Distributivity conditions and the order-skeleton of a lattice

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1 Distributivity conditions and the order-skeleton of a lattice Jianning Su, Wu Feng, and Richard J. Greechie Abstract. We introduce π-versions of five familiar conditions for distributivity by applying the various conditions to 3-element antichains only. We prove that they are inequivalent concepts, and characterize them via exclusion systems. A lattice L satisfies D 0π if a (b c) (a b) c for all 3-element antichains {a,b,c}. We consider a congruence relation whose blocks are the maximal autonomous chains and define the order-skeleton of a lattice L to be L := L/. We prove that the following are equivalent for a lattice L: (i) L satisfies D 0π, (ii) L satisfies any of the five π-versions of distributivity, (iii) the order-skeleton L is distributive. 1. Introduction Distributive lattices are perhaps the most familiar class of lattices. They may be defined via any of the following ternary relations on L: D(a,b,c) means a (b c) = (a b) (a c), D (a,b,c) means a (b c) = (a b) (a c), D m (a,b,c) means (a b) (b c) (c a) = (a b) (b c) (c a), and D 0 (a,b,c) means a (b c) (a b) c. In fact, a lattice L is distributive in case any one, and hence all, of the following equivalent conditions hold: (i) D(a,b,c) for all a,b,c L, (iii) D m (a,b,c) for all a,b,c L, (ii) D (a,b,c) for all a,b,c L, (iv) D 0 (a,b,c) for all a,b,c L. Recall that elements a, b of a lattice L are incomparable, written as a b, if they are not comparable. An antichain in L is a subset of L in which any two distinct elements are incomparable. We denote by π L the set of antichains in L and by πl n the set of n-element antichains in L, where n 1. In [8], one of us found that a π-version of distributivity proved to be of some importance in the study of when certain mappings are residuated. This motivated us to consider the π-version of each of the properties (i)-(iv), by replacing, in each case, for all a,b,c L with for all {a,b,c} πl 3. Presented by... Received...; accepted in final form Mathematics Subject Classification: Primary: 06D75. Key words and phrases: lattice, distributive lattice, π-distributive lattice, order-skeleton, residuated, exclusion systems.

2 2 J. Su, W. Feng, and R. J. Greechie Algebra univers. A lattice L is π-meet-distributive (resp. π-join-distributive) if D(a,b,c) (resp. D (a,b,c)) holds for all {a,b,c} πl 3. A lattice L is π-distributive if it is both π-meet-distributive and π-join-distributive. A lattice L satisfies the π-median law if D m (a,b,c) holds for all {a,b,c} πl 3. A lattice L satisfies D 0π if D 0 (a,b,c) holds for all {a,b,c} πl 3. We have resisted considering π-semi-distributivity because it is equivalent to semi-distributivity as defined in [4]. We observe that for a modular lattice, the five π-versions of distributivity are all equivalent to distributivity. But, in general, they are not equivalent to each other as we show in Section 3. Let C 1 and C 2 be two classes of algebras such that C 1 C 2 ; an exclusion system for C 1 C 2 is a class S C 2 C 1 such that, for L C 2, L / C 1 iff there exists S S isomorphic to a subalgebra of L. We denote by L, D, and M the classes of lattices, distributive lattices, and modular lattices, respectively. Let D 0π, D mπ, D π, D π, and D π be the classes of lattices satisfying D 0π, the π-median law, π-meet-distributivity, π-join-distributivity, and π-distributivity, respectively. RecallthatN 5 isthe 5-elementnon-modularlatticeandM 3 isthe 5-element modular non-distributive lattice. It is well known that {M 3,N 5 } is anexclusionsystemford L. InSection3, wecharacterizethefiveπ-versions of distributivity via exclusion systems; we show that [M 3,3 2 ), {L 15 }, {L 14 }, and {L 13 } are exclusion systems for D 0π L, D mπ D 0π, D π D mπ, and D π D mπ, respectively, where [M 3,3 2 ) is defined in Section 3. We study the notion of the order-skeleton L of a lattice L, which was first introduced in [8] and discussed in [9]. In Theorem2.8, we provethat L satisfies D 0π iff its order-skeleton L is distributive. Our main result, Corollary 3.6, is that L satisfies D 0π iff L satisfies any of the five π-versions of distributivity presented. In particular, if a lattice L is isomorphic to its order-skeleton, then all the five π-versions of distributivity are equivalent to distributivity. We conclude the paper by studying several other weakened distributivity conditions, giving the relation between them and our π-versions of distributivity. The authors wish to thank Dr. Jinko Kanno, Dr. Marcel Erné, and Dr. Peter Jipsen for comments which improved this paper. 2. Order-skeletons Let L be a lattice and a,b L. As usual, we write [a,b] := {x L a x b} and [a,b) := {x L a x < b}; we allow for the possibility that a b, in which case, of course, both sets are empty. Define π(a) := {b L b a}. The following definition plays an important role throughout this paper: a b means a b and π(a) = π(c) for all c [a,b] [b,a].

