An Infinite Family of Links
with Critical Bridge Spheres
Abstract
A closed, orientable, splitting surface in an oriented 3manifold is a topologically minimal surface of index if its associated disk complex is connected but not connected. A critical surface is a topologically minimal surface of index 2. In this paper, we use an equivalent combinatorial definition of critical surfaces to construct the first known critical bridge spheres for nontrivial links.
1 Introduction
In the 1960s Haken developed a framework for studying manifolds that contain incompressible surfaces. He demonstrated that for such manifolds, one can reduce the proofs of many theorems to induction arguments by using “hierarchies,” consecutively cutting a manifold into pieces along incompressible surfaces until a collection of balls is obtained. This powerful approach clearly demonstrated the utility of incompressible surfaces in the study of lowdimensional topology.
In 1987 Casson and Gordon introduced the idea of strongly irreducible surfaces [4]. Unlike incompressible surfaces which have no compressing disks, strongly irreducible surfaces have potentially many compressing disks to both sides, but any two compressing disks on opposite sides necessarily intersect. Somewhat surprisingly, it turns out that many theorems that are easy to prove for manifolds with incompressible surfaces also hold true for manifolds with strongly irreducible surfaces, although the proofs can be somewhat more involved. Casson and Gordon’s watershed result in the theory of strongly irreducible surfaces is that if a closed, connected, orientable 3manifold’s minimal genus Heegaard splitting is not strongly irreducible, then the manifold must contain an essential surface [4].
In 2002, Bachman introduced the notion of a critical surface [1], which he followed in 2009 with the introduction of the more general concept of a topologically minimal surface [2]. A closed, orientable, splitting surface in an oriented 3manifold is a topologically minimal surface of index if its associated disk complex is connected but not connected. This latter definition provides the framework into which incompressible, strongly irreducible, and critical surfaces all fit: They are topologically minimal surfaces of index 0, 1, and 2 respectively. Just like the incompressible and strongly irreducible surfaces before them, critical surfaces and topologically minimal surfaces in general have been used to prove longstanding conjectures that had remained unresolved for many years. For example, Bachman used these surfaces to prove The Gordon Conjecture^{1}^{1}1Gordon Conjecture: If the Heegaard splitting is a connect sum of two Heegaard splittings, and is stabilized, then one of its summands is stabilized. and to provide counterexamples to the Stabilization Conjecture^{2}^{2}2Stabilization Conjecture: Given any pair of Heegaard splittings, stabilizing the higher genus splitting once results in a stabilization of the other splitting. [3].
Most of the work done thus far regarding topologically minimal surfaces deals specifically with surfaces which are Heegaard splittings, twosided surfaces that split a manifold into two compression bodies. Less is known about topologically minimal bridge surfaces, which are a natural type of surface to consider when the 3manifold is a link complement. Lee has shown in [8] that all bridge spheres for the unknot with any number of bridges in are topologically minimal, and his results provide upper bounds for the indices of such bridge spheres. In particular, he concludes that the bridge sphere for an unknot in a 3bridge position has index exactly 2 (i.e., it is critical).
In this paper we provide the first known examples of critical bridge spheres for nontrivial links. Our construction is inspired by recent work of Johnson and Moriah [6] in which they construct links with bridge surfaces of arbitrarily high distance. The central result of this paper is the following:
Theorem 6.9. There is an infinite family of nontrivial links with critical bridge spheres.
In Section 2 we go over some of the foundational topological definitions upon which this paper relies, including an equivalent, combinatorial definition of a critical surface. Then in Section 2 we describe what it means for a link to be in a plat position. Following that, in Section 3 we embed a link in in a plat position with bridge sphere and discuss some of the specific details of the embedding as well as build some of the tools (certain arcs, loops, disks, and projection maps) which we will use throughout the rest of the paper. Sections 2 and 3 should be considered setup for the rest of the paper, and this is essentially the same setup as in Johnson and Moriah’s paper [6]. In particular we make use of Johnson and Moriah’s plat links and projection maps. In Section 4 we develop a link diagramatic way to visualize boundary loops of compressing disks for . Theorem 6.9 is proved in Sections 5 and 6.
