§ 6. Real and complex Lie groups

In this paragraph, K is assumed to be equal to $\mathbf{R}$ or $\mathbf{C}$ .

1. Passage from Lie algebra morphisms to Lie group morphisms

Lemma 1

Let G be a simply connected† topological group, W a symmetric connected open neighbourhood of e, H a group and f a mapping of $W^3$ into H such that

f(xyz) = f(x)f(y)f(z)

for x, y, z in W. There exists a morphism $f'$ of G into H such that $f' \mid W = f \mid W$ .

For $(g, h) \in G \times H$ and U an open neighbourhood of e in W, let $A(g, h, U)$ be the set of $(gu, hf(u)) \in G \times H$ for $u \in U$ . Then $(g, h) \in A(g, h, U)$ and $A(g, h, U_1) \cap A(g, h, U_2) = A(g, h, U_1 \cap U_2)$ . Let $(s, t) \in A(g, h, U)$ ; then $s = gu$ and $t = hf(u)$ for $u \in U$ ; there exists an open neighbourhood $U'$ of e in W such that $uU' \in U$ ; then, for $u' \in U'$ ,

(su', tf(u')) = (guu', hf(uu')) \in A(g, h, U)

† Cf. Chapter XI of General Topology (to appear). It is proved in this chapter that if $G_1, G_2$ are connected topological groups, $\phi$ is an open continuous homomorphism of $G_1$ onto $G_2$ with discrete kernel and $G_2$ is simply connected, then $\phi$ is a homeomorphism. Recall on the other hand that a simply connected space is connected.

and hence $A(s, t, U') \subset A(g, h, U)$ . It then follows that the $A(g, h, U)$ form the base of a topology on $G \times H$ . We shall denote by $Y$ the set $G \times H$ with this topology and denote by $p$ the canonical projection of $Y$ onto $G$ , which is open. The restriction of $p$ to $A(g, h, U)$ is a homeomorphism of $A(g, h, U)$ onto $gU$ . Hence $(Y, p)$ is a covering space of $G$ . Let $Y_0$ be the subgroup of $Y$ generated by $A(e, e, W)$ and let $\mathfrak{B}$ be the set of $A(e, e, U)$ . Clearly $\mathfrak{B}$ satisfies conditions (GV $_1'$ ) and (GV $_1''$ ) of General Topology, Chapter III, § 1, no. 2. The set $Y_0'$ of $y \in Y_0$ such that the mappings $z \mapsto yzy^{-1}$ and $z \mapsto y^{-1}zy$ of $Y_0$ into $Y_0$ are continuous at $(e, e)$ is a subgroup of $Y_0$ . Let $w \in W$ . The mapping $w' \mapsto ww'w^{-1}$ of $W$ into $G$ is continuous and hence the mapping

(w', f(w')) \mapsto (ww'w^{-1}, f(ww'w^{-1}))

of $A(e, e, W)$ into $Y$ is continuous at $(e, e)$ . Now $f(ww'w^{-1}) = f(w)f(w')f(w^{-1})$ and therefore $(ww'w^{-1}, f(ww'w^{-1})) = (w, f(w))(w', f(w'))(w, f(w))^{-1}$ .

As $w^{-1} \in W$ , we see that $(w, f(w)) \in Y_0'$ . Thus, $A(e, e, W) \subset Y_0'$ , so that $Y_0' = Y_0$ . The group $Y_0$ , with the filter base $\mathfrak{B}$ , therefore satisfies condition (GV $_{III}'$ ) of General Topology, Chapter III, § 1, no. 2. As

(g, h).A(e, e, U) = A(g, h, U),

$Y_0$ is a topological group which is connected since $A(e, e, W)$ is connected. Then $p(Y_0)$ is an open subgroup of $G$ , whence $p(Y_0) = G$ since $G$ is connected. The kernel of $p \mid Y_0$ is discrete. As $G$ is simply connected, $p \mid Y_0$ is a homeomorphism of $Y_0$ onto $G$ . Therefore $Y_0$ is the graph of a morphism $f'$ of $G$ into $H$ . For $g \in W$ , $(g, f(g)) \in A(e, e, W) \subset Y_0$ , whence $f'(g) = f'(g)$ .

Theorem 1

Let $G$ and $H$ be Lie groups and $h$ a continuous morphism of $L(G)$ into $L(H)$ . Suppose that $G$ is simply connected. Then there exists one and only one Lie group morphism $\phi$ of $G$ into $H$ such that $h = L(\phi)$ .

The existence of $\phi$ follows from Lemma 1 and § 4, no. 1, Theorem 1 (i). The uniqueness of $\phi$ follows from § 4, no. 1, Theorem 1 (ii) and the fact that $G$ is connected.

Corollary

Let $G$ be a finite-dimensional simply connected Lie group. There exists a finite-dimensional analytic linear representation of $G$ whose kernel is discrete.

There exist (Chapter I, § 7, Theorem 2) a finite-dimensional vector space $E$ and an injective morphism $h$ of $L(G)$ into the Lie algebra $\mathrm{End}(E)$ . By Theorem 1, there exists a morphism $\phi$ of $G$ into $\mathbf{GL}(E)$ such that $L(\phi) = h$ . Hence $\phi$ is an immersion and therefore its kernel is discrete.

Remark

(1) There exist finite-dimensional simply connected Lie groups which have no finite-dimensional injective analytic linear representation (Exercise 2).

(2) There exist finite-dimensional connected Lie groups $G$ such that every finite-dimensional analytic linear representation of $G$ has non-discrete kernel (Exercises 3 and 4).

2. Integral subgroups

Definition 1

Let G be a Lie group. An integral subgroup of G is a subgroup H with a connected Lie group structure such that the canonical injection of H into G is an immersion.

A one-parameter subgroup of G is a 1-dimensional integral subgroup of G.

Let H be an integral subgroup of G and i the canonical injection of H into G. Then L(i) defines an isomorphism of L(H) onto a Lie subalgebra of L(G) admitting a topological supplement. L(H) is identified with its image under L(i).

Example

(1) A connected Lie subgroup of G is an integral subgroup of G.

(2) Suppose that G is finite-dimensional. Let H be a subgroup of G; we give it the structure induced by the Lie group structure on G (§ 4, no. 5, Definition 3). Then its identity component H_0 is an integral subgroup of G and the subalgebra tangent to H at e is L(H_0) (§ 4, no. 5, Proposition 9 (ii)).

(3) Let G be a complex Lie group, H an integral subgroup of G and G_1 (resp. H_1) the underlying real Lie group of G (resp. H). Then H_1 is an integral subgroup of G_1 and L(H_1) is the underlying real Lie algebra of L(H).

Theorem 2

Let G be a Lie group.

(i) The mapping H ↦ L(H) is a bijection of the set of integral subgroups of G onto the set of Lie subalgebras of L(G) admitting a topological supplement.

(ii) Let H be an integral subgroup of G. Every connected Lie subgroup germ of G with Lie algebra L(H) is an open submanifold of H which generates H.

(a) Let h be a Lie subalgebra of L(G) admitting a topological supplement. Let H_1 be a Lie subgroup germ of G such that L(H_1) = h (§ 4, Theorem 3). H_1 can be chosen so that it is connected. Let H be the subgroup of G generated by H_1. There exists (§ 1, Corollary to Proposition 22) a Lie group structure on H such that H_1 is an open submanifold of H and the canonical injection of H into G is an immersion. As H_1 is connected, H is connected and is hence an integral subgroup of G. Then L(H) = L(H_1) = h. This proves that the mapping considered in (i) is surjective.

(b) Let H be an integral subgroup of G and N_1 a connected Lie subgroup germ of G with Lie algebra L(H). As the canonical injection of H into G is an immersion, there exists an open subgroup germ H_1 of H which is at the same time a submanifold of G and hence a Lie subgroup germ of G with Lie algebra L(H). On the other hand, let N be the subgroup of G generated by N_1; by part (a) of the proof, it has the structure of an integral subgroup of G such that N_1 is an open submanifold of N. By § 4, Theorem 3, H_1 ∩ N_1 is open in H_1 and N_1. Hence the subgroup of G generated by H_1 ∩ N_1 is equal on the one hand to H and on the other to N. Therefore the Lie groups H and N are equal. This proves (ii) and also proves that the mapping considered in (i) is injective.

Remark 1

Let H be an integral subgroup of G. Let Y be the left foliation of G associated with L(H). If $g \in G$ , let $gH$ be given the manifold structure derived from that on H by $\gamma(g)$ . By Differentiable and Analytic Manifolds, R, 9.3.2, the canonical injection of $gH$ into Y is a morphism. This morphism is étale. Hence the maximal connected leaves of Y are the left cosets modulo H.

Proposition 1

Let G and M be Lie groups, H an integral subgroup of G and $\phi$ a morphism of M into G such that $L(\phi)(L(M)) \subset L(H)$ . Suppose that M is connected. Then $\phi(M) \subset H$ and $\phi$ , considered as a mapping of M into H, is a Lie group morphism.

In the notation of Remark 1, $\phi$ is a morphism of M into Y (Differentiable and Analytic Manifolds, R, 9.3.2) and hence $\phi(M) \subset H$ since M is connected.

