Obstructing Sliceness in a Family of Montesinos Knots
Abstract.
Using Gauge theoretical techniques employed by Lisca for 2bridge knots and by GreeneJabuka for 3stranded pretzel knots, we show that no member of the family of Montesinos knots , with certain restrictions on , , and , can be (smoothly) slice. Our techniques use Donaldson’s diagonalization theorem and the fact that the 2fold covers of Montisinos knots bound plumbing 4manifolds, many of which are negative definite. Some of our examples include knots with signature 0 and square determinant.
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1. Introduction
In his recent breakthrough articles Lisca2007, Lisca20072, Lisca applies gauge theory, based on work of Donaldson Donaldson1987, to obstruct smooth sliceness for 2bridge knots. Lisca’s approach uses the observation that the 2fold branched cover of every 2bridge knot bounds a negative definite plumbing 4manifold. On the other hand, if a 2bridge knot is slice, its 2fold cover also bounds a rational homology 4ball. Gluing these two 4manifolds along their common boundary yields a smooth, closed, negative definite 4manifold . As such, according to Donaldson, its intersection form has to be standard. As Lisca goes on to show, this obstruction suffices to pin down all slice 2bridge knots and moreover, suffices to determine the smooth concordance orders of all 2bridge knots.
In GreeneJabuka, Greene and Jabuka use this approach, supplemented by Heegaard Floer homology techniques, to determine all slice knots among the 3stranded pretzel knots with odd. Both Lisca’s article, in the case of 2bridge knots, and GreeneJabuka’s article, in the case of 3stranded pretzel knots, resolve in the affirmative the sliceribbon conjecture kirby.
Building on Lisca’s work, we employ the same gauge theoretic techniques to address the question of smooth sliceness for a five parameter family of Montesinos knots with three rational tangles. Specifically, the family we consider is (see Section 3 for a precise definition)
with . Our main theorem is
Theorem 1.1.
Let subject to the condition
Moreover, assume that were chosen so that is a knot (see Proposition 4.1). Then is not smoothly slice.
The conditions on are realized on a multitude of examples. We only list a few but the interested reader will have little trouble finding many additional examples.
Example 1.2.
Before stating our next example, we remind the reader that the knot signature and knot determinant of a knot can used as obstructions to sliceness. If is slice then and is a square. With this in mind, consider the next example.
Theorem 1.3.
The family (with the restrictions on , and as outlined in Theorem 1.1) contains an infinite subfamily of knots whose signatures are 0 and whose determinants are square.
1.1. Organization
In Section 2, we formally outline the approach employed by Lisca to obstruct smooth sliceness. In Section 3, we give definitions of a Montesinos knot as well as outline plumbing 4manifolds whose boundaries are 2fold branched covers of such knots. Finally, in Sections 4 and 5, we prove Theorems 1.1 and 1.3 respectively.
2. Obstruction to Sliceness
This section outlines the technique used by Lisca Lisca2007. To start with, let be a knot in and let denote the 3maniold obtained by taking the the double branched cover of with branching set . We note that is always a rational homology sphere and moreover, if is slice^{1}^{1}1All instances of sliceness in the article will always refer to smooth sliceness., then bounds a rational homology 4ball . The manifold is obtained by taking a double branched cover of the 4ball branched over the slice disk of .
If is a knot whose double branched cover bounds a negative definite smooth 4manifold , one can form a closed, smooth, negative definite 4manifold by gluing to along . Recall Donaldson’s celebrated theorem
Theorem 2.1 (Donaldson’s Theorem A Donaldson1987).
Let be a closed, oriented, smooth 4manifold with negative definite intersection form , then is isomorphic to the standard negative definite form of the same rank.
Combining Donaldson’s theorem with the remarks preceding it, we arrive at
Theorem 2.2.
If is a slice knot whose double branched cover bounds the negative definite smooth 4manifold , then the intersection form of embeds into the standard negative definite form of the same dimension.
Example 2.3.
Consider the figureeight knot. bounds a plumbing 4manifold given by a single component of the plumbing given in Figure 1. If were slice, then there would be an embedding (see Section 4.2 for notation) such that . No such embedding exists since there are only two basis elements in the codomain. So, Theorem 2.2 implies that could not be slice.
We know that is slice. bounds the plumbing 4manifold in Figure 1. With the indicated basis, we can find an embedding :
Then, as expected, we cannot use Theorem 2.2 to obstruct from being slice.
3. Montesinos Knots and Their 2Fold Covers
Here, we outline the definition of a Montesinos knot and its 2fold branched cover. We follow Owens. Let , then we let be the continued fraction
Let be in lowest terms and let be a continued fraction representing , then a rational tangle corresponding to is given by Figure 2 where indicates halftwists.
With these conventions in place, we can state the definition of a Montesinos link and describe its 2fold cover.
Definition 3.1.
Let each be in lowest terms and let . Then the Montesinos link is given by Figure 3 where represents halftwists and represent a rational tangle corresponding to .
Theorem 3.2 (Montesinos).
The 2fold branched cover of branched along a Montesinos link is the boundary of the plumbing 4manifold given by Figure 4
where .
Neumann and Raymond NeumannRaymond give the following useful result to determine when such a plumbing could be blown down to a negative definite 4manifold.
Theorem 3.3 (NeumannRaymond).
A plumbing, , with no nonnegative weights in its plumbing graph and whose boundary is the 2fold branched cover of branched over the Montesinos link , is negative definite if and only if
4. Obstructing Sliceness in
In this section, we consider and prove Theorem 1.1. First, we outline the restrictions on found within Theorem 1.1. In order for to be a knot we need to carefully restrict the choices of parity on . The following proposition outlines these restrictions.
Proposition 4.1.
All but the following twelve combinations of parity for the , , and , result in knots in the family :

