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-

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?

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.

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).

[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.

viathe given $\mathbb Z$-action satisfies …"? Certainly the former is true, but I thought that the latter (which seems to be the kind of statement in which @kneidell is interested) wasn't. $\endgroup$ – LSpice Aug 6 '18 at 1:01