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It is widely known that $$ \frac{1}{n!}\sum_{\pi\in S_n}\chi_\lambda(\pi)\chi_\mu(\pi)=\delta_{\lambda,\mu},$$ where $S_n$ is the permutation group and $\chi$ are its irreducible characters.

In exercise 7.63 of his classic book Enumerative Combinatorics, Richard Stanley computes explicitly the value of $$\sum_{\pi\in D_n}\chi_\lambda(\pi),$$ where $\lambda$ is a hook and $D_n$ is the set of derangements (permutations without fixed points).

I would like to know the value of $$ \sum_{\pi\in D_n}\chi_\lambda(\pi)\chi_\mu(\pi),$$ at least when $\lambda$ and/or $\mu$ is a hook. Is anything known about this sum? (It is a generalization of the previous one, to which it reduces when $\mu=(n)$).

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  • $\begingroup$ @GeoffRobinson I can? If I decompose $\chi_\lambda\chi_\mu=\sum_\rho c_\rho \chi_\rho$ now the $\rho$'s are not necessarily hooks so the formula of Stanley does not help. $\endgroup$ – Marcel Apr 12 '18 at 11:47
  • $\begingroup$ Sorry, I did not read carefully enough to see that Stanley's computation was just for hooks- I was thinking it worked for all $\lambda.$ $\endgroup$ – Geoff Robinson Apr 12 '18 at 11:57
  • $\begingroup$ On the other hand, maybe Stanley's result, together with the Murnaghan-Nakayama rule, is enough to do the general $\lambda.$ $\endgroup$ – Geoff Robinson Apr 12 '18 at 13:11
  • $\begingroup$ @GeoffRobinson Maybe a bit more detail? $\endgroup$ – Marcel Apr 12 '18 at 16:15
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Using standard symmetric function notation, we have \begin{eqnarray*} \sum_{n\geq 0}\sum_{\lambda,\mu\vdash n} \frac{1}{n!}\left(\sum_{\pi\in D_n}\chi_\lambda(\pi)\chi_\mu(\pi)\right) s_\lambda(x)s_\mu(y) & = & \sum_{n\geq 0}\frac{1}{n!} \sum_{\pi\in S_n}\left.p_{\rho(\pi)}(x)p_{\rho(\pi)}(y)\right|_{p_1(x)=0}\\ & = & \sum_\nu \left.s_\nu(x)s_\nu(y)\right|_{p_1(x)=0}\\ & = & e^{-p_1(x)p_1(y)}\sum_\nu s_\nu(x)s_\nu(y), \end{eqnarray*} since $$\sum_\nu s_\nu(x)s_\nu(y)=\exp \sum_{m\geq 1}\frac {p_m(x)p_m(y)}{m}. $$ Thus your sum is obtained by expanding $e^{-p_1(x)p_1(y)}\sum_\nu s_\nu(x)s_\nu(y)$ in terms of Schur functions and taking $n!$ times the coefficient of $s_\lambda(x)s_\mu(y)$. To do this expansion you could write $$ e^{-p_1(x)p_1(y)} = \sum_{m\geq 0} (-1)^m\frac{s_1(x)^ms_1(y)^m}{m!} $$ and iteratively apply Pieri's formula for multiplying a Schur function by $s_1$.

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  • $\begingroup$ This looks very good indeed, thank you. $\endgroup$ – Marcel Apr 13 '18 at 0:17

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