Alexander Polynomials of 3-Braid Knots

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Mathematical Sciences Theses Mathematical Sciences

4-2020

Marwa Emad Alrefai

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Recommended Citation Emad Alrefai, Marwa, "Alexander Polynomials of 3-Braid Knots" (2020). Mathematical Sciences Theses. 4. https://scholarworks.uaeu.ac.ae/math_theses/4

This Thesis is brought to you for free and open access by the Mathematical Sciences at Scholarworks@UAEU. It has been accepted for inclusion in Mathematical Sciences Theses by an authorized administrator of Scholarworks@UAEU. For more information, please contact [email protected]. United Arab Emirates University

College of Science

Department of Mathematical Sciences

ALEXANDER POLYNOMIALS OF 3-BRAID KNOTS

Marwa Emad Alrefai

This thesis is submitted in partial fulﬁllment of the requirements for the degree of Master of Science in Mathematics

Under the Supervision of Dr. Nafaa Chbili

April 2020 ii

Declaration of Original Work

I, Marwa Emad Alrefai, the undersigned, a graduate student at the United Arab Emirates University (UAEU), and the author of this thesis entitled “Alexander Polynomials of 3- Braid Knots", hereby, solemnly declare that this thesis is my own original research work that has been done and prepared by me under the supervision of Dr. Nafaa Chbili, in the College of Science at UAEU. This work has not previously been presented or published, or formed the basis for the award of any academic degree, diploma or a similar title at this or any other university. Any materials borrowed from other sources (whether published or unpublished) and relied upon or included in my thesis have been properly cited and acknowledged in accordance with appropriate academic conventions. I further declare that there is no potential conﬂict of interest with respect to the research, data collection, authorship, presentation and/or publication of this thesis.

Student’s Signature Date 08-06-2020 iii

Advisory Committee

1) Advisor: Dr. Nafaa Chbili Title: Associate Professor Department of Mathematical Sciences College of Science

2) Co-advisor: Prof. Kanat Abdukhalikov Title: Professor Department of Mathematical Sciences College of Science v

Approval of the Master Thesis

This Master Thesis is approved by the following Examining Committee Members:

1) Advisor (Committee Chair): Dr. Nafaa Chbili Title: Associate Professor Department of Mathematical Sciences College of Science Signature Date 03-06-2020

2) Member: Ho Hon Leung Title: Assistant Professor Department of Mathematical Sciences College of Science Signature Date 04-06-2020

3) Member (External Examiner): Prof. Sadok Kallel Title: Professor Department of Mathematics and Statistics Institution: American University of Sharjah, United Arab Emirates Signature Date 07-06-2020 vi

This Master Thesis is accepted by:

Acting Dean of the College of Science: Professor Maamar Ben Kraouda

Signature Date June 8, 2020

Dean of the College of Graduate Studies: Professor Ali Al-Marzouqi

Signature Date June 8, 2020

Copy of vii

Abstract

A knot is an embedding of a circle S1 into the three-dimensional sphere S3. An n- ∏n 1 3 component link is an embedding of n disjoint circles i=1 S into S . The main objective of knot theory is to classify knots and links up to natural deformations called isotopies. While there is no simple algorithm that helps decide whether two given knots (or links) are equivalent, various topological invariants have been developed to help distinguish between non-equivalent knots and links. The Alexander polynomial ∆L(t) is one of the oldest such tools. It was originally deﬁned from the Seifert surface of the knot. A simpler recursive deﬁnition of this invariant has been introduced later using Conway skein relations on the link diagram.

The aim of this study is to compute the Alexander polynomial and obtain an explicit formula for some families of alternating knots of braid index 3. This formula is used to prove that ∆L(t) satisfies Fox’s trapezoidal conjecture. This conjecture states that the coefficients of the Alexander polynomial of an alternating knot are trapezoidal. In other words, these coefficients increase, stabilize then decrease in a symmetrical way. The main tool in this study is the Burau representation of the braid group.

Keywords: Alexander polynomial, Trapezoidal conjecture, alternating knots, 3-braids. viii

Title and Abstract (in Arabic)

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Acknowledgements

This thesis would not have been completed without the help and the support of many people around me.

I would like to take the opportunity to express my gratitude to my supervisor, Dr. Nafaa Chbili, who was the biggest source of motivation. I will never ﬁnd an inspirational, supportive, and wise supervisor like you. Thank you for your guidance, assistance, and constructive criticism throughout my research phase. Working with you allowed me to discover the beauty of math, as I was introduced to the Knot Theory, and since then I have been intrigued by this topic, and I started to learn more about it. Not to mention the Topology course, you made it easy to understand despite its difﬁculty. Dr. Nafaa, you spurred me on to greater accomplishments. I am tremendously grateful and proud to be your student.

My appreciation extended to Prof. Kanat Abdulkhalikov, who was my Co-advisor. Your way of teaching me the topic of Finite Field course made me love Algebra, which I linked a lot to my Thesis as there is a great connection between Topology and Algebra. Then I would like to thank the external examiner, Dr. Sadok Kallel, and the internal examiner, Dr. Ho Hon Leung, for their constructive remarks and for reviewing my thesis.

I would also like to thank Dr. Youssef El Khatib and Shaikha Al Neyadi for generously offering their knowledge of Latex, especially the Beamer. Without their help, I would not have been able to write this thesis and make a Beamer slideshow in conferences.

My sincere thanks also goes to the head of Mathematics Department, Dr. Mo- hamed Salem, and all my doctors, particularly, Prof. Muhammed Syam, Dr. Ahmed Rawashdeh, Dr. Adama Diene, Dr. Jianhua Gong, Dr. Mohamed Hajji, Dr. Abdessamad Tridan, Prof. Victor Bodi, and Prof. Qasem Almdallal. Thank you for solidifying and enriching my mathematical knowledge.

A very special thanks goes to my backbone, my parents, Dr. Emad Alrefai and Ne- x had Yahia, and my grandmothers for their unlimited support, continuous encouragement. Also, thank you, my twin sister, Safa, my sister Taqwa, and my brothers, Mohammad, and Ali and all my relatives for your love and for making this journey less stressful.

Last but not the least important, I would like to thank all my friends, particularly, Huda, Amena, Bayan, Asma, Sarah, Rasha, Shaikha and Namarig, who were my second family as they always supported and encouraged me to do my best. xi

Dedication

To my beloved parents and teachers xii

Table of Contents

Title...... i

Declaration of Original Work ...... ii

Advisory Committee ...... iv

Approval of the Master Thesis ...... v

Abstract ...... vii

Title and Abstract (in Arabic) ...... viii

Acknowledgments ...... ix

Dedication ...... xi

Table of Contents ...... xii

List of Figures ...... xiv

Chapter 1: Introduction ...... 1

Chapter 2: Knots and Links ...... 4 2.1 What are Knots and Links? ...... 4 2.2 Composition of Knots ...... 8 2.3 Reidemeister Moves ...... 11

Chapter 3: The Theory of Braids ...... 14 3.1 Braids in Mathematics ...... 14 3.2 The Braid Group ...... 17 3.3 Knots and Braids ...... 24 3.4 The Braid Index ...... 28 xiii

Chapter 4: The Alexander Polynomial ...... 29 4.1 The Alexander-Conway Polynomial ...... 29 4.2 Basic Properties of the Alexander Polynomial ...... 34 4.3 Burau Representation ...... 39

Chapter 5: Fox-Trapezoidal Conjecture for Closed 3-Braids ...... 43 5.1 Fox-Trapezoidal Conjecture ...... 43 5.2 Main Result ...... 44 5.3 Proof of the Main Theorem ...... 45 5.4 Conclusion and Future Work ...... 57

References ...... 58 xiv

List of Figures

Figure 2.1 Constructing a knot from a piece of string ...... 4 Figure 2.2 A regular projection of the trefoil knot...... 5 Figure 2.3 Examples of knot and link projections ...... 5 Figure 2.4 A 2-component split link ...... 6 Figure 2.5 Non-reduced diagram (a) and Reduced diagram (b) ...... 7 Figure 2.6 Positive and negative crossings ...... 7 Figure 2.7 A 2-component link with linking number 2 ...... 8 Figure 2.8 The composition J#K of two knots J and K ...... 8 Figure 2.9 Not the composition of J and K ...... 8 Figure 2.10 Orientations match (a), Orientations match (b), Orientations differ (c) ...... 9 Figure 2.11 Compositions (a) and (b) are the same ...... 10 Figure 2.12 The trefoil knot is invertible ...... 10 Figure 2.13 Two distinct composite knots with the same factors ...... 11 Figure 2.14 Planar isotopies ...... 11 Figure 2.15 Reidemeister moves ...... 12 Figure 2.16 A series of Reidemeister moves ...... 12 Figure 2.17 The ﬁgure-eight is equivalent to its mirror image ...... 13 Figure 3.1 The cube C ...... 14 Figure 3.2 Examples of braids (a) and (c), (b) is not a braid ...... 15 Figure 3.3 Two equivalent braids ...... 15 Figure 3.4 Regular diagrams of a braid in 2-dimensional plane ...... 16 Figure 3.5 The trivial n-braid ...... 16 Figure 3.6 The product of α and β ...... 18 Figure 3.7 The associative property for braids α,β and γ ...... 18 Figure 3.8 The unit element for the braid group B3 ...... 18 Figure 3.9 The mirror image α∗ of α ...... 19 Figure 3.10 The general elements of B2 ...... 19 Figure 3.11 The elementary braids ...... 20 −1 −1 −1 Figure 3.12 The braid α = σ1σ2σ1 σ3 σ1σ2 ...... 20 −1 −1 Figure 3.13 The braid β = σ3σ1 σ2σ3σ2 ...... 21 Figure 3.14 The braid relations ...... 22 Figure 3.15 The relation σ1σ2σ1 = σ2σ1σ2 in B5 ...... 22 Figure 3.16 The braids α and β are related by braid relations ...... 23 Figure 3.17 A knot K obtained from a braid α ...... 24 Figure 3.18 Alexander’s theorem ...... 25 Figure 3.19 Different braids give the same knot (trivial knot) ...... 25 Figure 3.20 Markov’s theorem ...... 27 Figure 4.1 Oriented 3 links that are identical except in a small region . . . . . 29 Figure 4.2 Three related links ...... 30 Figure 4.3 The resolving tree for the right-hand trefoil knot ...... 33 Figure 4.4 Resolving tree for the ﬁgure-eight knot ...... 33 Figure 4.5 The (−3,5,7)-pretzel knot ...... 37 Figure 4.6 µ-component link ...... 38 Figure 5.1 Case (1): p1 + p2 + q1 − 2 < q2 ...... 53 Figure 5.2 Case (2): p1 + p2 + q1 − 2 ≥ q2 and p1 + p2 + q1 + q2 even . . . . 53 Figure 5.3 Case (3): p1 + p2 + q1 − 2 ≥ q2 and p1 + p2 + q1 + q2 odd . . . . 54 xv

Figure 5.4 Example 1: All polynomials are trapezoidal ...... 56 Figure 5.5 Example 2: All polynomials are trapezoidal ...... 56 1

Chapter 1: Introduction

A knot is an embedding of a circle S1 into the three-dimensional sphere S3. An ∏n 1 3 n-component link is an embedding of n disjoint circles i=1 S into S . Knot theory is the branch of low dimensional topology concerned with the study of knots and links. This theory aims to classify these objects up to natural deformations called isotopies. The classiﬁcation of these objects is of central importance to the study of 3-dimensional manifolds as any oriented connected closed 3-manifold without boundary can be obtained from S3 by surgery along a link.

