Proposition 1

Let V be a $\mathfrak{g}$-module, $v$ a primitive element of V and $\omega$ the weight of $v$. Assume that V is generated by $v$ as a $\mathfrak{g}$-module.

(i) If $U(\mathfrak{n}_-)$ denotes the enveloping algebra of $\mathfrak{n}_-$, we have $V = U(\mathfrak{n}_-).v$.

(ii) For all $\lambda \in \mathfrak{h}^*,V^{\lambda}$ is the set of $x\in V$ such that $Hx=\lambda (H)x$ for all $H\in \mathfrak{h}$. We have $V =\bigoplus_{\lambda\in\mathfrak{h}^*}V^{\lambda}$, and each $V^{\lambda}$ is finite dimensional. The space

$V^{\omega}$ is of dimension 1, and every weight of V is of the form $\omega -\sum_{\alpha\in B}n_{\alpha}.\alpha$, where the $n_{\alpha}$ are integers $\geq 0$.

(iii) V is an indecomposable $\mathfrak{g}$-module, and its commutant reduces to the scalars.

(iv) Let $U(\mathfrak{g})$ be the enveloping algebra of $\mathfrak{g}$, and $\mathscr{Z}$ the centre of $U(\mathfrak{g})$. There exists a unique homomorphism $\chi$ from $\mathscr{Z}$ to $k$ such that, for all $z\in \mathscr{Z}$, $z_V$ is the homothety with ratio $\chi (z)$.

Let $U(\mathfrak{b}_+)$ be the enveloping algebra of $\mathfrak{b}_+$. We have $U(\mathfrak{g}) = U(\mathfrak{n}_-).U(\mathfrak{b}_+)$ (Chap. I, §2, no. 7, Cor. 6 of Th. 1). Hence

Denote by $\alpha_1, . . . , \alpha_n$ the distinct elements of $R_+$. Then

is a basis of $U(\mathfrak{n}_-)$, so

V = $\sum kX_{-\alpha_1}^{p_1}. . . X_{-\alpha_n}^{p_n}v$. (1)

$(p_1,...,p_n)\in \mathbf{N}^n$

For $\lambda \in \mathfrak{h}^*$, put

$(p_1,...,p_n)\in \mathbf{N}^n, \omega -p_1\alpha_1-\cdot \cdot \cdot -p_n\alpha_n=\lambda$

By Chap. VII, §1, no. 1, Prop. 2 (ii), if $h\in \mathfrak{h},h_V|T_{\lambda}$ is the homothety with ratio $\lambda (h)$. So $T_{\lambda}\subset V^{\lambda}$. On the other hand, (1) implies that

V = $\sum T_{\lambda}$.

$\lambda \in \omega -\mathbf{N}\alpha_1-\cdot \cdot \cdot -\mathbf{N}\alpha_n$

The sum of the $V^{\lambda}$ is direct (Chap. VII, §1, no. 1, Prop. 3). From these observations it follows that $V^{\lambda}= T_{\lambda}$, that V is the direct sum of the $V^{\lambda}$, and that $V^{\lambda}$ is the set of $x\in V$ such that $hx=\lambda (h)x$ for all $h\in \mathfrak{h}$. On the other hand, dim $V^{\lambda}$ is at most the cardinal of the set of $(p_1, . . . , p_n)\in \mathbf{N}^n$ such that $p_1\alpha_1+\cdots +p_n\alpha_n=\omega -\lambda$. This proves that $V^{\lambda}= 0$ if $\omega -\lambda \notin \sum_{\alpha\in B}\mathbf{N}\alpha$,

that dim $V^{\omega}= 1$, and that the $V^{\lambda}$ are all finite dimensional.

Let $c$ be an element of the commutant of V. For all $h\in \mathfrak{h}$,

so $c(v)\in V^{\omega}$; hence there exists $t\in k$ such that $c(v) =tv$. Now, for all $(p_1, . . . , p_n)\in \mathbf{N}^n$,

so that $c=t.1$. Hence, the commutant of V reduces to the scalars. This implies (iv) and the fact that V is indecomposable.