THE SLOPE CONJECTURE FOR MONTESINOS KNOTS

Transcription

1 THE SLOPE CONJECTURE FOR MONTESINOS KNOTS STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN Abstract. The Slope Conjecture relates the degree of the colored Jones polynomial of a knot to boundary slopes of incompressible surfaces. We develop a general approach that matches a state-sum formula for the colored Jones polynomial with the parameters that describe surfaces in the complement. We apply this to Montesinos knots proving the Slope Conjecture for Montesinos knots, with some restrictions. Contents. Introduction 2.. The Slope Conjecture and the case of Montesinos knots 2.2. Our results 3.3. Plan of the proof 5 2. Rational tangles 5 3. Essential surfaces of Montesinos knots Every essential surface is carried by a branched surface Essential surfaces for a rational knot Edge-paths and candidate surfaces for Montesinos knots The boundary slope of a candidate surface The Euler characteristic of a candidate surface 4. The colored Jones polynomial of pretzel knots 4.. Outline of the proof of Theorem Conventions for representing a Kauffman state Simplifying the state sum and pyramidal position for crossings Minimal states are taut and their degrees are δ(n, k) Enumerating all taut states Adding up all taut states in st(c, k) Proof of Theorem Quadratic integer programming Quadratic real optimization Quadratic lattice optimization Application: the degree of the colored Jones polynomial Matching the growth rate to the topology Proof of Theorem.2 for pretzel knots 3 7. The colored Jones polynomial of Montesinos knots The TR-move Montesinos state sum Special Montesinos knot case 33 Date: January 3, Mathematics Classification. Primary 57N0. Secondary 57M25. Key words and phrases: knot, Jones polynomial, Jones slope, quasi-polynomial, pretzel knots, incompressible surfaces.

2 2 STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN 7.4. The general case Matching the boundary slope and Euler characteristic Proof of Theorem.3 37 Appendix A. Background on the Temperley-Lieb algebra and the Jones-Wenzl idempotent 38 Acknowledgments 40 References 4. Introduction.. The Slope Conjecture and the case of Montesinos knots. The Slope Conjecture relates one of the most important knot invariants, the colored Jones polynomial, to incompressible surfaces in the knot complement [Garb]. More precisely, the growth of the degree as a function of the color determines boundary slopes. Understanding the topological information that the polynomial detects in the knot is a central problem in quantum topology. The conjecture suggests the polynomial can be studied through surfaces, which are fundamental objects in 3-dimensional topology. Our philosophy is that the connection follows from a deeper correspondence between terms in an expansion of the polynomial and surfaces. This would potentially lead to a purely topological definition of quantum invariants. The coefficients of the polynomial should count isotopy classes of surfaces, much like in the case of the 3d index [?] and work of [?]. As a first test of this principle, we focus on the Slope conjecture for Montesinos knots. In this case Hatcher-Oertel [HO89] provides a description of the set of incompressible surfaces of those knots, in particular, an effective algorithm to compute the set of boundary slopes of incompressible surfaces in such knots. We provide a state-sum formula for the colored Jones polynomial, that allows us to match the parameters of the terms of the sum that contribute to the degree of the colored Jones polynomial with the parameters that describe the locally incompressible surfaces. We interpret the curve systems on a Conway sphere enclosing a rational tangle in terms of skein elements. This is specialized to Montesinos knots in this paper, but may be done for all knots. In fact using this framework, one can determine the degree of the colored Jones polynomial and find candidates for corresponding incompressible surfaces in many new cases beyond Montesinos knots. The key innovation of our state sum is that we are able to identify the optimal configurations. They are organized in a way that makes their total contributions to the degree apparent. The resulting degree function is piecewise-quadratic, allowing application of quadratic integer programming methods. Although the local theory works in general, fitting together the surfaces in each tangle to obtain a (globally) incompressible surface has yet to be done. The behavior of the colored Jones polynomial under gluing of tangles has similar patterns, which may be explored in future work. The Montesinos knots are those which together with some well-understood algebraic knots have small Seifert fibered 2-fold branched cover [Mon73, Zie84]. For our purposes, we will not use this abstract definition, and instead use the description of Montesinos links as a

3 THE SLOPE CONJECTURE FOR MONTESINOS KNOTS 3 combination of 2-bridge knots and pretzel knots. More precisely, a Montesinos link is the closure of a list of rational tangles arranged as in Figure and concretely in Figure 2. r 0 r r m e Figure. A Montesinos link. Figure 2. The Montesinos link K( 3, 3 0, 4, 2 7 ). Rational tangles are parametrized by rational numbers, see Section 2, thus a Montesinos link K(r 0, r,..., r m ) is encoded by a list of rational numbers r j Q. Note that K(r 0, r,..., r m ) is a knot if and only if either there is only one even denominator, or, there is no even denominator and the number of odd numerators is odd. When r i = /q i is the inverse of an integer, the Montesinos knot K(/q 0,..., /q m ) is also known as the P (q 0,..., q m ) pretzel knot. Since the Strong Slope Conjecture is known for adequate knots [Garb, FKP, FKP3], we will ignore the Montesinos knots which are adequate. The remaining ones are the knots with precisely one negative tangle or precisely one positive tangle [LT88, p.529]. The Slope Conjecture and its refinement, the Strong Slope Conjecture [KT5], were established for many knots including alternating knots, adequate knots, torus knots, knots with at most 9 crossings, 2-fusion knots, graph knots, near-alternating knots, and most 3-tangle pretzel knots and 3-tangle Montesinos knots [Garb, FKP, GvdV6, LvdV6, MT7, Lee, LLY, How]. However the general case remains intractable and most proofs simply compute the quantum side and the topology side separately, comparing only the end results..2. Our results. Recall the colored Jones polynomial J K,n (v) Z[v ±2 ] of a (0-framed) knot K colored by the n-dimensional irreducible representation of sl 2 [Tur88]. Our variable v for the colored Jones polynomial is related to the skein theory variable A and to the Jones variable q [Jon87] by v = A = q 4. With our conventions, if 3 = P (,, ) denotes the left-hand trefoil, then J 3,2(v) = v 8 v 0 v 6 v 2. For the n-colored unknot we get J O,n = v2n v 2n v 2 v 2.

4 4 STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN Let δ K (n) denote the maximum v-degree of the colored Jones polynomial J K,n (v). It follows that δ K (n) is a quadratic quasi-polynomial [Gara]. In other words, for large n >> N K it can be written in the form where js K, jx K, and c K are periodic functions. δ K (n) = js K (n)n 2 + jx K (n)n + c K (n) () Conjecture.. (The Strong Slope Conjecture) For any knot K and any n there is an n and an incompressible surface S in S 3 \ K with boundary slope equal to js K (n) and 2χ(S) = jx #S K(n ). Here #S is the number of sheets of S. We call a value of the function js K a Jones slope and a value of the function jx K a normalized Euler characteristic, see [Garb] and [KT5] for additional background on the conjecture. By considering the mirror image K of K and the formula J K,n (v ) = J K,n (v), the Strong Slope Conjecture is equivalent to the statement in [KT5] that includes the behavior of the minimal degree. Before stating our main result on Montesinos knots we start with the special case of pretzel knots as they are the basis for our argument. In fact Theorem.2 is the bulk of our work. For P (q 0,..., q m ) to be a knot, at most one tangle has an even number of crossings, and if each tangle has an odd number of crossings, then the number of tangles has to be odd. In the theorem below, the condition on the parity of the q i s and the number of tangles may be dropped if one is willing to exclude an arithmetic sub-sequence of colors n. Theorem.2. Fix odd integers q 0,..., q m with m 2 even and q 0 < < < q,..., q m. Let P = P (q 0,..., q m ) denote the corresponding pretzel knot. Define rational functions s(q), s (q) Q(q): m s(q) = + q 0 + m i= (q i ), s i= (q) = (q i + q 0 2)(q i ) m i= (q. (2) i ) (a) If s(q) < 0, then the Strong Slope Conjecture holds with js P = 2s(q), jx P = 2s (q) + 4s(q) 2(m ). (3) (b) If s(q) = 0, then the Strong Slope Conjecture holds with { 2(m ) if s (q) 0 js P = 0, jx P = 2s (q) 2(m ) if s (q) < 0. (4) (c) If s(q) > 0, then the Strong Slope Conjecture holds with js P = 0, jx P = 2(m ). (5) Next, we consider the case of Montesinos knots. Recall that every rational number r has a unique positive continued fraction expansion r = [a 0,..., a lr ] of even length l r, see (6). This allows us to define r[j] = a j for j = 0,..., l r, and r[j] = 0 for j > l r. Let r e/o = l r j=3,j=even/odd r[j], so r = l r j=3 r[j] = r e + r o.

5 THE SLOPE CONJECTURE FOR MONTESINOS KNOTS 5 The next theorem is our main result. Here (r 0,..., r m ) Q m+ denotes a tuple of rational numbers and (q 0,..., q m ) Z m+ denotes the corresponding tuple of integers where q i = r i []. Again, the condition on the parity of the q i s and the number of tangles may be dropped if one is willing to exclude an arithmetic sub-sequence of colors n, thus proving a weaker version of the conjecture for all Montesinos knots. Theorem.3. Let K = K(r 0, r,..., r m ) be a Montesinos knot such that r 0 < 0, r i > 0 for all i m with m 2 even. Let { q i } m i=0 be defined by q 0 = q 0 and q i = r i [] + for i m. Suppose q 0 < < < q,..., q m are odd. Let P = P ( q 0, q,..., q m ) be the pretzel knot, and let ω(k) denote the writhe of K. Then the Strong Slope Conjecture holds with js K = js P r 0 [2] r 0 + (r i [2] ) + r i ω(p ) + ω(k), jx K = jx P r 0 o 2 Example.4. The Montesinos knot K( 46 = , 35 5 = i= (r i [2] ) 2 i=, 5 =, r i e. i= = , = ) is shown 5 4+ with the even-length continued fraction expansions, and has the associated pretzel knot K(,,,, ) The pretzel knot has a Jones slope js K = 72/7, and jx K = 22/7. Thus, the Montesinos knot has a rational Jones slope 00/7 = 72/ ( 3) }{{} ω(p ) and jx K = 22/ ( 4) 2( ) 2(2 + ) = 374/7. 43 }{{} ω(k) Note that when m 2, the Montesinos knot K(r 0, r,..., r m ) with precisely one negative/positive tangle is A/B-adequate, hence we need only to consider js K and jx K for a Montesinos knot with precisely one negative tangle..3. Plan of the proof. We first prove the pretzel knot case, Theorem.2. This is done in four steps. First we work out the relevant surfaces using the Hatcher-Oertel algorithm in Section 3. Next we use a mix of skein theory and fusion to find a formula for the degree of the dominant terms in the resulting state sum for the colored Jones polynomial in Section 4. Using quadratic integer programming techniques we determine the maximal degree of these dominant terms in Section 5. Finally in Section 6 we match the growth rate of the degree of the quantum invariant with the topology, using the analogy drawn between the parameters of the state sum and the parameters for the Hatcher-Oertel algorithm by Lemma 6.. The general case of Theorem.3 is then reduced to the pretzel case in Section Rational tangles Let us recall how to parametrize rational tangles by rational numbers and their continued fraction expansion. This material is well-known and may be found for instance in [KL04, BS]. Our building blocks of rational tangles are the horizontal and the vertical tangles.

