NORMAL AND JONES SURFACES OF KNOTS

Transcription

1 NORMAL AND JONES SURFACES OF KNOTS EFSTRATIA KALFAGIANNI AND CHRISTINE RUEY SHAN LEE Abstract. We describe a normal surface algorithm that decides whether a knot satisfies the Strong Slope Conjecture. We also establish a relation between the Jones period of a knot and the number of sheets of the surfaces that satisfy the Strong Slope Conjecture (Jones surfaces). 1. Introduction The Strong Slope Conjecture, as stated by the first named author and Tran in [18] refines the Slope Conjecture of Garoufalidis [7]. It has made explicit a close relationship between the degrees of the colored Jones polynomial and essential surfaces in the knot complement. The conjecture predicts that the asymptotics of the degrees determines the boundary slopes and the ratios of the Euler characteristic to the number of sheets of essential surfaces in the knot complement. Such surfaces are called Jones surfaces (see Section 2). Not much is known about the nature of these Jones surfaces, and it is unclear how they are distinguished from other essential surfaces of the knot complement. Our purpose in this paper is two-fold: On one hand we are interested in the information about the topology of Jones surfaces, and the topology of the knot complement, encoded in the period of the degree of the colored Jones polynomial. On the other hand we are interested in the question of how Jones surfaces behave with respect to normal surface theory in the knot complement. We organize the paper as follows: In Section 2 first we state the Slope Conjectures and briefly survey the cases where the conjectures have been proved. In Section 3, first we discuss knots of Jones period one: This class includes alternating knots and adequate knots. We discuss a characterization of alternating knots in terms of their Jones surfaces and propose a similar characterization of adequate knots (see Problem 3.3). Then we observe that, in general, there is a relation between the number of sheets of a Jones surface, the Euler characteristic and the period of the knot (see Lemma 3.5). Finally, we present numerical evidence suggesting that the number of sheets of a Jones surface should divide the period of the knot. See Examples and Question March 12, Kalfagianni is supported in part by NSF grant DMS Lee is supported in part by NSF grant DMS

2 2 In Section 4 we examine the Jones surfaces of knots from the viewpoint of normal surface theory in the knot complement. The question that we are concerned with is the following: If a knot satisfies the Strong Slope Conjecture, when can we find Jones surfaces that are fundamental in the sense of Haken [9]? As a result of our analysis we show that there is an algorithm to decide whether a given knot satisfies the Strong Slope Conjecture. We thank Josh Howie for bringing to our attention an oversight in the proof of Theorem 4.3 in an earlier version of the paper. 2. Jones slopes and surfaces 2.1. Definitions and statements. We recall the definition of the colored Jones polynomial; for more details the reader is referred to [21]: We first recall the definition of the Chebyshev polynomials of the second kind. For n 0, the polynomial S n (x) is defined recursively as follows: (1) S n+2 (x) = xs n+1 (x) S n (x), S 1 (x) = x, S 0 (x) = 1. Let D be a diagram of a knot K. For an integer m > 0, let D m denote the diagram obtained from D by taking m parallels copies of K. This is the m-cable of D using the blackboard framing. If m = 1 then D 1 = D. Let D m denote the Kauffman bracket of D m. This is a Laurent polynomial over the integers in the variable t 1/4, normalized so that unknot = (t 1/2 + t 1/2 ). Let c = c(d) = c + + c denote the crossing number and w = w(d) = c + c denote the writhe of D. For n > 0, we define J K (n) := (( 1) n 1 t (n2 1)/4 ) w ( 1) n 1 S n 1 (D), where S n 1 (D) is a linear combination of blackboard cables of D, obtained via equation (1), and the notation S n 1 (D) means extend the Kauffman bracket linearly. That is, for diagrams D 1 and D 2 and scalars a 1 and a 2, a 1 D 1 + a 2 D 2 = a 1 D 1 + a 2 D 2. For a knot K S 3 let d + [J K (n)] and d [J K (n)] denote the maximal and minimal degree of J K (n) in t, respectively. Garoufalidis [6] showed that the degrees d + [J K (n)] and d [J K (n)] are quadratic quasi-polynomials. This means that, given a knot K, there is n K N such that for all n > n K we have d [J K (n)] = a K (n)n 2 + b K (n)n + c K (n), d + [J K (n)] = a K(n)n 2 + b K(n)n + c K(n), where the coefficients are periodic functions from N to Q with integral period. Definition 2.1. The least common multiple of the periods of all the coefficient functions is called the Jones period p of K. For a sequence {x n }, let {x n } denote the set of its cluster points.

