Proposition 8

Let $l$ be the rank of $\mathfrak{g}$. There exists an open subgroup $G^*$ of G such that:

(i) the function $r^0_{Ad}$ is constant on $G^*$ and its value is $l$;

(ii) an element $a\in G^*$ is regular if and only if $\mathfrak{g}^1(a)$ is a Cartan subalgebra of $\mathfrak{g}$;

(iii) if $a\in G^*$, every Cartan subalgebra of $\mathfrak{g}^1(a)$ is a Cartan subalgebra of $\mathfrak{g}$.

(i) By Prop. 6, there exists an open neighbourhood U of 0 in $\mathfrak{g}$ and an exponential map exp from U to G satisfying the conditions of Lemma 4. In what follows, $G^*$ will denote the identity component of G if $k=\mathbf{R}$ or $\mathbf{C}$ and an open subgroup of G contained in exp(U) if $k$ is ultrametric. Since $r_{Ad}^0$ is locally constant and its value at any point of exp(U) is $l$ (Lemma 4 (i)), it follows that $r_{Ad}^0$ is constant and equal to $l$ on $G^*$.

(ii) Let $R^*$ (resp. $S^*$) be the set of regular elements of $G^*$ (resp. the set of elements $a\in G^*$ such that $\mathfrak{g}^1(a)$ is a Cartan subalgebra of $\mathfrak{g}$). Then $S^*\subset R^*$. Indeed, if $a\in S^*$, then $r_{Ad}(a) =l=r^0_{Ad}(a)$. We show that $R^*\subset S^*$. If $k$ is ultrametric, this follows from the inclusion $G^*\subset$ exp(U) and Lemma 4 (iii). Assume that $k=\mathbf{C}$. By the Cor. of Prop. 5, if $a\in R^*$, then for every $a'$ belonging to a neighbourhood of $a,\mathfrak{g}^1(a')$ is conjugate to $\mathfrak{g}^1(a)$ by an automorphism of $\mathfrak{g}$. This proves that $S^*$ and $R^*$ -- $S^*$ are open subsets of $G^*$. We have seen that $S^*$ contains all the regular elements in a neighbourhood of $e$ (Lemma 4 (iii)); consequently, $S^*$ is non-empty. Since $G^*$ is connected, so is $R^*$ (Prop. 1) and consequently $S^*= R^*$.

It remains to study the case $k=\mathbf{R}$. Assume first of all that $G^*$ is an integral subgroup of $\mathbf{G}\mathbf{L}(E)$ where E denotes a finite dimensional real vector space. Let $G^*_c$ be the integral subgroup of $\mathbf{G}\mathbf{L}(E\otimes_{\mathbf{R}}\mathbf{C})$ with Lie algebra $\mathfrak{g}_c=\mathfrak{g}\otimes \mathbf{C}$. There exists an analytic function on $G^*_c$ whose set of zeros is the complement of the open set of regular elements of $G^*_c$. By Differentiable and Analytic Manifolds, Results, 3.2.5, this function cannot vanish at every point of $G^*$. Consequently, $G^*$ contains a regular element of $G^*_c$. Let Ad$_c$ be the adjoint representation of $G^*_c$. For any $a\in G^*,\mathfrak{g}^1_c(a) =\mathfrak{g}^1(a)\otimes \mathbf{C}$, so $r_{Ad_c}(a) =r_{Ad}(a)$. If $a\in G^*$ is a regular element of $G^*_c$, this is a regular element of $G^*$ and $r_{Ad}^0(a) =r_{Ad}^0(a)$. The functions $r^0_{Ad}$ and $r_{Ad}^0$ being constant on $G^*_c$ and on $G^{^c*}$, respectively, it follows that the regular elements$^{^c}$ of $G^*$ are the regular elements of $G^*_c$ belonging to $G^*$. From the above, if $a$ is a regular element of $G^*,\mathfrak{g}^1_c(a) =\mathfrak{g}^1(a)\otimes \mathbf{C}$ is a Cartan subalgebra of $\mathfrak{g}_c$; this implies that $\mathfrak{g}^1(a)$ is a Cartan subalgebra of $\mathfrak{g}($§2, Prop. 3).

Assume now that G is simply connected. There exists a finite dimensional real vector space E and an étale morphism $f$ from G to an integral subgroup $G'$ of $\mathbf{G}\mathbf{L}(E)$ (Chap. III, §6, no. 1, Cor. of Th. 1). By Prop. 2, if $a\in G$ is regular, $f(a)$ is regular. By the preceding, ${\mathfrak{g}'}^1(f(a))$ is a Cartan subalgebra of the Lie algebra $\mathfrak{g}'$ of $G'$. Since ${\mathfrak{g}'}^1(f(a)) = (Tf)\mathfrak{g}^1(a)$ and $Tf$ is an isomorphism from $\mathfrak{g}$ to $\mathfrak{g}'$, this proves that $\mathfrak{g}^1(a)$ is a Cartan subalgebra of $\mathfrak{g}$.

We turn finally to the general case $(k=\mathbf{R})$. Let $\widetilde{G}$ be a universal covering of $G^*, \widetilde{\mathfrak{g}}= L( \widetilde{G})$, and $q$ the canonical map from $\widetilde{G}$ to $G^*$. Since the kernel of $q$ is contained in the centre of $\widetilde{G}$, if $a\in G^*$ is regular and if $a'\in q^{-1}(a)$, then $a'$ is regular (Prop. 2). By the preceding, $\widetilde{\mathfrak{g}}^1(a')$ is a Cartan subalgebra of $\widetilde{\mathfrak{g}}$. Since $\mathfrak{g}^1(a) = (Tq)\widetilde{\mathfrak{g}}^1(a')$ and since $Tq$ is an isomorphism from $\widetilde{\mathfrak{g}}$ to $\mathfrak{g}$, this proves that $\mathfrak{g}^1(a)$ is a Cartan subalgebra of $\mathfrak{g}$.

(iii) By Prop. 5, there exists a neighbourhood V of $a$ such that, for all $a'\in V,\mathfrak{g}^1(a')$ is conjugate to a subalgebra of $\mathfrak{g}^1(a)$ by an automorphism of $\mathfrak{g}$. Since every neighbourhood of $a$ contains a regular element of $G^*$, it follows from (ii) that $\mathfrak{g}^1(a)$ contains a Cartan subalgebra of $\mathfrak{g}$. Thus, by Prop. 3 of §3, every Cartan subalgebra of $\mathfrak{g}^1(a)$ is a Cartan subalgebra of $\mathfrak{g}$.