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 $\mu$?



Summing over normalized characters of the permutation group
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.


