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 $\mu=(n)$).



sum of character product over derangements
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 $s\_1$.


