By considering the left trivialization of $T(G)$ (§ 2, no. 3), it is seen immediately that the $g\mathfrak{h}$, for $g \in G$, are the fibres of a vector subbundle E of $T(G)$. Let $g \in G$. The set of $(L_a)_g$, where $a \in \mathfrak{h}$, is equal to $g\mathfrak{h}$. Now, if $a$ and $b$ belong to $\mathfrak{h}$, then $[L_a, L_b] = L_{[a, b]}$ and $[a, b] \in \mathfrak{h}$. Hence E is integrable (Differentiable and Analytic Manifolds, R, 9.3.3 (iv)). The argument is similar for the $\mathfrak{h}g$.

The integral foliation (Differentiable and Analytic Manifolds, R, 9.3.2) of the union of the $g\mathfrak{h}$ (resp. $\mathfrak{h}g$) is called the left (resp. right) foliation of G associated with $\mathfrak{h}$.

Let $V$ be an open neighbourhood of $e_G$ in $G$ such that, for $(s, t) \in V \times V$, the products $st$ and $\phi(s)\phi(t)$ are defined and $st \in U$. Consider the mappings $\alpha_1, \alpha_2$ of $V \times V$ into $H$ defined by

Then $\alpha_1(t, e) = \phi(t) = \alpha_2(t, e)$. On the other hand, let $t$ be fixed in $V$ and $\beta_i$ be the mapping $s \mapsto \alpha_i(s, t)$ of $V$ into $H$. Then, for all $s \in V$ and all $a \in L(G)$,

Hence (Differentiable and Analytic Manifolds, R, 9.3.7) $\alpha_1$ and $\alpha_2$ coincide on a neighbourhood of $(e_G, e_G)$. The restriction of $\phi$ to a sufficiently small symmetric open neighbourhood of $e_G$ is therefore a morphism of Lie group germs, whence (i).

Let $G_1, G_2, \phi_1, \phi_2$ be as in (ii) and let us prove that $\phi_1, \phi_2$ coincide on a neighbourhood of $e_G$. There exists an open neighbourhood $W$ of $e_G$ such that $\phi_1(ts) = \phi_1(t)\phi_1(s)$, $\phi_2(ts) = \phi_2(t)\phi_2(s)$ for all $s, t$ in $W$. Then, if $s \in W$ and $a \in L(G)$,

for $i = 1, 2$. As $\phi_1(e_G) = e_H = \phi_2(e_G)$, it follows from Differentiable and Analytic Manifolds, R, 9.3.7 that $\phi_1$ and $\phi_2$ coincide on a neighbourhood of $e_G$.

This follows from Theorem 1 and § 1, no. 10, Proposition 21.

The Lie algebra of the additive group $L(G)$ is isomorphic to $L(G)$. Hence it suffices to apply Corollary 1.

There exists an open Lie subgroup germ U of G which is commutative (Corollary 2). Let V be a neighbourhood of e such that V^2 ⊂ U. Then xy = yx for all x, y in V. Hence the subgroup of G generated by V is commutative; it is obviously open.

G is called the Lie group germ defined by L.

Suppose that K is ultrametric. Let $p$ be the characteristic of the residue field of K. If $p \neq 0$, let $\lambda = |p|^{1/(p-1)}$; if $p = 0$, let $\lambda = 1$.

G is called the Lie group defined by L.

The first assertion follows from Lemmas 2 and 3. The second assertion follows from Corollary 1 to Theorem 1 of no. 1.

We write L(G) = L. Choose a norm on L defining the topology on L and such that $\| [x, y] \| \leq \| x \| \| y \|$ for all x, y in L.

Suppose that K = R or C. Let G' be the Lie group germ defined by L. There exist an open ball U' of centre 0 in G' and an isomorphism $\phi$ of the Lie group germ U' onto an open neighbourhood U of e in G. Let V' = $\frac{1}{2} U'$, $V = \phi(V')$, H be a subgroup of G contained in V and $h \in H$. Let $x = \phi^{-1}(h) \in V'$. If $x \neq 0$, there exists an integer $n > 0$ such that $x, 2x, \ldots, nx$ are in V', $(n+1)x \in U'$, $(n+1)x \notin V'$. Then $h, h^2, \ldots, h^n$ are in V,

which is absurd. Hence H = {e}.

Suppose that K is ultrametric. It suffices to prove the corollary when G is the Lie group associated with L. If $g \in G$, the powers of g evaluated in G are the elements of $\mathbf{Z}g$ evaluated in L. These are distinct if $g \neq e$. Hence G contains no finite subgroup distinct from {e}.

