Distance two links
Abstract.
In this paper, we characterize all links in with bridge number at least three that have a bridge sphere of distance two. We show that a link has a bridge sphere of distance at most two then it falls into at least one of three categories:

The exterior of contains an essential meridional sphere.

can be decomposed as a tangle product of a Montesinos tangle with an essential tangle in a way that respects the bridge surface and either the Montesinos tangle is rational or the essential tangle contains an incompressible, boundaryincompressible annulus.

is obtained by banding from another link that has a bridge sphere of the same Euler characteristic as the bridge sphere for but of distance 0 or 1.
1. Introduction
Following Hempel’s definition of distance for Heegaard splittings [He], there have been a number of results showing that knots with high distance bridge surfaces are well behaved. For example, high distance knots are hyperbolic and do not contain essential surfaces of small euler characteristic [BS] and they have unique minimal bridge surfaces [To07]. Additionally, connect sums of knots with high distance bridge surfaces have multiple distinct bridge surfaces [JT]. Most recently, the Cabling Conjecture was established for all knots with a bridge surface of distance at least three [BCJTT, BCJTT2].
These results motivate us to study low distance knots and links. (We will use the term “link” below to mean both onecomponent and multiplecomponent links.) Links that have a bridge surface of distance two are of particular interest because a link with a minimal bridge sphere of distance one always contains an essential meridional surface [Th] and links with bridge surfaces of distance greater than two share many nice properties that often allow them to be addressed collectively as “high distance” links.
In this paper, we prove the following theorem. Precise definitions will be given later.
Theorem 1.1.
Let be a link in and let be a bridge sphere for of distance 2. Then at least one of the following holds:

There is an essential meridional sphere in the exterior of ,

can be decomposed as a tangle product of a bridgesplit Montesinos tangle with an essential tangle such that is a disk bridge surface for and either is rational or the exterior of in contains an incompressible, boundary incompressible annulus, or

can be obtained via banding from a link that has a bridge sphere with the same Euler characteristic as but has distance 0 or 1.
Conclusion (2) of Theorem 1.1 is illustrated in Figure 1. Conclusion (3) is illustrated in Figure 2. All of the conclusions are discussed in detail in Section 2.
We note that this paper is the analogue to recent results about 3manifolds with distancetwo Heegaard splittings. Hempel [He] and Thompson [Th3] (independently) classified all 3manifolds that have genus two, distancetwo Heegaard splittings. In work in preparation, Rubinstein and Thompson [RT] have proved a similar classification theorem for all manifolds that have a distancetwo Heegaard splitting of any genus. Portions of the proof of Theorem 1.1 are inspired by the work of Rubinstein and Thompson on distancetwo Heegaard splittings.
The analogy between the present work and these results about Heegaard splittings is most easily seen in the threebridge case. Theorem 1.1 implies the following:
Corollary 1.2.
If is a 3bridge knot with a minimal bridge sphere of distance two, then one of the following holds:

The exterior of contains an incompressible meridional 4punctured sphere.

is obtained by banding a 3bridge presentation of either the unknot, a 2bridge knot, or the connected sum of two 2bridge knots.

