Timeline for Representations of reductive groups over arbitrary fields
Current License: CC BY-SA 2.5
17 events
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| Nov 21, 2010 at 22:10 | comment | added | BCnrd | Kevin, for the finite group analogue there's no easy answer in general (concerning when the Brauer obstruction is trivial), and I suppose that for connected reductive groups using the naive Galois action it is similar. The existence of a descent to a splitting field of the torus doesn't seem to make the problem even harder, as far as I can see. (Analogue: for finite group of order $n$ we can descend its irreducible complex repns to $\mathbf{Q}(\zeta_n)$, if I remember correctly, but this doesn't seem to make the study of the Brauer obstruction to field of defn in general even more difficult.) | |
| Nov 21, 2010 at 8:27 | comment | added | Kevin Buzzard | Thanks Brian. In turn I thought about your version of the problem, which seems much harder! If I extend the base to trivialise the naive Galois action on the torus then of course I have split the torus so I have split the group and all the representations will be defined over the base! So one needs a much more delicate approach, as I'm sure you already realised. | |
| Nov 21, 2010 at 8:24 | history | edited | Kevin Buzzard | CC BY-SA 2.5 |
update on the mystery of the undefined Galois action
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| Nov 20, 2010 at 23:53 | comment | added | BCnrd | 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. | |
| Nov 20, 2010 at 21:10 | history | edited | Kevin Buzzard | CC BY-SA 2.5 |
summarised comments generated by this answer in an edit.
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| Nov 20, 2010 at 21:07 | comment | added | Kevin Buzzard | Brian: if you're right and it's the naive action, then my answer is not correct, as you say. One of us has to open the paper and look, I guess (but not me because I've got my hands full with real life at the minute). I think I'll edit the answer to clarify this. Thanks for the "iff" remark above: I hesitated to assert that, as you spotted. | |
| Nov 20, 2010 at 20:29 | comment | added | BCnrd | Kevin, let me try to clarify. I don't see why it's reasonable to expect $\mu_G$-action to be the "right" one in the question (analogy with finite gps & related Brauer obstruction suggests the naive one, which preserves the set of roots because the $T$-action on $\mathfrak{g}$ is defined over $k$, so I think question makes sense with naive action), and the conjugation ambiguity makes it unclear that your example satisfies hypotheses using the naive action. As an aside, $D$ is a matrix algebra iff $G$ is $k$-split, due to Galois cohom. classification of both sides. | |
| Nov 20, 2010 at 20:09 | comment | added | Kevin Buzzard | ...(because Galois acts trivially) and is not a map to any $GL_n$ in general if $D$ is not split. If however the OP really is talking about the naive Galois action on the torus then (a) I retract my answer and (b) I do not understand the question and am not even sure it makes sense (because I don't yet see why the roots will be preserved). I hope this clarifies precisely what I'm asserting. | |
| Nov 20, 2010 at 20:07 | comment | added | Kevin Buzzard | . 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... | |
| Nov 20, 2010 at 20:03 | comment | added | Kevin Buzzard | Hi Brian. Sorry for the delay. Ok so the OP says "everything has an action of Galois" but doesn't say what this action is. For the character group of the torus I think we have two natural choices. The first is the naive action, which depends on the arithmetic of the torus. I have no reason to believe that this will even preserve the roots in general! The second is the action called $\mu_G$ in Corvallis. This does preserve the roots. Now what follows is a guess because I'm at home: my guess is that the action the OP is talking about is $\mu_G$... | |
| Nov 20, 2010 at 19:39 | comment | added | BCnrd | Dear Kevin (& Lucio): I suppose you guys are using the *-action of Tits. But I thought that's just something which acts on the root datum, not nicely related to Galois descent for the representation space (unless I am missing something), so why is this the "right" action to be using? I also still don't understand why Kevin's example satisfies the desired starting hypotheses (and the canonicity of it remains puzzling to me). | |
| Nov 20, 2010 at 16:04 | comment | added | Lucio Guerberoff | I cannot make a comment, so I answer Brian here: the Galois action, as Kevin says, is different from the natural action on the Torus. I think it is described in Corvallis articles, and also in Borel-Tits. For this action, if G is a form of H, then it is an inner form if and only if the Galois actions on the root datum are the same. | |
| Nov 20, 2010 at 12:30 | comment | added | Kevin Buzzard | 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. | |
| Nov 20, 2010 at 4:02 | comment | added | BCnrd | 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. | |
| Nov 20, 2010 at 3:25 | comment | added | BCnrd | 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.) | |
| Nov 19, 2010 at 21:43 | history | edited | Kevin Buzzard | CC BY-SA 2.5 |
clarification
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| Nov 19, 2010 at 19:20 | history | answered | Kevin Buzzard | CC BY-SA 2.5 |