Related structures with involution

Acta Math. Hungar., 123 (1�2) (2009), 169�185.DOI: 10.1007/s10474-009-8081-6

RELATED STRUCTURES WITH INVOLUTIONR. PÖSCHEL1 and S. RADELECZKI2 ∗

1 Institut für Algebra, TU Dresden, 01062 Dresden, Germanye-mail: [email protected]

2 Institute of Mathematics, University of Miskolc, 3515 Miskolc-Egyetemváros, Hungarye-mail: [email protected]

(Received April 16, 2008; revised November 18, 2008; accepted November 19, 2008)

Abstract. We give a necessary and su�cient condition for an involutionlattice to be isomorphic to the direct square of its invariant part. This resultis applied to show relations between related lattices of an algebra. For instance,generalizing some earlier results of G. Czédli and L. Szabó it is proved that any al-gebra admits a connected compatible partial order whenever its quasiorder latticeis isomorphic to the direct square of its congruence lattice. Further, a majorityalgebra is lattice ordered if and only if the lattice of its compatible re�exive rela-tions is isomorphic to the direct square of its tolerance lattice. In the latter case,one can establish a bijective correspondence between factor congruence pairs ofthe algebra and its pairs of compatible lattice orders; several consequences of thisresult are given.

Introduction

It is well known that the compatible re�exive relations, the tolerances,the quasiorders and the congruences of an algebra (A,F ) form algebraic lat-

∗This paper is a result of a collaboration of the authors within DFG-MTA Grant no. 167, whosesupport is gratefully acknowledged by the authors. The partial support by Hungarian NationalResearch Fund (Grant no. T049433/05) is also acknowledged by the second author.

Key words and phrases: involution lattice, central element, quasiorder lattice and tolerancelattice of an algebra, compatible partial order of an algebra, lattice ordered majority algebra.

2000 Mathematics Subject Classi�cation: primary: 08A05, 06B05; secondary: 08A02, 06F99.

0236�5294/$ 20.00 c© 2009 Akadémiai Kiadó, Budapest

170 R. PÖSCHEL and S. RADELECZKI

tices with respect to inclusion j. We denote them by Refl (A,F ), Tol (A,F ),Quord (A, F ) and Con (A,F ), respectively. These lattices are not indepen-dent: Tol (A,F ) is a sublattice of the lattice Refl (A,F ), Con (A,F ) is asublattice of Quord (A,F ), moreover, the in�mum of a system {αi | i ∈ I} ofelements in each of these lattices is equal to the set theoretical intersection⋂{αi | i ∈ I}. Therefore, as algebraic structures with meet and join, we shalluse the notations

(Refl (A,F ),∩,t)

,(Tol (A,F ),∩,t)

,(Quord (A,F ),∩,∨)

and(Con (A,F ),∩,∨)

, respectively. Let α−1 ={

(a, b) | (b, a) ∈ α}

denotethe inverse of a relation α j A×A. Then the mapping ι : α 7−→ α−1 is aninvolution both in the lattice

(Refl (A,F ),∩,t)

and in(Quord (A,F ),∩,∨)

.As shown in [6] and [19], in the case of a partially ordered algebra (A,F,

5), there are some other relations between the above lattices. For instance,in [6] it is proved that for a lattice ordered majority algebra (A,F,5), thealgebra

(Quord (A,F ),∩,∨, ι

)is isomorphic to the algebra

(Con2(A,F ),∩,

∨, τ), where τ(θ1, θ2) = (θ2, θ1) for (θ1, θ2) ∈ Con2(A,F ).

In this paper several generalizations of this theorem are given. By usingTheorem 2.1 we prove necessary and su�cient conditions for the isomor-phism


) ∼=(Con2(A,F ),∩,∨, τ

), as well as for the iso-

morphism(Refl(A,F ),∩,t, ι

) ∼=(Tol2(A,F ),∩,∨, τ

)in the case of a general

algebra (A,F ). Moreover, we show that a majority algebra is lattice orderedif and only if we have the second isomorphism above (see Theorem 4.6),and we deduce several consequences of this fact. For instance, we prove thatthere exists a bijective correspondence between the pairs of factor congruencesof a lattice ordered majority algebra (A,F ) and its pairs of lattice orders(cf. Proposition 4.8(i)) and the conditions Refl (A,F ) = Quord (A, F ) andTol (A,F ) = Con (A, F ) are equivalent (cf. Proposition 4.10). The comple-mented elements of the lattice Refl (A,F ) are also characterized (see Propo-sition 5.2).

1. Preliminaries

1.1. Let α j A×A be a binary relation on a set A. If α is re�exive andsymmetric, then it is called a tolerance on the set A, and if it is re�exive andtransitive, then it is called a quasiorder on A. The re�exive relations, thetolerances, and the quasiorders on a set A form algebraic lattices with respectto j, denoted by Refl (A), Tol (A), and Quord (A), respectively. Their leastand greatest elements are 4 =

{(a, a) | a ∈ A

}and ∇ = A×A. A tolerance

of an algebra (A,F ) is a tolerance relation on the set A compatible with theoperations of the algebra (A,F ), and a quasiorder of an algebra (A,F ) is a

Acta Mathematica Hungarica 123, 2009

RELATED STRUCTURES WITH INVOLUTION 171

compatible quasiorder q j A×A. Obviously, we have

α ∈ Refl (A,F ) ⇐⇒ α−1 ∈ Refl (A,F ),

q ∈ Quord (A,F ) ⇐⇒ q−1 ∈ Quord (A, F ),

q is a partial order⇐⇒ q ∈ Quord (A,F ) and q ∩ q−1 = 4.