Knot theory has several applications that have been highlighted in many research areas. In addition to intrinsic mathematical interest, research work in biology, chemistry, and theoretical physics have had the light shed on the importance of knot theory. For a further reference on applications of knot theory in these areas, see chapter seven of Adams’s book [1].

A link diagram is a regular planar projection of the link together with hight information at each double point showing which of the 2 strands passes over the other. It is well known that the study of links up to isotopy is equivalent to the study of their diagrams up to local deformations known as Reidemeister moves. A link diagram is said to be alternating if the overpass and the underpass alternate as one travels along any strand of the diagram. A link L is alternating if it can be represented by an alternating diagram.

The braid group on n-strands (denoted by Bn) plays a central role in the study of knot theory and 3-dimensional topology in general. An example of braid group applications in knot theory is that any link in S3 can be represented as the closure of a braid b ∈ Bn (This result is known as Alexander’s theorem), it can be written as L = bˆ. This braid representation is not unique and the minimum value of n such that L is the closure of an n-braid is called the braid index of L. Links of braid index three represent an impor- 2 tant class of links in the 3-dimensional space, it has been widely studied and attracted the interest of many knot theorists. The 3-braid links have been classiﬁed [2], and a complete classiﬁcation of alternating 3-braid links has also been given [3].

In 1928, Alexander introduced an invariant of oriented links which associates with

±1/2 each link L a Laurent polynomial of integral coefficients ∆L(t) ∈ Z[t ]. In 1969, Con- way proved that this polynomial can be defined in a simple recursive way. Moreover, the Alexander polynomial can be defined through a relatively simple formula using Bu- rau representation, see [4]. Indeed, this invariant can be defined in several different (but equivalent) ways. More recently, Ozsva ´th and Szabo ´ [5] proved that this polynomial, up to the multiplication by some factor, is obtained as the Euler characteristic of link Floer homology. The Alexander polynomial is known to be symmetric as it satisfies

−1 ∆L(t) = ∆L(t ). Furthermore, for a knot K and up to the multiplication by a power of t 2n i it is always possible to write ∆K(t) = ∑ αit with α0 6= 0 and α2n 6= 0. i=0 Murasugi, studied the Alexander polynomial of alternating knots and proved that the co- efﬁcients of this polynomial alternate in sign and that the polynomial has no gaps. More- over, the degree of ∆K(t) is equal to twice the genus of the knot [6, 7].

One of the most curious properties of the Alexander polynomial of alternating knots is the "trapezoidal" property. In 1962, Fox conjectured that the coefﬁcients of the Alexander polynomial of an alternating knot are trapezoidal, as explained in the following conjecture [8].

2n i Conjecture 1.0.1. Let K be an alternating knot and ∆K(t) = ± ∑ αi(−t) , with αi > 0, i=0 its Alexander polynomial. Then there exists a non-negative integer m ≤ n such that:

α0 < α1 < ··· < αn−m = ··· = αn+m > ...... > α2n−1 > α2n.

This challenging conjecture has been proved for several classes of alternating knots such as two-bridge knots, algebraic knots, knots of genus 2, and stable alternating knots, see [9, 10, 5, 11, 12, 13]. 3

The purpose of this research work is to prove that the conjecture holds for some families of alternating knots of braid index 3. To achieve this desired aim, first, the Alexander polynomial is computed for some families of alternating knots of braid index 3, using the Burau representation of the braid group. Then, the coefficients of these polynomials are verified to satisfy the Fox-trapezoidal conjecture. 4

Chapter 2: Knots and Links

2.1 What are Knots and Links?

First of all, What is a knot? Imagine taking a piece of string and tying a knot on it, when you glue the two ends of the string together and have thus formed a knotted loop. This knotted loop is simply a knot, see Figure 2.1.

Figure 2.1: Constructing a knot from a piece of string

Now, what is a link? A link is a set of knots that do not intersect, but which may be tangled (or knotted) up together. Note that a knot is considered as a link of one component.

A knot K in R3(or S3) can be projected onto a plane R2(or S2). This projection is said to be regular if it is into everywhere, except at a ﬁnite number of crossing points, which are the projections of just two points of K, and these points are not collinear. For example, Figure 2.2 shows a regular projection of the trefoil knot.

Knots and links can also be formally deﬁned as follows:

Deﬁnition 2.1.1. A knot K is an embedding of a circle S1 into the three-dimensional sphere S3,

K : S1 −→ S3. 5

Figure 2.2: A regular projection of the trefoil knot. ∏n 1 Deﬁnition 2.1.2. An n-component link L is an embedding of n disjoint circles i=1 S

into S3, ∏n L : S1 −→ S3. i=1

Example 2.1.1. Figure 2.3 (a) shows three regular projections of knots. The ﬁrst is the trivial knot and the other two are non-trivial knots. Figure 2.3 (b) displays three regular projections of 2-component links. The ﬁrst is the 2-component trivial link and the other two are non-trivial 2-component links.

Figure 2.3: Examples of knot and link projections 6

A link diagram has some important properties that can be introduced by the below deﬁnitions.

Deﬁnition 2.1.3. A link L is alternating if it admits a link diagram (or projection) that has crossings alternate between over and under as one travels along any strand of the link diagram in a ﬁxed direction.

Example 2.1.2. The trefoil knot, ﬁgure-eight knot, Hopf link and Whitehead link in Fig- ure 2.3 are alternating.

Deﬁnition 2.1.4. A split link is a link that possesses a regular diagram L that can be separated into 2 pieces. i.e., a dotted circle (or closed curve) can be drawn in the regular diagram L that contains some part of diagram L inside it, and the other part of diagram

L is outside the dotted circle (or closed curve). For example, the link in Figure 2.4 is a

2-component split link.

Figure 2.4: A 2-component split link

Example 2.1.3. In Figure 2.3, the unlink is a split link, since its two components can be separated into 2 pieces by a plane. But the Hopf link is not a split link, since its two components do link each other.

Deﬁnition 2.1.5. A reduced diagram is a link diagram in which there are no reducible crossings (also called the nugatory or removable crossings). 7

Example 2.1.4. Figure 2.5(a) shows a non-reduced diagram, whereas Figure 2.5(b) shows a reduced diagram.

Figure 2.5: Non-reduced diagram (a) and Reduced diagram (b)

To measure how two components are linked up numerically, the following method is introduced.

Deﬁnition 2.1.6. Let L be an oriented link and choose a diagram D of L, then the linking number `k(L) is deﬁned as follows:

1 `k(L) = ∑sign(c) 2 c where c ranges over all crossings of D, where different components meet and   +1 if there is a ﬁrst type crossing, as in Figure 2.6(a). sign(c) :=  −1 if ther is a second type crossing, as in Figure 2.6(b).

Figure 2.6: Positive and negative crossings 8

Example 2.1.5. The linking number of the link pictured in Figure 2.7 is 2.

Figure 2.7: A 2-component link with linking number 2

2.2 Composition of Knots

Deﬁnition 2.2.1. Given two projections of knots J and K, the composition of J and K, denoted by J#K (it is also known as connected sum ), is a new knot that can be obtained by removing a small arc from each one and then connecting the four endpoints by two new arcs, see Figure 2.8. These new arcs do not overlap and do not intersect the remaining parts of J and K, see Figure 2.9.

Figure 2.8: The composition J#K of two knots J and K

Figure 2.9: Not the composition of J and K 9

Deﬁnition 2.2.2. A composite knot is a knot that can be expressed as the composition

J#K of two non-trivial knots. The knots K and J are called factor knots.

A knot is prime if it is not composite. For example, trefoil and ﬁgure-eight knots are prime knots. Prime knots with up to 10 crossings are tabulated in [14].

Note that the composition of a knot K with the unknot is again K, i.e., K#O = K.

It is not always obvious to know whether a knot is composite or not. Here is one of a simple questions, is the unknot composite? Actually, the unknot is not a composite knot, because there is no way to take the composition of two non-trivial knots and obtain the unknot. Note that the knot tables list only prime knots.

It is possible to obtain two different composites from the same factor knots J and K. It depends on the orientation placed along the factor knots. Note that the knot can be oriented by placing directed arrows along with the projection of the knot in a selected direction. This knot is called an oriented knot. There are two ways to compose two oriented knots J and K. One way is to orient each knot, and then connect the knots so that the orientations match up, see Figure 2.10 (a) and (b). Another way is to connect the knots J and K so that the orientations do not match up, see Figure 2.10 (b).

Figure 2.10: Orientations match (a), Orientations match (b), Orientations differ (c)

In this particular example, compositions using the ﬁrst way yield the same knot, no matter where the segments are chosen, see Figure 2.11. The third composition using the second way also gives the same knot as the previous two compositions, because one of the oriented knots is invertible. An invertible knots can be deﬁned as follows. 10

Figure 2.11: Compositions (a) and (b) are the same

Deﬁnition 2.2.3. A knot K in S3 is said to be invertible if there is an ambient isotopy of

S3, which deforms K into itself with orientation reversed.

