The function $v \rightarrow$ F((Ad $v$)$(x), y$) from K to $\mathbf{R}$ is continuous, and hence has a minimum at some point $k\in K$. Let $u\in L(K)$ and put

$h(t) =$ F((Ad exp($tu$)$.k$)$(x), y$)$,t\in \mathbf{R}$.

We have $h(t)\geq h(0)$ for all $t$; moreover, by Chap. III, §3, no. 12, Prop. 44,

$\frac{dh}{dt}(0) = F([u$,(Ad $k$)$(x)], y) = F(u$,[(Ad $k$)$(x), y])$,

hence the lemma (Functions of a Real Variable, Chap. I, §1, no. 7, Prop. 7).

Since $\mathfrak{g}$ is reductive, its Cartan subalgebras are commutative (Chap. VII, §2, no. 4, Cor. 3 of Th. 2). Conversely, let $\mathfrak{t}$ be a commutative subalgebra of $\mathfrak{g}$. By §1, no. 3, Prop. 1, ad $x$ is semi-simple for all $x\in \mathfrak{t}$; by Chap. VII, §2, no. 3, Prop. 10, there exists a Cartan subalgebra of $\mathfrak{g}$ containing $\mathfrak{t}$. This proves the first assertion of the theorem.

Now let $\mathfrak{t}$ and $\mathfrak{t}'$ be two Cartan subalgebras of $\mathfrak{g}$. We prove that there exists $u\in$ Int($\mathfrak{g}$) such that $u(\mathfrak{t}) =\mathfrak{t}'$. By Prop. 1 of §1, no. 3, we can assume that $\mathfrak{g}$ is of the form L(G), where G is a connected compact Lie group, and can choose a separating invariant symmetric bilinear form F on $\mathfrak{g}$. Let $x$ (resp. $x'$) be a regular element of $\mathfrak{g}$ such that $\mathfrak{t}=\mathfrak{g}^0(x)$ (resp. $\mathfrak{t}'=\mathfrak{g}^0(x')$) (Chap. VII, §3, no. 3, Th. 2). Applying Lemma 1 with K = G, we see that there exists $k\in G$ such that [(Ad $k$)$(x), x']$ is orthogonal to $\mathfrak{g}$ with respect to F, and hence is zero; then (Ad $k$)$(x)\in \mathfrak{g}^0(x') =\mathfrak{t}'$, so $\mathfrak{g}^0$((Ad $k$)$(x)$) $=\mathfrak{t}'$ since (Ad $k$)$(x)$ is regular. We conclude that (Ad $k$)$(\mathfrak{t}) =\mathfrak{t}'$, hence the theorem.

Put G = Int($\mathfrak{g}$), and consider the fixer $Z_G(\mathfrak{a})$ of $\mathfrak{a}$ in G; this is a Lie subgroup of G, whose Lie algebra $\mathfrak{z}_{\mathfrak{g}}(\mathfrak{a})$ consists of the elements of $\mathfrak{g}$ that commute with every element of $\mathfrak{a}$ (Chap. III, §9, no. 3, Prop. 7). Then $\mathfrak{t}$ and $u^{-1}(\mathfrak{t}')$ are two Cartan subalgebras of the compact Lie algebra $\mathfrak{z}_{\mathfrak{g}}(\mathfrak{a})$. By Th. 1, there exists an element $v$ of $Z_G(\mathfrak{a})$ such that $v(\mathfrak{t}) =u^{-1}(\mathfrak{t}')$; any such element has the desired properties.

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.

b) For all $n\geq 1$, the map $g \rightarrow g^n$ from G to itself is surjective.

Indeed, exp(L(G)) contains all the maximal tori of G, hence a). Assertion b) follows from the formula (exp $x$)$^n=$ exp $nx$ for $x$ in L(G).

Let $x$ be an element of the centre of G; by Th. $2c)$, there exists a maximal torus T of G containing $x$; then $x$ belongs to all the conjugates of T, hence to all the maximal tori of G. Conversely, if $x$ belongs to all the maximal tori of G, it commutes with every element of G by Th. $2c)$.

