# on paratopological gyrogroups

Ying-Ying Jin  (Y.Y. Jin) School of Mathematics and Computational Science,
Wuyi University, Jiangmen 529020, P.R. China<EMAIL_ADDRESS>and  Li-Hong
Xie  (L.H. Xie) School of Mathematics and Computational Science, Wuyi
University, Jiangmen 529020, P.R. China<EMAIL_ADDRESS>
###### Abstract.

The concept of gyrogroups is a generalization of groups which do not
explicitly have associativity. Recently, Atiponrat extended the idea of
topological (paratopological) groups to topological (paratopological)
gyrogroups. In this paper, we prove that every regular (Hausdorff) locally
gyroscopic invariant paratopological gyrogroup $G$ is completely regular
(function Hausdorff). These results improve theorems of Banakh and Ravsky for
paratopological groups. Also, we extend the Pontrjagin conditions of
(para)topological groups to (para)topological gyrogroups.

###### Key words and phrases:

Paratopological gyrogroup; separation axioms; completely regular; Pontrjagin
conditions

###### 2010 Mathematics Subject Classification:

22A30; 22A22,20N05,54H99

* The corresponding author.

This work is supported by Natural Science Foundation of China (Grant Nos.
11861018, 11871379, 11526158), the Natural Science Foundation of Guangdong
Province under Grant (No.2018A030313063), The Innovation Project of Department
of Education of Guangdong Province (No:2018KTSCX231), and key project of
Natural Science Foundation of Guangdong Province Universities (2019KZDXM025).

##  1\. Introduction

The concept of gyrogroup was discovered by A. A. Ungar when he studied the
Einstein velocity addition [15]. A gyrogroup is a generalization of a group in
the sense that it is a groupoid with an identity and inverses, but the
associative law is redefined by more general definitions which are the left
gyroassociative law and the left loop property. The gyrogroup does not form a
group since it is neither associative nor commutative. Nevertheless, Ungar
showed that gyrogroup is rich in structure and encodes a group-like structure,
namely the gyrogroup structure [15].

Many important characteristics of gyrogroups have been intensively studied in
[7], [8], [11], [12] and [13]. T. Suksumran and K. Wiboonton have been
studying some basic algebraic properties of gyrogroups, for example, the
isomorphism theorems, Cayley’s Theorem, Lagrange’s Theorem, the gyrogroup
actions, etc. in [11], [12] and [13]. Most of these properties are similar to
those in classical group theory.

As far as we known, Atiponrat [2] is the first scholar who extended the idea
of topological groups to topological gyrogroups as gyrogroups with a topology
such that its binary operation is jointly continuous and the operation of
taking the inverse is continuous. Some basic properties of topological
gyrogroups are studied by and studied by Atiponrat [2]. Specially, Atiponrat
[2] discovered that for a topological gyrogroup, $T_{0}$ and $T_{3}$ are
equivalent. It is worth noting that Z. Cai, S. Lin and W. He in [5] proved
that every Hausdorff first countable topological gyrogroup is metrizable.

Recently, Atiponrat and Maungchang [3] studied some separation axioms of
paratopological gyrogroups. They proved that:

######  Theorem 1.1.

[3, Theorem 3] Let $G$ be a micro-associative regular paratopological
gyrogroup such that for any neighborhoods $A,B$ of the identity $e$ of $G$,
$\overline{A}\subseteq\text{Int~{}}\overline{B\oplus A}$. Then $G$ is
completely regular.

Banakh and Ravsky [4] showed that every regular paratopological group is
completely regular. Hence they gave a generalization to Banakh and Ravsky’s
result.

In this paper, we introduce the concept of locally gyroscopic invariantness in
paratopological gyrogroup and prove that every regular (Hausdorff) locally
gyroscopic invariant paratopological gyrogroup $G$ is completely regular
(function Hausdorff). Also, we find that the condition“for any neighborhoods
$A,B$ of the identity $e$ of $G$,
$\overline{A}\subseteq\text{Int}\overline{B\oplus A}$” in Theorem 1.1 can be
dropped. Finally, we also extend the Pontrjagin conditions of paratopological
groups to paratopological gyrogroups.

All spaces are not assumed to satisfy any separation axiom unless otherwise
stated.

$\bullet$ A space $X$ is regular if for any element $x\in X$ and an open set
$U$ containing $x$, there is a neighborhood $V$ of $x$ such that
$\overline{V}\subseteq U$.

$\bullet$ A space $X$ is completely regular if for any element $x\in X$ and an
open set $U$ containing $x$, there is a continuous function
$f:X\rightarrow[0,1]$ such that $f(x)=0$ and $f^{-1}([0,1))\subseteq U$.

$\bullet$ A space $X$ is functionally Hausdorff if for any distinct points
$x,y\in X$ there exists a continuous function $f:X\rightarrow\mathbb{R}$ such
that $f(x)\neq f(y)$.

##  2\. Definitions and preliminaries

Let $G$ be a nonempty set, and let $\oplus:G\times G\rightarrow G$ be a binary
operation on $G$. Then the pair $(G,\oplus)$ is called a groupoid. A function
$f$ from a groupoid $(G_{1},\oplus_{1})$ to a groupoid $(G_{2},\oplus_{2})$ is
said to be a groupoid homomorphism if
$f(x_{1}\oplus_{1}x_{2})=f(x_{1})\oplus_{2}f(x_{2})$ for any elements
$x_{1},x_{2}\in G_{1}$. In addition, a bijective groupoid homomorphism from a
groupoid $(G,\oplus)$ to itself will be called a groupoid automorphism. We
will write $Aut(G,\oplus)$ for the set of all automorphisms of a groupoid
$(G,\oplus)$.

######  Definition 2.1.

[15, Definition 2.7] Let $(G,\oplus)$ be a nonempty groupoid. We say that
$(G,\oplus)$ or just $G$ (when it is clear from the context) is a gyrogroup if
the followings hold:

  1. ($G1$)

There is an identity element $e\in G$ such that

$e\oplus x=x=x\oplus e\text{~{}~{}~{}~{}~{}for all~{}~{}}x\in G.$

  2. ($G2$)

For each $x\in G$, there exists an inverse element $\ominus x\in G$ such that

$\ominus x\oplus x=e=x\oplus(\ominus x).$

  3. ($G3$)

For any $x,y\in G$, there exists an gyroautomorphism $\text{gyr}[x,y]\in
Aut(G,\oplus)$ such that

$x\oplus(y\oplus z)=(x\oplus y)\oplus\text{gyr}[x,y](z)$

for all $z\in G$;

  4. ($G4$)

For any $x,y\in G$, $\text{gyr}[x\oplus y,y]=\text{gyr}[x,y]$.

In this paper, $\text{gyr}[a,b]V$ denotes $\\{\text{gyr}[a,b]v:v\in V\\}$.

The following Proposition 2.2 below summarizes some algebraic properties of
gyrogroups

######  Proposition 2.2.

([14, 15]) Let $(G,\oplus)$ be a gyrogroup and $a,b,c\in G$. Then

  1. (1)

$\ominus(\ominus a)=a$ Involution of inversion

  2. (2)

$\ominus a\oplus(a\oplus b)=b$ Left cancellation law

  3. (3)

gyr$[a,b](c)=\ominus(a\oplus b)\oplus(a\oplus(b\oplus c))$ Gyrator identity

  4. (4)

$\ominus(a\oplus b)=\text{gyr}[a,b](\ominus b\ominus a)$ cf.
$(ab)^{-1}=b^{-1}a^{-1}$

  5. (5)

$(\ominus a\oplus b)\oplus\text{gyr}[\ominus a,b](\ominus b\oplus c)=\ominus
a\oplus c$ cf. $(a^{-1}b)(b^{-1}c)=a^{-1}c$

  6. (6)

$\text{gyr}[a,b]=\text{gyr}[\ominus b,\ominus a]$ Even property

  7. (7)

$\text{gyr}[a,b]=\text{gyr}^{-1}[b,a],\text{the inverse of gyr}[b,a]$
Inversive symmetry

######  Definition 2.3.

[2, Definition 1] A triple $(G,\tau,\oplus)$ is called a topological gyrogroup
if and only if

  1. (1)

$(G,\tau)$ is a topological space;

  2. (2)

$(G,\oplus)$ is a gyrogroup;

  3. (3)

The binary operation $\oplus:G\times G\rightarrow G$ is continuous where
$G\times G$ is endowed with the product topology and the operation of taking
the inverse $\ominus(\cdot):G\rightarrow G$, i.e. $x\rightarrow\ominus x$, is
continuous.

######  Definition 2.4.

[14, Definition 2.9] Let $(G,\oplus)$ be a gyrogroup with gyrogroup operation
(or, addition) $\oplus$. The gyrogroup cooperation (or, coaddition) $\boxplus$
is a second binary operation in $G$ given by the equation

$(\divideontimes)~{}~{}~{}~{}a\boxplus b=a\oplus\text{gyr}[a,\ominus b]b$

for all $a,b\in G$. The groupoid $(G,\boxplus)$ is called a cogyrogroup, and
is said to be the cogyrogroup associated with the gyrogroup $(G,\oplus)$.

Replacing $b$ by $\ominus b$ in $(\divideontimes)$, along with Identity
$(\divideontimes)$ we have the identity

$a\boxminus b=a\ominus\text{gyr}[a,b]b$

for all $a,b\in G$, where we use the obvious notation, $a\boxminus
b=a\boxplus(\ominus b)$.

######  Theorem 2.5.

[14, Theorem 2.14] Let $(G,\oplus)$ be a gyrogroup with cooperation $\boxplus$
given by Definition 2.4. Then,

$a\oplus b=a\boxplus\text{gyr}[a,b]b.$

If a triple $(G,\tau,\oplus)$ satisfies the first two conditions and its
binary operation is continuous, we call such triple a paratopological
gyrogroup [3]. Sometimes we will just say that $G$ is a topological gyrogroup
(paratopo-logical gyrogroup) if the binary operation and the topology are
clear from the context.

######  Proposition 2.6.

[3, proposition 1] Let $G$ be a paratopological gyrogroup and $A$ be an open
set. Then $B\oplus A$ is open for each $B\subseteq G$.

To study the separation axioms in paratopological gyrogroups, Atiponrat and
Maungchang [3, Definition 7] introduced the the following concept:

######  Definition 2.7.

[3, Definition 7] A paratopological gyrogroup $G$ micro-associative if for any
neighborhood $U\subseteq G$ of the identity $e$, there are neighborhoods
$W\subseteq V\subseteq U$ of $e$ such that $a\oplus(b\oplus V)=(a\oplus
b)\oplus V$ for any $a,b\in W$.