Corollary 1

Let G and H be Lie groups, $\phi$ a Lie group morphism of G into H, N the kernel of $\phi$ and $h = L(\phi)$ . Suppose that G is connected and that H is finite-dimensional.* (i) N is a Lie subgroup of G and $L(N) = \mathrm{Ker}\ h$ . (ii) Let H' be the integral subgroup of H with Lie algebra Im h. Then $\phi(G) = H'$ . (iii) *The mapping of G/N into H' derived from $\phi$ when passing to the quotient is a Lie group isomorphism.

(i) has already been proved (§ 3, no. 8, Proposition 28).

Let $\psi$ be the Lie group morphism of G/N into H derived from $\phi$ when passing to the quotient; it is an immersion (§ 3, no. 8, Proposition 28). By Proposition 1, $\psi$ is a Lie group morphism of G/N into H'. This morphism is étale and hence $\psi(G/N) = H'$ since H' is connected; this proves (ii). Then $\psi : G/N \to H'$ is bijective and is a Lie group isomorphism, which proves (iii).

Corollary 2

Let G be a Lie group and H_1 and H_2 integral subgroups of G. If $L(H_2) \subset L(H_1)$ , then H_2 is an integral subgroup of H_1.

Let $i_1 : H_1 \to G, i_2 : H_2 \to G$ be the canonical injections. Then

L(i_2)(L(H_2)) = L(H_2) \subset L(H_1).

By Proposition 1,

i_2

is an analytic mapping of

H_2

into

H_1

and even an immersion of

H_2

into

H_1

since

L(i_2)

is an isomorphism of

L(H_2)

onto a subalgebra of

L(H_1)

admitting a topological supplement.

Corollary 3

Let G be a finite-dimensional Lie group and $(H_i)_{i \in I}$ a family of Lie subgroups of G. Then $H = \bigcap_{i \in I} H_i$ is a Lie subgroup of G and*

L(H) = \bigcap_{i \in I} L(H_i).

There exists a finite subset J of I such that

\bigcap_{i \in J} L(H_i)

is equal to the intersection M of all the L(H_i). We know that H^* = \bigcap_{i \in J} H_i is a Lie subgroup such that L(H^*) = M (§ 3, no. 8, Corollary 2 to Proposition 29). Let H_0 be the identity component of H^*. It is a Lie subgroup of G and L(H_0) = M. By Corollary 2, H_0 \subset H_i for all i and hence H_0 \subset H \subset H^*, whence the corollary.

Corollary 4

Let G be a finite-dimensional connected Lie group. The following conditions are equivalent:* (i) G is unimodular (Integration, Chapter VII, § 1, no. 3, Definition 3); (ii) det Ad g = 1 for all g \in G; (iii) Tr ad a = 0 for all a \in L(G).

The mapping g \mapsto det Ad g is a morphism $\phi$ of G into K*. By § 3, Propositions 35 (no. 10) and 44 (no. 12), $L(\phi)a = Tr\ ad\ a$ for all $a \in L(G)$ . Clearly Im $L(\phi) = \{0\}$ or K. In the first (resp. second) case, Im $\phi = \{1\}$ (resp. Im $\phi = K^*$ ) by Corollary 1 and hence G is unimodular (resp. not unimodular) by § 3, no. 16, Corollary to Proposition 55.

Proposition 2

Let G be a finite-dimensional Lie group and H an integral subgroup of G. The following conditions are equivalent:* (i) H is closed; (ii) the topology on H is induced by that on G; (iii) H is a Lie subgroup of G. (i) $\Rightarrow$ (iii): this follows from § 1, Propositions 2 (iv) (no. 1) and 14 (iii) (no. 7). (iii) $\Rightarrow$ (ii): obvious. (ii) $\Rightarrow$ (i): if the topology on H is induced by that on G, H is closed because H is complete (§ 1, no. 1, Proposition 1).

Proposition 3

Let G be a Lie group, H an integral subgroup of G, M a non-empty connected analytic manifold, f a mapping of M into G and r \in N_K. Consider the following conditions:* (i) f is of class C^r and f(M) \subset H; (ii) f(M) \subset H and f, considered as a mapping of M into H, is of class C^r; (iii) f is of class C^r, f(M) meets H and the image of Tm(M) is contained in f(m).L(H) for all m \in M*.

Then (ii) $\Leftrightarrow$ (iii) $\Rightarrow$ (i). If the topology on H admits a countable base, the three conditions are equivalent. (ii) $\Rightarrow$ (i) and (ii) $\Rightarrow$ (iii): obvious. (iii) $\Rightarrow$ (ii): suppose that condition (iii) holds. By Differentiable and Analytic Manifolds, R, 9.2.8, f is a morphism of class C^r of M into the left foliation associated with L(H). As M is connected, f(M) \subset H.

If the topology on H admits a countable base, condition (i) implies that f is a mapping of class C^r of M into H (Differentiable and Analytic Manifolds, R, 9.2.8); hence (i) $\Rightarrow$ (ii).

Corollary 1

Let G be a finite-dimensional Lie group and H an integral subgroup of G. Then the Lie subalgebra tangent to H at e (§ 4, no. 5, Definitions 2 and 3) is L(H) and the Lie group structure on H is the structure induced by that on G.

As H is connected and finite-dimensional, its topology admits a countable base.

Corollary 2

Let G be a Lie group and H_1 and H_2 integral subgroups of G. Suppose that the topology on H_1 admits a countable base. Then

H_2 \subset H_1 \Leftrightarrow L(H_2) \subset L(H_1)

and, if these conditions hold, H_2 is an integral subgroup of H_1.

The last assertion and the implication $L(H_2) \subset L(H_1) \Rightarrow H_2 \subset H_1$ follow from Corollary 2 to Proposition 1. The converse implication follows from Proposition 3.

Corollary 3

Let G be a Lie group and H_1 and H_2 integral subgroups of G whose topology admits a countable base. If H_1 and H_2 have the same underlying set, the Lie group structures on H_1 and H_2 are the same.

This follows from Corollary 2.

Remark 2

Let G be a finite-dimensional Lie group. Let H be a subgroup of G. We shall say, by an abuse of language, that H is an integral subgroup of G if there exists a Lie group structure S on H such that H, together with S, is an integral subgroup of G. By Corollary 3 to Proposition 3, if S exists, S is unique.

Remark 3

Let V be a manifold of class C^r. Let M be a subset of V and x and y elements of M. Consider the following property:

P_{M, x, y}: there exist I, x_0, x_1, \ldots, x_n, f_1, \ldots, f_n such that: (a) I is a connected open subset of \mathbf{K}; (b) x_0, \ldots, x_n are in M, x_0 = x, x_n = y; (c) for 1 \leq i \leq n, f_i is a mapping of class C^r of I into V which takes the values x_{i-1} and x_i and f_i(I) \subset M.

We shall say that M is a C^r-connected subset of V if, for all elements x, y of M, the property P_{M, x, y} holds.

Proposition 4

Let G be a finite-dimensional Lie group and H a subgroup of G. Let r \in \mathbf{N}_\mathbf{K}. The following conditions are equivalent: (i) H is an integral subgroup of G; (ii) with the Lie group structure induced by that on G, H is connected; (iii) H is C^r-connected. (iii) \Rightarrow (i): obvious. (i) \Rightarrow (iii): suppose that H has a Lie group structure such that H is an integral subgroup of G. Using the notation of Remark 3, the set of y \in H such that property P_{H, e, y} holds is an open subgroup of H. As H is connected, this subgroup is equal to H and hence condition (iii) is satisfied.

(iii) ⇒ (ii): suppose that condition (iii) is satisfied and let H be given the structure induced by the Lie group structure on G. Let h be the subalgebra tangent to H at e. The identity component H_0 of H is an integral subgroup of G such that L(H_0) = h. We show that H = H_0. It suffices to prove the following: let I be a connected open subset of K, f a mapping of class C^r of I into G such that f(I) ⊂ H and λ and μ two points of I; if f(λ) ∈ H_0, then f(μ) ∈ H_0. But, for all v ∈ I, (T_v f)(K) ⊂ f(v)h by definition of h, so that our assertion follows from Proposition 3.

Remark 4

If K = R, the integral subgroups of G can also be characterized as the subgroups which, with the topology induced by that on G, are arcwise connected (§ 8, Exercise 4). However, subgroups may be connected but not integral (Commutative Algebra, Chapter VI, § 9, Exercise 2).

Corollary

Let G be a finite-dimensional Lie group and H_1 and H_2 two integral subgroups of G. The subgroup of G generated by H_1 and H_2 and the subgroup (H_1, H_2) of G are integral subgroups of G.

The subgroup (G, G) of G is not always closed (§ 9, Exercise 6).

Recall (§ 3, no. 11, Corollary 5 to Proposition 41) that if a is a finite-dimensional algebra, Aut(a) is a Lie subalgebra of GL(a) and that L(Aut(a)) is the Lie algebra of derivations of a.

Definition 2

Let a be a finite-dimensional Lie algebra. Let Ad(a) or Int(a) denote the integral subgroup of Aut(a) with Lie algebra ad(a). The elements of this group are called inner automorphisms of a.

By transport of structure, ad(a) is invariant under Aut(a) and hence Int(a) is normal in Aut(a). By § 4, no. 4, Corollary 1 to Proposition 8 and the fact that Int(a) is connected, the elements of Int(a) are the finite products of automorphisms of the form exp ad x where x ∈ a. In general, Int(a) is not a Lie subgroup of Aut(a) (Exercise 14).

