Conway’s Theorem

Emma Brakkee

July 19, 2013

Bachelor Thesis

Supervisor: dr. H.B. (Hessel) Posthuma

Korteweg-de Vries Instituut voor Wiskunde

Faculteit der Natuurwetenschappen, Wiskunde en Informatica

Universiteit van Amsterdam

Abstract

Conway’s theorem states that a compact, connected 2-orbifold represents a periodic tessellationof the Euclidean plane if and only if its orbifold Euler characteristic is 0.

The main part of this thesis is about orbifolds. We give basic definitions, do a few importantconstructions such as the orbifold fundamental group and give a classification of the compacttwo-dimensional orbifolds. Using this classification and some group theory, we prove Conway’stheorem and the corollary that there are 17 periodic tessellations of the plane.

Title: Conway’s TheoremAuthor: Emma Brakkee, emma.brakkee@student.uva.nl, 10026046Supervisor: dr. H.B. (Hessel) PosthumaSecond grader: dr. R.R.J. (Raf) BocklandtDate: July 19, 2013

Korteweg-de Vries Instituut voor WiskundeUniversiteit van AmsterdamScience Park 904, 1098 XH Amsterdamhttp://www.science.uva.nl/math

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Contents

Introduction 4

1. Background 61.1. Properly discontinuous actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2. Covering spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.2.1. Covering maps and liftings . . . . . . . . . . . . . . . . . . . . . . . . . . 111.2.2. The universal covering space . . . . . . . . . . . . . . . . . . . . . . . . . 14

2. Orbifolds 192.1. The orbifold structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.1.1. Definition and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.1.2. Developable orbifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.2. Algebraic topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.2.1. The orbifold fundamental group . . . . . . . . . . . . . . . . . . . . . . . 242.2.2. The universal orbifold covering . . . . . . . . . . . . . . . . . . . . . . . . 302.2.3. The orbifold Euler characteristic . . . . . . . . . . . . . . . . . . . . . . . 33

3. Classification of two-dimensional orbifolds 383.1. The orthogonal group O2(R) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.2. The compact 2-orbifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.3. Developability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4. Tessellations of the plane 514.1. Periodic tessellations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514.2. Representing tessellations with orbifolds . . . . . . . . . . . . . . . . . . . . . . . 53

5. Conway’s Theorem 56

Populaire samenvatting 58

Bibliography 61

A. Net convergence 62

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Introduction

People have been using tessellations in art for ages. A famous example of tessellations asdecoration are the patterns on the walls of the Alhambra palace in Granada in Spain, madeby the Moors in the 8th century. These decorations inspired the Dutch artist M.C. Escher(1898-1972) to draw many, many pictures of tessellations, not only of the flat plane, but also ofthe spherical and the hyperbolic plane.

Escher was also interested in the mathematical background of tessellations. He read apaper that he got from his brother, a crystallographer, which was written by the Hungarianmathematician George Polya. In this paper it was proved that there are seventeen periodictessellations of the plane. Polya was not the first one to prove this: Yevraf Fedorov, also acrystallographer, already showed it thirty years earlier. In this thesis, we will give yet anotherproof, using Conway’s theorem.

Orbifolds are topological spaces with some extra structure. They are a generalization ofmanifolds. They were first introduced in the 1950’s by Ichiro Satake, under the name V-manifolds.The name orbifold, which comes from ‘orbit-manifold’, was first used by Bill Thurston, whoreintroduced the spaces in the 1970’s. The part ‘orbit’ of the name refers to the fact thatorbifolds are locally homeomorphic to the orbit space Rn modulo a finite group action.

We will study orbifolds in chapters 2 and 3 of this thesis, first treating some general theory andthen going deeper into two-dimensional orbifolds. Chapter 1 contains some basic backgroundinformation needed to understand the rest of the thesis. In chapter 4, some group theory showsup when we are studying symmetries and tessellations of the plane. Finally, in chapter 5, wewill prove Conway’s theorem.

Conways theorem was named after the American mathematician John Horton Conway. Itrelates tessellations to orbifolds. The idea to do this comes, according to Conway, from Thurston:“The philosophy that geometrical groups should be studied through their orbifolds is BillThurston’s. I claim originality only for the simple and elegant notation introduced here.”

This quote comes from the article The orbifold notation for surface groups [12], written byConway in 1992. In this article, he introduces a new notation, orbifold notation, to distinguishbetween different tessellations, based on the orbifold that they correspond to. On the way, hegives an informal argument why Conway’s theorem (which he does not mention) holds.

More about orbifold notation can be found in the book The Symmetries of Things [3] byConway, Burgiel and Goodman-Strauss, which is also the resource of the picture on the titlepage of this thesis. I would like to recommend The Symmetries of Things to anyone who is atleast a bit interested in tessellations, because it is a very beautiful book, nice to read no matterwhat mathematical background you have.

When I was in my last year at school, I went to several open days of universities to getinformation about bachelor studies, in particular the study mathematics. The presentation thatI remember the best was given by Hessel Posthuma. It was about tessellations of the plane. Hesketched the proof that there are seventeen of them, using Conway’s theorem. Of course, thiswas in no sense a precise mathematical proof, because the listeners could not yet understand the

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theory behind it. However, it was a nice presentation and it made me very enthusiastic. Thiswas something that I wanted to learn.

And now I have learnt it. I understand orbifolds and the group theory necessary to proveConway’s theorem. I have proved the theorem myself, and the corollary that there are onlyseventeen periodic tessellations of the plane. I still like the subject a lot and I am happy to havewritten my bachelor thesis about it.

I would like to thank Hessel Posthuma for (maybe unconsciously) introducing me to, andleading me out of, my bachelor mathematics, for being a nice and motivated supervisor, foranswering (or at least trying to answer) all my questions, no matter if they were important ornot, and for telling me every now and then that it was going well. Also, I would like to thank mysecond grader, Raf Bocklandt, for his interest, his knowledge and his cutting-and-pasting skills.

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1. Background

We start with a basic chapter about two main topics that we need to understand the theory oforbifolds: properly discontinuous group actions and coverings of topological spaces. The first ofthese will be treated in the next section. We follow Tom Dieck [5] and use some results that canbe found in Munkres [15]. For the proof of one the of the propositions we need the concept ofnets and net convergence. Some theory about nets can be found in the appendix.

1.1. Properly discontinuous actions

A properly discontinuous action is a certain type of group action by a topological group. In thisthesis, all groups that appear will be topological groups. We give a few basic definitions.

Definition 1.1.1. A topological group is a group (G, ·) with a topology, such that the maps

G×G→ G, (g, h) 7→ g · h

andG→ G, g 7→ g−1

are continuous.

Note that every group becomes a topological group if we give it the discrete topology.

We can extend the notion of continuity to group actions.

Definition 1.1.2. Let G be a topological group. A continuous action of G on a space X is acontinuous map φ : G×X → X such that

(i) φ(e, x) = x for all x ∈ X,

(ii) φ(gh, x) = φ(g, φ(h, x)) for all g, h ∈ G and x ∈ X;

where e is the unit element of G. As usual, we will denote gx for φ(g, x).

From now on, whenever we talk about an action of a topological group, we will assume that theaction is continuous.

Two important types of group actions are effective and free actions.

Definition 1.1.3. An action of a group G on a set X is called effective if the only element in Gthat fixes all x ∈ X is the unit element. The action is free if no x in X is fixed by some g ∈ Gother than the unit element, i.e. Gx = {e} for all x ∈ X.

Clearly, every free action is effective. A free action is also properly discontinuous. Before wedefine this property, let us recall some facts about proper maps.

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Definition 1.1.4. A continuous map f : X → Y between two topological spaces is called properif for every space Z, the map

f × idZ : X × Z → Y × Z

is closed, or equivalently, if f is closed and for every y ∈ Y , the set f−1(y) is compact.

Proposition 1.1.5. If f : X → Y is a proper map then for any subset Z ⊆ Y , the map f |f−1(Z)

is proper.

Proposition 1.1.6. If f : X → Y is a continuous and injective map, then the following areequivalent:

(i) f is proper

(ii) f is closed

(iii) f is a homeomorphism onto a closed subspace of Y.

Proposition 1.1.7. Let f : X → Y and g : Y → Z be continuous maps between topologicalspaces.

1. If both f and g are proper, then g ◦ f is proper

2. If g ◦ f is proper and f is surjective, then g is proper,

3. If g ◦ f is proper and g is injective, then f is proper.

Definition 1.1.8. An action φ : G×X → X of a topological group G on a space X is calledproper if the map

Φ : G×X → X ×X(g, x) 7→ (x, gx)

is proper.

It turns out that for any space X and any group G acting properly on X, the quotient spaceX/G is Hausdorff. To prove this, we need two lemmas.

Lemma 1.1.9. Let f : X → Y be a map between two topological spaces which is continuous,open and surjective. Then Y is a Hausdorff space if and only if the set

∆f = {(x1, x2) ∈ X ×X | f(x1) = f(x2)}

is closed in X ×X.

Proof. Let ∆ = {(y, y) | y ∈ Y } be the diagonal of Y . Recall that Y is Hausdorff if and only if∆ is closed. The map (f, f) : X ×X → Y × Y is also continuous, open and surjective. Therefore∆ is closed if and only if (f, f)−1(∆) is closed. But (f, f)−1(∆) = ∆f .

Lemma 1.1.10. If G is a topological group acting on a space X, then the quotient mapp : X → X/G is open.

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Proof. Let U ⊆ X be an open set and consider p(U). This set is open if and only if p−1(p(U))is open. But

p−1(p(U)) = {gx | g ∈ G, x ∈ U}

=⋃g∈G

gU =⋃g∈G

(g−1)−1(U)

is open, since the map x 7→ g−1x is continuous.

Proposition 1.1.11. Let G be a topological group that acts properly on a space X. The spaceX/G, equipped with the quotient topology, is Hausdorff.

Proof. Because Φ is proper, it is a closed map and therefore the set Φ(G×X) = {(x, gx) | g ∈ G}is closed in X ×X.

By lemma 1.1.10, the quotient map p : X → X/G is open. Also, it is continuous and surjective.The set ∆p = {(x1, x2) ∈ X ×X | p(x1) = p(x2)} is equal to

{(x1, x2) ∈ X ×X | x1 = gx2 for some g ∈ G} = Φ(G×X),

which is closed by the first observation. It follows from lemma 1.1.9 that X/G is Hausdorff.

Proper group actions have another nice property.

Proposition 1.1.12. Let G be a group that acts properly on a space X. For every x ∈ X, themap ω : G→ X, g 7→ gx is proper, and the isotropy group Gx is compact.

Proof. Consider the set Φ−1({x} ×X) = G× {x}. By proposition 1.1.5, the map

Φx := Φ|G×{x} : G× {x} → {x} ×X(g, x) 7→ (x, gx)

is proper, so it is closed and for every (x, gx) ∈ {x} × X, the set Φ|−1x (x, gx) is compact. It

follows that ωx is closed and ω−1x (gx) is compact for every g ∈ G. So ωx is proper.

For (ii), note that Gx = ω−1x (x).

We will now give the definition of a properly discontinuous group action. As the nameindicates, this is just a proper group action with the extra condition of ‘discontinuity’. This isnot a restriction on the continuity of the action, but on the topology of the group.

Definition 1.1.13. A proper action φ : G × X → X of a discrete group G on a space X iscalled properly discontinuous.

Many authors, for example Bredon [2] and Munkres, define a properly discontinuous action ofG on X by the following property: every x in X has a neighbourhood U such that gU ∩ U = ∅for all g ∈ G except the unit element. As we will see, this requirement is strictly stronger thanour conditions. It is even stronger than the property of being free.

Even though this other definition is very common, we use Tom Dieck’s definition. This willbe convenient in the chapters about orbifolds.

In the rest of this section, we will prove two theorems that we will use in the chapter aboutorbifolds. The first gives a necessary and sufficient condition for a group action to be properlydiscontinuous.

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Theorem 1.1.14. An action of a discrete group G on a Hausdorff space X is proper if andonly if for all x, y ∈ X, there are neighbourhoods Vx of x and Vy of y such that the set

{g ∈ G | gVx ∩ Vy 6= ∅}

is finite.

Note that x is allowed to be equal to y.This theorem is a direct corollary of the next proposition, 1.1.17, for which we need two more

definitions and a lemma.

Definition 1.1.15. A space X is called locally compact if for every x ∈ X, there is a compactsubspace of X that contains an open neighbourhood of X.

A relatively compact or precompact subspace of a topological space is a subset whose closureis compact.

Lemma 1.1.16. Let G be a topological group acting on a Hausdorff space X. Then for anyx ∈ X the isotropy group Gx is closed.

Proof. Let g ∈ G\Gx, so gx 6= x. Let U and be an open neighbourhood of gx such that x /∈ U .Then the set φ−1(U) = {(h, y) ∈ G×X | hy ∈ U} 3 (g, x) is open. So there exist neighbourhoodsH of g and V of x such that H × V ⊂ φ−1(U). For all h ∈ H, we have hx 6= x so h ∈ G\Gx.This shows that G\Gx is open. So Gx is closed.

Proposition 1.1.17. Let G be a locally compact Hausdorff group that acts on a Hausdorff spaceX. The action is proper if and only if for all x, y ∈ X, there are neighbourhoods Vx of x and Vyof y such that the set Hx,y = {g ∈ G | gVx ∩ Vy 6= ∅} is relatively compact in G.

Proof, ⇐. Suppose that for all x, y ∈ X the set Hx,y is relatively compact in G. Let C ⊂ G×Xbe a closed subset and consider Φ(C) ⊆ X ×X. We will show that this set is equal to its closure.Let ((x, y)α)α∈A = (xα, gαxα) be a net in Φ(C) that converges to some point (x, y) (this point isin Φ(C)). Let Vx and Vy be neighbourhoods of x resp. y such that Hx,y is relatively compactin G. We can assume that xα ∈ Vx, gαxα ∈ Vy for all α ∈ A (if this is not the case, take someappropriate subnet). It follows that gα ∈ Hx,y ⊂ Hx,y for every α, and because Hx,y is compact,the net (gα) has a subnet converging to a point g ∈ Hx,y. Now (g, x) ∈ C because C is closed,and because Φ is continuous, Φ(g, x) = (x, gx) is equal to (x, y). So (x, y) ∈ Φ(C). This showsthat Φ is a closed map.

Now let (x, y) = (x, gx) ∈ Φ(G×X) and consider the set

Φ−1(x, gx) = {h ∈ G | hx = gx} × {x}= {h ∈ G | g−1hx = x} × {x} = gGx × {x}.

The group Gx is contained in Hx,x, which is compact by assumption. Also, Gx is closed bylemma 1.1.16. Therefore Gx is compact, and by the continuity of the group action, gGx iscompact as well. It follows that Φ−1(x, gx) is compact. Conclusion: Φ is proper.

For the other implication, we need a concept called the one-point compactification.

Definition 1.1.18. Let X be a topological space. A compactification of X is a compact spaceY such that X is a proper subset of Y and the closure of X in Y equals Y . If Y \X consists of asingle point, then Y is called a one-point compactification.

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Proposition 1.1.19. A space X has a one-point compactification if and only if X is a locallycompact Hausdorff space. Moreover, the one-point compactification of X is unique up toisomorphism.

Proof. See Munkres, p. 183.

The one-point compactification of a space X is usually denoted by X ∪ {∞}. Its topologyconsists of the open sets in X together with the sets Y \K with K ⊂ X compact. It is again aHausdorff space.

Proof of proposition 1.1.17, ⇒. Assume that Φ is a proper map. Since Φ is continuous andinjective, it follows from proposition 1.1.6 that Φ is a homeomorphism onto its image. Thereforethe map f : G×X → G×X ×X, (g, x) 7→ (g, x, gx) is also a homeomorphism onto its imageD ⊆ G×X ×X. So by proposition 1.1.7, the map h : D → X ×X defined by

h(g, x, gx) = Φ(f−1(g, x, gx)) = Φ(g, x) = (x, gx)

is proper.Let F be the one-point compactification of G, so F = G∪{∞}. We will show that D is closed

in F ×X×X. The set A = {(g, g) | g ∈ G} ⊆ F ×G is closed (the graph of the inclusion G→ Fis closed), so A×X ×X is closed in F ×G×X ×X. Therefore A′ = (A×X ×X)∩ (F ×D) isclosed in F ×D. Because h was proper, the map h′ : F ×D → F ×X×X that sends (f, g, x, gx)to (f, x, gx) = (idF ×h)(f, g, x, gx) is closed. So h′(A′) = D is closed in F ×X ×X.

Let x, y ∈ X. We have {∞} ×X ×X ∩D = ∅. Therefore, there are open neighbourhoodsU ⊂ F of the point ∞ and V ⊂ X ×X of (x, y) such that U × V ∩D = ∅. By definition of theone-point compactification, U is of the form F\K, with K ⊂ X compact. Let Vx and Vy beneighbourhoods of x and y such that Vx × Vy ⊂ V . Then ((G\K)× Vx × Vy) ∩D = ∅, whichmeans that g /∈ K implies gVx ∩ Vy = ∅.

Here is the second theorem that will prove useful when we are studying orbifolds.

Theorem 1.1.20. Let X be a Hausdorff space and let G act properly discontinuously on X.Then for every x ∈ X:

(i) The isotropy group Gx is finite,

(ii) There is an open neighbourhood U of x such that U is invariant under Gx and U ∩ gU = ∅for all g /∈ Gx,

(iii) There exists a U as in (ii), for which the canonical map U/Gx → X/G is a homeomorphismonto an open set.

Proof. (i) follows directly from proposition 1.1.12, since compact subspaces of discrete spacesare finite.

For (ii), let x ∈ X. By Theorem 1.1.14, there is a neighbourhood U0 of x such that the setH = {g ∈ G | gU0 ∩ U0 6= ∅} is finite. We have Gx ⊆ H. If H = Gx, let U = ∩g∈GxgU0. Thissatisfies the conditions in (ii). If Gx ( H, let H\Gx = {g1, . . . , gn} and let yi = gix for everyi. Then for all i, we have yi 6= x so because X is Hausdorff, there are open neighbourhoodsVi of x and V ′i of yi with Vi ∩ V ′i = ∅. Let Ui = Vi ∩ g−1

i V ′i , this is a neighbourhood of x thatsatisfies Ui ∩ giUi = ∅. Define U = U0 ∩ U1 ∩ · · · ∩ Un. This is a neighbourhood of x such thatU ∩ gU = ∅ for all g ∈ G\Gx. Finally, let U = ∩g∈GxgU . This is an open Gx-subspace thatcontains x and satisfies U ∩ gU = ∅ for g ∈ G\Gx.

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For (iii), consider the canonical map q : U/Gx → X/G. It is injective by construction. Namely,suppose that Gu = Gv for some u, v ∈ U and let h ∈ G such that v = hu. Since U ∩ gU = ∅ forg ∈ G\Gx, we have h ∈ Gx so Gxu = Gxv.

The following diagram commutes:

U

��

� � // X

��

U/Gxq// X/G

Because the other three maps are continuous and open, q is continuous and open as well. So q isa homeomorphism onto its image.

1.2. Covering spaces

In this section we develop some theory about covering spaces. This is an important subject in forexample homotopy theory and the study of Riemann surfaces. It is treated extensively in manybooks, for example in Bredon [2] and Munkres [15]. Therefore, we will not prove everything, butrefer to these books sometimes.