6 6 STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN A horizontal tangle has n horizontal half-twists (i.e., crossings) for n Z. 2 2 A vertical tangle has n vertical half-twists (i.e., crossings) for n Z. 2 2 Tangles can be added and multiplied (where addition is denoted by + and multiplication is denoted by ) as follows. T S T T + S S T S Tangle addition and multiplication follow the rules of addition and multiplication of rational numbers. Recall the notation of the (positive) continued fraction expansion [KL04, BS]: [a 0,..., a l ] = a 0 + a + a a l for integers a i. If r is a positive (resp. negative) rational number, then it has a unique positive continued fraction expansion where a i > 0 (resp. a i < 0) and a l > (resp. a l < ). In that case, we define a 2 + r[i] = a i, l r = l for r = [a 0,..., a l ]. (7) It will be useful to introduce the negative continued fraction expansion [BS, Ch.3] [[a 0,..., a l ]] = [a 0, a,..., ( ) l a l ] = a 0 a a 2 a 3 a l (6). (8)

7 THE SLOPE CONJECTURE FOR MONTESINOS KNOTS 7 Given a rational number r with the unique positive continued fraction expansion [0, a, a 2,..., a l, a l ] consider the rational tangle T r defined by: T r = (((a l ( a l ) + a l 2 ) a ) + 0. (9) Figure 3. The rational tangle T 3/36 corresponding to the continued fraction expansion [0, 2,, 3, 3] = 3/ Essential surfaces of Montesinos knots In this section we briefly describe the Hatcher-Oertel algorithm highlighting the features that will be important for the correspondence between the Jones slope and the boundary slope of an essential surface in the complement of a Montesinos knot. We will follow the conventions of [HO89]. 3.. Every essential surface is carried by a branched surface. A branched surface B, a notion originally due to Haken [Hak6], in a 3-manifold is a subspace locally homeomorphic to the space as shown in the following figure. Figure 4. Local picture of a branched surface, with the blue lines indicating the singularities. Every properly embedded surface S in a 3-manifold may be isotoped so that it runs nearly parallel to a branched surface B. In this case we say that S is carried by the branched surface. The number of parallels of the surface S along each component of the complement of the branched locus of B determines an integer weight. Conversely, it suffices to have the branched surface and integer weights in order to describe a surface carried by it. Floyd and Oertel [FO84] showed that essential surfaces are carried by finitely many branched surfaces in a Haken manifold.

8 8 STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN Theorem 3.. [FO84, Theorem ] Let M be a Haken 3-manifold with incompressible boundary. There are a finite number of branched surfaces B,..., B k, properly embedded in M such that (a) each surface carried with positive weights by one of the B i s is essential, and (b) every two-sided essential surface in M is isotopic to a surface carried by one of the B i s with positive weights. Given the finite list of branched surfaces B,..., B k, it is then possible to enumerate all essential surfaces by enumerating the set of positive weights on each of B,..., B k, and then determining which one of these gives a connected surface that is essential Essential surfaces for a rational knot. In [HT85], Hatcher and Thurston classify all essential surfaces for a rational knot K(r) where r Q by determining the set of branched surfaces of K(r) which carry essential surfaces. They show that these branched surfaces of S 3 \ K(r) correspond to continued fraction expansions of r of the form r = [[b 0, b, b 2,..., b k ]], b i Z, and that each such continued fraction expansion determines an edge-path on a one-simplex D. Here, D is the Farey ideal triangulation of H 2 on which PSL 2 (Z) is the group of orientationpreserving symmetries, see Figure 5. Recall that the vertices of D are Q and we will denote a typical vertex of D by p for coprime integers p, q with q nonnegative. There is an q edge between two vertices p and r, denoted by p r, whenever ps rq =, and q s q s an edge between 0/ and /0. An edge-path is simply a path on the -skeleton of D which may have endpoints on an edge rather than on a vertex. Given a continued fraction expansion of r, the vertices of the corresponding edge-path are the sequence of partial sums [[b 0, b,..., b k ]], [[b 0, b,..., b k ]],..., [[b 0, b ]], [[b 0 ]]. Such an edge-path determines an essential surface in the exterior of K(r) as follows. We isotope the 2-bridge knot presentation of K(r) so that it lies in S 2 [0, ], with the two bridges intersecting S 2 in two slope /0 arcs, and the arcs of slope r lying in S 2 0. See [HT85, p. Fig. (b)]. The slope here is determined by the lift of those arcs to R 2, where S 2 i \ K is identified with the orbit space of Γ, the isometry group of R generated by 80 -degree rotation about the integer lattice points. Each vertex v of an edge-path determines a curve system on S 2 i v, i v (0, ), i v i v if v v, with the specified number of sheets (the number of intersections of the curve system with the punctures). A surface is constructed whose intersections with S 2 i v coincide with the curve system via Morse theory by adding saddles. For details, see [HT85] Edge-paths and candidate surfaces for Montesinos knots. Hatcher and Oertel [HO89] give an algorithm that provides a complete classification of boundary slopes of Montesinos knots by decomposing K(r 0, r,..., r m ) via a system of Conway spheres {S 2 i } m i=, each of which contains a rational tangle T ri. Their algorithm determines the conditions under which the essential surfaces in the exterior of each rational tangle, as classified by [HT85] and put in the form as discussed in the previous section, may be glued together across the system of Conway spheres to form an essential surface in S 3 \ K(r 0, r,..., r m ).

9 THE SLOPE CONJECTURE FOR MONTESINOS KNOTS 9 2/ / 2/3 /2 /0 3/ 2/ / 0/ /2 Figure 5. The -simplex D. To describe the algorithm, it is now necessary to give coordinates to curve systems on a Conway sphere. Hatcher and Oertel determine that the curve system S Si 2 for a connected surface S S 3 \ K(r 0, r,..., r m ) may be described by homological coordinates A i, B i, and C i as shown in Figure 6. A i A i C i B i Figure 6. The Conway sphere containing the tangle corresponding to r i and the curve system on it. They also consider edge-paths in an augmented -simplex ˆD in the plane obtained by splitting open D along the slope /0 and adjoining constant edge-paths p p. See q q [HO89, Fig..3]. Again an edge-path in ˆD is a path in the -skeleton of ˆD which may or may not end on a vertex. A point on an edge p is denoted by q K M p q + M K M r s. The curve system coordinates (A, B, C) corresponding to this point is obtained by taking a linear combination of the A, B, C-coordinates of p and r. There are two cases: q s If p r, then the curve system is given by K(, q, p) + (M K)(, s, r). q s If p = r, then the curve system is given by (M K, K + M(q ), Mp). q s The algorithm, implemented by Dunfield [Dun0], is as follows. () For each fraction r i, pick an edge-path γ i in the -simplex ˆD corresponding to a continued fraction expansion r s r i = [[b 0, b,..., b k ]], b i Z.

10 0 STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN (2) For each edge p r in γ q s i, determine the integer parameters {K i } 0, {M i } 0 satisfying the following constraints. (a) A i = A j and B i = B j for all the A-coordinates A i and the B-coordinates B i of the point K i p M i q + M i K i r M i s. (b) m i=0 C i = 0 where C i is the C-coordinate of the point K i p M i q + M i K i r M i s. The edge-paths chosen in () with endpoints specified by the solutions to (a) and (b) determine a candidate edge-path system {γ i } m i=0, corresponding to a connected surface S in S 3 \ K(r 0, r,..., r m ). We call this the candidate surface associated to a candidate edge-path system. (3) Apply incompressibility criteria [HO89, Prop. 2., Cor. 2.4, Prop ] to determine if a candidate surface is an incompressible surface and actually gives a boundary slope. Below, we will write S = {γ i } m i=0 to indicate a candidate surface associated to a candidate edge-path system {γ i } m i=0. Note that for a candidate edge-path system, M i is identical for i = 0,..., m by condition (2a) in the algorithm, so we will simply write M for M i for a candidate surface S. Recall from Section 3.2 that M is the number of arcs coming out of each puncture and is therefore the number of sheets of S. We will mainly be applying [HO89, Corollary 2.4], which we restate here. Note that for an edge p r with 0 < q < s, the -value (called the r-value in [HO89]) is 0 if p = r q s q s or if the edge is vertical, and the -value is q s when p r. q s Theorem 3.2. [HO89, Corollary 2.4] A candidate surface S = {γ i } m i=0 is incompressible unless the cycle of -values for the final edges of the γ i s is of one of the following types: {0,,..., m }, {,,...,, m }, or {,...,, 2, m } The boundary slope of a candidate surface. The twist number tw(s) for a candidate surface S = {γ i } m i=0 is defined as tw(s) := 2 m (s i s + i M ) = 2 (e i e + i ), i=0 where s i is the number of slope-decreasing saddles of γ i, s + i is the number of slope-increasing saddles of γ i, and M is the number of sheets of S. Let an edge be given by p r, we say q s that the edge decreases slope if r < p, and that the edge increases slope if r > p. In terms s q s q of edge-paths, tw(s) can be written in terms of the number e i of edges of γ i that decreases slope and e + i, the number of edges of γ i that increases slope as shown. For an interpretation of the twist number in terms of the lifts of these arcs in R 2 /Γ, see [HO89, p. 460]. If γ i has endpoint K i p + M K i r. M q M s i=0

