(i) $\mathfrak{g}$ is isomorphic to the Lie algebra of a compact Lie group.

(ii) The group Int($\mathfrak{g}$) (Chap. III, §6, no. 2, Def. 2) is compact.

(iii) $\mathfrak{g}$ has an invariant bilinear form (Chap. I, §3, no. 6) that is symmetric, positive and separating.

(iv) $\mathfrak{g}$ is reductive (Chap. I, §6, no. 4, Def. 4); for all $x\in \mathfrak{g}$, the endomorphism ad $x$ is semi-simple, with purely imaginary eigenvalues.

(v) $\mathfrak{g}$ is reductive and its Killing form B is negative.

(i) $=\Rightarrow$ (ii): if $\mathfrak{g}$ is the Lie algebra of a compact Lie group G, the group Int($\mathfrak{g}$) is separating and isomorphic to a quotient of the compact group $G_0$ (Chap. III, §6, no. 4, Cor. 4), hence is compact.

(ii) $=\Rightarrow$ (iii): if the group Int($\mathfrak{g}$) is compact, there exists a symmetric bilinear form on $\mathfrak{g}$ that is positive, separating and invariant under Int($\mathfrak{g}$) (no. 1), hence also invariant under the adjoint representation of $\mathfrak{g}$.

(iii) $=\Rightarrow$ (iv): if (iii) is satisfied, the adjoint representation of $\mathfrak{g}$ is semi-simple (no. 1), hence $\mathfrak{g}$ is reductive; moreover, the endomorphisms ad $x$, for $x\in \mathfrak{g}$, have the indicated properties (no. 1).

(iv) $=\Rightarrow$ (v): for all $x\in \mathfrak{g}, B(x, x) =$ Tr((ad $x$)$^2$); consequently, $B(x, x)$ is the sum of the squares of the eigenvalues of ad $x$, and hence is negative if these are purely imaginary.

(v) $=\Rightarrow$ (i): assume that $\mathfrak{g}$ is reductive, hence the product of a commutative subalgebra $\mathfrak{c}$ and a semi-simple subalgebra $\mathfrak{s}$ (Chap. I, §6, no. 4, Prop. 5). The Killing form of $\mathfrak{s}$ is the restriction of the form B to $\mathfrak{s}$, hence is negative and separating if B is negative. The subgroup Int($\mathfrak{s}$) of $\mathbf{G}\mathbf{L}(\mathfrak{s})$ is closed (it is the identity component of Aut($\mathfrak{s}$), Chap. III, §10, no. 2, Cor. 2) and leaves the separating positive form $-B$ invariant; thus, it is compact, and $\mathfrak{s}$ is isomorphic to the Lie algebra of the compact Lie group Int($\mathfrak{s}$). Further, since $\mathfrak{c}$ is commutative, it is isomorphic to the Lie algebra of a torus T. Thus $\mathfrak{g}$ is isomorphic to the Lie algebra of the compact Lie group Int($\mathfrak{s}$)$\times T$.

b) A subalgebra of a compact Lie algebra is compact.

c) Let $\mathfrak{h}$ be an ideal of a compact Lie algebra $\mathfrak{g}$. Then the algebra $\mathfrak{g}/\mathfrak{h}$ is compact and the extension $\mathfrak{h}\rightarrow \mathfrak{g}\rightarrow \mathfrak{g}/\mathfrak{h}$ is trivial.

Assertions a) and b) follow from the characterization (iii) of Prop. 1. Part c) follows from a) and the fact that, in a reductive Lie algebra, every ideal is a direct factor (Chap. I, §6, no. 4, Cor. of Prop. 5).

(i) The Lie algebra L(G) is compact.

(ii) The group Ad(G) is compact.

(iii) There exists a separating positive symmetric bilinear form on L(G) invariant under the adjoint representation of G.

\*(iv) G has a riemannian metric invariant under left and right translations.\*

(i) $=\Rightarrow$ (ii): if L(G) is compact, the group Ad(G$_0$) $=$ Int(L(G)) is compact; since it has finite index in Ad(G), this latter group is also compact.

(ii) $=\Rightarrow$ (iii): this follows from no. 1.

(iii) $=\Rightarrow$ (i): since Int(L(G)) $\subset$ Ad(G), this follows from the characterization (iii) of Prop. 1.

\*(iii) $\Leftarrow \Rightarrow$ (iv): this follows from Chap. III, §3, no. 13.\*

Since G is semi-simple, its centre D is discrete. Moreover, the quotient group $G/D$ is isomorphic to Ad(G) (Chap. III, §6, no. 4, Cor. 4), hence compact (Prop. 3). Finally, the group $G/D$ is equal to its derived group (Chap. III, §9, no. 2, Cor. of Prop. 4). The theorem now follows from Integration, Chap. VII, §3, no. 2, Prop. 5.

$^4$ Note that a real topological vector space cannot be a compact topological space

unless it is reduced to 0. group S, a vector group V and a surjective morphism $f: V\times T\times S\rightarrow G$ with finite kernel. If G is compact, the group V is reduced to the identity element.

Let C (resp. S) be a simply-connected Lie group whose Lie algebra is isomorphic to the centre (resp. the derived algebra) of L(G). Then C is a vector group, S is a compact group with finite centre (Th. 1) and G can be identified with the quotient of $C\times S$ by a discrete subgroup D, which is central (Integration, Chap. VII, §3, no. 2, Lemma 4). Since the image of the projection of D onto S is central, hence finite, $D\cap C$ is of finite index in D. Let $C'$ be the vector subspace of C generated by $D\cap C$, and V a complementary subspace. Then the group $T = C'/(D\cap C)$ is a torus, and G is isomorphic to the quotient of the product group $V\times T\times S$ by a finite group.

