It can be assumed that $\phi(e) = 0$. Let $U' = \phi(U)$. Let $\psi : U' \to U$ be the inverse mapping of $\phi$. Let V be a symmetric open neighbourhood of e such that $V^2 \subset U$ and let $V' = \phi(V)$. We define mappings $\theta_1, \theta_2$ of $V' \times V'$ into $V' \times U'$ as follows:

It is immediately verified that $\theta_2(\theta_1(x, y)) = \theta_1(\theta_2(x, y)) = (x, y)$ for $x, y$ sufficiently close to 0. On the other hand, $\theta_1$ and $\theta_2$ are analytic and hence strictly differentiable at $(0, 0)$. Therefore (Differentiable and Analytic Manifolds, R, 1.2.2) there exist a neighbourhood $W'$ of 0 in $V'$ and constants $a > 0, b > 0$ such that

for all $x_1, x_2, y_1, y_2$ in $W'$. Writing $x_1 = x_2 = y_2$, we obtain

For $\delta > 0$, let $N_\delta$ be the set of ordered pairs $(x, y) \in W' \times W'$ such that $\| x - y \| \leq \delta$. The $N_\delta$ form a fundamental system of entourages in $W'$. We write $W = \psi(W')$. Let $M_\delta$ be the set of ordered pairs $(u, v) \in W \times W$ such that $\| \phi(uv^{-1}) \| \leq \delta$. The $M_\delta$ form a fundamental system of entourages in $W$ with the right uniform structure. But relation (2) proves that

and hence $W$ has the property of the lemma.

Since $e$ admits an open neighbourhood homeomorphic to an open ball of a normed space, $e$ admits a countable fundamental system of neighbourhoods whose intersection is $\{e\}$. Hence $G$ is metrizable (General Topology, Chapter III, § 1, Corollary to Proposition 2 and Chapter IX, § 3, Proposition 1). By Lemma 1, there exists a neighbourhood of $e$ which is complete under the right uniform structure and hence $G$ is complete (General Topology, Chapter III, § 3, Proposition 4).

Let $U$ be a neighbourhood of $e$. There exists an open neighbourhood $U_1$ of $e$ contained in $U$ and homeomorphic to an open ball of a normed space $E$ over $K$. If $K = \mathbf{R}$ or $\mathbf{C}$, $U_1$ is connected, which proves (i). Suppose that $K$ is ultrametric. There exists a neighbourhood $U_2$ of $e$ which is closed in $G$ and such that $U_2 \subset U_1$. Then there exists a neighbourhood $U_3$ of $e$ such that $U_3 \subset U_2$ and $U_3$ is open and closed relative to $U_1$. Then $U_3$ is closed relative to $U_2$ and hence $G$ and open relative to $U_1$ and hence $G$. This proves (ii). For $G$ to be locally compact, it is necessary and sufficient that $E$ be locally compact; if $K$ is locally compact, this amounts to saying that $E$ is finite-dimensional (Topological Vector Spaces, Chapter I, § 2, Theorem 3), whence (iii). Suppose that $G$ is generated by a subset $V$ and let $W = V \cup V^{-1}$; then

if there exists a sequence dense in $V$, we see that there exists a sequence dense in G and, as G is metrizable (Proposition 1), the topology on G admits a countable base.

Then there exists an open neighbourhood W of e in V and an automorphism θ of the manifold W such that θ(e) = e, θ(θ(x)) = x and m(x, θ(x)) = m(θ(x), x) = e for all x ∈ W.

m(e, y) = y for all y ∈ U and hence, by the implicit function theorem, there exists an open neighbourhood W_1 of e in V and a mapping θ_1 of class C^r of W_1 into V such that θ_1(e) = e, m(x, θ_1(x)) = e for all x ∈ W_1. Similarly, there exists an open neighbourhood W_2 of e in V and a mapping θ_2 of class C^r of W_2 into V such that θ_2(e) = e, m(θ_2(x), x) = e for all x ∈ W_2. For x ∈ W_1 ∩ W_2,

Let θ(x) be the common value of θ_1(x) and θ_2(x) for x ∈ W_1 ∩ W_2. Let W be the set of x ∈ W_1 ∩ W_2 such that θ(x) ∈ W_1 ∩ W_2. The set W is open. For x ∈ W,

and hence θ(x) ∈ W. We see that θ | W defines an automorphism of the manifold W.

