Related structures with involution

Transcription

1 Related structures with involution R. Pöschel 1 and S. Radeleczki 2 1 Institut für Algebra, TU Dresden, Dresden, Germany Reinhard.Poeschel@tu-dresden.de 2 Institute of Mathematics, University of Miskolc, 3515 Miskolc-Egyetemváros, Hungary matradi@gold.uni-miskolc.hu Abstract: We give a necessary and su cient condition for a bounded involution lattice to be isomorphic to the direct square of its invariant part. This result is 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 algebra admits a connected compatible partial order whenever its quasiorder lattice is isomorphic to the direct square of its congruence lattice. Further, a majority algebra is lattice ordered if and only if the lattice of its compatible re exive relations is isomorphic to the direct square of its tolerance lattice. In the latter case, one can establish a bijective correspondence between factor congruence pairs of the algebra and its pairs of compatible lattice orders; several consequences of this result 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 lattices with respect to inclusion. We denote them by Re (A; F ), Tol(A; F ), Quord(A; F ) and Con(A; F ), respectively. These lattices are not independent: Tol(A; F ) is a sublattice of the lattice Re (A; F ), Con(A; F ) is a sublattice of Quord(A; F ), moreover, the in mum of a system f i j i 2 Ig of elements in each of these lattices is equal to the set theoretical intersection \ f i j i 2 Ig. Therefore, as algebraic structures with meet and join, we shall use the notations (Re (A; F ); \; t), (Tol(A; F ); \; t), (Quord(A; F ); \; _) and (Con(A; F ); \; _), respectively. Let 1 = f(a; b) j (b; a) 2 g denote the inverse of a relation j A A. Then the mapping 0 Key words and phrases: Involution lattice, central element, quasiorder lattice and tolerance lattice of an algebra, compatible partial order of an algebra, lattice ordered majority algebra Mathematics Subject Classi cation: Primary: 08A05, 06B05; Secondary: 08A02, 06F99. 1

2 2 : 7! 1 is an involution both in the lattice (Re (A; F ); \; t) and in (Quord(A; F ); \; _). As shown in [6] and [19], in the case of a partially ordered algebra (A; F; ), there are some other relations between the above lattices. For instance, in [6] is proved that for a lattice ordered majority algebra (A; F; ), the algebra (Quord(A; F ); \; _; ) is isomorphic to the algebra (Con 2 (A; F ); \; _; ), where ( 1 ; 2 ) = ( 2 ; 1 ) for ( 1 ; 2 ) 2 Con 2 (A; F ). In this paper several generalizations of this theorem are given. By using Theorem 2.1 we prove necessary and su cient conditions for the isomorphism (Quord(A; F ); \; _; ) = (Con 2 (A; F ); \; _; ), as well as for the isomorphism (Re (A; F ); \; t; ) = (Tol 2 (A; F ); \; _; ) in the case of a general algebra (A; F ). Moreover, we show that a majority algebra is lattice ordered if 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 that there exists a bijective correspondence between the pairs of factor congruences of a lattice ordered majority algebra (A; F ) and its pairs of lattice orders (cf. Proposition 4.8(i)) and the conditions Re (A; F ) = Quord(A; F ) and Tol(A; F ) = Con(A; F ) are equivalent (cf. Proposition 4.10). The complemented elements of the lattice Re (A; F ) are also characterized (see Proposition 5.2). 1. Preliminaries 1.1 Let A A be a binary relation on a set A. If is re exive and symmetric, then it is called a tolerance on the set A, and if it is re exive and transitive, then it is called a quasiorder on A. The re exive relations, the tolerances, and the quasiorders on a set A form algebraic lattices with respect to, denoted by Re (A), Tol(A), and Quord(A), respectively. Their least and greatest elements are 4 = f(a; a) j a 2 Ag and r = A A. A tolerance of an algebra (A; F ) is a tolerance relation on the set A compatible with the operations of the algebra (A; F ), and a quasiorder of an algebra (A; F ) is a compatible quasiorder q j A A. Obviously, we have 2 Re (A; F ) () 1 2 Re (A; F ), q 2 Quord(A; F ) () q 1 2 Quord(A; F ), q is a partial order () q 2 Quord(A; F ) and q \ q 1 = 4. Let % denote the transitive closure of a relation % AA. A partial order r A A is called connected if r [ r 1 = r, and it is called locally bounded if any two elements a; b 2 A have an upper bound and a lower bound in the poset (A; r). Clearly, if 2 Re (A; F ), then

