Obstructing Sliceness in a Family of Montesinos Knots

# Obstructing Sliceness in a Family of Montesinos Knots

Luke Williams Department of Mathematics, Michigan State University, East Lansing, MI 48824
###### Abstract.

Using Gauge theoretical techniques employed by Lisca for 2-bridge knots and by Greene-Jabuka for 3-stranded 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 2-fold covers of Montisinos knots bound plumbing 4-manifolds, 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 Lisca-2007, Lisca-2007-2, Lisca applies gauge theory, based on work of Donaldson Donaldson-1987, to obstruct smooth sliceness for 2-bridge knots. Lisca’s approach uses the observation that the 2-fold branched cover of every 2-bridge knot bounds a negative definite plumbing 4-manifold. On the other hand, if a 2-bridge knot is slice, its 2-fold cover also bounds a rational homology 4-ball. Gluing these two 4-manifolds along their common boundary yields a smooth, closed, negative definite 4-manifold . 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 2-bridge knots and moreover, suffices to determine the smooth concordance orders of all 2-bridge knots.

In Greene-Jabuka, Greene and Jabuka use this approach, supplemented by Heegaard Floer homology techniques, to determine all slice knots among the 3-stranded pretzel knots with odd. Both Lisca’s article, in the case of 2-bridge knots, and Greene-Jabuka’s article, in the case of 3-stranded pretzel knots, resolve in the affirmative the slice-ribbon 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)

 M(0;[m1+1,n1+2],[m2+1,n2+2],q)

with . Our main theorem is

###### Theorem 1.1.

Let subject to the condition

 n1+2(m1+1)n1+2m1+1+n2+2(m2+1)n2+2m2+1+1q>0.

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.

Choose , , and . Then the conditions from Theorem 1.1 translate into

 n1,n2≥1andn1 % is even or n2 is odd.

Consequently there are infinitely many examples among which are obstructed from being smoothly slice by Theorem 1.1.

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 4-manifolds whose boundaries are 2-fold 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 Lisca-2007. To start with, let be a knot in and let denote the 3-maniold 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 slice111All instances of sliceness in the article will always refer to smooth sliceness., then bounds a rational homology 4-ball . The manifold is obtained by taking a double branched cover of the 4-ball branched over the slice disk of .

If is a knot whose double branched cover bounds a negative definite smooth 4-manifold , one can form a closed, smooth, negative definite 4-manifold by gluing to along . Recall Donaldson’s celebrated theorem

###### Theorem 2.1 (Donaldson’s Theorem A Donaldson-1987).

Let be a closed, oriented, smooth 4-manifold 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 4-manifold , then the intersection form of embeds into the standard negative definite form of the same dimension.

###### Example 2.3.

Consider the figure-eight knot. bounds a plumbing 4-manifold 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 4-manifold in Figure 1. With the indicated basis, we can find an embedding :

 φ(f1)=e2+e3+e4,φ(f2)=e1−e2,φ(f3)=e1+e2−e3,φ(f4)=e3−e4.

Then, as expected, we cannot use Theorem 2.2 to obstruct from being slice.

## 3. Montesinos Knots and Their 2-Fold Covers

Here, we outline the definition of a Montesinos knot and its 2-fold branched cover. We follow Owens. Let , then we let be the continued fraction

 [n1,n2,…,nk]=n1−1n2−1…−1nk.

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 half-twists.

With these conventions in place, we can state the definition of a Montesinos link and describe its 2-fold cover.

###### Definition 3.1.

Let each be in lowest terms and let . Then the Montesinos link is given by Figure 3 where represents half-twists and represent a rational tangle corresponding to .

###### Theorem 3.2 (Montesinos).

The 2-fold branched cover of branched along a Montesinos link is the boundary of the plumbing 4-manifold given by Figure 4

where .

Neumann and Raymond Neumann-Raymond give the following useful result to determine when such a plumbing could be blown down to a negative definite 4-manifold.

###### Theorem 3.3 (Neumann-Raymond).

A plumbing, , with no nonnegative weights in its plumbing graph and whose boundary is the 2-fold branched cover of branched over the Montesinos link , is negative definite if and only if

 e+n∑i=1βiαi>0.

