Université du Québec à MontréalDépartement de mathématiquesMarco A. Pérez B.marco.perez@cirget.ca

THE MAXIMAL TORUS OF A COMPACT LIEGROUP

April 2011

Contents

Introduction i

1 Lie Groups and Lie Algebras 1

1.1 Some basic notions on Lie groups . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Integration on Lie groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 The Mapping Degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Conjugacy Theorems 11

2.1 Maximal Tori and Weyl Groups . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.2 Conjugacy Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.3 Some consequences of the Conjugacy Theorems . . . . . . . . . . . . . . . . 25

Bibliography 29

Introduction

If G is a compact connected Lie group, we say that T is a torus if it is a connected abeliansubgroup of G. A torus T is said to be maximal if given another torus T ′ such that T ⊆ T ′,then T = T ′. An equivalent definition of a torus is that of a Lie subgroup of G isomorphicto Tk = Rk/Zk, for some k ∈ N. It is well known every compact connected Lie group has atorus and that every torus in contained in a maximal torus. The fact that G is compact is ofvital importance. A non compact Lie group need not have any nontrivial tori, for exampleRn.Maximal tori in a compact connected Lie group G have some interesting properties. Amongthem, the most important are known as the conjugacy theorems.

Theorem (First Conjugacy Theorem). In a compact connected Lie group G, every elementis conjugate to an element in any fixed maximal torus.

Theorem (Second Conjugacy Theorem). Any two maximal tori in a compact connected Liegroup G are conjugate.

The purpose of these notes is to give a self contained proof of the conjugacy theorems. Inthe first chapter we recall some basic notions on Lie group theory, such as the exponentialmap and integration. We also study the concept of mapping degree of a smooth map, whichshall play an important role in the proof of the conjugacy theorems.

We start the second chapter studying some notions related to maximal tori, such as Weylgroups. We prove that the Weyl group associated to a torus is finite. As a consequence ofthe Second Conjugacy Theorem, we prove that any two Weyl groups are isomorphic. Thenwe give the proof for both First and Second Conjugacy Theorems, which are based on [2].Finally, we show some consequences of these theorems. For example, the exponential map ofa compact connected Lie group is surjective. Other consequences are related to the notionof centralizer. We prove that the centralizer of any maximal torus is the torus itself.

i

ii

Chapter 1

Lie Groups and Lie Algebras

1.1 Some basic notions on Lie groups

Recall that a Lie Group G is a differentiable manifold with a group structure given by maps:

m : G×G −→ G (g, h) 7→ gh (multiplication),i : G −→ G g 7→ g−1 (inversion),

such that m and i are smooth maps. A Lie algebra is a finite dimensional vector space g(over R or C) with a multiplication [·, ·] : g× g −→ g, called the Lie bracket such that:

(1) [X, Y ] = −[Y,X],

(2) [X, [Y, Z]] + [Y, [Z,X]] + [Z, [X, Y ]] = 0.

A vector field X : G −→ TG is called left-invariant if for every g, h ∈ G we have

(dlg)(Xh) = Xlg(h),

where lg : G −→ G is the left translation h 7→ gh and (dlg) : ThG −→ Tlg(h)G is thedifferential map. We shall denote g the set of all left-invariant vector fields on a Lie group Gwith the Lie bracket [·, ·] for vector fields, i.e.,

[X, Y ]p(f) = Xp(Y (f))− Yp(X(f)),

where X and Y are vector fiends on G, p ∈ G, and f is a smooth function on a neighbourhoodof p.

1

Proposition 1.1.1.

(1) g is isomorphic to Te(G), where e denotes the identity element of G.

(2) [X, Y ] ∈ g, foe every X, Y ∈ g.

(3) g is a Lie algebra

Proof: See [4, Example 4.19 and Theorem 4.20].

A map F : G −→ H is a homomorphism of Lie groups if it is a smooth group homomor-phism. A map ψ : g −→ h is a homomorphism of Lie algebras if it is linear and satisfiesthe equality

ψ([X, Y ]) = [ψ(X), ψ(Y )].

A Lie subgroupH of G is a Lie group which is a subgroup of G and the inclusion i : H −→ Gis an injective immersion. A closed Lie subgroup of G is a Lie subgroup H such that i(H)is closed in G.

Theorem 1.1.1 (Closed Subgroup Theorem). If H is a subgroup of a Lie group G that isalso a closed subset of G, then H is an embedded Lie subgroup.

Proof: See [4, Theorem 20.10].

Let G be a Lie group and g its Lie algebra. Let V ∈ g and consider the local flow generatedby V , say ϕ. It is well known that every left-invariant vector field has a global flow (see [2,Chapter 1]). Then there exists a Lie group homomorphism ϕ : R −→ G such that dϕ(1) = V .The exponential map is the application exp : g −→ G given by V −→ ϕ(1).

Proposition 1.1.2 (Properties of exp). Let F : H −→ G be a homomorphism of Lie groups.Then the following diagram is commutative:

HF // G

hF∗

//

exp

OO

g

exp

OO

2

Proof: See [5, Theorem 3.32].

Let M be a differentiable manifold and let G be a Lie group. A smooth map α : G×M −→Msatisfying

(1) α(gh, x) = α(g, α(h, x)), and

(2) α(e, x) = x,

is called a left group action of G on M . Sometimes the map α is denoted simply as ·. Thenotion of right group action is defined similarly. Note that the product of a Lie groupG is a left group action. Such an action defines a diffeomorphism lg : G −→ G given bylg(h) = gh. Given a smooth form ω on G, we shall say that ω is left invariant if and onlyif l∗g(ω) = ω, for every g ∈ G, where l∗g is the pullback mapping of forms.

Example 1.1.1 (Adjoint Action). The map Ãd : G×G −→ G given by (g, h) 7→ ghg−1 is aleft action of G on itself. Note that the identity element e is fixed by the map Ãdg, for everyg ∈ G. Then

dÃdg : TeG ∼= g −→ TeG ∼= g

is a linear isomorphism of g onto itself since dÃdg is a diffeomorphism. Hence we have a map

Ad : G −→ Aut(g)g 7−→ dÃdg,

called the adjoint representation of G.

1.2 Integration on Lie groups

A volume form on a smooth n-manifold M is a nowhere vanishing n-form on M . Forinstance, dx1 ∧ · · · ∧ dxn is a volume form on Rn. Let {(Uα, ϕα)} be a smooth atlas of amanifold M . This atlas is called oriented if all transition functions ϕα ◦ ϕ−1β have positiveJacobian determinants everywhere they are defined. A manifold is called oriented if it hasan oriented atlas, or equivalently, if if has a volume form.

Now we recall how to integrate over a smooth oriented n-manifold M . Let {(Uα, ϕα)} be anoriented atlas of M . Consider a partition of unity subordinate to the cover {Uα}, i.e.,a collection of smooth functions ρα : M −→ R such that

3

(1) supp(ρα) = {x ∈M : ρα(x) 6= 0} ⊆ Uα.

(2) The collection {supp(ρα)} is locally finite (every point of M has a neighbourhood whichintersects finitely many members of {supp(ρα)}).

(3) 0 ≤ ρα ≤ 1 and∑

α ρα = 1.

