The Maximal Torus of a Compact Lie Group

Home , Compact group, Maximal torus, Weyl group

Universitédu Québec àMontréal Département de mathématiques Marco A. PérezB. [email protected]

THE MAXIMAL TORUS OF A COMPACT LIE GROUP

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 abelian subgroup of G. A torus T is said to be maximal if given another torus T 0 such that T ⊆ T 0, then T = T 0. An equivalent deﬁnition of a torus is that of a Lie subgroup of G isomorphic to Tk = Rk/Zk, for some k ∈ N. It is well known every compact connected Lie group has a torus and that every torus in contained in a maximal torus. The fact that G is compact is of vital importance. A non compact Lie group need not have any nontrivial tori, for example Rn. Maximal tori in a compact connected Lie group G have some interesting properties. Among them, the most important are known as the conjugacy theorems.

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

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

The purpose of these notes is to give a self contained proof of the conjugacy theorems. In the ﬁrst chapter we recall some basic notions on Lie group theory, such as the exponential map and integration. We also study the concept of mapping degree of a smooth map, which shall 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 Weyl groups. We prove that the Weyl group associated to a torus is ﬁnite. As a consequence of the Second Conjugacy Theorem, we prove that any two Weyl groups are isomorphic. Then we 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 of a compact connected Lie group is surjective. Other consequences are related to the notion of 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 the differential map. We shall denote g the set of all left-invariant vector fields on a Lie group G with 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 neighbourhood of 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 homomorphism. A map ψ : g −→ h is a homomorphism of Lie algebras if it is linear and satisﬁes the equality ψ([X,Y ]) = [ψ(X), ψ(Y )]. A Lie subgroup H of G is a Lie group which is a subgroup of G and the inclusion i : H −→ G is 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 is also 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 generated by 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:

F H / G O O exp exp

h / g F∗

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 −→ M satisfying

(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 ·. The notion of right group action is defined similarly. Note that the product of a Lie group G is a left group action. Such an action defines a diffeomorphism lg : G −→ G given by lg(h) = gh. Given a smooth form ω on G, we shall say that ω is left invariant if and only ∗ ∗ if lg(ω) = ω, for every g ∈ G, where lg is the pullback mapping of forms.

Example 1.1.1 (Adjoint Action). The map Adf : G × G −→ G given by (g, h) 7→ ghg−1 is a left action of G on itself. Note that the identity element e is ﬁxed by the map Adf g, for every g ∈ G. Then ∼ ∼ dAdf g : TeG = g −→ TeG = g is a linear isomorphism of g onto itself since dAdf g is a diﬀeomorphism. Hence we have a map

Ad : G −→ Aut(g)

g 7−→ dAdf g, 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. For n instance, dx1 ∧ · · · ∧ dxn is a volume form on R . Let {(Uα, ϕα)} be a smooth atlas of a −1 manifold M. This atlas is called oriented if all transition functions ϕα ◦ ϕβ have positive Jacobian determinants everywhere they are deﬁned. A manifold is called oriented if it has an 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 an oriented 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 which intersects ﬁnitely many members of {supp(ρα)}). P (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 the atlas {(Uα, ϕα)}. We define Z X Z ω = ραω, M α Uα where R = ρ ω means R (ϕ−1)∗(ρ ω) (a Riemann integral). Notice that the previous Uα α ϕα(Uα) α α sum has finitely many summands since supp(ω) is compact, and so it is covered by finitely many Uα. It is well known from elementary differential geometry that this definition of integral 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 state the following special case:

Theorem 1.2.1 (Stokes’s Theorem). Let M be an oriented smooth n-manifold. Let ω be a compactly supported (n − 1)-form on M. Then Z dω = 0. M

Proof: See [4, Theorem 14.9].

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

Theorem 1.2.2. Let G be a compact connected Lie group. Then there exists a symmetric bilinear form h·, ·iK on g, called the Killing form of G, which makes G into a metric space and satisﬁes

hAd(g)(X), Ad(g)(Y )iK = hX,Y iK , for every g ∈ G and X,Y ∈ g.

Proposition 1.2.1. Every connected and compact Lie Group is orientable. In fact, G has R a unique left-invariant orientation form ω with the property that G ω = 1.

