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\begin{document}
\title[Derangements]{On derangement polynomials of type $B$}
\author[C.-O.~Chow]{Chak-On~Chow}
\address{P.O. Box 91100, Tsimshatsui Post Office, Hong Kong}
\email{cchow@alum.mit.edu}
\begin{abstract}
Wachs [{\it Proc.\ Amer.\ Math.\ Soc.} {\bf 106} (1989), 273--278]
studied the $q$-enumeration of derangements in the symmetric group
$\mathfrak{S}_n$ by the major index and obtained a $q$-analogue of the
classical derangement number. We consider in this work the
$q$-enumeration of derangements in the hyperoctahedral group $B_n$ by
the flag major index and obtain a $q$-analogue of the type $B$
derangement number.  
\end{abstract}
\maketitle

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\markright{\its S\'eminaire Lotharingien de
Combinatoire \bfs 55 \rms (2006), Article~B55b\hfill}
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Let $n\geqslant 1$. Let $B_n$ be the $n$th hyperoctahedral group, which is a Coxeter group of rank $n$, consisting of signed permutations of $[n]:=\{1,2,\ldots,n\}$. We shall represent a signed permutation $\sigma\in B_n$ by the word $\sigma_1\sigma_2\cdots\sigma_n$, where $\sigma_i=\sigma(i)$, $i=1,2,\ldots,n$. The (type $A$) descent set of $\sigma$ is $D(\sigma):=\{i\in [n-1]\colon\sigma_i>\sigma_{i+1}\}$. A statistic derivable from $D(\sigma)$ is the usual (type $A$) major index: $\maj(\sigma):=\sum_{i\in D(\sigma)}i$. Let $N(\sigma):=\#\{i\in [n]\colon\sigma_i<0\}$ be the number of negative letters of $\sigma$.

Two candidates for the type $B$ major index, namely, the negative major index ($\nmaj$) and the flag major index ($\fmaj$), have recently been proposed and proven to be Mahonian, i.e.,
$$
\sum_{\sigma\in B_n}q^{\fmaj(\sigma)}=\sum_{\sigma\in B_n}q^{\nmaj(\sigma)}=\sum_{\sigma\in B_n}q^{l_B(\sigma)}
=[2]_q[4]_q\cdots[2n]_q,
$$
where $[k]_q:=1+q+q^2+\cdots+q^{k-1}$ is a $q$-integer, $\fmaj(\sigma):=2\maj(\sigma)+N(\sigma)$, and $l_B$ the length function on $B_n$. By pairing with the negative descent number ($\ndes$) and the flag descent number ($\fdes$), respectively, these major indices are shown to satisfy a $q$-rational generating function generalizing the classical Carlitz identity. See \cite{abr01,ar01} for the definitions of undefined terms and the above mentioned results.

In a recent work, Chow and Gessel \cite{cg03} studied the Euler-Mahonian pair $(\des_B,\fmaj)$, where $\des_B$ is the type $B$ descent number, on $B_n$ and computed its $q$-rational generating function. This $q$-rational generating function reduces to the rational generating function for the type $B$ Eulerian polynomial when $q\to 1$, thus providing a {\it natural} type $B$ generalization of the classical Carlitz identity. It is from this point of view that the flag major index ($\fmaj$) plays the same role on the hyperoctahedral group $B_n$ as the usual major index ($\maj$) does on the symmetric group $\mathfrak{S}_n$.

A number of statistical studies on $B_n$ sequel to \cite{abr01,ar01,cg03} have been pursued by various authors. See, e.g., \cite{fh05,fh06,hlr05} which considered the flag and negative major indices as well as other permutation statistics in the more general setting of signed words and wreath products.