There exists a maximal torus T of G containing $g$ (Th. $2c$)$)$, and hence contained in $C_0$. Moreover, the group $C_0$ is a connected compact Lie group, and hence the union of its maximal tori (Th. $2c$)$)$; these all contain $g$ (Cor. 2), hence are exactly the maximal tori of G containing $g$.

Since Ad $g$ is semi-simple, the dimension of the nilspace of Ad $g-1$ is also that of the centralizer C of $g$. By loc. cit., Prop. 8, and Th. $1,g$ is regular if and only if $C_0$ is a maximal torus of G. The conclusion now follows from Cor. 3.

b) Let $\mathfrak{s}$ be a commutative subalgebra of L(G). The fixer of $\mathfrak{s}$ in G is connected; it is the union of the maximal tori of G whose Lie algebras contain $\mathfrak{s}$.

To prove a), it suffices to prove that if an element $g$ of G centralizes S, there exists a maximal torus of G containing S and $g$. Now, if C is the centralizer of $g$, we have $g\in C_0$ (Cor. 3) and $S\subset C_0$; if T is a maximal torus of the connected compact Lie group $C_0$ containing S, we have $g\in T$ (Cor. 2), hence a). Assertion b) follows from a) applied to the closure of the integral subgroup with Lie algebra $\mathfrak{s}$, in view of Chap. III, §9, no. 3, Prop. 9.

Let T be a maximal torus of G and C its centralizer; since L(T) is a Cartan subalgebra of L(G), we have L(T) = L(C), hence C = T since C is connected (Cor. 5).

Let C be the centralizer of A. Then T and $s^{-1}(T')$ are two maximal tori of $C_0$; every element $g$ of $C_0$ such that (Int $g$)$(T) =s^{-1}(T')$ has the desired properties.

Let $h\in H$. Then $h^{-1}Th$ is a maximal torus of $H_0$, hence (Th. 2) there exists $g\in H_0$ such that $hg\in N_H(T)$; thus $h$ belongs to $N_H(T).H_0$, hence the first assertion. The second follows immediately.

a) Let $f: G\rightarrow G'$ be a surjective morphism of Lie groups. The maximal tori of $G'$ are the images under $f$ of the maximal tori of G. If the kernel of $f$ is central in G (for example discrete), the maximal tori of G are the inverse images under $f$ of the maximal tori of $G'$.

b) Let H be a connected closed subgroup of G. Every maximal torus of H is the intersection with H of a maximal torus of G.

c) Let H be a connected closed normal subgroup of G. The maximal tori of H are the intersections with H of the maximal tori of G.

a) Let T be a maximal torus of G; then L(T) is a Cartan subalgebra of L(G) (no. 2, Th. $2a$)$)$, so $L(f(T))$ is a Cartan subalgebra of $L(G')$ (Chap. VII, §2, no. 1, Cor. 2 of Prop. 4); it follows that $f(T)$ is a maximal torus of $G'$ (no. 2, Th. $2a$)$)$. If Ker $f$ is central in G, it is contained in T (Cor. 2 of Th. 2), so $T =f^{-1}(f(T))$.

Conversely, let $T'$ be a maximal torus of $G'$; we show that there exists a maximal torus T of G such that $f(T) = T'$. Let $T_1$ be a maximal torus of G; then $f(T_1)$ is a maximal torus of $G'$ and there exists $g'\in G'$ such that $T'=g'f(T_1)g^{'-1}$ (Th. $2b$)$)$; if $g\in G$ is such that $f(g) =g'$, we have $T'=f(T)$ with $T =gT_1g^{-1}$.

b) Let S be a maximal torus of H; this is a torus of G so there exists a maximal torus T of G containing S. Then $T\cap H$ is a commutative subgroup of H containing S, hence is equal to S (no. 2, Remark 2).

c) By §1, no. 3, Prop. $2c), L(G)$ is the direct product of L(H) with an ideal; thus, the Cartan subalgebras of L(H) are the intersections with L(H) of the Cartan subalgebras of L(G). Thus, for any maximal torus T of $G, T\cap H$ contains a maximal torus S of H and $S = T\cap H$ (no. 2, Remark 2).

a) The compact manifold $G/H$ is simply-connected.

b) The homomorphism $\pi_1(H)\rightarrow \pi_1(G)$, induced by the canonical injection of H into G, is surjective.

