Representations of reductive groups over arbitrary fields Let $k$ be a field and $G/k$ a connected reductive group. Fix a maximal torus $T$, and let $X$ denote the group of characters of $T_{\overline k}$, where $\overline k$ is a separable closure of $k$. Let $R\subset X$ denote the set of roots of $(G_{\overline k},T_{\overline k})$ and fix an ordering, with positive roots $R^+$ and dominant weights $X^+$. Everything has an action $\operatorname{Gal}(\overline k/k)$.
Then any $\lambda\in X^+$ gives an irreducible representation $V_{\lambda}$ of $G_{\overline k}$ with highest weight $\lambda$.
The question: if $k\subset F\subset\overline k$ is such that $\operatorname{Gal}(\overline k/F)$ stabilizes $\lambda$, is it true that $V_{\lambda}$ is actually the extension of scalars of a representation of $G_F$?
According to Théorème 3.3 of [Tits, Représentations linéaires irréductibles d'un groupe réductif], there exists a representation of $G_F$ with values in $GL_{m,D}$, where $D$ is a central division algebra over $F$, whose base change to $\overline k$ gives indeed $V_{\lambda}$. But $D$ is not necessarily equal to $F$.
 A: You've essentially answered your own question. Let $G$ be the units of a division algebra of dimension $n^2$. Then $G$ is an inner form of a general linear group so the Galois action on the root datum will be trivial. But the weight corresponding to the canonical $n$-dimensional representation of $G$ over the alg closure descends to an $n$-dimensional representation over $k$ iff $G$ is split.
EDIT: Bcnrd raises the issue that the questioner says "everything has an action of Galois" without saying what this action is. My answer implicitly assumes it is the action called $\mu_G$ in Corvallis (which has the property that it depends only on $G$ and not on $T$) and Bcnrd raises the issue that it could be the the "naive" action (which depends on the arithmetic of $T$). I do not know which action the OP means, and the validity of this answer is contingent upon my guess being the right one. UPDATE: BCnrd tells me that in the paper in question, the action seems to be the one I'm calling $\mu_G$ so this answer is probably OK, but it does leave open the question as to what happens if one uses the naive action.
