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TRANSACTIONS OF THE AMERICAN MATHMATICAL SOCIETY Volume 351 (1999), pages 2275--2294 DEHN S


TRANSACTIONS OF THE AMERICAN MATHMATICAL SOCIETY Volume 351 (1999), pages 2275–2294

DEHN SURGERY ON ARBORESCENT LINKS

Ying-Qing Wu
×?? ?? This paper studies Dehn surgery on a large class of links, called arborescent links. It will be shown that if an arborescent link L is su?ciently complicated, in the sense that it is composed of at least 4 rational tangles T (pi /qi ) with all qi > 2, and none of its length 2 tangles are of the form T (1/2q1 , 1/2q2 ), then all complete surgeries on L produce Haken manifolds. The proof needs some result on surgery on knots in tangle spaces. Let T (r/2s, p/2q) = (B, t1 ∪ t2 ∪ K) be a tangle with K a closed circle, and let M = B ? IntN (t1 ∪ t2 ). We will show that if s > 1 and p ≡ ±1 mod 2q, then ?M remains incompressible after all nontrivial surgeries on K. Two bridge links are a subclass of arborescent links. For such a link L(p/q), most Dehn surgeries on it are non-Haken. However, it will be shown that all complete surgeries yield manifolds containing essential laminations, unless p/q has a partial fraction decomposition of the form 1/(r ? 1/s), in which case it does admit nonlaminar surgeries.

0. Introduction In Dehn surgery theory, we would like to know what 3-manifolds are produced through certain surgeries on certain knots or links. More explicitly, we want to know how many surgeries yield Haken, hyperbolic, or laminar manifolds, and how many of them are “exceptional”, meaning that the resulting manifolds are reducible, or have cyclic or ?nite fundamental group, or are small Seifert ?bered spaces. There have been many results on these problems for surgery on knots. See [Gor] and [Ga] for surveys and frontier problems. These results, however, are not ready to be generalized to surgery on links of multiple components. The major di?culty is that surgery on one component of the link may change the property of the other components. An exception is Thurston’s hyperbolic surgery theorem [Th], which says that if L is a hyperbolic link, then except for ?nitely many slopes on each component of L, all other surgeries are hyperbolic. Another interesting result is Scharlemann’s simultaneous crossing change theorem, see [Sch]. There has been extensive study about surgery on a large class of knots called arborescent knots, also known as Conway’s algebraic knots [Co, BS], which include all Montesinos knots. The name “arborescent links” is ?rst used by Gabai [Ga2]. A knot or link is arborescent if it can be built by summing rational tangles together. See Section 1 for more detailed de?nitions. In [Oe] Oertel showed that surgeries on Montesinos knots of length ≥ 4 produce Haken manifolds. In [De1,De2] Delman showed that surgeries on all non-torus 2-bridge knots and most Montesinos knots are laminar, that is, they contain essential laminations. See [GO] for the de?nition and basic properties of laminar manifolds. In [Wu2] it was shown that if K is
1991 Mathematics Subject Classi?cation. Primary 57N10; Secondary 57M25, 57M50. 1

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a non-Montesinos arborescent knot, then all surgeries are laminar. Moreover, in most cases the surgeries are Haken and hyperbolic. Also, surgeries on 2-bridge knots have been classi?ed [BW], according to whether the resulting manifold is toroidal, Seifert ?bered, or “hyperbolike” in the sense that it would be hyperbolic if the hyperbolization conjecture is true. These results provide satisfactory understanding of surgery on all arborescent knots, except certain Montesinos knots of length 3. In particular, all arborescent knots have Strong Property P [Wu2], that is, manifolds obtained by surgery on such knots do not contain a fake 3-ball. Let L = l1 ∪ . . . ∪ ln be a link of n components; let γ = γ1 ∪ . . . ∪ γn , where γi is a slope on ?N (li ). We use L(γ) to denote the manifold obtained by γ surgery on L. More precisely, L(γ) is obtained by gluing n solid tori V1 , . . . , Vn to S 3 ? IntN (L) along their boundary so that each γi bounds a meridional disk in Vi . If some γi is a meridian of li , then L(γ) is the same as a surgery on a sub-link. We say that the γ surgery is a complete surgery if all γi are non-meridional slopes. In [Oe] Oertel studied closed incompressible surfaces in the exteriors of Montesinos links L. He also showed that if L is a knot with length ≥ 4 and is composed of rational tangles T (pi /qi ) with qi ≥ 3, then nontrivial Dehn surgeries on L always produce Haken manifolds. Note, however, that this is not true in general if L is a Montesinos link, as demonstrated in Remark 0.1(b) below. We need an extra condition. The following is one of the main theorems of the paper. It says that if we further require that none of the length 2 tangles of L is of the form T (1/2q1, 1/2q2 ), then all complete surgeries on L are Haken, and this is true even if L is an arborescent link. One is referred to section 1 for de?nitions of algebraic tangles, arborescent links, and their length. Theorem 3.4. Let L = k1 ∪ . . . ∪ kn be an arborescent link of length ≥ 4, so that (1) L is composed of rational tangles T (pi /qi ) with qi > 2, and (2) no length 2 tangle of L is a Montesinos tangle of the form T (1/2q1, 1/2q2 ) for some integers q1 , q2 . Then all complete surgeries on L produce Haken manifolds.

l2

l

1

Figure 0.1 Remark 0.1. (a) Consider the link L in Figure 0.1. It is easy to see that L is an arborescent link of two components. Surgery on k1 with coe?cient +1 yields S 3 , in which k2 is a trivial knot, so all surgeries on k2 yields non-Haken manifolds. The example can be modi?ed so that the length of L is arbitrarily large. It shows that condition (1) in the theorem is necessary. For another example one can take K to be a knot which is the union of two tangles T (1/2, p1/q1 ) and T (1/2, p2/q2 ). By choosing pi /qi properly K will satisfy condition (2) of the theorem, but it has been shown in [Wu2] that most integral surgeries on K are non-Haken.

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(b) Consider the Montesinos link L = L(1/2n1, . . . , 1/2nk ). There is a nonseparating planar surface P in S 3 with one boundary component on each component of L. Thus surgery along the boundary slope of P yields a reducible manifold. Therefore condition (2) in the theorem is necessary. However (2) can be removed if L has only two components. Theorem 4.2. Suppose L = l1 ∪ l2 is an arborescent link of two components. If L has length at least 4, and none of the rational tangles used to build L is a (1/2)-tangle, then all complete surgeries on L produce Haken manifolds. In section 5 we study surgeries on 2-bridge links. Since there are no closed essential surfaces in the complement of such links, most surgeries are non-Haken [Fl]. However, the following theorem shows that, like in the case of 2-bridge knots, surgeries on most 2-bridge links are laminar. Theorem 5.1. A non-torus 2-bridge link L = L(p/q) admits complete, nonlaminar surgeries if and only if p/q = 1/(r ? 1/s) for some odd integers r and s. This generalizes Delman’s result about surgery on 2-bridge knots [De1]. The theorem is proved by applying Delman’s construction of essential branched surfaces to certain “allowable paths” in the minimal diagram associated to p/q. It completely determines which 2-bridge link complements contain persistent laminations. The paper is organized as follows. In Section 1 we give some de?nitions, and a few easy lemmas. Section 2 is to study Dehn surgeries on a closed circle component of a tangle. The following result should be of independent interest. Theorem 2.7. Let T = T (r/2s, p/2q) = (B, t1 ∪ t2 ∪ K) be a tangle such that s > 1, and p ≡ ±1 mod 2q. Let K be the closed component of T , and let M = B ? IntN (t1 ∪ t2 ). Then ?M is incompressible in (M, K; γ) for all γ = m, where m is the meridional slope of K. This is used in Sections 3 and 4 to prove Theorems 3.4 and 4.2. Section 5 is to study essential laminations and surgery on 2-bridge links, and prove Theorem 5.1. 1. Preliminaries All 3-manifolds in this paper are assumed orientable and compact. We refer the reader to [He] for basic concepts about 3-manifolds. If X is a subset of a 3-manifold M , we use N (X) to denote a regular neighborhood of X, and use |X| to denote the number of components in X. For our purpose, we de?ne a tangle to be a pair (B, T ), where B is a 3-ball, and T is a properly embedded 1-manifold, containing two strings t1 , t2 and some circles. We use E(T ) to denote the tangle space B ? IntN (T ). If T is properly isotopic to a pair of arcs on ?B then (B, T ) is called a trivial tangle. A rational tangle is a triple (B, T ; D), where (B, T ) is a trivial tangle, and D is a disk on ?B containing two ends of T . We assign a rational number or ∞ to the tangle as follows. Let F be a torus which double branch covers ?B with branch set ?T . Let m be a component of the lifting of ?D, and let l be a curve on F intersecting m once. Orient m, l so that the intersection number of m with l is +1 with respect to the orientation of F induced from a ?xed orientation of ?B. (The pair (m, l) is called a coordinate system of F .) Let γ be a curve on ?B which bounds a disk ? in B separating