There exists a Lie group germ $G_1$ over K such that $L(G_1) = L'$ (Theorem 2). By Corollary 1 to Theorem 1 of no. 1, G and $G_1$, considered as Lie k-group germs, are locally isomorphic. Hence there exist an open neighbourhood G' of e in G and a Lie K-group germ structure on G', with Lie algebra L, which is compatible with the Lie group germ structure over k. Let V be a symmetric open neighbourhood of e in G such that $V^2 \subset G'$. Let $g \in G$. Then $\phi = \operatorname{Int} g$ is a k-isomorphism of a sufficiently small open Lie subgroup germ of G' onto an open Lie subgroup germ of G'; and $T_e(\phi)$ is K-linear and hence $T_x(\phi)$ is K-linear for x sufficiently close to e; therefore the restriction of Int g to a sufficiently small open neighbourhood of e in V is K-analytic (Differentiable and Analytic Manifolds, R, 5.14.6). By § 1, no. 9, Proposition 18, there exists on G an analytic K-manifold structure with which G is a Lie

K-group and V is an open sub-K-manifold of G. By translation, it is seen that the underlying k-manifold structure of G is the given structure. The Lie algebra of the Lie K-group G is the same as that of the open Lie K-subgroup germ V and hence is L'. Finally, the uniqueness stated in the corollary follows from § 3, no. 8, Proposition 32.

There exists a Lie group germ $H'$ with Lie algebra isomorphic to $\mathfrak{h}$ (Theorem 2). Shrinking $H'$ if necessary, it can be assumed that there exists a morphism $\phi$ of $H'$ into G such that $L(\phi)$ is an isomorphism of $L(H')$ onto $\mathfrak{h}$ (no. 1, Theorem 1). As $\mathfrak{h}$ admits a topological supplement, $\phi$ is an immersion at e. Hence, shrinking $H'$ further, it can be assumed that $\phi$ is an isomorphism of the manifold $H'$ onto a submanifold of G. This proves the existence of H. The second assertion follows from the following proposition:

If $H \cap H'$ is open in $H'$, then $L(H') = L(H \cap H') \subset L(H)$. Suppose that $L(H) \supset L(H')$. Let i, $i'$ be the canonical injections of H, $H'$ into G. By shrinking $H'$ if necessary, it can be assumed that there exists a morphism $\psi$ of $H'$ into H such that $L(\psi)$ is the canonical injection of $L(H')$ into $L(H)$ (no. 1, Theorem 1). Then $L(i \circ \psi) = L(i')$ and hence there exists a neighbourhood V of $e_{H'}$ in $H'$ such that $i \circ \psi$ and $i'$ coincide on V (Theorem 1). Therefore $V \subset H$, hence $V \subset H \cap H'$ and $H \cap H'$ is open in $H'$ (§ 1, no. 10).

There exists a Lie subgroup germ $H'$ of the Lie K-group G such that $L(H') = L(H)$ (Theorem 3). Consider G, H, $H'$ as Lie k-group germs; Theorem 3 then proves that $H \cap H'$ is open in H and $H'$. Hence there exists an open neighbourhood U of e in G such that $U \cap H$ is a submanifold of G over K. Therefore, H is a Lie subgroup of the Lie K-group G (§ 1, no. 3, Proposition 6).

(v) ⇒ (i): obvious, for (λb).(λ'b) = (λ + λ')b in W for |λ| and |λ'| sufficiently small. (i) ⇒ (ii): suppose that condition (i) is satisfied. Let b ∈ L. Let ψ be the restriction of φ to V ∩ Kb. By hypothesis, there exists a symmetric neighbourhood T of 0 in the additive Lie group Kb such that ψ|T is a morphism of the Lie group germ T into G. Hence

so that φ_*(b^n) is homogeneous of degree n in U(G). (ii) ⇒ (iii): this follows from the fact that TS^n(L) is the vector subspace of TS(L) generated by the n-th powers of the elements of L (Algebra, Chapter IV, § 5, Proposition 5). (iii) ⇒ (iv): the canonical mapping of TS(L) into the enveloping algebra of L is the unique morphism of graded cogebras mapping 1 to 1 and extending Id_L (Chapter II, § 1, no. 5, Remark 3). Now φ_* is a cogebra morphism and φ_*|L = Id_L by hypothesis. If condition (iii) holds, it is seen that condition (iv) also holds. (iv) ⇒ (v): suppose that condition (iv) is satisfied. Choose a norm on L defining the topology of L and such that \|[x, y]\| \leq \|x\|\|y\| for all x, y in L. Let H be the Lie group germ defined by the normed Lie algebra L. By Theorem 1, there exist an open subgroup germ S ⊂ V of H and an isomorphism φ' of S onto an open subgroup germ of G. As we know already that (v) ⇒ (iv), the mapping φ_* of TS(L) into U(G) is the canonical mapping of TS(L) into the enveloping algebra of L. Thus φ_*(t) = φ'_*(t) for all t ∈ T_0^{(∞)}(L). As φ and φ' are analytic, φ and φ' coincide on a neighbourhood of 0.

