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\usepackage[dvips]{epsfig}
\usepackage[francais,english]{babel}

\newtheorem{thm}{Theorem}[section]
\newtheorem{cor}[thm]{Corollary}
%\newtheorem{def}[theorem]{Definition}
%\newtheorem{ex}[theorem]{Example}
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%\newtheorem{conj}[theorem]{Conjecture}


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%% Commands
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%\renewcommand{\thesection}{\arabic{section}}
%\include{macros}

%Preliminaries
\newcommand{\br}[1]{[#1]}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Pos}{\mathbb{P}}
\newcommand{\pr}[1]{#1^{\prime}}
\newcommand{\permword}[1]{\sigma_1\sigma_2\cdots \sigma_{#1}}
\newcommand{\stack}[2]{\tiny\begin{array}{c} #1 , \\ #2\end{array}}
%\DeclareMathOperator{\inv}{inv}
\DeclareMathOperator{\cyc}{cyc} \DeclareMathOperator{\zsu}{zsu}
\DeclareMathOperator{\maj}{maj} \DeclareMathOperator{\des}{des}
\newcommand{\area}[1]{\mathrm{Area}(#1)}

%Boards
\newcommand{\Bcol}[1]{B(b_1, \ldots , b_{#1})}
\newcommand{\Bstep}[1]{B(h_1,d_1;\ldots ;h_{#1},d_{#1})}
\newcommand{\Bmstep}[2]{B(h_1+{#1}-1,d_1; \ldots ; h_{#2}, d_{#2}+{#1}-1)}
\newcommand{\tri}[1]{\mathbb{T}_{#1}}


%Basic Rook Theory
\newcommand{\rookset}[2]{{\cal R}_{#1}(#2)}
\newcommand{\hitset}[3]{{\cal H}_{#1,#2}(#3)}
\newcommand{\placeset}[2]{{\cal P}_{#1}(#2)}
\newcommand{\rooknum}[2]{r_{#1}(#2)}
\newcommand{\hitnum}[3]{h_{#1,#2}(#3)}


%q-Rook Theory
\newcommand{\inv}[2]{\mathrm{inv}_{#1}(#2)}
\newcommand{\qrooknum}[3][q]{R_{#2}(#1,#3)}
\newcommand{\qhitnum}[4][q]{A_{#2,#3}(#1,#4)}
\newcommand{\hagstat}[3]{h_{#1,#2}(#3)}
\newcommand{\hagstatprime}[2]{hag^{'}_{#1,h}(#2)}
\newcommand{\dworkinstat}[2]{s_{#1,d}(#2)}

%x,y-rook theory
\newcommand{\yrooknum}[2]{r_{#1}(y,#2)}
\newcommand{\xyhitnum}[3]{a_{#1,#2}(x,y,#3)}


%Cycle-Counting q-Rook Theory
\newcommand{\qyrooknum}[3][y]{R_{#2}(#1,q,#3)}
\newcommand{\qxyhitnum}[4][x,y]{A_{#2,#3}(#1,q,#4)}
\newcommand{\qyhitnum}[4][y]{A_{#2,#3}(#1,q,#4)}
\newcommand{\map}[3]{\phi_{#1,#2,#3}}
\newcommand{\mystat}[3]{b_{#1,#2}(#3)}
\newcommand{\combqyhitnum}[4][y]{\tilde{A}_{#2,#3}(#1,q,#4)}



%Cycle-Counting q-Eulerian numbers
\newcommand{\qEulernum}[2]{E_{#1,#2}(q)}
\newcommand{\combqyEulernum}[2]{\tilde{E}_{#1,#2}(y,q)}
\newcommand{\lrmin}[1]{\mathrm{rlmin}(#1)}


%p,q-Rook Theory
\newcommand{\brpq}[1]{[#1]_{p,q}}
\newcommand{\briggspstat}[2]{\zeta_{#1}(#2)}
\newcommand{\briggsqstat}[2]{\xi_{#1}(#2)}
\newcommand{\pstat}[2]{\hat{\zeta}_{#1}(#2)}
\newcommand{\qstat}[2]{\hat{\xi}_{#1}(#2)}
\newcommand{\pqrook}[2]{r_{#1}(p,q,#2)}
\newcommand{\pqhit}[3]{h_{#1,#2}(p,q,#3)}
\newcommand{\pqbinom}[2]{{#1 \brack #2}_{p,q}}
\newcommand{\combpqhit}[3]{\tilde{h}_{#1,#2}(p,q,#3)}
\newcommand{\mycombpqhit}[3]{\hat{h}_{#1,#2}(p,q,#3)}
\newcommand{\al}[2]{\alpha_{#1}(#2)}
\newcommand{\be}[2]{\beta_{#1}(#2)}
\newcommand{\ep}[2]{\epsilon_{#1}(#2)}


%Cycle-Counting p,q-Rook Theory
\newcommand{\ypqrook}[2]{R_{#1}(y,p,q,#2)}
\newcommand{\ypqhit}[3]{T_{#1,#2}(y,p,q,#3)}

\begin{document}

\newbox\Adr
\setbox\Adr\vbox{
\centerline{\sc Fred Butler}
\vskip18pt
\centerline{Department of Mathematics, West Virginia University}
\centerline{P.O. Box 6310, Morgantown, WV  26506-6310}
\vskip3pt
\centerline{\small \texttt{fbutler@math.wvu.edu}
}


\title{Symmetry and Unimodality in the $q,x,y$-Hit Numbers}
\author[Fred Butler]{\box\Adr}
        }
        \date{}


\begin{abstract}\noindent We prove symmetry, and in some cases
symmetry and unimodality, of polynomials related to the $q,x,y$-hit
numbers introduced by Haglund.  These results generalize theorems
proven by Haglund for the $q$-hit numbers. We also apply one of
these results to obtain a corollary concerning a generalization of
the Eulerian numbers.

\selectlanguage{francais}

\vskip10pt
\noindent{\sc R\'esum\'e.} 
Nous prouvons la sym\'etrie et dans certains cas la
sym\'etrie et l'unimodalit\'e des polyn\^omes relatifs aux $q,x,y$
nombres de contacts introduits par Haglund, g\'en\'eralisant ainsi
certains th\'eor\`emes. Un de ces r\'esultats nous permet d'obtenir
un corollaire \`a propos d'une g\'en\'eralisation des nombres
Eul\'eriens.

