Infinitely many roots of unity are zeros of some Jones polynomials
IINFINITELY MANY ROOTS OF UNITY ARE ZEROS OF SOMEJONES POLYNOMIALS
MACIEJ MROCZKOWSKI
Abstract.
Let N = 2 n − N = n + n −
1, for any n ≥
2. Let M = N − . We construct families of prime knots with Jones polynomials( − M (cid:80) Mk = − M ( − k t k . Such polynomials have Mahler measure equal to 1.If N is prime, these are cyclotomic polynomials Φ N ( t ), up to some shift inthe powers of t . Otherwise, they are products of such polynomials, includingΦ N ( t ). In particular, all roots of unity ζ N occur as roots of Jones poly-nomials. We also show that some roots of unity cannot be zeros of Jonespolynomials. Introduction
We study knots K with Jones polynomials that have Mahler measure equal to1, or M ( V K ( t )) = 1. Such knots were considered in [2, 3]. A Laurent polynomial P , with M ( P ) = 1, has the form: t a times a product of cyclotomic polynomialsΦ n , a ∈ Z . Following [3], we call such P cyclotomic .The motivation for studying such knots comes from some observed connectionsbetween the Mahler measure of the Jones polynomial of a knot and its hyperbolicvolume [2, 3]. The Mahler measure of Jones polynomials has also been studiedin [19, 20]. In a more general context, not much is known about the question:what polynomials are Jones polynomials? This is in contrast to the Alexanderpolynomial: there are simple conditions on a polynomial which are sufficient andnecessary for it to be the Alexander polynomial of a knot. Studying the locus ofthe zeros of Jones polynomials is part of the general question. It is shown in [8]that this locus is dense in C . Our result implies that the locus intersected with theunit circle is dense in the unit circle.Near the end of [2], after listing all knots up to 16 crossings with cyclotomicJones polynomials (there are only 17 such knots), the following problem is posed:“An interesting open question is how to construct more knots with M ( V K ( t )) =1.” In this paper, we construct four inifinite families of knots with cyclotomicJones polynomials of a particularly simple form. They include 4 and 9 with V ( t ) = t − − t − + 1 − t + t and V ( t ) = t − − t − + t − − t − t + t . TheJones polynomials of the other knots extend these two examples. Their coefficientsare finite sequences of alternating 1’s and − − m be odd. It is well known that, for any Mathematics Subject Classification 2020: 57K10, 57K14 a r X i v : . [ m a t h . G T ] F e b MACIEJ MROCZKOWSKI a ∈ N , t a − (cid:81) k | a Φ k ( t ). Hence, t m − (cid:89) k | m Φ k ( t ) = (cid:89) k | m Φ k ( t ) (cid:89) k | m Φ k ( t ) = ( t m − (cid:89) k | m Φ k ( t )It follows, that (cid:81) k | m Φ k ( t ) = t m + 1 = ( t + 1)( t m − − t m − + t m − + . . . − t + 1).Since Φ ( t ) = t + 1, we get: (cid:89) k | m,k> Φ k ( t ) = t m − − t m − + t m − + . . . − t + 1For m odd, we introduce the following notation: (cid:101) Φ m ( t ) := t − m − (cid:89) k | m,k> Φ k ( t ) = t − m − − t − m − +1 + . . . − t m − − + t m − We allow m = 1 with (cid:101) Φ ( t ) = 1. We will construct knots with cyclotomic Jonespolynomials equal to (cid:101) Φ m for infinitely many odd m .Notice, that ddt [ t n P ( t )] t =1 = P (cid:48) (1)+ n for any n ∈ Z , if P is a Laurent polynomialsatisfying P (1) = 1. If P is a Jones polynomial of a knot, it satisfies P (1) = 1 and P (cid:48) (1) = 0 (see [9]), hence t n P cannot be a Jones polynomial for n (cid:54) = 0. We saythat a Laurent polynomial P is palindromic , if P ( t − ) = t n P ( t ), for some n ∈ Z ; inparticular, if n = 0, we say that it is symmetric . One checks that P (cid:48) (1) = 0, if P issymmetric. Hence, a palindromic Jones polynomial of a knot must be symmetric.For n ≥
3, cyclotomic polynomials Φ n are palindromic of even degree, but notsymmetric (since they are not Laurent polynomials). In order to make them sym-metric, we multpily Φ n with t − ϕ ( n )2 (where ϕ is the Euler totient function and ϕ ( n )is the degree of Φ n ). For n ≥
3, we use the notation:Φ symn ( t ) = t − ϕ ( n )2 Φ n ( t )For example Φ sym ( t ) = t − − t − + 1 − t + t and Φ sym ( t ) = t − − t − + t − − t − t + t . Notice, that Φ symn and Φ n have the same roots. One checks that theformula for (cid:101) Φ m ( t ) given above, m odd, simplifies to: (cid:101) Φ m ( t ) := (cid:89) k | m,k> Φ sym k ( t )The paper is organized as follows: in section 2 the main theorems are stated,while the notion of arrow diagrams and some proofs are postponed to two lattersections 3 and 4. 2. Main results
Let W n,k , n, k ∈ Z , k ≥
0, be the knot shown for n = 2 and k = 3 in Figure 1,in the form of an arrow diagram . In general, there are n arrows on the left kink,and k arrows arranged on k strands, generalizing in an obvious way the case k = 3shown in this figure. When n <
0, there are | n | clockwise arrows on the left kink.In short, the arrows correspond to fibers in the Hopf fibration of S . A detailedexplanation of arrow diagrams is postponed to section 3. In section 4, we computethe Jones polynomials of the knots W n,k , denoted V W n,k : NFINITELY MANY ROOTS OF UNITY ARE ZEROS OF SOME JONES POLYNOMIALS 3
Figure 1. W , Theorem 1.
