Proposition 2

If the submodule V of E generated by the primitive element $e$ is finite dimensional, then:

(i) the weight $\lambda$ of $e$ is integral and equal to dim $V-1$;

(ii) $(e_0, e_1, . . . , e_{\lambda})$ is a basis of V, and $e_n= 0$ for $n > \lambda$;

(iii) the eigenvalues of $H_V$ are $\lambda , \lambda -2, \lambda -4, . . . ,-\lambda$; they are all of multiplicity 1;

(iv) every primitive element of V is proportional to $e$;

(v) the commutant of the module V is reduced to the scalars; in particular, V is absolutely simple.

Let $m$ be the largest integer such that $e_m\not= 0$. Then $0 =X_+e_{m+1}=$ $(\lambda -m)e_m$, so $\lambda =m$; since $(e_0, e_1, . . . , e_m)$ is a basis of V, this proves (i) and (ii). Assertion (iii) follows from the equality $He_n= (\lambda -2n)e_n$. We have $X_+e_n\not= 0$ for $1\leq n\leq m$, hence (iv). Let $c$ be an element of the commutant of the module V. Then $Hc(e) =cH(e) =\lambda c(e)$, so there exists $\mu\in k$ such that $c(e) =\mu e$; then

for all $q\geq 0$, so $c=\mu.1$, proving (v).