# Centraliser of regular semisimple element in $G^F$, for a connected reductive algebraic group $G$

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.

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

## 1 Answer

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 $$C_{G^{\prime F}}(s')\supsetneqq T^{\prime F}.$$

• 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! – Riju Jun 1 '19 at 9:30
• @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 $L$). – Mikhail Borovoi Jun 3 '19 at 0:34
• 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 $C_W(s)$. – Mikhail Borovoi Jun 3 '19 at 0:51
• 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, $n_0\in C_{G^F}(s)$. – Mikhail Borovoi Jun 3 '19 at 0:56