Let n, k ∈ Z , k ≥ . Then, V W n,k = t n ( n − + k ( k − − nk t − (cid:16) − t ( k +2) n +1 + t ( k +1)( n +1) ( t k +1 + 1) − t k ( n +3)+1 + t − (cid:17) Denote by D n,k the terms in the big parenthesis in the formula for V W n,k above.We want to check for which n, k the polynomial V W n,m is symmetric, i.e. V W n,m ( t − ) = V W n,m ( t ). It is easy to see that a necessary condion is that D n,k ( t − ) = − t a D n,k ( t )for some a ∈ Z . Say that such D n,k is antipalindromic .For k = 0, W n, is an oval with n ∈ Z arrows on it. Such knots are torus knots,trivial if and only if n ∈ { , , − , − } , see [15]. Thus, when W n, is non trivial, itsJones polynomial is not symmetric. Theorem 2.
Suppose that k > . The polynomial V n,k ( t ) is symmetric if and onlyif n = k − , k , k or k + 1 . Furthermore, let f ( k ) = k + k − and g ( k ) = 2 k − .Then, for k > , V W k − ,k ( t ) = (cid:101) Φ f ( k ) ( t ) V W k,k ( t ) = (cid:101) Φ f ( k +1) ( t ) V W k,k ( t ) = V W k +1 ,k ( t ) = (cid:101) Φ g ( k +1) ( t ) Proof.
We consider D n,k . First we check when t or − t cancels with a term − t a , if a = 1. This occurs when n = 0 or n = −
3. The term − n = − n = −
2. One checks, that D n,k is notantipalindromic in all these cases, except for ( n, k ) = (0 , W , is trivial isvery easy to check, see section 3. Notice, that (cid:101) Φ f (1) = (cid:101) Φ = 1.Suppose now that n and k are such that neither t nor − n ≥ n ≤ −
4. Let n = ( k + 2) n + 1, n = k ( n + 3) + 1, p = ( k + 1)( n + 1) and p = ( k + 1)( n + 2) be the exponents of the four remaining terms in D n,k (twonegative and two positive ones).Suppose that n ≤ −
4. The four exponents are negative with the exception n = 0 for ( n, k ) = ( − , t − t −
2, the lowest is − t n and the gap between n and any other exponent it at least 5. Hence, D n,k is notantipalindromic.Suppose that n ≥
1. The four exponents are greater or equal to 4. The four termscannot all cancel out, since otherwise V W n,k would not be a Laurent polynomial (onemay also check case by case that, if a pair of terms cancels, another pair does notcancel). Since there is a gap 1 in the powers of t and −
1, in order for D n,k to beantipalindromic, it should contain another pair of terms t a − t a − for some a ≥ p − n = k − n, p − n = n − k +1 , p − n = 2 k − n +1 , p − n = n − k MACIEJ MROCZKOWSKI
One of these four differences has to be equal to 1, which gives four cases: • p − n = k − n = 1. Then, p = n and these 2 terms cancel out. Now: D k − ,k = t n ( t −
1) + t − t k + k − + 1)( t − m = k + k −
1. One checks that: V W k − ,k = t − m − t − t m + 1)( t −
1) = (cid:101) Φ m = (cid:101) Φ f ( k ) • p − n = n − k + 1 = 1. Then, p = n and: D k,k = t n ( t −
1) + t − t k +3 k +1 )( t − m = k + 3 k + 1. Again, one checks that: V W k,k = (cid:101) Φ m = (cid:101) Φ f ( k +1) • p − n = 2 k − n + 1 = 1. Then p = n and: D k,k = t n ( t −
1) + t − t k +4 k +1 )( t − m = 2 k + 4 k + 1. One checks that: V W k,k = (cid:101) Φ m = (cid:101) Φ g ( k +1) • p − n = n − k = 1. Then, p = n and: D k +1 ,k = t n ( t −
1) + t − t k +4 k +1 )( t − m = 2 k + 4 k + 1. One checks that: V W k +1 ,k = (cid:101) Φ(2 m ) = (cid:101) Φ g ( k +1) (cid:3) As an immedaite consequence, we get:
Theorem 3.