3 Vol. 00, XX Distributivity conditions and the order-skeleton of a lattice 3 M 3 L 1 L 2 L 13 L 3 L 4 L 5 L 14 L 6 L 7 L 8 L 15 (a) A 9-element exclusion system for D 0π L (b) Figure 1. A 12-element exclusion system for D π L Following [16], a non-empty subset S of L is called (order-)autonomous in case, for all p / S, (1) if there is an s S with s p, then x p for all x S; and (2) if there is an s S with p s, then p x for all x S. Lemma 2.1. Let L be a lattice with a,b L. The following are equivalent. (i) a b. (ii) [a,b] or [b,a] is a chain and π(a) = π(b). (iii) [a,b] or [b,a] is an autonomous chain. Proof. (i) (ii) We may assume that a b and a b. We need only to show that [a,b] is a chain. For any c,d [a,b], we have π(c) = π(a) = π(d), so that c d. It follows that [a,b] is a chain. (ii) (iii) Since π(a) = π(b), we have a b. Assume a b, so that [a,b] is a chain. Let c [a,b] and p / [a,b]. If p c, then p b. Since π(a) = π(b), we have p a. Since p / [a,b], we have p a. Thus, p x for all x [a,b]. Dually, if c p, then x p for all x [a,b]. Therefore, [a,b] is autonomous. (iii) (i) We may assume that [a,b] is an autonomous chain. Let c [a,b] and x / [a,c]. Note that, x a iff x c, and a x iff c x. Thus, x a iff x c. It follows that π(a) = π(c), so that a b. Recall that an equivalence relation θ on a lattice L is a congruence relation iff for any a,b,c L, a θ b implies that (a c) θ (b c) and (a c) θ (b c) (cf. [5]). Lemma 2.2. The relation defined on a lattice L is a congruence relation.

4 4 J. Su, W. Feng, and R. J. Greechie Algebra univers. Proof. First, we claim that is an equivalence relation. The reflexivity and symmetry follow directly from the definition. The transitivity follows from the fact that the subsets of autonomous chains are autonomous chains and the union of two autonomous chains having a non-empty intersection is an autonomous chain. Now, we show that is a congruence relation. Let a,b,c L and suppose that a b. Since a b, we may assume that a b. We shall argue that a c b c by the following two cases. Case 1. Suppose that a c. Since π(a) = π(b), we have b c. Thus, {a,b,c} is a chain. If c a b, then a c = a b = b c. If a c b, then a c = c b = b c. If a b c, then a c = c c = b c. Therefore, in all cases, a c b c. Case 2. Suppose that a c. Since a b, we have a c b c. By Lemma 2.1, [a,b] is an autonomous chain. Since a a c, we have b a c, so that b c a c. Thus, a c = b c, and therefore, a c b c. By a dual argument, we have a c b c. Therefore, is a congruence relation. Define [a] := {b a b } and L := L/ = {[a] a L }. We call the quotient lattice L; L, L the order-skeleton of L. (The order-skeleton of a poset is introduced in [8].) Lemma 2.3. Let L be a lattice with a,b L. Then (i) [a] L [b] iff there exist a 1 [a] and b 1 [b] such that a 1 b 1 ; (ii) [a] < L [b] iff a < b and [a] [b]; (iii) [a] L [b] exists and equals [a b]; (iv) [a] L [b] exists and equals [a b]; (v) a b iff [a] L [b]; and (vi) π(a) = π(b) iff π L([a]) = π L([b]). Proof. (i), (ii), (iii), and (iv) follow directly from the fact that is a congruence relation (cf. [5]). Also, (v) follows from (i) and the definition of, and (vi) follows from (v). For a lattice L with a L, the element a is join-reducible if there exist b,c < a such that a = b c. A meet-reducible element is defined dually. An element a is doubly-reducible if it is both join-reducible and meet-reducible. Note that under this definition, 0 is not join-reducible and 1 is not meetreducible. For convenience of notation, we use a,b,c for elements in L and x,y,z for elements in L. The following lemma ensures that there is at most one joinreducible (resp., meet-reducible) element of L in each [a] L. Lemma 2.4. Let L be a lattice with a L and x L. (i) If a is join-reducible in L, then [a] exists and [a] = a. (ii) If x is join-reducible in L, then x exists and x x.

5 Vol. 00, XX Distributivity conditions and the order-skeleton of a lattice 5 (iii) If a is meet-reducible in L, then [a] exists and [a] = a. (iv) If x is meet-reducible in L, then x exists and x x. Proof. By duality, we need only prove (i) and (ii). (i) Let a L be join-reducible. We need only show that a is the lower bound of [a]. There exist b,c L such that b c and a = b c. Since b c and a c, we have [a] [b]. By Lemma 2.3 (i), we have [b] L [a], so that [b] < L [a]. Let u [a]. By Lemma 2.3 (ii), b < u. Similarly, c < u. Hence, a = b c u. (ii) Since x is join-reducible in L, there exist b,c L such that [b],[c] < L x and x = [b] L [c]. By Lemma 2.3 (ii), for all a x, we have b,c < a, so that b c a. By Lemma 2.3 (iii), x = [b c], so b c x. Therefore b c is the smallest element in x, i.e., x = b c x. Lemma 2.5. Let L be a lattice with x L. Then (i) x is a maximal autonomous chain in L; (ii) L := ( L) = {{x} x L}, i.e., L is equality on L. (iii) L = L iff L = {{a} a L}. Proof. (i) Let x L. For any b,c x, we have b c, so that b c; hence x is a chain. Let p / x and b,c x. If b p, then by Lemma 2.1, we have [b,c] or [c,b] is autonomous, so that c p. Similarly, p b implies p c. Hence, x is autonomous. We now show that x is maximal. Let S L be an autonomous chain containing x. For a x and s S, we have that [a,s] or [s,a] is autonomous, so that by Lemma 2.1, a s. Thus, S [a] = x, i.e., x is a maximal autonomous chain. (ii) Let x,y L and x L y. Since x L y, we may assume that x L y, and hence, there exist a x and b y such that a b. For any c L with a c b, we have x L [c] L y. Thus, π L(x) = π L([c]), and hence, by Lemma 2.3 (vi), π(a) = π(c). It follows that a b, so that x = y. Therefore, x = {x}. (iii) If L = {{a} a L}, then the function a [a] is easily seen to be an isomorphism. For the converse, assume that L = L via the isomorphism f: L L. Let a,b L with a b. We may assume that a b. Since f is an isomorphism, we have f(a) L f(b). Let x be an arbitrary element in [f(a),f(b)] and let c := f 1 (x). We have c [a,b], and thus, π(c) = π(a) since a b. Since f is an isomorphism, we have π L(x) = π L(f(a)). Therefore, by definition, f(a) L f(b). By (ii), we have f(a) = f(b), so that a = b. Therefore, is equality on L, and L = {{a} a L}. Let L and K be two lattices. Recall that a (lattice-)embedding f: L K is a one-to-one homomorphism from L to K. L is (lattice-)embeddable in K if there exists an embedding from L into K. The following lemma utilizes the Axiom of Choice. Lemma 2.6. Let L be a lattice. Consider the following conditions. (i) Every doubly-reducible element in L is a singleton subset of L.

