The Volume Conjecture, the Aj Conjectures and Skein Modules

THE VOLUME CONJECTURE, THE AJ CONJECTURES AND SKEIN MODULES

A Thesis Presented to The Academic Faculty

Anh Tuan Tran

In Partial Fulﬁllment of the Requirements for the Degree Doctor of Philosophy in the School of Mathematics

Georgia Institute of Technology August 2012 THE VOLUME CONJECTURE, THE AJ CONJECTURES AND SKEIN MODULES

Approved by:

Dr. Thang T.Q. Le, Advisor Dr. Patrick Gilmer School of Mathematics Department of Mathematics Georgia Institute of Technology Louisiana State University

Dr. John Etnyre Dr. Dan Margalit School of Mathematics School of Mathematics Georgia Institute of Technology Georgia Institute of Technology

Dr. Stavros Garoufalidis Date Approved: 1 July 2012 School of Mathematics Georgia Institute of Technology ACKNOWLEDGEMENTS

I am extremely grateful to my advisor, Dr. Thang T.Q. Le, for his valuable instruc- tion and constant support throughout my doctoral study at the Georgia Institute of

Technology. I would like to give a special thank to Dr. John Etnyre, with whom I

took many courses and readings, for valuable advice not only in teaching and research

but also in job search.

I would like to thank the faculty and staﬀ of the School of Mathematics, Georgia

Institute of Technology for providing me a great educational environment as well as constant ﬁnancial support. In particular, thanks go to Dr. Igor Belegradek, Dr. Jean

Bellissard, Dr. Luca Dieci, Dr. Stavros Garoufalidis, Dr. Mohammad Ghomi, Dr.

Dan Margalit, Dr. Doron Lubinsky, Ms. Klara Grodzinsky, Ms. Cathy Jacobson,

Ms. Sharon McDowell, and Ms. Genola Turner.

I would like to express my gratitude to the committee members, Dr. John Etnyre,

Dr. Stavros Garoufalidis, Dr. Patrick Gilmer, and Dr. Dan Margalit for their valuable advice and comments in my dissertation.

I am very grateful to my friends Meredith Casey, Phu Chung, Alan Diaz, Giang

Do, Yen Do, Luan Hoang, Huy Huynh, Amey Kaloti, James Krysiak, Phi Le, Nan

Lu, Quan Nguyen, Truyen Nguyen, Marc Sedjro, Hyunshik Shin, Bulent Tosun, Ar- men Vagharshakyan, Nham Vo, Thao Vuong, and Rebecca Winarski for their useful discussions not only in Mathematics but also in sharing life during the time I studied at the Georgia Institute of Technology.

I also would like to take this opportunity to express my sincere appreciation to my family, in particular my parents Tran Van Linh and Nguyen Thi Thu, and my wife Nhu Hoang for support and love.

iii TABLE OF CONTENTS

ACKNOWLEDGEMENTS ...... iii

LIST OF FIGURES ...... vii

SUMMARY ...... viii

I ON THE VOLUME CONJECTURE FOR CABLES OF KNOTS 1 1.1 Introduction ...... 1 1.1.1 The colored Jones polynomial and the Kashaev invariant of a link...... 1 1.1.2 Thevolumeconjectureforknotsandlinks...... 1 1.1.3 Mainresults ...... 2 1.1.4 Planofthechapter ...... 4 1.2 Cables, the colored Jones polynomial, and results ...... 4 1.2.1 Cables, the colored Jones polynomial ...... 4 1.2.2 General knot case and even m ...... 5 1.2.3 Figure8knotcase...... 6 1.3 Someelementarycalculations...... 7 1.3.1 Notationsandconventions ...... 7 1.3.2 The building blocks A(N,k) and S(k,l)...... 7 1.4 ProofofTheorem1.2.1 ...... 10 1.4.1 Habiroexpansion ...... 10 1.4.2 ProofofTheorem1.2.1 ...... 10 1.5 Proofoftheorem1.2.2...... 11 1.5.1 The case j

II ON THE AJ CONJECTURE FOR KNOTS ...... 20 2.1 Introduction ...... 20

iv 2.1.1 TheAJconjecture...... 20 2.1.2 Mainresults ...... 21 2.1.3 Otherresults...... 22 2.1.4 Planofthechapter ...... 22 2.2 Skein Modules and the colored Jones polynomial ...... 23 2.2.1 Skeinmodules ...... 23 2.2.2 The skein module of S3 and the colored Jones polynomial . . 24 2.2.3 Theskeinmoduleofthetorus ...... 24 2.2.4 Theorthogonalandperipheralideals ...... 25 2.2.5 Relation between the recurrence ideal and the orthogonal ideal 26 2.3 Character varieties and the A-polynomial ...... 27 2.3.1 Thecharactervarietyofagroup ...... 27 2.3.2 Theuniversalcharacterring ...... 28 2.3.3 The A-polynomial...... 28 2.3.4 The B-polynomial...... 29 2.3.5 Relation between the A-polynomial and the B-polynomial . . 29 2.3.6 Smallknots ...... 31 2.4 SkeinmodulesandtheAJconjecture ...... 33 2.4.1 Skein modules as quantizations of character varieties . . . . . 33 2.4.2 The peripheral polynomial and its role in the AJ conjecture . 34 2.4.3 The L-degree of the A-polynomial ...... 37 2.4.4 Localization and the L-degree of the peripheral polynomial . 38 2.4.5 ProofofTheorems5and6 ...... 40 2.5 The universal character ring of ( 2, 3, 2n +1)-pretzelknots . . . . . 40 − 2.5.1 Thecharactervariety ...... 41 2.5.2 Theuniversalcharacterring ...... 46

III THE SKEIN MODULE OF TWO-BRIDGE LINKS ...... 52 3.1 Introduction ...... 52

v 3.1.1 MainResults...... 52 3.1.2 Theuniversalcharacterring ...... 53 3.2 ProofofTheorem7andProposition1 ...... 54 3.2.1 Description of u ...... 56 3.2.2 The link complement ...... 58 3.2.3 Relative skein modules ...... 59 3.2.4 From (X ) to (X) through sliding ...... 62 S 1 S 3.2.5 ProofofTheorem7 ...... 64 3.2.6 ProofofProposition1...... 64

IV PROOF OF A STRONGER VERSION OF THE AJ CONJEC- TURE FOR TORUS KNOTS ...... 68 4.1 Introduction ...... 68 4.1.1 Mainresults ...... 68 4.1.2 Planofthechapter ...... 69 4.2 ProofoftheAJconjecturefortorusknots...... 69 4.2.1 The case a,b > 2...... 69 4.2.2 The case a =2...... 72 4.3 ProofofTheorem8 ...... 74 4.3.1 The case a,b > 2...... 74 4.3.2 The case a =2...... 77 4.3.3 ProofofTheorem8 ...... 78

V FUTURE PLANS ...... 81

REFERENCES ...... 83

VITA ...... 87

vi LIST OF FIGURES

1 The ( 2, 3, 2n +1)-pretzelknot ...... 41 − 2 The loops x, x′ and y ...... 53

3 The ball B1 ...... 55 4 The curves m and l...... 56 5 The total lifts m˜ ,˜l of m and l on the torus S˜ respectively...... 56

6 u0,l, u0,m, u0,r for 2p = 6 and q′ =5...... 57

7 The distribution of the points Aj, Fj, Ej and Aj′ , Fj′, Ej′ on the circles C and C′ respectively...... 59

8 ωl,ωm,ωr for 2p = 6 and q =5...... 59

9 ωs connects P, Q, and ωs′ connects P ′, Q′ ...... 60

10 The projections of a2, a3 and a4 onto the x1x3-plane...... 61

11 The projection of a2′ , a3′ and a4′ onto the x1x3-plane ...... 62

vii SUMMARY

This dissertation studies quantum invariants of knots and links, particularly the

colored Jones polynomials, and their relationships with classical invariants like the

hyperbolic volume and the A-polynomial.

In Chapter 1, we study the volume conjecture of Kashaev-Murakami-Murakami that relates the Kashaev invariant, a specialization of the colored Jones polynomial at a speciﬁc root of unity, and the hyperbolic volume of a link. We establish the conjecture for (m, 2)-cables of the ﬁgure 8 knot, when m is odd. For (m, 2)-cables of general knots where m is even, we show that the limit in the volume conjecture

depends on the parity of the color (of the Kashaev invariant). There are many cases

when the volume conjecture for cables of the ﬁgure 8 knot is false if one considers all

the colors, but holds true if one restricts the colors to a subset of the set of positive

integers. Chapter 2 studies the AJ conjecture of S. Garoufalidis that relates the linear recurrence relations of the colored Jones polynomials and the A-polynomial of a knot, using skein theory. We conﬁrm the AJ conjecture for those hyperbolic knots satisfying certain conditions. In particular, we show that the conjecture holds true for some classes of two-bridge knots and pretzel knots. This extends the previous result of T. Le where he established the AJ conjecture for a large class of two-bridge knots, including all twist knots. Along the way, we explicitly calculate the universal character ring of the knot group of the ( 2, 3, 2n + 1)-pretzel knot and show that it is reduced, i.e. has − no nilpotent elements, for all integer n.

viii Chapter 3 studies the Kauﬀman bracket skein module (KBSM) of the complement of all two-bridge links. For a two-bridge link, we show that the KBSM of its

1 complement is free over the ring C[t± ] and when reducing t = 1, it is isomorphic − to the ring of regular functions on the character variety of the link group.

In Chapter 4, we study a stronger version of the AJ conjecture, proposed by A.

Sikora, and conﬁrm it for all torus knots.

Results in chapters 1–3 of this dissertation are joint works with T. Le.

ix CHAPTER I

ON THE VOLUME CONJECTURE FOR CABLES OF

KNOTS

1.1 Introduction 1.1.1 The colored Jones polynomial and the Kashaev invariant of a link

Suppose K is framed oriented link with m ordered components in S3. To every

m-tuple (n1,...,nm) of positive integers one can associate a Laurent polynomial

1/4 J (n ,...,n ; q) Z[q± ], called the colored Jones polynomial, with n being the K 1 m ∈ j

color of the j-component of K. The polynomial JK (n1,...,nm; q) is the quantum link invariant, as deﬁned by Reshetikhin and Turaev [50, 57], associated to the Lie

algebra sl2(C), with the color nj standing for the irreducible sl2(C)-module Vnj of dimension nj. Here we use the functorial normalization, i.e. the one for which the colored Jones polynomial of the unknot colored by n is

n/2 n/2 q q− [n] := − . q1/2 q 1/2 − − When all the colors are 2, the colored Jones polynomial is the usual Jones polyno-

mial [28]. The colored Jones polynomials of higher colors are more or less the usual

Jones polynomials of cables of the link.

Following [44], we deﬁne the Kashaev invariant of a link K as the sequence K , N N =1, 2,... , by

JK (N,...,N; q) K := 1/4 . N [N] |q =exp(πi/2N)

1.1.2 The volume conjecture for knots and links

According to Thurston theory, by cutting the link complement S3 K along appro- \ priate disjoint tori one gets a collection of pieces, each is either Seifert ﬁbered or

1 hyperbolic; and Vol(K) is deﬁned as the sum of the hyperbolic volume of the hyperbolic pieces. It is known that Vol(K) = v S3 K , where v is the volume of the 3 || \ || 3 ideal regular tetrahedron, and S3 K is the Gromov norm. We can now formulate || \ || the volume conjecture of Kashaev-Murakami-Murakami [30, 44]:

Conjecture 1. Suppose K is a knot in S3, then

log K Vol(K) lim | N | = . N →∞ N 2π

For a survey on the volume conjecture, see [43]. Already in [44] it was noted that

the volume conjecture in the above form cannot be true for split links, since for split

links the Kashaev invariant vanishes. There are a few cases of links (of more than one components) when the volume conjecture had been conﬁrmed: in particular, the

volume conjecture was established for the Borromean rings [20], the Whitehead link

[60], and more general, for Whitehead chains [58].

When the Kashaev invariant vanishes, one might hope to remedy the conjecture

by renormalizing the colored Jones polynomial. One of consequences of the present

chapter is that the normalization alone is not good enough, we have also to distinguish between the cases N even and N odd.

1.1.3 Main results

For a knot K with framing 0, let K(m,p) be the (m, p)-cable of K, also with framing

0, see the precise deﬁnition in 1.2. Note that if m and p are co-prime, then K(m,p) § is again a knot. The two-component link K(0,2) is called the disconnected cable of K.

Note that we always have Vol(K(m,p)) = Vol(K).

In this chapter we study the volume conjecture for cables of a knot K. It turns

out that the case N even and the case N odd are quite diﬀerent.

Theorem 1. Suppose that K is a knot and K(0,2) the disconnected cable of K. Then

K(0,2) =0 for every even N. N

2 The case of odd N is quite diﬀerent, at least for the ﬁgure 8 knot:

Theorem 2. Suppose is the ﬁgure 8 knot and (0,2) its disconnected cable. Then E E the volume conjecture holds true for (0,2) if the colors are restricted to the set of odd E numbers: log (0,2) Vol( (0,2)) lim |E N | = E . N , N odd N 2π →∞

Thus for ﬁgure 8 knot, the sequence of Kashaev invariant (0,2) grows expo- |E N | nentially if N and N odd. While if N is even, then (0,2) = 0 (for any → ∞ |E N | knot).

However, when m = 0, i.e. when the two components of K(m,2) do have non- trivial linking number, the volume conjecture might still hold true even for even N.

For example, we have the following result.

Theorem 3. Suppose is the ﬁgure 8 knot and m 2 (mod 4). Then the volume E ≡ conjecture holds true for (m,2) if the colors are restricted to the set of numbers divisible E by 4: log (m,2) Vol( (m,2)) lim |E N | = E . N , N 0 (mod 4) N 2π →∞ ≡

According to the survey [43], the volume conjecture has been so far established

for the following knots:

4 (by Ekholm), • 1

5 , 6 , 6 (by Y. Yokota), • 2 1 2

torus knots (by Kashaev and Tirkkonen [31]), and •

Whitehead doubles of torus knots of type (2, b) (by Zheng [60]). •

3 We add to this the following result.

Theorem 4. Suppose is the ﬁgure 8 knot. Then the volume conjecture holds true E for the knot (m,2) for every odd number m. E

Actually, we will prove some generalizations of Theorems 1-4.

Remark 1.1.1. We arrived at the theorems through the symmetry principle studied in [32, 35], although we will not use the symmetry here. One important tool in our proof is the Habiro expansion of the colored Jones polynomial [22], which has been instrumental in integrality of the Witten-Reshetikhin-Turaev invariant of 3-manifolds

(see [2, 22, 37]) and in the proof of a generalization of the volume conjecture for small angles [20].

The odd colors correspond to the representations of the group SO(3), or representations of sl2 with highest weights in the root lattice.

1.1.4 Plan of the chapter

In Section 1.2 we exactly formulate the more general results that we want to prove.

Sections 1.3 contains some elementary calculations involving the building blocks in the Habiro expansion. Sections 1.4 and 1.5 contain the proof of the main theorems.

1.2 Cables, the colored Jones polynomial, and results 1.2.1 Cables, the colored Jones polynomial

Suppose K is a knot with 0 framing and m, p are two integers with d their greatest common divisor. The (m, p)-cable K(m,p) of K is the link consisting of d parallel copies of the (m/d, p/d)-curve on the torus boundary of a tubular neighborhood of K. Here an (m/d, p/d)-curve is a curve that is homologically equal to m/d times

the meridian and p/d times the longitude on the torus boundary. The cable K(m,p)

inherits an orientation from K, and we assume that each component of K(m,p) has

framing 0.

4 The colored Jones polynomial is a special case of tangle invariants deﬁned using ribbon Hopf algebras and their modules [50]. The ribbon Hopf algebra in our case is

the quantized enveloping algebra Uh(sl2), e.g. [46]. For each positive integer n, there is a unique irreducible Uh(sl2)-module Vn of rank n. In [46] our JK(n1,...,nm; q) is

sl ;V ,...,V denoted by Q 2 n1 nm (K).

The calculation of J (m,2) (N; q) is standard: one decomposes V V into irre- K N ⊗ N ducible components

N VN VN = l=1V2l 1. ⊗ ⊕ − Since the R-matrix commutes with the actions of the quantized algebra, it acts on

each component V2l 1 as a scalar l times the identity. The value of l is well-known: −

− 2 − N l 1 N l(l 1) =( 1) − q 2 q 2 . l −

Hence from the theory of quantum invariants, we have

N m J (m,2) (N; q)= J (2l 1; q). K l K − l=1 The symmetry of quantum invariant at roots of unity [32, 35] prompts us to combine the color N j with N + j. So we rewrite the above formula as follows − N 1 − m m JK(m,2) (N; q)= aN tj,N JK(N + j; q), (1) j=1 N,N j+1 even − − where 2 N 1 j (N+j) (3 4N 2)/8 tj,N = i − − q 8 and aN = q − .

1.2.2 General knot case and even m

Here we relate the Kashaev invariant of K(m,2) and the colored Jones polynomial of

Km/2, which is the same knot K, only with framing m/2. Increasing the framing by

(N 2 1)/4 1 has the eﬀect of multiplying the invariant by q − , hence

N2−1 p 4 JKp (N; q)= q JK (N; q).

5 Theorem 1.2.1. Suppose one of the following: (i) m 0 (mod 4) and N is even. ≡ (ii) m 2 (mod 4) and N 2 (mod 4). Then, with q1/4 = exp(πi/2N), one has ≡ ≡ N/2 (m,2) m/2 1/2 1/2 mN K = q (q q− ) J (2j 1; q). N − 4 Km/2 − j=1 In particular, if m =0 and N is even, then K(m,2) =0. N 1.2.3 Figure 8 knot case

Let be the ﬁgure 8 knot. We will show that the volume conjecture for (m,2) holds E E true under some restrictions.

For an integer m there are 4 possibilities, we list them here together with the deﬁnition of a set Sm:

(i) m is odd. Deﬁne Sm = N, the set of positive integers. (ii) m 0 (mod 8). Deﬁne S = N N, N 1 (mod 2) , the set of odd ≡ m { ∈ ≡ } positive integers.

(iii) m 2 (mod 4). Deﬁne S = N N, N 2 (mod 4) . ≡ m { ∈ ≡ } (iv) m 4 (mod 8). Deﬁne S = . ≡ m ∅

Theorem 1.2.2. Suppose is the ﬁgure 8 knot and m is in one of ﬁrst three cases E (i)–(iii) listed above. Then the volume conjecture for (m,2) holds true if the colors E are restricted to the corresponding set Sm, i.e. one has

log (m,2) Vol( (m,2)) lim |E N | = E . N ,N Sm →∞ ∈ N 2π

The proof of the theorem will be given in section 1.5. Theorems 2, 3, and 4 are parts of this theorem. We still don’t have any conclusion for the case (iv).

6 1.3 Some elementary calculations 1.3.1 Notations and conventions

We will work with the variable q1/4. Let v = q1/2. We will use the following notations. Here j,k,l,N are integers.

j j j := v v− , [j] := j / 1 , S(k,l) := j , S′(k,l) := j . { } − { } { } { } { } k j l k j l, j/ 0,N ≤ ≤ ≤ ≤ ∈{ }

(N+j)2 ′ N 1 j A(j, k) := j k j + k , a := a , and t := i − − q 8 . { − }{ } i j j,N i I i I j I i ∈ ∈ ∈\{ } 1 For f,g and h in Z[v± ], the equation f g (mod h) means that f g is divisible ≡ − by h.

