Lemma 6

Let $G$ be a Lie group germ, $\omega$ the canonical left differential form of $G$ (§ 3, no. 18.9), $M$ a manifold of class $C^r$ ($r \geq 2$) and $\alpha$ a differential form of class $C^{r-1}$ and degree 1 on $M$ with values in $L(G)$.

(i) The elements of $T(M \times G)$ at which the differential form

is zero constitute a vector subbundle $S$ of $T(M \times G)$ of class $C^{r-1}$.

(ii) For all $(x, g) \in M \times G$, $T(\mathrm{pr}_1)|S_{(x, g)}$ is an isomorphism of $S_{(x, g)}$ onto $T_x(M)$.

(iii) If $d\alpha + [\alpha]^2 = 0$ (cf. § 3, no. 14) the vector subbundle $S$ is integrable.

If $(x, g) \in M \times G$ and $(u, v) \in T_x(M) \times T_g(G)$, then

Hence the kernel of $\theta_{(x, g)}$ is the set $S_{(x, g)}$ of $(u, g\alpha(u))$ for $u \in T_x(M)$, whence (ii). We consider $T(M \times G)$ as the direct sum of two vector bundles $F$ and $F'$ with $F_{(x, g)} = T_x(M) \times \{0\}$ and $F'_{(x, g)} = \{0\} \times T_g(G)$ for all For $u \in T_x(M) \times \{0\}$, we write $\phi(u) = (0, g\alpha(u))$. Using the left trivialization of $T(G)$, it is seen that $\phi$ is a morphism of $F$ into $F'$, whence (i) (Lemma 5). Finally, if $d\alpha + [\alpha]^2 = 0$, then and hence $S$ is integrable (Differentiable and Analytic Manifolds, R, 9.3.6).