Twist Knots and Thickenings by Whitney Klaryn George (Under the Direction of Gordana Matic)

Twist Knots and Thickenings

Whitney Klaryn George

(Under the Direction of Gordana Mati´c)

Abstract

We prove that any neighborhood N of a positive twist knot Km, odd m ≥ 3 with extremal Eu- ler number can be thickened to a standard neighborhood of a maximal Thurston-Bennequin

number Legendrian representative of Km.

Index words: Contact Structure, Twist Knots, Bypass, Dividing Set, Legendrian, Thickenings Twist Knots and Thickenings

Whitney Klaryn George

B.A., University of Oregon, 2004 M.S., University of Oregon, 2006

A Dissertation Submitted to the Graduate Faculty of The University of Georgia in Partial Fulﬁllment of the Requirements for the Degree

Doctor of Philosophy

Athens, Georgia

Whitney Klaryn George

Approved:

Major Professor: Gordana Mati´c

Committee: Jason Cantarella David Gay William H. Kazez Michael Usher

Electronic Version Approved:

Maureen Grasso Dean of the Graduate School The University of Georgia August 2012 Acknowledgments

I would like to dedicate this work to my parents: Marcia Dooley, Joe Dooley, Boyd George, and Rhonda George. I would never have made it this far without their patience, love, and faith in me. A special thank you goes to my advisor Dr. Gordana Mati´c.You’ve not only taught me about contact topology, but you’re also taught me how to be an independent mathematician. I appreciate your patience when I needed it most. I also appreciate you holding me to your high standards. I would also like to thank John Etynre and David Gay for their useful comments and guidance in writing my thesis. Lastly, I would like to thank the person who has supported me through the most chal- lenging time in my life: Alex Rice. I don’t know why you chose to stand by me throughout all of the ups and downs, but I’m glad you did. Thank you for believing in me when I didn’t. You are my rock and I can’t imagine life without you.

iv Contents

Acknowledgments iv

List of Tables vii

List of Figures viii

1 Historical Background and Future Plans 1

2 An Introduction to Contact Topology 4 2.1 Contact 3-Manifolds ...... 4 2.2 Legendrian knots ...... 5 2.3 Classiﬁcation of Legendrian Knots and Invariants ...... 6 2.4 Convex Surface Theory ...... 8 2.5 Neighborhoods of Legendrian Knots ...... 12 2.6 Relative Euler Class ...... 14 2.7 Twist Knots ...... 14

3 Main Result 17 3.1 Introduction ...... 17 3.2 Set-Up ...... 18 3.3 Possible Dividing Sets on the Diagrams D ...... 21

v 3.4 Curves γ and γ0 ...... 23 3.5 Proof of Main Theorem ...... 25 3.6 Base Case Diagrams ...... 27 3.7 The diﬀerent eﬀects of the curves γ and γ0 ...... 29 3.8 Null Case ...... 30 3.9 Horizontal Cases ...... 32 3.10 Vertical Cases ...... 37

Bibliography 52

A The base case conﬁgurations 54 A.1 Null Cases ...... 55 A.2 Horiztonal Cases ...... 59 A.3 Vertical Cases ...... 71

vi List of Tables

3.1 The bypass configurations from Figure 3.14 along with the type of bypass found in Figure 3.15 ...... 32 3.2 The bypass configurations from Figure 3.17 along with the type of bypass found in Figure 3.18. When the slope s ≥ 3, bypasses of type C begin to occur. See appendix for examples...... 33 3.3 The bypass configurations from Figure 3.20 along with the type of bypass found in Figure 3.21 ...... 35 3.4 The bypass configurations from Figure 3.24, along with the type of bypass found in Figure 3.25. For slopes s ≥ 1, bypasses of type B occur...... 39 3.5 The bypass configurations from Figure 3.27 along with the type of bypass found in Figure 3.28. In this example, (11,12,1) is also of type A. For slopes s ≥ 1, bypasses of type C occur...... 41 3.6 The bypass configurations from Figure 3.30 along with the type of bypass found in Figure 3.31. In this example, bypass configuration (3,4,5) is also of typeC...... 44 3.7 The bypass configurations from Figure 3.32 along with the type of bypass found in Figure 3.33 ...... 46

vii List of Figures

3 2.1 (R , ξstd)...... 5 2.2 A front projection of a Legendrian unknot (left) and trefoil (right) ...... 6 2.3 The moves that will turn a topological planar diagram of a knot K into a Legendrian front projection of K ...... 6 2.4 Positive Stabilization corresponds to adding two Up cusps and Negative Sta- bilization corresponds to adding two Down cusps ...... 8

2.5 Two intersecting surfaces S1 and S2 before and after rounding edges. . . . . 11 2.6 A bypass for a convex surface S...... 11

2.7 The Eﬀect of a Bypass attachment to ΓS ...... 12 2.8 The standard tessellation of the hyperbolic unit disk ...... 13 2.9 Let D be a meridional disk of a solid torus. Suppose tb(∂D) = −3. In other

words, |∂D ∩ ΓD| = 6. Up to rotation, there are two conﬁgurations for ΓD.

To the left, we see the χ(R+) − χ(R−) = 2. Exchanging + and - would give

χ(R+) − χ(R−) = −2. To the right, we see χ(R+) − χ(R−) = 0...... 15 2.10 The eﬀect of a bypass to the dividing set of a disk. Note that the relative Euler number remains the same...... 15

2.11 The Twist Knot Km. The box contains m right-handed half twists if m ≥ 0 and |m| left-handed half twists if m < 0...... 16

viii 3.1 Model of a twist knot Km and unknot B in the complement of Km. If m > 0, we have m right-handed half-twists in the box. If m < 0, we have m left- handed twists in the box...... 17 3.2 Unknot U in the complement of N. The gray torus is S3 − N(U)...... 19

3.3 To the left, disks Di inside Q. To the right, Pi with boundary components ci,j 20 3.4 To the left, a diagram D with n = 4. To the right, a new diagram D0 with n = 3 along with U 0 which is topologically isotopic to the original U..... 21

3.5 After an isotopy, we can assume that ΓPi can be identiﬁed with one of the following conﬁgurations. From the left we have the horizontal type, null type, and vertical type...... 22 3.6 Slope convention for diagrams ...... 23 3.7 To the left, D along with γ. To the right, γ0 is the image of this curve after m half-Dehn twists along the curve t...... 24

3.8 Visualizing how γ bounds a disk Dout on the outside of the sphere S..... 24

0 3.9 Visualizing how γ bounds a disk Din on the inside of the sphere S. Here, we have shown one positive twist from the twist knot...... 25 3.10 To the left: a diagram D with dividing curves twisting one full time. To the right: a diagram with dividing curves that don’t twist. The intersections with the green curve γ that appear on the left but not on the right will reduce n. 28

0 3.11 To the left: a diagram D for the twist knot K3. The curve γ twists 3/2 times

around the pair of pants P2. To the right: a diagram for the twist knot K5.

0 0 This γ twist 5/2 times around P2. By adding an extra full twist to γ , the

0 new intersections between γ and ΓD will result in reducing n...... 29 3.12 A vertical diagram with three vertical components. The complexity for this diagram is (1, 2) and the slope is -1. presented is a bypass conﬁguration that increases n...... 30

ix 3.13 Nulll Diagram with complexity (n, l) with n = 2l and l ≥ 2...... 30

3.14 A Null Diagram D for the twist knot K3 so that s(ΓD) = 1...... 31 3.15 Null Case Configurations, up to rotation. Since the bypasses come from γ0, we “dig” the bypass...... 31 3.16 To the left, a horizontal Diagram with complexity (n, l) with n = 2l + 1 and l ≥ 2. To the right, a horizontal Diagram with complexity (n, l) with n = 2l + k and l ≥ 2 ...... 33 3.17 An example of a diagram D with one horizontal component and which has slope1...... 34 3.18 The types of bypass configurations that arise in the base case diagrams from γ when diagram D has one horizontal component and s > 0.Since the bypasses come from γ, we attach the bypass...... 34 3.19 The types of bypass configurations that arise in the base case diagrams from γ0 when diagram D has one horizontal component and s < 0. Since the bypasses come from γ0, we “dig” the bypass...... 35 3.20 An example of a diagram with 2 horizontal components. The slope of the dividing curves is 1. We use the curve γ to find a bypass that will reduce n. 36 3.21 (Isotoped) bypass configurations, up to rotation that result from γ when the s(D) > 0. Since the bypasses come from γ, we “attach” the bypass...... 36 3.22 Types of bypass configurations, up to rotation, that arise from γ0 for diagrams with two or more horizontal components and negative slope. Since the bypasses come from γ0, we “dig” the bypass...... 37 3.23 Vertical type diagrams...... 38 3.24 A vertical diagram with n = 1 and 3 vertical components and the curve γ.. 39

x 3.25 The types of bypass conﬁgurations from γ that occur for diagrams with complexity (1, l), l ≥ 2, and slope s(D) ≥ 0. Since the bypasses come from γ, we “attach” the bypass. Type B will result in a disconnected dividing set on

S once ci,j are filled in by convex disks with any dividing set configuration. Therefore, the only allowable bypass configurations will thicken N...... 40 3.26 The types of bypass configurations from γ0 that occur for diagrams with complexity (1, l), l ≥ 2, and slope s(D) < 0. Since the bypasses come from γ0, we “dig” the bypass. Type B will result in a disconnected dividing set on

S once ci,j are filled in by convex disks with any dividing set configuration. Therefore, the only allowable bypass configurations will thicken N...... 40 3.27 A diagram D with 2 vertical components and n = 2. The slope of the dividing curves of D is 0. We use the curve γ to find a bypass that will either thicken N or reduce n...... 41 3.28 The types of bypasses that arise when diagram D has at least two vertical components and s(D) ≥ 0. These are the bypass configurations for γ. Since the bypasses come from γ, we “attach” the bypass...... 42 3.29 The types of bypasses that arise when diagram D has at least two vertical components and s(D) < 0. These are the bypass configurations for γ0. Since the bypasses come from γ0, we “dig” the bypass...... 42 3.30 An example of a vertical diagram with 1 vertical component and slope 1. We use the curve γ to find a bypass that will either reduce n or decrease the slope of D...... 44 3.31 The base case bypass configurations for diagrams D with complexity (2l−1, l)

and slope s > 0 that arise when γ intersects ΓD. Since the bypasses come from γ, we “attach” the bypass. Types A and C will reduce n. Type B will thicken N. Type D will result in decreasing the slope s(D)by1...... 45

xi 3.32 A diagram with 1 vertical component which has slope -1. We use the curve γ0 to show the bypasses will either reduce n or increase the slope. Here, the

underlying twist knot is K3...... 46 3.33 The types of bypass conﬁgurations from γ0 that occur for diagrams with complexity (n, l), n = 2l − 1, and slope s(D) < 0. Since the bypasses come from γ0, we “dig” the bypass. Types A and C will reduce n. Type B will thicken N. However, type D will increase the slope of D by 1...... 47 3.34 A diagram with 1 vertical strand and with an odd number of twists...... 48

3.35 The two possible ways to ﬁll in the disks bounded by c1,3 and c2,3. In both cases, we have a disconnected dividing set on S...... 48

3.36 Filling in the disks bounded by c1,2, c1,3, and c2,3 when m is even, the dividing set is not disconnected...... 49 3.37 A Diagram D with slope 5. The disk below represents the disk bounded by

γ, Dout, with a speciﬁc dividing set in which all curves are nested. With

this conﬁguration of dividing curves on Dout, the slope of D will decrease by

1 after passing through each bypass from Dout. After 5 consecutive bypass attachments, the slope of D will be 0...... 51

A.1 Base case null diagrams when 0 < s(D) < 8 along with the curve γ0 when

the underlying twist knot Km has an odd number of twists. In this Figure, the slopes are −8 < s < 0, excluding s = −4 (which results in a disconnected

