# 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 $$\mu$$?

• Just a comment. If there were such a nice formula, then multiplying by the nice formula for $f^\lambda$ obtains a nice formula for $\chi_\lambda(\mu)$. In other words, there would be a nice, presumably meaning non-recursive, version of the Murnaghan--Nakayama rule. Jul 12, 2020 at 22:27
• @DavidCraven I was hoping for a nice formula only for the whole sum, not for individual terms Jul 13, 2020 at 0:12
• Ah sorry, I missed that. Jul 13, 2020 at 8:08

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.

• This is good enough, and suggests that no simple formula exists Jul 10, 2020 at 20:50

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...

• There is some information related to this problem in Exercise 7.68 of Enumerative Combinatorics, vol. 2. For instance, let $n$ be odd. Then the number of ways of writing a fixed $n$-cycle in the form $uvu^{-1}v^{-1}$ is $2n\cdot n!/(n+1)$. Jul 10, 2020 at 21:07