on the non-existence of compact lorentzian manifolds with constant positive … · 2016-03-16 ·...
TRANSCRIPT
On the Non-Existence of Compact
Lorentzian Manifolds with
Constant Positive Curvature
David Lundberg
Master's thesis
2015:E45
Faculty of Science
Centre for Mathematical Sciences
Mathematics
CENTRUMSCIENTIARUMMATHEMATICARUM
Abstract
In this Master’s Thesis we study Semi-Riemannian Geometry. More specifically, this is done bydiscussing some well-known classification theorems from Riemannian geometry and then showingsome counter-examples to these in the semi-Riemannian setting. In particular, we briefly discussthe theorem by Hopf and Rinow, and give a counter-example to this in the semi-Riemanniansetting.
Moreover, we study a well-known and important result by Klingler, which asserts that a compactLorentzian manifold of constant sectional curvature must be geodesically complete. We also discussa theorem by Calabi and Markus, which together with the theorem by Klingler asserts that in theLorentzian setting there can be no compact manifolds of positive constant curvature.
Throughout this work it has been my firm intention to give reference to the stated results andcredit to the work of others. All theorems, propositions, lemmas and examples left unmarked areassumed to be too well known for a reference to be given.
”Perhaps I could best describe my experience of doing mathematics in terms of enter-ing a dark mansion. You go into the first room and it’s dark, completely dark. Youstumble around, bumping into the furniture. Gradually, you learn where each piece offurniture is. And finally, after six months or so, you find the light switch and turn iton. Suddenly, it’s all illuminated and you can see exactly where you were. Then youenter the next dark room... ”
- Andrew Wiles
Acknowledgments
I start by thanking Arne Meurman for a very useful discussion on the notion of the cohomo-logical dimension and for providing an important argument from homotopy theory.
Furthermore I thank Yacin Ameur for providing useful comments on the text, as well as somevery useful questions on the subject, that certainly helped me with improving the overall quality.
Finally I thank my supervisor Sigmundur Gudmundsson for carefully reading through the text,patiently answering my questions and encouraging me during the project. His enormous love forMathematics and his kindness made this project even more enjoyable.
David LundbergLund, March 15 2016
Contents
1 Introduction 11.1 Completeness and the Hopf-Rinow Theorem . . . . . . . . . . . . . . . . . . . . . . 11.2 Historical background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
2 Semi-Riemannian Manifolds 52.1 Indefinite tangent spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Semi-Riemannian manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.3 Orbit manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3 Homogeneous spaces and (G,X)-manifolds 233.1 Homogeneous spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.2 Isometry groups and their properties . . . . . . . . . . . . . . . . . . . . . . . . . . 263.3 (G,X)-structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
4 Curvature in Lorentzian spaces 394.1 The Lorentzian Gauss-Bonnet Theorem . . . . . . . . . . . . . . . . . . . . . . . . 394.2 The Calabi-Markus Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464.3 The Klingler Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
A Covering spaces 63A.1 Homotopy theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63A.2 Covering maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
Bibliography 73
Chapter 1
Introduction
In this work we set out to study the subject of semi-Riemannian Geometry. This is different fromthe usual Riemannian Geometry in that the metric in this setting need not be positive definite.Moreover, many well-known classification theorems do no longer hold. To study these examples,we build up necessary pre-requisites and then give some examples or results illustrating this.
The text is written assuming that the reader has a good understanding of introductory Rie-mannian geometry, corresponding to the level e.g. in [15]. Furthermore a good understanding ofbasic group theory is assumed.
A review article, relevant to the current work, can be found in [21].
1.1 Completeness and the Hopf-Rinow Theorem
Let us start by reminding the reader of the notion of (geodesic) completeness.
Definition 1.1. Let (M, g) be a Riemannian manifold. A geodesic γ : [0, b) → M is said to becomplete if it can be extended to all t ∈ R. A Riemannian manifold (M, g) is called complete ifevery geodesic is complete.
We introduce some examples concerning completeness and incompleteness.
Example 1.2. As is well known, the geodesics in Em (Rm equipped with the standard Euclideanmetric) are the straight lines of the form γ(t) = p+ t · v. As this obviously is defined for all t ∈ Rthis makes Em into a (geodesically) complete Riemannian manifold.
Example 1.3. As a less trivial example, we consider the upper half-plane equipped with theLobachevsky metric. It is the set
R2+ =
(x, y) ∈ R2
∣∣ y > 0
(1.1)
equipped with the Riemannian metric
g(X,Y ) =1
y2〈X,Y 〉R2 . (1.2)
We now solve for the unique geodesic γ with γ(0) = p = (0, 1) and γ(0) = (0,−1). We begin bysimply stating the Christoffel symbols in the given metric. They are
Γ111 = 0, Γ1
12 = Γ121 = −1
y, Γ1
22 = 0,
Γ211 =
1
y, Γ2
21 = Γ221 = 0, Γ2
22 = −1
y.
These we use to solve the geodesic equation, which we state for recollection:
d2γk(t)
dt2+∑
i,j=1,2
γi(t)γj(t)Γkij(γ(t)) = 0. (1.3)
1
We assume that γ1(t) ≡ 0, and so we solve for γ2(t) only. We have
0 =d2γ2(t)
dt2+∑
i,j=1,2
γi(t)γj(t)Γ2ij(γ(t))
=d2γ2(t)
dt2+ γ2(t)γ2(t)Γ2
22(γ(t))
= γ2(t) + (− γ2(t)2
γ2(t)).
(1.4)
Given initial conditions on γ and γ, the general solution is
γ2(t) = exp(−t) γ(t) = (0, exp(−t)) (1.5)
and so we see that the geodesic is defined for all t.It can be shown, (see [8], p. 175), that the isometry group of the hyperbolic space is the group
GL2(R) =
(a bc d
) ∣∣∣∣ a, b, c, d ∈ R and ad− bc 6= 0
. (1.6)
For any two points p and q and any vector v ∈ TpR2+ there is an element f ∈ GL2(R) and a vector
u ∈ TqR2+ such that f(q) = p and df(u) = v. Hence any geodesic may be viewed as the image of a
geodesic on the above form and so must be complete.
Example 1.4. To show an example of incompleteness, we just take the punctured plane R2−0.The geodesic γ(t) = (1, 0) − t(1, 0) is defined on the interval (−∞, 1) but cannot be extended.Hence the punctured plane is not complete.
Having discussed some examples of geodesic completeness, we are now ready to state an im-portant classification result, regarding this property. The result was initially proven in [16].
Theorem 1.5. ([8], p. 146, ”The Hopf-Rinow Theorem”)Let (M, g) be a Riemannian manifold and p ∈M . Then the following assertions are equivalent:
1. M is complete as a metric space (or topologically complete).
2. M is geodesically complete.
3. For any q ∈M there exists a geodesic γ joining p and q with `(γ) = d(p, q).
We get the following important corollaries.
Corollary 1.6. ([8], p. 149) If M is compact, then M is (geodesically) complete.
Proof. By the compactness we have metric completeness by standard topological results. ByTheorem 1.5 this is equivalent to (geodesic) completeness.
Corollary 1.7. ([8], p. 149) Closed submanifolds of complete manifolds are complete in theinduced metric.
We will show, in Example 2.40, that there are compact semi-Riemannian manifolds equippedwith metrics that are not geodesically complete.
1.2 Historical background
The study of semi-Riemannian geometry dates far back and much of its interest has stemmed fromPhysics, more specifically the General Theory of Relativity. This theory is mathematically foundedon the sub-branch called Lorentzian Geometry - a notion that will be made precise in Chapter 2.
2
As is intuitively clear from the classical Gaussian geometry, it can be shown that the (geodesi-cally) complete Riemannian manifolds of constant positive curvature must be compact. The sim-plest example of this is surely the standard sphere Sn as a subset of the familiar Euclidean spaceEn+1 = (Rn+1, δ = diag(1, . . . , 1)). In the Riemannian setting, these spaces will be on the form
Mn ' Sn/Γ, n > 2, (1.7)
where Γ is a special type of group acting on Sn. The exact meaning of Γ will be made precise inSection 2.36 of Chapter 2.
In the Lorentzian setting, there is an analogous classification. The Lorentzian (geodesically)complete manifolds of poisitive constant curvature are on the form
Mn ' dSn/Γ, (1.8)
where dSn is the so-called de-Sitter space, which will be defined in Chapter 2, Section 2.3.However, the striking result of Theorem 4.19 by Calabi and Markus, that was proven 1962, showsthat the order of Γ in this case is finite. As an immediate consequence of this, (which is stated asCorollary 4.50), there can be no compact (geodesically) complete Lorentzian manifolds of constantpositive curvature for dimension n > 2.
The situation for dimension n = 2 is relatively straight-forward to discuss. In Section 4.1 wewill prove the Lorentzian analogue of the well-known Gauss-Bonnet theorem. The result in theLorentzian setting will have a slightly different form and the class of surfaces of interest for ourdiscussion is narrowed down quite severely - they are two-dimensional tori. The theorem will thenbe applied in the standard way: disjoining the surface M into components and then applyingthe Gauss-Bonnet formula to each one individually. The boundary integrals cancel out and whatremains is the integral ∫
M
κ = 0 (1.9)
from which we immediately get the remarkable result that compact Lorentzian surfaces must beflat. In particular, they cannot have constant positive curvature.
The question then remains, whether a compact Lorentzian manifold with constant positivecurvature can exist. From Corollary 4.50 it could not be complete. This question was open untilthe early 90s, where it was finally answered by Bruno Klingler in Theorem 4.21, which asserts thata compact Lorentzian manifold of constant curvature is (geodesically) complete. This theoremshows the assertion in the three relevant curvature cases, that is κ = +1, κ = 0 and κ = −1.
It was previously known that completeness would follow in the flat case, i.e. the case κ = 0.This was proven by Yves Carriere in 1989. Moreover, it was shown (with different methods) in thespherical case (κ = +1) to be true as well by Morrill 1996.
The work [17] by Klingler is very involved, and it is certainly the bulk part of the work that isthis thesis. It relies on the so-called (G,X)-structures, that were introduced by William Goldmanin [11] and [13] and many of the concepts in the proof are generalizations of some technical toolsused in [6]. The theorem is stated and proven in Section 4.3.
3
4
Chapter 2
Semi-Riemannian Manifolds
In the following we introduce the formalism used for manifolds with an indefinite metric. Suchmanifolds are called semi-Riemannian (pseudo-Riemannian). We prepare this by discussing someproperties of the tangent space TpM .
2.1 Indefinite tangent spaces
We now introduce the necessary formalism needed to understand the tangent space TpM of a semi-Riemannian manifold. Indeed, as has been mentioned, the topological structure is quite different.We follow the discussion in [20].
Definition 2.1. Let V be a real vector space. A symmetric bilinear form b is a mappingb : V × V → R such that b(v, w) = b(w, v) (symmetry) and b is linear in both its arguments. Theform is called non-degenerate if b(v, w) = 0 for all v implies w = 0.
Definition 2.2. Let V be a real vector space. A scalar product g is a non-degenerate symmetricbilinear form on V . If g is positive definite, it is called an inner product.
Furthermore a scalar product induces a quadratic form by q(v) = g(v, v). This can be negativeon some non-zero v, and so the span Rv generated by this vector is a negative definite vectorsubspace. This motivates the following definition.
Definition 2.3. Let V be a real vector space and g be a scalar product on V . The index ν of gis the maximal dimension of a vector subspace W in V such that the restriction g
∣∣W is negativedefinite. If ν = 1 the vector space is said to be Lorentzian.
Example 2.4. Let R2 be equipped with the metric g = diag(−1, 1). The set of vectors v = (v1, v2)such that v2
1 > v22 will span negative definite vector subspaces. Hence the index is at least ν ≥ 1.
But as the space is two-dimensional, and for instance the vector w = (0, 1) satisfies g(w,w) = 1 > 0,the index ν can be at most 1. Hence it is Lorentzian. Furthermore the subspaces R(1, 1) andR(1,−1) are degenerate as subspaces as the basis vectors are orthogonal to themselves.
Definition 2.5. Let V be a real vector space, equipped with a scalar product g. A vector v in Vis said to be
1. timelike if g(v, v) < 0,
2. spacelike if g(v, v) > 0 and
3. null if g(v, v) = 0.
We prove the analogue to the dimension theorem in linear algebra. First we need to specifywhat the orthogonal complement means.
5
Definition 2.6. Let V be a real vector space, equipped with a scalar product g, and let W be avector subspace. The set
W⊥ =v ∈ V
∣∣ g(v, w) = 0, w ∈W
(2.1)
is called the orthogonal complement of W .
The only non-trivial difference to the Euclidean case is that as although we assume the vectorspace to be non-degenerate, it can still have degenerate vector subspaces, as seen above. Thisallows for W = W⊥ to happen. The following two lemmas will still hold, though.
Lemma 2.7. ([20], p. 49) Let V be a real vector space, equipped with a scalar product g and W avector subspace. Then
1. dim W + dim W⊥ = dim V = n,
2. (W⊥)⊥ = W .
Proof. ([20], p. 49) Let e1, . . . , en be a basis for V such that e1, . . . , ek is a basis for W.We first show (1.). The elements w ∈ W⊥ are precisely those that satisfy g(w, ei) = 0 for
i = 1, . . . , k. Let w be expressed in terms of the basis vectors as w = w1e1 + · · ·+ wnen. In termsof the components of g, this yields the following equations:
0 = g(w, ei) =n∑j=1
gijwj , i = 1, . . . , k. (2.2)
Suppose the rows of the matrix g are linearly independent. Then we know from linear algebra thatthe equation system is invertible and solutions exist. It follows also that the space of solutionsmust be (n − k)-dimensional. Hence the assertion follows this from assumption. But the linearindependence follows naturally for any non-degenerate bilinear form: g(v, w) = 0 for all w if andonly if g(v, ei) = 0 for all ei. Hence
0 = g(v, ei) =
n∑j
g(vjej , ei) =
n∑j
vjbij , i = 1, . . . , n. (2.3)
This means that for some non-zero vj the system is singular and hence not invertible. Therefore itis equivalent that the bilinear form is non-degenerate and that the coefficient matrix is invertible.This proves (1.).
Obviously W ⊆ (W⊥)⊥. But from the first assertion, it follows that they must have the samedimension. Hence we have equality. This shows (2.).
Lemma 2.8. ([20], p. 49) A vector subspace W of V is non-degenerate if and only if V is thedirect sum V = W ⊕W⊥.
Proof. Regardless of the scalar product, one has for finite dimensional vector spaces that
dim(W ⊕W⊥) + dim(W ∩W⊥) = dim W + dim W⊥. (2.4)
If W is non-degenerate, then by the Lemma 2.7 it must hold that dim W + dim W⊥ = dim Vand so V is the direct sum. If V is the direct sum, the left hand side of the above implies
dim(W ∩W⊥) = 0, (2.5)
which is equivalent to W being non-degenerate.
Definition 2.9. A vector space V with a scalar product g has a norm defined through
||v|| =√|g(v, v)|. (2.6)
Definition 2.10. Let V be a Lorentzian vector space and z ∈ V . The orthogonal complementz⊥ to z is the orthogonal complement of the vector subspace Rz, as in Definition 2.6.
6
Lemma 2.11. ([20], p. 141) Let V be a Lorentzian vector space and suppose that z is time-like.Then
1. z⊥ is spacelike,
2. V = Rz ⊕ z⊥.
Proof. ([20], p. 141) As Rz is non-degenerate, it follows from Lemma 2.8 that, as Rz and z⊥
intersects trivially, and so V is the direct sum. This shows V = Rz ⊕ z⊥.As the index of Rz is 1, and V is Lorentz, z⊥ cannot have a positive index. If so, there would
be a time-like vector w ∈ z⊥, linearly independent of z and the vector subspace W = spanz, wwould be strictly time-like and have dimension 2.
Lemma 2.12. ([20], p. 52) Two scalar product spaces V and W have the same dimension andindex if and only if there exists a linear isometry between them.
Proof. See ([20], p. 52).
2.2 Semi-Riemannian manifolds
We are now equipped with some understanding of indefinite vector spaces. Hence we can constructthe manifolds of indefinite metrics.
Definition 2.13. Let M be a smooth topological manifold. A metric g is a non-degeneratesymmetric (0,2)-tensor field, of constant index ν, defined on M .
The metric assigns to each point p a non-degenerate scalar product to the tangent space TpM .The following conventions are necessary.
Definition 2.14. Let M be a smooth topological manifold equipped with a metric g. Then M iscalled a semi-Riemannian manifold. In the case of ν = 0, M is naturally called Riemannian.In the case of ν = 1 and dim M= n ≥ 2, M is called a Lorentzian manifold.
Theorem 2.15. ([20], p. 61, ”Levi-Civita”) Let (M, g) be a semi-Riemannian manifold. Thenthere exists a unique metric and torsion-free connection ∇.
Proof. See ([20], p. 61).
Definition 2.16. A complete and connected semi-Riemannian manifold of constant (sectional)curvature is called a space form. If, furthermore, the manifold is Lorentzian, it is said to be arelativistic space form. Space forms of positive constant curvature κ > 0 are called spherical,those of negative constant curvature κ < 0 are called hyperbolic and those with κ = 0 are calledflat.
Lorentz manifolds are of particular interest in semi-Riemannian geometry. Hence we studysome examples of Lorentz manifolds. The three Lorentzian manifolds defined in the following areall space forms.
Definition 2.17. Let Rn be equipped with the metric g = diag(−1, 1, . . . , 1). By convention, thismetric is denoted by g = η. The space (Rn, η) is called the Minkowski space and is denoted byMn.
Definition 2.18. Consider the (n + 1)-dimensional Minkowski space Mn+1 and the quadraticform Qn+1
1 (x, y), defined by the equation
Qn+11 (x, y) = −x1y1 + x2y2 + · · ·+ xn+1yn+1, (2.7)
for two points x = (x1, . . . , xn+1) and y = (y1, . . . , yn+1) in Mn+1. The subset defined throughQn+1
1 (x, x) = 1, that is the set of points x ∈Mn+1 that satisfies
1 + x21 = x2
2 + · · ·+ x2n+1, (2.8)
gives a semi-Riemannian submanifold equipped with the induced metric. This submanifold is calledthe de-Sitter space, or the dS-space for short.
7
Definition 2.19. Consider the set Rn+1 and the quadratic form
Qn+1−1 (x, y) = +x1y1 + x2y2 − (x3y3 + . . .+ xnyn + xn+1yn+1). (2.9)
The submanifold given by Qn+1−1 (x, x) = 1, i.e. the set of points x ∈ Rn+1 satisfying the equation
x21 + x2
2 = 1 + x23 + · · ·+ x2
n+1, (2.10)
combined with the induced metric from (Rn+1, g), where g = diag(−1,−1, 1, . . . , 1), is called theanti de-Sitter space, or the AdS-space for short.
Proposition 2.20. The Minkowski Mn space has the following properties:
1. Mn is simply connected,
2. Mn is complete and
3. Mn has constant curvature κ = 0.
Proof. Since Mn is convex Example A.10 shows that the simple connectivity follows. This shows(1.).
Since η = diag(−1, 1, . . . , 1) is constant, the Christoffel symbols vanish. This means both that
the sectional curvature vanishes and that the geodesic equation reduces to d2γdγ2 = 0. The solutions
to this differential equation are the straight lines, which are defined for all parameter values. Thisshows both (2.) and (3.).
Proposition 2.21. For n ≥ 2 consider dSn. Then
1. dSn is diffeomorphic to R× Sn−1,
2. dSn is simply connected for n ≥ 3,
3. dSn has constant curvature κ = +1.
Proof. We denote the coordinates (x1, . . . , xn+1) of the ambient Minkowski space by (t,−→x ).It is easily seen, that the mapping
ψ(t,−→x ) = (t,−→x√
1 + t2) (2.11)
is a diffeomorphism onto the space R× Sn−1. This shows (1.).For the first fundamental group, we have π1(dSn) ' π1(R× Sn−1) ' π1(R)× π1(Sn−1) ' e,
where we used Propositions A.14, A.13 and A.25. By the diffeomorphism we just showed, dSn ispath connected and the fundamental group is trivial. Hence dSn is simply connected.
For (3.), we show the case to be true in the n = 2 case first. Consider the parametrization
σ(t, φ) =(t,√
1 + t2 cos(φ),√
1 + t2 sin(φ)). (2.12)
The pull-back metric via σ satisfies
g
(∂
∂t,∂
∂t
)(t, φ) = − 1
1 + t2, g
(∂
∂φ,∂
∂φ
)(t, φ) = (1 + t2), g
(∂
∂φ,∂
∂t
)(t, φ) = 0. (2.13)
The corresponding Christoffel symbols are
Γttt = − t
1 + t2, Γttφ = Γtφt = 0, Γtφφ = t(1 + t2),
Γφφφ = 0, Γφφt = Γφtφ =t
1 + t2, Γφtt = 0.
8
The Gaussian curvature κ becomes, (after using Proposition 8.11 in [15]),
κ =
g
(R( ∂
∂φ ,∂∂t )
∂∂t ,
∂∂φ
)gφφgtt − g2
tφ
=1
gφφgttgφφ
(∂Γφtt∂φ−∂Γφφt∂t
+ (ΓφttΓφφφ − ΓφφtΓ
φtφ) + (ΓtttΓ
φtφ − ΓtφtΓ
φtt)
)= −(1 + t2)
(0− 1− t2
(1 + t2)2+ (0− t2
(1 + t2)2) + (− t2
(1 + t2)2− 0)
)= 1.
(2.14)
To prove the general case, we observe that if m < n, the mapping
ρ : dSn → dSn, by ρ(x, y) = (x,−y),
where x = (x1, . . . , xm+1) and y = (xm+2, . . . , xn+1),(2.15)
is an isometry with dSm as fix-point set. Hence dSm ⊂ dSn is totally geodesic (by Proposition7.25 in [15], modified to the semi-Riemannian setting). The metric η restricted to y is space-likeand so the Gauss-equation implies the curvature must be the same. Thus, the statement followsby induction.
Proposition 2.22. Let n ≥ 2 and consider dSn. Then
1. If, for two points x and y of dSn, Qn+11 (x, y) = 1, then there is a null geodesic in dSn, i.e.
a straight line in Mn+1, connecting x and y.
2. If, for two points x and y of dSn, −1 < Qn+11 (x, y) < 1, then there is a space-like geodesic
ellipse in dSn, connecting x and y.
3. If, for two non-antipodal points x and y of dSn, we have Qn+11 (x, y) > 1, there is a time-like
geodesic hyperbola connecting x and y.
4. If, for two non-antipodal points x and y of dSn, Qn+11 (x, y) ≤ −1, then there is no geodesic
in dSn connecting x and y.
5. The de-Sitter space is complete.
Proof. We denote the coordinates (x1, . . . , xn+1) of the ambient Minkowski space by (t,−→x ).We now discuss (1.). We have, as both x and y are in dSn, that Qn+1
1 (x, x) = Qn+11 (y, y) = 1.
If we can show that the line `(s) = x + s(y − x) is both contained in dSn and is null seen inMn+1, we are done, as the line then would be a null geodesic. As the space-like restriction of themetric has a rotational symmetry, we may assume that x = (u,
√1 + u2, 0, . . . , 0). Hence, we have
Qn+11 (y, y) = −y2
1 + y22 + · · · + y2
n+1 = 1 and Qn+11 (x, y) = −uy1 +
√1 + u2y2 = 1. The following
9
shows that the line is contained in dSn:
Qn+11 (`, `)(s) = −
(u+ s(y1 − u)
)2
+
(√1 + u2 + s(y2 −
√1 + u2)
)2
+ · · ·+(syn
)2
+
(syn+1
)2
= −(u2 + 2us(y1 − u) + s2(y1 − u)2
)+
((1 + u2) + 2
√1 + u2s(y2 −
√1 + u2) + s2(y2 −
√1 + u2)2
)+ · · ·+ s2y2
n + s2y2n+1
= −(u2 + 2usy1 − 2u2s+ s2y2
1 − 2s2y1u+ s2u2
)+
((1 + u2) + 2
√1 + u2sy2 − 2s(1 + u2)
+ s2y22 − 2s2y2
√1 + u2 + s2(1 + u2)
)+ · · ·+ s2y2
n + s2y2n+1
= 2s
(−uy1 +
√1 + u2︸ ︷︷ ︸
=1
)− 2s2(−uy1 +
√1 + u2︸ ︷︷ ︸
=1
)
+ s2
(−y2
1 + y22 + · · ·+ y2
n + y2n+1︸ ︷︷ ︸
=1
)
+
(− u2(1− s)2 + (1 + u2)(1− s)2
)= 2s− 2s2 + s2 + (1− s)2
= 1.
(2.16)
Furthermore, the tangent vector of ` is null:
η(x− y, x− y) = −(u− y1)2 + (√
1 + u2 − y2)2 + · · ·+ y2n + y2
n+1
= −(u2 − 2uy1 + y2
1
)+
((1 + u2)− 2
√1 + u2y2 + y2
2
)+ · · ·+ y2
n + y2n+1
= 1 +
(−y2
1 + y22 + · · ·+ y2
n + y2n+1︸ ︷︷ ︸
=1
)+ 2
(uy1 −
√1 + u2y2︸ ︷︷ ︸
=−1
)= 0.
(2.17)
This shows that the curve ` is a geodesic in Mn+1. The definition of the induced Levi-Civitaconnection to dSn guarantees that it is also a geodesic in the submanifold: ∇γ γ = (∇γ γ)> = 0.
Consider now the case −1 < Qn+11 (x, y) < 1. For the sake of brevity, we let Qn+1
1 (x, y) ≡Q(x, y). We consider the plane Π that contains the origin, x and y, i.e. Π = spanx, y. We findtwo orthogonormal elements e1 and e2 in Π, so as to parametrize the intersection. We claim thatthe curve
e(t) = sin(t)e1 + cos(t)e2
= sin(t)x+ cos(t)
(Q(x, y)√
1−Q2(x, y)x− 1√
1−Q2(x, y)y
).
(2.18)
is the desired parametrization. Clearly e(t) ∈ Π as Π = spanx, y. The following shows that the
10
basis vectors e1 and e2 are orthonormal. Clearly Q(e1, e1) = Q(x, x) = 1 and we have
Q(e1, e2) = Q
(x,
Q(x, y)√1−Q2(x, y)
x− 1√1−Q2(x, y)
y
)= Q(x, x)
Q(x, y)√1−Q2(x, y)
−Q(x, y)Q(y, y)√
1−Q2(x, y)
= 0.