0 dividing set on S. To see this, simply ﬁll in ci,j). We use γ to show that all

the possible boundary parallel dividing curves on Din result in a reduction of n. 55

xii A.2 Base case null diagrams when 0 < s(D) < 8 along with the curve γ0 when

the underlying twist knot Km has an even number or twists. In this Figure, the slopes are −8 < s < 0, excluding s = −4 (which results in a disconnected

0 dividing set on S. To see this, simply ﬁll in ci,j). We use γ to show that all

the possible boundary parallel dividing curves on Din result in a reduction of n...... 56 A.3 Base case null diagrams when −8 < s(D) < 0 along with the curve γ0 when

the underlying twist knot Km has an odd number or twists. Pictured is the

knot K3. In this Figure, the slopes are −8 < s < 0, excluding s = −4 (which

results in a disconnected dividing set on S. To see this, simply ﬁll in ci,j). We

0 use γ to show that all the possible boundary parallel dividing curves on Din result in a reduction of n...... 57 A.4 Base case null diagrams when −8 < s(D) < 0 along with the curve γ0 when

the underlying twist knot Km has an even number of twists. Pictured is the

knot K4. In this Figure, the slopes are −8 < s < 0, excluding s = −4 (which

results in a disconnected dividing set on S. To see this, simply ﬁll in ci,j). We

0 use γ to show that all the possible boundary parallel dividing curves on Din result in a reduction of n...... 58 A.5 Base Case Diagrams with one horizontal component. In this ﬁgure, 0 < s(D) < 8. We use γ to show that all boundary parallel dividing curves on

Dout will reduce n...... 59 A.6 Base Case Diagrams with one horizontal component. In this ﬁgure, s = 8, 9

. We use γ to show that all boundary parallel dividing curves on Dout will reduce n...... 60

xiii A.7 Base Case Diagrams with one horizontal component. In this ﬁgure, −7 <

0 s(D) < 0. We use γ , with the knot K3, to show that all boundary parallel

dividing curves on Dout will reduce n...... 61 A.8 Base Case Diagrams with one horizontal component. In this ﬁgure, s =

0 −8, −9. We use γ , with the knot K3, to show that all boundary parallel

dividing curves on Dout will reduce n...... 62 A.9 Base Case Diagrams with one horizontal component. In this ﬁgure, −7 < s <

0 0. We use γ , with the knot K4. to show that all boundary parallel dividing

curves on Din will reduce n...... 63 A.10 Base Case Diagrams with one horizontal component. In this ﬁgure, s =

0 −8, −9. We use γ , with the knot K4. to show that all boundary parallel

dividing curves on Din will reduce n...... 64 A.11 Base case horizontal diagrams with two or more horizontal components. In this Figure, the slopes are 0 < s < 6. We use γ to show that all the possible

boundary parallel dividing curves on Dout result in a reduction of n..... 65 A.12 Base case horizontal diagrams with two or more horizontal components. In this Figure, the slopes are 6 < s < 11. We use γ to show that all the possible

boundary parallel dividing curves on Dout result in a reduction of n..... 66 A.13 Base case horizontal diagrams with two or more horizontal components and

0 curve γ when the underlying twist knot Km has an odd number or twists.

Pictured is the knot K3 In this Figure, the slopes are −8 < s < 0, excluding s = −6 (which results in a disconnected dividing set on S. To see this, simply

0 ﬁll in ci,j). We use γ to show that all the possible boundary parallel dividing

curves on Din result in a reduction of n...... 67

xiv A.14 Base case horizontal diagrams with two or more horizontal components and

0 curve γ when the underlying twist knot Km has an odd number or twists.

Pictured is the knot K3. In this Figure, the slopes are −12 < s < −7. We

0 use γ to show that all the possible boundary parallel dividing curves on Din result in a reduction of n...... 68 A.15 Base case horizontal diagrams with two or more horizontal components and

0 curve γ when the underlying twist knot Km has an even number or twists.

Pictured is the knot K4. In this Figure, the slopes are −8 < s < 0, excluding s = −6 (which results in a disconnected dividing set on S. To see this, simply

0 ﬁll in ci,j).We use γ to show that all the possible boundary parallel dividing

curves on Din result in a reduction of n...... 69 A.16 Base case horizontal diagrams with two or more horizontal components and

0 curve γ when the underlying twist knot Km has an even number or twists.

Pictured is the knot K4. In this Figure, the slopes are −12 < s < −7. We

0 use γ to show that all the possible boundary parallel dividing curves on Din result in a reduction of n...... 70 A.17 Base case vertical type diagrams with one vertical component when the un-

derlying twist knot has an odd number of half-twists. Pictured here is K3. The slopes in these ﬁgures are −5 < s(D) < 0. We use γ0 to analyze the possible bypass conﬁgurations...... 71 A.18 Base case vertical type diagrams with one vertical component when the un-

derlying twist knot has an even number of half-twists. Pictured here is K4. The slopes in these ﬁgures are −5 < s(D) < 0. We use γ0 to analyze the possible bypass conﬁgurations...... 72

xv A.19 Base case vertical type diagrams with one vertical component. The slopes in these ﬁgures are 0 < s(D) < 5. We use γ to analyze the possible bypass conﬁgurations...... 73 A.20 To the left: base case diagrams of vertical types with complexity (2, 2) when

s ≥ 0. To the right, all possible boundary parallel dividing curves on Dout that correspond to the labels on γ on the left. The labels refer to Figures

3.28. Note, the disks do not show ΓDout ...... 74 A.21 To the left: base case diagrams of vertical types with complexity (2, 2) when

−3 < s < 0 and m = 3. Pictured here is K3. To the right, all possible

0 boundary parallel dividing curves on Din that correspond to the labels on γ

on the left. The labels refer to Figures 3.29. Note, the disks do not show ΓDin 75 A.22 To the left: base case diagrams of vertical types with complexity (2, 2) when −3 < s < 0 and m = 4 To the right, all possible boundary parallel dividing

0 curves on Din that correspond to the labels on γ on the left. Pictured here

is K4. The labels refer to Figures 3.29. Note, the disks do not show ΓDin .. 76 A.23 Base case diagrams of vertical types with complexity (1, 2) when 0 ≤ s ≤ 2.

We use γ to show all possible boundary parallel curve on Dout result in a thickening of N...... 77 A.24 Base case diagrams of vertical types when the underlying knot has an odd number of twists with diagram complexity (1, 2) when −2 < s < 0. Pictured

0 here is K3. We use γ to show all possible boundary parallel curve on Din result in a thickening of N...... 78 A.25 Base case diagrams of vertical types when the underlying knot has an even number of twists with diagram complexity (1, 2) when −2 < s < 0. Pictured

0 here is K4. We use γ to show all possible boundary parallel curve on Din result in a thickening of N...... 79

xvi Chapter 1

Historical Background and Future Plans

Contact topology is the study of odd dimensional manifolds equipped with a completely non- integrable plane field, ξ, called the contact structure. The field was introduced in 1896 by Sophus Lie, though certain aspects of the field can be traced to work by Christian Huygens in 1690 on geometric optics. In modern times, contact topology is often described as the odd dimensional sibling to symplectic geometry, and it is difficult to study one field without studying the other. Techniques from contact topology have been used to answer purely topological questions. One such example is the establishment of Property P for non-trivial knots by Peter Kronheimer and Tom Mrowka in 2004 [16]. While it certainly has applications to other areas of mathematics, contact topology stands on its own as a rich and interesting field. This work is focused towards knots and surfaces in 3-dimensional contact manifolds. One may ask that a knot K be everywhere transverse or everywhere tangent to the contact planes. Such knots are called transverse and Legendrian, respectively. If we were to also require that isotopies between two, say Legendrian, knots preserve the tangencies in the

1 intermediate stages, then we might have to rule out isotopies that worked in the general topological setting. It would then be reasonable to have two Legendrian knots that are topologically the same but not Legendrian isotopic. In fact, with even the simplest of knots, the unknot, this is precisely the case. There are 3 classical invariants that remain fixed under Legendrian isotopy: underlying topological knot type, rotation number, and Thurston-Bennequin number. If a Legendrian knot L can be identified up to Legendrian isotopy by these three invariants, then L is said to be simple. In 1995, Eliashberg and Fraser proved in [6] that Legendrian unknots are simple. These three invariants, however, are not enough for a complete classification. The first example of a non-simple Legendrian knot came in 1997 as a result of Chekanov, who showed that there exists two Legendrian 52 knots with rotation number 0 and Thurston- Bennequin number 1 that are not Legendrian isotopic. His proof utilized a combinatorial description of the theory of Contact Homology which Eliashberg was developing at the time. Since this discovery, new methods and techniques have been developed to help classify Legendrian knots up to Legendrian isotopy. For the interested reader, a running tabula- tion/classification for Legendrian knots, compiled by Ng and Chongchitmate, can be found in [3]. One important question is the determination of which knot types are simple and which ones are non-simple. The work for this thesis began as the study of twist knots and a knot invariant, developed by Etnyre and Honda in [7], called the Uniform Thickness Prop- erty (UTP). Roughly speaking, a knot K is uniformly thick if one is able to expand any neighborhood of K to a standard neighborhood (see Section 2.5),

Deﬁnition 1.0.1. A topological knot K is uniformly thick if every S1 × D2 representing K can be thickened to a standard neighborhood of a Legendrian knot L ∈ L(K) such that tb(K) = tb(L), where tb(K) is the maximal Thurston-Bennequin number for the knot K.1

1While this is not the original deﬁnition given in [7], the two deﬁnitions are equivalent. For more details,

2 The Uniform Thickness Property is a key ingredient in the following theorem which has been quite useful for identifying Legendrian simple knots.

Theorem 1.0.2. ([7],[17]) Let K be a knot type which is Legendrian simple and uniformly thick. Then, the (p,q)-cable of K is Legendrian simple, admits a classiﬁcation in terms of the classiﬁcation of K, and is uniformly thick.

In this same paper, [7], Etnyre and Honda showed that negative torus knots are uniformly thick, while positive torus knots fail to thicken. This result was later generalized to iterated torus knots by LaFountain [17] in his dissertation. The techniques used in the study of (iterated) torus knots and uniform thickness are specific to (iterated) torus knots. Therefore, new techniques are needed to study uniform thickness with respect to twist knots. The starting place to understanding the potential thickening of twist knots is a decomposition of S3 used in a paper by Etnyre, Ng, and Vértesititled Classification of Twist Knots [10]. Here we divided the problem into positive and negative twists knots, and in both cases there arose situations that resisted thickenings. However, an extra condition on the types of solid tori representing positive twist knots, namely extremal relative Euler number, allowed for us to show these solid tori thicken.

see [7].

3 Chapter 2

An Introduction to Contact Topology

2.1 Contact 3-Manifolds

A contact 3-manifold (M, ξ) is a 3-manifold M equipped with a smooth, completely nonin- tegrable two-plane distribution ξ called a contact structure. The contact structure ξ is given as the kernel of a non-degenerate 1-form α such that α ∧dα 6= 0 at each point of M. Locally,

3 all contact structures look like (R , ξstd), pictured in Figure 2.1, where ξstd = ker(dz − ydx), a result known as Darboux’s Theorem. Contact structures are either tight or overtwisted. We say the contact structure ξ is tight if there is no embedded disk D ⊂ M such that ξ is everywhere tangent to D along ∂D. If such a disk D exists, the contact structure ξ is said to be overtwisted. An example of an overtwisted

3 2 2 contact structure on R is given by the kernel of the one form αot = cos(πr )dz + sin(πr )dθ. 3 An example of a tight contact structure on R is ξstd, proven to be tight by Bennequin [1]. In 1971, Lutz and Marinet [19] showed that every closed orientable 3-manifold admits a contact structure. In the early 1990’s, Eliashberg showed that, up to isotopy, overtwisted contact structures are in one-to-one correspondence with homotopy classes of 2-plane ﬁelds

4 on M [5]. Two contact manifolds (M1, ξ1) and (M2, ξ2) are said to be contactomorphic if there is a diﬀeomorphism f : M1 → M2 such that f∗(ξ1) = ξ2.