6 6 J. Su, W. Feng, and R. J. Greechie Algebra univers. (ii) There exists an embedding β: L L such that β(x) x for every x L. (iii) L is embeddable in L. (iv) The cardinality of the set of doubly-reducible elements in L equals the cardinality of the set of doubly-reducible elements in L. Then (i) (ii) (iii) (iv). In particular, if L contains finitely many doubly-reducible elements, then the four conditions are equivalent to each other. Proof. (i) (ii) Assume that every doubly-reducible element in L is a singleton subset of L. Note that, by Lemma 2.4, if x is join-reducible (resp. meet-reducible) in L, then x (resp. x) exists and x x (resp. x x). By assumption, if x is doubly-reducible in L, then x = x. Thus, there exists a selection function β: L L such that (1) β(x) x, (2) if x is join-reducible, then β(x) = x, and (3) if x is meet-reducible, then β(x) = x. It is easy to see that β: L L is one-to-one. We now show that β is a homomorphism. Let [a],[b] L. If [a] L [b], then β([a] L[b]) = β([a]) β([b]). If [a] L [b], then both [a] L[b] and a b are join-reducible, so that by Lemma 2.3 (iii) and Lemma 2.4, β([a] L [b]) = β([a b]) = [a b] = β([a]) β([b]). Dually, we have β([a] L [b]) = β([a]) β([b]). Therefore, β is an embedding. (ii) (i) Let β: L L be an embedding such that β(x) x for every x L. Let x L be a doubly-reducible element. Since x is join-reducible in L, there exist y,z L such that y L z and x = y Lz. By Lemma 2.3 (v), β(y) β(z), so that β(x) = β(y) β(z) is join-reducible in L. By Lemma 2.4, β(x) = x. Dually, β(x) = x, so that x = x, i.e., x is a singleton subset of L. (ii) (iii) This is trivial. (iii) (iv) Let f: L L be an embedding. Let c L and c L be the cardinality ofthe doubly-reducibleelementsin Land Lrespectively. Note thatf mapsthe doubly-reducible elements in L to the doubly-reducible elements in L, so that c L c L. Also note that every doubly-reducible element a L corresponds to a doubly-reducible element [a] L, so that c L c L. Therefore, c L = c L. Now assume that L contains finitely many doubly-reducible elements. (iv) (i) Suppose that x is doubly-reducible in L but not a singleton subset of L. Since every doubly-reducible element a L corresponds to a doublyreducible element [a] L, we have c L c L. Also note that there is no doubly-reducible element in L that corresponds to x. It follows that c L < c L which contradicts (iv). Note that (i), (iii), and (iv) are not in general equivalent to each other. For example, let L be the horizontal sum of L 15 with countably many copies of the 3 2. Then L is the horizontal sum of countably many copies of the 3 2. Therefore L is embeddable in L, but L contains a horizontal summand isomorphic to L 15 so that L contains a doubly-reducible element which is not a singleton subset of L, i.e., (iii) (i). Now let M be the vertical sum of

7 Vol. 00, XX Distributivity conditions and the order-skeleton of a lattice 7 two 2 2 so that there is a bi-reducible element in M. Let Q be the horizontal sum of L 15 with countably many copies of M. In this example, both Q and Q have countably many bi-reducible elements, but Q is not embeddable in Q, i.e., (iv) (iii). Define a pentagon in a lattice L to be a quintuple a,b,c,u,v such that a,b,c,u,v L and v < b < a < u, c a = v, c b = u. Lemma 2.7. Let L be a lattice. If L is non-modular, then L contains a pentagon x,y,z,u,v and there exists an element w L such that either (i) x = y L w and {w,y,z} π 3 L; or (ii) y = x L w and {w,x,z} π 3 L. Proof. Suppose that L contains a pentagon x 1,y,z,u,v. Since x 1 L y, there exists x 2 L such that y < L x 2 L x 1 and π(y) π(x 2 ). Note that, x 2,y,z,u,v is a pentagon. Since π(y) π(x 2 ), we have either (i ) w L with w L y and w L x 2, or (ii ) w L with w L x 2 and w L y. By duality, we may assume that (i ) holds. Since y L x 2 and y L w, we have x 2 L w. Since x 2 L w and w L x 2, we have w L x 2. Let x = y L w. We have y < L x L x 2, and thus, x,y,z,u,v is a pentagon. Since w L y and x L z = v L y, we have w L x L z. Since w L x L z and w L x, we have w L z. Since w L x and z L x, we have z L w. Thus, w L z, and {w,y,z} π 3 L. Therefore, (i) holds. In the dual case (ii ), (ii) holds. Theorem 2.8. Let L be a lattice. The following are equivalent. (i) L satisfies D 0π. (ii) L satisfies D 0π. (iii) L is distributive. Proof. First, we prove that (i) and (ii) are equivalent. (i) (ii) If L satisfies D 0π, then {[a],[b],[c]} π 3 L implies {a,b,c} π 3 L and [a] L([b] L[c]) = [a (b c)] L [(a b) c] = ([a] L[b]) L[c], sothat Lsatisfies D 0π. Now we assume that L does not satisfy D 0π. Then {a,b,c} π 3 L such that a (b c) (a b) c. Let d = a (b c) and e = (a b) c, so that d e. Since {a,b,c} π 3 L, we have {[a],[b],[c]} π3 L. Since d a, c e, and c a, we have e d. Thus, d e. It follows, [d] L [e]. Since [a] L ([b] L [c]) = [d] L [e] = ([a] L [b]) L [c], we have D 0 ([a],[b],[c]) does not hold. Thus, L does not satisfy D 0π. Now we prove that (ii) and (iii) are equivalent. (ii) (iii) Assumethat Lisdistributive. Forany{x,y,z} π 3 L, x L(y Lz) = (x Ly) L(x Lz) L (x Ly) Lz. Thus, LsatisfiesD 0π. Nowwemayassume that L is not distributive. Then L contains a sublattice isomorphic to M 3 or N 5. It is easy to verify that M 3 does not satisfy D 0π. We may assume that L containsapentagon x,y,z,u,v. ByLemma2.7,wemayassumethat w L such that x = y Lw and {w,y,z} π 3 L. Since w Lz L x Lz = v L y, we have w L(z Ly) = w L y = (w Lz) Ly, i.e., L does not satisfy D 0π.

