We give $L(G)$ and $L(H)$ norms which define their topologies and such that $\| [x, y] \| \leq \|x\| \|y\|$ for all $x, y$. There exist an open ball V of centre 0 in $L(G)$ and an exponential mapping $\phi$ of G defined on V such that: (1) $\phi(V)$ is an open neighbourhood of $e$ in G; (2) $\phi$ is an isomorphism of the analytic manifold V onto the analytic manifold $\phi(V)$; (3) $\phi(nx) = \phi(x)^n$ for all $x \in V$ and all $n \in \mathbf{Z}$ such that $nx \in V$. We define similarly W and $\psi$ for H. By shrinking V if necessary, it can be assumed that $f(\phi(V)) \subset \psi(W)$. Then $g = \psi^{-1} \circ f \circ \phi$ is a continuous mapping of V into W.

We show that

It can be assumed that $\lambda \neq 0$. Let $\lambda = \frac{r}{q}$ with $r, q$ in $\mathbf{Z} - \{0\}$. Let $y = \frac{r}{q} x$.

If $K = \mathbf{R}$, we write $z = \frac{x}{q} = \frac{y}{r} \in V$. Then $x = qz, y = rz$, whence

We show that $\psi^{-1}(f(\phi(z))^q) = q \psi^{-1}(f(\phi(z))) = qg(z)$. It suffices to verify that, if $u \in \psi(W)$ is such that $u^q \in \psi(W)$, then $\psi^{-1}(u^q) = q \psi^{-1}(u)$; but if $u = \psi(v)$, $u^q = \psi(v^q)$, then $v^1/q \in W$ and $(\psi(v^1/q))^q = u^q$, hence $\psi(v^1/q) = u = \psi(v)$ and therefore $v^1 = qv$.

Similarly, $g(y) = rg(z)$, whence (1).

If $K = \mathbf{Q}_p$, we write $z = rx = qy \in V$, whence $g(z) = rg(x) = qg(y)$, whence again (1).

As $\mathbf{Q}$ is dense in $K$, (1) implies that

Let $x \in L(G)$ and $\lambda, \lambda'$ be elements of $K^*$ such that $\lambda x \in V, \lambda' x \in V$. Then by (2) and hence $\frac{1}{\lambda} g(\lambda x) = \frac{1}{\lambda'} g(\lambda' x)$. Thus an extension $h$ of $g$ to $L(G)$ is defined by writing $h(x) = \frac{1}{\lambda} g(\lambda x)$ for all $\lambda$ such that $\lambda x \in V$. Clearly $h$ is continuous. We show that Let $\lambda' \in K^*$ be such that $\lambda' x \in V$ and $\lambda' \lambda x \in V$. Then Let $x, y$ be in $L(G)$. Then, by Proposition 4 of § 4, no. 3, For $|\lambda|$ sufficiently small, $\lambda x \in V$ and $\lambda y \in V$ and hence the above expression is equal to Thus $h$ is continuous and linear, hence $g = h|V$ is analytic, hence $f$ is analytic on $\phi(V)$ and hence $f$ is analytic (§ 1, no. 10).

This follows immediately from Theorem 1.

Let $g \in G$. There exists an open neighbourhood $V'$ of $e$ in G such that $V' \cup gV'g^{-1} \subset V$. The mapping $v \mapsto gv g^{-1}$ of $V'$ into $V$ is a continuous and hence analytic morphism of the Lie group germ $V'$ into the Lie group germ $V$. It then suffices to apply Proposition 18 of § 1, no. 9.

Case (a). Let $G_0$ be the identity component of G. Then $f(G_0)$ is a connected subgroup of $G'$ and hence $f(G_0) = \{e\}$ and $G_0$ is open in G.

Case (b). Let $V'$ be a neighbourhood of $e$ in $G'$ such that every subgroup of $G'$ contained in $V'$ reduces to $\{e\}$ (§ 4, no. 2, Corollary 1 to Theorem 2). There exists a neighbourhood V of $e$ in G such that $f(V) \subset V'$. Then there exists an open subgroup $G_1$ of G such that $G_1 \subset V$ (§ 7, no. 1, Proposition 1). Then $f(G_1) = \{e\}$.