4 Proof: Let {e1, . . . , en} be an orthogonal basis of g with respect to the Killing form ∗ ∼ ∗ ∗ ∗ ∗ ∗ n ∗ h·, ·iK . Then g = Te G = spanR{e1, . . . , en}. Let ωe = e1 ∧ · · · ∧ en ∈ Λ (Te G). Note n that ωe is a nonzero alternating n-linear functional on TeG. Deﬁne ω : G −→ Λ (G) by ∗ ω := lg−1 (ωe).

(i) ω is smooth: We only prove the case n = 2. The rest follows similarly. Let X1,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) = lg−1 (e1 ∧ e2)(X1,...,X2) ∗ ∗ = e1 ∧ e2((lg−1 )∗X1, (lg−1 )∗X2) ∗ ∗ ∗ ∗ = e1((lg−1 )∗X1) · e2((lg−1 )∗X2) − e1((lg−1 )∗X2) · e2((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. Then ∗ 0 = lg−1 (ωe). Let X1,...,Xn ∈ TgG be linearly independent vectors. Since lg−1 is a diﬀeomorphism, we have that (lg−1 )∗X1,..., (lg−1 )∗Xn are linearly independent vectors in TeG. On the other hand, 0 = ωe((lg−1 )∗X1,..., (lg−1 )∗Xn). Since ωe is a nonzero alternating n-linear functional on TeG, we have that the last equality implies that (lg−1 )∗X1,..., (lg−1 )∗Xn are linearly dependent vectors on TeG, getting a contradiction.

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

∗ (lhω)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 −→ N be a smooth map and ω an n-form on N with integral 1. Deﬁne the mapping degree of f to be Z deg(f) = f ∗(ω). M Before proving that the degree of f is well deﬁned, i.e., does not depend on the choice of the n-form ω on N with integral 1, we recall the following result:

Theorem 1.3.1 (Poincar´eDuality Theorem). If N is a compact and oriented smooth n- manifold, deﬁne a map PD : Ωp(N) −→ Ωn−p(N)∗ by Z PD(ω)(η) = ω ∧ η. N

p n−p ∗ Then PD descends to a linear map PD : HdR(N) −→ HdR (N) , which is a diﬀeomorphism for each p.

Proof: See [1, Page 65].

Proposition 1.3.1. deg(f) is well deﬁned.

0 R R 0 Proof: Let ω and ω be two n-forms on N such that N ω = N ω = 1. Consider the case p = n in the Poincar´eDuality Theorem. We have an isomorphism

n 0 ∗ PD : HdR(N) −→ HdR(N)

given by Z PD([α])([f]) = fα, N n 0 for every [α] ∈ HdR(N) and every [f] ∈ HdR(N). Since N is connected, we have that 0 HdR(N) = {[f]: f is constant}. So we get Z Z Z Z PD([ω])([f]) = f · ω = f ω = f ω0 = f · ω0 = PD([ω0])([f]), N N N N

6 0 0 for every [f] ∈ HdR(N). It follows PD([ω]) = PD([ω ]). Since PD is an isomorphism, we get [ω] = [ω0], i.e., there exists a smooth (n − 1)-form η on N such that ω = ω0 + dη. Finally, we obtain Z Z Z f ∗(ω) = f ∗(ω0 + dη) = (f ∗(ω) + f ∗(dη)) M M M Z Z Z Z = f ∗(ω0) + f ∗(dη) = f ∗(ω0) + df ∗(η) M M M M Z = f ∗(ω0) + 0, by the Stokes’s Theorem M Z = f ∗(ω0). M

Proposition 1.3.2. Let f : M −→ N be a smooth map between oriented, connected and compact 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 exists a 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 . R R R Let ω be a smooth n-form on N with N ω = 1 and η = g ·ω. Let λ = N η = V g ·ω > 0. 0 R 0 0 Set ω = η/λ. We have N ω = 1 and ω = 0 outside of V . Let (x1, . . . , xn) be smooth 0 g coordinates for the neighbourhood V . On V we can write ω = λ dx1 ∧ · · · ∧ dxn. So we ∗ 0 g◦f get f (ω ) = λ d(x1 ◦ f) ∧ · · · ∧ d(xn ◦ f) = 0, since g(f(x)) = 0, for every x ∈ M. Hence Z Z deg(f) = f ∗(η0) = 0 = 0. M M

