Let $\chi_{\mu}^{\lambda}$ denote a value of an irreducible character of the symmetric group $\frak{S}_n$, where $\mu, \lambda\vdash n$. When $\mu=(n)$, then it's known that $$\sum_{\lambda\vdash n}\chi_{\mu}^{\lambda}=\delta_\text{odd}(n).$$ This time, I'm interested to know:

QUESTION 1. Is there anything known about this total sum: $$\frak{s}_n:=\sum_{\mu\vdash n}\sum_{\lambda\vdash n}\chi_{\mu}^{\lambda}=?$$

LeechLattice supplied the OEIS link here with values $\frak{s}_n=$$1, 2, 5, 13, 31, 89, 259, 842, 2810, \dots$

QUESTION 2. The following appears true. Is it? $\frak{s}_n$ is even iff $p(n)$ is even. Here, $p(n)$ stands for the number of partitions of $n$. Incidentally, one observes that it is not known "how often" $p(n)$ is even or odd.

  • 1
    $\begingroup$ This is closely related to the idea of a "Gelfand model" for the symmetric group. $\endgroup$ Sep 14 at 19:39
  • 3
    $\begingroup$ oeis.org/A082733 ? $\endgroup$ Sep 14 at 19:42
  • $\begingroup$ @SamHopkins: I always appreciate your quick reply. What do we get out of this idea? $\endgroup$ Sep 14 at 19:43
  • 3
    $\begingroup$ @T.Amdeberhan: A Gelfand model for a group just means the representation that is a direct sum of all the irreducibles, with multiplicity one (but the point is to give a nice, "tractable" model of this representation). So you are asking to sum the character of a Gelfand model over all permutations. I know there has been some work on Gelfand models for $\mathfrak{S}_n$, but I don't know of what specifically is known; probably someone more expert can say more. $\endgroup$ Sep 14 at 19:46
  • 1
    $\begingroup$ At a first glance of the numerics, it seems to be asymptotically close to the number of involutions in $S_n$ (the sum of the dimensions of irreps). $\endgroup$
    – Lucia
    Sep 14 at 21:45

The sum $\sum_\mu \chi^\lambda_\mu$ over partitions $\mu$ of $n$ is the multiplicity of the irreducible $\chi^\lambda$ in the character afforded by $\mathfrak S_n$ acting on itself by conjugation. If $\psi$ is the character for conjugation, then $\psi(g)$ is the size of the centralizer of $g$, so $$\langle \chi^\lambda,\psi\rangle =\sum_{\mu\vdash n} \frac{\chi^\lambda_\mu z_\mu}{z_\mu},$$ where $z_\mu$ is the size of the centralizer of any permutation with cycle type $\mu$. So your sum is counting how many irreducibles the conjugation representation breaks up into.

I should mention that the individual row sums $\sum_{\mu\vdash n} \chi^\lambda_\mu$ are not completely understood. It is already non-trivial to prove that they are greater than $0$ (this should be in "An explicit model for the complex representations of $S_n$" by Inglis, Richardson and Saxl.) This is unlike the situation for the group acting on itself by left multiplication. However, when you take your double sum, then there is a good combinatorial description. In fact, summing down any column we know: $$\sum_\lambda \chi^\lambda_\mu = \# \{g\in \mathfrak S_n : g^2=h\}$$ for any fixed $h\in\mathfrak S_n$ with cycle type $\mu$. This works because each irreducible character of $\mathfrak S_n$ is the character of a real representation. In general, there is the work of Frobenius--Schur. (This is what Lucia was already quickly and correctly observing. Indeed, the sum of degrees will dominate.)

For your second question, it comes from the paper linked to by Lucia in your previous post. Miller proved the statement you are observing (see Theorem 2 and its proof, particularly the last few lines). The number of even entries plus the number of odd entries (denoted $E_n$ and $O_n$) is congruent to the number of partitions $p(n)$, and $E_n$ is always even. So $O_n\equiv p(n)$ mod 2.


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.