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\title{The double Eulerian polynomial in terms of inversion tables}
\author{Erik Aas}
\date{}
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\begin{document}
\maketitle
\abstract{We prove a conjecture on double Eulerian polynomials due to Visontai \cite{V}, relating the number of descents and the number of inverse descents of permutations to the number of occupied rows and number of ascents of inversion tables.}

\section*{Introduction}

The double Eulerian polynomial $A_n(t,s)$ enumerates the number of descents and the number of inverse descents of a permutation,
$$
A_n(t,s) = \sum_{\pi\in\mbb{S}_n} t^{\des{\pi}} s^{\des{\pi^{-1}}}.
$$

It is a natural generalization of the classical Eulerian polynomial $A_n(t, 1)$.
One interesting property of the classical Eulerian polynomial is its {\it unimodality}, which is an easy consequence of being nonnegative in the basis $\{t^i(1+t)^{n-i}\}_{i=0} ^n$.
One way of proving this is to note that $A_n(t,1)$ is the $h$-polynomial of the Coxeter complex (see for instance \cite{BR}).
Foata and Strehl gave a bijective proof involving 'valley-hopping' - see \cite{P} for a nice exposition.
Gessel \cite{B} conjectured that $A_n(t,s)$ similarly has nonnegative coefficients when expressed in the basis $B_{ij} = (ts)^i(t+s)^j(1+ts)^{n-2i-j}$.
This conjecture has motivated some of the work on $A_n(t,s)$, including this.

For any $n \geq 0$, we denote the set of permutations of $[n] = \{1,\dots,n\}$ (the {\it $n$-permutations}) by $\mbb{S}_n$.
We think of permutations as {\it words}.
For example, the permutation mapping $1\mapsto 3$, $2\mapsto 1$, $3\mapsto 4$, $4\mapsto 2$ is identified with the word {\tt 3142}.
An {\it ($n$-)inversion table} is a sequence $e_1, \dots, e_n$ of integers satisfying $0 \leq e_i \leq i-1$ for each $i$.
The set of $n$-inversion tables is denoted $\mbb{I}_n$. We now define some statistics.

\begin{itemize}
\item  $\DES(\pi) = \{i : \pi(i) > \pi(i+1) \}$
\item $\IDES(\pi) = \{i : \pi^{-1}(i) > \pi^{-1}(i+1) \}$
\item  $\ASC(e) = \{i : e_i < e_{i+1} \}$ (note the strict inequality).
\item  $\ROW(e) = \{e_i : 1 \leq i \leq n\} \backslash \{0\}$.
\end{itemize}

Moreover, $\des = \#\DES, \ides=\#\IDES, \asc = \#\ASC, \row = \#\ROW$.

\subsection*{Examples}

For example, the inversion tables of length $5$ (written as words $e_1e_2e_3e_4e_5$) with $1$ ascent and two occupied rows are $00221, 00211, 00210,00021,00032,00031$.
The permutations of length $5$ with $1$ descent and two inverse descents are $24135,13524,23514,25134,35124,24513$.

We have $\DES(24135) = \{2\}$, $\IDES(24135) = \{1, 3\}$, $\ASC(00210) = \{2\}$, $\ROW(00210) = \{1,2\}$.

Given a standard Young tableau $P$, let $\des{P} = \{i : i$ is in a strictly higher row than $i+1\}$.
Well-known properties for the RSK correspondence give that $A_n(t,s) = \sum_{\lambda \langle n} f_{\lambda}(t) f_\lambda(s)$, where $f_{\lambda}(x) = \sum_{P{\textrm has shape }\lambda} x^{\des(P)}$.
For the $n = 5$, the polynomials $f_\lambda$ are, in no particular order, $1,2x+3x^2,4x,2x^3+3x^2,4x^3,x^4,3x^2+3x$.
These polynomials are easy to compute and using them we get $A_5(t,s)$.

\section*{The identity}

Visontai \cite{V} conjectured that
$$
\sum_{\pi \in \mbb{S}_n} t^{\des{\pi}} s^{\ides{\pi}} = \sum_{e\in \mbb{I}_n} t^{\asc{e}} s^{\row{e}}.
$$

In fact we will prove the following refined identity (which reduces to the conjecture by letting $t_1 = \dots = t_n = t$).
For a subset $S\subseteq [n]$, let $t^{S} = \prod_{i\in S} t_i$, a monomial in the indeterminates $t_1, \dots, t_n$.

\begin{theorem}
\label{main}
We have
$$
\sum_{\pi \in \mbb{S}_n} t^{\DES{\pi}} s^{\ides{\pi}} = \sum_{e\in \mbb{I}_n} t^{\ASC{e}} s^{\row{e}}.
$$

\end{theorem}

The following lemma is an immediate consequence of M\"{o}bius inversion (see \cite{S}).

\begin{lemma}
Suppose $\alpha, \beta$ are functions mapping subsets of $[n]$ to polynomials.
Then $\alpha = \beta$ if and only if for every $S\subseteq[n]$ we have
$$
\sum_{T\supseteq S} \alpha(T) = \sum _{T\supseteq S} \beta(T).
$$
\end{lemma}

We turn to the proof of Theorem \ref{main}.
Fix a subset $S\subseteq [n]$.
We will prove that 
$\sum _{\pi\in\mbb{S}_n : \DES(\pi) \supseteq S} v^{\ides(\pi)} = 
 \sum _{  e\in\mbb{I}_n : \ASC(  e) \supseteq S} v^{\row (e  )}$.


