Orthosystems of submodules of a module

Abstract Let M be a module over a ring. We first introduce a certain algebraic subsystem Σ of the lattice of all submodules of M (an orthosystem of submodules). We then show that any ortholattice can be represented as a Σ for a suitable module. Next, we introduce linear (resp. pre-Hilbert) ortholattices as those ortholattices that allow for a “linear” representation Σ (resp. for a meet-preserving “linear” representation Σ). These notions involve a type of splitting property of Σ. As an important example, we show that any Boolean algebra is pre-Hilbertian. We then find that linear orthosystems are orthomodular and that they satisfy the ortho-Arguesian law. In the rest, we consider complete orthosystems. We show that each complete orthosystem can be induced by an orthogonality relation on M. If is a linear orthospace on M then the collection of all -closed submodules is a complete orthosystem, and vice versa. Finally, we address a natural model theoretic question on the axiomatization of orthomodular orthosystems.—The results obtained may contribute to the algebraic foundation of quantum theory.


Introduction. Basic notions
In the introduction of the paper, let us recall two results of the theory of orthomodular structures. In the paper [8], R. Godowski and R.J. Greechie published a certain variant of the ortho-Arguesian law that is valid in any lattice L(H) of all topologically closed subspaces of a Hilbert space H but not in all orthomodular lattices. In the paper [9], R.I. Goldblatt showed that "the property of orthomodularity of the lattice of orthoclosed subspaces of a pre-Hilbert space P is not determined by any first-order properties of the relation ⊥ of orthogonality between vectors in P". The aim of this paper is to put the above cited results in a certain context. In the first three sections we study when (how) the ortholattices could be represented as orthosystems of submodules of a module. We then specify the representation by requiring a specific system of submodules. Accordingly, we classify the ortholattices as linear, pre-Hilbertian and Hilbertian, and we consider their properties. We then pass to an intrinsic approach-if M is a module, we introduce a relation ⊥ ⊥ on M so that all ⊥ ⊥-closed submodules form a complete orthosystem. We then consider special types of (M, ⊥ ⊥) and find several result that clarify the properties of the associated orthosystem (Proposition 4.4, Theorems 4.5 and 4.7).
It is worth noting that the previous efforts took place in linear spaces rather than in general modules [5,12,14,18,22,23] and main applications of the findings of this paper may presumably be directed toward linear spaces, too. However, since our general approach could e.g. involve certain links with rings and groups as examples of modules, the theme then extends itself over several algebraic structures (see also Section 4 of this paper).
Let us introduce basic notions as we shall use them in the sequel. Let L be a lattice and let x, y, z ∈ L such that x ≤ z. Let us write y [x,z] = (y ∧ z) ∨ x and y [x,z] = (y ∨ x) ∧ z. Evidently, y [x,z] ≤ y [x,z] . By the definition, the lattice L is said to be modular exactly when y [x,z] = y [x,z] for any x, y, z ∈ L, x ≤ z. The following characterization of modularity will be used later (Theorem 3.9). Lemma 1.1. Let L be a lattice. Then the following statements are equivalent: (1) L is modular, Proof. (1) ⇒ (2) Let L be modular. Let us choose elements x, y, z, w ∈ L. Then we have (2) ⇒ (3) Any lattice satisfies the inequality ( (3) ⇒ (1) Let us suppose that L satisfies the formula of condition (3). We are going to show that L is modular. Let x, y, z ∈ L be such elements that x ≤ z. Let us set w = z. [x,z] , and further Since L satisfies the condition (3), we have y [x,z] ≤ y [x,z] . Because the reverse inequality holds in any lattice, we have obtained y [x,z] = y [x,z] . It means that L is modular. Let us recall that by an ortholattice we mean a 6-tuple L = (X, ∧, ∨, , 0, 1) such that L 0 = (X, ∧, ∨, 0, 1) is a lattice with a least and a greatest element, 0 and 1, and with as an orthocomplementation on the lattice L 0 , i.e., : X → X is a mapping such that x ∧ x = 0, (x ) = x, and x ≤ y ⇒ y ≤ x for any x, y ∈ X. If L is an ortholattice and x, y ∈ L, we write x ⊥ y provided x ≤ y .

Lemma 1.2. Let L be an ortholattice. Then
Proof. It is a matter of a routine verification (see e.g. [18]).
An ortholattice L is said to be an orthomodular lattice if it satisfies the orthomodular law, x ≤ y ⇒ y = x ∨ (y ∧ x ). We shall only make use of basic properties of orthomodular lattices (see [3,17], etc.). It should be noted that the theory of orthomodular lattices has also been developed, in relation with theoretical physics, as the theory of quantum logics (see [21,23]).
Throughout the paper, the word ring will mean an associative nonzero ring with a unit (1 = 0), not necessarily commutative. Let M be a (left) module over a ring R. Let us denote by L(M) the set of all submodules of M. We will use the same symbol L(M) for the complete modular lattice (L(M), ∩, , 0, 1),

Representation of ortholattices by submodules
A lattice L is said to be "representable by R-modules" if L is embeddable in the lattice L(M) for some R-module M (see e.g. [16]). We will show in this section that any ortholattice is embeddable, in a certain sense, in the lattice L(M) for any ring R and an appropriate R-module M. Definition 2.1. Let M be a module over a ring R and let S ⊆ L(M). Let * : S → S be a mapping. Let us say that the couple = (S, * ) is an orthosystem of submodules of M (abbr., an orthosystem in L(M)) if, for any A, B ∈ S, the following conditions are fulfiled: In this case, let us write A∨ B = (A * ∩B * ) * for A, B ∈ S. Further, let us call an orthosystem = (S, * ) complete if i∈I A i ∈ S for any system {A i } i∈I ⊆ S. (2) The inclusion A + B ⊆ A ∨ B is obvious.
The submodule A ∨ B is the smallest submodule in S containing both the submodules A and B. But A + B ∈ S and A + B contains both the submodules A and B, too. It follows that A ∨ B ⊆ A + B. Making use of (2) we obtain A ∨ B = A + B.
If = (S, * ) is an orthosystem in L(M), then, by the preceding Lemma, the 6-tuple (S, ∩, ∨ , * , 0, 1) (where again 0 = {0 M } and 1 = M) is an ortholattice. We will denote it with the same symbol . In other words, every orthosystem of submodules will be identified with the corresponding ortholattice. If this ortholattice is orthomodular we can speak on an orthomodular orthosystem.
In the rest of this section we show that any ortholattice is isomorphic with some appropriate orthosystem of submodules. The proof of this fact is divided into a few statements.  Proof. The proof of (1) is obvious. Further, if a ≤ b (a, b ∈ L) then f (a) ⊆ f (b). Let us suppose that f (a) ⊆ f (b). If f (a) = ∅ then a = 0 and the inequallity a ≤ b obviously holds. If f (a) = ∅ then a ∈ f (a) and therefore a ∈ f (b). It follows that a ≤ b. Finally, let {a i } i∈I ⊆ L be some system of elements of L such that the meet i∈I a i exists in L. Then f ( Our next intention is to show how the mapping f of Lemma 2.3 transforms itself, for a module M, into a mapping that ranges in the lattice L(M). We shall use the following lemma. Lemma 2.4. Let M be a free module over a ring R and let B be a basis of M. Then the following statements hold: and therefore there are mutually distinct elements q 1 , . . . , q n ∈ B 2 and nonzero scalars α 1 , . . . , α n ∈ R such that p = α 1 q 1 + · · · + α n q n . If p ∈ {q 1 , . . . , q n } then {p, q 1 , . . . , q n } ⊆ B is a linearly dependent set, a contradiction. ( and therefore there are mutually distinct elements q 1 , . . . , q m ∈ B j and nonzero scalars β 1 , . . . , β m ∈ R such that x = β 1 q 1 + · · · + β m q m . As p n ∈ {q 1 , . . . , q m }, the equation α 1 p 1 + · · · + α n p n = β 1 q 1 + · · · + β m q m implies that the set {p 1 , . . . , p n } ∪ {q 1 , . . . , q m } is linearly dependent, a contradiction.

Definition 2.5.
Let L be an ortholattice and let M be a module. Let g : L → L(M) be a mapping. We say that g is a meet-embedding of L into L(M) if, for any a, b ∈ L, the following conditions are fulfiled: We say that g is a meet-complete embedding of L into L(M) if the mapping g fulfils conditions (i), (ii) and the following condition (iv): Proof. It amounts to a routinne verification.

Proposition 2.7. Let L be an ortholattice and let R be a ring. Then there is a module M over R and a meet-complete embedding of L into L(M).
Proof. Let us set X = L \ {0} and let f : L → exp(X) be the mapping of Lemma 2.3. Let M be a free module over R with the set X of free generators. Let us set, for a ∈ L, g(a) = [f (a)] (i.e., g(a) is the submodule of M generated by the set f (a) ⊆ X). We will show that g is a meet-complete embedding Finally, let {a i } i∈I ⊆ L be a system such that the meet i∈I a i exists in L. Then Lemma 2.4 (2) gives us the Theorem 2.8. Let L be an ortholattice and R be a ring. Then there is a module M over R and an orthosystem in L(M) such that the ortholattices L and are isomorphic. If, moreover, the ortholattice L is complete, there is a complete orthosystem in L(M) such that the ortholattices L and are isomorphic.
Proof. By Proposition 2.7 there exist a module M over R and a meet-complete embedding g : L → L(M). By Lemma 2.6, the ortholattices L and g = (S, * ) are isomorphic. Finally, let us suppose that the ortholattice L is complete. Let us choose some system {A i } i∈I ⊆ S. For any i ∈ I, let x i be an element of L with g(x i ) = A i . Since g "preserves" existing meets in L we obtain

Linear orthomodular lattices
Let us recall the well-known Amemiya-Araki theorem (see [1], for related considerations see also [13] and [14]). The theorem says that a pre-Hilbert space H is topologically complete (i.e., H is a Hilbert space) if and only if the ortholattice L(H) is orthomodular. The essential part of the proof of Amemiya-Araki theorem is the link of orthomodularity with the so called splitting property of elements of L(H). In this section we would like to investigate this link in a higher generality (for the specific situations in the pre-Hilbert spaces, see [5,12,22], etc.).  (1) is additive, It remains to show that A∨ B = A+B. Take an element x ∈ A∨ B. Since Conversely, let B = A + (B ∩ A * ) for any A, B ∈ S with A ⊆ B. Then, by taking B = M, we obtain the equality M = A + A * for any A ∈ S. Thus, is additive.
(2) Let us suppose that is additive. Let us take A, B ∈ S with A ⊆ B. By statement (1) we have At this stage it becomes natural to study the ortholattices that are isomorphic with additive orthosystems of submodules. Let us formulate the following definition. Proof. It follows from the fact that g is an isomorphism between ortholattices L and g (Lemma 2.6). We have adopted the adjective "Hilbertian" being motivated by [23]. The following proposition clarifies the meaning of additive orthosystems. Let us observe that Definition 3.6 determines a broader class of orthomodular lattices than the "projection" ones.

Proposition 3.8. Every Boolean algebra is pre-Hilbertian. As a consequence, every complete Boolean algebra is Hilbertian.
Proof. Let B = (X, ∧, ∨, , 0, 1) be a Boolean algebra. Let us denote by the operation of symmetric difference in B. So, for any x, y ∈ B, we have x y = (x ∧ y ) ∨ (x ∧ y) = (x ∨ y) ∧ (x ∧ y) . Let R B = (X, +, −, ·, 0, 1) be the Boolean ring associated with the Boolean algebra B. Let us recall that for x, y ∈ R B we have x + y = x y, −x = x and x · y = x ∧ y. Obviously, the ring R B could be viewed as a module over itself. Let us denote this module by M B . It is easy to show that, for any a ∈ X, the set [0, a] is a submodule of M B . Let as set g(a) = [0, a] for any a ∈ X. We will show that g is a meetcomplete embedding of B into L(M B ). Obviously, g(0) = {0} and g(1) g(b). Further, let {a i } i∈I ⊆ X be a system such that the meet i∈I a i exists in B. Then g( . It remains to show that g is additive. Let us take elements a, b ∈ B such that a ⊥ b. We have to show that g (a ∨ b) = g(a) + g(b). In other words, we have to check that The last result concerns the ortho-Arguesian law known to hold in L(H) (see [8]). We will extend it over linear orthomodular lattices. Theorem 3.9. Let L be a linear ortholattice. Then the following formula (called the ortho-Arguesian law) holds in L: Proof. We can suppose that there is a module, M, and an additive orthosystem of submodules = (S, * ) of M such that L = . Let us recall that in the lattice the infimum is performed by the intersection and the supremum is performed, for orthogonal submodules, by the summation (see Lemma 2.2). Let Since L(M) is modular, we can apply Lemma 1.1. Then Let us finally exhibit an example of a finite orthomodular lattice that does not fulfil the ortho-Arguesian law (and therefore this example cannot be Hilbertian). We will determine it by the Greechie diagram below (see [8] and [10]).¨¨¨s In this orthomodular lattice, it obviously holds that x 1 ⊥ y 1 and x 2 ⊥ y 2 . Further, ( Let us conclude this section by a few notes. First, in [11] R. Greechie indicates that the ortho-Arguesian law, without the precondition of orthogonality, would have implied modularity, as stated in Lemma 1.1 (3). The characterization of Lemma 1.1 (2), a new identity expression of modularity, further contributes to the understanding of the right side term in modular lattices. Second, in [20] R. Mayet writes that "the proof of the fact that an equation of this family holds in all HLs uses essentially the decomposition of a vector on orthogonal subspaces together with basic applications of the associativity and commutativity of the addition of vectors, and the inclusion M + N ⊆ M ∨ N = (M + N) ⊥⊥ for any M, N in a HL. " We believe that the notion of the linear orthomodular lattice introduced in this paper essentially captures the idea of the citation above and in turn may lead to a better understanding of the algebraic structure of L(H). In this connection, the following question is of importance in its own right and in the quantum logic interpretation as well.
Problem. Does there exist a characterization of the linear and Hilbertian orthomodular lattices in terms of the language of ortholattices?

Linear orthospaces
Let (S, * ) be a complete orthosystem of submodules of M. Let us exhibit a construction that enables us to extend the operation * over the entire L(M). In connection with this, the notion of orthogonality space (or simply an orthospace) may be recalled, see [2]. By an orthospace we mean a triple (X, ⊥, 0), where X is a set, 0 ∈ X, and ⊥ is an orthogonality relation on X, i.e., ⊥ is a binary relation on X that is symmetric and irreflexive on X \ {0} and, in addition, it satisfies 0 ⊥ x for all x ∈ X. It is known that each complete ortholattice can be represented by means of "closed" subsets of an orthospace. However, some properties of orthospaces that occur in the study of pre-Hilbert spaces cannot be directly expressed in general orthospaces (in this line of coping with these drawbacks, some attempts have been made, see [24]). The following definition indicates the way in which the "semantic" possibilities of orthospaces can be extended-some additional algebraic structure on the underlying set can be introduced.
Let us call the couple = (M, ⊥ ⊥) a linear orthospace (LOS) over R.
Obviously, the study of LOSs in this module formulation may involve several algebraic structures in their own right. For instance, in the paper [6] the author in fact reveals several examples of rings as LOSs (such are e.g. Baer rings and f-rings). In a similar vein, the papers [7] and [19] study groups (for example, lattice-ordered groups can be viewed as LOSs). Several open problems remain in this area, our quantum-logic paradigma rather lighten different kinds of questions. Let us also note that this notion of orthogonality on a module corresponds, in a certain sense, to the notion of "pre-orthogonality on a projective space" (see [15]). However, our concept of linear orthospaces allows us to situate the considerations into the theory of first order structures (compare with Theorem 4.7).
Let us recall the notions we shall use in the sequel. They follow the standard pattern of closure operators. Let = (M, ⊥ ⊥) be a LOS. If A is a subset (not necessarily a submodule) of M, let us write A ⊥ ⊥ = {x ∈ M; x ⊥ ⊥ y for all y ∈ A}. It is evident that A ⊥ ⊥ is a submodule of M.
Let us collect some properties of the operation ⊥ ⊥ on the lattice L(M).
Proof. The proof of (1) and (2) is an easy application of the conditions (⊥ ⊥ 1 )-(⊥ ⊥ 5 ). Let {A i } i∈I be subsystem of L(M). We will first show that ( On the other hand, let x ∈ i∈I A ⊥ ⊥ i and y ∈ i∈I A i . We will prove that x ⊥ ⊥ y. Since y ∈ i∈I A i , there exist indices i 1 , . . . , i n ∈ I and elements y 1 ∈ A i 1 , . . . , y n ∈ A i n such that y = y 1 + · · · + y n . Because x ∈ A ⊥ ⊥ i 1 ∩ . . . ∩ A ⊥ ⊥ i n , we obtain x ⊥ ⊥ y 1 , . . . , x ⊥ ⊥ y n . The condition of (⊥ ⊥ 2 ) could be applied to derive that x ⊥ ⊥ (y 1 + · · · + y n ). The following proposition gives us a generalized form of the classic result of pre-Hilbert spaces (see also [12,18,23], etc.). Proof. Let us again write A ⊥ ⊥ instead of A ⊥ ⊥ . Applying Proposition 4.2 (1), we see that {0 M } ∈ C( ) and M ∈ C( ). We show that the system C( ) is closed under the formation of intersections. Suppose that A i ∈ C( ), i ∈ I. By Proposition 4.2 (3), we see that We have shown that every linear orthospace determines a certain complete orthosystem of submodules. In the following two propositions we are going to show that every complete orthosystem of submodules can be obtained in this way. Moreover, we also obtain a natural correspondence between linear orthospaces and complete orthosystems of submodules. Let = (S, * ) be a complete orthosystem in L(M). Since (S, ⊆) is a complete lattice, an arbitrary system {A i } i∈I of submodules of S has its supremum in the lattice (S, ⊆). We will denote it by      As we indicated in the previous section, the linear orthospaces over a ring R could be viewed as firstorder structures. Indeed, any module M over a ring R could be viewed as an algebra in the language over R could be viewed as first-order structures in the language R ∪ {⊥ ⊥}. This enables us to talk about first-order axiomatizability of classes of linear orthospaces. Evidently, the class of all linear orthospaces over R is first-order axiomatizable. However, this is not the case for the class of all orthomodular and Hilbertian LOSs. This will be proved in Theorem 4.7. (The algebraic consideration then slightly overlaps with the notions of the model theory and the reader is assumed to be acquainted with them.) Theorem 4.7. Denote by AH the class of all atomic Hilbertian LOSs over the field R of reals numbers. Further, let us denote by OM the class of all orthomodular LOSs over R. Suppose that AH ⊆ T ⊆ OM. Then the class T is not axiomatizable.
Proof. It is sufficient to show that there are two LOSs 1 , 2 such that 1 is an elementary substructure of 2 , where 2 ∈ T but 1 ∈ T . The main technical tool for proving this is based on [9]. Let H be the separable Hilbert space l 2 of absolutely square-summable real sequences and P be the incomplete pre-Hilbert space of finitely-nonzero sequences of real numbers. It follows that the coresponding LOS (H, ⊥) is in AH, while the LOS (P, ⊥) is not in OM (Amemiya-Araki theorem [1]). It remains to show that (P, ⊥) is an elementary substructure of (H, ⊥). We will make use of the Tarski test [4]we will show that for any L 0 -formula ϕ(x 1 , . . . , x n , x) and any elements a 1 , . . . , a n from P such that (H, ⊥) | (∃x)ϕ[a 1 , . . . , a n ] there is an element a ∈ P with (H, ⊥) | ϕ[a 1 , . . . , a n , a]. Let us choose such elements a 1 , . . . , a n ∈ P. Since (H, ⊥) | (∃x)ϕ[a 1 , . . . , a n ] there is an element b ∈ H with (H, ⊥) | ϕ[a 1 , . . . , a n , b]. It can be easily shown (see also Theorem 1 in [9]) that there is an automorphism f of the Hilbert space H such that f (a i ) = a i (i = 1, . . . n) and f (b) ∈ P. Because the automorphism f preserves the relation ⊥, the proof is completed by setting a = f (b).

Funding
The second author acknowledges the support by the Austrian Science Fund (FWF): project I 4579-N and the Czech Science Foundation (GAČR): project 20-09869L.