There are infinitely many roots of unity that are zeros of Jones poly-nomials. Such roots are dense in the unit circle.Proof.
Since for any odd m , Φ m | (cid:101) Φ m , one has: for any k > ζ f ( k ) , ζ f ( k +1) , ζ g ( k +1) are zeros of some Jones polynomials.Since t m − t m − t + 1) (cid:101) Φ m t m − , the roots of (cid:101) Φ m are: ζ m , ζ m . . . ζ m − m , ζ m +22 m . . . ζ m − m It is clear that the roots of the (cid:101) Φ m ’s, that are Jones polynomials, are dense in theunit circle, since there are infinitely many such (cid:101) Φ m ’s. (cid:3) The knots appearing in Theorem 2 come in quadruplets for k = 1 , , ... InTable 2, are shown the first four quadruplets, their Jones polynomials and thecrossing numbers (together with identification for knots up to 15 crossings; also,the knot W , is 16 n ).Notice that 98 is the first index that is not twice a prime, hence (cid:101) Φ (cid:54) = Φ sym .The knots with k >
1, except W , , have more than 16 crossings (since their Jonespolynomials do not appear up to 16 crossings) and their crossing number seemsto increase rapidely. From Lemma 4 below, c ( W n,k ) < = k + ( n + k ) −
1. UsingKnotscape[7] (after removing the arrows in the diagrams, see section 3), the numberof crossings can sometimes be reduced by 1 or 2. Knotscape handles diagrams up to49 crossings and allows to check that the Alexander polynomial differentiates W , Table 1. W n,k with cyclotomic Jones polynomials up to k = 4 K V K c ( K ) K V K c ( K ) K V K c ( K ) K V K c ( K ) W , W , Φ sym W , Φ sym W , Φ sym W , Φ sym n W , Φ sym ≤ W , Φ sym ≤ W , Φ sym ≤ W , Φ sym ≤ W , Φ sym ≤ W , Φ sym ≤ W , Φ sym ≤ W , Φ sym ≤ W , Φ sym ≤ W , (cid:101) Φ ≤ W , (cid:101) Φ ≤ W , . Since W , has a diagram with 52 crossings, its Alexander polynomialcannot be computed with Knotscape (in order to check whether it is different from W , ). Though it seems unlikely, it is possible that some W k,k is the same knot as W k +1 ,k and/or some W k,k is the same knot as W k,k +1 .As an example, using Knotscape on a diagram of W , with 39 crossings one getsa reduction to 38 crossings with the following DT code:
38 1 6 -14 16 -28 26 -42 68 40 -50 -60 -34 36 -44 -54 52 62 -20 46-58 -4 2 56 64 -22 70 -76 8 -10 -66 -48 72 -74 24 12 38 18 -32 30 (bestavailable reduction)
We turn now to some properties of the knots W n,k . Proposition 1.
The knots W n,k are (1 , knots. In particular they have tunnelnumber , hence they are prime.Proof. We postpone the proof that W n,k are (1 ,
1) knots to section 3. Now (1 , (cid:3) Proposition 2.
The knots W n,k with cyclotomic Jones polynomials are non al-ternating except for and . There is no bound on the twist number of suchknots.Proof. Suppose that W n,k is non trivial and alternating with Jones polynomialequal to some (cid:101) Φ m . From [4], its twist number equals 2. Such knot can be eithera connected sum of torus knots of type (2 , m ) and (2 , n ) or a 2-bridge knot. FromProposition 1 W k,n is prime, so it has to be a 2-bridge knot. Using an explicitformula for Jones polynomials of 2-bridge knots with twist number 2 in [17], onechecks easily, that all such knots, except 4 , have Jones polynomials that are notequal to (cid:101) Φ m for any m .For the second part, it is shown in [3], that for a family of links with cyclotomicJones polynomials of unbounded span, there is no bound on the twist numbers ofthese links. (cid:3) We turn now to some obstructions for roots of unity being zeros of Jones poly-nomials.It is well known that the Jones polynomial has special values in 1, ζ , i and ζ , see [9, 10]. For a knot K , V K (1) = V K ( ζ ) = 1, V K ( ζ ) = ± V K ( ζ ) = ± ( i √ n , n ∈ N ∪ { } . This allows to exclude some roots of unity as zeros of Jonespolynomials: Theorem 4.
