# Idempotents in integral ring of dihedral quandle

P Narayanan and Saikat Panja<EMAIL_ADDRESS><EMAIL_ADDRESS>IISER
Pune, Dr. Homi Bhabha Road, Pashan, Pune 411 008, India

###### Abstract.

We provide the structure of some idempotents in the integral ring of a
dihedral quandles, for odd case. The solution is complete in case $n=5$. This
produces an example of a connected quandle $X$ such that
$\textup{Aut}(X)\not\cong\textup{Aut}(\mathbb{Z}[X])$.

###### Key words and phrases:

Quandle

###### 2020 Mathematics Subject Classification:

12F10, 16T05.

The first named author is partially supported by the IISER Pune research
fellowship and the second author has been partially supported by supported by
NBHM fellowship.

##  1\. Introduction

A quandle is a pair $(A,\cdot)$ such that ‘$\cdot$’ is a binary operation
satisfying

  1. (1)

the map $S_{a}:A\longrightarrow A$, defined as $S_{a}(b)=b\cdot a$ is an
automorphism for all $a\in A$,

  2. (2)

for all $a\in A$, we have $S_{a}(a)=a$.

The automorphisms $S_{a}$ are called the inner automorphisms of the quandle
$A$. A quandle is said to be connected  if the group of inner automorphisms
acts transitively on the quandle. Let $\mathbb{Z}_{n}$ denote the cyclic group
of order $n$. Then defining $a\cdot b=2b-a$ defines a quandle structure on
$A=\mathbb{Z}_{n}$. This is known as the dihedral quandle, to be denoted by
$Q_{n}$. Any dihedral quandle of odd order is connected. For other examples
see [BPS19].

To get a better understanding of the structure, a theory parallel to group
rings was introduced by Bardakov, Passi and Singh in [BPS19]. The quandle ring
of a quandle $A$ is defined as follows. Let $R$ be a commutative ring.
Consider

$R[A]=\left\\{\sum_{i}r_{i}a_{i}:r_{i}\in R,a_{i}\in A\right\\}.$

Then this is an additive group in usual way. Define multiplication as

$\left(\sum_{i}r_{i}a_{i}\right)\cdot\left(\sum_{j}s_{j}a_{j}\right)=\sum_{i,j}r_{i}s_{j}(a_{i}\cdot
a_{j}).$

The terms idempotents and ring automorphism are of usual meaning, as in ring
theory. Since $\mathbb{Z}[A]$ is an infinite ring, it is usually difficult to
conclude about the idempotent structure of the ring. Also since in quandle
$a*a=a$ for all $a\in A$, these are trivial idepotents in the quandle ring
$\mathbb{Z}[A]$.

Here we introduce the notion of an adjacency matrix of a quandle. This helps
us in finding some of the idempotents in integral quandle ring of a dihedral
quandle of odd order. Using this we further show that $\mathbb{Z}[Q_{5}]$ has
only trivial idempotents. This further helps in finiding the automorphism
group of $\mathbb{Z}[Q_{5}]$. This answers a question of Elhamdadi in
negation, whether $\textup{Aut}(\mathbb{Z}[X])\cong\textup{Aut}(X)$, for a
connected quandle $X$.

##  2\. Adjacency matrix of a quandle

The following definition is from [NH05]

######  Definition 2.1.

Let $(X,*)$ be a quandle of size $n$. Define the $n\times n$ matrix
$A=(a_{ij})$ with $a_{ij}=x_{i}*x_{j}$. This matrix will be called as the
adjacency matrix of the quandle.

######  Example 2.2.

Let $Q_{5}$ be the dihedral quandle of size $5$. Then the adjacency matrix of
$Q_{5}$ will be given by

$\displaystyle\begin{pmatrix}0&2&4&1&3\\\ 4&1&3&0&2\\\ 3&0&2&4&1\\\
2&4&1&3&0\\\ 1&3&0&2&4\end{pmatrix}.$

######  Lemma 2.3.

Let $n$ be an odd integer. Then the rows and columns of the adjacency matrix
corresponding to a dihedral quandle of order $n$, is determined by two
elements $\rho,\sigma\in S_{n}$. Here $S_{n}$ denotes the symmetric group on
letters $\\{0,1,\cdots,n-1\\}$. Furthermore, the first row is determined by

