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Hyperbolicity of 3-trip Lorenz knots
Topology and its Applications 212 (2016) 57–70
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Topology and its Applications www.elsevier.com/locate/topol
Hyperbolicity of 3-trip Lorenz knots Jiming Ma 1 School of Mathematical Sciences, Fudan University, Shanghai, 200433, PR China
a r t i c l e
i n f o
Article history: Received 12 November 2015 Received in revised form 8 September 2016 Accepted 8 September 2016 Available online 12 September 2016 MSC: 57M50 20F36 37C10 57M25
a b s t r a c t Using Xu’s conjugacy algorithm on the braid group B3 [6,31] and the hyperbolicity criterion of 3-braid links by Futer–Kalfagianni–Purcell [14], we classify Lorenz knots of trip number 3 into torus knots and hyperbolic knots. Moreover, we provide another approach to this problem. Modulo a conjectural pseudo-Anosov criterion, we can also classify Lorenz knots of trip number 3 based upon the Dehornoy floor theory. The author believes that the second approach is promising for the hyperbolicity of more classes of Lorenz knots. © 2016 Elsevier B.V. All rights reserved.
Keywords: Braid group Hyperbolic knot Lorenz knot Symbolic dynamics Pseudo-Anosov map
1. Introduction 1.1. Lorenz knots and hyperbolicity Lorenz knots are a kinds of knots which are periodic orbits of the Lorenz differential equation as follows [20]: ⎧ dx ⎪ ⎪ ⎨ dt = 10(y − x), dy dt = 28x − y − xz, ⎪ ⎪ ⎩ dz = xy − 8 z, dt
1
E-mail address: [email protected]. The author was partially supported by NSFC 11371094.
http://dx.doi.org/10.1016/j.topol.2016.09.005 0166-8641/© 2016 Elsevier B.V. All rights reserved.
3
(1)
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Fig. 1. The Lorenz template and a Lorenz braid.
which draw much attention recently [15,4,12]. For a three dimensional flow, the knot types of its periodic orbits are useful for understanding the flow. Let F be the modular surface, that is, F = H2 /P SL(2, Z). Ghys [15,24] showed if we view the unit tangent bundle T 1 F of the modular surface as the complement of the trefoil knot in S3 , then the periodic orbits of the geodesic flow of T 1 F are the same as the set of Lorenz knots. So Lorenz knots are also called modular knots. Moreover, primitive geodesics in the modular surface and the periodic orbits of the geodesic flow are highly interesting from the viewpoint of number theory [27,28,19,22,25]. An important question raised in [3] is to classifying knot types which can occur in Lorenz knots. For example, every torus knot is a Lorenz knot, and every Lorenz knot is a fibered knot [3]. Moreover, certain satellite knots are Lorenz knots [3,16]. From [7], for the 502 simplest hyperbolic knots, that is, knots with their complement in S3 admit ideal triangulations with at most eight tetrahedra, 249 of them are Lorenz knots. The complement of a lift of a periodic geodesic in the unit tangent bundle of F is a hyperbolic 3-manifold if the curve is filling, see [2] for the study of hyperbolic geometry of Lorenz knots in the trefoil complement. But the complete classification of Lorenz knot remains completely open. In this note, we classify the simplest Lorenz knots: trip number 3 Lorenz knots. From our result, we believe that the probability of a random Lorenz knot is a hyperbolic knot should be one, comparing to [21,17]. By [3,30], we can study Lorenz knots combinatorially as follows. Definition 1.1. Let T be the branched surface with a semi-flow in S3 as in the left-side of Fig. 1. T is called the Lorenz template. A Lorenz knot K is a simple closed curve in T which runs along the direction of the semi-flow. For a Lorenz knot K embedded in T , we associate a symbol x whenever the loop passes along the left-hand hole in T . Similarly, we associate a symbol y whenever the loop passes along the right-hand hole in T . We get a word in the semigroup generated by x and y, which is called a Lorenz word. For example, in Fig. 1, the curve corresponds to the word xy, which is the trivial knot in S3 . The trip number of a Lorenz knot K is the minimal number of syllables of type xa y b in all Lorenz words associated to K. So every 3-trip Lorenz knot corresponds to a word xa1 y b1 xa2 y b2 xa3 y b3 , and then we denote it as a Lorenz vector by (a1 , b1 , a2 , b2 , a3 , b3 ) ∈ Z6≥1 as in [1]. Let Bn be the braid group with n strands. Let K be a Lorenz knot in the template T . If we cut the two wings of the template T , from the Lorenz knot K, we get a positive braid in some Bn . It is called a Lorenz braid associated to K. Here our definition of the generating set {σ1 , σ2 . . . σn−1 } of the braid group Bn is different from the standard one. Here we take the meanings as in [6,31]. The difference is a sign change. See Fig. 2. Based on Bedient [1], we show Theorem 1.2. A 3-trip Lorenz knot K is either a torus knot or a hyperbolic knot: (1) If K is a torus knot, then it is a torus knot of type (3, 3b + 1) or (3, 3b + 2), this corresponds to Lorenz knot with Lorenz vector (1, b, 1, b, 1, b + 1) or (1, b, 1, b + 1, 1, b + 1) respectively;
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Fig. 2. Positive braids.
(2) If K is a hyperbolic knot, then it is the closure of 2(b −b)−1
Δ2(b+1) σ1−1 σ2
for b > b ≥ 1, this corresponds to Lorenz knot with Lorenz vector (1, b, 1, b , 2, b ); Or the closure of 2(b −b)−3
Δ2(b+1) σ1−1 σ2
for b − b ≥ 2 and b ≥ 1, this corresponds to Lorenz knot with Lorenz vector (1, b, 1, b , 1, b ); Or the closure of 2(b −b)−1
Δ2b+2 σ1−1 σ22a−3 σ1−1 σ2
for 1 ≤ b < b , 2 ≤ a ≤ b , this corresponds to Lorenz knot with Lorenz vector (a, b, a, b , 1, b ). 1.2. A question on pseudo-Anosov maps and the second approach We also provide another approach to the classification of 3-trip Lorenz knots in Section 3. We divide 3-trip Lorenz knots into two classes. The first class is easy to show the hyperbolicity using Dehornoy floor theory. Moreover, we divide the second class into six subclasses. By Lemmas 3.3–3.8, we have that for each subclass, as fibered knots, their monodromies are somehow similar. We use SnapPy to show the first Lorenz knot in each subclass is a hyperbolic knot. By Thurston’s theory, its monodromy is a pseudo-Anosov map. We have a question on pseudo-Anosov maps, that is, Question 1.3 below. Modular Question 1.3, we can show the hyperbolicity of Lorenz knots in each subclass. Even through for 3-trip Lorenz knots, the second approach is much more involving, but we believe that the second approach is promising for the hyperbolicity of more classes of Lorenz knots. The formulation of Question 1.3 is motivated by Theorem 1.1 and Corollary 3.1 of [8]. The author believes Question 1.3 has a positive answer, but he has difficulties to write down the proof completely. Let F be a compact oriented surface with non-empty boundary. We take a proper arc γ in F , a neighborhood of γ in F is γ × [−1, 1]. By adding a 1-handle to F along ∂γ × [−1, 1], we get a new compact oriented surface F . Any mapping class f in M CG(F ) extends naturally to a mapping class in M CG(F ), which is the identity in the attached 1-handle, we also denoted it by f . The union of the 1-handle and γ × [−1, 1] is an annulus A, the core curve of A is denoted by β. Let Dβ be the positive Dehn twist along β, then Dβ ◦ f in M CG(F ) is called a (positive) Hopf plumbing of f , see [8,11]. If the arc γ is essential in F , then we say the Hopf plumbing is essential. Colin–Honda showed in [8] if f is a pseudo-Anosov map, then for a carefully chosen essential arc γ, f is also a pseudo-Anosov map. Moreover, they showed if f is a pseudo-Anosov map with fractional Dehn twist coefficient at least four, then for any essential Hopf plumbing, the resulting map f is pseudo-Anosov.
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Question 1.3. Let F be an oriented compact surface with one boundary and genus g ≥ 3. Let c1 , c2 , · · · , ck be a set of circles in F . For each j ≤ k, cj intersects cj−1 in one point, and cj does not intersect ci for i ≤ j − 2 or i ≥ j + 1. We assume f = Dc1 ◦ Dc2 · · · Dck−1 ◦ Dck is a pseudo-Anosov map on F . Let T be a genus one oriented surface with one boundary, a and b be two simple closed curves in T which intersect in one point. Let Da and Db be the (positive) Dehn twists along a and b, then h = Da Db is a map with periodicity six in the mapping class group of T . We perform a plumbing to (F, f ) by (T, h), such that a intersects ck in one point, and a does not intersect ci for 1 ≤ i ≤ k − 1, b does not intersect ci for 1 ≤ i ≤ k. We get a map f = Dc1 ◦ Dc2 · · · Dck−1 ◦ Dck ◦ Da ◦ Db in a new surface F , F has genus g + 1 and one boundary component. In other words, we perform an essential trefoil plumbing to (F, f ). Is f a pseudo-Anosov map?
2. Hyperbolicity of 3-trip Lorenz knots
We need an old theorem of Schreier [26], see also [5]: Theorem 2.1 (Schreier [26]). Let w ∈ B3 be a three strands braid. Then w is conjugate in B3 to a braid in exactly one of the following forms: (1). (2). (3). (4). (5).
