Let $\mathfrak{g}$ be a complex simple Lie algebra and $x \in \mathfrak{g}$ be a regular element, i.e. its centralizer is of minimal dimension.

Consider the adjoint action of the adjoint group $G$ (with trivial center) on its Lie algebra $\mathfrak{g}$.

Is it true that the stabilizer of $x$ in $G$ is always connected?

If $x$ is semisimple, then it is known that the stabilizer of $x$ in the simply connected group is connected (see the book of Collingwood-McGovern, Theorem 2.3.3), and so by projection, also in the adjoint group.

If $x$ is regular and nilpotent (also called principal nilpotent), then one can check in the lists of all Lie algebras in the book of Collingwood-McGovern to check that the stabilizer is connected.

But what about a general regular element? And if it is true, is there a simple proof (without using the classification of simple Lie algebras)?

  • $\begingroup$ I don't know what is the adjoint group.. can I assume $G$ is a Lie group and $\mathfrak{g}$ be it's Lie algebra?? $\endgroup$ Oct 3, 2019 at 13:34
  • 1
    $\begingroup$ At least this is true in $\mathrm{PGL}_n(\mathbf{C})$, without assuming regular. Indeed, for $x\in\mathfrak{sl}_n(\mathbf{C})$, the centralizer $C_x$ of $x$ in $\mathrm{GL}_n(\mathbf{C})$ is connected (centralizers in $\mathrm{GL}_n(\mathbf{C})$ are connected, and this applies to $x+t$ for suitable scalar $t$). Hence the centralizer $C_x/\mathbf{C}^*$ of $x$ in $\mathrm{PGL}_n(\mathbf{C})$ is connected. (On the other hand, in $\mathrm{PGL}_2(\mathbf{C})$ there are regular elements inside the group itself, with non-connected centralizer.) $\endgroup$
    – YCor
    Oct 3, 2019 at 14:18
  • $\begingroup$ The centraliser of an arbitrary element is the centraliser in the centraliser of the semisimple part, of the (regular there) nilpotent part. $\endgroup$
    – LSpice
    Oct 3, 2019 at 14:43
  • $\begingroup$ @PraphullaKoushik: the adjoint group $G$ is the unique Lie group with Lie algebra $\mathfrak{g}$ which is centerless. If the centralizer is connected in a covering space, of course it is also connected in the adjoint group, but the converse is of course not true. $\endgroup$
    – AThomas
    Oct 3, 2019 at 15:55
  • $\begingroup$ By the way, your argument in the semisimple case doesn't work: it is true that the image in the adjoint group of the centraliser in the semisimple group is connected, but the centraliser in the adjoint group can, a priori, be strictly larger. (This happens with @YCor's example of $g = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}$, whose centraliser in $\operatorname{PGL}_2(\mathbb C)$ is bigger than the image of its centraliser in $\operatorname{SL}_2(\mathbb C)$; but that's on the group side.) The conclusion is still true (in characteristic 0), just not the proof. $\endgroup$
    – LSpice
    Oct 3, 2019 at 16:05

1 Answer 1


$\def\semi{_{\text{semi}}}\def\nil{_{\text{nil}}}$This is just an elaboration of my comment. If $X \in \mathfrak g$ is regular, with semisimple part $X\semi$ and nilpotent part $X\nil$, then $X\nil$ is regular nilpotent in $\operatorname{Lie}(\operatorname C_G(X\semi))$, so you have said that you already know that $\operatorname C_{\operatorname C_G(X\semi)^\circ}(X\nil)$ is connected. On the other hand, $\operatorname C_G(X\semi)$ is connected. (Your argument for this by reduction to the simply connected case doesn't work, but it is true in general, in characteristic 0 or even just in not-too-small positive characteristic. The reference that I know is Section 7 of Yu - Construction of tame supercuspidal representations, although that's clearly the wrong place to look for general questions of this sort; better to look in Steinberg. (Probably there's a precise reference in Collingwood–McGovern; the "simple connectedness implies connectedness" result you cite, which is valid on the group as well as the Lie-algebra level, is due to Steinberg.) Thus, $\operatorname C_{\operatorname C_G(X\semi)^\circ}(X\nil)$, which we have seen is connected, equals $\operatorname C_{\operatorname C_G(X\semi)}(X\nil) = \operatorname C_G(X)$.

  • $\begingroup$ Thank you very much! I felt that one should be able to reduce to the nilpotent case. But still, is there a direct argument without using the classification of Lie algebras? $\endgroup$
    – AThomas
    Oct 4, 2019 at 6:47
  • $\begingroup$ @AThomas, I almost never think of centralisers of nilpotents, so I don't know off the top of my head; but I'll think about it. $\endgroup$
    – LSpice
    Oct 4, 2019 at 17:14

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