Let $G$ be an connected reductive algebraic group over $k=\bar{\mathbb{F}_p}$. Suppose $G$ is defined over $\mathbb{F}_q$. Let $G^{F}$ be the corresponding finite group associated to $G$. Suppose $s\in G^{F}$ is a regular semisimple element. Now, $s$ is contained in a unique maximal Torus $T$, and $T$ is necessarily $F$-stable. Let $T^F$ denote the set of $F$-rational points of $T$.

It is clear that $T^F\subseteq C_{G^F}(s)$. My question is whether $T^F=C_{G^F}(s)$? In the case that this isn’t true is there a simple description of the quantity $C_{G^F}(s)$?

Thank you.

  • 2
    $\begingroup$ 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 $T^F$. $\endgroup$ May 30, 2019 at 19:53
  • $\begingroup$ 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 $\endgroup$ May 30, 2019 at 23:42
  • $\begingroup$ @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 $x$. $\endgroup$
    – Riju
    May 31, 2019 at 16:59
  • $\begingroup$ 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. $\endgroup$
    – Riju
    May 31, 2019 at 17:14
  • $\begingroup$ My question now is that what happen if $[G,G]$ is not simply connected. Is the claim of the question still holds true? $\endgroup$
    – Riju
    May 31, 2019 at 17:14

1 Answer 1


It is a good question! The answer is NO, see the counter-example below.

Take $p=3$; then $\mathbb F_3=\{0,1,-1\}$. Write $L=\mathbb F_3(i)$, where $i^2=-1$; then $L\simeq \mathbb F_9$.

Take $$G={\rm GL}_{2,L}\,,\quad G'=G/\{\pm 1\}.$$ Let $T\subset G$ denote the subgroup of diagonal matrices. Take $$ s={\rm diag}(i,-i)\in T(L)\subset G(L).$$ Then the centralizer of $s$ in $G$ is $T$, hence $s$ is a regular semisimple element of $G(L)$.

Write $$ n=\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}\in G(L). $$ Then $$ n s n^{-1} ={\rm diag}(-i,i)= -s.$$ This means that if we denote by $s'$ and $n'$ the images in $G'(L)$ of $s$ and $n$, respectively, then $$ n' s' (n')^{-1} = s'.$$ Thus $$ n'\in C_{G'}(s')(L),$$ but $n'\notin T'(L)$, where $T'$ denotes the image of $T$ in $G'$. We see that $$C_{G'(L)}(s')\supsetneqq T'(L).$$ In the notation of the question, we obtain that $$C_{G^{\prime F}}(s')\supsetneqq T^{\prime F}.$$

  • $\begingroup$ 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! $\endgroup$
    – Riju
    Jun 1, 2019 at 9:30
  • 1
    $\begingroup$ @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 $L$). $\endgroup$ Jun 3, 2019 at 0:34
  • $\begingroup$ 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 $C_W(s)$. $\endgroup$ Jun 3, 2019 at 0:51
  • $\begingroup$ 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, $n_0\in C_{G^F}(s)$. $\endgroup$ Jun 3, 2019 at 0:56

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