How to determine a highest weight corresponding to a parabolic subgroup?

Question

Let $G$ be a simply connected, semisimple algebraic group over $\mathbb C$ with maximal torus $T$ and Borel subgroup $B$ containing $T$. If $(V,\pi)$ is an irreducible representation of $G$, then $(V,d\pi)$ is an irreducible representation of the Lie algebra $\mathfrak g$ which has a unique highest weight $\lambda \in \mathfrak t^{\ast}$. I have read that if we identify the one-dimensional weight space $V_{\lambda} \subset V$ with a point in projective space $\mathbb P(V)$, then under the action $$ G \xrightarrow{\pi} \operatorname{GL}(V) \rightarrow \operatorname{Aut}(\mathbb P(V))$$ the stabilizer of $V_{\lambda}$ is a parabolic subgroup of $G$, and every parabolic subgroup of $G$ arises this way.

How does one go in the opposite direction? If $P$ is a (let's say maximal) parabolic standard subgroup of $G$, how does one find a dominant integral weight whose corresponding irreducible representation determines $P$ in the above sense? Can the highest weight occur as the highest root of $T$ in the unipotent radical of $P$?

Answer 1

This is done in e.g. Baston-Eastwood (1989, pp. 40, 55): with $P$ characterized as usual by a subset $\mathcal S_{\mathfrak p}$ of the simple roots (for a maximal parabolic, $\mathcal S_{\mathfrak p}$ is all but one simple root), take $\lambda=$ sum of the fundamental weights $\varpi_i$ corresponding to simple roots $\alpha_i$ not in $\mathcal S_{\mathfrak p}$. Then $G/P$ is the coadjoint orbit of $\lambda$ under the compact real form, and $V$ is its geometric quantization.

$\lambda$ can be the highest root (giving $V=$ adjoint representation) but then $P$ is usually not maximal.