2 Definitions
Suppose is a compact, orientable surface embedded in a compact, orientable 3manifold , and is a disk embedded in (not necessarily properly). is a compressing disk for if , and neither bounds a disk in nor is parallel to a boundary component of .
A compact, orientable surface embedded in a connected 3manifold is a splitting surface if has two components. We associate a simplicial complex called the disk complex to in the following way: Vertices of are isotopy classes of compressing disks for . A set of vertices will be filled in with an simplex if and only if the corresponding isotopy classes of compressing disks have pairwise disjoint representatives. is called a topologically minimal surface of index if is empty, or if it is connected but not connected. is called a critical surface if it is a topologically minimal surface of index 2. In [2], Bachman gives an alternative, combinatorial definition for critical surfaces and proves the two definitions are equivalent. This second definition, given below, is the one we will utilize in this paper.
A critical surface is a splitting surface whose isotopy classes of compressing disks can be partitioned into in a way that satisfies the following two conditions:

Whenever and for and on opposite sides of , must be nonempty.

For , there exists a pair of disjoint compressing disks, one on either side of , where each disk belongs to an isotopy class in .
Any link can be isotoped so that all of its maxima lie above all of its minima (with respect to the standard height function on ). After such an isotopy, is said to be in bridge position. Let be an open regular neighborhood of , and let be a level sphere in which separates all of the maxima of from all of the minima. Then the surface (a sphere with a finite number of open disks removed) is called a bridge sphere for . See Figure 2.
In our context, studying the bridge sphere is equivalent to studying the “sphere with marked points”, where the marked points are the points of . A disk in is considered a compressing disk for the sphere with marked points if and only if it is a compressing disk for in . This implies that a compressing disk for cannot intersect , and two compressing disks for are considered to be in the same isotopy class if and only if they are isotopic in .
A bridge arc is a component of , and each of these bridge arcs has exactly one critical point, so each is parallel into . An isotopy of a bridge arc into sweeps out a disk, called a bridge disk. (Note, bridge disks are not unique, even up to isotopy.)
A link is perturbed if there is a bridge disk above and another bridge disk below which intersect in a point of , and which are otherwise disjoint. We will say the link is perturbed at a bridge arc if corresponds to either of these two bridge disks. If a link is perturbed, then there is an isotopy of through the bridge disks which reduces the number of critical points of by one maximum and one minimum.
A cap for is a compressing disk such that bounds a disk in that contains exactly two marked points. The existence of caps is guaranteed by the existence of bridge disks. See Figure 3.
We will also find useful the notions of boundarycompressing along boundarycompressing disks. Suppose is a surface with boundary properly embedded in a 3manifold . A boundarycompressing disk (See Figure 4 for an example) is defined to be a disk with the following properties:

is an arc .

is an arc .^{3}^{3}3Note: Our definition here is looser than the standard definition since we don’t require to be an essential arc in . We drop this condition because we will want to perform boundarycompressions on disks, which have no essential arcs.

.

.
Next we define a half twist. Let be a twice punctured (marked) disk. The half twist can be simply understood to be the homeomorphism depicted in Figure 5. To be technical, the half twist of is the homeomorphism defined as follows: View as a union of the closed unit disk with an outer annulus whose intersection is an embedded circle containing the two marked points. See Figure 5. Give the coordinates , and identify with , with the coordinates . ( is glued to by the map taking in one component of to in .) Now define
For any punctured (marked) surface with more than one puncture (marked point), if is a loop which bounds an embedded twicepunctured (marked) disk we can identify with and define the half twist about to be the homeomorphism
Now we will define plat position for a link. We start by describing plat position for a braid. Fix integers and . We begin by constructing a set of vertical cylinders in which will provide a frame for our braid. (See Figure 6.) In the plane, define to be the circles of radius such that has center . Now define to be the cylinder obtained by crossing with the interval .
Next we define the twist regions, subcylinders of depicted in Figure 6. Let . For odd , let range from to , for even , let range from to . Then define
In this paper, superscripts will usually denote an object’s horizontal position (i.e., in the direction), and subscripts will usually denote an object’s vertical position (i.e., in the direction). This is reminiscent of typical matrix notation: an entry of a matrix is in the position vertically and the position horizontally. Here, however, unlike in a matrix, an array of objects is enumerated from bottom to top instead of from top to bottom.