(2.19)
Furthermore, there is
Q(e2, e2) = Q
(Q(x, y)√
1−Q2(x, y)x− 1√
1−Q2(x, y)y,
Q(x, y)√1−Q2(x, y)
x− 1√1−Q2(x, y)
y
)= Q(x, x)
Q2(x, y)
1−Q2(x, y)− 2Q(x, y)
Q(x, y)
1−Q2(x, y)+Q(y, y)
1
1−Q2(x, y)
=1 +Q2(x, y)
1−Q2(x, y)
= 1
(2.20)
so that e2 is a spacelike vector of unit norm. With this we have
Q(e(t), e(t)) = sin(t)e1 + cos(t)e2
= Q(e1, e1) sin2(t) + 2Q(e1, e2) sin(t) cos(t) +Q(e2, e2) cos2(t)
= 1
(2.21)
so that e(t) ∈ dSn.We only need to show that e(t) is a geodesic. To do this, we show e(t) ∈ Ne(t)dSn. This follows,
as for any e(t), we have that e(t) = −e(t). We choose coordinates so that, for some fixed t, e(t) ison the form e(t) = (s,
√1 + s2, 0, . . . , 0). We compute the derivative in the s-direction:
de(t)
ds= (1,
s√1 + s2
, 0, . . . , 0), (2.22)
from which it follows that
η(e(t),de(t)
ds) = −(1 · s) + (
√1 + s2 · s√
1 + s2) = 0. (2.23)
Hence e(t) ∈ Ne(t)dSn and so e is a geodesic. Since the parametrization σ is global, the geodesicconnecting x and y must be unique. This proves (2.).
We again consider the plane Π spanned by x and y and denote Qn+1−1 (x, y) ≡ Q(x, y). We claim
that the curve
h(t) = cosh(t)e1 + sinh(t)e2
= cosh(t)y + sinh(t)
(1√
Q2(x, y)− 1x− Q(x, y)√
Q2(x, y)− 1y
)(2.24)
is the geodesic that satisfies the claimed properties. Obviously h(t) ∈ Π. We have h(0) = y andfor some tx we have sinh(tx) =
√Q2(x, y)− 1, so that cosh(tx) = Q(x, y) and therefore h(tx) = x.
The vectors e1 and e2 form an orthonormal basis for Π:
Q(e1, e2) = Q
(y,
1√Q2(x, y)− 1
x− Q(x, y)√Q2(x, y)− 1
y
)= Q(y, x)
1√Q2(x, y)− 1
−Q(y, y)Q(x, y)√Q2(x, y)− 1
= 0
(2.25)
11
and
Q(e2, e2) = Q
(1√
Q2(x, y)− 1x− Q(x, y)√
Q2(x, y)− 1y,
1√Q2(x, y)− 1
x− Q(x, y)√Q2(x, y)− 1
y
)= Q(x, x)
1
Q2(x, y)− 1− 2Q(x, y)
Q2(x, y)
Q2(x, y)− 1+Q(y, y)
Q(x, y)2
Q2(x, y)− 1
=1−Q2(x, y)
Q2(x, y)− 1
= −1
(2.26)
so that e2 is a timelike unit vector orthogonal to e1.We can now show that h(t) ∈ dSn. We have
Q(h(t), h(t)) = Q
(cosh(t)e1 + sinh(t)e2, cosh(t)e1 + sinh(t)e2
)= Q(e1, e1) cosh2(t) + 2Q(e1, e2) sinh(t) cosh(t) +Q(e2, e2) sinh2(t)
= cosh2(t)− sinh2(t)
= 1
(2.27)
so that h(t) is a curve in dSn ∩Π.Furthermore, we have h(t) = h(t) and so by the calculation for the ellipse we already know
that such a curve is a geodesic. This shows (3.).For a geodesic the acceleration vector must be in the normal space. When we investigated the
properties of the elliptic geodesic, we showed that NxdSn = Rx. It follows that this happens for
every point along a geodesic if and only it is the intersection dSn ∩ Π for some plane Π (exceptin the case of the geodesic being a straight line, which we have already dealt with). Hence allgeodesics must be on the forms we have covered. This means on the one hand that for two (non-antipodal) points x and y on some geodesic, we must have Qn+1
1 (x, y) > −1 and on the other thatevery geodesic is complete. This shows (4.) and (5.).
Proposition 2.23. Let n ≥ 2 and consider AdSn. Then
1. AdSn is diffeomorphic to Rn−1 × S1,
2. AdSn is not simply connected,
3. AdSn has constant curvature κ = −1.
Proof. As in the proof of Proposition 2.21, we denote the coordinates (x1, x2, . . . , xn+1) = (t1, t2,−→x ).
One easily sees that the mapping
ψ(t1, t2,−→x ) =
( t1√1 +−→x · −→x
,t2√
1 +−→x · −→x,−→x)
(2.28)
is a diffeomorphism onto S1 × Rn−1.We have π1(AdSn) ' π1(Rn−1×S1) ' π1(Rn−1)×π1(S1) ' 〈Z,+〉, where we used Propositions
A.14, A.13 and A.25. As the fundamental group is not trivial, AdSn is not simply connected.We proceed as in the proof of (3.) in Proposition 2.21 and first work out the case of n = 2.
The space has the global parametrization
σ(s, φ) = (cos(φ)√
1 + s2, sin(φ)√
1 + s2, s). (2.29)
The metric, calculated through pull-back is
g
(∂
∂φ,∂
∂φ
)(s, φ) = −(1 + s2), g
(∂
∂s,∂
∂φ
)(s, φ) = 0,
g
(∂
∂s,∂
∂s
)(s, φ) =
1
1 + s2.
(2.30)
12
We get the Christoffel symbols
Γsss =s
1 + s2, Γssφ = Γsφs = 0, Γsφφ = s(1 + s2),
Γφφφ = 0, Γφφs = Γφsφ = − s
1 + s2, Γφss = 0.
The Gaussian curvature κ becomes, (after using Proposition 8.11 in [15]),
κ =
g
(R( ∂∂s ,
∂∂φ ) ∂
∂φ ,∂∂s
)gφφgss − g2
sφ
=1
gφφgssgφφ
(∂Γsφφ∂s−∂Γsφs∂φ
+ (ΓsφφΓsss − ΓsφsΓssφ) + (ΓφφφΓssφ − ΓφφsΓ
sφφ)
)=
1
1 + s2
(− (1 + 3s2) + 0 + (s2 − 0) + (0− (−s2))
)=
1
1 + s2
(− (1 + s2)
)= −1.
(2.31)
To show for the general case, we observe that if m < n, the mapping
ρ : AdSn → AdSn, by ρ(x, y) = (x,−y), (2.32)
is an isometry with AdSm as fix-point set. Hence AdSm ⊂ AdSn is totally geodesic. The metricη restricted to y is space-like and so the Gauss-equation implies the curvature must be the same.Thus, the statement follows by induction.
Proposition 2.24. Let n ≥ 2 and consider AdSn. Then
1. if, for two non-antipodal points x and y of AdSn, Qn+1−1 (x, y) = 1, then there is a null geodesic
in AdSn, i.e. a straight line in Mn+1, connecting x and y.
2. If, for two non-antipodal points x and y of AdSn, −1 < Qn+1−1 (x, y) < 1, then there is a
timelike geodesic ellipse in AdSn, connecting x and y.
3. If, for two non-antipodal points x and y of AdSn, we have Qn+1−1 (x, y) > 1, there is a spacelike
geodesic hyperbola connecting x and y.
4. If, for two non-antipodal points x and y of AdSn, Qn+1−1 (x, y) ≤ −1, then there is no geodesic
in AdSn connecting x and y.
5. The anti de-Sitter space is complete.
Proof. We do the analogous calculations to those we did in the dSn case. For the sake of brevity,we write Qn+1
−1 (x, y) = Q(x, y).For two non-antipodal points x and y that satisfies Q(x, y) = 1 we claim that the curve
`(t) = x + t(y − x) is a null geodesic. From the O(2)-symmetry of the timelike part and theO(n − 1)-symmetry of the spacelike part, we may assume that x = (0,
√1 + u2, u, 0, . . . , 0). We
denote y = (y1, y2, . . . , yn+1). The condition Q(x, y) = 1 gives√
1 + u2y2−uy3 = 1. The following
13
shows that the curve ` is contained in AdSn:
Q(`, `) = t2y21 +
(√1 + u2 + t(y2 −
√1 + u2)
)2
−(u+ t(y3 − u)
)2
− t2y24 − . . .− t2y2
n+1
= t2y21 +
((1 + u2)(1− t)2 + 2
√1 + u2(1− t)ty2 + t2y2
2
)−(u2(1− t)2 + 2u(1− t)ty3 + t2y2
3
)− . . .− t2y2
n+1
= t2(y21 + y2
2 − y23 − . . .− y2
n+1) + (1 + u2)(1− t)2 − u2(1− t)2
+ 2(1− t)t(√
1 + u2y2 − uy3
)= t2 + (1− t)2 + 2(1− t)t= 1.
(2.33)
Furthermore the tangent vector ˙ = x − y is null, as is shown in the following. With theAdSn-metric g = diag(−1,−1, 1, . . . , 1), we get
g(x− y, x− y) = −(−y1)2 −(√
1 + u2 − y2
)2
+ (u− y3)2 + . . .+ (−yn+1)2
= −y21 − y2
2 + y23 + . . .+ y2
n+1 − (1 + u2) + u2 + 2
(√1 + u2y2 − uy3
)= −1− 1 + 2
= 0.
(2.34)
This shows that the curve ` is a null straight line in AdSn. Since the metric g is contant, straightlines are geodesics in (Rn+1, g). The definition of the induced Levi-Civita connection to AdSn
guarantees that it is also a geodesic in the submanifold: ∇γ γ = (∇γ γ)> = 0. This shows (1.).We suppose now that −1 < Q(x, y) < 1. We claim that the curve
e(t) = cos(t)e1 + sin(t)e2
= cos(t)x+ sin(t)
(1√
1−Q2(x, y)y − Q(x, y)√
1−Q2(x, y)x
)(2.35)
is a geodesic. We have Q(x, x) = Q(y, y) = 1 and since Q is the metric g but with reverse sign,it follows that Π = spanx, y is a timelike plane. The basis vectors e1 and e2 form an orthonormalbasis:
Q(e1, e2) = Q(x,1√
1−Q2(x, y)y − Q(x, y)√
1−Q2(x, y)x)
= Q(x, y)1√
1−Q2(x, y)−Q(x, x)
Q(x, y)√1−Q2(x, y)
= 0
(2.36)
and
Q(e2, e2) = Q
(1√
1−Q2(x, y)y − Q(x, y)√
1−Q2(x, y)x,
1√1−Q2(x, y)
y − Q(x, y)√1−Q2(x, y)
x
)= Q(y, y)
1
1−Q2(x, y)− 2Q(x, y)
Q(x, y)
1−Q2(x, y)+Q(x, x)
Q2(x, y)
1−Q2(x, y)
=1−Q2(x, y)
1−Q2(x, y)
= 1.
(2.37)
14
Furthermore, we show that e(t) ∈ dSn. We have
Q(e(t), e(t)) = Q
(cos(t)e1 + sin(t)e2, cos(t)e1 + sin(t)e2
)= cos2(t)Q(e1, e1) + 2 sin(t) cos(t)Q(e1, e2) + sin2(t)Q(e2, e2)
= cos2(t) + sin2(t)
= 1.
(2.38)
Clearly, we have e(t) = −e(t). To show that e is a geodesic, we need only to show that theacceleration e(t) ∈ Ne(t)AdS
n. If x = (0,√
1 + u2, u, 0 . . . , 0) we get the vertical part of thetangent space to be
∂x
∂u= (0,
u√1 + u2
, 1, . . . , 0) (2.39)
and so
Q(x,∂x
∂u) =
√1 + u2
u√1 + u2
− u · 1 = 0. (2.40)
Therefore NxAdSn = Rx and so e is a geodesic. This shows (2.).
We now suppose Q(x, y) > 1. We claim that the curve
h(t) = cosh(t)e1 + sinh(t)e2
= cosh(t)y + sinh(t)
(1√
Q2(x, y)− 1x− Q(x, y)√
Q2(x, y)− 1y
)(2.41)
is a spacelike geodesic. We show that e1 and e2 form an orthonormal basis:
Q(e1, e2) = Q
(y,
1√Q2(x, y)− 1
x− Q(x, y)√Q2(x, y)− 1
y
)= Q(y, x)
1√Q2(x, y)− 1
−Q(y, y)Q(x, y)√Q2(x, y)− 1
= 0
(2.42)
and
Q(e2, e2) = Q
(1√
Q2(x, y)− 1x− Q(x, y)√
Q2(x, y)− 1y,
1√Q2(x, y)− 1
x− Q(x, y)√Q2(x, y)− 1
y
)= Q(x, x)
1
Q2(x, y)− 1− 2Q(x, y)
Q(x, y)
Q2(x, y)− 1+Q(y, y)
Q2(x, y)
Q2(x, y)− 1
=1−Q2(x, y)
Q2(x, y)− 1
= −1.
(2.43)
Finally we show that h(t) ∈ AdSn. We have
Q(h(t), h(t)) = Q
(cosh(t)e1 + sinh(t)e2, cosh(t)e1 + sinh(t)e2
)= cosh2(t)Q(e1, e1) + 2 sinh(t) cosh(t)Q(e1, e2) + sinh2(t)Q(e2, e2)
= cosh2(t)− sinh2(t)
= 1.
(2.44)
We clearly have h(t) = h(t) and by the previous this means that h(t) ∈ Nh(t)AdSn. Hence h is a
geodesic. This shows (3.).All these geodesics are clearly complete and since we know that NxAdS
n = Rx it follows thatall geodesics must be intersections AdSn∩Π, where Π is some plane. This shows that the geodesicswe have parametrized are all geodesics. Thus we have shown (4.) and (5.).
15
The following definitions will be of use.
Definition 2.25. On an n-dimensional semi-Riemannian manifold (M, g) we say that a smoothn-form ω is a volume element if for each orthonormal frame on M , we have ω(e1, . . . , en) = ±1.
Definition 2.26. For a smooth vector field V ∈ C∞(TM) we define
LV (f) = V (f), for all f ∈ C∞(M)
LV (X) = [V,X], for allX ∈ C∞(TM).(2.45)
The tensor derivation LV is called the Lie derivative with respect to V .
Lemma 2.27. ([20], p. 195) If ω is a local volume element on a semi-Riemannian manifold (M, g)and X is a smooth vector field, then LXω = (divX)ω.
Proof. ([20], p. 195) Let e1, . . . , en be a local orthonormal frame such that ω(e1, . . . , en) = 1. Wehave
(LXω)(e1, . . . , en) = Xω(e1, . . . , en)−n∑i
ω(e1, . . . , LXei, . . . , en)
= −n∑i
ω(e1, . . . , LXei, . . . , en),
(2.46)
as Xω(e1, . . . , en) = 0. If we write LXei = [X, ei] =∑j fijej we get
(LXω)(e1, . . . , en) = −n∑i
ω(e1, . . . ,∑j
fijej , . . . , en)
= −n∑i
ω(e1, . . . , fiiei, . . . , en)
= −n∑i
fiiω(e1, . . . , en),
(2.47)
where only the diagonal terms fii survive the summand, as the volume element is skew-symmetric.On the other hand we have
divX =∑i
g(∇eiX, ei)
=∑i
g([ei, X], ei) +∑i
g(∇Xei, ei)︸ ︷︷ ︸=0
= −∑i
fii,
(2.48)
where the second term vanishes as g(ei, ei) is constant and the connection ∇ is skew.
We finally briefly discuss the notion of a manifold with boundary. We denote the half-spacein Rn by
Hn = (x1, . . . , xn) ∈ Rn| xn ≥ 0 (2.49)
and equip it with the relative topology from the standard metric Euclidean topology in R, i.e. anopen set V in Hn is the intersection V = Hn ∩ U , where U is open in R.
Definition 2.28. An n-dimensional smooth manifold with boundary is a set M and a setof charts (Uα, φα) such that
1.⋃αUα = M ,
16
2. for all pairs α and β with φα(Uα) ∩ φβ(Uβ) = W 6= ∅ the sets φα(W ) and φβ(W ) are opensets in Hn and the transition maps φα φ−1
β and φβ φ−1α are smooth, and
3. the family (Uα, φα) is maximal relative the conditions (1.) and (2).
A point p in M is said to be a boundary point if for some chart p ∈ Uα and fα = (x1, . . . , 0).The set of boundary points is called the boundary and is denoted by ∂M . We make the followingassertion about manifolds with boundary.
Proposition 2.29. ([7], p. 62) The boundary ∂M of a smooth manifold with boundary is ann-dimensional manifold without boundary, i.e. ∂(∂M) = ∅.
Proof. See [7], p. 62.
2.3 Orbit manifolds
Definition 2.30. Let M be a topological manifold and Γ a group of diffeomorphisms on M . Fora fixed p ∈M , the set µ(p) | µ ∈ Γ is called a Γ-orbit of p. The sets of orbits is denoted M/Γ.
Definition 2.31. Let Γ be a group of diffeomorphisms on a differentiable manifold M . Then Γ isproperly discontinuous if:
1. for any point p ∈ M there is a neighbourhood U such that p ∈ U and µ(U) ∩ U 6= ∅, forµ ∈ Γ, if and only if µ = idM ,
2. if p, q ∈M are not in the same orbit, there are neighbourhoods U 3 p, V 3 q that are disjointand for all µ ∈ Γ we have µ(U) ∩ V = ∅.
We get the following direct consequence.
Lemma 2.32. Let Γ be a properly discontinuous group of diffeomorphisms on a differentiablemanifold M . Let p be a point in M and let Up be a neighbourhood of p, satisfying (1.) of Definition2.31. Then µ1(Up) ∩ µ2(Up) 6= ∅ if and only if µ1 = µ2.
Proof. We assume the converse. Then µ1(Up) ∩ µ2(Up) 6= ∅ and we denote by q some element inthe intersection. Then we have µ1(p1) = q = µ2(p2), for p1, p2 ∈ Up. But then µ−1
2 µ1 ≡ µ mapsp1 to p2 and so µ(Up) ∩ Up 6= ∅. Thus, µ = id which is the assertion.
Proposition 2.33. Let ϕ : M → N be a covering map between topological manifolds. Any decktransformation group D on M is properly discontinuous.
Proof. We first show (1.). Suppose that p ∈ N is arbitrary and consider the fiber ϕ−1(p). Sinceϕ covers N evenly, there is a neighbourhood U 3 p such that ϕ−1(U) decomposes as a union ofdisjoint sheets. As ϕ, restricted to a sheet, is a homeomorphism, any sheet contains precisely oneelement in ϕ−1(p). Since the deck transformations are diffeomorphisms, they must be bijectivebetween sheets. Hence the sheets around each element in M has the property in (1).
Suppose p, q are not in the same orbit. We consider ϕ(p) and ϕ(q) in N . Not being in thesame orbit guarantees that ϕ(p) 6= ϕ(q). N is Hausdorff, and so there are open disjoint setsU 3 ϕ(p) and V 3 ϕ(q). Since ϕ is an open cover, there are two evenly covered neighbourhoodsWp and Wq containing ϕ(p) and ϕ(q), respectively. Consider further the intersections Wp ∩ Uand Wq ∩ V . As U and V are disjoint, so must their inverse images ϕ−1(U) and ϕ−1(V ) be. Asthe deck transformations acts bijectively between sheets, so will they on the inverse intersections.Hence we have (2.).
Definition 2.34. A map ϕ : M → N is called a local isometry if, for each p ∈ M , there is aneighbourhood U , that contains p, such that ϕ|U is an isometry onto ϕ(U).
Definition 2.35. A covering map ϕ : M → M between a semi-Riemannian manifold M and aconnected semi-Riemannian manifold M is called a semi-Riemannian covering map if it alsois a local isometry.
17
The following theorem is of great importance.
Proposition 2.36. ([20], p. 188 and p. 191) Let M be a semi-Riemannian manifold and Γ bea properly discontinuous group of isometries acting on M . Then there is a unique way to makeM/Γ into a manifold such that the natural map π : M →M/Γ is a semi-Riemannian covering. Iffurthermore M is connected, the deck transformation group D is Γ.
Proof. See [20], p. 188 and p. 191.
Some care is required, for the notion of a properly discontiuous group of isometries in the semi-Riemannian setting. The metric now allows for a broader class of isometries on the manifold andit can happen that the standard metric results fail. The first comment regarding this we make isthe following proposition, which is not proven in the referred source.
Proposition 2.37. ([20], p. 214) Let (M, g) be a Riemannian manifold and Γ be a properlydiscontinuous group of isometries acting on M . Then (1.) from Definition 2.31 implies (2.).
Proof. We begin by showing that the Γ-orbits are closed sets. Since each point µ(p) ∈ Γ(p) hasa neighbourhood from (1.) of Definition 2.31 we know that they are isolated points. We mustshow that no sequence in Γ(p) converges. We therefore suppose that a convergent sequence exists,and denote it by pm. Since pm is in one Γ-orbit, we may identify pm = µm(p1) and pass toa subsequence so that all the µm are distinct. Denote the limit by q, i.e. we have µm(p1) → q.From (1.) of Definition 2.31 we have some neighbourhood Uq such that µm(Uq) ∩ Uq = ∅ for allm. Then there exists some N such that µN (p1) ∈ Uq and µN+1(p1) ∈ Uq. Then on the one hand,there must be some element µ ∈ Γ such that µ µN = µN+1 and so µ(pN ) = pN+1. On the otherhand, µ(Uq) ∩ Uq = ∅ and we have a contradiction.
We now show that two distinct orbits must have positive distance, i.e. that
d(Γ(p),Γ(q)) = inf︸︷︷︸p∈Γ(p),q∈Γ(q)
d(p, q) = r > 0. (2.50)
We assume not. Then there must be two sequences pn ⊂ Γ(p) and qn ⊂ Γ(q) such thatd(pn, qn) → 0 as n → ∞. But we have pn = µn(p1). Since Γ is a group of isomorphisms, theymust preserve distances and so it must follow that d(p1, µ
−1n (qn)) → 0, as n → ∞. This means
that there would be a sequence µ−1n (qn) ⊂ Γ(q) that converges to p1, an element outside of the
closed set Γ(q), which is a contradiction. Therefore the distance between orbits must be positive.For a positive distance r > 0 between two distinct orbits Γ(p) and Γ(q) we let ε = r/3. Since
every µ ∈ Γ is a Riemannian isometry, we have µ(Bε(p)) = Bε(µ(p)). Hence the sets U = Bε(p)and V = Bε(q) satisfies the required properties of (2.).
The following example shows that Proposition 2.37 is not true in general, i.e. that (2.) ofDefinition 2.31 is necessary in the general semi-Riemannian setting.
Example 2.38. ([20], p. 214) We consider the punctured plane M = R2−0 equipped with theLorentzian metric defined through
g =
(0 11 0
). (2.51)
We show that the mapping µ(u, v) = (2u, v/2) is an isometry. Let γ = (γu, γv) be a curve in M .Then
g(dµ(γ(t)), dµ(γ(t))) = 2µ(γu(t))µ(γv(t)) = 2(2γu(t)γv(t)/2) = g(γ(t), γ(t)). (2.52)
Let us now show that the cyclic isometry group Γ generated by µ is not properly discontinuous.We begin by making the following observation about how Γ behaves in the right- and left half-planes. Since no µ ∈ Γ changes sign it is clear that the quadrants are mapped into themselves.Furthermore, the right half-plane can be thought of as generated by the half-open strip
QR =
(x, y) ∈ R2 | x ∈ [1, 2)
(2.53)
18
under Γ. The image µn(QR) is the set
µn(QR) =
(x, y) ∈ R2 | x ∈ [2n, 2n+1). (2.54)
For m 6= n, the sets µn(QR) and µm(QR) are disjoint and homeomorhpic.Any point p in the right half-plane will be contained in some µn(QR). If p is in the interior,
any open ball Bε(p) that is contained in µn(QR) will satisfy (1.) of Definition 2.31. If p were tobe contained in the left boundary of µn(QR), the ball B2n−1(p) will satisfy (1.) of Definition 2.31.
Clearly, the positive y-axis is mapped into itself and may be thought of as generated by 0 ×[1, 2) under Γ in a similar fashion. The same argument holds for both the left half-plane as well asthe negative y-axis. Hence Γ satisfies (1.) of Definition 2.31.
However Γ does not fulfill (2.). This may be seen by the following. Let p and q be any twopoints on the positive y- and x-axes, respectively. Let Bεp(p) be any ball centered at p = (0, yp)on the y-axis. Let similarly Bεq (q) be some ball centered at q = (xq, 0) on the x-axis. The imageµn(Bεq (q)) is an ellipse with semi-minor axis length εq/2
n and semi-major axis length εq2n. Hence,
we may choose n big enough so that both xq/2n < εp and εq2
n > yp, and thus the balls will overlap.Hence we do not have (2.) and Γ is not properly discontinuous. This is illustrated in Figure 2.3.
Figure 2.1: Depiction of the base set QR together with Bε(p) and Bε(q) and the image of Bε(p)under µ−1.
We now show equivalence of completeness of a semi-Riemannian manifold and that of its cover.
Proposition 2.39. A semi-Riemannian manifold (M, g) is (geodesically) complete if and only ifits universal covering manifold M is (geodesically) complete (via the pull-back metric).
Proof. Suppose M is complete and γ : [0, b)→M is a geodesic. Pick any point e ∈ ϕ−1(γ(b)) andlet γ be the lifted geodesic with ϕ(γ) = γ. γ extends beyond b and by continuity we may defineγ(b) = ϕ(γ(b)). The point γ(b) has an evenly covered neighbourhood U that lifts to M such thatprecisely one sheet in the inverse set ϕ−1(γ(b)) contains γ(b). Since ϕ is a local diffeomorphism,the extension of γ on the sheet maps down isometrically to U and all is shown.
To show the converse, let M be complete and γ : [0, b)→ M be a geodesic. We map it down toM via ϕ and get the new geodesic ϕ(γ) ≡ γ : [0, b)→M . Since M is complete, we may extend itbeyond b. Let U be evenly covered in M and contain γ(b). Since U is open it must contain somepoint γ(b− ε) that is the image of γ(b− ε). Let e be such that ϕ(e) = γ(b) and Ue the unique sheetof U containing e. The geodesic ϕ−1(γ) : [b − ε, b + ε] lies in Ue and is mapped down to γ in U .There must be an element p in Ue such that ϕ(p) = γ(b − ε). In turn, there must be k ∈ D such
19
that k(p) = γ(b− ε). Since k is a diffeomorphism it will map Ue to the sheet that contains γ(b− ε)but this sheet will contain k(e) ≡ γ(b). Since k is an isometry it preserves geodesics. Hence γextends to b and since the sheet k(Ue) maps isometrically down to U γ extends beyond b as well.Hence M is complete.
