Pretzel knots up to nine crossings

R. Dı́az and P. M. G. Manchón ∗

November 20, 2020

The first author dedicates this paper to her father,who helped her with showing that the knot 77 is pretzel.

Abstract

There are infinitely many pretzel links with the same Alexander poly-nomial (actually with trivial Alexander polynomial). By contrast, in thisnote we revisit the Jones polynomial of pretzel links and prove that, givena natural number S, there is only a finite number of pretzel links whoseJones polynomials have span S.

More concretely, we provide an algorithm useful for deciding whetheror not a given knot is pretzel. As an application we identify all the pretzelknots up to nine crossings, proving in particular that 812 is the first non-pretzel knot.

Keywords: Pretzel link, Kauffman bracket, Jones polynomial, span.

MSC Class: 57M25, 57M27.

1 Introduction

The original motivation of this paper was to complete some tables appearing insome books (and knot atlas) deciding when a knot is or not of type pretzel (seefor example [2]). In a previous note [6] by the second author, it was given a closedformula for the Kauffman bracket of any pretzel diagram P = P (a1, . . . , an) (arecurrence formula was given in [4]), and based on this formula, the span of theJones polynomial of the pretzel link represented by P in terms of its entriesa1, . . . , an (see Theorem 1 and Theorem 9). In this paper, which can be seen as

∗The first author is partially supported by Project MTM2017-89420-P. The second authoris partially supported by MEC-FEDER grant MTM2016-76453-C2-1-P.

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a natural continuation of [6], we prove the following result (Theorem 14): givenan integer S, there is a finite number of pretzel links whose Jones polynomialshave span S. Moreover, the complete list of the pretzel links with span S canbe provided by using Theorem 9.

A by-product of our work is to provide the complete list of pretzel knots upto nine crossings (Theorems17 and 18). In particular, we discover that 812 isthe first non-pretzel knot in tables, and 942 is the first non alternating and nonpretzel knot.

The list of items in the statement of Theorem 9 is long and a bit cumbersome,and in this paper we fix a small exception that was missing in [6]. For this reasonand by completeness, we rewrite in an appendix the proof of Theorem 9, takingspecial care of item 5.

Pretzel links (and rational links) are special types of Montesinos links, andthere have been some important works trying to classify Montesinos links up tomutation and 5-move (see for example [3] or [7]). Sub-products of these workshave been certain (partial) tests for trying to decide whether or not a link isof Montesinos type, and the same for pretzel links. For example, if the test in[7] says no, the link is not a Montesinos link; if it says yes, the link could beof type Montesinos, and if so, would be obtained by applying mutations and 5-moves to an specific (representative of an equivalence class of) Montesinos link.Our contribution to this subject goes in another direction. When we face anarbitrary link L, our algorithm finds a (usually large but) finite list of pretzellinks such that, if L is pretzel, must be one of the list, the key points beingTheorems 14 and 9.

The paper is organized as follows: in Section 2 we recall the basic definitionsand the closed formula for the Kauffman bracket of pretzel links. In Section 3we state Theorem 9 and discuss its items with the help of some pictures. InSection 4 we prove Theorem 14 and precise an upper bound for the number ofpretzel links with a given span. Section 5 uses carefully the previous theoremsfor deciding which knots up to nine crossings are pretzel. The revisited proof ofTheorem 9 is left to the appendix.

2 Kauffman bracket of pretzel links

Given integers a1, ..., an, denote by P (a1, ..., an) the pretzel link diagram shownin Figure 1. Here ai indicates |ai| crossings, with signs ai/|ai| if ai 6= 0.

A pretzel link is a link that has a pretzel diagram.

For a link diagram D we denote by 〈D〉 its Kauffman bracket with normalization〈 〉 = δ = −A−2 − A2 (see [5]). Recall that 〈D〉 is a regular isotopy invariant

2

a1 a2 an

ai > 0 ai < 0

|ai|crossings

Figure 1: Pretzel link diagram P (a1, ..., an).

of diagrams, defined by the following additional relations:

(i) 〈 〉 = A〈 〉+A−1〈 〉, (ii) 〈D t 〉 = δ〈D〉.

Here is the diagram of the unknot with no crossings. In (i) the formula refersto three link diagrams that are exactly the same except near a point where theydiffer in the way indicated. In (ii) D t is a diagram consisting of the diagramD together with an extra closed curve that contains no crossings at all, notwith itself nor with D. From these relations we can deduce the effect on 〈D〉 ofa type I Reidemeister move on D:

(iii) 〈 〉 = −A3〈 〉, (iii’) 〈 〉 = −A−3〈 〉.

Theorem 1. [6] The Kauffman bracket of the pretzel link diagram P (a1, ..., an)is given by the formula

〈P (a1, . . . , an)〉 =n∏i=1

(Aaiδ + [ai]) + (δ2 − 1)

n∏i=1

[ai].

Here [0] = 0, and for any integer a 6= 0,

[a] = A−2a|a|−3a

|a|∑i=1

(−1)a+iA4ia|a| .

Note that [a] is a Laurent polynomial in the variable A. Its behaviour remindsthe quantum integers. Clearly [1] = A−1, [−1] = A, and we have the recurrenceformulas

[a] = A[a− 1] +A−1(−A−3)a−1 if a > 0,[a] = A−1[a+ 1] +A(−A3)−a−1 if a < 0.

In addition we have the following equalities (proofs can be found in [6]):

Lemma 2. δ[a] = −Aa + (−A−3)a.

Lemma 3. If the number of entries is greater than one, then

〈P (..., ai−1, a, ai+1, ...)〉 = Aa〈P (..., ai−1, 0, ai+1, ...)〉+ [a]〈P (..., ai−1, ai+1, ...)〉.

In particular 〈P (..., a, 0)〉 = (Aaδ + [a])〈P (..., 0)〉.

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Remark 4. 〈P (a)〉 = δ(−A−3)a since P (a) is the trivial knot diagram with |a|kinks, negative kinks if a > 0, positive kinks if a < 0.

3 Span of the Jones polynomial of pretzel links

If L is an oriented link, we denote by V (L) its Jones polynomial with normaliza-tion V ( ) = −t−1/2 − t1/2 (see [5]). Recall that V (L) = (−A)−3w(D)〈D〉 afterthe substitution A = t−1/4, where D is an oriented diagram of L and w(D) isits writhe. It follows that span(〈D〉) = 4 span(V (L)).

Remark 5. Denote by 〈D〉1 the Kauffman bracket of the diagram D definedthrough the same relations (i) and (ii) but with normalization 〈 〉1 = 1. Itfollows that 〈D〉 = δ〈D〉1 for every link diagram D. In parallel, denote by V1(L)the Jones polynomial of the oriented link L with normalization V1( ) = 1. Recallthat V1(L) = (−A)−3w(D)〈D〉1 after the substitution A = t−1/4, where D is anoriented diagram of L and w(D) is its writhe. It follows that V (L) = (−t−1/2−t1/2)V1(L) for every oriented link L, and hence span(V (L)) = 1 + span(V1(L)).

Notation 6. For a pretzel link diagram P (a1, . . . , an), we write r for the numberof ai > 1, s for the number of ai < −1, z for the number of ai = 0, α for thenumber of ai = 1 and β for the number of ai = −1. We also set λ = α − β.Finally, let Σ = Σ(P (a1, . . . , an)) =

∑|ai|>1 |ai|.

Remark 7. Recall that:

1. For any permutation σ of {1, . . . , n} the pretzel link defined by the di-agram P (aσ(1), . . . , aσ(n)) can be obtained applying a finite sequence ofmutations to the link defined by P (a1, . . . , an), and thus their Jones poly-nomials agree. That the Kauffman bracket of the corresponding diagramsagree follows directly from the formula in Theorem 1.

2. If L̄ is the mirror image of the oriented link L, then V (L̄) is obtainedfrom L by interchanging t and t−1, so both polynomials share span. Inparticular the span of the Jones polynomials of the pretzel links defined bythe diagrams P (a1, . . . , an) and P (−a1, . . . ,−an) is the same. Notice alsothat the symmetry ai 7→ −ai interchanges r and s and takes λ into −λ.

3. Any consecutive pair (1,−1) in (a1, . . . , an) can be canceled via a type IIReidemeister move without changing the link type. Moreover, after a π-rotation, any pair (1,−1) can be canceled, even if they are non-consecutive(see Figure 2). If there are only entries ±1, and in the same amount, thelink is just the split union of two trivial knots, , with Jones polynomial(−t−1/2 − t1/2)2 and span 2.

4. Similarly, if some entry ai is equal to 0, then any entry −1 or +1 can bedeleted by a π-rotation without changing the link type.

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ai ai+q ai ai+q

Figure 2: Canceling a pair (1,−1), consecutive or not

5. For n ≥ 2, the pretzel diagrams P (1,−2, a3, . . . , an) and P (2, a3, . . . , an)define the same pretzel link. Indeed, this can be achieved by a combinationof a type III plus a type I Reidemeister moves (see Figure 3).

R3 R1

Figure 3: P (1,−2, a3, . . . , an) is isotopic to P (2, a3, . . . , an)

Based on Remarks 7.3 and 7.4 we introduce the following definition:

Definition 8. We say that a pretzel diagram P is reduced if it is P = P (1,−1)or satisfies the following two conditions:

(i) αβ = 0, i.e., it does not have simultaneously entries equal to +1 and −1,and

(ii) if z 6= 0, then α = β = 0.

