ON TANGLE DECOMPOSITIONS OF TUNNEL NUMBER ONE LINKS

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1 ON TANGLE DECOMPOSITIONS OF TUNNEL NUMBER ONE LINKS HIROSHI GODA The Graduate School of Science and Technology, Kobe University, Rokkodai 1-1, Nada, Kobe , Japan goda@math.kobe-u.ac.jp MAKOTO OZAWA Department of Science, School of Education, Waseda University, Nishi-Shinjuku 1-6-1, Shinjuku-ku, Tokyo , Japan ozawa@mn.waseda.ac.jp MASAKAZU TERAGAITO Department of Mathematics and Mathematics Education, Hiroshima University, Kagamiyama 1-1-1, Higashi-Hiroshima , Japan mteraga@sed.hiroshima-u.ac.jp ABSTRACT We will discuss tangle decompositions of tunnel number one links, and show that any tangle composite tunnel number one link can be obtained as the union of a quasi Hopf tangle and a trivial tangle. We also consider the uniqueness of tangle decompositions of such links, and show that there are infinitely many tunnel number one links that admit non-isotopic essential tangle decompositions. Keywords: tunnel number, tangle, link. 1. Introduction In this paper, we will discuss tangle decompositions of tunnel number one links. Let L be a knot or link in S 3. We say that L has tunnel number one if there exists an arc t embedded in S 3, with t L = t, such that S 3 IntN(L t) isa genus 2 handlebody. Here N(X) denotes a regular neighborhood of X and IntX means the interior of X. Then the arc t is called a tunnel of L. Clearly a tunnel number one link has two components. An n-string tangle T, n 1, is a pair (B,s), where B is a 3-ball and s is a Typeset by AMS-TEX

2 H. Goda, M. Ozawa & M. Teragaito finite disjoint union of simple closed curves and n properly embedded arcs. These arcs are called the strings of the tangle. A tangle T is called a trivial tangle if it contains no closed curve components and the strings can be moved to arcs on T by an isotopy rel s. We say that T is essential if B IntN(s) is incompressible and boundary-incompressible in B IntN(s). A knot or link L is said to be n-string composite if the pair (S 3,L) can be expressed as the union of two essential n-string tangles, and tangle composite if it is n-string composite for some n. Note that a 1-string composite knot or link is composite in the usual sense. Let H n,n 1, be the n-string tangle shown in Fig.1. We call H n the n-string Hopf tangle. The loop e as shown in Fig.1 is called the equator of H n. Fig. 1 Hopf tangle Furthermore, the n-string quasi Hopf tangle is the (n 1)-string Hopf tangle with one additional trivial component (see Fig.2). Fig. 2 Quasi Hopf tangle For knots, Gordon and Reid [GR] showed that any tunnel number one knot is

3 Tangle decompositions of tunnel number one links tangle prime, that is, it has no essential tangle decomposition. This is a generalization of the fact that tunnel number one knots are prime [No,Sa]. There are also some results concerning essential tangle decompositions of tunnel number one links. In [GS] or [Jo], they treated the case of 1-string essential tangle decompositions of tunnel number one links. Indeed, they proved that a composite link has the 1-string Hopf tangle summand if the fundamental group of its complement is generated by two elements. Morimoto [Mo] proved that L is a tunnel number one composite link if and only if L is a connected sum of the Hopf link and a 2-bridge knot. (See [EU] also.) His result can be stated that a tunnel number one composite link has a decomposition into the union of the 2-string quasi Hopf tangle and the 2-string trivial tangle. Moreover, he decided the position of the tunnel of L. Gordon and Reid [GR] proved that if a tunnel number one link is n-string composite, then it can be decomposed into the m-string Hopf tangle for some m( n) and an essential tangle. However, they did not specify what form the essential tangle has. This is our motivation. Although we could not completely characterize which essential tangle appears as a summand of the decomposition given by the result of Gordon and Reid, we shall propose a conjecture in the last section. Our main result will give another decomposition of tangle composite tunnel number one links, which might be more explicit. Let L = K 1 K 2 be a tunnel number one link in S 3. Suppose that L is tangle composite. Then at least one component, K 2 say, is unknotted by the above result of Gordon and Reid. Therefore the wrapping number of K 1 with respect to the exterior E(K 2 )ofk 2 can be defined as the minimal geometric intersection number of K 1 and a meridian disk of the solid torus E(K 2 ). Theorem 1.1. Let L = K 1 K 2 be a tunnel number one link in S 3. Suppose that L is tangle composite. Then L can be decomposed into the union of the (n +1)- string quasi Hopf tangle and the (n+1)-string trivial tangle, where n is the wrapping number of K 1 with respect to the exterior of K 2. Note that this is a natural generalization of Morimoto s result in the 1-string composite link case. In Section 4, the uniqueness of the essential tangle decompositions of tunnel number one links will be discussed. We will give a class of tunnel number one links which admit unique essential tangle decompositions. Note that this class contains the infinite family of tangle composite, tunnel number one, hyperbolic links given by [GR]. Theorem 1.2. The links which are given by Gordon and Reid [GR, Theorem 1.3] have unique essential tangle decompositions. However, a tunnel number one link may have distinct (non-isotopic) essential tangle decompositions, in general. Theorem 1.3. There are infinitely many tunnel number one links in S 3 which have non-isotopic essential tangle decompositions.