They showed that for each neighborhood $U$ every micro-associative
paratopological gyrogroup $G$ of the identity $e$, there are neighborhoods
$W\subseteq V\subseteq U$ of $e$ such that $\text{gyr~{}}[V]\subseteq V$ [3,
Lemma 2]. This motives us introduce the following concept which plays an
important role in our study.

######  Definition 2.8.

Let $G$ be a paratopological gyrogroup and $\mathscr{B}$ a local base at the
identity $e$ of $G$. We say that $\mathscr{B}$ is locally gyroscopic invariant
if there is an open neighborhood $U$ of $e$ such that gyr$[a,b]V\subseteq V$
for each $V\in\mathscr{B}$ and $a,b\in U$. A paratopological gyrogroup $G$ is
called locally gyroscopic invariant if $G$ has a locally gyroscopic invariant
base at the identity.

######  Remark 2.9.

The Möbius gyrogroup $\mathbb{D}=\\{z\in\mathbb{C}:|z|<1\\}$ with the standard
topology is a topological gyrogroup(see [3, Example 2]). In fact, $\mathbb{D}$
is micro-associative (see [3, Example 8]). Also, one can easily show that
$\mathbb{D}$ is locally gyroscopic invariant. However, we do not what are
different for micro-associative paratopological gyrogroups and locally
gyroscopic invariant paratopological gyrogroups.

##  3\. Complete regularity in paratopological gyrogroups

In this section, we shall study the separation axioms in paratopological
gyrogroups. By the Urysohn’s technology we prove the following important
result.

######  Proposition 3.1.

Let $G$ be a locally gyroscopic invariant paratopological gyrogroup. Then for
any neighborhood $O\subseteq G$ of $e$ there exists a continuous function
$f:G\rightarrow[0,1]$ such that $f(e)=0$ and
$f^{-1}([0,1))\subseteq\overline{O}$.

###### Proof.

Since $G$ is a locally gyroscopic invariant paratopological gyrogroup,
according to Definition 2.8 $G$ has a locally gyroscopic invariant base
$\mathscr{B}$ at the identity $e$ of $G$ and an open neighborhood $U$ of $e$
such that gyr$[a,b]V\subseteq V$ for each $V\in\mathscr{B}$ and $a,b\in U$.
Without loss the generality, we can assume that $U\subseteq O$ and every
element of $\mathscr{B}$ is contained in $U$. Then we can find a family
$\\{U_{i}:i\in\omega\\}\subseteq\mathscr{B}$ of open neighborhoods of $e$ such
that $U_{i+1}\oplus U_{i+1}\subseteq U_{i}$ and $U_{0}\oplus U_{0}\subseteq
U\subseteq O$ for each $i\in\omega$ and gry$[a,b]U_{i}\subseteq U_{i}$ holds
for each $a,b\in U$ and each $i\in\omega$.

Put $V(1)=U_{0}$, fix $n\in\omega$, and assume that open neighborhoods
$V(m/2^{n})$ of $e$ are defined for each $m=1,2,...,2^{n}$. Put then
$V(1/2^{n+1})=U_{n+1}$, $V(2m/2^{n+1})=V(m/2^{n})$ for $m=1,...,2^{n}$, and

$(*)~{}~{}~{}~{}V((2m+1)/2^{n+1})=V(m/2^{n})\oplus U_{n+1}=V(m/2^{n})\oplus
V(1/2^{n+1}),$

for each $m=1,2,...,2^{n}-1$. By Proposition 2.6, one can easily show that
$V(r)$ is an open neighborhood of $e$ for every positive dyadic rational
number $r\leq 1$. We claim the following conditions are satisfied:

  1. (1)

$V(m/2^{n})\oplus V(1/2^{n})\subseteq V((m+1)/2^{n})\subseteq U\subseteq O$,
for all integers $m$ and $n$ satisfying $n>0$ and $2^{n}>m>0$.

  2. (2)

$V(r_{1})\subseteq V(r_{2})$ for any positive dyadic rational numbers $r_{1}$
and $r_{2}$ with $r_{1}<r_{2}\leq 1$

  3. (3)

$\overline{V(r_{1})}\subseteq\text{Int~{}}\overline{V(r_{2})}$ for any
positive dyadic rational numbers $r_{1}$ and $r_{2}$ with $r_{1}<r_{2}\leq 1$.

The proof of (1). Let us prove (1) by induction on $n$.

If $n=1$, then the only possible value for $m$ is also 1, and we have:

$V(1/2)\oplus V(1/2)=U_{1}\oplus U_{1}\subseteq U_{0}=V(1)\subseteq U\subseteq
O.$

Assume that $(1)$ holds for some $n$. Let us verify it for $n+1$.

If $m=2k$ for some $k\in\omega$, then we have that

$\displaystyle V(m/2^{n+1})\oplus V(1/2^{n+1})$
$\displaystyle=V(2k/2^{n+1})\oplus V(1/2^{n+1})$
$\displaystyle=V((2k+1)/2^{n+1})\quad\quad\text{by the definition~{}}(*)$
$\displaystyle=V((m+1)/2^{n+1}).$

This implies that $V(m/2^{n+1})\oplus V(1/2^{n+1})\subseteq V((m+1)/2^{n+1}).$
Also,

$\displaystyle V((m+1)/2^{n+1})$ $\displaystyle=V(2k/2^{n+1})\oplus
V(1/2^{n+1})\quad\quad\text{by the definition~{}}(*)$ $\displaystyle\subseteq
V(k/2^{n})\oplus U_{n}\quad\quad\quad\text{by~{}}V(1/2^{n+1})=U_{n+1}\subseteq
U_{n}$ $\displaystyle=V(k/2^{n})\oplus
V(1/2^{n})\quad\quad\quad\text{by~{}}V(1/2^{n})=U_{n}$
$\displaystyle=V((k+1)/2^{n})\subseteq U\subseteq O.\quad\quad\quad\text{by
the inductive assumption}$

Assume now that $0<m=2k+1<2^{n+1}$, for some integer $k$. Then

$\displaystyle V(m/2^{n+1})\oplus V(1/2^{n+1})$
$\displaystyle=V((2k+1)/2^{n+1})\oplus
U_{n+1}\quad\quad\quad\text{by~{}}V(1/2^{n+1})=U_{n+1}$
$\displaystyle=(V(k/2^{n})\oplus U_{n+1})\oplus U_{n+1}\quad\quad\text{by the
definition~{}}(*)$ $\displaystyle=\bigcup_{a\in V(k/2^{n}),b\in
U_{n+1}}a\oplus(b\oplus\text{gyr}[b,a]U_{n+1})\quad\text{by right
gyroassociative law}$ $\displaystyle\text{by locally gyroscopic invariant and
the inductive assumption~{}}V(k/2^{n})\subseteq U\subseteq O,$
$\displaystyle\text{clearly,~{}}U_{n+1}\subseteq U\subseteq O,\text{~{}then}$
$\displaystyle\subseteq\bigcup_{a\in V(k/2^{n}),b\in U_{n+1}}a\oplus(b\oplus
U_{n+1})$ $\displaystyle=V(k/2^{n})\oplus(U_{n+1}\oplus U_{n+1})$
$\displaystyle\subseteq V(k/2^{n})\oplus U_{n}$
$\displaystyle=V(k/2^{n})\oplus V(1/2^{n})$ $\displaystyle\subseteq
V((k+1)/2^{n})\quad\quad\quad\text{by the inductive assumption}$
$\displaystyle=V(2(k+1)/2^{n+1})$ $\displaystyle=V((m+1)/2^{n+1})$
$\displaystyle=V((k+1)/2^{n})\subseteq U\subseteq O\quad\quad\quad\text{by the
inductive assumption}$

This completes the proof of (1).

The proof of (2). If $n=1$, then the only possible value for $m$ is 1, and we
have:

$V(1/2)=U_{1}\subseteq U_{0}=V(2/2).$

Hence, to proof (2) it is enough to prove that

$V(m/2^{n+1})\subseteq V((m+1)/2^{n+1})\quad\quad\quad(**)$

holds for each $n\geq 1$ and $0<m,m+1\leq 2^{n}$.

If $m=2k$ for some $k$, then

$\displaystyle V(m/2^{n+1})=V(2k/2^{n+1})$ $\displaystyle=V(k/2^{n})\subseteq
V(k/2^{n})\oplus U_{n+1}$ $\displaystyle=V(k/2^{n})\oplus
V(1/2^{n+1})\quad\quad\quad\text{by~{}~{}}V(1/2^{n+1})=U_{n+1}$
$\displaystyle=V((2k+1)/2^{n+1})\quad\quad\text{by the definition~{}}(*)$
$\displaystyle=V((m+1)/2^{n+1}).$

If $m=2k+1$ for some $k$, then

$\displaystyle V(m/2^{n+1})=V((2k+1)/2^{n+1})$ $\displaystyle=V(k/2^{n})\oplus
U_{n+1}\quad\quad\text{by the definition~{}}(*)$ $\displaystyle\subseteq
V(k/2^{n})\oplus U_{n}$ $\displaystyle=V(k/2^{n})\oplus
V(1/2^{n})\quad\quad\quad\text{by~{}~{}}V(1/2^{n})=U_{n}$
$\displaystyle\subseteq V((k+1)/2^{n})\quad\quad\quad\text{by the fact }(1)$
$\displaystyle=V(2(k+1)/2^{n+1})$ $\displaystyle=V((m+1)/2^{n+1}).$

This completes the proof of (2).

The proof of (3). Since $r_{1}<r_{2}$, we can find positive integers $n,m$
such that $r_{1}\leq m/2^{n}<(m+1)/2^{n}\leq r_{2}$. From the fact that
$V(r_{1})\subseteq V(r_{2})$ if $0<r_{1}<r_{2}\leq 1$, it follows that:

$\displaystyle\overline{V(r_{1})}\subseteq\overline{V(m/2^{n})}\quad\quad\quad\quad\quad\quad\text{by
the fact~{}}(2)$ $\displaystyle\subseteq\overline{V(m/2^{n})}\oplus
V(1/2^{n})$ $\displaystyle\subseteq\overline{V(m/2^{n})\oplus
V(1/2^{n})}\quad\text{by~{}}\oplus\text{~{}being continuous on~{}}G\times G$
$\displaystyle\subseteq\overline{V((m+1)/2^{n})}\quad\quad\quad\quad\text{by
the fact~{}}(1)$
$\displaystyle\subseteq\overline{V(r_{2})}\quad\quad\quad\quad\quad\quad\text{by
the fact~{}}(2).$

Since the set $V(1/2^{n})=U_{n}$ is open, by [3, Proposition 1], the set
$\overline{V(m/2^{n})}\oplus V(1/2^{n})$ is open and contains
$\overline{V(r_{1})}$, and therefore, we obtain that
$\overline{V(r_{1})}\subseteq\text{Int~{}}\overline{V(r_{2})}$.

