Proposition 20

Let $G$ be a Lie group germ and $g \in G$. There exist an open neighbourhood $U$ of $e$ and an open neighbourhood $V$ of $g$ with the following properties: (a) $ug$ is defined for all $u \in U$; (b) $vg^{-1}$ is defined for all $v \in V$; (c) the mappings $u \mapsto ug,\ v \mapsto vg^{-1}$ are inverse analytic isomorphisms of one another of $U$ onto $V$ and of $V$ onto $U$.

§ 1.10

As the set of definition of the product is open in $G \times G$, there exist an open neighbourhood $U$ of $e$ and an open neighbourhood $V$ of $g$ with properties (a) and (b). Let $\eta(u) = ug$ for $u \in U$, $\eta'(v) = vg^{-1}$ for $v \in V$. By shrinking $U$ and $V$, it can be assumed that $(ug)g^{-1} = u$ and $(vg^{-1})g = v$ for $u \in U$ and $v \in V$. Then $\eta$ and $\eta'$ are injections. By shrinking $U$ further, it can be assumed that $\eta(U) \subset V$. Then $\eta'(V) \supset U$ and $\eta(U)$ is the inverse image of $U$ under $\eta'$ and hence is an open neighbourhood of $g$ in $V$. Replacing $V$ by $\eta(U)$, we finally arrive at the situation where $\eta$ and $\eta'$ are analytic inverse bijections.

Let $G_1, G_2$ be two Lie group germs with identity elements $e_1, e_2$. A mapping $f$ of $G_1$ into $G_2$ is called a morphism if $f$ satisfies the following conditions: (i) $f$ is analytic; (ii) $f(e_1) = e_2$; (iii) if $g, h$ are elements of $G_1$ such that $gh$ is defined, then $f(g)f(h)$ is defined and is equal to $f(gh)$.

Let $g \in G_1$. As $gg^{-1}$ is defined and is equal to $e_1$, $f(g)f(g^{-1})$ is defined and is equal to $e_2$ and hence

that is (11)

The composition of two morphisms is a morphism.

If $f : G_1 \to G_2$ and $f' : G_2 \to G_1$ are two inverse morphisms of one another, they are isomorphisms (using in particular formula (11)).

Let $G_1, G_2$ be two Lie group germs, and $f$ a mapping of $G_1$ into $G_2$ satisfying conditions (ii) and (iii) above, which is analytic in an open neighbourhood of $e_1$. Using Proposition 20, it can be proved as in Proposition 4 that $f$ is a morphism.

Let $(G, e, \theta, m)$ be a Lie group germ and $\Omega$ the set of definition of $m$. Let $H$ be a submanifold of $G$ containing $e$, which is stable under $\theta$. Suppose that the set $\Omega_1$ of $(x, y) \in \Omega \cap (H \times H)$ such that $m(x, y) \in H$ is open in $H \times H$. Then $(H, e, \theta|H, m|\Omega_1)$ is a Lie group germ. Such a Lie group germ is called a Lie subgroup germ of G. The canonical injection of $H$ into $G$ is a morphism. If $f : L \to G$ is a morphism of Lie group germs such that $f(L) \subset H$, then $f : L \to H$ is a morphism of Lie group germs.

Suppose that $K$ is of characteristic 0. If we replace the hypothesis that $H$ is a submanifold of $G$ by the hypothesis that $H$ is a quasi-submanifold of $G$, the results of the above paragraph remain true (cf. Differentiable and Analytic Manifolds, R, 5.8.5). $H$ is then called a Lie quasi-subgroup germ of $G$.

If $G$ is a Lie group germ with identity element $e$, every symmetric open neighbourhood of $e$ in $G$ is a Lie subgroup germ of $G$. (This applies in particular when $G$ is a Lie group.) Let $H$ be a Lie subgroup germ of $G$; if $H$ is a neighbourhood of $e$ in $G$, then $H$ is open in $G$ by Proposition 20.

The product Lie group germ of a finite number of Lie group germs is defined in an obvious way.