Proposition 6

Let V be an open subgroup of $\mathfrak{g}$ and let exp$: V\rightarrow G$ be an exponential map defined on V.

(i) There exists a neighbourhood W of 0 in V such that $\mathfrak{g}^1$(exp $x$) $=\mathfrak{g}^0(x)$ for all $x\in W$.

(ii) If $k=\mathbf{R}$ or $\mathbf{C},\mathfrak{g}^1$(exp $x$)$\supset \mathfrak{g}^0(x)$ for all $x\in \mathfrak{g}$.

By Cor. 3 of Prop. 8 of Chap. III, §4, no. 4, there exists a neighbourhood $V'$ of 0 in V such that, for all $x\in V'$, exp(ad($x$)) $=\sum_{n=0}^{\infty}\frac{1}{n!}$ ad($x$)$^n$ is defined and Ad(exp $x$) $=$ exp(ad($x$)). If $P\in k[X]$ and $\alpha \in$ End($\mathfrak{g}$), it is easy to check that $\mathfrak{g}^{\lambda}(\alpha )\subset \mathfrak{g}^{P(\lambda)}(P(\alpha ))$ for all $\lambda \in k$. Consequently,

$\mathfrak{g}^0$(ad($x$))$\subset \mathfrak{g}^1$(exp(ad($x$))) $=\mathfrak{g}^1$(Ad(exp $x$)) $=\mathfrak{g}^1$(exp $x$)

for all $x\in V'$. If $k=\mathbf{R}$ or $\mathbf{C}, V =\mathfrak{g}$ and we can take $V'$ = V, which proves (ii). We prove (i). Let U be a neighbourhood of 0 in End($\mathfrak{g}$) such that Log(1 $+\alpha$ ) $=\sum_{n>0}(-1)^{n+1 1}_{\overline{n}}\alpha^n$ is defined for all $\alpha \in U$. Then Log $\circ$ exp = 1 on a neighbourhood of 0 and $\mathfrak{g}^1(1 +\alpha )\subset \mathfrak{g}^0$(Log(1 $+\alpha$ )) for all $\alpha \in U$. Let W be the neighbourhood of 0 in $\mathfrak{g}$ consisting of those $x\in V'$ such that exp ad $x\in 1 + U$ and

Log(exp(ad($x$))) $=$ ad($x$).

Then, for all $x\in W$,

$\mathfrak{g}^1$(exp $x$) $=\mathfrak{g}^1$(Ad(exp $x$)) $=\mathfrak{g}^1$(exp(ad($x$)))

$\subset \mathfrak{g}^0$(Log(exp(ad($x$)))) $=\mathfrak{g}^0$(ad($x$)) $=\mathfrak{g}^0(x)$.

This shows that $\mathfrak{g}^1$(exp $x$) $=\mathfrak{g}^0(x)$ for all $x\in W$.