1.3.2 The building blocks A(N,k) and S(k,l)

The expressions A(N,k) and S(k,l) will be the building blocks in the Habiro expansion. We will prove here a few simple facts.

Lemma 1.3.1. One has

2 1 A(N j, k) A(N + j, k)+2 2j N N 1 / 1 (mod N ) in Z[v± ], − ≡ { }{ }{ − } { } { } j j N j = N + j + N (v + v− ), { − } −{ } { } m m mj m m N 2 1 t j,N = tj,N + tj,N N + tj,N (v + 1) Q, where Q Q[v± ]. − 2 { } ∈

Proof. The second equality follows directly from deﬁnition. For the ﬁrst one, by

j j k k noting that j = j and A(j, k)= q + q− q q− we have {− } −{ } − −

A(N j, k) A(N + j, k) = 2j 2N − − −{ }{ } N N 1 = 2j N (v + v− )(v v− )/ 1 −{ }{ } − { } 1 = 2j N [ N (v + v− ) 2 N 1 ]/ 1 −{ }{ } { } − { − } { } 2 2j N N 1 / 1 (mod N 2). ≡ { }{ }{ − } { } { }

7 To prove the last one, we note that for a positive integer k there is a polynomial P (a), whose coeﬃcients depend on k only, such that 1 ak = k(1 a)+(1 a)2P (a). − − − Hence if mj 0 then ≥

m m m N mj t j,N tj,N = tj,N [1 ( v− ) ] − − − − − m N N 2 1 = t [(1 + v− )mj +(1+ v− ) P (v− )] − j,N m mj N N 2 N 2 1 = t [ (v− (1 + v ) N )+(1+ v− ) P (v− )] − j,N 2 −{ } mj m m N 2 mj N 2N 1 = t N + t (v + 1) [ v− + v− P (v− )], 2 j,N { } j,N − 2 which asserts the third equality. Otherwise, i.e. if mj < 0, we have

m m m N mj t j,N tj,N = tj,N [1 ( v )− ] − − − − − = tm [ (1 + vN )mj +(1+ vN )2P (v)] − j,N − m mj N N 2 N 2 = t [ (v− (1 + v ) + N )+(1+ v ) P (v)] − j,N − 2 { } mj m m N 2 mj N = t N + t (v + 1) [ v− P (v)]. 2 j,N { } j,N 2 − This completes the proof of the lemma.

Lemma 1.3.2. Suppose vN = 1, then − l l m m tj,N [N + j]( A(N + j, k)) + t j,N [N j] A(N j, k) − − − k=1 k=1 = tm D(j, l), (2) [N] j,N where l l j j 2 ′ mj D(j, l)=(v + v− ) ( A(j, k))+2 j A(j, k) + j A(j, k). { } 2 { } k=1 1 k l k=1 ≤≤ Proof. From lemma 1.3.1, we have

l l A(N j, k) [A(N + j, k)+2 2j N N 1 / 1 ] (mod N 2) − ≡ { }{ }{ − } { } { } k=1 k=1 l A(N + j, k)+ ≡ k=1 + 2 2j N N 1 / 1 ′A(N + j, k) (mod N 2). { }{ }{ − } { } { } 1 k l ≤≤

8 This, together with the last two equalities in lemma 1.3.1, implies that when vN = 1 − the left hand side of (2) is equal to

l l mj m m j j t N + j ( A(N + j, k)) + t (v + v− )( A(N + j, k)) − 2 j,N { } j,N k=1 k=1 2 2j N 1 / 1 ′A(N + j, k)tm N + j − { }{ − } { } j,N { } 1 k l ≤≤ Hence (2) follows from the facts that A(N + j, k) = A(j, k) and N + j = j if { } −{ } vN = 1. −

Lemma 1.3.3. One has D(j, l)= D1(j, l)+ D2(j, l), where

mj vj +v−j l 1 2 + j +2 2j k=1 A(j,k) S(j l, j + l), if l< min(j, N j), D1(j, l)=  { } { } − −  0 if l min(j, N j), ≥ −  and 

2S′(j l, j + l) if j l < N j,  − ≤ −  D2(j, l)=  2S′(j l, j + l) if N j l < j, − − − ≤  0 if l< min(j, N j) or l max(j, N j).  − ≥ −   

Proof. This can be checked easily by direct calculations.

j Lemma 1.3.4. For j l N j, the sign of S′(j l, j + l) is ( 1) . For N j ≤ ≤ − − − − ≤ N j l j, the sign of S′(j l, j + l) is ( 1) − . ≤ − −

l+j l j j Proof. If j l N j then the sign of S′(j l, j + l) is i ( i) − = ( 1) . For ≤ ≤ − − − − N j l j, the proof is similar. − ≤ ≤

Lemma 1.3.5. Suppose vN = 1. For 1 j N 1 and 0 l N 1 one has − ≤ ≤ − ≤ ≤ − mN D(N j, l)+ D(j, l)= S(j l, j + l). (3) − 2 −

9 Proof. Let

l j j 2 ′ B(j, l)=(v + v− ) ( A(j, k))+2 j A(j, k) . { } k=1 1 k l ≤≤ Then by deﬁnition, D(j, l) = B(j, l)+ mj j l A(j, k). It is easy to see that if 2 { } k=1 vN = 1 then B(N j, l)+ B(j, l) = 0 and − − l l N j A(N j, k)= j A(j, k)= S(j l, j + l). { − } − { } − k=1 k=1 It implies that (3) holds true.

1.4 Proof of Theorem 1.2.1 1.4.1 Habiro expansion

1 By a deep result of Habiro [22], there are Laurent polynomials C (l; q) Z[q± ], K ∈ depending on the knot K, such that

N 1 l − JK(N; q) = [N] CK (l; q) A(N,k). l=0 k=1 From now on let q1/4 = exp(πi/2N). Then vN = 1. Using Eq. (2), we have for − 0 j N 1 ≤ ≤ − m m N 1 − tj,N JK(N + j; q)+ t j,N JK(N j; q) m − − = C (l; q) t D(j, l). [N] K j,N l=0 Hence Eq. (1) implies that

N 1 N 1 − − 1 ( 1)N K(m,2) = am C (l; q) ( tm D(j, l)) + − − tm D(0,l) . N N K j,N 4 0,N l=0 j=1,N j+1 even − (4)

1.4.2 Proof of Theorem 1.2.1

We assume that m, N satisfy the conditions of Theorem 1.2.1, i.e. m 0 (mod 4) ≡ and N is even; or m 2 (mod 4) and N 2 (mod 4). ≡ ≡

10 The symmetry [32, 35] hints that we should combine j and N j. Since N is even, − both j and N j are odd, so they both appear in the sum (4). Hence we rewrite the − expression in the big parenthesis in right hand side of Eq. (4) as follows N/2 1 − N/2 m m 1 ( 1) m ( tj,N D(j, l)+ tN j,N D(N j, l)) + − − tN/2,N D(N/2,l). − − 2 j=1,j odd m Under the assumption in the theorem on m and N, we can easily check that tN j,N = − m (m,2) m N 1 t . Therefore it follows from Eq. (3) that K is equal to a − C (l; q) j,N N N l=0 K times N/2 1 − mN 1 ( 1)N/2 mN ( tm S(j l, j + l)) + − − tm S(N/2 l,N/2+ l). j,N 2 − 2 N/2,N 4 − j=1,j odd m m(3 4N 2)/8 3m/8 Now, by noting that aN = q − = q and

m m mj2/8 tN j,N = tj,N = q , S(j l, j + l)= S(N j l, N j + l), − − − − − we obtain N 1 N 1 mN − − 2 K(m,2) = q3m/8 C (l; q) qmj /8S(j l, j + l) N 4 K − l=0 j=1,j odd N 1 − 2− m/2 mN m j 1 = q q 2 4 1 J (j; q) 4 { } K j=1,j odd N 1 mN − = qm/2 1 J (j; q), { } 4 Km/2 j=1,j odd where Km/2 is K with framing m/2. This proves theorem 1.2.1.

1.5 Proof of theorem 1.2.2

1/4 Let δ := exp(πi/4). We will write tj for tj,N . Then, with q = exp(πi/2N) one has

3N 2 j2/8 tj = δ − q .

For the ﬁgure 8 knot , we know that C (l, q) = 1, see [22]. From Eq. (4) we have E E N 1 N/2 1 N 1 N 1 (m,2) − − − − N mj2/8 mj2/8 E(3N 2)m m = ( q D(j, l))+( q D(j, l)) δ − a N l=0 j=1,N j+1 even l=0 j=N/2,N j+1 even − − N 1 1 ( 1)N − + − − D(0,l). (5) 4 l=0

11 1.5.1 The case j

We now consider the ﬁrst sum in the right hand side of Eq. (5). By lemma 1.3.3,

mj vj +v−j l 1 D1(j, l)= 2 + j +2 2j k=1 A(j,k) S(j l, j + l) if l < j  { } { } −  D(j, l)= D2(j, l)=2S′(j l, j + l) if j l < N j  − ≤ −  0 if l N j  ≥ −  We will consider two subcases when D(j, l) = 0: j l < N j and j

j N 1 By lemma 1.3.4, the sign of S′(j l, j + l) is ( 1) = ( 1) − . Note that k = − − − { } kπ N 1 2i sin , hence S′(j l, j + l)=( 1) − E(j, l), where N − − l j l+j − rπ rπ E(j, l)=( 2 sin )( 2 sin ). N N r=1 r=1 We will see that E(j, l) is maximized when j = 0 and l = 5N/6. Moreover, we have

the following result.

Proposition 1.5.1. There exists a nonzero number C such that for any α ( 1 , 2 ), ∈ 2 3 we have

N/2 1 N j − − mj2/8 N 1 5N 3α 2 q D (j, l)=( 1) − C E(0, )N(1 + O(N − )). 2 − 6 j=1,N j+1 even l=j −

Proof. By setting n jπ s = log 2 sin , n − | N | j=1 we have log E(j, l)= sl j sl+j. Consider the Lobachevsky function − − − x L(x) := log 2 sin u du. − | | 0 N nπ By a standard argument, see e.g. [20, 60], sn = π L( N )+ O(log N). Hence N (l j)π N (l + j)π N jπ lπ log E(j, l)= L( − ) L( )+ O(log N)= f( , )+ O(log N), − π N − π N π N N

12 where f(x, y)= L( x + y) L(x + y) for π y x 0 and π x + y. − − − ≥ ≥ ≥ ≥ It is easy to show that the function f attains its maximum at the unique point

5π 5π (x, y)=(0, 6 ) . Moreover, the Taylor expansion of f around (0, 6 ) is

5π 5π f(h, + k)= f(0, ) √3(h2 + k2)+ O( h 3 + k 3). 6 6 − | | | |

By the same argument as in the proof of theorem 1.2 in [60], there exists ǫ> 0 such

that

5N 2α 1 2 5N 2 2α < log E(0, 6 ) ǫN − + O(1) if j +(l 6 ) N , log E(j, l)  − − ≥  5N π√3 2 5N 2 3α 2 = log E(0, ) [j +(l ) ]+ O(N − ) otherwise 6 − N − 6  Let 

5N I = (j, l):1 j

Then we have

5N 2α 1 E(0, 6 ) exp( ǫN − )O(1) if (j, l) I1, E(j, l)=  − ∈  5N π√3 2 5N 2 3α 2 E(0, ) exp( [j +(l ) ])(1 + O(N − )) if (j, l) I2. 6 − N − 6 ∈  It implies that

mj2/8 5N 2 2α 1 q E(j, l)= E(0, )N exp( ǫN − )O(1), 6 − (j,l) I1 ∈ and qmj2/8E(j, l)= (j,l) I2 ∈ 2 πimj 5N π√3 2 5N 2 3α 2 = exp( )E(0, ) exp( [j +(l ) ])(1 + O(N − )) 4N 6 − N − 6 (j,l) I2 ∈ N πim 2 2 2 5N 3α 2 = exp( x π√3(x + y ))dxdyE(0, )(1 + O(N − )) 4 2 2 2α−1 4 − 6 x +y

13 Hence

N/2 1 N j − − mj2/8 N 1 mj2/8 mj2/8 q D (j, l) = 2( 1) − q E(j, l)+ q E(j, l) 2 −   j=1,N j+1 even l=j (j,l) I1 (j,l) I2 − ∈ ∈ N 1  5N 3α 2  = ( 1) − CE(0, )N(1 + O(N − )), − 6

where 1 πim C = exp( x2 π√3(x2 + y2))dxdy. 2 2 4 − R We can easily check that C is a nonzero number. This completes the proof the proposition 1.5.1.

From now on, we ﬁx the number α ( 1 , 2 ). ∈ 2 3 1.5.1.2 The subcase l < j

l+j j l 1 l By lemma 1.3.4, the sign of S(j l, j + l) is i /i − − = i( 1) . Hence we get − − S(j l, j + l)= i( 1)lF (j, l), where − − l+j j l 1 rπ − − rπ F (j, l)=( 2 sin )/( 2 sin ) N N r=1 r=1 for 0

Proposition 1.5.2. One has

N/2 1 − 2 N N qmj /8D (j, l)= N 3αF ( , )O(1). (6) 1 2 3 j=1,N j+1 even l

Proof. Let

N N I = (j, l):1 j

14 By the same argument as in the proof of the previous proposition, we have

N N 2α 1 = F ( 2 , 3 ) exp( ǫN − )O(1) if (j, l) I3, F (j, l)  − ∈  N N  F ( , ) if (j, l) I4. ≤ 2 3 ∈  To prove the proposition, we need the following three lemmas:

Lemma 1.5.3. One has

j j l 2 mj v + v 1 qmj /8 + − +2 2j S(j l, j + l)= 2 j { } A(j, k) − (j,l) I3 k=1 ∈ { }

5 N N 2α 1 = N F ( , ) exp( ǫN − )O(1). 2 3 −

Proof. It suﬃces to show that if (j, l) I then ∈ 3 mj vj + v j l 1 + − +2 2j = N 3O(1). (7) 2 j { } A(j, k) { } k=1 Indeed, since sin x 2 x for x [0, π ], we have ≥ π ∈ 2 vj + v j cos(jπ/N) π N N − = . | j | | sin(jπ/N) | ≤ 2 jπ ≤ 2 { } Note that k l j 1, hence ≤ ≤ − (j k)π (j + k)π π (2j 1)π A(j, k) = j k j + k = 4 sin − sin 4 sin sin − . | | |{ − }{ }| N N ≥ N N

It implies that 2j sin 2jπ { } N . π (2j 1)π |A(j, k)| ≤ − 2 sin N sin N N (2j 1)π 2jπ − If j > 4 then sin N > sin N > 0, therefore 2j 1 N { } π . |A(j, k)| ≤ 2 sin N ≤ 4 If j N then ≤ 4 2j 1 N 2 { } 2 π . |A(j, k)| ≤ 2 sin N ≤ 8 Hence (7) holds true and then the claim of the lemma is proved.

15 Lemma 1.5.4. One has

j j l 2 m N v + v 1 qmj /8 (j )+ − +2 2j S(j l, j + l)= 2 − 2 j { } A(j, k) − (j,l) I4 k=1 ∈ { } N N = N 3αF ( , )O(1). 2 3

Proof. Since (j N )2 +(l N )2 N 2α, we have − 2 − 3 ≤ j j v + v− jπ π jπ α 1 = cot = O(1) = N − O(1), | j | N | 2 − N | { } and 2jπ 2j 2j sin N α 1 { } { } = (j l)π (l+j)π = N − O(1), |A(j, k)|≤|A(j, l)| − 2 sin N sin N which proves the equality of the lemma.

Lemma 1.5.5. One has

mj2/8 3α 1 N N q S(j l, j + l)= N − F ( , )O(1). − 2 3 (j,l) I4 ∈

Proof. Denote by L the left hand side of the equality. We ﬁrst see that L is essentially equal to the following expression

1 mj2/8 L′ = q [S(j l, j + l)+ S(j l 1, j + l + 1)]. 2 − − − (j,l) I4 ∈ Note that for (j, l) I , we have S(j l, j + l)+ S(j l 1, j + l +1) = ∈ 4 − − − (j l 1)π (j + l + 1)π = S(j l, j + l)[1 4 sin − − sin ] − − N N 2jπ (2l + 1)π = S(j l, j + l)[(2 + 2 cos ) (1+2cos )] − N − N α 1 N N = N − F ( , )O(1). 2 3

3α 1 N N This implies that L′ = N − F ( 2 , 3 )O(1). Hence to complete the proof of the lemma, we need to estimate the diﬀerence between L and L′.

16 Let J = j : (j, l) I for some l . For each j J, let J = l : (j, l) I . We { ∈ 4 } ∈ j { ∈ 4} have

1 mj2/8 L′ L = q S(j l 1, j + l + 1) S(j l, j + l). (8) − 2 − − − − j J l J ∈ ∈ j For each j in J, it is easy to see that the set Jj is just a closed interval [aj, bj]. Hence

S(j l 1, j + l +1) S(j l, j + l)= S(j b 1, j + b +1) S(j a , j + a ) − − − − − j − j − − j j l J ∈ j N N has absolute value less than or equal to 2F ( , ). Eq. (8) then implies that L′ L 2 3 | − | ≤ N α2F ( N , N ). Since α< 3α 1, the conclusion of the lemma follows. 2 3 − We now come back to the proof of proposition 1.5.2. It is easy to see that lemmas

1.5.4 and 1.5.5 imply

j j l 2 mj v + v 1 N N qmj /8 + − +2 2j S(j l, j+l)= N 3αF ( , )O(1). 2 j { } A(j, k) − 2 3 (j,l) I4 k=1 ∈ { } Hence, by combining this with lemma 1.5.3, we obtain the equality of the proposition.

We can now estimate the ﬁrst sum in the right hand side of Eq. (5). To do that, we need one more lemma which allows us to express the right hand side of Eq. (6),

5N and hence the sum in its left hand side, in terms of E(0, 6 ).