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

n X deg(f) = sign(det(f)(xi)). i=1

Proof: Since f is smooth, there exists neighbourhoods Ui and Vi about xi and y, respectively, such that f(Ui) ⊆ Vi. Since M is Hausdorﬀ, we can choose the Ui such that n Ui ∩ Uj = ∅ if i 6= j. Let V = ∩i=1Vi. As we did in the proof of the previous proposition, 0 R 0 0 we can construct an n-form β from β such that N β = 1 and β = 0 outside of V . Let

fi := f|Ui , for each i = 1, . . . , n. Since the sets Ui are disjoint, we get Z n Z n ∗ 0 X ∗ 0 X deg(f) = f (β ) = (fi) (β ) = deg(fi). M i=1 Ui i=1

n Choose the Ui and V to be diﬀeomorphic to bounded open subsets of R . By the previous equality, it suﬃces to prove the formula for a smooth map f : U −→ V , where U and V are 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 coordinate neighbourhoods). 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) > 0 for 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., (b ◦ f)(x) · det(Jac(f)) det(f)(x) = . a(x) Since a(x) and b(x) are positive, we have sign(det(f)) = sign(det(Jac(f))). Now we use the hypothesis that det(Jac(f)) 6= 0, i.e, f is a local diﬀeomorphism of a neighbourhood

8 U 0 ⊆ U of 0 to a neighbourhood V 0 ⊆ V of 0. The derived form β0 is compactly supported since supp(β0) ⊆ V 0 is closed and bounded in Rn. Since f : U 0 −→ V 0 is a diﬀeomorphism, we have > 0 if f is orientation preserving, sign(det(Jac(f))) = < 0 if f is orientation reversing. So we get Z Z deg(f) = f ∗(β0) = sign(det(Jac(f))) β0 Rn Rn = 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 connected k abelian subgroup. Every maximal torus is isomorphic to a torus k = R .A generator of T Zk n a torus T is an element t ∈ T such that the set {t : n ∈ Z>0} is dense in T . Generators are characterized by the following result.

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

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

k k k 0 −→ Z −→ R −→ T −→ 0.

k ∼ k ∼ k k k Since R = T0R = T0modZk T , the exponential map exp : T0modZk T −→ T can be identiﬁed with the canonical quotient map ρ : Rk −→ Tk. By properties of the exponential map (see Proposition 1.1.2), we have the following commutative square:

exp Rk / Tk

f∗ f 1 R exp / S

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

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

f∗| k f∗ f Z / / / 1 / 0 Z j R exp S 0 where the i and j are inclusions. It follows f∗(v1, . . . , vk) = α1v1 + ··· + αkvk, for some α1, . . . , αk ∈ Z. Now we prove that the following statements are equivalent:

(i)1 , v1, . . . , vk are linearly dependent over Q. Pk k (ii) i=1 qivi ∈ Q for some k-tuple 0 6= (q1, . . . , qk) ∈ Q .

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

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

(v) vmodZk is not a generator of Tk.

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

k 1 • (iii) =⇒ (iv): Let v = (v1, . . . , vk). Let f : T −→ S be any nontrivial Lie group homomorphism such that f∗(v1, . . . , vk) = α1v1 + ··· + αkvk. Then we have

k f(vmodZ ) = 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 nontrivial homomorphism. 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 some α1, . . . , αk ∈ Z.

• (iv) =⇒ (v): Let vmodZk ∈ Ker(f : Tk −→ S1), where f is a nontrivial Lie group homomorphism. Then Ker(f) $ Tk, i.e., Ker(f) is a closed proper Lie subgroup of Tk. We have k {(vmodZk)n : n > 0} ⊆ Ker(f) = Ker(f) $ T . Hence vmodZk is not a generator.

12 • (v) =⇒ (iv): If vmodZk is not a generator then vmodZk is contained in a proper closed subgroup H $ Tk. It follows that the quotient Tk/H is a nontrivial connected abelian 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, where S1 × · · · × S1 has m factors. Let ρ : Tk −→ Tk/H be the quotient projection. 1 1 Consider the composition f = pr1 ◦ ψ ◦ ϕ ◦ ρ, where pr1 : S1 × · · · × S −→ S is the 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 given by the composite map pr1 ◦ ψ ◦ ϕ ◦ ρ.