For instance, the trefoil knot is the simplest non-trivial invertible knot, i.e. rotating the trefoil knot 180o in the 3-space around an axis on a plane diagram produces the same knot diagram, but with the direction of its arrows reversed, see Figure 2.12.

Figure 2.12: The trefoil knot is invertible

Note that if a knot K is invertible, then J#K is well deﬁned, no matter what orientations are used.

However, composing noninvertible knots with itself in the two different ways gives two distinct composite knots, see Figure 2.13. Up until now, no technique is es- tablished to detect whether or not a given knot is invertible. 11

Figure 2.13: Two distinct composite knots with the same factors

2.3 Reidemeister Moves

There are two different types of movements of a knot projection that cause no changes on the knotting information:

Type 1: Planar isotopy moves: are deformations of a knot projection, that do not interact with any crossings, i.e., change the projection plane as if it was made of rubber rather than moving the string freely in space, see Figure 2.14 that displays planar isotopies of a knot.

Type 2: Reidemeister moves: are three speciﬁc types of deformations of a knot projection, that interact with the crossings, but still do not change the information of how the curve is knotted.

Figure 2.14: Planar isotopies

The ﬁrst Reidemeister move (RI) enables us to insert or pull out a single twist in the knot, see Figure 2.15 (a). While the second Reidemeister move (RII) enables two crossings to be either added or removed, see Figure 2.15 (b). The third Reidemeister move (RIII) enables us to move a strand of the knot through a crossing from one side to another, see Figure 2.15 (c). 12

Figure 2.15: Reidemeister moves

Considering that these moves only change the relations of the crossings, and each Reidemeister move can be accomplished via an ambient planar isotopy of the knot, the resulting projection will be equivalent to the original one. Therefore, a deﬁnition of equivalence of projections of knots can be given as follows.

Deﬁnition 2.3.1. Two projections of knots are said to be equivalent if they are related by a ﬁnite series of Reidemeister moves and planar isotopies.

This was proved by the German mathematician Kurt Reidemeister in 1926.

Example 2.3.1. Figure 2.16 shows a series of Reidemeister moves that take us from one projection to the other.

Figure 2.16: A series of Reidemeister moves 13

Example 2.3.2. Figure 2.17 displays that the ﬁgure-eight is equivalent to its mirror image.

Figure 2.17: The ﬁgure-eight is equivalent to its mirror image

Note that the mirror image K∗ of a knot K can be obtained by changing every crossing in the knot to the opposite crossing. A knot that is equivalent to its mirror image is called amphicheiral (or achiral). 14

Chapter 3: The Theory of Braids

3.1 Braids in Mathematics

Deﬁnition 3.1.1. Assume taking a 3-dimensional cube C. On the top and base of C,

0 0 0 mark out n points, P1,P2,...,Pn and P1,P2,...,Pn, respectively. These points may be

1 i 0 arbitrary placed. However, for the convenience, assume that P = ( , ,1) and P = i 2 n + 1 i 1 i ( , ,0) as shown in Figure 3.1 below. 2 n + 1

Figure 3.1: The cube C

Then start drawing strands that connect the marked points in C and that subject to the following rules:

• The strands ﬂow from top to bottom, and each point Pi is joined to only one point

0 Pj and vice-versa.

• The strands are not allowed to turn back.

• The strands can be braided and linked together, but intersections are not allowed.

The n strands obtained by this construction form an n-braid. 15

Figure 3.2: Examples of braids (a) and (c), (b) is not a braid

Example 3.1.1. In Figure 3.2, (a) is an example of 1-braid, whereas (b) is not a 1-braid since the string turns back as one goes from top to the bottom. (c) is an example of

2-braid.

If given two n-braids, the elementary knot moves ( known as Reidemeister moves) can be applied to transform one into the other. However, the endpoints are kept fixed to avoid getting anything that does not match the given definition of a braid. This can be formulated as the following definition.

Deﬁnition 3.1.2. Two n-braids are equivalent if it is possible to deform one to the other without causing any of the strings to intersect each other and keeping the endpoints ﬁxed.

Figure 3.3: Two equivalent braids 16

Example 3.1.2. Figure 3.3 (a) and (b) are both example of two equivalent braids.

Braids are studied through their projections in a 2-dimensional plane, the yz- plane. A regular diagram of the braid in Figure 3.3 (b) is shown in Figure 3.4.

Figure 3.4: Regular diagrams of a braid in 2-dimensional plane

0 0 0 Note that if P1 to P1, P2 to P2,..., Pn to Pn are connected by parallel n strings, then a special type of braid can be obtained, known as the trivial n-braid. Its regular diagram is shown in Figure 3.5.

Figure 3.5: The trivial n-braid

Interestingly, to a n-braid, say α, a permutation can be assigned. Suppose that its strings are connected as follows: P to P0 , P to P0 ,..., P to P0 . Then the braid 1 i1 2 i2 n in permutation that is assigned to α is:   1 2 ··· n    . i1 i2 ··· in

For instance, the trivial braid corresponds to the identity permutation,

  1 2 ··· n    . 1 2 ··· n 17

Example 3.1.3. The braid permutation for Figure 3.3(b) is

  1 2 3     = (1 3).   3 2 1

Note that if two braids are equivalent, then their braid permutations are equal. However, the converse is not true since if two permutations are equal, then this does not necessarily guarantee that their corresponding braids are equal. In fact, the braid permutation is considered as the simplest braid invariant. In other words, one can deﬁne a surjective group homomorphism Bn −→ Sn from the n-braid group, Bn, onto the symmetric group, Sn. This homomorphism is not injective as explained above and its kernel is called the pure braid group.

3.2 The Braid Group

From deﬁnition 3.1.2, two braids are called equivalent if it is possible to deform one to the other by keeping the endpoints ﬁxed and not causing the strings to cross each other. Therefore, one braid can be represented in many different ways. Still, there are braids that can not be deformed one to the other. To be more precise, there are different equivalent classes formed by braids that can be transformed into each other. To determine a class, any of its representatives can be chosen.

Now, the group structure on the set of braids can be deﬁned. For n ≥ 1, let Bn be the set of all n-braids (or the set of all equivalence classes of braids). For any two elements of Bn, say α and β, it is possible to deﬁne a product. This just consists of putting one above the other vertically and connect the strands. An example is given in Figure 3.6.

In general, the two products αβ and βα, of the two braids α and β are not equivalent, i.e.αβ 6= βα. This can be shown easily by the braid permutations. Since the product of permutations is not commutative. Therefore, commutativity does not hold for braids.

Nevertheless, the multiplication of braids is associative, i.e., (αβ)γ = α(βγ). The associative property for braids α,β and γ is shown in Figure 3.7. 18

Figure 3.6: The product of α and β

Figure 3.7: The associative property for braids α,β and γ

Now the natural question is: Is Bn a group under the above-deﬁned product? To show this, a unit element, and an inverse element are needed to ﬁnd. Suppose given the trivial braid e, where all the strands are parallel. When any braid α is composed with e, as in Figure 3.8, the result is equivalent to the original braid α. Therefore, for any braid α, it is true that αe = α = eα.

Figure 3.8: The unit element for the braid group B3

If α is a braid, then α∗ is deﬁned as its mirror image. Let us take the products αα∗ and α∗α. First, the original braid α and then its mirror image α∗, as shown in Figure 19

3.9, or vice-versa, ﬁrst the mirror image, α∗, of α, and then the braid α. In both cases, they can be simpliﬁed and got the trivial braid e. Therefore, for every braid α, there is an inverse α∗, denoted by α−1, i.e., αα∗ = e = α∗α.

Figure 3.9: The mirror image α∗ of α

So far, it is known that an associative operation can be deﬁned on the set of braids on n strands. This operation has a neutral element, and for every element in the set, there is an inverse. A group structure is then deﬁned. This group is called the n-braid group, denoted by Bn.

Example 3.2.1. The 1-braid group, B1, contains only one element, which is the trivial braid, B1 = e. The 2-braid group, B2, has 2 types of elements, each of which is equivalent to one of the two elements displayed in Figure 3.10.

Figure 3.10: The general elements of B2

Therefore, the 2-braid group, B2, has an inﬁnite number of non-equivalent elements. So each n-braid group, Bn, with n ≥ 2 has an inﬁnite order. To deal with the 20 elements of the braid group, it will be useful to associate a word to each element in the group. There exists a very easy way of writing a general element in one of these groups.

The symbol 1 (or e) denotes the trivial braid because it is the neutral element for the product. The n-braid where the i’th and the i’th+1 strands are linked together by a clockwise twist, looking from the left, and the other remaining strands which form straight lines, is denoted as σi. If the i’th and the i’th+1 strands twist in the other direction, −1 denote the n-braid by σi , see Figure 3.11.

Figure 3.11: The elementary braids

In this way, (n − 1) special n-braids σ1,σ2,...,σn−1 can be created. Now use these elementary braids as bricks to construct any element in the braid group.

−1 −1 −1 Example 3.2.2. The braid α = σ1σ2σ1 σ3 σ1σ2 is drawn in Figure 3.12.

−1 −1 −1 Figure 3.12: The braid α = σ1σ2σ1 σ3 σ1σ2 . 21

−1 Conversely, any braid can be described by using the elementary braids σi and σi . Suppose given a braid β in Figure 3.13. Then it can be cut into levels so that each level is an elementary braid with just one crossing. If two crossings are at the same level, then one can be shifted slightly upwards and the other slightly downwards, so that in each level there is only one crossing. This procedure works on every braid, so to any braid a word

−1 can be associated, i.e. a sequence of σi and σi where i = 1,2,...,n − 1.

−1 −1 Figure 3.13: The braid β = σ3σ1 σ2σ3σ2

−1 Therefore, any element in Bn can be expressed as a ﬁnite product of σi and σi .

So σ1,σ2,...,σn−1 are called generators of Bn.

For instance, the elements of B2 are generated by one element, σ1 and any element in B2 can be written as :

m −m −1 −1 σ = σ1 ...σ1 or σ = σ1 ...σ1 , | {z } | {z } m times m times where m ≥ 0. Note that the group B2 is isomorphic to the inﬁnite cyclic group.

In fact, different words may represent the same braid.

