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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 $F$.

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    $\begingroup$ This looks identical to the analogous situation for absolutely irreducible representations of finite groups, where Schur's Lemma yields a Brauer obstruction to descending the field of definition to that of the character. (Serre's book on finite group representations gives many examples where the Brauer obstruction is nontrivial over interesting $F$.) In that spirit, there should be lots of examples where the answer is negative. $\endgroup$
    – BCnrd
    Commented Nov 19, 2010 at 19:12
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    $\begingroup$ My recollection of Tits' paper is that he doesn't provide abundant examples, though as Brian says there should be negative answers to the question. Indeed, it would be very surprising in this general context for the answer always to be yes. Meanwhile, a very minor notational comment: most people use $\overline{k}$ for an algebraic closure and $k_s$ for a separable closure. $\endgroup$
    – Jim Humphreys
    Commented Nov 19, 2010 at 19:24
  • $\begingroup$ P.S. Looking again at the actual paper, I realize that in Section 6 there are quite a few examples. But the examples as well as the general theorems get technical, so it's not easy at first to extract the information in Kevin's concise answer. $\endgroup$
    – Jim Humphreys
    Commented Nov 19, 2010 at 23:40

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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.

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    $\begingroup$ Kevin, in what sense is there a "canonical" $n$-dimensional representation of $G_{k_s}$? The choice of $k_s$-isom $G_{k_s}\simeq {\rm{GL}}_n$ is only well-defined up to conjugation and transpose-inverse. Even if you demand that it carry $T_{k_s}$ to the diagonal, there remains an ambiguity of $S_n$ and transpose-inverse. Can you please fill in your answer with more of the reasoning to justify it? (I had initially considered an example like this, but got myself confused over whether it really works. Also, probably you meant to use a central simple alg., not specifically division algebra.) $\endgroup$
    – BCnrd
    Commented Nov 20, 2010 at 3:25
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    $\begingroup$ Also, it seems that the Galois action on the root datum arises from the natural $k$-structure on $T_{k_s}$, and so the Galois action on the root datum is trivial if and only if $T$ is $k$-split (and likewise, the Galois action on the set of roots is trivial if and only if the maximal $k$-torus $T \cap \mathcal{D}(G)$ in $\mathcal{D}(G)$ is $k$-split. So I don't see how $G$ being an inner form of ${\rm{GL}}_n$ implies knowledge about the Galois action on the root datum. Please let me know if I am misunderstanding something. $\endgroup$
    – BCnrd
    Commented Nov 20, 2010 at 4:02
  • $\begingroup$ brian---i'm at the seaside on my phone. We're using different Galois actions. From the context I think mine is the right one. More later. $\endgroup$
    – Kevin Buzzard
    Commented Nov 20, 2010 at 12:30
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    $\begingroup$ . Maybe the OP wants to clarify. But if my guess is right then I think everything is fine. You worry about inverse-transpose but inverse-transpose is not an issue: $G$ is the units of some division algebra $D$ and you split $D$ not $G$. You worry about the representation only being defined up to conjugation but what representation isn't? The bottom line, I think, is this: if the OP is talking about the $\mu_G$ action then everything I've said makes sense. My representation of $G$ is the identity map $G\to D^\times$. This corresponds unambiguously to a Galois-stable weight... $\endgroup$
    – Kevin Buzzard
    Commented Nov 20, 2010 at 20:07
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    $\begingroup$ Kevin, thanks to the magic of the Internets, I just checked Tits' paper in section 3.1. Says that unless indications are given to the contrary, he always intends the Galois action on $X(T_{k_s})$ to be not the naive one but rather its twist by $W(G,T)(k_s)$ to preserve a chosen specified positive chamber. Perhaps this is also what is done in Corvallis (not within reach at the moment). So it seems that your guess is better than mine. But the finite group case is closer to the naive action, and when using the non-naive action I have no intuition for why the original question is reasonable. $\endgroup$
    – BCnrd
    Commented Nov 20, 2010 at 23:53

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