Title ON THE NUMBER OF p-COLORINGS OF KNOTS
Author(s) MIZUGUCHI, SHOU; TANAKA, TOSHIFUMI
Citation [岐阜大学教育学部研究報告. 自然科学] vol.[42] p.[9]-[13]
Issue Date 2018
Rights
Version Department of Mathematics, Faculty of Education, Gifu
University / Department of Mathematics, Faculty of Education, Gifu University
URL http://hdl.handle.net/20.500.12099/74996
※この資料の著作権は、各資料の著者・学協会・出版社等に帰属します。
SHOU MIZUGUCHI AND TOSHIFUMI TANAKA
Abstract. We show some examples of pairs of knots with the same number ofp-colorings for any positive integerp≥2. In particular, we show that there exists an infinite family of knots with only trivialp-colorings.
Key words: Knots; coloring; congruence.
1. Introduction
In the 1950’s, Fox introduced the concept of ap-coloringof a knot, where p≥2 is an integer. A p-coloring of a diagramD of a knotK is a map f :{arcs ofD} →Z/pZ such that at each crossing, if the three arcs are mapped tox, y and z, with the z corresponding to the overcrossing, then they satisfyx+y−2z ≡0 (mod p). We call the number of distinct p-colorings of a diagram of a knot K, the rank of thep-colorings ofK and we denote it by colp(K). We say that ap-coloring istrivial if it is a constant map. A diagram forK is p-colorableif it has a non-trivial p-coloring. The rank of thep-colorings of a knot and thep-colorabiliy are invariants of a knot. We know that if a knot is p-colorable, then it is alsopq-colorable for any positive integerq. Conversely, if a knot ispq-colorable for some positive integerspandq, then the knot is eitherp-colorable orq-colorable [5].
In this paper, we obtain the following results.
Theorem 1.1. Each pair of knots in the set {(41,51), (62,72), (74,3141)} (Figure1) is knots with the same rank of thep-colorings for any positive integerp≥2.
41 51
62 72
74 3 #1 41
Figure 1
1
2 SHOU MIZUGUCHI AND TOSHIFUMI TANAKA
In particular,
colp(41) =
5p (p= 5m(m∈N)) p (p= 5m(m∈N)) (1)
colp(62) =
11p (p= 11m (m∈N)) p (p= 11m (m∈N)) (2)
colp(74) =
⎧⎪
⎪⎨
⎪⎪
⎩
15p (p= 15n(n∈N))
3p (p= 3m, p= 5n(m, n∈N)) 5p (p= 3m, p= 5n(m, n∈N)) p (p= 3m,5n(m, n∈N))
. (3)
Theorem 1.2. There exists an infinite family of knots {Ki} such that colp(Ki) =pfor any positive integerp≥2.
In Sections 2, we shall give proofs of Theorems 1.1 and 1.2.
Acknowledgements. The second author is partially supported by the Ministry of Education, Science, Sports and Culture, Grant-in-Aid for Scientific Research(C), 2016-2018ʢ16K05145).
2. Proofs
Proof of Theorem 1.1. First we take ap-coloring of 41as in Figure 2.
z w
y x
Figure 2
Then we have the following congruences:
2z≡x+y (modp) 2x≡z+w(modp) 2w≡z+y (mod p) 2y≡w+x(modp)
Thus we have the following congruences:
x+y−2z≡0 (modp)
−5z+ 5w≡0 (modp) y+z−2w≡0 (modp)
In the case whenp= 5m, we havez≡wby the second congruence. Thus we havex≡zandy≡z by the first and the third congruences. Therefore we know that the rank isp.
In the case whenp= 5m, we have−z+w≡0 (modm) by the second congruence. Then we have 25mchoices for the pair (w, z). Thus we know that the rank is 25m.
Next we consider ap-coloring of 51as in Figure 3.
Then we have the following congruences:
2v≡x+y (modp) 2x≡z+v(mod p) 2z≡w+x(modp) 2w≡z+y (mod p)
z w
y
x
v Figure 3
2y≡v+w(modp)
Thus we have the following congruences:
x+ 3y−4z≡0 (modp) 5y−5z≡0 (modp) w−3y+ 2z≡0 (modp) v+y−2z≡0 (modp)
In the case whenp= 5m, we havey ≡z by the second congruence. Thus we havex≡z, w≡y andv≡z by the first, the third and the fourth congruences. Therefore we know that the rank isp.
In the case when p= 5m, we havey−z ≡0 (mod m) by the second congruence. Then we have 25mchoices for the pair (y, z). Thus we know that the rank is 25m.
Thus we know that the ranks of the p-colorings of 41 and 51 are equal. By the same way, we can show that 62 and 72have the same rank of thep-colorings.
