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Let $G$ be a complex semisimple (EDIT: simply-connected) 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.

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.

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.

Let $G$ be a complex semisimple (EDIT: simply-connected) 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.