# Quandle Coloring Quivers of general Torus links by dihedral quandles

Mohamed Elhamdadi  Department of Mathematics, University of South Florida,
Tampa, FL 33620<EMAIL_ADDRESS>,  Brooke Jones  Department of Mathematics,
University of South Florida, Tampa, FL 33620<EMAIL_ADDRESS>and
Minghui Liu  Florida College, Temple Terrace, FL 33617
<EMAIL_ADDRESS>
###### Abstract.

We completely characterize the coloring quivers of general torus links by
dihedral quandles by first exhausting all possible numbers of colorings,
followed by determining the interconnections between colorings in each case.
The quiver is obtained as function of the number of colorings. The quiver
always contains complete subgraphs, in particular a complete subgraph
corresponding to the trivial colorings, but the total number of subgraphs in
the quiver and the weights of their edges varies depending on the number of
colorings.

##  1\. Introduction

The concept of a quiver and its representations was introduced in [Gabriel] by
P. Gabriel in 1972. A quiver is an oriented graph; precisely, it is a pair
$Q=(V,E)$, where $V$ is a finite set of vertices and $E$ is a finite set of
arrows between them. Quivers have been used in many areas of mathematics such
as representation theory of finite dimensional algebras, ring theory
(Gabriel’s theorem), Lie algebras (Dynkin diagrams) and quantum groups among
others. A quiver structure on the set of quandle colorings of an oriented knot
was introduced in [CN] where some enhancements of the counting invariant were
obtained from the quiver structure. In [BC], quandle coloring quivers of
$(2,q)$-torus links by dihedral quandles were studied. The case of
$(3,q)$-torus links was investigated in [ZL]. The main goal of this paper is
to generalize the results of [BC, ZL] to any torus link $T(p,q)$. To give an
overview of the process, we decided to first cover the cases T(5,q) and T(7,q)
before the general case T(p,q). In investigating dihedral quandle coloring
quivers of general torus links, we first exhaust all possible numbers of
colorings of the torus links by dihedral quandles $R_{n}$. The quiver is
obtained as function of the number of colorings. Precisely, (1) if the number
of colorings is $n$ then the quiver is the complete graph on $n$ vertices with
each edge having weight $n$. (2) if the number of colorings is $pn$ then the
quiver graph is composed of two complete graphs, one graph on $n$ vertices and
the other on $(p-1)n$ vertices, such that there is a directed edge with weight
$\frac{n}{p}$ from one graph to the other. (3) if the number of colorings is
$2^{p-1}n$, then the quiver contains $2^{p-1}$ complete subgraphs of order
$n$. (4) if the number of colorings is $n^{p}$, we obtain
$\frac{n^{p}-n}{n(n-1)}$ disjoint subgraphs with vertices each having edges to
a complete subgraph on $n$ vertices. While some of the complete subgraphs of
the quivers are of different orders, we also find that the weights of the
edges between them vary depending on the number of colorings.

The paper is organized as follows. In Section 2 we review the basics of
quandles and link colorings. Section 3 gives the quandle coloring quivers of
the torus links $T(5,q)$ and $T(7,q)$. In Section 4, we give the main result
of the paper in which we characterize the quandle coloring quivers of general
torus links $T(p,q)$, where $p$ is prime.

##  2\. A review of Quandles and link colorings

In this section, we collect some basics about quandles that we will need
through the paper. We begin with the following definition taken from [EN,
Joyce, Matveev].

######  Definition 2.1.

A quandle is a set $X$ with a binary operation

$*:X\times X\longrightarrow X$ satisfying the following three axioms:

  * (i)

(right distributivity) for all $x,y,z\in X$, we have $(x*y)*z=(x*z)*(y*z)$;

  * (ii)

(invertibility) for all $x\in X$, the map $R_{x}:X\longrightarrow X$ sending
$y$ to $y*x$ is a bijection;

  * (iii)

(idempotency) for all $x\in X$, $x*x=x.$

Since axiom (ii) states that for each $y\in X$, the map $R_{y}$ is invertible,
we will denote $R_{y}^{-1}(x)$ by $x\bar{*}y$. When the binary operation
satisfies $(x*y)*y=x$ for all $x,y\in X$, then the quandles is called a kei
(or involutive quandle). Let $n$ be a positive integer. For
$x,y\in\mathbb{Z}_{n}$ (integers modulo $n$), define $x*y=2y-x\pmod{n}$. Then
the operation $*$ defines a quandle structure called dihedral quandle and is
denoted $R_{n}$.

Let $D$ be a diagram of an oriented link $K$ and let $X$ be a quandle. A
coloring of $D$ by the quandle $X$ is a map
${\mathcal{C}}:{\mathcal{A}}\rightarrow X$ from the set of arcs
${\mathcal{A}}$ of the diagram $D$ to $X$ such that the image of the map
satisfies the relation depicted in Figure 1 at each crossing. For more on
colorings of knots by quandles the reader can consult for example [EN, Joyce,
Matveev].

Figure 1. Colorings of arcs at positive and negative crossings

For any link $K$, there is an associated fundamental quandle $Q(K)$ of the
knot $K$ given by presentation with generators corresponding to the arcs of a
diagram $D$ of $K$ and quandle relations at crossings of $D$ as in Figure 1.
The set of colorings of a knot $K$ by a quandle $X$, then, is in one-to-one
correspondence with the set of quandle homomorphisms from the fundamental
quandle $Q(K)$ of $K$ to $X$. For example, the fundamental quandle of _figure
eight_ knot can be obtained from Figure 2. It is given by $Q(4_{1})=\langle
x,y,z,w;\;x\bar{*}z=y,z\bar{*}x=w,w*y=x,y*w=z\rangle=\langle
x,y,z;\;y*z=x,(z\bar{*}x)*y=x,y*(z\bar{*}x)=z\rangle$.

Figure 2. Oriented figure-8 knot with labeled arcs

The set of quandle homomorphisms from the fundamental quandle $Q(K)$ to the
quandle $X$ is denoted by $\textup{Hom}(Q(K),X)$. This set has been used to
define computable link invariants, for example, the cardinality of the set
$\textup{Hom}(Q(K),X)$ is known as the _quandle counting invariant_ [EN].

Notations: Using a similar notation as in [ZL], we denote the directed
complete graph on $n$ vertices with $\omega$ total number of edges between any
two vertices in the graph as $(\overleftrightarrow{K_{n}},\hat{\omega})$. When
two graphs $G_{1}$ and $G_{2}$ have disjoint vertex sets and edge sets, and
there are $d$ edges from each vertex of $G_{2}$ to each vertex of $G_{1}$, the
graph will be denoted as $G_{1}\overleftarrow{\nabla}_{\hat{d}}G_{2}$. Lastly,
the disjoint union of $m$ copies of the graph $G$ will be denoted as
$\bigsqcup\limits_{m}G$.

