# Set-theoretic solutions to the Yang–Baxter equation and generalized semi-braces

• Francesco Catino , Ilaria Colazzo and Paola Stefanelli
From the journal Forum Mathematicum

## Abstract

This paper aims to introduce a construction technique of set-theoretic solutions of the Yang–Baxter equation, called strong semilattice of solutions. This technique, inspired by the strong semilattice of semigroups, allows one to obtain new solutions. In particular, this method turns out to be useful to provide non-bijective solutions of finite order. It is well-known that braces, skew braces and semi-braces are closely linked with solutions. Hence, we introduce a generalization of the algebraic structure of semi-braces based on this new construction technique of solutions.

## 1 Introduction

The quantum Yang–Baxter equation appeared in the work of Yang [49] and Baxter [2]. It is one of the basic equations in mathematical physics, and it laid the foundations of the theory of quantum groups [33]. Solutions of the Yang–Baxter equation are instrumental in the construction of semisimple Hopf algebras [23, 43] and provide examples of coloring invariants in knot theory [41]. More recently, the Yang–Baxter solution popped up in the theory of quantum computation [34, 50], where solutions of the Yang–Baxter equation provide so-called universal gates. One of the central open problems is to find all solutions of the Yang–Baxter equation. Let V be a vector space over a field K. Then a solution of the Yang–Baxter equation is a linear map R:VVVV for which the following holds on V3:

(RidV)(idVR)(RidV)=(idVR)(RidV)(idVR).

The simplest solutions are the solutions R induced by a linear extension of a mapping r:X×XX×X, where X is a basis for V, satisfying the set-theoretic version of the Yang–Baxter equation, i.e., satisfying the following on X3:

(r×idX)(idX×r)(r×idX)=(idX×r)(r×idX)(idX×r).

In this case, r is said to be a set-theoretic solution of the Yang–Baxter equation (briefly, a solution). Drinfel’d, in [22], posed the question of finding these set-theoretic solutions. Denote for x,yX the element r(x,y)=(λx(y),ρy(x)). One says that a set-theoretic solution r is left non-degenerate if λx is bijective for every xS, right non-degenerate if ρy is bijective for every yS, and non-degenerate if r is both left and right non-degenerate. If a solution is neither left nor right non-degenerate, then it is called degenerate. The first papers on set-theoretic solutions are those of Etingof, Schedler and Soloviev [24] and Gateva-Ivanova and Van den Bergh [27]. Both papers considered involutive solutions, i.e., solutions r where r2=id. Rump [44] introduced a new algebraic structure, braces, that generalizes radical rings and provides an algebraic framework. We provide the equivalent definition formulated by Cedó, Jespers and Okniński [12]. A triple (B,+,) is called a left brace if (B,+) is an abelian group and (B,) is a group such that for any a,b,cB it holds that

($\diamond$)a(b+c)=ab-a+ac.

This new structure showed connections between the Yang–Baxter equation and ring theory, flat manifolds, orderability of groups, Garside theory, and regular subgroups of the affine group; see, for instance, [5, 6, 14, 15, 21, 25, 46]. Lu, Yan, and Zhu [38] and Soloviev [47] started the study of non-degenerate bijective solutions, not necessarily involutive. Almost all of the ideas used in the theory of involutive solutions can be transported to non-involutive solutions. We also mention Gateva-Ivanova and Majid [26] who studied and characterized most general set-theoretic solutions (braided sets) (X,r) in terms of the induced left and right actions of X on itself, and in terms of abstract matched pair properties of the associated braided monoid S(X,r). The algebraic framework now is provided by skew left braces [28]. Let (B,+) and (B,) be groups on the same set B. If, for any a,b,cB, condition ($\diamond$) holds, the triple (B,+,) is called a skew left brace. Skew left braces and some of their applications are intensively studied; see, for instance, [8, 13, 20, 35, 45].

In [37], Lebed drew the attention on idempotent solutions. Indeed, using idempotent solutions and graphical calculus from knot theory, she provides a unifying tool to deal with several diverse algebraic structures such as free monoids, free commutative monoids, factorizable monoids, plactic monoids and distributive lattice. Examples and classifications of these solutions have been provided by Matsumoto and Shimizu [39] and by Stanovskỳ and Vojtěchovskỳ [48]. Moreover, Cvetko-Vah and Verwimp [19] provided cubic solutions with skew lattices. A cubic solution r is a solution such that r3=r; hence, this class includes both involutive and idempotent solutions. More generally, a more systematic approach to the study of solutions with finite order can be found in the recent [9, 10, 11]. Catino, Colazzo, and Stefanelli [7] and Jespers and Van Antwerpen [32] introduced the algebraic structure called left semi-brace to deal with solutions that are not necessarily non-degenerate or that are idempotent. Let (B,+) be a semigroup and let (B,) be a group. Then (B,+,) is called a left semi-brace if, for any a,b,cB, it holds that

a(b+c)=ab+a(a-+c),

where a- denotes the inverse a in (B,). If (B,+) is a left cancellative semigroup, then we call (B,+,) a left cancellative left semi-brace. This was the original definition by Catino, Colazzo and Stefanelli [7]. It has been shown that left semi-braces, under some mild assumption, provide set-theoretic solutions of the Yang–Baxter equation. Moreover, the associated solution is left non-degenerate if and only if the left semi-brace is left cancellative.