3 2 Quord(A; F ). The supremum for q i 2 Quord(A; F ), i 2 I, in the lattice Quord(A; F ) is given by _ fqi j i 2 Ig = [ fq i j i 2 Ig. Thus Quord(A; F ) is a complete sublattice of Quord(A) (see e.g. [17]). A quadruple (L; ^; _; ) is called an involution lattice if is an automorphism of the lattice L = (L; ^; _) such that 2 (x) = x. For an involution lattice L = (L; ^; _; ), the set I = fx 2 L j (x) = xg of xed points of is a subalgebra (I; ^; _; ) which is called the invariant part of L. Of course, the restriction of 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 a bounded lattice, (i.e. a lattice with least element 0 and greatest element 1) then (L; ^; _; ) is called a bounded involution lattice and it follows (0) = 0, (1) = Example (1) Clearly, for every lattice L the algebra (L; ^; _; id) is an involution lattice, with invariant part I = L. (2) For any lattice L its direct square L 2 becomes an involution lattice (L 2 ; ^; _; ) if we de ne (x; y) = (y; x), for all (x; y) 2 L 2. (3) Let (A; F ) be any algebra. Then (Quord(A; F ); \; _; ) is a bounded involution lattice, where the involution is given by (q) = q 1, q 2 Quord(A; F ), and its invariant part is I = (Con(A; F ); \; _; id) (see e.g. [17] and [6]). (4) Analogously, the algebra (Re (A; F ); \; t; ), with () = 1, 2 Re (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 2 L is called a central element of L if a is complemented and for all x; y 2 L the sublattice generated by fa; x; yg is distributive. A complement of an element a 2 L (if it exists) will be denoted by a 0. The central elements of L form a Boolean sublattice denoted by Cen(L). Obviously, their complements are uniquely determined. If c 2 Cen(L), then c 0 2 Cen(L) and c; c 0 is called a central pair of L. Let (a] = fx 2 L j x 5 ag denote the principal ideal corresponding to an element a 2 L. It is well known that, for any direct product decomposition L = Q i2il i of a bounded lattice L there exist elements c i 2 Cen(L), i 2 I such that L i = (ci ] (see e.g. [18]). The central pairs of a bounded lattice have several characterizations. For instance, in [13] we can nd the following: 3

4 4 c; c 0 2 Cen(L) () x = (x ^ c) _ (x ^ c 0 ), 8x 2 L (1) x = (x _ c) ^ (x _ c 0 ), 8x 2 L (2) A lattice L with least element 0 is called 0-modular, if for any a; b; c 2 L the implication (M 0 ) (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 only if it does not contain an N 5 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 2 Cen(L) implies (c) 2 Cen(L) for any bounded involution lattice (L; ^; _; ). Hence, denoting by C the restriction of to Cen(L) we obtain that (Cen(L); ^; _; C ) is a subalgebra of (L; ^; _; ). 2. A structure theorem for involution lattices Our starting point was 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 isomorphisms given by formulas (3) and (4). 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 Example 1.2(3)), the present paper generalizes this in many ways and presents a lot of related results. Theorem 2.1. Let (L; ^; _; ) be a bounded involution lattice, I its invariant part and consider the mapping (x; y) = (y; x), for all (x; y) 2 I 2. Then (L; ^; _; ) = (I 2 ; ^; _; ) if and only if there exists an element c 2 Cen(L) such that (c) = c 0. Proof. Clearly, (I 2 ; ^; _; ) is an involution lattice. Assume that (c 0 ) = c, for some c 2 Cen(L) and consider the mappings f : L! I 2 and g : I 2! L de ned by f(x) := ((x ^ c) _ ((x) ^ c 0 ); (x ^ c 0 ) _ ((x) ^ c)), (3) g(x; y) := (x ^ c) _ (y ^ c 0 ). (4) It is easy to see that both f and g are order-preserving functions, hence to prove that they are lattice isomorphisms, it is enough to show that they are inverses each of other. Indeed, for any (x; y) 2 I 2 we get g(x; y) ^ c = [(x ^ c) _ (y ^ c 0 )] ^ c = (x ^ c ^ c) _ (y ^ c 0 ^ c) = x ^ c, (g(x; y))^c 0 = [((x)^c 0 )_((y)^c)]^c 0 = (x^c 0^c 0 )_(y^c^c 0 ) = x^c 0,