We will restrict ourselves to the theory needed to construct a so-called ‘universal’ coveringspace for topological spaces satisfying some connectedness conditions.

1.2.1. Covering maps and liftings

The intuitive idea of a covering space is that it is built up from ‘layers’ of copies of certain opensets of the original space, glued together in some way. We call these special open sets ‘evenlycovered’.

Definition 1.2.1. Let X and Y be topological spaces and p : Y → X a continuous surjectivemap. An open set U ⊆ X is called evenly covered by p if the inverse image p−1(U) is a disjointunion of open sets Vα, such that p|Yα is a homeomorphism onto U .

Note that if W ⊆ U is open, then W is also evenly covered by p.

Definition 1.2.2. Let X and Y be topological spaces and p : Y → X a continuous surjectivemap. If every x ∈ X has a neighbourhood U ⊆ X which is evenly covered by p, then Y is calleda covering space of X. The map p is a covering map.

Every covering map is open. Namely, let A ⊂ Y be open. Let x ∈ p(A) and let U be anevenly covered neighbourhood of x. There is an y ∈ A such that p(y) = x, let Vα be theconnected component of p−1(U) that contains y. The set Vα ∩A is open in Vα, and it is mappedhomeomorphically to p(Vα ∩A) ⊂ U . It follows that p(Vα ∩A) is an open neighbourhood of xcontained in p(A); this shows that p(A) is open.

If we have any map Z → Y , we can get a map Z → X by composing with p. It is an interestingquestion what happens if we start with a map Z → X. In some cases we can relate to it a mapZ → Y which is called a lifting.

Definition 1.2.3. Let X, Y and A be topological spaces and let p : Y → X be a continuousmap. If h : A→ X is continuous, a lifting, or lift, of h to Y is a continuous map h : A→ Y such

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that h = p ◦ h. In a diagram, this looks as follows:

Y

p

��

Ahoo

h~~

X

.

If p is a covering map and Z = [0, 1], such liftings always exist. Moreover, they are unique up tothe image of 0.

Proposition 1.2.4. Let p : (Y, y0)→ (X,x0) be a covering map. For every path γ : [0, 1]→ Xwith base point x0, there is a unique lifting γ : [0, 1]→ Y which is a path beginning at y0.

For the proof we need two lemmas. The first one is the pasting lemma:

Lemma 1.2.5 (Pasting lemma). Let X and Y be topological spaces and let A,B ⊆ X be closedsubsets of X, such that A ∪ B = X. Let f : A → Y and g : B → Y be continuous maps suchthat f(x) = g(x) for all x ∈ A ∩ B. Then the function h : X → Y defined by h(x) = f(x) forx ∈ A and h(x) = g(x) for x ∈ B is continuous.

Proof. See Munkres, p.108-109.

For the second lemma we need one more definition.

Definition 1.2.6. If (X, d) is a metric space and A ⊂ X is a bounded subspace, the diameterof A is the number sup{d(a1, a2) | a1, a2 ∈ A}.

Lemma 1.2.7 (Lebesgue number lemma). Let (X, d) be a metric space and U an open coveringof X. If X is compact, then there exists a δ > 0, called a Lebesgue number for U , such thatevery subset of X with diameter less than δ is contained in some U ∈ U .

Proof. If X ∈ U , the lemma holds for every δ > 0. Assume that X /∈ U . Because X is compact,we can choose a finite number of sets U1, . . . , Un ∈ U such that X =

⋃ni=1 Ui. Let Vi = X\Ui;

by assumption, every Vi is nonempty. So we can define a function f : X → R by

f(x) =1

n

n∑i=1

d(x, Vi),

where d(x, Vi) = inf{d(x, v) | v ∈ Vi}. Let x ∈ X and i ∈ {1, . . . , n} such that x ∈ Ui. Let ε > 0such that B(x, ε) ⊂ Ui. Then we have d(x, Vi) ≥ ε and therefore, f(x) ≥ ε

n . This shows thatf(x) > 0 for all x. Now f is a continuous function so since X is compact, f attains a minimum δ.This is the number we were looking for. Namely, let A ⊆ X be a subset with diameter less than δ.Let a ∈ A, then A ⊂ B(a, δ). Let m ∈ {1, . . . , n} such that d(a, Vm) = max{d(a, Vi) | i = 1 . . . n}.Then we have

δ ≤ f(a) ≤ 1

n

n∑i=1

d(a, Vi) = d(a, Vi)

so B(a, δ) ⊆ X\Vm = Um. We see that A ⊂ Um, which completes the proof.

Proof of proposition 1.2.4. Let U be a covering of X, such that every U ∈ U is evenly coveredby p. The set {γ−1(U) | U ∈ U } is an open covering of [0, 1]. This is a compact space, sothe Lesbesgue number lemma gives us a δ such that every subset with diameter less than δ

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is contained in some γ−1(U). We divide [0, 1] in a number of closed subintervals, say n, eachwith length less than δ. Call the endpoints of these intervals 0 = t0, t1, . . . , tn = 1. For everyi ∈ {0, . . . , n− 1}, we have γ([ti, ti+1]) ⊂ Ui for some Ui ∈ U .

We will now construct γ. Let γ(0) = y0. Suppose that γ(t) is defined for all t ∈ [0, ti], in sucha way that p ◦ γ(t) = γ(t). Consider the set Ui, which contains γ([ti, ti+1]). This set is evenlycovered by p, so p−1(Ui) is a disjoint union of open sets Vα such that p|Vα is a homeomorphismonto Ui. One of these, say V0, contains γ(ti). Let, for all t ∈ [ti, ti+1], γ(t) = (p|V0)−1(γ(t)). Inthis way we go on until 1 is reached.

Since p|V0 is a homeomorphism, γ is continuous on [ti, ti+1] for every i. By the Pasting lemma,γ is continuous on all of [0, 1], so it is a path. We have p ◦ γ = γ by construction; also, γ startsat y0. So this is the lifting we were looking for.

Now the uniqueness. Suppose that χ : [0, 1]→ Y is some lifting of γ beginning at y0. Thenγ(0) = χ(0) = y0. Suppose that γ(t) = χ(t) for all t ∈ [0, ti]. Let V0 be as above, thenγ(t) = (p|V0)−1(γ(t)) for t ∈ [ti, ti+1]. The set χ([0, 1]) is connected, and since the Vα are disjoint,it must be contained in one of the Vα. Because χ(ti) = γ(ti) ∈ V0, this must be V0. By definitionof lifting, we have p◦χ(t) = γ(t) for all t. Restricting t to [ti, ti+1], we can write p|V0 ◦χ(t) = γ(t)for all t ∈ [ti, ti+1]. This means (since p|V0 is a homeomorphism) that χ(t) = (p|V0)(t) = γ(t) forall t ∈ [ti, ti+1]. It follows that χ = γ. So γ is unique.

The next lemma states that homotopies of paths, with which we mean homotopy relative0, 1, in X can be lifted to homotopies in Y . The proof of the lemma is similar to the proof ofproposition 1.2.4. We will therefore not give it here, it can be found (for example) in Munkres.

Lemma 1.2.8. Let p : (Y, y0) → (X,x0) be a covering map. If F : [0, 1] × [0, 1] → X is acontinuous map starting at x0, then there is a unique lifting F : [0, 1]× [0, 1]→ Y of F , which isa continuous map starting at y0. If F is a path homotopy, then F is also a path homotopy.

Proof. See Munkres, p. 343/344.

Proposition 1.2.9. Let p : (Y, y0) → (X,x0) be a covering map such that p(y0) = x0. Let αand β be two paths in X with the same endpoint x1. Let α and β be the liftings of α and β to Ywith starting point y0. If α and β are path homotopic, then α and β have the same endpointand are path homotopic.

Proof. Let F : [0, 1] × [0, 1] → X be a path homotopy from α to β. Then F (0, 0) = x0. Bylemma 1.2.8, there is a unique path homotopy F : [0, 1]× [0, 1]→ Y starting at y0, which is alifting of F . Then F ({0} × [0, 1]) = {y0} and F ({1} × [0, 1]) = {y1}, where y1 is some point inY .

Now F |[0,1]×{0} is a lifting of F |[0,1]×{0} = α with starting point y0. By the uniqueness of path

liftings (lemma 1.2.4), we must have α = F |[0,1]×{0}. Similarly, β = F |[0,1]×{1}. So F is a path

homotopy from α to β; this also shows that α(1) = β(1) = y1.

Finally, we will define the notion of ‘number of sheets’ of a covering. For this to be well-defined,we need the covered space to be path connected. Some authors, for example Bredon and Massey[13], put in the definition of a covering map the requirement that both spaces are path connectedand locally path connected (and that all evenly covered sets are path connected, but this followsfrom local path connectedness and the fact that restrictions of covering maps to open sets arecovering maps). These conditions are necessary for most interesting results. In the next section,we will restrict to coverings that satisfy these conditions.

The following lemma was taken from Massey [13].

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Lemma 1.2.10. Let X be a path connected space and p : Y → X a covering map. For any twopoints x0, x1 in X, the sets p−1(x0) and p−1(x1) are in bijection.

Proof. Let γ : [0, 1]→ X be a path from x0 to x1. We define a map f : p−1(x0)→ p−1(x1) inthe following way. For any y ∈ p−1(x0), let γy be the lift of γ to Y with starting point y (whichis possible by proposition 1.2.4). Let f(y) = γy(1). Because p ◦ γy = γ, this is indeed an elementof p−1(x1). With the inverse path γ−1, we can define a map g : p−1(x1)→ p−1(x0) in the sameway. Clearly, g is the inverse of f and therefore f is a bijection.

Definition 1.2.11. If, in the situation above, |p−1(x0)| = k <∞, then k is called the numberof sheets of the covering. A covering with k sheets is called a k-fold covering.

1.2.2. The universal covering space

In this section we will, as mentioned, assume that if p : Y → X is a covering map, then both Xand Y are path connected and locally path connected. The main resource we use is Bredon [2].Also, we refer to Munkres [15] for some proofs.

Universal coverings are coverings that ‘factor through every other covering’. In this section,we will show that a covering space which is is simply connected, i.e. has trivial fundamentalgroup, satisfies this condition. First, however, we will show that simply connected coveringspaces do exist. To do this, we need some theory about fundamental groups.

Definition 1.2.12. Given a continuous map f : (X,x0)→ (Y, y0) between two pointed spaces,we can define a map f∗ : π1(X,x0)→ π1(Y, y0) by f∗([γ]) = [f ◦ γ]. (This map is well defined,because if H is a homotopy from γ to γ′, then f ◦H is a homotopy from f ◦ γ → f ◦ γ′. It isa homomorphism, because f ◦ (γ1 ? γ2) = (f ◦ γ1) ? (f ◦ γ2).) We call f∗ the homomorphisminduced by f .

Lemma 1.2.13. Let y0 be a base point in Y , let p : (Y, y0) → X be a covering map and letx0 = p(y0). The induced homomorphism p∗ : π1(Y, y0) → π1(X,x0) is injective. Its imageconsists of the classes of loops in X with base point x0, that lift to loops in Y with base points y0.

Proof. See Bredon, p. 141.

Definition 1.2.14. Let p : Y → X be a covering map, let x0 ∈ X and let y0 ∈ p−1(x0).For [α] ∈ π1(X,x0), denote the lifting of α to Y with starting point y0 by α. The liftingcorrespondence derived from p is the (set) map

φy0 : π1(X,x0)→ p−1(x0)

[α] 7→ α(1)

This definition depends on the choice of y0. If we let y0 vary, we get a map

p−1(x0)× π1(X,x0)→ p−1(x0)

(y, [α]) 7→ y[α] = φy(α).

It is easy to see that this defines a right group action of π1(X,x0) on p−1(x0), the monodromyaction (see for example Bredon, p. 146). The isotropy group of a point under this action satisfiesthe following equation:

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Lemma 1.2.15. For the isotropy group Jx0 of x0, the following equality holds:

Jx0 = p∗(π1(Y, y0)).

Proof. For a ∈ J , we have

a = [f ] ∈ Jx0 ⇐⇒ f lifts to a loop with base point x0 ⇐⇒ a ∈ p∗(π1(X,x0))

where the last equivalence holds by lemma 1.2.13.

Finally, we will give the construction of a simply connected covering space. We will followBredon’s proof, although his terminology is different from ours. For example, he puts strongerrequirements on a covering and his ‘locally relatively simply connected’ is our ‘semilocally simplyconnected’ (this is the term most authors use).

Definition 1.2.16. A topological space X is called semilocally simply connected if for everyx ∈ X, there is a neighbourhood U of x such that the homomorphism π1(U, x) → π1(X,x)induced by the inclusion U → X is trivial. If U is such a neighbourhood, it is sometimes calledrelatively simply connected. Another term for ‘semilocally simply connected’ is locally relativelysimply connected.

Theorem 1.2.17. A path connected, locally path connected and semilocally simply connectedspace X has a simply connected covering space X.

Proof. Again, whenever we talk about homotopy of paths, we mean homotopy relative {0, 1}.Choose a base point x0 ∈ X and let X be the set of homotopy classes of paths in X with startingpoint x0; let x0 ∈ X be the class of the constant path at x0. Let p : X → X be the map thatsends [γ] to γ(1).

We will first define a topology on X. Let

B = {U ⊂ X | U is open, path connected and relatively simply connected}.

By assumption on X, this is a basis for the topology on X. If γ is a path in X and γ(1) ∈ U ∈ B,define a subset U[γ] of X by

U[γ] = {[λ] ∈ p−1(U) | λ ' γ ? α for some path α in U}.

The sets U[γ] will form a basis for the topology on X. To show this, we prove a few properties:

(i) If λ ∈ U[γ], then U[λ] = U[γ].

Let α be a path in U such that λ ' γ ? α. Let [β] ∈ U[λ]. Then there is a path α′ in Usuch that β ' λ ? α′. Also, We find that β ' (γ ? α) ? α′ ' γ ? (α ? α′), which shows that[β] ∈ U[γ]. So U[λ] ⊆ U[γ]. The other inclusion holds because λ ? α−1 ' γ ? α ? α−1 ' γ, so[γ] ∈ U[λ].

(ii) If U, V ∈ B such that U ⊂ V , and γ(1) ∈ U , then U[γ] ⊂ V[γ].

This follows because every path α in U is a path in V .

Now consider the set {U[γ] | U ∈ B, γ(1) ∈ U}. Let [γ] ∈ X. Since B is a basis for the topologyon X, there is a U ∈ B such that γ(1) ∈ U . Then [γ] ∈ U[γ].

Suppose that λ ∈ U[γ1] ∩ V[γ2] = U[λ] ∩ V[λ] (by (i)). Then λ(1) ∈ U ∩ V so λ(1) ∈W ⊂ U ∩ Vfor some W ∈ B. So [λ] ∈ W[λ]. By (iii), we have W[λ] ⊂ U[λ] ∩ V[λ]. This shows that the set

{U[γ] | U ∈ B, γ(1) ∈ U} is a basis for a topology on X.

Next, we will show that p is a covering map.

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(iii) For all γ, p is a bijection between U[γ] and U .

Surjectivity is clear, since both X and U are path connected. Let [λ], [λ′] ∈ U[γ] suchthat λ(1) = λ′(1). By (i), there exists a path α in U[γ] such that λ ' λ′ ? α. Thenα(0) = λ(1) = λ′(1) = α(1) so α is a loop. Because U[γ] is relatively simply connected,i∗(α) is nullhomotopic, and therefore λ ' λ′. So [λ] = [λ′]. This shows that p is injectiveon U[γ].

(iv) The map p is open and continuous.

By the above, p(U[γ]) = U is open for all γ with γ(1) ∈ U . Because the sets U[γ] are abasis, it follows that p is open. For all U ∈ B, we have p−1(U) =

⋃{U[γ] | γ(1) ∈ U} is

open. Again, since B is a basis, it follows that p is continuous.

We see that every U ∈ B is evenly covered by p. So p is a covering map.

Finally, we need to show that X is simply connected, i.e., X is path connected and itsfundamental group is trivial.

(v) Let F : [0, 1] × [0, 1] → X be a homotopy relative {0} and let y0 = F (0, t). WriteF (s, t) = Ft(s), then Ft is a path starting at y0. Define a map γ ∈ X by γ(t) = [Ft]. Thenγ is a path in X which is a lifting of the path Ft(1) = F (1, t) in X.

It is clear that p ◦ γ(t) = F (1, t) for all t. We need to prove that γ is continuous. Lett0 ∈ [0, 1] and let U ∈ B be a neighbourhood of Ft0(1). For t ∈ [0, 1] near t0, we haveFt(1) ∈ U . Suppose, withou loss of generality, that t < t0. Then α = F |{1}×[t0−t] is a path inU such that Ft0 ' Ft ?α, where the homotopy is given by F . Therefore γ(t) = [Ft] ∈ U[Ft0 ].So for t near t0, p maps γ(t) to F (1, t) in U . Because F (1, t) is a continuous function andp maps U[Ft0 ] homeomorphically to U , it follows that γ is continuous at t0. This holds forevery t0 ∈ [0, 1], so γ is continuous.

(vi) X is path connected.

Let [γ] ∈ X and define, in X, a homotopy F from x0 to γ(1) by F (s, t) = γ(st). By (v),this gives us a path in X from x0 to [γ]. Because [γ] was arbitrary, it follows that X ispath connected.

(vii) π1(X, x0) = {0}.Let [γ] ∈ π1(X,x0), and define ft : [0, 1] → X by ft(s) = γ(st), for all t ∈ [0, 1]. Thenbecause p(γ(t)) = p([ft]) = ft(1) = γ(t), the map γ : [0, 1]→ Y with γ(t) = [ft] is a pathin Y which is a lift of γ. Now f0(t) = γ(0) = x0 for all t, so γ(0) = [f0] = x0 By definitionof the monodromy action, x0 · [γ] = γ(1) = [f1] = [γ]. Suppose that x0 · [γ] = x0. Then wesee that [γ] = x0 · [γ] = x0 so [γ] is the unit element 0 of π1(X,x0). So by lemma 1.2.15,we find that {0} = {[γ] ∈ π1(X,x0) | x0 · [γ] = x0} = Jx0 = p∗(π1(X, x0)). Because p∗ isinjective (see lemma 1.2.13), it follows that π1(X, x0) = {0}.

This finishes the proof.

In fact, the result is even stronger:

Theorem 1.2.18. Let X be a path connected, locally path connected and semilocally simplyconnected space and let x0 ∈ X. Then for every subgroup H of π1(X,x0) there exists a coveringspace Y , with covering map p : Y → X, and a point y0 ∈ p−1(x0) such that p∗(π1(Y, y0)) = H.

16

For the proof, see Munkres, p. 495.

Our next goal is proving that simply connected covering spaces are indeed universal. First,we prove a proposition about maps from connected spaces to covering spaces. The proof can befound in Bredon, however, we give a proof by Wilkins [19], because it is more general.

Recall that a topological space X is connected if and only if the only subsets that are bothopen and closed are X and the empty set.

Proposition 1.2.19. Let p : Y → X be a covering, let Z be a connected space and letf1, f2 : Z → Y be continuous maps such that p ◦ f1 = p ◦ f2. If there is a z ∈ Z such thatf1(z) = f2(z), then f1 = f2.

Proof. Let D = {z ∈ Z | f1(z) = f2(z). We will show that D = Z.Take any z ∈ Z, let w = p ◦ f1(z) = p ◦ f2(z) and let U be an open neighbourhood of w which

is evenly covered by p. Suppose that z ∈ D; let x = f1(z) = f2(z). Let V be the connectedcomponent of p−1(U) that contains x, then A = f−1

1 (V ) ∩ f−12 (V ) is a nonempty open set in Z.

