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I want to know given a connected (maybe we can assume it to be simply connected or linear) real semisimple lie group $G$ and one of its maximal parabolic group $P$, how can we embed the flag variety $G/P$ into some projective space?

If the lie algebra g is a split real form of its complexfication, then given any fundamental weight $\omega_i$ corresponding to the root $\alpha_i$, we can find the real irreducible highest weight module $V(\omega_i)$ such that $G$ can act on $\mathbb{P}(V(\omega_i))$, and the flag variety $G/P_{\alpha_i}$ can be identified with the orbit of the highest weight subspace $[\eta_{\omega_i}]$.

But if $g$ is not a split real form and what we know is a Cartan involution on it together with the restricted root space decomposition. I do not know how to construct a similar fundamental representation with respect to restricted roots.

There is one example one can deal with. Consider the lie group $\operatorname{SO}(p,q)$ where $p<q$ so its lie algebra is not a split real form. The cartan involution is sending $A\to A^{-T}$ on group level, so the maximal abelian is of rank $p$, and the restricted root system is of type $B_{p}$. The matrix group can act on $\mathbb{R}^{p,q}$. And indeed it acts transitively on isotropic lines in $\mathbb{R}^{p,q}$ and the stabilizer is the parabolic group corresponding to the first fundamental weight as usual. And after taking wedges of $k$ times ($k<p$), we can identify the flag varieties as isotropic $k$ grassmannians and embed it to the $kth$ wedge space which corresponds to the $kth$ fundamental representation.

But in general I do not know how to deal with it. And indeed what I concern is if we consider a real form of $E_6$ (or $E_7,E_8$) whose restricted root system is of type $F_4$, and pick $\alpha_1$ the first root in the restricted Dynkin diagram with $\omega_1$ the fundamental weight, can we find a representation $V(\omega_1)$ whose weight lattice can be described, and we can embed the first flag variety? Hope this clarifies.

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  • $\begingroup$ If $\alpha_i$ is a valid simple root to define a maximal parabolic in $G$ (i.e. a white node with no arrows on the Satake diagram) then $V(\omega_i)$ has a real highest weight space and its orbit is still the flag manifold. I wouldn't necessarily want to describe this in terms of the restricted roots instead though. Much easier to talk in terms of the full Dynkin diagram or the Satake diagram. $\endgroup$
    – Callum
    Commented Sep 23 at 12:05
  • $\begingroup$ @Callum Hi, can you show me a reference about this? Thanks! $\endgroup$
    – fffmatch
    Commented Sep 23 at 13:21
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    $\begingroup$ I've been meaning to write a more full answer for this but haven't had the time yet. So I will just correct my comment for the time being. I think we might not be guaranteed a real representation in fact but we are at least guaranteed a quaternionic one and we can take the exterior square or symmetric square if needs be to get a real representation. I don't have a good reference for any of these though I'm afraid $\endgroup$
    – Callum
    Commented Sep 27 at 12:45

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If $P$ is any parabolic subgroup in a semisimple real Lie group $G$, one can construct a $G$-equivariant embedding of the partial flag manifold $G/P$ into some (high dimensional) real projective space $\mathbb R\mathrm P^n$ as follows. Consider the adjoint representation $\operatorname{ad}$ of $G$ on its Lie algebra $\mathfrak g$. Let $\mathfrak p\subseteq\mathfrak g$ denote the Lie algebra of $P$. Since parabolics are self-normalizing, $P$ is the $G$-stabilizer of the linear subspace $\mathfrak p\subseteq\mathfrak g$. Now put $V=\bigwedge^k\mathfrak g$, where $k$ is the dimension of $\mathfrak p$. Then $G$ acts via $\bigwedge^k\operatorname{ad}$ on $V$ and the stabilizer of the one-dimensional linear subspace $L=\bigwedge^k\mathfrak p\subseteq\bigwedge^k\mathfrak g=V$ is precisely $P$.

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    $\begingroup$ Neat trick! This is in the spirit of Chevalley's proof that every Zariski-closed subgroup of $G$ arises as the stabilizer of a line in some finite-dimensional representation. $\endgroup$
    – LSpice
    Commented Sep 27 at 13:48
  • $\begingroup$ You can make this slightly lower dimensional by considering the nilradical $\mathfrak{p}^\perp \leq \mathfrak{p}$ instead $\endgroup$
    – Callum
    Commented Sep 28 at 14:10

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