Since H is connected, we have an exact sequence (General Topology, Chap. XI, in preparation)

where $\overline{e}$ is the image in $G/H$ of the identity element of G. Since $G/H$ is connected, this immediately implies the equivalence of assertions a) and b). Moreover, if $f: G'\rightarrow G$ is a surjective morphism of connected compact Lie groups whose kernel is central, proving the proposition (in the form $a$)$)$ for G is the same as proving it for $G'$ (Prop. 2). Thus, we can first of all replace G by Ad(G), then assume that G is semi-simple, and then by replacing G by a universal covering (§1, no. 4, Cor. 2), assume that G is simply-connected. But then assertion b) is trivial.

Indeed, the Lie algebra of H contains a Cartan subalgebra of L(G). Thus, by Chap. VII, §2, no. 1, Cor. 4 of Prop. 4, H is the identity component of N. Since N is compact, H is of finite index in N.

b) The intersections with T of the conjugacy classes of G are the orbits of the Weyl group.

Let $g\in G$; by Th. 2 of no. 2, there exists $h\in G$ such that $g\in hTh^{-1}$, hence a). By definition of the Weyl group, any two elements in the same orbit of $W_G(T)$ on T are conjugate in G; conversely, let $a, b$ be two elements of T conjugate under G. There exists $h\in G$ such that $b=hah^{-1}$; applying Cor. 7 of Th. 2 (no. 2) with $A =\{a\},s=$ Int $h, T'= T$, we see that there exists $g\in G$ such that Int $hg$ maps T to T and $a$ to $b$. The class of $hg$ in $W_G(T)$ then maps $a$ to $b$, hence the proposition.

Indeed, this is a bijective continuous map between two compact spaces (cf. General Topology, Chap. III, p. 29, Cor. 1).

This follows from Cor. 1 and the fact that the canonical maps $\mathrm{T}\to \mathrm{T}/\mathrm{W}_{G}(\mathrm{T})$ and $\mathrm{G}\to \mathrm{G}/\mathrm{Int}(\mathrm{G})$ are open (General Topology, Chap. III, p. 10, Lemma 2).

Denote the Lie algebra of $\mathrm{G}$ by $\mathfrak{g}$, and that of $\mathrm{T}$ by $\mathfrak{t}$. The operation of $\mathrm{W}_{G}(\mathrm{T})$ on $\mathrm{T}$ induces a representation, called the canonical representation, of the group $\mathrm{W}_{G}(\mathrm{T})$ on the $\mathbf{R}$-vector space $\mathfrak{t}$.

b) The intersections with $\mathfrak{t}$ of the orbits of $\mathrm{G}$ are the orbits of $\mathrm{W}_{G}(\mathrm{T})$ on $\mathfrak{t}$.

Assertion a) follows from Th. 1 (no. 1). Let $x,y$ be two elements of $\mathfrak{t}$ conjugate under $\mathrm{Ad}(\mathrm{G})$, and let $h\in\mathrm{G}$ be such that $(\mathrm{Ad}\,h)(x)=y$. Applying the corollary of Th. 1 (no. 1) with $\mathfrak{a}=\{x\}$, $u=\mathrm{Ad}\,h$, $\mathfrak{t}'=\mathfrak{t}$, we see that there exists $g\in\mathrm{G}$ such that $\mathrm{Ad}\,hg$ maps $\mathfrak{t}$ to $\mathfrak{t}$ and $x$ to $y$. Then $hg\in\mathrm{N}_{G}(\mathrm{T})$ (Chap. III, §9, no. 4, Prop. 11), and the class of $hg$ in $\mathrm{W}_{G}(\mathrm{T})$ maps $x$ to $y$, hence the proposition.

