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Aug 7, 2023 at 23:10 comment added LSpice @Riju, re, $C_G(s)$ is not always generated by $T$ and certain root groups; those are both connected, so we can only access $C_G(s)^\circ$ that way. As @‍MikhailBorovoi's comments 1 2 indicate, $C_G(s)$ will generally fail to be connected (although it is so for $G$ simply connected!), and the failure is measured, algebraically (for any field) and rationally (for any finite field), by a quotient of $C_W(s)$.
Jun 3, 2019 at 0:56 comment added Mikhail Borovoi Now if $w=nT$ and $wsw^{-1}=s$, and moreover $w\in W^F$, then by Lang's theorem there exists $n_0\in N^F$ such that $w=n_0 T$. Clearly, $n_0\in C_{G^F}(s)$.
Jun 3, 2019 at 0:51 comment added Mikhail Borovoi Namely, write $N=N_G(T)$, $W=N/T$. Then $C_G(s)\subset N$. For any $w=nT\in W$ we can define $wsw^{-1}$, and for a given regular semisimple element $s$ you have to compute the centralizer $C_W(s)$.
Jun 3, 2019 at 0:34 comment added Mikhail Borovoi @Riju: A standard approach is first to compute $C_G(s)$, and after that to compute the set of $L$-points $C_{G^F}(s)$ using Galois cohomology (which is not difficult over a finite field $L$).
Jun 2, 2019 at 9:18 vote accept Riju
Jun 1, 2019 at 9:30 comment added Riju Fine example! Btw is there any way to find a description of this quantity $C_{G^F}(s)$, in terms of $T^F$ and something more, just like we have a description of $C_{G}(s)$ , which says that it is generated by $T$ and “certain” root subgroups!
Jun 1, 2019 at 2:18 history edited Mikhail Borovoi CC BY-SA 4.0
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Jun 1, 2019 at 2:01 history answered Mikhail Borovoi CC BY-SA 4.0