Article
Kyungpook Mathematical Journal 2020; 60(2): 239-253
Published online June 30, 2020
Copyright © Kyungpook Mathematical Journal.
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
Abstract
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.
Keywords: determinant of a matrix, Seifert matrix of a link, Alexander polynomial of a link, signatures of a link, pretzel link
1. Introduction
A pretzel link
Suppose that
In this paper, we will use Seifert matrices to find a formula for the Alexander polynomials of pretzel links
2. Preliminaries
The authors have previously developed techniques for the calculation of the Alexander polynomial. See [1, 2, 3, 4] for details.
A
Let
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
The
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
A
where
Let
See [4, 6] for further details.
For
Proposition 2.1.([1])
Let
From now on, we suppose that the Seifert surface of
To calculate the Alexander polynomial of a pretzel link
Proposition 2.2.([Viète’s formula])
The Alexander polynomial Δ
Notice that the signs of the coefficents are always positive, e.g., Δ
3. Seifert Matrices of Pretzel Links and Related Invariants
Lemma 3.1
If we choose the oriented simple closed curves
Theorem 3.2
From Lemma 3.2, we know that
Hence
by the formula (3) in
Theorem 3.3
Suppose that
By the formula (1) in
since
Suppose that
By the formula (1) in
since
The proof is completed.
Corollary 3.4
From the definition of a link and Lemma 3.1, we can prove it by using the formula (3) in
To prove the main theorem, we show the following lemma.
Lemma 3.5
Δ
P (p 1)(t ) = ΔO (t ) = 1.Δ
P (p 1, −p 1)(t ) = ΔOO (t ) = 0.Δ
P (p 1, −p 1, −p k )(t ) = ΔP (p 1, −p 1, −p 1)(t ),for any k = 1, 2, · · ·,n. Δ
P (p 1,p 2, ···,p n )(t ) = ΔP (p 1,p i )(t )ΔP (p 1,p 2, ···,p i −1,p i +1, ···,p n )(t ) +ΔP (p 1,p 2, ···,p i −1, −p i ,p i +1, ···,p n )(t ).
(1) and (2) are trivial.
(3) Suppose that
(4) The basic idea of determining the determinant of matrix is as
The proof is completed.
The following is the main theorem of the paper.
Theorem 3.6
We divide our proof into two cases (Case 1) All of
All of
We proceed by the mathematical induction on
Assume that the formula is true for
By applying the identity Lemma 3.5(4) to the last polynomial Δ
All of
Example 3.7
Let
And by using the main theorem, one can get the same result.
Figures
References
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