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?

$\begingroup$ 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 CohenCooperstein, "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 $F_4$. $\endgroup$– Jason StarrSep 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. $\endgroup$– Robert BryantSep 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 $SO(16)$. $\endgroup$– Robert BryantSep 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. $\endgroup$– Jason StarrSep 17, 2011 at 3:05