On the minimum number of Fox Colorings of Knots

Abstract

We investigate Fox colorings of knots that are 17-colorable. Precisely, we prove that any 17-colorable knot has a diagram such that exactly 6 among the seventeen colors are assigned to the arcs of the diagram.

Full text

On the minimum number of Fox colorings of knots
H. Abchir
Hassan II University. Casablanca, Morocco.
e-mail: hamid.abchir@univh2c.ma
M. Elhamdadi
University of South Florida, Tampa FL.
e-mail: emohamed@usf.edu
S. Lamsifer
Hassan II University. Casablanca, Morocco.
e-mail: soukainalamsifer@gmail.com
September 29, 2020
Abstract
We investigate Fox colorings of knots that are 17-colorable. Precisely, we prove
that any 17-colorable knot has a diagram such that exactly 6 among the seventeen
colors are assigned to the arcs of the diagram.
1 Introduction
Knot theory has been a very active area of research for more than a century. In the
last fifty years, Knot theory has been successfully applied to chemistry and molecular
biology. Recently some algebraic structures, related to quandles which generalize Fox n-
colorability, have been used to classify topological structures of proteins were introduced
in [1] and labelled bondles. The 3-coloring invariant is the simplest invariant that distin-
guishes the trefoil knot from the trivial knot. The idea of 3-coloring and more generally
m-coloring was developed by R. Fox around 1960 (see [5]). He introduced a diagram-
matic definition of colorability of a knot by
Z
m(the integers modulo m). Precisely, for
any natural number m, a knot diagram is said to be m-colorable if we can assign to each
of its arcs an element of
Z
m, called the color of that arc, such that, at each crossing, the
sum of the colors of the under-arcs is twice the color assigned to the over arc modulo m
(see Figure 1 below). A knot is said to be m-colorable if it has an m-colorable diagram.
For obvious reasons mwill be restricted to the odd primes. A coloring that uses only
one color is usually called a trivial coloring. For an explicit example of a Fox 3-coloring
of the knot 819 consult example 60 on page 82 of [4].
Let pbe an odd prime integer. Let Kbe a p-colorable knot and let Cp(K) denote the
minimal number of colors needed to color a diagram of K. The problem of finding the
minimum number of colors for p-colorable knots with primes up to 13 was investigated
1
arXiv:2005.11336v2 [math.GT] 26 Sep 2020

by many authors. In 2009, S. Satoh showed in [9] that C5(K) = 4. In 2010, K. Oshiro
proved that C7(K) = 4 [8]. In 2016, T. Nakamura, Y. Nakanishi and S. Satoh showed
in [7] that C11(K) = 5. In 2017, M. Elhamdadi and J. Kerr [3] and independently F.
Bento and P. Lopes [2] proved that C13(K) = 5. In what follows we investigate the case
of the prime number p= 17.
pCp(K)
3 3
5 4
7 4
11 5
13 5
First, using the result of T. Nakamura, Y. Nakanishi and S. Satoh [6] which states
that for any knot Kand any prime p,Cp(K)≥ blog2pc+2, we obtain that C17(K)≥6.
The main result of this article is to show that C17(K) = 6. It is still an open question
whether for any p-colorable knot K, the equality Cp(K) = blog2pc+2 holds. In addition
to the results already known, regarding the problem of finding the minimum number of
colors for p-colorable knots, the result of this article reinforces the validity that equality
holds.
2 Any 17-colorable knot can be colored by six colors
Through this article we will adopt the same notations as in [3]. So we will use {a|b|c}
to denote a crossing, as in Figure 1 where bis the color of the over-arc and aand care
the colors of the under-arcs with a+c= 2bmodulo 17. When the crossing is of the type
{a|a|a}(trivial coloring), we will omit over and under-arcs and draw them crossing each
other.
Figure 1: The coloring {a|b|c}.
Our main result is the following
Theorem 2.1. Any 17-colorable knot has a 17-colored diagram with exactly six colors.
Proof. Let Dbe a non-trivially 17-colored knot diagram of a knot K. We will show that
the integers {0,2,3,4,8,12}are enough to color K. To do this, we will proceed by steps.
2

