TITLE: How parabolic subgroups are determined by roots?
QUESTION [6 upvotes]: 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)?

REPLY [4 votes]: 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 $(\alpha, 0) \ge 0$.