11 THE SLOPE CONJECTURE FOR MONTESINOS KNOTS Then the final edge of γ i is called a fractional edge and counted as a fraction K i M. Finally, the boundary slope bs(s) of a candidate surface S is given by bs(s) = tw(s) tw(s 0 ) where S 0 is a Seifert surface that is a candidate surface from the Hatcher-Oertel algorithm The Euler characteristic of a candidate surface. We compute the Euler characteristic of a candidate surface S associated to an edge-path system {γ i } m i=0, where none of the γ i s are constant or end in /0 as follows. M is again the number of sheets of the surface S. We begin with 2M disks which intersect S 2 i 0 in slope r i = p i q i arcs in each B i. From left to right in an edge-path γ i, each non-fractional edge p r is constructed q s by gluing M number of saddles that change 2M arcs of slope p (representing the q intersections with Si 2 i p ) to slope r (representing the intersections with q s S2 i i r ), s therefore decreasing the Euler characteristic by M. A fractional final edge of γ i of the form p K p + M K r changes 2(M K) q M q M s out of 2M arcs of slope p to 2(M K) arcs of slope r via M K saddles, thereby q s decreasing the Euler characteristic by M K. This takes care of the individual contribution of an edge-path {γ i }. Now the identification of the surfaces on each of the 4-punctured sphere will also affect the Euler characteristic of the resulting surface. In terms of the common (A, B, C)-coordinates of each edge-path, there are two cases: The identification of hemispheres between neighboring balls B i and B i+ identifies 2M arcs and B i half circles. Thus it subtracts 2M + B i from the Euler characteristic for each identification. The final step of identifying hemispheres from B 0 and B m on a single sphere adds B i to the Euler characteristic. 4. The colored Jones polynomial of pretzel knots We will consider the standard diagram of the pretzel knot K = K(/q 0,..., /q m ), with q i >. Throughout the section the integer n 2 is fixed, and we will illustrate graphically using the example K( /5, /3, /3, /3, /5). We will also assume the standard material summarized in the Appendix on Kauffman state sums and the Temperley-Lieb algebra. To compute the colored Jones for a fixed n we have to take the n cable of K, insert a JW (Jones-Wenzl) idempotent and then take the Kauffman bracket. We write the colored Jones polynomial as J K,n+ = ( ) n ( v) ω(k)n(n+2) K n. Instead of computing the usual bracket or fusion state sum we use a customized state sum reflecting the splitting K = K K + where K is the negative twist region of q 0 crossings and K + is the rest of the knot viewed as a 2-2 tangle.

12 2 STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN K n + K n + K n + K n K n 2k 0 n n I k0 n Figure 7. From left to right: K n = K n K n +, doubling the idempotents, and I k0. It is convenient to compute the bracket of these 2-2 tangles first. For any tangle T write T n to mean cabling each component by a JW idempotent of order n and evaluating in the Temperley-Lieb algebra TL 2n 2n. We may write K n = k 0 G k0 (v)i k0 for tangles I k0 with four JW idempotents of size n connected in the middle to a JW idempotent of size 2k 0 arranged in an I-shape using the fusion and untwisting formulas, see the Appendix for more information. The other tangle does get computed in the standard Kauffman way, leaving the four JW idempotents of size n: K+ n = σ vsgn(σ) Tσ n. The state sum we consider consists of pairs (k 0, σ) and we write K n = G k0 (v)v sgn(σ) I k0 Tσ n (0) (k 0,σ) where the product means identifying the JW idempotents, see Figure 7. Using the notion of through strands, we collect like terms together in our state sum. Definition 4.. Consider the Temperley-Lieb algebra TL n n with n inputs and n outputs. Let T be an element of TL n n with no crossings. Viewing D2 as a square, an arc in T with one endpoint on the top boundary of the disc D 2 defining TL n n and another endpoint on the bottom boundary is called a through strand of T. We can organize states (k 0, σ) according to the number of through strands at various levels. The global number of through strands of σ, denoted by c = c(σ), is the number of through strands of the whole Temperley-Lieb element Tσ n in TL 2n 2n inside the box framed by four idempotents in K+, n see Figure 8 for an example. Figure 8. T n σ with c(σ) = 4. When restricting σ to the 4th twist region, we have c 4 (σ) = k 4 (σ) = 2. On the far right we show an example of a state σ where c 4 (σ) = and therefore k 4 (σ) =.

13 THE SLOPE CONJECTURE FOR MONTESINOS KNOTS 3 We will also define c i (σ) to be the number of ith local through strands when restricting σ to the ith twist region, that are also global through strands. The parameter for each twist region, k i, will be defined as k i (σ) = c i(σ). 2 With the notation k = (k 0,..., k m ) we set G c,k = G k0 (v)v sgn(σ) I k0 Tσ n. () k 0 σ:k i (σ)=k i,c(σ)=c We prove the following theorem. Theorem 4.2. Assume q i >. Referring to the above state sum K n = c,k G c,k we have the following. Note 0 k i n and define the parameters c, k to be tight if k 0 = k + +k m = c. For tight c, k we have G 2 c,k = ( ) q 0(n k 0 )+n+k 0 + P m i= (n k i)(q i ) v δ(n,k) + l.o.t. and δ(k, n) = ( ) 2 (q 0 + )k0 2 + (q i )ki 2 n(n + 2) + ( 2 + q 0 + q i )k i q i + (m )n (2) 2 i= i= If c, k is not tight then deg v G c,k < max c, k tight δ( k, n). This theorem will be used in the next section to find the actual degree using quadratic integer programming. 4.. Outline of the proof of Theorem 4.2. Let st(c, k) be the set of states (k 0, σ) with c(σ) = c and k i (σ) = k i for all i such that the parameters c, k are tight. A state in st(c, k) is said to be taut if its term G k0 (v)v sgn(σ) I k0 T n σ in () maximizes the v-degree within st(c, k). For any fixed tight c, k we plan to construct all taut states. The first examples of we construct will be minimal states, from which we will derive all taut states. A state in st(c, k) is minimal if it has the least number of A-resolutions. We will first show that minimal states are characterized by having a certain configuration on the set of crossings where they choose the A-resolution, called pyramidal. This will also be used to show that c, k not tight implies deg v G c,k < max c, k tight δ( k, n). Then, with the construction of all taut states from minimal states, we show that δ(n, k) is the maximal degree of a taut state with parameters k, and G taut c,k tight i=0 = ( ) q 0(n k 0 )+n+k 0 + P m i= (n k i)(q i ) v δ(n,k) + l.o.t., where Gc,k taut tight is the double sum of G c,k only over taut states with tight c, k. This will lead to G c,k tight = ( ) q 0(n k 0 )+n+k 0 + P m i= (n k i)(q i ) v δ(n,k) + l.o.t. and conclude Theorem 4.2. Conventions for representing a Kauffman state. Throughout the rest of Section 4, we will indicate schematically a crossing-less skein element S σ, resulting from applying a Kauffman state to a skein element S with crossings, by the following convention. Let S B be the result of applying the all-b state on the crossings of S. For a Kauffman state σ let A σ be the set of crossings of S on which σ chooses the A-resolution. The skein element S σ is represented by S B with colored edges, such that the edge in S B corresponding to a crossing The abbreviation l.o.t. means lower order terms in v.

14 4 STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN in A σ is colored red, and all other edges remain black. The skein element S σ may then be recovered by a local replacement of two arcs with a dashed segment. See Figure 9 below. Figure 9. A red edge indicates the state where the B-resolution replaces the A-resolution for a Kauffman state σ Simplifying the state sum and pyramidal position for crossings. We will denote by S(k 0, σ) the skein element I k0 T n σ as in (). Lemma 4.3. Fix (k 0, σ) determining a skein element S(k 0, σ) with k i = k i (σ) and c = c(σ). If k 0 > m i= k i, then S(k 0, σ) = 0. Proof. Note that m i= k i c. Thus if k 2 0 > m i= k i, then k 0 > c, and the lemma follows 2 from [Lee, Lemma 3.2]. With the information of through strands c(σ) and {k i (σ)}, we describe the structure of A σ for a Kauffman state σ. It is necessary to introduce a labeling of the crossings with respect to their positions in the all-b Kauffman state graph S(k 0, B) = I k0 TB n, where T B n is the all-b state on T n. We first further decompose T n = S t S w S b where is the multiplication by stacking in TL, and let the crossings contained in those skeins be denoted by C t, C w, and C b, respectively. See Figure 0 for an example. S t K n + 2k 0 S w S b n Figure 0. Skein element S = I k0 K( /5, /3, /3, /3, /5). (S t S w S b ) of the pretzel knot See Figure for a guide to the labeling. The skein element TB n consists of n arcs on top in the region defining S t, n arcs on the bottom in the region defining S b, and q i sets of n circles for the ith twist region in the region defining S w. The n upper arcs are labeled by S u,..., Sn, u and the n lower arcs are labeled by S, l..., Sn, l respectively. Cj u is the set of crossings whose corresponding segments in TB n lie between the arcs Su j and Sj+. u Similarly we define Cj l by reflection. For the crossings in the region defining S w, we divide each state circle into upper and lower half arcs as also shown in Figure 0, and use an additional label s for s q i. Thus, the notation C l,s i,j, where s q i for each twist region with q i crossings and j n

15 THE SLOPE CONJECTURE FOR MONTESINOS KNOTS 5 s = s = C u 4 C u C2 u C3 u S u S2 u S3 u S4 u... S l, S l,,4 S l,,3,2 S l,, S u,q, S u,q,2... C l, 2,3 C l, 2,2 C l, 2, s = q s = q 2 C l 3 C l 2 C l S4 l S3 l S2 l S l Figure. Labeling of crossings, arcs, and circles from applying the all-b state on T n. In this example n = 4. indicating a circle in the n-cable, means the crossings between the state circles S l,s i,j S l,s i,j+, see Figure. It is helpful to see a local picture at each n-cabled crossing in T n. and n S u 2 S3 u S3 l S u S l C u C 2 u Upper C3 u = C3 l Lower C 2 l C l x n S l 2 Figure 2. Local labeling of n 2 crossings on the all-b state of an n-cabled crossing. In this example n = 3. The goal of this subsection is to prove the following theorem. Theorem 4.4. Suppose a skein element S(k 0, σ) has parameters k i = k i (σ) and c = c(σ). Then the Kauffman state σ chooses the A-resolution on a set of crossings A σ A σ such that (a) A σ = c2 4 c 2 + m i= (k2 i + k i ) + m i= (q i 2)k 2 i.