Let ω be a compactly supported n-form in M , where M has the orientation given by theatlas {(Uα, ϕα)}. We define ∫

M

ω =∑α

∫Uα

ραω,

where∫Uα

= ραω means∫ϕα(Uα)

(ϕ−1α )∗(ραω) (a Riemann integral). Notice that the previous

sum has finitely many summands since supp(ω) is compact, and so it is covered by finitelymany Uα. It is well known from elementary differential geometry that this definition ofintegral does not depend on the oriented atlas {(Uα, ϕα)} and partition of unity either.The most important theorem about integrals on manifolds is the Stokes’s theorem. We statethe following special case:

Theorem 1.2.1 (Stokes’s Theorem). Let M be an oriented smooth n-manifold. Let ω be acompactly supported (n− 1)-form on M . Then∫

M

dω = 0.

Proof: See [4, Theorem 14.9].

Now we study integration on Lie groups. We shall only consider the case where G is acompact connected Lie group. Recall the following result, which gives rise to a Riemannianstructure on every Lie group G.

Theorem 1.2.2. Let G be a compact connected Lie group. Then there exists a symmetricbilinear form 〈·, ·〉K on g, called the Killing form of G, which makes G into a metric spaceand satisfies

〈Ad(g)(X),Ad(g)(Y )〉K = 〈X, Y 〉K ,

for every g ∈ G and X, Y ∈ g.

Proposition 1.2.1. Every connected and compact Lie Group is orientable. In fact, G hasa unique left-invariant orientation form ω with the property that

∫Gω = 1.

4

Proof: Let {e1, . . . , en} be an orthogonal basis of g with respect to the Killing form〈·, ·〉K . Then g∗ ∼= T ∗eG = spanR{e∗1, . . . , e∗n}. Let ωe = e∗1 ∧ · · · ∧ e∗n ∈ Λn(T ∗eG). Notethat ωe is a nonzero alternating n-linear functional on TeG. Define ω : G −→ Λn(G) byω := l∗g−1(ωe).

(i) ω is smooth: We only prove the case n = 2. The rest follows similarly. LetX1, X2 ∈ TgG. We have

(lg−1)∗X1 = a1(g)e1 + a2(g)e2, where a1 and a2 depend smoothly on g,

(lg−1)∗X2 = b1(g)e1 + b2(g)e2, where b1 and b2 depend smoothly on g,

ωg(X1, X2) = l∗g−1(e

∗1 ∧ e∗2)(X1, . . . , X2)

= e∗1 ∧ e∗2((lg−1)∗X1, (lg−1)∗X2)= e∗1((lg−1)∗X1) · e∗2((lg−1)∗X2)− e∗1((lg−1)∗X2) · e∗2((lg−1)∗X1),= a1(g)b2(g)− a2(g)b1(g) is a smooth function.

Since ω is smooth on a neighbourhood of each point of G, we have that ω is smooth.

(ii) ω is nowhere vanishing: Suppose there exists g ∈ G such that ωg = 0. Then0 = l∗g−1(ωe). Let X1, . . . , Xn ∈ TgG be linearly independent vectors. Since lg−1is a diffeomorphism, we have that (lg−1)∗X1, . . . , (lg−1)∗Xn are linearly independentvectors in TeG. On the other hand, 0 = ωe((lg−1)∗X1, . . . , (lg−1)∗Xn). Since ωe isa nonzero alternating n-linear functional on TeG, we have that the last equalityimplies that (lg−1)∗X1, . . . , (lg−1)∗Xn are linearly dependent vectors on TeG, gettinga contradiction.

(iii) ω is left-invariant: Let h ∈ G and X1, . . . , Xn ∈ TgG. We have:

(l∗hω)g(X1, . . . , Xn) = ωlh(g)((lh)∗X1, . . . , (lh)∗Xn)

= (l(hg)−1)∗ωe((lh)∗X1, . . . , (lh)∗Xn)

= ωe((l(hg)−1)∗(lh)∗X1, . . . , (l(hg)−1)∗(lh)∗Xn)

= ωe((lg−1)∗(lh−1)∗(lh)∗X1, . . . , (lg−1)∗(lh−1)∗(lh)∗Xn)

= ωe((lg−1)∗X1, . . . , (lg−1)∗Xn)

= (lg−1)∗ωe(X1, . . . , Xn)

= ωg(X1, . . . , Xn).

5

1.3 The Mapping Degree

Let M and N be two oriented, connected and compact smooth n-manifolds. Let f : M −→ Nbe a smooth map and ω an n-form on N with integral 1. Define the mapping degree of fto be

deg(f) =

∫M

f ∗(ω).

Before proving that the degree of f is well defined, i.e., does not depend on the choice of then-form ω on N with integral 1, we recall the following result:

Theorem 1.3.1 (Poincaré Duality Theorem). If N is a compact and oriented smooth n-manifold, define a map PD : Ωp(N) −→ Ωn−p(N)∗ by

PD(ω)(η) =

∫N

ω ∧ η.

Then PD descends to a linear map PD : HpdR(N) −→ Hn−pdR (N)

∗, which is a diffeomorphismfor each p.

Proof: See [1, Page 65].

Proposition 1.3.1. deg(f) is well defined.

Proof: Let ω and ω′ be two n-forms on N such that∫Nω =

∫Nω′ = 1. Consider the

case p = n in the Poincaré Duality Theorem. We have an isomorphism

PD : HndR(N) −→ H0dR(N)∗

given by

PD([α])([f ]) =

∫N

fα,

for every [α] ∈ HndR(N) and every [f ] ∈ H0dR(N). Since N is connected, we have thatH0dR(N) = {[f ] : f is constant}. So we get

PD([ω])([f ]) =

∫N

f · ω = f∫N

ω = f

∫N

ω′ =

∫N

f · ω′ = PD([ω′])([f ]),

6

for every [f ] ∈ H0dR(N). It follows PD([ω]) = PD([ω′]). Since PD is an isomorphism, weget [ω] = [ω′], i.e., there exists a smooth (n − 1)-form η on N such that ω = ω′ + dη.Finally, we obtain∫

M

f ∗(ω) =

∫M

f ∗(ω′ + dη) =

∫M

(f ∗(ω) + f ∗(dη))

=

∫M

f ∗(ω′) +

∫M

f ∗(dη) =

∫M

f ∗(ω′) +

∫M

df ∗(η)

=

∫M

f ∗(ω′) + 0, by the Stokes’s Theorem

=

∫M

f ∗(ω′).

Proposition 1.3.2. Let f : M −→ N be a smooth map between oriented, connected andcompact smooth n-manifolds. If f is not surjective, then deg(f) = 0.

Proof: Note that the set f(M) is compact in N . Let y ∈ N − f(M). Then there existsa neighbourhood V of y such that V ∩ f(M) = ∅. Let B be a closed set contained in V .There exists a bump function for B supported in V , say g. Thus, g is zero outside of V .Let ω be a smooth n-form on N with

∫Nω = 1 and η = g ·ω. Let λ =

∫Nη =

∫Vg ·ω > 0.

Set ω′ = η/λ. We have∫Nω′ = 1 and ω′ = 0 outside of V . Let (x1, . . . , xn) be smooth

coordinates for the neighbourhood V . On V we can write ω′ = gλdx1 ∧ · · · ∧ dxn. So we

get f ∗(ω′) = g◦fλd(x1 ◦ f)∧ · · · ∧ d(xn ◦ f) = 0, since g(f(x)) = 0, for every x ∈M . Hence

deg(f) =

∫M

f ∗(η′) =

∫M

0 = 0.