Let $k_i$ be the canonical injection of $L_i$ into $L_1 \times L_2 \times \cdots \times L_n$. Then, for all $b \in L_i$, $(T_{(0, \ldots, 0)}\theta)(T_0k_i)(b) = (T_0\phi)(b) = b$ and hence $(T_{(0, \ldots, 0)}\theta)|L_i$ is the canonical injection of $L_i$ into L.

In particular, $\theta$ is étale at $(0, 0, \ldots, 0)$. Its restriction to a sufficiently small

LIE GROUPS

open neighbourhood U of $(0, 0, \ldots, 0)$ has an open image in G and is an isomorphism of the manifold U onto the manifold $\theta(U)$. The inverse mapping $\eta$ of $\theta(U)$ onto U is called a canonical chart of the second species of G, associated with the given decomposition of L as a direct sum. If further G is finite-dimensional and each $L_i$ is generated by a non-zero vector $e_i$, the coordinate system on $\theta(U)$ defined by $\eta$ and the $e_i$ is called a canonical coordinate system of the second species.

(1)

(2) (note that $\phi^{-1}(\phi(\lambda x) \phi(\lambda y))$ and $\phi^{-1}(\phi(\lambda x) \phi(\lambda y) \phi(-\lambda x) \phi(-\lambda y))$ are defined for $|\lambda|$ sufficiently small).

Let $L = L(G)$ be given a norm defining the topology of L and such that $\| [x, y] \| \leq \| x \| \| y \|$ for all $x, y$ in L. Using Theorems 2 and 4, it can be assumed that G is the Lie group germ defined by L and that $\phi = \mathrm{Id}_G$. Let $(x, y) \mapsto x \cdot y$ denote the product in the group G. The formulae to be proved can then be written (3)

(4)

There exists an open neighbourhood V of 0 in K such that the function

is defined and analytic on V. By Chapter II, § 6, no. 4, Remark 2, the expansion of $f$ as an integral series about the origin is and this proves (3). On the other hand, for $u, v$ in G and $\|u\|, \|v\|$ sufficiently small, $u \cdot v$ is an analytic function of $(u, v)$ and the terms of degrees 1 and 2 in the expansion of this function as an integral series about the origin are $u + v + \frac{1}{2}[u, v]$. By Differentiable and Analytic Manifolds, R, 3.2.7 and 4.2.3, the terms of degrees 1 and 2 in the expansion of the function $f(\lambda) \cdot f(-\lambda)$ as an integral series about the origin are the terms of degrees 1 and 2 in or also in and this proves (4).

EXponential mappings

§ 4.3

$\phi$ satisfies hypothesis (i) of Theorem 4 relative to G' and is therefore an exponential mapping of G'.

Let L be given a norm compatible with its topology and such that $\| [x, y] \| \leq \| x \| \| y \|$ for all $x, y \in L$. It suffices to consider the case where G is the Lie group germ defined by L and $\phi = \operatorname{Id}_G$. By definition, $\varpi(x)$ is then the tangent mapping at x to the mapping $y \mapsto y . x^{-1}$ of G into G. If $H(X, Y)$ denotes the Hausdorff series, $\varpi(x)$ is therefore, for $\| x \|$ sufficiently small, the tangent mapping at 0 to the mapping $y \mapsto H(x, +y, -x)$ of G into G. In $H(X + Y, -X)$, the sum of terms of the first degree in Y is

(Chapter II, § 6, no. 5, Proposition 5). The proposition then follows from Differentiable and Analytic Manifolds, R, 3.2.4 and 4.2.3.

Let G be a Lie group germ and $t \in K$. A t-th power mapping of G is any mapping, defined and analytic on an open neighbourhood of e, with values in G and coinciding on a neighbourhood of e with a mapping

where $\phi$ is an injective exponential mapping of G.

(ii) The tangent mapping at e to a t-th power mapping is the homothety of ratio t.

(iii) If h is a t-th power mapping and h' a t'-th power mapping of G, $h \circ h'$ is a (tt')-th power mapping and $g \mapsto h(g)h'(g)$ is a (t + t')-th power mapping.

(iv) If h is a t-th power mapping and $u \in U^n(G)$, then $h_*(u) = t^n u$.

It suffices to prove the proposition when G is the Lie group germ defined by a complete normed Lie algebra and when the t-th power mappings considered are constructed using the exponential mapping $\phi = \operatorname{Id}_G$. But in that case everything is obvious.

