Tunnel number degeneration under the connected sum of prime knots
TTUNNEL NUMBER DEGENERATIONUNDER THE CONNECTED SUM OF PRIME KNOTS
JO ˜AO MIGUEL NOGUEIRA
Abstract.
We study 2-string free tangle decompositions of knots with tunnel number two. Asan application, we construct infinitely many counter-examples to a conjecture in the literaturestating that the tunnel number of the connected sum of prime knots doesn’t degenerate bymore than one: t ( K K ) ≥ t ( K ) + t ( K ) −
1, for K and K prime knots. Introduction
Given a knot K in S , an unknotting tunnel system for K is a collection of arcs t , t , . . . , t n ,properly embedded in the exterior of K , with the complement of a regular neighborhood of K ∪ t ∪ · · · ∪ t n being a handlebody . The minimum cardinality of an unknotting tunnel systemfor a knot K is a knot invariant, referred to as the tunnel number of K and is denoted by t ( K ).A natural question of study on knot invariants is their behavior under the connected sum ofknots. In the particular case of the tunnel number, it is known, by Norwood [21], that tunnelnumber one knots are prime. This result is now consequence of more general work. For instance,Scharlemann and Schultens prove in [26] that the tunnel number of the connected sum of knotsis bigger than or equal to the number of summands: t ( K · · · K n ) ≥ n, where K · · · K n represents the connected sum of the knots K , . . . , K n . Also, in [4] Gordonand Reid prove that tunnel number one knots are, in fact, n -string prime for any positive integer n . On the tunnel number behavior under connected sum, it is a consequence from the definitionof connected sum of knots that for two knots K and K in S we have: t ( K K ) ≤ t ( K ) + t ( K ) + 1 . For some time the only examples known had an additive behavior: t ( K K ) = t ( K ) + t ( K ) . However, in the early nineties, Morimoto [13] constructed connected sum examples of primeknots K with 2-bridge knots K whose tunnel number degenerates by one : t ( K K ) = t ( K ) + t ( K ) − . Every knot has an unknotting tunnel system obtained from the knot exterior triangulation. A knot is n -string prime if it has no n -string essential tangle decomposition. For definitions on tangledecompositions see Definitions 2 and 3 in section 2 (Preliminaries). In [16], without mentioning it, Morimoto gives also the first examples of knots that when connected sumwith themselves the tunnel number degenerates (by one): all tunnel number two 3-bridge knots with a 2-stringfree tangle decomposition (as the knot 10 from Rolfsen’s list in [24]). a r X i v : . [ m a t h . G T ] J un hortly afterwards, Moriah and Rubinstein in [12], and also independently Morimoto, Sakumaand Yokota in [17], gave examples of knots with supper-additive behavior: t ( K K ) = t ( K ) + t ( K ) + 1 . Furthermore, about the same time, Kobayashi in [7] constructed examples of knots that degen-erate arbitrarily under connected sum: for any positive integer n , there are knots K and K where t ( K K ) ≤ t ( K ) + t ( K ) − n. However, Kobayashi’s examples to show arbitrarily hight degeneration of the tunnel numberunder connected sum are composite knots.In this paper we study further the tunnel number degeneration under connected sum of primeknots. For this study we use the work of Morimoto in [16] that relates n -string free tangledecompositions of knots and high tunnel number degeneracy under the connected sum of primeknots. Within this setting, we study 2-string free tangle decompositions of knots with tunnelnumber two and we obtain Theorem 1, and its Corollary 1.1, for which statement we need thefollowing definition. Definition 1.
Let s be a properly embedded arc in a ball B . Suppose the knot obtained bycapping off s along ∂B has tunnel number at most one. We say that s is µ -primitive if there isan unknotted arc t properly embedded in B , disjoint from s , such that the tangle ( B, s ∪ t ) isfree. Remark . Note that a string s is µ -primitive if, and only if, the knot obtained by capping off s along ∂B is the unknot or a µ -primitive tunnel number one knot . Theorem 1.
Let K be a tunnel number two knot with a -string free tangle decomposition. Thenboth strings of some tangle are µ -primitive. Corollary 1.1.
Let K be a knot with a -string free tangle decomposition where no tangle hasboth strings being µ -primitive. Then t ( K ) = 3 . The only examples of prime knots whose tunnel number degenerates under connected sum arethe ones given by Morimoto, and in this case the tunnel number only degenerates by one. Also,Kobayashi and Rieck in [8], and also Morimoto in [15], proved that the tunnel number of theconnected sum of m -small knots doesn’t degenerate. With this and other results in perspective,Moriah conjectured in [11] that the tunnel number of the connected sum of prime knots doesn’tdegenerate by more than one: t ( K K ) ≥ t ( K ) + t ( K ) −
1, for K and K prime knots.In this paper, we construct infinitely many counter-examples to this conjecture as in Theorem 2and its Corollary 2.1. Theorem 2.
There are infinitely many tunnel number three prime knots K such that, for any -bridge knot K , t ( K K ) ≤ . Corollary 2.1.
There are infinitely many prime knots K and K where t ( K K ) = t ( K ) + t ( K ) − . For a definition of µ -primitive knot see Definition 5.13 of the survey paper [11] by Moriah. The correspondent result to Theorem 1 for links exists and is also proved by the author in [20]. A knot is said m -small if there is no incompressible surface with meridional boundary components in itscomplement. n [27], Scharlemann and Schultens introduced the concept of degeneration ratio for theconnected sum of two prime knots, K and K : d ( K , K ) = t ( K ) + t ( K ) − t ( K K ) t ( K ) + t ( K ) . If the knots K and K behave additively we have d ( K , K ) = 0.In case the knots K and K have supper-additive behavior then − ≤ d ( K , K ) <
0. The min-imum is achieved by the examples of Morimoto, Sakuma and Yokota in [17]. From the examplesof Moriah and Rubinstein in [12] we can choose a sequence of pairs of prime knots ( K , K ),with super-additive behavior, where d ( K , K ) converges to zero.For the sub-additive behavior of the tunnel number, the degeneration ratio is not so well under-stood. Naturally d ( K , K ) >
0, and from Corollary 9.2 in [27] , d ( K , K ) ≤ . The examplesof Morimoto in [13] have degeneration ratio . The examples from Corollary 2.1 have degenera-tion ratio . If K is a knot as in the statement of the Theorem 2 and K is any 3-bridge knotwith tunnel number one, from the main theorem of Morimoto in [14], t ( K K ) = 3. Hence,the degeneration ratio for these knots is . So, for sub-additive behavior, from the results inthis paper we have the lowest known degeneration ratio for the connected sum of prime knots , , and also the highest, .The proof of Theorem 2 is a consequence of Morimoto’s work in [16] and Theorem 1, andis explained in Section 8. For the proof of Theorem 1, we present the setting in Section 2. InSections 3 and 4 we prove some auxiliary technical lemmas that are used along the paper. InSections 5 and 6 we present the main lemmas that together give an outline of the proof. Andfinally in Section 7 we organize all the information to prove Theorem 1. For this proof, newand deeper arguments of innermost-arc type are developed to study the 2-string free tangledecomposition of K with respect to a minimal unknotting tunnel system of K .AcknowledgmentI would like to express deep gratitude to my thesis advisor, Cameron Gordon, for the excellentsupport and guidance. I also thank John Luecke for several useful discussions on the subject ofthis paper, and The University of Texas at Austin for the outstanding institutional support.2. Preliminaries
The subject of this paper is mainly on tunnel number two knots with 2-string free tangledecompositions. So, we introduce the following definitions.
Definition 2. A n -string tangle , ( B, T ), is collection T of n mutually disjoint arcs properlyembedded in a ball B .We say that a tangle ( B, T ) is: essential (resp., inessential ), if the planar surface ∂B − ∂ T isessential (resp., inessential) in B − T ; free , if π ( B − T ) is a free group or, equivalently, if thecomplement of a regular neighborhood of T in B is a handlebody; trivial , if T can be ambientisotoped into ∂B .A string s of ( B, T ) is said to be: trivial if there is an embedded disk in B , disjoint from T − s ,with boundary s and an arc in ∂B ; unknotted if it is trivial in the tangle ( B, s ). In [28], Schirmer proves that if K is a m -small knot then t ( K K ) ≥ max t ( K ) , t ( K ), which implies thatfor these knots d ( K , K ) ≤ . From work of Morimoto in [16], there are pairs of prime knots with tunnel number two that also havedegeneration ratio of . In this paper, the same degeneration ratio is obtained with a tunnel number three knotand a tunnel number one knot. he tangle ( B, T ) is said to be the product tangle with respect to a disk D in ∂B , if T canbe isotoped to a tangle ( B, p × I, . . . , p n × I ), with B = D × I , for a collection of n points, p , . . . , p n , in intD . Definition 3.
Consider a 2-sphere S in S in general position with a knot K , bounding theballs B and B . Let T i = B i ∩ K , for i = 1 ,
2. If T i is a collection of n arcs we say that S defines a n -string tangle decomposition of K : ( S , K ) = ( B , T ) ∪ S ( B , T ).If both tangles ( B i , T i ) are free (resp., essential), we say that the tangle decomposition of K defined by S is free (resp., essential ); if the tangle decomposition of K defined by S is notessential, we say that it is inessential .If the tangles ( B i , T i ), for i = 1 ,
2, are trivial then we say that K has an n -bridge decomposition,defined by S .Two tangle decompositions of a knot K defined by the 2-spheres S , S (cid:48) are said to be isotopic , ifthere is an ambient isotopy of S ∪ K to S (cid:48) ∪ K .Let K be a tunnel number two knot in S with a 2-string essential free tangle decompositiondefined by the 2-sphere S . We represent this tangle decomposition by ( S , K ) = ( B , T ) ∪ S ( B , T ). As the tangles are free, their strings have no local knots . This property and the nextlemma will be frequently used along this paper. Lemma 2.1.
The two strings of a -string essential free tangle are not parallell .Proof. Let (
B, s ∪ s ) be a 2-string essential free tangle. Suppose that s and s are parallel,and let D be a disk in B with boundary the strings s ∪ s and two arcs in ∂B connecting theends of these strings. As s and s are parallel, from Theorem 1’ of [2], the strings are knottedin B .Let N be a regular neighborhood of D in B . Hence, N is a regular neighborhood of s and of s . We have that B − int N is embedded in B − s ∪ s and ∂N is a proper essential surface in B − s ∪ s . So, π ( B − intN ), that is not a free group, injects into π ( B − ( s ∪ s )), that isfree, which is a contradiction. So, the strings s ∪ s are not parallel. (cid:3) As in the statement of Theorem 1, we want to prove that the two strings of ( B , T ) or ( B , T )are µ -primitive. With this purpose, it is useful to consider the following characterization of µ -primitive string. Lemma 2.2.
Let s be a string properly embedded in a ball B . Then s is µ -primitive if and onlyif s is trivial in a solid torus T in B intersecting ∂B in a single disk and whose complement in B is also a solid torus.Proof. Assume s is µ -primitive in B . Then there is a trivial string t in B , disjoint from s andwhere ( B, s ∪ t ) is a free tangle. Let T (cid:48) = B − intN ( t ) . As t is trivial in B we have that T (cid:48) isa solid torus and, from Theorem 1’ in [2], s is trivial in T (cid:48) . Consider the annulus A = ∂B ∩ ∂T (cid:48) .Let D (cid:48) be a disk in A that is a regular neighborhood of an arc in A connecting the two boundarycomponents of A . We have that A − intD (cid:48) is also a disk D . Consider a regular neighborhood of D (cid:48) in T (cid:48) and isotope ∂T (cid:48) , along the neighborhood of D (cid:48) , away from D (cid:48) . We are left with a solid Along the following sections we are assuming the tangle decomposition is essential. The inessential case isobserved in the proof of Theorem 1. We say that a tangle ( B, T ) contains a local knot , if there is a ball in B intersecting a single string of T at aknotted arc. We say that the strings of a tangle ( B ; s , s ) are parallel if there is an embedded disk D in B with boundarythe strings s ∪ s and two arcs in ∂B connecting the ends of these strings. For a manifold X smoothly embedded in the manifold Y , we denote by N ( X ) the regular neighborhood of X in Y . orus T in B , intersecting ∂B at the disk D . Furthermore, the complement of T in B is also asolid torus, it is a 1-handle attached to a ball, and s is trivial in T .Assume now that s is a trivial string in a solid torus T in B intersecting ∂B in a single disk andwhose complement in B is also a solid torus. Take a meridian disk L of the complement of T in B not intersecting S . Add the 2-handle with core L to T . We have that R = N ( L ) ∪ T is aball intersecting ∂B in a single disk. So, the complement of R in B is a ball. We isotope ∂R to ∂B along this ball, and from T we obtain the solid torus T (cid:48) , and from the disk L we obtain thedisk L (cid:48) . We have that ∂T (cid:48) ∩ ∂B is an annulus and the complement of T (cid:48) in B is the cylinder N ( L (cid:48) ), where N ( L (cid:48) ) intersects ∂B in two disks. Let t be the co-core arc of N ( L (cid:48) ). Hence, as T (cid:48) is a solid torus, t is a trivial string in B . Also, N ( t ) = N ( L (cid:48) ) and s is trivial in the complementof N ( t ). Therefore, ( B, t ∪ s ) is a free tangle, and s is µ -primitive. (cid:3) Consider an unknotting tunnel system of K , { t , t } , and the respective union of regularneighborhoods to be V = N ( K ) ∪ ( N ( t ) ∪ N ( t )). So, W = S − intV is a handlebodyand S = V ∪ W is a genus three Heegaard decomposition of S . Taking K ∪ t ∪ t ingeneral position with respect to S , we can assume that S ∩ V is a collection of essential disks: S ∩ V = D ∗ ∪ · · · ∪ D ∗ n ∪ D ∪ · · · ∪ D n , where D ∗ i , i = 1 , . . . , n , are the disks of S ∩ V intersecting K . Let D ∗ = D ∗ ∪ · · · ∪ D ∗ n and D = D ∪ · · · ∪ D n . Lemma 2.3. (a)
There is no -sphere in V defining a tangle decomposition of K isotopic to the onedefined by S . (b) Let C be a component of V − V ∩ S that intersects K . Then C ∩ K is parallel to theboundary of C .Proof. As V = N ( K ) ∪ N ( t ) ∪ N ( t ) there is an annulus A in V with ∂A = K ∪ b , where b isa simple closed curve in ∂V in general position with S ∩ V . As K ∩ D ∗ is non-empty, A ∩ S isalso non-empty. Assume that | A ∩ S | is minimal.First assume that some arc γ of A ∩ S has both ends in a string s from the tangle decomposition,and also that γ co-bounds a disk in A together with the string s , that intersects S ∩ V only at γ .As γ is in S , we have that s is trivial in the respective tangle decomposition, which contradictsthe tangle decomposition being essential.Suppose A ∩ S contains a simple closed curve c essential in A . Then K is isotopic to c . As c is asimple closed curve in S it bounds a disk in S . Therefore, in this case, K would be unknotted,which is a contradiction. Therefore, if c is a simple closed curve of A ∩ S then c bounds a disk in A .(a) Suppose there is a 2-sphere in V defining a tangle decomposition isotopic to the onedefined by S , and, abusing notation, denote it also by S . Hence, S ⊂ V . So, there cannot bearcs of A ∩ S between K and b . Let B be the ball bounded by S in V . Suppose A ∩ S containssome simple closed curve c . As observed before, c bounds a disk D in A . Suppose that c is aninnermost simple closed curve of A ∩ S in A . Then, D intersects S only at c . As S ⊂ V , if c bounds a disk S disjoint from K we can reduce | A ∩ S | , which is a contradiction to the minimalityof | A ∩ S | . Otherwise, both disks bounded by c in S intersect K , which contradicts the surface S − intN ( K ) being essential in S − N ( K ). Then, A ∩ S contains no simple closed curves. Then,from the previous observations, the components of A ∩ B are two disks co-bounded by the stringsof the tangle in B and two arcs of A ∩ S . As each disk of A ∩ B intersects S only at a singlearc in its boundary, we have that both strings of the tangle ( B, B ∩ K ) are trivial, which is acontradiction to the tangle decomposition defined by S being essential. igure 1 : An annulus A in V with boundary being K and a curve b in the boundaryof V . The disk ∆ is a disk in the component C of V − V ∩ S , with boundary beingthe string s and a curve in the boundary of C . (b) To prove part (b) of this lemma we just need to prove that A ∩ S contains no simple closedcurves, and that no arc of A ∩ S has both ends in ends of strings from the tangle decomposition.Assume now A ∩ S contains a simple closed curve, c . As observed before, c bounds a disk D in A ; suppose that it is an innermost curve with this property. Let L be the disk bounded by c in S ∩ V . If L doesn’t intersect K then we can reduce | A ∩ S | , which contradicts the minimality of | A ∩ S | . If L intersects K in less than four points then D contradicts the tangle decompositiondefined by S being essential. If L intersects K in four points then the tangles decompositiondefined by D ∪ L ⊂ V and S are isotopic, which is a contradiction to (a). Then A ∩ ( S ∩ V )contains no simple closed curve.From the previous arguments all arcs of A ∩ S either have both ends in b or one end in b andthe other at an end of a string in C . Also, as A is in general position with S ∩ V , each stringend is attached to a single arc of A ∩ S . Let C be a component of V − V ∩ S that intersects K . Then each string s of C belongs to the boundary of a properly embedded disk componentof A − A ∩ S in C , disjoint from the other string components in C , as in Figure 1. Therefore,all components of C ∩ K are independently parallel to the boundary of C , which gives us thestatement (b) of the lemma. (cid:3) Considering the previous lemma and that all 2-spheres in S intersect K an even number oftimes, no disk of D is parallel to a disk of D ∗ in V .From the work of Ozawa [23], we know that if a knot has an essential 2-string free tangledecomposition then this decomposition is unique up to isotopy , and, furthermore, K is n -stringprime for n (cid:54) = 2. (In particular, K is prime.) This is a result frequently throghout this paper,and we refer to it as Ozawa’s unicity theorem.We assume the unknotting tunnel system and the tangle decomposition defined by S up toisotopy are such that S ∩ V is a collection of disks with minimum cardinality | S ∩ V | . FromLemma 2.3 and the minimality of | S ∩ V | , we can assume that all disks S ∩ V are essential in V . As S decomposes K in two 2-string tangles we have n ≤
4. If n ≥
3, we denote the stringwith one end in D ∗ i and the other end in D ∗ j by s ij .Let P denote the planar surface S ∩ W . By the minimality of | S ∩ V | and the incompressibilityof S − int ( N ( K )) in S − int ( N ( K )), we have that P is essential in W . For a topological space X , | X | denotes the number of connected components of X . or a complete system of meridian disks of W , { E , E , E } , we write E = E ∪ E ∪ E .Considering E and P in general position, we choose E such that | P ∩ E | is minimal between thecomplete systems of meridian disks of W .By the incompressibility of P and the minimality of | P ∩ E | , no component of P ∩ E is a simpleclosed curve. Also, if an arc component of P ∩ E is a loop co-bounding a disk in P disjoint from P ∩ E , using this disk, we can change the complete system of meridian disks of W to E (cid:48) with | P ∩ E | > | P ∩ E (cid:48) | . This is a contradiction, and therefore P ∩ E is a collection of essential arcs in P .With the arcs of P ∩ E we define a graph in S that we denote by G P : the vertices are thedisks from S ∩ V , each of which corresponds to a boundary component of P , and the edges arethe arcs P ∩ E . The graph G P is connected: in fact, if the graph G P is not connected thenby cutting along a complete system of meridian disks of W we can find a compressing disk for P in W , which is a contradiction as P is essential. As G P is a connected graph in a 2-sphere,from the arcs P ∩ E in E we can create a sequence of isotopies of type A over a sequence ofarcs α , α , . . . , α m of P ∩ E such that the closure of the components of P − α ∪ · · · ∪ α m is acollection of disks. Figure 2 : An illustration of some arc components of E ∩ P in P . The arc α (resp., α ) is a sk-arc (resp., st-arc). The arc α is a d ∗ -arc that is also a k-arc, and α isboth a d-arc and a d ∗ -arc. Note also that the arc α is an example of a type I d-arcthat is not a st-arc. The vertices of G P associated to D , resp. D ∗ , are referred to as d-vertices , resp. d ∗ -vertices ,and are illustrated as white disks, resp. dark disks. Between the edges of G P it is useful todefine some types of arcs as follows. (See Figure 2.) Type I : is an edge connected to a single vertex.
