Proposition 24

Let G be a Lie group and $(U, \phi, E)$ a chart on G such that $\phi(e) = 0$. Let V be an open neighbourhood of e such that $V^2 \subset U$. Let m be the analytic mapping $(a, b) \mapsto \phi(\phi^{-1}(a)\phi^{-1}(b))$ of $\phi(V) \times \phi(V)$ into E. Let

be the expansion of m as an integral series about (0, 0), where $m_{i,j}$ is a bihomogeneous continuous-polynomial of bidegree $(i, j)$ on $E \times E$ with values in $E$.

(i) $m_{i,0} = m_{0,j} = 0$ for all $i \neq 1$ and $j \neq 1$. (ii) $m_{1,0}(a, b) = a$ and $m_{0,1}(a, b) = b$ for all $a \in E, b \in E$. (iii) *Let $\psi : T_e(G) \to E$ be the differential of $\phi$ at e. For all $u, v$ in $T_e(G)$,

$m(a, 0) = a,\ m(0, b) = b$ for all $a, b$ in $\phi(V)$, which proves (i) and (ii). Let $u, v$ be in $T_e(G)$. Let $T_0(E)$ be identified with $E$ and hence $\psi$ with $T_e(\phi)$: The images of $u$ and $v$ under $T_e(\phi)$ are $\psi(u)$ and $\psi(v)$. The tensor product point distribution of these images is the symmetric product of $(\psi(u), 0)$ and $(0, \psi(v))$ in $TS(E \times E) = TS(E) \times TS(E)$, that is

Hence $\phi * (u * v)$ is the image of the above element under the mapping $m$ of $\phi(V) \times \phi(V)$ into $E$. Its component of degree 1 in $TS(E)$ is, by Lemma 1,

We define a bilinear mapping $n : (E \times E)^2 \to E$ by

Then $n((a, b), (a, b)) = m_{1,1}(a, b)$ and hence

Similarly, $\phi * (v * u)$ admits $m_{1,1}(\phi(v), \phi(u))$ as component of degree 1 in $TS(E)$. As $\phi([u, v])$ is of degree 1, this proves (iii).