I'm trying to understand a proof sketch found here: https://mathoverflow.net/questions/10635/why-are-the-characters-of-the-symmetric-group-integer-valued

If $g$ is an element of order $m$ in a group $G$, and $V$ a complex representation of $G$, then $\chi_V(g)$ lies in $F=\mathbb{Q}(\zeta_m)$. Since the Galois group of $F/\mathbb{Q}$ is $(\mathbb{Z}/m)^\times$, for any $k$ relatively prime to $m$ the elements $\chi_V(g)$ and $\chi_V(g^k)$ differ by the action of the appropriate element of the Galois group.

If $G$ is a symmetric group and $g$ an element as above, then $g$ and $g^k$ are conjugate: they have the same cycle decomposition. So $\chi_V(g)=\chi_V(g^k)$ whenever $(k,m)=1$, and thus $\chi_V(g)\in \mathbb{Q}$.

I'm struggling with understanding how the Galois group acts on the character. The character $\chi$ of a representation $\phi(g)$ is $\chi _{\phi} (g) = \textrm{Tr} \phi (g)$ so $\chi _{\phi} (g^k) = \textrm{Tr} \phi (g)^k$. I know that the character is the sum of roots of unity, but I don't see why $\chi _{\phi} (g^k)=\sigma _k (\chi _{\phi} (g))$, where $\sigma _k$ is the element of the Galois group that takes $\sigma _k (\zeta _m)=\zeta _m ^k$.

Is this the way the automorphism is acting? If so (or not), how is it acting and how can I see this? Thanks!

  • $\begingroup$ Yes, this is the way. $\endgroup$ – Berci May 9 '13 at 23:51

A simple result in linear algebra is:

Let $A$ be a linear map from $\mathbb{C}^n\rightarrow \mathbb{C}^n$ with eigenvalues $\{ \lambda_1,\cdots, \lambda_n \}$ counted with multiplicity. Then the eigenvalues of $A^k$ is $\{ \lambda_1^k,\cdots, \lambda_n^k\}$ also counted with multiplicity.

In your problem, $\phi(g)$ is a linear map from finite dimensional vector space over $\mathbb{C}$, so the above holds for the set of eigenvalues of $\phi(g)$. Note that all the eigenvalues of $\phi(g)$ should be $\zeta_m^i$ for some $i$.

Also, we have that trace of a linear map is sum of eigenvalues.

Hence, $\chi _{\phi} (g^k)=\sigma _k (\chi _{\phi} (g))$ should be true.

  • $\begingroup$ Clear answer, delivered very quickly. Thank you! I need to brush up on my linear algebra :) $\endgroup$ – curiousmathlete May 9 '13 at 23:53

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