## 4. Obstructing Sliceness in M(0;[m1+1,n1+2],[m2+1,n2+2],q)

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 :

1. , , and even and no other restrictions,

2. , , and even and no other restrictions,

3. , , , odd and no other restrictions,

4. , , , odd and no other restrictions,

5. , , , , and odd.

###### Proof.

The proof is a simple matter of an exhaustive check of the thirty-two possible combinations of parity for the , and . ∎

To apply Theorem 2.2, we need the 2-fold cover of to bound a negative definite 4-manifold. In light of Theorem 3.3, we require that , and satisfy

 n1+2(m1+1)n1+2m1+1+n2+2(m2+1)n2+2m2+1+1q>0. (1)

### 4.1. Choosing a Plumbing 4-manifold

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.

Let , then .

###### Proof.

This fact is easily proven by induction. ∎

This, combined with Theorem 3.2, gives that the 2-fold cover of bounds the plumbing 4-manifold in Figure 5. Blowing down appropriate vertices results in the negative definite 4-manifold 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 4-manifold 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

 φ(f1)⋅φ(f2)=e1⋅ei−e1⋅ei+1−e2⋅ei+e2⋅ei+1.

Since is injective, we cannot have . If , then the above quantity is zero, so it follows that and thus . Similar arguments show that

 φ(fi)=ei−ei+1,1≤i≤n1. (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

 φ(fi)=ei−ei+1,n1+2≤i≤n1+m1+m2.

Similarly, we can assume that and thus

 φ(fi)=ei−ei+1,n1+m1+m2+2≤i≤n1+m1+m2+n2+1.

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

 λi+1−λi=0,

for , , and . It follows that

 λ1=λ2=…=λn1,λn1+3=λn1+4=…=λn1+m1+m2+1,
 λn1+m1+m2+3=λn1+m1+m2+4=…=λn1+m1+m2+n2+2.

Let , and . Now, . Therefore, . Similarly, . So,

 φ(fn1+1) = =n1∑i=1λei+(λ±1)en1+1+(η−1)en1+2+n1+m1+m2+1∑i=n1+3ηei+n1+m1+m2+n2+2∑i=n1+m1+m2+2γei. (4)

Direct calculation along with (3) gives that

 φ(fn1+1)2=−(n1λ2+(λ±1)2+(η−1)2+(m1+m2−1)η2+(n2+1)γ2)=−3.

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

 φ(fn1+1)=n1∑i=1λei+(λ±1)en1+1+(η−1)en1+2+n1+m1+m2+1∑i=n1+3ηei.

The same argument applies to where the pairings are given by

 φ(fi)⋅φ(fn1+m1+m2+1)=⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩±1i=n1+m1+m2+2,−3i=n1+m1+m2+1,1i=n1+m1+m2,0otherwise. (5)

Then, there exist such that

 φ(fn1+m1+m2+1) = =n1+m1+m2∑i=n1+2βei+(β+1)en1+m1+m2+1+(α±1)en1+m1+m2+2+n1+m1+m2+n2+2∑i=n1+m1+m2+3αei. (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

 (m1+m2)βη−β+η=0.

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

 φ(fn1+1)=n1∑i=1λei+(λ±1)en1+1−en1+2,
 φ(fn1+m1+m2+1)=en1+m1+m2+1+(α±1)en1+m1+m2+2+n1+m1+m2+n2+2∑i=n1+m1+m2+3αei.

Moreover, and , and , so

 φ(fn1+1)=φ(f3)=∓e1∓e2−e4,
 φ(fn1+m1+m2+1)=φ(fm1+m2+3)=em1+m2+3∓e+m1+m2+5∓em1+m2+6.

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,

 φ(f1) = e1−e2 φ(f2) = e2−e3 φ(f3) = ∓e1∓e2−e4 φ(f4) = e4−e5 ⋮ φ(fm1+m2+2) = em1+m2+3−em1+m2+2 φ(fm1+m2+3) = em1+m2+3∓e+m1+m2+5∓em1+m2+6 φ(fm1+m2+4) = em1+m2+5−em1+m2+4 φ(fm1+m2+5) = em1+m2+6−em1+m2+5

Now, suppose that for . Then, we get the following system of equations arising from the pairings indicated in Figure 6

Then , , , and