For $n\geqslant 1$, let $\mathscr{D}_n:=\{\sigma\in\mathfrak{S}_n\colon\textrm{$\sigma(i)\ne i$ for all $i\in [n]$}\}$ be the set of all derangements in $\mathfrak{S}_n$. Wachs \cite{w89} considered derangement polynomials defined by $d_n(q):=\sum_{\sigma\in\mathscr{D}_n}q^{\maj(\sigma)}$ and showed combinatorially that
\begin{equation}\label{eqn:1}
d_n(q)=[n]_q!\sum_{k=0}^n\frac{(-1)^kq^{\binom{k}{2}}}{[k]_q!},
\end{equation}
where $[n]_q!:=[1]_q[2]_q\cdots[n]_q$ is a $q$-factorial. Letting $q\to 1$, the above formula reduces to the formula of the $n$th classical derangement number: $d_n=n!\sum_{k=0}^n(-1)^k/k!$. It is from this point of view that Wachs' derangement polynomials are $q$-analogues of the derangement numbers.

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\markboth{\SMALL CHAK-ON CHOW}{\SMALL ON DERANGEMENT POLYNOMIALS OF
TYPE $B$}
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We show in this work (Theorem~\ref{thm:2}) that $d_n^B(q):=\sum_{\sigma\in\mathscr{D}_n^B}q^{\fmaj(\sigma)}$ can be written as
$$
d_n^B(q)=[2]_q[4]_q\cdots[2n]_q\sum_{k=0}^n\frac{(-1)^kq^{2\binom{k}{2}}}{[2]_q[4]_q\cdots[2k]_q},
$$
where $\mathscr{D}_n^B:=\{\sigma\in B_n\colon\textrm{$\sigma(i)\ne i$ for all $i\in [n]$}\}$ is the set of derangements in $B_n$. The first four $d_n^B(q)$ are given as follows:
\begin{equation*}
\begin{split}
d_1^B(q)&=q,\\
d_2^B(q)&=q+2q^2+q^3+q^4,\\
d_3^B(q)&=q+3q^2+4q^3+5q^4+5q^5+4q^6+4q^7+2q^8+q^9,\\
d_4^B(q)&=q+4q^2+8q^3+13q^4+18q^5+22q^6+26q^7+28q^8+28q^9+25q^{10}\\
&\qquad+21q^{11}+17q^{12}+11q^{13}+7q^{14}+3q^{15}+q^{16}.
\end{split}
\end{equation*}

Following Wachs,
for any signed permutation $\alpha\in B_A$, where $A=\{a_1<a_2<\cdots<a_k\}$, define the {\it reduction} of $\alpha$ to be the signed permutation in $B_k$
obtained from $\alpha$ by replacing each letter $\alpha_j$ by $(\sgn\alpha_j)i$ if $|\alpha_j|=a_i$, $i=1,2,\ldots,k$. The {\it derangement part} of a signed permutation $\sigma\in B_n$, denoted $\dep(\sigma)$, is the reduction of the subword of non-fixed points of $\sigma$. For example, $\dep(5\bar 3\bar 1476\bar 2)=4\bar 3\bar 15\bar 2$. Note that the derangement part of a signed permutation is a signed derangement, and that conversely, any derangement in $\mathscr{D}_k^B$ and $k$-element subset of $[n]$ determine a signed permutation in $B_n$ with $n-k$ fixed points. Hence, the number of signed permutations in $B_n$ with a given derangement part in $\mathscr{D}_k^B$ is $\binom{n}{k}$.

Now let $\sigma=\sigma_1\sigma_2\cdots\sigma_n\in B_n$. A letter $\sigma_i$ of $\sigma$ is an {\it excedant} (respectively {\it subcedant}) of $\sigma$ if $\sigma_i>i$ (respectively $\sigma_i<i$). Let $s(\sigma)$ and $e(\sigma)$ be the number of subcedants and excedants of $\sigma$, respectively. It is clear that excedants of $\sigma$ are necessarily positive. We now fix $n$ and let $k\leqslant n$. For $\sigma\in B_k$, let $\tilde\sigma$ be the signed permutation of $k$ letters obtained from $\sigma$ by replacing its $i$th smallest (in absolute value) subcedant $\sigma_j$ by $(\sgn\sigma_j)i$, $i=1,2,\ldots,s(\sigma)$, its $i$th smallest fixed point by $s(\sigma)+i$, $i=1,2,\ldots,k-s(\sigma)-e(\sigma)$, and its $i$th largest excedant by $n-i+1$, $i=1,2,\ldots,e(\sigma)$. The map $\sigma\to\tilde\sigma$ restricted to the symmetric group $\mathfrak{S}_k$ is precisely the descent set preserving map used in \cite{w89}. If $k=n$ then $\tilde\sigma\in B_n$. If $\sigma$ is a signed derangement, then $\tilde\sigma\in\mathscr{D}_A^B$, where $A=\{1,2,\ldots,s(\sigma)\}\cup\{n-e(\sigma)+1,n-e(\sigma)+2,\ldots,n\}$.

