Timeline for Lifting $\mathfrak{sl}_2$-triples
Current License: CC BY-SA 4.0
14 events
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| Jun 7, 2021 at 14:41 | history | edited | David Stewart | CC BY-SA 4.0 |
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| Jun 7, 2021 at 11:22 | comment | added | David Stewart | @LSpice. Agreed, that's reasonable. The B-T thm says that any closed subgroup $H$ of reductive $G$ lives in a parabolic $P$ of $G$ such that $R_u(H)\subseteq R_u(P)$. | |
| Jun 7, 2021 at 0:02 | comment | added | LSpice | I'm not sure what the Borel–Tits theorem is (maybe just the statement that any connected unipotent subgroup can be so conjugated?), but conjugacy of $U$ into the unipotent radical of a Borel is elementary: the chosen Borel of $H$ is a connected soluble subgroup of $H$, hence of $G$, hence is contained in a maximal connected soluble subgroup $B$ of $G$, which is a Borel; and, since it is unipotent, it has trivial projection on the torus $B/\operatorname{Rad}_u(B)$, so is contained in $\operatorname{Rad}_u(B)$. | |
| Jun 7, 2021 at 0:01 | comment | added | LSpice | An argument that avoids the Springer isomorphism and associated cocharacters is certainly good, since those are both comparatively large-$p$ phenomena, and here we're interested in the small-$p$ case. (I was suspicious of "the same must be true", but it's just saying that, since the derivative at the identity of $U \to U_\alpha$ is non-trivial, so is $U \to U_\alpha$ itself, and that's perfectly reasonable. ($U \to U_\alpha$ comes from $U \subseteq R := \operatorname{Rad}_u(B) \to R/[R, R] \cong \prod U_\alpha$, product possibly including some non-simple roots for small $p$.)) | |
| Jun 6, 2021 at 18:35 | comment | added | David Stewart | @spin. There must be a better way to do this, maybe using a Springer isomorphism and/or associated cocharacters, but here is an answer that I think works. Let $U$ be the unipotent radical of a Borel of $H$. By the Borel–Tits theorem, we can conjugate $U$ into the unipotent radical of a Borel of $G$. Since $e$ is regular, it must be supported on all the simple roots. Since $Lie(U)=\langle e\rangle$, the same must be true of $1\neq u\in U$. Hence $u$ is a regular unipotent of $U$. | |
| Jun 6, 2021 at 18:00 | history | edited | David Stewart | CC BY-SA 4.0 |
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| Jun 6, 2021 at 10:43 | history | edited | LSpice | CC BY-SA 4.0 |
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| Jun 6, 2021 at 4:11 | comment | added | spin | If $e$ is a regular nilpotent element, why does it follow that unipotent elements of $H$ are regular unipotent? | |
| Jun 6, 2021 at 0:12 | comment | added | LSpice | Thank you for this very interesting answer, and the pointer to your paper! | |
| Jun 5, 2021 at 20:57 | vote | accept | LSpice | ||
| Jun 5, 2021 at 20:36 | history | edited | David Stewart | CC BY-SA 4.0 |
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| Jun 5, 2021 at 20:28 | comment | added | David Stewart | One reference is [Liebeck-Seitz, Unipotent and Nilpotent Classes in Simple Algebraic Groups and Lie Algebras] and look in the index for 'Orders …'. This probably goes back to the Order Formula 0.4 in [Testerman, A_1-type overgroups of elements of order p]. (Will clarify where I base-change to char p.) | |
| Jun 5, 2021 at 19:58 | comment | added | LSpice | It seems that at some point you switch from characteristic $0$ to characteristic $p$, but I guess characteristic $0$ was just motivation. Do you have a reference for the fact that about regular unipotent elements when $p$ is less than the Coxeter number? | |
| Jun 5, 2021 at 19:41 | history | answered | David Stewart | CC BY-SA 4.0 |