3. Passage from Lie algebras to Lie groups

Theorem 3

(i) If L is a finite-dimensional Lie algebra, there exists a simply connected Lie group G such that L(G) is isomorphic to L.

(ii) Let G_1 and G_2 be two connected Lie groups, with G_1 simply connected. Let f be an isomorphism of L(G_1) onto L(G_2), φ the morphism of G_1 into G_2 such that L(φ) = f and N the kernel of φ. Then N is a discrete subgroup of the centre of G_1 and the morphism of G_1/N into G_2 derived from φ is a Lie group isomorphism. If G_2 is simply connected, φ is an isomorphism.

Let L be a finite-dimensional Lie algebra. There exists a finite-dimensional vector space E such that L can be identified with a Lie subalgebra of End(E) (Chapter I, § 7, Theorem 2). Let H be the integral subgroup of GL(E) with

Lie algebra L. Let $\hat{H}$ be its universal covering ( $§ 1$ , no. 9, Remark). Then $L(\hat{H})$ is isomorphic to L, whence (i).

Let $G_1, G_2, f, \phi, N$ be as in (ii). Then $\phi$ is étale, hence $\phi(G_1)$ is an open subgroup of $G_2$ and hence $\phi(G_1) = G_2$ . On the other hand, N is discrete and hence contained in the centre of $G_1$ (Integration, Chapter VII, $§ 3$ , Lemma 4). Clearly the morphism of $G_1/N$ onto $G_2$ derived from $\phi$ is a Lie group isomorphism. If $G_2$ is simply connected, every étale mapping of $G_1$ onto $G_2$ is injective and hence $N = \{e\}$ .

Proposition 5

Let G be a connected real Lie group. Suppose that $L(G)$ has a complex normable Lie algebra structure $L'$ compatible with its real normable Lie algebra structure. There exists on G one and only one complex Lie group structure compatible with the real Lie group structure and with Lie algebra $L'$ .

By $§ 4$ , no. 2, Corollary 2 to Theorem 2, it suffices to prove that the structure of $L'$ is invariant under Ad G. Let $\phi$ be an exponential mapping of G. By $§ 4$ , no. 4, Corollary 3 (i) to Proposition 8, there exists a neighbourhood V of 0 in $L(G)$ such that the structure of $L'$ is invariant under Ad $\phi(V)$ . But Ad $\phi(V)$ generates G because G is connected.

The conclusion of Proposition 5 is not necessarily true if G is not assumed to be connected (Exercise 7).

Proposition 6

Let G be a connected complex Lie group. If G is compact, G is commutative.

The holomorphic mapping $g \mapsto \mathrm{Ad}\,g$ of G into $\mathscr{L}(L(G))$ is constant (Differentiable and Analytic Manifolds, R, 3.3.7) and hence $\mathrm{ad}\,a = 0$ for all $a \in L(G)$ ( $§ 3$ , no. 12, Proposition 44). Hence G is commutative ( $§ 4$ , Corollary 3 to Theorem 1).

4. Exponential mapping

Theorem 4

Let G be a Lie group. There exists one and only one exponential mapping of G defined on $L(G)$ .

There exist a convex open neighbourhood U of 0 in $L(G)$ and an exponential mapping of G defined on U. It can be assumed, by choosing U sufficiently small, that

\phi((\lambda + \lambda')a) = \phi(\lambda a)\phi(\lambda'a)

for

a \in L(G)

\lambda, \lambda'

in K,

\lambda a, \lambda'a, (\lambda + \lambda')a

in U.

Let $a \in L(G)$ . There exists an integer $n > 0$ such that $\frac{1}{n} a \in U$ . If m is another integer such that $\frac{1}{m} a \in U$ , then $\frac{1}{nm} a \in U$ and relation (1) implies

\phi\left(\frac{1}{n} a\right) = \left(\phi\left(\frac{1}{nm} a\right)\right)^m, \quad \phi\left(\frac{1}{m} a\right) = \left(\phi\left(\frac{1}{nm} a\right)\right)^n;

hence $\left( \phi \left( \frac{1}{n} a \right) \right)^n = \left( \phi \left( \frac{1}{m} a \right) \right)^m$ . There exists an extension $\psi : L(G) \to G$ of $\phi$ such that $\psi(a) = \left( \phi \left( \frac{1}{n} a \right) \right)^n$ for $a \in L(G)$ and $n$ an integer $> 0$ such that $\frac{1}{n} a \in U$ . Clearly $\psi$ is analytic and is an exponential mapping of $G$ . If $\psi' : L(G) \to G$ is an exponential mapping of $G$ , $\psi$ and $\psi'$ coincide on a neighbourhood of 0 and hence are equal since $L(G)$ is connected.

Henceforth when we speak of the exponential mapping of $G$ , we shall mean the mapping considered in Theorem 4. It will be denoted by $\exp_G$ or $\exp$ if there is no risk of confusion.

Example

Let $A$ be a complete normed unital associative algebra. Then $\exp_{A^*}$ is the exponential mapping defined in Chapter II, § 7, no. 3.

Proposition 7

Let $G$ be a Lie group and $a$ an element of $L(G)$ . The mapping $\lambda \mapsto \exp(\lambda a)$ of $K$ into $G$ is the unique morphism $\phi$ of the Lie group $K$ into $G$ such that $(T_0 \phi)1 = a$ .

The mappings $(\lambda, \lambda') \mapsto \exp(\lambda a) \exp(\lambda' a)$ and $(\lambda, \lambda') \mapsto \exp(\lambda + \lambda') a$ of $K \times K$ into $G$ are analytic and coincide on a neighbourhood of $(0, 0)$ . As $K \times K$ is connected, these mappings are equal. Hence $\phi : \lambda \mapsto \exp(\lambda a)$ is a Lie group morphism of $K$ into $G$ . The tangent mapping at 0 to $\lambda \mapsto \lambda a$ is the mapping $\lambda \mapsto \lambda a$ ; and $T_e(\exp) = \mathrm{Id}_{L(G)}$ ; hence $(T_0 \phi)1 = a$ . The uniqueness assertion of the proposition follows from Theorem 1.

Proposition 8

Let $G$ be a Lie group. For all $x, y$ in $L(G)$ and $n$ an integer,

\exp(x + y) = \lim_{n \to + \infty} \left( \left( \exp \frac{1}{n} x \right) \left( \exp \frac{1}{n} y \right) \right)^n

\exp[x, y] = \lim_{n \to + \infty} \left( \left( \exp \frac{1}{n} x \right) \left( \exp \frac{1}{n} y \right) \left( \exp \frac{1}{n} x \right)^{-1} \left( \exp \frac{1}{n} y \right)^{-1} \right)^{n^2}.

By Proposition 7, this follows from Proposition 4 of § 4, no. 3, taking $\lambda = \frac{1}{n}$ .

Proposition 9

Let $G$ be a complex Lie group and $G'$ the underlying real Lie group. Then $\exp_G = \exp_{G'}$ .

This follows from Proposition 5 of § 4, no. 3 and the analyticity of $\exp_G$ and $\exp_{G'}$ .

Proposition 10

Let $G$ and $H$ be Lie groups and $\phi$ a morphism of $G$ into $H$ . (i) $\phi \circ \exp_G = \exp_H \circ L(\phi)$ . (ii) *If $G$ is an integral subgroup of $H$ , then $\exp_G = \exp_H | L(G)$ .

EXponential mapping

The two sides of the equality (i) are analytic mappings of $L(G)$ into $H$ which coincide in a neighbourhood of 0 ( $§ 4$ , Proposition 8, no. 4) and hence are equal. Assertion (ii) is a special case of (i).

Corollary 1

Let $G$ be a Lie group, $G'$ a Lie subgroup of $G$ and $a \in L(G)$ . The following conditions are equivalent*: (i) $a \in L(G')$ ; (ii) $\exp(\lambda a) \in G'$ for all $\lambda \in K$ and $|\lambda|$ sufficiently small; (iii) $\exp(\lambda a) \in G$ for all $\lambda \in K$ . The argument is as in $§ 4$ , no. 4, Corollary 1 to Proposition 8.

Corollary 2

Let $G$ be a Lie group, $H$ an integral subgroup of $G$ and $a \in L(G)$ . Consider the following conditions*: (i) $a \in L(H)$ ; (ii) $\exp_G(\lambda a) \in H$ for all $\lambda \in K$ . Then (i) $\Rightarrow$ (ii). If the topology of $H$ admits a countable base, then (i) $\Leftrightarrow$ (ii). Let $i$ be the canonical injection of $H$ into $G$ . If $a \in L(H)$ , then

\exp_G(\lambda a) = (\exp_G \circ L(i))(\lambda a) = (i \circ \exp_H)(\lambda a) \in H.

Hence (i)

\Rightarrow

(ii). The converse when

H

has a countable base follows from Proposition 3.

Corollary 3

Let $G$ be a Lie group, $\rho$ an analytic linear representation of $G$ , $x \in L(G)$ and $g \in G$ . (i) $\rho(\exp x) = \exp L(\rho)x$ ; (ii) $\mathrm{Ad}(\exp x) = \exp \mathrm{ad}\, x$ ; (iii) $g(\exp x)g^{-1} = \exp(\mathrm{Ad}\, g.x)$ . The argument is as in $§ 4$ , no. 4, Corollaries 2 and 3 to Proposition 8.