Out of algebraic interest, Brzeziński introduced left trusses [4] and left semi-trusses [3]. A quadruple (B,+,,λ) is called a left semi-truss if both (B,+) and (B,) are semigroups and λ:B×BB is a map such that a(b+c)=(ab)+λ(a,c). Clearly, the class of left semi-trusses contains all left semi-braces, rings, associative algebras and distributive lattices. This entails that it will prove difficult to present deep results on this class. However, one may examine large subclasses. In particular, Brzeziński [4] focused on left semi-trusses with a left cancellative semigroup (B,+) and a group (B,), and showed that such a left semi-truss is equivalent with a left cancellative semi-brace, and thus providing set-theoretic solutions of the Yang–Baxter equation, albeit known ones. In [40], Miccoli introduced almost left semi-braces, a particular instance of left semi-trusses, and constructed set-theoretic solutions associated with this algebraic structure. In [18], Colazzo and Van Antwerpen continued this study focusing on the subclass of brace-like left semi-trusses, i.e., left semi-trusses in which the multiplicative semigroup is a group and which includes almost left semi-braces. Concerning solutions, they showed that the solution one can associate with an almost left semi-brace is already the associated solution of a left semi-brace. In particular, this shows that brace-like left semi-trusses will not yield a universal algebraic structure that produces set-theoretic solutions.

In this paper, we focus on a new algebraic structure that includes left semi-braces and is an instance of left semi-trusses, which is on a different path with respect to brace-like left semi-trusses, called generalized left semi-brace. A triple (S,+,) is called a generalized left semi-brace if (S,+) is a semigroup, (S,) is a completely regular semigroup (or union of groups), and such that, for any a,b,cS, it holds

a(b+c)=ab+a(a-+c),

where a- denotes the (group) inverse of a in (B,). We prove that, under some mild assumptions, generalized left semi-braces provide solutions. In particular, elementary examples of generalized left semi-braces produce cubic solutions that cannot be obtained by skew lattices and left semi-braces. Also, we introduce a construction technique that provides generalized left semi-braces called the strong semilattice of generalized left semi-braces. This technique is inspired by the description of semigroups which are unions of groups due to Clifford [16].

Furthermore, we introduce a construction technique for solutions called the strong semilattice of solutions. This technique takes a family of disjoint sets {XααY} indexed by a semilattice Y and solutions defined on these sets. Then, under some assumptions of compatibility, it allows one to construct a solution on the union of the sets Xα. We prove that the solutions provided by the strong semilattice of left semi-braces are a particular instance of a strong semilattice of solutions.

Finally, we prove that the strong semilattice of solutions is a useful tool to provide solutions of finite order. Indeed, the strong semilattice of solutions of finite order is a solution of finite order. Moreover, a solution r is of finite order if there exist a non-negative integer i and a positive integer p such that rp+i=ri, and the minimal integers that satisfy such relation are said to be index and period, respectively. We show that it is possible to determine the index and the period of the semilattice of solutions {rααY} as a function of the indexes and periods of rα. As a corollary of this result, we prove that solutions associated with strong semilattices of left semi-braces are not bijective, so they are clearly different from solutions obtained by left semi-braces.

## 2 Basic tools on left semi-braces

Let us briefly present some basic background information regarding left semi-braces. Most of the content of this section appears in [32]. In particular, we provide a different proof of [32, Corollary2.9] based on a result in semigroup theory due to Hickey [29] that gives a clear description of completely regular semigroups with middle units. Moreover, we add further information on the behavior of middle units of the additive semigroup of a left semi-brace. Finally, we present concrete examples of left semi-braces.

Let us start by recalling the definition of left semi-braces.

## Definition 2.1.

Let B be a set with two operations + and such that (B,+) is a semigroup and (B,) is a group. Then (B,+,) is said to be a left semi-brace if

a(b+c)=ab+a(a-+c)

for all a,b,cB, where a- is the inverse of a in (B,).

Throughout, 0 denotes the identity of the group (B,). Moreover, we call (B,+) and (B,) the additive semigroup and the multiplicative group of the left semi-brace (B,+,), respectively. Furthermore, if the semigroup (B,+) has a pre-fix, pertaining to some property of the semigroup, we will also use this pre-fix with the left semi-brace. Hence, the left semi-braces introduced in [7], where one works under the restriction that the semigroup (B,+) is left cancellative, will be called left cancellative left semi-braces.

Now, we recall that an element u of an arbitrary semigroup (S,+) is a middle unit of S if a+u+b=a+b for all a,bS. Thus, u+u is idempotent, but u itself need not be idempotent (see [17, p. 98]). This is not the case for the element 0 in the additive semigroup of a left semi-brace: the following proposition shows that 0 is an idempotent middle unit.

## Proposition 2.2.

Let B be a left semi-brace. Then the following assertions hold:

1. 0 is a middle unit of (B,+).

2. 0 is an idempotent of (B,+).

3. B=B+B.

4. B+0 is a subgroup of (B,).

5. 0+B is a subsemigroup of (B,).

## Proof.

(i) See [32, Lemma 2.4 (1)].

(ii) Since, by (i), 0+0+0=0+0, we have that

0=(0+0)-(0+0)
=(0+0)-0+(0+0)-(0+0+0)
=(0+0)-+(0+0)-(0+0)
=(0+0)-+0.

Thus, since 0 is a middle unit, we obtain

0+0=(0+0)-+0+0=(0+0)-+0=0.

(iii) If bB, by (ii) we have that

b=b0=b(0+0)=b0+b(b-+0)B+B.

(iv) By (ii), it is clear that B+0 is not empty. Moreover, if a,bB then, using (i), we get the equalities

(a+0)-(b+0)=(a+0)-b+(a+0)-(a+0+0)
=(a+0)-b+(a+0)-(a+0)
=(a+0)-b+0B+0.