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the strings of T . Then the lifting of γ represents some pm + ql in H1 (F ). We say that (B, T ; D) is a p/q rational tangle, and use T (p/q) to denote it. Because of the ambiguity of the choice of l, the number p/q is de?ned mod Z. Thus T (r) = T (r′ ) if and only if r = r′ mod Z. One can check that if a tangle is a (p, q) rational tangle in the usual sense (see e.g. [HT]), and if we choose the left hand side disk as the disk D, then it is a T (p/q) according to our de?nition. Given two tangles (B1 , T1 ) and (B2 , T2 ), we can choose a disk Di on ?Bi , then glue the two disks Di together to form a new tangle (B, T ). We say that (B, T ) is a sum of (B1 , T1 ; D1 ) and (B2 , T2 ; D2 ), and write (B, T ) = (B1 , T1 ) ∪D (B2 , T2 ), where D = D1 = D2 . This process depends on the choice of Di and the gluing map. When Di is not important, we will simply say that (B, T ) is the sum of (B1 , T1 ) and (B2 , T2 ). If neither of (Bi , Ti ; Di ) is T (0) or T (∞), we say that the sum is a nontrivial sum. A tangle is called an algebraic tangle if it is obtained by nontrivially summing rational tangles together in various ways. Thus a nontrivial sum of algebraic tangles is still an algebraic tangle. A Montesinos tangle T (r1 , . . . , rn ) is obtained by gluing rational tangles T (ri ) = (Bi , Ti ; Di ) together so that Di is glued to the disk ?Bi+1 ? IntDi+1 , where ri are non integral rational numbers. When further gluing Dn to ?B1 ? IntD1 , we get a Montesinos link L(r1 , . . . , rn ), also known as star link [Oe]. The number n is called the length of L. When n = 1, L is a 2-bridge link. Given two tangles (B1 , T1 ) and (B2 , T2 ), we may glue the boundaries of the Bi together so that T1 ∪T2 becomes a link L in S 3 = B1 ∪B2 . A link L is an arborescent link if either it is a Montesinos link, or (S 3 , L) is the union of two algebraic tangles (B1 , T1 ) and (B2 , T2 ), each of which has length at least 2. Note that if (B1 , T1 ) has ′ ′ ′′ ′′ length ≥ 3 and (B2 , T2 ) has length = 1 then (B1 , T1 ) = (B1 , T1 ) ∪ (B1 , T1 ), where ′ ′ 3 ′ ′ length of (B1 , T1 ) is ≥ 2, so (S , L) can be rewritten as a union of (B1 , T1 ) and ′ ′ ′ ′ ′′ ′′ (B2 , T2 ), where (B2 , T2 ) is a sum of (B1 , T1 ) and (B2 , T2 ). The above restriction ′ ′ ′′ ′′ about the length of (Bi , Ti ) is to make sure that (B2 , T2 ) = (B1 , T1 ) ∪ (B2 , T2 ) is a nontrivial sum. Let L = l1 ∪ . . . ∪ ln be a link in a 3-manifold M . A slope γ is a set of curves γ1 ∪ . . . ∪ γn , where γi is an essential curve on ?N (li ). We use (M, L; γ) to denote the manifold obtained from M by surgery on L along γ, that is, (M, L; γ) = (M ? IntN (L)) ∪ (V1 ∪ . . . ∪ Vn ), where γi bounds a disk in the solid torus Vi . When M = S 3 , the surgered manifold (M, L; γ) is simply denoted by L(γ). Lemma 1.1. Let T (p/q) = (B, t1 ∪ t2 ; D) be a rational tangle. If ?D bounds a disk ? in E(T ), then q = 0. Proof. The disk ? separates the two strings t1 and t2 . Since both ti are trivial strings in B, t1 is rel ?t1 isotopic to a string on D or ?B ? D without crossing t2 . Lemma 1.2. Let T (p/q) = (B, T ; D) be a rational tangle. If ?E(T ) ? ?D is compressible in E(T ), then q ≤ 1. Proof. Let W be the manifold obtained from E(T ) by attaching a 2-handle to E(T ) along ?D. Note that W is the exterior of a 2-bridge link L(p/q). We use the fact that if q ≥ 2 then L is a nontrivial link. In particular, ?W is incompressible. Let ? be a compressing disk of ?E(T ) ? ?D in E(T ). Then ? is a compressing disk of ?W unless ?? is a trivial curve on ?W , which happens only if (1) ??

