On the Slice-Ribbon Conjecture for pretzel knots

# On the Slice-Ribbon Conjecture for pretzel knots

## Abstract.

We give a necessary, and in some cases sufficient, condition for sliceness inside the family of pretzel knots with one even. The three stranded case yields two interesting families of examples: the first consists of knots for which the non-sliceness is detected by the Alexander polynomial while several modern obstructions to sliceness vanish. The second family has the property that the correction terms from Heegaard-Floer homology of the double branched covers of these knots do not obstruct the existence of a rational homology ball; however, the Casson-Gordon invariants show that the double branched covers do not bound rational homology balls.

###### Key words and phrases:
Pretzel knots, slice–ribbon conjecture, rational homology balls
This work is partially supported by the Spanish GEOR MTM2011-22435.

## 1. Introduction

Pretzel knots have been thoroughly studied since they were first introduced by Reidemeister in [b:Re]. In recent work Greene and Jabuka have determined the order in the smooth knot concordance group of –stranded pretzel knots with all odd [b:GJ, Theorem 1.1]. A corollary of their result is that the slice–ribbon conjecture proposed by Fox in [b:Fo] holds true for this family of knots. In this paper we address the question of sliceness in the family of pretzel knots with one even. Our main result, Theorem 1, gives a necessary condition for sliceness in this family. The condition obtained is not sufficient, but we propose a conjecture of what constraints should be added to Theorem 1 to obtain a complete characterisation of ribbon pretzel knots.

As a byproduct of our research we have found two interesting families of pretzel knots. The first one is a one parameter family of knots for which most of the known slice obstructions vanish: the signature, the determinant, the Arf invariant, the -invariant and the invariant among others. However, the Alexander polynomial is able to show that more than three quarters of the knots in this family are not slice. This property makes this set of knots an excellent source to test future slice invariants. The second interesting family consists of a set of pretzel knots whose double branched covers do not bound rational homology balls. The non-existence of these balls is determined via the Casson-Gordon invariants which, for this particular family of -manifolds, turn out to be a more powerful tool than the -invariants from Heegaard Floer homology in obstructing the existence of a rational ball.

Given nonzero integers the pretzel link is obtained by taking pairs of parallel strands, introducing half twists on the -th pair, with the convention for right–hand crossings and for left–hand crossings, and connecting the strands with pairs of bridges. As an example, the first knot in Figure 2 corresponds to . If more than one of the is even or if is even and none of the is even then is a link. In all other cases it is a knot. Inside the family of pretzel knots we limit our considerations to those with one even parameter and moreover from now on we fix and for all . Note that if or if and one of the satisfies , then the pretzel knot is a –bridge, already studied in [b:Li, b:Li2].

This paper addresses the question of sliceness of pretzel knots. A knot in is said to be slice if it bounds a disc smoothly embedded in the -ball. The double cover of branched over a slice knot bounds a rational homology 4-ball. This property is one of the main obstructions we shall study. The strategy we follow is straightforward: use different obstructions to rule out the non-slice pretzel knots and show that the remaining knots are slice by explicitly constructing the slice discs in . The discs we find have the property of being ribbon, i.e. they have no local maxima for the radial function in . The slice–ribbon conjecture says that all slice knots are ribbon and the results in the present work support this conjecture.

It is well known [b:BZ, Theorem 12.19] (recall that pretzel knots are a particular case of the more general family of Montesinos links) that among the permutations of the parameters , the of them which correspond to cyclic permutations, order reversing permutations and compositions of these leave invariant the knot . Two pretzel knots and which are not isotopic but who share the same set of parameters are mutants, that is can be obtained from by removing a -ball from that meets in two proper arcs and gluing it back via an involution of its boundary , where is orientation preserving and leaves the set invariant. All pretzel knots defined by the same set of parameters have the same double branched cover.

Our main result, Theorem 1, is stated for pretzel knots up to reordering of the parameters because the obstructions to sliceness that we analyze live in the double branched cover of these knots. Up to reordering we are able to establish the sliceness of pretzel knots with one even except for the following set

 E={a,−a−2,−(a+1)22,q1,−q1,…,qm,−qm},

where , odd and (mod 60). Our main result is the following.