(v) See [32, Lemma 2.6 (iii)]. ∎

To show the following theorem, let us recall that an arbitrary semigroup (S,+) is said to be a rectangular group if it is isomorphic to the direct product of a group and a rectangular band. For more background and details on this topic, we refer the reader to [17].

## Theorem 2.3.

Let B be a completely simple left semi-brace. Then the additive semigroup (B,+) of B is a rectangular group.

## Proof.

The thesis follows by [29, Corollary 3.5], which states that any completely simple semigroup with a middle unit is a rectangular group. ∎

A special case in which the additive semigroup is completely simple is when 0+B is a subgroup of (B,) (see [32, Theorem 2.8]). For instance, this is the case when B is finite.

The set of idempotents of a semigroup S will be denoted by E(S). As a consequence of Theorem 2.3, we have that the additive semigroup (B,+) can be written as

B=I+G+Λ,

that is, the direct sum of the left zero semigroup I:=E(B+0), the group G:=0+B+0, and the right zero semigroup Λ:=E(0+B). Moreover, the set of idempotents is E(B)=I+Λ and it is a rectangular band.

For the sake of completeness, let us introduce a further property of middle units of left semi-braces that holds without restrictions on the additive semigroup. At first, we recall that the additive semigroup of a left semi-brace B does not contain a zero element if B has at least two elements (see [32, Lemma 2.3]).

In the following, we prove that middle units of the additive structure of an arbitrary left semi-brace are idempotents.

## Proposition 2.4.

Let B be a left semi-brace. Then every middle unit eB is an idempotent of the semigroup (B,+).

## Proof.

Since 0 is idempotent, we have that

e=e0=e(0+0)=e0+e(e-+0)=e+e(e-+0).

Since e is a middle unit, it follows that

e+e=e+e+e(e-+0)=e+e(e-+0)=e,

which is our assertion. ∎

Following Ault’s paper [1, Theorem 1.8], by Proposition 2.4, we have that the additive semigroup of any left semi-brace contains a subsemigroup MB, called the semigroup of middle units of B, that is explicitly given by

MB={xxB,x has inverse x with x+x,x+x middle units},

where the inverse x of x means that x=x+x+x and x=x+x+x.

Then MB is a subsemigroup of (B,+) that is a rectangular group. This is an interesting substructure of a left semi-brace, which is beyond the purpose of this paper and shall be studied elsewhere.

Now, having as reference [7, Example 2], we provide the following class of examples of completely simple left semi-braces that allow one to obtain solutions.

## Example 2.5.

Let (B,) be a group with identity 0, and let f,g be idempotent endomorphisms of (B,) such that fg=gf. Let us consider the following operation:

a+b:=bfg(b-)f(a)

for all a,bB. It is easy to check that (B,+,) is a completely simple left semi-brace. Now, observe that the map ρ is an anti-homomorphism from the group (B,) into the monoid BB, where BB denotes the monoid of the functions from B into itself. Indeed, ρb is given by

ρb(a)=(a-+b)-b=(bfg(b-)f(a-))-b=f(a)fg(b).

Moreover,

ρbρa(c)=ρb(f(c)fg(a))=f2(c)f2g(a)fg(b)=f(c)fg(ab)=ρab(c)

for all a,b,cB. Thus, by [32, Proposition 2.14], the semigroup (B,+) is completely simple.

In addition, since ρ is an anti-homomorphism of the group (B,), we obtain that the map r:B×BB×B defined by r(a,b)=(λa(b),ρb(a)) in [32, Theorem 5.1] is a solution. Note that λa(b)=abfg(b-)f(a-). Therefore, r is explicitly given by

r(a,b)=(abf(g(b-)a-),f(ag(b)))

for all a,bB.

Let us examine special cases of the previous class of examples. Firstly, observe that if fid and g is not the constant map of value 0, then the semigroup (B,+) is neither left nor right cancellative. Moreover, note the following three cases:

Case 1. If g is the constant map of value 0, then

a+b=bf(0)f(a)=bf(a)

for all a,bB, i.e., B coincides with the left cancellative left semi-brace provided in [7, Example 2]. Moreover, the solution associated to B is given by

r(a,b)=(abf(a-),f(a)).

Case 2. If f=id, then

a+b=bg(b-)a

for all a,bB. In this case, the semigroup (B,+) is right cancellative. Indeed, if a+b=c+b, it follows that bg(b-)a=bg(b-)c, and so a=c. Moreover, it is easy to check that B is both a right and left semi-brace. In addition, note that the solution associated to B is given by

r(a,b)=(abg(b-)a-,ag(b)).

Case 3. If f=g, then

a+b=bf2(b-)f(a)=bf(b-)f(a)

for all a,bB. Note that if aB, it holds

a+a=af(a-)f(a)=a,

i.e., every element is idempotent with respect to the sum. In this case, the semigroup (B,+) is a rectangular band where kerf=0+B and imf=B+0. Moreover, the solution r associated to B, given by

r(a,b)=(abf(b-a-),f(ab)),

is an idempotent solution, consistently with [32, Theorem 5.1] in the case in which G={0}.

The following is a class of examples of completely simple left semi-braces that, under suitable assumptions, give rise to solutions.