\end{abstract}

\maketitle

\selectlanguage{english}

%First page headline in AmS-LaTeX for S\'eminaire Lotharingien de Combinatoire
%--first part
\thispagestyle{myheadings}
\font\rms=cmr8
\font\its=cmti8
\font\bfs=cmbx8
\markright{\its S\'eminaire Lotharingien de
Combinatoire \bfs 54A \rms (2006), Article~B54Ae\hfill}
\def\thepage{}
%
%

\section{Introduction}

\subsection{Preliminaries}

\begin{figure}
\begin{center}
\includegraphics{Bstep.eps}
\caption{The Ferrers board $\Bstep{t}.$\label{fig:Bstep}}
\end{center}
\end{figure}

We will use the notation $SQ_n$ to denote the $n\times n$ square
chess board.  We will number the columns of $SQ_n$ with $1$ through
$n$ going from left to right across the bottom, and the rows of
$SQ_n$ with $1$ through $n$ going from bottom to top.  We will label
a square on $SQ_n$ in column $i$ row $j$ with $(i,j)$.

More generally, a {\itshape board} will be any subset of $SQ_n$
for some $n \in\N$.  A \textit{Ferrers board} is a board with
non-decreasing column heights from left to right, or more
precisely a board of the form $\{ (i,j)\in SQ_n \vert \ 1\leq
j\leq b_i, \ 1\leq i\leq n \}$ where $b_1\leq b_2\leq \cdots \leq
b_n$.  We will denote the Ferrers board with column heights $b_1,
b_2, \ldots , b_n$ by $\Bcol{n}$. We will also specify a Ferrers
board by its step heights and depths. The Ferrers board
$\Bstep{t}$ is shown in Figure~\ref{fig:Bstep}. We will call
$B=\Bcol{n}=\Bstep{t}$ a \textit{regular Ferrers board} if
$b_i\geq i$ for $1\leq i\leq n$, or equivalently if $h_1+\cdots +
h_i \geq d_1+\cdots +d_i$ for $1\leq i\leq t$ as was defined
in~\cite{hyper}. In this paper we will focus on regular Ferrers
boards.

A {\itshape rook placement} on a board $B\subseteq SQ_n$ is a subset
of squares of $B$ such that no two of these squares lie in the same
row or the same column.  As the name suggests, these squares
represent positions on an $n\times n$ chess board where
non-attacking rooks can be placed.  Let $\rooknum{k}{B}$ denote the
number of $k$ rook placements on $B$, and let $\hitnum{n}{k}{B}$
denote the number of $n$ rook placements on $SQ_n$ such that exactly
$k$ rooks lie on $B$.  These are known as the \textit{$k$th rook
number} and the \textit{$k$th hit number}, respectively, of the
board $B$.
%Classical rook theory is concerned with studying the
%relationships between these two numbers.

\subsection{Cycle-counting $q$-rook theory\label{sec:qyrooktheory}}

The cycle-counting $q$-rook numbers were first introduced in the
unpublished work of Ehrenborg, Haglund, and Readdy~\cite{EHR},
defined only for Ferrers boards.  These rook numbers generalize both
the $q$-rook numbers $\qrooknum{k}{B}$ of Garsia and
Remmel~\cite{GaRe}, and the cycle-counting rook numbers
$\yrooknum{k}{B}$ of Chung and Graham~\cite{ChGr}. In order to
describe them, we need to define the following three statistics.

The first statistic is denoted $\mathrm{inv}_B$, a generalization of
the number of inversions of a permutation.  Given a placement $P$ of
rooks on a Ferrers board $B\subseteq SQ_n$, let each rook cancel all
squares to the right in its row and below in its column.  We can
then define $\inv{B}{P}$ to be the number of squares of $B$ which
neither contain a rook from $P$ nor are cancelled.

%First page headline in AmS-LaTeX for S\'eminaire Lotharingien de Combinatoire
%--restoring the headers and pagenumbering
\pagenumbering{arabic}
\addtocounter{page}{1}
\markboth{\SMALL FRED BUTLER}{\SMALL SYMMETRY AND UNIMODALITY IN THE
$q,x,y$-HIT NUMBERS}
%
%

The second statistic is denoted $\cyc$, and is a generalization of
the number of cycles of a permutation.  Given a rook placement $P$
on a board $B\subseteq SQ_n$, it is possible to associate to $P$ a
simple directed graph $G_P$ on $n$ vertices. This fact was first
noted in~\cite{Gessel} (see also~\cite{ChGr} and~\cite{DwoCyc}).
There is an edge from $i$ to $j$ in $G_P$ if and only if there is
a rook from $P$ on the square $(i,j)$.  We can then define
$\cyc(P)$ to be the number of cycles in $G_P$.

The third statistic, denoted $E$, depends on the following fact.
Given any placement $P$ of $j$ non-attacking rooks in columns $1$
through $i-1$ of a Ferrers board $B$ (where $j\leq i-1$), it is an
easy exercise to see that if $b_i\geq i$ then there is exactly one
square in column $i$ where placement of a rook will complete a new
cycle in the digraph $G_P$.  If $b_i<i$ then there is no square
where placing a rook will complete a new cycle. Note that a regular
Ferrers board will have such a square in each of its columns (since
$b_i\geq i$ for all $1\leq i\leq n$). Now for $i$ with $b_i\geq i$
we can define $s_i(P)$ to be the unique square which, considering
only rooks from $P$ in columns $1$ through $i-1$ of $P$, completes a
new cycle. Then let $E(P)$ be the number of $i$ such that $b_i\geq
i$ and there is no rook from $P$ in column $i$ on or above square
$s_i(P)$.  For the rook placement $P$ pictured in
Figure~\ref{fig:GPEx}, we see that $\inv{B}{P}=4$, $\cyc(P)=2$, and
$E(P)=2$ (corresponding to $i=4$ and $i=5$).

