5
$\begingroup$

Questions a bit similar to this one have already appeared I think on the forum but I couldn't find the answer to my question using those answers. I must say from the beginning that my knowledge of group theory is very very basic so please adapt your answers.

Let $G$ be a reductive group over an algebraically closed field (say of reasonable characteristic). Fix a torus $T$ and let $w$ be an element of the Weyl group. Consider the subtorus of fixed points of $w$, say $S:=(T^w)^\circ$ and put $M=Z_G(S)^\circ$ (I guess this is already connected if $G$ is simply-connected).

From the general theory we know that $M$ is reductive. Question: is $M$ always a Levi subgroup? If no, can we say when does this happen? I'm only interested in the simply-connected case actually but the question makes sense for a general $G$.

Any references and comments are appreciated!

$\endgroup$
1
  • 1
    $\begingroup$ By the way, you don't need simple connectedness of $G$ to conclude that $M$ is connected; it is always so for $G$ reductive. The simple connectedness hypothesis comes in when you try to prove that the centraliser of a single semisimple element is connected. $\endgroup$
    – LSpice
    Commented Sep 16, 2019 at 21:07

2 Answers 2

7
$\begingroup$

Yes, it is true. You can look at the book of Digne-Michel Proposition 1.22 for a proof.

$\endgroup$
8
$\begingroup$

While zyxel has provided a concise answer and reference, it's worth filling in more details about the original source of this kind of result. Unfortunately, it wasn't clearly articulated in textbooks before Digne-Michel (who were especially interested in the structure of groups over finite fields following the work of Deligne and Lusztig).

1) First, I'd emphasize that simply connected only makes sense here for a semisimple group, and in any case doesn't influence the answer to your question. The basic object of study is a connected reductive group $G$, but one might as well focus here on the connected semisimple derived group when it has positive dimension. This is where the combinatorics of root systems matters.

2) In the question, $G$ can be taken to be defined (and automatically split) over any algebraically closed field $k$ of arbitrary characteristic. But the detailed structure theory involving centralizers of tori and parabolics was first undertaken over a general field of definition by Borel and Tits in their foundational 1965 paper here. The relevant material (applicable whenever $G$ is $k$-isotropic) is contained in sections 3 and 4, with a fairly explicit general statement in Theorem 4.15. What Digne and Michel do is essentially extracted from this source.

3) The ideas of Borel-Tits were intentionally coordinated at the time with Chapter VI of Bourbaki's treatise Groupes et algebres de Lie, treating root systems axiomatically. This was only published later in 1968, so references in Borel-Tits are numbered tentatively. In the published Bourbaki volume the crucial results are in section 1.7, especially Prop. 23, 24.

4) Its easy to get lost in the details, especially when working over arbitary $k$, so it may be helpful to outline briefly what goes into the basic proof that the centralizer $H:=C_G(S)$ of an arbitrary torus $S$ is a Levi factor of some parabolic subgroup of $G$. First, it follows readily from the Borel-Chevalley structure theory that $H$ is connected and reductive (and this is contained in standard textbooks). Taking an arbitrary basis for the root system of the derived group of $H$, the key point is to embed this basis in a basis of the full root system of $G$ relative to a maximal torus $T$ containing $S$. (This is where Bourbaki's abstract arguments come into play.) Then the Borel-Tits study of parabolics leads to a parabolic subgroup in which $H$ is a Levi factor. (Of course, this will turn out to be one of the standard parabolics of $G$ relative to $T$.)

$\endgroup$
1
  • $\begingroup$ Thank you Jim for this nice answer and for the precise reference to Borel Tits which I was always afraid to open. $\endgroup$ Commented Oct 26, 2013 at 16:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .