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Given a multi-variable polynomial $F$, denote the number of monomials by $N(F)$. Take for instance, \begin{align*}N(x(x+y)+(x+y)^2-(x-y)^2)=N(x^2+5xy)&=2 \qquad \text{and} \\ N((x+z)(x+y)^2)=N(x^3 + 2x^2y + x^2z + xy^2 + 2xyz + y^2z)&=6.\end{align*} Denote the symmetric group on $n$ letters $\{1,2,\dots,n\}$ by $\mathfrak{S}_n$. Define the action of $\mathfrak{S}_n$ on a function $F(x_1,\dots,x_n)$ in a natural way: given $w\in\mathfrak{S}_n$, then $w\cdot F(x_1,\dots,x_n)=F(x_{w(1)},\dots,x_{w(n)})$. Introduce the (multi-variable) rational functions $$G(\mathbf{x},\mathbf{z})=\prod_{k=1}^n\frac{x_1z_1+x_2z_2+\cdots+x_kz_k}{x_k-x_{k+1}}.$$

Assuming that the symmetric groups act only on the $x$-variables, let's compute the polynomial $$G_{\mathfrak{S}_{n+1}}=\sum_{w\in\frak{S}_{n+1}}w\cdot G.$$ I would like to ask:

QUESTION. Let $C_n=\frac1{n+1}\binom{2n}n$ be the Catalan numbers. Is there a combinatorial proof that $N(G_{\mathfrak{S}_{n+1}})=C_n$?

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  • $\begingroup$ Are you sure the $G(\mathbf{x},\mathbf{z})$ are polynomials (as opposed to rational functionals)? I guess the point is that after symmetrization the $G_{\mathfrak{S}_{n+1}}$ become polynomials... $\endgroup$ Apr 22 at 21:07
  • $\begingroup$ Edited. Thank you. $\endgroup$ Apr 22 at 22:07
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    $\begingroup$ I checked findstat for the distribution of coefficients, to no avail :-( $\endgroup$ Apr 23 at 9:18

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Any monomial $P:=\prod x_i^{c_i}$ of degree $\sum c_i=n$ maps to a non-zero constant after symmetrization $$ P\to \Phi(P):=G_{\mathfrak{S}_{n+1}}\frac{P}{(x_1-x_2)(x_2-x_3)\ldots (x_n-x_{n+1})}. $$ Indeed, $\Phi(P)$ is a constant by a degree consideration, to prove that this constant is non-zero you may use, for example, Theorem 2 here.

Thus any monomial $\prod (x_iz_i)^{c_i}$ maps to a non-zero constant times $\prod z_i^{c_i}$.

Thus, you simply count the number of monomials which may arise, when you multiply the linear forms $\prod_{k=1}^n(x_1z_1+x_2z_2+\cdots+x_kz_k)$. The only condition is that $c_1+\ldots+c_i\geqslant i$ for all $i$. Such sequences are indeed enumerated by Catalan numbers, a bijection with lattice paths below diagonal is straightforward.

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  • $\begingroup$ This is now findstat.org/StatisticsDatabase/St001786 $\endgroup$ Apr 25 at 8:04
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    $\begingroup$ Does your Theorem yield a direct interpretation of the coefficients? Is it true that the coefficient equals 1 if and only if the path is a bounce path? $\endgroup$ Apr 25 at 10:36
  • $\begingroup$ @MartinRubey the coefficient equals (upto sign) to the number of enumerations of $n+1$ vertices of the path which satisfy $n$ inequalities which in turn correspond to the $n$ edges. The inequality $\pi(x)<\pi(y)$ for an edge $xy$ corresponds to our monomial $C$ being $y$-biased in the following sense: remove the edge $xy$, let $k$ denote the number of vertices in the piece containing $y$; then the total degree of $C$ in these $k$ variables is at least $k$. So, the number of monomials with coefficient $\pm 1$ (happens when all inequalities have the same direction) is twice Catalan number. $\endgroup$ Apr 25 at 12:30
  • $\begingroup$ That's strange, possibly I have a bug in my code, but my coefficients are all positive. $\endgroup$ Apr 25 at 14:33
  • $\begingroup$ @MartinRubey this is quite possible, I simply did not care about the sign writing my comment $\endgroup$ Apr 25 at 15:48

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