Because p ◦ f1 = p ◦ f2 and p restricted to V is a homeomorphism, we have f1(a) = f2(a) for alla ∈ A. So A ⊂ D. We showed that D is open.

Now suppose that f1(z) 6= f2(z). Let V1 be the component of p−1(U) containing f1(z), and V1

the one containing f1(z). Then V1 ∩ V2 is empty. Let B be the open set f−11 (V1)∩ f−1

2 (V2), thatcontains z. We have f1(B) ⊂ V1 and f2(B) ⊂ V2 so f1(B) ∩ f2(B) = ∅. Therefore B ⊂ Z\D.This shows that Z\D is open.

It follows that D is both open and closed. Because Z is connected and D is nonempty byassumption, D must be all of Z.

Lemma 1.2.20. Let X, Y and Z be path connected and locally path connected spaces and letp : Y → X, q : Y → Z and r : Z → X be continuous maps such that r ◦ q = p. If p and r arecovering maps, then q is also a covering map.

Proof. We first show that q is surjective. Let z ∈ Z and let α be a path from z0 to z. Letβ = r ◦ α, this is a path in X beginning at x0. Lift β to a path β in Y beginning at y0, thenq ◦ β is a lift of β to Z that starts at z0. By the uniqueness of liftings starting at the same point,we have q ◦ β = α. Now q(β(1)) = α(1) = z. This shows that q is surjective.

Let z ∈ Z. We will give a neighbourhood V of z evenly covered by q. Let x = r(z). Let U1

and U2 be neighbourhoods of x that are evenly covered by p resp. r, then U = U1 ∩U2 is evenlycovered by both. We can assume that U is path connected. Let V be the connected componentof r−1(U) that contains z. Let {Wi} be the set of connected components of p−1(U). Then forevery i, q(Ui) is contained in a connected component of r−1(U). Therefore, the components ofq−1(V ) are those components Uj of p−1(U) for which q(Uj) ⊂ V . Now consider the followingcommutative diagram:

Uj

p|Uj��

q|Uj// V

r|V��

U

.

Because p|Uj and r|V are homeomorphisms, q|Uj = (r|V )−1 ◦ p|Uj is also a homeomorphism. SoV is evenly covered by q.

Lemma 1.2.21. Let p : (Y, y0)→ (X,x0) be a covering map and let f : (W,w0)→ (X,x0) becontinuous. There exists a lifting f : (W,w0)→ (Y, y0) of f if and only if

f∗(π1(W,w0)) ⊂ p∗(π1(Y, y0)).

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Proof. See Munkres, p. 478.

Corollary 1.2.22. Let p : (Y, y0) → (X,x0) and r : (Z, z0) → (X,x0) be covering maps suchthat Y is simply connected. Then there is a unique covering map q : (Y, y0)→ (Z, z0) such thatr ◦ q = p.

Proof. Because Y is simply connected, we automatically have p∗(π1(Y, y0)) ⊂ r∗(π1(Z, z0)). Bythe preceding lemma, p can be lifted to a map q : (Y, y0) → (Z, z0) such that r ◦ q = p. Bylemma 1.2.20, q is a covering map.

Uniqueness of q follows immediately from proposition 1.2.19.

Now universality of a covering p : Y → X is exactly the condition that for any coveringr : Z → X, there exists a covering map q : Y → Z. So we showed that any simply connectedcovering is universal. It follows that simply connected coverings are equivalent in the followingsense:

Definition 1.2.23. Let p : Y → X and p′ : Y ′ → X be covering maps. The covering spacesY and Y ′ are called equivalent if there is a homeomorphism f : Y → Y ′ with p = p′ ◦ f . Thishomeomorphism is called an equivalence of covering spaces (or an equivalence of covering maps).

For this reason, we often talk about the universal covering space.

The equivalence follows immediately from corollary 1.2.22 if we take both Y and Z simplyconnected. We get covering maps p′ : (Y, y0) → (Z, z0) and r′ : (Z, z0) → (Y, y0) such thatr◦p′ = p and p◦r′ = r. We see that p◦r′◦p′ = p and r◦p′◦r′ = r. Now r′◦p′(y0) = y0 = idY (y0),so it follows from proposition 1.2.19 that r′ ◦ p′ = idY . Similarly, p′ ◦ r′ = idZ .

Finally, we will state a theorem that relates the fundamental group of the universal coveringspace to a certain subgroup of the automorphisms of Y , the deck transformations.

Definition 1.2.24. Let p : Y → X be a covering map. A homeomorphism f : Y → Y such thatp ◦ f = p is called a deck transformation or covering transformation. We denote the set of decktransformations of Y with respect to p by Aut(Y, p). It is clear that (Aut(Y, p), ◦) is a group.

Note that a deck transformation of a covering is an equivalence of this covering with it-self. Note also, that if f : Y → Y ′ is an equivalence of covering spaces, then the mapf : Aut(Y, p)→ Aut(Y ′, p′) that sends g to f ◦ g ◦ f−1 is an isomorphism.

Theorem 1.2.25. If p : Y → X is a covering map and Y is simply connected, then Aut(Y, p)is isomorphic to π1(X).

Proof. Let x0 ∈ X be a base point. Because simply connected coverings are equivalent wecan assume that Y is the space we constructed in the proof of Theorem 1.2.17, consisting ofthe homotopy classes of paths starting at x0. For any [α] ∈ π1(X,x0), let f[α] : Y → Y bethe continuous map that sends [γ] ∈ Y to [α ? γ]. Clearly, this map is well-defined and it is ahomeomorphism: its inverse is f[α−1]. Moreover, p([α ? γ]) = α ? γ(1) = γ(1) = p([γ]), so f[α] isa deck transformation.

This gives us a map f : π1(X,x0) → Aut(Y, p). Clearly, this is a homomorphism withtrivial kernel, so it is a monomorphism. It is surjective as well. Namely, let g : Y → Y bea deck transformation, let [γ] ∈ Y and let [γ′] = g([γ]). Then p([γ′]) = p([γ]), which meansthat γ′(1) = γ(1). So γ′ ? γ−1 is a loop, based at x0. Now the map f[γ′?γ−1] maps [γ] to[γ′ ? γ−1 ? γ] = [γ′] = g([γ]). By proposition 1.2.19, it follows that f[γ′?γ−1] = g.

Conclusion: f is an isomorphism.

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2. Orbifolds

Orbifolds are generalizations of manifolds. Just like manifold atlases, we have orbifold atlases,and most constructions on manifolds can be extended such that they work for orbifolds as well.

In the first part of this chapter, we will give the definition of an orbifold and derive somegeneral properties. In the second part, we will do some algebraic topology on orbifolds.

2.1. The orbifold structure

2.1.1. Definition and examples

Definition 2.1.1. Let X be a Hausdorff space. An orbifold chart is a tuple (U, U ,Γ, ϕ) whereU ⊂ X is open, U is an open subset of Rn, Γ is a finite topological group acting on U andϕ : U → U is a continuous map that factors through a homeomorphism ϕ : U/Γ→ U .

We will usually denote (U, U ,Γ, ϕ) just by ϕ, if it is clear which chart is meant.

If the action of Γ on U is effective, then the chart is called reduced. In this thesis, we willassume that all orbifold charts are effective. See Dragomir [6] for details on orbifolds withnon-reduced charts.

Definition 2.1.2. Let (Ui, Ui,Γi, ϕi) and (Uj , Uj ,Γj , ϕj) be two orbifold charts such thatUi ⊆ Uj . We say that the first embeds in the second if there exists a pair (fij , ϕij), called anembedding of charts, where

(i) fij : Γi → Γj is an injective homomorphism;

(ii) ϕij : Ui → Uj is an embedding which is equivariant with respect to fij , i.e. for all γ ∈ Γi,we have ϕij(γx) = fij(γ)ϕij(x),

such that the following diagram commutes:

Ui

��

ϕij// Uj

��

Ui/Γi

ϕi ∼=

��

ϕij// Uj/Γi

fij��

Uj/Γj

∼= ϕj

��

Ui� � // Uj

Here the action of Γi on Uj is given by γy = fij(γ)y = ϕij(γϕ−1ij (y)) (so the upper part commutes

by definition) and fij sends [x] ∈ Uj/Γi to [x] ∈ Uj/Γj .In particular, ϕj ◦ ϕij = ϕi.

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The maps fij and ϕij in this definition are defined up to conjugation, resp. composition,with elements of Γj . If Ui ⊂ Uj ⊂ Uk, there exists a γ ∈ Γk such that γϕik = ϕjk ◦ ϕij andγfikγ

−1 = fjk ◦ fij . For the proof, see Dragomir [6], p. 33.

Definition 2.1.3. An orbifold atlas on a Hausdorff space X is a collection

U = {(Ui, Ui,Γi, ϕi) | i ∈ I}

of orbifold charts such that {Ui} is a covering of X closed under finite intersection and for allcharts ϕi, ϕj in U , if Ui ⊂ Uj then Ui embeds in Uj .

Definition 2.1.4. An orbifold is a Hausdorff space X with an orbifold atlas.

We will denote an orbifold by O and its underlying space by |O|.

Let U and V be two orbifold atlases on O. We say that V refines U if for every chart(U, U ,ΓU , ϕ) ∈ U , there is a chart (V, V ,ΓV , ψ) ∈ V such that U ⊂ V and there exists anembedding (fUV , ϕUV ) of U into V . We call two orbifold atlases equivalent if they have acommon refinement. It is easy to check that this indeed defines an equivalence relation. Manyauthors define an orbifold to be a Hausdorff space with an equivalence class of orbifold atlases. Itcan be showed, in the same way as for manifolds, that for each such class there exists a maximalatlas.

We need one more definition before we give some examples.

Definition 2.1.5. The local group Γx of a point x ∈ O is defined as follows: let (U, U ,Γ, ϕ) bea chart around x, then Γx ⊂ Γ is the isotropy group of any point in ϕ−1(x). It follows fromthe existence of embeddings that the local group is defined up to isomorphism. If Γx is trivial,then x is called regular, otherwise, x is singular. The set of all singular points of O is called thesingular locus of O. It is denoted by ΣO.

It can be proved, see Dragomir [6], that the singular locus is closed and nowhere dense.

Example 2.1.6.

(i) Every manifold is an orbifold. The local groups are all trivial.

(ii) A manifold with boundary is an orbifold. For every point outside the boundary, the localgroup is trivial. For a boundary point, the local group is Z2, acting on Rn by reflection ina hyperplane.

(iii) Consider the action of Zn on R2 by rotation over 2πn . The orbit space R2/Zn is an orbifold.

The only singular point is [0], its local group is Zn.

(iv) An interesting example is the so-called teardrop, showed in the following picture.

This a 2-sphere with one singular point, modeled on the quotient of a neighbourhood of theorigin in R2 by the action of Zn by rotation, as above. We will denote this space by Tn.

20

(v) An example similar to the teardrop is the rugby ball Rm,n, which is a sphere with twosingular points. For one of those points the local group is Zm, for the other it is Zn. Theyact on R2 by rotation around the origin over 2π

m resp. 2πn .

If (U, U ,Γ, ϕ) is an orbifold chart and V ⊂ U , then we can define a chart on V by restriction.Let V = ϕ−1(U). We will show that V is invariant under Γ. Let x ∈ V and let γ ∈ Γ. Thenϕ(γx) = ϕ([γx]) = ϕ([x]) = ϕ(x), so γx ∈ V . It follows that (V, V ,Γ, ϕ|V ) is an orbifold chart.We will denote this chart by ϕ|V .

Just like for manifolds, we have transition maps on orbifolds. Let U be an orbifold atlasand (Ui, Ui,Γi, ϕi) and (Uj , Uj ,Γj , ϕj) charts in U , such that Ui ∩ Uj is nonempty. Because themaps ϕi : Ui → Ui and ϕj : Uj → Uj are not necessarily injective, we cannot directly definetransition maps ϕ−1

i (Ui ∩ Uj)→ Uj and ϕ−1j (Ui ∩ Uj)→ Ui as in the manifold case. However,

there is a chart (Uk = Ui ∩ Uj , Uk,Γk, ϕk) in U with two embeddings (fki, ϕki) and (fkj , ϕkj).We use these to construct transition maps.

Definition 2.1.7. In the situation above, the transition maps are given by

τij = ϕkj ◦ ϕ−1ki : ϕki(Uk) ⊂ Ui → Uj

τji = ϕki ◦ ϕ−1kj : ϕkj(Uk) ⊂ Uj → Ui

λij = fkj ◦ f−1ki : fki(Γk) ⊂ Γi → Γj

λij = fki ◦ f−1kj : fkj(Γk) ⊂ Γj → Γi.

Many authors (e.g. Davis [4] and Boileau, Maillot, Porti [1]) give a slightly different definitionof orbifold atlases. Instead of putting the conditions that the orbifold atlas is closed under finiteintersections and that there are embeddings for charts contained in another chart, they onlyrequire the existence of ‘local transition maps’ on Rn for overlapping charts. To be precise,let (Ui, Ui,Γi, ϕi) and (Uj , Uj ,Γj , ϕj) be overlapping charts. The condition set is that for allx ∈ Ui and y ∈ Uj with ϕi(x) = ϕj(y), there are neighbourhoods Vx of x and Vy of y and ahomeomorphism (diffeomorphism, in the differentiable case) τ : Vx → Vy such that ϕj ◦ τ = ϕi.

Clearly, our definition implies this definition. Pronk [16] (p. 88-89) proves that given anorbifold atlas satisfying the weaker condition, there exists an orbifold atlas satisfying ourconditions.

We end this section with the definition of a continuous map between orbifolds.

Definition 2.1.8. An orbifold map O1 and O2 is a continuous map f : |O1| → |O2| that satisfiesthe following: for every x ∈ O1, there exist charts ϕ1 : U1 → U1 around x and ϕ2 : U2 → U2

around f(x) such that f maps U1 into U2, and there is a map f : U1 → U2 such that f ◦ϕ1 = ϕ2◦f .Two orbifolds O1 and O2 are called homeomorphic if there exist orbifold maps f : O1 → O2 andg : O2 → O1 such that g ◦ f = idO1 and f ◦ g = idO2 .

2.1.2. Developable orbifolds

The first important result about orbifolds is that they can arise as the quotient of a manifold.

Theorem 2.1.9. If M is a manifold and Γ a group acting properly discontinuously on M , thenM/Γ has an orbifold structure.

21

Proof. Step 1. First, we will construct a set of charts that covers M .Let q : M →M/Γ be the quotient map. Let x ∈M/Γ and let x ∈ q−1(x). Then by Theorem

1.1.20, there is a neighbourhood Ux of x which is invariant under the finite isotropy group Γxand is disjoint from the translations γUx for γ /∈ Γx, such that the map px : Ux/Γx →M/Γ is ahomeomorphism onto its image.

Let (Vx, Vx, φx) be a chart of M with x ∈ Vx. Let U ′x =⋂γ∈Γx

γ(Ux ∩ Vx) ⊂ Vx. BecauseΓx is finite and the maps x 7→ γx are homeomorphisms, U ′x is open. It is easy to verify thatU ′x is also invariant under Γx, and that γU ′x ∩ U ′x = ∅ for γ /∈ Γx. The map px sends U ′x/Γxhomeomorphically to Ux := px(U ′x/Γx) ⊂M/Γ.

Let Wx = φ−1x (U ′x) ⊂ Rn. The group Γx acts on Wx by γ · w = φ−1

x (γ · φx(w)). We have thefollowing commutative diagram:

Wxφx∼=

//

��

U ′x

��

Wx/Γxφx∼=// U ′x/Γx

px ∼=��

Ux

Let ψx : Wx → Ux be defined by ψx(y) = px([φx(y)]) = p(φx(y)). Then (Ux,Wx,Γx, ψx) is achart for M/Γ.

The set A = {Ux | x ∈M/Γ} is an open covering of M/Γ.

Step 2 Closure of A under finite intersections.Suppose that U = Ux1 ∩ · · · ∩ Uxk is nonempty for some x1, . . . , xk ∈M/Γ. For every i, the

inverse image of Uxi under q is the set {γU ′xi | γ ∈ Γ}. If Γxi Γ, this set consists of more thanone connected component, one of which is U ′xi (without loss of generality we assume that this setis connected). Consider the set q−1(U) =

⋂i=1,...,k q

−1(Uxi). One of its connected components iscontained in U ′x1 , call it Vx1 . The other components are Vxi ⊂ U ′xi with Vxi = γiVx1 for someγi ∈ Γ.

We had a manifold chart (Wx1 , U′x1 , φx1) on M . Let W = φ−1

x1 (Vx1). The group

Λ = Γx1 ∩ γ2Γx2γ−12 ∩ · · · ∩ γkΓxkγ

−1k

acts on W by λw = φx1(λ · φ−1x1 (w)). Let ψ : W → U be defined by y 7→ px1([φx1(y)]). With a

diagram similar to the one in step 1, we see that the tuple (U,W,Λ, ψ) is an orbifold chart.Let A be the closure of A under finite intersection. We will take U = {(U, U ,Λ, ϕ) | U ∈ A}

as the set of charts for the orbifold atlas.

Step 3. Construction of embeddings of charts.Suppose that (Ui,Wi,Γi, ϕi) and (Uj ,Wj ,Γj , ϕj) are charts in U such that Ui ⊂ Uj . By

construction, there are U ′i ⊂M and U ′j ⊂M , which are connected components of p−1(Ui) and

p−1(Uj) with homeomorphisms φi : Wi → U ′i and φj : Wj → U ′j . The connected components of

p−1(Ui) are contained in those of p−1(Uj), therefore, there is a γ0 ∈ Γ such that γ0U′i ⊂ U ′j . Let

f0 : U ′i → U ′j be the map that sends y to γ0y. We have the following commutative diagram:

Wiφi∼=//

ϕi

U ′if0//

p

��

U ′j

p

��

Wj∼=

φjoo

ϕj~~

Ui� � // Uj

22

Let ϕij be the embedding φ−1j ◦ f0 ◦ φi. An injective homomorphism fij : Γi → Γj is given by

fij(γ) = γ0γγ−10 . This is a well-defined map because fij(γ) · y ∈ U ′j for all y ∈ U ′j , and because

U ′j ∩ τU ′j = ∅ for all τ ∈ γ\Γj . For γ ∈ Γi and x ∈Wx, we have

fij(γ)ϕij(x) = γ0γγ−10 φ−1

j (γ0φi(x))

= φ−1j (γ0γγ

−10 φj(φ

−1j (γ0φi(x))))

= φ−1j (γ0γφi(x))

and this is equal to φ−1j (γ0φi(γx)) = ϕij(γx) because φi(γx) = φi(φ

−1i (γφi(x))) = γφi(x). This

shows that ϕij is equivariant with respect to fij .Finally, consider the diagram

Wi

��

ϕij//Wj

��

Wi/Γi

ϕi ∼=

��

ϕij//Wj/Γi

fij��

Wj/Γj

∼= ϕj

��

Ui� � // Uj

As mentioned before, the upper part commutes by definition. Let x ∈Wi, then ϕi([x]) = ϕi(x).We have

ϕj ◦ fij ◦ ϕij([x]) = ϕj([φ−1j ◦ f0 ◦ φi(x)])

= ϕj(φ−1j ◦ f0 ◦ φi(x))

= p ◦ φj(φ−1j ◦ f0 ◦ φi(x))

= p(f0 ◦ φi(x)) = p(φi(x),

where the last three steps follow from the commutativity of the first diagram on this page. Thisshows that the lower part of the diagram above commutes as well. So we have an embedding.