We consider a p-coloring of 74 as in Figure 3.
x t
u z
y w
v
Figure 4
Then we have the following congruences:
2t≡x+y (modp) 2x≡u+t (modp) 2u≡x+w(mod p) 2y≡u+v (mod p) 2v≡t+z (modp) 2z≡v+w (modp) 2w≡y+z (modp)
Thus we have the following congruences:
−5x+ 5z≡0 (modp)
−w−2x+ 2y+z≡0 (modp)
−15x+ 15y≡0 (modp) u+ 2x−4y+z≡0 (modp) t−4x+ 4y−z≡0 (modp)
−4x+ 3y+z≡0 (modp)
In the case whenp= 3m,5n, we know thatx≡y≡zby the first and the third congruences. Thus we have x≡t≡u≡v≡w. Hence the rank is equal top.
4 SHOU MIZUGUCHI AND TOSHIFUMI TANAKA
In the case when p= 5nand p= 3m, we have x≡z (mod n) by the first congruence. Thus we have 25n choices for the pair (x, z). We also know thatx≡y (mod p) by the assumption and the third congruence. Hence we know that the rank is equal to 25nby the rest congruences.
In the case whenp= 3mandp= 5n, we have 5x−5z≡0 (modm) wheremis not divisible by 5.
Thus we have x≡z (mod m) and we have 9mchoices for the pair (x, z). Hence the rank is equal to 9mby the rest congruences.
In the case when p= 15n, we know that x≡y (mod n) by the third congruence. Thus we have 225nchoices for the pair (x, y). Hence we know that the rank is equal to 225nby the rest congruence.
We consider a p-coloring of 3141 as in Figure 3.
t
v
u w
y x
z
Figure 5
Then we have the following congruences:
2t≡u+v (modp) 2u≡t+v (modp) 2v≡t+z (modp) 2w≡x+u(modp) 2x≡y+w(mod p) 2y≡w+z(mod p) 2z≡x+y (modp)
Then we have the following congruences:
5y−5z≡0 (modp) 3v−3z≡0 (modp) x+v−2z≡0 (modp) u−z≡0 (modp) w−2y+z≡0 (modp) x+y−2z≡0 (modp)
In the case when p= 3m,5n, we know that v ≡y ≡z by the first and the second congruences.
Thus we havev≡t≡u≡w≡x. Hence the rank is equal top.
In the case when p= 5n andp= 3m, we havey ≡z (mod n) by the first congruence. Thus we have 25n choices for the pair (y, z). We also know that v ≡y (mod p) by the assumption and the second congruence. Hence we know that the rank is equal to 25nby the rest congruences.
In the case whenp= 3mandp= 5n, we have v≡z (modm) by the second congruence. Thus we have 9mchoices for the pair (v, z). We also know thaty≡z(modp) by the assumption and the first congruence. Hence the rank is equal to 9mby the rest congruences.
In the case whenp= 15n, we know that y ≡z (mod 5n) by the first congruence. We also know that v ≡ z (mod 3n) by the second congruence. Thus we have 225n choices for the triad (v, y, z).
Hence we know that the rank is equal to 225nby the rest congruence.
Proof of Theorem 1.2. Let Kc be a knot as in Figure 6. (The number c denotes the number of full-twists.) We know that det(Kc)=1 [3], where det(Kc) is the knot determinant of Kc . If p is a prime number, then the knot Kc is not p-colorable since p cannot divide det(Kc) [1]. Suppose that Kc isp-colorable for some positive integer p≥2. Then Kc is d-colorable for some prime numberd [5], where dis a prime factor ofp. This is a contradiction. Thus we know thatKc is notp-colorable for any positive integerp≥2. Hence colp(Kc) =pfor any positive integerp≥2.
TheJones polynomialV(t) ofKcis easily calculated as follows. (See [2] for the definition. We may use [Theorem 1.1, [4]] for the calculation.)
V(t) = (−1)ct−c+ (1−(−1)ct−c)(−t4+ 2t3−3t2+ 4t−3 + 4t−1−3t−2+ 2t−3−t−4).
Then we know that Kc1 is not equivalent toKc2 ifc1=c2. This completes the proof.
c
Figure 6
References
1. C. Livingston,Knot Theory, The Mathematical Association of America, Washington DC, 1993.
2. W. B. R. Lickorish,An introduction to knot theory, Graduate Texts in Mathematics, 175. Springer-Verlag, New York, 1997.
3. C. Lamm,Symmetric unions and ribbon knots, Osaka J. Math., Vol. 37 (2000), 537–550.
4. T. Tanaka,The Jones polynomial of knots with symmetric union presentations, J. Korean Math. Soc. 52 (2015), no. 2, 389–402.
5. S. Ganzell and C. Vanblargan,A note on composite colorings, preprint.
Department of Mathematics, Faculty of Education, Gifu University, Yanagido 1-1, Gifu, 501-1193, Japan.
Department of Mathematics, Faculty of Education, Gifu University, Yanagido 1-1, Gifu, 501-1193, Japan.
E-mail address:[email protected]