Type II : is an edge connected to two distinct vertices. d-arc : is an edge with at least one end attached to a d-vertex. d ∗ -arc : is an edge with at least one end attached to a d ∗ -vertex. t-arc : is an edge of type II, α i , in a sequence of isotopies of type A as above, connected to somed-vertex D and where α j , j < i , is disjoint from D . (See Remark 2 and also Lemma 1of [22] by Ochiai.) A complete system of meridian disks of a handlebody H is a collection of disks in H whose complement isa ball. An arc α in P is essential if the components closure of P − α doesn’t contain any disk component. See chapter 2 of [1] by Jaco for a definition of an isotopy of type A , and section 2 of [22] by Ochiai for adefinition of the latter and also of an inverse isotopy of Type A . -arc : is a type II arc connecting two d ∗ -vertices. st-arc : is a type I d-arc separating P into two planar surfaces, each with some boundary com-ponent of D ∗ . sk-arc : is a type I d ∗ -arc separating P into two planar surfaces, each with some boundarycomponent of D ∗ . Remark . Suppose α is a type II d-arc with one end in the d-vertex associated with D . If oneof the disks separated by α from E intersects the disk D only at the end of α in D , then all arcsof E ∩ P in this disk have no end in D . This implies that α is a t-arc. (See Figure 3.) In Lemma2.4 we prove that such arcs cannot exist. Figure 3 : An illustration of arc components of E ∩ P in some component of E . If anarc of E ∩ P in E has one end in the disk D i , resp. D ∗ j , then we label the end of thearc in E by i , resp. by j ∗ . The ends of the arc α in the figure exemplify this notation.If all arcs in one of the disks separated by α from E have no ends being i , then α is at-arc. We say that an arc δ of P ∩ E is an outermost arc , if δ separates a disk component ∆ of E − E ∩ S from E . We have ∆ ∩ P = δ and ∂ ∆ = δ ∪ β with β ⊂ ∂E . The disk ∆ is saidto be an outermost disk . (See Figure 3.) An outermost disk is said to be over a component of V − V ∩ S if the correspondent arc β is in the (boundary) of the component.In the next lemma, we study the arcs P ∩ E in P and in E and obtain properties that givethe base setting for these arcs along the work in this paper. Lemma 2.4. (a)
All outermost arcs are of type I. (b) If n is the number of vertices of G P then either n = 1 and the graph G P has no edges,or n ≥ and at least two vertices of G P are not adjacent to edges of type I. (c) No arc of E ∩ P is a t-arc. (d) The outermost d-arcs of E ∩ P in E are of type I. (e) Each type I arc of E ∩ P is a st-arc or sk-arc. (f) All d-vertices are adjacent to a st-arc. (g)
Each outermost arc of E ∩ P in E is a st-arc or a sk-arc.Proof. (a) Suppose some outermost arc is of type II. Then, proceeding with an isotopy of type A alongthe respective outermost disk we can reduce | S ∩ V | , which is a contradiction to the minimalityof | S ∩ V | . b) If n = 1 , G P , the outermost loop co-bounds a disk in P .Furthermore, if G P has no loops and n = 2 then the outermost arcs of E ∩ P in E are of typeII, which is a contradiction to (a). If n ≥ P . In both cases we contradict the fact that alledges of G P are essential in P .(c) If there is a t-arc, then by a sequence of isotopies of type A followed by a sequence ofinverse isotopies of type A, as in Lemma 1 of [22] by Ochiai, we can ambient isotope S , in theexterior of K , to some 2-sphere S (cid:48) where | S ∩ V | > | S (cid:48) ∩ V | . This is a contradiction to theminimality of | S ∩ V | .(d) If an outermost d-arc of E ∩ P in E is of type II then it is a t-arc, which is a contradictionto (c). Therefore, the outermost d-arcs of E ∩ P in E are of type I.(e) Let α be a type I arc of E ∩ P . As α is essential in P it separates P into two componentsthat are not disks. If one of these components, say F , only contains boundary componentscorresponding to d-vertices there is some arc of E ∩ P in F that is a t-arc, which contradicts (c).(f) Assume there is a d-vertex D that is only adjacent to edges of type II. Then there is at-arc with respect to D (choose an outermost arc, in E , between the edges of type II attachedto D ), which is a contradiction to (c). Hence, there is at least one edge of type I with ends in D , and from (e) it is a st-arc.(g) From (a) the outermost arcs are of type I, and from (e) the type I arcs are st or sk-arcs. (cid:3) Outermost disks over ball components of V − V ∩ S In this section we study the case when there is an outermost disk over some ball componentof V − V ∩ S , as in Lemma 3.2. We also have presented other crucial lemmas relating ballcomponents of V − V ∩ S and certain disks of E − E ∩ P , together with the next lemma wherewe show several properties of tangles obtained from balls in B or B . Lemma 3.1.
Suppose there is a ball Q in one of the tangles defined by S that intersects eachstring of the tangle in a single arc. (a) Let Q c denote the complement of Q in S . The tangle ( Q c , Q c ∩ K ) is essential. (b) If one of the strings of Q ∩ K is unknotted in Q then either the tangle ( Q, Q ∩ K ) istrivial or some string of some tangle defined by S is unknotted. (c) Suppose both strings of the tangle are in Q and have ends in one or two disk componentsof Q ∩ S . Then the tangle ( Q, Q ∩ K ) is essential. (d) If a ball component of V − S ∩ V contains a string with both ends in the same componentof D ∗ , then some string of some tangle is unknotted.Proof. Assume that the tangle containing Q is ( B , T ).(a) Suppose that ( Q c , Q c ∩ K ) is inessential. As this tangle contains only two strings, bothstrings are trivial in it. Let s (cid:48) be an arc component of Q c ∩ K , and D (cid:48) be a disk in Q c withinterior disjoint from K and with boundary being the union of s (cid:48) and an arc in ∂Q c . Let s bethe string from the tangle decomposition defined by S that is a subset of s (cid:48) . So, s is a string ofthe tangle ( B , T ). As s (cid:48) contains only the string s of K − S ∩ K , we have that ∂D (cid:48) intersects S only at two points, which are the end points of s . Considering a minimal collection D (cid:48) ∩ S and following an innermost curve or arc type of argument, we can prove that D (cid:48) ∩ S is a singlearc a with ends being the ends of s ⊂ ∂D (cid:48) . Let D be the disk in D (cid:48) with boundary defined bythe arcs a ⊂ S and s . Then D is in the tangle ( B , T ) and the interior of D doesn’t intersect S .Therefore, the string s is trivial in ( B , T ), which is a contradiction to the tangle decompositiondefined by S being essential. b) Assume that one of the strings of Q ∩ K is unknotted in B . If the tangle ( Q, Q ∩ K )is essential then, from (a), the 2-sphere ∂Q defines a 2-string essential tangle decomposition of K . By Ozawa’s unicity theorem, the tangle decompositions given by S and ∂Q are isotopic.Hence, as one string of ( Q, Q ∩ K ) is unknotted, some string of some tangle defined by S is alsounknotted. Otherwise, the tangle ( Q, Q ∩ K ) is trivial.(c) Suppose the tangle ( Q, Q ∩ K ) is trivial. Let Q (cid:48) be obtained from Q after we isotope awayfrom S in B the components of Q ∩ S that don’t contain any string ends. If Q (cid:48) intersects S in a disk with all string ends in it, as the strings are trivial in Q (cid:48) they are both unknotted in( B , T ). From Theorem 1’ in [2], this is a contradiction to the tangle ( B , T ) being essential.Otherwise, if Q (cid:48) intersects S in two disks that also contain the strings ends in them. As thetangle ( B , T ) is free, following an argument as in Lemma 2.1, the complement of Q (cid:48) in B is asolid torus. Then ∂Q (cid:48) is ambient isotopic to S in S − K , which is also a contradiction to thetangle ( B , T ) being essential. So, the tangle ( Q, Q ∩ K ) is essential.(d) Suppose there is a ball component C of V − V ∩ S containing a string s with both endsin the same component of D ∗ . From Lemma 2.3, the tangle ( C, C ∩ K ) is trivial. Consequently,the string s is trivial in C . As the ends of s are in the same disk of C ∩ S , it is also unknottedin the respective tangle defined by S . (cid:3) Lemma 3.2.
If there is an outermost disk over a ball component of V − ( S ∩ V ) then somestring of some tangle is unknotted.Proof. Suppose there is an outermost disk ∆ over a ball component C of V − ( S ∩ V ), and let δ be the respective outermost arc. Without loss of generality assume that C ⊂ B . Let A bethe annulus in the intersection of C ⊂ V with the 2-sphere obtained after an isotopy, along aregular neighborhood of ∆, of a regular neighborhood of δ in S into V . The component C iseither disjoint from K or contains one or both strings of T . Then, a core of A bounds a disk D in ∂C that is either disjoint or intersects K in one or two points. We isotope int D into C ,slightly, such that D ∩ S = ∂D .Assume D is disjoint from K . The arc δ union an arc component of ∂P − ∂δ is a simple closedcurve parallel, in S − K , to a core of A . Also, the arc δ separates P into two planar surfacescontaining boundary components of D ∗ . Therefore, D separates the strings of the tangle ( B , T )and intersects S only at ∂D , which is a contradiction to the tangle decomposition defined by S being essential.Assume that | D ∩ K | = 1. Let D (cid:48) be the disk in S with ∂D (cid:48) = ∂D and | D (cid:48) ∩ K | = 1, and Q be the ball in B bounded by D ∪ D (cid:48) . Then Q ∩ T is a single trivial arc in Q . So, consideringthe 2-sphere S (cid:48) = ( S − D (cid:48) ) ∪ D , the tangle decompositions defined by S and S (cid:48) are isotopic with | S (cid:48) ∩ V | < | S ∩ V | , which is a contradiction to the minimality of | S ∩ V | .Assume now that | D ∩ K | = 2. Then D splits the tangle ( B , T ) into two 2-string tangles:( B (cid:48) , T (cid:48) ) and ( B (cid:48)(cid:48) , T (cid:48)(cid:48) ). If D intersects K in the same string of T then one string of this tangle,say s , is either in ( B (cid:48) , T (cid:48) ) or in ( B (cid:48)(cid:48) , T (cid:48)(cid:48) ). Assume, without loss of generality, that s is in( B (cid:48) , T (cid:48) ). From Lemma 3.1 (a), if the tangle ( B (cid:48) , T (cid:48) ) is essential then ∂B (cid:48) defines an essential2-string tangle decomposition of K with | ∂B (cid:48) ∩ V | < | S ∩ V | , which contradicts the minimality of | S ∩ V | . Hence, the tangle ( B (cid:48) , T (cid:48) ) is trivial. So, s is trivial in ( B (cid:48) , T (cid:48) ) and therefore unknottedin ( B , T ). Otherwise, assume that D intersects K in different strings of T . By a similarargument as when D intersects K in the same string we can prove that the tangles ( B (cid:48) , T (cid:48) ) and( B (cid:48)(cid:48) , T (cid:48)(cid:48) ) are trivial. Then the string s ∩ B (cid:48) is trivial in B (cid:48) and the string s ∩ B (cid:48)(cid:48) is trivial in B (cid:48)(cid:48) , which implies that s is unknotted in ( B , T ). (cid:3) emark . From Lemma 3.2, if some outermost disk is over a ball component of V − S ∩ V then we have Theorem 1. So, we can assume that all outermost disks are over components of V − S ∩ V other than balls.We say that two arcs of E ∩ P are parallel in E if the union of these arcs cuts a disk componentof E − E ∩ P from E . An arc outermost in E between the arcs of E ∩ P not in a sequence ofparallel arcs to a outermost arc is said to be a second-outermost arc . A disk of E − E ∩ P in theoutermost side of a second-outermost arc is called a second-outermost disk . The arcs α and γ inFigure 3 are examples of second-outermost arcs.Let γ and γ (cid:48) be two type I arcs of E ∩ P parallel in E attached to disks D and D (cid:48) of S ∩ V ,resp., parallel in V . (See Figure 4(a).) Denote by Γ the disk cut by γ ∪ γ (cid:48) from E , and by C theball component of S − S ∩ V cut by D ∪ D (cid:48) from V . Suppose that C and Γ are in the same ballcomponent bounded by S , say B . Then Γ is a proper surface in the complement of the solidtorus B ∪ D ∪ D (cid:48) C in S , which is B − intC . The curve ∂ Γ is inessential in the boundary of B ∪ D ∪ D (cid:48) C , and as we are in S , it bounds a disk in ∂ ( B ∪ D ∪ D (cid:48) C ). Let L be the disk boundedby ∂ Γ in ∂ ( B − intC ). Note that L intersects S in two disks and C in a disk band from D to D (cid:48) . Let R be the ball bounded by Γ ∪ L in B − intC . (See Figure 4(b).) For the next lemma,denote by q a core arc of C in B , this is a proper arc in B with regular neighborhood C . Thisconstruction and the following lemma will be frequently used throughout this paper. Figure 4
Lemma 3.3.