Let % denote the transitive closure of a relation % j A×A. A partialorder r j A×A is called connected if r ∪ r−1 = ∇, and it is called locallybounded if any two elements a, b ∈ A have an upper bound and a lower boundin the poset (A, r). Clearly, if α ∈ Refl (A, F ), then α ∈ Quord (A,F ). Thesupremum for qi ∈ Quord (A,F ), i ∈ I, in the lattice Quord (A,F ) is given by

∨{qi | i ∈ I} =

⋃{qi | i ∈ I}.

Thus Quord (A,F ) is a complete sublattice of Quord (A) (see e.g. [17]).A quadruple (L,∧,∨, ι) is called an involution lattice if ι is an automor-

phism of the lattice L = (L,∧,∨) such that ι2(x) = x. For an involutionlattice L = (L,∧,∨, ι), the set I =

{x ∈ L | ι(x) = x

}of �xed points of ι is a

subalgebra (I,∧,∨, ι) which is called the invariant part of L. Of course, therestriction of ι to the set I is the identity mapping id, moreover if (L,∧,∨)is a complete lattice, then (I,∧,∨) is a complete sublattice of it. If L is abounded lattice (i.e. a lattice with least element 0 and greatest element 1),then (L,∧,∨, ι) is called a bounded involution lattice and it follows ι(0) = 0,ι(1) = 1.

1.2. Example (1) Clearly, for every lattice L the algebra (L,∧,∨, id) isan involution lattice, with invariant part I = L.

(2) For any lattice L its direct square L2 becomes an involution lattice(L2,∧,∨, τ) if we de�ne τ(x, y) = (y, x), for all (x, y) ∈ L2.

(3) Let (A,F ) be any algebra. Then(Quord (A,F ),∩,∨, ι

)is a bounded

involution lattice, where the involution is given by ι(q) = q−1, q ∈Quord (A, F ), and its invariant part is I =

(Con (A, F ),∩,∨, id

)(see e.g.

[17] and [6]).(4) Analogously, the algebra

(Refl (A,F ),∩,t, ι

), with ι(α) = α−1, α ∈

Refl (A,F ), is a bounded involution lattice with invariant part

I =(Tol (A, F ),∩,t, id

).

1.3. Let L be a bounded lattice. An element a ∈ L is called a centralelement of L if a is complemented and for all x, y ∈ L the sublattice gen-erated by {a, x, y} is distributive. A complement of an element a ∈ L (if



it exists) will be denoted by a′. The central elements of L form a Booleansublattice denoted by Cen (L). Obviously, their complements are uniquelydetermined. If c ∈ Cen (L), then c′ ∈ Cen (L) and c, c′ is called a central pairof L. Let (a] = {x ∈ L | x 5 a} denote the principal ideal corresponding toan element a ∈ L. It is well known that, for any direct product decompo-sition L ∼= ∏

i∈I Li of a bounded lattice L there exist elements ci ∈ Cen (L),i ∈ I such that Li

∼= (ci] (see e.g. [18]). The central pairs of a bounded latticehave several characterizations. For instance, in [13] we can �nd the following:

(1)

(2)c, c′ ∈ Cen (L) ⇐⇒

{x = (x ∧ c) ∨ (x ∧ c′), ∀x ∈ L

x = (x ∨ c) ∧ (x ∨ c′), ∀x ∈ L.

A lattice L with least element 0 is called 0-modular, if for any a, b, c ∈ L theimplication

(M0) (a ∧ c = 0 and b 5 c) =⇒ (a ∨ b) ∧ c = b

holds. In view of Varlet's result [21], a lattice with 0 is 0-modular if and onlyif it does not contain an N5 sublattice including 0. It was proved in [13] that,for any 0-modular lattice, (1) implies (2).

As the centre of a bounded lattice is preserved by its automorphisms,c ∈ Cen (L) implies ι(c) ∈ Cen (L) for any bounded involution lattice (L,∧,∨, ι). Hence, denoting by ιC the restriction of ι to Cen (L) we obtain that(Cen (L),∧,∨, ιC

)is a subalgebra of (L,∧,∨, ι).

2. A structure theorem for involution lattices

Our starting point is the following result of [5]:Theorem F. If L is a bounded distributive lattice with involution, then

L is isomorphic with the square of its invariant part via the isomorphismsgiven by formulas (2.3) and (2.4) below.

First, we generalize this result to arbitrary bounded involution lattices.While Theorem F is used for Quord (L) of a lattice L in [5] (see Example1.2(3)), the present paper generalizes this in many ways and presents a lot ofrelated results.

Theorem 2.1. Let (L,∧,∨, ι) be a bounded involution lattice, I its invari-ant part and consider the mapping τ(x, y) = (y, x), for all (x, y) ∈ I2. Then(L,∧,∨, ι) ∼= (I2,∧,∨, τ) if and only if there exists an element c ∈ Cen (L)such that ι(c) = c′.



