AbstractIn this thesis we construct explicitly a unique hyperbolic geome-try on the complements of a certain class of links. We reduce theproblem of deciding whether a link is part of this class to finding adecomposition of a hyperbolic polyehdron in regular tetrahedra.

DetailsTitle: The unique hyperbolic geometryof regular link complementsAuthor: Kyndylan NienhuisE-mail: [email protected].: 5773377Supervisor: Roland van der VeenSecond supervisor: Ale Jan HomburgEnd date: July 16, 2010

Korteweg de Vries Instituut voor WiskundeUniversiteit van AmsterdamPlantage Muidergracht 24, 1018 TV Amsterdamhttp://www.science.uva.nl/math

Contents

1 Introduction 2

2 Knots, links and diagrams 42.1 Knots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Links . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.3 Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.4 Alternating links . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.5 Balanced links . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3 Link complements 123.1 Contracting a space . . . . . . . . . . . . . . . . . . . . . . . . . 123.2 Gluing spaces together . . . . . . . . . . . . . . . . . . . . . . . . 133.3 Expanding a space . . . . . . . . . . . . . . . . . . . . . . . . . . 173.4 Transforming the complement . . . . . . . . . . . . . . . . . . . . 18

4 Hyperbolic geometry 274.1 The hyperbolic space . . . . . . . . . . . . . . . . . . . . . . . . . 274.2 Hyperbolic polyhedra . . . . . . . . . . . . . . . . . . . . . . . . 30

5 Hyperbolic manifolds 365.1 Gluing hyperbolic tetrahedra . . . . . . . . . . . . . . . . . . . . 365.2 Regular Links . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

6 Conclusion 41

7 Populaire samenvatting 427.1 Knopen en schakels . . . . . . . . . . . . . . . . . . . . . . . . . . 427.2 Hyperbolisch complement . . . . . . . . . . . . . . . . . . . . . . 437.3 Veelvlakken en betegelingen . . . . . . . . . . . . . . . . . . . . . 447.4 Conclusie . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

1

Chapter 1

Introduction

Some say that knot theory started in the 1860s when Lord Kelvin thought thatatoms were knots in the aether (see [12]). Others say that it began with Gauss,as almost everything did. Nowadays, knot theory is linked to many subjectsand it is researched for various reasons.

Knots and links, which are knots with more components, can be studiedby looking at their complement. A link complement is a 3-manifold, in somecases it also has a unique geometric structure. This geometric structure helpsus to understand the link and gives us topological invariants to distinguishdifferent links. An example of these invariants is the hyperbolic volume of thecomplement of a link. To investigate the geometric structure of a manifold is inline with Thurston’s geometrization conjecture. This conjecture has been provedby Grigori Perelman in 2003, it states that there are eight relevant geometriesfor 3-manifold. Hyperbolic geometry is one of them.

We call a link whose complement has a hyperbolic geometry a hyperboliclink. William Menasco proved in [7] that every prime, non-split alternating linkthat is not a torus link is a hyperbolic link.

In this thesis we consider a subclass of alternating balanced links, calledthe regular links, and construct explicitly a unique hyperbolic geometry on thecomplements of these links. We know then that the regular links are hyperboliclinks, but more importantly, we can use this explicit construction to do hyper-bolic geometry on link complements. We will see that the hyperbolic volume ofthe complement of a link is a topological invariant of links.

Alan Reid proved in [11] that the figure-eight knot is the only regular linkwith one component. We will find an infinite family of nonequivalent regularlinks with more components.

In the first chapter after this we define everything we need about knots,links and their diagrams. Then in Chapter 3 we consider the complementsof links. We develop topological tools that will enable us to transform thecomplement of an alternating balanced link to a space obtained by gluing twoabstract polyhedra at their faces.

We introduce hyperbolic geometry in Chapter 4, and we define the hyper-bolic polyhedra. In particular, we consider polyhedra that are decomposableinto ideal regular tetrahedra, these are tetrahedra with vertices at infinity andcorners of 1

3π. We construct an infinite family of decomposable polyhedra, calledthe n-tetrahedra.

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In Chapter 5 we show how to make complete hyperbolic manifolds by gluingideal regular tetrahedra together. In general this can be done by non-regulartetrahedra, but by only regarding regular tetrahedra the complete hyperbolicstructure is almost an immediate consequence.

Then we refer to Chapter 3 where we showed that the complement of everyalternating balanced link is homeomorphic with two abstract polyhedra glued attheir faces. If we can give these abstract polyhedra a hyperbolic shape such thatthey are decomposable in ideal regular tetrahedra, we call the link regular andwe show that is has a complete hyperbolic structure. With Mostow’s rigiditytheorem we show that this structure is unique.

To see if an alternating balanced link is regular, one must decompose thecorresponding abstract polyhedra into regular tetrahedra. In Chapter 5 wealso show that this can be done the other way around, every decomposablepolyhedron give rise to a regular link. Since we have found an infinite familyof decomposable polyhedra, namely the n-tetrahedra, we have found an infinitefamily of hyperbolic links. We show that these links are not equivalent.

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Chapter 2

Knots, links and diagrams

In this chapter we give the definition of knots and links and we will see thatthey behave (mostly) as ordinary links and knots. The diagrams introduced inthis chapter will give us a convenient way of describing particular knots andlinks and more importantly, they enable us the associate polyhedra with linksin later chapters.

The definitions given coincide with the usual definitions in knot theory, ex-cept at one point. For our goal we do not need to differentiate between knotsthat are mirrored versions of each other, so we see them as equal.

2.1 Knots

Let Sn denote the n-dimensional sphere. We will only encounter the circleS1, the ordinary sphere S2 and the 3-sphere S3. The first two can easily bevisualized, you can think of S3 as R3 ∪∞.

Throughout this thesis we use the following convention. The sphere S2

divides R3∪∞ = S3 in two parts, we call the part containing∞ the upper part,and the other part the lower.

Definition 2.1 (Knot). A knot is the image of an embedding of the circle S1

in the 3-sphere S3.

Recall that a map f : X → Y is called an embedding if f is a homeomorphismto its image f(X), where f(X) has the subspace topology.

With this definition knots can still exhibit very strange behavior, the knot

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in Figure 2.1 for example is wild at the point on the right1. To avoid this wewill only consider tame knots.

Figure 2.1: A wild knot

Definition 2.2 (Tame knot). A tame knot is the image of an embedding i :S1 → S3, such that i(S1) is a simple closed polygon. In other words, i(S1)consists of finite many straight line segments.

As always, we want to know when we can consider two knots as the same.

Definition 2.3 (Equivalence). Two knots K1,K2 ∈ S3 are equivalent if thereis a homeomorphism ϕ : S3 → S3 with ϕ(K1) = K2.

Remark. Usually, the homeomorphism is required to be orientation preserving.Then the definition would coincide with definitions using ambient isotopies, orReidemeister moves. See [3] for a proof.

However, for our purposes it is more convenient to see mirrored knots asequivalent, so we do not require that.

Example. The knots in Figure 2.2 are equivalent. The first knot is the unknot ,it is the image of the inclusion S1 ↪→ S3. If a knot is equivalent to the unknot,we say that it is trivial.

(a) The unknot (b) A trivial knot

Figure 2.2

From now on, all knots will be tame. Although they are simple closedpolygons, we will still draw them curved with the justification that there willalways be a homeomorphism of S3 that transforms it to a simple closed polygon.

1R. Brode proved in 1981 in his article ”Uber wilde Knoten und ihre ’Anzahl’” that almostall knots are wild at every point.

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2.2 Links

A link consists of multiple knots tied together.

Definition 2.4 (Link). A link is the image of an embedding i of a disjointunion of n circles qnj=1S1

j in the 3-sphere S3, such that i(Sj) is a simple closedpolygon for j = 1, . . . , n.

We call n the multiplicity of the link, and for j = 1, . . . , n we call i(Sj) acomponent of the link.

The equivalence of links is essentially the same as with knots, we include itfor the sake of completeness.

Definition 2.5 (Equivalence). Two links L1,L2 ∈ S3 are equivalent if there isa homeomorphism ϕ : S3 → S3 with ϕ(L1) = L2.

Example. The links in Figure 2.3 have multiplicity 2 and each component istrivial.

Figure 2.3: Links consisting of two unknots

2.3 Diagrams

Describing a link can be complicated, diagrams will make this easier. A diagramof a link is a projection of the link on S2 in such a way that all the relevantinformation to reconstruct the link is maintained, see Figure 2.4 for example.To avoid the need to explain what we mean by ‘relevant information’ we willbegin with diagrams, and then show how to construct a link from a diagram.

(a) A diagram (b) The corresponding link

Figure 2.4: The figure-eight knot

2.3.1 Embedded graphs

We take an embedding of a graph in S2 as the basis for a diagram.

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Definition 2.6 (Embedded graph). Let V be a finite, non-empty subset of S2.Let E be a finite set of simple curves in S2 that start and end at a v ∈ V, suchthat the interiors of the curves are disjoint from V and from each other.

Then the pair G = (V, E) is an embedded graph. We call V the vertices ofthe graph and E the edges.

By removing the edges e ∈ E from S2 we get n disjoint sets. These sets arecalled the faces of the graph.

We do not care about the orientation of the edges. If an edge e goes fromv to v′ we say that v and v′ are its endpoints and that e is an edge between vand v′. If there is only one edge between v and v′, we can identify e = {v, v′}.Sometimes we will abuse this, ’an edge {v, v′}’ means one of the edges betweenv and v′.

The vertices of a graph G will represent the crossings of a link. An edge{v, v′} will represent a part of the link between the crossings v and v′.

Because a link does not stop at a crossing, we want for every edge {v, w} atthe crossing w that there is a edge {w, v′} where that part of the link continues.So we want an even number of edges at a vertex and we want to know whichedges connect to each other.

Definition 2.7 (Connecting edges). Let G be an embedded graph. We saythat the edges e and e′ connect at the vertex v if e and e′ are distinct edgesof v and there are the same number of edges between e and e′ clockwise ascounterclockwise.

If the number of edges of v is even, then for every edge e of v there is aunique edge e′ that connects to e at v.

To make things easier we do not want more than two parts of the link crossingat the same point, so we want the graph to be 4-regular.

2.3.2 Crossing functions

An embedded graph gives no information about which part of the crossing liesabove the other part. For this we introduce crossing functions.

Definition 2.8 (Crossing function). Let G be an embedded, 4-regular graph,with vertices V and edges E . A crossing function on G is a function γ :V → {{e1, e2} | e1, e2 ∈ E} such that if γ(v) = {e1, e2} then e1, e2 are edgesconnecting at v.

For each v we have two choices for γ(v), the converse of γ is the crossingfunction that always takes the other choice. We will see this when we discussalternating links.

Definition 2.9 (Converse crossing function). Let γ be a crossing function.The converse crossing function −γ is the crossing function with e ∈ −γ(v) iffe /∈ γ(v) for every edge e at v.

Proof. Let v be a vertex and e1, . . . , e4 the edges around v with γ(v) = {e1, e2}.By definition −γ(v) = {e3, e4}. Since e1 and e2 connect, e3 and e4 lie betweene1 and e2 on opposite sides. So e3 and e4 connect, and we conclude that −γ isindeed a crossing function.

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2.3.3 Diagrams

Definition 2.10 (Diagram). A diagram is a pair (G, γ) with G an embedded,4-regular graph and γ a crossing function on G.