The ball R contains a single string of T , and this string is parallel to q in B .Proof. Denote by O and O (cid:48) the disks of L ∩ S , which are the disks cut by γ and γ (cid:48) , resp., in S − int ( D ∪ D (cid:48) ). As γ , and γ (cid:48) , is a st or sk-arc we have that O and O (cid:48) contain some componentof D ∗ . This means that R contains some string(s) of T .Suppose that T is in R . From Lemma 3.1(a) and (c), we have that ∂R defines a 2-string essentialtangle decomposition of K . As the 2-string essential tangle decomposition of K is unique, wehave that the tangle decompositions defined by S and ∂R are isotopic. But | ∂R ∩ V | < | S ∩ V | ,which is a contradiction to the minimality of | S ∩ V | .Then R contains a single string, s , of T . As there are no local knots, s is trivial in R . Theintersection of R with C is the disk L − O ∪ O (cid:48) , that intersects each D and D (cid:48) at an arc. Let a be an arc in L − O ∪ O (cid:48) with one end in D ∪ O and other in D (cid:48) ∪ O (cid:48) . Then, s (resp., q ) isparallel to a in R (resp., C ) through a disk with boundaries being s ∪ a (resp., a ∪ q ) and twoarcs in O ∪ O (cid:48) (resp., D ∪ D (cid:48) ). As R intersects C in L − O ∪ O (cid:48) , we have that s and q are parallelthrough a disk with boundaries being s ∪ q and two arcs in S . Consequently, s and q are parallelin B . (cid:3) Corollary 2.2.
The disks D and D (cid:48) cannot be disks of D ∗ . roof. As no disk of D is parallel to a disk of D ∗ in V , suppose both D and D (cid:48) are disks of D ∗ .Then C contains some string(s) of T . As R contains a string of T we have that C containsa single string of T , and from Lemma 2.3(b) this string is also a core of C . Therefore, fromLemma 3.3, the strings of T in R and in C are parallel in B , which is a contradiction to Lemma2.1. (cid:3) Lemma 3.4.
Let D k , D ∗ i and D ∗ j be disks of S ∩ V where D k ∪ D ∗ i ∪ D ∗ j cuts a ball component C of V − V ∩ S from V ; assume that C intersects K at a single string, with one end at D ∗ i andthe other at D ∗ j . Suppose there is a disk component of E − E ∩ P , in the same tangle componentas C , that intersects S in arcs where all but one of these arcs have both ends in D k , and theremaining arc has either at least one end in D k , or one end in D ∗ i and the other in D ∗ j . Thensome string of some tangle is unknotted.Proof. Denote by Γ the disk component of E − E ∩ P referred to in the statement, and by γ thearc of Γ ∩ S that doesn’t have by assumption both ends in D k , as in Figure 5(a). Without lossof generality, suppose C is in B . Let s and s (cid:48) be the strings in this tangle with s in C , and C i denote the cylinder obtained from C after an isotopy pushing D ∗ j away from S in B . Consideralso the solid torus T i defined by B ∪ D ∗ i ∪ D k C i .Assume that γ also has both ends in D k . Figure 5 : (a) Arc γ after the outermost arcs attached to D k . The label k at an endof an arc means the end is at the disk D k . (b) The ball C cut by D ∗ i ∪ D ∗ j ∪ D k from V , and the string s ij in it. The curve ∂ Γ is inessential in T i and it bounds a disk in ∂T i that we denote by L . The disk L intersects D ∗ . In fact, suppose L is disjoint from D ∗ , and consider a disk D of L ∩ C i or L ∩ S with boundary intersecting ∂L at a single component. Then, if D ⊂ C i it is also a disk in C and we get a contradiction to the minimality of | P ∩ E | , and if D ⊂ S we obtain a contradictionwith Lemma 2.4(e).Consider the ball R in B bounded by Γ ∪ L . As L contains some component of D ∗ the ball R intersects T ; it contains at most two arcs, the string s (cid:48) or a portion of the string s .Suppose R contains the string s (cid:48) only. As there are no local knots in the tangle, s (cid:48) is parallel to L . By pushing L to S from ∂C i we have that the string s (cid:48) is unknotted in ( B , T ).Suppose R contains also a portion of the string s . From Lemma 3.1(a), we have that the tangle( R c , R c ∩ K ), where R c is the complement of R in S , is essential. As | ∂R ∩ V | < | S ∩ V | , if thetangle ( R, R ∩ K ) is essential, we have a contradiction to the minimality of | S ∩ V | . Therefore,( R, R ∩ K ) is a trivial tangle. Then, the string s (cid:48) is unknotted in R and parallel to the disk L .By an isotopy of L from T i to S we have that the string s (cid:48) is also unknotted in ( B , T ).So, we can assume that R ∩ K is only a portion of the string s . Consider the solid torus T (cid:48) i = T i ∪ R , and the complement in B of the ball obtained by cutting T (cid:48) i along D ∗ i that wedenote by Q . Then, Q is a ball in B containing s (cid:48) and a portion of s . The 2-sphere ∂Q isisotopic to S rel. Q ∩ S in B . Then, if s is unknotted in Q it is also unknotted in B . As ∂Q ∩ V | < | S ∩ V | , following a similar reasoning as when R contains two arcs, we also have thatsome string of some tangle is unknotted.Assume now that γ has only one end at D k .Suppose, without loss of generality, that the other end of γ is in D ∗ i . We isotope S along aregular neighborhood of a disk in C intersecting K once, intersecting the disk D k along a singlearc, and separating D ∗ i from D ∗ j in ∂C . In this way, we split D k into two disks D k and D k (cid:48) , and C into two cylinders from D k to D ∗ i , C k,i ∗ , and from D k (cid:48) to D ∗ j , C k (cid:48) ,j ∗ . The boundary of Γ liesin S , and in the boundaries of the balls C k,i ∗ and C k (cid:48) ,j ∗ . The arcs of ∂ Γ ∩ C k (cid:48) ,j ∗ have both endsattached to D k (cid:48) . Hence, we can isotope these arcs to S . Also, all but one arc of ∂ Γ ∩ C k,i ∗ hasboth ends in D k . The other arc has one end in D ∗ i and the other in D k . We isotope all arcsof ∂ Γ ∩ ( C k,i ∗ ∪ C k (cid:48) ,j ∗ ) with both ends in D k or both ends in D k (cid:48) from ∂C k,i ∗ or ∂C k (cid:48) ,j ∗ to S ,respectively. We are left with the disk Γ with boundary defined by one arc in S and other arc inthe boundary of C k,i ∗ from D ∗ i to D k . Using this disk we can isotope C k,i ∗ through S . After thisisotopy of V we obtain a new 2-string tangle decomposition of K , that contains in each tanglea string from the original tangle decomposition defined by S . We also reduced | S ∩ V | . So, thenew tangle decomposition cannot be essential, which implies that some string of the originaltangle decomposition defined by S is unknotted.Assume at last that γ has one end in D ∗ i and the other end in D ∗ j .Then each arc of Γ ∩ S co-bounds a disk in S − intD k , with ∂D k , containing D ∗ i ∪ D ∗ j , and otherdisk containing none of these disks. Hence, we can isotope ∂ Γ to lie in ∂C with the exceptionof γ . So, after the isotopy ∂ Γ is defined by γ and an arc in C from D ∗ i to D ∗ j . The string s istrivial in C and therefore it is parallel to the arc ∂ Γ ∩ C . Therefore, the string s is unknotted in( B , T ). (cid:3) Outermost disks over torus components of V − V ∩ S In this section, we prove several lemmas related with the existence of outermost disks overtori components of V − V ∩ S disjoint or intersecting K at a single arc. These lemmas arefundamental on the proof of Theorem 1. Lemma 4.1 ((c.f. Morimoto [14, Lemma 2.4])) . There is no outermost disk over a solid toruscomponent of V − V ∩ S disjoint from K and containing a single disk of V ∩ S .Proof. Denote by T the solid torus as in the statement and by D the disk T ∩ S . Let δ bethe outermost arc co-bounding an outermost disk ∆ as in the statement. Consider, also, thecorresponding arc β and a disk O cut by δ in S − intD . Let L = O ∪ ∆. The disk L is a meridianfor the complement of T and intersects a meridian of T once. Consider a regular neighborhoodof ∆ in W , N (∆). So, N (∆) ∩ S is a regular neighborhood of δ in S , N ( δ ), and N (∆) ∩ ∂T is aregular neighborhood of β in ∂T , N ( β ). We isotope the annulus N ( δ ) ∪ D through N (∆) to theannulus N ( β ) ∪ D . As β intersects a meridian of T once, the annulus A = N ( β ) ∪ D is such that T = A × I . Therefore, we can isotope A ⊂ S through T to ∂T − A and out of V . Let S (cid:48) be the2-sphere obtained from S after this isotopy. The tangle decomposition of K obtained from S (cid:48) isthe same as the one given by S . However, | S (cid:48) ∩ V | < | S ∩ V | and we contradict the minimalityof | S ∩ V | . (cid:3) Lemma 4.2.
Assume V − V ∩ S has a solid torus component T intersecting K in a single stringand with T ∩ D ∗ being a single disk. If there is an outermost disk over T then some string ofsome tangle is unknotted.Proof. Suppose T is in the tangle ( B , T ). Let s be the string T ∩ K , that is a component of T , and D ∗ be the component of T ∩ S that intersects K . Then, both ends of s are in D ∗ ⊂ ∂T . et ∆ be an outermost disk over T and δ the respective outermost arc in E attached to thedisk D of T ∩ S . Consider also the disk O cut by δ in S − intD and disjoint from D ∗ . Let L = O ∩ ∆. Isotope the disks (if any) of T ∩ O away from S in B , and denote the resultingsolid torus by T (cid:48) . By adding the 2-handle with core L to T (cid:48) we define a ball Q that intersects S at disk components.Suppose D = D ∗ . As δ is a sk-arc, and two ends of strings are in D , we have that O intersects D ∗ in a a single disk. In this case, the ball Q contains the string s , and also an unknottedportion of the other string of T . From Lemma 3.1(b), some string of some tangle defined by S is unknotted or the tangle ( Q, Q ∩ K ) is trivial. So, we can assume that ( Q, Q ∩ K ) is trivial,and consequently that s is trivial in this tangle. As s has both ends at the same disk componentof Q ∩ S , we have that s is unknotted in ( B , T ).Suppose D (cid:54) = D ∗ . If O intersects K at a single point, then following the argument used inthe previous case we have that some string of some tangle is unknotted. So, assume that O intersects K at a collection of two points. In this case, the ball Q contains the string s , and alsotwo portions of the other string, s (cid:48) that are unknotted in Q . So, ∂Q defines a 3-string tangledecomposition of K . Let Q c denote the complement of Q in S . From Ozawa’s unicity theorem,either the tangle ( Q, Q ∩ K ) or the tangle ( Q c , Q c ∩ K ) is inessential. As there are no local knotsin the tangles defined by S and the tangles ( Q, Q ∩ K ) and ( Q c , Q c ∩ K ) are 3-string tangles,the tangle that is inessential has a trivial string. If the tangle ( Q c , Q c ∩ K ) is inessential then,following an argument as in the proof of Lemma 3.1, either some string of the tangle ( B , T ) istrivial, which is a contradiction, or the string Q c ∩ s (cid:48) is trivial in Q c . In the latter case isotope Q from S in such a way that Q ∩ S is only D ∪ O , and denote by Q (cid:48) the ball after the isotopy. Then,( Q (cid:48) , Q (cid:48) ∩ s (cid:48) ) is the product tangle with respect to D ∪ O , and Q (cid:48) c ∩ s (cid:48) is isotopic to ∂Q (cid:48) − ∂Q (cid:48) ∩ S .Therefore, after the isotopy of Q (cid:48) c ∩ s (cid:48) to Q (cid:48) we have that s (cid:48) is unknotted in ( Q (cid:48) , Q (cid:48) ∩ K ). As s (cid:48) has both ends in the same disk component of Q (cid:48) ∩ S we have that s (cid:48) is unknotted in the tangle( B , T ). Suppose now that the 3-string tangle ( Q, Q ∩ K ) is inessential. Then one of the strings Q ∩ K is trivial in this tangle. If such a string is s then the string s is unknotted in ( B , T ). Ifsuch a string is one of the arcs obtained from Q ∩ s (cid:48) , then consider the compressing disk C for ∂Q in the interior of Q and the ball Q (cid:48)(cid:48) , containing the string s , obtained after cutting Q along C . From Lemma 3.1(b), either some string of ( B , T ) is unknotted or the tangle ( Q (cid:48)(cid:48) , Q (cid:48)(cid:48) ∩ K )is trivial. Then s is trivial in Q (cid:48)(cid:48) and unknotted in Q (that is obtained from Q (cid:48)(cid:48) after gluing aball along a disk). As s has both strings in the disk T ∩ D ∗ we also have that s is unknotted inthe tangle ( B , T ). (cid:3) Lemma 4.3.
Let T be a solid torus component of V − V ∩ S with more than one component from V ∩ S in its boundary. Then there is no ball Q , in the tangles defined by S , with the followingproperties, (1) T ⊂ Q , ∂Q ∩ ∂T is the disks T ∩ S union an annulus A (that contains at least twocomponents of T ∩ S ); (2) ( ∂Q ∩ S ) ∪ A is an annulus A (cid:48) , and A (cid:48) − A is a collection of disks, attached to somecomponents of T ∩ S , containing the disks of T ∩ S not in A ; (3) the two strings of a tangle are in Q and the tangle in Q with these two strings is essential.Proof. Suppose there is a ball Q as in the statement. From Lemma 3.1(a) the complement of Q in S contains an essential tangle. As ( Q, Q ∩ K ) is an essential tangle, from Ozawa’s unicitytheorem, we have that the tangle decomposition of K defined by S and ∂Q are isotopic. Notethat as A (cid:48) − A is a collection of disks in S attached to ∂T ∩ S , A − A ∩ S = A (cid:48) − A (cid:48) ∩ S .Consider an arc a in A − A ∩ S connecting the two components of ∂A . Then a is also an arc in A (cid:48) − A (cid:48) ∩ S connecting the two different components of ∂A (cid:48) . Consider S (cid:48) after an isotopy of ∂Q long a regular neighborhood of the arc a in the complement of Q . Then, all disks of S ∩ T thatare in A are now in a single disk component of S (cid:48) ∩ T . The sphere S (cid:48) defines the same tangledecomposition of K than S does. And also, as A contains at least two components of T ∩ S , wehave | S ∩ V | > | S (cid:48) ∩ V | , which contradicts the minimality of | S ∩ V | . (cid:3) For the next lemmas assume that n ≥ T of V − V ∩ S intersecting K at a single arc component. Suppose there is an outermost disk ∆ over T , and let δ be the respective outermost arc attached to the disk D of V ∩ S . Assume also without loss ofgenerality that T is in ( B , T ). Denote by s and s the strings of T , and by s and s thestrings of T . Suppose that s is the string of T that T contains. Lemma 4.4.