Corollary 4

Let $G$ be a finite-dimensional connected Lie group*. (i) $\mathrm{Int}(L(G)) = \mathrm{Ad}(G)$ . (ii) Let $Z$ be the centre of $G$ . Then $Z$ is a Lie subgroup of $G$ whose Lie algebra is the centre of $L(G)$ . The mapping of $G/Z$ into $\mathrm{Int}\, L(G)$ derived from $g \mapsto \mathrm{Ad}\, g$ when passing to the quotient is a Lie group isomorphism. Assertion (i) follows from Corollary 3 (ii) and the remarks after Definition 2. Let $g \in G$ . In order that $\mathrm{Ad}\, g = \mathrm{Id}_{L(G)}$ , it is necessary and sufficient that $\mathrm{Int}\, g$ coincide with $\mathrm{Id}_G$ on a neighbourhood of $e$ ( $§ 4$ , no. 1, Theorem 1 (ii)) and hence on the whole of $G$ ; in other words, it is necessary and sufficient that $g \in Z$ . Then (ii) follows from Corollary 1 to Proposition 1.

Definition 3

Let $G$ be a finite-dimensional connected Lie group. The Lie group $\mathrm{Int}(L(G)) = \mathrm{Ad}(G)$ is called the adjoint group of $G$ *.

Proposition 11

Let $G$ be a connected commutative Lie group*. (i) exp is an étale morphism of the additive Lie group $L(G)$ onto $G$ .

(ii) *If K = \mathbf{R} and G is finite-dimensional, G is isomorphic to a Lie group of the form $\mathbf{R}^p \times \mathbf{T}^q$ ( $p, q$ integers $\geqslant 0$ ).

By the Hausdorff formula, $(\exp x)(\exp y) = \exp(x + y)$ for $x, y$ sufficiently close to 0 and hence for all $x, y$ in $L(G)$ by analytic continuation. Hence $\exp$ is a group homomorphism and is étale since

T_e(\exp) = \mathrm{Id}_{L(G)}.

Hence (i). Assertion (ii) follows from (i) and General Topology, Chapter VII, § 1, Proposition 9.

Proposition 12

Let G be a Lie group and $L = L(G)$ . For all $x \in L$ , let $T_x(L)$ be identified with $L$ , so that the right differential $\varpi(x)$ of $\exp$ at $x$ is a linear mapping of $L$ into $L$ . For all $x \in L$ ,

\varpi(x) = \sum_{n \geqslant 0} \frac{1}{(n + 1)!} (\mathrm{ad}\ x)^n.

The two sides are analytic functions of $x$ and are equal for $x$ sufficiently close to 0 (§ 4, no. 3, Proposition 6).

Remark

$\varpi(x) \cdot (\mathrm{ad}\ x) = \exp \mathrm{ad}\ x - 1$ . We write, by an abuse of notation,

\varpi(x) = \frac{\exp \mathrm{ad}\ x - 1}{\mathrm{ad}\ x}.

Corollary

Let G be a complex Lie group and $x \in L(G)$ . The tangent mapping at $x$ to $\exp_G$ has kernel $\bigoplus_{n \in \mathbf{Z} - \{0\}} \mathrm{Ker}(\mathrm{ad}\ x - 2i\pi n)$ .

The integral function $z \mapsto \sum_{n \geqslant 0} \frac{1}{(n + 1)!} z^n$ , equal to $\frac{e^z - 1}{z}$ for $z \neq 0$ , admits as zeros the points of $2\pi i \mathbf{Z} - \{0\}$ , which are all simple zeros. The corollary then follows from Proposition 12 and the following lemma:

Lemma 2

Let E be a complex Banach space, u an element of $\mathscr{L}(E)$ , S the spectrum of u in $\mathscr{L}(E)$ (Spectral Theories, Chapter I, § 1, no. 2) and f a holomorphic complex function on an open neighbourhood $\Omega$ of S. Suppose that f admits in $\Omega$ only a finite number of distinct zeros $z_1, \ldots, z_n$ , of multiplicities $h_1, \ldots, h_n$ . Then $\mathrm{Ker}\ f(u)$ is the direct sum of the $\mathrm{Ker}(u - z_i)^{h_i}$ for $1 \leqslant i \leqslant n$ .

(For the definition of $f(u)$ , see Spectral Theories, Chapter I, § 4, no. 8.)

There exists a holomorphic function g on $\Omega$ , everywhere non-zero, such that $f(z) = (z - z_1)^{h_1} \ldots (z - z_n)^{h_n} g(z)$ . Then $g(u)g^{-1}(u) = g^{-1}(u)g(u) = 1$ and hence $\mathrm{Ker}\ f(u) = \mathrm{Ker} \prod_{i=1}^n (u - z_i)^{h_i}$ . Considering $\mathrm{Ker}\ f(u)$ as a $\mathbf{C}[X]$ -module by means of the external law $(h, x) \mapsto h(u)x$ for $h \in \mathbf{C}[X]$ , $x \in \mathrm{Ker}\ f(u)$ , we see that Ker $f(u)$ is the direct sum of the $\mathrm{Ker}(u - z_i)^{n_i}$ , using Algebra, Chapter VII, § 2, no. 1, Proposition 1.

5. Application to linear representations

Proposition 13

Let $G$ be a connected Lie group and $\varphi$ an analytic linear representation of $G$ on a complete normable space $E$ . Let $E_1, E_2$ be two closed vector subspaces of $E$ such that $E_2 \subset E_1$ . The following conditions are equivalent:

(i) $\varphi(g)x \equiv x \pmod{E_2}$ for all $g \in G$ and all $x \in E_1$ ; (ii) $L(\varphi)(L(G))$ maps $E_1$ into $E_2$ .

\begin{align*} &\varphi(g)x \equiv x \quad (\text{mod } E_2) \quad \text{for all } g \in G \text{ and all } x \in E_1 \\ \Leftrightarrow &\varphi(\exp a)x \equiv x \quad (\text{mod } E_2) \quad \text{for all } a \in L(G) \text{ and all } x \in E_1 \\ \Leftrightarrow &(\exp L(\varphi)a)x \equiv x \quad (\text{mod } E_2) \quad \text{for all } a \in L(G) \text{ and all } x \in E_1. \end{align*}

On the other hand, if $u \in \mathscr{L}(E)$ , then

\begin{align*} &\exp(\lambda u)x \equiv x \quad (\text{mod } E_2) \quad \text{for all } \lambda \in K \text{ and all } x \in E_1 \\ \Leftrightarrow &u(E_1) \subset E_2 \end{align*}

whence the proposition.

Corollary 1

For $E_1$ to be stable under $\varphi$ , it is necessary and sufficient that $E_1$ be stable under $L(\varphi)$ .

It suffices to take $E_1 = E_2$ in Proposition 13.

Corollary 2

Suppose that $\varphi$ is finite-dimensional. For $\varphi$ to be simple (resp. semi-simple), it is necessary and sufficient that $L(\varphi)$ be simple (resp. semi-simple).

This follows from Corollary 1.

Corollary 3

Let $x \in E$ . For $x$ to be invariant under $\varphi(G)$ , it is necessary and sufficient that $x$ be annihilated by $L(\varphi)(L(G))$ (that is that $x$ be invariant under $L(\varphi)$ in the sense of Chapter I, § 3, Definition 3).

It suffices to take $E_1 = Kx$ and $E_2 = 0$ in Proposition 13.

Corollary 4

Let $\varphi'$ be another analytic linear representation of $G$ on a complete normable space $E'$ . Let $T \in \mathscr{L}(E, E')$ . The following conditions are equivalent:

(i) $T\varphi(g) = \varphi'(g)T$ for all $g \in G$ ; (ii) $TL(\varphi)(a) = L(\varphi')(a)T$ for all $a \in L(G)$ .

Let $\sigma$ be the linear representation of $G$ on $\mathscr{L}(E, E')$ derived from $\varphi$ and $\varphi'$ (§ 3, no. 11, Corollary 1 to Proposition 41). Condition (i) means that $T$ is invariant under $\sigma(G)$ . Condition (ii) means that $T$ is annihilated by $L(\sigma)(L(G))$ . It then suffices to apply Corollary 3.

Corollary 5

Suppose that $\rho$ and $\rho'$ are finite-dimensional. For $\rho$ and $\rho'$ to be equivalent, it is necessary and sufficient that $L(\rho)$ and $L(\rho')$ be equivalent.

This is a special case of Corollary 4.

Corollary 6

Suppose that $G$ is finite-dimensional. Let $t \in U(G)$ . For $L_t$ (resp. $R_t$ ) to be right (resp. left) invariant, it is necessary and sufficient that $t$ belong to the centre of $U(G)$ .

For $L_t$ (resp. $R_t$ ) to be right (resp. left) invariant, it is necessary and sufficient that $\varepsilon_g * t = t * \varepsilon_g$ for all $g \in G$ , that is that $(\mathrm{Int}\, g)_* t = t$ . There exists an integer $n$ such that $t \in U_n(G)$ . By Corollary 3 and Proposition 45 of § 3, no. 12, $(\mathrm{Int}\, g)_* t = t$ for all $g \in G$ if and only if $[a, t] = 0$ for all $a \in L(G)$ , that is if and only if $t$ commutes with $U(G)$ .

