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### Article

Kyungpook Mathematical Journal 2020; 60(2): 239-253

Published online June 30, 2020

### On Alexander Polynomials of Pretzel Links

Yongju Bae* and In Sook Lee

Department of Mathematics, Kyungpook National University, Daegu 41566, Republic of Korea
e-mail : ybae@knu.ac.kr and insooki1109@knu.ac.kr

Received: February 28, 2020; Accepted: June 15, 2020

In this paper, we will find a Seifert matrix for a class of pretzel links with a certain symmetry. Using the symmetry, we find formulae for the Alexander polynomials, determinants and signatures of the pretzel links.

### 1. Introduction

A pretzel link P(p1, p2, p3, · · ·, pn) is defined by an n-tuple (p1, p2, p3, · · ·, pn), n ≥ 3, such that each pi is nonzero integer. The absolute value of pi is the number of half twists and the sign of pi is either positive or negative as seen in Fig. 1. Pretzel links are a well-known family of links in knot theory, and they have been studied extensively. J. Ge and L. Zhang [5] used graph theory to study the determinants of pretzel links and Y. Shinohara [9] used the Goreitx matrix to study their signatures. In [8], Y. Nakagawa studied the Alexander polynomials of pretzel links where at least two pis are even, while D. Kim and J. Lee [7] studied the Conway polynomials of pretzel links. Even though the Alexander polynomial of a link can be obtained from its Conway polynomial, the practical calculation of the Alexander polynomial of a link is very difficult.

Suppose that P(p1, p2, p3, · · ·, pn) is oriented so that the induced orientation of the tangle pi is either parallel or opposite, as seen in Fig. 2.

In this paper, we will use Seifert matrices to find a formula for the Alexander polynomials of pretzel links P(p1, p2, p3, · · ·, pn) all of whose tangles have opposite orientation. We will also use Seifert matrices to calculate the determinant and the signature of P(p1, p2, p3, · · ·, pn).

### 2. Preliminaries

The authors have previously developed techniques for the calculation of the Alexander polynomial. See [1, 2, 3, 4] for details.

A Seifert surface for an oriented link L in S3 is a connected compact oriented surface contained in S3 which has L as its boundary. The following Seifert algorithm is one way to get a Seifert surface from a diagram D of L.

Let D be a diagram of an oriented link L. In a small neighborhood of each crossing, make the following local change to the diagram;

Delete the crossing and reconnect the loose ends in the only way compatible with the orientation.

When this has been done at every crossing, the diagram becomes a set of disjoint simple loops in the plane. It is a diagram with no crossings. These loops are called Seifert circles. By attaching a disc to each Seifert circle and by connecting a half-twisted band at the place of each crossing of D according to the crossing sign, we get a Seifert surface F for L.

The Seifert graph Γ of F is constructed as follows.

Associate a vertex with each Seifert circle and connect two vertices with an edge if their Seifert circles are connected by a twisted band.

Note that the Seifert graph Γ is planar, and that if D is connected, so is Γ. Since Γ is a deformation retract of a Seifert surface F, their homology groups are isomorphic: H1(F) ≅ H1(Γ). Let T be a spanning tree for Γ. For each edge eE(Γ) E(T), the graph T ∪ {e} contains the unique simple closed circuit Te which represents an 1-cycle in H1(F). The set {Te | eE(Γ) E(T)} of these 1-cycles is a homology basis for F. For such a circuit Te, let $Te+$ denote the circuit in S3 obtained by lifting slightly along the positive normal direction of F. For E(Γ) E(T) = {e1, · · ·, en}, the linking number between Tei and $Tej+$ is defined by

$lk(Tei,Tej+)=12∑crossing c∈Tei∩Tej+sign(c).$

A Seifert matrix of L associated to F is the n × n matrix M = (mij) defined by

$mij=lk(Tei,Tei+),$

where E(Γ) E(T) = {e1, · · ·, en}. A Seifert matrix of L depends on the Seifert surface F and the choice of generators of H1(F).

Let M be any Seifert matrix for an oriented link L. The Alexander polynomial ΔL(t) ∈ ℤ[t, t−1], the determinant det(L) and the signature σ(L) of L are defined by

$ΔL(t) =˙det(t12M-t-12MT)det(L) =|det(M+MT)|σ(L) =σ(M+MT).$

See [4, 6] for further details.

