# Rota–Baxter operators on Clifford semigroups and the Yang–Baxter equation

Francesco CATINO<EMAIL_ADDRESS>Marzia MAZZOTTA
<EMAIL_ADDRESS>Paola STEFANELLI<EMAIL_ADDRESS>Dipartimento di Matematica e Fisica “Ennio De Giorgi”, Università del Salento,
Via Provinciale Lecce-Arnesano, 73100 Lecce (Italy)

###### Abstract

In this paper, we introduce the theory of Rota–Baxter operators on Clifford
semigroups, useful tools for obtaining dual weak braces, i.e., triples
$\left(S,+,\circ\right)$ where $\left(S,+\right)$ and $\left(S,\circ\right)$
are Clifford semigroups such that $a\circ\left(b+c\right)=a\circ b-a+a\circ c$
and $a\circ a^{-}=-a+a$, for all $a,b,c\in S$. To each algebraic structure is
associated a set-theoretic solution of the Yang–Baxter equation that has a
behaviour near to the bijectivity and non-degeneracy. Drawing from the theory
of Clifford semigroups, we provide methods for constructing dual weak braces
and deepen some structural aspects, including the notion of ideal.

###### keywords:

Rota–Baxter operator , Quantum Yang–Baxter equation , Set-theoretic solution ,
Inverse semigroup, Clifford semigroup , Skew brace , Brace , Weak brace

###### MSC:

[2020] 17B38 , 16T25, 81R50, 20M18

mytitlenotemytitlenotefootnotetext: This work was partially supported by the
Dipartimento di Matematica e Fisica “Ennio De Giorgi” - Università del
Salento. The authors are members of GNSAGA (INdAM).

## Introduction

Rota–Baxter operators on commutative algebras firstly appeared in 1960 in G.
Baxter’s probability studies [4] and were subsequently explored by several
authors, including Atkinson [1], Cartier [8], and Rota [30, 31], among others.
An entire monograph has been dedicated by Guo [23] on this topic. Moreover,
Guo, Lang, and Sheng [24] have introduced the notion of Rota–Baxter operator
on groups, pursued further by Bardakov and Gubarev [2]. Recently, the latter
have found in [3] a strict connection between Rota–Baxter operators on groups
and the Yang–Baxter equation, a fundamental identity of theoretical physics
that takes the name from Yang [33] and R. Baxter [5], and of which Drinfel’d
[21] posed the problem of finding all set-theoretic solutions. Specifically,
if $S$ is a non-empty set, a map $r:S\times S\to S\times S$ is a _set-
theoretic solution of the Yang–Baxter equation_ , or shortly a _solution_ , if
it satisfies the relation

$\displaystyle\left(r\times\operatorname{id}_{S}\right)\left(\operatorname{id}_{S}\times
r\right)\left(r\times\operatorname{id}_{S}\right)=\left(\operatorname{id}_{S}\times
r\right)\left(r\times\operatorname{id}_{S}\right)\left(\operatorname{id}_{S}\times
r\right).$

Writing a solution $r$ as
$r\left(x,y\right)=\left(\lambda_{x}\left(y\right),\rho_{y}\left(x\right)\right)$,
with $\lambda_{x},\rho_{x}$ maps from $S$ into itself, then $r$ is _left non-
degenerate_ if $\lambda_{x}\in\operatorname{Sym}_{S}$, _right non-degenerate_
if $\rho_{x}\in\operatorname{Sym}_{S}$, for every $x\in S$, _non-degenerate_
if it is both left and right non-degenerate.

As proved in [3, Proposition 3.1], every Rota–Baxter operator on a group
determines a skew brace, an algebraic structure introduced by Guarnieri and
Vendramin in 2017 that always gives rise to a bijective non-degenerate
solution (see [22, Theorem 3.1]).

In this regard, Caranti and Stefanello [7] have recently described gamma
functions that characterise skew braces obtained from Rota-Baxter operators on
groups.

A generalisation of the skew brace that determines a solution has been newly
introduced in [13], and it is named weak brace. To give the definition, we
firstly recall that a _Clifford semigroup_ $(S,\cdot)$ is an inverse
semigroup, i.e., a semigroup for which every $a\in S$ admits a unique
$a^{-1}\in S$ satisfying $aa^{-1}a=a$ and $a^{-1}a^{-1}=a^{-1}$, having
central idempotents. Into detail, a triple $(S,+,\circ)$ is a _weak (left)
brace_ if $\left(S,+\right)$ and $\left(S,\circ\right)$ both are inverse
semigroups and they hold

$\displaystyle a\circ\left(b+c\right)=a\circ b-a+a\circ
c\qquad\text{and}\qquad a\circ a^{-}=-a+a,$

for all $a,b,c\in S$, where $-a$ and $a^{-}$ denote the inverses of $a$ with
respect to $+$ and $\circ$, respectively, and we assume that the
multiplication has higher precedence than the addition. As demonstrated in
[13, Thereom 10], the additive semigroup of any weak brace is a Clifford
semigroup. Moreover, if both $(S,+)$ and $(S,\circ)$ are groups, then
$(S,+,\circ)$ is a _skew brace_ ; in addition, if $(S,+)$ is abelian, then
$(S,+,\circ)$ is a _brace_ [32]. Furthermore, weak braces form a subclass of
_inverse (left) semi-braces_ [14], namely, triples $(S,+,\circ)$ with $(S,+)$
a semigroup and $(S,\circ)$ an inverse semigroup such that

$a\circ(b+c)=a\circ b+a\circ\left(a^{-}+b\right),$

for all $a,b\in S$. More specifically, weak braces are all the inverse semi-
braces having $(S,+)$ as a Clifford semigroup and in which $-a+a\circ
b=a\circ\left(a^{-}+b\right)$, for all $a,b\in S$. According to [13, Theorem
11], given a weak brace $(S,+,\circ)$, the map $r:S\times S\to S\times S$
defined by

$\displaystyle r\left(a,b\right)=\left(-a+a\circ b,\left(-a+a\circ
b\right)^{-}\circ a\circ b\right),$

for all $a,b\in S$, is a solution. Such a map $r$ has a behaviour close to
bijectivity, since there exists the solution $r^{op}$ associated to the
_opposite weak brace_ $\left(S,+^{op},\circ\right)$ of $S$, where
$a+^{op}b:=b+a$, for all $a,b\in S$, such that

$\displaystyle r\,r^{op}\,r=r,\qquad
r^{op}\,r\,r^{op}=r^{op},\qquad\text{and}\qquad rr^{op}=r^{op}r,$

namely, $r$ is a completely regular element in $\operatorname{Map}(S\times
S)$. In this context, one can note that weak braces are instances of
semitrusses that produces solutions but they are not necessarily
YB–semitrusses, see [6] and [20].

The aim of this paper is to show how to obtain weak braces from Rota–Baxter
operators defined on Clifford semigroups. For this reason, we introduce the
notion of _Rota–Baxter operator on a Clifford semigroup $(S,+)$_, i.e., a map
$\mathfrak{R}:S\to S$ satisfying the relations

$\displaystyle\mathfrak{R}\left(a\right)+\mathfrak{R}\left(b\right)=\mathfrak{R}\left(a+\mathfrak{R}\left(a\right)+b-\mathfrak{R}\left(a\right)\right)\qquad\text{and}\qquad
a+\mathfrak{R}\left(a\right)-\mathfrak{R}\left(a\right)=a,$

for all $a,b\in S$. Clearly, such definition includes the original one given
for groups, cf. [24, Definition 2.2]. Any Rota–Baxter operator $\mathfrak{R}$
on a Clifford semigroup $(S,+)$ determines a dual weak brace
$S_{\mathfrak{R}}=(S,+,\circ)$ with

$\displaystyle a\circ
b=a+\mathfrak{R}\left(a\right)+b-\mathfrak{R}\left(a\right),$

for all $a,b\in S$, that belongs to the class of weak braces having also
$(S,\circ)$ as a Clifford semigroup. We name such particular weak braces _dual
weak braces_. The solution associated to a dual weak brace has a behaviour
close to the non-degeneracy since also the maps $\lambda_{a}$ and $\rho_{b}$
are completely regular in $\operatorname{Map}(S)$. In addition, we prove that
the opposite solution $r^{op}$ of the solution $r$ associated to a dual weak
brace $S_{\mathfrak{R}}$ derives from the dual weak brace obtained through the
Rota–Baxter operator $\mathfrak{R}^{op}=-a+\mathfrak{R}(a)$, for every $a\in
S$.
This fact has motivated us to deepen the structure of the dual weak brace on
which we study the notion of ideal that allows us to define its quotient
structure. In particular, we specialise ideals of dual weak braces obtained
from Rota–Baxter operators. Besides, we introduce the _socle_ of a dual weak
brace, a special ideal already investigated for braces [32] and other
connected structures [22, 11, 16, 10]. Moreover, we show how to construct new
ideals starting from two given ones.
Furthermore, we provide a method to construct Rota–Baxter operators on a
Clifford semigroup seen as strong semilattices of groups, each of which admits
a Rota–Baxter operator. Besides, we give a description of all endomorphisms
$\mathfrak{R}$ of Clifford semigroups that are Rota–Baxter operators for which
$\operatorname{Im}\mathfrak{R}$ is commutative. Consistently with [27,
Proposition 5.1], these special operators determine _bi-weak braces_ , i.e.,
special dual weak braces in which the roles of the sum and the multiplication
can be reversed. Finally, we characterise all Rota–Baxter endomorphisms of
groups that are also idempotent.

##  1 Basic results on weak braces

The aim of this preliminary section is to give essential results on the
structures of weak braces, recently introduced in [13] to find solutions of
the Yang–Baxter equation. To this purpose, we initially recall some basics on
Clifford semigroups, covered in detail in the books [19, 25, 28].

A semigroup $S$ is called an _inverse semigroup_ if, for each $a\in S$, there
exists a unique element $a^{-1}$ of $S$ such that $a=aa^{-1}a$ and
$a^{-1}=a^{-1}aa^{-1}$. We call such an element $a^{-1}$ the _inverse_ of $a$.
Evidently, every group is an inverse semigroup. The behavior of inverse
elements in an inverse semigroup $S$ is similar to that in a group, since
$(ab)^{-1}=b^{-1}a^{-1}$ and $(a^{-1})^{-1}=a$, for all $a,b\in S$.
The set $\operatorname{E}(S)$ of the idempotents of an inverse semigroup $S$
is a commutative subsemigroup of $S$. Clearly, $e=e^{-1}$, for every
$e\in\operatorname{E}(S)$. In addition, the idempotents of $S$ are exactly the
elements $aa^{-1}$ and $a^{-1}a$, with $a\in S$.

If $S$ is an inverse semigroup such that $aa^{-1}=a^{-1}a$, for any $a\in S$,
then $S$ is called _Clifford semigroup_. In this case, we briefly write

$\displaystyle a^{0}:=aa^{-1}=a^{-1}a,$

for every $a\in S$. Equivalently, Clifford semigroups are the inverse
semigroups in which the idempotents are central.
Moreover, if $\phi:S\to T$ is a homomorphism from a Clifford semigroup $S$
into a semigroup $T$, then $\operatorname{Im}\phi$ is a Clifford semigroup,
$\phi(a^{-1})=\phi(a)^{-1}$, and $\phi(a^{0})=\phi(a)^{0}$, for any $a\in S$
(cf. [25, Lemma 2.4.4]).
According to [25, Theorem 4.2.1], an inverse semigroup $S$ is a Clifford
semigroup if and only if it is a strong (lower) semilattice $Y$ of disjoint
groups $G_{\alpha}$, for every $\alpha\in Y$. Specifically, for each pair
$\alpha,\beta$ of elements of $Y$ such that $\alpha\geq\beta$, let
$\phi_{\alpha,\beta}:G_{\alpha}\to G_{\beta}$ be a homomorphism of groups such
that

  1. 1.

$\phi_{\alpha,\alpha}$ is the identical automorphism of $G_{\alpha}$, for
every $\alpha\in Y$,

  2. 2.

$\phi_{\beta,\gamma}{}\phi_{\alpha,\beta}{}=\phi_{\alpha,\gamma}{}$, for all
$\alpha,\beta,\gamma\in Y$ such that $\alpha\geq\beta\geq\gamma$.

Then, $S=\displaystyle\bigcup_{\alpha\in Y}\,G_{\alpha}$ endowed with the
operation defined by

$\displaystyle a\,b:=\phi_{\alpha,\alpha\beta}(a)\phi_{\beta,\alpha\beta}(b),$

for every $a\in G_{\alpha}$ and $b\in G_{\beta}$, is a _strong semilattice $Y$
of groups $G_{\alpha}$_, usually written as
$S=[Y;G_{\alpha};\phi_{\alpha,\beta}]$.