By Lemma 2, G is a neighbourhood of the identity element. For all g ∈ G, the mapping x ↦ m(g, x) is an automorphism of the manifold X. Hence the image of G under this mapping is a neighbourhood of g, obviously contained in G. Therefore G is open in X. Clearly conditions (GL_1) and (GL_2) hold. Condition (GL_3) holds by Lemma 2.

Examples of Lie groups

(1) Let E be a complete normable space over K. The mapping (x, y) ↦ x − y of E × E into E is continuous and linear and hence analytic.

Hence E, with its additive group and analytic manifold structures, is a Lie group.

In particular, K is a Lie group.

(2) Let A be a complete normable unital associative algebra over K. The multiplication $(x, y) \mapsto xy$ of $A \times A$ into A is bilinear and continuous and hence analytic. Proposition 3 shows that the group $A^*$ of invertible elements of A is open in A (which also follows from General Topology, Chapter IX, § 3, Proposition 13) and that $A^*$ is a Lie group.

For example, let E be a complete normable space over K and let $A = \mathscr{L}(E)$ (General Topology, Chapter IX, § 3, Proposition 5). Then $A^*$ is the automorphism group $\mathbf{GL}(E)$ of E. This group therefore has canonically a Lie group structure over K. More particularly, $\mathbf{GL}(n, K)$, with the manifold structure induced by that on $\mathbf{M}_n(K)$, is a Lie group. For $n = 1$, we see that the multiplicative group $K^*$ is a Lie group with the manifold structure induced by that on K.

(3) Let G be a Lie group over K. Let $K' = \mathbf{R}$ or $\mathbf{C}$ or a non-discrete complete ultrametric field and $\sigma$ an isomorphism of the valued field $K'$ onto a valued subfield of K. Then the group G, with the $K'$-manifold structure obtained by restriction of scalars, is a Lie group over $K'$, which is said to be derived from the Lie group G by restriction of scalars (from K to $K'$ by means of $\sigma$). For example, every complex Lie group has canonically a real Lie group structure. Again, with every complex Lie group G is associated a complex Lie group called the conjugate of G, derived from G by means of the automorphism $z \mapsto \overline{z}$ of $\mathbf{C}$.

The condition is obviously necessary. Suppose that it holds. For all $x_0 \in G$, $f(x_0x) = f(x_0)f(x)$ for all $x \in U$ and hence $f|_{x_0U}$ is analytic. But the sets $x_0U$, where $x_0 \in G$, form an open covering of G.

By Differentiable and Analytic Manifolds, R, 5.8.3, H is locally closed. Hence H is closed (General Topology, Chapter III, § 2, Proposition 4). Assertion (ii) is obvious. Assertion (iii) follows from Differentiable and Analytic Manifolds, R, 5.8.5.

The condition is obviously necessary. Suppose that it holds. For all h' ∈ H, the translation γ(h'h^{-1}) is an automorphism of the manifold G and maps the submanifold H ∩ U of U into the submanifold (h'h^{-1}H) ∩ (h'h^{-1}U) of h'h^{-1}U. As h'h^{-1}H = H and h'h^{-1}U is an open neighbourhood of h' in G, we see that every point of H has an open neighbourhood V such that V ∩ H is a submanifold of G. Hence H is a submanifold of G.

Let G be a Lie group and H a Lie subgroup of G. If L is a Lie subgroup of H, L is a Lie subgroup of G by Differentiable and Analytic Manifolds, R, 5.8.6. Let M be a Lie subgroup of G such that M ⊂ H. Then M is a Lie subgroup of H, for the canonical injection of M into H is obviously an immersion.

Let k be a non-discrete closed subfield of K. A Lie k-subgroup of G is a Lie subgroup of the underlying Lie k-group of G.

For $l, l'$ in $L$ and $m, m'$ in $M$,

whence the proposition.

If the conditions of Proposition 7 hold, the Lie group $S$ is called the (external) semi-direct product Lie group of $L$ by $M$ relative to $\sigma$.

