1
$\begingroup$

Let $G$ be a real semisimple Lie group; we suppose $G$ is connected and centerless. Let $\mathfrak{g}$ be its Lie algebra, $\mathfrak{a}$ a Cartan subspace of $\mathfrak{g}$ (i. e. a maximal abelian subalgebra made of hyperbolic elements); $\mathfrak{a}^+$ an open Weyl chamber of $\mathfrak{a}$; and $a_0$ some element of $\mathfrak{a}^+$.

I would like to have the following property:

Let $g \in G$. Suppose that $\operatorname{Ad} g$ fixes $a_0$. Then $\operatorname{Ad} g$ fixes the whole subspace $\mathfrak{a}$.

(In other words, the stabiliser of $a_0$ is equal to the stabiliser of $\mathfrak{a}$, commonly called $L$ or $MA$.)

Since all Cartan subspaces are conjugate to each other, an equivalent statement would be:

Any two distinct open Weyl chambers in $\mathfrak{g}$ (not necessarily lying in the same Cartan subspace) are disjoint.

This looks like it should be a classical result, but I can neither prove it myself nor find a reference. Can anyone help me?

If the proof is longer than a couple of lines, I would rather quote it than rewrite it, so a reference would be appreciated!

$\endgroup$
3
$\begingroup$

The answer to your basic question is yes, though the notion of "Weyl chamber" comes up in somewhat different contexts (so it has to be defined carefully). Abstractly, the Bourbaki development of "root systems" leads to a simple geometric picture in a real vector space: you have hyperplanes orthogonal to finitely many pairs of roots, and the resulting Weyl chambers are defined to be the (open) connected components in the complement of the union of hyperplanes. In turn, the Weyl group generated by reflections in the hyperplanes will act simply transitively on the Weyl chambers, which are in particular pairwise disjoint.

Concretely, all of this arises in the traditional theory of Lie groups, most transparently in the extreme cases where the group is compact semisimple and where it is "split"; so its complexification is one of the usual semisimple groups over $\mathbb{C}$ and can also be regarded as a semisimple algebraic group. In general, for a real semisimple Lie group, you get a relative root system and relative Weyl group, along with its "Weyl chambers". But in any case the Weyl chambers are mutually disjoint.

There are quite a few sources for the standard Lie group structure theory. Look for example at the modern book by A.W. Knapp, Representation Theory of Semisimple Lie Groups. Here Chapter 3 works out the structure theory for noncompact groups, and in section 3 he treats the Weyl group and Weyl chambers in a notational framework similar to yours.

$\endgroup$
  • $\begingroup$ The Weyl group does act simply transitively on the Weyl chambers, but only on those that lie in the same Cartan subspace. I would like to show that two Weyl chambers are always disjoint, even when they lie in different Cartan subspaces. I added a comment in the original statement to clarify this. $\endgroup$ – Ilia Smilga Apr 15 '14 at 9:21
  • $\begingroup$ @Ilia: I was probably oversimplifying the framework you work in, since I don't usually think about Lie groups directly. $\endgroup$ – Jim Humphreys Apr 24 '14 at 22:36
2
$\begingroup$

Let $H$ be the centralizer of $a_0$. $H$ certainly contains the centralizer $MA$ of $A$. We now compare their Lie algebras, $\mathfrak{h}$ and $\mathfrak{m}+\mathfrak{a}$. Since $H$ contains $A$ it is normalized by it and we can decompose $\frak{h}$ under this action. Now $\frak{m}+\frak{a}=\frak{g}_0\subset\frak{h}$ is the subspace corresponding to the trivial character of $A$, so the rest of $\frak{h}$ must transform under other characters (that is, under the roots). However, no root vanishes on $a_0$ (since $a_0$ is not on any wall), so no root can occur in this action. It follows that $\frak{h} = \frak{a}+\frak{m}$.

This shows that the connected components of $H$ and $MA$ agree.

In the algebraic category, the centralizer of $A$ is Zariski-connected, and $A$ is the maximal split torus in the center of the centralizer, hence characteristic. It follows that $A$ is normal in $H$, so $H$ is contained in the normalizer of $A$. But $N_G(A) / Z_G(A)$ is the Weyl group, which acts simply transitively on the chambers. It follows that $H = Z_G(A)$.

In the smooth category, $\frak{a}$ is the unique Cartan subalgebra of $MA$, so again $A$ is characteristic there and $H \subset N_G(A)$, at which point the argument can proceed in the same fashion.

[Edited: erroneous claim about connectedness of $H$ replaced with correct discussion. Reference added]

[Edited: I think I have a reference: Lemma 6.4.3 in Springer's "Linear Algebraic Groups", 2nd edition. The result is not exactly the same (it concerns regular elements of $A$, not of $\frak{a}$), and the arguments depend on an embedding in $\mathrm{GL}_n$, but it may be enough for you.]

$\endgroup$
  • 1
    $\begingroup$ Concerning your remark toward the end: In the algebraic setting, it isn't true in all cases that the centralizer of a semisimple element is connected. Can you clarify? (And I guess a reference is really needed here.) $\endgroup$ – Jim Humphreys Apr 3 '14 at 12:44
  • $\begingroup$ I added a reference. $\endgroup$ – Lior Silberman Apr 3 '14 at 18:36
  • $\begingroup$ At least one point does not seem clear to me. Why do you treat separately the algebraic and the smooth category? The equality $H = Z_G(A)$ is either true or false, regardless of the category we are working in; why would you need to show it twice? $\endgroup$ – Ilia Smilga Apr 15 '14 at 14:37
  • $\begingroup$ Because "real semisimple Lie group" can mean one of two things. It can mean the group of $\mathbb{R}$-points of a semisimple linear algebraic group defined over $\mathbb{R}$, or it can mean a Lie group whose Lie algebra is a direct sum of simple Lie algebras. $\endgroup$ – Lior Silberman Apr 16 '14 at 2:41
  • $\begingroup$ Sorry for coming back to this two years later. If you are still around, could you please explain to me the logical step at the beginning of the third paragraph? You know that $A$ is characteristic in $MA$, and you want to show it is normal in $H$. Let $h \in H$; then if $h(MA)h^{-1} = MA$, certainly this implies that $hAh^{-1} = A$. But how do you obtain the first equality? $\endgroup$ – Ilia Smilga Oct 9 '16 at 18:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.