For k ∈ N ∪ { } , let N = p k , p k , p k , with p prime; or N = 6 p k with p (cid:54) = 3 prime. Then Φ N cannot divide any Jones polynomial. MACIEJ MROCZKOWSKI
Proof.
We check the values of cyclotomic polynomials in 1, ζ , i and ζ . From [1],we have for p prime:Φ p k (1) = p for k > (1) = 0) | Φ p k ( ζ ) | = Φ p k ( i ) = | Φ p k ( ζ ) | = p We see that none of these polynomials can divide a Jones polynomial V K , since V K (1) = V K ( ζ ) = | V K ( i ) | = 1 and | V K ( ζ ) | = √ n , except if p = 3 in the case ofΦ p k , which we excluded in our assumptions. (cid:3) We can also exclude easily some (cid:101) Φ k as divisors of Jones polynomial: Proposition 3.
Let k be odd. If (cid:101) Φ k divides a Jones polynomial, then k is notdivisible by .Proof. If 3 | k , then Φ | (cid:101) Φ k , hence (cid:101) Φ k ( ζ ) = 0, which is impossible for a divisor ofa Jones polynomial. (cid:3) Notice, that all roots of unity appearing as zeros of Jones polynomials in Theo-rem 2 are of the form ζ k , with k odd, 3 (cid:45) k . It is natural to ask what other roots ofunity ζ N can be zeros of Jones polynomials of knots. Using Theorem 4, the smallestpossible N for such roots are 18 , , , , , , , , , ,
60. Let us sum thisup:
Question 1.
Is there a knot with Jones polynomial having a zero in ζ N such that: | N ; N is odd; | N ; or N = 2 k , k odd, (cid:45) k but N not coming from Theorem 2? One may also ask the question, whether there are infinitely many primes suchthat Φ sym p are Jones polynomials of some knots. A positive answer would follow,if there were infinitely many primes in the image of f or g from Theorem 2 (twospecial cases of the Bunyakovsky conjecture).Since a Mersenne prime 2 p −
1, with p >
2, satisfies 2 p − p − ) − g (2 p − ), we get: Corollary 1.
Let N = 2 p − , p > , be a Mersenne prime. Then Φ sym N is theJones polynomial of a knot. Arrow diagrams
Arrow diagrams where introduced in [12] for links in F × S , where F is anorientable surface. They were subsequently extended for links in Seifert manifolds(see [6, 13, 14]). In [15], they were applied for links in S : it was shown there,that projections of links under the Hopf fibration from S to S can be encodedwith arrow diagrams in a disk: such a diagram is like a usual diagram of a link,except that it is in a disk and there may be some arrows on it, outside crossings.Two arrow diagrams represent the same link if and only if one diagram can betransformed into the other with a series of six Reidemeister moves, see Figure 2.For the Ω ∞ move in this figure, the boundary of the disk is drawn in thick. Forsimplicity we can also omit this boundary when picturing arrow diagrams (as wehave done in Figure 1).A detailed interpretation of the arrow diagrams and Reidemeister moves can befound in [15]. One can picture easily a link L from its arrow diagram D in thefollowing way: pick a solid torus T = C × S , C a disk, consisting of some oriented NFINITELY MANY ROOTS OF UNITY ARE ZEROS OF SOME JONES POLYNOMIALS 7 Ω ←→ Ω ←→ Ω ←→ Ω ←→ Ω ←→ Ω ←→ Ω ∞←→ Figure 2.
Reidemeister movesfibers p × S , p ∈ C , in the Hopf fibration of S . Let S = I ∪ I (cid:48) consist oftwo intervals glued along their endpoints. Then T = B ∪ B (cid:48) , where B = C × I and B (cid:48) = C × I (cid:48) are two balls. If there are no arrows in D , L lies entirely in B .Otherwise it lies in B except for some neighborhoods of the arrows where it goesthrough B (cid:48) along an oriented fiber and the orientation of the arrow agrees with theorientation of the fiber.We turn now to the proof of Proposition 1. We want to show that the knots W n,k are (1 ,
1) knots. Recall from [5] that a link L admits a ( g, b ) decomposition,if there is a genus g Heegard splitting ( V , V ) of S such that V i intersects L in b trivial arcs, for i ∈ { , } . To show that W n,k is a (1 ,
1) knot, we need to show thatit intersects each T i in a trivial arc, for a Heegard spliting of S into two solid tori T i , i ∈ { , } .We say that an arrow diagram of a knot is annulus monotonic , if there is anannulus A = S × I , containing the diagram and such that the curve of the diagramhas exactly one minimum and one maximum w.r.t. I . Applying Ω ∞ on the left kinkof W n,k (see Figure 1), we obtain a diagram consiting of a spiral with some arrowson it. Such a diagram is clearly annulus monotonic, see Figure 3. Proposition 1now follows directly from the following: Lemma 1.
