Let $G$ be a complex semisimple Lie group with Lie algebra $\mathfrak{g}$. In a particular [paper](https://arxiv.org/abs/math/0209053), 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.



Centralizers of regular elements are abelian
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](http://www.ams.org/mathscinet-getitem?mr=158024), [Prop. 14](https://books.google.com/books?id=Ae_Ayk5OitsC&pg=PA353)). *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.


