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Let $G$ be a complex semisimple Lie group with Lie algebra $\mathfrak{g}$. In a particular paper, the following statement is made:

If $X\in\mathfrak{g}$ is regular (i.e. has centralizer of minimal dimension) then $$ Z_G(X) := \{g\in G:\mathrm{Ad}_gX=X\}, $$ is abelian.

However, no justification is given, and I was wondering if anybody knows how to prove it or can point a reference discussing this.

The Lie algebra of $Z_G(X)$ is the centralizer $Z_{\mathfrak{g}}(X)$ of $X$ in $\mathfrak{g}$, and since $X$ is regular, $Z_{\mathfrak{g}}(X)$ is a Cartan subalgebra of $\mathfrak{g}$. In particular, $Z_{\mathfrak{g}}(X)$ is abelian and hence the identity component of $Z_G(X)$ is abelian. But the problem is that $Z_G(X)$ might not be connected.

For example, if $G=\mathrm{SL}(2,\mathbb{C})$, then $X=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ is regular and $$Z_G(X)=\left\{\begin{pmatrix}1&z\\0&1\end{pmatrix}:z\in\mathbb{C}\right\}\cup\left\{\begin{pmatrix}-1&z\\0&-1\end{pmatrix}:z\in\mathbb{C}\right\}$$ is abelian, but not connected.

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    $\begingroup$ A small correction: The centraliser $Z_{\mathfrak{g}}(X)$ is not always a Cartan subalgebra when X is regular (it's not self-normalising when X isn't semisimple). But for regular X, this centraliser is still abelian, so the argument that the identity component of $Z_G(X)$ is abelian still holds, without assuming simply-connectedness. $\endgroup$ – Peter McNamara Oct 31 '16 at 0:03
  • $\begingroup$ @PeterMcNamara Right, thanks. The correct argument is that $Z_{\mathfrak{g}}(X)$ is contained in the generalized $0$-weight space of $\mathrm{ad}(X)$ which is a Cartan subalgebra since $X$ is regular. $\endgroup$ – SHP Oct 31 '16 at 9:15
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    $\begingroup$ @SHP: I don't think (Lie algebra) regular centralizers (which have the same dimension as a Cartan subalgebra) are contained in Cartan subalgebras unless $X$ is itself regular semisimple (so that its centralizer is a Cartan). For example the Lie algebra centralizer of a regular nilpotent element (such as the $X$ in your post) always consists entirely of nilpotent matrices. $\endgroup$ – David Ben-Zvi Oct 31 '16 at 18:56
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Here is an attempted proof. We have that $Z_G(X)$ is the full preimage of $Z_H(X)$ under the covering central extension $\operatorname{Ad}$: $$ 1\longrightarrow C\longrightarrow G\overset{\operatorname{Ad}}\longrightarrow H\longrightarrow 1 \tag{*} $$ where $C$ is the (discrete) center and $H=\operatorname{Ad}(G)$ the adjoint group of $G$. Now:

Theorem (Kostant 1963, Prop. 14). If $X$ is regular, then $Z_H(X)$ is connected abelian.

To deduce that $Z_G(X)$ is also abelian, pick $g_0$ and $g$ there, projecting to $h_0$ and $h_1$. Since $Z_H(X)$ is connected, it contains a continuous path $h_t$ from $h_0$ to $h_1$. Lift that to a continuous path $g_t$ in $Z_G(X)$ starting at $g_0$. Since $Z_H(X)$ is abelian, the path $c_t=g_tg_0g_t^{-1}g_0^{-1}$ satisfies $$ \operatorname{Ad}(c_t)=h_th_0h_t^{-1}h_0^{-1}=1. $$ So $c_t$ lies in the (discrete) center $C$, hence constantly equals $c_0=1$. So $g_0$ commutes with $g_1$. But $\operatorname{Ad}(g)=h_1=\operatorname{Ad}(g_1)$ shows that $g=g_1z$ for some $z\in C$. So $g_0$ commutes with $g$, too.