Theorem 9. Let P (a1, . . . , an) be an unoriented, reduced pretzel diagram of anoriented pretzel link L, with r, s, z, λ and Σ as in Notation 6. Let S be thespan of the Jones polynomial V (L) of L. Since we are interested in the calculusof this span, we can assume that a1 ≥ · · · ≥ an by Remark 7.1. Then:

1. S = Σ + z if z > 0.

In the remaining cases we assume z = 0.

2. S = Σ −min{1, r + λ, s − λ} + 1 if r + λ 6= 1 and s − λ 6= 1, except thecase P (1,−1) for which S = 2.

In the remaining cases we consider z = 0 and r + λ = 1 (taking the mirrorimage the case s− λ = 1 can be reduced to this one by Remark 7.2).

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3. In this item we assume r > 1.

3.1. S = Σ− 1 if (r, λ, s) 6= (2,−1, 0).

In the rest of the cases in this item we assume that (r, λ, s) = (2,−1, 0).

3.2. S = Σ− 2 = 2 if a2 = 2 and a1 = 2.3.3. S = Σ− 4 = 1 if a2 = 2 and a1 = 3.3.4. S = Σ− 3 = a1 − 1 if a2 = 2 and a1 > 3.3.5. S = Σ− 2 = a1 + a2 − 2 if a2 > 2.

4. In this item we assume r = 1 and s ≤ 1 (so a1 > 1 and, if s = 1, −1 > a2).

4.1. S = 1 = Σ− a1 + 1 if s = 0.4.2. S = 2 if s = 1 and |a1 + a2| = 0.4.3. S = 1 if s = 1 and |a1 + a2| = 1.4.4. S = 1 + |a1 + a2| if s = 1 and |a1 + a2| > 1.

5. In this item we assume r = 1 and s > 1 (so a1 > 1 and −1 > a2 ≥ a3 ≥. . . ).

5.1. S = Σ−min{a1, |a2| − 1} if a1 6= |a2| − 1.5.2. S = Σ−min{|a2|, |a3| − 1} if a1 = |a2| − 1 and |a2| 6= |a3| − 1.5.3. S = 2a1 if s = 2, a1 = |a2| − 1 and |a2| = |a3| − 1, except the case

P (2,−3,−4), for which S = 3.5.4. S = Σ − a1 − 3 if a1 = |a2| − 1, |a2| = |a3| − 1 and |a3| < |a4| − 1,

except the cases P (2,−3,−4, a4) with a4 < −6, for which S = Σ− 6.5.5. S = Σ− a1 − 2 if a1 = |a2| − 1, |a2| = |a3| − 1 and |a3| = |a4| − 1.5.6. S = Σ− a1 − 1 if a1 = |a2| − 1, |a2| = |a3| − 1 and |a3| = |a4|.

6. In this item we assume r = 0 and a2 6= −2 (so a1 = 1, a2 < −2 or s = 0).

6.1. S = Σ + 1 = 1 if s = 0.

6.2. S = Σ = −a2 if s = 1.6.3. S = Σ− 2 if s = 2.6.4. S = Σ− 1 if s > 2.

7. In this item we assume r = 0 and a2 = −2 (so a1 = 1 and a2 = −2).

7.1. S = Σ− 1 = 1 if s = 1.7.2. S = Σ− 2 = 2 if s = 2 and a3 = −2.7.3. S = Σ− 4 = 1 if s = 2 and a3 = −3.7.4. S = Σ− 3 = −1− a3 if s = 2 and a3 < −3.7.5. S = Σ− 1 = 3− a4 if s = 3 and a3 = −2.

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7.6. S = Σ− 2 = 6 if s = 3, a3 = −3 and a4 = −3.7.7. S = Σ− 6 = 3 if s = 3, a3 = −3 and a4 = −4.7.8. S = Σ− 3 = 2− a4 if s = 3, a3 = −3 and a4 < −4.7.9. S = Σ− 2 = −a3 − a4 if s = 3 and a3 < −3.

7.10. S = Σ− 1 if s > 3 and a3 = −2.7.11. S = Σ− 2 if s > 3, a3 = −3 and a4 = −3.7.12. S = Σ− 3 if s > 3, a3 = −3, a4 = −4 and a5 = −4.7.13. S = Σ− 4 if s > 3, a3 = −3, a4 = −4 and a5 = −5.7.14. S = Σ − 5 if s > 3, a3 = −3, a4 = −4 and a5 < −5, except for the

pretzel P (1,−2,−3,−4, a5) with a5 < −6, for which S = Σ− 6.7.15. S = Σ− 3 if s > 3, a3 = −3 and a4 < −4.7.16. S = Σ− 2 if s > 3 and a3 < −3.

Remark 10. The exception stated in item 5.4 of the previous theorem amend-ments the original statement of Theorem 2 in [6]. The proof of this theorem,including this correction and more details of item 5 is given in a final appendix.

Remark 11. Item 1 of Theorem 9 analyzes when there is an entry equal tozero. The rest of cases assume that there are no entries equal to zero. Item 2considers the generic case: (r, λ, s) is neither in the plane r + λ = 1 nor ins− λ = 1. The symmetry ai 7→ −ai interchanges r and s and takes λ into −λ,so (r, λ, s) 7→ (s,−λ, r). It follows that it is enough to give a formula for thespan assuming that the point (r, λ, s) is in the plane r + λ = 1 (light blue planein Figure 4). We then distinguish cases r > 1, r = 1 and r = 0. We deal withcase r > 1, except for the point (r, λ, s) = (2,−1, 0), in 3.1. Note that r+λ = 1,r > 1 and s − λ = 1 implies (r, λ, s) = (2,−1, 0). By symmetry this case canbe studied as the case (r, λ, s) = (0, 1, 2). We deal with case r = 1 and s ≤ 1in 4. The difficult case, when r = 1 and s > 1, is treated in 5. Finally items6 and 7 consider the cases when r = 0, depending on whether or not a2 = −2(see Figure 5). Note that item 7 can be directly deduced from items 4 and 5 byRemark 7.5, but it has been included in the statement for practical computationpurposes.

4 There are finitely many pretzel links with afixed span of the Jones polynomial

Suppose that D is a connected diagram with n crossings of an oriented link Lwith Jones polynomial V (L). Recall that span(V (L)) ≤ n+ 1 (see [5]).

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Plane r + λ = 1

Planes− λ = 1

Symmetry planer + 2λ− s = 0

r

λ

s

(0, 1, 0)

(0, 1, 1)

(0, 1, 2)

(1, 0, 0)

(1, 0, 1)

(2,−1, 0)

Figure 4: Cases in Theorem 9 according to the values of (r, λ, s)

Theorem 12. Let P (a1, . . . , an) be a pretzel diagram of a pretzel link L. LetV (L) be the Jones polynomial of L, where L has been arbitrarily oriented. Sup-pose that L is not a torus link T (2, n) with two strands. Then

span (V (L)) ≥∑|ai|>1

|ai| −min{|ai| / |ai| > 1} − 4.

Proof. Let Σ =∑|ai|>1 |ai| and M = min{|ai|/|ai| > 1} if r+s > 0. If r = s = 0

let Σ = M = 0. Let S = span(V (L)). We want to prove that S ≥ Σ−M − 4.

Clearly we may assume that the pretzel diagram is reduced (note that if P =P (1,−1), then S = 2 and Σ − M − 4 = −4). We may also assume thata1 ≥ a2 ≥ · · · ≥ an by Remark 7.1, hence the pretzel diagram P (a1, . . . , an) isunder the hypothesis of Theorem 9.

If z > 0 then S = Σ + z > Σ > Σ−M − 4 by Theorem 9.1, since M ≥ 0. Hence

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r

s

Item 6.4a2 6= −2

Expand andcalculate

Item 6.3a2 6= −2

Item 7.1a2 = −2

Items 7.5-16a2 = −2

Items 7.2-4a2 = −2

Items 4.2-4

Item 6.2a2 6= −2

Item 1z > 0

Item 5

Item 3.1

Item 3.2

Item 6.1 Item 4.1 s− λ = 1

Figure 5: Plane r + λ = 1 coordinated by (r, s). An oval means that thecorresponding knot is the unknot

we may assume that ai 6= 0 for any i = 1, . . . , n.

Assume that r + λ 6= 1 and s− λ 6= 1. Then by Theorem 9.2

S = Σ−min{1, r + λ, s− λ}+ 1 ≥ Σ− 1 + 1 = Σ > Σ−M − 4.

Suppose that s−λ = 1. Consider then the pretzel diagram P ′ = P (−a1, . . . ,−an).On one hand, the values for Σ, M and the bound Σ −M − 4 are unchanged,since Σ and M are defined in terms of absolute values. On the other hand,the span of the new pretzel link is the same, by Remark 7.2. And for this newpretzel diagram P ′ we have that r′ + λ′ = s + (−λ) = 1. In other words, inorder to conclude the proof we may assume that r + λ = 1 (and z = 0).

• Suppose that r > 1. Then, by Theorem 9.3 we have S ≥ Σ− 4 ≥ Σ−M − 4.

• Suppose that r = 1, hence λ = 0. Recall that we have already deleted thepairs +1,−1, and in addition we have a1 ≥ 2, ai ≤ −2 for i = 2, . . . , n and|a2| ≤ |a3| ≤ . . . |an|.

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If s = 0 then S = 1 = Σ− a1 + 1 = Σ−M + 1 > Σ−M − 4 by Theorem 9.4.1.