Δ2k σ1−p1 σ2q1 σ1−p2 σ2q2 . . . σ1−ps σ2qs , where pi and qi are positive integers, k ∈ Z and Δ = σ1 σ2 σ1 ; Δ2k σ1p , where p, k ∈ Z; Δ2k σ1 σ2 , where k ∈ Z; Δ2k σ1 σ2 σ1 , where k ∈ Z; Δ2k σ1 σ2 σ1 σ2 , where k ∈ Z.
A 3-braid w is called generic if w is conjugate to Δ2k σ1−p1 σ2q1 σ1−p2 σ2q2 . . . σ1−ps σ2qs . The following is proved in Theorem 5.5 of [14]: Theorem 2.2 (Futer–Kalfagianni–Purcell [14]). Let w be a braid in B3 , and let K → S3 be the link obtained as the closure of w. Then S3 − K is hyperbolic if and only if w is generic, and w is not conjugate to σ1p σ2q for any integers p and q. Note that the notations σ1 and σ2 in [14] are different from ours here, we take the meanings as in [6,31]. If there is an orientation preserving self-homeomorphism h of S3 , which maps the knot K1 to the knot K2 , then we say K1 and K2 are equivalent. We need Theorem 1 from [1]: Theorem 2.3 (Bedient [1]). Every 3-trip Lorenz knot is equivalent to exactly one of the following knots: Type Type Type Type
Ia, (1, b, 1, b, 1, b + 1); Ib, (1, b, 1, b + 1, 1, b + 1); 2(b −b)−1 II, (1, b, 1, b , 2, b ), which is the closure of the 3-braid w = Δ2(b+1) σ1−1 σ2 , where 1 ≤ b < b ; 2(a−1)+1 2(b −b)+1 III, (a, b, a, b , 1, b ), which is the closure of the 3-braid w = Δ2b σ1 σ2 , where 1 ≤ a ≤ b , 1 ≤ b < b , and if a = 1, we require b − b ≥ 2.
Moreover, Type Ia and Type Ib represent torus knots of knot type (3, 3b + 1) and (3, 3b + 2) respectively.
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Remark 2.4. (1). There is no the condition “if a = 1, we require b − b ≥ 2” of Type III case in [1]. But if a = 1 and b = b + 1, it is of Type Ib, so we add this condition. (2). Note that there is a typo in the proof of Theorem 1 of [1], so we use a instead of a in [1]. Lemma 2.5. The braids in Type II and Type III of Theorem 2.3 are generic in the sense of Theorem 2.2. Proof. Note that a braid of Type II in Theorem 2.3 is already in a generic form. We now consider a Type III braid. 2(b −b)+1 by σ2−2 . Note that Δ2 lies in the center of the braid group, If a = 1, we conjugate w = Δ2b σ1 σ2 2(b −b)+1 we have w is conjugate to w = Δ2b σ22 σ1 σ2 σ2−2 , which is 2(b −b)+1 −2 σ2 .
Δ2b+2 σ22 (σ2−1 σ1−1 σ2−1 σ1−1 σ2−1 σ1−1 )σ1 σ2 Using the simple relation σ2 σ1 σ2 = σ1 σ2 σ1 , it then equals to 2(b −b) −2 σ2
Δ2b+2 σ22 σ2−1 σ1−1 σ2−1 σ1−1 σ2
2(b −b)−3
= Δ2b+2 σ1−1 σ2
,
which is generic if b − b ≥ 2. 2(a−1)+1 2(b −b)+1 σ2 is conjugate to If a ≥ 2, w = Δ2b σ1 2(b −b)+1 −1 σ2 .
w = Δ2b σ2 σ12a−1 σ2 We rewrite it as
2(b −b)−1
Δ2b+2 σ2 (σ2−1 σ1−1 σ2−1 σ1−1 σ2−1 σ1−1 )σ12a−1 (σ2 σ1 σ1−1 )σ2
,
it equals to 2(b −b)−1
Δ2b+2 (σ1−1 σ2−1 σ1−1 σ2−1 )(σ12a−2 σ2 σ1 )σ1−1 σ2
.
Using 2a − 2 times the simple relation σ2 σ1 σ2 = σ1 σ2 σ1 , we have 2(b −b)−1
w = Δ2b+2 (σ1−1 σ2−1 σ1−1 σ2−1 )(σ2 σ1 σ22a−2 )σ1−1 σ2
,
then we get 2(b −b)−1
w = Δ2b+2 σ1−1 σ22a−3 σ1−1 σ2
,
which is generic. 2 For the proof of Lemma 2.6, we introduce the band presentation of the 3-braid group B3 as in [6,31]: B3 = a1 , a2 , a3 |a2 a1 = a3 a2 = a1 a3 , where a1 = σ1 , a2 = σ2 , and a3 = σ1−1 σ2 σ1 . Now let δ = a2 a1 = a3 a2 = a1 a3 . We have the following simple relations: if k = 1, we have a−1 = δ −1 ai+1 , i
(2)
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and if k ≥ 2, we have a−k = δ −k ai−k+2 ai−k+3 · · · ai ai+1 , i
(3)
where the indices are given mod 3. Note that there are exactly k many of aj in the equation (3). Moreover, we have aj δ m = δ m aj+m
(4)
for all m ∈ Z. Note that there is also a typo for the above relations in [6]. Lemma 2.6. The braids in Type II and Type III of Theorem 2.3 are not conjugate to σ1p σ2q for any p, q ∈ Z. Proof. From (2), (3) and (4) above, for any word w in a1 , a2 and a3 , we can rewrite it as δ n P , where n ∈ Z, P is a positive word in a1 , a2 , a3 , and there is no adjacent letters ai+1 ai in P (here the subscripts are defined mod 3). Any such form of w is called a normal form, and n is the power of the normal form, the syllable length of P is called the syllable length of this normal form. Here by syllable length we mean the minimal number of syllables of type ani . For a word w in terms of a1 , a2 and a3 in B3 , the summit set of w is the subset of elements in the conjugacy class of w which has the maximal power. The maximal power is finite and the summit set of w is finite, see [6,31]. 2(b −b)−1 For a Type II Lorenz knot, which has a 3-braid representative w = Δ2(b+1) σ1−1 σ2 . We rewrite w as 2(b −b)−1
w = (a1 a2 )3(b+1) a−1 1 a2
2(b −b)−1
= (a1 a2 )3(b+1) δ −1 a2 a2
.
Then w equals to 2(b −b)
a1 δ 3b+2 a2 δ −1 a2
2(b −b)
= a1 δ 3b+2 δ −1 a1 a2
2(b −b)
= a1 δ 3b+1 a1 a2
2(b −b)
= δ 3b+1 a2 a1 a2
2(b −b)
= δ 3b+2 a2
.
2(b −b)
Now δ 3b+2 a2 is in a normal form and its power is 3b + 2 ≥ 5. 2(a−1)+1 2(b −b)+1 For a Type III Lorenz knot, which has a 3-braid representative w = Δ2b σ1 σ2 . Then w equals to 2(b −b)+1
= (a1 a2 )3b a2a−1 a2 1
.
We perform a tail move of w, which is just a conjugacy of the word w by a1 in this case, see [6,31]. We get that w is conjugate to 2(b −b)+1
w = (a2 a1 )3b−1 a2 a2a−1 a2 1
a1 ,
and then 2(b −b)
w = (a2 a1 )3b−1 a2 a2a−1 a2 1
2(b −b)
δ = δ 3b a3 a2a−1 a3 2
by (4). Then we have 2(b −b)
w = δ 3b+1 a2a−2 a3 2
.
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2(b −b)
Now δ 3b+1 a2a−2 a3 is in a normal form and its power is 3b + 1 ≥ 4. 2 For σ1p σ2q = ap1 aq2 , with p and q be two non-negative integers. It is in a normal form already. If p ≥ 1 and q ≥ 1, we perform a tail move on it, we get a new word aq2 ap1 = aq−1 (a2 a1 )ap−1 = aq−1 δap−1 = δaq−1 ap−1 . 2 1 2 1 3 1 q−1 p−1 Now δa3 a1 is in a normal form, its power is one, the syllable length is two. If we perform a tail move on δaq−1 ap−1 , we get δap−1 aq−1 , its power is one, the syllable length is two. Since the power does not increase 3 1 1 2 and the syllable length does not decrease. From [6,31], δaq−1 ap−1 lies in the summit set of ap1 aq2 . Then for 3 1 any element in the summit set of ap1 aq2 , the power of it is one. If, say p = 0, then easily, we have the power of any element in the summit set of σ2q = aq2 is zero. q For σ1−p σ2q = a−p 1 a2 , with p and q be two non-negative integers. From (2) and (3), we have q −p a−p a3−p a4−p · · · a0 a1 a2 aq2 , 1 a2 = δ
which has power −p and syllable length 2 − (3 − p) + 1 + 1 = p + 1. Now we perform a tail move on it, that is, we conjugate it by δ −p a3−p δ p , we get a new word w = δ −p a4−p · · · a0 a1 aq+1 (δ −p a3−p δ p ). 2 By (4), it equals to δ −p a4−p · · · a0 a1 aq+1 a3 . 2 q It also has power −p and syllable length p + 1. So from [6,31], for any element in the summit set of a−p 1 a2 , the power of it is −p ≤ 0. For σ1p σ2−q = ap1 a−q 2 , with p and q be two non-negative integers, p −q ap1 a−q a2−q+2 a2−q+3 · · · a2 a3 = δ −q ap1−q a2−q+2 a2−q+3 · · · a2 a3 . 2 = a1 δ
Note that a1−q = a2−q+2 , so δ −q ap1−q a2−q+2 a2−q+3 · · · a2 a3 has power −q and syllable length 3 − (4 − q) + 2 = q + 1. Now we perform a tail move on it, we get the new word q −q δ −q a5−q a6−q · · · a2 a3 (δ −q ap+1 a5−q a6−q · · · a2 a3 ap+1 , 1−q δ ) = δ 1
which also has power −q and syllable length q + 1. As above, for any element in the summit set of ap1 a−q 2 , the power of it is −q ≤ 0. Now, since the power of elements in the summit set of w is a conjugacy invariant of a braid w ∈ B3 , see q p −q [6,31]. We have a braid of Type II or Type III in Theorem 2.3 is not conjugate to ap1 aq2 , a−p 1 a2 , a1 a2 , for any non-negative integers p and q. Moreover, the exponent in a band representative of a braid of Type II or −q Type III in Theorem 2.3 is positive, it is not conjugate to a−p 1 a2 , for any non-negative integers p and q. 2 Now we can prove Theorem 1.2, we restate it as: Theorem 2.7. A 3-trip Lorenz knot is either a torus knot or a hyperbolic knot. It is hyperbolic if and only if it is of Type II or Type III. Proof. By Lemma 2.5, Lemma 2.6, Theorem 2.2 and Theorem 2.3, a Lorenz knot of Type II or Type III in Theorem 2.3 is hyperbolic, and a Lorenz knot of Type I in Theorem 2.3 is a torus knot. 2
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Corollary 2.8. The volume of 3-trip hyperbolic Lorenz knot is bounded above by 8ν8, where ν8 = 3.6638 . . . is the volume of a regular ideal octahedron in H3 . Proof. By Theorem 1.2, if K is a 3-trip hyperbolic Lorenz knot, then K is the closure of 2(b −b)−1
Δ2(b+1) σ1−1 σ2
,
or 2(b −b)−3
Δ2(b+1) σ1−1 σ2
,
or 2(b −b)−1
Δ2b+2 σ1−1 σ22a−3 σ1−1 σ2
.
