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

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 $$P$$?

## 1 Answer

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.