I have asked this in MSE 8 days ago, even offered a bounty, and got nothing, so will try here.

I would like to understand the value of the skew characters of the symmetric group, $\chi_{\lambda/\mu}$ in the particular case when both $\lambda$ and $\mu$ are hooks, i.e. $\lambda=(a,1^{n-a})$ and $\mu=(b,1^{m-b})$ with $n>m$ and $a>b$.

As far as I can see, there would be two approaches to their calculation, but I'm a beginner and I cant work any of them through. I have looked around, but to no avail.

1) First, the Murnaghan–Nakayama rule says $$\chi_{\lambda/\mu}(\nu)=\sum_T (-1)^{{\rm ht}(T)},$$ where the sum is taken "over all border-strip tableaux of shape $\lambda/\mu$ and type $\nu$", according to Wikipedia. I do not really understand these tableaux. I mean, I think $\lambda/\mu$ is never a border strip if both are hooks; can I still sum over border strips of format $\lambda/\mu$?

2) I could also write $$\chi_{\lambda/\mu}(\nu)=\sum_\rho c^\lambda_{\mu\rho}\chi_\rho(\nu),$$ where $c^\lambda_{\mu\rho}$ are the Littlewood-Richardson coefficients. I looked around to see if they are known when $\lambda$ and $\mu$ are both hooks, but found nothing.

  • 2
    $\begingroup$ Border-strip tableaux are more than just a single border-strip. Anyway, when $\lambda$ and $\mu$ are hooks as in your post, the Schur function $s_{\lambda / \mu}$ is just the product $h_{a-b} e_{n-m-a+b}$, and you should be easily able to get the character values from there. $\endgroup$ Nov 13 '17 at 17:40
  • $\begingroup$ @darijgrinberg Thanks. Do you have a reference for this? $\endgroup$
    – thedude
    Nov 13 '17 at 17:48
  • 1
    $\begingroup$ Hmm, I would expect this to be in Sagan's "The symmetric group" or Fulton's "Young tableaux" or whatever source connects representations of $S_n$ with symmetric functions. $\endgroup$ Nov 13 '17 at 17:58

As Darij pointed out, if both $\lambda$ and $\mu$ are hooks, your diagram will be the disjoint union of a row of size $r$ and a column of size $c$, say.

To compute the character value at $\nu$, you sum over all ways to partition the parts, $\nu = \rho \cup \eta$, such that $\rho \vdash r$, $\eta \vdash c$ and evaluate the corresponding characters: $$ \sum_{\rho \cup \eta = \nu} \chi_{(r)}(\rho) \chi_{1^c}(\eta). $$ Now, the Murnaghan-Nakayama rule implies that $\chi_{(r)}(\rho)=1$, as you can fill the row in a unique way with border-strips, so you end up with $$ \sum_{\rho \cup \eta = \nu} \chi_{1^c}(\eta). $$ However, Murnaghan-Nakayama again says that $\chi_{1^c}(\eta) = (-1)^{ep(\eta)}$, where $ep(\eta)$ denotes the number of parts in $\eta$ which are even.

This can probably be simplified further, and be expressible in the parity of the length of $\eta$, and $c$.

I have typed down some info on the Murnaghan-Nakayama rule here.


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.