       On symmetric unions and alternating knots
Toshifumi Tanaka
Key words: symmetric union, alternating knot, ribbon knot.
Institution: Department of Mathematics, Faculty of Education, Gifu University, Yanagido 1-1, Gifu, 501-1193, Japan.

Abstract. In this paper, we introduce the partial alternating number for a symmetric union andshow that for any positive integer n, there exists a symmetric union with partial alternating number≥ n. We also show that there exist inﬁnitely many symmetric unions with partial alternating numberone.

1Introduction

The union of knots was introduced by S. Kinoshita and H. Terasaka [2]. A symmetric union [3] is a knot which
is obtained from the connected sum of a knot and its mirror image by inserting some vertical twists along thesymmetry axis to the diagram. A symmetric union is known to be a ribbon knot [4]. In this paper, we study
symmetric unions for alternating knots. For any symmetric union, we have only ﬁnitely many alternating partial
knots for the symmetric union presentations (Proposition 3.1). We deﬁne the partial alternating number for asymmetric union as the number of alternating knots (up to mirror images). Then we have the following.
Theorem 1.1. For any positive interger n, there exists a symmetric union with partial alternating number ≥ n.
The notation for prime knots up to 10 crossings is due to Rolfsen’s book [1]. We denote the mirror imageof a knot K by K ∗ . In Section 2, we shall deﬁne a symmetric union. In Section 3, we shall deﬁne the partial
alternating number for a symmetric union and prove Theorem 1.1. In Section 4, we shall show that there exist
inﬁnitely many symmetric unions with partial alternating number one.

2Deﬁnition

We deﬁne a symmetirc union [3] as follows. We denote the tangles made of half twists by integers n ∈ Z andthe horizontal trivial tangle by ∞ as in Figure 1.
0n===n>0n<0Figure 1: Tangles.

1 ∗Deﬁnition 2.1. Let DK be an unoriented diagram of knot K and DK, the diagram DK reﬂected at an axis
in the plane. We take k 0-tangles Ti (i = 0, . . . , k) on the symmetry axis as in Figure 2(a). Then we replace
the tangles Ti with T0 = ∞ and Ti = ni ∈ Z for i = 1, . . . , k as in Figure 2(b). We call the resultant diagram∗(n1 , . . . , nk ) and the diagram is called a symmetric union presentation.
a symmetric union and write DK ∪ DK
The knot K is called the partial knot for the symmetric union presentation. We say that a knot K is a symmetricunion if K has a symmetric union presentation.

T0n1

T1
DK

T2

D*KDKn2

Tknk(a)(b)

D*KFigure 2: A symmetric union.

3

Symmetric unions and alternating knots

Proposition 3.1. For any symmetric union, we have only ﬁnitely many alternating partial knots for thesymmetric union presentations.
Proof. Let K be a knot with a symmetric union presentation with K as a partial knot. Then by [3, Theorem 2.6],
we have det(K) = det(K)2 , where det(K) is the determinant of K. Suppose that K is a (non-trivial) alternatingknot. Then by [6] and [5, Chapter 8], we have degQK (x) = c(K) − 1, where degQK is the maximal degree ofthe Q-polynomial of K. Since any alternating knot is quasi-alternating, by [7], we have degQK (x) < det(K).
Thus we obtain that c(K) − 1 ≤ (c(K) − 1)2 < (det(K))2 = det(K).
Deﬁnition 3.2. Let K be a symmetric union. Then the number of alternating partial knots (excluding mirrorimages) for the symmetric union presentations for K is called the partial alternating number for K.
Remark 3.3. Clearly, the partial alternating number is an invariant of a symmetric union. For example, theconnected sum of a prime alternating knot and its mirror image is a symmetric union with partial alternating
number ≥ 1. In particular, the partial alternating number of the connected sum of 3 1 and its mirror image isequal to one. However we do not know if the equality holds in general.
Proof of Theorem 1.1. For a positive integer m ≥ 2, let K m be a knot which has the symmetric union
presentation Dm as shown in Figure 3.