If s > 1, we apply Theorem 9.5:

1. If a1 6= |a2| − 1 then by Theorem 9.5.1

S = Σ−min{a1, |a2| − 1} ≥ Σ−min{a1, |a2|} = Σ−M > Σ−M − 4.

In the following cases a1 = |a2| − 1 hence M = a1. We have:

2. If a1 = |a2| − 1 and |a2| 6= |a3| − 1 then by Theorem 9.5.2

S = Σ−min{|a2|, |a3| − 1} ≥ Σ− (M + 1) > Σ−M − 4.

3. If s = 2, a1 = |a2| − 1 and |a2| = |a3| − 1 but the diagram is notP (2,−3,−4), then

Σ−M − 4 = (a1 − a2 − a3)− a1 − 4 = −a2 − a3 − 4 = 2a1 − 1 < 2a1 = S

and for P (2,−3,−4) the bound is sharp: Σ−M − 4 = 9− 2− 4 = 3 = S.

4. If a1 = |a2| − 1, |a2| = |a3| − 1 and |a3| < |a4| − 1 then by Theorem 9.5.4

S = Σ− a1 − 3 > Σ−M − 4

excepting the cases P (2,−3,−4, a4) with a4 < −6, where S = Σ − 6 =Σ− 2− 4 = Σ−M − 4, again a case for which the bound is sharp.

5. If a1 = |a2| − 1, |a2| = |a3| − 1 and |a3| = |a4| − 1 then by Theorem 9.5.5

S = Σ− a1 − 2 = Σ−M − 2 > Σ−M − 4.

6. If a1 = |a2| − 1, |a2| = |a3| − 1 and |a3| = |a4| then by Theorem 9.5.6

S = Σ− a1 − 1 = Σ−M − 1 > Σ−M − 4.

We do not have to consider the case s = 1 since P (a, b) ∼ P (0, a+b) = T (2, a+b)is a torus link with two strands.

• Finally suppose that r = 0. Then the bound can be directly checked by simpleinspection of items 6 and 7 of Theorem 9, having in mind that M = 2 for item 7(7.7 and the exception in 7.14 are cases of sharp bound).

Remark 13. We must discard torus links T (2, n) with two strands by the fol-lowing: if 5 < a < −b − 1 in P = P (a, b) ∼ P (0, a + b) = T (2, a + b), then onone hand the Jones polynomial of the link has span S = 1 + |a+ b| = 1− a− bby Theorem 9.1 and on the other hand Σ−M − 4 = (a− b)− a− 4 = −b− 4.And 1 − a − b < −b − 4 since 5 < a. This shows in particular that there areinfinitely many pretzel diagrams with the same span. In the example all thesediagrams correspond to the same torus link with two strands. Next we will seethat these are the only exceptions.

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Theorem 14. Given a natural number S, there are finitely many oriented pret-zel links whose Jones polynomials have span S.

Proof. This follows from Theorem 12, items 1 and 2 in Theorem 9 and the factthat the span of V (L) is |q| + 1 if L is the torus link T (2, q) (see Remark 13).Let us see the details of the proof. Let L be an oriented pretzel link, representedby the unoriented pretzel diagram P = P (a1, . . . , an), with span(V (L)) = S.By Remarks 7.3 and 7.4, we may choose this diagram to be reduced. And wemay assume that P is under the hypothesis of Theorem 9, having in mind thatthere are n! possible reorderings that could give different pretzel links, althoughall of them would have the same Jones polynomial, hence the same span.

If L is a torus link with two strands, it is necessarily T (2, S − 1) or its mirrorimage, only two possibilities. Next suppose that L is not a torus link with twostrands. Then by Theorem 12, and with its notation,

S = span(V (L)) ≥ Σ−M − 4 ≥ 2(r + s− 1)− 4 = 2(r + s− 3)

from which we deduce that r + s ≤ S2 + 3. Also, by Theorem 9.1, z ≤ S.

Taking into account that r+s has an upper bound, the possible value of λ = α−βhas also lower and upper bounds by Theorem 9.2, assumed that z = 0 (inProposition 15 we will determine specific bounds for λ). This proves that thenumber n of entries has necessarily an upper bound as long as z = 0.

Assume now that r + s > 1 (and z = 0). Then any entry ai satisfy that|ai| ≤ S + 4 since, if for example M = |ak| and |aj | > 1, j 6= k, then

S ≥ Σ−M − 4 =∑|ai|>1

|ai| − |ak| − 4 =∑

|ai|>1,i6=k

|ai| − 4 ≥ |aj | − 4.

And, indeed, |ak| ≤ |aj | ≤ S + 4. This leaves a finite number of possibilities forthe numbers ai with |ai| > 1 and proves the statement if r + s > 1.

Consider now the case r + s = 1 (and z = 0). By symmetry we may assumer = 1, s = 0. If λ 6= −1, 0, then r + λ 6= 1 and s − λ 6= 1, and by Theorem 9.2S = |a| −min{1, r+ λ, s− λ}+ 1 ≥ |a|, hence |a| < S where a is the only entrywith |a| > 1. So, a finite number of possibilities. If λ = −1, P = P (a,−1) is atorus link. If λ = 0, then P = P (a) is the unknot.

Finally assume z > 0. Since P is reduced, α = β = 0. Then by Theorem 9.1 wehave that n ≤ S since

S = Σ + z ≥ 2(n− z) + z = 2n− z ≥ 2n− n = n

and |ai| < S, since |ai| ≤ Σ < S. This completes the proof.

We now collect specific upper bounds in the following result. As usual, weconsider a reduced pretzel diagram P (a1, . . . , an) with n entries representing a

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pretzel link whose Jones polynomial (with normalization as in Section 3) hasspan S, and let z, r, s and λ be integers as in Notation 6.

Proposition 15. We have that

z ≤ S, r + s ≤ S2

+ 3 and |λ| ≤ max{S2

+ 2, S − 1}.

Moreover, |ai| ≤ S + 4 for i = 1, . . . , n, and

n ≤ max{2S + 5, 52S + 2}.

Proof. The bounds for z, r+s and |ai| were obtained in the proof of Theorem 14.We now prove that |λ| ≤ max{S2 + 2, S− 1} (note that the maximum is S− 1 ifS > 5). If λ = −1, 0 or 1 this is obvious. Now assume that λ ≥ 2 (a symmetricargument works if λ ≤ −2). First note that if λ > S2 + 2 then s− λ < 1, since

λ >S

2+ 2 =

S

2+ 3− 1 ≥ r + s− 1 ≥ s− 1.

Moreover, s − λ 6= 1 and r + λ 6= 1 (since λ ≥ 2), hence by Theorem 9.2 wehave S = Σ−min{1, r + λ, s− λ}+ 1 = Σ− (s− λ) + 1 and, since s− Σ ≤ 0,λ = S − 1 + (s− Σ) ≤ S − 1.

We finally check the upper bound for the number n of entries:

n = r + s+ z + |λ|≤ S2 + 3 + S + max{

S2 + 2, S − 1} ≤ max{2S + 5,

52S + 2}.

Looking closer at the proof of Theorem 14, we actually prove the followingcorollary.

Corollary 16. a) Given a natural number S, there are finitely many reducedpretzel diagrams and not representing a torus link with two strands, suchthat the Jones polynomials of the pretzel links represented by these dia-grams have span S.

b) A pretzel link L has a finite number of reduced pretzel diagrams if and onlyif L is not a torus link with two strands.

5 Pretzel knots up to nine crossings

In this section we provide the complete list of pretzel knots up to nine crossings.By abuse of language we will say that a link L has span S if its Jones polynomialV (L), normalized as in Section 3, has span S (note that the span is not affected

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by the orientation of the link); also, a pretzel diagram P = P (a1, . . . , an) hasspan S if the link L defined by P has span S. In the same line we will speakabout V (L) as simply the Jones polynomial of the pretzel diagram P .

Theorem 17. The following table gives the information about pretzel and non-pretzel knots up to 8 crossings. For the pretzel ones, we provide a pretzel dia-gram. In particular, the first non-pretzel knot in tables is the knot 812.

31 = P (1, 1, 1)41 = P (1, 1, 2)51 = P (1, 1, 1, 1, 1)52 = P (1, 1, 3)61 = P (1, 1, 4)62 = P (1, 2, 3)63 = P (2, 1,−3, 1)71 = P (1, 1, 1, 1, 1, 1, 1)72 = P (5, 1, 1)73 = P (1, 1, 1, 4)74 = P (3, 1, 3)75 = P (2, 1, 1, 3)76 = P (2, 1, 1,−3, 1)77 = P (1, 1, 1,−3,−3)

81 = P (1, 1, 6)82 = P (1, 2, 5)83 = P (1, 1, 1, 1, 4)84 = P (1, 3, 4)85 = P (2, 3, 3)86 = P (1, 1, 1, 2, 3)87 = P (4, 1,−3, 1)88 = P (2, 1, 1, 1,−3, 1)89 = P (3, 1,−4, 1)810 = P (−3,−2, 3,−1)811 = P (3, 1, 1,−3, 1)812 is not pretzel813 = P (1, 1, 1,−3,−4)814 is not pretzel815 = P (−3, 1, 2, 1,−3)816 is not pretzel817 is not pretzel818 is not pretzel819 = P (3, 3,−2)820 = P (−3, 2, 3,−1)821 = P (−3,−3, 1, 2)

Some of the above knots were known to be pretzel (see for example tables in [2]or [8]), and other have been recently matched as pretzel in [1]. That the abovelist is now exhaustive can be deduced exactly in the same way as in the proofof Theorem 18 below for knots with nine crossings.