Clearly the canonical injection of $L$ (resp. $M$) into $S$ is an isomorphism of $L$ (resp. $M$) onto a Lie subgroup of $S$ which we identify with $L$ (resp. $M$). The canonical mapping of $S$ onto $L$ is a Lie group morphism.

Conversely, let $G$ be a Lie group and $L$, $M$ two Lie subgroups such that the group $G$ is (algebraically) the semi-direct product of $L$ by $M$ (Algebra, Chapter I, § 6, no. 1). We write $\sigma(l)m = lml^{-1}$ for $l \in L$ and $m \in M$. Then $\sigma$ satisfies the conditions of Proposition 7. We can therefore form the semi-direct product Lie group $S$ of $L$ by $M$ relative to $\sigma$. The mapping $j: (m, l) \mapsto ml$ of $S$ onto $G$ is a group isomorphism and is analytic. If $j$ is a Lie group isomorphism, the Lie group $G$ is called the (internal) semi-direct product of $L$ by $M$ and $S$ and $G$ are identified. For all $g \in G$, we write $g = p(g)q(g)$, where $p(g) \in M$ and $q(g) \in L$. For the Lie group $G$ to be the semi-direct product of $L$ by $M$, it is necessary and sufficient that one of the mappings $p: G \to M$ and $q: G \to L$ be analytic, in which case both are analytic; or alternatively, it is necessary and sufficient that $T_e(G)$ be the topological direct sum of $T_e(M)$ and $T_e(L)$ (for, if this condition holds, $j$ is étale at $e_S$).

$T_e(p) \circ T_e(s) = \mathrm{id}_{T_{e(H)}}$ and hence $p$ (resp. $s$) is a submersion (resp. an immersion). By Differentiable and Analytic Manifolds, R, 5.10.5, N is a Lie subgroup of G. On the other hand, $s$ is a homeomorphism of H onto $s(H)$ and hence $s$ is an isomorphism of H onto a Lie subgroup of G (Differentiable and Analytic Manifolds, R, 5.8.3). Finally, for all $g \in G$, $g = (s \circ p)(g) \cdot n$ for some $n \in N$; as $s \circ p$ is analytic, the Lie group G is the semi-direct product of $s(H)$ by N.

(i) If $\rho(x)$ is an immersion (resp. a submersion, a subimmersion) at $g_0$, then, for all $g \in G$, $\rho(gx)$ is an immersion (resp. a submersion, a subimmersion).

(ii) If $\rho(x)$ is of rank k at $g_0$, then for all $g \in G$, $\rho(gx)$ is of constant rank equal to k.

This follows immediately from formulae (4) and (5) since $T_g(\delta(g)), T_x(\tau(g))$ and $T_g(\gamma(g^{-1}))$ are isomorphisms.

This follows from Proposition 9 and Differentiable and Analytic Manifolds, R, 5.10.6.

Observe that, if $\eta$ denotes the mapping $(g, x) \mapsto gx$ of $G \times X$ into X, then, for $g \in G, x \in X, u \in T_g(G), v \in T_x(X)$,

that is

(a) the group G operates properly and freely on X;

(b) for all $x \in X$, $\rho(x)$ is an immersion (which is a consequence of (a) if K is of characteristic 0 and X is finite-dimensional).

Then the equivalence relation defined by G on X is regular (Differentiable and Analytic Manifolds, R, 5.9.5). There exists on the quotient set X/G one and only one manifold structure such that the canonical mapping $\pi : X \to X/G$ is a submersion. The underlying topology of this manifold structure is the quotient topology of that on X; it is Hausdorff. Finally, $(X, G, X/G, \pi)$ is a principal left fibre bundle.\footnote{† The principal fibre bundles defined in Differentiable and Analytic Manifolds, R, 6.2.1 are principal right fibre bundles. The definition of principal left fibre bundles can be deduced from this in an obvious way.}

Let $\theta$ be the mapping $(g, x) \mapsto (x, gx)$ of $G \times X$ into $X \times X$. This mapping is of class $C^\tau$. We show that it is an immersion. For $u \in T_g(G)$ and $v \in T_x(X)$, by (6),