Given a connected, compact and oriented smooth n-manifold M with orientation ω, for everyvolume form α and for every x ∈M there exists a scalar λx ∈ R such that αx = λxωx. We saythat α is associated with the orientation in M if λx > 0, for every x ∈ M . If f : M −→ Nis a smooth map between connected, compact and oriented smooth n-manifolds, and if β isa volume form on M , then f ∗(β) is a smooth n-form on N . For every x ∈ M , there existsλx ∈ R such that f ∗(β)x = λx · αx. We shall denote det(f)(x) = λx.

7

Proposition 1.3.3. Let f : M −→ N be a smooth map between oriented, connected andcompact smooth n-manifolds, and let α and β be two volume forms on M and N , respectively,associated with the orientations on M and N . If y is a point with a finite pre-image f−1(y) ={x1, . . . , xn} and xi is a regular point of f for all i, then

deg(f) =n∑i=1

sign(det(f)(xi)).

Proof: Since f is smooth, there exists neighbourhoods Ui and Vi about xi and y, re-spectively, such that f(Ui) ⊆ Vi. Since M is Hausdorff, we can choose the Ui such thatUi ∩Uj = ∅ if i 6= j. Let V = ∩ni=1Vi. As we did in the proof of the previous proposition,we can construct an n-form β′ from β such that

∫Nβ′ = 1 and β′ = 0 outside of V . Let

fi := f |Ui , for each i = 1, . . . , n. Since the sets Ui are disjoint, we get

deg(f) =

∫M

f ∗(β′) =n∑i=1

∫Ui

(fi)∗(β′) =

n∑i=1

deg(fi).

Choose the Ui and V to be diffeomorphic to bounded open subsets of Rn. By the previousequality, it suffices to prove the formula for a smooth map f : U −→ V , where U and Vare bounded open subsets of Rn, and a point y ∈ Rn with a single pre-image f−1(y) = {x}.Without loss of generality, we can assume x = 0 and y = 0 (choose centred coordinateneighbourhoods). In this particular case, the volume forms α and β can be written as

αx = a(x)dx1 ∧ · · · ∧ dxn,βx = b(x)dx1 ∧ · · · ∧ dxn.

Since α and β are associated with the orientation of Rn, we have a(x) > 0 and b(y) > 0for every x ∈ U , and y ∈ V . We have

f ∗(β)(x) = f ∗(bdx1 ∧ · · · ∧ dxn)(x)= (b ◦ f)(x)d(x1 ◦ f) ∧ · · · ∧ d(xn ◦ f)= (b ◦ f)(x) · det(Jac(f))dx1 ∧ · · · ∧ dxn.

Also, f ∗(β) = det(f)α. So we have

det(f)(x)a(x)dx1 ∧ · · · ∧ dxn = (b ◦ f)(x) · det(Jac(f))dx1 ∧ · · · ∧ dxn,

i.e.,

det(f)(x) =(b ◦ f)(x) · det(Jac(f))

a(x).

Since a(x) and b(x) are positive, we have sign(det(f)) = sign(det(Jac(f))). Now we usethe hypothesis that det(Jac(f)) 6= 0, i.e, f is a local diffeomorphism of a neighbourhood

8

U ′ ⊆ U of 0 to a neighbourhood V ′ ⊆ V of 0. The derived form β′ is compactly supportedsince supp(β′) ⊆ V ′ is closed and bounded in Rn. Since f : U ′ −→ V ′ is a diffeomorphism,we have

sign(det(Jac(f))) =

{> 0 if f is orientation preserving,< 0 if f is orientation reversing.

So we get

deg(f) =

∫Rnf ∗(β′) = sign(det(Jac(f)))

∫Rnβ′

= sign(det(Jac(f)))

= sign(det(f)).

9

10

Chapter 2

Conjugacy Theorems

2.1 Maximal Tori and Weyl Groups

Let G be a compact connected Lie group. A maximal torus T ⊆ G is maximal connectedabelian subgroup. Every maximal torus is isomorphic to a torus Tk = RkZk . A generator ofa torus T is an element t ∈ T such that the set {tn : n ∈ Z>0} is dense in T . Generators arecharacterized by the following result.

Theorem 2.1.1 (Kronecker’s Theorem). A vector v ∈ Rk represents a generator of Tk ifand only if 1 and the components v1, . . . , vk of v are linearly independent over the rationalnumbers Q.

Proof: Consider Zk, Rk and Tk as additive groups. We have an exact sequence

0 −→ Zk −→ Rk −→ Tk −→ 0.

Since Rk ∼= T0Rk ∼= T0modZkTk, the exponential map exp : T0modZkTk −→ Tk can beidentified with the canonical quotient map ρ : Rk −→ Tk. By properties of the exponentialmap (see Proposition 1.1.2), we have the following commutative square:

Rkexp //

f∗

��

Tk

f

��R exp // S1

where f : Tk −→ S1 is any Lie group homomorphism. This gives rise to the followingcommutative diagram with exact rows:

11

0 // Zk i //

f∗|Zk

��

Rkexp //

f∗

��

Tk

f

��

// 0

0 // Zj

// R exp // S1 // 0

where the i and j are inclusions. It follows f∗(v1, . . . , vk) = α1v1 + · · · + αkvk, for someα1, . . . , αk ∈ Z. Now we prove that the following statements are equivalent:

(i) 1, v1, . . . , vk are linearly dependent over Q.

(ii)∑k

i=1 qivi ∈ Q for some k-tuple 0 6= (q1, . . . , qk) ∈ Qk.

(iii)∑k

i=1 αivi ∈ Z for some k-tuple 0 6= (α1, . . . , αk) ∈ Zk.

(iv) vmodZk is in the kernel of a nontrivial homomorphism f : Tk −→ S1, wherev = (v1, . . . , vk).

(v) vmodZk is not a generator of Tk.

The equivalence (i)⇐⇒ (ii)⇐⇒ (iii) is clear.

• (iii) =⇒ (iv): Let v = (v1, . . . , vk). Let f : Tk −→ S1 be any nontrivial Lie grouphomomorphism such that f∗(v1, . . . , vk) = α1v1 + · · ·+ αkvk. Then we have

f(vmodZk) = exp(f∗(v1, . . . , vk)) = exp(α1v1 + · · ·+ αkvk)= ei2π(α1v1+···+αkvk)

= 1, since α1v1 + · · ·+ αkvk ∈ Z.

Then, vmodZk ∈ Ker(f).

• (iv) =⇒ (iii): Suppose vmodZk ∈ Ker(f), where f : Tk −→ S1 is a nontrivialhomomorphism. We have

f(exp(v1, . . . , vk)) = 0 =⇒ exp(f∗(v1, . . . , vk)) = 0.

It follows f∗(v1, . . . , vk) ∈ Z. But f∗(v1, . . . , vk) = α1v1 + · · · + αkvk, for someα1, . . . , αk ∈ Z.

• (iv) =⇒ (v): Let vmodZk ∈ Ker(f : Tk −→ S1), where f is a nontrivial Lie grouphomomorphism. Then Ker(f) $ Tk, i.e., Ker(f) is a closed proper Lie subgroup ofTk. We have

{(vmodZk)n : n > 0} ⊆ Ker(f) = Ker(f) $ Tk.Hence vmodZk is not a generator.

12

• (v) =⇒ (iv): If vmodZk is not a generator then vmodZk is contained in a properclosed subgroup H $ Tk. It follows that the quotient Tk/H is a nontrivial connectedabelian Lie group, i.e., a torus. So there is an ismomorphism ϕ : Tk/H −→ Tm,for some m > 0. Also, we have an isomorphism ψ : Tm −→ S1 × · · · × S1, whereS1 × · · · × S1 has m factors. Let ρ : Tk −→ Tk/H be the quotient projection.Consider the composition f = pr1 ◦ ψ ◦ ϕ ◦ ρ, where pr1 : S1 × · · · × S1 −→ S1 isthe projection onto the first factor. Since vmodZn ∈ H, we have f(vmodZk) = 1.Hence vmodZk ∈ Ker(f), where f is the nontrivial Lie group homomorphism givenby the composite map pr1 ◦ ψ ◦ ϕ ◦ ρ.