, , and even and no other restrictions,

, , and even and no other restrictions,

, , , odd and no other restrictions,

, , , odd and no other restrictions,

, , , , and odd.
Proof.
The proof is a simple matter of an exhaustive check of the thirtytwo possible combinations of parity for the , and . ∎
To apply Theorem 2.2, we need the 2fold cover of to bound a negative definite 4manifold. In light of Theorem 3.3, we require that , and satisfy
(1) 
4.1. Choosing a Plumbing 4manifold
Rather than applying Theorem 3.2 directly, we note that can be expressed as a different continued fraction expansion  one which is more compatible with Lisca’s approach.
Proposition 4.2.
Let , then .
Proof.
This fact is easily proven by induction. ∎
This, combined with Theorem 3.2, gives that the 2fold cover of bounds the plumbing 4manifold in Figure 5. Blowing down appropriate vertices results in the negative definite 4manifold in Figure 6.
4.2. Proving Theorem 1.1
Before we present the proof of Theorem 1.1, we define our notation for embeddings of intersection forms over the integers. For an integer , we view the matrix as an intersection pairing on with respect to the standard basis . Similarly, we view as the standard negative definite intersection pairing on another copy of with respect to its standard basis, . We abbreviate to , and similarly for . Finally, an embedding of into is a monomorphism such that .
Proof of Theorem 1.1.
Let be the intersection form associated to the plumbing 4manifold in Figure 6, we show that does not embed into the standard negative definite form of equal rank.
Suppose to the contrary that does embed into . Then, there must exist an embedding , where . Taking the indicated in Figure 6 as our basis for , we explore the structure of .
First, note that up to a change of basis for , any vertex with self intersection 2 or 3 has essentially a unique image under . For instance, we can assume that, . Similarly, we can assume that for some (up to a change of basis). Now ; therefore . By definition
Since is injective, we cannot have . If , then the above quantity is zero, so it follows that and thus . Similar arguments show that
(2) 
It is worth noting that if , we could take while still satisfying the pairings and . However, with such a definition, we still need , implying that must contain either or , but not both. However, this would imply that would pair with nontrivially  which contradicts the fact that is an embedding. It follows that, up to a change of basis for , the images defined in (2) are the only possibilities.
Next, we consider the image of under . As above, up to a change of basis we can take and by the same argument used for , we arrive at
Similarly, we can assume that and thus
Next, we consider . Since is a homomorphism, we know that for some . Given that , we have
(3) 
Note that the (rather than just 1) arises from the fact that we may have had to change the basis of to get images of the previous vertices in their correct forms. Thus, we may have caused the pairing of and to become negative.
Since for each , we have that for the same . Then, from (3) we have that
for , , and . It follows that
Let , and . Now, . Therefore, . Similarly, . So,
(4) 
Direct calculation along with (3) gives that
Clearly, if , then . If and , then , , , and must be be identically zero  clearly an impossibility; therefore if , . Moreover, if and , then three of , , , and must be zero  which is again impossible. It follows that is necessarily 0. Therefore
The same argument applies to where the pairings are given by
(5) 
Then, there exist such that
(6) 
Therefore, we have explicit forms for the images of all but the last basis element in under . Before we consider this last image, we extract as much information out of the pairings in (3) and (5) as we can. (5) gives that . Therefore,
and thus
Noting that the coefficient of each basis element in the image of a square 3 vertex is necessarily in the set , we have that ; so, or and or .
If , then and if , then and . In this latter case, is forced to be which, in turn, forces . Moreover, is forced to be which makes . We’ll come back to this case.
Now, suppose that , then
Moreover, and , and , so
Therefore, if and are not identically 2, then no such embedding exists.
Corresponding to the case when , we have and . Then, we know that up to a change of basis,
Now, suppose that for . Then, we get the following system of equations arising from the pairings indicated in Figure 6
Then , , , and