\begin{lemma}\label{lem:1}
Let $\sigma\in B_k$, $k\leqslant n$. Then $D(\sigma)=D(\tilde\sigma)$ and $N(\sigma)=N(\tilde\sigma)$.
\end{lemma}
\begin{proof}
It is clear from the construction of $\tilde\sigma$ that
$N(\sigma)=N(\tilde\sigma)$. Let
$\sigma=\sigma_1\sigma_2\cdots\sigma_k$ and
$\tilde\sigma=\tilde\sigma_1\tilde\sigma_2\cdots\tilde\sigma_k$. For
each $i\in [k-1]$, we shall show that $i\in D(\sigma)$
if and only if $i\in D(\tilde\sigma)$, by considering the nine possible designations of subcedant $(s)$, excedant $(e)$, and fixed point $(f)$ to $\sigma_i$ and $\sigma_{i+1}$. Since the letters $\tilde\sigma_i$ are determined according to whether $\sigma_i$ is a subcedant $(e)$, a fixed point $(f)$, or an excedant $(e)$, and in this order, $\tilde\sigma_i<\tilde\sigma_{i+1}$ if the designation of $\sigma_i$ precedes that of $\sigma_{i+1}$.

Suppose that $(\sigma_i,\sigma_{i+1})$ is an $(f,e)$ or $(f,f)$ pair. Then $\sigma_i=i<i+1\leqslant\sigma_{i+1}$ so that $\tilde\sigma_i<\tilde\sigma_{i+1}$. If $(\sigma_i,\sigma_{i+1})$ is an $(f,s)$, then $\sigma_i=i>i-1\geqslant\sigma_{i+1}$ so that $\tilde\sigma_i>s(\sigma)\geqslant|\tilde\sigma_{i+1}|\geqslant\tilde\sigma_{i+1}$.

Suppose now that $(\sigma_i,\sigma_{i+1})$ is an $(s,f)$ or $(s,e)$
pair. Then $\sigma_i<i<i+1\leqslant\sigma_{i+1}$ so that
$\tilde\sigma_i<\tilde\sigma_{i+1}$. If $(\sigma_i,\sigma_{i+1})$ is
an $(s,s)$, then since $\sgn\sigma_i=\sgn\tilde\sigma_i$,
$\sigma_i\gtrless\sigma_{i+1}\gtrless 0$ if and only if 
$\tilde\sigma_i\gtrless\tilde\sigma_{i+1}\gtrless 0$, and
$\sigma_i\gtrless 0\gtrless\sigma_{i+1}$ if and only if
$\tilde\sigma_i\gtrless 0\gtrless\tilde\sigma_{i+1}$.

Suppose finally that $(\sigma_i,\sigma_{i+1})$ is an $(e,f)$
pair. Then $\sigma_i\geqslant i+2>i+1=\sigma_{i+1}$ so that
$\tilde\sigma_i>\tilde\sigma_{i+1}$; if $(\sigma_i,\sigma_{i+1})$ is
an $(e,s)$ pair, then $\sigma_i>i\geqslant\sigma_{i+1}$ so that
$\tilde\sigma_i>\tilde\sigma_{i+1}$; consider now
$(\sigma_i,\sigma_{i+1})$ being an $(e,e)$ pair. If $\sigma_i$
(respectively 
$\sigma_{i+1}$) is the $j$th (respectively $k$th) largest excedant of
$\sigma$, then $\sigma_i\lessgtr\sigma_{i+1}$ if and only if
$j\gtrless k$ if and only if
$\tilde\sigma_i=n-j+1\lessgtr n-k+1=\tilde\sigma_{i+1}$.
\end{proof}

