Cusp Shapes of Hyperbolic Link Complements and Dehn Filling

CUSP SHAPES OF HYPERBOLIC LINK COMPLEMENTS AND DEHN FILLING

a dissertation submitted to the department of mathematics and the committee on graduate studies of stanford university in partial fulfillment of the requirements for the degree of doctor of philosophy

Jessica S. Purcell May 2004 c Copyright by Jessica S. Purcell 2004

ii I certify that I have read this dissertation and that, in my opinion, it is fully adequate in scope and quality as a dissertation for the degree of Doctor of Philosophy.

Steven P. Kerckhoﬀ (Principal Adviser)

I certify that I have read this dissertation and that, in my opinion, it is fully adequate in scope and quality as a dissertation for the degree of Doctor of Philosophy.

Ralph Cohen

I certify that I have read this dissertation and that, in my opinion, it is fully adequate in scope and quality as a dissertation for the degree of Doctor of Philosophy.

Gregory Brumﬁel

Approved for the University Committee on Graduate Studies.

iii Abstract

For every knot or link with hyperbolic complement, each cusp of the complement has a geometric shape given by the Euclidean similarity class of structures on horoball neighborhoods of the cusp. We give estimates on the geometric shapes of cusps of “complicated” links using only combinatorial information on a diagram of the link, namely, the number of twists and twist regions in a prime, twist-reduced diagram of the link. We use this to show that there are numbers N and M so that any hyperbolic knot having at least N twists in each of M twist regions has the property that every non-trivial Dehn ﬁlling is hyperbolic.

iv Acknowledgements

I would like to thank my advisor, Steve Kerckhoff, for at least three years’ worth of helpful conversations. It’s lucky for me that he knows everything [at least for all practical purposes]. I would also like to thank my husband, Tim Purcell. This dissertation wouldn’t have happened without him. Or at the very least, it wouldn’t have happened at Stanford. Thanks to my officemates, especially Dana Rowland, Josh Sabloff, Yu Yan, and Pierre Albin, for many enlightening conversations. Thanks to the all the people who read relevent papers with me, including Dave Futer and Jian He, but especially thanks to Henry Segerman, who was willing to read about harmonic deformations of cone manifolds. Oh and thanks again to my advisor for writing about harmonic deformations of cone manifolds. Thanks to my reading committee, for very carefully reading every singel word of this dissertation. Thanks to the assorted non-mathematician people who tried to help keep life real, especially Jennie and Jonathan Clark, Todd and Johanna Neller, Dan and Deborah Butler. And thanks to my high school teachers, Mrs. Swensen, Mr. Jensen, Mrs. Keir, and Mr. Pehrson for making math fun, even if Alison always was the better student.

v Contents

Abstract iv

Acknowledgements v

1 Introduction 1 1.1 Hyperbolic Knot and Link Complements ...... 1 1.2 Cusp Shapes of Hyperbolic 3-Manifolds ...... 3 1.3 Twist Regions, Twisting, and Main Results ...... 4 1.4 DehnFilling ...... 7 1.5 OrganizationofThesis ...... 11

2 Initial Cusp Shape Estimates 12 2.1 ObtainingFlatAugmentedLink ...... 12 2.2 Flat Augmented Links are Hyperbolic ...... 15 2.3 Estimates on Cusp Shapes of Flat Augmented Links ...... 20 2.4 Initial Cusp Shape Estimates with Half Twists ...... 29

3 Deforming Through Cone Manifolds 32 3.1 HyperbolicConeManifolds ...... 33 3.2 ObtaininganE-valued1-form ...... 34 3.2.1 Inﬁnitesimal Isometries ...... 35 3.2.2 Deﬁning the 1-form ω ...... 37 3.2.3 Properties of ω ...... 39 3.3 Deforming the Augmented Link Complement ...... 41

vi 4 Analyzing Boundary Terms 45 4.1 Analysis of the Boundary Term from the Cusp ...... 45 4.1.1 The Area of ∂U ...... 48 4.1.2 The Scale Term (a2 + b2) ...... 51 b 4.2 Analysis of the Boundary Term from the Singular Locus ...... 55 4.3 Solving the diﬀerential inequality ...... 60

5 Finding Numerical Values for Constants 63 5.1 Computing number of twists per region ...... 65 5.2 AdditionalConsequences ...... 67

6 Some Analysis of Results 69 6.1 Analysis of Constants for Dehn Filling ...... 69 6.2 AnExample:2-BridgeKnots ...... 70 6.3 Conclusions ...... 76

Bibliography 77

vii List of Tables

viii List of Figures

1.1 Left: A satellite knot. Right: A torus knot...... 2 1.2 A horosphere and horoball region, its projection to S3 K, and the − lift of a meridian and longitude to the horosphere in H3...... 4 1.3 The above diagram has 3 twist regions, N(K) = 3, with twisting 1 1 amounts in each twist region equal to 2, 2 , and 1 2 , respectively. . . . 5 1.4 (a.) Left: a prime diagram. (b.) Right: A twist-reduced diagram. . . 5 1.5 When N(K)= N and the twist amount in each twist region is at least N , then L (0.8154)2N ...... 6 2 ≥ 1.6 Dehn ﬁlling: Glue a solid torus to the boundary of M in such a way that the slope s boundsadisk...... 7

2.1 Primeandtwist-reduceddiagrams...... 13 2.2 Creatingtheﬂataugmentedlink...... 14 2.3 Decomposing S3 L into ideal polyhedra...... 14 − 2.4 Create a trivalent graph by replacing each twist region with an edge. 16 2.5 Form of dual graph to a triangulation with an edge with a single vertex atbothendpoints...... 16 2.6 Form of original link diagram if triangulation has an edge with a single vertexatbothendpoints...... 17 2.7 Form of dual graph to a triangulation with two vertices connected by twodistinctedges...... 17 2.8 Form of the dual graph to a triangulation with two vertices connected bytwodistinctedges...... 18

ix 2.9 One possible form of the original link diagram if the triangulation γ contains two vertices connected by two distinct edges...... 19 2.10 Another possible form of the original link diagram if the triangulation γ contains two vertices connected by two distinct edges...... 19 2.11 The cusp diagram corresponding to a crossing circle ...... 21 2.12 The cusp diagram for the link on the projection plane ...... 21 2.13 The possible configurations of the four circles closest to a point of tangency...... 22 2.14 Result of taking a point of tangency to infinity ...... 23 2.15 A fundamental region for a three punctured sphere. Projections of horospheres are labeled by the short dashed lines. Note there is reflective symmetry through the dashed line from 0 to infinity...... 26 2.16 On the left is the standard gluing of the twice-punctured disk. On the right is the new gluing creating a half-twist...... 29 2.17 Adding a half twist to a flat augmented link shifts the gluing along the cusp...... 30

3.1 A cone of angle α...... 33

4.1 Top edge of quadrilateral maps to shortest path on ∂U under the cov- eringprojection...... 50 b 5.1 Left: Rectangular shape of a crossing circle cusp without a half twist. Right: Shape of a cusp in which a half twist has been added. . . . . 66 5.2 Top: Dashed line is the 1/n curve on a crossing circle without a half twist. Bottom: Dashed line is the 1/n curve on a crossing circle with ahalftwist...... 66

6.1 The ﬂat augmented link of a 2-bridge knot ...... 71 6.2 Trivalent graph associated with a 2-bridge knot ...... 71 6.3 Circle packing associated to a 2-bridge knot ...... 72 6.4 Left: The vertex shapes corresponding to crossing circle cusps at circles B and B . Right: The cusp shape at these crossing circles is then 2 1. 72 1 2 ×

x 6.5 The shape of the vertex corresponding to crossing circle cusps at circles

Ci, possibly with roles of A and D switched...... 73 6.6 Four types of vertices of the ﬂat augmented link of a 2-bridge knot, which glue to give the initial cusp shape in the projection plane. . . . 74

xi Chapter 1

Introduction

In this thesis, we determine geometric constraints on the cusp shapes of a hyperbolic knot or link complement, based only on combinatorial information on the manifold. These results allow us to distinguish classes of hyperbolic knots and links based on their twist number, and to determine information about the geometry of such objects with only a combinatorial description. The results also have applications to hyperbolic Dehn filling on complicated knots and links. This chapter is organized as follows. In the first four sections, we include definitions of terms which will be used throughout the thesis, motivations for our results, statements of our main results and background and information about additional applications to Dehn filling. The final section contains information on the organization of the thesis.

1.1 Hyperbolic Knot and Link Complements

A hyperbolic 3-manifold is a 3-manifold in which each point has a neighborhood modeled on a ball in hyperbolic 3-space, H3. Overlap maps are restrictions of isometries of H3. Examples of hyperbolic 3-manifolds include “most” knot complements: Thurston has shown that any knot K which is not a satellite or torus knot has a complement S3 K admitting a complete hyperbolic structure (see [20]). −

1 CHAPTER 1. INTRODUCTION 2

Here a satellite knot is a knot obtained by embedding a circle in a small solid torus neighborhood of another knot. A torus knot is a knot which winds around the outside of an ordinary torus in S3. Examples of these knots are shown in Figure 1.1.

Figure 1.1: Left: A satellite knot. Right: A torus knot.

By the Mostow-Prasad rigidity theorem, the hyperbolic structure on a knot complement is unique. Speciﬁcally, the Mostow-Prasad rigidity theorem asserts that if

M1 and M2 are complete hyperbolic manifolds of ﬁnite volume and dimension at least 3, and there is an isomorphism φ : π (M ) π (M ), then there is an isometry 1 1 → 1 2 F : M M which induces the isomorphism φ up to conjugacy. (The original theo- 1 → 2 rem doesn’t include the case of noncompact manifolds and was given by Mostow [16]. Prasad completed the proof for noncompact manifolds [18].) By the work of Gordon and Luecke [8], two knots with homeomorphic complement are equivalent. Hence associated to the diagram of a hyperbolic knot, which is a very combinatorial object, is a unique geometric structure. Thus in theory, to study a hyperbolic knot, geometric information such as volumes or geodesic lengths might be very useful. However, in practice, it seems to be diﬃcult to gain information on the geometric hyperbolic structure of classes of knot complements given only the combinatorial information of their knot diagrams. Results along these lines, which use only combinatorial diagrams of knots to determine geometric information, seem to be scarce. Lackenby [14] has given estimates on volumes for alternating knots and links. In this CHAPTER 1. INTRODUCTION 3

thesis, we give estimates on the cusp shape of a hyperbolic knot (or link) for sufficiently complicated links, where we will give a definition of “sufficiently complicated” in Section 1.3 that depends only on the combinatorics of a knot diagram.

1.2 Cusp Shapes of Hyperbolic 3-Manifolds

Let K be a knot such that S3 K admits a complete hyperbolic structure. Note that − the knot K does not exist in the manifold S3 K. Hence in a complete structure, − approaching the knot is equivalent to approaching a point at infinity. Rather than study the knot itself, we take a tubular neighborhood of the knot in S3 K and − examine its geometry. The universal cover of S3 K is hyperbolic 3-space H3. We take the upper half − space model of H3, and conjugate so that the knot K lifts to the point . Then ∞ the planes z = c for c large project to non-intersecting tori about the knot K. These planes z = c are called horospheres. Under the hyperbolic metric, these planes inherit a Euclidean structure. The regions z>c above the horospheres in H3 are called horoballs. Their projection to S3 K give solid torus neighborhoods about the knot, called horoball − neighborhoods. The boundaries of these horoball neighborhoods are flat tori about the knot K. If we let m be the meridian of the knot K, and we let ` be any longitude, then for fixed c the curves m and ` are homotopic to closed geodesics in the Euclidean metric on the projection of the horosphere z = c to the knot complement S3 K. These lift − to give a tiling of the horosphere z = c by parallelograms in H3. Changing the value of the constant c doesn’t affect the Euclidean similarity structure. Hence for each knot we have a point in (T 2), the Teichmüller space of the torus. T In general, any cusps of a finite volume hyperbolic 3-manifold are foliated by tori with a unique Euclidean similarity structure. We investigate the shapes of these tori when the 3-manifold is a link complement. CHAPTER 1. INTRODUCTION 4

z > c

m z = c l l

Figure 1.2: A horosphere and horoball region, its projection to S3 K, and the lift of a meridian and longitude to the horosphere in H3. −

1.3 Twist Regions, Twisting, and Main Results

Following Lackenby [13], define the twist number of a knot or link as follows: Given a diagram of a knot or link, the projection of the diagram to a plane in S3 gives a four-valent graph with over-under crossing information at each vertex. A bigon is a complementary region of this graph having two edges in its boundary. A twist region of the knot or link is a string of bigons arranged end to end, attached at vertices, which is maximal in the sense that there are no bigons attached to either end of the bigon string. A single crossing adjacent to no bigons is also a twist region. The twist number N(D) of a diagram D is the number of twist regions in its graph. We are also concerned with the amount of twisting that occurs in each twist region. A full twist is where two strands wrap once entirely around each other. Note it consists of two crossings in a link diagram. A half twist is a single crossing. For a given twist region, we define the twisting amount to be the number of full twists in that twist region. See Figure 1.3 for an example. Before stating our main results, we need two more definitions concerning properties of a diagram of a knot or link. Recall we can view the link diagram as a 4-valent graph with over-under crossing information. The diagram lies on a plane S2 in S3. A link diagram is said to be prime if, for any simple closed curve in this plane S2 which intersects the 4-valent graph transversely in two points in the interior of edges, the simple closed curve bounds a subdiagram containing no crossings of the original diagram. See Figure 1.4 (a). CHAPTER 1. INTRODUCTION 5

Figure 1.3: The above diagram has 3 twist regions, N(K) = 3, with twisting amounts 1 1 in each twist region equal to 2, 2 , and 1 2 , respectively.

We also require the diagram to be twist reduced, in the sense of Lackenby [14]. That is, whenever a simple closed curve in S2 intersects the link projection transversely in four points, with two points adjacent to one crossing and the other two points adjacent to another crossing, then the simple closed curve bounds a subdiagram consisting of a (possibly empty) collection of bigons strung end to end between these crossings. See Figure 1.4 (b).

A or B is A or B is

A B = A B = . ⇒ ⇒ .

Figure 1.4: (a.) Left: a prime diagram. (b.) Right: A twist-reduced diagram.