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is parallel to ?D, or (2) ?? bounds a torus on ?E(T ) containing ?D as a nonseparating curve. In the ?rst case, ?D bounds a disk in E(T ), so by Lemma 1.1 we have q = 0. In the second case, ? cuts E(T ) into two solid tori, because E(T ) is a handlebody of genus 2. A meridian disk of the solid torus which does not contain ?D would then be a compressing disk of ?W . 2. Surgery on knots in tangle spaces Let (B, T ) = (B, t1 ∪ t2 ∪ K) be a tangle, where ti are arcs, and K is a closed circle in B. After γ surgery on K, we get a manifold B ′ = (B, K; γ). We want to know whether the manifold M = B ′ ? IntN (t1 ∪ t2 ) is irreducible and ?-irreducible. Recall that a manifold M is ?-irreducible if its boundary is incompressible. The main result of this section is Theorem 2.7, which solves this problem for Montesinos tangles of length 2. The result will be used in the next section to prove Theorem 3.4. We use m to denote a meridian of the knot K. Lemma 2.1. Let F be a compressible surface on the boundary of a 3-manifold M , let K be a knot in M . If F is compressible in (M, K; γ) for some γ = m, then it is compressible in (M, K; γ ′ ) for some γ ′ with ?(γ ′ , m) = 1. Proof. We may assume that F is incompressible in M ?K, and M ?IntN (K) is not a T 2 × I, otherwise F is compressible in (M, K; γ) for all γ. If there is no essential annulus A in M ? IntN (K) with one boundary on each of F and ?N (K), then by [Wu3, Theorem 1], F is incompressible in (M, K; γ) unless ?(γ, m) ≤ 1, so we may choose γ ′ = γ. Now assume that there is an essential annulus A in M ? IntN (K) with one boundary on each of F and ?N (K), and let γ0 be the slope A ∩ ?N (K). If γ0 = m, then again F is compressible in (M, K; γ) only if ?(γ, m) ≤ 1 [CGLS, Theorem 2.4.3], so we can choose γ ′ = γ. If γ0 = m, by [CGLS, Theorem 2.4.3] F is compressible in (M, K; γ) if and only if ?(γ0 , γ) ≤ 1. In particular, ?(γ0 , m) = 1. Thus if we choose γ ′ = γ0 + m (homologically), then ?(m, γ ′ ) = 1. Since ?(γ0 , γ ′ ) = ?(γ0 , γ0 + m) = 1, F is compressible in (M, K; γ ′ ). The following result is due to Starr [St]. Lemma 2.2. Let C be a simple closed curve on the boundary of a handlebody M . Then ?M ? C is incompressible in M if and only if there exists a set of disks D1 , . . . , Dk in M , such that (1) ?D1 ∪ . . . ∪ ?Dk cuts ?M into P1 , . . . , Pr , each of which is a pair of pants, (2) C ∩ Pi are essential arcs, with at least one arc connecting any two components of ?Pi . Lemma 2.3. Let T (p/q) = (B, T ; D) = (B, t1 ∪ t2 ; D). Let T (1/0) = (B, T ′ ; D) = (B, t′ ∪ t′ ; D). Let D′ be another disk on ?B such that (B, T ; D′ ) is a T (1/0). 1 2 Then (B, T ′ ; D′ ) = T (?s/q) with ps ≡ 1 mod q. In particular, if p ≡ ±1 mod q, then s ≡ ±1 mod q. Proof. Let F → ?B be a double branched cover with branch set ?T . Let (m, l) be a coordinate system of F such that m covers ?D. Similarly, let (m′ , l′ ) be a coordinate system with m′ covering ?D′ . Since (B, T ; D′ ) is a 1/0 tangle, ?D′ bounds a disk ?′ in B which separates the strings of T . So by our de?nition m′

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represents pm + ql in H1 (F ). Thus m′ l′ = p r q s m l

for some r, s such that ps ? qr = 1. So m l = s ?r ?q p m′ l′ .

In particular, m = sm′ ? ql′ . Since m covers ?D, which bounds a disk in B separating the strings of T ′ , this implies that (B, T ; D′ ) is a T (?s/q). From ps ? qr = 1 we see that ps ≡ 1 mod q. Lemma 2.4. Let T (p/2q) = (B, T ; D) = (B, t1 ∪ t2 ; D). Let P be the punctured torus on ?E(T ) bounded by ?D, containing Q = D ∩ E(T ). Let C be a circle in P intersecting both Q and A = P ? IntQ in k ≥ 1 essential arcs. If p ≡ ±1 mod 2q, then ?E(T ) ? C is incompressible in E(T ). Proof. Without loss of generality we may assume that t1 is the component of T with ends on D. Let γ be an arc in D with ?γ = ?t1 . Choose another disk D′ on ?B such that (B, T ; D′ ) = T (1/0). Then by Lemma 2.3 γ is an arc on ?B having slope ?r/2q, and rp ≡ 1 mod 2q. Since p ≡ ±1 mod 2q, we may assume that 1 < ?r < 2q ? 1. Let ? be a properly embedded disk in B containing T . Then ? ∩ E(T ) consists of three disks D1 , D2 , D3 , cutting E(T ) into two 3-balls W1 and W2 , see Figure 2.1(a). Now ?? intersects ?E(T ) in four arcs e1 , . . . , e4 , as shown in Figure 2.1(b). Let Pi be the pair of pants ?E(T ) ∩ Wi . Note that the two vertical arcs e1 and e2 on ?E(T ) belong to di?erent component ?D1 , ?D3 of ?Pi , and the two horizontal arcs e3 and e4 belong to the other component ?D2 of ?Pi .
e3

D1

D2

D3

e1

e2

e4

Figure 2.1 By an isotopy of C we may assume that C ∩ Pi consists of essential arcs. Each component αj of C ∩ Q is a copy of γ ∩ Q, so it is an arc of slope ?r/2q on ?E(T ), as shown in Figure 2.1(b), where ?r/2q = 3/8. Since 1 < ?r < 2q ? 1, αj ∩ Pi has some arcs connecting e1 to e2 and some arcs connecting e3 ∪ e4 to each of e1 and e2 . Therefore, for each i, and each pair of boundaries ?j , ?k of ?Pi , there are arcs of C ∩ Pi connecting ?j to ?k . It follows from Lemma 2.2 that ?E(T ) ? C is incompressible.

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Lemma 2.5. Let P, Q, A, and T (p/2q) be as in Lemma 2.4. (1) If q = 0, then P and ?E(T ) ? P are incompressible. (2) If p = ±1 mod 2q, there is no compressing disk of ?E(T ) intersecting P at just one arc. Proof. (1) Assuming the contrary, let ? be a compressing disk of P in E(T ). If ?? is a separating curve in P , then ?D = ?P bounds a disk in B ? t1 ∪ t2 , which would imply that T is a 1/0 tangle. If ?? is non-separating, then ?P , being coplanar to ?D, also bounds a disk in E(T ), which again would imply that q = 0. Similarly one can show that ?E(T ) ? P is incompressible. (2) Let ? be a compressing disk of ?E(T ) intersecting P in an essential arc γ. If γ were disjoint from A then by [Wu1, Lemma 2.1] T would be a 1/2 tangle. So assume k = |γ ∩ A| > 0, and k is minimal up to isotopy of γ. Let γ ′ be an arc on ?P with ?γ ′ = ?γ. Pushing γ ∪ γ ′ o? γ and ?P , we get a circle C in P which intersects both Q and A in k essential arcs. Since ? is a compressing disk of ?E(T ) ? C, this contradicts Lemma 2.4. Lemma 2.6. Let T = T (1/2, p/2q) = (B, t1 ∪ t2 ∪ K) be a tangle such that p ≡ ±1 mod q. Put M = B ? IntN (t1 ∪ t2 ), and M (γ) = (M, K; γ). Let m1 be a meridional circle of t1 on ?N (t1 ) ? ?M . Then (1) M (γ) is a handlebody for all γ; (2) ?M (γ)?m1 is incompressible in M (γ) for all γ = m, where m is the meridian slope of K. Proof. (1) T = T (1/2, p/2q) is a sum of T1 = T (1/2) = (B1 , t1 ∪ t′ , D1 ) and 1 T2 = T (p/2q) = (B2 , t′ ∪ t2 , D2 ), and K = t′ ∪ t′ . Notice that B1 ? IntN (t1 ) is a 2 1 2 solid torus, and t′ is the arc shown in Figure 2.2(a). Thus gluing B1 ? IntN (t1 ) to 1 (B2 , t′ ∪ t2 ) is the same as adding a 1-handle D2 × I to B2 so that D2 × ?I is glued 2 to a neighborhood of ?t′ in ?B2 , and the ends of t′ = 0 × ?I are glued to ?t′ . See 2 1 2 Figure 2(b). Since (B2 , t′ ∪ t2 ) is homeomorphic to the trivial tangle, K = t′ ∪ t′ 2 1 2 is a central curve of the solid torus V = (B1 ? IntN (t1 )) ∪ B2 = B ? IntN (t1 ), and t2 is properly isotopic to a trivial arc in V ? K, as shown in Figure 2(c). It is now clear that M (γ), the manifold obtained from V ? IntN (t2 ) by surgery on K, is a handlebody of genus 2. This proves (1).

t'1 D
1

K t
2

(a)

(b)

(c)