As an example, the Jones polynomial of the non-alternating knot with eightcrossings 821 is V1(821) = 2t−2t2+3t3−3t4+2t5−2t6+t7 hence span(V (821)) =6 + 1 = 7. Then we list all the pretzel links whose Jones polynomial have spanequal to 7 and observe that the Jones polynomial of P (3, 3,−1,−2) coincideswith that of 821. Although there is no such coincidences for small knots, wecheck by hand that both knots are the same (see Figure 6).

Theorem 18. The following table gives the information about pretzel and non-pretzel knots with 9 crossings. For the pretzel ones, we provide a pretzel diagram.In particular the first non-alternating and non-pretzel knot is the knot 942. Infact, all the non-alternating knots with nine crossings are non-pretzel except 946and 948.

13

7→ 7→ 7→

7→ 7→ 7→

Figure 6: The knot 821 is the pretzel knot P (2, 1,−3,−3).

91 = P (1, 1, 1, 1, 1, 1, 1, 1, 1)92 = P (1, 1, 7)93 = P (1,−4, 5)94 = P (1,−5, 4)95 = P (1, 3, 5)96 = P (1, 1,−3,−6)97 = P (−3,−1,−1,−1,−1,−2)98 = P (2, 1, 1, 1, 1,−3, 1)99 = P (−4,−1,−1,−3)910 = P (−3,−1,−1,−1,−3)911 = P (5,−2,−1,−1,−1)912 = P (4, 1, 1,−3, 1)913 = P (3, 1, 1,−4, 1)914 = P (1, 1, 1,−3,−5)915 is not pretzel916 = P (2, 3, 1, 3)917 = P (−3, 1, 1, 1, 1, 1,−3)918 is not pretzel919 is not pretzel920 = P (1, 1, 1, 1,−3,−4)921 is not pretzel922 is not pretzel923 is not pretzel924 = P (3, 1,−3, 1, 2)925 is not pretzel

926 is not pretzel927 is not pretzel928 = P (1, 1,−3, 2, 1,−3)929 is not pretzel930 is not pretzel931 is not pretzel932 is not pretzel933 is not pretzel934 is not pretzel935 = P (3, 3, 3)936 is not pretzel937 = P (1,−3, 3,−3, 1)938 is not pretzel939 is not pretzel940 is not pretzel941 is not pretzel942 is not pretzel943 is not pretzel944 is not pretzel945 is not pretzel946 = P (3, 3,−3)947 is not pretzel948 = P (−3, 1,−3, 1,−3)949 is not pretzel

14

The following definition will be useful in the proof. Fix an integer S. A set LSof pretzel diagrams with span S is said to be complete if any pretzel knot Kwith span S has a diagram which can be obtained from a diagram in LS bymirror image and/or by reordering its entries. Results in Section 4 (preciselyTheorem 14 and Corollary 16) guaranty that there exists a finite complete set LSfor each natural number S. In Lemma 19 we will find a finite complete set LSfor S = 10, which is the key ingredient to prove Theorem 18.

Proof. The Jones polynomial of the alternating knots 91 to 941 have indeedspan S = 10. Let L10 be the complete set of pretzel diagrams with span 10determined in Lemma 19 below. We compute their Jones polynomials by usingTheorem 1 and with the help of Sage. Now, let K be an alternating knot with 9crossings. Let L10(K) be the subset of L10 consisting on those pretzel diagramswith Jones polynomial equal or symmetric to V (K). If L10(K) is empty wecan confirm that K is not pretzel. Otherwise K could be still non-pretzel,but if it is pretzel, must have a diagram that can be obtained from a diagramP = P (a1, . . . , an) ∈ L10(K) by reordering its entries (K would be a mutant ofthe knot with diagram P ) and/or by taking its mirror image. In all the cases inwhich L10(K) was non-empty we directly checked by hand, using Reidemeistermoves, that K was pretzel (since they are in general small knots with the sameJones polynomial, this was quite expected).

For non-alternating knots 942 to 949 we have repeated the same strategy, butin these cases the considered set LS corresponds to a span S < 10 (for S < 10the set LS is even smaller and can be found in the same way as in Lemma 19).This completes the proof.

Lemma 19. The following set L10 of reduced pretzel knot diagrams with spanS = 10 is complete (“Type” indicates the corresponding item of Theorem 9 satis-fied; a pretzel diagram P (a1, . . . , an) will be shorted just by writing (a1, . . . , an)):

• Type 1:

(3, 3, 3, 0), (3, 3, 0,−3), (9, 0).

• Type 2:

15

(7, 2), (7, 1,−2), (2, 1,−7), (1, 1,−2,−7),(3, 3, 3), (3, 1, 1,−3,−3),(6, 3), (6, 1,−3), (3, 1,−6), (1, 1,−3,−6),(5, 4), (5, 1,−4), (4, 1,−5), (1, 1,−4,−5),(3, 3, 2, 1), (3, 3, 1, 1,−2), (3, 2, 1, 1,−3),(3, 1, 1, 1,−2,−3), (2, 1, 1, 1,−3,−3), (1, 1, 1, 1,−2,−3,−3),(5, 3, 1), (1, 1, 1,−3,−5),(8, 1), (1, 1,−8),(5, 2, 1, 1), (5, 1, 1, 1,−2), (2, 1, 1, 1,−5), (1, 1, 1, 1,−2,−5),(4, 3, 1, 1), (4, 1, 1, 1,−3), (3, 1, 1, 1,−4), (1, 1, 1, 1,−3,−4),(7, 1, 1),(3, 3, 1, 1, 1), (1, 1, 1, 1, 1,−3,−3),(6, 1, 1, 1), (1, 1, 1, 1,−6),(3, 2, 1, 1, 1, 1), (3, 1, 1, 1, 1, 1,−2), (2, 1, 1, 1, 1, 1,−3), (1, 1, 1, 1, 1, 1,−2,−3),(5, 1, 1, 1, 1),(4, 1, 1, 1, 1, 1), (1, 1, 1, 1, 1, 1,−4),(3, 1, 1, 1, 1, 1, 1),(2, 1, 1, 1, 1, 1, 1, 1), (1, 1, 1, 1, 1, 1, 1, 1,−2),(1, 1, 1, 1, 1, 1, 1, 1, 1).

• Type 3:(3, 3, 3, 2,−1,−1,−1), (3, 3, 3,−1,−1,−2), (3, 3, 2,−1,−1,−3),(3, 3,−1,−2,−3), (3, 2,−1,−3,−3),(5, 3, 3,−1,−1),(11, 2,−1),(9, 3,−1),(7, 5,−1).

• Type 5:(3,−3,−6), (6,−3,−3),(3,−2,−3,−3),(4,−3,−5), (5,−3,−4),(2,−5,−5), (3,−5,−5),(4,−5,−5),(3,−4,−7),(5,−6,−7).

• Type 6:

(1,−3,−9), (1,−5,−7).

Proof. We say that a pretzel diagram P = P (a1, . . . , an) is of type i.j if itsatisfies the hypothesis of Theorem 9, item i.j. In particular, we will say that Pis generic if it is of type 2; otherwise is called non-generic. Recall that all thesediagrams are reduced pretzel diagrams and in addition a1 ≥ a2 ≥ · · · ≥ an.We will run along all the items of Theorem 9. For each one, i.j, we will findthe pretzel knot diagrams of type i.j with span equal to S = 10, discarding a

16

diagram if it can be obtained by mirror image or reordering the entries of aprevious selected one. Another important remark that will greatly reduce thelist (and our effort) is that we will avoid to take pretzel diagrams with more thanone even entry, since they correspond to links with two or more components.For the same reason, from the final list we will also delete the pretzel diagramswith n even and all entries ai odd. We will use notation r, s, z, λ and Σ as inTheorem 9.

• If P is of type 1 then z > 0 and S = Σ + z. Since we want P to representa knot, necessarily z = 1, hence Σ = 9. On the other hand, since P isreduced, it is λ = 0. Then the decompositions of Σ = 9 as sum of positivenumbers greater than 1 and none of them even (0 is already one evenentry) are 3 + 3 + 3 and 9. We then need to assign different signs to theaddends in order to obtain the different possibilities (in the rest of proofwe will refer to this process as assigning signs), although avoiding to havediagrams that differ just by mirror image and reordering:

(3, 3, 3, 0), (3, 3, 0,−3), (9, 0).

Note that z = 0 in the rest of items in Theorem 9. Also, along the rest of proof,a decomposition of an integer Σ ≥ 2 will mean a non-decreasing sequence ofintegers greater than one, at most one of them even, whose sum is exactly Σ.

• Suppose that P is of type 2, i.e., it is generic. Then r+λ 6= 1, s−λ 6= 1 andS = Σ−min{1, r+λ, s−λ}+ 1 ≥ Σ− 1 + 1 = Σ. As S = 10, the possiblevalues of Σ are 10, 9, 8, 7, 6, 5, 4, 3, 2 and 0. Let m = min{1, r + λ, s− λ}.

– If Σ = 10 then m = 1 and, since r + λ 6= 1 and s − λ 6= 1, it mustbe r + λ ≥ 2 and s− λ ≥ 2, thus r + s ≥ 4. On the other hand, themaximum number of addends in a decomposition of Σ = 10 occursat 10 = 2 + 3 + 5, which means that r+ s ≤ 3, and we conclude thatno P satisfies this case.

