Take a finite extension of characteristic zero fields $L/F$ which is Galois. Let $G$ be a finite group and $(V,\rho)$ a finite-dimensional representation of $G$ over $L$. Assume that the representations $V$ and $V^{\sigma}$ are isomorphic for every $\sigma\in\text{Gal}(L/F)$. Does this mean that $V$ is realizable over $F$?
Remark: I write $V^{\sigma}$ for the same set as $V$, but with each $g\in G$ acting by $\sigma(\rho(g))$ rather than by $\rho(g)$. By $\sigma(\rho(g))$ I refer to the entry-wise conjugation of the matrix $\rho(g)$.
EDIT: It would also be nice to know what happens when the extension $L/F$ is not finite (but still Galois).
Best Answer
No, that the representations are isomorphic means they have the same character $\chi$ so it must be $F$-valued, there exists some representation over $F$ such that $tr \ \Pi = d \chi$ but sometimes we need a larger field to get $d=1$.
That's the difference between the extension $K$ (of $\Bbb{Q}$) generated by the character table and a splitting field for $G$ : a finite extension (of $K$) where all the irreducible representations can be realized.
The character table of the quaternions $Q_8$ is in $\Bbb{Q}$ but some of the representations need fields like $\Bbb{Q}(i)$ (or $\Bbb{Q}(\sqrt{-1-m^2})$ to be realized.