$\displaystyle\rho=\left(0\quad 2\quad\cdots\quad n-1\quad 1\quad
3\quad\cdots\quad n-2\right),$

and the first column is determined by

$\displaystyle\sigma=(0\quad 1\quad 2\quad\cdots\quad n-1)^{-1}.$

###### Proof.

We first prove that all the rows of the adjacency matrix is determined by one
row. To achieve this let us assume $x_{i}*x_{j}=x_{k}$ and
$x_{i}*x_{j+1}=x_{l}$. We need to show that $x_{m}*x_{r}=x_{k}$ implies
$x_{m}*x_{r+1}=x_{l}$. From the given three conditions we have

$\displaystyle 2j-i=k,\quad 2j+2-i=l,\quad 2r-m=k\pmod{n}.$

Hence we have $2r+2-m=l\pmod{n}$. Now we compute the first row of the matrix.
This can be easily shown to be equal to $\rho$. The comment about $\sigma$
follows similarly. ∎

######  Lemma 2.4.

Let $n$ be an even integer. Then the rows and columns of the adjacency matrix
corresponding to a dihedral quandle of order $n$, is determined by three
elements $\rho_{1},\rho_{2},\sigma\in S_{n}$. Here $S_{n}$ denotes the
symmetric group on letters $\\{0,1,\cdots,n-1\\}$. Furthermore, the row is
determined by

$\displaystyle\rho_{1}$ $\displaystyle=\left(0\quad 2\quad\cdots\quad
n-2\right)$ $\displaystyle\rho_{2}$ $\displaystyle=\left(1\quad
3\quad\cdots\quad n-1\right),$

and the column is determined by

$\displaystyle\sigma=(0\quad 1\quad 2\quad\cdots\quad n-1)^{-1}.$

###### Proof.

Same as the last lemma and hence left to the reader.

∎

######  Lemma 2.5.

Let $Q_{n}$ denote the dihedral quandle of odd order $n$. The
$\sum\limits_{i=0}^{n-1}\alpha_{i}x_{i}\in\mathbb{Z}[Q_{n}]$ is an idempotent
if and only if $(\alpha_{0},\alpha_{1},\cdots,\alpha_{n-1})$ is a solution to
the system of equations given by

$\displaystyle B\cdot T=T,\quad\sum\limits_{i=0}^{n-1}t_{i}=1,$

where $T=(t_{0},t_{1},\cdots,t_{n-1})^{t}$ and $b_{ij}=t_{a_{ij}}$ with
$A=(a_{ij})$ being the adjacency matrix of $Q_{n}$.

###### Proof.

Let $\sum_{i=0}^{n-1}\alpha_{i}x_{i}\in\mathbb{Z}[Q_{n}]$ is an idempotent
element. Then we get that

$\displaystyle\left(\sum_{i=0}^{n-1}\alpha_{i}x_{i}\right)^{2}=\sum_{i=0}^{n-1}\alpha_{i}x_{i}$
$\displaystyle\implies$
$\displaystyle\sum\limits_{i=0}^{n-1}\alpha_{i}^{2}x_{i}+\sum\limits_{0\leq
i<j\leq
n}\alpha_{i}\alpha_{j}(x_{i}*x_{j}+x_{j}*x_{i})=\sum_{i=0}^{n-1}\alpha_{i}x_{i}.$

Now we want to compare coefficient of $x_{k}$ of both side. Now note that for
all $0\leq i\leq n-1$, we have the equality

$\displaystyle(x_{k}*x_{i})*x_{i}$ $\displaystyle=x_{k}.$

Hence comparing the both side we get that

$\displaystyle\alpha_{k}=\sum_{i=0}^{n-1}\alpha_{k*i}\alpha_{i}.$

Since we know that $\sum\limits_{i=0}^{n-1}\alpha_{i}=1$, we get that
$(\alpha_{0},\alpha_{1},\cdots,\alpha_{n-1})$ should satisfy the given system
of equations. A mutatis mutandis produce the other direction for the proof of
the other direction. ∎

######  Definition 2.6.

Let $a=\sum\alpha_{i}x_{i}\in\mathbb{Z}[X]$. We define the length of $a$ to be
$\ell(a)=|\\{i:\alpha_{i}\neq 0\\}|$.

######  Lemma 2.7.

Let $n$ be odd. Then there is no idempotent $a\in\mathbb{Z}[Q_{n}]$ satisfying
$\ell(a)=2$.