Case (c). By § 7, Theorem 4 and the Corollary to Proposition 8, there exists a neighbourhood $V'$ of $e$ in $G'$ such that, for all $x' \in V' - \{e\}$, ${x'}^{p^n}$ does not tend

† See for example D. Montgomery and L. Zippin, Topological transformation groups, Interscience tracts in pure and applied mathematics, no. 1, Interscience publishers, New York 1955 (in particular pp. 169 and 184).

to $e$ as $n$ tends to $+\infty$. There exists a neighbourhood $V$ of $e$ in $G$ such that $f(V) \subset V'$. By § 7, Theorem 4 and Proposition 9, there exists an open subgroup $G_1$ of $G$ such that $G_1 \subset V$ and such that, for all $x \in G_1$, $x^{p^n}$ tends to $e$ as $n$ tends to $+\infty$. Then $f(G_1) = \{e\}$.

Let $\mathfrak{h}$ be the Lie subalgebra tangent to $H$ at $e$ (§ 4, no. 5, Definition 2). There exists a Lie subgroup germ $H_0$ of $G$ with Lie algebra $\mathfrak{h}$ and contained in $H$. We prove that $H_0$ is open in $H$ with the topology induced by that on $G$. This will prove that $H$ is an analytic submanifold of $G$ and will establish the theorem.

There exist a vector subspace $\mathfrak{k}$ supplementary to $\mathfrak{h}$ in $L(G)$, symmetric open neighbourhoods $V_1, V_2$ of zero in $\mathfrak{h}$ and $\mathfrak{k}$ respectively and an exponential mapping $\phi$ of $G$ defined on $V_1 + V_2$ and possessing the following properties:

(a) the mapping $(a_1, a_2) \mapsto \phi(a_1)\phi(a_2)$ is an analytic isomorphism of $V_1 \times V_2$ onto an open subset $V$ of $G$; (b) $\phi(V_1) \subset H_0$; (c) $V^2 \subset U$.

We shall prove (and this will achieve the proof) that there exists an open neighbourhood $V'_2$ of 0 in $V_2$ such that $H \cap (\phi(V_1)\phi(V'_2)) = \phi(V_1)$.

Suppose that the assertion is false. Then we can find a sequence $(x_n)$ in $V_1$ and a sequence $(y_n)$ in $V_2 - \{0\}$ tending to 0, such that $\phi(x_n)\phi(y_n) \in H$ for all $n$. Then $\phi(y_n) \in H$ by (c).

If $K = \mathbf{Q}_p$, it can be assumed that $V_2$ is an additive subgroup of $\mathfrak{k}$ and that $\phi(pa) = \phi(a)^p$ for all $a \in V_2$ and all $p \in \mathbf{Z}$. Then $\phi(\lambda y_1) \in H$ for all $\lambda \in \mathbf{Z}$ and hence by continuity for all $\lambda \in \mathbf{Z}_p$. The mapping $f : \lambda \mapsto \phi(\lambda y_1)$ of $\mathbf{Z}_p$ into $G$ is analytic and takes its values in $H$ and $(T_0f)(1) = y_1$. Hence $y_1 \in \mathfrak{h}$, which is absurd. The theorem is therefore established in the case of $\mathbf{Q}_p$.

If $K = \mathbf{R}$, it can be assumed that $V_2$ is connected and that $y_n$ belongs to $\frac{1}{2}V_2 - \{0\}$. By taking a subsequence of $(y_n)$ if necessary, we can find a sequence $(\lambda_n)$ of non-zero scalars such that $\lambda_n^{-1}y_n$ tends to an element $y$ of $V_2 - \{0\}$. The sequence $(\lambda_n)$ tends to 0. Let $\lambda \in \mathbf{R}$ be such that $\lambda y \in \frac{1}{2}V_2$; we prove that $\exp(\lambda y) \in H$. It can be assumed that $\lambda \lambda_n^{-1}y_n \in \frac{1}{2}V_2$ for all $n$. Let $k_n \in \mathbf{Z}$ be such that $|\lambda - k_n \lambda_n|$ tends to 0. For $n$ sufficiently large, $(\lambda - k_n \lambda_n)\lambda_n^{-1}y_n \in \frac{1}{2}V_2$ and hence $k_n y_n \in \frac{1}{2}V_2$. Hence $\exp(h y_n) \in H$ for $h$ an integer and $0 \leq |h| \leq |k_n|$ (as is seen by induction on $|h|$). Then