8 8 J. Su, W. Feng, and R. J. Greechie Algebra univers. We conclude this section by observing that a simple induction proves that a lattice L is π-meet-distributive iff x ( (A {x})) = (x a) for a A {x} every finite π-set A L and x A. And the dual statement holds for π-joindistributivity. Note that this is not true when A is not finite. For example, let N 0 := N {0} and consider the lattice N 0 ;lcm,gcd. Note that 0 is the top element and 1 is the bottom element. This is a distributive lattice, and thus, a π-meet-distributive lattice. Let O be the set of all odd integers, A = O {2}, and x = 2. We have 2 ( O) = 2 0 = 2 but (2 a) = 1 = 1. a O a O 3. Exclusion systems In this section, we characterize the five π-versions of distributivity introduced in Section 1 by the exclusion systems. We observe that they are not equivalent to each other. Note that in Figure 1, L 13 is π-meet-distributive but not π-join-distributive, while L 14 is π-join-distributive but not π-meetdistributive. L 15 satisfies D 0π but does not satisfy the π-median law. Also, both L 13 and L 14 satisfy D 0π and the π-median law, but do not satisfy π- distributivity. For a lattice L with a,b,c L, we use Γ{a,b,c} to denote the sublattice of L generated by a,b,c. We write 1 K (resp. 0 K ) for the top (resp. bottom) element of a sublattice K of L. Lemma 3.1. Let L be a lattice with {a,b,c} πl 3. The following statements hold. (i) If a (b c) / {a b,a c}, then {a (b c),b,c} π 3 L. (ii) If a (b c) / {a b,a c}, then {a (b c),b,c} π 3 L. (iii) If a b < 1 Γ{a,b,c} and a c < 1 Γ{a,b,c}, then {(a b) (a c),b,c} π 3 L. (iv) If 0 Γ{a,b,c} < a b and 0 Γ{a,b,c} < a c, then {(a b) (a c),b,c} π 3 L. Proof. By duality, we need only prove (i) and (iii). (i) Let a 1 := a (b c) and assume that a 1 / {a b,a c}. Since a 1 a b and a b a 1, we have a b < a 1. Since a 1 b = a (b c) b = a b < a 1 and a 1 b = a b < b, we have a 1 b. By symmetry, a 1 c. It follows that {a 1,b,c} πl 3. (iii) Let a u := (a b) (a c). Since a a u and a b, we also have a u b. We have b a u, otherwise, b a u = (a b) (a c) a c, which implies 1 Γ{a,b,c} = a b c = a c, contradicting a c < 1 Γ{a,b,c}. Hence, a u b. By symmetry, a u c. Therefore, {a u,b,c} πl 3. Recall that D 0 (a,b,c) means a (b c) (a b) c. Dually, we can define D 0 (a,b,c) to mean (a b) c a (b c). Note that D 0 (a,b,c) = D 0(c,b,a). Lemma 3.2. Let L be a lattice with {a,b,c} π 3 L. The following statements are equivalent.

9 Vol. 00, XX Distributivity conditions and the order-skeleton of a lattice 9 (i) D 0 (a,b,c) holds. (ii) c (a (b c)) = (a b) c. (iii) a (b c) = a (c (a b)). (iv) {a (b c),b,(a b) c} / π 3 L. Proof. It is easy to see that (i), (ii), and (iii) are equivalent. It is trivial that (i) implies (iv). We need only to show that (iv) implies (i). Let a 1 := a (b c) and c 1 := (a b) c. Note that a 1 a and c c 1. Suppose that {a 1,b,c 1 } / π 3 L. We have a 1 b, b c 1, or a 1 c 1. Note that, since b a and a 1 a, we have b a 1. Therefore, if a 1 b, then since b a 1, we have a 1 < b, and thus, a 1 a b (a b) c = c 1. Similarly, since c b and c c 1, we have c 1 b. Therefore, if b c 1, then since c 1 b, we have b < c 1, and thus, a 1 = a (b c) b c c 1. Thus, we may assume that a 1 c 1. Since c c 1, a 1 a, and c a, we have c 1 a 1. Thus, a 1 c 1. Lemma 3.3. Let L be a lattice. The following conditions are equivalent. (i) L does not satisfy D 0π. (ii) {a,b,c} πl 3 such that a b c and a b c. (iii) {a,b,c} πl 3 such that a b = 0 Γ{a,b,c} and b c = 1 Γ{a,b,c}. Proof. Let a 1 := a (b c) and c 1 := (a b) c. Note that a 1 a and c c 1. (i) (ii) Since L does not satisfy D 0π, {a,b,c} π 3 L such that a 1 c 1. By Lemma 3.2 (iv), {a 1,b,c 1 } π 3 L. Thus, a 1 b = a b (a b) c = c 1. Similarly, a 1 b c 1. (ii) (iii) It is trivial. (iii) (i) a (b c) = a c = (a b) c, i.e., D 0 (a,b,c) does not hold. As mentioned in the introduction, distributivity implies D 0π, but D 0π does not imply distributivity. Moreover, for {a,b,c} πl 3, if either D(a,b,c) or D (c,b,a) holds, then D 0 (a,b,c) holds. But the converse is not true. Figure 2 is an example of a lattice satisfying D 0π, but not D(a,b,c) or D (c,b,a) (this can be verified by applying the above lemma). The following lemma is well known. Lemma 3.4. {M 3 } is an exclusion system for D M. In [6], the following lemma is proved. Lemma 3.5. Let L be a lattice containing a pentagon a,b,c,u,v and an element d such that a = b d and {b,c,d} πl 3. Then L contains a sublattice isomorphic to L 1, L 3, L 4, L 6, L 7, or L 8. We follow the notation from [14, 11, 12]. Note that M 3, L 1, L 2, L 3, L 4, L 5, L 13, L 14 and L 15 are subdirectly irreducible lattices and each L i is the order-skeleton of the corresponding lattice L i for i = 6,7,8 as found in these references. Note that L 1 and L 2 are dual; L 4 and L 5 are dual. By Lemma 2.7, Lemma 3.5 and its dual, and the fact that the eight lattices L 1,L 2,L 3,L 4,L 5, L 6, L 7, and L 8 are not modular, we have the following corollary.

