Theorem 1

A simple $\mathfrak{g}$-module is finite dimensional if and only if it has a highest weight belonging to $P_{++}$.

We denote by V a simple $\mathfrak{g}$-module and by $\mathscr{X}$ its set of weights.

a) Assume that V is finite dimensional. Then $\mathscr{X}$ is finite and non-empty (Prop. 1) and so has a maximal element $\omega$. Let $\alpha \in B$. Then $\omega +\alpha \notin \mathscr{X}$, which proves that $\omega$ is the highest weight of V (§6, no. 2, Lemma 2). On the other hand, there exists a homomorphism from $\mathfrak{s}\mathfrak{l}(2, k)$ to $\mathfrak{g}$ that takes $H$ to $H_{\alpha}$; by §1, Prop. 2 (i), $\omega (H_{\alpha})$ is an integer $\geq 0$, so $\omega \in P_{++}$.

b) Assume that V has a highest weight $\omega \in P_{++}$. Let $\alpha \in B$ and let $e$ be a primitive element of weight $\omega$ in V. Put $e_j=X_{-\alpha}^je$ for $j\geq 0$,

$m=\omega (H_{\alpha})\in \mathbf{N}$, and N = $\sum_{j=0}^mke_j$. By §1, no. 2, Prop. $1,X_{\alpha}e_{m+1}$ = 0.

If $\beta \in B$ and $\beta \not=\alpha$, then $[X_{\beta}, X_{-\alpha}] = 0$ so $X_{\beta}e_{m+1}=X_{\beta}X_{-\alpha}^{m+1}e$ = $X_{-\alpha}^{m+1}X_{\beta}e= 0$. If $e_{m+1}\not= 0$, we conclude that $e_{m+1}$ is primitive, which is absurd (§6, Prop. 3 (i)); so $e_{m+1}= 0$. Thus, by §1, no. 2, Cor. of Prop. 1, N is stable under the subalgebra $\mathfrak{s}_{\alpha}$ generated by $H_{\alpha}, X_{\alpha}$ and $X_{-\alpha}$. Now $\mathfrak{s}_{\alpha}$ is reductive in $\mathfrak{g}$, so the sum of the finite dimensional subspaces of V that are stable under $\mathfrak{s}_{\alpha}$ is a $\mathfrak{g}$-submodule of V (Chap. I, §6, no. 6, Prop. 7); since this sum is non-zero, it is equal to V. It follows from this that $(X_{\alpha})_V$ and $(X_{-\alpha})_V$ are locally nilpotent. In view of Lemma 1 (ii), $\mathscr{X}$ is stable under $s_{\alpha}$, and this holds for all $\alpha$. Hence $\mathscr{X}$ is stable under W. Now every orbit of W on P meets $P_{++}$ (Chap. VI, §1, no. 10). On the other hand, if $\lambda \in \mathscr{X}\cap P_{++}$, then $\lambda =\omega -\sum_{\alpha\in B}n_{\alpha}\alpha =\sum_{\alpha\in B}n'_{\alpha}\alpha$ with $n_{\alpha}\in \mathbf{N}$ and $n'_{\alpha}\geq 0$ for all $\alpha \in B$

(Chap. V, §3, no. 5, Lemma 6). So $\mathscr{X}\cap P_{++}$ is finite and hence so is $\mathscr{X}$. Since each weight has finite multiplicity (§6, no. 1, Prop. 1 (ii)), V is finite dimensional.