If one of the disks separated by δ in S − intD contains just one disk of V ∩ S andit intersects K once, some string of some tangle is unknotted.Proof. Suppose one of the disks cut by δ in S − intD , say O , contains a single disk of V ∩ S . As δ is a st or sk-arc the disk of V ∩ S in O is a disk D ∗ of D ∗ , which from the statement intersects K once.Consider the disk L = ∆ ∪ O . As we are in S , by attaching the 2-handle with core L to T weobtain a ball. Consequently, as D ∗ is a disk of V ∩ S (separating or non-separating in V ), byattaching a regular neighborhood of the annulus A = L − intD ∗ to V we have a handlebody V (cid:48) also of genus three. Furthermore, as A is incompressible and non-separating in W , by cutting W along A we obtain a handlebody W (cid:48) of genus three. Altogether, by cutting W along A andsimultaneously adding a regular neighborhood of A to V , we obtain a Heegaard decompositionof S , V (cid:48) ∪ W (cid:48) , of the same genus as the one defined by ∂V .Let T (cid:48) be a solid torus obtained from T by an ambient isotopy of B ∩ V taking D ∗ awayfrom S in B . We denote by Q the ball obtained by attaching a regular neighborhood of L to T (cid:48) . As T intersects K at a single arc and as L intersects K at a single point, we have that Q intersects K at two arcs, with one being unknotted.Let T (cid:48)(cid:48) be a solid torus obtained from T by an ambient isotopy of ( T ∩ S ) − D away from S in B . We denote by R the ball obtained by attaching a regular neighborhood of L to T (cid:48)(cid:48) . As T intersects K at a single arc and as L intersects K at a single point, we have that R intersects K at two arcs, with one being unknotted.Suppose n = 4.(1) Suppose D is in D and D ∗ is not in T . Then Q intersects each string of T at a singlearc. Then by Lemma 3.1(b) some string of some tangle defined by S is unknotted or thetangle ( Q, Q ∩ K ) is trivial. So, we can assume the latter. Each disk of Q ∩ ( V (cid:48) − intQ )intersects K at most at a single point. Therefore, the arcs Q ∩ K can be isotoped to ∂Q intersecting Q ∩ ( V (cid:48) − intQ ) only at the end points. From Lemma 2.3, all the other componentsof ( V (cid:48) − V (cid:48) ∩ S ) − Q intersecting K have the same property. Furthermore, if two consecutivearcs are in adjoint components of V (cid:48) − V (cid:48) ∩ S then, after the isotopy to the boundary of the arcsin the respective components, we can choose that the common ends are at the same point ofthe disk of intersection between the components. (In this case, this is a consequence from eachcomponent of V (cid:48) ∩ S intersecting K at most once and the tangle in each component of V (cid:48) − V (cid:48) ∩ S being trivial.) So, with V (cid:48) being the union of components with these properties, K is parallel to ∂V (cid:48) . We also note that there is a meridian disk of V (cid:48) intersecting K once. Altogether, we havethat ( V (cid:48) − N ( K )) ∪ W (cid:48) is a genus three Heegaard decomposition of the knot K exterior. But | S ∩ V (cid:48) | < | S ∩ V | , which is a contradiction to the minimality of | S ∩ V | .
2) Suppose D is in D and D ∗ is in T . Note that Q intersects T at s in two arcs, withone of the arcs being unknotted in Q . If the tangle ( Q, Q ∩ K ) is trivial then following a similarargument as in (1), we obtain contradiction with the minimality of | S ∩ V | . So, we can assumethat ( Q, Q ∩ K ) is essential. Consider the complement of Q in S , Q c . If the tangle ( Q c , Q c ∩ K )is also essential then the tangle decomposition defined by S is isotopic to the one defined by ∂Q ;as ( Q, Q ∩ K ) contains an unknotted string, this means that some string of some tangle definedby S contains an unknotted string. So, we can assume that the tangle ( Q c , Q c ∩ K ) is trivial.Let s be the intersection of Q c with s , and s the other string of Q c ∩ K . As ( Q c , Q c ∩ K ) istrivial and K is prime, s or s are trivial in Q c . Suppose that s is trivial in Q c . By followinga similar argument as in the proof of Lemma 3.1(a), we have that either s and s are trivialin ( B , T ), or s is trivial in ( B , T ), which is a contradiction to these tangles being essential.Suppose s is knotted in Q c . As ( Q c , Q c ∩ K ) is trivial, there is a proper disk in Q c separating s and s ; let B be the ball separated by this disk containing s . Then s is knotted in B . As K is prime, the string in the complement of B in S , B c ∩ K , is trivial. We have B c ∩ K being s and Q ∩ K . Then, following a similar argument as in Lemma 3.1(a), we have that one ofthe strings of Q ∩ K is trivial in Q , which contradicts ( Q, Q ∩ K ) being essential, or the string s is trivial in Q c and the strings of Q ∩ K are parallel in Q . But one of the strings Q ∩ K isunknotted in Q , which is a contradiction to the assumption that ( Q, Q ∩ K ) is essential.(3) Suppose D is in D ∗ and D ∗ is not in T . The ball R intersects each string of T at a singlearc component, with one of them being unknotted in R . From Lemma 3.1(a), some string of sometangle defined by S is unknotted or the tangle ( R, R ∩ K ) is trivial. Let R be the complementof R in B , and R c the complement of R in S . Suppose the tangle ( R , R ∩ K ) is trivial then,as there are no local knots in ( B , T ), R ∩ s is unknotted in R . As R ∩ s is also unknottedin R we have that s is unknotted in B . So, we can assume that ( R , R ∩ K ) is essential.Again from Lemma 3.1(a), we have that the tangle ( R c , R c ∩ K ) is essential. Therefore, thetangle decompositions defined by S and ∂R are isotopic. This means that the tangle ( R, R ∩ K )is the following product tangle: it is ambient isotopic to the tangle in the ball ( D ∪ O ) × I , thatis R , with strings being (( D ∪ O ) ∩ K ) × I . Let V (cid:48) be obtained from V by replacing T (cid:48)(cid:48) by R , as in (1), and W (cid:48) = S − intV (cid:48) . Then, the arcs R ∩ K can be isotoped to ∂R intersecting R ∩ ( V (cid:48) − intR ) only at the end points. Also, if two arcs are in adjoint components of V (cid:48) − V (cid:48) ∩ S then, after the isotopy to the boundary of the respective components, we can assume that thecommon ends are at the same point of the disk of intersection between the components. (In thiscase, this is a consequence from ( R, R ∩ K ) being the product tangle described, the tangle ineach component of V (cid:48) − V (cid:48) ∩ S being trivial and each component of V ∩ S intersecting K atmost once.) So, as in (1), ( V (cid:48) − intN ( K )) ∪ W (cid:48) is a Heegaard decomposition of the knot exteriorwith | S ∩ V (cid:48) | < | S ∩ V | , and we have a contradiction to the minimality of | S ∩ V | .(4) Suppose D is in D ∗ and D ∗ is in T . So, the ball R intersects s at two arcs, and R intersects K at a portion of s and the string s . If the tangle ( R , R ∩ K ) is trivial wehave that the string s is trivial in R , and as it has ends in the same disk component of R ∩ S it is unknotted in ( B , T ). So, we can assume that ( R , R ∩ K ) is essential. FromLemma 3.1(a), the tangle ( R c , R c ∩ K ) is essential. Then the tangle decompositions definedby S and ∂R are isotopic. This means that the tangle ( R, R ∩ K ) is the product tangle as in(3). Following a similar argument as in (3), we obtain a contradiction to the minimality of | S ∩ V | .Suppose n = 3.Assume that the ends of s are at the same disk of T ∩ S . Then Q intersects each string of T at a single component. Therefore, from Lemma 3.1(b) some string of some tangle is unknotted orthe tangle ( Q, Q ∩ K ) is trivial. In the latter case we have that s is trivial in Q and unknotted n B . In case the ends of s are in distinct components of T ∩ S , we can follow a similarargument as in case (4). (Note that, as n = 3 and the genus of V is three the solid torus T cannot contain two disks of D ∗ and components of D ; so, in this case we have necessarily D in D ∗ and D ∗ in T .) (cid:3) Lemma 4.5.
Suppose T intersects D ∗ at two disks, D and D (cid:48) , and is disjoint from D . Thensome string of some tangle is unknotted, or there is a ball Q , in B , where (1) Q ∩ S is a disk intersecting D ∗ in two components; (2) Q ∩ K is a collection of two arcs each with one end in Q ∩ S ; (3) ( Q, Q ∩ K ) is a product tangle with respect to the disk Q ∩ S and its intersection with K ; (4) the complement of Q in B intersects T either in a cylinder between D (cid:48) and a diskparallel to it in V , or in a cylinder between two disks parallel to D (cid:48) in V .Proof. As T contains a single component from the intersection with K , we have that D and D (cid:48) intersects K once. As D intersects K at one point, one of the disks separated by δ in S − intD intersects K once; denote by O this disk. Let T (cid:48) be the solid torus obtained by an isotopy of T taking D (cid:48) away from S in B . Consider the ball Q defined by adding the 2-handle with core L = O ∪ ∆ to T (cid:48) . Denote by Q the complement of Q in B .First assume that O ∩ D ∗ is a disk not in T . Then Q ∩ T is a cylinder between D (cid:48) and a diskparallel to it in V . The arc Q ∩ s is unknotted in Q . From Lemma 3.1(b), either some stringin the tangle ( B , T ) is unknotted or the tangle ( Q, Q ∩ K ) is trivial. So, we can assume thelatter. Also, from Lemma 3.1(a), the tangle in the complement of Q in S is essential. If thetangle defined in ( Q , Q ∩ K ) is also essential then the tangle decompositions defined by S and ∂Q are isotopic. Then the tangle in Q is the product tangle as in the statement. Otherwise, ifthe tangle ( Q , Q ∩ K ) is inessential then, as the strings of Q ∩ K are trivial in Q , both stringsof the tangle ( B , T ) are unknotted. So, we either have that one string of ( B , T ) is unknottedor that ( Q, Q ∩ K ) is the product tangle described with Q intersecting T in a cylinder between D (cid:48) and a disk parallel to it in V .Assume now that O ∩ D ∗ is a disk in T . In this case, Q ∩ T is a cylinder having intersectionwith Q in two disks parallel to D (cid:48) in V . From Lemma 3.1(a), the tangle in the complement of Q ,or of Q , in S is essential. If the tangle ( Q , Q ∩ K ) is essential then the tangle decompositionsdefined by ∂Q and S are isotopic. This implies that the tangle ( Q, Q ∩ K ) is the product tangleas in the statement. Otherwise, the tangle ( Q , Q ∩ K ) is trivial. As the string s is in Q withends in the disk Q ∩ S , it is also unknotted in ( B , T ). Hence, we either have that one stringof ( B , T ) is unknotted or that ( Q, Q ∩ K ) is the product tangle described with Q intersecting T in a cylinder between two disks parallel to D (cid:48) in V . (cid:3) Outermost disks over components of V − V ∩ S when n = 3In this section we consider the several cases when n = 3 with respect to the existence of agenus two or a genus one component of V − V ∩ S .So assume n = 3 and let D ∗ , D ∗ and D ∗ be the disk components of S ∩ V that intersect K .As | S ∩ K | = 4, without loss of generality, we assume that | D ∗ ∩ K | = 2 and | D ∗ i ∩ K | = 1, for i = 2 ,
3. As no 2-sphere is non-separating in S , we have that D ∗ is not parallel to D ∗ or D ∗ in V . So, D ∗ isn’t parallel in V to any other disk of S ∩ V . Lemma 5.1. If V − V ∩ S has a genus two component then some string of some tangle isunknotted.Proof. Assume there is a component of V − S ∩ V with genus two that we denote by V . As thegenus of V is three, S ∩ V is a collection of at most two disks. f S ∩ V is a collection of two disks or a single disk disjoint from K , then, as the genus of V isthree, some disk of D is parallel to a disk of D ∗ , or D ∗ is parallel to D ∗ or D ∗ in V . This isimpossible as observed before. Therefore, S ∩ V is a single disk intersecting K .As S ∩ V is also separating, we can only have S ∩ V = D ∗ , as in Figure 6. So, the disks D ∗ Figure 6 and D ∗ are necessarily parallel in the solid torus separated by D ∗ in V , and we have n = 0.Also, as V is the only non-ball component of V − V ∩ S , from Remark 3, all outermost disksare over V and attached to D ∗ .Let C be the ball component of V − V ∩ S cut from V by D ∗ ∪ D ∗ ∪ D ∗ and suppose it lies inthe tangle ( B , T ). The ball C contains both strings of the tangle ( B , T ): the string s withone end in D ∗ and the other in D ∗ , and the string s with one end in D ∗ and the other end in D ∗ , and from Lemma 2.3 both strings are mutually trivial in C .Between the arcs of E ∩ P with end in D ∗ or D ∗ we choose one that is outermost in E , say γ , asin Figure 7(a). We note that γ cannot have one end in D ∗ and the other in D ∗ , as, otherwise δ wouldn’t be essential in P . (See Figure 7(b).) So, without loss of generality, assume that γ hasone end in D ∗ . The disk Γ is in the complement of C in B and its boundary intersects D ∗ only Figure 7 : In (a) the arc γ represents an arc of E ∩ P outermost in E between theones with at least one end in either D ∗ or D ∗ . The label 2 ∗ | ∗ at an end of the arc γ means that this end is either at the disk D ∗ or at the disk D ∗ . once. So, D ∗ is a primitive disk with respect to the complement of C in B , which is a genustwo handlebody. Then, by an isotopy of C along D ∗ away from S in B , we are left with theball C ∗ , ∗ that intersects S at D ∗ and D ∗ , whose complement in B is a solid torus and withthe string s as a core. Hence, the string s is unknotted in ( B , T ). (cid:3) Lemma 5.2.
