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I'm interested in the following question: let $k$ be a field of characteristic zero (just for simplicity), $G$ a connected semi-simple $k$-group, $P\subsetneq G$ a parabolic $k$-subgroup, and $H\subsetneq G$ a connected $k$-subgroup. Assume that $H\nsubseteq P$, then does the intersection $H\cap P$ becomes a parabolic $k$-subgroup for $H$? One might even assume that $H$ is reductive.

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This is true precisely when the $H$-orbit of the base point of $G/P$ is closed which is usually not the case. (If it is the case for all conjugates of a given $P$ then all the $H$-orbits are closed which is very rarely the case. –  Torsten Ekedahl Dec 27 '10 at 15:56
    
What Torsten points out is true but of course hard to test even if the field is algebraically closed. The problem with arbitrary closed subgroups $H$ (reductive or not) is that they can be extremely small relative to $G$ and not well-placed in terms of Lie-theoretic structure. Think of $G$ as a huge special linear group, for instance. In other words, your question is too loosely formulated to have an interesting answer. –  Jim Humphreys Dec 27 '10 at 16:11

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