######  Definition 1.

[13, Definition 5] Let $S$ be a set endowed with two operations $+$ and
$\circ$ such that $\left(S,+\right)$ and $\left(S,\circ\right)$ are inverse
semigroups. Then, $(S,+,\circ)$ is a _weak (left) brace_ if the following
relations

$\displaystyle a\circ\left(b+c\right)=a\circ b-a+a\circ
c\qquad\text{and}\qquad a\circ a^{-}=-a+a$

hold, for all $a,b,c\in S$, where $-a$ and $a^{-}$ denote the inverses of $a$
with respect to $+$ and $\circ$, respectively. We call $(S,+)$ and $(S,\circ)$
the _additive semigroup_ and the _multiplicative semigroup_ of
$\left(S,+,\circ\right)$, respectively, and, since the sets of idempotents
$E(S,+)$ and $E(S,\circ)$ coincide, we denote it simply by $E(S)$.

Given a weak brace $(S,+,\circ)$, let $\lambda_{a}:S\to S$ be the map defined
by

$\displaystyle\lambda_{a}\left(b\right)=-a+a\circ b,$

for all $a,b\in S$. The map $\lambda_{a}$ can be also written as it is usual
in the context of semi-braces [11, 26], namely,
$\lambda_{a}\left(b\right)=a\circ\left(a^{-}+b\right)$, for all $a,b\in S$.
Furthermore, the map
$\lambda:S\to\operatorname{End}\left(S,+\right),a\mapsto\lambda_{a}$ is a
homomorphism of the inverse semigroup $\left(S,\circ\right)$ into the
endomorphism semigroup of $\left(S,+\right)$.
The operations of sum and multiplication of any weak brace are closely related
by the following relation

$\displaystyle a\circ b=a+\lambda_{a}\left(b\right),$ (1)

for all $a,b\in S$, as proved in [13, Lemma 1]. In particular, they coincide
on $E(S)$. Indeed, if $e_{1},e_{2}\in E(S)$, we get

$\displaystyle e_{1}+e_{2}$
$\displaystyle=\left(e_{1}+e_{2}\right)\circ\left(e_{1}+e_{2}\right)=\left(e_{1}+e_{2}\right)\circ
e_{1}-\left(e_{1}+e_{2}\right)+\left(e_{1}+e_{2}\right)\circ e_{2}$
$\displaystyle=e_{1}\circ\left(e_{1}+e_{2}\right)+e_{1}+e_{2}+e_{2}\circ\left(e_{2}+e_{1}\right)=e_{1}+\lambda_{e_{1}}\left(e_{2}\right)+e_{2}+\lambda_{e_{2}}\left(e_{1}\right)$
$\displaystyle=e_{1}\circ e_{2}+e_{2}\circ e_{1}$ by (1)
$\displaystyle=e_{1}\circ e_{2}.$

As proved in [13, Theorem 8], the additive semigroup of any weak brace $S$ is
a Clifford semigroup. Moreover, the multiplicative semigroup of a weak brace
need not be a Clifford semigroup (see, for instance, [13, Example 3]). Thus,
the following question arises.

######  Question 1.

Describe the structure of the multiplicative semigroup of a weak brace.

However, one can find some instances of weak braces having also a Clifford
semigroup as a multiplicative structure among [12, 13, 14].

######  Definition 2.

A weak brace $(S,+,\circ)$ is said to be a _dual weak brace_ if
$\left(S,\circ\right)$ is a Clifford semigroup.

In the case of a dual weak brace $S$, we highlight that $a^{0}$ is also
$a^{-}\circ a$, for every $a\in S$.
Note that any Clifford semigroup $\left(S,\circ\right)$ determines two dual
weak braces, by setting $a+b:=a\circ b$ or $a+b:=b\circ a$ for all $a,b\in S$.
The first weak brace is the _trivial weak brace_ and the second one the
_almost trivial weak brace_ (see [13, Example 1]).

Below, we show that in any weak brace the sum can be written in terms of the
multiplication as it is already known in the context of skew braces, see [22,
Remark 1.8].

######  Proposition 3.

Let $S$ be a weak brace. Then, $a+b=a\circ\lambda_{a^{-}}\left(b\right),$ for
all $a,b\in S$.

###### Proof.

Let $a,b\in S$. Observing that by [13, Lemma 1] $a\circ(-b)=a-a\circ b+a$, we
obtain

$\displaystyle\left(a\circ\lambda_{a^{-}}\left(b\right)\right)^{-}$
$\displaystyle=\left(a+b\right)^{-}\circ\left(a-a\right)=\left(a+b\right)^{-}\circ
a-\left(a+b\right)^{-}+\left(a+b\right)^{-}\circ\left(-a\right)$
$\displaystyle=\left(a+b\right)^{-}\circ a-\left(a+b\right)^{-}\circ
a+\left(a+b\right)^{-}$
$\displaystyle=\left(a+b\right)^{-}-\left(a+b\right)^{-}\circ
a+\left(a+b\right)^{-}\circ a$
$\displaystyle=\left(a+b\right)^{-}+\lambda_{\left(a+b\right)^{-}}\left(-a\right)-\left(a+b\right)^{-}+\left(a+b\right)^{-}\circ
a$ by (1)
$\displaystyle=\left(a+b\right)^{-}+\left(a+b\right)^{-}\circ\left(a+b-a+a\right)$
$\displaystyle=\left(a+b\right)^{-}+\left(a+b\right)^{-}\circ\left(a+b\right)$
$\displaystyle=\left(a+b\right)^{-}-\left(a+b\right)^{-}+\left(a+b\right)^{-}$
$\displaystyle=\left(a+b\right)^{-}.$

Therefore, the claim follows. ∎

The trivial and the almost trivial weak braces lie in a special class of dual
weak braces that we call _bi-weak braces_. We underline that such a definition
has been originally introduced by Childs in [18] for the particular case of
skew braces.

######  Definition 4.

Let $(S,+,\circ)$ be a weak brace. Then, $S$ is called a _bi-weak brace_ if
$(S,\circ,+)$ is a weak brace, i.e., the relations

$\displaystyle a+(b\circ c)=\left(a+b\right)\circ
a^{-}\circ\left(a+c\right)\qquad\text{and}\qquad a-a=a^{-}\circ a$

hold, for all $a,b,c\in S$.

The following is a general way to construct weak braces involving strong
semilattices of inverse semi-braces provided in [14, Proposition 4] and also
used in the case of generalized left semi-braces in [12].

######  Proposition 5.

Let $Y$ be a (lower) semilattice, $\left\\{S_{\alpha}\ \left|\ \alpha\in
Y\right.\right\\}$ a family of disjoint weak braces. For each pair
$\alpha,\beta$ of elements of $Y$ such that $\alpha\geq\beta$, let
$\phi_{\alpha,\beta}:S_{\alpha}\to S_{\beta}$ be a homomorphism of weak braces
such that

  1. 1.

$\phi_{\alpha,\alpha}$ is the identical automorphism of $S_{\alpha}$, for
every $\alpha\in Y$,

  2. 2.

$\phi_{\beta,\gamma}{}\phi_{\alpha,\beta}{}=\phi_{\alpha,\gamma}{}$, for all
$\alpha,\beta,\gamma\in Y$ such that $\alpha\geq\beta\geq\gamma$.

Then, $S=\bigcup\left\\{S_{\alpha}\ \left|\ \alpha\in Y\right.\right\\}$
endowed with the addition and the multiplication defined by

$\displaystyle
a+b:=\phi_{\alpha,\alpha\beta}(a)+\phi_{\beta,\alpha\beta}(b)\quad\text{and}\quad
a\circ b:=\phi_{\alpha,\alpha\beta}(a)\circ\phi_{\beta,\alpha\beta}(b),$

for all $a\in S_{\alpha}$ and $b\in S_{\beta}$, is a weak brace, called
_strong semilattice $S$ of weak braces $S_{\alpha}$_ and is denoted by
$S=[Y;S_{\alpha};\phi_{\alpha,\beta}]$.

###### Proof.

Since $S$ is an inverse semi-brace and $(S,+)$ clearly is an inverse
semigroup, by [13, Proposition 16], it is sufficient to show that
$a\circ\left(a^{-}+b\right)=-a+a\circ b$, for all $a,b\in S$. If $a\in
S_{\alpha}$ and $b\in S_{\beta}$, then

$\displaystyle a\circ\left(a^{-}+b\right)$
$\displaystyle=a\circ\left(\phi_{\alpha,\alpha\beta}\left(a^{-}\right)+\phi_{\beta,\alpha\beta}\left(b\right)\right)$
$\displaystyle=\phi_{\alpha,\alpha\beta}\left(a\right)\circ\phi_{\alpha\beta,\alpha\beta}\left(\phi_{\alpha,\alpha\beta}\left(a^{-}\right)+\phi_{\beta,\alpha\beta}\left(b\right)\right)$
$\displaystyle=\phi_{\alpha,\alpha\beta}\left(a\right)\circ\left(\phi_{\alpha,\alpha\beta}\left(a\right)^{-}+\phi_{\beta,\alpha\beta}\left(b\right)\right)$
$\displaystyle=-\phi_{\alpha,\alpha\beta}\left(a\right)+\phi_{\alpha,\alpha\beta}\left(a\right)\circ\phi_{\beta,\alpha\beta}\left(b\right)$
$S_{\alpha\beta}$ is a weak brace
$\displaystyle=\phi_{\alpha,\alpha\beta}\left(-a\right)+\phi_{\alpha\beta,\alpha\beta}\left(\phi_{\alpha,\alpha\beta}\left(a\right)\circ\phi_{\beta,\alpha\beta}\left(b\right)\right)$
$\displaystyle=-a+\phi_{\alpha,\alpha\beta}\left(a\right)\circ\phi_{\beta,\alpha\beta}\left(b\right)$
$\displaystyle=-a+a\circ b.$

Therefore, the claim follows. ∎

######  Corollary 6.

A strong semilattice $S=[Y;S_{\alpha};\phi_{\alpha,\beta}]$ is a dual weak
brace if and only if $S_{\alpha}$ is a dual weak brace, for every $\alpha\in
Y$, as well. Specifically, $S$ is a bi-weak brace if and only if $S_{\alpha}$
is a bi-weak brace.

##  2 Rota–Baxter operators on Clifford semigroups

In this section, we present and investigate the Rota–Baxter operators on a
Clifford semigroup, consistently with that introduced for groups in [24].
Moreover, we provide some examples.

######  Definition 7.

Let $(S,+)$ be a Clifford semigroup. A map $\mathfrak{R}:S\to S$ satisfying

$\displaystyle\mathfrak{R}\left(a\right)+\mathfrak{R}\left(b\right)=\mathfrak{R}\left(a+\mathfrak{R}\left(a\right)+b-\mathfrak{R}\left(a\right)\right)\qquad\text{and}\qquad
a+\mathfrak{R}\left(a\right)^{0}=a,$

for all $a,b\in S$, is called _Rota–Baxter operator_ on $(S,+)$.

If $(S,+)$ and $(T,+)$ are Clifford semigroups, then two Rota–Baxter operators
$\mathfrak{R}$ and $\mathfrak{T}$ on $S$ and $T$, respectively, are said to be
_equivalent_ if there exists an isomorphism $\varphi$ from $(S,+)$ to $(T,+)$
such that $\mathfrak{R}=\varphi^{-1}\mathfrak{T}\varphi$.

######  Example 1.

Let $\left(S,+\right)$ be a Clifford semigroup and $\varphi$ an idempotent
endomorphism of $\left(S,+\right)$ such that $\varphi\left(e\right)=e$, for
every $e\in E\left(S,+\right)$. Then, the map $\mathfrak{R}:=-\varphi$ is a
Rota–Baxter operator on $S$.

Easy examples of operators belonging to the previous class are the _elementary
Rota–Baxter operators_ , namely, the maps $\mathfrak{E}$ and $\mathfrak{O}$
from a Clifford semigroup $(S,+)$ into itself given by $\mathfrak{E}(a)=a^{0}$
and $\mathfrak{O}(a)=-a$, for every $a\in S$. Such a terminology is analogue
to that used in [2]. One can observe that in the context of groups such
Rota–Baxter operators are not equivalent. In addition, if $(S,+)$ is a
Clifford monoid with identity $0$, the map $\mathfrak{R}(a)=0$, for every
$a\in S$, is a Rota–Baxter operator on $(S,+)$.

If $\mathfrak{R}$ is a Rota–Baxter operator on a Clifford semigroup $(S,+)$,
the map $\mathfrak{R}^{op}:S\to S$, given by
$\mathfrak{R}^{op}(a)=-a+\mathfrak{R}(-a)$, for every $a\in S$, is still a
Rota–Baxter operator on $(S,+)$, which we call _opposite Rota–Baxter operator_
of $\mathfrak{R}$. We observe that the elementary Rota–Baxter operators can be
obtained one from the other since $\mathfrak{O}=\mathfrak{E}^{op}$ and
$\mathfrak{E}=\mathfrak{O}^{op}$.