We now construct two (rather large) rooted labelled trees. 

The first tree $T_1$ has as node set all permutations $\pi$ (of any length) such that if $\pi$ is an $r$-permutation then $\DES(\pi)\supseteq S \cap [r]$.

Suppose $r<s$, and that $\pi\p$ is an $s$-permutation.
Denote by $\pi\p_{(r)}$ the $r$-permutation obtained by relabeling the first $r$ values of $\pi\p$ by $[r]$, retaining the relative ordering of these values.
There is an edge from $\pi$ to $\pi\p$ if $\{r, r+1, \dots, s\} \cap S = \{r+1, \dots, s-1\}$.
All edges are given in this way.
The node $\pi\in\mbb{S}_n$ in $T_1$ is labelled by the pair $(\ides(\pi), n)$

The other tree $T_2$ has as node set all inversion tables $e$ (of any length) such that if $e$ is an $r$-inversion table then $\ASC(e) \supseteq S \cap [r]$.

Suppose $r < s$, $e$ is an $r$-inversion table, $e\p$ is an $s$-inversion table.
There is an edge from $e$ to $e\p$ if the restriction of $e\p$ (considered as a function from $[r]$ to $\mbb{Z}_{\geq 0}$) to $[r]$.
The node $e\in\mbb{I}_n$ in $T_2$ is labelled $(\row(e), n)$.

By the lemma it suffices to prove that $T_1$ and $T_2$ are isomorphic as rooted labelled trees.
We prove this by constructing a bijection $\Phi$ from $T_1$ to $T_2$ inductively.
Although we do not construct $\Phi$ completely explicitly, this can be done, though we have not found a particularly nice choice.

We let $\Phi$ map the root of $T_1$ (the empty permutation) to the root of $T_2$ (the empty inversion table).
Now suppose $\pi$ is an $r$-permutation and that $e = \Phi(\pi)$ has been defined, where $e$ is an $r$-inversion sequence.

Let $F(\pi, s, p)$ be the set of children $\pi\p$ of $\pi$ such that $\pi\p$ has length $s$ and $\ides(\pi\p) = \ides(\pi)+p$.
Similarly, let $G(e, s, p)$ be the set of children $e\p$ of $e$ such that $e\p$ has length $s$ and $\row(e\p) = \row(e)+p$.

To prove that $\Phi$ can be extended to the children of $\pi$ it suffices to prove that $|F(\pi,s,p)| = |G(e,s,p)|$.
This will finish the proof of the theorem.

It is easy to see that elements $\pi\p$ of $F(\pi,s,p)$ correspond in a natural way to nonnegative integer $(s+1)$-tuples $(x_1,\dots,x_{s+1})$ with sum $r$ such that $\sum_{i=1} ^{t+1} x_i + \sum_{i = t+2} ^{s+1} (x_i-1)_+ = p$; $x_i$ represents the number of letters in $\pi\p$ less than the $i$'th largest and the $(i+1)$'st largest among the first $r$ letters in $\pi\p$.

Similarly elements $e\p$ of $G(e, s, p)$ can naturally be represented by $r$-subsets $S$ of $[r+s]$ with $|S \cap \{1,\dots,t+1\}| = p$; $S$ represents the values of $e\p$ on $[s]\backslash[r]$ (which are distinct by the construction of $T_2$).

From these remarks one easily finds $|F(\pi,s,t,p)| = \sum_{a=0}^p {t+1\choose r-p} {r-a-1\choose a}{a+s-t-1\choose a}$. 
Similarly, $|G(e,s,t,p)| = {t+1\choose r-p}{r+s-t-1\choose p}$. Verifying that these sums are in fact equal is routine.

\begin{thebibliography}{9}
\bibitem{BR} Matthias Beck and Sinai Robins: Computing the continuous discretely: Integer-point enumeration in polyhedra, Undergraduate Texts in Mathematics, Springer, New York, 2007.
\bibitem{B} Petter Br\"{a}nd\'{e}n: Actions on permutations and unimodality of descent polynomials. European J. Combin., 29(2):514-531, 2008.
\bibitem{P} Kyle Petersen: Two-sided Eulerian numbers via balls in boxes. Math. Mag. (to appear). Preprint available at arXiv:1209.6273v1, 2012.
\bibitem{S} Richard Stanley: Enumerative Combinatorics, Vol 2, Cambridge, 1999.
\bibitem{V} Mirk\'{o} Visontai: Some remarks on the joint distribution of descents and inverse descents, Electron. J. Combin., Volume 20, Issue 1, 2013, Research article 52, 12pp.
\end{thebibliography}
\end{document}