10 10 J. Su, W. Feng, and R. J. Greechie Algebra univers. a b c Figure 2. A lattice L satisfying D 0π but satisfying neither D(a,b,c) nor D (c,b,a). Corollary 3.6. Let L be a lattice. The order-skeleton L is modular iff L contains no sublattice isomorphic to L 1,L 2,L 3,L 4,L 5, L 6, L 7, or L 8. Note that the lattices M 3,L 1,L 2,L 3,L 4,L 5, L 6, L 7, and L 8 satisfy the condition (i) in Lemma 2.6. Therefore, we have the following corollary. Corollary 3.7. For F {M 3,L 1,L 2,L 3,L 4,L 5, L 6, L 7, L 8 }, if L contains a sublattice isomorphic to F, then L contains a sublattice isomorphic to F. Let L F ; be the poset of all finite lattices with the ordering defined by order-embedding, i.e., L 1 L 2 iff there exists a one-to-one mapping f: L 1 L 2 such that x y iff f(x) f(y) for all x,y L 1. We define the half open interval in L F by [L 1,L 2 ) := {L L F : L 1 L < L 2 }. One can verify that [M 3,3 2 ) = {M 3,L 1,L 2,L 3,L 4,L 5, L 6, L 7, L 8 }. We now characterize the condition D 0π. Theorem 3.8. [M 3,3 2 ) is an exclusion system for D 0π L. Proof. It is easy to verify that the nine lattices in [M 3,3 2 ) do not satisfy D 0π. We argue that, if L does not satisfy D 0π, then L contains a sublattice isomorphic to a lattice in [M 3,3 2 ). Suppose that L does not satisfy D 0π. By Theorem 2.8, L is not distributive. If L is modular, then by Lemma 3.4, it contains a sublattice isomorphic to M 3. If L is non-modular, then by Corollary 3.6, it contains a sublattice isomorphic to one of L 1,L 2,L 3,L 4,L 5, L 6, L 7, L 8. Hence, by Corollary 3.7, L contains a sublattice isomorphic to one of M 3,L 1,L 2,L 3,L 4,L 5, L 6, L 7, L 8. Recall that a lattice L satisfies the π-median law iff D m (a,b,c) holds for all {a,b,c} π 3 L. For convenience of notation, for a,b,c L, define a u := (a b) (a c), b u := (a b) (b c), and c u := (a c) (b c). Dually, define

11 Vol. 00, XX Distributivity conditions and the order-skeleton of a lattice 11 a l := (a b) (a c), b l := (a b) (b c), and c l := (a c) (b c). Also, define m u := (a b) (b c) (c a) and m l := (a b) (b c) (c a). It is easy to verify that the nine lattices in [M 3,3 2 ) do not satisfy the π-median law, so that D mπ D 0π. Lemma 3.9. Let L be a lattice and L = 3 2. If L does not satisfy the π-median law, then L 15 is embeddable in L. Proof. Note that L = 3 2 is generated by its 3-element antichain. Since L does not satisfy the π-median law, there exists {a,b,c} π 3 L such that D m(a,b,c) does not hold. Since {[a],[b],[c]} π 3 L, we have L = Γ L{[a],[b],[c]}. Without loss of generality, we may assume that [a] and [c] are complements in L. Since D m (a,b,c) does not hold, we have m l = (a b) (b c) (c a) < (a b) (b c) (c a) = m u. Since m l = b l b b u = m u and m l m u, we have [m l ] = [b] = [m u ]. Consider the inverse image of : L L in L, one can verify that {0,1,a,b u,b l,c,a b,b c,a b,b c} = L 15. Lemma {L 15 } is an exclusion system for D mπ D 0π. Proof. It is easy to verify that L 15 does not satisfy the π-median law. Now assume that L satisfies D 0π, but does not satisfy the π-median law. There exists {a,b,c} πl 3 such that D m(a,b,c) does not hold, so that m l < m u. We claim that at least two of a, b, and c are incomparable with m u. Otherwise, we may assume that m u a and m u b. If m u a and m u b, then m u a b m l, contradicting m l < m u. If a < m u and b < m u, then a m l and b m l, so that m u a b m l, contradicting m l < m u. If a m u and m u b, then a b, contradicting a b. If b m u and m u a, then b a, contradicting a b. Therefore, two of a, b, and c are incomparable with m u. We may assume that m u a and m u b, so that {a,b,m u } πl 3. By Theorem 2.8, L is distributive, so that m l m u. Observe that a b m u a b, i.e., a and b are complement elements in Γ{a,b,m u }. One can verify that Γ L{[a],[b],[m u ]} = 3 2, so that by Lemma 3.9, L 15 is a sublattice of L. By Theorem 3.8 and Lemma 3.10, we have the following theorem. Theorem [M 3,3 2 ) {L 15 } is an exclusion system for D mπ L. It is easy to verify that the ten lattices in [M 3,3 2 ) {L 15 } are not π-meetdistributive, so that D π D mπ. Lemma {L 14 } is an exclusion system for D π D mπ. Proof. ItiseasytoverifythatL 14 isnotπ-meet-distributive. Nowassumethat L satisfies the π-median law, but does not satisfy the π-meet-distributivity. There exists {a 1,b,c} π 3 L such that D(a 1,b,c) does not hold. Let a := a 1 (b c). Since a 1 b (a 1 b) (a 1 c) < a 1 (b c) = a, we have a a 1 b. By symmetry, a a 1 c. By Lemma 3.1 (i), {a,b,c} π 3 L. We have (a b) (a c) = (a 1 b) (a 1 c) < a 1 (b c) = a (b c), i.e., D(a,b,c) does not hold. Let F = Γ{a,b,c}. We have b c = a b c = 1 F.

