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\def\al{\alpha}


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\newtheorem{proposition}[lemma]{Proposition}

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\textwidth=15cm \textheight=22cm \topmargin=0.5cm \oddsidemargin=0.5cm
\evensidemargin=0.5cm \advance\headheight1.15pt


\begin{document}
\title{A $q$-Catalan-numbers-determinant}
\author[\"O. E\u gecio\u glu and C. Krattenthaler]{\"O. E\u gecio\u glu$^*$
and C. Krattenthaler$^{\dagger}$} 

\address{$^*$Department of Computer Science, University of
California at Santa Barbara, CA 93106-5110\newline
WWW: \tt http://www.cs.ucsb.edu/\lower0.5ex\hbox{\~{}}omer/.
}

\address{$^{\dagger}$Fakult\"at f\"ur Mathematik, Universit\"at Wien,
Nordbergstrasze~15, A-1090 Vienna, Austria.
WWW: {\tt http://www.mat.univie.ac.at/\lower0.5ex\hbox{\~{}}kratt}.}

\thanks{$^\dagger$Research partially supported by the Austrian
Science Foundation FWF, grants Z130-N13 and S9607-N13,
the latter in the framework of the National Research Network
``Analytic Combinatorics and Probabilistic Number
Theory"}

\subjclass[2010]{Primary 15A15;
Secondary 05A15 05A19 05A30}

\keywords{Carlitz $q$-Catalan numbers, non-intersecting lattice paths}


\begin{abstract}
We compute a determinant involving Carlitz's $q$-Catalan numbers.
\end{abstract}
\maketitle

Let $C_n(q)$ be defined by $C_0(q)=1$ and
\begin{equation} \label{eq:rek} 
C_n(q)=
\sum _{k=0} ^{n-1}q^k C_k(q)C_{n-k-1}(q),\quad \quad n\ge1.
\end{equation}
From \cite[Eq.~(2.1)]{FuHoAA},\footnote{Our $C_n(q)$ is denoted by
$\tilde C_n(q)$ in \cite{FuHoAA}. The relation between these
$q$-Catalan numbers and the ones in \cite[Eq.~(2.1)]{FuHoAA} is
described in \cite[Eq.~(2.3)]{FuHoAA}.} one can derive the expansion 
\begin{equation} \label{eq:exp} 
1=
\sum _{n=0} ^{\infty}(-1)^{n}q^{-\binom n2}C_n(q)z^n(-z;q^{-1})_{n+1}.
\end{equation}

The purpose of this note is to prove the following theorem.
It has recently been reproved by Cigler \cite{CiglZZ} using completely
different means. In fact, Cigler is able to embed the result below
into a wider framework and is thereby able to establish several
similar theorems --- some already known, some new.

\begin{theorem} \label{thm:1}
For all non-negative integers $n$, we have
\begin{equation} \label{eq:1} 
\det_{0\le i,j\le n}\left(
\sum _{m=0} ^{i+j}C_{i+j-m}(q)x^m\right)=
q^{n(n-1)(4n+1)/6}
\sum _{k=0} ^{n}(-1)^kq^{(n-k)^2}\begin{bmatrix} n+k\\n-k\end{bmatrix}x^k.
\end{equation}
\end{theorem}