We draw a diagram (G, γ) as the embedded graph G, with at each vertextwo edges drawn above the other two according to the crossing function. See forexample Figure 2.4a. A diagram has faces, these are the faces of the embeddedgraph G.

From a diagram you can immediately see the link it is meant to describe,but nevertheless, we will make it explicit how to do this.

(a) The crossing in the diagram (b) The crossing in S3

Figure 2.5: Transforming a crossing

Definition 2.11 (Corresponding link). Given a diagram (G, γ), we constructthe corresponding link in the following way. We consider the embedded graphG in S2 ⊂ S3. The sphere S2 divides S3 in two parts. At each crossing v we liftthe two edges γ(v) to one side of S3 and the other two edges ending at v to theother side, see Figure 2.5.

In this way we get n non-intersecting, simple closed polygons in S3. Theyare the image of an embedding i : qnj=1S1

j → S3 so together they form a link.

If L is the corresponding link of (G, γ) we also say that (G, γ) is a diagramof L.

Proposition 2.12. For every link L, there is a diagram D such that L and thecorresponding link of D are equivalent.

This is proved in [3] by projecting the link to S2. Note that to find adiagram of the unknot (or a trivial link) we need to introduce a crossing, sincewe demanded that an embedded graph has at least one vertex.

Definition 2.13 (Equivalence). Two diagrams are equivalent if their corre-sponding links are equivalent.

2.4 Alternating links

A link alternates if it has a diagram where a line is alternatively the upper andthe lower part of a crossing. We make this more precise.

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Definition 2.14 (Alternating link). We call a crossing function γ alternatingif for each edge e = {v1, v2} we have e ∈ γ(v1) or e ∈ γ(v2) and not both.

We call a diagram (G, γ) alternating if γ is alternating. A link L is alternatingif it has an alternating diagram.

Example. The diagram of the figure-eight knot in Figure 2.4a is alternating,so the figure-eight knot is an alternating link.

In most cases, knowing that a diagram is alternating gives the same infor-mation as the crossing function. We can see this as follows.

Algorithm 2.15. Let G be a finite, undirected, planar, 4-regular and connectedgraph. We construct an alternating crossing function γ.

1. Pick a v ∈ V and an edge e at v. Let f be the edge connecting with e at v(see Definition 2.7). Define γ(v) = {e, f}.

2. For each edge e = {v1, v2} where we have defined γ(v1) but not γ(v2), wedefine γ(v2) so that γ becomes alternating. In other words, let f be theedge connecting with e at v2 and let g, h be the other two edges at v2. Ife ∈ γ(v1) define γ(v2) = {g, h}, else γ(v2) = {e, f}.

3. Repeat 2.

Proof. Since G is finite, this construction will end. Because G is connected, γwill be defined for all v ∈ V. And by construction, γ is alternating.

Proposition 2.16. Let G be an embedded, 4-regular and connected graph. Thereare precisely two alternating crossing functions γ and γ′ on G and we haveγ′ = −γ.

Proof. Take a vertex v. Let {e, f} and {g, h} be the two pairs of connectingedges at v. Let γ and γ′ be the alternating crossing functions obtained bypicking v and e, and v and g in the above algorithm.

Let δ be an alternating crossing function of G. By choosing v and one of theedges in δ(v) in step 1, we can make δ by the above construction. Since step 2leaves no choice, we see that either δ = γ or δ = −γ, so there are at most two.

Because either γ′ = γ or γ′ = −γ and γ′(v) 6= γ(v) we must have γ′ =−γ.

Notation. We call these two alternating functions γalt and −γalt.

Corollary 2.17. Let G be an embedded, 4-regular and connected graph. Thereare two alternating links with G the graph of one of their diagrams. They areequivalent.

Proof. The two links are the corresponding links of (G, γalt) and (G,−γalt). Theyare equivalent by reflecting S3 in S2.

In an alternating diagram we can give the faces an orientation.

Definition 2.18 (Face-orientation). Let (G, γ) be an alternating diagram, let fbe a face. Since γ is alternating, the crossings around f look either like Figure2.6a or like Figure 2.6b looking from above to S2 (recall our convention that thepart of S3 = R3 ∪∞ that contains ∞ is the upper part).

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In the first case we say that the face is clockwise, in the other case that f iscounterclockwise.

(a) Clockwise (b) Counterclockwise

Figure 2.6: The orientation of a face

Note. It may seem strange to choose the orientation this way in stead of theother way around. But this definition saves us from confusion when we willwork with oriented edges that point exactly in this direction.

Proposition 2.19. Let (G, γ) be an alternating diagram, let f, f ′ be adjacentfaces. Then f and f ′ have different orientations.

Proof. To see this, observe that in Figure 2.6a and 2.6b the faces adjacent tothe face in the middle have a different orientation.

2.5 Balanced links

Balanced diagrams are diagrams that have not too many or too little faces thatare bounded by two edges, like in Figure 2.7. We will need this property whenwe will transform the complement of a link.

Figure 2.7: A face bounded by two edges (a digon)

Definition 2.20 (Balanced link). Let G be an embedded graph with verticesV. We call v, v′ a balanced pair if v and v′ are distinct, there are exactly twoedges between v and v′, and for every w different from v and v′ there is at mostone edge between v and w, and at most one edge between v′ and w.

We call an embedded graph balanced if its vertices V can be partitioned intobalanced pairs.

We call a diagram (G, γ) balanced if G is balanced. A link L is balanced if ithas an balanced diagram.

A way to obtain a balanced link is by using matchings.

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Definition 2.21 (Matching). Let G be an embedded graph with vertices V andedges E . A matching M is a subset of E such that for every vertex v ∈ V thereis exactly one edge e ∈M that has v as endpoint and that edge is not a loop.

Note. Edges are not oriented, an edge has two endpoints.

Proposition 2.22. Let G be an embedded, 3-regular and connected graph thatis also simple, i.e. there are no different edges that have the same endpoints.Let M be a matching of G.

The pairs (G,M) are in one-to-one correspondence with pairs of alternating,balanced diagrams {(H, γ), (H,−γ)} with H connected. Furthermore, if (G,M)corresponds to {(H, γ), (H,−γ)} then the graph H is the same as the graph Gwhere the edges in M are doubled.

Proof. Let G be an embedded, 3-regular, connected and simple graph, let Mbe a matching of G. We will construct a pair of alternating, balanced diagrams{(H, γ), (H,−γ)}. We obtain the new graph G′ from G by doubling all the edgesin the matching M, we push the doubled edges slightly away from each otherto make sure that G′ is an embedded graph. G′ is 4-regular, since there is ateach vertex exactly one edge doubled, because M is a matching.

The vertices V of G are the vertices of G′. We partition them in the followingway. For every edge e ∈ M we make the pair {ve, v′e} where ve and v′e are theendpoints of e. Because for every vertex v there is exactly one edge e ∈M thathas v as endpoint, the set P = {{ve, v′e} | e ∈M} is a partition of V.

Let {v, v′} ∈ P. Since M does not contain loops we have that v and v′ aredistinct. Because G is simple and the edge {v, v′} is doubled in G′, we see thatthere are exactly two edges between v and v′.

Furthermore, for every w different from v and v′ there is at most one edgebetween v and w. This is because G is simple, and if there is an edge {v, w} itis not in the matching so it is not doubled. We see that {v, v′} is a balancedpair, and we conclude that G′ is a balanced graph.

Because G′ is an embedded, 4-regular and connected graph, by Proposition2.16 there are two alternating crossing functions on G′ and these are γalt and−γalt. We define the corresponding pair of alternating, balanced diagrams as{(G′, γalt), (G′,−γalt)}.

Let {(H, γ), (H,−γ)} be a pair of alternating, balanced diagrams with Hconnected. Let G be the graph obtained from H by collapsing all double edgesto single edges, we have that G is simple. Since only double edges are collapsed,G is connected. Because H is 4-regular and balanced, there is one pair of edgescollapsed at each vertex, so G is 3-regular.

LetM be the set of edges of G that are the result of collapsing a double edge.Since H is balanced, M is a matching. We define (G,M) as the correspondingpair.

If (G,M) 7→ {(H, γ), (H,−γ)} 7→ (G′,M′) then G = G′ and M = M′. Tosee this observe that the edges in G are doubled according to M, and thencollapsed with M′ the set of collapsed edges.

In the same way, we see that if

{(H, γ), (H,−γ)} 7→ (G,M) 7→ {(H′, γ′), (H′,−γ′)}

then H = H′. By Proposition 2.16 there are only two alternating crossingfunctions on a connected graph, and they are each others converse. Hence,{γ,−γ} = {γ′,−γ′}. We conclude that the correspondence is one-to-one.

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Chapter 3

Link complements

In this chapter we will transform the complement of a link to a combinatorialobject. To do that we will develop some tools in the first sections that allow usto manipulate topological spaces.

In the next chapter we will see that this combinatorial object is homeomor-phic to hyperbolic polyhedra glued together.

3.1 Contracting a space

From a topological space you can obtain a new space by contracting certainparts. The way to do this is by making a quotient space.

Notation. Let X be a set and ∼ an equivalence relation on X. Then [x] isthe equivalence class of x ∈ X, X/ ∼ is the set of equivalence classes andπ : X → X/ ∼ the canonical projection x 7→ [x].

We can make X/ ∼ a topological space in the following way.

Definition 3.1 (Quotient space). Let X be a topological space and ∼ an equiv-alence relation on X. Define the topology on X/ ∼ by calling a set U ⊆ X/ ∼open if its inverse image π−1(U) is open in X. With this topology X/ ∼ is thequotient space of X by ∼.

Figure 3.1: The quotient [0, 1]/ ∼

Proposition 3.2. Let f : X/ ∼ → Y be a map. We have that f is continuousif and only if f ◦ π is continuous.

Proof. If f is continuous, then f ◦π is a composition of continuous function andtherefore continuous.

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Suppose f ◦ π is continuous. Let U ⊆ Y be an open set, then (f ◦ π)−1 =π−1(f−1(U)) is open in X. Then by definition, f−1(U) is open in X/ ∼. Hence,f is continuous.

The following proposition tells us that a homeomorphism X → Y gives us ahomeomorphism from X/ ∼ to a quotient space of Y . This enables us to forgetthe equivalence relation for a while: suppose Z ∼= X/ ∼. Instead of searchingfor spaces homeomorphic with X/ ∼ we can search for spaces homeomorphicwith X. Then this proposition gives us a space homeomorphic with X/ ∼.

Proposition 3.3. Let ϕ : X → Y be a homeomorphism. Let ∼X be an equiva-lence relation on X, and ∼Y on Y with

a ∼X b iff ϕ(a) ∼Y ϕ(b). (3.1)

Then X/ ∼X and Y/ ∼Y are homeomorphic.

Proof. Define ϕ′ : X/ ∼X → Y/ ∼Y by [x]X → [ϕ(x)]Y . By (3.1) there is norepresentation problem.

Since ϕ′ ◦ πX = πY ◦ ϕ and the latter is a composition of continuous func-tions, we see that ϕ′ ◦ πX is continuous. By Proposition 3.2 we have that ϕ′ iscontinuous.

Analogously, we see that the inverse [y]Y 7→ [ϕ−1(y)]X is continuous and weconclude that X/ ∼X and Y/ ∼Y are homeomorphic.

When there is an equivalence relation defined on a quotient space, we canswitch the order of the equivalence relations under certain conditions.