At the step number iwe will prove that one can do without the color ci, which is the
i-th number in the ordered list {16,15,9,10,6,7,5,1,11,14,13}.
We will start by proving that we can modify Dto get an equivalent colored diagram
D1where the color c1= 16 is not used. The step i,i≥2, consists in showing that if
one begins with a colored diagram Di−1in which none of the already discarded colors
{c1, . . . , ci−1}is used, then one can modify Di−1to get a new equivalent colored diagram
Diwhere none of the colors {c1, . . . , ci}appear. Note that any color ccan occur in D
in three ways:
•at a crossing of the form {c|c|c},
•or on an over-arc at a crossing {a|c|2c−a}for some color a
•or as the color of an under-arc that connects two crossings of the type {2a−c|a|c}
and {c|b|2b−c}for some colors aand b.
Then at each step, we will show that in each one of these three cases, one can modify
the diagram such that the color cwill be eliminated.
In all the figures we will use, we denote by cthe color we want to drop. To make things
clear, we start by dealing with the first step when c= 16. We will show that there is a
non-trivially equivalent 17-colored diagram D1with no arc colored by 16.
Case 1: Assume that Dhas a crossing of type {16|16|16}. Then Dwill necessarily
have one of the two crossings, {2a+ 1|a|16}or {a|16|15 −a}for some a6= 16. Since
2a+ 1 6= 16 and 15 −a6= 16, we deform the arc colored by aas shown in Figure 2 in the
case of the first crossing, or as shown in Figure 3 in the case of the second crossing. Each
of those two deformations provides an equivalent diagram where the crossing {16|16|16}
disappeared.
Figure 2: Transformation of {c|c|c}when ais the color of an over-arc.
In the case of the second crossing, we do the deformation described in Figure 3.
3

Figure 3: Transformation of the crossing {c|c|c}when ais the color of an under-arc.
Case 2: Assume that Dhas a crossing whose over-arc has the color 16, i.e. it is
of the type {a|16|15 −a}for some a6= 16. Then we deform Das shown in Figure
4. We easily check that the generated colors 2a+ 1 and 3a+ 2 are both distinct from
16. Furthermore there is no more over-arc with color 16 in the region concerned by the
modification.
Figure 4
Case 3: Assume that Dhas a crossing whose under arc is colored by 16. Then
this under-arc will connect a crossing of the type {2a+ 1|a|16}to a crossing of type
{16|b|2b+ 1}for some aand bdistinct from 16. If a=b, the deformation shown in
Figure 5 allows to eliminate the color 16. If a6=b, we do the deformation described in
Figure 6 and then the color 16 disappears unless when 2a−b= 16 i.e. b= 2a+ 1. In this
case we apply to Dthe transformation shown in Figure 7. Finally, we get an equivalent
diagram D1in which no arc has the color 16.
Figure 5
4

Figure 6
Figure 7
Now we will deal with a general step i,i≥2. Assume that we have a diagram
Di−1that is equivalent to Dwhere the colors {c1, . . . , ci−1}are not used. We want
to show that there exists an equivalent colored diagram Diwhich does not use colors
{c1, . . . , ci−1, ci}. Here, ciwill be denoted by cas in the figures. Like in the first step,
we will consider the three cases:
Case 1 Assume that Di−1has a crossing of the type {c|c|c}. Then there exists a crossing
of type {2a−c|a|c}or {a|c|2c−a}for some adistinct from cand a /∈ {c1, . . . , ci−1}.
In the case of the first crossing we deform the arc colored by aas indicated in Figure 2
which results in the crossing {c|c|c}disappearing.
In the case of the second crossing, we do the deformation described in Figure 3. The
obtained color 2a−cwill be different from cand ckiff a6=cand a6= 9(c+ck), for each
ksuch that 1 ≤k≤i−1.
If a= 9(c+ck) we resolve the problem by making the deformation of Figure 8, unless if
(c, ck) = (7,16) or (c, ck) = (7,15) which occur in the sixth step (i.e. i= 6). For those
cases we will apply to the diagram Di−1=D5one of the deformations described in the
Figure 9 according to the value of a. So, we eliminate all crossings of the type {c|c|c}.
5

Figure 8
Figure 9
Case 2 Assume that Di−1has a crossing whose over-arc is of color c=ci, i.e. it is of
the type {a|c|2c−a}for some adifferent from cand ck, for each k, 1 ≤k≤i−1. We
deform the diagram Di−1as shown in Figure 4.
This deformation provides the two new colors 2a−cand 3a−2c, which are different from
cand ckiff a6=c,a6= 9(c+ck) and a6= 6(ck+ 2c). If a= 9(c+ck) or a= 6(ck+ 2c) for
some k, the deformation of Figure 10 resolves the problem except when (c, ck) = (7,16)
or (c, ck) = (7,15) wich occur in the sixth step (i.e. i= 6). For the two remaining cases
we resolve the problem by applying to Di−1=D5one of the deformations described
Figure 11 according to the value of a.
6