Now we define a real-valued function $f$ on $G$ as follows:

$f(y)=\left\\{\begin{array}[]{r@{\quad\quad}l}1&\text{if~{}}y\in
G\setminus\overline{V(1)},\\\
\text{inf~{}}\\{r:y\in\overline{V(r)}\\}&\text{if~{}}y\in\overline{V(1)}.\end{array}\right.$

Clearly, the function $f$ is well-defined and $f(e)=0,$
$f^{-1}([0,1))\subseteq\overline{V(1)}=\overline{U_{0}}\subseteq\overline{O}$.
Thus it is enough to prove the following:

(4) $f$ is a continuous function on $G$.

Proof of (4). Take any $y\in G$ and an open neighborhood $W$ of $f(y)$ in
$[0,1]$.

(i) If $f(y)=1$, then one can find a positive dyadic rational number $r<1$
such that $[r,1]\subseteq W$. According to the definition of $f$ it is obvious
that $y\in G\setminus\overline{V(r)}$ and
$f(G\setminus\overline{V(r)})\subseteq[r,1]\subseteq W$, which implies that
$f$ is continuous at $y$.

(ii) If $0<f(y)<1$, then one can find positive dyadic rational numbers
$r_{1}<r_{2}<1$ such that $f(y)\in(r_{1},r_{2})\subseteq[r_{1},r_{2}]\subseteq
W$. By (3) and the definition of $f$, one can easily show that
$y\in(\text{Int~{}}\overline{V(r_{2})})\setminus\overline{V(r_{1})}$. Indeed,
$y\notin\overline{V(r_{1})}$ is obvious. By $f(y)<r_{2}$ one can find a
positive dyadic rational numbers $r_{3}$ such that $f(y)<r_{3}<r_{2}$. Hence,
$y\in\overline{V(r_{3})})\subseteq\text{Int~{}}\overline{V(r_{2})}$ by (3) and
the definition of $f$. Therefore, the set
$(\text{Int~{}}\overline{V(r_{2})})\setminus\overline{V(r_{1})}$ is an open
neighbourhood of $y$ such that
$f((\text{Int~{}}\overline{V(r_{2})})\setminus\overline{V(r_{1})})\subseteq[r_{1},r_{2}]\subseteq
W$, which implies that $f$ is continuous at $y$.

(iii) If $f(y)=0$, then one can find a positive dyadic rational number $r<1$
such that $[0,r]\subseteq W$. By (3) and the definition of $f$, one can easily
show that $y\in\text{Int~{}}\overline{V(r)}$ and
$f(\text{Int~{}}\overline{V(r)})\subseteq[0,r]\subseteq W$, which implies that
$f$ is continuous at $y$. Hence we have proved that $f$ is continuous at each
point of $G$. ∎

Now we obtain our main results:

######  Theorem 3.2.

Every regular locally gyroscopic invariant paratopological gyrogroup $G$ is
completely regular.

###### Proof.

Since $G$ is homogeneous, it is enough to show that there is a continuous
real-valued function which can separate the identity $e$ in $G$ from the
closed set $F$ such that $e\notin F$. Take any open neighborhood $O$ of $e$.
Since $G$ is regular, one can find an open neighborhood $U$ of $e$ such that
$\overline{U}\subseteq O$. According to Proposition 3.1, we can find a
continuous function $f:G\rightarrow[0,1]$ such that $f(e)=0$ and
$f^{-1}([0,1))\subseteq\overline{U}\subseteq O$. This implies that $f(y)=1$
for each $y\in G\setminus O$. Thus $G$ is completely regular. ∎

######  Theorem 3.3.

Every Hausdorff locally gyroscopic invariant paratopological gyrogroup $G$ is
functionally Hausdorff.

###### Proof.

Take any distinct points $x,y\in G$. Since $G$ is homogeneous, we can assume
that $x$ is the identity $e$ in $G$. Since $G$ is Hausdorff, one can find an
open neighborhood $O$ of $x$ such that $y\notin\overline{O}$. According to
proposition 3.1, we can find a continuous function $f:G\rightarrow[0,1]$ such
that $f(x)=0$ and $f^{-1}([0,1))\subseteq\overline{O}$. This implies that
$f(y)=1\neq 0=f(x)$. Thus $G$ is functionally Hausdorff. ∎

By Proposition 2.6, we can prove the following result:

######  Lemma 3.4.

Let $G$ be a paratopological gyrogroup and $A\subseteq G$ be an open
neighborhood of the identity $e$. Then
$A\subseteq\text{Int~{}}\overline{B\oplus A}$ for any set $B\subseteq G$
containing $e$.

###### Proof.

Indeed, one can easily show that $A\subseteq B\oplus
A\subseteq\overline{B\oplus A}$. Since $A$ is open in $G$, $B\oplus A$ is open
by Proposition 2.6. Hence, we have that $A\subseteq B\oplus
A\subseteq\text{Int~{}}\overline{B\oplus A}$. ∎

Indeed, Atiponrat and Maungchang [3] have shown that the condition “for any
neighborhoods $A,B$ of the identity $e$ of $G$” can be replaced by “for any
open neighborhoods $A,B$ of the identity $e$ of $G$” in Theorem 1.1. Thus, by
Lemma 3.4 we have the following result:

######  Theorem 3.5.

Every micro-associative regular (Hausdorff) paratopological gyrogroup is
completely regular (function Hausdorff).

##  4\. properties of (para)topological gyrogroups

In this section, we will show that every topological gyrogroup has many self-
homeomorphisms. Let $(G,\oplus)$ be a gyrogroup, $x\in G$ and $A,B\subseteq
G$. We write $A\oplus B=\\{a\oplus b:a\in A,b\in B\\}$, $x\oplus A=\\{x\oplus
a:a\in A\\}$ and $A\oplus x=\\{a\oplus x:a\in A\\}$.

######  Proposition 4.1.

[2, Proposition 3 and Corollary 5] Let $G$ be a topological gyrogroup, $x\in
G$ and $A,B\subseteq G$.

  1. (1)

the inverse mapping $inv:G\rightarrow G$, where $inv(x)=\ominus x$ for every
$x\in G$, is a homeomorphism;

  2. (2)

the left translation $L_{x}:G\rightarrow G$, where $L_{x}(y)=x\oplus y$ for
every $y\in G$, is a homeomorphism;

  3. (3)

the right translation $R_{x}:G\rightarrow G$, where $R_{x}(y)=y\oplus x$ for
every $y\in G$, is a homeomorphism;

  4. (4)

if $A$ is open in $G$, then $x\oplus A$, $A\oplus x$, $A\oplus B$, $B\oplus A$
and $\ominus A$ are all open in $G$.

Recall that a topological space $X$ is said to be homogeneous if for every
$x\in X$ and every $y\in X$, there exists a homeomorphism $f$ of the space $X$
onto itself such that $f(x)=y$.

In the following theorem, we characterize the families of subsets of a
gyrogroup $G$ which can appear as neighborhood bases of the neutral element in
some paratopological gyrogroup and consider the Pontrjagin conditions, which
are generalization of Ravsky’s results [9].

######  Theorem 4.2.

Let $G$ be a Hausdorff paratopological gyrogroup and $\mathcal{U}$ an open
base at the neutral element $e$ of $G$. Then the following conditions hold:

  1. (1)

for every $U\in\mathcal{U}$, there exists an element $V\in\mathcal{U}$ such
that $V\oplus V\subseteq U$;

  2. (2)

for every $U\in\mathcal{U}$, and every $x\in U$, there exists
$V\in\mathcal{U}$ such that $x\oplus V\subseteq U$;

  3. (3)

for every $U\in\mathcal{U}$ and $x\in G$, there exists $V\in\mathcal{U}$ such
that $\ominus x\oplus(V\oplus x)\subseteq U$;

  4. (4)

for $U,V\in\mathcal{U}$, there exists $W\in\mathcal{U}$ such that $W\subseteq
U\cap V$;

  5. (5)

for every $U\in\mathcal{U}$ and $a,b\in G$, there exists an element
$V\in\mathcal{U}$ such that gyr$[a,b]V\subseteq U$;

  6. (6)

for every $U\in\mathcal{U}$ and $b\in G$, there exists an element
$V\in\mathcal{U}$ such that $\bigcup_{v\in V}\text{gyr}[v,b]V\subseteq U$;

  7. (7)

$\\{e\\}=\bigcap_{U\in\mathcal{U}}(U\boxminus U)$.

Conversely, let $G$ be a gyrogroup and let $\mathcal{U}$ be a family of
subsets of $G$ satisfying conditions (1)-(7). Then the family
$\mathcal{B}_{\mathcal{U}}=\\{a\oplus U:a\in G,U\in\mathcal{U}\\}$ is a base
for a Hausdorff topology $\mathcal{T}_{\mathcal{U}}$ on $G$. With this
topology, $G$ is a paratopological gyrogroup.

###### Proof.

Let $U\in\mathcal{U}$.

(1) For $G$ is a paratopological gyrogroup, then $op_{2}:G\times G\rightarrow
G$ defined by $op_{2}(x,y)=x\oplus y$ is continuous. Because $e\oplus e=e$,
and $U\in\mathcal{U}$, there exist neighborhood $O$ and $W$ of $e$ such that
$O\oplus W\subseteq U$. We choose $V\in\mathcal{U}$ such that $V\subseteq
O\cap W$. Then $V\oplus V\subseteq W$.

(2) Let $x\in U$. We define $R_{x}:G\rightarrow G$ by $R_{x}(y)=x\oplus y$.
Since $R_{x}(e)=x$ and $R_{x}$ is continuous at $e$, there exists
$V\in\mathcal{U}$ such that $x\oplus V=R_{x}(V)\subseteq U$.