An example of an orbifold that arises in this way is the cone R2/Zn. The teardrop, however,is not a quotient of a manifold, as we will see later this chapter.

Definition 2.1.10. A covering of an orbifold O is an orbifold O′ with a continuous mapp : |O′| → |O|, called a projection, satisfying the following condition. For each x ∈ O there is aneighbourhood U ∼= U/Γ, such that every connected component Vα of p−1(U) is homeomorphicto U/Γα for some subgroup Γα of Γ. This homeomorphism must respect p, i.e. if ϕ : U → U isa chart for U , then the following diagram commutes:

U/Γα∼= //

��

Vα

p

��

U

aa ??

ϕ��}}

U/Γ ∼=// U

23

Note that in general, an orbifold covering is not an ‘ordinary’ covering of topological spaces.However, if x is a regular point, then it has a neighbourhood U which is evenly covered (in theclassical sense) by p|p−1(U).

Like topological covering maps, all orbifold covering maps are open.

Just like for topological coverings, we can define the number of sheets of an orbifold covering.

Definition 2.1.11. Let p : O → O be an orbifold covering, where O is (path) connected. Letx be a regular point in O. The number of sheets of the covering is |p−1(x)|, if this is finite. Ak-fold orbifold covering is an orbifold covering with k many sheets.

We wrote ‘(path) connected’ here, because just like for manifolds, the properties of connected-ness and path connectedness are equivalent for orbifolds.

We have come to explaining the title of this section.

Definition 2.1.12. An orbifold is called developable (or good) if it has a covering orbifold whichis a manifold.

Developability is a nice property. A lot of information about a developable orbifold can bederived from the manifold that covers it, together with the covering map. In section 2.2.2 wewill show that the developable orbifolds are exactly those that are the quotient of a manifold,like in the beginning of this section.

An example of an orbifold which is not developable is the teardrop. There are many ways toprove this. The proof below is the one given by Guerreiro [10], p. 15.

Proposition 2.1.13. The teardrop Tn is not developable.

Proof. Suppose that Tn has a covering p : M → Tn such that M is a manifold. Because Tn isconnected, we can choose M as well. Let U be a neighbourhood of the singular point. ThenU ∼= U/Zn for some U ⊂ R2 which is homeomorphic to the open disc D2. Therefore p−1(U)must be a disjoint union of open discs. Now Tn\U is homeomorphic to the closed disc D2.Because this space is simply connected, it has only trivial coverings: disjoint copies of D2. Itfollows that M exists of pairs of open and closed discs, joined at the boundaries. Now becauseM is connected, it must be one pair, so M ∼= S2.

Let V ⊂ Tn such that V ∩ U 6= ∅ but V does not contain the singular point. Let x ∈ V ∩ U .Because x ∈ V , the set p−1(x) contains one point. On the other hand, p−1(x) exists of n pointsbecause x ∈ U . Contradiction.

2.2. Algebraic topology

With the right extensions and extra conditions, concepts in algebraic topology that work formanifolds, work for orbifolds as well. Three important examples are the fundamental group,the universal covering space and the Euler characteristic, which we will study in the followingthree sections. From now on, we will assume that every orbifold is connected (and thereforepath connected).

2.2.1. The orbifold fundamental group

There are many ways to approach the fundamental group of obifolds. We will do it as in theclassical construction, with loops, following Boileau, Maillot, Porti [1]. Orbifold paths and

24

orbifold homotopy are defined just like for topological spaces, but with some extra requirements.Thanks to these requirements, we can work with the orbifold fundamental group in the sameway as we are used to.

Definition 2.2.1. A path in an orbifold O is a map α : [0, 1]→ |O| such that α([0, 1]) containsat most finitely many singular points, with for every t0 ∈ [0, 1] for which α(t0) is singular, aso-called local lift of α around t0. This is a tuple (ϕ,W, l), where ϕ : U ⊂ Rn → U is a chart,W ⊂ [0, 1] is a neighbourhood of t0 such that for every t ∈W\{t0}, α(t) is a regular point in Uand l : α−1(U)→ U is a lift of α|α−1(U) to U , so the following diagram commutes:

Uϕ

��

α−1(U)

l

<<

α // U

The local lift can be seen as a choice for the direction of the path when it is lifted to Rn.Note that it is not unique. In fact, if n is the order of the local group at α(t0) then there are n2

choices for l. This is because l is determined by its starting point and its endpoint. Paths withthe same underlying map but different local lifts are not equal. However, we will see that theyare homotopic in certain cases.

Example 2.2.2. Consider the map in the next picture. The orbifold is the disc D2, dividedout by the action of Z3 by rotation around the origin by 2π

3 . We denote this orbifold by Kn.

On the left, we see the underlying map. It goes through the singular point once. In the middle,we see the 9 possible local lifts around the singular point: every combination of a path to andone from the origin is a possibility. On the right, we see our choice for the lift.

If p : O → O is an orbifold covering and α is a path in O, then we can define a path in O withunderlying map p ◦ α in the following way. Suppose that p ◦ α(t) is singular, for some t ∈ [0, 1].There are two possibilities.

If α(t) is singular, then there is a local lift (ϕ,W, l) of α around t. If we choose W smallenough, then (p ◦ ϕ, U , p(U),Γt) is a chart around p(α(t)) (this follows immediately from theproperties of p) and (p ◦ ϕ,W, l) is a local lift of p ◦ α around t.

If α(t) is regular, then there is a chart φ : U → U with α(t) ∈ U , such that (p◦φ, U , U,Γp(α(t)))is a chart around p ◦ α(t). The map φ is a homeomorphism. The tuple (p ◦ φ,U, φ−1 ◦ α) is alocal lift of p ◦ α around t. In fact, we could consider φ−1 ◦ α to be a local lift for α around t aswell (namely, the only possible local lift). Then this is the same construction as when α(t) issingular.

25

Definition 2.2.3. The map defined above is called the projection of α. We denote it by p(α).If α is a path in O and β a path in O such that p(α) = β, then α is called a lift of β.

Note that a path and its projection have the same local lifts. The local lifts are also part ofthe definition of projections. If two paths α and α′ have the same underlying map, but differentlocal lifts, they will have different projections (namely, p(α) and p(α′) will also have the sameunderlying map but different local liftings). The same holds for lifts of α and α′.

The next proposition is the equivalent of proposition 1.2.4 for orbifolds.

Proposition 2.2.4. Let p : O′ → O be a covering map and x0 ∈ O′ a regular point such thatp(x0) is regular. Any path α in O starting in p(x0) can be lifted in a unique way to a path α′ inO′ beginning at x0.

Proof. We start in the same way as for ordinary topological spaces: let U be a covering ofO, such that every U ∈ U is a chart covered by p. The set {α−1(U) | u ∈ U } is an opencovering of [0, 1]. Let δ > 0 such that every subset with diameter less than δ is contained in someα−1(U), which is possible by the Lebesgue number lemma. Divide [0, 1] in closed subintervals[t0 = 0, t1], . . . [tn−1, tn = 1] of length less than δ. Let, for every i ∈ {0, . . . , n− 1}, Ui ∈ U suchthat α([ti, ti+1]) ⊂ Ui.

If α does not cross any singular points, we can lift it just like for topological coverings.Otherwise, let Ik = [tk, tk+1] be the first subinterval containing a point y0 such that α(y0) issingular. Without loss of generality we can assume that y0 is the only such point in Ik, if not,divide Ik into smaller intervals. Let (ϕ,W, l) be the local lift of α around y0. We can assumethat Ik ⊂W , again, if this is not satisfied we just make Ik a bit smaller.

For [0, tk], construct a lift α′ of α as in the normal case. For [tk, tk+1], we do the following.Let (U, U ,Γ, ϕ) be the chart of the local lift (ϕ,W, l). Let V0 be the component of p−1(U) thatcontains α′(tk). Then there is a chart (V0, U ,Γ

′ ⊂ Γ, ϕ0) on V0 such that the right part of thefollowing diagram commutes.

U

��

ϕ0//

ϕ

��

V0

p

��

U/Γ

!!

Ik

l

AA

α // U

The left part commutes by definition of the local lift. If ϕ0 ◦ l(tk) = α′(tk), then we defineα′(t) = ϕ0 ◦ l(t) for t ∈ Ik. Otherwise, there is a γ ∈ Γ such that α′(tk) = γϕ0 ◦ l(tk). Then wedefine α′(t) = ϕ0 ◦ γl(t) for t ∈ Ik (clearly, replacing l by γl does not affect the commutativity ofthe diagram). The rest of the construction of α′ is clear: if α|[ti,ti+1] does not meet any singularpoints, we follow the procedure for ordinary paths, if it does, we use the construction above.

Finally, to prove that α is a path in O′, we need to define local lifts. But this is simple.Suppose that t0 ∈ [0, 1] such that α′(t0) is singular. Then α(t0) must be singular as well. Sowe have a local lift (ϕ,W, l) of α around t0. Now by construction, α′ = ϕ0 ◦ γl on some smallinterval [a, b] around t0 and some γ ∈ Γ (possibly the identity). So (ϕ0, [a, b], γl|[a, b]) is a locallift of α′ around t0.

Now the uniqueness. Suppose that β : [0, 1]→ O′ is another lift of α. On the interval [0, tk],where α has no singular points, β must be equal to α′ by uniqueness of topological liftings. Nowbecause α = p(β), β must have the same local lifts as α, and as α′. From the construction above,

26

we see that on subintervals with a singular point β must also be equal to α′. This shows that α′

is unique.

We will now generalize the concept of path homotopy to orbifolds. Note that if (U, U ,Γ, ϕ) isa chart, then ϕ : U → U is an orbifold covering. Let α be a path and let [a, b] ⊂ [0, 1] such thatα([a, b]) ⊂ U . If α(a) is regular, then by the proposition above, there exists a lift of α|[a,b] to U .This lift is unique up to the starting point. If α(a) is not regular, then a 6= 0 so because U isopen, we can choose a′ < a such that α(a′) is regular and α([a′, b]) ⊂ U . Lift this path to U andrestrict it to the part that lifts α|[a,b].

Definition 2.2.5. Let α be a path in O. Let (U, U ,Γ, ϕ) be a chart in O and let [a, b] ⊂ [0, 1]such that α([a, b]) ⊂ U . Let β be a lift of α|[a,b] to U . Let β′ be a path in U which is homotopicto β relative the endpoints and replace, in α, the part α|[a,b] by the projection of β′. Then weget a new path in O which is elementary homotopic to α. We define orbifold homotopy of pathsas the equivalence relation generated by elementary homotopies.

We have to say ‘generated by’ because otherwise transitivity does not hold. The existenceof elementary homotopies from α1 to α2 and from α2 to α3 does not imply the existence of anelementary homotopy from α1 to α3, just because there may not be a chart that contains boththe images of α1 and α3.

We look again at the example we had above. The path we had is the one on the left in thenext picture. It is orbifold homotopic to the path on the right, because the lifts of the singularpoints are path homotopic in R2.

These paths are also homotopic to the following path, where the starting point for the local liftwas chosen differently.

However, they are not homotopic to the following path.

27

Here we also took a different local lift, but in contrast to the one of the second picture, this liftcannot be obtained from the ones above by composition with a group element. Clearly, thisgeneralizes to the following:

Two orbifold paths with the same underlying map are orbifold homotopic if and only if theirlocal lifts are equal up to composition with elements of the local groups.

It follows that the paths going through a singular point with local group of order n fall apart inn homotopy classes.

We prove three more important properties of orbifold path homotopy.

(i) If the singular locus is empty, then orbifold homotopy is the same as homotopy of theunderlying space.

Let α and α′ be orbifold homotopic paths as in definition 2.2.5. Let F : [0, 1]× [0, 1]→ Ube a homotopy from β to β′. We can define a homotopy F : [0, 1]× [0, 1]→ O from α toα′ by

G(y, t) =

{ϕ−1 ◦ F (y, t) ◦ ϕ if y ∈ [a, b]

α(t) = α′(t) otherwise.

Note that this is well-defined, because ϕ is a homeomorphism.

Suppose that α and α′ are path homotopic in the classical sense. Again using the fact thatall charts are homeomorphisms, we can show that they are orbifold homotopic by makinga sequence of paths obtained from each other by elementary homotopies, starting with αand ending with α′.

(ii) If p : O′ → O is an orbifold covering and α, α′ are orbifold homotopic paths in O′, thenthe projections p(α) and p(α′) are orbifold homotopic.

(iii) If p : O′ → O is an orbifold covering, α and α′ are orbifold homotopic paths in O, and α,α′ are lifts of α resp. α′ with the same initial point, then α, α′ are orbifold homotopic.

Properties (i) and (ii) hold because elementary homotopy only depends on homotopy inRn, not on the group acting on it, and because we used the local lifts to define projectionsand lifts of paths.

Definition 2.2.6. Let x0 ∈ O be a regular point. A path α in O with α(0) = α(1) = x0 iscalled a loop based at x0. The orbifold fundamental group πorb

1 (O, x0) is the set of homotopyclasses op loops based at x0, with the usual group operation ?.

Just like for the ordinary fundamental group, one can check that πorb1 (O, x0) is indeed a group.

Also, a change of the base point gives an isomorphic group.

We will calculate the fundamental group of a few spaces. As we have seen, every path thatgoes through a singular point is homotopic to a path that does not go through this point, becausewe can ‘pull it away’ in Rn. So we only have to calculate the homotopy classes of the pathsoutside the singular locus.

Intuitively, the fundamental group of an orbifold should be at least as complicated as thefundamental group of its underlying space, the space obtained by replacing all singular pointsby regular points. On the other hand, it should be at most as complicated as the fundamentalgroup of the manifold that we get by removing all singular points, thus creating a number ofholes. Namely, a loop around a hole is different from a loop ‘outside’ a hole, while a loop arounda singular point can be homotopic to a loop outside it. We will use this observation to get a firstidea what the fundamental group of a certain orbifold could be.

28

Example 2.2.7.

(i) The teardrop Tn

Let x be the singular point of Tn. Intuitively, π1(Tn) should be ‘something between’ π1(S2)and π1(Tn\{x}) = π1(D2). As both are trivial, this would mean that π1(Tn) is trivial.Clearly, this is true: every loop outside the singular locus is homotopic (‘via the bottom ofTn’) to the constant loop at its base point.

(ii) The cone Kn = D2/ZnThe underlying space of Kn is D2, which has trivial fundamental group; if we removethe singular point we get an annulus, which has fundamental group Z. It turns out thatπorb

1 (Kn) is isomorphic to Zn. It is not difficult to see why.

Fix a base point and take a loop that goes around the singular point once. Lift it to R2.Without loss of generality, we can assume that the lift starts on the positive x-axis (thesingular point is mapped to zero) and that it goes around the origin counterclockwise.Then the lift ends on the line that crosses the x-axis at angle 2π

n . If we take a loop thatgoes around the singular point twice, its lift ends on the line that crosses the x-axis atangle 4π

n . In general, a loop that goes around the singular point m times lifts to a pathstarting on the x-axis and ending on the line that crosses the x-axis at angle 2mπ

n . So aloop going around the singular point n times is lifted to a loop, which is nullhomotopic inR2. Therefore the homotopy class of this loop is the trivial one.

With this idea in mind, it is clear that loops with m windings around the singular point inone direction are homotopic to loops with n−m windings around the singular point in theother direction. The following picture shows this for n = 3.

(iii) The rugby ball Rn,n

Just like for Kn, we have πorb1 (Rn,n) ∼= Zn. Namely, consider loops around the orbifold,

outside the singular locus. We consider them as loops around one of the singular points(it does not matter which one, because their local groups are the same). As in the lastexample, we see that there are n homotopy classes, corresponding to the group Zn.

29

2.2.2. The universal orbifold covering

In this section we will construct a simply connected orbifold covering. The idea is similar tothe proof for ordinary topological spaces. However, we need to define an orbifold atlas on thecovering space. We will simultaneously define charts and a topology on the covering orbifold.

Theorem 2.2.8. A connected orbifold (O, x0) has a covering (O′, x′0) such that πorb1 (O′, x′0) is

trivial.

Proof. Just like in the classical construction, we let O′ be the set of orbifold homotopy classesof paths in O starting at x0, and x′0 the homotopy class of the constant path at x0. We definep : O′ → O by [α] 7→ α(1). This map is surjective because O is path connected.

Next, we define maps from open subsets of Rn to O′, that will give both a topology and anatlas on O′. Let {(Ui, Ui,Γi, ϕi)} be an atlas for O. Without loss of generality we can assumethat every Ui is simply connected (otherwise, we divide it into smaller open sets). For each i,choose a base point yi ∈ Ui and a point yi ∈ ϕ−1(yi). Let Fi = p−1(yi), so this is the set oforbifold homotopy classes of paths in O from x0 to yi.

We define a map ψi : Ui × Fi → p−1(Ui) as follows. Let z ∈ Ui and [α] ∈ Fi. Let λ be apath in Ui from yi to z, and project λ to a path λ in Ui from yi to ϕ(z). Because Ui is simplyconnected, the homotopy class of α ? λ does not depend on the choice of λ; clearly, it does notdepend on α either. So we can define ψi(z, [α]) = [α?λ]. The situation is sketched in the picturebelow.

This map is surjective. Namely, let [γ] ∈ p−1(Ui), so γ(1) ∈ Ui. Let z ∈ ϕ−1i (γ(1)), let θ be a

path in Rn from z to yi and project θ to a path θ in O from γ(1) to yi. Then ψi(z, [γ ? θ]) = [γ].

We give every Fi the discrete topology and Ui×Fi the product topology. The sets ψi(A) withA ⊂ Ui × Fi open form a basis for a topology on p−1(Ui). The connected components of p−1(Ui)are the sets ψi(Ui × {[α]}) for [α] ∈ Fi. Let Λi be the group

{γ ∈ Γi | ψi(γx, [α]) = ψi(x, [α]) for all x ∈ Ui, [α] ∈ Fi}.

We give O′ the atlas generated by the tuples (ψi(Ui × {[α]}), Ui,Λi, ψi|Ui×{[α]}) for all i and all

[α] ∈ Fi.It is clear that by this construction, p becomes an orbifold covering map. We still need to show

that πorb1 (O′, x′0) is trivial. Let η : [0, 1]O′ be a loop, so η(0) = η(1) = x′0, the homotopy class

of the constant path at x0. We define a homotopy H : [0, 1]times[0, 1]→ O′ from the constantloop c : [0, 1]→ O′, c(t) = x′0 to η as follows. Let H(s, u) : [0, 1]→ O be the map that sends tto η(s)(tu). Clearly, H is continuous. It satisfies the conditions for an ‘ordinary’ homotopy:

• H(s, 0)(t) = η(s)(0) = x0 for all t, because η(s) ∈ O′

• H(s, 1)(t) = η(s)(t) for all t.

30

• H(0, u) = η(0)(u) = η(1)(u) = H(1, u) = x0 for all u.

It is easy to see that the extra requirements of orbifold homotopy are satisfied as well.Conclusion: every path in O′ is nullhomotopic.

