TTHE AJ CONJECTURE FOR CABLES OF TWO-BRIDGE KNOTS
NATHAN DRUIVENGA
Abstract.
The AJ -conjecture for a knot K ⊂ S relates the A -polynomial and thecolored Jones polynomial of K . If a two-bridge knot K satisfies the AJ -conjecture,we give sufficient conditions on K for the ( r, C to also satisfy the AJ -conjecture. If a reduced alternating diagram of K has η + positive crossings and η − negative crossings, then C will satisfy the AJ -conjecture when ( r +4 η − )( r − η + ) > Introduction
The AJ -conjecture [3] is a proposed relationship between two different invariantsof a knot K in S , the A -polynomial and the colored Jones polynomial. The A -polynomial is determined by the fundamental group of the knot complement, whilethe colored Jones polynomial is a sequence of Laurent polynomials J K ( n ) ∈ Z [ t, t − ]that have no apparent connection to classical knot invariants. The relationship positedby the AJ -conjecture allows us to extract information about one invariant from theother. For example, if the AJ -conjecture is true, then the fact that the A -polynomialrecognizes the unknot implies the colored Jones function does as well.The AJ -conjecture has been verified for the trefoil and figure eight knots [3], alltorus knots [6, 15], some classes of two-bridge knots and pretzel knots [7, 8], theknot 7 [4], and most cable knots over torus knots, the figure eight knot, and twistknots [10, 11, 12, 13]. Early methods of proving the AJ -conjecture relied on explicitformulas for each invariant. Recently, methods have been developed in [12] to verifythe conjecture for cables of knots even when explicit formulas are unknown. In thecurrent paper, a closed formula for the A -polynomial is needed.The main result of this paper is an extension of work done by Dennis Ruppe andAnh Tran [10, 12, 13]. Theorem 4.1
Let K = b ( p, m ) be a two-bridge knot with a reduced alternatingdiagram D that has η + positive crossings and η − negative crossings. Let A K ( M, L ) bethe A -polynomial of K and let C be the ( r, -cable of K . Assume the following:i. K satisfies the AJ -conjecture with an associated degree p annihilator α K ( t, M, L ) ii. A K ( M,L ) L − is irreducible and A K ( M,L ) L − (cid:54) = A K ( M, − L ) L − .iii. The matrix N ( − , M ) described in Section 3 has nonzero determinant.Then for odd integers r with ( r +4 η − )( r − η + ) > the cable knot, C , satisfies the AJ -conjecture. a r X i v : . [ m a t h . G T ] M a r NATHAN DRUIVENGA
The need for condition iii will be apparent in Section 3 as it leads to a solution ofa system of equations.In Section 2, we review each invariant, state the AJ -conjecture, and provide otherpreliminary material. In Section 3, we establish an important lemma needed to provethe main result in Section 4. Applications of the main result are given in section 5.The author would like to Anh Tran for his substantial influence on the completionof this paper. Gratitude is also extended to Cody Armond, Ben Cooper and StevenLandsburg for helpful conversations.2. Preliminaries
The A-Polynomial.
The A -polynomial is the defining polynomial of an alge-braic curve in C ∗ × C ∗ where C ∗ are the nonzero complex numbers [1]. Let K ⊂ S be a knot and M be the complement of a regular neighborhood of K . Then M isa compact manifold with boundary homeomorphic to a torus, ∂ M = T . Denote byRep( T ) ⊂ (cid:81) ni =1 SL ( C ) the space of all representations of π ( T ) into SL ( C ).The fundamental group π ( T ) is a free abelian group with two generators. Let { µ, λ } be the standard basis for π ( T ). Consider the subset Rep ∆ ( π ( M )) of Rep( π ( M ))consisting of upper triangular SL ( C ) representations. Set ρ ( µ ) = (cid:18) m (cid:63) m − (cid:19) and ρ ( λ ) = (cid:18) l (cid:63) l − (cid:19) and let (cid:15) : Rep ∆ ( π ( M )) → C ∗ × C ∗ be the eigenvalue map defined by (cid:15) ( ρ ) = ( m, l ).Let Z be the Zariski closure of (cid:15) (Rep ∆ ( π ( M ))) in C ∗ × C ∗ . Each of the components of Z are one dimensional [2]. The components are hyper-surfaces and can be cut out bya single polynomial unique up to multiplication by a constant. The A -polynomial , A K ( m, l ), is the product of all such defining polynomials. The A -polynomial can betaken to have relatively prime integer coefficients and is well defined up to a unit. Theabelian component of Z will have defining polynomial l − A -polynomialcan be factored as A K ( m, l ) = ( l − A (cid:48) K ( m, l ) [1].2.2. The Colored Jones Function.