(3) For every $x\in G$, we define left translation map $L_{\ominus
x}:G\rightarrow G$ by $L_{\ominus x}(y)=\ominus x\oplus y$. By the continuous
of $L_{\ominus x}$, for every $U\in\mathcal{U}$, there exists the neighborhood
$V^{\prime}$ of $x$ such that $L_{\ominus x}(V^{\prime})\subseteq U$, that is
$\ominus x\oplus V^{\prime}=L_{\ominus x}(V^{\prime})\subseteq U$. We also
define the right translation map $R_{x}:G\rightarrow G$ by $R_{x}(y)=y\oplus
x$. Then $R_{x}(e)=x$. Because $R_{x}$ is continuous at $e$, for the
neighborhood $V^{\prime}$ of $x$, there exists $V\in\mathcal{U}$ such that
$R_{x}(V)\subseteq V^{\prime}$, that is $V\oplus x=R_{x}(V)\subseteq
V^{\prime}.$ So we get $\ominus x\oplus(V\oplus x)\subseteq\ominus x\oplus
V^{\prime}\subseteq U$.

(4) It is clear since $\mathcal{U}$ is an open base at $e.$

(5) For every $a,b\in G$, we define $f_{a,b}:G\rightarrow G$ by
$f_{a,b}(x)=\text{gyr}[a,b]x$. Since $f_{a,b}(e)=e$ and $f_{a,b}$ is
continuous at $e$, for every $U\in\mathcal{U}$, there exists $V\in\mathcal{U}$
such that $f_{a,b}(V)\subseteq U$, that is, gyr$[a,b]V\subseteq U$.

(6) Take $W\in\mathcal{U}$ such that $W\oplus W\subseteq U$. Then
$b\oplus(W\oplus W)$ is an open set containing $b$. Since $G$ is a
paratopological gyrogroup, one can find $V\in\mathcal{U}$ such that

$(b\oplus V)\oplus V\subseteq b\oplus(W\oplus W)\quad\quad\quad(*)$

Note that

$(b\oplus V)\oplus V=b\oplus(V\oplus\bigcup_{v\in
V}\text{gyr}[v,b]V)\quad\quad\quad(**)$

By (*) and (**) we have $b\oplus(V\oplus\bigcup_{v\in
V}\text{gyr}[v,b]V)\subseteq b\oplus(W\oplus W)$, which means
$V\oplus\bigcup_{v\in V}\text{gyr}[v,b]V\subseteq W\oplus W$. So we can get
$\bigcup_{v\in V}\text{gyr}[v,b]V\subseteq W\oplus W\subseteq U$.

(7) We assume that $G$ is Hausdorff. If $\bigcap_{U\in\mathcal{U}}(U\boxminus
U)\neq\\{e\\}$, then there is $x\in\bigcap_{U\in\mathcal{U}}(U\boxminus U)$
such that $x\neq e$. Since $G$ is Hausdorff, there are an open set $V_{1}$
containing $x$ and a open set $V\in\mathcal{U}$ such that $V_{1}\cap
V=\emptyset$. Since $V_{1}$ is a neighborhood of $x$, $x\oplus e=x$ and $G$ is
a paratopological gyrogroup, one can find $U\in\mathcal{U}$ such that
$(x\oplus U)\subseteq V_{1}$, hence we have that $(x\oplus U)\cap
V=\emptyset$. We choose $W\in\mathcal{U}$ such that $W\subseteq U\cap V$. Then
$(x\oplus W)\cap W=\emptyset$, that is, $x\notin W\boxminus W$, which is a
contradiction.

To prove the converse, let $\mathcal{U}$ be a family of subsets of $G$ such
that conditions (1)-(7) hold. Let $\mathcal{T}=\\{W\subseteq G:\text{for
every~{}}x\in W~{}\text{there exists~{}}U\in\mathcal{U}~{}\text{such
that~{}}x\oplus U\subseteq W\\}.$

Claim 1. $\mathcal{T}$ is a topology on $G$. It is clear that
$G\in\mathcal{T}$ and $\emptyset\in\mathcal{T}$. It also easy to see that
$\mathcal{T}$ is closed under unions. To show that $\mathcal{T}$ is closed
under finite intersections, let $V,W\in\mathcal{T}$. Let $x\in V\cap W$. Since
$x\in V\in\mathcal{T}$ and $x\in W\in\mathcal{T}$, there exist
$O,Q\in\mathcal{U}$ such that $x\oplus O\subseteq V$ and $x\oplus Q\subseteq
W$. From (5) it follows that there exists $U\in\mathcal{T}$ such that
$U\subseteq O\cap Q$. Then, we have $x\oplus U\subseteq V\cap W$. Hence,
$V\cap W\in\mathcal{T}$, and $\mathcal{T}$ is a topology on $G$.

Claim 2. If $O\in\mathcal{U}$ and $g\in G$, then $g\oplus O\in\mathcal{T}$.

Take any $x\in g\oplus O$, then $\ominus g\oplus x\in O$. By property (2),
there exists $V^{\prime}\in\mathcal{U}$ such that $\ominus g\oplus x\oplus
V^{\prime}\subseteq O$. For $V^{\prime}$ and $\ominus g,x\in G$, there exists
$V\in\mathcal{U}$ such that $\text{gyr}[\ominus g,x]V\subseteq V^{\prime}$ by
condition (5). So we have $\ominus g\oplus(x\oplus V)=(\ominus g\oplus
x)\oplus\text{gyr}[\ominus g,x]V\subseteq O$, that is $x\oplus V\subseteq
g\oplus O$. Hence $g\oplus O\in\mathcal{T}$.

Claim 3. The family $\mathcal{B}_{\mathcal{U}}=\\{a\oplus U:a\in
G,U\in\mathcal{U}\\}$ is a base for the topology $\mathcal{T}$ on $G$.

Indeed, it follows from Claim 2.

Claim 4. The multiplication in $G$ is continuous with respect to the topology
$\mathcal{T}$.

Let $a,b$ be arbitrary elements of $G$, and $O$ be any element of
$\mathcal{T}$ such that $a\oplus b\in O$. Then there exists $W\in\mathcal{U}$
such that $(a\oplus b)\oplus W\subseteq O$. There exists $U\in\mathcal{U}$
such that $a\oplus b\oplus\text{gyr}[a,b]U\subseteq(a\oplus b)\oplus W$ by
condition (5). For $U$ there exists $U_{1}\in\mathcal{U}$ such that
$U_{1}\oplus U_{1}\subseteq U$. For $b$ and $U_{1}$ there exists
$U_{2}\in\mathcal{U}$ such that $\bigcup_{v\in
U_{2}}\text{gyr}[v,b]U_{2}\subseteq U_{1}$ by condition (6). By condition (4),
we can get $U_{3}\subseteq U_{1}\cap U_{2}$. For $U_{3}\in\mathcal{U}$, apply
(3) to choose $U_{4}\in\mathcal{U}$ such that $U_{4}\oplus b\subseteq b\oplus
U_{3}$. Using condition (6) we can get $U_{5}\in\mathcal{U}$ such that
$\bigcup_{v\in U_{5}}\text{gyr}[v,b]U_{5}\subseteq U_{3}$. By the condition
(4), we get $U_{6}\subseteq U_{4}\cap U_{5}$. We have

$\displaystyle a\oplus U_{6}\oplus(b\oplus U_{6}))\quad$
$\displaystyle=a\oplus U_{6}\oplus\text{gyr}[a,e](b\oplus U_{6}))\quad$
$\displaystyle\subseteq a\oplus U_{6}\oplus\bigcup_{v\in
U_{6}}\text{gyr}[a,v](b\oplus U_{6}))\quad$
$\displaystyle=a\oplus(U_{6}\oplus(b\oplus U_{6}))\quad$
$\displaystyle=a\oplus((U_{6}\oplus b)\oplus\bigcup_{v\in
U_{6}}\text{gyr}[v,b]U_{6})))\quad$ $\displaystyle\subseteq
a\oplus((U_{4}\oplus b)\oplus\bigcup_{v\in U_{5}}\text{gyr}[v,b]U_{5})))\quad$
$\displaystyle\subseteq a\oplus((U_{4}\oplus b)\oplus
U_{3})\quad\quad\quad\quad\quad\quad$ $\displaystyle\subseteq a\oplus((b\oplus
U_{3})\oplus U_{3})\quad\quad\quad\quad\quad\quad$
$\displaystyle=a\oplus(b\oplus(U_{3}\oplus\bigcup_{v\in
U_{3}}\text{gyr}[v,b]U_{3})))\quad$ $\displaystyle\subseteq
a\oplus(b\oplus(U_{1}\oplus\bigcup_{v\in U_{2}}\text{gyr}[v,b]U_{2})))\quad$
$\displaystyle\subseteq a\oplus(b\oplus(U_{1}\oplus U_{1}))\quad$
$\displaystyle\subseteq a\oplus(b\oplus U)\quad\quad\quad\quad\quad\quad$
$\displaystyle=a\oplus b\oplus\text{gyr}[a,b]U\quad\quad\quad$
$\displaystyle\subseteq(a\oplus b)\oplus W\quad\quad\quad$

Since $U_{6}\in\mathcal{U}$, $a\oplus U_{6},b\oplus U_{6}$ are the
neighborhood of $a,b$. Thus, the multiplications in $G$ is continuous with
respect to the topology $\mathcal{T}$. This proves Claim 4.

Claim 5. The gyrogroup $G$ with topology $\mathcal{T}$ is Hausdorff.

For every $x,y\in G$ and $x\neq y$, then $\ominus y\oplus x\neq e$. There
exist $U\in\mathcal{U}$ such that $\ominus y\oplus x\notin U\boxminus U$,
which implies $\ominus y\oplus x\oplus U\cap U=\emptyset$. For $x,y\in G$ and
$U\in\mathcal{U}$, there exists $V\in\mathcal{U}$ such that
$\text{gyr}[\ominus y,x]V\subseteq U$ by condition (5). Then we claim that
$x\oplus V\cap y\oplus U=\emptyset$, which implies that the gyrogroup $G$ with
the topology $\mathcal{T}$ is Hausdorff.

In fact, if $x\oplus V\cap y\oplus U\neq\emptyset$, then $\ominus
y\oplus(x\oplus V)\cap U\neq\emptyset$. Hence, we have that $(\ominus y\oplus
x)\oplus\text{gyr}[\ominus y,x]V\cap U\neq\emptyset$. Since
$\text{gyr}[\ominus y,x]V\subseteq U$, we have that $(\ominus y\oplus x)\oplus
U\cap U\neq\emptyset$. This is a contradiction.

