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

Graduate J. Math. 555 (2020), 122 – 137
Hamid Abchir, Mohamed Elhamdadi, Soukaina Lamsifer
Abstract
We investigate Fox colorings of knots that are 17-colorable. Pre-
cisely, we prove that any 17-colorable knot has a diagram such that
exactly 6among the seventeen colors are assigned to the arcs of the
diagram.
MSC 2010. Primary: 57M25. Secondary: 57K12.
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 pro-
teins 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 devel-
oped by R. Fox around 1960 (see [5]). He introduced
a diagrammatic definition of colorability of a knot by
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 m,
called the color of that arc, such that, at each cross-
ing, the sum of the colors of the under-arcs is twice
the color assigned to the over arc modulo m(see Fig-
ure 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 num-
ber 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
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. Elham-
dadi 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. Nakan-
ishi 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 ques-
tion 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 equal-
ity 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 a
and care the colors of the under-arcs with a+c= 2b
modulo 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.
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 dia-
gram 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. At the step number iwe will
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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}
Fig. 1: The coloring {a|b|c}.
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 Din 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 col-
ored by aas shown in Figure 2 in the case of the first
crossing, or as shown in Figure 3 in the case of the sec-
ond crossing. Each of those two deformations provides
an equivalent diagram where the crossing {16|16|16}
disappeared.
Fig. 2: Transformation of {c|c|c}when ais the color of an over-
arc.
In the case of the second crossing, we do the defor-
mation described in Figure 3.
Fig. 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.
Fig. 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.
Fig. 5
Now we will deal with a general step i,i≥2. As-
sume 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
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Fig. 6
Fig. 7
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 de-
form 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 defor-
mation described in Figure 3. The obtained color 2a−c
will 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}.
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−c
and 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
Fig. 8
Fig. 9
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 de-
scribed Figure 11 according to the value of a.
Fig. 10
Fig. 11
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Case 3. Assume that Di−1has a crossing whose under-
arc is colored by c=ci. Then cconnects 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 defor-
mation shown in Figure 5. We get the two new colors
3a−2cand 4a−3c. They are different from cand ck
iff a6=c a 6= 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.
Step (c, ck)a=
6(ck+ 2c)
Required
deforma-
tion
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
deforma-
tion
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
Tab. 1: List of the remaining cases at each step and the corre-
sponding deformations.
Fig. 12
Fig. 13
Fig. 14
125

Fig. 15
Fig. 16
Fig. 17
Fig. 18
126

Fig. 19
Fig. 20
Fig. 21
Fig. 22
127

Fig. 23
Fig. 24
Fig. 25
Fig. 26
128

Fig. 27
Fig. 28
Fig. 29
Fig. 30
Fig. 31
129

Fig. 32
Fig. 33
If a6=b, we do the deformation described in Figure 6. We get
the two new colors 2a−band 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 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−1
different deformations which will be indicated in tables 2to 5
next. 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.
Fig. 34
130

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)
Tab. 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
deformation
(a, b) =
(2ck−c, 10ck+ 8c)
Required
deformation
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
Tab. 3: Table of (a, b) = (10ck+ 8c, 2ck−c)
or (a, b) = (2ck−c, 10ck+ 8c).
Step (c, ck, cl) (a, b) = (9cl+ck+
8c, cl+ck−c)
Required
deformation
(c, ck, cl) (a, b) = (cl+ck−
c, cl+ 9ck+ 8c)
Required
deformation
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)
Tab. 4: Table of (a, b) = (9cl+ck+ 8c, cl+ck−c)
or (a, b) = (cl+ck−c, cl+ 9ck+ 8c).
131

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
Tab. 5: Table of (a, b) = (6cl+ 12ck,12cl+ 6ck).
Fig. 35
Fig. 36 Fig. 37
132

Fig. 38
Fig. 39
Fig. 40
Fig. 41
Fig. 42
Fig. 43
133

Fig. 44
Fig. 45
Fig. 46
Fig. 47
Fig. 48
134

Fig. 49
Fig. 50
Fig. 51
Fig. 52
135

Fig. 53
Fig. 54
Fig. 55
Fig. 56
Fig. 57
136

This completes the proof of our Theorem 2.1.
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Hamid Abchir
Hassan II University,
Casablanca, Morocco
E-mail address:hamid.abchir@univh2c.ma
Mohamed Elhamdadi
University of South Florida, Tampa, Florida, USA.
E-mail address:emohamed@usf.edu
Soukaina Lamsifer
Hassan II University,
Casablanca, Morocco
E-mail address:soukainalamsifer@gmail.com
137