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  • $\begingroup$ A couple of other belated comments. Kostant's 1959 and 1963 papers are pioneering but also somewhat improvised. His notion of "regular" shifted from (semisimple) elements having a Cartan subalgebra as centralizer to the more general notion which Steinberg then generalized to an algebraic setting. Since only semisimple Lie groups are mentioned in the question, Ad has finite kernel equal to the center, so your analytic arguments are unnecessary. On the other hand, the theorem of Kostant you cite deals just with "nilpotent" elements, but the question includes all $X$. $\endgroup$ – Jim Humphreys Nov 1 '16 at 20:48
  • $\begingroup$ @JimHumphreys : Kostant's cited and linked Prop. 14 (p. 362) deals with all (regular) $X$, not just nilpotent! See his definition of $\mathfrak r$ (3.4.3, p. 358). As to "necessity" of the above argument: I was happy with sufficient. If, as the last § of your answer suggests, there is an algebraic proof that $(*)$ actually splits over $Z_H(X)$, so that $Z_G(X)=C\times Z_H(X)$, I haven't seen it -- but would love to. $\endgroup$ – Francois Ziegler Nov 2 '16 at 6:18
  • $\begingroup$ Yes, I misread Kostant's symbol. However, in this paper he is getting closer to the algebraic group viewpoint soon developed by Steinberg and Springer. He starts here with a reductive Lie algebra, though in the question it's semisimple, along with its adjoint group $G$. To get to an arbitrary semisimple group over $\mathbb{C}$ is then easy, since the center of such a group is finite. No need for a "discrete center" or other steps here. Note that Kostant's argument is essentially algebraic. $\endgroup$ – Jim Humphreys Nov 2 '16 at 13:29
  • $\begingroup$ @JimHumphreys : Oh, $(∗)$ does not split over $Z_H(X)$ when $G=SL_2(\mathbf C)$ and $X=\operatorname{diag}(1,−1)$. For then $Z_G(X)=\{\operatorname{diag}(z,1/z):z\in\mathbf C^\times\}\simeq\mathbf C^\times$ covers $Z_H(X)\simeq\mathbf C^\times/\{\pm1\}$ nontrivially. So I guess you only meant the splitting for $X$ nilpotent. Where does one find algebraic proofs of 1) this, and 2) how commutativity follows in general? $\endgroup$ – Francois Ziegler Nov 2 '16 at 23:52
  • $\begingroup$ Sorry to have overlooked this question. My edited answer may be more precise, since I am indeed thinking about regular nilpotent elements as the crucial case. $\endgroup$ – Jim Humphreys Nov 11 '16 at 16:56
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EDIT: The question and some of the responses are out of focus, so it's worth clarifying a few of the issues here. First, the intended notion of "regular element" is unclear. In Kostant's early papers, for example here, only regular semisimple elements in semisimple Lie groups are directly considered. But later he broadened the definition. In the general Lie group framework, there is no notion of "Jordan-Chevalley decomposition", so a broader definition of regularity was given by Steinberg for a connected semisimple algebraic group $G$ here. An element $x \in G$ is called regular if its centralizer has dimension equal to the rank $\ell$ of $G$ (which is the least possible).

At the 1966 ICM in Moscow, Steinberg surveyed questions about conjugacy classes and regular elements here. Among the problems he raised were (14) and (15), asking whether regular $x \in G$ is characterized by having a commutative centralizer. This and more was later confirmed by two of his students B. Lou and S.V. Keny, even for fields of "bad" characteristic.

For $X \in \mathfrak{g}$, the Lie algebra of $G$, there are actually two relevant centralizers: the subgroup $G_X$ of $G$ fixing $X$ under the adjoint representation Ad, and the subalgebra $\mathfrak{g}_X$ of $\mathfrak{g}$ fixing $X$ under ad. The latter always contains the Lie algebra of $G_X$ and equals it in characteristic 0. Springer refined and extended Steinberg's ideas in a 1966 paper here: see $\S4$ and $\S5$. He further showed in Proposition 2 here that when $x \in G$ is regular, the identity component of $G_x$ is commutative. The work of Lou and Keny showed in fact that $G_x$ itself is commutative, while $G_X$ is commutative for a regular element $X\in \mathfrak{g}$ provided $G$ is of adjoint type.