## Example 2.6.

Let G, H be two groups, let B:=G×H and consider the group (B,) where

(a,u)(b,v)=(aub,ubv)

for all (a,u),(b,v)G×H, i.e., the classical Zappa–Szép product of G and H (see [36]) that has identity (1,1). Let φ be a map from G into H such that φ(1)=1 and define the following operation on B:

(a,u)+(b,v)=(a,uφ(b)v)

for all (a,u),(b,v)G×H. It is easy to check that the structure (B,+,) is a left semi-brace. Let us note that (B,+) is a left group. Moreover, (a,u) in G×H is idempotent with respect to the sum if and only if φ(a)=u-1. In addition, if (a,u),(b,v)E(B), we have that

(a,u)+(b,v)=(a,uφ(b)v)=(a,u).

Hence E(B) is a sub-semigroup of (B,+) and it is also a left zero semigroup. Note also that

(a,u)+(1,1)=(a,uφ(1)1)=(a,u),

i.e., (1,1) is a right identity with respect to the sum. Furthermore, we have that

ρ(b,v)(a,u)=((aub)(φ(b)v)-1,(((φ(b)v)-1)au)bv)

for all (a,u),(b,v),(c,w)B. One can check that ρ is an anti-homomorphism if and only if it holds

(2.1)φ(b)vφ(c)=φ(bvc)vc

for all a,bG and uH. Moreover, by the characterization [10, Theorem 3], one can verify that the map r in [32, Theorem 5.1] associated to the left semi-brace B is a solution if and only if

(2.2)φ(a)uφ(b)=φ(b)vφ((φ(b)v)-1(aub))(((φ(b)v)-1)au)b

holds for all a,bG and u,vH. On the other hand, note that, if a,bG and uH, considering v=φ(b)-1 in (2.2), we obtain

φ(a)uφ(b)=v-1vφ((v-1v)-1(aub))(((v-1v)-1)au)b=φ(aub)ub.

Hence (2.1) is satisfied.

Now, let G be the cyclic group C2 of 2 elements, let H be the cyclic group C3 of 3 elements, and let φ be the constant map of value 1 from G into H. Hence, if (B,) is the cyclic group C2×C3, then condition (2.1) trivially holds, and hence r is a solution. Instead, if (B,) is the symmetric group C2C3, then (2.1) is not satisfied; equivalently, (2.2) does not hold, and hence r is not a solution.

## 3 Definitions and examples

Braces, skew braces, and semi-braces were introduced to study set-theoretic solutions of the Yang–Baxter equation. The following definition generalizes these structure to the case in which the multiplicative structure is no more a group.

At first, we recall that a semigroup (S,) is completely regular if for any element a of S there exists a (unique) element a- of S such that

(3.1)a=aa-a,a-=a-aa-,aa-=a-a.

Conditions (3.1) imply that a0:=aa-=a-a is an idempotent element of (S,).

## Definition 3.1.

Let S be a set with two operations + and such that (S,+) is a semigroup (not necessarily commutative) and (S,) is a completely regular semigroup. Then we say that (S,+,) is a generalized left semi-brace if

(3.2)a(b+c)=ab+a(a-+c)

for all a,b,cS. We call (S,+) and (S,) the additive semigroup and the multiplicative semigroup of S, respectively.

A generalized right semi-brace is defined similarly, replacing condition (3.2) by

(a+b)c=(a+c-)c+bc

for all a,b,cS.

A generalized two-sided semi-brace is a generalized left semi-brace that is also a generalized right semi-brace with respect to the same pair of operations.

Let us note that if S is a generalized left semi-brace and aS, then the map

λa:SS,ba(a-+b),

is an endomorphism of the semigroup (S,+) and λab(x)=(ab)0+λaλb(x) for all a,b,xS. Indeed, if a,b,x,yS, we have that

λa(x+y)=a(a-+x+y)=a(a-+x)+a(a-+y)=λa(x)+λa(y)

and

λab(x)=(ab)((ab)-+x)
=a(b(ab)-+b(b-+x))
=ab(ab)-+a(a-+λb(x))
=(ab)0+λaλb(x).

Of course, left semi-braces [7, 32] are examples of generalized left semi-braces. Moreover, a generalized left semi-brace can be obtained from every completely regular semigroup.

## Example 3.2.

If (S,) is an arbitrary completely regular semigroup and (S,+) is a right zero semigroup (or a left zero semigroup), then (S,+,) is a generalized two-sided semi-brace.

Unlike left semi-braces, a generalized left semi-brace S can have a zero element even if S has more than one element. Examples of such generalized left semi-braces can be easily obtained by any Clifford semigroup.

## Example 3.3.

If (S,) is a Clifford semigroup, which is a completely regular semigroup where all idempotent elements are central, then (S,+,), where a+b=ab for all a,bS, is a generalized two-sided semi-brace.

More generally, the previous generalized left semi-braces can be obtained through the following construction.

## Proposition 3.4.

Let Y be a (lower) semilattice, let {SααY} be a family of disjoint generalized left semi-braces. For each pair α,β of elements of Y such that αβ, let ϕα,β:SαSβ be a homomorphism of generalized left semi-braces such that the following conditions hold:

1. ϕα,α is the identical automorphism of Sα. for every αY.

2. ϕβ,γϕα,β=ϕα,γ for all α,β,γY such that αβγ.

Then S={SααY} endowed by the addition and the multiplication defined by

a+b=ϕα,αβ(a)+ϕβ,αβ(b),
ab=ϕα,αβ(a)ϕβ,αβ(b)

for any aSα and bSβ, is a generalized left semi-brace. Such a generalized left semi-brace is said to be the strong semilattice Y of the generalized left semi-brace Sα and is denoted by S=[Y;Sα,ϕα,β].

