Let us call a partition odd if all its parts are odd, and let $Odd(n)$ be the set of all odd partitions of $n$, e.g. $Odd(6)=\{(5\,1),(3\, 3),(3\,1^3),(1^6)\}$.

Let $H(n)$ denote the set of all hook partitions of $n$. I have made the following surprising observation (to me, at least): For every $\mu\in H(m)$, $$\frac{1}{2^{\ell(\nu)}}\sum_{\lambda\in H(n+m)} \chi_{\lambda\backslash\mu}(\nu)=\begin{cases} 0, \text{ if }\nu\notin Odd(n)\\ 1, \text{ if }\nu\in Odd(n),\end{cases}$$ where $\chi_{\lambda\backslash\mu}$ are skew characters of the symmetric group.

I think a proof would come from the combinatorial description of skew characters in terms of tableaux (although I didn't really develop this all the way through). My question is: is this result already in the literature somewhere?

EDIT: Upon a moment of reflection it becomes clear that my sum is indeed independent of $\mu$. So we could use any $\mu$ to compute it, even the trivial one $\mu=(0)\vdash0=m$.

However, user61318 mentioned a paper in his comment which proves precisely that $$ \frac{1}{2^{\ell(\nu)-1}}\sum_{\lambda\in H(n)} \chi_{\lambda}(\nu)=\begin{cases} 0, \text{ if }\nu\notin Odd(n)\\ 1, \text{ if }\nu\in Odd(n),\end{cases}$$ which differs from the $m=0$ case of my sum by a factor of $2$. I have checked that both formulas are right as they stand, so this difference is a puzzle.

  • $\begingroup$ Yeah, sorry, I forgot about the sum. $\endgroup$
    – Jay Taylor
    Mar 7, 2016 at 16:27
  • $\begingroup$ Hooks and characters... the following might be useful: arxiv.org/pdf/1108.3170.pdf $\endgroup$
    – user35313
    Mar 7, 2016 at 17:05
  • $\begingroup$ Your formula is independent of $\mu$ so long as $m > 0$. I think there is a difference when $m=0$ due to the way that hooks are computed in the skew diagram versus a proper Young diagram. For instance a hook Young diagram of a partition of $n$ has a hook of length $n$ but a hook skew diagram whose shape is a partition of $n$ has no hook of length $n$. The Murnaghan–Nakayama formula indicates that these situations are therefore different. $\endgroup$
    – Jay Taylor
    Mar 7, 2016 at 19:48
  • $\begingroup$ @JayTaylor Funny, huh? So the proof for $m> 0$ cannot rely on the $m=0$ case; it has to be a different proof... $\endgroup$
    – Marcel
    Mar 7, 2016 at 20:15
  • $\begingroup$ @JayTaylor: Funny I thought the Regev paper would be useful. Instead you ended up deriving results therein. Nice! $\endgroup$
    – user35313
    Mar 15, 2016 at 1:07

1 Answer 1


I know you were asking for a reference, and there may be better approaches, but just to offer one proof of your statement based on the Murnaghan–Nakayama formula. Assume $m>0$ then any skew tableaux $\lambda/\mu$ has shape $(a,1^b)$ with $a+b=n$. This means $a$ boxes in the first row and $b$ boxes in the first column but the row and column are disconnected. We want to evaluate the sum

$$ \Lambda_n = \sum_{a=0}^n \chi_{(a,1^{n-a})} $$

If $1 \leqslant k \leqslant n$ then the skew partition $(a,1^b)$ has at most two $k$-hooks, one of leg length 0 if $k \leqslant a$ and one of leg length $k-1$ if $k \leqslant b$. Now assume $\nu\vdash n$ is a partition with part equal to $k$ and let $\hat{\nu} \vdash n-k$ be the partition obtained by removing this part. By the Murnaghan–Nakayama formula for skew partitions we have

\begin{align*} \Lambda_n(\nu) &= \sum_{a=0}^{k-1}\chi_{(a,1^{n-a})}(\nu) + \sum_{a=k}^{n-k}\chi_{(a,1^{n-a})}(\nu) + \sum_{a=n-k+1}^n\chi_{(a,1^{n-a})}(\nu)\\ &= \sum_{a=0}^{k-1}(-1)^{k-1}\chi_{(a,1^{n-a-k})}(\hat{\nu}) + \sum_{a=k}^{n-k}(\chi_{(a-k,1^{n-a})}(\hat{\nu}) + (-1)^{k-1}\chi_{(a,1^{n-a-k})}(\hat{\nu})) + \sum_{a=n-k+1}^n\chi_{(a-k,1^{n-a})}(\hat{\nu})\\ &= \sum_{a=0}^{n-k}\chi_{(a,1^{n-k-a})}(\hat{\nu}) + (-1)^{k-1}\sum_{a=0}^{n-k}\chi_{(a,1^{n-k-a})}(\hat{\nu})\\ &= (1+(-1)^{k-1})\Lambda_{n-k}(\hat{\nu}). \end{align*}

Hence if $k$ is even then $\Lambda_n(\nu) = 0$ and an easy induction shows that $\Lambda_n(\nu) = 2^{\ell(\nu)}$ if $\nu \in \mathrm{Odd}(n)$.

Maybe I would just add that the same argument essentially works when $m=0$. For this case let $\tilde{\chi}_{(a,1^{n-a})}$ be the irreducible character labelled by a hook partition and let $\tilde{\Lambda}_n = \sum_{a=1}^n \tilde{\chi}_{(a,1^{n-a})}$. The calculation above goes through in this case except when $\nu = (n)$ in which case we have

\begin{equation*} \tilde{\Lambda}_n(\nu) = \sum_{a=1}^n \chi_{(a,1^{n-a})}(\nu) = \sum_{a=1}^n (-1)^{n-a} \end{equation*}

which is 0 if $n$ is even and 1 if $n$ is odd. This explains why one gets $\tilde{\Lambda}_n(\nu) = 2^{\ell(\nu)-1}$ when $\nu \in \mathrm{Odd}(n)$.

EDIT: It should also be possible to show that $\Lambda_n = 2\tilde{\Lambda}_n$ directly using the fact that $\chi_{\lambda/\mu} = \sum_{\tau \vdash n} c^{\lambda}_{\mu\tau}\tilde{\chi}_{\tau}$, where $c^{\lambda}_{\mu\tau}$ is the Littlewood–Richardson coefficient.

  • $\begingroup$ I have this otther question (232631) on the analogue of Murnaghan–Nakayama for skew zonal spherical functions. Maybe you can contribute an answer there? $\endgroup$
    – Marcel
    Mar 8, 2016 at 21:40

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