5 g(x; y) ^ c 0 = [(x ^ c) _ (y ^ c 0 )] ^ c 0 = y ^ c 0, and (g(x; y)) ^ c = [((x) ^ c 0 ) _ ((y) ^ c)] ^ c = (y) ^ c = y ^ c, hence, we obtain: (f g)(x; y) = = ((g(x; y) ^ c) _ ((g(x; y)) ^ c 0 ); (g(x; y) ^ c 0 ) _ ((g(x; y)) ^ c)) = = ((x ^ c) _ (x ^ c 0 ); (y ^ c 0 ) _ (y ^ c)), and 1.3 (2) implies (f g)(x; y) = (x; y). Further, for any x 2 L we obtain (g f)(x) = ([(x ^ c) _ ((x) ^ c 0 )] ^ c) _ ([(x ^ c 0 ) _ ((x) ^ c)] ^ c 0 ) = [(x^c^c)_((x)^c 0^c)]_[(x^c 0^c 0 )_((x)^c^c 0 )] = (x^c)_(x^c 0 ) = x, proving that f and g are inverses each of other. Now, it is easy to check that f((x)) = (f(x)), for any x 2 L and g((x; y)) = (g(x; y)), for any x; y 2 I, thus we obtain that f and g are involution lattice isomorphisms, and hence (L; ^; _; ) = (I 2 ; ^; _; ). Conversely, assume that (L; ^; _; ) = (I 2 ; ^; _; ). It is easy to see that (1; 0) and (0; 1) is a central pair in the lattice (I 2 ; ^; _) and (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 2 Cen(L) and c ^ c = 0, c _ c = 1 by isomorphism, thus c and c is also a central pair in L and, by uniqueness of the complement, c = c 0. As (1; 0) = (0; 1) in I 2, we obtain by isomorphism (c) = c = c 0 in L, completing the proof. Now, consider a bounded involution lattice (L; ^; _; ) with invariant part I. Then (Cen(L); ^; _; C ) is also an involution lattice and it is easy to see that its invariant part is I C = I \ Cen(L). With the notations of Theorem 2.1 we have: Corollary 2.2. The following conditions are equivalent: (i) (L; ^; _; ) = (I 2 ; ^; _; ), (ii) (Cen(L); ^; _; C ) = (IC 2 ; ^; _; ), (iii) There exists a lattice K = (K; ^; _) such that (Cen(L); ^; _; C ) = (K 2 ; ^; _; ). Proof. (i))(ii): By Theorem 2.1, (i) is equivalent to the condition: () 9c; c 0 2 Cen(L) : (c) = c 0. But then () is also satis ed by (Cen(L); ^; _; C ) since the centre of the lattice Cen(L) is Cen(L) itself. Hence, by Theorem 2.1, we get (ii). (ii))(iii) is obvious. 5

6 6 (iii))(i): As Cen(L) is a bounded lattice, K must be also bounded. Let 0 K and 1 K stand for the least and the greatest element of K, respectively. Then (0 K ; 1 K ) and (1 K ; 0 K ) are complements of each other in the lattice (K 2 ; ^; _) and ((0 K ; 1 K )) = (1 K ; 0 K ). Denote by c and c the elements in Cen(L) corresponding by isomorphism to (0 K ; 1 K ) and (1 K ; 0 K ), respectively. Then by the same argument as in the proof of Theorem 2.1, we get C (c) = c and c = c 0. Hence, (c) = C (c) = c 0, proving that condition () holds for (L; ^; _; ), 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 general algebras. Let (A; F ) be any algebra. It is easy to see that, q; q 1 2 Quord(A; F ) are complements of each other in Quord(A; F ) if and only if q is a connected partial order. Similarly, % 0 = (%) = % 1, for some % 2 Re (A; F ), if and only if % j A A is an antisymmetric relation and % t % 1 = r. From these observations and Theorem 2.1 it follows immediately Proposition 3.1. Let (A; F ) be any algebra. Then we have: (i) (Quord(A; F ); \; _; ) = (Con 2 (A; F ); \; _; ) if and only if there exists a connected partial order % j A A such that % 2 Cen(Quord(A; F )). (ii) (Re (A; F ); \; t; ) = (Tol 2 (A; F ); \; t; ) if and only if there exists an antisymmetric re exive relation % j A A such that % 2 Cen(Re (A; F )) and % t % 1 = r. (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 2 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 complemented pairs of elements, we obtain: Corollary 3.2. Let (A; F ) be a majority algebra. Then (Quord(A; F ); \; _; ) = (Con 2 (A; F ); \; _; ) 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 i 2 Re (A; F ), i 2 I we have F f i j i 2 Ig = S f i j i 2 Ig. Therefore, (Re (A; F ); \; t) and consequently (Tol(A; F ); \; t) are distributive lattices. Hence, by applying Proposition 3.1, we obtain