With these deﬁnitions, we have the following theorem:

Theorem 1. There is a number N2 such that if K is any hyperbolic knot with a prime, twist-reduced diagram with twist number N, and in each twist region, a twisting amount at least N2, and if the meridian of K is scaled to have length 1, then any longitude of K will have length L (0.8154)2N. ≥ CHAPTER 1. INTRODUCTION 6

Schematically, the theorem is illustrated by Figure 1.5. If the twist amount per twist region is higher than the universal constant N2, then we obtain a lower bound on the length of a longitude in terms of the twist number of the knot. This forces the cusp shape, or the shape of the parallelogram associated with the cusp, to be “long and skinny”.

Twisting amount of at least N2 full twists per twist region.

N twist regions

Figure 1.5: When N(K) = N and the twist amount in each twist region is at least N , then L (0.8154)2N 2 ≥

This theorem can be extended to apply to links as follows.

Theorem 2. There is a number N2 such that if L is any hyperbolic knot with a prime, twist-reduced diagram, and K is a component of L which meets N twist regions of the diagram of L, and in each of those twist regions there is a twisting amount at least

N2, then if the meridian of the cusp corresponding to K is scaled to have length 1, any longitude of that cusp will have length L (0.8154)2N. ≥ CHAPTER 1. INTRODUCTION 7

1.4 Dehn Filling

One major application of our results on cusp shapes is to Dehn filling. In this section, we define Dehn filling and outline some previous results relating Dehn filling to 3- manifolds. Let M be a 3-manifold with torus boundary ∂M. Let s be a slope on ∂M, that is, s is an isotopy class of non-trivial simple closed curves on the torus ∂M. We obtain a new manifold by gluing a solid torus S1 D2 to ∂M along the common torus × boundary, in such a way that the slope s bounds a disk in the resulting manifold.

Figure 1.6: Dehn ﬁlling: Glue a solid torus to the boundary of M in such a way that the slope s bounds a disk.

This procedure is called Dehn filling along the slope s, and the resulting manifold is denoted M(s). More generally, if M is a 3-manifold with multiple torus boundary components and along each component we have a slope si, we obtain a manifold by Dehn filling along these slopes. The new manifold is denoted M(s1,...,sv). Note that if M is taken to be S3 K, where K is a knot, then Dehn filling along − a meridian of K will always give S3. This Dehn filling is called the trivial filling. All other Dehn fillings are non-trivial. Conventions: Using a meridian-longitude basis µ,λ for the fundamental group { } of the torus, slopes on cusps are parameterized by Q . Thus a slope corresponds ∪{∞} to a/b if and only if the slope is equivalent to aµ + bλ. Note = 1/0. ∞ In the early 1960’s, Wallace [22] and then Lickorish [15] proved that every closed orientable 3-manifold is obtained by Dehn filling on a link in S3. Moreover, we can CHAPTER 1. INTRODUCTION 8

take the link complement to admit a complete hyperbolic structure. Thus by these results, an understanding of Dehn filling on hyperbolic links would give a better understanding of 3-manifolds. Specifically concerning the geometry of these manifolds, Thurston [19] showed that “most” Dehn fillings are hyperbolic. That is, for each link component, only a finite number of Dehn fillings need to be excluded, and the remaining fillings will yield a 3-manifold admitting a complete hyperbolic structure. Since this result of Thurston, there have been several results giving more explicit information on which Dehn fillings will and will not yield hyperbolic manifolds. Much progress has been made in determining properties of slopes which cannot produce a manifold with a hyperbolic structure. Techniques along these lines are very topological or combinatorial, and have included identifying incompressible tori or spheres that are created under Dehn filling, or Dehn fillings that yield finite or cyclic fundamental groups of the resulting manifolds. Some of the results in this area are due to Gordon, Luecke, Wu, Culler, Shalen, Boyer, Zhang, and many others (see, for example, survey articles [6], [7]). There has also been progress in determining which slopes are guaranteed to produce a manifold that admits a hyperbolic structure, or at least a structure that will be hyperbolic given the geometrization conjecture. These results involve finding bounds on the length of a slope. The setup for these theorems is the same. Let M be a hyperbolic manifold with

ﬁnite volume, and let P1,...,Pv be disjoint horoball neighborhoods of the cusps of

M. Let si be a slope on ∂Pi represented by a geodesic αi in the Euclidean metric.

The 2π-Theorem (Gromov, Thurston [9]). If the length of each αi is at least

2π, then M(s1,...,sv) admits a metric of negative sectional curvatures.

The 2π theorem was extended recently in two diﬀerent ways. Independently, Agol [2] and Lackenby [13] were able to reduce the length from 2π to 6, if algebraic properties of the fundamental group were required. CHAPTER 1. INTRODUCTION 9

The 6-Theorem (Agol, Lackenby). If the length of each αi is at least 2π, then

M(s1,...,sv) is irreducible, atoroidal, and not Seifert ﬁbered, with inﬁnite word hyperbolic fundamental group.

Following Agol, we call such manifolds “hyperbolike”. Such manifolds are known to be hyperbolic given the Geometrization Conjecture.

Note that in the theorem above, the slope si can be given a longer length if the horoball neighborhoods about the cusps are chosen to be as large as possible. Hence manifolds admitting large horoball neighborhoods are likely to have more hyperbolike Dehn ﬁllings. A second extension of the 2π theorem was by Hodgson and Kerckhoﬀ [10]. By putting further restrictions on the lengths of the αi, they were able to prove the resulting manifold admitted a genuine hyperbolic metric.

The 7.515-Theorem (Hodgson, Kerckhoﬀ). If α1 has normalized length at least

7.515, and α2 through αv have normalized length at least 10.6723, then M(s1,...,sv) is hyperbolic.

Here the normalized length of the curve αi is deﬁned to be the length of αi divided by the square root of the area of the torus ∂Pi:

length(αi) Normalized length(αi)= Area(∂Pi) p Note the normalized length is independent of the choice of horoball neighborhood of the cusp, so unlike the 6-theorem, we can’t improve results by taking a horoball neighborhood as large as possible. Now, combining the 7.515-Theorem or the 6-Theorem with Theorem 1 gives the following results.

Theorem 3. If K is a hyperbolic knot with a prime, twist reduced diagram which has at least 84 twist regions and a twist amount of at least 73 full twists per region, then any non-trivial Dehn ﬁlling on S3 K will result in a manifold admitting a complete − CHAPTER 1. INTRODUCTION 10

hyperbolic structure.

Or, if we are only concerned with hyperbolike manifolds:

Theorem 4. If K is a hyperbolic knot with a prime, twist reduced diagram with at least 10 twist regions and a twist amount of at least 73 full twists per region, then any non-trivial Dehn filling on S3 K will result in a manifold which is irreducible, − atoroidal, not Seifert fibered, with infinite word hyperbolic fundamental group.

These results should be compared with previously known results. Lackenby [13] defined a combinatorial length for slopes on alternating knots, and showed that when a slope had a long combinatorial length, Dehn filling along that slope yields a hyperbolike manifold. He was able to show that for alternating knots with twist number N(K) 9, all slopes had long combinatorial length, and hence all non-trivial fillings ≥ are hyperbolike. The above theorems extend his work, in that the knots we consider are no longer restricted to be alternating. Additionally, the methods in this thesis are more geometrically constructive. Rather than dealing with combinatorial lengths, we estimate the actual geometric length of slopes in the knot complements. Others have made much progress in the classification of Dehn fillings on particular classes of knots. For example, Wu has made significant progress in the classification of non-hyperbolic, or so called exceptional Dehn fillings on arborescent knots and links [24], [23]. Brittenham and Wu have classified exceptional Dehn fillings on 2- bridge knots [4]. These results are obtained by topological arguments, involving, for example, determining which Dehn fillings create an incompressible torus. By contrast, the results of this thesis apply very geometric arguments, involving the deformation of metrics. Additionally, we don’t require the restriction to any particular class of knots or links. However, it is an interesting problem to determine whether there are large classes of knots whose Dehn fillings can be classified neatly by these results. CHAPTER 1. INTRODUCTION 11

1.5 Organization of Thesis

The proof of Theorems 1 or 2 goes as follows. Starting with a hyperbolic knot or link, replace that link with a link which is geometrically simpler, called a flat augmented link. We show these links are hyperbolic, and we obtain estimates on the cusp shapes of these links based on the initial twist number of the original link. Now, these flat augmented links have a structure we understand well. To return to the original link, we must deform the hyperbolic structure on the flat augmented links through cone manifolds, bounding the geometry under the deformation. In order for the bounds and the deformation to go through, we need to ensure the twist amount in each twist region is high, hence we pick up requirements on the twist amount in each region. By tracing through the deformation, we obtain a differential inequality. Its analysis leads to Theorem 2. This thesis is organized as follows. In Chapter 2, we replace our initial knot or link by the flat augmented link and obtain initial estimates on the cusp shapes. In Chapter 3, we deform through cone manifold structures, obtaining an inequality relating certain boundary terms. These boundary terms are then analyzed in Chapter 4, and the implications of the inequality relating them are studied. Using the analysis of Chapter 4, we find the constants for Theorems 2, 3, and 4 in Chapter 5. Finally, in Chapter 6 we analyze our results in comparison to related results, and make suggestions for future directions of this research. Chapter 2

Initial Cusp Shape Estimates from Flat Augmented Links

In this chapter, we show how to replace a link by a new link, which we call a flat augmented link. We show that these flat augmented links are always hyperbolic, provided we started with a prime, twist reduced diagram of the original link, and we give estimates on the cusp shapes of these flat augmented links. Many ideas from this section are from the appendix by I. Agol and D. Thurston to Lackenby’s paper on volumes ([14]). In this appendix, Agol and D. Thurston create a flat augmented link from a given link and then find a polyhedral decomposition of the complement of the flat augmented link with totally geodesic faces. They use this polyhedral decomposition to give estimates on the volume of the initial manifold. We will use the polyhedral decomposition to make precise statements about the cusp shapes of the resulting manifold.

2.1 Obtaining Flat Augmented Link

We describe here how the ﬂat augmented link and its polyhedral decomposition are created. Begin with a diagram of a hyperbolic knot or link. We consider the diagram as a 4-valent graph in S2 with “under-over” crossing information assigned to each

12 CHAPTER 2. INITIAL CUSP SHAPE ESTIMATES 13

vertex. We require the diagram to be prime. That is, any simple closed curve in S2 which intersects the graph transversely in two points in the interior of edges bounds a subdiagram containing no crossings of the original diagram. Note this ensures the diagram contains no monogons. We also require the diagram to be twist reduced in the sense of Lackenby ([14]). That is, whenever a simple closed curve in S2 intersects the link projection transversely in four points, with two points adjacent to one crossing and the other two points adjacent to another crossing, then this curve bounds a subdiagram consisting of a (possibly empty) collection of bigons arranged in a row between these two crossings. See Figure 2.1.

A or B is A or B is

A B = A B = . ⇒ ⇒ .

Figure 2.1: Left: A prime diagram; Right: A twist-reduced diagram.

Each twist region of the diagram consists of two strands twisting around each other. Around these two strands, add a simple closed curve to the diagram, known as a crossing circle. Note that the complement of this new link with crossing circles added is homeomorphic to the complement of the link obtained by either removing all the crossings in the twist region, if the number of crossings is even; or removing all of its crossings except one if the number of crossings is odd. We further alter the link projection by removing all remaining single crossings in twist regions, and call the resulting link L. See Figure 2.2. The link L now has components lying flat in the projection plane bound together by crossing circles. It is a flat augmented link. Note that the flat augmented link L is the limit of the knots obtained by per- forming 1/n Dehn filling on each crossing circle of the link as n goes to infinity. See Thurston [19] for details. CHAPTER 2. INITIAL CUSP SHAPE ESTIMATES 14

Figure 2.2: Left to right: The original knot with twist regions marked; the knot with crossing circles added; the homeomorphic link; the ﬂat augmented link L

Following Agol and D. Thurston [14], obtain the decomposition of the ﬂat augmented link in to two polyhedra as follows. Consider the planar surface lying in the projection plane and the two-punctured disks bounded by each crossing circle. We take three edges per two-punctured disk, each edge lying on the intersection of these two-punctured disks with the planar surface. Now slice along the planar surface. We have cut the manifold S3 L into two identical pieces, one on either side of the plane. − Each two-punctured disk has been sliced in half along the three edges. To complete the decomposition, slice each remaining half of the two-punctured disks, opening them up into two faces. See Figure 2.3.

A A

C C

Figure 2.3: Decomposing S3 L into ideal polyhedra: First slice along the projection − plane, then split remaining halves of two-punctured disks. Obtain polygon on right.

This gives an ideal polyhedral decomposition of the manifold. The edges of the polyhedra are the edges given by intersection of two-punctured disks with the planar surface. On each of the two polyhedra, we have one face per planar region of the CHAPTER 2. INITIAL CUSP SHAPE ESTIMATES 15

diagram of L, and two triangular faces per two-punctured disk. We shade the faces arising from two-punctured disks, giving each polyhedron a checkerboard coloring. The edges meet in four-valent vertices. At each vertex two white faces are separated by two shaded faces. The gluing pattern on the polyhedra is given by following the above process in reverse: First, on each polyhedron fold the pairs of shaded faces together at their common vertex. Then glue each white face to its corresponding face on the opposite polyhedron via the identity map.

2.2 Flat Augmented Links are Hyperbolic

We show that the ﬂat augmented link created in the previous section is hyperbolic, and that the two polyhedra described above that give a polyhedral decomposition of the link have totally geodesic faces in H3. To do so, we will use Andreev’s theorem [19]:

Theorem 5 (Andreev). Let γ be a triangulation of S2 such that each edge has distinct ends and no two vertices are joined by more than one edge. Then there is a packing of circles in S2 whose nerve is isotopic to γ. This circle packing is unique up to Moebius transformation.

The reason we use Andreev’s theorem is that geodesic planes in H3 extend to give circles on the boundary S2 at inﬁnity. Hence if our polyhedra were totally geodesic in H3, the planes containing its faces would extend to give circles in S2. Conversely, if we can show a circle packing exists on S2 in such a way that corresponding hemispheres in H3 can be cut away to give one of our polyhedra, we will be done.

Theorem 6. Given a prime, twist-reduced link diagram, the ﬂat augmented link obtained from this diagram, as described in Section 2.1, is hyperbolic. Moreover, the polyhedral decomposition of this ﬂat augmented link is totally geodesic in H3.

