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.



Skew character with hooks
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](https://www.math.upenn.edu/%7Epeal/polynomials/murnaghanNakayama.htm).