Example 3.2.3. The two braids σ1σ3 and σ3σ1, in Figure 3.14 (a), are equivalent 4-braids.

Therefore, the relation σ1σ3 = σ3σ1 holds in B4. Moreover, the two braids σ1σ2σ1 and

σ2σ1σ2 , in Figure 3.14 (b), are equivalent 3-braids and so the relation σ1σ2σ1 = σ2σ1σ2 holds in B3. 22

Figure 3.14: The braid relations

Actually, these two relations in Figure 3.14, (a) and (b), hold for general n-braids (n ≥ 3) where the regular diagram of n-braid has some extra non-crossed lines (or straight lines) added to it, see Figure 3.15.

Figure 3.15: The relation σ1σ2σ1 = σ2σ1σ2 in B5

In fact, all the movements on braids are easily translated into manipulations of the words. But how all possible manipulation of the words can be found? This seems an end- less job, but mathematicians have already solved this problem. The ﬁrst mathematician who described the structure of the group on the set of braids and noted that the group can be described by explicit presentations was Emil Artin in 1947. 23

Generalizing the above examples to n strands, it can be easily seen the following two types of relations hold Bn:

(1) σiσ j = σ jσi if |i − j| ≥ 2.

(2) σiσi+1σi = σi+1σiσi+1 ∀1 ≤ i ≤ n − 2

Putting generators and relations together gives the presentation of the Artin braid group Bn in the following theorem:

Theorem 3.2.1. (Artin, 1947) For n ≥ 2, the braid group Bn admits the following presentation:

Bn = σ1,...,σn−1 | σiσi+1σi = σi+1σiσi+1,σiσ j = σ jσi for 1 ≤ i ≤ n − 2, |i − j| ≥ 2 .

For instance:

B2 = hσ1 | −i,

B3 = hσ1,σ2 | σ1σ2σ1 = σ2σ1σ2i,

B4 = hσ1,σ2,σ3 | σ1σ3 = σ3σ1,σ1σ2σ1 = σ2σ1σ2,σ2σ3σ2 = σ3σ2σ3i. Note that the lack of relations is denoted by −.

As a matter of fact, the braid relations can be used several times to ensure that two n-braids are equivalent.

Figure 3.16: The braids α and β are related by braid relations

Example 3.2.4. From Figure 3.16, a word can be assigned for each braid, α and β in B4 24 as follows: −1 −1 −1 α = σ1σ2σ1 σ3σ1σ2 and β = σ1σ3 σ2σ3

Let us show that these two braids are equivalent by applying the braid relations

−1 several times. Note that the trivial relation can also be used, namely σiσi = e and the changes in the braid relations inside the () are as follows:

−1 −1 −1 −1 −1 −1 −1 α = σ1σ2σ1 (σ3σ1)σ2 = σ1σ2σ1 (σ1σ3)σ2 = σ1σ2σ3σ2 = σ1σ3 (σ3σ2σ3)σ2 =

−1 −1 −1 σ1σ3 (σ2σ3σ2)σ2 = σ1σ3 σ2σ3 = β.

−1 −1 Example 3.2.5. The braids β1 = σ1σ2σ1 and β2 = σ2 σ1σ2 are equivalent. Since

−1 −1 −1 −1 −1 −1 −1 −1 β1β2 = σ1σ2σ1 (σ2 σ1σ2) = σ1σ2σ1 σ2 σ1 σ2 = σ1σ2(σ1σ2σ1) σ2 =

−1 −1 −1 −1 σ1σ2(σ2σ1σ2) σ2 = σ1σ2σ2 σ1 σ2 σ2 = e. Hence, β1 = β2. Similarly, The braids

−1 −1 −1 −1 −1 α1 = σ2σ1 σ2 and α2 = σ1 σ2 σ1 are equivalent (check that α1α2 = e).

3.3 Knots and Braids

When given a braid, the strand ends can be tied together and formed into a regular

0 diagram of a knot (or link), K. More precisely, each pair of points, Pi and Pi , is joined by a simple arc lying outside the cube without creating any crossings, see Figure 3.17.

Figure 3.17: A knot K obtained from a braid α

In this way, the closure of the braid α can be formed, denoted by αˆ . An orientation can be assigned to each strand in a braid α. Hence, from a braid α, an oriented knot K 25

(or link) can be formed, and vice-versa an oriented closed braid can be obtained from an oriented knot as stated in the following Alexander’s theorem [15].

Theorem 3.3.1. Every oriented knot (or link) can be represented as the closure of a braid.

Example 3.3.1. Figure 3.18, from (a) to (c), shows the procedure of obtaining a braid for the oriented right-hand trefoil.

Figure 3.18: Alexander’s theorem

So far, it is known that two equivalent braids give two equivalent knots by closing their corresponding braids. But the question now is, when two braids close to the same knot? An overview of the answer appears in Figure 3.19, where all these different braids close to the same knot, which is the trivial knot.

Figure 3.19: Different braids give the same knot (trivial knot)

Therefore, equivalent knots can also be obtained from the closure of non-equivalent braids. 26

To better understand the relationship between braids and knot theory, it must be determined which braids give the same knot. A Russian mathematician named Markov has already solved this problem. He noticed that two closed braids give the same knot if and only if they are related by a sequence of moves known as Markov moves, which are deﬁned as follows. It is worth mentioning that starting from an n-braid, one may attach an additional trivial strand to obtain an (n + 1)-braid. Thus, an injective homomorphism is deﬁned from Bn to Bn+1. Let B∞ denotes the resulting direct limit.

Deﬁnition 3.3.1. Let β be an n-braid, i.e.,β ∈ Bn. The following two operations may be applied in B∞:

−1 (1) M1 (conjugation): is the operation that transforms β into the n-braid γβγ , where

−1 γ ∈ Bn. Here γβγ is called a conjugate of β. See Figure 3.20 (a).

(2) M2 (stabilization): is the operation that transforms β into either of two (n+1)-braids

−1 βσn or βσn , see Figure 3.20 (b).

These two operations M1 and M2 are called Markov moves.

−1 Example 3.3.2. Figure 3.20 (a) shows that the braid β = σ1σ2σ1 (element of B3), be-

−1 comes γβγ by applying M1, where γ = σ1σ2. Also, the converse can be done to get

−1 −1 β from γβγ by applying M1 . Figure 3.20 (b) shows that the braid β(element of B3),

−1 becomes βσ3 or βσ3 by applying M2. Similarly, the converse can be done to get β from

−1 −1 βσ3 or βσ3 by performing M2 .

Now, the concept of M-equivalence shall be introduced between two braids.

Deﬁnition 3.3.2. (Markov equivalence) Two braids, α and β (elements of B∞), are said to be Markov equivalent (M-equivalent) if one can be transformed into the other by applying

−1 −1 the operations (Markov moves) M1, M2, and their inverses M1 , M2 a ﬁnite number of 27 times. They are written as α ∼ β (orβ ∼ α).

Figure 3.20: Markov’s theorem

Example 3.3.3. The braids β and γβγ−1, Figure 3.20 (a), are M-equivalent. Note that if the closure is formed of braid γβγ−1, its lateral pieces γ and γ−1 are shifted so that they

−1 cancel out since one is the inverse of the other. Then a knot K2 (closure of γβγ ) is obtained, which is equivalent to the knot K1(closure of β). Moreover, the braid β and braid

−1 βσ3 (or βσ3 ) , Figure 3.20 (b), are M-equivalent. Note that if the closure is formed of

−1 βσ3 (or βσ3 ), K2, a knot equivalent to it is obtained, which is K1 (closure of β). In this case, it is only needed to undo the loop in the K2 that resulted from adding a strand to the right of the braid β.

Based on deﬁnition 3.3.2, Markov’s theorem can be formulated as follows.

Theorem 3.3.2. Let K1 be the closure of braid β1 and K2 be the closure of braid β2, then

∼ K1 = K2 ⇐⇒ β1 ∼ β2. 28

3.4 The Braid Index

Deﬁnition 3.4.1. The braid index of a knot (or link) K is the least number of strings required to construct a closed braid presentation of K and is denoted by b(K).

Example 3.4.1. The unknot has braid index 1. Conversely, a knot of braid index 1 is the unknot.

Note that the braid index is a knot invariant, because it is invariant under Reide- meister moves.

Example 3.4.2. The trefoil knot has braid index 2, see Figure 3.18. Note that the knots with braid index 2 are only the elementary torus knots of type (n;2), where n 6= 0,±1. 29

Chapter 4: The Alexander Polynomial

4.1 The Alexander-Conway Polynomial

The Alexander polynomial is the very ﬁrst polynomial knot invariant discovered

1 −1 in 1928 [16]. It is a Laurent polynomial, ∆L(t) ∈ Z[t 2 ,t 2 ], of oriented knots (or links) with a variable t. In 1928, James Alexander deﬁned this polynomial in relatively abstract mathematical concepts. But in 1969 John Conway provided a simple recursive procedure to compute the Alexander polynomial. He showed that the Alexander polynomial can be calculated using only the following two axioms.

Axiom 1: ∆O(t) = 1 where O denotes the trivial knot. 1 −1 2 2 Axiom 2: ∆L+ (t) − ∆L− (t) = (t −t )∆L0 (t).

Where L+,L− and L0 are three oriented link diagrams which are identical except in a small region where they are as pictured in Figure 4.1.

Figure 4.1: Oriented 3 links that are identical except in a small region

Note that the three regular diagrams L+,L− and L0 shown in Figure 4.1 above are called skein diagrams, and the second relation in the axiom 2, between the Laurent polynomials of L+,L− and L0 is called the skein relation. Also, an operation that replaces one of the three regular diagrams L+,L− and L0 by the other two is called a skein operation.

Example 4.1.1. Let us compute the Alexander polynomial of the 2-component unlink. 30

Consider the 2-component unlink as L0, with a crossing surrounded by a dotted circle as shown in Figure 4.1. If the skein operation is applied, the skein diagrams are shown in

Figure 4.2.

Figure 4.2: Three related links

Where

∆L+ (t) = ∆L− (t) = ∆O(t) = 1

and

1 −1 2 2 0 = 1 − 1 = ∆L+ (t) − ∆L− (t) = t −t ∆L0 (t), then

∆L0 (t) = ∆OO(t) = 0.