Below we give the example that shows that the Hopf-Rinow Theorem, (Theorem 1.5), does nothold in the semi-Riemannian setting.
Example 2.40. ([20], p. 193, ”The Clifton-Pohl Torus”) Consider M = R2 − 0 equipped withthe metric
g =1
u2 + v2
(0 11 0
). (2.55)
Define the mapping µ2(u, v) = (2u, 2v). The following shows that µ is an isometry. Let γ(t) =(γu(t), γv(t)) be a curve through M . Then
g(dµ2(γ(t)), dµ2(γ(t)))µ2(γ(t)) =2dµ2(γu(t))dµ2(γv(t))
µ2(γu(t))2 + µ2(γv(t))2
=2(2γu(t))(2γv(t))
(2γu(t))2 + (2γv(t))2
=2γu(t)γv(t)
γu(t)2 + γv(t)2
= g(γ(t), γ(t))γ(t).
(2.56)
Hence the mapping µ2 generates a cyclic isometry group Γ = µn2∣∣ n ∈ Z, and the isometries are
obviously properly discontinuous, from the discussion in Example 2.38. The orbits under Γ are thesets defined through the congruence relation
(u, v) ≡ (x, y) mod Γ ⇔ u
x=v
y= 2n, (2.57)
for some integer n. Hence the quotient set M/Γ can be identified with the set
M/Γ =
(u, v) ∈M∣∣1 ≤ (u2 + v2) < 2
, (2.58)
which, after identifying the boundary circles, is a torus and therefore also compact.We now want to show that M/Γ is not complete. From Proposition 2.36 we know that the map
π : M →M/Γ is a local isometry and therefore we need only to show that (M, g) is not complete.This we do by solving for the geodesic equations. The Christoffel symbols for the given metric are
Γuuu =∂uguvguv
= − 2u
u2 + v2, Γvuu = 0, Γuuv = 0,
Γvvv =∂vguvguv
= − 2v
u2 + v2, Γvuv = 0, Γuvv = 0,
(2.59)
from which it follows that the geodesic equations are
u− 2u
u2 + v2u2 = 0,
v − 2v
u2 + v2v2 = 0.
(2.60)
The curve γ(t) is a solution:
γ(t) = (1
1− t, 0), (2.61)
but cannot be extended beyonds its domain (−∞, 1). The projected geodesic π(γ) is a geodesic inM/Γ that is not complete. Hence M/Γ is a compact Lorentzian manifold that is not complete.
We end this section by stating an important result.
20
Corollary 2.41. Suppose (M, g) is a complete semi-Riemannian manifold of constant (sectional)curvature κ = ±1 or 0. Then
M ' M/Γ, (2.62)
where Γ is a properly discontinuous group of isometries and M is the simply connected universalcovering space, which is unique up to isometry. If furthermore the curvature is positive, i.e. κ = 1,then M = dSn.
Proof. From Theorem A.33 we know that M has some universal cover and we use the notationϕ : M → M . By Theorem 2.39 we know, that as M is complete, so is also M . We know fromTheorem A.24 that there is a bijection between Γ(p) and ϕ−1(p), for each p ∈ M . But thenϕ−1(p) = Γ(p). If we consider the orbit manifold M/Γ it is obvious that M and M/Γ are inbijection, through Γ(p) ' p. By Theorem 2.36, the quotient M/Γ is a semi-Riemannian manifold,and since the metric on M/Γ is the pull-back metric through a local isometry, it must have thesame curvature.
From Proposition A.35 we have that since the curvature is positive, the covering space mustbe dSn. This follows as it has constant curvature κ = 1, is complete and simply connected byPropositions 2.21 and 2.22.
21
22
Chapter 3
Homogeneous spaces and(G,X)-manifolds
In this chapter we begin by discussing the concept of a smooth homogeneous space and a Lie-groupaction on a manifold. This is a necessity as we need some properties of special isometry groupsto the spaces Mn, dSn and AdSn. Moreover we discuss the important notion of the so-called(G,X)-manifolds, which is a tool that will be essential for the proof of Theorem 4.21 in Chapter6.
3.1 Homogeneous spaces
The following discussion stems from [1], [20] and [22]. We begin by reminding the reader of thenotion of a Lie group.
Definition 3.1. Let G be a smooth manifold and suppose the following holds:
1. G is a group, and
2. the mapping ρ : G×G→ G by ρ(x, y) = xy−1 is smooth.
Then G is called a Lie group.
Definition 3.2. Let G be a Lie group and H be a subgroup of G under the given group structure,such that H is an immersed submanifold of G. Then H is said to be a Lie subgroup.
Definition 3.3. Let H be a Lie subgroup of a Lie group G. If H is a (topologically) closed subsetof G H is called a closed subgroup.
A priori, a closed subgroup and a Lie subgroup are distinct things. It turns out, however, thatbeing a closed subgroup implies being a Lie subgroup.
Theorem 3.4. ([22], p. 110 Theorem 3.42) Let G be a Lie group and H be a closed subgroup.Then H is a Lie subgroup and the submanifold structure of H is unique.
Proof. See ([22], p. 110 Theorem 3.42).
We make the following definition:
Definition 3.5. Let G be a Lie group and K be a closed subgroup. The set G/K, consisting ofall left cosets gK, g ∈ G, is called the coset space.
The coset space may be turned into a manifold as is shown in the following theorem:
Theorem 3.6. ([1], [20], [22]) Let G be a Lie group and K a closed subgroup. Then there isa unique way to make G/K into a manifold, so that the natural projection π : G → G/K withg → gK is a smooth submersion.
23
Proof. See [22].
Definition 3.7. Let M be a manifold and G be a Lie group. A smooth map
λ : G×M →M, (3.1)
such that λ(e, x) = x and λ(a, λ(b, x)) = λ(a · b, x), is called a left action by G on M . The Liegroup G is said to act on M through λ.
Definition 3.8. Let G be a Lie group acting on M . If for each x, y ∈M there is an element µ ∈ Gsuch that µ(x) = y, we say that G acts transitively on M .
Definition 3.9. Let G be a Lie group acting on a smooth manifold M . The subgroup Kx of G,defined through
Kx = g ∈ G | λ(g, x) = x, (3.2)
is called the isotropy subgroup of G at x.
With the above at hand, we may state the following:
Proposition 3.10. [1] Let G be a Lie group acting transitively on M and Kx be the isotropy groupat x. Then we have the following:
1. Kx is closed in G, and
2. the map ρ : G/Kx →M with ρ(gKx) = λ(g, x) is a diffeomorphism.
Proof. See [20].
We are now ready to define the notion of a homogeneous space.
Definition 3.11. A (smooth) manifold M is said to be a homogeneous space if there exists agroup G acting transitively on it.
With the above, we may identify any homogeneous space M with some (not necessarily unique)coset manifold G/K, where G is a Lie group and K is a closed subgroup of G.
In the above we only assume that M is a smooth manifold, but now we continue assuming thatM is equipped with some metric.
Definition 3.12. Let (M, g) be a semi-Riemannian manifold. The set of isometric diffeomorphismsform a group under composition. The isometry group on (M, g) is denoted by I(M).
Definition 3.13. Let (M, g) be a semi-Riemannian manifold with isometry group I(M). If I(M)acts transitively, then (M, g) is said to be a semi-Riemannian homogeneous manifold.
The following theorem shows that our previous discussion of this chapter may be applied to asemi-Riemannian homogeneous space.
Theorem 3.14. ([20], p. 255) Let (M, g) be a semi-Riemannian homogeneous space together withan isometry group I(M). Then there is a unique way to make I(M) into a manifold such that:
1. I(M) is a Lie group, and
2. the natural action I(M)×M →M is smooth.
Definition 3.13 is made for the general semi-Riemannian setting, but there are some resultsthat hold only in the Riemannian case. The following theorem is such an example.
Theorem 3.15. ([20], p. 257) Let (M, g) be a homogeneous Riemannian manifold. Then (M, g)is complete.
24
Proof. ([20], p. 257) We suppose we have a geodesic γ : [0, b) → M and we want to show that itcan be extended beyond [0, b). As (M, g) is Riemannian, every point p is contained in some normalneighbourhood U . Since U is open, there is an open ball Bε(p) ⊂ U with ε < b/2. Let φ be someelement mapping γ(0) to the point γ(b− ε/2). Denote by v the tangent vector γ(b− ε/2). Since Gis transitive any differential is surjective, and there is an element v ∈ Tγ(0)M such that dφ(v) = v.
The unique geodesic γ with γ(0) = γ(0) and ˙γ(b − ε/2) = v extends γ by φ γ. Since the Bε(p)contains γ(b) as an interior point, the geodesic extends beyond γ(b) and (M, g) is complete.
Theorem 3.15 does not hold in the semi-Riemannian setting, as the following example shows.
Example 3.16. [20] We consider the set
R2+ =
(x, y) ∈ R2
∣∣ x > 0, (3.3)
equipped with the metric defined through
g =
(0 11 0
). (3.4)
This space has isometries on the form µa(u, v) = (u/a, av), a > 0, and φa(u, v) = (u, v−a). Letting(a, b) be arbitrary, we have φab µa(a, b) = φab(1, ab) = (1, 0). Conversely, if (c, d) is arbitrary,then µ1/c φ−cd(1, 0) = µ1/c(1, cd) = (c, d) and so any two points can be joined by isometries.This makes (M, g) a semi-Riemannian homogeneous manifold. As (M, g) is flat, the geodesics arestraight lines, and so the line γ(t) = (1, 0)−t(1, 0) is a geodesic. γ is defined on (−∞, 1) but cannotbe extended. Hence (M, g) is a homogeneous semi-Riemannian manifold that is not geodesicallycomplete.
We want to state and prove Theorem 3.21 of Marsden, that tells us that a compact andhomogeneous semi-Riemannian manifold is geodesically complete. For this we need the followinglemmas.
Lemma 3.17. ([20], p. 30) Let γ : [0, b)→M, b <∞ be an integral curve of some vector field Xin C∞(TM). Then the following are equivalent:
1. γ is extendible as an integral curve of X to [0, b+ ε),
2. γ is extendible,
3. γ lies in a compact subset of M , and
4. there exists a sequence tn → b such that γ(tn) converges.
Proof. See ([20], p. 30).
Definition 3.18. A Killing vector field X on a semi-Riemannian manifold is a vector field suchthat LXg = 0.
Lemma 3.19. ([20], p. 252, ”The conservation lemma”) Let X be a Killing vector field on asemi-Riemannian manifold M and let γ be a geodesic in M . Then the restriction of X to γ is aJacobi field and g(γ, X) is constant along γ.
Proof. See ([20], p. 252).
Lemma 3.20. ([20], p. 258) Each tangent vector to a homogeneous semi-Riemannian manifoldM extends to a Killing vector field on M .
Proof. See ([20], p. 258).
We may now prove Marsdens Theorem.
Theorem 3.21. ([20], p. 258, ”Marsdens Theorem”) Let (M,g) be a compact homogeneous semi-Riemannian manifold. Then M is (geodesically) complete.
25
Proof. [20] One needs to show extendability for a geodesic γ(t) defined on [0, b). Let tn be asequence in [0, b) that converges increasingly towards b. Then, by the compactness, there is asubsequence γ(tnk) that converges to p := γ(b). By continuity of γ, the limit is independent ofsubsequence and so γ can be extended to the closed interval [0, b]. Let v1, . . . , vn be a basis forTpM . By Lemma 3.20, each vector vi extends to a Killing vector field Vi on M . Furthermore,by Lemma 3.19, each component ci = g(γ, Xi) is constant along γ. From this it follows that thesequence γ(tn) converges in TM to a vector v ∈ TpM .
It can be shown, that the vector field γ may be extended to a vector field on TM . From Lemma3.17 it follows that since the sequence γ(tn) converges as in (4.), we have that the integral curveextends and so completeness follows.
3.2 Isometry groups and their properties
We continue by discussing isometry groups. Example 3.22 shows that for a space with indefinitescalar product, i.e. a space with non-zero index ν, the isometry group will be different from thestandard O(n) isometry group of Euclidean space.
Example 3.22. Let us consider the two-dimensional Lorentzian space M2. The metric η of thisspace is, as we are familiar with from above, the tensor η = diag(−1, 1). For two vectors v = (v1, v2)and u = (u1, u2), we thus have
g(u, v) = η(u, v) = −u1v1 + u2v2. (3.5)
This is obviously not invariant under SO(2)-rotations (u → A(φ)u, v → A(φ)v where A(φ) ∈SO(2)), as the following calculation shows:
(A(φ)u)T ηA(φ)v = (u1, u2)T(
cos(φ) sin(φ)− sin(φ) cos(φ)
)(−1 00 1
)(cos(φ) − sin(φ)sin(φ) cos(φ)
)(v1
v2
)= (u1, u2)T
(sin2(φ)− cos2(φ) 2 cos(φ) sin(φ)
2 cos(φ) sin(φ) cos2(φ)− sin2(φ)
)(v1
v2
)6= (u1, u2)T
(−1 00 1
)(v1
v2
)= uT ηv.
(3.6)
However, the failure for this symmetry might be remedied - the diagonal terms look like thehyperbolic identity. Hence, one might guess that hyperbolic functions might preserve the scalarproduct. Indeed, they do:
(B(φ)u)T ηB(φ)v
= (u1 u2)T(
cosh(φ) sinh(φ)− sinh(φ) cosh(φ)
)(−1 00 1
)(cosh(φ) − sinh(φ)sinh(φ) cosh(φ)
)(v1
v2
)= (u1, u2)T
(sinh2(φ)− cosh2(φ) cosh(φ) sinh(φ)− cosh(φ) sinh(φ)
cosh(φ) sinh(φ)− cosh(φ) sinh(φ) cosh2(φ)− sinh2(φ)
)(v1
v2
)= (u1, u2)T
(−1 00 1
)(v1
v2
)= uT ηv.
(3.7)
Definition 3.23. The group of linear isometries on (Rn, g), where g = diag(−1, . . . ,−1, 1, . . . , 1)is the diagonal metric with ν negative signs and n−ν positive signs, is called the semi-orthogonalgroup. The group of linear isometries on Mn = (Rn, η), where η = diag(−1, 1, . . . , 1), is denotedby O1(n), and is called the Lorentz group.
From Example 3.22 we know that O1(n) is not compact. In order to better understand itsstructure, let us calculate its Lie algebra. The discussion is similar to that of ([15], Example 3.11).
26
Lemma 3.24. The tangent space TeO1(n) at the neutral element is given by
TeO1(n) =X ∈ Rn×n |XT η + ηX = 0
, (3.8)
where η = diag(−1, 1, . . . , 1).
Proof. Let γ : I → O1(n) be a curve in O1(n) with γ(0) = e. Then γ(t)T ηγ(t) = η for all t ∈ I.We differentiate and get
γ(0)T η + ηγ(0) = 0, (3.9)
after evaluation at t = 0.We also need to show the reverse inclusion. Suppose that X satisfies XT + ηXη = 0. Then
XT = −ηXη and (XT )n = (−ηXη)n = η(−1)nXnη. It follows that, for A = Exp(X), we get
AT = Exp(X)T
= Exp(XT )
= E +XT +1
2(XT )2 +
1
3!(XT )3 + · · ·+ 1
n!(XT )n + . . .
= ηEη + (−ηXη) +1
2(ηX2η)− 1
3(ηX3η) + · · ·+ (−1)n
n!(ηXnη) + . . .
= η
(E −X +
1
2(X2η)2 − 1
3(ηX3η) + · · ·+ (−1)n
n!(ηXnη) + . . .
)η
= ηExp(−X)η,
(3.10)
from which it follows that
AT ηA = Exp(X)T ηExp(X)
= Exp(XT )ηExp(X)
= ηExp(−X)ηηExp(X)
= ηExp(X −X)
= η.
(3.11)
Hence A = Exp(X) ∈ O1(n) and we have shown the assertion.
Lemma 3.25. Let X ∈ O1(n) be an element of the Lie algebra of O1(n). Then X has the (unique)decomposition
X =
(0 00 B
)+
(0 uuT 0
), (3.12)
where B ∈ O(n− 1) and u = (u2, . . . , un) is a row-vector.
Proof. From Lemma 3.24 we know that XT = −ηXη. Since η = diag(−1, 1, . . . , 1) the right-lower(n− 1)× (n− 1) components Y of X satisfy Y + Y T = 0 and from ([15], Proposition 3.12) weknow that this is precisely TeO(n− 1).
By simply looking at the rows u and uT we see that they make X satisfy XT = −ηXη.
Let X,Y ∈ Rn×n. We let 〈 , 〉 denote the standard Euclidean scalar product on the space Rn×n,which is explicitly given by 〈X,Y 〉 = Re
∑ij XijYij . The following Lemma will be of use.
Lemma 3.26. ([20], p. 303) Let G be a Lie subgroup of GLn(C). Then
1. Ada(X) = aXa−1 for all a ∈ G and X ∈ g0, and
2. if X ∈ g0 implies that XT ∈ g0, then B(X,Y ) = Re Trace (XY ) = 〈 XT , Y 〉 is an Ad-invariant scalar product.
Proof. See [20], p. 303.
The following definitions, stemming from ([18], p. 446), will be useful.
27
Definition 3.27. A bijective mapping θ from a set into itself is called an involution if it is itsown inverse.
Definition 3.28. Let M = G/H be a homogeneous space. Then M is said to be reductive ifthe Lie algebra g0 of G has a decomposition g0 = h0 ⊕ m0 where h0 is the Lie algebra of H andm0 is Ad(H)-invariant.
Definition 3.29. Let g0 be a Lie algebra. Then the set
(g0)C = u+ iv | u, v ∈ g0 (3.13)
is called the complexification of g0.
We remind ourselves that an automorphism on a Lie algebra is an invertible linear map Lthat preserves Lie brackets, i.e.
L([X,Y ]) = [L(X), L(Y )]. (3.14)
We suppose for simplicity that g0 is real. It follows that the Ad-mapping in Proposition 3.26 is aLie algebra-automorphism and may thus be viewed as a Lie group homomorphism
Ad : G→ AutR(g0) ⊆ GLR(g0) by Ad(g) : X → gXg−1. (3.15)
Since Ad is a map between Lie groups, it has a differential between the associated Lie algebras:
ad : g0 → EndR(g0) by ad : X → [X, ·], (3.16)
where X ∈ g0 and EndR(g0) is the Lie algebra of GLR(g0). From the Jacobi-identity of Lie-brackets,it follows that the ad-map is a Lie-algebra homomorphism. Hence we may view the image ad(g0)as a Lie subalgebra. We may now make the following definition.
Definition 3.30. Let g0 be a Lie algebra. By Int g0 we mean the subgroup that has ad(g0) as Liealgebra. An automorphism L is called inner if L ∈ Int g0.
Clearly Int g0 ⊆ AutR with equality if AutR is connected.
Definition 3.31. The bilinear form B(X,Y ) = Trace(Ad(X)Ad(Y )) is called the Killing form.
Definition 3.32. ([18], p. 446) A reductive Lie group is a 4-tuple (G,K, θ,B), where G is aLie group, K is a compact Lie subgroup, θ is an involution of the Lie algebra g0 of G and B is anon-degenerate, Ad(G)-invariant, θ-invariant bilinear form on g0, such that
1. g0 is a reductive Lie algebra,
2. the decomposition of g0 into θ-eigenspaces with eigenvalues ±1 is g0 = l0⊕p0 where l0 is theLie algebra of K,
3. l0 and p0 are ortogonal under B and B is positive definite on p0 and negative definite on l0(i.e. B(P, P ) > 0 and B(L,L) < 0 for P ∈ p0 and L ∈ l0),
4. multiplication as a map from K × exp p0 into G, is a diffeomorphism onto, and
5. every automorphism Ad(g) of g = (g0)C is inner for g ∈ G.
With this, θ is called the Cartan involution, g0 = l0⊕ p0 is called the Cartan decomposition,K is called the maximal compact subgroup and B is called the invariant bilinear form.
Observe that p0 need not be a Lie subalgebra of g0.We show that the isometry group O1(n+1) naturally has the structure of a reductive Lie group.
Proposition 3.33. The 4-tuple (O1(n+ 1), O(n), θ, B), where θ(X) = −XT and B is the Killingform, is a reductive Lie group.
28
Proof. That O1(n) is a Lie group follows from Theorem 3.14. O(n) is compact and hence a closedsubgroup of O1(n + 1). It follows that O(n) is also a Lie subgroup and from construction, it isobvious that θ is an involution.
We want to show (1.). From Lemma 3.25 we get the decomposition g0 = l0 ⊕ p0, where
l0 =
(0 00 B
)|B +BT = 0
, and
p0 =
(0 uuT 0
)| u = (u2, . . . , un)
(3.17)
and that l0 is precisely the Lie algebra (at the neutral element) of K = O(n− 1). In order for g0
to be reductive, we must therefore show that p0 is Ad(K)-invariant. Let P ∈ p0 and L ∈ l0. Then
AdL(P ) = [P,L] =
(0 uB
BuT 0
)∈ p0 (3.18)
and so g0 is reductive. This shows (1.).Let P ∈ p0 and L ∈ l0 be on the form in Lemma 3.25. We have
θ(L) = −(
0 00 B
)T=
(0 00 −BT
)=
(0 00 B
)= L, and
θ(P ) = −(
0 uuT 0
)T= −
(0 uuT 0
)= −P.
(3.19)
Thus we see that l0 and p0 are the corresponding eigenspaces of θ. This shows (2.).We need to show the properties of B in (3.). We start by showing that B is symmetric. Let
e1, . . . , en+1 be a basis for g0. There is [ei, ej ] =∑k c
ijk ek, where cijk are the structure constants.
We have
[ei, [ej , ek]] = [ei,∑m
cjkm em]
=∑m
cjkm [ei, em]
=∑m
cjkm∑l
ciml el
=∑m
∑l
cjkm ciml el,
(3.20)
from which we see that Bij =∑k
∑m c
jkm c
imk and thus obviously Bij = Bji. Hence B is symmetric.
We now show (3.). From (2.) of Lemma 3.26 we know that B(X,Y ) = XT · Y . From this weimmediately get
B(P, P ) = PT · P = P · P =∑ij
|Pij |2 > 0,
B(L,L) = LT · L = −L · L = −∑ij
|Lij |2 < 0 and
B(L,P ) = LT · P =∑ij
PijLij = 0,
(3.21)
where the last line follows as every term PijLij = 0. From this we also get the θ-invariance of B,and from Lemma 3.26 we already know that B is Ad-invariant. Hence we have (3.).
Property (4.) follows from Proposition 7.14 in [18].Property (5.) follows as a special case of Example (4) in [18] p. 474.
Proposition 3.34. ([18], p. 459, ”KAK-decomposition”) Suppose G is a reductive Lie group anda0 is a maximal abelian vector subspace of p0. Then G = KAK, where K is the maximal compactsubgroup in the 4-tuple (G,K, θ,B), and A is the subgroup of G with a0 as Lie algebra.
29
Proof. See ([18], p. 459).
Proposition 3.35. [17] Let g be an element of O1(n+ 1) and B be a closed Euclidean ball. ThengB is an ellipsoid with axis lengths e−t, 1, . . . , 1, et, for some real t ≥ 0.
Proof. [17] We consider p0 and define the vector subspace a0 to be the matrices at ∈ p0 whereu = (u2 = t, 0, . . . , 0). It follows that allowing any other ui to be non-zero ruins the closure underthe Lie bracket. Moreover the matrices in a0 obviously commute. Hence a0 is a maximal abelianvector subspace of l0. The exponentiation
Exp
(0 αα 0
)=
(1 00 1
)+
(0 αα 0
)+
1
2!
(0 αα 0
)2
+1
3!
(0 αα 0
)3
+ · · ·+ 1
n!
(0 αα 0
)n+ . . .
=
(1 00 1
)+
(0 αα 0
)+
1
2
(α2 00 α2
)+
1
6
(0 α3
α3 0
)+ . . .
=
(cosh(α) sinh(α)sinh(α) cosh(α)
)(3.22)
shows that
At = Exp(at) =
cosh(t) sinh(t) 0 . . . 0
sinh(t) cosh(t) 0 . . ....
0 0 1 . . ....
......
.... . . 0
0 . . . . . . 0 1
. (3.23)
We therefore define A = Exp(a0) and from Proposition 3.34 get that
O1(n+ 1) = O(n)AO(n), (3.24)
in the sense that any g ∈ G can be written as g = k1Atk2, where At ∈ A and
ki =
(1 00 Bi
), Bi ∈ O(n). (3.25)
To investigate the geometrical interpretation of a rotation of such a matrix, we consider thetwo-dimensional case and let B be the closed unit ball centered around the origin. The boundary,which is identified with S1, may be parametrized as (t, x) = (sin(θ), cos(θ)). We get, after thehyperbolic rotation, that(
cosh(t) sinh(t)sinh(t) cosh(t)
)(sin(θ)cos(θ)
)=
(cosh(t) sin(θ) + sinh(t) cos(θ)sinh(t) sin(θ) + cosh(t) cos(θ)
)=
(y1
y2
). (3.26)
Suppose for simplicity that t is positive. A straightforward calculation shows that
y21 + y2
2 = cosh(2t) + sin(2θ) sinh(2t) (3.27)
and so we see that the Euclidean distance is at most a2 = cosh(2t) + sinh(2t) = e2t and at leastb2 = cosh(2t) − sinh(2t) = e−2t, for the values θ = kπ + π/4 and θ = kπ − π/4, respectively. Itremains only to show that the rotated ball is indeed an ellipse with these axes. We claim that thenew semi-major and semi-minor axes are the lines t = x and t = −x, respectively. The ellipticequation is (
u
a
)2
+
(v
b
)2
= 1, (3.28)
where a and b are the semi-major and semi-minor axes. We change basis into mentioned axesu↔ x = t and v ↔ x = −t by rotating counter clock-wise by φ = π/4. The new elliptic equation
30
is (u
a
)2
+
(v
b
)2
=
( y1+y2√2
et
)2
+
( y1−y2√2
e−t
)2
=1
2
(y2
1 + 2y1y2 + y22
e2t
)+
1
2
(y2
1 − 2y1y2 + y22
e−2t
)=
1
2
(y2
1 + y22
)(1
e2t+
1
e−2t
)+
(y1y2
)(1
e2t− 1
e−2t
)=
(cosh(2t) + sin(2θ) sinh(2t)
)cosh(2t)
−(
sinh(2t) + sin(2θ) cosh(2t)
)sinh(2t)
= 1.