{thm}

Let be a slice pretzel knot with one even parameter and such that . Then, the –tuple of integers can be reordered so that it has the form

• if is even;

• if is odd.

{rem}

Ligang Long [b:Long] has independently obtained Theorem 1 for the case . Moreover, he has several partial results concerning the sliceness of pretzel knots without the restriction of having one even parameter.

As further explained in Section 2 not all possible orders of the parameters in Theorem 1 yield slice knots. In Corollary 2 we show that for certain orders of the parameters the knots in the above Theorem are actually slice. Moreover we conjecture that the orders proposed in Conjecture 2 are all the possible orders of the parameters in Theorem 1 yielding slice knots.

As detailed in Section 4.2 most of the pretzel knots in the family , odd, are not slice. However establishing that none of them is slice is still an open challenge. There is a great amount of evidence supporting the following conjecture

{conj}

If then the pretzel knot is not slice.

Note that a pretzel knot of the form is independent of the order of the parameters, since cyclic permutations, order reversing permutations and compositions of these comprise all possible permutations of three elements. For -stranded pretzel knots whose defining parameters are not in an easy corollary of Theorem 1 is the validity of the slice–ribbon conjecture.

In the following Corollary 1 the results for with three odd parameters were already proved in [b:GJ]. Our work proves the statement for –stranded pretzel knots with one even parameter and leaves out the case , odd and (mod 60).

{cor}

The slice–ribbon conjecture holds true for pretzel knots of the form where and .

In order to prove Theorem 1 we start using the approach of [b:Li], which is also followed in [b:GJ]: if a pretzel knot is slice then its double branched cover is the boundary of a rational homology ball [b:Ka, Lemma 17.2]. Moreover, up to considering the mirror image of , the –manifold is also the boundary of a negative definite –manifold obtained by plumbing together disc bundles over spheres. We can build a closed, oriented, negative definite, –manifold as . By Donaldson’s celebrated theorem [b:Do] the intersection form of must be diagonalizable and therefore, since , there must exist a monomorphism such that for every .

The existence of is enough to guarantee sliceness among –bridge knots [b:Li]. In the case of pretzel knots with all odd, this obstruction shows that there must exist some such that [b:GJ, Proposition 3.1]. However, not all these pretzel knots are slice. In fact, using the Ozsváth–Szabó correction terms for rational homology spheres Greene and Jabuka conclude that only result in slice knots.

In our case, for pretzel knots with one even parameter, even in the case of three strands, the existence of is a weaker obstruction to sliceness than in the cases studied in [b:GJ, b:Li]. For example, contrary to what happens for –bridge knots and pretzel knots of the form with all odd, in our case the existence of does not imply that the knot signature is zero. The proof of Theorem 1 has three main steps: first we determine the pretzel knots with vanishing signature such that exists. Not all the resulting knots are slice. In a second step we use the correction terms from Heegaard-Floer homology to further restrain the family of candidates to slice knots. This leaves us with two one parameters families to further study. The sliceness of one of these families is ruled out using Casson-Gordon invariants while the other one is partially treated studying Alexander polynomials.

The rest of the paper is organized as follows. The proof of Corollary 1 is carried out in Section 2 assuming Theorem 1. An easy algorithm is given to detect ribbon pretzel knots and we show that many of the knots from Theorem 1 are actually slice. In Section 3 we recall some properties of the Seifert spaces associated to pretzel knots and of the knot signature. Section 4 deals with the two interesting families of pretzel knots whose properties were described above. Finally, Sections 5 and 6 treat the general case combining Donaldson’s theorem with the knot signature and the correction terms from Heegaard Floer homology.

Acknowledgments. This project began while I was a Ph.D. student under the supervision of Paolo Lisca to whom I am very grateful and from whom I still learn so much. I have had the opportunity to discuss this problem with many people who have all contributed to my better understanding of its different aspects: Benjamin Audoux, David Cimasoni, Jose F. Galvan, Jose M. Gamboa, Farshid Hajir, Lukas Lewark, Marco Mazzucchelli, Brendan Owens and Stepan Orevkov. Very special thanks go to Josh Greene and Slaven Jabuka, not only for their inspiring paper on this subject that introduced me to pretzel knots, but especially for their great generosity and the many hours we have shared thinking about this problem together.

