How parabolic subgroups are determined by roots?

Question

I am interested in understanding the general construction of important subgroups in reductive groups, and how they are parametrized (Borel, Levi, parabolic, etc.). But for simplicity I take the example of $SL(3)$ over a field $k$.

There are six roots, three positive roots and thus three "root" hyperplanes (do they have a name?). I saw many times the statement that parabolic subgroups correspond to subsets of positive roots, and I would like to understand how this correspondence work. Give a parabolic or a subset of roots, how can I construct the corresponding element on the other side?

More precisely, let $M$ be a standard minimal Levi subgroup. I would like to understand why

• the set of parabolic having Levi component $M$ is in bijection with the (6) open Weyl chambers (i.e. the connected sectors between hyperplanes)
• the set of Levi subgroups containing M is in bijection with the (5) "subspaces" in the diagram: the plane, the three line, and the origin
• the set of parabolics containing $M$ is in bijection with the (13) "half-subspaces": the half-lines, the chambers and the origin.

I am really troubled by these statements but I would like to develop a geometric version of this reductive structure. What is a good reference for it (with lots of examples)?

Answer 1

I should say right away that I am only familiar with the case that $k=\mathbb C$, I still hope that the answer is helpful.

Parabolics do not correspond to subsets of positive roots but to subsets of simple roots. There is a description in terms of saturated subsets of positive roots (i.e. subsets $\Phi\subset\Delta^+$ such that if $\alpha,\beta\in\Phi$ and $\alpha+\beta$ is a root, then $\alpha+\beta\in\Phi$), but it turns out that such a subset is uniquely determined by the simple roots it contains. This leads to a rather simple description of parabolic subalgebras in the Lie algebra, but it sounds like this is not quite what you are after.

On a more group theoretic level, the best description that comes to my mind is to start from an irreducible representation $V$ of your group and consider the associated action on the projective space $\mathcal P(V)$. Then the highest weight line of $V$ determines a point in $\mathcal P(V)$ and the stabilizer of that point is a parabolic subgroup. Indeed, the orbit of that point $\mathcal P(V)$ turns out to be the unique closed orbit in there. In appropriate conventions this is the group corresponding to the subset of all those simple roots which are not orthgonal to the highest weight of $V$. So maximal parabolic subgroups can be obtained from fundamental representations.

In the case of $SL(3,\mathbb C)$ you get 3 parabolics, the stabilizer of a line, the stabilizer of a plane and the stabilizer of a full flag in the standard representation.

Answer 2

I'll amend what Andreas Cap describes as the rather simple description of parabolic subalgebras in the Lie algebra $\mathfrak{sl}_3(k)$. Actually, I would have thought that most of this translates over to the group level, but I'm not an expert in that. I will use both terminologies in what follows and hope to be corrected by experts if wrong.

If your minimal Levi is the (maximal torus / Cartan subalgebra) $M = \pmatrix{*&0&0\\0&*&0\\0&0&*}$, and we call $\alpha_1$ the root whose (one-parameter unipotent group / root space) is $\pmatrix{(1/0)&*&0\\0&(1/0)&0\\0&0&(1/0)}$, $\alpha_2$ the one with $\pmatrix{(1/0)&0&0\\0&(1/0)&*\\0&0&(1/0)}$, then the entire root system is $R= \{\pm \alpha_1, \pm \alpha_2, \pm(\alpha_1+\alpha_2) \}$, and we have:

• The parabolics with Levi component $M$ (i.e. minimal parabolics = Borels containing $M$) are

$$ \pmatrix{*&*&*\\0&*&*\\0&0&*}, \pmatrix{*&0&*\\*&*&*\\0&0&*}, \pmatrix{*&0&*\\*&*&0\\*&0&*}, \pmatrix{*&0&0\\*&*&0\\*&*&*}, \pmatrix{*&*&0\\0&*&0\\*&*&*}, \pmatrix{*&*&*\\0&*&0\\0&*&*} $$ which correspond to the six possible root bases $$\{\alpha_1, \alpha_2\}, \{\alpha_1 +\alpha_2, -\alpha_1 \}, \{\alpha_2, -\alpha_1-\alpha_2\}, \{ -\alpha_1, -\alpha_2\}, \{-\alpha_1-\alpha_2, -\alpha_2 \}, \{ -\alpha_2, \alpha_1+\alpha_2\},$$ respectively. Such root bases, a.k.a. sets of simple roots, are in bijection with Weyl chambers, as every serious source on root systems will tell.

