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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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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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.

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