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 $\mu$?

  • $\begingroup$ 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. $\endgroup$ – David A. Craven Jul 12 at 22:27
  • $\begingroup$ @DavidCraven I was hoping for a nice formula only for the whole sum, not for individual terms $\endgroup$ – Marcel Jul 13 at 0:12
  • $\begingroup$ Ah sorry, I missed that. $\endgroup$ – David A. Craven Jul 13 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.

| cite | improve this answer | |
  • $\begingroup$ This is good enough, and suggests that no simple formula exists $\endgroup$ – Marcel Jul 10 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...

| cite | improve this answer | |
  • 3
    $\begingroup$ 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 $2n\cdot n!/(n+1)$. $\endgroup$ – Richard Stanley Jul 10 at 21:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.