3 Definition 2.2. An element of the sets js K := { 4n 2 d + [J K (n)] }, js K := { 4n 2 d [J K (n)] } is called a Jones slope of K. Also let jx K := { 2n 1 ld + [J K (n)] } = {2bK (n)}, jx K := { 2n 1 ld [J K (n)] } = {2b K (n)}, where ld + [J K (n)] and ld [J K (n)] denote the linear term of d + [J K (n)] and d [J K (n)], respectively. Given a knot K S 3, let n(k) denote a tubular neighborhood of K and let M K := S 3 \ n(k) denote the exterior of K. Let µ, λ be the canonical meridian longitude basis of H 1 ( n(k)). A properly embedded surface (S, S) (M K, n(k)), is called essential if it is π 1 -injective and it is not a boundary parallel annulus. An element a/b Q {1/0} with gcd(a, b) = 1 is called a boundary slope of K if there is an essential surface (S, S) (M K, n(k)), such that S represents [aµ + bλ] H 1 ( n(k)). Hatcher showed that every knot K S 3 has finitely many boundary slopes [11]. The Slope Conjecture [7, Conjecture 1] asserts that the Jones slopes of any knot K are boundary slopes. The Strong Slope Conjecture [18, Conjecture 1.6] asserts that the topology of the surfaces realizing these boundary slopes may be predicted by the linear terms of d + [J K (n)], d [J K (n)]. Strong Slope Conjecture. Given a Jones slope of K, say a/b js K, with b > 0 and gcd(a, b) = 1, there is an essential surface S M K with S boundary components such that each component of S has slope a/b, and χ(s) S b jx K. Similarly, given a /b js K, with b > 0 and gcd(a, b ) = 1, there is an essential surface S M K with S boundary components such that each component of S has slope a /b, and χ(s ) S b jx K. Definition 2.3. With the notation as above, a Jones surface of K is an essential surface S M K such that, either S represents a Jones slope a/b js K, with b > 0 and gcd(a, b) = 1, and we have χ(s) S b jx K; or S represents a Jones slope a /b js K, with b > 0 and gcd(a, b ) = 1, and we have χ(s) S b jx K. 3

4 4 The number S b is called the number of sheets of the Jones surface What is known. The Strong Slope Conjecture is known for the following classes of knots. Alternating knots [7]. Adequate knots [4, 5], which is a generalization of alternating knots. Iterated torus knots [18]. Families of 3-tangle pretzel knots [20]. Knots with up to 9 crossings [7, 13, 18]. Graph knots [25]. An infinite family of arborescent non-montesinos knots [3]. Near-adequate knots [19] constructed by taking Murasugi sums of an alternating diagram with a non-adequate diagram. Knots obtained by iterated cabling and connect sums of knots from any of the above classes, since the conjecture was shown to be closed under these operations [18, 25]. The Slope Conjecture is also known for a family of 2-fusion knots, which is a 2- parameter family K(m 1, m 2 ) of closed 3-braids, where K(m 1, m 2 ) is obtained by the ( 1/m 1, 1/m 2 ) Dehn filling on a 3-component link K [8]. 3. Jones period and Jones surfaces In this section we discuss some properties of Jones surfaces of knots with Jones period one, and we establish some relations between the Jones period and the number of sheets of a Jones surface of a knot. We also state some open questions Knots of period one. A large class of knots with Jones period one is the class of adequate knots which contains the class of alternating knots. The property of having period one does not characterize adequate knots. That is, there are knots of period one that are not adequate. See Example 3.4 below. Nevertheless there exists a characterization of adequate knots in terms of the degree of their colored Jones polynomial. Before we can describe it, we need to introduce some notation and terminology. Let D be a link diagram, and x a crossing of D. Associated to D and x are two link diagrams, called the A resolution and B resolution of the crossing. See Figure 1. A Kauffman state σ is a choice of A resolution or B resolution at each crossing of D. The result of applying a state σ to D is a collection s σ of disjointly embedded circles in the projection plane. We can encode the choices that lead to the state σ in a graph G σ as follows. The vertices of G σ are in 1 1 correspondence with the state circles of s σ. Every crossing x of D corresponds to a pair of arcs that belongs to circles of s σ ; this crossing gives rise to an edge in G σ whose endpoints are on the state circles containing those arcs. Definition 3.1. A link diagram D is called A adequate if the state graph G A corresponding to the all A state contains no 1 edge loops. Similarly, D is called