Then by Theorem 5.6 of [14], we have the volume of a 3-trip hyperbolic Lorenz knot is bounded above by 4ν8 in the first two cases, and it is bounded above by 8ν8 in the third case. 2 3. Monodromy and the second approach In this section, we provide another approach to the classification of 3-trip Lorenz knots. The Dehornoy ordering
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If K is a Type III Lorenz knot corresponding to the Lorenz vector (a, b, a, b , 1, b ), which is the closure 2(a−1)+1 2(b −b)+1 of the 3-braid w = Δ2b σ1 σ2 , then the Dehornoy floor of w is at least b by the positivity of 2(a−1)+1 2(b −b)+1 σ1 σ2 . So the Dehornoy floor is at least three by our assumption. It is well-known that a 3-braid w acts on the 4-punctured sphere (see, for example, [14]). In other words, the braid group B3 is the mapping class group of a 3-punctured disk, so consider the projection map from B3 to M CG(F04 ), where the mapping class group of the 4-punctured sphere is just SL(2, Z). The projection map is the following: σ1 →
1 0 1 1
,
σ2−1
→
1 1 0 1
,
and Δ2 projects to the identity in SL(2, Z). −1 1−2(b −b) 1 0 1 1 is 1 + 2(b − b), it is at least 3. Which in The trace of the matrix 1 1 0 1 turn means w projects to a pseudo-Anosov element in the mapping class group of the 4-punctured 2(b −b)−1 sphere. So w = Δ2(b+1) σ1−1 σ2 is a pseudo-Anosov braid. Similarly, the corresponding matrix of 2(a−1)+1 2(b −b)+1 w = Δ2b σ1 σ2 in SL(2, Z) has trace 1 + (b − b)(4a − 2) + 2a, so w is also a pseudo-Anosov braid. Now by Theorem 1.3 of [18], if w is a pseudo-Anosov braid with Dehornoy at least three, then the closure of w is a hyperbolic knot. 2 Remark 3.2. Even though the generators {σ1 , σ2 , · · · , σn−1 } of the braid group Bn in our note are different from the usual ones, such as in [18], this does not affect the result above. For example, if 2(b −b)−1 w = Δ2(b+1) σ1−1 σ2 , then Δ2b
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Fig. 3. Lorenz braid associated to the Lorenz vector (1, 1, 1, b, 2, b).
Proof. The corresponding Lorenz word of the Lorenz vector (1, 1, 1, b, 2, b) is w = xyxy b x2 y b . Let s(w) be the shift of w, that is, if w = x1 x2 · · · xn , xi equals x or y, then s(w) = x2 x3 · · · xn x1 . If ν is obtained from μ by a shift, we denote this process by μ → ν. Consider the set of shifts of w, which are {x2 y b xyxy b → xy b xyxy b x → y b xyxy b x2 → y b−1 xyxy b x2 y → · · · → yxyxy b x2 y b−1 → xyxy b x2 y b → yxy b x2 y b x → xy b x2 y b xy → y b x2 y b xyx → y b−1 x2 y b xyxy → · · · → yx2 y b xyxy b−1 → x2 y b xyxy b }. We denote the set of shifts of w by {w1 < w2 < · · · < w2b+4 < w2b+5 }, here we use the lexicographical rule x < 0 < y as in [3]. There is only one word in the set of all shifts of w which begins with x2 . It is w1 = x2 y b xyxy b , which is the smallest one in the set of shifts of w. There are three words which begins with xy, which are w2 = xyxy b x2 y b , w3 = xy b x2 y b xy and w4 = xy b xyxy b x. There are three words which begins with y, which are w5 = yx2 y b xyxy b−1 , w6 = yxyxy b x2 y b−1 and w7 = yxy b x2 y b x. There are two words which begins with y 2 , which are w8 = y 2 x2 y b xyxy b−2 and w9 = y 2 xyxy b x2 y b−2 . There are two words which begins with y b , which are w2b+4 = y b x2 y b xyx and w2b+5 = y b xyxy b x2 . So the permutations between the shifts of w are {w1 → w4 → w2b+5 → w2b+3 → w2b+1 → · · · w9 → w6 → w2 → w7 → w3 → w2b+4 → w2b+2 · · · → w10 → w8 → w5 → w1 }. From above, by Algorithm 2.4.3 of [3], the Lorenz braid of w is shown in Fig. 3. Since a Lorenz braid is a simple braid, it is determined by the permutations on its strands. We adopt the convention of drawing the figure such that the over-crossing strands are thicker than the under-crossing strands. Now the corresponding Young diagram is shown in the left-side of Fig. 4, with height (2b, 4, 2). From Fig. 3, it is easy to see the trip number is three. 2 Now we consider a Lorenz knot of Type III, which corresponding to the Lorenz vector (a, b, a, b , 1, b ) with b equals to 1, that is, (a, 1, a, b , 1, b ). We rewrite it as (a, 1, a, b, 1, b) with 1 ≤ a ≤ b, and b is at least two. We first consider the case that a ≥ 2.
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Fig. 4. Young diagrams: the left-side with height (4, 4, 2) and the right-side with height (4, 3, 2, 1).
Fig. 5. Lorenz braid associated to the Lorenz vector (a, 1, a, b, 1, b).
Lemma 3.4. The Lorenz vector (a, 1, a, b, 1, b) with 2 ≤ a ≤ b corresponds to the Young diagram with height (2b, 3, 2, 1, · · · , 1), where there are 2a − 3 many of 1. Proof. The corresponding Lorenz word is w = xa yxa y b xy b . Consider the set of all shifts of w, we denote it by {w1 < w2 < · · · < w2a+2b+1 < w2a+2b+2 }. There are two words which begins with xa , which are w1 = xa yxa y b xy b and w2 = xa y b xy b xa y. There are two words which begins with xi , for each 2 ≤ i ≤ a − 1, which are w3 , w4 , · · · , w2a−2 . There are three words which begins with xy, which are w2a−1 = xyxa y b xy b xa−1 , w2a = xy b xa yxa y b and w2a+1 = xy b xy b xa yxa−1 . There are three words which begins with yx, which are w2a+2 = yxa yxa y b xy b−1 , w2a+3 = yxa y b xy b xa and w2a+4 = yxy b xa yxa y b−1 . There are two words which begins with y i , for each 2 ≤ i ≤ b, which are w2a+5 = y 2 xa yxa y b xy b−2 , · · · , w2a+2b+1 = y b xa yxa y b x, and w2a+2b+2 = y b xy b xa yxa . So the permutations between the shifts of w are {w1 → w3 → w5 → · · · → w2a−1 → w2a+3 → w2 → w4 → · · · → w2a−2 → w2a+1 → w2a+2b+2 → w2a+2b → · · · → w2a+4 → → w2a → w2a+2b+1 · · · → w2a+5 → w2a+2 → w1 }. By Algorithm 2.4.3 of [3], the Lorenz braid is shown in Fig. 5. The corresponding Young diagram has height (2b, 3, 2, 1, · · · , 1), there are 2a − 3 many of 1. In the right-side of Fig. 4, we show the case a = b = 2, so the Young diagram has height (4, 3, 2, 1). 2 The proof of the following lemmas just like the proof of Lemma 3.3 and Lemma 3.4. We omits the details. Lemma 3.5 handle Type III case with a = b = 1, so the Dehornoy floor is 1. We have (1, 1, 1, b , 1, b ) and b ≥ 3 by the assumption of Theorem 2.3. We rewrite it as (1, 1, 1, b, 1, b) and b ≥ 3. Lemma 3.5. The Lorenz vector (1, 1, 1, b, 1, b) with b ≥ 3 corresponds to the Young diagram with height (2b, 4).