For e, fE(Γ)E(T), the intersection TeTf is either the empty set, a single vertex, or a simple path in the spanning tree T. If TeTf is a simple path, and v0 and v1 are two ends of TeTf, we may assume that the neighborhood of v0 looks like Fig. 3. In other words, the cyclic order of edges incident to v0 is given by TeTf, Te, Tf with respect to the positive normal direction of the Seifert surface. Also we may assume that the directions of Te and Tf are given so that v0 is the starting point of TeTf. For, if the direction is reversed, one can change the direction to adapt to our setting so that the resulting linking number changes its sign. In [1], the authors showed the following proposition which is the key tool to calculate the linking numbers for Seifert matrix of a link.

### Proposition 2.1.([1])

For e, fE(Γ)E(T), let p and q denote the numbers of edges in TeTf corresponding to positive crossings and negative crossings, respectively. Suppose that the local shape of TeTf in F looks like Fig. 3. Then,

$lk(Te,Tf+)={-12(p-q),if p+q is even;-12(p-q+1),if p+q is odd, andlk(Tf,Te+)={-12(p-q),if p+q is even;-12(p-q-1),if p+q is odd.$

Let P(p1, p2, p3, · · ·, pn) denote the pretzel link whose all tangles have opposite orientation. Then the Seifert surface of P(p1, p2, p3, · · ·, pn) is drawn in Fig. 4. In this case of orientation, we will see that the Seifert matrix has very nice symmetry, in which Viète’s formula (from algebra) can be applied.

From now on, we suppose that the Seifert surface of P(p1, p2, p3, · · ·, pn) is depicted as in Fig. 4. In order for a Seifert surface of P(p1, p2, p3, · · ·, pn) to be drawn as Fig. 4, the orientations of pi for all i, 1 ≤ in must be opposite. To do this, the pis must be either all odd or all even. Because if there exist i ∈ {1, 2, · · ·, n − 1} such that pi is odd and pi+1 is even, then the Seifert circles of P(p1, p2, p3, · · ·, pn) are depicted as in Fig. 5.

To calculate the Alexander polynomial of a pretzel link P(p1, p2, p3, · · ·, pn), we introduce that Viète’s formula:

### Proposition 2.2.([Viète’s formula])

Let f(x) = xn−1 +Cn−2xn−2 +· · ·+C1x+C0be a polynomial of degree n − 1 and let x1, x2, · · ·, xn−1be roots of the equation f(x) = 0. Then the relation between coefficients of f(x) and its roots are related to symmetric polynomial expression:

$∏k(x1,x2,⋯,xn-1)=x1x2⋯xk+⋯+xn-kxn-k+1⋯xn-1=(-1)kCn-1-k.$

The Alexander polynomial ΔP (t) of a pretzel link P(p1, p2, p3, · · ·, pn) can be expressed as f(x) = xn−1+Cn−2xn−2+· · ·+C1x+C0 where we think of ΔP(p1, p2)(t), ΔP(p1, p3)(t), · · ·, ΔP(p1, pn)(t) as roots and think of $ΔP(p1,-p1,⋯,-p1︸k)(t)$ as xk.

Notice that the signs of the coefficents are always positive, e.g., ΔP(3, −5,5)(t) = ΔP(3, −3, −3)(t) + {ΔP(3, −5)(t) + ΔP(3,5)(t)}ΔP(3, −3)(t) + ΔP(3, −5)(tP(3,5)(t).

### Lemma 3.1

Let P = P(p1, p2, · · ·, pn) be a pretzel link. Suppose that the Seifert surface of the pretzel link P looks like Fig. 4. Then there exist a Seifert matrix M of the pretzel link P such that if p1is odd, then a Seifert matrix M of the pretzel link P is given by

$M=12(p1+p2p1-1p1-1⋯p1-1p1+1p1+p3p1-1⋯p1-1p1+1p1+1p1+p4⋯p1-1⋮⋮⋮⋱⋮p1+1p1+1p1+1⋯p1+pn)(n-1)×(n-1),$

and if p1is even, then a Seifert matrix M of the pretzel link P is given by

$12(p1+p2p1p1⋯p1p1p1+p3p1⋯p1p1p1p1+p4⋯p1⋮⋮⋮⋱⋮p1p1p1⋯p1+pn)(n-1)×(n-1).$
Proof

If we choose the oriented simple closed curves f1, f2, · · ·, fn−1 shown in Fig. 4 as the basis of H1(F, ℤ) where F is the Seifert surface of the pretzel link P(p1, p2, p3, · · ·, pn), then by using Proposition 2.1 one can calculate the linking numbers to get the result. The proof is completed.

