BOOLEAN SUBALGEBRAS OF ORTHOALGEBRAS

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1 OOLEAN SUALGERAS OF ORTHOALGERAS JOHN HARDING, CHRIS HEUNEN, AND MIRKO NAVARA Abstract. Earlier work had shown that every nontrivial orthomodular lattice is characterized by its poset of oolean subalgebras in that two such orthomodular lattices with isomorphic posets of oolean subalgebras are isomorphic. In this note we give a method to reconstruct a non-trivial orthoalgebra A from its poset of oolean subalgebras P. Additionally, we characterize those posets that arise as the posets of oolean subalgebras of some orthoalgebra. The technique is to reconstruct A from what we call the directions of P, which are certain partial maps d P P 2. When specialized to the oolean setting, these results are used to show there is a functor from the category of oolean algebras and their homomorphisms to a category of oolean domains with appropriate morphisms that is nearly an equivalence. It remains for future work to extend such categorical considerations to the orthoalgebra setting. 1. Introduction In [12] Sachs showed that every oolean algebra with at least 4 elements is determined by its lattice Sub() of oolean subalgebras. The restriction to oolean algebras with at least 4 elements is necessary since the 1-element and 2-element oolean algebra both have a 1-element lattice as their lattice of subalgebras. Grätzer et. al. [5] characterized those lattices that arise as the lattice Sub() for a oolean algebra as those algebraic lattices where each compactly generated principal ideal is isomorphic to a finite partition lattice. We call these oolean domains. Grätzer et. al. also produced a method to reconstruct a oolean algebra with at least 4 elements from Sub() via a direct limit construction. Here we present a different method to construct a oolean algebra with at least 4 elements from the poset Sub(). The atoms of Sub() are the subalgebras {0, x, x, 1} where x 0, 1. Calling an element of a bounded lattice basic if it is either the bottom or an atom, the basic elements of Sub() correspond to complementary pairs of elements of. A direction for a basic element of Sub() says which of these two complementary elements this basic element corresponds, and is given in the oolean case by a certain ordered pair of elements of Sub(). It is a property of oolean domains that each basic element has two directions. We create a oolean algebra that is isomorphic to from the set of directions of Sub(). This reconstruction of an isomorphic copy of from its subalgebra lattice Sub() is more direct and tractable than the construction of Grätzer et. al. via a direct limit. This allows us to employ this in establishing a categorical connection between oolean algebras and oolean domains. A priori there are limits on what can be expected. A 1-element and 2-element oolean algebra have isomomorphic subalgebra lattices, and the 4-element oolean algebra has 2 automorphisms while its subalgebra lattice, a 2-element lattice, has only a single automorphism. With these considerations in mind, we come near to an equivalence to the category oo of oolean algebras and a suitable category ood of oolean domains. We show there is a functor Sub oo ood that is dense, faithful except for homomorphisms whose image has 4 elements or less, and full except with respect to morphisms whose image is a 2-element lattice. 1

2 2 JOHN HARDING, CHRIS HEUNEN, AND MIRKO NAVARA For an orthomodular lattice L, its poset of oolean subalgebras Sub(L) is not a lattice unless L is oolean since there will be oolean subalgebras of L whose union is not contained in a oolean subalgebra of L. Sachs results were extended in [7] to show that an orthomodular lattice L with more than two elements is determined by the poset Sub(L). This result has implication for the topos approach to quantum mechanics of utterfield and Isham [9]. Further results along this line were given by Döring and Harding [3] where it was shown that a von Neumann algebra with no type-2 factor is determined up to Jordan isomorphism by its poset of unital abelian subalgebras, and by Hamhalter [6] with corresponding results for C algebras. A thorough account of matters surrounding this area with many results is found in the Ph.D. thesis of ert Lindenhovius [?]. The fundamental reason why an orthomodular lattice L is determined by its poset Sub(L) of oolean subalgebras is that it is built in a very tight way by gluing together its oolean subalgebras [10]. There are other types of structures that generalize orthomodular lattices and are also built by gluing together oolean algebras. Orthomodular posets [10], their generalization to orthoalgebras [4], and the partial oolean algebras of Kochen and Specker [11] are the most notable examples. One can glue two 8-element oolean algebras together by identifying an atom of each and a coatom of each, or by identifying an atom of each with a coatom of the other. oth give examples of partial oolean algebras, but only the first is an orthoalgebra. These two partial oolean algebras have the same poset of oolean subalgebras, so partial oolean algebras are not determined by their posets of oolean subalgebras, and this is for an essential reason. We show that an orthoalgebra A with at least 4 elements is determined by its poset of oolean subalgebras Sub(A). Further, we characterize the posets, that we call orthodomains, that arise as Sub(A) for an orthoalgebra A. Here matters are somewhat delicate since such posets will in general have neither finite joins or finite meets, and the combinatorial properties of even finite orthoalgebras are notoriously complicated. They do have directed joins, are atomistic, and have the property that each principal ideal is a oolean domain. Several other properties are required to characterize these posets, with the main one being the existence of enough directions. Here directions are a generalization of the notion of direction from the oolean domain setting. The situation is somewhat analogous to the characterization of lattices isomorphic to open set lattices of topological spaces as the complete meet-distributive lattices with enough points. This paper is arranged in the following way. The second section describes directions in the oolean setting, and the third section gives categorical results in the oolean setting. The fourth section gives background on orthostructures and their posets of oolean subalgebras. The fifth section discusses directions for orthoalgebras and establishes that an orthoalgebra A can be reconstructed from the directions of its poset P of oolean subalgebas. The sixth section characterizes those posets P that arise as Sub(A) for some orthoalgebra A. At the later stages of preparation of this manuscript we were contacted by ert Lindenhovius who had a manuscript that had overlap with ours. In particular, he had a direct reconstruction of a oolean algebra using essentially the same directions for oolean algebras as we use, and with a different method than taken here, used oolean directions to reconstruct an orthomodular poset from its poset of oolean subalgebras under the assumption of a non-trivial center. The results obtained by us and by Lindenhovius were obtained independently at approximately the same time.