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The following lemma treat Type II case with Dehornoy floor 2, that is, (1, b, 1, b , b, b ) in Type II of Theorem 2.3 with b = 2 and b ≥ 3. We rewrite it as (1, 2, 1, b, 2, b) with b ≥ 3. Lemma 3.6. The Lorenz vector (1, 2, 1, b, 2, b) with b ≥ 3 corresponds to the Young diagram with height (2b + 1, 7, 2). The following two Lemmas treat Type III case with Dehornoy floor 2. If a = 1 and b = 2 in Type III of Theorem 2.3, that is, (1, 2, 1, b , 1, b ) with b ≥ 4 by the assumption of Theorem 2.3. We rewrite it as (1, 2, 1, b, 1, b) with b ≥ 4. Lemma 3.7. The Lorenz vector (1, 2, 1, b, 1, b) with b ≥ 4 corresponds to the Young diagram with height (2b + 1, 7). If a ≥ 2 and b = 2 in Type III of Theorem 2.3, that is, (a, 2, a, b , 1, b ) with b ≥ 3 by the assumption of Theorem 2.3. We rewrite it as (a, 2, a, b, 1, b) with b ≥ 3. Lemma 3.8. The Lorenz vector (a, 2, a, b, 1, b) with a ≥ 2 and b ≥ 3 corresponds to the Young diagram with height (2b + 1, 6, 2, 1, 1, · · · , 1), where there are 2a − 3 many of 1. Remark 3.9. It is easy to see Lorenz braids corresponding to Young diagrams in Lemmas 3.3–3.8 have trip number three, see [3,4]. The second approach of Theorem 1.2 modulo Question 1.3: It is easy to see the Lorenz knot with Young diagram (4, 4, 2) is the closure of the braid w = σ4 σ3 σ5 σ2 σ4 σ6 σ1 σ3 σ5 σ7 σ2 σ4 σ6 σ8 σ3 σ5 σ7 σ6 , which is hyperbolic with volume approximate 2.82812208833 by SnapPy [9]. Now if the Lorenz knot with Young diagram (2b, 4, 2) is hyperbolic, then the fiber monodromy f of the knot complement is pseudo-Anosov by Thurston’s theorem [29,23]. The fiber monodromy f of the Lorenz knot with Young diagram (2(b + 1), 4, 2) is obtained from f by two times of essential positive non-trivial Hopf plumbings, or by an essential trefoil plumbing, which is easy to see from [11,12]. So if we assume Question 1.3 is true, then f is pseudo-Anosov, and the Lorenz knot with Young diagram (2(b + 1), 4, 2) is hyperbolic. By induction, every Lorenz knot corresponding to the Young diagram (2b, 4, 2) for any b at least two is hyperbolic. The case that the Lorenz knot corresponding to the Young diagram (2b, 3, 2, 1, · · · , 1) is similar. The Lorenz knot with the Young diagram (4, 3, 2, 1) is the closure of the braid w = σ5 σ4 σ6 σ3 σ5 σ7 σ2 σ4 σ6 σ8 σ1 σ3 σ5 σ7 σ2 σ4 σ6 σ9 σ8 , which is hyperbolic with volume approximate 5.916745735 by SnapPy [9]. The monodromy of a Lorenz knot with the Young diagram (2b, 3, 2, 1, · · · , 1), where there are 2a −3 many of 1, is obtained from the monodromy of the Lorenz knot with the Young diagram (4, 3, 2, 1) by b − 2 times of essential trefoil plumbings, and then a − 2 times of essential trefoil plumbings, which are independent of the performed b − 2 times of essential trefoil plumbings. By induction and by Question 1.3, every Lorenz knot corresponding to the Young diagram (2b, 3, 2, 1, · · · , 1) is hyperbolic for any a and b at least two. For the Lorenz knot corresponding to the Young diagram (2b, 4) with b ≥ 3, the first one is (6, 4), which is the closure of the braid w = σ3 σ2 σ4 σ1 σ3 σ5 σ2 σ4 σ6 σ3 σ5 σ7 σ4 σ6 σ8 σ5 σ7 σ9 σ8 .
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It is hyperbolic with volume approximate 2.82812208833 by SnapPy [9]. For the Lorenz knot with Young diagram (2b + 1, 7, 2) and b ≥ 3, the first one in this list is (7, 7, 2), which is the closure of the braid w = σ4 σ3 σ5 σ2 σ4 σ6 σ1 σ3 σ5 σ7 σ2 σ4 σ6 σ8 σ3 σ5 σ7 σ9 σ6 σ8 σ10 σ7 σ9 σ11 σ8 σ10 σ9 . It is hyperbolic with volume approximate 3.474 by SnapPy [9]. For the Lorenz knot with Young diagram (2b + 1, 7) and b ≥ 4, the first one is (9, 7), which is the closure of the braid w = σ3 σ2 σ4 σ1 σ3 σ5 σ2 σ4 σ6 σ3 σ5 σ7 σ4 σ6 σ8 σ5 σ7 σ9 σ6 σ8 σ10 σ7 σ9 σ11 σ8 σ10 σ12 σ11 . It is hyperbolic with volume approximate 3.47424776131 by SnapPy [9]. For the Lorenz knot with Young diagram (2b + 1, 6, 2, 1, 1, · · · , 1, 1) with a ≥ 2 and b ≥ 3, the first one is (7, 6, 2, 1), which is the closure of the braid w = σ5 σ4 σ6 σ3 σ5 σ7 σ2 σ4 σ6 σ8 σ1 σ3 σ5 σ7 σ9 σ2 σ4 σ6 σ8 σ10 σ7 σ9 σ11 σ8 σ10 σ12 σ9 σ11 . It is hyperbolic with volume approximate 7.0285041168 by SnapPy [9]. Now if we assume Question 1.3 has a positive answer, then by Theorem 2.3, Lemma 3.1, Lemmas 3.3–3.8, we have a trip-3 Lorenz knot is either a torus knot or a hyperbolic knot. Acknowledgements The author would like to thank the referee for the detailed comments of the paper, which certainly helped to improve the manuscript. The author would also like to thank Youlin Li for the help on CorelDRAW and SnapPy. References [1] Richard E. Bedient, Classifying 3-trip Lorenz knots, Topol. Appl. 20 (1) (1985) 89–96. [2] Maxime Bergeron, Tali Pinsky, Lior Silberman, An upper bound for the volumes of complements of periodic geodesics, arXiv:1412.2446 [math.GT], 2014. [3] J.S. Birman, R. Williams, Knotted periodic orbits in dynamical systems, I: Lorenz equations, Topology 22 (1983) 47–82. [4] J.S. Birman, Ilya Kofman, A new twist on Lorenz links, J. Topol. 2 (2) (2009) 227–248. [5] J.S. Birman, William.W. Menasco, Studying links via closed braids, III: classifying links which are closed 3-braids, Pac. J. Math. 161 (1993) 25–113. [6] Joan S. Birman, William W. Menasco, A note on closed 3-braids, Commun. Contemp. Math. 10 (Suppl. 1) (2008) 1033–1047. [7] Abhijit Champanerkar, IIya Kofman, Timothy Mullen, The 500 simplest hyperbolic knots, J. Knot Theory Ramif. 23 (12) (2014) 1450055. [8] Vincent Colin, Ko Honda, Stabilizing the monodromy of an open book decomposition, Geom. Dedic. 132 (2008) 95–103. [9] Marc Culler, Nathan Dunfield, Jeff Weeks, SnapPy. http://www.math.uic.edu/t3m/SnapPy/. [10] Patrick Dehornoy, Dynnikov Patrick, Ivan Dynnikov, Dale Rolfsen, Bert Wiest, Ordering Braids, Mathematical Surveys and Monographs, vol. 148, American Mathematical Society, Providence, RI, 2008. [11] Pierre Dehornoy, On the zeroes of the Alexander polynomial of a Lorenz knot, Ann. Inst. Fourier (Grenoble) 65 (2) (2015) 509–548. [12] Pierre Dehornoy, Small dilatation homeomorphisms as monodromies of Lorenz knots, arXiv:1402.3765 [GT]. [13] Benson Farb, Margalit Dan, A Primer on Mapping Class Groups, Princeton Mathematical Series, vol. 49, Princeton University Press, Princeton, NJ, 2012. [14] David Futer, Efstratia Kalfagianni, Jessica S. Purcell, Cusp areas of Farey manifolds and applications to knot theory, Int. Math. Res. Not. (23) (2010) 4434–4497. [15] Étienne Ghys, Knots and dynamics, in: International Congress of Mathematicians, vol. I, Eur. Math. Soc., Zürich, 2007, pp. 247–277. [16] Paulo Gomes, Nuno Francoa, Luis Silva, Partial classification of Lorenz knots: syllable permutations of torus knots words, Physica D 306 (2015) 16–24.