## 2. Slice pretzel knots

In this section we prove that, for certain orders of the parameters, the knots in Theorem 1 are actually slice. Moreover, assuming this theorem we prove Corollary 1 which deals with the slice-ribbon conjecture for -stranded pretzel knots. We start with the following proposition which explains an easy algorithm to determine whether a pretzel knot is ribbon.

{prop}

[Ribbon Algorithm] Let be a pretzel knot and let . While for some it holds , we reduce the number of parameters to and repeat with the the knot . If at the end of the sequence of reductions we are left with a pretzel knot with exactly one parameter or with two parameters and satisfying , then is ribbon.

###### Proof.

On a pretzel knot , whenever there are two adjacent strands and with the same number of crossings but of opposite signs, we can perform the ribbon move shown in Figure 2, which simplifies the pretzel knot yielding the disjoint union of an unknot and a new pretzel knot . The knot is equal to without and . Therefore, if is odd and after the sequence of reductions the set of parameters defining consists of only one integer, we have that after performing ribbon moves on we obtain the disjoint union of unknots. Thus, is ribbon. On the other hand, if is even and after the sequence of reductions the set of parameters defining consists of exactly two integers and satisfying , then after performing ribbon moves on , we obtain, since is the unknot, the disjoint union of unknots. Thus again, is ribbon. ∎

{cor}

Let be a pretzel knot satisfying the assumptions of Theorem 1. Then the above Ribbon Algorithm shows that for certain orderings of the parameters is slice.

###### Proof of Corollary 1.

Given as in the assumptions, if any of or equals , then is a –bridge knot (see [b:GJ, Figure 2] for a proof) and by [b:Li, Corollary 1.3], the slice–ribbon conjecture holds in this case. On the other hand, if for every , then the parameters satisfy either (mod 2) or there is exactly one even parameter. For the first possibility [b:GJ, Theorem 1.1] holds and the statement follows. In the second case, if is slice then Theorem 1 holds and we obtain that is of the form for some ordering. Since –stranded pretzel knots are independent from the ordering of the parameters, we have and Proposition 2 shows that is ribbon. ∎

{rem}

The only -stranded pretzel knots for which the slice-ribbon conjecture remains open are a subset of the family . We conjecture that none of these knots are slice and thus that the slice-ribbon conjecture holds for all -stranded pretzel knots. Given a pretzel knot with for all , the necessary condition for sliceness that we establish in this paper, namely that for some permutation of the parameters the knot can be shown to be ribbon using Proposition 2, is not sufficient. For instance, in [b:H, Section 11] it is shown that the mutant of the slice pretzel knot is not slice. Therefore, one may ask what constraints on the ordering of the parameters need to be added in order to obtain a sufficient condition for sliceness. For all the examples we know (including in particular the knots and above) the ribbon algorithm of Proposition 2 establishes that the knot is actually ribbon. On the basis of these considerations we propose: {conj} The pretzel knots with for all that are ribbon are precisely those detected by the algorithm in Proposition 2.

## 3. Preliminaries

### 3.1. Double branched covers of pretzel knots

Let be a plumbing graph, that is, a graph in which every vertex carries an integer weight , . Associated to each vertex is the –dimensional disc bundle with Euler number . If the vertex has edges connected to it in the graph , we choose disjoint discs in the base of and call the disc bundle over the th disc . When two vertices are connected by an edge, we identify with by exchanging the base and fiber coordinates and smoothing the corners. This pasting operation is called plumbing (for a more general treatment we refer the reader to [b:GS]), and the resulting smooth –manifold is said to be obtained by plumbing according to .

The group has a natural basis represented by the zero-sections of the plumbed bundles. We note that all these sections are embedded –spheres, and they can be oriented in such a way that the intersection form of will be given by the matrix with the entries

 qij=⎧⎪⎨⎪⎩aiif i=j;1if i is connected to j by an edge;0otherwise.

We will call the intersection lattice associated to .