Thanks to the fact that the centralizer of a semisimple element is reductive (the almost-direct product of a central torus and a semisimple group), the main problems are reduced to $x \in G$ unipotent or $X \in \mathfrak{g}$ nilpotent. But $G_x$ is not necessarily connected for $x$ semisimple unless $G$ is simply connected. (An algebraic proof due to Digne-Michel is discussed in Chapter 2 of my 1995 book on conjugacy classes; the first proof was by Springer and Steinberg.) A similar reduction to the regular nilpotent case works for $X \in \mathfrak{g}$ and $G_X$ when $G$ is simply connected.

In the case of $G_X$ for $X$ regular nilpotent, the outcome in characteristic 0 is fairly straightforward: here the centralizer is the product of the center of $G$ and a (connected!) commutative unipotent group. But for bad prime characteristics, it took some case-by-case work to describe $G_X$.

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    $\begingroup$ Which "some" of the two other extant responses are out of focus? ;-) The OP works over $k=\mathbf C$ and spells out detailed definitions in agreement with current terminology for both centralizer ($Z_G(X)$, $X\in\mathfrak g$) and regular. $\endgroup$ – Francois Ziegler Oct 31 '16 at 14:57
  • $\begingroup$ @Francois: I've added a comment to your answer. Working over the complex field isn't significant for this kind of question, only the fact that the field is algebraically closed of characteristic 0. The examples mentioned are not very helpful. Also, the question here is algebraic and doesn't require analysis or differential geometry. (It's of course not necessary here to get into the more difficult problems in prime characteristic, but I wanted to emphasize that Steinberg's setting for regular elements is more natural.) $\endgroup$ – Jim Humphreys Nov 1 '16 at 20:52
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In $PGL_2$, the centralizer of the diagonal element $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ consists of all elements of the form $\begin{pmatrix}* & 0 \\ 0 & * \end{pmatrix}$ or $\begin{pmatrix} 0 & * \\ * & 0 \end{pmatrix}$, and so has the minimal dimension (one, after modding out by scalars) but is not connected.

So I believe this is false without additional assumptions.

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    $\begingroup$ In the paper they consider $G$ simply-connected, which I forgot to add to the question. $\endgroup$ – SHP Oct 30 '16 at 17:09
  • $\begingroup$ No, a matrix $\left(\begin{smallmatrix}0&*\\*&0\end{smallmatrix}\right)$ won't centralize $\left(\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\right)$ unless it is zero. $\endgroup$ – Francois Ziegler Oct 31 '16 at 5:28
  • $\begingroup$ Actually, given an element of $\mathfrak{sl}_n$, its centralizer in the whole algebras of matrices is a subalgebra, so its centralizer in $GL_n$ is an Zariski open subset of this algebra and in particular is Zariski connected, and in turn the image of the latter in $PGL_n$ is Zariski connected. So the result is true in $PGL_n$ (and connectedness of the centralizer of any Lie algebra element does not make use of regularity). $\endgroup$ – YCor Oct 31 '16 at 5:54
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    $\begingroup$ To summarize erased comments: the 1st sentence of the post is true, but is not the point: this is the centralizer in $PGL_2$ of the diagonal matrix $(1,-1)$ viewed as element of $PGL_2$. But one has to consider instead, the centralizer in $PGL_2$ of this matrix viewed as an element of the Lie algebra. This is the same as the image in $PGL_2$ of its genuine centralizer in $GL_2$, so is reduced to diagonal matrices. And thus this does not yield a counterexample. $\endgroup$ – YCor Oct 31 '16 at 6:39
  • $\begingroup$ @YCor Ah, good point. $\endgroup$ – Will Sawin Oct 31 '16 at 22:07

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