6. Normal integral subgroups

Lemma 3

Let $G$ be a Lie group, $H_1$ and $H_2$ integral subgroups whose topology admits a countable base and $g \in G$ . Then

g H_1 g^{-1} = H_2 \Leftrightarrow (\mathrm{Ad}\, g)(L(H_1)) = L(H_2).

$\mathrm{Ad}\, g = T_e(\mathrm{Int}\, g)$ . Hence, by transport of structure, $(\mathrm{Int}\, g)(H_1)$ has Lie algebra $(\mathrm{Ad}\, g)(L(H_1))$ . As $H_1$ and $H_2$ have countable bases, to say that the sets $H_2$ and $(\mathrm{Int}\, g)(H_1)$ are equal amounts to saying that the integral subgroups $H_2$ and $(\mathrm{Int}\, g)(H_1)$ are equal (no. 2, Corollary 3 to Proposition 3). Then the lemma follows from Theorem 2 (i).

Proposition 14

Let $G$ be a Lie group and $H$ an integral subgroup whose topology admits a countable base. The following conditions are equivalent: (i) $H$ is normal in $G$ ; (ii) $L(H)$ is invariant under $\mathrm{Ad}(G)$ . If further $G$ is connected, these conditions are equivalent to the following: (iii) $L(H)$ is an ideal of $L(G)$ . If further $G$ is simply connected and $L(H)$ is of finite codimension in $L(G)$ , these conditions imply that $H$ is a Lie subgroup of $G$ and that $G/H$ is simply connected.

The equivalence (i) $\Leftrightarrow$ (ii) follows from Lemma 3. If further $G$ is connected, condition (ii) is equivalent to saying that $L(H)$ is stable under $\mathrm{ad}\, L(G)$ (no. 5, Corollary 1 to Proposition 13 and § 3, no. 12, Proposition 44).

Suppose that $G$ is simply connected and that $L(H)$ is an ideal of finite codimension in $L(G)$ . By Theorem 3 of no. 3, there exists a simply connected Lie group $G'$ such that $L(G')$ is isomorphic to $L(G)/L(H)$ . There exists a continuous morphism $f$ of $L(G)$ onto $L(G')$ with kernel $L(H)$ . By Theorem 1 of no. 1, there exists a morphism $\phi$ of $G$ into $G'$ such that $L(\phi) = f$ . This morphism is a submersion and hence its kernel $N$ is a Lie subgroup of $G$ such that $L(N) = \mathrm{Ker}\, f = L(H)$ . Hence $H$ is the identity component of $N$ and is therefore a Lie subgroup of G. Let $\psi$ be the morphism of $G/H$ into $G'$ derived from $\phi$ when passing to the quotient. This morphism is étale; since G is simply connected, $\psi$ is an isomorphism of $G/H$ onto $G'$ .

Corollary 1

Let G be a finite-dimensional simply connected Lie group. Let m, h be Lie subalgebras of L(G) such that L(G) is the semi-direct product of m by h. Let M, H be the corresponding integral subalgebras of G. Then M and H are simply connected Lie subgroups of G and, as a Lie group, G is the semi-direct product of M by H.

By Proposition 14, H is a normal Lie subgroup of G and the Lie group G/H is simply connected. Let $\pi$ be the canonical morphism of G onto G/H. There exists a morphism $\theta$ of G/H into M such that $L(\theta)$ is the canonical isomorphism of $L(G)/L(H) = L(G)/h$ onto $L(M) = m$ . Then

L(\pi \circ \theta) = L(\pi) \circ L(\theta) = \mathrm{Id}_{L(G/H)}

and hence $\pi \circ \theta = \mathrm{Id}_{G/H}$ . By no. 1, Corollary 1 to Proposition 1, $\theta(G/H) = M$ . By Proposition 8 of § 1, no. 4, M is a Lie subgroup of G and the Lie group G is the semi-direct product of M by H.

Corollary 2

Let G be a finite-dimensional simply connected Lie group, H a normal connected Lie subgroup of G and $\pi$ the canonical morphism of G onto G/H.

(i) There exists an analytic mapping $\varphi$ of G/H into G such that $\pi \circ \varphi = \mathrm{Id}_{G/H}$ .

(ii) For every mapping $\varphi$ with the properties of (i), the mapping $(h, m) \mapsto h \varphi(m)$ of $H \times (G/H)$ into G is an isomorphism of analytic manifolds.

(iii) H and G/H are simply connected.

Let $n = \dim G - \dim H$ . The corollary is obvious for $n = 0$ . We argue by induction on n.

Suppose that there exists an ideal of L(G) containing L(H) distinct from L(G) and L(H). Let H' be the corresponding connected Lie subgroup of G. Let $\pi_1 : G \to G/H'$ and $\pi_2 : H' \to H'/H$ be the canonical morphisms. By the induction hypothesis, there exist analytic mappings $\rho_1 : G/H' \to G$ , $\rho_2 : H'/H \to H'$ such that $\pi_1 \circ \rho_1 = \mathrm{Id}_{G/H'}$ , $\pi_2 \circ \rho_2 = \mathrm{Id}_{H'/H}$ . Let

\pi_3 : G/H \to G/H'

be the canonical morphism. If $x \in G/H$ and y is a representative of x in G, y and $\rho_1(\pi_3(x))$ have the same canonical image modulo H' and hence $x^{-1} \pi(\rho_1(\pi_3(x))) \in H'/H$ . Let

\rho(x) = \rho_1(\pi_3(x)) \rho_2(\pi(\rho_1(\pi_3(x))))^{-1} x \in G.

Clearly $\rho$ is an analytic mapping of G/H into G and

\pi(\rho(x)) = \pi(\rho_1(\pi_3(x))) \pi_2(\rho_2(\pi(\rho_1(\pi_3(x))))^{-1} x)) \\ = \pi(\rho_1(\pi_3(x))) \pi(\rho_1(\pi_3(x)))^{-1} x = x.

If now the only ideals of L(G) containing L(H) are L(G) and L(H), the Lie algebra $L(G)/L(H)$ is either 1-dimensional or simple. In both cases, $L(G)$ is the semi-direct product of a subalgebra by $L(H)$ ; this is obvious in the first case and in the second this follows from Chapter I, § 6, Corollary 3 to Theorem 5. Assertion (i) then follows from Corollary 1.

Assertion (ii) is obvious. Assertion (iii) follows from (i) and (ii).

The conclusions of Corollary 2 are no longer necessarily true when G is infinite-dimensional or when H is not normal (Exercises 8 and 15).

Corollary 3

Let G be a finite-dimensional connected Lie group and H a normal connected Lie subgroup of G. The canonical morphism of $\pi_1(H)$ into $\pi_1(G)$ is injective.

Let $G_1$ be the universal covering of G and $\lambda$ the canonical mapping of $G_1$ onto G. The Lie algebra of $G_1$ is identified with $L(G)$ . The Lie subgroup $\lambda^{-1}(H)$ of $G_1$ is normal in $G_1$ and its Lie algebra is $L(H)$ . Let $H_1$ be the identity component of $\lambda^{-1}(H)$ and $\lambda_1 = \lambda | H_1$ . Then $L(H_1) = L(H)$ and hence $\lambda_1$ is an étale morphism of $H_1$ onto H. On the other hand, $H_1$ is simply connected (Corollary 2) and hence is identified with the universal covering of H. Then, by General Topology, Chapter XI, the canonical morphism of $\pi_1(H)$ into $\pi_1(G)$ is identified with the canonical injection of Ker $\lambda_1$ into Ker $\lambda$ .

7. Primitives of differential forms with values in a Lie algebra

Proposition 15

Let G be a Lie group, 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 $L(G)$ , such that $d\alpha + [\alpha]^2 = 0$ . Suppose that M is simply connected. For all $x \in M$ and all $s \in G$ , there exists one and only one mapping f of class $C^{r-1}$ of M into G such that $f(x) = s$ and $f^{-1} \cdot df = \alpha$ .

The uniqueness of f follows from § 3, no. 17, Corollary 2 to Proposition 59 and the fact that M is connected. We prove the existence of f. There exist an open covering $(U_i)_{i \in I}$ of M and, for all $i \in I$ , a mapping $g_i : U_i \to G$ of class $C^{r-1}$ such that $g_i^{-1} \cdot dg = \alpha$ on $U_i$ (§ 4, no. 6, Theorem 5). By § 3, no. 17, Corollary 2 to Proposition 59, $g_i g_j^{-1}$ is locally constant on $U_i \cap U_j$ . Let $g_i g_j^{-1} = g_{ij}$ . Let $G_d$ be the group G with the discrete topology. The $g_{ij} : U_i \cap U_j \to G_d$ are continuous and $g_{ij} g_{jk} = g_{ik}$ on $U_i \cap U_j \cap U_k$ . Since M is simply connected, there exist continuous mappings $\lambda_i : U_i \to G_d$ such that $g_i g_j^{-1} = \lambda_i \lambda_j^{-1}$ on $U_i \cap U_j$ . Let g be the mapping of M into G whose restriction to $U_i$ is $\lambda_i^{-1} g_i$ for all $i \in I$ . This mapping is of class $C^{r-1}$ and $g^{-1} dg = \alpha$ . The mapping f of M into G defined by $f = s(g(x))^{-1} g$ satisfies the conditions $f^{-1} \cdot df = \alpha$ and $f(x) = s$ .