A star–shaped graph is a connected tree with a distinguished vertex (called the central vertex) such that the degree of any vertex other than the central one is . A leg of a star–shaped graph is any connected component of the graph obtained by removing the central vertex. If is a star–shaped graph then the boundary is a Seifert space (see [b:Ran] for a proof) with as many singular fibers as legs of the graph .

Given a pretzel knot , let denote the –manifold obtained as the –fold cover of branched along . These –manifolds are Seifert fibered spaces with singular fibers [b:Mo], which can be described as the boundary of the –manifold obtained by plumbing according to the graph in Figure 3.1.

The order of the first homology group of can be computed via the incidence matrix of any graph such that and it holds (see [b:NR])

 |H1(YΓ)|=|detQΓ|=(n∑i=11pi)n∏i=1pi. (1)

By [b:NR, Theorem 5.2], the Seifert space can be written as the boundary of a negative definite –plumbing as long as

 1p1+⋯+1pn>0. (2)

If the inequality holds, there is a canonical negative plumbing tree, which from now on will be denoted , satisfying and is negative definite. All the vertices in have weight except for the central vertex which has weight . The tree is obtained as follows: take the graph in Figure 3.1: for every such that substitute its corresponding length–one leg with a –chain with vertices and subtract from the weight of the central vertex. In this way, we obtain a new four manifold, which is negative definite, and has the same boundary as before the substitutions. Formally this is done by a series of blow downs and blow ups (see [b:Ne]). An example is shown in Figure 3.1. We call the –chain corresponding to the parameter .

Recall that if a knot is slice then its mirror image is also slice. In the case of pretzel knots we have that for the mirror image satisfies . Therefore, when studying sliceness of pretzel knots, up to taking mirror images, we can always suppose that the double branched cover is the boundary of a negative definite –plumbing or equivalently that the defining parameters satisfy inequality (2). From now on we will only consider pretzel knots satisfying (2) and we shall divide them in the following three families

• is even and all except one of the are odd.

• is odd, all except one of the are odd and the only even parameter is positive.

• is odd, all except one of the are odd and the only even parameter is negative.

Since the Seifert space does not depend on the order of , from now on we adopt the following convention for the ordering and notation of the parameters. We write

 Y=Y(a1,...,as;c1,...,ct),s,t≥0,n=s+t,

where and . Note that the central vertex in has weight .

We label the vertices of the graph as follows: the central vertex will be called ; the vertices corresponding to the negative parameters will be called ; the vertices of the –chain , , will be called , where is connected to and is connected to for all . The number of vertices in , which will be called , coincides with the rank of . It is immediate to check that

 m=|Ψ|=s+1+t∑i=1(ci−1).

Let be the standard negative diagonal lattice with the elements of a fixed basis labeled as . As an abbreviation in notation let us write to denote . If the intersection lattice admits an embedding into , then we will omit the in the notation, i.e. instead of writing we will directly write . If exists we will call, for every ,

 US:={ekj∈\catE|ekj⋅v≠0 for some v∈S}.

### 3.2. Signature of pretzel knots

Let be a pretzel knot and let be its double branched cover described as the boundary of a –dimensional plumbing manifold. Since is a knot the determinant of the intersection form is odd and the equation

 QΓ(w,x)≡QΓ(x,x)(mod 2)∀x∈H2(MΓ;\Z)

has exactly one solution in . This solution admits a unique integral lift such that its coordinates are or in the natural basis of given by the vertices of the graph . The homology class is called the Wu class. There is a well defined subset such that

 w=∑j∈Jvj∈H2(MΓ;\Z)

and we define the Wu set as . In order to calculate the Wu set for a given plumbing graph with odd determinant, we apply the algorithm described in [b:NR, Theorem 7.1]: start by reducing the graph to a collection (possibly empty) of isolated points with odd weights by a sequence of moves of type and below. Consider a leaf connected to the vertex .

• Move 1: If the weight on is even, then erase and from .

• Move 2: If the weight of is odd, then erase and change the parity of the weight on .

In order to determine which vertices belong to the Wu–set we undo the sequence of movements starting with the isolated vertices until we reobtain , taking the following into account:

• All the isolated vertices with odd weight obtained in the final step of the reduction of belong to the Wu set.