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

Proposition 2.1.1. If T and T 0 are two maximal tori in G, then W (T ) and W (T 0) are isomorphic.

Proof: Since T and T 0 are conjugate, there exists g ∈ G such that T 0 = gT g−1. Let ϕ : N(T 0) −→ N(T )/T be the map given by ϕ(x) = g−1xgT , for every x ∈ N(T 0).

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

(g−1xg)T (g−1xg)−1 = g−1x(gT g−1)x−1g = g−1xT 0x−1g = g−1T 0g = 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 0). In fact,

(gyg−1)T 0(gyg−1)−1 = gy(g−1T 0g)y−1g−1 = gyT y−1g−1 = gT g−1 = T 0.

Also, yT = g−1(gyg−1)gT = ϕ(gyg−1). (iii) Ker(ϕ) = T 0: Suppose ϕ(x) = eT . Then (g−1xg)T = eT , i.e., g−1xg ∈ T . Hence x = gtg−1 ∈ gT g−1 = T 0. Now suppose x ∈ T 0. We have ϕ(x) = g−1xgT = eT since g−1xg ∈ T . Therefore, Ker(ϕ) = T 0.

By the First Fundamental Theorem of Isomorphisms, we have N(T )/Ker(ϕ) ∼= Im(ϕ), i.e. N(T )/T ∼= N(T 0)/T 0.

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

i exp 0 / Zk / Rk / Tk / 0

ϕ∗| k ϕ∗ ϕ Z / k / k / k / 0 Z i R exp T 0

k k Lemma 2.1.1. Let Φ : Aut(T ) −→ Aut(Z ) be the map ϕ 7→ ϕ∗|Zk . Then Φ is an isomorphism.

Proof: It is clear that Φ is a group homomorphism. Let {e1, . . . , ek} denote the canonical basis 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) k = v1e1 + ··· + vkek, since each ej ∈ Z = (v1, . . . , vk).

k k So we get ϕ(vmodZ ) = ϕ ◦ exp(v1, . . . , vk) = exp(v1, . . . , vk) = vmodZ . Hence ϕ = Id.

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

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

Now deﬁne ϕ : Tk −→ Tk by

k ϕ((v1, . . . , vk)modZ ) = exp ◦Fe(v1, . . . , vk).

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

exp ◦Fe(v1, . . . , vk) = exp ◦Fe(v1e1 + ··· + vkek)

= exp(v1F (e1) + ··· + vkF (ek))

= exp(w1F (e1) + ··· + wkF (ek) + m1F (e1) + ··· + mkF (ek))

= exp(w1F (e1) + ··· + wkF (ek))

= exp(Fe(w1, . . . , wk)).

The map ϕ is surjective since exp and Fe are. Suppose ϕ(vmodZk) = 0modZk. Then exp(Fe(v)) = 0. Since exp is the canonical quotient map Rk −→ Rk/Zk, we have that Fe(v) ∈ Zk. So v ∈ Z, i.e., vmodZk = 0modZk. Hence ϕ is surjective.

Therefore, ϕ is an automorphism. Also, Φ(ϕ) = ϕ∗|Zk = Fe|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 prove N0 = 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 is ﬁnite, 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 ) on ∼ T as a continuous map f : N(T ) −→ Aut(T ) = Gl(k, Z). Since N0 is connected we have that f(N0) is connected in Gl(k, Z). On the other hand, Gl(k, Z) is a −1 discrete space, it follows f(N0) = {Id}. We get ntn = t, for every n ∈ N0 and every 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 are connected, we have that α(R) · T is connected in N0. Then α(R) · T is a connected abelian subgroup of G (i.e., a torus) containing T . By maximality of T , we have α(R) · T = T . Hence α(R) ⊆ T . Now, the image α(R) covers a neighbourhood of 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 follows that α( ) generates N , i.e., N = S α( )n, (see [5, Proposition 3.18]). Then, R 0 0 n∈N R 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. Let xN0 be a point of N(T )/N0. We know xN0 is open in N(T )/N0 is and only if −1 ρ (xN0) is open in N, where ρ : N(T ) −→ N(T )/N0 is the canonical quotient map. We have