### Theorem 3.2

Let $P=P(p1,-p1,⋯,-p1︸n-1)$be a pretzel link. Suppose that the Seifert surface of the pretzel link P looks like Fig. 4. Then the determinant det(P) and the signature σ(P) of the pretzel link P are given by

$det(P) =(n-2)∣p1∣n-1,σ(P) ={-n+3,if p1>0;n-3,if p1<0.$
Proof

From Lemma 3.2, we know that

$M+MT=(0p1p1⋯p1p10p1⋯p1p1p10⋯p1⋮⋮⋮⋱⋮p1p1p1⋯0)(n-1)×(n-1).$

Hence $det(M+MT)=(-1)n(n-2)p1n-1$ by the formula (1) in Appendix A. Since n − 2 > 0, det(P) = (n − 2)|p1|n−1. The characteristic equation of M +MT to be

$det((M+MT)-λI)=det(-λp1p1⋯p1p1-λp1⋯p1p1p1-λ⋯p1⋮⋮⋮⋱⋮p1p1p1⋯-λ)(n-1)×(n-1)=(-λ-p1)n-2{p1(n-1)+(-λ-p1)}=0$

by the formula (3) in Appendix A. Thus the eigenvalues of M + MT are λ1 = p1(n − 2) and λ2 = λ3 = · · · = λn−1 = −p1. If p1 is positive, then −p1 is negative and p1(n − 2) is positive since n > 3. Hence the signature of M + MT is 3 − n. Similarly, if p1 is negative, then the signature of M + MT is n − 3. The proof is completed.

### Theorem 3.3

Let $P=P(p1,-p1,⋯,-p1︸n-1)$be a pretzel link. Suppose that the Seifert surface of the pretzel link P looks like Fig. 4. Then the Alexander polynomials ΔP (t) of P are given by

$ΔP(t)={(-1)n-1ΔP(p1,-p1,-p1)(t)×{(t12p1-12-t-12p1+12)n-2-(t12p1+12-t-12p1-12)n-2}(t12p1-12-t-12p1+12)-(t12p1+12-t-12p1-12),if p1 is odd;(-1)n-1(n-2)ΔP(p1,-p1,-p1)(t) (t12p12-t-12p12)n-3,if p1 is even.$
Proof

Suppose that p1 is odd. From Lemma 3.1, we know that $t12M-t-12MT=(0t12p1-12-t-12p1+12⋯t12p1-12-t-12p1+12t12p1+12-t-12p1-120⋯t12p1-12-t-12p1+12⋮⋮⋱⋮t12p1+12-t-12p1-12t12p1+12-t-12p1-12⋯0)(n-1)×(n-1).$.

By the formula (1) in Appendix A,

$det(t12M-t-12MT)=(-1)n-2(t12p1-12-t-12p1+12) (t12p1+12-t-12p1-12)×{(t12p1-12-t-12p1+12)n-2-(t12p1+12-t-12p1-12)n-2}(t12p1-12-t-12p1+12)-(t12p1+12-t-12p1-12)=(-1)n-1ΔP(p1,-p1,-p1)(t)×{(t12p1-12-t-12p1+12)n-2-(t12p1+12-t-12p1-12)n-2}(t12p1-12-t-12p1+12)-(t12p1+12-t-12p1-12)$

since

$ΔP(p1,-p1,-p1)(t)=det(0t12p1-12-t-12p1+12t12p1+12-t-12p1-120)=(-1) (t12p1-12-t-12p1+12) (t12p1+12-t-12p1-12).$

Suppose that p1 is even. From Lemma 3.1, we know that $t12M-t-12MT=(0t12p12-t-12p12⋯t12p12-t-12p12t12p12-t-12p120⋯t12p12-t-12p12⋮⋮⋱⋮t12p12-t-12p12t12p12-t-12p12⋯0)(n-1)×(n-1).$.

By the formula (1) in Appendix A,

$det(t12M-t-12MT)=(-1)n-2(n-2) (t12p12-t-12p12)n-1=(-1)n-1(n-2)ΔP(p1,-p1,-p1)(t) (t12p12-t-12p12)n-3$

since

$ΔP(p1,-p1,-p1)(t)=det(0t12p12-t-12p12t12p12-t-12p120)=(-1) (t12p12-t-12p12)2.$

The proof is completed.