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Kazuhiro Ichihara, Jiming Ma, A random link via bridge position is hyperbolic, Topol. Appl., in press. Tetsuya Ito, Braid ordering and the geometry of closed braid, Geom. Topol. 15 (1) (2011) 473–498. Kelmer Dubi, Quadratic irrationals and linking numbers of modular knots, J. Mod. Dyn. 6 (4) (2012) 539–561. E. Lorenz, Deterministic nonperiodic flow, J. Atmos. Sci. 20 (1963) 130–141. Jing Ma, The closure of a random braid is a hyperbolic link, Proc. Am. Math. Soc. 142 (2014) 695–701. Charles J. Mozzochi, Linking numbers of modular geodesics, Isr. J. Math. 195 (1) (2013) 71–95. Jean-Pierre Otal, Le théorème d’hyperbolisation pour les variétés fibrées de dimension 3, in: Astérisque, vol. 235, 1996, Translated into English as Jean-Pierre Otal, Leslie D. Kay, The Hyperbolization Theorem for Fibered 3-Manifolds, SMF/AMS Texts and Monographs, American Mathematical Society, 2001. Tali Pinsky, Templates for geodesic flows, Ergod. Theory Dyn. Syst. 34 (2014) 211–235. Peter Sarnak, Linking numbers of modular knots, Commun. Math. Anal. 8 (2) (2010) 136–144. O. Schreier, Uber die Gruppen AaBb = 1, Abh. Math. Semin. Univ. Hamb. 3 (1924) 167–169. Caroline Series, The modular surface and continued fractions, J. Lond. Math. Soc. (2) 31 (1) (1985) 69–80. Mark Pollicott, Distribution of closed geodesics on the modular surface and quadratic irrationals, Bull. Soc. Math. Fr. 114 (4) (1986) 431–446. William P. Thurston, Hyperbolic structures on 3-manifolds, II: surface groups and 3-manifolds which fiber over the circle, arXiv:math.GT/9801045, 1998. W. Tucker, A rigorous ODE solver and Smale’s 14th problem, Found. Comput. Math. 2 (2002) 53–117. P.J. Xu, The genus of closed 3-braids, J. Knot Theory Ramif. 1 (1992) 303–326.
Contents lists available at ScienceDirect
Topology and its Applications www.elsevier.com/locate/topol
Hyperbolicity of 3-trip Lorenz knots Jiming Ma 1 School of Mathematical Sciences, Fudan University, Shanghai, 200433, PR China
a r t i c l e
i n f o
Article history: Received 12 November 2015 Received in revised form 8 September 2016 Accepted 8 September 2016 Available online 12 September 2016 MSC: 57M50 20F36 37C10 57M25
a b s t r a c t Using Xu’s conjugacy algorithm on the braid group B3 [6,31] and the hyperbolicity criterion of 3-braid links by Futer–Kalfagianni–Purcell [14], we classify Lorenz knots of trip number 3 into torus knots and hyperbolic knots. Moreover, we provide another approach to this problem. Modulo a conjectural pseudo-Anosov criterion, we can also classify Lorenz knots of trip number 3 based upon the Dehornoy floor theory. The author believes that the second approach is promising for the hyperbolicity of more classes of Lorenz knots. © 2016 Elsevier B.V. All rights reserved.
Keywords: Braid group Hyperbolic knot Lorenz knot Symbolic dynamics Pseudo-Anosov map
1. Introduction 1.1. Lorenz knots and hyperbolicity Lorenz knots are a kinds of knots which are periodic orbits of the Lorenz differential equation as follows [20]: ⎧ dx ⎪ ⎪ ⎨ dt = 10(y − x), dy dt = 28x − y − xz, ⎪ ⎪ ⎩ dz = xy − 8 z, dt
1
E-mail address: [email protected]. The author was partially supported by NSFC 11371094.
http://dx.doi.org/10.1016/j.topol.2016.09.005 0166-8641/© 2016 Elsevier B.V. All rights reserved.
3
(1)
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Fig. 1. The Lorenz template and a Lorenz braid.
which draw much attention recently [15,4,12]. For a three dimensional flow, the knot types of its periodic orbits are useful for understanding the flow. Let F be the modular surface, that is, F = H2 /P SL(2, Z). Ghys [15,24] showed if we view the unit tangent bundle T 1 F of the modular surface as the complement of the trefoil knot in S3 , then the periodic orbits of the geodesic flow of T 1 F are the same as the set of Lorenz knots. So Lorenz knots are also called modular knots. Moreover, primitive geodesics in the modular surface and the periodic orbits of the geodesic flow are highly interesting from the viewpoint of number theory [27,28,19,22,25]. An important question raised in [3] is to classifying knot types which can occur in Lorenz knots. For example, every torus knot is a Lorenz knot, and every Lorenz knot is a fibered knot [3]. Moreover, certain satellite knots are Lorenz knots [3,16]. From [7], for the 502 simplest hyperbolic knots, that is, knots with their complement in S3 admit ideal triangulations with at most eight tetrahedra, 249 of them are Lorenz knots. The complement of a lift of a periodic geodesic in the unit tangent bundle of F is a hyperbolic 3-manifold if the curve is filling, see [2] for the study of hyperbolic geometry of Lorenz knots in the trefoil complement. But the complete classification of Lorenz knot remains completely open. In this note, we classify the simplest Lorenz knots: trip number 3 Lorenz knots. From our result, we believe that the probability of a random Lorenz knot is a hyperbolic knot should be one, comparing to [21,17]. By [3,30], we can study Lorenz knots combinatorially as follows. Definition 1.1. Let T be the branched surface with a semi-flow in S3 as in the left-side of Fig. 1. T is called the Lorenz template. A Lorenz knot K is a simple closed curve in T which runs along the direction of the semi-flow. For a Lorenz knot K embedded in T , we associate a symbol x whenever the loop passes along the left-hand hole in T . Similarly, we associate a symbol y whenever the loop passes along the right-hand hole in T . We get a word in the semigroup generated by x and y, which is called a Lorenz word. For example, in Fig. 1, the curve corresponds to the word xy, which is the trivial knot in S3 . The trip number of a Lorenz knot K is the minimal number of syllables of type xa y b in all Lorenz words associated to K. So every 3-trip Lorenz knot corresponds to a word xa1 y b1 xa2 y b2 xa3 y b3 , and then we denote it as a Lorenz vector by (a1 , b1 , a2 , b2 , a3 , b3 ) ∈ Z6≥1 as in [1]. Let Bn be the braid group with n strands. Let K be a Lorenz knot in the template T . If we cut the two wings of the template T , from the Lorenz knot K, we get a positive braid in some Bn . It is called a Lorenz braid associated to K. Here our definition of the generating set {σ1 , σ2 . . . σn−1 } of the braid group Bn is different from the standard one. Here we take the meanings as in [6,31]. The difference is a sign change. See Fig. 2. Based on Bedient [1], we show Theorem 1.2. A 3-trip Lorenz knot K is either a torus knot or a hyperbolic knot: (1) If K is a torus knot, then it is a torus knot of type (3, 3b + 1) or (3, 3b + 2), this corresponds to Lorenz knot with Lorenz vector (1, b, 1, b, 1, b + 1) or (1, b, 1, b + 1, 1, b + 1) respectively;
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Fig. 2. Positive braids.
(2) If K is a hyperbolic knot, then it is the closure of 2(b −b)−1
Δ2(b+1) σ1−1 σ2
for b > b ≥ 1, this corresponds to Lorenz knot with Lorenz vector (1, b, 1, b , 2, b ); Or the closure of 2(b −b)−3
Δ2(b+1) σ1−1 σ2
for b − b ≥ 2 and b ≥ 1, this corresponds to Lorenz knot with Lorenz vector (1, b, 1, b , 1, b ); Or the closure of 2(b −b)−1
Δ2b+2 σ1−1 σ22a−3 σ1−1 σ2
for 1 ≤ b < b , 2 ≤ a ≤ b , this corresponds to Lorenz knot with Lorenz vector (a, b, a, b , 1, b ). 1.2. A question on pseudo-Anosov maps and the second approach We also provide another approach to the classification of 3-trip Lorenz knots in Section 3. We divide 3-trip Lorenz knots into two classes. The first class is easy to show the hyperbolicity using Dehornoy floor theory. Moreover, we divide the second class into six subclasses. By Lemmas 3.3–3.8, we have that for each subclass, as fibered knots, their monodromies are somehow similar. We use SnapPy to show the first Lorenz knot in each subclass is a hyperbolic knot. By Thurston’s theory, its monodromy is a pseudo-Anosov map. We have a question on pseudo-Anosov maps, that is, Question 1.3 below. Modular Question 1.3, we can show the hyperbolicity of Lorenz knots in each subclass. Even through for 3-trip Lorenz knots, the second approach is much more involving, but we believe that the second approach is promising for the hyperbolicity of more classes of Lorenz knots. The formulation of Question 1.3 is motivated by Theorem 1.1 and Corollary 3.1 of [8]. The author believes Question 1.3 has a positive answer, but he has difficulties to write down the proof completely. Let F be a compact oriented surface with non-empty boundary. We take a proper arc γ in F , a neighborhood of γ in F is γ × [−1, 1]. By adding a 1-handle to F along ∂γ × [−1, 1], we get a new compact oriented surface F . Any mapping class f in M CG(F ) extends naturally to a mapping class in M CG(F ), which is the identity in the attached 1-handle, we also denoted it by f . The union of the 1-handle and γ × [−1, 1] is an annulus A, the core curve of A is denoted by β. Let Dβ be the positive Dehn twist along β, then Dβ ◦ f in M CG(F ) is called a (positive) Hopf plumbing of f , see [8,11]. If the arc γ is essential in F , then we say the Hopf plumbing is essential. Colin–Honda showed in [8] if f is a pseudo-Anosov map, then for a carefully chosen essential arc γ, f is also a pseudo-Anosov map. Moreover, they showed if f is a pseudo-Anosov map with fractional Dehn twist coefficient at least four, then for any essential Hopf plumbing, the resulting map f is pseudo-Anosov.
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Question 1.3. Let F be an oriented compact surface with one boundary and genus g ≥ 3. Let c1 , c2 , · · · , ck be a set of circles in F . For each j ≤ k, cj intersects cj−1 in one point, and cj does not intersect ci for i ≤ j − 2 or i ≥ j + 1. We assume f = Dc1 ◦ Dc2 · · · Dck−1 ◦ Dck is a pseudo-Anosov map on F . Let T be a genus one oriented surface with one boundary, a and b be two simple closed curves in T which intersect in one point. Let Da and Db be the (positive) Dehn twists along a and b, then h = Da Db is a map with periodicity six in the mapping class group of T . We perform a plumbing to (F, f ) by (T, h), such that a intersects ck in one point, and a does not intersect ci for 1 ≤ i ≤ k − 1, b does not intersect ci for 1 ≤ i ≤ k. We get a map f = Dc1 ◦ Dc2 · · · Dck−1 ◦ Dck ◦ Da ◦ Db in a new surface F , F has genus g + 1 and one boundary component. In other words, we perform an essential trefoil plumbing to (F, f ). Is f a pseudo-Anosov map?
