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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.

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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 $-\alpha>0$.

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  • $\begingroup$ 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 $G/P$. $\endgroup$
    – Vít Tuček
    Commented Oct 30, 2013 at 8:26
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    $\begingroup$ 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 $\lambda$. $\endgroup$
    – Peter Crooks
    Commented Oct 30, 2013 at 13:27
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    $\begingroup$ Peter's formulation is the most natural and useful one for geometric purposes, emphasizing the role of the standard parabolic subgroup (or Lie subalgebra). Note that all of this works uniformly for any semisimple algebraic group (or its Lie algebra) over an algebraically closed field, since the highest weight theory is the same up to a point. Also, a good reference is Bourbaki, Groupes et algebfes de Lie, Chap. VIII, 7.2, Cor. 3 of Prop. 3. Jantzen's book on representations of algebraic groups has similar ideas. $\endgroup$
    – Jim Humphreys
    Commented Oct 30, 2013 at 23:49
  • $\begingroup$ @Peter Crooks: You are right of course. I apologize for my sloppiness. $\endgroup$
    – Vít Tuček
    Commented Oct 31, 2013 at 23:51
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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.

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