Let $G$ be an adjoint group over an algebraically closed field $k$ and $s\in G$ a semisimple element.

Let $H=C_{G}(s)^{0}$ the neutral component of the centralizer of $s$. Do we have that the center of $H$ is itself connected?

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    $\begingroup$ No, we do not have that. Take $G=PSO_8$, take for $s$ the image of ${\rm diag}(1,1,1,1,-1,-1,-1,-1)\in SO_8$, then $H=(SO_4\times SO_4)/\mu_2$, and the center of $H$ is isomorphic to $\mu_2$, hence not connected. $\endgroup$ Jan 8, 2013 at 20:30
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    $\begingroup$ The general study of this and related phenomena is "Borel-de Siebenthal theory". $\endgroup$ Jan 8, 2013 at 20:47

1 Answer 1


{EDIT] Since I was discussing a slightly different question than the one actually asked, I'll start over with your connected group $H$ and look at its center.

The question illustrates some of the fine points in the structure theory of reductive groups, which are worth considering beyond Mikhail's counterexample (which I guess has to require good characteristic, here different from 2). In the formulation of the question, I'm assuming that $G$ is meant to be connected as well as semisimple of adjoint type over an algebraically closed field. Here the general result for a semisimple element $s \in G$ is that the centralizer is reductive but not necessarily connected (unless $G$ itself is simply connected). Your group $H$ is both connected and reductive.

More precisely, it's a familiar result that the centralizer $C_G(s)$ of $s$ is generated by some connected subgroups of $G$ (including a maximal torus) along with perhaps part of the Weyl group. Moreover, those connected subgroups are enough to generate the identity component $H$, which contains all unipotent elements of the centralizer (a relevant issue in characteristic $p>0$).

I think the basic problem you encounter is that while $G$ itself is of adjoint type, there is no reason why the connected semisimple derived group of $H$ should also be of adjoint type. So the center of $H$ might well be the direct product of the nontrivial finite center of this derived group and a torus. Beyond Mikhail's proposed example, I expect that a transparent example might be seen when $G$ has type $G_2$: this group is both simply connected and of adjoint type, but will contain subgroups isomorphic to $\mathrm{SL}_2$ or $\mathrm{SL}_3$ that aren't of adjoint type.

P.S. To answer the original question more systematically, I should refer to the notion of pseudo-Levi subgroup defined by my colleague Eric Sommers IMRN 1998: see 2.1. (He and I disagree about applying this label as he does to actual Levi subgroups of parabolics.) This comes from the idea of Borel and de Siebenthal, in line with Allen's comment. It is well known that pseudo-Levi subgroups coincide with the connected centralizers of semisimple elements, as written down by Eric's thesis adviser George Lusztig in 5.5 of his paper IMRN 1995. For example, the simple group of type $G_2$ (which is both adjoint and simply connected) has such a subgroup isomorphic to $\mathrm{SL}_3$ (which visibly has a nontrivial center unless the characteristic is 3).

  • $\begingroup$ Dear Prof. Humphreys, could you give a more precise reference for your 2nd paragraph "More precisely, Borel-Tits show..."? $\endgroup$
    – Alexander
    Apr 5, 2018 at 22:39
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    $\begingroup$ @Alexander: Sorry to be slow in seeing your question. I don't recall an exact reference now, but for example they cover a lot of ground in $\S3$ of their 1965 IHES paper Groupes reductifs (freely available at numdam.org) and I gave a concise trreatment in my 1995 AMS book Conjugacy Classes in Semisimple Algebraic Groups (see Thm. 2.2). I'll do some further checking of their work and Steinberg's. $\endgroup$ Apr 9, 2018 at 0:27
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    $\begingroup$ P.S. See my rewritten version above, with added references. While the basic idea may come from Borel and Tits, their main efforts went into the study of structure theory when the field of definition isn't algebraically closed. Anyway, I'm still unsure about the original source of the description of centralizers given here; but it's fairly straightforward and written down in a number of places including Carter's 1985 book and mine in 1995. $\endgroup$ Apr 10, 2018 at 1:16

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