5 5 Figure 1. Crossing with the A-resolution and the B-resolution. The dashed line indicates the corresponding edge of state graphs. B adequate if the all B graph G B contains no 1 edge loops. A link diagram is adequate if it is both A and B adequate. A link that admits an adequate diagram is also called adequate. We recall that the T uraev genus of a knot diagram D = D(K) is defined by (2) g T (D) = (2 v A (D) v B (D) + c(d))/2, where v A (D) = s A is the number of disjointly embedded circles in the all-a state of D, and similarly v B (D) = s B is the number of disjointly embedded circles in the all-b state of D. The Turaev genus of a knot K is defined by (3) g T (K) = min {g T (D) D = D(K)}. Abe [1, Theorem 3.2] and Manturov [23] have shown that if D is an adequate diagram of a knot K, then we have g T (K) = g T (D) = (2 v A (D) v B (D) + c(d))/2. The following characterization of adequate knots is shown in [17]. Theorem 3.2. For a knot K let js K, js K and jx K, jx K be the sets associated to J K (n) as in Definitions 2.2 and 2.3. Also let c(k) and g T (K) denote the crossing number and the Turaev genus of K. Then, K is adequate if and only if (1) There are Jones slopes s js K and s js K, with s s = 2c(K), and (2) there are x jx K and x jx K with x x = 2(2 2g T (K) c(k)). Assuming the Strong Slope Conjecture, Theorem 3.2 leads to a characterization of adequate knots in terms of Jones slopes and essential spanning surfaces. More specially, as discussed in [17], Theorem 3.2 implies the following: A knot K is adequate if and only if it admits Jones slopes s js K and s js K, that are realized by essential spanning surfaces S, S, such that (4) s s = 2c(K) and χ(s) + χ(s ) + c(k) = 2 2g T (K). The proof of the Strong Slope Conjecture for adequate knots [18] shows they satisfy equations (4). Alternating knots are exactly adequate knots with Turaev genus zero. In this case, the characterization of (4) reduces to the following: A knot K is alternating

6 6 if and only if it admits Jones slopes s, s, that are realized by essential spanning surfaces S, S, such that (5) (s s )/2 + χ(s) + χ(s ) = 2 and s s = 2c(K). The proof of the Slope Conjecture for alternating knots shows that they satisfy equation (5). Conversely, work of Howie [12] implies that knots that satisfy equation (5) are alternating, providing additional evidence supporting the Strong Slope Conjecture. The following problem is still open. Problem 3.3. Show that a knot K is adequate if and only if it admits Jones slopes s, s, that are realized by essential spanning surfaces S, S, such that s s = 2c(K) and χ(s) + χ(s ) + c(k) = 2 2g T (K). We finish this subsection with examples of non-adequate knots of period one. Example 3.4. From [20], we have that if a pretzel knot K = P (1/r, 1/s, 1/t) is such that r < 0, s, t > 0 odd, 2 r < s and 2 r < t, then p = 1 and js K = {0}, where 0 = 2r +2r = 2c +2r, and js K = { 2(s+t)}, where 2(s+t) = 2c +. In [22] it is shown that the pretzel diagram specified by P (1/r, 1/s, 1/t) is a minimum-crossing diagram for the knot [22, Lemma 8], so c(k) = r +s+t. By Theorem 3.2, K admits an adequate diagram if and only if there are Jones slopes s js K and s js K with s s = 2c(K) = 2(r + s + t). However, since js K = { 2(s + t)} and js K = {0}, we have that s s = 0 ( 2(s + t)) = 2c(K) 2r. This shows that there do not exist Jones slopes s js K, s js K where s s = 2c(K), so K is not adequate Period and number of sheets. We show that the Strong Slope Conjecture implies a relationship between the number of sheets of a Jones surface for a knot K, its Euler characteristic, and the period of the colored Jones polynomial of the knot. Lemma 3.5. Suppose that K S 3 is a knot of Jones period p. Let a/b js K js K be a Jones slope and let S be a corresponding Jones surface. Then b divides p 2 and b S divides 2pχ(S). In particular, if p = 1 then all the Jones slopes of K are integral and for every Jones surface we have 2χ(S) Z. S Proof. Suppose, for notational simplicity, that a/b js K and thus S corresponds to the highest degree 4d + [J K (n)] = 4a(n)n 2 + 4b(n)n + 4c(n). The case a/b js K is completely analogous. The claim that b divides p 2 is shown in [7, Lemma 1.11]. By above discussion we can assume that for n >> 0 and for every integer k > 0 we have 4a(n) = 4a(n + kp) = a b and 4b(n) = 4b(n + kp) = 2χ(S) S b,