Denote this map by $j$; it is bijective and continuous (Prop. 6). We have a commutative diagram

where $p$ and $q$ are quotient maps, and $i$ is the canonical injection. Since $i$ and $q$ are proper (General Topology, Chap. I, §10, no. 1, Prop. 2 and General Topology, Chap. III, §4, no. 1, Prop. 2 c)) and since $p$ is surjective, it follows that $j$ is proper (General Topology, Chap. I, §10, no. 1, Prop. 5), and hence is a homeomorphism.

a) Denote by $\mathrm{W}_{H}(\mathrm{T})$ the subgroup $\mathrm{N}_{H}(\mathrm{T})/\mathrm{T}$ of $\mathrm{W}_{G}(\mathrm{T})$; the group $H/H_{0}$ is isomorphic to the quotient group $\mathrm{W}_{H}(\mathrm{T})/\mathrm{W}_{H_{0}}(\mathrm{T})$.

b) $H$ is connected if and only if every element of $\mathrm{W}_{G}(\mathrm{T})$ that has a representative in $H$ belongs to $\mathrm{W}_{H_{0}}(\mathrm{T})$.

Assertion a) follows from Cor. 8 of Th. 2 (no. 2), and assertion b) is a particular case of a).

Put $Z = C(G)_0$; by §1, no. 4, Cor. 1 of Prop. 4, the morphism of Lie groups $g: Z\times \widetilde{D}(G)\rightarrow G$ such that $g(z, x) =z^{-1}p(x)$ is a covering; its kernel consists of the pairs $(z, x)$ such that $p(x) =z$, for which $x\in p^{-1}(Z)\subset \widetilde{T}$. Since the morphism $(z, x) \rightarrow f_T(z^{-1}) \widetilde{f}(x)$ from $Z\times \widetilde{D}(G)$ to H maps Ker $g$ to $\{e\}$, there exists a morphism $f$ from G to H such that $f\circ p= \widetilde{f}$ and $f(z) =f_T(z)$ for $z\in Z$. But we also have $f(t) =f_T(t)$ for $t\in p( \widetilde{T})$; since $T = Z.p( \widetilde{T})$, the restriction of $f$ to T is indeed $f_T$.

If $f: G\rightarrow H$ is a morphism of Lie groups such that $L(f) =\varphi$, the restriction $f_T$ of $f$ to T is the unique morphism from T to H such that $L(f_T) =\varphi |L(T)$. Conversely, let $f_T: T\rightarrow H$ be a morphism of Lie groups such that $L(f_T) =\varphi |L(T)$. Let $\widetilde{D}(G)$ and $p$ be as above; the map $L(p)$ induces an isomorphism from $L( \widetilde{D}(G))$ to the derived algebra $\mathfrak{b}$ of L(G). There exists a morphism of Lie groups $\widetilde{f}: \widetilde{D}(G)\rightarrow H$ such that $L( \widetilde{f}) = (\varphi |\mathfrak{b})\circ L(p)$ (Chap. III, §6, no. 1, Th. 1). The morphisms $t \rightarrow \widetilde{f}(t)$ and $t \rightarrow f_T(p(t))$ from $\widetilde{T}$ to H induce the same homomorphism of Lie algebras, and hence coincide. Applying Lemma 2, we deduce the existence of a morphism $f: G\rightarrow H$ such that $L(f)$ and $\varphi$ coincide on L(T) and $\mathfrak{b}$. Since $L(G) =\mathfrak{b}+ L(T)$, we have $L(f) =\varphi$.

Indeed, by Th. 2 (no. 2) the normal subgroup Ker $f$ of G reduces to the identity element if and only if its intersection with T reduces to the identity element.