Proof. Clearly, (I2,∧,∨, τ) is an involution lattice. Assume thatι(c′) = c, for some c ∈ Cen (L) and consider the mappings f : L → I2 andg : I2 → L de�ned by

f(x) := ((x ∧ c) ∨ (ι(x) ∧ c′

), (x ∧ c′) ∨ (

ι(x) ∧ c)),(2.3)

g(x, y) := (x ∧ c) ∨ (y ∧ c′).(2.4)

It is easy to see that both f and g are order-preserving functions, henceto prove that they are lattice isomorphisms, it is enough to show that theyare inverses each of other. Indeed, for any (x, y) ∈ I2 we get

g(x, y) ∧ c =[(x ∧ c) ∨ (y ∧ c′)

] ∧ c = (x ∧ c ∧ c) ∨ (y ∧ c′ ∧ c) = x ∧ c,

ι(g(x, y)

) ∧ c′ =[(ι(x) ∧ c′) ∨ (ι(y) ∧ c)

] ∧ c′

= (x ∧ c′ ∧ c′) ∨ (y ∧ c ∧ c′) = x ∧ c′,

g(x, y) ∧ c′ =[(x ∧ c) ∨ (y ∧ c′)

] ∧ c′ = y ∧ c′,

and

ι(g(x, y)

) ∧ c = [(ι(x) ∧ c′) ∨ (ι(y) ∧ c

)] ∧ c = ι(y) ∧ c = y ∧ c.

Hence, we obtain

(f ◦ g)(x, y)

=((

g(x, y) ∧ c) ∨ (ι

(g(x, y)

) ∧ c′),(g(x, y) ∧ c′

) ∨ (ι(g(x, y)

) ∧ c))

=((x ∧ c) ∨ (x ∧ c′), (y ∧ c′) ∨ (y ∧ c)

),

and 1.3 (2) implies (f ◦ g)(x, y) = (x, y).Further, for any x ∈ L we obtain

(g ◦ f)(x) = ([(x ∧ c) ∨ (ι(x) ∧ c′)

] ∧ c) ∨ ([(x ∧ c′) ∨ (ι(x) ∧ c)

] ∧ c′)

=[(x ∧ c ∧ c) ∨ (ι(x) ∧ c′ ∧ c)

] ∨ [(x ∧ c′ ∧ c′) ∨ (ι(x) ∧ c ∧ c′)

]

= (x ∧ c) ∨ (x ∧ c′) = x,

proving that f and g are inverses of each other.



Now, it is easy to check that f(ι(x)

)= τ

(f(x)

), for any x ∈ L and

g(τ(x, y)

)= ι

(g(x, y)

), for any x, y ∈ I, thus we obtain that f and g are

involution lattice isomorphisms, and hence (L,∧,∨, ι) ∼= (I2,∧,∨, τ).Conversely, assume that (L,∧,∨, ι) ∼= (I2,∧,∨, τ). It is easy to see that

(1, 0) and (0, 1) are central pairs in the lattice (I2,∧,∨) and τ(1, 0) = (0, 1).Denote by c and c∗ the elements in L corresponding (by isomorphism) to(1, 0) and (0, 1), respectively. Then we have c, c∗ ∈ Cen (L) and c ∧ c∗ = 0,c ∨ c∗ = 1 by isomorphism, thus c and c∗ are also central pairs in L and, byuniqueness of the complement, c∗ = c′. As τ(1, 0) = (0, 1) in I2, we obtainby isomorphism ι(c) = c∗ = c′ in L, completing the proof. ¤

Now, consider a bounded involution lattice (L,∧,∨, ι) with invariantpart I. Then

(Cen (L),∧,∨, ιC

)is also an involution lattice and it is easy to

see that its invariant part is IC = I ∩ Cen (L). With the notations of Theo-rem 2.1 we have

Corollary 2.2. The following conditions are equivalent:(i) (L,∧,∨, ι) ∼= (I2,∧,∨, τ),(ii)

(Cen (L),∧,∨, ιC

) ∼= (I2C ,∧,∨, τ),

(iii) There exists a lattice K = (K,∧,∨) such that

(Cen (L),∧,∨, ιC

) ∼= (K2,∧,∨, τ).

Proof. (i) ⇒ (ii). By Theorem 2.1, (i) is equivalent to the condition

(∗) ∃c, c′ ∈ Cen (L) : ι(c) = c′.

But then (∗) is also satis�ed by(Cen (L),∧,∨, ιC

)since the centre of the

lattice Cen (L) is Cen (L) itself. Hence, by Theorem 2.1, we get (ii).(ii) ⇒ (iii) is obvious.(iii) ⇒ (i). As Cen (L) is a bounded lattice, K must be also bounded. Let

0K and 1K stand for the least and the greatest element of K, respectively.Then (0K, 1K) and (1K, 0K) are complements of each other in the lattice (K2,

∧,∨) and τ((0K,1K)

)= (1K,0K). Denote by c and c∗ the elements in Cen (L)

corresponding by isomorphism to (0K, 1K) and (1K, 0K), respectively. Thenby the same argument as in the proof of Theorem 2.1, we get ιC(c) = c∗

and c∗ = c′. Hence, ι(c) = ιC(c) = c′, proving that condition (∗) holds for(L,∧,∨, ι), too. Now, by applying Theorem 2.1, we obtain (i). ¤



3. Conclusions for general algebras

In this section, we collect several conclusions from Theorem 2.1 for gen-eral algebras. Let (A,F ) be any algebra. It is easy to see that, q, q−1 ∈Quord (A,F ) are complements of each other in Quord (A,F ) if and only if q isa connected partial order. Similarly, %′ = ι(%) = %−1, for some % ∈ Refl(A,F ),if and only if % j A×A is an antisymmetric relation and %t %−1 = ∇. Fromthese observations and Theorem 2.1 it follows immediately

Proposition 3.1. Let (A,F ) be any algebra. Then we have:(i)


) ∼=(Con2(A,F ),∩,∨, τ

)if and only if there ex-

ists a connected partial order % j A×A such that % ∈ Cen(Quord (A,F )

).