It can be proved, in a similar way as for topological coverings, that simply connected orbifoldcoverings are universal. From this, we can derive another proof that Tn is not developable.Namely, we saw in example 2.2.7 that Tn is simply connected. Therefore, it is its own universalcovering. So any other covering of Tn is equivalent to Tn and is therefore an orbifold.

Just like for topological coverings, we have the notion of a deck transformation of an orbifoldcovering.

Definition 2.2.9. Let p : O′ → O be an orbifold covering. A deck transformation of thiscovering is an orbifold homeomorphism (see definition 2.1.8) f : O′ → O′ such that p ◦ f = p.

We can now prove that all developable orbifolds are quotients of manifolds, and the other wayround.

Proposition 2.2.10. A connected orbifold O is developable if and only if it is the quotient of amanifold by a properly continuous group action.

In the proof, we use the following proposition.

Proposition 2.2.11. Let (Ui, Ui,Γi, ϕi) and (Uj , Uj ,Γj , ϕj) be orbifold charts such that Ui ⊂ Uj.If φij , ψij : Ui → Uj are two embeddings such that ϕj ◦ϕij = ϕi = ϕj ◦ψij, then there is a uniqueλ ∈ Γj such that ψij = λ ◦ ϕij.

Proof. See Moerdijk, Pronk [14], p. 20.

Proof of proposition 2.2.10. Suppose that O = M/Γ, where Γ is a discrete group acting properlyon M . We will show that the quotient map q : M →M/Γ is a projection.

Let x ∈ M/Γ and let (Ux, Ux,Γx, ϕx) be the chart as constructed in Theorem 2.1.9. Wedefined Ux = p(U ′x) where U ′x was a neighbourhood of some x ∈ q−1(x) which was invariantunder the isotropy group Γx, such that γU ′x ∩ U ′x = ∅ for γ /∈ Γx. We had a manifold chartφ : Ux → U ′x and ϕx was defined as q ◦ φ. As mentioned in step 2 of the proof of Theorem 2.1.9,q−1(Ux) = {γU ′x | γ ∈ Γ}. Each component of this set is homeomorphic to Ux.

It follows immediately from the above that q : M → M/Γ is a projection. This proves thefirst implication.

Suppose that p : M → O is an orbifold covering, where O is connected and M is a manifold.Let x ∈ M . Let U ∼= U/Γ be a neighbourhood of p(x) which is evenly covered by p and letV be the connected component of p−1(U) that contains x. Suppose that f1, f2 : M → M betwo deck transformations such that f1(V ) ∩ V 6= ∅ 6= f2(V ). Because p ◦ f1 = p and f1 is ahomeomorphism, f1 permutes the components of p−1(U). From f1(V ) ∩ V 6= ∅, it follows thatf1(V ) = V . The same holds for f2.

There is a homeomorphism ψ : U → V such that ϕ ◦ ψ−1 = p, where ϕ : U → U is the map ofthe chart on U . Now ψ−1 ◦ f1 ◦ψ and ψ−1 ◦ f2 ◦ψ are embeddings of U in itself. By proposition2.2.11, there is a unique γ ∈ Γ such that ψ−1 ◦ f1 ◦ ψ = γ ◦ ψ−1 ◦ f2 ◦ ψ, so f1 = ψ ◦ γ ◦ ψ−1 ◦ f2.Because Γ is finite, this shows that the set {f ∈ Aut(M,p) | f(V ) ∩ V 6= ∅} is finite. It followsfrom Theorem 1.1.14 that Aut(M,p) acts properly discontinuously on M .

31

Because O is connected, we can assume that M is connected as well. So M has a universalcovering M ′, which is a manifold. Let N be the universal orbifold covering of O, as constructedabove. Then N is equivalent to M ′ and is therefore a manifold.

Let q : N → O be the projection, that maps [γ] to γ(1). We will prove that O ∼= N/Aut(N, q).Let q : N/Aut(N, q) → O be the map that sends [x] to q(x) (where x is itself an equivalenceclass). Because q ◦ f = q for all f ∈ Aut(M, q), this map is well defined. The inverse map isgiven by y 7→ [z] for a z ∈ q−1(y). For this to be well-defined, Aut(N, q) must act transitivelyon the elements in p−1(y), i.e., for all [γ1], [γ2] ∈ q−1(y) there must be a deck transformation fsending [γ1] to [γ2].

Just like we did in the proof of Theorem 1.2.25, we can define a map f : π1(O)→ Aut(N, q)by letting f[α] = f([α]) be the map that sends [γ] ∈ N to [α ? γ]. If [γ1], [γ2] ∈ q−1(y),

then γ1(1) = γ2(1) so γ2 ? γ−11 is a loop. The deck transformation f[γ2?γ

−11 ] sends [γ1] to

[γ2 ? γ−11 ? γ1] = [γ2]. This shows that Aut(N, q) acts transitively on the elements in q−1(y). It

follows that q is bijective.Because q and the quotient map π : N → N/Aut(N, q) are open and continuous and q = q ◦π,

the map q is a homeomorphism. Also, it is clear that it satisfies the conditions of an orbifoldhomeomorphism on the charts, see definition 2.1.8. We conclude that O ∼= N/Aut(N, q).

Just like in he classical case, it can be showed that the map f : π1(O)→ Aut(N, q) defined inthe proof is an isomorphism. This proves the following:

Theorem 2.2.12. If p : (O′, x′0)→ (O, x0) is a covering map such that πorb1 (O′, x′0) is trivial,

then Aut(O′, p) is isomorphic to πorb1 (O, x0).

With this theorem, we can easily compute the fundamental groups of the orbifolds in example2.2.7(ii) and (iii). However, does not help us to compute, for example, π1(Rm,n) with m 6= n,because this orbifold is not developable.

It turns out that the Seifert-van Kampen theorem holds for the orbifold fundamental group.This result is very useful for computing the fundamental groups of more complicated orbifolds.

There are two ways of proving Seifert-van Kampen for the classical fundamental group. Thefirst one is directly from the definitions, see for example Munkres [15]. The second one, which isless basic but much more efficient, is by using universal coverings. Both proofs can be adaptedfor orbifolds.

We will not prove the theorem. Dragomir [6] refers to Haefliger [11] for a proof.

Recall the definition of the amalgamated product:

Definition 2.2.13. Let

K

j2��

j1// G1

i1��

G2 i2// Γ

be a commutative diagram of groups and group homomorphisms. We say that (Γ, i1, i2) is theamalgamated product of G1 and G2 via j1 and j2, if for every commutative diagram of groups

K

j2��

j1// G1

f1��

G2f2// H

32

there is a unique homomorphism ϕ : Γ→ H such that the following diagram commutes:

K

j2��

j1// G1

f1

��

i1��

G2

f2 ,,

i2// Γ

ϕ

H

The amalgamated product of G1 and G2 via j1 and j2 is denoted by G1 ∗j1,K,j2 G2, or byG1 ∗K G2, if it is clear what the maps j1 and j2 are.

An open subset A of an orbifold O has a unique orbifold structure U such that the inclusionA→ O extends to an embedding of orbifolds. The orbifold (A,U ) is an open suborbifold of O.

Theorem 2.2.14 (Seifert-van Kampen for orbifolds). Let O be an orbifold and O1,O2 two opensuborbifolds (definieren!), such that O1, O2 and O1 ∩ O2 are connected and O = O1 ∪ O2. Letx0 ∈ O1 ∩ O2 and let G = πorb(O1 ∩ O2, x0). Then we have the following isomorphism:

πorb1 (O, x0) ∼= πorb

1 (O1, x0) ∗G πorb1 (O2, x0).

Here the homomorphisms G → πorb1 (O1, x0) and G → πorb

1 (O2, x0) are (j1)∗ and (j2)∗, wherej1 : O1 ∩ O2 → O1 and j2 : O1 ∩ O2 → O2 are the inclusion maps.

With Seifert-van Kampen we can compute, for example, the fundamental group of the rugbyball Rm,n.

Example 2.2.15. Let O1∼= (Km)◦ be an open neighbourhood around the point with local

group Zm, and O2∼= (Kn)◦ an open neighbourhood around the point with local group Zn,

such that their intersection is an annulus contained in the regular part of Rm,n. We haveπorb

1 (O1) ∼= Zm and πorb1 (O2) ∼= Zn, as we saw in example 2.2.7, and piorb

1 (O1 ∩ O2) ∼= Z. BySeifert-van Kampen, we get that πorb

1 (Rm,n) ∼= Zm ∗Z Zn. Now recall the following property ofthe amalgamated product:

Proposition 2.2.16. Suppose that in the situation of definition 2.2.13, the map j1 : K → G1

is surjective. Then G1 ∗ KG2∼= G2/N , where N ⊂ G2 is the normal subgroup generated by

j2(ker(j1)).

A loop in O1 ∩ O2 that goes around the hole of he annulus one time is a generator of G. Theinclusion of this generator into O1 (or O2) is a loop that goes around the singular point once, soit is a generator of πorb

1 (O1) (or πorb1 (O2)). It follows that πorb

1 (Rm,n) is a quotient of Zn, or ofZm, if we interchange O1 and O2.

2.2.3. The orbifold Euler characteristic

The Euler characteristic is an important topological invariant (it is invariant under homotopyequivalence). It was originally defined for polyhedrons. Nowadays, in algebraic topology, theEuler characteristic is usually defined using the homology group of a space, see Bredon [2]. Thisdefinition makes use of the fundamental theorem of abelian groups.

33

Theorem 2.2.17 (Fundamental theorem of abelian groups). Let G be a finitely generated abeliangroup. Then there are unique integers r ≥ 0 and n1, n2, . . . , nt with ni > 1 for all i, such thatn1|n2| · · · |nt and

G ∼= Zr × Zn1 × Zn2 × · · · × Znt .

The number r is called the rank of G.

Definition 2.2.18. Let X be a topological space such that Hi(X) is finitely generated for all i,and unequal to 0 for only finitely many i. The Euler characteristic of X is

χ(X) =∑i

(−1)i rank(Hi(X)).

Example 2.2.19.

(i) The space R2 is contractible and therefore its only nontrivial homology group is H0(R2) ∼= Z.The rank of Z is 1. We see that χ(R2) = 1.

(ii) The 0th and 2nd homology of S2 are Z, in all other cases, the homology group is trivial.Therefore, the Euler characteristic of S2 is 2.

(iii) For T 2, we have H0(T 2) ∼= Z ∼= H2(T 2) and H1(T2) ∼= Z2. So χ(T 2) = 1− 2 + 1 = 0.

(iv) The homology of P 2(R) is Z in the 0th case, Z2 in the 1st case and trivial in all other cases.The rank of Z2 is 0, and it follows that χ(P 2(R)) = 1.

If a space has a CW -structure, then its Euler characteristic can be computed in a very simpleway:

Theorem 2.2.20 (Euler-Poincare). Let K be a finite CW-complex and, for every i, ai thenumber of i-cells. Then χ(K) is defined and

χ(K) =∑i

(−1)iai.

Proof. Let, for all i, Ci = ⊕σ i-cell{zσ | z ∈ Z} ∼= ⊕σi-cellZ be the ith cellular group. Thenai = rank(Ci). Denote by Zi ⊂ Ci the subset of cycles and let Bi = ∂Ci+1. Let Hi be the groupZi/Bi = Hi(C∗(K)) ∼= Hi(K). We have the following short exact sequences:

0→ Zi → Ci → Bi+1 → 0

0→ Bi → Hi → Bi → 0.

Claim. If 0→ A→ B → C → 0 is a short exact sequence of finitely generated abelian groups,then rank(B) = rank(A) + rank(C).

We will first finish the proof and then prove the claim.

It follows from the claim that

rank(Ci) = rank(Zi) + rank(Bi−1)

rank(Zi) = rank(Bi) + rank(Hi).

34

Therefore∑i

(−1)i(rank(Bi) + rank(Hi)) =∑i

(−1)i(rank(Ci)− rank(Bi−1))

=∑i

(−1)i rank(Ci) +∑i

(−1)i−1 rank(Bi−1)

The numbers rank(Bi) on the left and the right cancel each other. This gives that

χ(K) =∑i

(−1)i rank(Hi) =∑i

(−1)i rank(Ci) =∑i

(−1)iai.

Proof of the Claim. Recall the Splitting lemma:

Lemma 2.2.21 (Splitting lemma). Let 0 → Fi→ G

q→ H → 0 be a short exact sequence ofabelian groups. The following are equivalent:

(i) There exists a homomorphism r : G→ F such that r ◦ i = idF ;

(ii) There exists a homomorphism s : H → G such that q ◦ s = idH ;

(iii) There is an isomorphism ϕ : F ⊕H → G such that the following diagram commutes:

0 // Ff 7→(f,0)

//

i''

A⊕Gϕ

��

(f,h)7→h// H // 0

G

q

77 .

If F , G and H are vector spaces, then the properties of the lemma always hold (in fact, itis enough if either F and G, or G and H are vector spaces). Namely, choose a basis F for F ,then F := i(F ) is a basis for i(F). Expand this to a basis G for G. Let r : G→ F be the linearextension of the map that sends x ∈ G to i−1(x) if x ∈ F and to 0 if x ∈ G\F . Then r is ahomomorphism such that r ◦ i = idF .

Let 0f→ A

g→ B → C → 0 be a short exact sequence of finitely generated abelian groups. The

sequence A⊗Q f⊗id−→ B ⊗Q g⊗id−→ C ⊗Q→ 0 is also exact, and because Q is torsion-free the mapf ⊗ idQ is injective. So we have a short exact sequence 0→ A⊗Q→ B ⊗Q→ C ⊗Q→ 0.

Write A ∼= Zra × Zk1 × · · · × Zkt , B ∼= Zrb × Zm1 × · · · × Zmu and C ∼= Zrc × Zn1 × · · · × Zmv .We find that

A⊗Q ∼= (Zra × Zk1 × · · · × Zkt)⊗Q∼= (Z⊗Q)ra × Zk1 ⊗Q× · · · × Zkt ⊗Q∼= Qra .

In the same way, B ⊗Q ∼= Qrb and C ⊗Q ∼= Qrc . So we get a short exact sequence of vectorspaces 0 → Qra → Qrb → Qrc → 0. By what we saw above, it follows that Qrb ∼= Qra × Qrc .This shows that rb = ra + rc. So rank(B) = rank(A) + rank(C).

The proofs of the following two statements can be found in Bredon, [2].

35

Lemma 2.2.22. Let p : (Y, y0)→ (X,x0) be a covering map, let (W,w0) be a path connected,locally path connected and simply connected space and let f : (W,w0) → (X,x0). For anyy ∈ p−1(x0) there exists a uniquely determined lift g : (W,w0)→ (Y, y0) with g(w0) = y.

Corollary 2.2.23. A covering space of a CW-complex is a CW-complex (in particular, it hasthe weak topology).

With these statements, we can prove a useful proposition.

Proposition 2.2.24. Let X be a connected space and p : Y → X a covering map with k <∞sheets. If X is a finite CW-complex, then Y is also a finite CW-complex and χ(Y ) = kχ(X).

Proof. Let σ be an n-cell and fσ : Dn → X a characteristic map. Because Dn is path connected,locally path connected and simply connected, it follows from lemma 2.2.22 that there are exactlyk lifts of fσ to Y . These lifts give Y the structure of a CW-complex, with in each dimension ktimes as many cells as X. Therefore χ(Y ) = kχ(X).

We will now define the orbifold Euler characteristic. This number depends on the underlyingspace of an orbifold, but also on the orbifold structure. It is defined by way of a formula derivedfrom the Euler-Poincare formula for the underlying space.

Definition 2.2.25. Let O be an orbifold such that |O| has a finite CW-complex decompositionwhere the local group is constant in each cell, i.e. within each cell, the local groups of the pointsare isomorphic. The orbifold Euler characteristic of O is defined as

χorb(O) =∑ci

(−1)dim(ci)1

|Γ(ci)|

where the ci are the cells of the CW-complex and |Γ(ci)| is the order of the local group at ci.

Note that if O is a manifold, the definition simplifies to the Euler-Poincare formula. Notealso that the Euler characteristic of the underlying space can be different from the orbifoldEuler characteristic of O. In particular, the Euler characteristic is always an integer, while theorbifold Euler characteristic is not, in general. Moreover, different orbifold structures on thesame underlying space lead to a different orbifold Euler characteristic.

Example 2.2.26.

(i) Consider the space Kn = D2/Zn. An example of a CW-structure for Kn is given by thefollowing picture:

It is built from two points, two lines and a disc. The points of the disc and those on theline lying in the interior of Kn are regular. The line that forms the boundary of Kn haslocal group Z2 acting by reflection, just like the point y on ∂(Kn). The point x in theinterior of Kn has local group Zn. We find that χorb(Kn) = 1

2 + 1n − 1− 1

2 + 1 = 1n .

36

(ii) For Tn, we can build a CW-structure with a point with local group Zn and a disc of regularpoints, joining the boundary of the disc to the point. The orbifold Euler characteristic ofTn is 1 + 1

n .

(iii) The rugby ball Rm,n has a CW-structure existing of two points, one with local group Znand the other with local group Zm, one regular line connecting these two points and a discof regular points, joined to the 1-skeleton as in the picture.

This gives that χorb(Rm,n) = 1m + 1

m − 1 + 1 = 1m + 1

n .

For the orbifold Euler characteristic, we can derive a similar result as for the topological Eulercharacteristic.

Proposition 2.2.27. Let O be an orbifold with a CW-structure as in definition 2.2.25. Letp : O′ → O be an orbifold covering with k <∞ sheets. Then χorb(O′) = kχorb(O).

Proof. Let (U, U ,Γ, ϕ) be an orbifold chart such that U is evenly covered by p. Write U = tαVα,and let Γ′ ⊂ Γ such that Vα ∼= U/Γ′. Consider the following diagram:

Vα

p

��

U/Γ′∼=oo

U U/Γ

OO

∼=oo Uoo

ϕll

``

ϕ′

dd

Let y ∈ U be a regular point. Then the number of points in ϕ−1(y) is #Γ. For each point

s ∈ p−1(y)∩Vα, the set ϕ−1(s) contains |Γ′| many elements. Therefore, there are |Γ||Γ′| elements of

p−1(y) in Vα. As the set p−1(y) contains k elements, we find that∑

α|Γ||Γ′| = k, so

∑α

1|Γ′| = k

|Γ| .

Let {ci | i ∈ I} be the cells in the CW-structure of O. Just like in corollary 2.2.23, thepreimages under p of these cells form a CW-structure for O′. Denote these by {dj | j ∈ I ′}.Then we get

kχ(O) =∑ci

k(−1)dim(ci)

|Γ(ci)|=∑ci

∑p−1(ci)

(−1)dim(ci)

|Γ(p−1(ci))|

=∑dj

(−1)dim(dj)

|Γ(dj)|= χ(O′).

This is an important result that turns out to be very useful if one wants to determine if anorbifold is developable or not. We will see this in the next chapter.

37

3. Classification of two-dimensionalorbifolds

In this chapter we will classify the compact 2-dimensional orbifolds on the basis of the Eulercharacteristic. We will see which types of singular points a 2-orbifold can have and derive whichof the 2-orbifolds are developable. Before we can do this, we need to prove a few results aboutthe group O2(R).

3.1. The orthogonal group O2(R)

We start by recalling some basic theory about O2(R) and semidirect products, that can forexample be found in Van der Geer [8]. In the end of this section, we will classify the finitesubgroups of O2(R).

