# Moving Between Weight Spaces in Highest-Weight Representations

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.

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 $-\alpha>0$.
• Let me just add that the subalgebra of $\mathfrak{g}$ that stabilizes the highest weight vector is generated precisely by these roots and hence it is a standard parabolic subalgebra whose Levi part is given precisely by the roots orthogonal to $\lambda$. Since $\mathfrak{g}$ acts transitively on highest weight vectors (pretty much by definition), it's only a small step to see that the orbit of highest weight vectors in $\mathbb{P}V(\lambda)$ is $G/P$. Oct 30, 2013 at 8:26
• I think it is the Lie algebra of the $G$-stabilizer of $[v_{\lambda}]\in\mathbb{P}(V(\lambda))$ that is this standard parabolic. I ask because the $\frak{g}$-stabilizer of $v_{\lambda}\in V(\lambda)$ does not contain all of the Cartan subalgebra. Only the kernel of $\lambda$ belongs to this stabilizer. I definitely agree with your conclusion that the $G$-orbit of $[v_{\lambda}]\in\mathbb{P}(V(\lambda))$ is $G/P$, where $P$ is the parabolic determined by the negative roots orthogonal to $\lambda$. Oct 30, 2013 at 13:27
Nope, it's exactly the ones where $\lambda+\alpha$ is not a weight. Think about this in terms of how the $\mathfrak{sl}_2$ generated by $\mathfrak{g}_{\pm \alpha}$ acts; $\mathfrak{g}_{\alpha}$ is only going to kill the highest weight vector if it has weight 0 for this $\mathfrak{sl}_2$, which is exactly when $\lambda+\alpha$ is not a weight.