When Σ ≤ 9, m < 1 hence m = r + λ or m = s− λ. One can also easilycheck that if P is generic and m = r + λ, then its mirror image is alsogeneric and satisfies that m = s−λ. Thus, for Σ ≤ 9, we can assume thatm = s− λ and therefore r + λ ≥ s− λ.

– If Σ = 9 then m = s−λ = 0. Since r+λ 6= 1 it follows that r+s 6= 1and the admissible decompositions of Σ = 9 are 2 + 7, 3 + 3 + 3, 3 + 6and 4 + 5. Assigning signs to the addends and, since λ = s, addingas many 1’s as negative addends, we obtain

(7, 2), (7, 1,−2), (2, 1,−7), (1, 1,−2,−7)(3, 3, 3), (3, 3, 1,−3), (3, 1, 1,−3,−3), (1, 1, 1,−3,−3,−3)(6, 3), (6, 1,−3), (3, 1,−6), (1, 1,−3,−6),(5, 4), (5, 1,−4), (4, 1,−5), (1, 1,−4,−5).

17

For 0 < Σ ≤ 8 we have m = s−λ ≤ −1, hence λ ≥ s+1, and the relationsr + λ 6= 1, r + λ ≥ −1 are automatic.

– If Σ = 8, then s − λ = −1. Decompositions of Σ = 8 are 2 + 3 + 3,3 + 5 and 8. Assigning signs and adding λ = s+ 1 entries equal to 1,we obtain

(3, 3, 2, 1), (3, 3, 1, 1,−2), (3, 2, 1, 1,−3),(3, 1, 1, 1,−2,−3), (2, 1, 1, 1,−3,−3), (1, 1, 1, 1,−2,−3,−3),(5, 3, 1), (5, 1, 1,−3), (3, 1, 1,−5), (1, 1, 1,−3,−5),(8, 1), (1, 1,−8).

– If Σ = 7 then s− λ = −2. Decompositions of 7 are 2 + 5, 3 + 4 and7. Assigning signs and adding λ = s+ 2 entries equal to 1, we obtain

(5, 2, 1, 1), (5, 1, 1, 1,−2), (2, 1, 1, 1,−5), (1, 1, 1, 1,−2,−5),(4, 3, 1, 1), (4, 1, 1, 1,−3), (3, 1, 1, 1,−4), (1, 1, 1, 1,−3,−4),(7, 1, 1), (1, 1, 1,−7).

– If Σ = 6 then s − λ = −3. Decompositions of 6 are 3 + 3 and 6.Assigning signs and adding λ = s+ 3 entries equal to 1, we obtain

(3, 3, 1, 1, 1), (3, 1, 1, 1, 1,−3), (1, 1, 1, 1, 1,−3,−3),(6, 1, 1, 1), (1, 1, 1, 1,−6).

– If Σ = 5 then s − λ = −4. Decompositions of 5 are 2 + 3 and 5.Assigning signs and adding λ = s+ 4 entries equal to 1, we obtain

(3, 2, 1, 1, 1, 1), (3, 1, 1, 1, 1, 1,−2), (2, 1, 1, 1, 1, 1,−3), (1, 1, 1, 1, 1, 1,−2,−3),(5, 1, 1, 1, 1), (1, 1, 1, 1, 1,−5).

– If Σ = 4 then s−λ = −5. The only decomposition of 4 is 4. Assigningsigns and adding λ = s+ 5 entries equal to 1, we obtain

(4, 1, 1, 1, 1, 1), (1, 1, 1, 1, 1, 1,−4).

– Similarly, for Σ = 3 and Σ = 2 we obtain the lists

(3, 1, 1, 1, 1, 1, 1), (1, 1, 1, 1, 1, 1, 1,−3),(2, 1, 1, 1, 1, 1, 1, 1), (1, 1, 1, 1, 1, 1, 1, 1,−2).

– Finally, for Σ = 0 we have s − λ = −9. In this case r = s = 0 andwe obtain the pretzel diagram

(1, 1, 1, 1, 1, 1, 1, 1, 1).

• If P is of type 3.1 then r + λ = 1, r > 1 and (r, λ, s) 6= (2,−1, 0), thusr + s ≥ 3 and S = Σ− 1. The decompositions of Σ = S + 1 = 11 with at

18

least 3 numbers are 2 + 3 + 3 + 3 and 3 + 3 + 5. Assigning signs takinginto account that r > 1 and adding r − 1 entries equal to −1, we obtain

(3, 3, 3, 2,−1,−1,−1), (3, 3, 3,−1,−1,−2), (3, 3, 2,−1,−1,−3),(3, 3,−1,−2,−3), (3, 2,−1,−3,−3),(5, 3, 3,−1,−1), (5, 3,−1,−3), (3, 3,−1,−5).

Pretzel diagrams of types 3.2 and 3.3 have span equal to 2 and 1, respec-tively, so they do not belong to L10.

• If P is of type 3.4 then (r, λ, s) = (2,−1, 0), a2 = 2, a1 > 3 and S = a1−1.The only diagram is

(11, 2,−1).

• If P is of type 3.5 then (r, λ, s) = (2,−1, 0), a2 > 2 and S = Σ − 2 =a1 + a2 − 2. Decompositions of Σ = 12 with two numbers are 3 + 9 and5 + 7. We obtain

(9, 3,−1), (7, 5,−1).

Pretzel diagrams of type 4 are either trivial or torus links with 2 strands,which have representative pretzel diagram of type 1.

In all diagrams of type 5, r = 1, λ = 0 and s > 1, so they have at least threeentries, only the first one positive, none of them ±1.

• If P is of type 5.1 then a1 6= −a2 − 1 and S = Σ−min{a1,−a2 − 1}. Wedistinguish two possibilities:

If a1 > −a2 − 1 then S = Σ− (−a2 − 1) hence S − 1 = a1 − a3 − . . . Thedecompositions of S − 1 = 9 with at least two numbers are 2 + 7, 3 + 6,3 + 3 + 3 and 4 + 5. Assigning signs (recall that r = 1) and taking intoaccount the restrictions on a2 we obtain

(3,−3,−6), (6,−3,−3),(3,−2,−3,−3),(3,−3,−3,−3),(4,−3,−5), (5,−3,−4).

If a1 < −a2 − 1 then S = Σ − a1, i.e., S = −a2 − a3 − . . . The onlydecomposition of S = 10 with at least two numbers and all of them greaterthan 3 (since −a2 > a1 + 1 ≥ 3) is 5 + 5. Assigning signs (r = 1) andtaking into account the restriction 2 ≤ a1 < −a2 − 1, we obtain

(2,−5,−5), (3,−5,−5).

19

• If P is of type 5.2 then a2 = −a1 − 1, a3 6= a2 − 1(= −a1 − 2) andS = Σ−min{a1 + 1,−a3 − 1}. Again we distinguish two possibilities:If a3 > −a1− 2 then S = Σ− (−a3− 1) hence S − 1 = a1− a2− a4− . . . .The only decomposition of S − 1 = 9 with at least two numbers, and twoof them consecutive (since −a2 = a1 + 1), is 4 + 5. Assigning signs (recallthat r = 1) and having into account the restriction a2 ≥ a3 > a2 − 1, i.e.,a3 = a2, we obtain the only possibility

(4,−5,−5).

If a3 < −a1−2 then S = Σ−(a1+1) = Σ+a2 hence S = a1−a3−. . . . Theonly decomposition of S = 10 with at least two numbers, the differencebetween the two smallest ones greater than 2 (since −a3 > a1+2), is 3+7.Assigning signs and adding a2 = −a1 − 1, we obtain

(3,−4,−7).

• If P is of type 5.3 then P = P (a1,−a1 − 1,−a1 − 2). Exceptional casesin this item have two even entries, so they do not define knots. Since10 = S = 2a1, the only possibility is

(5,−6,−7).

• If P is of type 5.4, P = P (a1,−a1−1,−a1−2, a4, . . . ) with a4 < −a1−3.As before, exceptional cases do not define knots. Notice that a1 must beodd, otherwise there would be two even entries. So a2 is even. In this case,the theorem gives S = Σ− a1 − 3, so S + 3 = −a2 − a3 − a4 − . . . . Thuswe have to decompose S + 3 = 13 as sum of at least 3 positive numbers,a ≤ b ≤ c with a > 3, b = a+ 1 and c ≥ a+ 2. But this is not possible.All diagrams of type 5.5 have two even entries, so they are not in L10.

• If P is of type 5.6 then P = P (a1,−a1 − 1,−a1 − 2,−a1 − 2, . . . ) withspan S = Σ− a1 − 1. Thus S + 1 = −a2 − a3 − a4 − . . . and we have todecompose S + 1 = 11 as sum of at least 3 positive numbers a, b, c witha+ 1 = b = c, a ≥ 3 and b odd. But this is not possible.Pretzel diagrams of types 6.1 and 6.2 are either trivial or torus links withtwo strands, which have representative pretzel diagrams of type 1.

• If P is of type 6.3 then P = P (1, a2, a3) with a2, a3 negative, a2 6= −2,and S = Σ − 2. The decompositions of S + 2 = 12 with two numbersgreater than 2 are 3 + 9 and 5 + 7. We have diagrams

(1,−3,−9), (1,−5,−7).

• If P is of type 6.4 then P = P (1, a2, a3, a4, . . . ) with a2 < −2 and S =Σ− 1. The only decomposition of S + 1 = 11 with at least three numbersgreater than 2 is 3 + 3 + 5, which gives the diagram

(1,−3,−3,−5).