Proposition 3.4. Let ∼1 and ∼3 be equivalence relations on X, and ∼2 anequivalence relation on X/ ∼1, such that

x ∼3 y iff [x]1 ∼2 [y]1. (3.2)

Then (X/ ∼1) / ∼2 and X/ ∼3 are homeomorphic.

Proof. Define the functions ϕ and ψ between (X/ ∼1) / ∼2 and X/ ∼3 by

ϕ ([[x]1]2) = [x]3ψ ([x]3) = [[x]1]2.

By (3.2) there is no representation problem, so ϕ and ψ are well defined.Since ϕ ◦ π2 ◦ π1 = π3 is continuous, we have by Proposition 3.2 that ϕ is

continuous. Because ψ ◦ π3 = π2 ◦ π1, we have that ψ is continuous.We see that ϕ ◦ ψ = id and ψ ◦ ϕ = id and we conclude that ϕ is a homeo-

morphism.

3.2 Gluing spaces together

Using the construction of quotient spaces we can glue topological spaces to-gether. First we make a space that consists of the two spaces without anyinteraction between them.

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Definition 3.5 (Topological sum). Let X,Y be topological spaces. The sumX + Y of X and Y is the set X ×{0} ∪ Y ×{1} with the following topology. Aset U ⊆ X + Y is open if {x | (x, 0) ∈ U} is open in X and {y | (y, 1) ∈ U} isopen in Y .

When no confusion can arise, we identify X ×{0} with X and Y ×{1} withY .

Proposition 3.6. A function f : X + Y → Z is continuous iff f restricted toX and f restricted to Y are continuous.

Proof. By definition of the sum topology.

Now we can glue the copies of X and Y in X + Y together, we use a home-omorphism to tell us what to glue to what.

Definition 3.7 (Generated equivalence relation). Let X0 ⊆ X, Y0 ⊆ Y bespaces and ψ : X0 → Y0 a homeomorphism. Define the equivalence relation ∼ψon X + Y generated by ψ as

a ∼ψ b iff

a = b ora ∈ X0, b ∈ Y0 and b = ψ(a) ora ∈ Y0, b ∈ X0 and a = ψ(b).

Definition 3.8 (Gluing spaces). Let X0 ⊆ X, Y0 ⊆ Y be spaces and ψ : X0 →Y0 a homeomorphism. The space X ∪ψ Y obtained by gluing X to Y by ψ isthe space X + Y/ ∼ψ.

Figure 3.2: [0, 1] glued to R by ϕ

The following proposition tells us that homeomorphisms and gluing work welltogether. This enables us to forget gluings for a while: suppose W ∼= X ∪ψ Y .In stead of searching for spaces homeomorphic with X ∪ψ Y we can search forspaces homeomorphic with X. Then by the following Proposition we get a spacehomeomorphic with X ∪ψ Y .

Proposition 3.9. Let X0 ⊆ X, Y0 ⊆ Y and Z0 ⊆ Z be spaces. Let ψ : X0 →Y0, ψ′ : Z0 → Y0 and ϕ : X → Z be homeomorphisms. If Z0 = ϕ(X0) andψ′ = ψ ◦ ϕ−1 then X ∪ψ Y is homeomorphic to Z ∪ψ′ Y .

Proof. Let ∼ψ be the equivalence relation on X +Y generated by ψ, let ∼ψ′ bethe equivalence relation on Z + Y generated ψ′. We extend ϕ canonically to ahomeomorphism ϕ′ : X + Y → Z + Y .

Suppose a ∼ψ b, we will prove ϕ′(a) ∼ψ′ ϕ′(b). Since ∼ψ is generated by ψthere are three cases.

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Case 1. a = b. Then clearly ϕ′(a) ∼ψ′ ϕ′(b).

Case 2. a ∈ X0, b ∈ Y0 and b = ψ(a). We have ϕ′(a) = ϕ(a) and ϕ′(b) = b.Since b = ψ(a) = ψ′(ϕ(a)) we have ϕ(a) ∼ψ′ b.

Case 3. a ∈ Y0, b ∈ X0 and a = ψ(b). Same as the previous case.

The other way, if ϕ′(a) ∼ψ′ ϕ′(b) then a ∼ψ b, goes analogously. ApplyingProposition 3.3 we conclude that X+Y/ ∼ψ is homeomorphic to Z+Y/ ∼ψ′ .

If an equivalence relation only contracts points in the gluing surface, we canswitch the order of gluing and contracting.

Proposition 3.10. Let X0 ⊆ X, Y0 ⊆ Y and ϕ : X0 → Y0 a homeomorphism.Let ∼, ∼X , ∼Y be equivalence relations on respectively X ∪ϕ Y , X and Y , suchthat ∼ only contracts points in the gluing surface, i.e.

a ∼ b only if

{a = b ora, b ∈ X0 ∪ϕ Y0,

(3.3)

and the equivalence relations are compatible, i.e.

x ∼X x′ iff [x]ϕ ∼ [x′]ϕy ∼Y y′ iff [y]ϕ ∼ [y′]ϕ. (3.4)

Let ϕ′ : X ′0 → Y ′0 be the function with

X ′0 = {[x]X ∈ X/ ∼X | x ∈ X0}Y ′0 = {[y]Y ∈ Y/ ∼Y | y ∈ Y0}

ϕ′ ([x]X) = [ϕ(x)]Y . (3.5)

Then ϕ′ is a well defined homeomorphism, and

(X ∪ϕ Y ) / ∼ ∼= (X/ ∼X) ∪ϕ′ (Y/ ∼Y ) .

Proof. Suppose [x]X = [x′]X with x, x′ ∈ X0. By (3.4) we have [x]ϕ ∼ [x′]ϕ.Since x, x′ ∈ X0, we have by definition of the generated equivalence relation ∼ϕthat [x]ϕ = [ϕ(x)]ϕ and [x′]ϕ = [ϕ(x′)]ϕ. Hence, [ϕ(x)]ϕ ∼ [ϕ(x′)]ϕ. Then by(3.4) we have [ϕ(x)]Y = [ϕ(x′)]Y and we conclude that ϕ′ is well defined.

We define the function ψ′ : Y ′ → X ′ by ψ′ ([y]Y ) = [ϕ−1(y)]X . Analogouslyto ϕ′, we can show that ψ′ is well defined.

We have that ϕ′ ◦ πX = πY ◦ ϕ. Since the latter is continuous, we have byProposition 3.2 that ϕ′ is continuous. For the same reason we conclude fromψ′ ◦ πY = πX ◦ϕ−1 that ψ′ is continuous. Since ψ′ ◦ϕ′ = id and ϕ′ ◦ψ′ = id weconclude that ϕ′ is a homeomorphism.

Define the equivalence relation ∼Z on X + Y by

z ∼Z z′ iff [z]ϕ ∼ [z′]ϕ. (3.6)

By Proposition 3.4 we have

(X ∪ϕ Y ) / ∼∼= (X + Y ) / ∼Z .

15

We abbreviate

Z = X + Y

Z ′ = Z/ ∼ZX ′ = X/ ∼XY ′ = Y/ ∼Y

We will construct a homeomorphism from Z ′ to X ′ ∪ϕ′ Y ′, which will com-plete the proof. Define ρ : Z → X ′ + Y ′ by

ρ(z) =

{[z]X if z ∈ X[z]Y if z ∈ Y

.

Since the restrictions of ρ to X and Y are πX and πY we have by Proposition3.6 that ρ is continuous.

We have that πϕ′ ◦ ρ is constant on [z]Z . To see this, let z ∼Z z′. Then by(3.6) we have [z]ϕ ∼ [z′]ϕ. We assume without loss of generality that z ∈ X.

Case 1. z ∈ X − X0. Then [z]ϕ = {z} /∈ X0 ∪ϕ Y0. So by (3.3), {z} = [z′]ϕ.Hence, z = z′.

Case 2. z ∈ X0, z′ ∈ X. By (3.4) [z]x = [z′]x, hence ρ(z) = ρ(z′).

Case 3. z ∈ X0, z′ ∈ Y . Then [ϕ(z)]ϕ = [z]ϕ ∼ [z′]ϕ, so by (3.4) [ϕ(z)]Y = [z′]Y .By (3.5), ϕ′ (ρ(z)) = ρ(z′).

Since πϕ′◦ρ is constant on [z]Z , there is a ρ′ : Z ′ → X ′∪ϕ′Y ′ with ρ′◦πZ = πϕ′◦ρ.By Proposition 3.2 ρ′ is continuous. For completeness, ρ′ is described below

ρ′ ([z]Z) =

{[[z]X ]ϕ′ if z ∈ X[[z]Y ]ϕ′ if z ∈ Y.

(3.7)

Define σX : X ′ → Z ′ and σY : Y ′ → Z ′ by

σX ([x]X) = [x]ZσY ([y]Y ) = [y]Z .

These functions are well defined: if [x]X = [x′]X then by (3.4) and (3.6)[x]Z = [x′]Z and the same for σY . They are both continuous (again by Propo-sition 3.2), so their sum τ : X ′ + Y ′ → Z ′ is continuous by Proposition 3.6.

We have that τ is constant on [a]ϕ′ . To see this, let a ∼ϕ′ a′ with a 6= a′.We can assume without loss of generality that ϕ′(a) = a′. So, a = [x]X ∈ X ′and a′ = [y]Y ∈ Y ′, with [ϕ(x)]Y = [y]Y .

By (3.4) we have [x]ϕ = [ϕ(x)]ϕ ∼ [y]ϕ. Then by (3.6) we have [x]Z = [y]Z ,hence τ(x) = τ(y).

Since τ is constant on [a]ϕ′ , there is a τ ′ : X ′ ∪ϕ′ Y ′ → Z ′ with τ ′ ◦ πϕ′ = τ .By Proposition 3.2 τ ′ is continuous. We have τ ′ described below

τ ′ ([a]ϕ′) =

{[x]Z if a = [x]X ∈ X ′

[y]Z if a = [y]Y ∈ Y ′.(3.8)

16

By comparing (3.7) and (3.8) we see that ρ′ ◦ τ ′ = id and τ ′ ◦ ρ′ = id, sothey are homeomorphisms. We conclude that

(X ∪ϕ Y ) / ∼ ∼= (X/ ∼X) ∪ϕ′ (Y/ ∼Y ) .

3.3 Expanding a space

Expanding a space is the converse of contracting. In stead of identifying points,we double certain points.

Definition 3.11 (Expansion space). Let X be a topological space, and A ⊆ Xa subset. The expansion of X obtained by doubling A is the space Y = X∪ϕX,where ϕ : X −A→ X −A is the identity.

Figure 3.3: Expanding (1, 2] ⊂ [0, 2]

Proposition 3.12. Let A ⊆ X, let X ∪idX−AX be the expansion of X by

doubling A and let ∼A the equivalence relation on X ∪idX−AX that identifies

A× {0} with A× {1}. Then

X ∼=(X ∪idX−A

X)/ ∼A .

Proof. Abbreviate Y = X ∪idX−AX. By Definition 3.8 we have

Y = (X +X)/ ∼X−A,

where ∼X−A be the equivalence relation on X + X generated by idX−A (seeDefinition 3.7).

Let [(x, i)] be the equivalence class of (x, i) ∈ X + X by ∼X−A, and πX−Athe projection (x, i) 7→ [(x, i)]. Let 〈y〉 be the equivalence class of y ∈ Y by ∼A,and πA the projection y 7→ 〈y〉.