Figure 10
Figure 11
Case 3 Assume that Di−1has a crossing whose under-arc is colored by c=ci. Then c
connects two crossings of the type {2a−c|a|c}and {c|b|2b−c}for some aand bboth
distinct from cand ck, for each k, 1 ≤k≤i−1.
If a=b, we apply to the diagram Di−1the deformation shown in Figure 5. We get
the two new colors 3a−2cand 4a−3c. They are different from cand ckiff a6=c
a6= 6(ck+ 2c) and a6= 13(ck+ 3c), for each k, 1 ≤k≤i−1.
For the remaining cases, if a= 6(ck+2c) or a= 13(ck+3c) (obviously a6=cand a6=ck),
some other transformations are required. They are listed in the following table.
7

Step (c, ck)a=
6(ck+ 2c)
Required
deformation
2 (15,16) 4 Fig. 12
3 (9,16) 0 Fig. 14
(9,15) 11 Fig. 12
4 (10,16) 12 Fig. 12
(10,15) 6 Fig. 15
5 (6,10) 13 Fig. 12
6 (7,15) 4 Fig. 14
(7,9) 2 Fig. 15
(7,6) 1 Fig. 20
7 (5,16) 3 Fig. 21
(5,10) 1 Fig. 16
(5,6) 11 Fig. 22
(5,7) 0 Fig. 23
9 (11,7) 4 Fig. 29
11 (13,14) 2 Fig. 35
Step (c, ck)a=
13(ck+ 3c)
Required
deformation
2 (15,16) 11 Fig. 13
3 (9,15) 2 Fig. 13
4 (10,16) 3 Fig. 13
(10,15) 7 Fig. 17
(10,9) 14 Fig. 13
5 (6,16) 0 Fig. 13
(6,15) 4 Fig. 15
(6,10) 7 Fig. 16
6 (7,16) 5 Fig. 18
(7,10) 12 Fig. 19
7 (5,16) 12 Fig. 16
8 (1,15) 13 Fig. 24
(1,7) 11 Fig. 25
(1,5) 2 Fig. 26
9 (11,15) 12 Fig. 27
(11,6) 14 Fig. 28
10 (14,9) 0 Fig. 30
(14,10) 13 Fig. 31
(14,7) 8 Fig. 32
11 (13,10) 8 Fig. 33
(13,11) 4 Fig. 34
Table 1: List of the remaining cases at each step and the corresponding deformations.
Figure 12
Figure 13
8

Figure 14
Figure 15
Figure 16
9

Figure 17
Figure 18
Figure 19
10

Figure 20
Figure 21
Figure 22
11

Figure 23
Figure 24
Figure 25
12

Figure 26
Figure 27
Figure 28
13

Figure 29
Figure 30
Figure 31
14

Figure 32
Figure 33
15

Figure 34
Figure 35
16

If a6=b, we do the deformation described in Figure 6. We get the two new colors 2a−b
and 2a−2b+c. They are different from cand ckiff b6= 2a−c,b6= 2a−ckand
b6=a+ 9c−9ck, for each k, 1 ≤k≤i−1. Then the color cdisappears and none of the
the colors ckappears.
Now if b= 2a−cor b= 2a−ckor b=a+ 9c−9ck, then we apply to Di−1the
transformation shown in Figure 7. We obtain the new colors 2b−aand 2b−2a+c. They
are different from c,ckand clwhere l6=kfor each l,k, 1 ≤l, k ≤i−1, iff (a, b) is distinct
from (6ck+ 12c, 12ck+ 6c), (6c+ 12ck,12c+ 6ck), (10ck+ 8c, 2ck−c), (2ck−c, 10ck+ 8c),
(6cl+ 12ck,12cl+ 6ck), (cl+ck−c, cl+ 9ck+ 8c) and (9cl+ck+ 8c, cl+ck−c). when
(a, b) is one of those pairs, we will apply to the diagram Di−1different deformations
which will be indicated in the following tables. Finally we get a diagram Diequivalent
to Di−1in which no arc has the color ci.
We remark that in all those cases, the colors aand bplay symmetric roles. Then the
adequate figures are similar. In such cases, we fill just one box in the table and the other
is left blank. For example, in the first table, when (c, ck) = (15,16), we get (a, b) = (4,10)
and (a, b) = (10,4). The deformation in Figure 36 allows to resolve the problem in the
two cases in a similar way.
Step (c, ck) (a, b) =
(6ck+ 12c, 12ck+ 6c)
Required
deformation
(a, b) =
(6c+ 12ck,12c+ 6ck)
Required
deformation
2 (15,16) (4,10) (10,4) Fig. 36
3 (9,16) (0,8) (8,0) Fig. 36
(9,15) (11,13) (13,11) Fig. 36
4 (10,16) (12,14) (14,12) Fig. 40
(10,15) (6,2) (2,6) Fig. 37
5 (6,10) (13,3) (3,13) Fig. 36
6 (7,15) (4,1) Fig. 42 (1,4)
(7,9) (2,14) (14,2) Fig. 37
(7,6) (1,12) (12,1) Fig. 43
7 (5,16) (3,1) (1,3) Fig. 46
(5,6) (11,0) (0,11) Fig. 45
(5,7) (0,12) (12,0) Fig. 50
9 (11,7) (4,14) (14,4) Fig. 53
11 (13,14) (2,8) Fig. 56 (8,2)
Table 2: Table of (a, b) = (6ck+ 12c, 12ck+ 6c) or (a, b) = (6c+ 12ck,12c+ 6ck).
Step (c, ck) (a, b) =
(10ck+ 8c, 2ck−c)
Required
deforma-
tion
(a, b) =
(2ck−c, 10ck+ 8c)
Required
deforma-
tion
2 (15,16) (8,0) (0,8) Fig. 36
3 (9,16) (11,6) (6,11) Fig. 36
4 (10,16) (2,5) (5,2) Fig. 38
(10,9) (0,8) (8,0) Fig. 36
5 (6,10) (12,14) Fig. 36 (14,12)
6 (7,6) (14,5) Fig. 36 (5,14)
7 (5,15) (3,8) (8,3) Fig. 36
(5,9) (11,13) (13,11) Fig. 36
Table 3: Table of (a, b) = (10ck+ 8c, 2ck−c) or (a, b) = (2ck−c, 10ck+ 8c).
17