3 Figure 2.1: (R , ξstd)

2.2 Legendrian knots

3 We are interested in a type of knot inside (R , ξstd) called a Legendrian knot, which is a knot 3 that is everywhere tangent to ξstd. In other words, if L is a knot in (R , ξstd) such that

TxL ⊂ ξx for all x ∈ L, then L is a Legendrian knot. If we are trying to visualize a Legendrian knot, it is often useful to do so in the front projection given by projecting the knot onto the xz−plane. Let L be

1 3 a Legendrian knot parametrized by the function φ : S → R sending θ 7→ (x(θ), y(θ), z(θ)). 1 2 Then, φxz : S → R sending θ 7→ (x(θ), z(θ)) parametrizes πxz(L). Front projections of Legendrian knots have no vertical tangencies, the only non-smooth points are generalized cusps, and at each crossing the slope of the over crossing is smaller

5 than that of the undercrossing. In fact, any planar S1 is the front projection for some Legendrian knot, provided it has no vertical tangencies and is immersed except at cusps with horizontal tangencies. This can be seen by recovering the y coordinate via the equation z0(θ) − y(θ)x0(θ) = 0.

Figure 2.2: A front projection of a Legendrian unknot (left) and trefoil (right)

Legendrian knots are not hard to ﬁnd. Any topological knot can be C0 approximated by a Legendrian knot. This is obtained by taking a planar diagram of any topological knot and turning the vertical tangencies into cusps, then adjusting the crossings so that the slope of the over crossing is smaller than that of the undercrossing. Refer to Figure 2.3

Figure 2.3: The moves that will turn a topological planar diagram of a knot K into a Legendrian front projection of K

2.3 Classiﬁcation of Legendrian Knots and Invariants

Legendrian knots are classiﬁed up to isotopy through Legendrian knots. We say two Legen-

drian knots L1 and L2 are Legendrian isotopic if there is a continuous family Lt, t ∈ [1, 2],

of Legendrian knots starting at L1 and ending at L2.

6 The three classical invariants of a Legendrian knot are underlying topological knot type, rotation number, and Thurston-Bennequin number. If a Legendrian knot L in the topological class of K can be identified up to Legendrian isotopy by these three invariants, then K is said to be Legendrian simple. These three invariants, however, are not enough for a complete classification of Legendrian knots up to Legendrian isotopy, which is to say not all knots are simple. While a formal definition of the Thurston-Bennequin number requires the introduction of a few concepts, it is useful to think of this invariant as measuring the amount of twisting of the contact planes along the knot L. Let ` be the line bundle given by ξx ∩ νx for all x ∈ L, where ν is the normal bundle of L. Let F be the framing given by the Seifert surface of L. The Thurston-Bennequin number is defined as tb(L) = tw(`, F), where tw(`, F) is the twisting of ` with respect to F. The other classical invariant is the rotation number. Let L = ∂Σ, where Σ is an embedded orientable surface. The contact planes restricted to Σ form a trivial two-dimensional bundle

2 which induces a trivialization ξ|L = L × R . Let v be a non-zero vector ﬁeld tangent to L pointing in the direction of the orientation on L. We may use this trivialization to think of

2 v as a path of non-zero vectors in R . The rotation number of L is deﬁned as the winding number of v, and is denoted r(L). Given a Legendrian knot L, we may use the front projection of L to compute the Thruston-Bennequin number and the rotation number. In the front projection, r(L) =

(1/2)(D − U) and tb(L) = writhe(πxz(L)) − 1/2(D + U). where D is the number down cusps and U is the number of up cusps. Refer to Figure 2.4 for an example of up and down cusps. If K is a topological knot type, let L(K) be the set of Legendrian representatives for K. If K ∈ L(K) and tb(K) = n, there always exists a K0 ∈ L(K) such that tb(K0) = n − 1. The Legendrian knot K0 is found by a process called Stabilization.

7 There are two types of stabilization, a positive and negative form denoted by S+ and

0 S−, respectively. In other words, K = S±(K). It is a natural question to ask if one can

0 0 destabilize: Given K, does there exist a K such that K = S±(K )? If such a knot exists, we say that K destabilizes.

S+

S−

Figure 2.4: Positive Stabilization corresponds to adding two Up cusps and Negative Stabi- lization corresponds to adding two Down cusps

Closely related to the Thurston-Bennequin number is a knot invariant called the maximal Thurson-Bennquin number, deﬁned by tb(K) = max tb(L). Bennequin proved the ﬁrst L∈L(K) upper bound on tb(K) in terms of the genus of K [1]. Rephrased by Eliashberg in [4], we have the following Bennequin inequality:

Theorem 2.3.1. Let (M, ξ) be a tight contact 3-manifold. Let L be a Legendrian knot in M with Seifert surface ΣL. Then, tb(L)+|rot(L)| ≤ −χ(ΣL).

2.4 Convex Surface Theory

The theory of convex surfaces in contact topology was developed by Giroux in [11], and has proven to be a very useful tool in the field. A surface S ⊂ M is convex if there exists a vector field v transverse to S whose flow preserves the contact structure ξ. Such a vector field v is called a contact vector field. Given a convex surface S, the dividing set is defined to be

ΓS := {x ∈ S|v(x) ∈ ξstd}. If S is closed, ΓS is a union of pairwise disjoint, smooth closed

8 curves called dividing curves. We will denote the number of curves in ΓS by |ΓS|. If S has boundary, then arcs that begin or end on ∂S are also allowed [11]. The union of dividing curves divide S into positive and negative regions R±. R+ ⊂ S is the set of points where the orientation of ξ agrees with the orientation of S. Similarly, R− is where the orienation of ξ disagrees with the orientation of S. If we have an ambient metric and v normal to S, the dividing set is where the contact planes are normal to the tangent planes on S. While it might seem that convex surfaces could be hard to ﬁnd, any surface S ⊂ M can be perturbed by a C∞-small isotopy to a convex surface [11]. The Legendrian realization principle is a criterion for when a collection of curves and arcs on a convex surface S can be made Legendrian. Let C be a collection of closed curves and arcs on a convex surface S with Legendrian boundary. The collection C is called nonisolating if all of the following hold:

1. C is transverse to ΓS

2. Every arc begins and ends on ΓS

3. Every component of S\(ΓS ∪ C) has a boundary component which intersects ΓS

An isotopy φs, s ∈ [0, 1], of a convex surface with contact vector ﬁeld v is called admissible if φs(S) is transverse to v for all s.

Theorem 2.4.1 (Legendrian Realization Principle (LeRP)). Consider a nonisolating collection of disjoint properly embedded closed curves and arcs, C, on a convex surface S with

Legendrain boundary. Then, there exists an admissible isotopy φs, s ∈ [0, 1], such that

(1.) φ0 = id

(2.) φs(S) are all convex

(3.) φ1(ΓS) = Γφ1(S)

9 (4.) φ1(C) is Legendrian

A corollary to this theorem was observed by Kanda [14]

Corollary 2.4.2. A closed curve C on Σ which is transverse to ΓΣ can be realized as a

Legendrian curve if C ∩ ΓΣ 6= ∅. Giroux’s Criterion allows one to determine when a convex surface S has a tight neighborhood. A tight neighborhood of a surface is one that supports a tight contact structure.

Theorem 2.4.3 (Giroux’s Criterion). If Σ 6= S2 is a convex surface (closed or compact with Legendrian boundary) in a contact manifold (M, ξ), then Σ has a tight neighborhood if and

2 only if ΓΣ has no homotopically trivial curves. If Σ = S , Σ has a tight neighborhood if and only if |ΓΣ| = 1.

In particular, when Σ is a torus, ΓΣ consists of 2n non-intersecting closed curves. We may assign a slope to the curves of Γ in the following way. Let γ be a closed curve in Γ that intersects the meridian of Σ p times and intersects the longitude q times. Then, we say γ has slope p/q.

Suppose we have two convex surfaces S1 and S2 with Legendrian boundary that intersect along a common Legendrian boundary curve. Honda described a process called Edge Rounding that allows one to describe the dividing curves of the new surface that is obtained from intersection [12]. Locally, the intersection of S1 and S2 is isomorphic to

2 2 2 {x + y ≤ } ⊂ R × R/Z with the contact 1-form α = sin(2πnz)dx + cos(2πnz)dy with + n ∈ Z . The two surfaces intersect along the curve x = y = 0. After rounding this common

edge, the dividing curves of S1 and S2 are connected by “turning to the right”, when viewed from the outside of the surface, which is enforced by the existence of a positive contact structure. Figure 2.4 seems to “round to the left” because the view is from the inside of the surface.

10 S2

Figure 2.5: Two intersecting surfaces S1 and S2 before and after rounding edges.

Bypasses

Introduced by Honda in [12], a bypass (see Figure 2.6) for a convex surface S consists of a Legendrian arc α connecting 3 dividing curves in S, a Legendrian arc β joining ∂α, and a convex half disk in M with boundary equal to α ∪ β which contains a single dividing curve.

α S

Figure 2.6: A bypass for a convex surface S.

We can think of forming a neighborhood of the original convex surface by adding a neighborhood of this half-disk to a contact product neighborhood of our surface. After isotoping the convex surface through this topological product neighborhood (but not contact product neighborhood), the dividing curves change as described in Figure 2.7.

11 Figure 2.7: The Eﬀect of a Bypass attachment to ΓS

Theorem 2.4.4 (Honda’s Bypass Attachment, [12]). Assume D is a bypass for a convex surface S. There exists a nieghborhodd of S ∪ D ⊂ M diﬀeomorphic to S × [0, 1] such that

Si = S × {i}, i = {0, 1} are convex. S × [0, ] is I−invariant, S = S × {}, and ΓS1 is obtained from ΓS0 by the operation found in Figure 2.7 While establishing the existence of bypasses requires work, they are usually quite abun- dant. Honda proved the following theorem [12]:

Theorem 2.4.5. Let Σ = D2 be a convex surface with Legendrian boundary inside a tight contact manifold and t(∂Σ, F rΣ) = −n < 0. Then, every component of ΓΣ is an arc which begins and ends on ∂Σ, and there exists a bypass along ∂Σ if tb(∂Σ) < −1.

Honda also showed the existence of trivial bypasses in Lemma 1.8 in [13], which are bypasses that do not change the dividing curve conﬁguration.

2.5 Neighborhoods of Legendrian Knots

We will make the assumption that our ambient manifold is S3 with the standard contact structure. Let L be a Legendrian knot. We call a neighborhood N(L) a standard neighborhood if N(L) satisiﬁes the following:

12 (i) ∂N(L) is convex

(ii) |Γ∂N(L)| = 2

(iii) The slopes of Γ∂N(L) = 1/tb(L)

We use the following convention for defining slopes on ∂N(L). The slope of the meridian is s = 0 and the slope of the longitude, given by the Seifert surface, is s = ∞. Looking near a Legendrian knot L, we can always find a standard neighborhood. By Giroux’s Criterion, we know that given any neighborhood N of a Legendrian knot with convex boundary (in a toght contact manifold, standard contact S3), the boundary must have no homotopically trivial curves, and hence must consist of an even number of parallel dividing curves. By cutting N along a convex meridional disk with Legendrian boundary on ∂N, if there are more than two dividing curves, we can find a bypass that is a part of this disk, as described in Theorem 2.4.5. The effect of attaching/digging a nontrivial bypass to ∂N when |Γ∂N | > 2 is a reduction of the number of dividing curves by 2. We can interpret the Bypass Attachment Theorem in terms of the standard Farey tessellation of the hyperbolic unit disk, H (Figure 2.8).