(ii)(Refl (A,F ),∩,t, ι

) ∼=(Tol2(A,F ),∩,t, τ

)if and only if there exists

an antisymmetric re�exive relation % j A×A such that % ∈ Cen(Refl(A,F )

)

and % t %−1 = ∇.(A,F ) is called a majority algebra if it has a ternary term function m such

that m(x, x, y) = m(x, y, x) = m(y, x, x) = x, for all x, y ∈ A. In [8] and [14],it was proved that the quasiorder lattice of a majority algebra is distributive.As the central pairs in a distributive lattice are exactly the complementedpairs of elements, we obtain

Corollary 3.2. Let (A, F ) be a majority algebra. Then(Quord (A,F ),∩,∨, ι

) ∼=(Con2(A,F ),∩,∨, τ

)

if and only if (A,F ) admits a connected compatible partial order.Let us observe that for a unary algebra (A,F ) and any system

αi ∈ Refl (A,F ), i ∈ I we have⊔{αi | i ∈ I} =

⋃{αi | i ∈ I}. Therefore,(Refl (A,F ),∩,t)

and consequently(Tol (A,F ),∩,t)

are distributive lat-tices. Hence, by applying Proposition 3.1, we obtain

Corollary 3.3. If (A,F ) is a unary algebra, then(Refl (A,F ),∩,t, ι

) ∼=(Tol2(A,F ),∩,t, τ

)

if and only if (A,F ) admits an antisymmetric re�exive compatible relation% j A×A such that % t %−1 = ∇.

Now, let A be a nonempty set and F = {idA}. Then obviously, Refl (A)= Refl (A,F ) and Tol (A) = Tol (A,F ). Since on any nonempty set thereexists a linear order R j A×A and since R ∪R−1 = ∇, from Corollary 3.3we obtain the following

Corollary 3.4. For any nonempty set A we have(Refl (A),∩,∪, ι

) ∼=(Tol2(A),∩,∪, τ

).



Let (A, %) be a partially ordered set. An equivalence relation θ j A×Ais called a congruence of the poset (A, %) if it is the kernel of a %-preservingmapping, i.e. if there exists a poset (B, %∗) and a mapping f : A → B suchthat (a, b) ∈ % =⇒ (

f(a), f(b)) ∈ %∗ and θ = Ker f =

{(x, y) ∈ A×A | f(x)

= f(y)}. In [7] it was proved that an equivalence θ is a congruence if and only

if θ = (θ ∨ ρ)∩ (θ ∨ ρ−1) holds in the lattice(Quord (A),∩,∨)

(see also [12]).Let % j A×A be a compatible partial order of an algebra (A,F ). An

order-congruence of the partially ordered algebra (A,F, %) (see [7]) is a con-gruence of (A,F ), which is at the same time a congruence of the poset (A,%).Now, using these notions we have:

Corollary 3.5. Let (A,F ) be an algebra. If(Quord (A,F ),∩,∨, ι

) ∼=(Con2(A,F ),∩,∨, τ

), then there exists a connected compatible partial order %

of (A,F ) such that every θ ∈ Con (A,F ) is an order-congruence of (A,F, %).Proof. Assume that


) ∼=(Con2(A,F ),∩,∨, τ

).

Then, by Proposition 3.1(i), there exists a connected partial order % jA×A such that % ∈ Cen

(Quord (A,F )

). Hence %−1 = %′ also belongs to

Cen(Quord (A,F )

), and so the relation θ = (θ ∨ %) ∩ (θ ∨ %−1) holds in

Quord (A, F ), for all θ ∈ Con (A,F ). As Quord (A,F ) is a sublattice ofQuord (A), the same equality holds in Quord (A), too. ¤

Corollary 3.6. Let (A, F, R) be a linearly ordered algebra.(i) If Quord (A, F ) is a 0-modular lattice, then

(Quord (A, F ),∩,∨, ι

) ∼=(Con2(A,F ),∩,∨, τ

).

(ii) If Refl (A, F ) is a 0-modular lattice, then(Refl (A,F ),∩,t, ι

) ∼=(Tol2(A,F ),∩,t, τ

).

(iii) If A is �nite, then(Quord (A,F ),∩,∨, ι

) ∼=(Con2(A,F ),∩,∨, τ

)if

and only if Quord (A,F ) (and Con (A,F )) is a distributive lattice.Proof. (i) Clearly, we have % = (% ∩R) ∪ (% ∩R−1), for any relation

% j A×A, whence we get q = (q ∩R) ∨ (q ∩R−1), for all q ∈ Quord (A,F ).Since Quord (A,F ) is a 0-modular lattice, R and R−1 form a central pairin the lattice Quord (A,F ) (as mentioned in 1.3, cf. also [13]). Hence, byProposition 3.1(i), we get


) ∼=(Con2(A,F ),∩,∨, τ

).

(ii) is proved analogously.(iii) If


) ∼=(Con2(A, F ),∩,∨, τ

), then according to

Corollary 3.5, any θ ∈ Con (A,F ) is a congruence of the linearly ordered set(A,R). In [12] it was proved that the congruences of a �nite linearly ordered



set form a distributive sublattice of the lattice Equ (A) of all equivalence rela-tions de�ned on the set A. Since for any algebra (A,F ) the lattice Con (A,F )is a sublattice of Equ (A), in our case Con (A,F ) is also a sublattice of thelattice of all congruences of (A,R), hence Quord (A,F ) (and also Con (A,F ))is distributive.

