Summing over normalized characters of the permutation group Let $\chi_\lambda(\mu)$ be the usual characters of the irreducible representations of the permutation group $S_n$. The normalized character is the quotient $\chi_\lambda(\mu)/f^\lambda$, where $f^\lambda=\chi_\lambda(1)$ is the dimension of the representation.
Can I hope for a nice formula expressing their sum
$$ \sum_{\lambda\vdash n}\frac{\chi_\lambda(\mu)}{f^\lambda},$$
in terms of the parts of $\mu$?
 A: The quantity you are asking about is in fact a well-known expression: When multiplied by $n!$, it is the number of ordered pairs $\sigma, \tau \in S_{n}$ such that
$[\sigma, \tau] = \mu$, where $[\sigma, \tau] = \sigma^{-1}\tau^{-1}\sigma \tau$ is the commutator of $\sigma$ and $\tau$. However, I do not know how to relate this to the disjoint cycle structure of $\mu$, except to say that this quantity is clearly zero if $\mu$ is an odd permutation.
A: This is just an observation. I normalize your problem by $n!$, to get rid of denominators.
Let $A_n(\mu) := n! \sum_{\lambda \vdash n} \chi^{\lambda}(\mu)/f^\lambda$.
Define $B_n(x) := n! \sum_{\lambda \vdash n} \frac{p_\lambda(x)}{f^\lambda}$.
Then $A_n(\mu) = \langle B_n(x), s_\mu \rangle$.
That is, $A_n(\mu)$ is the coefficient of $s_\mu$ when expanded in the Schur basis.
The Schur expansion of $B_n(x)$ for $n=1,2,\dotsc$
are
\begin{array}{l}
s_{1} \\ 
4 s_{2} \\
15 s_{3}+6 s_{21}+9 s_{111} \\
76 s_{4}+64 s_{22}+44 s_{31}+76 s_{211}+12 s_{1111} \\
368 s_{5}+628 s_{32}+416 s_{41}+580 s_{221}+792 s_{311}+344 s_{2111}+200 s_{11111} 
\end{array}
Perhaps there is some pattern...