Definition 3.1.1. Let (X, d) be a metric space. An isometry on X is a map f : X → X that‘preserves distance’, i.e. for all x, y ∈ X we have d(f(x), f(y)) = d(x, y). If X = R2 and d is theEuclidean metric, then we call f a Euclidean plane geometry or just a Euclidean geometry.

In section 4.1, we will study all Euclidean geometries. For now, we restrict to those that fixthe origin.

Definition 3.1.2. The orthogonal group O2(R) is the group of Euclidean isometries that fixthe origin. The elements of O2(R) are called orthogonal maps.

The group operation on O2(R) is composition of maps. Any isometry is injective, butsurjectivity is not obvious. However, it turns out that O2(R) exists of only two types of elements,which are clearly invertible.

Lemma 3.1.3. An orthogonal map is either a rotation around the origin or a reflection in aline through the origin.

As a corollary, every orthogonal map is linear. We can therefore represent it by a matrix. Arotation by an angle α corresponds to the following matrix(

cos(α) − sinαsinα cos(α)

).

It has determinant cos2(α)−− sin2(α) = 1.Let us denote the map ‘reflection in the x-axis’ by s, and ‘rotation by an angle α’ by rα. Take

any reflection t in a line l through the origin. Let α be the angle between l and the x-axis, thent = rαs. We see that every reflection is a product of a rotation and s. Now s is represented bythe matrix (

1 00 −1

),

it has determinant −1. So every reflection t = rαs has determinant det(rα) det(s) = −1.

38

Definition 3.1.4. The special orthogonal group SO2(R) is the subgroup of O2(R) consisting ofthe rotations or, equivalently, the orthogonal maps with determinant 1.

As SO2(R) is the kernel of the determinant homomorphism det : O2(R)→ {±1}, it is a normalsubgroup of O2(R).

From now on, we will just write O2 and SO2. The group O2 is not commutative: For areflection σ and a rotation r, the map rσ is a reflection so rσ = (rσ)−1 = σ−1r−1 = σr−1. Wewill see that O2 is a semidirect product.

Definition 3.1.5. Let N and H be two groups, with a homomorphism φ : H → Aut(N). Thesemidirect product N oφ H is the cartesian product N ×H with the following group operation:

(n1, h1) · (n2, h2) = (n1φ(h1)(n2), h1h2).

Instead of N oφ H, the notation H nφ N is sometimes used. If it is clear what φ is, we willomit the subscript.

Proposition 3.1.6. Let G be a group with subgroups N and H, such that

(i) N is a normal subgroup;

(ii) N ∩H = {e};

(iii) G = NH.

Let φ : H → Aut(N) be the homomorphism defined by φ(h)(n) = hnh−1. Then the mapπ : N oφ H → G, (n, h) 7→ nh is an isomorphism.

Proof. Note that φ is well-defined because N is a normal subgroup. We check that π is ahomomorphism:

π((n1, h1) · (n2, h2)) = π((n1h1n2h−11 , h1h2)) = n1h1n2h

−11 h1h2

= n1h1n2h2 = π((n1, h1))π((n2, h2)).

Surjectivity of π follows by (iii). Suppose that (n, h) ∈ N oφ H such that nh = e, so h = n−1.By (ii), it follows that n = h = e. So π is injective.

Corollary 3.1.7. The orthogonal group O2 is isomorphic to the semidirect product SO2 o Z2,where Z2 is generated by the reflection s in the x-axis.

An important example of a semidirect product is the dihedral group.

Definition 3.1.8. The dihedral group Dn of order 2n is the group of symmetries of the regularpolygon with n sides.

We will describe Dn as the group of symmetries of a particular polygon, namely ∆n, theregular polygon in R2 with n sides, one vertex in the point (1, 0) and center the origin. In thisway, Dn is a subgroup of O2. It consists of the rotations rα with α ∈ {2πk

n | k = 0, 1, . . . , n− 1}and the reflections in the lines through either the origin and two vertices, or the origin, onevertex and the middle of the opposite side of ∆n. Let αn = 2πk

n . We can write

Dn = {rkαn | k = 0, 1, . . . , n− 1} ∪ {rkαns | k = 0, 1, . . . , n− 1},

where s again denotes the reflection in the x-axis.The subgroup of Dn consisting of the rotations {rkαn | k = 0, 1, . . . , n− 1}, generated by rα1 ,

is isomorphic to Zn. It is a normal subgroup, because it has index 2.

We derived another corollary of proposition 3.1.6:

39

Corollary 3.1.9. The dihedral group Dn is isomorphic to Zn o Z2, where Z2 is generated by sand Zn is the subgroup of SO2 generated by rotation by an angle 2π

n .

Before we classify the finite subgroups of O2, we first look at SO2.

Lemma 3.1.10. All finite subgroups of SO2 are isomorphic to Zn for some n.

Proof. We can represent every element in SO2 by a complex number e2πiα, with α ∈ [0, 1], where2πα is the angle of the rotation. Let G = {e2πα1 , e2πα2 , . . . , e2παn} be a finite subgroup of SO2.Let αi be the least of α1, . . . , αn which is bigger than 0. Suppose that there is a j ∈ {1, . . . , n}such that αj is not a multiple of αi, in particular, αj > 0. Let k = bαjαi c and β = αj − kαi. Then

0 < β < αi and e2πβ = eαje−kαi ∈ G. This is in contradiction with the choice of αi.This shows that G is generated by one element x. Since G is finite, x has finite order n. Then

G ∼= 〈x〉 ∼= Zn.

Theorem 3.1.11. The finite subgroups of O2 are Z2 generated by s, Zn generated by r 2πn

and

the dihedral groups Dn.

Proof. We have already seen that all finite subgroups of SO2 are isomorphic to Zn. We only needto show that any other finite subgroup of O2 is Z2 or a dihedral group. Let G ⊂ O2 be a finitesubgroup which is not a subgroup of SO2. Then it contains a reflection σ and the homomorphismdet : G→ {±1} ∼= 〈σ〉 is surjective. The kernel of this homomorphism is N = G ∩ SO(2). Wecan visualize this by a short exact sequence 0→ N → G→ 〈σ〉 → 0. By the first isomorphismtheorem, G/N ∼= 〈σ〉. So G = N〈σ〉. Since N ∩ 〈σ〉 = {idR2}, it follows by proposition 3.1.6that G ∼= N o 〈σ〉. Now N is a finite subgroup of SO2, so it is cyclic. If N is trivial, thenG ∼= 〈σ〉 ∼= Z2; if not, then G ∼= Zn o Z2 for some n. By corollary 3.1.9, this is Dn.

3.2. The compact 2-orbifolds

The goal of this section is to give a list of the compact 2-orbifolds with nonnegative Eulercharacteristic. For the first theorem, differentiability of the orbifolds is necessary. This means,just like for manifolds, that the transition maps are C∞. From now on, we will assume ourorbifolds to be differentiable. We will only give a sketch of the proof. For a detailed proof, seeGuerreiro [10], p. 17.

Theorem 3.2.1. Let O be a 2-dimensional orbifold and let x ∈ O. The local group at x isisomorphic to a finite subgroup of O2(R).

Sketch of the proof. We use the Slice Theorem (see [7], p. 98,99 for a precise formulation anda proof): Let Γ be a Lie group acting properly on a manifold M . For every x ∈ M there is aneighbourhood Ux ⊂M of x and an open set Vx ⊂ TxM such that there exists a Γx-equivariantdiffeomorphism Ux → Vx.

Let Γ be a group acting on a manifold M by diffeomorphisms, i.e., the map φγ : M → Msending x to γx is a diffeomorphism for all γ ∈ Γ. The derivative dxφγ is a linear map fromTxM to TγxM . If γ ∈ Γx, then TγxM = TxM so dxφγ is bijective. This gives us a linear actionof Γx on TxM .

Let U ∼= U/Γ be a neighbourhood of x ∈ O. We can construct a Γx-invariant metric g′ on TxUby taking the standard Euclidean metric g and defining g′(y, z) = 1

|Γx|∑

γ∈Γx(dφγ(y), dφγ(z).

With this metric, the determinant of every γ ∈ Γx is 1 or −1 and therefore Γx is isomorphic to afinite subgroup of O2(R).

40

Because Γ is finite, it is a Lie group and its action is properly discontinuous. By the SliceTheorem, there is a Γx-equivariant diffeomorphism f : Vx → Wx, with φ−1(x) ∈ Vx ⊂ U andWx ⊂ TxU . Via this diffeomorphism, the actions of Γx on Vx and Wx are equivalent.

It follows from Theorem 3.1.11 that

Corollary 3.2.2. A two-dimensional orbifold can have three types of singular points:

(i) Points with local group Z2, acting on R2 by reflection in a line (the x-axis, for example);

(ii) Points with local group Zn, acting on R2 by rotation around the origin by an angle 2πn ;

(iii) Points with local group Dn, generated by reflection in a line (for example the x-axis) androtations by 2π

n .

Singular points of the first type are called mirror points, points of the second type ellipticpoints of index n and points of the third type corner reflectors of index n.

Because each of these points has an open neighbourhood homeomorphic to an open subsetof the (half-)plane, it follows that the underlying space of a 2-orbifold is always a topological2-manifold, possibly with boundary.

A connected 2-manifold is usually called a surface. Two important theorems, the ClassificationTheorem for Compact Surfaces and the Classification Theorem for Compact Surfaces withBoundary, show that there are only a few types of surfaces, and we know exactly which ones.We will state the theorems after the following definition. For the proofs, see Massey, [13].

Definition 3.2.3. Let S1 and S2 be disjoint surfaces. Let D1 ⊂ S1 and D2 ⊂ S2 be closedsubsets which are homeomorphic to the closed disc D2. Let S′1 = S1\D◦1 and S′2 = S2\D◦2 andlet h : ∂D1 → ∂D2 be a homeomorphism. The connected sum S1#S2 is the quotient spaceS′1∪S′2/ h, where h is the equivalence relation formed by identifying x and h(x), for all x ∈ ∂D1.

The space S1#S2 is again a surface. It can be proved that the definition is independent of thechoices of D1, D2 and h.

Theorem 3.2.4 (Classification Theorem for Compact Surfaces). A compact surface is isomorphicto one of the three following spaces:

(i) The sphere S2

(ii) A connected sum of tori T 2# · · ·#T 2

(iii) A connected sum op projective planes P 2(R)# · · ·#P 2(R).

Proof. See Massey.

Theorem 3.2.5 (Classification Theorem for Compact Surfaces with Boundary). A compactsurface N with boundary is of the form M\(D1 t · · · tDk) where M is a compact surface andD1, . . . , Dk are disjoint open discs in M .

Proof. See Massey.

With these two theorems and the next lemma, we can compute the Euler characteristic of thepossible underlying spaces of a compact, connected 2-orbifold.

41

Proposition 3.2.6. If S1 and S2 are disjoint surfaces, then

χ(S1#S2) = χ(S1) + χ(S2)− 2.

Proof. Take a triangulation of S1 and let a0, a1 and a2 be number of points, lines and triangles.A triangulation is a CW-complex, and therefore χ(S1) = a0 + a2 − a1. In the same way, we findthat χ(S2) = b0 + b2 − b1, where b0, b1 and b2 are the number of points, lines and triangles insome triangulation of S2. Now S1#S2 can be formed by letting D1 and D2 be closed trianglesof the triangulations. The space S1#S2 then has a triangulation with a0 + b0 − 3 many points,a1 + b1 − 3 many lines and a2 + b2 − 2 many triangles. So

χ(S1#S2) = a0 + b0 − 3 + a2 + b2 − 2− (a1 + b1 − 3) = χ(S1) + χ(S2)− 2.

Corollary 3.2.7. The Euler characteristic of a compact surface K is

(i) 2 if K ∼= S2

(ii) 2− 2n if K ∼= T 2# · · ·#T 2︸ ︷︷ ︸n times

(iii) 2− n if K ∼= P 2(R)# · · ·#P 2(R)︸ ︷︷ ︸n times

.

Proposition 3.2.8. Let N ∼= M\(D1 t · · · tDk) be a compact surface with boundary. Thenχ(N) = χ(M)− k.

Proof. The idea is as follows: take a triangulation of M , again, this is a CW-structure for M .Then N is homeomorphic to M with the interior of k of the triangles removed. This decreasesχ(M) with k.

We can now give a formula for the Euler characteristic of an orbifold that only depends onthe indices of its elliptic points and corner reflectors.

Proposition 3.2.9. Let O be a 2-dimensional orbifold with n elliptic points of index a1, . . . , anand m corner reflectors of index b1, . . . , bm, then

χorb(O) = χ(|O|)−n∑i=1

(1− 1

ai)− 1

2

m∑j=1

(1− 1

bj).

Proof. Take a cell division for |O| such that in O, the local groups in the cells are constant. Tocalculate χ(|O|), we add 1 for each 0-cell and −1 for each 1-cell. To calculate χorb(O), firstconsider the elliptic points. For each point, instead of adding 1, we add 1

ai. This decreases χ(|O|)

by 1 − 1ai

. At a corner reflector, we have a 0-cell with local group Dn, plus a line, a 1-cell of

mirror points. For the 0-cell, we add 12bj

instead of 1; for the 1-cell, we add −12 instead of −1.

So a corner reflector decreases χ(|O|) by 1− 12bj− 1

2 = 12(1− 1

bj).

We have come to the classification of the compact, connected 2-orbifolds. There are onlyfinitely many of those with Euler characteristic zero. Proving this is just a combinatoric puzzle.

42

We start with a surface with nonnegative Euler characteristic and add elliptic points and cornerreflectors until the Euler characteristic is zero.

The only compact surfaces without boundary with nonnegative Euler characteristic are S2

(characteristic 2), P 2(R) (characteristic 1), T 2 (characteristic 0) and P 2(R)#P 2(R), which ishomeomorphic to the Klein bottle and has Euler characteristic 0. An orbifold O with a surfacewithout boundary as underlying space cannot have corner reflectors. Namely, these only occurin the presence of a reflection, which is a boundary in the surface.

Let a1 ≤ · · · ≤ an, ai > 1 for all i, be the indices of the elliptic points in O. If the coveringspace of O is S2, then by proposition 3.2.9: χ(O) = 2−

∑ni=1(1− 1

ai). Setting this to zero gives

the equation∑n

i=1(1− 1ai

) = 2. Because 1− 1ai∈ [1

2 , 1), there are no solutions for n = 1, n = 2and n ≥ 5. We get the following:

• n = 3: 3− 1a1− 1

a2− 1

a3= 2, so 1

a1+ 1

a2+ 1

a3= 1. By trying possible values for a1 we find

the three solutions (2, 3, 6), (2, 4, 4) and (3, 3, 3).

• n = 4: 4 − 1a1− 1

a2− 1

a3− 1

a4= 2, so 1

a1+ 1

a2+ 1

a3+ 1

a4= 2. The only solution is

a1 = a2 = a3 = a4 = 2.

If the covering space of O is P 2(R), then χ(O) = 1−∑n

i=1(1− 1ai

). Again, because 1− 1ai∈ [1

2 , 1),

the equation∑n

i=1(1 − 1ai

) = 1 has no solutions for n = 1 and n > 2. So the only solution isn = 2 and a1 = a2 = 2.

The compact surfaces with boundary with nonnegative Euler characteristic are S2 minusa disc, which is homeomorphic to the closed disc D2 and has characteristic 1, S2 minus twodisjoint discs, which is an annulus A and has characteristic 0, and P 2(R) minus a disc, which ishomeomorphic to the Mobius band and has Euler characteristic 0.

The Euler characteristic of an orbifold O with underlying space D2, with indices of ellipticpoints a1 ≤ · · · ≤ an and indices of corner reflectors b1 ≤ · · · ≤ bm is

1−n∑i=1

(1− 1

ai)− 1

2

m∑j=1

(1− 1

bj).

Just like before, the equation∑n

i=1(1 − 1ai

) + 12

∑mj=1(1 − 1

bj) = 1 has no solutions for n > 2.

Also, 12 −

12bj∈ [1

4 ,12) so m ≤ 4.

• If n = 0, then the equation reduces to∑m

j=1(1− 1bj

) = 2 which we solved above: for m = 3

there are three solutions (2, 3, 6), (2, 4, 4) and (3, 3, 3). For m = 4 there is one solutionb1 = · · · = b4 = 2.

• If n = 1, then m ≥ 1 so∑m

j=1(1− 1bj

) ≥ 14 . Therefore 1− 1

a1≤ 3

4 so a1 ≤ 4.

If a1 = 2, the equation becomes 12 + 1

2

∑mj=1(1− 1

bj) = 1 so

∑mj=1(1− 1

bj) = 1. This is also

an equation we solved before: m = 2 and b1 = b2 = 2.

If a1 = 3, then 23 + 1

2

∑mj=1(1− 1

bj) = 1 so

∑mj=1(1− 1

bj) = 2

3 . This has only one solution:

m = 1 and b1 = 3.

If a1 = 4, then 34 + 1

2

∑mj=1(1− 1

bj) = 1 so

∑mj=1(1− 1

bj) = 1

2 . The only solution is m = 1

and b1 = 2.

• If n = 2, then∑n

i=1(1− 1ai

) ≥ 1. So m = 0 and the only solution for 2− 1a1− 1

a2= 1 is

a1 = a2 = 2.

43

For the orbifolds with positive Euler characteristic, we get a similar puzzle. However, thereare also 2-orbifolds with positive Euler characteristic that are not the quotient of a surface by aproperly discontinuous group action. These are the teardrop Tn and the rugby ball Rm,n withm 6= n. They have Euler characteristic 1 + 1

n and 1m + 1

n , respectively. In the next section wewill see that these are the only nondevelopable compact 2-orbifolds.

We do not find finitely many, but finitely many types of compact two-dimensional orbifoldswith positive Euler characteristic. They are listed on the next page. All compact, connected2-orbifolds that are not in the list have negative Euler characteristic.

44

Euler characteristic 0

Underlying space Indices of elliptic points Indices of corner reflectors

S2 2,3,62,4,43,3,32,2,2,2

P 2(R) 2,2T 2

P 2(R)#P 2(R)

S2\D ∼= D2 2,22 2,23 34 2

2,3,62,4,43,3,32,2,2,2

AP 2(R)\D

Euler characteristic > 0

Underlying space Indices of elliptic points Indices of corner reflectors

S2

n,n2,2,n2,3,32,3,42,3,5

P 2(R)n

S2\D ∼= D2

n2 m3 2

n,n2,2,n2,3,32,3,42,3,5

Tn nRm,n m,n

45

3.3. Developability

As announced, in this section we will determine exactly which 2-orbifolds are developable. Westart with the ones that are not. We have already seen two proofs that the teardrop is notdevelopable. Here, we give yet another one.

Proposition 3.3.1. The teardrop Tn and the rugby ball Rm,n with m,n > 1, m 6= n are notdevelopable.

Proof. As seen before, χorb(Tn) = 1 + 1n = n+1

n . Suppose that Tn is the quotient of a manifoldM by a properly discontinuous group action. In the proof of proposition 2.1.13, we saw thatthis manifold has to be a sphere S. Because a sphere has finite area, the number of sheets ofthe covering must be finite. Call it k. By proposition 3.2.9, χ(S) = kχorb(Tn) = k(n+1)

n . Butχ(S) = 2 and n+1

n ∈ (1, 32 ] does not divide 2. Contradiction.

