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Does there exist (as elementary as possible) a generalized flag manifold (e.g. sphere, projective space, Grassmanian, etc.) of a simple classical Lie group (SL, SO or Sp over real or complex field), such that $F_4$ is acting transitively on it with stabilizer given by parabolic subgroup? In other words, is there a Lie group G (equal to SL, SO or Sp) and a parabolic subgroup $P< G$, such that G/P is equal to $F_4/Q$, where Q is parabolic subgroup of $F_4$ given by intersection of $F_4$ with P?

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  • $\begingroup$ I don't think so. Probably this can be deduced from the tables in Bourbaki. But one reference which I happen to have on my desk is Cohen-Cooperstein, "Lie Incidence Systems ..." If you look at their table, it seems that the type of coincidence you are suggesting does occur for $G_2$, but does not occur for $F_4$. $\endgroup$ Commented Sep 6, 2011 at 18:14
  • $\begingroup$ Your 'in order words' reinterpretation is not really a reinterpretation. For example, $F_4^{(-20)}$ acts transitively on the $15$-sphere (the boundary of the octonic hyperbolic plane), which is a generalized flag manifold of $G$ any one of $SL(16,\mathbb{R})$, $SO(16)$, or $Sp(8,\mathbb{R})$, and the stabilizer subgroup of this action is a (maximal) parabolic subgroup. However, none of these groups $G$ contain $F_4$ as a subgroup, so there is no equality $F_4/Q = G/P$ in the sense of your second sentence. $\endgroup$ Commented Sep 8, 2011 at 0:35
  • $\begingroup$ I'm sorry about the typos in the comment above, but there doesn't appear to be a way to fix them: "in order words" $\mapsto$ "in other words", "$F_4$" means "$F_4^{(-20)}$" near the end, and I should have pointed out that this group also does not appear as a subgroup of $SO(17,1)$, which is the other maximal, connected, finite dimensional Lie subgroup of $Diff(S^{15})$ that contains $SO(16)$. $\endgroup$ Commented Sep 9, 2011 at 14:56
  • $\begingroup$ In the original question, was everything complex? I remember this as being a question about complex Lie groups from when I commented earlier. $\endgroup$ Commented Sep 17, 2011 at 3:05

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