16 6 STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN (b) Moreover, we have A σ = A t σ A w σ A b σ denoting the crossings in the regions determining S t, S w and S b, respectively, such that (i) A t σ = A b σ = c2 /4 c/2+ P m i= (k2 i +k i). The set A t 2 σ = n j=n c/2+ u j is a union of crossings u j Cj u, and the set A b σ = n j=n c/2+ l j is a union of crossings l j Cj l satisfying: For n c + j n, u 2 j (resp. l j ) has j n + c crossings. 2 For each n c + 2 j n and a pair of crossings x, 2 x in u j (resp. l j ) whose corresponding segments e, e in TB n are adjacent (i.e., there is no other crossing in u j whose corresponding segment is between e and e ), there is a crossing x in u j (resp. l j ), where the end of the corresponding segment e on Sj u (resp. Sj) l lies between the ends of e and e. (ii) A w σ = m i= (q i 2)ki 2. The set A w σ = m i= q i s= k i j= (us i,j l s i,j) is a union of crossings with u s i,j C u,s i,j and l s i,j C l,s i,j, such that For each n k i + j n, u s i,j, l s i,j each has j n + k i crossings. For each n k i + 2 j n and a pair of crossings x, x in u s i,j (resp. l s i,j) whose corresponding segments e, e in TB n are adjacent (i.e., there is no other edge in u s i,j between e and e ), there is a crossing x in u s i,j (resp. l s i,j ), where the end of the corresponding segment e on Si,j s lies between the ends of e and e. The set of crossings A σ is said to be in pyramidal position. Proof. (ii) is a direct application to every set of n-cabled crossings in each twist region of S w of the following result from [Lee]. Lemma 4.5. [Lee, Lem. 3.7] Let S be a skein element in TL 2n 2n consisting of a single n-cabled positive crossing x n with labels as shown in Figure 2. If S σ for a Kauffman state σ on x n has 2k through strands, then σ chooses the A-resolution on a set of k 2 crossings C σ of x n, where C σ = n j=n k (u j l j ) is a union of crossings u j C u j and l j C l j, such that For each n k + j n, u j, l j each has j n + k crossings. For each n k + 2 j n, and a pair of crossings x, x in u j (resp. l j ) whose corresponding segments c, c in the all-b state of x n are adjacent (i.e., there is no other edge in C σ between c and c ), there is a crossing x in u j (resp. l j ), where the end of the corresponding segment c on Sj u (resp. Sj) l lies between the ends of c and c. The same proof applies to the crossings in the strip S t, see Figure 3. Vertical reflection to S b will show (i). We will now apply what we know about the crossings on which a state σ chooses the A- resolution from Theorem 4.4 to construct degree-maximizing states for given global through strands c(σ) and parameters {k i (σ)}. See Figure 4 for an example of a pyramidal position of crossings.

17 THE SLOPE CONJECTURE FOR MONTESINOS KNOTS 7 n n n... n n n n n Figure 3. The arrow indicates the direction from left to right of the crossings in S t. 2k 0 2k 0 Figure 4. A minimal state τ is shown with n = 3 and c(τ) = 6 global through strands. From the representation on the left one can see the pyramidal position of the crossings A τ as described by Theorem 4.4. The skein element S(k 0, τ) with k = (k 0, 0, 0, 2, ) resulting from applying τ is shown on the right Minimal states are taut and their degrees are δ(n, k). The contribution of the state (k 0, σ) to the state sum is G k0 (v)v sgn(σ) I k0 T n σ as in (). We denote its v-degree by d(k 0, σ). Recall the skein element S(k 0, σ) = I k0 T n σ. Also recall A σ denotes the set of crossings on which σ chooses the A-resolution, and A σ is the number of crossings in A σ. A minimal state with tight parameters c, k (k 0 = k + +k m = c 2 ) has the least A σ. Let o(a σ ) denote the number of circles of S(k 0, σ), which is the skein element obtained by replacing all the JW idempotents in S(k 0, σ) by the identity, respectively. Lemma 4.6. A minimal state (k 0, τ) with c(τ) through strands and tight c, k has A τ in pyramidal position as specified in Theorem 4.4 and distance A τ from the all-b state given by ( ) A τ = 2 ( k i ) ( m i= k i ) k i (k i + ) + + (q i 2)ki Moreover, i= i= G k0 (v)v sgn(σ) I k0 T n σ = ( ) q 0(n k 0 )+n+k 0 + P m i= (n k i)(q i ) v δ(k,n) + l.o.t. (3) i=

18 8 STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN Proof. Observe that minimal states τ have corresponding crossings A τ in pyramidal position. Moreover, if A τ is pyramidal, then A τ determines the number of circles o(a τ ). The skein element S(k 0, τ) is adequate as long as k 0 m i k i, thus by [Arm3, Lem. 4], we have deg v sgn(τ) S(k 0, τ) = deg v sgn(τ) S(k 0, τ), and we simply need to determine the number of circles in S(k 0, τ) and sgn(τ) in order to compute the degree of the Kauffman bracket. This is completely specified by the pyramidal configuration of A τ by just applying the Kauffman state. With the assumption that k 0 = m i= k i = c since c, k is tight, the degree is then 2 ) ( ( m i= k i) (( m i= k i) ) 2 d(k 0, τ) = q i n 2 k i (k i + ) 2(2 + + (q i 2)ki 2 ) 2 i= i= i= }{{} ( ( m ) + 2 2n ( k i k 0 ) + i= sgn(τ) ) (n k i )(q i ) i= } {{ } 2o(A τ ) + q 0 (2n 2k 0 + 2n2 4k0 2 ) + 2k 0 2n. }{{ 2 } fusion and untwisting The sign of the leading term is given by q 0 (n k 0 ) + n + k }{{} 0 fusion and untwisting ( ) +o(a τ ) = ( ) q 0(n k 0 )+n+k 0 + P m i= (n k i)(q i ). Lemma 4.7. Minimal states are taut. In other words, given c, k tight, we have max σ:c(σ)=c,k i (σ)=k i d(k 0, σ) = d(k 0, τ), where τ is a minimal state with c(τ) = c and k i (τ) = k i. Proof. Note that for any state σ with corresponding skein element S(k 0, σ) A τ A σ for a minimal state τ with the same parameter set (c, k) by Theorem 4.4 and d(k 0, τ) = d(k 0, τ ) for two minimal states τ, τ with the same parameters c(τ) = c(τ ) and k i (τ) = k i (τ ) by Lemma 4.6. This implies d(k 0, σ) d(k 0, τ) Constructing minimal states. Lemma 4.8. A minimal state exists for any tight c, k, where c is an even integer between 0 and 2n and k 0 = m i= k i = c 2.

19 THE SLOPE CONJECTURE FOR MONTESINOS KNOTS 9 Proof. It is not hard to see that at an n-cabled crossing x n in a twist region with q i crossings, for any 0 k i n there is always a minimal state giving 2k i through strands. For an n- cabled crossing x n in S t or S b, it is also not hard to see that we may take the pyramidal position P for the minimal state for the bottom half (or upper half, for S b ) of the crossings in x n in Cn u and C l, i,j for each twist region. What remains to be shown is that a minimal state always exists, given the set of parameters {k i } and c total through strands for crossings in the top and bottom strips delimited by {Sj u } n j= and {Sj} l n j=. To see this, we take the leftmost configuration with {k i } through strands for the bottom half of the crossings in x n for each twist region, which we already know to exist. Given two crossings x and x in Cn u whose corresponding segment in S(k 0, B) has ends on Sn u we can always find another crossing x in Cn u, the end of whose corresponding segment on Sn u lies between those of x and x, because the previously chosen crossings in Cn u are leftmost. Pick the leftmost possible and repeat to choose crossings in Cj u for n k + j n 2. We pick crossings in the bottom strip by reflection. Lemma 4.9. Let σ be a state with c = c(σ) and k i = k i (σ) which is not tight, that is, m i= k i > c 2 or k 0 < c 2, then d(k 0, σ) < d(k 0, τ), where τ is a minimal state with c(τ) = c through strands. Proof. For the case m i= k i > c, we can apply Theorem 4.4 to conclude that there is a 2 minimal state τ (there may be multiple such states) such that A τ A σ, with k i (τ) k i (σ) for each i. There must be some i for which k i (τ) < k i (σ). Applying the B-resolution to the additional crossings to obtain a sequence of states from τ to σ, we see that it must contain two consecutive terms that merge a pair of circles. If k 0 < c with similar arguments we can see that o(a 2 σ) < o(a τ ) Enumerating all taut states. By Lemma 4.7, we have shown that every taut state contains a minimal state. Next we show that every taut state is obtained from a unique such minimal state τ by changing the resolution from B-to A-on a set of crossings F τ. We show that any taut state σ with c(σ) = c(τ) and k i (σ) = k i (τ) containing τ as the leftmost minimal state, to be defined below, satisfies A σ = A τ p, where p is any subset of F τ. All the circles here in the definitions and theorems are understood with possible extra labels u, l, s, i, j indicating where they are in the regions defining S t, S w, and S b. To simplify notation we do not show these extra labels. Definition 4.0. For each x A τ between S j and S j, let R x be the set of crossings to the right of x between S j and S j, but to the left of any x A τ between S j 2 and S j, and any x A τ between S j and S j+. We define the following possibly empty subset F τ of crossings of K n. See Figure 5 and 6 for examples. F τ := x Aτ R x.

