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.



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> **Question** Does there exist $h\in G$ such that $\mathrm{C}\_G(h)=C^\circ$?
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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.



Is the connected centralizer of a semisimple element in a connected reductive group also a centralizer?
So after thinking about this for a few extra days, I think I found a counterexample. I'm recording it here, as community wiki, in case it would be of interest for anyone in the future.




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Let $k$ be an algebraically closed field of characteristic not $2$ and $G=\mathrm{PGSp}\_{2n}(k)=\mathrm{GSp}\_{2n}(k)/k^\times$, where $\mathrm{GSp}\_{2n}(k)$ is the group of similitudes of the standard symplectic form, i.e. 
$$\mathrm{GSp}\_{2n}(k)=\{x\in\mathrm{GL}\_{2n}(k):x^tJx=\lambda J\text{ for some }\lambda\in k^\times\}\text{ where }J=\begin{pmatrix}0&I\_n\\-I\_n&0\end{pmatrix},$$
(I don't know if this notation is common).


$G$ is simple of adjoint type with maximal torus $T=\lbrace d(t\_1,t\_2):=\left[\begin{smallmatrix}t\_1\\&t\_2\\&&t\_1^{-1}\\&&&t\_2^{-1}\end{smallmatrix}\right]:t\_1,t\_2\in k^\times\rbrace$, (here $[\cdot]$ denotes the class mod $k^\times$ of a matrix; note that $d(\lambda t\_1,\lambda t\_2)=\lambda d(t\_1,t\_2)$ implies $\lambda=\pm 1$) and root system with simple roots:
$$\alpha(d(t\_1,t\_2))=t\_1/t\_2\text{ and } \beta(d(t\_1,t\_2))=t\_2^2$$
(the other positive roots are $\alpha+\beta$ and $2\alpha+\beta$).


Consider the subsystem $\Sigma=\lbrace\pm \beta,\pm(2\alpha+\beta)\rbrace$ (viz. the long roots). Then $\Sigma$ is the root system of a pseudo-Levi subgroup of $G$ which is isomorphic to $(\mathrm{GL}\_n(k)\times\mathrm{GL}\_n(k))/k^\times$, and one can easily verify that 
$$(\star)\quad \beta(d(t\_1,t\_2))=(2\alpha+\beta)(d(t\_1,t\_2))=1\:\iff\: d(t\_1,t\_2)\in\lbrace\left[\begin{smallmatrix}1\\&1\\&&1\\&&&1\end{smallmatrix}\right],\left[\begin{smallmatrix}1\\&-1\\&&1\\&&&-1\end{smallmatrix}\right]\rbrace.$$
Let $g$ be the non-central element in this set. Then, a standard computation, taking into account that $g=-g$ in $G$, shows that $C\_G(g)$ is disconnected (the non-identity connected component is generated by the coset of the Weyl group element permuting $t\_1$ and $t\_2$). On the other hand, $Z(C\_G(g))$ consists of precisely the two elements on the RHS of $(\star)$, so there exists no $g\ne h\in Z(C\_G(g))$ such that $C\_G(h)\subseteq C\_G(g)$, and, in particular, the question above has a negative answer in this case.


