Lemma 1

Let V be a $\mathfrak{g}$-module and $v\in V$. The following conditions are equivalent:

(i) $\mathfrak{b}_+v\subset kv$;

(ii) $\mathfrak{h}v\subset kv$ and $\mathfrak{n}_+v= 0$;

(iii) $\mathfrak{h}v\subset kv$ and $\mathfrak{g}^{\alpha}v= 0$ for all $\alpha \in B$.

(i) $=\Rightarrow$ (ii): Assume that $\mathfrak{b}_+v\subset kv$. There exists $\lambda \in \mathfrak{h}^*$ such that $v\in V^{\lambda}$. Let $\alpha \in R_+$. Then $\mathfrak{g}^{\alpha}.v\subset V^{\lambda}\cap V^{\lambda+\alpha}= 0$. Hence $\mathfrak{n}_+v= 0$.

(ii) $=\Rightarrow$ (iii): This is clear.

(iii) $=\Rightarrow$ (i): This follows from the fact that $(X_{\alpha})_{\alpha\in B}$ generates $\mathfrak{n}_+($§3, no. 3, Prop. 9 (iii)).