Step (c, ck, cl) (a, b) =
(9cl+ck+8c, cl+ck−c)
Required
defor-
mation
(c, ck, cl) (a, b) =
(cl+ck−c, cl+9ck+8c)
Required
defor-
mation
3 (9,16,15) (2,5) (9,15,16) (5,2) Fig. 36
(9,15,16) (10,5) Fig. 36 (9,16,15) (5,10)
4 (10,9,15) (3,14) (10,15,9) (14,3) Fig. 38
(10,15,9) (6,14) (10,9,15) (14,6) Fig. 38
5 (6,16,10) (1,3) (6,10,16) (3,1) Fig. 38
(6,9,15) (5,1) (6,15,9) (1,5) Fig. 36
6 (7,9,16) (5,1) Fig. 40 (7,16,9) (1,5)
(7,10,15) (14,1) Fig. 41 (7,15,10) (1,14)
(7,9,10) (2,12) (7,10,9) (12,2) Fig. 44
7 (5,16,7) (0,1) (5,7,16) (1,0) Fig. 39
(5,7,16) (4,1) (5,16,7) (1,4) Fig. 47
(5,16,6) (8,0) Fig. 49 (5,6,16) (0,8)
(5,6,16) (3,0) (5,16,6) (0,3) Fig. 39
(5,7,10) (1,12) Fig. 48 (5,10,7) (12,1)
(5,10,7) (11,12) (5,7,10) (12,11) Fig. 51
8 (1,9,5) (11,13) (1,5,9) (13,11) Fig. 52
9 (11,5,9) (4,3) (11,9,5) (3,4) Fig. 37
(11,9,16) (3,14) (11,16,9) (14,3) Fig. 38
(11,10,15) (12,14) (11,15,10) (14,12) Fig. 54
10 (14,16,15) (8,0) Fig. 55 (14,15,16) (0,8)
(14,9,5) (13,0) (14,5,9) (0,13) Fig. 38
11 (13,16,14) (8,0) (13,14,16) (0,8) Fig. 37
(13,6,9) (4,2) Fig. 57 (13,9,6) (2,4)
Table 4: Table of (a, b) = (9cl+ck+ 8c, cl+ck−c) or (a, b)=(cl+ck−c, cl+ 9ck+ 8c).
Step (c, ck, cl) (a, b) =
(6cl+ 12ck,12cl+ 6ck)
Required deformation
6 (7,10,16) (12,14)
(7,16,10) (14,12) Fig. 36
7 (5,10,6) (3,13)
(5,6,10) (13,3) Fig. 37
Table 5: Table of (a, b) = (6cl+ 12ck,12cl+ 6ck).
Figure 36
Figure 37
18

Figure 38
Figure 39
Figure 40
19

Figure 41
Figure 42
Figure 43
20

Figure 44
Figure 45
Figure 46
21

Figure 47
Figure 48
Figure 49
22

Figure 50
Figure 51
Figure 52
23

Figure 53
Figure 54
Figure 55
24

Figure 56
Figure 57
25

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