We prove that $G$ is a Hausdorff paratopological gyrogroup with the topology
$\mathcal{T}$.

∎

######  Lemma 4.3.

Let $G$ be a topological gyrogroup and $x\in G$. Then
$L_{x}^{\boxplus}(\cdot):G\rightarrow G$ is homeomorphisms, where
$L_{x}^{\boxplus}(\cdot)$ is defined as: $L_{x}^{\boxplus}(y)=x\boxplus y$ for
each $y\in G$.

###### Proof.

According to the definition, we have that

$\displaystyle x\boxplus y$ $\displaystyle=x\oplus\text{gyr}[x,\ominus y]y$
$\displaystyle=x\oplus(\ominus(x\ominus y)\oplus(x\oplus(\ominus y\oplus y)))$
$\displaystyle=x\oplus(\ominus(x\ominus y)\oplus x)$

Hence, $L_{x}^{\boxplus}(y)=L_{x}(R_{x}(\ominus(L_{x}(\ominus(y)))))$. Since
the operations $L_{x},R_{x}$ and $\ominus$ are homeomorphisms, so is their the
compositions. ∎

######  Theorem 4.4.

Let $G$ be a Hausdorff topological gyrogroup and $\mathcal{U}$ an open base at
the neutral element $e$ of $G$. Then the conditions (1)-(7) in Theorem 4.2
hold. In addition, the following conditions hold:

  1. (8)

for every $U\in\mathcal{U}$ and $x\in G$, there exists $V\in\mathcal{U}$ such
that $V\boxplus x\subseteq x\oplus U$ and $x\oplus V\subseteq x\boxplus U$;

  2. (9)

for every $U\in\mathcal{U}$, there exists $V\in\mathcal{U}$ such that $\ominus
V\subseteq U$.

Conversely, let $G$ be a gyrogroup and let $\mathcal{U}$ be a family of
subsets of $G$ satisfying conditions (1)-(9). Then the family
$\mathcal{B}_{\mathcal{U}}=\\{a\oplus U:a\in G,U\in\mathcal{U}\\}$ is a base
for a Hausdorff topology $\mathcal{T}_{\mathcal{U}}$ on $G$. With this
topology, $G$ is a topological gyrogroup.

###### Proof.

Let $G$ be a Hausdorff topological gyrogroup and $\mathcal{U}$ an open base at
the neutral element $e$ of $G$. Then the conditions (1)-(7) hold by Theorem
4.2.

For (8). Since $G$ is a topological gyrogroup, it is obvious that $(x\oplus
U)\ominus x$ is an open set containing the the neutral element $e$ of $G$.
Hence there is a $V_{1}\in\mathcal{U}$ such that $V_{1}\subseteq(x\oplus
U)\ominus x$, which is equivalent to $V_{1}\boxplus x\subseteq x\oplus U$. By
Lemma 4.3, we have that $x\boxplus U$ is an open set containing $x$. Since the
operation $L_{x}$ is continuous and $L_{x}(e)=x$, one can find
$V_{2}\in\mathcal{U}$ such that $L_{x}(V_{2})=x\oplus V_{2}\subseteq x\boxplus
U$. Take a $V\in\mathcal{U}$ such that $V\subseteq V_{1}\cap V_{2}$. Then the
set $V$ is the required.

For (9). Since $G$ is a topological gyrogroup, the operation $\ominus$ is
continuous. Clearly, $U$ is an open set containing the neutral element $e$, so
one can find $V\in\mathcal{U}$ such that $\ominus V\subseteq U$.

To prove the converse, let $\mathcal{U}$ be a family of subsets of $G$ such
that conditions (1)-(9) hold. Let $\mathcal{T}=\\{W\subseteq G:\text{for
every~{}}x\in W~{}\text{there exists~{}}U\in\mathcal{U}~{}\text{such
that~{}}x\oplus U\subseteq W\\}.$ Then by Theorem 4.2 we have that $G$ with
the topology $\mathcal{T}$ is a paratopological gyrogroup. Thus It is enough
to show that the inverse operation
$\ominus:(G,\mathcal{T})\rightarrow(G,\mathcal{T})$ is continuous.

Take any $x\in G$ and any $U\in\mathcal{U}$. By the condition (8), there is
$U_{1}\in\mathcal{U}$ such that $U_{1}\boxplus(\ominus x)\subseteq\ominus
x\oplus U$. For $U_{1}$, applying the condition (9), one can find
$U_{2}\in\mathcal{U}$ such that $\ominus U_{2}\subseteq U_{1}$. For $U_{2}$,
applying the condition (8) again, one can find $V\in\mathcal{U}$ such that
$x\oplus V\subseteq x\boxplus U_{2}$. Then we have that

$\displaystyle\ominus(x\oplus V)$ $\displaystyle\subseteq\ominus(x\boxplus
U_{2})$ $\displaystyle=\ominus U_{2}\boxplus(\ominus x)$
$\displaystyle\subseteq U_{1}\boxplus(\ominus x)$
$\displaystyle\subseteq\ominus x\oplus U$

Thus we have proved that the inverse operation $\ominus$ is continuous. ∎

Clearly, every topological gyrogroup $G$ is a paratopological gyrogroup. In
general, paratopological gyrogroups need not be topological gyrogroups. There
exists a gyrogroup $G$ and a family $\mathcal{U}$ of subsets of $G$ satisfying
conditions (1)-(7) of Theorem4.2. And the family
$\mathcal{B}_{\mathcal{U}}=\\{a\oplus U:a\in G,U\in\mathcal{U}\\}$ is a base
for a topology $\mathcal{T}_{\mathcal{U}}$ on $G$. With this topology, $G$ is
a paratopological gyrogroup.

######  Example 4.5.

The M$\ddot{o}$bius gyrogroup equipped with the standard topology
$\mathcal{T}$ is a topological gyrogroup (see [2, Example 2]).

Put $\mathcal{U}=\\{U_{n}:n\in\mathbb{N}\\}$, where
$U_{n}=\\{x\in\mathbb{D}:|x|<\frac{1}{n}\\}$ for each $n\in\mathbb{N}$. Then
the family
$\mathcal{B}_{\mathcal{U}}=\\{a\oplus_{M}U_{n}:a\in\mathbb{D},U_{n}\in\mathcal{U}\\}$
is a base for a Hausdorff topology $\mathcal{T}_{\mathcal{U}}$ on
$\mathbb{D}$. With this topology, $\mathbb{D}$ is a topological gyrogroup. In
particular, we have that $\mathcal{T}_{\mathcal{U}}=\mathcal{T}$.

###### Proof.

Clearly, $\mathcal{U}$ is a local base at the identity 0 when the
M$\ddot{o}$bius gyrogroup $\mathbb{D}$ and any $n\in\omega$, let
$U_{n}=\\{x\in\mathbb{D}:|x|<\frac{1}{n}\\}$. Let’s verify
$\mathcal{U}=\\{U_{n}:n\in\omega\\}$ satisfying conditions (1)-(9).

(1) For each $n\in\omega$, we put $m=[n+\sqrt{n^{2}+1}]+1$ such that $|x\oplus
y|=|\frac{x+y}{1+\overline{x}y}|\leq\frac{|x|+|y|}{1-|\overline{x}y|}\leq\frac{2/m}{1-1/m^{2}}<1/n$
for every $x,y\in U_{m}$, that is $x\oplus y\subseteq U_{n}$. Thus we verify
$U_{m}\oplus U_{m}\subseteq U_{n}$.

(2) For each $n\in\omega$ and $x\in U_{n}$, we put $m=[N]+1$, where
$1/{N}=\frac{{1}/{n}-|x|}{1+({1}/{n})|x|}$, such that for every $y\in U_{m}$,
we have $|x\oplus
y|=|\frac{x+y}{1+\overline{x}y}|\leq\frac{|x|+|y|}{1-|\overline{x}y|}<\frac{|x|+1/m}{1-|\overline{x}(1/m)|}<\frac{|x|+1/N}{1-|\overline{x}(1/N)|}=\frac{|x|+\frac{{1}/{n}-|x|}{1+({1}/{n})|x|}}{1-|\overline{x}\frac{{1}/{n}-|x|}{1+({1}/{n})|x|}|}=1/n$.
So we get $x\oplus y\in U_{n}$, that is $x\oplus U_{m}\subseteq U_{n}$.

(3) For each $n\in\omega$ and $x\in\mathbb{D}$, we put $m=[T]+1$, where
$1/{T}=\frac{{1}/{n}(1-|x|^{2})}{\sqrt{|x|^{4}+(2-4/n^{2})|x|^{2}+1}}$, such
that for every $v\in U_{m}$ and $x\in\mathbb{D}$, we have

$|\ominus x\oplus(v\oplus x)|=|\ominus
x\oplus\frac{x+v}{1+\overline{v}x}|=|\frac{-x+\frac{x+v}{1+\overline{v}x}}{1-\overline{x}\frac{x+v}{1+\overline{v}x}}|$
$=|\frac{-\overline{v}x^{2}+v}{1+\overline{v}x-v\overline{x}-x\overline{x}}|<\frac{1}{n}$,

which implies $\ominus x\oplus(U_{m}\oplus x)\subseteq U_{n}$.

In fact, for the above $v$ and $x$ we have

$(|x|^{4}+1-(2/n^{2})|x|^{2})|v|^{2}+2(1-1/n^{2})|v|^{2}|x|^{2}<1/n^{2}(1-|x|^{2})^{2}.$

The above formula implies

$(|x|^{4}+1-(2/n^{2})|x|^{2})|v|^{2}-(1-1/n^{2})2Re(\overline{v}^{2}x^{2})<1/n^{2}(1-|x|^{2})^{2}$

if and only if

$(-\overline{v}x^{2}+v)\overline{(-\overline{v}x^{2}+v)}<1/n^{2}(1+\overline{v}x-v\overline{x}-\overline{x}x)(\overline{1+\overline{v}x-v\overline{x}-\overline{x}x})$

if and only if

$|-\overline{v}x^{2}+v|^{2}<1/n^{2}|1+\overline{v}x-v\overline{x}-\overline{x}x|^{2}$

if and only if

$\frac{|-\overline{v}x^{2}+v|}{|1+\overline{v}x-v\overline{x}-x\overline{x}|}<\frac{1}{n}.$

(4) For $U_{m},U_{n}\in\mathcal{U}$, we put $n_{1}=\text{max}\\{m,n\\}+1$ such
that $W_{n_{1}}\subseteq U_{m}\cap U_{n}$.