The following are two examples of Rota–Baxter operators on a Clifford monoid
which admits an exact factorization. We recall that a Clifford semigroup
$(S,+)$ admits an _exact factorization_ $(U,V)$ if $U$ and $V$ are Clifford
subsemigroups of $S$ and any element $a\in S$ can be written in a unique way
as $a=u_{a}+v_{a}$, with $u_{a}\in U$ and $v_{a}\in V$. In such a case, it
holds that $U\cap V=\\{0\\}$, with $0$ both right identity of $U$ and left
identity of $V$, cf. [9, Theorem 2]. As a consequence, $(S,+)$ is a Clifford
monoid.

######  Example 2.

Let $(U,V)$ be an exact factorization of a Clifford monoid $(S,+)$. Then, the
maps $\mathfrak{R}:S\to S$ and $\mathfrak{T}:S\to S$ given by

$\displaystyle\mathfrak{R}(a)=-v_{a}\qquad\text{and}\qquad\mathfrak{T}(a)=u_{a}^{0}-v_{a}$

for every $a\in S$, respectively, are two Rota–Baxter operators on $S$.
We underline that the map $\mathfrak{R}$ is a Rota–Baxter operator also in the
case of a right univocal factorizable Clifford semigroup. To this purpose, we
recall that, in general, a factorization $\left(U,V\right)$ of a semigroup $S$
is _right univocal_ if

$\displaystyle\forall\ u,u^{\prime}\in U,\ v,v^{\prime}\in V\quad
u+v=u^{\prime}+v^{\prime}\implies v=v^{\prime},$

see, for instance, [9, Definition 1].

In the next result, we introduce a method for constructing a new Rota–Baxter
operator on a Clifford semigroup seen as strong semilattice $Y$ of groups
$G_{\alpha}$ starting from given Rota–Baxter operators on each group
$G_{\alpha}$. Hereinafter, we denote by $0_{\alpha}$ the identity of
$G_{\alpha}$, for every $\alpha\in Y$.

######  Theorem 8.

Let $(S,+)$ be a Clifford semigroup, $S=[Y;G_{\alpha};\phi_{\alpha,\beta}]$
and assume that $\mathfrak{R}_{\alpha}$ is a Rota–Baxter operator on the group
$(G_{\alpha},+)$, for every $\alpha\in Y$. Then, the map $\mathfrak{R}:S\to S$
given by $\mathfrak{R}\left(a\right)=\mathfrak{R}_{\alpha}\left(a\right)$, for
every $a\in G_{\alpha}$, is a Rota–Baxter operator on $(S,+)$ if and only if
the condition

$\displaystyle\mathfrak{R}_{\beta}\phi_{\alpha,\beta}=\phi_{\alpha,\beta}\mathfrak{R}_{\alpha},$
(2)

is satisfied, for all $\alpha,\beta\in Y$ such that $\alpha\geq\beta$. We call
such an operator $\mathfrak{R}$ _strong Rota–Baxter operator on $(S,+)$_.

###### Proof.

Initially, assume that (2) holds. If $a\in G_{\alpha}$ and $b\in G_{\beta}$,
we have that

$\displaystyle a+\mathfrak{R}\left(a\right)+b-\mathfrak{R}\left(a\right)$
$\displaystyle=\phi_{\alpha,\alpha\beta}\left(a+\mathfrak{R}_{\alpha}\left(a\right)\right)+\phi_{\beta,\alpha\beta}\left(b\right)-\mathfrak{R}_{\alpha}\left(a\right)$
$\displaystyle=\phi_{\alpha,\alpha\beta}\left(a\right)+\phi_{\alpha,\alpha\beta}\mathfrak{R}_{\alpha}\left(a\right)+\phi_{\beta,\alpha\beta}\left(b\right)-\phi_{\alpha,\alpha\beta}\mathfrak{R}_{\alpha}\left(a\right)$
$\displaystyle=\phi_{\alpha,\alpha\beta}\left(a\right)+\mathfrak{R}_{\alpha\beta}\phi_{\alpha,\alpha\beta}\left(a\right)+\phi_{\beta,\alpha\beta}\left(b\right)-\mathfrak{R}_{\alpha\beta}\phi_{\alpha,\alpha\beta}\left(a\right)$
by (2)

Thus, we obtain

$\displaystyle\mathfrak{R}\left(a+\mathfrak{R}\left(a\right)+b-\mathfrak{R}\left(a\right)\right)$
$\displaystyle=\mathfrak{R}_{\alpha\beta}\left(\phi_{\alpha,\alpha\beta}\left(a\right)+\mathfrak{R}_{\alpha\beta}\phi_{\alpha,\alpha\beta}\left(a\right)+\phi_{\beta,\alpha\beta}\left(b\right)-\mathfrak{R}_{\alpha\beta}\phi_{\alpha,\alpha\beta}\left(a\right)\right)$
$\displaystyle=\mathfrak{R}_{\alpha\beta}\phi_{\alpha,\alpha\beta}\left(a\right)+\mathfrak{R}_{\alpha\beta}\phi_{\beta,\alpha\beta}\left(b\right)$
$\displaystyle=\phi_{\alpha,\alpha\beta}\mathfrak{R}_{\alpha}\left(a\right)+\phi_{\beta,\alpha\beta}\mathfrak{R}_{\beta}\left(b\right)$
by (2) $\displaystyle=\mathfrak{R}\left(a\right)+\mathfrak{R}\left(b\right).$

Therefore, since the second condition in 7 is trivially satisfied,
$\mathfrak{R}$ is a Rota–Baxter operator on $S$.
Conversely, assume that $\mathfrak{R}$ is a Rota–Baxter operator on $(S,+)$.
Then, if $a\in G_{\alpha}$, by [2, Lemma 2.5] we get

$\displaystyle\mathfrak{R}\left(a\right)+\mathfrak{R}\left(0_{\beta}\right)=\mathfrak{R}_{\alpha}\left(a\right)+\mathfrak{R}_{\beta}\left(0_{\beta}\right)=\phi_{\alpha,\alpha\beta}\mathfrak{R}_{\alpha}\left(a\right)+0_{\alpha\beta}=\phi_{\alpha,\alpha\beta}\mathfrak{R}_{\alpha}\left(a\right).$

On the other hand,

$\displaystyle\mathfrak{R}\left(a+\mathfrak{R}\left(a\right)+0_{\beta}-\mathfrak{R}\left(a\right)\right)$
$\displaystyle=\mathfrak{R}_{\alpha\beta}\left(\phi_{\alpha,\alpha\beta}\left(a\right)+\phi_{\alpha,\alpha\beta}\mathfrak{R}_{\alpha}\left(a\right)+\phi_{\beta,\alpha\beta}\left(0_{\beta}\right)-\phi_{\beta,\alpha\beta}\mathfrak{R}_{\alpha}\left(a\right)\right)$
$\displaystyle=\mathfrak{R}_{\alpha\beta}\phi_{\alpha,\alpha\beta}\left(a\right),$

therefore (2) holds. ∎

We observe that not all Rota-Baxter operators on Clifford semigroups can be
constructed as in 8. Indeed, if $(S,+)$ is a Clifford monoid with $|E(S)|>1$,
the map $\mathfrak{R}:S\to S$ given by $\mathfrak{R}(a)=0$, for every $a\in
S$, is such an example.

Strong Rota–Baxter operators on a Clifford semigroup
$S=[Y;G_{\alpha};\phi_{\alpha,\beta}]$ can be easily obtained starting from
arbitrary Rota–Baxter operators on groups $G_{\alpha}$.

######  Example 3.

Let $S=[Y;G_{\alpha};\phi_{\alpha,\beta}]$ be a Clifford semigroup with
$\phi_{\alpha,\beta}(a)=0_{\beta}$, for every $a\in G_{\alpha}$ and
$\alpha,\beta\in Y$ such that $\alpha\geq\beta$. If $\mathfrak{R}_{\alpha}$ is
an arbitrary Rota–Baxter operator on $G_{\alpha}$, for every $\alpha\in Y$,
then the map $\mathfrak{R}:S\to S$ given by
$\mathfrak{R}(a)=\mathfrak{R}_{\alpha}(a)$, for every $a\in G_{\alpha}$, is a
strong Rota–Baxter operator on $S$.

Finally, let us show some useful properties of Rota–Baxter operators on
Clifford semigroups.

######  Proposition 9.

Let $\mathfrak{R}$ be a Rota–Baxter operator on a Clifford semigroup $(S,+)$.
Then,

$\displaystyle\mathfrak{R}\left(E\left(S\right)\right)\subseteq
E\left(S\right).$

In particular, if $(S,+)$ is a Clifford monoid with identity $0$, it holds
that $\mathfrak{R}\left(0\right)=0$.

###### Proof.

If $e\in E\left(S\right)$, we have that

$\displaystyle\mathfrak{R}\left(e\right)=\mathfrak{R}\left(e+\mathfrak{R}\left(e\right)-\mathfrak{R}\left(e\right)\right)=\mathfrak{R}\left(e+\mathfrak{R}\left(e\right)+e-\mathfrak{R}\left(e\right)\right)=\mathfrak{R}\left(e\right)+\mathfrak{R}\left(e\right).$

Besides , if $(S,+)$ is a Clifford monoid with identity $0$, since
$\mathfrak{R}(0)$ is idempotent, we get

$\displaystyle
0=0+\mathfrak{R}\left(0\right)-\mathfrak{R}\left(0\right)=\mathfrak{R}\left(0\right)+\mathfrak{R}\left(0\right)=\mathfrak{R}\left(0\right),$

thus the claim follows. ∎

In general, it is not true that
$E(S)\subseteq\mathfrak{R}\left(E\left(S\right)\right)$. In fact, it is enough
to take the Rota-Baxter operator of constant value $0$ on a Clifford monoid
$(S,+)$ with $|E(S)|>1$.

######  Proposition 10.

Let $\mathfrak{R}$ be a Rota–Baxter operator on a Clifford semigroup $(S,+)$.
Then, the following statements hold:

  1. 1.

$\mathfrak{R}(a)=\mathfrak{R}(a)+\mathfrak{R}\left(a^{0}\right)$,

  2. 2.

$a=a+\mathfrak{R}\left(a^{0}\right)$,

  3. 3.

$-\mathfrak{R}\left(a\right)=\mathfrak{R}\left(-\mathfrak{R}\left(a\right)-a+\mathfrak{R}\left(a\right)\right)$,

for every $a\in S$.

###### Proof.

If $a\in S$, we have that

$\displaystyle\mathfrak{R}(a)=\mathfrak{R}\left(a+\mathfrak{R}\left(a\right)^{0}\right)=\mathfrak{R}\left(a+\mathfrak{R}(a)+a^{0}-\mathfrak{R}(a)\right)=\mathfrak{R}\left(a\right)+\mathfrak{R}\left(a^{0}\right),$

i.e., $1.$ is verified. Moreover, by $1.$,

$\displaystyle
a=a-\mathfrak{R}(a)+\mathfrak{R}(a)=a-\mathfrak{R}(a)+\mathfrak{R}(a)+\mathfrak{R}\left(a^{0}\right)=a+\mathfrak{R}\left(a^{0}\right),$

and so $2.$ holds. To prove $3.$, we note that

$\displaystyle\mathfrak{R}(a)+\mathfrak{R}\left(-\mathfrak{R}\left(a\right)-a+\mathfrak{R}\left(a\right)\right)=\mathfrak{R}\left(a^{0}\right).$
(3)

Thus, we obtain that

$\displaystyle\mathfrak{R}(a)+\mathfrak{R}\left(-\mathfrak{R}\left(a\right)-a+\mathfrak{R}\left(a\right)\right)+\mathfrak{R}(a)$
$\displaystyle=\mathfrak{R}\left(a^{0}\right)+\mathfrak{R}(a)$ by (3)
$\displaystyle=\mathfrak{R}(a)+\mathfrak{R}\left(a^{0}\right)$ by 9
$\displaystyle=\mathfrak{R}(a)$ by $1.$

and

$\displaystyle\mathfrak{R}\left(-\mathfrak{R}\left(a\right)-a+\mathfrak{R}\left(a\right)\right)+\mathfrak{R}(a)+\mathfrak{R}\left(-\mathfrak{R}\left(a\right)-a+\mathfrak{R}\left(a\right)\right)$
$\displaystyle\quad=\mathfrak{R}\left(-\mathfrak{R}\left(a\right)-a+\mathfrak{R}\left(a\right)\right)+\mathfrak{R}\left(a^{0}\right)$
by (3)
$\displaystyle\quad=\mathfrak{R}\left(a^{0}\right)+\mathfrak{R}\left(-\mathfrak{R}\left(a\right)-a+\mathfrak{R}\left(a\right)\right)$
by 9
$\displaystyle\quad=\mathfrak{R}\left(a^{0}+\mathfrak{R}\left(a^{0}\right)-\mathfrak{R}\left(a\right)-a+\mathfrak{R}\left(a\right)-\mathfrak{R}\left(a^{0}\right)\right)$
$\displaystyle\quad=\mathfrak{R}\left(-\mathfrak{R}\left(a\right)-a+\mathfrak{R}\left(a\right)+\mathfrak{R}\left(a^{0}\right)\right)$
by 9
$\displaystyle\quad=\mathfrak{R}\left(-\mathfrak{R}\left(a\right)-a+\mathfrak{R}\left(a\right)\right)$
by $1.$

hence $3.$ follows. ∎

######  Remark 1.

If $\mathfrak{R}$ is a Rota–Baxter operator on a Clifford semigroup $(S,+)$,
by $3.$ in 10, we obtain that $\operatorname{Im}\mathfrak{R}$ is a Clifford
subsemigroup of $(S,+)$. Furthermore, we can make explicit the statement in 9,
since, using $3.$ in 10 and (3), we get

$\displaystyle\mathfrak{R}\left(a^{0}\right)=\mathfrak{R}\left(a\right)^{0},$
(4)

for every $a\in S$.

