Theorem 4

Let G be a Lie group. There exists one and only one exponential mapping of G defined on $L(G)$.

There exist a convex open neighbourhood U of 0 in $L(G)$ and an exponential mapping of G defined on U. It can be assumed, by choosing U sufficiently small, that

for $a \in L(G)$, $\lambda, \lambda'$ in K, $\lambda a, \lambda'a, (\lambda + \lambda')a$ in U.

Let $a \in L(G)$. There exists an integer $n > 0$ such that $\frac{1}{n} a \in U$. If m is another integer such that $\frac{1}{m} a \in U$, then $\frac{1}{nm} a \in U$ and relation (1) implies

hence $\left( \phi \left( \frac{1}{n} a \right) \right)^n = \left( \phi \left( \frac{1}{m} a \right) \right)^m$. There exists an extension $\psi : L(G) \to G$ of $\phi$ such that $\psi(a) = \left( \phi \left( \frac{1}{n} a \right) \right)^n$ for $a \in L(G)$ and $n$ an integer $> 0$ such that $\frac{1}{n} a \in U$. Clearly $\psi$ is analytic and is an exponential mapping of $G$. If $\psi' : L(G) \to G$ is an exponential mapping of $G$, $\psi$ and $\psi'$ coincide on a neighbourhood of 0 and hence are equal since $L(G)$ is connected.

Henceforth when we speak of the exponential mapping of $G$, we shall mean the mapping considered in Theorem 4. It will be denoted by $\exp_G$ or $\exp$ if there is no risk of confusion.