The Weyl group of G associated to a maximal torus T is defined as the quotient groupW (T ) = N(T )/T , where N(T ) is the normalizer of T , i.e., N(T ) = {g ∈ G/gTg−1 = T}.We shall prove bellow that any two maximal tori are conjugate. Assume this for a momentto prove the following result.

Proposition 2.1.1. If T and T ′ are two maximal tori in G, then W (T ) and W (T ′) areisomorphic.

Proof: Since T and T ′ are conjugate, there exists g ∈ G such that T ′ = gTg−1. Letϕ : N(T ′) −→ N(T )/T be the map given by ϕ(x) = g−1xgT , for every x ∈ N(T ′).

(i) ϕ is well defined: In other words, g−1xg ∈ N(T ). If fact,

(g−1xg)T (g−1xg)−1 = g−1x(gTg−1)x−1g = g−1xT ′x−1g = g−1T ′g = T.

Also, it is clear that ϕ is a group homomorphism.

(ii) ϕ is surjective: Let yT ∈ N(T )/T . Then gyg−1 ∈ N(T ′). In fact,

(gyg−1)T ′(gyg−1)−1 = gy(g−1T ′g)y−1g−1 = gyTy−1g−1 = gTg−1 = T ′.

Also, yT = g−1(gyg−1)gT = ϕ(gyg−1).

(iii) Ker(ϕ) = T ′: Suppose ϕ(x) = eT . Then (g−1xg)T = eT , i.e., g−1xg ∈ T . Hencex = gtg−1 ∈ gTg−1 = T ′. Now suppose x ∈ T ′. We have ϕ(x) = g−1xgT = eTsince g−1xg ∈ T . Therefore, Ker(ϕ) = T ′.

By the First Fundamental Theorem of Isomorphisms, we have N(T )/Ker(ϕ) ∼= Im(ϕ),i.e. N(T )/T ∼= N(T ′)/T ′.

13

Note that the normalizer N(T ) acts on T via conjugation

N(T )× T −→ T(n, t) 7→ ntn−1

For each n ∈ N(T ), we have the adjoint automorphism Ãdn : T −→ T given by Ãdn(t) =ntn−1. So we get a continuos map N(T ) −→ Aut(T ). Since T is isomorphic to some Tk,we have Aut(T ) ∼= Aut(Tk). Recall from the proof of the Kronecker’s theorem that givenan automorphism ϕ : Tk −→ Tk one can construct the following commutative diagram withexact rows:

0 // Zk i //

ϕ∗|Zk��

Rkexp //

ϕ∗

��

Tk

ϕ

��

// 0

0 // Zk i// Rk exp // Tk // 0

Lemma 2.1.1. Let Φ : Aut(Tk) −→ Aut(Zk) be the map ϕ 7→ ϕ∗|Zk . Then Φ is anisomorphism.

Proof: It is clear that Φ is a group homomorphism. Let {e1, . . . , ek} denote the canonicalbasis of Rk.

(i) Φ is injective: Suppose ϕ∗|Zk = Id. Then, ϕ∗ = Id. In fact,

ϕ∗(v1, . . . , vk) = ϕ∗(v1e1 + · · ·+ vkek)= v1ϕ∗(e1) + · · ·+ vkϕ∗(ek)= v1e1 + · · ·+ vkek, since each ej ∈ Zk

= (v1, . . . , vk).

So we get ϕ(vmodZk) = ϕ ◦ exp(v1, . . . , vk) = exp(v1, . . . , vk) = vmodZk. Henceϕ = Id.

(ii) Φ is surjective: Let F : Zk −→ Zk be an automorphism. Define F̃ : Rk −→ Rk by

F̃ (v1e1 + · · ·+ vkek) = v1F (e1) + · · ·+ vkF (ek).

Now define ϕ : Tk −→ Tk by

ϕ((v1, . . . , vk)modZk) = exp ◦F̃ (v1, . . . , vk).

14

We show ϕ is well defined. Suppose (v1, . . . , vk)modZk = (w1, . . . , wk)modZk. Thenvi = wi +mi, for some mi ∈ Z, for every i = 1, . . . , k. We have

exp ◦F̃ (v1, . . . , vk) = exp ◦F̃ (v1e1 + · · ·+ vkek)= exp(v1F (e1) + · · ·+ vkF (ek))= exp(w1F (e1) + · · ·+ wkF (ek) +m1F (e1) + · · ·+mkF (ek))= exp(w1F (e1) + · · ·+ wkF (ek))= exp(F̃ (w1, . . . , wk)).

The map ϕ is surjective since exp and F̃ are. Suppose ϕ(vmodZk) = 0modZk.Then exp(F̃ (v)) = 0. Since exp is the canonical quotient map Rk −→ Rk/Zk, wehave that F̃ (v) ∈ Zk. So v ∈ Z, i.e., vmodZk = 0modZk. Hence ϕ is surjective.Therefore, ϕ is an automorphism. Also, Φ(ϕ) = ϕ∗|Zk = F̃ |Zk = F .

Theorem 2.1.2. The Weyl group associated to T is finite.

Proof: Let N0 be the connected component of the identity in N(T ). We shall proveN0 = T . Then W (T ) = N(T )/N0. This quotient is compact since N(T ) is compact.We shall also show that N(T )/N0 is discrete. Since every compact and discrete space isfinite, we have W (T ) = N(T )/N0 is finite.

(i) N0 = T : By the previous lemma, we can consider the adjoint action of N(T ) onT as a continuous map f : N(T ) −→ Aut(T ) ∼= Gl(k,Z). Since N0 is connectedwe have that f(N0) is connected in Gl(k,Z). On the other hand, Gl(k,Z) is adiscrete space, it follows f(N0) = {Id}. We get ntn−1 = t, for every n ∈ N0 andevery t ∈ T . In other words, N0 acts trivially on T . Let α : R −→ N0 be a one-parameter group, i.e., a continuous group homomorphism. Since α(R) and T areconnected, we have that α(R) · T is connected in N0. Then α(R) · T is a connectedabelian subgroup of G (i.e., a torus) containing T . By maximality of T , we haveα(R) · T = T . Hence α(R) ⊆ T . Now, the image α(R) covers a neighbourhoodof the identity in N0. Notice that N0 is closed since it is a connected component.By the Closed Submanifold Theorem, we have that N0 is a Lie group. It followsthat α(R) generates N0, i.e., N0 =

⋃n∈N α(R)n, (see [5, Proposition 3.18]). Then,

T ⊆ N0 ⊆ T . Therefore N0 = T .

15

(ii) N(T )/N0 is discrete: We have to show that every point in N(T )/N0 is open. LetxN0 be a point of N(T )/N0. We know xN0 is open in N(T )/N0 is and only ifρ−1(xN0) is open in N , where ρ : N(T ) −→ N(T )/N0 is the canonical quotientmap. We have

ρ−1(xN0) = {g ∈ N(T ) : gN0 = xN0} = {g ∈ N(T ) : g = xn, ∃!n ∈ N0}.