7 while 4c(n + kp) = 4c(n). Furthermore we have d + [J K (n)] and d + [J K (n + kp)] are integers for all k > 0 as above. If S b = 1 then there is nothing to prove. Otherwise, we choose n so that it is a multiple of b S, say n = mb S. The difference 4d + [J K (n + p)] 4d + [J K (n)] is then 4d + [J K (n + p)] 4d + [J K (n)] = 2a S mp + ap2 b + 2pχ(S) b S, which must be an integer. Since a S mp is an integer, the number ap 2 b + 2pχ(S) b S is also an integer. Multiplying through by b gives that S divides 2pχ(S), and b divides p 2 gives that ap2 b is an integer so b S divides 2pχ(S). It turns out that for all knots where the Strong Slope Conjecture is known and the Jones period is calculated, for each Jones slope we can find a Jones surface where the number of sheets of a Jones surface b S actually divides the period of the knot. This leads us to give the following definition. Definition 3.6. We call a Jones surface S of a knot K characteristic if the number of sheets of S divides the Jones period of K. Example 3.7. An adequate knot has Jones period equal to 1, two Jones slopes and two corresponding Jones surfaces each a single boundary component [4]. By the proof of [18, Theorem 3.9], this property also holds for iterated cables of adequate knots. Thus in all the cases, we can find characteristic Jones surfaces. Note, that for adequate knots the characteristic Jones surfaces are spanning surfaces that are often non-orientable. In these cases the double surface is also a Jones surface but it is no-longer characteristic since it has two boundary components. Example 3.8. Table 1 gives the Jones period, the Jones slopes, and the numbers of sheets of a corresponding characteristic Jones surfaces, for all non-alternating knots up to nine crossings. The Jones slopes and Jones period in the table are compiled from [7]. The Jones surface data for all examples, but 9 47 and 9 49, are obtained from [18]. The proof that the knots 9 47 and 9 49 satisfy the Strong Slope Conjecture was recently completed by Howie [13]. Example 3.9. By [18], the Jones slopes of a (p, q)-torus knot K = T (p, q) are pq and 0, with Jones surfaces an annulus and a minimum genus Seifert surface, respectively. The Jones period of K is 2 and thus both Jones surfaces are characteristic. By the proof of [18, Theorem 3.9], this property also holds for iterated torus knots. 7

8 8 Knot js K S χ(s) b S js K S χ(s ) b S p 8 19 {12} {0} {8/3} {-10 } {1} {-12} {6} {-8} {32/3} {-4} {14/3} {-10} {1} {-14} {2} {-12} {9} {-6} {11} {-4} {15} {0} Table 1. Eight and nine crossing non-alternating knots Example Consider the pretzel knot K = P (1/r, 1/s, 1/t) where r, s, t are odd, r < 0, and s, t > 0. If 2 r < s and 2 r < t, then we can find Jones surfaces which are spanning surfaces of K [20] for each Jones slope. Thus their numbers of sheets is 1 and clearly divides p. If r > s or r > t, the Jones period is equal to p = 2 + s + t, the Jones ( ) 2 1 st slopes are given by s = s + t r and s = 2(s + t). A Jones surface with boundary slope equal to s has number of sheets the least common multiple of fractions with denominator p = 2 + s + t, reduced to lowest terms, and a Jones 2 surface with boundary slope equal to s is a spanning surface of K. In both cases the number of sheets divides the period and hence the Jones surfaces are characteristic. For example, the pretzel knot P ( 1/101, 1/35, 1/31) has a Jones slope s = 1345/8 and realized by a Jones surface with number of sheets 32. This means that the number of boundary components is 4. The Jones period is 32. This is an interesting example where both b and S are not equal to 1. Yet another interesting example comes from this family the pretzel P ( 1/101, 1/61, 1/65), which has p = 62. It has a Jones slope js K = 4280/31 from a Jones surface with number of sheets 31, which divides the Jones period 62, but is not equal to it. We note that currently there are no examples of knots which admit multiple Jones slopes for either d + [J K (n)] or d [J K (n)]. That is, in all the known cases the functions a(n), a (n) are both constant. One may ask the following Question Are there knots for which a(n), a (n) are not constant functions? That is, is there a knot K that admits multiple Jones slopes for d + [J K (n)] or d [J K (n)]?

