Lemma 4

Let U be a neighbourhood of 0 in $\mathfrak{g}$ and exp an exponential map from U to G, étale at every point of U and such that $\mathfrak{g}^1$(exp $x$) $=\mathfrak{g}^0(x)$ for all $x\in U$.

(i) The function $r^0_{Ad}$ is constant and equal to the rank of $\mathfrak{g}$ on exp(U).

(ii) If $x\in U$, exp $x$ is regular if and only if $x$ is a regular element of $\mathfrak{g}$.

(iii) An element $a\in$ exp(U) is regular if and only if $\mathfrak{g}^1(a)$ is a Cartan subalgebra of $\mathfrak{g}$.

Let $l=$ rk($\mathfrak{g}$). If $x\in U$ is a regular element of $\mathfrak{g}$,

$r_{Ad}$(exp $x$) $=$ dim$\mathfrak{g}^1$(exp $x$) $=$ dim$\mathfrak{g}^0(x) =l$.

Since the regular elements of $\mathfrak{g}$ belonging to U constitute a neighbourhood of $x$ and exp is étale at $x$, this shows that exp $x$ is regular and that $r^0_{Ad}$(exp $x$) $=l$. The regular elements of $\mathfrak{g}$ belonging to U being dense in U, we have $r^0_{Ad}(a) =$ $l$ for all $a\in$ exp(U). Let $a\in$ exp(U) be a regular element of G and let $x\in U$ be such that $a=$ exp $x$. Since $\mathfrak{g}^0(x) =\mathfrak{g}^1(a)$, dim $\mathfrak{g}^0(x) =r^0_{Ad}(a) =l$. Consequently, $x$ is a regular element of $\mathfrak{g}$ and $\mathfrak{g}^1(a)$ is a Cartan subalgebra of $\mathfrak{g}$. Finally, if $a\in$ exp(U) and $\mathfrak{g}^1(a)$ is a Cartan subalgebra of $\mathfrak{g}$,

$r_{Ad}(a) =$ dim$\mathfrak{g}^1(a) =l=r_{Ad}^0(a)$,

so $a$ is regular.