(5) For every $U_{n}\in\mathcal{U}$ and $a,b\in\mathbb{D}$, we put $m=n+1$
such that for each $v\in U_{m}$ we have
$|\text{gyr}[a,b]v|=|v|<\frac{1}{m}<\frac{1}{n}$, that is
$\text{gyr}[a,b]U_{m}\subseteq U_{n}$.

(6) For every $U_{n}\in\mathcal{U}$ and $b\in\mathbb{D}$, we put $m=n+1$ such
that for each $v\in U_{m}$ we have
$|\text{gyr}[v,b]v|=|v|<\frac{1}{m}<\frac{1}{n}$, that is $\bigcup_{v\in
U_{m}}\text{gyr}[v,b]U_{m}\subseteq U_{n}$.

(7) Let $v\in\bigcap_{n\in\omega}(U_{n}\boxminus U_{n})$ and $v\neq 0$. We put
$t=[\frac{|v|}{-1+\sqrt{1+|v|^{2}}}]+1$ such that $|x\boxminus
y|=|x\ominus\text{gyr}[x,y]y|=|\frac{x-\frac{1+x\overline{y}}{1+\overline{x}y}y}{1-\overline{x}\frac{1+x\overline{y}}{1+\overline{x}y}y}|\leq\frac{|x|+|y|}{1-|\overline{x}y|}\leq\frac{2/t}{1-1/t^{2}}<|v|$
for every $x,y\in U_{t}$, since $|\text{gyr}[x,y]|=1$. This implies that
$v\notin U_{t}\boxminus U_{t}$, which is a contradiction. Thus we verify
$\bigcap_{n\in\omega}(U_{n}\boxminus U_{n})=\\{0\\}$.

Then the family $\mathcal{B}_{\mathcal{U}}=\\{a\oplus
U_{n}:a\in\mathbb{D},U_{n}\in\mathcal{U}\\}$ is a base for the topology
$\mathcal{T_{U}}$ on $\mathbb{D}$. In fact, $\mathcal{T_{U}}$ is equal to the
standard topology. We know that $\mathbb{D}$ equipped with the standard
topology is a topological gyrogroup, which further confirms our conclusion.

(8) For each $n\in\omega$ and $x\in\mathbb{D}$, we put
$m=\text{max}\\{[T]+1,3n\\}$, where
$1/{T}=\sqrt{\frac{-B+\sqrt{B^{2}-4AC}}{2A}}$, $A=(1-|x|^{2})^{2}|x|^{2}$,
$B=(1-|x|^{2})^{2}(1-\frac{1}{n^{2}}|x|^{2}+\frac{2|x|}{3n})$ and
$C=\frac{1}{n^{2}}(1-|x|^{2})^{2}(\frac{2|x|}{3n}-1)$ such that for every
$y\in U_{m}$ and $x\in\mathbb{D}$, we have

$|\ominus x\oplus(y\boxplus x)|=|\ominus x\oplus(y\oplus\text{gyr}[y,\ominus
x]x)|=|\ominus
x\oplus(y\oplus\frac{1-\overline{x}y}{1-\overline{y}x}x)|=|\ominus
x\oplus\frac{(1-|x|^{2})y+(1-|y|^{2})x}{1-|x|^{2}|y|^{2}}|=$
$\frac{|(|x|^{2}-1)|y|^{2}x+(1-|x|^{2})y|}{|(1-|x|^{2})(1-\overline{x}y)|}<\frac{1}{n}$,
which implies $\ominus x\oplus(U_{m}\boxplus x)\subseteq U_{n}$, furthermore,
$U_{m}\boxplus x\subseteq x\oplus U$.

In fact, for the above $y$ and $x$ we have

$(1-|x|^{2})^{2}|y|^{4}+(1-|x|^{2})^{2}(1-\frac{1}{n^{2}}|x|^{2})|y|^{2}+(1-|x|^{2})^{2}\frac{2|x|}{3n}|y|^{2}+\frac{1}{n^{2}}(1-|x|^{2})^{2}\frac{2|x|}{3n}$

$<\frac{1}{n^{2}}(1-|x|^{2})^{2}$

The above formula implies

$(1-|x|^{2})^{2}|y|^{4}+(1-|x|^{2})^{2}(1-\frac{1}{n^{2}}|x|^{2})|y|^{2}-(1-|x|^{2})^{2}\cdot
2\text{Re}(\overline{x}y)|y|^{2}+\frac{1}{n^{2}}(1-|x|^{2})^{2}\cdot
2\text{Re}(\overline{x}y)$

$<\frac{1}{n^{2}}(1-|x|^{2})^{2}$

if and only if

$[(|x|^{2}-1)|y|^{2}x+(1-|x|^{2})y]\overline{[(|x|^{2}-1)|y|^{2}x+(1-|x|^{2})y]}$

$<1/n^{2}[(1-|x|^{2})(1-\overline{x}y)](\overline{[(1-|x|^{2})(1-\overline{x}y)]}$

if and only if

$|(|x|^{2}-1)|y|^{2}x+(1-|x|^{2})y|^{2}<1/n^{2}|(1-|x|^{2})(1-\overline{x}y)|^{2}$

if and only if

$\frac{|(|x|^{2}-1)|y|^{2}x+(1-|x|^{2})y|}{|(1-|x|^{2})(1-\overline{x}y)|}<\frac{1}{n}.$

For every $U_{n}\in\mathcal{U}$ and $x\in\mathbb{D}$, we get $V=\bigcup_{u\in
U_{n}}\text{gyr}[x,\ominus x](\text{gyr}[x,\ominus u]u)$. Since for every
$v\in V$, $|v|=|\text{gyr}[x,\ominus x](\text{gyr}[x,\ominus u]u)|=|u|<1/n$,
we have $V\in\mathcal{U}$. Then, for every $v\in V$ and some $u\in U_{n}$ we
have

$\displaystyle\ominus x\oplus(x\oplus v)$ $\displaystyle=(\ominus x\oplus
x)\oplus\text{gyr}[\ominus x,x]v$ $\displaystyle=\text{gyr}[\ominus x,x]v$
$\displaystyle=\text{gyr}[\ominus x,x](\text{gyr}[x,\ominus
x](\text{gyr}[x,\ominus u]u))$ $\displaystyle=\text{gyr}[x,\ominus u]u$

So we get $\ominus x\oplus(x\oplus v)=\text{gyr}[x,\ominus u]u$, that is
$x\oplus v=x\oplus\text{gyr}[x,\ominus u]u=x\boxplus u$. We have proved
$x\oplus V\subseteq x\boxplus U$.

(9) For every $U\in\mathcal{U}$, we put $V=U$ such that $U=\ominus U=\ominus
V\subseteq U$, since $\ominus U=U$.

At last, we prove $\mathcal{T}_{\mathcal{U}}=\mathcal{T}$. In fact, for each
$r\in(0,1)$ we can get $m=[T]+1$, where
$1/{T}=\frac{-a(1-r^{2})+r(1-a^{2})}{1-r^{2}a^{2}}$, such that for every $u\in
U_{m}$ and $a\in\mathbb{D}$, we have $|a\oplus
u|=|\frac{a+u}{1+\overline{a}u}|<r$.

Since

$(1-r^{2}|a|^{2})|u|^{2}+2(1-r^{2})|a\overline{u}|+|a|^{2}-|r^{2}|<0$

for every $u\in U_{m}$ and $a\in\mathbb{D}$.

The above formula implies

$(1-r^{2}|a|^{2})|u|^{2}+2(1-r^{2})2\text{Re}a\overline{u}+|a|^{2}-|r^{2}|<0$

if and only if

$(a+u)\overline{(a+u)}<r^{2}(1+\overline{a}u)\overline{(1+\overline{a}u)}$

if and only if

$|a+u|^{2}<r^{2}|1+\overline{a}u|^{2}$

if and only if

$\frac{|a+u|}{|1+\overline{a}u|}<r.$

We prove $a\oplus U_{m}\subseteq U(0,r)$, that is
$\mathcal{T}_{\mathcal{U}}=\mathcal{T}$. ∎

######  Lemma 4.6.

Let $G$ be a paratopological gyrogroup. For all $a,b\in G$ and every
nontrivial normal subgyrogroup $H$ of $G$, the following holds:

$\text{gyr}[a,b]H=H.$

###### Proof.

For $H$ is a nontrivial normal subgyrogroup, there exists a gyrogroup
homomorphisms $\varphi:G\rightarrow K$ between gyrogroups $G$ and $K$ such
that $H=\text{ker}\varphi$. For every $x\in\text{gyr}[a,b]\text{ker}\varphi$,
there exists $y\in\text{ker}\varphi$ such that $x=\text{gyr}[a,b]y$. So
$\varphi(x)=\varphi(\text{gyr}[a,b]y)=\text{gyr}[\varphi(a),\varphi(b)]\varphi(y)=\text{gyr}[\varphi(a),\varphi(b)]0=0$
(see [13, Proposition 5.1]), which implies $x\in\text{ker}\varphi$. Thus, we
have proved $\text{gyr}[a,b]\text{ker}\varphi\subseteq\text{ker}\varphi$ for
all $a,b\in G$. By Proposition 2.6 [13], we have
$\text{gyr}[a,b]\text{ker}\varphi=\text{ker}\varphi$. That is
$\text{gyr}[a,b]H=H$ For all $a,b\in G$. ∎

######  Example 4.7.

Let $G$ be a gyrogroup and

$\mathcal{U}=\\{U:U\subseteq G,U\text{~{}is a nontrivial normal subgyrogroup
of G}\\}$

Then $\mathcal{U}$ satisfying conditions (1)-(9) and the family
$\mathcal{B}_{\mathcal{U}}=\\{a\oplus U:a\in G,U\in\mathcal{U}\\}$ is a base
for the topology $\mathcal{T_{U}}$ on $G$. With this topology, $G$ is a
topological gyrogroup.

###### Proof.

(1) For each $U\in\mathcal{U}$, we can take $V=U$ such that $V\oplus V=U\oplus
U=U$ by the closure property of groups.

(2) For every $U\in\mathcal{U}$, and every $x\in U$, we can take $V=U$ such
that $x\oplus V=x\oplus U=U$ by the closure property of groups.

(3) For every $U\in\mathcal{U}$ and $x\in G$, we can take $V=U$ such that
$\ominus x\oplus(V\oplus x)=\ominus x\oplus(U\oplus x)=\ominus x\oplus(x\oplus
U)=(\ominus x\oplus x)\oplus\text{gyr}[\ominus x,x]U=U$.

