Timeline for $\mathbb{F}_q$-rational elements in unipotent classes of a finite group of Lie-type
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 11, 2019 at 13:20 | history | edited | YCor |
edited tags
|
|
| Sep 18, 2018 at 8:57 | comment | added | Mikko Korhonen | By the way, it is true that every unipotent class of split $G$ has a representative of the form $\prod_{\alpha \in \Phi^+} x_{\alpha}(c_{\alpha})$, where $c_{\alpha} \in \mathbb{F}_q$, and we can even arrange $c_{\alpha} \in \{1,-1\}$. Consequently $G$ has only finitely many conjugacy classes of unipotent elements, which is a non-trivial theorem due to Lusztig. To prove that such representatives exist I suppose you need to know the unipotent conjugacy classes and proceed case-by-case (at least I am not aware of any other proof). | |
| Sep 18, 2018 at 8:09 | comment | added | Mikko Korhonen | See Corollary 6, p. 6 in the book of Liebeck-Seitz mentioned by Jay. | |
| Sep 18, 2018 at 7:37 | history | edited | kneidell | CC BY-SA 4.0 |
added 74 characters in body
|
| Sep 17, 2018 at 13:02 | comment | added | kneidell | Excellent, thank you! I was unaware of this reference, I'll check it out. | |
| Sep 17, 2018 at 12:49 | comment | added | Jay Taylor | In any case, you should look at the book "Unipotent and nilpotent classes in simple algebraic groups and Lie algebras" by Liebeck and Seitz. It has all the information you will need. | |
| Sep 17, 2018 at 12:48 | comment | added | Jay Taylor | As you say, you just need to check whether a class is invariant under the automorphism induced by the Frobenius. In good characteristic it is enough to check that the weighted Dynkin diagram of the class is invariant under the automorphism. If $G$ is simple then this is true unless $G$ is type $\mathsf{D}_{2n}$. In this case there are a family of classes which are interchanged by the automorphism. | |
| Sep 17, 2018 at 12:32 | history | asked | kneidell | CC BY-SA 4.0 |