Figure 2.8: The standard tessellation of the hyperbolic unit disk

13 Theorem 2.5.1 (Honda, [12]). Let T be a torus with two dividing curves and let s =

0 slope(ΓT ). If a bypass is attached along a closed Legendrian curve of slope s , then the resulting slope s00 is obtained as follows: Let (s0, s) ⊂ ∂H be the counterclockwise interval from s0 to s. Then, s00 is the point on (s0, s) which is closest to s0 and has an edge to s.

2.6 Relative Euler Class

Let ξ be a tight contact structure on M so that ∂M is convex and ξ|∂M is trivializable. Choose

2 a nowhere zero section s of ξ on ∂M. The relative Euler class e(ξ, s) ∈ H (M, ∂M; Z) is deﬁned as: H2(∂M) → H2(M, ∂M) → H2(M) → H2(∂M) e(ξ, s) 7→ e(ξ) 7→ 0

Let S be a closed convex surface in M with positive and negative regions R+ and R−. Then,

1 2 he(ξ, S)i = χ(R+) − χ(R−). In particular, taking M to be a solid torus S × D and S to be a meridional disk with Legendrian boundary allows us to compute the relative Euler class of S1 × D2 [12]. An example is given in Figure 2.9. We say that a solid torus has extremal relative Euler class if a meridional disk has only boundary parallel dividing curves. If a solid torus has 2n dividing curves, then −n + 1 ≤ e(ξ, S) ≤ n − 1. If the relative Euler number is extreme, then |e(ξ, S)| = n − 1. It should be noted that the relative Euler number is well-deﬁned. If the meridional disk is isotoped, the dividing set of the disk is potentially changed by a bypass, but the relative Euler number is not aﬀected:

2.7 Twist Knots

A twist knot is a twisted Whitehead double of the unknot, specifically any knot of the form in Figure 2.11. In 2010, Etnyre, Ng, and Vértesigave a complete classification for Legendrian

14 +

+ - + +- -

Figure 2.9: Let D be a meridional disk of a solid torus. Suppose tb(∂D) = −3. In other words, |∂D ∩ ΓD| = 6. Up to rotation, there are two conﬁgurations for ΓD. To the left, we see the χ(R+) − χ(R−) = 2. Exchanging + and - would give χ(R+) − χ(R−) = −2. To the right, we see χ(R+) − χ(R−) = 0.

+ -+- + -

Figure 2.10: The eﬀect of a bypass to the dividing set of a disk. Note that the relative Euler number remains the same.

twist knots in [10]. The techniques used in the classiﬁcation involved convex surface theory, Legendrian ruling invariants, and Heegaard Floer homology. The classiﬁcation is summarized in Theorem 2.7

Theorem 2.7.1 (Etnyre, Ng, Vertesi). Let K = Km be a twist knot as in Figure 2.11 with m half-twists.

1. For m ≥ −2 even, there is a unique representative of Km with maximal Thurston- Bennequin number, tb=-m-1. This representative has rot =0, and all other Legendrian

15 Left-handed

m Right-handed

Figure 2.11: The Twist Knot Km. The box contains m right-handed half twists if m ≥ 0 and |m| left-handed half twists if m < 0.

knots of type Km destabilze to the one with maximal Thurston-Bennquin number.

2. For m ≥ 1 odd, there are exactly two representatives with maximal Thurston-Bennequin number, tb=-m-5. These representatives are distinguished by their rotation number, rot=±1 and negative stabilization of the rot=1 knot is isotopic to a postive stabilization of the rot=-1 knot. All other Legendrian knots destabilize to at least one of these two.

m + 1 3. For m ≤ −3 odd, K has − Legendrian representatives with (tb, rot) = (−3, 0). m 2 All other Legendrian knots destabilize to one of these. After any number of positive number of stabilizations (with a fixed number of positive and negative stabilizations), m + 1 these − representatives all become isotopic. 2 n2 4. For m ≤ −2 even with m = −2n, K has different Legendrian representatives m 2 with (tb, rot) = (1, 0). All other Legendrian knots destabilize to one of these. These lnm Legendrian knots fall into different Legendrian isotopy classes after any given 2 lnm positive number of positive stabilizations, and different Legendrian isotopy classes 2 after any given positive number of negative stabilization. After at least one positive and one negative stabilization (with a fixed number of each), the knots become Legendrian isotopic.

16 Chapter 3

Main Result

3.1 Introduction

In the proof of the classiﬁcation of twist knots found in [10], Etnyre, Ng, and V´ertesishow that in the complement of a standard neighborhood of a Legendrian twist knot Km for m 6= −1, there exists a Legendrian unknot U with tb(U) = −1 that is isotopic to the curve B found in Figure 3.1

Figure 3.1: Model of a twist knot Km and unknot B in the complement of Km. If m > 0, we have m right-handed half-twists in the box. If m < 0, we have m left-handed twists in the box.

The proof of the existence of U in [10] is by contradiction: they assume there exists some

Legendrian knot L in the topological class B, disjoint from Km, so that tb(L) = −n, n > 1,

17 is maximal for Legendrians in this class. By analyzing the dividing curve on two disks, Din

and Dout, in the complement of Km, the authors derive their contradiction. In the proof of Theorem 3.5.2, we use the same construction and disks used in [10].

However, the analysis of the dividing curves on Din and Dout is diﬀerent because we don’t

assume our neighborbood of Km is standard. This results in many more types of bypass

conﬁgurations. We use Din and Dout to understand the complement of a neighborhood of a

twist knot Km and prove the following:

Theorem 3.1.1 (Main Theorem). Every solid torus with extremal relative Euler number

representing a twist knot Km with odd m ≥ 3 can be thickened to a standard neighborhood of a maximal Thurston-Bennequin representative of Km. ∼ 1 2 For each N = S × D that represents Km, we have an associated convex surface, D, with Legendrian boundary which has an assigned complexity (n, l) (refer to Definition 3.2), where n, l ≥ 1 and n, l ∈ Z . We first argue that the complexity of a diagram can be reduced to (1, l) for l ≥ 1 and then use the disks Din and Dout to find thickenings for N. We thicken to reduce the complexity to (1, 1). Then we show that N is a standard neighborhood of a

Legendrian representative of Km. Lastly, we use results from [10] to thicken to a standard neighborhood of a maximal Thurston-Bennequin representative.

3.2 Set-Up

We use the decomposition of S3 following the work of Etnyre, Ng, and V´ertesiin [10]. Let

3 N be a solid torus representing the twist knot Km ⊂ (S , ξstd) so that ∂N is convex and

3 |Γ∂N | = 2l . Inside (S , ξst), there exists a Legendrian realization of an unknot, U, such that tb(U) = −n, n > 0, and U ∩ N = ∅ (refer to Figure 3.2). Take a standard neighborhood

of U, N(U), such that |Γ∂N(U)| = 2, s(Γ∂N(U)) = −1/n, and N(U) ∩ N = ∅.

18 Q

Figure 3.2: Unknot U in the complement of N. The gray torus is S3 − N(U).

3 Let Q = S − N(U). Then, ∂Q is convex and s(Γ∂Q) = −n. (Note that N ⊂ Q). Take two meridians, c1,1 and c2,1, on ∂Q that bound disks D1 and D2, respectively, which separate

∂Q into two cylinders, A1 and A2 so that the clasp of N is contained in D1 ∪ A2 ∪ D2 and the m twists of N are contained in D1 ∪ A1 ∪ D2. The curves c1,1 and c2,1 can be made

Legendrian by Corollary 2.4. Isotope D1 and D2 so that they are convex and each intersect

N in 2 convex disks bounded by Legendrian meridians of ∂N. Let ci,j = Di ∩ N for i = 1, 2 and j = 2, 3 (Refer to Figure 3.3). Deﬁne the pair of pants Pi = Di − N. We are now ready to deﬁne one of the key objects used in the proof of Theorem 3.5.2.

Deﬁnition 3.2.1. Given an embedded solid torus N representing Km and a Legendrian unknot U, as described above, an associated diagram D is the convex 4-punctured sphere with Legendrian boundary given by (after rounding corners)

D := P1 ∪ A1 ∪ P2

19 Q ci,3

D1 D2 ci,2

Pi ci,1

Figure 3.3: To the left, disks Di inside Q. To the right, Pi with boundary components ci,j

along with its dividing curves. Note that if we want to emphasize the underlying N and U,

U then we will denote D by DN We will use the diagram D to help us understand the complement of N in S3. It will serve as a record of the bypass attachments we will perform to try to thicken N. Let us introduce a complexity for the diagram D:

U Deﬁnition 3.2.2. The complexity of a diagram DN is a pair (n, l) ∈ Z × Z where 2n =

U Proposition 3.2.1. Let DN be a diagram with complexity (n, l). If D can be modiﬁed via bypass attachment/digging so that n is reduced by one, there exists a Legendrian unknot U 0 in the complement of N which is topologically isotopic to U and has larger tb.

Proof. Let U be a Legendrian unknot with tb(U) = −n. Then, the dividing curves on a

U corresponding diagram D will intersect ci,1 in 2n points (if the intersection is eﬃciently minimal, which we will always assume). Suppose the number of intersections can be reduced

0 0 0 to 2n − 2. Let U be a parallel copy of ci,1 so that U is disjoint from N. Note that U is in

20 the topological class of U and has larger tb than U. In particular, tb(U 0) = −(n − 1). Refer to Figure 3.4.

Figure 3.4: To the left, a diagram D with n = 4. To the right, a new diagram D0 with n = 3 along with U 0 which is topologically isotopic to the original U.

We will use this fact repeatedly in the proof of our main result to show the existence of a Legendrian unknot U with tb(U) = −1.

3.3 Possible Dividing Sets on the Diagrams D

Consider the dividing curves on A1 and Pi - they determine the dividing set on D after

rounding corners. For ΓA1 , recall that A1 ⊂ Q, where s(ΓQ) = −n. Therefore, ΓA1 consists of 2n non-intersecting arcs connecting c1,1 to c2,1. Next, we can assume that there are no boundary parallel curves on Pi. If there were, we could push Pi through the corresponding bypasses, reducing n, which contradicts the maximality of tb.

n > 2l, n = 2l, and n < 2l. In the case that n > 2l,ΓPi consists of 2l arcs joining ci,1 to ci,2,

2l arcs joining ci,1 to ci,3, and n − 2l arcs joining ci,1 to itself. Note that the arcs from ci,1 to itself must be between ci,2 and ci,3 so that there are no boundary parallel curves. When

n = 2l,ΓPi consists of 2l arcs joining ci,1 to ci,2 and 2l arcs joining ci,1 to ci,3. Lastly, when

21 n < 2l,ΓPi consists of n arcs from ci,1 to ci,2, n arcs from ci,1 to ci,3 and 2l − n arcs from ci,2 to ci,3.

After an isotopy of Pi, we can assume that there exists an annular collar, Ci,j, about all three ci,j for j = 1, 2, 3, so that in the complement of these collars in Pi, which we will

0 denote by Pi , the dividing set has one of the conﬁgurations found in Figure 3.5. For obvious reasons, we will call the case n ≥ 2l horizontal , the case n = 2l null, and the case n < 2l vertical.

Figure 3.5: After an isotopy, we can assume that ΓPi can be identiﬁed with one of the following conﬁgurations. From the left we have the horizontal type, null type, and vertical type.

0 0 Our next goal is to describe a type of coordinate system on ΓA where A = D − (P1 ∪ P2).