Let $L(G)$ and $L(H)$ be given norms defining their topologies and such that $\| [x, y] \| \leq \| x \| \| y \|$ for all x and y. It can be assumed that G (resp. H) is the Lie group germ defined by $L(G)$ (resp. $L(H)$), so that $\phi_G$ (resp. $\phi_H$) coincides with $\mathrm{Id}_G$ (resp. $\mathrm{Id}_H$) in a neighbourhood of 0. On the other hand, there exists a symmetric open neighbourhood W of 0 in $L(G)$ such that $L(h)$ is a morphism of the Lie group germ W into H. By Theorem 1, $L(h)$ coincides with h on a neighbourhood of 0, whence the proposition.

Loosely speaking, if G and H are identified in a neighbourhood of the identity element with $L(G)$ and $L(H)$ by means of exponential mappings, every morphism of G into H is linear in a neighbourhood of 0.

(i) There exists an open neighbourhood V of 0 in $L(G')$ such that $\phi|V$ is an isomorphism of the manifold V onto an open neighbourhood of e in G'.

(ii) Let $x \in L(G)$. The following conditions are equivalent: (a) $x \in L(G')$; (b) $\phi(\lambda x) \in G'$ for $|\lambda|$ sufficiently small.

(i) is obtained by applying Proposition 8 to the canonical injection of G' into G and (ii) follows from (i).

This follows from Proposition 8 and Example 2 of no. 3.

(i) There exists a neighbourhood V of 0 in $L(G)$ such that

for all $x \in V$.

(ii) If $g \in G$, there exists a neighbourhood W of 0 in $L(G)$ such that

for all $x \in W$.

(i) follows from Corollary 2 and § 3, no. 12, Proposition 44. (ii) follows from Proposition 8 applied to Int g.

(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).

(i) There exists on $H$ one and only one analytic manifold structure with the following property: for all $r$ between 1 and $\omega$, for every manifold $V$ of class $C^r$ and for every mapping $f$ of $V$ into $H$, $f$ is of class $C^r$ as a mapping of $V$ into $H$ if and only if $f$ is of class $C^r$ as a mapping of $V$ into $G$.

(ii) With this structure, $H$ is a Lie group, the canonical injection $i$ of $H$ into $G$ is an immersion and $L(i)(L(H))$ is the Lie subalgebra tangent at $e$ to $H$.

In (i) the uniqueness is obvious. We prove the existence. Let $\mathfrak{h}$ be the Lie algebra tangent at $e$ to $H$. Let $H'$ be a Lie subgroup germ of $G$ with Lie algebra $\mathfrak{h}$. By replacing $H'$ by an open subgroup germ of $H'$, it can be assumed that $H' \subset H$ (Lemma 4 (iv)). For all $x \in H$, $xH'x^{-1}$ is a Lie subgroup germ of $G$ with Lie algebra $x\mathfrak{h}x^{-1} = \mathfrak{h}$. Hence $H' \cap (xH'x^{-1})$ is open in $H'$ (no. 2, Theorem 3) and the mapping $y \mapsto xyx^{-1}$ is an isomorphism of $H' \cap x^{-1}H'x$ onto $xH'x^{-1} \cap H'$. Using Proposition 18 of § 1, no. 9, there exist an open Lie subgroup germ $W$ of $H'$ and a Lie group structure on $H$ with the following properties: $W$ is open in $H$ and the manifold structures on $H$ and $H'$ induce the same structure on $W$. It then follows that the canonical injection $i$ of $H$ into $G$ is an immersion and that $L(i)(L(H)) = L(H') = \mathfrak{h}$. Moreover, let $V$ and $f$ be as in (i). If $f : V \to H$ is of class $C^r$, $i \circ f : V \to G$ is of class $C^r$. Suppose that $i \circ f : V \to G$ is of class $C^r$; then we prove that $f : V \to H$ is of class $C^r$. By translation, it suffices to consider the case where there exists $v_0 \in V$ such that $f(v_0) = e$ and to prove that $f : V \to H$ is of class $C^r$ in an open neighbourhood of $v_0$. Now, by Lemma 4 (iii), $f(v) \in H'$ for $v$ sufficiently close to $v_0$, whence our assertion. Thus we have proved (i) and (ii) has been obtained on the way.

Let $\theta$ and $\theta'$ be the mappings of $F \oplus F'$ into itself defined as follows: if $(u, v) \in F_x \oplus F'_x$, then

By Differentiable and Analytic Manifolds, R, 7.7.1, $\theta$ and $\theta'$ are morphisms of $F \oplus F'$ into itself. Clearly $\theta \circ \theta' = \theta' \circ \theta = \mathrm{Id}_{F \oplus F'}$. Hence $\theta$ and $\theta'$ are automorphisms of $F \oplus F'$. Therefore, $S = \theta(F \oplus \{0\})$ is a vector subbundle of $F \oplus F'$.