A similar argument holds for Rm,n. Suppose that it has a covering M which is a mani-fold. Because Rm,n is connected, we can choose M to be connected and therefore M has auniversal covering M ′ which is simply connected. Let p : M ′ → Rm,n be a covering. ThenAut(M,p) ∼= πorb

1 (Rm,n) and in example 2.2.15 we saw that this is a subgroup of Zn, so it isfinite. So the number of sheets of this covering is finite, call it k.

The Euler characteristic of Rm,nis 1m + 1

n = m+nmn . Again by proposition 2.1.13, we have

χ(M ′) = kχorb(Rm,n) = k(m+n)mn and this must be an integer because M ′ is a manifold. Let

d = gcd(m,n), m′ = md and n′ = n

d . Then at least one of m′ and n′ is greater than 1. Ifk(m+n)mn = k(m′+n′)

m′n′d is to be an integer, then m′n′d divides k. It follows that χ(M ′) ≥ m+ n > 2.Now Rm,n is compact and because the number of sheets is finite, M ′ is compact as well. Bythe Classification Theorems all compact 2-manifolds have Euler characteristic at most 2. So wederived a contradiction.

Proposition 3.3.2. The disc with one corner reflector and the disc with 2 corner reflectors ofdifferent index are not developable.

Proof. Let us denote these spaces by Yn and Zm,n, with m,n > 1 and m 6= n. Let X be oneof Yn and Zm,n. Suppose that X has a covering M which is a manifold. Again, because X isconnected we can assume that M is connected and therefore has a universal covering M ′. Thisspace covers every other orbifold that covers X.

We can construct such a covering by taking two copies of X and joining them along theboundary. If X = Yn, this gives us the teardrop Tn. If X = Zm,n, we get the rugby ball Rm,n.We derived that these spaces can be covered by a manifold, which is in contradiction with thelast proposition.

It is easy to see that the arguments used in these proofs do not work for other underlyingspaces or more singular points. It turns out that all other 2-orbifolds are developable. The ideaof the proof is to remove all singular points in steps, by covering the orbifold with a simplerorbifold in each step until a manifold is reached.

First, we consider the orbifolds with underlying space a surface without boundary. These can,as we noted before, only contain elliptic points. We consider four cases in which we can removesuch points.

1. Two elliptic points of the same index.

46

Let O have two elliptic points a, b of index n. Cover O by n copies (sheets) of itself. Draw aline ab from a to b that does not cross any other elliptic points (which is possible becausethese are nowhere dense in O) and cut each sheet open along the inverse image of this line.Then join the sheets cyclically along the boundaries just created, as in the following picture:

At a and b, there are n sheets coming together, with the result that these singular points are‘unfolded’ to regular points. The result is a connected orbifold which is an n-fold cover of Owith two singular points less.

2. Three elliptic points of the same odd index.

Let a, b and c be three elliptic points of index 2n+ 1. Draw lines ac and bc that do not crossany elliptic points, cover O by 2n+ 1 copies of itself and cut each of the sheets open alongthe inverse images of the lines. We call the sides acl, acr, bcl and bcr, where l stands for leftand r for right. Schematically, it looks as follows:

Enumerate the sheets by s1, . . . , sn+1. We join them in the following way.In the first step, start joining acr of s1 to acr of s2. Then join bcr of s2 to bcr of s3. Thenjoin s3 and s4 along acr and so forth, alternatively joining along abr and bcr until sn+1 isreached (sn is joined to it along bcr). Then join acr of sn+1 to acl of s1.In the second step, we go through the sheets again, alternatively joining along bcl and acl.The first joining, s1 to s3, is along bcl. We take the following path:

s1 − s3 − s5 − · · · − sn+1 − s2 − s4 − · · · − sn.

Here the last joining, sn−2 to sn, is along acl. Finally, we join bcl of sn to bcr of S1 and thenall sides are joined.By joining the sheets in this order, the points a, b and c are joined to one point each. Therefore,the resulting (2n+ 1)-fold connected cover of O does not contain the singular points a, b andc anymore. An example with 5 sheets:

Step 1.

47

Step 2.

3. Several pairs and triples of the same index

Suppose that we have several pairs (a1, b1), . . . , (ak, bk) of elliptic points with index n1, . . . , nkand triples (p1, q1, r1), . . . , (pl, ql, rl) of points with index m1, . . .ml. We can remove themsimultaneously in the following way. Let a1b1, . . . , akbk be lines between the points of the pairsand p1r1, . . . plrl, q1r1, . . . , qlrl lines between two of the three points of the triples, such thatthe lines do not cross each other or any elliptic points. Let t = lcm(n1, . . . , nk,m1, . . . ,ml,the least common multiple of all the indices. Cover O by t copies of itself and cut the copiesalong the inverse images of all the lines. For any pair (ai, bi), divide the sheets into t

nigroups

and apply for each group the method of joining as in 1. The result will be tni

times the

unfolding of the points a1 and b1. For a triple (pj , qj , rj), divide the sheets in groups of tmj

and use method 2 for each of the groups, so that the points pj , qj and rj are each unfoldedtmj

many times.

Because t is the least common multiple of the indices, the resulting covering will be connected.

From this, we can already derive that any orbifold O without boundary with underlying spaceunequal to the sphere is developable. Namely, by the Classification Theorems |O| is a connectedsum of tori or projective planes and we can draw therefore on |O| a simple closed non-separatingcurve (‘a nontrivial loop that does not divide the surface in two parts’) γ. Moreover, we canmake sure that γ does not meet any elliptic points. Cover O by two copies and cut these alongthe inverse image of γ. Because γ was non-separating, the sheets stay connected. If we join thesheets along the boundaries just created, we get a connected orbifold with every elliptic point inO doubled. By the above method, we can remove all these points. What is left is a manifold.

4. More than one elliptic point of even index.

Suppose that O has two points a,b of index 2n resp. 2m. Let d = gcd(n,m). Cover O with2d copies, draw a line ab which does not meet any elliptic points and cut the sheets openalong the inverse images of ab. Join the sheets cyclically along the created boundaries. Thismethod divides the indices of a and b by 2d, so at least one of them now has odd index.If O contained elliptic points other than a and b, there are now 2d multiples of these. Theycan be grouped in pairs and removed by the first method. So what is left is at most twoelliptic points, at most one of which has even index.

Now we will show that a sphere with more than two elliptic points is developable. To do this,we need what Stillwell calls the ‘(p, q, r)-Hypothesis’ (see Stillwell): For any integers p, q, r ≥ 2there is a permutation σp of order p and a permutation σq of order q whose product σqσp hasorder r. Note that the number of elements that are permuted is not fixed, it depends on p, qand r. The proof of the hypothesis can be found in Stillwell.

48

Theorem 3.3.3. A 2-sphere with more than two elliptic points is developable.

Proof. By point 4 above, we can assume that at most one of all elliptic points has even index. Ifthere is one, call it a, otherwise, let a be an arbitrary elliptic point of O. By assumption, thereare (at least) two more elliptic points, call them b and c. Let p, q and r be the indices of a, band c and let σp, σq and σr be permutations as above. Let t be the number of elements thatthey permute.

Draw lines ac and bc that do not cross any elliptic points, cover O by t copies and cut each ofthe sheets open along the inverse images of the lines. Name the sheets and their boundaries asin 2. We join the sheets as follows. Let x be one of the t elements permuted by σp, σq and σr.Then acl of x is joined to acr of σ−1

p (x) and acr of x is joined to acl of σp(x). Similarly, bcl of xis joined to bcr of σ−1

q (x) and bcr of x is joined to bcl of σq(x).Here is an example with p = 2, q = 5, r = 3, σp = (12)(45), σq = (21543) and σr = (253).

Because σp has order p, this permutation is the product of cycles with the least common multipleof their lengths p1, . . . , pA equal to p. With the first cycle, we joined p1 sheets at a. Here weget an elliptic point a1 of index p

p1. With the other cycles, we created elliptic points a2, . . . , aA

of index pp2, . . . p

pA. Similarly, we have elliptic points b1, . . . bB of index q

q2, . . . q

qBand, exactly

because σqσp = σr, points c1, . . . cC of index rr2, . . . r

rC.

Suppose that O had more elliptic points than a, b and c. Then we created t copies of each ofthese points. Because t ≥ 2, we can divide these copies in groups of two and three. Becauseby assumption all points other than a had odd index, we can remove these groups of copies bymethod 3 described above. Now we note that in the methods of 1 and 2, we could as well havetaken twice as many sheets, making the cycles that we follow during the joining twice as long. Ifwe do this when removing the groups of copies, we create an even number of copies of the ai, bjand ck. We can remove these in the end with method 3 and then we are done.

Suppose that a, b and c were the only elliptic points of O. Then the t-fold covering we createdhas only the elliptic points. If these are more than three points, we can remove them as describedabove. If a1, b1 and c1 are the only ones, then they have index 1 so they are regular.

We conclude that except Tn and Rm,n with m 6= n, all 2-orbifolds with underlying space asurface without boundary are developable.

Suppose that the underlying space of our orbifold O has a boundary. Then O has a mirrorline. We cover O by two copies that we join along the boundary. The result is a connected 2-foldcovering without boundary. This orbifold does not contain any mirror points. If O containedcorner reflectors, what is left in the covering are elliptic points of the same index. as the cornerreflectors. If O is not a disc with one corner reflector of two corner reflectors of different index,then the constructed covering will not be the teardrop or the rugby ball Rm,n with m 6= n so wecan remove the remaining elliptic points with the methods described before.

49

Conclusion: except for the disc with one corner reflector of two corner reflectors of differentindex, all 2-orbifolds with underlying space a surface with boundary are developable.

Remark. It is important to note that in the ‘unfolding’ of our orbifolds, we used only finitecoverings. As we started with a compact orbifold, the manifolds we constructed are thereforecompact as well. Moreover, they are connected, so they have a universal covering. We haveproved the following theorem:

Theorem 3.3.4. Any compact, connected 2-orbifold unequal to the teardrop Tn, the rugby ballRm,n with m 6= n, the disc with one corner reflector and the disc with two corner reflectors ofdifferent index, has a universal orbifold covering with finitely many sheets, such that the coveringspace is a compact surface without boundary.

50

4. Tessellations of the plane

Tessellations or tilings of surfaces are an interesting topic. The book Tilings and Patterns byGrunbaum an Shephard [9] is a very detailed and extensive work on tilings of the plane. Wewill develop some theory about one specific type of tiling, the wallpaper pattern, which is thesubject of Conway’s theorem.

4.1. Periodic tessellations

There are many different definitions of a tessellation. We will use the definitions given byGrunbaum and Shephard. Their definition of a tiling is more general than the definition mostother authors give: they usually require at least what Grunbaum and Shephard call normality.

Definition 4.1.1. A tiling or tessellation of the plane is a countable family of closed setsT = {T1, T2, . . . }, called tiles, such that T is a covering of the plane and the interiors of thesets Ti are pairwise disjoint.

Clearly, this definition extends to all surfaces.

In principle, tiles can have any shape, for example, they could be disconnected or encloseholes. We will only be concerned with so-called normal tilings.

Definition 4.1.2. A normal tiling is a tiling satisfying the following conditions:

(i) Every tile is homeomorphic to the open disc D2;

(ii) The intersection of two tiles is either empty or a connected set;

(iii) The tiles are uniformly bounded.

We will now go into the concept of symmetry of a tessellation. A symmetry is an isometry,so we can only talk about symmetry in metric spaces. When we talk about R2, we will alwaysmean R2 with the Euclidean metric, the Euclidean plane.

If X is a subset of a metric space, we define a symmetry of X to be an isometry that maps Xonto itself (note that any isometry is injective, so symmetries are bijections). For a tile T , wedenote the set of symmetries of T , which is a group under composition of maps, by S(T ). Asymmetry of a tessellation is defined as follows:

Definition 4.1.3. Let T be a tessellation of R2. An isometry of the plane is a symmetry of Tif it maps every tile of T onto a tile of T .

The set of symmetries of T also forms a group, that we denote by S(T ). This group couldof course be trivial, containing only the identity. If S(T ) is nontrivial, we call T a symmetrictiling.

Before we go on about tessellations, first some theory about the symmetries of R2. In section3.1, we gave the definition of a Euclidean isometry : this is a map f : R2 → R2 such thatd(f(x), f(y)) = d(x, y) for all x, y ∈ R2, where d is the Euclidean metric on R2.

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Clearly, the composition of two isometries is again an isometry. It can easily be showed, seefor example Stillwell [17], that there are only four types of Euclidean isometries:

(i) Translation: for all a, b ∈ R, the translation t(a,b) is given by t(a,b)(x, y) = (x+ a, y + b).

(ii) Rotation around a point in R2 by an angle θ.

(iii) Reflection in a line l .

(iv) Glide reflection: a reflection composed with a translation.

Because these maps have inverses (that are isometries as well), the set of Euclidean isometriesforms a group Iso(R2) with as group law composition of maps.

Note that if x = (x1, x2) ∈ R2, then rotating the plane around x by an angle θ is the same asfirst translating the origin to x, then rotating by θ and then translating back. So rotation aroundx by θ is given by t(−x1,−x2)rθt(x1,x2). Here rθ is the map we saw in section 3.1 represented bythe matrix (

cos(α) − sinαsinα cos(α)

).

In the same way, we can represent reflection in a line l by more elementary isometries. Supposethat l is a line through the origin. Then rθ will carry the x-axis to l for the right θ. Reflection inl can be represented by r2π−θsrθ, where s is the map we saw in section 3.1 given by the matrix(

1 00 −1

).

If l does not go through the origin, let t be some translation that moves the origin to l . Thenreflection in l is represented by t−1r2π−θsrθt.

We see that the full group Iso(R2) is generated by the maps s, rθ and the translations. Anotherway to see this is using the following proposition:

Proposition 4.1.4. Let T be the subgroup of Iso(R2) consisting of the translations. ThenIso(R2) ∼= T oO2.

Proof. Let us first show that Iso(R2) = T O2. Take an arbitrary λ ∈ Iso(R2) and let (x, y) = λ(0, 0).Then λ′ = t(−x,−y) ◦ λ is an isometry mapping (0, 0) to t(−x,−y)(x, y) = (0, 0) so λ′ ∈ O2. Nowλ = t(x,y) ◦ λ′ so λ ∈ T O2.

Now we show that T is a normal subgroup of Iso(R2). Let (p, q) ∈ R2 and let t(a,b) be sometranslation. Let g ∈ O2, then g is linear so g◦t(a,b)◦g−1(p, q) = g(g−1(p, q)+(a, b) = (p, q)+g(a, b).So g ◦ t(a,b) ◦ g−1 is a translation. Now let λ ∈ Iso(R2) and represent it by t(x,y) ◦ λ′ as above.Then

λ ◦ t(a,b) ◦ λ−1 = t(x,y) ◦ λ′ ◦ t(a,b) ◦ (λ′)−1 ◦ t(−x,−y)

= t(x,y) ◦ t ◦ t(−x,−y)

where t is the translation λ′ ◦ t(a,b) ◦ (λ′)−1. We see that λ ◦ t(a,b) ◦ λ−1 is again a translation.Conclusion: T is a normal subgroup of Iso(R2).

Because T ∩O2 = {idR2}, it follows from proposition 3.1.6 that Iso(R2) ∼= T oO2.

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The group of symmetries of a tessellation is a subgroup of Iso(R2). We will call a tessellationperiodic if its symmetry group contains two translations over linearly independent vectors. Suchtilings are also called wallpaper patterns, and their groups of symmetries are wallpaper groups.We will call two wallpaper patterns equivalent if the corresponding groups are isomorphic.

The following picture is an example of a wallpaper pattern.

4.2. Representing tessellations with orbifolds

Let T be a periodic tessellation of the plane. The space R2/S(T ) is a representation of T . Wewill show that this space is an orbifold, by proving that S(T ) acts properly discontinuously onR2.

We introduce one more notion. Consider the following pattern. It is not immediately clearwhat we mean by a tile of this pattern, for example, any of the sets next to it forms a tile.

If me make the first tile a bit smaller, it will get too small to reconstruct the whole tessellationwith; the same holds for the fourth tile. They are both fundamental domains for this tiling.

Definition 4.2.1. A closed subset F of the plane is a fundamental domain for a tessellation Tif

1. F is connected;

2. F contains a representative of every orbit S(T )x;

3. in the interior of F , there is only one representative of every orbit;

4. F is minimal with these properties, i.e., it is not a proper subset of a set with properties(i)− (iii).

53

In the example, a fundamental domain is given by the first tile.

A fundamental region of course not unique: whenever F is a set satisfying the above conditionsand λ is an isometry, λ(F ) again satisfies the conditions. Still, we will usually talk about ‘the’fundamental domain, as it is in some sense uniquely determined. Namely, it is the smallestregion of the plane such that with this region and the symmetry group of the tiling, we canreconstruct the original tiling. Dividing out R2 by S(T ) gives the same result as ‘gluing together’the boundaries of a fundamental domain for T according to the symmetries of the tiling.

Now we will show that the group S(T ) acts properly discontinuously on R2. We start byshowing that S(T ) is discrete. We saw that Iso(R2) ∼= T o O2, so the underlying space ofIso(R2) is T × O2. The group O2 can be viewed as a subspace of R4, as its elements can berepresented by matrices. We give O2 the subspace topology. The group T is isomorphic to R2

via the map t(a,b) 7→ (a, b). We give T the topology induced by this map. The set Iso(R2) hasthe product topology.

The group S(T ) has a subgroup T (T ) of translations that, by assumption, contains twotranslations over linearly independent vectors. If (a, b) ∈ R2 such that the translation t(a,b) is inS(T ), let us call (a, b) a translation vector of T . Let d > 0 be the smallest distance such thatthere is a translation vector (a, b) of T with distance d from the origin. Clearly, (−a,−b) is alsosuch a translation vector. If there is yet another translation vector of T , linearly independent of(a, b), with distance d from the origin, call it (p, q). If not, consider the set of translation vectorsof T which are linearly independent of (a, b), and let (p, q) be the one that has the smallestdistance to the origin, as in the following picture.

The group T (T ) is generated by t(a,b) and t(p,q). It is a subgroup of T and can therefore beseen as a subgroup of R2. It exists of the images of the origin under the maps tn(p,q) ◦ t

m(a,b) for

m,n ∈ Z. Clearly, this is a discrete subspace (it is homeomorphic to Z2 ⊂ R2). This shows thatT (T ) is a discrete subspace of T .

The group T (T ) is a normal subgroup of S(T ). Just like for Iso(R2), it can be proved thatS(T ) ∼= T (T )oH, where H ⊂ S(T ) is a subgroup of O2 isomorphic to S(T )/T (T ). Dividingout R2 by S(T ) has the same result as first dividing out by T (T ) and then by H.

Consider set of the images of the origin (or any fixed point) under the maps tn(p,q) ◦ tm(a,b). It is

a so-called lattice of points in the plane. The points can be viewed as the vertices of a tiling bycongruent parallelograms. If we have just one of these parallelograms, say P , we can rebuild thewhole tiling by adding the tiles tn(p,q) ◦ t

m(a,b)(P ). So we can choose a fundamental domain F of T

which is contained in P . The space R2/T (T ) can be seen as the space obtained by joining theboundaries of P according to the translations. This gives a torus, tiled by a tessellation withsymmetry group H. The tiles of this tessellation are copies of F . Now the torus has finite areaand F has positive area, so the number of tiles is finite. It follows that H is finite.

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So H is a finite subgroup of O2, and it must therefore be discrete. So the underlying set ofS(T ) is a product T (T )×H of discrete sets. This shows that S(T ) is discrete.