## Proof.

First note that (S,+) is a semigroup and (S,) is a completely regular semigroup. Now, let aSα, bSβ, and cSγ. Set δ:=αβ, ε:=βγ, ζ:=αγ, and η:=αβγ. It follows that

a(b+c)=a(ϕβ,ε(b)+ϕγ,ε(c))
=ϕα,η(a)ϕε,η(ϕβ,ε(b)+ϕγ,ε(c))(since αε=η)
=ϕα,η(a)(ϕε,ηϕβ,ε(b)+ϕε,ηϕγ,ε(c))
=ϕα,η(a)(ϕβ,η(b)+ϕγ,η(c))(by (ii))
=ϕα,η(a)ϕβ,η(b)+ϕα,η(a)((ϕα,η(a))-+ϕγ,η(c)),

where the last equality holds since Sη is a generalized left semi-brace. Moreover,

ab+a(a-+c)=ϕα,δ(a)ϕβ,δ(b)+a((ϕα,ζ(a))-+ϕγ,ζ(c))
=ϕα,δ(a)ϕβ,δ(b)+ϕα,ζ(a)((ϕα,ζ(a))-+ϕγ,ζ(c))
=ϕδ,η(ϕα,δ(a)ϕβ,δ(b))+ϕζ,η(ϕα,ζ(a)((ϕα,ζ(a))-+ϕγ,ζ(c)))(since δζ=η)
=ϕα,η(a)ϕβ,η(b)+ϕα,η(a)((ϕα,η(a))-+ϕγ,η(c))(by (ii)).

Therefore, S is a generalized left semi-brace. ∎

## Remark 3.5.

If S=[Y;Sα,ϕα,β] is a strong semilattice Y of left semi-braces Sα, then (S,) is a strong semilattice of groups, and hence, by [30, Theorem 4.2.1], (S,) is a Clifford semigroup.

## 4 Solutions related to generalized left semi-braces

This section is devoted to provide a sufficient condition to obtain solutions through a generalized left semi-brace. To this end, we recall that if S is a left cancellative left semi-brace, then the map r:S×SS×S given by

(4.1)r(a,b):=(a(a-+b),(a-+b)-b)

for all a,bS is a solution. Moreover, [32, Theorem 5.1] gives a sufficient condition to obtain that the map in (4.1) is still a solution for a left semi-brace, not necessarily left cancellative. In addition, in [10, Theorem 3], we state a necessary and sufficient condition to ensure that r is a solution.

Specifically, if (S,+,) is a left semi-brace, then the map r:S×SS×S, defined by

r(a,b):=(a(a-+b),(a-+b)-b)for all a,bS,

is a solution if and only if

(4.2)a+λb(c)(0+ρc(b))=a+b(0+c)

holds for all a,b,cS.

Let us remark that if S is a generalized left semi-brace with (S,+) being a right zero semigroup, then the map r as in (4.1) is a solution if and only if (ab)0=(a0b)0 holds for all a,bS. Observe that semigroups (S,) satisfying such a condition lie in the wide class of right cryptogroups; see [42]. In this way, we get new idempotent solutions that are of the form r(a,b)=(a0,ab), different from those obtained in [37, 39, 19, 48].

Moreover, note that if (S,+,) is the generalized left semi-brace of Example 3.3, we obtain that

r(a,b)=(a0b,b-ab)

is a solution. In particular, if (S,) is commutative, clearly r(a,b)=(a0b,b0a) and it is easy to verify that r is a cubic solution, i.e., r3=r.

Our aim is to show that if S=[Y;Sα,ϕα,β] is a strong semilattice of generalized left semi-braces such that every Sα satisfies condition (4.2), then the map in (4.1) is a solution. This result is a consequence of a more general construction technique on solutions we introduce in the following theorem.

## Theorem 4.1.

Let Y be a (lower) semilattice, let {(Xα,rα)αY} be a family of disjoint solutions indexed by Y such that for each pair α,βY with αβ there is a map ϕα,β:XαXβ. Let X be the union

X={XααY}

and let r:X×XX×X be the map defined by

r(x,y):=rαβ(ϕα,αβ(x),ϕβ,αβ(y))

for all xXα and yXβ. Then (X,r) is a solution if the following conditions are satisfied:

1. ϕα,α is the identity map of Xα for every αY.

2. ϕβ,γϕα,β=ϕα,γ for all α,β,γY such that αβγ.

3. (ϕα,β×ϕα,β)rα=rβ(ϕα,β×ϕα,β) for all α,βY such that αβ.

We call the pair (X,r) a strong semilattice of solutions (Xα,rα) indexed by Y.

The proof of Theorem 4.1 is technical, and for the sake of clarity, we present it in the next section. Now, as a consequence of this theorem, we obtain the following result.

## Theorem 4.2.

Let S=[Y;Sα,ϕα,β] be a strong semilattice of generalized left semi-braces. Then, if Sα satisfies (4.2) for every αY, then the map rS:S×SS×S defined by

r(a,b):=(a(a-+b),(a-+b)-b)

for all a,bS is a solution.

## Proof.

For any αY, let rα:Sα×SαSα×Sα be the solution associated to the left semi-brace Sα, i.e., the map defined by rα(x,y)=(x(x-+y),(x-+y)-y). Since S is a strong semilattice of left semi-braces, by Proposition 3.4, ϕα,α is the identical automorphism of Sα and ϕβ,γϕα,β=ϕα,γ for all α,β,γY such that αβγ. Hence, conditions (i) and (ii) in Theorem 4.1 are satisfied. Moreover, let a,bY such that αβ. Since, ϕα,β is a homomorphism of left semi-braces, for all x,ySα it follows that

(ϕα,β×ϕα,β)rα(x,y)=(ϕα,β(x(x-+y)),ϕα,β((x-+y)-y))
=(ϕα,β(x)((ϕα,β(x))-+ϕα,β(y)),((ϕα,β(x))-+ϕα,β(y))-ϕα,β(y))
=rβ(ϕα,β(x),ϕα,β(y))
=rβ(ϕα,β×ϕα,β)(x,y).