is a small Montesinos knot.
Recall that every genustwo 3manifold is the doublebranched cover of over a threebridge knot and that every genustwo Heegaard surface is the lift of a 6punctured bridge sphere to the doublebranched cover. Additionally, every essential loop in a genustwo Heegaard surface is the lift of an essential loop in the corresponding bridge sphere. This fact implies that the distance of a 6punctured bridge sphere is equal to the distance of the corresponding genustwo Heegaard surface in the doublebranched cover. Thus, as expected, the Heegaard splittings in the Hempel/Thompson classification are precisely the double branched covers of the links in Corollary 1.2.
2. Definitions and Constructions
2.1. Bridge surfaces
Let be a link in the 3sphere and a regular neighborhood of . Let be the standard height function from to whose level surfaces are concentric 2spheres. A level sphere is called a bridge sphere for if all maxima of are above and all minima of are below . In particular, if is a bridge sphere for , then intersects each of the 3balls bounded by in boundary parallel arcs called bridges. The disks of parallelism are called bridge disks. More precisely, a bridge disk is a disk with interior disjoint from and the bridge surface, and whose boundary is the union of an arc in and an arc in the bridge surface. For technical reasons, we will only discuss links whose bridge spheres have bridge number at least three (so at least 6 punctures). However, twobridge knots and links are well understood [HT, STo] and every onebridge knot is an unknot.
Given a Morse function where is any 3manifold, an arc or a surface properly embedded in is vertical if does not have any critical points in the interior of .
2.2. Distance
The curve complex of a compact surface is the simplicial complex with vertices that correspond to isotopy classes of essential (including not boundary parallel) simple closed curves in . An edge connects two vertices if the corresponding isotopy classes of curves have disjoint representatives. The distance of a bridge sphere for a link in is the length of the shortest path in the curve complex for from the boundary of a compressing disk for in above to the boundary of a compressing disk for in below . The notion of distance measured in the curve complex was first introduced by Hempel in [He]. There is also a similar notion of distance measured in the arc and curve complex introduced by Bachman and Schleimer in [BS].
For a bridge sphere of distance two, every compressing disk on one side of the bridge sphere intersects every compressing disk on the other side nontrivially, but there is an essential simple closed curve in that is disjoint from at least one compressing disk on each side of . For each of the possible conclusions of our main theorem, we consider the possibility of a converse.
2.3. Incompressible spheres
Recall that a meridional surface for a link is a surface properly embedded in whose boundary is a set of meridional curves in . A meridional surface is compressible if there is a disk in with interior disjoint from whose boundary is contained in and is essential in . A meridional surface is incompressible if it is not compressible.
The first possible conclusion of our Main Theorem is that our link has an incompressible, boundaryincompressible planar meridional surface. In general, it is likely that there are knots of distance at least 3 which have essential meridional planar surfaces in their exterior. However, if there is an essential planar surface C for the link which intersects the bridge surface in a single simple closed curve which is essential on both surfaces, such that is disjoint from compressing discs for on both sides of , then the link will have distance at most 2. The intersection curve is the middle vertex in a length two path in the curve complex of . In some cases, the planar surface we construct in the proof of the Main Theorem will be of this type.
2.4. Disk Bridge Surfaces and Montesinos Tangles
In what follows, it will be natural to study tangles from the perspective of the height function on . Let be the projection map . Given a tangle properly embedded in such that , we will say that is in bridge position if all of the maxima of project into and all of the minima of project into . A disk is a disk bridge surface for if is a regular value of the height function , and all the minima of lie below and all the maxima lie above it.
Recall that an strand tangle is rational if each of the strands can be simultaneously isotoped into while fixing their endpoints. An strand tangle is essential if is incompressible when viewed as a meridional, planar surface. Let for be a rational tangle and let be an incompressible punctured disc for some . The result of gluing and along and is a Montesinos tangle if the image of and is not boundaryparallel in . The disc in is a Montesinos disk for . Given a Montesinos tangle properly embedded in such that , we say that is bridgesplit if has both maxima and minima, the foliation induced by on consists of parallel arcs and separates the maxima of from the minima of , as in Figure 3. A knot in is a tangle product of tangles and if is the result of gluing to along their boundary and .
Suppose a link has a tangle decomposition such that one of the tangles is a bridgesplit Montesinos tangle. If the bridge surface meets in a disk bridge surface for , then has distance at most two. A disk bridge surface for a bridgesplit Montesinos tangle is depicted in green in Figure 3.
2.5. Banding
A third way to construct a bridge surface of distance at most two is as follows: Let be any link such that some local maximum lies below some local minimum with respect to the standard height function . Notice that if we monotonically isotope all maxima of above all minima of , there will be a compressing disk for one of the resulting bridge spheres that surrounds and a second disjoint compressing disk around , on the other side of . Thus, the level sphere corresponding to a bridge sphere for will have distance at most 1. We can also construct a distanceone bridge surface for any link by perturbing a given bridge surface, i.e. adding a new pair of canceling bridges. This new bridge surface will no longer be minimal.
Choose any vertical arc with one endpoint at and the other endpoint at a local maximum above , as in Figure 4. Let be a band along , i.e. the image of a square such