Note that the polynomial of a splittable link is always zero (This property of the Alexander polynomial will be proved in Section 4.2).

Example 4.1.2. The Alexander polynomial of the Hopf link can also be obtained as follows.

1 −1 ∆ (t) − ∆ (t) = t 2 −t 2 ∆ (t) where 31

∆ (t) = ∆OO(t) = 0 and ∆ (t) = ∆O(t) = 1 then

1 −1 ∆ (t) = t 2 −t 2 .

Note that if the orientation is reversed of one of these circles, then

∆ (t) = −∆ (t).

Example 4.1.3. The Alexander polynomial of the right-hand trefoil knot can be obtained as follows.

1 −1 ∆ (t) − ∆ (t) = t 2 −t 2 ∆ (t) where

1 −1 ∆ (t) = ∆O(t) = 1 and by Example 4.1.2 ∆ (t) = ∆ (t) = t 2 −t 2 .

Then

2 1 −1 −1 ∆ (t) = t 2 −t 2 + 1 = t − 1 +t.

In fact, there is a more effective way to compute the Alexander polynomial using the skein tree diagram (also called the resolving tree). To make this way easier to understand, see Example 4.1.4. First of all, to facilitate the next calculations, let us rewrite the skein relation in the axiom 2, as follows:

∆L+ (t) = ∆L− (t) + x∆L0 (t) (4.1)

∆L− (t) = ∆L+ (t) − x∆L0 (t) 32

1 −1 where x = t 2 −t 2 . Suppose that L is a regular diagram of knot K (or link) and follow the steps below:

(1) Choose any crossing in L that corresponds to either L+ or L−. Assume L here is L+ as in Example 4.1.4 for the right-hand trefoil knot, see Figure 4.3.

(2) Apply a skein operation to the chosen crossing so that L+ is transformed to the other

diagrams: one is L− and the other is L0. Then connect L+ and L− by drawing a left

arrow and assign +1 to it (and, similarly, connect L+ and L0 by drawing another right arrow and assign x to it) .

(3) Repeat the same steps on the resulting knots (or links) that are not trivial knots (or links) until each branch of the tree ends with a trivial knot (or link).

(4) Compute ∆L(t) as the sum of the Alexander polynomial of the each terminating trivial knot (or link) multiplied by the product of the coefﬁcients (on the arrows) along the branch path that starts with the original regular diagram L of K and ends with the regular diagram of this trivial knot (or link).

Note that the process of repeatedly choosing a crossing, and then applying the skein operation to get two simpler knots (or links) gives a tree of knots (or links) called the resolving tree (or skein tree diagram). At the top is the original knot (or link) and at the bottom, all of the trivial links that resulted from repeatedly applying the skein relation are found.

Example 4.1.4. The resolving tree for the right-hand trefoil knot is given in Figure 4.3.

From this resolving tree, the result is:

2 2 ∆L(t) = 1∆O(t) + x∆OO(t) − x ∆O(t) = 1 + x .

Therefore,

1 −1 2 −1 ∆L(t) = 1 + (t 2 −t 2 ) = t − 1 +t. 33

Figure 4.3: The resolving tree for the right-hand trefoil knot

Example 4.1.5. The resolving tree for the ﬁgure-eight knot is given in Figure 4.4.

Figure 4.4: Resolving tree for the ﬁgure-eight knot

From this resolving tree, the result is:

2 2 ∆L(t) = 1∆O(t) + x∆OO(t) − x ∆O(t) = 1 − x .

Therefore,

1 −1 2 −1 ∆L(t) = 1 − (t 2 −t 2 ) = −t + 3 −t. 34

4.2 Basic Properties of the Alexander Polynomial

In this section, the light will be shed on several important properties of the Alexander polynomial. But before that lets mention some deﬁnitions that will help in understanding the upcoming properties of the Alexander polynomial. For the proofs of theorems and propositions in this section, see Murasugi’s book [15].

Deﬁnition 4.2.1. A Seifert surface Σ for a knot K is a connected, orientable surface em- bedded in R3 with boundary K.

Deﬁnition 4.2.2. The genus of a knot (or non-split link) K is the smallest integer g(K) such that there exists an oriented surface, Σ, of genus g(Σ) whose boundary is K.

g(K) = ming(Σ); Σ is a Seifert surface for K,∂Σ =∼ K

Recall that the linking number lk(α1,α2) of two-component link with components

α1 and α2 is half the sum of the signs of the crossings at which one strand is from α1 and the other is from α2.

Deﬁnition 4.2.3. Let K be an oriented knot (or link) and Σ be a Seifert surface for K.

The orientation of F is induced naturally from the orientation of the knot K (or link) that forms its boundary. If the genus of the Seifert surface Σ of a knot (or link) is g(Σ), then on

Σ there are 2g(Σ) + µ(K) − 1(= m) closed curves α1,α2,...,αm, where µ(K) is just the number of components of the link K. The positive push-off can be formed of each closed

# curve αi where i = 1,2,...,m, denoted by αi , which runs parallel to αi and lies slightly above the surface Σ. The orientation may be assigned to the closed curves α1,α2,...,αm.

These orientations induce, in a natural manner orientations on the positive push-offs, see chapter 5 of Murasugi’s book [15]. This now enables us to calculate the linking number 35

# # lk(αi,α j ) of two closed curves,αi and α j , on Σ. The Seifert matrix of K is the integer matrix M of size mxm given as follows.

M = lk( , #) . αi α j i, j=1,2,...,m

Note that the Seifert matrix of a knot (or link) is not an invariant, as it is not unique. Different matrices may be obtained by changing the order of the closed curves

α1,α2,...,αm on the surface, or by using a different orientation.

Theorem 4.2.1. The Alexander polynomial of a knot (or link) K, ∆K(t), with Seifert matrix M, coming from a connected Seifert surface is given by the formula

− k T t 2 det(M −tM )

where k denotes the size of M and MT denotes the transpose of M. For a link which has a disconnected Seifert surface, the Alexander polynomial is deﬁned to be zero.

The ﬁrst property of Alexander polynomial is related to the following proposition.

Proposition 4.2.2. If K is a knot, then ∆K(1) = 1.

Note that if L is a µ-component (µ ≥ 2) link, then ∆L(1) = 0.

The next is an important property of the Alexander polynomial.

Theorem 4.2.3. Suppose that K is a knot, then ∆K(t) is a symmetric Laurent polynomial, i.e.,

−n −(n−1) n−1 n ∆K(t) = a−nt + a−(n−1)t + ··· + an−1t + ant 36 and

a−n = an,a−(n−1) = an−1,...,a−1 = a1

µ−1 Note that if L is a µ-component link, then ∆L(x) = x g(x) may be written,

2 1 −1 where g(x) is an integer polynomial in x and x = t 2 −t 2 .

Now, if Proposition 4.2.2 and Theorem 4.2.3 are connected, the Alexander polynomial can be characterized as in the following theorem.

Theorem 4.2.4. Suppose that f (t) is a Laurent polynomial that satisﬁes the following two conditions:

(1)f (1) = 1

(2)f (t) = f (t−1).

Then there exists a knot that has f (t) as its Alexander polynomial.

The Alexander polynomial is a useful invariant for classiﬁcation of knots (or links). However, it can not distinguish between the knot and its mirror image, and whether the knot is invertible or not.

Recall that an invertiable knot is a knot that can be continuously deformed to itself, but with its orientation reversed.

Theorem 4.2.5. Let K be a knot.

(1) If K∗ is the mirror image of K, then

∆K∗ (t) = ∆K(t). 37

(2) If −K is the knot obtained from K by reversing the orientation on K, then

∆K(t) = ∆−K(t).

Also, the Alexander polynomial cannot always distinguish a given knot from the trivial one. There are known examples of nontrivial knots with Alexander polynomial equal to 1. For example, the (−3,5,7)-pretzel knot given in Figure 4.5 has Alexander polynomial 1.

Figure 4.5: The (−3,5,7)-pretzel knot

In Example 4.1.1, the Alexander polynomial was obtained of the 2-component link is zero by applying the skein relation. In the same manner, the following proposition can be proven.

Proposition 4.2.6. The Alexander polynomial of the trivial link with µ-component (µ >

2) is 0.

Proof. If the skein relation is applied on a µ-component link, the skein diagrams L+ and

L− can be easily noticed that are trivial (µ − 1)-component links, see Figure 4.6. The 38 skein relation is

1 −1 2 2 ∆L+ (t) − ∆L− (t) = (t −t )∆L0 (t).

Since, ∆L+ (t) = ∆L− (t), therefore, ∆L0 (t) = 0.

Figure 4.6: µ-component link

In general, the proposition 4.2.6 may be generalized.

Recall that a split link is a link whose components can be deformed so that they lie on different sides of a plane in the 3-space.

Proposition 4.2.7. If L is a split link, then ∆L(t) = 0

Note that the converse is not true, ∆L(t) = 0 does not necessarily imply that L is a split link.

Theorem 4.2.8. Suppose that K1#K2 is the connected sum of two knots (or links) K1 and

K2, then

∆K1#K2 (t) = ∆K1 (t)∆K2 (t). 39

In fact, it is quite difﬁcult to determine the genus of a knot K. However, the Alexander polynomial can provide us with an approximate value of genus or an exact value in some cases. This remarkable property of the Alexander polynomial is stated by the following theorem.

Theorem 4.2.9. Suppose that the genus of a knot K is g(K), then the maximum degree of t in the Alexander polynomial cannot exceed g(K).

Proposition 4.2.10. The maximum degree of t in the Alexander polynomial is exactly g(K) if and only if the det(M) 6= 0.

Using Proposition 4.2.10 Murasugi proved the following theorem [6].

Theorem 4.2.11. Given an alternating knot K. Then, degree(∆K(t)) = g(K).

Also, he proved that the Alexander polynomial for an alternating knot has a special form [7]:

m i Theorem 4.2.12. Suppose that K is an alternating knot and ∆K(t) = ∑−m ait , with am and a−m nonzero. Then

(1)a −m,a−m+1,...,am are never equal to zero,

(2) The sign of two consecutive coefﬁcients alternates, i.e., aiai+1 < 0,∀ − m ≤ i ≤

m − 1.

