Corollary 3

Let $v\in V$ be primitive of weight $\omega$. Let $\Sigma$ be the set of $\alpha \in B$ such that $\omega (H_{\alpha}) = 0$. The stabilizer in $\mathfrak{g}$ of the line $kv$ is the parabolic subalgebra $\mathfrak{p}_{\Sigma}$ associated to $\Sigma ($§3, no. 4, Remark).

Replacing V by the $\mathfrak{g}$-submodule generated by $v$, if necessary, we can assume that V is simple. Let $\mathfrak{s}$ be the stabilizer. We have $(\mathfrak{n}_+)_Vv$ = 0, $(\mathfrak{h})_Vv\subset kv$. Let $\alpha \in B$ be such that $\omega (H_{\alpha}) = 0$. We have $\omega +\alpha \notin \mathscr{X}$, hence $\omega -\alpha \notin \mathscr{X}$ (Prop. 3 (i)) and consequently $(\mathfrak{g}^{-\alpha})_Vv= 0$. The preceding proves that $\mathfrak{p}_{\Sigma}\subset \mathfrak{s}$. If $\mathfrak{p}_{\Sigma}\not=\mathfrak{s}$, then $\mathfrak{s}=\mathfrak{p}_{\Sigma'}$, where $\Sigma '$ is a subset of B strictly containing $\Sigma$. Let $\beta \in \Sigma '$ -- $\Sigma$. Then $\mathfrak{g}^{-\beta}$ stabilizes $kv$, and hence annihilates $v$. But, since $\omega (H_{\beta})>0$, this contradicts Prop. 3 (iii). Q.E.D.

A subset $\mathscr{X}$ of P is called R-saturated if it satisfies the following condition: for all $\lambda \in \mathscr{X}$ and all $\alpha \in R$, we have $\lambda -t\alpha \in \mathscr{X}$ for all integers $t$ between 0 and $\lambda (H_{\alpha})$. Since $s_{\alpha}(\lambda ) =\lambda -\lambda (H_{\alpha})\alpha$, we see that an R-saturated subset of P is stable under W. Let $\mathscr{Y}\subset P$. An element $\lambda$ of $\mathscr{Y}$ is called R-extremal in $\mathscr{Y}$ if, for all $\alpha \in R$, either $\lambda +\alpha \notin \mathscr{Y}$ or $\lambda -\alpha \notin \mathscr{Y}$.