Is the connected centralizer of a semisimple element in a connected reductive group also a centralizer?

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.

Question Does 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.

• Note that $\Sigma$ is not an arbitrary closed subsystem, but is the root system of a pseudo-Levi. As such it corresponds to a subset of the simple affine roots. 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. Apr 12, 2020 at 18:28
• You write: $\langle{s_\alpha:\alpha\in \Phi}\rangle\subseteq S\subseteq W$. However, it seems that $\langle{s_\alpha:\alpha\in \Phi}\rangle= W$. Is there a typo in the formula? Apr 12, 2020 at 18:33
• @Sam Gunningham Thank you, I was unaware of this terminology, but I think I do know what you mean (I called these Deriziotis root systems, but i guess that's uncommon). In any case, one thing that comes into play in my proof is that the root system of a Levi (as opposed to a pseudo Levi) is not contained in any subsystem of the same rank.. I would be very interested if you could supplement this. Apr 12, 2020 at 18:37
• @Mikhail Borovoi: sorry, you're correct. I meant $\alpha\in\Sigma$.. Apr 12, 2020 at 18:38

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.