20

Finally, since P (1,−2, a3, . . . , an) and P (2, a2, . . . , an) are isotopic by Re-mark 7.5, and r = 1 and λ = 0 for P (2, a3, . . . , an), any pretzel linkrepresented by a diagram of type 7 has also a diagram of type 4 or type 5.It follows that type 7 diagrams are not necessary in order to fill a completeset L10. This completes the proof.

6 Appendix

Here we detail the proof of Theorem 9.

Proof. Let P = P (a1, . . . , an) and �i = (−1)ai for every i ∈ {1, . . . , n}. Indeed,

spantV (L) =1

4spanA〈P 〉. (1)

By Theorem 1 and Lemma 2

δn〈P 〉 =n∏i=1

A−3ai

(n∏i=1

(�i + (δ2 − 1)A4ai) +(δ2 − 1)

n∏i=1

(�i −A4ai)

).

Let B = A4. Then δ2 − 1 = B−1 + 1 +B and we have that

4n+ spanA〈P 〉 = spanA(δn〈P 〉) = 4 spanB(p(B)) (2)

where

p(B) =

n∏i=1

(�i + (B

−1 + 1 +B)Bai)

+ (B−1 + 1 +B)

n∏i=1

(�i −Bai). (3)

Putting together 1 and 2, we deduce that

spantV (L) = spanB(p(B))− n. (4)

In the rest of the proof we fix our attention in the polynomial p(B) and calculateits span in the variable B. Let F (B) =

∏ni=1(�i+(B

−1+1+B)Bai) and S(B) =(B−1+1+B)

∏ni=1(�i−Bai) be the first and second summands of p(B). Let hF

and lF be respectively the highest and lowest degree of F (B), and let hS and lSbe respectively the highest and lowest degree of S(B). Let h and l be respectivelythe highest and lowest degree of p(B). By definition spanB(p(B)) = h− l, andclearly h = max{hF , hS} if hF 6= hS and l = min{lF , lS} if lF 6= lS . Thestrategy will be then to calculate hF , lF , hS and lS , and whenever hF = hSlook carefully at the possible cancellations of the highest degree terms of the

21

summands F (B) and S(B). Under the hypothesis of the theorem lF and lS willbe found to be different. Suppose that z > 0. Then p(B) = F (B) and

spanB(p(B)) =

n∑i=1

spanB(�i + (B−1 + 1 +B)Bai)

=∑|ai|>1

(|ai|+ 1) +∑|ai|=1

1 +∑|ai|=0

2

=∑|ai|>1

|ai|+ r + s+ α+ β + 2z

=∑|ai|>1

|ai|+ n+ z

and item 1 follows.

Assume now that z = 0. It is easy to see that

hF = r +∑al>1

al + 2α− β, lF = −s+∑aj1

al + α, lS = −1 +∑aj 1hS if r + λ < 1

, l =

lF if s− λ > 1lS if s− λ < 1

and item 2 follows (if r + λ < 1 and s − λ < 1 then r = s = λ = 0 and, sinceP is reduced, P = (1,−1) which represents the split union of two trivial knotswhose Jones polynomial has span two).

We now prove item 3.1. Assume r+λ = 1, r > 1 and (r, λ, s) 6= (2,−1, 0). Firstnote that s− λ > 1 hence l = lF . By Remark 7.1 we may assume that al > 1,l = 1, . . . , r and aj < −1, j = r + 1, . . . , r + s. We have that hF = hS ,

F (B) = �r+1 . . . �r+sBhF + (r + α+ β + d)�r+1 . . . �r+sB

hF−1

+ monomials of degree < hF − 1

and

S(B) = (−1)r�r+1 . . . �r+s(−1)α(−1)βBhS − (α+ β + 1)�r+1 . . . �r+sBhS−1+ monomials of degree < hS − 1

where d is the number of ak = −2 (note that �k = +1 if ak = −2). Sincer + λ = 1 the first summands cancel in p(B). Since r > 1, (r − 1 +d)�r+1 · · · �r+sBhF−1 is the highest degree term of p(B). Hence h = hF − 1and 3.1 follows.

22

The exceptional cases 3.2 to 3.5, when (r, λ, s) = (2,−1, 0), will be proved laterby using items 6 and 7.

In the rest of the proof we will write P ∼ P ′ if both diagrams P and P ′ definethe same link.

We now prove item 4. Item 4.1 corresponds to the trivial knot P (a1), withspan 1. Since P (a, b) ∼ P (0, a+ b), in the item 4.3 we have also the trivial knotwith span one, and for items 4.2 and 4.4 we may apply item 1.

Item 5 is the difficult one, and we leave it for last. We now prove item 6:

6.1. It is P (1), the trivial knot, with S = 1.

6.2. Since P (1, a2) ∼ P (0, 1 + a2), then S = 1 + |1 + a2| = 1− (1 + a2) = −a2by item 1 (note that |1 + a2| > 1 since a2 < −2).

6.3. The diagram is then P (1, a2, a3). By expanding p(B) we find that spanB(p(B)) =h− l = 1− (a2 + a3).

6.4. It is the case P (1, b1, . . . , bs) with s > 2 and bj < −2, j = 1, . . . , s. Then

p(B) = (B +B2)∏sj=1(B

bj−1 +Bbj +Bbj+1 + �j)

+(B−1 + 1 +B)(−1−B)∏sj=1(−Bbj + �j)

has span s− b1 − . . .− bs.

Remark 7.5 is used constantly along the proof of item 7. Items 7.1 to 7.4 arederived from items 4.1 to 4.4 respectively. Items 7.5 to 7.15 are derived fromitem 5.

7.1. P (1,−2) ∼ P (2), the trivial knot, S = 1.

7.2. P (1,−2,−2) ∼ P (2,−2). Then S = 2 by item 4.2.

7.3. P (1,−2,−3) ∼ P (2,−3). Then S = 1 by item 4.3.

7.4. P (1,−2, a3) ∼ P (2, a3) with a3 < −3. Then S = 1+|2+a3| = 1−(2+a3) =−1− a3 by item 4.4.

7.5. P (1,−2,−2, a4) ∼ P (2,−2, a4) and we use item 5.1.

7.6. P (1,−2,−3,−3) ∼ P (2,−3,−3) and we use item 5.2.

7.7. P (1,−2,−3,−4) ∼ P (2,−3,−4) and we use the exceptional case of item5.3.

7.8. By item 5.2.

7.9. By item 5.1.

23

7.10. By item 5.1.

7.11. By item 5.2.

7.12. By item 5.6.

7.13. By item 5.5.

7.14. By item 5.4.

7.15. By item 5.2.

7.16. By item 5.1.

Exceptional cases of item 3 correspond to pretzel diagrams P (a1, a2,−1) witha1 ≥ a2 ≥ 2. After taking the mirror image and reordering the entries we obtainP (1, b2, b3) = P (1,−a2,−a1) with −2 ≥ −a2 ≥ −a1, hence (r′, λ′, s′) = (0, 1, 2).

3.2 By item 7.2 since if a2 = 2 and a1 = 2 then b2 = b3 = −2.

3.3 By item 7.3 since if a2 = 2 and a1 = 3 then b2 = −2 and b3 = −3.

3.4 By item 7.4 since if a2 = 2 and a1 > 3 then b2 = −2 and b3 < −3.

3.5 By item 6.3 since if a2 > 2 then b2 6= −2.

Finally we concentrate in the difficult cases, item 5. Since r+ λ = 1, r = 1 ands > 1 it follows that s − λ > 1 and l = lF . But as in item 3.1, hF = hS andthe terms with this degree cancel. Moreover, other previous terms cancel too.The point is how many steps we have to go down in order to find the first nocancellation. This require some laborious calculations, that are shown carefullyafterwards (for the item 5.4 the case a1 = 2 must be considered separately). Inthis process more and more inner coefficients have to be considered, having theimpression that the process has not an end. But it has!

Recall that �i = (−1)ai for every i ∈ {1, . . . , n}, a1 > 1, aj < −1 if j ∈{2, . . . , n} and |a2| ≤ . . . ≤ |an|. Also spantV (L) = spanB(p(B)) − n wherep(B) = F (B) + S(B). It will be convenient to write both polynomials F (B)and S(B) in such a way that the degree of the monomials in each factor increaseswhen going right:

F (B) = (�1 +Ba1−1 +Ba1 +Ba1+1)

n∏j=2

(Baj−1 +Baj +Baj+1 + �j)

and

S(B) = (B−1 + 1 +B)(�1 −Ba1)∏nj=2(−Baj + �j)

= (�1B−1 + �1 + �1B −Ba1−1 −Ba1 −Ba1+1)

∏nj=2(−Baj + �j).

24

The proof of item 5.3 is direct: it is enough to expand P (a1,−a1 − 1,−a1 − 2)to find that spanB(p(B)) = 2a1 + 3 if a1 6= 2, and it is 6 if a1 = 2.

Under the hypothesis of item 5 we have that hF − lF =∑|ai|>1 |ai|+n, so it is

enough to show that the value of h is the following for each case of item 5:

5.1. h =

{hF − a1 = 1 if a1 < |a2| − 1,hF − (|a2| − 1) = 2 + a1 + a2 if a1 > |a2| − 1.