the interior of is disjoint from ,

is vertical with respect to ,

is the image of and

the arcs and are contained in small neighborhoods of .
Let be the result of replacing the arcs and of with the arcs and . Note that even after we choose , there are infinitely many different possible bands , related by twisting around . In particular, if is a knot, but our initial choice of produces a link, we can add a half twist to ensure that is a knot. We will say that is obtained from by banding.
More generally, the banding operation described here is also known as integersloped rational tangle replacement. However, for our purposes it suffices to note that the resulting link has a bridge surface of distance at most two. To see this, let be the level sphere immediately above the highest minimum of and let be the boundary of a regular neighborhood in of . Any maximum of disjoint from gives rise to a compressing disk for contained above and disjoint from . Similarly, any minimum of that does not cobound (with ) a monotone subarc of will give rise to a compressing disk for contained below and disjoint from . Thus, has distance at most two. By choosing a “sufficiently complicated” arc , we expect that one can always construct a link of distance exactly two, but proving this is beyond the scope of the present paper.
3. Simplifying bridge disks
In this section, we introduce several results that will be needed in the proof of the main theorem. We will have several occasions to consider a tangle embedded in a cylinder and a disjoint collection of boundary compressing disks for its disk bridge surface. In the next lemma, we show that we can always assume that such boundary compressing disks for the disk bridge surface are vertical with respect to the natural height function on given by projection onto the second component. We can in fact assume so while also assuming that the tangle is polygonal and composed only of vertical and horizontal components.
Lemma 3.1.
Let be a tangle properly embedded in a cylinder such that all endpoints of are in . Suppose that is a disk bridge surface for . Let and be collections of pairwise disjoint boundary compressing disks for above and below respectively such that . Then there is an isotopy during which the tangle remains transverse to and after which all disks in and are disjoint and vertical. Furthermore, we may assume that after the isotopy, the tangle is a set of polygonal curves composed only of horizontal and vertical arcs such that all horizontal arcs are contained in or and each component of contains at most one horizontal arc.
Proof.
Let and . The proof follows the following basic structure. We begin by using standard techniques to construct an ambient isotopy so that , and subsequently have the desired properties. We then analyze the intersection between and the vertical bridge disks for to construct an ambient isotopy that results in being vertical and disjoint from while preserving the desired properties of , and .
First, perform an isotopy supported in a neighborhood of after which each arc of that is parallel into consists of two vertical arcs and one horizontal arc in and all other arcs of are fibers of . Because the boundary of each disk of consists of two vertical and two horizontal arcs, each disk is isotopic to a vertical disk. By the Isotopy Extension Theorem, these isotopies extend to a single ambient isotopy in a neighborhood of that fixes and results in being vertical. Note that this ambient isotopy preserves .
Next, perform an isotopy supported in (where is a regular neighborhood of ) after which each arc of that is parallel into consists of two vertical arcs and one horizontal arc in and every other arc of is an fiber of . This isotopy does not affect or . After these isotopies, meets the conclusion of the theorem, with the exception that it remains to be shown that we can now perform an isotopy that guarantees that is a collection of vertical disks without affecting the conditions on and that we have already imposed.
Let be a complete collection of vertical bridge disks for . Consider . After an isotopy of supported in the interior of which does not affect the tangle or , we may assume that this intersection consists only of arcs. Let be an outermost arc of intersection on some disk . This arc cuts off a subdisk whose interior is disjoint from all disks in and a subdisk from some disk in . Let be the ball cobounded by , and a subdisk of .
By standard transversality arguments, we can assume that there is so that is vertical. Let be the set of all bridges contained in . Since is disjoint from , the bridge disk of each is disjoint from so there is a vertical isotopy of relative to and supported in the interior of after which arcs in are contained in . Now the interior of is a ball that is completely disjoint from the link and thus there is an isotopy shrinking so it lies in a neighborhood of , as in Figure 5.
Let be the cylinder together with the ball after the above isotopies. The cylinder may, of course, intersect both in its boundary and in its interior. Perform a horizontal isotopy of , first shrinking it horizontally until is contained in a neighborhood of and then pushing through , thus reducing , as in Figure 5. This sequence of isotopies preserves the property that is composed of horizontal and vertical arcs and that is vertical. Note also that these ambient isotopies preserve the property that .
Thus, we may assume that . However, if the collection of disks is contained in the complement of all vertical bridge disks then they are in fact contained in a product region and thus can be isotoped to be vertical via an isotopy that is supported in , does not affect , and preserves the fact that . ∎
The following is a standard sweepout argument similar to that of [G].
Lemma 3.2.
Let be a tangle with such that and let be a bicompressible disk bridge surface for . Let be the natural projection map. Suppose is a compressing disk for in such that cannot be isotoped to be disjoint from . Then there is an isotopy of that fixes and the height and number of saddles of after which one of the following occurs:

There exists a value of such that is isotopic to and there exists a compressing disk for that is disjoint from a compressing disk or boundary compressing disk for on the opposite side of ,

there exists a regular value of such that is isotopic to and there are two disjoint boundary compressing disks for with boundary contained in and on opposite sides of , or

there exists a critical value of such that is isotopic to and there are two disjoint boundary compressing disks for on opposite sides of with the property that the boundaries of these disks meet in the boundary of a regular neighborhood of the component of that contains a saddle.
Proof.
Consider the projection map . We have for each and we will assume . Isotope all maxima of up above all critical values of into a neighborhood of and isotope all minima of down below all critical values of into a neighborhood of . This isotopy fixes and fixes the height and number of saddles of . After this isotopy, each component of the intersection between and a neighborhood of is either a vertical boundary compressing disk or a compressing disk with level boundary and a single critical point corresponding to a maximum. Similarly, meets a neighborhood of in a collection of vertical boundary compressing disks and compressing disks with level boundary and a single critical point corresponding to a minimum.
Let be the level of the highest local minimum of (with respect to ) and let be the level of the lowest local maximum. Then for every , the disk is isotopic to by an isotopy transverse to . If for any meets in a collection of arcs and loops that are all inessential in , then can be isotoped to be disjoint from . Hence, we will assume that for every meets in at least one component that is essential in .
For each fixed , an innermost loop essential in or an outermost arc essential in of in will bound a subdisk of that can be isotoped to be either a compressing or boundary compressing disk for . For near , at least one of these disks will be below , while for near , at least one of these will be above . If there is a value with loops or arcs of intersection defining disks both above and below then these disks are disjoint (since is embedded) and we have one of conditions (1) or, in the case that both disks are boundary compressing disks, conclusion (2). Otherwise, there must be a critical point of at which the disks switch from below to above. However, in this case the loops or arcs above and below the critical point correspond to disjoint arcs or loops in , so we obtain conclusion (1) or, in the case that both disks are boundary compressing disks, conclusion (3). ∎
Lemma 3.3.
Let be a tangle with such that and let be a bicompressible disk bridge surface for . Let be the natural projection map. Suppose is a compressing disk for in such that can be isotoped to be disjoint from . Assuming , there is an isotopy of that fixes and the height and number of saddles of after which one of the following occurs:

There exists a regular value of such that is isotopic to and there are two disjoint compressing disks for on opposite sides of ,

There exists a critical value of such that is isotopic to and there are two disjoint compressing disks for with boundaries contained in the boundary of a regular neighborhood of the component of that contains a saddle and on opposite sides of ,