• If we undo Move , then the vertex does not belong to the Wu set whereas the vertex will belong to the Wu set only if the weight on the vertex and the number of adjacent vertices to which already belong to the Wu set do not have the same parity.

• If we undo Move , then the vertex will belong to the Wu set if and only if does not belong to the Wu set.

In Figure 3.2 the encircled vertices form the Wu–set of the canonical negative plumbing graphs associated to pretzel knots in families (p1), (p2) and (p3). In family (p2) we assume that the only even parameter is , while in family (p3) the only even parameter is .

The following formula, due to Saveliev [b:Sa, Theorem 5], expresses the signature of the pretzel knot as

 σ(P(p1,...,pn))=sign(QΓ)−w⋅w, (3)

where stands for . Notice that the expression equals , where is Neumann’s invariant. It is well known, [b:Ka, Theorem 8.3], that slice knots have vanishing signature. In Section 5, we will find constraints on the parameters defining the Seifert spaces which arise as double branched covers over slice pretzel knots combining equality (3) with Donaldson’s theorem on the intersection form of definite -manifolds.

## 4. Two interesting families of pretzel knots

In this section we study the sliceness of two subfamilies of pretzel knots of type (p3). Both of them stem from the detailed general study on pretzel knots developed in the remaining sections but need specific arguments to show that the knots in these families are not slice. In the first subsection we study the family with odd. We shall prove that the double branched covers of these knots do not bound rational homology balls, in spite of the fact that the obstruction given by the -invariants from Heegaard-Floer homology (see Section 6.1) vanishes. The main tool we use are the Casson-Gordon invariants. The second subsection deals with the family with odd. A major difficulty presented by this family is that the double branched covers of these knots are all integer homology spheres. As further explained in Section 4.2 many of the recent obstructions to knot sliceness defined from Heegard-Floer homology or Khovanov homology vanish for this family while, perhaps surprisingly, the Alexander polynomial is able to detect the non-sliceness of many (perhaps all) the knots in this family.

### 4.1. Casson-Gordon invariants and the family P(a,−a−2,−a−a2+92)

In 1975 Casson and Gordon introduced some knot invariants that allowed to show that not all algebraically slice knots are smoothly slice. The invariants depend on a knot and on the choice of a character defined on the first homology group of the double branched cover of . In [b:CG] two different invariants, denoted and , are defined from the difference of the twisted signatures of some 4-manifolds associated to the couple . We shall only deal with the properties of a specific version of that suits our purposes.

Consider a slice knot with double branched cover . Let be a prime, and be a character of order . Suppose that the covering induced by the character satisfies . Let be the double cover of branched over a slicing disc for and be the kernel of the map induced by the inclusion.

{thm}

[Casson-Gordon] With the above assumptions, for every character of prime power order vanishing on we have .

We shall compute the Casson-Gordon invariants of some of the pretzel knots in family (p3) via the formula given in [b:CiF, Theorem 6.7]. This formula computes from a surgery presentation of regarded as a colored link and the term stands for the colored signature of .

{thm}

[Cimasoni-Florens] Let be the 3-manifold obtained by surgery on a framed link with components and linking matrix . Let be the character mapping the meridian of the -th component of to with and coprime to . Consider as an -colored link and set . Then,

 σ(K,χ)=σL(ω)−∑i

In order to use this formula to compute the Casson-Gordon invariants for the pretzel knots , odd, we will use the surgery presentation of the Seifert space given by the framed link in Figure 4.1. It is obtained from the diagram associated to the canonical negative plumbing tree by blowing down the central vertex and subsequently blowing down every new -framed unknot.

From this surgery presentation of we can read the following presentation of its first homology group:

 H1(Ya;\Z)=⟨μ1,μ2|−a2+92μ1+aμ2,aμ1−2μ2⟩=⟨μ1,μ2|a+92μ1−μ2,9μ1⟩.

It follows that with being a generator of the group and . We now proceed to define a character on vanishing on the subgroup , which is cyclic of order generated by (cf. Remark LABEL:r:order). The character such that , where is a generator of , has the desired properties.