20 20 STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN S j 2 x S j S j S j+ Figure 5. Only the blue edge is in R x because of the presence of the top and bottom red edges. Figure 6. An example of F τ state τ shown as red edges. with edges shown in blue with the minimal Definition 4.. Given a set of crossings C of K n, a crossing x C, and j n, define the distance x C of a crossing x C from the left to be x C := For x C j, the # of edges in S(k 0, B) to the left of x between S j and S j+, the distance of the set C from the left is x C. x C Given any state σ with tight parameters c, k, we extract the leftmost minimal state τ σ where A τσ A σ, i.e., there is no other minimal state τ such that A τ A σ, and the distance of A τ from the left is less than the distance of A τσ from the left. Lemma 4.2. A Kauffman state σ with tight parameters c(σ), {k i (σ)} is taut if and only if A σ may be written as A σ = A τσ p where τ σ is the leftmost minimal state from σ such that A τσ A σ, and p is a subset of F τσ. See Figure 7 for an example of a taut state that is not a minimal state, and how it is obtained from the leftmost minimal state that it contains. Proof. By construction, if a state σ is such that A σ = A τσ p where p is a subset of F τσ, then σ is a taut state. Conversely, suppose by way of contradiction that σ is taut, which means that it has the same parameters (c, k) as its leftmost minimal state τ σ with the same degree, but that there is a crossing x A σ and x / F τσ. Then there are two cases () x is to the left or to the right of all the edges in A τσ.

21 THE SLOPE CONJECTURE FOR MONTESINOS KNOTS 2 2k 0 2k 0 Figure 7. On the left, a taut state having the same degree as a minimal state but is not equal to it. We have c = 6, k = 0, k 2 = 0, k 3 = 2 and k 4 = as the minimal state in Figure 4, and the thickened red edges indicate the difference from a minimals state with the same parameters. Choosing the B-resolution at each of the thickened red edges splits off a circle. (2) x C j is between x, x C j in A τσ for some j. In both cases we consider the state σ where A σ = A τ {x}, and we assume that taking the A-resolution on x splits off a circle from the skein element S(k 0, σ) otherwise by Lemma A.6, deg v sgn(σ) S(k 0, σ) < deg v sgn(τσ) S(k 0, τ σ ), a contradiction to σ being taut. In case (), the state σ has parameters (c, k ) such that m i= k i < m i k i. If each step of a sequence from σ to σ splits a circle in order to maintain the degree, then the parameters for σ, and hence the number of global through strands of S(k 0, σ) will differ from S(k 0, τ σ ), a contradiction. In case (2), we have that x / F τσ must be an edge of the following form between a pair of edges x, x as indicated in the generic local picture shown in Figure 8, since τ σ is assumed to be leftmost. x x x Figure 8. The crossing x corresponds to the green edge.

22 22 STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN Choosing the A-resolution at x merges a pair of circles which means that d(k 0, σ) < d(k 0, τ σ ), a contradiction Adding up all taut states in st(c, k). Note that in general there may be many taut states σ with fixed parameters k. Theorem 4.3. Let c, k = {k i } m i= be tight. The sum v sgn(σ) S(k 0, σ) = ( ) q0(n k0)+n+k0+ P m i= (n k i)(q i ) v d(k0,τ) + l.o.t., (4) σ taut:c(σ)=c,k i (σ)=k i where τ is a minimal state in the sum. We are finally ready to prove Theorem 4.3. Proof. Every minimal state with parameters c, k may be obtained from the leftmost minimal state of the entire set of minimal states M by transposing to the right. Now we organize the sum (4) by putting it into equivalence classes of states indexed by the leftmost minimal state τ σ. We may write v sgn(σ) S(k 0, σ) = σ taut:c(σ)=c,k i (σ)=k i τ minimal σ : τ σ=τ v sgn(σ) S(k 0, σ). By Lemma 4.2, this implies σ taut:c(σ)=c,k i (σ)=k i v sgn(σ) S(k 0, σ) = If F τ, then by a direct computation, F τ ( ) deg n v sgn(τ) 2j ( v 2 v 2 ) j j=0 F τ τ minimal j=0 o(aτ )+j ( ) n v sgn(τ) 2j ( v 2 v 2 ) o(aτ )+j. j = sgn(τ) + 2o(A τ ) 4 F τ < deg ( v sgn(τ) S(k, τ) ) = δ(n, k) by Lemma 4.6. Every taut state can be grouped into a nontrivial canceling sum except for the rightmost minimal state. Thus it remains and determines the degree of the sum Proof of Theorem 4.2. Recall that J K,n+ = c,k G c,k and G c,k = G k0 (v) v sgn(σ) I k0 Tσ n k 0 σ:k i (σ)=k i,c(σ)=c By the fusion and untwisting formulas we have 2k0 G k0 (v) = ( ) q 0(k 0 +n) θ(n, n, 2k 0 ) vq 0(2n 2k 0 +n2 k2) 0.

23 THE SLOPE CONJECTURE FOR MONTESINOS KNOTS 23 We apply the previous lemmas to compute for each c, k the v-degree of the sum v sgn(σ) I k0 T σ. σ:k i (σ)=k i,c(σ)=c When c, k is tight the top degree part of the sum is Gc,k taut. By Theorem 4.3, we have that the coefficient and the degree of the leading term are given by a minimal state τ with parameters c, k. The degree is computed to be δ(n, k) in Lemma 4.6, which also determines the leading coefficient. When σ is a state such that c, k is not tight, and k 0 c(σ) or k 0 m i= k i(σ), Lemma 4.3 says that S(k 0, σ) is zero. Otherwise, Lemma 4.9 says that there exists a taut state corresponding to a tight c, k that has strictly higher degree. 5. Quadratic integer programming In this section we collect some facts regarding real and lattice optimization of quadratic functions. 5.. Quadratic real optimization. We begin with considering the well-known case of real optimization. Lemma 5.. Suppose that A is a positive definite m m matrix and b R m. Then, the minimum min x R m 2 xt Ax + b x (5) is uniquely achieved at x = A b and equals to 2 bt Ab. Proof. The function is proper with the only critical point at x = A b which is a local minimum since the Hessian A is positive definite. For a vector v R m, we let v i for i =,..., m denote its ith coordinate, so that v = (v,..., v m ). When v is are nonzero for all i, we set v = (v,..., vm ). The next lemma concerns optimization of convex separable functions f(x), that is, functions of the form f(x) = f i (x i ), f i (x i ) = a i x 2 i + b i x i (6) i= where a i > 0 and b i are real for all i. The terminology follows Onn [Onn0, Sec.3.2]. Lemma 5.2. (a) Fix a separable convex function f(x) as in (6) and a real number t R. Then the minimum min{f(x) x i = t, x R m } (7) i is uniquely achieved at x (t) where x i (t) = a i t + 2 j (b j b i )a j a j i a j (8)

24 24 STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN and equals to a t2 + b a a t + s 0(a, b) (9) where Z m denotes the vector with all coordinates equal to. (b) If t 0, then the minimum min{f(x) i x i = t, x R m, 0 x i, i =,..., m} (20) is uniquely achieved at (8) and given by (9). Note that the coordinates of the minimizer x (t) are linear functions of t for t 0; we will call such minimizers linear. It is obvious that the minimal value is then quadratic in t for t 0. Proof. Let f(x) = j a jx 2 j + b j x j and g(x) = j x j and use Lagrange multipliers. { f = λ g g = t. So, 2a j x j + b j = λ for all j, hence x j + b j /(2a j ) = λ/(2a j ) for all j and summing up, we get t + j b j/(2a j ) = λ j /(2a j). Solving for λ, we get λ = 2t+P j b ja j and using Pj a j x i = λ b i = 2t + j (b j b i )a j = a i t + 2 j (b j b i )a i a j, 2a i 2a i j a j j a j Equation (8) follows. Observe that x (t) is an affine linear function of t. It follows that f(x (t)) is a quadratic function of t. An elementary calculation gives (9) for an explicit rational function s 0 (a, b). If in addition t 0 observe that x (t) = t a + O(), therefore x (t) is in the simplex a x i 0 for all i and i x i = t. The result follows Quadratic lattice optimization. In this section we discuss the lattice optimization problem min{f(x) Ax = t, x Z m, 0 x t} (2) for a nonnegative integer t, where A = (,,..., ) is a m matrix and f(x) is a convex separable function (6) with a, b Z m with a > 0, and t Z is a nonnegative integer. We will follow the terminology and notation from Onn s book [Onn0]. Lemma 3.8 of Onn [Onn0] gives a necessary and sufficient condition for a lattice vector x to be optimal. In the next lemma, suppose that a feasible x Z m is non-degenerate, that is, x i < t and x j > 0 for all i, j. Note that this is not a serious restriction since otherwise the problem reduces to a lattice optimization problem of the same shape in one dimension less. Lemma 5.3. [Onn0] Fix a feasible x Z m which is non-degenerate. Then it is optimal (i.e., a lattice optimizer for the problem (2)) if and only if it satisfies the certificate 2(a i x i a j x j ) (a i + a j ) (b i b j ). (22)

25 THE SLOPE CONJECTURE FOR MONTESINOS KNOTS 25 Proof. Lemma 3.8 of Onn [Onn0] implies that x is optimal if and only if f(x) f(x + g) for all g G(A) where G(A) is the Graver basis of A. In our case, the Graver basis is given by the roots of the A m lattice, i.e., by G((,,..., )) = {e j e i i, j m, i j}. Let g = e j e i G(A) and f(x) as in (6). Then f(x) f(x + g) is equivalent to (22). Below, we will call a vector quasi-linear if its coordinates are linear quasi-polynomials. Proposition 5.4. (a) Every non-degenerate lattice optimizer x (t) of (2) is quasi-linear of the form x i (t) = for some ϖ-periodic functions c i, where ϖ = i a i j a j t + c i (t) (23) a j. (24) (b) When t 0 is an integer, the minimum value of (2) is a quadratic quasi-polynomial a t2 + b a a t + s 0(a, b)(t) (25) where s 0 (a, b) is a ϖ-periodic function. Note that in general there are many minimizers of (2). Comparing with (8) it follows that any lattice minimizer of (2) is within O() from the real minimizer. Proof. Let A i = j j a j = a... â j... a m, then ϖ = A + + A m. Suppose x satisfies the optimality criterion (22) and Ax = t where A = (,,..., ). Let x = x + (A,..., A m ). Since a i A i a j A j = 0 for i j, it follows that j i 2(a i x i a j x j) = 2(a i x i a j x j ). Hence x satisfies the optimality criterion (22) if and only if x does. Moreover, Ax = Ax + ϖ = t + ϖ. Since a i /( j a ) = A i /ϖ, it follows that every minimizer x (t) satisfies the property that x i (t) a P i j a j j t is a ϖ-periodic function of t. Part (a) follows. For part (b), write x (t) = t a + c(t) and use the fact that Ac(t) = 0 to deduce that f(x (t)) is a a quadratic quasi-polynomial of t with constant quadratic and linear term given by (2) 5.3. Application: the degree of the colored Jones polynomial. Recall that our aim is to compute the maximum of the degree function δ(k) = δ(k, n) of the states in the state sum of the colored Jones polynomial with tight parameters k 0 = m i= k i, see Theorem 4.2. Here we make use of the assumption that q i is odd for all 0 i m. We will compute the maximum in two steps, Step : We will apply Proposition 5.2 to the function δ(k) (divided by 2, and ignoring the terms that depend on n and q but not on k): m 2 δ(k) = ( m ) 2 m (q i )ki 2 + (q 0 + ) k i + k i ( 2 + q 0 + q i ). (26) i= i= i=