Proof. We begin by ﬁnding a triangulation of S2 associated with one of the polyhedra described above. For each white face, take a vertex. Connect two vertices by an edge CHAPTER 2. INITIAL CUSP SHAPE ESTIMATES 16

if and only if the two corresponding white faces meet at a vertex of the polyhedron. This gives a triangulation γ of S2. Note if we draw vertices of γ on top of white faces of the polyhedron, and edges of γ through the vertices of the polyhedron, that each triangle of γ circumscribes a shaded face of the polyhedron. To apply Andreev’s theorem, we need to show this triangulation γ satisﬁes the conditions: each edge has distinct ends, and no two vertices are joined by more than one edge. To show these conditions, we consider the graph dual to γ, as follows. Consider the original link diagram. We view it as a 4-valent graph with over-under crossing information at each vertex. For each twist region, replace all twist bigons by a single edge. If the twist region contains no bigons, insert a single edge. See Figure 2.4 below.

Figure 2.4: Create a trivalent graph by replacing each twist region with an edge.

Since each crossing is part of some twist region, this gives a trivalent graph Γ. Note the dual graph to Γ is isotopic to the triangulation γ. We use properties of the dual Γ to show the triangulation γ satisﬁes the conditions necessary for Andreev’s theorem.

Lemma 7. No edge of γ has a single vertex at both endpoints.

Proof. Consider the graph Γ. If its dual γ as an edge with one vertex at both endpoints, then the graph Γ must be of the form in Figure 2.5.

Figure 2.5: Form of dual graph to a triangulation with an edge with a single vertex at both endpoints. CHAPTER 2. INITIAL CUSP SHAPE ESTIMATES 17

Note then if the edge e was an original edge of the 4-valent link diagram, then we have a simple closed curve intersecting the link diagram transversely in a single point. This is impossible. If instead the edge e replaced a twist region in the original 4-valent link diagram, then the original diagram had the form of Figure 2.6.

DD0 ···

Figure 2.6: Form of original link diagram if triangulation has an edge with a single vertex at both endpoints.

However, note one of the subdiagrams D and D0 must contain crossings else the link is not hyperbolic. Without loss of generality, say D contains crossings. Then the dashed simple closed curve enclosing D intersects the link diagram transversely in two points, but both its interior and its exterior contain crossings. This contradicts the fact that the original diagram was prime.

Lemma 8. No two vertices of γ are joined by more than one edge.

Proof. Suppose two vertices of γ are joined by two distinct edges. Then consider the dual graph Γ. Γ must be of the form shown in Figure 2.7, where region B and region B0 are the same. B ··· ··· a e1 d

. A . . .

be2 c B0 ··· ···

Figure 2.7: Form of dual graph to a triangulation with two vertices connected by two distinct edges. CHAPTER 2. INITIAL CUSP SHAPE ESTIMATES 18

Give the edge e1 an orientation, say e1 points in the direction of the edge a. Then this gives the region A an orientation, and hence assigns to the edge e2 an orientation as well. Now, trace a path of edges around the region B, listing the edges in the order they occur. Our list begins with e1, then a, etc. Note e2 must appear in the list before the edge d, else region B would be cut oﬀ from region B0. Note also that edge b must occur before edge c, by the orientation on e2. Hence the graph Γ has the form illustrated in Figure 2.8.

B e1

Figure 2.8: Form of the dual graph to a triangulation with two vertices connected by two distinct edges.

Case 1: Edges e1 and e2 both were edges of the original 4-valent link diagram.

Then there exists a simple closed curve intersecting e1 and e2 transversely in the original link diagram which doesn’t intersect the diagram in any other points. Since the original diagram must have crossings on either side of the region A, this contradicts the fact that we started with a prime diagram.

Case 2: Edges e1 and e2 both replaced twist regions in the original link diagram. Then the original link diagram must have been of the form in Figure 2.9. Then since the original diagram was twist reduced, the either the subdiagram D or D0 must be a string of bigons. But then rather than two distinct twist regions, our diagram must have had only one twist region. Thus edges e1 and e2 would not have been distinct.

Case 3: One of edge e1 or e2 is from the original link diagram, the other replaced a twist region. Then the original link diagram must have been of the form shown in Figure 2.10. However, now note that the simple closed curve bounding subdiagram D intersects CHAPTER 2. INITIAL CUSP SHAPE ESTIMATES 19

··· D D0

···

Figure 2.9: One possible form of the original link diagram if the triangulation γ contains two vertices connected by two distinct edges.

DD0

···

Figure 2.10: Another possible form of the original link diagram if the triangulation γ contains two vertices connected by two distinct edges. the link transversely in only three points. This is impossible. Since these are the only three cases, these contradictions prove the lemma.

Now Lemmas 7 and 8 are enough to show that Andreev’s theorem applies, and there is a circle packing of S2 whose nerve is γ. Now, complete the proof by slicing off geodesic hemispheres in H3 corresponding to each circle in the circle packing. These give the geodesic white faces of the polyhedron. The shaded faces are obtained by slicing off hemispheres in H3 corresponding to each circle of the dual circle packing. Thus the polyhedron for the flat augmented link is totally geodesic in H3, and hence the flat augmented link is hyperbolic. CHAPTER 2. INITIAL CUSP SHAPE ESTIMATES 20

2.3 Estimates on Cusp Shapes of Flat Augmented Links

Using the setup described above, we can determine estimates on the shapes of the cusps of the ﬂat augmented links.

Proposition 9. The shape of any cusp of a ﬂat augmented link is rectangular.

Proof. To determine the shapes of cusps, we intersect horospheres about ideal vertices with the polyhedra. Note that a horosphere around any ideal vertex will meet two white faces and two shaded faces. Since we know that the two-punctured spheres are perpendicular to the planar projection surface on which lie the white faces, it follows that the horospheres intersect the polyhedra in rectangles. These rectangles are glued according to the gluing pattern on the polyhedra to form cusps. We need only show these rectangles are glued without shearing. Consider a single such rectangle. Two opposite sides of the rectangle come from the intersection of the horosphere with white faces on the polyhedron. Recall that each white face is glued by the identity map to its corresponding white face on the opposite polyhedron. Hence this rectangle is glued along each “white” side to an identical rectangle obtained from a horosphere intersecting the same vertex on the opposite polyhedron. Thus a cycle has been formed after only these two rectangles have been glued. We let the closed curve running along a shaded edge of both rectangles be a generator for the fundamental group of the cusp. In the case of the cusps corresponding to crossing circles, we ﬁnd the other generator of the fundamental group of the cusp without much more eﬀort. Recall that in our polyhedra, each pair of shaded faces was folded across its common vertex. Each such vertex and no other vertex in the polyhedron corresponds to a crossing circle. Hence in rectangles from horospheres about these vertices, one shaded side of the rectangle is folded over and glued to the opposite shaded side of the same rectangle. Thus a white edge of a single rectangle forms the other generator of the fundamental group of this cusp. In fact, this generator is a meridian of the crossing circle. See Figure 2.11. Since the two generators are orthogonal, the cusp is rectangular. CHAPTER 2. INITIAL CUSP SHAPE ESTIMATES 21

A A

C C

Figure 2.11: The cusp diagram corresponding to a crossing circle

Now consider the remaining cusps. To see that their shapes are also rectangular, we consider the gluing pattern along the shaded sides of the rectangles given by horospheres meeting vertices of the polyhedra. Recall that shaded faces of a polyhedron are glued to shaded faces in the same polyhedron. Thus a shaded side of a rectangle will be glued to the shaded side of another rectangle from a horosphere about a vertex of the same polyhedron, and so on. Hence eventually a shaded edge will be glued to a shaded edge of the original rectangle. We obtain the full cusp by adding the rectangles from the opposite polyhedron glued along white sides. Since no rectangle is repeated in this picture, it follows that the rectangle made up of these rectangles is a fundamental region for the cusp. Hence each remaining cusp is also rectangular, proving the theorem.

a d a b c d a

b c a' b' c' d' a'

Figure 2.12: The cusp diagram for the link on the projection plane

Now let L be a ﬂat augmented link. By the above theorem, each of its cusps has a rectangular shape. Consider in particular the cusp shapes of each of the links lying ﬂat in the projection plane. The following result gives information about the shapes of these cusps. CHAPTER 2. INITIAL CUSP SHAPE ESTIMATES 22

Proposition 10. Let L be a ﬂat augmented link, and consider the cusp shapes of the link components lying ﬂat in the projection plane. If we normalize so that the meridian in each of these cusps has length one, then the sum of the lengths of the longitudes of these link components will be greater than or equal to the number of crossing circles in L.

Proof. We know each of the cusps from the link components lying ﬂat in the projection plane are made up of small rectangles glued two across along the white sides. A curve running along a shaded side gives a meridian. We normalize so that this has length 1. Consequently, each of the small rectangles in the diagram making up the cusp will have shaded side of length 1/2. We need to determine the possible lengths of the white sides of these rectangular pieces. Consider again the circle packing given by the white faces of one of the two polyhedra in the polyhedral decomposition of the ﬂat augmented link complement. Consider a point of tangency of two circles. This corresponds to an ideal vertex in the polyhedron. Tangent to both circles on either side of this point of tangency are two more circles. This collection of four circles will be in one of the two forms shown in Figure 2.13.

Figure 2.13: The possible conﬁgurations of the four circles closest to a point of tangency CHAPTER 2. INITIAL CUSP SHAPE ESTIMATES 23

We take the point of tangency to infinity by a Möbius transformation. This has the effect of sending the two circles tangent to the point of interest to two parallel lines. The circles tangent to the original two are sent to circles of the same radius tangent to the two parallel lines. The rectangle given by the intersection of the horosphere about this vertex and the polyhedron is now exactly the rectangle formed by the two parallel lines and the lines through the points of tangency of the other two circles, as in Figure 2.14. The “shaded” side is marked by the dotted line in Figure 2.14. We scale so that the shaded side has length 1/2.

Figure 2.14: Result of taking a point of tangency to inﬁnity

Note that the length of the “white” side of the rectangle can be completely determined by the distance between the two circles in the picture. Because of our scaling, each circle has radius 1/4. Since the two circles cannot intersect, the length of the white side must be greater than or equal to 1/2. Thus the total length of all planar cusps will be greater than 1/2 times the total number of all rectangles occurring in the cusp diagram. The total number of such rectangles occurring in all planar cusps is equal to the number of vertices in one of our polyhedra minus the number of vertices corresponding to crossing circles. Consider again the nerve of the circle packing corresponding to only the white faces of a polyhedron. Each edge of the nerve passes through a vertex, so the number of edges E in the nerve is equal to the number of vertices in the polyhedron. There is just one vertex in the polyhedron corresponding to each crossing circle, so the number of rectangles is equal to E C, where C is the number of crossing circles in L. − Each triangle in the nerve bounds a shaded face in the polyhedron. Since there are two shaded faces in the polyhedron per crossing circle, the total number T of such triangles is equal to 2C. Finally, since the nerve is a triangulation of S2, 3T = 2E. Hence E = 3C, and so the total number of rectangles corresponding to planar cusps CHAPTER 2. INITIAL CUSP SHAPE ESTIMATES 24

is 2C. Then the total length of the rectangles corresponding to planar cusps is greater than 1/2 2C = C. ∗ So far we have only given information about the similarity class of the cusps of flat augmented links. In fact, we can determine explicitly the lengths of curves contained in the intersection of twice-punctured disks with a maximal cusp. First, recall that we obtain the maximum Euclidean size of the rectangular cusps when we expand a horosphere about the cusp until it becomes tangent to itself. Following [1], we call such a horosphere a maximal cusp. At this point, it can no longer be expanded without intersecting itself. Perhaps this is best understood by considering the universal cover H3 of the complete manifold. In the universal cover, horospheres about the cusp correspond to spheres tangent to the points at infinity that project to the cusp under covering transformations. As the radius of any particular horosphere is increased, the radii of the horospheres of all its images under covering transformations also increase. The maximum expansion of the horosphere is reached when two of these horospheres abut each other. Given two abutting horospheres, conjugate so that one goes to the horizontal plane z = 1 in H3. That is, send one of the cusp points to infinity in the upper half space model of H3, sending the horosphere to the plane z = 1. Then the other horosphere, which was tangent to this horosphere, maps under this transformation to a horosphere of diameter 1, tangent to the plane z = 1. We call such a horosphere a full-sized horosphere. We can find the Euclidean size of the torus now by examining the lattice given by the Z Z subgroup of the covering transformations preserving the plane z = 1. × Lengths of the meridian and longitude as measured on this plane under the metric inherited from H3 are the same as the usual Euclidean lengths on a plane. Moreover, because the horospheres cannot be expanded any further, these lengths are maximum. In our flat augmented links, we have several different cusps to consider and expand. Note that the maximum size of a cusp will depend on the order in which we expand the horospheres about any particular cusp. For example, if we expand one cusp after we have already expanded another to its maximum size, the new cusp may abut CHAPTER 2. INITIAL CUSP SHAPE ESTIMATES 25

another horosphere corresponding to the other expanded cusp rather than abutting a horosphere corresponding to the same cusp. To avoid this difficulty, we fix the order in which we will expand the cusps. For ease of exposition, first assume there is one unique link component lying flat in the projection plane. For the remainder of the section, we will expand the horosphere about this cusp first, expanding until it abuts itself. Then expand horospheres about the crossing circles. We will see below that the order in which the horospheres about the crossing circles are expanded doesn’t matter once the horosphere about the planar cusp is maximal. However, for now assume that the crossing circle horospheres are expanded simultaneously, such that each horoball neighborhood of the cusp has the same volume. Recall that a meridian of the planar cusp runs along a curve corresponding to the intersection of the horosphere with a “shaded face”, which glues to give a twice- punctured disk, or thrice-punctured sphere, in the 3-manifold. It is from considering these thrice-punctured spheres that we are able to obtain results on the lengths of these meridians. Note that in the case of only one planar link component, two of the three punctures in each of our thrice-punctured spheres lie at cusps corresponding to the planar link component. The other puncture will correspond to a crossing circle.

Lemma 11. Let S be any thrice-punctured sphere in the complement M of a flat augmented link with a unique planar link component. Consider the universal cover H3 of M, and conjugate so that three points projecting to the three punctures of S in some lift of S into H3 lie at the points 0, 1, and infinity, with 1 and infinity projecting to the cusp corresponding to the planar link. If there is a horosphere of (Euclidean) radius r about the point 1, then there will be a horosphere of (Euclidean) height 1/2r about the point infinity.