\begin{figure}
\begin{center}
\includegraphics{GPex.eps}
\caption{The placement $P$ on $B$ and the associated digraph
$G_P$.\label{fig:GPEx}}
\end{center}
\end{figure}

We will use the common notation $\br{x}=(1-q^x)/(1-q)$ to denote the
\textit{$q$-analog} of the real number $x$,
%Note that
%when $x=n\in\N$,
%$$\br{n}=\frac{1-q^n}{1-q}=1+q+\cdots +q^{n-1}$$
%is a polynomial in $q$.
and $\br{n}!$ to denote the product $\br{n}\br{n-1}\cdots
\br{2}\br{1}$, the $q$-analog of $n!$. For $n,k\in\N$ we denote by
${n \brack k}$ the $q$-analog of the binomial coefficient
$\binom{n}{k}$, equal to
$$\frac{\br{n}!}{\br{k}!\br{n-k}!}=\frac{\br{n}\br{n-1}\cdots\br{n-k+1}}{\br{k}!}$$
for $k\leq n$ and equal to $0$ for $k>n$.  It is a well known fact
that ${n \brack k}$ is a polynomial in $q$. More generally for
$z\in\C$ we will write ${z \brack k}$ for
$$\frac{\br{z}\br{z-1}\cdots \br{z-k+1}}{\br{k}!}.$$

As in~\cite{EHR}, we now define the \textit{$k$th cycle-counting
$q$-rook number} of a Ferrers board $B$ by the equation
\begin{equation}\qyrooknum{k}{B}=\sum_{P \ k \ \mathrm{rooks \ on} \ B}\br{y}^{\cyc(P)}q^{\inv{B}{P}+(y-1)E(P)},\label{eq:qyrooknumdef}\end{equation}
where the sum is taken over all placements $P$ of $k$
non-attacking rooks on $B$.  Letting $y=1$ in
(\ref{eq:qyrooknumdef}) yields the $q$-rook numbers
of~\cite{GaRe}, and letting $q\to 1$ gives the cycle-counting rook
numbers of~\cite{ChGr}.  The $\qyrooknum{k}{B}$ satisfy the useful
equation
\begin{multline} \label{eq:qyfacttheorem}
\sum_{k=0}^{n}\qyrooknum{n-k}{B}\br{z}\br{z-1}\cdots\br{z-k+1}\\=
\prod_{i \ \textrm{with} \ b_i\geq i} \br{z+b_i-i+y} \prod_{i \
\textrm{with} \
b_i<i}\br{z+b_i-i+1},
\end{multline}
a
version of the well-known factorization theorems proven for the
$\rooknum{k}{B}$~\cite{GJW}, $\qrooknum{k}{B}$~\cite{GaRe}, and
$\yrooknum{k}{B}$~\cite{ChGr}.

Haglund~\cite[p.~449]{hyper} further extended this model by defining
the \textit{$q,x,y$-hit numbers} algebraically by the equation
\begin{multline} \label{eq:qxyhitnumdef}
\sum_{k=0}^{n}\qxyhitnum{n}{k}{B}z^k\\=
\sum_{k=0}^{n}\qyrooknum{n-k}{B}\br{x}\br{x+1}\cdots\br{x+k-1}z^k\prod_{i=k+1}^{n}(1-zq^{x+i-1}).
\end{multline}
The $\qxyhitnum{n}{k}{B}$ generalize the $\xyhitnum{n}{k}{B}$ also
discussed in~\cite{hyper} (obtained by letting $q\to 1$ in
(\ref{eq:qxyhitnumdef})), along with the $q$-hit numbers of Garsia
and Remmel~\cite{GaRe} (letting $x=y=1$) and the cycle-counting hit
numbers in the model of Chung and Graham~\cite{ChGr} (when $x=1$ and
$q\to 1$).

The case $x=y$ is studied in~\cite{mythesis}, where for a regular
Ferrers board $B$ the combinatorial interpretation
$$\qxyhitnum[y,y]{n}{k}{B}=\sum_{\stack{P \ n \ \mathrm{rooks \ on} \ SQ_n}{n-k \ \mathrm{rooks \ on} \
B}}\br{y}^{\cyc(P)} q^{(n-\cyc(P))(y-1)+\mystat{n}{B}{P}+E(P)}$$ is
given.  Here the sum is taken over all placements of $n$
non-attacking rooks on $SQ_n$ such that exactly $n-k$ of the rooks
lie on $B$.  The statistic $E$ is as defined above, and
$\mystat{n}{B}{P}$ is the number of squares on $SQ_n$ which neither
contain a rook from $P$ nor are cancelled, after applying the
following cancellation scheme:
\begin{enumerate}
 \item each rook cancels all squares to the right in its row;
 \item each rook on $B$ cancels all squares above it in its column (squares both on $B$ and strictly above $B$);
 \item each rook on $B$ which is also on a square which completes a cycle cancels all squares below it in its column as well;
 \item each rook off $B$ cancels all squares below it but above $B$.
\end{enumerate}
While no combinatorial interpretation is known for the
$\qxyhitnum{n}{k}{B}$ when $x\ne y$, the author suspects that one
exists similar to that given for the $\xyhitnum{n}{k}{B}$
in~\cite[p.~418]{hyper}. Such an interpretation would enhance the
results that follow.

In Section~\ref{Sec:SUAk} we sketch an easy proof of the symmetry
and unimodality of\break $\qxyhitnum[a,b]{n}{k}{B}$ for $a,b\in\N$.  Our
proof for regular Ferrers boards is a simplified version of that
given in~\cite{mat}, for an analogous result concerning the $q$-hit
numbers. We also deduce two corollaries in this section.
%We then apply this theorem to prove a symmetry and unimodality
%property of the \textit{cycle-counting $q$-Eulerian numbers}
%introduced in~\cite{mythesis}.
In Section~\ref{Sec:SAk/di}, we
prove symmetry of the polynomial
$\qxyhitnum[a,b]{n}{k}{B}/\prod_{i=1}^{t}\br{d_i}!$ for any regular
Ferrers board $B=\Bstep{t}$.  Finally in Section~\ref{Sec:SUAk/di},
we prove unimodality of
$\qxyhitnum[a,b]{n}{k}{B}/\prod_{i=1}^{t}\br{d_i}!$ for a certain
class of regular Ferrers boards.