4 H. Goda, M. Ozawa & M. Teragaito To prove Theorem 1.1, we will prove a stronger theorem from which Theorem 1.1 follows immediately. Theorem 1.4. Let L = K 1 K 2 be a tunnel number one link. Suppose that at least one component of L, K 2 say, is unknotted. Then L can be decomposed into the union of the (n +1)-string quasi Hopf tangle and the (n +1)-string trivial tangle, where n is the wrapping number of K 1 with respect to the exterior of K 2. Furthermore the position of a tunnel of L is as illustrated in Fig. 2. Corollary 1.5. Let L = K 1 K 2 be a tunnel number one link with K 2 unknotted. Then the bridge index of K 1 is less than or equal to n +1, where n is the wrapping number of K 1 with respect to the exterior of K 2. In Section 5 we will show that this estimate is best possible (see Example 5.1). In general, tunnel number one links can be classified into three classes : both components are unknotted, exactly one component is unknotted, or both components are knotted. It is known that if both are unknotted then L is a 2-bridge link [Ku]. Our theorem can be considered to give an explicit description of the second case. We remark that there exist infinitely many examples of tunnel number one links whose components are knotted [Be]. 2. Quasi Hopf tangle Throughout this paper, a link L will be assumed to be nontrivial. In this section, we will prove the following. Proposition 2.1. Let L = K 1 K 2 be a tunnel number one link such that K 2 is unknotted, and let t be a tunnel of L. Let D be a disk bounded by K 2 such that D K 1 + D t is minimal among all disks bounded by K 2 whose interiors meet K 1 and t transversely. Then IntD t =. By considering a regular neighborhood of the union of such a disk D and t, we have : Corollary 2.2. Let L = K 1 K 2 be a tunnel number one link such that K 2 is unknotted. Then L has the (n +1)-string quasi Hopf tangle summand, where n is the wrapping number of K 1 with respect to the exterior of K 2. First, note that the exterior E(L) is irreducible, since L is nontrivial and so non-splittable. Let V 1 = N(L t) and V 2 = cl(s 3 V 1 ), and let F = V 1 = V 2. Then both V 1 and V 2 are genus two handlebodies. We can regard V 1 as N(K 1 ) N(t) N(K 2 ). Let D be a disk bounded by K 2. By an isotopy, we may assume that IntD meets K 1 and t transversely. Then we may also assume that D V 1 consists of a parallel family of non-separating disks D 1,...,D n and a parallel family of separating disks D1,...,D l and an annulus A (see Fig.3). Since L is non-splittable, we have n>0. Suppose that D is chosen so that n + l is minimal among all disks bounded by K 2 whose interiors meet K 1 and t

5 Tangle decompositions of tunnel number one links transversely. Let D = D V 2. Then D is incompressible in V 2 by the minimality of n + l. Let E be a meridian disk of V 2. If E D =, then D is an annulus. Thus we have n = 1 and l = 0 as desired. Hence suppose that E D. We assume that l>0, and ultimately obtain a contradiction. Since D is incompressible in V 2, we can also assume that E D consists of properly embedded arcs on E. Finally, we may assume that E meets D i, Di and A transversely, and that E meets the closure of each component of V 1 ( n 1 D i ) ( l 1 Di ) A in essential arcs. Fig. 3 D V 1 As in Fig.3, the components of D are numbered consecutively on V 1. Label the points of E D with the number of the corresponding component of D, i.e., if a point of E D is in D i ( Di resp.), then we label the point with i (i resp.), and if a point of E D is in A, then we label it with 0. Since (V 1,V 2 ) gives a Heeggard splitting of S 3, we may assume that E runs over N(K 1 ) at least once. Then, if E does not meet Dl, then E would be a disk Seifert surface of K 1 and so L is splittable. Therefore we may assume that all i and i appear on E at least once. Let α be an arc of E D.Wesayα is of type (x, x ) if the labels at the endpoints of α are x and x. The alternatives of the type of each outermost arc of E D on E are 9 cases as follows. (a) (i, i + 1) for i =1, 2,...,n 1, (b) (1,n), (c) (1, 1), (n, n), (d) (1,l ), (n, l ),

6 H. Goda, M. Ozawa & M. Teragaito (e) (i, (i +1) ) for i =1, 2,...,l 1, (f) (1, 0), (g) (l,l ) for l>1, (h) (0, 0), (i) (1, 1 ). Let α be an outermost arc of E D in E, the outermost 2-gon which is cut off by α and β = cl( α). In cases (a) and (b), we can use to reduce n + l. In case (c), let P be the component of V 1 cut along D which contains D 1, D n and Dl. Then P is a 2-sphere with 3 holes. Suppose that α is of type (1, 1). The arc β appears on P as an essential arc. Then the number of intersections of E with D n would be less than that with D 1. Similarly for type (n, n). In cases (d), (e) and (f), we can use to reduce n + l. In case (g), let γ be an arc properly embedded in Dl with γ = α = β (see Fig. 4). Note that β γ is isotopic in V 1 to a meridian of K 1.Thusβ γbounds a disk with Int D = and K 1 = one point. By cut and paste D along, we can obtain a disk D with D = K 2, D K 1 D K 1 and D t < D t, which is a contradiction. Fig. 4 Case (g) In case (h), let γ be an arc in A with γ = α = β as in Fig. 5. Since β γ is isotopic to a meridian of t in V 1, β γ bounds a disk with Int D = and t = one point. Consider the disk δ bounded by α γ in D. Then δ