4.3 Burau Representation

In the 1930s, Werner Burau discovered an interesting connection between a representation of the braid group Bn and the Alexander polynomial. This representation

−1 ψn,t : Bn −→ GL(n,Z[t,t ]) known as the Burau representation is deﬁned as follows. Let

σi ;1 ≤ i ≤ n − 1 be the standard generators of the braid group Bn. The image, ψn,t(σi), 40 is given by the following matrix:

  Ii−1 OO O      O 1 −t t O    ψn,t(σi) =  ,  O O O   1    O OO In−(i+1)

for 1 ≤ i ≤ n − 1, where Ii denotes the ixi identity matrix. This representation, ψn,t, is re-

0 −1 ducible, and is reduced to an irreducible representation ψn,t : Bn −→ GL(n − 1,Z[t,t ]). For n ≥ 3 the reduced Burau representation is given by the following matrices:

  −t 1 O   0   ψ (σ1) =  0 1 O , n,t     OO In−3   Ii− OOO O  2     O O   1 0 0  0     ψn,t(σi) =  O t −t 1 O , 2 ≤ i ≤ n − 2,      O 0 0 1 O      O OOO In−(i+2)   In−3 OO   0   ψ (σn−1) =  O 1 0 , n,t     O t −t 0 while for n = 2, ψ2,t(σ1) = (−t).

The reduced Burau representation of the braid group offers another insight to study the Alexander polynomial. The Alexander polynomial can be computed using a relatively simple formula. If a link L is the closure of a braid α in Bn, and eα is the exponent sum of α as a word in the elementary braids σi, for 1 ≤ i ≤ n − 2, then the 41

Alexander polynomial ∆L(t) of the link L = αˆ is given by the following formula.

0 √ det(In−1 − ψ (α)) eα −2 n,t ∆L(t) = (−1/ t) (1 −t) 0 , det(In−1 − ψn,t(σ1σ2 ...σn−1))

0 whereψn,t(α) is the reduced Burau representation of the braid α.

In the case of B3, this representation is faithful and can be described simply as follows. Let b be a given 3-braid and eb the exponent sum of b as a word in the elementary 0 −1 braids σ1 and σ2. Let ψ3,t : B3 −→ GL(2,Z[t,t ]) be the reduced Burau representation    

0 −t 1 0 1 0 B ( ) =   ( ) =   deﬁned on the generators of 3 by ψ3,t σ1   and ψ3,t σ2  . 0 1 t −t The Alexander polynomial of the link L = bˆ is given by:

√ 1 −t 0 ∆ (t) = (−1/ t)eb−2 det(I − ψ (b)), L 1 −t3 2 3,t

where I2 is the identity matrix. Given that this representation is 2-dimensional, then 0 0 0 det(I2 −ψ3,t(b)) = 1−tr(ψ3,t(b))+det(ψ3,t(b)), where tr denotes the usual matrix-trace function deﬁned as the sum of diagonal entries of the matrix.

Example 4.3.1. Let α = σ1σ2 in B3. The matrices above can be used to compute the

Alexander polynomial of L = αˆ . The reduced Burau representation of α is given by:

     −t 1 1 0 0 −t 0 0 0      ψ (σ1σ2) = ψ (σ1)ψ (σ2) =    =  , 3,t 3,t 3,t      0 1 t −t t −t

and   1 t 0   2 det(I2 − ψ (σ1σ2)) = det( ) = 1 +t +t . 3,t   −t 1 +t 42

1 −t 0 1 −t ∆ (t) = det(I − ψ (σ σ )) = (1 +t +t2) = 1. L 1 −t3 2 3,t 1 2 1 −t3

Note that the closure of α = σ1σ2 is the unknot whose Alexander polynomial is 1.

2 Example 4.3.2. Let α = (σ1σ2) in B3. The matrices above can be used to compute the

Alexander polynomial of L = αˆ . The reduced Burau representation of α is given by:

0 2 0 0 0 0 ψ3,t((σ1σ2) ) = ψ3,t(σ1)ψ3,t(σ2)ψ3,t(σ1)ψ3,t(σ2)        −t 1 1 0 −t 1 1 0 −t2 t2        =      =  ,        0 1 t −t 0 1 t −t −t2 0

and   1 +t2 −t2 0   2 4 det(I2 − ψ (σ1σ2)) = det( ) = 1 +t +t . 3,t   t2 1

So 2 1 1 −t 0 1 −t 2 4 ∆L(t) = √ det(I − ψ (σ σ )) = (1 +t +t ) t 1 −t3 2 3,t 1 2 t(1 −t3)

1 −t = (1 −t +t2)(1 +t +t2) = t−1 − 1 +t. t(1 −t3)

2 Note that the closure of α = (σ1σ2) is the trefoil knot whose Alexander polynomial is

−1 ∆L(t) = t − 1 +t, as it was computed using the skein relations in Example 4.1.4. 43

Chapter 5: Fox-Trapezoidal Conjecture for Closed 3-Braids

5.1 Fox-Trapezoidal Conjecture

In 1962, R. Fox conjectured that the coefﬁcients of the Alexander polynomial of an alternating knot are trapezoidal [8]. In other words, these coefﬁcients, increase, stabilize then decrease in a symmetrical way. This conjecture can be stated as follows.

2n i Conjecture 5.1.1. Let K be an alternating knot and ∆K(t) = ± ∑ αi(−t) , with αi > 0, i=0 its Alexander polynomial. Then there exists a non-negative integer m ≤ n such that:

α0 < α1 < ··· < αn−m = ··· = αn+m > ...... > α2n−1 > α2n.

This conjecture, known as Fox’s trapezoidal conjecture, has been confirmed for several classes of alternating knots. The case of two-bridge knots has been proved by Hartley [9]. Murasugi, confirmed the conjecture for all alternating algebraic knots [10]. Using knot Floer homology, Ozsva ´th and Szabo ´ [5] proved that the conjecture holds for knots of genus 2. The same result has been obtained by Jong using combinatorial methods [11, 12]. More recently, Hirasawa and Murasugi [13] proved that Conjecture 5.1.1 holds for stable alternating knots and suggested a refinement of this conjecture by stating that the length of the stable part is less than or equal the signature of the knot σ(K), which means m ≤ |σ(K)|/2. In [17], Chen confirmed that this refinement holds for two-bridge knots.

Note that σ(L) denotes the signature of link L, which can be brieﬂy deﬁned as a topological invariant σ(L) ∈ Z that can be computed from the Seifert matrix of L. If L is the closure of a 3-braid, then σ(L) can be computed using Erle’s formula [18]. For

p −q p −q instance, the signature of the closed braid σ1 1 σ2 1 ...σ1 n σ2 n is q − p, where p = n n ∑i=1 pi and q = ∑i=1 qi. 44

5.2 Main Result

In this section, the Fox trapezoidal conjecture for alternating knots of braid index 3 will be discussed, and the main result in this master thesis will be introduced.

A full classiﬁcation of alternating links of braid index 3 is given by Stoimenow [3].

Theorem 5.2.1. Let L be an alternating link of braid index 3. Then L is either:

(1) the connected sum of two (2,k)-torus links (with parallel orientation), or

(2) the closure of an alternating 3-braid, including split unions of a (2,k)-torus link

and an unknot and the 3 component unlink), or

(3) a pretzel link P(1, p,q,r) with p,q,r ≥ 1 (oriented so that the twists corresponding

to p,q,r are parallel).

Conjecture 5.1.1 holds for cases 1 and 3 above, as in both cases the knots are algebraic. This study focuses on the other case and proves that for a certain family of alternating knots described in case 2 above the Alexander polynomial is trapezoidal.

It is well known that 3-braids have been completely classiﬁed, up to conjugation, by Murasugi [2].

3 Theorem 5.2.2. Let b be a 3-braid and let h = (σ1σ2) be a full positive twist. Then b is conjugate to exactly one of the following:

n p1 −q1 pr −qr (1)h σ1 σ2 ...σ1 σ2 , where r, pi and qi are positive integers.

n m (2)h σ2 where m ∈ Z.

n m −1 (3)h σ1 σ2 , where m ∈ {−1,−2,−3}.

According to Murasugi’s classiﬁcation of 3-braids, an alternating 3-braid whose

p −q p −q closure is non-split can be written in the form α = σ1 1 σ2 1 ...σ1 r σ2 r , where r ≥ 1 45

and pi,qi ≥ 1 for all i. The length of the braid will be called r, and it is written l(α) = r. The main theorem can be stated as follows.

Theorem 5.2.3. Let α be an alternating 3-braid of length l(α) ≤ 3 and L = αˆ . Then the Alexander polynomial ∆L(t) is trapezoidal. Moreover, the length of the stable part in

∆L(t) is 2m ≤ σ(L).

The proof of Theorem 5.2.3 is discussed in details in Section 5.3.

5.3 Proof of the Main Theorem

This section will be devoted to the proof of Theorem 5.2.3, which conﬁrms that

p −q p −q p −q Fox’s trapezoidal conjecture holds for the closure of the σ1 1 σ2 1 , σ1 1 σ2 1 σ1 2 σ2 2 p −q p −q p −q and σ1 1 σ2 1 σ1 2 σ2 2 σ1 3 σ2 3 . Since the Alexander polynomial of these links is alternating, It will be more convenient to make the variable change s = −t.

p −q First, it is worth mentioning that the closure of the braid σ1 1 σ2 1 is actually a connected sum of the torus links of type (2, p1) and (2,−q1). Without loss of generality, assume that p1 ≤ q1. Hence, its Alexander polynomial up to the multiplication by a power of s is,

p1−1 q1−1 i i p1−1 q1−1 q1 p1+q1−2 ( ∑ s )( ∑ s ) = 1+2s+···+ p1s +···+ p1s +(p1 −1)s +···+s . i=0 i=0 (5.1)

This polynomial is clearly trapezoidal and the length of the stable part is 2m =

(q1 − 1 − (p1 − 1)) = q1 − p1 = σ(L).