8. Passage from laws of infinitesimal operation to laws of operation

Lemma 4

Let G be a connected topological group, X a Hausdorff topological space and $f_1, f_2$ laws of left (resp. right) operation of G on X such that, for all $x \in X$ , the mappings* $s \mapsto f_1(s, x),\ s \mapsto f_2(s, x)$ of $G$ into $X$ are continuous. Suppose that there exists a neighbourhood $V$ of $\{e\} \times X$ in $G \times X$ such that $f_1$ and $f_2$ coincide on $V$ . Then $f_1 = f_2$ .

Let $x \in X$ and $A$ be the set of $g \in G$ such that $f_1(g, x) = f_2(g, x)$ . Then $A$ is closed in $G$ . On the other hand, let $g \in A$ ; we write $y = f_1(g, x) = f_2(g, x)$ . There exists a neighbourhood $U$ of $e$ in $G$ such that $f_1(t, y) = f_2(t, y)$ for $t \in U$ , in other words such that $f_1(t', x) = f_2(t', x)$ for $t' \in Ug$ (resp. $gU$ ). Hence $A$ is open in $G$ and therefore $A = G$ .

Proposition 16

Let G be a connected Lie group, X a Hausdorff manifold of class C^r and f_1, f_2 laws of left (resp. right) operation of class C^r of G on X. If the laws of infinitesimal operation associated with f_1 and f_2 are equal, then f_1 = f_2.

By § 4, no. 7, Corollary to Proposition 11, there exists a neighbourhood V of \{e\} \times X in G \times X such that f_1 and f_2 coincide on V. Hence f_1 = f_2 (Lemma 4).

Lemma 5

Let G be a simply connected topological group, X a Hausdorff topological space, U an open neighbourhood of e in G and \psi a continuous mapping of U \times X into X such that \psi(e, x) = x and \psi(s, \psi(t, x)) = \psi(st, x) for all x \in X and s, t in U such that st \in U. Let W be a symmetric connected open neighbourhood of e such that W^3 \subset U. There exists one and only one law of continuous left operation \psi' of G on X such that \psi' and \psi coincide on W \times X. If G is a Lie group and X a manifold of class C^r and \psi is of class C^r, then \psi' is of class C^r.

The uniqueness on \psi follows from Lemma 4. Let P be the permutation group of X. For u \in W^3, the mapping x \mapsto \psi(u, x) is an element f(u) of P and

f(u_1u_2u_3) = f(u_1)f(u_2)f(u_3)

for u_1, u_2, u_3 in W. Applying Lemma 1 of no. 1, we obtain a morphism f' of G into P which extends f|W. Let \psi'(g, x) = f'(g)(x) for all (g, x) \in G \times X. Then \psi' is a law of left operation of G on X which coincides with \psi on W \times X. As \psi'(g, \psi'(g', x)) = \psi'(gg', x) for (g, g', x) \in G \times G \times X, the continuity of \psi on W \times X implies the continuity of \psi' on gW \times X for all g \in G. Hence \psi' is continuous. If \psi is of class C^r, it is seen similarly that \psi' is of class C^r.

Theorem 5

Let G be a simply connected Lie group, X a compact manifold of class C^r (r \geq 2) and a \mapsto D_a a law of left (resp. right) infinitesimal operation of class C^{r-1} of L(G) on X. There exists one and only one law of left (resp. right) operation of class C^{r-1} of G on X such that the associated law of infinitesimal operation is a \mapsto D_a.

The uniqueness follows from Proposition 16. By § 4, no. 7, Corollary 1 to Theorem 6, there exist a neighbourhood V of \{e\} \times X in G \times X and a law chunk of left (resp. right) operation of class C^{r-1} of G on X, defined on V, such that the associated law of infinitesimal operation is a \mapsto D_a. As X is compact, V can be assumed to be of the form U × X, where U is an open neighbourhood of e in G. It then suffices to apply Lemma 5.

9. Exponential mapping in the linear group

Proposition 17

Let Δ be the set of z ∈ C such that −π < J(z) < π and Δ' the set of z ∈ C which are not real ≤ 0. Let E be a complete normable space over C and A (resp. A') the set of x ∈ L(E) whose spectrum Sp(x) is contained in Δ (resp. Δ'). Then A (resp. A') is an open subset of L(E) (resp. GL(E)) and the mappings exp: A → A' and log: A' → A (Spectral Theories, Chapter I, § 4, no. 9) are inverse isomorphisms of analytic manifolds.

This follows from Spectral Theories, Chapter I, § 4, Proposition 10 and no. 9.

Theorem 6

Let E be a real or complex Hilbert space and U the unitary group of E.

(i) The set H of Hermitian elements of L(E) is, with the real normed space structure, a closed vector subspace of L(E) admitting a topological supplement.

(ii) The set H' of elements ≥ 0 of GL(E) is a real analytic submanifold of GL(E).

(iii) The restriction to H of the mapping exp is an isomorphism of real analytic manifolds of H onto H'.

(iv) The mapping (h, u) ↦ (exp h)u of H × U into GL(E) is an isomorphism of real analytic manifolds.

Recall that, if x ∈ L(E), x* denotes the adjoint of x. Let H_1 be the set of x ∈ L(E) such that x* = −x. The formula x = $\frac{1}{2}(x + x^*) + \frac{1}{2}(x - x^*)$ proves that, with its real normed space structure, L(E) is the topological direct sum of H and H_1, whence (i).

Suppose that K = C. In the notation of Proposition 17, H' is the set of h ∈ H ∩ A' such that Sp h ⊂ R^*_+. As exp(R) = R^*_+, (ii) and (iii) follow from Proposition 17 and Spectral Theories, Chapter I, § 4, Proposition 8 and § 6, no. 5. The mapping (h, u) ↦ y = (exp h)u of H × U into GL(E) is bijective by Spectral Theories, Chapter I, § 6, Proposition 15. It is real analytic by the above. The mapping y ↦ h = $\frac{1}{2} \log(y y^*)$ is real analytic and so also is the mapping y ↦ u = (exp h)^{-1}y. Hence (iv).

Suppose that K = R. Let Ē be the complexified Hilbert space of E and J the mapping ξ + iη ↦ ξ − iη (for ξ, η in E) of Ē into Ē. Let Ũ, Ũ', Ũ denote the sets defined for Ē as were H, H', U for E. Then H (resp. H', U) is identified with the set of x ∈ Ũ (resp. Ũ', Ũ) such that JxJ^{-1} = x. Properties (ii), (iii) and (iv) then follow easily from (i) and the analogous properties in the complex case.

Proposition 18

Let E be a complete normable space over C, v ∈ L(E) and g = exp v. Suppose that Sp(v) contains none of the points 2iπn with n ∈ Z − {0}. Then, for all x ∈ E, the conditions vx = 0 and gx = x are equivalent.

This follows from Lemma 2 of no. 4, applied to the function z ↦ e^z − 1.

Corollary 1

Let E be a complete normable space over $\mathbf{C}$ and F the space of continuous n-linear mappings of $E^n$ into E. For all $v \in \mathscr{L}(E)$ , let $\sigma(v)$ be the element of $\mathscr{L}(F)$ defined by

(\sigma(v)f)(x_1, \ldots, x_n) = v(f(x_1, \ldots, x_n)) - \sum_{i=1}^n f(x_1, \ldots, vx_i, \ldots, x_n).

For all $g \in \mathbf{GL}(E)$ , let $\rho(g)$ be the element of $\mathbf{GL}(F)$ defined by

(\rho(g)f)(x_1, \ldots, x_n) = g(f(g^{-1}x_1, \ldots, g^{-1}x_n)).

Let $u \in \mathscr{L}(E)$ be such that every $z \in \mathrm{Sp}\,u$ satisfies $|\mathscr{J}(z)| < \frac{2\pi}{n+1}$ . Then, for all $f \in F$ , the conditions $\sigma(u)f = 0$ and $\rho(\exp u)f = f$ are equivalent.

$\mathbf{L}(\rho) = \sigma$ (§ 3, no. 11, Corollary 1 to Proposition 41) and hence

\rho(\exp u) = \exp \sigma(u)

(no. 4, Corollary 3 to Proposition 10). By Proposition 18 it then suffices to prove that $\mathrm{Sp}\,\sigma(u)$ does not meet $2i\pi(\mathbf{Z} - \{0\})$ . But this follows from the following lemma:

Lemma 6. If $v \in \mathscr{L}(E)$ , then $\mathrm{Sp}\,\sigma(v) \subset \mathrm{Sp}\,v + \mathrm{Sp}\,v + \cdots + \mathrm{Sp}\,v$ , where the sum comprises $n+1$ terms.

We define elements $v_0, v_1, \ldots, v_n$ of $\mathscr{L}(F)$ by writing, for all $f \in F$ ,

\begin{align*} (v_0f)(x_1, \ldots, x_n) &= v(f(x_1, \ldots, x_n)) \\ (v_if)(x_1, \ldots, v_n) &= -f(x_1, \ldots, vx_i, \ldots, x_n) \quad \text{for } 1 \leq i \leq n. \end{align*}

Then $\sigma(v) = \sum_{i=0}^n v_i$ and the $v_i$ are pairwise permutable. Let A be the total closed subalgebra of $\mathscr{L}(F)$ generated by the $v_i$ ; it is commutative (Spectral Theories, Chapter I, § 1, no. 4) and $\mathrm{Sp}_{\mathscr{L}(F)} v' = \mathrm{Sp}_A v' \subset \sum_{i=0}^n \mathrm{Sp}\,v_i$ (Spectral Theories, Chapter I, § 3, Proposition 3 (ii)). Now, if $\lambda \in \mathbf{C}$ is such that $v - \lambda$ is invertible, clearly the $v_i - \lambda_i$ are invertible and hence $\mathrm{Sp}\,v_i \subset \mathrm{Sp}\,v$ for all $i$ .