Note that ρ−1(xN0) is diffeomorphic to N0. On the other hand, N0 is open inN(T ) since it is a connected component of a locally path connected space. Henceρ−1(xN0) is open.

2.2 Conjugacy Theorems

We need some more structure before proving the conjugacy theorems. Let H be a closedLie subgroup of G. As a consequence of the Quotient Manifold Theorem (see [4, Theorem9.16]), the quotient group G/H is a smooth manifold of dimension dim(G) − dim(H), andthe canonical quotient map ρ : G −→ G/H is smooth. As we did in Theorem 1.2.1, we shallconstruct a nowhere vanishing left invariant form on G/T .

Theorem 2.2.1. Let H and G as above. If H is connected, then G/H is orientable.

Proof: Let τe be a nonzero alternating (n − k)-linear functional on the tangent spaceTeHG/H. Define a map τ : G/H −→ Λn−k(G/H) by τ(gH) = L∗g−1(τe), where Lg−1 :G/H −→ G/H is the map Lg−1(xH) = g−1xH. We only proof that τ is well defined.The rest of the proof (τ is smooth, nowhere vanishing and left invariant) is similar tothe proof of Theorem 1.2.1. Suppose gH = g′H. Then g′ = gh for some h ∈ H. Wehave to show L∗(gh)−1(τe) = L

∗g−1(τe). Note that L

∗(gh)−1(τe) = L

∗h−1g−1(τe) = L

∗g−1L

∗h−1(τe).

Then L∗(gh)−1(τe) = L∗g−1(τe) if and only if τe = L

∗h−1(τe). Also, note that L

∗h−1(τe) ∈

Λn−k(T ∗eHG/H), i.e., L∗h−1 is an alternating (n − h)-linear functional on TeHG/H. If

we follow the construction done in the proof of Theorem 1.2.1, we have that τe is thedeterminant function. From linear algebra we know L∗h−1(τe) = λ(h)τe, for some λ(h) ∈ R(see [3, Theorem 2, page 152]). By properties of pullbacks, we have that λ is a grouphomomorphism from H to the multiplicative group R∗ = R − {0}. Moreover, λ is acontinuous map since L∗h−1 is smooth. Since H is connected and λ(e) = 1, we have thatλ(h) > 0 for every h ∈ H. Now suppose that there exists h ∈ H such that λ(h) 6= 1. We

16

can assume λ(h) > 1. Then λ(h)n = λ(hn) ∈ λ(H) = [a, b]. Note that λ(H) is a closedinterval since H is connected and compact (it is a closed subgroup of a compact groupG). Since λ(h) > 1, the fact that λ(h)n ∈ [a, b] for every n ∈ N is a contradiction. Itfollows λ(h) = 1 for every h ∈ H. Hence L∗h−1(τe) = τe.

We shall denote the forms ω and τ on G and G/H by dg and d(gH), respectively. Let g bethe Lie algebra if G. Let T be a maximal torus of G. Since T is a Lie subgroup of G, wehave that t, the Lie algebra of T , is a Lie subalgebra of g. Consider the Killing form 〈·, ·〉Kon g. We know this product is invariant under the action of the adjoint representation Ad(g)for all g ∈ G. Let m be the orthogonal complement to t with respect to 〈·, ·〉K . Notice thatg = t⊕ m. We have the following result:

Proposition 2.2.1. The adjoint map restricted to the torus Ad|T acts trivially on everyvector in t, and non-trivially on every nonzero vector in m. Moreover, for every t ∈ T , themap Ad(t) : g −→ g preserves g = t⊕ m.

Proof: Let V ∈ t. We show that Ad(t)(V ) = V , for every t ∈ T . We know Ad(t)(V ) =dÃd(t)(V ). Let f be a smooth function about e. Then dÃd(t)(V )(f) = V (f ◦ Ad(t)).Let α be an integral curve of V passing through e. We have

dÃd(t)(V )(f) = α′(0)(f ◦ Ãd(t)) = dds

∣∣∣∣s=0

(f ◦ Ãd(t) ◦ α(s))

=d

ds

∣∣∣∣s=0

(f(tα(s)t−1))

=d

ds

∣∣∣∣s=0

f(α), since t, α(s) ∈ T and T is abelian

= V (f).

Hence Ad(t)(V ) = V . Now suppose that V is a vector in m such that Ad(t)(V ) = V . Letα be the maximal integral curve of V passing through e. Consider β(s) = tα(s)t−1. Wehave

V (f) = Ad(t)(V )(f) = dÃd(t)(V )(f) =d

ds

∣∣∣∣s=0

(f ◦ Ãd(t) ◦ α(s))

=d

ds

∣∣∣∣s=0

(f ◦ β(s)) = β′(0)(f).

Then β is a maximal integral curve of V . By the uniqueness of α, we get tα(s) = α(s)t,i.e., α(R) commutes with T . So α(R) · T is a connected abelian group containing T .

17

By maximality of T , we get T = α(R) · T . Thus, α(R) ⊆ T . It follows V ∈ t. HenceV ∈ t ∩ m = {0} and so V = 0.Finally, we prove that Ad(t) preserves g = t⊕ m. We already know that Ad(t)(t) ⊆ t. Itis only left to show that Ad(t)(m) ⊆ m. Let X ∈ m and Y ∈ t. We have

〈Y,Ad(t)(X)〉K =〈Ad(t−1)(Y ),Ad(t−1)(Ad(t)(X))

〉K

=〈Ad(t−1)(Y ), X

〉K

= 0.

Hence Ad(t)(X) ∈ m.

We have that m is an invariant subspace of Ad(t), for every t ∈ T . Then we can define anaction of T on m, denoted

AdG/T : T −→ Aut(m)t 7→ Ad(t)|m.

We know that G/T is a smooth manifold of dimension n − k, where n = dim(G) andk = dim(T ). Since the canonical quotient map ρ : G −→ G/T is smooth, it induces a mapρ∗ : g = t⊕m −→ TeTG/T . It is known that ρ∗ maps m isomorphically onto TeTG/T . We shallidentify m with TeTG/T via this isomorphism. Since T is a Lie group, we can find a volumeform dt which is left invariant. Then d(gT )dt is a nowhere vanishing left invariant form onG/T × T . Note that (d(gT )dt)(eT, e) and dg(e) are nonzero alternating n-linear functionalson g. We know that, up to multiplication by a constant, there is a unique alternating n-linearfunctional on t⊕ m. So we get (d(gT )dt)(eT, e) = c · dg(e), where c > 0 or c < 0. Replacing−d(gT ) by d(gT ) if necessary, we may assume that c > 0. From now on, we consider themanifolds G/T ×T and G with the orientations given by d(gT )dt and dg. Consider the mapq : G/T × T −→ G given by (gT, t) 7→ gtg−1. Note that q is a well defined smooth map. Inorder to prove the conjugacy theorems, we need the following result:

Lemma 2.2.1 (Main Lemma). Let G be a connected and compact Lie group and T amaximal torus in G. Then the map q has mapping degree deg(q) = |W (T )|, where |W (T )|is the order of the Weyl group associated to T .

Theorem 2.2.2 (First Conjugacy Theorem). In a connected compact Lie group G, everyelement is conjugate to an element in any fixed maximal torus.

Proof: By the previous lemma, deg(q) = |W (T )| 6= 0, and Proposition 1.3.2 implies q issurjective. Let g ∈ G. Then there exists hT ∈ G/T and t ∈ T such that g = q(hT, t) =hth−1, i.e., g is conjugate to t, where t ∈ T .

