A fresh perspective on canonical extensions for bounded lattices

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A fresh perspective on canonical extensions for bounded lattices Mathematical Institute, University of Oxford Department of Mathematics, Matej Bel University Second International Conference on Order, Algebra and Logics Kraków, Poland 8 June 2011

Let L be a lattice and C a complete lattice with L isomorphic to a sublattice of C. Then C is a completion of L and C is a dense completion if any element of C can be expressed as both a meet of joins and join of meets of elements of L, C is a compact completion if for any filter F, and ideal I, of L, F I = there exist finite subsets F F and I I such that F I. A dense and compact completion C of L is called a canonical extension of L. Theorem (Gehrke & Harding, 2001) Every bounded lattice L has a canonical extension L δ, and this is unique up to an isomorphism which fixes L.

Construction of the canonical extension Gehrke & Harding construction uses a polarity between FiltL and IdlL. The original construction for Boolean algebras (Jónsson & Tarski, 1951) used Stone Duality. The canonical extension of a Boolean algebra, B, is the power set of its Stone space. We present a novel construction of the canonical extension of a bounded lattice L, based on hom-functors. This method has the advantage of simplifying the extension of lattice homomorphisms. The hom-functors act on morphisms by composition.

Canonical extensions for distributive lattices Let D be the variety of bounded distributive lattices. Canonical extensions for distributive lattices make use of Priestley duality. The underlying set of the dual space of L D is the set of prime filters. Let 2 = {0, 1};,. The set of prime filters of L can be identified with the set of distributive lattice homomorphisms from L into 2. Denote this D(L, 2). Theorem (Gehrke & Jónsson, 1994) For L D, the set of order-preserving maps from the Priestley dual space of L into 2 is a canonical extension of L. The variety of bounded lattices L is not finitely generated, so there is no natural duality available. However, we use a similar approach, and get hom-functors from L to the dual space and to the canonical extension.

Hom-functors for Priestley duality Let L D and consider the set D(L, 2). Define a topology T on D(L, 2) by subbasic open sets of the form V a = { f D(L, 2) f (a) = 1 } and V a = { f D(L, 2) f (a) = 0 } for a L. Let X be the category of compact totally order-disconnected (Priestley) spaces with continuous order-preserving maps as morphisms. The functor D: D X is defined by D(L) = (D(L, 2), T, ). Let 2 T be the two-element ordered set with the discrete topology. Given a Priestley space X = (X, T, ), the functor E: X D is defined by E(X) = (X (X, 2 T ),, ) where and are defined pointwise.

Hom-functors in the distributive case Let L D and let P be the category of posets with order-preserving maps as morphisms. The functor F : D P is defined by F(L) = (D(L, 2), ) Note: F is the composition of D and the forgetful functor. The functor G : P D is defined by G(P) = (P(P, 2 ),, ) L E D G F L = e L (L) = ED(L) L δ

Maximal partial E-preserving maps Let G be the category of graphs with edge-preserving maps and X = (X, E) a graph where E is a reflexive relation on X. Let ϕ: X 2 be a maximal partial E-preserving map. Then ϕ 1 (0) = { x X y ϕ 1 (1), (y, x) / E } ϕ 1 (1) = { x X y ϕ 1 (0), (x, y) / E } We order the set G mp (X, 2 ) by ϕ ψ ϕ 1 (1) ψ 1 (1) ψ 1 (0) ϕ 1 (0) Theorem The ordered set G(X) = ( G mp (X, 2 ), ) is a complete lattice.

Ploščica s representation of bounded lattices Let L L and 2 = {0, 1};,, 0, 1. Denote by L mp (L, 2) the maximal partial homomorphisms from L into 2. For f, g L mp (L, 2), (f, g) E ( a dom f dom g) (f (a) g(a)). Example: L = M 3. For x, y, z L, define 1 a b c 0 1 if z x f xy (z) = 0 if z y otherwise

Example of the relation E: L mp (M 3, 2) = { f ab, f ac, f bc, f ba, f ca, f cb }. f ab f cb f ca f ba f ac f bc Note: E not transitive, since (f bc, f ac ) E and (f ac, f ab ) E but (f bc, f ab ) / E.

