1)q^{D+1}}{t(q-1)}\right)\\
&=\left(1+ \frac{q(t-1)(q^{D}-1)}{t(q-1)}\right)\\
&=\frac{q(t-1)}{t(q-1)}q^D +(1 - 
\frac{q(t-1)}{t(q-1)}).
\end{align*}
It then follows that 
\begin{equation}\label{genH22}
W(H(2)) = 
\frac{q(t-1)}{t(q-1)}W(bbH(2)) + t\left(1- \frac{q(t-1)}{t(q-1)}\right)W(rH/row1(2)).
\end{equation}

By part 2 of Theorem~\ref{thm:rrcon2},
$$\frac{(q-1)}{t^{r(rH/row1)}}= W(rH/row1)=W(rH/row1(1))+W(rH/row1(2)).$$ 
Since  
$r(rH/row1) = r(H)-1$, we get that   
\begin{multline}
W(H(1))+W(H(2)) =
\frac{q(t-1)}{t(q-1)}(W(bbH(1)) + W(bb(H(2))))\\
+t\left(1- \frac{q(t-1)}{t(q-1)}\right)\frac{(q-1)}{t^{r(H)-1}} + 
(t-q)\left(1-\frac{q(t-1)}{t(q-1)}\right)W(B_\beta).
\label{gen1+2}
\end{multline}



\bigskip
\noindent 
{\bf Case 3.} Cell $u$  is blank and cell $v$ is  red-blue. 
See Figure \ref{br3corner3}.

\bigskip

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{br3corner3.pdf}
$$
\caption{Blue-red $\geq$ 3 lower corners shape: Case 3}
\label{br3corner3}
\end{figure}
 

We shall refer to the sum of the weights of all the colorings in this case as 
$W(H(3))$. Similarly, we shall refer to the similar sum of the weights 
of coloring for $bbH$ as $W(bbH(3))$. 
Then it is easy to see that 
\begin{align}
W(H(3)) &= W(bV(A)) W(B^{b}_\beta)\frac{1}{t^{C+1}}q^D\frac{q(t-1)}{t} \nonumber \\
&=\left((t-q) + \frac{(q-1)}{t^A}\right)
\left(\frac{q^{D+1}(t-1)}{t^{C+2}}\right)W(B^b_\beta).
\label{genH3}
\end{align}
and 
\begin{align}
W(bbH(3)) &= W(bV(A)) W(B^{b}_\beta) \frac{1}{t^{C+1}}q^D (q-1)\nonumber \\
&=\left((t-q) + \frac{(q-1)}{t^A}\right)\frac{t(q-1)q^D}{t^{C+2}} W(B^b_\beta).
\label{genH3bb}
\end{align}
Thus 
\begin{equation}\label{genH13}
W(H(3)) = 
\frac{q(t-1)}{t(q-1)}W(bbH(3)).
\end{equation}


\bigskip
\noindent 
{\bf Case 4.} Both $u$ and $v$ are red-blue. 
See Figure \ref{br3corner4}.

\bigskip

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=1.25in]{br3corner4.pdf}
$$
\caption{Blue-red $\geq$ 3 lower corners shape: Case 4}
\label{br3corner4}
\end{figure}
 

We shall refer to the sum of the weights of all the colorings in this case as 
$W(H(4))$. Similarly, we shall refer to the similar sum of the weights 
of coloring for $bbH$ as $W(bbH(4))$. 
Then it is easy to see that 
\begin{equation}\label{genH4}
W(H(4)) =  (-1)\frac{1}{t^A}q^{B+1}W(B^{ab}_\beta)\frac{1}{t^{C+1}}q^D\frac{q((t-1)}{t}.
\end{equation}
and 
\begin{equation}\label{genH4bb}
W(bbH(4)) =  (-1)\frac{1}{t^A}q^{B+1}W(B^{ab}_\beta) 
\frac{1}{t^{C+1}}q^D (q-1).
\end{equation}
Thus again we have 
\begin{equation}\label{genH14}
W(H(4)) = 
\frac{q(t-1)}{t(q-1)}W(bbH(4)).
\end{equation}