Suppose 〈[(x, i)]〉 = 〈[(y, j)]〉. By definition of ∼A we either have [(x, i)] =[(y, j)], or x, y ∈ A, i 6= j and x = y. In the first case, we have by definition of∼X−A that (x, i) = (y, j) or x, y ∈ X −A, i 6= j and x = y. In all cases we havex = y.

Therefore, the function ϕ : Y/ ∼A→ X with 〈[(x, i)]〉 7→ x is well-defined.Since the function ϕ′ : X + X → X with (x, i) 7→ x is continuous and ϕ′ =ϕ ◦ πA ◦ πX−A, we have by Proposition 3.2 that ϕ is continuous.

Define the function ψ : X → Y/ ∼ by x 7→ 〈[(x, 0)]〉. Since ψ is the compo-sition of x 7→ (x, 0), πX−A and πA, we have that ψ is continuous.

Because ψ ◦ ϕ = idY/∼Aand ϕ ◦ ψ = idX we conclude that X and Y/ ∼A

are homeomorphic.

17

3.4 Transforming the complement

We will transform the complement of a link to a combinatorial object. In alater chapter we will see that it is homeomorphic to hyperbolic polyhedra gluedtogether, for this reason we shall call this combinatorial object a polyhedroid.

To avoid confusion, we write vertices of a graph with small letters a, . . . , zand points in S3 with capital letters A, . . . , Z. Then we have to upgrade the cap-ital letters we used for spaces to calligraphic letters A, . . . ,Z. Links, diagramsand graphs of diagrams already have calligraphic letters.

3.4.1 Digon tolerant polyhedroids

In a hyperbolic polyhedron it is not possible that two distinct edges have thesame endpoints, for example the digon in Figure 3.4 does not exists. So if thediagram contains multiple edges between two vertices, we cannot lift it to apolyhedron.

Since diagrams can contain multiple edges between two vertices, we willdefine a digon tolerant version of the polyhedroids to formulate our intermediateresult.

Figure 3.4: A digon

Definition 3.13 (Digon tolerant polyhedroid). Let (G, γ) be an alternatingdiagram, with V ⊂ S2 the vertices of G. The digon tolerant polyhedroid P+

d.t.

of (G, γ) is the space S2 − V united with the part of S3 above S2. The digontolerant polyhedroid P−d.t. is the space S2 − V united with the lower part of S3.P+d.t. and P−d.t. have edges and faces, these are the edges and faces of the

graph G.

We shall abbreviate digon tolerant to d.t. When we glue the d.t. polyhe-droids together in a certain way, they form the complement of a link.

Lemma 3.14. Let (G, γ) be an alternating diagram, with V the vertices of G, letL be the corresponding link. There is a homeomorphism ϕrot : S2−V → S2−Vand an equivalence relation ∼E on P+

d.t. ∪ϕrot P−d.t. such that

• ϕrot rotates the clockwise faces of G including their bounding edges, oneedge clockwise, and the counterclockwise faces of G without their boundingedges, one edge counterclockwise;

• for each v ∈ V, ∼E contracts the equivalence classes [e]ϕrotand [f ]ϕrot

where −γ(v) = {e, f}, to one edge;

• the complement C = S3 − L is homeomorphic to(P+d.t. ∪ϕrot

P−d.t.)/ ∼E .

18

Proof. We look at a crossing of the corresponding link L.

(a) In the diagram (b) In the corresponding link

Figure 3.5: A crossing

Let v be a vertex of G with edges {v, a}, {v, b}, {v, c} and {v, d} such thatγ(v) = {{v, a}, {v, c}}, see Figure 3.5a. It is possible that some of a, b, c, d areequal. Let AC and BD be parts of the link that correspond to the crossing vin the region pqrs, see Figure 3.5b. Note in the line EF ⊂ C that joins the twoparts of the link, the endpoints are part of the link and therefore not part of C.

We double this line EF for each crossing. More precisely, let J =⋃v∈G EvFv

be all these lines without the endpoints, we obtain the space D by doubling Jin C. We have

D = C ∪idC−J C.With ∼J the equivalence relation that contracts J ×{0} and J ×{1} to J

we have by Proposition 3.12 that

C ∼= D/ ∼J . (3.9)

Now we cut D along a surface Z ⊂ D into two parts D+ and D− that lieroughly above and below S2. Near the crossing v, Z consists of the curvedsurfaces AEFBQ, BFECR, CEFDS and DFEAP . Near other crossings Z isdefined in the same way. Away from crossings Z coincides with S2.

By construction L lie on S2 except near crossings. The curved faces and S2

therefore join up nicely on their borders. We define D+ as the part of D aboveZ (and including Z), and D− as the part below Z. We have

D ∼= D+ ∪idZ D−. (3.10)

Let F+ be S2 united with the part of S3 above S2 minus some parts of thelink, as shown in Figure 3.7b. We will construct a homeomorphism ϕ+ : D+ →F+.

Figure 3.6: The crossing v in D+

19

We look at D+ near the crossing v. In Figure 3.6 the boundaries AEFBand BFEC are colored grey, the boundaries CEFD and DFEA look similar.We see that in D+, the only way to go underneath AC is through the doublededge EF .

We can therefore make a homeomorphism ϕ+ : D+ → F+ that moves onecopy of EF to the left, and the other copy to the right, see Figure 3.7. It maylook as if the endpoint E still needs to be split into E′ and E′′, but E is partof the link and therefore not in D+.

(a) The crossing in D+ (b) Flattened crossing in F+

Figure 3.7: Flattening the crossings

For the lower part D− we can do something similar. We observe that the onlyway to go over BD is through the doubled line EF , so there is a homeomorphismϕ− : D− → F− that moves the copies of EF away from each other.

In the homeomorphisms ϕ± we have not chosen yet which copy of EF movesto which side, we will do this now. Two opposite faces f ′ and f ′′ of v have aclockwise orientation, see Definition 2.18. In both ϕ+ as ϕ− we move the copyE′F ′to the face f ′ and the copy E′′F ′′ to the face f ′′. This is illustrated inFigure 3.8.

(a) The lines EF ′ and EF ′′ in F+ (b) The lines E′F and E′′F in F−

Figure 3.8: The configuration of the doubled line EF

Now we can shrink the remaining parts of the link to points. We follow thepart AC of the link in F+ to its endpoints (where it turns into a line ExFx ofa crossing x). Since L is alternating, AC ends at the first crossings next to v,see Figure 3.9a. Since AC is not part of F+, we can shrink it to a point with ahomeomorphism.

20

Let H+ be S2 united with the upper part of S3 minus the points in thegraph G. We construct the homeomorphism χ+ : F+ → H+ by shrinking eachpart of the link to a point. See Figure 3.9b. We can do the same in F−, thehomeomorphism χ− : F− → H− shrinks the parts of the link to points.

(a) The part of the link in F+ (b) The shrunken part in H+

Figure 3.9: Shrinking the link to points

The shrunk parts of the link lie exactly on the crossings. So with V thevertices of the graph, we see that H+ is S2−V united with the part of S3 aboveS2 and H− is S2 − V united with the part of S3 below S2.

Furthermore, we see that the (copies of the) lines ExFx coincide with theedges of the graph G. Therefore, we have that H+ is the d.t. polyhedroid P+

d.t.

of the diagram (G, γ) and H− is the d.t. polyhedroid P−d.t..We have the homeomorphisms (χ± ◦ ϕ±) : D± → P±d.t. and by (3.10) that

D ∼= D+ ∪idZ D−. Proposition 3.9 gives us a homeomorphism ψ : P+d.t. → P

−d.t.

ψ = χ− ◦ ϕ− ◦ idZ ◦(ϕ+)−1 ◦

(χ+)−1 , such that

D ∼= P+d.t. ∪ψ P

−d.t.. (3.11)

Let f be a face of the diagram, we will apply ϕ± and χ± on the part of Dcorresponding to the face f to see how ψ acts. If f is a clockwise face, then ψglues the face in P+

d.t. to the face in P−d.t. by rotating the face by one edge inclockwise direction. If f is a counterclockwise face, then ψ rotates by one edgein counterclockwise direction. This is illustrated in Figure 3.10.

(a) The image of Z in P+d.t.

(b) The cutting surface Z ⊂ D (c) The image of Z in P−d.t.

Figure 3.10: The gluing between P+d.t. and P−d.t.

21

In the definition of ϕ± we said that ϕ± moves the same copies of the edgesEF to the clockwise faces. So, ψ rotates the face including the edges on clock-wise faces and it rotates the face without the edges on counterclockwise faces.We see that ψ is precisely the function ϕrot (ϕrot is defined in the statement ofthis theorem).

We now have a homeomorphism ω : D → P+d.t. ∪ϕrot

P−d.t.. In Figure 3.10we see that the lines ExFx in J (see the definition a few lines above (3.9))are moved by ω to the edges of the polyhedroids. The equivalence relation ∼Jcontracts the edges with similar types of arrows. By looking at Figure 3.7 we seethat in the upper part P+

d.t. the edges {e, f} that go underneath are contracted,these are the edges {e, f} = −γ(v). So after gluing with ϕrot, the edges [e]ϕrot

and [f ]ϕrotare identified by ∼J , for all v and −γ(v) = {e, f}.

Hence, the equivalence relations ∼J and ∼E (which is defined in the state-ment of this theorem) satisfy the requirement of Proposition 3.3 and we concludethat

C =(P+d.t. ∪ϕrot P−d.t.

)/ ∼E .

3.4.2 Polyhedroids

Now we will remove the digons of the d.t. polyhedroids. We begin by defining(digon intolerant) polyhedroids.

Definition 3.15 (Polyhedroid). Let (G, γ) be an alternating, balanced diagram,with V ⊂ S2 the vertices of G. The polyhedroid P+ of (G, γ) is the space S2−Vunited with the part of S3 above S2. The polyhedroid P− is the space S2 − Vtogether with the part of S3 below S2.P+ and P− have edges and faces. If Q is a partition of V in balanced pairs

(see Definition 2.20), then one of the two edges of G that are between a pair{v, v′} ∈ Q is an edge of P+ and P−, and all the edges that are not between abalanced pair are edges of P+ and P−.

Consequently, the faces of P+ and P− are the faces of G except those thatare bounded by precisely two edges.

(a) A diagram (b) The polyhedroid

Figure 3.11: The figure-eight knot

Example. In Figure 3.11a is an alternating, balanced diagram of the figure-eight knot, in Figure 3.11b the corresponding Polyhedroid (with the space aboveand below the paper, we get P+ and P−).

22

Note that the vertices are not part of the polyhedroid, so they are drawnas open circles. We also see that only one of the two edges between a balancedpair is an edge of the polyhedroid, so that is has no faces bounded by two edges.

Definition 3.16 (Face gluing). Let (G, γ) be an alternating, balanced diagram.Let P± be the polyhedroids of (G, γ) and let f+

1 , . . . , f+n and f−1 , . . . , f

−n be the

faces of respectively P+ and P−. Then the face gluings ϕ1, . . . , ϕn are the home-omorphisms ϕi : f+

i → f−i that rotate fi including bounding edges one edgeclockwise if fi is a clockwise face, and counterclockwise if fi is counterclockwise.

We can combine the equivalence relations ∼ϕi P+ + P− in one equivalencerelation ∼Φ. We say that ∼Φ is the face gluing of P+ and P− generated by(G, γ).