2. Hyperbolicity of 3-trip Lorenz knots
We need an old theorem of Schreier [26], see also [5]: Theorem 2.1 (Schreier [26]). Let w ∈ B3 be a three strands braid. Then w is conjugate in B3 to a braid in exactly one of the following forms: (1). (2). (3). (4). (5).
Δ2k σ1−p1 σ2q1 σ1−p2 σ2q2 . . . σ1−ps σ2qs , where pi and qi are positive integers, k ∈ Z and Δ = σ1 σ2 σ1 ; Δ2k σ1p , where p, k ∈ Z; Δ2k σ1 σ2 , where k ∈ Z; Δ2k σ1 σ2 σ1 , where k ∈ Z; Δ2k σ1 σ2 σ1 σ2 , where k ∈ Z.
A 3-braid w is called generic if w is conjugate to Δ2k σ1−p1 σ2q1 σ1−p2 σ2q2 . . . σ1−ps σ2qs . The following is proved in Theorem 5.5 of [14]: Theorem 2.2 (Futer–Kalfagianni–Purcell [14]). Let w be a braid in B3 , and let K → S3 be the link obtained as the closure of w. Then S3 − K is hyperbolic if and only if w is generic, and w is not conjugate to σ1p σ2q for any integers p and q. Note that the notations σ1 and σ2 in [14] are different from ours here, we take the meanings as in [6,31]. If there is an orientation preserving self-homeomorphism h of S3 , which maps the knot K1 to the knot K2 , then we say K1 and K2 are equivalent. We need Theorem 1 from [1]: Theorem 2.3 (Bedient [1]). Every 3-trip Lorenz knot is equivalent to exactly one of the following knots: Type Type Type Type
Ia, (1, b, 1, b, 1, b + 1); Ib, (1, b, 1, b + 1, 1, b + 1); 2(b −b)−1 II, (1, b, 1, b , 2, b ), which is the closure of the 3-braid w = Δ2(b+1) σ1−1 σ2 , where 1 ≤ b < b ; 2(a−1)+1 2(b −b)+1 III, (a, b, a, b , 1, b ), which is the closure of the 3-braid w = Δ2b σ1 σ2 , where 1 ≤ a ≤ b , 1 ≤ b < b , and if a = 1, we require b − b ≥ 2.
Moreover, Type Ia and Type Ib represent torus knots of knot type (3, 3b + 1) and (3, 3b + 2) respectively.
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Remark 2.4. (1). There is no the condition “if a = 1, we require b − b ≥ 2” of Type III case in [1]. But if a = 1 and b = b + 1, it is of Type Ib, so we add this condition. (2). Note that there is a typo in the proof of Theorem 1 of [1], so we use a instead of a in [1]. Lemma 2.5. The braids in Type II and Type III of Theorem 2.3 are generic in the sense of Theorem 2.2. Proof. Note that a braid of Type II in Theorem 2.3 is already in a generic form. We now consider a Type III braid. 2(b −b)+1 by σ2−2 . Note that Δ2 lies in the center of the braid group, If a = 1, we conjugate w = Δ2b σ1 σ2 2(b −b)+1 we have w is conjugate to w = Δ2b σ22 σ1 σ2 σ2−2 , which is 2(b −b)+1 −2 σ2 .
Δ2b+2 σ22 (σ2−1 σ1−1 σ2−1 σ1−1 σ2−1 σ1−1 )σ1 σ2 Using the simple relation σ2 σ1 σ2 = σ1 σ2 σ1 , it then equals to 2(b −b) −2 σ2
Δ2b+2 σ22 σ2−1 σ1−1 σ2−1 σ1−1 σ2
2(b −b)−3
= Δ2b+2 σ1−1 σ2
,
which is generic if b − b ≥ 2. 2(a−1)+1 2(b −b)+1 σ2 is conjugate to If a ≥ 2, w = Δ2b σ1 2(b −b)+1 −1 σ2 .
w = Δ2b σ2 σ12a−1 σ2 We rewrite it as
2(b −b)−1
Δ2b+2 σ2 (σ2−1 σ1−1 σ2−1 σ1−1 σ2−1 σ1−1 )σ12a−1 (σ2 σ1 σ1−1 )σ2
,
it equals to 2(b −b)−1
Δ2b+2 (σ1−1 σ2−1 σ1−1 σ2−1 )(σ12a−2 σ2 σ1 )σ1−1 σ2
.
Using 2a − 2 times the simple relation σ2 σ1 σ2 = σ1 σ2 σ1 , we have 2(b −b)−1
w = Δ2b+2 (σ1−1 σ2−1 σ1−1 σ2−1 )(σ2 σ1 σ22a−2 )σ1−1 σ2
,
then we get 2(b −b)−1
w = Δ2b+2 σ1−1 σ22a−3 σ1−1 σ2
,
which is generic. 2 For the proof of Lemma 2.6, we introduce the band presentation of the 3-braid group B3 as in [6,31]: B3 = a1 , a2 , a3 |a2 a1 = a3 a2 = a1 a3 , where a1 = σ1 , a2 = σ2 , and a3 = σ1−1 σ2 σ1 . Now let δ = a2 a1 = a3 a2 = a1 a3 . We have the following simple relations: if k = 1, we have a−1 = δ −1 ai+1 , i
(2)
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and if k ≥ 2, we have a−k = δ −k ai−k+2 ai−k+3 · · · ai ai+1 , i
(3)
where the indices are given mod 3. Note that there are exactly k many of aj in the equation (3). Moreover, we have aj δ m = δ m aj+m
(4)
for all m ∈ Z. Note that there is also a typo for the above relations in [6]. Lemma 2.6. The braids in Type II and Type III of Theorem 2.3 are not conjugate to σ1p σ2q for any p, q ∈ Z. Proof. From (2), (3) and (4) above, for any word w in a1 , a2 and a3 , we can rewrite it as δ n P , where n ∈ Z, P is a positive word in a1 , a2 , a3 , and there is no adjacent letters ai+1 ai in P (here the subscripts are defined mod 3). Any such form of w is called a normal form, and n is the power of the normal form, the syllable length of P is called the syllable length of this normal form. Here by syllable length we mean the minimal number of syllables of type ani . For a word w in terms of a1 , a2 and a3 in B3 , the summit set of w is the subset of elements in the conjugacy class of w which has the maximal power. The maximal power is finite and the summit set of w is finite, see [6,31]. 2(b −b)−1 For a Type II Lorenz knot, which has a 3-braid representative w = Δ2(b+1) σ1−1 σ2 . We rewrite w as 2(b −b)−1
w = (a1 a2 )3(b+1) a−1 1 a2
2(b −b)−1
= (a1 a2 )3(b+1) δ −1 a2 a2
.
Then w equals to 2(b −b)
a1 δ 3b+2 a2 δ −1 a2
2(b −b)
= a1 δ 3b+2 δ −1 a1 a2
2(b −b)
= a1 δ 3b+1 a1 a2
2(b −b)
= δ 3b+1 a2 a1 a2
2(b −b)
= δ 3b+2 a2
.
2(b −b)
Now δ 3b+2 a2 is in a normal form and its power is 3b + 2 ≥ 5. 2(a−1)+1 2(b −b)+1 For a Type III Lorenz knot, which has a 3-braid representative w = Δ2b σ1 σ2 . Then w equals to 2(b −b)+1
= (a1 a2 )3b a2a−1 a2 1
.
We perform a tail move of w, which is just a conjugacy of the word w by a1 in this case, see [6,31]. We get that w is conjugate to 2(b −b)+1
w = (a2 a1 )3b−1 a2 a2a−1 a2 1
a1 ,
and then 2(b −b)
w = (a2 a1 )3b−1 a2 a2a−1 a2 1
2(b −b)
δ = δ 3b a3 a2a−1 a3 2
by (4). Then we have 2(b −b)
w = δ 3b+1 a2a−2 a3 2
.
