Let $G$ be a connected reductive algebraic group defined over an algebraically closed field and let $g\in G$ be semisimple. Write $C=\mathrm{C}_G(g)$ and $C^\circ=\mathrm{C}_G(g)^\circ$ for the centralizer of $g$ and for its identity component, respectively.

QuestionDoes there exist $h\in G$ such that $\mathrm{C}_G(h)=C^\circ$?

The statement is true in the case where $C^\circ$ is the Levi factor of a parabolic subgroup of $G$. The proof I know is by computing dimensions.

Fix $T$ to be a maximal torus containing $g$, and let $\Sigma$ (resp $\Phi$) denote the root system of $C^\circ$ (resp of $G$) with respect to $T$. Then, for $h\in T$, it holds that $\mathrm{C}_G(h)=C^\circ$ if and only if $\{\alpha \in \Phi:\alpha(h)=1\}=\Sigma$ and $\mathrm{Stab}_W(h)$ (the stabilizer of $h$ in the Weyl group of $G$) is generated by the reflections $s_\alpha$ for $\alpha\in \Sigma$.

Given a root systems $\Sigma\subseteq \Psi\subseteq\Phi$ and subgroups $\langle{s_\alpha:\alpha\in \Sigma}\rangle\subseteq S\subseteq W$, put
$$T_\Psi^S=\{h\in T: \{\alpha\in \Phi:\alpha(h)=1\}=\Psi\text{ and }\mathrm{Stab}_W(h)\supseteq S\}.$$
Then, using the assumption that $\Sigma$ is the root system of a Levi subgroup in several key steps, one can show that $\dim T_\Psi^S$ attains a maximum if *and only if* $\Psi=\Sigma$ and $S=\langle{s_\alpha:\alpha\in \Sigma}\rangle$. Therefore $T_\Sigma^{\langle s_\alpha:\alpha\in \Sigma\rangle}\setminus (\bigcup T_\Psi^S)$ is non-empty, and contains the $h$ we are seeking.
However, this proof falls apart completely if we take $\Sigma$ to be an arbitrary (closed) subsystem of $\Phi$.

I would very much appreciate if anyone could either suggest a different proof for this statement, which hopefully extends to general centralizers, or otherwise provide a counterexample for my question. Thank you.

affineroots. I think this should lead to a similar proof in general - I will try to think this through carefully later if there are no other answers. $\endgroup$ – Sam Gunningham Apr 12 '20 at 18:28