(3.29)
So we see that the image of the rotated unit circle At(S1) is an ellipse with semi-major and semi-
minor axis lengths et and e−t.The result now follows as for higher dimensions, the O(n) group rotates the spatial parts
isometrically and hence axis lengths are preserved. Furthermore, since O1(n+1) is a linear isometrygroup, we may use the decomposition
g(Bε(p)) = g(p+Bε(0)) = g(p) + g(Bε(0)) (3.30)
to see that the action on any Euclidean closed ball centered at p is an ellipsoid with mentionedprincipal axes, centered at p.
We now state the isometry groups of the three Lorentzian space forms.
Proposition 3.36. ([20], p. 239-240) The Lorentzian space forms have the following isometrygroups:
1. The Minkowski space Mn has as isometry group the semi-direct product O1(n) o Tn, whereTn is the group of translations on Mn,
2. dSn has the isometry group I(dSn) = O1(n+ 1), and
3. AdSn has the isometry group I(AdSn) = O2(n+ 1).
Proof. See [20], p. 239-240.
The group I(Mn) is called the Poincare group. From the proof to Proposition 3.35 we getthe following Corollary:
Corollary 3.37. The isotropy subgroups Kp of O1(n+ 1) is O1(n).
Proof. Suppose p = (0, . . . , 0, 1). We know that O1(n + 1) = O(n)AtO(n). Clearly, the isometrysubgroup preserving p must be O(n− 1)AtO(n− 1), but this is just O1(n).
We must show that any two isotropy groups are conjugate. Let p and q be two points inMn+1
and h(p) = q for some h ∈ O1(n + 1). Let kp ∈ Kp be an element in the isotropy subgroup at p,and similarly kq ∈ Kq. Since both h−1kph ∈ Kq and hkqh
−1 ∈ Kp it follows that hKph−1 = Kq
and so two arbitrary isotropy subgroups are conjugate. The result now follows.
Let (X, d) be some metric space and let F denote collection of the closed and bounded sets inthis metric. The following definition equips F with a metric.
Definition 3.38. Let (X, d) be a metric space and let X and Y be two closed and bounded subsetsof X. The Hausdorff distance dH(X,Y ) between X and Y is defined to be
dH(X,Y ) = max
supx∈X
infy∈Y
d(x, y), supy∈Y
infx∈X
d(x, y)
. (3.31)
31
The 2-tuple (F , dH) is called the Hausdorff metric space. A sequence of sets An is said tobe Hausdorff convergent to A if dH(An, A) → 0 as n → ∞. We show the following immediateproperty.
Lemma 3.39. The metric space (F , dH) is complete if and only if (X, d) is complete. Furthermore,for a Cauchy sequence An in (F , dH) let A be the set of limit points of all Cauchy sequencesxn in (X, d) with xn ∈ An and let A be its closure. Then dH(An, A)→ 0 as n→∞.
Proof. We suppose first that (F , dH) is complete. Then a Cauchy-sequence xn in (X, d) may beviewed as a sequence in (F , dH) and clearly the limit x ∈ F serves as limit in (X, d). Hence wehave shown that (X, d) is complete.
Conversely, we suppose that (X, d) is complete. We let An be a Cauchy sequence in (F , dH)and let A be the set as stated in the assertion. We suppose that An 6→ A. Then we may assume,by perhaps passing to a subsequence, that dH(An, A) ≥ c > 0. By again passing to a subsequence,we may assume that d(An, An+1) < 1/2n. From the definition of dH the first condition meansthat, for each n, there exists either
(1) xn ∈ A s.t. infy∈An
d(xn, y) ≥ c, or
(2) xn ∈ An s.t. infy∈A
d(xn, y) ≥ c.(3.32)
Possibility (1) is ruled out immediately by definition. Any element y ∈ A is by definition a limitof some Cauchy sequence and so infy∈An d(xn, y) can only be bounded below by 0 over n. Hence(1) is ruled out.
We now consider (2). Let n be big enough so that dH(An, An+1) < c/22 and let y ∈ An be anelement as in (2). We denote y by x1. We know that
infz∈An+1
d(x1, z) ≤ maxx∈An
infz∈An+1
d(x, z) ≤ dH(An, An+1) < c/22. (3.33)
Hence there exists x2 ∈ An+1 such that d(x1, x2) < c/22. For eack k we get, by iterating thisprocedure, that d(xk, xk+1) < c/2k+1. This defines a sequence xk in (X, d) with xk ∈ An+k.This sequence is on the one hand clearly Cauchy and as (X, d) is complete, it has a limit xk → x.But
d(x1, x) ≤ d(x1, x2) + d(x2, x3) + . . . < c
(1
22+
1
23+ . . .
)= c
1
2< c (3.34)
so that x 6∈ A. This is a contradiction. Therefore we must have dH(An, A) → 0. By taking theclosure of A and observing that dH(A, A) = 0 it follows that dH(An, A) → 0 and so (F , dH) iscomplete.
With the constructions in this work, we need to modify this definition to the following:
Definition 3.40. A sequence Fn of closed sets is said to converge to F if, for each closedEuclidean ball B, the sequence Fn ∩B converges to F in the Hausdorff sense.
Definition 3.41. InMn, a hyperplane H is a subset H = v+W , were v is some vector inMn
and W is a (n − 1)-dimensional vector subspace. If v ∈ W , H is said to be vectorial, otherwiseH is said to be affine. If the subspace W contains its orthogonal complement, i.e. W⊥ ⊂ W , His said to be coisotropic.
With these definitions at hand, we get the following corollary to Proposition 3.35.
Corollary 3.42. [17] Suppose gn is a sequence in O1(n + 1), Bε(p) is a closed Euclidean balland the sequence gnBε(p) converges with limit E. Then E is either of codimension 0 or 1, andin the latter case is contained in a coisotropic hyperplane. If gn lie in the isotropy group Kp, thehyperplane is vectorial and contains p.
Proof. Since gn ⊂ O1(n+ 1), every gn is a linear mapping. Hence we may consider the decom-position
gn(Bε(p)) = gn(p+Bε(0)) = gn(p) + gn(Bε(0)). (3.35)
32
From Proposition 3.35 it is clear that the limit may have codimension 1 or 0. If the sequence gnis in the isotropy group Kp we have gn(p) = p and moreover as gn is linear, it must fix the lineRp. The segment of this line is contained in Bε(p) must therefore also be contained in the limit E .Thus E is a vector subspace.
3.3 (G,X)-structures
In this section we briefly describe the so-called (G,X)-structures. This construction will be crucial,when proving Theorem 4.21. We follow [11], [12] and [13].
Definition 3.43. Let X be a real analytic manifold and G be a Lie-group acting transitively onX. A smooth manifold M is called a (G,X)-manifold if there exists a maximal atlas of localcharts
(φα, Uα)
such that φα : Uα → X where Uα ⊂M , with the following properties:
1. M =⋃α∈A
Uα,
2. φα is a diffeomorphism onto is image φα(Uα) and
3. whenever Uα ∩ Uβ 6= ∅ there is, for each connected component C of the intersection, anelement gC,α,β ∈ G such that gC,α,β φα = φβ .
The atlas is denoted by (Φ, U), where Φ = φα | α ∈ A and U = Uα | α ∈ A. Figure 3.3illustrates the construction. The following result is of enormous importance.
Theorem 3.44. ([12], p. 69, ”The Development Theorem”) Let M be a (G,X)-manifold withuniversal covering space M and deck group D ' π1(M). Then there exists a pair (D,h) where D isa local diffeomorphism D : M → X, called the developing map, and h is a group homomorphismh : D→ G, such that the following diagram is commutative:
MD−−−−→k
MyD yDX
Γ−−−−→h(k)
X
(3.36)
or equivalently, that the developing map is equivariant:
D k = h(k) D, for all k ∈ D. (3.37)
The image h(D) = Γ is called the holonomy subgroup of G.Furthermore, the pair (D,h) is unique up to holonomy-conjugation, i.e. if (D′, h′) is another
pair, then there exists g ∈ Γ such that
D′ = g D, g ∈ Γ, and
h′(k) = Inn(g) h(k).(3.38)
Proof. See [12], p. 69.
Conversely, it can be shown that the existence of a developing pair (D,h) is equivalent toexistence of a (G,X)-structure on M . See e.g. [11] for details.
It will become apparent that it is easier to work with the definition in Theorem 3.44 as thisdefinition is, in some sense, more concrete.
The following insight is of great importance.
33
Figure 3.1: Depiction of the action of two charts (Uα, φα) and (Uβ , φβ) on a (G,X)-manifold M .
Theorem 3.45. ([12], p. 72) Let M be a (G, X) manifold and let (G,X) the pair where X is theuniversal cover of X and G is the lift of G to X with group homomorphism ψ : G→ G such thatthe following diagram commutes:
XG−−−−→g
Xyϕ yϕX
G−−−−→ψ(g)
X
i.e. ψ(g) ϕ = ϕ g. (3.39)
Let the pair (D, h) be the development pair for (G, X). Then the pair (ϕD, ψh) is a developmentpair for (G,X).
Proof. Since D and ϕ are both local diffeomorphisms their composition is also a local diffeomor-phism. Similarly the composition ψ h is a group homomorphism. We need only to show theequivariance condition. For any p ∈ M we have
ϕ D(g(p)) = ϕ(h(g) D(p)) = ψ h(g) ϕ(D(p)) (3.40)
so that the development pair (ϕ D, ψ h) is also holonomy-equivariant.
Theorem 3.45 shows that if we model M on some space X that is not simply connected we knowfrom Theorem 3.44 that the development pair (D,h) on (G,X) will be ”holonomy-equivalent” tothe pair (ϕ D, ψ h) and hence this pair may be the one of choice.
Theorem 3.46. [12] Let (M, g) be a semi-Riemannian manifold of constant (sectional) curvature.Then there exists a (G,X)-structure on M .
Proof. See [11].
We get the following properties for the (G,X)-manifolds.
34
Lemma 3.47. Suppose (M, g) is a (G,X)-manifold and Γ = h(D) is the holonomy subgroup of G.Then Γ acts properly discontinuously on the image D(M) in X.
Proof. We first show property (1.) from Definition 2.31. Suppose x is in D(M) and it is the imageof D(x). Then there is a neighbourhood Ux in X, containing x such that D−1, restricted to Ux isa diffeomorphism. If x = D(x), we denote the lift Ux to the neighbourhood that contains x. ButD acts properly discontinuously on M and so there is a neighbourhood Vx around x that satisfies(1.) of Definition 2.31. Let Wx = Ux ∩ Vx denote the intersection.
We claim that the image D(Wx) = Wx of the intersection satisfies (1.) of Definition 2.31. Wesuppose not. The converse is that for any open ball Bn there must exist some µn ∈ Γ such thatµ Bn ∩Bn 6= ∅. We let Bn be a sequence of balls, of decreasing radius and succesive inclusion,centered at x and contained in Wx. By hypothesis, for each n we have some zn = µn(yn) in theintersection. The respective sequences zn and yn must obviously converge to x, as n → ∞.Since D is a diffeomorphism on Wx, we may lift both sequences up to M . We denote the liftedsequences by yn and zn and they converge to the limit x. On the one hand, they are twosequences in Wx that converges to x. On the other hand there must for each n exist a kn ∈ D suchthat h(kn) = µn and therefore also kn(yn) = zn. This contradicts the proper discontinuity of D ifthe kn are non-trivial.
We now show property (2.) of Definition 2.31. Again we assume this not to be true and derivea contradiction. The converse is that for some two points x and y that are not congruent, i.e.x 6≡ y mod Γ, there are no two neighbourhoods Vx 3 x and Vy 3 y satisfying µ(Vx) ∩ Vy = ∅. Welet Bn,x and Bn,y be decreasing, nested sequences of balls centered around x and y, respectively.By hypothesis we have µn(Bn,x)∩Bn,y 6= ∅ for each n. We let yn = µn(xn) be in the intersections.Clearly yn converges to y and xn converge to x.
We let x and y be elements in M such that x = D(x) and y = D(y). We must have x 6≡ ymodD,as otherwise
x = D(x) = D(k(y)) = h(k) D(y) = h(k)(y), (3.41)
which would contradict our hypothesis that x and y were not congruent under Γ. Since D actsproperly discontinuously we therefore have Vx and Vy such that k(Vx) ∩ Vy = ∅ for all non-trivial
k ∈ D. We lift the sequences xn to Vx and denote the lifted sequence by xn. This is done bythe same procedure as the discussion on proving (1.) - we assume D is a diffeomorphism here asotherwise we take intersections. Similarly, yn is lifted to yn in Vy.
For each n there must exist kn ∈ D such that h(kn) = µn. It now follows that kn(yn) = xn.We have
yn = D(yn) = D(kn(xn)) = h(kn)D(xn) = µn(xn). (3.42)
This contradicts the proper discontinuity of D acting on M .Hence Γ must act in a properly discontinuous manner on D(M) ⊂ X.
We now want to show that if M is compact the developing map D is a covering onto its image.
Lemma 3.48. Let (M, g) be a compact (G,X)-manifold. Then every fibre D−1(x) is finite.
Proof. We assume that the assertion is false, i.e. that there exists a sequence xn where we haveD(xn) = x for each n. We first demonstrate that this set is closed. In fact we demonstrate thatit is the union of isolated points, i.e. that there is no convergent subsequence in D−1(x). If sucha convergent subsequence existed, we would have yn → y. Then any neighbourhood containing ywould contain some yN and this breaks the local injectivity.
Now we use the compactness of M . xn has no convergent subsequence by what we justdeduced. Let un be the projected sequence in M , i.e. un = ϕ(xn) for each n. By compactnessof M , there is a convergent subsequence, which for notational convenience, we assume is un.The sequence converges to some limit u and this point has an evenly covered neighbourhood Uu.We pick any of the sheets ϕ−1(Uu) and denote it by Uu. We may assume that all the un arecontained in the neighbourhood Uu as we otherwise discard the first N elements and keep the”tail”. Denote by yn = ϕ−1(un) ∩ Uu. Since ϕ is a local diffeomorphism, this sequence mustconverge, i.e. yn → y = ϕ−1(u) ∩ Uu.
We now consider how the sequences yn and xn are related. If, for each n, we were to havek(xn) = yn, where k ∈ D, it would follow that xn would be a convergent sequence. Hence no
35
k ∈ D may relate any yn to any xn for more than finitely many n. From this it follows that, as wecan always pass to a subsequence, we may assume that there is kn(xn) = yn where all kn ∈ D aredistinct.
From the developing map equation it now follows that
D(yn) = D(kn(xn)) = h(kn)D(xn) = h(kn)(x) (3.43)
and hence that the image sequence D(yn) must both lie in the Γ-orbit Γx and be convergent.By the proof in Proposition 2.37 this cannot happen. Hence D−1(x) is finite.
In particular, Lemma 3.48 shows that if M is compact, then Ker(h) is a finite subgroup of D.The following theorem is crucial in the proof of Theorem 4.21.
Theorem 3.49. Let M be a compact (G,X)-manifold. Then the developing map D is a coveringmap onto the image Ω = D(M).
Proof. From Lemma 3.48 we know that the fibers D−1(x) have finite cardinality. If the fibres havethe same cardinality N , then for each x ∈ X we may construct an evenly covered neighbourhoodUx in the following way. Since M is Hausdorff we may take finitely many intersections so as tohave disjoint neighbourhoods Uxi around each element xi in D−1(x) such that D, restricted toUxi , is a diffeomorphism. Then we just take the intersection
W =
N⋂i
D(Uxi) (3.44)
and in turn let D−1(W )∩ Uxi be the sheet containing xi. This makes D into a covering map ontothe image Ω = D(M).
However, the cardinalities need not be the same. We consider the sets
On =x ∈ Ω
∣∣ |D−1(x)| = n. (3.45)
Clearly, we have
Ω =
∞⋃n=1
On (3.46)
and in principle infinitely many On may be non-empty. We now want to show that the path-liftingproperty will still hold in this setting. We show both existence and uniqueness of such a lift1.
We suppose that γ : [0, 1]→ X is a curve and that γ1 and γ2 are two lifts with starting pointsγ1(0) = γ2(0) = x0. We define the set
I = t ∈ [0, 1] | γ1(t) = γ2(t). (3.47)
Since [0, 1] is connected and 0 ∈ I, so that I is non-empty, it is sufficient to show that I ⊂ [0, 1]is both open and closed.
We show that I is open by considering any element t0 ∈ I. We need to show that there existsan ε > 0 such that the interval (t0− ε, t0 + ε) ⊂ I. The developing map D is a local diffeomorphismand so there is some neighbourhood U around γ1(t0) = γ2(t0) such that D, restricted to U , is adiffeomorphism. Since γ is continuous there exists an ε > 0 such that γ((t0 − ε, t0 + ε)) ⊂ D(U).This shows that I is open.
To show that I is closed we consider the set
∆X = (x, x) ∈ X ×X |x ∈ X. (3.48)
The function γ1 × γ2 : [0, 1] × [0, 1] → M × M is continuous, and as ∆X is closed we have that(γ1 × γ2)−1(∆X) = I is also closed. Hence I = [0, 1) and by continuity of the lifts this extends toI = [0, 1]. We have now showed that γ1 = γ2 so that any lift is unique.
It remains to show that the lift exists. We define
t = sups ∈ [0, 1] | there exists a lift γ : [0, s]→ M. (3.49)
1The following arguments are essentially due to Arne Meurman.
36
Since D is a local homeomorphism we must have t > 0. By the preceding any such lift willnecessarily be unique. We suppose now that t < 1. Let tn ⊂ [0, 1] be a monotonically increasingsequence with tn → t. We define the curves γn : [0, tn]→ M as the unique lifts of γ : [0, tn]→ X.This defines a lift γ : [0, t)→ M and we define the sequence xn via xn = γ(tn).
Now either xn contains a convergent subsequence or not. If not, then by the proof of Lemma3.48 we may assume that there exists a sequence kn ⊂ D of distinct elements such that knxn =yn → y. But D(xn) = γ(tn) → γ(t) as γ is continuous. Hence xn must have a convergentsubsequence and by the local continuity of D we must have D(γ(t)) = γ(t). Again by localcontinuity D is homeomorphic on some U 3 γ(t) and there exists some ε > 0 such that γ may beextended to be defined on [0, t + ε]. Hence we are in contradiction to the assumption that t < 1and the path lifting exists.
We can now show the covering property. We suppose that some x ∈ On and for W as above,containing x, there is some element y ∈ Om, where m > n. Now on the one hand as by LemmaA.30 X is locally path-connected, we have that there is a path γ : [0, 1] → W with γ(0) = y andγ(1) = x. Let x denote the lift of x to some of the sheets Ui constructed above, with image W ,and similarly y. We let z be some elements not in a sheet but with D(z) = y. Now on the onehand, we have just deduced existence of a unique path-lift of γ that connects z and x. But on theother hand D is locally diffeomorphic onto W and so this lift must lie in the lifted sheet. This isa contradiction to the assumption that z does not. Hence y ∈ On and D is a covering onto itsimage.
It follows that when the developing map D is a covering map onto the image D(M) the numberof sheets equal Ker h. This, together with Proposition 2.36, gives the manifold diffeomorphism
M/Ker (h) ' Ω, Ω = D(M). (3.50)
The special case when X is simply connected restricts the possibilities even further.
Proposition 3.50. Let (M, g) be a (G,X)-manifold where X is simply connected. If D is acovering, then D is injective.
Proof. We assume that D(x) = D(y) = x. Since M is path-connected by Lemma A.28 there isa curve γ : [0, 1] → M such that γ(0) = x and γ(1) = y. This would mean that the curve D(γ)is a loop in X and as X is simpy connected, it is contractible to the trivial one-point curve. Wedenote the associated homotopy map by F (s, t). By Lemma A.21 there must exist a homotopymap G(s, t) in M with G(0, 0) = γ(0) = x and D(G) = F . Since G is not a loop we are incontradiction. Hence x = y.
An immediate consequence of this insight is that when D is a covering and X is simply con-nected, the group homomorphism h : D → G is an isomorphism onto its image h(D) = Γ. Withthis we get the following diffeomorphisms:
M ' M/D ' Ω/Γ, Ω = D(M). (3.51)
We invoke the following definition of completeness.
Definition 3.51. A (G,X)-manifold M is complete if the developing map D is a covering.
Proposition 3.52. Let (M, g) be a (G,X)-manifold, where X is a (geodesically) complete semi-Riemannian manifold. If M is complete in the sense of Definition 3.51, it is (geodesically) complete.
Proof. In light of Proposition 2.39, we need only to show that any geodesic γ : [0, b)→ M extendsbeyond b. But this follows since D is a diffeomorphism and X is (geodesically) complete, the imageD(γ) is defined for all parameters and so geodesic completeness of M follows.
37
38
Chapter 4
Curvature in Lorentzian spaces
This chapter is devoted to the understanding of constant curvature in Lorentzian spaces. Indeed,such an understanding is quite different from the Riemannian case, as will be shown. We beginby stating the well-known Gauss-Bonnet theorem for two-dimensional Lorentzian manifolds. Weproceed by proving the Theorem of Calabi and Markus and proceed by studying the Theorem ofKlingler.
4.1 The Lorentzian Gauss-Bonnet Theorem
In this section we prove the Gauss-Bonnet Theorem for Lorentzian surfaces. In order to do so, wemust make precise the notion of an angle. This was initially done in [3]. The same authors thenshowed the Gauss-Bonnet Theorem in the Lorentzian setting in [2].
We begin by defining the notion of orientability. Let Rn be equipped with two bases e1, . . . , enand e1, . . . , en. If there is a matrix A such that ei = Aij ej with det(A) > 0 these bases are saidto have the same orientation. If detA < 0 they are said to have opposite orientation. Itfollows that the set of bases with the same orientation form an equivalence class and that preciselytwo such equivalence classes exist. We denote by
λ = [e1, . . . , en] (4.1)
the orientation of a given basis.We take this definition to the manifold setting in the following:
Definition 4.1. Let (M, g) be an n-dimensional semi-Riemannian manifold and p any point inM . Suppose (Up, φp) is a local chart containing p. Then let λφp be an orientation to each basis inTpM , i.e.
λφp =
[∂
∂x1, . . . ,
∂
∂xn
]. (4.2)
An orientation of M is a mapping λ that assigns an orientation at each point p in M and issmooth in the sense that for each coordinate system λφp we have λ = λφp . A manifold with anorientation is said to be oriented.
With this definition, we may discuss the vector space geometry of the Lorentzian surfaces. Wewill denote the two-dimensional Minkowski space by M2, and assume it to be oriented. For anytimelike unit vector z in TpM2, we will always denote by z⊥ the unique spacelike unit vector thatboth satisfies η(z, z⊥) = 0 and is such that the ordered basis z, z⊥ has positive orientation.We will say that a coordinate system (x1, x2) in TpM2 is allowable if (0, 1) is a future-pointingtimelike unit vector and (0, 1)⊥ = (1, 0).
The concept of orientability extends to a more elaborate notion in Lorentzian geometry, namelythat of time-orientability. To make this notion precise, we invoke the following definition.
Definition 4.2. Let V be a Lorentzian vector space with scalar product g and consider a timelikevector v ∈ V . The set
C(v) = u ∈ V | g(u, v) < 0, and u is timelike (4.3)
39
is called the timecone containing v. The opposite timecone is denoted by C(−v) = −C(v).
It follows straightforwardly that if v and u lie in the same timecone, then C(v) = C(u). Withthis definition we may discuss the following.
Definition 4.3. Let τ be a function on M that for each point p in M assigns a timecone τp inTpM . τ is smooth if for each p in M there is a smooth vector field V on some neighbourhoodUp of p such that V (q) ∈ τq for all q ∈ Up. τ is called a time-orientation of M . If M admits atime-orientation, M is said to be time-orientable. If such an orientation is chosen, M is said tobe time-oriented.
Clearly this definition is very analogous to Definition 4.1. We state an immediate consequence.
Lemma 4.4. ([20], p. 145) A connected Lorentzian manifold M is time-orientable if and only ifthere exists a timelike vector field X ∈ C∞(TM).
Proof. ([20], p. 145) If such an X exists, it will for each point p assign a timecone τp. Since X isglobally defined and M is connected this gives the time-orientation.
We suppose now that M is time-oriented. From Definition 4.3 we know that every point phas a neighbourhood Up and a vector field V such that V |q lies in τq for all q ∈ Up. We now letfα |α ∈ A be a smooth partition of unity subordinate to the covering of neighbourhoods Up. Allfα are non-negative and each fα is contained in some Upα . Since timecones are convex any finitesum
∑α vα of timelike vectors vα must be timelike and contained within the same timecone. It
follows that the sum ∑α
fαVqα (4.4)
is a globally defined timelike vector field.
Depending on which timecone the vector field V lies in, V is called either future-pointing orpast-pointing in the natural sense.
With this preparation, we are ready to impose the following definition.
Definition 4.5. Let (M, g) be an oriented Lorentzian surface and let γ : I → M be a smoothcurve in M . Then
kg = g(∇γ γ, γ⊥) (4.5)
is called the geodesic curvature along γ.
One might say, that the geodesic curvature kg measures the deviation of γ from being a geodesic.For a geodesic, ∇γ γ = 0 and so kg = 0.
We can now state the Gauss-Bonnet theorem for a Riemannian manifold.
Theorem 4.6. (”The Gauss-Bonnet Theorem”) Let (M, g) be a Riemannian surface and D be adomain in M with piece-wise smooth boundary ∂D. Then∫
∂D
kgds+
∫∫D
KdA+∑i
θi = 2π (4.6)
where θi are the inner angles made by the ”polygon” ∂D.
We now wish to generalise this to the Lorentzian setting, following [2]. We make the followingdefinition.
Definition 4.7. Let (M, g) be a connected, oriented and time-oriented Lorentzian manifold andX,Y be two unit time-like future-pointing (or past-pointing) vectors at TpM . Then the angle αbetween X and Y is defined by(
cosh(α) sinh(α)sinh(α) cosh(α)
)(x1
x2
)=
(y1
y2
)(4.7)
where X = (x1, x2) and Y = (y1, y2) in any allowable coordinate system. We will use the notationα(X,Y ).