Let L be a link in S . The colored Jonesfunction is a quantum link invariant that assigns to each n ∈ N a Laurent polynomial J L ( n ) ∈ Z [ t ± ]. Like the classic Jones polynomial, J L ( n ) can be defined using theKauffman bracket. For a complete definition of the colored Jones function, see forexample [7].2.3. The AJ Conjecture.
To set up the AJ conjecture, we follow [3] and use no-tation from [7]. Given a discrete function f : N → C [ t ± ], define operators M and L acting on f by M ( f )( n ) = t n f ( n ) and L ( f )( n ) = f ( n + 1) HE AJ CONJECTURE FOR CABLES OF TWO-BRIDGE KNOTS 3
It can be seen that these operators satisfy LM = t M L . Let T = C [ t ± ] (cid:10) M ± , L ± (cid:11) / ( LM − t M L )This non-commutative ring, T , is called the quantum torus.If there exists a P ∈ T such that P ( f ) = 0, then P is called a recurrence relation for f . The recurrence ideal of a discrete function f is a left ideal A f of T consistingof recurrence relations for f . A f = { P ∈ T | P ( f ) = 0 } When A f (cid:54) = { } , the discrete function f is said to be q-holonomic . Also, denoteby A K the recurrence ideal of J K ( n ).The ring T is not a principal ideal domain, so a non-trivial recurrence ideal is notguaranteed to have a single generator. Fix this by localizing at Laurent polynomialsin M [3]. The process of localization works since T satisfies the Ore condition. Theresulting ring, (cid:101) T , has a Euclidean algorithm and is therefore a principal ideal domain.Let C [ t ± ]( M ) be the fraction field of the polynomial ring C [ t ± ][ M ]. Then, (cid:101) T = (cid:40)(cid:88) i ∈ Z a i ( M ) L i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a i ( M ) ∈ C [ t ± ]( M ) , a i ( M ) = 0 for almost every i (cid:41) . Extend the recurrence ideal A K to an ideal (cid:102) A K of (cid:101) T by (cid:102) A f = (cid:101) T A K which can bedone since T embeds as a subring of (cid:101) T . The extended ideal will then have a singlegenerator (cid:102) α K ( t, M, L ) = n (cid:88) i =0 α i ( M ) L i This generator has only positive powers of M and L , and the degree of L is assumedto be minimal since it generates the ideal. The coefficients α i ( M ) can be assumedto be co-prime and α i ( M ) ∈ Z [ t ± , M ]. The operator (cid:102) α K is called the recurrencepolynomial of K and is defined up to a factor of ± t j M k for j, k ∈ Z . Garoufalidisand Le showed that for every knot K , J K ( n ) satisfies a nontrivial recurrence relation[5].Let f, g ∈ C [ M, L ]. If the quotient f /g does not depend on L , then f and g aresaid to be M -essentially equivalent , denoted f M = g . In other words, the functions f and g are equal up to a factor only depending on M . Let (cid:15) be the map were thesubstitution t = − The AJ Conjecture [3]: If A K ( M, L ) is the A -polynomial of a knot K and (cid:102) α K ( t, M, L ) is as above, then (cid:15) ( (cid:102) α K ) M = A K ( M, L ) NATHAN DRUIVENGA
Cabling Formulas and the Resultant.
Let r, s ∈ Z with greatest commondivisor d . The ( r, s )-cable, C , of a zero framed knot K is the link formed by taking d parallel copies of the ( rd , sd ) curve on the torus boundary of a tubular neighborhoodof K . Homologically, one can think of this curve as rd times the meridian and sd timesthe longitude on the torus boundary. Notice that if r and s are relatively prime, C isa knot.Let A K ( M, L ) be the A -polynomial of a knot K in S and let C be the ( r, K . Assume that A (cid:48) K ( M, L ) is irreducible. Write A C ( M, L ) in terms of A K ( M, L ) using the following cabling formula given by Ni and Zhang, c.f. [10]. Let F r ( M, L ) = (cid:26) M r L + 1 if r > L + M − r if r < A C ( M, L ) = ( L − F r ( M, L )Res λ (cid:0) A (cid:48) K ( M , λ ) , λ − L (cid:1) where Res λ is the resultant defined below. Use the fact that the product of theresultants is the resultant of the product and Res λ ( λ − , λ −
1) = L − A C ( M, L ) = F r ( M, L )Res λ (cid:0) A K ( M , λ ) , λ − L (cid:1) Let F be a field. Let f ( x ) = (cid:80) ni =0 f i x i and g ( x ) = (cid:80) mi =0 g i x i be polynomials in F [ x ]. The resultant of f and g is the determinant of the following ( m + n ) × ( m + n )matrix. Res( f, g ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) f . . . g . . . f f g g f f . . . 0 g g . . . 0... ... . . . ... ... ... . . . ... f n f n − . . . f g k ... . . . g f n . . . f g k +1 g k . . . g ... ... . . . ... ... ... . . . ...0 . . . f n f n − . . . g m g m − . . . f n . . . g m (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) The coefficients of f ( x ) constitute the first m columns while the coefficients of g ( x )are in the last n columns.There is also a formula relating the colored Jones polynomial of the cable to asubsequence of the colored Jones polynomial of K . Since we only consider the ( r, C , the cabling formula given in [11] simplifies as(2.2) ( M r L + t − r M − r ) J C ( n ) = J K (2 n + 1) . HE AJ CONJECTURE FOR CABLES OF TWO-BRIDGE KNOTS 5
In the next section, we seek a homogeneous annihilator (cid:101) β ( t, M, L ) of the odd sub-sequence J K (2 n + 1). With this in hand, the above cabling formula implies that (cid:101) β ( t, M, L )( M r L + t − r M − r ) annihilates J C ( n ).3. The Annihilator
Lemma 3.1.