###### Proof.

Suppose on the contrary assume that there exist $i\neq j$ with
$\alpha_{i},\alpha_{j}\neq 0$ and $\alpha_{k}=0$ for al $k\neq i,j$. Consider
the $i$-the column of the adjacency matrix. Then as seen above it will be the
permutation $\sigma$. Therefore there exist a $k$ such that $k$-th entry of
the column is $\alpha_{j}$. Moreover $k$ cannot be equal to $i$. If $k$ were
equal to $j$, then $x_{j}*x_{i}=x_{k}$ since $(x_{j}*x_{i})*x_{i}=x_{j}$. This
will force $i=j$, contrary to our assumptions. Hence $k\neq i,j$. So from the
$k$-th row we get an equation of the form $\alpha_{i}.\alpha_{j}=0$ or
$2\alpha_{i}.\alpha_{j}=0$ again contradicting our assumption
$\alpha_{i},\alpha_{j}\neq 0$. ∎

##  3\. Idempotents in $\mathbb{Z}[Q_{5}]$

######  Theorem 3.1.

The idempotents of $\mathbb{Z}[Q_{5}]$ are the trivial idempotents.

###### Proof.

We will be using Lemma 2.5. Note that the adjacency matrix is given by

$\displaystyle\begin{pmatrix}0&2&4&1&3\\\ 4&1&3&0&2\\\ 3&0&2&4&1\\\
2&4&1&3&0\\\ 1&3&0&2&4\\\ \end{pmatrix}$

Hence we need to find $\mathbb{Z}$-solutions of the system of equations given
by

$\displaystyle\begin{pmatrix}t_{0}&t_{2}&t_{4}&t_{1}&t_{3}\\\
t_{4}&t_{1}&t_{3}&t_{0}&t_{2}\\\ t_{3}&t_{0}&t_{2}&t_{4}&t_{1}\\\
t_{2}&t_{4}&t_{1}&t_{3}&t_{0}\\\ t_{1}&t_{3}&t_{0}&t_{2}&t_{4}\\\
\end{pmatrix}\begin{pmatrix}t_{0}\\\ t_{1}\\\ t_{2}\\\ t_{3}\\\
t_{4}\end{pmatrix}=\begin{pmatrix}t_{0}\\\ t_{1}\\\ t_{2}\\\ t_{3}\\\
t_{4}\end{pmatrix},\sum\limits_{i=0}^{4}t_{i}=1.$

We solve this using the Groebner basis, since a common zero of the above set
will be a common zero of the Groebner basis. The Groebner basis of the above
equations is given by

$\displaystyle
t_{4}^{3}-t_{4}^{2},t_{1}^{2}-t_{1},t_{1}t_{2}+3t_{4}^{2}-3t_{4},t_{2}^{2}-3t_{4}^{2}-t_{2}+3t_{4},t_{1}t_{3}-t_{1}^{2}+t_{1},t_{2}t_{3},$
$\displaystyle
t_{3}^{2}-4t_{2}^{2}-t_{3}+4t_{2},t_{1}t_{4}+t_{4}^{2}-t_{4},t_{2}t_{4},t_{3}t_{4},5t_{4}^{2}-5t_{4}.$

This system can be easily seen to have only solutions $t_{i}=1$ and $t_{j}=0$
for all $i\neq j$, $i,j\in\\{0,1,2,3,4\\}$. Hence the idempotents of
$\mathbb{Z}[Q_{5}]$ are trivial. ∎

######  Corollary 3.2.

We have that $\textup{Aut}(\mathbb{Z}[Q_{5}])\cong S_{5}$.

###### Proof.

Note that an automorphism $\varphi$ of $\mathbb{Z}[Q_{5}]$ is determined by
the images of $x_{i}$. Since $\varphi(x_{i})$ is an idempotent and
$\mathbb{Z}[Q_{5}]$ ahs only trivial idempotents, it follows that
$\varphi(x_{i})=x_{j}$ for some $j$. This finishes the proof. ∎

## References

  * [BPS19] Valeriy G. Bardakov, Inder Bir S. Passi, and Mahender Singh.  Quandle rings.  J. Algebra Appl., 18(8):1950157, 23, 2019.
  * [EFT19] Mohamed Elhamdadi, Neranga Fernando, and Boris Tsvelikhovskiy.  Ring theoretic aspects of quandles.  J. Algebra, 526:166–187, 2019.
  * [NH05] Sam Nelson, Benita Ho.  Matrices and Finite quandles.  Homotopy and Homology Appl. 7, 197-208, 2005.