\begin{proof}
We start by subtracting $x$ times row $i-1$ from row $i$,
$i=n,n-1,\dots,1$ (in this order!), in the matrix of which we want to
compute the determinant. As a result, we obtain that the determinant
on the left-hand side of \eqref{eq:1} is equal to
\begin{equation} \label{eq:2} 
\det_{0\le i,j\le n}\begin{pmatrix} 
\sum _{m=0} ^{j}C_{j-m}(q)x^m&i=0\\
C_{i+j}(q)&i\ge1\end{pmatrix}.
\end{equation}
Next we replace column $j$ by
$$
(\text{column $j$})-
\sum _{s=0} ^{j-1}q^{j-s-1}C_{j-s-1}(q)\cdot (\text{column $s$}),
$$
$j=0,1,\dots,n$.
Clearly, this can be achieved by elementary column
operations, which do not change the value of the determinant. 
Thereby, the determinant in \eqref{eq:2} is transformed into
\begin{align} \notag
&\det_{0\le i,j\le n}\begin{pmatrix} 
\sum _{m=0} ^{j}C_{j-m}(q)x^m-\sum _{s=0} ^{j-1}q^{j-s-1}C_{j-s-1}(q)
\sum _{m=0} ^{s}C_{s-m}(q)x^m&i=0\\
C_{i+j}(q)-\sum _{s=0}
^{j-1}q^{j-s-1}C_{j-s-1}(q)C_{i+s}(q)&i\ge1\end{pmatrix}\\ 
\notag
&=\det_{0\le i,j\le n}\begin{pmatrix} 
\sum _{m=0} ^{j}C_{j-m}(q)x^m-\sum _{m=0} ^{j-1}
x^m\sum _{s=m} ^{j-1}q^{j-s-1}C_{j-s-1}(q)
C_{s-m}(q)&i=0\\
C_{i+j}(q)-\sum _{s=0}
^{j-1}q^{j-s-1}C_{j-s-1}(q)C_{i+s}(q)&i\ge1\end{pmatrix}\\ 
\notag
&=\det_{0\le i,j\le n}\begin{pmatrix} 
x^j&i=0\\
C_{i+j}(q)-\sum _{s=0}
^{j-1}q^{j-s-1}C_{j-s-1}(q)C_{i+s}(q)&i\ge1\end{pmatrix},
\end{align}
where we used \eqref{eq:rek} to obtain the last line.
We may expand this determinant along the first row to arrive at
\begin{multline} \label{eq:3}
\sum _{k=0} ^{n}(-1)^kx^k\det_{1\le i\le n,\ 0\le j\le n-1}
\bigg(
C_{i+j+\chi(j\ge k)}(q)\\
-\sum _{s=0}
^{j+\chi(j\ge k)-1}q^{j+\chi(j\ge k)-s-1}C_{j+\chi(j\ge
k)-s-1}(q)C_{i+s}(q)
\bigg),
\end{multline}
where $\chi(\mathcal S)=1$ if $\mathcal S$ is true and
$\chi(\mathcal S)=0$ otherwise.

We now focus our attention 
on the coefficient of $(-x)^k$ in this expansion,
that is, on the determinant
\begin{equation} \label{eq:4} 
\det_{0\le i,j\le n-1}
\bigg(
C_{i+j+\chi(j\ge k)+1}(q)\\
-\sum _{s=0}
^{j+\chi(j\ge k)-1}q^{j+\chi(j\ge k)-s-1}C_{j+\chi(j\ge
k)-s-1}(q)C_{i+s+1}(q)
\bigg).
\end{equation}
(The reader should note that we did the replacement $i\to i+1$
compared to \eqref{eq:3}.)
By using column operations, working from the left to the right, this
determinant can be reduced to
$$
\det_{0\le i,j\le n-1}\begin{pmatrix} 
C_{i+j+1}(q)&j<k\\
C_{i+j+2}(q)
-\sum _{s=k}
^{j}q^{j-s}C_{j-s}(q)C_{i+s+1}(q)&j\ge k
\end{pmatrix}.
$$
Now, for $j=k,k+1,n-1$, we replace column~$j$ by
$$
\sum _{\ell=k} ^{j}C_{j-\ell}(q)(\text{column $\ell$}).$$
Again, this can be achieved by elementary column operations.
In this manner, we see that the determinant in \eqref{eq:4} is equal to
\begin{align*}
&\det_{0\le i,j\le n-1}\begin{pmatrix} 
C_{i+j+1}(q)&j<k\\
\begin{matrix}
\sum _{\ell=k} ^{j}C_{j-\ell}(q)C_{i+\ell+2}(q)\hfill\\
\hfill-\sum _{\ell=k} ^{j}C_{j-\ell}(q)\sum _{s=k}
^{\ell}q^{\ell-s}C_{\ell-s}(q)C_{i+s+1}(q)
\end{matrix}&j\ge k
\end{pmatrix}\\
&=\det_{0\le i,j\le n-1}\begin{pmatrix} 
C_{i+j+1}(q)&j<k\\
\begin{matrix}
\sum _{\ell=k} ^{j}C_{j-\ell}(q)C_{i+\ell+2}(q)\hfill\\
\hfill
-\sum _{s=k} ^{j}C_{i+s+1}(q)\sum _{\ell=s} ^{j}C_{j-\ell}(q)
q^{\ell-s}C_{\ell-s}(q)
\end{matrix}&j\ge k
\end{pmatrix}\\
&=\det_{0\le i,j\le n-1}\begin{pmatrix} 
C_{i+j+1}(q)&j<k\\
\sum _{\ell=k} ^{j}C_{j-\ell}(q)C_{i+\ell+2}(q)
-\sum _{s=k} ^{j}C_{i+s+1}(q)C_{j-s+1}
&j\ge k
\end{pmatrix}\\
&=\det_{0\le i,j\le n-1}\begin{pmatrix} 
C_{i+j+1}(q)&j<k\\
C_{i+j+2}(q)
-C_{i+k+1}(q)C_{j-k+1}
&j\ge k
\end{pmatrix},
\end{align*}
where we used \eqref{eq:rek} to arrive at the next-to-last line.