Let (cid:101) α K ( t, M, L ) = (cid:80) di =0 P i ( t, M ) L i be a minimal degree homogeneousrecurrence polynomial for J K ( n ) of degree d ≥ such that (cid:101) α K ( − , M, L ) M = A K ( M, L ) .Also assume deg L ( (cid:101) α K ) = deg l ( A K ) . If the matrix N defined later has nonzero deter-minant at t = − , then J K (2 n + 1) has the homogeneous recurrence polynomial givenby (cid:101) β ( t, M, L ) = d (cid:88) i =0 ( − i Q i ( t, M ) L i with Q i ( t, M ) = − det(A i+1 ) where the A i +1 are matrices to be defined below. Proof : The process is similar to the case of the figure eight knot [10], except herea homogeneous annihilator is considered. Since (cid:101) α K is a homogeneous annihilator, (cid:101) α K ( t, M, L ) J K ( n ) = 0. Since M acts on discrete functions as multiplication by t n ,change M to t n yielding (cid:32) d (cid:88) i =0 P i ( t, t n ) L i (cid:33) J K ( n ) = d (cid:88) i =0 P i ( t, t n ) J K ( n + i ) = 0For each 0 ≤ j ≤ d , substitute 2 n + j + 1 for n yielding d + 1 equations of the form d (cid:88) i =0 P i ( t, t n + j +1) ) J K (2 n + j + 1 + i ) = 0A degree d homogeneous recurrence relation of J K (2 n + 1) has the form (cid:32) d (cid:88) i =0 Q i ( t, t n ) L i (cid:33) J K (2 n + 1) = d (cid:88) i =0 Q i ( t, t n ) J K (2( n + i ) + 1) = 0To construct (cid:101) β solve(3.1) d (cid:88) j =0 c j d (cid:88) i =0 P i ( t, t n +2 j +2 ) J K (2 n + i + j + 1) − d (cid:88) i =0 Q i ( t, t n ) J K (2 n + 2 i + 1) = 0Setting the coefficients of each J K ( m ) equal to zero, where m satisfies 2 n + 1 ≤ m ≤ n + 2 d + 1, gives 2 d + 1 equations in the 2 d + 2 unknowns c , c , ..., c d , Q , Q , ...Q d . NATHAN DRUIVENGA
Form a (2 d + 1) × (2 d + 2) matrix, C , where the columns are labeled by the unknownsin the order given above. In the following matrix, P i ( r ) is used as shorthand for P i ( t, t n + r ). C = P (2) 0 0 ··· ··· − ··· P (2) P (4) 0 ··· ··· ··· P (2) P (4) P (6) ··· ··· − ··· P (2) P (4) P (6) ... ··· ··· P (2) P (4) P (6) ... ··· − ··· P (2) P (4) P (6) ... ··· ··· ... ... ... ... ... ... ... ... ... ... ... ... ... ... P d − (2) P d − (4) P d − (6) ··· ··· ··· P d (2) P d − (4) P d − (6) ··· P (2 k +2) 0 0 0 0 0 0 ··· P d (4) P d − (6) ··· P (2 k +2) ... ··· P d (6) ··· P (2 k +2) ... ··· ··· P (2 k +2) ... ··· ... ... ... ··· ... ... ... ... ... ... ... ... ... ... ··· P d − (2 k +2) ... P k (2 d ) P k − (2 d +2) 0 0 0 0 ··· ··· P d (2 k +2) ... P k +1 (2 d ) P k (2 d +2) 0 0 0 0 ··· ··· ... P k +2 (2 d ) P k +1 (2 d +2) 0 0 0 0 ··· ... ... ... ... ... ... ... ... ... ... − ··· ··· P d (2 d ) P d − (2 d +2) 0 0 0 0 ··· ··· ··· P d (2 d +2) 0 0 0 0 ··· − If < c , ..., c d , Q , ..., Q d > T = (cid:126)X ∈ R d +2 is the vector of unknowns, then (3.1) isequivalent to C (cid:126)X = (cid:126)