Lemma 1.5.6. One has N N 1 5N F ( , )= E(0, )O(1). 2 3 N 6

N 1 kπ − Proof. The follows from the fact that k=1 (2 sin N )= N. From proposition 1.5.2 and lemma 1.5.6, we have

N/2 1 − mj2/8 3α 1 5N q D (j, l)= N − E(0, )O(1). 1 6 j=1,N j+1 even l

N 1 N/2 1 − − mj2/8 N 1 5N 3α 2 q D(j, l)=( 1) − CE(0, )N(1 + O(N − )). (9) − 6 l=0 j=1,N j+1 even −

17 1.5.2 The case j N/2 ≥ Again, by lemma 1.3.3 we have

mj vj +v−j l 1 D1(j, l)= 2 + j +2 2j k=1 A(j,k) S(j l, j + l) if l < N j  { } { } − −  D(j, l)= D2(j, l)= 2S′(j l, j + l) if N j l < j  − − − ≤  0 if l j  ≥   In this case all the estimations we have done in section 1.5.1 also work for the second sum in the right hand side of Eq. (5) except two things. The ﬁrst one is that for

N j N j l < j, by lemma 1.3.4, the sign of S′(j l, j + l) is ( 1) − = 1 and the − ≤ − − − other is

mj2/8 m(N j)2/8 q = q − exp(πim(2j N)/4), − where

β = exp(πim(N + 2)/4), if N j +1 0 (mod4), exp(πim(2j N)/4) =  − ≡ −  γ = exp(πim(N 2)/4), if N j +1 2 (mod4). − − ≡  Therefore by similar arguments as in the proof of Eq. (9), we get

N 1 N 1 − − mj2/8 1 5N 3α 2 q D(j, l)= (β + γ)CE(0, )N(1 + O(N − )). (10) 2 6 l=0 j=N/2,N j+1 even − 1.5.3 Proof of theorem 1.2.2

From Equations (9) and (10), we have

N 1 N 1 − − mj2/8 1 N 1 5N 3α 2 q D(j, l)= (β + γ + 2( 1) − )CE(0, )N(1 + O(N − )). 2 − 6 l=0 j=1,N j+1 even − (11)

Moreover, it is easy to see that

N 1 − 5N D(0,l)= N αE(0, )O(1). (12) 6 l=0 Therefore, to complete the proof of theorem 1.2.2, we need the following lemma

18 N 1 Lemma 1.5.7. We have β + γ + 2( 1) − = 0 if and only if one of the following − holds:

(i) m 0 (mod 4) and N is even. ≡ (ii) m 2 (mod 4) and N 2 (mod 4). ≡ ≡ (iii) m 4 (mod 8) and N is odd. ≡ N 1 N 1 Moreover, if β + γ + 2( 1) − =0 then β + γ + 2( 1) − 2. − | − | ≥

N 1 Proof. Suppose that β + γ + 2( 1) − = 0. If N is even then β + γ = 2. Since − β = γ = 1, it implies that β = γ = 1 which means that m 0 (mod 4); or m 2 | | | | ≡ ≡ (mod 4) and N 2 (mod 4). If N is odd then similarly, we have β = γ = 1, which ≡ − is equivalent to the condition that m 4 (mod 8). ≡ N 1 Note that if m is odd then β + γ = 0, hence β + γ + 2( 1) − = 2. Now let us | − | N 1 N 1 consider the case m is even. Then β = γ and β + γ + 2( 1) − = 2(β +( 1) − ), − − N 1 m (N+2) so it is remaining to show that β +( 1) − 1. Since β = i 2 1, i | − | ≥ ∈ {± ± } N 1 N 1 and β +( 1) − = 0, it implies that the angle between β and ( 1) − is less than − − π or equal to 2 . Hence

N 1 2 2 N 1 2 β +( 1) − β + ( 1) − =2, | − | ≥| | | − | which completes the proof of the lemma.

N 1 Under the assumption of theorem 1.2.2, by lemma 1.5.7, β + γ +2( 1) − 2. | − | ≥ Hence from Equations (11) and (12) we get

(m,2) N 1 N 1 5N 3α 2 − − E(3N 2)m m = (β + γ + 2( 1) )CE(0, )N(1 + O(N )), δ − aN 2 − 6 which, together with the simple fact that 5N 2N 5π 2N π log E(0, )= L( )+ O(log N)= L( )+ O(log N), 6 − π 6 π 6 conﬁrms the volume conjecture for (m,2): E log (m,2) π 2π lim |E N | =4L( ) = Vol( (m,2)). N ,N Sm →∞ ∈ N 6 E

19 CHAPTER II

ON THE AJ CONJECTURE FOR KNOTS

2.1 Introduction 2.1.1 The AJ conjecture

3 1 For a knot K in S , let J (n) Z[t± ] be the colored Jones polynomial of K colored K ∈

by the (unique) n-dimensional simple representation of sl2 [28, 50], normalized so that for the unknot U, 2n 2n t t− J (n) = [n] := − . U t2 t 2 − − The color n can be assumed to take negative integer values by setting J ( n) = K − J (n). In particular, J (0) = 0. It is known that J (1) = 1, and J (2) is the − K K K K ordinary Jones polynomial.

Deﬁne two linear operators L, M acting on the set of discrete functions f : Z → 1 := C[t± ] by R (Lf)(n)= f(n + 1), (Mf)(n)= t2nf(n).

2 1 1 It is easy to see that LM = t ML. Besides, the inverse operators L− , M − are well-deﬁned. One can consider L, M as elements of the quantum torus

1 1 2 = L± , M ± /(LM t ML), T R −

which is not commutative, but almost commutative.

Let

= P PJ =0 , AK { ∈T | K } which is a left ideal of , called the recurrence ideal of K. It was proved in [19] that T for every knot K, the recurrence ideal is non-zero. Partial results were obtained AK

20 earlier by Frohman, Gelca, and Lofaro through their theory of non-commutative A- ideal [15, 21]. An element in is called a recurrence relation for the colored Jones AK polynomial of K.

The ring is not a principal left-ideal domain, i.e. not every left-ideal of is T T generated by one element. In [17], by adding the inverses of polynomials in t, M to

one gets a principal left-ideal domain ˜, and a generator α of the extension ˜ T T K AK of . The element α can be presented in the form AK K d j αK(t; M, L)= αK,j(t, M) L , j=0 where the degree in L is assumed to be minimal and all the coeﬃcients α (t, M) K,j ∈ 1 Z[t± , M] are assumed to be co-prime. The polynomial αK is deﬁned up to a polyno-

1 mial in Z[t± , M]. Moreover, one can choose α , i.e. it is a recurrence relation K ∈AK

for the colored Jones polynomial. We call αK the recurrence polynomial of K.

Garoufalidis [17] formulated the following conjecture (see also [15, 21]).

Conjecture 2. (AJ conjecture) For every knot K, αK t= 1 is equal to the A- | − polynomial, up to a factor depending on M only.

Here in the deﬁnition of the A-polynomial [10], we also allow the abelian compo-

nent of the character variety, see Section 2.

2.1.2 Main results

Conjecture 2 was established for a large class of two-bridge knots, including all twist

knots, by T. Le using skein theory [36]. Here we have the following stronger results.

Theorem 5. Suppose K is a knot satisfying all the following conditions: (i) K is hyperbolic,

(ii) The SL -character variety of π (S3 K) consists of 2 irreducible components 2 1 \ (one abelian and one non-abelian),

21 (iii) The universal SL -character ring of π (S3 K) is reduced, 2 1 \ (iv) The recurrence polynomial of K has L-degree greater than 1.

Then the AJ conjecture holds true for K.

Theorem 6. The following knots satisfy all the conditions (i)–(iv) of Theorem 5 and hence the AJ conjecture holds true for them.

(a) All pretzel knots of type ( 2, 3, 6n 1), n Z. − ± ∈ (b) All two-bridge knots for which the character variety has exactly 2 irreducible components; these includes all twist knots, double twist knots of the form J(k,l) with

k = l in the notation of [24], all two-bridge knots b(p, m) with p prime or m = 3. Here we use the notation b(p, m) from [8].

Remark 2.1.1. Besides the inﬁnitely many cases of two-bridge knots listed in Theo-

rem 6, explicit calculation seems to conﬁrm that ”most two-bridge knots” satisfy the

conditions of Theorem 5 and hence AJ conjecture holds for them. In fact, among 155

b(p, m) with p< 45, only 9 knots do not satisfy the conditions of Theorem 5.

2.1.3 Other results

In our proof of Theorem 6, it is important to know whether the universal character

ring of a knot group is reduced, i.e. whether its nil-radical is 0. Although it is diﬃcult

to construct a group whose universal character ring is not reduced (see [38]), so far

there are a few groups for which the universal character ring is known to be reduced:

free groups [52], surface groups [11], two-bridge knot groups [48], torus knot groups

[39], two-bridge link groups (see Chapter 3). In the present chapter, we show that the universal character ring of the ( 2, 3, 2n + 1)-pretzel knot is reduced for all integer n. − 2.1.4 Plan of the chapter

We review skein modules and their relation with the colored Jones polynomial in

Section 2.2. In Section 2.3 we prove some properties of the character variety and

22 the A-polynomial a knot. We discuss the role of the quantum peripheral polynomial in the AJ conjecture and give the proof of Theorem 5 and Theorem 6 in Section

2.4. In Section 2.5, we prove the reducedness of the universal character ring of the

( 2, 3, 2n + 1)-pretzel knot for all integer n. − 2.2 Skein Modules and the colored Jones polynomial

In this section we will review skein modules and their relation with the colored Jones polynomial. The theory of Kauﬀman bracket skein module (KBSM) was introduced by Przytycki [47] and Turaev [56] as a generalization of the Kauﬀman bracket [29] in

S3 to an arbitrary 3-manifold. The KBSM of a knot complement contains a lot, if not all, of information about the colored Jones polynomial.

2.2.1 Skein modules

1 Recall that = C[t± ]. A framed link in an oriented 3-manifold Y is a disjoint R union of embedded circles, equipped with a non-zero normal vector ﬁeld. Framed links are considered up to isotopy. Let be the set of isotopy classes of framed links L in the manifold Y , including the empty link. Consider the free -module with basis R , and factor it by the smallest submodule containing all expressions of the form L 1 2 2 t t− and +(t + t− ) , where the links in each expression are − − ∅ identical except in a ball in which they look like depicted. This quotient is denoted

by (Y ) and is called the Kauﬀman bracket skein module, or just skein module, of S Y .

For an oriented surface Σ we deﬁne (Σ) = (Y ), where Y = Σ [0, 1], the S S × cylinder over Σ. The skein module (Σ) has an algebra structure induced by the S operation of gluing one cylinder on top of the other. The operation of gluing the cylinder over ∂Y to Y induces a (∂Y )-left module structure on (Y ). S S

23 2.2.2 The skein module of S3 and the colored Jones polynomial

When Y = S3, the skein module (Y ) is free over of rank one, and is spanned by S R the empty link. Thus if ℓ is a framed link in S3, then its value in the skein module (S3) is ℓ times the empty link, where ℓ , known as the Kauﬀman bracket of S ∈ R ℓ [29], and is just the Jones polynomial of framed links in a suitable normalization.

Let Sn(z) be the Chebychev polynomials deﬁned by S0(z) = 1, S1(z) = z and

3 Sn+1(z)= zSn(z) Sn 1(z) for all n Z. For a framed knot K in S and an integer − − ∈ n 0, we define the n-th power Kn as the link consisting of n parallel copies of K. ≥ Using these powers of a knot, S (K) is defined as an element of (S3). We then n S define the colored Jones polynomial JK (n) by the equation

J (n + 1) := ( 1)n S (K) . K − × n

The ( 1)n sign is added so that for the unknot U, J (n) = [n]. Then J (1) = 1, − U K J (2) = K . We extend the deﬁnition for all integers n by J ( n) = J (n) K − K − − K and JK(0) = 0. In the framework of quantum invariants, JK(n) is the sl2-quantum

invariant of K colored by the n-dimensional simple representation of sl2.

2.2.3 The skein module of the torus

Let T2 be the torus with a ﬁxed pair (,λ) of simple closed curves intersecting at

exactly 1 point. For co-prime k and l, let λk,l be a simple closed curve on the torus homologically equal to k+lλ. It is not diﬃcult to show that the skein algebra (T2) S of the torus is generated, as an -algebra, by all λ . In fact, Bullock and Przytycki R k,l [7] showed that (T2) is generated over by 3 elements ,λ and λ , subject to S R 1,1 some explicit relations.

1 1 2 Recall that = M ± , L± /(LM t ML) is the quantum torus. Let σ : T R − T → T k l k l be the involution deﬁned by σ(M L ) = M − L− . Frohman and Gelca [14] showed

24 that there is an algebra isomorphism Υ : (T2) σ given by S → T

k+l kl k l k l Υ(λ )=( 1) t (M L + M − L− ). k,l −

The fact that (T2) and σ are isomorphic algebras was also proved by Sallenave S T [51].

2.2.4 The orthogonal and peripheral ideals

Let N(K) be a tubular neighborhood of an oriented knot K in S3, and X the closure of S3 N(K). Then ∂(N(K)) = ∂(X)= T2. There is a standard choice of meridian \ and longitude λ on T2 such that the linking number between the longitude and the knot is 0. We use this pair (,λ) and the map Υ in the previous subsection to identify (T2) with σ. S T The torus T2 = ∂(N(K)) cut S3 into two parts: N(K) and X. We can consider (X) as a left (T2)-module and (N(K)) as a right (T2)-module. There is a S S S S bilinear bracket

3 , : (N(K)) (T2) (X) (S ) S ⊗S S → S ≡ R given by ℓ′,ℓ′′ = ℓ′ ℓ′′ , where ℓ′ and ℓ′′ are links in respectively N(K) and X. ∪ Note that if ℓ (T2), then ∈ S

ℓ′ ℓ,ℓ′′ = ℓ′,ℓ ℓ′′ .

In general (X) does not have an algebra structure, but it has a unit: The empty S link serves as the unit. The map

Θ: (T2) (X), Θ(ℓ)= ℓ S → S ∅ is (T2)-linear. Its kernel = ker Θ is called the quantum peripheral ideal, ﬁrst S P introduced in [15]. In [15, 21], it was proved that every element in gives rise to a P recurrence relation for the colored Jones polynomial.

25 The orthogonal ideal in [15] is deﬁned by O

:= ℓ (∂X) ℓ′, Θ(ℓ) = 0 for every ℓ′ (N(K)) . O { ∈ S | ∈ S }

It is clear that is a left ideal of (∂X) σ and . In [15], was called O S ≡ T P ⊂ O O the formal ideal. According to [36], if = for all knots then the colored Jones P O polynomial distinguishes the unknot from other knots.

2.2.5 Relation between the recurrence ideal and the orthogonal ideal

As mentioned above, the skein algebra of the torus (T2) can be identiﬁed with S σ via the -algebra isomorphism Υ sending ,λ and λ to respectively (M + T R 1,1 − 1 1 1 1 M − ), (L + L− ) and t(ML + M − L− ). −

Proposition 2.2.1. One has

n ( 1) Sn 1(λ), Θ(ℓ) = Υ(ℓ)JK(n) − − for all ℓ (T2). ∈ S

Proof. We know from the properties of the Jones-Wenzl idempotent (see e.g. [46]) that

2n 2n Sn 1(λ) , = (t + t− ) Sn 1(λ), − ∅ − ∅

Sn 1(λ) λ, = Sn(λ)+ Sn 2(λ), − ∅ − ∅ 2n+1 2n+1 Sn 1(λ) λ1,1, = t Sn(λ)+ t− Sn 2(λ), . − ∅ − − ∅

n 1 2n By deﬁnition JK(n)=( 1) − Sn 1(λ), and (MJK )(n) = t JK(n), (LJK)(n) = − − ∅

JK(n + 1). Hence the above equations can be rewritten as

n 1 ( 1) Sn 1(λ), Θ() = (M + M − )JK(n)=Υ()JK(n), − − − n 1 ( 1) Sn 1(λ), Θ(λ) = (L + L− )JK(n)=Υ(λ)JK(n), − − − n 1 1 ( 1) Sn 1(λ), Θ(λ1,1) = t(ML + M − L− )JK(n)=Υ(λ1,1)J(n). − −

26 Since (T2) is generated by ,λ and λ , we conclude that S 1,1

n ( 1) Sn 1(λ), Θ(ℓ) = Υ(ℓ)JK(n) − −

for all ℓ (T2). ∈ S

Noting that S (λ) generates the skein module (N(K)), we get { n }n S

Corollary 2.2.2. One has = σ. O AK ∩ T

Remark 2.2.3. Corollary 2.2.2 was already obtained in [16] by another method. Our proof here uses the properties of the Jones-Wenzl idempotent only.

2.3 Character varieties and the A-polynomial

For non-zero f,g C[M, L], we say that f is M-essentially equal to g, and write ∈ f =M g, if the quotient f/g does not depend on L. We say that f is M-essentially

divisible by g if f is M-essentially equal to a polynomial divisible by g.

2.3.1 The character variety of a group

The set of representations of a ﬁnitely presented group G into SL2(C) is an algebraic

set deﬁned over C, on which SL2(C) acts by conjugation. The naive quotient space, i.e. the set of orbits, does not have a good topology/geometry. Two representations in the same orbit (i.e. conjugate) have the same character, but the converse is not true in general. A better quotient, the algebro-geometric quotient denoted by χ(G)

(see [12, 38]), has the structure of an algebraic set. There is a bijection between

χ(G) and the set of all characters of representations of G into SL2(C), hence χ(G) is usually called the character variety of G. For a manifold Y we use χ(Y ) also to denote χ(π1(Y )). Suppose G = Z2, the free abelian group with 2 generators. Every pair of generators

2 2 λ, will deﬁne an isomorphism between χ(G) and (C∗) /τ, where (C∗) is the set of

1 1 non-zero complex pairs (L, M) and τ is the involution τ(M, L)=(M − , L− ), as

27 follows: Every representation is conjugate to an upper diagonal one, with L and M being the upper left entry of λ and respectively. The isomorphism does not change

1 1 if one replaces (λ,) with (λ− ,− ).

2.3.2 The universal character ring

For a finitely presented group G, the character variety χ(G) is determined by the traces of some fixed elements g , ,g in G. More precisely, one can find g , ,g 1 k 1 k in G such that for every element g G there exists a polynomial P in k variables ∈ g such that for any representation r : G SL (C) we have tr(r(g)) = P (x , , x ) → 2 g 1 k

where xj := tr(r(gj)). The universal character ring of G is then deﬁned to be the quotient of the ring C[x , , x ] by the ideal generated by all expressions of the 1 k form tr(r(v)) tr(r(w)), where v and w are any two words in g , ,g which are − 1 k equal in G. The universal character ring of G is actually independent of the choice

of g , ,g . The quotient of the universal character ring of G by its nil-radical is 1 k equal to the ring of regular functions on the character variety of G.

2.3.3 The A-polynomial

Let X be the closure of S3 minus a tubular neighborhood N(K) of a knot K. The

boundary of X is a torus whose fundamental group is free abelian of rank 2. An orientation of K will deﬁne a unique pair of an oriented meridian and an oriented

longitude such that the linking number between the longitude and the knot is 0, as

2 in Subsection 2.2.4. The pair provides an identiﬁcation of χ(∂X) and (C∗) /τ which actually does not depend on the orientation of K.

The inclusion ∂X ֒ X induces the restriction map →

2 ρ : χ(X) χ(∂X) (C∗) /τ −→ ≡

2 2 Let Z be the image of ρ and Zˆ (C∗) the lift of Z under the projection (C∗) ⊂ → 2 2 2 2 (C∗) /τ. The Zariski closure of Zˆ (C∗) C in C is an algebraic set consisting ⊂ ⊂

28 of components of dimension 0 or 1. The union of all the 1-dimension components is defined by a single polynomial A Z[M, L], whose coefficients are co-prime. Note K ∈ that A is defined up to 1. We call A the A-polynomial of K. By definition, A K ± K K does not have repeated factors. It is known that A is always divisible by L 1. The K − A-polynomial here is actually equal to L 1 times the A-polynomial defined in [10]. − 2.3.4 The B-polynomial

It is also instructive and convenient to see the dual picture in the construction of the

A-polynomial. For an algebraic set V (over C) let C[V ] denote the ring of regular

2 σ functions on V . For example, C[(C∗) /τ] = t , the σ-invariant subspace of t :=

1 1 k l k l C[L± , M ± ], where σ(M L )= M − L− . The map ρ in the previous subsection induces an algebra homomorphism

θ : C[χ(∂X)] tσ C[χ(X)]. ≡ −→

We will call the kernel p of θ the classical peripheral ideal; it is an ideal of tσ. Let

pˆ := tp be the ideal extension of p in t. The set of zero points of pˆ is the closure of

Zˆ in C2.

1 1 The ring t = C[M ± , L± ] embeds naturally into the principal ideal domain ˜t :=

1 C(M)[L± ], where C(M) is the fractional field of C[M]. The ideal extension of pˆ in ˜t, which is ˜t pˆ = ˜tp, is thus generated by a single polynomial B Z[M, L] which has K ∈ co-prime coefficients and is defined up to a factor M k with k Z. Again B can ± ∈ K be chosen to have integer coefficients because everything can be defined over Z. We will call BK the B-polynomial of K.