Hence Theorem 4.1 (iii) holds. Therefore, according to Theorem 4.1, we shall consider the strong semilattice Y of solutions rα, i.e., the map r defined by

r(x,y)=rαβ(ϕα,αβ(x),ϕβ,αβ(y))

for all xSα, ySβ. Finally, note that, by Proposition 3.4,

r(x,y)=(ϕα,αβ(x)((ϕα,αβ(x))-+ϕβ,αβ(y)),((ϕα,αβ(x))-+ϕβ,αβ(y))-ϕβ,αβ(y))
=(x(x-+y),(x-+y)-y)

for all xSα, ySβ. ∎

## 5 Strong semilattices of set-theoretical solutions

This section aims to provide a proof of Theorem 4.1 and to give some examples of strong semilattices of solutions. Furthermore, we analyze strong semilattices of solutions with finite order.

## Proof of Theorem 4.1.

At first, note that if λx[ω] and ρy[ω] are the maps from Xω into itself that define every solution rω, i.e., rω is written as

rω(x,y)=(λx[ω](y),ρy[ω](x))

for all x,yXω, then condition (iii) is equivalent to the following equalities:

(5.1)ϕω,ιλx[ω](y)=λϕω,ι(x)[ι]ϕω,ι(y),
(5.2)ϕω,ιρy[ω](x)=ρϕω,ι(y)[ι]ϕω,ι(x)

for all ω,ιY such that ωι and x,yXω. In addition, let us observe that if xXω and yYι, then the two components of the map r, i.e.,

λx(y)=λϕω,ωι(x)[ωι]ϕι,ωι(y),ρy(x)=ρϕι,ωι(y)[ωι]ϕω,ωι(x),

lie in Xωι, consistently with the second part of the subscript of the maps ϕ. To avoid overloading the notation, hereinafter we will write the previous elements as

λx(y)=λϕω,ωι(x)ϕι,ωι(y),ρy(x)=ρϕι,ωι(y)ϕω,ωι(x).

Now, we verify that r is a solution proving that the relations

L1:=λxλy(z)=λλx(y)λρy(x)(z)=:L2,
C1:=λρλy(z)(x)ρz(y)=ρλρy(x)(z)λx(y)=:C2,
R2:=ρρy(z)ρλz(y)(x)=ρzρy(x)=:R1

are satisfied for all x,y,zX. For this purpose, let x,y,z be elements of Xω, Xι, Xκ, respectively, and assume ν:=ωικ and

𝒳:=ϕω,ν(x),𝒴:=ϕι,ν(y),𝒵:=ϕκ,ν(z)  (in Xν).

Setting

U:=ρϕι,ωι(y)ϕω,ωι(x),V:=λϕω,ωι(x)ϕι,ωι(y)  (in Xωι),

we have that

L2=λλx(y)λρy(x)(z)
=λVλU(z)
=λVλϕωι,ν(U)ϕκ,ν(z)(ν=ωικ)
=λϕωι,ωιν(V)ϕν,ωινλϕωι,ν(U)(𝒵)
=λϕωι,ν(V)ϕν,νλϕωι,ν(U)(𝒵)(ωιν=ν)
=λϕωι,ν(V)λϕωι,ν(U)(𝒵)(ϕν,ν=idXν).

Since

ϕωι,ν(V)=λϕωι,νϕω,ωι(x)ϕωι,νϕι,ωι(y)(by (5.1))
=λϕω,ν(x)ϕι,ν(y)(ϕωι,νϕω,ωι=ϕω,ν,ϕωι,νϕι,ωι=ϕι,ν)
=λ𝒳(𝒴)

and

ϕωι,ν(U)=ρϕωι,νϕι,ωι(y)ϕωι,νϕω,ωι(x)(by (5.2))
=ρϕι,ν(y)ϕω,ν(x)(ϕωι,νϕι,ωι=ϕι,ν,ϕωι,νϕω,ωι=ϕω,ν)
=ρ𝒴(𝒳),

it follows that

L2=λλ𝒳(𝒴)λρ𝒴(𝒳)(𝒵)(in Xν)
=λ𝒳λ𝒴(𝒵)(rν is a solution).

Moreover, it holds

L1=λxλy(z)=λλϕι,ικ(y)xϕκ,ικ(z)
=λϕω,ν(x)ϕικ,νλϕι,ικ(y)ϕκ,ικ(z)(ν=ωικ)
=λ𝒳λϕικ,νϕι,ικ(y)ϕικ,νϕκ,ικ(z)(by (5.1))
=λ𝒳λϕι,ν(y)ϕκ,ν(z)(ϕικ,νϕι,ικ=ϕι,ν,ϕικ,νϕκ,ικ=ϕκ,ν)
=λ𝒳λ𝒴(𝒵).