9 The discussion above and examples also raise the following question. Question Is it true that for every Jones slope of a knot K we can find a characteristic Jones surface? 4. Haken sums for Jones surfaces In this section we show that there is a normal surface theory algorithm to decide whether a given knot satisfies the Strong Slope Conjecture. Here we will briefly recall a few facts about normal surfaces. For more background and terminology on normal surface theory, the reader is referred to [24], [16], or the introduction of [15]. Let M be a 3-manifold with a triangulation T consisting of t tetrahedra. A property embedded surface S is called normal if for every tetrahedron, the intersection S consists of triangular or quadrilateral discs each intersecting each edge of the tetrahedron in at most one point and away from vertices of T. There are seven normal isotopy classes of normal discs, four are triangular and three are quadrilateral; these are called disc type. Thus we have total of 7t normal discs in T. Fixing an order of these normal discs, D 1,..., D 7t, then S is represented by a unique (up to normal isotopy) 7t-tuple of non-negative integers n(s) = (y 1,..., y 7t ), where y i is the number of the discs D i contained in S. Definition 4.1. Two normal surfaces S 1, S 2 are called compatible if they do not contain quadrilateral discs of different types. Given compatible normal surfaces S 1, S 2 one can form their Haken sum S 1 S 2 : This is a geometric sum along each arc and loop of S 1 S 2 and it is uniquely determined by the requirement that the resulting surface S 1 S 2 be normal in T. See Figure 2. 9 Figure 2. The Haken sum of compatible normal surfaces S 1 and S 2. If n(s 1 ) = (y 1,..., y 7t ) and n(s 2 ) = (y 1,..., y 7t), then n(s 1 S 2 ) = n(s 1 ) + n(s 2 ) = (y 1 + y 1,..., y 7t + y 7t),

10 10 and χ(s 1 S 2 ) = χ(s 1 ) + χ(s 2 ). Figure 3. The resulting normal curves from the Haken sum. Conversely, given a 7t-tuple of non-negative integers n, we can impose constraints on the y i s so that it represents a unique up to isotopy normal surface in T. These constraints are known as normal surface equations. Definition 4.2. A normal surface S is called fundamental if n(s) cannot be written as a sum of two solutions to the normal surface equations. There are only finitely many fundamental surfaces and there is an algorithm to find all of them. Furthermore, all normal surfaces can be written as a finite sum of fundamental surfaces [9]. Theorem 4.3. Given a knot K with known sets js K js K, jx K jx K and Jones period p, there is a normal surface theory algorithm that decides whether K satisfies the Strong Slope Conjecture. Proof. There is an algorithm to determine whether M K = S 3 \K is a solid torus and thus if K is the unknot [9, 16]. If K is the unknot then the Strong Slope Conjecture is known and we are done. If K is not the unknot then we can obtain a triangulation T 1 of the complement M K = S 3 \ K together with a meridian of M K that is expressed as a path that follows edges of T 1 on M K. A process for getting this triangulation is given in [10, Lemma 7.2]. Apply the algorithm of of Jaco and Rubinstein [14, Proposition 5.15 and Theorem 5.20] to convert T 1 to a triangulation T that has a single vertex (a onevertex triangulation) and contains no normal embedded 2-spheres. The algorithm assures that the only vertex of the triangulation lies on M K. Then we can apply the process known as layering a triangulation to alter the edges on M K till the meridian becomes a single edge in the triangulation (see [15]). We will continue to denote this last triangulation by T and we will use µ to denote the single edge corresponding to the meridian of K.