Suppose that a knot K has an annulus monotonic arrow diagram D .Then K is a (1 , -knot.Proof. Let A = S × I be an annulus containing D and such that D has exactlyone minimum and one maximum w.r.t. I . The closure of S \ ( A × S ) consists oftwo solid tori T and T , chosen so that T i ∩ ( A × S ) = ( S × i ) × S , i ∈ { , } .Cut I into I = [0 , a ] and I = [ a,
1] for some a ∈ (0 , S × I ) ∩ D is a small trivial arc. A decomposes into two annuli A i = S × I i , i ∈ { , } . Let T (cid:48) i = T i ∪ ( A i × S ) be two solid tori, i ∈ { , } , so that S = T (cid:48) ∪ T (cid:48) .Then T (cid:48) ∩ K is clearly a trivial arc in T (cid:48) . We claim that T (cid:48) ∩ K is also a trivial arc MACIEJ MROCZKOWSKI in T (cid:48) . Let I = [ b, b > a , be such that ( S × I ) ∩ D is a small trivial arc. Let A = S × I . Since D is annulus monotonic, the pair ( A × S , K ∩ ( A × S )) canbe isotoped to ( A × S , K ∩ ( A × S )), by removing the tori ( S × c ) × S for c from a to b . Such isotopy clearly extends to ( T (cid:48) , K ∩ T (cid:48) ), so K ∩ T (cid:48) is a trivial arcin T (cid:48) . Thus K is a (1 ,
1) knot. (cid:3)
Figure 3.
An annulus monotonic diagram, drawn inside the annulusWe remark here, that for some n ’s, hypothetical knots with Jones polynomialsΦ symn would only admit a ( g, b ) decomposition with large g + b . Indeed, in [1]the values of cyclotomic polynomials in ζ are computed. It is shown there, that | Φ n ( ζ ) | can be arbitrarily large for some n ’s. For example it grows very fastwith the number of primes in the decomposition of n , when n is a product of anodd number of distinct primes congruent to 2 or 3 modulo 5. For instance, for n = 2 3 7 13 17, one checks that this module is approximately 2207. On theother hand, it follows from [11] that, if a knot K admits a ( g, b ) decomposition, itsJones polynomial V K satisfies | V K ( ζ ) | ≤ α g β b − , where α > β > g + b .It was shown in [15], that the usual blackboard framing for links obtained fromtheir diagrams extends to arrow diagrams and that such framing is invariant underall Reidemeister moves except Ω . In particular, to compute the writhe of a framedlink represented by an arrow diagram, one may eliminate all arrows without usingΩ , then sum the signs of all crossings in the arrowless diagram.We present now a formula for the writhe of any arrow diagram of a knot. Thisformula holds also for oriented links.Let D be an oriented arrow diagram. Let r be an arrow in D . The sign of r ,denoted (cid:15) ( r ), is defined as follows: (cid:15) ( r ) = 1 (resp. (cid:15) ( r ) = − r points in thesame (resp. opposite) direction as the orientation of the diagram. We also saythat r is positive (resp. negative ). The winding number of r , denoted ind ( r ), isby definition the winding number ind D ( P ), where D is the diagram considered asan oriented curve and P is a point close to r , to the right of D according to theorientation of D .For example, consider W , in Figure 1. Orient it so that the left kink is orientedclockwise. Then the 3 arrows on the right are positive and the two arrows on theleft are negative. Also the winding numbers of the arrows on the right are 0, 1 and2, wheras the 2 arrows on the left have winding number − w ( D ) the writhe of the framed knot represented by the arrow diagram D . Denote by ¯ w ( D ) the writhe, when all arrows in D are ignored (it is sum of thesigns of crossings in D ). We have the following formula for the writhe: Lemma 2.
Let D be an oriented arrow diagram. Let n = (cid:88) r (cid:15) ( r ) , the sum takenover all arrows of D . Then: NFINITELY MANY ROOTS OF UNITY ARE ZEROS OF SOME JONES POLYNOMIALS 9 w ( D ) = ¯ w ( D ) + (cid:88) r (cid:15) ( r ) ind ( r ) + n ( n + 1) Proof.