18

Theorem 2.2.3 (Second Conjugacy Theorem). Any two maximal tori in a connected com-pact Lie group G are conjugate.

Proof: Let T and T ′ be two maximal tori in G. We have to show T ′ = gTg−1 forsome g ∈ G. Let t be a generator of T , which exists by the Kronecker’s theorem. By theprevious theorem, there exists g ∈ G such that gtg−1 ∈ T ′. It follows gtng−1 = (gtg−1)n ∈T ′, for every n > 0. Consider the set {gtng−1 : n > 0}. Note that g{tn : n > 0}g−1 is aclosed set containing {gtng−1 : n > 0}. Then {gtng−1 : n > 0} ⊆ g{tn : n > 0}g−1. Theother inclusion follows similarly. Hence

{gtng−1 : n > 0} = g{tn : n > 0}g−1 = gTg−1.

Since {gtng−1 : n > 0} ⊆ T ′, we get gTg−1 ⊆ T ′, i.e., T ⊆ g−1T ′g, where g−1T ′g is atorus in G. By maximality of T , we get T = g−1T ′g, i.e., gTg−1 = T ′.

The proof of the main lemma is quite difficult and technical. We prove the following twohelper lemmas. Using them, the proof of the main lemma follows easily.

Lemma 2.2.2. sign(det(q)(gT, t)) = sign(det(AdG/T (t−1)−IdG/T )), where IdG/T is the iden-

tity on m.

Proof: Let V1, . . . , Vn be vector fields on G/T × T . Then

q∗dg(V1(gT, t), . . . , Vn(gT, t)) =

= dg(q∗(V1(gT, t)), . . . , q∗(Vn(gT, t)))

= l∗q(gT,t)−1dg(q∗(V1(gT, t)), . . . , q∗(Vn(gT, t))), since dg is left invariant

= dg((lq(gT,t)−1)∗q∗(V1(gT, t)), . . . , (lq(gT,t)−1)∗q∗(Vn(gT, t)))

= dg((lq(gT,t)−1 ◦ q)∗(V1(gT, t)), . . . , (lq(gT,t)−1 ◦ q)∗(Vn(gT, t))).

Since d(gT )dt is left invariant we also have

(d(gT )dt)(V1(gT, t), . . . , Vn(gT, t)) =

= (d(gT )dt)((l(gT,t)−1)∗(V1(gT, t)), . . . , (l(gT,t)−1)∗(Vn(gT, t))),

Notice that

lq(gT,t)−1 ◦ q(gT, t) = L(gtg−1)−1(gtg−1) = (gtg−1)−1 · (gtg−1) = e,l(gT,t)−1(gT, t) = e,

then the arguments of the expressions above are elements of TeG ∼= t⊕ m.

19

Now we use the equality (d(gT )dt)(eT, e) = cdg(e), where c > 0. Write Wi(gT, t) =(L(gT,t)−1)∗Vi(gT, t), for simplicity. We obtain

(d(gT )dt)(V1(gT, t), . . . , Vn(gT, t)) = c · dg(W1(gT, t), . . . ,Wn(gT, t)).

Thus,

q∗dg(V1(gT, t), . . . , Vn(gT, t)) =

= dg((lq(gT,t)−1 ◦ q)∗V1(gT, t), . . . , (lq(gT,t)−1 ◦ q)∗Vn(gT, t))= dg((lq(gT,t)−1 ◦ q)∗(l(gT,t)−1)−1∗ W1(gT, t), . . . , (lq(gT,t)−1 ◦ q)∗(l(gT,t)−1)−1∗ Wn(gT, t))= dg((lq(gT,t)−1 ◦ q)∗(l(gT,t))∗W1(gT, t), . . . , (lq(gT,t)−1 ◦ q)∗(l(gT,t))∗Wn(gT, t))= dg((lq(gT,t)−1 ◦ q ◦ l(gT,t))∗W1(gT, t), . . . , (lq(gT,t)−1 ◦ q ◦ l(gT,t))∗Wn(gT, t))= det(Jac(lq(gT,t)−1 ◦ q ◦ l(gT,t))∗)dg(W1(gT, t), . . . ,Wn(gT, t)).

It follows

q∗dg =det(Jac(lq(gT,t)−1 ◦ q ◦ l(gT,t))∗)

c· d(gT )dt.

By definition of det(q)(gT, t), we obtain

det(q)(gT, t) =det(Jac(lq(gT,t)−1 ◦ q ◦ l(gT,t))∗)

c.

We compute lq(gT,t)−1 ◦ q ◦ l(gT,t):

lq(gT,t)−1 ◦ q ◦ l(gT,t)(hT, s) = lq(gT,t)−1 ◦ q(ghT, ts) = lq(gT,t)−1((gh)(ts)(gh)−1)= lgt−1g−1(ghtsh

−1g−1) = gt−1g−1ghtsh−1g−1

= gt−1htsh−1g−1.

On the other hand, Ãd(g)(Ãd(t−1)(h)sh−1) = Ãd(g)(t−1htsh−1) = gt−1htsh−1g−1.

Hence lq(gT,t)−1 ◦ q ◦ l(gT,t)(hT, s) = Ãd(g)(Ãd(t−1)(h)sh−1). Now consider the functionf : G/T × T −→ G given by f(hT, s) = Ãd(t−1)(h)sh−1. So we get

lq(gT,t)−1 ◦ q ◦ lq(gT,t) = Ãd(g) ◦ f,(lq(gT,t)−1 ◦ q ◦ lq(gT,t))∗ = (Ãd(g) ◦ f)∗ = Ãd(g)∗ ◦ f∗

= d(Ãd(g)) ◦ f∗ = Ad(g) ◦ f∗,

det(q)(gT, t) =1

cdet(Jac(lq(gT,t)−1 ◦ q ◦ l(gT,t))∗)

=1

cdet(Ad(g) ◦ f∗)

=1

cdet(Ad(g)) · det(f∗).

20

Recall that Ad(g) is orthogonal with respect to the Killing form 〈·, ·〉K . Thendet(Ad(g)) = ±1. Since G is connected and det(Ad(e)) = 1, we get det(Ad(g)) = 1,for every g ∈ G. Hence det(q)(gT, t) = 1

cdet(f∗). In order to compute the matrix of f∗,

it suffices to determine the matrices of f∗|t and f∗|m, since g = t⊕ m and

f∗ =

(f∗|t 00 f∗|m

).

Let V ∈ t and let α : R −→ G be the maximal integral curve of V passing through e.Then β(r) = (eT, α(r)) is also an integral curve of V . Let g be a smooth function aboute. We have

f∗|t(V )(g) = f∗ ◦ β∗(d

dr

∣∣∣∣r=0

)(g) = (f ◦ β)∗

(d

dr

∣∣∣∣r=0

)(g)

=d

dr

∣∣∣∣r=0

(g ◦ f ◦ β(r)) = ddr

∣∣∣∣r=0

(g(f(eT, α(r))))

=d

dr

∣∣∣∣r=0

(g(Ãd(t−1)(e)α(r)e−1)) =d

dr

∣∣∣∣r=0

(g(α(r)))

= α∗

(d

dr

∣∣∣∣r=0

)(g)

= V (g).