We have all the necessary ingredients to compute via the formula (4). Notice however that if (mod 3) then the above setting yields which is not a valid value for in the hypothesis of Theorem 4.1. Therefore we shall first compute for (mod 3) and later we will use yet another surgery presentation of to deal with this remaining case.

The linking matrix for the link in Figure 4.1 is given by

 Qa=\cbra−a2+92aa−2

and formula (4) yields

 σ(Ka,χ)= σLa(1,n2)−a+2+29(−a2+922+a2n2+a(3−n2)−2(3−n2)n2)) = σLa(1,n2)−a+2+29(−a2−13+a(n2+3)), (5)

where has two possible values, and , depending on whether (mod 3) or (mod 3).

Our aim is to show that the knots are not slice and to do so it suffices to estimate in (5) and show it is greater than 1. Indeed, all the assumptions of Theorem 4.1 are satisfied since is cyclic and the character is of order , which implies that the first rational homology of the induced covering vanishes [b:CG2, Lemma 4.4]. The term is the coloured signature of the link and it is defined as the signature of a hermitian matrix where is the rank of the first homology group of a -complex for (see [b:CiF] for the definitions and details). The -complex for depicted in Figure 4.1 allows us to determine .

Since the term is negative for all and we have the following estimate

 |σ(Ka,χ)|≥29(a2+13−a(n2+3))−1>1for all a>1 odd, a≢0(mod 3).

This last inequality shows that the knots with (mod 3) are not slice.

To deal with the remaining case, the knots with (mod 3), we shall use the surgery presentation of given in Figure 4.1.

It is obtained from the diagram in Figure 4.1 by sliding the left handle over the handle with framing . The new framings are obtained applying the rules of Kirby calculus (see [b:GS] for details). This surgery diagram yields the following presentation of the first homology group of :

 H1(Ya;\Z)=⟨μ1,μ2|−(a+2)2+92μ1−(a+2)μ2,−(a+2)μ1−2μ2⟩=⟨μ1,μ2|a−72μ1+μ2,9μ1⟩.

As before is generated by and this time we have . We choose the character defined by which satisfies the assumptions of Theorem 4.1 and since we can use the formula (4) to compute . This time the link of the surgery presentation in Figure 4.1 has linking matrix

 Qa=\cbra−(a+2)2+92−a−2−a−2−2

 σ(Ka,χ)= σ~La(1,2)+a+2+2+29(−(a+2)2+922−(a+2)4−(a+2)−4)= σ~La(1,2)+a+4−29(27+9a+a2)).

Similar arguments to the ones used before allow us to estimate

 |σ(Ka,χ)|≥−2a−5+6+2a+2a29=2a29+1>1

and therefore by Theorem 4.1 we have the following statement.

{thm}

For all odd the knots in the family are not slice.

{rem}

In fact we have shown something stronger than the non sliceness of the pretzel knots in the family : the arguments in the proof of Theorem 4.1 imply that the Seifert spaces do not bound rational homology balls.

### 4.2. Alexander polynomials and the family P(a,−a−2,−(a+1)22)

This section is devoted to the study of the sliceness of the pretzel knots of the form with odd. All the knots in this family have determinant and therefore the double branched covers are integer homology spheres. It follows that the Casson-Gordon invariants cannot be used to study the existence of rational homology balls bounded by the Seifert manifolds . We will pursue the study of the sliceness of the knots leaving open the question of the existence of rational homology balls bounded by their double branched covers.

There are several well-known obstructions to sliceness that we have computed for the knots of the form but all of them vanished. In the sequel we shall not use any of the following facts but we have decided to include them for completeness. In the remaining sections we will show that each knot has vanishing signature, its determinant is a square and Donaldson’s theorem does not obstruct sliceness. Moreover, we have checked that the only -invariant of their double branched covers, which are homology spheres, vanishes. The hat version of the knot Floer homology of pretzel knots is known. The family lies within the hypothesis of [b:Ef, Theorem 2] which combined with the Alexander polynomials computed below suffices to determine that the Ozsváth-Szabó invariant is zero for all the knots in this family. Moreover, the Rasmussen -invariant of the knot is known to be zero and a crossing change argument implies that the first knot in our family, namely the knot , also has vanishing Rasmussen invariant.