##  3 Rota–Baxter endomorphisms

This section focuses on endomorphisms of Clifford semigroups that are
Rota–Baxter operators, namely Rota–Baxter endomorphisms. Specifically, we give
a description of such maps $f$ for which $\operatorname{Im}f$ is commutative.
In addition, we characterize Rota–Baxter endomorphisms of groups that are also
idempotent.

######  Definition 11.

Let $(S,+)$ be a Clifford semigroup. Any endomorphism $\mathfrak{R}:S\to S$
that is a Rota–Baxter operator is called a _Rota–Baxter endomorphism_ of
$(S,+)$.

Clearly, the constant map of value $0$ of a Clifford monoid $(S,+)$ is an easy
example of Rota–Baxter endomorphism.

From now on, given two Clifford semigroups $(S,+)$ and $(T,+)$, we will denote
by $\operatorname{Hom}\left(S,T\right)$ the set of all the homomorphisms from
$(S,+)$ to $(T,+)$. If $f\in\operatorname{Hom}\left(S,T\right)$ is such that
$\operatorname{Im}f$ is commutative, we say that $f$ is a _commutative
homomorphism_. In the context of groups, these maps are named _abelian
homomorphisms_ [17].
If $G$ is a group, one has that the abelian endomorphisms of $G$ are
Rota–Baxter operators. Instead, if $\left(S,+\right)$ is a Clifford semigroup,
not every commutative endomorphism of $(S,+)$ is a Rota–Baxter operator.
Indeed, if $|E(S)|>1$ and $e\in E(S)$, the constant map of value $e$ is such
an example. Thus, our aim is to show how to construct all the commutative
Rota–Baxter endomorphisms on a Clifford semigroup
$S=[Y;G_{\alpha};\phi_{\alpha,\beta}]$ starting from given abelian
homomorphisms. To this purpose, we recall the following result.

######  Lemma 12 (cf. Proposition II.2.8, [28]).

Let $S=[Y;G_{\alpha};\phi_{\alpha,\beta}]$ be a Clifford semigroup, $h:Y\to Y$
an endomorphism, and consider a family of maps
$f_{\alpha}\in\operatorname{Hom}\left(G_{\alpha},G_{h\left(\alpha\right)}\right)$,
for every $\alpha\in Y$, such that

$\displaystyle
f_{\beta}\phi_{\alpha,\beta}=\phi_{h\left(\alpha\right),h\left(\beta\right)}f_{\alpha}$

holds, for all $\alpha,\beta\in Y$ such that $\alpha\geq\beta$. Then, the map
$f:S\to S$ given by $f(a)=f_{\alpha}(a)$, for every $a\in G_{\alpha}$, is an
endomorphism of $S$. Conversely, every endomorphism of $S$ can be so
constructed.

######  Theorem 13.

Let $S=[Y;G_{\alpha};\phi_{\alpha,\beta}]$ be a Clifford semigroup, $h:Y\to Y$
an endomorphism such that $\alpha\leq h\left(\alpha\right)$, and consider a
family of abelian homomorphisms
$f_{\alpha}\in\operatorname{Hom}\left(G_{\alpha},G_{h\left(\alpha\right)}\right)$,
for every $\alpha\in Y$. If the following condition is satisfied

$\displaystyle
f_{\beta}\phi_{\alpha,\beta}=\phi_{h\left(\alpha\right),h\left(\beta\right)}f_{\alpha},$
(5)

for all $\alpha,\beta\in Y$ such that $\alpha\geq\beta$, then the map
$\mathfrak{R}:S\to S$ given by $\mathfrak{R}(a)=f_{\alpha}(a)$, for every
$a\in G_{\alpha}$, is a commutative Rota–Baxter endomorphism of $S$.
Conversely, any commutative Rota–Baxter endomorphism on $S$ can be so
constructed.

###### Proof.

Initially, by Lemma 12, note that $\mathfrak{R}$ is an endomorphism of $S$ and
obviously $\mathfrak{R}$ is commutative since $f_{\alpha}$ is abelian, for
every $\alpha\in Y$. Now, if $a\in G_{\alpha}$ and $b\in G_{\beta}$, with
$\alpha,\beta\in Y$ such that $\alpha\geq\beta$, we have that

$\displaystyle a$
$\displaystyle+\mathfrak{R}\left(a\right)+b-\mathfrak{R}\left(a\right)$
$\displaystyle=\left(\phi_{\alpha,\alpha
h\left(\alpha\right)}\left(a\right)+\phi_{h\left(\alpha\right),\alpha
h\left(\alpha\right)}f_{\alpha}\left(a\right)\right)+\left(\phi_{\beta,\beta
h\left(\alpha\right)}\left(b\right)-\phi_{h\left(\alpha\right),\beta
h\left(\alpha\right)}f_{\alpha}\left(a\right)\right)$
$\displaystyle=a+\phi_{h\left(\alpha\right),\alpha}f_{\alpha}\left(a\right)+\phi_{\beta,\beta
h\left(\alpha\right)}\left(b\right)-\phi_{h\left(\alpha\right),\beta
h\left(\alpha\right)}f_{\alpha}\left(a\right)$
$\displaystyle=\phi_{\alpha,\alpha\beta
h\left(\alpha\right)}\left(a+\phi_{h\left(\alpha\right),\alpha}f_{\alpha}\left(a\right)\right)+\phi_{\beta
h\left(\alpha\right),\alpha\beta}\left(\phi_{\beta,\beta
h\left(\alpha\right)}\left(b\right)-\phi_{h\left(\alpha\right),\beta
h\left(\alpha\right)}f_{\alpha}\left(a\right)\right)$
$\displaystyle=\phi_{\alpha,\alpha\beta}\left(a\right)+\phi_{\alpha,\alpha\beta}\phi_{h\left(\alpha\right),\alpha}f_{\alpha}\left(a\right)+\phi_{\beta,\alpha\beta}\left(b\right)-\phi_{h\left(\alpha\right),\alpha\beta}f_{\alpha}\left(a\right)$
$\displaystyle=\phi_{\alpha,\alpha\beta}\left(a\right)+\phi_{h\left(\alpha\right),\alpha\beta}f_{\alpha}\left(a\right)+\phi_{\beta,\alpha\beta}\left(b\right)-\phi_{h\left(\alpha\right),\alpha\beta}f_{\alpha}\left(a\right)$

and, consequently,

$\displaystyle\mathfrak{R}$
$\displaystyle\left(a+\mathfrak{R}\left(a\right)+b-\mathfrak{R}\left(a\right)\right)$
$\displaystyle=f_{\alpha\beta}\left(\phi_{\alpha,\alpha\beta}\left(a\right)+\phi_{h\left(\alpha\right),\alpha\beta}f_{\alpha}\left(a\right)+\phi_{\beta,\alpha\beta}\left(a\right)-\phi_{h\left(\alpha\right),\alpha\beta}f_{\alpha}\left(a\right)\right)$
$\displaystyle=f_{\alpha\beta}\phi_{\alpha,\alpha\beta}\left(a\right)+f_{\alpha\beta}\phi_{\beta,\alpha\beta}\left(b\right)$
$f_{\alpha\beta}$ is abelian
$\displaystyle=\phi_{h\left(\alpha\right),h\left(\alpha\beta\right)}f_{\alpha}\left(a\right)+\phi_{{}_{,}h\left(\beta\right),h\left(\alpha\beta\right)}f_{\beta}\left(b\right)$
by (5)
$\displaystyle=f_{\alpha}\left(a\right)+f_{\beta}\left(b\right)=\mathfrak{R}\left(a\right)+\mathfrak{R}\left(b\right)$

Moreover, it holds that

$\displaystyle a+\mathfrak{R}\left(a\right)^{0}$
$\displaystyle=a+f_{\alpha}\left(0_{\alpha}\right)=\phi_{\alpha,\alpha
h\left(\alpha\right)}\left(a\right)+0_{\alpha
h\left(\alpha\right)}=\phi_{\alpha,\alpha}\left(a\right)+0_{\alpha}=a.$

Therefore, $\mathfrak{R}$ is a commutative Rota–Baxter endomorphism of $S$.
Conversely, assume that $\mathfrak{R}$ is a commutative Rota–Baxter
endomorphism of $S$. Let $h:Y\to Y$ be the map such that
$h\left(\alpha\right)=\chi$ if $\mathfrak{R}\left(0_{\alpha}\right)=0_{\chi}$.
Note that $h\left(\alpha\right)\geq\alpha$, since if $a\in G_{\alpha}$, by
$2.$ in 10, we obtain

$\displaystyle a=a+\mathfrak{R}\left(a^{0}\right)=a+0_{\chi}\in
G_{\alpha\chi},$

thus $\alpha\chi=\alpha$, and so $h(\alpha)\geq\alpha$. Now, denoted by
$f_{\alpha}:=\mathfrak{R}_{|_{G_{\alpha}}}$, the map
$\mathfrak{R}_{\alpha}:=\phi_{h\left(\alpha\right),\alpha}f_{\alpha}$ is
clearly an abelian endomorphism of $G_{\alpha}$, for every $\alpha\in Y$.
Moreover, if $a\in G_{\alpha}$ and $\alpha,\beta\in Y$ with $\alpha\geq\beta$,
we get

$\displaystyle\mathfrak{R}_{\beta}\phi_{\alpha,\beta}\left(a\right)$
$\displaystyle=\phi_{h\left(\beta\right),\beta}f_{\beta}\phi_{\alpha,\beta}\left(a\right)=\phi_{h\left(\beta\right),\beta}\mathfrak{R}\left(a+0_{\beta}\right)=\phi_{h\left(\beta\right),\beta}\left(\mathfrak{R}\left(a\right)+\mathfrak{R}\left(0_{\beta}\right)\right)$
$\displaystyle=\phi_{h\left(\beta\right),\beta}\phi_{h\left(\alpha\right),h\left(\alpha\beta\right)}\mathfrak{R}\left(a\right)=\phi_{h\left(\alpha\right),\beta}f_{\alpha}\left(a\right)=\phi_{\alpha,\beta}\phi_{h\left(\alpha\right),\alpha}f_{\alpha}\left(a\right)=\phi_{\alpha,\beta}R_{\alpha}\left(a\right).$

Therefore, the claim follows. ∎

Obviously, any Rota–Baxter endomorphism is a strong Rota–Baxter operator only
in the case $h=\operatorname{id}_{Y}$.

######  Remark 2.

Let us note that if the endomorphism $h$ in 13 is surjective, then
$\mathfrak{R}\left(E\left(S\right)\right)=E\left(S\right)$. Indeed, if
$\beta=h\left(\alpha\right)$ for a certain $\alpha\in Y$, observing that
$\alpha\leq\beta$, we have that

$0_{\beta}=0_{h\left(\alpha\right)}=\phi_{h\left(\beta\right),h\left(\alpha\right)}\left(0_{h\left(\beta\right)}\right)\underset{\eqref{eq:Worawiset}}{=}f_{\alpha}\phi_{\beta,\alpha}\left(0_{\beta}\right)=f_{\alpha}\left(0_{\alpha}\right)=\mathfrak{R}\left(0_{\alpha}\right),$

hence, by 9, the claim follows.

To conclude this section, we give a description of all idempotent Rota–Baxter
endomorphisms of not necessarily abelian groups $\left(G,+\right)$. In this
case, $G$ is exactly factorizable in $\ker\mathfrak{R}$ and
$\operatorname{Im}\mathfrak{R}$.