• The possible Levis which contain $M$ are $$\pmatrix{*&0&0\\0&*&0\\0&0&*} = M, L_1 := \pmatrix{*&*&0\\*&*&0\\0&0&*}, L_2:= \pmatrix{*&0&0\\0&*&*\\0&*&*}, L_3:= \pmatrix{*&0&*\\0&*&0\\*&0&*}, \pmatrix{*&*&*\\*&*&*\\*&*&*} = G $$

They generally correspond to subsets $S_L$ of the root system which are saturated (see Andreas Cap's answer) and symmetric ($S_L =-S_L$). Here, the above possibilities obviously correspond to $S_L=$

$$ \emptyset, \{\pm \alpha_1\}, \{\pm \alpha_2\}, \{\pm(\alpha_1+\alpha_2)\}, R,$$

respectively, and you can mattch that with those "subspaces".

• The set of all parabolics containing $M$ consists of:

• • the Borels $$ \pmatrix{*&*&*\\0&*&*\\0&0&*}, \pmatrix{*&0&*\\*&*&*\\0&0&*}, \pmatrix{*&0&*\\*&*&0\\*&0&*}, \pmatrix{*&0&0\\*&*&0\\*&*&*}, \pmatrix{*&*&0\\0&*&0\\*&*&*}, \pmatrix{*&*&*\\0&*&0\\0&*&*} $$ from the first question, corresponding to the Weyl chambers;
• • $$ \pmatrix{*&*&*\\*&*&*\\0&0&*}, \pmatrix{*&*&0\\*&*&0\\*&*&*}, \pmatrix{*&*&*\\0&*&*\\0&*&*}, \pmatrix{*&0&0\\*&*&*\\*&*&*}, \pmatrix{*&*&*\\0&*&0\\*&*&*}, \pmatrix{*&0&*\\*&*&*\\*&0&*},$$

of which the first two contain $L_1$, the second and third contain $L_2$, the last two contain $L_3$ as Levi factor; and finally, of course, the full group / algebra

• • $$\pmatrix{*&*&*\\*&*&*\\*&*&*}. $$

In general, these correspond e.g. to subsets $S_P \subseteq R$ which are saturated and contain a system of simple roots. This contains of course, first, the six root bases / Weyl chambers /Borels listed in the first part; then, in our case (and I am not sure how that would generalise), for each of the next six, we could make it correspond to a half-plane, which one can make correspond to a single vector, namely e.g. for the first one, $S_P = \{\pm \alpha_1, \alpha_2, \alpha_1+\alpha_2 \}$ which is $\{ \alpha \in R: (\alpha, 2\alpha_2+\alpha_1) \ge 0\}$; ...; for the sixth one, $S_P = \{\pm (\alpha_1 +\alpha_2), \alpha_2, -\alpha_1 \}$ which is $\{ \alpha \in R: (\alpha, -\alpha_1+\alpha_2) \ge 0\}$. I do not see how one would naturally parametrise these via one of the roots. (Of course one can via some convention, after all there's six such parabolics and six roots; but the "natural" choice of parameter here seems to be, rather, a vector which is perpendicular to a certain root.)

If one does that, -- i.e. classify the Borels by Weyl chambers $C$ as belonging to those $S \subset R$ where $S = \{\alpha \in R: (\alpha, c) \ge 0 \text{ for all } x \in C\}$, and the six next bigger parabolics via those $S$ where $S = \{\alpha \in R: (\alpha, v) \ge 0\}$ for certain vectors $v$ perpendicular to one root -- then at least the last one falls into place as well, consisting of the entire root system, which we might phrase as those roots such that $(\alpha, 0) \ge 0$.