Generalized cycle index polynomial for the symmetric group

Question

The answer to a particular calculation in quantum information theory gives me the following expression:

Given $M$ specific elements of the symmetric group $S_n$, define the polynomial

$$Z_n(\pi_1, \dots , \pi_M ; X_i) = \sum_{\pi \in S_n} X_1^{C(\pi \cdot \pi_1)} X_2^{C(\pi \cdot \pi_2)} \cdots X_M^{C(\pi \cdot \pi_M)}$$

where $C(\pi)$ counts the number of disjoint cycles in the permutation $\pi$.

For $M=1$, this is related to the ordinary cycle index $Z(S_n; a_1, a_2, \dots)$ by

$$Z_n(\pi_1;X) = n! Z(S_n ; X,X, \dots)$$

My question is whether there is some useful discussion of this object $Z_n$ in the literature, or whether anyone can recommend an approach for calculating $Z_n$ for specific elements $(\pi_1,\dots, \pi_M)$ (a simple example of interest would be $\pi_1$ equal to the identity and $\pi_2$ equal to a cyclic permutation).

Answer 1

For $M=2$ we get a "solution" as follows. For given $\pi_1$ and $\pi_2$ we want to know how many permutations $\pi$ are there such that $\pi\pi_1$ has $c_1$ cycles and $\pi\pi_2$ has $c_2$ cycles. Let $\rho_1=\pi\pi_1$ and $\rho_2=\pi\pi_2$.

Writing $\pi\pi_2=\pi\pi_1\pi_1^{-1}\pi_2=\rho_1\pi_1^{-1}\pi_2=\rho_2$, we arrive at a factorization problem: Given $\pi_1^{-1}\pi_2$, find in how many ways it can be written as a product $\rho_1^{-1}\rho_2$ such that $\rho_1$ has $c_1$ cycles and $\rho_2$ has $c_2$ cycles.

Such quantities are related to connection coefficients of the symmetric group, and can be computed in terms of characters. There is a large literature on factorization problems, but simple explicit formulas are not easy to come by.

You can find somre references here: G. Berkolaiko, J. Irving, Inequivalent Factorizations of Permutations, http://arxiv.org/abs/1405.5255.

For general $M$ we can take $\pi_1$ and insert it in the other equations, arriving at $M-1$ simultaneous factorizations, $$\pi_1^{-1}\pi_i=\rho_1^{-1}\rho_i,\quad 2\le i\le M.$$ The problem here is that the first factor is common to all of them, so they are coupled. I have never seen anything about this.

Answer 2

For $M=2$, the following can be shown using the case $k=3$ of Exercise 7.70 in Enumerative Combinatorics, vol. 2. I use notation from this reference. We may assume that $\pi_1$ is the identity permutation. Suppose that $\pi_2$ has cycle type $\lambda$. Write $$ F_\lambda(x,y) = Z_n(\mathrm{id},\pi_2;x,y). $$ Define the symmetric function $$ G_n = \sum_{\lambda\vdash n}\frac{n!}{z_\lambda}F_\lambda(x,y)p_\lambda. $$ For instance, $$ G_3 = (x^3y^3+3x^2y^2+2xy)p_1^3 + 3(x^3y^2+x^2y^3+2x^2y+2xy^2)p_2p_1 + 2(x^3y+xy^3+3x^2y^2+xy)p_3. $$ Then $$ G_n = \sum_{\lambda\vdash n} f^\lambda \prod_{u\in\lambda}(x+c(u))(y+c(u))\cdot s_\lambda. $$ To extract the coefficient of $p_\mu$ from $G_n$, use $$ s_\lambda =\sum_{\mu\vdash n}z_\lambda^{-1} \chi^\lambda(\mu) p_\mu. $$ This gives an explicit formula for $F_\lambda(x,y)$, but it is not so elegant since it involves irreducible character values of $S_n$.