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

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

2.2 Conjugacy Theorems

We need some more structure before proving the conjugacy theorems. Let H be a closed Lie subgroup of G. As a consequence of the Quotient Manifold Theorem (see [4, Theorem 9.16]), the quotient group G/H is a smooth manifold of dimension dim(G) − dim(H), and the canonical quotient map ρ : G −→ G/H is smooth. As we did in Theorem 1.2.1, we shall construct 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 space n−k ∗ TeH G/H. Deﬁne a map τ : G/H −→ Λ (G/H) by τ(gH) = Lg−1 (τe), where Lg−1 : −1 G/H −→ G/H is the map Lg−1 (xH) = g xH. We only proof that τ is well deﬁned. The rest of the proof (τ is smooth, nowhere vanishing and left invariant) is similar to the proof of Theorem 1.2.1. Suppose gH = g0H. Then g0 = gh for some h ∈ H. We ∗ ∗ ∗ ∗ ∗ ∗ have to show L(gh)−1 (τe) = Lg−1 (τe). Note that L(gh)−1 (τe) = Lh−1g−1 (τe) = Lg−1 Lh−1 (τe). ∗ ∗ ∗ ∗ Then L(gh)−1 (τe) = Lg−1 (τe) if and only if τe = Lh−1 (τe). Also, note that Lh−1 (τe) ∈ n−k ∗ ∗ Λ (TeH G/H), i.e., Lh−1 is an alternating (n − h)-linear functional on TeH G/H. If we follow the construction done in the proof of Theorem 1.2.1, we have that τe is the ∗ determinant function. From linear algebra we know Lh−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 group homomorphism from H to the multiplicative group R∗ = R − {0}. Moreover, λ is a ∗ continuous map since Lh−1 is smooth. Since H is connected and λ(e) = 1, we have that λ(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 closed interval since H is connected and compact (it is a closed subgroup of a compact group G). Since λ(h) > 1, the fact that λ(h)n ∈ [a, b] for every n ∈ N is a contradiction. It ∗ follows λ(h) = 1 for every h ∈ H. Hence Lh−1 (τe) = τe.

We shall denote the forms ω and τ on G and G/H by dg and d(gH), respectively. Let g be the Lie algebra if G. Let T be a maximal torus of G. Since T is a Lie subgroup of G, we have that t, the Lie algebra of T , is a Lie subalgebra of g. Consider the Killing form h·, ·iK on 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 h·, ·iK . Notice that g = t ⊕ m. We have the following result:

Proposition 2.2.1. The adjoint map restricted to the torus Ad|T acts trivially on every vector in t, and non-trivially on every nonzero vector in m. Moreover, for every t ∈ T , the map 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 ) = dAd(f t)(V ). Let f be a smooth function about e. Then dAd(f t)(V )(f) = V (f ◦ Ad(t)). Let α be an integral curve of V passing through e. We have

0 d dAd(f t)(V )(f) = α (0)(f ◦ Ad(f t)) = (f ◦ Ad(f t) ◦ α(s)) ds s=0

d −1 = (f(tα(s)t )) ds s=0

d = f(α), since t, α(s) ∈ T and T is abelian ds s=0 = V (f).

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

d V (f) = Ad(t)(V )(f) = dAd(f t)(V )(f) = (f ◦ Ad(f t) ◦ α(s)) ds s=0

d 0 = (f ◦ β(s)) = β (0)(f). ds s=0 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. Hence V ∈ 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. It is only left to show that Ad(t)(m) ⊆ m. Let X ∈ m and Y ∈ t. We have

−1 −1 −1 hY, Ad(t)(X)iK = Ad(t )(Y ), Ad(t )(Ad(t)(X)) K = Ad(t )(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 an action 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) and k = dim(T ). Since the canonical quotient map ρ : G −→ G/T is smooth, it induces a map ρ∗ : g = t⊕m −→ TeT G/T . It is known that ρ∗ maps m isomorphically onto TeT G/T . We shall identify m with TeT G/T via this isomorphism. Since T is a Lie group, we can ﬁnd a volume form dt which is left invariant. Then d(gT )dt is a nowhere vanishing left invariant form on G/T × T . Note that (d(gT )dt)(eT, e) and dg(e) are nonzero alternating n-linear functionals on g. We know that, up to multiplication by a constant, there is a unique alternating n-linear functional 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 the manifolds G/T × T and G with the orientations given by d(gT )dt and dg. Consider the map q : G/T × T −→ G given by (gT, t) 7→ gtg−1. Note that q is a well deﬁned smooth map. In order 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 a maximal 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, every element 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 is surjective. 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 compact Lie group G are conjugate.

