Rectangle condition and its applications
Abstract.
In this paper, we define the rectangle condition on the bridge sphere for a bridge decomposition of a knot whose definition is analogous to the definition of the rectangle condition for Heegaard splittings of manifolds. We show that the satisfaction of the rectangle condition for a bridge decomposition can guarantee that the Hempel distance for the bridge decomposition is greater than or equal to . In particular, we give an interesting family of alternating 3bridge knots by using the rectangle condition and a modified train track argument.
1. Introduction
Casson and Gordon [2] introduced the rectangle condition on Heegaard surfaces to show strong irreducibility of Heegaard splittings of manifolds. A Heegaard splitting is strongly irreducible if for any pair of essential disks and , meets in . In other words, the Hempel distance of the Heegaard splitting is greater than or equal to two.
We consider the Heegaard surface as the branched double covering of a punctured sphere, denoted by . Then the natural question is if there is a similar criterion for bridge decompositions of a knot to check whether or not the Hempel distance is greater than or equal to two. Currently, K. Takao [11] defined the wellmixed condition for a bridge decomposition of a knot which is a variation of the rectangle condition and the satisfaction of the condition guarantees the Hempel distance to be at least two. In this paper, we define the rectangle condition on the bridge sphere for a bridge decomposition of a knot based on the classical definition and show that the Hempel distance is greater than or equal to two if two pants decompositions related to the bridge decomposition satisfy the rectangle condition. To check the satisfaction of the wellmixed condition, two pants decompositions obtained from the definition need adjacent subarcs in a conditional diagram on . The adjacent subarcs come from different simple arcs connecting two punctures in . The adjacent subarcs are possibly not parallel sides of a rectangle. Actually, the wellmixed condition needs more conditions to be checked than the conditions for the rectangle condition we define. For example, the wellmixed condition needs 18 conditions to be checked for a bridge decomposition of a knot but the rectangle condition needs 9 conditions(rectangles) for one. So, the obvious advantage of the rectangle condition would be less conditions to be checked. However, the parallel sides of each rectangle for the rectangle condition are usually longer than the adjacent arcs for wellmixed condition in . In other words, there are more obstacles to have rectangles than adjacent arcs. So, it is hard to say that the rectangle condition is a better criterion than the wellmixed condition. Moreover, we use the same technique to prove the main result, Proposition 3.2. In order to give another effectiveness of the rectangle condition I would give an algorithm to check whether or not two pants decompositions on the bridge sphere for a given bridge decomposition of a knot satisfy the rectangle condition in Section 5. The figure 4 is a diagram that satisfies the rectangle condition but not the wellmixed condition.
Hempel distance is a measurement of complexity for Heegaard splittings of manifolds [5]. Bachman and Schleimer transfered the definition from Heegaard surfaces to bridge spheres to compute the complexity of bridge decompositions. Actually, Hempel distance for bridge decompositions of knots in bridge sphere is one of nice tools to detect nonperturbed knots which are in bridge position. We show that if a knot satisfies the rectangle condition for bridge decomposition of then the Hempel distance for the bridge decomposition is greater than or equal to in Section 2. This implies that if satisfies the rectangle condition on the bridge sphere of a bridge decomposition of then is not perturbed. Especially, Otal [8] showed that any bridge presentation of any bridge knot is perturbed for . If has a bridge presentation but the bridge sphere of a bridge decomposition does not allow any perturbation then is 3bridge knot. Therefore, for a knot having bridge decomposition , if then is a 3bridge knot. Coward [3] gave a theoretical method to calculate the bridge number of hyperbolic knots. However, we hardly even know how to find the bridge numbers of alternating knots having a bridge presentation. Especially, we wonder whether or not a knot having a reduced alternating bridge presentation is a bridge knot if the presentation is obtained from a reduced alternating presentation of a bridge knot by adding more crossings without violating the alternating condition. We conjecture that they are all bridge knots. In order to support our conjecture, we investigate the following families. In Section 6, we construct special families of alternating bridge links and . Then, we show that they are bridge links if by using the “Hexagon parameterization” and a modified “train track” diagrams based on the Rectangle condition. Since the number of crossings of a reduced alternating knot diagram is the minimal number of crossings of the knot, we know that each family has an infinitely many elements.
