Lemma 4

Let G be a finite-dimensional Lie group, $\Omega$ a symmetric open neighbourhood of e in G and H a subset of $\Omega$ containing e such that the conditions $x \in H, y \in H, xy^{-1} \in \Omega$ imply $xy^{-1} \in H$. Let $r \in N_K$. For all $x \in H$, let $h_x$ be the set of $a \in T_x(G)$ with the following property: there exist an open neighbourhood I of 0 in K and a mapping f of class $C^r$ of I into G such that $f(0) = x, f(I) \subset H, (T_0f)(1) = a$.

(i) Let $h_e = h$. Then h is a Lie subalgebra of $L(G)$ which is invariant under $\mathrm{Ad}_{L(G)}(H)$.

(ii) $h_x = xh = hx$ for all $x \in H$ (where $xh$ and $hx$ are evaluated in $T(G)$).

(iii) Let V be a manifold of class $C^r$, $v_0$ a point of V and f a mapping of class $C^r$ of V into G such that $f(v_0) = e$ and $f(V) \subset H$. For every Lie subgroup germ $H'$ of G with Lie algebra h, $f(v) \in H'$ for v sufficiently close to $v_0$.

(iv) For every Lie subgroup germ $H'$ of G with Lie algebra h, $H' \cap H$ is a neighbourhood of e in $H'$.

(v) For all $x \in H$ and all $a \in h_x$, there exist an open neighbourhood I of 0 in K and a mapping f of class $C^a$ of I into G such that $f(0) = x, f(I) \subset H, (T_0f)(1) = a$. Clearly $Kh = h$ and $xh_yz = h_{xyz}$ for $x, y, xy, xyz$ in H. This implies (ii) and the fact that h is invariant under $\mathrm{Ad}_{L(G)}(H)$.

Let $a_1, a_2$ be in h. Let I be an open neighbourhood of 0 in K and $f_1, f_2$ mappings of class $C^r$ of I into G such that $f_j(0) = e, f_j(I) \subset H, (T_0f_j)(1) = a_j$ ($j = 1, 2$). We define $f : I \to G$ by $f(\lambda) = f_1(\lambda)f_2(\lambda)$. Then f is of class $C^r$ and $f(0) = e$. By shrinking I if necessary, we have $f(I) \subset H$. On the other hand, the mapping of $T_e(G) \times T_e(G)$ into $T_e(G)$ tangent to the mapping $(g, g') \to gg'$ is addition; hence $(T_0f) = a_1 + a_2$. Hence $a_1 + a_2 \in h$ and h is a vector subspace of $L(G)$. Since $xhx^{-1} = h$ for all $x \in H$, $(\mathrm{Ad}_{f_1(\lambda)}) . a_2 \in h$ for all $\lambda \in I$. The tangent mapping at 0 to the mapping $\lambda \mapsto \mathrm{Ad}f_1(\lambda)$ is, by Proposition 44 of § 3, no. 12, the mapping $\lambda \mapsto \mathrm{ad}(a_1\lambda)$; hence

since h is closed in $L(G)$. Hence we have proved (i). In the rest of the proof we fix a Lie subgroup germ $H'$ of G with Lie algebra h.

Let V, $v_0, f$ be as in (iii). Let Y be the left foliation of G associated with h (no. 1). For all $y \in H'$, $T_y(H') = yh$. On the other hand, for all $v \in V$, the image of $T_v(V)$ under $T_v(f)$ is contained in $h_{f(v)} = f(v)h$ (by definition of $h_{f(v)}$). By Differentiable and Analytic Manifolds, R, 9.3.2, f is a morphism of V into Y. As $H'$ is a leaf of Y (Differentiable and Analytic Manifolds, R, 9.2.8), $f(v) \in H'$ for v sufficiently close to $v_0$.

Let $(a_1, \ldots, a_s)$ be a basis of h. There exist an open neighbourhood I of 0 in K and mappings $f_1, \ldots, f_s$ of class $C^r$ of I into G such that $f_j(0) = e, f_j(I) \subset H, (Tf_j)1 = a_j$ for all j. By (iii), $f_j(\lambda) \in H'$ for $|\lambda|$ sufficiently small. Hence the $f_1(\lambda_1)f_2(\lambda_2)\ldots f_s(\lambda_s)$ constitute, for $|\lambda_1|, |\lambda_2|, \ldots, |\lambda_s|$ sufficiently small, a neighbourhood of $e$ in $H'$; and this neighbourhood is contained in $H$. Hence (iv).

If $a \in \mathfrak{h}$, there exist an open neighbourhood $I$ of $0$ in $K$ and a mapping $f$ of class $C^\omega$ of $I$ into $G$ such that $f(0) = e, f(I) \subset H', (T_0 f)1 = a$. This, with (iv), implies (v).