Known results: The only cases where quandle coloring quivers for torus links
are completely known are the _two_ cases of $T(2,q)$ and $T(3,q)$. In [BC] the
structure of the quandle coloring quiver of a $T(2,q)$ with respect to a
dihedral quandle of order $n$, denoted $R_{n}$, was given. In [ZL] the
coloring quiver was studied for $T(3,q)$. We give the known results for
$\mathcal{Q}_{R_{n}}(T(p,q))$ for $p=2,3$ in the table below.

Number of colorings $N$ | Quivers when $p=2$ | Quivers when $p=3$
---|---|---
When $N=n$ | $(\overleftrightarrow{K_{n}},\hat{n})$ | $(\overleftrightarrow{K_{n}},\hat{n})$
When $N=pn$ | $(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}_{\hat{\frac{n}{2}}}\left(\overleftrightarrow{K_{n}},\hat{\frac{n}{2}}\right)$ | $(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}_{\hat{\frac{n}{3}}}\left(\overleftrightarrow{K_{2n}},\hat{\frac{n}{3}}\right)$
When $N=n\cdot 2^{p-1}$ | $(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}_{\hat{\frac{n}{2}}}\left(\overleftrightarrow{K_{n}},\hat{\frac{n}{2}}\right)$ | $(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}_{\hat{\frac{n}{2}}}\big{[}\bigsqcup\limits_{3}\left(\overleftrightarrow{K_{n}},\hat{\frac{n}{2}}\right)\big{]}$
When $N=n^{p}$ | $(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}\left(\overleftrightarrow{K_{n(n-1)}},\hat{1}\right)$ | $(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}\big{[}\bigsqcup\limits_{n+1}\left(\overleftrightarrow{K_{n(n-1)}},\hat{1}\right)\big{]}$

Table 1. Quandle coloring quivers of $T(2,q)$ [BC] and $T(3,q)$ [ZL] by
$R_{n}$.

Our results: We solve the problem of determining the quandle coloring quivers
of a general torus links $T(p,q)$ by dihedral quandles. We give a summary of
the result in the following chart. For the details see the sections below. We
begin by deriving the quandle coloring quivers in the $p=5$ and $p=7$ cases to
give an overview of the process, then we generalize to $T(p,q)$ for any prime
$p$.

Number of colorings $N$ | Quivers when $p$ is a prime
---|---
When $N=n$ | $(\overleftrightarrow{K_{n}},\hat{n})$
When $N=pn$ | $(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}_{\hat{\frac{n}{p}}}\left(\overleftrightarrow{K_{(p-1)n}},\hat{\frac{n}{p}}\right)$
When $N=n\cdot 2^{p-1}$ | $(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}_{\hat{\frac{n}{2}}}\big{[}\bigsqcup\limits_{2^{p-1}-1}\left(\overleftrightarrow{K_{n}},\hat{\frac{n}{2}}\right)\big{]}$
When $N=n^{p}$ | $(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}\big{[}\bigsqcup\limits_{\frac{n^{p}-n}{n(n-1)}}\left(\overleftrightarrow{K_{n(n-1)}},\hat{1}\right)\big{]}$

Table 2. Quandle coloring quivers of $T(p,q)$ by $R_{n}$ (our main result).

##  3\. Quandle Coloring Quivers of $(5,q)$ and $(7,q)$-Torus links with
Dihedral Quandles

To further the understanding of deriving the quandle coloring quivers of
$T(p,q)$-torus links with dihedral quandles, we go through the process of
determining the number of colorings of $T(p,q)$ by the dihedral quandle
$R_{n}$ and the quivers when $p=5$ and when $p=7$.

Quandle Coloring Quivers of $(5,q)$-Torus links with Dihedral Quandles: Using
the braid form $(\sigma_{1}\sigma_{2}\sigma_{3}\sigma_{4})^{q}$ of the torus
link $T(5,q)$ as shown in Figure 3, we calculate $N$, where $N$ is the number
of colorings of $T(5,q)$ by the dihedral quandle $R_{n}$, for varying values
of $q$.

Figure 3. Coloring of the braid
$(\sigma_{1}\sigma_{2}\sigma_{3}\sigma_{4})^{q}$ whose closure is $T(5,q)$

Let $f$ be a coloring of the torus link $T(5,q)$ with $f(x_{i})=y_{i}$ for
$i=1,...,5$ as in Figure 3. Coloring the braid
$(\sigma_{1}\sigma_{2}\sigma_{3}\sigma_{4})^{q}$ by the dihedral quandle and
using the fact that $f(x_{i})=y_{i}$ gives the following equation:

$y_{i}=y_{5}+(-1)^{q+1}y_{q}+(-1)^{q}y_{i+q}$ (1)

Now, using equation (1), we calculate $N$ for the following values of $q$:

  1. (1)

Let $q=10k$. A straightforward computation gives that all $y_{i}$ are free
variables and thus the number of colorings is $N=n^{5}$.

  2. (2)

Let $q=10k+1$. The computation gives $y_{1}=y_{2}=y_{3}=y_{4}=y_{5}$.
Therefore, $N=n$.

  3. (3)

Let $q=10k+2$. The computation gives $5(y_{i}-y_{j})=0\mod n$. Therefore when
$\gcd(5,n)=1$, we have $N=n$, and when $\gcd(5,n)=5$, we have $N=5n$.

  4. (4)

Let $q=10k+3$. The computation gives $y_{1}=y_{2}=y_{3}=y_{4}=y_{5}$.
Therefore, $N=n$.

  5. (5)

Let $q=10k+4$. The computation gives $5(y_{i}-y_{j})=0\mod n$. Therefore when
$\gcd(5,n)=1$, we have $N=n$, and when $\gcd(5,n)=5$, we have $N=5n$.

  6. (6)

Let $q=10k+5$. The computation gives $2(y_{i}-y_{j})=0\mod n$. Therefore when
$\gcd(2,n)=1$, we have $N=n$, and when $\gcd(2,n)=2$, we have $N=2^{4}n$.

  7. (7)

Let $q=10k+6$. The computation gives $5(y_{i}-y_{j})=0\mod n$. Therefore when
$\gcd(5,n)=1$, we have $N=n$, and when $\gcd(5,n)=5$, we have $N=5n$.

  8. (8)

Let $q=10k+7$. The computation gives $y_{1}=y_{2}=y_{3}=y_{4}=y_{5}$, and so
$N=n$.

  9. (9)

Let $q=10k+8$. The computation gives $5(y_{i}-y_{j})=0\mod n$. Therefore when
$\gcd(5,n)=1$, we have $N=n$, and when $\gcd(5,n)=5$, we have $N=5n$.

  10. (10)

Let $q=10k+9$. The computation gives $y_{1}=y_{2}=y_{3}=y_{4}=y_{5}$, and so
$N=n$.

We summarize these results in the following table below.