Theorem 3.17. Let (G, γ) be an alternating, balanced diagram and L the cor-responding link. Let P+ and P− be the polyhedroids of (G, γ) and ∼Φ the facegluing generated by (G, γ).

Then the complement C = S3 − L is homeomorphic to (P+ + P−) / ∼Φ.

Proof. Let Q be a partitioning of the vertices V in balanced pairs, let {v, v′} ∈ Qbe a balanced pair. Define the lines E′F ′ and E′′F ′′ that join the parts of thelinks that cross at respectively v and v′, see Figure 3.12a.

We define an equivalence relation ∼∆ that contracts the face ∆ = E′F ′E′′F ′′

to a single edge EF , for every balanced pair {v, v′} ∈ Q. Since the face ∆ isa closed disk in S3, we can contract it to a line segment without changing thestructure of S3. This is true also for the complement C ⊂ S3. Hence, we have

C ∼= C/ ∼∆ .

(a) A digon (b) The contracted digon

Figure 3.12: Contracting a digon

By Lemma 3.14, we have the homeomorphism

ψ : C →(P+d.t. ∪ϕrot P−d.t.

)/ ∼E .

The digon ∆ is transformed by ψ to the region ∆′ in Figure 3.13. Theequivalence relation is therefore transformed by ψ to the relation ∼∆′ thatcontracts ∆′ along the grey lines in Figure 3.13. Then by Proposition 3.3 wehave

C ∼=((P+d.t. ∪ϕrot

P−d.t.)/ ∼E

)/ ∼∆′ . (3.12)

23

Figure 3.13: The digon in(P+d.t. ∪ϕrot

P−d.t.)/ ∼E

Now we will ’push’ the relation ∼∆′ through the expression to P±d.t. to obtainthe polyhedroids P±.

The relation ∼E contracts the double edges in P+d.t. ∪ϕrot P−d.t.. In Figure

3.14a the edges e, e′ are contracted to one edge, and the edges f , f ′. We willcontract the digons in P+

d.t. ∪ϕrotP−d.t. before we identify these edges. We abuse

notation a little by calling that contraction also ∼∆′ , but its essentially thesame.

By letting ∼∆′ work on P+d.t. ∪ϕrot

P−d.t. , we see that the edges e′ and f ′ willform one edge, with the head of e′ identified with the tail of f ′ and vice versa.We call this new edge g, it has the direction of e. Now we are looking for arelation ∼E′ on

(P+d.t. ∪ϕrot

P−d.t.)/ ∼∆′ that contracts this new edges so that

[x]∆′ ∼E′ [y]∆′ iff [x]E ∼∆′ [y]E . (3.13)

Because e, e′ and f , f ′ are contracted and e′ and f ′ form a new edge g, therelation ∼E′ must contract e, f and g. Since g got the direction of e′ which is theopposite direction of f ′, we must reverse the direction of f and then contract itto g. See Figure 3.14b.

(a) Edge contraction with the digon (b) Edge contraction with the collapsed digon

Figure 3.14: Changing the order of contractions

We define ∼E′ as follows. Let {v, v′} be a balanced pair in Q. Let g be theedge between v, v′ (that originally was a digon), let e, f be edges of respectivelyv and v′, with e /∈ γ(v) and f /∈ γ(v′). Give e, g an orientation towards v, and fan orientation away from v′. Then ∼E′ contracts the edges e, f, g to one edge,respecting their orientation. Since (G, γ) is alternating and balanced, every edgeis contracted precisely once with two other edges.

By Proposition 3.4 and (3.13) we have((P+d.t. ∪ϕrot

P−d.t.)/ ∼E

)/ ∼∆′

∼=((P+d.t. ∪ϕrot

P−d.t.)/ ∼∆′

)/ ∼E3 . (3.14)

24

The relation ∼∆′ contracts the digons in P+d.t. ∪ϕrot

P−d.t., we want to con-tract the digons first and then glue the spaces together. Since contracting thedigons in P±d.t. goes the same as in P+

d.t. ∪ϕrotP−d.t., we call the relations on P±d.t.

again ∼∆′ . Because all the digon lie in the surface of the d.t. polyhedroidswhere we glue them together, we can apply Proposition 3.10. We define thehomeomorphism ψrot : P+

d.t./ ∼∆′→ P−d.t./ ∼∆′ as the ϕ′ in Proposition 3.10,we get (

P+d.t. ∪ϕrot

P−d.t.)/ ∼∆′

∼=(P+d.t./ ∼∆′

)∪ψrot

(P−d.t./ ∼∆′

). (3.15)

By definitionψrot ([x]∆′) = [ϕrot(x)]∆′ ,

so we see that ψrot rotates all the faces of P+d.t./ ∼∆′ that are not digons in

the same way as ϕrot. By definition of ϕrot we see the following. If f is a non-digon face, ψrot rotates it with bounding edges clockwise if f is clockwise, andit rotates it without bounding edges counterclockwise if f is counterclockwise.If e is an edge between two counterclockwise faces (e used to be a digon), thenψrot reverses it.

Now consider P±d.t./ ∼∆′ . Since P±d.t. is S2−V plus the space above or belowS2, contracting digons does not change the topology. If we forget the edges andfaces of P±d.t. and P± we have

P±d.t./ ∼∆′∼= P±d.t. = P±.

The edges and faces of P±d.t. are changed however. By applying ∼∆′ every twoedges between a balanced pair is turned into one edge, and the faces boundedby two edges are therefore removed. We see that this are exactly the edges andfaces of the polyhedroids P±. So the function ψrot acts exactly the same onP+d.t./ ∼∆′and P+. Hence,(

P+d.t./ ∼∆′

)∪ψrot

(P−d.t./ ∼∆′

) ∼= P+ ∪ψrotP−.

With (3.15), (3.14) and (3.12) we conclude that

C ∼=(P+ ∪ψrot P−

)/ ∼E′ . (3.16)

We can combine the relations ∼ψrot and ∼E′ working after each other to onerelation ∼Ψ on P+ + P− with

x ∼Ψ y iff [x]ψrot ∼E′ [y]ψrot . (3.17)

Let x, y be in the inside of faces. Since ∼E′ does not change faces, and ∼Φ

and ∼ϕrot rotate faces in the same way, we have x ∼Ψ y iff x ∼Φ y.To complete the case that x, y lie on edge we will look how ∼Ψ and ∼Φ

identify edges. Consider a region with a digon, see Figure 3.15a. The homeo-morphism ψrot rotates the clockwise faces, and it rotates the collapsed edge ofthe digon in the middle. So the edges with arrows ’→’ in P+ are glued to theedge with arrows ’ 7→’ in P−, see Figure 3.15b. Then the relation ∼E′ identifiesthese three edges (for an illustration of the definition of ∼E′ , see Figure 3.14b).

25

(a) Edges in P+ (b) Edges in P−

Figure 3.15: Edge identifications by ∼Ψ

With the face gluing ∼Φ all faces are rotated with their bounding edges. Sothe edges in P+ in Figure 3.16a will be glued to the edges in P− in Figure 3.16baccording to there arrow types. We see that edges with different arrow typesin P+ are glued to the same edge in P−, the edges with these arrow types aretherefore identified by ∼Φ. By looking at Figure 3.16b we see that all the threearrow types are identified, and we conclude that ∼Φ contracts the same edgesas ∼Ψ.

(a) Edges in P+ (b) Edges in P−

Figure 3.16: Edge identifications by ∼Φ

We see that ∼Φ and ∼Ψ are equal. Since (3.17) we can apply Proposition3.4 to see that (

P+ ∪ψrot P−)/ ∼E′ ∼=

(P+ + P−

)/ ∼Ψ .

With (3.16) we conclude

C ∼=(P+ + P−

)/ ∼Φ .

Corollary 3.18. In (P+ + P−) / ∼Φ the edges are identified in groups of six.

Proof. The face gluings as defined in Definition 3.16 glue the edges as follows.The three edges with arrow types→, 7→ and � in P+ in Figure 3.16a are gluedto edges with corresponding arrow types in Figure 3.16b by rotating in thedirection that is shown in Figure 3.16a. We see that the six edges with arrowtypes in P+ and P− are glued together.

26

Chapter 4

Hyperbolic geometry

Figure 4.1: A regular heptagonal tessellation of H2

4.1 The hyperbolic space

The hyperbolic space is one example of a non-Euclidean space. Given a line `and a point P , there are infinite lines through the point P parallel to `.

We use the Poincare half-plane model to work with the hyperbolic space.For the proofs in this section, we refer to [2].

Consider the half-space

H3 = {(x, y, u) ∈ R3 | u > 0}.

Let ρ be a piecewise differentiable curve in H3 given by the parametrization

ρ(t) = (x(t), y(t), u(t)) , t ∈ [a, b].

Definition 4.1 (Hyperbolic length). The hyperbolic length of the curve ρ is

`hyp(ρ) =ˆ b

a

√x′(t)2 + y′(t)2 + u′(t)2

u(t)dt.

27

Definition 4.2 (Hyperbolic volume). The hyperbolic volume of a region X ⊆H3 ⊂ R3 is defined by

Vhyp(X) =ˆX

1u3

dxdydu.

Then we can define a metric on H3 in the following way.

Definition 4.3 (Hyperbolic distance). Let P,Q ∈ H3. The hyperbolic distancefrom P to Q is

dhyp(P,Q) = inf{`hyp(ρ) | ρ is a curve from P to Q}.

Definition 4.4 (Hyperbolic space). The hyperbolic space is the metric space(H3, dhyp).

The hyperbolic space has a natural extension to infinity, namely the plane{(x, y, u) ∈ R3 | u = 0} beneath the half-space H3 together with a single point{∞}. We identify this with the Riemann sphere C = C ∪ {∞}.

4.1.1 Isometries

An isometry of (H3, dhyp) is a bijection ϕ : H3 → H3 that preserves distance,i.e.

dhyp(ϕ(P ), ϕ(Q)) = dhyp(P,Q).

The following theorem tells us that isometries of H3 are determined by howthey behave on the sphere C at infinity.

Theorem 4.5. Every linear or antilinear fractional map ϕ : C → C has aunique continuous extension ϕ : H3 ∪ C→ H3 ∪ C whose restriction to H3 is anisometry of (H3, dhyp).

Conversely, every isometry of (H3, dhyp) is obtained in this way.

Recall that a linear fractional map ϕ : C→ C is of the form

ϕ(z) =az + b

cz + d

with a, b, c, d ∈ C and ad− bc = 1.An antilinear fractional map ϕ : C→ C is of the form

ϕ(z) =az + b

cz + d

with a, b, c, d ∈ C and ad− bc = 1.

4.1.2 Geodesics

Definition 4.6. A geodesic is a curve ρ that is locally the shortest curve, i.e.for every P,Q ∈ ρ sufficiently close to each other, the section of ρ joining P toQ is the shortest curve from P to Q.

Since isometries preserve distance, they send geodesics to geodesics.

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The following proposition tells us how geodesics look. For simplicity we seea line as a circle with radius ∞.

Proposition 4.7. Let P,Q ∈ H3 be distinct points. There is one geodesicstarting at P and ending at Q. It is the circle arc joining P to Q that iscontained in the vertical circle passing through P and Q and centered on thexy-plane.

Geodesics can be extended, when a geodesic cannot be extended to a largergeodesic we call it a complete geodesic. Using Proposition 4.7 we see whatcomplete geodesics look like.