• Suppose first that a1 < |a2| − 1. If in F we take a Baj+1 of one of the lastn − 1 factors, we obtain degree < 1 since (a1 + 1) + (a2 + 1) < 1. If in Swe take a −Baj of one of the last n − 1 factors, we obtain degree < 0 since(a1 + 1) + a2 < 0. Then

F = �2 · · · �nBa1+1 + �2 · · · �nBa1 + �2 · · · �nBa1−1 + monomials of degree < 1

and

S = −�2 · · · �nBa1+1 − �2 · · · �nBa1 − �2 · · · �nBa1−1 + �1 · · · �nB+ monomials of degree < 1

and F + S = �1 · · · �nB + monomials of degree < 1, hence h = 1 = hF − a1.

• Suppose that a1 > |a2| − 1 and let h0 = 2 + a1 + a2 = hF − (|a2| − 1). If inS we take a −Baj of one of the last n− 1 factors, we obtain degree < h0 since(a1 + 1) + a2 < h. Then, noting that h0 > 1, we have that

F = �2 · · · �nBa1+1 + �2 · · · �nBa1 + �2 · · · �nBa1−1 + (1 + k)�3 · · · �nBa1+1+a2+1+ monomials of degree < h0

if a2 = a3 = . . . = ak+2 > ak+3 (note that if a2 = a3, �2 = �3 hence �3 · · · �n =�2�4 · · · �n, etc.) and

S = −�2 · · · �nBa1+1 − �2 · · · �nBa1 − �2 · · · �nBa1−1+ monomials of degree < h0.

It follows that F +S = (1 + k)�3 · · · �nBh0 + monomials of degree < h0, henceh = h0 = 2 + a1 + a2 and item 5.1 is proved.

5.2. h =

{hF − |a2| = 0 if a1 = |a2| − 1 and |a2| < |a3| − 1,hF − a1 = 1 if a1 = |a2| − 1 and |a2| = |a3|.

• Suppose that a1 = |a2| − 1 and |a2| < |a3| − 1, hence a1 + a2 + 1 = 0 anda1 + a3 + 2 < 0. Then h = hF − |a2| = 1 + a1 + a2 = 0. Then

F = �2 · · · �nBa1+1 + �2 · · · �nBa1 + �2 · · · �nBa1−1 + �1 · · · �n+�3 · · · �nBa1+1+a2+1=1 + �3 · · · �nBa1+a2+1=0+�3 · · · �nBa1+1+a2=0 + monomials of degree < 0

25

and

S = −�2 · · · �nBa1+1 − �2 · · · �nBa1 − �2 · · · �nBa1−1 + �1 · · · �nB + �1 · · · �n+�3 · · · �nBa1+1+a2=0 + monomials of degree < 0.

Since �1�2 = −1 we have that F +S = �3 · · · �n + monomials of degree < 0 andh = 0 = 1 + a1 + a2 = hF − |a2|.

• Suppose that a1 = |a2| − 1 and |a2| = |a3| hence a1 + a2 + 1 = 0. If inF we take two Baj+1 of the last n − 1 factors, we obtain degree < 1 since(a1 + 1) + (a2 + 1) + (a3 + 1) = a3 + 2 < 1. If in S we take a −Baj of one of thelast n− 1 factors, we obtain degree < 1 since (a1 + 1) + a2 = 0. Then

F = �2 · · · �nBa1+1 + �2 · · · �nBa1 + �2 · · · �nBa1−1+(2 + k)�3 · · · �nBa1+1+a2+1=1 + monomials of degree < 1

if a2 = a3 = . . . = ak+3 > ak+4 and

S = −�2 · · · �nBa1+1 − �2 · · · �nBa1 − �2 · · · �nBa1−1 + �1 · · · �nB+ monomials of degree < 1.

Since �1�2 = −1 we have that F+S = (1+k)�3 · · · �nB+ monomials of degree < 1and h = 1 = hF − a1.

5.4. h =

hF − (|a3|+ 1) = −2

if a1 = |a2| − 1, |a2| = |a3| − 1, |a3| < |a4| − 1 and a1 > 2or (a1, a2, a3, a4) = (2,−3,−4,−6),

−3 if (a1, a2, a3, a4) = (2,−3,−4, a4) with a4 < −6.

• Suppose that a1 = |a2| − 1, |a2| = |a3| − 1, |a3| < |a4| − 1 and a1 > 2, hencea1 +a2 +1 = 0, a1 +a3 +2 = 0 and a1 +a4 +3 < 0. If in F we take two B

aj+1 ofthe last n−1 factors we obtain degree < −2 since (a1 +1)+(a2 +1)+(a3 +1) =−a1 < −2. If in S we take two −Baj of the last n− 1 factors we obtain degree< −2 since (a1 + 1) + a2 + a3 = a3 < −2. Then

F = �2 · · · �nBa1+1 + �2 · · · �nBa1 + �2 · · · �nBa1−1 + �1 · · · �n+�3 · · · �nBa1+1+a2+1=1 + �2�4 · · · �nBa1+1+a3+1=0

+(1 + k)�2�3�5 · · · �nBa1+1+a4+1≤−2+�3 · · · �nBa1+a2+1=0 + �2�4 · · · �nBa1+a3+1=−1+�3 · · · �nBa1−1+a2+1=−1 + �2�4 · · · �nBa1−1+a3+1=−2+�3 · · · �nBa1+1+a2=0 + �2�4 · · · �nBa1+1+a3=−1+�3 · · · �nBa1+a2=−1 + �2�4 · · · �nBa1+a3=−2+�3 · · · �nBa1−1+a2=−2+�3 · · · �nBa1+1+a2−1=−1 + �2�4 · · · �nBa1+1+a3−1=−2+�3 · · · �nBa1+a2−1=−2+ monomials of degree < −2

26

if a4 = a5 = . . . = ak+4 > ak+5 and

S = −�2 · · · �nBa1+1 − �2 · · · �nBa1 − �2 · · · �nBa1−1 + �1 · · · �nB+�1 · · · �n + �1 · · · �nB−1+�3 · · · �nBa1+1+a2=0 + �2�4 · · · �nBa1+1+a3=−1+�3 · · · �nBa1+a2=−1 + �2�4 · · · �nBa1+a3=−2+�3 · · · �nBa1−1+a2=−2+ monomials of degree < −2.

Since �1 = −�2 = �3 all terms of degree greater than −2 cancel. For degree−2, the two addends provided by S cancel, and the last four addends providedby F cancel. Then if a1 + 1 + a4 + 1 < −2 we have F + S = �2�4 · · · �nB−2;otherwise a1 + 1 + a4 + 1 = −2 hence �3 = �1 = �4 and it follows that F + S =(2 + k)�2�3�5 · · · �nB−2. In both cases h = −2 = hF − (|a3|+ 1).

• Suppose that (a1, a2, a3, a4) = (2,−3,−4,−6) hence a1 +a2 +1 = 0, a1 +a3 +2 = 0, a1+a4+4 = 0, �1 = �3 = 1 and �2 = −1. If in F we take two Baj+1 otherthan Ba2+1 and Ba3+1 among the last n − 1 factors, we obtain degree < −2since (a1 +1)+(a2 +1)+(a4 +1) = −4 while (a1 +1)+(a2 +1)+(a3 +1) = −2.Instead, if in S we take two −Baj of the last n − 1 factors we obtain degree< −2 since (a1 + 1) + a2 + a3 = −4 < −2. Then

F = �2 · · · �nBa1+1 + �2 · · · �nBa1 + �2 · · · �nBa1−1 + �1 · · · �n+�3 · · · �nBa1+1+a2+1=1 + �2�4 · · · �nBa1+1+a3+1=0

+(1 + k)�2�3�5 · · · �nBa1+1+a4+1=−2+�3 · · · �nBa1+a2+1=0 + �2�4 · · · �nBa1+a3+1=−1+�3 · · · �nBa1−1+a2+1=−1 + �2�4 · · · �nBa1−1+a3+1=−2+�1�3 · · · �nB0+a2+1=−2+�3 · · · �nBa1+1+a2=0 + �2�4 · · · �nBa1+1+a3=−1+�3 · · · �nBa1+a2=−1 + �2�4 · · · �nBa1+a3=−2+�3 · · · �nBa1−1+a2=−2+�3 · · · �nBa1+1+a2−1=−1 + �2�4 · · · �nBa1+1+a3−1=−2+�3 · · · �nBa1+a2−1=−2+�4 · · · �nBa1+1+a2+1+a3+1=−2+ monomials of degree < −2

if a4 = a5 = . . . = ak+4 > ak+5, and

S = −�2 · · · �nBa1+1 − �2 · · · �nBa1 − �2 · · · �nBa1−1 + �1 · · · �nB+�1 · · · �n + �1 · · · �nB−1

+�3 · · · �nBa1+1+a2=0 + �2�4 · · · �nBa1+1+a3=−1+�3 · · · �nBa1+a2=−1 + �2�4 · · · �nBa1+a3=−2+�3 · · · �nBa1−1+a2=−2−�1�3 · · · �nB1+a2=−2+ monomials of degree < −2.

Monomials of degree greater that −2 cancel, and it turns out that F + S =(1 + k)�2�3�5 · · · �nB−2 hence h = −2 = hF − (|a3|+ 1).