The next step is to expand the last determinant by linearity in
columns $k,k+1,\dots,n-1$. In theory, this produces $2^{n-k}$
terms. However, since choosing the term $C_{i+k+1}(q)C_{j-k+1}$ for
two different $j$'s will result in a vanishing determinant, only very
few terms survive. To be precise, these are
\begin{align}
\notag
&\det_{0\le i,j\le n-1}\left(
C_{i+j+\chi(j\ge k)+1}(q)
\right)
-
\sum _{\ell=k} ^{n-1}C_{\ell-k+1}(q)
\det_{0\le i,j\le n-1}\begin{pmatrix} 
C_{i+j+\chi(j\ge k)+1}(q)&j\ne\ell\\
C_{i+k+1}(q)
&j=\ell
\end{pmatrix}\\
\notag
&\kern2cm
=
\sum _{\ell=k-1} ^{n-1}(-1)^{\ell-k+1}C_{\ell-k+1}(q)
\det_{0\le i,j\le n-1}\left( 
C_{i+j+\chi(j\ge \ell+1)+1}(q)
\right)\\
&\kern2cm
=
\sum _{\ell=k} ^{n}(-1)^{\ell-k}C_{\ell-k}(q)
\det_{0\le i,j\le n-1}\left( 
C_{i+j+\chi(j\ge \ell)+1}(q)
\right).
\label{eq:5}
\end{align}


\begin{figure}
$$
\Einheit.5cm
\Gitter(11,11)(-10,0)
\Koordinatenachsen(11,11)(-10,0)
\Pfad(-2,0),34\endPfad
\Pfad(-4,0),333444\endPfad
\Pfad(-6,0),333334444434\endPfad
\Pfad(-8,0),3333333444344444\endPfad
\Pfad(-10,0),33333333334444444444\endPfad
\DickPunkt(-2,0)
\DickPunkt(-4,0)
\DickPunkt(-6,0)
\DickPunkt(-8,0)
\DickPunkt(-10,0)
\DickPunkt(0,0)
\DickPunkt(2,0)
\DickPunkt(6,0)
\DickPunkt(8,0)
\DickPunkt(10,0)
\Label\u{A_0}(-2,0)
\Label\u{A_1}(-4,0)
\Label\u{A_2}(-6,0)
\Label\u{A_3}(-8,0)
\Label\u{A_4}(-10,0)
\Label\u{E_0}(0,0)
\Label\u{E_1}(2,0)
\Label\u{E_2}(6,0)
\Label\u{E_3}(8,0)
\Label\u{E_4}(10,0)
\PfadDicke{.5pt}
$$
\caption{}
\label{fig:1}
\end{figure}

By the Lindstr\"om--Gessel--Viennot theorem
(\cite[Lemma~1]{LindAA}, \cite[Theorem~1]{GeViAB}), the determinant in
\eqref{eq:5} can be given a combinatorial interpretation: it is equal
to the generating function $
\sum _{\mathbf P} ^{}w(\mathbf P)$, where the sum is over all families
$\mathbf P=(P_0,P_1,\dots,P_{n-1})$ of nonintersecting paths
consisting of up-steps and down steps, where $P_i$ is a path 
from $(-2i-2,0)$ to $(2i+2\chi(j\ge\ell))$
never running below the $x$-axis, $i=0,1,\dots,n-1$, and where the
weight $w(\mathbf P)$ is $q^{A(P_0)+\dots+A(P_{n-1})}$, with $A(P)$
denoting the area under the path $P$. An example with $n=5$ and
$\ell=2$ is given in Figure~\ref{fig:1}.