2.3.5 Relation between the A-polynomial and the B-polynomial

From the deﬁnitions one has immediately that the polynomial BK is M-essentially divisible by A . Moreover, their zero sets B =0 and A =0 are equal, up to K { K } { K } some lines parallel to the L-axis in the LM-plane.

29 Proposition 2.3.1. The B-polynomial BK does not have repeated factors.

Proof. We ﬁrst note that the ring C[χ(X)] has a tσ-module structure via the algebra

σ 1 σ homomorphism θ : C[χ(∂X)] t C[χ(X)], hence a C[M ± ] -module structure ≡ → 1 σ σ since C[M ± ] is a subring of t .

1 σ The extension from C[M ± ] to C(M) can be done in two steps: The ﬁrst one is

1 σ 1 from C[M ± ] to its ﬁeld of fractions C(x) where x := M + M − ; the second step is from C(x) to C(M). Each step is a ﬂat extension. The extension from tσ to ˜t can be

1 σ viewed as the extension from C[M ± ] to C(M): One can easily check that

σ ˜t = t C ±1 σ C(M), p˜ = p C ±1 σ C(M). ⊗ [M ] ⊗ [M ]

We want to show that p˜ is radical, i.e. p˜ = p˜. Here p˜ denotes the radical of p˜.

Let

σ t = t C ±1 σ C(x), p = p C ±1 σ C(x). ⊗ [M ] ⊗ [M ] Note that p, the kernel of θ : tσ C[χ(X)], is radical since the ring C[χ(X)] is → reduced. We claim that p is also radical. Indeed, suppose γ t and γ2 p. Then ∈ ∈ γ2 = δ/f for some δ p and f C[x]. It implies that (fγ)2 = fδ is in p. Hence ∈ ∈ fγ √p = p which means γ p. ∈ ∈ 1 Since t = C(x)[L± ] is a principal ideal domain, the radical ideal p can be generated

1 by one element, say γ(L) C(x)[L± ], which does not have repeated factors. Note ∈ that the polynomial γ(L) and δ(L) := γ′(L), the derivative of γ(L) with respect to L,

1 are co-prime. Since C(x)[L± ] is an Euclidean domain, there are f,g C(x) such that ∈ 1 fγ + gδ =1. It follows that γ(L) and δ(L) are also co-prime in C(M)[L± ]. Hence the

1 ideal p˜ = p C C(M) in C(M)[L± ] is radical. This means that the B-polynomial ⊗ (x)

BK does not have repeated factors.

Corollary 2.3.2. For every knot K one has

A B = K . K its M-factor

30 Here the M-factor of AK is the maximal factor of AK depending on M only; it is deﬁned up to a non-zero complex number.

Remark 2.3.3. Corollary 2.3.2 implies that the A-polynomial and B-polynomial are

equal for a knot K if the A-polynomial of K has trivial M-factor.

2.3.6 Small knots

A knot K is called small if its complement X does not contain closed essential surfaces.

It is known that all 2-bridge knots and all 3-tangle pretzel knots are small [26, 45].

Proposition 2.3.4. Suppose K is a small knot. Then the A-polynomial AK has trivial M-factor. Hence the A-polynomial and B-polynomial of a small knot are equal.

Proof. The A-polynomial A always contains the factor L 1 coming from characters K − of abelian representations [10]. Hence we write A = (L 1)An where An is a K − ab ab polynomial in C[M, L].

Suppose the polynomial Anab of a knot has non-trivial M-factor, then the Newton

polygon of Anab has the slope inﬁnity. It is known that every slope of the Newton

polygon of Anab is a boundary slope of the knot complement [10]. Hence the knot complement has boundary slope inﬁnity. The complement of a small knot does not

have boundary slope inﬁnity by a result in [9], hence its polynomial Anab has trivial M-factor.

Remark 2.3.5. There exists a non-small knot whose A-polynomial has non-trivial

M-factor; it is the knot 938 in the Rolfsen table, see [27].

A knot K is called SL2-small if the character variety χ(X) has dimension 1. It is

known that every small knot is SL2-small, see e.g. [10].

31 Proposition 2.3.6. Suppose K is SL2-small. Then both C[χ(X)] and the universal

1 character ring have ﬁnite rank over C[M + M − ].

Proof. By [38], the universal character ring of K is the quotient of the polynomial

1 ring R := C[x, x ,...,x ] by an ideal C[x, x ,...,x ], where x = M + M − is 1 k I ⊂ 1 k

the trace of the meridian and x1,...,xk are traces of some fixed elements in the knot group. Then the character ring of K is C[χ(X)] = C[x, x ,...,x ]/√ . 1 k I We first claim that the C[x]-module √ / has finite rank. Indeed, since R is I I Noetherian √ can be generated by a finite number of elements f ’s. Choose N such I j that f N for all j. Then (√ )N and hence the rank of the C[x]-module √ / j ∈ I I ⊂ I I I is less than or equal to N.

Consider the exact sequence of C[x]-modules

0 √ / R/√ R/ 0. → I I → I → I →

One has rankC R/ = rankC R/√ + rankC √ / . Hence to prove the theorem, [x] I [x] I [x] I I it suffices to show that the C[x]-module C[χ(X)] has finite rank for a SL2-small knot K. Let V , 1 j l, be irreducible components of χ(X). Since C[χ(X)] embeds into j ≤ ≤ the product C[V ] C[V ], it suffices to show that each C[V ] has finite rank, as 1 ×× l j a C[x]-module.

Let V be an irreducible component of the character variety. Let pr : Ck+1 C → be the projection onto the ﬁrst coordinate axis. Then pr(V ) is either (i) one point, or

(ii) the Zariski closure of pr(V ) is C, i.e. pr(V ) is C minus a finite number of points. In case (ii): the map pr is a finite map. Hence C[V ] has finite rank over C[x].

In case (i): V is determined by the ideal (J, x x ), where J C[x ,...,x ] and − 0 ⊂ 1 k x C. It follows that C[V ] is a torsion C[x]-module: it is annihilated by x x . In 0 ∈ − 0 this case C[V ] has rank 0, although as a C[x]-module it is inﬁnitely generated.

32 2.4 Skein modules and the AJ conjecture

Our proof of the main theorems is more or less based on the ideology that the KBSM

is a quantization of the SL2(C)-character variety [5, 48] which has been exploited in the work of Frohman, Gelca, and Lofaro [15] where they deﬁned the non-commutative

A-ideal. In this section we will discuss the quantum peripheral polynomial and its role in our approach to the AJ conjecture, and then prove Theorems 5 and 6.

2.4.1 Skein modules as quantizations of character varieties

Let ε be the map reducing t = 1. An important result [5, 48] in the theory of skein − modules is that ε( (Y )) has a natural C-algebra structure and is isomorphic to the S universal SL2-character algebra of the fundamental group of Y . The product of 2 links in ε( (Y )) is their union. Using the skein relation with t = 1, it is easy to see that S − the product is well-deﬁned, and that the value of a knot in the skein module depends only on the homotopy class of the knot in Y . The isomorphism between ε( (Y )) and S the universal SL -character algebra of π (Y ) is given by K(r) = tr r(K), where 2 1 − K is a homotopy class of a knot in Y , represented by an element, also denoted by

K, of π (Y ), and r : π (Y ) SL (C) is a representation of π (Y ). The quotient 1 1 → 2 1 of ε( (Y )) by its nilradical is canonically isomorphic to C[χ(Y )], the ring of regular S functions on the SL2-character variety of π1(Y ). In many cases ε( (Y )) is reduced, i.e. its nilradical is 0, and hence ε( (Y )) is S S exactly equal to the ring of regular functions on the character variety of π1(Y ). For example, this is the case when Y is a torus, or when Y is the complement of a two- bridge knot/link (see [36, 48] and Chapter 3), or when Y is the complement of the

( 2, 3, 2n + 1)-pretzel knot for any integer n (see Section 2.5 below). We conjecture − that

Conjecture 3. For every knot K the universal character ring is reduced.

33 2.4.2 The peripheral polynomial and its role in the AJ conjecture

Suppose for a knot K, the nilradical of ε( (X) is trivial. One has the following S commutative diagram (∂X) σ Θ (X) S ≡ T −−−→ S ε ε

 θ  C[χ(∂X)] tσ C[χ(X)]. ≡ −−−→ Recall that the classical peripheral ideal p is the kernel of θ; it is an ideal of tσ. The

σ 1 ring t embeds into the principal ideal domain t = C(M)[L± ]. The ideal extension p = tp of p in t is generated by the B-polynomial B . K Let us now adapt the construction of the polynomial B to the quantum setting. K By definition the quantum peripheral ideal is the kernel of Θ; it is a left-ideal of P σ. The ring σ embeds into the principal left-ideal domain which can be formally T T T defined as follows. Let (M) be the fractional field of the polynomial ring [M]. Let R R ˜ be the set of all Laurent polynomials in the variable L with coefficients in (M): T R

˜ = f (M)Lj f (M) (M), f = 0 almost everywhere , T { j | j ∈ R j } j Z ∈ and define the product in ˜ by f(M)Lk g(M)Ll = f(M) g(t2kM)Lk+l. T The left-ideal extension := of in is then generated by a polynomial P T P P T d ′ j βK (t; M, L)= βK,j(t, M) L , j=0 1 where d ′ is assumed to be minimum and all the coefficients β (t, M) Z[t± , M] K,j ∈ are co-prime. Note that the polynomial β is defined up to tkM l with k,l Z. We K ± ∈

will call βK the peripheral polynomial of K. The polynomial βK was introduced in [36].

Proposition 2.4.1. a) ε(βK) is M-essentially divisible by BK , and hence is M-

essentially divisible by AK.

34 b) βK is divisible by the recurrence polynomial αK in the sense that there are polynomials g(t, M) Z[t, M] and γ(t, M, L) such that ∈ ∈ T 1 β (t, M, L)= γ(t, M, L) α (t, M, L). (13) K g(t, M) K

Moreover g(t, M) and γ(t, M, L) can be chosen so that ε(g) =0.

Proof. See [36]. Note that the ﬁrst part follows from the fact that ε( ) p which can P ⊂ be easily deduced from the above commutative diagram, while the second fact follows the fact that the peripheral ideal is contained in the recurrence ideal by Proposition

2.2.2.

From Proposition 2.4.1, we see that both ε(αK) and AK divide ε(βK) for a knot K. This observation gives us some important information for studying the AJ conjecture for K. Indeed, we have the following.

Proposition 2.4.2. Suppose K is a knot satisfying all the following conditions:

(i) The L-degrees of the polynomials AK and ε(βK) are equal, (ii) The A-polynomial has exactly two irreducible factors,

(iii) The recurrence polynomial αK has L-degree greater than 1. Then the AJ conjecture holds for K.

M Proof. Since ε(βK) is M-essentially divisible by AK, conditon (i) implies that ε(βK) =

AK. Combining this with (13), we get

M ε(γ)ε(αK) = AK . (14)

It is known that A always contains the factor L 1 coming from characters of K − abelian representations [10]), and ε(α ) is also divisible by L 1 [36, Proposition K − 2.3]. Hence we can rewrite (14) as follows:

ε(α ) A ε(γ) K =M K . (15) L 1 L 1 − −

35 By (ii), A has exactly two irreducible factors. One of them is L 1, hence the K − AK other factor L 1 is irreducible. Equation (15) then implies that − ε(α ) ε(α ) A K =1M , or K =M K . L 1 L 1 L 1 − − − ε(αK ) M If L 1 = 1, then, by Lemma 2.4.3 below, the recurrence polynomial αK has L-degree − ε(αK ) M AK 1. This contradicts (iii), hence we must have L 1 = L 1 . In other words, the AJ − − conjecture holds true for K.

Lemma 2.4.3. The polynomial ε(α ) is M-essentially equal to L 1 if and only if K − the L-degree of the recurrence polynomial αK is 1.

Proof. The backward direction is obvious since ε(α ) is always divisible by L 1. K − Now suppose the polynomial ε(α ) is M-essentially equal to L 1, i.e. ε(α )= K − K 1 g(M)(L 1) for some non-zero g(M) C[M ± ]. Then we have − ∈ d α = g(M)(L 1)+(1+ t) f (M)Lj (16) K − j j=0 1 where f (M)’s are Laurent polynomials in [M ± ] and d is the L-degree of α . j R K By a result in [16], the recurrence ideal is invariant under the involution AK σ. Hence σ(α ) is contained in . Since α is the generator, it follows that K AK K α = h(M)σ(α )Ld for some h(M) (M). Equation (16) implies that K K ∈ R d g(M)(L 1)+(1+ t) f (M)Lj − j j=0 d 1 1 d 1 n j = h(M)g(M − )(L− 1)L +(1+ t) h(M)f (M − )L − . − j j=0 If d > 1 then by comparing the coeﬃcients of L0 in both sides of the above

1 equation, we get g(M)+(1+ t)f (M)=(1+ t)h(M)f (M − ), i.e. − 0 d

1 g(M)=(1+ t) f (M) h(M)f (M − ) (17) 0 − d Since g(M) is a Laurent polynomial in M with coeﬃcients in C, equation (17) implies that g(M) must be equal to 0. This is a contradiction. Hence we must have d = 1.

36 2.4.3 The L-degree of the A-polynomial

Suppose K is a hyperbolic knot. Then it has discrete faithful SL2(C)-representations.

Let χ0 denote an irreducible component of χ(X) containing the character of a discrete faithful representation. By a result of Thurston [54], χ0 has dimension 1 since X has one boundary component.

.(Recall that the inclusion ∂X ֒ X induces the restriction map ρ : χ(X) χ(∂X → → Dunﬁeld [13] showed that the map ρ : χ χ(∂X) is a birational isomorphism |χ0 0 → onto its image. Suppose K is hyperbolic and the character variety χ(X) consists of

2 irreducible conponents (one abelian and one non-abelian containing the characters of discrete faithful representations). Then the restriction map ρ : χ(X) χ(∂X) → is a birational isomorphism onto its image. i.e. χ(X) and its image are equal up to

adding a ﬁnite number of points, since χ(X) has dimension 1. It implies that the map

1 θ˜ : C(M)[L± ]= C[χ(∂X)] C ±1 σ C(M) C[χ(X)] C ±1 σ C(M), ⊗ [M ] → ⊗ [M ] induced by ρ, is surjective. Hence the polynomial BK , the generator of the kernel of

θ˜, has L-degree equal to the dimension of the C(M)-vector space C[χ(X)] C ±1 σ ⊗ [M ] 1 C(M), which is equal to the rank of the C[M + M − ]-module C[χ(X)]. Since the A-polynomial is M-essentially equal to the B-polynomial by Corollary 2.3.2, we have

Proposition 2.4.4. Suppose K is a knot satisfying all the following conditions:

(i) K is hyperbolic,

(ii) The SL -character variety of π (S3 K) consists of 2 irreducible components 2 1 \ (one abelian and one non-abelian).

1 Then the L-degree of the A-polynomial of K is equal to the rank of the C[M + M − ]- module C[χ(X)].

37 2.4.4 Localization and the L-degree of the peripheral polynomial 2.4.4.1 Localization

Recall that is the quantum torus and ε is the map reducing t = 1. Let D := T − 1 [M ± ] and consider the localization of D at the ideal (1+ t): R f D := f,g D, ε(g) =0 . { g | ∈ }

1 Note that D, being a localization of D, is ﬂat over D. The ring D = [M ± ] is R 1 σ 1 σ ﬂat over [M ± ] , since it is free over [M ± ] : R R

1 1 σ 1 σ [M ± ]= [M ± ] M [M ± ] . R R ⊕ R

Tensoring with D, we get T

:= f (M)Lj f (M) D, f = 0 almost everywhere T { j | j ∈ j } j Z ∈ with commutation rule: f(M)Lk g(M)Ll = f(M)g(t2kM)Lk+l. Note that if in the deﬁnition of we allow f (M) to be in the fractional ﬁeld (M) of D then we get T j R . T For a left-ideal I of (or σ) let I and I be its extensions in and respectively. T T T T 2.4.4.2 The L-degree of the peripheral polynomial

The involution σ acts on D. Let Dσ denote the σ-invariant part of D. There is a

σ 1 natural D = [¯x]-module structure on (X), herex ¯ (M + M − ) is a meridian R S ≡ − thus belongs to the boundary of X. Hence by tensoring (X) with D, (X) := S S 2 2 (X) σ D is an D-module. Note that (T ) := (T ) σ D is a free D-module S ⊗D S S ⊗D with basis Lj : j Z . { ∈ } Consider the following exact sequence of Dσ-modules

(σ Θ (X ֒ 0 → P → T −→ S

38 Since D is a ﬂat over Dσ, the following sequence is also exact

σ id Θ (D σ ֒ D σ ⊗ D σ (X 0 → ⊗D P → ⊗D T −→ ⊗D S i.e. the sequence of D-modules

(Θ (X) (18 ֒ 0 → P → T −→ S

is exact. Note that (X) is a module over the principal ideal domain D. S Suppose (X) C(M) is a ﬁnite dimensional vector space over C(M). Consider S ⊗D the exact sequence (18). The third module (X) is a D-module of ﬁnite rank and S hence l k (X)= D D/(f ). S j j=1 Here each f is a power of (1+ t). The middle module of (10) is free D-module j T with basis Lj : j Z . Hence the image of k + l + 1 elements 1, L, L2,...,Lk+l are { ∈ } linearly dependent. Hence there must be a non-trivial element δ in the kernel of P L-degree less than or equal to k + l. This element δ is in , hence it divides β , the P K generator of . From this, we conclude that β is non-trivial and has L-degree less P K that or equal to k + l, which is the dimension of the vector space (X) C(M) over S ⊗D C(M) since (X) C(M)= C(M)k+l. S ⊗D

It is easy to check that (X) C(M) = ε( (X)) C ±1 σ C(M). Since the S ⊗D S ⊗ [M ]

dimension of the C(M)-vector space ε( (X)) C ±1 σ C(M) is equal to the rank of S ⊗ [M ] 1 the C[M + M − ]-module ε( (X)), we have the following. S

1 Proposition 2.4.5. Suppose for a knot K, ε( (X)) is a ﬁnite rank C[M + M − ]- S

module. Then the L-degree of the peripheral polynomial βK is less than or equal the

1 rank of the C[M + M − ]-module ε( (X)). S

39 2.4.5 Proof of Theorems 5 and 6 2.4.5.1 Proof of Theorem 5

Suppose the knot K satisﬁes all the conditions of the theorem. By assumptions (i), (ii) and Proposition 2.4.4, the L-degree of the A-polynomial AK is equal to the rank of the

1 C[M + M − ]-module C[χ(X)]. Since the universal SL2-character variety is reduced, ε( (X)) = C[χ(X)]. Hence the L-degree of the A-polynomial A is also equal to the S K 1 the rank of the C[M + M − ]-module ε( (X)). This, together with Proposition 2.4.5, S implies that the L-degree of βK is less than or equal to that of AK. From this, one can easily check that the knot K satisﬁes all the conditions of Proposition 2.4.2 and hence the AJ conjecture holds true for K.

2.4.5.2 Proof of Theorem 6

It is known that two-bridge knots and ( 2, 3, 2n + 1)-pretzel knots, excluding torus − knots, are hyperbolic. (Note that the AJ conjecture holds true for torus knots by [25] and Chapter 4). Their universal character rings are reduced by [36] and Theorem 2.5.6 below, respectively. The L-degrees of their recurrence polynomials are greater than

1 according to the results in [36, Proposition 2.2] and [18, Section 4.7] respectively.