26 26 STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN under the usual assumptions that q 0 < 0, q i > 0 for i =,..., m. We assume that k = (k,..., k m ) Z m. Restricting δ(k) to the simplex k i 0 and k + + k m = t and using Proposition 5.4, it follows that min P k i 0 δ(k) = Q 0 (t), where Q 0 (t) = s(q)t 2 + s (q)t + s 0 (q)(t), (27) i k i=t and s(q), s (q) are given by (2) and s 0 (q) is a ϖ-periodic function where ϖ is the denominator of s(q). Step 2: Since min δ(k) = min Q 0(t), P k i 0 0 t n i k i n it remains to compute the minimum min Q 0(t) 0 t n of a quadratic function of t (the fact that this is a quasi-polynomial whose constant term is a periodic function of t does not affect the argument, since we can work in a fixed congruence). It follows that Q 0 (t) is positive definite, degenerate, or negative definite if and only if s(q) > 0, s(q) = 0, or s(q) < 0, respectively. Case : s(q) < 0. Then Q 0 (t) is negative definite and the minimum is achieved at the boundary t = n (since this has lower value than that of t = 0). It follows that min δ(k) = s(q)n 2 + s (q)n + s 0 (q)(n). P k i 0 i k i n Case 2a: s(q) = 0, s (q) 0. Then Q 0 (t) is a linear function of t and the minimum is achieved at t = 0 or t = n depending on whether s (q) 0 or s (q) < 0, so we have { s 0 (q)(n) if s (q) 0 min δ(k) = P k i 0 s (q)n + s 0 (q)(n) if s (q) < 0. i k i n Case 2b: s(q) = 0 = s (q). Now t = 0 and t = n both contribute equally so cancellation may occur. It does not because the sign of the leading term is constant due to the parity of the q i s. Case 3: s(q) > 0. Then Q 0 (t) is positive definite and Proposition 5.4 implies that the lattice minimizers are near s (q)/(2s(q)) or at 0, when s (q) < 0 or s (q) 0 and the minimum value is given by: min δ(k) = P k i 0 i k i n { s (q) 2 2s(q) if s (q) < 0 s 0 (q)(n) if s (q) 0. Again cancellation of multiple lattice minimizers is ruled out because the signs of the leading terms are always the same due to the assumption on the parity of the q i s. For future reference it may be of interest to note that there are very few pretzel knots with s(q) 0 and s (q) = 0. These are cases 2b and 3 above where cancellations might occur if we had no control on the sign of the leading coefficients. The case P ( 3, 5, 5) is mentioned in [LvdV] for its colored Jones polynomial with growing leading coefficient.

27 THE SLOPE CONJECTURE FOR MONTESINOS KNOTS 27 Lemma 5.5. (Exceptional Pretzel knots) The only pretzel knots with q 0 2 < 3 q,... q m for which s(q) 0 and s (q) = 0 are () P ( 3, 5, 5), P ( 3, 4, 7), P ( 2, 3, 5, 5), with s(q) = 0. (2) P ( 2, 3, 7), with s(q) = 2. Proof. Changing variables to f i = q i turns the two equations s(q) 0 and s (q) = 0 into: f 0 (f + + fm ) + m = 0 and 2 + f 0 + f = c for some c 0. Solving for + +fm f 0 yields f 0 = (c 2) m. Since f m 0 3 we must have 0 c 2 3 m. This means m there can only be such c when m = 2 or 3. Suppose m = 2 then c = 0 or c =. In the first 2 case we find f 2 = 2f so the positive integer solutions are (f f 2, f 2 ) {(3, 6), (4, 4), (6, 3)}. In the case c = we find f 2 2 = 3f so (f 2f 3, f 2 ) {(2, 6), (3, 3), (6, 2)}. Finally the case m = 3, c = 0, f 0 = 3 yields (f, f 2, f 3 ) {(2, 4, 4), (2, 3, 6), (3, 3, 3)} and permutations. 6. Matching the growth rate to the topology We consider two candidate surfaces from the Hatcher-Oertel algorithm whose boundary slope, Euler characteristic, and number of sheets match the growth rate of the degree of the colored Jones polynomial from the previous section as predicted by the Strong Slope Conjecture The surface S(M, x ). Let n = M be the least common multiple of the denominators of {x i (M)} as defined by (9), reduced to lowest terms. Write x i (M) = x i,m + x i,0, so x i, = a i j a j where a i = q i and b i = q 0 + q i 2. and x i,0 = 2 j (b j b i )a j a j Lemma 6.. There is a candidate surface S(M, x ) from the Hatcher-Oertel algorithm with M sheets and C-coordinates { M, Mx,, Mx 2,,..., Mx m,}. Proof. Directly from the proof of Lemma 5., the parameters {x i,(m)} m i= satisfy the following equations. x i,(q i ) = x j,(q j ), for i j, and x i, =. (28) i= Therefore, letting K i = Mx i,, the parameters {Mx i,} m i=0 satisfy the equations coming from (a) and (b) of Step (2) of the algorithm, with edge-path systems determined by the following choice of continued fraction expansions for {/q i } m i=0 /q 0 = [[, 2, 2,..., 2]] }{{} q 0 /q i = [[0, q i ]], for i 0. i a j.

28 28 STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN Thus, there is a candidate surface with { M, Mx,, Mx 2,,..., Mx m,} as the C-coordinates in the tangle corresponding to r i. Explicitly, with K i = Mx i, and K 0, q determined by solving for 0 K 0 M, q 0 q 2 such that K 0 + M(q 2) = K (q ), (29) the explicit description of S(M, x ) in terms of edge-paths are: The edge-path for q 0 is For i 0, we have the edge-path q 0 q 0 + K 0 M q + M K 0 M q +. q i K i M q i + M K i M 0, Note that there could be different pairs K 0, q satisfying (29), but the resulting edge-path systems all have the same boundary slope. The twist number of S(M, x ). With the given edge-path system and applying the formula for computing the boundary slope in Section 3.4, the twist number of S(M, x ) is given by tw(s(m, x )) = 2( q 0 x,(q ) + m 2). (30) The Euler characteristic of S(M, x ). With the given edge-path system and applying the formula for computing the Euler characteristic in Section 3.5, the Euler characteristic over the number of sheets of S(M, x ) is given by 2χ(S(M, x )) #S(M, x ) = 4 tw(s(m, x )) 2(m )(x, )(q ). (3) The reference surface R. Note that the set of parameters {0} m i=0 also trivially satisfy the equations from Step 2(a) and 2(b) of the Hatcher-Oertel algorithm with the choice of continued fraction expansion /q i = [[0, q i ]] for 0 i m, and therefore defines a connected candidate surface in the complement of K(/q 0,..., /q m ). We will call this surface the reference surface R. This is a state surface for K(/q 0,..., /q m ) obtained from a Kauffman state σ, where for each twist region consisting of adjacent bigons, σ chooses the resolution on each crossing in the twist region such that the bigons become state circles. By the Hatcher-Oertel algorithm, the reference surface is essential except the one for K(,, ), K(,, ), and K(,, ) In the framework of the Hatcher-Oertel algorithm, the edge-path corresponding to the reference surface has the following form for each q i : q i 0. The twist number of R. With the exception of γ 0, each γ i is slope-increasing of length, thus the twist number of the reference surface R is tw(r) = 2(m ). (32)

29 THE SLOPE CONJECTURE FOR MONTESINOS KNOTS 29 The Euler characteristic of R. From the state surface that gives R, we have that the number of sheets of R is and the Euler characteristic, and therefore χ(r)/#r, is χ(r) #R = m. (33) Matching the Jones slope. Note that both S(M, x ) and R are essential by an immediate application of Theorem 3.2. Let τ be a minimal state. Write Define d(k 0, τ) = δ(n, k) = s 2 (n, k)n 2 + s (n, k)n + s 0 (n, k). js(s(k 0, τ)) = ω(k) + s 2 (n, k) where ω(k) is the writhe of K. We associate to S(M, x ) the skein element S(M, τ ), where τ is a minimal state such that δ(m, τ ) maximizes δ(m, k) as in Lemma 5.2, and we associate to R the skein element S(0, τ 0 ), where τ 0 is the Kauffman state that chooses the B-resolution on all the crossings in K n +. Let bs(r) denote the boundary slope of R and bs(s(m, x )) denote the boundary slope of S(M, x ). Note that js(s(0, τ 0 )) = bs(r) by [FKP, Lemma 4]. Lemma 6.2. Let R be the reference surface associated to S(0, τ 0 ), and S(M, x ) the surface associated to the unique degree-maximizing skein element S(M, τ ) from the minimal state τ with boundary slope bs(s(m, x )) and bs(r), respectively. If js(s(m, τ )) js(s(0, τ 0 )) = tw(s(m, x )) tw(r), then js(s(m, τ )) is the boundary slope of the surface S(M, x ). Proof. It is easy to check that js(s(0, τ 0 )) = tw(r) tw(s 0 ) = bs(r), where S 0 is a Seifert surface from the Hatcher-Oertel algorithm. Then by assumption, js(s(m, τ )) js(s(0, τ 0 )) = tw(s(m, x )) tw(r) js(s(m, τ )) = tw(s(m, x )) tw(r) + tw(r) tw(s 0 ) js(s(m, τ )) = tw(s(m, x )) tw(s 0 ) = bs(s(m, x )). Theorem 6.3. We have: js(s(m, τ )) js(s(0, τ 0 )) = tw(s(m, x )) tw(r). Proof. We have js(s(m, τ )) js(s(0, τ 0 )) = ω(k) + s 2 (M, x ) (ω(k) + s 2 (M, 0)) = s 2 (M, x ) s 2 (M, 0).