0. Because C is not a square matrix, Cramer’s rule can not beapplied. However, a slight modification can be made to get a square matrix.Let D be the (2 d + 1) × (2 d + 1) matrix obtained from C by removing the lastcolumn. Let (cid:126)Y ∈ R d +1 be the vector obtained from (cid:126)X by removing the last entry Q d .Finally, let (cid:126)b = < , , ..., , Q d > T ∈ R d +1 . It can be seen that the equation C (cid:126)X = (cid:126) D (cid:126)Y = (cid:126)b .Assume det( D ) (cid:54) = 0, then by Cramer’s rule c k = det( D k +1 )det( D ) and Q k = det( D d + k +2 )det( D )where D j is the same as the matrix D except the j -th column is replaced by (cid:126)b . Atthis point it is convenient to make the choice Q d ( t, M ) = det( D ). In order to verifythe AJ-conjecture later, it will be helpful to simplify these determinants. Let us first HE AJ CONJECTURE FOR CABLES OF TWO-BRIDGE KNOTS 7 examine det( D ). The last d columns of D have only one nonzero entry(that is, − d columns yields,(3.2) det( D ) = det( X )where X is a ( d + 1) × ( d + 1) matrix made up of the first d + 1 columns of D withthe odd rows removed except the 2 d + 1 row. There is no factor of − − X ) = P d ( t, t n +2 d ) P d ( t, t n +2 d +2 )det( N )where N is the upper left ( d − × ( d −
1) block sub-matrix of X . It is the matrix N that will be important in verifying the AJ-conjecture. Example : Let d = 3. Then we have D = P (2) 0 0 0 − P (2) P (4) 0 0 0 0 0 P (2) P (4) P (6) 0 0 − P (2) P (4) P (6) P (8) 0 0 00 P (4) P (6) P (8) 0 0 −
10 0 P (6) P (8) 0 0 00 0 0 P (8) 0 0 0 N = (cid:18) P (2) P (4) P (2) P (4) (cid:19) Now simplify the determinants corresponding to the c j for 0 ≤ j ≤ d . For example, c = det( D )det( D ) . Simplify det( D ) by expanding along the last d columns to get a factorof 1 and then expand along the first column. Recall that the first column of D is thevector (cid:126)b with only one nonzero entry, Q d , in the last row. Therefore,det( D ) = ( − d Q d ( t, M ) · det( B )where B is the ( d + 1 , X . In general,det( D j ) = ( − d + j − Q d ( t, M ) · det( B j )where B j is the ( d + 1 , j ) cofactor of X and 1 ≤ j ≤ d + 1. This gives c j = ( − d + j det( B j +1 ) for 0 ≤ j ≤ d. Finally, simplify the determinants corresponding to the Q j for 0 ≤ j ≤ d − D d +2 ) simplifies by expanding along the last d columns, the first ofwhich is the vector (cid:126)b . This givesdet( D d +2 ) = − Q d · det( A ) NATHAN DRUIVENGA where A is a ( d + 1) × ( d + 1) sub-matrix of D . The cofactor expansion implies that A is formed from the first d + 1 columns of D while removing all the odd rows exceptthe first. Similarly, det( D d + j +2 ) = − Q d · det( A j )where A j is a ( d + 1) × ( d + 1) sub-matrix of D . Form A j from the first d + 1 columnsof D while removing all the odd rows except the 2 j − Example : In the case d = 3, A = P (2) 0 0 0 P (2) P (4) 0 0 P (2) P (4) P (6) P (8)0 0 P (6) P (8) B = P (4) 0 0 P (4) P (6) P (8)0 P (6) P (8) A = P (2) P (4) 0 0 P (2) P (4) P (6) 0 P (2) P (4) P (6) P (8)0 0 P (6) P (8) B = P (2) 0 0 P (2) P (6) P (8)0 P (6) P (8) A = P (2) P (4) 0 0 P (2) P (4) P (6) P (8)0 P (4) P (6) P (8)0 0 P (6) P (8) B = P (2) P (4) 0 P (2) P (4) P (8)0 0 P (8) To complete the proof of the lemma, it remains to show that (cid:101) β ( t, M, L ) = d (cid:88) i =0 ( − i Q i ( t, M ) L i defines a nontrivial operator which can be accomplished with the following claim. Claim : det( B j +1 ( − , M )) = P j ( − , M ) · det( N ( − , M )) where N is the ( d − × ( d −
1) matrix defined above.