Hence the mapping $f : \lambda \mapsto \exp \lambda y$, where $\lambda y \in \frac{1}{4}V_2$, takes its values in $H$ and $(T_0f)(1) = y$. Hence $y \in \mathfrak{h}$, which is absurd. The theorem is thus established in the case of $\mathbf{R}$.

Theorem 2 is no longer true if $G$ is not assumed to be finite-dimensional (Exercise 12).

There exists a compact neighbourhood $V$ of $e$ in $G'$ such that $f|V$ is a homeomorphism of $V$ onto a compact subspace of $G$. If $U$ is a sufficiently small open neighbourhood of $e$ in $G$, the hypotheses of Theorem 2 are satisfied with $H = f(V) \cap U$. Hence $H$ is a Lie subgroup germ of $G$. Let $W$ be the inverse image of $H$ under $f|V$. Then $W$ is a neighbourhood of $e$ in $G'$. Let $W$ be given the analytic manifold structure transported from that on $H$ by $(f|W)^{-1}$. For all $z \in G'$, the mapping $x \mapsto f(z)x f(z)^{-1}$ of $G$ into $G$ is analytic; hence there exists an open neighbourhood $W'$ of $e$ in $W$ such that the mapping $x' \mapsto zx'z^{-1}$ of $W'$ into $W$ is analytic. By Proposition 18 of § 1, no. 9, there exists on $G'$ a Lie group structure which induces on a sufficiently small open neighbourhood of $e$ the same analytic structure as $W$ and hence the same topology as the initial topology on $G'$.

(i) If $K = \mathbf{C}$, $H$ is a Lie subgroup of $G$. (ii) If $K$ is a finite extension of $\mathbf{Q}_p$ and $I$ is finite, $H$ is a Lie subgroup of $G$.

(i) Suppose that $K = \mathbf{C}$. We consider $G$ as a real Lie group. Then $H$ is a real Lie subgroup of $G$ (Theorem 2). Let $a \in L(H)$. There exists a connected open neighbourhood $W$ of $0$ in $\mathbf{C}$ such that $\exp \lambda a \in V$ for all $\lambda \in W$. Let $i \in I$. Then $f_i(\exp \lambda a) = f_i(e)$ if $\lambda \in \mathbf{R} \cap W$. Hence $f_i(\exp \lambda a) = f_i(e)$ if $\lambda \in W$ by analytic continuation. Thus, $\exp \lambda a \in H$ for $\lambda \in W$ and therefore $\mu a \in L(H)$ for all $\mu \in \mathbf{C}$. Therefore $H$ is a Lie subgroup of the complex Lie group $G$ (§ 4, no. 2, Proposition 2).

(ii) Suppose that $K$ is a finite extension of $\mathbf{Q}_p$. We consider $G$ as a Lie group over $\mathbf{Q}_p$. It is finite-dimensional and Theorem 2 implies that $H$ is a $p$-adic Lie subgroup of $G$. Since $I$ is finite, $\prod_{i \in I} M_i$ is a manifold and it can be assumed that the family $(f_i)$ reduces to a single mapping $f$. Let $a \in L(G)$. Let $\phi$ be an exponential mapping of $G$. Then $f(\phi(\lambda a)) = f(e)$ for $\lambda \in \mathbf{Q}_p$ and $|\lambda|$ sufficiently small. Since $f$ is K-analytic, it follows that $f(\phi(\lambda a)) = f(e)$ for $\lambda \in K$ and $|\lambda|$ sufficiently small. Hence $\phi(\lambda a) \in H$ for $\lambda \in K$ and $|\lambda|$ sufficiently small and therefore $\mu a \in L(H)$ for all $\mu \in K$. The proof is completed as in (i).

Corollary 2 (ii) is no longer true if we omit the hypothesis that I is finite.