Timeline for Centraliser of regular semisimple element in $G^F$, for a connected reductive algebraic group $G$
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 7, 2023 at 23:03 | comment | added | LSpice | @MikhailBorovoi, re, isn't that usually called strongly regular semisimple (whereas just rss, as @Riju says and your answer does, usually drops the connectedness hypothesis)? | |
| Jun 3, 2019 at 9:41 | comment | added | Riju | @JimHumphreys Probably the fact that if $s$ is regular, semisimple then $C_{G}(s)^{\circ}=T$, made me believe that the conclusion in the question has more possibility to hold than just a semisimple element. | |
| Jun 3, 2019 at 1:00 | comment | added | Mikhail Borovoi | @JimHumphreys: Simply connectedness does make a difference, see my answer. | |
| Jun 2, 2019 at 9:18 | vote | accept | Riju | ||
| Jun 1, 2019 at 21:50 | comment | added | Jim Humphreys | @Riju: Your two questions are related but not the same. I'm sorry if I implied the reverse. I'm also sorry for oversimpllifying the questions too much. (By trhe way, simply connected makes little difference to the question.) Is there a reason to concentrate just on regular semisimple elements? | |
| Jun 1, 2019 at 2:01 | answer | added | Mikhail Borovoi | timeline score: 2 | |
| May 31, 2019 at 17:20 | comment | added | Riju | @JimHumphreys, I have been following the references that you have given, but I couldn’t find a exact answer to my question there. Maybe I have missed something. As far as the reference to the math overflow question, that question was also asked by me, but I don’t seem to find much of resemblence, of this question to that! Again may be, I am missing something. | |
| May 31, 2019 at 17:14 | comment | added | Riju | My question now is that what happen if $[G,G]$ is not simply connected. Is the claim of the question still holds true? | |
| May 31, 2019 at 17:14 | comment | added | Riju | Moreover, it is known that $[G,G]$ is simply connected then the centraliser of a semisimple element is connected, in which case your claim that $C_{G}(x)=T$, holds, and my claim holds. | |
| May 31, 2019 at 16:59 | comment | added | Riju | @MikhailBorovoi the definition of regular element is that $x$ will be called regular if dim($C_{G}(x))$ is minimal. Since, it is known that dim($C_{G}(x)) \geq rank(G)$, it turns out that $x$ is regular if dim($C_{G}(x))$ is equal to $rank(G)$. Now, since my consideration is $x$ is regular semisimple element, it is clear that $C_{G}(x)^{\circ}=T$, where $T$ is the unique maximal torus containing $x$. | |
| May 30, 2019 at 23:42 | comment | added | Jim Humphreys | Two comments: 1) See Chapter 3 in the 1985 book of R.W. Carter.or the long article by Springer-Steinberg in Lecture Notes in Mathematis 131 (1970). 2) This question has considerable overlap with a recent question mathoverflow.net/questions/332689 | |
| May 30, 2019 at 19:53 | comment | added | Mikhail Borovoi | You need a definition of a regular semisimple element. I think that a semisimple element $s$ of $G$ is called regular if its centralizer in $G$ is a (maximal) torus. Then in your case the centralizer of $s$ in $G$ is $T$, and hence, the centralizer of $s$ in $G^F$ is $T^F$. | |
| May 30, 2019 at 16:20 | history | asked | Riju | CC BY-SA 4.0 |