7 Tangle decompositions of tunnel number one links meets K 1 at least twice. Because δ is a 2-sphere and L is unsplittable. By cut and paste D along, we have a disk D such that D = K 2 and D (K 1 t) < D (K 1 t), which is a contradiction. Fig. 5 Case (h) In case (i), β runs on either N(K 1 )or N(K 2 ). Note that the former case occurs only if l = 1, and we can get a contradiction by the same argument as in case (g). Thus we consider the latter case. By performing isotopy of type A along α (cf. [Ja]), we obtain an annulus from D 1. Boundary components of the annulus and a component of A bound annuli on F (= V 1 = V 2 ). We call them F and F. Here F ( F resp.) consists of a component of A and ζ (ζ resp.). Without loss of generality, we may suppose that ζ is located in the inside of ζ on D. Fig. 6 Case (i) If ζ bounds a disk in D, then there would be a disk Q such that Q = K 2 and

8 H. Goda, M. Ozawa & M. Teragaito Q K 1 =, which is a contradiction. Therefore ζ does not bound a disk in D.By cut and paste D along F, we obtain a disk D such that D = K 2, D K 1 D K 1 and D t < D t, which is a contradiction. This completes the proof of Proposition Proof of Theorem 1.4 Let D be a disk bounded by K 2. We may assume that D intersects K 1 transversely and n = D K 1 is minimal among all such disks. Hence n is the wrapping number of K 1 with respect to the exterior of K 2. Furthermore, we can assume that IntD t = by Proposition 2.1. As in Section 2, let D = D V 2. Let B 1 = N(D) and let B 2 = cl(s 3 B 1 ). Then B 1 is the n-string Hopf tangle. Put S = B 1 = B 2. We can suppose that S intersects V 1 in a parallel family of non-separating disks D 1 D 2 D 2n of V 1 and a separating disk D1, where the labels 1, 2,...,2n appear consecutively along K 1. We may suppose that the subarc of K 1 between D 1 and D 2 is contained in B 1. (If n = 1, then this holds.) See Fig.7. Let S 1 = S V 1 and let S 2 = S V 2. Then S 2 is a 2-sphere with (2n + 1)-holes. Fig. 7 S V 1 Lemma 3.1. There is a meridian disk E of V 2 such that E intersects D i (i = 1, 2,...,2n) exactly once. Proof. Since K 2 is unknotted, E(K 2 ) is a solid torus. Furthermore, E(K 2 ) IntN(K 1 t) is a genus 2 handlebody V 2. This means that the solid torus E(K 2 ) is obtained from V 2 by attaching a 2-handle along D 1, say. Therefore we have the conclusion by [Go, Theorem 1]. Lemma 3.2. S 2 is incompressible in B 1 IntV 1. Proof. If there is a compressing disk, this disk would divide B 1 into two balls. This means that a component of K 1 B 1 and K 2 lie in disjoint balls. This contradicts that any string of K 1 B 1 is linked with K 2.

9 Tangle decompositions of tunnel number one links By an isotopy, we can assume that E S 2 consists of simple closed curves and arcs, and that E S 2 is minimal. Then E is divided into subdisks by the arc components of E S 2. Let be such a subdisk of E. Then consists of an alternating sequence of arcs in E S 2 and ones in E. In particular, the latter are referred to as corners of. Since S 2 is separating in V 2, either every corner of is contained in B 1 or every corner is contained in B 2. In the former case, is called a black region, and in the latter, is called a white region. Lemma 3.3. There is no innermost circle component of E S 2 on E that bounds a disk in B 1. Proof. If there is an innermost circle component that bounds a disk in B 1, we can reduce E S 2 by Lemma 3.2. Let G be a disk properly embedded in B 1 IntV 1 as illustrated in Fig 8. We can see that (B 1 IntV 1 ) IntN(G) is homeomorphic to D [0, 1]. An annulus properly embedded in B 1 IntV 1 is said to be vertical if it is incompressible in B 1 IntV 1 and of the form (a circle in D {0}) [0, 1]. Fig. 8 G in B 1 IntV 1 Since V 2 is irreducible, any circle component of E D which bounds a disk in D can be removed by an isotopy. Then we can assume that any component of E B 1 which does not meet E is a vertical annulus. Lemma 3.4. Let be a white region of E. Then Int S 2 =. Proof. Assume that Int S 2. Then there exists at least one circle component of E S 2 in Int. Every innermost circle bounds a disk in B 2 by Lemma 3.3.

10 H. Goda, M. Ozawa & M. Teragaito Therefore there are circle components γ 1 and γ 2 of E S 2 such that γ 1 and γ 2 bound a vertical annulus H in. Let γ = H D. Then we obtain a new disk D such that D K 1 < D K 1 by cut and paste D along γ. Note that D does not meet the tunnel t, since E is a disk in V 2. Thus this contradicts the minimality of D. By an isotopy, we may assume that E meets the closure of each component of F ( D 1 D 2n D1 ) in essential arcs. Let us consider the arc components of E S 2 in E. Each endpoint of an arc is a point in D i or D1. As in Section 2, assign the label i to the endpoint if it lies in D i, and label it with if it lies in D1. Thus, each arc of E S 2 has a pair of labels at its endpoints. We say that an arc α of E S 2 is of type (x, x )ifα has the labels x and x as its endpoints. Note that as we read around E, we see a label sequence 1, 2,...,2n exactly once, and the other labels are by Lemma 3.1. Then a corner is called [i, i +1]([, 1], [2n, ], [, ] resp.) if its endpoints have the labels {i, i +1} ({, 1}, {2n, }, {, } resp.). Let κ i be a subarc of K 1 between the disks D i and D i+1 in V 1, and κ 0 the subarc of K 1 between D 1 and D 2n. Note that if i is odd, κ i B 1, and if i is even, κ i B 2. An arc of E S 2 is said to be good if it has the label, bad otherwise. Lemma 3.5. Let α be an outermost arc of E S 2 in E. Suppose that α is bad. Then the outermost 2-gon cut off by α from E is a white region. Proof. Assume that α is of type (i, i + 1). If the outermost 2-gon cut off by α is black, then we can move the subarc κ i of K 1 to α. This contradicts the minimality of n. Lemma 3.6. Any bad arc of E S 2 is outermost in E. Proof. Suppose not. Then there is a bad arc α which cut off a disk δ from E such that Intδ contains only outermost bad arcs of E S 2 (see Fig. 9).

