3
$\begingroup$

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.

$\endgroup$
9
  • 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

1
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.