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

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

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

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

−1 Lemma 2.2.2. sign(det(q)(gT, t)) = sign(det(AdG/T (t )−IdG/T )), where IdG/T is the identity 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))) ∗ = lq(gT,t)−1 dg(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

−1 −1 −1 −1 lq(gT,t)−1 ◦ q(gT, t) = L(gtg−1)−1 (gtg ) = (gtg ) · (gtg ) = 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)) −1 −1 = dg((lq(gT,t)−1 ◦ q)∗(l(gT,t)−1 )∗ W1(gT, t),..., (lq(gT,t)−1 ◦ q)∗(l(gT,t)−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 det(Jac(l −1 ◦ q ◦ l ) ) q∗dg = q(gT,t) (gT,t) ∗ · d(gT )dt. c By deﬁnition of det(q)(gT, t), we obtain

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

We compute lq(gT,t)−1 ◦ q ◦ l(gT,t):

−1 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 −1 −1 −1 −1 −1 = lgt−1g−1 (ghtsh g ) = gt g ghtsh g = gt−1htsh−1g−1.

On the other hand, Ad(f g)(Ad(f t−1)(h)sh−1) = Ad(f g)(t−1htsh−1) = gt−1htsh−1g−1. −1 −1 Hence lq(gT,t)−1 ◦ q ◦ l(gT,t)(hT, s) = Ad(f g)(Ad(f t )(h)sh ). Now consider the function f : G/T × T −→ G given by f(hT, s) = Ad(f t−1)(h)sh−1. So we get

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

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

= d(Ad(f g)) ◦ f∗ = Ad(g) ◦ f∗, 1 det(q)(gT, t) = det(Jac(l −1 ◦ q ◦ l ) ) c q(gT,t) (gT,t) ∗ 1 = det(Ad(g) ◦ f ) c ∗ 1 = det(Ad(g)) · det(f ). c ∗

20 Recall that Ad(g) is orthogonal with respect to the Killing form h·, ·iK . Then det(Ad(g)) = ±1. Since G is connected and det(Ad(e)) = 1, we get det(Ad(g)) = 1, 1 for every g ∈ G. Hence det(q)(gT, t) = c det(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∗|t 0 f∗ = . 0 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 about e. We have d d f∗|t(V )(g) = f∗ ◦ β∗ (g) = (f ◦ β)∗ (g) dr r=0 dr r=0

d d = (g ◦ f ◦ β(r)) = (g(f(eT, α(r)))) dr r=0 dr r=0

d −1 −1 d = (g(Ad(f t )(e)α(r)e )) = (g(α(r))) dr r=0 dr r=0 d = α∗ (g) dr r=0 = 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 be the maximal integral curve of V passing through e. Then β(r) = (α(r)T, e) is an integral curve of V . We have

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

d d = (g ◦ f ◦ β(r)) = g ◦ f(α(r)T, e) dr r=0 dr r=0

d −1 −1 = g(Ad(f t )(α(r))eα(r) ) dr r=0

d −1 = g(Ad(f t )(α(r))α(−r)) dr r=0 d = γ∗ (g), dr r=0 where γ(r) = Ad(f t−1)(α(r))·α(−r). Since G is a Lie group, the diﬀerential of the product is the sum of the diﬀerentials. Then γ∗ = dAd(f −t) ◦ α∗ − α∗.

21 We have d f∗|m(V )(g) = γ∗ (g) dr r=0 −1 d d = dAd(f t ) ◦ α∗ (g) − α∗ (g) dr r=0 dr r=0 −1 −1 = Ad(t )V (g) − V (g) = AdG/T (t )(V ) − IdG/T (V ) (g).

−1 Hence f∗|m = AdG/T (t ) − IdG/T . Finally, we get IdT 0 f∗ = −1 , 0 AdG/T (t ) − IdG/T

and then 1 det(q)(gT, t) = det(f ) c ∗ 1 = det(Ad (t−1) − Id ) · det(Id ) c G/T G/T T 1 = det(Ad (t−1) − Id ). c G/T G/T

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

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

(i) |q−1(t)| = |W (T )|;

−1 (ii) AdG/T (t ) 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−1T g. Since g−1tg ∈ T , we have {g−1tng : n > 0} ⊆ T . It follows T ⊆ gT g−1. Also, gT g−1 is a torus since it is a connected abelian Lie subgroup of G. So we get T = gT g−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 well deﬁned. We also show it is a bijection.