\section{Symmetry and Unimodality of $\qxyhitnum[a,b]{n}{k}{B}$\label{Sec:SUAk}}

If $B=\Bstep{t}\subseteq SQ_n$ is a Ferrers board, let us denote by
$B-h_p-d_p$ the Ferrers board $B(h_1,d_1; \ldots ; h_p-1,d_p-1;
\ldots h_t, d_t)\subseteq SQ_{n-1}$, obtained from $B$ by decreasing
the $p$th step by $1$.  We will write $\area{B}$ for the number of
squares in the board $B$.

Suppose
$$f(q)=\sum_{i=M}^{N}a_iq^i,$$
is a polynomial in $q$ with $a_M,a_N\ne 0$. We call $M+N$ the {\it
virtual degree} of $f$. We will say the polynomial $f(q)$ is
$\zsu(d)$ if either
\begin{enumerate}
\item $f(q)$ is  identically zero, or
\item $f(q)$ is in $\N[q]$, symmetric, and unimodal with virtual degree $d$.
\end{enumerate}

Note that for $s\in\N$, $q^s$ is $\zsu(2s)$ and $\br{s}$ is
$\zsu(s-1)$.  It is also easy to see that if $f$ and $g$ are
polynomials which are both $\zsu(d)$, then $f+g$ is also $\zsu(d)$.
We will use the following lemmas to prove the main proposition of
this section. A proof of Lemma~\ref{lem:zsu*} can be found
in~\cite{stanleylog}.

%\begin{lem} If $f$ and $g$ are polynomials which are both $\zsu(d)$, then $f+g$ is $\zsu(d)$.\label{lem:zsu+}\end{lem}

\begin{lem} If $f$ is $\zsu(d)$ and $g$ is $\zsu(e)$, then $fg$ is $\zsu(d+e)$.\label{lem:zsu*}\end{lem}

%Combining Lemmas~\ref{lem:zsu+} and~\ref{lem:zsu*} with
%(\ref{eq:qyfacttheorem}) and (\ref{eq:qxyhitnumdef}), we can easily
%prove the following.

%\begin{lem}
%Let $a,b \in \N$.  For any regular Ferrers board $B\subseteq
%SQ_n$, $\qxyhitnum[a,b]{n}{0}{B}$ is
%$\zsu(Area(B)+n(b-1)-\binom{n+1}{2})$.\label{lem:A0zsu}
%\end{lem}
%\begin{proof}
%Letting $z=0$ in (\ref{eq:qxyhitnumdef}), we see that
%$\qxyhitnum{n}{0}{B}=\qyrooknum{n}{B}$. Now letting $z=0$ in
%(\ref{eq:qyfacttheorem}) and using the fact that $B$ is regular (so
%there is no $i$ with $b_i<i$), we get that
%$$\qyrooknum{n}{B}=\prod_{i=1}^{n}\br{b_i-i+y},$$
%hence
%$$\qxyhitnum[a,b]{n}{0}{B}=\prod_{i=1}^{n}\br{b_i-i+b}.$$
%Since each $b_i \geq i$, we get that $\br{b_i-i+b}$ is
%$zsu(b_i-i+b-1)$. Finally, by Lemma~\ref{lem:zsu*} we see that
%$\qxyhitnum[a,b]{n}{0}{B}$ is $zsu(\sum_{i=1}^{n}(b_i-i+b-1)) =
%zsu(\sum_{i=1}^{n}b_i - \sum_{i=1}^{n}i +n(b-1)) =
%zsu(Area(B)+n(b-1)-\binom{n+1}{2})$, since $Area(B)=b_1+\cdots +b_n$
%when $B$ is a Ferrers board with column heights $b_1$, \ldots
%$b_n$.\end{proof}

\begin{lem}Let $B=\Bstep{t}\subseteq SQ_n$ be a regular Ferrers board, $B-h_t-d_t\subseteq SQ_{n-1}$ as described earlier.
Then
\begin{multline*}
\qxyhitnum{n}{k}{B}=\br{k+y+d_t-1}\qxyhitnum{n-1}{k}{B-h_t-d_t}\\+
q^{k+y+d_t-2}\br{n+x-y-d_t-k+1}\qxyhitnum{n-1}{k-1}{B-h_t-d_t}
\end{multline*}
for any $1\leq k\leq n$.\label{lem:qxyhitrecurrence}
\end{lem}

\begin{proof}
Let $p=t$ in Lemma 5.7 of~\cite{hyper}.
\end{proof}

The following is now a simple corollary of the above lemmas.  We
offer a brief sketch of the proof.

\begin{prop}
Let $B=\Bstep{t}\subseteq SQ_n$ be a regular Ferrers board, $a,b \in
\N$.  If $n+a+1\geq b+d_t+k$, then $\qxyhitnum[a,b]{n}{k}{B}$ is
$\zsu(\area{B}+n(b+k-1)+k(a-1)-\binom{n+1}{2})$ for $0\leq k\leq n$.
\label{thm:qxysymunimodal}\end{prop}
\begin{proof}
The proof is by induction on $\area{B}$.  When $k=0$, we use
Lemma~\ref{lem:zsu*} with (\ref{eq:qyfacttheorem}) and
(\ref{eq:qxyhitnumdef}) to prove that
$$\qxyhitnum[a,b]{n}{0}{B}=\prod_{i=1}^{n}\br{b_i-i+b},$$ which is
$\zsu(\area{B}+n(b-1)-\binom{n+1}{2})$.  We then use
Lemma~\ref{lem:qxyhitrecurrence}, Lemma~\ref{lem:zsu*}, and the fact
that two polynomials which are $\zsu(d)$ sum to another polynomial
which is $\zsu(d)$ for the case when $k>0$.  Note the assumption
$n+a+1\geq b+d_t+k$ is necessary to ensure that the factor
$\br{n+a-b-d_t-k+1}$ in the recurrence is a polynomial in $q$.