Figure 2.2 (2) Clearly, F = ?M ? m1 is compressible in M , so by Lemma 2.1, we need only show that F is incompressible in M (γ) for all γ such that ?(m, γ) = 1. As above, B ? IntN (t1 ) can be obtained by gluing a 1-handle H = D2 × I to B2 . Let W be the solid torus H ∪ N (t′ ), let X = B2 ? IntN (t′ ∪ t2 ) = E(T2 ). 2 2 Then B ? IntN (t1 ∪ t2 ) = W ∪A X, where A = W ∩ X is an annulus identi?ed to a meridional annulus on W and to ?N (t′ ) on X. 2

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Let α be an arc of slope 1/0 on ?B2 , connecting the ends of t′ . Let α′ = α ∩ X. 2 Then m1 is the union of α′ and an arc β on the boundary of the 1-handle H. Since ?(m, γ) = 1, W ′ = (W, K; γ) is a solid torus such that A is a longitudinal annulus. Therefore M (γ) = W ′ ∪A X is homeomorphic to X. The homeomorphism sends β to an arc β ′ on ?N (t′ ). Therefore (M (γ), m1 ) ? (X, α′ ∪ β ′ ). Now we can = 2 apply Lemma 2.4, and conclude that ?M ? m1 = ?X ? α ∪ β is incompressible in M (γ) = X. Theorem 2.7. Let T = T (r/2s, p/2q) = (B, t1 ∪ t2 ∪ K) be a tangle such that s > 1, and p ≡ ±1 mod 2q. Let K be the closed component of T , and let M = B ? IntN (t1 ∪ t2 ). Then ?M is incompressible in (M, K; γ) for all γ = m, where m is the meridional slope of K. Proof. By Lemma 2.1, we need only prove the theorem for γ with ?(m, γ) = 1. Let T1 = T (r/2s) = (B1 , t1 ∪ t′ .D1 ), and T2 = T (p/2q) = (B, t′ ∪ t2 , D2 ). 1 2 Let D = D1 = D2 be the disk in B cutting (B, T ) into (B1 , T1 ) and (B2 , T2 ). Let Pi = Qi ∪ Ai be the punctured torus on ?E(Ti ) bounded by ?Di , where Qi = Di ∩ E(Ti ), and Ai = N (t′ ) ∩ ?E(Ti ). i P2 cuts M into two pieces X and Y , where X = (B1 ∪ IntN (t1 )) ∪ N (t′ ), and 2 Y = B2 ? IntN (t′ ∪ t2 ) = E(T2 ). By the argument in the proof of Lemma 2.6, 2 we see that when ?(m, γ) = 1, (X, K; γ) ? E(T1 ), with P2 on ?X identi?ed to P1 = on ?E(T1 ). Hence, (M, K; γ) = (X, K; γ) ∪P2 Y ? E(T1 ) ∪P E(T2 ), where P is = identi?ed to Pi on ?E(Ti ). By Lemma 2.5, P is incompressible in (M, K; γ), and if r ≡ ±1 mod 2s, then P is also ?-incompressible in (M, K; γ), so P is an essential surface in (M, K; γ). By a standard innermost circle – outermost arc argument, one can show that ?M is incompressible in (M, K; γ) in this case. Now we may assume without loss of generality that r = 1. Let V be the solid torus B1 ? IntN (t1 ). Since T1 is a 1/2s tangle, t′ is rel ?t′ isotopic to an arc α 1 1 on ?V such that α ∩ (?V ? D1 ) is a single arc. Let X be a regular neighborhood of α ∪ D1 in V containing t′ , and let Y = V ? IntX. Thus V = Y ∪A X, and 1 A = Y ∩ X is an annulus running s > 1 times along the longitude of the solid torus Y. Now consider M ′ = V ∪D (B2 ? IntN (t2 )) ? M . Notice that if we glue a 2handle D2 × I to V so that ?D2 × I is glued to A, then (V ∪ D2 × I, t′′ ∪ t′ , D1 ) is 1 1 a 1/2 tangle, where t′′ = 0 × I ? D2 × I. Thus if we put (B ′ , T ′ ) = T (1/2, p/2q) = 1 (B ′ , u1 ∪u2 ∪K ′ ), then M ′ is homeomorphic to B ′ ?IntN (u1 ∪u2 ), with K identi?ed to K ′ , and A to ?N (u1 ). Therefore by Lemma 2.6, ?M ′ ? A is incompressible in (M ′ , K; γ) for all γ = m. As A is an annulus, this implies that A is incompressible. Also, if ? is a compressing disk of ?M ′ in (M ′ , K; γ) intersecting A in an essential arc, then the boundary of a regular neighborhood of ?? ∪ A on ?M ′ would bound a disk in (M, K; γ), contradicting the incompressibility of ?M ′ ? A. We conclude that there is no compressing disk ? of ?M ′ in (M ′ , K, γ) such that |?? ∩ A| ≤ 1. Since A runs s > 1 times along the longitude of the solid torus Y , any compressing disk of ?Y also intersects A at least twice. Notice that (M, K; γ) ? Y ∪A (M ′ , K; γ). The above results show that A is an = essential surface in (M, K; γ), and there is no compressing disk of ?M in (M, K; γ) disjoint from A. It is now a standard innermost circle – outermost arc argument to show that ?M is incompressible in (M, K; γ). Remark 2.8. The two conditions of Theorem 2.7 are necessary. If s = 1 or p = 1, by Lemma 2.6 we see that (M, K; γ) is a handlebody for all γ. If r ≡ ±1 mod 2s

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and p ≡ ±1 mod 2q, then K is isotopic to a curve on ?M , so ?M is compressible in(M, K; γ) for in?nitely many γ. Theorem 2.7 covers the remaining cases. 3. Dehn surgery on sufficiently complicated arborescent links In this section we will prove Theorem 3.4, which says that if L is an arborescent link composed of algebraic tangles satisfying certain conditions, then all surgeries on L are Haken. The main di?culty in dealing with surgery on links is that surgery on di?erent components of L will interact each other. The idea here in proving Theorem 3.4 is to ?nd a branched surface in the link complement, cutting the link complement into simpler pieces, so that we can deal with each piece separately. Given a branched surface F in a 3-manifold M , we use E(F) to denote the exterior of F, i.e. E(F) = M ? IntN (F). We refer the reader to [GO] for de?nitions of essential laminations, essential branched surfaces, and related concepts such as horizontal and vertical surfaces. Recall that the vertical surfaces are annuli on ?E(F). We call them cusps. Lemma 3.1. Let L be an arborescent link of length ≥ 4. Then there is a branched surface F in E(L) = S 3 ? IntN (L) such that each component M of E(F) = S 3 ? IntN (F) is of one of the following types. (a) M = B ? IntN (t1 ∪ t2 ), where (B, t1 ∪ t2 ) = T (p1 /q1 , p2 /q2 ), and at least one of the qi is odd; ?M has no cusps; M ∩ L = ?. (b) M = B ? IntN (t1 ∪ t2 ), where (B, t1 ∪ t2 ∪ K) = T (p1 /q1 , p2 /q2 ), both qi are even; ?M has no cusps; M ∩ L = K. (c) M = B ? IntN (t1 ∪ t2 ), where (B, t1 ∪ t2 ; D) = T (p/q) is a rational tangle with q ≥ 2; ?M has a cusp at ?D; M ∩ L = ?. (d) M is a solid torus, ?M has 3 cusps, each of which is a longitude of M ; M ∩ L = ?. (e) M = N (Ki ), Ki is a component of L; M has at least two meridional cusps; M ∩ L = Ki . Proof. By assumption (S 3 , L) is the union of two tangles (B1 , T1 ) and (B2 , T2 ), each of which is an algebraic tangle of length ≥ 2. Let F be a branched surface in B, where (B, T ) is a tangle. A component M of E(F) is called an inner component if it is disjoint from ?B, otherwise it is an outer component. All components of S 3 ? IntN (F) are inner components. For each algebraic tangle (B, T ) of length ≥ 2 and for each arborescent link (S 3 , L) of length ≥ 4, we construct inductively a branched surface F in B or S 3 , satisfying (*) Each inner component of E(F) is of one of the types listed in the lemma, and each outer component is a neighborhood of an arc of T , with possibly some meridional cusps. CASE 1. (B, T ) has length = 2. Then (B, T ) = T (p1 /q1 , p2 /q2 ) with qi ≥ 2. If at least one of the qi is odd, then T = t1 ∪ t2 , and we choose F = ?E(T ). If both qi are even, T = t1 ∪ t2 ∪ K, where K is a circle component of T . Let F = ?(B ? IntN (t1 ∪ t2 )). In both cases F is a genus 2 surface containing the punctured sphere ?B ? IntN (T ).