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2(b −b)
Now δ 3b+1 a2a−2 a3 is in a normal form and its power is 3b + 1 ≥ 4. 2 For σ1p σ2q = ap1 aq2 , with p and q be two non-negative integers. It is in a normal form already. If p ≥ 1 and q ≥ 1, we perform a tail move on it, we get a new word aq2 ap1 = aq−1 (a2 a1 )ap−1 = aq−1 δap−1 = δaq−1 ap−1 . 2 1 2 1 3 1 q−1 p−1 Now δa3 a1 is in a normal form, its power is one, the syllable length is two. If we perform a tail move on δaq−1 ap−1 , we get δap−1 aq−1 , its power is one, the syllable length is two. Since the power does not increase 3 1 1 2 and the syllable length does not decrease. From [6,31], δaq−1 ap−1 lies in the summit set of ap1 aq2 . Then for 3 1 any element in the summit set of ap1 aq2 , the power of it is one. If, say p = 0, then easily, we have the power of any element in the summit set of σ2q = aq2 is zero. q For σ1−p σ2q = a−p 1 a2 , with p and q be two non-negative integers. From (2) and (3), we have q −p a−p a3−p a4−p · · · a0 a1 a2 aq2 , 1 a2 = δ
which has power −p and syllable length 2 − (3 − p) + 1 + 1 = p + 1. Now we perform a tail move on it, that is, we conjugate it by δ −p a3−p δ p , we get a new word w = δ −p a4−p · · · a0 a1 aq+1 (δ −p a3−p δ p ). 2 By (4), it equals to δ −p a4−p · · · a0 a1 aq+1 a3 . 2 q It also has power −p and syllable length p + 1. So from [6,31], for any element in the summit set of a−p 1 a2 , the power of it is −p ≤ 0. For σ1p σ2−q = ap1 a−q 2 , with p and q be two non-negative integers, p −q ap1 a−q a2−q+2 a2−q+3 · · · a2 a3 = δ −q ap1−q a2−q+2 a2−q+3 · · · a2 a3 . 2 = a1 δ
Note that a1−q = a2−q+2 , so δ −q ap1−q a2−q+2 a2−q+3 · · · a2 a3 has power −q and syllable length 3 − (4 − q) + 2 = q + 1. Now we perform a tail move on it, we get the new word q −q δ −q a5−q a6−q · · · a2 a3 (δ −q ap+1 a5−q a6−q · · · a2 a3 ap+1 , 1−q δ ) = δ 1
which also has power −q and syllable length q + 1. As above, for any element in the summit set of ap1 a−q 2 , the power of it is −q ≤ 0. Now, since the power of elements in the summit set of w is a conjugacy invariant of a braid w ∈ B3 , see q p −q [6,31]. We have a braid of Type II or Type III in Theorem 2.3 is not conjugate to ap1 aq2 , a−p 1 a2 , a1 a2 , for any non-negative integers p and q. Moreover, the exponent in a band representative of a braid of Type II or −q Type III in Theorem 2.3 is positive, it is not conjugate to a−p 1 a2 , for any non-negative integers p and q. 2 Now we can prove Theorem 1.2, we restate it as: Theorem 2.7. A 3-trip Lorenz knot is either a torus knot or a hyperbolic knot. It is hyperbolic if and only if it is of Type II or Type III. Proof. By Lemma 2.5, Lemma 2.6, Theorem 2.2 and Theorem 2.3, a Lorenz knot of Type II or Type III in Theorem 2.3 is hyperbolic, and a Lorenz knot of Type I in Theorem 2.3 is a torus knot. 2
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Corollary 2.8. The volume of 3-trip hyperbolic Lorenz knot is bounded above by 8ν8, where ν8 = 3.6638 . . . is the volume of a regular ideal octahedron in H3 . Proof. By Theorem 1.2, if K is a 3-trip hyperbolic Lorenz knot, then K is the closure of 2(b −b)−1
Δ2(b+1) σ1−1 σ2
,
or 2(b −b)−3
Δ2(b+1) σ1−1 σ2
,
or 2(b −b)−1
Δ2b+2 σ1−1 σ22a−3 σ1−1 σ2
.
Then by Theorem 5.6 of [14], we have the volume of a 3-trip hyperbolic Lorenz knot is bounded above by 4ν8 in the first two cases, and it is bounded above by 8ν8 in the third case. 2 3. Monodromy and the second approach In this section, we provide another approach to the classification of 3-trip Lorenz knots. The Dehornoy ordering
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If K is a Type III Lorenz knot corresponding to the Lorenz vector (a, b, a, b , 1, b ), which is the closure 2(a−1)+1 2(b −b)+1 of the 3-braid w = Δ2b σ1 σ2 , then the Dehornoy floor of w is at least b by the positivity of 2(a−1)+1 2(b −b)+1 σ1 σ2 . So the Dehornoy floor is at least three by our assumption. It is well-known that a 3-braid w acts on the 4-punctured sphere (see, for example, [14]). In other words, the braid group B3 is the mapping class group of a 3-punctured disk, so consider the projection map from B3 to M CG(F04 ), where the mapping class group of the 4-punctured sphere is just SL(2, Z). The projection map is the following: σ1 →
1 0 1 1
,
σ2−1
→
1 1 0 1
,
and Δ2 projects to the identity in SL(2, Z). −1 1−2(b −b) 1 0 1 1 is 1 + 2(b − b), it is at least 3. Which in The trace of the matrix 1 1 0 1 turn means w projects to a pseudo-Anosov element in the mapping class group of the 4-punctured 2(b −b)−1 sphere. So w = Δ2(b+1) σ1−1 σ2 is a pseudo-Anosov braid. Similarly, the corresponding matrix of 2(a−1)+1 2(b −b)+1 w = Δ2b σ1 σ2 in SL(2, Z) has trace 1 + (b − b)(4a − 2) + 2a, so w is also a pseudo-Anosov braid. Now by Theorem 1.3 of [18], if w is a pseudo-Anosov braid with Dehornoy at least three, then the closure of w is a hyperbolic knot. 2 Remark 3.2. Even though the generators {σ1 , σ2 , · · · , σn−1 } of the braid group Bn in our note are different from the usual ones, such as in [18], this does not affect the result above. For example, if 2(b −b)−1 w = Δ2(b+1) σ1−1 σ2 , then Δ2b
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Fig. 3. Lorenz braid associated to the Lorenz vector (1, 1, 1, b, 2, b).
Proof. The corresponding Lorenz word of the Lorenz vector (1, 1, 1, b, 2, b) is w = xyxy b x2 y b . Let s(w) be the shift of w, that is, if w = x1 x2 · · · xn , xi equals x or y, then s(w) = x2 x3 · · · xn x1 . If ν is obtained from μ by a shift, we denote this process by μ → ν. Consider the set of shifts of w, which are {x2 y b xyxy b → xy b xyxy b x → y b xyxy b x2 → y b−1 xyxy b x2 y → · · · → yxyxy b x2 y b−1 → xyxy b x2 y b → yxy b x2 y b x → xy b x2 y b xy → y b x2 y b xyx → y b−1 x2 y b xyxy → · · · → yx2 y b xyxy b−1 → x2 y b xyxy b }. We denote the set of shifts of w by {w1 < w2 < · · · < w2b+4 < w2b+5 }, here we use the lexicographical rule x < 0 < y as in [3]. There is only one word in the set of all shifts of w which begins with x2 . It is w1 = x2 y b xyxy b , which is the smallest one in the set of shifts of w. There are three words which begins with xy, which are w2 = xyxy b x2 y b , w3 = xy b x2 y b xy and w4 = xy b xyxy b x. There are three words which begins with y, which are w5 = yx2 y b xyxy b−1 , w6 = yxyxy b x2 y b−1 and w7 = yxy b x2 y b x. There are two words which begins with y 2 , which are w8 = y 2 x2 y b xyxy b−2 and w9 = y 2 xyxy b x2 y b−2 . There are two words which begins with y b , which are w2b+4 = y b x2 y b xyx and w2b+5 = y b xyxy b x2 . So the permutations between the shifts of w are {w1 → w4 → w2b+5 → w2b+3 → w2b+1 → · · · w9 → w6 → w2 → w7 → w3 → w2b+4 → w2b+2 · · · → w10 → w8 → w5 → w1 }. From above, by Algorithm 2.4.3 of [3], the Lorenz braid of w is shown in Fig. 3. Since a Lorenz braid is a simple braid, it is determined by the permutations on its strands. We adopt the convention of drawing the figure such that the over-crossing strands are thicker than the under-crossing strands. Now the corresponding Young diagram is shown in the left-side of Fig. 4, with height (2b, 4, 2). From Fig. 3, it is easy to see the trip number is three. 2 Now we consider a Lorenz knot of Type III, which corresponding to the Lorenz vector (a, b, a, b , 1, b ) with b equals to 1, that is, (a, 1, a, b , 1, b ). We rewrite it as (a, 1, a, b, 1, b) with 1 ≤ a ≤ b, and b is at least two. We first consider the case that a ≥ 2.
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Fig. 4. Young diagrams: the left-side with height (4, 4, 2) and the right-side with height (4, 3, 2, 1).
Fig. 5. Lorenz braid associated to the Lorenz vector (a, 1, a, b, 1, b).
Lemma 3.4. The Lorenz vector (a, 1, a, b, 1, b) with 2 ≤ a ≤ b corresponds to the Young diagram with height (2b, 3, 2, 1, · · · , 1), where there are 2a − 3 many of 1. Proof. The corresponding Lorenz word is w = xa yxa y b xy b . Consider the set of all shifts of w, we denote it by {w1 < w2 < · · · < w2a+2b+1 < w2a+2b+2 }. There are two words which begins with xa , which are w1 = xa yxa y b xy b and w2 = xa y b xy b xa y. There are two words which begins with xi , for each 2 ≤ i ≤ a − 1, which are w3 , w4 , · · · , w2a−2 . There are three words which begins with xy, which are w2a−1 = xyxa y b xy b xa−1 , w2a = xy b xa yxa y b and w2a+1 = xy b xy b xa yxa−1 . There are three words which begins with yx, which are w2a+2 = yxa yxa y b xy b−1 , w2a+3 = yxa y b xy b xa and w2a+4 = yxy b xa yxa y b−1 . There are two words which begins with y i , for each 2 ≤ i ≤ b, which are w2a+5 = y 2 xa yxa y b xy b−2 , · · · , w2a+2b+1 = y b xa yxa y b x, and w2a+2b+2 = y b xy b xa yxa . So the permutations between the shifts of w are {w1 → w3 → w5 → · · · → w2a−1 → w2a+3 → w2 → w4 → · · · → w2a−2 → w2a+1 → w2a+2b+2 → w2a+2b → · · · → w2a+4 → → w2a → w2a+2b+1 · · · → w2a+5 → w2a+2 → w1 }. By Algorithm 2.4.3 of [3], the Lorenz braid is shown in Fig. 5. The corresponding Young diagram has height (2b, 3, 2, 1, · · · , 1), there are 2a − 3 many of 1. In the right-side of Fig. 4, we show the case a = b = 2, so the Young diagram has height (4, 3, 2, 1). 2 The proof of the following lemmas just like the proof of Lemma 3.3 and Lemma 3.4. We omits the details. Lemma 3.5 handle Type III case with a = b = 1, so the Dehornoy floor is 1. We have (1, 1, 1, b , 1, b ) and b ≥ 3 by the assumption of Theorem 2.3. We rewrite it as (1, 1, 1, b, 1, b) and b ≥ 3. Lemma 3.5. The Lorenz vector (1, 1, 1, b, 1, b) with b ≥ 3 corresponds to the Young diagram with height (2b, 4).