For some , ,

There is a strict subdisk of that is bicompressible.
Proof.
Without loss of generality, we can isotope to be contained in via an isotopy supported in a regular neighborhood of . Note that there can be no arcs of intersection between and for any . Isotope all maxima of up above all critical points of into a neighborhood of and isotope all minimum of down below all critical points of into a neighborhood of . This isotopy fixes and fixes the height and number of saddles of . After this isotopy, the collection of components of the intersection between and a neighborhood of is the union of a single vertical annulus with compressing disks, each having level boundary and a single critical point which is a maximum. Similarly, meets a neighborhood of in a collection of compressing disks with level boundary and a single critical point corresponding to a minimum.
If the intersection between and a neighborhood of contains a compressing disk, then by the proof of Lemma 3.2 we can conclude (1) or (2) holds.
If the intersection between and a neighborhood of is a single vertical annulus , then meets the portion of below in a single compressing disk . If is isotopic to , then (3) holds, so we can assume that is not isotopic to . Since is bicompressible, let be a compressing disk for above and let be a compressing disk for below . Boundary compress along and boundary compress along to produce two new compressing disks for each of which is disjoint from . If the boundaries of these disks lie in distinct components of the complement of in , then conclusion (1) holds. If the boundaries of these disks lie in a common component of the complement of in , then conclusion (4) holds. ∎
The following is known as the popover lemma and has been used in a number of papers. We refer the reader who is interested in the proof to Lemma 2.3 of [BT1].
Lemma 3.4.
Suppose that a tangle has and, under the projection to only has maxima. Then there is an isotopy fixing so that has the same number of maxima as and so that there is a strand of that contains the highest maximum of , meets a collar neighborhood of in two vertical arcs and a horizontal arc and meets the complement of in in a vertical arc as in Figure 6.
4. Distance 2 links
Proof of Theorem 1.1.
Let and be the two balls bounded by in . Let be a sequence of essential simple closed curves in such that is disjoint from for and such that bounds a compressing disk in and bounds a compressing disk in . The curve separates into two disks. If and are in distinct disk components of then they are disjoint and . Thus and must be contained in a single disk with boundary . Let be the closure of the complement of in . In general, there may be many different distancetwo paths for and we will assume that , and have been chosen amongst all such triples of curves so that is minimal.
Let be the boundary of a regular neighborhood of . Isotope relative to so as to minimize without creating any new minima or maxima, as in Figure 7. Let be the closure of the complement in of . Note that and are both balls bounded by the sphere such that contains and is a tangle in . Moreover, we can naturally identify with so that is a disk bridge surface for .
Claim 0: is incompressible in .
Proof of Claim 0: Suppose that is a compressing disk for . Since both and are disjoint from then there is a compressing disk for that is disjoint from and on the opposite side of from . Hence, is distance at most one, a contradiction. ∎(Claim 0)
Claim 1: is incompressible in and is incompressible in .
Proof of Claim 1: Assume for contradiction is a compressing disk for in . Then also bounds a disk in . The union is a ball and each arc of in contains exactly one maximum since is a bridge sphere for . Hence, is a trivial tangle in . Moreover, the endpoints of each strand of are in , so we can isotope a strand of into . After pushing just past and into or out of , will meet in two fewer points, contradicting our minimality assumption. The proof that is incompressible in is essentially the same. ∎(Claim 1)
Claim 2: As a properly embedded surface in , does not have a compressing disk on one side that is disjoint from a compressing or a boundary compressing disk on the opposite side.
Proof of Claim 2: If there are two disjoint compressing disks on opposite sides of then is a bridge sphere of with distance at most one, contrary to our hypothesis.
Suppose there is a compressing disk for on one side, say in , that is disjoint from a boundary compressing disk for in , as in Figure 8. Let be the compressing disk that bounds in . By Claim 1, we may assume that this disk is disjoint from and therefore contained in . We can eliminate intersections between and by compressing and boundary compressing along . One of the resulting components will be a compressing disk for in that is disjoint from . If is on the opposite side of from , then and are disjoint so the distance of is at most one, contradicting the assumption. Thus, and are on the same side of .
Consider the curve that is contained in the boundary of a regular neighborhood of in and separates from , as in Figure 8. Let be the disk in bounded by that contains and . Then , contradicting the minimality assumption on the path . ∎(Claim 2)
Remark 4.1.
Note that if has disjoint boundary compressing disks and on opposite sides of , then we know where the critical points of the tangle occur with respect to and . If the tangle has maxima on both sides of then we can find compressing disks associated to each of these maxima that are disjoint from and occur on both sides of by taking the frontier of regular neighborhoods of bridge disks that have been chosen to be disjoint from . Since is on one side or the other of in , this implies that at least one of these compressing disks is disjoint from . Thus by Claim 2, every maximum of the tangle must be on the same side of as . Symmetrically, every minimum of the tangle must be on the same side of as .