11 Tangle decompositions of tunnel number one links Fig. 9 The disk δ Let be the black region cut off from E by α and the outermost bad arcs β 1,β 2,...,β s. Then =α [i, i+1] β 1 [i+2,i+3] β 2 β s [i+2s, i+2s+1], where α is of type (i, i +2s + 1). Let j be the outermost (white) 2-gon cut off by β j. By using 1 s, we can move the subarc κ i κ i+2s of K 1 between D i and D i+2s+1 to α and push off from S into B 2. This contradicts the minimality of n. We push S into B 2 along the tunnel t by an isotopy as illustrated in Fig. 10. Let S be the resulting new 2-sphere. Let B and W be the 3-balls bounded by S in S 3. We may assume that B 1 Band W V 2 = B 2 V 2. Then (B, B L) is the (n + 1)-string quasi Hopf tangle, and (W, W L) =(W, W K 1 ) gives a (n + 1)-string tangle. Fig. 10 The new 2-sphere S Lemma 3.7. (W, W L) is a trivial tangle. Proof. First, suppose that every arc of E S 2 is good. That is, each arc is of type (, ) or(,i) for i =1,...,2n. Then, for each (white) corner [, 1], [2, 3], [4, 5],..., [2n 2, 2n 1], [2n, ], there exists the unique white region whose boundary contains it. For instance, let be the white region meeting [, 1]. Let =[, 1] a 1 b 1 a 2 b 2 a m 1 b m 1 a m, where a j are arcs of E S 2 and b k are corners [, ]. Recall that Int S 2 = by Lemma 3.4. By using, the string of (W, W L) meeting D 1 can be pushed into W. Similarly for the other strings in (W, W L). This means that (W, W L) is equivalent to a trivial tangle. Next, suppose that there exists a bad arc. Let us consider the case where n = 1. Then there exists the only one bad arc α of E S 2 that is of type (1, 2), and therefore α is outermost in E. Let be the outermost 2-gon cut off from E by α. Then =α [1, 2], and so is a black region. This is impossible by Lemma 3.5.

12 H. Goda, M. Ozawa & M. Teragaito Thus we can suppose that n>1. By Lemma 3.6, every bad arc of E S 2 is outermost in E. For each (white) corner [i, i +1] {[2, 3], [4, 5],...,[2n 2, 2n 1]}, either there is a bad arc of type (i, i + 1) or there is the unique white region whose boundary contains [i, i + 1]. In the former case, let α be a bad arc of type (i, i + 1) for some i. Since the outermost 2-gon cut off by α is a white region by Lemma 3.5, can be used to move the string κ i into W. In the latter case, the argument in the first paragraph works. For the remaining two strings meeting D 1 and D1 respectively, the white regions whose boundaries contain the corners [, 1], and [, 2n] respectively give the isotopies of the strings into W. This completes the proof of Theorem 1.4, and therefore Theorem Uniqueness of essential tangle decompositions In this section, we will consider the uniqueness of essential tangle decompositions of tunnel number one links. Let s be a 1-manifold properly embedded in a 3-manifold M, and let F be a surface properly embedded in M. Suppose that F intersects s transversely. By an isotopy of F in (M,s), we always mean an isotopy φ : F I M of F in M such that φ((f s) I) s. Let L be a link in S 3 which admits an essential tangle decomposition by a 2- sphere S. For any 2-sphere S giving an essential tangle decomposition of L, if there is an isotopy taking S to S, then we say that the essential tangle decomposition of L is unique. Proof of Theorem 1.3. Let us consider the link L m (m 1) as illustrated in Fig. 11. It is straightforward to verify that L m has tunnel number one, and that two 2-spheres S and S shown in Fig. 11 give essential tangle decompositions respectively by [Li,Nan]. Furthermore it is easy to see that two 2-spheres are not isotopic. Finally, an calculation of Jones polynomial shows that all L m are distinct. This completes the proof of Theorem 1.3.