Given all this vanishing of obstructions one might be tempted to think that the knots in the family are actually slice. However, we conjecture that this is not the case for any parameter and we will show that indeed for (mod 60) the knot does not bound a disk embedded in the -ball. The invariant capable of detecting the non-sliceness of these knots is the Alexander polynomial, in its most classical version. Perhaps it is surprising that one of the first invariants used in the slice problem is still capable of distinguishing subtleties that more modern invariants fail to see.

We have postponed the tedious computation of the Alexander polynomials of the knots to Appendix A. In it we show that it holds

 Δa(t) ≐ta+2+1t+1ta+1t+1−(a+1)24ta−1(t−1)2 =∏d|a+2d≠1Φd(−t)∏δ|aδ≠1Φδ(−t)−(a+1)24ta−1(t−1)2,

where stands for the -th cyclotomic polynomial.

If the knots in family were slice then it would follow, by Fox and Milnor’s theorem [b:FM], that for some polynomial . Our goal is to show that this is not the case and to this end several strategies are possible. If we showed that the polynomials are irreducible in , we would be done. Stepan Orevkov has kindly taken a look at this problem and informed us that he checked with the computer and up to the polynomials are in fact irreducible. However we have not found a proof to show the irreducibility of every polynomial in our family. Another possible approach is to look at these polynomials modulo for some prime and study their irreducibility in . However, by [b:AV, Theorem 12] the number of irreducible factors of (mod ) is even for every odd , which implies that many (perhaps all) the polynomials are irreducible over the integers but reducible over for every .

We are not able to show that none of the polynomials satisfies the Fox-Milnor factorisation but there is a lot of evidence pointing that this might be the case. Taking advantage of the rich literature on the reducibility of cyclotomic polynomials, we will prove the following statement.

{thm}

For (mod 60) the polynomials do not have a Fox-Milnor factorisation.

{cor}

For (mod 60) the pretzel knots are not slice.

###### Proof of Theorem 4.2.

Since the Alexander polynomial is a self reciprocal polynomial, i.e. , its irreducible factors are all self reciprocal or come in reciprocal pairs, that is if and is not self reciprocal then is also a factor of . Suppose that there exists a polynomial such that then we have that the irreducible self reciprocal factors of all have even multiplicity. The idea of the proof is to show that the reduction mod , for suitable primes , of has an odd number of self reciprocal irreducible factors. We start with a prime number dividing and consider the polynomial

 ¯¯¯¯¯Δpa(t):=Δa(t) (mod p)=∏d|a+2d≠1¯¯¯¯Φpd(−t)∏δ|aδ≠1¯¯¯¯Φpδ(−t). (6)

Notice that since gcdgcdgcd all the irreducible factors of the polynomials and in appearing in (6) have multiplicity one and are all distinct. Since cyclotomic polynomials are self reciprocal, if satisfies the Fox-Milnor condition, it follows that every cyclotomic polynomial in (6) has an even number of irreducible factors. It is well known that the number of irreducible factors of equals the quotient between and the order of mod , where is Euler’s totient function. Let us call this quotient . It follows that if has a Fox-Milnor factorisation then for every dividing and every , , and every , , the numbers and are even. The rest of the proof will consist of suitable choices of and that force or to be odd.

Let us consider the odd parameter modulo 12. If (mod 12) then divides either or and is even. We obtain . Moreover, if (mod 12) then divides and we choose dividing such that (mod 3). This choice is always possible since under these assumptions , , and (mod 3). Again in this case we obtain . It follows that for (mod 12) the polynomial does not have a Fox-Milnor factorisation.

We can push further this same argument to study the cases (mod 12) with less success. If for some then is even and we choose . The difficulty comes in the choice of , since in this case there is not a common divisor for all or . Adding the hypothesis (mod 5) we obtain that divides or and . Similarly, if and (mod 5) then again divides either or . We choose as dividing a prime such that (mod 5). This choice is always possible since under the current assumptions (mod 5). In this case it follows that