Centralizers of regular elements are abelian

Question

Let $G$ be a complex semisimple Lie group with Lie algebra $\mathfrak{g}$. In a particular paper, the following statement is made:

If $X\in\mathfrak{g}$ is regular (i.e. has centralizer of minimal dimension) then $$ Z_G(X) := \{g\in G:\mathrm{Ad}_gX=X\}, $$ is abelian.

However, no justification is given, and I was wondering if anybody knows how to prove it or can point a reference discussing this.

The Lie algebra of $Z_G(X)$ is the centralizer $Z_{\mathfrak{g}}(X)$ of $X$ in $\mathfrak{g}$, and since $X$ is regular, $Z_{\mathfrak{g}}(X)$ is a Cartan subalgebra of $\mathfrak{g}$. In particular, $Z_{\mathfrak{g}}(X)$ is abelian and hence the identity component of $Z_G(X)$ is abelian. But the problem is that $Z_G(X)$ might not be connected.

For example, if $G=\mathrm{SL}(2,\mathbb{C})$, then $X=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ is regular and $$Z_G(X)=\left\{\begin{pmatrix}1&z\\0&1\end{pmatrix}:z\in\mathbb{C}\right\}\cup\left\{\begin{pmatrix}-1&z\\0&-1\end{pmatrix}:z\in\mathbb{C}\right\}$$ is abelian, but not connected.

Answer 1

Here is an attempted proof. We have that $Z_G(X)$ is the full preimage of $Z_H(X)$ under the covering central extension $\operatorname{Ad}$: $$ 1\longrightarrow C\longrightarrow G\overset{\operatorname{Ad}}\longrightarrow H\longrightarrow 1 \tag{*} $$ where $C$ is the (discrete) center and $H=\operatorname{Ad}(G)$ the adjoint group of $G$. Now:

Theorem (Kostant 1963, Prop. 14). If $X$ is regular, then $Z_H(X)$ is connected abelian.

To deduce that $Z_G(X)$ is also abelian, pick $g_0$ and $g$ there, projecting to $h_0$ and $h_1$. Since $Z_H(X)$ is connected, it contains a continuous path $h_t$ from $h_0$ to $h_1$. Lift that to a continuous path $g_t$ in $Z_G(X)$ starting at $g_0$. Since $Z_H(X)$ is abelian, the path $c_t=g_tg_0g_t^{-1}g_0^{-1}$ satisfies $$ \operatorname{Ad}(c_t)=h_th_0h_t^{-1}h_0^{-1}=1. $$ So $c_t$ lies in the (discrete) center $C$, hence constantly equals $c_0=1$. So $g_0$ commutes with $g_1$. But $\operatorname{Ad}(g)=h_1=\operatorname{Ad}(g_1)$ shows that $g=g_1z$ for some $z\in C$. So $g_0$ commutes with $g$, too.

Answer 2

EDIT: The question and some of the responses are out of focus, so it's worth clarifying a few of the issues here. First, the intended notion of "regular element" is unclear. In Kostant's early papers, for example here, only regular semisimple elements in semisimple Lie groups are directly considered. But later he broadened the definition. In the general Lie group framework, there is no notion of "Jordan-Chevalley decomposition", so a broader definition of regularity was given by Steinberg for a connected semisimple algebraic group $G$ here. An element $x \in G$ is called regular if its centralizer has dimension equal to the rank $\ell$ of $G$ (which is the least possible).

At the 1966 ICM in Moscow, Steinberg surveyed questions about conjugacy classes and regular elements here. Among the problems he raised were (14) and (15), asking whether regular $x \in G$ is characterized by having a commutative centralizer. This and more was later confirmed by two of his students B. Lou and S.V. Keny, even for fields of "bad" characteristic.

For $X \in \mathfrak{g}$, the Lie algebra of $G$, there are actually two relevant centralizers: the subgroup $G_X$ of $G$ fixing $X$ under the adjoint representation Ad, and the subalgebra $\mathfrak{g}_X$ of $\mathfrak{g}$ fixing $X$ under ad. The latter always contains the Lie algebra of $G_X$ and equals it in characteristic 0. Springer refined and extended Steinberg's ideas in a 1966 paper here: see $\S4$ and $\S5$. He further showed in Proposition 2 here that when $x \in G$ is regular, the identity component of $G_x$ is commutative. The work of Lou and Keny showed in fact that $G_x$ itself is commutative, while $G_X$ is commutative for a regular element $X\in \mathfrak{g}$ provided $G$ is of adjoint type.

Thanks to the fact that the centralizer of a semisimple element is reductive (the almost-direct product of a central torus and a semisimple group), the main problems are reduced to $x \in G$ unipotent or $X \in \mathfrak{g}$ nilpotent. But $G_x$ is not necessarily connected for $x$ semisimple unless $G$ is simply connected. (An algebraic proof due to Digne-Michel is discussed in Chapter 2 of my 1995 book on conjugacy classes; the first proof was by Springer and Steinberg.) A similar reduction to the regular nilpotent case works for $X \in \mathfrak{g}$ and $G_X$ when $G$ is simply connected.

In the case of $G_X$ for $X$ regular nilpotent, the outcome in characteristic 0 is fairly straightforward: here the centralizer is the product of the center of $G$ and a (connected!) commutative unipotent group. But for bad prime characteristics, it took some case-by-case work to describe $G_X$.

Answer 3

In $PGL_2$, the centralizer of the diagonal element $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ consists of all elements of the form $\begin{pmatrix}* & 0 \\ 0 & * \end{pmatrix}$ or $\begin{pmatrix} 0 & * \\ * & 0 \end{pmatrix}$, and so has the minimal dimension (one, after modding out by scalars) but is not connected.

So I believe this is false without additional assumptions.