%When $Area(B)=1$ the only regular Ferrers board is the $1\times 1$
%square $SQ_1$.  A quick calculation from the definition shows that
%$\qxyhitnum{1}{0}{SQ_1}=\br{y}$,
%$\qxyhitnum{1}{1}{SQ_1}=q^y\br{x-y}$, and $\qxyhitnum{1}{k}{SQ_1}=0$
%for $k>1$.  Thus $\qxyhitnum[a,b]{1}{0}{B}=\br{b}$ and
%$\qxyhitnum[a,b]{1}{1}{B}=q^b\br{a-b}$ are $\zsu(b-1)$ and
%$\zsu(a+b-1)$ respectively, and the result holds for the case
%$Area(B)=1$.

%Now assume the result holds for all regular Ferrers boards with
%$Area<A$, and let $B$ be such a board with $Area(B)=A$.  We know by
%Lemma~\ref{lem:A0zsu} that the result holds for
%$\qxyhitnum[a,b]{n}{0}{B}$, so assume $k>0$.  Then by
%Lemma~\ref{lem:qxyhitrecurrence} with $x=a$ and $y=b$, we have that
%$$\qxyhitnum[a,b]{n}{k}{B}=\br{k+b+d_t-1}\qxyhitnum[a,b]{n-1}{k}{B-h_t-d_t}+$$
%\begin{equation}q^{k+b+d_t-2}\br{n+a-b-d_t-k+1}\qxyhitnum[a,b]{n-1}{k-1}{B-h_t-d_t}.\label{eq:abhitrec}\end{equation}

%Now we know that $\br{k+b+d_t-1}$ is $\zsu(k+b+d_t-2)$, and by the
%induction hypothesis, $\qxyhitnum[a,b]{n-1}{k}{B-h_t-d_t}$ is
%$\zsu(Area(B-h_t-d_t)+(n-1)(b+k-1)+k(a-1)-\binom{n}{2})$.  Note
%here that $Area(B-h_t-d_t)=Area(B)-n-d_t+1$.  Then by
%Lemma~\ref{lem:zsu*}, the first term on the right side of
%(\ref{eq:abhitrec}) is
%$\zsu(Area(B)+n(b+k-1)+k(a-1)-\binom{n+1}{2})$.

%For the second term on the right side of (\ref{eq:abhitrec}), we
%know that $q^{k+b+d_t-2}$ is $\zsu(2k+2b+2d_t-4)$,
%$\br{n+a-b-d_t-k+1}$ is $\zsu(n+a-b-d_t-k)$ (since we have assumed
%$n+a+1\geq b+d_t+k$), and by the induction hypothesis
%$\qxyhitnum[a,b]{n-1}{k-1}{B-h_t-d_t}$ is
%$\zsu(Area(B-h_t-d_t)+(n-1)(b+k-2)+(k-1)(a-1)-\binom{n}{2})$.
%Finally, applying Lemma~\ref{lem:zsu*} one last time we get that
%the second term on the right side of (\ref{eq:abhitrec}) is
%$\zsu(Area(B)+n(b+k-1)+k(a-1)-\binom{n+1}{2})$, and
%Lemma~\ref{lem:zsu+} gives us the result for
%$\qxyhitnum[a,b]{n}{k}{B}$ as well.
\end{proof}

An immediate corollary is the following.

\begin{cor}
For any regular Ferrers board $B\subseteq SQ_n$ and $m\in\N$, the
polynomial
$$\sum_{\stack{P \ n \ \mathrm{rooks \ on} \ SQ_n}{n-k \ \mathrm{rooks \ on} \
B}}\br{m}^{\cyc(P)} q^{(m-\cyc(P))(y-1)+\mystat{n}{B}{P}+E(P)}$$ is
$\zsu(\area{B}+n(m+k-1)+k(m-1)-\binom{n+1}{2})$.
\end{cor}
\begin{proof}
Let $x=y=m$ in Proposition~\ref{thm:qxysymunimodal}, and use the
combinatorial interpretation for $\qxyhitnum[y,y]{n}{k}{B}$ given in
Section~\ref{sec:qyrooktheory}.
 Note that assuming $B$ is a regular Ferrers board, we always have
$n+m+1\geq m+d_t+k$.  This is because the last $d_t$ columns of $B$
have height n, so in a placement of $n$ non-attacking rooks on
$SQ_n$ with $k$ rooks off $B$, we must have $n-d_t\geq k$.
\end{proof}

A cycle-counting version of the Eulerian numbers is given
in~\cite{mythesis}, defined by the equation
\begin{equation}\combqyEulernum{n}{k}=\sum_{\sigma \in S_n, \ \des(\sigma)=k-1}
\br{y}^{\lrmin{\sigma}}q^{(n-\lrmin{\sigma})(y-1) +
\maj(\sigma)}.\label{eq:qyEulerNumDef}\end{equation} Here
$\des(\sigma)$ denotes the number of descents, and $\lrmin{\sigma}$
the number of right-to-left minima, in the permutation $\sigma$. A
\emph{right-to-left minimum} of $\permword{n}$ is an entry
$\sigma_i$ which is smaller than $\sigma_j$ for all $j>i$ (so for
example $\lrmin{51243}=3$, corresponding to the right-to-left minima
$3$, $2$, and $1$). Note that right-to-left minima have the same
overall distribution as cycles in $S_n$, justifying the term
``cycle-counting.''  It was proven in~\cite{mythesis} that
\begin{equation}\combqyEulernum{n}{k}=\qxyhitnum[y,y]{n}{k-1}{\tri{n}},\label{eq:combqyEulernum=qyhitnum}\end{equation}
where $\tri{n}=B(1,2,\ldots , n)$ denotes the triangular Ferrers
board.  In light of (\ref{eq:combqyEulernum=qyhitnum}) and
Proposition~\ref{thm:qxysymunimodal}, the following can be easily
proven.

\begin{cor}For $m\in\N$, the polynomial
$$\sum_{\sigma \in S_n, \ \des(\sigma)=k-1}
\br{m}^{\lrmin{\sigma}}q^{(n-\lrmin{\sigma})(m-1) + \maj(\sigma)}$$
is $\zsu(n(m+k-2)+(k-1)(m-1))$.
\end{cor}