By part 3 of Theorem~\ref{thm:rbcon},
$$W(bbH(1))+W(bbH(2)) + W(bbH(3)) + W(bbH(4))= W(bbH) =
(t-q)q^{c(H)-1}.$$ Thus 
\begin{align}
 \nonumber
W(H) &= W(H(1))+W(H(2))+W(H(3))+W(H(4)) \\
&= \frac{q(t-1)}{t(q-1)}W(bbH)+ t\left(1- \frac{q(t-1)}{t(q-1)}\right)\frac{(q-1)}{t^{r(H)-1}} + 
(t-q)\left(1-\frac{q(t-1)}{t(q-1)}\right)W(B_\beta) \nonumber\\
&= \frac{q(t-1)}{t(q-1)}((t-q)q^{c(H)-1})+ t\left(1- \frac{q(t-1)}{t(q-1)}\right)\frac{(q-1)}{t^{r(H)-1}} \nonumber\\
&\kern6cm
+ 
(t-q)\left(1-\frac{q(t-1)}{t(q-1)}\right)W(B_\beta).
\label{gen1+2+3+4}
\end{align}
Using the fact that 
\begin{equation*}
1-\frac{q(t-1)}{t(q-1)}= \frac{t(q-1) -q(t-1)}{t(q-1)} = \frac{(q-t)}{t(q-1)},
\end{equation*}
we see that 
$$
W(H) = -\frac{(q-t)(t-1)}{t(q-1)}q^{c(H)} + \frac{(q-t)}{t^{r(H)-1}} - 
\frac{(q-t)^2}{t(q-1)}W(B_\beta).
$$
Since $W(B_\beta) = 1 +(t-1)\sum_{i=1}^{\ell(\beta)}\frac{q^{\beta_i}-1}{t^i}$,
it follows that 
$$
W(H) = \frac{(q-t)(t-1)}{t(q-1)}\left(\frac{(q-1)}{(t-1)t^{r(H)-2}} - q^{c(H)} - \frac{(q-t)}{(t-1)}-(q-t) \sum_{i=1}^{\ell(\beta)}\frac{q^{\beta_i}-1}{t^i}  \right). 
$$
Finally using the fact that 
\begin{align*}
\frac{(q-1)}{(t-1)t^{r(H)-2}} - \frac{(q-t)}{(t-1)} &=
\frac{(q-1)-(q-t)t^{r(H)-2}}{(t-1)t^{r(H)-2}}  \\
&= \frac{(t^{r(H)-1}-1)-q(t^{r(H)-2}-1)}{(t-1)t^{r(H)-2}} \\
&=\frac{[r(H)-1]_t -q[r(H)-2]_t}{t^{r(H)-2}},
\end{align*}
we see that  
\begin{equation}\label{final}
W(H) = 
\frac{(q-t)(t-1)}{t((q-1)}\left(
\frac{[r(H)-1]_t -q[r(H)-2]_t}{t^{r(H)-2}} - q^{c(H)} -  (q-t) \sum_{i=1}^{\ell(\beta)}\frac{q^{\beta_i}-1}{t^i}.  \right) 
\end{equation}
as claimed.

\end{proof}

\section{Conclusion}

In this paper, we have proved the following theorem.
\begin{thm} \label{main}
Suppose that $\la = (\la_1 \geq \cdots \geq \la_n)$ is a partition of $n$. Then
\begin{equation}
D^1_n(s_\la(x_1, \ldots ,x_n)) = \sum_{\gamma \vdash n} d_{\la,\gamma}(q,t) 
s_\gamma(x_1,\ldots, x_n)
\end{equation}
where 
\begin{description}
\item[(a)] $d_{\la,\la} = \sum_{i=1}^n t^{n-i}q^{\la_i}$, 

\item[(b)] $d_{\la,\gamma} = 0$ unless $\gamma$ satisfies (*) where

\bigskip
\noindent
(*) $\gamma$ arises from $\la$ by first removing a broken rim hook $R$ 
of $\la$ from $\lambda$ to get a partition $\mu$ and then 
adding a broken rim hook 
$B$ of $\gamma$ on the outside of $\mu$ to obtain obtain $\gamma$ so that 
rows occupied by $B$ lie strictly above the rows occupied by $R$ and 
$|B| =|R|$.
\bigskip

\item[(c)] If $\gamma$ satisfies (*), we let $D$ be the diagram consisting of the squares of $\la \cup \gamma$ and we color the squares of $R$ with red and the squares of $B$ with blue. We then let $H_1, \ldots ,H_m$ be the 
connecting rim hooks determined by $\la$ and $\gamma$, ${\mathcal B} = 
\{b_1, \ldots, b_p\}$ be 
the blue squares which are not part of any connecting rim hook, and $
{\mathcal R} = \{r_1, \ldots,r_q\}$ be the red squares which are not part of any connecting rim hook. Then 
\begin{equation}
d_{\la,\gamma}(q,t) = \frac{t^n}{(t-1)}\left(\prod_{i=1}^m W(H_i)\right)
\left(\prod_{j=1}^p w_b(b_j)\right)\left(\prod_{k=1}^q w_r(r_k)\right)
\end{equation}
where for any square $s \in {\mathcal B} \cup {\mathcal R}$, 
\begin{equation}\label{a-weight2}
w_b(s) = \begin{cases} t & \text{if $s$ has a blue square in $D$ to its right},\\
-1  & \text{if $s$ has a blue square in $D$ below it},\\
(t-1) & \text{otherwise},
\end{cases}
\end{equation}
  
\begin{equation}\label{r-weight2}
w_r(s) = \begin{cases} \frac{q}{t} & \text{if $s$ has a red square in $D$ to its right},\\
\frac{-1}{t} & \text{if $s$ has a red square in $D$ below it},\\
\frac{(q-1)}{t} & \text{otherwise}, 
\end{cases}
\end{equation}
and for connecting rim hook $H_i$, 
\begin{enumerate}
\item $W(H_i)=q^{c(H_i)}$ if  $H_i$ is a blank-blue connecting rim hook, 