Corollary 2

Let E be a complete normable algebra over $\mathbf{C}$ and $w \in \mathscr{L}(E)$ . Suppose that every $z \in \mathrm{Sp}\,w$ satisfies $|\mathscr{J}(z)| < \frac{2\pi}{3}$ . The following conditions are equivalent:* (i) w is a derivation of E; (ii) exp w is an automorphism of E. This follows from Corollary 1 with $n = 2$ and $f$ the multiplication of E.

Proposition 19

Let E be a complete normable space over $\mathbf{C}$ , $v \in \mathscr{L}(E)$ and $g = \exp v$ . Suppose that every $z \in \mathrm{Sp}\, v$ satisfies $-\pi < \mathscr{J}(z) < \pi$ . Then, for every closed vector subspace $E'$ of $E$ , the conditions $v(E') \subset E'$ and $g(E') = E'$ are equivalent.

The condition $v(E') \subset E'$ implies $g(E') \subset E'$ and $g^{-1}(E') \subset E'$ and hence $g(E') = E'$ . Suppose that $g(E') = E'$ . We use the notation $\Delta, \Delta'$ of Proposition 17. Since $\mathrm{Sp}\, v$ is a compact subset of $\Delta$ , there exists a compact rectangle $Q = (a, b) \times (a', b')$ such that $\mathrm{Sp}\, v \subset Q \subset \Delta$ . The set $\Delta - Q$ is connected. Hence $\mathrm{Sp}\, g \subset \exp Q \subset \Delta'$ , the set $\exp Q$ is compact and the set $\Delta' - \exp Q$ is connected. The closure of the latter contains $]-\infty, 0]$ and hence

(\Delta' - \exp Q) \cup ]-\infty, 0] = \mathbf{C} - \exp Q

is connected. Then $\exp Q$ is polynomially convex (Spectral Theories, Chapter I, § 3, Corollary 2 to Proposition 9) and hence the function $\log$ , defined on $\Delta'$ , is a limit in $\mathscr{O}(\exp Q)$ of polynomial functions (Spectral Theories, Chapter I, § 4, Proposition 3). Hence $v = \log g$ is the limit in $\mathscr{L}(E)$ of elements of the form $P(g)$ , where $P$ is a polynomial (Spectral Theories, Chapter I, § 4, Theorem 3). As $P(g)(E') \subset E'$ , it follows that $v(E') \subset E'$ .

Corollary

Let E be a complete normed space over $\mathbf{C}$ , $v \in \mathscr{L}(E)$ and $g = \exp v$ . Suppose that every $z \in \mathrm{Sp}\, v$ satisfies $-\frac{\pi}{2} < \mathscr{J}(z) < \frac{\pi}{2}$ . Then, for every closed vector subspace M of $\mathscr{L}(E)$ , the conditions $gMg^{-1} = M$ and $[v, M] \subset M$ are equivalent.

Let $F = \mathscr{L}(E)$ , $g'$ be the mapping $f \mapsto gfg^{-1}$ of F into F and $v'$ be the mapping $f \mapsto [v, f]$ of F into F. Then $g' = \exp v'$ (no. 4, Corollary 3 to Proposition 10 and § 3, no. 11, Corollary 1 to Proposition 41). Lemma 6 proves that $-\pi < \mathscr{J}(z) < \pi$ for all $z \in \mathrm{Sp}\, v'$ . It then suffices to apply Proposition 19.

10. Complexification of a finite-dimensional real Lie group

Lemma 7

Let B be a group, A a normal subgroup of B, C the group $B/A$ and $i : A \to B$ and $p : B \to C$ the canonical morphisms. Let $A'$ be a group and $f$ a homomorphism of A into $A'$ . Let $\omega$ be a morphism of B into the automorphism group of $A'$ . Suppose that, for $a \in A'$ , $a' \in A'$ , $b \in B$ ,

f\left(bab^{-1}\right) = \omega(b)f(a), \quad \omega(a)a' = f(a)a'f(a)^{-1}.

Let $B''$ be the semi-direct product of B by $A'$ relative to $\omega$ and q the canonical morphism of $B''$ onto B.

(i) The mapping $a \mapsto (f(a^{-1}), i(a))$ of A into $B''$ is a morphism of A onto a normal subgroup D of $B''$ . Let $B' = B''/D$ and $i' : A' \to B'$ , $g : B \to B'$ be the morphisms of $A'$ and B into $B'$ derived when taking quotients from the canonical injections of $A'$ and B into $B''$ .

(ii) The morphism $p \circ q$ of $B''$ into C defines when passing to the quotient a morphism of $B'$ into C.

(iii) $i'$ is injective, $p'$ is surjective, $\mathrm{Ker}(p') = \mathrm{Im}(i')$ and the following diagram is commutative

\begin{array}{ccc} A & \xrightarrow{i} & B \xrightarrow{p} C \\ f \downarrow & & g \downarrow \mathrm{Id}_C \\ A' & \xrightarrow{i'} & B' \xrightarrow{p'} C. \end{array}

(iv) If $b \in B$ and $a' \in A'$ , then

g(b)i'(a')g(b)^{-1} = i'(\omega(b)a').

(i) For $a_1, a_2$ in $A$ , we have, in $B''$ ,

(f(a_1^{-1}), i(a_1))(f(a_2^{-1}), i(a_2)) = (f(a_1^{-1})(\omega(a_1)f(a_2^{-1})), i(a_1)i(a_2)) = (f(a_1^{-1})f(a_1a_2^{-1}a_1^{-1}), i(a_1a_2)) = (f((a_1a_2)^{-1}, i(a_1a_2))

and hence

a \mapsto (f(a^{-1}), i(a))

is a homomorphism

h

A

into

B''

. Let

a \in A

a' \in A'

b \in B

; then, in

B''

bh(a)b^{-1} = bf(a^{-1})ab^{-1} = (\omega(b)f(a^{-1}))(bab^{-1}) = f(ba^{-1}b^{-1})(bab^{-1}) = h(bab^{-1})

a'h(a){a'}^{-1} = a'f(a^{-1})a{a'}^{-1} = a'f(a^{-1})(\omega(a){a'}^{-1})a = a'f(a^{-1})f(a){a'}^{-1}f(a^{-1})a = h(a)

and hence

h(A) = D

is normal in

B''

(ii) For $a \in A$ ,

(p \circ q)(h(a)) = p(q(f(a^{-1})a)) = p(a) = e

and hence

p \circ q

is trivial on

D

(iii) Let $a' \in A'$ be such that $i'(a') = e$ ; then $a' \in D$ and hence there exists $a \in A$ such that $a' = f(a^{-1})a$ ; this implies $a = e$ , whence $a' = e$ ; thus $i'$ is injective. As $p$ and $q$ are surjective, $p'$ is surjective.

Let $r$ denote the canonical morphism of $B''$ onto $B'$ . Let $a' \in A'$ , $b \in B$ and $b' = r(a'b)$ . If $b' \in \mathrm{Im}(i')$ , there exists $a'_1 \in A'$ such that $b' = r(a'_1)$ ; then there exists $a \in A$ such that $a'b = a'_1f(a^{-1})a$ , whence $b = a \in A$ and

p'(b') = p(q(a'b)) = p(b) = e;

thus,

\mathrm{Im}(i') \subset \mathrm{Ker}(p')

. We preserve the notation

a', b, b'

but assume that

b' \in \mathrm{Ker}(p')

; then

e = p'(b') = p(q(a'b)) = p(b)

and hence

b \in A

, whence

b' = r(a'f(b)f(b^{-1})b) = r(a'f(b)) \in \mathrm{Im}(i');

thus

\mathrm{Ker}(p') \subset \mathrm{Im}(i')

If $a \in A$ , then

i'(f(a)) = r(f(a)) = r(f(a)f(a^{-1})a) = r(a) = g(i(a)).

If $b \in B$ , then

p'(g(b)) = p(b)

and hence diagram (4) is commutative.

(iv) Let $b \in B, a' \in A'$ . Then

g(b)i'(a')g(b)^{-1} = r(b)r(a')r(b)^{-1} = r(ba'b^{-1}) = r(\omega(b)a') = i'(\omega(b)a').

Proposition 20

Let G be a finite-dimensional real Lie group.

(i) There exist a complex Lie group $\tilde{G}$ and an $\mathbf{R}$ -analytic morphism $\gamma$ of G into $\tilde{G}$ with the following properties: for every complex Lie group H and every $\mathbf{R}$ -analytic morphism $\phi$ of G into H, there exists one and only one $\mathbf{C}$ -analytic morphism $\psi$ of $\tilde{G}$ into H such that $\phi = \psi \circ \gamma$ .

(ii) If $(\tilde{G}', \gamma')$ has the same properties as $(\tilde{G}, \gamma)$ , there exists one and only one isomorphism $\theta$ of $\tilde{G}$ onto $\tilde{G}'$ such that $\theta \circ \gamma = \gamma'$ .