Claim 3: If is compressible in , then is obtained from a link via banding where has a bridge sphere with the same bridge number as , but with distance at most one.
Proof of Claim 3: Suppose is a compressing disk for contained in that intersects minimally amongst all such disks. If contains a simple closed curve, then an innermost loop of intersection in bounds a compressing disk for that is disjoint from and , so the distance of would be at most one, contradicting our hypothesis. Also, by Claim 1, . Thus, we can conclude that is a nonempty collection consisting entirely of arcs.
Let be any compressing disk for or for that decomposes or into two 3balls and such that both and have nontrivial intersection with . In the current situation, we could let be a disk cut from by an outermost arc of . However, we will need the full generality of the definition of in Claim 5.
Choose a collection of bridge disks for the arcs of above which is disjoint from . Use these bridge disks to lower all bridges of above below any critical points of . Now we can isotope to have a single maximum by an isotopy that does not affect the link. We can then isotope to its original position without introducing any new critical points in . Therefore we may assume that after an isotopy has a single maximum in its interior.
Let and be two copies of with the point at infinity between them, as in the top row of Figure 9. After thickening and an isotopy, and are the balls indicated in the top left of the Figure bounded by and , respectively, and disjoint disks in . Then is a tangle that only has maxima so we can apply Lemma 3.4 to pull one strand with at least one endpoint in out of the braid in each of and as indicated in the upper right picture of Figure 9. Our definition of guarantees that such strands exist. These two strands then form a band as indicated in the last picture. The link produced by undoing this band, the inverse of the process depicted in Figure 4, continues to have bridge surface . However, now has an obvious pair of disjoint compressing disks on opposite sides. Therefore has a bridge surface with the same bridge number as , but distance at most one. (Note that this bridge surface may be the result of perturbing a bridge surface with strictly lower bridge number.) ∎(Claim 3)
Both Claim 4 and Claim 5 derive consequences from the assumption that is compressible in . It is necessary to use the conclusions of these claims in tandem to prove the theorem.
Claim 4: If is compressible in , then is a bridgesplit Montesinos tangle or is obtained from a link via banding where has a bridge sphere with the same bridge number as , but with distance at most one.
Proof of Claim 4: Suppose is a compressing disk for in . If can be isotoped to be disjoint from , then we can apply Lemma 3.3 and we arrive at a contradiction to Claim 2 if conclusion (1) or (2) holds. If (3) of Lemma 3.3 holds, then we contradict Claim 0. If (4) of Lemma 3.3 holds, then we contradict how we chose the curves by finding a smaller bicompressible subdisk of .
Thus, we can assume that cannot be isotoped to be disjoint from . Apply Lemma 3.2 to conclude that there are two cases to consider. If conclusion (1) of Lemma 3.2 holds, then we contradict Claim 2. Thus, we may assume conclusion (2) or (3) holds. Hence, has a pair of disjoint boundary compressing disks on opposite sides. Let and be these disks such that lies above and lies below . By Lemma 3.1, we may assume that these disks are vertical and by Remark 4.1, every compressing disk disjoint from must be contained in the component of that contains and vice versa. Let be the result of boundary compressing along . By the above argument, some component of must separate all the minima from all the maxima of the tangle.
Assume is boundary parallel in . This boundary parallelism induces a product structure of the form on at least one of the two components of . Since is bicompressible, some strand of contains a maximum and some strand contains a minimum. Without loss of generality, the component of that contains maxima, inherits the product structure. Let be a strand of that contains a maximum. The product structure of allows for a level preserving isotopy of that fixes away from and results in being pulled out of and into , as in leftmost isotopy of Figure 10. Apply Lemma 3.4 to pull one strand with at least one endpoint in out of , as in the middle isotopy in Figure 10. This strand and cobound a band, as illustrated in the right most isotopy in Figure 10. Hence, is obtained from a link via banding where has a bridge sphere with the same bridge number as , but with distance at most one. Since we arrive at the same conclusion independent of which component of illustrates the boundary parallelism, we can assume that is not boundary parallel in .
We can think of as a subdisk of together with vertical flaps corresponding to and , as in Figure 11. Hence, after a small tilt of the portion of in , the foliation of induced by the restriction of the projection of onto its factor is a collection of parallel arcs. To demonstrate that is a bridgesplit Montesinos tangle, it only remains to show that is incompressible in .
Suppose is a compressing disk for . After an isotopy, becomes a compressing disk for with boundary disjoint from and . Without loss of generality, assume is above . Since is bicompressible in by construction, there is a compressing disk for contained below . By compressing and boundary compressing this compressing disk along subdisks of , we obtain a compressing disk for for contained below , with boundary disjoint from . must be contained in the same component of as , otherwise we contradict Claim 2. Since the boundary of and both are contained in a subdisk of that meets in strictly fewer points, we arrive at a contradiction to how we chose . Thus, is incompressible.
∎(Claim 4)
Claim 5: If is compressible in , then at least one of the following holds:

the exterior of contains an incompressible meridional planar surface,

is rational,

is obtained from a link via banding where has a bridge sphere with the same bridge number as , but with distance at most one,

the exterior of in contains an incompressible, boundary incompressible annulus.
The proof of the following claim is inspired by the work of Rubinstein and Thompson [RT] on the classification of distancetwo Heegaard splittings and the proof of Lemma 3.2 in [CR].
Proof of Claim 5: By Claim 3, we can assume that is incompressible in . Let be a maximal collection of compressing disks for in . Let be the regular neighborhood of . If any of the boundary components of are 2spheres bounding balls in disjoint from or are boundary parallel twice punctured 2spheres, then fill these components with the corresponding 3balls or 3balls containing unknotted arcs to form the 3manifold . Let . If , then is rational and the claim holds. Hence, we can assume that . Since was maximal, is incompressible to the side not containing . is a punctured 3ball that meets in a collection of arcs isotopic into or connecting to . Thus, is incompressible in .
Suppose that is compressible in the exterior of with compressing disk . Isotope to be transverse to so that consists of loops and 3valent graphs in . Assume that we have isotoped to meet in a minimal number of edges, where we think of each loop component as consisting of a single vertex and a single edge. Since is incompressible in , . A component of in is innermost in if all but one of the boundary components of a closed regular neighborhood of in bound disks in that are disjoint from . Let be an innermost component of in .
Case 1: Suppose that is a loop. Since is innermost it bounds a disk in which is disjoint from . If is contained in , then is disjoint from since it is a loop and not a 3valent graph. If is contained in , then we obtain a contradiction to Claim 0. Thus, we can assume that . If is contained in , then is a compressing disk for in or in , a contradiction to Claim 1. Hence, is contained in . Given the natural identification of with and the fact that is disjoint from , then we can assume without loss of generality that has boundary in . Apply Lemma 3.3 to and analyze each possible conclusion. If conclusions (1) or (2) of Lemma 3.3 hold, then we contradict the fact that is distance two. If conclusion (3) of Lemma 3.3 holds, then we contradict Claim 0. If conclusion (4) of 3.3 holds, we contradict how we chose . Thus, can not be a loop.
Case 2: Suppose that is a 3valent graph. Since is embedded in the interior of the disk , the formula for Euler characteristic gives where is the number of vertices of , is the number of edges of and counts the number of components of the complement of in . Since is 3valent, . Thus, . From this equality follows the inequality
Since every edge of borders exactly two faces and since the closure of each component of has boundary a cycle in consisting of an even number of edges, then the above inequality implies the existence of a disk component of with closure such that consists of exactly two edges and two vertices of ( is a bigon), or consists of exactly four edges and four vertices of ( is a square).
Suppose is a bigon properly embedded in . Since is a bigon, meets in exactly one arc that is essential in . Identify with and isotope to consist of two vertical arcs in and two horizontal arcs in . Hence meets every level surface in exactly one arc. Apply Lemma 3.2 and the proof of Claim 4 to the compressing disk to conclude that has a pair of disjoint boundary compressing disks on opposite sides, and such that the arc is isotopic to the arc .
Since is bicompressible in , we can boundary compress along to find a compressing disk for above which is disjoint from . Similarly, we can find a compressing disk for below which is disjoint from . If these compressing disks have boundary on the same side of in , then we contradict how we chose the curves . If these compressing disks have boundary on opposite sides of in , then has distance at most one, a contradiction. Hence, is not a bigon properly embedded in .
Suppose is a bigon properly embedded in or . Then satisfies the definition of in the proof of Claim 3 and we conclude that is obtained from a link via banding where has a bridge sphere with the same bridge number as , but with distance at most one.
Suppose is a square properly embedded in . Identify with and, since is a square, we can isotope to consist of four vertical arcs in , two horizontal arcs in and two horizontal arcs in . Recall that is the natural height function on . At any regular value of , consists of exactly two arcs and some number of loops. Since is connected, there must be some saddle singularity of at which two arcs are banded together.