$q$ | Number of colorings $N$ when $p=5$
---|---
When $q\neq 0\mod p$ is odd | $N=n$
When $q\neq 0\mod p$ is even | $N=n\cdot\gcd(5,n)$
When $q=p\mod 2p$ | When $n$ is even, $N=2^{4}n$. When $n$ is odd, $N=n$
When $q=0\mod 2p$ | $N=n^{5}$

Table 3. Number of colorings of $T(5,q)$ by $R_{n}$

We characterize in this section all the quandle coloring quivers of the torus
links $T(5,q)$ with the dihedral quandle $R_{n}$.

In the following four theorems we let $T(5,q)$ be a torus link and $R_{n}$ be
the dihedral quandle.

######  Theorem 3.1.

If $|Hom(T(5,q),R_{n})|=n$, then the full quandle coloring quiver is the
complete directed graph:

$\mathcal{Q}_{R_{n}}(T(5,q))=(\overleftrightarrow{K_{n}},\hat{n}).$

###### Proof.

If $|Hom(T(5,q),R_{n})|=n$, this corresponds to the $n$ trivial colorings. Now
assume that $f$ and $g$ are both trivial colorings given respectively by
$f(x_{i})=j$ and $g(x_{i})=k$ for all $x_{i}\in Q(T(5,q))$. Since any
endomorphism of the dihedral quandle, $R_{n}$, has the form

$\phi(f)=af+b$ (2)

where $a,b\in R_{n}$ (see [EMR]), then there are exactly $n$ solutions of
endomorphisms $\phi$ making equation (2) hold. Thus we obtain the complete
graph $\mathcal{Q}_{R_{n}}(T(5,q))=(\overleftrightarrow{K_{n}},\hat{n})$. ∎

######  Theorem 3.2.

If $|Hom(T(5,q),R_{n})|=5n$, then the full quandle coloring quiver is the
directed graph:

$\mathcal{Q}_{R_{n}}(T(5,q))=(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}_{\hat{d}}\left(\overleftrightarrow{K_{4n}},\hat{d}\right),$

where $d=\frac{n}{5}$.

###### Proof.

Let $f,g\in Hom(Q(T(5,q)),R_{n})$ be two vertices of the quiver. Since any
endomorphism of the dihedral quandle $R_{n}$ is given by equation (2), we need
to find the number of endomorphisms $\phi$ making equation (2) hold. We know
when there are $N=5n$ colorings, then $gcd(5,n)=5$. Now, since there are no
edges from a trivial coloring to a non-trivial coloring, we consider the
following $3$ cases

  1. (1)

Case 1. It is clear that if $f$ and $g$ are trivial colorings, then there are
$n$ possible solutions to equation (2). Therefore, we have $n$ edges between
trivial colorings

  2. (2)

Case 2. When $f$ is a nontrivial coloring, and $g$ is a trivial coloring, so
$g(x_{i})=k$ for all $x_{i}$, equation (2) becomes $a(f(x_{i})-f(x_{j}))=0\mod
n$ for all $i,j$.Thus, we have $\frac{n}{5}$ edges from nontrivial colorings
to trivial colorings.

  3. (3)

Case 3. When $f$ and $g$ are both non-trivial colorings, then $f$ and $g$ are
given by $g(x_{1})=g(x_{3})=g(x_{5})\neq g(x_{2})=g(x_{4})$, and
$5(g(x_{1})-g(x_{2}))=0$ modulo $n$. Since the equation
$(g(x_{1})-g(x_{2}))-a(f(x_{1})-f(x_{2}))=0$ in $R_{n}$ has $\frac{n}{5}$
solutions for $a$, we get the result.

∎

######  Theorem 3.3.

If $|Hom(T(5,q),R_{n})|=16n$, then the full quandle coloring quiver is the
directed graph:

$\mathcal{Q}_{R_{n}}(T(5,q))=(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}_{\hat{d}}\big{[}\bigsqcup\limits_{15}\left(\overleftrightarrow{K_{n}},\hat{d}\right)\big{]},$

where $d=\frac{n}{2}$.

###### Proof.

In order to have $N=16n$ colorings, we know $n$ must be even and for any
coloring $f$, $2(f(x_{i})-f(x_{j}))=0$ for any $i,j$. The $16n$ colorings
consist of $n$ trivial colorings and $15n$ nontrivial colorings. By Theorem
3.1, the $n$ trivial colorings correspond to a subgraph
$(\overleftrightarrow{K_{n}},\hat{n})$ of $\mathcal{Q}_{R_{n}}(T(5,q))$. Now,
there are $15n$ nontrivial colorings, so let $f$ be a nontrivial coloring,
thus we have for some $j$, $f(x_{i})=f(x_{1})+m_{i-1}$ for $1<i\leq 5$ where
$m_{i-1}=0$ or $\frac{n}{2}$. Since $n$ is even and $f$ is nontrivial, there
exists $i$ such that $m_{i-1}=\frac{n}{2}$. We know for some coloring $g\in
Hom(Q(T(5,q)),R_{n})$, $g(x_{i})=g(x_{1})+m^{\prime}_{i-1}$ where
$m^{\prime}_{i-1}=0$ or $\frac{n}{2}$ . By satisfying equation (2), we have
$am_{i-1}=m^{\prime}_{i-1}$ for $1<i\leq 5$. If any $m_{i-1}=0$, then
$m^{\prime}_{i-1}=0$. Thus, consider the $m_{i-1}$ such that
$m_{i-1}=\frac{n}{2}$ and so we have $a\frac{n}{2}=0$ or $n/2$. If
$a\frac{n}{2}=0$ then $a$ is even, i.e. $g$ must be trivial, and we have $n/2$
edges between nontrivial colorings to trivial colorings. If
$a\frac{n}{2}=\frac{n}{2}$ then $a$ is odd, so
$m_{i-1}=m^{\prime}_{i-1}=\frac{n}{2}$, and we have $n/2$ edges between
nontrivial colorings to nontrivial colorings. For each choice of
$m_{1},m_{2},m_{3},m_{4}$, there are $n$ colorings which divides the
nontrivial colorings into 15 disjoint
$\left(\overleftrightarrow{K_{n}},\hat{\frac{n}{2}}\right)$.

∎

######  Theorem 3.4.

If $|Hom(T(5,q),R_{n})|=n^{5}$, where $n$ is prime, then the full quandle
coloring quiver is the directed graph:

$\mathcal{Q}_{R_{n}}(T(5,q))=(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}\big{[}\bigsqcup\limits_{m}\left(\overleftrightarrow{K_{n(n-1)}},\hat{1}\right)\big{]},$

where $m=\frac{n^{5}-n}{n(n-1)}$.