Conversely, assume that Quord (A,F ) is a distributive lattice. Then it is0-modular, too. Hence, by applying (i) we obtain the required isomorphism.¤

4. Lattice ordered majority algebras

In this section we prove necessary and su�cient conditions for the iso-morphism


) ∼=(Tol2(A,F ),∩,t, τ

)in case of a majority

algebra (A,F ). It is well-known that all compatible relations of a majorityalgebra are determined by compatible binary re�exive relations (see e.g. [16,Proposition 4.1.13]). In our proofs we will use the following result of [20]:

Fact 4.1 [20, Proposition 6.14]. If (A,F,5) is a partially ordered ma-jority algebra such that 5 is locally bounded, then 5 is a lattice order and themeet and join operations with respect to 5 are local polynomial functions of(A,F ).

Namely, if x, y ∈ A, and u and v are a lower bound and an upper boundof {x, y}, respectively, then x ∧ y = m(x, y, u) and x ∨ y = m(x, y, v), wherem(x, y, z) denotes the majority term of (A,F ).

Corollary 4.2. If (A,F,5) is a lattice ordered majority algebra, thenany α ∈ Refl (A,F ) is preserved by the lattice operations ∨ and ∧.

Proof. Take any (x1, x2), (y1, y2) ∈ α and let u := x1 ∧ x2 ∧ y1 ∧ y2 andv := x1 ∨ x2 ∨ y1 ∨ y2. Then, in view of Fact 4.1,

(x1 ∧ y1, x2 ∧ y2) =(m(x1, y1, u),m(x2, y2, u)

) ∈ α

and(x1 ∨ y1, x2 ∨ y2) =

(m(x1, y1, v),m(x2, y2, v)

) ∈ α. ¤

A lattice L with 0 is called pseudocomplemented if for each x ∈ L thereexists an element x∗ ∈ L such that y ∧ x = 0 ⇔ y 5 x∗, for any y ∈ L. Let(A,F ) be a majority algebra. Then, according to [9], Tol (A,F ) is a 0-modularand pseudocomplemented lattice and α∗ ∈ Con(A,F ), for any α ∈ Tol (A,F ).(The particular case where (A,F ) is a lattice goes back to Bandelt [1].) ForRefl (A,F ) the same lattice properties were proved in [4], and it was shown



that α ∈ Refl (A,F ) implies α∗ ∈ Quord (A,F ). It was also proved that forany α, β, γ ∈ Tol (A,F )

α ∩ γ = 4 =⇒ (α ∩ β) t γ = (α t γ) ∩ (β t γ),

α ∩ β ∩ γ = 4 =⇒ α ∩ (β t γ) = (α ∩ β) t (α ∩ γ).

We say that two elements a, b ∈ L of a bounded lattice L form a semi-central pair if a ∧ b = 0 and x = (x ∧ a) ∨ (x ∧ b) for all x ∈ L (see also [2]).Clearly, if a, b ∈ L form a semicentral pair, then a ∨ b = 1, hence a and bare complements of each other. A pair θ1, θ2 ∈ Con (A,F ) is called a pairof factor congruences of the algebra A = (A,F ), if A ∼= A/θ1 ×A/θ2. Itis well-known, that this is equivalent to the conditions θ1 ∩ θ2 = 4, θ1 ◦ θ2

= θ2 ◦ θ1 = ∇.Remark 4.3. Note that in a 0-modular pseudocomplemented lattice L

a complement x′ of an element x ∈ L (if it exists) equals x∗, and hence it isunique. Indeed, if x′ exists, then x∧ x′ = 0, hence x′ 5 x∗. As x∨ x′ = 1, weget x ∨ x∗ = 1. Now, using 0-modularity, we obtain x∗ = (x ∨ x′) ∧ x∗ = x′.

The next lemma, which is an easy consequence of the de�nition of α t βin Refl (A,F ), will be frequently used in our proofs:

Lemma 4.4. Let (A, F ) be an algebra, α, β ∈ Refl (A,F ) and let 〈α ∪ β〉denote the subalgebra generated by the union α ∪ β in (A,F )2. Then

(i) α t β = 〈α ∪ β〉,(ii) α t β j (α ◦ β) ∩ (β ◦ α).

Proposition 4.5. Let (A,F ) be a majority algebra and let % ∈ Refl(A,F ).Then the following conditions are equivalent:

(i) %, %−1 ∈ Refl (A,F ) are complements of each other.(ii) % is a compatible lattice order of (A, F ).(iii) %, %−1 ∈ Cen

(Refl (A,F )

)and %′ = %−1.

Proof. (i) ⇒ (ii). Assume %−1 = %′. By Remark 4.3 we have %′ = %∗.Therefore, %−1 = %∗ ∈ Quord (A,F ), and hence % = (%∗)−1 is a compatiblequasiorder. Since %∩%−1 =4, % is a partial order. As %t%−1 =∇, by Lemma4.4(ii) we get (% ◦ %−1) ∩ (%−1 ◦ %) k % t %−1 = ∇, i.e. % ◦ %−1 = %−1 ◦ % = ∇.Now, the latter equalities imply that % is a locally bounded (compatible) par-tial order. Since (A,F ) is a majority algebra, according to Fact 4.1, we obtainthat % is a compatible lattice order of (A,F ).