(4) For every $U,V\in\mathcal{U}$, we can take $W=U\cap V\in\mathcal{U}$ such
that $W\subseteq U\cap V$, since intersection of normal subgroups is a normal
subgroup.

(5) For every $U\in\mathcal{U}$ and $a,b\in G$, there exists $V=U$ such that
$\text{gyr}[a,b]V=\text{gyr}[a,b]U=U$, since the normal subgyrogroup is
invariant under the gyroautomorphisms of $G$.

(6) For every $U\in\mathcal{U}$ and $b\in G$, there exists $V=U$ such that
$\bigcup_{v\in V}\text{gyr}[v,b]V=\bigcup_{v\in U}\text{gyr}[v,b]U=U$.

(7) Let $v\in\bigcap_{U\in\mathcal{U}}(U\boxminus U)$ and $v\neq 0$. Then we
have $v\in\bigcap(\text{ker}\varphi\boxminus\text{ker}\varphi)$ for every
gyrogroup homomorphisms $\varphi:G\rightarrow H$ between gyrogroups $G$ and
$H$, since $U$ is a nontrivial normal subgyrogroup. We put
$\varphi=\text{gyr}[a,\ominus v]$, where $a\in G$ and $a\neq v$. It is obvious
that $v\notin\text{ker}\varphi$, which is a contradiction. Thus we verify
$\bigcap_{U\in\mathcal{U}}(U\boxminus U)=\\{0\\}$.

(8) For every $U\in\mathcal{U}$ and $x\in G$, since $U$ is a nontrivial normal
subgyrogroup, there exists a gyrogroup homomorphisms $\psi:G\rightarrow H$
between gyrogroups $G$ and $H$ such that $U=\text{ker}\psi$ and
$\psi^{-1}(\psi(x))=x\oplus U$. For every $u\in U$ and $x\in G$, we have
$\psi(u\boxplus x)=\psi(u)\boxplus\psi(x)=\psi(x)$ (see [13, Proposition
5.1]), which means that $u\boxplus x\in x\oplus U$. So we have proved
$U\boxplus x\subseteq x\oplus U$.

Next we prove for every $U\in\mathcal{U}$ and $x\in G$, $x\oplus U=x\boxplus
U$. Let $x\in G$ and $u\in U$. By Definition 2.4, $x\boxplus
u=x\oplus\text{gyr}[x,\ominus u]u$. Since $U$ is a normal subgyrogroup,
$\text{gyr}[x,\ominus u]u\in U$ by Lemma 4.6. That is $x\boxplus u\in x\oplus
U$. This proves $x\boxplus U\subseteq x\oplus U$.

On the other hand, let $x\in G$ and $u\in U$. By Theorem 2.5 $x\oplus
u=x\boxplus\text{gyr}[x,u]u$. By Lemma 4.6, $\text{gyr}[x,u]u\in U$. Hence,
$x\oplus u\in x\boxplus U$. So $x\oplus U\subseteq x\boxplus U$. We have
proved $x\oplus U=x\boxplus U$.

(9) For every $U\in\mathcal{U}$, we put $V=U$ such that $U=\ominus U=\ominus
V\subseteq U$, since $U$ is a nontrivial normal subgyrogroup of $G$. ∎

Let $G$ be a gyrogroup and $\mathcal{U}=\\{0\\}$. It is obvious that
$\mathcal{U}$ satisfies conditions (1)-(9). Then the family
$\mathcal{B}_{\mathcal{U}}=\\{a\oplus U:a\in G,U\in\mathcal{U}\\}$ is a base
for the topology $\mathcal{T_{U}}$ on $G$. In fact, $\mathcal{T_{U}}$ is the
discrete topology. With this topology, $G$ is a topological gyrogroup.

In addition, let $G$ be a gyrogroup and $\mathcal{U}=\\{G\\}$. $\mathcal{U}$
also satisfies conditions (1)-(9). Then the family
$\mathcal{B}_{\mathcal{U}}=\\{a\oplus U:a\in G,U\in\mathcal{U}\\}$ is a base
for the topology $\mathcal{T_{U}}$ on $G$. In fact, $\mathcal{T_{U}}$ is the
trivial topology. With this topology, $G$ is a topological gyrogroup.

## References

  * [1] A.V. Arhangel’ski, M. Tkachenko, Topological Groups and Related Structures, Atlantis Press and World Sci., 2008.
  * [2] W. Atiponrat, Topological gyrogroups: generalization of topological groups, Topol. Appl. 224 (2017) 73-82.
  * [3] W. Atiponrat, R. Maungchang, Complete regularity of paratopological gyrogroups, Topol. Appl. 270 (2020) 106951.
  * [4] T. Banakh, A. Ravsky, Each regular paratopological group is completely regular, Proc. Am. Math. Soc. 145 (3) (2017) 1373-1382.
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  * [11] T. Suksumran, The Algebra of Gyrogroups: Cayley’s Theorem, Lagrange’s Theorem, and Isomorphism Theorems, Springer International Publishing, Cham, (2016) 369-437.
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  * [15] A.A. Ungar, Analytic Hyperbolic Geometry and Albert Einstein’s Special Theory of Relativity, World Scientific, 2008.


# SYNERGY OF MACHINE AND DEEP LEARNING MODELS FOR MULTI-PAINTER RECOGNITION

###### Abstract

The growing availability of digitized art collections has created the need to
manage, analyze and categorize large amounts of data related to abstract
concepts, highlighting a demanding problem of computer science and leading to
new research perspectives. Advances in artificial intelligence and neural
networks provide the right tools for this challenge. The analysis of artworks
to extract features useful in certain works is at the heart of the era. In the
present work, we approach the problem of painter recognition in a set of
digitized paintings, derived from the WikiArt repository, using transfer
learning to extract the appropriate features and classical machine learning
methods to evaluate the result. Through the testing of various models and
their fine tuning we came to the conclusion that RegNet performs better in
exporting features, while SVM makes the best classification of images based on
the painter with a performance of up to 85%. Also, we introduced a new large
dataset for painting recognition task including 62 artists achieving good
results.

Index Terms—  Painter recognition, Deep learning, Machine learning, Feature
extraction

##  1 Introduction

Art as an integral part of human culture has entered the digital age,
following the trend of recent years to digitize our world. Large collections
of artworks, mainly from museums and galleries, but also private collections,
are digitized, published and made available on the internet with the aim not
only of preserving and presenting them as creations of human expression, but
also as data for processing and analysis using machine learning methods.

The large volume of this type of data, as well as the advanced tools of
artificial intelligence impose the need to automate the work of analysis,
extraction of features and categorization of certain components of artworks. .
Art historians, curators and art experts, in general, are able to identify the
particular characteristics of a painting, its creator, the genre, the art
movement and the artistic style to which it belongs. Their ability is based
both on their perception and memory, as well as on their artistic experience
that allows them to connect the various artistic features with each other and
to draw relevant conclusions. Therefore, the aim is to imitate the human
ability, experience and knowledge to automate the above tasks.

We are interested in the problem of automatic painter recognition. Approaching
this problem, we consider that the artworks of a painter present common
features related to the style and preferences of each artist. However, it has
been observed that an artist does not follow from the beginning to the end of
his course the same art movement, the same genre, the same theme, the same
style or color palette, as well as may have been influenced by other creators,
for example by their teachers, in which case their paintings have common
features. Therefore, the present work is called to deal with the complexity of
recognizing a painter from a painting by finding appropriate methods in
machine learning.

Recognizing the creator, the genre, the art movement and the style of a
painting are problems of increasing complexity, the solution of which is a
research challenge in the field of machine learning. Deep learning techniques
are mainly employed, but also classical machine learning techniques are used
to deal with these complex problems. On the one hand, classical methods of
image processing are used for feature extraction and combined with typical
machine learning classifiers. On the other hand, feature extraction and
classification is done by an end-to-end training of convolutional neural
networks and a use of transfer learning. We combined these two approaches in
order to yield better results to the painter recognition task. Having image
data from the WikiArt repository and knowing the metadata that accompanies the
painter in each painting, we fine-tuned pre-trained deep neural networks to
extract the appropriate features from images and then used machine learning
classifiers on those features to get our predictions.

The advantage of our method lies in its application to a new large-scale
dataset of 62 artists. There are no other studies, as we know, that have
approached the specific problem to such an extent, achieving such good results
without the use of end-to-end deep neural networks. Thus, we managed to
approach the task of painter recognition using a different from the state-of-
the-art methodologies that gives satisfactory results for usual datasets of
about 20 artists and excellent for the problem of increased complexity of 62
artists.

##  2 Background

Automation in the processing, analysis and categorization of artworks has been
a major challenge in recent decades and machine learning has made a
significant contribution to progress in this field. There are many studies
regarding the distinction of the type and style of a painting, as well as the
identification of its artist.

Early studies addressed these problems with traditional machine learning
methods based on low-level features and a relatively small number of artworks
[1, 2, 3, 4, 5, 6, 7]. These studies use features that capture shape, texture,
edge and color properties and are extracted with the use of classical computer
vision methods, like SIFT, GIST, HOG, GLCM, and HSV color histograms. Then,
they train classical classifiers of machine learning, like SVM, K-NN, Random
Forest, MLP, to make the predictions.

Following the studies, CNN was introduced as extractors of features. The first
large-scale such study [8] has shown that features derived from the layers of
a CNN pre-trained in non-artistic images achieve high performance in the tasks
of art classification. Then, many studies [9, 10, 11, 12, 13, 14, 15, 16] have
confirmed the effectiveness of the methods based on the extraction of features
using CNN, as well as the combination of them with the handmade features of
the images.