Now we define a (nonclosed) braid to be in plat position if it satisfies the following conditions:

is a strand braid embedded in whose bottom endpoints’ coordinates are for and whose top endpoints’ coordinates are for .

Arcs of are partitioned into two types: Vertical arcs which lie in the intersection of with the plane, and Twisting arcs which are not contained in the plane but are properly embedded in the twist regions.

Every Twisting arc is strictly increasing in a twist region, and as it ascends, it moves either strictly clockwise around the cylinder or strictly counterclockwise.
These conditions guarantee that each twist region contains exactly two arcs of . Observe that the endpoints of every Twisting arc must lie in the plane. Since Twisting arcs move strictly clockwise or counterclockwise, they intersect the plane minimally. That is, there is no isotopy of a Twisting arc in a twist region, relative its endpoints, which would decrease the intersection of that Twisting arc and the plane. Looking down at the link from above, if the Twisting arc moves counterclockwise as it ascends, we define to be plus the number of times the Twisting arc intersects the plane. Similarly, if the Twisting arc moves clockwise as it ascends, we define to be plus the number of times the Twisting arc intersects the plane. In other words, to each twisting arc we assign an integer such that the twisting arc travels around the twist region through an angle of . Notice that the other twisting arc in the same twist region must twist through an equal angle (or else the two arcs would intersect). Thus we can define the number to be the twist number corresponding to that twist region. We will use to denote the twist number for .
There is a standard way to create a link from a braid in plat position. See Figure 7. Along the bottom of , we have a row of endpoints at a height of . For each odd between and , connect the endpoints and with an lower semicircle in the plane. The way we connect the top endpoints of depends on the parity of . Notice that if is even, then the highest row of twist regions consists of twist regions. Along the top of we have a row of endpoints at a height of . For each odd between and , connect the endpoints and with an upper semicircle in the plane. If is odd, there are twist regions at the top. For each even between and , connect the endpoints and with an upper semicircle in the plane. Then connect the points and with a larger upper semicircle in the plane. See Figure 7. Thus we obtain a link from .
Any link constructed in this way from a braid in plat position will be said to be a link in plat position. Observe that such a link is in a bridge position with bridge number . For , a link in plat position is called twisted if for all . See Figure 8 for an example of a twisted link in plat position.
3 Setting
Let be a height function on , and let be a strictly increasing arc in with endpoints being the two critical points of . Then is homeomorphic to . Let be a homeomorphism that respects the height function on and the standard height function on . That is, if is the level plane in of height , and is the level sphere in of height , then is minus the point . Throughout the rest of the paper, we adopt the point of view of someone standing on the side of the plane. This gives meaning to words like “left,” “right”, “up,” “down,” “horizontal,” and “vertical.” (See Figure 9.)
The Link : Let be a twisted link in plat position with bridge sphere . Suppose has twist numbers such that for all , the twist numbers are positive, and are negative. We will work with this link for the rest of the paper.
Consider the diagram for obtained by projection to the plane. Our choice of signs for the twist numbers makes an alternating diagram, and we know that is a split link if and only if is a split diagram (see Theorem 4.2 of [9]). Since is clearly a nonsplit diagram, it follows that is a nonsplit link.
The Bridge Disks: Refer to Figure 10. Above in the plane lie the upper bridge arcs. We will name these arcs consecutively from left to right, . Vertical projection of the bridge arcs into gives us straight line segments at level which we will name consecutively from left to right, , , , and . We will refer to these as arcs. Observe that and form a loop. Let be the disk in the plane bounded by this loop. Notice each is a bridge disk. Define to be the straight line segment connecting the points and . We will refer to these arcs as arcs.