Topology on L mp (L, 2): For a L, V a = { f L mp (L, 2) f (a) = 0 }, W a = { f L mp (L, 2) f (a) = 1 }. Let T be the topology having subbasic closed sets of the form V a, W a for a L.

Topological graphs: A graph X = (X, E) with a topology T is called a topological graph. For L L, let D(L) = (L mp (L, 2), T, E) and 2 T = {0, 1}, T,. Consider G T, the category of topological graphs with continuous E-preserving maps. Given D(L) and 2 T above, we will be interested in the maximal partial continuous E-preserving maps from D(L) to 2 T. We denote this set by G mp T (D(L), 2 T ).

Representation theorem: For f L mp (L, 2) and a L, define e a : L mp (L, 2) 2 T 1 if f (a) = 1 e a (f ) = 0 if f (a) = 0 otherwise by Theorem (Ploščica, 1995) The bounded lattice L is isomorphic to G mp T (D(L), 2 T ) via the isomorphism a e a (a L).

Canonical extensions using maximal partial maps: Let L L and let F(L) = (L mp (L, 2), E). Theorem The complete lattice G(F(L)) = ( G mp (F(L), 2 ), ) is a canonical extension of L.

Hom-functors acting on morphisms We want to replicate the distributive case. For L, K D, and u D(L, K), we have F(u): F(K) F(L), f f u. However, this will not work for non-surjective lattice homomorphisms.

Example of non-surjective homomorphism Let L = 3, K = M 3 and u : L K (u L (L, K)) with u(a) = b. 1 a 0 1 b c d 0 Consider f cd L mp (K, 2). Now f u : L 2 is defined as 1 if x = 1 (f u)(x) = if x = a 0 if x = 0 Clearly, f u / L mp (L, 2).

How do we solve this problem? We have to use an enlarged first dual in order to have well-defined hom-functors on non-surjective homomorphisms. This is used by Allwein & Hartonas (1993) for a duality theorem for bounded lattices. For L L, consider the set of all (not necessarily maximal) partial homomorphisms L p (L, 2). We define F : L G by F (L) = ( L p (L, 2), E ). Also, with C the category of complete lattices with complete lattice homomorphisms, consider G : G C G(X) = ( G mp (X, 2 ),, ). Proposition Let L be a bounded lattice with F(L) = (L mp (L, 2), E) and F (L) = (L p (L, 2), E). Then the complete lattices G mp (F(L), 2 ) and G mp (F (L), 2 ) are order-isomorphic.

Hom-functors acting on morphisms Let u : L K be a lattice homomorphism. Then for x L p (K, 2), we define ( F (u) ) (x) = x u. It is easy to show that for any x L p (K, 2,), we have x u L p (L, 2). Similarly for graphs X and Y, and v G (X, Y), define for ϕ G mp (Y, 2 ), ( G(v) ) (ϕ) = ϕ v.

Extending lattice homomorphisms For { ϕ i i I } G mp (F (L), 2 ) meets and joins are defined for x L p (L, 2) by ( ( i I ϕ i i I ϕ i ) (x) = {1 if x i I ϕ 1 i (1), 0 if there is no y i I ϕ 1 i (1) with (y, x) E; ) (x) = {1 if there is no y i I ϕ 1 i (0) with (x, y) E, 0 if x i I ϕ 1 i (0). Theorem Let u : L K be a lattice homomorphism. Then GF (u) : L δ K δ is a complete lattice homomorphism.

Proof. Let x L p (K, 2) and { ϕ i i I } G mp (F (L), 2). Then x ( i I (G F (u)(ϕ i )) ) 1 (1) x (G F (u)(ϕ i )) 1 (1) i I ( i I ) ϕ i (F (u)(x)) = 1 ( i I ) ϕ i (x u) = 1 x u i ) i I(ϕ 1 (1) x u ( 1(1) i I ϕ i) ( i I ϕ i) (x u) = 1 ( i I ϕ i) (F (u)(x)) = 1 x ( G F (u)( i I ϕ i) ) 1 (1).