30 30 STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN The reference surface R comes from the Kauffman state that chooses the A-resolution on all the crossings in the n-cabled negative twist region with q 0 crossings and the B-resolution everywhere else. Therefore, s 2 (M, 0) = q i. i=0 The quadratic term for js(s(m, τ )) is s 2 (M, x ) = 4 i j ( m ) x i,x j, 2q i (x i,) 2 q 0 + i= q i. i= Recall that (q i )x i, = (q j )x j, and m i= x i, =, so by Equation (30) and (32) for the twist numbers of R and S(M, x ), respectively, s 2 (M, x ) s 2 (M, 0) = 2 2q 0 + 2( q + )x,( = tw(s(m, x )) tw(r) Matching the Euler characteristic. Again we write and define i= d(k 0, τ) = s 2 (n, k)n 2 + s (n, k)n + s 0 (n, k) jx(s(k 0, τ)) = s (n, k) 2s 2 (n, k). It is also immediate from the description of the reference surface R as a state surface and [FKP, Lemma 4] that jx(s(0, τ 0 )) = χ(r) = χ(r) #R. For the proof, see [Lee]. Lemma 6.4. We have jx(s(m, τ )) = 2 χ(s(m, x )) #S(M, x ), where χ(s(m, x )) is the Euler characteristic and #S(M, x ) is the number of sheets M of the surface S(M, x ). Proof. We have by (3), 2χ(S(M, x )) #S(M, x ) x i,) = 4 tw(s(m, x )) 2(m )(x, )(q ).

31 THE SLOPE CONJECTURE FOR MONTESINOS KNOTS 3 The quantity jx(s(m, τ )) = s (M, x ) 2s 2 (M, x ) is given by s (M, x ) 2s 2 (M, x ) = 2(m ) 2(q 0 2) 2 (q i )2x i,x i,2 2 i= 2(tw(S(M, x )) tw(r) + s 2 (M, 0)) = 2(m ) + 4 2q 0 4(q )x, i= x i,2 }{{} 0 2 q i x i, + 2 i= i=0 (q i )x i, 2 (tw(s(m, x )) 2(m )) ( 2 2q 0 + 2( q + )x, + 2 = 4 2(m )(q )(x, ) tw(s(m, x )). i= q i i= x i, }{{} q i ) i=0 +2 i=0 q i 6.. Proof of Theorem.2 for pretzel knots. Now we prove Theorem.2. Fix odd integers q 0,..., q m with q 0 < < < q,..., q m. Let P = P (q 0,..., q m ) denote the pretzel knot K( q 0, q,..., q m ). By Theorem 3.2, both of the surfaces S(M, x ) and R are incompressible by examining their edge-paths and computing their -values. In Section 6, Theorem 6.3, Lemma 6.4, and previous work of [FKP] say that js(s(m, τ )) = bs(s(m, x )), js(s(0, τ 0 )) = bs(r), jx(s(m, τ )) = 2 χ(s(m,x )), and jx(s(0, τ #S(M,x ) 0)) = 2 χ(r). #R From Section 5.3, we have the following cases for the degree of the colored Jones polynomial J P,n (v). The choice of the surface detected by the Jones slope swings between the surface S(M, x ) and the reference surface R. Case : s(q) < 0. We have that the maximum of δ(n, k) is given by δ P (n) = 2s(q)n 2 2s (q)n 2(m )n + (n 2 + 2n) q i 2s 0 (q)(n), where recall that s(q) and s (q) are explicitly defined by (2) and s 0 (q)(n) is a periodic function. By Lemma 5.2, we see that s(q) and s (q) for any n are actually the same as when n is equal to the multiple of M, where there is a unique minimal state τ with parameters M, x realizing δ P (M). Thus the fact that js(s(m, τ )) = bs(s(m, x )) and jx(s(m, τ )) = verifies the Strong Slope Conjecture in this case. Case 2a: s(q) = 0, s (q) 0. If s (q) > 0, the maximum 2s 0 (q)(n) of δ(n, k) has no quadratic or linear term, and the reference surface R verifies the conjecture. If s (q) < 0. Then the maximum 2 χ(s(m,x )) #S(M,x ) 2s (q)n 2(m )n + (n 2 + 2n) i=0 q i 2s 0 (q)(n) of δ(n, k) is found at maximizers τ with parameters n, k, again all satisfying n = k 0 = k + + k m. Thus the surface S(M, x ) verifies the conjecture. i=0

32 32 STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN Case 2b: s(q) = s (q) = 0. There is no quadratic or linear term of the maximum of δ(n, k), thus the reference surface R verifies the conjecture. Case 3: s(q) > 0. In this case the maximum of δ(n, k) also does not have quadratic/linear terms, and the reference surface R verifies the conjecture. Remark 6.5. With the analogy between the C-curve system coordinates K i and the real maximizers x as established by Lemma 6., it is interesting to note that for n M, the degrees of the terms in the state sum of the colored Jones polynomial seem to correspond to disconnected surfaces with the same C-curve system coordinates. The boundary slope and normalized Euler characteristic of the disconnected surfaces approximate the connected one associated to the real maximizers when n = M. 7. The colored Jones polynomial of Montesinos knots In this section we will extend our proof of the Strong Slope Conjecture of pretzel knots to the class of Montesinos knots. To do so, we introduce the tangle replacement move (in short, TR-move), and study its effect on the state-sum formula for the colored Jones polynomial, as well as on the Hatcher-Oertel algorithm. 7.. The TR-move. The TR-move is a local modification of a link diagram D. Suppose D contains a twist region T. Viewing T as a rational tangle T = for some integer t we may t consider a new diagram D obtained by replacing T by the rational tangle T = r for some t non-zero integer r with the same sign as t. Alternatively, viewing T as an integer tangle t we replace it with T 2 = + t, also with r having the same sign. Collectively these two r operations are referred to as the TR-move. Iteration of the TR-move leads to replacement by a rational tangle, see Figure 9 and 20. t r t r t t + t r r t Figure 9. Two types of TR-move.

33 THE SLOPE CONJECTURE FOR MONTESINOS KNOTS 33 r > 0 r > 0 > 0 r > 0 t Figure 20. Iteration of the TR-move leads to a rational tangle. In the picture shown, we have performed three TR-moves: first on /t, then r, then /r. Note that a Montesinos knot K(r 0, r,..., r m ) may be put in the standard form where each r i satisfies 0 < r i <, unless all the r is have the same sign [BZ03], thus we need only to consider rational numbers r i s where r i [0] = 0. We will use the TR-moves to reduce a Montesinos knot to a pretzel knot Montesinos state sum. As in the case of pretzel knots we use a customized state sum to compute the colored Jones polynomial, splitting K = K K +. In this case K is the single twist region /r 0 [] and K + is the union of all other twist regions. As before we apply fusion and untwisting to K and the usual Kauffman state sum to K + after cabling with the Jones-Wenzl idempotent of size n Special Montesinos knot case. We start by generalizing the pretzel knot case to the case where l ri = 2 for all i >. This includes the pretzel knots by allowing a continued fraction expansion with r i [2] =. We will prove the main theorem for such special Montesinos knots where q i = r i [] are even, q i >, and q 0 < 2 are odd. Figure 2. The special case K(,, ) The methods used previously on the pretzel knots also apply to this case with minor modifications. In particular the notion of global through strands c(σ) for a Kauffman state σ on K n + still makes sense and k i (σ) is still well defined by restricting σ to the ith-tangle. In this case c i (σ) means the number of through strands of the ith tangle of K n + that are also

34 34 STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN global through strands, and as before k i = c i 2. Let G c,k = G k0 (v)v sgn(σ) I k0 T σ. k 0 σ:k i (σ)=k i,c(σ)=c We prove the following theorem.,..., q m+ ). Assume q i >, q q m i > 0, and let Theorem 7.. Consider K = K( q 0, q + q q i = q i + for i m while q 0 = q 0. Referring to the above state sum K n = c,k G c,k we have the following. For a state σ, define the parameters c = c(σ), k = k(σ) to be tight if k 0 = k + +k m = c 2. For tight c, k we have G c,k = ( ) q 0(n k 0 )+n+k 0 + P m i= (n k i)( q i ) v δ(n,k) +l.o.t. 2 and δ(k,n) = 2 (q 0 +)k0+ 2 ( q i )ki 2 + ( 2+q 0 + q i )k i i= i= n(n + 2) 2 If c, k is not tight then deg v G c,k < max c, k tight δ( k, n). i=0 q i +(m )n n2 2 ( q i ). (34) Proof. The proof is analogous to that of Theorem 4.2 for pretzel knots. As in the pretzel case we identify the minimal states and show that they maximize the degree and do not cancel out. Since these arguments are exactly the same we focus on describing the minimal states, one for each tight parameters of through strands c, k. The minimal states are produced by choosing a minimal state for the pretzel knot P = K( q 0,..., q m ) and extending it to a Kauffman state of K+ n by choosing a pyramidal configuration on the remaining twist regions. The new pyramidal configuration has exactly ki 2 extra A-states for each i > 0, so the degree of the minimal pretzel state is increased by m i= q in 2 2ki 2 in the new state sum. The number of additional circles in the pyramidal configuration is m i= n k i. Adjusting the degree accordingly concludes the proof The general case. Given K = K K + we further split K + into K + = D V where D is the union of the first two twist regions of each rational tangle in K + and V is the remaining tangle. When V is empty K is a special Montesinos knot. We approach the general case as insertion of V into this special knot, where V is constructed by applying TR moves. The essential feature of V is that the all-b state acts like the identity plus some closed loops, see Figure 22. Lemma 7.2. The standard diagram of a Montesinos knot K is of the form K = K D V, where K s = K D is a special Montesinos knot. If q 0 < is odd and q i > is even for every i > 0 then we have deg v K n = deg v K n s + c(v )n 2 + 2n o(v B ), where o(v B ) is the number of disjoint circles resulting from applying the all-b state to V. Proof. Applying quadratic integer programming to the formula of Theorem 7. for the degreemaximizing states of K n s, discarding any terms that depend only on q i and n, we see that there are minimal states of the state sum of any special Montesinos knot that attain the 2 The abbreviation l.o.t. means lower order terms in v. i=