3 OOLEAN SUALGERAS OF ORTHOALGERAS 3 2. Subalgebras of oolean algebras and their directions Here, and throughout the paper, we use standard terminology from posets as found in for instance [1]. In particular, for a poset P and x P we use x = {a P a x} and x = {a P x a} for the principal ideal and principal filter generated by x. Definition 2.1. For a oolean algebra, let Sub() be the set of subalgebras of partially ordered by set inclusion. This notion was introduced by Sachs [12], and the details of his study of its finer properties will form the basis of our investigation. First, some more general comments about its nature are in order. Since the intersection of oolean subalgebras is a subalgebra, Sub() is a complete lattice, and since finitely generated oolean algebras are finite, the compact elements of this lattice are the finite oolean subalgebras of. Since every oolean algebra is the union of its finite subalgebras, it follows that Sub() is an algebraic lattice. The following characterization of those algebraic lattices that arise as Sub() was given later by Grätzer et. al. [5]. We recall that a partition lattice is one that is isomorphic to the lattice of partitions of a set. Theorem 2.2. A poset L is isomorphic to Sub() for some oolean algebra iff (1) L is an algebraic lattice (2) for each compact c L, c is isomorphic to a finite partition lattice Lattices with these properties are called oolean domains. If I is an ideal of a oolean algebra, the elements belonging to I, and their complements, form a subalgebra of. Properties of these subalgebras, and in particular a way to characterize these subalgebras order theoretically in Sub(), was given in Sachs [12]. We review the details since they will form the basis of our study. Definition 2.3. A subalgebra m of a oolean algebra is called an (principal) ideal subalgebra when m = I I for some (principal) ideal I of and I = {x x I}. 1 x x 0 Definition 2.4. An element m of a lattice is dual modular if (m n) o = m (n o) for every o m and (n m) p = n (m p) for every p n. Lemma 2.5 ([12, Thm. 1]). The dual modular elements of Sub() are the ideal subalgebras. For a oolean algebra, the least element of the oolean domain Sub() is {0, 1}, the largest element is, and that the atoms of Sub() are the elements {0, x, x, 1} where x 0, 1. So there is a bijection between pairs {x, x } of complementary elements of and elements of Sub() that are either or an atom. We introduce terminology for this.

4 4 JOHN HARDING, CHRIS HEUNEN, AND MIRKO NAVARA Definition 2.6. Let P be a poset with a least element. An element of P is basic if it is either an atom of P or is. So pairs {x, x } of complementary elements of correspond to basic elements of Sub(). Pairs where x 0, 1 correspond to atoms of Sub(). We will also need to characterize those pairs {x, x } where neither of x, x is an atom of. Since {0, x, x, 1} is an ideal subalgebra iff one of x, x is either 0 or an atom of, the following is immediate from Lemma 2.5 Lemma 2.7. For a oolean algebra, the basic elements of Sub() that are dual modular are the least element = {0, 1} and those subalgebras {0, x, x, 1} where either x, x is an atom of. We come to the key notion, that of a direction for a basic element. The definition is broken into two parts to reflect that basic elements come in two flavors, those that are dual modular and those that are not. Definition 2.8. Let D be a oolean domain and r be an atom of D that is not dual modular. An ordered pair (p, q) of elements of D is a direction for r if (1) p, q are dual modular elements (2) p q = r (3) p q = The directions for a dual modular basic element r are (r, ) and (, r). We say that (p, q) is a direction of D if it is a direction for the basic element r = p q of D. Remark 2.9. In the definition of a direction, the directions (r, ) and (, r) for a dual modular basic element r satisfy conditions (1) (3). However, there may be other pairs (p, q) that satisfy conditions (1) (3) for r. Consider a 16-element oolean algebra and let b, c be elements of height 2 with a = b c an atom. Then r = {0, a, a, 1} is basic. For p = b b and q = c c both p, q are dual modular, p q = r, and p q =. Lemma Let be a oolean algebra and x. Then the directions for the basic element r = {0, x, x, 1} are (p, q) and (q, p) where p = x x and q = x x, and if has more than 4 elements, these are distinct. x q p 1 0 p q x Proof. Let x with r = {0, x, x, 1} a dual modular basic element. y definition (r, ) and (, r) are the directions for r, and if has more than 4 elements, these are distinct. If x = 0, then r = and (p, q) = (, ). If x = 1, then r = and (p, q) = (, ). If x is an atom of, then r = {0, x, x, 1} and (p, q) = (r, ), and if x is a coatom of, then r = {0, x, x, 1} and (p, q) = (, r). Suppose that x is neither 0, 1, an atom, or a coatom of. Then r = {0, x, x, 1} is a basic element that is not dual modular, and there are y, z such that 0 < y < x < z < 1. It is clear that p, q are ideal subalgebras, hence are dual modular elements. Also p q = r. For any b we have

5 OOLEAN SUALGERAS OF ORTHOALGERAS 5 b = (b x) (b x ). Since b x p and b x q, then b is in the subalgebra p q generated by p, q. So p q =. So (p, q) and (q, p) are directions for r. Since y p and y / q, these directions are distinct. We next show that in the case when x is neither 0, 1, an atom, or a coatom of, that these are the only directions for r. From our assumptions of x there are 0 < y < x < z < 1. Suppose that (s, t) is a direction for r. Since s, t are dual modular, they are ideal subalgebras. So there are ideals I, J of with s = I I and t = J J. Note that r = s t gives {0, x, x, 1} = (I J) (I J ) (I J) (I J ) It cannot be the case that x I J since I and J being ideals would imply that y I J, and it cannot be the case that x I J since I and J being filters would imply that z I J. So one of x, x belongs to I J and the other to I J. Assume that x I J. Then since J is a filter, there cannot be an element of I other than 1 that is larger than x since it would belong to I J, and since I is an ideal and x < z < 1 it cannot be that 1 I since this would imply that z I. So x is the largest element of I, and dual reasoning shows that x is the least element of J. So (s, t) = (p, q). If it had been the case that x I J, then symmetry gives (s, t) = (q, p). Definition Let Dir(D) be the set of all directions on D. The idea is as follows. If is a oolean algebra with more than 4 elements and D = Sub(), then by Lemma 2.10 each x gives a direction (p, q) where p = x x and q = x x. So the basic element r = {0, x, x, 1} of D has 2 directions, one from x and one from x. Directions allow us to recreate elements of from the order structure of D. We will require structure on these directions, and the relation and unary operation are easiest. If (p, q) is the direction for r given by x, it is easily seen that (q, p) is the direction for r given by x. If (p 1, q 1 ) is the direction for x and (p 2, q 2 ) that for y, then x y implies p 1 p 2 and q 2 q 1. These relationships can hold in some cases without x y. If x is an atom of, so r = {0, x, x, 1} is a dual modular atom of D, then the direction for x is (r, ) and that of x is (, r), but of course x / x. Excluding this situation leads to the following definitions. Definition Let D be a oolean domain and let (p, q), (p 1, q 1 ), (p 2, q 2 ) be directions for the basic elements r, r 1, r 2. Set (p, q) = (q, p). In the case when r 1 is not a dual modular atom set (p 1, q 1 ) (p 2, q 2 ) p 1 p 2 and q 1 q 2 When r 1 is a dual modular atom, let (p 1, q 1 ) (p 2, q 2 ) when the condition above holds and additionally (p 2, q 2 ) (p 1, q 1 ). Proposition Let be a oolean algebra with more than 4 elements. Dir(Sub()) be given by ϕ(x) = ( x x, x x) Then ϕ is an isomorphism with respect to the structures given by and. Let ϕ Proof. Let r be a basic element of Dir(). Then there is x with r = {0, x, x, 1}. Lemma 2.10 gives that there are two directions for r, and these are given by ϕ(x) and ϕ(x ). So ϕ is onto. To see that ϕ is one-one, suppose that x, y and ϕ(x) = ϕ(y). Lemma 2.10 provides that ϕ(x) is a direction for {0, x, x, 1} and ϕ(y) is a direction for {0, y, y, 1}. So {0, x, x, 1} = {0, y, y, 1}.

6 6 JOHN HARDING, CHRIS HEUNEN, AND MIRKO NAVARA So either y = x or y = x. Since has more than 4 elements, Lemma 2.10 gives that ϕ(x) and ϕ(x ) are distinct. Thus y = x, showing that ϕ is also one-one, hence a bijection. It is obvious that ϕ preserves the operation. Let x y. Then x x y y and y y x x. y Lemma 2.10, ϕ(x) is a direction for r = {0, x, x, 1}. Then Lemma 2.7 gives that r is a dual modular atom of Dir() iff one of x, x is an atom of. So if neither x, x is an atom of, then ϕ(x) ϕ(y). If one of x, x is an atom of, then to show that ϕ(x) ϕ(y) we must additionally show that ϕ(y) ϕ(x), or in terms of what has already been shown, that y x. ut this follows since x 0 and x y. So in either case, x y implies that ϕ(x) ϕ(y). Suppose that x, y with ϕ(x) ϕ(y). Then x x y y and y y x x To show that x y we argue by contradiction. Suppose that x / y, hence y / x. The first containment implies x y and the second containment implies y x. So y x and x y, giving y = x. Since ϕ(x) ϕ(y) and ϕ(y) = ϕ(x), the definition of implies that r = {0, x, x, 1} cannot be a dual modular atom of Dir(), hence neither of x, x is an atom of. Since x y we have x 0. So there is 0 < z < x. Then z x x, but z / x x = y y, a contradiction. Theorem Let be a oolean algebra with more than 4 elements and D be a oolean domain with more than 2 elements. Then (1) Sub() is a oolean domain (2) Dir(D) is a oolean algebra (3) is isomorphic to Dir(Sub()) (4) D is isomorphic to Sub(Dir(D)) Proof. y Theorem 2.2, if is a oolean algebra then Sub() is a oolean domain, and if D is a oolean domain then there is a oolean algebra with D Sub(). This provides (1). Proposition 2.13 shows that with the structures given by and, the map ϕ is an isomorphism from to Dir(Sub()). Since a oolean algebra may be defined through such structure, Dir(Sub()) is a oolean algebra and ϕ is a oolean algebra isomorphism from to it. This provides (3). Since D is a oolean domain with more than 2 elements, by Theorem 2.2 there is a oolean algebra with more than 4 elements with D Sub(). Then Dir(D) is isomorphic to Dir(Sub()), and by (3) this is oolean. This provides (2). To establish (4), there is a oolean algebra with D Sub(). Then using (3) we have Sub(Dir(D)) Sub(Dir(Sub())) Sub() D 3. Categorical aspects in the oolean setting In this section we consider categorical ties between oolean algebras and oolean domains. The idea is to take Sub() as the object part of a functor from oolean algebras to a suitable category of oolean domains, and to take a oolean homomorphism f 1 2 to the direct image map f[ ] Sub( 1 ) Sub( 2 ). From the outset, there are limitations on what can be achieved. The 2-element and 1-element oolean algebras both have 1-element oolean domains as their subalgebra lattices, and the 4-element oolean algebra has 2 automorphism, while its subalgebra lattice is a 2-element lattice, and has only the trivial automorphism. So there is no

7 OOLEAN SUALGERAS OF ORTHOALGERAS 7 possibility of obtaining an equivalence of categories. arring these obstacles, we come as close as one could hope to an equivalence. Definition 3.1. Let oo be the category of oolean algebras and homomorphisms between them, and let oo* be its full subcategory consisting of oolean algebras having more than 1 element, i.e. the non-trivial oolean algebras. In the following we note that if D is a oolean domain and m D, then m is a oolean domain. Thus it makes sense to talk of a direction in m. Definition 3.2. Let ood be the category whose objects are oolean domains with morphisms being functions α D 1 D 1 that preserve arbitrary joins and for each m D 1 (1) α maps m onto α(m) (2) if (p, q) is a direction in m, then (α(p), α(q)) is a direction in α(m) (3) if (p 1, q 1 ) (p 2, q 2 ) in Dir( m), then (α(p 1 ), α(q 1 )) (α(p 2 ), α(q 2 )) in Dir( α(m)) Remark 3.3. The third condition in Definition 3.2 is not so elegant, and carries less weight than it may seem. Since α preserves joins, it is order preserving. This provides that (p 1, q 1 ) (p 2, q 2 ) implies that (α(p 1 ), α(q 1 )) (α(p 2 ), α(q 2 )) except for the case where (α(p 1 ), α(q 1 )) is a dual modular atom of α(m). The content of condition (3) is to provide that in this case (α(p 2 ), α(q 2 )) (α(p 1 ), α(q 1 )). I do not know if (3) is needed, or whether it follows from (1) and (2). There may also be more elegant ways to get this result, perhaps in terms of α preserving meets of certain pairs of dual modular elements. Proposition 3.4. There is a functor Sub oo ood taking a oolean algebra to Sub() of subalgebras, and a homomorphism f 1 2 to f[ ]. Proof. In general, the image under a homomorphism of the subalgebra generated by the union of a family of sets is equal to the subalgebra generated by the union of the images of the sets. So f[ ] preserves arbitrary joins. Let m Sub( 1 ) and n = f[m]. We verify (1) (3) of Definition 3.2 for m. If t Sub( 2 ) with t n, then s = f 1 [t] m is a subalgebra of 1 that is contained in m and f[s] = t. So f[ ] maps m onto f[m], giving (1). Suppose that (p, q) is a direction in m. y Lemma 2.10 there is x m with p = m x m x and q = m x m x Here m x = {w m w x} is the downset of x in the subalgebra m and similarly for m x. Since f maps m onto n, we have f[ m x] = n f(x) where n f(x) is the downset in n of f(x), and similarlyf[ m x ] = n f(x). So f[p] = n f(x) m f(x ) and f[q] = n f(x) n f(x). y Lemma 2.10 this is a direction in n, so (2) holds. For (3), if (p 1, q 1 ) (p 2, q 2 ) are directions in m, then there are x y in m with p 1 = m x m x, q 1 = m x m x, p 2 = m y m y, and q 2 = m y m y. From the argument in (2) we have (f[p 1 ], f[q 1 ]) is the direction in f[m] for f(x) and (f[p 2 ], f[q 2 ]) is the direction in f[m] for f(y). Since f(x) f(3) item (3) follows. Thus f[ ] is a morphism of domains. That this assignment on morphisms preserves composition is clear. Remark 3.5. At the opening of this section we mentioned several obstacles to an equivalence between oo* and ood. Here we mention others. Suppose that 1 and 2 are oolean algebras with 2 having more than 1 element. Then every prime ideal of 1 gives a distinct homomorphism f 1 2 whose image has 2 elements. For each of these homomorphisms, f[ ] is the map sending each element of Sub( 1 ) to the of Sub( 2 ). For another pathology,

8 8 JOHN HARDING, CHRIS HEUNEN, AND MIRKO NAVARA suppose f 1 2 is a oolean algebra homomorphism whose image has 4 elements. Then composing with the non-trivial automorphism of this image gives a homomorphism g 1 2 distinct from f with f[ ] and g[ ] equal. Thus the functor Sub is not injective with respect to homomorphisms having four or fewer elements in their image. A functor F C D gives an equivalence of categories if it is full, faithful, and dense, i.e. each object of D is isomorphic to one in the image of F. The functor Sub oo ood does not have these properties, but it comes close. efore establishing this, we collect several observations that are needed. Lemma 3.6. Let be a oolean algebra with more than 4 elements, and let (p, q), (p 1, q 1 ), (p 2, q 2 ) be directions for the basic elements r, r 1, r 2 in Sub(). (1) If r is dual modular, one of p, q is basic and the other equals (2) If r is not dual modular, neither of p, q is basic or equal to (3) If p 1 = p 2 and either p 1 or r 1 = r 2, then q 1 = q 2 (4) The join (p 1, q 1 ) (p 2, q 2 ) in Dir() has first component p 1 p 2 Proof. y Lemma 2.10 there are elements x, x 1, x 2 with (p, q) = ( x x, x x) and with (p i, q i ) = ( x i x i, x i x i) for i = 1, 2. (1) This is from Definition 2.8. (2) y Lemma 2.7 x, x are not 0, 1, atoms, or coatoms. (3) Since x 1 x 1 = x 2 x 2 either x 1 = x 2, which implies q 1 = q 2, or p 1 = p 2 =. If p 1 = p 2 =, then (1) gives (p i, q i ) = (, r i ) for i = 1, 2 and the result follows. (4) Theorem 2.14 shows the first component of (p 1, q 1 ) (p 2, q 2 ) is S = (x 1 x 2 ) (x 1 x 2 ). Clearly S contains p 1, p 2 hence S contains the subalgebra p 1 p 2 they generate. If z (x 1 x 2 ), then z = (z x 1 ) (z x 2 ). It follows that S p 1 p 2. Lemma 3.7. Let 1 and 2 be oolean algebras with 2 having more than 4 elements, and let f 1 2 be a function that preserves order and complementation. If f(x y) = f(x) f(y) whenever f(x) f(y) is not a coatom or 1, then f is a homomorphism. Proof. Since f preserves order and complementation it preserves orthogonality. We will show that f preserves orthogonal joins. From this it follows that f preserves binary joins, so is a homomorphism. Indeed, for any x, y there are pairwise orthogonal u, v, w with x = u v and y = v w. Then if f preserves orthogonal joins f(x y) = f(u v w) = f(u) f(v) f(w) = (f(u) f(v)) (f(v) f(w)) = f(x) f(y). We must show that f preserves orthogonal joins. Suppose x and y are orthogonal. y assumption f preserves their join if f(x) f(y) is not a coatom or 1, and since f is order preserving it clearly preserves their join if f(x) f(y) = 1. It remains to consider the case when f(x) f(y) is a coatom c of 2. Since f(x) and f(y) are orthogonal, they cannot both be coatoms of 2, and we assume that f(y) is not a coatom. We argue by contradiction. Assume f(x y) f(x) f(y), hence f(x y) = 1. Since f preserves complementation f(x y ) = 0. So f(x y ) f(y) = f(y) is not a coatom or 1. This implies that f(y) = f(x y ) f(y) = f((x y ) y) = f(x y). Then f(x ) f(y) c and f(x) f(x) f(y) = c. This contradicts that f preserves complementation. Lemma 3.8. Suppose that 1 and 2 are oolean algebras and α Sub( 1 ) Sub( 2 ) is a oolean domain morphism with α( 1 ) = T having more than 4 elements. Then there is a unique homomorphism f 1 2 with α = f[ ]. Proof. Note first that α( 1 ) = T implies by the definition of a morphism of oolean domains that α maps Sub( 1 ) onto Sub(T ). Since T has more than 4 elements, it follows that 1 has more than 4 elements. Then by Theorem 2.14 there are isomorphisms

9 OOLEAN SUALGERAS OF ORTHOALGERAS 9 1 Dir(Sub( 1 )) x ( x x, x x) T Dir(Sub(T )) y ( T y T y, T y T y) For (p, q) a direction of Sub( 1 ), Definition 3.2 gives that (α(p), α(q)) is a direction in α( 1 ) = T = Sub(T ). So there is a function (α, α) Dir(Sub( 1 )) Dir(Sub(T )) taking (p, q) to (α(p), α(q)). Let f 1 T be the unique function making the square below commute. Dir(Sub( 1 )) (α, α) Dir(Sub(T )) 1 f T To give f explicitly, for each x 1 set f(x) = y where y T is the unique element with α( x x ) = T y T y α( x x) = T y T y Let g 1 2 be a homomorphism with α = g[ ]. Then g( 1 ) = α( 1 ) = T, so the image of g is T. Thus g[ x x ] = T g(x) T g(x), and g[ x x] = T g(x) T g(x), and this gives g(x) = f(x). So there is at most one homomorphism from 1 to 2 whose image mapping is α, and the candidate for this homomorphism is the function f. It remains to show that f is a homomorphism and the image map f[ ] is equal to α. To show that f is a homomorphism, it is enough to show that (α, α) is a homomorphism, and for this we rely on Lemma 3.7. Condition (3) of Definition 3.2 shows that (α, α) is order preserving. It would be nice not to have to use condition (3). For (p, q) Dir(Sub( 1 )) with (α(p), α(q)) = (s, t), we see that (α, α) preserves complementation since (α, α)((p, q) ) = (α(q), α(p)) = (α(p), α(q)) = ((α, α)(p, q)) To show the final condition of Lemma 3.7 required to show that (α, α), and hence f, is a homomorphism, suppose that (p 1, q 1 ) and (p 2, q 2 ) are such that (α(p 1 ), α(q 1 )) (α(p 2 ), α(q 2 )) is not 1 or a coatom of Dir(Sub(T )). This means that the first coordinate of this join is not T. Making use of Lemma 3.6 and the fact that α preserves joins, we have the following where indicates an irrelevant second component. (α, α)((p 1, q 1 ) (p 2, q 2 )) = (α, α)(p 1 p 2, ) = (α(p 1 ) α(p 2 ), ) (α, α)(p 1, q 1 ) (α, α)(p 2, q 2 ) = (α(p 1 ), α(q 1 )) (α(p 2 ), α(q 2 )) = (α(p 1 ) α(p 2 ), ) These are directions in Sub(T ) with the same first component that is different from T. So they are equal. This shows that (α, α) satisfies the hypotheses of Lemma 3.7, hence it, and therefore f, is a homomorphism. Finally, it remains to show that f[ ] = α. Let x 1. Then ( x x, x x) is a p.p. for r = {0, x, x, 1}. y the definition of a oolean domain morphism, (α( x x ), α( x x)) is a p.p. for α(r). From the definition of f, this p.p. is equal to ( f(x) f(x), f(x) f(x)),

10 10 JOHN HARDING, CHRIS HEUNEN, AND MIRKO NAVARA and this is a p.p. for {0, f(x), f(x), 1}. Thus α(r) = f[r]. So α and f[ ] agree on the basic elements of Sub( 1 ). Since every element of Sub( 1 ) is the join of basic elements and both α and f[ ] preserve arbitrary joins, α = f[ ]. Theorem 3.9. The functor Sub oo ood is dense, faithful except with regards to oolean algebra homomorphisms whose image has 4 elements or less, and full except possibly with regards to oolean domain morphisms whose image has 2 elements. Proof. Density is given by Sach s result saying that each oolean domain D is isomorphic to Sub() for a oolean algebra. This further determines the oolean algebra up to isomorphism except when D has a single element. Lemma 3.8 says that if α Sub( 1 ) Sub( 2 ) is a morphism of oolean domains where the image of α has more than 2 elements (so α(1) has height more than 1), then there is a unique oolean algebra homomorphism f 1 2 with α = f[ ]. y considering a trivial homomorphism and a homomorphism induced by a prime ideal, this result is seen to hold (without uniqueness) also in the case that the image of α is just the of Sub( 2 ). This gives fullness. Suppose that f 1, f with f 1 [ ] = α and f 2 [ ] = α. Then if the image of α has more that 2 elements, so equivalently if the image of either f 1, f 2 has more than 4 elements, then Lemma 3.8 provides that f 1 = f Orthoalgebras In this section, we introduce the basics of orthoalgebras and their subalgebras. For further details, the reader should consult [4]. Definition 4.1. An orthoalgebra (abbreviated oa) (A,,, 0, 1) is a set A with a partially defined binary operation with domain of definition, unary operation operation and constants 0, 1 that satisfies (1) is commutative and associative in the usual sense for a partial operation (2) x 0 is defined and equal to x for each x A (3) x is the unique element with x x defined and equal to 1 (4) x x is defined iff x = 0 For any oolean algebra, restricting the binary join operation of to pairs of orthogonal elements produces an orthoalgebra, and we call an orthoalgebra a oolean orthoalgebra if it arises this way. For any orthoalgebra A, there is a partial ordering on A given by x z iff there exists y with x y and x y = z. An orthoalgebra is oolean iff this partial ordering is the partial ordering of a oolean algebra. Definition 4.2. Let A be an oa and S A. Then S is a sub-oa of A if (1) 0, 1 S (2) x S x S (3) x, y S and x y x y S A sub-oa S that is a oolean oa is called a oolean subalgebra of A. Definition 4.3. A maximal oolean subalgebra of an oa A is a block of A. We call a block small if it has 4 or fewer elements.

11 OOLEAN SUALGERAS OF ORTHOALGERAS 11 It is known that small blocks of an oa are what are known as horizontal summands of A. y removing a small block of A, we meant removing the two elements of the block that are not 0, 1 from A. It is known that removing small blocks from A leaves an oa A with no small blocks, and that the original oa A can be obtained from A by taking the horizontal sum of A and an appropriate number of 4-element oolean algebras. Definition 4.4. For an oa A, let Sub(A) be the poset of oolean subalgebras of A, partially ordered by set inclusion. Our aim is to reconstruct an oa A from Sub(A), and to characterize the posets that arise as Sub(A) for some oa A. This will be the content of the following sections. We conclude this section with a brief explanation of our choice of oas as our ambient setting. Remark 4.5. For an oa A, each x A belongs to the oolean subalgebra {0, x, x, 1} of A. So an oa is constructed by gluing together a family of oolean oas. One can work from a general perspective. A family F of oolean oas is compatible if for each, C F (1) and C have the same 0 and 1. (2) If x C, then x in equals x in C. (3) for x, y C, x y exists in iff it exists in C, and when defined they are equal. For a compatible family F, its pasting P(F) is the structure (A,,, 0, 1) where A is the union of the underlying sets of the oolean oas in F, and letting,, 0, 1 be the unions of the corresponding operations on the members of F. A structure (A,,, 0, 1) that is the pasting of a compatible family of oolean oas a weak orthostructure. This is an extension to oolean oas of the notion of a compatible family of oolean algebras [2]. It is a general setup that includes oas and the partial oolean algebras of Kochen and Specker. A oolean subalgebra of a weak orthostructure A is a A that is closed under 0, 1,, and forms a oolean oa. One might hope to reconstruct A from its poset Sub(A) of oolean subalgebras, but this is not possible. The poset at left below is Sub(A) where A is the oa MO 2 2 shown in the middle. Here a = {0, p, p, 1}, b = {0, q, q, 1}, c = {0, r, r, 1}, d = {0, s, s, 1}, and e = {0, t, t, 1}, while m and n are the two 8-element blocks of A. However, we can also produce a weak orthostructure A by taking the two 8-element oolean algebras shown at right. It should be noted that this weak orthostructure is not an oa. However, Sub(A ) is isomorphic to the poset at left. Thus the poset at left does not determine a unique weak orthostructure, but as we will see, it does determine a unique, to isomorphism, oa m n p q r s t p q r r s t a b c d e p q r s t p q r r s t Orthodomains and orthodirections In this section, we define orthodomains which abstract basic properties of the posets Sub(A) for an oa A. We define orthodirections, abbreviated to o-directions, on orthodomains, and

12 12 JOHN HARDING, CHRIS HEUNEN, AND MIRKO NAVARA show that an orthoalgebra A can be reconstructed from the o-directions on the orthodomain Sub(A). We begin with an example that shows there is some counterintuitive behavior in the posets Sub(A). Example 5.1. There is an orthoalgebra F called the Fraser cube that is constructed from the usual 3-dimension cube with 8 vertices and 6 faces. One considers each face to give a 16-element oolean algebra whose atoms are the four vertices of the face. Consider the element a b. Its orthocomplement in the oolean algebra corresponding to the bottom of the cube is c d, and its orthocomplement in the oolean algebra corresponding to the front of the cube is e f. Thus c d = e f. Similar reasoning shows that the intersection of the oolean subalgebras for the top and bottom of the cube consists of the elements 0, a b, b c, c d, a d, 1. Thus the intersection of two oolean subalgebras of F is not oolean. This implies that in Sub(F ), two elements need not have a meet, and two elements that have an upper bound need not have a least upper bound. These are contrary to the situation for oolean domains. h g e f d c a b Definition 5.2. In a poset P we use for the covering relation. Thus x z if x < z and there is no y with x < y < z. Definition 5.3. An orthodomain is a poset P with least element such that (1) every directed subset of P has a join (2) P is atomistic and the atoms are compact (3) for each x P, x Sub() for a oolean algebra (4) if r, s are distinct atoms of P and r, s u, then r s = u Note that condition (2) in this definition is not implied by condition (3) even though every oolean domain is atomistic. For instance, the poset with a bottom, three atoms, and two elements that both cover all three atoms satisfies (1) and (3), but not (2). We next examine more closely condition (4). Definition 5.4. For atoms r, s of a oolean domain P, we say that r, s are near if they are distinct, their join r s = u exists, and r, s u. From condition (4), atoms r, s are near iff they are distinct and have an upper bound u of height 2, since this implies that u is their join. We come now to a property that will be key for us, a property similar to the exchange property of geometry. Proposition 5.5 (Exchange property). If r, s are near atoms of an orthodomain with r s = u, then there is exactly one atom t that is distinct from r, s and with t u. Further, any two of r, s, t are near.

13 OOLEAN SUALGERAS OF ORTHOALGERAS 13 Proof. y the definition of near, r s = u exists and covers both r, s. y the definition of an orthodomain, u is a oolean domain. Since the top of this oolean domain covers an atom in it, the oolean domain u must be isomorphic to the subalgebra lattice of an 8-element oolean algebra. Then u must have 3 distinct atoms, so there is a third atom t distinct from r, s with t u. Then r, s, t u. It follows from the definition of an orthodomain that u is the join of any two of r, s, t, hence any two of r, s, t are near. Proposition 5.6. If A is an oa, then Sub(A) is an orthodomain where directed joins are given by unions. Proof. Take a directed family S of oolean subalgebras of A and let = S. Using directedness, it is easy to see that is closed under,, 0, 1, hence is a sub-oa. To show is a oolean oa it is enough to show that if D is a finite subset of, then there is a finite jointly orthogonal family F in with D { E E F }. This follows from the directedness of S. Thus each directed family in Sub(A) has a join that is given by its union. The atoms of Sub(A) are the oolean subalgebras {0, x, x, 1} where x 0, 1. Since directed joins are given by unions, it follows that the compact elements of Sub(A) are exactly the finite oolean subalgebras, and hence every atom is compact. For any Sub(A) we have that is the union of the atoms beneath it, hence is the join of the atoms beneath it. So Sub(A) is atomistic. Finally, for a oolean oa, its oolean subalgebras are exactly its sub-oas that are oolean, so (3) holds in Sub(A). For (4), suppose r, s are distinct atoms of Sub(A) with r, s u. Then r = {0, x, x, 1} and s = {0, y, y, 1} for some x, y A with 0, x, x, y, y, 1 all distinct. Since u covers an atom of P and u is a oolean domain, then u is an 8-element oolean subalgebra of A that contains r, s. So there is z A with u = {0, x, x, y, y, z, z, 1} an 8-element oolean subalgebra of A. One of x, x is an atom of u, one of y, y is an atom of u, and one of z, z is an atom of u. We may assume that x, y, z are atoms of u. Then in u we have x y = z. So in A we have x y = z. Then if v is a oolean subalgebra of A that contains r, s, we have x, y v, hence x y = z v. Thus u = {0, x, x, y, y, z, z, 1} v. So r s = u. Thus Sub(A) is an orthodomain. Definition 5.7. For an orthodomain P and basic element r P, an orthodirection, written o-direction, for r is a map d r P 2 such that for each k, n r (1) d(k) is a direction for r in the oolean domain k (2) if k n and d(n) = (p, q), then d(k) = (k p, k q) (3) if r k, n, d(k) = (r, k) and d(n) = (n, r) then k n exists Let odir(p ) be the set of all o-directions for basic elements of P. We note that if d is an o-direction of some basic element r of P, then r can be determined from the partial mapping d as the least element of its domain. Proposition 5.8. Let r be a basic element of an orthodomain P and d be an o-direction for r. (1) each element of P lies beneath a maximal element (2) for any r k n the value of d(k) is determined by d(n) (3) for any r < k n, the value of d(n) is determined by d(k) Proof. (1) Zorn s lemma produces a maximal directed set containing a given element k of an orthodomain P. Taking the join of this maximal directed set provides a maximal element of P above k. (2) This is part of the definition of a o-direction. (3) Let p, q n be such that (p, q)

14 14 JOHN HARDING, CHRIS HEUNEN, AND MIRKO NAVARA and (q, p) are the two directions for r in n. Then d(n) = (p, q) or d(n) = (q, p). In the first case, d(k) = (k p, k q) and in the second d(k) = (k q, k p). We claim that k p k q, thus d(k) determines d(n). y Definition 2.8, p q = r. So if k p = k q, then r = k p q = k p = k q. This would give d(k) = (r, r), contrary to d(k) being a direction for r in k. Proposition 5.9. A basic element r in an orthodomain P has at most two o-directions. Proof. First consider the case when r =, and let d be an o-direction for r. For any k with r k there are two directions for r in k, (k, ) and (, k). Suppose there is a maximal element m above r with d(m) = (m, ). We claim that for any maximal element m above r that d(m ) = (m, ). It follows from this, and the definition of an o-direction, that d(k) = (k, ) for all k. Suppose m is maximal above r with d(m ) = (, m ). Then m m. Since P is atomistic and m, m are both maximal, we can find atoms k, n with k m and k m, and n m and n m. Then r k, n, d(k) = (k, r), and d(n) = (r, n). So by Definition 5.7, k n exists. Having d(k n) = (k n, ) would contradict d(n) = (, n), and having d(k n) = (, k n) would contradict d(k) = (k, ). Next, consider the case where r is an atom of P that is maximal in P and d is an o-direction for r. Since the domain of d is r = {r} and the only direction for r in r is (r, r), there is only one o-direction for r in P. What remains is the main case when r is an atom of P that is not maximal. Any o- direction for r is determined by its value on the maximal elements of P above r. The value of an o-direction at a maximal element m above r is a direction for r in m, so can take at most two possible values. Suppose there are three distinct o-directions for r in P, say d 1, d 2, d 3. Choose any maximal m above r. Then two of d 1, d 2, d 3 must agree at m. Say d 1 and d 2 agree at m. Since d 1 d 2, there is a maximal m with d 1 (m ) d 2 (m ). Choose n, k with r k m and r n m. Since d 1 (m) = d 2 (m) and k m, we have d 1 (k) = d 2 (k), and since k has height 2 this value is either (k, r) or (r, k). Suppose without loss of generality that d 1 (k) = d 2 (k) = (r, k). Since r < n m and d 1 (m ) d 2 (m ), it follows from Proposition 5.8 that d 1 (n) d 2 (n). So one of d 1 (n), d 2 (n) is (n, r). Suppose that d 1 (n) = (n, r). Then by the definition of an o-direction k n exists. Take m maximal with k n m. Since d 1 (k) = d 2 (k), Proposition 5.8 gives d 1 (m ) = d 2 (m ), hence d 1 (n) = d 2 (n). This contradiction shows that the assumption of there being more than 2 directions for r in P is invalid. We next begin the process of putting structure on the set of o-directions of an orthodomain. The following is easily verified. Proposition Let P be an orthodomain and d be an o-direction for a basic r P. Then there is a direction d for r given by d (k) = (q, p) if d(k) = (p, q). Further, there are directions 0 and 1 for the basic element P, given by 0(k) = (, k) 1(k) = (k, ) for all k P for all k P Proposition Let P be an orthodomain with no maximal elements that are atoms or. Then these are equivalent. (1) each basic element of P has an o-direction (2) each basic element of P has exactly two o-directions. Proof. (1) (2) If d is an o-direction for r, then so is d given by Proposition If d = d, then for a maximal element m above r we have that d(m) = d (m), and this implies that m is

15 OOLEAN SUALGERAS OF ORTHOALGERAS 15 an atom or, contrary to our assumptions on P. Thus each basic element of P has at least two directions, so by Proposition 5.9 has exactly 2 directions. (2) (1) is trivial. Definition An orthodomain P is proper if it has no maximal elements that are or atoms, and has enough o-directions if it is proper and each basic element has an o-direction. Note that the situation is somewhat analogous to that of spatial frames, which are defined through the existence of a sufficient supply of points. In the following, we restrict our attention to orthodomains with enough o-directions. Definition Let P be an orthodomain and d, e odir(p ) where d is an o-direction for r and e is an o-direction for s. Set d e if one of the following conditions applies. (1) at least one of d, e is the o-direction 0 (2) d = e (3) r, s are near and for r s = u we have d(u) = (r, u) and e(u) = (s, u) Definition Let P be an orthodomain with enough o-directions. Define a partial binary operation on odir(p ) with domain as follows. (1) d 0 = d = 0 d (2) d d = 1 (3) d e is the o-direction for t with (d e)(u) = (u, t). In case (3) we assume d is an o-direction for r, e is an o-direction for s, that r, s are near with r s = u, and that t is the third atom distinct from r, s beneath u. Theorem Let A be an OA with no small blocks. Then the orthodomain Sub(A) has enough o-directions and A is isomorphic to odir(sub(a)) with its given structure. Proof. We have seen that P = Sub(A) is an orthodomain. Let x A and r = {0, x, x, 1}. Define d x r P 2 by setting for each h r d x (h) = ( h x h x, h x h x) To see that d x is a direction for r, we must verify the three conditions in Definition 5.7. For (1) we have seen that for any r k that d x (k) is a direction in k. The definition of d x gives condition (2). For (3) we consider first the case that r =, hence x is either 0 or 1. If x = 0 then d x (h) = (, h) for all h, and if x = 1 then d x (h) = (h, x) for all h, so (3) holds vacuously. Suppose r is an atom of P. Suppose r k, n and that d x (k) = (r, k) and d x (n) = (n, r). This means that k, n are 8-element oolean algebras, that x is an atom in k, and x is a coatom in n. Let y 1, y 2 be the atoms of k distinct from x and z 1, z 2 be the atoms of n distinct from x. y 2 y 1 x z 2 z 1 x x y 1 y 2 x z 1 z 2 k n

16 16 JOHN HARDING, CHRIS HEUNEN, AND MIRKO NAVARA Then in A we have that z 1 z 2 = x and y 1 y 2 = x. This implies that y 1, y 2, z 2, z 2 are the atoms of a oolean subalgebra m of A, and clearly m = k n. This shows condition (3) of Definition 5.7, so d x is a direction of P = Sub(A). Since this is valid for each x A, the orthodomain P has enough directions. Consider ϕ A odir(p ) defined by ϕ(x) = d x. Every basic element of P is of the form {0, x, x, 1} and each basic element has 2 directions. Since d x and d x are directions for {0, x, x, 1} we have that ϕ is onto. If ϕ(x) = ϕ(y), then since d x is a direction for x and d y is a direction for y, we must have that y = x or y = x. ut d x d x, so ϕ is one-one. To show that ϕ is an isomorphism, it is easily seen that ϕ maps 0, 1 of A to the directions 0, 1 of P, and that ϕ(x ) = ϕ(x). It remains to show that x y iff ϕ(x) ϕ(y) and that when these hold, we have ϕ(x y) = ϕ(x) ϕ(y). We consider various possibilities to have x y in A. For any x we have x 0, ϕ(x) ϕ(0), and ϕ(x 0) = ϕ(x) = ϕ(x) ϕ(0). For any x we have x x and x x = 1. Since ϕ(x ) = ϕ(x), then ϕ(x) ϕ(x ) and ϕ(x x ) = 1 = ϕ(x) ϕ(x ). The remaining possibility to have x y in A is to have x, y distinct atoms of an 8-element oolean subalgebra u of A. In this case, r = {0, x, x, 1} and s = {0, y, y, 1} are basic elements that are near, r s = u, t = {0, x y, (x y), 1} is the third atom beneath u, and d x (u) = (r, u), d y (u) = (s, u). Thus ϕ(x) ϕ(y), and as d x d y is the direction for t with (d x d y )(u) = (u, t) we have that ϕ(x) ϕ(y) = ϕ(x y). Conversely, suppose that ϕ(x) ϕ(y) via the remaining condition (3) of Definition Since d x is a direction for r = {0, x, x, 1} and d y is a direction for s = {0, y, y, 1}, the assumptions of this condition provide that x, y are r, s are near and generate an 8-element oolean subalgebra of A. Further, since d x (u) = (r, u) and d y (u) = (s, u), we have that x, y are atoms of u, hence x y in A. Finally, d x d y is the direction for the third atom t = {0, x y, x y, 1} beneath u with (d x d y )(u) = (u, t), so d x d y = d x y. Remark This result provides one of our primary aims, a means to reconstruct an oa A from its poset of oolean subalgebras. This result applies only in the case when A has no small blocks, meaning blocks of 4 or fewer elements. If A does have blocks of 4 elements, these are horizontal summands of A and appear in Sub(A) as maximal atoms. Stripping these blocks from A yields an oa A with Sub(A ) being the result of stripping maximal atoms from Sub(A). Since we can reconstruct A as odir(sub(a )), we can then reconstruct A since the number of horizontal summands we must add is the number of maximal atoms of Sub(A). 6. Characterizing orthodomains of the form Sub(A) For any oa A with no blocks of 4 or fewer elements, results of the previous section have shown that poset Sub(A) is an orthodomain with enough o-directions and A odir(sub(a)). In this section we characterize those orthodomains with enough o-directions that are isomorphic to Sub(A) for some oa A. Definition 6.1. For an orthodomain P, let P be the downset of P consisting of elements m of height 3 or less. We say S P is a shadow of P if (1) S is a downset of P (2) S is closed under existing joins in P Proposition 6.2. Let P be an orthodomain, S be a shadow of P, r be a basic element of P that belongs to S, and d be an o-direction of P for r. (1) S is an orthodomain (2) The restriction d S of d to r S is an o-direction of S

17 OOLEAN SUALGERAS OF ORTHOALGERAS 17 So if P has enough o-directions and S has no maximal atoms, then S has enough o-directions. Proof. (1) Since P, and hence S, has finite height, every directed set has a maximal element and hence a join. Further, each element is compact. Since P is atomistic and S is a downset of P, it is atomistic. Since S is a downset of P, for each x S we have that x is a oolean domain. Finally, if r, s are atoms of S and r, s u, then r s = u in P, hence r s = u in S as well. So S is an orthodomain. (2) To see that the restriction of d to r S is an o-direction, we must verify the three conditions of Definition 5.7. The first two are trivial consequences of restricting, and the the third holds since r k, n with k, n S implies that k n S since k n has height at most 3, so is in P, and S is closed under existing joins in P. For the further comment, if P has enough o-directions, then each basic r S has an o-direction in P that restricts to an o-direction of r in S. Then if S has no maximal atoms, by definition it has enough o-directions. Definition 6.3. Let P be an orthodomain with enough directions and S be a shadow of P with no maximal atoms. Let odir S (P ) be the directions of P for basic elements r S. Then by Proposition 6.2 we can define a map µ S odir S (P ) odir(s) by setting µ S (d) = d S. Proposition 6.4. Let P be an orthodomain with enough directions and S be a shadow of P with no no maximal atoms. (1) odir S (P ) contains 0, 1 and is closed under the operations and (2) µ S is a bijection from odir S (P ) to odir(s) (3) µ S preserves 0, 1 and (4) d e iff µ S (d) µ S (e) and in this case µ S (d e) = µ S (d) µ S (e) Proof. (1) Since 0, 1 are directions for and S, we have 0, 1 odir S (P ). If d is a direction for r, then d is also a direction for r, and this gives closure under. For closure under, suppose d, e odir S (P ) with d a direction for r S, e a direction for s S, and d e. There are several cases for. If one of d, e is 0, then d e equals d or e, and if e = d, then d e = 1, so these cases are trivial. For the remaining case we have that r, s are near. Say r s = u with t the third atom beneath u. Then u S since S is closed under joins in P, and so t S since S is a downset of P. Since d e is a direction for t, we have d e odir S (P ). (2) For a basic r S, the two directions for r in P are d and d. These restrict to directions of S for r and their restrictions are orthocomplements. Then as S has enough directions, these restrictions are distinct and are the only two directions for r in S. (3) This is trivial (4) Suppose d, e odir S (P ) with d a direction for r and e a direction for s. Note that one of d, e is 0 iff one of µ S (d), µ S (e) is 0, and in this case µ S (d e) = µ S (d) µ S (e). Next, e = d iff µ S (e) = µ S (d), and in this case µ S (d e) = 1µ S (d) µ S (e). For the remaining case we have d e iff r, s are near and d(u) = (r, u), e(u) = (s, u) where r s = u. ut these conditions are equivalent to having the restrictions µ S (d) µ S (e). When these conditions hold, d e is the direction for the third atom t beneath u with (d e)(u) = (u, t), and thus its restriction is a direction for t taking value (u, t) at u, and hence is µ S (d) µ S (e). A specific case of this result is of particular interest. It is easily seen that P is a shadow of P and that if P has no maximal atoms, then P has none. Further, since every basic element of P belongs to P, we have that odir P (P ) = odir(p ). The following is then immediate. Corollary 6.5. If P is an orthodomain with enough directions, then so is P and restriction gives an isomorphism odir(p ) odir(p ). We next turn attention to establishing that for any orthodomain P with enough directions, odir(p ) is an oa. We begin with the following.