Double twist knots of the form J(k,l) with k = l, two-bridge knots of the form b(p, m) with p prime or m = 3, and ( 2, 3, 6n 1)-pretzel knots satisfy assumption (ii) of − ± Theorem 5 by [42], [23] and [6], and [40] respectively. Hence the theorem follows.

2.5 The universal character ring of ( 2, 3, 2n + 1)-pretzel knots −

In this section we explicitly calculate the universal character ring of the ( 2, 3, 2n+1)- − pretzel knot and prove its reducedness for all integer n.

40 2.5.1 The character variety

For the ( 2, 3, 2n + 1)-pretzel knot K , we have − 2n+1

π (X)= a, b, c cacb = acba, ba(cb)n = a(cb)nc , 1 | where X = S3 K and a, b, c are meridians depicted in Figure 1. \ 2n+1

Figure 1: The ( 2, 3, 2n + 1)-pretzel knot −

Let w = cb then the ﬁrst relation of π1(X) becomes caw = awa. It implies that

1 1 1 1 1 1 c = awaw− a− and b = c− w = awa− w− a− w. The second relation then has the form

1 1 1 n n 1 1 awa− w− a− waw = aw awaw− a−

i.e.

n 1 1 1 1 1 1 n w awa− w− a− = a− w− awaw− w .

Hence we obtain a presentation of π1(X) with two generators and one relation

π (X)= a, w wnE = Fwn 1 |

1 1 1 1 1 1 where E := awa− w− a− and F := a− w− awaw− . The character variety of the free group F = a, w in 2 letters a and w is isomor- 2 phic to C3 by the Fricke-Klein-Vogt theorem, see [38]. For every element ω F ∈ 2

there is a unique polynomial Pω in 3 variables such that for any representation r : F SL (C) we have tr(r(ω)) = P (x, y, z) where x := tr(r(a)), y := tr(r(w)) 2 → 2 ω

41 and z := tr(r(aw)). The polynomial Pω can be calculated inductively using the following identities for traces of matrices A, B SL (C): ∈ 2

1 1 tr(A) = tr(A− ), tr(AB)+tr(AB− ) = tr(A)tr(B). (19)

Thus for every representation r : π (X) SL (C), we consider x, y, and z as func- 1 → 2 tions of r. The character variety of π1(X) is the zero locus of an ideal in C[x, y, z], which we describe explicitly in the next theorem.

Theorem 2.5.1. The character variety of the pretzel knot K2n+1 is the zero locus of

2 polynomials P := P P and Q := P n P n . Explicitly, E − F n w Ea − Fw a P = x xy +( 3+ x2 + y2)z xyz2 + z3, (20) − − − 2 Qn = Sn 2(y)+ Sn 3(y) Sn 4(y) Sn 5(y) Sn 2(y) x (21) − − − − − − − − 2 + Sn 1(y)+ Sn 3(y)+ Sn 4(y) xz Sn 2(y)+ Sn 3(y) z − − − − − − where Sn(y) are the Chebychev polynomials deﬁned by S0(y)=1, S1(y) = y and

Sn+1(y)= ySn(y) Sn 1(y) for all integer n. − − Proof. The explicit formulas (20) and (21) follow from an easy calculation of the trace polynomials using (19).

n n n Because E and F are conjugate (by w ) and w Ea = Fw a in π1(X), we have P = Q = 0 for every representation r : π (X) SL (C). n 1 → 2

We will prove the converse: ﬁx a solution (x, y, z) of P = Qn = 0, we will ﬁnd a representation r : π (X) SL (C) such that x = tr(r(a)), y = tr(r(w)) and 1 → 2 z = tr(r(aw)). We consider the following 3 cases:

2 1 Case 1: y = 4. Then there exist s,u,v C such that s + s− = y, u + v = ∈ 1 sk+1 s−k−1 x, su + s− v = z. Since Sk(y)= s −s−1 for all integer k, we have −

3 P = s− (s 1)P ′, − 3 n 2n Q = s− − (s u sv)P ′ (1 + s)( 1+ uv)Q′ , n − − − n 42 where

3 4 5 2 2 2 3 2 4 2 5 2 P ′ = s u s u s u + v + sv s v s u v s u v + s u v + s u v − − − − − uv2 suv2 + s2uv2 + s3uv2, − − 5 2n 2+2n 2 4+2n 2 3 5 2n 2+2n 2 3 2 Q′ = s + s s u + s u + s uv s uv s uv + s uv + sv s v . n − − − −

Since s = 1, P = Q = 0 is equivalent to P ′ =( 1+ uv)Q′ = 0. We consider the ± n − n following 2 subcases:

u 1 s 0 Subcase 1.1: Qn′ = 0. Choose r(a) = and r(w) = .    1  uv 1 v 0 s−  −    It is easy to check x = tr(r(a)), y = tr(r(w)), z = tr(r(aw)) and the calculations  in the following 2 lemmas.

Lemma 2.5.2. One has

2 2 3 1 s− H11 s− H12 s− H22 s− H21 r(E)= − , r(F )= − −  2 2   3 1  s− ( 1+ uv)H21 s− H22 s− ( 1+ uv)H12 s− H11  − −   −      where

H = s2u s4u + v s2u2v + s4u2v uv2 + s2uv2, 11 − − − H = 1 s2u2 + s4u2 uv + s2uv, 12 − − H = s4 s2uv + s4uv v2 + s2v2, 21 − − − H = s4u + v s2v s2u2v + s4u2v uv2 + s2uv2. 22 − − − −

Lemma 2.5.3. One has

3+n 2 n s− P ′ s− − Qn′ r(wnE Fwn)= − . −  3 n 2 n  s− − ( 1+ uv)Qn′ s− − P ′  − − −   

43 n n n Since P ′ = Q′ = 0, Lemma 2.5.3 implies that r(w E Fw ) = 0, i.e. r(w E)= n − r(Fwn).

1 Subcase 1.2: 1+ uv = 0 then v = u− . In this case the equation P ′ = 0 becomes − 2 1 2 2 s u− (s u ) = 0 i.e. s = u . Let − u 0 u2 0 r(a)= , r(w)= .  1   2  0 u− 0 u−         Then it is easy to check that x = tr(r(a)), y = tr(r(w)), z = tr(r(aw)) and r(Ewn)= r(wnF ). (Note that r(a) and r(w) commute in this case).

Case 2: y = 2. Then Sk(y)= k for all integer k. Hence

P = (x z)( 1+ xz z2), − − − Q = 4 (n 1)x2 + (3n 5)xz (2n 3)z2. n − − − − −

1 2 4 Hence (x, z)=( 2, 2), (2, 2) or (x = z + z− and 1 n +(1+ n)z z = 0). − − − − If x = z = 2 we choose

1 0 1 0 r(a)=   , r(w)=   . 0 1 0 1         If x = z = 2 we choose − 1 0 1 0 − r(a)=   , r(w)=   . 0 1 0 1  −    1  2 4   If x = z + z− and 1 n +(1+ n)z z = 0 we choose − − z 0 1 1 r(a)= , r(w)= .  1 1    z− z− 0 1  −        Lemma 2.5.4. One has

1 2 4 0 z− ( 1+ n (1 + n)z + z ) n n − − r(w E Fw )=   − 0 0    

44 3 1 z 2z + z z z− z Proof. By direct calculations we have r(E)= − , r(F )= −  1   1  0 z− 0 z−         1 n n and r(w )=  . The lemma follows. 0 1     Hence x = tr(r(a)), y = tr(r(w)), z = tr(r(aw)) and r(wnE)= r(Fwn).

Case 3: y = 2. Then S (y)=( 1)kk for all integer k. Hence − k −

P = 3x + z + x2z +2xz2 + z3,

n Q = ( 1) xP (x + z)Q′′ /2, n − − n 2 2 where Qn′′ = x +2nx +2z + x z + xz .

Hence the system P = Qn = 0 is equivalent to P =(x + z)Qn′′ = 0. We consider the following 2 subcases:

Subcase 3.1: x + z = 0. Then it is easy to see that P = 0 is equivalent to x = z = 0. In this case we choose

i 0 1 0 − r(a)=   , r(w)=   0 i 0 1  −   −      where i is the imaginary number.

Subcase 3.2: x + z = 0. Then Q′′ = 0. Choose n

x/2 (1 x2/4)/(x + z) 1 1 − − − r(a)=   , r(w)=   . x z x/2 0 1  − −   −      Lemma 2.5.5. One has

nP Qn′′ Qn′′ /2 r(wnE Fwn)=( 1)n − . − −   0 (n 1)P + Qn′′  − −   

45 1 n n n Proof. By direct calculations, we have r(w )=( 1)   and − 0 1     2 3 2 2 (x +2z + x2z + xz2)/2 4+3x +4xz+x z+x z − − 4(x+z) r(E) =   , (x + z)(1 + xz + z2) (3x +2z + x2z + xz2)/2     (x +2z + x2z + xz2)/2 4+5x2+10xz+3x3z+4z2+5x2z2+2xz3 − 4(x+z) r(F ) =   (x + z)(1 + xz + z2) ( 5x 4z 3x2z 5xz2 2z3)/2.  − − − − −    The lemma follows.

Hence x = tr(r(a)), y = tr(r(w)), z = tr(r(aw)) and r(wnE) = r(Fwn) in all

cases. It implies that the character variety of the pretzel knot K2n+1 is exactly equal to the algebraic set P = Q =0 . { n } 2.5.2 The universal character ring

In this subsection, we will prove the following theorem.

Theorem 2.5.6. The universal character ring of K2n+1 is reduced and is equal to the

ring C[x, y, z]/(P, Qn).

Proof. Suppose we have shown that the ring C[x, y, z]/(P, Qn) is reduced, then it is exactly the character ring C[χ(X)] of K2n+1. Recall that π (X) = a, w wnE = Fwn where F = a, w is the free group on 1 | 2

two generators a, w. It is known that the universal character ring of F2 is the ring C[x, y, z] where x = tr(r(a)), y = tr(r(w)) and z = tr(r(aw)) as above. The quotient

map h : F2 π1(X) induces the epimorphism h : C[x, y, z] ε( (X)). Since P, Qn → ∗ → S come from traces, they are contained in ker h . ∗ Since C[χ(X))] is the quotient of ε( (X)) by its nilradical, we have the quotient S homomorphism φ : ε( (X)) C[χ(X))] = C[x, y, z]/(P, Q ). Then S → n

φ h : C[x, y, z] ε( (X)) C[χ(π)] = C[x, y, z]/(P, Qn) ◦ ∗ → S →

46 is a homomorphism. It follows that ker h (P, Qn). Hence we must have ker h = ∗ ⊆ ∗ (P, Q ), which implies ε( (X)) = C[x, y, z]/(P, Q ) C[χ(X)]. n S ∼ n ≡

In the remaining part of this section we will show that the ring C[x, y, z]/(P, Qn) is reduced, i.e. the ideal In := (P, Qn) is radical. The proof of this fact will be divided into several steps.

2.5.2.1 C[x, y, z]/In is free over C[x].

Lemma 2.5.7. For every x =0, 2, the polynomial P is irreducible in C[y, z]. 0 ± |x=x0

Proof. Assume that P can be decomposed as |x=x0

z3 x yz2 +(y2 + x2 3)z + x (1 y)= z + f z2 (x y + f )z + f , (22) − 0 0 − 0 − 1 − 0 1 2 where f C[y]. Equation (22) implies that f f (x y + f ) = y2 + x2 3 and j ∈ 2 − 1 0 1 0 − f f = x (1 y). 1 2 0 − If f is a constant then f = x (1 y)/f has y-degree 1. Hence f f (x y + f ) 1 2 0 − 1 2 − 1 0 1 has y-degree 1 also. It implies that f f (x y + f ) = y2 + x2 3. 2 − 1 0 1 0 − If f is a constant then f = x (1 y)/f . Hence 2 1 0 − 2

x0 x0 x0 x0 f2 f1(x0y + f1) = f2 y y + x0y − − f2 − f2 f2 − f2 2 2 2 x0 1 2 x0 2 x0 = 1 y 1 y + f2 2 . f2 − f2 − f2 − f2 − f2

2 2 2 2 x0 1 x0 2 Then since f2 f1(x0y + f1)= y + x 3, we have 1 = 1, 1 = 0, − 0 − f2 − f2 f2 − f2 x2 0 2 and f2 f 2 = x0 3. This implies x0 =0 or x0 = 2. − 2 − ±

Lemma 2.5.8. For every x , the polynomials P and Q are co-prime in 0 |x=x0 n |x=x0 C[y, z].

Proof. If x = 0, 2 then, by Lemma 2.5.7, P is irreducible in C[y, z]. Lemma 0 ± |x=x0 2.5.8 then follows since P and Q have z-degrees 3 and 2 respectively. |x=x0 n |x=x0

47 At x = 0, we have P = z( 3+ y2 + z2) and Q = a + b z2 where 0 − n n n

an = Sn 2(y)+ Sn 3(y) Sn 4(y) Sn 5(y), − − − − − −

bn = Sn 2(y) Sn 3(y). − − − −

2 In this case, it suﬃces to show that Qn z2=3 y2 = an + bn(3 y ) = 0. This is true by | − − Lemma 2.5.11 below.

2 2 At x = 2, we have P = z +1 y z (1+ y)z +2 and Q = a′ + b′ z + c′ z 0 − − n n n n where a′ , b′ ,c′ C[y]. When z = y 1, we have Q = 1 and Q = y 1 and n n n ∈ − 0 1 −

Qn+1 = yQn Qn 1. It implies that Qn z=y 1 is a polynomial of y-degree n and hence − − | − 2 2 is not identically 0. It remains to show that Q = a′ +b′ z+c′ z = c′ (z (1+y)z+2). n n n n n −

It suﬃces to show that bn′ y= 1= 0. Indeed, when x0 = 2 and y = 1 we have | − − bn′ = 2 Sn 1( 1) + Sn 3( 1) + Sn 4( 1) . It is easy to check that Sk( 1) = 1 if − − − − − − − k = 0 (mod 3), S ( 1) = 1 if k = 1 (mod 3) and S ( 1) = 0 otherwise. Hence k − − k − bn′ =2 Sn 1( 1) + Sn 3( 1) + Sn 4( 1) = 0. − − − − − − The case x = 2 is similar. 0 −

Proposition 2.5.9. C[x, y, z]/In is a free C[x]-module.

Proof. Since √In = √I, Proposition 2.3.6 implies that C[x, y, z]/In has ﬁnite rank over C[x]. It suﬃces to show that C[x, y, z]/In is a torsion-free C[x]-module. Suppose S C[x, y, z] and (x x )S I for some x C. We will show that S I . ∈ − 0 ∈ n 0 ∈ ∈ n Indeed, we have (x x )S = fP gQ for some f,g C[x, y, z]. Hence − 0 − n ∈ (fP ) = (gQ ) which implies that f is divisible by Q , since |x=x0 n |x=x0 |x=x0 n |x=x0 P and Q are co-prime in C[y, z] by Lemma 2.5.8. Hence f = |x=x0 n |x=x0 x=x0 h Q for some h C[y, z]. From this, we may write f = h Q +(x x )Q n |x=x0 ∈ n − 0 for some Q C[x, y, z]. Hence ∈

(x x )S = fP gQ = hQ +(x x )Q P gQ =(x x )QP +(hP g)Q − 0 − n n − 0 − n − 0 − n hP g which implies that hP g is divisible by x x and S = QP + − Q I . 0 x x0 n n − − − ∈

48 2.5.2.2 Reduction to a special case

Proposition 2.5.10. I is radical if I is radical for some x C. n n |x=x0 0 ∈

Proof. We ﬁrst note that I = I C C, where C is considered as an C[x]-module n |x=x0 n⊗ [x] by reducing x = x0. Let R = C[x, y, z]. Consider the exact sequence of C[x]-modules

0 I /I R/ I R/I 0. → n n → n → n → Since R/In is free by Proposition 2.5.9, the sequence splits and hence √In/In is isomorphic to free sub-module of R/In. Let k be the rank of the C[x]-module √In/In then the rank of the C-module (√I /I ) is always k for every x C. Hence if n n |x=x0 0 ∈ I is radical for some x C then k = 0 which implies that √I = I . n |x=x0 0 ∈ n n

2.5.2.3 In x=0 is radical By Lemma 2.5.8, P x=0 and Qn x=0 are co-prime. This means In x=0 is a zero- C dimensional ideal of [y, z]. By Seidenberg’s lemma (see [33, Proposition 3.7.15]), if there exist two non-zero free-square polynomials in I C[y] and I C[z] n x=0 ∩ n x=0 ∩ respectively, then I is radical. n x=0 From now on we ﬁx x = 0. Then P = z( 3+ y2 + z2) and Q = a + b z2 where − n n n

an = Sn 2(y)+ Sn 3(y) Sn 4(y) Sn 5(y), − − − − − −

bn = Sn 2(y) Sn 3(y). − − − −

2 Let Un = an + bn(3 y ). Then U0 =1, U1 = y + 1 and Un+1 = yUn Un 1. Hence − − −

Un = Sn(y)+ Sn 1(y). −

Lemma 2.5.11. One has

n j2π U = (y 2 cos ). n − 2n +1 j=1

49 Proof. It is easy to see that Un is a polynomial of degree n in y. Note that if y =

k+1 −k−1 j2π j2π 1 s s i 2n+1 i 2n+1 j2π s + s− = 2 then Sk(y)= s −s−1 . We now take y = e + e− = 2 cos 2n+1 ± − where 1 j n. Then ≤ ≤ j2π j2π sin((n + 1) 2n+1 ) sin(n 2n+1 ) Sn(y)= = = Sn 1(y). j2π j2π − sin( 2n+1 ) − sin( 2n+1 ) − The lemma follows.

Lemma 2.5.12. One has

n 3 − (2k + 1)π a = (y 2 cos ). n − 2n 5 k=0 − Proof. The proof is similar to that of the previous lemma.

Note that

b2 zP = b z2 ( 3+ y2)b + b z2 =(Q a )(Q U ). n n − n n n − n n − n Hence a U = a Q Q2 + Q U + b2 zP is contained in I . But a U is a n n n n − n n n n n |x=0 n n polynomial in y, hence it is actually contained in I C[y]. It is easy to see that n |x=0 ∩ an Un is square-free, i.e. does not have repeated factors. Let

n n 3 j2π − (2k + 1)π V = z ( 3 + 4 cos2 + z2) ( 3 + 4 cos2 + z2). n − 2n +1 − 2n 5 j=1 k=0 − Then it is easy to show that V C[z] is square-free. Moreover, since n ∈ n j2π V = z ( 3+ y2 + z2 + (4 cos2 y2)) n − 2n +1 − j=1 n 3 − (2k + 1)π ( 3+ y2 + z2 + (4 cos2 y2)) × − 2n 5 − k=0 − n n 3 j2π − (2k + 1)π z (4 cos2 y2) (4 cos2 y2) (mod P ), ≡ 2n +1 − 2n 5 − j=1 k=0 − 0 (mod(P, a U )) ≡ n n

50 it is contained in I . Hence V is in I C[z] and is square-free. n |x=0 n n |x=0 ∩ Since both a U I C[y] and V I C[z] are square-free, I is n n ∈ n |x=0 ∩ n ∈ n |x=0 ∩ n |x=0

a radical ideal by Seidenberg’s lemma. Hence by Proposition 2.5.10, In is also radical.