######  Theorem 14.

Let $(G,+)$ be a group. If $N\unlhd G$ is such that $G/N$ is abelian and
$\mathcal{S}$ is a set of representatives of $G/N$ that is a subgroup of $G$,
then any map $\mathfrak{R}:G\to G$ such that
$\operatorname{Im}\mathfrak{R}=\mathcal{S}$ and $\mathfrak{R}\left(g\right)\in
N+g$, for every $g\in G$, is an idempotent Rota–Baxter endomorphism of $G$.
Conversely, if $\mathfrak{R}$ is an idempotent Rota–Baxter endomorphism of
$G$, then $G/\ker\mathfrak{R}$ is abelian, $\operatorname{Im}\mathfrak{R}$ is
a subgroup of $G$ that is also a set of representatives of
$G/\ker\mathfrak{R}$, and $\mathfrak{R}\left(g\right)\in\ker\mathfrak{R}+g$,
for every $g\in G$.

###### Proof.

Initially, assume that $N\unlhd G$ and that $G/N$ is abelian, and let
$\mathfrak{R}:G\to G$ be a map such that
$\operatorname{Im}\mathfrak{R}=\mathcal{S}$ and $\mathfrak{R}\left(g\right)\in
N+g$, for every $g\in G$. Then, we clearly have that $\mathfrak{R}$ is a
homomorphism from $G$ to $\mathcal{S}$. Besides, if $g,h\in G$, by [31,
Theorem 2.23], since $G^{\prime}\subseteq N$,

$\displaystyle\mathfrak{R}\left(g+\mathfrak{R}\left(g\right)+h-\mathfrak{R}\left(g\right)\right)$
$\displaystyle\in
N+g+\mathfrak{R}\left(g\right)+h-\mathfrak{R}\left(g\right)=N+g+g+h-g$
$\displaystyle=N+g+[g,h]+h\subseteq N+g+h.$

Moreover, since $\mathfrak{R}\left(g\right)=n+g$, for a certain $n\in N$, we
get
$\mathfrak{R}^{2}\left(g\right)=\mathfrak{R}\left(n\right)+\mathfrak{R}\left(g\right)=\mathfrak{R}\left(g\right)$,
hence the first part of the claim follows.
Conversely, assume that $\mathfrak{R}:G\to G$ is an idempotent Rota–Baxter
endomorphism of $G$. Then, obviously, $\operatorname{Im}\mathfrak{R}$ is a
subgroup of $G$. Moreover, if $g,h\in G$, it holds that

$\displaystyle\mathfrak{R}\left(g+h-g-h\right)$
$\displaystyle=\mathfrak{R}\left(g\right)-\mathfrak{R}\left(h\right)+\mathfrak{R}\left(h\right)+\mathfrak{R}^{2}\left(h\right)+\mathfrak{R}\left(-g\right)-\mathfrak{R}^{2}\left(h\right)$
$\displaystyle=\mathfrak{R}\left(g\right)-\mathfrak{R}\left(h\right)+\mathfrak{R}\left(h+\mathfrak{R}\left(h\right)-g-\mathfrak{R}\left(h\right)\right)$
$\displaystyle=\mathfrak{R}\left(g\right)-\mathfrak{R}\left(h\right)+\mathfrak{R}\left(h\right)-\mathfrak{R}\left(g\right)=0,$

hence $g+h-g-h\in\ker\mathfrak{R}$, i.e., $G/\ker\mathfrak{R}$ is abelian.
Now, if $g\in G$, since

$\displaystyle\mathfrak{R}\left(\mathfrak{R}\left(g\right)-g\right)=\mathfrak{R}^{2}\left(g\right)-\mathfrak{R}\left(g\right)=0,$

we obtain that
$\mathfrak{R}\left(g\right)=\mathfrak{R}\left(g\right)-g+g\in\ker\mathfrak{R}+g$.
Thus,
$\mathfrak{R}\left(g\right)\in\mathfrak{R}\left(G\right)\cap\left(\ker\mathfrak{R}+g\right)$.
In addition, if
$\mathfrak{R}\left(h\right)\in\mathfrak{R}\left(G\right)\cap\left(\ker\mathfrak{R}+g\right)$,
it follows that
$\mathfrak{R}\left(g-h\right)=\mathfrak{R}\left(g\right)-\mathfrak{R}\left(h\right)\in\ker\mathfrak{R}$,
hence $\mathfrak{R}\left(g-h\right)=\mathfrak{R}^{2}\left(g-h\right)=0$,
namely $g-h\in\ker\mathfrak{R}$. It follows that $g=n+h$, with
$n\in\ker\mathfrak{R}$, thus

$\displaystyle\mathfrak{R}\left(g\right)=\mathfrak{R}\left(n\right)+\mathfrak{R}\left(h\right)=\mathfrak{R}\left(h\right).$

Therefore, $\operatorname{Im}\mathfrak{R}$ is a set of representatives of
$G/\ker\mathfrak{R}$. ∎

##  4 Dual weak braces obtained by Rota–Baxter operators

In this section, we illustrate a method for obtaining a dual weak brace $S$
starting from a given Rota–Baxter operator on a Clifford semigroup $(S,+)$.
This construction is inspired by that of skew braces due to Bardakov and
Gubarev [3, Proposition 3.1]. Furthermore, we show that every commutative
Rota–Baxter endomorphism determines a bi-weak brace.

######  Lemma 15.

Let $(S,+)$ be a Clifford semigroup and $\mathfrak{R}:S\to S$ a Rota–Baxter
operator on $(S,+)$. Set $a\circ
b:=a+\mathfrak{R}\left(a\right)+b-\mathfrak{R}\left(a\right)$, for all $a,b\in
S$. Then, $(S,\circ)$ is a Clifford semigroup.

###### Proof.

It is a routine computation to verify that $(S,\circ)$ is a semigroup.
Moreover, if $a\in S$, set
$x:=-\mathfrak{R}\left(a\right)-a+\mathfrak{R}\left(a\right)$, we get

$\displaystyle a\,\circ x\circ a$
$\displaystyle=\left(a+\mathfrak{R}\left(a\right)^{0}-a+\mathfrak{R}\left(a\right)^{0}\right)\circ
a$ $\displaystyle=a^{0}\circ a$ by 9
$\displaystyle=a^{0}+a+\mathfrak{R}\left(a^{0}\right)-\mathfrak{R}\left(a^{0}\right)$
by 9 $\displaystyle=a^{0}+a=a.$

To prove the claim we use [25, Theorem 4.2.1] and thus we show that the
idempotents of $(S,\circ)$ are central. To this end, observe that
$E(S,\circ)\subseteq E(S,+)$. Indeed, if $e\in E(S,\circ)$, then

$\displaystyle\mathfrak{R}\left(e\right)+\mathfrak{R}\left(e\right)=\mathfrak{R}\left(e+\mathfrak{R}\left(e\right)+e-\mathfrak{R}\left(e\right)\right)=\mathfrak{R}\left(e\circ
e\right)=\mathfrak{R}\left(e\right),$

i.e., $\mathfrak{R}\left(e\right)\in E(S,+)$, and thus

$\displaystyle e=e\circ
e=e+\mathfrak{R}\left(e\right)+e-\mathfrak{R}\left(e\right)=e+e+\mathfrak{R}\left(e\right)^{0}=e+e.$

Hence, if $a\in S$, we get

$\displaystyle a\circ e$
$\displaystyle=a+\mathfrak{R}\left(a\right)+e-\mathfrak{R}\left(a\right)=a+\mathfrak{R}\left(a\right)^{0}+e=a+e$
$\displaystyle=a+e+\mathfrak{R}\left(e\right)^{0}=e+\mathfrak{R}\left(e\right)+a-\mathfrak{R}\left(e\right)=e\circ
a.$

Therefore, $(S,\circ)$ is a Clifford semigroup and $a^{-}$ coincides with $x$.
∎

######  Remark 3.

As it happens in the context of groups (see [24, Proposition 3.1]), if
$\mathfrak{R}$ is a Rota–Baxter operator on a Clifford semigroup $(S,+)$, then
clearly $\mathfrak{R}$ is a homomorphism from $(S,\circ)$ to $(S,+)$.
Moreover, $\mathfrak{R}$ is also a Rota–Baxter operator on $(S,\circ)$, indeed
if $a,b\in S$ we have that

$\displaystyle\mathfrak{R}\left(a\right)\circ\mathfrak{R}\left(b\right)$
$\displaystyle=\mathfrak{R}\left(a\right)+\mathfrak{R}^{2}\left(a\right)+\mathfrak{R}\left(b\right)-\mathfrak{R}^{2}\left(a\right)=\mathfrak{R}\left(a\circ\mathfrak{R}\left(a\right)\circ
b\right)+\mathfrak{R}\left(\mathfrak{R}\left(a\right)^{-}\right)$
$\displaystyle=\mathfrak{R}\left(a\circ\mathfrak{R}\left(a\right)\circ
b\circ\mathfrak{R}\left(a\right)^{-}\right)$

and, by (1),

$\displaystyle
a\circ\mathfrak{R}\left(a\right)\circ\mathfrak{R}\left(a\right)^{-}$
$\displaystyle=a+\lambda_{a}\mathfrak{R}\left(a\right)-\lambda_{a}\mathfrak{R}\left(a\right)=a\circ\mathfrak{R}\left(a\right)-a\circ\mathfrak{R}\left(a\right)+a$
$\displaystyle=a^{0}\circ\mathfrak{R}\left(a\right)^{0}+a=a+a^{0}+\mathfrak{R}\left(a\right)^{0}=a.$

######  Theorem 16.

Let $(S,+)$ be a Clifford semigroup. If $\mathfrak{R}:S\to S$ is a Rota–Baxter
operator on $(S,+)$, then $(S,+)$ endowed with the operation $\circ$ defined
by

$\displaystyle a\circ
b:=a+\mathfrak{R}\left(a\right)+b-\mathfrak{R}\left(a\right),$ (6)

for all $a,b\in S$, is a dual weak brace, which we call _weak brace associated
to $\mathfrak{R}$_ and denote by $S_{\mathfrak{R}}$.

###### Proof.

Initially, by Lemma 15, $\left(S,\circ\right)$ is a Clifford semigroup where,
for any $a\in S$,
$a^{-}=-\mathfrak{R}\left(a\right)-a+\mathfrak{R}\left(a\right)$. Moreover, if
$a,b,c\in S$, we get

$\displaystyle a\circ b-a+a\circ c$
$\displaystyle=a+\mathfrak{R}\left(a\right)+b-\mathfrak{R}\left(a\right)+a^{0}+\mathfrak{R}\left(a\right)+c-\mathfrak{R}\left(a\right)$
$\displaystyle=a+\mathfrak{R}\left(a\right)+b+\mathfrak{R}\left(a\right)^{0}+c-\mathfrak{R}\left(a\right)$
$\displaystyle=a+\mathfrak{R}\left(a\right)+b+c-\mathfrak{R}\left(a\right)=a\circ\left(b+c\right),$

In addition, it holds $a\circ
a^{-}=a+\mathfrak{R}(a)^{0}-a+\mathfrak{R}\left(a\right)^{0}=a^{0}$.
Therefore, the statement is proved. ∎

For example, the dual weak braces associated to the elementary Rota–Baxter
operators $\mathfrak{E}$ and $\mathfrak{O}$ on a Clifford semigroup $(S,+)$
are the trivial and the almost trivial weak braces, respectively. Moreover, if
$(S,+)$ is commutative, then any weak brace $S_{\mathfrak{R}}$ related to an
arbitrary Rota–Baxter operator $\mathfrak{R}$ on $S$ is a trivial weak brace.

Let us note that if $\mathfrak{R}$ and $\mathfrak{T}$ are two equivalent
Rota–Baxter operators on two Clifford semigroups $(S,+)$ and $(T,+)$,
respectively, then, $S_{\mathfrak{R}}$ and $T_{\mathfrak{T}}$ are isomorphic.
Indeed, if $\varphi$ is the isomorphism from $(S,+)$ to $(T,+)$ such that
$\mathfrak{R}=\varphi^{-1}\mathfrak{T}\varphi$, then

$\displaystyle\varphi(a\circ b)$
$\displaystyle=\varphi(a+\mathfrak{R}(a)+b-\mathfrak{R}(a))=\varphi(a)+\varphi\mathfrak{R}(a)+\varphi(b)-\varphi\mathfrak{R}(a)$
$\displaystyle=\varphi(a)+\mathfrak{T}\varphi(a)+\varphi(b)-\mathfrak{T}\varphi(a)=\varphi(a)\circ\varphi(b),$

for all $a,b\in S$.
It is natural to ask whether the converse holds. The answer is negative, since
if we consider a commutative Clifford semigroup $(S,+)$, with $E(S)\neq S$,
the trivial weak brace on $(S,+)$ can be obtained starting from arbitrary
Rota–Baxter operators on $(S,+)$.
Analogously to [3, Question 7.4], the following question arises.