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It is clear that in this case an inner component M of B ? IntN (F) is of type (a) or (b) in the lemma. An outer component of B ? IntN (F) is a neighborhood of ti . Therefore (*) holds. CASE 2. (B, T ) = (B1 , T1 ) ∪D (B2 , T2 ), where (B2 , T2 , D) is a rational tangle T (p/q) for some q ≥ 2. By Case 1, we may assume that (B1 , T1 ) is not a rational tangle. So by induction we have a branched surface F1 in B1 satisfying (*). In particular, F1 ∩ ?B1 = ?B1 ? IntN (T1 ). De?ne F = F1 ∪ ?E(T2 ). Since F1 ∩ ?E(T2 ) is a pair of pants P , there are three new branch loci ?0 ∪ ?1 ∪ ?2 = ?P . We smooth F so that the cusp at ?0 = ?D is inside of E(T2 ), and the cusps on ?1 and ?2 are outside of both E(T1 ) and E(T2 ), so they are in N (T ), see Figure 3.1.

E(T1 )

E(T )
2

Figure 3.1 Let M be an inner component of B ? IntN (F). Then one of the following holds. (i) M is an inner component of B1 ? IntN (F1 ); (ii) M = E(T2 ), with a cusp at ?D; (iii) M is the union of N (t2 ) and an outer component of B1 ? IntN (F1 ), where t2 is a string of T2 . In case (i) M is of one of the types in the lemma by induction. In case (ii) M is of type (c). In case (iii) M is of type (e), where the two meridional cusps are the cusps at ?1 and ?2 in the above construction. Note that (iii) happens only if both Ti has a string with both ends on D. It is also easy to see that an outer component of B ? IntN (F) is a neighborhood of a string of T . Therefore (*) holds for F.

E(T1 )

E(T )
2

Figure 3.2 CASE 3. (B, T ) = (B1 , T1 ) ∪D (B2 , T2 ), and neither of (Bi , Ti ) is a rational tangle. By induction we may assume that we have constructed branched surfaces Fi in Bi satisfying (*), so F ∩?Bi = ?Bi ? IntN (Ti ). Construct F as in Figure 3.2. More explicitly, let ?0 = ?D be the outer boundary of P = D ? IntN (T ) in B1 ∪ B2 . We

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can think of B as the union of B1 ∪ B2 and a solid torus N which has a longitudinal annulus A′ glued to a neighborhood of ?0 on ?(B1 ∪ B2 ). Then F = F1 ∪ F2 ∪ A, where A is the annulus ?N ? IntA′ . It has ?ve new branch loci. We smooth it so that at the two inner boundary components of P it is smoothed in the same way as in Case 2, and the three cusps on ?N are on the side of N . An inner component M of B ? IntN (F) is of one of the following types. (i) M is an inner component of Bi ? IntN (Fi ); (ii) M is the solid torus in N above, with three longitudinal cusps, and M ∩T = ?; (iii) M = N (K) with at least two meridional cusps, where K is the union of two strings, one in each Ti . By induction we see that M is of one of the types in the lemma. Note that in case (ii) M is of type (d). One can check that an outer component is a neighborhood of a string of T , with possibly some meridional cusps. CASE 4. (S 3 , L) = (B1 , T1 ) ∪ (B2 , T2 ), where (Bi , Ti ) are algebraic tangles of length ≥ 2. Let Fi be the branched surface in Bi constructed above. De?ne F = F1 ∪ F2 , smoothed so that the four new cusps lie between F1 and F2 . In other words, the new cusps are in the solid tori whose intersection with Bi is a regular neighborhood of the arcs in Ti . A component M of S 3 ? IntN (F) is either an inner component of Bi ? IntN (Fi ), or is the union of two or four outer components of Bi ? IntN (Fi ). In the later case it is an N (Ki ) for some component Ki of L, with at least two meridional cusps on its boundary, so it is of type (e). Since each inner component of Bi ? IntN (Fi ) is of one of the types listed in the lemma, the result follows. Lemma 3.2. Let F be a branched surface such that its branch loci are mutually disjoint simple closed curves, cutting F into an orientable surface S. Then F fully carries a lamination. Proof. N (F) is obtained from S × I by gluing ?S × I together in certain way. Laminating S × I by S × C for a Cantor set C in I, and choosing the gluing map to preserve the lamination, we get the required lamination. See for example the construction of [GO, Example 5.1]. Corollary 3.3. Let F1 be a branched surface in a 3-manifold M . Let F be a two sided surface in F1 having a collar N = F × I in M such that F1 ∩ (F × I) = F × 0 = F . Let F be a branched surface obtained by gluing a surface G to F1 so that G ∩ N = ?G × I. If F1 fully carries a lamination, so does F. Proof. We can split N (F) along F into N (F1 ) and N (F2 ), where F2 is a branched surface homeomorphic to F ∪ G. By assumption and Lemma 3.2, each N (Fi ) fully carries a lamination λi . Thus λ = λ1 ∪ λ2 is a lamination fully carried by F. Theorem 3.4. Let L = k1 ∪. . .∪kn be an arborescent link of length ≥ 4, so that (1) L is composed of rational tangles T (pi /qi ) with qi > 2, and (2) no length 2 tangle of L is of the form T (1/2q1, 1/2q2 ) for some integers q1 , q2 . Then all complete surgeries on L produce Haken manifolds. Proof. Let W be a manifold obtained from S 3 by a complete surgery on L. Let F be the branched surface constructed in Lemma 3.1. We want to show that F is an essential branched surface in W . By [GO], W will then be irreducible. Note that ?h F has some closed surface components. Since each leaf of a lamination carried