Hence we obtain that L1=L2. Now, setting

W:=ρϕκ,ικ(z)ϕι,ικ(y),Z:=λϕι,ικ(y)ϕκ,ικ(z)  (in Xικ),

observe that

C1=λρλy(z)(x)ρz(y)
=λρλϕι,ικ(y)ϕκ,ικ(z)(x)ρϕκ,ικ(z)ϕι,ικ(y)
=λρZ(x)(W)
=λρϕικ,ν(Z)ϕω,ν(x)(W)(ν=ωικ)
=λϕν,νικρϕικ,ν(Z)(𝒳)ϕικ,νικ(W)
=λϕν,νρ(𝒳)ϕικ,ν(Z)ϕικ,ν(W)(νικ=ν)
=λρϕικ,ν(Z)(𝒳)ϕικ,ν(W)(ϕν,ν=idXν).

Since

ϕικ,ν(Z)=λϕικ,νϕι,ικ(y)ϕικ,νϕκ,ικ(z)(by (5.1))
=λϕι,ν(y)ϕκ,ν(z)(ϕικ,νϕι,ικ=ϕι,ν,ϕικ,νϕκ,ικ=ϕκ,ν)
=λ𝒴(𝒵)

and

ϕικ,ν(W)=ρϕικ,νϕικ,νϕκ,ικ(z)ϕι,ικ(y)by (5.2))
=ρϕκ,ν(z)ϕι,ν(y)(ϕικ,νϕκ,ικ=ϕκ,ν,ϕικ,νϕι,ικ=ϕι,ν)
=ρ𝒵(𝒴),

it follows that

C1=λρλ𝒴(𝒵)(𝒳)ρ𝒵(𝒴)(in Xν).

Furthermore, since U and V lie in Xωι, we have

C2=ρλρy(x)(z)λx(y)
=ρλρϕι,ωι(y)ϕω,ωι(x)(z)λϕω,ωι(x)ϕι,ωι(y)
=ρλU(z)(V)
=ρλϕωι,ν(U)ϕκ,ν(z)(V)(ν=ωικ)
=ρϕν,ωινλϕωι,ν(U)(𝒵)ϕωι,ωιν(V)
=ρϕν,νλϕωι,ν(U)(𝒵)ϕωι,ν(V)(ωιν=ν)
=ρλϕωι,ν(U)(𝒵)ϕωι,ν(V)(ϕν,ν=idXν).

As seen before, ϕωι,ν(U)=ρ𝒴(𝒳) and ϕωι,ν(V)=λ𝒳(𝒴). Thus

C2=ρλ𝒳λρ𝒴(𝒳)(𝒵)(𝒴)(in Xν).

Consequently, since rν is a solution, we obtain that C1=C2.

Finally, since W and Z lie in Xικ, note that

R2=ρρy(z)ρλz(y)(x)
=ρρϕκ,ικ(z)ϕι,ικ(y)ρλϕι,ικ(y)ϕκ,ικ(z)(x)
=ρWρZ(x)
=ρWρϕικ,ν(Z)ϕω,ν(x)(ν=ωικ)
=ρϕικ,νικ(W)ϕν,νικρϕικ,ν(Z)(𝒳)
=ρϕικ,ν(W)ϕν,νρϕικ,ν(Z)(𝒳)(νικ=ν)
=ρϕικ,ν(W)ρϕικ,ν(Z)(𝒳)(ϕν,ν=idXν).

As seen before, ϕικ,ν(W)=ρ𝒵(𝒴). In addition, we have

ϕικ,ν(Z)=λϕικ,νϕι,ικ(y)ϕικ,νϕκ,ικ(z)(by (5.1))
=λϕι,ν(y)ϕκ,ν(z)(ϕικ,νϕι,ικ=ϕι,ν,ϕικ,νϕκ,ικ=ϕκ,ν)
=λ𝒴(𝒵).

It follows that

R2=ρρ𝒵(𝒴)ρλ𝒴(𝒵)(𝒳)(in Xν)
=ρ𝒵ρ𝒴(𝒳)(rν is a solution).

Moreover, it holds

R1=ρzρy(x)=ρzρϕι,ωι(y)ϕω,ωι(x)
=ρϕκ,ν(z)ϕωι,νρϕι,ωι(y)ϕω,ωι(x)
=ρ𝒵ρϕωι,νϕι,ωι(y)ϕωι,νϕω,ωι(x)(by (5.2))
=ρ𝒵ρϕι,ν(y)ϕω,ν(x)(ϕωι,νϕι,ωι=ϕι,ν,ϕωι,νϕω,ωι=ϕω,ν)
=ρ𝒵ρ𝒴(𝒳),

and hence R1=R2. Therefore, the map r is a solution. ∎

One can use strong semilattices of solutions (X,r) with rα’s of finite order to produce examples of new solutions with finite order.

## Examples 5.1.

(i) Let X be a semilattice of sets indexed by Y such that conditions (i) and (ii) of Theorem 4.1 are satisfied and let rα be the twist map on Xα for every αY. Then, if αβ, we have that

ϕα,βλx(y)=ϕα,β(y)=λϕα,β(x)ϕα,β(y),
ϕα,βρy(x)=ϕα,β(x)=ρϕα,β(y)ϕα,β(x)

for all x,yXα, i.e., condition (iii) of Theorem 4.1 holds. Hence, the strong semilattice of solutions (X,r) is such that r3=r. Indeed, if xXα and yXβ, assuming ν:=αβ, we have that r(x,y)=rν(ϕα,ν(x),ϕβ,ν(y)). Hence r2(x,y)=(ϕα,ν(x),ϕβ,ν(y)), and so

r3(x,y)=rν(ϕα,ν(x),ϕβ,ν(y))=r(x,y).

Consequently, r3=r.