Proposition 4.8. A complete geodesic is the intersection of H3 with a verticalcircle centered on the xy-plane.

Strictly speaking, a complete geodesic ρ lies in H3 and has no points in C.Nevertheless, we will speak of the endpoints α, β ∈ C of the geodesic ρ. The fol-lowing corollary of Proposition 4.8 says that complete geodesics are determinedby two distinct points in H3 ∪ C, so we can speak of the geodesic from α ∈ C toβ ∈ C.

Corollary 4.9. Let P,Q ∈ H3∪C be two distinct points. There is one completegeodesic going through or ending in P and Q.

The angles in (H3, dhyp) coincide with Euclidean angles. For our purpose,the following definition is useful.

Definition 4.10 (Angle). Let ρ, σ be geodesics intersecting at P ∈ H3, byProposition 4.7 they are circle arcs through P . These circle arcs define 2, 3 or4 regions around P , depending on whether ρ and σ end in P or go through P .

Define the angle between ρ and σ viewed from one of these regions as theEuclidean angle in this region between the circle arcs. When we do not specifya region, then the region is used that gives the smallest angle.

Let ρ, σ be geodesics with a shared endpoint α ∈ C. Define the angle aszero.

4.1.3 Planes

Definition 4.11 (Hyperbolic plane). Let P ∈ H3 be a point and ρ a geodesicthrough P . The hyperbolic plane through P orthogonal to ρ is the union of allthe geodesics σ through P such that the angle between ρ and σ is 1

2π.

Proposition 4.12. A hyperbolic plane is the intersection of H3 with a (Eu-clidean) sphere centered at the xy-plane or with a vertical Euclidean plane.

Corollary 4.13. If P,Q are distinct points in the hyperbolic plane H, then thecomplete geodesic through P and Q lie in H.

Corollary 4.14. The intersection of two intersecting planes is a completegeodesic.

Proposition 4.15. Every isometry of (H3, dhyp) sends hyperbolic planes tohyperbolic planes.

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4.1.4 Horospheres

Horospheres will enable us to look at the neighbourhood of a point α ∈ Cat infinity. When we want the space obtained by gluing polyhedra to have acomplete, hyperbolic structure, we will truncate polyhedra by intersecting themwith horospheres to see if the space ’behaves’ at infinity.

Definition 4.16. A horosphere centered at α ∈ C is the intersection of H3 witha Euclidean sphere which is tangent to the xy-plane at z, and lies above thisxy-plane.

A horosphere centered at∞ is a horizontal Euclidean plane contained in H3.

Proposition 4.17. Every isometry of (H3, dhyp) sends horospheres to horo-spheres.

4.2 Hyperbolic polyhedra

(a) Tetrahedron (b) Cube (c) Icosahedron

Figure 4.2: Hyperbolic polyhedra

Definition 4.18 (Polygon). Let H be a hyperbolic plane, and H the plane Hunited with its endpoints in C. Let n ≥ 3 and P1, . . . , Pn distinct points in H,such that the geodesics from Pi to Pi+1 for i = 1, . . . , n − 1 and the geodesicfrom Pn to P1 only intersect at endpoints. By Corollary 4.13 these geodesics liein H.

A polygon is the subset of H delimited by (and including) these geodesics.The endpoints P1, . . . , Pn are called the vertices of the polygon, the geodesicsare called the edges of the polygon.

Since there is a unique geodesic between two points P,Q ∈ H3 ∪ C (seeCorollary 4.9) there are no hyperbolic digons, which are polygons with preciselytwo edges.

When a vertex lies in C, we call this vertex an ideal vertex . When all thevertices of a polygon are ideal, we call the polygon an ideal polygon.

Definition 4.19 (Polyhedron). Let F1, . . . , Fn be polygons such that an edge ofa polygon Fi is an edge of exactly one other polygon Fj , and that the intersectionof two polygons Fi and Fj consists only of edges and vertices of Fi.

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A polyhedron is the region delimited by (and including) F1, . . . , Fn. Thepolygons F1, . . . , Fn are called the faces of the polyhedron, the edges of thepolyhedron are the edges of its faces and the vertices of the polyhedron are thevertices of its faces.

When all the vertices of a polyhedron ∆ are ideal, we call the polyhedronan ideal polyhedron.

The tetrahedron is the simplest example of a polyhedron.

Definition 4.20 (Tetrahedron). A tetrahedron is a polyhedron with four ver-tices.

Proposition 4.21. An (ideal) tetrahedron has a finite volume.

Proof. In [9] a direct formula is given to compute the volume of a tetrahedronwith zero to four ideal vertices.

4.2.1 Skeleton

We can construct a graph that represents the edges and vertices of a polyhedron.

Definition 4.22 (Skeleton). The skeleton of a polyhedron ∆ is the graph Gwith a vertex in G for every vertex of ∆, and an edge between two graph verticesif the two corresponding vertices in ∆ are the endpoints of an edge in ∆.

Since there do not exists digons, a skeleton is a simple graph (a graph withat most a single edge between two vertices).

Example. In Figure 4.3 are the skeletons of a tetrahedron and a cube.

(a) The skeleton of a tetrahedron (b) The skeleton of a cube

Figure 4.3: Skeletons

Theorem 4.23 (Steinitz). The skeleton of a convex polyhedron is planar and3-connected. Every planar and 3-connected graph is the skeleton of a convexpolyhedron.

See [14] (page 103) for a proof.

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4.2.2 Dihedral angles

When we will define a geometry on glued polyhedra together, we must checkthat they align nicely. In particular, the polyhedra around an edge should fillthe space around the edge. We can check this by measuring the angles of thepolyhedra around that edge.

The angle of a polyhedron along an edge is defined in the same way is inEuclidean geometry.

Definition 4.24 (Dihedral angle). Let ∆ be a polyhedron, E an edge and F1,F2 the two faces meeting at E. Choose a point P ∈ E, let H be the planetrough P orthogonal to E. The intersections σi = Fi ∩H are geodesics endingat P . The dihedral angle of ∆ along E is the angle between σ1 and σ2 viewedfrom the region that is inside the polyhedron.

Note. The angle does not depend on the choice of P ∈ E.

Proposition 4.25. Let ∆ be a polyhedron and P an ideal vertex. Let E1, . . . , Enbe the edges of the vertex P . Then the dihedral angles of ∆ along Ei add up to(n− 2)π.

Proof. Let P = α ∈ C be the ideal vertex. Let ϕ be an isometry that sends αto ∞ (see Theorem 4.5), let Λ = ϕ(∆) and Di = ϕ(Ei). Since ϕ is an isometry,the angle of Λ along Di is the same as the angle of ∆ along Ei.

Let Fi be the face between Di and Di+1, i = 1, . . . , n where we set n+1 ≡ 1for convenience. Since the edges Di have ∞ as endpoint, they are vertical linesand the faces Fi lie in vertical planes. Let Π be a horizontal (Euclidean) planeand Pi the intersection Di ∩Π.

We measure the dihedral angle of Λ along Di in these points Pi. Let Hi

be the plane through Pi orthogonal to Di, then Hi is a half-sphere with Pi ashighest point. Therefore, the dihedral angle of Λ along Di, which is the anglebetween between the intersections Fi−1∩Hi and Fi∩Hi viewed from the regionthat is inside Λ, is the same as the Euclidean angle between the intersectionsFi−1 ∩Π and Fi ∩Π viewed from inside Λ.

The Euclidean line segments Fi∩Π form an Euclidean polygon with n edges,so the angles between Fi−1 ∩Π and Fi ∩Π with i = 1, . . . , n viewed from insideΛ add up to (n− 2)π.

Figure 4.4: The opposite edges of an ideal tetrahedron

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Lemma 4.26. Let F , F ′ be opposite edges in an ideal tetrahedron ∆. Then theangles of ∆ along F and F ′ are equal.

Proof. Let F1, . . . , F4 be the faces of ∆. Let αij be the angle of ∆ along theedge between Fi and Fj . Since the skeleton is 3-regular, Proposition 4.25 givesthat the angles along each of the vertices add up to π. Hence,

α23 + α34 + α42 = π

α34 + α41 + α13 = π

α41 + α12 + α24 = π

α12 + α23 + α31 = π.

Hence, α12 = α34, α13 = α24 and α14 = α23.

Proposition 4.27. Ideal tetrahedra (up to orientation preserving isometry) arein one-one corresponding with complex triples {z, 1

1−z , 1−1z}, z ∈ C, Im(z) > 0.

Proof. Let {z, 11−z , 1 −

1z}, Im(z) > 0. Each triple in the set {0, 1,∞, z} ⊂ C

defines a different ideal polygon, these polygons together give an ideal tetra-hedron. The orientation preserving isometries z 7→ 1

1−z and z 7→ 1 − 1z take

{0, 1,∞, z} to {0, 1,∞, 11−z} and {0, 1,∞, 1 − 1

z}, so it does not matter whichz ∈ {z, 1

1−z , 1−1z} we choose.

Let ∆ be an ideal tetrahedron. Apply an orientation preserving isometrysuch that three of the vertices of ∆ go to 0, 1 and ∞, and the fourth to z withIm(z) > 0. Define the triple {z, 1

1−z , 1 −1z}. Another orientation preserving

isometry sending the vertices of ∆ to 0, 1,∞ and z′, gives the same triple. Tosee this observe the following.

At a vertex v, there is one edge from each opposite pair (Figure 4.4). Sinceby Lemma 4.26 opposite edges have the same angle, there is a triple α1, α2, α3

with α1 +α2 +α3 = π that are the angles of the edges at v. The triangle 0, 1, zhas angles α1, α2 and α3. There are only two other z′ with Im(z′) > 0 suchthat the triangle 0, 1, z′ has angles α1,α2 and α3. These are 1

1−z and 1− 1z and

these give the same triple.The mappings from ideal tetrahedra to triples and back are each others

converse. We conclude that they are in one-one correspondence.

Definition 4.28 (Complex parameter). Let ∆ be an ideal tetrahedron, and{z, 1

1−z , 1 −1z} the triple it is bijective with. The triple is called the complex

parameter of ∆.

4.2.3 Regular polyhedra

Definition 4.29 (Regular). A polyhedron is regular if the angles along its edgesare equal. A polyhedron is x-regular if the angles along its edges are x.

Proposition 4.30. The skeleton of a regular, ideal polyhedron ∆ is 3-regularif and only if ∆ is 1

3π-regular.

Proof. Let ∆ be a x-regular, ideal polyhedron, let G be its skeleton. Let v be avertex of G, let n be the number of edges at v. The vertex v corresponds withan ideal vertex α of ∆, there are n edges Ei of ∆ ending at α.

Since ∆ is x−regular, the angles along the edges Ei add up to nx. ByProposition 4.25 they add up to (n− 2)π. Hence, x = 1

3π iff n = 3.

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Corollary 4.31. An ideal regular tetrahedron is 13π-regular.

Proof. The skeleton of a tetrahedron is 3-regular.

Corollary 4.32. The complex parameter of an ideal regular tetrahedron is{eπi/3

}.

Proof. Since it is 13π-regular, it follows from Proposition 4.27.

4.2.4 Decomposable polyhedra

Definition 4.33 (Decomposition). A decomposition of a polyhedron ∆ is afamily polyhedra Λ1, . . . ,Λn such that

⋃ni=1 Λi = ∆ and the intersection of two

polyhedra Λi, Λj consists only of faces, edges and vertices of Λi.