We get f∗|t(V ) = V for every V ∈ t, i.e., f∗|t = IdT . Let V ∈ m and let α : R −→ G bethe maximal integral curve of V passing through e. Then β(r) = (α(r)T, e) is an integralcurve of V . We have

f∗|m(V )(g) = V (g ◦ f) = β′(0)(g ◦ f)

=d

dr

∣∣∣∣r=0

(g ◦ f ◦ β(r)) = ddr

∣∣∣∣r=0

g ◦ f(α(r)T, e)

=d

dr

∣∣∣∣r=0

g(Ãd(t−1)(α(r))eα(r)−1)

=d

dr

∣∣∣∣r=0

g(Ãd(t−1)(α(r))α(−r))

= γ∗

(d

dr

∣∣∣∣r=0

)(g),

where γ(r) = Ãd(t−1)(α(r))·α(−r). Since G is a Lie group, the differential of the productis the sum of the differentials. Then γ∗ = dÃd(−t) ◦ α∗ − α∗.

21

We have

f∗|m(V )(g) = γ∗(d

dr

∣∣∣∣r=0

)(g)

= dÃd(t−1) ◦ α∗(d

dr

∣∣∣∣r=0

)(g)− α∗

(d

dr

∣∣∣∣r=0

)(g)

= Ad(t−1)V (g)− V (g) =[AdG/T (t

−1)(V )− IdG/T (V )]

(g).

Hence f∗|m = AdG/T (t−1)− IdG/T . Finally, we get

f∗ =

(IdT 00 AdG/T (t

−1)− IdG/T

),

and then

det(q)(gT, t) =1

cdet(f∗)

=1

cdet(AdG/T (t

−1)− IdG/T ) · det(IdT )

=1

cdet(AdG/T (t

−1)− IdG/T ).

Since c > 0, we obtain sign(det(q)(gT, t)) = sign(det(AdG/T (t−1)− IdG/T )).

Lemma 2.2.3. There exists a generator t of T such that:

(i) |q−1(t)| = |W (T )|;

(ii) AdG/T (t−1) has no real eigenvalues. Consequently, dim(G/T ) is even.

(iii) det(q) > 0 at each point of q−1(t).

Proof:

(i) Let t be any generator of T . Let (gT, s) ∈ q−1(t). Then gsg−1 = t, i.e., g−1tg =s ∈ T . As we did in the proof of the Second Conjugacy Theorem, we have that{g−1tng : n > 0} = g−1Tg. Since g−1tg ∈ T , we have {g−1tng : n > 0} ⊆ T . Itfollows T ⊆ gTg−1. Also, gTg−1 is a torus since it is a connected abelian Liesubgroup of G. So we get T = gTg−1, by maximality of T . Hence g ∈ N(T ).Define a map ϕ : q−1(t) −→ W (T ) by ϕ(gT, s) = gT . It is clear that ϕ is welldefined. We also show it is a bijection.

22

– ϕ is injective: Suppose ϕ(g1T, s1) = ϕ(g2T, s2). Then g1T = g2T and henceg2 = g1t

′, for some t′ ∈ T . Since T is abelian, we have

s1 = g−11 (g2s2g

−12 )g1 = g

−11 g1t

′s2(g1t′)−1g1 = t

′s2(t′)−1g−11 g1 = t

′s2(t′)−1 = s2.

Hence (g1T, s1) = (g2T, s2).

– ϕ is surjective: Let gT ∈ W (T ) and s = g−1tg. Then s ∈ T since g ∈ N(T ).Also, (gT, s) ∈ q−1(t) since q(gT, s) = gsg−1 = t. Moreover, gT = ϕ(gT, s).Hence ϕ is onto.

Since ϕ is a bijection, we get |q−1(t)| = |W (T )|.

(ii) Recall that AdG/T (t−1) is an orthonormal linear transformation on m with respect

to 〈·, ·〉K . Then the real eigenvalues of AdG/T (t−1) (if any) must be ±1. By theKronecker’s Theorem t0 = t

2 is also a generator of T . Replace t by t0 and supposethat AdG/T (t

−1) has a real eigenvalue λ = ±1. We have AdG/T (t−1)(V ) = λV ,where V is an eigenvector associated to λ. Also,

AdG/T (t−10 )(V ) = AdG/T (t

−1t−1)(V ) = AdG/T (t−1) · AdG/T (t−1)(V )

= AdG/T (t−1)(λV ) = λ2V = V.

Hence 1 is a real eigenvalue of AdG/T (t−10 ). It follows AdG/T (t

−n0 )(V ) = V , for every

n > 0. Since t0 is a generator of T , we have T = {t−n0 : n > 0}. Consider the setEV = {s ∈ T : AdG/T (s)(V ) = V }. Note that EV is a closed set containing {t−n0 :n > 0}. It follows T = {t−n0 : n > 0} ⊆ EV and so AdG/T (s)(V ) = V , for everys ∈ T . On the other hand, Proposition 2.2.1 states that AdG/T (s) acts nontriviallyon every nonzero vector of m, getting a contradiction since AdG/T (s)(V ) = V andV 6= 0. Therefore, the automorphism AdG/T (t−1) : m ∼= TeTG/T −→ m ∼= TeTG/Thas no real eigenvalues. It follows the characteristic polynomial of AdG/T (t

−1) haseven degree, since every polynomial of odd degree has at least one real root. Hencedim(G/T ) = dim(TeTG/T ) is even.

(iii) Let (gT, s) ∈ q−1(t). By part (ii), AdG/T (t−1)− IdG/T has no real eigenvalues. LetA = AdG/T (t

−1)− IdG/T . Suppose that A has negative determinant. Let

p(x) = det(xI − A) = xn + · · ·+ a1x+ a0be the characteristic polynomial of A. Note that a0 = (−1)ndet(A) = det(A),since n is even. Thus, p(0) < 0. On the other hand, limx→+∞ p(x) = +∞. Sothere exists a > 0 such that p(a) > 0. By the Intermediate Value Theorem, thereexists b ∈ (0, a) such that p(b) = 0, i.e., b is a real eigenvalue of A, getting acontradiction. Therefore, det(AdG/T (t

−1)− IdG/T ) > 0. By the previous lemma, wehave det(q)(gT, s) > 0.

23

Proof of the Main Lemma. We know that q∗dg = det(q)(d(gT )dt). By the previ-ous lemma, let t be a generator of T satisfying (i), (ii) and (iii). Then q−1(t) ={(g1T, s1), . . . , (gmT, sm)}, where m = |W (T )|. By Proposition 1.3.3, we have deg(q) =∑m

i=1 sign(det(q)(giT, si)). By (iii), we have sign(det(q)(giT, si)) = 1. Hence

deg(q) =m∑i=1

1 = |W (T )|.

Now we give some examples of maximal tori and Weyl groups. If you are interested in thedetails, see [2, Chapter 4, Section 3].

Example 2.2.1.

(1) Consider the group U(1) of unitary n× n matrices. Let ∆(n) ⊆ U(1) be the subgroupof diagonal matrices

D =

z1 · · · 0... . . . ...0 · · · zn

, zi ∈ S1,and let S∆(n) = ∆(n) ∩ SU(n) be the subgroup of diagonal matrices as above withz1 · · · zn = 1. The groups ∆(n) and S(n) are maximal tori in U(n) and SU(n), respec-tively. The Weyl group of the unitary group U(n) is the symmetric group Sn.

(2) Let SO(2n) be the orthogonal special group, i.e., the orthogonal matrices whose de-terminant in 1. The subgroup T (n) = SO(2) × · · · × SO(2) (n times) is a maximaltorus in SO(2n) and SO(2n + 1). Let Gn be the group of permutations σ of the set{−n, . . . ,−1, 1, . . . , n} for which σ(−i) = −σ(i). The Weyl group of SO(2n + 1) isGn. Also, the Weyl group of SO(2n) is SGn, the subgroup of Gn consisting of evenpermutations.