If there is a solid torus component of V − V ∩ S then both strings of some tangleare µ -primitive.Proof. As the genus of V is three, and one component of V − V ∩ S is a solid torus, the componentsof V − V ∩ S are balls or solid tori. From Remark 3, all outermost disks are over solid toruscomponents of V − V ∩ S . Let T be a torus component of V − S ∩ V with a outermost disk overit, and suppose T is in B . The collection of disks T ∩ S cannot be bigger than four as the genusof V is three. If the number of disks in T ∩ S is four then D ∗ is parallel to some other disk of V ∩ S , which is impossible as previously observed. So, | T ∩ S | is at most three. uppose T ∩ S is a single disk. If T ∩ S is disjoint from K we get a contradiction to Lemma4.1. If T ∩ S intersects K , from Lemma 4.2 some string of some tangle is unknotted.In case T ∩ S is a collection of two disks then we have several cases to consider. If these twodisks don’t intersect K then D ∗ is necessarily separating. Furthermore, one string from a tanglelies in a ball of V − V ∩ S cut by T ∩ S and D ∗ with the two ends in D ∗ . Then, from Lemma2.3 this string is trivial in the respective tangle, which is a contradiction to the tangle beingessential. If only one disk of T ∩ S intersects K then it is necessarily D ∗ , because K intersects T ∩ S an even number of times. In this situation, T intersects K at a single arc and from Lemma4.2 some string of some tangle is unknotted.If the two disks of T ∩ S intersect K then T ∩ S = D ∗ ∪ D ∗ . In this case, D ∗ ∪ D ∗ separate V Figure 8 into two solid tori components, T and V . The disk D ∗ is in V and is necessarily separating.(See Figure 8.) We also have n = 0. Then, for the respective outermost arc of an outermostdisk over T we are always under the statement of Lemma 4.4, which means that some string ofsome tangle defined by S is unknotted.Assume now that T ∩ S is a collection of three disks. At least some disk of T ∩ S intersects K ,as otherwise D ∗ would have to be parallel in V to some other disk of V ∩ S , which is impossibleas previously observed.If only one disk of T ∩ S intersects K then this disk is D ∗ , and from Lemma 4.2 some string ofsome tangle is unknotted.If two disks of T ∩ S intersect K then these disks have to be D ∗ and D ∗ . As the genus of V isthree either T ∩ S or D ∗ ∪ D ∗ cuts a ball from V . In either case, D ∗ would have to be parallelto some other disk, which is impossible as previously observed.The last case is when T ∩ S = D ∗ ∪ D ∗ ∪ D ∗ . The disk D ∗ can be separating or non-separating.In the latter case D ∗ ∪ D ∗ ∪ D ∗ separates a ball from V , and in the former case the disks D ∗ and D ∗ are parallel and the disk D ∗ separates a solid torus V in V . (See Figure 9.) So, fromLemma 2.3 and the fact that no disk of D is parallel to a disk of D ∗ , we can assume that n = 0.From | S ∩ V | = 3 and Lemma 2.4(b), there is only one disk attached to outermost arcs.Assume D ∗ is non-separating, then D ∗ ∪ D ∗ ∪ D ∗ separates a ball C from V , which is in thetangle ( B , T ). If there is a string in C with both ends in D ∗ then, from Lemma 3.1(d), this Figure 9 string is unknotted in ( B , T ). So, we can assume that each string in C has only one end in D ∗ . onsider a second-outermost disk Γ ∗ , and the respective second-outermost arc γ ∗ . Then Γ ∗ is inthe complement of C in B . If γ ∗ has equal ends then, following the proof Lemma 3.4, we havethat some string of some tangle is unknotted. If the ends of γ ∗ are distinct, then D ∗ , D ∗ or D ∗ is primitive in the complement of C in B . Suppose D ∗ (or D ∗ ) is primitive with respect to thecomplement C in B . After an isotopy of C along D ∗ (resp., D ∗ ) away from S , we have thatthe complement of a regular neighborhood of the string s (resp., s ) is a solid torus, whichimplies that this string is unknotted in ( B , T ). Suppose D ∗ is primitive with respect to thecomplement of C in B . As the complement of C in B is a handlebody, after an isotopy of C along D ∗ away from S , we obtain a cylinder from D ∗ to D ∗ , with core t , whose complement in B is a solid torus. Then t is unknotted in B . As s and s are trivial in C , we have that C is the union of the regular neighborhoods of t ∪ s , and also of t ∪ s . Consequently, both s and s are µ -primitive.Assume now that D ∗ is separating. Suppose D ∗ and D ∗ are the only disks attached to outermostarcs. As D ∗ is parallel to D ∗ by the finiteness of outermost arcs, if we consider a second-outermostarc we have that both disks have loops attached in G P , which contradicts Lemma 2.4(b). So, D ∗ has outermost arcs attached and all second-outermost arcs are after outermost arcs attachedto D ∗ . If there is an outermost disk over V , from Lemma 4.2 some string of some tangle isunknotted. So, we can assume that all outermost disks are over T . Let Γ ∗ be a second-outermostdisk, then Γ ∗ is in the complement of V in B . Suppose ∂ Γ ∗ is essential in ∂V ∪ D ∗ S . Thenthe complement of V in B is also a solid torus (intersecting S at a single disk). From Lemma2.3 the string s is trivial in V . Then, from Lemma 2.2, s is µ -primitive. We note also that B ∩ V is V together with the cylinder cut by D ∗ ∪ D ∗ in V , C ∗ ∗ , where the string s isa core. As the complement of B ∩ V in B is a handlebody we have that s is trivial in thecomplement of V in B . Therefore, from Lemma 2.2, s is also µ -primitive. Suppose now that ∂ Γ ∗ is inessential in ∂V ∪ D ∗ S . Then ∂ Γ ∗ bounds a disk L in ∂V ∪ D ∗ S . Let R be the ball in B bounded by Γ ∗ ∪ L . By similar arguments as in the proof of Lemma 3.4, we have that s isin R and is parallel to L . So, s is trivial in the complement of V in B . As the complementof B ∩ V in B is a handlebody, this implies that the complement of V in B is a solid torus.Then, as when ∂ Γ ∗ is essential in ∂V ∪ D ∗ S , we have that both s and s are µ -primitive. (cid:3) Outermost disks over components of V − V ∩ S when n = 4Along this section we consider the several cases when n = 4 with respect to the componentsof V − V ∩ S topology and their intersection with S ∩ V .So assume that n = 4. As | S ∩ K | = 4 we have | D ∗ i ∩ K | = 1, for i = 1 , , ,
4. Therefore, D ∗ i is a non-separating disk in V .Denote by γ ∗ i the outermost arcs of E ∩ P , in E , among the arcs with at least one end in D ∗ i ,for i = 1 , , ,
4. Also, let Γ ∗ i denote the disk of E − E ∩ P co-bounded by γ ∗ i in the outermostside of this arc in E , for i = 1 , , , Lemma 6.1. If V − V ∩ S contains a genus two component then some string in some tangle isunknotted.Proof. Assume that V − V ∩ S contains a genus two component, V . As the genus of V is three S ∩ V is a collection of at most two disks. If S ∩ V is a collection of two disks then S ∩ V are allparallel disks in V and, from Remark 3, all outermost disks are over V . Therefore, n = 0 andall disks of D ∗ are parallel, as in Figure 10(a). Consequently, by the finiteness of outermost arcs,we have parallel type I d ∗ -arcs in E , as in Figure 11(a1) or (a2), in contradiction to Corollary2.2. Then, S ∩ V is a single disk. As each disk of D ∗ intersects K once, S ∩ V is a disk of D .Then, all disks of D ∗ are parallel in the solid torus cut from V by S ∩ V , and all disks of D are .. Figure 10Figure 11 parallel to S ∩ V in V , as in Figure 10(b). Let D , . . . , D n be the disks of D , with S ∩ V being D . The outermost disks are all adjacent to D and are over V . Consider a second-outermostarc γ , as in Figure 11(b). If the arc γ has at least one end in D n , or has one end in D ∗ and theother in D ∗ , by Lemma 3.4, some string in the tangle decomposition defined by S is unknotted.Otherwise, if all second-outermost arcs have both ends in D ∗ or both ends in D ∗ , as when S ∩ V is two disks, by the finiteness of outermost arcs we have a contradiction to Corollary 2.2. (cid:3) Assume now that V − V ∩ S has a solid torus component T with some outermost disk overit. Hence, as the genus of V is three, the components of V − V ∩ S are solid tori or balls, andthe solid torus T components intersect S at most in four disks. As each disk of D ∗ intersects K once, the solid torus T intersects D ∗ at an even number of disks. Lemma 6.2.
Suppose V − V ∩ S contains a solid torus component intersecting D ∗ at the fourdisks. Then some string in some tangle is unknotted.Proof. Let T be the solid torus component of V − V ∩ S as in the statement, and suppose it liesin the tangle ( B , T ). As the genus of V is three and T intersects D ∗ at the four disks, we havethat the disks of D ∗ are parallel two-by-two in V , say D ∗ parallel to D ∗ and D ∗ parallel to D ∗ .So, n = 0, and V ∩ S is as in Figure 12. Also, from Remark 3, we can assume all outermostdisks are over T . From Lemma 2.4(b), at most two disks are adjacent to outermost arcs.If the outermost arcs are attached to a single disk or if they are attached to two non-paralleldisks, by the finiteness of outermost arcs of E ∩ P in E we have a contradiction to Corollary 2.2.Then, the outermost arcs are attached to two parallel disks. Without loss of generality, assume Figure 12 that the only disks adjacent to outermost arcs are D ∗ and D ∗ . Consider the second outermost igure 13 arc γ of E ∩ P in E , after the outermost arcs attached to D ∗ and D ∗ , and the disk componentof E − E ∩ P , Γ, co-bounded by γ on the outermost side of γ in E . Let C and C be thecylinders cut from V by D ∗ ∪ D ∗ and D ∗ ∪ D ∗ , resp.. We have that Γ is in B − intC . If Γ isessential in S ∪ ∂C , as in Figure 13(a), then Γ is a meridian disk to B − intC , which impliesthat the string s , a core of C , is unknotted in the tangle B . Otherwise, if Γ is inessential in S ∪ ∂C , we have that ∂ Γ bounds a disk L in the torus S ∪ D ∗ ∪ D ∗ ∂C . (See Figure 13(b), (c).)Let R be the ball in B bounded by Γ ∪ L . The string s , as a core of C , is in R and, as thereare no local knots, it is trivial in R and parallel to L . Hence, as the complement of C ∪ C in B is a handlebody, we have that the complement of C in B is a solid torus. Therefore, inthis case, the string s is also unknotted. (cid:3) Lemma 6.3.
Suppose V − V ∩ S contains a solid torus that intersects D ∗ in a collection of twodisks and D in a single separating disk. Then both strings of some tangle are µ -primitive.Proof. Let T be the solid torus component of V − V ∩ S as in the statement, and suppose it liesin the tangle ( B , T ). Assume that T ∩ D ∗ = D ∗ ∪ D ∗ and that D ∩ T = D , and denote by V the solid torus separated by D in V . As D ∩ T is separating, and K is connected, the fourdisks of D ∗ have to be parallel in V . If all outermost arcs are attached to D ∗ or to D ∗ then,by the finiteness of outermost arcs, we have a contradiction to Corollary 2.2. Hence, there is anoutermost arc attached to D . The set D contains a collection of separating disks, D , . . . , D k in V , and might also contain a collection of non-separating parallel disks D k +1 , . . . , D n in V ,as in Figure 14(a), (b). From Remark 3, the outermost disks are over T or V , and from Lemma4.1 there are no outermost disks over V . So, all outermost disks are over T , attached to D ∗ , D ∗ or D , with no sequence of parallel arcs of E ∩ P in E after an outermost arc attached to D ∗ or D ∗ . Case 1.
Assume that all disks of D are parallel and separating as in Figure 14(a). If n >
1, by the finiteness of outermost arcs, there is a sequence of parallel arcs of E ∩ P in E , δ , . . . , δ n , as in the Figure 15(a), where δ i has both ends in D i and δ is an outermost arcattached to D . Denote the outermost disk that δ co-bounds by ∆, and the disk between δ i and δ i +1 by ∆ i . Considering the disks ∆ i and the cylinder cut from V by D i ∪ D i +1 we definea ball R i as in Lemma 3.3. The balls R i intersect S at disks O i and O i +1 , co-bounded by δ i .. ... ... Figure 14 and δ i +1 resp., each containing an end of the string in R i . Then, in particular, O contains asingle disk of D ∗ . If n = 2, as R contains a single string, we have that O intersects S ∩ V at a single disk, that is of D ∗ . Assume n ≥
3. If D ∗ , or D ∗ , is in O then R contains T andconsequently two strings of the tangle, which is a contradiction as R contains a single string.Therefore, without loss of generality, we can assume that D ∗ is in O . Suppose that some diskof D , say D i , is in O . Then O i ⊂ O and D ∗ ⊂ O i . Consequently, following the strings in thesequence of balls R j , we have T ⊂ R i − and D in O i , which is a contradiction as D i is in O .Therefore, D ∗ is the only disk of S ∩ V in O . Then, by Lemma 4.4, if n > n = 1. As before we denote by δ an outermost arc attached to D . If a disk cutfrom S − intD by δ intersects D ∗ at a single disk from Lemma 4.4 some string of some tangleis unknotted. Therefore, we can assume that all outermost arcs δ separate S − intD into twodisks each intersecting D ∗ at two disks. Consequently they are all parallel in P . Let Γ be asecond-outermost disk. The disk Γ is in the complement of V in B . If ∂ Γ is inessential inthe solid torus B ∪ D V then Γ bounds a disk L in S ∪ D ∂V . Let R be the ball boundedby Γ ∪ L in B . By similar arguments as in the proof of Lemma 3.4, we have that the strings s and s are in R and are parallel to L . Hence, the complement of V in B is a solid torusintersecting S at a single disk. Altogether, from Lemma 2.2 we have that both strings s and s are µ -primitive. Suppose now that ∂ Γ is essential in the solid torus B ∪ ∂ D V . Then thecomplement of V in B is also a solid torus. Consider an outermost arc among the arcs withone end in D ∗ or D ∗ , and denote these arcs by γ ∗ . Suppose there are arcs γ ∗ with both endsin D ∗ and also in D ∗ . Then there are arcs γ ∗ and γ ∗ of Type II outermost among the d ∗ -arcs,and the disks Γ ∗ and Γ ∗ are in B and intersect D ∗ and D ∗ , resp., exactly once. Then, D ∗ and D ∗ are primitive with respect to the complement of V ∩ B in B . Let T (cid:48) be the solid torusobtained by an isotopy of T along D ∗ ∪ D ∗ away from S . We also have that an outermost disk∆ intersects a meridian of T (cid:48) once. Altogether, the complement of the cylinder from D ∗ to D ∗ in B is a solid torus; as the core of this cylinder is the string s , this string is unknotted in( B , T ). Otherwise, without loss of generality, suppose there is an arc γ ∗ with only one end in D ∗ . This means γ ∗ is an γ ∗ arc, and we can consider the respective disk Γ ∗ . Let C (resp., C )be the cylinder from D ∗ to D ∗ (resp., D ∗ to D ∗ ) in V . As D ∗ is primitive with respect to thecomplement of V ∩ B in B , a core of C , as the string s , is trivial in the complement of V in B . If the other end of γ ∗ is in D ∗ then D ∗ is also primitive with respect to the complementof V ∩ B in B , and similarly a core of C , as the string s , is trivial in the complement of V in B . Otherwise, if the other end of γ ∗ is not in D ∗ , using the disk Γ ∗ , we have that a coreof C , as the string s , is trivial in the complement of V in B . Then, from Lemma 2.2, bothstrings s and s are µ -primitive. Case 2.
Assume now that D also has a collection of non-separating disks in V , as in Figure14(b). laim 6.3.1. If the outermost arcs attached to D are not parallel in P then some string ofsome tangle is unknotted.Proof of Claim 6.3.1. In fact, let δ and δ (cid:48) be outermost arcs attached to D , non-parallel in P . Consider the disjoint disks O and O (cid:48) co-bounded, respectively, by δ and δ (cid:48) in S − D , andalso the respective outermost disk ∆ , ∆ (cid:48) . Consider the disks L = O ∪ ∆ and L (cid:48) = O (cid:48) ∪ ∆ (cid:48) .Let Q be the ball obtained by attaching a regular neighborhood of L and L (cid:48) to T and addinga ball to the respective boundary component disjoint from S . If D ∗ ⊂ O ∪ O (cid:48) then the arcs δ and δ (cid:48) are parallel. If ( O ∪ O (cid:48) ) ∩ D ∗ is only D ∗ and D ∗ , then ∂Q − ∂Q ∩ S is a compressingdisk for P . Otherwise, D ∗ ∪ D ∗ is in O ∪ O (cid:48) and the string s is in Q . From Lemma 4.3 thetangle ( Q, Q ∩ K ) is trivial. Therefore, the string s is trivial in Q . As the ends of s are inthe same disk component of Q ∩ S we have that s is unknotted in ( B , T ). (cid:52) From the previous claim, we assume that the outermost arcs attached to D are parallel in P . If k >
1, by the finiteness of outermost arcs we have a sequence of arcs, δ i for i = 1 , . . . , k ,after an outermost arc, δ , as in the Figure 15(b). Following the construction at the beginningof Case 1, from each sequence of parallel arcs after an outermost arc δ we have a sequence ofballs R i , i = 1 , . . . , k . Also, as there are no t-arcs, the outermost arc, δ k +1 , after these arcs is ast-arc. If some arc δ k +1 has both ends in D k , following an argument as in Lemma 3.4, we havethat some string of some tangle is unknotted. Then, the arcs δ k +1 have both ends in D k +1 , asin Figure 15(c), or in D n .For any k , suppose we have both situations, that there are arcs δ k +1 and δ (cid:48) k +1 with both endsin D k +1 and D n , resp.. Consider the component disks ∆ k and ∆ (cid:48) k of E − E ∩ P co-boundedby δ k +1 and δ (cid:48) k +1 , resp., in the outermost side of these arcs in E . As the outermost arcs δ areparallel in P , and the balls R i , i = 1 , . . . , k −
1, contain only one string, the arcs of ∂ ∆ k and ∂ ∆ (cid:48) k that have both ends in D k are parallel in P . Let C be the ball cut from V by D k ∪ D k +1 ∪ D n ,and C k,k +1 (resp., C k,n ) be the ball obtained from C by an isotopy of D n (resp., D k +1 ) awayfrom S . Let L k and L (cid:48) k be the disks bounded by ∂ ∆ k and ∂ ∆ (cid:48) k , resp., in ∂C k,k +1 ∪ D k ∪ D k +1 S Figure 15 and ∂C k,n ∪ D k ∪ D n S , resp.. Consider the balls R k and R (cid:48) k bounded by L k ∪ ∆ k and L (cid:48) k ∪ ∆ (cid:48) k ,not containing S . Similarly, as observed in Case 1, the balls R k and R (cid:48) k contain only one string.Suppose none of these balls contains the other, as in Figure 16(a). Hence, each of the disks L k and L (cid:48) k intersection with S contains a disk component, O k and O (cid:48) k resp., co-bounded, with ∂D k ,by a single arc of ∂ ∆ k ∩ S , δ k , and ∂ ∆ (cid:48) k ∩ S , δ (cid:48) k . Each of the arcs δ k and δ (cid:48) k is in a sequence of arcs fter an outermost arc, as in Figure 15(b). As observed before in this Claim, we are assumingthat these arcs are parallel in P . Then one of the disks O k or O (cid:48) k has to be contained in theother, which is a contradiction with the assumption that R k and R (cid:48) k are disjoint. So, assumethat, say, R (cid:48) k is contained in R k , as in Figure 16(b). Then L k contains D n and L (cid:48) k . Therefore,from the minimality of | E ∩ P | and from the arcs of ∂ ∆ k ∩ S and ∂ ∆ (cid:48) k ∩ S that have both endsin D k being parallel in P , we have that ∂ ∆ k intersects S in two arcs, one with two ends in D k and the other with two ends in D k +1 ; similarly, ∂ ∆ (cid:48) k intersects S in two arcs, one with twoends in D k and the other with two ends in D k +1 . Let O k +1 , resp. O (cid:48) k +1 , be the disks cut from S − intD k +1 , resp. S − intD n , by δ k +1 , resp. δ (cid:48) k +1 , disjoint from D k . As there are no localknots, the string in R (cid:48) k ⊂ R k is trivial. Then, from the minimality of | S ∩ V | , we have | O (cid:48) k +1 ∩ V | equal to | O (cid:48) k ∩ V | . Also, O k ∩ ( V ∩ S ) is the same as O (cid:48) k ∩ ( V ∩ S ). Therefore, | O k +1 ∩ V | isbigger than | O k ∩ ( V ∩ S ) | . So, we can isotope D k +1 ∪ O k +1 along R k union the ball C k,k +1 toreduce | S ∩ V | , which is a contradiction.So, assume without loss of generality that all arcs δ k +1 have both ends in D k +1 . By the finiteness Figure 16 : The disks L and L k when R (cid:48) k is disjoint from R k and when R (cid:48) k is containedin R k , resp.: the arcs δ k won’t be parallel in P as previously observed. of outermost arcs we have a sequence of parallel arcs, δ k +2 , . . . , δ n , as in Figure 15(d), and therespective sequence of balls R k +2 , . . . , R n − . Then, we have a sequence of arcs parallel to anoutermost arc, δ , . . . , δ n , and the respective balls R , . . . , R n − . Following a similar argumentas in Case 1, we have that δ is as in Lemma 4.4, which means that some string of some tangleis unknotted. (cid:3) Lemma 6.4.