Let $n_1,\ldots,n_r$ be non-negative integers such that $\sum_{i=1}^rn_i=n$. Recall the $q$-multinomial coefficient
$$
\left[\begin{matrix}
n\\
n_1,\ldots,n_r\end{matrix}\right]_q:=\frac{[n]_q!}{[n_1]_q!\cdots[n_r]_q!}.
$$
Now let $(N_1,\ldots,N_r)$ be a partition of the set $[n]$. For each $1\leqslant i\leqslant r$ let $\pi_i$ be a permutation of the elements of $N_i$. Recall that a permutation $\sigma\in\mathfrak{S}_n$ is a shuffle of $\pi_1,\pi_2,\ldots,\pi_r$ if, for every $i$, the letters of $N_i$ appears in $\sigma$ in the same order as the corresponding letters in $\pi_i$. The generating function of shuffles of permutations by their major index has been computed by Garsia and Gessel \cite[Theorem 3.1]{gg79}.

\begin{theorem}[Garsia-Gessel]\label{thm:3}
Let $\sh(\pi_1,\ldots,\pi_r)$ be the collection of all shuffles of given permutations $\pi_1,\pi_2,\ldots,\pi_r$. Then
$$
\sum_{\sigma\in\sh(\pi_1,\ldots,\pi_r)}q^{\maj(\sigma)}=\left[\begin{matrix}
n\\
n_1,\ldots,n_r\end{matrix}\right]_qq^{\maj(\pi_1)+\cdots+\maj(\pi_r)},
$$
where $n_i$ is the length of $\pi_i$ $(1\leqslant i\leqslant r)$.
\end{theorem}

Theorem~\ref{thm:3} remains true if $\pi_1,\ldots,\pi_r$ are signed words of distinct letters from the totally ordered alphabet $\{-n<-n+1<\cdots<-1<1<2<\cdots<n\}$. It is clear in this case that $N(\sigma)=N(\pi_1)+\cdots+N(\pi_r)$. Replacing $q$ by $q^2$, followed by multiplication by $q^{N(\sigma)}$ and noting that $\fmaj(\sigma)=2\maj(\sigma)+N(\sigma)$, we have the following identity:
\begin{equation}\label{eqn:2}
\sum_{\sigma\in\sh(\pi_1,\ldots,\pi_r)}q^{\fmaj(\sigma)}=\left[\begin{matrix}
n\\
n_1,\ldots,n_r\end{matrix}\right]_{q^2}q^{\fmaj(\pi_1)+\cdots+\fmaj(\pi_r)},
\end{equation}
where $n_i$ is the length of $\pi_i$ $(1\leqslant i\leqslant r)$.

\begin{theorem}\label{thm:1}
Let $\alpha\in\mathscr{D}_k^B,$ $k\leqslant n,$ and $\gamma=(s(\alpha)+1)(s(\alpha)+2)\ldots(n-e(\alpha))$.
Then the map $\varphi\colon\{\sigma\in B_n\colon\dep(\sigma)=\alpha\}\to\sh(\tilde\alpha,\gamma)$ defined by $\varphi(\sigma)=\tilde\sigma$
is a flag major index preserving bijection, i.e., $\fmaj(\sigma)=\fmaj(\varphi(\sigma))$.
\end{theorem}
\begin{proof}
The flag major index preserving property of $\varphi$ is clear from Lemma~\ref{lem:1}. Since $\#\{\sigma\in B_n\colon\dep(\sigma)=\alpha\}=\#\sh(\tilde\alpha,\gamma)=\binom{n}{k}$, to prove that $\varphi$ is a bijection, it suffices to prove that $\varphi(\{\sigma\in B_n\colon\dep(\sigma)=\alpha\})=\sh(\tilde\alpha,\gamma)$.