(3) Now consider the symplectic group Sp(n) of unitary matrices of the form

(A −BB A

).

We have a canonical inclusion U(n) −→ Sp(n) given by A 7→(A 00 A

). Let T (n)

be the image of the torus ∆(n) under this inclusion. Then T (n) is a maximal torus inSp(n). The Weyl group of Sp(n) is Gn.

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2.3 Some consequences of the Conjugacy Theorems

We collect a few consequences of the Second Conjugacy Theorem. From now on, any dis-cussion of the maximal torus or operation of the Weyl group shall refer to a particular toruschosen once and for all. As we saw above, any other choice leads to conjugate tori and iso-morphic Weyl groups. Hence, the following result do not depend of the choice of a particulartorus.

Corollary 2.3.1. The exponential map of a compact connected Lie group G is surjective.

Proof: Let g ∈ G, there exists h ∈ G such that g ∈ hTh−1. Notice that hTh−1 is atorus in G. If there is no torus T ′ such that hTh−1 ⊆ T ′, then hTh−1 is a maximal torus.Now if there exists a maximal torus T ′ such that hTh−1 ⊆ T ′, then g ∈ T ′. In any case,g is in a maximal torus in G. Let T be a maximal torus containing g. The exponentialmap exp : t −→ T is surjective since it can be identified with the canonical quotient mapρ : Rk −→ R/Zk ∼= T , where k = dim(T ). Since ρ is surjective, so is exp : t −→ T . Hencethere exists V ∈ t such that g = exp(V ).

The centralizer of a subgroup H ⊆ G is the subgroup Z(H) = {g ∈ G : gh = hg ∀h ∈ H}.The center is the centralizer of the entire group G.

Corollary 2.3.2. Let G be a compact connected Lie group and T its maximal torus.

(i) If S ⊆ G is a connected abelian subgroup, then Z(S) is the union of the maximal toricontaining S.

(ii) Z(T ) = T .

(iii) The center of G is the intersection of all maximal tori in G. In particular, Z(G) iscontained in any maximal torus.

Proof:

(i) Note that S is a connected abelian subgroup of G, i.e., a torus. Let g ∈ Z(S). Thengs = sg, for every s ∈ S. It folows gs = sg, for every s ∈ S. Hence Z(S) ⊆ Z(S).The other inclusion is obvious. We get Z(S) = Z(S). So we may assume thatS = S is a torus. Let x ∈ Z(S) and let 〈x, S〉 be the subgroup generated by x andS. Let B = 〈x, S〉. Then B is a compact abelian subgroup of G. Let B0 denote the

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connected component of the identity of B. We have that B0 is a torus, since it isconnected and abelian. Consider the quotient B/B0. We show that xB0 generatesB/B0. Consider yB0 ∈ B/B0. If y ∈ 〈x, S〉 then y = xnsk11 · · · skmm , for somen,m ∈ Z. Since S is connected and the identity e is in S, we have S ⊆ B0 since B0is a component. Then, yB0 = x

nB0. It follows xB0 generates B/B0. As we did inthe proof of Theorem 2.1.2, we have that B/B0 is finite. Then B/B0 is a finite cyclicgroup and hence there exists l ∈ N such that B/B0 ∼= Zl. It follows B ∼= B0 × Zl.It is known that a compact Lie group contains a dense cyclic subgroup if and onlyif is isomorphic to Tk × Zl, for some k, l ∈ N. (See [2, Corollary 4.14]). Thus, Bcontains a dense cyclic subgroup {gn : n ∈ Z}. We get B = {gn : n ∈ Z}. Let T bea maximal torus containing g. Then gn ∈ T for every n ∈ Z and so B ⊆ T . Thenx ∈ T and S ⊆ T . We obtain Z(S) ⊆

⋃{T : T is a maximal torus and T ⊇ S}.

The other inclusion follows easily since T is abelian.

(ii) Z(T ) =⋃{T ′ : T ′ is a maximal torus and T ′ ⊇ T} = T , since T is maximal.

(iii) Suppose x is in the center of G. There exists a maximal torus T in G such thatx ∈ T . Let T ′ be any other maximal torus in G. By the Second Conjugacy Theorem,there exists g ∈ G such that T ′ = gTg−1. Then x = gg−1x = gxg−1 ∈ T ′. Hencex is in any maximal torus of G, and Z(G) ⊆

⋂{T : T is a maximal torus in G}.

Now suppose that x is in every maximal torus. Let g ∈ G. There exists a maximaltorus T such that g ∈ T . Since x, g ∈ T and T is abelian, we get xg = gx. Thenx ∈ Z(G) and the other inclusion follows.

The Weyl group W (T ) of G acts on T via the map product xT · t = xtx−1. Consider thegroup homomorphism ρ : W (T ) −→ Aut(T ) given by

ρ(xT ) : T −→ T ρ(xT )(t) = xtx−1.

Corollary 2.3.3. The Weyl group acts effectively on the maximal torus, i.e., the homo-morphism ρ is injective.

Proof: Suppose ρ(xT ) = IdT . Then xtx−1 = t, i.e., xt = tx for every t ∈ T . Hence

x ∈ Z(T ) = T and so xT = eT .

Recall that the orbit of t ∈ T under the action ρ is the set W (T )t = {gT · t : gT ∈ W (T )}.

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Corollary 2.3.4. Two elements of the maximal torus are conjugate in G if and only if theylie in the same orbit under the action of the Weyl group.

Proof: Let x, y ∈ T such that y = gxg−1, for some g ∈ G. Let Z(x) and Z(y) be thecentralizers of x and y. Consider the map Ãdg : G −→ G. Let h ∈ Z(x). We have

Ãdg(h)y = (ghg−1)y = (gh)(g−1y) = (gh)(xg−1) = g(hx)g−1

= g(xh)g−1 = (gx)(hg−1) = (yg)(hg−1) = y(ghg−1)

= yÃdg(h).

Then Ãdg(h) ∈ Z(y). We have Ãdg(T ) ⊆ Z(y) since T ⊆ Z(x). Let T ′ = Ãdg(T ). Wehave that T and T ′ are maximal tori in the connected component Z(y)0 of the identityof Z(y). Then there exists z ∈ Z(y)0 such that zT ′z−1 = T , i.e.,

T = Ãdz(T′) = Ãdz ◦ Ãdg(T ) = Ãdzg(T ).

We obtain T = zgT (zg)−1 and hence zg ∈ N(T ). Thus, zgT ∈ W (T ). Moreover,

(zgT ) · x = (zg)x(zg)−1 = z(gxg−1)z−1 = zyz−1 = y, since z ∈ Z(y).

Therefore, x and y are in the same orbit. Now suppose that y = (gT ) · x, for somegT ∈ W (T ). Then y = gxg−1 and hence x and y are conjugate.

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Bibliography

[1] Bott, R; Tu, L. Differentials forms in Algebraic Topology. Springer-Verlag. New York(1983).

[2] Bröcker, T; Dieck, T. Representations of Compact Lie Groups. Springer-Verlag. NewYork (1985).

[3] Hoffman, K; Kunze, R. Linear Algebra. Prentice-Hall, Inc. New Jersey (1971).

[4] Lee, J. Introduction to Smooth Manifolds. Springer-Verlag. New York (2003).

[5] Warner, F. Foundations of Differentiable Manifolds and Lie Groups. Springer-Verlag.New York (1983).

[6] Wright, A. The Conjugacy Theorems.

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