In other words, A consists of A1 along with the annular collars C1,1 and C2,1 of c1,1 and

1 1 c2,1 with corners rounded. Identify A with S × [1, 2] so that S × {1} is the boundary

0 1 0 component of C1,1 inside P1 and S ×{2} is the boundary component of C2,1 inside P2. Since

1 1 |ΓA ∩ (S × ∂[1, 2])| = 2n, there is a natural way to associate the length of S with 2n. In other words, view A = ([0, 2n] × [1, 2])/(0, t) ∼ (2n, t).

This coordinate system is well-deﬁned (use the longitude given by the Siefert surface) and allows us to deﬁne the slope of ΓA, accounting for the twisting inside Ci,1 that arises from the

above mentioned isotopy on Pi, as well as the twisting of ΓA1 . It should be noted that while

22 the slope of ΓA does depend on the slope of ΓA1 , the twisting from both P1 and P2 could give rise to a slope of ΓA that is much diﬀerent than the slope on ΓA1 . We will denote the slope of the dividing set Γ by s(Γ). If ΓA connects the points (i, 1) to (i + k, 2) for i ∈ [1, 2n], k ∈ Z, and i + k is identiﬁed with (i + k) mod 2n, then s(Γ) = k.

s = 0 s = 1

s = −1

Figure 3.6: Slope convention for diagrams

3.4 Curves γ and γ0

0 Deﬁne curves γ and γ on D as shown in Figure 3.7. Let S be the sphere given by D1 ∪A1 ∪D2 with corners rounded. Note D ⊂ S. Note that γ bounds a disk outside S and γ0 bounds a disk inside S which are both disjoint from N(Km). Call these disks Dout and Din, respectively.

Figure 3.8 and Figure 3.9 show how to visualize the disks Din and Dout. Legendrian realize

0 0 γ and γ , and make Din and Dout convex. If γ (or γ ), do not have maximal tb, ie if they intersect more than two dividing curves of D, then Theorem 2.4.5 shows that a boundary

parallel curve on Dout (or Din) corresponds to a bypass. Since the structure of ΓDout (or

ΓDin ) is not given to us a priori, we want to examine every possible boundary parallel curve

0 on Dout (or Din). On our diagram D, this corresponds to segments of γ (or γ ) that intersect

ΓD in 3 consecutive points. We will call such segments bypass conﬁgurations. It should be

23 noted that bypass conﬁgurations may or may not represent actual bypasses that exist in S3. For instance, a conﬁguration might lead to a disconnected dividing set and hence to a contradiction of tightness of S3, showing that such a bypass could not possibly exist.

Figure 3.7: To the left, D along with γ. To the right, γ0 is the image of this curve after m half-Dehn twists along the curve t.

Figure 3.8: Visualizing how γ bounds a disk Dout on the outside of the sphere S.

24 0 Figure 3.9: Visualizing how γ bounds a disk Din on the inside of the sphere S. Here, we have shown one positive twist from the twist knot.

3.5 Proof of Main Theorem

In this section, we sketch a proof our main theorem, Theorem 3.5.2, using results proven in later sections. Recall that a solid torus is said to have extremal relative Euler number if the dividing curves on a meridional disk are all boundary parallel . Before we prove the main theorem, we establish (modulo results proved later) the following lemma.

Lemma 3.5.1. Let N be a neighborhood of a twist knot Km for odd m ≥ 3 with extremal relative Euler number. There exists a Legendrian unknot U that is isotopic to B in Figure 3.1 with tb(U) = −1.

Proof. Let N be a neighborhood of the twist knot Km. We can assume that N is convex. Let U be a Legendrian unknot that is isotopic to B in Figure 3.1 which has maximal Thurston- Bennequin number within its topological knot class. If tb(U) = −n, we will show that the assumption n > 1 leads to a contradiction.

U Let DN be an associated diagram of complexity (n, l) with l ≥ 1, n > 1. If n = 2l, then our diagram is a null type diagram. In Section 3.8, we show that in all null type diagrams, n can be reduced, which is a contradiction. If n > 2l, then the diagram is a horizontal diagram. In Section 3.9, we show that in all horizontal type diagrams, n can be reduced,

25 which is a contradiction. If n < 2l the diagram is of vertical type. In Section 3.10, we show that in all vertical type diagrams, n can be reduced, which is a contradiction. We impose the extremal relative Euler number on vertical diagrams with one vertical component to achieve this result. Therefore, when n > 1, all the possible conﬁgurations will result in a reduction of n. Hence, there exists a maximal Thurston-Bennequin unknot isotopic to B in Figure 3.1. We are now ready to prove our main theorem:

Theorem 3.5.2 (Main Theorem). Every solid torus with extremal Euler number representing a twist knot Km with odd m ≥ 3 can be thickened to a standard neighborhood of a maximal

Thurston-Bennequin representative of Km.

Proof of Theorem 3.5.2. Let N be a neighborhood of the twist knot Km. We may assume that N is convex. By Lemma 3.5, there exists a Legendrian unknot U isotopic to B in Figure 3.1 with tb(U) = −1. The associated diagram for N and U is of vertical type with complexity (1, l) with l ≥ 1. If l ≥ 2, then by Proposition 3.10.1 there is a thickening of N.

We may repeat this process until l = 1. When l = 1, s(Γ∂N ) = 1/r for some r ∈ Z, thus the Legendrian divides also have slope 1/r. Let d be one such Legendrian divide.

We claim tb(d) = r. We follow methods found in [7]. Let CKm be the coordinate system on ∂N(Km), where the longitude (given by the Seifert surface) has slope ∞ and the meridian

0 has slope 0. Let CKm be the coordinate system on ∂N(d) where the meridian has slope 0 and slope ∞ is given by A ∩ ∂N(d) where A is an annulus who boundary lies on ∂N(d) such that (∂N(d))\A consists of two disjoint annuli Σ1 and Σ2 with A ∪ Σi, i = 1, 2, isotopic to

∂N(Km).

0 We calculate tb(d) by relating the two coordinate systems Cd and CKm , where Cd is the standard coordinate system described above. Let Σ(Km) be the Seifert surface for the knot Km. The Seifert Surface Σ(d) is obtained in the following way: Let ∂Σ(Km) be a

26 longitudinal curve of slope ∞ on ∂N. Take r parallel meridional disks whose boundaries lie on ∂N. Replace each point of intersection between the meridional disks and Σ(Km) by a

0 band, and the resulting surface is Σ(d). The framings from CKm and Cd diﬀers by r. In other words,

0 t(d, CKm ) + r = t(d, Cd) = tb(d)

0 Note that t(d, CKm ) = 0 which gives us the desired result. Push d inside N via the ﬂow of the contact vector ﬁeld ~v. Take a tubular neighborhood

−1 1 2 1 2 0 Nd of φ (d) inside S × D so that Nd ∩ {S × ∂D } = ∅. Let φ be a reparametrization of S1 × D2 ﬁxing the boundary so that φ0(S1 × {0}) = φ−1(d). Then, φ ◦ φ0 : S1 × D2 ,→ S3 is an embedded solid torus whose core curve, d, is Legendrian with tb(d) = n. Therefore, we have a standard neighborhood of a Legendrian representative of

Km. Lastly, we destabilize N by Theorem 2.7 to a maximal Thurston-Bennequin representative of Km.

3.6 Base Case Diagrams

In the following sections, we refer to a type of diagram as being base case. The idea behind base case diagrams is that if the dividing curves of D twist multiple times around the core curve, t, of A1, then most of the bypass configurations will behave the same. In a similar way, if the twist knot has more than 4 half twists, and the curve γ0 is used to find bypasses, then most of the bypass configurations will behave the same way. Figure 3.10 demonstrates such an example.

0 It is not hard to see that the interesting intersections between ΓD and either γ or γ occur when the diagram is, in some sense, minimal. The diagrams of twist knots with either 3 or

27 Figure 3.10: To the left: a diagram D with dividing curves twisting one full time. To the right: a diagram with dividing curves that don’t twist. The intersections with the green curve γ that appear on the left but not on the right will reduce n.

4 twists which have minimal twisting of their dividing curves and small complexity will be considered base case diagrams. A base case diagram D with complexity (n, l) is a diagram with slope −2n < s < 2n. Let D be a diagram with slope s. If s < 0, then it can be written as s = t(2n) + r with t ≤ 0 and −2n < r ≤ 0. If s > 0, then it can be written as s = t(2n) + r with t ≥ 0 and

0 ≤ r < 2n. When s > 0, we may think of the dividing curves from c1,1 to c2,1 as “looking” like a base case diagram with slope r with t extra right-handed Dehn twists about the core curve of A1. When s < 0, we may think of the dividing curves from c1,1 to c2,1 as “looking” like a base case diagram with slope r with t extra left-handed Dehn twists about the core curve of A1. A similar argument can be made for the the twisting of γ0, which is a direct consequence of the parity of the integer m. If m ≥ 3 is odd, then the diagrams with γ0 will “look” like the diagram when m = 3 with (m − 3)/2 right-handed Dehn twists about the core curve of

0 A1. If m ≥ 4 is even, then the diagrams with γ will “look” like the diagram when m = 4 with (m − 4)/2 right-handed Dehn twists about the core curve of A1. The types of bypass

28 conﬁgurations that arise in the generalized cases will all occur away from the pairs of pants and hence will reduce n. Figure 3.11 demonstrates such an example.

0 Figure 3.11: To the left: a diagram D for the twist knot K3. The curve γ twists 3/2 times 0 around the pair of pants P2. To the right: a diagram for the twist knot K5. This γ twist 0 0 5/2 times around P2. By adding an extra full twist to γ , the new intersections between γ and ΓD will result in reducing n.

3.7 The diﬀerent eﬀects of the curves γ and γ0

Throughout the proofs in the following sections, we specify one of the two curves γ or γ0. Recall that γ and γ0 bound disks that are disjoint from the neighborhood N of the twist knot Km. The curve γ bounds a disk Dout on the outside of the underlying 2-sphere S, while

0 γ bounds a disk Din on the inside of S. Therefore, the resulting bypasses from the two disks

0 Din and Dout are dug and attached, respectively. The choice of γ or γ depends on the slopes of the dividing curves for a given diagram. In some cases, using one curve will not result in a reduction of n nor a thickening of N, but the use of the other will. Figure 3.12 shows a bypass conﬁguration from γ on a vertical diagram with a slope of -1 that will increase n. This is not the desired eﬀect , and so we use the curve γ0.

29 Figure 3.12: A vertical diagram with three vertical components. The complexity for this diagram is (1, 2) and the slope is -1. presented is a bypass conﬁguration that increases n.

3.8 Null Case

In this section, we show that n can always be reduced by ﬁnding bypasses from the curve γ0 for null type diagrams. Recall that a null type diagram is one in which the pair of pants has dividing curves of the form in Figure 3.13.

Figure 3.13: Nulll Diagram with complexity (n, l) with n = 2l and l ≥ 2

Example 3.8.1. We present an example to demonstrate the types of bypasses that arise on

0 null type diagrams from γ . Figure 3.14 shows a null diagram D for K3 with s(D) = 1. We

0 label the intersection points of ΓD and γ from 1 to 22. To three consecutive points along γ0, say i, i + 1, i + 2, we associate a bypass conﬁguration. We label this conﬁguration as the

30 tuple (i, i + 1, i + 2). Table 3.1 show the correspondence between the bypass conﬁgurations and the type of bypasses that arise.

22 5 14 21 6 13 20 7 12

19 1 8 2 18 9 3 17 10 11 4 16 15

Figure 3.14: A Null Diagram D for the twist knot K3 so that s(ΓD) = 1.

ABCD

Figure 3.15: Null Case Conﬁgurations, up to rotation. Since the bypasses come from γ0, we “dig” the bypass.

U Proposition 3.8.1. Let N be a neighborhood of a twist knot Km for m ≥ 3. Let DN be an associated diagram for N with complexity (n, l) = (2l, l), l ≥ 2. Any bypass conﬁguration from γ0 will reduce n.

