Theorem 2

Let G be a connected compact Lie group.

a) The Lie algebras of the maximal tori of G are the Cartan subalgebras of L(G).

b) Let $T_1$ and $T_2$ be two maximal tori of G. There exists $g\in G$ such that $T_2=gT_1g^{-1}$.

c) G is the union of its maximal tori.

Let $\mathfrak{t}$ be a Cartan subalgebra of L(G); the integral subgroup of G whose Lie algebra is $\mathfrak{t}$ is closed (Chap. VII, §2, no. 1, Cor. 4 of Prop. 4) and commutative (Th. 1), and hence is a torus of G. If T is a maximal torus of G, its Lie algebra is commutative, hence is contained in a Cartan subalgebra of L(G) (Th. 1). It follows that the maximal tori of G are exactly the integral subgroups of G associated to the Cartan subalgebras of L(G), hence a). Assertion b) follows from Th. 1, since the canonical homomorphism $G\rightarrow$ Int(L(G)) is surjective (Chap. III, §6, no. 4, Cor. 4 of Prop. 10).

Denote by X the union of the maximal tori of G, and let T be a maximal torus of G. The continuous map $(g, t) \rightarrow gtg^{-1}$ from $G\times T$ to G has image X, which is therefore closed in G; thus, to prove c), it suffices to prove that X is open in G; since X is invariant under inner automorphisms, it suffices to show that, for all $a\in T, X$ is a neighbourhood of $a$. We argue by induction on the dimension of G and distinguish two cases:

$1)a$ is not central in G. Let H be the identity component of the centralizer of $a$ in G; this is a connected compact subgroup of G distinct from G, which contains T, and hence $a$. Since Ad $a$ is semi-simple (§1, no. 1), the Lie algebra of H is the nilspace of Ad $a-1$; it now follows from Chap. VII, §4, no. 2, Prop. 4, that the union Y of the conjugates of H is a neighbourhood of $a$. By the induction hypothesis, $H\subset X$, and hence $Y\subset X$; thus, X is a neighbourhood of $a$.

$2)a$ is central in G. It suffices to prove that $a$ exp $x$ belongs to X for all $x$ in L(G). Now every element $x$ of L(G) belongs to a Cartan subalgebra of G (Th. 1); the corresponding integral subgroup $T'$ contains exp $x$; since it is conjugate to T, it contains $a$ and hence $a$ exp $x$, as required.