###### Proof.

In this case, $y_{1},...,y_{5}$ are free elements of $R_{n}$, so there are
$n^{5}$ colorings (vertices), and we need to determine the edges in this
quiver. By Theorem 3.1 we will have the $n$ trivial colorings corresponding to
the subgraph $(\overleftrightarrow{K_{n}},\hat{n})$. Now, there are $n^{5}-n$
nontrivial colorings. Let $f$ be a nontrivial coloring, thus we have

$f(x_{i})=f(x_{1})+m_{i-1}$

where $1<i\leq 5$ and $0\leq m_{i-1}\leq n-1$. Since $f$ is nontrivial, there
exists an index $j$ such that $m_{j}\neq 0$. Let $g$ be another coloring, thus
we have

$g(x_{i})=g(x_{1})+m^{\prime}_{i-1}$

where $1<i\leq 5$ and $0\leq m^{\prime}_{i-1}\leq n-1$. To satisfy equation
(2), we get for $1<i\leq 5$,

$am_{i-1}=m^{\prime}_{i-1}$

If any $m_{i-1}=0$, then $m^{\prime}_{i-1}=0$. Thus, we need only consider the
$i-1$ such that $m_{i-1}\neq 0$. Let $\rho(n)$ be the Euler function of $n$,
which is prime by hypothesis, and so $m_{i-1}^{\rho(n)}=1$. Therefore, we have
a unique solution (an edge) for $a$ if the following equation is satisfied for
all $i-1$ such that $m_{i-1}\neq 0$ :

$m^{\prime}_{1}m^{\rho(n)-1}_{1}=m^{\prime}_{2}m^{\rho(n)-1}_{2}=m^{\prime}_{3}m^{\rho(n)-1}_{3}=...=m^{\prime}_{4}m^{\rho(n)-1}_{4}\
$ (3)

Fix $m_{j}$ for each $j$ such that $m_{j}\neq 0$, so $1\leq m_{j}\leq n-1$.
Now, we consider the following $3$ cases

  1. (1)

Case 1: if $m_{j}=m^{\prime}_{j}$ for all $j$:
It is clear that equation (3) is satisfied in this case. Therefore there is an
edge between the $n$ vertices for each choice of $m_{1},...,m_{4}$. This
divides the nontrivial colorings into $\frac{n^{5}-n}{n}$
$(\overleftrightarrow{K_{n}},\hat{1})$ subgraphs.

  2. (2)

Case 2: if $m_{j}=m^{\prime}_{j}$ for some $j$ and $m_{k}\neq m^{\prime}_{k}$
for some $k$:
In this case, it is impossible to satisfy equation (3), since
$1=m^{\prime}_{j}m_{j}^{\rho(n)-1}\neq m^{\prime}_{k}m_{k}^{\rho(n)-1}$.
Therefore there are no edges between colorings in this case.

  3. (3)

Case 3: if $m_{j}\neq m^{\prime}_{j}$ for all $j$:
We have $n-2$ options for $m^{\prime}_{j}$, which determines the other
$m^{\prime}_{k}$’s to satisfy equation (3). Therefore there is an edge between
each vertex of the $\frac{n^{5}-n}{n}$ $(\overleftrightarrow{K_{n}},\hat{1})$
subgraphs from Case 1, to each vertex of $n-2$ other $\frac{n^{5}-n}{n}$
$(\overleftrightarrow{K_{n}},\hat{1})$ subgraphs. This means we now have
subgraphs $(\overleftrightarrow{K_{n(n-1)}},\hat{1})$. We thus have
$\frac{n^{5}-n}{n(n-1)}$ $(\overleftrightarrow{K_{n(n-1)}},\hat{1})$
subgraphs.

Case 2 tells us that the $(\overleftrightarrow{K_{n(n-1)}},\hat{1})$ subgraphs
are disjoint. However, we have an edge from each nontrivial coloring to each
trivial coloring since equation (3) is always satisfied if
$m^{\prime}_{i-1}=0$ for all $i-1$. This gives the desired result.

∎

We give the following examples of the full quandle coloring quiver for varying
values of $N$.

######  Example 3.5.

When $N=n$, we have
$\mathcal{Q}_{R_{n}}(T(5,q))=(\overleftrightarrow{K_{n}},\hat{n})$ by Theorem
3.1. In Figure 4, we show the graph $(\overleftrightarrow{K_{5}},\hat{5})$
corresponding to the full $R_{5}$-quandle coloring quiver of $T(5,q)$, while
Figure 5 shows
$\mathcal{Q}_{R_{20}}(T(5,q))=(\overleftrightarrow{K_{20}},\hat{20})$. For
visual simplicity, we leave out arrows on the graph edges and remind the
reader that the weight of each directed edge is $n$.

Figure 4. $(\overleftrightarrow{K_{5}},\hat{5})$

Figure 5. $(\overleftrightarrow{K_{20}},\hat{20})$

######  Example 3.6.

By Theorem 3.2, when $N=5n$, we have

$\mathcal{Q}_{R_{n}}(T(5,q))=(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}_{\hat{d}}\left(\overleftrightarrow{K_{4n}},\hat{d}\right),$

where $d=\frac{n}{5}$. The full $R_{5}$-quandle coloring quiver of $T(5,q)$ is
thus,

$\mathcal{Q}_{R_{5}}(T(5,q))=(\overleftrightarrow{K_{5}},\hat{5})\overleftarrow{\nabla}_{\hat{1}}\left(\overleftrightarrow{K_{20}},\hat{1}\right).$

Notice that the subgraph components of the quiver
$\mathcal{Q}_{R_{5}}(T(5,q))$ can be seen in Figures 4 and 5. In Figure 6 we
draw $\mathcal{Q}_{R_{5}}(T(5,q))$ as a simplified graph where the blue (left)
and red (right) dots represent $(\overleftrightarrow{K_{5}},\hat{5})$ and
$(\overleftrightarrow{K_{20}},\hat{20})$, respectively, and the directed edge
indicates a directed edge with weight 1 from every vertex of
$(\overleftrightarrow{K_{20}},\hat{20})$ to every vertex of
$(\overleftrightarrow{K_{5}},\hat{5})$.

Figure 6.
$\mathcal{Q}_{R_{5}}(T(5,q))=(\overleftrightarrow{K_{5}},\hat{5})\overleftarrow{\nabla}_{\hat{1}}\left(\overleftrightarrow{K_{20}},\hat{1}\right)$

######  Example 3.7.

By Theorem 3.3, when $N=16n$, we have

$\mathcal{Q}_{R_{n}}(T(5,q))=(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}_{\hat{d}}\big{[}\bigsqcup\limits_{15}\left(\overleftrightarrow{K_{n}},\hat{d}\right)\big{]},$

where $d=\frac{n}{2}$. Thus, if $n=6$, we know
$\mathcal{Q}_{R_{6}}(T(5,q))=(\overleftrightarrow{K_{6}},\hat{6})\overleftarrow{\nabla}_{\hat{3}}\big{[}\bigsqcup\limits_{15}\left(\overleftrightarrow{K_{6}},\hat{3}\right)\big{]}$.
In Figure 7 we again use the simplified technique of illustrating this quiver,
where the blue dot (center) represents $(\overleftrightarrow{K_{6}},\hat{6})$
and each of the 15 green dots represent
$(\overleftrightarrow{K_{6}},\hat{3})$. Each of the directed edges from a
green dot to a blue dot indicates that there is a directed edge from each
vertex of $(\overleftrightarrow{K_{6}},\hat{3})$ to every vertex of
$(\overleftrightarrow{K_{6}},\hat{6})$.