12 12 J. Su, W. Feng, and R. J. Greechie Algebra univers. Since L satisfies the π-median law, we have m l = m u. Since b c = 1 F, m u = (a b) (b c) (c a) = (a b) (a c) = a u. Since a l = (a b) (a c) < a (b c) = a 1 F = a a u and a l (b c) = (a b) (b c) (c a) = m l = m u = a u, we have 0 F < b c. Since a a u and a b, we have a u b. Since a u b and b l b, we have a u b l. Since a u b l and b l (a c) = (a b) (b c) (c a) = m l = m u = a u, we have 0 F < a c. By symmetry, 0 F < a b. By Lemma 3.1 (iv), {a l,b,c} πl 3, {a,b l,c} πl 3, and {a,b,c l } πl 3. Since b l c and c l c, we have b l c l. By symmetry, c l b l. Thus, b l c l. It follows, {a,b l,c l } πl 3. Since a a u = m u = m l = b l c l, (a b l ) (a c l ) (a b) (a c) < a = a (b l c l ), i.e., D(a,b l,c l ) does not hold. Since a b a l a and a b b l b, we have a b a l b l a b, so that a l b l = a b. Similarly, b l c l = b c and a l c l = a c. Observe that a l b l = b l c l = a l c l = m l, so that Γ{a l,b l,c l } = 2 3. Since a l a m l, we have Γ{a,b l,c l } = L 14. By Theorem 3.11, Lemma 3.12, and their dual statements, we have the following theorem. Theorem (i) [M 3,3 2 ) {L 14,L 15 }is an exclusion system for D π L. (ii) [M 3,3 2 ) {L 13,L 15 } is an exclusion system for D π L. (iii) [M 3,3 2 ) {L 13,L 14,L 15 } is an exclusion system for D π L. Remark By definition, π-distributivity implies both π-meet- and π- join-distributivity. By comparing the exclusion systems of the π-versions of distributivity, one can also see that either one of π-meet-distributivity and π-join-distributivity implies the π-median law, and the π-median law implies D 0π. In particular, π-distributivity implies D 0π. By using the exclusion systems with Theorem 2.8, we have the following corollary. Corollary Let L be a lattice. The following statements are equivalent. (i) L satisfies D 0π. (ii) L contains no sublattice isomorphic to a lattice in [M 3,3 2 ). (iii) L is distributive. (v) L is π-meet-distributive. (vii) L satisfies the π-median law. (iv) L is π-distributive. (vi) L is π-join-distributive. (viii) L satisfies D 0π. This corollary tells us that, if a lattice L is isomorphic to its own orderskeleton, then all these π-properties are equivalent to distributivity. We conclude this section by observing that no two of the properties (iii) - (viii) are equivalent for general lattices.

13 Vol. 00, XX Distributivity conditions and the order-skeleton of a lattice Motivation and related results In another paper ([9]), we study the residuated approximation to orderpreserving maps on complete lattices. A mapping between two posets is residuated if the inverse image of every principle ideal in Q is a principle ideal in L (see [2]). If the posets are complete lattices, then the mapping is residuated iff it preserves all joins. Any function f: L Q between two complete lattices dominates a largest residuated mapping, called ρ f, the residuated approximation of f. In [1], Andréka, Greechie, and Strecker introduced the shadow σ f of f in order to efficiently approximate ρ f by iterating σ f where σ f is defined by σ f (x) := {q Q x f 1 ( q)}. They proved the following theorem. Theorem 4.1. For a complete lattice Q, Q is completely distributive iff for every complete lattice L and every order-preserving mapping f: L Q, the shadow σ f is residuated. Also, Greechie and Janowitz[10, 9, 8] proved a result implying the following corollary. Corollary 4.2. Let L and Q be two complete lattices and f: L Q be an order-preserving mapping. If L is completely distributive, then σ f is residuated. Notice that both the theorem and the corollary are concerned with distributivity conditions. But even if both L and Q are non-distributive, it is still possible that the shadow is residuated. In [9], we study the structure of L in order to make the shadow σ f residuated and found the following theorem. Theorem 4.3. Let L be a lattice with no infinite chains. The following statements are equivalent. (i) σ f is residuated whenever Q is a complete lattice and f: L Q is an order-preserving mapping. (ii) L satisfies D 0π. (iii) L is distributive. This theorem motivated us to study the π-versions of distributivity and the order-skeleton of a lattice. Now we discuss the relation between the π-versions of distributivity in this paper to some weakened conditions found elsewhere [7, 13]. Note that both 2 and N 5 are π-distributive lattices, but 2 N 5 does not satisfy D 0π. Thus, by Comment 3.14, both 2 and N 5 satisfy the five π-versions of distributive conditions, but 2 N 5 does not satisfy any of these five conditions. Therefore, the five classes of lattices satisfying the various π-versions of distributivity described in this paper are not lattice varieties since they are not closed under products. A lattice L is semi-distributive whenever, for every a,b,c L, (SD1) a b = a c implies a b = a (b c); and (SD2) a b = a c implies a b = a (b c).

