Proposition 42

Let $G$ be a Lie group, $X$ an analytic manifold, $(g, x) \mapsto gx$ (resp. $xg$) a law of analytic left (resp. right) operation of $G$ on $X$ and $x_0$ a point of $X$ which is invariant under $G$. For all $g \in G$, let $\tau(g)$ be the automorphism $x \mapsto gx$ (resp. $xg$) of $X$ and let $\pi(g)$ be the automorphism of $T_{x_0}(X)$ tangent at $x_0$ to $\tau(g)$. (i) $\pi$ is an analytic representation of $G$ (resp. $G^\vee$) on $T_{x_0}(X)$. (ii) For all $a \in L(G)$ and all $\xi_0 \in T_{x_0}(X)$, $L(\pi)a.\xi_0$ can be calculated as follows: let $D_a$ be the vector field defined by $a$ on $X$ and $\xi$ a vector field of class $C^1$ in an open neighbourhood of $x_0$ such that $\xi(x_0) = \xi_0$; then

$\tau(gg') = \tau(g)\tau(g')$ (resp. $\tau(g')\tau(g)$) and hence $\pi(gg') = \pi(g)\pi(g')$ (resp. $\pi(g')\pi(g)$). On the other hand, since $TX$ is a vector G-bundle of class $C^\omega$ (§ 1, no. 8, Proposition 16), $\pi$ is analytic, whence (i).

To prove (ii), suppose that G operates on the left. There exists an open neighbourhood I of 0 in K and an analytic mapping $\gamma$ of I into G such that $\gamma(0) = e, T_0(\gamma)1 = a$. Then $D_a$ is the vector field on X defined by the mapping $\phi : (\lambda, x) \mapsto \gamma(\lambda)x$ of $I \times X$ into X (§ 2, no. 2). If $\phi_\lambda$ denotes the bijection $x \mapsto \gamma(\lambda)x$ of X into X, then

As the mappings $\lambda \mapsto \gamma(\lambda)^{-1}$ and $\lambda \mapsto \gamma(-\lambda)$ are tangent at 0, this is also equal to