Figure 7.
$\mathcal{Q}_{R_{6}}(T(5,q))=(\overleftrightarrow{K_{6}},\hat{6})\overleftarrow{\nabla}_{\hat{3}}\big{[}\bigsqcup\limits_{15}\left(\overleftrightarrow{K_{6}},\hat{3}\right)\big{]}$

Quandle Coloring Quivers of $(7,q)$-Torus links with Dihedral Quandles We
calculate the quiver of $T(7,q)$ when $q=14k,14k+1,14k+2,\ldots,14k+13$. Let
$N$ be the number of colorings of $T(7,q)$ by $R_{n}$.

Using the braid form $(\sigma_{1}\sigma_{2}\ldots\sigma_{6})^{q}$ of $T(7,q)$,
we calculate $N$ for varying values of $q$. Let $f$ be a coloring of $T(7,q)$
by $R_{n}$, and $f(x_{i})=y_{i}$ for $i=1,...,7$. Coloring the braid
$(\sigma_{1}\sigma_{2}\sigma_{3}\sigma_{4}\sigma_{5}\sigma_{6})^{q}$ by the
dihedral quandle and using the fact that $f(x_{i})=y_{i}$ gives the following
equation:

$y_{i}=y_{7}+(-1)^{q+1}y_{q}+(-1)^{q}y_{i+q}$ (4)

Now, using equation (4) and following a similar process as in the $p=5$ case,
we summarize the results for $p=7$ in the following table below:

$q$ | Number of colorings $N$ when $p=7$
---|---
When q=p | When $n=7k$ for $k\geq 2$, $N=2^{6}n$
When q=p | When gcd(7,n)=1 or n=7, $N=n$
When q is even | $N=n\cdot\gcd(7,n)$
When q is odd, not p | $N=n$

Table 4. Number of colorings of $T(7,q)$ by $R_{n}$

######  Theorem 3.8.

If $|Hom(T(7,q),R_{n})|=n$, then the full quandle coloring quiver is the
complete directed graph:

$\mathcal{Q}_{R_{n}}(T(7,q))=(\overleftrightarrow{K_{n}},\hat{n}).$

###### Proof.

If $|Hom(T(7,q),R_{n})|=n$ (this corresponds to trivial colorings). Now assume
that $f$ and $g$ are both trivial colorings given respectively by
$f(x_{i})=j\;\textit{and}\;g(x_{i})=k$ for all $x_{i}\in Q(T(7,q))$. Since any
endomorphism of the dihedral quandle $R_{n}$ has the form $\phi(f)=af+b$,
where $a,b\in R_{n}$, then there are exactly $n$ solutions of functions $\phi$
making the equation $\phi(f)=g$ hold. Thus we obtain the complete graph
$\mathcal{Q}_{R_{n}}(T(7,q))=(\overleftrightarrow{K_{n}},\hat{n})$. ∎

######  Theorem 3.9.

If $|Hom(T(7,q),R_{n})|=7n$, then the full quandle coloring quiver is the
directed graph:

$\mathcal{Q}_{R_{n}}(T(7,q))=(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}_{\hat{d}}\left(\overleftrightarrow{K_{6n}},\hat{d}\right),$

where $d=\frac{n}{7}$.

###### Proof.

Let $f,g\in Hom(Q(T(7,q)),R_{n})$ be two vertices of the quiver. Since any
endomorphism of the dihedral quandle $R_{n}$ is given by $\phi(f)=af+b$ (see
[EMR]), we need to find the number of endomorphisms $\phi$ making the equation
(2) hold. We know when there are $N=7n$ colorings, then $gcd(7,n)=7$. Now,
since there are no edges from a trivial coloring to a non-trivial coloring, we
consider the following $3$ cases:

  1. (1)

Case 1. It is clear that if $f$ and $g$ are trivial colorings, i.e. constant
maps, then there are $n$ possible solutions to equation (2), and we have $n$
edges between trivial colorings.

  2. (2)

Case 2. When $f$ is a non-trivial coloring, then for some trivial coloring $g$
we have $g(x_{i})=k$ for all $x_{i}$, and equation (2) becomes
$a(f(x_{i})-f(x_{j}))=0\mod n$ for all $i,j$. Since
$7(f(x_{i})-f(x_{j}))=0\mod n$, we have $\frac{n}{7}$ from nontrivial
colorings to trivial colorings.

  3. (3)

Case 3. When $f$ and $g$ are both non-trivial colorings, then $f$ and $g$ are
given by $7(f(x_{i})-f(x_{j}))=7(g(x_{i})-g(x_{j}))=0$ modulo $n$. Since the
equation $(g(x_{i})-g(x_{j}))-a(f(x_{i})-f(x_{j}))=0$ in $R_{n}$ has
$\frac{n}{7}$ solutions for $a$, we get the result.

∎

######  Theorem 3.10.

If $|Hom(T(7,q),R_{n})|=64n$, then the full quandle coloring quiver is the
directed graph:

$\mathcal{Q}_{R_{n}}(T(7,q))=(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}_{\hat{d}}\big{[}\bigsqcup\limits_{63}\left(\overleftrightarrow{K_{n}},\hat{d}\right)\big{]},$

where $d=\frac{n}{2}$.

###### Proof.

In order to have $N=64n$ colorings, we know $n$ must be even and for any
coloring $f$, $2(f(x_{i})-f(x_{j}))=0$ for any $i,j$. The $64n$ colorings
consist of $n$ trivial colorings and $63n$ nontrivial colorings. By Theorem
3.8, the $n$ trivial colorings correspond to a subgraph
$(\overleftrightarrow{K_{n}},\hat{n})$ of $\mathcal{Q}_{R_{n}}(T(7,q))$. Now,
there are $63n$ nontrivial colorings, so let $f$ be a nontrivial coloring,
thus we have, $f(x_{i})=f(x_{1})+m_{i-1}$ for $1<i\leq 7$ where $m_{i-1}=0$ or
$\frac{n}{2}$. Since $n$ is even and $f$ nontrivial, there exists $i$ such
that $m_{i-1}=\frac{n}{2}$. We know for some coloring $g\in
Hom(Q(T(7,q)),R_{n})$, $g(x_{i})=g(x_{1})+m^{\prime}_{i-1}$ where
$m^{\prime}_{i-1}=0$ or $\frac{n}{2}$ . By satisfying equation (2), we have
$am_{i-1}=m^{\prime}_{i-1}$ for $1<i\leq 7$. If any $m_{i-1}=0$, then
$m^{\prime}_{i-1}=0$. Thus, consider the $m_{i-1}$ such that
$m_{i-1}=\frac{n}{2}$ and so we have $a\frac{n}{2}=0$ or $n/2$. If
$a\frac{n}{2}=0$ then $a$ is even, i.e. $g$ must be trivial, and we have $n/2$
edges between nontrivial colorings to trivial colorings. If
$a\frac{n}{2}=\frac{n}{2}$ then $a$ is odd
($m_{i-1}=m^{\prime}_{i-1}=\frac{n}{2}$ and we have $n/2$ edges between
nontrivial colorings to nontrivial colorings. For each choice of
$m_{1},m_{2},...,m_{6}$, there are $n$ colorings which divides the nontrivial
colorings into 63 disjoint
$\left(\overleftrightarrow{K_{n}},\hat{\frac{n}{2}}\right)$. ∎

######  Theorem 3.11.

If $|Hom(T(7,q),R_{n})|=n^{7}$, then the full quandle coloring quiver is the
directed graph:

$\mathcal{Q}_{R_{n}}(T(7,q))=(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}\big{[}\bigsqcup\limits_{m}\left(\overleftrightarrow{K_{n(n-1)}},\hat{1}\right)\big{]},$

where $m=\frac{n^{7}-n}{n(n-1)}$.