14 14 J. Su, W. Feng, and R. J. Greechie Algebra univers. It is easy to show that the π-version of semi-distributivity (again by applying the conditions only to {a,b,c} πl 3 ) is equivalent to semi-distributivity. Davey, Poguntke, and Rival proved that {M 3,L 1,L 2,L 3,L 4,L 5 } is an exclusion system for SD L where SD is the class of all semi-distributive lattice [4]. A lattice L is near-distributive whenever, for every a,b,c L, (ND1) a (b c) = a (b (a (c (a b)))) and (ND2) a (b c) = a (b (a (c (a b)))). As with semi-distributivity, it is not difficult to show that the π-version of near-distributivity is equivalent to near-distributivity. It is also easy to show that near-distributivity implies semi-distributivity. We now show that D 0π implies near-distributivity. Lemma 4.4. Let L be a lattice. If L satisfies D 0π, then L is near-distributive. Proof. Since near-distributivity and its π-version are equivalent, we need only to show that the π-version holds. Let {a,b,c} πl 3. By Lemma 3.2 (iii), we know that a (b c) = a (c (a b)). Let r := a (b (a (c (a b)))) be the right hand side of (ND1). Since a (c (a b)) a and a (c (a b)) b (a (c (a b))), we have a (c (a b) r. Also, we have r a (b (a (c b))) a (b (c b)) = a (b c). Therefore, a (b c) = r = a (b (a (c (a b)))), i.e., (ND1) holds. By duality, (ND2) holds. Therefore, L is near-distributive. A lattice L is almost-distributive if it is near-distributive and for every x,y,z,u,v L, (AD1) v (u c) u (c (v a)); and (AD2) v (u c ) u (c (v a )), where a = (x y) (x z), c = x (y (x z)), a = (x y) (x z), and c = x (y (x z)). Note that a is the dual of a, c is the dual of c, and (AD2) is the dual of (AD1). Lemma 4.5. Let L be a lattice. If L satisfies D 0π, then L is almost distributive. Proof. Let L be a lattice satisfying D 0π. By Lemma 4.4, L is near-distributive. Thus, by duality, we need only to show that (AD1) holds. Since L satisfies D 0π, by Theorem 2.8, L is distributive. Recall that is a congruence relation on L. In L, [a] = [(x y) (x z)] = ([x] L[y]) L([x] L [z]) = [x] L ([y] L [z]) = [x (y z)] = [c]. Thus, a c and clearly, a c. If a v, then c v and by Lemma 2.4, v a = [v a] = [v c] = v c, so that v (u c) u c = u (c (v c)) = u (c (v a)). If v a, then v (u c) v a u a = u (c (v a)). If c v, then v (u c) u c = u (c (v a)). If a v c, then v (u c) = v u v = u (c (v a)). Thus, (AD1) holds.

15 Vol. 00, XX Distributivity conditions and the order-skeleton of a lattice 15 Note that the converse of this lemma is not true. For example, L 6 is almost distributive but does not satisfy D 0π. In [15], Rose proved that for any subdirectly irreducible lattice L, L is almost distributive iff L = D[d] for some distributive lattice D and d D (see also [11, 13]). Here D[d] is the doubling construction introduced by Day [3]. Note that D[d] = D. If D = D, then the order-skeleton D[d] is isomorphic to D, and every block of the order-skeleton is a singleton subset except one block which is a doubleton subset. Recall the definition of a pentagon a,b,c,u,v preceding Lemma 2.7. The following lemma follows from the definition of the doubling construction. Lemma 4.6. Let L = D[d] for some distributive lattice D with d D. If L contains a pentagon a,b,c,u,v and θ is the smallest congruence relation that identifies a and b, then L/θ = D and θ is the congruence relation that identifies only a and b. Lemma 4.7. Let L be a π-distributive lattice with x,y,z L and L = D[d] for some distributive lattice D with d D. If L contains a pentagon a,b,c,u,v and x v = y v, then (x z) v = (y z) v. Proof. Assume that (x z) v (y z) v. Let θ be the smallest congruence relation that identifies a and b. By Lemma 4.6, L/θ is distributive, so that [(x z) v] θ = ([x] θ θ [z] θ ) θ [v] θ = ([x] θ θ [v] θ ) θ ([z] θ θ [v] θ ) = [(x v) (z v)] θ = [(y v) (z v)] θ = [(y z) v] θ. Since (x z) v θ (y z) v, we have {(x z) v,(y z) v} = {a,b} and we may assume that (x z) v = a and (y z) v = b. Sincea b,wehavex y. Notethatx a = x v = y v = y b. Since a x a = y b and a b, we have y b. Since x x a = y v and x y, we have b y. Thus, y b. Since b y b = y v y c and b c, we have y c. Since y < y b = y v y c, we have c y. Hence, y c, so that {y,b,c} πl 3. Since b y < b and a y θ b y, we have a y = b y. Since a x a = y v y c and L is π-distributive, we have a = a (y c) = (a y) (a c) = (b y) (b c) b, contradicting b < a. Therefore, (x z) v = (y z) v. Lemma 4.8. Let L be a subdirectly irreducible π-distributive lattice. If L contains a pentagon a,b,c,u,v, then u = 1 and v = 0. Proof. Let L be a subdirectly irreducible π-distributive lattice with a pentagon a,b,c,u,v. Assume that u 1 or v 0. By duality, we may assume v 0. Define a relation α on L by x α y iff x v = y v. It is easy to see that α is an equivalent relation such that for any x,y,z L with x α y, we have x z α y z and, by Lemma 4.7, x z α y z, so that α is a congruence relation. Let θ be the smallest congruence relation that identifies a and b. By Lemma 4.6, θ identifies only a and b. Note that, since a v = a b = b v, we have a α b, so that θ α. Let β be a congruence relation with β α and β θ, and suppose that c,d L with c β d. Since c θ d, we have c = d or