Regarding the problem of painter recognition from a painting, in which we are
interested in this work, in 2013, Cetinic et al. [17] studied the style of
individual artists by extracting specific features, like color, light,
texture, and then several classifiers, such as MLP, SVM, Naïve Bayes, Random
Forest and Adaboost, were applied. In 2015, Saleh et al. [12] investigated the
applicability of metric learning approaches and performance of different
visual features (feature fussion) coupled with SVM for learning similarity
between artistic items. In 2016, Tan et al. [18] used CNN as feature extractor
and SVM for classification or CNN as an end-to-end fine-tuned architecture. In
2017, Viswanathan [19] trained from scratch a ResNet18 to resolve the problem
of painter recognition. In 2018, Cetinic et al. [20], also, focused on fine-
tuned networks based on VGG, Resnet, Googlenet and CaffeNet models. In 2019,
Kelek et al. [21] used pre-trained architectures such as GoogleNet,
Inceptionv3, ResNet50, ResNet101 and DenseNet. Zhong et al. [22] proposed a
dual path classification scheme, including RBG and brush stroke information
channels, based on architecture of ResNet131. In 2020, Choundhury [23]
compared the results of feature extraction methods to train Random Forest and
SVM classifiers and deep convolutional networks with transfer learning, such
as basic CNN, ResNet18 and ResNet50. In 2021, Cömert et al. [24] focused on
the fine-tuning of MobileNet v2, ResNet, Inception v2 and NasNet to identify
the painter of an artwork. The same year, Zhao et al. [25] compared the
architectures ResNet, RegNet and EfficientNet for painter identification.
Finally, Nevo et al. [26] proposed a novel dual-stream architecture, based on
pretrained EfficientNet model, for capturing in parallel both global elements
and local structures in painting’s images. It is important to mention that
most of the related work has used datasets derived form WikiArt.

##  3 Dataset and Features

Our data were retrieved from Kaggle111https://www.kaggle.com/c/painter-by-
numbers/overview. Most of these digitized images were taken from the WikiArt
repository. In total, 103.250 paintings by 2.319 artists are available in this
dataset. It is noted that for our convenience in managing our data and in the
application of the various techniques and algorithms that we have chosen, we
built two different datasets. We did that in order to test our algorithms in
different scales and difficulty of the problem. The first one corresponds to
20 distinct painters with 9,986 paintings in total (Medium Dataset) and the
second one to 62 with 26,263 (Large Dataset). The selection of data was based
on the minimum number of paintings for each artist in order to have adequate
number of artworks per artist. For this reason, we selected for the study
those painters who have at least 270 artworks in the original dataset. In
addition, in order to do equal comparison with state-of-the-art works, we
applied our experiments to the dataset used by the state-of-the art (SOTA)
methods, which contains 19,050 paintings for 23 painters
222https://github.com/cs-chan/ArtGAN and is derived from WikiArt repository.
From this point we will refer to this dataset as SOTA Dataset.

Furthermore, for the best management of this two dimensional data, resizing of
the images (3x256x256) was applied, so that the dimensions of the input to our
algorithms are specific. For the better generalization of our models we
performed data-augmentation, we used random cuts and reflections of the
images. We also applied normalization to the data based on the mean value and
the covariance of the features, which helped to accelerate the training of our
neural networks.

##  4 Methodology

Regarding the feature extraction from images, none of the classic techniques
were applied, but it was preferred to use SOTA pre-trained convolutional
neural networks that are distinguished for their ability to extract high-level
features from images. This is also a challenge of our study, as it is
interesting to investigate to what extent convolutional networks can offer to
classification networks those appropriate features that will help to solve the
problem of painter recognition as effectively as possible.

To construct a deep-neural-network-based feature extractor we conducted many
experiments with different SOTA architectures and datasets. We tried different
sizes of ResNet[27] and RegNet[28] architectures in different datasets. We
started by experimenting on the Medium dataset in order to select an
architecture. The RegNet architecture outperformed the ResNet, so we continued
experimenting with it on the Large dataset.

Further experiments focused on determining the best hyper-parameters for the
RegNet architecture. We experimented with different ways of training. We tried
different values of model depth and learning rate in order to a model of the
right capacity. Additionally we tried adding a dropout layer. Also, we used
new techniques such as label smoothing[29], warm-up layers and frozen layers
in order to avoid over-fitting and catastrophic forgetting [30].

After choosing the best architecture for each sub-task we applied various
machine learning algorithms. We applied heuristic algorithms for finding the
best combination of hyper-parameters for each model and ended up selecting the
best one. We followed the same procedure for the Large Dataset and we tried
one extra SOTA classifier, XGBoost. Also in order to compare with SOTA model’s
we applied our best model on the SOTA Dataset. Our aim was to use classifiers
that don’t follow the stantard MLP model with fully connected layers.

A crucial factor for achieving good results was the hyper-parameter tuning.
Specifically for our best classifier SVM we tuned the hyper-parameters: C,
$\gamma$ and kernel-type. We performed gradually grid-search in order to find
the best combination for our model, using the sklearn python package.

##  5 Experiments

Our models managed to be comparable to SOTA models in the painter recognition
task, especially for the works due to 2021, where started approaching the
problem with different manner using dual-stream architectures for feature
extraction. In Table 2 the performance of our best models is given compared to
previous works. For the implementation of our experiments we provide a GitHub
repository 333https://github.com/jliartis/art-recognition. Below our
experiments are described analytically.

###  5.1 Recognition of 20 artists

The sub-task of recognizing paintings from 20-artists was based in the Medium
Dataset. We conducted a number of experiments in order to choose the best
architecture for feature extraction. We split the dataset in three parts: 20%
validation set, 20% test set and 60% train set. We experimented with different
sizes of the layers of ResNet (34, 101, 152) and of the number of parameters
of RegNet (400mf, 800mf). We had approximately 20 experiments and ended up
choosing RegNet_Y_800MF and ResNet-152. After choosing the two best
architecture we did fine tuing to choose one of them. RegNet_y_800MF with
learning rate 1e-5, trained in 20 epochs with 3 frozen layers had the best
results achieving 0.84 accuracy in the validation set, so we chose it for our
feature extractor.

After using the extractor to retrieve the desired features, we applied 6
different ML classifiers: Logistic Regression , Naive Bayes, K-Nearest
Neighbors (K-NN), Decision Trees, AdaBoost and Support Vector Machines (SVM).
We applied cross-validation with 5 folds. In order to compare them fairly, we
applied hyper-parameter optimization to all of them, using the gridsearch
algorithm. After conducting all the experiments SVM prevailed all the others
reaching 85% accuracy, while Decision Trees having the worst results. An
overview of the results is shown in Table 1.

From Fig. 1, we can conclude that in general the performance of our method was
good for all classes. We had the worst results for the painter Ilya Repin and
the best for the painter Battista Piranesi. It is also noted that there was a
confusion between the painters Boris Kustodiev and Ilya Repin. This confusion
may be due to the fact that both artists belong to the same artistic movement,
Realism. In addition, both are from Russia and studied at the same time at the
Imperial Academy of Arts of Russia.

Table 1: Comparing our ML classifiers Classifier |  Medium Dataset |  Large Dataset
---|---|---
Plain RegNet |  - |  79%
Decision Trees |  63% |  -
Random Forest |  - |  69%
AdaBoost |  79% |  -
Naive Bayes |  82% |  -
Logistic Regression |  84% |  -
K-NN |  84% |  72%
XGBoost |  - |  75%
SVM |  85% |  84%

###  5.2 Recognition of 62 artists

The second sub-task with the 62-artists is based on the Large Dataset. We
split the dataset in three parts: 20% validation set, 20% test set and 60 %
train set. Since the best model in our previous experiments (see Section 5.1)
was RegNet architecture we chose it a priori. In order to chose the best
hyper-parameters for RegNet we had 30 different experiments. We tried various
sizes of the model’s parameters (400mf, 800mf, 1.6gf) and ended up choosing
RegNet_Y_1_6GF. The hyper-parameters of the best model that achieved 0.8
accuracy in the validation set are: 1000 epochs, 128 batch size, 4 warm up
layers, 2 freeze layers, 0.015 label smoothing and 1.6 learning rate.

Since we chose the best feature extractor for this task, we tried four machine
learning classifiers for the task: Random Forest, K-Nearest Neighbors, Support
Vector Machines and XGBoost. We used cross-validation for the training and
tuning of each classifier with 5 folds. For the hyper-parameter tuning of the
classifiers we used Optuna [31], a python library. The best classifier was
proven to be SVM achieving 84% accuracy and the worst was Random Forest
achieving 69%. The results are displayed in Table 1. It is important to point
that plain RegNet achieved 79%, which means that SVM boosted our performance
with 5 percentage points.

Fig. 1: Confusion matrix for the best SVM trained in a subset of the Medium
Dataset

###  5.3 Comparison with state-of-the-art models

In order to compare our methodology with SOTA [12], [20], [25], [26], we
applied our experiments to the same dataset, the performance in which can be a
mark for the efficiency of our approaching system. We used our RegNet model
that was fine-tuned on the Large Dataset for extracting representative
features of the dataset and then performed gridsearch for SVM models.

We achieved results close to SOTA model’s performance as shown in Table 2. Our
approach deviates from the so far proposed end-to-end deep convolutional
network architectures to solve the task of painter recognition and proves that
traditional classifiers have still the potential to have comparable results to
pure deep-learning models.

We consider that that the complexity of the problem lies in the increase in
the number of artists to be recognized. Regarding Fig. 2 we observe that our
methodology achieves appreciable performance for the ordinary datasets and
much better for the extended dataset of 62 artists. The references of Fig. 2
are briefly descibed in Section 2.

Table 2: Comparing our results with previous state-of-the-art models on SOTA
Dataset

Work | Year | Methodology | Accuracy
---|---|---|---
Saleh et al. [12] | 2015 | Feature fussion & SVM | 63%
Cetinic et al. [20] | 2018 | CaffeNet | 82%
Zhao et al.[25] | 2021 | EfficientNet | 92%
Nevo et al.[26] | 2022 | Dual-stream (EfficientNet) | 94%
Our approach | 2023 | RegNetY-1.6MF & SVM | 85%

Fig. 2: Scatter plot for comparing the change in accuracy with the number of
artists (blue points represent previous works and red ones ours - all points
are annotated with the corresponding references)

##  6 Conclusions and future work

In this work, we had a large number of experiments trying a variety of SOTA
deep learning and machine learning architectures achieving very good results
in the painters recognition task. A new methodology was applied, as a synergy
of old and new techniques, in order to reach high accuracy in a large-scale
and complex multi-class classification problem, like this of 62 artists. It is
proven that traditional machine learning classifiers can have close to SOTA
results, assembling more versatile model-structures. The use of a pre-trained
deep neural network for feature extraction and an SVM for classification
instead of the use of an end-to-end deep neural network has advantages in the
performance and can lead to a different comparable approach with SOTA ones.

Future work will be the extension of our experiments in other art recognition
tasks in order to success the generalization of our model. One other
interesting future direction, in order to tackle the problem of
generalization, is to try multi-task learning. Finally, it will be useful to
include explainability to make the artist identification process more
understandable and detect possible bias of model’s parameters.

## References

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