Suppose V − V ∩ S contains a solid torus that intersects D ∗ at two disks and D at a single non-separating disk. Then both strings of some tangle are µ -primitive.Proof. Let T be the solid torus component of V − V ∩ S as in the statement, and suppose it liesin the tangle ( B , T ). Assume that T ∩ D ∗ = D ∗ ∪ D ∗ and that T ∩ D = D . The disks D ∗ and D ∗ are not parallel, otherwise D would be separating. Then, D ∪ D ∗ ∪ D ∗ separate a ball rom V , as in Figure 17, and all outermost disks are over T with corresponding outermost arcsattached to D ∗ , D ∗ or D . The disks D , D , . . . , D n are all parallel and non-separating in V . ... Figure 17
Claim 6.4.1.
If the disks of D ∗ are parallel two-by-two then some string of some tangle isunknotted.Proof of Claim 6.4.1. Assume that D ∗ is parallel to D ∗ and that D ∗ is parallel to D ∗ in V , as in Figure 17(a). If D ∗ or D ∗ are the only disks with outermost disks attached then bythe finiteness of outermost arcs we have parallel sk-arcs in E , as in Figures 11(a1), (a2), whichis a contradiction to Corollary 2.2. So, D has an outermost arc attached. Furthermore, fromCorollary 2.2, even if D ∗ , or D ∗ , has outermost arcs attached we cannot have a sequence ofparallel sk-arcs in E after such outermost arcs. So, by the finiteness of outermost arcs only someoutermost arc attached to D is before a sequence of parallel arcs of E ∩ P in E , as in Figure15(a). Consider a second-outermost arc γ , and the disk component of E − E ∩ P , Γ, co-boundedby γ in the outermost side of this arc in E , as in Figure 11(b). The boundary of Γ intersects S in γ and arcs with both ends in D n . If γ has at least one end in D n , or one end in D ∗ andthe other in D ∗ , then from Lemma 3.4 we have that some string in some tangle is unknotted.Otherwise, the ends of all second outermost arcs are both in D ∗ or both in D ∗ , and by thefiniteness of outermost arcs we have a contradiction to Corollary 2.2. (cid:52) From this claim, we can assume that the disks of D ∗ are not parallel two-by-two in V .Therefore, as no disk of D ∗ can be parallel in V to a disk of D , without loss of generality, weassume that the disks D ∗ and D ∗ are parallel to D ∗ , as in Figure 17(b). Under this setting, wecontinue the lemma’s proof in several steps with respect to which disks are attached to outermostarcs and to the value of n . Claim 6.4.2.
The disks D or D ∗ have outermost arcs attached; and the disks D ∗ and D ∗ cannot have simultaneously outermost arcs attached.Proof of Claim 6.4.2. If all outermost arcs are attached to D ∗ then there is a sequence ofparallel sk-arcs, as in Figure 11(a2), which is a contradiction to Corollary 2.2. Then D or D ∗ have outermost arcs attached.Suppose D ∗ and D ∗ have simultaneously outermost arcs attached. Let δ ∗ i be an outermost arcattached to D ∗ i , and ∆ ∗ i the respective outermost disk, for i = 1 ,
4. Consider also the disjointdisks O ∗ and O ∗ , in S − int { D ∗ ∪ D ∗ } , co-bounded by δ ∗ and δ ∗ , respectively. Let L ∗ i = ∆ ∗ i ∪ O ∗ i ,for i = 1 ,
4. As the arcs δ ∗ i are sk-arcs, D ∗ ∪ D ∗ is in O ∗ ∪ O ∗ . Taking a regular neighborhood ofthe disks L ∗ i together with T , and by capping off the boundary component of N ( T ) ∪ i =1 , N ( L ∗ i )disjoint from S with the ball it bounds, we get a ball Q in the tangle ( B , T ) containing bothstrings s and s . Each string of T in Q has ends in two distinct disk components of ∂Q ∩ S , D ∗ ∪ O ∗ and D ∗ ∪ O ∗ . Then with the tangle ( Q, T ) we have a contradiction between Lemma4.3 and Lemma 3.1(c). (cid:52) laim 6.4.3. If D or D ∗ is not attached to outermost arcs then n ≤ .Proof of Claim 6.4.3. If D or D ∗ is not attached to outermost arcs then all outermost d-arcshave either both ends in D n or in D . Then by the finiteness of outermost arcs there is a se-quence of parallel arcs, δ i , as in Figure 15(a), parallel to some outermost d-arc attached to D n or D . As in Case 1 of Lemma 6.3, using the disks ∆ i between the arcs δ i and δ i +1 in E , attachedto the disks D i and D i +1 , resp., and the disk that ∂ ∆ i bounds in the torus C i,i +1 ∪ D i ∪ D i +1 S ,we define a ball R i . Each of these balls contains a single string of the tangle decomposition andit is regular neighborhood of it. If n ≥ V − S ∩ V are contained insome ball R i ∪ C i,i +1 . We note that these balls are either disjoint or intersect at a disk, wetherthe strings they contain are disjoint or intersect at an end. Then, taking the union of the largestballs R i ∪ C i,i +1 for each string, we have a solid torus with K as its core, V in its interior andboundary essential in W , which is a contradiction as W is a handlebody. So, given that n isodd, n ≤ D and D ∗ have outermost arcs attached, in Claim 6.4.5 we also prove that n ≤ (cid:52) Claim 6.4.4. If D ∗ and D are the only disks with outermost arcs attached then some string ofsome tangle is unknotted.Proof of Claim 6.4.4. Suppose both disks D and D ∗ are attached to outermost arcs, δ and δ , resp., . If n = 1 then either δ or δ are as in Lemma 4.4, which means that some string ofsome tangle is unknotted.So, from Claim 6.4.3, we can assume that n = 3. From Corollary 2.2 there are no parallelsk-arcs after an outermost arc attached to D ∗ , as in Figure 11(a2). Consequently, from thefiniteness of outermost arcs, we have such a sequence of parallel arcs after an outermost arcattached to D , as in Figure 15(a), and consider the respective balls R i , for i = 1 ,
2. Let O i and O i +1 be the disk components of R i ∩ S that are co-bounded by δ i and δ i +1 , resp., for i = 1 , R contains a single string, O intersects D ∗ at a single disk. Then, as D ∗ has a type I arcattached, this disk is not in O . If D ∗ or D ∗ are in O then R contains T and, consequently,two strings of the tangle, which we know is impossible. Then D ∗ is in O , the string s is in R and the string s is in R . So, if D or D is in O , also O or O will be, and consequentlythe same for D ∗ or D ∗ , which is impossible as observed before. Then, O ∩ ( S ∩ V ) is only D ∗ .So, δ is an outermost arc as in Lemma 4.4. Then, some string of some tangle is unknotted. (cid:52) Claim 6.4.5. If D ∗ and D are both attached to outermost arcs then both strings of some tangleare µ -primitive.Proof of Claim 6.4.5. Suppose that both D and D ∗ have outermost arcs attached, denotedby δ and δ ∗ resp.. Let O and O ∗ be the disjoint disks in S − int ( D ∪ D ∗ ) co-bounded by δ and δ ∗ , resp.. Consider also the disks L ∗ = ∆ ∗ ∪ O ∗ and L = ∆ ∩ O . Let Q be the ballobtained by adding a regular neighborhood of L ∗ and L to T , together with the ball that theboundary component of N ( T ) ∪ N ( L ) ∪ N ( L ∗ ), disjoint from S , bounds. As δ ∗ and δ are ansk-arc and an st-arc, we have that O ∗ and O intersect D ∗ . As D ∗ is not in O ∗ ∪ O , in thisparticular case D ∗ ∪ D ∗ is necessarily in O ∗ ∪ O , and the string s is also in Q . The disk D ∗ may or not be in O ∗ ∪ O .If D ∗ is in O ∗ ∪ O then Q intersects S in two components: D ∗ ∪ O ∗ and D ∪ O . From Lemma4.3, the tangle ( Q, T ) is trivial, which is a contradiction to Lemma 3.1(c).So, we can assume that D ∗ is not in O ∗ ∪ O and Q intersects S in three component disks: D ∗ , D ∗ ∪ O ∗ and D ∪ O . Also, O ∩ D ∗ , and O ∗ ∩ D ∗ , is either D ∗ or D ∗ . Furthermore, from emma 4.3, both strings s and s are trivial in Q .If n = 1 then δ is as in Lemma 4.4, which means that some string of some tangle is unknotted.So we can assume that n ≥ E after an outermost arc δ , δ , . . . , δ n , as inFigure 15(a), and consider the balls R i as in Case 1 of Lemma 6.3. Then, as O ∩ D ∗ is either D ∗ or D ∗ we have that the ball R contains two strings, which is a contradiction to the balls R i containing a single string. Consequently, there is no sequence of parallel arcs in E , δ , . . . , δ n ,after an outermost arc δ .Consider an arc parallel to an outermost arc δ ∗ or otherwise a second-outermost arc, γ , anddenote by Γ the disk of E − E ∩ S , co-bounded by γ , in the outermost side of this arc in E .(See Figure 18(a).) As there is no sequence δ , . . . , δ n after the outermost arcs δ , as in Figure18(a), we have that Γ intersects S in γ and outermost arcs δ ∗ . Note that γ cannot have only oneend in D n , otherwise γ would be a t-arc, which is a contradiction to Lemma 2.4(c). If γ hastwo ends in D ∗ or one end in D ∗ and the other end in D ∗ , following reasoning as in the proofof Lemma 3.4, we have that some string in some tangle is unknotted. If the ends of all arcs γ are both in D ∗ , by the finiteness of outermost arcs, we have a contradiction to Corollary 2.2.Hence, we can assume that some arc γ has both ends at D n .Let O n be the disk in S − intD n cut by γ , disjoint from D ∗ . Denote by C the ball cut from Figure 18 V by D ∗ ∪ D ∗ ∪ D n , and by C ∗ ,n the cylinder obtained from C by an isotopy of C along D ∗ away from S . Note that C is in B . Consider the disk L bounded by ∂ Γ in the torus ∂C ∗ ,n ∪ D ∗ ∪ D n S . Let R be the ball bounded by Γ ∪ L in B . If R intersects K in two com-ponents, then we can prove that γ is parallel to δ ∗ in E . By taking R together with C ∗ ,n wedefine a cylinder containing the two strings of T with ends in the disks D ∗ ∪ O ∗ and D n ∪ O n .Then from Lemma 3.1(a), (c), and because ∂C ∗ ,n − L ∩ ∂C ∗ ,n is a single disk containing D ∗ ∪ D n , we obtain a contradiction to the minimality of | S ∩ V | . So, we have that R intersects T at a single component. Naturally O n ⊂ L , and also O n ∩ D ∗ is D ∗ . In fact, if D ∗ is in O n then, s is in R . As R intersects T at a single component, and O ∗ intersects D ∗ , we have D ∗ in O ∗ , which contradicts our assumption that O ∗ ∩ D ∗ is only D ∗ or D ∗ . If D ∗ is in O n then, following a similar reasoning, D ∗ is in O ∗ and D ∗ is in O . As before, with the existenceof parallel arcs to γ or δ in E we can define the balls R n − or R . But then, in this case, R or R n contain two strings, which is a contradiction. Then, D ∗ is in O n . As O n is disjoint from O ∗ , and O ∗ ∩ D ∗ is either D ∗ or D ∗ , we have that D ∗ is in O ∗ . Then, D ∗ is in O and if R exists it has two strings, which is impossible. So, we can assume that there is a sequence of balls R n − , . . . , R exists, related to a sequence of parallel arcs of E ∩ P to γ in E , δ n − , . . . , δ .As O n contains D ∗ , if n ≥ R n − contains T and consequently twostrings, which is a contradiction. Therefore n = 3, and the ball R contains the string s . But cannot contain T , otherwise it would contain two strings. Hence, O ⊂ O ∗ and O ⊂ O .(See Figure 18(b).)Consider an arc α outermost after the outermost arcs δ and parallel arcs to γ . Then α has endsin D ∪ D . If the arcs α have one end in D and the other in D then we get a contradiction to D ⊂ O ∗ and O being disjoint from O ∗ . Then α has equal ends. If the ends of α are in D then α is in O ∗ (because D is in O ∗ ). All loops attached to D , as α , have to be parallel in P to thearc parallel to γ in P attached to D . Otherwise, D ∗ is contained in O ∗ , which contradicts theassumption that it is not. Let A be the disk of E − E ∩ P co-bounded by α in the outermost sideof the arc in E . Suppose α is attached to D or is parallel to δ in P . The boundary of A boundsa disk in S ∪ D ∪ D ∂C , that contains O , and the union of these two disks bounds a ball, R (cid:48) ,in B . The ball R (cid:48) has similar properties to the balls R i ; including containing a single string of T , which is a consequence of Lemma 3.1 (a), (c), the arcs ∂A ∩ S − γ with both ends in D and D being parallel in P resp., and also from the minimality of | S ∩ V | . But has R (cid:48) contains O , italso contains two strings, which a contradiction to the previous observation. Then, α is attachedto D and is not parallel to δ . In this case, R (cid:48) contains the string s as a core, that is parallelto the core of the cylinder C , . Consider the outermost arcs γ (cid:48) among the arcs of E ∩ P withdistinct ends in D ∗ ∪ D . Given the configuration of G P , as in Figure 18(b), the only possibleends for γ (cid:48) are one end in D ∗ and the other in D , one end in D ∗ and the other in D and oneend in D and the other in D . The only possible case, because the disks involved belong to thesame component of V − V ∩ S , is having γ (cid:48) with one end in D ∗ and the other in D . Let Γ (cid:48) bethe disk, of E − E ∩ S , co-bounded by γ (cid:48) , in the outermost side of γ (cid:48) in E . Then Γ (cid:48) is over Q and S , in B . All the arcs of ∂ Γ (cid:48) ∩ S that intersect D ∗ are either γ (cid:48) or have both ends in D ∗ andare parallel to δ ∗ in P . By an isotopy of these arcs to Q we get that D ∗ ∪ O ∗ is primitive withrespect to the complement of Q in B , that is a handlebody. Then the core of the cylinder from D ∗ to D is unknotted. As the string s is parallel to the core from D ∗ to D ∗ in Q and thestring s is parallel to the core from D ∗ to D in Q , we have that both strings are µ -primitive. (cid:52) Claim 6.4.6.