7 Corollary 3.3. If (A; F ) is a unary algebra, then (Re (A; F ); \; t; ) = (Tol 2 (A; F ); \; t; ) if and only if (A; F ) admits an antisymmetric re exive compatible relation % j A A such that % t % 1 = r. Now, let A be a nonempty set and F = fid A g. Then obviously, Re (A) = Re (A; F ) and Tol(A) = Tol(A; F ). Since on any nonempty set there exists a linear order R j A A and since R [ R 1 = r, from Corollary 3.3 we obtain the following Corollary 3.4. For any nonempty set A we have (Re (A); \; [; ) = (Tol 2 (A); \; [; ). Let (A; %) be a partially ordered set. An equivalence relation j A A is called a congruence of the poset (A; %) if it is the kernel of a %-preserving mapping, i.e. if there exists a poset (B; % ) and a mapping f : A! B such that (a; b) 2 % =) (f(a); f(b)) 2 % and = Kerf = f(x; y) 2 A A j f(x) = f(y)g. In [7] it was proved that an equivalence is a congruence if and only if = ( _ ) \ ( _ 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 congruence 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 ); \; _; ) = (Con 2 (A; F ); \; _; ), then there exists a connected compatible partial order % of (A; F ) such that every 2 Con(A; F ) is an order-congruence of (A; F; %). Proof. Assume that (Quord(A; F ); \; _; ) = (Con 2 (A; F ); \; _; ). Then, by Proposition 3.1(i), there exists a connected partial order % j A A such that % 2 Cen(Quord(A; F )). Hence % 1 = % 0 also belongs to Cen(Quord(A; F )), and so the relation = ( _ %) \ ( _ % 1 ) holds in Quord(A; F ), for all 2 Con(A; F ). As Quord(A; F ) is a sublattice of Quord(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 ); \; _; ) = (Con 2 (A; F ); \; _; ). (ii) If Re (A; F ) is a 0-modular lattice, then (Re (A; F ); \; t; ) = (Tol 2 (A; F ); \; t; ). 7

8 8 (iii) If A is nite, then (Quord(A; F ); \; _; ) = (Con 2 (A; F ); \; _; ) 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 2 Quord(A; F ). Since Quord(A; F ) is a 0-modular lattice, R and R 1 form a central pair in the lattice Quord(A; F ) (as mentioned in 1.3, cf. also [13]). Hence, by Proposition 3.1(i), we get (Quord(A; F ); \; _; ) = (Con 2 (A; F ); \; _; ). (ii) is proved analogously. (iii) If (Quord(A; F ); \; _; ) = (Con 2 (A; F ); \; _; ), then according to Corollary 3.5, any 2 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 relations 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 the lattice 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 is 0-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 isomorphism (Re (A; F ); \; t; ) = (Tol 2 (A; F ); \; t; ) in case of a majority algebra (A; F ). It is well-known, that all compatible relations of a majority algebra are determined by compatible binary re exive relations (see e.g. [16, Prop ]). In our proofs we will use the following result of [20]: Fact 4.1. [20, Proposition 6.14]: If (A; F; ) is a partially ordered majority algebra such that is locally bounded, then is a lattice order and the meet and join operations with respect to are local polynomial functions of (A; F ). Namely, if x; y 2 A; and u and v are a lower bound and an upper bound of fx; yg, respectively, then x ^ y = m(x; y; u) and x _ y = m(x; y; v), where m(x; y; z) denotes the majority term of (A; F ). Corollary 4.2. If (A; F; ) is a lattice ordered majority algebra, then any 2 Re (A; F ) is preserved by the lattice operations _ and ^.