We remove with Reidemeister moves all arrows in D keeping track of thesigns of the crossings that appear. We do not use Ω , thus the writhe is unchanged.Consider an arrow r in D . We push it next to the boundary of the diagramin such a way that the orientation of the arc next to the arrow agrees with thecounterclockwise orientation of the boundary of the diagram (see Figure 4 (left),where 3 arrows have been pushed and the arcs are oriented as wished). Figure 4.
Two positive, one negative arrow (left); pushing anarrow through the arc next to it (right)To achieve this, we use Ω and Ω moves repeatedly. When r crosses an arc, twopositive or two negative crossings appear. One checks that the total contribution,when r is next to the boundary, is 2 (cid:15) ( r ) ind ( r ). Notice that when r is next to theboundary, but the orientation of the arc is not the desired one, then ind ( r ) = − r has to be pushed once through a piece of arc next to it, see Figure 4 (right).After all the arrows have been pushed, so they are as in Figure 4 (left), the sum ofthe signs of all crossings is ¯ w ( D ) + (cid:88) r (cid:15) ( r ) ind ( r ).Suppose now, that there are a positive arrows and b negative ones, so that n = a − b . Push every positive arrow through an arc next to it as in Figure 4(right). This adds 2 a positive crossings. Now any arrow r can be eliminated withΩ ∞ followed by Ω . We push the remaining arrows through the arc created byΩ ∞ . One checks that if an arrow r (cid:48) is pushed through the arc coming from r , thisadds two positive (resp. negative) crossings if (cid:15) ( r ) = (cid:15) ( r (cid:48) ) (resp. (cid:15) ( r ) = − (cid:15) ( r (cid:48) )).Then one repeats the process with the arrow r (cid:48) (eliminating it and pushing allother arrows through it). Hence, at the end any pair of arrows r and r (cid:48) contributes2 (cid:15) ( r ) (cid:15) ( r (cid:48) ) to the writhe. The total contribution to the writhe of this second part isthus: 2 a + a ( a −
1) + b ( b − − ab = ( a − b )( a − b + 1) = n ( n + 1)Combined with the first part, this gives the required formula. (cid:3) Applying Lemma 2 to W n,k we get: Lemma 3.
Let W n,k stand for the diagram in Figure1, as well as for the framedknot represented by this diagram. Then: w ( W n,k ) = n + n + 2 k + k − nk Proof.
Orient W n,k so that the k arrows are positive. If n > n arrowsare negative. If n < | n | arrows are positive. The k arrows have wind-ing numbers 0 , , , . . . , k −
1. The n arrows have all winding number −
1. Also,¯ w ( W n,k ) = k . Hence: w ( W n,k ) = k + 2( − n )( −
1) + 2(1 + 2 + . . . + k −
1) + ( k − n )( k − n + 1) = k + 2 n + k ( k −
1) + ( k − n )( k − n + 1) = n + n + 2 k + k − nk (cid:3) For an upper estimate of the number of crossings, c ( W n,k ), we use Lemma 1from [15]. It states that, if a diagram D has k crossings and all its arrows are next tothe boundary, with a of them removable (i.e. one can remove them with Ω ∞ followedby Ω ) and b > c ( K ) ≤ k + b − a + b )( a + b − Lemma 4.
For n ≥ , k ≥ and k + n > , one has: c ( W n,k ) ≤ k + ( n + k ) − Proof.
Starting with the diagram of W n,k shown in Figure 1, we push k − k +2(1+2+ . . . + k −
1) = k + k ( k −
1) = k crossings. Then, we can apply Lemma 1 from [15]. Since n ≥
0, allarrows will be non removable. Since n + k >
0, there is at least one non removablearrow. Thus, c ( W n,k ) ≤ k + ( n + k ) − n + k )( n + k −
1) = k + ( n + k ) − (cid:3) We end this section with a visualization of any knot W n,k . Such knot is obtainedby a small modification from a pair of torus knots lying on the boundary of athickened Hopf link. It was shown in [15] how to get some simple arrow diagramsof torus knots: one checks that W n, is the torus knot T ( n, n + 1) and W ,k is thetorus knot T ( k, k + 1). Consider the diagram of W n,k in Figure 1. Let W sn,k be thediagram of a 2-component link, obtained from W n,k by smoothing vertically thecrossing next to the n arrows. The components of W sn,k are torus knots T ( n, n + 1)and T ( k, k + 1). Let D and D (cid:48) be two disjoint disks, such that D contains the n arrows, D (cid:48) contains the k arrows and W sn,k is contained in D ∪ D (cid:48) . Let T , resp. T (cid:48) ,be two solid tori consisting of fibers intersecting D , resp. D (cid:48) , in the Hopf fibrationof S . Then T and T (cid:48) form a thickened Hopf link. The torus ( n, n + 1) componentof W sn,k can be pushed onto ∂T and the torus ( k, k + 1) component can be pushedonto ∂T (cid:48) . Then W n,k is obtained from such two linked torus knots by reverting thesmoothing back to the crossing.4. Jones polynomials of the knots W n,k Let G n , G (cid:48) n and G (cid:48) a,b , n, a, b ∈ Z , be the arrow diagrams shown in Figure 5. Inthe box is an arrow tangle G (a tangle with, possibly, some arrows on it). Let g n , g (cid:48) n and g (cid:48) a,b be the Kauffman brackets of, respectively, G n , G (cid:48) n and G (cid:48) a,b . We wantto express g (cid:48) n with some g k ’s. In order to do it, we will use g (cid:48) a,b ’s. n G n G a b G Figure 5. G n , G (cid:48) n and G (cid:48) a,b NFINITELY MANY ROOTS OF UNITY ARE ZEROS OF SOME JONES POLYNOMIALS 11
It is useful to define for n ≥ S n = A n g − n + A n − g − n +2 + . . . + A − n +2 g n − + A − n g n = n (cid:88) i =0 A n − i g − n +2 i Extend S n for negative n , by defining S − = 0 and, for n < − S n = − S | n |− Lemma 5.