%\begin{proof}By (\ref{eq:qyEulerNumDef}) and (\ref{eq:combqyEulernum=qyhitnum}),
%$$\sum_{\sigma \in S_n, \ des(\sigma)=k-1}
%\br{m}^{\lrmin{\sigma}}q^{(n-\lrmin{\sigma})(m-1) + maj(\sigma)}=
%\qxyhitnum[m,m]{n}{k-1}{\tri{n}}.$$ The board $\tri{n}$ is a regular
%Ferrers board with area $\binom{n+1}{2}$ and $d_t=1$. For
%$\qxyhitnum[m,m]{n}{k-1}{\tri{n}}$ we have $x=y=m\in\N$, and we can
%assume that $n\geq k-1$ since $\qxyhitnum[y,y]{n}{k-1}{\tri{n}}=0$
%for $k-1>n$. This implies that $n+m+1\geq m+1+k-1$, so the
%hypotheses of Theorem~\ref{thm:qxysymunimodal} hold. Hence we have
%that
%$$\sum_{\sigma \in S_n, \ des(\sigma)=k-1}
%\br{m}^{\lrmin{\sigma}}q^{(n-\lrmin{\sigma})(m-1) + maj(\sigma)}$$
%is $zsu(n(m+k-2)+(k-1)(m-1))$ as desired.
%\end{proof}


\section{Symmetry of $\qxyhitnum[a,b]{n}{k}{B}/\prod_{i=1}^{t}\br{d_i}!$}\label{Sec:SAk/di}
In this section we prove a more general symmetry result for all
regular Ferrers boards, namely the symmetry of the polynomial
$$\frac{\qxyhitnum[a,b]{n}{k}{B}}{\prod_{i=1}^{t}\br{d_i}!},$$
where $B=\Bstep{t}$.  Throughout the rest of the paper we will use
the notation $H_i$ for the partial sum $h_1 + \cdots +h_i$, and
$D_i$ for $d_1+ \cdots + d_i$. We have the following lemmas.

\begin{lem}
Let $B=\Bcol{n}=\Bstep{t}$ be a regular Ferrers board, $j\in\N$.
Then
\begin{equation*}
\frac{\prod_{i=1}^{n}\br{j+b_i-i+y}}{
\prod_{i=1}^{t}\br{d_i}!}=\prod_{i=1}^{t}{j+H_i-D_{i-1}+y-1 \brack
d_i}.
\end{equation*}\label{lem:prod/d_i=prodchoosed_i}\end{lem}

\begin{proof}We see that
\begin{multline*}
\prod_{i=1}^{n}\br{j+b_i-i+y}=
\prod_{i=1}^{t}\Big(\br{j+H_i-D_{i-1}+y-1}\br{(j+H_i-D_{i-1}+y-1)-1}\\
\cdots\br{(j+H_i-D_{i-1}+y-1)-d_i+1}\Big).
\end{multline*}
Thus
\begin{multline*}
\frac{\prod_{i=1}^{n}\br{j+b_i-i+y}}{\prod_{i=1}^{t}\br{d_i}!}\\=
\prod_{i=1}^{t}\frac{\br{j+H_i-D_{i-1}+y-1}\cdots\br{(j+H_i-D_{i-1}+y-1)-d_i+1}}{\br{d_i}!},
\end{multline*}
which is
$$\prod_{i=1}^{t} {j+H_i-D_{i-1}+y-1 \brack d_i}$$
by definition.
\end{proof}

\begin{lem}
Let $B=\Bcol{n}=\Bstep{t}\subseteq SQ_n$ be a regular Ferrers board.
Then 
$$\frac {\qxyhitnum{n}{k}{B}} {\prod_{i=1}^{t}\br{d_i}!}=
\sum_{j=0}^{k}{n+x \brack k-j}{x+j-1 \brack
j}(-1)^{k-j}q^{\binom{k-j}{2}}\prod_{i=1}^{t} {j+H_i-D_{i-1}+y-1
\brack d_i}.$$\label{lem:prodA_k/prodd_i!}\end{lem}
\begin{proof}By Lemma 5.1 of~\cite{hyper}, we have
$$\qxyhitnum{n}{k}{B}=\sum_{j=0}^{k}{n+x \brack k-j}{x+j-1 \brack
j}(-1)^{k-j}q^{\binom{k-j}{2}} \prod_{i=1}^{n}\br{j+b_i-i+y}.$$ The
lemma now follows trivially from
Lemma~\ref{lem:prod/d_i=prodchoosed_i}.
\end{proof}

We can now prove the following.

\begin{thm}
Let $B=\Bstep{t}$ be a regular Ferrers board (so $H_i\geq D_i$ for
$1\leq i\leq t$).  Let $a,b \in \N$ with $a\geq b\geq 1$, and set
$$L_k^{a,b}(B)=\area{B}+n(b-1)+k(n+a-1)-\sum_{i=1}^{t}d_iD_i.$$
Then $\qxyhitnum[a,b]{n}{k}{B}/\prod_{i=1}^{t}\br{d_i}!$ is either
zero or symmetric with virtual degree
$L_k^{a,b}(B)$.\label{thm:A_k/\br{d_i}!sym}
\end{thm}

\begin{proof}
By Lemma~\ref{lem:prodA_k/prodd_i!},
$\qxyhitnum[a,b]{n}{k}{B}/\prod_{i=1}^{t}\br{d_i}!=$
$$\sum_{j=0}^{k}{n+a \brack k-j}{a+j-1 \brack
j}(-1)^{k-j}q^{\binom{k-j}{2}}\prod_{i=1}^{t} {j+H_i-D_{i-1}+b-1
\brack d_i},
$$
which is a polynomial in $q$ (the first two $q$-binomial
coefficients in each summand are clearly polynomials, and the third
is since $H_i\geq D_i\geq D_{i-1}$ and $b\geq 1$).  Using the fact
that ${r \brack s}$ is symmetric with virtual degree $s(r-s)$ and
Lemma~\ref{lem:zsu*}, we see that each term on the right side above
has virtual degree
$(k-j)(n+a-k+j)+j(a-1)+(k-j)(k-j-1)+\sum_{i=1}^{t}d_i(j+H_i-D_i+b-1)$
(which is exactly $L_k^{a,b}(B)$).  We then conclude that if
$\qxyhitnum[a,b]{n}{k}{B}/\prod_{i=1}^{t}\br{d_i}!$ is non-zero,
then it is symmetric with virtual degree $L_k^{a,b}(B)$.
\end{proof}

\section{Unimodality of $\qxyhitnum[a,b]{n}{k}{B}/\prod_{i=1}^{t}\br{d_i}!$\label{Sec:SUAk/di}}

In this section we give some sufficient conditions on the regular
Ferrers board $B$ for the polynomial of the previous section to also
be unimodal.  Let us first define some more notation.