Proof. Because the points 1 and infinity both project to the same cusp of M, there will be a covering transformation T of H3 taking 1 to infinity. Under this transformation, the points 0 and infinity must go to puncture points of another lift of S, and so will lie on the complex plane at infinity. We know that any other lift of S will correspond to a shaded face lying over one of the shaded edges of the cusp diagram Figure 2.12. CHAPTER 2. INITIAL CUSP SHAPE ESTIMATES 26

0 1

Figure 2.15: A fundamental region for a three punctured sphere. Projections of horospheres are labeled by the short dashed lines. Note there is reﬂective symmetry through the dashed line from 0 to inﬁnity.

In particular, the images of 0 and infinity under T will still be Euclidean distance 1 apart. Hence T takes 0 to 0+c and infinity to 1+c where c is some complex number. Thus T is given by the composition of Möbius transformations:

1 c 1 0 T = 0 1 ! 1 1 ! − Note the parabolic translation by c won’t affect the Euclidean height of a point in the upper half space. Hence we need only consider the effect of the leftmost Möbius transformation making up T . A simple calculation checks that under this transformation, the horosphere of radius r about 1 will go to the plane of Euclidean height 1/2r. Since this plane must be the horosphere about infinity, the lemma is proved.

Lemma 12. Let M be the complement of a flat augmented link with a unique planar cusp. Expand the horosphere about the planar cusp until it becomes tangent to itself, and consider the universal cover of M, with the planar cusp at infinity. Let S be the lift of any thrice-punctured sphere in M such that one of the punctures of S is at infinity. Then over the other puncture corresponding to the planar cusp will lie a full-sized horosphere.

Proof. Without loss of generality, we can assume that S contains points 0, 1 and inﬁnity, with inﬁnity and 1 projecting to the planar cusp of M, and 0 projecting to a crossing circle as in Lemma 11. Then the horosphere over 1 will have radius r and CHAPTER 2. INITIAL CUSP SHAPE ESTIMATES 27

the horosphere about infinity will have Euclidean height 1/2r. Suppose the horosphere over 1 is not full-sized. Then the diameter of the horosphere over 1 is strictly less than the height of the horosphere about infinity, or 2r < 1/2r, and hence r < 1/2. So the height of the horosphere about infinity is strictly greater than 1. Thus, any full-sized horosphere must have diameter strictly greater than 1. However, recall M is has a reflective symmetry through the projection plane of the flat augmented link. In the universal cover, this corresponds to a symmetry under reflection through the planes perpendicular to S, running through any two points projecting to punctures in the thrice-punctured sphere of M. I.e., there is a reflection through the white edge in Figure 2.12, and all its translates. Lifts of these white faces give a vertical plane through 0 perpendicular to S, as well as one through 1. Now note that under a reflection in one of these planes, a horosphere of diameter greater than 1 must intersect one of its translates. This contradicts the fact that the horospheres are only expanded until they are tangent. So the horosphere over 1 must be full-sized.

Proposition 13. Let M be the complement of a ﬂat augmented link with a unique planar link component. Take a maximal horoball neighborhood about the cusp in M corresponding to the planar link component, and let T be the boundary torus of this horoball neighborhood. Then the meridian of T has length exactly equal to 2.

Proof. Consider the universal cover of M, and conjugate so that the cusp corresponding to the planar link component lies at infinity. Conjugate again so that some lift of a thrice-punctured sphere S has points 0, 1 and infinity projecting to punctures in M. By Lemma 12, there is a full-sized horosphere lying over the point 1. If this horosphere has radius r, then because it is full-sized, Lemma 11 implies 2r = 1/2r, and hence r = 1/2. So the horosphere about infinity is at height 1. This is convenient: Distances measured on the plane of Euclidean height 1 in H3 are the same as the Euclidean distances along this plane. Thus we need to show a meridian as measured along this plane has length 2. Recall that a meridian runs along one of the shaded edges of Figure 2.12. That CHAPTER 2. INITIAL CUSP SHAPE ESTIMATES 28

is, a meridian runs the length of a thrice-punctured sphere in M. Consider again our lift of the thrice-punctured sphere S. The points 0, 1 and infinity in H3 project to punctures of S. However, the triangle with those vertices in H3 projects to only half of S: that part of S lying in one of the two polyhedra in the proof of Proposition 9. The other half is obtained by reflection through the white faces of the polyhedra. In the universal cover, this reflection is reflection through the plane containing the points 0 and infinity, and perpendicular to the plane containing 0, 1 and infinity which projects to S. Hence a fundamental region for S is given by the union of the two triangles with vertices 0, 1 and infinity and 0, 1, and infinity. Thus a meridian of the planar link − component runs from 1 to 1 in the plane of height 1 in H3, so has length 2. − Remark. All the above results apply immediately for a flat augmented link with more than one planar link component. Simply increase the horospheres corresponding to two different planar link components in such a way that the horospheres about the punctures of any thrice-punctured sphere maintain the same (Euclidean) radius.

Corollary 14. Expand the horospheres about the planar link components as above. Then expand the horosphere about any crossing circle link component until it abuts another horosphere. The horospheres it abuts will correspond to those from the planar link component, and the crossing circle will have longitude exactly length 2.

Proof. Recall (Figure 2.11) that the longitude of a crossing circle runs along a thrice- punctured sphere. In this case, if we put the crossing circle cusp at inﬁnity, then the other punctures of the thrice punctured sphere will correspond to the planar cusp. We know from Lemma 12 that horospheres about these punctures must abut each other. The proof of Lemma 12 also applies to show that these horospheres are full-sized. Hence the longitude has length exactly 2. CHAPTER 2. INITIAL CUSP SHAPE ESTIMATES 29

2.4 Initial Cusp Shape Estimates with Half Twists

Recall that to obtain a flat augmented link, we added crossing circles at twist regions. The new manifold was homeomorphic to one with either zero or one crossings at each twist region. If a single crossing was left, we removed it. However, in general we’ll want to keep the single crossings to apply the results of the next sections, so we need to extend the estimates in the previous section to include flat augmented links with single crossings, or half-twists, added at crossing circles. When half-twists are present, the decomposition of the link into two polyhedra is the same as when the augmented link is completely flat. To obtain half-twists, only the gluing pattern changes along the twice-punctured disks. This change in gluing pattern is illustrated by Figure 2.16.

Figure 2.16: On the left is the standard gluing of the twice-punctured disk. On the right is the new gluing creating a half-twist.

In particular, in the ﬂat case the half of the twice-punctured disc lying above the projective plane will be glued to a piece lying above the projective plane. When we put in a half-twist, now the piece lying above the projective plane will be glued to one lying below the projective plane. The eﬀect on the cusp diagram will be to shift the gluing of the small rectangular pieces by one step along the shaded faces, as illustrated in Figure 2.17. Hence the cusp shapes may be sheared. In particular, if only one half-twist is added, the longitude of the cusp in the projection plane will no longer be perpendicular CHAPTER 2. INITIAL CUSP SHAPE ESTIMATES 30

Figure 2.17: Adding a half twist to a ﬂat augmented link shifts the gluing along the cusp. to the meridian, which still runs along a twice-punctured disk. However, note that if the link component passes through an even number of half- twists then the actual shape of the cusp doesn’t change. That is, given two shifts as in Figure 2.17, the lattice corresponding to the torus is unaﬀected. Hence it is still rectangular, although the curve running perpendicular to the meridian is no longer a true longitude. In particular, this must occur if there is only one link component lying on the projection plane. Hence we have shown:

Proposition 15. If L is a hyperbolic augmented link with only one link component in the projection plane, and this component lies ﬂat on the projection plane except possibly for half-twists at twice-punctured discs, then the cusp corresponding to this link component will be rectangular. Moreover, when the meridian is scaled to have length 1, the length of a longitude will be at least equal to the number of crossing circles in L.

Finally, note that regardless of the parity of half-twists through which a link component passes, in the hyperbolic structure any twice-punctured disk will still have length 2, giving an analogue to Proposition 13 for half-twists.

Proposition 16. Let L be a hyperbolic augmented link, possibly with half-twists. Then the length of a meridian of a cusp corresponding to a link component lying on CHAPTER 2. INITIAL CUSP SHAPE ESTIMATES 31

the projection plane will be exactly 2. Similarly, the length of a longitude of a crossing circle will be exactly 2.

Proof. Because the link complement can be broken into the two hyperbolic polyhedra described above, which still exhibit reﬂective symmetry through the white faces, the proof of Lemma 12 still holds without modiﬁcation. Hence any full-sized horospheres in a cusp occur over punctures of a thrice-punctured sphere. But then the proof of Proposition 13 continues to apply in this case, proving that the length of a curve running along a thrice-punctured sphere is exactly 2. Chapter 3

Deforming Through Cone Manifolds

In this chapter, we will modify the geometry of augmented link complement we obtained in Chapter 2, deforming smoothly until the augmented link complement becomes the original knot complement with a complete hyperbolic structure. We analyze changes in geometry under this deformation.

Recall that our original link is obtained from the augmented link by a 1/ni Dehn

filling along each crossing circle cusp, where ni is the twisting amount in the i-th twist region. We would like to control the geometric change in the cusp in the projection plane under this topological Dehn filling, and thus preserve geometric estimates of cusp shapes for the original manifold. To obtain this geometric control, we deform through hyperbolic cone manifold structures. In the first section of this chapter, we’ll give background on cone manifolds and the process of deforming through cone manifolds. In the second section, we explain in some detail how the deformation leads to a 1-form. This 1-form has been analyzed in detail by Hodgson and Kerckhoff, and we refer the reader to their papers for more information ([11], [10], [12]). In the third section, we apply some of the analysis of Hodgson and Kerckhoff to the situation at hand. That is, we will analyze the 1-form we obtain when our manifold starts as a link complement with cusps coming from crossing circle components and a complete hyperbolic structure, and then is modified

32 CHAPTER 3. DEFORMING THROUGH CONE MANIFOLDS 33

through cone manifolds to give a complete hyperbolic structure on the original knot complement we began with in Chapter 2.

3.1 Hyperbolic Cone Manifolds

A hyperbolic cone manifold is a manifold deﬁned as follows. Let X be a 3-manifold. Let Σ be a closed 1-manifold in X. Then X is a hyperbolic cone manifold if M = X Σ − admits an incomplete hyperbolic structure, such that in the completion any point in

Σ has a neighborhood of the form Cα I, where Cα is a 2-dimensional hyperbolic × cone with cone angle α, and I is an open interval. 2 That is, Cα can be obtained by taking a wedge of angle α in H and identifying the sides of the wedge, as in Figure 3.1.

Figure 3.1: A cone of angle α.

Σ is called the singular locus of the cone manifold X. Note that when α = 2π we have a complete, non-singular hyperbolic metric on the manifold X. When α approaches 0, the cones become more and more narrow and sharp. In the limit, we obtain a complete hyperbolic structure on M = X Σ with a cusp along Σ. − Thurston [19] showed that one can always deform a hyperbolic metric through cone manifold structures in some neighborhood of a complete hyperbolic structure with cusps. That is, it is always possible to increase the cone angle along Σ of the cone manifold X from 0 to some > 0. CHAPTER 3. DEFORMING THROUGH CONE MANIFOLDS 34

If the cone angle increases from 0 to 2π, we have performed hyperbolic Dehn surgery on the manifold. Hodgson and Kerckhoﬀ ([11], [10]) found constraints on the change in geometric structure of the manifold when the cone angle is deformed. Applying their results to our situation with cusp shapes, we are able to ﬁnd bounds on the change in geometry of the cusp shapes as we deform from the augmented link through cone manifold structures with singular loci the cores of the crossing circles to the original knot or link with cone angle 2π along the crossing circles. For simplicity of exposition, we will assume throughout the rest of this chapter that our initial manifold was a knot complement, rather than a link complement. The arguments extend easily to the case of a link complement.

3.2 Obtaining an E-valued 1-form

In this section, we explain notation and show how an E-valued 1-form describes the change in geometry of the manifold under the deformation through cone manifolds. The ideas are from the work of Hodgson and Kerckhoﬀ, and more details can be found in their papers ([11], [10], [12]). We begin with a hyperbolic 3-manifold M with a complete hyperbolic structure with cusps. That is, M can be viewed as the cone manifold X = M (cusps) with ∪ singular locus Σ = (cusps) and cone angle 0 along Σ. M is the initial manifold in our deformation. We will deform the hyperbolic structure on M smoothly through structures with cone singularities along components of Σ. Recall that any (not necessarily complete) hyperbolic structure on M is given by local charts, each modeled on a ball in H3, such that overlap maps are restrictions of hyperbolic isometries. Given such a structure, we obtain a developing map Φ : M → H3 as follows. f Fix a basepoint x0 on M. We consider M, the universal cover of M, to be the space of homotopy classes of paths in M that start at x0. Select an initial chart f 3 whose domain contains x0. Along each path in M, we can map into H by analytic

f CHAPTER 3. DEFORMING THROUGH CONE MANIFOLDS 35

continuation. This gives the developing map

Φ : M H3. →

It is unique up to composition by anf isometry of H3, the result of changing the basepoint x0 or the initial chart.