First, we claim that if $\sigma\in B_n$ is such that $\dep(\sigma)=\alpha$, then $\tilde\sigma$ is obtained from $\sigma$ by replacing the subword of non-fixed points of $\sigma$ by $\tilde\alpha$ and the subword of fixed points of $\sigma$ by $\gamma$. Indeed, the subword of fixed points of $\sigma$ is replaced by the word $(s(\sigma)+1)(s(\sigma)+2)\cdots(n-e(\sigma))$, which is precisely $\gamma$ since $s(\sigma)=s(\alpha)$ and $e(\sigma)=e(\alpha)$. Also since $\alpha$ is the reduction of the subword of non-fixed points of $\sigma$, the position of the $i$th smallest subcedant of $\alpha$ is the same as the position of the $i$ smallest subcedant of $\sigma$ in its subword of non-fixed points. The same is true for the $i$th largest excedant. Hence each subcedant and excedant of $\sigma$ is replaced by the same letter that replaces the corresponding subcedant and excedant of $\alpha$, which means that the subword of subcedants and excedants of $\sigma$ is replaced by $\tilde\alpha$. Thus $\varphi(\sigma)=\tilde\sigma\in\sh(\tilde\alpha,\gamma)$.

To finish the proof, we show that $\varphi$ is surjective. Let $\tau\in\sh(\tilde\alpha,\gamma)$. Replace the $\tilde\alpha$ subword by the signed permutation, of the subword positions, whose reduction is $\alpha$, and the letters of the $\gamma$ subword by their positions, we obtain a unique signed permutation $\sigma\in B_n$ such that $\dep(\sigma)=\alpha$ and $\varphi(\sigma)=\tau$.
\end{proof}

\begin{proposition}
Let $\alpha\in\mathscr{D}_k^B$ and $0\leqslant k\leqslant n$. Then
$$
\sum_{\dep(\sigma)=\alpha,\,\sigma\in B_n}
q^{\fmaj(\sigma)}=q^{\fmaj(\alpha)}\left[\begin{matrix}
n\\
k\end{matrix}\right]_{q^2}.
$$
\end{proposition}
\begin{proof}
This follows from
\begin{equation*}
\sum_{\dep(\sigma)=\alpha,\,\sigma\in B_n}q^{\fmaj(\sigma)}=\sum_{\sigma\in\sh(\tilde\alpha,\gamma)}q^{\fmaj(\sigma)}
=q^{\fmaj(\tilde\alpha)}\left[\begin{matrix}
n\\
k\end{matrix}\right]_{q^2}=q^{\fmaj(\alpha)}\left[\begin{matrix}
n\\
k\end{matrix}\right]_{q^2},
\end{equation*}
where the first equality follows from Theorem~\ref{thm:1}, the second from \eqref{eqn:2}, and the third from Lemma~\ref{lem:1}.
\end{proof}

Recall that the two $q$-exponential functions defined by
\begin{equation*}
e(u;q):=\sum_{n\geqslant 0}\frac{u^n}{(q;q)_n},\qquad
E(u;q):=\sum_{n\geqslant 0}\frac{q^{\binom{n}{2}}u^n}{(q;q)_n},
\end{equation*}
satisfy $E(-u;q)e(u;q)=1$
(which is what the so-called Gau{\ss} inversion \cite[p.~96]{a79} amounts to),
where
$$
(u;q)_n:=\begin{cases}
1&\textrm{if $n=0$},\\
(1-u)(1-uq)\cdots(1-uq^{n-1})&\textrm{if $n\geqslant 1$}.\end{cases}
$$

