Theorem 4

Let E be a finite dimensional $\mathfrak{g}$-module. The following conditions are equivalent:

(i) E has an admissible lattice;

(ii) every element of $\mathscr{H}$ defines a diagonalizable endomorphism of E with integer eigenvalues.

(i) $=\Rightarrow$ (ii): this follows from Remark 3.

(ii) $=\Rightarrow$ (i): we assume that condition (ii) is satisfied and prove (i). By Th. 4 of Chap. I, §6, no. 5, we can assume that the elements of $\mathfrak{c}$ define homotheties of E, and that E is a simple $\mathscr{D}\mathfrak{g}$-module. Let B be a basis of R, and $\mathfrak{g}=\mathfrak{n}_-\oplus \mathfrak{h}\oplus \mathfrak{n}_+$ the corresponding decomposition of $\mathfrak{g}$. Let $\lambda$ be the highest weight of the $\mathscr{D}\mathfrak{g}$-module E, and let $e\in E^{\lambda}$ -- $\{0\}$. Put $\mathscr{E}=\mathscr{U}.e$. It is clear that $\mathscr{U}.\mathscr{E}\subset \mathscr{E}$. Since E is simple, $U(\mathfrak{g}).e= E$ and hence $\mathscr{E}$ generates E as a $\mathbf{Q}$-vector space. For $h\in \mathscr{H}$ and $n\in \mathbf{N}$, we have $(^h_n)e=(^{\lambda(h)}_n)e\in \mathbf{Z}e$, so

Since $U(\mathfrak{n}_+).e= 0$, we have $\mathscr{E}= (\mathscr{U}\cap U(\mathfrak{n}_-)).e$ by Prop. 3. It now follows from Th. 3 (iii) that $\mathscr{E}$ is a $\mathbf{Z}$-module of finite type.