Suppose we have integers $h_1, \ldots , h_t$, $d_1, \ldots , d_t$,
and $e_1, \ldots , e_t$ with $d_i \in \Pos$, $h_i \in \N$, and $0
\leq e_i \leq d_i$.  We will denote the vector $(e_1,e_2,\ldots ,
e_t)$ by $\vec e$.  We will continue to denote the partial sum
$h_1+\cdots +h_i$ by $H_i$, $d_1+\cdots +d_i$ by $D_i$, and we will
also let $E_i=e_1+\cdots +e_i$.  We make the convention that
$H_0=D_0=E_0=0$. For fixed $h_1, \ldots , h_t$ and $d_1, \ldots ,
d_t$ we can define
$$P(\vec e,x,y)=\prod_{i=1}^{t}{H_i-D_{i-1}+E_{i-1}+y-1 \brack d_i-e_i}
 {D_i+D_{i-1}-H_i-E_{i-1}+x-y \brack e_i}$$
and prove the following lemmas.

\begin{lem}Let $B=B(h_1,d_1;\ldots h_{t-1},d_{t-1};h_t,d_t)\subseteq SQ_n$ be a regular Ferrers board,
$\pr{B}=\Bstep{t-1}\subseteq SQ_{H_{t-1}}$.  Then
\begin{multline*}
\qxyhitnum{n}{k}{B}=\br{d_t}!\sum_{s=k-d_t}^{k}\qxyhitnum{H_{t-1}}{s}{\pr{B}}{y+d_t+s-1
\brack d_t-k+s}\\
\times {n-y-d_t+x-s \brack k-s}q^{(k-s)(y+k-1)}.
\end{multline*}
\label{lem:AkxyJim}\end{lem}
\begin{proof}Let $p=t$ in Corollary 5.10
of~\cite{hyper} and note that because $B$ is a regular Ferrers
board, $H_t=D_t=n$.\end{proof}

\begin{lem}
Let $B=\Bstep{t}$ be a regular Ferrers board.  Then
\begin{equation}  \label{eq:Akxyexp}
\qxyhitnum{n}{k}{B}=
\prod_{i=1}^{t}\br{d_i}!\sum_{e_1+\cdots +e_t=k, \ 0\leq e_i\leq
d_i}^{} P(\vec
e,x,y)\prod_{i=1}^{t}q^{e_i(H_i-D_i+E_i+y-1)}.
\end{equation}
\label{lem:A_kzsu}
\end{lem}

\begin{proof}
By induction on $t$.  When $t=1$ we have that $d_1=n$, and
Lemma~\ref{lem:AkxyJim} gives us
\begin{multline}\qxyhitnum{n}{k}{B}=\br{d_1}!\sum_{s=k-n}^{k}\qxyhitnum{0}{s}{\emptyset}{y+n+s-1
\brack d_1-k+s}\label{eq:t=1}\\
\times{n-y-n+x-s \brack k-s}\times q^{(k-s)(y+k-1)}.
\end{multline}
In this case we have that $H_1=D_1=d_1=n$ and $D_0=H_0=0$, so we
get that the $s=0$ term in (\ref{eq:t=1}) is equal to
\begin{equation}\br{d_1}!{H_1-D_0+y-1\brack d_1-k}{D_1+D_0-H_1+x-y\brack k}\times
q^{k(H_1-D_1+k+y-1)}.\label{eq:2ndt=1}\end{equation} Note that by
definition
$$\qxyhitnum{0}{s}{\emptyset}=\delta_{s,0},$$
so the only nonzero summand in (\ref{eq:t=1}) occurs when $s=0$ and
hence (\ref{eq:2ndt=1}) is actually equal to (\ref{eq:t=1}).
Finally if we recall that $E_1=e_1$ and $E_0=0$, we can rewrite
(\ref{eq:2ndt=1}) as
\begin{multline*}
 \br{d_1}!\sum_{e_1=k, \ 0\leq e_1\leq d_1}{H_1-D_0+E_0+y-1\brack d_1-e_1}\\
\times{D_1+D_0-H_1-E_0+x-y\brack e_1}\times q^{e_1(H_1-D_1+E_1+y-1)},
\end{multline*}
which is exactly of the form of (\ref{eq:Akxyexp}).

For $t>1$, Lemma~\ref{lem:AkxyJim} gives that
\begin{multline} \label{eq:induction}
\qxyhitnum{n}{k}{B}=\br{d_t}!\sum_{E_{t-1}=E_t-d_t}^{E_t}\qxyhitnum{H_{t-1}}{E_{t-1}}{\pr{B}}{y+d_t+E_{t-1}-1 \brack d_t-e_t}\\
\times {n-y-d_t+x-E_{t-1}\brack e_t}\times q^{e_t(y+E_t-1)}.
\end{multline}
Here we are letting $E_{t-1}=s$ and defining $e_t=k-s$ and
$E_t=E_{t-1}+e_t=k$.  Since $B$ is regular $H_t=D_t=n$, so
$H_t-D_{t-1}=D_t-D_{t-1}=d_t$ and (\ref{eq:induction}) can be
rewritten as
\begin{multline*}
\qxyhitnum{n}{k}{B} = \br{d_t}!\sum_{e_t=0}^{d_t}\qxyhitnum{H_{t-1}}{E_{t-1}}{\pr{B}}
{H_t-D_{t-1}+E_{t-1}+y-1 \brack d_t-e_t}\\
\times {D_t+D_{t-1}-H_t-E_{t-1}+x-y \brack e_t} \times q^{e_t(H_t-D_t+E_t+y-1)}.
\end{multline*}
By the inductive hypothesis, the above is equal to
\begin{multline*}
\br{d_t}!\sum_{e_t=0}^{d_t} \bigg{\{} \prod_{i=1}^{t-1}\br{d_i}!\sum_{e_1+\cdots +e_{t-1}=E_{t-1}, \ 0\leq e_i\leq d_i} \prod_{i=1}^{t-1} {H_i-D_{i-1}+E_{i-1}+y-1 \brack
d_i-e_i}\\
\times {D_i+D_{i-1}-H_i-E_{i-1}+x-y \brack e_i} q^{e_i(H_i-D_i+E_i+y-1)} \bigg{\}}\\
\times {H_t-D_{t-1}+E_{t-1}+y-1 \brack d_t-e_t}
 {D_t+D_{t-1}-H_t-E_{t-1}+x-y \brack e_t}
 q^{e_t(H_t-D_t+E_t+y-1)},
\end{multline*}
which is
$$\prod_{i=1}^{t}\br{d_i}!\sum_{e_1+\cdots +e_t=k, \ 0\leq e_i\leq d_t}^{}
P(\vec e,x,y)\prod_{i+1}^{t}q^{e_i(H_i-D_i+E_i+y-1)},$$ as desired.
\end{proof}

