Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\lambda\vdash n$ be a partition of $n\in\mathbb N$ and $\chi=\chi_\lambda$ the corresponding irreducible character of the symmetric group $S_n$. Denote by $\lambda^t$ be the transpose of $\lambda$ and $\ell(\lambda):=\lambda^t_1$ the length of $\lambda$. Let $\tau\in S_n$ be any transposition. In [R, Theorem 3.5], the following formula is quoted: $$ \binom n2 \cdot \frac{\chi(\tau)}{\chi(1)} = \sum_{i=1}^{\ell(\lambda)} \left( \binom{\lambda_i}2 - \binom{\lambda^t_i}2 \right). $$

I would like to see a proof for this formula. Roichman refers to the paper [I], which itself refers to a German paper by Frobenius, "Über die Charaktere der symmetrischen Gruppe". I was unable to obtain a copy of said paper, but I believe this must somehow follow from the Murnaghan-Nakayama rule (or directly from the Frobenius character formula).

Frankly, I have not invested a lot of time into an attempt to prove this myself because I was hoping that there is a more accessible and detailed reference for such an early formula. I would be very grateful if someone could provide such reference. Of course, if you happen to know a short proof, don't refrain from posting it.

share|cite|improve this question
So, for $\lambda = \lambda^t$ we get zero on the RHS. That seems strange. Also, what is $\chi(1)$ supposed to mean? My characters use to have $\chi(1) = 1$. Unless I am misunderstanding something trivial, this formula is very weird... – Marek Feb 8 '13 at 14:33
@Marek $\chi(1)$ is the degree of the representation afforded by $\chi$. – Alexander Gruber Feb 8 '13 at 15:26
@Alexander: ah, right, thanks. I confused this with characters that are homomorphisms (as opposed to just functions, like here). – Marek Feb 8 '13 at 15:43
up vote 3 down vote accepted

The paper Über die Charaktere der symmetrischen Gruppe by Frobenius is in the Sitzungsberichte der königlich preußischen Akademie der Wissenschaften zu Berlin of $1900$, available at in various formats. It begins on p. $516$; the formula you cite seems to correspond to the last formula of the paper on p. $534$. The notation $f^{(\kappa)}=\chi^{(\kappa)}_0$ is introduced on p. $522$, and $h_\varrho=n!/(c(n-c)!)$ is introduced on p. $533$. I'm not sure though how the $a_i$ and $b_i$ are related to your $\lambda_i$ and $\lambda_i^t$. Let me know if you need any help with the German.

share|cite|improve this answer
Oh that's great! The German will be the least of my problems. I will check it out right away. – Jesko Hüttenhain Feb 12 '13 at 10:10
@Jesko: Sorry, I didn't even check your name, let alone your location :-) – joriki Feb 12 '13 at 11:49
As anticipated, this is quite tough to read. Distilling the proof from it would probably require more time than I am willing to invest; Maybe someone knows a reference which is about 100 years younger? Otherwise, I'll accept your answer and just quote the result =). – Jesko Hüttenhain Feb 12 '13 at 14:31
Alright then, I shall accept this answer. Thanks a bunch for the research, I am quite happy to know that the Sitzungsberichte can be accessed via arXiv. – Jesko Hüttenhain Feb 18 '13 at 18:47
@Jesko: This is; there's no relation to (other than the name). And you're welcome :-) – joriki Feb 18 '13 at 18:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.