2. bridge decompositions and Hempel distance
Suppose is a link in and is a sphere which divides into two balls and .
Assume that intersects transversely. Let , where are the components of .
We note that is decomposed into and by . The triple is called an nbridge decomposition of if each is a rational tangle. is said to be rational if there exists a homeomorphism of pairs , where . Also,
we say that is in
nbridge position with respect to if has a bridge decomposition . Then consider the projection of onto the plane so that the projection of is a horizontal line and the projection of has maxima and minima. Then we say that the diagram is bridge presentation of .
Let . Then we say that a simple closed curve on is essential if neither it bounds a disk nor it is boundary parallel to a puncture. Also, we say that is an essential disk of if and is essential in . The essential simple closed curves on form a complex which is called the curve complex of . If , the vertices of are the isotopy classes of essential simple closed curves on and a pair of vertices spans an edge of if the corresponding isotopy classes can be realized as disjoint curves. We define that is the minimal distance between and measured in with the path metric. Bachman and Schleimer defined the Hempel distance (or just the distance) of is defined by
is an essential disk of for .
We note that the distance is a finite nonnegative integer since the curve complex is connected.
Now, consider a disk in the ball for so that and , where is a simple arc between the two endpoints of in . The disk is called a bridge disk. Then let be a collection of bridge disks for if are pairwise disjoint. We note that there exist such disks since are rational tangles. Let be a maximal collection of pairwise disjoint, nonisotopic essential disks in . This is called a collection of cut disks. Similarly, we have a collection of cut disks for . We note that there exists a collection of cut disks for so that for . Suppose is in bridge position with respect to for . If there exist essential disks and of and respectively such that are cut disks and , then is separated by the sphere into an bridge sublink and an bridge sublink of . We note that since is an essential disk in . So, if is a knot then there are no such disks and . Let be a knot which is in bridge position with respect to a sphere . Suppose there is a pair of bridge disks and so that the arcs and intersect precisely at one end. Then is said to be perturbed with respect to (and vice verse), and are called cancelling disks for We note that if there are cancelling disks for then we can construct collections of cut disks so that as in the proof of Lemma 2.1.
Lemma 2.1 (Ozawa, Takao [9]).
Suppose that a knot () in bridge position has Hempel distance greater than equal to . Then is not perturbed with respect to the bridge sphere .
3. Rectangle condition on the bridge sphere for a bridge decomposition of a knot
A ntangle is the disjoint union of properly embedded arcs in the unit 3ball; the embedding must send the endpoints of the arcs to marked (fixed) points on the ball’s boundary. Without loss of generality, consider the marked points on the 3ball boundary to lie on a great circle (or a horizontal line ). The tangle can be arranged to be in general position with respect to the projection onto the flat disk or the upper plane in the plane bounded by (or ). The projection then gives us a tangle diagram , where we make note of over and undercrossings as with knot diagrams.
Recall that a tangle in a 3ball , denoted by , is rational if
there exists a homeomorphism of pairs
, where .
Now, let be a knot which has a bridge decomposition . Then, let and be maximal collections of cut disks for and respectively. Recall that there exists a collection of bridge disks for so that for . For instance, is a maximal collection of essential cut disks for the trivial rational 3tangle as in Figure 1, where trivial means that the projection of the tangle on plane has no crossing. Consider a cut disk in which may intersect with . We need to assume that intersects transversely and minimally. A subarc of cut by is a wave for the cut disk in if there exists an outermost arc in and a corresponding outermost disk of with .
Lemma 3.1.
Suppose that is a maximal collection of cut disks for a rational tangle and is an essential disk in . Then if is not isotopic to any for then contains a wave.
Proof.
If is not isotopic to any and then is not essential since is a maximal collection of cut disks. Therefore, . Then by taking an outermost arc of them we have a wave since intersects minimally.
∎
Suppose that be a maximal collection of cut disks for a rational tangle . A cut disk is essential if cuts into a punctured disk and punctured disk. Then there is a maximal collection of essential cut disks . Then they cut into 2punctured disks and a planer surface with boundary components. We may assume that the collection of essential cut disks are the first elements of the maximal collection of cut disks . Then has a pants decomposition with pairs of pants. Suppose that and are pairs of pants of the pants decompostions of and respectively, where is a maximal collection of essential cut disks for a rational tangle . We assume that and intersect transversely and minimally. Then we say that the pairs of pants and are tight if for each tuple of nine combinations as below there is a rectangle embedded in and such that the interior of is disjoint from and the four edges of are subarcs of the tuple, where and are the three boundary components of and respectively.