\item $W(H_i)= (t-q)q^{c(H_i)-1}$, if $H_i$ is a blue-blue connecting rim hook, 
\item $W(H_i)= \frac{(q-t)}{t^{r(H_i)}}$ if  $H_i$ is a red-red connecting rim hook, 

\item $W(H_i)= \frac{(q-1)}{t^{r(H_i)}}$ if  $H_i$ is a red-blank connecting rim hook, and 


\item $W(H_i) = \frac{(t-1)(q-t)}{t(q-1)}\left( 
\frac{[r(H) -1]_t -q[r(H)-2]_t}{t^{r(H) -2}} -q^{c(H)} - (q-t)
\sum_{i=1}^{\ell(\beta)} \frac{q^{\beta_i} -1}{t^i}\right)$ if $H_i$ is a blue-red connecting rim hook and $\beta$ is the partition whose diagram results by removing the first 
row and column of the smallest shape $\alpha$ which contains $H_i$ and by definition, the sum 
$\sum_{i=1}^{\ell(\beta)} \frac{q^{\beta_i} -1}{t^i}$ is equal to 0 if $\beta$ is empty.
\end{enumerate}

\end{description}
\end{thm}

For example, suppose that $\la = (2^2,1)$.  Then the two $\gamma$ that 
satisfy condition (*) for $\la$ are pictured in Figure \ref{221} along 
with the corresponding connecting rim hooks $H_i$, blue squares $b_j$ and 
red squares $r_k$. Note that as usual, we do not consider the red square 
at the bottom of $H_1$ to be part of $H_1$.

\begin{figure}[tbp]
\centering
%\vspace{.1in}
$$
\includegraphics[height=2.25in]{221.pdf}
$$
\caption{The $\gamma$ for $\la =(2^2,1)$}
\label{221}
\end{figure}
 
Note by Theorem~\ref{main}, $d_{(2^2,1),(2^2,1)}(q,t) = t^4q^2+t^3q^2 + t^2 q + t + 1$. For $\gamma = (2,1^3)$, we have 
\begin{align*} 
W(H_1) &= \frac{(t-1)(q-t)}{t(q-1)}\left(\frac{[3-1]_t -q[3-2]_t}{t^{3-2}} -q\right) \\
&= \frac{(t-1)(q-t)}{t(q-1)}\left(\frac{1+t-q}{t} -q\right)\\
&= \frac{(t-1)(q-t)}{t(q-1)}\frac{1+t-q -qt}{t} \\
&= \frac{(t-1)(q-t)}{t(q-1)}\frac{(q-1)(t+1)}{t} \\
&=\frac{(t-1)(q-t)(t+1)}{t^2}
\end{align*}
and $W(H_2) = \frac{(q-1)}{t^2}$ so that
\begin{align*}
d_{(2^2,1),(2,1^3)}(q,t) &= \frac{t^5}{(t-1)} \frac{(t-1)(q-t)(t+1)}{t^2} \frac{(q-1)}{t^2}\\
&= t(q-t)(t+1)(q-1).
\end{align*}

For $\gamma = (1^5)$, we have $w_b(b_1) = -1$, $w_r(r_1) = \frac{-1}{t}$, 
$w_r(r_2) = \frac{(q-1)}{t}$, and $W(H_1) = \frac{(t-1)(q-t)(t+1)}{t^2}$ as above. Thus 
\begin{align*}
d_{(2^2,1),(1^5)}(q,t) &= \frac{t^5}{(t-1)} (-1) \frac{-1}{t} \frac{(q-1)}{t}\frac{(t-1)(q-t)(t+1)}{t^2}\\
&= t(q-t)(t+1)(q-1).
\end{align*}

Finally, we should note that Macdonald \cite{Macb} defined a whole 
sequence of operators $D_n^i$ via the equation
\begin{equation}
D_n(X;q,t) = \sum_{i=1}^n D_n^i X^i = \frac{1}{\Delta(x_1,\ldots,x_n)}
\sum_{w \in S_n} \epsilon(w) x^{w\delta} \prod_{i=1}^n (1+Xt^{(w\delta)_i}T_q^{(i)})
\end{equation}
where $X$ is another indeterminate, $\delta = (n-1, n-2, \ldots, 1,0)$, $\Delta(x_1,\ldots,x_n)$ is the Vandermonde determinant, $x^{(p_1,\ldots,p_n)} = x_1^{p_1} \cdots x_n^{p_n}$, 
$\epsilon(w)$ is the sign of $w$, and $(w\delta)_i$ is the $i$-th component of 
$w\delta$. A $\la$-ring  expression for $D_n^i$ can be found in \cite{BGHT}, but we have not tried of find a combinatorial description of the coefficients that arise in Schur function expansion of $D_n^i(s_\la(x_1, \ldots,x_n))$ for 
$i \geq 2$. 






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