11 11 For notational simplicity we will work with js K and jx K as the argument for js K and jx K is completely analogous. Fix a Jones slope a/b js K, with b > 0 and gcd(a, b) = 1, and suppose that we have Jones surfaces corresponding to it. Let S be such a surface with β := χ(s) S b jx K. By Lemma 3.5, S b divides 2pχ(S), where p is the Jones period of K. Thus (6) 2pχ(S) λ S b = 0 where λ = 2pβ Z. Lemma 4.4. Suppose that S is a Jones surface with boundary slope s := a/b js K, Jones period p, and with β = χ(s) S b jx K and λ as defined in (6). Then, exactly one of the following is true: (1) There is a Jones surface Σ with boundary slope a/b js K and with (7) χ(σ) Σ b = β = λ 2p, that is also a normal fundamental surface with respect to T. (2) There is a nonempty set EF s of essential surfaces that are normal fundamental with respect to T, have boundary slope s and such that we have for every F EF s. 2pχ(Σ) λ Σ b 0, Proof. Let S be a Jones surface as above. Any essential surface in M K may be isotoped to a normal surface with respect to above fixed T. Moreover, this normal surface S may be be taken to be minimal in the sense of [24, Definition 4.1.6]: This means that the surface minimizes the number of intersections with the 1-skeleton T 1 of T in the (normal) isotopy class of the surface. The significance of this minimality condition is the following: By [24, Corollary ], applied to (M K, µ), if S can be written as a Haken sum of non empty normal surfaces then each of these summands is essential in M K. (See also [15, Theorem 5.1]). Suppose that S is not fundamental. Then S can be represented as a Haken sum (8) S = Σ 1... Σ n F 1... F k, where each Σ i is a fundamental normal surface with boundary, and each F i is a closed fundamental normal surface. A theorem of Jaco and Sedgwick [15] states that each Σ i has the same boundary slope as S. As said earlier we have (9) χ(s) = χ(σ 1 ) χ(σ n ) + χ(f 1 ) χ(f k ). Recall that the number of sheets of a surface S, that is properly embedded in M K, is the number of intersections of S with the edge µ. We also recall that the boundary of a Haken sum S 1 S 2 is obtained by resolving the double points in

12 12 S 1 S 2 so that the resulting curves are still normal. In particular, the homology class of (S 1 S 2 ) is the sum of the homology classes of S 1 and S 2 in H 1 ( M K ). This implies that the number of intersections of (S 1 S 2 ) with µ is the sum of the numbers of intersection of S 1 and S 2 with µ. Thus by (8) we obtain (10) S b = Σ 1 b Σ n b. As said above, [24, Corollary ] shows that Σ i must be essential, for all i = 1,..., n. If for some i we have 2pχ(Σ i ) λ Σ i b = 0, then Σ := Σ i is a Jones surface as claimed in (1) in the statement of the lemma. Otherwise we have (11) 2pχ(Σ i ) λ Σ i b 0, for all 1 i n and option (2) is satisfied. To continue suppose that there exist Jones surfaces S, with boundary slope s := a/b, with β = χ(s), and λ defined as in (6), but there are no such surfaces that S b are fundamental with respect to T. Then, by Lemma 4.4, we have a set EF s of properly embedded essential surfaces in M K such that for every Σ i EF s we have: Σ i has boundary slope s and is a normal fundamental surface with respect to T ; and we have 2pχ(Σ i) λ Σ i b 0. By the proof of Lemma 4.4, a Jones surface S as above is a Haken sum of essential fundamental surfaces (12) S = ( i n i Σ i) ( j m j F j ), and where Σ i EF s, the F j s are closed surfaces and n i, m j 0 are integers. We have χ(s) = i χ(σ i)n i + j χ(f j )m j, S b = i Σ i bn i. Multiplying the first equation by 2p, the second by λ and subtracting we obtain (13) x(σ i)n i + 2p χ(f j )m j = 0 i j where x(σ i) := 2pχ(Σ i) λ Σ i b 0. Thus the vector n : = (n 1,..., m 1,...) corresponds to a solution of the homogeneous equation (13), with non-negative integral entries. We recall that a solution

