Lemma 9

With the notations of Prop. 8, put $\mathfrak{g}^p=$ Ker(ad $h^0-p$) for $p\in \mathbf{Z}$. Let $\mathfrak{g}^2_*$ be the set of elements of $\mathfrak{g}^2=\sum_{\alpha\in B}\mathfrak{g}^{\alpha}$ that have a non-zero component

in each $\mathfrak{g}^{\alpha}$. Let $R_+$ be the set of positive roots relative to B$,\mathfrak{n}_+=\sum_{\alpha\in R_+}\mathfrak{g}^{\alpha}$, and $x\in \mathfrak{g}^2_*$. Then $e^{ad\mathfrak{n}_+}.x=x+ [\mathfrak{n}_+,\mathfrak{n}_+]$.

It is clear that $e^{ad\mathfrak{n}_+}.x\subset x+ [\mathfrak{n}_+,\mathfrak{n}_+]$. We prove that, if $v\in [\mathfrak{n}_+,\mathfrak{n}_+]$, then $x+v\in e^{ad\mathfrak{n}_+}.x$. Put $\mathfrak{n}^{(p)}=\sum_{r\geq p}\mathfrak{g}^{2r}$; it suffices to prove that

for all $p\geq 2$. This is clear for $p= 2$ since $\mathfrak{n}^{(2)}$ $= [\mathfrak{n}_+,\mathfrak{n}_+] ($§3, no. 3, Prop. 9 (iii)). Assume that we have found $z\in \mathfrak{n}_+$ such that $v+x-e^{adz}.x\in$ $\mathfrak{n}^{(p)}$. Since there exists an $\mathfrak{s}\mathfrak{l}_2$-triplet of the form $(x, h^0, y)$ (Prop. 8), §1, no. 2, Cor. of Prop. 2 proves that $[x,\mathfrak{g}^{2p-2}] =\mathfrak{g}^{2p}$; hence, there exists $z'\in \mathfrak{g}^{2p-2}\subset \mathfrak{n}_+$ such that

So $v+x\in e^{ad(z+z')}.x+\mathfrak{n}^{(p+1)}$, and our assertion is established by induction.