Every polyhedron has a decomposition, since we are only interested in de-compositions with ideal, regular tetrahedra, we reserve the word decomposablefor this case.

Definition 4.34 (Decomposable). A polyhedron is decomposable if it has adecomposition of ideal, regular tetrahedra.

The n-tetrahedra are a family of ideal, decomposable, 13π-regular polyhedra.

They are recursively defined.

Definition 4.35 (n-tetrahedron). A n-tetrahedron is an ideal, decomposable,13π-regular polyhedron defined as follows.

An 1-tetrahedron is an ideal, regular tetrahedron.A (n + 1)-tetrahedron is obtained by pasting several n-tetrahedrons in the

following way. Start with a n-tetrahedron ∆. For each face F of ∆, mirror ∆ inF and obtain an n-tetrahedron ∆F . Let Λ be the union of ∆ with all the ∆F .

By construction we have that Λ is ideal and decomposable. We will modifyΛ so that it becomes 1

3π-regular.Let E be an edge of ∆. There are two faces F1, F2 in ∆ around E. Therefore

there are three polyhedra ∆, ∆F1 and ∆F2 around E in Λ. Since all thesepolyhedra are 1

3π-regular, the angle of Λ along E is π, so the two faces H1 andH2 of Λ around E lie in one plane. Delete the edge E and the faces H1, H2

from Λ and add the face H = H1 ∪H2.Repeat this for every edge of E, the resulting polyhedron Λ′ is an n + 1-

tetrahedron. Since we did not change the shape of Λ, Λ′ is also ideal anddecomposable, but now Λ′ is also 1

3π-regular. To see this E be an edge of Λ′.We deleted all edges from ∆, so E is an edge of ∆F that is not the edge ofanother polyhedron in Λ′. Since ∆F is 1

3π-regular, the angle of Λ′ along E isalso 1

3π.

Example. The hyperbolic cube is the 2-tetrahedron. The 10-tetrahedron hasabout 10154 faces, edges and vertices. By looking at the construction, the func-tions fn and en that give the number of faces and edges of the n-tetrahedron,are defined by

f1 = 4e1 = 6

fn+1 = (fn − 1)fn − enen+1 = (fn − 1)en − en.

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Since a n-tetrahedron is 13π-regular and therefore has a 3-regular skeleton, its

number of vertices is given by vn = 23en.

Proposition 4.36. A n-tetrahedron has a finite volume. A n-tetrahedron anda m-tetrahedron have a different volume if m 6= n.

Proof. The volume of an ideal regular tetrahedron is finite (see Proposition4.21). Since a n-tetrahedra consists of finite many tetrahedra it also has a finitevolume.

Assume m > n. Then the m-tetrahedron is obtained by gluing multiplen-tetrahedra together.

Using the number of faces fi defined in the previous example we can computethe hyperbolic volume of a n-tetrahedron. Let V be the volume of an idealregular tetrahedron. Then the volume Vn of a n-tetrahedron is given by

Vn = V ·n−1∏i=1

(fi + 1) .

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Chapter 5

Hyperbolic manifolds

In this chapter we show that some link complements have a hyperbolic metricdefined on them. With this metric one can derive a new kind of property of linkcomplements, for instance the hyperbolic volume of a complement.

With Mostow’s theorem we have that this metric is unique. Hence, theseproperties do not depend on the chosen metric, but are properties of the com-plement itself.

Definition 5.1 (Hyperbolic manifold). A n-manifold M has a hyperbolic struc-ture if it has an atlas of charts ϕ : U → Hn such that for every (U,ϕ) and (V, ψ)in the atlas with U ∩ V 6= ∅ we have that

ϕ ◦ ψ−1 : ψ(U ∩ V ) → ϕ(U ∩ V )ψ ◦ ϕ−1 : ϕ(U ∩ V ) → ψ(U ∩ V )

are isometries of Hn. In this way, M inherits the metric of Hn. We also saythat M is a hyperbolic n-manifold .

If the inherited metric on M is complete, we say that M has a completehyperbolic structure and that M is a complete hyperbolic n-manifold .

Theorem (Mostow’s rigidity theorem). Suppose M and N are complete finite-volume hyperbolic n-manifolds with n > 2. If M and N are homeomorphic thenM and N are isometric.

For a proof, see [8].

5.1 Gluing hyperbolic tetrahedra

Under certain conditions, we can obtain a complete hyperbolic manifold bygluing ideal polyhedra together.

Definition 5.2 (Truncated tetrahedron). Let ∆ be an ideal tetrahedron. Atevery ideal vertex α ∈ C we make equally large horospheres Hi (see Definition4.16).

The truncated tetrahedron ∆′ is the inside of ∆ bounded by, and not includ-ing, these horospheres. The intersections Hi ∩∆ are called the truncated facesof ∆′, although strictly speaking there are not part of ∆′.

36

Proposition 5.3. The truncated faces of an ideal regular tetrahedron are iso-metric with Euclidean equilateral triangles.

Proof. Let v be an ideal vertex. By moving v with an isometry to ∞, thehorosphere H at v is a horizontal plane, so H ∩ ∆′ is an Euclidean triangle.Since ∆ is 1

3π-regular (Corollary 4.31), this triangle is equilateral.

Theorem 5.4. Let M be a 3-manifold such that ∂M is a collection of tori.Suppose there are ideal truncated tetrahedra ∆1, . . . ,∆n such that M − ∂M canbe obtained by gluing their (non-truncated) faces in pairs. At the i-th edge ofthis gluing, let wi1, . . . , wik ∈ C be a certain number in the complex parameterof the tetrahedra around that edge.

Then M − ∂M inherits a hyperbolic structure if and only if for each edge i

k∏j=1

wij = 1.

For a proof, see [6]. In [6] it specified which certain number wij is. For ourpurposes however, we only need to know that wij is in the complex parameter.

Theorem 5.5. Let M be a 3-manifold such that ∂M is a collection of tori andM−∂M has a hyperbolic structure. Suppose there are ideal truncated tetrahedra∆1, . . . ,∆n such that M − ∂M can be obtained by gluing their (non-truncated)faces in pairs.

If every Euclidean quadrilateral in ∂M consisting of truncated faces is aparallelogram, then M has a complete hyperbolic structure.

For a proof, see [13].

Proposition 5.6. Let ∆1, . . . ,∆n be ideal regular tetrahedra. Let M be obtainedby gluing the faces of the tetrahedra in pairs, such that precisely six tetrahedraare glued around each edge, and such that ∂M is a collection of circles.

Then M is a complete hyperbolic 3-manifold and up to isometries, the hy-perbolic metric on M is unique.

Proof. Truncate ∆i, let ∆′i be the truncated tetrahedra. Let N be obtainedby gluing the truncated tetrahedra ∆′i in the same way as ∆i. Since ∂M is acollection of circles, we have that ∂N is a collection of tori.

Let wi1, . . . , wik ∈ C be defined as in the statement of Theorem 5.4. Thecomplex parameter of a ideal, regular tetrahedron is {eπi/3} (by Corollary 4.32),so wij = eπi/3 for every i, j. Because precisely six tetrahedra are glued aroundevery edge, we have k = 6, hence

k∏j=1

wij =(e

13πi)6

= 1.

Then by Theorem 5.4 we have that N − ∂N has a hyperbolic structure.Every truncated face of a ∆′i is (isometric to) an Euclidean equilateral tri-

angle, by Proposition 5.3. So every quadrilateral in ∂N is a parallelogram. ByTheorem 5.5 we have that N has a complete hyperbolic structure.

By contracting the collection of closed tori to a collection of circles, we canshow that M has a complete hyperbolic structure.

37

Since ideal tetrahedra have a finite volume (see Proposition 4.21), and Mis obtained by gluing n tetrahedra, we have that the volume of M is finite.Suppose M ′ is the same space as M with a different metric, we can applyMostow’s theorem to see that M and M ′ are isometric. Hence, the metric isunique up to isometries.

5.2 Regular Links

The regular links are a subset of the links that have a complete hyperboliccomplement. We obtain the regular links by gluing regular tetrahedra.

5.2.1 Gluing hyperbolic polyhedra

The polyhedroids of Chapter 3 give us a way of constructing a complement bygluing hyperbolic polyhedra.

Definition 5.7 (Corresponding polyhedron). Let (G, γ) be a diagram. A poly-hedron ∆ is a corresponding polyhedron of (G, γ) if its skeleton G is the same asG with all the double edges collapsed to a single edge.

Proposition 5.8. Every alternating, balanced diagram (G, γ), with G a 3-connected graph has a corresponding polyhedron.

Proof. By Proposition 2.22 there is an embedded, 3-regular graph G that is thesame as G where double edges are collapsed. Because G is 3-connected, G is3-connected. By Theorem 4.23 there is a polyhedron ∆ with G as skeleton.

Proposition 5.9. Let (G, γ) be an alternating, balanced diagram, with G a 3-connected graph, let L be the corresponding link and let ∆ be a correspondingpolyhedron.

Then there is a gluing ∼Φ that glues every face of ∆ to the same face of acopy of ∆, such that edges are glued together in groups of six and

S3 − L ∼= (∆ + ∆) / ∼Φ .

Proof. Let G be the skeleton of ∆. By definition of corresponding polyhedron,we have that G is the same as G where all the double edges are collapsed to asingle edge.

Let P+ and P− be the polyhedroids defined by (G, γ) (see Definition 3.15).The edges of P+ are the edges of G with double edges collapsed, so these areprecisely the edges of G. Hence, the edges of ∆ and P+ correspond with eachother.

The vertices V of G are not in P+ and since ∆ is ideal, the vertices V of theskeleton G are also not part of ∆. Because the vertices G and the skeleton arethe same, the ’missing’ vertices of P+ and ∆ correspond with each other.

Since the spaces P+ ∪ V and ∆ ∪ V are three dimensional, closed, convexsubsets of R4 there is a homeomorphism between them. We can arrange thishomeomorphism so that it sends V to V . Hence, P+ and ∆ are homeomorphic.Analogously, P− ∼= ∆.

Because the edges between corresponding vertices correspond with eachother, we can choose the homeomorphism so that it sends edges of P+ to edges

38

of ∆. The face gluing ∼Φ of P+ to P− (see Definition 3.16) is therefore a facegluing of ∆ to ∆ that glues the a face Fi of one ∆ to the face Fi of the other∆. We have (

P+ + P−)/ ∼Φ

∼= (∆ + ∆) / ∼Φ .

By Theorem 3.17 we then have

S3 − L ∼= (∆ + ∆) / ∼Φ .

By Corollary 3.18 we have that edges are glued together in groups of six.

5.2.2 Gluing regular tetrahedra

The gluing of polyhedra we considered in the previous section do not necessarilygive rise in a complete hyperbolic manifold, but decomposable polyhedra do.

Definition 5.10 (Regular link). A diagram (G, γ) is called regular , if it is analternating, balanced diagram, G is 3-connected and it has a corresponding poly-hedron ∆ that is 1

3π-regular and decomposable (into ideal, regular tetrahedra).A link is regular if it has a regular diagram.

Example. The figure eight-knot has an alternating, balanced diagram (G, γ)with G 3-connected. See Figure 2.4. The ideal regular tetrahedron is a corre-sponding polyhedron, and it is 1

3π-regular (see Corollary 4.31) and decompos-able.

We can make regular links in the following way.