13 Tangle decompositions of tunnel number one links Fig. 11 The link L m We define a class of links which includes the examples of tangle composite, tunnel number one, hyperbolic links given by Gordon and Reid [GR, Theorem 1.3], and prove that any link in this class has the unique essential tangle decomposition. Let H n =(C, k) be the n-string Hopf tangle with the loop component l, and let A be a horizontal annulus bounded by l and the equator e of H n (see Fig. 1) such that IntA k =. Let (B,s) be an essential free 2-string tangle. That is, cl(b N(s)) is a genus two handlebody. Let m be a simple closed curve in B s which divides B into two disks, each meeting both of strings. If we replace each string of s by a bunch of its parallel copies, then the resulting tangle (B, s) is essential and free again. We call it a parallelized tangle of (B,s). Assume that (B, s) has n strings. Then we can construct a link as the union of H n and (B, s), where the equator e is identified with the loop m. The set of such links is denoted by O. We remark that any link in O has at least one essential tangle decomposition, but is not necessarily tunnel number one. Theorem 4.1. Let L be a link in the class O defined above. Suppose that L has no essential tangle decomposing sphere intersecting l. Then the essential tangle decomposition of L is unique. We remark that there exists a link in O which admits an essential tangle decomposing sphere meeting l, and has two non-isotopic essential tangle decomposing spheres. It is obvious to see that the examples of tangle composite, tunnel number one, hyperbolic links given by Gordon and Reid [GR, Theorem 1.3] belong to our class O. Furthermore, any essential tangle decomposing sphere of tunnel number one links intersects its knotted component. (From the argument in [GR, Proof of Theorem 1.4], any essential tangle decomposing sphere of a tunnel number one link L meets only one component K 1,say,ofL. Then the argument in [GR, Proof of Theorem 1.5] implies that another component is unknotted. By [Ku], at least one component of L is knotted, and we have the conclusion.) Therefore their links satisfy the assumption of Theorem 4.1. This gives the proof of Theorem 1.2. The next lemma was originally proved by [GR] in a slightly different form, but it was revised by Morimoto [Mo2]. Lemma 4.2 [GR, Mo2]. Let (V,W) be a genus 2 Heegaard splitting of a closed 3-manifold M and let P be an incompressible planar surface properly embedded in V such that each component of P is non-separating on V and bounds a disk in W. Then one of the following conclusions holds. (1) M has a lens space or S 2 S 1 summand. (2) P is isotopic into V.

14 H. Goda, M. Ozawa & M. Teragaito Lemma 4.3 [Oz, Lemma 3.2]. Let (B,s) be a free 2-string tangle, and let P be either a 2-sphere embedded in B or a disk properly embedded in B such that P B s is disjoint from m, where m is a simple closed curve in B s which divides B into two disks, each meeting both of strings. Suppose that P intersects s transversely. If P IntN(s) is incompressible in B IntN(s), then one of the following conclusions holds. (1) P is a 2-sphere which bounds a trivial 1-string tangle. (2) P is a 2-sphere which is isotopic to B. (3) P is a disk which is isotopic into B. Proof. Since (B,s) is a free 2-string tangle, V = B IntN(s) is a genus two handlebody. Suppose that B is in S 3 and put C = cl(s 3 B). Then C N(s) is also a genus two handlebody W.Thus(V,W) gives a Heegaard splitting of S 3. Put P = P V.IfPis a 2-sphere, it is clear that P satisfies the assumption of Lemma 4.2. If P is a disk, the condition of P implies that. Therefore, P is isotopic into V in either case. Since V is genus two and each component of P is non-separating in V, P is either an annulus, a 2-sphere with 3 holes, or a 2-sphere with 4 holes. If P is an annulus, then P is a 2-sphere of type (1) when P N(s), or P is a disk of type (3) otherwise. If P is a 2-sphere with 3 holes, then two components of P are contained in N(s) and the other in B. Then P is a disk of type (3). If P is a 2-sphere with 4 holes, then P is a 2-sphere of type (2) when P N(s), and P is a disk of type (3) otherwise. Lemma 4.4 [Wa]. Let M = F I be the product of an orientable surface F S 2 and the interval I =[0, 1]. Letp : M F be the projection onto the factor F.Let G be a system of incompressible surfaces properly embedded in M. Suppose G is contained in F {1}. Then there is an isotopy, constant on M, taking G to a system G such that p G is homeomorphic on each component of G. The next lemma immediately follows from Lemma 4.4. Lemma 4.5. Let D be a disk and ρ a finite set of distinct points in IntD. Let F be a surface properly embedded in D I such that F intersects ρ I transversely, F is contained in (D ρ) {1}, and F IntN(ρ I) is incompressible in D I IntN(ρ I). Then there is an isotopy which takes F to a surface in D {1}. Lemma 4.6. Let H n =(C, k) be the n-string Hopf tangle, and let P be either a 2-sphere in C or a disk properly embedded in C such that P e =, where e is the equator of H n. Suppose that P intersects s transversely, P l =, and P IntN(k) is incompressible in C IntN(k). Then one of the following conclusions holds. (1) P is a 2-sphere which bounds a trivial 1-string tangle. (2) P is a 2-sphere which is isotopic to C. (3) P is a disk which is isotopic to a disk in C. Proof. Let D be a disk in C bounded by l such that D intersects each string in exactly one point. Recall that A is the horizontal annulus in H n cobounded by l