(ii) ⇒ (iii). Assume that % is a compatible lattice order of (A,F ), andlet ∧(%),∨(%) stand for the corresponding lattice operations. Then %, %−1 ∈Refl (A,F ) and % ∩ %−1 = 4. First, we prove that %, %−1 is a semicentral



pair in the lattice Refl (A,F ) (see the de�nition before 4.3). Take any α ∈Refl (A,F ). In view of Lemma 4.4(i), we have to show that

α = (α ∩ %) t (α ∩ %−1) =⟨(α ∩ %) ∪ (α ∩ %−1)

⟩.

Clearly,⟨(α ∩ %) ∪ (α ∩ %−1)

⟩j α, hence we have to prove only the con-

verse inclusion. Take any (x, y) ∈ α. Now let u := x ∧(%) y be the greatestlower bound and v := x ∨(%) y the least upper bound of {x, y}. Then, byCorollary 4.2, we get

(x, v) = (x, x) ∨(%) (x, y) ∈ α and (v, y) = (x, y) ∨(%) (y, y) ∈ α.

Hence, by the de�nition of v, we have (x, v) ∈ α∩ % and (v, y) ∈ α ∩ %−1.Thus we get (x, v), (v, y) ∈ (α∩ %)∪ (α∩ %−1) j

⟨(α∩ %)∪ (α∩ %−1)

⟩. Now,

again by Corollary 4.2, we obtain (x, y) = (x, v) ∧(%) (v, y) ∈ ⟨(α ∩ %) ∪

(α ∩ %−1)⟩proving α j

⟨(α∩ %)∪ (α∩ %−1)

⟩. Hence % and %−1 form a semi-

central pair in Refl (A,F ). Then clearly %′ = %−1, and since Refl (A,F ) is a0-modular lattice, we obtain (cf. 1.3) %, %−1 ∈ Cen

(Refl (A,F )

).

(iii) ⇒ (i) is obvious. ¤Theorem 4.6. Let (A,F ) be a majority algebra. Then the following are

equivalent:(i) (A,F ) has a compatible lattice order.(ii)


) ∼=(Tol2(A,F ),∩,t, τ

).

Proof. (i) ⇒ (ii). Suppose that (A,F ) has a compatible lattice order %.Then, by Proposition 4.5, we get %, %−1 ∈ Cen

(Refl (A,F )

)and %′ = %−1 =

ι(%). Hence, by applying Theorem 2.1 to the involution lattice(Refl (A,F ),

∩,t, ι), we obtain (ii).

(ii) ⇒ (i). Assume(Refl (A, F ),∩,t, ι

) ∼=(Tol2(A,F ),∩,t, τ

). Now,

Proposition 3.1(ii) implies that there exists % ∈ Cen(Refl (A,F )

)such that

%−1 = %′ ∈ Cen(Refl (A,F )

). Then, in view of Proposition 4.5, % is a com-

patible lattice order of (A,F ). ¤Corollary 4.7. If L is a distributive lattice, then Refl (L) is a distribu-

tive lattice.Proof. In view of [3], any distributive lattice has a distributive tolerance

lattice. As any lattice is a lattice ordered majority algebra, by Theorem 4.6,Refl (L) ∼= Tol2(L) is also a distributive lattice. ¤

The next proposition is a generalisation of the corresponding results givenonly for lattices in [10].



Proposition 4.8. Let (A,F, %) be a lattice ordered majority algebra.Then we have:

(i) There exists a bijection between the pairs of factor congruences of(A,F ) and the pairs of compatible lattice orders of (A,F ). In particular, thepair %, %−1 corresponds to the pair ∇,4 of factor congruences.

(ii) (A,F ) is directly irreducible if and only if (A,F ) has no compatiblelattice-order di�erent from % and %−1.

Proof. (i) If (A,F, %) is a lattice ordered majority algebra, then by The-orem 4.6 we have

(Tol2(A, F ),∩,t, τ

) ∼=(Refl (A,F ),∩,t, ι

), and in view

of Theorem 2.1, the isomorphism can be given by the mapping

g : Tol2(A, F ) → Refl (A,F ), g(α, β) := (α ∩ %) t (β ∩ %′).

Let θ1, θ2 be a pair of factor congruences of (A,F ). In view of [9], this isequivalent to the fact that θ1 and θ2 are complements of each other in thelattice Tol (A,F ). Now, let σ1 = g(θ1, θ2) and σ2 = g(θ2, θ1). Then (θ2, θ1) isthe complement of (θ1, θ2) in Tol2(A,F ), and hence by isomorphism we getσ2 = σ′1 and σ2 = g

(τ(θ1, θ2)

)= ι

(g(θ1, θ2)

)= ι(σ1) = σ−1

1 . Thus σ′1 = σ−11

in Refl (A,F ), and applying Proposition 4.5 we get that σ1 and σ2 = σ−11 are

compatible lattice orders of (A,F ).Conversely, if σ is a compatible lattice order of (A,F ), then in view

of Proposition 4.5, σ−1 = σ′ in the lattice Refl (A,F ). Take (θ1, θ2) :=g−1(σ). As g−1 is an isomorphism of involution algebras, we get g−1(σ−1) =g−1

(ι(σ)

)= τ

(g−1(σ)

)= (θ2, θ1) and (θ2, θ1) = (θ1, θ2)′ in Tol2(A,F ). Then

θ1 ∩ θ2 = 4 and θ1 t θ2 = ∇ in Tol (A,F ), and hence, in view of [9], θ1, θ2 isa pair of factor congruences of (A,F ). Therefore, restricting g we obtain abijection between the pairs of factor congruences of (A,F ) and the pairs ofcompatible lattice orders of (A, F ). It can be readily seen that g(∇,4) = %

and g(4,∇) = %′ = %−1.(ii) is clear. ¤In view of [19] a linearly ordered majority algebra (A,F, %) has no lo-

cally bounded compatible partial orders di�erent from % and %−1. Hence, byProposition 4.8(ii), we obtain

Corollary 4.9. Any linearly ordered majority algebra is directly irre-ducible.