Now, I would like to define the rectangle condition for bridge spheres with an analogous definition to the rectangle condition for Heegaard surfaces. For two pants decompositions and of and for and respectively. Then we say that and satisfy the rectangle condition if all the nonessential pairs of pants and are tight for . Then the following proposition is the main result about the rectangle condition.
Proposition 3.2.
Suppose and are two pants decompositions of and for and respectively. If and satisfy the rectangle condition, then the Hempel distance .
Proof.
Suppose and are two pants decompositions of and for and respectively.
Let be the disjoint, nonisotopic disks in so that .
Suppose that . Then we have since is a knot.
So, there exist essential disks and in and respectively so that and . We assume that meets transversely and minimally if they intersect.
First, assume that . Then we note that is isotopic to a boundary of . Otherwise, we should have more than pairs of pants for . Then, we isotope so that for some without changing the condition of disjointness with . Let be the boundary component of which is isotopic to without loss of generality. Then we note that there is no path to connect and , and and in without meeting , where and are the other boundary components of . Then there are two subcases for as follows.
If is isotopic to one of then should meet to satisfy the rectangle condition. This contradicts the condition that .
If is isotopic to none of then has a wave by Lemma 3.1. We note that the interior of does not intersect with and the two endpoints of are in the same component of for some . Let be the boundary component of which meets .
We note that separates into two nonempty sets of boundary components since is an essential disk in . Especially, there is a pair of pants so that the boundary components of are separated by since is connected. Let and be two components of which are separated by .
So, needs to meet the all for and especially there are two points in the interior of so that , and the arc between and in only meets at and by the rectangle condition. Since , we can take a path which is closely parallel to so that connects and in without meeting . This makes a contradiction.
Now, assume that . Take a component of whose closure is a bigon. Then assume that for . Then separates into two nonempty sets since is an essential disk in . So, there exists so that the boundary components of are separated by . Let and be the components of so that they are separated by . Then there is no path to connect and in without meeting or . We note that is isotopic to one of the boundary components .
If is isotopic to one of we switch the indices and to have a contradiction with the same reason as above.
If is isotopic to none of then has a wave . Then, by using a similar argument above, there are two points in the interior of so that , and the arc between and in only meets at and by the rectangle condition. Since , we can take a path which is closely parallel to so that connects and in without meeting or . This contradicts the existence of and so that there is no path to connect and in without meeting or . This completes the proof.
∎
Now, we will discuss how to check whether or not given two pants decompositions satisfy the rectangle condition. First of all, we parameterize the boundary of an essential disk in in Section 4.
4. Dehn’s parameterization of
Let be a simple closed curve in . We consider the standard essential disks as in Figure 1. Consider the pair of pants , where is the two punctured disk in so that as in Figure 2.
Figure 2 shows standard arcs in the pair of pants . We notice that we can isotope into in so that each component of is isotopic to one of the standard arcs and . Then we say that subarc of is carried by if some component of is isotopic to . The closed arc is called a window. Let . Then can have many parallel arcs which are the same type in . Let be the number of parallel arcs of the type which is called the of
Lemma 4.1.
determine the weights for .
Proof.
We have two subcases for this.
First, suppose that for all distinct .
We claim that . If not, then for some .
We notice that . So we have , and .
This shows that . This makes a contradiction. So, .
Now, we have . This implies that
.
Now, suppose that for some . Then we note that has , and . This implies that and .
∎
Now, we discuss the arc components of . Let and be the simple arcs as in Figure 3.
We assume that has no bigon in . We note that separates into two semidisks and as in Figure 3.
Let be the number of subarcs of from to in . Also, let be the number of subarcs of from to in and let be the number of subarcs of from to in .
Let and in . We note that each component of meets exactly once. Also, we know that each such component is essential in
. The components of are determined by three parameters as in Figure 3,
where in and .
In order to define and , consider and .
Then we know that .
So, . Therefore, .