The fundamental group, π1(M) acts on M by covering transformations. Note that if γ π1(M) is an element of the fundamental group, then Φ γ = ρ(γ) Φ, where ∈ f ◦ ◦ ρ(γ) is an isometry of H3. The map

ρ : π (M) Isom(H3) 1 → is called the holonomy representation. See Thurston [21] for more details. Now, when the hyperbolic structure on a 3-manifold is deformed smoothly, this deformation determines a smooth 1-parameter family of developing maps Φt : M → 3 3 H . Note that for any ﬁxed m in M, Φt(m) is a path in H . If we let Φ˙ be the f derivative at time t = 0, then Φ(˙ m) is the initial tangent vector to this path. This f gives us a function Φ˙ : M T H3. → f 3.2.1 Inﬁnitesimal Isometries

We will connect Φ˙ with the bundle E of infinitesimal isometries of M. First, we describe the bundle E and also the bundle E of infinitesimal isometries of M. e f If M admits a complete hyperbolic structure, then its universal cover, M, will be e H3. We know the group of isometries of H3 is G = SL(2, C). The Lie algebra to this f group is = sl(2, C). We view as the tangent space to G at the identity, TeG. G G From this viewpoint, elements of correspond to the infinitesimal isometries of H3. G Thus the bundle of infinitesimal isometries of H3 is H3 sl(2, C). × In general, M will not be H3, since during our deformation we pass through incomplete hyperbolic structures. However, in the general case, any one parameter f family of isometries ϕt of M, for small t, can be composed with the initial developing −1 3 map, Φ0 ϕt Φ0 to give a path of local isometries of H , and similarly given such a path e f e CHAPTER 3. DEFORMING THROUGH CONE MANIFOLDS 36

of isometries of H3, the initial developing map will give us a path of local isometries of M. Hence taking derivatives of these paths of isometries, we find again that an infinitesimal isometry of M corresponds to an element of = sl(2, C). Since M is f G simply connected, the bundle of infinitesimal isometries of M is E = M sl(2, C). f × f It also makes sense to speak of the bundle E of infinitesimal isometries of M. f e f Again because M is hyperbolic, the fiber of the bundle E is = sl(2, C), with G the correspondence given by lifting (projecting) paths of isometries to (from) M in small neighborhoods. However, we now have added relations on the bundle E, for f if γ is an element of π1(M), then ϕ is a lift of some isometry of M to M if and −1 only if γ ϕ(γm) = ϕ(m). Hence any one parameter family ϕt of isometries of M f corresponds to a family of isometriese of H3 satisfying: e e −1 −1 −1 ρ(γ) Φ0(ϕt) Φ0 ρ(γ) = Φ0 ϕt Φ0

Hence E is isomorphic to (M )/e , where (m, ζ) e (γm, Ad(ρ)ζ) for γ in π (M). × G ∼ ∼ 1 Here recall ρ : π1(M) is the holonomy representation. → G f Now, note that both E and E are flat vector bundles. That is, they both have a trivialization relative to which the transition functions are locally constant. Thus e we can define the de Rham cohomology with coefficients in E or E. Elements of H0(M,E) are sections on M with values in E. Elements of H1(M,E) are E-valued e 1-forms. See Bott and Tu [3] for details. We connect this with the deformation of a 3-manifold through cone manifold structures. First, note any element ζ of the Lie algebra gives a vector field on H3 as follows: G Exponentiate ζ to give a path expζ(t) through the identity in SL(2, C) with tangent 3 vector ζ at zero. For any time t, the element expζ (t) acts on H . In particular, for a 3 3 fixed m in H , expζ(t)(m) is a path through m in H . When we take the derivative 3 of this path at time t = 0, we get a vector in TmH . Thus we obtain a vector field V (called a Killing vector field) on H3. CHAPTER 3. DEFORMING THROUGH CONE MANIFOLDS 37

This Killing vector field V gives all the information necessary to recover the original Lie algebra element. For at each point, sl(2, C) splits into a direct sum of infinitesimal translations and infinitesimal rotations. The infinitesimal translation part at m 3 can be identified with TmH . This is picked up by the vector V (m). The infinitesimal rotation part can be identified with so(3) = su(2). To recover the infinitesimal rotation from V , take curl(V )(m), where the curl of a vector field at a point is defined to be 1/2 times the skew symmetric part of its covariant derivative at the point x. − In fact, given any vector field V on H3 we can lift to a section s of H3 by letting ×G s(m) be the element of whose infinitesimal translational part agrees with V (m) and G whose infinitesimal rotational part agrees with curl(V )(m). If V is a Killing field, it lifts to the constant section. In our current situation, rather than having a vector field on H3, we have a function Φ˙ : M T H3. We can lift this to a section s of E as in the previous paragraph. → Again choose s(m) to be the unique element of whose infinitesimal translational f G e part agrees with Φ(˙ m) and whose infinitesimal rotational part agrees with curl(Φ)(˙ m). Thus we have obtained s Ω0(M, E): sections on M with values in E. ∈ f e f e 3.2.2 Defining the 1-form ω

Now, from our deformation of the hyperbolic structure we obtained first a vector valued function Φ˙ on M, and then a section s in Ω0(M, E). Note first that in general, s does not project to give an element of Ω0(M,E). To f f e give an element of Ω0(M,E), we’d need s(γm) to agree with Adρ(γ)s(m) for each γ in π (M). However, s(γm) Adρ(γ)s(m) will be a constant infinitesimal isometry, 1 − as shown by the following argument: Recall

Φt(γm)= ρt(γ)Φt(m) where ρt denotes the holonomy at time t. So:

d Φ(˙ γm)= (ρt(γ)Φt(m)) t dt | =0 CHAPTER 3. DEFORMING THROUGH CONE MANIFOLDS 38

The right hand side equals:

d ρt(γ) t (Φ (m)) + dρ Φ(˙ m) dt | =0 0 0

d Here ρt(γ) t (Φ (m)) is the vector tangent to the path ρt(γ)(Φ (m)). When dt | =0 0 0 we lift to E, it is equal toρ ˙ , translated to ρ (γ). This is independent of Φ (m), ∈ G 0 0 hence independent of m, so is a constant infinitessimal isometry. e As for the term dρ0Φ(˙ m), this lifts to give the action of dρ0 on s(m), which is the adjoint action: Ad(ρ(γ)). Thus the lift s of Φ at Φ (m) equalsρ ˙ + Ad(ρ(γ))s(m) and we have s(γm) 0 − Ad(ρ(γ))s(m) is a constant infinitesimal isometry. But then consider ds. This is an exact 1-form in H1(M, E) which satisfies ds(γm) = Adρ(γ)ds(m), hence it descends to a closed, not generally exact 1-form f e ω in H1(M,E). The form ω encodes the information we need on the deformation. The only issue we have not yet dealt with is that of choice: We chose a particular family developing maps. Post multiplication by a family of isometries of H3 would give the same hyperbolic structure. Similarly, if we first apply a lift of a diffeomorphism of M to M we get the same hyperbolic structures on M. That is, a family of developing maps giving the same hyperbolic structure as Φt is anything of the form ktΦtft, where f 3 kt are isometries of H and ft are lifts of diffeomorphisms of M to M. Assume k0 and f0 are the identity. We want to ensure that ktΦtft give the same closed 1-form ω in f H1(M,E).

Tracing through the above procedure, note the isometries kt will add a constant infinitessimal isometry k˙ to s : M E. Hence ds will remain unchanged. As for → the diffeomorphisms ft, their effect is to add a section dΦ0 f˙ to Φ˙ which satisfies f e ◦ dΦ f˙(γm) = dΦ f˙(m). Hence this lifts to modify the section s by addition of 0 ◦ 0 ◦ a section s0 which satisfies s0(γm) = Adρ(γ)s0(m), hence descends to a well defined element of Ω0(M,E). But then its exterior derivative ds0 is exact in H1(M,E), so it has no effect on the class ω. Thus, given a deformation of hyperbolic structures on the manifold M, we have obtained a closed E-valued 1-form ω whose cohomology class is well-defined. CHAPTER 3. DEFORMING THROUGH CONE MANIFOLDS 39

3.2.3 Properties of ω

Note in choosing our one-parameter family of developing maps above we had a lot of flexibility. It would be ideal to choose a family with nice analytic properties. Hodgson and Kerckhoff ([11]) showed this could be done, and in fact, that we can assume ω is a particularly nice representative for its de Rham cohomology class in H1(M,E). We recall some properties of the Lie algebra, of the group of isometries of H3. G Recall that , the fiber of E, splits at each point into a direct sum of translational and G rotational parts. In H3, these have a particularly nice structure. The infinitesimal rotations fixing a point p can be identified with so(3) ∼= su(2), and the infinitesi- H3 mal translations at p correspond to i su(2) ∼= T . Here geometrically, if τ is an infinitesimal translation, then iτ is an infinitesimal rotation with axis in the direction of τ. The infinitesimal translational part of E can be identified with T M via the developing map. Thus the bundle E can be identified with the complexified tangent bundle T M C. ⊗ By the work of Hodgson and Kerckhoff, [11], we see that ω is determined by its translational part η, which determines the infinitesimal change in metric. In particular, ω can be chosen so that η is a harmonic T M-valued 1-form which satisfies:

D∗η = 0,

D∗Dη = η, (3.1) − where D is the exterior covariant derivative on T M-valued 1-forms and D∗ is its adjoint. Also, η and Dη are symmetric and traceless, and

ω = η + i Dη. ∗

Additionally, Hodgson and Kerckhoﬀ showed that if U is a neighborhood of the singular locus of M, then inside U we have a decomposition:

η = η0 + ηc (3.2) CHAPTER 3. DEFORMING THROUGH CONE MANIFOLDS 40

where η0 gives the inﬁnitesimal change in holonomy of the meridian and longitude on the torus ∂U, and ηc is a “correction term”. 2 Similarly, ω = ω0 + ωc where ω0 changes the holonomy and ωc is L . In fact,

Hodgson and Kerckhoﬀ further showed that ω0 decomposes:

ω0 = ωm + ωl, (3.3)

where ωm = ηm + i Dηm is an inﬁnitesimal deformation which changes the cone ∗ angle but doesn’t change the real part of the complex length of the meridian, and

ωl = ηl +i Dηl stretches the singular locus, but leaves the holonomy of the meridian ∗ unchanged. Now, let α and β be E-valued 1-forms. Each can be written as e σ where σ is ⊗ a real-valued 1-form and e is an element of E. We let

α β =(eα σα) (eβ σβ)=(eα eβ) (σα σβ) ∧ ⊗ ∧ ⊗ · ⊗ ∧ where here (eα eβ) denotes the dot product on E induced from the geometric inner · 2 product after identifying Ex = TxM TxM. Then we obtain an L inner product on ⊕ E-valued forms deﬁned by (α, β)= α β, ∧ ZM when the integral exists (see [11]). Now recall that ω = η + i Dη, where η satisﬁes: ∗

D∗Dη + η = 0.

Taking the inner product of the above formula and integrating by parts over any submanifold N of M with boundary ∂N, as in [11] (Proposition 1.3 and p. 36), and [10] (Lemma 2.3), we obtain

0= η 2 + Dη 2 + Dη η, (3.4) k kN k ∗ kN ∗ ∧ Z∂N where the boundary ∂N is oriented by the outward normal. CHAPTER 3. DEFORMING THROUGH CONE MANIFOLDS 41

Using similar notation as that of [10], we deﬁne:

bV (α, β)= Dα β ∗ ∧ Z∂V where V is a neighborhood of the singular locus or cusp of M, and ∂V is oriented ∂ with ∂r , the tangent vector orthogonal to ∂V in V , as outward normal when viewed from V . Then letting N = M V in equation 3.4, note ∂N = ∂V , so we obtain − −

2 2 2 bV (η,η)= η − + Dη − = ω − . (3.5) k kM V k ∗ kM V k kM V

Hodgson and Kerckhoﬀ have also analyzed the eﬀect of the decomposition of equation 3.2 on bV (η,η). They showed ([10], Lemma 2.5 and 2.6) that

bV (η,η)= bV (η0,η0)+ bV (ηc,ηc) (3.6) and

bV (ηc,ηc) 0. (3.7) ≤

3.3 Deforming the Augmented Link Complement

In our particular situation, we take the manifold M to be the augmented link complement. That is, M is the 3-manifold with torus boundary components. One of these components corresponds to the component of the augmented link in the projection plane. The others correspond to the crossing circle link components. We showed in Chapter 2 that M admits a complete hyperbolic structure with cusps at the crossing circles. This will be our initial cone manifold structure on M. That is, at the initial point of our deformation the singular locus of M consists of cusps, with cone angle zero. We will deform this structure through manifolds with increasing cone angles along the crossing circles, until we reach cone angle 2π, at which point we have performed hyperbolic Dehn surgery and obtained the original hyperbolic structure on the knot CHAPTER 3. DEFORMING THROUGH CONE MANIFOLDS 42

complement S3 K. − As in [10], we’ll parameterize the deformation by the square of the cone angle at the singular locus. That is, if Φt is the 1-parameter family of developing maps giving the change in hyperbolic structure, t = α2 where α is the cone angle along the singular locus. This choice of parameterization becomes useful in the analysis of Chapter 4. We will analyze η and ω in the following neighborhoods of the manifold M: Let U be a horoball neighborhood of the cusp that is on the projection plane at t = 0, and has become the cusp corresponding to the knot K at t = (2π)2. Thus U b is a horoball throughout the deformation, with boundary ∂U a torus. It is the shape b of this boundary ∂U which we would like to analyze. b Let U be a tubular neighborhood of the crossing circles. That is, at time t = 0, b U is a collection of horoball neighborhoods of crossing circle cusps. At time t> 0, U is made up of components consisting of tubular neighborhoods of each component of the singular locus at the crossing circles. So U = U U UN K where each Uj 1 ∪ 2 ∪···∪ ( ) is a tubular neighborhood of the j th component of the singular locus, and we let − N(K) denote the twist number of the original knot complement. At any fixed time t > 0, we require each component Uj to have the same tube radius. That is, if Rj is the distance from ∂Uj to the j-th component of the singular locus, then Rj = Ri = R for each i, j. We define R to be the tube radius of the deformation. Before stating our first results, recall that inside a neighborhood of the singular locus, we could decompose η into η = η0 + ηc = ηm + ηl + ηc (equations 3.2 and 3.3). In our current case, we have neighborhoods U and U. We can decompose η in each of them. b In U, we write η = ηm +ηl +ηc. In fact, in U this decomposition of η becomes even simpler, for ηm encodes the change in cone angle under the deformation. However, b b in U, the neighborhoode of thee cuspe in the projection plane, the cone angle remains 0 e throughout the deformation. Thus ηm = 0. b In U, write η = η0 + ηc. We are now ready to state our firste results.