### Corollary 3.4

Let P = P(p1, p2, p3, · · ·, pn) be a pretzel link (n ≥ 3). If the Seifert surface of pretzel link P is shown in Fig 4, then the determinant det(P) of P is given by

$det(P)=|p2p3⋯pn{p1pn(n-2)+p1p2+1}|.$
Proof

From the definition of a link and Lemma 3.1, we can prove it by using the formula (3) in Appendix A. The proof is completed.

To prove the main theorem, we show the following lemma.

### Lemma 3.5

Suppose that the Seifert surface of the pretzel link P looks like Fig. 4.

• ΔP(p1)(t) = ΔO(t) = 1.

• ΔP(p1, −p1)(t) = ΔOO(t) = 0.

• ΔP(p1, −p1, −pk)(t) = ΔP(p1, −p1, −p1)(t), for any k = 1, 2, · · ·, n.

• ΔP(p1, p2, ···, pn)(t) = ΔP(p1, pi)(tP(p1, p2, ···, pi−1, pi+1, ···, pn)(t) +ΔP(p1, p2, ···, pi−1, −pi, pi+1, ···, pn)(t).

Proof

(1) and (2) are trivial.

(3) Suppose that p1 is odd. Then a Seifert matrix ML of P(p1, −p1, −pk) and a Seifert matrix MR of P(p1, −p1, −p1) are given by $ML=(0p1-12p1+12p1+pk2)$ and $MR=(0p1-12p1+120)$ if p1 is odd. For p1 is even, it is similar for p1 is odd.

(4) The basic idea of determining the determinant of matrix is as

$det(a11a12⋯a1i-1a1ia1i+1⋯a1na21a22⋯a2i-1a2ia2i+1⋯a2n⋮⋮⋱⋮⋮⋮⋱⋮ai-11ai-12⋯ai-1i-1ai-1iai-1i+1⋯ai-1nai1ai2⋯aii-1aiiaii+1⋯ai-1nai+11ai+22⋯ai+1i-1ai+1iai+1i+1⋯ai+1n⋮⋮⋱⋮⋮⋮⋱⋮an1an2⋯ani-1aniani+1⋯ann)=aiidet(a11a12⋯a1i-1a1i+1⋯a1na21a22⋯a2i-1a2i+1⋯a2n⋮⋮⋱⋮⋮⋱⋮ai-11ai-12⋯ai-1i-1ai-1i+1⋯ai-1nai+11ai+22⋯ai+1i-1ai+1i+1⋯ai+1n⋮⋮⋱⋮⋮⋱⋮an1an2⋯ani-1ani+1⋯ann)+det(a11a12⋯a1i-1a1ia1i+1⋯a1na21a22⋯a2i-1a2ia2i+1⋯a2n⋮⋮⋱⋮⋮⋮⋱⋮ai-11ai-12⋯ai-1i-1ai-1iai-1i+1⋯ai-1nai1ai2⋯aii-10aii+1⋯ai-1nai+11ai+22⋯ai+1i-1ai+1iai+1i+1⋯ai+1n⋮⋮⋱⋮⋮⋮⋱⋮an1an2⋯ani-1aniani+1⋯ann).$

The proof is completed.

The following is the main theorem of the paper.

### Theorem 3.6

Let P = P(p1, p2, p3, · · ·, pn) be a pretzel link (n ≥ 3). If the Seifert surface of the pretzel link P is shown in Fig. 4, then the Alexander polynomial ΔP (t) of P is given by

$∑k=1n{ΔP(p1,-p1,⋯,-p1︸k-1)(t)×∏n-k(ΔP(p1,p2)(t),ΔP(p1,p3)(t),⋯,ΔP(p1,pn)(t))}.$
Proof

We divide our proof into two cases (Case 1) All of pi are odd, and (Case 2) All of pi are even.