Below in the plane lie the lower bridge arcs, which we will call consecutively from left to right, , , , and . We will also define to be the straight line segment between the endpoints of . (See Figure 11.) is a loop which bounds a bridge disk in the plane which we will call .
loops and disks: For , define to be the circle of radius in the horizontal plane centered at the point . For , define to be the circle of radius in the horizontal plane centered at the point . Then each bounds a twicepunctured disk which we call . Each disk lies below a corresponding twist region . See Figure 8.
projection: It will be convenient to be able to talk about simple vertical projections. If denotes the horizontal plane in of height , let be the vertical projection map defined by .
projection: We will also need a more subtle type of projection which respects . We closely follow Johnson and Moriah [6]: Note that intersects each in the same number of points and these points vary continuously as varies from to . This can be thought of as an isotopy of these marked points in which extends to an ambient isotopy of . To be precise, there is a projection map for each that sends each arc component of the plat braid to a point for some and defines a homeomorphism for each . These homeomorphisms are canonical up to isotopy fixing the points , and the induced homeomorphism is the identity. Further, for each , the homeomorphism induced by is a composition of half twists about the loops, which can be expressed as . (Note that since the disks at any given level are pairwise disjoint, the corresponding half twists all commute with each other.) Therefore the homeomorphism induced by can be expressed as .
Disk partition: Our goal is to show that is critical, so we need to exhibit a partition of the isotopy classes of the compressing disks for . (Recall the definition of critical from the beginning of Section 2.) Above , let be the frontier of a regular neighborhood of . Below , let be the frontier of a regular neighborhood of . and are depicted in Figure 12. Let , and let all of the other isotopy classes of compressing disks be in . It will be convenient to refer to a compressing disk as red if and blue if .
4 The Labyrinth
We are interested in the image of under , the homeomorphism induced by . As takes up from level 1 to level 4, undergoes a series of twists, following the strands of the link. As explained in Section 3, can be expressed as a product of half twists around the loops. Explicitly, . In Figure 13 we show exactly what looks like under in a case where for all .
Of course, we want to understand what will look like in general for any set of twist numbers . To do so, we will devise a convenient pictorial notation for loops if which undergo half twists about loops. Suppose is a loop in with one arc of intersection with which passes between the two marked points, as in the first picture in Figure 14. The second and third pictures in Figure 14 are homotopic pictures of . Observe that the third picture is reminiscent of knot diagram smoothing. In fact, using the smoothing rule in Figure 15, we can “unsmooth” into the fourth picture, which consists of a link diagram on with two components: 1) the original and 2) . This suggests a convenient algorithm to draw any such loop after a (positive) half twist about an loop: We start by drawing . Next we draw the loop so that it passes under . Last we smooth the crossings using the rule in Figure 15. If we want to perform a negative half twist instead, the procedure is the same, except that we draw the loop crossing over .
We can generalize this process to the situation where there are strands of passing through . In Figure 16, the top row is an example in which , and the bottom row depicts the general case. As in Figure 14, the second and third pictures of each row in Figure 16 are homotopic pictures of , and we can “unsmooth” into the fourth picture, which consists of a link diagram on . This time, the link diagram contains components: and parallel copies of . Thus our generalized procedure for drawing a half twist of about is the following: First determine the number of times that passes between the two punctures of , then we draw parallel, disjoint copies of so that at each crossing, passes under for a positive half twist, or over for a negative half twist, and then we smooth the crossings using the rule from Figure 15.
Note that to perform consecutive half twists around , we simply perform this process times, taking care that every time we add new parallel copies of , they are nested inside the ones previously drawn. Thus if is halftwisted around a total of times, and passes between the punctures of a total of times, then to build the link diagram which depicts , we will add a total of parallel copies of (all passing under if or over if ). Thus after any sequence of half twists about loops, we obtain a link diagram representing . Instead of drawing all of the parallel copies of each loop, we will draw only one of each and label it with the number of strands it represents.
In the resulting link diagram, we will have crossings involving two strands which are labeled with different numbers. Figure 17 depicts an example in which we show how to recover what actually looks like near such a crossing, where a strand marked with a 7 crosses a strand marked with a 3. Figure 18 shows the general case.