Proof of Claim : Break the claim into two cases; d is even or d is odd. The casewhere d is odd is shown, the even case is analogous. All calculations are done at t = − P k is shorthand for P k ( − , M ). HE AJ CONJECTURE FOR CABLES OF TWO-BRIDGE KNOTS 9
When d is odd, the matrix N is formed from the two vectors (cid:126)v = P P ... P d − P d and (cid:126)w = P P ... P d − P d − where (cid:126)v, (cid:126)w ∈ R d − . For example, when d = 7 (cid:126)v = P P P P and (cid:126)w = P P P P Let S be the ( d − × ( d −
1) permutation matrix that shifts each vector entry bythe permutation (1 2 3 4 . . . d − N is the ( d − × ( d − N = (cid:16) (cid:126)v (cid:126)w (cid:126)Sv (cid:126)Sw . . . (cid:126)S d − v (cid:126)S d − w (cid:17) . Let i : R d − → R d be inclusion. Then i ( (cid:126)v ) is almost the same as (cid:126)v except thereis an extra zero in the last row of i ( (cid:126)v ). Let T be the permutation matrix that shiftseach vector entry by the permutation (1 2 3 4 . . . d ). Form any of the matrices B j bychoosing d vectors (all but the j -th vector) from the following set of d + 1 vectors; { (cid:126)i ( v ) , (cid:126)i ( w ) , (cid:126)T i ( v ) , (cid:126)T i ( w ) , · · · , (cid:126)T d − i ( v ) , (cid:126)T d − i ( w ) } For example, B d +1 = (cid:16) (cid:126)i ( v ) (cid:126)i ( w ) (cid:126)T i ( v ) (cid:126)T i ( w ) . . . (cid:126)T d − i ( v ) (cid:126)T d − i ( w ) (cid:126)T d − i ( v ) (cid:17) . Notice that det( B d +1 ) = P d · det( N ). This is clear since B d +1 is a block matrix. Toshow the claim for the rest of the B j , consider the equation B d +1 · (cid:126)X = (cid:126)T d − i ( w ) This equation has the solution (cid:126)X = 1 P d P − P ... − P d − P d − The fact that (cid:126)X is the solution to this equation follows from the successive shiftingof the vectors (cid:126)i ( v ) and (cid:126)i ( w ). Notice that because (cid:102) α K was a minimal degree annihila-tor, P , P d (cid:54) = 0, implying (cid:126)X is defined and nonzero.By Cramer’s rule, P i P d = det( B d +1 ) i +1 det( B d +1 )where ( B d +1 ) i +1 is the matrix B d +1 with the i + 1 column replaced by T d − (cid:126)i ( w ). Notethat placing the vector T d − (cid:126)i ( w ) in the i + 1 column and then exchanging appropriatecolumns cancels the negative signs from the solution (cid:126)X . After this replacement andshifting of columns, the matrix B i +1 is obtained. Therefore, P i P d = det( B i +1 )det( B d +1 ) = det( B i +1 ) P d · det( N )Conclude that det( B i +1 ) = P i · det( N ) for 0 ≤ i ≤ d which proves the claim. (cid:3) For general d , an important observation isdet( A i ) = (cid:88) j + k =2 i − ( − k + i − P k (2 j ) · det( B j )where 0 ≤ k ≤ d and 1 ≤ j ≤ d + 1. This can be calculated directly by expandingalong the extra row included in A i that B i does not contain.The proof of Lemma 3.1 can be completed by combining the above observation andthe previous claim. − Q i ( − , M ) = det( A i +1 ) = (cid:88) j + k =2 i +1 ( − k + i P k · P j − · det( N )where 0 ≤ k ≤ d and 1 ≤ j ≤ d + 1. Now shift the j indexing by 1 to getdet( A i +1 ) = (cid:88) j + k =2 i ( − k + i P k · P j · det( N ) HE AJ CONJECTURE FOR CABLES OF TWO-BRIDGE KNOTS 11 where 0 ≤ k ≤ d and 0 ≤ j ≤ d . But then,(3.3) ( − i +1 Q i ( − , M ) = ( − i (cid:88) j + k =2 i ( − k P k · P j · det( N ) . The fact that (cid:101) α K ( t, M, L ) = (cid:80) di =0 P i ( t, M ) L i is a minimal degree annihilator and deg L ( (cid:101) α K ) = deg l ( A K ) implies that P ( − , M ) and P d ( − , M ) are nonzero. Since