The proof of Theorem 5.2.3 in the case of the closure of the braid

p −q p −q σ1 1 σ2 1 σ1 2 σ2 2 is based on elementary calculations of the Alexander polynomial of closed 3-braids using the reduced Burau representation.

p −q p −q Proposition 5.3.1. Let L be the closure of the 3-braid σ1 1 σ2 1 σ1 2 σ2 2 . Then, up to 46 the multiplication by a power of s, the Alexander polynomial of L is,

−s−q1−q2 ∆ (s) = (s(1−sp1 )(1−sp2 )(1−sq1 )(1−sq2 )+(1−sp1+p2 )(1−sq1+q2 )(1−s)2). L (1 − s)4

Proof. For any positive integers pi and qi, elementary computations show that:

    1 − spi spi 1 0      1 − s    0 pi   0 −qi   ψ (σ ) =  , and ψ (σ ) =  . 3,s 1   3,s 2          −s(1 − s−qi )  0 1 s−qi 1 − s

Consequently,   1 − spi −s(1 − s−qi ) 1 − spi spi + s−qi  1 − s 1 − s 1 − s  0   pi −qi   ψ (σ σ ) =  . 3,s 1 2      −s(1 − s−qi )  s−qi 1 − s

p1 −q1 p2 −q2 Thus, for α = σ1 σ2 σ1 σ2 , the reduced Burau representation of α is,

0 0 p1 −q1 p2 −q2 0 p1 −q1 0 p2 −q2 ψ3,s(α) = ψ3,s(σ1 σ2 σ1 σ2 ) = ψ3,s(σ1 σ2 )ψ3,s(σ1 σ2 ) =

   1 − sp1 −s(1 − s−q1 ) 1 − sp1 1 − sp2 −s(1 − s−q2 ) 1 − sp2 sp1 + s−q1 sp2 + s−q2     1 − s 1 − s 1 − s  1 − s 1 − s 1 − s       .        −s(1 − s−q1 )  −s(1 − s−q2 )  s−q1 s−q2 1 − s 1 − s

  A(s) B(s)       This product of matrices can be put in the form  , where       C(s) D(s) 47

1 − sp1 −s(1 − s−q1 ) −s(1 − s−q2 ) A(s) = sp2 sp1 + + 1 − s 1 − s 1 − s 1 − sp1 1 − sp2 1 − sp1 −s(1 − s−q1 ) s−q1 + sp1 + 1 − s 1 − s 1 − s 1 − s 1 − sp1 1 − sp2 1 − sp1 −s(1 − s−q1 ) B(s) = s−q2 s−q1 + sp1 + 1 − s 1 − s 1 − s 1 − s −s(1 − s−q1 ) −s(1 − s−q2 ) 1 − sp2 −s(1 − s−q1 ) C(s) = sp2 + s−q1 + and 1 − s 1 − s 1 − s 1 − s 1 − sp2 −s(1 − s−q1 ) D(s) = s−q2 s−q1 + . 1 − s 1 − s

The determinant and the trace of the resulting matrix above are given by:

0 p1+p2−q1−q2 • det(ψ3,s(α)) = s . p −q −q p 0 1 − s 1 −s(1 − s 1 ) −s(1 − s 2 ) 1 − s 1 tr(ψ (α)) = sp2 [sp1 + ] + [s−q1 + 3,s 1 − s 1 − s 1 − s 1 − s 1 − sp2 1 − sp1 −s(1 − s−q1 ) • [sp1 + ]] + s−q2 [s−q1 1 − s 1 − s 1 − s 1 − sp2 −s(1 − s−q1 ) + ]. 1 − s 1 − s Thus,

0 0 0 det(1 − ψ3,s(α)) = 1 −tr(ψ3,s(α) + det(ψ3,s(α)) s−q1−q2 = − (1 − s + s2)(1 − s + s2 − sp1+1 − sp2+1 − sp1+p2 + (s − 1)4

3sp1+p2+1 − sp1+p2+2 − sq1+1 + sp1+q1+1 + sp2+q1+1−

sp1+p2+q1+1 − sq2+1 + sp1+q2+1 + sp2+q2+1 − sp1+p2+q2+1−

sq1+q2 + 3sq1+q2+1 − sq1+q2+2 − sp1+q1+q2+1 − sp2+q1+q2+1+

sp1+p2+q1+q2 − sp1+p2+q1+q2+1 + sp1+p2+q1+q2+2).

0 1 + s Now, the det(1 − ψ (α)) will be multiplied by to get the Alexander polynomial 3,s 1 + s3 −1eα −2 of the link L = αˆ . It is worth mentioning here that the term corresponding to √ t is ignored as it will not affect the trapezoidal behavior of the polynomial. 48

1 + s 0 ∆ (s) = det(1 − ψ (α)) L 1 + s3 3,s s−q1−q2 = − (1 − s + s2 − sp1+1 − sp2+1 − sp1+p2 + 3sp1+p2+1 − sp1+p2+2− (s − 1)4

sq1+1 + sp1+q1+1 + sp2+q1+1 + sp1+p2+q1+1 − sq2+1 + sp1+q2+1 + sp2+q2+1−

sp1+p2+q2+1 − sq1+q2 + 3sq1+q2+1 − sq1+q2+2 − sp1+q1+q2+1 − sp2+q1+q2+1+

sp1+p2+q1+q2 − sp1+p2+q1+q2+1 + sp1+p2+q1+q2+2).

Finally, by doing routine elementary calculations, the result of these calculations is as follows: s−q1−q2 ∆ (s) = − (1 − s(1 + sp1 + sp2 − 3sp1+p2 + sq1 − sp1+q1 − sp2+q1 + sp1+p2+q1 + L (1 − s)4

sq2 − sp1+q2 − sp2+q2 + sp1+p2+q2 − 3sq1+q2 + sp1+q1+q2 + sp2+q1+q2 +

sp1+p2+q1+q2 ) − s2(sp1+p2 + sq1+q2 − sp1+p2+q1+q2 − 1) − sp1+p2 −

sq1+q2 + sp1+p2+q1+q2 )

s−q1−q2 = − (1 − s(−1 + sp1 + sp2 − sp1+p2 + sq1 − sp1+q1 − sp2+q1 + sp1+p2+q1 + (1 − s)4

sq2 − sp1+q2 − sp2+q2 + sp1+p2+q2 − sq1+q2 + sp1+q1+q2 + sp2+q1+q2 −

sp1+p2+q1+q2 ) + (−2s + 2sp1+p2+1 + 2sq1+q2+1 − 2sp1+p2+q1+q2+1)−

s2(sp1+p2 + sq1+q2 − sp1+p2+q1+q2 − 1) − sp1+p2 − sq1+q2 + sp1+p2+q1+q2 )

s−q1−q2 = − (−s(−(1 − sp1 )(1 − sp2 )(1 − sq1 )(1 − sq2 )) − 2s(1 − sp1+p2 − (1 − s)4

sq1+q2 + sp1+p2+q1+q2 ) + s2(1 − sp1+p2 − sq1+q2 + sp1+p2+q1+q2 )+

(1 − sp1+p2 − sq1+q2 + sp1+p2+q1+q2 ))

s−q1−q2 = − (s(1 − sp1 )(1 − sp2 )(1 − sq1 )(1 − sq2 )) + (1 − sp1+p2 − sq1+q2 + (1 − s)4

sp1+p2+q1+q2 )(−2s + s2 + 1))

s−q1−q2 = − (s(1 − sp1 )(1 − sp2 )(1 − sq1 )(1 − sq2 ) + (1 − sp1+p2 )(1− (1 − s)4

sq1+q2 )(1 − s)2). 49

Lemma 5.3.2. The polynomial h(s) = (1 + ··· + sp1−1)(1 + ··· + sp2−1)(1 + ··· + sq1−1)

p1+p2+q1+q2−4 q −1 k (1 + ··· + s 2 ) is trapezoidal. It can be written ∑ cks , where 0

(1) if p1 + p2 + q1 − 2 < q2 then

• ck+1 > ck for k < p1 + p2 + q1 − 3

• ck+1 = ck for p1 + p2 + q1 − 3 ≤ k < q2 − 1

• ck+1 < ck otherwise

(2) if p1 + p2 + q1 − 2 ≥ q2 and p1 + p2 + q1 + q2 even, then

p1+p2+q1+q2−4 • ck+1 > ck for k < 2

• ck+1 < ck otherwise

(3) if p1 + p2 + q1 − 2 ≥ q2 and p1 + p2 + q1 + q2 odd, then

p1+p2+q1+q2−5 • ck+1 > ck for k < 2

p1+p2+q1+q2−5 • ck+1 = ck for k = 2

• ck+1 < ck otherwise.

Proof. Without loss of generality, It can be assumed that p1 ≤ p2 ≤ q1 ≤ q2. The ﬁrst two cases will be proved in the lemma above. Similar arguments can be used to prove the third case.

Case 1. Assume that p1 + p2 + q1 − 2 < q2. Let

f (s) = (1 + ··· + sp1−1)(1 + ··· + sp2−1)

p −1 p −1 p p +p −2 = (1 + 2s + ··· + p1s 1 + ··· + p1s 2 + (p1 − 1)s 2 + ··· + s 1 2 )

and 50

g(s) = (1 + ··· + sq1−1)(1 + ··· + sq2−1)

q −1 q −1 q q +q −2 = (1 + 2s + ··· + q1s 1 + ··· + q1s 2 + (q1 − 1)s 2 + ··· + s 1 2 ).

i i For simplicity, write f (s) = ∑ ais and g(s) = ∑ bis . Then, i=0 i=0 p1+p2+q1+q2−4 k k h(s) = ∑ cks where ck = ∑ aibk−i. Note here that ai = 0 for i > p1 + p2 − 2 0 i=0 k and bi = 0 for i > q1 + q2 − 2. Thus, ck+1 − ck = ∑ ai(bk+1−i − bk−i) + ak+1b0. i=0

It is clear that if k < p1 + p2 + q1 − 3 then k + 1 < q2. Consequently, bk+1−i − bk−i ≥ 0.

Moreover, ai ≥ 0 and one may check easily that at least one of the terms ai(bk+1−i −bk−i) is nonnegative. Indeed, if k ≤ p1 + p2 − 2, then ak(b1 − b0) > 0. Otherwise, the last

term in the sum will be ap1+p2−2(bk+1−(p1+p2−2) − bk−(p1+p2−2)). It is easy to check that k + 1 − (p1 + p2 − 2) = k − p1 − p2 + 3 < p1 + p2 + q1 − 3 − p1 − p2 + 3 = q1 and then

(bk+1−(p1+p2−2) − bk+1−(p1+p2−2)) > 0. Finally, ck+1 − ck > 0.