For n ∈ Z : g (cid:48) n = ( A − − A ) A n S n − A n − g − n Proof.
One checks easily that the formula holds for n = 0 and n = − A + ( A − − A )Using this relation and Ω and Ω moves, we get:(*) g (cid:48) a,b = A g (cid:48) a − ,b − + ( A − − A ) g a + b Suppose that n ≥
1. Iterating equation (*) until g (cid:48) , − n = − A g − n , we get: g (cid:48) n, = A g (cid:48) n − , − + ( A − − A ) g n = A g (cid:48) n − , − + A ( A − − A ) g n − + ( A − − A ) g n = . . . = − A A n g − n + ( A − − A ) (cid:0) g n + A g n − + . . . + A n − g − n +2 (cid:1) = − A n − g − n + ( A − − A ) (cid:0) g n + A g n − + . . . + A n g − n (cid:1) = ( A − − A ) A n S n − A n − g − n Suppose now that n ≤ −
2. Rewriting equation (*) and replacting a by a + 1 and b by b + 1 one gets: g (cid:48) a,b = A − g (cid:48) a +1 ,b +1 − A − ( A − − A ) g a + b +2 Iterating until g (cid:48) n + | n | , | n | = g (cid:48) , − n = − A g − n , we get: g (cid:48) n, = A − g (cid:48) n +1 , − A − ( A − − A ) g n +2 = A − g (cid:48) n +2 , − A − ( A − − A ) g n +4 − A − ( A − − A ) g n +2 = . . . = − A A n g − n − A − ( A − − A ) (cid:0) g n +2 + A − g n +4 + . . . + A n +2 g − n (cid:1) = − A n − g − n − A − ( A − − A ) (cid:0) g n +2 + A − g n +4 + . . . + A n +4 g − n − (cid:1) = − A n − g − n − A − ( A − − A ) A n +2 (cid:0) A − n − g n +2 + . . . + A n +2 g − n − (cid:1) = − A n − g − n − ( A − − A ) A n S | n |− = ( A − − A ) A n S n − A n − g − n (cid:3) We now prove Theorem 1 by induction on k . W n, is an oval with n ∈ Z arrowson it. This is the torus knot T ( n, n + 1) if n ≥ n < W n, = W − − n, (use Ω ∞ and Ω moves). One checks that for k = 0 the formula in Theorem 1 isthe correct formula for such torus knots (see also [15]).We restate Theorem 1 in terms of the Kauffman bracket using the formula V K ( t ) = ( − A ) − w ( K ) < K > , where w ( K ) is the writhe of K and t = A − . FromLemma 3, we have w ( W n,k ) = n + n + 2 k + k − nk , hence ( − w ( W n,k ) = ( − k and: < W n,k > = ( − k A n +1) n +3 k (2 k +1 − n ) V W n,k One checks, that Theorem 1 can be restated as:
Proposition 4. < W n,k > ( A − −
1) = ( − k A n − kn +2 k + n − k − (cid:0) (1 − A ) A kn +4 n +8 k +4 − A n − k +4 + A k +4 − A k − n +4 + 1 (cid:1) Proof.
For k = 0 the formula is correct, since it is correct for the Jones polynomialand we just restate it with the Kauffman bracket using the writhe.Let k ≥
0. Assume that the formula holds for < W n,k > , n ∈ Z . We useLemma 5, with W n,k +1 = G (cid:48) n . One has to identify G a (in that lemma), for any a ∈ Z : it has a diagram shown in Figure 6 (for k = 2 as an example). Usinga single Ω ∞ move on the strand with the a + 1 arrows, one gets W − a − ,k (oneextra arrow comes from the move). Since this move does not change the writhe, < G a > = < W − a − ,k > . a + 1 Figure 6.