thecondition det( N ( − , M )) (cid:54) = 0 has been assumed, − Q ( − , M ) = (cid:88) j + k =0 ( − k P k ( − , M ) · P j ( − , M ) · det( N ( − , M ))= P ( − , M ) · P ( − , M ) · det( N ( − , M )) (cid:54) = 0and − Q d ( − , M ) = (cid:88) j + k =2 d ( − k P k ( − , M ) · P j ( − , M ) · det( N ( − , M ))= P d ( − , M ) · P d ( − , M ) · det( N ( − , M )) (cid:54) = 0 . Therefore, the operator (cid:101) β ( t, M, L ) = (cid:80) di =0 ( − i Q i ( t, M ) L i is a nontrivial degree d annihilator of the odd sequence J K (2 n + 1) completing the proof of Lemma 3.1. (cid:3) Until now, is has been assumed that det( N ) (cid:54) = 0 when evaluated at t = −
1. Itwould be nice if this determinant was nonzero in general. Generically though, thereare cases where it is clearly zero. For example, assume that a knot K has associatedannihilator (cid:102) α K ( t, M, L ) = r (cid:88) i =0 P i ( t, M ) · L i for some positive integer r . This annihilator has only even power of L and thus eachcolumn of N corresponding to the odd powers of L is the zero column. Then clearlydet( N ) = 0. Therefore, knots where A K ( M, L ) has at least one nonzero odd degree L coefficient should be considered to apply Lemma 3.1. Remark : For twist knots K m , Anh Tran uses skein theory to show that a variantof N has a nonzero determinant, [12]. In fact, he shows that under a suitable basis,the columns of N are linearly independent. It should be noted that other authorshave encountered this matrix obstruction while proving the AJ -conjecture for cablesof knots. 4. Verifying the AJ conjecture
Let K be a knot that satisfies the AJ -conjecture with homogeneous annihilator (cid:102) α K ( t, M, L ) such that deg L ( (cid:101) α K ) = deg l ( A K ). Let C be the ( r,
2) cable knot of K .In the previous section, we found the annihilator (cid:101) β of the colored Jones function J K (2 n + 1). From the cabling formula (2.2) for the colored Jones function, we havean annihilator of J C ( n ) given by (cid:101) β ( t, M, L )( M r L + t − r M − r )Recall (2.1) the A -polynomial of the ( r, A C ( M, L ) = F r ( M, L )Res λ (cid:0) A K ( M , λ ) , λ − L (cid:1) where F r ( M, L ) = M r L + 1 if r > L + M − r if r < AJ -conjecture it must be shown that (cid:101) β ( − , M, L ) = R ( M )Res λ (cid:0) A K ( M , λ ) , λ − L (cid:1) where R ( M ) is some function of M . Lemma 4.1.
Let P ( L, M ) = (cid:80) di =0 P i ( M ) L i be a degree d polynomial. Then,Res λ ( P ( M , λ ) , λ − L ) = d (cid:88) i =0 (cid:32) (cid:88) k + j =2 i ( − k P k ( M ) P j ( M ) (cid:33) L i Proof : The proof uses induction on the L degree of P ( L, M ) and the fact thatRes λ ( P ( M , λ ) , λ − L ) = P ( M , √ L ) · P ( M , −√ L ) . When d = 2, calculate the determinant of the resultant matrix to getRes λ ( P ( M , λ ) , λ − L ) = P ( M ) L + (2 P ( M ) P ( M ) − P ( M )) L + P ( M ) . Now assume that the statement holds for degree less than d and let P ( M, L ) be adegree d polynomial in L . ThenRes λ ( P ( M , λ ) , λ − L ) = P ( M , √ L ) · P ( M , −√ L )= ( − d P d L d + P d √ L d (cid:32) d − (cid:88) i =0 ( − d P i √ L i + P i ( −√ L ) i (cid:33) + d − (cid:88) i =0 P i √ L i · d − (cid:88) i =0 P i √− L i = ( − d P d L d + 2( − d P d √ L d d − (cid:88) i =0 i ≡ d mod 2 P i √ L i + Res λ ( d − (cid:88) i =0 P i λ i , λ − L )= ( − d P d L d + 2 d − (cid:88) i =0 i ≡ d mod 2 ( − d P d P i √ L i + d + d − (cid:88) i =0 (cid:32) (cid:88) k + j =2 i ( − k P k P j (cid:33) L i = d (cid:88) i =0 (cid:32) (cid:88) k + j =2 i ( − k P k P j (cid:33) L i (cid:3) HE AJ CONJECTURE FOR CABLES OF TWO-BRIDGE KNOTS 13