Lemma 17. The boundary terms bUb (ηl, ηl) and bU (η0,η0) are related by the following

e e CHAPTER 3. DEFORMING THROUGH CONE MANIFOLDS 43

inequality:

b b (ηl, ηl) bU (η ,η ) (3.8) − U ≤ 0 0 e e Proof. By equation 3.5, we know:

2 2 2 ω − = Dη − + η − = bU (η,η) k kM U k kM U k kM U

2 Here M−U denotes the L -norm on M U. k · k − We know further by equations 3.6 and 3.7 that

bU (η,η)= bU (η0,η0)+ bU (ηc,ηc)

where bU (ηc,ηc) is non-positive. Thus

2 ω − bU (η ,η ). (3.9) k kM U ≤ 0 0

As for the neighborhood U, in equation 3.4, let the submanifold N be the horoball neighborhood U and we obtain: b

b 2 2 0= η b + Dη b + Dη η, k kU k ∗ kU b ∗ ∧ Z∂U or 2 2 2 b b (η,η)= η b + Dη b = ω b . − U k kU k ∗ kU k kU

Note the sign on bUb (η,η) is negative due to the orientation of ∂U. Thus inside of U, using equation 3.6, we can write: b

b 2 ω b = b b (η,η)= b b (ηl, ηl) b b (ηc, ηc). k kU − U − U − U

Again by equation 3.7, b b (ηc, ηc) 0. So we obtain the estimate: U ≤ e e e e

2 e eb b (ηl, ηl) ω b (3.10) − U ≤ k kU e e CHAPTER 3. DEFORMING THROUGH CONE MANIFOLDS 44

Finally, 2 2 2 ω b = ω − ω b . k kU k kM U − k kM−U−U Putting this with equations 3.9 and 3.10, we obtain our result:

2 2 b b (ηl, ηl) ω b ω − = bU (η ,η ). − U ≤ k kU ≤ k kM U 0 0 e e

This relationship between the boundary terms of the manifold M U U under − − the deformation through cone manifold structures leads to bounds on the change of b geometry of our cusp. In the next chapter, we will study each boundary term from the lemma above, and determine what the inequality implies in terms of the deformation of cusp shapes. Chapter 4

Analyzing Boundary Terms

In this chapter, we analyze both sides of the inequality 3.8. In Section 4.1, we study the left hand side, or the boundary term coming from the horoball neighborhood U of the cusp which becomes the original knot at the end of the deformation. In Section b 4.2, we study the right hand side, obtaining estimates on the boundary term given by the tubular neighborhood of the singular locus. Putting these together, we obtain a diﬀerential inequality, under certain conditions, which we analyze in Section 4.3. Throughout this section, we will obtain estimates which depend on the size of the tube radius R. We will keep a running list of these conditions, then ensure they are all met when we compute actual values for constants in the next chapter.

4.1 Analysis of the Boundary Term from the Cusp

In this section, we analyze the left hand side of the inequality 3.8, or the boundary term b b (ηl, ηl). − U Recall that U is a horoball neighborhood of the cusp of the augmented link which lies in the projectione e plane at time t = 0. Over the deformation through cone manifold b structures, U remains a horoball neighborhood of a cusp. At time t = (2π)2, U has

b b 45 CHAPTER 4. ANALYZING BOUNDARY TERMS 46

become a horoball neighborhood of the original knot K. As in Chapter 3, we let ω H1(M,E) be a harmonic 1-form encoding the de- ∈ formation through cone manifold structures. Recall that we could decompose ω in a neighborhood of the singular locus into the parts ω = ωm + ωl + ωc (equation 3.3).

In the neighborhood U, we write this decomposition as ω = ωm + ωl + ωc. Recall that ωm = 0, since it gives the infinitesimal change in cone angle, and in U our cone b angle is constant, equal to 0, throughout the deformation. e e e e b In general, ωl is a complex multiple of a standard form, which Hodgson and Kerckhoff computed ([11], p. 32). We review their results. First, note anye E-valued 1-form σ can be viewed as an element of Hom(TM,TM ⊕ i T M), for σ takes a vector v T M and gives an element of E = T M i T M. ∈ ⊕ Then in the neighborhood U, we use cylindrical coordinates, and let the vectors e1, e2, and e3 be an orthonormal frame, with e2 tangent to the meridian of K, e3 tangent b to the longitude, and e1 pointing in the radial direction. We let ω1, ω2, and ω3 be dual to these vectors. Then an element of Hom(TM,TM i T M) can be described ⊕ as a matrix whose (i, j) coordinate is the coefficient of ei ωj. ⊗ In these coordinates, the standard form of ωl is given by the matrix:

1 − 0 0 cosh2 R      i sinh(R)   0 1 −   cosh(R)   −         2   i sinh(R) cosh R + 1   0 −   cosh(R) cosh2(R)      Here R is the size of the tube radius of the neighborhood of U. Since U is the horoball neighborhood of a cusp throughout, in this case R is inﬁnite. Hence we take the limit b b of the above matrix as R goes to inﬁnity and obtain the standard form for a cusp CHAPTER 4. ANALYZING BOUNDARY TERMS 47

neighborhood: 0 0 0  0 1 i  − −  0 i 1   −    Thus we know: 0 0 0

ωl =(a + ib)  0 1 i  − −  0 i 1   −  e   Since in general, ωl = ηl + i Dηl, gives the real and imaginary parts of ωl (see ∗ Section 3.2.3), this implies:

0 0 0 0 0

ηl =  0 a b  =  a  ω2 +  b  ω3 (4.1) − −  0 b a   b   a        e       and: 0 0 0 0 0

Dηl =  0 b a  =  b  ω2 +  a  ω3 (4.2) ∗ − − − −  0 a b   a   b   −      e       We can then compute b b (ηl, ηl) explicitly in terms of the constants a and b: − U e e b b (ηl, ηl)= ηl Dηl, − U b ∧ ∗ Z∂U e e e e CHAPTER 4. ANALYZING BOUNDARY TERMS 48

where

0 0 0 0

ηl Dηl =  a  ω2 +  b  ω3  b  ω2 +  a  ω3 ∧ ∗ − ∧ − −  b   a    a   b             e e          

0 0 0 0

=  a   a  ω2 ω3 +  b   b  ω3 ω2 − · − ∧ · − ∧  b   b   a   a                  = 2(a2 + b2)ω ω . 2 ∧ 3

Since ω ω is the area form for the torus ∂U, we have: 2 ∧ 3

2 2 b 2 2 b b (ηl, ηl)= 2(a + b )dA = 2(a + b )Area(∂U). (4.3) − U b Z∂U b The term (a2 + be2)e above gives information in the change in the cusp shape of the torus. We want to bound the area below. Then the area bound along with the boundary calculation will give bounds on (a2 + b2), or the change of shape of cusp torus.

4.1.1 The Area of ∂U

In this section, we bound the area of the torus ∂U below. b The area of the torus ∂U is equal to its base times its height. We let the base b length be the length of the meridian at time t, m(t), and denote by H(t) the height b of the torus at time t. To bound the area term, we ﬁnd bounds on the length of the meridian m(t) for all time t. Note we showed in Proposition 16 that at time t = 0, the meridian of ∂U had length exactly 2. At time t = (2π)2, when we have reached the complete hyperbolic b structure on the knot complement, it is known that the meridian has length at least 1 (see Adams [1]). For time t between these values we need to ensure that U and the

b CHAPTER 4. ANALYZING BOUNDARY TERMS 49

tubular neighborhood of the singular locus U do not intersect, and so the length of the meridian of ∂U will depend on the tube radius R of the tubular neighborhood U. In particular, we have the following result: b Lemma 18. Suppose the tube radius R about the singular locus remains greater than or equal to log(2/√3) throughout the deformation. Then the meridian of ∂U has length greater than or equal to 1 e−2R throughout the deformation. − b Proof. Step 1: For any ﬁxed time t between 0 and (2π)2, expand the tube U about the singular locus Σ until the tube hits itself. Denote this maximal tube radius by R. Step 2: Expand the horoball U about the cusp until it either becomes tangent to itself or to the tube about Σ. b Take the shortest nontrivial simple closed curve on ∂U starting at a point on ∂U which is closest to Σ. Lift the path to the universal cover X of the manifold b S3 K Σ= X. b− − e Now, we obtain a region D by continuing to expand U as follows. First, choose a 3 particular lift U0 of U in the universal cover. Because U is embedded in H by the b developing map on the universal cover of X, we can assume the developing map is b b3 one to one on U0 and takes ∂U0 to the plane z =1 in H . Take: D = U S 0 ∪ where S is the set consisting of all points x in X such that dist(x, ∂U0) < dist(x, ∂U1) for U1 any other lift of U, and also such that dist(φ(x),∂U) < dist(∂U, Σ), where φ e is the covering projection. Note that D is embedded in H3 by the developing map. b b b That is, the developing map is one to one on D.

Then check: Let p be a point on ∂U closest to Σ, and let p0 and p1 be lifts of p such that the path from p0 to p1 is shortest on ∂U. Let a0 be the point in the b universal cover on a lift of Σ such that the path from p0 to a0 has length equal to b the minimum distance from p0 to Σ. Similarly, let a1 be the point on a lift of Σ such that the path from p1 to a1 has length equal to the minimum distance from p1 to Σ.

Note the paths between p0 and a0 and between p1 and a1 are contained in D. CHAPTER 4. ANALYZING BOUNDARY TERMS 50

Then note the geodesic γ from a0 to a1 is also in D, along with all the points in the quadrilateral with sides given by γ and the paths from p0 to p1, from p0 to a0, and from p1 to a1. This is because, first, all points x in this quadrilateral (except the endpoints of γ) have dist(φ(x),∂U) < dist(∂U, Σ). Also, if any x were closer to some other lift U1 of U, then there would be points in the quadrilateral equidistant to two b b different lifts of U. All points equidistant to two different lifts of U lie on some totally b geodesic pieces of H3. Hence some geodesic must intersect γ but not the paths from b b p0 to p1, p0 to a0, or p1 to a1. This is impossible.

Now let H be the length of the paths from p0 to a0 and from p1 to a1. Let L be the length of γ. We know that H R and L 2R since the distance between any ≥ ≥ two components of the lift of Σ is greater than or equal to R. We let d be the length of the path between p0 and p1. Since D is embedded by the developing map, we can assume the quadrilateral is of the form shown in Figure 4.1.

p0 p1

a0 a1

Figure 4.1: Top edge of quadrilateral maps to shortest path on ∂U under the covering projection. b Then solving for d in terms of H and L, d = 2e−H sinh(L/2). Case 1: H = R. Then d 2e−R sinh(R) = 1 e−2R. ≥ − Case 2: H > R Then U must be tangent to itself, and not necessarily tangent to the tube about Σ. Then the image of D under the developing map must b contain some portion of the region above the geodesic surface equidistant from a horizontal plane and some other horosphere. Note if D were to contain all the points above this horosphere, then d 1. This occurs as long as any such point x satisﬁes ≥ dist(φ(x),∂U) < dist(∂U, Σ), or as long as H is large enough.

b b CHAPTER 4. ANALYZING BOUNDARY TERMS 51

A calculation shows as long as H log(2/√3), d 1. Since H > R, we can ≥ ≥ ensure this will occur by restricting to R log(2/√3). ≥ Corollary 19. If R log(2/√3), then Area(∂U) (1 e−2R) H(t), where H(t) is ≥ ≥ − the height of the torus at time t. b We will want to apply this corollary below. To do so, we obtain our ﬁrst condition on the tube radius R.

Condition 1. The tube radius R must be at least log(2/√3) 0.143841. ≈

4.1.2 The Scale Term (a2 + b2)

2 2 We now return to considering the 1-form ωl, and analyze how the scale term (a + b ) relates to the change in torus shape of ∂U. e By equations 4.1 and 4.2, we see that ωl has no effect on or contribution to the b change in structure in the radial direction of U. Hence we will disregard the first row e and column of the matrix for ωl and consider only the 2 2 matrix which gives the b × change in the longitude and meridional directions in U. Further, only the real part ofe the form ω gives the change in metric of the manifold. b The imaginary part encodes an infinitesimal rotation (see Hodgson and Kerckhoff [11]), whose effect is to keep the torus parabolic. Hence when considering the effect of ωl on the metric of the torus, we need only consider its real part. So in particular, we know from equation 4.1 that the change in metric of ∂U is encodede by the form: a b b A = − (4.4) " b a #

∂ The matrix A takes vectors in the space spanned by e2 = ∂m , a unit vector ∂ pointing in the meridional direction, and e2 = ∂` , an orthogonal vector pointing in the longitudinal direction. Multiplication by A gives a new vector, encoding the change in metric, or in other words, the change in shape of the torus. Now, note that A is traceless. Therefore, it is area preserving. Then it becomes simpler to evaluate the eﬀect of the matrix A on the normalized length of a longitude CHAPTER 4. ANALYZING BOUNDARY TERMS 52

than on actual length. Recall that if α is a geodesic representative of a slope s on the torus T , then we deﬁne the normalized length of s to be

Length(α) Normalized Length(s)= . Area(T ) p We are interested in determining how the normalized length of a slope changes over the deformation. Let L0 be the initial normalized length of a geodesic α on the torus ∂U. Let Lt be the normalized length of α at time t. We will assume without loss of generality that the area of the torus is 1, since the actual area doesn’t aﬀect the b scaled normalized length. Let v0 be a unit vector tangent to α. Under the deformation through cone manifolds, v0 will change with time, becoming the vector vt at time t. Then considering the matrix A, we know that to the ﬁrst order,

2 vt = v0 + Av0t + O(t )

1 a 2 b 2 = v0 + v − + v t + O(t ), " b # " a #! where v1 v0 = . " v2 # And therefore,

2 2 1 a 2 b 2 vt = v0 + 2v0 v − + v t + O(t ). k k k k · " b # " a #!

Thus when we consider the change in its length,

d 2 1 a 2 b ( vt ) t=0= 2v0 v − + v . dt k k | · " b # " a #!

d Writing vt t = v˙ , the above equation becomes: dt k k | =0 k k

2 v˙ v = 2vT Av , k k k 0k 0 0 CHAPTER 4. ANALYZING BOUNDARY TERMS 53

or, using the fact that v0 is a unit vector,

2 v˙ = 2vT Av . (4.5) k k 0 0

We diagonalize the matrix A, and ﬁnd A is similar to the matrix

λ 0 B = − , " 0 λ # where λ = √a2 + b2. Thus equation 4.5 becomes:

λ 0 v˙ = wT − w, k k " 0 λ # for some unit vector w, and we ﬁnd

v˙ √a2 + b2. k k ≥ −

This is true for every initial unit vector v0 along the geodesic α. Thus, since we are assuming the area of the torus is 1, the change in normalized length must satisfy

L˙ = v˙ = v˙ Lt, k k k k Zα for v˙ is constant along the curve α. k k So then, 2 2 L˙ √a + b Lt. (4.6) ≥ − The above result holds for any closed curve on the torus. We are particularly interested in the second shortest curve and how it changes. Note that at time t = 0, the cusp shape is rectangular and so the second shortest curve is perpendicular to the meridian. At time t, let pt denote the arc perpendicular to the meridian. For time t = 0, this arc will generally not be a closed curve. However, the second shortest 6 curve will always have length at least as long as the height of the torus, which is the CHAPTER 4. ANALYZING BOUNDARY TERMS 54

length of the arc pt.

Let ht be the height of the torus over time t, where the torus is still assumed to have unit area. So ht = `t(pt), where `t denotes the length at time t under the metric on the torus at time t. Then taking the derivative at time t = 0, we ﬁnd:

d h˙ = `˙(p )+ (` (pt)) t 0 dt 0 | =0

d The term (` (pt)) t gives the change in length on the initial torus of the arcs dt 0 | =0 pt. Note that at time t = 0, p0 is the closed curve perpendicular to the meridian.