Our purpose in developing this link diagramatic representation of a twisted loop is to understand what looks like after performing (in order) half twists about all of the loops for any set of twist numbers . With this method in hand, this is a straightforward task. Refer to Figure 19 as we describe this process. We start with the simple circle in at Level 1, and then push it up to Level 2 (i.e., we apply ). had to pass through the twist region , so this means it underwent half twists about . Before twisting (i.e., at Level 1), passed one time (i.e., ) between the punctures of ; therefore what looks like at Level 2 is a link diagram consisting of the original circle , plus parallel copies of . Since we specified in Section 3 that is positive for all , each copy of will be drawn to pass under . Now the original and the copies of together with the crossing information make a link diagram which represents the image of under . Since the value of is arbitrary, we cannot smooth the crossings to see exactly what looks like. But that is not a problem; we can still push this loop up from Level 2 to Level 3. As we do, we add copies of and to the link diagram. This time, since for all , these loops will all be drawn with overcrossings, which gives us a link diagram representing . Finally we push this up to Level 4, adding copies of and (with undercrossings) to the link diagram. In this way we see that can be represented by the diagram of the fourcomponent unlink in Figure 20.
Each circle in the diagram in Figure 20 is marked with a number indicating how many parallel copies of that circle are present, where , , , and . Recall that for all , , and . No matter what values the twist numbers take on, provided they follow this rule, it can be shown that for each , and these three inequalities hold true: , , and . This means that at each of the six crossings, the strand labeled with the higher number passes under the strand labeled with the lower number, so all six crossings look like the top picture in Figure 18.
Proposition 4.1.
The position of described by the link diagram in Figure 20 is minimal with respect to .
Proof.
The Bigon Criterion in [5] tells us that as long as cobounds no bigons with , then they are in minimal position. Consider as being cut into component strands by . To bound a bigon, one of these strands would have to have both endpoints on a single arc. However, each strand has endpoints on distinct arcs. ∎
Let Lab be the disk in which contains and whose complement is a regular neighborhood of . (Lab is depicted in Figures 13 and 20, and also, Lab is isotopic in to the top grey rectangle in Figure 12.) We will refer to Lab as the Labyrinth.
Define , and to be the dotted arcs depicted in Figures 13 and 20. We will call these the brown, orange, and purple gates of the labyrinth, respectively. Each gate is an arc in with endpoints on .
Proposition 4.2.
and the three gates cut into five components: three oncepunctured disks, an annulus, and a twicepunctured disk.
Proof.
Since each crossing in our link diagram in Figure 20 looks like the top picture of Figure 18, it is apparent that there is an annulus component, which we will call . has as one boundary component, and the other component of alternates between the three gates and three subarcs of . Observe that is not punctured. cuts off the twicepunctured disk from by definition, and clearly none of the gates are in this disk. is a thricepunctured annulus in which the three gates are properly embedded but not nested; If any subset of the gates were nested, then at least one of them would not be a boundary component of . None of the gates can be parallel to because that would either force to be punctured or force the other gates to be nested. Thus the only possible configuration is (isotopic to) Figure 21. ∎
We will refer to the three oncepunctured disks in Proposition 4.2 as the brown, orange, and purple punctured disks, according to the color of the corresponding gate, and we will call the marked point contained therein a brown, orange, or purple marked point. In the brown disk, there is a unique arc (up to isotopy) connecting the brown point to , which we will call the brown escape route. We define the orange and purple escape routes similarly. (The three escape routes are depicted in the second picture of Figure 32.) We can think of the colored points as escaping from a maze whose walls are (), and the escape routes are the paths they take to get to the exits (the gates).
5 Red disks enter the labyrinth
Refer to Figure 10.
Lemma 5.1.
If is a red disk above , then must intersect or .
Proof.
cuts into two 3balls: above and below . divides into two 3balls which we will call and and refer to as the two sides of .
Case 1: , and are not all on the same side of . Then for or , and are on opposite sides of . Since connects an endpoint of and an endpoint of , must intersect .
Case 2: All of the bridge arcs , and are on the same side of . Without loss of generality, say they are in . See Figure 23. Notice that cannot also be in because that would imply is nullhomotopic in , contradicting the fact that is a compressing disk. Therefore, in Case 2, separates from the other bridge arcs, which implies is a cap for . Let , and isotope to minimize .
Subcase 2.1: is disjoint from .