###### Proof.

We know the quiver will have a subgraph $(\overleftrightarrow{K_{n}},\hat{n})$
corresponding to the $n$ trivial colorings. Let $f$ be a nontrivial coloring,
thus we have

$f(x_{i})=f(x_{1})+m_{i-1}$

where $1<i\leq 7$ and $0\leq m_{i-1}\leq n-1$. Since $f$ is nontrivial, there
exists an index $j$ such that $m_{j}\neq 0$. Let $g$ be another coloring, thus
we have

$g(x_{i})=g(x_{1})+m^{\prime}_{i-1}$

where $1<i\leq 7$ and $0\leq m^{\prime}_{i-1}\leq n-1$. To satisfy equation
(2), we get for $1<i\leq 7$,

$am_{i-1}=m^{\prime}_{i-1}$

If any $m_{i-1}=0$, then $m^{\prime}_{i-1}=0$. Thus, we need only consider the
$i-1$ such that $m_{i-1}\neq 0$. Let $\rho(n)$ be the Euler function of $n$,
which is prime by hypothesis, and so $m_{i-1}^{\rho(n)}=1$. Therefore, we have
a unique solution (an edge) for $a$ if the following equation is satisfied for
all $i-1$ such that $m_{i-1}\neq 0$:

$m^{\prime}_{1}m^{\rho(n)-1}_{1}=m^{\prime}_{2}m^{\rho(n)-1}_{2}=m^{\prime}_{3}m^{\rho(n)-1}_{3}=...=m^{\prime}_{6}m^{\rho(n)-1}_{6}\
$ (5)

Fix $m_{i}$ for each $i$ such that $m_{i}\neq 0$, so $1\leq m_{i}\leq n-1$.
Now, we consider the following $3$ cases

  1. (1)

Case 1: if $m_{i}=m^{\prime}_{i}$ for all $i$:
It is clear that equation (5) is satisfied in this case. Therefore there is an
edge between the $n$ vertices for each choice of $m_{1},...,m_{6}$. This
divides the nontrivial colorings into $\frac{n^{7}-n}{n}$
$(\overleftrightarrow{K_{n}},\hat{1})$ subgraphs.

  2. (2)

Case 2: if $m_{i}=m^{\prime}_{i}$ for some $i$ and $m_{j}\neq m^{\prime}_{j}$
for some $j$:
In this case, it is impossible to satisfy equation (5), since
$m^{\prime}_{j}m_{j}^{\rho(n)-1}\neq 1=m^{\prime}_{i}m_{i}^{\rho(n)-1}$.
Therefore there are no edges between colorings in this case.

  3. (3)

Case 3: if $m_{i}\neq m^{\prime}_{i}$ for all $i$:
We have $n-2$ options for $m^{\prime}_{i}$, which determines the other
$m^{\prime}_{j}$’s to satisfy equation (5). Therefore there is an edge between
each vertex of the $\frac{n^{7}-n}{n}$ $(\overleftrightarrow{K_{n}},\hat{1})$
subgraphs from Case 1, to each vertex of $n-2$ other $\frac{n^{7}-n}{n}$
$(\overleftrightarrow{K_{n}},\hat{1})$ subgraphs. This means we now have
subgraphs $(\overleftrightarrow{K_{n(n-1)}},\hat{1})$. We thus have
$\frac{n^{7}-n}{n(n-1)}$ $(\overleftrightarrow{K_{n(n-1)}},\hat{1})$
subgraphs.

Case 2 tells us that the $(\overleftrightarrow{K_{n(n-1)}},\hat{1})$ subgraphs
are disjoint. However, we have an edge from each nontrivial coloring to each
trivial coloring since equation (5) is always satisfied if $m^{\prime}_{i}=0$
for all $i$. This gives the desired result.

∎

##  4\. Quandle Coloring Quivers of General $(p,q)$-Torus links with Dihedral
Quandles

Using the braid form $(\sigma_{1}\sigma_{2}...\sigma_{p-1})^{q}$ of $T(p,q)$,
we calculate the number of colorings, $N$, for varying values of $q$ when $p$
is an odd prime. Let $f$ be a coloring of $T(p,q)$ by $R_{n}$ and
$f(x_{i})=y_{i}$ for $i=1,...,p$. Coloring the braid
$(\sigma_{1}\sigma_{2}...\sigma_{p-1})^{q}$ by the dihedral quandle and using
the fact that $f(x_{i})=y_{i}$ gives the following equation:

$y_{i}=y_{p}+(-1)^{q+1}y_{q}+(-1)^{q}y_{i+q}$ (6)

Thus, we have the following:

  * •

When $q=2pk$, $N=n^{p}$.

  * •

When $q=2pk+i$ where $i<2p$ is an odd number but $i\neq p$, $N=n$.

  * •

When $q=2pk+j$ where $j<2p$ is an even number,

    * –

When $n=p$ or $\gcd(n,p)=1$, $N=n$;

    * –

When $n\geq 2p$ and $\gcd(n,p)=p$, $N=pn$.

  * •

When $q=2pk+p$,

    * –

When $\gcd(2,n)=1$, $N=n$.

    * –

When $\gcd(2,n)=2$, $N=2^{p-1}n$.