(iii) The $\mathbf{C}$ -linear mapping of $L(G) \otimes \mathbf{C}$ into $L(\tilde{G})$ which extends $L(\gamma)$ is surjective; in particular $\dim_{\mathbf{C}}(\tilde{G}) \leq \dim_{\mathbf{R}}(G)$ .

Assertion (ii) is obvious. We prove the existence of an ordered pair $(\tilde{G}, \gamma)$ with properties (i) and (iii).

(a) Suppose first that G is connected. Let $g = L(G)$ , $g_C = g \otimes_{\mathbf{R}} \mathbf{C}$ be the complexification of g, S (resp. S') the simply connected real (resp. complex) Lie group with Lie algebra g (resp. $g_C$ ) and $\sigma$ the unique $\mathbf{R}$ -analytic morphism of S into S' such that $L(\sigma)$ is the canonical injection of g into $g_C$ . Let $\pi$ be the unique $\mathbf{R}$ -analytic morphism of S onto G such that $L(\pi) = \mathrm{Id}_{L(G)}$ and F = Ker $\pi$ .

\begin{array}{ccc} S & \xrightarrow{\sigma} & S' \\ \downarrow \pi & & \downarrow \lambda \\ G & \xrightarrow{\gamma} & G \\ & \searrow \phi & \swarrow \psi \\ & & H \end{array}

For every complex Lie group H and every $\mathbf{R}$ -analytic morphism $\phi$ of G into H, $L(\phi): g \to L(H)$ has a unique $\mathbf{C}$ -linear extension to $g_C$ and this extension is of the form $L(\phi^*)$ , where $\phi^*$ is a $\mathbf{C}$ -analytic morphism of S' into H. Then

L(\phi \circ \pi) = L(\phi) \circ L(\pi) = L(\phi) = L(\phi^*) \circ L(\sigma) = L(\phi^* \circ \sigma)

and hence

\phi \circ \pi = \phi^* \circ \sigma

. Therefore

\phi^*(\sigma(F)) = \phi(\pi(F)) = \{e\}

, whence

\sigma(F) \subset \mathrm{Ker}\, \phi^*.

Let P be the intersection of the

\mathrm{Ker}\, \phi^*

for variable

\phi

. This is a normal Lie subgroup of S' (no. 2, Corollary 3 to Proposition 1). Let

\tilde{G} = S'/P

and λ: S' → Ġ be the canonical morphism. Then σ(F) ⊂ P and hence there exists one and only one R-analytic morphism γ of G into Ġ such that γ ∘ π = λ ∘ σ. If ψ: Ġ → H denotes the morphism derived from φ* when passing to the quotient, then

(\varphi \circ \gamma) \circ \pi = \psi \circ (\lambda \circ \sigma) = \varphi^* \circ \sigma = \varphi \circ \pi

whence ψ ∘ γ = φ. Clearly L(ψ), and hence ψ, are determined uniquely by the equality ψ ∘ γ = φ. Thus we have proved that the ordered pair (\tilde{G}, γ) has properties (i) and (iii).

(b) We pass now to the general case. Let F be the identity component of G, M = G/F and i: F → G and p: G → M be the canonical morphisms. We apply part (a) of the proof to F. We obtain an ordered pair (\tilde{F}, δ). For all g ∈ G, Int g|F = ω'(g) is an automorphism of F. By the universal property of \tilde{F}, there exists one and only one automorphism ω(g) of the complex Lie group \tilde{F} such that δ ∘ ω'(g) = ω(g) ∘ δ. Clearly ω is a morphism of G into Aut(\tilde{H}). If g ∈ G and f ∈ F, then

\delta(gf g^{-1}) = (\delta \circ \omega'(g))(f) = (\omega(g) \circ \delta)(f) = \omega(g)(\delta(f)).

If f ∈ F, then δ ∘ (Int_F f) = (Int_\tilde{F} δ(f)) ∘ δ and Int_\tilde{F} δ(f) is an automorphism of the complex Lie group \tilde{F}; hence Int_\tilde{F} δ(f) = ω(f).

Hence Lemma 7 can be applied, which gives a diagram

\begin{array}{ccccc} F & \xrightarrow{i} & G & \xrightarrow{p} & M \\ \downarrow{\delta} & & \downarrow{\gamma} & & \downarrow{\mathrm{Id}} \\ \tilde{F} & \xrightarrow{\tilde{i}} & \tilde{G} & \xrightarrow{\tilde{p}} & M. \end{array}

Let \tilde{F} be identified with a normal subgroup of \tilde{G} by means of ī. The group \tilde{G} is generated by \tilde{F} and γ(G); hence the automorphisms of \tilde{F} defined by the elements of \tilde{G} are automorphisms of the complex Lie group structure. By § 1, no. 9, Proposition 18, there exists one and only one complex Lie group structure on \tilde{G} such that \tilde{F} is an open Lie subgroup of \tilde{G}. Henceforth \tilde{G} will have this structure. As δ is R-analytic, γ is R-analytic.

The ordered pair (\tilde{G}, γ) has property (iii) of the proposition. We show that it has property (i). Let H be a complex Lie group and ψ an R-analytic morphism of G into H. There exists a C-analytic morphism η of \tilde{F} into H such that η ∘ δ = φ|F. Let g ∈ G. The mappings

f \mapsto \eta(\omega(g)f), \quad f \mapsto \phi(g)\eta(f)\phi(g)^{-1}

of \tilde{F} into H are C-analytic morphisms; they coincide on δ(F), for, if f' ∈ F, then

\phi(g)\eta(\delta(f'))\phi(g)^{-1} = \phi(g)\phi(f')\phi(g)^{-1} = \phi(gf'g^{-1})

= \eta(\delta(gf'g^{-1})) = \eta(\omega(g)\delta(f'));

therefore $\eta(\omega(g)f) = \phi(g)\eta(f)\phi(g)^{-1}$ for all $g \in G$ and all $f \in \tilde{F}$ . If $G'$ denotes the semi-direct product of $G$ by $\tilde{F}$ relative to $\omega$ , there then exists a morphism $\zeta$ of the group $G'$ into $H$ which coincides with $\phi$ on $G$ and with $\eta$ on $\tilde{F}$ . For $f \in F$ ,

\zeta(\delta(f^{-1})f) = \eta(\delta(f^{-1}))\phi(f) = \phi(f^{-1})\phi(f) = e.

Hence

\zeta

defines on passing to the quotient a morphism

\psi

\tilde{G}

into

H

. Then

\psi \circ \gamma = \phi

and

\psi \circ \tilde{i} = \eta

; the latter inequality implies that

\psi

\mathbf{C}

-analytic.

Finally, let $\psi'$ be a $\mathbf{C}$ -analytic morphism of $\tilde{G}$ into $H$ such that $\phi = \phi' \circ \gamma$ . Then

\psi' \circ \tilde{i} \circ \delta = \psi' \circ \gamma \circ i = \phi \circ i = \psi \circ \tilde{i} \circ \delta

and hence

\psi' \circ \tilde{i} = \psi \circ \tilde{i}

. As

\tilde{G}

is generated by

\tilde{i}(\tilde{F})

and

\gamma(G)

\psi' = \psi

Definition 4

$(G, \gamma)$ , or simply $\tilde{G}$ , is called the universal complexification of $G$ .

Remark

(1) Let $(\tilde{G}, \gamma)$ be the universal complexification of $G$ . Let $G_0$ (resp. $\tilde{G}_0$ ) be the identity component of $G$ (resp. $\tilde{G}$ ). By the proof of Proposition 20, $(\tilde{G}_0, \gamma|G_0)$ is the universal complexification of $G_0$ and the composite morphism

G \to \tilde{G} \to \tilde{G}/\tilde{G}_0

defines on passing to the quotient an isomorphism of

G/G_0

onto

\tilde{G}/\tilde{G}_0

(2) Suppose that $G$ is simply connected. Let $g = L(G)$ , $g_C$ be the complexification of $g$ , $S'$ the simply connected complex Lie group with Lie algebra $g_C$ and $\sigma$ the morphism of $G$ into $S'$ such that $L(\sigma)$ is the canonical injection of $g$ into $g_C$ . We again use the notation of the proof of Proposition 20, part (a). If $H = S'$ and $\phi = \sigma$ , then $\phi^* = \mathrm{Id}_{S'}$ . Hence $(S', \sigma)$ is the universal complexification of $G$ . Note that $\sigma$ is not in general injective (Exercise 16); however its kernel is discrete since $L(\sigma)$ is injective. On the other hand, let $\theta$ be the involution of $g_C$ defined by $g$ and let $\eta$ be the corresponding automorphism of the underlying real Lie group of $S'$ ; let ${S'}^\eta$ be the set of points of $S'$ invariant under $\eta$ ; it is a real Lie subgroup of $S'$ with Lie algebra $g$ (§ 3, no. 8, Corollary 1 to Proposition 29). By no. 1, Corollary 1 to Proposition 1, $\sigma(G)$ is a real integral subgroup of $S'$ with Lie algebra $g$ and hence $\sigma(G)$ is the identity component of ${S'}^\eta$ ; in particular $\sigma(G)$ is a real Lie subgroup of $S'$ .

Exercises

See the exercises for § 6.