U Proof. We use the base cases when the diagram DN is of null type with complexity (n, l) = (4, 2), 0 < |s| < 8, and m = 3, 4. The complexity (4, 2) is the smallest complexity in this case that fails to represent a maximal unknot U and standard nieghorhood N of a Legendrian twist knot. Refer to Figures A.1 - A.4 for the base case conﬁgurations along with γ0. We make

the observation that when s = 0 mod n,ΓS is disconnected which contradicts the tightness of S3. From these base cases we make generalizations to null diagrams with complexity (n, l) with n > 4 and l > 2, |s| > 8, or m > 4.

31 Table 3.1: The bypass conﬁgurations from Figure 3.14 along with the type of bypass found in Figure 3.15 Bypass Type Bypass Type (22,1,2) D (11,12,13) D (1,2,3) A (12,13,14) A (2,3,4) A (13,14,15) A (3,4,5) B (14,15,16) B (4,5,6) C (15,16,17) C (5,6,7) A (16,17,18) A (6,7,8) C (17,18,19) A (7,8,9) B (18,19,20) B (8,9,10) A (19,20,21) C (9,10,11) A (20,21,22) A (10,11,12) D (21,22,1) D

The bypass configurations that occur from γ0 in the base cases are summarized in Figure 3.15. It is easily checked that all the bypass configurations that occur will reduce n. The base cases are generalized to all null type diagrams by increasing n (and hence also l), |s|, and m. The new bypass configurations that arise from the generalizations are the same type as in the base cases (there are just more of them). Therefore, all null type diagrams will reduce n.

3.9 Horizontal Cases

In this section, we show that n can be reduced for horizontal diagrams. Recall that a horizontal diagram is one such that the dividing curves of D are like those in Figure 3.16. We work with two cases: the first is when there is exactly one horizontal component and the second is when there are more than one horizontal components. The two cases are treated differently because we use different methods for finding the necessary bypasses for the reduction of n. When there is one horizontal component and the slope of the dividing

32 Figure 3.16: To the left, a horizontal Diagram with complexity (n, l) with n = 2l + 1 and l ≥ 2. To the right, a horizontal Diagram with complexity (n, l) with n = 2l + k and l ≥ 2

curves for the diagram are positive, we use the curve γ to ﬁnd bypasses. When the slope is negative, we use the curve γ0. On the other hand, when we have more than one horizontal component, we can get away with using only the curve γ0 to ﬁnd bypasses. An example of each case is provided before each proof. For the interested reader, more detailed illustrations of the base cases are provided in the appendix.

Example 3.9.1. We present an example of a diagram D that has one horizontal arc. Here, the slope of ΓD is 1 (refer to Figure 3.17). The intersections between γ and ΓD are labeled 1-14. Table 3.2 shows the type of bypass to which each bypass conﬁgurations corresponds. The type of bypass, A-E, is shown in Figure 3.18.

Table 3.2: The bypass conﬁgurations from Figure 3.17 along with the type of bypass found in Figure 3.18. When the slope s ≥ 3, bypasses of type C begin to occur. See appendix for examples. Bypass Type Bypass Type (14,1,2) B (7,8,9) B (1,2,3) A (8,9,10) A (2,3,4) A (9,10,11) A (3,4,5) B (10,11,12) B (4,5,6) D (11,12,13) E (5,6,7) E (12,13,14) E (6,7,8) E (13,14,1) E

33 12 13 11 2

10 9 3 4 5

6 7

Figure 3.17: An example of a diagram D with one horizontal component and which has slope 1.

ABCDE

Figure 3.18: The types of bypass conﬁgurations that arise in the base case diagrams from γ when diagram D has one horizontal component and s > 0.Since the bypasses come from γ, we attach the bypass.

U Proposition 3.9.1. Let N be a neighborhood of a twist knot, Km for m ≥ 3. Let DN be an associated diagram for N with complexity (n, l) with l ≥ 2 and n = 2l + 1. When s(D) > 0, the bypass conﬁgurations from γ will reduce n. When s(D) < 0, the bypass conﬁgurations from γ0 will reduce n.

Proof. We rule out the slopes s that are s = 0 mod n because these slopes result in a disconnected dividing set on S which contradicts the tightness of S3. We analyze in detail

U the base cases when the diagram DN has one horizontal component with complexity (5, 2) and slope 0 < |s| < 10. The complexity (5, 2) is the smallest complexity in this case that

34 ABCDE

Figure 3.19: The types of bypass conﬁgurations that arise in the base case diagrams from γ0 when diagram D has one horizontal component and s < 0. Since the bypasses come from γ0, we “dig” the bypass.

fails to represent a maximal unknot U and standard nieghorhood N of a Legendrian twist knot. Refer to Figures A.5- A.10 for the base cases. We summarize the bypass configurations found from these images in Figures 3.18 and 3.19. All the bypass configurations will reduce n. We generalize the base cases by increasing l (and hence n). By doing so, the bypass configurations are the same as those found in the base cases. However, we have more bypass configurations of type E.

Example 3.9.2. In this next example, we have a diagram with two horizontal components which has slope 1, as shown in Figure 3.20. We use the curve γ and investigate all possible bypass attachments. Table 3.3 shows the bypass conﬁgurations along with the type of bypass. The types of bypasses refer to those in Figure 3.21

Table 3.3: The bypass conﬁgurations from Figure 3.20 along with the type of bypass found in Figure 3.21 Bypass Type Bypass Type (16,1,2) D (8,9,10) D (1,2,3) B (9,10,11) B (2,3,4) B (10,11,12) D (3,4,5) C (11,12,13) B (4,5,6) D (12,13,14) D (5,6,7) E (13,14,15) A (6,7,8) A (14,15,16) A (7,8,9) A (15,16,1) A

35 14 15 13 2

1 12 3 11 4 10 5 6 7 8

Figure 3.20: An example of a diagram with 2 horizontal components. The slope of the dividing curves is 1. We use the curve γ to ﬁnd a bypass that will reduce n.

ABCDE

Figure 3.21: (Isotoped) bypass conﬁgurations, up to rotation that result from γ when the s(D) > 0. Since the bypasses come from γ, we “attach” the bypass.

U Proposition 3.9.2. Let N be a neighborhood of a twist knot, Km for m ≥ 3. Let DN be an associated diagram for N with complexity (n, l), l ≥ 2 and n = 2l + k, k ≥ 2. Every allowable bypass conﬁguration from γ0 will reduce n.

Proof. We first rule out the slopes on D which cannot occur. Note that if s = 0, we have a disconnected dividing set on S which contradicts the tightness of S3. Next, suppose s(D) > 0. By examining base cases in Figures A.11 and A.12 in the Appendix, we see every bypass configuration from γ can be isotoped away from the pairs of pants Pi to one of the configurations found in Figure 3.21. These bypass configurations will all reduce n.

36 ABC

DEF

Figure 3.22: Types of bypass conﬁgurations, up to rotation, that arise from γ0 for diagrams with two or more horizontal components and negative slope. Since the bypasses come from γ0, we “dig” the bypass.

We use the base cases when diagram D has complexity (2l + k, l) for k ≥ 2, l ≥ 2, and −11 < s < 0. Refer to Figures A.13- A.16 for images of the base cases along with the curve γ0 when m = 3, 4. The types of bypass configurations that occur from γ0 in the base cases are summarized in Figure 3.22. It is easily checked that the bypass configurations A-F in Figure 3.22 reduce n. We generalize the base cases to diagrams of horizontal type with a bigger complexity than the base cases, with slope 11 < |s|, and for twist knots Km with m > 4. We find that the new bypass configurations that result from the generalizations are the same as those found in Figure 3.22.

3.10 Vertical Cases

This section serves several purposes and is centered around vertical diagrams. Recall that a vertical diagram is one of the form given in Figure 3.23. The ﬁrst purpose is to show that once n has been reduced to 1, we can ﬁnd a thickening of the underlying twist knot until there is only one vertical component. When there is only one vertical component and n = 1, the underlying neighborhood of the twist knot represents a neighborhood of a Legendrian

37 Figure 3.23: Vertical type diagrams.

twist knot. We handle this situation in Proposition 3.10.1. The second purpose of this section, using the remaining propositions, is similar to the previous sections: when n 6= 1, we can find a bypass which will reduce n. It should be noted that, for the first time, we need to use the maximal relative Euler number condition imposed on our solid tori. Before each case, an example is provided to the reader. For more examples (of the base cases), refer to the appendix. The first example and proposition involve the types of diagrams which have n = 1.

Example 3.10.1. Consider a Vertical diagram when n = 1. In particular, consider the diagram D with complexity (1, 2). This means that there are two intersections between the dividing curves of Pi and the boundary component ci,1, and there are three vertical strands between ci,2 and ci,3. In Figure 3.24, the intersection points of ΓD and γ are labeled 1-14. Table 3.4 indicates the type of each bypass conﬁguration, where the types are shown in Figure 3.25.

38 14

11 1213 123 8 9 10 456

Figure 3.24: A vertical diagram with n = 1 and 3 vertical components and the curve γ.

Table 3.4: The bypass conﬁgurations from Figure 3.24, along with the type of bypass found in Figure 3.25. For slopes s ≥ 1, bypasses of type B occur. Bypass Type Bypass Type (14,1,2) A (7,8,9) A (1,2,3) A (8,9,10) A (2,3,4) A (9,10,11) A (3,4,5) A (10,11,12) A (4,5,6) A (11,12,13) A (5,6,7) A (12,13,14) A (6,7,8) C (13,14,1) C

39 ABC

Figure 3.25: The types of bypass configurations from γ that occur for diagrams with complexity (1, l), l ≥ 2, and slope s(D) ≥ 0. Since the bypasses come from γ, we “attach” the bypass. Type B will result in a disconnected dividing set on S once ci,j are filled in by convex disks with any dividing set configuration. Therefore, the only allowable bypass configurations will thicken N.

ABC

Figure 3.26: The types of bypass configurations from γ0 that occur for diagrams with complexity (1, l), l ≥ 2, and slope s(D) < 0. Since the bypasses come from γ0, we “dig” the bypass. Type B will result in a disconnected dividing set on S once ci,j are filled in by convex disks with any dividing set configuration. Therefore, the only allowable bypass configurations will thicken N.

U Proposition 3.10.1. Let N be a neighborhood of a twist knot Km for m ≥ 3. Let DN be an associated diagram for N with complexity (n, l) with n = 1 and l ≥ 2. When s ≥ 0, every allowable bypass conﬁguration from γ will thicken N(Km). When s < 0, every allowable

0 bypass conﬁguration from γ will thicken N(Km).

Proof. We use the base case when D has complexity (1, 2). The complexity (1, 2) is the smallest complexity in this case that fails to represent a maximal unknot U and standard nieghorhood N of a Legendrian twist knot. Refer to Figures A.23-A.25. We summarize the type of bypass conﬁgurations that occur from γ and γ0 in Figure 3.25 and 3.26. Note that type B is disallowed as it would result in a disconnected dividing set on S which contradicts

40 the tightness of S3. Therefore, the only allowable bypass conﬁgurations will result in a thickening of N. When we generalize to diagrams with complexity (1, l) with l > 2, we see the same type of bypass conﬁgurations as in the base cases. The next example and proposition involve the type of vertical diagram with more than

one vertical component and more than 2 intersections of the dividing curves with ci,1.

Example 3.10.2. The next example is for a vertical diagram with n > 1 and more than one vertical component. In particular, n = 2 and l = 2, as shown in Figure 3.27. We have

labeled the intersection points between γ and ΓD 1-12. Next, we indicate the type of each bypass conﬁguration in Table 3.5

10 12 9 7 8 34

5 6

Figure 3.27: A diagram D with 2 vertical components and n = 2. The slope of the dividing curves of D is 0. We use the curve γ to ﬁnd a bypass that will either thicken N or reduce n.