######  Question 2.

Given two Rota–Baxter operators $\mathfrak{R}$ and $\mathfrak{T}$ on a
Clifford semigroup $(S,+)$, under which conditions the associated weak braces
$S_{\mathfrak{R}}$ and $S_{\mathfrak{T}}$ are isomorphic?

If $(S,+)=[Y;G_{\alpha};\phi_{\alpha,\beta}]$ is a Clifford semigroup in which
each group $G_{\alpha}$ has a Rota–Baxter operator $\mathfrak{R}_{\alpha}$,
the strong semilattice of the skew braces $G_{\mathfrak{R}_{\alpha}}$
coincides with the dual weak brace $S_{\mathfrak{R}}$, where $\mathfrak{R}$ is
the strong Rota–Baxter operator on $(S,+)$, as we show below.

######  Remark 4.

Let $Y$ be a (lower) semilattice, $\\{G_{\alpha}\ |\ \alpha\in Y\\}$, a family
of disjoint groups each one having a Rota–Baxter operator
$\mathfrak{R}_{\alpha}$. Thus, by 16, $\\{G_{\alpha}\ |\ \alpha\in Y\\}$ is a
family of disjoint skew braces associated to $\mathfrak{R}_{\alpha}$, for
every $\alpha\in Y$. Moreover, considered a strong semilattice
$S=[Y;G_{\alpha};\phi_{\alpha,\beta}]$ of the groups $G_{\alpha}$, by 5, the
structure $\left(S,+,\circ\right)$ is a dual weak brace, with
$a+b=\phi_{\alpha,\alpha\beta}(a)+\phi_{\beta,\alpha\beta}(b)$ and

$\displaystyle a\circ
b=\phi_{\alpha,\alpha\beta}(a)\circ\phi_{\beta,\alpha\beta}(b)=\phi_{\alpha,\alpha\beta}\left(a\right)+\mathfrak{R}_{\alpha\beta}\phi_{\alpha,\alpha\beta}\left(a\right)+\phi_{\beta,\alpha\beta}\left(b\right)-\mathfrak{R}_{\alpha\beta}\phi_{\alpha,\alpha\beta}\left(a\right),$

for all $a\in G_{\alpha}$ and $b\in G_{\beta}$. On the other hand, by 8, we
have that the map $\mathfrak{R}:S\to S$ given by
$\mathfrak{R}\left(a\right)=\mathfrak{R}_{\alpha}\left(a\right)$, for every
$a\in G_{\alpha}$, is a Rota–Baxter operator on the Clifford semigroup
$\left(S,+\right)$ if and only if (2) is satisfied. Hence, in this case, by
16, $S_{\mathfrak{R}}=(S,+,\circ_{\mathfrak{R}})$ is a weak brace and

$\displaystyle a\circ_{\mathfrak{R}}b$
$\displaystyle=a+\mathfrak{R}_{\alpha}\left(a\right)+b-\mathfrak{R}_{\alpha}\left(a\right)$
$\displaystyle=\phi_{\alpha,\alpha\beta}\left(a\right)+\phi_{\alpha,\alpha\beta}\mathfrak{R}_{\alpha}\left(a\right)+\phi_{\beta,\alpha\beta}\left(b\right)-\phi_{\alpha,\alpha\beta}\mathfrak{R}_{\alpha}\left(a\right)$
$\displaystyle=\phi_{\alpha,\alpha\beta}\left(a\right)+\mathfrak{R}_{\alpha\beta}\phi_{\alpha,\alpha\beta}\left(a\right)+\phi_{\beta,\alpha\beta}\left(b\right)-\mathfrak{R}_{\alpha\beta}\phi_{\alpha,\alpha\beta}\left(a\right)$
by (2) $\displaystyle=a\circ b,$

for all $a\in G_{\alpha}$ and $b\in G_{\beta}$.

Instances of bi-weak braces can be obtained through special Rota–Baxter
operators, as we show in the following result that is related to [27,
Proposition 5.1].

######  Proposition 17.

Let $(S,+)$ be a Clifford semigroup and consider a commutative Rota–Baxter
endomorphism $\mathfrak{R}$ of $(S,+)$. Then, the weak brace
$S_{\mathfrak{R}}$ is a bi-weak brace.

###### Proof.

If $a,b,c\in S$. Since
$\mathfrak{R}\left(a^{-}\right)=-\mathfrak{R}\left(a\right)$, we get

$\displaystyle\left(a+b\right)\circ a^{-}\circ\left(a+c\right)$
$\displaystyle=\left(a+b\right)\circ\left(-\mathfrak{R}\left(a\right)+a^{0}+c+\mathfrak{R}\left(a\right)\right)$
$\displaystyle=a+b+\mathfrak{R}\left(a\right)+\mathfrak{R}\left(b\right)-\mathfrak{R}\left(a\right)+a^{0}+c+\mathfrak{R}\left(a\right)-\mathfrak{R}\left(b\right)-\mathfrak{R}\left(a\right)$
$\displaystyle=a+b+\mathfrak{R}\left(a\right)^{0}+\mathfrak{R}\left(b\right)+c-\mathfrak{R}\left(b\right)+\mathfrak{R}\left(a\right)^{0}$
$\displaystyle=a+b+\mathfrak{R}\left(b\right)+c-\mathfrak{R}\left(b\right)$
$\displaystyle=a+(b\circ c).$

Moreover, since $S_{\mathfrak{R}}$ is a dual weak brace, it trivially holds
$a^{-}\circ a=a-a$. Therefore, the claim follows. ∎

##  5 Solutions associated to Rota–Baxter operators

In this section, we make explicit the solutions associated to dual weak braces
obtained through Rota-Baxter operators. In particular, such maps have a
behaviour near to bijectivity and non-degeneracy.

Let us recall that every weak brace $(S,+,\circ)$ gives rise to a solution
$r:S\times S\to S\times S$ defined by

$\displaystyle r\left(a,b\right)=\left(-a+a\circ b,\ \left(-a+a\circ
b\right)^{-}\circ a\circ b\right),$

for all $a,b\in S$ (see [13, Theorem 11]). In particular, the images of $r$
can be written by the maps $\lambda_{a}$ and $\rho_{b}:S\to S$ given by

$\displaystyle\rho_{b}\left(a\right)=\lambda_{a}\left(b\right)^{-}\circ a\circ
b,$

for all $a,b\in S$. The map $\rho:S\to\operatorname{Map}(S),b\mapsto\rho_{b}$
is a semigroup anti-homomorphism of the inverse semigroup
$\left(S,\circ\right)$ into the monoid $\operatorname{Map}(S)$ of the maps
from $S$ into itself.
Any solution $r$ associated to an arbitrary weak brace $(S,+,\circ)$ has a
behaviour close to bijectivity, namely $r$ is a completely regular element in
$\operatorname{Map}(S\times S)$. Indeed, considered the solution $r^{op}$
associated to the _opposite weak brace_ $S^{op}$ of $S$, i.e., the structure
$\left(S,+^{op},\circ\right)$ where $a+^{op}b:=b+a$, for all $a,b\in S$, they
hold

$\displaystyle r\,r^{op}\,r=r,\qquad
r^{op}\,r\,r^{op}=r^{op},\qquad\text{and}\qquad rr^{op}=r^{op}r.$

In general, the maps $\lambda_{a}$ and $\rho_{b}$ are not completely regular
in $\operatorname{Map}(S)$ (see [13, Example 3]), unless $S$ is a dual weak
brace. Indeed, in this case, $r$ has also a behaviour close to the non-
degeneracy because they hold

$\displaystyle\lambda_{a}\lambda_{a^{-}}\lambda_{a}=\lambda_{a},\qquad\lambda_{a^{-}}\lambda_{a}\lambda_{a^{-}}=\lambda_{a^{-}},\quad$
$\displaystyle\text{and}\qquad\lambda_{a}\lambda_{a^{-}}=\lambda_{a^{-}}\lambda_{a},$
$\displaystyle\rho_{b}\rho_{b^{-}}\rho_{b}=\rho_{b},\qquad\rho_{b^{-}}\rho_{b}\rho_{b^{-}}=\rho_{b^{-}},\qquad$
$\displaystyle\text{and}\qquad\rho_{b}\rho_{b^{-}}=\rho_{b^{-}}\rho_{b},$

for all $a,b\in S$. In particular, if $S$ is a skew brace, $r$ is bijective
with $r^{-1}=r^{op}$ and $r$ is non-degenerate.

Note that if $(S,+,\circ)$ is a dual weak brace and the map $\lambda_{a}$ is
bijective, for every $a\in S$, then $S$ is a skew brace. Hence, since the maps
$\lambda_{a}$ are completely regular, we obtain that
$\lambda_{a}^{-1}=\lambda_{a^{-}}$, for every $a\in S$. Thus,
$\lambda_{e}=\operatorname{id}_{S}$, for every $e\in E(S)$, and if
$e_{1},e_{2}\in E(S)$, we get

$\displaystyle e_{1}=\lambda_{e_{2}}(e_{1})=-e_{2}+e_{2}\circ
e_{1}=e_{2}+e_{2}+e_{1}=e_{2}+e_{1}.$

Similarly, $e_{2}=e_{1}+e_{2}$, and so $e_{1}=e_{2}$. Therefore, all the
idempotents are equal and $|E(S)|=1$.

Now, let us make explicit the solution $r$ associated to a dual weak brace
$S_{\mathfrak{R}}$ associated to a Rota–Baxter operator on a Clifford
semigroup $(S,+)$, that is

$\displaystyle r\left(a,b\right)=\left(\lambda_{a}\left(b\right),\
-\mathfrak{R}\lambda_{a}\left(b\right)-\lambda_{a}\left(b\right)+a+\lambda_{a}\left(b\right)+\mathfrak{R}\lambda_{a}\left(b\right)\right),$
(7)

for all $a,b\in S$. In fact,

$\displaystyle\rho_{b}\left(a\right)$
$\displaystyle=\lambda_{a}\left(b\right)^{-}\circ\left(a\circ
b\right)=\lambda_{a}\left(b\right)^{-}+\mathfrak{R}\left(\lambda_{a}\left(b\right)^{-}\right)+a\circ
b-\mathfrak{R}\left(\lambda_{a}\left(b\right)^{-}\right)$
$\displaystyle=\lambda_{a}\left(b\right)^{-}-\mathfrak{R}\lambda_{a}\left(b\right)+a+\lambda_{a}(b)+\mathfrak{R}\lambda_{a}\left(b\right)$
by (1)
$\displaystyle=-\mathfrak{R}\lambda_{a}\left(b\right)-\lambda_{a}(b)+\mathfrak{R}\lambda_{a}\left(b\right)^{0}+a+\lambda_{a}(b)+\mathfrak{R}\lambda_{a}\left(b\right)$
$\displaystyle=-\mathfrak{R}\lambda_{a}\left(b\right)-\lambda_{a}\left(b\right)+a+\lambda_{a}\left(b\right)+\mathfrak{R}\lambda_{a}\left(b\right),$

for all $a,b\in S$.

Now, given a Rota–Baxter $\mathfrak{R}$ on a Clifford semigroup $(S,+)$, we
show that the solution $r^{op}$ associated to the opposite weak brace
$S_{\mathfrak{R}}^{op}$ of $S_{\mathfrak{R}}$ is strictly linked to the
solution associated to $S_{\mathfrak{R}^{op}}$, where $\mathfrak{R}^{op}$ is
the opposite Rota–Baxter operator of $\mathfrak{R}$ defined by
$\mathfrak{R}^{op}(a)=-a+\mathfrak{R}(-a)$, for every $a\in S$. To this
purpose, we recall that two solutions $r,s$ on a set $S$ are said to be
_equivalent_ (or _isomorphic_) if there exists a bijection $f$ on $S$ such
that $\left(f\times f\right)r=s\left(f\times f\right)$, see [15, p. 105].

######  Proposition 18.

Let $S_{\mathfrak{R}}$ be the dual weak brace associated to a Rota–Baxter
operator $\mathfrak{R}$ on a Clifford semigroup $\left(S,+\right)$ and $r$ the
solution associated to $S_{\mathfrak{R}}$. Then,
$S_{\mathfrak{R}^{op}}=(S,+,\tilde{\circ})$ and $S_{\mathfrak{R}}^{op}$ are
isomorphic and the opposite solution $r^{op}$ of $r$ is equivalent to the
solution associated to $S_{\mathfrak{R}^{op}}$. In particular, it is given by

$\displaystyle r^{op}(a,b)$
$\displaystyle=\left(\lambda_{a}\left(b\right)^{op},\
-\mathfrak{R}\left(\lambda_{a}\left(b\right)^{op}\right)-\lambda_{a}\left(b\right)^{op}+a+\lambda_{a}\left(b\right)^{op}+\mathfrak{R}\left(\lambda_{a}\left(b\right)^{op}\right)\right),$

where
$\lambda_{a}(b)^{op}=a+\mathfrak{R}\left(a\right)+b-\mathfrak{R}\left(a\right)-a$,
for all $a,b\in S$.