40
If Y is past-pointing time-like and X is future-pointing time-like, we make the following defi-nition. Observe that (−Y ) is future-pointing. Hence α(X,−Y ) is(
cosh(α) sinh(α)sinh(α) cosh(α)
)(x1
x2
)=
(−y1
−y2
)(4.8)
from which it follows that(−1 00 −1
)(cosh(−α) sinh(−α)sinh(−α) cosh(−α)
)(y1
y2
)=
(x1
x2
). (4.9)
We now ignore the (−E) matrix and define the angle as −α, i.e. α(X,Y ) = −α(X,−Y ).
Before stating any properties, we make the following remark.
Remark 4.8. The angle α is independent of the choice of the allowable coordinate system.
The angle has the following properties.
Lemma 4.9. [2] Let X and Y be unit time-like vectors on a M2. Then
1. α(X,X) = 0,
2. α(X,−X) = 0,
3. α(X,Y ) + α(Y,Z) = α(X,Z),
4. α(X,Y ) = −α(Y,X),
5. α(X,−Y ) = α(X,Y ).
Proof. We denote the matrix in Definition 4.7 by B(α). (1.) follows immediately, as the B(α) = Eif and only if α = 0.
If X is future-pointing, −X is past-pointing. From the definition we have
α(−X,X) = −α(−(−X), X) = −α(X,X) = 0. (4.10)
This shows (2.).Suppose first that X,Y and Z have the same time-orientation, say future-pointing. Denote the
angles by θXZ = α(X,Z), θXY = α(X,Y ) and θY Z = α(Y,Z). Then we have
Z = B(θY Z)Y = B(θY Z)B(θXY )X. (4.11)
The matrix product satisfies B(θY Z)B(θXY ) = B(θY Z + θXY ), which can be seen, using standardtrigonometric formulas:(
cosh(θY Z) sinh(θY Z)sinh(θY Z) cosh(θY Z)
)(cosh(θXY ) sinh(θXY )sinh(θXY ) cosh(θXY )
)=
(cosh(θY Z + θXY ) sinh(θY Z + θXY )sinh(θY Z + θXY ) cosh(θY Z + θXY )
).
(4.12)
But the product matrix can also be identified with B(θXZ) and so the assertion follows.Suppose that X is past-pointing, the others future-pointing. Then (4.), whose proof is inde-
pendent of this, may be used and the same argument follows mutatis mutandis. The same goes forthe case of two past-pointing vectors.
From the matrix identityB(α)B(β) = B(α+β) that we just showed, it follows that B(θXY )−1 =B(−θXY ). From this, in turn, it follows that since B(θXY )X = Y we have B(θXY )−1Y = X andthus B(−θXY )Y = X, which is (4.).
Using (3.) and (2.), we get α(X,−Y ) = α(X,Y )+α(Y,−Y )︸ ︷︷ ︸=0
= α(X,Y ). and so (5.) follows.
The following lemma will be useful.
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Lemma 4.10. Suppose X is a future-pointing timelike vector in the oriented manifold M2 thatmakes an angle α(X, z) with the timelike basis vector in some allowable coordinate system z, z⊥.Then
X = cosh(α)z − sinh(α)z⊥. (4.13)
Proof. We letX = c1z+c2z⊥ in the allowable coordinate system. ThatX is unit implies c22−c21 = 1.
The angle assumption implies (cosh(α) sinh(α)sinh(α) cosh(α)
)(c2c1
)=
(01
). (4.14)
In total, we have the three conditions
1 =c22 − c210 = cosh(α)c2 + sinh(α)c1
1 = sinh(α)c2 + cosh(α)c1,
(4.15)
from which it immediately follows that c1 = cosh(α) and c2 = − sinh(α).
With Example 4.11 we give an illustration of Lorentzian trigonometry.
Example 4.11. Let v1, v2 ∈ TpM2 be time-like and future-pointing. Then v1 +v2 is also time-likeand future-pointing. We consider the triangle spanned by these three vectors (v1 at the origin, v2
at the tip of v1 and −(v1 + v2) starts at the tip of v2 and takes us back to the origin). From (3.)and (4.) above, the following holds:
α(v1, v2) + α(v2, v1 + v2) + α(v1 + v2, v1) = α(v1, v2) + α(v2, v1)
= α(v1, v2)− α(v1, v2)
= 0.
(4.16)
This has an interesting geometric interpretation, as we try to illustrate in Figure 4.1.
The following lemma generalizes this example.
Lemma 4.12. Let D be a region in M2, bounded by a polygon ∂D of piece-wise time-like straightlines, denoted γi, i = 1, 2, . . . , k. Then
α(v1, v2) + · · ·+ α(vn−1, vn) + α(vn, v1) = 0. (4.17)
Proof. We use induction over n corners. We have already shown the case to be true for n = 3.Suppose the statement holds for polygons with n corners. We have∑
i
α = α(v1, v2) + · · ·+ α(vn−1, vn) + α(vn, vn+1) + α(vn+1, v1)
= α(v1, v2) + · · ·+ α(vn−1, vn) + α(vn, v1)︸ ︷︷ ︸=0
− α(vn, v1) + α(vn, vn+1) + α(vn+1, v1)
= −α(vn, v1) + α(vn, vn+1) + α(vn+1, v1)
= α(v1, vn) + α(vn, vn+1) + α(vn+1, v1)
= 0.
(4.18)
Here the sum of the first n terms vanishes by the induction hypothesis and the last three byExample 4.11. The assertion now follows by induction.
Lemma 4.13. [2] Let γ(t) be a smooth time-like curve on M2 and Z be a parallel future-pointingvector field on γ. Then we have
d
dtα(γ(t), Z(t)) = −kg. (4.19)
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Figure 4.1: The addition of two future pointing time-like vectors in M2. The angle α(v, w) isdenoted by a, α(w, v + w) is denoted by b and α(v + w, v) is denoted by c.
Proof. [2] We first observe that if Z is parallel then so is Z⊥. This follows from
0 =d
dtg(Z,Z⊥) = g(∇γZ︸ ︷︷ ︸
=0
, Z⊥) + g(Z,∇γZ⊥) = g(Z,∇γZ⊥) (4.20)
together with
0 =d
dtg(Z⊥, Z⊥) = 2g(Z⊥,∇γZ⊥). (4.21)
Therefore ∇γZ⊥ is orthogonal to both Z and Z⊥ and by non-degeneracy of the vector space wemust have ∇γZ⊥ = 0.
In the case of γ(t) being future-pointing, we have from Lemma 4.10 that
γ = cosh(α)Z − sinh(α)Z⊥, γ⊥ = − sinh(α)Z + cosh(α)Z⊥. (4.22)
If γ(t) is a past-pointing vector, −γ(t) is future-pointing, and we have from Lemma 4.10 that
γ = − cosh(α)Z + sinh(α)Z⊥, γ⊥ = sinh(α)Z − cosh(α)Z⊥. (4.23)
We differentiate the angle α with respect to the time, t. We get, in the future pointing case, that
∇γ γ = ∇γ(
cosh(α)Z − sinh(α)Z⊥)
= cosh(α)∇γZ + sinh(α)dα
dtZ
− sinh(α)∇γZ⊥ − cosh(α)dα
dtZ⊥
= sinh(α)dα
dtZ − cosh(α)
dα
dtZ⊥
(4.24)
43
and in the past-pointing case, by similar calculation, that
∇γ γ = − sinh(α)dα
dtZ + cosh(α)
dα
dtZ⊥. (4.25)
In either case, we get
kg = g(∇γ γ, γ⊥) =
(sinh2(α)− cosh2(α)
)dα
dt= −dα
dt(4.26)
which was the assertion.
Before stating and proving the Lorentzian Gauss-Bonnet Theorem, we need the followinglemma. As M2 is time-orientable there exists by Lemma 4.4 a unit, time-like and future pointingvector field X on M2. Hence, we can define the connection form:
ωX(V ) = g(∇VX,X⊥). (4.27)
Remark 4.14. Since we have the formula 0 = V g(X,X⊥) = g(∇VX,X⊥) + g(X,∇VX⊥), itfollows that ωX(V ) = −g(X,∇VX⊥).
Lemma 4.15. [2] The connection form satisfies
dω = κdA, (4.28)
where κ is the Gaussian curvature.
Proof. [2] Let V and W be arbitrary vector fields. Then
dωX(V,W ) = V ωX(W )−WωX(V )− ωX([V,W ]) =
= V g(∇WX,X⊥)−Wg(∇VX,X⊥)− g(∇[V,W ]X,X⊥)
= g(∇V∇WX,X⊥) + g(∇WX,∇VX⊥)− g(∇W∇VX,X⊥)
− g(∇VX,∇WX⊥)− g(∇[V,W ]X,X⊥)
= g(R(V,W )X,X⊥) + g(∇WX,∇VX⊥)− g(∇VX,∇WX⊥).
(4.29)
From the metric compatibility of the Levi-Civita connection, it follows that 0 = V g(X,X) =2g(∇VX,X) and so ∇VX is orthogonal to X. Similarly one gets that ∇VX⊥ is orthogonal to X⊥.Thus the last two terms vanish, and for V = X and W = X⊥, we get
dωX(X,X⊥) = g(R(X,X⊥)X,X⊥) = κ. (4.30)
The corresponding volume element satisfies dA(X,X⊥) = 1 and so the assertion follows.
We can now state and prove the famous Gauss-Bonnet Theorem in the Lorentzian setting.
Theorem 4.16. [2] Let (M, g) be an oriented Lorentzian surface. Let D be a domain in M withpiece-wise smooth boundary curve ∂D consising of time-like curve segments. Then∫
∂D
kgdt+∑i
θi −∫∫
D
KdA = 0. (4.31)
Proof. [2] Let Z be a unit and timelike vector field parallel along the boundary γ. As γ wasonly piece-wise smooth, we denote the partial smooth curves γi and the start- and endpointsγ(t0), γ(t1), . . . , γ(tk) for the k curve-segments, so that θi = α(γi(ti), γi+1(ti)). Again kg =g(γ(t), Z). Along the first curve γ1, we have by Lemma 4.13
−∫γ1
kgdt =
∫γ1
(dα
dt
)= α(γ1(t1), Z(t1))− α(γ1(t0), Z(t0)) (4.32)
44
so that
−∫γ1
kgdt− θ1 = α(γ1(t1), Z(t1))− α(γ1(t0), Z(t0))
− α(γ1(t1), γ2(t1))
= α(γ1(t1), Z(t1))− α(γ1(t0), Z(t0))
+ α(γ2(t1), γ1(t1))
= α(γ2(t1), Z(t1))− α(γ1(t0), Z(t0))
(4.33)
where we first used (4.) and then (3.) of Lemma 4.9. With the same argument, we get for thegeneral curve segment γi that
−∫γi
kgdt− θi = α(γi+1(ti), Z(ti))− α(γk(ti−1), Z(ti−1)) (4.34)
for the i : th curve. For the final segment, we identify θk = α(γk(tk), γ1(tk)) = α(γk(t0), γ1(t0)).We now sum up our k terms:∑
i
(−∫γi
kgdt− θi)
= −∫γ
kgdt−∑i
θi = α(Z(tk), Z(t0)), (4.35)
where we used the notation α(Z(t0), Z(tk)) for the angle between Z before and after the paralleltransportation along γ. Let X be some future-pointing time-like vector field defined on M2 anddenote α(Z,X) = β. Then Z can be decomposed, following Lemma 4.10, as
Z = cosh(β)X − sinh(β)X⊥. (4.36)
With this decomposition we get
0 = ∇γZ = ∇γ(
cosh(β)X − sinh(β)X⊥)
= sinh(β)dβ
dtX + cosh(β)∇γX
− cosh(β)dβ
dtX⊥ − sinh(β)∇γX⊥
= sinh(β)dβ
dtX + cosh(β) g(∇γX,X)︸ ︷︷ ︸
=0
X
+ cosh(β)g(∇γX,X⊥)X⊥ − cosh(β)dβ
dtX⊥
− sinh(β) g(∇γX⊥, X⊥)︸ ︷︷ ︸=0
X⊥ − sinh(β)g(∇γX⊥, X)X
= sinh(β)
(dβ
dt− ω(γ)
)X + cosh(β)
(ω(γ)− dβ
dt
)X⊥,
(4.37)
where the terms are zero as 0 = ddtg(X,X) = 2g(∇γX,X) and similar for X⊥. The above
calculation implies, as X and X⊥ are linearly independent, that ω(γ) = dβdt .
Now we can relate this to the angle:
α(Z(tk), Z(t0)) = α(Z(tk), X) + α(X,Z(t0))
= α(Z(tk), X)− α(Z(t0), X)
=
∫γ
(dβ
dt
)dt
=
∫γ
ω(γ)dt
=
∫∫D
dω
=
∫∫D
κdA,
(4.38)
45
where we use Greens theorem in the next-to-last line. This completes the proof.
Before stating Corollary 4.18, we need the following result.
Proposition 4.17. ([20], p. 149) Let M be a smooth manifold. Then the following is equivalent:
1. M admits a Lorentzian metric,
2. M admits a time-orientable metric,
3. there exists a non-vanishing vector field on M, and
4. either M is non-compact, or M is compact with Euler characteristic χ(M) = 0.
Proof. See [20], p. 149.
With this at hand, we are ready to state the following remarkable result.
Corollary 4.18. A compact, oriented Lorentzian surface with constant curvature κ must be flat.
Proof. From Proposition 4.17 it follows that the only oriented, connected and compact Lorentziansurfaces are tori, T2. On T2 we apply the standard procedure of dividing T2 into disjoint regionsbounded by time-like curves. The Gauss-Bonnet formula is applied to each of these. The curveintegrals will go in reverse directions and so must add up to zero. The angle sum must add up tozero by Lemma 4.12. What remains is the integral∫
T2
κ dA = 0 (4.39)
from which the result follows immediately.
4.2 The Calabi-Markus Theorem
In this section we prove Theorem 4.19 of Calabi and Markus. This is important when classifyingspherical Lorentzian space forms.
Let (Mn, g) be an n-dimensional Lorentzian space form with positive constant (sectional) curva-ture. Let furthermore Γ be a properly discontinuous group of isometries on Mn. From Proposition2.36 we know that the quotient Mn/Γ admits a manifold structure and the ”push-forward” met-ric on Mn/Γ preserves curvature, so that Mn/Γ has positive constant curvature as well. FromCorollary 2.41 we know that such a space form must be on the form
Mn ' dSn/Γ. (4.40)
The following theorem shows finiteness of Γ.
Theorem 4.19. ([5], ”The Calabi-Markus Theorem”) Let Γ be a properly discontinuous group ofisometries on dSn. Then Γ is finite.
Proof. We follow the discussions in ([5], p. 67), and ([20], p. 245).From Proposition 3.36 we know that I(dSn) = O1(n+ 1), which is a group of linear isometries.
Therefore Γ is also a group of linear isometries. We denote by Sn−1 the sphere
Sn−1 =
(x1, x2, . . . , xn+1) ∈Mn+1 | x1 = 0, x22 + · · ·+ x2
n+1 = 1. (4.41)
We assume Γ is infinite. Let p be a point on the unit sphere Sn−1, i.e. p = (0, x2, . . . , xn+1). Weclaim that there is a neighbourhood Up of p such that µ(U) ∩ Sn−1 6= ∅ only for finitely manyµ ∈ Γ. If not, then for every neighbourhood V that contains p, there would exist an infinitesequence µm such that µm(V ) ∩ Sn−1 6= ∅. We may assume the µm all to be distinct.
We now claim that it is possible to construct a sequence pm in Sn−1 that both converges top and for each m satisfies µm(pm) ∈ Sn−1. Let N be such that the open ball B1/N (p) satisfies (1.)of Definition 2.31. By hypothesis, we must for each m > N , have µm(B1/m(p)) ∩ Sn−1 6= ∅. Let
46
pm be an element in this intersection. Clearly pm → p and µm(pm) ⊂ Sn−1.Since Sn−1 is compact, the sequence µm(pm) must have a convergent subsequence, which we
for the sake of notational convenience assume to be the same. We denote the limit by q. Now wecan have either q ≡ p mod Γ or not. We show that in both cases this will contradict the properdiscontinuity of Γ.
We first suppose not. Then, by (2.) of Definition 2.31 there are open balls Bε(p) and Bε(q)that satisfy
µ(Bε(p)) ∩Bε(q) = ∅, µ ∈ Γ. (4.42)
The ball Bε(p) must contain some tail of pm. Similarly, the ball Bε(q) must contain some tailof the sequence µm(pm). But this is a contradiction to (2.), as then there is some N such thatboth pN ∈ Bε(q) and µN (pN ) ∈ Bε(q). Hence, we are contradicted.
Therefore, we assume that q ≡ p mod Γ, i.e. that µ(q) = p. By construction, we must haveµµm(pm)→ p. By (1.) of Definition 2.31 there is an open ball Bε(p) such that µ(Bε(p))∩Bε(p) =∅. Now, for some N , there must be pN ∈ Bε(p), but then on the other hand µ µN (pN ) ∈ Bε(p)as well and we are contradicted.
Therefore, we may assume existence of some neighbourhood U containing p, such that onlyfinitely many µ ∈ Γ satisfy µ(U) ∩ Sn−1 6= ∅.
We now want to demonstrate a consequence of the linearity of Γ. Let W be the vector subspacedefined by x1 = 0. Then Sn−1 ⊂ W . Since Γ is linear, the image µ(W ) must, for any µ ∈ Γ be avector subspace of the same dimension. By Lemma 2.7 we have
dim(W ∩ µ(W )
)+ dim
(W + µ(W )
)= dimW︸ ︷︷ ︸
n
+ dim µ(W )︸ ︷︷ ︸n
= 2n. (4.43)
Furthermore, µ(W ) must be spacelike. Since(W + µ(W )
)is at most the full ambient space, we
have dim(W +µ(W )
)≤ n+ 1 from which it follows that dim
(W ∩µ(W )
)≥ n− 1 > 0. Therefore,
any µ ∈ Γ must satisfy µ(Sn−1) ∩ Sn−1 6= ∅.The neighbourhoods Up form an open cover of Sn−1 which is a compact set. Therefore we have
Sn−1 ⊂N⋃i
Upi . (4.44)
For each Upi we have only finitely many µ1, . . . , µKi such that Upi intersects Sn−1 non-trivially.For each N such neighbourhoods we remove these finitely many elements. As Γ is infinite theremust remain elements in Γ that would satisfy µ(Sn−1) ∩ Sn−1 = ∅. This is a contradiction and soΓ must be finite.
Corollary 4.20. [5] A spherical relativistic space form Mn is not compact.
Proof. By Corollary 2.41, any spherical space form Mn must be on the form dSn/Γ. The spacedSn is not compact by Proposition 2.21. By Theorem 4.19, Γ is a finite group.
Denote by ϕ : dSn → M the covering map. Any point p ∈ Mn has an evenly coveredneighbourhood Up through ϕ, and the finiteness of D implies there are k disjoint sheets in ϕ−1(Up).
The sets⋃
p∈Mn
Up forms an open cover, and compactness implies Mn ⊂N⋃i
Upi . It follows that
Mn ⊂N⋃i
ϕ−1(Up), i.e. that Mn is contained within k · N open neighbourhoods. Since dSn is
not compact, there is a sequence xn ⊂ dSn with no convergent subsequence. Since dSn, byhypothesis, is supposed to be finitely covered, some sheet V must contain infinitely many pointsof xn. Since ϕ(V ) ⊂ Mn and Mn is compact, there is a convergent subsequence. We liftthis through ϕ−1 and as ϕ is a local diffeomorphism, this must be mapped onto a convergentsubsequence yn of xn, which is a contradiction and so Mn ' dSn/D cannot be compact.
4.3 The Klingler Theorem
In this section we prove Theorem 4.21 of Klingler. It shows that a compact Lorentzian manifoldwith constant sectional curvature must be (geodesically) complete:
47
Theorem 4.21. ([17], ”Klinglers Theorem”) Every compact Lorentzian manifold of constant cur-vature, with dimension n > 2, must be geodesically complete.
This will, together with Calabis and Markus’ Theorem 4.19 and its consequence, Corollary4.20, show our main classification result Corollary 4.50.
Since the proof of Theorem 4.21 is rather considerable, we discuss it in parts below. Throughout,(M, g) is assumed to be a smooth, compact Lorentzian manifold of constant (sectional) curvature.
For the respective cases κ = ±1, 0, we identify (M, g) with the spaces
κ = +1 : Mn = (O1(n+ 1), dSn)
κ = 0 : Mn = (O1(n),Mn)
κ = −1 : Mn = (O2(n+ 1), AdSn).
(4.45)
From Propositions 2.20, 2.21, 2.22, 2.23 and 2.24 we know that these spaces are (geodesically)complete spaces with the desired sectional curvatures. From the discussion in Section 3.3 we al-ready know that in the κ = 1 and κ = 0 cases, as X is simply connected, D is injective.
Before moving on to discuss the proof, we first describe the method of attack. By Theorem 3.49we know that D : M → X will be a covering map onto its image. The overall idea is then to showthat it is a surjection. As, in all three cases the space X is (geodesically) complete, this wouldimply (geodesic) completeness of M . Proposition 2.39 then would imply that (M, g) is complete.
We will assume that D is not surjective so as to derive a contradiction. If M were not complete,the image D(M) = Ω would be a strict subset of X. It is possible to equip M with a Riemannianmetric. This is explicitly done by equipping X, in each three cases, with the standard Euclideanmetric and pulling it back via D−1. Via this metric we would be able to discuss the metric com-pletion of M .
If D : M → X were not surjective, it can be shown that the boundary of D(M) in X mustidentify with a hyperplane. This is Corollary 4.46. This is shown more specifically by studyinghow the holonomy group Γ ⊂ G treats a sequence of Euclidean balls that converge. It will beshown that these ”collapse” into subsets of hyperplanes in X from which Corollary 4.46 is ratherdirect.
Having established that the ”new boundary” in the completion of M must identify with hyper-planes in X via D we arrive at a concrete problem in the well-known spaces Mn, dSn and AdSn.In Propositions 4.47, 4.48 and 4.49 it is shown that there can be at most one such hyperplanefor the flat, spherical and hyperbolic cases respectively. In the proof of Theorem 4.21 it is finallyshown that existence of one such component results in contradiction and so the metric completionof D(M) is empty. This is to say that D is surjective onto the (geodesically) complete space Xwhich in turn implies (geodesic) completeness and all is shown.
We start by making the following definition.
Definition 4.22. Let X be any of the space forms Mn, dSn or AdSn. For Mn we define asegment between two points x and y to be the (closed) geodesic (a straight line) that connectsthe two points. For dSn and AdSn we define a segment to be a geodesic connecting the two pointsx and y that does not contain any anti-podal points. If a segment connecting x and y exists, wesay that x sees y and it is denoted by [x, y].
For a fixed point x ∈ X, there is a subset of X consisting of the set of points connected to xby some segment. This stimulates for the following definition:
Definition 4.23. Let x be a point in X. The set of points seen by x is denoted Xx and is calledthe star of X.
From the definition of a segment in X, we get the following definition for a segment in M :
Definition 4.24. A segment in M is a curve γ such that identifies locally through D(γ) witha segment in X. As D is a local isometry via the pull-back metric, the segment in M is also ageodesic. If it exists, it is denoted by [p, q].
The stars Mp and Mp are defined in a similar manner. The following shows what the stars are,in the three (G,X)-manifolds:
48
Example 4.25. For an arbitrary point p, the following sets are the stars in the correspondingrelativistic space forms:
1. Mn+1x =Mn+1
2. dSnx = y ∈ dSn |Qn+11 (x, y) > −1
3. AdSnx = y ∈ AdSn |Qn+1−1 (x, y) > −1.
The first case is obvious, as the straight line between two points is a segment. The other two followfrom Propositions 2.22 and 2.24.
Corollary 4.26. The sets Xx, Mp and Mp are open.
Proof. For the case of Xx with κ = ±1, Example 4.25 shows that Xx is the inverse image of anopen set under the continuous functions Qn+1
1 (p, ·) and Qn+1−1 (p, ·) and thus is open. Since both
D : M → X and ϕ : M →M are continuous it follows that also Mp and Mp are open.
Definition 4.27. A subset C of M (or M) is said to be simply convex if for any two of itspoints p and q the segment [p, q] exists and is contained within C.
Definition 4.28. Suppose C1 and C2 are two subsets of X such that C1 ⊂ C2. Then C1 is saidto be relatively convex in C2 if, when two points x, y ∈ C1 are connected by a segment in C2,the segment is contained in C1.
A simply convex set must be relatively convex subset of any set containing it. Note however,that a relatively convex subset C1 ⊂ C2 of X, can still have two points seing one another but thesegment is outside of C2.
Lemma 4.29. Let x be a point in X. Then there exists a simply convex neighbourhood in X,containing x.
Proof. The assertion is, by Example 4.25, trivial in the case κ = 0. For the cases κ = ±1, weassume not and derive a contradiction.
We start by showing that there is some neighbourhood containing x where every point seesone another. To show this, we construct Euclidean balls B1/n(x) containing zn and yn that are
points in X that satisfy Qn+1±1 (zn, yn) ≤ −1. Clearly both zn and yn converge to x. But
Qn+1(x, x) = 1 as x ∈ X. This contradicts the continuity of Qn+1±1 .
Now let x = (0, 1, . . . , 0) in dSn or x = (1, 0, . . . , 0) in AdSn. By what we just deduced, theremust be some neighbourhood Ux where all elements see one another. If ε is small enough, theintersection Bε(x) ∩ X = Wx will both be contained in Ux and must contain all segments [y, z]where y, z ∈Wx.
Since G = O1(n + 1) or G = O2(n + 1) acts transitively and continuously on X, we may forany element y ∈ X find the element g ∈ G that maps gx = y and then g(Wx) is simply convexaround y.
The following two properties are immediate:
Lemma 4.30. Let D(x) = x. Then D(Mx) ⊂ Xx.
Proof. If x sees y there is a segment in Mx joining them. But this segment is the image ofD−1([x, y]) by definition, so there is a segment [x, y] joining x and y, (with x = D(x) and y = D(y))and so y ∈ Xx.
Lemma 4.31. D restricted to Mx is injective.
Proof. We assume that D restricted to Mx is not injective, i.e. that there are two points in Mx
such that D(y) = y, D(z) = z and y = z. Hence the segments [x, y] and [x, z] have the sameendpoints. From Propositions 2.23 and 2.21, it follows that the segments are unique. So since Dis a local diffeomorphism, it follows that z = y.
The following proposition is of great importance.
49
Proposition 4.32. [17] The image D(Mx) is relatively convex in Xx.
