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Ordering the Order of a Distributive Lattice by Itself View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by Elsevier - Publisher Connector Available online at www.sciencedirect.com The Journal of Logic and Algebraic Programming 76 (2008) 198–208 www.elsevier.com/locate/jlap Ordering the order of a distributive lattice by itself Michel Krebs ∗, Jürg Schmid Institute of Mathematics, University of Bern, CH-3012 Bern, Switzerland Available online 18 April 2008 Abstract We order the ordering relation of an arbitrary poset P component-wise by itself, obtaining a poset (P) extending P. In particular, the effects of on L ∈ DLAT01, the category of all bounded distributive lattices, are studied, mainly with the aid of Priestley duality. We characterize those L ∈ DLAT01 which occur as (K) for some K ∈ DLAT01, decide this situation in polynomial time for finite L, characterize fixpoints of within DLAT01 and relate them to free objects in DLAT01. © 2008 Elsevier Inc. All rights reserved. AMS classification: Primary 06A07, 06D05, 06D50 Keywords: Partially ordered set; Distributive lattice; Priestley duality; Free distributive lattice 1. Introduction In January 2005, at the end of a long seminar day devoted to ordered sets and lattices, van der Zypen wondered what happens if one orders the order relation of a poset component-wise – that is, by itself. Since no interesting fixed points of this operation came up, the topic was dropped quickly. Soon after Krebs fired up the AlgebraWorkBench (AWB) (see [9,8]) – a software package developed by Sprenger for the handling of finite (universal) algebras in an e-learning environment – fed it the two-element chain and performed the above operation on the latter repeatedly. To general surprise, AWB returned the first few freely generated distributive 0–1-lattices in succession, and eventually led to Krebs’ Ph.D. thesis [6], in whose title stands – for lack of a better idea – for the operation sketched above. This note collects some results, partially contained in [6], on the effects of “ordering the order by itself” within bounded distributive lattices. The paper is organized as follows: Section 2 collects, mostly without proofs, the basic properties of we need; for worked-out proofs and details one might consult [7,6]. Section 3 contains the germinating example which sparked our interest in , with a brute force direct proof (tailored to the finite case) showing that n(2) is nothing else than FD01(n). In Section 4, the effects of on a bounded distributive lattices L are studied in topological terms, based on Priestley duality, the main result being that (L) is the free product of L with the three-element chain. Returning to the finite case, we show in Section 5 how to decide in polynomial time whether a given finite distributive lattice occurs ∗ Corresponding author. E-mail addresses: [email protected] (M. Krebs), [email protected] (J. Schmid). 1567-8326/$ - see front matter ( 2008 Elsevier Inc. All rights reserved. doi:10.1016/j.jlap.2008.02.003 M. Krebs, J. Schmid / Journal of Logic and Algebraic Programming 76 (2008) 198–208 199 as the -image of another one. Section 6 characterizes those L which stay fixed under and relates them to free bounded distributive lattices on infinitely many generators. The final Section 7 considers the special case of Boolean lattices and lists some open questions. The authors’ sincere thanks go to referee who with his constructive and detailed criticism substantially improved the presentation of this paper. 2. Set-up A partially ordered set – for short: a poset – is a pair P = (P, RP ) consisting of a carrier set P and a bi- nary relation RP or just R on P which is reflexive, antisymmetric and transitive, that is, an order (relation). We will freely confuse P and P if the order is clear from the context; also, instead of RP we will mostly use the standard infix notation ≤P or just ≤. Write POSET for the category of all posets together with order-preserving maps. Let P × P be the cartesian square of P, having carrier P × P and the product order ≤P ×P given by (a, a ) ≤P ×P (b, b ) iff a ≤ b and a ≤ b , for all (a, a ), (b, b ) ∈ P × P .NowRP as a binary relation is actually a subset of P × P , consisting of all pairs (a, a) ∈ P × P satisfying a ≤ a, and thus carries the canonical order induced by ≤ × = ≤ × P P . For some of the properties of (P) (RP , P P RP ) for