Timeline for homomorphism into reductive groups
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Aug 9, 2012 at 18:36 | comment | added | user22479 | @rvarma: Since the answer addresses perfect fields, you might wonder about imperfect fields. A dichotomy arises: should we only try to embed in connected unipotent groups that are split over the ground field? (Over perfect fields the "split" condition is automatic.) Tits focused on the split ones, calling unipotent $u \in G(k)$ "good" when it lies in one and "bad" otherwise (with connected reductive $G$). Over local function fields bad $u$ abound if $G$ isn't simply connected. Check out Gille's article "Unipotent subgroups of reductive groups in characteristic $p > 0$". | |
| Aug 8, 2012 at 14:53 | vote | accept | rvarma | ||
| Aug 8, 2012 at 14:53 | comment | added | rvarma | @quasicoherent : yes I see that.. thanks a lot! | |
| Aug 8, 2012 at 8:49 | comment | added | user22479 | @rvarma: Focusing on finite $p$-groups does not seem to make the problem any simpler, in the sense that the key construction in the Borel-Tits proof (see $U \mapsto L(U)$ in my answer below) applied to a finite unipotent group will typically immediately leave the framework of finite unipotent groups when it is iterated. So it seems likely that a simpler proof in the finite case would require an entirely different idea. | |
| Aug 8, 2012 at 8:40 | answer | added | user22479 | timeline score: 4 | |
| Aug 8, 2012 at 4:34 | answer | added | YCor | timeline score: 5 | |
| Aug 8, 2012 at 3:04 | comment | added | Will Sawin | @rvama: Do you mean minimal parabolic / Borel subgroup? | |
| Aug 7, 2012 at 23:08 | answer | added | Will Sawin | timeline score: 0 | |
| Aug 7, 2012 at 16:52 | comment | added | user22479 | (In the preceding, "no nontrivial $p$-power torsion" meant of course in the sense of geometric points, not closed subgroup schemes such as $\mu_p$ which abound inside nontrivial tori.) | |
| Aug 7, 2012 at 16:50 | comment | added | user22479 | There is no difference between the (connected) reductive and semisimple cases because (letting $G'$ be the derived group of a connected reductive $G$) (i) the quotient $G/G'$ is a torus in characteristic $p > 0$ and hence contains no nontrivial $p$-power torsion (so your $P$ automatically lands inside $G'$), (ii) the Borel subgroups of $G$ and $G'$ are in natural bijection correspondence via $B \mapsto B \cap G'$ and $B' \mapsto B' \cdot Z$ for the maximal central torus $Z$ in $G$, and likewise for their unipotent radicals (which are pairwise conjugate in any connected reductive $k$-group). | |
| Aug 7, 2012 at 15:40 | comment | added | rvarma | Someone pointed out to me the paper by Borel,Tits in inventiones, where they prove that a subgroup $H \subset G $ consisting of only unipotent elements with $G$ reductive can be realised inside the unipotent radical of a parabolic subgroup $P$ of $G$. I haven't read the paper. Hence I don't really know about the issues involved. But Does assuming the group $H$ being finite make the problem any easier? | |
| Aug 7, 2012 at 15:04 | comment | added | Jim Humphreys | Probably not much can be said in this generality. Keep in mind that general linear groups often embed in much larger reductive groups as well. So the possibilities are open-ended. | |
| Aug 7, 2012 at 14:12 | history | asked | rvarma | CC BY-SA 3.0 |