Let $G$ be a connected, simply-connected complex semisimple linear algebraic group with Lie algebra $\frak{g}$. Fix a maximal torus $T\subseteq G$ and let $\Delta\subseteq Hom(T,\mathbb{C}^\*)$ be the resulting collection of roots. Choose positive and negative roots, $\Delta\_{+}$ and $\Delta\_{-}$, respectively. Let $\Pi\subseteq\Delta\_{+}$ be the set of simple roots. Let $V(\lambda)$ be the irrep of $G$ with highest (dominant) weight $\lambda$. We know that $$V(\lambda)=U(\frak{n}\_-) v\_{\lambda},$$ where $v\_{\lambda}\in V(\lambda)$ is a high-weight vector. 


Is there some characterization of those $\alpha\in\Delta\_{-}$ for which $\{0\}=\frak{g}\_{\alpha}\cdot v\_{\lambda}$? (ie. Which negative root spaces annihilate $v\_{\lambda}$?) I think we might generally obtain more than just those $\alpha\in\Delta\_{-}$ for which $\lambda+\alpha$ is not a weight of $V(\lambda)$. If this is documented in the literature, I would appreciate any references. If not, I would appreciate any thoughts. 



Moving Between Weight Spaces in Highest-Weight Representations
They're the ones perpendicular to $\lambda$.


To see this, note, first off, that $\mathfrak g\_\alpha v\_\lambda \in V\_{\lambda+\alpha}$ so if $g\_\alpha v\_\lambda \neq 0$ then $\lambda+\alpha$ must be a weight of $V$. But then so too would be $s\_\alpha(\lambda+\alpha)=s\_\alpha(\lambda)-\alpha$. Now if $\langle \lambda, \alpha \rangle = 0$, then $s\_\alpha(\lambda)=\lambda$, and we'd get a weight $\lambda-\alpha$ in $V$ higher than $\lambda$. Thus, $\langle \lambda, \alpha \rangle = 0$ implies that $\mathfrak g\_\alpha v\_\lambda = 0$.


Conversely, assume that $\langle \lambda, \alpha \rangle \neq 0$. Let $H\_\alpha \in \mathfrak t$ be such that $\lambda(H\_\alpha) = \langle \lambda, \alpha \rangle$ and let $X\_{\pm \alpha} \in \mathfrak g\_{\pm\alpha}$ be such that $[X\_{-\alpha},X\_\alpha]=H\_\alpha$, as usual (except keep in mind that here $\alpha <0$!). Then I claim that $X\_\alpha v\_\lambda \neq 0$ and so $\mathfrak g\_\alpha v\_\lambda \neq 0$. Indeed, for otherwise we'd get $$ 0 = X\_{-\alpha}X\_\alpha v\_\lambda = [X\_{-\alpha},X\_\alpha]v\_\lambda + X\_\alpha X\_{-\alpha}v\_\lambda = H\_\alpha v\_\lambda = \langle \lambda, \alpha \rangle v\_\lambda \neq 0, $$
where I've used the fact that $X\_{-\alpha}v\_\lambda = 0$, because $-\alpha>0$.