13 vector n with non-negative integer entries, for equation (13), is called fundamental if it cannot be written as a non-trivial sum of solution vectors with non-negative integer entries. For any system of linear homogeneous equations, there is a finite number of fundamental solutions that can be found algorithmically, and every solution is linear combination of fundamental ones (see, for example, [24, Theorem 3.2.8]). Lemma 4.5. Suppose that there is a Jones surface S corresponding to boundary slope s which satisfies equation (6). Suppose moreover that there are no normal fundamental surfaces with respect to T that are Jones surfaces satisfying (7). Then, there is a Jones surface Σ, with boundary slope s and χ(σ ) Σ b = β = λ, such that 2p Σ = i k i Σ i ( j l j F j ), where k = (k 1,..., l 1,..., ) is a fundamental solution of equation (13). Proof. By assumption we have a Jones surface S of the form shown in equation (12) corresponding to a solution n with non-negative integer entries of equation (13). If n is not fundamental then, n = k + m where k is fundamental and m a non-trivial solution with non-negative integer entries of equation (13), corresponding to normal surfaces Σ and Σ via equation (12). We have S = Σ Σ. In order for Σ to be a Jones surface it is enough to see that Σ is essential. But this follows by [24, Corollary ] as noted earlier. Now we finish the proof of the theorem: Given a knot K with known sets js K js K and jx K jx K, to check whether it satisfies the Strong Slope Conjecture we need to check that the elements in js K js K are boundary slopes and to find Jones surfaces for all these slopes. To use Lemma 4.4, we need to know the fundamental normal surfaces with respect to triangulation T fixed in the beginning of the proof. There are finitely many fundamental surfaces in M K and there is an algorithm to find them [9]. Let F = {F 1,..., F k } denote the list of all fundamental surfaces. There is an algorithm to compute χ(f ) for all surfaces F F, and to compute their boundary slopes of the ones with boundary [16]. Let A = {a 1 /b 1,..., a s /b s } denote the list of distinct boundary slopes of the surfaces in F, where (a i, b i ) = 1 and b i > 0. Now proceed as follows: (1) Check whether js K A and js K A. If one of the two inclusions fails then K does not satisfy the Slope Conjecture. (2) If F contains no closed surfaces move to the next step. If we have closed surfaces we need to find any incompressible ones among them. There is an algorithm that decides whether a given 2-sided surface is incompressible and boundary incompressible if the surface has boundary. See [24, Theorem ] or [2, Alorithm 3]. Apply the algorithm to each closed surface in F 13

14 14 to decide whether they are incompressible. Let C F denote the set of incompressible surfaces found, that have genus bigger than one. (3) For every s := a/b js K A consider the set F s F that have boundary slope a/b. By [15] we know that F s. Decide whether F s contains essential surfaces and find them. Note that the surfaces in F s may not be 2-sided. To decide that an 1-sided surface is essential one applies the incompressibility and -algorithm to the double of the surface. Let EF s denote the set of essential surfaces found. If EF s = then K fails the conjecture. (4) For every λ 2pjx K and every F EF s calculate the quantity x(f ) := 2pχ(F ) λb F. Suppose that there is F EF s with x(f ) = 0. Then any such F is a Jones surface corresponding to s. (5) Suppose EF s := {Σ 1,..., Σ r} and that we have x(f ) 0, for all F EF s. Then consider equation (13) x(σ 1)n x(σ r)n r + 2pχ(C 1 )m pχ(C t )m t = 0, where C i runs over all the surfaces in C. Find and enumerate all the fundamental solutions of the equation. Among these solutions pick the admissible ones: That is solutions for which, for any incompatible pair of surfaces in C EF s, at most one of the corresponding entries in the solution should be non-zero. Hence pairs of non-zero numbers correspond to pairs of compatible surfaces. Every admissible fundamental solution represents a normal surface. By Lemma 4.5, we need only to check if one of these surfaces is essential. If a surface in this set is essential, then it is a Jones surface, otherwise, K fails the Strong Slope Conjecture. (6) For every a/b js K A repeat steps (3)-(5) above and run the analogous process for the Jones slopes in js K. Remark 4.6. Suppose that Question 3.12 has an affirmative answer: That is for every Jones slope s := a/b there is a Jones surface S, with χ(σ) = β Σ b, for some β jx K jk, that is characteristic (i.e. Σ b divides the period p). Thus Σ b p and we obtain (14) χ(σ) + Σ b (1 β)p. Now [24, Theorem ], applied to (M K, µ) implies that there are finitely many essential surfaces in M K that satisfy (14) and they can be found algorithmically. Using this observation, one can see that a positive answer to Question 3.12 will lead to an alternative algorithm for finding Jones surfaces than the one outlined above.

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