2

 

K1K2 
  K2m

Ki =Figure 3: K m .

Figure 4: A ﬂype.

Note that Dm has 2mi=1 31 , which is the connected sum of 2m 31 s, as the partial knot. Then for an integer p(1 ≤ p ≤ m), by ﬂyping K1 , . . . , Kp to Dm as shown in Figure 4, we have a symmetric union presentation withthe partial knot (pi=1 3∗1 )(2m−pi=1 31 ) denoted by Hp .
Then it is easily seen that the Jones polynomials [5] of Hp1 and Hp2 are distinct if p1 = p2 . So we have atleast m distinct alternating partial knots (up to mirror images) for K m .

4Examples

Proposition 4.1. There exist inﬁnitely many symmetric unions with partial alternating number one.
Proof. For a positive integer n, let K̂n be a knot which has the symmetric union presentation D ∪ D∗ (n) as
shown in Figure 5. Suppose that K is a partial knot for a symmetric union presentation for K̂n . By using
the same method as in the proof of Theorem 1.1, we know that c(K) − 1 < det(K) = 3. Since c(K) ≤ 3 anddet(K) = 3, K is either 31 or 3∗1 . Thus we know that the partial alternating number of K̂n is equal to one. By
using the Jones polynomial, we know that K̂n1 = K̂n2 if n1 = n2 .
Example 4.2. Let K be the connected sum of 41 and 4∗1 and let K be the partial knot for a symmetric union
presentation for K. Then as in the proof of Proposition 4.1, we know that c(K) − 1 < det(K) = 5. There are
only four alternating knots 41 , 51 , 4∗1 and 5∗1 such that the crossing numbers ≤ 5 and the determinants are equalto 5. Thus we know that the partial alternating number of K is ≤ 2.
3 Let K̃n be a knot which has the symmetric union presentation D ∪ D∗ (n) as shown in Figure 6. Supposethat K is a partial knot for a symmetric union presentation for K̃n . As in the proof of Proposition 4.1, weknow that c(K) − 1 < det(K) = 9. There are only four alternating knots, 61 , 91 , 31 31 and 31 3∗1 , such that thecrossing numbers ≤ 9 and the determinants are equal to 9, up to mirror images. Thus we know that the partial
alternating number of K̃n is ≤ 4. By using the same method as in the proof of Theorem 1.1, we know that31 31 and 31 31∗ are the partial knots for K̃n . So we know that the partial alternating number of K̃n is ≥ 2.

nFigure 5: K̂n .

nFigure 6: K̃n .
Question. Can the knots 61 and 91 be partial knots for K̃n ?
Acknowledgements. The author is partially supported by the Ministry of Education, Science, Sports andCulture, Grant-in-Aid for Scientiﬁc Research(C), 2022-2024(22K03310).

References[1] D. Rolfsen, Knots and links, Publish or Perish. Inc. (1976).
[2] S. Kinoshita and H.Terasaka, On unions of knots, Osaka J. Math. Vol. 9 (1957), 131-153.
[3] C. Lamm, Symmetric unions and ribbon knots, Osaka J. Math., Vol. 37 (2000), 537-550.
[4] L. H. Kauﬀman, On knots, Ann. of Math. Stud., 115, Princeton University Press, Princeton, NJ, 1987,
xvi+481 pp.
[5] A. Kawauchi, A survey of knot theory, Translated and revised from the 1990 Japanese original by the author.
Birkhäuser Verlag, Basel, 1996.

4    [6] M. E. Kidwell, On the degree of the Brandt-Lickorish-Millett-Ho polynomial of a link, Proc. Amer. Math.
Soc.100 (1987), no.4, 755–762.
[7] K. Qazaqzeh, and N. Chbili, A new obstruction of quasialternating links, Algebr. Geom. Topol.15 (2015),
no.3, 1847–1862.

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