Proof. ([17]) We do proof by contradiction and suppose that so is not the case. The converse tothe statement of D(Mx) being relatively convex in Xx is that there is a segment in Xx, joiningtwo points y and z of D(Mx) not contained within D(Mx). We suppose that these exist and thatthe segment [y, z] is contained in Xx.
We now consider some of the geometry induced by x, y and z. The idea is to construct a”triangle” in Xx with end-points being x, y, z, which is done separately for the zero and non-zerocurvature cases.
For κ = 0, we consider the affine (or vectorial) plane S spanned by x, y and z. The triangle isdenoted by Txyz and is the set contained by the straight lines (segments) connecting x, y and z.Txyz is clearly well-defined, i.e. it is a triangle and not a straight line; if that were the case, then
the segment [y, z] would be contained in D(Mx), which we assumed not to be the case.For the case of κ = ±1, we define it as follows: we define S to be the intersection S =
X ∩ spanx, y, z. The span must be three-dimensional; if it were not, it would be a planespanx, y, z = Π. By Propositions 2.22 and 2.24 the intersection S would be a geodesic andas this geodesic contains x, y and z it would follows that y sees z. This is contrary to our hypothe-sis that [y, z] 6⊂ D(Mx). The constraint of being an intersection with X reduces the dimensionalityso that S is two-dimensional.
Given now what S is, we construct Txyz by considering the three planes spanned by the individ-ual segments and the origin. The segments are intersections of X with some plane that intersectsthe origin, and the three such planes will define a ”half-cone”. The planes, being pair-wise inter-secting in precisely one line (spanned between the origin and one of the three points) will producea cone. The intersection of the half of the cone containing x, y, z with X and the interior of thisclosed and bounded set will be Txyz.
It follows immediately that Txyz is simply convex: any two points u, v can be joined by asegment contained within; two points and the origin span a plane that must intersect two otherplanes and so the full geodesic containing the segment [u, v] must intersect the boundary of Txyzin two distinct segments. Hence Txyz is simply convex.
Another property of Txyz is that it contains no antipodal points. If it were so, there wouldbe a plane containing both the origin and u and −u. By the simple convexivity that we justshowed this segment [u,−u] will be contained in Txyz. On the other hand Qn±1(u,−u) = −1 andby the discussion of Example 4.25 these elements cannot see one another. Hence Txyz contains noanti-podal points.
We now consider the segment [y, z] in Xx. By hypothesis this was not included in D(Mx). SinceD(Mx) is open and contains y, there must be some half-open segment [y, v) in D(Mx). Moreover,the segment [x, v) will be contained in D(Mx).
Let now vn be a sequence in [x, v) that converges to v. We denote by vn the lift throughD of this sequence to M , which by Lemma 4.31 exists uniquely. A depiction of this is given inFigure 4.3. We make use of the compactness of M . Let un be the projected sequence in M ,i.e. un = ϕ(vn) for each n. By compactness of M , there is a convergent subsequence, which fornotational convenience, we assume is un. The sequence converges to some limit u and this pointhas an evenly covered neighbourhood Uu. We pick any of the sheets ϕ−1(Uu) and denote it byUu. We may assume that all the un are contained in the neighbourhood Uu, otherwise we discardthe first N elements and keep the ”tail”. Denote by wn the set ϕ−1(un) ∩ Uu. Since ϕ is a localdiffeomorphism, this sequence must converge, i.e. wn → w = ϕ−1(u) ∩ Uu. We now consider howthe sequences wn and vn are related. If, for each n, we were to have k(vn) = wn, where k ∈ D,it would follow that vn would be a convergent sequence. This in turn would mean that the geodesic[x, v) extends to [x, v]. But this contradicts the inclusion of v in D(Mx). Hence no k ∈ D mayrelate any wn to any vn for more than finitely many n. From this it follows that, as we can alwayspass to a subsequence, we may assume that there is kn(vn) = wn where all the kn ∈ D are distinct.
We now consider Euclidean balls in Rn+1 centered at wn = D(wn). We define the followingprojection
φ : x→
x if κ = 0,
x/√Qn+1±1 (x, x) if κ = ±1.
(4.46)
50
If X = dSn φ is, restricted to the spacelike part of Rn+1, clearly a projection onto dSn that mapsstraight lines to geodesics. If X = AdSn then φ is, restricted to the timelike part of Rn+1, aprojection onto AdSn. Furthermore, if κ = ±1, then we have
φ(g(x)) = g(x)/√Qn+1±1 (g(x), g(x)) = g(x)/
√Qn+1±1 (x, x) = g(φ(x)), (4.47)
where the G-invariance of Q follows as G = O1(n + 1) is a linear isometry. Therefore φ is G-equivariant for all three cases κ = 1, ±1.
We now choose a compact neighbourhood B, containing w in M , such that k(B) ∩ B = ∅ fornon-trivial k ∈ D. Such a ball will exist, as by (1.) in Definition 2.31 the neighbourhood U willcontain w. We take any coordinate chart and restrict to its intersection with U , map down to Rn,take a closed epsilon-ball centered at the image of w and contained in the image of the intersection,and map it back up. Clearly this definition of B will have the property. By local continuity of D,the image D(B) will also be compact in X.
We now choose r small enough, so that the projected image φ(B2r(w)) ⊂ D(B), where B2r(w)is the Euclidean ball of radius 2r centered at w in Rn+1. We let Bn = Br(wn). When n is bigenough, we have Bn ⊂ B2r(w) and therefore also φ(Bn) ⊂ φ(B2r(w)) ⊂ D(B). Denote furtherBn = B ∩D−1(φ(Bn)) and Cn = k−1
n (Bn). Clearly the Cn are compact and by Lemma 2.32 theyare also disjoint.
We define gn = h(k−1n ). With this, it follows from the equivariance equation D k = h(k) D
that
vn = D(vn) = D(k−1n (wn)) = h(k−1
n )(D(wn)) = h(k−1n )(wn) = gn(wn) (4.48)
and
D(Cn) = D(k−1n (Bn)) = h(k−1
n )(D(Bn)) = gn(φ(Bn)) = φ(gn(Bn)), (4.49)
where we used the G-equivariance of φ in the last line.We now define three sequences in G. The sequences an, bn and cn are defined so that
wn = a−1n (w), vn = bn(v) and cn = bng, (4.50)
where g ∈ G is an element such that v = g(w). With the equality wn = gn(vn) we get the successivecomputations
vn = gn(wn) = gna−1n (w)
v = b−1n (vn) = b−1
n gna−1n (w)
w = g−1(v) = g−1b−1n gna
−1n (w) = c−1
n gna−1n (w)
(4.51)
so that c−1n gna
−1n ∈ Kw ' O1(n) (or Kw ' O2(n)). Thus gn = cnonan, where on ∈ Kw.
Figure 4.2: Depiction of the triangle Txyz and an element of the convergent sequence vn. The
boldface denotes the intersection Txyz ∩D(Mx). The picture is taken from [17].
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We now consider the sequence of closed sets gn(Bn). As Bn is a Euclidean ball and gn ∈O1(n + 1), Proposition 3.35 shows that gn(Bn) is an ellipsoid for each n. Furthermore, as φ isG-invariant, the Cn are disjoint and vn → v, it must follow that the φ(gn(Bn)) are disjoint.The φ(gnBn) are disjoint and so it follows that the gn(Bn) are disjoint and so the limit setgn(Bn)→ E must be degenerate. By Corollary 3.42 we have that E is a coisotropic hyperplane ofcodimension 1. Clearly φ(gn(Bn))→ φ(E) and since the φ(gnBn) are disjoint in spanx, y, z itfollows that φ(E) ∩ spanx, y, z must be two-dimensional.
We consider now the the intersections of gn(Bn) and Txyz. Since S is a surface, the tangentspaces TvnS must be affine planes in Rn+1, more specifically in spanx, y, z. Since the anand the cn lie in compact subsets, there is a real number r′ such that r > r′ > 0 and r′ issmaller than every principal axis but one. Therefore, it follows that each gnBn must contain someaffine segment σn of length 2r′, centered at vn. We define δn = φ(σn). Clearly, as the σn are allstraight lines the projected images δn must be contained in vectorial planes and by Propositions2.20 , 2.22 and 2.24 therefore be geodesics. Since φ maps spanx, y, z to itself the geodesics δnmust be contained in S.
Let us decompose σn = vn+σ0n. Then the closed ball B2r′(0) will contain all segments σ0
n. Sinceσn ∈ gn(Bn) ∩ Txyz, the segments σ0
n must converge to some segment σ0 that in turn is containedin some limit hyperplane E ′. As furthermore vn → v, we must have σn = vn + σ0
n → v + σ0 = σ,which is an affine segment of length 2r′ that is centered at v and contained in TvS. We let δ = φ(σ)and clearly δn → δ.
Since every δn intersects Txyz, the limit segment δ must intersect Txyz. This can happen intwo distinct ways; either δ ∩ [y, v) 6= ∅, or δ ∩ int(Txyz) 6= ∅. At any case, let s be an element inthis intersection. By Lemma 3.39 there exists a sequence sn of elements converging to s, andwith sn ∈ δn. We may (uniquely) lift the sequence sn to Mx, and we denote this sequence sn.Since also s ∈ D(Mx) there is a (unique) corresponding lifted point s ∈ Mx ∩D−1(s) and sn → s.
Each element in the sequence sn is contained in Cn. By construction, Cn = k−1n (Bn) and
so kn(sn) ∈ Bn for each n. Since Bn ⊂ B we have kn(sn) ⊂ B. As B was compact, there is aconvergent subsequence in kn(sn) and the limit t is contained in B.
We have now reached a contradiction to the proper discontinuity of D; there must on the onehand be some element k ∈ D such that k(s) = t. Since k is a diffeomorphism, it maps convergentsequences to convergent sequences, and so k(sn)→ t. But on the other hand, this cannot happenas by construction k(B) ∩ B = ∅ and so we have a contradiction.
Hence, D(Mx) is relatively convex in Xx.
Figure 4.3: Depiction of the triangle Txyz and the ellipsoids, containing the segments δn con-
verging along the convergent sequence vn. The boldface denotes the intersection Txyz ∩D(Mx).The picture is taken from [17].
Before stating the important consequence of Proposition 4.32, we state and prove Lemma 4.33:
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Lemma 4.33. [17] Let C1 and C2 be two subsets of M and C be a subset of X such that D(C1) ⊂ Cand D(C2) ⊂ C. Suppose
1. D restricted to C1 is injective,
2. D(C1) ⊂ C is relatively convex,
3. C2 is simply convex,
4. C2 ⊂ D−1(C),
5. C1 ∩ C2 6= ∅.
Then D, restricted to C1 ∪ C2 is injective.
Proof. [17] By (1.) D is injective on C1. We show that it follows from (3.) that D restricted toC2 is also injective. If any two points x and y in C2 can be joined by a segment [x, y] included inC2 the segment D([x, y]) = [x, y] exists in X. If x = y the corresponding segment [x, y] is triviallyone point and therefore also the lift D−1([x, y]) = [x, y]. Hence we have injectivity also on C2.
We now suppose x ∈ C1 and y ∈ C2 such that D(x) = D(y) and that neither lies in theintersection C1 ∩ C2. Since by (5.) the intersection C1 ∩ C2 is not empty, there exists z ∈ C1 ∩ C2.Since C2 is simply convex, there exists a segment [y, z] ⊂ C2, and since by (4.) C2 ⊂ D−1(C) themapped segment satisfies [x, z] ⊂ C. By the hypothesis that x = y, the segment [x, z] exists andmust coincide with [y, z] as segments are unique. Now by (2.) D(C1) is relatively convex, so that[x, z] ⊂ D(C1). Hence D−1([x, z]) is a segment in C1 and z sees both x and y. This means thatboth x and y lie in Mz. By Lemma 4.31 it follows that x = y. Hence D restricted to the unionC1 ∩ C2 is injective.
Corollary 4.34. [17] Denote by ∂Mx the boundary of the star of x in M , and similarly ∂Xx =Xx −Xx. Then D(∂Mx) ⊂ ∂Xx.
Proof. [17] Let y ∈ ∂Mx and let By be a simply convex open neighbourhood of y in M that must
exist by Lemma 4.29. The point y = D(y) is in the closure of Xx as D(Mx) ⊂ Xx and D iscontinuous. We need to show that y is not in the interior of Xx, so we assume the opposite, i.e.that y ∈ Xx. Since Xx is open by Corollary 4.26 we may assume D(By) ⊂ Xx.
Now By ∩ Mx 6= ∅, By is simply convex and by Proposition 4.32 D(Mx) is relatively convex in
Xx. Lemma 4.33 implies that D restricted to the union Mx ∪ By is injective.Since we assumed that y ∈ Xx the segment [x, y] must exist inXx. Therefore we may lift [x, y] to
[x, y] in M and thus y ∈ Mx, i.e. x sees y and we have a contradiction. Hence D(∂Mx) ⊂ ∂Xx.
We now equip X with a Riemannian metric as follows. If X =Mn, we simply change metricfrom η, the Minkowski metric, to the Euclidean metric g0. If X = dSn, or X = AdSn, we equip itsambient space Rn+1 with the standard Euclidean metric and let the Riemannian metric g0 on Xbe the induced metric. With this (X, g0) becomes a Riemannian metric space in all three κ = 0,±1cases. Furthermore (X, g0) will be complete as a metric space. This follows trivially in the flatcase, as then (X, g0) = (Rn, dE) which is just the standard Euclidean space. In the curved cases(X, g0) is merely a closed metric subspace of a complete space and will thus also be metricallycomplete. (3.) from Hopf’s and Rinow’s Theorem 1.5 now implies that any two points x, y ∈ Xmay be connected with a geodesic γ such that `(γ) = d0(x, y).
By pullback, we get a Riemannian metric g0 = D−1(g0) on M . We can thereby discuss themetric completion of M , which we denote by M . We make the following definitions:
Definition 4.35. Let x ∈ Mx and y ∈ Mx and suppose γ : [0, 1] → M is some curve withendpoints x and y. γ is called a segment if γ([0, 1)) ⊂ Mx and D(γ([0, 1])) is a segment in X. Asbefore, if it exists, it is denoted by [x, y].
The definition of a star generalizes in the natural manner, with notation Mx. The following,however, is new:
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Definition 4.36. Let N be a subset of M . By ∂N we denote the boundary of N in M , and it isthe disjoint union of the sets
∂N = ∂N ∩ (M − M), ∂N = ∂N ∩ M. (4.52)
We state some properties of this construction with the following lemmas.
Lemma 4.37. [17] The stars Mx in the completion M are the disjoint union
Mx = Mx
⋃(∂Mx ∩D−1(Xx)
). (4.53)
Proof. [17] We show both inclusions. On the one hand, we have the obvious inclusion Mx ⊂ Mx.
We suppose y ∈ Mx, but y 6∈ Mx. Then there must be y ∈ ∂Mx. From the definition there mustbe a segment [x, y] in Xx such that its lift through D is [x, y) ⊂ Mx. Therefore y ∈ D−1(Xx).
Hence we have the inclusion Mx ⊂ Mx
⋃(∂Mx ∩D−1(Xx)).
To show the other inclusion, we must show ∂Mx ∩ D−1(Xx) ⊂ Mx. We suppose y ∈ ∂Mx ∩D−1(Xx) and let yn be a sequence in Mx converging to y. Since the points yn are seen by x thereare associated segments [x, yn] ⊂ Xx. Since yn → y, and by hypothesis y ∈ Xx, these segmentsconverge to [x, y] ⊂ Xx. By Proposition 4.32, D(Mx) is relatively convex in Xx. Hence, if z ∈ [x, y]is an interior point in the segment, the segment [x, z] ⊂ D(Mx). Therefore the half-open segmentmust satisfy [x, y) ⊂ Mx.
This shows the assertion.
Lemma 4.38. D restricted to Mx is injective.
Proof. The proof is much similar to Proposition 4.31. We suppose D(y) = D(z) for y, z ∈ Mx.Then there exist segments in M that join the points: [x, y] and [x, z]. By construction, we have thatthe segments [x, y] and [x, z] exist in X, (with usual notation D(x) = x, D(y) = y and D(z) = z).In X we have that, as D(y) = D(z), the segments [x, z] and [x, y] are the same. Since, by Lemma4.31 we have D is injective restricted to Mx and so the segments [x, z) and [x, y) are the same,and by the metric completion [x, z] = [x, y] from which it follows that z = y.
Lemma 4.39. [17] The stars Mx are open sets.
Proof. [17] From Corollary 4.26 and Proposition 4.37, it suffices to show that if v ∈ ∂Mx∩D−1(Xx)then v is contained in some open ball in Mx.
Let V1 and V2 be two open balls in M , centered at v, and with radii r1 = ε and r2 = 3ε,respectively. The distance is that of the Riemannian metric d induced by the pull-back metricD−1(g0) and ε is chosen so that the ball V2 is contained within the open set D−1(Xx). Since
v ∈ ∂M , the segment [x, v) lies in Mx. The ball V1 intersects the segment [x, v) with some pointv0.
Now for each element y in V1 we construct a curve βy, lying in the intersection V2 ∩ M , such
that βy(0) = v0 and βy(1) = y. To do this, we let yn be a sequence in M such that yn → y. Wemay assume that both yn and d(yn, yn+1) < ε/2n so that the sum
`(β) =
∞∑n
d(yn, yn+1) <
∞∑n
ε
2n= ε <∞ (4.54)
is convergent. The latter statement is easy to see: let zn be any sequence tending towards y.We first pick N1 > 0 so that d(zn, zm) < ε/2 for all m,n > N1 and let y1 = zN1
. We proceedby choosing N2 so that d(zn, zm) < ε/22 for all m,n > N2 and let y2 = zN2 . Proceeding in thisfashion we choose Nk such that d(zn, zm) < ε/2k for m,n > Nk and let yk = zNk . Clearly thisprocedure is well-defined and we have d(yn, yn+1) < ε/2n for all n.
With this at hand, we construct a curve β, contained in V1 such that it has end-points y1 andy. This we do by pair-wise path-connecting the points in yn and then define β(1) = y. We nowclaim that the successive pair of elements yn and yn+1 may in fact be geodesically connected. Welet U be an open neighbourhood of D(y) such that D, restricted to U ∩D(M), is a local isometry.
54
We may assume that the full sequence yn is contained in D−1(U ∩D(M)). By the Hopf-RinowTheorem there must exist a geodesic joining D(yn) to D(yn+1) and D is isometric between thesetwo points we may by continuity lift this geodesic to M . Geodesics can be parametrized by arc-length and so we have already a parametrization for β. Clearly it follows that `(β) < ε.
Since furthermored(y1, v0) ≤ d(y1, v) + d(v, v0) < ε+ ε = 2ε (4.55)
we may construct a curve δ with δ(0) = v0 and δ(1) = y1. Clearly, as both v0, y0 ∈ V1 we haveEuclidean length `(δ) < 2ε. Now the product β δ suffices as βy. Clearly we have βy([0, 1)) ⊂ M .
We consider the product αy = βy [x, v] that for each y ∈ V1 defines a mapping αy : [0, 1]→ M .
Now we have D(αy) ⊂ Xx; this follows as [x, v0] is already in Mx and β and δ are both contained
within V2 that, in turn, was chosen so that D(V2) ⊂ Xx. Clearly we have αy([0, 1)) ⊂ M and we
need only to verify that αy([0, 1)) ⊂ Mx.
By construction we have αy([0, 1)) ⊂ M . Therefore, if the element αy(t0) 6∈ Mx for some
t0 ∈ [0, 1) then αy(t0) ∈ ∂Mx. This is a contradiction to Corollary 4.34. Hence V1 ⊂ Mx so that
Mx is open.
Lemma 4.40. [17] The stars Mx form an open cover of M i.e. we have
M =⋃x∈M
Mx. (4.56)
Proof. [17] We need to show, that every v ∈ M is seen by some x ∈ M . Again, we define V1 andV2 to be two open balls of radius r1 = ε and r2 = 3ε, such that they are centered at v. Again,the metric is induced from the pull-back D−1(g0). ε is chosen, such that the open image D(V2) issimply convex in Xx. This implies, that for every y ∈ V2 we have D(V2) ⊂ Xy, where y = D(y).
Now, we take some element x from the intersection V1 ∩ M and construct a curve α : [0, 1] → V2
with endpoint x and v. Again, it must follow that α([0, 1)) ⊂ Mx as otherwise we would have acontradiction to Proposition 4.34. Hence the assertion follows.
In the two curved Lorentzian space forms in our discussion, we will define a hyperplane. Inorder to do so we first need the following insight.
Lemma 4.41. Let H be a coisotropoic vectorial hyperplane in Rn+1 and n > 2. Then
1. dimH⊥ = 1,
2. dSn ∩H is path-connected, and
3. AdSn ∩H consists of two path connected components.
Proof. Statement (1.) follows immediately from Lemma 2.7, as dimH = n− 1.We now show (2.). We may always use the O(n)-symmetry to rotate the spatial vectors so that
the null vector of H lies in the (x1, x2)-plane. That x lies in the intersection dSn ∩H is equivalentto it satisfying the equation system
1 = −x21 + x2
2 + . . . x2n+1
x1 = x2
⇒ 1 = x2
3 + · · ·+ x2n+1, (4.57)
from which it follows that the intersection is path-connected.For the intersection AdSn ∩H, we may use the rotational symmetry so that H is given by the
equation x2 = x3. Hence, we have
1 = x21 + x2
2 − (x23 + · · ·+ x2
n+1)
x2 = x3
⇒ 1 + x2
4 + · · ·+ x2n+1 = x2
1. (4.58)
Here we see that the intersection AdSn ∩ H will have two connected components - one for eachsign of x1.
55
With Lemma 4.41 at hand, we may state the following.
Definition 4.42. Consider dSn and AdSn as subsets of Rn+1. In dSn a coisotropic hyperplaneis a connected component of the intersection dSn ∩H with some coisotropic vectorial hyperplaneH in Rn+1. In AdSn we call one of the two connected components in AdSn ∩ H a coisotropichalf-hyperplane.
Clearly, a coisotropic hyperplane H in Mn+1 must contain one null vector but no timelikevectors in the Mn and dSn cases, and two null orthogonal vectors but no timelike vectors in theAdSn case. For convenience with our next result, we first make the following definition.
Definition 4.43. A subset C ⊂ M is called totally geodesically coisotropic if D(C) identifieslocally with a coisotropic hyperplane in X.
We state the following important result:
Proposition 4.44. [17] The boundary ∂M is totally geodesically coisotropic, or empty.
Proof. [17] Let v be an element in ∂M . By Lemma 4.40 and its proof, we know that there is anelement x that sees v and some neighbourhood V , centered at v such that its image D(V ) is simplyconvex. We denote the intersection V ∩ M by V .
Let α : [0, 1]→ M be the parametrization of the segment [x, v]. By construction, α([0, 1)) ⊂ M .We let tn be a sequence in [0, 1) tending towards 1. The corresponding sequence α(tn) tendsto v. We denote α(tn) by vn. From the proof of Proposition 4.32 we have existence of a sequence ofdistinct elements kn ⊂ D such that the sequence wn = kn(vn) converges to w in M . By thesame procedure as in the proof of Proposition 4.32, we construct the compact balls Bn, centeredat wn and Cn = k−1
n (Bn), centered at vn. The Cn are compact and as they are all subsets ofMx they are disjoint.
The balls Cn are contained within V and thus the intersection V ∩ Cn is closed in Mx. Thesequence of sets is identified with φ(gn(Bn)) ∩D(V ) in X, where Bn is the Euclidean closed ballof radius r, centered at wn in Rn+1. The sequence vn converges to v, and so we must haveconvergence of φ(gn(Bn)) ∩D(V ) to some limit ellipsoid E , centered at v. Furthermore, as in theproof of Proposition 4.32, we may identify gn = cnonan, where on ∈ Kw ' O1(n) and an andcn are contained in compact subsets of G. Since the φ(gn(Bn)) are disjoint, the limit must bedegenerate. By Corollary 3.42 it follows that E must have codimension 1 and thus be contained ina coisotropic hyperplane H.
By this it follows that φ(gn(Bn)) ∩D(V ) converges, and the limit E has φ(E) ⊂ D(V ).We need to show that φ(E) ⊂ D(V − V ), i.e. that D(V ) ∩ φ(E) = ∅. Let u be any point in
φ(E). From the definition of convergence, we know that there is a sequence un in X such thatun → u and for each n we have un ∈ gn(Bn). Since the D−1(φ(gn(Bn))) ⊂ Mx we know that thereis a uniquely lifted sequence un in Mx and that for each n we have un ∈ Cn.
Now, we must on the one hand have kn(un) ⊂ B for each n. This means in particular, as B iscompact, that there must be a convergent subsequence of kn(un) that converges to t in B.
On the other hand, the sequences un and kn(un) are projected down to M to the samesequence, and so if un → u, we must have k(u) = t. This means that u ∈ t(B) but the k(B) areall disjoint and so there can be no convergence of un within M .
By our deduction it now follows that every element in φ(E) is a limit point to D(M), which iscontained in some coisotropic hyperplane H in Rn+1.
In the case κ = 0, this is already the assertion. In the case of κ = ±1, we need to show thatthe hyperplane H is vectorial. In order to show this, we consider the metric boundary V − V inM . Clearly, this set has no interior point. Since D is a local diffeomorphism it must follow thatD(V − V ) has no interior point either. But D(V − V ) = φ(E) and if E were not vectorial, theprojected image φ(E) would have an interior point. Hence H must be vectorial and so X ∩H is acoisotropic hyperplane.
In the space M we make the following definitions.
Definition 4.45. In M we define a hyperplane H to be a connected subset of M such thatD(H) is a hyperplane in X.
56
We get the following result.
Corollary 4.46. [17] If the boundary ∂M is non-empty, then every connected component is ahyperplane in M .
Proof. This assertion follows, as from Proposition 2.29 we know that every connected componentin ∂M is an n−1-dimensional manifold without boundary. But on the other hand we know that the
image D(M) = D(M) is closed and from this the only remaining possibility is that the connectedcomponents must identify with hyperplanes.
With this in hand, we formulate a very important result. For convenience, we state and proveit for the three cases κ = 0,±1 respectively. For the sake of simplicity, we start with the simplestcase, namely that of κ = 0.
Proposition 4.47. [17] Suppose D : M → Mn is not surjective. Then D(M) will be identifiedwith one of the two connected components in Mn −H, for some coisotropic hyperplane H.
