Let $\newcommand{\GG}{\mathbf{G}}\newcommand{\g}{\mathfrak{g}}\GG$ be a connected semisimple algebraic group over the algebraically closed field $k=\overline{\mathbb{F}\_q}$, and let $\g$ be its Lie-algebra. Let $x\in\g$ be a semisimple element. My question is the following-



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> Assuming that the characteristic of $k$ is good for $\GG$, is it true that the centralizer $\newcommand{\CC}{\mathbf{C}}\CC\_\GG(x)$ for the adjoint action of $\GG$ on $\g$ is connected? 
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Of course, the analogous statement for the case where $x$ is a semisimple element of $\GG$ is false. For example, if $\GG=\mathrm{PGL}\_2(k)$ and $x=\left(\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\right)$ (or rather, its class mod $k^\times$) then it is easy to verify that $\CC\_{\GG}(x)$ has two connected components. However, for $\mathrm{PGL}\_2(k)$ acting on its Lie algebra, which consists of traceless $2\times 2$ matrices, the centralizer of the element represented by the same matrix is simply the diagonal torus in $\GG$.


The statement does hold in the case where $\GG$ is simply-connected, essentially, by the same argument as in Steinberg's Connectedness Theorem (see II, 3.19 in [1]). In the case where $\GG$ is not simply-connected I was wondering if maybe the following argument shows the connectedness of the centralizer:


*Proof(?)*. Let $\GG$ be as above, not necessarily simply connected, and let $\renewcommand{\sc}{\mathrm{sc}}\pi:\GG\_{\sc}\to \GG$ be the simply-connected cover of $\GG$. The differential map $d\pi:\newcommand{\Lie}{\mathrm{Lie}}\Lie(\GG\_\sc)\to\g$ is an isomorphism of Lie-algebras which intertwines the actions of both groups on their Lie-algebras. In particular, the centralizer of $x$ in $\GG\_{\sc}$ is mapped by $\pi$ in to its centralizer in $\GG$. On the other hand, the adjoint action of $\GG\_\sc$ on its Lie-algebra factors through the adjoint quotient, and hence the kernel of $\pi$ acts trivially on $\g$. It follows that for any $g\in \GG\_{\sc}$, the operators $\newcommand{\Ad}{\mathrm{Ad}}\Ad(g)$ and $\Ad(\pi(g))$ coincide, and hence $g\in \CC\_{\GG\_\sc}(x)$ if and only if $g\in \CC\_{\GG}(x)$. 


Thus $\CC\_\GG(x)=\pi(\CC\_{\GG\_\sc}(x))$ is the image of a connected group and hence connected. 




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I don't see where the argument above fails, in fact it seems surprisingly simple. However, it could very well be that I'm missing out on something crucial here. Also, I haven't seen this statement written out explicitly in any text, which could be because it is false, or because it is trivial.


In any case- I've been looking at this argument for a while, and I can't seem to be confident that I'm not missing any details, so I would be very grateful to anyone who can point out any mistakes I might have made, or otherwise, corroborate the statement (preferably, but not necessarily, with a reference).




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[1]*Springer, T. A.; Steinberg, R.*, Conjugacy classes, Sem. algebr. Groups related finite Groups Princeton 1968/69, Lect. Notes Math. 131, E1-E100 (1970). [ZBL0249.20024](https://zbmath.org/?q=an:0249.20024).



Connectedness of centralizers of semisimple Lie-algebra elements under the action of a semisimple algebraic group
This is true for $p$ not a torsion prime of $G$; it is exactly Theorem 3.14 (p. 88) of [Steinberg - Torsion in reductive groups](http://www.sciencedirect.com/science/article/pii/0001870875901255) ([MSN](http://mathscinet.ams.org/mathscinet-getitem?mr=354892)). In fact, as you probably know, it's always good to look in Steinberg (often in the torsion paper or the [endomorphisms](http://www.ams.org/books/memo/0080) paper) for questions of this type. (If you'll permit another philosophical statement, it's not a bad heuristic that "whatever is true in characteristic $0$ is true in large characteristic" (in the $p$-adic setting, "… in large residual characteristic"), where the meaning of "large" depends on the group $G$ in question, but only through its root datum—my colleagues and I often start our discussions with "let $p$ be huge", and worry about sorting out "how huge is huge" only later; but maybe you were interested in the precise bound on $p$. In some sense, the concepts around motivic integration make precise this heuristic for statements in a certain logical language. See, for the latest on this, [Cluckers, Gordon, and Halupczok - Uniform analysis on local fields and applications to orbital integrals](https://arxiv.org/abs/1703.03381).)


I originally said the following wrong thing:



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> A prime is good for a semisimple group $G$ if it is not torsion for the dual root system of $G$ ([Springer–Steinberg](https://link.springer.com/chapter/10.1007/BFb0081546), Definition 4.1) …. I guess you can cook up an example of a non-self-dual semisimple group and a good prime $p$ for which some centraliser is disconnected; but I haven't tried.
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(see below for the text elided with the ellipsis). The first sentence is wrong, as I would have found if I had read just one more sentence ([Springer–Steinberg](https://link.springer.com/chapter/10.1007/BFb0081546), Caution 4.2). However, the good primes for $G$ are *among* the non-torsion primes for $G$, at least for $G$ semisimple (see [Springer–Steinberg](https://link.springer.com/chapter/10.1007/BFb0081546), §§4.3–4.4), so that a result for non-torsion primes is stronger than one for good primes. The ellipsis indicates the following text, which is correct:



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> For reductive groups, the definition is more complicated ([Steinberg - Torsion …](http://www.sciencedirect.com/science/article/pii/0001870875901255), Definition 2.1). The results in Steinberg are quite sharp (in fact the connectedness condition is proven to be *equivalent* to $p$ not being a torsion prime for $G$).
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