Corollary 1

Put $\mathscr{G}=\mathscr{U}\cap \mathfrak{g}$. Then $\mathscr{G}$ is an order in the Lie algebra $\mathfrak{g}$, stable under $\theta$. We have $\mathscr{G}=\mathscr{H}+\sum_{\alpha\in R}(\mathscr{G}\cap \mathfrak{g}^{\alpha})$. For all $\alpha \in B$ and all $n\in \mathbf{N}$, the maps (ad $x_{\alpha}$)$^n/n$!, (ad $y_{\alpha}$)$^n/n$! leave $\mathscr{U}$ and $\mathscr{G}$ stable.

The first assertion is clear. The second follows by considering the graduation of type Q(R) on $U(\mathfrak{g})$ and $\mathscr{U}$. The third follows from Lemma 2 of no. 5.