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by an essential surface is π1 -injective, a closed surface component of ?h F will be an incompressible surface in W , so W will be a Haken manifold. To prove that F is essential in W , according to [GO] we need to prove the following. (i) F has no disk of contact; (ii) ?h F is incompressible, has no sphere components, and has no monogons; (iii) Each component M of W ? IntN (F) is irreducible; (iv) F contains no Reeb branched surface; (v) F fully carries a lamination. By the construction of F one can see that each branch locus is a homotopically nontrivial curve in F, so (i) and (iv) hold. It is also easy to see that ?h F has no sphere components. Let M be a component of S 3 ? IntN (F). Then it is of one of the ?ve types in Lemma 3.1. We want to show that after surgery it is irreducible, and ?h F ∩ ?M is incompressible and has no monogons. If M is of type (a), then M = E(T ) for some T = T (p1 /q1 , p2 /q2 ), and at least one of the qi is odd. Since we have assumed qi = 2, by [Wu1, Lemma 3.3] ?M = ?h F ∩ M is incompressible. It has no monogons because there is no cusps on ?M . As a tangle space, M is irreducible. Since M ∩ L = ?, these properties persist after surgeries. If M is of type (b), then M = B ? IntN (t1 ∪ t2 ) for some (B, t1 ∪ t2 ∪ K) = T (p1 /2q1 , p2 /2q2 ), where K is the circle component of the tangle, and is a component of L. Our assumption says that pi ≡ ±1 mod 2qi . By Theorem 2.7, the manifold (M, K; γ) obtained from M by some nontrivial surgery on K is irreducible and ?-irreducible. If M is of type (c), M = E(T ) for some rational tangle T (p/q) = (B, t1 ∪ t2 , D), and ?M ∩ ?h F = ?M ? ?D, which is incompressible by Lemma 1.2. Since ?M has genus > 1, this implies that there is no monogon. As a tangle space, M is irreducible. In case (d), M is a solid torus with 3 longitudinal cusps, and M ∩ L = ?, so the result is obvious. In case (e), M = N (ki ) for some component ki of L, and ?M has at least two meridional cusps. After surgery on ki , M remains a solid torus, but the cusps becomes non-meridional. Since there are at least two cusps, ?h F ∩ ?M is incompressible and has no monogons after surgery. In summary, (ii) and (iii) hold for all components of W ? IntN (F). It remains to show that F fully carries a lamination λ. We follow the steps in the construction of F in Lemma 3.1. In Case 1, F is a surface. We can simply take λ to be F × ?I in N (F) = F × I. In Case 2, F = F1 ∪ ?E(T2 ) = F1 ∪ G, where G is the closed up surface of ?E(T2 ) ? F1 . Let F be the surface ?E(T1 ) in F1 . Then F1 , F and G satisfy the conditions of Corollary 3.3, so by that corollary and induction, F fully carries a lamination. In Case 3, F = F1 ∪ F2 ∪ A. Let Si be the surface ?E(Ti ) on Fi . Then N (F) can be split into three pieces homeomorphic to N (F1 ), N (F2 ), and N (S1 ∪ S2 ∪ A), respectively. By induction N (Fi ) fully carries a lamination λi . By Lemma 3.2 N (S1 ∪ S2 ∪ A) also fully carries a lamination λ3 . Thus λ = λ1 ∪ λ2 ∪ λ3 is a lamination fully carried by N (F).

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In Case 4, N (F) can be split into N (F1 ) and N (F2 ). If λi is a lamination in N (Fi ) fully carried by Fi , then λ = λ1 ∪ λ2 is one fully carried by F. 4. Surgery on two component arborescent links Lemma 4.1. Let (B, t1 ∪t2 ∪K) = (B1 , t1 ∪t′ )∪D (B2 , t2 ∪t′ ) be an algebraic tangle 1 2 with K = t′ ∪ t′ , such that neither (Bi , ti ∪ t′ ; D) is the (1/2)-tangle T (1/2). Put 1 2 i M = B ? IntN (t1 ∪ t2 ), and M (γ) = (M, K; γ). Let mi be a meridional circle of ti on ?N (ti ) ? ?M . Then M (γ) is irreducible, and ?M ? m1 ∪ m2 is incompressible in M (γ). Proof. Let Xi = Bi ? IntN (ti ∪ t′ ) be the exterior of the tangle (Bi , ti ∪ t′ ). Let i i W be a regular neighborhood of D ∪ K. The frontier of W consists of two oncepunctured tori, denoted by Q1 and Q2 . They cut B into three pieces. One of them is W . The other two are homeomorphic to X1 and X2 , and will still be denoted by X1 and X2 respectively. Up to relabeling we may assume that Qi = Xi ∩ W . Since each (Bi , ti ∪ t′ ; D) a nontrivial algebraic tangle, Qi is incompressible in E(Ti ). i Let W (γ) denote that manifold obtained from W by Dehn surgery on K along the slope γ. We can consider W as the union of N (K) and P × I along two annuli A1 ∪ A2 , where P = D ? IntN (K) is a twice punctured disk. After surgery, the solid torus V = N (K) becomes a new solid torus V (γ), on which Ai becomes nonmeridional annuli. By a standard innermost circle outermost arc argument one can show that W (γ) = (P × I) ∪A1 ∪A2 V (γ) is irreducible, and Qi is incompressible in W (γ). Since M (γ) = X1 ∪Q1 W (γ) ∪Q2 X2 , we see that M (γ) is irreducible. Now suppose F = ?M ? m1 ∪ m2 is compressible in M (γ), with ? a compressing disk. Let Fi = F ∩ ?Xi , i = 1, 2, and let A be the annulus F ∩ W . Since (Bi , ti ∪ t′ ) i are nontrivial algebraic tangles, Fi are incompressible in Xi . Also, A is incompressible, because it is an annulus with boundary the same as the incompressible surface Q = Q1 ∪ Q2 . Therefore, ? ∩ Q = ?. By minimizing |? ∩ Q| we may assume that ? ∩ Q has no circle components. An outermost arc of ? ∩ Q on ? cuts o? a disk ?′ with ??′ = c1 ∪ c2 , where c1 is an arc on Q, and c2 is an essential arc on one of the F1 , F2 and A. The arc c2 can not be on A, because ?c2 = ?c1 must be on the same boundary component of A, but there is no such essential arc on an annulus. If c2 lies on Fi , then (Bi , ti ∪ t′ ) would be a T (1/2) by Lemma 4.1 of [Wu], contradicting i our assumption. Therefore, F is incompressible in M (γ). Remark. If (B, t1 ∪ t2 ∪ K) = T (1/2, 1/2n) then ?M ? m1 ∪ m2 is compressible in M (γ) for some γ = m. Theorem 4.2. Suppose L = l1 ∪ l2 is an arborescent link of two components. If L has length at least 4, and none of the rational tangles used to build L is a (1/2)-tangle, then all complete surgeries on L produce Haken manifolds. Proof. Write (S 3 , L) = (B1 , T1 ) ∪ (B2 , T2 ), where (Bi , Ti ) are algebraic tangles of length ≥ 2. We separate the proof into two cases. CASE 1. (Neither of Ti has a closed circle component.) Let S be the 2-sphere B1 ∩ B2 . Let W be a regular neighborhood of S ∪ L in S 3 . Let Q = ?W . It cuts E(L) into three pieces. One of them is W , and the other two are homeomorphic to E(T1 ) and E(T2 ) respectively, where E(Ti ) = Bi ? IntN (Ti ) are the tangle spaces. Let W (γ) be the manifold obtained from W by γ surgery on L, where γ = γ1 ∪ γ2 , γi a non-meridional slope on ?N (li ). As in the proof of Lemma 4.1, one can show