It implies that R/In is reduced. Hence the ring C[x, y, z]/(P, Qn) is reduced and is equal to the universal character ring of K2n+1. This completes the proof of Theorem 2.5.6.

51 CHAPTER III

THE SKEIN MODULE OF TWO-BRIDGE LINKS

3.1 Introduction

The theory of Kauﬀman bracket skein module (KBSM) was introduced by Przytycki

[47] and Turaev [56] as a generalization of the Kauﬀman bracket [29] in S3 to an arbitrary 3-manifold. The KBSM of a knot complement contains a lot, if not all, of information about the colored Jones polynomial. It also contains a lot of information about classical geometric invariants such as the character variety, and has been instrumental in the study of the AJ conjecture which relates the colored Jones polynomial and the A-polynomial of a knot, see [15, 21, 17, 36] and Chapter 2. The

calculation of the KBSM of a knot complement is a diﬃcult task. At the moment, the

KBSM has been calculated only for two-bridge knots [36] (with earlier work for twist

knots [3]) and torus knots [39] (with earlier work for (2, 2m + 1)-torus knots [4]). In

this chapter, we calculate the KBSM of the complement of all two-bridge links (see

Chapter 2 for a deﬁnition of the skein module of a 3-manifold). Applications to the theory of AJ conjecture for links will be discussed in a future work.

3.1.1 Main Results

A two-bridge link is a two-component link L S3 such that there is a 2-sphere ⊂ S2 S3 separating S3 into 2 balls B and B , and the intersection of L and each ⊂ 1 2 ball is isotopic to 2 trivial arcs in the ball. The branched double covering of S3 along a two-bridge link is a lens space L(2p, q), which is obtained by doing a 2p/q surgery on the unknot. Such a two-bridge link is denoted by b(2p, q). Here gcd(q, 2p) = 1, and one can always assume that 2p > q 1. It is known that b(2p′, q′) is isotopic to ≥ 1 b(2p, q) if and only if p′ = p and q′ q± (mod 2p), see [8]. ≡

52 Assume the 3-ball B is presented as a vertical cylinder B = D [0, 1], where 1 1 ×

D is a 2-dimensional disk, and the two arcs of L inside B1 are two vertical line

segments U [0, 1] and U ′ [0, 1], where U and U ′ are 2 interior points of D. Let × ×

D = D U, U ′ , then B1 L = D [0, 1]. Hence (B1 L)= (D ) is an algebra. ∗∗ \{ } \ ∗∗ × S \ S ∗∗ Let x, x′ D are respectively small loops around U, U ′, and y = ∂D D is the ⊂ ∗∗ ⊂ ∗∗ boundary of D. We consider x, x′, and y as elements of the algebra (B L). Using S 1 \ 3 a b c 3 the embedding (B L) (S L) we will consider x (x′) y as an element of (S L). 1\ ⊂ \ S \

Figure 2: The loops x, x′ and y

Theorem 7. For the two-bridge link L = b(2p, q), the skein module (S3 L) is free S \ 1 a b c over C[t± ] with basis x (x′) y 0 a, b, 0 c p . { | ≤ ≤ ≤ } 3.1.2 The universal character ring

Let ε( (Y )) be the quotient of (Y ) by the relation t = 1. An important result S S − [5, 48] in the theory of skein modules is that ε( (Y )) has a natural C-algebra structure S

and is isomorphic to the universal SL2-character algebra of the fundamental group of Y . For a deﬁnition of the universal character algebra, see [1, 38]. The product of 2 links in ε( (Y )) is their union. Using the skein relation with t = 1, it is S − easy to see that the product is well-deﬁned, and that the value of a knot in the

skein module depends only on the homotopy class of the knot in Y . The isomorphism

53 between ε( (Y )) and the universal SL -character algebra of π (Y ) is given by K(ρ)= S 2 1 tr ρ(K), where K is a homotopy class of a knot in Y , represented by an element, − also denoted by K, of π (Y ), and ρ : π (Y ) SL (C) is a representation of π (Y ). 1 1 → 2 1 The quotient of ε( (Y )) by its nilradical is canonically isomorphic to C[χ(π (Y ))], S 1 the ring of regular functions on the SL2-character variety of π1(Y ). The above fact has been exploited in the work of Frohman, Gelca, and Lofaro [15] where they deﬁned the non-commutative A-ideal of a knot, and in our proof of the AJ conjecture for some classes of two-bridge knots and pretzel knots in [36] and Chapter

2. In our work on the AJ conjecture, it is important to know whether the universal

character algebra ε( (Y )) is reduced, i.e. whether its nilradical is 0. Although it is S diﬃcult to construct a group whose universal character algebra is not reduced (see

[38]), so far there are a few groups for which the universal character algebra is known

to be reduced: free groups [52], surface groups [11], two-bridge knot groups [48], torus knot groups [39], and some pretzel knot groups (see Chapter 2).

As a consequence of Theorem 7, we will show the following.

Proposition 1. For a two-bridge link L, the universal SL -character algebra ε( (S3 2 S \ L)) is reduced, and hence ε( (S3 L)) is canonically isomorphic to the ring of regular S \ functions on the SL -character variety of π (S3 L). 2 1 \ 3.2 Proof of Theorem 7 and Proposition 1

We change the picture and will present the ball B R3 as the closed ball of radius √2 1 ⊂ centered at the origin, i.e. B = (x , x , x ) R3 x2+x2+x2 2 . We suppose that 1 { 1 2 3 ∈ | 1 2 3 ≤ }

the two-bridge link L = b(2p, q) intersects the interior of B1 in two straight intervals

UV and U ′V ′ in the x x -plane, where U =( 1, 1, 0), U ′ = (1, 1, 0),V =( 1, 1, 0) 1 2 − − − and V ′ = (1, 1, 0), see Figure 3. After an isotopy, we assume that the part of L −

outside the interior of B1 are 2 non-intersecting arcs u and u′ on the sphere S = ∂B1,

where u connects U and V , and u′ connects U ′ and V ′. If one cuts S along the arc

54 u, then one obtains a disk, hence the other arc u′, is uniquely determined by u, up to isotopy.

Figure 3: The ball B1

For a set Z R3 let Z[α, β] be the part of Z in the strip α x β , i.e. ⊂ { ≤ 1 ≤ } Z[α, β] := Z (x , x , x ) α x β . ∩{ 1 2 3 | ≤ 1 ≤ } Let S˜ be the 2-fold covering of S branched along the 4 points U, U ′,V,V ′. Note that S˜ is a torus, with the following preferred meridian and longitude. The plane passing through U, U ′,V,V ′ (i.e. the x x plane) intersects S[ √2, 1] in an arc 1 2 − − m that connects U and V . In other words, m is the shortest arc on the sphere S

connecting U and V , see Figure 4. The total lift m˜ of m is a closed curve on the the

torus S˜ which will serve as the meridian, see Figure 5. Let l be the shortest arc on S

connecting U and U ′. The total lift ˜l of l is a closed curve serving as the longitude.

It is easy to see that m˜ and ˜l form a basis of H1(S,˜ Z).

According to [8, Chapter 12], the isotopy class of the pair of arcs (u, u′) in the ball

B2 is uniquely determined by the homology class of the total lift u˜ of the curve u in

H (S,˜ Z). Moreover, the homology class of u˜ is equal to 2p m˜ + q′˜l for some q′ Z 1 ∈ 1 satisfying the condition q′ q± (mod 2p). We will describe explicitly the arc u in ≡

55 Figure 4: The curves m and l.

Figure 5: The total lifts m˜ ,˜l of m and l on the torus S˜ respectively.

the next subsection.

3.2.1 Description of u

We will present u by describing 3 parts of it: the left part ul, the middle part um, and the right part u , which are respectively the intersection of u with S := S[ √2, 1], r l − − S := S[ 1, 1], and S := S[1, √2]. For two non-antipodal points A, B on the sphere m − r S let γ(AB) be the shortest geodesic on S connecting A and B.

The boundary Cl := ∂Sl is a circle containing U and V . On the circle C mark 2p

points A0 = V, A1,...,A2p 1 which are: (i) counter-clockwise in that order if viewing − from the origin of the coordinate system, and (ii) uniformly distributed on the circle

Then Ap = U, and for 1 j p 1, the segment Ap jAp+j is parallel to the ≤ ≤ − −

x3-axis. The shortest geodesic γ(Ap jAp+j) lies in Sl. Let u0,l be the union of all the −

56 disjoint γ(Ap jAp+j), 1 j p 1. See Figure 6. − ≤ ≤ −

Let Ej be the midpoint of the arc AjAj+1 on the circle C (indices are taken modulo

2p). In other words, Ej is the image of Aj under the rotation by 2π/4p about the x1-axis, counter-clockwise if viewing from the origin.

Let Ej′ be the reﬂection of Ej through the x2x3-plane. Note that all the points Ej′

are on the circle C′ := ∂Sr. The p geodesics γ(Ep′ jEp′ +j 1), j =1,...,p, are disjoint − −

and are in Sr. Let u0,r be the union of the p geodesics γ(Ep′ jEp′ +j 1), j =1,...,p. − −

On Sm let u0,m be the union of 2p geodesics γ(AjEj′+(q 1)/2), j = 0, 1,..., 2p 1 − −

(indices taken modulo 2p). Note that the 2p components of u0,m are obtained from

each other by rotations by 2jπ/2p about the x1-axis.

Figure 6: u0,l, u0,m, u0,r for 2p = 6 and q′ =5

Let u0 be the arc on S obtained by combining u0,l, u0,m and u0,r; it connects U

and V . Up to isotopy there is a unique arc u0′ on S connecting V and V ′, and disjoint

from u0.

Lemma 3.2.1. The pair (u0, u0′ ) is isotopic, relative endpoints, to (u, u′) in the ball

B2.

2 Proof. It is easy that the homology class of the total lift of the arc u0 in H1(S˜ , Z)

is equal to 2p m˜ + q′˜l, which is exactly equal to the homology class of the total lift

57 ˜2 of the arc u in H1(S , Z) . According to [8, Chapter 12], (u0, u0′ ) is isotopic, relative endpoints, to (u, u′) in the ball B2.

From now on we identify (u, u′) and (u0, u0′ ). Without loss of generality we also

assume that q′ = q.

3.2.2 The link complement

Let ω be the boundary curve of a small normal neighborhood of the arc u in S = ∂B1.

Let X := B (UV U ′V ′), which is homeomorphic to the cylinder over a two- 1 1 \ ∪ punctured disk D in Subsection 3.1.1. Then the complement X of the link L is ∗∗

obtained from X1 by gluing a 2-handle to X1 along ω. Up to isotopy, ω can be described as follows.

On the circle C = ∂Sl mark 4p points F0, F1,...,F4p 1 which are: (i) counter- − clockwise in that order if viewing from the origin of the coordinate system, and (ii)

uniformly distributed on the circle C, and such that (iii) V is the midpoint of the arc

F4p 1F0 on C. We can also say that F2j is the midpoint of the arc AjEj, and F2j+1 −

is the midpoint of the arc EjAj+1 on C, see Figures 7 and 8.

For 1 j 2p, the segment F2p jF2p+j 1 is parallel to the x3-axis. The ≤ ≤ − −

shortest geodesic γ(F2p jF2p+j 1) lies in Sl. Let ωl be the union of all the disjoint − −

γ(F2p jF2p+j 1), 1 j 2p. − − ≤ ≤

Let Fj′ be the reﬂection of Fj through the x2x3-plane. We can also say that F2′j

is the midpoint of the arc Aj′ Ej′ , and F2′j+1 is the midpoint of the arc Ej′ Aj′ +1 on C,

where Aj′ is the reﬂection of Aj through the x2x3-plane. Note that all the points Fj′

are on the circle C′ = ∂Sr. For 1 j 2p, the segment F2′p jF2′p+j 1 is parallel to ≤ ≤ − −

the x3-axis. The shortest geodesic γ(F2′p jF2′p+j 1) lies in Sr. Let ωr be the union of − −

all the disjoint γ(F2′p jF2′p+j 1), 1 j 2p. − − ≤ ≤

Note that we can also say that ωr is the reﬂection of ωl through the x2x3-plane.

On S , let ω is the union of 4p geodesics γ(F F ′ ), j =0, 1,..., 4p 1 (indices m m j q+j −

58 taken modulo 4p). Note that the 4p components of ωm are obtained from each other by rotations by 2jπ/4p about the x1-axis.

Then, up to isotopy, ω is obtained by combining ωl,ωm and ωr.

Figure 7: The distribution of the points Aj, Fj, Ej and Aj′ , Fj′, Ej′ on the circles C and C′ respectively

Figure 8: ωl,ωm,ωr for 2p = 6 and q =5

o Let ψ be the rotation by 180 about the x1-axis. One has ψ(B1) = B1. Up to isotopy, we can assume that ψ(ω)= ω.

Let P = Fp+(q+1)/2,P ′ = F3p+(q+1)/2 and Q = Fp′+1 (q+1)/2, Q′ = F3′p+1 (q+1)/2. See − −

Figure 9. Note that ψ(P )= P ′ and ψ(Q)= Q′.

3.2.3 Relative skein modules

Let us recall the deﬁnition of the relative skein module (X ; P, Q′) (see [3, 36]). A S 1 type 1 tangle is the disjoint union of a framed link and a framed arc in X1 such that the parts of the arc near the two end points are on the boundary ∂X1, and the framing on these parts are given by vectors normal to ∂X1. Type 1 tangles are considered up

1 to isotopy relative the endpoints. Then (X ; P, Q′) is the C[t± ]-module generated S 1

59 by type 1 tangles with endpoints at P, Q′ modulo the usual skein relations, like in the definition of (X). One defines in a similar way the relative Kauffman bracket skein S module (∂X ; P, Q′) := (∂X [0, 1]; P, Q′), where we identify ∂X [0, 1] with a S 1 S 1 × 1 ×

collar of ∂X1 in X1.

There is a natural bilinear map (∂X ; P, Q′) (X ) (X ; P, Q′), where S 1 ⊗ S 1 → S 1 ℓ ℓ′ ℓ⋆ℓ′, which is the disjoint union of ℓ and ℓ′. ⊗ → The pair P, Q divide ω into two arcs, the one that is fully drawn in Figure 9 (and

that goes around the points U and V ′ exactly once) is denoted by ωs. Similarly, the

pair P ′, Q′ divide ω into two arcs, the one that is fully drawn in Figure 9 (and that

goes around the points U ′ and V exactly once) is denoted by ωs′ .

Let Pc = F3p+1 (q+1)/2,Pc′ = Fp+1 (q+1)/2 and Qc = F ′ , Qc′ = F ′ . − − 3p+(q+1)/2 p+(q+1)/2

Then ωs consists of 3 parts: the left part is an arc on Sl connecting P and Pc, the middle part is an arc on Sm connecting Pc and Qc, and the right part is an arc on Sr connecting Qc and Q. Similarly, ωs′ also consists of 3 parts: the left part is an arc on

Sl connecting P ′ and Pc′, the middle part is an arc on Sm connecting Pc′ and Qc′ , and the right part is an arc on Sr connecting Qc′ and Q′.

Figure 9: ωs connects P, Q, and ωs′ connects P ′, Q′

Let γin(P Q′),γin(P ′Q) be respectively the shortest arcs on the surface S = ∂B1

connecting P and Q′, P ′ and Q, whose interiors are slightly pushed inside the interior

of B1 (to avoid intersections with other arcs on S) and whose framings are given by

60 vectors normal to S.

Let din(PP ′), din(QQ′) be respectively the straight intervals connecting P and P ′,

Q′ and Q, whose interiors are slightly pushed into the interior of B [ √2, 1] and 1 − −

the interior of B1[1, √2] respectively (to avoid intersections with the straight lines

UV and U ′V ′ respectively).

Let a1 be γin(P Q′); a2 be ωs followed by din(QQ′); a3 be din(PP ′) followed by ωs′ ; and a4 be ωs followed by γin(QP ′) then followed by ωs′ .

Figure 10: The projections of a2, a3 and a4 onto the x1x3-plane

4 Lemma 3.2.2. The relative skein (X ; P, Q′) is equal to the union (a ⋆ (X )). S 1 ∪i=1 i S 1

Proof. Let a1′ = a1. Let a2′ be γ(PPc) folowed by γin(PcQ′); a3′ be γin(P Qc′ ) followed by γ(Qc′ Q′); and a4′ be γ(PPc) followed by γin(PcQc′ ) then followed by γ(Qc′ Q′). Here

γin(PcQ′), γin(P Qc′ ), γin(PcQc′ ) are respectively the shortest arcs on S connecting Pc and Q′, P and Qc′ , Pc and Qc′ , whose interiors are slightly pushed inside the interior of B1 (to avoid intersections with other arcs on S) and whose framings are given by vectors normal to S.

Using the skein relations one can simplify the arc part of elements in (X ; P, Q′), S 1 showing that the arc part is one of the four ai′ , i =1, 2, 3, 4. More precisely, by similar arguments as in the proof of [3, Lemma 3.1] one can show that the relative skein

4 (X ; P, Q′) is equal to the union (a′ ⋆ (X )). S 1 ∪i=1 i S 1

61 Figure 11: The projection of a2′ , a3′ and a4′ onto the x1x3-plane

It is easy to see that a is isotopic to a′ for all 1 i 3. Using the skein relations i i ≤ ≤ to resolve all the crossings of a one can easily show that the set a ⋆ (X ) is equal 4 4 S 1 3 to a′ ⋆ (X ), modulo the union (a′ ⋆ (X )). The lemma follows. 4 S 1 ∪i=1 i S 1 3.2.4 From (X ) to (X) through sliding S 1 S

Recall that X is obtained from X1 by attaching a 2-handle along the curve ω. Note that (X1) = (D ) is isomorphic to the commutative algebra [x, x′,y], see [47]. S S ∗∗ R The embedding of X into X gives rise to a linear map from (X ) [x, x′,y] to 1 S 1 ≡ R (X). It is known that the map is surjective, and its kernel , see [47, 3], can be S K described through slides as follows. Suppose a is a type 1 tangle whose 2 endpoints are on ω such that outside a small

neighborhood of the 2 endpoints a is in the interior of X1 and in a small neighborhood

of the endpoints a is on the boundary S = ∂B1. The two end points of a divide ω into 2 arcs ω1 and ω2. The loop ω partitions S, which is a sphere, into 2 parts; the one not containing U, U ′ is called the outside one. Let us isotope a (relatively to the endpoints) to a′ so that in a small neighborhood of the endpoints, a′ is in the outside part of ω.

Let sl(a) be a′ ω a′ ω , considered as an element of the skein module (X ). 1 − 2 S 1 Here a′ ω is the framed link obtained by combining a′ and ω . Note that sl(a) is 1 1 deﬁned up to a factor t3n, n Z. The exchange ω ω changes the sign, and ± ∈ 1 ↔ 2 isotopies in neighborhoods of the endpoints change the framing, which results in a

62 factor equal to a power of ( t3). − It is clear that as framed links in X, a′ ω is isotopic to a′ ω , since one is obtained 1 2 from the other by sliding over the 2-handle attached to the curve ω. Hence we always have sl(a) . It was known that is spanned by all possible sl(a), where a can be ∈K K chosen among all type 1 tangles with pre-given two endpoints on ω.

From the description of (X ; P, Q′) in Lemma 3.2.2 we have S 1

1 Lemma 3.2.3. The kernel is equal to the C[t± ]-span of sl(a ) ⋆ (X ), i = K { i S 1 1, 2, 3, 4 . }

Lemma 3.2.4. One has

sl(a1) = sl(γin(P Q′)),

sl(a2) = sl(din(PP ′)),

sl(a3) = sl(din(QQ′)),

sl(a4) = sl(γin(P ′Q)).