As t changes, the endpoints of the arc pt will change on the initial torus. Because pt is perpendicular to the meridian of the initial torus at time t = 0, this will be a local minimum for the function `0(pt). Hence its derivative is zero. So then the above becomes:

h˙ = `˙(p0) which is just the change in normalized length of the second shortest curve on the initial torus. We plug this into equation 4.6 and ﬁnd that:

2 2 h˙ √a + b ht (4.7) ≥ −

The actual height H(t) of the torus from Corollary 19 is related to the normalized height ht by the equation H(t) ht = Area(∂U) q so solving for H(t) in terms of ht, and pluggingb into the results of Corollary 19, we obtain −2R Area(∂U) (1 e )ht, (4.8) ≥ − q provided Condition 1 holds, i.e. R log(2b /√3). ≥ CHAPTER 4. ANALYZING BOUNDARY TERMS 55

4.2 Analysis of the Boundary Term from the Sin- gular Locus

In this section, we analyze the boundary term bU (η0,η0) from the tubular neighborhood U of the singular locus. Recall that U is made up of components Uj, each of which is a tubular neighborhood of the j-th component of the singular locus. Since the singular locus corresponds to the crossing circles in the augmented link, there are a total of N(K) such components, where N(K) is the twist number of the knot K.

Let R be the tube radius of U. That is, for t > 0 we let Rj be the distance from ∂Uj to the component of the singular locus at the core of Uj, and we require

Rj = Ri = R for any i, j. We let m denote the length of a meridian on any of the tori ∂Uj. Here by “meridian” we mean a curve given by the intersection of ∂Uj with a plane cutting through Uj orthogonal to the singular locus when time t> 0. Notice that at time t > 0, the meridian curve about a component of the singular locus is the deformed slope we originally chose to fill on the cusp at time t = 0, for at time t = (2 π)2, when we have completed hyperbolic Dehn filling, the meridian bounds a ∗ disk in the filled manifold, hence must be our original slope.

Note that m depends on the size of the Uj: m = α sinh R, with α the cone angle. With this notation, we apply the following lemma, due to Hodgson and Kerckhoﬀ ([12] equation (17)).

Lemma 20 (Hodgson and Kerckhoﬀ).

Area(∂U) bU (η ,η ) 0 0 ≤ 8m4

Then using Lemma 20, we ﬁnd that

N(K) Area(∂U) Area(∂U ) b (η ,η ) j (4.9) U 0 0 8m4 8m4 ≤ ≤ j X The rest of this section will be devoted to proving: CHAPTER 4. ANALYZING BOUNDARY TERMS 56

Lemma 21. Let the tube radius R about the singular locus Σ be bounded below by some constant R1. Then

N(K) Area(∂U ) j C N(K) (4.10) 8m4 0 j ≤ X where C0 is some constant depending on R1.

Note: We will determine conditions on the lower bound on tube radius, R1, during the course of the proof, and add these to our list of conditions the tube radius must satisfy.

Proof. First, suppose we are ﬁlling only one component of U. The summation in equation 4.9 becomes simply Area(∂U) . 8m4 We let 8m4 Φ= A where A = Area(∂U). We want to bound Φ below by a constant. We deﬁne some notation. Let h be the height of the torus ∂U. So A = mh. Let ` be the length of the singular locus. Thus h = ` cosh(R). We let α denote the cone angle. Using this notation, note:

8m4 8m4 8m 8α sinh(R) Φ= = = m2 = α2 sinh2(R). A mh h ` cosh(R)

Now deﬁne α u = . ` Then 8α sinh(R) Φ= α2 sinh(R) = 8u tanh(R) α2 sinh2(R). (4.11) ` cosh(R) To compute our bounds, we now need bounds on the tube radius which were determined by Hodgson and Kerckhoﬀ [10]. Namely, we have the following results. CHAPTER 4. ANALYZING BOUNDARY TERMS 57

Proposition 22 (Hodgson, Kerckhoﬀ). Deﬁne the function Υ(R) by

tanh R Υ(R) = 3.3957 . 2 cosh(2R)

Then the tube radius R satisﬁes

R Υ−1(α`) ≥

when α` Υmax 1.019675 and R 0.531. ≤ ≈ ≥ To use this proposition below, we add a condition on the tube radius.

Condition 2. R 0.531. ≥ Note that if R satisﬁes Condition 2, it trivially satisﬁes Condition 1, so we need not concern ourselves anymore with Condition 1. We let ρ ρ = Υ−1(α`), for some ρ . Then we will ensure R ρ throughout 0 ≥ 0 ≥ 0 the deformation.

Let z = tanh(R), z0 = tanh(ρ0). We replace R in equation 4.11 by z.

z3 z3 Φ = 8 u α2 8 u α2 0 1 z2 ≥ 1 z2 − − 0 By Proposition 22, α` Υ(R). Writing Υ(R) in terms of z, we have ≤ z (1 z2) Υ(R) = 3.3957 − . 2(1+ z2)

Now consider 1 1 2(1+ z2) = = , α` Υ(ρ) 3.3957 z (1 z)2 − where z = tanh(ρ). e e e We deﬁne K(z) to be the function e 2(1+ z2) K(z)= 3.3957 z (1 z)2 − CHAPTER 4. ANALYZING BOUNDARY TERMS 58

The fact that Υ(ρ) is decreasing with ρ implies K(z) is increasing with ρ, hence K(z) is increasing with z. Thus: e z3 Φ = 8 u α2 1 z2 − 2(1+ z2) 3.3957 z = 8 u α2 z3 3.3957 z (1 z2) 2(1+ z2) − 3.3957 z = 8 u α2 z3K(z) 2(1+ z2) 1 3.3957z4 8 uα2 ≥ α` 2(1 + z2)

And so ﬁnally: 3.3957 z4 Φ 8 u2 . (4.12) ≥ 2(1+ z2) We now analyze what happens to u during the deformation. Hodgson and Kerck- hoﬀ found the following result (see [10] Proposition 5.6).

Proposition 23 (Hodgson, Kerckhoﬀ). For any smooth family of hyperbolic cone- manifolds whose initial tube radius is at least 0.531, the following diﬀerential inequality holds as long as z z0: ≥ du G(z) dt ≥ − where (1 + z2) G(z)= . 3.3957 z3

Then 2 2 (2π) (2π) 1+ z2 ∆u G(z)dt = dt ≥ − −3.3957z3 Zt=0 Z0 Also, at time t = 0, u = L2, where L is the normalized length of the curve along the crossing circle cusp which we are ﬁlling, i.e., the normalized length of the 1/n b b curve where n is the number of twists we’re putting back into the knot (see [10]). Hence, u2 (L2 +∆u)2 throughout the deformation, as long as (L2 +∆u) is ≥ b b CHAPTER 4. ANALYZING BOUNDARY TERMS 59

greater than or equal to zero, for

(τ)2 ∆u(τ) = G(t)dt | | − Z0 is increasing in τ. As long as (L2 +∆u) 0, we can put this with equation 4.12, we obtain: ≥

2 2 b (2π) 1+ z2 3.3957 z4 Φ 8 L2 (4.13) ≥ − 3.3957 z3 2(1+ z2) Z0 ! b If we choose L large so that R R for a ﬁxed constant R throughout the ≥ 1 1 2 deformation, then z z1 for z1 constant. Hence we obtain an estimate on (L +∆u): b ≥ 2 2 2 2 1+ z1 b L +∆u L (2π) 3 . (4.14) ≥ − 3.3957 z1 b b So to ensure that (L2 +∆u) remains greater than zero, we add the next condition on R. b Condition 3. The tube radius R must be at least R1 where R1 is large enough that the quantity on the right hand side of equation 4.14 is greater than 0, where z1 = tanh R1.

Assuming this condition, equation 4.13 above becomes:

1+ z2 2 3.3957 z4 Φ 8 L2 (2π)2 1 1 (4.15) ≥ − 3.3957 z3 2(1+ z2) 1 1 b This will be a constant, say equal to C0. Then for ﬁlling a single crossing circle component, the above implies: f Area(∂U) 1 1 4 = . 8m Φ ≤ C0

We deﬁne C0 = 1/C0. Now C0 is a constant dependingf on R1, which must satisfy conditions 2 and 3. f CHAPTER 4. ANALYZING BOUNDARY TERMS 60

Applying the above result to each crossing circle cusp component, we have:

N(K) Area(∂U ) j N(K) C 8m4 ≤ 0 j=1 X

4.3 Solving the diﬀerential inequality

In the previous two sections, we studied the boundary terms b b (ηl, ηl) and bU (η ,η ). − U 0 0 In this section, we will put the results of those two sections together. First, by equations 3.8, 4.3, 4.9, and 4.10, we obtain the estimeatee

2(a2 + b2)Area(∂U) C N(K), ≤ 0 b for C0 a constant, provided Conditions 2 and 3 hold. We can make C0 small. Taking the square root of both sides and dividing by 2 times the area, we obtain

C N(K) √a2 + b2 0 (4.16) ≤ s2 Area(∂U)

By equation 4.7, we have b 2 2 h˙ √a + b ht. ≥ − So putting this with equation 4.16,

C0 N(K) h˙ ht. ≥ − s2 Area(∂U)!

Finally, combining this with our area estimate,b equation 4.8 (assuming still Con- ditions 2 and 3, which imply Condition 1 on R) we obtain

C N(K)/2 h˙ 0 . (4.17) ≥ − 1 e−2R p − CHAPTER 4. ANALYZING BOUNDARY TERMS 61

We calculate: t t C N(K)/2 h˙ 0 , ≥ − 1 e−2R Z0 Z0 p − and so C0 N(K)/2 ht h t . − 0 ≥ − 1 e−2R p − Extending to t = (2π)2:

2 C0 N(K)/2 h 2 h (2π) . (4.18) (2π) ≥ 0 − 1 e−2R p −

We are interested in the normalized length Lt of the second shortest curve on the torus, rather than the normalized height of the torus. However, note that (for a torus of area 1) the second shortest curve is always at least as long as the height of the torus. In particular, L 2 h 2 . (2π) ≥ (2π) Also, the initial normalized height h0 is the normalized length of the closed curve perpendicular to the meridian in the initial cusp. Recall this is equal to L0: the normalized length of the longitude of the initial cusp corresponding to the link component in the projection plane, normalized so that the area is 1. We know by our analysis of this initial cusp from Chapter 2 that the actual length of this longitude is at least 2N(K), with a meridian of length 2. Hence its normalized length will be L N(K). 0 ≥ Thusp equation 4.18 becomes:

2 C0 N(K)/2 L 2 L (2π) . (4.19) (2π) ≥ 0 − 1 e−2R p −

Also, L(2π)2 is the normalized length of the second shortest curve on the cusp of the ﬁnal knot complement. Since the meridian of the ﬁnal cusp can be assumed to 2 be of length at least 1, (L(2π)2 ) will give a lower bound for the length of the second shortest curve on the torus ∂U. Thus if we denote the length of the second shortest curve on the torus ∂U by , b L b CHAPTER 4. ANALYZING BOUNDARY TERMS 62

then we have: 2 2 C0 N(K)/2 N(K) (2π) −2R L ≥ − p1 e ! p − or:

Proposition 24. Assume the tube radius R remains large enough throughout the deformation to ensure Conditions 1, 2, and 3 hold above. Then if is the length of L the second shortest curve on the cusp of S3 K, i.e. the cusp of the manifold resulting − from hyperbolic Dehn ﬁlling, then

2 C /2 N(K) 1 (2π)2 0 (4.20) L ≥ − 1 e−2R p− !

Note that the entire expression in parentheses in equation 4.20 is a constant, independent of the original knot complement. In the next chapter we ﬁnd values for R and this constant, concluding the proof of Theorem 1. Chapter 5

Finding Numerical Values for Constants

In this chapter, we find conditions which assure that our values of R are large enough that the calculations of the previous chapter can be made explicitly. We compute values of the constants obtained in the previous chapter, which concludes the proof of Theorems 1, 2, 3, and 4 introduced in the introduction. We need to ensure Conditions 1, 2, and 3 on the tube radius R hold, that is, that the tube radius stays large. By results of Hodgson and Kerckhoff [10], restated below, to ensure R stays large, we need to ensure the normalized lengths Li of the curves on the crossing circles which we fill stay large. Li determines how many twists must be b put back into each twist region for our constants above to work. To ensure our knot b cusp stays long enough that every non-trivial Dehn filling along it is hyperbolic, we will also need to find bounds below on the number of twist regions. This is all done in this section. L is related to z = tanh R by equation (47) from Hodgson and Kerckhoff’s paper [10]: b 3.3957 z t L2(1 z) exp F (w)dw ≥ 2 − − Z1 b

63 CHAPTER 5. FINDING NUMERICAL VALUES FOR CONSTANTS 64

Here (1+4z + 6z2 + z4) F (z)= − (z + 1)(1 + z2)2 Or: 2(2π)2 z1 L2 exp F (w)dw . (5.1) ≥ 3.3957(1 z ) − 1 Z1 Recall here that zb1 = tanh(R1), where R1 is a ﬁxed lower bound for the tube radius which guarantees our tube radius R satisﬁes Conditions 1, 2 and 3 from the previous chapter. Of these conditions, we will see that Conditions 1 and 2 become trivially true once we satisfy Condition 3. So we restrict our attention to Condition 3. For Condition 3, we needed R R , where R is large enough that the right hand ≥ 1 1 side of equation 4.14 is greater than zero. That is, we need

1+ z2 (L2 +∆u) L2 (2π)2 1 > 0 ≥ − 3.3957 z3 1 b b Putting this with the estimate on L2, equation 5.1, we need to ensure

2(2π)2 z1 b 1+ z2 exp F (w)dw (2π)2 1 = C > 0 (5.2) 3.3957(1 z ) − 3.3957 z3 1 − 1 Z1 1

We can ﬁx a value for R1 and compute whether equation 5.2 holds. If we start by

ﬁxing R1 = 0.531, the minimal value required by Condition 2, then the value of C1, the left hand side of equation 5.2 becomes negative. However, note it is increasing with R1. So we need R1 to be greater than 0.531.

Plugging in values for R1, we ﬁnd that C1, the left hand side of equation 5.2, is positive, equal to approximately 1.4621, when R1 is as low as 0.56. Now we can use these values to solve for the constants in Proposition 24.

