Timeline for homomorphism into reductive groups
Current License: CC BY-SA 3.0
14 events
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| Aug 8, 2012 at 4:30 | comment | added | Will Sawin | What sort of insight does the OP want? The only thing we know for sure is he wants to know if it is easier for finite subgroups. You stated in your second comment that it is not much easier even in the case of just a cyclic subgroup. Perhaps you could explain why this is so in an answer? | |
| Aug 8, 2012 at 3:17 | comment | added | user22479 | @Will: The point of my 2nd comment is twofold: (i) the notion of "maximal solvable subgroup" (sans connectedness) is a poor one, so it is not clear why "maximal unipotent subgroup" is a useful notion to introduce by itself (and as you know, it is rendered moot by the Borel-Tits result from Inventiones that the OP has noted), (ii) it suggests a suitable viewpoint if one might want to consider the question over other fields (e.g., finite fields). Anyway, the real content is the Borel-Tits Inventiones result, and the OP is seeking insight into that fact (applied to finite subgroups). | |
| Aug 8, 2012 at 2:32 | comment | added | Will Sawin | While the second comment is interesting, I don't see how it adds anything, for the purposes of answering this question or correcting mistakes, to rvama's comment. | |
| Aug 8, 2012 at 2:31 | comment | added | Will Sawin | Then I do not understand what you are saying. For the purposes of this question, the indubitably correct analogue is a maximal $p$-subgroup of $G$, thus a maximal unipotent subgroup. The correct analogue is not the unipotent radical of a Borel subgroup unless the maximul unipotent subgroup is connected, in which case the distinction is irrelevant. The minimal parabolic subgroup construction is irrelevant to the question, and is something I already know. | |
| Aug 8, 2012 at 2:21 | comment | added | user22479 | @Will: Every single phenomenon which I am raising as an objection to your answer occurs over algebraically closed fields (and I only included general fields in my comments to provide you with some additional perspective on the situation). If you delete your preceding comment (which is entirely wrong) then I will delete this one. | |
| Aug 8, 2012 at 2:12 | comment | added | Will Sawin | In the question the field is assumed to be algebraically closed, so that's not a problem. | |
| Aug 8, 2012 at 2:12 | comment | added | user22479 | @Will: To give perspective, let's see that "maximal solvable subgroup" is a poor notion. Away from characteristic 2, special orthogonal groups (for split quadratic spaces of dimension at least 3) contain finite abelian 2-torsion subgroups that are too big to lie in a torus and consequently do not lie in any Borel subgroup (see the end of 11.17 in Borel's textbook). In characteristic $> 0$, the fact that a single unipotent element in a connected linear algebraic group lies in a Borel (equivalently, connected unipotent) subgroup is a serious theorem -- try to prove it using whatever you know. | |
| Aug 8, 2012 at 2:01 | comment | added | user22479 | @Will: Connectedness aspects of linear algebraic groups deserve more respect. Your "answer" seems not to be informed by enough experience, and your "correct" analogue of the upper-triangular group in ${\rm{GL}}_n$ for a general (connected) reductive $G$ over a field $k$ is incorrect. For $k$-split $G$ the correct analogue is the unipotent radical of a Borel $k$-subgroup, and in general it is the unipotent radical of a minimal parabolic $k$-subgroup (all of which are $G(k)$-conjugate). As you gain experience, you will learn to appreciate this very important conjugacy theorem of Borel-Tits. | |
| Aug 8, 2012 at 1:11 | comment | added | Will Sawin | It's clear that $U$ is not connected in general. Just take $G$ a finite group. On the other hand if $G$ is any nice example the larger statement is true, or if $U$ is defined over a finite field. I think that perhaps someone who knows about reductive groups should figure out the last part. | |
| Aug 8, 2012 at 0:27 | comment | added | rvarma | @ Will Sawin , I think there is a problem with you argument. Let us say we embedd $f : G \rightarrow GL_n$. Now we know the image of the composite map $f\circ \rho$ lies in $B_u$ (The unipotent radical of a Borel). Set $U$ to be $B_u \cap G$. Now why should $U$ be connected? If it is connected then your argument works. else we will have to work with the connected component of $U$, but then why should it contain $im(\rho)$? | |
| Aug 8, 2012 at 0:14 | comment | added | Will Sawin | That is correct. That was lazy thinking on my part. I replaced it with a different argument. | |
| Aug 8, 2012 at 0:14 | history | edited | Will Sawin | CC BY-SA 3.0 |
added 277 characters in body
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| Aug 7, 2012 at 23:29 | comment | added | Olivier Benoist | I think your claim is not true : there may be elements of order $p$ not defined over a finite field. Consider for example the $p$-subgroup of $SL_2$ generated by an upper triangular matrix with $1$ as diagonal coefficients and a transcendental element as upper right coefficient. | |
| Aug 7, 2012 at 23:08 | history | answered | Will Sawin | CC BY-SA 3.0 |