Proof. The ﬁrst identity is a tautology. The last three follows from trivially a simple isotopy of the links involved.

Lemma 3.2.5. For every ℓ (X ), one has ψ(ℓ)= ℓ. ∈ S 1

Proof. This is because x, x′ and y are invariant under the rotation ψ.

Lemma 3.2.6. One has

sl(γ (P Q′)) ⋆ (X ) = sl(γ (P ′Q)) ⋆ (X ), in S 1 in S 1

sl(d (PP ′)) ⋆ (X ) = 0, in S 1

sl(d (QQ′)) ⋆ (X ) = 0. in S 1

63 Proof. Since ψ(P ) = P ′ and ψ(Q) = Q′, we have ψ(sl(γin(P Q′))) = sl(γin(P ′Q)).

Hence sl(γ (P Q′)) ⋆ (X )= sl(γ (P ′Q)) ⋆ (X ) by Lemma 3.2.5. in S 1 in S 1 Since both d (PP ′) and ω is invariant under ψ, we have ψ(d (PP ′) ω (P,P ′)) = in in 1 d (PP ′) ω (P,P ′), where ω (P,P ′) and ω (P,P ′) are the two arcs of ω obtained by in 2 1 2 dividing ω using the two points P,P ′. It implies that

sl(d (PP ′)) ⋆ (X )=(d (PP ′) ω (P,P ′) d (PP ′) ω (P,P ′)) ⋆ (X )=0. in S 1 in 1 − in 2 S 1

This completes the proof of the lemma.

3.2.5 Proof of Theorem 7

1 Let = C[t± ]. We have (X) = [x, x′,y]/ , where is the -span of sl(a ) ⋆ R S R K K R 1 [x, x′,y], by Lemmas 3.2.3, 3.2.4 and 3.2.6. Note that there is a natural [x, x′]- R R module structure on (X): Here x, x′ are meridians, thus belong to the boundary S 2 of X. Over [x, x′], [x, x′,y] is spanned by 1,y,y ,... . Hence , as an [x, x′]- R R K R module, is spanned by sl(a ) ⋆yk =(a ω a ω ) ⋆yk,k =0, 1, 2,... . 1 1 1 − 1 2 Note that a ω is the closure in the sense of [36, Section 1.5] of a braid on 1 1 (2p + 2) strands, while a ω is the closure of a braid on (2p 2) strands. Moreover, 1 2 − (a ω ) ⋆yk is the closure of a braid on (2p +2)+2k strands, while (a ω ) ⋆yk 1 1 1 2 is the closure of of a braid on (2p 2)+2k strands. Lemma 1.1 in [36] then shows − k that (a ω a ω ) ⋆y , as an element of [x, x′,y], has y-degree (p +1)+ k, with 1 1 − 1 2 R highest coeﬃcient invertible and of the form a power of t. Hence when we factor out

l [x, x′,y] by , we get a free [x, x′]-module with representatives y ,l =0, 1, 2,...,p R K R as a basis.

3.2.6 Proof of Proposition 1

From Theorem 7 it follows that ε( (X)) is the quotient of the ring C[¯x, x¯ , y¯] by the S ′ ideal I generated by ε(sl(a1)) ⋆ C[¯x, x¯′, y¯], wherez ¯ denotes the negative of the trace of the loop z.

64 Note that ε( (X)) has a natural C-algebra structure and ⋆ is just the multipli- S cation of this algebra. It implies that I can be generated by only one element which is ε(sl(a )). Hence ε( (X)) = C[¯x, x¯ , y¯]/(ϕ), where ϕ = ε(sl(a )) C[¯x, x¯ , y¯]. Note 1 S ′ 1 ∈ ′ that ϕ is a polynomial ofy ¯-degree p + 1 with leading coeﬃcient 1. ± We claim that ϕ has no repeated factors. Since ϕ is a polynomial ofy ¯-degree p+1

with leading coeﬃcient 1, it suﬃces to show that ϕ(0, 0, y¯) has no repeated factors. ±

Lemma 3.2.7. One has

2 ϕ(0, 0, y¯)= (¯y 4)Sp 1(¯y), ± − −

where Sn(¯y) are the Chebyshev polynomials deﬁned by S0(¯y)=1,S1(¯y)=y ¯ and

Sn+1(¯y)=¯ySn(¯y) Sn 1(¯y) for all integer n. − −

Proof. By [8], the fundamental group of the two-bridge link L = b(2p, q) is

π (L)= x,˜ x˜ xw˜ = wx˜ , 1 ′ |

kq ε ε ε − ε − where w =(x˜ ) 1 (˜x) 2 (˜x) 2p 2 (x˜ ) 2p 1 and ε =( 1)⌊ 2p ⌋. Herex, ˜ x˜ are meridians ′ ′ k − ′ of the link L, and are conjugate to x, x′ respectively.

3 The character variety of the free group in 2 lettersx ˜ and x˜′ is isomorphic to C , by the Fricke-Klein-Vogt theorem. For every word z, the trace of z is a polynomial

in 3 variables trx ˜ = x,¯ tr x˜ = x¯ and tr(˜xx˜ )= y.¯ − ′ − ′ ′ − 1 1 1 Note that the traces of the words (˜x)− wx˜(x˜′)− and w(x˜′)− are equal. Hence

1 1 1 η = tr (˜x)− wx˜(x˜ )− w(x˜ )− ′ − ′ ¯ is divisible by ϕ in C[¯x, x′, y¯].

1 Suppose from now onx ¯ = x¯ = 0. We have (˜x)− +x ˜ = tr˜x = x¯ = 0, i.e. ′ − 1 (˜x)− = x˜, by the Cayley-Hamilton theorem applying for matrices in SL (C). Here − 2 1 we identifyx ˜ with its representation matrix in SL (C). Similarly, (x˜ )− = x˜ . 2 ′ − ′

65 1 Let k be the the number of times the power 1 appears in the word w(x˜ )− = − ′ (x˜ )ε1 (˜x)ε2 (˜x)ε2p−2 . Then it is easy to see that the number of times the power ′ 1 1 1 ε1 ε2 ε2p−2 ε2p−1 1 1 appears in the word (˜x)− wx˜(x˜ )− = (˜x)− (x˜ ) (˜x) (˜x) (x˜ ) x˜(x˜ )− is − ′ ′ ′ ′ 1 1 1 k +2. If we replace (˜x)− and (x˜′)− in w(x˜′)− byx ˜ and x˜′ respectively then we pick

k 1 k p 1 1 1 up the sign ( 1) , i.e. we have w(x˜ )− =( 1) (x˜ x˜) − . Similarly, (˜x)− wx˜(x˜ )− = − ′ − ′ ′ k+2 p+1 k p+1 p 1 ( 1) (˜xx˜ ) . It implies that η(0, 0, y¯)=( 1) tr (˜xx˜ ) (x˜ x˜) − . − ′ − ′ − ′ n+1 n 1 1 Let δ = tr (˜xx˜ ) (x˜ x˜) − . By the Cayley-Hamiton, x ˜x˜ +(˜xx˜ )− = tr(˜xx˜ )= n ′ − ′ ′ ′ ′ 2 y¯. This implies that δn+1 = yδ¯ n δn 1. It is easy to check that δ1 =y ¯ 4, δ2 = − − − − − 2 n 1 2 (¯y 4)¯y. Hence δn =( 1) − (¯y 4)Sn 1(¯y) where Sn(¯y) are the Chebyshev poly- − − − − −

nomials deﬁned by S0(¯y)=1,S1(¯y)=y ¯ and Sn+1(¯y)=yS ¯ n(¯y) Sn 1(¯y) for all − − integer n.

k k+p 1 2 We have η(0, 0, y¯)=( 1) δp = ( 1) − (¯y 4)Sp 1(¯y), which is a polynomial − − − − k+p 1 of degree p + 1 iny ¯ with leading coeﬃcient ( 1) − . Since η is divisible by ϕ, and − ϕ is also a polynomial ofy ¯-degree p + 1 with leading coeﬃcient 1, we must have ± 2 ϕ(0, 0, y¯)= (¯y 4)Sp 1(¯y) as desired. ± − −

p 1 πj 2 It is known that Sp 1(¯y) = − (¯y 2 cos ) and hence (¯y 4)Sp 1(¯y) has no − j=1 − p − − repeated factors. By Lemma 3.2.7, it follows that ϕ has no repeated factors either.

Hence the nil-radical of ε( (X)) is zero, which means that ε( (X)) is exactly equal S S

to C[χ(π1(X))]. This completes the proof of Proposition 1.

Corollary 3.2.8. The character ring of the two-bridge link b(2p, q) is the quotient of

1 1 the ring C[¯x, x¯ , y¯] by the ideal generated by the polynomial η = tr((˜x)− wx˜(x˜ )− ) ′ ′ − 1 tr(w(x˜′)− ), where x,¯ x¯′, y,¯ x,˜ x˜′ and w are deﬁned as in the proof of Lemma 3.2.7.

Proof. We still use the notations in the proof of Lemma 3.2.7.

kq ε ε ε − ε − Since w = (x˜ ) 1 (˜x) 2 (˜x) 2p 2 (x˜ ) 2p 1 and ε = ( 1)⌊ 2p ⌋ = 1, it is easy to ′ ′ k − ± 1 1 1 show that the traces of the words (˜x)− wx˜(x˜′)− and w(x˜′)− havey ¯-degrees equal to p + 1 and p 1 respectively, with leading coeﬃcients 1. It implies that the − ±

66 polynomial η hasy ¯-degree p + 1 with leading coeﬃcient 1. Since η is divisible by ± ϕ, we must have η = ϕ. Hence, by Proposition 1, the character ring of b(2p, q) is ± equal to the quotient of the ring C[¯x, x¯′, y¯] by the ideal generated by the polynomial

1 1 1 η = tr((˜x)− wx˜(x˜ )− ) tr(w(x˜ )− ). ′ − ′

Remark 3.2.9. Corollary 3.2.8 was already obtained in [49] although it was not

completely written in form of traces. The proof we present here essentially follows

directly from Theorem 7.

One can easily show that the characters of abelian representations (into SL2(C)) of the two-bridge link b(2p, q) is determined by the polynomial

1 1 2 2 2 ηab = tr(˜xx˜ (˜x)− (x˜ )− )=¯y +¯x + x¯ +¯yx¯x¯ 4. ′ ′ ′ ′ −

Hence, by Corollary 3.2.8, the characters of non-abelian representations of b(2p, q)

is determined by the polynomial ηnab = η/ηab. The polynomial ηnab hasy ¯-degree

p 1 with leading coeﬃcient 1. After a suitable change of variables, it is exactly − ± the polynomial Φ in [49, Lemma 2], up to 1. πL ±

67 CHAPTER IV

PROOF OF A STRONGER VERSION OF THE AJ

CONJECTURE FOR TORUS KNOTS

4.1 Introduction

The AJ Conjecture was veriﬁed for the trefoil and ﬁgure 8 knots by Garoufalidis [17], and was partially checked for all torus knots by Hikami [25]. It was established

for some classes of two-bridge knots and pretzel knots, including all twist knots and

( 2, 3, 6n 1)-pretzel knots, see [36] and Chapter 2. In this chapter we provide a − ± full proof of the AJ conjecture for all torus knots. Moreover, we show that a stronger

version of the conjecture, due to A. Sikora, holds true for all torus knots.

4.1.1 Main results

Recall from Chapter 2 that is the recurrence ideal and p is the classical peripheral A ideal of a knot K. The involution σ acts on the quantum torus by σ(M kLl) = T k l σ M − L− . Let be the σ-invariant part of the recurrence ideal ; it is an ideal A A of σ. Let ε is the map reducing the quantum parameter t to 1. A. Sikora [52] T − proposed the following conjecture.

Conjecture 4. Suppose K is a knot. Then ε( σ)= p. A Here ε( σ) denotes the radical of the ideal ε( σ) in the ring tσ = ε( σ). A A T It is easy to see that Conjecture 4 implies the AJ conjecture. Conjecture 4 was veriﬁed for the unknot and the trefoil knot by Sikora [52]. In the present chapter we

conﬁrm it for all torus knots.

Theorem 8. Conjecture 4 holds true for all torus knots.

68 4.1.2 Plan of the chapter

We provide a full proof of the AJ conjecture for all torus knots in Section 4.2 and

prove Theorem 8 in Section 4.3.

4.2 Proof of the AJ conjecture for torus knots

We will always assume that knots have framings 0.

We consider the two cases: a,b > 2 and a = 2 separately. Lemmas 4.2.1 and 4.2.5

below were ﬁrst proved in [25] using formulas for the colored Jones polynomial and

Alexander polynomial of torus knots given in [41]. We present here direct proofs.

4.2.1 The case a,b > 2

Lemma 4.2.1. For the (a, b)-torus knot, we have

2 2 4ab(n+1) 2ab(n+1) t λ(a+b)(n+1) t− λ(a b)(n+1) J(n +2)= t− J(n)+ t− − − , t2 t 2 − − 2k 2k where λk := t + t− .

Proof. For the (a, b)-torus knot, by [41], we have

n−1 2 ab(n2 1) 4bj(aj+1) J(n)= t− − t [2aj + 1]. (23) − j= n 1 − 2 Hence

n+1 2 ab((n+2)2 1) 4bj(aj+1) J(n +2) = t− − t [2aj + 1] j= n+1 − 2 n−1 2 ab((n+2)2 1) 4bj(aj+1) ab((n+2)2 1) = t− − t [2aj +1]+ t− − − j= n 1 − 2 b(n+1)(a(n+1)+2) b(n+1)(a(n+1) 2) t [a(n + 1) + 1] t − [a(n + 1) 1] × − − 2 2 4ab(n+1) 2ab(n+1) t λ(a+b)(n+1) t− λ(a b)(n+1) = t− J(n)+ t− − − , t2 t 2 − − 2k 2k 2 2 since [k]=(t t− )/(t t− ). − −

69 Lemma 4.2.2. The colored Jones polynomial of the (a, b)-torus knot is annihilated

3 2 by the operator αa,b = c3L + c2L + c1L + c0 where

2 2(a+b) a+b 2(a+b) (a+b) 2 2(a b) a b 2(a b) (a b) c = t t M + t− M − t− t − M − + t− − M − − , 3 − 2ab 2 4(a+b) a+b 4(a+b) (a+b) 2 4(a b) a b 4(a b) (a b) c = t− t t M + t− M − + t− t − M − + t− − M − − , 2 − 8ab 2ab c = t− M − c , 1 − 3 4ab 2ab c = t− M − c . 0 − 2

4ab(n+2) 4ab(n+1) Proof. It is easy to check that c3t− + c1 = c2t− + c0 = 0 and

2 2 2ab 2 2 c3 t λ(a+b)(n+2) t− λ(a b)(n+2) + c2t t λ(a+b)(n+1) t− λ(a b)(n+1) = 0. − − − − Hence, by Lemma 4.2.1, αa,bJ(n) is equal to

c3J(n +3)+ c2J(n +2)+ c1J(n +1)+ c0J(n) 2 2 4ab(n+2) 2ab(n+2) t λ(a+b)(n+2) t− λ(a b)(n+2) − − − = c3 t J(n +1)+ t 2 − 2 t t− 2 − 2 4ab(n+1) 2ab(n+1) t λ(a+b)(n+1) t− λ(a b)(n+1) − − − + c2 t J(n)+ t 2 − 2 t t− − + c1J(n +1)+ c0J(n)

4ab(n+2) 4ab(n+1) = (c3t− + c1)J(n +1)+(c2t− + c0)J(n) 2 2 2 2 2ab(n+1) t λ(a+b)(n+2) t− λ(a b)(n+2) 2ab t λ(a+b)(n+1) t− λ(a b)(n+1) − − − + t c3 2 − 2 + c2t 2 − 2 t t− t t− − − = 0.

This proves Lemma 4.2.2.

Let and denote the recurrence ideal of the (a, b)-torus knot and its Aa,b Aa,b extension in respectively. T

Proposition 4.2.3. For the (a, b)-torus knot, with a,b > 2, we have = α . Aa,b a,b

70 2 Proof. By Lemma 4.2.2 it suﬃces to show that if an operator P = P2L + P1L + P0,

1 where Pj’s are polynomials in C[t± , M], annihilates the colored Jones polynomial then P =0.

Indeed, suppose PJ(n) = 0. Then, by Lemma 4.2.1,

0 = P2J(n +2)+ P1J(n +1)+ P0J(n) 2 2 t λ(a+b)(n+1) t− λ(a b)(n+1) 4ab(n+1) 2ab(n+1) − = P2 t− J(n)+ t− 2 − 2 t t− − + P1J(n +1)+ P0J(n) 2 2 4ab(n+1) 2ab(n+1) t λ(a+b)(n+1) t− λ(a b)(n+1) = (t− P + P )J(n)+ P J(n +1)+ P t− − − . 2 0 1 2 t2 t 2 − − 2 −2 t λ t λ − 4ab(n+1) 2ab(n+1) (a+b)(n+1)− (a b)(n+1) Let P2′ = t− P2 + P0 and P0′ = P2t− t2 t−2 . Then −

P2′J(n)+ P1J(n +1)+ P0′ =0. (24)

1 Note that P2′ and P0′ are polynomials in C[t± , M].

Lemma 4.2.4. The lowest degree in t of J(n) is

2 1 n 1 l = abn + ab + (1 ( 1) − )(a 2)(b 2). n − 2 − − − −

Proof. From (23), it follows easily that l = abn2 +ab if n is odd, and l =( abn2 + n − n − ab)+(ab 2b 2a + 4) if n is even. − −

Suppose P ′,P = 0. Let r and s be the lowest degrees (in t) of P ′ and P 2 1 n n 2 1 respectively. Note that, when n is large enough, rn and sn are polynomials in n of degrees at most 1. Equation (24) then implies that rn + ln = sn + ln+1, i.e.

r s = l l = ab(2n + 1) ( 1)n(a 2)(b 2). n − n n+1 − n − − − − −

This cannot happen since the LHS is a polynomial in n, when n is large enough, while

the RHS is not (since (a 2)(b 2) > 0). Hence P ′ = P = P ′ = 0, which means − − 2 1 0 P =0.

71 2ab a a b b It is easy to see that ε(α ) = M − (M M − )(M M − )A where A = a,b − − a,b a,b (L 1)(L2M 2ab 1) is the A-polynomial of the (a, b)-torus knot when a,b > 2. This − − means the AJ conjecture holds true for the (a, b)-torus knot when a,b > 2.

4.2.2 The case a =2

Lemma 4.2.5. For the (2, b)-torus knot, we have

(4n+2)b 2nb J(n +1) = t− J(n)+ t− [2n + 1]. −

Proof. For the (2, b)-torus knot, by (23), we have

n−1 2 2b(n2 1) 4bj(2j+1) J(n)= t− − t [4j + 1]. − j= n 1 − 2 Hence n 2 2b((n+1)2 1) 4bk(2k+1) J(n +1) = t− − t [4k + 1] k= n − 2 Set k = (j + 1 ). Then − 2 n+1 − 2 2b((n+1)2 1) 4bj(2j+1) J(n +1) = t− − t [ (4j + 1)] − − j= n 1 2 n−1 2 2b((n+1)2 1) 4bj(2j+1) 2bn(n+1) = t− − t [4j +1]+ t [2n + 1] − − j= n 1 − 2 (4n+2)b 2nb = t− J(n)+ t− [2n + 1]. − This proves Lemma 4.2.5.