(ii) Let X:=XαXβ be a semilattice of sets such that α>β, let c be a fixed element of Xβ, and let ϕα,β(x)=c for every xXα. Let rα be the twist map on Xα and let rβ be the idempotent solution on Xβ defined by rβ(x,y):=(x,c) for all x,yXβ. Then, if x,yXα, we have that

ϕα,βλx(y)=c=ϕα,β(x)=λϕα,β(x)ϕα,β(y),
ϕα,βρy(x)=c=ρϕα,β(y)ϕα,β(x).

Hence the assumptions of Theorem 4.1 are satisfied. Moreover, the strong semilattice of solutions (X,r) is such that r3=r. Indeed, if xXα and yXβ, since r(x,y)=rβ(ϕα,β(x),y)=rβ(c,y)=(c,c), it follows that

r2(x,y)=rβ(c,c)=(c,c)=r(x,y).

Therefore, we obtain that r3=r.

(iii) Let X:=XαXβ be a semilattice of sets such that α>β, let c be a fixed element of Xβ, and let ϕα,β(x)=c for every xXα. Let f be an idempotent map from Xβ into itself, fidXβ, and let rα be the map from Xα×Xα into itself defined by rα(x,y):=(f(x),x) for all x,yXα. Thus, rα is a solution such that rα3=rα2. Let rβ be the idempotent solution defined by rβ(x,y)=(x,c) for all x,yXβ. Then, if x,yXα, we obtain that

ϕα,βλx(y)=c=ϕα,β(x)=λϕα,β(x)ϕα,β(y),
ϕα,βρy(x)=c=ρϕα,β(y)ϕα,β(x).

Thus the hypotheses of Theorem 4.1 are satisfied. Moreover, the strong semilattice of solutions (X,r) is such that r3=r2. Indeed, if xXα and yXβ, since r(x,y)=rβ(ϕα,β(x),y)=rβ(c,y)=(c,c), it follows that

r2(x,y)=rβ(c,c)=(c,c)=r(x,y),

and clearly r3(x,y)=r2(x,y). Therefore, r3=r2.

To investigate strong semilattices of solutions with finite order, we need the notions of the index and the period of a solution r that are

i(r):=min{jj0,there exists l such that rj=rl,jl},
p(r):=min{kk,ri(r)+k=ri(r)},

respectively. These definitions of the index and the order are slightly different from the classical ones (cf. [31, p. 10]), but they are functional to distinguish bijective solutions from non-bijective ones. For more details, we refer the reader to [10].

In the following theorem, we show that, given a semilattice Y of finite cardinality, the strong semilattice of solutions (X,r) indexed by Y is of finite order if and only if solutions rα are. Furthermore, it allows for establishing the order of the strong semilattice of solutions (X,r) if the index and the period of solutions rα are known. Conversely, the index and the period of a strong semilattice of solutions (X,r) give us upper bounds of the indexes and periods of solutions rα.

## Theorem 5.2.

Let (X,r) be a strong semilattice of solutions indexed by a finite semilattice Y. Then rα is a solution with finite order on Xα for every αY if and only if r is a solution with finite order. More precisely, the index of r is

i(r)=max{1,i(rα)αY}

and the period is

p(r)=lcm{p(rα)αY}.

## Proof.

At first, suppose that rα is a solution with finite order for every αY. Let n:=lcm{p(rα)αY} and i:=max{i(rα)αY}. If xXα and yXβ, setting ν:=αβ, we have that

rn+i(x,y)=rνn+i(ϕα,ν(x),ϕβ,ν(y))=rνi(ϕα,ν(x),ϕβ,ν(y)).

Consequently, if i=0, i.e., rα is bijective for every αY, we obtain that

rn+1(x,y)=rν(ϕα,ν(x),ϕβ,ν(y))=r(x,y).

Therefore, rn+1=r and clearly the index of r is 1 by the assumption on i. Now, assume that i0 and that i=i(rγ) for a certain γY. If rn=rh, for a positive integer h, in particular, we have that rγn=rγh. Since n=p(rγ)q+i for a certain natural number q, it follows that

rγp(rγ)+i=rγi=rγp(rγ)q+i=rγn=rγh.

Hence ih, and so i(r)=i.

Now, we proceed to determine the period of r. If rm=ri(r) for a natural number m, then rαm=rαi(r) for every αY. Consequently, p(rα) divides m-i(r) for every αY. Thus lcm{p(rα)αY} divides m-i(r), i.e., n-i(r) divides m-i(r). Therefore, n-i(r)m-i(r), and hence p(r)=n-i(r). Conversely, suppose that the solution r is with finite order and set i:=i(r) and p:=p(r). If αY, since ϕα,α=idXα×Xα, we obtain that rαp+i=rαi. Therefore, rα is a solution with finite order. Clearly, we have that i(rα) is less than i and p(rα) divides p. ∎

Let us note that if (X,r) is a strong semilattice of non-bijective solutions rα such that i(rα)=i and p(rα)=n for every αY, then r is still a solution of index i and period n, also in the case of an infinite semilattice Y. Indeed, one can prove this statement by similar computations used for the non-bijective case in the proof of Theorem 5.2.

Communicated by Manfred Droste

Funding statement: This work was partially supported by the Dipartimento di Matematica e Fisica “Ennio De Giorgi” – Università del Salento. The second author is supported in part by Onderzoeksraad of Vrije Universiteit Brussel and Fonds voor Wetenschappelijk Onderzoek (Flanders), Grant G016117. The authors are members of GNSAGA (INdAM).

## Acknowledgements

We would like to thank the referee for the accurate review.

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