22 – ϕ is injective: Suppose ϕ(g1T, s1) = ϕ(g2T, s2). Then g1T = g2T and hence 0 0 g2 = g1t , for some t ∈ T . Since T is abelian, we have −1 −1 −1 0 0 −1 0 0 −1 −1 0 0 −1 s1 = g1 (g2s2g2 )g1 = g1 g1t s2(g1t ) g1 = t s2(t ) g1 g1 = t s2(t ) = 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 )|.

−1 (ii) Recall that AdG/T (t ) is an orthonormal linear transformation on m with respect −1 to h·, ·iK . Then the real eigenvalues of AdG/T (t ) (if any) must be ±1. By the 2 Kronecker’s Theorem t0 = t is also a generator of T . Replace t by t0 and suppose −1 −1 that AdG/T (t ) has a real eigenvalue λ = ±1. We have AdG/T (t )(V ) = λV , where V is an eigenvector associated to λ. Also, −1 −1 −1 −1 −1 AdG/T (t0 )(V ) = AdG/T (t t )(V ) = AdG/T (t ) · AdG/T (t )(V ) −1 2 = AdG/T (t )(λV ) = λ V = V. −1 −n Hence 1 is a real eigenvalue of AdG/T (t0 ). It follows AdG/T (t0 )(V ) = V , for every −n n > 0. Since t0 is a generator of T , we have T = {t0 : n > 0}. Consider the set −n EV = {s ∈ T : AdG/T (s)(V ) = V }. Note that EV is a closed set containing {t0 : −n n > 0}. It follows T = {t0 : n > 0} ⊆ EV and so AdG/T (s)(V ) = V , for every s ∈ T . On the other hand, Proposition 2.2.1 states that AdG/T (s) acts nontrivially on every nonzero vector of m, getting a contradiction since AdG/T (s)(V ) = V and −1 ∼ ∼ V 6= 0. Therefore, the automorphism AdG/T (t ): m = TeT G/T −→ m = TeT G/T −1 has no real eigenvalues. It follows the characteristic polynomial of AdG/T (t ) has even degree, since every polynomial of odd degree has at least one real root. Hence dim(G/T ) = dim(TeT G/T ) is even.

−1 −1 (iii) Let (gT, s) ∈ q (t). By part (ii), AdG/T (t ) − IdG/T has no real eigenvalues. Let −1 A = AdG/T (t ) − IdG/T . Suppose that A has negative determinant. Let n p(x) = det(xI − A) = x + ··· + a1x + a0 n be the characteristic polynomial of A. Note that a0 = (−1) det(A) = det(A), since n is even. Thus, p(0) < 0. On the other hand, limx→+∞ p(x) = +∞. So there exists a > 0 such that p(a) > 0. By the Intermediate Value Theorem, there exists b ∈ (0, a) such that p(b) = 0, i.e., b is a real eigenvalue of A, getting a −1 contradiction. Therefore, det(AdG/T (t ) − IdG/T ) > 0. By the previous lemma, we have det(q)(gT, s) > 0.

23 Proof of the Main Lemma. We know that q∗dg = det(q)(d(gT )dt). By the previous 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) = Pm i=1 sign(det(q)(giT, si)). By (iii), we have sign(det(q)(giT, si)) = 1. Hence

m X deg(q) = 1 = |W (T )|. i=1

Now we give some examples of maximal tori and Weyl groups. If you are interested in the details, 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 subgroup of diagonal matrices   z1 ··· 0  . .. .  1 D =  . . .  , zi ∈ S , 0 ··· zn and let S∆(n) = ∆(n) ∩ SU(n) be the subgroup of diagonal matrices as above with z1 ··· zn = 1. The groups ∆(n) and S(n) are maximal tori in U(n) and SU(n), respectively. 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 determinant in 1. The subgroup T (n) = SO(2) × · · · × SO(2) (n times) is a maximal torus 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) is Gn. Also, the Weyl group of SO(2n) is SGn, the subgroup of Gn consisting of even permutations. A −B (3) Now consider the symplectic group Sp(n) of unitary matrices of the form . B A A 0 We have a canonical inclusion U(n) −→ Sp(n) given by A 7→ . Let T (n) 0 A be the image of the torus ∆(n) under this inclusion. Then T (n) is a maximal torus in Sp(n). The Weyl group of Sp(n) is Gn.