If G is compact, so is $V\times T\times S ($General Topology, Chap. III, §4, no. 1, Cor. 2 of Prop. 2), hence so is V, which implies that $V =\{e\}$.

With the notation in Prop. 4, we have $V =\{e\}$ and the subgroups $f(T)$ and $f(S)$ of G are compact, hence closed. Thus it suffices to show that $f(T) =$ $C(G)_0,f(S) = D(G)$. Now, $L(G) = L(f(T))\times L(f(S))$; since S is semi-simple and T is commutative, this implies that $L(f(T)) =\mathscr{C}(L(G)) = L(C(G)_0)$ (Chap. III, §9, no. 3, Prop. 8) and $L(f(S)) =\mathscr{D}L(G) = L(D(G))$ (Chap. III, §9, no. 2, Cor. of Prop. 4), hence the stated assertion.

With the notation in Prop. 4, the groups V and T are reduced to the identity element; thus S is a universal covering of G, and the fundamental group of G is isomorphic to Ker $f$, hence finite. The centre D of G is discrete since G is semi-simple, so D is finite.

Indeed, with the notations in Cor. 1, the fundamental group of $C(G)_0$ is isomorphic to $\mathbf{Z}^n$, with $n=$ dim $C(G)_0$, and the fundamental group of D(G) is finite (Cor. 2).

(i) G is semi-simple;

(ii) C(G) is finite;

(i) $\pi_1(G)$ is finite.

If G is simply-connected, it is semi-simple.

This follows from Cor. 1 to 3.

Let $f\in$ Aut(G)$_0$. Then $f$ induces an automorphism $f_1$ of $C(G)_0$ and an automorphism $f_2$ of D(G), and we have $f_1\in$ Aut(C(G)$_0$)$_0, f_2\in$ Aut(D(G))$_0$. Since Aut(C(G)$_0$) is discrete (General Topology, Chap. VII, §2, no. 4, Prop. 5), we have $f_1=$ Id; since D(G) is semi-simple, by Chap. III, §10, no. 2, Cor. 2 of Th. 1 there exists an element $g$ of D(G) such that $f_2(x) =gxg^{-1}$ for all $x\in$ D(G). For all $x\in C(G)_0$, we have $gxg^{-1}=x$ = $f_1(x)$; since $G = C(G)_0.D(G)$, it follows that $gxg^{-1}=f(x)$ for all $x\in G$, so $f=$ Int $g$.

a) Assume that G is connected. Then G has a largest compact subgroup K; it is connected. There exists a closed central vector subgroup (no. 2) N of G such that G is the direct product $N\times K$.

b) Assume that the group of connected components of G is finite. Then:

(i) Every compact subgroup of G is contained in a maximal compact subgroup.

(ii) If $K_1$ and $K_2$ are two maximal compact subgroups of G, there exists $g\in G$ such that $K_2=gK_1g^{-1}$.

(iii) Let K be a maximal compact subgroup of G. Then $K\cap G_0$ is equal to $K_0$; it is the largest compact subgroup of $G_0$.

(iv) There exists a closed central vector subgroup N of $G_0$, normal in G, such that, for any maximal compact subgroup K of G, $G_0$ is the direct product of $K_0$ by N, and G is the semi-direct product of K by N.

a) We retain the notations of Prop. 4. The projection of Ker $f$ onto V is a finite subgroup of the vector group V, hence is reduced to the identity element. It follows that Ker $f$ is contained in $T\times S$, hence that G is the direct product of the vector group $N =f(V)$ and the compact group $K =f(T\times S)$. Every compact subgroup of G has a projection onto N that is reduced to the identity element, hence is contained in K. This proves a).

b) Assume now that $G/G_0$ is finite. By $a), G_0$ is the direct product of its largest compact subgroup M and a vector subgroup P; the subgroup M of G is clearly normal. Let $\mathfrak{n}$ be a vector subspace complement of L(M) in L(G), stable under the adjoint representation of G (no. 1 and no. 3, Prop. 3); this is an ideal of L(G) and we have $L(G) = L(M)\times \mathfrak{n}$. Let N be the integral subgroup of G with Lie algebra $\mathfrak{n}$; by Chap. III, §6, no. 6, Prop. 14, it is normal in G. The projection of L(G) onto L(P) with kernel L(M) induces an isomorphism from $\mathfrak{n}$ to L(P); it follows that the projection of $G_0$ onto P induces an étale morphism from N to P; since P is simply-connected, this is an isomorphism, and N is a vector group. The morphism $(x, y) \rightarrow xy$ from $M\times N$ to $G_0$ is an injective étale morphism (since $M\cap N$ is reduced to the identity element), hence an isomorphism. It follows that N is a closed subgroup of G and that the quotient $G/N$ is compact, since $G_0/N$ is compact and $G/G_0$ is finite (General Topology, Chap. III, §4, no. 1, Cor. 2 of Prop. 2).

By Integration, Chap. VII, §3, no. 2, Prop. 3, every compact subgroup of G is contained in a maximal compact subgroup, these are conjugate, and for any maximal compact subgroup K of G, G is the semi-direct product of K by N. Since $G_0$ contains N, it is the semi-direct product of N by $G_0\cap K$; it follows that $G_0\cap K$ is connected, hence equal to $K_0$, since $K/(G_0\cap K)$ is isomorphic to $G/G_0$, hence finite; finally, $K_0$ is clearly the largest compact subgroup of $G_0$ by a).

Indeed, by (ii) there exists an element $g\in G$ such that $gK_1g^{-1}= K_2$; by (iv), there exists $n\in N$ and $k\in K_1$ such that $g=nk$. The element $n$ then has the required properties.