Lemma 4.2.6. The colored Jones polynomial of the (2, b)-torus knot is annihilated

2 by the operator α2,b = d2L + d1L + d0 where

2 2 2 2 d = t M t− M − , 2 − 2b 4b 2b 2 2 2 2 6 2 6 2 d = t− t− M − (t M t− M − ) (t M t− M − ) , 1 − − − 4b 2b 6 2 6 2 d = t− M − (t M t− M − ). 0 − −

72 Proof. From Lemma 4.2.5 we have

(4n+2)b 2nb J(n +1) = t− J(n)+ t− [2n + 1], − 8(n+1)b 6(n+1)b 2(n+1)b J(n +2) = t− J(n) t− [2n +1]+ t− [2n + 3]. −

It is easy to check that

8(n+1)b (4n+2)b t− d t− d + d = 0, 2 − 1 0 6(n+1)b 2(n+1)b 2nb d t− [2n +1]+ t− [2n + 3] + d t− [2n +1] = 0. 2 − 1 Hence

α2,bJ(n) = d2J(n +2)+ d1J(n +1)+ d0J(n)

8(n+1)b 6(n+1)b 2(n+1)b = d t− J(n) t− [2n +1]+ t− [2n + 3] 2 − (4n+2)b 2nb + d t− J(n)+ t− [2n + 1] + d J(n) 1 − 0 8(n+1)b (4n+2)b = t− d t− d + d J(n) 2 − 1 0 6(n+1)b 2(n+1)b 2nb + d t− [2n +1]+ t− [2n + 3] + d t− [2n + 1] 2 − 1 = 0.

This proves Lemma 4.2.6.

Proposition 4.2.7. For the (2, b)-torus knot, we have = α . A2,b 2,b

Proof. By Lemma 4.2.6, it suﬃces to show that if an operator P = P1L + P0, where

1 Pj’s are polynomials in C[t± , M], annihilates the colored Jones polynomial then P =0.

Indeed, suppose PJ(n) = 0. Then

0 = P1J(n +1)+ P0J(n)

(4n+2)b 2nb = P t− J(n)+ t− [2n + 1] + P J(n) 1 − 0 (4n+2)b 2nb = t− P + P J(n)+ t− [2n + 1]P . − 1 0 1 73 (4n+2)b 2nb Let P ′ = t− P + P and P ′ = t− [2n + 1]P . Then P ′,P ′ are polynomials 1 − 1 0 0 1 1 0 1 in C[t± , M] and P1′J(n)+ P0′ = 0. This implies that P1′ = P0′ = 0 since the lowest degree in t of J(n) is 2bn2 +2b, which is quadratic in n, by Lemma 4.2.4. Hence − P =0.

2b 2 2 2b It is easy to see that ε(α )= M − (M M − )A where A =(L 1)(LM +1) 2,b − 2,b 2,b − is the A-polynomial of the (2, b)-torus knot. This means the AJ conjecture holds true

for the (2, b)-torus knot.

4.3 Proof of Theorem 8

As in the previous section, we consider the two cases: a,b > 2 and a = 2 separately.

4.3.1 The case a,b > 2

We claim that

Proposition 4.3.1. The colored Jones polynomial of the (a, b)-torus knot is annihilated by the operator P Q where

10ab 3 2ab 3 2ab 2(a b) 2(b a) 4ab 2 2ab 2 2ab P = t− (L M + L− M − ) (t − + t − )t− (L M + L− M − ) − 2ab 2ab 1 2ab 2ab 2ab 1 + t (LM + L− M − ) (t + t− )(L + L− ) − 2(a b) 2(b a) 4ab 4ab +(t − + t − )(t + t− ),

6ab 3 2ab 3 2ab 2(a+b) 2(a+b) ab 2 2ab 2 2ab Q = t− (L M + L− M − ) (t + t− )q− (L M + L− M − ) − 2ab 2ab 1 2ab 2ab 2ab 1 2(a+b) 2(a+b) + t− (LM + L− M − ) (t + t− )(L + L− )+2(t + t− ). −

Proof. We ﬁrst prove the following two lemmas.

Lemma 4.3.2. For the (a, b)-torus knot, we have

2abn 2abn 4ab 2 t λ(a b)(n+1) t− λ(a b)(n 1) QJ(n)= t − (λa+b λa b) − − − − . − − t2 t 2 − −

74 Proof. Let 2 2 2abn t λ(a+b)n t− λ(a b)n g(n)= t− − − . t2 t 2 − − 4ab(n+1) Then, by Lemma 4.2.1, J(n +2) = t− J(n)+ g(n + 1). Hence QJ(n) is equal to

6ab 4ab(n+3) 4ab(n 3) t− t J(n +3)+ t− − J(n 3) − 2(a+b) 2(a+b) 4ab 4ab(n+2) 4ab(n 2) (t + t− )t− t J(n +2)+ t− − J(n 2) − − 2ab 4ab(n+1) 4ab(n 1) + t− t J(n +1)+ t− − J(n 1) − 2ab 2ab 2(a+b) 2(a+b) (t + t− ) J(n +1)+ J(n 1) + 2(t + t− )J(n) − − 6ab 4ab 2ab(n+5) 2ab(n 5) = t− t J(n +1)+ J(n 1) + t g(n + 2) t− − g(n 2) − − − 2(a+b) 2(a+b) 4ab 4ab 2ab(n+4) 2ab(n 4) (t + t− )t− 2t J(n)+ t g(n + 1) t− − g(n 1) − − − 2ab 4ab 2ab(n+3) 2ab(n 3) + t− t (J(n 1) + J(n +1))+ t t− − g(n) − − 2ab 2ab 2(a+b) 2(a+b) (t + t− ) J(n +1)+ J(n 1) + 2(t + t− )J(n) − − 6ab 2ab(n+5) 2ab(n 5) = t− t g(n + 2) t− − g(n 2) − − 2(a+b) 2(a+b) 4ab 2ab(n+4) 2ab(n 4) t + t− t− t g(n + 1) t− − g(n 1) − − − 2ab 2ab(n+3) 2ab(n 3) + t− t t− − g(n). − Using the deﬁnition of g(n), we can rewrite

2 2 2 2 4ab 2abn t λ(a+b)(n+2) t− λ(a b)(n+2) 2abn t λ(a+b)(n 2) t− λ(a b)(n 2) QJ(n) = t t − − t− − − − − t2 t 2 − t2 t 2 − − − − 2(a+b) 2(a+b) 4ab (t + t− )t − × 2 2 2 2 2abn t λ(a+b)(n+1) t− λ(a b)(n+1) 2abn t λ(a+b)(n 1) t− λ(a b)(n 1) − − − − − t 2 − 2 t 2 − 2 t t− − t t− − 2 2 − 4ab 2abn 2abn t λ(a+b)n t− λ(a b)n + t (t t− ) − − . − t2 t 2 − −

Now applying the equality λk+l + λk l = λkλl, we then obtain − 2abn 2abn 4ab 2 t λ(a b)(n+1) t− λ(a b)(n 1) QJ(n)= t − (λa+b λa b) − − − − . − − t2 t 2 − − This proves Lemma 4.3.2.

75 2abn 2abn Let h(n)= t λ(a b)(n+1) t− λ(a b)(n 1). − − − − Lemma 4.3.3. The function h(n) is annihilated by the operator P , i.e. P h(n)=0.

Proof. Let c = a b. Then P h(n) is equal to −

10ab 4ab(n+3) 4ab(n 3) t− t h(n +3)+ t− − h(n 3) − 2(a b) 2(b a) 4ab 4ab(n+2) 4ab(n 2) (t − + t − )t− t h(n +2)+ t− − h(n 2) − − 2ab 4ab(n+1) 4ab(n 1) + t t h(n +1)+ t− − h(n 1) − 2ab 2ab 2(a b) 2(b a) 4ab 4ab (t + t− ) h(n +1)+ h(n 1) +(t − + t − )(t + t− )h(n) − − 2ab(3n+4) 2ab(n 2) 2ab(n+2) 2ab(3n 4) = t λc(n+4) t − λc(n+2) + t− λc(n 2) t− − λc(n 4) − − − − 2ab(3n+4) 2abn 2abn 2ab(3n 4) λc t λc(n+3) t λc(n+1) + t− λc(n 1) t− − λc(n 3) − − − − − 2ab(3n+4) 2ab(n+2) 2ab(n 2) 2ab(3n 4) + t λc(n+2) t λcn + t− − λcn t− − λc(n 2) − − − 2ab 2ab 2ab(n+1) 2ab(n+1) 2ab(n 1) 2ab(n 1) (t + t− ) t λc(n+2) t− λcn + t − λcn t− − λc(n 2) − − − − 4ab 4ab 2abn 2abn + λc(t + t− ) t λc(n+1) t− λc(n 1) . − −

Note that λk+l + λk l = λkλl. Hence P h(n) is equal to −

2ab(n 2) 2ab(n+2) 2abn 2abn t − λc(n+2) + t− λc(n 2) λc t λc(n+1) + t− λc(n 1) − − − − − 2ab(n+2) 2ab(n 2) + t λ + t− − λ − cn cn 2ab 2ab 2ab(n+1) 2ab(n+1) 2ab(n 1) 2ab(n 1) (t + t− ) t λc(n+2) t− λcn + t − λcn t− − λc(n 2) − − − − 4ab 4ab 2abn 2abn + λc(t + t− ) t λc(n+1) t− λc(n 1) − − 4ab 4ab 2abn 4ab 4ab 2abn = (t + t− + 1)t λc(n+2) +(t + t− + 1)t− λc(n 2) − − 4ab 4ab 2abn 2abn (t + t− + 1)(t t− )λ − − cn 4ab 4ab 2abn 2abn + λc(t + t− + 1) t λc(n+1) t− λc(n 1) − − 4ab 4ab 2abn = (t + t− + 1)t λ + λ λ λ − c(n+2) cn − c c(n+1) 4ab 4ab 2abn +(t + t− + 1)t− λc(n 2) + λcn λcλc(n 1) − − − = 0.

76 This proves Lemma 4.3.3.

Proposition 4.3.1 follows directly from Lemmas 4.3.2 and 4.3.3.

4.3.2 The case a =2

We claim that

Proposition 4.3.4. The colored Jones polynomial of the (2, b)-torus knot is annihilated by the operator

4b 2 2b 2 2b 2b 2b 1 R = t− (L M + L− M − )+(t + t− )(L + L− )

4 4 2b 2b 1 2b 2b 2b 4 4 (t + t− )t− (LM + L− M − )+(M + M − ) 2(t + t− ). − −

Proof. From Lemma 4.2.5 we have

(4n+2)b 2nb J(n +1) = t− J(n)+ t− [2n + 1], − 8(n+1)b 6(n+1)b 2(n+1)b J(n +2) = t− J(n) t− [2n +1]+ t− [2n + 3], − (4n 2)b 2nb J(n 1) = t − J(n)+ t [2n 1], − − − 8(n 1)b 6(n 1)b 2(n 1)b J(n 2) = t − J(n) t − [2n 1] + t − [2n 3]. − − − −

77 Hence RJ(n) is equal to

4b 4(n+2)b 4(n 2)b 2b 2b t− t J(n +2)+ t− − J(n 2) +(t + t− ) J(n +1)+ J(n 1) − − 4 4 2b 4(n+1)b 4(n 1)b (t + t− )t− t J(n +1)+ t− − J(n 1) − − 4nb 4nb 4 + (t + t− ) 2(t + t− ) J(n) − 4b 2(n 1)b 2(n+3)b 2(n+1)b 2(n 3)b = t− t− − [2n +1]+ t [2n + 3] t [2n 1] + t− − [2n 3] − − − − 2b 2b 2nb 2nb +(t + t− ) t− [2n +1]+ t [2n 1] − 4 4 2b (2n+4)b ( 2n+4)b (t + t− )t− t [2n +1]+ t − [2n 1] − − 4 2b 2nb 2nb (t + t− )t t [2n +1]+ t− [2n 1] − − 2b 2nb 4 4 = t t [2n + 3] + [2n 1] (t + t− )[2n + 1] − − 2b 2nb 4 4 + t t− [2n 3] + [2n + 1] (t + t− )[2n 1] − − − = 0,

2l 2l since [k + l] + [k l]=(t + t− )[k]. − 4.3.3 Proof of Theorem 8

We ﬁrst note that the A-ideal p, the kernel of θ : tσ C[χ(X)], is radical i.e. −→ √p = p. This is because the character ring C[χ(X)] is reduced, i.e. has nil-radical 0, by deﬁnition.

Lemma 4.3.5. Suppose δ(t, M, L) . Then there are polynomials g(t, M) ∈ AK ∈ 1 C[t± , M] and γ(t, M, L) such that ∈ T 1 δ(t, M, L)= γ(t, M, L) α (t, M, L). (25) g(t, M) K Moreover, g(t, M) and γ(t, M, L) can be chosen so that εg =0. Proof. By deﬁnition α is a generator of , the extension of in the principal K AK AK left-ideal domain . Since δ , it is divisible by α in . Hence (25) follows. T ∈AK K T We can assume that t + 1 does not divide both g(t, M) and γ(t, M, L) simultane- ously. If ε(g) = 0 then g is divisible by t + 1, and hence γ is not. But then from the

78 equality gδ = γαK , it follows that αK is divisible by t + 1, which is impossible, since

all the coeﬃcients of powers of L in αK are supposed to be co-prime.

Showing ε( σ) p. For torus knots, by Section 4.2, we have ε(α )= f(M)A , A ⊂ K K 1 where f(M) C[M ± ]. For every δ , by Lemma 3.2.7, there exist g(t, M) ∈ ∈ AK ∈ 1 1 C[t± , M] and γ such that δ = γα and εg = 0. It implies that ∈ T g(t,M) K 1 1 ε(γ)= ε(γ) ε(α )= ε(γ) f(M)A . (26) εg(M) K εg(M) K

The A-polynomial of a torus knot does not contain any non-trivial factor de-

1 1 pending on M only. Since ε(γ) t = C[L± , M ± ], equation (26) implies that ∈ h := 1 ε(γ) f(M) is an element of t. Hence ε(γ) A t, the ideal of t generated εg(M) ∈ K by A . It follows that ε( ) A t and thus ε( σ) (A t)σ = p. Hence K AK ⊂ K A ⊂ K ε( σ) √p = p. A ⊂ Showing p ε( σ). For a,b > 2, by Proposition 4.3.1 the colored Jones poly- ⊂ A nomial of the (a, b)-torus knot is annihilated by the operator P Q. Note that

1 2 2 2ab 2 2ab 2 ε(P Q) = (L + L− 2) (L M + L− M − 2) − − 2 1 ab 2 2ab 4 = L− L− M − (L 1)(L M 1) . − − 1 ab 2 2ab 1 ab If u p then u = vA′ , where A′ = L− M − (L 1)(L M 1) = L− M − A ∈ a,b a,b − − a,b 1 1 and v C[M ± , L± ]. It is easy to see that σ(v)= Lv since σ(u)= u and σ(A′ )= ∈ a,b 1 2 2 1 2 L− Aa,b′ . This implies that σ(v L)= σ(v) L− = v L. We then have

′ u4 = v4A 4 = ε(v4L2P Q) ε( σ), a,b ∈ A hence u ε( σ). ∈ A For a = 2, by Proposition 4.3.4 the colored Jones polynomial of the (2, b)-torus

knot is annihilated by the operator R. Note that σ(R)= R and

1 2b 1 2b ε(R) = (L + L− 2)(LM + L− M − + 2) − 1 b 2b 2 = L− M − (L 1)(LM + 1) . − 79 1 b 2b 1 b If u p then u = vA′ , where A′ = L− M − (L 1)(LM +1)= L− M − A and ∈ 2,b 2,b − 2,b 1 1 2 2 v C[M ± , L± ]. It is easy to see that σ(v)= v and hence σ(v )= σ(v)σ(v)= v . ∈ − We then have

′ u2 = v2A 2 = ε(v2R) ε( σ), 2,b ∈ A hence u ε( σ). ∈ A In both cases p ε( σ). Hence ε( σ)= p for all torus knots. ⊂ A A

80 CHAPTER V

FUTURE PLANS

There are several problems we want to study in the future:

Problem 1. Study the volume conjecture for all cables of a knot.

The colored Jones polynomials of cables of a knot are available in [59], so we hope to apply the method in Chapter 1 to study the volume conjecture for them, especially to extend the results of Theorems 1, 2, 3 and 4 to all cables of a knot.

Problem 2. Study the AJ conjecture for those knots whose character varieties have at least 3 irreducible components.

All the knots for which the AJ conjecture has been so far conﬁrmed have character varieties consisting of 2 irreducible components. Hence we want to investigate the conjecture for those knots whose character varieties have 3 or more irreducible components. One way to study this is to understand the action of the skein algebra of the boundary torus on the skein module of the knot complement.

Problem 3. Study the AJ conjecture for links.

By the result of Chapter 3, for a two-bridge link when reducing t = 1, the skein − module is isomorphic to the ring of regular functions on the character variety. This is the starting point for considering the AJ conjecture for links. In a joint work in progress with T. Le, we have formulated the AJ conjecture for links and conﬁrmed it for (2, 2m)-torus links. The next step we plan to do is to extend Theorems 5, 6 and

2.5.6 to the case of hyperbolic links, in particular to some classes of two-bridge links and pretzel links.

81 Problem 4. Study the σ = conjecture. A P

Although the recurrence ideal is deﬁned algebraically, it has a beautiful geo- AK metric meaning relating to skein modules. Let N(K) be a tubular neighborhood and

X be the complement of a knot K. There is a bilinear pairing

3 1 (N(K)) (X) (S )= C[t± ] S ⊗ S → S deﬁned by l l′ = l,l′ := l l′ , where l and l′ are framed links in N(K) and X ⊗ ∪ respectively. The inclusion ∂X ֒ X induces a map Θ : (∂X) (X). The kernel → S → S of Θ is the non-commutative A-ideal of K. The orthogonal ideal of K is P O

= l′ (∂X) l, Θ(l′) = 0 for all l (N(K)) . O { ∈ S | ∈ S }

It is known that = σ, the σ-invariant part of under the involution O A AK a b a b σ(M L ) = M − L− , see [21, 16] and Chapter 2. It implies that is contained in P σ = . If = σ for all knots then according to [36] the colored Jones polynomial A O P A distinguishes the unknot from other knots.

Problem 5. Study the stronger version of the AJ conjecture.

In the future we hope to conﬁrm the conjecture for all the knots for which the AJ conjecture has been established.

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86 VITA

Contact Information

Anh T. Tran

School of Mathematics, Georgia Institute of Technology

Oﬃce 117, Skiles building, 686 Cherry ST, Atlanta, GA 30332-0160

Email: [email protected]

Education

2006–2012: PhD in Mathematics, Georgia Institute of Technology, Atlanta GA, USA.

Advisor: Dr. Thang T.Q. Le

2000–2004: BSc in Mathematics and Computer Science, Vietnam National University at Hochiminh city, Vietnam. Advisors: Dr. Duong Minh Duc and Dr. Dang Duc Trong.

Publications and Preprints

1. (with T. Le) On the volume conjecture for cables of knots, J. Knot Theory Rami-

ﬁcations 19 (2010), no. 12, 1673–1691, arXiv:0907.0172. 2. (with T. Le) On the AJ conjecture for knots, arXiv:1111.5258.

3. (with T. Le) The skein module of two-bridge links, to appear in Proceedings of the

American Mathematical Society, arXiv:1111.0332.

4. Proof of a stronger version of the AJ conjecture for torus knots, arXiv:1111.5065.