# Connectedness of centralizers of semisimple Lie-algebra elements under the action of a semisimple algebraic group

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 ). 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).

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.

• Concerning your attempted argument, it's not correct to say that the differential of the covering map is an isomorphism of Lie algebras in characteristic $p>0$. This fails in many cases (for arbitrarily large $p$), as seen in the thesis work of G.M.D. Hogeweij summarized here mathscinet.ams.org/mathscinet-getitem?mr=683531 (or in 0.13 of my 1995 monograph on conjugacy classes). Aug 5, 2018 at 16:01
• @JimHumphreys, does "arbitrarily large $p$" mean "for arbitrarily large $p$, there is a power $q$ of $p$ and an $\mathbb F_q$-group $\mathbf G$ such that …" or "for some root datum with $\mathbb Z$ action, there are arbitrarily large $p$ such that, for some power $q$ of $p$, the $\mathbb F_q$-group $\mathbf G$ with that root datum and on which Frobenius acts via the 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. Aug 6, 2018 at 1:01
• @LSpice: I don't mean anything sophisticated here, just a comment on why the proposed "proof" doesn't work. For example, even for $\mathrm{SL}_n$ when $p$ divides $n$ there is an obvious problem about getting an isomorphism of Lie algebras. (I'm just using the older language of the Springer-Steinberg reference given here by kneidell.) Aug 6, 2018 at 12:27
• @JimHumphreys, right, although such conditions may, I think, always be avoided by imposing a lower bound on $p$ depending only on the absolute root datum of $\mathbf G$, which seems to be in the spirit of @‍kneidell's question. (That is, that the proof might still work if we replace the "good prime" condition by a still stronger one; say, also that the algebraic fundamental group has no $p$-torsion might be enough.) Aug 6, 2018 at 14:22

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 (MSN). In fact, as you probably know, it's always good to look in Steinberg (often in the torsion paper or the endomorphisms 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.)
A prime is good for a semisimple group $G$ if it is not torsion for the dual root system of $G$ (Springer–Steinberg, 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.
(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, 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, §§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:
For reductive groups, the definition is more complicated (Steinberg - Torsion …, 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 $G$).
• Great, thank you! I'm interested mainly in things happening for almost all primes'', so the precise bound is not incredibly important. I wasn't aware of the Torsion paper, thank you for the reference; I'll be sure to check it out. Aug 3, 2018 at 13:33