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.

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    $\begingroup$ 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. $\endgroup$ Apr 12, 2020 at 18:28
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    $\begingroup$ 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? $\endgroup$ Apr 12, 2020 at 18:33
  • $\begingroup$ @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. $\endgroup$
    – kneidell
    Apr 12, 2020 at 18:37
  • $\begingroup$ @Mikhail Borovoi: sorry, you're correct. I meant $\alpha\in\Sigma$.. $\endgroup$
    – kneidell
    Apr 12, 2020 at 18:38

1 Answer 1


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.

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.

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    $\begingroup$ I just realized this counterexample too! Glad you figured that out, and sorry if I sent you down the wrong path. $\endgroup$ Apr 16, 2020 at 21:22

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