######  Theorem 4.1.

If $|Hom(T(p,q),R_{n})|=n$, then the full quandle coloring quiver is the
complete directed graph:

$\mathcal{Q}_{R_{n}}(T(p,q))=(\overleftrightarrow{K_{n}},\hat{n}).$

###### Proof.

We know if $|Hom(T(p,q),R_{n})|=n$, this corresponds to the $n$ trivial
colorings. Now assume that $f$ and $g$ are both trivial colorings given
respectively by $f(x_{i})=j\;\textit{and}\;g(x_{i})=k$ for all $x_{i}\in
Q(T(p,q))$. Since any endomorphism of the dihedral quandle $R_{n}$ is given by
$\phi(f)=af+b$ (see [EMR]), we need to find the number of endomorphisms $\phi$
making the equation

$\displaystyle\phi(f)=af+b=g$ (7)

hold, where $a,b\in R_{n}$. there are exactly $n$ solutions of functions
$\phi$ making the equation $\phi(f)=g$ hold. Thus we obtain the complete graph
$\mathcal{Q}_{R_{n}}(T(p,q))=(\overleftrightarrow{K_{n}},\hat{n})$. ∎

######  Theorem 4.2.

If $|Hom(T(p,q),R_{n})|=pn$, then the full quandle coloring quiver is the
directed graph:

$\mathcal{Q}_{R_{n}}(T(p,q))=(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}_{\hat{d}}\left(\overleftrightarrow{K_{(p-1)n}},\hat{d}\right),$

where $d=\frac{n}{p}$.

###### Proof.

Let $f,g\in Hom(Q(T(p,q)),R_{n})$ be two vertices of the quiver. Since any
endomorphism of the dihedral quandle $R_{n}$ is given by $\phi(f)=af+b$, we
need to find the number of endomorphisms $\phi$ making the equation (7) hold.
We know when there are $N=pn$ colorings, then $gcd(p,n)=p$. Now, since there
are no edges from a trivial coloring to a non-trivial coloring, we consider
the following $3$ cases:

  1. (1)

Case 1. It is clear that if $f$ and $g$ are trivial colorings, i.e. constant
maps, then there are $n$ possible solutions to equation (7), and we have $n$
edges between trivial colorings.

  2. (2)

Case 2. When $f$ is a non-trivial coloring, then for some trivial coloring $g$
we have $g(x_{i})=k$ for all $x_{i}$, and equation (7) becomes
$a(f(x_{i})-f(x_{j}))=0\mod n$ for all $i,j$. Since
$p(f(x_{i})-f(x_{j}))=0\mod n$, we have $\frac{n}{p}$ edges from nontrivial
colorings to trivial colorings.

  3. (3)

Case 3. When $f$ and $g$ are both non-trivial colorings, then $f$ and $g$ are
given by $p(f(x_{i})-f(x_{j}))=p(g(x_{i})-g(x_{j}))=0$ modulo $n$. Since the
equation $(g(x_{i})-g(x_{j}))-a(f(x_{i})-f(x_{j}))=0$ in $R_{n}$ has
$\frac{n}{p}$ solutions for $a$, we get the result.

∎

######  Theorem 4.3.

If $n$ is even and $|Hom(T(p,q),R_{n})|=2^{p-1}n$, then the full quandle
coloring quiver is the directed graph:

$\mathcal{Q}_{R_{n}}(T(p,q))=(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}_{\hat{d}}\big{[}\bigsqcup\limits_{2^{p-1}-1}\left(\overleftrightarrow{K_{n}},\hat{d}\right)\big{]},$

where $d=\frac{n}{2}$.

###### Proof.

When determining the number of colorings, $N$, we get the equation
$2(f(x_{i})-f(x_{j}))=0$ for any coloring $f$, and any $i,j.$ Since $f(x_{i})$
takes $n$ possible values, and for $j\neq i$, $f(x_{j})$ takes 2 values, thus
giving $N=n\cdot 2^{p-1}$. This includes $n$ trivial colorings and
$(2^{p-1}-1)n$ nontrivial colorings. By Theorem 4.1, the $n$ trivial colorings
correspond to a subgraph $(\overleftrightarrow{K_{n}},\hat{n})$ of
$\mathcal{Q}_{R_{n}}(T(p,q))$. Now, there are $(2^{p-1}-1)n$ nontrivial
colorings, so let $f$ be a nontrivial coloring, thus we have,
$f(x_{i})=f(x_{1})+m_{i-1}$ for $1<i\leq p$ where $m_{i-1}=0$ or
$\frac{n}{2}$. Since $n$ is even and $f$ nontrivial, there exists $i$ such
that $m_{i-1}=\frac{n}{2}$. We know for some coloring $g$ of $T(p,q)$ by
$R_{n}$, we have $g(x_{i})=g(x_{1})+m^{\prime}_{i-1}$ where
$m^{\prime}_{i-1}=0$ or $\frac{n}{2}$ . By satisfying equation (7), we have
$am_{i-1}=m^{\prime}_{i-1}$ for $1<i\leq p$. If any $m_{i-1}=0$, then
$m^{\prime}_{i-1}=0$. Thus, consider the $m_{i-1}$ such that
$m_{i-1}=\frac{n}{2}$ and so we have $a\frac{n}{2}=0$ or $n/2$. If
$a\frac{n}{2}=0$ then $a$ is even, i.e. $g$ must be trivial, and we have $n/2$
edges between nontrivial colorings to trivial colorings. If
$a\frac{n}{2}=\frac{n}{2}$ then $a$ is odd
($m_{i-1}=m^{\prime}_{i-1}=\frac{n}{2}$ and we have $n/2$ edges between
nontrivial colorings to nontrivial colorings. For each choice of
$m_{1},m_{2},...,m_{p-1}$, there are $n$ colorings which divides the
nontrivial colorings into $2^{p-1}-1$ disjoint
$\left(\overleftrightarrow{K_{n}},\hat{\frac{n}{2}}\right)$. ∎

######  Theorem 4.4.

If $n$ is prime and $|Hom(T(p,q),R_{n})|=n^{p}$, then the full quandle
coloring quiver is the directed graph:

$\mathcal{Q}_{R_{n}}(T(p,q))=(\overleftrightarrow{K_{n}},\hat{n})\overleftarrow{\nabla}\big{[}\bigsqcup\limits_{m}\left(\overleftrightarrow{K_{n(n-1)}},\hat{1}\right)\big{]},$

where $m=\frac{n^{p}-n}{n(n-1)}$.

###### Proof.

We know the quiver will have a subgraph $(\overleftrightarrow{K_{n}},\hat{n})$
corresponding to the $n$ trivial colorings. Let $f$ be a nontrivial coloring,
so

$f(x_{i})=f(x_{1})+m_{i-1}$

for some $0\leq m_{1},m_{2},...,m_{p-1}\leq n-1$. Now, let $g\neq f$ be
another coloring, so

$g(x_{i})=g(x_{1})+m^{\prime}_{i-1}$

for some $1\leq m^{\prime}_{1},m^{\prime}_{2},...,m^{\prime}_{p-1}\leq n$. To
satisfy equation (7) , we get

$am_{i-1}=m^{\prime}_{i-1}.$

If any $m_{i}=0$, then $m^{\prime}_{i}=0$. Thus, we need only consider the $i$
such that $m_{i}\neq 0$. Let $\rho(n)$ be the Euler function of $n$, which is
prime by hypothesis, and so $m_{i-1}^{\rho(n)}=1$. Therefore, we have a unique
solution (an edge) if the following equation is satisfied for all $i$ such
that $m_{i}\neq 0$:

$m^{\prime}_{1}m^{\rho(n)-1}_{1}=m^{\prime}_{2}m^{\rho(n)-1}_{2}=m^{\prime}_{3}m^{\rho(n)-1}_{3}=...=m^{\prime}_{p-1}m^{\rho(n)-1}_{p-1}\
$ (8)

Fix $m_{i}$ for each $i$ such that $m_{i}\neq 0$, so $1\leq m_{i}\leq n-1$.
Now, we consider the following $3$ cases

  1. (1)

Case 1: if $m_{i}=m^{\prime}_{i}$ for all $i$:
It is clear that equation (8) is satisfied in this case. Therefore there is an
edge between the $n$ vertices for each choice of $m_{1},...,m_{p-1}$. This
divides the nontrivial colorings into $\frac{n^{p}-n}{n}$
$(\overleftrightarrow{K_{n}},\hat{1})$ subgraphs.

