Let $G$ be a finite group and $\chi$ a character of $G$. The values of $\chi$ generate an abelian Galois extension $K$ of $\mathbb{Q}$, and so one can consider the conjugate $\sigma(\chi)$ of $\chi$ by any element $\sigma$ of the Galois group. What's the shortest way to prove that $\sigma(\chi)$ is also a character of $G$?
This follows from the fact that the corresponding representation $V$ is realizable over a finite extension of $K$, but this fact is somewhat annoying to prove, and on an exam I don't want to prove it from scratch if I can avoid it. Are there any shorter arguments?
Motivation: I'm looking at a representation theory exam question (for practice) which asks me to fill out a character table. If I could assume the above statement, I could conclude that the missing characters are integer-valued.
Best Answer
OK, I've thought about it a bit more and I'll give an alternate proof below the first horizontal line. However, I highly deny that this proof is better, and I am not sure it is actually shorter. To my mind, the morally correct proof is to show that, if $K \subseteq L$ with $L$ algebraically closed, and a system of polynomial equations has a root in $L$, then it has a root in a finite extension of $K$.
The point, which I am sure Qiaochu understands, is that he only knows a priori that the representation is defined over $\mathbb{C}$. Once he knows that the representation is definable over an algebraic extension $K'$ of $K$, he can replace $K'$ by its normal closure, note that $\mathrm{Gal}(K', \mathbb{Q}) \to \mathrm{Gal}(K, \mathbb{Q})$ is surjective, lift any element $\sigma$ of $\mathrm{Gal}(K, \mathbb{Q})$ to some $\tilde{\sigma}$ in $\mathrm{Gal}(K', \mathbb{Q})$, and apply $\tilde{\sigma}$ to the entries of his matrices.
Part of the problem is that the representation may honestly not be defined over $K$. For example, the two dimensional representation of the quaternion eight group has character with values in $\mathbb{Q}$, but can't be defined over $\mathbb{Q}$.
Fix $G$ and a representation $V$ of $G$. For $g \in G$, let $\lambda_1(g)$, $\lambda_2(g)$, ..., $\lambda_n(g)$ be the multiset of eigenvalues of $g$ acting on $V$. These are necessarily roots of unity, since $g^N=1$ for some $N$. For any symmetric polynomial $f$, with integer coefficients, define $\chi(f,g) = f(\lambda_1(g), \ldots, \lambda_n(g))$.
Lemma: With notation as above, $g \mapsto \chi(f,g)$ is a virtual character.
Proof: If $f$ is the elementary symmetric function $e_k$, then this is the character of $\bigwedge^k V$. Any symmetric function is a polynomial (with integer coefficients) in the $e_k$'s; take the corresponding tensor product and formal difference of virtual characters.
Any Galois symmetry $\sigma$ of $\mathbb{Q}(\zeta_N)$ is of the form $\zeta_N \mapsto \zeta_N^s$, for $s$ relatively prime to $N$. Consider the power sum symmetric function $p_s := \sum x_i^s$. So $\chi(p_s, \ )$ is the Galois conjugate $\chi^{\sigma}$, and we now know that it is a virtual character.
But $\langle \chi^{\sigma}, \chi^{\sigma} \rangle = \langle \chi, \chi \rangle =1$, because the inner product is built out of polynomial operations and complex conjugation, and complex conjugation is central in the Galois group. So this virtual character must correspond to $\pm W$, for some representation $W$. Since $\chi^{\sigma}(e) = \chi(e) = \dim V$, we conclude that the positive sign is correct.
It just occurred to me that actually writing this out for some specific small values of $s$ makes some nonobvious statements about representation theory. For example, if $G$ has odd order and $V$ is a $G$-irrep, then $\bigwedge^2 V$ has a $G$-equivariant injection into $\mathrm{Sym}^2 V$. Proof: The difference of their characters is the character of $V^{\sigma}$, where $\sigma: \zeta \mapsto \zeta^2$.