Timeline for Follow-up to Steinberg's problem (12) in his 1966 ICM talk?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 7, 2020 at 3:42 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
link updated
|
| Feb 21, 2017 at 14:45 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
added 11 characters in body
|
| Feb 20, 2017 at 21:44 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
added 216 characters in body
|
| Feb 15, 2017 at 23:51 | history | edited | YCor | CC BY-SA 3.0 |
added "below" to help the reader
|
| Feb 15, 2017 at 19:24 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
deleted 269 characters in body
|
| Jul 20, 2016 at 13:31 | comment | added | Jim Humphreys | @David: Yes, that's apparently the status at present. However, Lusztig's less direct proof of finiteness for the number of unipotent classes makes it tempting to look for an analogous proof. Of course, the "number" of classes or orbits can vary for bad $p$, which complicates matters. | |
| Jul 20, 2016 at 11:09 | comment | added | David Stewart | I don't think there's any proof of the finiteness of the list of nilpotent orbits in bad characteristic which doesn't involve (i) working out conjugacy class dimensions for a putative list, (ii) using those to write down the number of elements in each orbit in $\mathfrak{g}(\mathbb F_q)$ in terms of a polynomial in $q=p^r$, (iii) then checking you have all of them by adding it up and praying that you get $q^{|\Phi|}$ elements. | |
| Jul 19, 2016 at 15:52 | comment | added | Jim Humphreys | Yes, these later theoretical refinements are more precise. But Steinberg himself worked out many details for good prime characteristics, leaving some analogues for bad $p$ as problems. (Recent computations have filled in a lot more of the specifics for bad primes. But for example it's hard to find a nice proof that there are only finitely many nilpotent orbits without just pointing to case-by-case work.) | |
| Jul 18, 2016 at 21:33 | comment | added | David Stewart | I don't know if it's subsequently referred to again, but it's probably worth a comment that at least in the exceptional types, Liebeck and Seitz's book gives a list of the centraliser dimensions dim $G_x$ for x unipotent. Together with Lawther you can presumably get a list of all conjugacy classes where the answer is no. I guess there are a few. On the other hand work of Bate, Martin, Roehrle tell you that all subgroups are separable in very good characteristic---uses an argument of Richardson. So the answer will be yes there (well, also there's no centre $Z(L)$ to worry about!). | |
| Jul 18, 2016 at 20:24 | history | asked | Jim Humphreys | CC BY-SA 3.0 |