Assume (for contradiction) that is also disjoint from and . Then we have a single straight arc disjoint from because . So cuts into two disks, one of which contains , and the other of which contains . But of course, there is only one such loop in , which is . Thus , so the red disk is blue, a contradiction. We conclude that in this subcase, must intersect or .
Subcase 2.2: is not disjoint from .
For some , is nonempty. cannot contain loop intersections because is minimal, and is irreducible. cannot contain an arc with either endpoint on because that would imply that intersects . Thus the components of intersection in must all be arcs with both endpoints on . See Figure 24.
Take an arc of intersection outermost on and call it . Let be the subarc of that shares endpoints with . Then let be the subdisk of cobounded by and . See Figures 24 and 25. for either or . Whichever is the case, cuts into two balls which we call the sides of . Recall that we assume , so therefore . cannot be in with . If it were, then since cannot intersect or , would have to be completely contained on one side or the other of . Then the other side of would be an empty 3ball through which we can isotope , removing at least one component of intersection with , which is a contradiction since is already minimal. Thus is in with , and .
If , and were on the same side of , that would mean that the other side of was an empty 3ball through which we could isotope , removing the arc of intersection , contradicting minimality. Thus it must be the case that there are some arcs in either side of .
Then for or , and are on opposite sides of . An endpoint of and an endpoint of are connected by the arc in . None of the arcs have interiors that intersect , so must pass through in order to connect to . Thus we have proved that in Subcase 2.2, .
Now we have proved both subcases, concluding Case 2 and finishing the proof of Lemma 5.1. ∎
Corollary 5.2.
Every red disk intersects at least one of , , , and .
Proof.
Recall that by Proposition 4.2, has five components: three (colored) oncepunctured disks, one twicepunctured disk (), and an annulus. See Figure 21. Let be a red compressing disk above . The boundary of a compressing disk cannot bound a oncepunctured disk in . Thus cannot be contained in one of the three colored oncepunctured disks. If lies in the twicepunctured disk, then must be isotopic to . But then after the isotopy, would be a splitting sphere for , a contradiction since is nonsplit. We see in Figures 13 and 20 that and are disjoint from the annulus component of , so if is contained in the annulus, would fail to intersect or , contradicting Lemma 5.1. We conclude cannot lie in a single component of ; therefore it must intersect . ∎
6 All Red Disks Above Intersect All Blue Disks Below
Recall the disks and arcs defined in Section 3 and pictured in Figures 10 and 11. Let as in Section 5. Let be the set of red compressing disks above which are disjoint from . Throughout Section 6, whenever is nonempty, assume that is a disk in such that for all , and assume is in minimal position with respect to and with respect to the gates. cuts into multiple subarcs, which we call lanes.
Lemma 6.1.
Assume is nonempty. If two points of lie in the same lane, then they cannot be the endpoints of a common arc of .
Proof.
Suppose there exists an arc of whose endpoints lie in the same lane. We may assume the arc is outermost on . It cuts off a small subdisk which doesn’t intersect . Boundarycompress along . The result is two disks, whose boundaries lie in , and which are disjoint from . Suppose is trivial. Then it represents an isotopy through which we can move to decrease its intersection with the bridge disks. But that implies was not in minimal position with respect to the bridge disks, a contradiction. Thus is not trivial, and similarly, neither is .
This means both and are compressing disks. By construction, neither of them intersects , and for both , . Thus by our choice of as a minimal representative from , . Then by the definition of , both must be blue. There is only one blue disk above , which is , so and must be parallel copies of . This implies is a band sum of parallel disks, but any band sum of parallel disks is trivial, so is a trivial disk, which is a contradiction. ∎
By definition, implies , so can also be made disjoint from since is the boundary of a regular neighborhood of . It follows from Corollary 5.2, that must intersect at least one gate. Figure 21 makes clear something not at all obvious in Figures 13 and 20, which is that if an arc of “enters” the labyrinth through a particular gate, it must subsequently “exit” the labyrinth from the same gate. We will call the components of the intersection of and the brown disk brown tracks, and we will define orange tracks and purple tracks similarly. Observe that these tracks are pairwise disjoint, and each track has endpoints on the gate of the corresponding color.