It is left to prove that the action of S(T ) on R2 is proper. Recall Theorem 1.1.14: an actionof a discrete group G on a Hausdorff space X is proper if and only if for all x, y ∈ X, there areneighbourhoods Vx of x and Vy of y such that the set Hx,y = {g ∈ G | gVx ∩ Vy 6= ∅} is finite.

Let x, y ∈ R2. Let P and Q be parallelograms with as vertices four images of the origin bytranslations, such that x ∈ P and y ∈ Q.

For any parallelogram Z with images of the origin by translations as vertices, there are exactly9 translations t such that t(Z) ∩Q 6= ∅. Namely, these are the ones that map Z to Q, to a tileadjacent with (sharing an edge with) Q, or to a tile that has a vertex in common with Q. Leth ∈ H, the subgroup S(T ) ∩O2 of S(T ). Then h(P ) is a tile, and there are 9 translations thatmap it to a tile that is not disjoint with Q. Because every element in S(T ) is of the form t ◦ hwith t ∈ T (T ) and h ∈ H, this shows that |{λ ∈ S(T ) | λ(P ) ∩ Q 6= ∅}| ≤ 9|H| and this isfinite, as we saw above.

Conclusion: S(T ) acts properly discontinuously on R2. It follows from Theorem 2.1.9 thatR2/S(T ) is an orbifold.

We end this chapter with a short remark about normal tessellations of the sphere. Just like wesaw for the torus, such a tessellation must have a finite number of tiles, because the fundamentaldomain has positive area and the area of S2 is finite. Therefore the symmetry group of the tilingmust be finite and it follows immediately that it acts properly discontinuously on S2. So for anynormal tessellation T of the sphere, the space S2/S(T ) is an orbifold.

The following picture is an example of a spherical pattern.

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5. Conway’s Theorem

Finally, we have arrived to the goal of this thesis: proving Conway’s theorem about the relationbetween wallpaper patterns and orbifolds. John Conway never formulated the statement as amathematical theorem himself. He was not the first one to prove it, either (he attributes it toThurston, see [12]). Still, we devote the following theorem to Conway because it was he whobrought it to the attention of a bigger public.

Theorem 5.1 (Conway’s theorem for planar patterns). A two-dimensional compact orbifoldrepresents a periodic tessellation of the plane if and only if its Euler characteristic is 0.

Now that we have all our theory about orbifolds, the proof is very simple.

Proof. Suppose that we have a periodic tessellation T of the plane. As we saw in the lastchapter, the space R2/S(T ) can be obtained by first dividing out the subgroup T (T ). Thisgives a torus T 2, finitely tessellated by T ′, which consists of copies of the fundamental domainof T and has symmetry group H = S(T )∩O2. Now R2/S(T ) ∼= T 2/H. By proposition 2.2.27,the Euler characteristic of this space divides χ(T 2) = 0. So χorb(R2/S(T )) = 0.

Let O be a compact, connected, 2-dimensional orbifold with Euler characteristic 0. As weproved in chapter 3, O is developable and can be covered by a compact surface M without bound-ary, with a finite number k of sheets. Again by proposition 2.2.27, we have χ(M) = kχorb(O) = 0.The only compact surfaces without boundary that satisfy this are the torus and the Klein bottleP 2(R)#P 2(R). But P 2(R)#P 2(R) is also covered by the torus, for example via the 2-foldcovering sketched in the following picture:

So there is a finite covering of O by the torus. The sheets form a tessellation of T 2; let F be itsfundamental domain. The torus is covered by R2 via the map R2 7→ R2/Z2 ∼= T 2. The sheetsare the 1× 1 squares bounded by the lines {x = z | z ∈ Z} and {y = z | z ∈ Z}. The coveringgives us a tiling T of R2 by copies of F . The group S(T ) contains the translations over thelinear independent vectors (1, 0) and (0, 1).

Informally, the construction above goes as follows: We cut the tiled T 2 open along two loops,one ‘through the hole’ and one ‘around the hole’ (generators of π1(T 2)), and get a parallelogramP . For all (x, y) ∈ Z × Z, we take a copy of P , ‘squeeze’ it into a square and ‘glue’ it on thesquare with bottom left point (x, y).

As mentioned in chapter 4, we call two tessellations equivalent if their symmetry groups areisomorphic. Clearly, this means that they correspond to the same orbifold. We conclude fromthe classification of 2-orbifolds that there are, up to equivalence, 17 wallpaper patterns.

56

We only stated Conway’s theorem for tessellations of the Euclidean plane. In fact, the theoremconsists of two more parts:

Theorem 5.1a (Conway’s theorem for spherical patterns). A two-dimensional compact orbifoldrepresents a tessellation of the sphere if and only if it is developable and its Euler characteristicis positive.

Theorem 5.1b (Conway’s theorem for hyperbolic patterns). A two-dimensional compact orbifoldrepresents a periodic tessellation of the hyperbolic plane if and only if its Euler characteristic isnegative.

We will not go into the hyperbolic case. For the sphere, the proof is similar to the proof forthe Euclidean plane. We already saw that for any tessellation T of S2, the space S2/S(T ) is anorbifold because S(T ) is finite. Therefore the number of sheets of the covering S2 → S2/S(T )

must be finite. Call this number k. The Euler characteristic of S2/S(T ) is χ(S2)k = 2

k > 0.

For the other implication, let O be a developable, compact, connected 2-orbifold with positiveEuler characteristic. Then it has a covering space which is a compact surface without boundary,with Euler characteristic greater than zero. The only surfaces that satisfy this are S2 and theprojective plane P 2(R). The projective plane is covered by the sphere, for example, as in thefollowing picture:

We see that O is finitely covered by the sphere. The sheets form a tessellation.

It follows from our classification of 2-orbifolds that the number of spherical patterns is 14. Thereare infinitely many tessellations of the hyperbolic plane.

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Populaire samenvatting

Periodieke betegelingen zijn patronen die het platte vlak vullen op zo’n manier dat een kleinstukje genoeg is om het hele patroon te construeren. Dit kan omdat de betegeling symmetriebezit. Hier zien we twee voorbeelden van periodieke betegelingen.

Er zijn vier soorten symmetrieen in het vlak: translaties (verschuivingen), rotaties (draaiin-gen), spiegelingen en glijspiegelingen (translaties gevolgd door spiegelingen). We noemen tweebetegelingen equivalent als ze dezelfde symmetrieen hebben. Van de volgende drie betegelingenzijn de eerste twee equivalent, maar de derde is niet equivalent met de eerste twee.

Dit is bijvoorbeeld omdat de eerste twee wel gespiegeld kunnen worden in een diagonale lijn,maar de derde niet.

Om betegelingen beter te kunnen bestuderen, gooien we zoveel mogelijk overbodige informatieweg. We nemen een stukje uit de betegeling dat groot genoeg is om het patroon mee tereconstrueren, en dat kiezen we zo klein mogelijk: als we er nog wat vanaf halen kunnen wehet patroon niet meer reconstrueren. Zo’n stukje heet een fundamentaalgebied. In het volgendevoorbeeld is het fundamentaalgebied een ruit.

We plakken nu de randen van het fundamentaalgebied op elkaar volgens de symmetrieenvan de betegeling. De betegeling in dit voorbeeld heeft als symmetrieen twee translaties. De

58

ene translatie beeldt het ene paar tegenovergestelde randen van de ruit op elkaar af, de anderetranslatie doet dat met het tweede paar tegenovergestelde randen. Als we ze aan elkaar plakken,ziet het er als volgt uit.

De figuur die ontstaat heet een torus. Het is een voorbeeld van een orbifold.

Een orbifold is een ruimte die niet overal mooi glad hoeft te zijn (zoals de torus, die eenheel eenvoudig orbifold is). Behalve gaten kunnen er randen aan zitten, puntjes, en andereuitsteeksels. Punten waar het orbifold niet glad is heten singuliere punten. Bij elk orbifold hoorteen zeker getal genaamd de orbifold-Eulerkarakteristiek. Dit getal is vernoemd naar de belangrijkeachttiende-eeuwe Duitse wiskundige Leonhard Euler, hoewel die zelf nooit van orbifolds gehoordhad (hij kende alleen de ‘gewone’ Eulerkarakteristiek). De orbifold-Eulerkarakteristiek hangt afvan de singuliere punten van het orbifold, maar ook van bijvoorbeeld de dimensie en het aantalgaten. We noteren hem met χorb.

Hieronder zien we een aantal voorbeelden van tweedimensionale orbifolds, met hun Eulerka-rakteristiek.

χ = 2 χ = 0 χ = 1 + 1n χ = 1

n

De Amerikaanse wiskundige John Horton Conway heeft in 1992 bewezen dat elk orbifold datafkomstig is van een periodieke betegeling, door de randen van een fundamentaalgebied aanelkaar te plakken, Eulerkarakteristiek 0 heeft. De omkering is ook waar: als een tweedimensionaalorbifold Eulerkarakteristiek 0 heeft, dan kan je er een periodieke betegeling bij vinden. Dezetwee beweringen vormen samen de stelling van Conway.

Conway is niet de eerste die dit bewezen heeft. Hij is wel de eerste die het netjes heeftopgeschreven, en daarom hebben we de stelling naar hem vernoemd. Met behulp van de stellingvan Conway kan je bewijzen dat er op symmetrie na maar 17 verschillende periodieke betegelingenzijn. Er zijn namelijk precies 17 tweedimensionale orbifolds met orbifold-Eulerkarakteristiek 0.

59

Bibliography

[1] Boileau, M., Maillot, S., Porti, J.: Three-dimensional orbifolds and their geometric structures,vol. 10. Societe mathematique de France (2003). URL http://mat.uab.es/~porti/main.

pdf

[2] Bredon, G.E.: Topology and geometry, vol. 139. Springer Verlag (1993)

[3] Conway, J.H., Burgiel, H., Goodman-Strauss, C.: The Symmetries of Things. AK Peters(2008)

[4] Davis, M.: Lectures on orbifolds and reflection groups. Transformation Groups and ModuliSpaces of Curves pp. 63–93 (2010). URL http://www.math.osu.edu/~davis.12/papers/

lectures%20on%20orbifolds.pdf

[5] tom Dieck, T.: Transformation groups, vol. 8. De Gruyter (1987)

[6] Dragomir, G.C.: Closed geodesics on compact developable orbifolds (2011)

[7] Duistermaat, J.J., Kolk, J.A.: Lie groups. Springer Berlin (2000)

[8] Geer, G.v.d.: Syllabus Algebra 1 (2013). URL http://www.science.uva.nl/~geer/

algebra-input.pdf

[9] Grunbaum, B., Shephard, G.: Tilings and patterns. New York: WH Freeman and Company(1987)

[10] Guerreiro, J.: Orbifolds and wallpaper patterns (2009). URL http://www.math.columbia.

edu/~guerreiro/orbifolds_notes.pdf

[11] Haefliger, A.: Groupoıdes d’holonomie et classifiants. Universite de Geneve-Section demathematiques (1982)

[12] Liebeck, M.W., Saxl, J.: Groups, combinatorics and geometry, vol. 165. CambridgeUniversity Press (1992)

[13] Massey, W.S.: Algebraic topology: an introduction. Springer-Verlag (New York) (1977)

[14] Moerdijk, I., Pronk, D.A.: Orbifolds, sheaves and groupoids. K-theory 12(1), 3–21 (1997)

[15] Munkres, J.R.: Topology (Second Edition). Prentice-Hall Inc., Engla-wood Cliffs, NewJersey (2000)

[16] Pronk, D.A.: Groupoid representations for sheaves on orbifolds. Universiteit Utrecht,Faculteit Wiskunde en Informatica (1995)

[17] Stillwell, J.: Geometry of surfaces. Springer (1992)

[18] Thurston, W.: The geometry and topology of three-manifolds (chapter 13). PrincetonUniversity (2002). URL http://library.msri.org/books/gt3m/PDF/13.pdf

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[19] Wilkins, D.: Course 421: Algebraic topology. section 3: Covering maps and the monodromytheorem (2008). URL http://www.maths.tcd.ie/~dwilkins/Courses/421/421S3_0809.

pdf

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A. Net convergence

This chapter is meant to be a concise introduction to nets and net convergence, needed for theproof of proposition 1.1.17.

Definition A.1. A partial order is a set A with a relation � which is reflexive, anti-symmetricand transitive. A directed set is a partial order (A,�) such that for all α, β ∈ A, there is a γ ∈ Asuch that α � γ and β � γ.

Definition A.2. Let X be a topological space. A net in X is a function f : A → X from adirected set (A,�) to X.

It is common to write xα for f(α) and (xα)α∈A for f(A) (or (xα), if it is clear what the indexset is).

Definition A.3. A net (xα) converges to x ∈ X if for every open neighbourhood U of x, thereis an α ∈ A such that for all β ∈ A, α � β implies that xβ ∈ U .

Proposition A.4. Let B ⊆ X be a subset of a topological space X. Then x ∈ B if and only ifthere is a net (xα) in B that converges to x.

Proof. The implication ⇒ is trivial.Suppose that x ∈ B. Let U = {Uα} be the set of all open neighbourhoods of x, and let

Uα � Uβ iff Uβ ⊆ Uα. Then (U,�) is a directed set, as is easy to see. Since x ∈ B, Uα ∩ B isnonempty for every α. So we can define a net (xα) by choosing, for every α, some xα ∈ Uα ∩B.Clearly, this is a net in B that converges to x.

Theorem A.5. A function f : X → Y is continuous if and only if for every net (xα) in X thatconverges to a point x ∈ X, the net (f(xα)) converges to f(x).

Proof. Again, the left to right implication is trivial.Suppose that for every net (xα) in X that converges to a point x ∈ X, the net (f(xα))

converges to f(x). Let U ⊆ Y be open. We will show that X\f−1(U) = f−1(Y \U) is closed.Let y ∈ f−1(Y \U), then there is a net (yα)α∈A that converges to y. By assumption, the net(f(yα)) converges to f(y) ∈ Y \U . This means that f(y) ∈ Y \U = Y \U and it follows thaty ∈ f−1(Y \U).

Definition A.6. A subset A′ of a partial order (A,�) is called cofinal if for every α ∈ A, thereis an α′ ∈ A′ such that α � α′.

Note that if (A,�) is a directed set and A′ ⊆ A is cofinal, then (A′,�) is again a directed set.

Definition A.7. Let (A,�A) be a directed set and f : A→ X a net in a space X. Let (B,�B)is a directed set and g a function B → A. If the following two conditions hold:

(i) if β �B β′, then g(β) �A g(β′),

(ii) g(B) is cofinal in A

62

then the function f ◦ g : B → X is called a subnet of f .

It is easy to see that if a net converges to a point x, then every subnet converges to x.

Definition A.8. Let (xα)α∈A be a net in a topological space X. An element x ∈ X is anaccumulation point of (xα) if for each open neighbourhood U of x, the set {α ∈ A | xα ∈ U} iscofinal in A.

Proposition A.9. A point x is an accumulation point of (xα)α∈A = {f(α) | α ∈ A} if and onlyif there is a subnet of (xα) that converges to x.

Proof. First, suppose that (xα) has a subnet f ◦ g, where g : B → A is as in definition A.7,that converges to x. Let U be an open neighbourhood of x, then there is a β0 ∈ B suchthat for all β ∈ B with β0 �B β, xg(β) ∈ U . So the set M = {α ∈ A | xα ∈ U} containsL = {g(β) | β0 �B β}. Let α ∈ A. Since g(B) is cofinite, there is a γ ∈ B such that α �A g(γ).By definition of a directed set, there is a δ ∈ B such that β0 �B δ, so δ ∈ L, and γ �B δ. Nowα �A g(γ) �A g(δ). This shows that L is cofinal in A and therefore M is cofinal in A. So x isan accumulation point of (xα).

Now suppose that x is an accumulation point of (xα). Let B be the set of pairs (α,U) suchthat α ∈ A and U is an open neighbourhood of x that contains xα. Define a relation �B on Bby (α,U) �B (β, V ) if α �A β and V ⊆ U . It is easy to check (use that x is an accumulationpoint) that (B,�B) is a directed set. Define g : B → A by g(α,U) = α. By definition of �B,(α,U) �B (β, V ) implies g(α,U) �A g(β, V ). Also, g(B) is cofinal in A. Namely, let α ∈ A andU an open neighbourhood of x that contains xα. Since B is directed, there is a (β, V ) ∈ B forwhich (α,U) �B (β, V ), and we have α �A g(β, V ).

So f ◦ g is a subnet of (xα). We will now show that f ◦ g converges to x. Let U be an openneighbourhood of x. Because x is an accumulation point of (xα), the set M = {α ∈ A | xα ∈ U}is cofinal in A and therefore nonempty. Let α in M , then (α,U) ∈ B. If (α,U) �B (β, V ), thenf ◦ g(β, V ) = xβ ∈ V ⊆ U . This shows that f ◦ g converges to x.

Recall the following definition and proposition:

Definition A.10. Let X be a topological space and {Zα | α ∈ A} a collection of subsets. Thiscollection has the finite intersection property if for every finite collection {Zα1 , . . . , Zαn}, theintersection

⋂ni=1 Zαi is nonempty.

Proposition A.11. A space X is compact if and only if every for every collection {Cα | α ∈ A}of closed subsets that has the finite intersection property, the intersection

⋂α∈ACα is nonempty.

Theorem A.12. A topological space X is compact if and only if every net (xα) in X has asubnet that converges.

Proof. Suppose that X is compact and let (xα)α∈A) be a net. Let Bα = {β ∈ A | α � β} for allα ∈ A and let B = {Bα | α ∈ A}. Because A is directed, for each α ∈ A there exists a β ∈ Awith α � β, so all of the Bα are nonempty.

We will show that B = {Bα | α ∈ A} has the finite intersection property. Let {Bα1 , . . . , Bαn}be a finite subset of B. Because A is a directed set, there is a γ ∈ A such that αi � γ fori = 1, . . . , n. By transitivity of �, we have Bαi ∩ Bγ = Bγ for each i. So

⋂ni=1Bαi ⊇ Bγ 6= ∅.

This shows that B has the finite intersection property. Now C = {Bα | α ∈ A} is a collection ofclosed subsets of X with the finite intersection property. By proposition A.11 and the assumptionthat X is compact, ∩C

⋂α∈ABα is nonempty.

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Let x ∈ ∩C and let U be an open neighbourhood of x. Consider the set M = {α ∈ A | xα ∈ U .Let α0 ∈ A, then x ∈ Bα0 so U ∩ Bα0 contains some element xβ. For this β, we have β ∈ Mand α0 � β. This shows that M is cofinal. It follows that x is an accumulation point of (xα).But by proposition A.9, this means that (xα) has a subnet which converges to x. This provesthe ⇒ implication.

For the ⇐ implication, suppose that there exists an open covering {Uα | α ∈ A} of X thathas no finite subcovering. Let M = {B ⊂ A | B is finite }, it is easy to see that (M,⊆) is adirected set. For every B ∈M , there is an xB ∈ X which is not in

⋃α∈B Uα. Let f : B → X be

the map that sends B to xB. This defines a net in X. Now let x ∈ X, then there is an α0 ∈ Asuch that x ∈ Uα0 . We have {α0} ∈ M . Let B ∈ A, then there is a B′ ∈ A such that B ⊆ B′

and {α} ⊆ B′. We have xB′ /∈ Uα, and it follows that (xB) cannot have a convergent subnet.This proves, by contraposition, the ⇐ implication.

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