Lemma 2

(i) Let $\mathfrak{h}_r$ be the set of $x\in \mathfrak{h}$ such that $\mathfrak{g}^0(x) =\mathfrak{g}^0(\mathfrak{h})$; this is the set of $x\in \mathfrak{h}$ such that $\lambda (x)\not= 0$ for all $\lambda \in R$, and $\mathfrak{h}_r$ is open and dense in $\mathfrak{h}$ in the Zariski topology.

(ii) Put $R =\{\lambda_1, \lambda_2, . . . , \lambda_p\}$ where the $\lambda_i$ are mutually distinct. Let F be the map from $\mathfrak{g}^0(\mathfrak{h})\times \mathfrak{g}^{\lambda_1}(\mathfrak{h})\times \cdots \times \mathfrak{g}^{\lambda_p}(\mathfrak{h})$ to $\mathfrak{g}$ defined by the formula

Then F is a dominant polynomial map (App. I).

Assertion (i) is clear. We prove (ii). Let $n=$ dim $\mathfrak{g}$. If $\lambda \in R$ and $x\in \mathfrak{g}^{\lambda}(\mathfrak{h})$, we have (ad $x$)$^n= 0$. It follows that $(y, x) \rightarrow e^{adx}y$ is a polynomial map from $\mathfrak{g}\times \mathfrak{g}^{\lambda}(\mathfrak{h})$ to $\mathfrak{g}$; it follows by induction that F is polynomial. Let $h_0\in \mathfrak{h}_r$ and let DF be the tangent linear map of F at $(h_0,0, . . . ,0)$; we show that DF is surjective. For $h\in \mathfrak{g}^0(\mathfrak{h}), F(h_0+h,0, . . . ,0) =h_0+h$, so DF($h,0, . . . ,0$) $=h$ and Im(DF) $\supset \mathfrak{g}^0(\mathfrak{h})$. On the other hand, for $x\in \mathfrak{g}^{\lambda_1}(\mathfrak{h})$,

$F(h_0, x,0, . . . ,0) =e^{adx}h_0=h_0+$ (ad $x$)$.h_0+$ (ad2!$x$)$^2h_0+\cdots$

so DF(0$, x,0, . . . ,0$) $=$ (ad $x$)$.h_0=-$(ad $h_0$)$x$; since ad $h_0$ induces an automorphism of $\mathfrak{g}^{\lambda_1}(\mathfrak{h})$, Im(DF) $\supset \mathfrak{g}^{\lambda_1}(\mathfrak{h})$. Similarly,

Im(DF) $\supset \mathfrak{g}^{\lambda_i}(\mathfrak{h})$

for all $i$, hence the surjectivity of DF. Prop. 4 of App. I now shows that F is dominant.