If only D ∗ is attached to outermost arcs then some string of some tangle isunknotted.Proof of Claim 6.4.6. Denote by δ ∗ the outermost arcs attached to D ∗ . Consider a secondoutermost arc, γ , and let Γ be the disk of E − E ∩ V co-bounded by γ in the outermost side ofthis arc in E . (See Figure 18(a).) The curve ∂ Γ bounds a disk L in the torus S ∪ D ∗ ∪ D n ∂C ∗ ,n .Following a similar argument as in Claim 6.4.5, we can assume γ has both ends in D n and wedefine similarly the ball R in B with boundary Γ ∪ L . So, either the string s or a portion thestring s with end in D ∗ is in R , and therefore, this string is parallel to the core of the cylinder C ∗ ,n . Let O ∗ and O be the disjoint disks in S − int { D ∗ ∪ D n } co-bounded by δ ∗ and γ ,resp.. Note that O is in L ⊂ ∂R . As R contains a single string, we have that O intersects D ∗ ata single disk. From Claim 6.4.3, we have n ≤
3; also, when n = 3 we consider the balls R , R and the respective disks of intersection with S , O , O and O , attached to D , D and D , resp. .Assume R contains the string s . In this case O ∗ is in R , and each O and O ∗ contain a singledisk of D ∗ , D ∗ or D ∗ . Then, if n = 3 one of the balls R or R contains two strings of a tangle,which is impossible. Hence, n = 1. As O ∗ is disjoint from D we have that O ∗ intersects S ∩ V at a single disk of D ∗ . Therefore, some arc δ ∗ is as in Lemma 4.4, which means that some stringof some tangle is unknotted.Assume now that R contains a portion of the string s . uppose n = 3. We have O ∩ D ∗ = D ∗ and consequently s is in R and s is in R , whichmeans that O ∩ D ∗ = D ∗ and O ∩ D ∗ = D ∗ , as in Figure 19(a). Consider an outermost arc, α ,between the arcs with ends in distinct disk components of D ∗ ∪ D , and A the disk of E − E ∩ P co-bounded by α in its outermost side in E . Note that α can only have ends in disks in the samecomponent of V − V ∩ S . So, α can only have ends in D ∗ and D , D and D , D and D , oralso, D ∗ and D , as in Figure 19(b).If the ends of α are in D ∗ and D , D and D , or D and D , then the strings s , s or s are unknotted, respectively.So, assume that all arcs α have ends in D ∗ and D . Consider now the outermost arc α (cid:48) between Figure 19 the ones with ends in distinct components of D ∗ ∪ D − D or that have ends in distinct componentsof D . Let A (cid:48) be the disk of E − E ∩ S co-bounded by α (cid:48) in the outermost side of the arc in E .(See Figure 19(c).) The arc α (cid:48) can only connect components of V − V ∩ S with the disks D ∗ and D in them . Hence, the disk A (cid:48) is in the tangle with the strings s , s . Using the disk A (cid:48) and depending on the ends of α (cid:48) we can prove that s or s is unknotted.Suppose n = 1. Suppose that O ∗ ∩ D ∗ is either D ∗ or D ∗ . As O ∗ and D are disjoint, δ ∗ isas in Lemma 4.4, which means that some string of some tangle is unknotted.Suppose, now, that O ∗ intersects D ∗ in D ∗ ∪ D ∗ , as in Figure 20(c). Consider the arcs γ ∗ and Figure 20 γ ∗ , and the respective disks Γ ∗ and Γ ∗ . From Lemma 2.4(b), the two disks D ∗ or D ∗ cannot ave simultaneously loops attached in G P . Then, all arcs γ ∗ or all arcs γ ∗ have distinct ends.Assume that all arcs γ ∗ have distinct ends. Suppose also that some Γ ∗ intersects D ∗ as in Figure20(a). Then the disks D ∗ and D ∗ are primitive with respect to the complement of V ∩ B in B .Consequently, the complement of C ∗ , in B is a solid torus. As s is parallel to the core ofthe ball C ∗ , we have that s is unknotted. Otherwise, suppose that all disks Γ ∗ intersect D ∗ as in Figure 20(b). Then, the disks D ∗ and D ∗ are primitive with respect to the complement of V ∩ B in B . Consider an outermost arc α between the arcs with one end in D ∗ and the otherend in D . Let A be the disk, of E − E ∩ V , co-bounded by α in the outermost side of α in E , as in Figure 20(d). Suppose that A is in B . The components of ∂A ∩ S that intersect D are α and eventually arcs with both ends in D parallel to γ . The disk D ∗ is primitive in thecomplement of V ∩ B in B . Then after adding the 2-handle with core D ∗ to the complementof V ∩ B in B we are left with the complement of C ∗ , ∪ C ∗ , ∗ . We isotope the arcs of A ∩ S parallel to γ , through O , to the boundary of the cylinder C ∗ , . After this isotopy, A intersects Figure 21 D geometrically once. Then, the complement of C ∗ , ∗ in B is a solid torus, which means thatthe string s is unknotted. Otherwise, assume that A is in B . The components of ∂A ∩ S that intersect D are α and eventually arcs with both ends in D parallel to γ , or arcs with oneend in D and the other in D ∗ . As Γ ∗ is in B , we have that A doesn’t intersect any arc γ ∗ .Then we can proceed as follows. Take T union with a regular neighborhood of O . Isotope to N ( D ∪ O ) the arcs of A ∩ S parallel to γ . Then, the disk A intersects D ∪ O geometricallyonce, and Γ ∗ intersects D ∗ geometrically once. As A is disjoint from any γ ∗ , cut T ∪ N ( O ) along D ∗ and, afterwards, we isotope T ∪ N ( O ) along D ∪ O away from S . Denote the solid torusafter the isotopy as T (cid:48) . Then, the complement of T (cid:48) in B is a handlebody. Let O ∗ c be thecomplement of O ∗ in S − intD ∗ . Denote by Q (cid:48) the ball obtained by adding the two handle withcore O ∗ c ∪ ∆ ∗ to T (cid:48) . The ball Q (cid:48) intersects S in D ∗ ∪ O ∗ c and D ∗ , and its complement in B isa solid torus. The ball Q (cid:48) contains s and intersects the string s at an unknotted component.Then, by Lemma 3.1(b), either one string of the tangle decomposition given by S is unknottedor the tangle in ( Q (cid:48) , Q (cid:48) ∩ T ) is trivial. Hence, we can assume the latter and that the string s is trivial in Q (cid:48) . As the string s has one end in each of the two components of Q (cid:48) ∩ S , it is acore of the cylinder Q (cid:48) . Consequently, the string s is unknotted.Suppose now that some γ ∗ has identical ends. Then, all arcs γ ∗ have distinct ends, and fromFigure 21(a), the other end of γ ∗ is in D ∗ . As γ ∗ is the outermost d ∗ -arc with one end in D ∗ and γ ∗ has one end in D ∗ , we have that Γ ∗ intersects D ∗ once. (See Figure 21(b).) This meansthat Γ ∗ is in B and that D ∗ and D ∗ are primitive with respect to the complement of V ∩ B in B . Then, considering the arc α and disk A and proceeding as before, we have that some stringof some tangle is unknotted. (cid:52) laim 6.4.7. If only D is attached to outermost arcs then some string of some tangle isunknotted.Proof of Claim 6.4.7. Let δ denote the outermost arcs attached to D . If n = 3 by thefiniteness of outermost arcs there is a sequence of parallel arcs to some δ , that is δ and δ , asin Figure 15(a), and with this sequence we can consider the balls R i as in Case 1 of Lemma 6.3.Let γ be a second-outermost arc of E ∩ P in E , as in Figure 11(b). From Lemma 3.4, if γ hasone end in D n or one end in D ∗ and the other in D ∗ then some string of some tangle definedby S is unknotted. If all arcs γ have both ends in D ∗ , then, by the finiteness of outermost arcs,we have a contradiction to Corollary 2.2. Then some arc γ has both ends in D ∗ . Consider thisarc γ and let Γ be the disk component of E − E ∩ P co-bounded by γ in the outermost side ofthis arc in E . The disk Γ bounds a disk L in ∂C ∗ ,n ∪ D ∗ ∪ D n S that together with Γ bounda ball R in B . As in the previous claim, we have that either a portion of the string s withend in D ∗ is in R , or the string s is in R . Let O ∗ be the disk co-bounded by γ in S − intD ∗ ,disjoint from D n . Then O ∗ ⊂ L .Assume R contains a portion of the string s . In this case, D ∗ is the only disk of D ∗ in L and D ∗ ⊂ O ∗ . Consider the ball C ∗ obtained by an isotopy of the ball C , cut from V by D ∗ ∪ D ∗ ∪ D n , along D ∗ ∪ D n away from S in B . From Lemma 2.3, the arc C ∗ ∩ s is trivialin C ∗ . We also have that the portion of s in the complement of C ∗ in B is unknotted. Infact, we can assume that this arc is R ∩ s . As there are no local knots, R ∩ s is trivial in R , and therefore, it is parallel to L . We can isotope the components of L ∩ S − O ∗ from S to ∂C ∗ ,n . With the isotopy we verify that the arc R ∩ s is parallel to the boundary of C ∗ .Altogether, we have that s is unknotted in ( B , T ).Assume now that R contains the string s . Following along an argument of the similarsituation in Claim 6.4.6, we have some string of some tangle is unknotted. (cid:52) (cid:3) Lemma 6.5.
Suppose V − V ∩ S contains a solid torus that intersects D ∗ and D at two disks.Then some string of some tangle is unknotted.Proof. Let T be the solid torus component of V − V ∩ S as in the statement, and suppose itlies in the tangle ( B , T ). As the genus of V is three, all disks of D ∗ are parallel in V , and thesame is true for the disks of D . Assume that ∂T ∩ D ∗ = D ∗ ∪ D ∗ and ∂T ∩ D = D ∪ D n , as inFigure 22). From Remark 3, we can assume that all outermost disks are over T with respectiveoutermost arcs attached to D ∗ , D ∗ , D or D n . If all outermost arcs are attached to D ∗ or D ∗ ,then by the finiteness of outermost arcs there are parallel sk-arcs in contradiction to Corollary2.2. Then, some outermost arc is attached to D or D n . Furthermore, even if D ∗ or D ∗ isattached to outermost arcs the only sequences of arcs parallel to outermost arcs in E are withrespect to outermost arcs attached to D or D n . Without loss of generality, we assume that D is always attached to some outermost arc, and denote by δ an outermost arc attached to D . ... Figure 22 laim 6.5.1. If D n is not attached to outermost arcs then some string of some tangle isunknotted.Proof of Claim 6.5.1. Assume that D n is not attached to outermost arcs. By the finiteness ofoutermost arcs and Lemma 2.2, there is a sequence of arcs of E ∩ P in E parallel to an outermostarc δ , that is δ , . . . , δ n , as in Figure 15(a). As in Case 1 of Lemma 6.3 we define the balls R i ;consider also the respective disks O i and O i +1 . As R contains a single string we have that O intersects D ∗ at a single disk. If n = 2 then as O and O are disjoint, we have that O intersects S ∩ V at a single disk. Hence, δ is as in Lemma 4.4, which means that some string of some tangleis unknotted. Suppose n ≥
4. (Note that n is necessarily even.) If D ∗ or D ∗ are in O then R contains two strings, which is impossible. Then, without loss of generality, we can assume that D ∗ is in O . Suppose some disk of D is in O , say D i . Then D ∗ is also in O i . This means that T is in R i − , and consequently, D is in O i , which is a contradiction as D i is in O . Therefore, δ isunder the conditions of Lemma 4.4, which means that some string of some tangle is unknotted. (cid:52) Claim 6.5.2. If D n is attached to outermost arcs then some string of some tangle is unknotted.Proof of Claim 6.5.2. Assume that both D and D n have outermost arcs attached, denotedby δ and δ n resp.. Let the outermost disk co-bounded by δ (resp., δ n ) be denoted by ∆ (resp., ∆ n ) and let O (resp., O n ) be the disk in S − int ( D ∪ D n ) separated by δ (resp., δ n ).By adding a regular neighborhood of O n ∪ ∆ n and O ∪ ∆ to T , and the ball bounded by theboundary component that is disjoint from S , we define a ball Q . If Q contains both strings of T , from Lemma 4.3, we have that the tangle ( Q, Q ∩ K ) is trivial. If D ∗ is in O ∪ O n then weget a contradiction between Lemma 3.1(c) and Lemma 4.3. Then, O or O n intersects D ∗ at asingle disk.If n = 2, O and O are disjoint, and δ or δ n are as in Lemma 4.4, which means that somestring of some tangle is unknotted.Assume n ≥ O ∪ O n intersect D ∗ in three disks and, without loss of generality, that O intersects D ∗ at a single disk. If there is any arc of E ∩ P parallel to δ n in E , the respectiveball R n − contains two strings, which is impossible as observed in Lemma 3.3. Then, there isa sequence of parallel arcs of E ∩ P in E , δ , . . . , δ n − and we can consider the respective balls R , . . . , R n − . If O ∩ D ∗ is D ∗ or D ∗ , then R contains two strings, which is impossible. Then, O n ∩ D ∗ is D ∗ ∪ D ∗ . The string s is trivial in Q and has ends in the same disk component of Q ∩ S , then s is unknotted in ( B , T ).Suppose that O ∪ O n intersect D ∗ in two disks. Assume D ∗ ∪ D ∗ is in O ∪ O n . If thereare two consecutive balls R i after O or O n , then some ball R i contains two strings, which isimpossible. Then, n = 4 and both δ and δ have a parallel arc of E ∩ P in E , δ and δ , resp.,from where we define the balls R and R . As R and R have a single string, we have that O is disjoint from D , D and D . Hence, δ is as in Lemma 4.4 and some string of some tangle isunknotted. Assume now that D ∗ ∪ D ∗ is in O ∪ O n . If there is a sequence of parallel arcs to δ (or to δ n ) attached to all disks D , . . . , D n , as in Figure 15(a), following the same argument asin Claim 6.5.1, we prove that some string of some tangle is unknotted. Otherwise, the sequencesof parallel arcs from δ go up to some arc δ i and from δ n go up to some arc δ i +1 . If n = 4,then as before δ or δ n are as in Lemma 4.4, which means that some string of some tangle isunknotted. Suppose n ≥
6. Then, again using arguments as in Claim 6.5.1, if O intersects D , it is in D i or D i +1 . From the sequences of parallel arcs we can consider the respective balls R , . . . , R i − and R i +1 , . . . , R n − . Denote by C k,k +1 the cylinder in V between D k and D k +1 .If D i and D i +1 are not in O then δ resp., is as in Lemma 4.4, which means that some string f some tangle is unknotted. Otherwise, without loss of generality, suppose that D i is in O .Then, then as R i − cannot be in Q, we have that C i,i +1 is in Q . Each string of the tanglesdefined by S is in Q or is some ball R k . Following as in the previous claim, consider Q unionwith R k ∪ C k,k +1 , for k = 1 , . . . , i − , i + 1 , . . . , n , we define a solid torus that is a neighborhoodof K , containing V , and with boundary essential in W , which is a contradiction to W being ahandlebody. (cid:52) (cid:3) Lemma 6.6.
Suppose V − V ∩ S contains a solid torus component disjoint from D and inter-secting D ∗ at two disks. Then both strings of some tangle are µ -primitive.Proof. Let T be a solid torus component as in the statement and suppose D ∗ ∩ T = D ∗ ∪ D ∗ .Assume that T is in the tangle ( B , T ). From Remark 3, all outermost disks are over solid toruscomponents of V − V ∩ S . Suppose some outermost disk is attached to some disk of D . As thegenus of V is three, this outermost disk is over a solid torus disjoint from K intersecting S ∩ V at a single disk, which is a contradiction to Lemma 4.1. Then all outermost disks are attachedto disks of D ∗ . Claim 6.6.1.
If the disks of D ∗ are parallel two-by-two then some string in some tangle isunknotted.Proof of Claim 6.6.1. Suppose only one disk or two non-parallel disks are adjacent to out-ermost arcs. By the finiteness of outermost arcs we have parallel sk-arcs, as in Figure 11(a1),(a2), and we get a contradiction to Corollary 2.2.Otherwise, we are left with the case when the outermost arcs are only adjacent to two paralleldisks of D ∗ . Following an argument of a similar situation in Lemma 6.2, we have that somestring in some tangle is unknotted. (cid:52) Claim 6.6.2.