(Case 1)

All of pi are odd. By the definition of the Alexander polynomial of a link and by Lemma 3.1, we have the Alexander polynomial of P is given by

$det((t12-t-12)p1+p22t12p1-12-t-12p1+12⋯t12p1-12-t-12p1+12t12p1+12-t-12p1-12(t12-t-12)p1+p32⋯t12p1-12-t-12p1+12⋮⋮⋱⋮t12p1+12-t-12p1-12t12p1+12-t-12p1-12⋯(t12-t-12)p1+pn2).$

We proceed by the mathematical induction on n(n ≥ 3). For n = 3,

$ΔP(p1,p2,p3)(t)=det(t12(p1+p2)-t-12(p1+p2)2t12(p1-1)-t-12(p1+1)2t12(p1+1)-t-12(p1-1)2t12(p1+p3)-t-12(p1+p3)2)=det(t12(p1+p2)-t-12(p1+p2)2)det(t12(p1+p3)-t-12(p1+p3)2)+det(t12(p1+p2)-t-12(p1+p2)2t12(p1-1)-t-12(p1+1)2t12(p1+1)-t-12(p1-1)20),by Lemma 3.5(4)=ΔP(p1,p2)(t)ΔP(p1,p3)(t)+ΔP(p1,p2,-p1)(t)=ΔP(p1,p2)(t)ΔP(p1,p3)(t)+ΔP(p1,-p1,-p1)(t) by Lemma 3.5(3).=ΔP(p1,p2)(t)ΔP(p1,p3)(t)+{ΔP(p1,p2)(t)+ΔP(p1,p3)(t)}ΔP(p1,-p1)(t)+ΔP(p1,-p1,-p1)(t), by Lemma 3.5(2).$

Assume that the formula is true for n − 1.

$ΔP(p1,p2,⋯,pn)(t)=ΔP(p1,pn)(t)ΔP(p1,p2,⋯,pn-1)(t)+ΔP(p1,p2,⋯,pn-1,-p1)(t)=ΔP(p1,pn)(t)ΔP(p1,p2,⋯,pn-1)(t)+ΔP(p1,pn-1)(t)ΔP(p1,p2,⋯,pn-2,-p1)(t)ΔP(p1,p2,⋯,pn-2,-p1,-p1)(t),by Lemma 3.5(4).$

By applying the identity Lemma 3.5(4) to the last polynomial ΔP(p1, p2, ···, pn−2, −p1, −p1)(t) repeatedly, we get the following result.

$ΔP(p1,p2,⋯,pn)(t)=∑i=2nΔP(p1,pi)(t)ΔP(p1,p2,⋯,pi-1,-p1,⋯,-p1︸n-i)(t)+ΔP(p1,-p1,⋯,-p1︸n-1)(t).=∑i=2nΔP(p1,pi)(t)∑k=1n-1{ΔP(p1,-p1,⋯,-p1︸k-1)(t)×∏n-1-k(ΔP(p1,p2)(t),⋯,ΔP(p1,pi-1)(t),ΔP(p1,-p1)(t),⋯,ΔP(p1,-p1)(t)︸n-i)}+ΔP(p1,-p1,⋯,,-p1︸n-1)(t)=∑k=1n-1{ΔP(p1,-p1,⋯,-p1︸k-1)(t)∑i=2nΔP(p1,pi)(t)×∏n-1-k(ΔP(p1,p2)(t),⋯,ΔP(p1,pi-1)(t),ΔP(p1,-p1)(t),⋯,ΔP(p1,-p1)(t)︸n-i)}+ΔP(p1,-p1,⋯,,-p1︸n-1)(t)=∑k=1n-1{ΔP(p1,-p1,⋯,-p1︸k-1)(t)×∏n-k(ΔP(p1,p2)(t),⋯,ΔP(p1,pn)(t))}+ΔP(p1,-p1,⋯,,-p1︸n-1)(t)=∑k=1n-1{ΔP(p1,-p1,⋯,-p1︸k-1)(t)×∏n-k(ΔP(p1,p2)(t),ΔP(p1,p3)(t),⋯,ΔP(p1,pn)(t))}.$
(Case 2)

All of pi are even, is similar to the proof of (Case 1). The proof is completed.

Let P(3, −5, 5) be a pretzel link. Then a Seifert matrix of P(3, −5, 5) is given by $M=(-1124)$. By direct calculation, one can see that ΔP(3, −5,5)(t) = −6t2 + 13t − 6.

And by using the main theorem, one can get the same result.

$ΔP(3,-5,5)(t)=ΔP(3,-5)(t)ΔP(3,5)(t)+ΔP(3,-3)(t){ΔP(3,-5)(t)+ΔP(3,5)(t)}+ΔP(3,-3,-3)(t)=(-t+1) (4t-4)+(-1)2t+1 ((3-1)t-(3+1)2) ((3+1)t-(3-1)2)×{((3-1)t-(3+1)2)-((3+1)t-(3-1)2)}=-6t2+13t-6.$
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