Recall that C0 was equal to 1/C0, a formula for which was given in equation 4.15.

We substitute C1 into this equation. f 3.3957 z4 C = 8C2 1 0 1 2(1+ z2) 1 f CHAPTER 5. FINDING NUMERICAL VALUES FOR CONSTANTS 65

When R = 0.56, C0 is approximately 1.5368, giving C0 approximately 0.65068. Then we can put this value, as well as R> 0.56 into equation 4.20 of Proposition f 24. However, with this choice of R, the expression in parentheses on the right hand side of equation 4.20 becomes negative. Since we have the option of making R large, we increase R until this gives us a better bound. Using R = 1.0, we ﬁnd L2 must be greater than or equal to 144.45, and we obtain 2 2 2 (L +∆u) (102.861) 10580.54. Then C0 0.00003268 which implies ≥ ≈b ≤ b (0.8154)2N(K) . (5.3) ≤ L

Thus if we choose N(K) large enough and L2 on each crossing circle larger than 144.45, the second longest curve on the cusp of S3 K will have normalized length b − greater than or equal to √56.4696, so by [10], all nontrivial Dehn ﬁllings along K will produce a hyperbolic manifold. We need: (0.8154)2N(K) √56.4696 ≥ so this will happen if N(K)p 84.93, or in other words, if: ≥

N(K) 84. (5.4) ≥

So if we have at least 84 twist regions with enough twists per region to ensure L is long, and therefore R> 1.0 on each region, all ﬁllings except the trivial one will be b hyperbolic. We now analyze how many twists per region we must have to ensure that L2 ≥ 144.45. b

5.1 Computing number of twists per region

We know quite a bit about the shape of a crossing circle cusp. If no half twist is added at the crossing circle, we know its torus is rectangular with longitude length exactly 2. If a half twist is added, the result on the torus shape is a shearing. The CHAPTER 5. FINDING NUMERICAL VALUES FOR CONSTANTS 66

resulting torus still has longitude exactly 2, however rather than being rectangular its shape is as illustrated in Figure 5.1.

Figure 5.1: Left: Rectangular shape of a crossing circle cusp without a half twist. Right: Shape of a cusp in which a half twist has been added.

To put twists into the knot complement, we perform a 1/n ﬁlling along the crossing circle. That is, we take a curve that travels n times along the longitude and once along the meridian. Let m denote the length of the side edge of one of the rectangles in Figure 5.1, so in particular, m is the length of the meridian when no half twists are added, and √m2 + 1 is the length of the meridian in the half twist case. We know that m 1, but m could be very large. ≥ Then the 1/n curve on the crossing circle torus has length √m2 + 4n2 in the case of no half twists, and length m2 + (2n 1)2 when half twists are included. See − Figure 5.2. p

. . .

Figure 5.2: Top: Dashed line is the 1/n curve on a crossing circle without a half twist. Bottom: Dashed line is the 1/n curve on a crossing circle with a half twist.

In either case, the area of the torus is 2m. So in the case of no half twists, the CHAPTER 5. FINDING NUMERICAL VALUES FOR CONSTANTS 67

normalized length of the 1/n curve is:

L = √m2 + 4n2/√2m, and in the case of a half twist,b the normalized length is:

L = m2 + (2n 1)2/√2m. − p Note that for constant nb, in the first case L is at its minimum for m = 2n, and that minimum value is √2n. In the second case, L is minimized for m = 2n 1, with b − value √2n 1. So regardless of m, if we choose n large enough that √2n or √2n 1 − b − is greater than or equal to √144.45, our above results will go through. In either case, if n 73 then L √144.45. ≥ ≥ Hence our final results can be stated: b Theorem 25. If K is a hyperbolic knot with at least 84 twist regions and at least 73 twists per twist region, then any non-trivial Dehn filling on S3 K will result in a − manifold admitting a complete hyperbolic structure.

Note: The number 73 came from analysis of the worst possible case. In general, the number of required twists can be reduced if more is known about the knot diagram. Recall from Chapter 2 that associated to each flat augmented link is a circle packing. The circle packing gives the shape of the crossing circle cusps. In many cases, the meridian of these cusps is actually much shorter than in the worst case scenario. For example, all but two of the crossing circle cusps in a flat augmented link coming from a two-bridge knot have a meridian of length 2, and a longitude of length 2. In this case, the number of twists required to ensure the normalized length of the filling curve is at least as long as √144.45 is reduced to 12.

5.2 Additional Consequences

Agol [2] and Lackenby [13] have shown that if the length of a curve on a cusp in a hyperbolic manifold is greater than 6, then the manifold obtained by Dehn ﬁlling CHAPTER 5. FINDING NUMERICAL VALUES FOR CONSTANTS 68

along that curve is irreducible, with infinite word hyperbolic fundamental group. Using equation 5.3, we can find estimates on the number of twist regions required to guarantee all curves but the trivial curve on the knot have length greater than 6. As long as the tube radius R is at least 1.0, or in other words, as long as there are at least 73 twists per twist region in the most general case, the results of equation 5.3 hold. Then: 6 < (0.8154)2N(K) . ≤ L Or in other words, N(K) > 9.02 implies > 6. Thus we have: L Theorem 26. If K is a hyperbolic knot with at least 10 twist regions and at least 73 twists per region, then any non-trivial Dehn filling on S3 K will result in a manifold − that is irreducible with infinite word hyperbolic fundamental group.

Again the number of twists required per region can be greatly reduced if more information about the knot diagram is known. However, 73 twists per region is the most general result that we can prove. Chapter 6

Some Analysis of Results

In this chapter, we analyze the constants obtained in the previous chapter in comparison with other known results. We present an example showing how our numbers can be improved for certain known classes of knots, and suggest directions for future research.

6.1 Analysis of Constants for Dehn Filling

In the previous chapter, we saw that there are universal constants M and N such that if K is a hyperbolic knot with a prime, twist-reduced diagram with at least N twist regions and a twisting amount of at least M full twists per region, then any non-trivial Dehn filling on that knot is hyperbolic. The numbers we obtained for M and N are M = 73 and N = 84. If we are only interested in hyperbolike fillings, N is reduced to 10. Although this work is unique in that it links Dehn fillings to cusp shapes of the hyperbolic knot, there are many results on Dehn fillings which we present as a comparison to our results. First, a theorem of Lackenby [13] relates the number of twist regions of an alternating knot to the number of nontrivial Dehn fillings.

Theorem 27 (Lackenby). If K is an alternating knot with a prime, twist reduced diagram with at least 9 twist regions, then all non-trivial Dehn ﬁllings on S3 K are − 69 CHAPTER 6. SOME ANALYSIS OF RESULTS 70

hyperbolike.

Note Lackenby’s result makes no restriction on the twisting amount in each twist region. Thus an alternating knot with 9 twist regions but just a half-twist, or single crossing, in each twist region meets the criteria of his theorem. Although we would hope for such a result for general knots, i.e., no restriction on the twisting amount in each twist region of such knots, this will not be possible due to the following theorem, which has been observed for diﬀerent classes of knots by Eudave-Mu˜noz, Luecke, and others [5].

Theorem 28 (Eudave-Muñoz, Luecke). There are infinitely many knots which admit non-trivial Dehn fillings which are not hyperbolic. Moreover, the volume of such knot complements can be arbitrarily large.

Since the volume of the knots in Theorem 28 can be arbitrarily large, this implies the number of crossing circles in the augmented diagram of such a knot can also be taken to be arbitrarily large. Hence there are hyperbolic knots admitting non- trivial Dehn ﬁllings which have more than N twist regions, for any constant N. By Lackenby’s theorem, these knots are not alternating. As a consequence of the results in this dissertation, such knots must have twist regions with small twisting amounts. It is an interesting question to ask how small the twisting amount can be. We conjecture that the number 73, obtained in this paper, is unnecessarily large, and we hope to reduce it using topological techniques.

6.2 An Example: 2-Bridge Knots

In this section, we work through a particular class of knots, namely, 2-bridge knots, and show how our results can be applied to more specific classes of knots to obtain better constants. It is well known that a 2-bridge knot has a corresponding flat augmented link of the form in Figure 6.1 (see, for example [17]). This flat augmented link corresponds to a trivalent graph of the form shown in Figure 6.2. CHAPTER 6. SOME ANALYSIS OF RESULTS 71

···

Figure 6.1: The ﬂat augmented link of a 2-bridge knot

··· B1 C2 Cm

C1 ··· D Cm−1 B2

A ···

···

Figure 6.2: Trivalent graph associated with a 2-bridge knot

Note each region of the graph will give a circle in our circle packing, with tangencies across edges. Thus each circle will be tangent to the circles corresponding to regions A and D. The circle packing associated with this graph is of the form in Figure 6.3. Recall that the tangencies of circles correspond to pieces of the cusps. Those tangencies corresponding to crossing circle cusps have been marked with an x. Now we analyze the shapes of these crossing circle cusps and show that the twisting amount required at the twist region corresponding to these crossing circles is small.

Consider ﬁrst the cusps between circles B1 and A, and B2 and D. In both of these cases, conjugating the point of tangency to gives a square vertex shape, as ∞ in Figure 6.4. Recall that for a crossing circle cusp, these vertex shapes were glued in pairs, with a meridian running along a solid edge. Scaling this to have length 1, we ﬁnd a longitude has length exactly 2 in this case. CHAPTER 6. SOME ANALYSIS OF RESULTS 72

B D 1 C1

A . .

Cm B2

Figure 6.3: Circle packing associated to a 2-bridge knot

D C1

A Cm

Figure 6.4: Left: The vertex shapes corresponding to crossing circle cusps at circles B and B . Right: The cusp shape at these crossing circles is then 2 1. 1 2 × CHAPTER 6. SOME ANALYSIS OF RESULTS 73

At these crossing circles, for the tube radius to stay longer than 1.0 throughout a cone deformation, we need the normalized length of the 1/n curve to be longer than √144.45. Such curves have normalized lengths 1/2 + 2n2, so if n > 8.48 we have suﬃcient length. p Now consider the cusp shapes at the circles Ci in Figure 6.3. Conjugating these vertices to , we ﬁnd each is of the shape shown in Figure 6.5. ∞ A C0 = B1

Ci−1 D Ci+1 Cm+1 = B2

Figure 6.5: The shape of the vertex corresponding to crossing circle cusps at circles Ci, possibly with roles of A and D switched.

In this case, the cusp shape will be square: 2 2. Then the normalized length of × the 1/n curve on this cusp will be √1+ n2. For the tube radius to stay longer than 1.0 throughout a cone deformation, we need √1+ n2 > √144.45, or n> 11.98. Thus, putting this with equation (5.3) from Chapter 5, we find the second shortest curve on a 2-bridge knot has length at least (0.8154)2N(K) when there is a twisting amount of at least 12 in each twist region. This is significantly better than our previous result requiring a twisting amount of at least 73. In fact, we can also improve the constant (0.8154)2, since we can determine explicitly the shape of the cusp in the projection plane for the flat augmented link, which corresponds to the initial cusp shape of the deformation. To determine this initial cusp shape, we determine the shapes of the vertices that are not marked by x’s in Figure 6.3. There are four types of vertex shapes. These are illustrated in Figure 6.6. Since the dashed edge of any vertex above has length 1, we can compute the length of a longitude by adding the lengths of the vertices in the figure. There is just one vertex as in Figure 6.6 (a), of length N(K). There are N(K) 2 − of the form 6.6 (c), of length 2. The remaining vertices are of length 1, and there are CHAPTER 6. SOME ANALYSIS OF RESULTS 74

(a.) A

B C C Cm B 1 1 2 ··· 2

(b.) D A

A C1 D C2

B1 B2

(c.) A

Ci−1 D Ci+1

(d.)

A D

Ci+1

Figure 6.6: Four types of vertices of the ﬂat augmented link of a 2-bridge knot, which glue to give the initial cusp shape in the projection plane. CHAPTER 6. SOME ANALYSIS OF RESULTS 75

N(K) + 1 of these. Thus the total length of a longitude is N(K) + 2(N(K) 2) + − (N(K) + 1) = 4N(K) 3. − Then the normalized length of the second shortest curve in this particular case will initially be 2N(K) 3/2. We put this into equation 4.19 from Chapter 4, as − well as our boundpR> 1.0, and calculate as in Chapter 5. We ﬁnd that the length of the second shortest curve at the end of the deformation, is at least: L

( 2N(K) 3/2 0.18456 N(K))2. L ≥ − − p p This gives a better estimate on the shape of the cusp. Using this estimate in place of that of equation 5.3 in the slope length theorems for Dehn filling, we find the normalized length of the second shortest curve in cusp of the original knot complement is at least 7.515 as long as N(K) > 38.21. Thus the numbers in Theorem 25 for a 2-bridge knot become the following: For any 2-bridge knot with at least 39 twist regions and a twisting amount of at least 12 twists per region, any non-trivial Dehn filling is hyperbolic. Compare this with our original results of 84 and 73, respectively. For hyperbolike Dehn fillings, the number 39 twist regions is further reduced to 5. These Dehn filling results in this case do not give any additional information, since Dehn fillings on 2-bridge knots have been completely classified by Brittenham and Wu [4]. They have been able to show that any 2-bridge knot with at least 3 twist regions admits only hyperbolic non-trivial Dehn fillings. However, the methods of Brittenham and Wu are purely topological, and give no information on the geometry of the cusp shape of the 2-bridge knot. Additionally, although using the methods of this dissertation to obtain Dehn filling results for the 2-bridge case gives nothing new, this example does illustrate the potential of this method for giving information on Dehn fillings on other classes of knots with associated circle packings that are easy to understand. CHAPTER 6. SOME ANALYSIS OF RESULTS 76

6.3 Conclusions

In this dissertation, we have been able to relate purely combinatorial data of a knot diagram to specific geometric information on the knot. In particular, we have shown that the number of twist regions and the twisting amount per region gives estimates on the shape of the cusp of a hyperbolic knot. These results give geometric applications to problems concerning Dehn filling, through Hodgson and Kerckhoff’s 7.515-theorem and Agol and Lackenby’s 6-theorem. In particular, we showed that there are universal constants N and M such that if a hyperbolic knot has a diagram with N twist regions and a twisting amount of at least M twists per region, then any Dehn filling on the knot was hyperbolic or hyperbolike. Finally, these results have applications to the shapes of cusps of particular classes of knots, such as 2-bridge knots, and to Dehn fillings along those knots. Bibliography

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