Proof. [17] By Corollary 4.46 the boundary ∂M must contain a hyperplane H. We denote the
image D(H) = H the hyperplane in Mn. Let h be a point in H. For each y ∈ H we define, inMn, the curve
αy(t) = y + t(x− h), (4.59)
where y = D(y), h = D(h) and x = D(x). Since these curves are straight lines in Mn theyare geodesics and may be lifted through D to segments in M . The curves may be defined for allt ∈ [0,∞) or not. If not, they must terminate at some hyperplane Hy. Clearly Hy is distinct fromH. This happens if and only if they are parallel.
In this manner, if the geodesic αy is not defined for all positive t, the points y are all mappedto some plane Hy. This defines a mapping ϕ : y → Hy, were we take Hy to be the empty set ∅ ifthe curve αy is defined for all positive t. Since this is locally constant and the hyperplanes are allconnected components, the mapping ϕ is globally constant, i.e. ϕ(y) ≡ Hy.
Since M is connected and D is a local homeomorphism it follows that D(M) = Ω is the enclosedinterior if Hy is not empty and the open affine half-space in Mn if Hy is empty. Theorem 2 in [9]
p. 771 asserts that D(M) cannot lie between two hyperplanes and hence the former possibility isruled out. The assertion follows.
Proposition 4.48. [17] Suppose D : M → dSn is not surjective. Then D(M) will be identifiedwith one of the two connected components in dSn −H, for some coisotropic hyperplane H.
Proof. [17] We suppose D : M → dSn is not surjectiv and so the boundary ∂M is non-empty,
containing some element h. Again, by Corollary 4.46 the connected components of ∂M must behyperplanes, and we denote D(H) = H. From Lemma 4.41 we know that there is an orthonormalbasis e1, . . . , en+1 were H ⊥ e1 and in these coordinates the metric is represented by
B(x, y) = x1y2 + y1x2 + x3y3 + . . .+ xn+1yn+1. (4.60)
Again, by Lemma 4.37, there is some element x ∈ M that sees h. We let e1 be such that connectedcomponent Ω of dSn − H containing x = D(x) identifies with the set of points y such thatB(e1, y) > 0.
We define the curve
γy(t) =
sinh(t)B(y,e2)e2 + exp(−t)y, if B(y, e2) 6= 0 and
y + te2, if B(y, e2) = 0.(4.61)
57
If B(y, e2) 6= 0, we have
B(γy, γy) = B
(sinh(t)
B(y, e2)e2 + exp(−t)y, sinh(t)
B(y, e2)e2 + exp(−t)y
)=
(sinh(t)
B(y, e2)
)2
B(e2, e2) + 2 exp(−t) sinh(t)
B(y, e2)B(e2, y)
+ exp2(−t)B(y, y)
= 2 exp(−t) sinh(t) + exp(−2t)
= 1,
(4.62)
and if B(e2, y) = 0 we have
B(γy, γy) = B(y + te2, y + te2)
= B(y, y) + 2tB(y, e2) + t2B(e2, e2)
= 1
(4.63)
so that in both cases γy ⊂ dSn.Furthermore γ is a geodesic: from Proposition 2.22 we know that any geodesic is on the form
dSn ∩Π, where Π is a vector sub-plane. Clearly γy ⊂ spane2, y. Since, in the case B(e2, y) 6= 0,we have γy = γy it follows that γy is a correctly parametrized geodesic. If B(e2, y) = 0 it isimmediately obvious that γy is a straight line and so since Π is a degenerate plane with indexν = 0 it is also a geodesic.
We now consider the image of the mapping ϕ : y → H, just as in the proof of Proposition 4.47;for each fixed y we have that if γy is not defined for all t, it must hit the boundary ∂M , which byProposition 4.46 identifies with a hyperplane Hy. (Otherwise, just as before, we take the image tobe the empty set). We again denote this mapping ϕ : y → Hy. Since ϕ is locally constant and thehyperplanes are connected components, it is constant, i.e. ϕ ≡ Hy.
If Hy is non-empty it is a full coisotropic hyperplane in dSn. But by construction a hyperplanein dSn is the intersection H ∩ dSn, for a vectorial coisotropic hyperplane H in Mn+1. Two suchhyperplanes have the origin in common and their intersection must be a vectorial hyperplane ofdimension n − 1. The intersection will obviously intersect dSn and hence H ∩ Hy 6= ∅. Thiscontradicts Proposition 4.46 and therefore Hy must be empty.
We need only to verify that D(M) is surjective onto Ω. To do this we consider the curveγ(t) = exp(t)e1 + 1
2 exp(t)e2. We have B(γ(t), γ(t)) = 2 exp(t) 12 exp(t) = 1, so that γ ⊂ dSn.
Furthermore, γ is also on the form dSn ∩ Π, where Π = spane1, e2 is a Lorentzian plane. ByProposition 2.22 its image is a geodesic. Clearly we also have ¨γ = γ, and by the proof in Proposition2.22 we then know that γ ∈ NγdSn so that γ has the correct parametrization. Clearly, if y ∈ His a boundary element the corresponding curve γy must be defined for all parameter values t > 0.
The image of all γy is Ω − γ. Since D(M) is closed and D is continuous it follows that γ mustnecessarily also be contained in the image Ω. Hence the assertion must follow.
Proposition 4.49. [17] Suppose D : M → AdSn is not surjective. Then D(M) will be identifiedwith one of the two connected components in AdSn −H, for some coisotropic hyperplane H.
Proof. [17] Since AdSn is not simply connected, we consider its universal covering space, which is
AdSn = Rn. We denote the covering map ϕ : Rn → AdSn and it is given explicitly
ϕ(θ, x1, . . . , xn−1) = (√x2 + 1 cos(θ),
√x2 + 1 sin(θ), x1, . . . , xn−1), (4.64)
where we use x2 =∑n−1k x2
k. By Theorem 3.45 we have that the following diagram commutes:
M
ϕ
D
""
D // AdSn
ϕ
M AdSn
(4.65)
58
i.e. there is ϕ D = D. D will be a covering map so that by Proposition 3.50 it will be an isometryand hence we have the diffeomorphisms M/D ' Ω. We denote by Γ the image h(D), i.e. the
holonomy group on AdSn, and Γ the holonomy group on AdSn.Let H be a hyperplane in AdSn and e = (f1, f2, e1, . . . , en−1) be a vector orthogonal to it,
(where f1 =√e2 cos(θe) and f2 =
√e2 sin(θe)). We solve for an equation for ϕ−1(H): suppose
ϕ(θ, x) ∈ H. Then
0 = Qn+1−1 (e, x) = f1
√x2 + 1 sin(θ) + f2
√x2 + 1 cos(θ)− (e1x1 + · · ·+ en−1xn−1)
=√e2 cos(θe)
√x2 + 1 cos(θ) +
√e2 sin(θe)
√x2 + 1 sin(θ)− 〈e, x〉
=√x2 + 1
√e2 cos(θ − θe)− 〈e, x〉
⇒ cos(θ − θe) =〈e, x〉
√x2 + 1
√e2,
(4.66)
where we defined 〈e, x〉 =∑n−1i eixi.
The lift ϕ−1(H) will consist of disjoint connected components, which we denote
Hk = (θ, x) ∈ ϕ−1(H) | (θ − θe) ∈ (kπ, (k + 1)π), k ∈ Z. (4.67)
These subsets are called coisotropic half-hyperplanes. From the definition it follows that the
disunion AdSn − Hk consists of two disjoint open sets; one contains Hk+1 and the other Hk−1.The set Hk is furthermore diffeomorphic to Rn−1. To show this we observe first that as 〈x, e〉
is an n− 1-dimensional Euclidean scalar product we have
−√x2√e2 ≤ 〈x, e〉 ≤
√x2√e2, (4.68)
from which it follows that −1 < cos(θ − θe) < 1. Hence we may define the following maps:
ψ2k(x1, . . . , xn−1) =
(θ =
(arccos
( 〈e, x〉√x2 + 1
√e2
)+ θe
)+ 2k · π, x1, . . . , xn−1
)ψ2k+1(x1, . . . , xn−1) =
(2π −
(θ = arccos
( 〈e, x〉√x2 + 1
√e2
)+ θe
)+ 2k · π, x1, . . . , xn−1
) (4.69)
that clearly are diffeomorphisms from Rn−1 onto Hk.Clearly we have Hk = Hk+2 and for even and odd k Hk is mapped down to one of the two
associated half-hyperplanes in AdSn −H.Suppose h belongs to H0 ⊂ ∂M , with image D(H0) = H0. By Lemma 4.40 it is seen by some
element x. If D(x) belongs to the connected component containing H1 we say that H0 is positiveand otherwise negative.
We now proceed with the actual proof. We suppose that D : M → AdSn, so that Proposition4.46 implies that there must be a hyperplane H0 in ∂M , which identifies with D(H0) = H0. Wesuppose H0 is positive.
For each element y ∈ H0 we define the curve αy(θ) = y + (θ, 0, . . . , 0). If αy is not lifted to M
through D to a curve that is defined for all θ ∈ M , it is because it hits a hyperplane Hy ⊂ ∂M .
This defines a mapping y → Hy, where Hy is taken to be the empty set if the lifted curve of αyis defined for all positive t. Furthermore, this mapping is locally constant. Since the connectedcomponents of ∂M are all connected, the mapping is constant. It must equal either the empty setor one half-hyperplane.
In the case of the image Hy being empty, we construct a subset Ω ⊂ M that is diffeomorphic
through D to the connected component Ω of AdSn − H0, containing H1 (as H0 is positive).If Hy is not empty, we let D(Hy) = Hy denote the half-hyperplane. Clearly H0 and Hy will
be the boundary of some connected open set Ω and as such there is a subset Ω ⊂ M such thatD(Ω) = Ω.
In both cases, as AdSn is simply connected, the discrete, infinite subgroup h(D) = Γ ⊂ G acts ina properly discontinuous manner by Lemma 3.47. Since Ω/Γ 'M is compact the cohomological
59
dimension cd Γ of Γ, must satisfy cd Γ = dimM = n, following ([4], p. 209).If Hy is not empty but Hy 6= H1, we must have non-empty intersection ϕ−1(H0)∩Ω 6= ∅. This
intersection will contain a half-hyperplane, of dimension n− 1, which we denote by Ω′. We denoteby Γ′ the subgroup of Γ that fixes Ω′ and we have [Γ : Γ′] = N <∞.
We now show that the subquotient Ω′/Γ′ is also compact. We let Γ′x′n be a sequence inΩ′/Γ′. Clearly x′n ⊂ Ω′. We consider the sequence Γx′n in Ω/Γ that by the compactness has aconvergent subsequence, which for notational convenience we assume is itself and the limit is Γx′.By Proposition 2.36 there exists a covering map ψ : Ω → Ω/Γ and so we may choose a sequencezn ⊂ Ω such that ψ(zn) = Γx′n and zn → z with ψ(z) = Γx′. Furthermore, as [Γ : Γ′] = N <∞we have the disjoint union
Γ = Γ′ ∪ g1Γ′ ∪ . . . ∪ gN−1Γ′, gi ∈ Γ. (4.70)
Clearly zn and x′n are congruent mod Γ, i.e. we have zn = gnx′n. But as we have only finitely
many cosets giΓ′ in Γ, one of these must contain infinitely many elements of the gn. We may
pass to this subsequence, or assume that already gn ⊂ gKΓ′ for notational convenience. Butthen gn = gKg
′n, where g′n ∈ Γ′ and so
zn = gn(x′n) = gKg′n(x′n). (4.71)
Since zn → z and g−1K is a diffeomorphism, we must have g−1
K (zn) → x = g−1K (z). On the other
hand we have g−1K (zn) = g−1
K gKg′n(xn) = g′n(x′n) ∈ Ω′, as g′n ∈ Γ′ that fixes Ω′. Since Ω′ ⊂ Ω
was closed it follows that x ∈ Ω′. Hence we may project down the convergent sequence x′n intoΓ′x′n ∈ Ω′/Γ′ and it converges to Γ′x. Thus Ω′/Γ′ is compact.
Now on the on hand, we have from ([4], p. 187) that cdΓ′ = cdΓ. But on the other we have thatas Ω′/Γ′ is compact we must have cd Γ′ = dim Ω′ = n− 1 by ([4], p. 209). This is a contradiction.Hence we have Hy = H1, which is the assertion.
With the above results in hand, we are finally ready to deduce the full Theorem 4.21.
Proof. ([6], [17]) We deal with the flat and non-flat cases separately.For the case of κ = 0, Proposition 4.46 implies that D(M) = Ω, a connected component
of Mn − H, for a hyperplane H. The holonomy group Γ must leave a hyperplane parallel to Hinvariant. This contradicts ([14], Theorem 2.8 p. 644). Hence D(M) =Mn and asMn is completeby Proposition 2.20 it follows that (M, g) must be complete. This argument is initially due to [6].
The following is due to [17]. Suppose κ = ±1. Let H be the coisotropic hyperplane in theambient space Rn+1 and e1 a vector orthogonal to H. Clearly then e1 is null. Let furthermore GIdenote the isometry subgroup of G fixing Ω. We denote HI the isotropy subgroup, so that we mayidentify Ω = GI/HI . We define the vector field
Yy =e1
Qn+1±1 (e1, y)
− y. (4.72)
It follows that Yy ∈ TyΩ as
Qn+1±1 (Yy, y) = Qn+1
±1 (e1
Qn+1±1 (e1, y)
− y, y)
=Qn+1±1 (e1, y)
Qn+1±1 (e1, y)
−Qn+1±1 (y, y)
= 0,
(4.73)
where we used the fact that if y belongs to dSn or AdSn, we have Qn+1±1 (y, y) = 1.
Furthermore we show that Yy is GI -invariant, in the sense that if g ∈ GI , then gYy = Ygy.Clearly GI must act on e1 by scalar multiplication, i.e. ge1 = c · e1. Furthermore, as it is a groupof isometries, it must preserve the scalar product, i.e. Qn+1
±1 (ge1, gy) = Qn+1±1 (e1, y). We therefore
60
get
gYy = g
(e1
Qn+1±1 (e1, y)
− y)
= ge1
Qn+1±1 (e1, y)
− gy
=ge1
Qn+1±1 (ge1, gy)
− gy
=c · e1
Qn+1±1 (c · e1, gy)
− gy
=e1
Qn+1±1 (e1, gy)
− gy
= Ygy,
(4.74)
where we used linearity and scalar product invariance.We now calculate the divergence of Yy for both the de-Sitter space and the anti de-Sitter
separately. For the de-Sitter space, we let (e1, . . . , en+1) in Rn+1 be a basis such that e1 is thevector orthogonal to H and so that the quadratic form Qn+1
+1 is given by
B+1(x, y) = x1y2 + y1x2 + x23 + · · ·+ x2
n+1. (4.75)
in this basis. We get that Ω is the set of points x in Rn+1 such that both B+1(x, x) = 1 and x2 > 0and the hyperplane H is orthogonal to Re1. A parametrization ρκ=1 : Rn+ → Ω for Ω in the dSn
is given by
ρκ=1(x2, . . . , xn+1) =
(1− x2
3 − . . .− x2n+1
2x2, x2, . . . , xn+1
), (4.76)
where Rn+ is the open half-space in Rn defined by x2 > 0. We have
B+1(ρ, ρ) = 21− x2
3 − · · · − x2n+1
2x2· x2 + x2
3 + · · ·+ x2n+1 = 1, (4.77)
so that ρ ∈ Ω. The metric components in this basis are
gκ=1 =
− 1−x2
3−...x2n+1
x22
−x3
x2. . . −xn+1
x2
−x3
x21 . . . 0
......
. . . 0−xn+1
x20 0 1
(4.78)
so that the determinant becomes
g =√|det g| = 1
x2. (4.79)
We haveY (x2, . . . , xn+1) = −(x2, . . . , xn+1) (4.80)
and with this we may calculate the divergence div Y following ([20], p. 213):
div Y =1
g
∑i
∂
∂xi
(g Yi
)
= x2
(∂
∂x2
−x2
x2+
n+1∑i=3
∂
∂xi
(−xix2
))
= x2
(0 +
n+1∑i=3
−1
x2
)= −(n− 1).
(4.81)
61
With this we may, together with Lemma 2.27, interpret LY ω = −(n− 1)ω.We now do same calculation in the anti de-Sitter space. The quadratic form Qn+1
−1 becomes
B−1(x, y) = 2x1x3 + x22 − (x2
4 + . . .+ x2n+1) (4.82)
in the new basis. A parametrization ρκ=−1 : Rn+ → Ω is
ρκ=−1(x2, . . . , xn+1) =
(1− x2
2 + (x24 + . . .+ x2
n+1)
2x3, x2, . . . , xn+1
). (4.83)
Clearly, ρ ∈ AdSn as
B−1(ρ, ρ) = 2 ·1− x2
2 + (x24 + . . .+ x2
n+1)
2x3· x3 + x2
2 − (x24 + . . .+ x2
n+1) = 1 (4.84)
and similarly if x ∈ AdSn it must be on the form ρ(x2, . . . , xn+1). The metric components are
gκ=−1 =
1 0 0 . . . 0
0 − 1−x22+(x2
4+...+x2n+1)
x23
−x3
x3. . . −xn+1
x3
0 −x3
x31 . . . 0
......
.... . . 0
0 −xn+1
x30 0 1
(4.85)
so that the determinant becomes
g =√|det g| = 1
x3. (4.86)
We haveY (x2, . . . , xn+1) = −(x2, . . . , xn+1) (4.87)
so that we may calculate the divergence div Y :
div Y =1
g
∑i
∂
∂xi
(g Yi
)
= x3
( n+1∑i=2
∂
∂xi
(−xix3
))= −(n− 1).
(4.88)
With this we may, together with Lemma 2.27, interpret LY ω = −(n− 1)ω.We now look on this from another angle. The volume element ωI is also GI -invariant as GI
is an isometry group. From the proof of Lemma 2.27 we know that LY ωI = λωI pointwise. TheGI -invariance implies that the λ is constant on dSn or AdSn. Furthermore, Y induces a vectorfield YM and a volume element ωM on M . We may therefore apply Stokes Theorem ([7], p. 63):∫
M
LYMωM =
∫M
λωM = λ
∫M
ωM = λ
∫∂M
dωM = 0, (4.89)
where the integral must vanish as M has no boundary. Hence, λ = 0 6= −(n− 1).We have now showed that there can be no boundary of D(M) in X and hence D is surjective.
Completeness follows.
We get the the following consequence.
Corollary 4.50. There are no compact Lorentzian manifolds of constant positive curvature.
Proof. Suppose (Mn, g) is an n-dimensional, compact Lorentzian manifold with positive constantcurvature. If n = 2, this would be a contradiction to Corollary 4.18. If n > 2, Theorem 4.21 showsthat (M, g) is geodesically complete. But then it would be a compact spherical space form. Thisis a contradiction to Corollary 4.20. Hence (Mn, g) cannot exist.
62
Appendix A
Covering spaces
In this Appendix we discuss what is called covering spaces. It will be shown that every topologicalspace, and hence any manifold, has such an associated space. The covering space may be viewedas a simplified version of the original space, where the information of the original space is storedin the corresponding allowable symmetry groups to the covering space. This is of fundamentalimportance when proving Theorem 4.21.
A.1 Homotopy theory
We now introduce what is known as homotopy theory. We follow the discussion in [10].
Definition A.1. Let (M, τ) be a topological manifold. Suppose γ1 : [0, 1]→M and γ2 : [0, 1]→Mare two curves with γ1(0) = γ2(0) = a and γ1(1) = γ2(1) = b. Then they are said to be homotopicwith endpoints fixed if there is a map F : [0, 1]× [0, 1]→M such that
F (0, t) = γ1(t), ∀t,F (1, t) = γ2(t), ∀t,F (s, 0) = a, ∀s,F (s, 1) = b, ∀s.
(A.1)
F is called the homotopy map. In this case we write γ1 ' γ2.
The above definition shows, that if γ1 ' γ2 the former curve may be continuously deformedinto the latter.
Lemma A.2. ([10], p. 114) Let (M, τ) be a topological space. Then the relation ”'”, definedabove, is an equivalence relation.
Proof. ([10], p. 114) We need to prove reflexivity, symmetry and transitivity. For reflexivity, weobserve that if γ is a curve through M , then the map F (s, t) = γ(t) for all s, gives a map with thenecessary properties.
For symmetry, we observe that if γ1 ' γ2 through F , i.e. F (0, t) = γ1(t) and F (1, t) = γ2(t),then F (s, t) = F (1− s, t) has F (1, t) = γ2(t) and F (0, t) = γ1(t). Hence we have symmetry.
For reflexivity, suppose γ1 ' γ2 and γ2 ' γ3. Let F12 and F23 be the corresponding homotopymaps. Then the map
F (s, t) =
F12(2s, t) 0 ≤ s ≤ 1/2,F23(2s− 1, t) 1/2 ≤ s ≤ 1
(A.2)
is a homotopy map with the fixed endpoints a and b and has the properties F (0, t) = γ1(t) andF (1, t) = γ3(t). Hence γ1 ' γ3 and ”'” is an equivalence relation.
The equivalence classes of γ in the above definition are denoted [γ], i.e. if γ1 ' γ2 then[γ1] = [γ2]. We need the following lemmas.
63
Lemma A.3. ([10], p. 114) Let f : [0, 1] → [0, 1] be a map such that f(0) = 0 and f(1) = 1.Then [γ f ] = [γ].
Proof. ([10], p. 114) The explicit map
F (s, t) = γ((1− s)t+ sf(t)) (A.3)
is a homotopy map with F (0, t) = γ(t) and F (1, t) = γ(f(t)).
We now want to introduce something akin towards a product of equivalence classes. First, wedefine the product of two curves.
Definition A.4. Let (M, τ) be a topological manifold, a, b, c ∈M and suppose α(t) and β(t) arecurves in M with α(0) = a, α(1) = β(0) = b and β(1) = c. Then the curve product is(
αβ)(t) =
α(2t) 0 ≤ t ≤ 1/2,β(1− 2t) 1/2 ≤ t ≤ 1.
(A.4)
Lemma A.5. ([10], p. 115) The curve product is well-defined and associative up to reparametriza-tion, where the former is meant in the sense that if α1 ' α2 and β1 ' β2 then [α1β1] = [α2β2],and the latter in the sense of (αβ)γ(t) = α(βγ)(f(t)).
Proof. ([10], p. 115) The construction is obvious from the definition.Let α, β and γ be curves such that the endpoints of the former is the starting point of the
latter, as above. We look at the different combinations explicitly:
((αβ)γ)
=
α(4s), 0 ≤ s ≤ 1/4,β(4s− 1), 1/4 ≤ s ≤ 1/2,γ(2s− 1), 1/2 ≤ s ≤ 1.
(A.5)
and (α(βγ))
=
α(2s), 0 ≤ s ≤ 1/2,β(4s− 2), 1/2 ≤ s ≤ 3/4,γ(4s− 3), 3/4 ≤ s ≤ 1.
(A.6)
Comparing we see that with the reparametrization
f(t) =
s/2, 0 ≤ 1/2,s− 1/4, 1/2 ≤ s ≤ 3/4,2s− 1, 3/4 ≤ s ≤ 1.
(A.7)
the curves are homotopic with endpoints fixed.
Definition A.6. Let [γ] be a homotopy class. The inverse class is defined as [γ(t)]−1 = [γ(t)−1] =[γ(1− t)].
Lemma A.7. ([10], p. 117) Let α be a curve that starts at a and ends at b. Then
[α][α]−1 = [a], (A.8)
where a is the constant curve at a.
Proof. ([10], p. 117) The homotopy map
F (s, t) =
α(2t), 0 ≤ t ≤ s/2,α(s), s/2 ≤ t ≤ 1− s/2,α(1− 2t), 1− s/2 ≤ t ≤ 1
(A.9)
has F (1, t) running along α at double speed and then back, whereas the curve F (0, t) is constantat a.
Proposition A.8. ([10], p. 118) Let (M, τ) be a topological manifold and p ∈ M . Then the setof equivalence classes of curves, that start and end at p, form a group under the above compositionrule.
64
Proof. ([10], p. 118) The trivial curve γ(t) ≡ a serves as an identity element. We saw in LemmaA.5 that the product composition rule is well-defined and associative. Lemma A.7 shows that everyequivalence class [γ] has an inverse element [γ]−1 such that [γ][γ]−1 becomes the identity element.Hence we have group structure.
Definition A.9. Let p ∈ M be any point. The group defined in Proposition A.8, is denotedπ1(M,p) and is called the fundamental group.
Let us study an example.
Example A.10. We take any convex subset A ⊂ R2 and let p ∈ A. Let furthermore γ be anyloop from p to p in A. As A is star shaped, any point on γ can be joined to p by a straight line,and thus γ is homotopic to the trivial curve, constant at p. The fundamental group will thus betrivial, i.e. π1(R2, p) ' e.
The following results show that for parth-connected sets the fundamental group does not dependon p.
Lemma A.11. ([10], p. 119) Let p and q be points in M and let γ be a curve such that γ(0) = qand γ(1) = p. The mapping
φγ([α]) = [γ][α][γ]−1 (A.10)
is an isomorphism from π1(M,p) to π1(M, q).
Proof. ([10], p. 119) φγ is a homomorphism as
φγ([α])φγ([β]) = [γ][α] [γ]−1[γ]︸ ︷︷ ︸id
[β][γ]−1
= [γ][α][β][γ]−1
= [γ][αβ][γ]−1
= φγ([αβ]),
(A.11)
where we used Lemma A.7 and Lemma A.5.If φγ([α]) = φγ([β]), then [γ][α][γ]−1 = [γ][β][γ]−1 and by left- and right multiplication with
[γ]−1 and [γ] respectively, it follows that [α] = [β] and so φγ is injective.If [α] ∈ π1(M, q), then the curve class [β] = [γ]−1[α][γ] is mapped φγ([β]) = [α] and so φγ is
surjective. Hence φγ is an isomorphism.
If M is path connected any two points can be joint with curves and hence any two fundamentalgroups π1(M,p) and π1(M, q) are isomorphic by Lemma A.1. Hence we need not include the pointin the notation, i.e. π1(M,p) = π1(M, q) = π1(M).
Definition A.12. Let M be a path-connected manifold. If the fundamental group π1(M) istrivial, then M is called simply connected.
Proposition A.13. Let M and N be a path-connected manifolds. Then
π1(M ×N) ' π1(M)× π1(N). (A.12)
Proof. A curve in M ×N is on the form γ(t) = (α(t), β(t)). Hence the mapping
φ([γ]) = [α]× [β] (A.13)
maps π1(M ×N) into π1(M)×π1(N). That φ is onto is obvious from construction. The followingshows that φ is a homomorphism:
φ([γ1]) ∗ φ([γ2]) =[α1]× [β1] ∗ [α2]× [β2]
=[α1α2]× [β1β2]
=φ([γ1γ2]).