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that W (γ) is irreducible and ?-irreducible. Since T1 and T2 have length at least 2, by Lemma 3.3 of [Wu1] E(Ti ) are also irreducible and ?-irreducible. Therefore L(γ) = E(T1 ) ∪ W (γ) ∪ E(T2 ) is a Haken manifold. CASE 2. (T2 = t2 ∪ t′ ∪ K has a circle component K.) Suppose (B2 , T2 ) = 2 (B3 , T3 ) ∪ (B4 , T4 ). If T3 still has a circle component, then (S 3 , L) = (B3 , T3 ) ∪ ((B4 , T4 ) ∪ (B1 , T1 )), so we can consider (B3 , T3 ) instead. Therefore without loss of generality we may assume that T3 = t3 ∪ t′ and T4 = t4 ∪ t′ have no circle 3 4 components. Up to relabeling we may assume that l1 = t′ ∪ t′ is the component of 3 4 L contained in T2 . Let M be the manifold B2 ? IntN (t2 ∪ t′ ). Let M (γ1 ) be the manifold obtained 2 from M by γ1 surgery on K. Since neither of (B3 , T3 ) and (B4 , T4 ) is a (1/2)-tangle, by Lemma 4.1 M (γ1 ) is irreducible, and ?M ? m2 ∪ m′ is incompressible in M (γ1 ), 2 where m2 and m′ are the meridians of t2 and t′ on ?M . 2 2 Let Y = B2 ∪ N (L) = M ∪A V , where A is a union of two annuli, identi?ed to a neighborhood of m2 ∪ m′ on ?M , and to a pair of meridional annuli on boundary of 2 the solid torus V = ?N (l2 ). After γ surgery on L, we get Y (γ) = M (γ1 ) ∪A V (γ2 ). Clearly A is incompressible in both M (γ1 ) and V (γ2 ), and there is no compressing disk of ?V (γ2 ) intersecting A just once. If there is a compressing disk D of ?M (γ1 ) intersecting A just once, then the boundary of a regular neighborhood of A ∪ ?D in ?M (γ1 ) would bound a disk in M (γ1 ), contradicting the fact that ?M (γ1 )?m2 ∪m′ 2 is incompressible. Thus A is essential in Y (γ). By an innermost circle outermost arc argument it follows that Y (γ) is irreducible and ?-irreducible. As in Case 1, the tangle space E(T1 ) is also irreducible and ?-irreducible. Therefore L(γ) = E(T1 ) ∪ Y (γ) is a Haken manifold. 5. Surgery on 2-bridge links An essential lamination in the complement of a link L is persistent if it remains essential after all complete surgeries on the link. In this section we study the problem of which 2-bridge link complement contains a persistent lamination. This has been done by Delman [De1,De2] for all 2-bridge knots and most Montesinos knots. It has also been shown in [Wu2] that all non-Montesinos arborescent knot complements contain persistent laminations. In this section we will solve this problem for 2-bridge links. Two rational numbers p/q and r/s are considered equivalent, denoted by p/q ≡ r/s, if |q| = |s|, and p ≡ ±r mod Z. Recall that L(p/q) and L(r/s) are equivalent if p/q ≡ r/s. Denote by [a1 , . . . , ak ] the partial fraction expansion 1/(a1 ? 1/(. . . ? ak ) . . . ). In particular, [r, s] = 1/(r ? 1/s). Theorem 5.1. A non-torus 2-bridge link L = L(p/q) admits complete, nonlaminar surgeries if and only if p/q ≡ [r, s] for some odd integers r and s. The proof of the theorem relies heavily on Delman’s construction [De2]. Before proving the theorem, let us have a brief review of some basic concepts of [De1,De2]. For each rational number p/q, there is associated a diagram D(p/q), which is the minimal sub-diagram of the Hatcher-Thurston diagram [HT, Figure 4] that contains all minimal paths from 1/0 to p/q. See [HT, Figure 5] and [De1]. See also below in the proof of Lemma 5.4 for a construction of D(p/q). To each vertex vi of D(p/q) is associated a rational number ri /si . It has one of the three possible parities: odd/odd, odd/even, or even/odd, denoted by o/o, o/e,

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and e/o, respectively. Note that the three vertices of any simplex in D(p/q) have mutually di?erent parities. Let ?1 , ?2 be two simplices in D(p/q) with an edge in common. Assume that the two vertices which are not on the common edge are of parity o/o. Then the arcs indicated in Figure 5.1(a) and (b) are called channels. A path γ in D(p/q) is a union of arcs, each of which is either an edge of D(p/q) or a channel.

o/o

o/o

o/o

o/o

Figure 5.1 De?nition 5.2. A path γ in D(p/q) is an allowable path if (1) γ passes any point of D(p/q) at most once; (2) When ignoring the middle points of channels, γ intersects the interior of at most one edge of any given simplex; and (3) γ contains at least one channel. Lemma 5.3. Let L = L(p/q) be a 2-bridge link of two components. If D(p/q) has an allowable path γ which contains at least two channels, then S 3 ? L has a persistent lamination. Proof. Put (S 3 , L) = (B1 , T (1/0)) ∪ (B2 , T (p/q)). Let S be the sphere B1 ∩ B2 . A con?guration on S is a train track containing a circle around each point of L ∩ S, and two arcs of slope 1/0 joining the circles. The tangencies at the branch points determines the type of the con?guration. See [De2, Figure 3.1] for the types of con?gurations. The one in Figure 5.2 was said to be in group I. One can check that with respect to this con?guration, a channel by our de?nition is also a channel in the sense of Delman.

Figure 5.2 Thus given an allowable path γ in D(p/q), we can apply Delman’s construction in [De2] to obtain a branched surface F ′ in B2 , such that F ′ ∩?B2 is the con?guration in Figure 5.2. F ′ can be extended to a branched surface F in S 3 ? L by adding to

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it a trivial branched surface in B1 , as in [De2, Figure 3.10]. By Proposition 3.1 of [De2], F is an essential branched surface in S 3 ?L. The components of S 3 ?IntN (F) containing L form a regular neighborhood of L, denoted by N (L) = V1 ∪ V2 , each Vi being a solid torus. Since each channel contributes two cusps on ?N (L), there are 2k cusps on ?N (L) if γ has k channels. We claim that each Vi has k cusps. Actually, when examining the construction of the part of F corresponding to a channel [De2, Section 3], one can see that the two cusps appear around two points of L on some level sphere which have the same orientation. Since each component of L intersects the sphere in two points of di?erent orientation, there must be one cusp for each component of L, so the claim follows. Thus if k ≥ 2, then after complete surgeries on L, Vi becomes a solid torus with at least two non-meridional cusps. Since these are the only components of S 3 ? IntN (F) which will change after the surgery, it follows that F remains an essential branched surface after complete surgery, so any lamination fully carried by F is persistent. Lemma 5.4. Suppose q is an even number. Then there is an allowable path in D(p/q) with two channels, unless p/q ≡ [r, s] for some r, s. Proof. Replacing p/q by p′ /q = (q ? p)/q if necessary, we may assume that 0 < p < (q/2). Let [a1 , . . . an ] be the partial fraction expansion of p/q such that all ai ’s are even. Since q is even, n is odd. We may assume n ≥ 3, otherwise p/q = [q] = [q + 1, 1], and we are done. The condition 0 < p < (q/2) implies that (*) either a1 > 2 or a1 a2 < 0.

To each ai is associated a “fan” Fai consisting of ai simplices in D(p/q), see Figure 5.3(a) and (b) for the fans F4 and F?4 . The edges labeled e1 are called initial edges, and the ones labeled e2 are called terminal edges. The diagram D(p/q) can be constructed by gluing the Fai together in such a way that the terminal edge of Fai is glued to the initial edge of Fai+1 . Moreover, if ai ai+1 < 0 then Fai and Fai+1 have one edge in common, and if ai ai+1 > 0 then they have a 2-simplex in common. See Figure 5.3(c) for the diagram of [2, ?2, ?4, 2].

e1

e2

e1

e2

(a)

(b)
Figure 5.3

(c)

Note that a vertex on initial and terminal edges of Fai always has parity o/e or e/o. We use ? to indicate vertices with parity o/o. If ai ai+1 < 0, there is a channel in Fai ∪ Fai+1 which starts and ends with boundary edges of D(p/q), see Figure 5.4(a) for the channel in F2 ∪F?2 . If ai ai+1 > 0 and ai > 2, there is a channel which starts with a boundary edge and ends with

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an interior edge, but its union with a boundary edge of D(p/q) is an allowable path, see Figure 5.4(b) for the path in F4 ∪ F2 .