35 THE SLOPE CONJECTURE FOR MONTESINOS KNOTS 35 maximal degree. Fix one such minimal state τ. Denote the skein element resulting from applying such a state to Ks n by S(k 0, τ), and the degree by δ(n, k) = d(k, τ). Now we consider the effect of adding V. Note deg G k0 (v)v sgn(σ)+sgn(bv) S(k 0, σ) V B > deg G k0 (v)v sgn(σ)+sgn(σ ) S(k 0, σ) V σ, where σ is any other state on V and B V indicates the all-b state on V. Taking the all-b state on V also preserves the states of Ks n. Thus for a minimal state τ maximizing the degree in the state sum Ks n, the term G k0 (v)v sgn(τ) S(k 0, τ) V B also maximizes the degree in the new Montesinos state sum. The leading terms all have the same sign because of the assumption on the parity of the q i and Theorem 7.. Thus there is no cancellation of these maximal term, and we can determine deg v K n by counting the number of disjoint circles o(v B ), giving the formula in the lemma. It is useful to reformulate the above lemma in a more relative sense, pinpointing how the degree changes as a result of applying a TR-move. For our purposes it is more convenient to work with the composite moves TR 2 (T ) = ( r + r 2 ) T, and TR + (T ) = (r r 2 ) + T. r 2 r t r r 2 t Figure 22. Examples of applying the all-b state and the resulting disjoint circles for moves sending tangles to ( t r + r 2 ) and sending t to (r t r 2 ) + t. TR Lemma 7.3. Suppose two standard diagrams K, L of Montesinos links satisfying the conditions of Lemma 7.2 where L is obtained from K by applying one of the moves TR, TR 2, TR +, locally replacing tangle T by T, then the degree of the colored Jones polynomial changes as follows. See Figure 22 for examples of the moves TR 2, TR +. move: Suppose r, t < 0, T = t is a vertical twist region, and T = r t, then deg L n = deg K n rn 2 + 2( r )n. TR 2 move: Suppose r, r 2, t < 0, T = t is a vertical twist region, and T = ( r + r 2 ) t, then deg L n = deg K n (r + r 2 )n 2 2r 2 n. TR + move: Suppose r, r 2, t > 0, T = t is a horizontal twist region, and T = (r r 2 ) + t, then deg L n = deg K n + (r + r 2 )n 2 + 2r 2 n. Proof. Applying Lemma 7.2 we may simply count the number of crossings and state circles added to the degree in applying the all-b state to the newly added tangle V in each of these cases.

36 36 STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN 7.5. Matching the boundary slope and Euler characteristic. Similar to the case of pretzel knots, we define the reference surface R for K(r 0, r,..., r m ) to be the state surface from the Kauffman state which chooses the A-resolution on the negative twist region q 0 and the B-resolution everywhere else. The surface R is associated to the skein S(0, τ 0 ), where τ 0 is the all-b state on K+. n The other surface S(M, x ) corresponds to a real maximizer S(M, τ ) of δ(n, k) as in Theorem 7., when we apply the method of Lagrange multipliers as in Lemma 5.2 to (34). The edge-path systems of S(M, x ) and R are explicitly given below. Note both of these surfaces are also essential by a direct application of Proposition For the surface S(M, x ). The edge-path system of S(M, x ) is described as follows. For i = 0, say r 0 = [0, a, a 2,..., a lr0 ] for a i < 0, we take the following continued fraction expansion [, 2,..., 2, a }{{} 2, 2,..., 2, a }{{} 2j, 2,..., 2,...], (35) }{{} a times a 3 times a 2j+ times with corresponding edge-path [0, a, a 2,..., a lr0 ] [, 2, 2] [, 2]. For i m, say r i = [0, a, a 2,..., a lri ] for a i > 0, we take the following continued fraction expansion [0, a, 2,..., 2, a }{{} 3, 2,..., 2, a }{{} 2j+, 2,..., 2,...], (36) }{{} a 2 times a 4 times a 2j+2 times with corresponding edge-path [0, a, a 2,..., a lri ] [0, a, 2] [0, a ] 0. We similarly let n = M be the least common multiple of the denominators of {x i (M)} as given below, reduced to lowest terms. Write x i (M) = x i,m + x i,0, so x i, = and x i,0 = j (b j b i )a i a j. 2 a i j a j j a j where a i = q i and b i = q 0 + q i. S(M, x ) is the candidate surface from the Hatcher-Oertel algorithm with M sheets and C-coordinates { M, Mx,, Mx 2,,..., Mx m,} The reference surface R. For the reference surface R, we have for each r i, the edgepath system corresponding to the following continued fraction expansion For r 0 = [0, a, a 2,..., a lr0 ] for a i < 0, we take the following continued fraction expansion. [0, a, a 2, 2,..., 2, a }{{} 4, 2,..., 2, a }{{} 2j, 2,..., 2,...], (37) }{{} a 3 times a 5 times a 2j+ times with corresponding edge-path [0, a, a 2,..., a ] lr0 [0, a ] 0.

37 THE SLOPE CONJECTURE FOR MONTESINOS KNOTS 37 For i m, say r i = [0, a, a 2,..., a lri ] for a i > 0, we take the following continued fraction expansion. [0, a, 2,..., 2, a }{{} 3, 2,..., 2, a }{{} 2j+, 2,..., 2,...], (38) }{{} a 2 times a 4 times a 2j+2 times with corresponding edge-path [0, a, a 2,..., a ] lri [0, a, 2] [0, a ] 0. Write Define δ(n, k) = s 2 (n, k)n 2 + s (n, k)n + s 0 (n, k). js(s(k, τ)) = ω(k) + s 2 (n, k) and jx(s(k, τ)) = s (n, k) 2s 2 (n, k), where ω(k) is the writhe of K. Lemma 7.4. We have and js(s(0, τ 0 )) = bs(r), jx(s(0, τ 0 )) = 2χ(R) #R, js(s(m, τ )) = bs(s(m, x )), jx(s(m, τ )) = 2χ(S(M, x )) #S(M, x ). Proof. It is easy to verify for the state surface R that js(s(0, τ 0 )) = bs(r) and jx(s(0, τ 0 )) = 2χ(R) #R using [FKP]. For showing js(s(m, τ )) = bs(s(m, x )), it suffices then to verify that js(s(m, τ )) js(s(0, τ 0 )) = tw(s(m, k)) tw(r). and apply Lemma 6.2. Notice that the edge-path systems of the two surfaces coincide beyond the first segments of their edge-path systems, which define candidate surfaces S P (M, x ) and R P for the pretzel P = P (r 0 [], r [] +, r 2 [] +,..., r m [] + ). Now by Theorem 6.3, we have js(s P (M, τ )) js(s P (0, τ 0 )) = tw(s P (M, x )) tw(r P ) = tw(s(m, x )) tw(r). Now Theorem 7. says that js(s P (M, τ )) js(s P (0, τ 0 )) = js(s(m, τ )) js(s(0, τ 0 )), and this finishes matching js(s(m, τ )) to bs(s(m, x )). The proof that jx(s(m, τ )) = 2χ(S(M,x )) is similar using jx(s #S(M,x ) P (M, τ )) = 2χ(S P (M,x )) #S P, which is verified in Section (M,x ) 7.6. Proof of Theorem.3. Putting everything together we prove Theorem.3. Proof. Given a Montesinos knot K(r 0, r,..., r m ) where r 0 < 0 < r,..., r m and r i <, take the unique positive continued fraction expansion r i = [[0, r i [], r i [2],... r i [l ri ]]]. Then for each i if l ri is odd, replace with the even-length continued fraction expansion r i = [[0, r i [], r i [2],..., r i [l ri ], ]]. Otherwise, use the original continued fraction expansion. Let K be the diagram for the Montesinos knot corresponding to this choice of continued fraction expansions for each rational tangle r i, then we may obtain it from the special

38 38 STAVROS GAROUFALIDIS, CHRISTINE RUEY SHAN LEE, AND ROLAND VAN DER VEEN Montesinos knot diagram K s = K([[0, r 0 []]], [[0, r [], r [2]]],..., [[0, r m [], r m [2]]]) by the combination of TR, TR 2, and TR + -moves. Lemma 7.3 shows how the degree of the colored Jones polynomial changes and Theorem 7. gives the base case. The resulting formulas are matched with the boundary slope and normalized Euler characteristic of incompressible surfaces by Lemma 7.4. Appendix A. Background on the Temperley-Lieb algebra and the Jones-Wenzl idempotent We consider the skein module of link diagrams on an oriented surface with a finite (possibly empty) collection of points specified on the boundary F. For the original reference for skein modules see [Prz9]. We will follow the approach of Lickorish [Lic97] except for the variable substitution (our v is his A to avoid confusion with the A for a Kauffman state). Definition A.. Let v be a fixed complex number. The linear skein module S(F ) of F is a vector space of formal linear sums over C, of (unoriented) link diagrams in F, considered up to isotopy of F fixing F, and quotiented by the relations (i) D = ( v 2 v 2 )D, and (ii) = v + v. We consider the linear skein module S(D 2, n, n ) of the disc D 2 with n+n -points specified on its boundary, where the boundary is viewed as a rectangle with n marked points above and n marked points below. For D S(D 2, n, n ), and D 2 S(D 2, n, n), there is a natural multiplication operation D D 2 defined by identifying the top boundary of D with the bottom boundary of D 2 and matching the boundary points. This makes S(D 2, n, n ) into an algebra TL n n, called Temperley-Lieb algebra. For the original reference see [TL7]. A Kauffman state [Kau87], which we will denote by σ, is a choice of the A- or B-resolution at a crossing of a link diagram. A-resolution B-resolution Figure 23. A- and B-resolutions of a crossing. Definition A.2. Let σ be a Kauffman state on a skein element with crossings, define sgn(σ) = (# of B-resolutions of σ) (# of A-resolutions of σ). Definition A.3. Given a skein element S with crossings in R 2, the σ-state graph denoted by S σ is the set of disjoint circles resulting from applying a Kauffman state σ to S along with segments recording the original location of the crossing. For the precise definitions of semi-adequacy (A/B-adequacy) of a link based on the Kauffman state graphs of its link diagrams, see the original reference [LT88] and [FKP3].