###### Proof.

Let $a,b\in S$, then

$\displaystyle a\ \tilde{\circ}\
b=a^{0}+\mathfrak{R}\left(-a\right)+b-\mathfrak{R}\left(-a\right)+a=\mathfrak{R}\left(-a\right)+b-\mathfrak{R}\left(-a\right)+a=-\left(\left(-a\right)\circ\left(-b\right)\right).$

Besides, on can check that $f:S\to S$ defined by $f\left(a\right)=-a$, for
every $a\in S$, is an isomomorphism of weak braces from
$S_{\mathfrak{R}^{op}}$ to $S^{op}_{\mathfrak{R}}$.
Now, it is a routine computation to calculate the solution $r^{op}$ as in [13,
Remark 2] and, if $\tilde{r}$ is the solution associated to
$S^{op}_{\mathfrak{R}}$ as in (7), then $f$ is a bijection of $S$ satisfying
$\left(f\times f\right)r^{op}=\tilde{r}\left(f\times f\right)$, namely,
$r^{op}$ and $\tilde{r}$ are equivalent. ∎

##  6 Ideals of dual weak braces

In this section, we deal with the ideals of a dual weak brace, by extending
the already known theory for skew braces. In particular, we introduce the
notion of the socle of a dual weak brace. Finally, we describe the ideals of
dual weak braces associated to a given Rota–Baxter operator.

Inverse semigroups are a widely studied class of semigroups for which
congruences are well understood. Indeed, in this context it is possible to
mimic the group-theoretical treatment of congruences since one can find
special subsemigroups that are analogue of normal subgroups of groups. To this
end, we recall the definition of such structures below.

######  Definition 19.

[29, Definition VI.1.2] Let $S$ be a Clifford semigroup. A subset $N$ of $S$
is _normal_ if it satisfies the following conditions:

  1. 1.

$E(S)\subseteq N$,

  2. 2.

$\forall\ a\in S\quad a\in N\Longrightarrow a^{-1}\in N$,

  3. 3.

$\forall\ a,b\in S\quad a,\;a^{0}b\in N\Longrightarrow ab\in N$,

  4. 4.

$\forall\ a,b\in S\quad ab\in N\Longrightarrow ba\in N$.

Note that a subset $N$ of a group $G$ is a normal subgroup of $G$ if and only
if it is a normal subset of $G$ (see [29, Exercises VI.1.22(i)]. Given a
Clifford semigroup $S=[Y;G_{\alpha};\phi_{\alpha,\beta}]$, for each $\alpha\in
Y$, let $K_{\alpha}$ be a normal subgroup of $G_{\alpha}$ and assume that
$\phi_{\alpha,\beta}(K_{\alpha})\subseteq K_{\beta}$ if $\alpha>\beta$. Set
$\psi_{\alpha,\beta}={\phi_{\alpha,\beta}}_{|_{K_{\alpha}}}$ if
$\alpha\geq\beta$ and $K=[Y;K_{\alpha};\psi_{\alpha,\beta}]$, then $K$ is a
normal subset of $S$; conversely, every normal subset of $S$ is of this form
(see [28, Exercises III.1.9(ii)]).
Clearly, if $S$ is a Clifford semigroup, then $E(S)$ is a normal subset of
$S$. Indeed, the conditions (1), (2), and (3) are trivially satisfied;
moreover, the condition (4) follows by [29, Lemma II.2.2(iii)].

######  Remarks 5.

Let $N$ be a normal subset of a Clifford semigroup $S$.

  1. 1.

Observe that $N$ is a subsemigroup of $S$. Indeed, if $a,b\in N$, then, by
$1.$ of 19, we get $b^{0}a^{0}\in N$ and, by $3.$, $ba^{0}\in N$, thus
$a^{0}b\in N$. Hence, by $3.$, it follows that $ab\in N$.

  2. 2.

If $n\in N$ and $a\in S$, then $a^{-1}na\in N$. In fact, since $a^{0}\in N$,
by $1.$, we have that $\left(na\right)a^{-1}\in N$ and, by $4.$ in 19,
$a^{-1}na\in N$.

The following definition is essential for obtaining a quotient structure and
it is consistent with that given for skew braces (see [22, Definition 2.1]).

######  Definition 20.

Let $(S,+,\circ)$ be a dual weak brace. Then, a subset $I$ of $S$ is an
_ideal_ of $S$ if the following hold:

  1. 1.

$I$ is a normal subset of $\left(S,+\right)$,

  2. 2.

$\lambda_{a}\left(I\right)\subseteq I$, for every $a\in S$,

  3. 3.

$I$ is a normal subset of $\left(S,\circ\right)$.

Observe that if $S$ is a dual weak brace, then $S$ and $E(S)$ are trivial
ideals of $S$.

######  Theorem 21.

Let $I$ be an ideal of a dual weak brace $(S,+,\circ)$. Then, the relation
$\sim_{I}$ on $S$ given by

$\displaystyle\forall\ a,b\in S\quad a\sim_{I}b\ \iff\ a^{0}=b^{0}\ \
\text{and}\ \ -a+b\in I,$

is a congruence of $(S,+,\circ)$. Moreover, $S/\sim_{I}$ is a dual weak brace
with semilattice of idempotents isomorphic to $E\left(S\right)$.

###### Proof.

Initially, we prove that

$\displaystyle a\sim_{I}b\ \iff\ a^{0}=b^{0}\ \ \text{and}\ \ a^{-}\circ b\in
I,$

for all $a,b\in S$. Indeed, if $a,b\in S$ are such that $a\sim_{I}b$, then,
clearly $a^{0}=b^{0}$, and, since $\lambda_{x}\left(x^{-}\right)=-x$, for
every $x\in S$ (see [13, Lemma 2]), we get

$a^{-}\circ
b=a^{-}+\lambda_{a^{-}}\left(b\right)=\lambda_{a^{-}}\left(-a+b\right)\in I.$

Conversely, if $a,b\in S$ are such that $a^{0}=b^{0}$ and $a^{-}\circ b\in I$,
then

$\displaystyle-a+b=\lambda_{a}\left(a^{-}\right)+\lambda_{b^{0}}\left(b\right)=\lambda_{a}\left(a^{-}\right)+\lambda_{a^{0}}\left(b\right)=\lambda_{a}\left(a^{-}+\lambda_{a^{-}}\left(b\right)\right)=\lambda_{a}\left(a^{-}\circ
b\right)\in I.$

Now, by [29, Lemma VI.3.1], [29, Exercise VI.2.13(i)], and [25, Theorem
4.2.1], $\sim_{I}$ is a congruence of the dual weak brace
$\left(S,+,\circ\right)$ whose both quotient structures are Clifford
semigroups. The remaining parts of the statement are for immediate
verification, so the proof is complete. ∎

The following result describes the ideals of dual weak braces associated to
Rota–Baxter operators and it consistent with [3, Proposition 5.4(a)]. Into the
specific, the request of $\lambda$-invariance is redundant.

######  Proposition 22.

Let $S_{\mathfrak{R}}$ be the dual weak brace obtained by a Rota–Baxter
operator $\mathfrak{R}$ on a Clifford semigroup $(S,+)$ and $I\subseteq S$.
Then, $I$ is an ideal of $S_{\mathfrak{R}}$ if and only if $I$ both is a
normal subset of the Clifford semigroups $(S,+)$ and $(S,\circ)$.

###### Proof.

Assume that $I$ is a normal subset both of $(S,+)$ and $(S,\circ)$. Thus,
$-a+x+a\in I$, for all $x\in I$ and $a\in S$. Indeed, since $a^{0}\in I$, it
follows that $a^{0}+x=a+(-a+x)\in I$ and so, by $4.$ in 19, $-a+x+a\in I$.
Hence, if $a\in S$ and $x\in I$, we obtain that

$\displaystyle\lambda_{a}\left(x\right)=-a+a\circ
x=a^{0}+\mathfrak{R}\left(a\right)+x-\mathfrak{R}\left(a\right)\in I.$

Therefore, since the converse is trivial, the claim follows. ∎

If $S$ is a dual weak brace, we call _socle_ of $S$ the following set

$\displaystyle\operatorname{Soc}\left(S\right)=\\{a\,|\,a\in S,\,\,\forall\
b\in S\quad a+b=a\circ b\quad\text{and}\quad a+b=b+a\\}.$

Clearly, the previous definition includes that already known in the context of
skew braces in [22] originally introduced in [32]. Moreover, by the following
result, we can always consider the quotient structure of a dual weak brace by
its socle.

######  Proposition 23.

Let $(S,+,\circ)$ be a dual weak brace. Then, $\operatorname{Soc}(S)$ is an
ideal of $S$.

###### Proof.

Initially, note that $E(S)\subseteq\operatorname{Soc}(S)$. Indeed, if $e\in
E(S)$ and $a\in S$, it is sufficient to show that

$\displaystyle e+a=e\circ a.$ (8)

In fact, $e\circ a=e+\lambda_{e}\left(a\right)=e-e+e\circ
a=\lambda_{e}\left(a\right),$ and, by 3,

$\displaystyle e+a=e\circ\lambda_{e^{-}}\left(a\right)=e\circ e^{-}\circ
a=e\circ a.$

Now, let us prove that
$\lambda_{x}\left(\operatorname{Soc}\left(S\right)\right)\subseteq\operatorname{Soc}\left(S\right)$,
for every $x\in S$. If $a\in\operatorname{Soc}(S)$ and $b\in S$, we obtain

$\displaystyle\lambda_{x}(a)\circ b$
$\displaystyle=x\circ\left(a+x^{-}\right)\circ b=x\circ\left(a\circ x^{-}\circ
b\right)=x\circ\left(a+x^{-}\circ b\right)$ $\displaystyle=x\circ
a-x+x^{0}\circ b$ $\displaystyle=x\circ a-x+x^{0}+b$ by (8)
$\displaystyle=x\circ(a+x^{-})+b=\lambda_{x}(a)+b$

and, on the other hand, by the previous computations,

$\displaystyle\lambda_{x}(a)+b$ $\displaystyle=\lambda_{x}(a)\circ b=x\circ
a\circ x^{-}\circ b=b^{0}\circ x\circ\left(a\circ x^{-}\circ b\right)$
$\displaystyle=b^{0}\circ x\circ\left(a+x^{-}\circ
b\right)=b\circ\left(b^{-}\circ x\circ\left(x^{-}\circ b+a\right)\right)$
$\displaystyle=b\circ\lambda_{b^{-}\circ x}\left(a\right)$
$\displaystyle=b+\lambda_{x}(a)$ by 3

Now, it is a routine computation to prove that $\operatorname{Soc}(S)$ is
closed with respect to the $\circ$. Hence, if $a,a^{0}\circ
b\in\operatorname{Soc}(S)$, then $a\circ b=a\circ a^{0}\circ b\in Soc(S)$.
Moreover, if $a\in\operatorname{Soc}(S)$, then
$a^{-}\in\operatorname{Soc}(S)$. Indeed, we have

$\displaystyle a^{-}+b$ $\displaystyle=a^{-}\circ\left(-a+a\circ b\right)$ by
3 $\displaystyle=a^{-}\circ\left(-a+a+b\right)$ $\displaystyle=a^{-}\circ
a^{0}\circ b$ by (8) $\displaystyle=a^{-}\circ b.$

Considering $b=a$, it follows that

$\displaystyle a^{-}+a=a^{0}.$ (9)

Thus, we get

$\displaystyle b+a^{-}$
$\displaystyle=b+a^{-}\circ\left(a^{-}\right)^{0}=b+a^{-}\circ a^{0}$
$\displaystyle=b+a^{-}+a^{0}$ by (8) $\displaystyle=a^{-}\circ a+b+a^{-}$
$\displaystyle=a^{-}+a+b+a^{-}$ by (9) $\displaystyle=a^{-}+b+a+a^{-}$
$\displaystyle=a^{-}+b+a^{0}$ by (9) $\displaystyle=a^{-}+b,$

and so $a^{-}\in\operatorname{Soc}(S)$. It immediately follows that
$-a=\lambda_{a}\left(a^{-}\right)\in\operatorname{Soc}(S)$. Finally, if
$a\circ b\in\operatorname{Soc}(S)$,

$\displaystyle b\circ a=b\circ\left(a\circ b\circ
b^{-}\right)=b\circ\left(a\circ b+b^{-}\right)=b\circ\left(b^{-}+a\circ
b\right)=\lambda_{b}(a\circ b),$

thus $b\circ a\in\operatorname{Soc}(S)$, i.e., $\operatorname{Soc}(S)$ is a
normal subset of $(S,\circ)$.
It is easy to prove that $\operatorname{Soc}(S)$ is closed with respect to the
$+$, hence if $a,a^{0}+b\in\operatorname{Soc}(S)$, clearly
$a+b\in\operatorname{Soc}(S)$. To conclude the proof, we observe that if
$a+b\in\operatorname{Soc}(S)$, then

$\displaystyle b+a=-a+(a+b)+a=-a+a+(a+b)=a+b\in\operatorname{Soc}(S).$

Therefore, the statement is proved. ∎

The following proposition provides a method to obtain ideals of dual weak
braces starting from two given ones.