an arbitrary poset P, the reader may wish to consult [6]. This paper is about (P) in the special case when P is a bounded distributive lattice. Such lattices will be written L = (L;∧, ∨, 0, 1), and DLAT01 stands for the category of all bounded distributive lattices satisfying 0 =/ 1, together with all 0–1-preserving lattice homomorphisms. Here’s the formal definition of for this case: Definition 2.1. For L ∈ DLAT01 with order ≤L ⊆ L × L, let (L) be the 0–1-sublattice of L × L with carrier ≤L. Given f : K −→ L in DLAT01, define (f ) : (K) −→ (L) by (f )(a, a):=(f (a), f (a)) for all (a, a) ∈ (K), that is, (f ) = f × f . Fact 2.2. is a functor from DLAT01 to DLAT01. Proof. This follows easily since, for L ∈ DLAT01, the sup of (a, a), (b, b) ∈ (L) is given by (a ∨ b, a ∨ b), and their inf by (a ∧ b, a ∧ b). Define L : L −→ (L) by L(a):=(a, a) for a ∈ L. The subscript L will be omitted if clear from the context. Fact 2.3. (L) is a 0–1-sublattice of (L) isomorphic to L for any L ∈ DLAT01, thus L =∼ (L) ⊆ (L) ⊆ L × L (as 0–1-sublattices). Proof. Routine. One may check that preserves, e.g., injectivity of morphisms (if K is a 0–1-sublattice of L, then (K) a0– =∼ ∈ 1-sublattice of (L)), arbitrary direct products (( i∈I Li) ∈I (Li) for any index set I and lattices Li DLAT01), and even completeness (if L is a complete lattice, so is (L)), but we do not need these facts for the purposes of this paper. 3. Unleashing So far, we have not seen doing any real work. The simplest nontrivial candidate for testing will be the two-element chain 2 with carrier {0, 1} and order {(0, 0), (0, 1), (1, 1)}. The following figures show 0(2):=2, 1(2) = (2), 2(2) and 3(2). 200 M. Krebs, J. Schmid / Journal of Logic and Algebraic Programming 76 (2008) 198–208 Especially the last diagram will immediately ring a bell for anyone having been exposed to an introductory “Order and Lattices” course: It depicts the free algebra in DLAT01 on three generators, and invites us to recognize the other three as the diagrams of the free algebras on 0, 1 and 2 generators. To formulate the expected theorem, we write FD01(n) for the lattice freely generated in DLAT01 by n generators. ∼ n Theorem 3.1. FD01(n) = (2) for all n ∈ ω. ∼ Proof. Put Ln := FD01(n). We show (Ln) = Ln+1 for all n ∈ ω by brute force, transforming an arbitrary n- element set freely generating Ln into a (n + 1)-element set freely generating (Ln). The procedure is based on a peculiar property of free generating sets in DLAT01: G is such for L ∈ DLAT01 iff G generates L and H ≤ K implies H ∩ K/=∅, for any nonempty subsets H,K ⊆ G (see, e.g., [2, section V.3]). So assume G ={g1,...,gn} freely generates Ln (if n = 0, G is empty), and put G :=[G]∪{(0, 1)}. Consider an arbitrary element (a, b) ∈ (Ln), thus a,b ∈ Ln and a ≤ b. We see that (a, b) = ((0, 1) ∧ (b)) ∨ (a) since meets and joins are computed component-wise in (Ln). Since a ∈ Ln, we may write a = p(g1,...,gn) for some M. Krebs, J. Schmid / Journal of Logic and Algebraic Programming 76 (2008) 198–208 201 (lattice) polynomial p, and thus (a) = p((g1),...,(gn)). Analogously, (b) = q((g1),...,(gn)) for a suitable polynomial, hence, (a, b) is in the 0–1-sublattice of (Ln) generated by G . It remains to show that G freely generates (Ln). Assume that, for some H ,K ⊆ G ,wehave H ≤ K . Put H :=−1(H ) and K :=−1(K), thus H,K ⊆ G.If(0, 1) ∈ H ∩ K, we are done, so we may assume that (0, 1) occurs in at most one of H resp. K . Suppose(0, 1) ∈ H but (0, 1)/∈K . Hence K/=∅and K = ( K, K). It follows that H =/ (0, 1) since (0, 1) ≤ ( K, K) implies 1 ≤ K which is not possible for any nonempty K⊆ G, G being a free generating set for Ln. But H =/(0, 1) is only possible if H/=∅ , in which case H = (0, H). We conclude that (0, H) = H ≤ K = ( K, K) and thus H ≤ K. The last comparability is in Ln and we infer that H ∩ K/=∅, G being a free generating set. Pick gi ∈ H ∩ K, then (gi,gi) ∈ H ∩ K and we are done. The dual argument works for (0, 1)/∈ H but (0, 1) ∈ K. Finally, if (0, 1)/∈ H ∪ K, the desired conclusion H ∩ K/=∅is obtained directly. So the orbit of on 2 gives the complete list of all finitely generated free algebras in DLAT01. The obvious question at this point is what other – if any – lattices K ∈ DLAT01 are of the form (L) for some L ∈ DLAT01. As it turns out, the best way to characterize these lattices is to use Priestley’s topological duality theory for DLAT01.
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