Now, the next case p1 + p2 + q1 − 3 ≤ k < q2 − 1 is checked. Note that in the product ai(bk+1−i −bk−i) either i > p1 + p2 −2 or k −i ≥ p1 + p2 +q1 −3− p1 − p2 +2 = q1 −1.

If i > p1 + p2 − 2 then ai = 0 otherwise bk+1−i − bk−i = 0 as q1 − 1 ≤ k − i < q2 − 1.

The case k ≥ q2 is treated in the same way as the case k ≤ p1 + p2 +q1 −3. This ends the proof of case 1.

Case 2. Assume that p1 + p2 + q1 − 2 ≥ q2 and p1 + p2 + q1 + q2 is even.

p1+p2+q1+q2−4 Note that if k < 2 , there are two cases to consider. If k < q2 − 1, then

(bk+1−i − bk−i) ≥ 0 and then ck+1 − ck > 0 as one of the terms ai(bk+1−i − bk−i) is non- negative. 51

p1+p2+q1+q2−4 Assume now that q2 − 1 ≤ k < 2 . Then,

k ck+1 − ck = ∑ ai(bk+1−i − bk−i) + ak+1b0 i=0

= a0(bk+1 − bk) + ··· + ak−(q2−1)(bq2 − bq2−1) + ...

+ak−q1+2(bq1−1 − bq1−2) + ··· + ak(b1 − b0) + ak+1b0.

Obviously, for all j the difference b j+1 − b j is either -1, 0 or 1. Thus ck+1 − ck = −(a0 +

a1 +···+ak−(q2−1))+(ak−q1+2 +···+ak)+ak+1b0. It is clear that the number of negative terms is k − q2 + 2 while the number of positive terms is either (p1 + p2 − 2 − (k − q1 +

2)) = p1 + p2 + q1 − k − 3 or q1. One may check easily that in both cases the number of positive terms is larger than the number of negative terms. Moreover both terms are consecutive coefﬁcients of f (t) which are trapezoidal and satisfy (ai+1 −ai) ∈ {−1,0,1}.

This implies that ck+1 − ck > 0.

p1+p2+q1+q2−4 The proof for k ≥ 2 is done similarly using symmetries of the polynomials f and g. This ends the proof of case 2.

According to Proposition 5.3.1, the Alexander polynomial of L, −s−q1−q2 ∆ (s) = (s(1 − sp1 )(1 − sp2 )(1 − sq1 )(1 − sq2 ) + (1 − sp1+p2 )(1 − sq1+q2 )(s − L (s − 1)4 p1+p2+q1+q2−2 2 −q −q k 1) ), can be written in the form −s 1 2 [sh(s) + r(s)] where r(s) = ∑ dks . k=0

Lemma 5.3.3. The polynomial w(s) = sh(s) + r(s) is trapezoidal.

Proof. By Lemma 5.3.2 and equation (5.1), the two polynomials h(s) and r(s) are both 52 trapezoidal. Let us prove now that w(s) is trapezoidal.

w(s) = sh(s) + r(s)

p1+p2+q1+q2−4 p1+p2+q1+q2−2 k k = s ∑ cks + ∑ dks k=0 k=0 p1+p2+q1+q2−4 p1+p2+q1+q2−2 k+1 k = ∑ cks + ∑ dks k=0 k=0 p1+p2+q1+q2−3 p1+p2+q1+q2−3 k k p1+p2+q1+q2−2 = ∑ ck−1s + ∑ dks + s k=1 k=0 p1+p2+q1+q2−3 k p1+p2+q1+q2−2 = 1 + ∑ (ck−1 + dk)s + s . k=1

p1+p2+q1+q2−3 k Now, let αk = ck−1 + dk. It should be proven that ∑ αks is trapezoidal. As k=1 in the previous lemma, there are 3 cases to consider. The second case will be just proven, and the two other cases can be done similarly. Suppose that p1 + p2 + q1 − 2 ≥ q2 and p1 + p2 + q1 + q2 even, then

p1+p2+q1+q2−4 • if k ≤ 2 , then ck > ck−1 and dk+1 > dk. Hence, αk+1 > αk.

p1+p2+q1+q2−4 • if k > 2 , then ck < ck−1 and dk+1 < dk. Hence, αk+1 < αk.

Therefore, the polynomial w(s) is trapezoidal. Consequently, the Alexander poly-

p −q p −q nomial of the closure of the 3-braid σ1 1 σ2 1 σ1 2 σ2 2 is trapezoidal as well. More- over, it is easy to see that in all cases discussed above, the length of the stable part in the

Alexander polynomial satisﬁes 2m ≤ σ(L) = q2 +q1 − p1 − p2. This completes the proof of Theorem 5.2.3 in the case l(α) = 2.

Example 5.3.1. A graphical illustration of the coefﬁcients of the polynomials sh(s),r(s) and w(s) for the three cases using Wolfram Mathematica.

• For the ﬁrst case ( Figure 5.1 ): p1 = 5, p2 = 8,q1 = 6, and q2 = 24.

• For the second case ( Figure 5.2 ): p1 = 4, p2 = 5,q1 = 6, and q2 = 7. 53

• For the third case ( Figure 5.3 ): p1 = 4, p2 = 5,q1 = 6, and q2 = 8.

All the polynomials sh(s),r(s), and w(s) that are found in the three cases are trape-

zoidal.

Figure 5.1: Case (1): p1 + p2 + q1 − 2 < q2

Figure 5.2: Case (2): p1 + p2 + q1 − 2 ≥ q2 and p1 + p2 + q1 + q2 even 54

Figure 5.3: Case (3): p1 + p2 + q1 − 2 ≥ q2 and p1 + p2 + q1 + q2 odd

The case of alternating braids of length 3 is done similarly. The details will be omitted, and the idea will be brieﬂy described. Indeed, using the same elementary computations of the matrix trace function, a formula for the Alexander polynomial of the clo-

p −q p −q p −q sure of the braid σ1 1 σ2 1 σ1 2 σ2 2 σ1 3 σ2 3 , is obtained, as stated in the proposition below.

p1 −q1 p2 −q2 p3 −q3 Proposition 5.3.4. Let L be the closure of the 3-braid σ1 σ2 σ1 σ2 σ1 σ2 .

Then, up to the multiplication by a power of s, the Alexander polynomial of L is,

−s−q1−q2−q3 ∆ (s) = [s2(1 − sp1 )(1 − sp2 )(1 − sp3 )(1 − sq1 )(1 − sq2 )(1 − sq3 ) L (1 − s)6

+ s(1 − s)2((1 − sp1 )(1 − sp2+p3 )(1 − sq1 )(1 − sq2+q3 )

+ (1 − sp2 )(1 − sp3 )(1 − sq2 )(1 − sp1+q1+q3 )

+ (1 − sp1 )(1 − sp3 )(1 − sq2 )(sp2+q1 − sq3 ))

+ (1 − s)4(1 − sp1+p2+p3 )(1 − sq1+q2+q3)]. 55

Now, a brief explanation will be provided to explain why the polynomial ∆L(s) in Proposition 5.3.4 is trapezoidal. In the same way, as discussed above, this polynomial can be written in the following form.

−q −q −q 2 ∆L(s) = −s 1 2 3 [s A(s) + s(B(s) +C(s) + D(s)) + E(s)] , where

p1+p2+p3+q1+q2+q3−6 k A(s) = ∑ aks k=0 p1+p2+p3+q1+q2+q3−4 k B(s) = ∑ bks k=0 p1+p2+p3+q1+q2+q3−4 k C(s) = ∑ cks k=0 p1+p2+p3+q1+q2+q3−4 k D(s) = ∑ dks and k=0 p1+p2+p3+q1+q2+q3−2 k E(s) = ∑ eks . k=0 The proof of Lemma 5.3.2 can be reproduced here to prove that the polynomial A(s) is trapezoidal.

p −1 Lemma 5.3.5. For any p1, p2, p3,q1,q2,q3 ≥ 1 the polynomial A(s) = (1+···+s 1 )(1+

··· + sp2−1)(1 + ··· + sp3−1)(1 + ··· + sq1−1)(1 + ··· + sq2−1)(1 + ··· + sq3−1) is trapezoidal.

Lemma 5.3.6. The polynomial T(s) = s2A(s) + s(B(s) +C(s) + D(s)) + E(s) is trapezoidal.

By Lemma 5.3.2, It is easy to show that the polynomial B(s) + C(s) + D(s) is trapezoidal. So is the polynomial E(s) by equation (5.1). An argument similar to that used in Lemma 5.3.3 will enable us to prove that T(s) is trapezoidal, which allows us to conclude that ∆L(s) is trapezoidal. Finally, one can easily check that the length of the sable part in the Alexander polynomial satisﬁes 2m ≤ σ(L) = q3 +q2 +q1 − p1 − p2 − p3. This ends the proof of Theorem 5.2.3.

Example 5.3.2. A graphical illustration of the coefﬁcients of the polynomials s2A(s),G(s) = s(B(s) + C(s) + D(s)),E(s) and T(s) is demonstrated in the following examples using 56

Wolfram Mathematica.

• Example 1 (Figure 5.4): p1 = 3, p2 = 4, p3 = 6,q1 = 7,q2 = 9 and q3 = 13.

• Example 2 (Figure 5.5): p1 = 3, p2 = 4, p3 = 5,q1 = 7,q2 = 9 and q3 = 10.

All the polynomials s2A(s),G(s),E(s), and T(s) that are found are trapezoidal.

Figure 5.4: Example 1: All polynomials are trapezoidal

Figure 5.5: Example 2: All polynomials are trapezoidal 57

5.4 Conclusion and Future Work

In this thesis, It was proved that the Alexander polynomial of the closure of the alternating 3-braid of length less than or equal three is trapezoidal. By now, it is believed that the trapezoidal conjecture for the Alexander polynomial of alternating knots of braid

p1 −q\1 pr −qr index 3 (in the form L = σ1 σ2 ...σ1 σ2 , where r ≥ 1 and pi,qi ≥ 1, according to Murasugi’s classiﬁcation of 3-braids) might be proved. This will require introducing a general formula of the Alexander polynomial of knots of braid index 3, and test whether or not it satisﬁes the trapezoidal conjecture. 58

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