Timeline for Normalizer of a split torus
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 2, 2018 at 21:20 | comment | added | LSpice | (Also "non-$0$" should be "non-central".) | |
| Oct 2, 2018 at 11:47 | comment | added | LSpice | (This assumes that centralisers of Lie-algebra elements are Levis, which is a large- or zero-characteristic thing.) | |
| Oct 2, 2018 at 11:47 | comment | added | LSpice | If $M$ is a Levi subgroup of $G$, then (by looking at the action of simple roots outside $M$ on elements of $Z(M)^\circ$), for every $M$-regular semisimple element $t \in M$, there is a $G$-regular semisimple element of $t Z(M)^\circ$. If $w$ fixes a non-$0$ element of $\operatorname{Lie}(T)$, then we can take $M = \operatorname C_G(\operatorname{Lie}(T)^w)$ and work inductively. | |
| Oct 2, 2018 at 2:56 | comment | added | Shawn | @LSpice: I don't see how this induction process can be carried out in the first case, could you elaborate a little bit? The second case is true for GL_n and indeed all lifts work, by an elementary matrix calculation. | |
| Oct 2, 2018 at 0:25 | comment | added | LSpice | If $w$ fixes a non-$0$ element of $\operatorname{Lie}(T)$, then we should be done by induction on the semisimple rank. Thus the difficult case is when $w$ fixes no such element, i.e., when it's elliptic. In this case it seems to me heuristically that either no lift should work or all should, but (a) I can't prove this and (b) I can't prove that it's not the case that no lift works. | |
| Oct 1, 2018 at 23:35 | comment | added | LSpice | Not if $k$ is finite and its characteristic divides the order of $w$. In that case the order of $n$ would be of finite order divisible by the characteristic, which is impossible for a semisimple element. | |
| Oct 1, 2018 at 22:25 | history | asked | Shawn | CC BY-SA 4.0 |