9 9 Proof. Take any (x 1 ; x 2 ), (y 1 ; y 2 ) 2 and let u := x 1 ^ x 2 ^ y 1 ^ y 2 and v := x 1 _ x 2 _ y 1 _ y 2. Then, in view of Fact 4.1, (x 1 ^ y 1 ; x 2 ^ y 2 ) = (m(x 1 ; y 1 ; u); m(x 2 ; y 2 ; u)) 2 and (x 1 _ y 1 ; x 2 _ y 2 ) = (m(x 1 ; y 1 ; v); m(x 2 ; y 2 ; v)) 2. A lattice L with 0 is called pseudocomplemented if for each x 2 L there exists an element x 2 L such that y ^ x = 0, y x, for any y 2 L. Let (A; F ) be a majority algebra. Then, according to [9], Tol(A; F ) is a 0-modular and pseudocomplemented lattice and 2 Con(A; F ), for any 2 Tol(A; F ). (The particular case where (A; F ) is a lattice goes back to Bandelt [1].) For Re (A; F ) the same lattice properties were proved in [4], and it was shown that 2 Re (A; F ) implies 2 Quord(A; F ). It was also proved that for any ; ; 2 Tol(A; F ) \ = 4 =) ( \ ) t = ( t ) \ ( t ), \ \ = 4 =) \ ( t ) = ( \ ) t ( \ ). We say that two elements a; b 2 L of a bounded lattice L form a semicentral pair, if a ^ b = 0 and x = (x ^ a) _ (x ^ b), for all x 2 L (see also [2]). Clearly, if a; b 2 L form a semicentral pair, then a _ b = 1, hence a and b are complements of each other. A pair 1, 2 2 Con(A; F ) is called a pair of factor congruences of the algebra A = (A; F ), if A = A= 1 A= 2. It is well-known, that this is equivalent to the conditions 1 \ 2 = 4, 1 2 = 2 1 = r. Remark 4.3. Note that in a 0-modular pseudocomplemented lattice L a complement x 0 (if it exists) of an element x 2 L equals x, and hence it is unique. Indeed, if x 0 exists, then x^x 0 = 0, hence x 0 5 x. As x_x 0 = 1, we get x _ x = 1. Now, using 0-modularity, we obtain x = (x _ x 0 ) ^ x = x 0. The next lemma, which is an easy consequence of the de nition of t in Re (A; F ),.will be frequently used in our proofs: Lemma 4.4. Let (A; F ) be an algebra, ; 2 Re (A; F ) and let h[i denote the subalgebra generated in by the union [ in (A; F ) 2. Then (i) t = h [ i. (ii) t j ( ) \ ( ).