If the disks of D ∗ are not parallel two-by-two then both strings of some tangle are µ -primitive.Proof of Claim 6.6.2. Assume, without loss of generality, that no other disk of S ∩ V is parallelto D ∗ . If there are disks (of D ∗ ) parallel to D ∗ , and D ∗ or one disk parallel to it are the onlydisks with outermost arcs attached, then we get a contradiction to Corollary 2.2. So, withoutloss of generality, assume there is some outermost arc attached to D ∗ , and that it is over T .Under these conditions, we define a ball Q as in Lemma 4.5, using the outermost disk attachedto D ∗ over T . From Lemma 4.5, the tangle ( Q, Q ∩ K ) is the product tangle with respect to Q ∩ S . So, we can isotope S through Q , and we replace D ∗ with a disk parallel to D ∗ . So,if n = 0 we reduce this case to either the case when there is a genus two component, as inLemma 6.1 or to the case when V − V ∩ S contains a solid torus component intersecting D ∗ atthe four disks as in Lemma 6.2. If n > V − V ∩ S containsa solid torus component intersecting D ∗ in a collection of two disks and D in one or two disks,as in Lemmas 6.3, 6.4 and 6.5. From these lemmas we get that both strings in some tangle are µ -primitive. (cid:52) (cid:3) Lemma 6.7.
Suppose V − V ∩ S contains a solid torus component disjoint from K . Then bothstrings of some tangle are µ -primitive.Proof. As the genus of V is three and no disk of D ∗ is parallel to a disk of D , D ∩ T is a collectionof at most three disks. If there is some solid torus component of V − V ∩ S intersecting D ∗ we follow as in the previous lemmas to get the conclusion that two strings of some tangle are -primitive. Otherwise, the solid torus components of V − V ∩ S are disjoint from K . FromRemark 3, without loss of generality, we can assume that some outermost disk is over T .If D ∩ T is a single disk we get a contradiction to Lemma 4.1. Then, we have that T intersects D at more than one disk, in which case T is the only solid torus component of V − V ∩ S andall outermost disks are over T .Assume that D ∩ T is a collection of two disks, D and D (cid:48) . The outermost arcs are attached ...... ...... V ... Figure 23 to D or to D (cid:48) , with outermost disks over T . Let D , D , . . . , D k be the disks of D parallel to D in V , in case there exists such a sequence. Without loss of generality, assume there is anoutermost arc δ attached to D , and that there is a sequence of arcs, δ i , after an outermostarc, δ , as in Figure 15(a). Let ∆ be the outermost disk bounded by δ , in E , and also, ∆ i bethe disk of E − E ∩ S between δ i and δ i +1 . As S has no lens space summand, we have that ∂ ∆intersects a meridian of T geometrically once. So, we can perform an isotopy of the annulus in S , A = D ∪ ( S ∩ N (∆)) through N (∆) to the annulus A (cid:48) = D ∪ ( ∂T ∩ N (∆)). As ∂ ∆ intersects ameridian of T geometrically once, we isotope A (cid:48) through T to a disk in T parallel to D (cid:48) , that wealso denote by D . Using the disk ∆ ∪ ∆ we can perform a similar isotopy, and from the disk D of E ∩ S we get a disk in T parallel to the new disk D (c.f. Morimoto [14, Lemma 3.3]). Inthis way we can perform a sequence of isotopies of S to get from the disks D , D , . . . , D k newdisks in T parallel to D (cid:48) . With this isotopy we reduce this case to other cases: If the disks of D ∗ are not parallel in V we can reduce this case to the case when T ∩ D ∗ is a collection of twodisks and T ∩ D is one non-separating disk, as in Lemma 6.4. So, we are left with the situationwhen the disk components of D ∗ are parallel. The disk components of D in V can be parallelto D or to D (cid:48) , or can be separating. Assume there is a disk of D that is separating in V , asin Figure 23(a). By the previous isotopy we reduce this case to the case, considered next, when D ∗ ∩ T is empty and D ∩ T is a collection of three disks. Otherwise, suppose that no disk of D is separating, as in Figure 23(b). Similarly, we reduce this case to the case when D ∗ ∩ T and D ∩ S is a collection of two disks, as in Lemma 6.5.At last, suppose that D ∩ T is a collection of three disks. We have a collection of parallel non- ...... Figure 24 separating disks of D , and a collection of separating disks of D in V , as in Figure 24. As inLemma 3.4, let C be the ball component of V − V ∩ S cut from V by D ∗ ∪ D ∗ ∪ D n . Everyoutermost arc of E ∩ S in E attached to D n has both ends attached to it (otherwise, it would be t-arc, which don’t exist from Lemma 2.4(c)). By the finiteness of outermost arcs, we consideran outermost arc γ after outermost arcs with both ends in D n . From Lemma 3.4, if γ hasat least one end in D n or one end in D ∗ and the other in D ∗ , some string of some tangle isunknotted. In case, all arcs γ have both ends in D ∗ or in D ∗ , by the finiteness of outermostarcs, we have a contradiction to Corollary 2.2. (cid:3) Proof of Theorem 1
For the proof of Theorem 1, we study all cases of S ∩ V with respect to the value n . Proposition 1. If n = 1 then both strings of some tangle are µ -primitive.Proof. Suppose n = 1. If n > n = 0, P is a disk and | P ∩ E | = 0.Let D ∗ = S ∩ V . The 2-sphere S = D ∗ ∪ P is separating, then D ∗ is a separating disk in V . Figure 25
As the handlebody V has genus three, the disk D ∗ separates V in a solid torus V and a genustwo handlebody V , as in Figure 25. Let ( B , T ) denote the tangle containing V . The stringsof this tangle lie in V , have end points in D ∗ and, by Lemma 2.3, are simultaneously parallelto ∂V . Also, the complement of V in B is a torus. Hence, from Lemma 2.2, both strings ofthe tangle ( B , T ) are µ -primitive. (cid:3) Proposition 2. n (cid:54) = 2 .Proof. Suppose n = 2. We denote by D ∗ and D ∗ the components of D ∗ . From Lemma 2.4(b),(g) n > Figure 26 laim. If n = 2 there is no ball C of V − V ∩ S containing strings of a tangle.Proof of Claim. Suppose that there is a ball component of V − V ∩ S , C , containing strings ofa tangle.Suppose, the ball C contains two strings. From Lemma 2.3, the strings are parallel to ∂C .Therefore, the tangle ( C, C ∩ K ) is trivial, which is a contradiction to Lemma 3.1(c).Otherwise, suppose that C contains a single string. As D ∗ ∪ D ∗ intersects K in four points onlyone of these disks can be in ∂C , and both ends of the string in C are in this disk. Then, thisstring is trivial in ∂C . Furthermore, as this is the only string in C it is also trivial in the respec-tive tangle, which is a contradiction to the tangle decomposition defined by S being essential. (cid:52) If D ∗ and D ∗ are parallel in V then the ball component of V − S ∩ V cut by D ∗ ∪ D ∗ is incontradiction to Claim.Suppose now that D ∗ and D ∗ are not parallel, as in the examples of Figure 26. Then, thecomponents of V − D ∗ ∪ D ∗ are solid tori. As n >
0, the disks of D are in some of these solidtori. Then, some ball component of V − V ∩ S contains D ∗ , D ∗ , or both, which is a contradictionto Claim. (cid:3) Proposition 3. If n = 3 then both strings of some tangle are µ -primitive.Proof. Consider the components of V − V ∩ S . From Remark 3 we can assume that somecomponent of V − S ∩ V is not a ball.If there is a genus two component of V − V ∩ S then, by Lemma 5.1 , some string of some tangleis unknotted. Otherwise, there is some solid torus component of V − V ∩ S , and from Lemma5.2 two strings of some tangle are µ -primitive. (cid:3) Proposition 4. If n = 4 then both strings of some tangle are µ -primitive.Proof. As in Proposition 3, we consider the components of V − V ∩ S and we assume that somecomponent of V − S ∩ V is not a ball.If V − V ∩ S has a genus two component then, by Lemma 6.1, some string of some tangle isunknotted.Now, assume that V − V ∩ S has no genus two component. This means at least one of itscomponents is a solid torus, T . The collection of disks D ∗ ∩ T is always even, because ∂T is aseparating torus in S . We consider several cases with respect to D ∗ ∩ T .If D ∗ ⊂ T , from Lemma 6.2, some string of some tangle is unknotted.Suppose D ∗ ∩ T is a collection of two disks. As the genus of V is three, there are at most twodisks of D in ∂T . Then we are under Lemmas 6.3, 6.4, 6.5 and 6.6, and we have that bothstrings of some tangle are µ -primitive.At last, suppose D ∗ ∩ T = ∅ . From Lemma 6.7, we also have that both strings of some tangleare µ -primitive. (cid:3) We can now prove Theorem 1 and its Corollary 1.1. of Theorem 1. If K has an inessential 2-string free tangle decomposition then the tunnel numberof K is one. This is a contradiction with the assumption that the tunnel number of K is two.Hence, the 2-string free tangle decomposition of K is essential.We have that 0 ≤ n ≤
4. If n = 0 then, as S ∩ K ⊂ S ∩ V we have n = 0. Hence, S ⊂ V which is a contradiction to Lemma 2.3(a). In case n (cid:54) = 0, from Propositions 1, 2, 3 and 4, wehave that two strings of some tangle are µ -primitive. (cid:3) f Corollary 1.1. Let K be a knot with a 2-string free tangle decomposition where at least astring of each tangle is not µ -primitive.From Corollary 2.4 in [16] by Morimoto, if a knot K has a n -string free tangle decomposition,then t ( K ) ≤ n −
1. Hence, in this case t ( K ) ≤ K has both strings being µ -primitive,from Theorem 1 we have t ( K ) ≥ t ( K ) = 3. (cid:3) On the tunnel number degeneration under the connected sum of prime knots
In this section, we construct an infinite class of knots with a 2-string free tangle decompositionwhere no tangle has both strings being µ -primitive. With these collection of knots, Theorem 1and the work of Morimoto [16] we prove Theorem 2.A particular, simplified, version of Theorem 3.4 in [16] by Morimoto gives us the followingproposition, which is relevant to the proof of Theorem 2. Proposition 5 ([16], Morimoto) . Let K be a knot which has a -string free tangle decompositionand K be a knot with a -bridge decomposition. Then t ( K K ) ≤ . Figure 27 : The knot K ( m ) and one unknotting tunnel, with m a natural number. For the construction of knots, K , as in Theorem 2 we consider 2-string free tangle decom-positions. Suppose there are two 2-string free tangles ( B , T ) and ( B , T ) where one of thestrings in each tangle is not µ -primitive. Identify ( ∂B , ∂ T ) to ( ∂B , ∂ T ), such that no stringof T has its end identified to the ends of the same string of T . Then ( B ∪ B , T ∪ T ) is a knot( S , K ) under the conditions of Proposition 5. Furthermore, from Corollary 1.1, t ( K ) = 3.Hence, this procedure gives us a knot as in the statement of Theorem 2.So, we need 2-string free tangles with one of the strings not µ -primitive. As observed in Remark1, if a string s properly embedded in a ball B is µ -primitive, then by capping s along ∂B weget a µ -primitive knot. Then, for the construction of a 2-string free tangle where at least one ofthe strings is not µ -primitive we consider a tunnel number one knot K that is not µ -primitive,and one of its unknotting tunnels. For such a knot K , let s be a string in a ball B , that when apped off along ∂B we obtain K , together with one unknotting tunnel for K . If we slide theends of the unknotting tunnel from s to ∂B we get an essential 2-string free tangle where one ofthe strings is not µ -primitive.Then, we want tunnel number one knots that are not µ -primitive. Existence results of suchknots are known by work Johnson and Thompson in [6] and also Moriah and Rubinstein in [12].On the other hand, explicit or constructive examples of knots with tunnel number one that arenot µ -primitive is given by work Eudave-Mu˜noz in [18] and [19], Ram´ırez-Losada and Valdez-S´anchez in [9], Minsky, Moriah and Schleimer in [10] and also Morimoto, Sakuma and Yokota in[17]. With any of these examples it is possible to construct knots as in the statement of Theorem2. As an example of such construction we consider the class of knots K (7 ,
17; 10 m −
4) from [17],where m is an integer, together with an unknotting tunnel. We denote these knots by K ( m ), asin Figure 27.As previously described, from the knots K ( m ) and an unknotting tunnel we construct tangles T ( m ) where at least one of the strings is not µ -primitive, as in Figure 28. From the construction Figure 28 : A possible construction of a tangle T ( m ) from the knot K ( m ) and one ofits unknotting tunnels. we have that the tangles T ( m ) are free. With the tangles T ( m ) and T ( m (cid:48) ) we construct a knot K ( m, m (cid:48) ), as explained before, that has a 2-string free tangle decomposition where no tanglehas both strings being µ -primitive. With this construction we can now prove Theorem 2 and itsCorollary 2.1. of Theorem 2. Consider the collection of knots { K ( m, m (cid:48) ) : m, m (cid:48) ∈ N , m ≤ m (cid:48) } . From Corol-lary 1.1, we have that t ( K ( m, m (cid:48) )) = 3. From Ozawa’s unicity theorem, the knot K ( m, m (cid:48) ) isprime. And, from Proposition 5, for any 3-bridge knot K , t ( K ( m, m (cid:48) ) K ) ≤ (cid:3) of Corollary 2.1. Consider the collection of knots { K ( m, m (cid:48) ) : m, m (cid:48) ∈ N , m ≤ m (cid:48) } . Let K beany 3-bridge prime knot with tunnel number two. From Proposition 5, t ( K ( m, m (cid:48) ) K ) ≤ t ( K ( m, m (cid:48) ) K ) ≥
3. Then, t ( K ( m, m (cid:48) ) K ) = 3 = t ( K ( m, m (cid:48) )) + t ( K ) − (cid:3) eferences [1] W. Jaco, Lectures on three-manifold topology, CMBS 43 by Amer. Math. Soc., Providence, Rhode Island,1997 reprint.[2] C. McA. Gordon, On the primitive sets of loops in the boundary of a handlebody, Topology and it Applica-tions 27 (1987), 285-299.[3] C. McA. Gordon, Combinatorial methods in Dehn surgery, Lectures at knots ’96 (Tokyo), World Sci. Pub-lishing (1997), 263-290.[4] C. MCA. Gordon and A. W. Reid, Tangle decompositions of tunnel number one knots and links, J. KnotTheory and its Ramifications Vol. 4 No. 3 (1995), 389-409.[5] J. Hempel, 3-manifolds, AMS Chelsea Publishing vol. 349, reprint 2004.[6] J. Johnson and A. Thompson, On tunnel number one knots which are not (1 , n ), J. Knot Theory and itsRamifications 20 No. 4 (2011), 609-615.[7] T. Kobayashi, A construction of arbitrarily high degeneration of tunnel numbers of knots under connectedsum, J. Knot Theory and its Ramifications 3 No. 2 (1994), 179-186.[8] T. Kobayashi and Y. Rieck, Heegaard genus of the connected sum of m -small knots, Comm. Anal. Geom.14 No. 5 (2006), 1037-1077.[9] E. Ramirez-Losada and L. Valdez-Snchez, Hyperbolic (1 , S with crosscap number two tunnelnumber one, Topology and its Applications 156 (2009), 1463-1481.[10] Y. Minsky, Y. Moriah and S. Schleimer, High distance knots, Algebraic and Geometric Topology 7 (2007),1471-1483.[11] Y. Moriah, Heegaard splittings of knot exteriors, Geometry and Topology Monographs 12 (2007), 191-232.[12] Y. Moriah and H. Rubinstein, Heegaard structures of negatively curved 3-manifolds, Comm. Anal. Geom. 5(1997), 375-412.[13] K. Morimoto, There are knots whose tunnel numbers go down under connected sum, Proc. Amer. Math.Soc. Vol. 123 No. 11 (1995), 3527-3532.[14] K. Morimoto, On the additivity of tunnel number of knots, Topology and its Applications 53 (1993), 37-66.[15] K. Morimoto, Tunnel number, connected sum and meridional essential surfaces, Topology 39 (2000), 469-485.[16] K. Morimoto, On the degeneration ratio of tunnel numbers and free tangle decompositions of knots, Geometryand Topology Monographs 12 (2007), 265-275.[17] K. Morimoto, M. Sakuma and Y. Yokota, Examples of tunnel number one knots which have the property’1 + 1 = 3’, Math. Proc. Camb. Phil. Soc. 119 (1996), 113-118.[18] M. Eudave-Mu˜noz, Incompressible surfaces and (1 , , n knots is at least n , Topology38 (1999), 265-270.[27] M. Scharlemann and J. Schultens, Annuli in generalized Heegaard splittings and degeneration of tunnelnumber, Math. Ann. 317 (2000), 783-820.[28] T. Schirmer, Lower bounds on the Heegaard genus of amalgamated 3-manifolds, preprint, arXiv:1211.4568. CMUC, Department of Mathematics, University of Coimbra, Apartado 3008 EC Santa Cruz, 3001-501 Coimbra, Portugal
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