Since is simply connected, there is at most one saddle where two arcs are banded together to form two new arcs. Let be the height of this singularity, then the graph consists of a single “X” and possibly a collection of loops. If any of the loops of intersection of are inessential in , then we can eliminate them via an isotopy of that leaves the saddle singularity fixed. As in the proof of Claim 4, we can apply Lemma 3.2 to find an isotopy of after which conclusion (2) or (3) of Lemma 3.2 holds. Let be the value for which has a pair of disjoint boundary compressing disks on opposite sides, and .
Suppose conclusion (2) of Lemma 3.2 holds. If , then the arc separates into two disks. Let be the possibly immersed disk that is the union of and the portion of that is cut off by and lies above when near , see Figure 12. Since , consists of two vertical arcs in , one horizontal arc in and one horizontal arc in . Since the selfintersections of correspond to the transverse intersections of embedded disks, then after surgering along any loops of self intersection, we can assume is an embedded compressing disk. In this case is a bigon in and we can apply the above argument to derive a contradiction. Similarly, we derive a contradiction if . Hence . However, implies conclusion (3) rather that conclusion (2) of Lemma 3.2.
Suppose conclusion (3) of Lemma 3.2 holds. If let for small. We arrive at a contradiction when we consider the immersed disk that is the union of below and the portion of that is cut off by and lies above when near exactly as before. Similarly, we derive a contradiction if . If , then there are no loops in as we have already isotoped to eliminate loops that are inessential in and if any such loops are essential in we derive a contradiction as in Claim 4. Hence, consists of a single “X.”
Identify with and let be the four properly embedded arcs in that lie in the boundary of a regular neighborhood of in . By the above argument, each of these arcs bound a boundary compressing disk for . In particular, and bound boundary compressing disks above and and bound boundary compressing disks below . By Remark 4.1, all maxima of must be contained in the 3ball between the upper boundary compressing disks in and all minima of must be contained in the 3ball between the lower boundary compressing disks in . Hence decomposes into two rational tangles glued together along an unpunctured disk. See Figure 13. Thus, is a rational tangle.
Suppose is a square properly embedded in or . We consider the case that is a square properly embedded in and the case when is a square properly embedded in will follow by a symmetric argument. Recall that is the closure of the complement of in . Since is a square, it meets in exactly two essential arcs. If these arcs are not isotopic in then meets the definition of in the proof of Claim 3 and we conclude that is obtained from a link via banding where has a bridge sphere with the same bridge number as , but with distance at most one. Hence, we can assume that the two arcs in together with two arcs in bound a disk in which is disjoint from . After a small isotopy pushing into the interior of , is an annulus properly embedded in with boundary in . If is compressible in the exterior of in , then, after compressing , each component of bounds a compressing disk for in , a contradiction to the incompressibility of in . Hence, is incompressible. is not boundary parallel since partitions into three subsurfaces, each of which has nontrivial intersection with . Lastly, is boundary incompressible since boundary compressing would result in a compressing disk for in which was ruled out as a possibility at the beginning of the proof of this claim.
∎(Claim 5)
To conclude, we note that if is incompressible, then conclusion (1) holds; if is compressible into , then, by Claim 4, conclusion (3) holds; and if is compressible into but incompressible into , then, by Claim 3 and Claim 5, at least one of conclusions (1), (2) or (3) holds. ∎
Proof of Corollary 1.2.
Let be a 3bridge knot with a 6punctured bridge sphere of distance two. Let be the boundary of a regular neighborhood of as defined at the beginning of the proof of Theorem 1.1. Recall that after isotopying relative to so as to minimize , is at most since is compressible in and in . Since is essential in , 2, 3 or 4.
If , then and is distance zero, a contradiction.
If , then