  2. (2)

Case 2: if $m_{i}=m^{\prime}_{i}$ for some $i$ and $m_{j}\neq m^{\prime}_{j}$
for some $j$:
In this case, it is impossible to satisfy equation (8), since
$m^{\prime}_{j}m_{j}^{\rho(n)-1}\neq 1=m^{\prime}_{i}m_{i}^{\rho(n)-1}$.
Therefore there are no edges between colorings in this case.

  3. (3)

Case 3: if $m_{i}\neq m^{\prime}_{i}$ for all $i$:
We have $n-2$ options for $m^{\prime}_{i}$, which determines the other
$m^{\prime}_{j}$’s to satisfy equation (8). Therefore there is an edge between
each vertice of the $\frac{n^{p}-n}{n}$ $(\overleftrightarrow{K_{n}},\hat{1})$
subgraphs from Case 1, to each vertice of $n-2$ other $\frac{n^{p}-n}{n}$
$(\overleftrightarrow{K_{n}},\hat{1})$ subgraphs. This means we now have
subgraphs $(\overleftrightarrow{K_{n(n-1)}},\hat{1})$. We thus have
$\frac{n^{p}-n}{n(n-1)}$ $(\overleftrightarrow{K_{n(n-1)}},\hat{1})$
subgraphs.

Case 2 tells us that the $(\overleftrightarrow{K_{n(n-1)}},\hat{1})$ subgraphs
are disjoint. However, we have an edge from each nontrivial coloring to each
trivial coloring since equation (8) is always satisfied if $m^{\prime}_{i}=0$
for all $i$. This gives the desired result.

∎

### Acknowledgements

Mohamed Elhamdadi was partially supported by Simons Foundation collaboration
grant 712462.

## References

  * [1]
BasiJagdeepCaprauCarmenQuandle coloring quivers of $(p,2)$-torus linksJ. Knot
Theory Ramifications3120229Paper No. 2250057, 14ISSN 0218-2165Review
MathReviewsDocument@article{BC, author = {Basi, Jagdeep}, author = {Caprau,
Carmen}, title = {Quandle coloring quivers of $(p,2)$-torus links}, journal =
{J. Knot Theory Ramifications}, volume = {31}, date = {2022}, number = {9},
pages = {Paper No. 2250057, 14}, issn = {0218-2165}, review = {\MR{4475496}},
doi = {10.1142/S0218216522500572}}

  * [3]
ElhamdadiMohamedMacquarrieJenniferRestrepoRicardoAutomorphism groups of
quandlesJ. Algebra Appl.11201211250008, 9ISSN 0219-4988Review
MathReviewsDocument@article{EMR, author = {Elhamdadi, Mohamed}, author =
{Macquarrie, Jennifer}, author = {Restrepo, Ricardo}, title = {Automorphism
groups of quandles}, journal = {J. Algebra Appl.}, volume = {11}, date =
{2012}, number = {1}, pages = {1250008, 9}, issn = {0219-4988}, review =
{\MR{2900878}}, doi = {10.1142/S0219498812500089}}

  * [5]
ElhamdadiMohamedNelsonSamQuandles—an introduction to the algebra of
knotsStudent Mathematical Library74American Mathematical Society, Providence,
RI2015x+245ISBN 978-1-4704-2213-4Review MathReviews@book{EN, author =
{Elhamdadi, Mohamed}, author = {Nelson, Sam}, title = {Quandles—an
introduction to the algebra of knots}, series = {Student Mathematical
Library}, volume = {74}, publisher = {American Mathematical Society,
Providence, RI}, date = {2015}, pages = {x+245}, isbn = {978-1-4704-2213-4},
review = {\MR{3379534}}}

  * [7]
GabrielPeterUnzerlegbare darstellungen. iGerman, with English
summaryManuscripta Math.6197271–103; correction, ibid. 6 (1972), 309ISSN
0025-2611Review MathReviewsDocument@article{Gabriel, author = {Gabriel,
Peter}, title = {Unzerlegbare Darstellungen. I}, language = {German, with
English summary}, journal = {Manuscripta Math.}, volume = {6}, date = {1972},
pages = {71–103; correction, ibid. 6 (1972), 309}, issn = {0025-2611}, review
= {\MR{332887}}, doi = {10.1007/BF01298413}}

  * [9]
JoyceDavidA classifying invariant of knots, the knot quandleJ. Pure Appl.
Algebra231982137–65ISSN 0022-4049Review MathReviewsDocument@article{Joyce,
author = {Joyce, David}, title = {A classifying invariant of knots, the knot
quandle}, journal = {J. Pure Appl. Algebra}, volume = {23}, date = {1982},
number = {1}, pages = {37–65}, issn = {0022-4049}, review = {\MR{638121
(83m:57007)}}, doi = {10.1016/0022-4049(82)90077-9}}

  * [11]
MatveevSergei V.Distributive groupoids in knot theoryRussianMat. Sb.
(N.S.)119(161)1982178–88, 160ISSN 0368-8666Review MathReviews@article{Matveev,
author = {Matveev, Sergei V.}, title = {Distributive groupoids in knot
theory}, language = {Russian}, journal = {Mat. Sb. (N.S.)}, volume =
{119(161)}, date = {1982}, number = {1}, pages = {78–88, 160}, issn =
{0368-8666}, review = {\MR{672410 (84e:57008)}}}

  * [13]
ChoKarinaNelsonSamQuandle coloring quiversJ. Knot Theory
Ramifications28201911950001, 12ISSN 0218-2165Review
MathReviewsDocument@article{CN, author = {Cho, Karina}, author = {Nelson,
Sam}, title = {Quandle coloring quivers}, journal = {J. Knot Theory
Ramifications}, volume = {28}, date = {2019}, number = {1}, pages = {1950001,
12}, issn = {0218-2165}, review = {\MR{3910948}}, doi =
{10.1142/S0218216519500019}}

  * [15]
  * [16]
ZhouBoxinLiuXiminQuandle coloring quivers of ($p$,3)-torus linksJ. Knot Theory
Ramifications3220233Paper No. 2350016, 23ISSN 0218-2165Review
MathReviewsDocument

  * [18]