By looking at the last up-step of paths $\ell,\ell+1,\dots,n-1$, it is
not difficult to see that this generating function equals
\begin{multline*}
q^{
\sum _{t=1} ^{n-1}
\sum _{s=1} ^{t}(4s-1)}
q^{2\sum _{s=\ell} ^{n-1}(s-\ell)}
\sum _{0\le k_\ell\le\dots\le k_{n-1}} ^{}
q^{k_\ell+\dots+k_{n-1}}\\
=
q^{n(n-1)(4n+1)/6}q^{(n-\ell)(n-\ell-1)}\begin{bmatrix} n+\ell+1\\n-\ell\end{bmatrix}.
\end{multline*}
Therefore, if we substitute this in \eqref{eq:5}, and if we
subsequently replace $\ell$ by $\ell+k$, then we obtain the expression
\begin{equation} \label{eq:6} 
\sum _{\ell=0} ^{n-k}(-1)^{\ell}
q^{n(n-1)(4n+1)/6}q^{(n-k-\ell)(n-k-\ell-1)}
C_{\ell}(q)\begin{bmatrix} n+k+\ell+1\\n-k-\ell\end{bmatrix}.
\end{equation}
Using the $q$-binomial theorem, we have
$$
\coef{z^m}(-z;q)_M=q^{\binom m2}\begin{bmatrix} M\\m\end{bmatrix},
$$
where $(\alpha;q)_M:=(1-\al )(1-\al q)\cdots(1-\al q^{M-1})$ for
$M\ge1$ and $(\al;q)_0:=1$, and
where $\coef{z^{m}}h(\mathbf z)$
denotes the coefficient of $z^{m}$ in the formal
Laurent series $h(z)$. Hence, we can rewrite \eqref{eq:6} as
\begin{align*} %\label{eq:7} 
\sum _{\ell=0} ^{n-k}&(-1)^{\ell}
q^{n(n-1)(4n+1)/6}q^{\binom {n-k-\ell}2}
C_{\ell}(q)\coef{z^{n-k-\ell}}(-z;q)_{n+k+\ell+1}\\
&=\sum _{\ell=0} ^{n-k}(-1)^{\ell}
q^{n(n-1)(4n+1)/6}q^{\binom {n-k-\ell}2}
C_{\ell}(q)\\
&\kern5cm
\cdot
q^{(n-k-\ell)(n+k+\ell)}
\coef{z^{n-k-\ell}}(-zq^{-n-k-\ell};q)_{n+k+\ell+1}\\
&=\coef{z^{n-k}}(-z;q^{-1})_{n+k}
q^{n(n-1)(4n+1)/6}q^{\binom n2+n(n-k)-\binom k2}\\
&\kern5cm
\times
\sum _{\ell=0} ^{n-k}(-1)^{\ell}
q^{-\binom \ell2-\ell(n+k)}
C_{\ell}(q)
z^\ell(-zq^{-n-k};q^{-1})_{\ell}\\
&=\coef{z^{n-k}}(-z;q^{-1})_{n+k}
q^{n(n-1)(4n+1)/6}q^{\binom n2+n(n-k)-\binom k2}\\
&=q^{n(n-1)(4n+1)/6}q^{(n-k)^2}\begin{bmatrix} n+k\\n-k\end{bmatrix},
\end{align*}
where we used \eqref{eq:exp} to obtain the next-to-last line.
To summarize: we have proved that the coefficient of $(-x)^k$ in
\eqref{eq:3} is equal to this last expression. In its turn,
expression~\eqref{eq:3} is equal to our determinant on the left-hand
side of \eqref{eq:1}. This completes the proof of the theorem.
\end{proof}




\begin{thebibliography}{9}

\bibitem{CiglZZ} J. Cigler, {\em Hankel determinants of some polynomial sequences},
preprint, {\tt ar$\chi$iv:1211.0816}.

\bibitem{FuHoAA} J.    F\"urlinger and J.~Hofbauer,
{\em $q$-Catalan numbers}, J. Combin.\ Theory Ser.~A {\bf 40} (1985), 
248--264.

\bibitem{GeViAB} I. M. Gessel and X. Viennot,  
{\em Determinants, paths, and plane partitions},
preprint, 1989, available at 
{\tt http://www.cs.brandeis.edu/\~{}ira}.

\bibitem{LindAA} B.    Lindstr\"om,
{\em On the vector representations of induced matroids}, 
Bull.\ London Math.\ Soc.\ {\bf 5} (1973), 85--90.

\end{thebibliography}



\end{document}