27

• If (a1, a2, a3, a4) = (2,−3,−4, a4) with a4 < −6 then a1 + a2 + 1 = 0, a1 +a3 + 2 = 0, a1 +a4 < −4, �1 = �3 = 1 and �2 = −1. In F there are four addendsof degree ≥ −3 formed with two terms of non-extremal degree among the lastn− 1 factors, since (a1 + 1) + (a2 + 1) + (a3 + 1) = −2 and

a1 + (a2 + 1) + (a3 + 1) = (a1 + 1) +a2 + (a3 + 1) = (a1 + 1) + (a2 + 1) +a3 = −3

but (a1 + 1) + (a2 + 1) + (a4 + 1) = a4 + 2 < −6 + 2 = −4. Instead, if inS we take two −Baj of the last n − 1 factors we obtain degree < −3 since(a1 + 1) + a2 + a3 = −4 < −3. Then

F = �2 · · · �nBa1+1 + �2 · · · �nBa1 + �2 · · · �nBa1−1 + �1 · · · �n+�3 · · · �nBa1+1+a2+1=1 + �2�4 · · · �nBa1+1+a3+1=0

+(1 + k)�2�3�5 · · · �nBa1+1+a4+1≤−3+�3 · · · �nBa1+a2+1=0 + �2�4 · · · �nBa1+a3+1=−1+�3 · · · �nBa1−1+a2+1=−1 + �2�4 · · · �nBa1−1+a3+1=−2+�1�3 · · · �nB0+a2+1=−2 + �1�2�4 · · · �nB0+a3+1=−3+�3 · · · �nBa1+1+a2=0 + �2�4 · · · �nBa1+1+a3=−1+�3 · · · �nBa1+a2=−1 + �2�4 · · · �nBa1+a3=−2+�3 · · · �nBa1−1+a2=−2 + �2�4 · · · �nBa1−1+a3=−3+�1�3 · · · �nB0+a2=−3+�3 · · · �nBa1+1+a2−1=−1 + �2�4 · · · �nBa1+1+a3−1=−2+�3 · · · �nBa1+a2−1=−2 + �2�4 · · · �nBa1+a3−1=−3+�3 · · · �nBa1−1+a2−1=−3+�4 · · · �nBa1+1+a2+1+a3+1=−2+�4 · · · �nBa1+a2+1+a3+1=−3+�4 · · · �nBa1+1+a2+a3+1=−3 + �4 · · · �nBa1+1+a2+1+a3=−3+ monomials of degree < −3

if a4 = a5 = . . . = ak+4 > ak+5, and

S = −�2 · · · �nBa1+1 − �2 · · · �nBa1 − �2 · · · �nBa1−1 + �1 · · · �nB+�1 · · · �n + �1 · · · �nB−1+�3 · · · �nBa1+1+a2=0 + �2�4 · · · �nBa1+1+a3=−1+�3 · · · �nBa1+a2=−1 + �2�4 · · · �nBa1+a3=−2+�3 · · · �nBa1−1+a2=−2 + �2�4 · · · �nBa1−1+a3=−3−�1�3 · · · �nB1+a2=−2 − �1�2�4 · · · �nB1+a3=−3−�1�3 · · · �nB0+a2=−3+ monomials of degree < −3.

Note that the monomials of S in the fourth and fifth rows cancel, what corre-sponds for a1 = 2) to the fact that the first factor in S can be simplified, beingS = (�1B

−1 + �1 −Ba1 −Ba1+1)(−Ba2 + �2) · · · (−Ban + �n).

Since �1 = �3 = 1 and �2 = −1, the monomials of degree greater than −3cancel. Let us see the degree −3. If a1 + 1 + a4 + 1 < −3 it turns out thatF + S = �4 . . . �nB

−3; otherwise a1 + 1 + a4 + 1 = −3, equivalently a4 = −7

28

hence �4 = −1, and

F + S = (1 + k)�2�3�5 · · · �nB−3 + �4 · · · �nB−3 = (2 + k)�4 · · · �nB−3

since �2�3 = −1 = �4. In both cases h = −3 since k is a non-negative integer.

This completes the proof of the formula in item 5.4. For example, the spanof the Jones polynomial of the pretzel link represented by the pretzel diagramP (2,−3,−4,−7) is (2 + 3 + 4 + 7)− 6 = 10. Note that, by applying the formula∑|ai|>1 |ai| − a1 − 3 of item 5.4 it would be 11. This amendments the case

(iv)(c) in Theorem 2 of [6].

5.5. h = hF − |a3| = −1 if a1 = |a2| − 1, |a2| = |a3| − 1 and |a3| = |a4| − 1, and

• Suppose that a1 = |a2|−1, |a2| = |a3|−1 and |a3| = |a4|−1 hence a1+a2+1 =0, a1+a3+2 = 0 and a1+a4+3 = 0. If in F we take two B

aj+1 of the last n−1factors, we obtain degree < −1 since (a1 + 1) + (a2 + 1) + (a3 + 1) = a3 + 2 =−a1 < −1. If in S we take two −Baj of the last n− 1 factors, we obtain degree< −1 since (a1 + 1) + a2 + a3 = a3 < −1. Then

F = �2 · · · �nBa1+1 + �2 · · · �nBa1 + �2 · · · �nBa1−1 + �1 · · · �n+�3 · · · �nBa1+1+a2+1=1 + �2�4 · · · �nBa1+1+a3+1=0

+(1 + k)�2�3�5 · · · �nBa1+1+a4+1=−1+�3 · · · �nBa1+a2+1=0 + �2�4 · · · �nBa1+a3+1=−1+�3 · · · �nBa1−1+a2+1=−1+�3 · · · �nBa1+1+a2=0 + �2�4 · · · �nBa1+1+a3=−1+�3 · · · �nBa1+a2=−1+�3 · · · �nBa1+1+a2−1=−1 + monomials of degree < −1

if a4 = a5 = . . . = ak+4 > ak+5, and

S = −�2 · · · �nBa1+1 − �2 · · · �nBa1 − �2 · · · �nBa1−1+�1 · · · �nB + �1 · · · �n + �1 · · · �nB−1+�3 · · · �nBa1+1+a2=0 + �2�4 · · · �nBa1+1+a3=−1+�3 · · · �nBa1+a2=−1 + monomials of degree < −1.

Since �1 = −�2 = �3 there is no terms of degree greater that −1 and F +S = (1 + k)�2�3�5 · · · �nB−1 + monomials of degree < −1, with k ≥ 0. Henceh = −1 = hF − |a3|.

5.6. h = hF − |a2| = 0 if a1 = |a2| − 1, |a2| = |a3| − 1 and |a3| = |a4|.

• Suppose that a1 = |a2|−1, |a2| = |a3|−1 and |a3| = |a4| hence a1+a2+1 = 0,a1 + a3 + 2 = 0 and a1 + a4 + 2 = 0. If in F we take two B

aj+1 of the last n− 1factors, we obtain degree < 0 since (a1 + 1) + (a2 + 1) + (a3 + 1) = a3 + 2 =

29

−a1 ≤ −2. If in S we take two −Baj of the last n− 1 factors, we obtain degree< 0 since (a1 + 1) + a2 + a3 = a3 < 0. Then

F = �2 · · · �nBa1+1 + �2 · · · �nBa1 + �2 · · · �nBa1−1 + �1 · · · �n+�3 · · · �nBa1+1+a2+1=1 + �2�4 · · · �nBa1+1+a3+1=0

+(1 + k)�2�3�5 · · · �nBa1+1+a4+1=0+�3 · · · �nBa1+a2+1=0+�3 · · · �nBa1+1+a2=0 + monomials of degree < 0

if a4 = a5 = . . . = ak+4 > ak+5, and

S = −�2 · · · �nBa1+1 − �2 · · · �nBa1 − �2 · · · �nBa1−1 + �1 · · · �nB + �1 · · · �n+�3 · · · �nBa1+1+a2=0 + monomials of degree < 0.

It follows that F + S = (1 + k)�2�3�5 · · · �n + monomials of degree < 0 since�1 = −�2 = �3. In particular h = 0 = hF − |a2|. This proves item 5.6 andcompletes the proof of Theorem 9.

References

[1] Asensio-Mart́ınez, C.: Cálculo automático del polinomio de Jones, DegreeFinal Work, 2018.

[2] Cromwell, P. R.: Knots and links. Cambridge University Press., 2004.

[3] Dabkowski, M. K., Ishiwata, M. and Przytycki, J. H.: 5-move equivalenceclasses of links and their algebraic invariants. J. Knot Theory Ramifications16 (10) (2007) 1413–1449.

[4] Landvoy, R. A.: The Jones polynomial of pretzel knots and links. Topologyand its applications 83 (1998) 135–147.

[5] Lickorish, W. B. R.: An introduction to Knot Theory. Graduate texts inMathematics, 175. Springer-Verlag (1997).

[6] Manchón, P. M. G.: Kauffman bracket of pretzel links. Marie Curie Fellow-ships Annals, Second Volume. 118–122 (2003).

[7] Stoimenow, A.: 5-moves and Montesinos links. J. Math. Soc. Japan 59, no.3 (2007), 729–749.

[8] The Knot Atlas, http://katlas.org/

Raquel Dı́azDepartment of Algebra, Geometry and Topology

Facultad de Matemáticas, Universidad Complutense de Madridradiaz@ucm.es

30

http://katlas.org/

Pedro M. G. ManchónDepartment of Applied Mathematics to Industrial Engineering

ETSIDI, Universidad Politécnica de Madridpedro.gmanchon@upm.es

31

1 Introduction2 Kauffman bracket of pretzel links3 Span of the Jones polynomial of pretzel links4 There are finitely many pretzel links with a fixed span of the Jones polynomial5 Pretzel knots up to nine crossings6 Appendix