10 10 Proposition 4.5. Let (A; F ) be a majority algebra and let % 2 Re (A; F ). Then the following conditions are equivalent: (i) %,% 1 2 Re (A; F ) are complements of each other, (ii) % is a compatible lattice order of (A; F ), (iii) %; % 1 2 Cen(Re (A; F )) and % 0 = % 1. Proof. (i))(ii): Assume % 1 = % 0. By Remark 4.3 we have % 0 = %. Therefore, % 1 = % 2 Quord(A; F ), and hence % = (% ) 1 is a compatible quasiorder. Since % \ % 1 = 4, % is a partial order. As %t% 1 = r, by Lemma 4.4(ii) we get (%% 1 )\(% 1 %) k %t% 1 = r, i.e. % % 1 = % 1 % = r. Now, the latter equalities imply that % is a locally bounded (compatible) partial order. Since (A; F ) is a majority algebra, according to Fact 4.1, we obtain that % is a compatible lattice order of (A; F ). (ii))(iii): Assume that % is a compatible lattice order of (A; F ), and let ^(%) ; _ (%) stand for the corresponding lattice operations. Then %; % 1 2 Re (A; F ) and %\% 1 = 4. First, we prove that %; % 1 is a semicentral pair in the lattice Re (A; F ) (see de nition before 4.3). Take any 2 Re (A; F ). In view of Lemma 4.4(i), we have to show that = ( \ %) t ( \ % 1 ) = h( \ %) [ ( \ % 1 )i. Clearly, h( \ %) [ ( \ % 1 )i j, hence we have to prove only the converse inclusion. Take any (x; y) 2. Now let u := x ^(%) y be the greatest lower bound and v := x _ (%) y the least upper bound of fx; yg. Then, by Corollary 4.2, we get (x; v) = (x; x) _ (%) (x; y) 2 and (v; y) = (x; y) _ (%) (y; y) 2. Hence, by the de nition of v, we have (x; v) 2 \% and (v; y) 2 \% 1. Thus we get (x; v); (v; y) 2 ( \ %) [ ( \ % 1 ) j h( \ %) [ ( \ % 1 )i. Now, again by Corollary 4.2, we obtain (x; y) = (x; v) ^(%) (v; y) 2 h( \ %) [ ( \ % 1 )i proving j h( \ %) [ ( \ % 1 )i. Hence % and % 1 form a semicentral pair in Re (A; F ). Then clearly % 0 = % 1, and since Re (A; F ) is a 0-modular lattice, we obtain (cf. 1.3) %; % 1 2 Cen(Re (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) (Re (A; F ); \; t; ) = (Tol 2 (A; F ); \; t; ).

11 Proof. (i))(ii): Suppose that (A; F ) has a compatible lattice order %. Then, by Proposition 4.5, we get %; % 1 2 Cen(Re (A; F )) and % 0 = % 1 = (%). Hence, by applying Theorem 2.1 to the involution lattice (Re (A; F ); \; t; ), we obtain (ii). (ii))(i). Assume (Re (A; F ); \; t; ) = (Tol 2 (A; F ); \; t; ). Now, Proposition 3.1(ii) implies that there exists % 2 Cen(Re (A; F )) such that % 1 = % 0 2 Cen(Re (A; F )). Then, in view of Proposition 4.5, % is a compatible lattice order of (A; F ). Corollary 4.7. If L is a distributive lattice, then Re (L) is a distributive 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, Re (L) = Tol 2 (L) is also a distributive lattice. The next proposition is a generalisation of the corresponding results given only 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, the pair %; % 1 corresponds to the pair r; 4 of factor congruences. (ii) (A; F ) is directly irreducible if and only if (A; F ) has no compatible lattice-order di erent from % and % 1. Proof. (i) If (A; F; %) is a lattice ordered majority algebra, then by Theorem 4.6 we have (Tol 2 (A; F ); \; t; ) = (Re (A; F ); \; t; ), and in view of Theorem 2.1, the isomorphism can be given by the mapping g : Tol 2 (A; F )! Re (A; F ), g(; ) := ( \ %) t ( \ % 0 ). Let 1 ; 2 be a pair of factor congruences of (A; F ). In view of [9], this is equivalent to the fact that 1 and 2 are complements of each other in the lattice Tol(A; F ). Now, let 1 = g( 1 ; 2 ) and 2 = g( 2 ; 1 ). Then ( 2 ; 1 ) is the complement of ( 1 ; 2 ) in Tol 2 (A; F ), and hence by isomorphism we get 2 = 0 1 and 2 = g(( 1 ; 2 )) = (g( 1 ; 2 )) = ( 1 ) = 1 1. Thus 0 1 = 1 1 in Re (A; F ), and applying Proposition 4.5 we get that 1 and 2 = 1 1 are compatible lattice orders of (A; F ). Conversely, if is a compatible lattice order of (A; F ), then in view of Proposition 4.5, 1 = 0 in the lattice Re (A; F ). Take ( 1 ; 2 ) := g 1 (). As g 1 is an isomorphism of involution algebras, we get g 1 ( 1 ) = g 1 (()) = (g 1 ()) = ( 2 ; 1 ) and ( 2 ; 1 ) = ( 1 ; 2 ) 0 11