Now use the set up from the previous section to finish verifying the AJ -conjecture.Recall (3.3), ( − i +1 Q i ( − , M ) = ( − i (cid:88) j + k =2 i ( − k P k · P j · det( N ) . Therefore, by Lemma 4.1, (cid:101) β ( − , M, L ) = d (cid:88) i =0 ( − i Q i ( t, M ) L i = − d (cid:88) i =0 ( − i (cid:88) j + k =2 i ( − k P k · P j · det( N ) · L i = − det( N ) d (cid:88) i =0 (cid:88) j + k =2 i ( − k P k · P j · L i = − det( N ) · Res λ ( (cid:102) α K ( − , M , λ ) , λ − L )= − det( N ) · Res λ (cid:0) R ( M ) · A K ( M , λ ) , λ − L (cid:1) = − det( N ) · R ( M ) · Res λ (cid:0) A K ( M , λ ) , λ − L (cid:1) The R ( M ) in the last two expressions arises because K satisfies the AJ -conjecture.Therefore, it has been shown that this annihilator of the cable knot C evaluated at t = − M -essentially equivalent to the A -polynomial A C ( L, M ) when det( N ) (cid:54) = 0.The only thing left to show is that there does not exist an annihilator of lower L degree than (cid:101) β ( t, M, L )( M r L + t − r M − r ). To this end, it is sufficient to put somerestrictions on the A -polynomial and then use following lemmas and propositions. Lemma 4.2. If P ( M, L ) ∈ C [ M, L ] is an irreducible polynomial that contains onlyeven powers of M , then P ( M , L ) is irreducible over C [ M, L ] . Proof : Let u = M . Since P ( M, L ) contains only even powers of M , P ( u, L ) ∈ C [ M, L ]. Assume for contradiction that P ( u , L ) is reducible in C [ M, L ]. By symme-try, if h ( M, L ) is a factor of P ( u , L ) then so is h ( − M, L ). Since P ( u, L ) is irreducible, P ( u , L ) = h ( M, L ) h ( − M, L ) where h ( M, L ) is irreducible in C [ M, L ]. If h ( M, L )contains a term of odd degree in M , let d be the smallest of those degrees. Then P ( u , L ) contains an term of degree d in u , a contradiction. This means that all theterms in h ( M, L ) are of even degree in M . Therefore, P ( u, L ) = h ( u, L ) h ( − u, L ) is apolynomial factorization, contradicting the assumption that P ( M, L ) is irreducible.
Lemma 4.3 (Tran [12]) . Let P ( M, L ) ∈ C [ M, L ] be an irreducible polynomial with P ( M, L ) (cid:54) = P ( M, − L ) . Then, R K ( M, L ) =
Res λ ( P ( M, λ ) , λ − L ) ∈ C [ M, L ] is irreducible and has L degree equal to that of P ( M, L ) . Lemma 4.4.
Let K = b ( p, m ) be a two-bridge knot. Assume K satisfies the AJ -conjecture. Also assume A (cid:48) K ( m, l ) is irreducible in C [ M, L ] and has L degree p − .Then (cid:101) α K ( t, M, L ) has L degree p +12 . Proof : By [7] Proposition 6.1, (cid:101) α K ( t, M, L ) has L degree at most p +12 . Since (cid:101) α K ( − , M, L ) M = ( L − A (cid:48) K ( M, L ) has L degree p +12 , conclude that (cid:101) α K ( t, M, L ) has L degree p +12 . (cid:3) Proposition 4.1 (Tran [13]) . Suppose K is a non-trivial alternating knot. Then theannihilator of the odd sequence J K (2 n + 1) has L -degree greater than 1. Proposition 4.2 (Tran [13]) . For any non-trivial annihilator (cid:101) δ ( t, M, L ) of the oddsequence J K (2 n + 1) , (cid:15) ( (cid:101) δ ) is divisible by L − . Proposition 4.3 (Tran [12]) . Suppose K is a non-trivial knot with reduced alternat-ing diagram D . Let (cid:101) δ ( t, M, L ) be the minimal degree annihilator of J K (2 n + 1) and (cid:101) ∆ C ( t, M, L ) be the minimal degree annihilator of J C ( n ) . Then for odd integers r with ( r + 4 η − )( r − η + ) > , (cid:101) ∆ C ( t, M, L ) = (cid:101) δ ( t, M, L )( L + t − r M − r ) . Theorem 4.1.
