Lemma 2

Let V be a simple $\mathfrak{g}$-module, $\omega$ a weight of V. The following conditions are equivalent:

(i) every weight of V is of the form $\omega -\mu$ where $\mu$ is a radical weight $\geq 0$;

(ii) $\omega$ is the highest weight of V;

(iii) for all $\alpha \in B,\omega +\alpha$ is not a weight of V;

(iv) there exists a primitive element of weight $\omega$.

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

(iii) $=\Rightarrow$ (iv): Assume that condition (iii) is satisfied. For all $h\in \mathfrak{h}$,

Ker($h_V-\omega (h)$)

is non-zero, contained in $V^{\omega}$, and stable under $\mathfrak{h}_V$. By induction on dim $\mathfrak{h}$, we see that there exists a non-zero $v$ in $V^{\omega}$ such that $\mathfrak{h}v\subset kv$. Condition (iii) implies that $\mathfrak{n}_+v= 0$, so $v$ is primitive.

(iv) $=\Rightarrow$ (i): Let $v$ be a primitive element of weight $\omega$. Since V is simple, V is generated by $v$ as a $\mathfrak{g}$-module. Assertion (i) now follows from Prop. 1.

Q.E.D.

Thus, for any simple $\mathfrak{g}$-module, the existence of a primitive element is equivalent to that of a highest weight, or to that of a maximal weight. There exist simple $\mathfrak{s}\mathfrak{l}(2,\mathbf{C}$)-modules V that have no weights for any Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{s}\mathfrak{l}(2,\mathbf{C}) ($§1, Exerc. $14f))$. These modules are of infinite dimension over $\mathbf{C}($§1, no. 3, Th. 1).