Timeline for Is the normalizer of a reductive subgroup reductive?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jun 10, 2016 at 0:24 | history | suggested | LSpice | CC BY-SA 3.0 |
Added link to Mostow paper
|
| Jun 9, 2016 at 23:43 | review | Suggested edits | |||
| S Jun 10, 2016 at 0:24 | |||||
| Nov 25, 2012 at 21:14 | comment | added | George McNinch | I guess I should add that $\dim T(p) = 2p$ so that $G = \operatorname{GL}_{2p}$ in the previous comment... | |
| Nov 25, 2012 at 21:12 | comment | added | George McNinch | Concerning the rank 1 situation: given (say) a $p$-nilpotent element $X$ of $\mathfrak{g}$, if the char is "very good" for $G$, there is indeed a rank 1 subgroup $H$ of $G$ with $X \in \operatorname{Lie}(H)$ for which $C_G(H)$ is reductive. But not every rank 1 subgroup has reductive centralizer, even when $p$ is "not too tiny". e.g. if $H = \operatorname{SL}_2$ and $V$ is the $H$-representation $V = T(p)$ where $T(p)$ is the tilting module of highest weight $p$ (say) the argument sketched in my answer shows that $C_G(H)$ is not reductive, where $G=\operatorname{GL}(V)$. | |
| Nov 25, 2012 at 14:29 | comment | added | Jim Humphreys | @Aakumadula: That seams reasonable, though I still wonder whether a more elementary proof in characteristic 0 (say in the equivalent Lie algebra setting) might exist. Mostow's theorem relies heavily on Lie group theory even though the desired result is purely algebraic. (Though Mostow is an old acquaintance of mine.) | |
| Nov 25, 2012 at 3:59 | comment | added | Venkataramana | @Jim: Since $G,H$ and the normaliser are all defined and split over a finitely generated field $k$ over the prime field, the question of "reductivity" does not change over extensions of the field $k$. We may thus assume that in characteristic zero, the field $k$ is embedded in ${\mathbb C}$ and then argue using Mostow's theorem. | |
| Nov 24, 2012 at 21:46 | comment | added | Jim Humphreys |
@Martin: The question is natural but doesn't seem to come up in the standard structure theory. Aside from motivation, it makes a big difference whether you are interested only in the complex field or want an arbitrary algebraically closed field of char 0: the methods may differ. I still hope that a fairly "elementary" algebraic proof (based on Bourbaki's Chapter 1) will be found here, by translating to the Lie algebra setting. In char $p>0$ there might be counterexamples for small primes, even though the rank 1 case I outlined is encouraging for $p\gg 0$.
|
|
| Nov 24, 2012 at 16:15 | comment | added | Martin Orr | I confess that the question had no great motivation at all. I was considering the conjugacy class of $H$ in $G$, and hence the normalizer comes up. I noticed that when I picked examples, the normalizer is always reductive, but I couldn't prove that this was true in general. | |
| Nov 23, 2012 at 22:41 | history | answered | Jim Humphreys | CC BY-SA 3.0 |