But $T_g(\rho(x))$ is injective by hypothesis (b) and hence $T_{(g, x)}(\theta)$ is injective. Its image is the topological direct sum of the subspace $H_{g, x}$ consisting of the vectors $(v, T_x(\tau(g))v)$ for $v \in T_x(X)$ and the subspace

By hypothesis (b), $T_g(\rho(x))(T_g(G))$ admits a topological supplement $J_{g, x}$ in $T_{gx}(X)$. Hence the image of $T_{(g, x)}(\theta)$ admits the topological supplement $\{0\} \times J_{g, x}$. Hence we have proved that $\theta$ is an immersion of $G \times X$ in $X \times X$.

As G operates freely on X, $\theta$ is injective. Let C be the graph of the equivalence relation R defined by G on X. As G operates properly, $\theta$ is a homeomorphism of $G \times X$ onto C (General Topology, Chapter I, § 10, Proposition 2). By Differentiable and Analytic Manifolds, R, 5.8.3, it is a submanifold of $X \times X$ and $\theta$ is an isomorphism of the manifold $G \times X$ onto the manifold C. The tangent space $T_{(x, gx)}(C)$ is identified with

Let pr$_1$ and pr$_2$ be the canonical projections of $X \times X$ onto the two factors. It is immediate that $T_{(x, gx)}(\mathrm{pr}_1)$ maps $H_{g, x}$ onto $T_x(X)$ and that the kernel of $T_{(x, gx)}(\mathrm{pr}_1) | T_{(x, gx)}(C)$ is $I_{g, x}$. Thus $\mathrm{pr}_1 | C$ is a submersion of C onto X. By Differentiable and Analytic Manifolds, R, 5.9.5, R is regular. By definition, there therefore exists on the quotient set X/G one and only one manifold structure such that $\pi$ is a submersion. The underlying topology of X/G is the quotient topology of that of X (Differentiable and Analytic Manifolds, R, 5.9.4). This topology is Hausdorff (General Topology, Chapter III, § 4, Proposition 3).

For all $b \in X/G$, there exist an open neighbourhood W of b and a morphism $\sigma : W \to X$ such that $\pi \circ \sigma = \mathrm{id}_W$ (Differentiable and Analytic Manifolds, R, 5.9.1). Let $\phi$ be the bijection $(g, w) \mapsto g\sigma(w)$ of $G \times W$ onto $\pi^{-1}(W)$. It is of class $C^\tau$. Then $\pi(g\sigma(w)) = w$ and

and hence the inverse bijection of $\phi$ is of class $C^r$. Clearly $\phi(gg', w) = g\phi(g', w)$ for $w \in W,\ g \in G,\ g' \in G$. Hence $(X, G, X/G, \pi)$ is a principal left fibre bundle.

(i) There exists on the homogeneous set $X/G$ one and only one analytic manifold structure such that the canonical projection $\pi$ of $X$ onto $X/G$ is a submersion. The law of operation of $X$ on $X/G$ is analytic. For all $x \in X$, the kernel of $T_x(\pi)$ is obtained from $T_e(G)$ by $T_e(\gamma(x))$.

(ii) If $G$ is normal in $X$, $X/G$ is a Lie group with its group structure and the manifold structure defined in (i). The mapping $\pi$ is a Lie group morphism.

By General Topology, Chapter III, § 4, no. 1, Example 1, $G$ operates properly and freely on $X$ by right translation. Hence the first assertion of (i) follows from

Proposition 10 of no. 5. The second follows from the Remark of no. 5. Since $\pi$ is a submersion, the kernel of $T_x(\pi)$ is the tangent space at $x$ to

and hence is obtained from $T_e(G)$ by $T_e(\gamma(x))$.

Suppose that $G$ is normal. Let $m$ be the mapping $(x, y) \mapsto xy^{-1}$ of $(X/G) \times (X/G)$ into $X/G$. Then $(m \circ (\pi \times \pi))(x, y) = \pi(xy^{-1})$ for all $x, y$ in $X$. Hence $m \circ (\pi \times \pi)$ is analytic. As $\pi \times \pi$ is a surjective submersion, $m$ is analytic (Differentiable and Analytic Manifolds, R. 5.9.5), whence (ii).