12 12 in Tol 2 (A; F ). Then 1 \ 2 = 4 and 1 t 2 = r in Tol(A; F ), and hence, in view of [9], 1 ; 2 is a pair of factor congruences of (A; F ). Therefore, restricting g we obtain a bijection between the pairs of factor congruences of (A; F ) and the pairs of compatible lattice orders of (A; F ). It can be readily seen that g(r; 4) = % and g(4; r) = % 0 = % 1. (ii) is clear. In view of [19] a linearly ordered majority algebra (A; F; %) has no locally bounded compatible partial orders di erent from % and % 1. Hence, by Proposition 4.8(ii), we obtain Corollary 4.9. Any linearly ordered majority algebra is directly irreducible. Proposition Let A = (A; F; %) be a lattice ordered majority algebra. Then the following are equivalent: (i) Re (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 2 Re (A; F ) has the form = ( 1 \ %) t ( 2 \ % 1 ), for some 1 ; 2 2 Con(A; F ). First, we prove (]) ( 1 \ %) t ( 2 \ % 1 ) = ( 1 \ %) ( 2 \ % 1 ) = ( 2 \ % 1 ) ( 1 \ %): By Lemma 4.4(ii) we have ([) ( 1 \ %) t ( 2 \ % 1 ) j ( 1 \ %) ( 2 \ % 1 ). In order to show (]), take any (a; b) 2 ( 1 \ %) ( 2 \ % 1 ). Then there is a v 2 A such that (a; v) 2 1 \ % and (v; b) 2 2 \ % 1. Thus we have (a; v); (v; b) 2 ( 1 \ %) t ( 2 \ % 1 ). Denoting by ^(%) and _ (%) the meet and join operation of the lattice (A; %), in view of Corollary 4.2, we obtain (a; b) = (a; v) ^(%) (v; b) 2 ( 1 \ %) t ( 2 \ % 1 ). Thus we get ( 1 \ %) ( 2 \ % 1 ) j ( 1 \ %) t ( 2 \ % 1 ), and now (]) follows from ([) by the ( 1 ; %) - ( 2 ; % 1 ) symmetry. Since the relations 1 \ % and 2 \ % 1 are permutable quasiorders of (A; F ), we obtain ( 1 \ %) _ ( 2 \ % 1 ) = ( 1 \ %) ( 2 \ % 1 ) = ( 1 \ %) t ( 2 \ % 1 ) =. Hence 2 Quord(A; F ), and this proves Re (A; F ) = Quord(A; F ).

13 13 5. Compatible re exive relations on a lattice ordered majority algebra It was proved in [4] that for any majority algebra, Re (A; F ) is a pseudocomplemented and 0-modular lattice. Now, as an immediate consequence of Theorem 4.6 we obtain Corollary 5.1. If (A; F; %) is a lattice ordered majority algebra and ; ; 2 Re (A; F ), then (1) \ = 4 =) ( \ ) t = ( t ) \ ( t ), (2) \ \ = 4 =) \ ( t ) = ( \ ) t ( \ ). Proof. As (A; F ) is a majority algebra, the implications (1) and (2) are satis ed in Tol(A; F ), for any ; ; 2 Tol(A; F ) (see [4]). By Theorem 4.6, (Re (A; F ); \; t; ) = (Tol 2 (A; F ); \; t; ), thus the same implications are satis ed by any ; ; 2 Re (A; F ). Proposition 5.2. Let (A; F ) be a majority algebra and is a complemented element in the lattice Re (A; F ). Then (i) = = t holds for every 2 Re (A; F ), (ii) if (A; F ) is lattice ordered, is a central element of Re (A; F ). Proof. (i) Let 0 be the complement of in the lattice Re (A; F ) ( 0 is unique since Re (A; F ) is 0-modular and pseudocomplemented cf. [4]). Then by Lemma 4.4 (ii) we get r = t 0 j ( 0 ) \ ( 0 ), whence 0 = 0 = r. By symmetry, it su ces to show t =. The j part is evident (according to Lemma 4.4 (ii)). To show the reverse inclusion t k, assume that (a; b) 2. Then there is an x 2 A with (a; x) 2 and (x; b) 2. Since 0 = 0 = r, there exist u; t 2 A with (a; u); (t; b) 2 and (u; b); (a; t) 2 0. Since (m(x; u; b); b) = (m(x; u; b); m(b; u; b)) 2, (m(x; u; b); b) = (m(x; u; b); m(x; b; b)) 2 0 and \ 0 = 4, we get m(x; u; b) = b. Hence (a; b) = (m(a; a; b); m(x; u; b)) 2 t. This proves part (i). (ii) If (A; F ) is a lattice ordered majority algebra, then by Corollary 5.1(2) we obtain = \ ( t 0 ) = ( \ ) t ( \ 0 ), for any 2 Re (A; F ). Since Re (A; F ) is a 0-modular lattice, in view of [13], (and 0 ) is a central element of it. Corollary 5.3. Let j A A be a compatible lattice order or a factor congruence of a lattice ordered majority algebra (A; F ). Then is a central element of Re (A; F ) and = = t holds for every 2 Re (A; F ).