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The following lemma treat Type II case with Dehornoy floor 2, that is, (1, b, 1, b , b, b ) in Type II of Theorem 2.3 with b = 2 and b ≥ 3. We rewrite it as (1, 2, 1, b, 2, b) with b ≥ 3. Lemma 3.6. The Lorenz vector (1, 2, 1, b, 2, b) with b ≥ 3 corresponds to the Young diagram with height (2b + 1, 7, 2). The following two Lemmas treat Type III case with Dehornoy floor 2. If a = 1 and b = 2 in Type III of Theorem 2.3, that is, (1, 2, 1, b , 1, b ) with b ≥ 4 by the assumption of Theorem 2.3. We rewrite it as (1, 2, 1, b, 1, b) with b ≥ 4. Lemma 3.7. The Lorenz vector (1, 2, 1, b, 1, b) with b ≥ 4 corresponds to the Young diagram with height (2b + 1, 7). If a ≥ 2 and b = 2 in Type III of Theorem 2.3, that is, (a, 2, a, b , 1, b ) with b ≥ 3 by the assumption of Theorem 2.3. We rewrite it as (a, 2, a, b, 1, b) with b ≥ 3. Lemma 3.8. The Lorenz vector (a, 2, a, b, 1, b) with a ≥ 2 and b ≥ 3 corresponds to the Young diagram with height (2b + 1, 6, 2, 1, 1, · · · , 1), where there are 2a − 3 many of 1. Remark 3.9. It is easy to see Lorenz braids corresponding to Young diagrams in Lemmas 3.3–3.8 have trip number three, see [3,4]. The second approach of Theorem 1.2 modulo Question 1.3: It is easy to see the Lorenz knot with Young diagram (4, 4, 2) is the closure of the braid w = σ4 σ3 σ5 σ2 σ4 σ6 σ1 σ3 σ5 σ7 σ2 σ4 σ6 σ8 σ3 σ5 σ7 σ6 , which is hyperbolic with volume approximate 2.82812208833 by SnapPy [9]. Now if the Lorenz knot with Young diagram (2b, 4, 2) is hyperbolic, then the fiber monodromy f of the knot complement is pseudo-Anosov by Thurston’s theorem [29,23]. The fiber monodromy f of the Lorenz knot with Young diagram (2(b + 1), 4, 2) is obtained from f by two times of essential positive non-trivial Hopf plumbings, or by an essential trefoil plumbing, which is easy to see from [11,12]. So if we assume Question 1.3 is true, then f is pseudo-Anosov, and the Lorenz knot with Young diagram (2(b + 1), 4, 2) is hyperbolic. By induction, every Lorenz knot corresponding to the Young diagram (2b, 4, 2) for any b at least two is hyperbolic. The case that the Lorenz knot corresponding to the Young diagram (2b, 3, 2, 1, · · · , 1) is similar. The Lorenz knot with the Young diagram (4, 3, 2, 1) is the closure of the braid w = σ5 σ4 σ6 σ3 σ5 σ7 σ2 σ4 σ6 σ8 σ1 σ3 σ5 σ7 σ2 σ4 σ6 σ9 σ8 , which is hyperbolic with volume approximate 5.916745735 by SnapPy [9]. The monodromy of a Lorenz knot with the Young diagram (2b, 3, 2, 1, · · · , 1), where there are 2a −3 many of 1, is obtained from the monodromy of the Lorenz knot with the Young diagram (4, 3, 2, 1) by b − 2 times of essential trefoil plumbings, and then a − 2 times of essential trefoil plumbings, which are independent of the performed b − 2 times of essential trefoil plumbings. By induction and by Question 1.3, every Lorenz knot corresponding to the Young diagram (2b, 3, 2, 1, · · · , 1) is hyperbolic for any a and b at least two. For the Lorenz knot corresponding to the Young diagram (2b, 4) with b ≥ 3, the first one is (6, 4), which is the closure of the braid w = σ3 σ2 σ4 σ1 σ3 σ5 σ2 σ4 σ6 σ3 σ5 σ7 σ4 σ6 σ8 σ5 σ7 σ9 σ8 .
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It is hyperbolic with volume approximate 2.82812208833 by SnapPy [9]. For the Lorenz knot with Young diagram (2b + 1, 7, 2) and b ≥ 3, the first one in this list is (7, 7, 2), which is the closure of the braid w = σ4 σ3 σ5 σ2 σ4 σ6 σ1 σ3 σ5 σ7 σ2 σ4 σ6 σ8 σ3 σ5 σ7 σ9 σ6 σ8 σ10 σ7 σ9 σ11 σ8 σ10 σ9 . It is hyperbolic with volume approximate 3.474 by SnapPy [9]. For the Lorenz knot with Young diagram (2b + 1, 7) and b ≥ 4, the first one is (9, 7), which is the closure of the braid w = σ3 σ2 σ4 σ1 σ3 σ5 σ2 σ4 σ6 σ3 σ5 σ7 σ4 σ6 σ8 σ5 σ7 σ9 σ6 σ8 σ10 σ7 σ9 σ11 σ8 σ10 σ12 σ11 . It is hyperbolic with volume approximate 3.47424776131 by SnapPy [9]. For the Lorenz knot with Young diagram (2b + 1, 6, 2, 1, 1, · · · , 1, 1) with a ≥ 2 and b ≥ 3, the first one is (7, 6, 2, 1), which is the closure of the braid w = σ5 σ4 σ6 σ3 σ5 σ7 σ2 σ4 σ6 σ8 σ1 σ3 σ5 σ7 σ9 σ2 σ4 σ6 σ8 σ10 σ7 σ9 σ11 σ8 σ10 σ12 σ9 σ11 . It is hyperbolic with volume approximate 7.0285041168 by SnapPy [9]. Now if we assume Question 1.3 has a positive answer, then by Theorem 2.3, Lemma 3.1, Lemmas 3.3–3.8, we have a trip-3 Lorenz knot is either a torus knot or a hyperbolic knot. Acknowledgements The author would like to thank the referee for the detailed comments of the paper, which certainly helped to improve the manuscript. The author would also like to thank Youlin Li for the help on CorelDRAW and SnapPy. References [1] Richard E. Bedient, Classifying 3-trip Lorenz knots, Topol. Appl. 20 (1) (1985) 89–96. [2] Maxime Bergeron, Tali Pinsky, Lior Silberman, An upper bound for the volumes of complements of periodic geodesics, arXiv:1412.2446 [math.GT], 2014. [3] J.S. Birman, R. Williams, Knotted periodic orbits in dynamical systems, I: Lorenz equations, Topology 22 (1983) 47–82. [4] J.S. Birman, Ilya Kofman, A new twist on Lorenz links, J. Topol. 2 (2) (2009) 227–248. [5] J.S. Birman, William.W. Menasco, Studying links via closed braids, III: classifying links which are closed 3-braids, Pac. J. Math. 161 (1993) 25–113. [6] Joan S. Birman, William W. Menasco, A note on closed 3-braids, Commun. Contemp. Math. 10 (Suppl. 1) (2008) 1033–1047. [7] Abhijit Champanerkar, IIya Kofman, Timothy Mullen, The 500 simplest hyperbolic knots, J. Knot Theory Ramif. 23 (12) (2014) 1450055. [8] Vincent Colin, Ko Honda, Stabilizing the monodromy of an open book decomposition, Geom. Dedic. 132 (2008) 95–103. [9] Marc Culler, Nathan Dunfield, Jeff Weeks, SnapPy. http://www.math.uic.edu/t3m/SnapPy/. [10] Patrick Dehornoy, Dynnikov Patrick, Ivan Dynnikov, Dale Rolfsen, Bert Wiest, Ordering Braids, Mathematical Surveys and Monographs, vol. 148, American Mathematical Society, Providence, RI, 2008. [11] Pierre Dehornoy, On the zeroes of the Alexander polynomial of a Lorenz knot, Ann. Inst. Fourier (Grenoble) 65 (2) (2015) 509–548. [12] Pierre Dehornoy, Small dilatation homeomorphisms as monodromies of Lorenz knots, arXiv:1402.3765 [GT]. [13] Benson Farb, Margalit Dan, A Primer on Mapping Class Groups, Princeton Mathematical Series, vol. 49, Princeton University Press, Princeton, NJ, 2012. [14] David Futer, Efstratia Kalfagianni, Jessica S. Purcell, Cusp areas of Farey manifolds and applications to knot theory, Int. Math. Res. Not. (23) (2010) 4434–4497. [15] Étienne Ghys, Knots and dynamics, in: International Congress of Mathematicians, vol. I, Eur. Math. Soc., Zürich, 2007, pp. 247–277. [16] Paulo Gomes, Nuno Francoa, Luis Silva, Partial classification of Lorenz knots: syllable permutations of torus knots words, Physica D 306 (2015) 16–24.
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