Proposition 4.10. Let A = (A,F, %) be a lattice ordered majority alge-bra. Then the following are equivalent:

(i) Refl (A, F ) = Quord (A,F ).(ii) Tol (A,F ) = Con (A,F ).Proof. (i) ⇒ (ii). Clearly, (i) implies that any tolerance of (A,F ) is a

congruence, i.e. Tol (A,F ) = Con (A,F ).



(ii) ⇒ (i). Suppose Tol (A,F ) = Con (A,F ). Then, by Theorem 4.6,any σ ∈ Refl (A,F ) has the form σ = (θ1 ∩ %) t (θ2 ∩ %−1), for some θ1, θ2

∈ Con (A,F ). First, we prove

(]) (θ1 ∩ %) t (θ2 ∩ %−1) = (θ1 ∩ %) ◦ (θ2 ∩ %−1) = (θ2 ∩ %−1) ◦ (θ1 ∩ %).

By Lemma 4.4(ii) we have

([) (θ1 ∩ %) t (θ2 ∩ %−1) j (θ1 ∩ %) ◦ (θ2 ∩ %−1).

In order to show (]), take any (a, b) ∈ (θ1 ∩ %) ◦ (θ2 ∩ %−1). Then there is av ∈ A such that (a, v) ∈ θ1∩% and (v, b) ∈ θ2∩%−1. Thus we have (a, v), (v, b)∈ (θ1∩%)t (θ2∩%−1). Denoting by ∧(%) and ∨(%) the meet and join operationof the lattice (A, %), in view of Corollary 4.2, we obtain (a, b) = (a, v) ∧(%)

(v, b) ∈ (θ1 ∩ %)t (θ2 ∩ %−1). Thus we get (θ1 ∩ %) ◦ (θ2 ∩ %−1) j (θ1 ∩ %)t (θ2

∩ %−1), and now (]) follows from ([) by the (θ1, %)�(θ2, %−1) symmetry.

Since the relations θ1 ∩ % and θ2 ∩ %−1 are permutable quasiorders of(A,F ), we obtain

(θ1 ∩ %) ∨ (θ2 ∩ %−1) = (θ1 ∩ %) ◦ (θ2 ∩ %−1) = (θ1 ∩ %) t (θ2 ∩ %−1) = σ.

Hence σ ∈ Quord (A,F ), and this proves Refl (A,F ) = Quord (A,F ). ¤

5. Compatible re�exive relations on a lattice ordered majorityalgebra

It was proved in [4] that for any majority algebra, Refl (A, F ) is a pseu-docomplemented and 0-modular lattice. Now, as an immediate consequenceof Theorem 4.6 we obtain

Corollary 5.1. If (A,F, %) is a lattice ordered majority algebra andα, β, γ ∈ Refl (A,F ), then

α ∩ γ = 4 =⇒ (α ∩ β) t γ = (α t γ) ∩ (β t γ),(1)

α ∩ β ∩ γ = 4 =⇒ α ∩ (β t γ) = (α ∩ β) t (α ∩ γ).(2)

Proof. As (A,F ) is a majority algebra, the implications (1) and (2) aresatis�ed in Tol (A,F ), for any α, β, γ ∈ Tol (A,F ) (see [4]). By Theorem 4.6,(Refl (A,F ),∩,t, ι

) ∼=(Tol2(A,F ),∩,t, τ

), thus the same implications are

satis�ed by any α, β, γ ∈ Refl (A,F ). ¤



Proposition 5.2. Let (A,F ) be a majority algebra and ν a complementedelement in the lattice Refl (A,F ). Then

(i) ν ◦ α = α ◦ ν = α t ν holds for every α ∈ Refl (A,F ),(ii) if (A,F ) is lattice ordered, ν is a central element of Refl (A, F ).Proof. (i) Let ν ′ be the complement of ν in the lattice Refl (A,F )

(ν ′ is unique since Refl (A,F ) is 0-modular and pseudocomplemented, cf. [4]).Then by Lemma 4.4(ii) we get ∇ = ν t ν ′ j (ν ◦ ν ′) ∩ (ν ′ ◦ ν), whence ν ◦ ν ′

= ν ′ ◦ ν = ∇. By symmetry, it su�ces to show α t ν = α ◦ ν. The j partis evident (according to Lemma 4.4(ii)). To show the reverse inclusion α t νk α ◦ ν, assume that (a, b) ∈ α ◦ ν. Then there is an x ∈ A with (a, x) ∈ αand (x, b) ∈ ν. Since ν ◦ ν ′ = ν ′ ◦ ν = ∇, there exist u, t ∈ A with (a, u), (t, b)∈ ν and (u, b), (a, t) ∈ ν ′. Since (m(x, u, b), b) =

(m(x, u, b),m(b, u, b)

) ∈ ν,(m(x,u, b), b

)=

(m(x,u, b),m(x, b, b)

) ∈ ν ′ and ν ∩ν ′ =4, we get m(x,u, b)= b. Hence (a, b) =

(m(a, a, b),m(x, u, b)

) ∈ α t ν. This proves part (i).(ii) If (A,F ) is a lattice ordered majority algebra, then by Corollary 5.1(2)

we obtain α = α∩ (ν t ν ′) = (α∩ ν)t (α∩ ν ′), for any α ∈ Refl (A,F ). SinceRefl (A,F ) is a 0-modular lattice, in view of [13], ν (and ν ′) is a centralelement of it. ¤

Corollary 5.3. Let ν j A×A be a compatible lattice order or a factorcongruence of a lattice ordered majority algebra (A,F ). Then ν is a central el-ement of Refl (A,F ) and ν ◦α = α◦ν = αtν holds for every α ∈ Refl (A,F ).