16 16 J. Su, W. Feng, and R. J. Greechie Algebra univers. {c,d} = {a,b}. Since c α d, we have c = d. Therefore, β is equality, which implies that L is not subdirectly irreducible, contradicting the assumption. Theorem 4.9. Let L be a subdirectly irreducible lattice with L N 5. Then L is distributive iff L is π-distributive. Proof. Let L be a subdirectly irreducible π-distributive lattice with L N 5. ByComment3.14, LsatisfiesD 0π, andbylemma4.5, Lis almostdistributive, so that L = D[d] for some distributive lattice D and d D. Assume that L is not distributive, so that L contains a sublattice isomorphic to M 3 or N 5. Since M 3 does not satisfy D 0π, L contains a pentagon a,b,c,u,v. By Lemma 4.8, u = 1 and v = 0. Since L N 5, there exists e L such that e / {a,b,c,0,1}. We have a e, for if a < e, then a,b,c e,b (c e),0 is a pentagon, so that, by Lemma 4.8 again, 1 = b (c e) e, contradicting e 1. Similarly, e b. Since L is subdirectly irreducible, e / [b,a], so that e a and e b. Since {a,b,e,a e,a e} is a pentagon, by Lemma 4.8, a e = b e = 1 and a e = b e = 0. Let θ be the smallest congruence relation that identifies a and b. By Lemma 4.6, θ identifies only a and b. Since [c] θ = [c (a e)] θ = [(c a) (c e)] θ = [c e] θ, we have c θ c e, so that c = c e. Similarly, c = c e, so that c = e, contradicting the assumption. Theorem Let L be a subdirectly irreducible lattice. Then L is almost distributive iff L satisfies D 0π. Proof. By Lemma 4.5, D 0π implies almost distributivity. We now prove the sufficiency. Let L be a subdirectly irreducible almost distributive lattice. There exists a distributive lattice D and an element d D such that L = D[d]. Notice that the order-skeleton L = D[d] = D is distributive, by Corollary 3.15, L satisfies D 0π. In [7], Erné introduced n-zipper-distributivity and the conditions of H n where n 3. It turns out that the π-version of n-zipper-distributivity is also equivalent to n-zipper-distributivity. In Figure 3, we present a diagram indicating the implications between the various weakened distributive conditions discussed above. We observe that nothing collapses except the five π-versions of distributivity discussed in Corollary 3.15 even if a lattice L is isomorphic to its own order-skeleton. Details can be found in [17]. References [1] Andréka H., Greechie R. J., Strecker G. E.: On residuated approxiations. Categorical Methods in Computer Science With Aspects from Topology, 393, (1989) [2] Blyth T. S., Janowitz M. F.: Residuation theory. Pergamon Press, (1972) [3] Day A.: A simple solution to the word problem for lattices. Canadian Mathemtical Bulleton, 13, (1970) [4] Davey B. A., Poguntke W., Rival I.: A characterization of semi-distributivity. Algebra Universalis, 5, (1975) [5] Davey B. A., Priestley H. A.: Introduction to lattices and order. Cambridge University Press, New York (1980)

17 Vol. 00, XX Distributivity conditions and the order-skeleton of a lattice 17 meet-semi-distributivity 4-zipper-distributivity semi-distributivity meet-near-distributivity near-distributivity almost-distributivity 3-zipper-distributivity modularity D 0π H 3 π-median law π-meet-distributivity π-join-distributivity H 4 π-distributivity distributivity Figure 3. The implications between various weakened distributive conditions. [6] Davey B. A., Rival I.: Finite sublattices of three-generated lattices. Journal of Australian Mathematical Society (Series A), 21, (1976) [7] Erné M.: Weak distributive laws and their role in lattices of congruences and equational theories. Algebra Univeralis, 25, (1980) [8] Feng W.: On calculating residuated approximations and the structure of finite lattices of small width. PhD dissertation, Louisiana Tech University (2010) [9] Feng W., Su J., Greechie R. J.: On calculating residuated approximations (submitted to Algebra Universalis) [10] Greechie R. J., Janowitz M. F.: Personal Communication. [11] Jónsson B., Rival I.: Lattice varieties covering the smallest non-modular variety. Pacific Journal of Mathematics, 82, (1979) [12] Jipsen P., Rose H.: Varieties of lattices. Springer Verlag, (1992) [13] Lee J. G.: Almost distributive lattice varieties. Algebra Univeralis, 21, (1985) [14] McKenzie R.: Equational bases and nonmodular lattice varieties. Transactions of the American Mathematical Society, 174, 1-43 (1972) [15] Rose H.: Nonmodular lattice varieties. Memoirs of American Mathematical Society, 292 (1984) [16] Schröder B. S. W.: Ordered sets: an introduction, Birkhäuser (2002) [17] Su J.: Results in lattices, ortholattices, and graphs. PhD dissertation, Louisiana Tech University (2011) Jianning Su Program of Mathematics and Statistics, Louisiana Tech University, Ruston 71272, USA sujianning@hotmail.com Wu Feng

18 18 J. Su, W. Feng, and R. J. Greechie Algebra univers. The Center for Secure Cyberspace, Louisiana Tech University, Ruston 71272, USA Richard J. Greechie Program of Mathematics and Statistics, Louisiana Tech University, Ruston 71272, USA URL: greechie/