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For a permutation $\pi\in\frak{S}_n$, define the number of descents of $\pi$ as $$\text{des}(\pi)=\vert\{i: \pi(i)>\pi(i+1)\}\vert.$$ The following is a well-known (and interesting) identity: $$\binom{k\ell+n-\text{des$(\pi)$}-1}n=\sum_{\sigma\tau=\pi} \binom{k+n-\text{des$(\sigma)$}-1}n\binom{\ell+n-\text{des$(\tau)$}-1}n.$$ This motivated me to ask:

QUESTION. Is there a similar identity for the major index? What about other "statistic" on $\frak{S}_n$?

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    $\begingroup$ Can you please give the reference for this well-known identity? Thanks $\endgroup$ Apr 28, 2021 at 19:24
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    $\begingroup$ @VladimirDotsenko Here is one place to look: Mielnik, B. and Plebanski, J., Combinatorial approach to Baker-Campbell-Hausdorff exponents,Annalesde l’ I.H.P., Section A12(1970), 215–254. $\endgroup$ Apr 28, 2021 at 19:38
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    $\begingroup$ Not exactly what you are looking for, but you might be interested in "shuffle compatible" permutations statistics in the sense of arxiv.org/abs/1706.00750. $\endgroup$ Apr 28, 2021 at 20:10
  • $\begingroup$ Thanks for this info, Sam. $\endgroup$ Apr 28, 2021 at 20:51
  • $\begingroup$ The binomials in the right-hand side are exactly what one gets if one evaluates the fundamental quasisymmetric functions at 1,1,...,1. So, your example should have a good chance opf have a quasisymmetric lift and a q-analog with q-binomials involved.... $\endgroup$ Apr 29, 2021 at 7:05

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Richard's identity $(*)$ can be found as Theorem 11 (though not stated exactly this way) in my paper Multipartite  P-partitions and inner products of skew Schur functions, Combinatorics and algebra (Boulder, Colo., 1983), 289–317,Contemp. Math., 34, Amer. Math. Soc., Providence, RI, 1984. (It's actually $\tau\sigma=\pi$ rather than $\sigma\tau=\pi$.) You can find this paper at http://people.brandeis.edu/~gessel/homepage/papers/multipartite.pdf.

I don't think that setting $x_i=q^{i-1}$ and $y_j=t^{j-1}$ works except when $\pi$ is an identity permutation or the reverse of an identity permutation when $F_{D(\pi)}$ is symmetric.

However we can get a $q$-analogue if we keep the descent number in addition to the major index, thought it's not as simple as one might like: we take $x_i = q^{i-1}$ for $i=1,2,\dots, M$ with $x_i=0$ for $i>M$ and $y_j= q^{m(j-1)}$ for $j=1,2,\dots, N$ (where $N$ may be $\infty$). See Section 4 of T. Kyle Petersen, Cyclic descents and P-partitions, Journal of Algebraic Combinatorics 22 (2005) 343-375, https://arxiv.org/abs/math/0405479.

Some related formulas can be found in D. Krob, B. Leclerc, and J.-Y.Thibon, Noncommutative symmetric functions. II. Transformations of alphabets, Internat. J. Algebra Comput. 7 (1997), no. 2, 181–264.

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  • $\begingroup$ thank you for expanding on Richard's claim and adding valuable pointers. Upvoted. $\endgroup$ Apr 30, 2021 at 15:22
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Let $F_{S,n}(x)$ denote the fundamental quasisymmetric function in the variables $x_1,x_2,\dots$ indexed by the set $S\subseteq [n-1]$, so $F_{S,n}(x_1,\dots,x_k)$ is homogeneous of degree n. Let $F_{S,n}(xy)$ denote the fundamental quasisymmetric function indexed by $S\subseteq [n-1]$, in the variables $x_iy_j$, $i,j\geq 1$, where the order of the variables is lexicographic, i.e., $x_iy_j<x_hy_m$ if $i<h$ or if $i=h$ and $j<m$. Let $\pi\in\mathfrak{S}_n$, with descent set $D(\pi)$. Then one can conjecture the following identity: $$ F_{D(\pi),n}(xy) =\sum_{\sigma\tau=\pi} F_{D(\sigma),n}(x) F_{D(\tau),n}(y). \qquad\qquad (*)$$ (This might not be exactly right. Perhaps a different order of the variables $x_iy_j$ is necessary, or some other small modification.) We can then set $x_1=\cdots=x_k=1$, $x_h=0$ if $h>k$, $y_1=\cdots= y_\ell=1$, $y_m=0$ if $m>\ell$, to get the stated identity. I would not be surprised if ($*$) is already known.

If we set $x_i=q^{i-1}$ and $y_j=t^{j-1}$, then we should get a major index analogue, using $$ F_{D(\pi),n}(1,q,q^2,\dots) = \frac{q^{\mathrm{comaj}(\pi)}} {(1-q)(1-q^2)\cdots (1-q^n)}. $$

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  • $\begingroup$ Your conjectured identity sounds like the internal product on quasisymmetric functions. If so, this might be in Malvenuto-Reutenauer. $\endgroup$
    – user35313
    Apr 29, 2021 at 17:41
  • $\begingroup$ @Richard: thank you for initiating a fine conjecture. Upvoted. $\endgroup$ Apr 30, 2021 at 15:21

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