(A.14)
It remains only to show that φ is injective. But this is clear as if φ([γ1]) = φ([γ2]) then there mustbe [α1]× [β1] = [α2]× [β2]. This happens if and only if α1 ' α2 and β1 ' β2. But then it followsthat γ1 ' γ2 and so [γ1] = [γ2]. Thus φ is an isomorphism.
65
Proposition A.14. If two path-connected manifolds M and N are homeomorphic, then π1(M) 'π1(N).
Proof. Let φ : M → N be the homeomorphism. Suppose we have γ1 ' γ2 in M . Denote thehomotopy map F (s, t) with F (0, t) = γ1(t) and F (1, t) = γ2(t). Then the map F (s, t) = φ(F (s, t))is a homotopy in N that shows that the curves φ(γ1) and φ(γ2) are homotopic with end-pointsfixed. Thus the mapping φ∗([γ]) = [φ γ] is a well-defined mapping from π1(M) to π1(N). Thefollowing shows that φ∗ is a group homomorphism:
φ∗([γ1]) ∗ φ∗([γ2]) =[φ γ1] ∗ [φ γ2]
=[(φ γ1)(φ γ2)]
=[φ (γ1γ2)]
=φ∗([γ1][γ2]).
(A.15)
It remains to show that φ∗ is bijective. The surjectivity follows from the fact that φ is a homeo-morphism, and so the mapping φ−1
∗ is just as well-defined as φ∗, so that φ∗ indeed is onto. Theinjectivity is follows in the same manner, i.e. if φ∗([γ1]) = φ∗([γ2]) = [γ], then φ−1
∗ ([γ]) = [γ1] = [γ2]and so the injectivity follows. Hence φ∗ is a group isomorphism.
A.2 Covering maps
Definition A.15. Let ϕ : M → N be a map between two topological manifolds. An open setU ⊂ N is said to be evenly covered by ϕ if ϕ−1(U) is a number of disjoint sets such that ϕ,restricted to any of them, is a homeomorphism onto U . If ϕ is onto N and every p ∈ N has anevenly covered neighbourhood U , then ϕ is said to be a covering map. The disjoint open sets inϕ−1(U) are called sheets.
Example A.16. We consider the standard exponential map exp : R→ S1. For an arbitrary pointp = exp(s0i2π) ∈ S1 the open set
Um =t ∈ R
∣∣ |t− (s0 +m)| < 1/2, m ∈ Z
(A.16)
is a union⋃m∈Z
Um in R of disjoint sheets Um = (m− ε,m+ ε) and the exponential map restricted
to each Um is a homeomorphism. Since s0 was arbitrary, S1 is evenly covered by exp.
Example A.17. We consider the mapping φn : (C−0)→ (C−0) by φn(z) = zn, n ∈ Z. Foreach z = r exp(is) ∈ C, the inverse image is
φ−1n (z) = n
√r exp(is/n), n
√r exp(is/n) exp(2πi/n), . . . , n
√r exp(is/n) exp((n− 1)πi/n)︸ ︷︷ ︸
n
. (A.17)
The open balls Bk, centered at z0 = n√r exp(is/n) exp(kπi/n), k = 0, 1, . . . , n − 1, and with
radius ε = n√r sin(π/n), are the corresponding sheets. They are mapped homeomorphically onto
their common image. Since z is arbitrary, φn is a covering map.
For later purposes the following definition is useful.
Definition A.18. Let ϕ : M → N be a covering map between topological manifolds, and suppose,for L a topological manifold, ψ : L → N is a diffeomorphism. If there is a map π : L → M suchthat π = ψ φ, such a map is called a lift.
The situation is depicted in Figure A.2. Lemma A.19 shows that such a lift is determined bya single point.
Lemma A.19. ([10], p. 125) Let ϕ : M → N be a covering map between topological manifolds,ψ : L→ N be a diffeomorphism from a connected manifold L and π1 : L→M and π2 : L→M betwo lifts. If π1(y) = π2(y) for some y ∈ L, then π1 = π2.
66
Figure A.1: Depiction of path lift.
Proof. ([10], p. 126) We define the following sets:
T =y ∈ L
∣∣ π1(y) = π2(y),
S =y ∈ L
∣∣ π1(y) 6= π2(y).
(A.18)
Since both T ∪ S = L and T ∩ S = ∅ we have that if both T and S are open, the connectedness ofL implies either T = L or S = L. Hence we show that both sets are open.
Let x ∈ T and U be an open evenly covered neighbourhood of ψ(x). Then ψ−1(U) will consistof disjoint sheets, containing π1(x) and π2(x). Since x ∈ T , π1(x) = π2(x) and π1(x) = π2(x) is insome sheet, which we denote Vx. As π1 and π2 are continuous, the inverse images of open sets areopen. The inverse open sets of the sheet Vx intersects to form an open set: π−1
1 (Vx)∩π−12 (Vx) = W
and y ∈ W . Furthermore, π1 and π2 restricted to W must coincide, i.e. π1 = π2|W . Indeed itmust be so, as otherwise there would be z ∈W such that π1(z) 6= π2(z) but still π1(z), π2(z) ∈ Vx.Restricted to each sheet, ϕ is a homemorphism, and so ϕ π1(z) 6= ϕ π2(z) from the bijectivityof ϕ. But π1 and π2 are lifts, and so must satisfy ϕ π1(z) = ϕ π2(z) = ψ(z) and so we wouldhave a contradiction. Hence W ⊂ T so that T is open.
To show that S is open, we let x ∈ S. We again let U contain x and be evenly covered.As π1(x) 6= π2(x) they must be in distinct sheets, V1 and V2. Again, π1 and π2 are continuous,and so have open inverse images. Both π−1
1 (V1) and π−12 (V2) contain x and their intersection
π−11 (V1) ∩ π−1
2 (V2) = W is open. Now it must be so that W ⊂ S. Otherwise, for some z ∈ Wwe would have π1(w) = π2(w). But then this element would be in the intersection of V1 and V2.These are two disjoint sets by the properties of the covering map, and so we have contradition.Hence also S is open.
Theorem A.20. ([10], p. 126, ”The path-lifting Theorem”) Let ϕ : M → N be a covering mapbetween topological manifolds and suppose γ : [0, 1] → N is a continuous curve with γ(0) = p.Then, if q ∈ ϕ−1(p), there is a unique path lift curve γ : [0, 1] → M such that γ(0) = q andϕ(γ) = γ.
Proof. ([10], p. 126) Let x ∈ X be a point and Ux be an evenly covered neighbourhood of x. Thenthe sets
Uxx∈X form an open covering of X. As γ is continuous, the inverse images γ−1(Ux) are
also open and form an open covering of [0, 1]. The interval [0, 1] is compact, and hence we maychoose finitely many γ−1(Uxi) to cover it. We choose points 0 = s0 < s1 < · · · < sN = 1 such thateach si lie in the intersection γ−1(Uxi−1
) ∩ γ−1(Uxi). This means that γ([si−1, si]) ⊂ Uxi .We now lift the curve piece wise. We start at γ(0) = γ(s0). Since γ(s0) is evenly covered by Ux0
we consider the sheet ϕ−1(Ux0) that contains q and denote it Wq. Since ϕ is a homeomorphismon Wq, the curve segment γ([s0, s1]) is mapped to a curve in Wq, starting at q. Inductively, wecontinue with the next curve segment γ([s1, s2]). Again, it is contained in the evenly coveredneihbourhood Ux1
. We consider the sheet that meets the endpoint of ϕ−1(γ(s0, s1)). Again ϕ is ahomeomorphism on the sheet and we may extend the curve to be defined on the interval [s0, s2].This process is repeated N times, and hence the curve is lifted to the topological space M .
To show uniqueness we simply apply Lemma A.19. The curve γ may be seen as the lift, andγ = ϕ γ and hence if the lifts were ever to coincide, e.g. in the starting point, they would be thesame.
Not only paths may be uniquely lifted, as the above theorem showed. In fact, entire homotopymaps may be lifted.
67
Theorem A.21. ([10], p. 127) Let ϕ : M → N be a covering map between topological manifolds.Suppose F (s, t) : [0, 1] × [0, 1] → N is a homotopy map with F (0, 0) = p. Then, for q ∈ ϕ−1(p),there exists a unique homotopy map G(s, t) : [0, 1]×[0, 1]→M such that G(0, 0) = q and ϕ(G) = F .
Proof. ([10], p. 127) The uniqueness follows from Lemma A.19.By Theorem A.20 there is a unique path γ : [0, 1] → M with γ(0) = e such that ϕ(γ)(t) =
F (0, t). In the same manner, for each t, we may lift F in the other variable, i.e. there is a mapG(s, t) such that G(0, t) = γ(t) and ϕ(γ)(s) = F (s, t). This defines the desired lift. We need onlyto demonstrate continuity.
We define the curve γ(s) = F (s, 0). We use the same notation as in the proof of Theorem A.20.We define Ux to be some evenly covered neighbourhood, for each x ∈ M . We know that γ−1(Ux)is open in [0, 1], and more precisely an open neighbourhood of [sj−1, sj ]. It follows that F−1(Uj)must include [sj−1, sj ] × [0, ε]. That is, we have F ([sj−1, sj ] × [0, ε]) ⊂ Uj . We have e ∈ V1 andtherefore γ(s) ∈ V1 for s ∈ [0, ε]. From what we have constructed we now must have
G = (ϕ∣∣V1)−1 F, on [0, s1]× [0, ε] (A.19)
with G continuous on [0, s1] × [0, ε]. We proceed in this fashion to find that G is continuous onevery rectangle [sj−1, sj ]× [0, ε].
The same argument holds for any t0 ∈ (0, 1], so that G is continuous both on any [sj−1, sj ] ×[t0 − ε, t0 + ε] and on any [sj−1, sj ]× [1− ε, 1] and so G is continuous on [0, 1]× [0, 1].
Theorems A.20 and A.21 have important consequences. Suppose that ϕ : M → N is a coveringmap between connected topological manifolds. Let γ : [0, 1] → N be a loop with γ(0) = γ(1) = pand consider a lifted curve γ : [0, 1] → M with γ(0) = q and ϕ(q) = p. Although the curve maystart at q, in order for it to be mapped into a loop through ϕ it must not end where it started, onlysomewhere in ϕ−1(p). If such a curve ends at q, then the loop will be in the trivial equivalenceclass of π1(M), and so by Theorem A.21 will mapped down into the trivial equivalence class ofπ1(N). This suggests the following mapping:
ϕq : π1(N, p)→ φ−1(p), by ϕq([γ]) = γ(1), (A.20)
i.e. ϕq maps the equivalence class [γ] to the end-point of the lifted curve γ that starts at q. Weget the following result:
Lemma A.22. ([10], p. 128) If M is simply connected, the map ϕq is a bijection.
Proof. ([10], p. 128) For any y ∈ ϕ−1(p) we let γ be a curve from q to y. Then the compositionγ = ϕ γ is a curve in N that starts and ends at p, i.e. it is a loop. Therefore ϕq([γ]) = γ(1) = yand so ϕq is onto.
To show injectivity, we suppose ϕq([γ1]) = ϕq([γ2]), where γ1 and γ2 are loops based at p. Welet γ1 and γ2 be lifts of γ1 and γ2 respectively, that start at q. Since γ1 and γ2 have the sameterminal point, we may construct the curve γ1γ
−12 , which is a loop based at q. Since M is simply
connected, there is a map F : [0, 1] × [0, 1] that contracts it down to the trivial curve. Thereforewe have [γ1][γ2]−1 = [q] and so [γ1] = [γ2] so that ϕq is injective.
Hence ϕq is a bijection.
Definition A.23. Let ϕ : M → N be a covering map between topological manifolds. A diffeo-morphism k : M →M is called a deck transformation if ϕ k = ϕ. If, for any two points p andq in ϕ−1(x), there exists some k ∈ D such that k(p) = q, D is called normal.
Since the identity mapping is trivially a deck transformation. If k is a deck transformation,then clearly k−1 is also. Compositions are also deck transformations and so it follows that the setof deck transformations form a group under composition.
The following is an immediate consequence of Theorem A.24.
Corollary A.24. Let ϕ : M → N be a covering map between two topological manifolds and supposethere exists a normal deck group D on M . If M is simply connected, then π1(N) ' D.
68
Proof. In light of Theorem A.24, we need only to show that the map ϕq may be extended tobecome a homomorphism. We define
ψq : π1(N, p)→ D, by ψq : [γ]→ ϕq([γ]) = γ(1)→ k,
where k ∈ D such that k(q) = γ(1).(A.21)
Since D i normal, the map ψq is well-defined. We consider two curve classes [α] and [β] of π1(N, p).
Denote the associated lifts, starting at q, by α and β respectively. The curve product [α][β] = [αβ]has representative αβ whose lift αβ starts at q and ends up at αβ(1). We denote by kα(q) = α(1)and kβ(q) = β(1) the elements of the deck group D that maps the starting point to the respectiveendpoints. We must only show that kα kβ = kαβ , where kαβ(q) = αβ(1).
The curve αβ is α for 0 ≤ t < 1/2 and β for 1/2 < t ≤ 1. Any lifted representative γremains projected into the same equivalence class when shifted through D. Hence, we must haveϕ(kα(β)) = β. But then the lifted curve αβ starts at q, passes through α(1) and moves throughkα(β) to kα(β(1)). This end point is both αβ(1) and kα kβ(q). Hence there must be
ψq([α][β]) =
αβ(1) = kαβ(q),
kα β(1) = kα kβ(q)(A.22)
and so ψq is a homomorphism.Since, by Lemma A.22, ψq is also bijective, it must be an isomorphism.
Proposition A.25. The fundamental group of S1 is π1(S1) ' 〈Z,+〉, i.e. the infinite cyclic group.
Proof. We have already shown that the map exp : R → S1 is a covering map. In order for adeck transformation to satisfy exp k = exp we must have exp(2πis0) = exp(2πik(s0)). Thishappens if and only if k(s0) = s0 + n, n ∈ Z. Denote kn(s0) = s0 + n. Compositions of suchmappings have kn km(s0) = s0 + (m+n). Thus, the mapping ϕ(n) = kn is an isomorphism from〈Z,+〉 → D. Since R is simply connected, the fundamental group π1(S1) must be isomorphic tothe deck transformation group, which was just the infinite cyclic group.
Proposition A.26. The fundamental group of Sn, n > 1, is trivial, i.e. Sn is simply connected.
Proof. We let γ : [0, 1]→ Sn be a loop based at x0 ∈ Sn. Let x be a point in Sn that is not in theimage of γ. From rotational symmetry we may assume that x = (0, . . . , 0, 1), i.e. the north pole.We use the stereographic projection P : Sn − x → Rn, explicitly given by
P : (x1, . . . , xn+1)→(
x1
1− xn+1,
x2
1− xn+1, . . . ,
xn1− xn+1
). (A.23)
We show that P is a map onto Rn. Clearly the components are continuous. We show that P isinjective. Let y = (y1, . . . , yn+1) and z = (z1, . . . , zn+1). If P (y) = P (z), we have(
z1
1− zn+1,
z2
1− zn+1, . . . ,
zn1− zn+1
)=
(y1
1− yn+1,
y2
1− yn+1, . . . ,
yn1− yn+1
). (A.24)
Furthermore, P (y) and P (z) have the same norm:
n∑i
y2i
(1− yn+1)2=
n∑i
z2i
(1− zn+1)2(A.25)
and since both y, z ∈ Sn we have
n+1∑i
x2i = 1 and
n+1∑i
y2i = 1. (A.26)
With this we getn∑i
y2i
(1− yn+1)2=
1− y2n+1
(1− yn+1)2(A.27)
69
and similarly for z, which in turn gives the equality
1− y2n+1
(1− yn+1)2=
1− z2n+1
(1− zn+1)2(A.28)
from which it follows that yn+1 = zn+1. Since we have equality of components
zi1− zn+1
=yi
1− yn+1(A.29)
the injectivity now follows.To show surjectivity, we suppose z = (z1, . . . , zn) ∈ Rn. Then
x =
(2
z1
z21 + . . .+ z2
n + 1, . . . , 2
znz2
1 + . . .+ z2n + 1
,z2
1 + . . .+ z2n − 1
z21 + . . .+ z2
n + 1
)(A.30)
belongs to Sn and satisfies P (x) = z. The first assertion follows, as
||x||2 =
n∑i
4z2i
(z21 + . . .+ z2
n + 1)2+
(z12 + . . .+ z2
n − 1)2
(z21 + . . .+ z2
n + 1)2
=
n∑i
4z2i
(z21 + . . .+ z2
n + 1)2+
((z12 + . . .+ z2
n)2 − 2∑ni 2z2
i + 1)
(z21 + . . .+ z2
n + 1)2
=((z1
2 + . . .+ z2n)2 + 2
∑ni 2z2
i + 1)
(z21 + . . .+ z2
n + 1)2
=(z2
1 + . . .+ z2n + 1)2
(z21 + . . .+ z2
n + 1)2
= 1.
(A.31)
The second follows as, component-wise, we have
1− xn+1 = 1− z21 + . . .+ z2
n − 1
z21 + . . .+ z2
n + 1=
2
z21 + . . .+ z2
n + 1, (A.32)
so that
P (x) · ei =xi
1− xn+1= 2
z1
z21 + . . .+ z2
n + 1
z21 + . . .+ z2
n + 1
2= zi (A.33)
and so P is surjective.With this we have the homemorphism Sn − x ' Rn and Lemma A.14 shows that π1(Sn −
x) = π1(Rn). Example A.10 shows that π1(Rn) = e and so it follows that the loop γ iscontinuously contractible to the trivial loop based at x0. Hence Sn is simply connected.
Proposition A.27. The fundamental group of R2 − 0 is π1(R2 − 0) ' 〈Z,+〉.
Proof. We work with the topologically equivalent space C−0 instead and consider the mappingexp : z → ez. For an arbitrary complex number w, we have exp−1(w) = ln(w) + 2πin, n ∈ Z. Thisshows, that if we choose the sheets
U =⋃n∈Z
Un, Un = Bπ(
ln(w) + 2πin), (A.34)
the exponential becomes a covering map from C→ C−0. A deck transformation k must satisfyexp k = exp and so k is a deck transformation if and only if kn(z) = z + 2πin. Composition isadditive, i.e. km kn(z) = z + 2πi(m+ n). The full complex plane is simply connected and hencethe deck transformation group is isormophic to 〈Z,+〉.
With the following lemmas we aim to prove existence of a universal cover.
Lemma A.28. Every connected topological manifold M is path-connected.
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Proof. We pick a point p ∈ M and consider the set subset N ⊂ M consisting of the points thatare path-connected to p. We want to show that this set is the full of M , and to do so it wouldsuffice to show that N is both open and closed.
We first show that N is open: We pick an arbitrary point q in N and take a chart (Uq, φ)containing it. Then φ(Uq) ⊂ Rn is open and Rn is path connected. Hence, for some ball Bε(φ(q)) ⊂φ(Uq), all points contained can be joined to q with a straight line. Since φ−1(Bε(φ(q)) ⊂ φ(Uq)) ⊂Uq it follows that this set is now connected to q, and since q is path connected to p, so is the point.Hence q is an interior point of N and so N is open.
We show that N is closed: We show the complement M −N is open. Let q /∈ N . Then, again,there is a chart (Uq, φ) containing q and the open set φ(Uq) ⊂ Rn contains some open ball Bε(φ(q)).This open ball is of course path connected in Rn, and hence so is the open ball φ−1(Bε(φ(q))).Since q was not path connected to p, nor can any element in the ball φ−1(Bε(φ(q))) be, and henceφ−1(Bε(φ(q))) 6⊂ N . Thus the complement of N is open.
Definition A.29. For a topological space (M, τ), we say that it is locally path-connected if,for each p ∈M and neighbourhood U containing p, there is a subset V ⊂ U that is path-connected.
Lemma A.30. Every topological manifold M is locally path-connected.
Proof. It suffices to show that for p ∈M and neighbourhood U containing p, U is path-connected.This follows by similar arguments to those in the proof of Lemma A.28. Every point in U is con-tained in some chart (Uα, φα). Since any open and connected subset of Rn is also path-connected,we know that φα(Uα) is path-connected, and hence that Uα is. Since U is the union of intersectionsof the type U ∩ Uα, that are path-connected, and since U is connected, it follows that also U ispath-connected.
Definition A.31. A topological space (M, τ) is called semi-locally 1-connected if every pointp ∈ M has some neighbourhood U such that any loop, contained in U , starting at p, may becontinuously deformed (within U) to the trivial map.
Lemma A.32. A topological manifold (M, τ) is semi-locally 1-connected.
Proof. We use the fact that open sets in Rn have this property. Indeed, if U is open in Rn andx ∈ U , then there is some ball Bε(p) ⊂ U . Let γ(t) be a loop in Bε(x) starting and ending at x.Then the homotopical map F (s, t) = x + s(γ(t) − x) continuously deforms γ down to the trivialcurve at x.
This follows naturally to the manifold setting. For p ∈ M we take any chart (Uα, φα) thatcontains p, and consider the image. The image φα(Uα) is an open subset in Rn and so there is someball Bε(φα(p)) that deforms loops within it to the trivial curve. The inverse image φ−1
α (Bε(φα))has this property since φα is a homeomorphism.
We are now ready to prove existence of a universal covering space. The following theorem isstated, but not proven in the sources.
Theorem A.33. ([20],[22]) Every connected topological manifold M has a simply connected cover.
Proof. We consider all curves on M and their equivalence classes. Denote the space of suchequivalence classes by M . We now show that this space has a naturally inherited topology, issimply connected and there exists a covering map.
We show existence of a topology. For a curve class [γ] in M and neighbourhood U containingγ(1), we define U([γ]) as the elements [α], that are homotopically equivalent to some curve thatfollows γ and then stays in U . We denote the topology of such open sets τ .
This is first of all well defined; if U([γ]) contains [α], then U([γ]) = U([α]). This follows, as byconstruction α, (which we pick as representative of [α]), first follows γ and then some curve γ thatlies in U . But then conversely [γ] has a representative γ that follows α and then γ−1 and so thedefinition is well defined.
We show that this space becomes a topological space by showing that these neighbourhoodsform a basis for a topology. By construction, they form a cover. We need only to show, that foreach two non-trivially intersecting sets U1([γ1]) and U2([γ2]) we have some element W ([α]) in the
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intersection. We now show that the full intersection U1([γ1]) ∩ U2([γ2]) already has this property:suppose [α] ∈ U1([γ1])∩U2([γ2]). Then on the on hand, [α] has a represenative that follows γ1 andthen stays in U1, but on the other hand some representative follows γ2 and then stays in U2. Hencethe endpoint α(1) ∈ U1 ∩ U2 ≡ W . We need only to show, that any other element [β] ∈ W ([α])has a representative that follows α and then stays in W . But if [β] lies in U2([γ2]), then it has arepresentative β that follows γ2 and then stays in U2, i.e. β = β γ2. Meanwhile the same goes forα; α = αγ2. Finally since Lemma A.30 there is a curve δ in W with δ(0) = α(1) and δ(1) = β(1).We want to show that δ α = β. But this follows as the loops α−1 α γ2 and β−1 β γ2 areboth homotopically trivial, from which it follows that the loop β−1 δ α is homotopically trivialand therefore δ α = β.
Hence we have a basis for a topology and so (M, τ) is a topological space.We now show that the mapping ϕ : M → M , ϕ([γ]) = γ(1) is a covering map. We show
existence of evenly covered neighbourhood of any p ∈M . By Lemma A.32 there is a neighbourhoodUp for which any loop at p in Up is homotopically trivial. Let [γ] be some class with γ(1) = p. Thenwe claim that ϕ restricted to Up([γ]) is bijective. The surjectivity follows, as by Lemma A.30 anyopen connected neighbourhood is path connected. To show injectivity, we suppose ϕ([α]) = ϕ([β])in Up([γ]). Then each class has a representative α and β respectively, that first follow γ and then
α and β respectively. Since these classes are not the same, the compositions α γ and β γ cannotbe equivalent. Hence β and α cannot be equivalent. But they share both end-points and startingpoints (p) and so the composition β−1 α is a loop in Up([γ]). By hypothesis any loop could be
deformed down to the trivial one, and this is equivalent to α ' β and we are contradicted. HenceUp is evenly convered.
It remains only to show simple connectivity of M . To do so we let [σ] ∈ π1(M, p) be anequivalence class of loops with base p in M . We let ϕ(p) = p and let p be the trivial curve at p inM . We want to show that [σ] is trivial. We let σ represent [σ] and since σ is a curve in M , wehave that ϕ(σ) is a curve in M . ϕ(σ) starts and ends at p. Both σ and the lift ϕ−1(ϕ σ) thatstarts at p are curves in M . By Lemma A.21, they are the same.
On the one hand, we have σ ∈ π1(M, p) so that σ(1) = p, the constant curve at p. But on theother we have σ(1) = [ϕ(σ)]. Then ϕ(σ) must be trivial in π1(M,p) and so [σ] must also be trivial.Hence π1(M, p) ' e and M is simply connected.
The simply connected covering space is usually called the universal cover. By taking com-positions of the covering map with the local charts of M , we naturally equip M with an atlas sothat it becomes a manifold. The following properties are immediate.
Proposition A.34. The deck group of the universal covering space M is normal.
Proof. ([20], p. 444) Let p and q be in M such that ϕ(p) = ϕ(q) and let x be any point not p or qsuch that ϕ(x) 6= ϕ(p). Then there is a curve γ that travels from p to x, by path-connectedness.Clearly ϕ(γ) is a curve that travels from p to ϕ(x) = x. Since q is in ϕ−1(p) we may by LemmaA.20 lift this curve back up to M but with starting point q. We denote this curve by β. We definek(p) = β(1). Since this procedure is continuous in each step and M is simply connected, it followsthat k is a diffeomorphism and the assertion is shown.
Proposition A.35. The universal cover is unique up to diffeomorphism.
Proof. See [20], p. 444.
We state a constraint given by topological properties of this universal cover.
Lemma A.36. Suppose M is a manifold and M its universal cover. If M is compact, then thedeck group D is finite.
Proof. We let ϕ : M → M and suppose the assertion is false. Then, for p ∈ M , we have thatϕ−1(p) ⊂ M would consist of infinitely many points. Since M is compact, there is a convergentsubsequence pn converging to some p ∈ M . This contradicts the fact that the orbit ϕ−1(p) isthe closed union of isolated points.
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