15 Tangle decompositions of tunnel number one links and e. We can take D so that D A = D = l. Then the union of A and D forms a disk which divides C into two 3-balls C + and C. We can regard C ± as the product (D, D k) I respectively. By an isotopy of P, we may assume that P is chosen so that P (D A) consists of loops and P D + P A is minimal. Note that for each component F of P C ±, F IntN(k) is incompressible in C ± IntN(k) by the minimality of P. First, suppose that P (D A) =. IfP is a 2-sphere, it is not hard to see that P is a 2-sphere of type (1). If P is a disk, then it follows from Lemma 4.5 that P is a disk of type (3). Therefore, we suppose that P (D A). Then there is an innermost disk E in P bounded by a loop in P (D A). Without loss of generality, we may assume that E is contained in C. If E is contained in D, then E is isotopic to a subdisk in D by Lemma 4.5. Then there is an isotopy of P which reduces P D, which contradicts the minimality of P. Hence E is contained in A, and essential in A. Then E divides C into two 3-balls C 1 and C 2 with D C 1. We can regard (C i,c i (k l)) as (E, E k) I by Lemma 4.5. Let A i = A C i for i =1, 2. First, we consider P C 1. Claim A. There is no component F of P C 1 such that F D. Proof. Let F be a component of P C 1, and suppose that F D. Then F is isotopic into D by Lemma 4.5. Thus there is an isotopy of P, which reduces P D, a contradiction. (Claim A) Claim B. Let F be a component of (P C 1 ) E such that F A 1. Then F is a disk isotopic to E. Proof. By Lemma 4.5, F is either an annulus isotopic into A 1 or disk isotopic to E. In the former case, there is an isotopy of P, reducing P A, which is a contradiction. (Claim B) Therefore we can assume that E is chosen so that there is no disk in C 1 isotopic to E. Claim C. There is an isotopy in (C, k) of P, keeping the minimality of P, which takes each component of (P C 1 ) E to a parallel copy of an annulus N(l) C 1. Proof. Let F be a component of (P C 1 ) E. Then F D and F A 1 by Claim A and the choice of E. By Lemma 4.5, F is isotopic to a surface F in A 1 D, and so F A 1 = 1. Furthermore, we may assume that F is chosen so that there is no other component of P C 1 between F and F. Let A 0 be a parallel copy of N(l) C 1. If F is parallel to A 0 in C 1, then we may ignore this component. Hence we assume that F is not parallel to A 0 in C 1.

16 H. Goda, M. Ozawa & M. Teragaito Let A 1 be an annulus in A 1 cobounded by A 0 A 1 and F A 1. Put Q = A 0 A 1, and push it slightly into IntC 1, keeping Q D fixed. Then Q F is a simple closed curve parallel to F A 1 in F. See Fig. 12. We deform F along Q by an isotopy in (C, k) so that F C 1 becomes a disjoint union of a parallel copy of A 0 and a connected surface F with F D. As in the proof of Claim A, we can push out F into C + by an isotopy of P. During these isotopies, the complexity P D + P A does not change. (In fact, it turns out that F is an annulus.) By repeating the above argument, we have the conclusion. (Claim C) Fig. 12 The annulus Q Second, we consider P C 2. Claim D. Let F be a component of (P C 2 ) E. Then, (1) F is a disk which is isotopic to E, if F C =. (2) F is an annulus which is isotopic into C 2 E, if F C. Proof. Suppose that F C =. Then F A 2, and so F is either an annulus isotopic into A 2, or a disk isotopic to E. In the former case, there is an isotopy of P which reduces P A, a contradiction. Therefore we have the conclusion (1). Next, suppose that F C. Then F A 2, since P A. By Lemma 4.5, F is isotopic into C 2 E. Then F is an annulus, and we have the conclusion (2). (Claim D) Finally, we consider P C +. Claim E. There is no component of P C + which is isotopic to the annulus N(l) C + in C + k. Proof. Suppose that there is a component F of P C + which is isotopic to the annulus N(l) C + in C + k. Thus there is an annulus F D A such that F F bounds a solid torus J in C + k. By the incompressibility and the minimality, any component of (P J) F is isotopic to F, and meets both A and D. Thus we may assume that F is chosen as the innermost one. Then there is an annulus F in C such that F = F by Claim C. Hence we obtain a torus component of P, which is absurd. (Claim E)

17 Claim F. Let F be a component of P C +. Then, Tangle decompositions of tunnel number one links (1) F is a disk with F A, which is isotopic into C +,if F C =. (2) F is an annulus cobounded by P and an essential loop in A, which is isotopic into C + D, if F C. Proof. Suppose that F C =. If F A, then F is either an annulus isotopic into A or a disk isotopic into C +. In the former case, there is an isotopy of P which reduces P A, a contradiction. Thus we have the conclusion (1). Therefore suppose that F is not contained in A. If F D, then F is isotopic into D. Then there is an isotopy of P which reduces P D, a contradiction again. Therefore F meets both D and A. Then F is isotopic to the annulus N(l) C +, but this is impossible by Claim E. Next, suppose that F C. By Lemma 4.5, F is isotopic into C +. Since F P consists of mutually parallel loops in C + k, F is an annulus. If F P D, then there does not exist a surface of type (1) in this Claim. Hence P C + = F. But this is impossible, because P A. Therefore F P A, and we have the conclusion (2). (Claim F) By Claim F, we have P D =. Suppose that there is a disk F of type (1) in Claim F. If we choose F so that F is innermost in A (that is, nearest to l in A), then E = F by the choice of E (see the remark before Claim C). Therefore P = F E is a 2-sphere, which is isotopic to C. This is the conclusion (2) of Lemma 4.6. Suppose that there is no disk of type (1) in Claim F. Then P C + is an annulus F of type (2) in Claim F. Hence P C is the disk E, and we have the conclusion (3) of Lemma 4.6. This completes the proof of Lemma 4.6. Proof of Theorem 4.1. Let S be the 2-sphere which gives the decomposition of L into the n-string Hopf tangle H n =(C, k) and a parallelized (free) tangle (B, s). Consider a 2-sphere P giving an essential tangle decomposition of L. We will show that P is isotopic to S. We may assume that P S consists of simple closed curves, which are disjoint from S K 1. In addition, we may assume that ( ) P e = by an isotopy of P, where e is the equator of H n, because P does not intersect l.