Identifying G a Recall the notations used in Lemma 5: g (cid:48) n = < G (cid:48) n >, g a = < G a >, S n = n (cid:88) i =0 A n − i g − n +2 i Also, by definition: S − = 0 and S n = − S | n |− for n < − S (cid:48) n =( − k A n − kn − n +2 k − k − (cid:0) − A kn +12 n +12 k +16 + A kn +8 n +4 k + A n +8 k +8 − A n (cid:1) One checks that S (cid:48)− = 0 and S (cid:48) n = − S (cid:48)− n − for any n ∈ Z .We claim that for any n ∈ Z :(**) S n ( A − −
1) = S (cid:48) n Because of the skew-symmetry of both S n and S (cid:48) n around −
1, it is sufficient toprove (**) for n ≥ − n = −
1. Now S = g = < W − ,k > . By induction on k : < W − ,k > ( A − −
1) = ( − k A k − k − ( − A k +16 + A k +8 + A k −
1) = S (cid:48) One has obviously: S n +2 = S n + A n +2 g − n − + A − n − g n +2 = S n + A n +2 < W n,k > + A − n − < W − n − ,k > Thus, to prove (**), we need to show that: S (cid:48) n +2 = S (cid:48) n + ( A − − (cid:0) A n +2 < W n,k > + A − n − < W − n − ,k > (cid:1) One checks: S (cid:48) n +2 = ( − k A n − kn +2 k − n − k − (cid:0) − A kn +12 n +20 k +40 + A kn +8 n +12 k +16 − A n +16 + A n +8 k +16 (cid:1) By induction on k : S (cid:48) n + ( A − − (cid:0) A n +2 < W n,k > + A − n − < W − n − ,k > (cid:1) =( − k A n − kn − n +2 k − k − (cid:0) − A kn +12 n +12 k +16 + A kn +8 n +4 k + A n +8 k +8 − A n (cid:1) + ( − k A n − kn +2 k +2 n − k − (cid:0) (1 − A ) A kn +4 n +8 k +4 − A n − k +4 + A k +4 − A k − n +4 + 1 (cid:1) + ( − k A n +2 kn +2 k +6 n +7 k +2 (cid:0) (1 − A ) A − kn − n − k − − A − n − k − + A k +4 − A k +4 n +20 + 1 (cid:1) = ( − k A n − kn +2 k − n − k − (cid:0) − A kn +8 n +16 k +24 + A kn +4 n +8 k +8 + A k +16 − A n +4 k +8 + (1 − A ) A kn +8 n +12 k +16 − A n +16 + A n +8 k +16 − A k +16 + A n +4 k +12 + (1 − A ) A n +4 k +8 − A kn +4 n +8 k +8 + A kn +8 n +16 k +24 − A kn +12 n +20 k +40 + A kn +8 n +12 k +20 (cid:1) = S (cid:48) n +2 Since g − n = < W n − ,k > , from Lemma 5 we get: < W n,k +1 > = ( A − − A ) A n S n − A n − < W n − ,k > Hence: < W n,k +1 > ( A − −
1) = ( A − − A ) A n S (cid:48) n − A n − < W n − ,k > ( A − − − k ( A − − A ) A n − kn − n +2 k − k − (cid:0) − A kn +12 n +12 k +16 + A kn +8 n +4 k + A n +8 k +8 − A n (cid:1) − ( − k A n − kn +2 k − n +3 k − (cid:0) (1 − A ) A kn +4 n − − A n − k − + A k +4 − A k − n +12 + 1 (cid:1) = ( − k A n − kn +2 k − n +3 k − (cid:0) − (1 − A ) A kn +8 n +8 k +12 + (1 − A ) A kn +4 n − + (1 − A ) A k +4 − (1 − A ) A n − k − − (1 − A ) A kn +4 n − + A n − k − − A k +4 + A k − n +12 − (cid:1) = ( − k +1 A n − kn +2 k − n +3 k − (cid:0) (1 − A ) A kn +8 n +8 k +12 − A n − k + A k +8 − A k − n +12 + 1 (cid:1) = ( − k +1 A n − k +1) n +2( k +1) + n − ( k +1) − (cid:16) (1 − A ) A k +1) n +4 n +8( k +1)+4 − A n − k +1)+4 + A k +1)+4 − A k +1) − n +4 + 1 (cid:17) Thus the formula holds for < W n,k +1 > and we are done. (cid:3) References [1] B. Bzdega, A. Herrera-Poyatos, P. Moree,
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