Proof. If ν is a compatible lattice order of (A,F ), then our assertion fol-lows directly from Proposition 4.5 and Proposition 5.2. Now, let ν and ν ′ bea pair of factor congruences of (A,F ). Then, by [9], ν and ν ′ are complementsof each other in the lattice Tol (A,F ), i.e. ν ∩ ν ′ = 4 and ν t ν ′ = ∇. Henceν is also a complemented element in Refl (A,F ), so applying Proposition 5.2we get the required result. ¤

Now, let A =∏n

i=1Ai be a �nite direct product of algebras Ai andπi : A → Ai, πi(x1, . . . , xn) = xi be the corresponding natural projections,i ∈ {1, . . . , n}, n = 1. For any system αi j A2

i , i ∈ {1, . . . , n} of compatiblebinary relations of the algebras Ai de�ne the relation α j A2 by

(a, b) ∈ α :⇐⇒ (πi(a), πi(b)

) ∈ αi, for all i ∈ {1, . . . , n}.

Then α is a compatible relation of A, denoted by∏n

i=1 αi, which is called thedirect product of the relations αi, i ∈ {1, . . . , n}. Obviously, if each αi j A2

i isre�exive, then

∏ni=1 αi is also re�exive. Notice also, that each Ai is a lattice

ordered majority algebra, whenever A =∏n

i=1Ai is a lattice ordered major-ity algebra. We will use also the notation x := (x1, . . . , xn) for elements of∏n

i=1Ai.



The following statement is a generalization of the so called Fraser�Hornproperty of the congruences of a congruence-distributive algebra (see [11,Corollary 1]).

Theorem 5.4. Let A =∏n

i=1Ai be a lattice ordered majority algebra.Then any α ∈ Refl (A) is equal to a direct product of some αi ∈ Refl (Ai),i ∈ {1, . . . , n}, and Refl (A) ∼= ∏n

i=1 Refl (Ai).Proof. Let α ∈ Refl (A). Then it is easy to see that the relations

πi(α) :={

(x, y) ∈ A2i | ∃(u, v) ∈ α : x = πi(u), y = πi(v)

}

are compatible re�exive relations on the algebras Ai, i ∈ {1, . . . , n}. Let νi :=kerπi j A2, i ∈ {1, . . . , n}. In order to prove α =

∏ni=1 πi(α), observe �rst

thatn∏

i=1

πi(α) =n⋂

i=1

(νi ◦ α ◦ νi).

Indeed, for any a, b ∈ A we have

(a, b) ∈n∏

i=1

πi(α) ⇐⇒ ∀i ∈ {1, . . . , n} :(πi(a), πi(b)

) ∈ πi(α)

⇐⇒ ∀i ∈ {1, . . . , n} : ∃(u(i), v(i)) ∈ α with

πi(a) = πi(u(i)), πi(b) = πi(v(i)).

However, the latter formula is equivalent to (a, b) ∈ νi ◦ α ◦ νi, for alli ∈ {1, . . . , n}, i.e. to (a, b) ∈ ⋂n

i=1(νi ◦ α ◦ νi).Now, since each νi is a factor congruence of A, by Corollary 5.3 we get

νi ◦ α ◦ νi = α ◦ νi ◦ νi = α ◦ νi = α t νi. Hence, we obtain

(+)n∏

i=1

πi(α) =n⋂

i=1

(α t νi).

Further, according to Corollary 5.3, every νi is a central element in thelattice Refl (A). Therefore, for any α ∈ Refl (A) the sublattice of Refl (A)generated by α, ν1, . . . , νn is distributive. Since

⋂ni=1 νi = 4, we get

(++) α = α t( n⋂

i=1

νi

)=

n⋂

i=1

(α t νi).



Summarizing (+) and (++), we obtain α =⋂n

i=1(α t νi) =∏n

i=1 πi(α), andthis completes the proof of the �rst assertion.

To prove Refl (A) ∼= ∏ni=1 Refl (Ai), consider the maps

Φ : Refl (A) →n∏

i=1

Refl (Ai), Ψ :n∏

i=1

Refl (Ai) → Refl (A),

de�ned by

Φ(α) =(π1(α), . . . , πn(α)

)and Ψ(α1, . . . , αn) =

n∏

i=1

αi.

Then Φ(Ψ(α1, . . . , αn)

)= (α1, . . . , αn) by de�nition, and Ψ

(Φ(α)

)= α, ac-

cording to the previous proof, i.e. Φ and Ψ are inverses of each other. Itis easy to check that both Φ and Ψ are order-preseving, therefore they arelattice isomorphisms. Thus Refl (A) ∼= ∏n

i=1 Refl (Ai). ¤Acknowledgment. The authors thank the anonymous referee for valu-

able remarks and helpful suggestions which made the paper more readable.

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