Proposition 6

Assume that $k$ is $\mathbf{R},\mathbf{C}$ or a non-discrete complete ultrametric field of characteristic 0. Let $G$ be a finite dimensional Lie group over $k,e$ its identity element, $\mathfrak{g}$ its Lie algebra, $\mathfrak{h}$ a Cartan subalgebra of $\mathfrak{g},\mathfrak{h}_r$ the set of regular elements of $\mathfrak{g}$ belonging to $\mathfrak{h}$.

(i) Let $\mathfrak{s}$ be a vector space complement of $\mathfrak{h}$ in $\mathfrak{g},\mathfrak{s}_0$ a neighbourhood of 0 in $\mathfrak{s}$ on which an exponential map is defined, and $h_0\in \mathfrak{h}_r$. The map $(s, h) \rightarrow F(s, h) =$ (exp ad $s$)$.h$ from $\mathfrak{s}_0\times \mathfrak{h}$ to $\mathfrak{g}$ is étale at $(0, h_0)$.

(ii) The map $(g, h) \rightarrow F'(g, h) =$ (Ad $g$)$.h$ from $G\times \mathfrak{h}_r$ to $\mathfrak{g}$ is a submersion. In particular, its image $\Omega$ is open. For all $x\in \Omega ,\mathfrak{g}^0(x)$ is a Cartan subalgebra of $\mathfrak{g}$ conjugate to $\mathfrak{h}$ under Ad(G).

(iii) Let $h_0\in \mathfrak{h}_r$. For any neighbourhood U of $e$ in G, the set $\bigcup_{a\in U}$(Ad $a$)$(\mathfrak{h}_r)$

is a neighbourhood of $h_0$ in $\mathfrak{g}$.

Let $h_0$ and $\mathfrak{s}$ be as in (i). Let T be the tangent linear map of F at $(0, h_0)$. Then $F(0, h) =h$ for all $h\in \mathfrak{h}$, so $T(0, h) =h$ for all $h\in \mathfrak{h}$. On the other hand, for $\mathfrak{s}_0$ sufficiently small, the tangent linear map at 0 of the map $s \rightarrow$ exp ad $s$ from $\mathfrak{s}_0$ to End($\mathfrak{g}$) is the map $s \rightarrow$ ad $s$ from $\mathfrak{s}$ to End($\mathfrak{g}$). Thus $T(s,0) = [s, h_0]$ for all $s\in \mathfrak{s}$. Now the map from $\mathfrak{g}/\mathfrak{h}$ to $\mathfrak{g}/\mathfrak{h}$ induced by ad $h_0$ by passage to the quotient is bijective. It follows that T is bijective, hence (i). Since exp ad $s=$ Ad exp $s$ for all $s\in \mathfrak{s}$ sufficiently close to 0, (iii) and the first assertion of (ii) follow. Every $x\in \Omega$ is of the form (Ad $a$)$(h)$ with $a\in G$ and $h\in \mathfrak{h}_r$, so $\mathfrak{g}^0(x) =$ (Ad $a$)$(\mathfrak{g}^0(h)) =$ (Ad $a$)$(\mathfrak{h})$ is a subalgebra of $\mathfrak{g}$ conjugate to $\mathfrak{h}$ under Ad(G).