18 H. Goda, M. Ozawa & M. Teragaito Fig. 13 Two ribbon disks O 1 and O 2 By the definition of (B, s), there are two ribbon disks O 1 and O 2, each containing a bunch of parallel strings. That is, O i is the image of an embedding from [0, 1] [0, 1] to B. The images of the segments [0, 1] {0} and [0, 1] {1} are contained in B. An arc in O i is called vertical (horizontal, resp.) if it corresponds to the image of the segment {p} [0, 1] ([0, 1] {p}, resp.) for some p [0, 1]. We may assume that each string in O i is vertical. See Fig. 13. Since P gives an essential tangle decomposition of L, we can isotope P so that ( ) P intersects O i (i =1, 2) in vertical or horizontal arcs. Suppose that P S is minimal, up to isotopy of P, under the condition that P satisfies ( ) and ( ) above. If P S = 0, then P is contained in B or C. First assume that P B. In this case, P O i contains no vertical arcs. Then by regarding each O i as a string, it follows from Lemma 4.3 that P is isotopic to S = B. Secondly, assume that P C. In this case, we have the desired conclusion by Lemma 4.6. Therefore, we can assume that P S > 0 hereafter. By the minimality of P S, for any component F of P B (P C, resp.), F IntN(L) is incompressible in B IntN( s) (C IntN(k), resp.). Let E be an innermost disk in P, that is, E P S and IntE S =. Suppose that E B. Since E m = (recall that m is identified with e), E O i contains no vertical arcs. By regarding each O i as a string, it follows from Lemma 4.3 that E is isotopic into B. Therefore there exists an isotopy of P which reduces P S by Lemma 4.5, a contradiction. Suppose that E C. Note that E does not meet the equator e of H n. Therefore the similar argument as above works, by using Lemma 4.6 instead of Lemma

19 4.3. This completes the proof of Theorem Some examples Tangle decompositions of tunnel number one links Under the situation of Theorem 1.4, the bridge index of K 1 is at most n +1. We show that this estimate is best possible. Example 5.1. Let L = K 1 K 2 be a link as illustrated in Fig. 14. Then L has tunnel number one, and there is a disk D bounded by K 2 and D K 1 = n. Since K 1 is the torus knot of type (n +1,n+ 2), its bridge index is equal to n + 1 [Su]. In Theorem 1.1, we showed that any tangle composite, tunnel number one link can be decomposed into the union of a quasi Hopf tangle and a trivial tangle. However, the converse is not true. Example 5.2. Let L be the link as illustrated in Fig. 15. Then L can be decomposed into the 3-string quasi Hopf tangle and a trivial tangle. Since its knotted component is non-prime, L has tunnel number greater than one. By [GR, Theorem 1.5], any tangle composite tunnel number one link has at least one unknotted component. Our Theorem 1.4 deals with tunnel number one links which have at least one unknotted component, which give a broader class as shown in the following example, indeed. Example 5.3. Let K 1 be a non-trivial torus knot in S 3. Then E(K 1 ) is a Seifert fibered manifold with orbit-manifold a disk and two exceptional fibers. Let K 2 be one of the two exceptional fibers. Then L = K 1 K 2 has tunnel number one, but does not admit an essential tangle decomposition. For, E(L) is Seifert fibered, and there is no incompressible planar surface properly embedded in E(L) (see [Ja, VI.34]).

20 H. Goda, M. Ozawa & M. Teragaito Fig. 14 The link L Fig. 15 The link L Finally, we show that the number of strings stated in Theorems 1.1 and 1.4 is crucial. Proposition 5.4. Let L = K 1 K 2 be a two component link in S 3, and let t be an arc such that t L = t and that one component of t is in K 1 and the other in K 2. If K 2 is unknotted, then there exists an integer n such that L can be decomposed into the n-string quasi Hopf tangle and the n-string trivial tangle. Proof. Since K 2 is unknotted, K 2 bounds a disk D. By an isotopy of D, wemay suppose that IntD intersects K 1 and t transversely. Let V 1 be a regular neighborhood of L t. Then V 1 D consists of disks D i (i =1,...,n), Di (i =1,...,l) and an annulus as illustrated in Fig. 3. Let δ be a regular neighborhood of Dl t in Dl. By moving δ by an isotopy rel δ toward K 1, we can reduce one component of D t. Instead, K 1 D increases by two. We can repeat this operation l times, so that we may suppose that IntD does not intersect t. Let B 1 = N(D t) and B 2 = cl(s 3 B 1 ). Then (B 1,B 1 L) is a quasi Hopf tangle, but (B 2,B 2 L) is not necessarily trivial. Choose a point p in IntB 2 K 1, and let h be a projection map B 2 p ( = S 2 [0, 1)) [0, 1). We put K 1 B 2 in Morse position with respect to h. Let us consider a minimal point q of K 1 B 2. Move q toward B 2 vertically by an isotopy, and push into B 1. Furthermore, move q through K 2 in B 1 and put it back in B 2. As a result, the number of strings in (B 1,B 1 L) increased by two, and (B 2,B 2 L) lost one minimal point and gained one maximal point. If we repeat this operation for each minimal point of (B 2,B 2 L), then we obtain a quasi Hopf tangle and the tangle (B 2,B 2 L) whose critical points are maximal points only. This means that L can be decomposed into a quasi Hopf tangle and a trivial tangle.

