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Suppose that we have a closed embedding $G_1\hookrightarrow G_2$ of reductive groups (say over $\mathbb{Q}$), and suppose that we have a maximal parabolic sub-group $P_2\subset G_2$, and a minimal parabolic $P_1\subset G_1$. Is it possible to have two different maximal parabolic sub-groups of $G_1$ contained in $P_2$ and containing $P_1$?

Actually, is it even possible that there are two different maximal parabolics of $G_1$ contained in $P_2$?

Even more optimistically, if there is a maximal parabolic sub-group of $G_1$ contained in $G_1\cap P_2$, does that make $G_1\cap P_2$ a parabolic sub-group and thus equal to the maximal parabolic it contains?

EDIT: As Jim Humphreys points out below, there is one pathological case, where $G_1\cap P_2$ might be all of $G_1$. In which case, all parabolic sub-groups of $G_1$ will be contained in $P_2$! This is the case, for example, when $G_1$ is a Levi sub-group of $G_2$. But Angelo's answer below shows that, excluding this possibility, the answer to my third question is 'yes'.

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To clarify, my last example was intended to provide a negative answer to your second question (there are lots of similar examples involving higher rank groups). It would help to restate your questions in a more precise way. (And what if any role would be played by a field of definition if the groups are not split over that field?) –  Jim Humphreys Dec 9 '11 at 22:53
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2 Answers

up vote 4 down vote accepted

If $P$ is a parabolic subgroup of a reductive group $G$ and $H$ is a closed subgroup of $G$ containing $P$, then $G/H$ is a quotient of $G/P$, so it is projective, and $H$ is parabolic. It follows that a maximal parabolic subgroup is maximal among all proper closed subgroups, which would seem to imply that the answer to all of your questions is positive.

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Great. Thank you! –  Keerthi Madapusi Pera Dec 8 '11 at 18:02
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There are several questions being asked (and an unexplained reference to a field of definition), but the answer to at least one of them is no: Take $G_1 = G_2 = \mathrm{SL}_3(\mathbb{C})$, with a given minimal = maximal parabolic subgroup involving a single simple root subgroup relative to some choice of positive roots; this parabolic clearly won't lie in two distinct maximal ones. (I'm assuming "minimal" excludes a Borel subgroup and "maximal" means proper, though a reductive group might be just a torus.)

So it's a good idea to separate out the more precise question you have in mind and specify whether a field of definition is really involved for the various groups and subgroups involved.

P.S. I've only pointed to the trivial case where there is just one reductive group. But I think your line of questioning will also run into obstacles in situations where the group and subgroup share a common maximal torus but the subgroup is of "pseudo-Levi" type. An example with both $G_1, G_2$ simple of rank 2 occurs when you consider the reductive subgroup of a group of Lie type $G_2$ (in root system notation!) which involves the long roots and has Lie type $A_2$. These pseudo-Levi subgroups are described in terms of proper subsets of vertices in the extended Dynkin diagram which fail to define Levi subgroups of parabolics.

To summarize, there are pairs $G_1 \subset G_2$ and respective parabolic subgroups (all defined over a given subfield of $\mathbb{C}$ such as $\mathbb{Q}$) for which the answers to the various questions asked can be either yes or no. Another example: Take a reductive subgroup $G_1 \cong \mathrm{SL}_2(\mathbb{C})$ of $G_2 = \mathrm{SL}_3(\mathbb{C})$ with a Borel subgroup $P_1$ (involving one simple root $\alpha$) as minimal = maximal parabolic of $G_1$; then $P_1$ and the opposite Borel both lie in the standard maximal parabolic $P_2$ of $G_2$ having $-\alpha$ as a root.

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I suspect that the term "minimal parabolic" is in this case not supposed to rule out a Borel. Since the OP wants to work over $\mathbf{Q}$, in general $G_1$ needn't have a Borel defined over $\mathbf{Q}$, but one can speak of a "minimal $\mathbf{Q}$-parabolic". –  George McNinch Dec 8 '11 at 20:50
    
Sorry, I realize now that it wasn't so clear, but what I wanted to know is that the situations in my first two questions never happen. This would of course follow from an affirmative answer to my last question. And George McNinch is right, 'minimal' is Borel inclusive, but 'maximal' is 'maximal proper'. –  Keerthi Madapusi Pera Dec 8 '11 at 22:44
    
And yes all sub-groups are defined over $\mathbb{Q}$. –  Keerthi Madapusi Pera Dec 8 '11 at 22:46
    
@Keerthi: I'm not yet sure I've understood what your precise questions are, but there seem to be positive and negative answers even in the case of split groups. I've added another type of example, since you allow Borel subgroups to be viewed as minimal parabolics. –  Jim Humphreys Dec 9 '11 at 13:54
    
Thanks for that last example. I think the issue there is that the intersection of $P_2$ with $G_1$ is all of $G_1$. So, as long as we exclude that possibility, the answer to my last question should be 'yes', I think. –  Keerthi Madapusi Pera Dec 9 '11 at 14:17
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