Table 3.5: The bypass conﬁgurations from Figure 3.27 along with the type of bypass found in Figure 3.28. In this example, (11,12,1) is also of type A. For slopes s ≥ 1, bypasses of type C occur. Bypass Type Bypass Type (12,1,2) D (6,7,8) D (1,2,3) D (7,8,9) D (2,3,4) D (8,9,10) D (3,4,5) D (9,10,11) D (4,5,6) B (10,11,12) B (5,6,7) B (11,12,1) B

41 BA C D

Figure 3.28: The types of bypasses that arise when diagram D has at least two vertical components and s(D) ≥ 0. These are the bypass conﬁgurations for γ. Since the bypasses come from γ, we “attach” the bypass.

BA C D

Figure 3.29: The types of bypasses that arise when diagram D has at least two vertical components and s(D) < 0. These are the bypass conﬁgurations for γ0. Since the bypasses come from γ0, we “dig” the bypass.

U Proposition 3.10.2. Let N be a neighborhood of a twist knot Km for m ≥ 3. Let DN be an associated diagram for N with complexity (n, l) with n ≥ 2 and 2l > n + 1. In other words, the diagram has more than 1 vertical component. When s ≥ 0, all boundary parallel dividing curves on Dout will reduce n or thicken N. When s < 0, all boundary parallel dividing curves on Din will reduce n or thicken N.

U Proof. We use the base case when the diagram DN is of vertical type with complexity (n, l) = (2, 2), 0 ≤ |s| ≤ 3, and m = 3, 4. The complexity (2, 2) is the smallest complexity in this case that fails to represent a maximal unknot U and standard nieghorhood N of a Legendrian twist knot. From these base cases we make generalizations to vertical diagrams with complexity tuple (n, l)D with n > 2 and l ≥ 2, |s| > 3, or m > 4. Assume diagram D is a base case. These ﬁgures can be found in the appendix, speciﬁcally

42 Figures A.20- A.22. When s(D) ≥ 0, we use γ to either thicken or reduce n. When s(D) < 0, we use γ0 to either thicken or reduce n. Figures 3.29 and 3.28 show the 8 types of bypass conﬁgurations that arise from either γ or γ0. Of these 8 cases, type D and B could result in only a thickening and not a reduction of n.

We generalize the base cases by (1) increasing only n (by adding dividing arcs from c1,1

to c2,1), (2) increasing both n and l (by adding vertical components between ci,2 and ci,3), (3) increasing m, and (4) increasing |s|. When only n is increased, the bypass conﬁgurations are the same as those found in Figure 3.28 and 3.29 with the exception that γ or γ0 will

now intersect three dividing curves that originate from the same ci,j, resulting in a bypass conﬁguration of type A. This bypass conﬁguration will result in a reduction of n. When

both n and l are increased by adding vertical components between ci,2 and ci,3, the new bypass conﬁgurations are of type D, resulting in a thickening. When m or |s| is increased, there are more bypass conﬁgurations of type A and C, which will reduce n. The next two examples and two propositions are for vertical diagrams with only one vertical component. This is the case where we must assume the solid torus representing the twist knot has extremal relative Euler number. We use this assumption in the case that the slope of the diagram is 0, which is saved for the last proposition, Proposition 3.10.5.

Example 3.10.3. Take the diagram D with one vertical component and n = 2 so that the slope of ΓD is 1, as shown in Figure 3.30. We label the 14 intersection points between the dividing curves of Pi and γ, 1-14. Next, we classify each type of bypass in Table 3.6. The types of bypasses refer to the ones shown in Figure 3.31.

43 12

13 11 2 14

10 1 9 8 3

6 7

Figure 3.30: An example of a vertical diagram with 1 vertical component and slope 1. We use the curve γ to ﬁnd a bypass that will either reduce n or decrease the slope of D.

Table 3.6: The bypass conﬁgurations from Figure 3.30 along with the type of bypass found in Figure 3.31. In this example, bypass conﬁguration (3,4,5) is also of type C. Bypass Type Bypass Type (14,1,2) D (7,8,9) D (1,2,3) B (8,9,10) B (2,3,4) D (9,10,11) D (3,4,5) B (10,11,12) B (4,5,6) A (11,12,1) A (5,6,7) A (12,13,14) A (6,7,8) B (13,14,1) B

U Proposition 3.10.3. Let N be a neighborhood of a twist knot Km for m ≥ 3. Let DN be an associated diagram for N with complexity (n, l) with l ≥ 2 and n = 2l − 1. In other words, D is a diagram with 1 vertical component. When s(D) > 0, all the possible boundary

parallel dividing curves on Dout will reduce n, with the exception of two boundary parallel curves that will each reduce the slope s(D) by 1.

44 ABCD

Figure 3.31: The base case bypass conﬁgurations for diagrams D with complexity (2l − 1, l) and slope s > 0 that arise when γ intersects ΓD. Since the bypasses come from γ, we “attach” the bypass. Types A and C will reduce n. Type B will thicken N. Type D will result in decreasing the slope s(D) by 1.

Proof. We analyze the base cases when the diagram D has complexity (3, 2) and the slope 0 < s(D) < 6. The complexity (3, 2) is the smallest complexity in this case that fails to represent a maximal unknot U and standard nieghorhood N of a Legendrian twist knot. Refer to Figures A.19 for the base cases. We summarize the types of bypass conﬁgurations resulting from γ0 in Figure 3.31. Types A, B, and C will reduce n. However, type D will decrease the slope by 1. We generalize the base cases by increasing n, m, and s. By doing so, we gain bypass conﬁgurations of type A, B and C, all of which reduce n.

Example 3.10.4. Consider the diagram D with one vertical component and n = 2 so that the slope of the dividing curves of D is -1. This diagram is shown in Figure 3.32. We label

0 the 22 intersection points between the dividing curves of Pi and γ , 1-22. Because we’re using

0 the curve γ , the underlying twist knot is relevant. In this example the knot is K3. Recall that γ0 is obtained by following one of the components of the twists inside D. For the knot

0 K3, γ twists 3/2 times around the pair of pants P2 . Next, we classify the types of bypasses in Table 3.7.

45 5 20 14 6 13 19

18 7 12

22 11

17 8 2 9 16 3 10 15 4 21

Figure 3.32: A diagram with 1 vertical component which has slope -1. We use the curve γ0 to show the bypasses will either reduce n or increase the slope. Here, the underlying twist knot is K3.

Table 3.7: The bypass conﬁgurations from Figure 3.32 along with the type of bypass found in Figure 3.33 Bypass Type Bypass Type (22,1,2) B (11,12,13) B (1,2,3) A (12,13,14) A (2,3,4) A (13,14, 15) A (3,4,5) C (14,15,16) C (4,5,6) A (15,16,17) A (5,6,7) A (16,17,18) A (6,7,8) C (17,18,19) C (7,8,9) A (18,19,20) A (8,9,10) A (19,20,21) A (9,10,11) B (20,21,22) C (10,11,12) D (21,22,1) D

U Proposition 3.10.4. Let N be a neighborhood of a twist knot Km for m ≥ 3. Let DN be an associated diagram for N with complexity (n, l) with l ≥ 2 and n = 2l − 1. In other words, D is a diagram with 1 vertical component. When s(D) < 0, all the possible boundary

parallel dividing curves on Din will reduce n, with the exception of two boundary parallel curves which will increase the slope s(D) by 1.

46 B DAC

Figure 3.33: The types of bypass conﬁgurations from γ0 that occur for diagrams with complexity (n, l), n = 2l − 1, and slope s(D) < 0. Since the bypasses come from γ0, we “dig” the bypass. Types A and C will reduce n. Type B will thicken N. However, type D will increase the slope of D by 1.

Proof. We use the base case when the diagram D has complexity (3, 2) and slope −6 < s(D) < 0. The complexity (3, 2) is the smallest complexity in this case that fails to represent a maximal unknot U and standard nieghorhood N of a Legendrian twist knot. Refer to Figures A.17 and A.18 for the base cases. We summarize the types of bypass configurations resulting from γ0 in Figure 3.33. Types A and C will reduce n. Type B will thicken N. However, type D will increase the slope by 1. We generalize the base cases by increasing n and m and by decreasing s. By doing so, we gain bypass configurations of type A and C, all of which reduce n. In this next example, we establish the fact that for a solid torus N with extremal relative Euler number, a diagram D for N cannot have 1 vertical component and slope 0. We use the interpretation of extremal relative Euler number as the meridional disks being boundary parallel. We establish this fact by filling in parts of the twist knot. In particular, we fill in one of the strands from the clasp and one of the strands from the twist and derive a contradiction to the tightness of S3 by showing the underlying S2 has more than one dividing curve.

Notice that there is a strand from the clasp connecting the curves c1,2 and c1,3. By ﬁlling

in this clasp, we ﬁll in the disks bounded by these curves c1,2 and c1,3. The disks must have the same dividing set conﬁguration, but with the R+/R− regions switched. Now, if m is

47 odd, there is a strand from the twist of the twist knot connecting c1,2 to c2,3. The same argument holds for why the R+/R− regions of the disks bounded by c1,2 and c2,3 must be

switched. Therefore, the disks bounded by c1,3 and c2,3 must have the same Euler number. This concept is demonstrated in Figure 3.34.

c1,2 A

c1,3 A c2,3

Figure 3.34: A diagram with 1 vertical strand and with an odd number of twists.

Now, because the disks have only boundary parallel dividing curves, there are only two ways to connect them (refer to Figure 3.35). In both cases, we see a disconnected dividing set on the underlying sphere S which contradicts the tightness of S3. This can be generalized to diagrams with bigger n and is proven in the next proposition.

+ +- - + - - + + + + +

Figure 3.35: The two possible ways to ﬁll in the disks bounded by c1,3 and c2,3. In both cases, we have a disconnected dividing set on S.

If m were even, then we would not have have a disconnected dividing set on the underlying 2-sphere S. For instance, in Figure 3.36, one can see that the dividing set is not disconnected as in the case when m is odd.

A A + - A - + - +

Figure 3.36: Filling in the disks bounded by c1,2, c1,3, and c2,3 when m is even, the dividing set is not disconnected.

Proposition 3.10.5. Let l ≥ 2, and let m ≥ 3 be odd. Let N be a neighborhood of the twist

knot Km with extremal Euler number. If D is any diagram associated to N with complexity (n, l) with n = 2l − 1, then s(D) 6= 0.

Proof. Suppose s(D) = 0. Let Di,j be the meridional disks of N bounded by ci,j for i ∈ {1, 2}, j ∈ {2, 3}, and consider the 2-sphere S deﬁned by

( ) [ S = D ∪ Di,j . i,j The assumption that N has extremal Euler number implies that the dividing curves on the

2l 2l disks Di,j are boundary parallel. Let {pi}i=1 = ΓD1,3 ∩ c1,2 and {qi}i=1 = ΓD2,3 ∩ c2,2 be labeled in such a way that p1 is connected to q1 by ΓD. The assumption that m is odd allows us to determine that the Euler number of D1,3 and D2,3 are the same. Inside D1,3, if p1 is connected to p2l, then p2 must be connected to p3. This implies that, inside D2,3, q2

49 3 is connected to q3, which leads to a contradiction of the tightness of S . Inside D1,3, if p1 is connected to p2, then q1 must be connected to q2, which again is a contradiction of the tightness of S3.

Corollary 3.10.5. Let N be a neighborhood of a twist knot Km with m = 2k + 1 ≥ 3 that has extremal Euler number. If s(D) > 0, all the bypass conﬁgurations from γ will reduce n. If s < 0, all the bypass conﬁgurations from γ0 will reduce n.