24 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 torus chosen once and for all. As we saw above, any other choice leads to conjugate tori and isomorphic Weyl groups. Hence, the following result do not depend of the choice of a particular torus.

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 ∈ hT h−1. Notice that hT h−1 is a torus in G. If there is no torus T 0 such that hT h−1 ⊆ T 0, then hT h−1 is a maximal torus. Now if there exists a maximal torus T 0 such that hT h−1 ⊆ T 0, then g ∈ T 0. In any case, g is in a maximal torus in G. Let T be a maximal torus containing g. The exponential map exp : t −→ T is surjective since it can be identiﬁed with the canonical quotient map k k ∼ ρ : R −→ R/Z = T , where k = dim(T ). Since ρ is surjective, so is exp : t −→ T . Hence there 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 tori containing S.

(ii) Z(T ) = T .

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

Proof:

(i) Note that S is a connected abelian subgroup of G, i.e., a torus. Let g ∈ Z(S). Then gs = 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 that S = S is a torus. Let x ∈ Z(S) and let hx, Si be the subgroup generated by x and S. Let B = hx, Si. Then B is a compact abelian subgroup of G. Let B0 denote the

25 connected component of the identity of B. We have that B0 is a torus, since it is connected and abelian. Consider the quotient B/B0. We show that xB0 generates n k1 km B/B0. Consider yB0 ∈ B/B0. If y ∈ hx, Si then y = x s1 ··· sm , for some n, m ∈ Z. Since S is connected and the identity e is in S, we have S ⊆ B0 since B0 n is a component. Then, yB0 = x B0. It follows xB0 generates B/B0. As we did in the proof of Theorem 2.1.2, we have that B/B0 is ﬁnite. Then B/B0 is a ﬁnite cyclic ∼ ∼ group 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 only k if is isomorphic to T × Zl, for some k, l ∈ N. (See [2, Corollary 4.14]). Thus, B contains a dense cyclic subgroup {gn : n ∈ Z}. We get B = {gn : n ∈ Z}. Let T be a maximal torus containing g. Then gn ∈ T for every n ∈ Z and so B ⊆ T . Then x ∈ T and S ⊆ T . We obtain Z(S) ⊆ S{T : T is a maximal torus and T ⊇ S}. The other inclusion follows easily since T is abelian.

(ii) Z(T ) = S{T 0 : T 0 is a maximal torus and T 0 ⊇ T } = T , since T is maximal.

(iii) Suppose x is in the center of G. There exists a maximal torus T in G such that x ∈ T . Let T 0 be any other maximal torus in G. By the Second Conjugacy Theorem, there exists g ∈ G such that T 0 = gT g−1. Then x = gg−1x = gxg−1 ∈ T 0. Hence x is in any maximal torus of G, and Z(G) ⊆ T{T : T is a maximal torus in G}. Now suppose that x is in every maximal torus. Let g ∈ G. There exists a maximal torus T such that g ∈ T . Since x, g ∈ T and T is abelian, we get xg = gx. Then x ∈ 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 the group 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 homomorphism ρ is injective.

−1 Proof: Suppose ρ(xT ) = IdT . Then xtx = 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 )}.

26 Corollary 2.3.4. Two elements of the maximal torus are conjugate in G if and only if they lie 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 the

centralizers of x and y. Consider the map Adf g : G −→ G. Let h ∈ Z(x). We have

−1 −1 −1 −1 Adf g(h)y = (ghg )y = (gh)(g y) = (gh)(xg ) = g(hx)g = g(xh)g−1 = (gx)(hg−1) = (yg)(hg−1) = y(ghg−1)

= yAdf g(h).

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

0 T = Adf z(T ) = Adf z ◦ Adf g(T ) = Adf zg(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 some gT ∈ W (T ). Then y = gxg−1 and hence x and y are conjugate.

27 28 Bibliography

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

[2] Br¨ocker, T; Dieck, T. Representations of Compact Lie Groups. Springer-Verlag. New York (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.