Let K = b ( p, m ) be a two-bridge knot with a reduced alternating dia-gram D that has η + positive crossings and η − negative crossings. Assume A K ( M, L ) has L degree p +12 and let C be the ( r, -cable of K . Assume the following properties;i. K satisfies the AJ -conjecture and J K ( n ) has a minimal degree d homogeneous an-nihilator α K ( t, M, L ) with (cid:101) α K ( − , M, L ) M = A K ( M, L ) , where d = p +12 .ii. The A (cid:48) polynomial, A (cid:48) K ( M, L ) of K is irreducible and A (cid:48) K ( M, L ) (cid:54) = A (cid:48) K ( M, − L ) .iii. The matrix N ( − , M ) described in Section 3 has nonzero determinant.Then for odd integers r with ( r + 4 η − )( r − η + ) > the cable knot C satisfies the AJ -conjecture. Proof : The proof is similar to that of Theorem 1 given by Anh Tran [12]. Con-ditions ( i ) and ( iii ) allow for the application of Lemma 3.1 to find a non-trivialoperator (cid:101) β ( t, M, L ) such that (cid:101) β ( t, M, L ) J K (2 n + 1) = 0 . From the discussion above, (cid:15) ( (cid:101) β ) M = ( L − R K ( M , L ) where R K ( M , L ) = Res( A (cid:48) K ( M , λ ) , λ − L ). Let (cid:101) δ ( t, M, L )be the minimal degree annihilator of J K (2 n + 1). Then (cid:15) ( (cid:101) δ ) left divides (cid:15) ( (cid:101) β ). Since K is alternating Propositions 4.1 and 4.2 imply (cid:15) ( (cid:101) δ ) has L degree ≥ L −
1. It is well known that the A -polynomial has only even powers of M . Therefore,the resultant, R K ( M , L ), is irreducible over C [ M, L ] by condition ( ii ) and Lemmas4.2 and 4.3. Since (cid:15) ( (cid:101) δ ) L − has L degree at least 1 and R K ( M , L ) is irreducible, conclude HE AJ CONJECTURE FOR CABLES OF TWO-BRIDGE KNOTS 15 that (cid:15) ( (cid:101) δ ) L − M = R K ( M , L ). Let (cid:101) ∆ C ( t, M, L ) be the minimal degree annihilator of J C ( n ).All that is left to show is A C ( M, L ) M = (cid:15) ( (cid:101) ∆ C ). Since ( r + 4 η − )( r − η + ) >
0, Proposi-tion 4.3 implies that (cid:101) ∆ C = (cid:101) δ ( L + M − r ). By these remarks and the cabling formula(2.1) for A -polynomials, A C ( M, L ) = ( L − R K ( M , L )( L + M − r ) M = (cid:15) ( (cid:101) δ )( L + M − r )= (cid:15) ( (cid:101) ∆ C ) (cid:3) Results
To apply Theorem 4.1, first find a two-bridge knot with reduced alternating dia-gram that satisfies the AJ -conjecture. In [8], conditions were given in order for a knotto satisfy the AJ conjecture. Specifically, all two-bridge knots for which the SL ( C )character variety has exactly two irreducible components satisfy the AJ conjecture.We will apply the main theorem to ( r, The knot = b (11 , A -polynomial of the knot 6 (without the L − A (cid:48) ( M, L ) = − L M + (3 M + 2 M + M − M − M − M + M ) L +( − M + M − M + 3 M + 8 M + M − M − M + 5 M + 3 M ) L +(8 M + 3 M − M + M − M − M + 3 M − M + 5 M + M ) L + ( M +3 M − M − M + 1 − M + 2 M ) L − M The degree of A (cid:48) ( M, L ) is 5 = − . Lemma 4.4 implies the annihilator is degree 6.The only condition to verify in order to apply Theorem 4.1 is det( N ) (cid:54) = 0. Since thedegree of A ( M, L ) is 6, N will be a 5 × P i = P i ( M )is used. N = P P P P P P P P P P P P P P P P P With the aid of Maple the determinant of N is found using the coefficients of A ( M, L ). The determinant of N ( − , M ) will be a nonzero multiple of the belowpolynomial and therefore still nonzero. The knot 6 has a reduced alternating diagram with four positive and two negativecrossings. Therefore, when ( r − r + 8) >
0, the ( r, will satisfythe AJ conjecture.Follow the same process to show that the ( r, AJ conjecture when r is in the proper range. It would be toocumbersome and take up to much space to list all of the determinants. However, ithas been verified with Maple that each relevant determinant is non-zero. Below is atable of some two-bridge knots whose cables satisfy the AJ conjecture by Theorem 4.1.The needed A -polynomials were gathered from KnotInfo.Rolfsen Notation Two-Bridge Notation6 b (11 , b (13 , b (13 , b (17 , b (19 , b (17 , b (17 , b (19 , b (23 , b (23 , b (25 , b (25 , b (29 , b (29 , b (31 , b (19 ,
13) Rolfsen Notation Two-Bridge Noation9 b (19 , b (21 , b (29 , b (31 , b (31 , b (37 , b (37 , b (39 , b (39 , b (41 , b (41 , b (41 , b (43 , b (47 , b (49 , References [1] D. Cooper, M. Culler, H. Gillet, D. Long and P. Shalen,
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HE AJ CONJECTURE FOR CABLES OF TWO-BRIDGE KNOTS 17 [2] N. Dunfield and S. Garoufalidis,
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Proof of a stronger version of the AJ conjecture of torus knots , Algebra. Geom. Topol. (2013), no. 1, 609-624. Mathematics Department, University of Iowa, Iowa City, IA
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