######  Proposition 24.

Let $S$ be a dual weak brace and $I,J$ ideals of $S$. Then, $I+J$ and $I\circ
J$ are ideals of $S$.

###### Proof.

Initially, by $2.$ in 5, one can check that $I+J=\\{i+j\ |\ i\in I\
\text{and}\ j\in J\\}$ is closed with respect to $+$. Moreover,
$\lambda_{a}(i+j)=\lambda_{a}(i)+\lambda_{a}(j)\in I+J$, for every $a\in S$.
Hence, by (1), it also follows that $I+J$ is closed with respect to $\circ$
since, if $a,b\in I+J$, $a\circ b=a+\lambda_{a}\left(b\right)\in I+J$. In
addition, it is clear that $E(S)\subseteq I+J$.
Now, let us show that $I+J$ is a normal subset of $(S,+)$. If $a=i+j\in I+J$,
then $-a=\left(-j-i+j\right)-j\in I+J$. Moreover, if $a\in I+J$ and
$a^{0}+b\in I+J$, then

$\displaystyle a+b=a+a^{0}+b\in I+J$

and if $a+b=i+j\in I+J$, it follows that

$\displaystyle b+a=a^{0}+b+a=-a+\left(i+j\right)+a=-a+i+a-a+j+a\in I+J.$

Thus, $I+J$ is a normal subset of $(S,+)$. Finally, we prove that $I+J$ is a
normal subset of $(S,\circ)$. Now, observe that
$a^{-}=-\lambda_{a^{-}}\left(a\right)\in I+J$ and, if $a,a^{0}\circ b\in I+J$,
then

$\displaystyle a\circ b=a\circ a^{0}\circ b\in I+J.$

Furthermore, if $a\circ b=i+j\in I+J$, by (1) and 3, we obtain that

$\displaystyle b\circ a=a^{-}\circ(i+j)\circ a=a^{-}\circ
i\circ\lambda_{i^{-}}\left(j\right)\circ a=\left(a^{-}\circ i\circ
a\right)\circ\left(a^{-}\circ\lambda_{i^{-}}\left(j\right)\circ a\right)\in
I+J,$

therefore $I+J$ is an ideal of $S$.
To prove the second part of the statement, note that $I\circ J=\\{i\circ j\ |\
i\in I\ \text{and}\ j\in J\\}$ is closed with respect to $\circ$. Besides, if
$a,b\in I\circ J$, it follows that $a+b=a\circ\lambda_{a^{-}}\left(b\right)\in
I\circ J$, hence $I\circ J$ is closed with respect to $+$. Moreover, by (1),
we have that $\lambda_{a}(i\circ j)=\lambda_{a}(i)+\lambda_{a\circ i}(j)\in
I\circ J$, for every $a\in S$. In addition, it is clear that $E(S)\subseteq
I+J$.
Now, let us show that $I\circ J$ is a normal subset of $(S,+)$. If $a=i\circ
j\in I\circ J$, then
$-a=\lambda_{i}\left(-j\right)-i=-i+\left(i+\lambda_{i}\left(-j\right)-i\right)\in
I\circ J$. Moreover, if $a\in I\circ J$ and $a^{0}+b\in I\circ J$, then
$a+b=a+a^{0}+b\in I\circ J$ and if $a+b=i\circ j\in I\circ J$, it follows that

$\displaystyle b+a$ $\displaystyle=a^{0}+b+a=-a+i\circ
j+a=-a+i+\lambda_{i}\left(j\right)+a$
$\displaystyle=\left(-a+i+a\right)+\left(-a+\lambda_{i}\left(j\right)+a\right)\in
I\circ J.$

Thus, $I\circ J$ is a normal subset of $(S,+)$. Finally, we prove that $I\circ
J$ is a normal subset of $(S,\circ)$. Now, observe that $a^{-}=j^{-}\circ
i^{-}=i^{-}\circ\left(i\circ j^{-}\circ i^{-}\right)\in I\circ J$ and, if
$a,a^{0}\circ b\in I\circ J$, then

$\displaystyle a\circ b=a\circ a^{0}\circ b=a+\lambda_{a}\left(a^{0}\circ
b\right)\in I\circ J$

and, if $a\circ b=i\circ j\in I\circ J$,

$\displaystyle b\circ a=a^{-}\circ(i\circ j)\circ a=\left(a^{-}\circ i\circ
a\right)\circ\left(a^{-}\circ j\circ a\right)\in I\circ J,$

which completes the proof. ∎

In light of the results obtained in this section, it is interesting a further
deepening of the dual weak brace theory. Our intent is to continue such an
investigation in future work.

## Acknowledgement

We would like to thank the referee for the question related to 3 that allows
us to extend the result to weak braces that are not necessarily dual.

## References

  * [1] F. V. Atkinson, Some aspects of Baxter’s functional equation, J. Math. Anal. Appl. 7 (1963) 1–30.
URL https://doi.org/10.1016/0022-247X(63)90075-1

  * [2] V. G. Bardakov, V. Gubarev, Rota-Baxter operators on groups, arXiv preprint arXiv:2103.01848.
  * [3] V. G. Bardakov, V. Gubarev, Rota-Baxter groups, skew left braces, and the Yang-Baxter equation, J. Algebra 596 (2022) 328–351.
URL https://doi.org/10.1016/j.jalgebra.2021.12.036

  * [4] G. Baxter, An analytic problem whose solution follows from a simple algebraic identity, Pacific J. Math. 10 (1960) 731–742.
URL http://projecteuclid.org/euclid.pjm/1103038223

  * [5] R. J. Baxter, Partition function of the eight-vertex lattice model, Ann. Physics 70 (1972) 193–228.
URL https://doi.org/10.1016/0003-4916(72)90335-1

  * [6] T. Brzeziński, Towards semi-trusses, Rev. Roumaine Math. Pures Appl. 63 (2) (2018) 75–89.
  * [7] A. Caranti, L. Stefanello, Skew braces from Rota–Baxter operators: a cohomological characterisation and some examples, Ann. Mat. Pura Appl. (2022).
URL https://doi.org/10.1007/s10231-022-01230-w

  * [8] P. Cartier, On the structure of free Baxter algebras, Advances in Math. 9 (1972) 253–265.
URL https://doi.org/10.1016/0001-8708(72)90018-7

  * [9] F. Catino, Factorizable semigroups, Semigroup Forum 36 (2) (1987) 167–174.
URL https://doi.org/10.1007/BF02575013

  * [10] F. Catino, F. Cedó, P. Stefanelli, Nilpotency in left semi-braces, J. Algebra 604 (2022) 128–161.
URL https://doi.org/10.1016/j.jalgebra.2022.04.004

  * [11] F. Catino, I. Colazzo, P. Stefanelli, Semi-braces and the Yang-Baxter equation, J. Algebra 483 (2017) 163–187.
URL https://doi.org/10.1016/j.jalgebra.2017.03.035

  * [12] F. Catino, I. Colazzo, P. Stefanelli, Set-theoretic solutions to the Yang-Baxter equation and generalized semi-braces, Forum Math. 33 (3) (2021) 757–772.
URL https://doi.org/10.1515/forum-2020-0082

  * [13] F. Catino, M. Mazzotta, M. M. Miccoli, P. Stefanelli, Set-theoretic solutions of the Yang-Baxter equation associated to weak braces, Semigroup Forum 104 (2) (2022) 228–255.
URL https://doi.org/10.1007/s00233-022-10264-8

  * [14] F. Catino, M. Mazzotta, P. Stefanelli, Inverse semi-braces and the Yang-Baxter equation, J. Algebra 573 (2021) 576–619.
URL https://doi.org/10.1016/j.jalgebra.2021.01.009

  * [15] F. Cedó, E. Jespers, J. Okniński, Braces and the Yang-Baxter equation, Comm. Math. Phys. 327 (1) (2014) 101–116.
URL https://doi.org/10.1007/s00220-014-1935-y

  * [16] F. Cedó, A. Smoktunowicz, L. Vendramin, Skew left braces of nilpotent type, Proc. Lond. Math. Soc. (3) 118 (6) (2019) 1367–1392.
URL https://doi.org/10.1112/plms.12209

  * [17] L. N. Childs, Fixed-point free endomorphisms and Hopf Galois structures, Proc. Amer. Math. Soc. 141 (4) (2013) 1255–1265.
URL https://doi.org/10.1090/S0002-9939-2012-11418-2

  * [18] L. N. Childs, Bi-skew braces and Hopf Galois structures, New York J. Math. 25 (2019) 574–588.
URL http://nyjm.albany.edu/j/2019/25-26p.pdf

  * [19] A. H. Clifford, G. B. Preston, The algebraic theory of semigroups. Vol. I, Mathematical Surveys, No. 7, American Mathematical Society, Providence, R.I., 1961\.
  * [20] I. Colazzo, E. Jespers, A. Van Antwerpen, C. Verwimp, Left non-degenerate set-theoretic solutions of the Yang-Baxter equation and semitrusses, J. Algebra 610 (2022) 409–462.
URL https://doi.org/10.1016/j.jalgebra.2022.07.019

  * [21] V. G. Drinfel′ d, On some unsolved problems in quantum group theory, in: Quantum groups (Leningrad, 1990), vol. 1510 of Lecture Notes in Math., Springer, Berlin, 1992, pp. 1–8.
URL https://doi.org/10.1007/BFb0101175

  * [22] L. Guarnieri, L. Vendramin, Skew braces and the Yang-Baxter equation, Math. Comp. 86 (307) (2017) 2519–2534.
URL https://doi.org/10.1090/mcom/3161

  * [23] L. Guo, An introduction to Rota-Baxter algebra, vol. 4 of Surveys of Modern Mathematics, International Press, Somerville, MA; Higher Education Press, Beijing, 2012.
  * [24] L. Guo, H. Lang, Y. Sheng, Integration and geometrization of Rota-Baxter Lie algebras, Adv. Math. 387 (2021) Paper No. 107834, 34.
URL https://doi.org/10.1016/j.aim.2021.107834

  * [25] J. M. Howie, Fundamentals of semigroup theory, vol. 12 of London Mathematical Society Monographs. New Series, The Clarendon Press, Oxford University Press, New York, 1995, Oxford Science Publications.
  * [26] E. Jespers, A. Van Antwerpen, Left semi-braces and solutions of the Yang-Baxter equation, Forum Math. 31 (1) (2019) 241–263.
URL https://doi.org/10.1515/forum-2018-0059

  * [27] A. Koch, Abelian maps, bi-skew braces, and opposite pairs of Hopf-Galois structures, Proc. Amer. Math. Soc. Ser. B 8 (2021) 189–203.
URL https://doi.org/10.1090/bproc/87

  * [28] M. Petrich, Inverse semigroups, Pure and Applied Mathematics (New York), John Wiley & Sons, Inc., New York, 1984, a Wiley-Interscience Publication.
  * [29] M. Petrich, N. R. Reilly, Completely regular semigroups, vol. 23 of Canadian Mathematical Society Series of Monographs and Advanced Texts, John Wiley & Sons, Inc., New York, 1999, a Wiley-Interscience Publication.
  * [30] G.-C. Rota, Baxter algebras and combinatorial identities. I, II, Bull. Amer. Math. Soc. 75 (1969), 325–329; ibid. 75 (1969) 330–334.
URL https://doi.org/10.1090/S0002-9904-1969-12156-7

  * [31] J. J. Rotman, An introduction to the theory of groups, vol. 148 of Graduate Texts in Mathematics, 4th ed., Springer-Verlag, New York, 1995.
URL https://doi.org/10.1007/978-1-4612-4176-8

  * [32] W. Rump, A decomposition theorem for square-free unitary solutions of the quantum Yang-Baxter equation, Adv. Math. 193 (1) (2005) 40–55.
URL https://doi.org/10.1016/j.aim.2004.03.019

  * [33] C. N. Yang, Some exact results for the many-body problem in one dimension with repulsive delta-function interaction, Phys. Rev. Lett. 19 (1967) 1312–1315.
URL https://doi.org/10.1103/PhysRevLett.19.1312