\begin{lem}
Let $B=\Bstep{t}$ be a regular Ferrers board, $a,b \in \N$ with $a
\geq b\geq 1$.  Let $e_i$, $d_i$, $h_i$, $E_i$, $D_i$, and $H_i$ be
as in the definition of $P(\vec e,x,y)$.  Assume that $B$ is such
that $d_{i-1}+d_i \geq h_i$ for $1\leq i\leq t$ (where $d_0:=0)$.
If any of the numerators of the $q$-binomial coefficients in
$$P(\vec e,a,b) = \prod_{i=1}^{t} {H_i-D_{i-1}+E_{i-1}+b-1 \brack
d_i-e_i}{D_i+D_{i-1}-H_i-E_{i-1}+a-b \brack e_i}$$ are negative,
then $P(\vec e,a,b)=0$. \label{lem:P(e,x,y)=0}
\end{lem}

\begin{proof}
First note that $H_i-D_{i-1}+E_{i-1}+b-1 \geq 0$ for $1\leq i\leq
t$, since $H_i\geq D_i\geq D_{i-1}$ and $b\geq 1$, so none of the
numerators in the first $q$-binomial coefficient of the product are
ever negative.

Now suppose that $D_k+D_{k-1}-H_k-E_{k-1}+a-b < 0$ for some $k$ with
$0\leq k\leq t$.  Note $D_1+D_0-H_1-E_0+a-b=d_1-h_1+a-b$, and since
we assumed $d_{i-1}+d_i\geq h_i$ (and in particular $d_1\geq h_1$)
and $a\geq b$, we have that $d_1-h_1+a-b\geq 0$.  Thus we see that
such a $k$ must be greater than $2$.

Now choose $j$ such that $D_i+D_{i-1}-H_i-E_{i-1}+a-b \geq 0$ for
$1\leq i <j$, but $D_j+D_{j-1}-H_j-E_{j-1}+a-b <0$ (such a $j$
exists because of the remarks in the previous paragraph).  Then
$D_j+D_{j-1}-H_j-E_{j-1}+a-b<0$ implies
$D_j+D_{j-1}-H_j-E_{j-2}+a-b<e_{j-1}$, which is equivalent to
$d_j+d_{j-1}-h_j+D_{j-1}+D_{j-2}-H_{j-1}-E_{j-2}+a-b<e_{j-1}$, which
implies $D_{j-1}+D_{j-2}-H_{j-1}-E_{j-2}+a-b<e_{j-1}$ (since
$d_j+d_{j-1}\geq h_j$).  Hence
$${D_{j-1}+D_{j-2}-H_{j-1}-E_{j-2}+a-b \brack e_{j-1}} = 0,$$
since the numerator is non-negative by definition of $j$ but less
than the denominator, so the product $P(\vec e,a,b)$ is $0$ as well.
\end{proof}

We are now ready to prove the main theorem of this section, the
unimodality of the
$\qxyhitnum[a,b]{n}{k}{B}/\prod_{i=1}^{t}\br{d_i}!$.

\begin{thm}
Let $B=\Bstep{t}$ be a regular Ferrers board such that
$d_{i-1}+d_i\geq h_i$ for $1\leq i\leq t$.  Let $a,b \in \N$ with
$a\geq b\geq 1$, and set
$$L_k^{a,b}(B)=\area{B}+n(b-1)+k(n+a-1)-\sum_{i=1}^{t}d_iD_i$$
as before. Then $\qxyhitnum[a,b]{n}{k}{B}/\prod_{i=1}^{t}\br{d_i}!$
is $\zsu(L_k^{a,b}(B))$. \label{thm:A_k/\br{d_i}!zsu}
\end{thm}

\begin{proof}
We apply Lemma~\ref{lem:A_kzsu}, which says that
$$\frac{\qxyhitnum[a,b]{n}{k}{B}}{\prod_{i=1}^{t}\br{d_i}!}=
\sum_{e_1+\cdots +e_t=k, \ 0\leq e_i \leq d_i}^{}P(\vec
e,a,b)\prod_{i=1}^{t}q^{e_i(H_i-D_i+E_i +b-1)},$$ and all of the
terms on the right hand side above are in $\N[q]$ by
Lemma~\ref{lem:P(e,x,y)=0}.  Using the fact that ${r \brack s}$ is
$\zsu(s(r-s))$ (for a proof of the unimodality see~\cite{GO}) along
with Lemma~\ref{lem:zsu*}, each term above is
$\zsu(\sum_{i=1}^{t}\{(d_i-e_i)(H_i-D_i+E_i+b-1)+e_i(D_i+D_{i-1}-H_i-E_i+a-b)+2e_i(H_i-D_i+E_i+b-1)\})$.
A simple calculation shows this is the same $\zsu(L_k^{a,b}(B))$.
Thus $\qxyhitnum[a,b]{n}{k}{B}/\prod_{i=1}^{t}\br{d_i}!$ is
$\zsu(L_k^{a,b}(B))$ as well. \end{proof}

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\end{document}