\begin{theorem}\label{thm:2}
For $n\geqslant 1,$ we have
\begin{enumerate}
\item[(i)]
$\displaystyle d_n^B(q)=[2]_q[4]_q\cdots[2n]_q\sum_{k=0}^n\frac{(-1)^kq^{2\binom{k}{2}}}{[2]_q[4]_q\cdots[2k]_q},$
\item[(ii)]
$\displaystyle \sum_{n\geqslant 0}d_n^B(q)\frac{u^n}{[2]_q[4]_q\cdots[2n]_q}=\frac{E(-u(1-q);q^2)}{1-u},$
\item[(iii)]
$d_{n+1}^B(q)=[2n+2]_qd_n^B(q)+(-1)^{n+1}q^{2\binom{n+1}{2}}$.
\end{enumerate}
\end{theorem}
\begin{proof}
By virtue of the preceding proposition, we have
\begin{equation}\label{eqn:3}
\begin{split}
[2]_q[4]_q\cdots[2n]_q
&=\sum_{\sigma\in B_n}q^{\fmaj(\sigma)}\\
&=\sum_{k=0}^n\sum_{\alpha\in\mathscr{D}_k^B}\sum_{\dep(\sigma)=\alpha}q^{\fmaj(\sigma)}\\
&=\sum_{k=0}^n\sum_{\alpha\in\mathscr{D}_k^B}q^{\fmaj(\alpha)}\left[\begin{matrix}
n\\
k\end{matrix}\right]_{q^2}\\
&=\sum_{k=0}^n\left[\begin{matrix}
n\\
k\end{matrix}\right]_{q^2}d_k^B(q).
\end{split}
\end{equation}
Since $[2]_q[4]_q\cdots [2n]_q=(1+q)^n[n]_{q^2}!$, \eqref{eqn:3} can be simplified as
\begin{equation*}
(1+q)^n=\sum_{k=0}^n\frac{d_k^B(q)}{[k]_{q^2}![n-k]_{q^2}!}.
\end{equation*}
Multiplying through by $u^n/(1-q^2)^n$, followed by summing over $n\geqslant 0$, we get
\begin{equation*}
e(u;q^2)\sum_{n\geqslant 0}d_n^B(q)\frac{u^n}{(q^2;q^2)_n}
=\sum_{n\geqslant 0}\frac{u^n}{(1-q)^n},
\end{equation*}
since $(q^2;q^2)_n=(1-q^2)^n[n]_{q^2}!$. Multiplying the preceding equation by $E(-u;q^2)$, followed by extracting the coefficients of $u^n$, we have
$$
d_n^B(q)=\sum_{k=0}^n\frac{(-1)^kq^{2\binom{k}{2}}(q^2;q^2)_n}{(q^2;q^2)_k(1-q)^{n-k}}=\sum_{k=0}^n\frac{(-1)^kq^{2\binom{k}{2}}[2]_q[4]_q\cdots[2n]_q}{[2]_q[4]_q\cdots[2k]_q},
$$
which is (i). Since $(q^2;q^2)_n=(1-q)^n[2]_q[4]_q\cdots[2n]_q$, we have
$$
\sum_{n\geqslant 0}d_n^B(q)\frac{u^n}{(1-q)^n[2]_q[4]_q\cdots[2n]_q}=\frac{E(-u;q^2)}{1-u/(1-q)}.
$$
Replacing $u$ by $u(1-q)$, we get (ii). (iii) is immediate from (i).
\end{proof}

Theorem~\ref{thm:2}(i) is the type $B$ analogue of \eqref{eqn:1}, in
the sense that $[2]_q[4]_q\cdots[2n]_q$ 
(respextively $[n]_q!$) is the Poincar\'e series of $B_n$
(respectively $\mathfrak{S}_n$). (See \cite[Chapter 7]{bb05}.) By letting $q\to 1$, $E(-u(1-q);q^2)\to e^{-u/2}$ and Theorem~\ref{thm:2} specializes to
\begin{equation*}
\begin{split}
\sum_{n\geqslant 0}d_n^B\frac{u^n}{2^nn!}&=\frac{e^{-u/2}}{1-u},\\
d_{n+1}^B&=2(n+1)d_n^B+(-1)^{n+1},\\
d_n^B&=n!\sum_{k=0}^n\frac{(-1)^k2^{n-k}}{k!},
\end{split}
\end{equation*}
where $d_n^B=d_n^B(1)$ is the derangement number of $B_n$; the last formula can also be obtained by a routine application of the principle of inclusion-exclusion \cite[Chapter 2]{s97}.


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\end{document}