So, we know . Now, we define and as follows. If then (mod ) and , and and if then (mod ) and , and .
Then is called the in .
Let be the three parameters to determine the arcs in . Similarly, we have the three parameters for ().
Therefore, is determined by a sequence of nine parameters by Lemma 4.1.
Let be the set of isotopy classes of simple closed curves in . For a given simple closed curve in , we define , and in as above. Let for . Then we have the following Dehn’s Theorem.
Theorem 4.2 (Special case of Dehn’s Theorem ).
There is an onetoone map so that . i.e., it classifies isotopy classes of simple closed curves.
When then if the simple closed curve is isotopic to and if . Refer [10] to see the general Dehn’s theorem.
5. Detecting bridge knots with bridge presentations
Theorem 5.1.
Let be the 3bridge decomposition of a knot . Suppose and are two pants decompositions of and for and respectively. If and satisfy the rectangle condition, then is a 3bridge knot.
To use the rectangle condition to check if a bridge presentation link is bridge link, we need to choose a level sphere to have the bridge decomposition of . Especially we can choose to have a bridge decomposition of by the following lemma, where . (Refer to Figure 1.) However, it does not mean that for a given pants decomposition for there exists a pants decomposition for to satisfy the rectangle condition even if there are two pants decompositions for the two rational tangles and to satisfy the rectangle condition.
Lemma 5.2.
Suppose and are bridge decompositions of a link . Then there exist minimal collections of essential cut disks and for and respectively so that there exist two pants decompositions and for and respectively which satisfy the rectangle condition if and only if there exist minimal collections of essential cut disks and for and respectively so that there exist two pants decompositions and for and respectively which satisfy the rectangle condition.
Proof.
First, we note that the pants decompositions have only one pair of pants. We consider the selfhomeomorphism of as an extension of a homeomorphims from to which is obtained from a combination of half Dehn twists to get from to by following up the bridge presentation. We note that the homeomorphism of preserves rectangles to satisfy the rectangle condition. Let . Then it is enough to show that and bound the minimal collections of cut disks for and . By construction of , it is clear that bounds essential cut disks. (Refer to [6].)
∎
Unfortunately, there are infinitely many different maximal collection of essential cut disks in . In other words, we have infinitely many different pants decompositions of for . So, it is impossible to check the all combinations of two pants decompositions. In this paper, we will especially discuss the case that is the collection of maximal essential cut disks for . (Refer to Figure 1.) Then, we note that is the boundaries of which is pants decomposition. In order to get a pants decomposition for , we consider three bridge disks for . Then let which is called essential arc. Let be the regular neighborhoods of so that they are pairwise disjoint. Then by deleting from we have a pants decomposition for .
For an easier argument, now we consider three essential arcs which connect two punctures in instead of the three simple closed curves for the boundary components of .
Then, we want to modify the rectangle condition as follows.
Let be a bridge decomposition of a link .
Let be the collection of bridge disks for two rational tangles and respectively.
Then we take the collection of arcs .
Then we say that the collections of arcs and satisfy the rectangle condition if there is a rectangle embedded in such that the interior of is disjoint from and the four edges of are subarcs of the four entries of each combination as below respectively.
Proposition 5.3.
Suppose that is a bridge decomposition of a link . For given collections of bridge disks for two rational tangles and respectively, take the collection of arcs . If the collections of arcs and satisfy the rectangle condition in then is a bridge link.
Now, we will discuss how to check whether or not two pants decompositions satisfy the rectangle condition.
The figure 4 shows that and satisfy the rectangle condition, where is the arc from to , is the arc from to and is the arc from to , and are the straight line which connects two punctures in each two punctured disk . The alphabetic sequences are the orders of arcs, where.
Then we can check that each sequence contains all the possible adjacent pairs and .
We recall that determines the arc pattern in a two punctured disk as Figure 5. In order to analyize the arcs, take the boundaries of the regular neighborhood of the essential arcs as in Figure 5. We note that is the same with the intersection number between the three essential arcs and the window . Except the two arcs start from the punctures, each component meets twice with . So, we name the arcs by using positive integers to . For example, we name to the arc components as in Figure 5.
Lemma 5.4.
and determine the pairs of intersection points in which are the two endpoints of each arc component in
a two punctured disk as follows.

If then replace by .

If