21 Tangle decompositions of tunnel number one links 6. Problems (I) By [GR, Theorem 1.5], any tunnel number one link can be decomposed into the union of a Hopf tangle and some essential tangle. Our motivation was to specify which essential tangle can appear as the summand. Here we will propose a conjecture, which gives a complete description of tangle composite, tunnel number one links. Let L = K 1 K 2 be a tunnel number one link in S 3. Suppose that L is tangle composite. Then L can be decomposed into the n-string Hopf tangle and an essential tangle T, where n is the wrapping number of K 1 with respect to the exterior E(K 2 ) of K 2. An essential tangle (B,s) is called an annulus 1-bridge tangle if s consists of arcs s 1,s 2,...,s n and there is an unknotted annulus A properly embedded in B which satisfies the following conditions: (1) s i A for i =1, 2,...,n 1, (2) s n is lying in A except a 1-bridge, that is, s n consists of three arcs α, β, α, where α A, α A, and β is an unknotted arc which is contained in a meridian disk of the solid torus divided by A in B. Then (B,s) is a parallelized tangle of the essential free tangle (B,s 1 s n ). Conjecture 6.1. The above essential tangle T has the form of (i) an annulus 1-bridge tangle, or (ii) the sum of an annulus 1-bridge tangle and a rational tangle. Furthermore, in case (i), L admits the unique essential tangle decomposition, and in case (ii), there are exactly three non-isotopic essential tangle decompositions. We can see that an annulus 1-bridge tangle is a parallelized tangle of the 2-string free tangle (B,s 1 s n ). Therefore, in case (i), the uniqueness of essential tangle decompositions holds by Theorem 4.1. (II) Let L = K 1 K 2 be a tangle composite, tunnel number one link in S 3, with K 2 unknotted. Let us take a disk D bounded by K 2 which intersects K 1 transversely. In particular, we choose D so that D K 1 is minimal among such disks. Let S = N(D). Question 6.2. Does S give an essential tangle decomposition of L? Conjecture 6.1 implies a positive answer to Question 6.2. (III) For 1-string composite, tunnel number one links L, Morimoto [Mo] determined the positions of tunnels. Indeed, such a link has the unique essential tangle decomposition, but it can have non-isotopic tunnels. Let L be a tunnel number one link which is n-string composite for n 2.

22 H. Goda, M. Ozawa & M. Teragaito Problem 6.3. Determine the tunnels of L. (IV) As stated in Section 1, tunnel number one links are classified into three classes, according to the number of its unknotted component. Recall that a tunnel number one link has exactly two components. If both components are unknotted, then the link is 2-bridge [Ku]. Our discussions in this paper give mainly a description of the case where such a link has exactly one unknotted component. Finally, there exist tunnel number one links whose both components are knotted. (cf. [Be]). Problem 6.4. Find an explicit description of tunnel number one links whose both components are knotted. Acknowledgements The first author is partially supported by Grant-in-Aid for Encouragement of Young Scientists , The Ministry of Education, Science and Culture, and Hyogo Science and Technology Association. The third author is partially supported by Grant-in-Aid for Encouragement of Young Scientists , The Ministry of Education, Science and Culture. References [Be] J. Berge, Embedding the exteriors of one-tunnel knots and links in the 3-sphere, preprint. [EU] M. Eudave-Muños and Y. Uchida, Non-simple links with tunnel number one, Proc. Amer. Math. Soc. 124 (1996) [GS] F. González-Acuña and H. Short, Knot surgery and primeness, Math. Proc. Cambridge Phil. Soc. 90 (1986) [Go] C. McA. Gordon, On primitive sets of loops in the boundary of a handlebody, Topology Appl. 27 (1987) [GR] C. McA. Gordon and A. Reid, Tangle decompositions of tunnel number one knots and links, J. Knot Theory and its Rami. 4 (1995) [Ja] W. Jaco, Lectures on three manifold topology, CBMS Regional Conf. Ser. Math. 43, CBMS, Washington DC (1980). [Jo] A. Jones, Composite two-generator links have a Hopf link summand, Topology Appl. 67 (1995) [Ku] M. Kuhn, Tunnels of 2-bridge links, J. Knot Theory and its Rami. 5 (1996) [Li] W. B. R. Lickorish, Prime knots and tangles, Trans. Amer. Math. Soc. 267 (1987) [Mo] K. Morimoto, On composite tunnel number one links, Topology Appl. 59 (1994) [Mo2] K. Morimoto, Planar surfaces in a handlebody and a theorem of Gordon-Reid, inproc. Knots 96, ed. S. Suzuki, World Sci. Publ. Co., Singapore (1997) [Nan] Y. Nakanishi, Primeness of links, Math. Sem. Notes Kobe Univ. 9 (1981) [No] F.H. Norwood, Every two generator knot is prime, Proc. Amer. Math. Soc. 86 (1982) [Oz] M. Ozawa, On uniqueness of essential tangle decompositions of knots with free tangle decompositions, in Proc. Appl. Math. Workshop 8, ed. G.T.Jin and K.H.Ko, KAIST, Taejon (1998)

23 Tangle decompositions of tunnel number one links [Sa] M. Scharlemann, Tunnel number one knots satisfy the Poenaru conjecture, Topology Appl. 18 (1984) [Su] H. Schubert, Über eine numerische Knoteninvariante, Math Z. 61 (1954) [Wa] F. Waldhausen, On irreducible 3-manifolds which are sufficiently large, Ann. of Math. 87 (1968)