The homogeneous set $X/G$ with the manifold structure defined in (i) is called the quotient (left) Lie homogeneous space of $X$ by $G$. The (right) Lie homogeneous space $G \setminus X$ is defined analogously. When $G$ is normal, the Lie group $X/G$ defined in (ii) is called the quotient Lie group of $X$ by $G$.

As the canonical mapping of $X$ onto $X/X_y$ is a submersion, the equivalences (i) $\Leftrightarrow$ (ii), (i') $\Leftrightarrow$ (ii') are immediate. (i) $\Leftrightarrow$ (i') by Proposition 9 of no. 5. The equivalence (i') $\Leftrightarrow$ (iii) follows from formula (7) of no. 5.

Under the conditions of Proposition 12, $Y$ is called a (left) Lie homogeneous space of $X$. Right Lie homogeneous spaces of $X$ are defined analogously.

(i) The law of left operation $(h, x) \mapsto hx$ of $H$ on $X$ satisfies conditions (a) and (b) of Proposition 10 (so that we can consider the quotient manifolds $X/G$ and $X/H$).

(ii) The law of left operation of $G$ on $X$ defines on taking quotients a law of left operation of class $C^r$ of $G/H$ on $X/H$; this law satisfies conditions (a) and (b) of Proposition 10 (so that we can consider the quotient manifold $(X/H)/(G/H)$).

(iii) The canonical mapping of $X$ onto $X/H$ defines on taking quotients a bijection of $X/G$ onto $(X/H)/(G/H)$. This bijection is an isomorphism of manifolds of class $C^r$.

Clearly $H$ operates freely on $X$; it operates properly by General Topology, Chapter III, § 4, no. 1, Example 1. The orbital mappings of $H$ on $X$ are immersions since the canonical injection of $H$ into $G$ is an immersion. This proves (i).

The law of left operation of $G$ on $X$ obviously defines, on taking quotients, a law of left operation of $G/H$ on $X/H$. This law is of class $C^r$ by Differentiable and Analytic Manifolds, R, 5.9.6. Let $g \in G$ and $x \in X$ be such that $(Hg)(Hx) = Hx$; then $H(gx) = Hx$ and hence $gx \in Hx$ and $g \in H$; this proves that $G/H$ operates freely on $X/H$. The mapping $\theta : (g, x) \mapsto (x, gx)$ of $G \times X$ into $X \times X$ is closed; on the other hand, $\theta(Hg \times Hx) = Hx \times H(gx)$; it follows immediately that the mapping

of $(G/H) \times (X/H)$ into $(X/H) \times (X/H)$ is closed; as moreover $G/H$ operates freely on $X/H$, Theorem 1 (c) of General Topology, Chapter I, § 10, no. 2 proves that $G/H$ operates properly on $X/H$.

Let $\pi$ be the canonical mapping of $X$ onto $X/H$, $\sigma$ the canonical mapping of $G$ onto $G/H$, $x$ an element of $X$ and $y = \pi(x)$.

Then $\pi \circ \rho(x) = \rho(y) \circ \sigma$ and hence

Let $u \in T_e(G/H)$ be such that $T_e(\rho(y))u = 0$. There exists $v \in T_e(G)$ such that $u = T_e(\sigma)v$. Then $T_x(\pi)(T_e(\rho(x))v) = 0$, hence $T_e(\rho(x))v$ is tangent to $Hx$ (Differentiable and Analytic Manifolds, R, 5.10.5) and therefore is of the form $T_e(\rho(x) | H)v'$ for some $v' \in T_e(H)$. As $T_e(\rho(x))$ is injective, it follows that $v = v'$, whence $v \in T_e(H)$ and therefore $u = 0$. Thus $T_e(\rho(y))$ is injective. The image of $T_e(\rho(y))$ is equal to that of $T_x(\pi) \circ T_e(\rho(x))$; now the image of $T_e(\rho(x))$ admits a topological supplement in $T_x(X)$ and contains the kernel of $T_x(\pi)$. It is therefore seen that $\rho(y)$ is an immersion, which completes the proof of (ii).

Assertion (iii) follows from the above and Differentiable and Analytic Manifolds, R, 5.9.7.