Proposition 5.11. Let ∆ be a 13π-regular and decomposable polyhedron. For

every matching M on the skeleton G of ∆ there is a different pair of regularlinks L± that has ∆ as corresponding polyhedron.

Proof. Let ∆ be a 13π-regular and decomposable polyhedron, let G be its skele-

ton. Since there are no hyperbolic digons, G is simple. By Theorem 4.23 G isconnected and can be embedded. By Proposition 4.30 G is 3-regular.

Let M be a matching on G. By Proposition 2.22 there are two alternatingbalanced diagrams (G, γ) and (G,−γ) where G is the same as the graph Gwhere are the edges in M are doubled. Then ∆ is a corresponding polyhedronof (G,±γ). Since ∆ is 1

3π-regular and decomposable, the diagram is regular.The corresponding link L± of (G,±γ) is therefore a regular link with ∆ ascorresponding polyhedron.

Let M′ be a different matching than M. The diagrams (H,±δ) obtainedfrom M′ are different from the diagrams (G,±γ) since other edges of G aredoubled. Therefore, the corresponding links are different. (Note that the corre-sponding links could be equivalent.)

Example. The n-tetrahedra together with matchings on their skeletons giverise to a family of regular links.

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5.2.3 Complete hyperbolic structure

Theorem 5.12. Let L be a regular link. The complement S3 −L is a completehyperbolic 3-manifold and up to isometries, the metric on S3 − L is unique.

Proof. Let L be a regular link, let ∆ be the corresponding 13π-regular and

decomposable polyhedron. By Proposition 5.9

S3 − L ∼= (∆ + ∆) / ∼Φ .

Let ∆ decompose into the ideal regular tetrahedra ∆1, . . . ,∆n. Let ∼ gluethe ∆i together to form ∆, and then glue them further according to ∼Φ. So

(∆ + ∆) / ∼Φ= (∆1 + . . .+ ∆n + ∆1 + . . .+ ∆n) / ∼ .

We abbreviate this with M . The face gluing ∼Φ of ∆ glues edges in groupsof six. Since ∆ is 1

3π-regular, there is only one ideal regular tetrahedron at anedge, so ∼ glues precisely six tetrahedra around every edge. Since ∂M = L is acollection of circles, we can apply Proposition 5.6 and we conclude that S3 −Lis a complete hyperbolic 3-manifold, and that the metric on S3−L is unique upto isometries.

Corollary 5.13. The hyperbolic volume of the complement of a regular link isa topological invariant, i.e. if L and L′ are equivalent regular links, then thevolumes of their complements are equal.

Corollary 5.14. Let L and L′ be links obtained from a n-tetrahedron ∆ and am-tetrahedron ∆′ with m 6= n (see Proposition 5.11). Then L and L′ are notequivalent.

Proof. By Proposition 5.9 the hyperbolic volumes of S3−L and ∆ differ with afactor 2, and the same for the volumes of S3 − L′ and ∆′. By Proposition 4.36the volumes of ∆ and ∆′ are different. Hence, L and L′ are not equivalent.

40

Chapter 6

Conclusion

We considered a subclass of the alternating balanced links, called the regularlinks, and we constructed a hyperbolic structure on the complements of theseregular links. This explicit hyperbolic structure enables us to do geometryon the complements, we can compute for instance the hyperbolic volume of acomplement.

We showed that this hyperbolic geometry is unique. Therefore, all the ge-ometric properties derived using this geometry are properties of the link itself.This gives us new tools to distinguish links, we saw for example that the hyper-bolic volume of a complement is a topological invariant.

To make regular links, we showed that every hyperbolic polyhedron that isdecomposable into ideal regular tetrahedra gives rise to several regular links.We constructed an infinite family of decomposable polyhedra, namely the n-tetrahedra. By using the hyperbolic volume invariant we showed that the linksderived from different n-tetrahedra are not equivalent. The question remainswhether links derived from the same n-tetrahedron are equivalent or not. Sincethe hyperbolic geometries of these links are very similar, hyperbolic geometryis probably not suitable to distinguish them.

A way to find more links with an explicit hyperbolic geometry is by classify-ing decomposable polyhedra. Another way is to generalize the construction toother links than the alternating balanced links we considered. We can then usegeometry to find invariants of a larger class of links.

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Chapter 7

Populaire samenvatting

In deze scriptie gaan we op zoek naar knopen en schakels met een bepaaldemooie eigenschap. Deze knopen en schakels noemen we welgevormd, ze hebbeneen relatief eenvoudige structuur en zijn daardoor makkelijker te begrijpen.

Het vermoeden is dat als je met welgevormde schakels begint, je alle andereschakels daaruit kunt construeren. Als je iets weet over de welgevormde schakels,en dit blijft behouden onder de constructie, dan weet je automatisch iets overalle schakels. Aangezien welgevormde schakels makkelijker te bestuderen zijndan schakels in het algemeen, zou dit erg handig zijn.

De mooie eigenschap waar we het over hebben is het volgende. Als het com-plement van een schakel hyperbolisch is en betegeld kan worden met regelmatigetetraeders, dan is de schakel welgevormd. De wiskundige begrippen zijn schuingedrukt en worden in de volgende paragrafen behandeld. Maar eerst beginnenwe met knopen en schakels zelf.

7.1 Knopen en schakels

Een wiskundige knoop lijkt erg op een normale knoop in een stuk touw. Normaleknopen zijn echter (in theorie) altijd te ontwarren en dat zou flauwe wiskundeopleveren, dus we eisen dat de uiteinden van het touw aan elkaar worden geplakt.In Figuur 7.1a is de gewone achtknoop te zien, door de uiteindes naar boven tebuigen en aan elkaar te plakken krijgen we in Figuur 7.1b de wiskunde acht-knoop.

(a) Gewone achtknoop (b) Wiskundige achtknoop

Figure 7.1

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Een schakel is een knoop die uit meerdere stukken kan bestaan. In Figuur7.2 zijn enkele voorbeelden gegeven.

Figure 7.2: Schakels

7.2 Hyperbolisch complement

Het complement van een schakel is de ruimte om de schakel heen. Oftewel, hetis de hele ruimte behalve de schakel. In het complement zit dus een soort gatwaar eerst de schakel was, maar je kan er ook over nadenken alsof de schakelaangeeft waar oneindig is. Je ziet het gat dan nooit omdat het oneindig ver wegligt.

We kunnen het complement meer structuur geven door te bepalen hoe meetkundein het complement werkt. Meetkundes komen in drie smaken. De bekenste isde Euclidische meetkunde, dit is de meetkunde van het platte vlak (zie Figuur7.3b). Dan bestaat er nog de bolmeetkunde, die gaar zoals de naam doet ver-moeden over meetkunde op een bol (zie Figuur 7.3a). Als laatste hebben we dehyperbolische meetkunde, waar we in geınteresseerd zijn (zie Figuur 7.3c).

(a) Bolmeetkunde (b) Vlakke meetkunde (c) Hyperbolische meetkunde

Figure 7.3: Verschillende soorten meetkunde

Bolmeetkunde is de meest compacte van de drie. Als je een vel A4 op eenbal plat probeert te drukken zie je dat het gaat golven, er zit meer ruimte inhet A4-tje dan op de bal past.

In een hyperbolisch vlak is juist meer ruimte dan op het platte vlak. Als jedie plat probeert te slaan op een tafel gaat het ook golven, zoals in Figuur 7.4te zien is.

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Figure 7.4: (Een deel van) Het hyperbolische vlak

7.3 Veelvlakken en betegelingen

In Figuur 7.5 is een voorbeeld van een betegeling te zien. In het algemeenbestaat een betegeling uit een voorwerp dat volgens een patroon is neergelegd,zodat de voorwerpen niet overlappen en alle ruimte wordt opgevuld.

Figure 7.5: Een betegeling

Behalve een plat vlak kan je ook een hyperbolische ruimte betegelen. InFiguur (7.6) is een regelmatige tetraeder te zien. Bij bepaalde hyperbolischeruimtes kan je deze tetraeders zo neerleggen dat ze niet overlappen en dat ze dehele ruimte overdekken.

Een tetraeder is een voorbeeld van een veelvlak. Als je meerdere tetraedersop elkaar plakt krijg je andere veelvlakken.

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(a) De ribben van een tetraeder (b) Een (opgevulde) tetraeder

Figure 7.6

7.4 Conclusie

In deze scriptie hebben we schakels gevonden wiens complementen betegelbaarzijn met regelmatige tetraeders. Om dit te kunnen doen hebben we het vol-gende verband gevonden tussen de diagrammen van schakels en diagrammenvan veelvlakken.

Als het diagram van een schakel op bepaalde punten na hetzelfde is als hetdiagram van een veelvlak, dan weten we dat het complement van de schakel tebetegelen is met het veelvlak.

Als voorbeeld is in (7.7a) is het diagram van de achtknoop te zien, en inFiguur (7.7b) het diagram van de tetraeder. Als je de dubbele lijnen in hetdiagram van de achtknoop dichtklapt krijg je het diagram van de tetraeder. Ditdichtklappen is precies het ’op bepaalde punten na’, dus we weten nu dat hetcomplement van de achtknoop te betegelen is met tetraeders.

(a) Een diagram van de achtknoop (b) Een diagram van detetraeder

Figure 7.7

Door veelvlakken te bedenken die zelf uit tetraeders bestaan en daar daarde bijbehorende schakels bij te zoeken, vinden we schakels die welgevormd zijn.

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Index

alternating crossing function, 9alternating diagram, 9alternating link, 9angle

of intersecting geodesics, 29antilinear fractional map, 28

balanced diagram, 10balanced graph, 10balanced link, 10balanced pair vertices, 10

clockwise face, 10complete hyperbolic manifold, 36complete hyperbolic structure, 36complex parameter, 33connecting edge, 7converse crossing function, 7corresponding link, 8corresponding polyhedron, 38counterclockwise face, 10crossing function, 7

alternating, 9converse, 7

d.t. polyhedroid, 18decomposable, 34decomposition, 34diagram, 8

alternating, 9balanced, 10corresponding link, 8equivalence, 8regular, 39

dihedral angle, 32

edgeconnecting, 7of a polygon, 30of a polyhedroid, 22of a polyhedron, 31

embedded graph, 7

embedding, 4

faceof a diagram, 8of a graph, 7of a polyhedroid, 22of a polyhedron, 31

face gluingof polyhedroids, 23

geodesic, 28complete, 29

gluing, 14graph

balanced, 10

horosphere, 30hyperbolic distance, 28hyperbolic length, 27hyperbolic manifold, 36hyperbolic plane, 29hyperbolic space, 28hyperbolic structure, 36hyperbolic volume, 28

ideal polygon, 30ideal polyhedron, 31ideal vertex, 30isometry, 28

knot, 4equivalence, 5tame, 5wild, 5

linear fractional map, 28link, 6

alternating, 9balanced, 10equivalence, 6of a graph, 9regular, 39

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matching, 11

n-tetrahedron, 34

polygon, 30ideal, 30

polyhedroid, 22digon tolerant, 18

polyhedron, 31ideal, 31regular, 33

quotient space, 12

regular diagram, 39regular link, 39regular polyhedron, 33

skeleton, 31

tetrahedron, 31truncated, 36

topological sum, 14truncated face, 36truncated tetrahedron, 36

unknot, 5

vertexideal, 30of a polygon, 30of a polyhedron, 31

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Bibliography

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The unique hyperbolic geometry of regular link complements

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