14 14 Proof. If is a compatible lattice order of (A; F ), then our assertion follows directly from Proposition 4.5 and Proposition 5.2. Now, let and 0 be a pair of factor congruences of (A; F ). Then, by [9], and 0 are complements of each other in the lattice Tol(A; F ), i.e. \ 0 = 4 and t 0 = r. Hence is also a complemented element in Re (A; F ), so applying Proposition 5.2 we get the required result. Now, let A = n Q A i be a nite direct product of algebras A i and i : A! A i, i (x 1 ; :::; x n ) = x i be the corresponding natural projections, i 2 f1; :::; ng, n = 1. For any system i j A 2 i, i 2 f1; :::; ng of compatible binary relations of the algebras A i de ne the relation j A 2 by (a; b) 2 :() ( i (a); i (b)) 2 i, for all i 2 f1; :::; ng. Q Then is a compatible relation of A, denoted by n i, which is called the direct product of the relations i, i 2 f1; :::; ng. Obviously, if each Q i j A 2 i is re exive, then n i is also re exive. Notice also, that each Q A i is a lattice ordered majority algebra, whenever A = n A i is a lattice ordered majority algebra. We will use also the notation x := (x 1 ; :::; x n ) nq for elements of A i. The following statement is a generalization of the so called Fraser- Horn property of the congruences of a congruence-distributive algebra (see [11, Corollary 1]). Q Theorem 5.4. Let A = n A i be a lattice ordered majority algebra. Then any 2 Re (A) is equal to a direct product of some i 2 Re (A i ) i 2 f1; :::; ng, and Re (A) Q = n Re (A i ). Proof. Let 2 Re (A). Then it is easy to see that the relations i () := f(x; y) 2 A 2 i j 9(u; v) 2 : x = i (u); y = i (v)g are compatible re exive relations on the algebras A i, i 2 f1; :::; ng. Let Q i := ker i j A 2, i 2 f1; :::; ng. In order to prove = n i (),

15 15 observe rst that nq T i () = n ( i i ): Indeed, for any a; b 2 A we have Q (a; b) 2 n i () () 8i 2 f1; :::; ng: ( i (a); i (b)) 2 i () () 8i 2 f1; :::; ng: 9(u (i) ; v (i) ) 2 with i (a) = i (u (i) ), i (b) = i (v (i) ). However, the latter formula is equivalent to (a; b) 2 i i, for all T i 2 f1; :::; ng, i.e. to (a; b) 2 n ( i i ). Now, since each i is a factor congruence of A, by Corollary 5.3 we get i i = i i = i = t i. Hence, we obtain nq T (+) i () = n ( t i ): Further, according to Corollary 5.3, every i is a central element in the lattice Re (A). Therefore, for any 2 Re (A) the sublattice of T Re (A) generated by ; 1 ; :::; n is distributive. Since n i = 4, we get n T T (++) = t i = n ( t i ). T Summarising (+) and (++), we obtain = n Q ( t i ) = n i (), and this completes the proof of the rst assertion. To prove Re (A) Q = n Re (A i ), consider the maps Q : Re (A)! n nq Re (A i ), : Re (A i )! Re (A), de ned by Q () = ( 1 (); :::; n ()) and ( 1 ; ::; n ) = n i. Then ( ( 1 ; ::; n )) = ( 1 ; ::; n ) by de nition, and (()) =, according to the previous proof, i.e. and are inverses of each other. It is easy to check that both and are order-preseving, therefore they are lattice isomorphisms. Thus Re (A) Q = n Re (A i ).

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