*

*
(a)
Figure 5.4

*
(b)

*

We call i a channel index for [a1 , . . . , an ] if either ai ai+1 < 0 or ai ai+1 > 4. Claim. If there are two channel indices for [a1 , . . . , an ], then D(p/q) has an allowable path from 1/0 to p/q with two channels. The idea is to join the channels given above with some boundary edges to form the required path. But we have to be careful, for example, there are no such paths for [2, 4, 2]. (However, condition (*) above excludes this from the set of p/q we are considering.) We prove the claim case by case. Assume that i and j are channel indices, and i < j. CASE 1: Both ai ai+1 and aj aj+1 are negative. We may choose i and j to be the ?rst two such indices. Then ai > 0, and aj < 0. The channel for i starts with a bottom edge and ends with a top edge of D(p/q), while the channel for j starts with a top edge and ends with a bottom edge, so they can be connected by boundary edges of D(p/q) to become an allowable path. See Figure 5.5 for the case p/q = [4, ?2, ?4, ?2, 2]. Note that this works even if j = i + 1.

*

*

*

*
Figure 5.5

*

CASE 2. ai ai+1 > 4, and aj aj+1 < 0. By Case 1 we may assume that ai , ai+1 and aj have the same sign. Then the path constructed above in Fai ∪ Fai+1 start and end with bottom edges, and the channel in Faj ∪ Faj+1 starts with bottom edge. So they can be joined with boundary edges

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of D(p/q) to form an allowable path from 1/0 to p/q. See Figure 5.6 for the paths in the diagrams of [2, 4, ?2] and [4, 2, ?2].
* *

* (a)

*

* (b)

*

Figure 5.6 CASE 3. ai ai+1 < 0, and aj aj+1 > 4. The proof is similar to that of Case 2. CASE 4. ai ai+1 > 4, and aj aj+1 > 4. By the above cases we may assume that all ak are positive. Since we have assumed that either a1 > 2 or a1 a2 < 0, we may further assume that i = 1, and a1 > 2. The channels in Fai ∪ Fai+1 and Faj ∪ Faj+1 starts and ends on bottom edges of D(p/q), so they can be joined by bottom edges of D(p/q) to form the required allowable path. This can be done even if j = i + 1. See Figure 5.7 for the cases of [4, 4, 2] and [4, 2, 4].

*

* (a)

*

*

* (b)

*

Figure 5.7 This completes the proof of the claim. The lemma is proved unless there is only one channel index for [a1 , . . . , an ]. Condition (*) above says that 1 is a channel index. Therefore [a1 , . . . , an ] is either [a1 , 2, . . . , 2] or [a1 , ?2, . . . , ?2]. One can check that in the ?rst case p/q = [a1 ? 1, n], and in the second case p/q = [a1 + 1, n]. This completes the proof of Lemma 5.4. Proof of Theorem 5.1. The “only if” direction follows from Lemma 5.3 and Lemma 5.4, noticing that if q is even and p/q = [r, s] then r and s must be odd numbers. So we need only show that if p/q ≡ [2t1 + 1, 2t2 + 1] then L admits a complete surgery producing a non-laminar manifolds. Without loss of generality we may assume that t1 > 0. The link L = l1 ∪ l2 can be drawn as in Figure 5.8(a), where t1 = t2 = 3. Let t = t1 ? t2 + 1. Let α be a curve on ?N (l2 ) representing tm + l, where m, l are the meridian and longitude of l2 . Denote by M the manifold obtained from S 3 by

DEHN SURGERY ON ARBORESCENT LINKS

19

′ t surgery on l2 . Thus α bounds a disk in the Dehn ?lled solid torus, so if l1 is ′ a band sum of l1 and α, then l1 is isotopic to l1 in M . See Figure 5.8(b). Note ′ ′ that l1 is a (2, u) cable of a knot K on ?N (l2 ), for some u, so the surgery on l1 along the cabling slope produces a manifold containing a lens space summand. The combination of these two steps yields a complete surgery on L which has a lens space summand, and hence is non-laminar by [GO].

Remark 5.5. Surgeries on torus links are easy to understand. Most of them are Seifert ?bered spaces. The exceptional ones are reducible, containing a lens space summand. If p/q = [r, s], r, s = ±1, then there is an allowable path in D(p/q) with one channel. Let M (γ1 , γ2 ) be the manifold obtained from S 3 by γi surgery on li . By the argument in Lemma 5.3, the corresponding branched surface has one cusp for each component of L. Therefore if γi are non-integral slopes on ?N (li ), then M (γ1 , γ2 ) is laminar. It is not known how many M (γ1 , γ2 ) are laminar if at least one of the γi is an integral slope.

(a) Figure 5.8 References

(b)

M. Brittenham and Y-Q. Wu, The classi?cation of Dehn surgeries on 2-bridge knots, preprint. [BS] F. Bonahon and L. Siebenmann, Geometric splittings of knots, and Conway’s algebraic knots, preprint. [CGLS] M. Culler, C. Gordon, J. Luecke and P. Shalen, Dehn surgery on knots, Annals of Math. 125 (1987), 237–300.

[BW]

20 [De1] [De2] [Fl]

Y-Q. WU C. Delman, Essential laminations and Dehn surgery on 2-bridge knots, Topology and its Appl. 63 (1995), 201–221. , Constructing essential laminations which survive all Dehn surgeries, preprint. W. Floyd, Incompressible surfaces in 3-manifolds: The space of boundary curves, Lowdimensional topology and Kleinian groups, London Math. Soc. LN. 112, 1986, pp. 131– 143. D. Gabai, Eight problems in the geometric theory of foliations and laminations on 3manifolds, Proceedings of Georgia Topology Conference, Part II, 1997, pp. 1–33. , Genera of the arborescent links, Mem. Amer. Math. Soc. 339 (1986), 1–98. D. Gabai and U. Oertel, Essential laminations in 3-manifolds, Ann. Math. 130 (1989), 41–73. C. Gordon, Dehn surgery on knots, Proceedings of the International Congress of Mathematicians, Kyoto (1990), 631–642. C. Gordon and J. Luecke, Knots are determined by their complements, Jour. Amer. Math. Soc. 2 (1989), 371–415. A. Hatcher and W. Thurston, Incompressible surfaces in 2-bridge knot complements, Invent. Math. 79 (1985), 225–246. J. Hempel, 3-manifolds, Ann. Math. Studies 86, Princeton Univ. Press, 1976. U. Oertel, Closed incompressible surfaces in complements of star links, Paci?c J. Math. 111 (1984), 209–230. M. Scharlemann, Unlinking via simultaneous crossing changes, Trans. Amer. Math. Soc. 336 (1993), 855–868. E. Starr, Curves in handlebodies, Thesis UC Berkeley (1992). W. Thurston, Three dimensional manifolds, Kleinian groups and hyperbolic geometry, Bull. Amer. Math. Soc. 6 (1982), 357–381. Y-Q. Wu, The classi?cation of nonsimple algebraic tangles, Math. Ann. 304 (1996), 457–480. , Dehn surgery on arborescent knots, J. Di?. Geo. 43 (1996), 171–197. , Imcompressibility of surfaces in surgered 3-manifolds, Topology 31 (1992), 271– 279.
? ??? ?? ? ? ? ? ? ×? ?? ? ?× ?? ? ??? ? ??? ??? ? ?? ?

[Ga] [Ga2] [GO] [Gor] [GL] [HT] [He] [Oe] [Sch] [St] [Th] [Wu3] [Wu2] [Wu3]

E-mail address: wu@math.uiowa.edu


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