Proof. Propositions 3.10.3 (3.10.5) imply that all the bypass conﬁgurations from γ (γ0) will reduce (increase) s with the exception of two bypass conﬁgurations. Assume the dividing curves on Dout (Din) are nested in such a way that the only two boundary parallel curves are precisely those two. Then, passing through each consecutive bypass will eventually result in the slope of ΓD reaching 0. We demonstrate this fact in the next example. This is a contradiction.

Example 3.10.6. In this example, we demonstrate when the bypasses from Dout will reduce the slope of ΓD to eventually reach 0. The specific dividing set configuration on Dout is one with nested dividing curves. We illustrate this in Figure 3.37. The specific bypasses are labeled in the disk Dout, as well as in each diagram. After performing a bypass attachment, the resulting configuration, up to isotopy, is the same as the original but with a slope that is one less than the original.

50 1 2

3 4

Figure 3.37: A Diagram D with slope 5. The disk below represents the disk bounded by γ, Dout, with a speciﬁc dividing set in which all curves are nested. With this conﬁguration of dividing curves on Dout, the slope of D will decrease by 1 after passing through each bypass from Dout. After 5 consecutive bypass attachments, the slope of D will be 0.

51 Bibliography

[1] D. Bennequin, Entrelacements et équationsde Pfaff, Astérisque 107-108 (1983), 87-161

[2] Yuri Chekanov. Diﬀerential algebra of Legendrian links. Invent. Math., 150(3):441483, 2002.

[3] W. Chonchitmate, L. Ng, An atlas of Legendrian knots, arXiv: 1010.3997, (To appear in Exp. Math)

[4] Y. Eliashberg, Contact 3-manifolds twenty years since J. Marinet’s work, Ann. Inst. Fourier (Grenoble) 42 (1992), no1-2, 165-192

[5] Y. Eliashberg, Classiﬁcation of overtwisted contact structures on 3-manifolds, Invent. Math 98 (1989), 623-637

[6] Y. Eliashberg and M. Fraser, Legendrian and transverse knots in tight contact 3- manifolds, in Geometry, topology, and dynamics (Montreal, PQ, 1995), 17-51, CRM Proc. Lecture Notes 15, Amer. Math. Soc., Providence, RI, 1998

[7] J. Etnyre and K. Honda, Cabling and transverse simplicity. Annals of Mathematics, 162 (2005), 1305-1333

[8] J. Etnyre and K. Honda. Knots and contact geometry. I. Torus knots and the ﬁgure eight knot. J.Symplectic Geom., 1(1):63120, 2001.

52 [9] J. Etnyre and K. Honda. Knots and contact geometry. II. Connected Sums arXiv: 0205310v1

[10] J. Etnyre, L. Ng, V. V´ertesi, Legendrian and Transverse Twist Knots. arXiv:1002.2400

[11] E. Giroux, Convexite‘ en topologie de contact, Comment. Math. Helv, 66 (1991), 637- 677

[12] K. Honda, On the Classiﬁcation of tight contact structures I. Geom. Topol. 4 (2000), 309-368

[13] K. Honda, Gluing tight contact structures. Duke Math J. 115 (2002),no. 3, 435-478

[14] Y. Kanda, On the Thurston-Bennequin invariant of Legendrina knots and non exactness of Bennequin’s inequality Invent. Math 133 (1998), 227-242

[15] W. Kazez, A cut and paste approach to contact topology. arXiv:math.GT/0210091

[16] P. Kronheimer and T. Mrowka, Wittens conjecture and Property P, Geom. Topol. 8 (2004), 295310.

[17] D. LaFountain, Studying uniform thickness I: Legendrian simple iterated torus knots, arXiv: 0905.2760

[18] D. LaFountain, Studying uniform thickness II: Transversely non-simple iterated torus knots., arXiv: 0909.1452

[19] J. Martinet, Formes de contact sur les vari´et´esde dimension 3, Springer Lect. Notes in Math 209 (1971), 142-163

[20] P. Ozsv´ath,A. Stipsicz, Contact Surgeries and the transverse invariant in knot Floer homology, arXiv: 0803.1252, 2008

53 Appendix A

The base case conﬁgurations

The diagrams in this Appendix are the base case diagrams referred to in the proof of our main result. The figures are separated, by section, by the type of diagram: null, horizontal, and vertical. Recall that the base case diagrams are diagrams which show the minimal amount of twisting of the dividing curves and that have the fewest curves in the dividing set. For more detail, refer to Section 3.6. The types of diagrams are further separated by the slopes being positive and negative. In most cases, we use the different curves, γ and γ0, when the slopes are positive or negative. We establish which curve is used in the captions, though it should be apparent to the reader by examining the illustrations. It is also noted in the proofs which curve is used. Furthermore, a summary of the types of bypasses are given in the proofs. The behavior of γ and γ0 with the dividing curves of a diagram D is very much affected by the slopes and number of curves in ΓD. Recall that we “attach” bypasses coming from γ, and we “dig” bypasses coming from γ0.

54 A.1 Null Cases

Figure A.1: Base case null diagrams when 0 < s(D) < 8 along with the curve γ0 when the underlying twist knot Km has an odd number of twists. In this Figure, the slopes are −8 < s < 0, excluding s = −4 (which results in a disconnected dividing set on S. To see 0 this, simply ﬁll in ci,j). We use γ to show that all the possible boundary parallel dividing curves on Din result in a reduction of n.

55 Figure A.2: Base case null diagrams when 0 < s(D) < 8 along with the curve γ0 when the underlying twist knot Km has an even number or twists. In this Figure, the slopes are −8 < s < 0, excluding s = −4 (which results in a disconnected dividing set on S. To see 0 this, simply ﬁll in ci,j). We use γ to show that all the possible boundary parallel dividing curves on Din result in a reduction of n.

56 Figure A.3: Base case null diagrams when −8 < s(D) < 0 along with the curve γ0 when the underlying twist knot Km has an odd number or twists. Pictured is the knot K3. In this Figure, the slopes are −8 < s < 0, excluding s = −4 (which results in a disconnected 0 dividing set on S. To see this, simply ﬁll in ci,j). We use γ to show that all the possible boundary parallel dividing curves on Din result in a reduction of n.

57 Figure A.4: Base case null diagrams when −8 < s(D) < 0 along with the curve γ0 when the underlying twist knot Km has an even number of twists. Pictured is the knot K4. In this Figure, the slopes are −8 < s < 0, excluding s = −4 (which results in a disconnected 0 dividing set on S. To see this, simply ﬁll in ci,j). We use γ to show that all the possible boundary parallel dividing curves on Din result in a reduction of n.

58 A.2 Horiztonal Cases

Figure A.5: Base Case Diagrams with one horizontal component. In this ﬁgure, 0 < s(D) < 8. We use γ to show that all boundary parallel dividing curves on Dout will reduce n.

59 Figure A.6: Base Case Diagrams with one horizontal component. In this ﬁgure, s = 8, 9 . We use γ to show that all boundary parallel dividing curves on Dout will reduce n.

60 Figure A.7: Base Case Diagrams with one horizontal component. In this ﬁgure, −7 < 0 s(D) < 0. We use γ , with the knot K3, to show that all boundary parallel dividing curves on Dout will reduce n.

61 Figure A.8: Base Case Diagrams with one horizontal component. In this ﬁgure, s = −8, −9. 0 We use γ , with the knot K3, to show that all boundary parallel dividing curves on Dout will reduce n.

62 Figure A.9: Base Case Diagrams with one horizontal component. In this ﬁgure, −7 < s < 0. 0 We use γ , with the knot K4. to show that all boundary parallel dividing curves on Din will reduce n.

63 Figure A.10: Base Case Diagrams with one horizontal component. In this ﬁgure, s = −8, −9. 0 We use γ , with the knot K4. to show that all boundary parallel dividing curves on Din will reduce n.

64 Figure A.11: Base case horizontal diagrams with two or more horizontal components. In this Figure, the slopes are 0 < s < 6. We use γ to show that all the possible boundary parallel dividing curves on Dout result in a reduction of n.

65 Figure A.12: Base case horizontal diagrams with two or more horizontal components. In this Figure, the slopes are 6 < s < 11. We use γ to show that all the possible boundary parallel dividing curves on Dout result in a reduction of n.

66 Figure A.13: Base case horizontal diagrams with two or more horizontal components and 0 curve γ when the underlying twist knot Km has an odd number or twists. Pictured is the knot K3 In this Figure, the slopes are −8 < s < 0, excluding s = −6 (which results in a 0 disconnected dividing set on S. To see this, simply ﬁll in ci,j). We use γ to show that all the possible boundary parallel dividing curves on Din result in a reduction of n.

67 Figure A.14: Base case horizontal diagrams with two or more horizontal components and 0 curve γ when the underlying twist knot Km has an odd number or twists. Pictured is the 0 knot K3. In this Figure, the slopes are −12 < s < −7. We use γ to show that all the possible boundary parallel dividing curves on Din result in a reduction of n.

68 Figure A.15: Base case horizontal diagrams with two or more horizontal components and 0 curve γ when the underlying twist knot Km has an even number or twists. Pictured is the knot K4. In this Figure, the slopes are −8 < s < 0, excluding s = −6 (which results in a 0 disconnected dividing set on S. To see this, simply ﬁll in ci,j).We use γ to show that all the possible boundary parallel dividing curves on Din result in a reduction of n.

69 Figure A.16: Base case horizontal diagrams with two or more horizontal components and 0 curve γ when the underlying twist knot Km has an even number or twists. Pictured is the 0 knot K4. In this Figure, the slopes are −12 < s < −7. We use γ to show that all the possible boundary parallel dividing curves on Din result in a reduction of n.

70 A.3 Vertical Cases

Figure A.17: Base case vertical type diagrams with one vertical component when the underlying twist knot has an odd number of half-twists. Pictured here is K3. The slopes in these ﬁgures are −5 < s(D) < 0. We use γ0 to analyze the possible bypass conﬁgurations.

71 Figure A.18: Base case vertical type diagrams with one vertical component when the underlying twist knot has an even number of half-twists. Pictured here is K4. The slopes in these ﬁgures are −5 < s(D) < 0. We use γ0 to analyze the possible bypass conﬁgurations.

72 Figure A.19: Base case vertical type diagrams with one vertical component. The slopes in these ﬁgures are 0 < s(D) < 5. We use γ to analyze the possible bypass conﬁgurations.

73 Figure A.20: To the left: base case diagrams of vertical types with complexity (2, 2) when s ≥ 0. To the right, all possible boundary parallel dividing curves on Dout that correspond to the labels on γ on the left. The labels refer to Figures 3.28. Note, the disks do not show

ΓDout

74 Figure A.21: To the left: base case diagrams of vertical types with complexity (2, 2) when −3 < s < 0 and m = 3. Pictured here is K3. To the right, all possible boundary parallel 0 dividing curves on Din that correspond to the labels on γ on the left. The labels refer to

Figures 3.29. Note, the disks do not show ΓDin

75 Figure A.22: To the left: base case diagrams of vertical types with complexity (2, 2) when −3 < s < 0 and m = 4 To the right, all possible boundary parallel dividing curves on Din 0 that correspond to the labels on γ on the left. Pictured here is K4. The labels refer to

Figures 3.29. Note, the disks do not show ΓDin

76 Figure A.23: Base case diagrams of vertical types with complexity (1, 2) when 0 ≤ s ≤ 2. We use γ to show all possible boundary parallel curve on Dout result in a thickening of N.

77 Figure A.24: Base case diagrams of vertical types when the underlying knot has an odd number of twists with diagram complexity (1, 2) when −2 < s < 0. Pictured here is K3. We 0 use γ to show all possible boundary parallel curve on Din result in a thickening of N.

78 Figure A.25: Base case diagrams of vertical types when the underlying knot has an even number of twists with diagram complexity (1, 2) when −2 < s < 0. Pictured here is K4. We 0 use γ to show all possible boundary parallel curve on Din result in a thickening of N.