5
$\begingroup$

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 $\mu$?

Improve this question
$\endgroup$
3
  • $\begingroup$ 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
    Commented Jul 12, 2020 at 22:27
  • $\begingroup$ @DavidCraven I was hoping for a nice formula only for the whole sum, not for individual terms $\endgroup$
    – Marcel
    Commented Jul 13, 2020 at 0:12
  • $\begingroup$ Ah sorry, I missed that. $\endgroup$
    – David A. Craven
    Commented Jul 13, 2020 at 8:08

2 Answers 2

Reset to default
7
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ This is good enough, and suggests that no simple formula exists $\endgroup$
    – Marcel
    Commented Jul 10, 2020 at 20:50
4
$\begingroup$

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...

Improve this answer
$\endgroup$
1
  • 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
    Commented Jul 10, 2020 at 21:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .