CCategories of orthogonality spaces
Jan Paseka and Thomas Vetterlein [email protected] Department of Knowledge-Based Mathematical Systems, Johannes Kepler University LinzAltenberger Straße 69, 4040 Linz, Austria
Abstract
An orthogonality space is a set equipped with a symmetric and irreflexive binaryrelation. We consider orthogonality spaces with the additional property thatany collection of mutually orthogonal elements gives rise to the structure of aBoolean algebra. Together with the maps that preserve the Boolean structures,we are led to the category
N OS of normal orthogonality spaces.Moreover, an orthogonality space of finite rank is called linear if for any twodistinct elements e and f there is a third one g such that exactly one of f and g isorthogonal to e and the pairs e, f and e, g have the same orthogonal complement.Linear orthogonality spaces arise from finite-dimensional Hermitian spaces. Weare led to the full subcategory LOS of N OS and we show that the morphismsare the orthogonality-preserving lineations.Finally, we consider the full subcategory
EOS of LOS whose members arisefrom positive definite Hermitian spaces over Baer ordered (cid:63) -fields with a Eu-clidean fixed field. We establish that the morphisms of
EOS are induced bygeneralised semiunitary mappings.
Keywords:
Orthogonality spaces; undirected graphs; categories; Boolean sub-algebras; linear orthogonality spaces; generalised semilinear map; generalisedsemiunitary map
MSC:
In quantum mechanics, physical processes are described in a way assigning an es-sential role to the observer. Rather than predicting on the basis of complete initial1 a r X i v : . [ m a t h . R A ] M a r onditions the unambiguous development of some physical system, the theory as-signs probabilities to pairs consisting of a preparation procedure and the outcome ofa subsequent measurement. Why the formalism has proved successful is by and largetoday still unanswered; we could admit that we rather got used to it. But even at themost basic level, there are unresolved issues. A key ingredient of the model is a cer-tain inner-product space – a Hilbert space over the field of complex numbers –, andthe deeper reasons for this choice are a matter of ongoing discussions.The probably oldest approach aiming to clarify the basic principles on which quantumtheory is based is due to Birkhoff and von Neumann [BiNe]. The keyword “quantumlogic” is often used in this context but might be misleading. What in our eyes rathermatters is the idea of increasing the degree of abstraction: the question is whetherthe Hilbert space can be recovered from a considerably simpler structure. Numeroustypes of algebras, including partial ones, have been proposed and investigated, thebest-known example being orthomodular lattices, which describe the Hilbert spaceby means of the inner structure of its closed subspaces. For an overview of possibledirections, we may refer, e.g., to the handbooks [EGL1, EGL2].Increasing the degree of abstraction means to restrict the structure to the necessaryminimum. An approach that was proposed in the 1960s by David Foulis and hiscollaborators goes presumably to the limits of what is possible. They coined thenotion of an orthogonality space, which is simply a set endowed with a symmetricand irreflexive binary relation [Dac, RaFo, Wlc]. The prototypical example is thecollection of one-dimensional subspaces of a Hilbert space together with the usualorthogonality relation.The notion of an orthogonality space is in the centre of the present work and themain motivation behind our work is to elaborate on its role within the basic quantum-physical model. We generally deal with the case of a finite rank, meaning that thereare only finitely many pairwise orthogonal elements. We should certainly be awareof the fact that orthogonality spaces are as general as undirected graphs, which inturn are rarely put into context with inner-product spaces. As has been shown in[Vet3], however, the relationship between the two types of structures is close. Anorthogonality space of finite rank is called linear if, for any distinct elements e and f ,there is a further one g such that exactly one of f and g is orthogonal to e and theset of elements orthogonal to e and f coincides with the set of elements orthogonalto e and g . Linearity characterises the orthogonality spaces that arise from finite-dimensional Hermitian spaces.In physics, symmetries of the model generally play a fundamental role. It mightthus not come as a surprise that orthogonality spaces associated with complex Hil-bert spaces are describable by the particular properties of their automorphisms [Vet1,Vet2]. Here, we explore this issue further, but we adopt a more general perspectivethan in the previous works.The present paper is devoted to the investigation of structure-preserving maps betweenorthogonality spaces. We do so first in a general context, taking into account fea-tures inherent to orthogonality spaces, and in a second step, we turn to the narrowerclass of linear orthogonality spaces. We start with the question how to reasonablydefine morphisms. It certainly seems to make sense to require nothing more than2he preservation of the single binary relation on which the structures are based. Wecall orthogonality-preserving maps homomorphisms. To choose homomorphisms asmorphisms, however, is inexpedient when the context that we ultimately have inmind is given by inner-product spaces. Indeed, for linear orthogonality spaces, weexpect a morphism to preserve, in some sense, linear dependence. The followingsituation illustrates the difficulties [ ˇSem], even though we otherwise deal with thefinite-dimensional case only. Consider the complex projective space over three di-mensions P ( C ) as well as over ℵ dimensions P ( C ℵ ) ; then any injective mapfrom P ( C ) to P ( C ℵ ) such that the image consists of mutually orthogonal elementsis a homomorphism of orthogonality spaces, but in no way related to the preservationof linear dependence.We note that these problems do not arise in approaches that consider the orthogonalityrelation not as basic but as an additional structure. Projective geometries enhanced byan orthogonality relation were studied, e.g., in [FaFr, StSt]. Here, we try an alternativeway. Having in mind the Hilbert space model of quantum physics, we focus on anadjusted kind of orthogonality spaces, ruling out structural peculiarities that we mustconsider as inappropriate. In quantum mechanics, observables correspond to Booleanalgebras. In a finite-state system, measurement outcomes correspond to mutuallyorthogonal subspaces, which in turn generate a Boolean subalgebra of the lattice ofclosed subspaces. We require to have an analogue of this situation in our more abstractsetting and we take it into account for our definition of morphisms.To be more specific, let us first recall that orthogonality spaces lead us straightfor-wardly to the realm of lattice theory. A subset A of an orthogonality space ( X, ⊥ ) is called orthoclosed if A = B ⊥ for some B ⊆ X , where B ⊥ is the set of e ∈ X orthogonal to all elements of B . The set of orthoclosed subsets form a completeortholattice C ( X, ⊥ ) . Now, consider a collection E = { x , . . . , x k } of mutually or-thogonal elements of X . Then the subsets of E generate a subortholattice of C ( X, ⊥ ) .This subortholattice is, in general, not isomorphic to the Boolean algebra of subsetsof E ; in case it always is, we call ( X, ⊥ ) normal . We moreover name homomorph-isms in the same way if they preserve, in a natural sense, Boolean subalgebras of C ( X, ⊥ ) . We thus arrive at the category N OS of normal orthogonality spaces andnormal homomorphisms.We take up in this way an often-discussed issue. Indeed, for the aim of recovering aHilbert space or, more generally, an orthomodular lattice from suitable substructures,it has been a guiding motive to consider the lattice as being glued together from itsBoolean subalgebras; see, e.g., [Nav, Section 4]. Moreover, deep results have beenachieved on the question how to reconstruct orthomodular lattices or related quantumstructures from the poset of their Boolean subalgebras [HaNa, HHLN].Any linear orthogonality space is normal and thus our next step is to consider normalhomomorphisms between linear orthogonality spaces. That is, we investigate the fullsubcategory
LOS of N OS , consisting of linear orthogonality spaces. It turns outthat the morphisms in
LOS do have the most basic property to be expected: they aremaps between projective spaces that preserve the triple relation of being contained ina line, that is, they are lineations. In fact, we show that the morphisms are exactly theorthogonality-preserving lineations. 3ur final objective is to describe the morphisms in
LOS as precisely as possible.Generalisations of the fundamental theorem of projective geometry show that anylineation is induced by a generalised semilinear transformation – provides it is non-degenerate [Mach, Fau]. Here, non-degeneracy means two additional conditions tohold: (1) the image is not contained in a -dimensional subspace, and (2) the image ofa line is never -element. Provided that the rank is at least , condition (1) is ensured.Condition (2), however, leads us to an issue dealt with in the discussions aroundthe peculiarities of quantum physics: we show that a violation of (2) implies theexistence of two-valued measures. The exclusion of two-valued measures is in turna consequence of Gleason’s Theorem in case that the skew field is C or R . Althoughthe case of specific further skew fields has been discussed [Dvu], not much seems tobe known about the general case. Here, we show that if the skew field of scalars is aEuclidean subfield of the reals, two-valued measures do not exist. Consequently, thesame applies if a (cid:63) -field possesses a subfield of this type.Moreover, we deal with lineations that in addition preserve an orthogonality relation.It seems natural to ask whether the representing generalised semilinear map can bechosen to preserve in some sense the inner product. We establish that this is the caseunder particular conditions: the skew field is commutative, that is, a field, and thereis a basis of vectors of equal length. These conditions apply for positive definiteHermitian spaces over Baer ordered (cid:63) -fields whose fixed field is Euclidean. In thiscase, morphisms are induced by what we call generalised semiunitary maps.The paper is organised as follows. In the following Section 2, we fix the basic notationused in this paper. Moreover, we introduce and discuss normal orthogonality spaces,including a characterisation of normality as an intrinsic property, without referenceto the associated ortholattice. In Section 3, we investigate the category N OS of nor-mal orthogonality spaces and normal homomorphisms. In Section 4, we prepare theground for the discussion of those orthogonality spaces that arise from inner-productspaces; in particular, we discuss lineations between projective spaces and discuss theirrepresentation in the presence of an inner product. Then, in Section 5, we recall thenotion of a linear orthogonality space and we show that linearity implies normality.Finally, in Section 6, we study the full subcategory
LOS of N OS that consists oflinear orthogonality spaces, focussing especially on the morphisms induced by gen-eralised semiunitary maps. Some concluding remarks are found in the final Section 7.
We deal in this paper with the following relational structures.
Definition 2.1. An orthogonality space is a non-empty set X equipped with a sym-metric, irreflexive binary relation ⊥ , called the orthogonality relation . The supremumof the cardinalities of sets of mutually orthogonal elements of X is called the rank of ( X, ⊥ ) .We may observe that orthogonality spaces are essentially the same as undirectedgraphs, understood such that the edges are -elements subsets of the set of nodes.4he rank of an orthogonality space is under this identification the supremum of thesizes of cliques. The present work, however, is not motivated by graph theory, ourguiding example rather originates in quantum physics. Example 2.2.
Let H be a Hilbert space. Then the set P ( H ) of one-dimensionalsubspaces of H , together with the usual orthogonality relation, is an orthogonalityspace, whose rank coincides with the dimension of H . The (orthogonal) complement of a subset A of an orthogonality space X is A ⊥ = { x ∈ X : x ⊥ a for all a ∈ A } . The map P ( X ) → P ( X ) , A (cid:55)→ A ⊥⊥ is a closure operator on X . We call the closedsubsets orthoclosed and we denote the collection of orthoclosed subsets by C ( X, ⊥ ) .Endowed with the set-theoretical inclusion and the orthocomplementation ⊥ , C ( X, ⊥ ) becomes a complete ortholattice. The ortholattice ( C ( X, ⊥ ); ∩ , ∨ , ⊥ , ∅ , X ) will beour primary tool to investigate ( X, ⊥ ) . Example 2.3.
Let ( P ( H ) , ⊥ ) be the orthogonality space arising from the Hilbertspace H according to Example 2.2. Then we may identify C ( P ( H ) , ⊥ ) with the set C ( H ) of closed subspaces of H , endowed with the set-theoretical inclusion and theorthocomplementation. In this paper, we will focus exclusively on the case of a finite rank. Our guidingexample is, accordingly, the orthogonality space associated with a finite-dimensionalHilbert space. From now on, all orthogonality spaces are tacitly assumed to be offinite rank.We will next introduce a condition on orthogonality spaces that mimics a key fea-ture of the quantum-physical formalism. In quantum mechanics, a physical system ismodelled by means of a Hilbert space and observables correspond to Boolean subal-gebras of the lattice of its closed subspaces. We will require that orthogonality spacespossess substructures of the corresponding type.
Definition 2.4.
An orthogonality space ( X, ⊥ ) is called normal if, for any mutuallyorthogonal elements e , . . . , e k of X , where k (cid:62) , the subalgebra of the ortholattice C ( X, ⊥ ) generated by { e } ⊥⊥ , . . . , { e k } ⊥⊥ is Boolean.We may understand normality also as a coherence condition. By a subset A of anorthogonality space to be orthogonal, we mean that A consists of mutually orthogonalelements. Lemma 2.5.
For an orthogonality space ( X, ⊥ ) , the following are equivalent: (1) ( X, ⊥ ) is normal. (2) For any maximal orthogonal set { e , . . . , e n } ⊆ X , there is a finite Booleansubalgebra of C ( X, ⊥ ) whose atoms are { e } ⊥⊥ , . . . , { e n } ⊥⊥ . (3) For any maximal orthogonal set { e , . . . , e n } ⊆ X and any (cid:54) k < n , if f ⊥ e , . . . , e k and g ⊥ e k +1 , . . . , e n , then f ⊥ g . roof. (1) ⇒ (2): Let ( X, ⊥ ) be normal and let { e , . . . , e n } be a maximal orthogonalsubset of X . By normality, the subalgebra B of C ( X, ⊥ ) generated by { e } ⊥⊥ , . . . , { e n } ⊥⊥ is Boolean. Moreover, { e } ⊥⊥ , . . . , { e n } ⊥⊥ are mutually orthogonal ele-ments and we have { e } ⊥⊥ ∨ . . . ∨ { e n } ⊥⊥ = { e , . . . , e n } ⊥⊥ = ∅ ⊥ = X . Thus B is a finite Boolean subalgebra of C ( X, ⊥ ) , its atoms being { e } ⊥⊥ , . . . , { e n } ⊥⊥ .(2) ⇒ (3): Let { e , . . . , e n } be a maximal orthogonal subset of X and assume that { e } ⊥⊥ , . . . , { e n } ⊥⊥ are the atoms of a finite Boolean subalgebra of C ( X, ⊥ ) . Let (cid:54) k < n . Then f ⊥ e , . . . , e k means f ∈ { e , . . . , e k } ⊥ = { e k +1 , . . . , e n } ⊥⊥ , andsimilarly, g ⊥ e k +1 , . . . , e n means g ∈ { e , . . . , e k } ⊥⊥ . If both f ⊥ e , . . . , e k and g ⊥ e k +1 , . . . , e n holds, we hence conclude f ⊥ g .(3) ⇒ (1): Let D = { e , . . . , e k } , k (cid:62) , be an orthogonal subset of X . Thenwe may extend D to a maximal orthogonal subset E = { e , . . . , e n } of X , where n (cid:62) k . For any A ⊆ E , we have (cid:87) {{ e } ⊥⊥ : e ∈ A } = A ⊥⊥ ; for any A, B ⊆ E ,we have A ⊥⊥ ∨ B ⊥⊥ = ( A ∪ B ) ⊥⊥ ; and E ⊥⊥ = X . Let ∅ (cid:54) = A (cid:40) E . Then ( E \ A ) ⊥⊥ ⊆ A ⊥ . Moreover, if f ∈ A ⊥ and g ∈ ( E \ A ) ⊥ , we have by assumption f ⊥ g ; hence f ∈ ( E \ A ) ⊥⊥ . We conclude that A ⊥⊥ = ( E \ A ) ⊥ . We haveshown that { e } ⊥⊥ , . . . , { e n } ⊥⊥ generate a Boolean subalgebra of C ( X, ⊥ ) ; hence sodo { e } ⊥⊥ , . . . , { e k } ⊥⊥ .The following notation will be useful. Let e , . . . , e k be mutually orthogonal elementsof a normal orthogonality space ( X, ⊥ ) . Then the closure of {{ e } ⊥⊥ , . . . , { e k } ⊥⊥ } under joins in C ( X, ⊥ ) has the structure of a Boolean algebra, whose top element is { e , . . . , e k } ⊥⊥ . We will denote this Boolean algebra by B ( e , . . . , e k ) .The property of normality applies to our canonical example. We write [ x , . . . , x k ] for the linear hull of non-zero vectors x , . . . , x k of a linear space. Example 2.6.
Let x , . . . , x k , k (cid:62) , be mutually orthogonal non-zero vectors of aHilbert space H . Then the subalgebra of C ( H ) generated by [ x ] , . . . , [ x k ] consistsof the joins of subspaces among [ x ] , . . . , [ x k ] , [ x , . . . , x k ] ⊥ . This algebra is Booleanand we conclude that ( P ( H ) , ⊥ ) is normal. For later considerations, we introduce a further, particularly simple example.
Example 2.7.
For n ∈ N \ { } , we denote by n an n -element set and we considerthe binary relation (cid:54) = on n . Then ( n , (cid:54) =) is an orthogonality space and C ( n , (cid:54) =) is thepowerset of n . Since C ( n , (cid:54) =) is Boolean, we have that ( n , (cid:54) =) is normal. In general, however, an orthogonality space need not be normal. The subsequentexamples of finite orthogonality spaces will be graphically depicted as follows: theelements of the space are represented by points, and two elements are orthogonal ifthe points are connected by a straight line. For instance, in the Example 2.8 belowwe have that a, b, c are mutually orthogonal and moreover d ⊥ a as well as e ⊥ b, c .We note that this representation might remind of Greechie diagrams. It must be keptin mind, however, that an element of an orthogonality space does not necessarilyrepresent an atom of the associated ortholattice. In Example 2.8, for instance, { e } ⊥⊥ properly contains { a } ⊥⊥ . 6 xample 2.8. Consider the orthogonality space X = { a, b, c, d, e } given by the fol-lowing scheme: { a, b, c } is a maximal orthogonal set. Furthermore, we have { a } ⊥⊥ = { a } and { b } ⊥⊥ ∨ { c } ⊥⊥ = { b, c } ⊥⊥ = { b, c } . Since { b, c } ⊥ = { a, e } , X is not normal. Given a normal orthogonality space ( X, ⊥ ) , we call an orthoclosed subset A of X together with the inherited orthogonality relation, which we usually still denote by ⊥ ,a subspace of ( X, ⊥ ) .The following proposition and example show that a subspace of a normal orthogon-ality space is not in general normal, but a subspace that is the closure of any maximalorthogonal subset is so. Proposition 2.9.
Let ( X, ⊥ ) be a normal orthogonality space and let A ∈ C ( X, ⊥ ) be such that, for any maximal orthogonal subset D of A , we have D ⊥⊥ = A . Thenthe subspace ( A, ⊥ ) is normal.Proof. We shall use criterion (3) of Lemma 2.5. Let { e , . . . , e n } be a maximal or-thogonal subset of A , let (cid:54) k < n , and assume that there are f, g ∈ A suchthat f ⊥ e , . . . , e k and g ⊥ e k +1 , . . . , e n . Then we may choose e n +1 , . . . , e m ∈ X such that { e , . . . , e m } is a maximal orthogonal subset of X . By assumption, A = { e , . . . , e n } ⊥⊥ , hence g ⊥ e n +1 , . . . , e m . Thus we have f ⊥ e , . . . , e k and g ⊥ e k +1 , . . . , e m and the normality of X implies f ⊥ g . We conclude that ( A, ⊥ ) isnormal. Example 2.10.
Let ( X, ⊥ ) be the -element orthogonality space given as follows:(Here, the sets { e, f, g, h } and { g, h, i, j } are meant to be orthogonal; but, none of e or f is orthogonal to i or j .)By criterion (3) of Lemma 2.5, we may check that X is normal. However, the subspace { f, i } ⊥ = { a, g, h, n } is not.
7e might expect that normality of an orthogonality space is closely related to theorthomodularity of the associated ortholattice. This is indeed the case but the twoproperties do not coincide.A
Dacey space is an orthogonality space ( X, ⊥ ) such that C ( X, ⊥ ) is an orthomodularlattice. We have the following characterisation of Dacey spaces [Dac, Wlc]. Lemma 2.11.
An orthogonality space ( X, ⊥ ) is a Dacey space if and only if, for any A ∈ C ( X, ⊥ ) and any maximal orthogonal subset D of A , we have that D ⊥⊥ = A . Example 2.12.
Let H be a Hilbert space. Then C ( H ) is an orthomodular lattice andhence ( P ( H ) , ⊥ ) a Dacey space. Example 2.13.
By means of Lemma 2.11, we observe that the orthogonality space ( X, ⊥ ) from Example 2.8 is not a Dacey space. Indeed, A = { b, c, d } ∈ C ( X, ⊥ ) , { b, c } is a maximal orthogonal subset of A , and { b, c } ⊥⊥ = { b, c } (cid:40) A . The following proposition and example show that the Dacey spaces form a strictsubclass of the normal orthogonality spaces.
Proposition 2.14.
A Dacey space is a normal orthogonality space.Proof.
Let ( X, ⊥ ) be a Dacey space and { e , . . . , e k } be an orthogonal subset of X .Then { e i } ⊥⊥ , i = 1 , . . . , k , are pairwise orthogonal and hence pairwise commutingelements of the orthomodular lattice C ( X, ⊥ ) . It follows that they generate a Booleansubalgebra [BrHa, Prop. 2.8]. Example 2.15.
Consider the following orthogonality space ( X, ⊥ ) :The maximal orthogonal subsets are the elements along a straight line, e.g., { a, b, c } .By criterion (3) of Lemma 2.5, we observe that ( X, ⊥ ) is normal. We may also checkthat each subspace of ( X, ⊥ ) is normal.Moreover, the set { a, e } is orthoclosed. But { a } is a maximal orthogonal subset of { a, e } and { a } ⊥⊥ = { a } . Hence by Lemma 2.11, ( X, ⊥ ) is not a Dacey space. N OS of normal orthogonality spaces
We discuss in this section structure-preserving maps between orthogonality spaces.We shall introduce a category consisting of normal orthogonality spaces and invest-igate its basic properties. 8or orthogonality spaces X and Y , we call a map ϕ : X → Y a homomorphism if ϕ is orthogonality-preserving, that is, if, for any e, f ∈ X , e ⊥ f implies ϕ ( e ) ⊥ ϕ ( f ) .In this case, ϕ induces the map ¯ ϕ : C ( X, ⊥ ) → C ( Y, ⊥ ) , A (cid:55)→ { ϕ ( a ) : a ∈ A } ⊥⊥ . Obviously, ¯ ϕ is order- and orthogonality-preserving. It seems that in general, how-ever, we cannot say much more about ¯ ϕ . We will be interested in homomorphismsfulfilling the following additional condition. Definition 3.1.
Let ϕ : X → Y be a homomorphism between the normal orthogonal-ity spaces X and Y . We will call ϕ normal if, for any orthogonal set e , . . . , e k ∈ X , k (cid:62) , ¯ ϕ maps B ( e , . . . , e k ) isomorphically to B ( ϕ ( e ) , . . . , ϕ ( e k )) .The following lemma might help to elucidate the condition of normality for homo-morphisms. Lemma 3.2.
Let ϕ : X → Y be a homomorphism between normal orthogonalityspaces. Then the following are equivalent: (1) ϕ is normal. (2) For any orthogonal subset { e , . . . , e k } of X , where k (cid:62) , we have ¯ ϕ ( { e , . . . , e k } ⊥⊥ ) = { ϕ ( e ) , . . . , ϕ ( e k ) } ⊥⊥ . (3) For any orthogonal subset { e , . . . , e k } of X , where k (cid:62) , we have ϕ ( { e , . . . , e k } ⊥⊥ ) ⊆ { ϕ ( e ) , . . . , ϕ ( e k ) } ⊥⊥ . (4) For any maximal orthogonal subset { e , . . . , e n } of X , we have ϕ ( X ) ⊥⊥ = { ϕ ( e ) , . . . , ϕ ( e n ) } ⊥⊥ . (5) For any maximal orthogonal subset { e , . . . , e n } of X , we have ϕ ( X ) ⊆{ ϕ ( e ) , . . . , ϕ ( e n ) } ⊥⊥ .Proof. (1) ⇒ (2): Let ϕ be normal and let { e , . . . , e k } ⊆ X be orthogonal. Then ¯ ϕ maps the top element of B ( e , . . . , e k ) to the top element of B ( ϕ ( e ) , . . . , ϕ ( e k )) , thatis, ¯ ϕ ( { e , . . . , e k } ⊥⊥ ) = { ϕ ( e ) , . . . , ϕ ( e k ) } ⊥⊥ .(2) ⇒ (1): Let (2) hold and let { e , . . . , e k } ⊆ X be orthogonal. Recall that theBoolean algebra B ( e , . . . , e k ) consists of the elements A ⊥⊥ ∈ C ( X, ⊥ ) , where A ⊆{ e , . . . e k } . By assumption, ¯ ϕ ( A ⊥⊥ ) = ( ϕ ( A )) ⊥⊥ . Thus ¯ ϕ establishes an isomorph-ism between B ( e , . . . , e k ) and B ( ϕ ( e ) , . . . , ϕ ( e k )) .The equivalence of (2) and (3) as well as the equivalence of (4) and (5) are clear.Moreover, (2) clearly implies (4). We conclude the proof by showing that (5) implies(3).Assume that (5) holds. Let { e , . . . , e k } be an orthogonal subset of X . We extendit to a maximal orthogonal set E = { e , . . . , e k , e k +1 , . . . , e m } . Furthermore, f = ϕ ( e ) , . . . , f m = ϕ ( e m ) are pairwise orthogonal elements of Y . We extend ϕ ( E ) to amaximal orthogonal subset F of Y . 9et A ⊆ E . We shall show that ϕ ( A ⊥⊥ ) ⊆ ϕ ( A ) ⊥⊥ , then in particular (3) willfollow. As ϕ is orthogonality-preserving, ϕ ( A ⊥⊥ ) ⊥ ϕ ( E \ A ) ⊥⊥ . Moreover, byassumption, ϕ ( A ⊥⊥ ) ⊆ ϕ ( X ) ⊆ ϕ ( E ) ⊥⊥ ⊥ ( F \ ϕ ( E )) ⊥⊥ . It follows ϕ ( A ⊥⊥ ) ⊥ ϕ ( E \ A ) ⊥⊥ ∨ ( F \ ϕ ( E )) ⊥⊥ = ( F \ ϕ ( A )) ⊥⊥ and hence, by the normality of Y , ϕ ( A ⊥⊥ ) ⊆ ( F \ ϕ ( A )) ⊥ = ϕ ( A ) ⊥⊥ .We observe that normal homomorphisms are, in a restricted sense, linearity-preserv-ing. Lemma 3.3.
Let X and Y be normal orthogonality spaces and let ϕ : X → Y bea normal homomorphism. Let e, f ∈ X be such that e ⊥ f . If g ∈ { e, f } ⊥⊥ , then ϕ ( g ) ∈ { ϕ ( e ) , ϕ ( f ) } ⊥⊥ .Proof. The assertion holds by Lemma 3.2, property (3).An automorphism of an orthogonality space ( X, ⊥ ) is a bijection ϕ : X → X suchthat, for any e, f ∈ X , e ⊥ f if and only if ϕ ( e ) ⊥ ϕ ( f ) . Automorphisms are alwaysnormal homomorphisms, in particular the identity is normal. Lemma 3.4.
Let X be a normal orthogonality space and let ϕ : X → X be anautomorphism. Then ϕ is normal.Proof. ¯ ϕ is an automorphism of C ( X, ⊥ ) .We see next that normal homomorphisms are closed under composition. Lemma 3.5.
Let X , Y , and Z be normal orthogonality spaces and let ϕ : X → Y and ψ : Y → Z be normal homomorphisms. Then also ψ ◦ ϕ is a normal homomorphism.Proof. Clearly, ψ ◦ ϕ is orthogonality-preserving. Moreover, the normality followsby means of property (3) in Lemma 3.2.We define the category N OS to consist of the normal orthogonality spaces (of finiterank) and the normal homomorphisms.We first check whether an inclusion map between normal orthogonality spaces isnormal. The following example shows that this is not in general the case.
Example 3.6.
Consider again the orthogonality space ( X, ⊥ ) from Example 2.15,which is normal but not Dacey, and let A = { a, e } . Then A ∈ C ( X, ⊥ ) and ( A, ∅ ) is a subspace of ( X, ⊥ ) , which is normal. Let now i A : A → X be the inclusion map.We have that { a } is a maximal orthogonal subset of A and i A ( A ) ⊥⊥ = { a, e } ⊥⊥ = { a, e } (cid:54) = { a } = { a } ⊥⊥ = { i A ( a ) } ⊥⊥ . Hence, by Lemma 3.2, property (4) , i A is not normal. Theorem 3.7.
Let ( X, ⊥ ) be a normal orthogonality space. Then X is a Dacey spaceif and only if, for any A ∈ C ( X, ⊥ ) , the subspace ( A, ⊥ ) is normal and the inclusionmap ι : A → X is a morphism in N OS . roof. Assume first that X is a Dacey space. Let A ∈ C ( X, ⊥ ) . By Lemma 2.11 andProposition 2.9, ( A, ⊥ ) is a normal subspace. Moreover, the inclusion map ι : A → X, x (cid:55)→ x is clearly orthogonality-preserving. Let { e , . . . , e n } be a maximal or-thogonal subset of A . Then A = { e , . . . , e n } ⊥⊥ by Lemma 2.11. By Lemma 3.2,property (4), we conclude that ι is actually a normal homomorphism.Conversely, assume that, for any A ∈ C ( X, ⊥ ) , ( A, ⊥ ) is normal and the inclusionmap ι : A → X, x (cid:55)→ x is a morphism of N OS . Let { e , . . . , e n } be a maximal or-thogonal subset of some A ∈ C ( X, ⊥ ) . Then again by Lemma 3.2, property (4),we have that ι ( A ) ⊥⊥ = { ι ( e ) , . . . , ι ( e n ) } ⊥⊥ , that is, A = { e , . . . , e n } ⊥⊥ . ByLemma 2.11, we conclude that X is a Dacey space.We note that for a normal orthogonality space to be a Dacey space, it is not enoughto assume that all subspaces are normal. Indeed, Example 2.15 provides a counter-example. We turn now our attention to orthogonality spaces arising from inner-product spaces.In this section, we compile the necessary background material.We first consider linear spaces without any additional structure. We are interested inthe representation of maps between projective spaces that preserve the collinearity ofpoint triples. The most general results in this area are, to our knowledge, due to Faure[Fau]. Here, we will follow the work of Machala [Mach]. The reader is referred toeither of these papers for more detailed information.By an sfield , we mean a skew field (i.e., a division ring). Let V be a linear space overan sfield K . We write V • = V \ { } and in accordance with Example 2.2, we define P ( V ) = { [ x ] : x ∈ V • } to be the projective space associated with V . For x, y, z ∈ V • ,we write (cid:96) ([ x ] , [ y ] , [ z ]) if [ x ] , [ y ] , [ z ] are on a line of P ( V ) , that is, if x, y, z are linearlydependent.Let V and V (cid:48) be linear spaces over the sfields K and K (cid:48) , respectively. We call a map ϕ : P ( V ) → P ( V (cid:48) ) a lineation if:(L1) For any x, y, z ∈ V • , (cid:96) ([ x ] , [ y ] , [ z ]) implies (cid:96) ( ϕ ([ x ]) , ϕ ([ y ]) , ϕ ([ z ])) .Thus a lineation is a map between projective spaces that preserves the collinearity ofpoint triples. Obviously, (L1) is equivalent to:(L1’) For any x, y, z ∈ V • such that ϕ ([ x ]) (cid:54) = ϕ ([ y ]) and [ z ] ⊆ [ x ] + [ y ] , we have ϕ ([ z ]) ⊆ ϕ ([ x ]) + ϕ ([ y ]) .Thus a lineation can also be understood as follows: if the point [ z ] lies on the linethrough [ x ] and [ y ] and if [ x ] and [ y ] are not mapped to the same point, then ϕ ([ z ]) is on the line through ϕ ([ x ]) and ϕ ([ y ]) . It is natural to ask whether a lineation isinduced by a suitable map between the underlying linear spaces.11et K be an sfield. We denote by K × = K \ { } the multiplicative group of K .A valuation ring F K of K is a subring of K such that, for each α ∈ K × , at leastone of α or α − is in F K . In this case, the subgroup U ( F K ) of K × consisting ofthe units of F K is called the group of valuation units . Obviously, F K is a local ring, I K = F K \ U ( F K ) = { α ∈ F K : α = 0 or α − / ∈ F K } being its unique maximalleft (right) ideal. Let K (cid:48) be a further sfield; then a ring homomorphism (cid:37) : F K → K (cid:48) with kernel I K is called a place from K to K (cid:48) . Note that in this case, (cid:37) induces anembedding of the sfield F K /I K into K (cid:48) .Let V be a linear space over an sfield K . Let F K be a valuation ring of K and let F V be a submodule of V over F K such that any one-dimensional subspace of V containsa non-zero element of F V . Let V (cid:48) be a further linear space over an sfield K (cid:48) . Let (cid:37) : F K → K (cid:48) be a place from K to K (cid:48) and let A : F V → V (cid:48) be such that (i) anyone-dimensional subspace of V contains a vector in F V that A does not map to ,(ii) A is additive, and (iii) for any x ∈ F V and α ∈ F K , we have A ( αx ) = (cid:37) ( α ) A ( x ) .Then A is called a generalised semilinear map from V to V (cid:48) w.r.t. the place (cid:37) . Theorem 4.1.
Let A : F V → V (cid:48) be a generalised semilinear map between the linearspaces V and V (cid:48) . Then the prescription ϕ A : P ( V ) → P ( V (cid:48) ) , [ x ] (cid:55)→ [ A ( x )] , where x ∈ F V and A ( x ) (cid:54) = 0 ,defines a lineation.Sketch of proof; for full details see the proof of [Mach, Satz 5] . Each one-dimension-al subspace of V contains by assumption an element x ∈ F V such that A ( x ) (cid:54) = 0 .Moreover, let y ∈ [ x ] ∩ F V such that A ( y ) (cid:54) = 0 . Then either y = αx or x = αy forsome α ∈ F K \ { } . In the former case, we have A ( y ) = (cid:37) ( α ) A ( x ) ; in the latter case,we have A ( x ) = (cid:37) ( α ) A ( y ) . Here, (cid:37) is the place associated with A . It follows that [ A ( x )] = [ A ( y )] . We conclude that we can define ϕ A as indicated.Let now x, y, z ∈ F V such that A ( x ) , A ( y ) , A ( z ) (cid:54) = 0 and (cid:96) ([ x ] , [ y ] , [ z ]) . We haveto show that (cid:96) ( ϕ A ([ x ]) , ϕ A ([ y ]) , ϕ A ([ z ])) . We may assume that ϕ A ([ x ]) , ϕ A ([ y ]) , and ϕ A ([ z ]) are mutually distinct. Let α, β ∈ K be such that z = αx + βy . Then α, β (cid:54) = 0 . Moreover, either α − β ∈ F K or β − α ∈ F K . In the former case, we set z (cid:48) = α − z = x + α − β y ; then z (cid:48) ∈ F V and A ( z (cid:48) ) = A ( x ) + (cid:37) ( α − β ) A ( y ) (cid:54) = 0 because ϕ A ([ x ]) = [ A ( x )] and ϕ A ([ y ]) = [ A ( y )] are distinct. Hence ϕ A ([ z ]) =[ A ( z (cid:48) )] = [ A ( x ) + (cid:37) ( α − β ) A ( y )] ⊆ [ A ( x )] + [ A ( y )] = ϕ A ([ x ]) + ϕ A ([ y ]) and theassertion follows. In the latter case, we set z (cid:48) = β − z and proceed similarly.Let A : F V → V (cid:48) be a generalised semilinear map between the linear spaces V and V (cid:48) . We will then write I V = { x ∈ F V : A ( x ) = 0 } . Note that, for any x ∈ F V \ I V ,we have [ x ] ∩ F V = F K · x . Moreover, let α ∈ F K ; then αx ∈ I V if and only if α ∈ I K , or in other words, αx ∈ F V \ I V if and only if α ∈ U ( F K ) .For a converse of Theorem 4.1, we need to take into account additional conditions.A lineation ϕ : P ( V ) → P ( V (cid:48) ) is called non-degenerate if the following conditionshold:(L2) For any linearly independent vectors x, y ∈ V • , { ϕ ([ z ]) : z (cid:54) = 0 , z ∈ [ x, y ] } contains at least three elements. 12L3) The image of ϕ is not contained in a -dimensional subspace of V (cid:48) .We arrive at the main theorem of [Mach]. Theorem 4.2.
Every non-degenerate lineation between projective spaces is, in thesense of Theorem 4.1, induced by a generalised semilinear map.
We next consider linear spaces that are equipped with an inner product. For, the mapsbetween projective spaces that we are going to study are supposed to respect first andforemost an orthogonality relation.A (cid:63) -sfield is an sfield equipped with an involutorial antiautomorphism (cid:63) . An (aniso-tropic) Hermitian space is a linear space H over a (cid:63) -sfield K that is equipped withan anisotropic, symmetric sesquilinear form ( · , · ) : H × H → K . For x, y ∈ H , wewrite x ⊥ y if ( x, y ) = 0 , and for x, y ∈ H • , [ x ] ⊥ [ y ] means x ⊥ y .Let H and H (cid:48) be Hermitian spaces over the (cid:63) -sfield K and K (cid:48) , respectively. We call U : H → H (cid:48) a generalised semiunitary map if U is a generalised semilinear mapw.r.t. some place (cid:37) from K to K (cid:48) and there are a λ ∈ K and a λ (cid:48) ∈ K (cid:48) such that ( U ( x ) , U ( y )) = (cid:37) (( x, y ) λ ) λ (cid:48) for any x, y ∈ F H . The question arises whether orthogonality-preserving lineationsare induced by semiunitary maps. Under particular circumstances, we can give anaffirmative answer. Theorem 4.3.
Let H and H (cid:48) be finite-dimensional Hermitian spaces over the (cid:63) -fields K and K (cid:48) , respectively. Assume that H possesses an orthogonal basis consisting ofvectors of equal length. Then any non-degenerate orthogonality-preserving lineation ϕ : P ( H ) → P ( H (cid:48) ) is induced by a generalised semiunitary map.Proof. By Theorem 4.2, ϕ is induced by a generalised semilinear map U : F H → H (cid:48) w.r.t. a place (cid:37) : F K → K (cid:48) .We proceed by showing several auxiliary lemmas.(a) For any a, b ∈ F H \ I H , a ⊥ b implies U ( a ) ⊥ U ( b ) .Proof of (a): Assume a ⊥ b . Then [ a ] ⊥ [ b ] and hence [ U ( a )] = ϕ ([ a ]) ⊥ ϕ ([ b ]) =[ U ( b )] as ϕ is orthogonality-preserving. It follows U ( a ) ⊥ U ( b ) .(b) There is an orthogonal basis b , . . . , b n ∈ F H \ I H of H consisting of vectors ofequal length.Proof of (b): By assumption, H possesses an orthogonal basis b , . . . , b n consistingof vectors of equal length. In view of condition (i) of the definition of a generalisedsemilinear map, we may assume that b ∈ F H \ I H . Let (cid:54) i (cid:54) n ; we claim that b i ∈ F H \ I H as well. Assume that b i ∈ I H . Then U ( b + b i ) = U ( b − b i ) = U ( b ) (cid:54) = 0 but b + b i ⊥ b − b i , in contradiction to (a). Assume that b i / ∈ F H . Let λ ∈ K be suchthat λb i ∈ F H \ I H . Then λ − ∈ F K would imply that b i = λ − · λb i ∈ F H contraryto the assumption; hence λ ∈ I K . It follows λb , λb i ∈ F H and U ( λb i + λb ) = U ( λb i − λb ) = U ( λb i ) (cid:54) = 0 but λb i + λb ⊥ λb i − λb , again a contradiction to (a).13or the rest of the proof, we fix a basis b , . . . , b n of H as specified in (b).(c) U ( b ) , . . . , U ( b n ) are vectors of equal length.Proof of (c): Let (cid:54) i (cid:54) n . From b + b i ⊥ b − b i it follows by (a) that U ( b + b i ) ⊥ U ( b − b i ) , that is, ( U ( b ) + U ( b i ) , U ( b ) − U ( b i )) = 0 . By (a), it follows ( U ( b ) , U ( b )) = ( U ( b i ) , U ( b i )) . The assertion is shown.(d) F K and I K are closed under (cid:63) . Moreover, for any α ∈ F K , we have (cid:37) ( α (cid:63) ) = (cid:37) ( α ) (cid:63) .Proof of (d): Let α ∈ I K \ { } . Assume that α (cid:63) / ∈ F K . Then ( α − ) (cid:63) = ( α (cid:63) ) − ∈ I K .Moreover, αb − b and ( α − ) (cid:63) b + b are orthogonal vectors in F H \ I H and hence − U ( b ) = U ( αb − b ) ⊥ U (( α − ) (cid:63) b + b ) = U ( b ) , a contradiction. We concludethat α (cid:63) ∈ F K . Furthermore, αb − b and b + α (cid:63) b are orthogonal vectors in F H \ I H and hence U ( b ) = − U ( αb − b ) ⊥ U ( b + α (cid:63) b ) = U ( b ) + (cid:37) ( α (cid:63) ) U ( b ) . Weconclude that (cid:37) ( α (cid:63) ) = 0 , that is, α (cid:63) ∈ I K . We have shown that I K is closed under (cid:63) .Let now α ∈ K \ { } be such that α (cid:63) / ∈ F K . Then ( α (cid:63) ) − ∈ I K and hence also α − ∈ I K . This means α / ∈ F K . It follows that also F K is closed under (cid:63) .Finally, let α ∈ F K . We have that αb − b and b + α (cid:63) b are orthogonal vectors in F H \ I H . It follows that (cid:37) ( α ) U ( b ) − U ( b ) ⊥ U ( b ) + (cid:37) ( α (cid:63) ) U ( b ) , that is, (cid:37) ( α )( U ( b ) , U ( b )) − (cid:37) ( α (cid:63) ) (cid:63) ( U ( b ) , U ( b )) = 0 . Thus, by (c), the assertion follows.(e) Let α , . . . , α n ∈ K . Then there is an α ∈ K \ { } such that α − α , . . . , α − α n ∈ F K and α − α i / ∈ I K for at least one i .The proof of (e) can be found in [Rad, Lemma 6].(f) Let x = α b + . . . + α n b n ∈ F H . Then α , . . . , α n ∈ F K .Proof of (f): Assume to the contrary that one of the coefficients is not in F K . By(e), there is an α ∈ K such that α − α , . . . , α − α n ∈ F K and α − α i / ∈ I K for some i . Then α / ∈ F K and hence α − ∈ I K . Hence (cid:37) ( α − ) U ( x ) = U ( α − x ) = (cid:37) ( α − α ) U ( b ) + . . . + (cid:37) ( α − α n ) U ( b n ) (cid:54) = 0 , because (cid:37) ( α − α i ) U ( b i ) (cid:54) = 0 and thesummed vectors are mutually orthogonal. The assertion follows.Let now x = α b + . . . + α n b n and y = β b + . . . + β n b n be elements of F H . By(f), α , . . . , α n , β , . . . , β n ∈ F K . Using (c) and (d), we get ( U ( x ) , U ( y )) = (cid:37) ( α )( U ( b ) , U ( b )) (cid:37) ( β ) (cid:63) + . . . + (cid:37) ( α n )( U ( b n ) , U ( b n )) (cid:37) ( β n ) (cid:63) = (cid:37) ( α β (cid:63) + . . . + α n β (cid:63)n )( U ( b ) , U ( b ))= (cid:37) (cid:0) ( α ( b , b ) β (cid:63) + . . . + α n ( b n , b n ) β (cid:63)n )( b , b ) − (cid:1) ( U ( b ) , U ( b ))= (cid:37) (( x, y )( b , b ) − )( U ( b ) , U ( b )) , thus the theorem is proved.We may observe that the requirement in Theorem 4.3 regarding the existence of basisvectors of equal length is related to the orderability of the scalar (cid:63) -field. In the re-mainder of this section, we review orders on fields and (cid:63) -fields and we will indicateexamples for our above results. For further information, we refer the reader, e.g.,to [Fuc, Pre, Hol]. 14e recall that an ordered field is a field K equipped with a linear order such that theadditive reduct becomes a linearly ordered group and the positive elements are closedunder multiplication [Fuc, Chapter 8]. In this case, K n , n (cid:62) , endowed with thestandard inner product, (cid:32)(cid:32) α ... α n (cid:33) , (cid:32) β ... β n (cid:33)(cid:33) = α β + . . . + α n β n , is an n -dimensional Hermitian space. Moreover, K n is positive definite and possessesan orthonormal basis.The elements of K may be roughly classified by means of the linear order. We call F K = { α ∈ K : | α | (cid:54) n for some n ∈ N } the set of finite elements, which we partition in turn into the two sets M K = { α ∈ K : n (cid:54) | α | (cid:54) n for some n ∈ N } ,I K = { α ∈ K : | α | (cid:54) n for all n ∈ N } , consisting of the medial and the infinitesimal elements, respectively. Then F K is avaluation ring of K . We have that U ( F K ) = M K is the group of valuation unitsand, in accordance with the notation used above, the unique maximal ideal of F K is I K = F K \ M K .If is the only infinitesimal element, that is, if all non-zero elements are medial, then K is called Archimedean . This is the case if and only if K is isomorphic to an orderedsubfield of R ; see, e.g., [Fuc].Any positive definite Hermitian space over a non-Archimedean ordered field K thatpossesses an orthonormal basis gives rise to an example for Theorem 4.3. Namely, thequotient of the module over F K of the vectors of finite length modulo the submoduleof the vectors of infinitesimal length results in a Hermitian space over R . For thedetails, see, e.g., [Hol]; let us here consider the particular case of non-standard reals. Example 4.4.
Let R (cid:63) be the ordered field of hyperreal numbers. The set F R (cid:63) of finitehyperreals is a valuation ring. Moreover, F R (cid:63) /I R (cid:63) is isomorphic to R , hence thequotient map induces a place (cid:37) : F R (cid:63) → R .Consider now the linear space ( R (cid:63) ) n , n (cid:62) , endowed with the standard innerproduct. Then F ( R (cid:63) ) n = { x ∈ ( R (cid:63) ) n : ( x, x ) ∈ F R (cid:63) } = (cid:40)(cid:32) α ... α n (cid:33) : α , . . . , α n ∈ F R (cid:63) (cid:41) is a submodule of ( R (cid:63) ) n over F R (cid:63) . Furthermore, A : F ( R (cid:63) ) n → R n , (cid:32) α ... α n (cid:33) (cid:55)→ (cid:32) (cid:37) ( α ) ... (cid:37) ( α n ) (cid:33) is a generalised semiunitary map from ( R (cid:63) ) n to R n , inducing a non-degenerate orthog-onality-preserving lineation. K be a (cid:63) -field and let S K = { α ∈ K : α (cid:63) = α } be its fixed field. Assume that S K , as anadditive group, is linearly ordered such that (cid:62) and, for any β ∈ K , α (cid:62) implies βαβ (cid:63) (cid:62) . Then K is called Baer ordered .All above considerations allow analogues in this broader context. In particular, apositive definite Hermitian space over a Baer ordered (cid:63) -field allows the constructionof a quotient space over R or C ; see [Hol]. Example 4.4 can be adapted in a way thatthe reals are replaced with the field of complex numbers. The orthogonality spaces to which we turn in this section are directly related to thosethat arise from Hermitian spaces. We will show that they are Dacey spaces and hencebelong to the normal orthogonality spaces.
Definition 5.1.
An orthogonality space ( X, ⊥ ) is called linear if, for any two distinctelements e, f ∈ X , there is a third element g such that { e, f } ⊥ = { e, g } ⊥ and exactlyone of f and g is orthogonal to e .In other words, for ( X, ⊥ ) to be linear means that (i) for distinct, non-orthogonalelements e, f ∈ X there is a g ⊥ e such that { e, f } ⊥ = { e, g } ⊥ and (ii) for orthogonalelements e, f ∈ X , there is a g (cid:54)⊥ e such that { e, f } ⊥ = { e, g } ⊥ . Note that in bothcases g is necessarily distinct from e and f . Example 5.2.
Let H be a Hilbert space and let ( P ( H ) , ⊥ ) again be the orthogon-ality space arising from H according to Example 2.2. Then we readily check that ( P ( H ) , ⊥ ) is linear. We start with the following observation. We call an orthogonality space ( X, ⊥ ) irre-dundant if, for any e, f ∈ X , { e } ⊥ = { f } ⊥ implies e = f . Moreover, we call ( X, ⊥ ) strongly irredundant if, for any e, f ∈ X , { e } ⊥ ⊆ { f } ⊥ implies e = f . Obviously,strong irredundancy implies irredundancy. We may express strong irredundancy alsoclosure-theoretically; cf., e.g., [Ern]. Indeed, ( X, ⊥ ) is strongly irredundant exactlyif the specialisation order associated with the closure operator ⊥⊥ is the equality. Lemma 5.3.
Linear orthogonality spaces are strongly irredundant.Proof.
Let ( X, ⊥ ) be a linear orthogonality space.We first show that X is irredundant. Let e and f be distinct elements of X . If e and f are orthogonal, then f ⊥ e but e (cid:54)⊥ e . If not, there is by the linearity some g ⊥ e suchthat { e, f } ⊥ = { e, g } ⊥ . Then g / ∈ { e, g } ⊥ = { e, f } ⊥ , hence g ⊥ e but g (cid:54)⊥ f . Hence { e } ⊥ (cid:54) = { f } ⊥ in either case.Let now e, f ∈ X be such that { e } ⊥ ⊆ { f } ⊥ . We shall show that then actually { e } ⊥ = { f } ⊥ ; by irredundancy, it will follow that X is strongly irredundant. Assumeto the contrary that { e } ⊥ (cid:40) { f } ⊥ . Then e (cid:54) = f and e (cid:54)⊥ f . Hence, by the linearityof X , there is a g ⊥ e such that { e, g } ⊥ = { e, f } ⊥ . But this means g ∈ { e, g } ⊥⊥ = { e, f } ⊥⊥ = { e } ⊥⊥ , a contradiction. 16he following correspondence between linear orthogonality spaces and linear spaceswas shown in [Vet3]. Theorem 5.4.
Let H be a Hermitian space of finite dimension n . Then ( P ( H ) , ⊥ ) isa linear orthogonality space of rank n .Conversely, let ( X, ⊥ ) be a linear orthogonality space of finite rank n (cid:62) . Thenthere is a (cid:63) -sfield K and an n -dimensional Hermitian space H over K such that ( X, ⊥ ) is isomorphic to ( P ( H ) , ⊥ ) . Clearly, the assumption regarding the rank cannot be omitted in Theorem 5.4. For lowranks, linear orthogonality spaces may be of a much different type than those arisingfrom inner-product spaces.
Example 5.5.
For n (cid:62) , let D n = { , , . . . , n , n } , endowed with the ortho-gonality relation such that i and i , for each i = 1 , . . . , n , are orthogonal and nofurther pair. We easily see that ( D n , ⊥ ) is linear. Note that C ( D n , ⊥ ) is isomorphicto MO n , the horizontal sum of n four-element Boolean algebras, which is a modularortholattice with n + 2 elements. Each linear orthogonality space is a Dacey space and hence normal. The exact rela-tionship is as follows.Here, an orthogonality space ( X, ⊥ ) is called irreducible if X cannot be partitionedinto two non-empty subsets A and B such that e ⊥ f for any e ∈ A and f ∈ B . Theorem 5.6.
An orthogonality space ( X, ⊥ ) is linear if and only if X is an irredu-cible, strongly irredundant Dacey space. In particular, X is in this case normal.Proof. Let ( X, ⊥ ) be linear. By [Vet3, Theorem 3.7], C ( X, ⊥ ) is orthomodular, thatis, a Dacey space. By Proposition 2.14, X is hence normal. By Lemma 5.3, X isstrongly irredundant. Assume now that X = A ∪ B , where A and B are disjointnon-empty subsets of X and e ⊥ f for any e ∈ A and f ∈ B . By linearity, for any e ∈ A and f ∈ B , there is a g (cid:54)⊥ e such that { e, f } ⊥ = { e, g } ⊥ . Then g / ∈ B andconsequently g ∈ A and thus g ⊥ f . It follows { g } ⊥⊥ ⊆ { e, f } ⊥⊥ ∩ { f } ⊥ = { e } ⊥⊥ by orthomodularity and hence f ∈ { f } ⊥⊥ ⊆ { e, f } ⊥⊥ = { e } ⊥⊥ ∨ { g } ⊥⊥ = { e } ⊥⊥ ,in contradiction to f ∈ { e } ⊥ . We conclude that X is irreducible.Conversely, let ( X, ⊥ ) be an irreducible, strongly irredundant Dacey space. By thestrong irredundancy, { e } ⊥⊥ is, for any e ∈ X , an atom of C ( X, ⊥ ) and it followsthat ( X, ⊥ ) is atomistic. Furthermore, C ( X, ⊥ ) is a complete orthomodular latticeof finite length. It follows that C ( X, ⊥ ) is in fact a modular lattice and hence fulfilsthe covering property and the exchange property. Moreover, C ( X, ⊥ ) is irreducible.Indeed, if the centre of C ( X, ⊥ ) contained an element ∅ (cid:40) A (cid:40) X , then each atomwould be below A or below A ⊥ , that is, we would have X = A ∪ A ⊥ and X wouldnot be irreducible.Let e, f ∈ X be distinct, non-orthogonal elements. Then { e } ⊥⊥ and { f } ⊥⊥ are dis-tinct atoms and hence { e, f } ⊥⊥ = { e } ⊥⊥ ∨{ f } ⊥⊥ covers { e } ⊥⊥ . By orthomodularity,there is an element g ⊥ e such that { e, f } ⊥⊥ = { e } ⊥⊥ ∨ { g } ⊥⊥ = { e, g } ⊥⊥ , that is, { e, f } ⊥ = { e, g } ⊥ . 17et e, f ∈ X be distinct, orthogonal elements. Since C ( X, ⊥ ) is irreducible, thejoin of { e } ⊥⊥ and { f } ⊥⊥ contains a third atom, that is, there is a g (cid:54) = e, f such that g ∈ { e, f } ⊥⊥ . By the exchange property, it follows { e, f } ⊥⊥ = { e, g } ⊥⊥ . Thus { e, f } ⊥⊥ = { e, g } ⊥⊥ , and g (cid:54)⊥ e because otherwise g = f . The proof of the linearityof X is complete. Example 5.7.
We observe from Theorem 5.6 that not every Dacey space is linear.The probably simplest counterexample is ( , (cid:54) =) , the orthogonality space consistingof two orthogonal elements, cf. Example 2.7. Obviously, is Dacey but not linear.More generally, the same applies, for any n (cid:62) , to ( n , (cid:54) =) . We have seen above that subspaces of normal orthogonality spaces are not necessarilynormal. In the present context, the situation is different.
Proposition 5.8.
Any subspace of a linear orthogonality space is linear.Proof.
Let ( X, ⊥ ) be a linear orthogonality space and let ( A, ⊥ A ) be a subspace of X . For B ⊆ A , we have B ⊥ A = B ⊥ ∩ A . By Theorem 5.6, C ( X, ⊥ ) is orthomodularand hence B ⊥ A ⊥ A = ( B ⊥ ∩ A ) ⊥ ∩ A = B ⊥⊥ .Let e and f be distinct elements of A . Then there is a g ∈ X such that { e, f } ⊥ = { e, g } ⊥ and exactly one f and g is orthogonal to e . Since g ∈ { e, g } ⊥⊥ = { e, g } ⊥ A ⊥ A ,we have that g ∈ A . Furthermore, we have { e, f } ⊥ A = { e, f } ⊥ ∩ A = { e, g } ⊥ ∩ A = { e, g } ⊥ A . The linearity of ( A, ⊥ A ) is shown. LOS of linear orthogonality spaces
We denote by
LOS the full subcategory of
N OS consisting of linear orthogonalityspaces.Our aim is to describe the morphisms in
LOS . We restrict our considerations toorthogonality spaces that arise from Hermitian spaces. In view of Theorem 5.4, theresults hence apply to linear orthogonality spaces whose rank is at least . Theorem 6.1.
Let H and H (cid:48) be finite-dimensional Hermitian spaces. Then a map ϕ : P ( H ) → P ( H (cid:48) ) is a morphism in LOS if and only if ϕ is an orthogonality-preserving lineation.Proof. Let ϕ : P ( H ) → P ( H (cid:48) ) be a morphism in LOS . By definition, ϕ preservesthe orthogonality relation. Let x, y, z ∈ H • be such that ϕ ([ x ]) (cid:54) = ϕ ([ y ]) and z ∈ [ x, y ] . Let y (cid:48) ⊥ x be such that [ x, y ] = [ x, y (cid:48) ] . By Lemma 3.3, ϕ ([ y ]) , ϕ ([ z ]) ⊆ ϕ ([ x ]) + ϕ ([ y (cid:48) ]) . By assumption, ϕ ([ x ]) (cid:54) = ϕ ([ y ]) , so that ϕ ([ z ]) ⊆ ϕ ([ x ]) + ϕ ([ y (cid:48) ]) = ϕ ([ x ]) + ϕ ([ y ]) . By criterion (L1’), ϕ is a lineation.Conversely, let ϕ be an orthogonality-preserving lineation. Then ϕ is a homomorph-ism of orthogonality spaces. Let x , . . . , x n be an orthogonal basis of H . We have toverify that ϕ ([ x ]) ∈ { ϕ ([ x ]) , . . . , ϕ ([ x n ]) } ⊥⊥ , that is, ϕ ([ x ]) ⊆ ϕ ([ x ])+ . . . + ϕ ([ x n ]) for any x ∈ H ; then it will follow by Lemma 3.2, property (5), that ϕ is normal andhence a morphism. 18ssume that x ∈ [ x , x ] . We have that ϕ ([ x ]) ⊥ ϕ ([ x ]) and hence ϕ ([ x ]) (cid:54) = ϕ ([ x ]) . As ϕ is a lineation, it follows ϕ ([ x ]) ⊆ ϕ ([ x ]) + ϕ ([ x ]) . The assertionfollows thus by an inductive argument.A morphism of LOS being a lineation, the question seems natural whether it is non-degenerate. We consider the conditions (L2) and (L3), which define non-degeneracy,separately.The latter condition is automatic, provided that we assume dimensions of at least . Lemma 6.2.
Let H and H (cid:48) be finite-dimensional Hermitian spaces and assume thatthe dimension of H is at least . Then any morphism in LOS from P ( H ) to P ( H (cid:48) ) is a lineation fulfilling (L3) .Proof. A morphism ϕ : P ( H ) → P ( H (cid:48) ) in LOS is by Theorem 6.1 an orthogonality-preserving lineation. Since H is at least -dimensional, the image of ϕ contains threemutually orthogonal elements. It follows that ϕ fulfils (L3).In the next lemma, ( , (cid:54) =) is, in accordance with Example 2.7, the orthogonality spaceconsisting of three mutually orthogonal elements. Lemma 6.3.
Let H and H (cid:48) be Hermitian spaces of finite dimension (cid:62) over the (cid:63) -sfields K and K (cid:48) , respectively. Assume that there is a morphism in LOS from P ( H ) to P ( H (cid:48) ) that does not fulfil (L2) . Then there is a -dimensional subspace H of H and a morphism in N OS from ( P ( H ) , ⊥ ) to ( , (cid:54) =) .Proof. For convenience, we will formulate this proof in the language of orthogonalityspaces rather than linear spaces.Let ϕ : P ( H ) → P ( H (cid:48) ) be a morphism in LOS that violates (L2). By Theorem 6.1, ϕ is an orthogonality-preserving lineation. Moreover, there are e, f ∈ P ( H ) suchthat e ⊥ f and the image of { e, f } ⊥⊥ under ϕ contains exactly two elements. Thus ϕ ( { e, f } ⊥⊥ ) = { e (cid:48) , f (cid:48) } , where e (cid:48) = ϕ ( e ) and f (cid:48) = ϕ ( f ) . We choose a g ∈ P ( H ) besuch that g ⊥ e, f . Then e (cid:48) , f (cid:48) , and g (cid:48) = ϕ ( g ) are mutually orthogonal.Let H be the -dimensional subspace of H spanned by e , f , and g . Then we have that { e, f, g } ⊥⊥ = P ( H ) , the orthogonality relation being induced by the inner producton H . Similarly, let H (cid:48) be the subspace of H spanned by e (cid:48) , f (cid:48) , and g (cid:48) , so that { e (cid:48) , f (cid:48) , g (cid:48) } ⊥⊥ = P ( H (cid:48) ) . As ϕ is normal, we conclude from Lemma 3.2, property (3),that the image of P ( H ) under ϕ is contained in P ( H (cid:48) ) . In other words, ϕ | P ( H ) is anorthogonality-preserving lineation from P ( H ) to P ( H (cid:48) ) . ( { e, f, g } ⊥⊥ , ⊥ ) is, by Proposition 5.8, a linear orthogonality space. Furthermore, { e (cid:48) , f (cid:48) , g (cid:48) } , together with the orthogonality relation of P ( H (cid:48) ) , is an orthogonalityspace isomorphic to ( , (cid:54) =) . In particular, ( { e (cid:48) , f (cid:48) , g (cid:48) } , ⊥ ) is normal. Our aim is toshow that there is an orthogonality-preserving map ψ : { e, f, g } ⊥⊥ → { e (cid:48) , f (cid:48) , g (cid:48) } .Then it will follow that ψ is a morphism of N OS and the lemma will be proved. For,such a map is a homomorphism of orthogonality spaces, and since the image of anyset of three mutually orthogonal elements in { e, f, g } ⊥⊥ is { e (cid:48) , f (cid:48) , g (cid:48) } , the normalityholds by Lemma 3.2, property (4). 19y Lemma 3.3 we observe that, for any x ∈ { e, f } ⊥⊥ such that ϕ ( x ) = e (cid:48) , wehave ϕ ( { g, x } ⊥⊥ ) ⊆ { e (cid:48) , g (cid:48) } ⊥⊥ , and for any x ∈ { e, f } ⊥⊥ such that ϕ ( x ) = f (cid:48) , wehave ϕ ( { g, x } ⊥⊥ ) ⊆ { f (cid:48) , g (cid:48) } ⊥⊥ . Furthermore, for any y ∈ { e, f, g } ⊥⊥ , there is an x ∈ { e, f } ⊥⊥ such that y ∈ { g, x } ⊥⊥ . We conclude that ϕ ( { e, f, g } ⊥⊥ ) ⊆ { e (cid:48) , g (cid:48) } ⊥⊥ ∪ { f (cid:48) , g (cid:48) } ⊥⊥ . (1)We now distinguish three cases. Case 1.
There is an h ∈ { e, g } ⊥⊥ such that h (cid:48) = ϕ ( h ) (cid:54) = e (cid:48) , g (cid:48) . Note that h (cid:48) ∈{ e (cid:48) , g (cid:48) } ⊥⊥ . We claim that ϕ ( { f, h } ⊥⊥ ) = { f (cid:48) , h (cid:48) } . (2)Let x ∈ { f, h } ⊥⊥ . Since x (cid:54) = g , there is a unique t ∈ { e, f } ⊥⊥ such that x ∈{ g, t } ⊥⊥ . Depending on whether ϕ ( t ) = e (cid:48) or ϕ ( t ) = f (cid:48) , we have that ϕ ( x ) ∈{ e (cid:48) , g (cid:48) } ⊥⊥ or { f (cid:48) , g (cid:48) } ⊥⊥ . Furthermore, since ϕ ( x ) ∈ { f (cid:48) , h (cid:48) } ⊥⊥ , we conclude thateither ϕ ( x ) = h (cid:48) or ϕ ( x ) = f (cid:48) . Thus (2) is shown.We next claim that ϕ ( { e, f, g } ⊥⊥ ) ⊆ { e (cid:48) , g (cid:48) } ⊥⊥ ∪ { f (cid:48) } . (3)Let x ∈ { e, f, g } ⊥⊥ such that ϕ ( x ) / ∈ { e (cid:48) , g (cid:48) } ⊥⊥ . Note that then ϕ ( x ) ∈ { f (cid:48) , g (cid:48) } ⊥⊥ by (1). Moreover, there is a unique y ∈ { f, h } ⊥⊥ ∩ { e, x } ⊥⊥ . Then x ∈ { e, y } ⊥⊥ and, by (2), either ϕ ( y ) = h (cid:48) or ϕ ( y ) = f (cid:48) . If ϕ ( y ) = h (cid:48) , then ϕ ( x ) ∈ { e (cid:48) , h (cid:48) } ⊥⊥ = { e (cid:48) , g (cid:48) } ⊥⊥ , in contradiction to our assumption. Hence we have ϕ ( y ) = f (cid:48) and ϕ ( x ) ∈{ e (cid:48) , f (cid:48) } ⊥⊥ and we conclude ϕ ( x ) = f (cid:48) . Thus (3) is shown.Finally, let τ : { e (cid:48) , g (cid:48) } ⊥⊥ → { e (cid:48) , g (cid:48) } be any orthogonality-preserving map. We define ψ : { e, f, g } ⊥⊥ → { e (cid:48) , f (cid:48) , g (cid:48) } , x (cid:55)→ (cid:40) τ ( ϕ ( x )) if ϕ ( x ) ∈ { e (cid:48) , g (cid:48) } ⊥⊥ , f (cid:48) if ϕ ( x ) = f (cid:48) .Then ψ is orthogonality-preserving, as desired. Case 2.
There is a h ∈ { f, g } ⊥⊥ such that ϕ ( h ) (cid:54) = f (cid:48) , g (cid:48) . Then we argue similarly toCase 1. Case 3. ϕ ( { e, g } ⊥⊥ ) = { e (cid:48) , g (cid:48) } and ϕ ( { f, g } ⊥⊥ ) = { f (cid:48) , g (cid:48) } . Let then x ∈ { e, f, g } ⊥⊥ such that x / ∈ { e, g } ⊥⊥ ∪ { f, g } ⊥⊥ . By (1), ϕ ( x ) ∈ { e (cid:48) , g (cid:48) } ⊥⊥ or ϕ ( x ) ∈ { f (cid:48) , g (cid:48) } ⊥⊥ .In the former case, let y ∈ { e, g } ⊥⊥ be such that x ∈ { f, y } ⊥⊥ ; then ϕ ( y ) = e (cid:48) andhence also ϕ ( x ) = e (cid:48) , or ϕ ( y ) = g (cid:48) and hence ϕ ( x ) = g (cid:48) . In the latter case, wesimilarly see that ϕ ( x ) = f (cid:48) or ϕ ( x ) = g (cid:48) .We conclude that ϕ ( { e, f, g } ⊥⊥ ) = { e (cid:48) , f (cid:48) , g (cid:48) } . Taking ψ = ϕ , we again have that ψ is orthogonality-preserving.We may formulate the conclusion of Lemma 6.3 measure-theoretically. By a measure on a finite-dimensional Hermitian space H , we mean a map µ from C ( H ) to the realunit interval such that (i) µ ( A + B ) = µ ( A ) + µ ( B ) for any orthogonal subspaces A and B of H and (ii) µ ( H ) = 1 . We call a measure two-valued if its image consists of and only. 20 emma 6.4. Let H and H (cid:48) be Hermitian spaces of finite dimension (cid:62) over the (cid:63) -sfields K and K (cid:48) , respectively. Assume that there is a morphism in LOS from P ( H ) to P ( H (cid:48) ) that does not fulfil (L2) . Then there is a -dimensional Hermitian spaceover K that possesses a two-valued measure.Proof. By Lemma 6.3, there is a -dimensional subspace H of H and a map from P ( H ) to the -element set such that any three mutually orthogonal elements of P ( H ) are mapped to distinct elements. Assigning to one of the elements of and to the remaining two, we may construct a two-valued measure as asserted.Combining the results that we have achieved so far, we arrive at the following state-ment. Proposition 6.5.
Let H and H (cid:48) be Hermitian spaces of finite dimension (cid:62) over (cid:63) -fields. Assume moreover that (1) H possesses a basis of vectors of equal lengthand (2) no -dimensional subspace of H possesses a two-valued measure. Then anymorphism in LOS between P ( H ) and P ( H (cid:48) ) is induced by a generalised semiunitarymap.Proof. Let ϕ : P ( H ) → P ( H (cid:48) ) be a morphism in LOS . By Theorem 6.1, ϕ is anorthogonality-preserving lineation and by Lemma 6.2, ϕ fulfils (L3).Assume now that ϕ does not fulfil (L2). By Lemma 6.4, H possesses a two-valuedmeasure, in contradiction to condition (2). We conclude that ϕ does fulfil (L2) and ishence non-degenerate.Taking into account condition (1), the assertion now follows by Theorem 4.3.The question arises when a Hermitian space fulfils the conditions of Proposition 6.5,which are admittedly rather technical. It would in particular be desirable to find aformulation referring to the scalar fields only. Whereas an answer seems to be difficultto find in general, the situation is somewhat more transparent in the case of Baerordered (cid:63) -fields. Theorem 6.6.
Let H be a positive definite Hermitian space of finite dimension (cid:62) over the Baer ordered (cid:63) -field K , and let H (cid:48) be a further finite-dimensional Hermitianspace. Assume moreover that K has the following properties: (1) For any α ∈ S K such that α (cid:62) , there is a β ∈ K such that α = ββ (cid:63) . (2) K , endowed with the standard inner product, does not possess a two-valuedmeasure.Then any morphism in LOS between P ( H ) and P ( H (cid:48) ) is induced by a generalisedsemiunitary map.Proof. Let x ∈ H • . As H is positive definite, there is, by (1), a β ∈ K such that ( x, x ) = ββ (cid:63) and hence ( β x, β x ) = 1 . That is, every subspace of H contains a unitvector and H possesses an orthonormal basis.21t furthermore follows that H can be identified with K , endowed with the standardinner product. Thus, by (2), H does not possess a two-valued measure. The assertionfollows now from Proposition 6.5.Theorem 6.6 leads in turn to the question how to characterise the Baer ordered (cid:63) -fields K fulfilling condition (2), that is, how to exclude the existence of two-valuedmeasures on K . We will give a sufficient criterion; see Lemma 6.11 below. We needseveral preparatory steps, some of which might be interesting in their own right.We recall that an ordered field is called Euclidean if any positive element is a square.In the proof of the next lemma, we follow the lines of Piron’s proof of Gleason’sTheorem [Pir, p. 75–78].
Theorem 6.7. A -dimensional positive definite Hermitian space over a Euclideansubfield of R does not possess a two-valued measure.Proof. Let R be a Euclidean subfield of the reals and let H be a -dimensional posit-ive definite Hermitian space over R .We claim that each one-dimensional subspace of H possesses a unit vector. Indeed,the only automorphism of R is the identity, hence (cid:63) = id. Hence the assertion followslike in the proof of Theorem 6.6.Let us assume that there is a two-valued measure µ on H , that is, a map µ : P ( H ) →{ , } such that, among any three orthogonal elements [ x ] , [ y ] , [ z ] ∈ P ( H ) , exactlyone is mapped to . Pick b ∈ H • such that µ ([ b ]) = 1 and let b , b , b be anorthogonal basis of H . By the previous paragraph, we can suppose that b , b , b areunit vectors. We may hence identify H with R , endowed with the standard innerproduct.We have that µ ([ (cid:16) (cid:17) ]) = 1 and consequently µ ([ (cid:16) αβ (cid:17) ]) = 0 for any elements α, β ∈ R that are not both . The map ι : R → P ( R ) , ( α, β ) (cid:55)→ [ (cid:16) αβ (cid:17) ] establishes a one-to-one correspondence between R and the set of those elements of P ( R ) that are notorthogonal to b . We shall write ¯ µ for µ ◦ ι . Let ¯0 be the origin of R ; then ¯ µ (¯0) = 1 .We proceed by showing several auxiliary statements.(a) Let L ⊆ R be a line and let r ∈ L be the element closest to ¯0 . Then ¯ µ ( r ) (cid:62) ¯ µ ( s ) for any s ∈ L .Proof of (a): We may assume that s (cid:54) = r . Let [ l ] ∈ P ( R ) be parallel to L . Then [ l ] ⊥ ιr and µ ([ l ]) = 0 . Moreover, let t ∈ L be such that ιt ⊥ ιs . Then ιr and [ l ] spanthe same -dimensional subspace of R as ιs and ιt . Hence ¯ µ ( r ) = ¯ µ ( r ) + µ ([ l ]) =¯ µ ( s ) + ¯ µ ( t ) (cid:62) ¯ µ ( s ) .We will denote by (cid:107) r (cid:107) the (Euclidean) distance between ¯0 and some r ∈ R .(b) For any r ∈ R and τ ∈ R such that < τ (cid:54) , we have ¯ µ ( r ) (cid:54) ¯ µ ( τ · r ) .Proof of (b): Let r ⊥ arise from rotating r by π . Consider s = τ · r + (cid:112) τ (1 − τ ) · r ⊥ . (cid:94) ¯0 s r = π , hence ¯ µ ( r ) (cid:54) ¯ µ ( s ) by (a). Likewise, we have (cid:94) s τ r ¯0 = π , hence ¯ µ ( s ) (cid:54) ¯ µ ( τ r ) again by (a).In what follows, D ω : R → R denotes the rotation by ω .(c) Let r ∈ R , n (cid:62) , and s = cos π n D π n r . Then s ∈ R and ¯ µ ( r ) (cid:54) ¯ µ ( s ) .Proof of (c): From the fact that, for any x ∈ R , we have cos x = (1 + cos x ) ,we conclude that cos π n , sin π n ∈ R for all n . It follows that s ∈ R . Moreover, (cid:94) r s ¯0 = π . Hence the last assertion follows from (a).(d) lim n →∞ cos n πn = 1 .Proof of (d): By L’Hospital’s rule, we have lim x → πxx = − π lim x → tan πx = 0 .Hence lim n →∞ ln cos n πn = lim n →∞ n ln cos πn = 0 as well and the assertion follows.(e) Let r ∈ R such that (cid:107) r (cid:107) > . Then ¯ µ ( r ) (cid:54) ¯ µ ( − (cid:107) r (cid:107) r ) .Proof of (e): Because of (d), we may choose an m large enough such that cos m π m > (cid:107) r (cid:107) . Let ω = π m and define a sequence r ( i ) ∈ R , i = 0 , . . . , m , as follows: r (0) = r, r ( i +1) = cos ω · D ω r ( i ) for (cid:54) i < m . By (c), r ( i ) ∈ R for all i , and ¯ µ ( r ) = ¯ µ ( r (0) ) (cid:54) ¯ µ ( r (1) ) (cid:54) . . . (cid:54) ¯ µ ( r (2 m ) ) . Moreover, r (2 m ) = − cos m ω · r and − cos m ω < − (cid:107) r (cid:107) . Hence, by (b), it follows ¯ µ ( r (2 m ) ) (cid:54) ¯ µ ( − (cid:107) r (cid:107) r ) .(f) Let r ∈ R be such that (cid:107) r (cid:107) > . Then ¯ µ ( r ) = 0 .Proof of (f): Assume that ¯ µ ( r ) = 1 . By (e), ¯ µ ( r ) (cid:54) ¯ µ ( − (cid:107) r (cid:107) r ) , hence ¯ µ ( − (cid:107) r (cid:107) r ) = 1 .But ιr and ι ( − (cid:107) r (cid:107) r ) are perpendicular, a contradiction.(g) There are r, s, t ∈ R such that ιr, ιs, ιt are mutually orthogonal and (cid:107) r (cid:107) , (cid:107) s (cid:107) , (cid:107) t (cid:107) > .Proof of (g): Consider (2 , , ( − , , and ( − , − ) .Whereas Theorem 6.7 might in the present context be of limited applicability, itsfollowing corollary is more useful. Lemma 6.8.
Let R be a subfield of the Baer ordered (cid:63) -field K . Assume moreover that R is isomorphic to a Euclidean subfield of R . Then K , endowed with the standardinner product, does not possess a two-valued measure.Proof. The inclusion map R → K induces a map P ( R ) → P ( K ) , which isinjective and orthogonality-preserving. Hence, if there is an orthogonality-preservingmap from P ( K ) to , there is an orthogonality-preserving map from P ( R ) to .Assume that there is a two-valued measure on K . Then it follows that there is atwo-valued measure on R . This in turn is impossible by Theorem 6.7.Let R be a subfield of an ordered field S . Equipped with the inherited order, R is anordered field again. We note that we may speak about the infinitesimal and the medialelements of R without the need to specify whether we refer to R or S . Indeed, we23ave that M R = M S ∩ R and I R = I S ∩ R because R contains the rational subfieldof S . Lemma 6.9.
Let S be a Euclidean field and let R be an Archimedean subfield of S .Then any quadratic extension of R is Archimedean as well.Proof. Let γ ∈ R + not possess a square root in R . We have to show that R ( √ γ ) isArchimedean. Since R is Archimedean, we have that R ⊆ M S ∪ { } . In particular, γ ∈ M S and this implies that also √ γ ∈ M S .Furthermore, R ( √ γ ) = { α + β √ γ : α, β ∈ R } . Assume that α, β ∈ R are suchthat α + β √ γ is a non-zero infinitesimal element. Note that then β (cid:54) = 0 . Since α − β √ γ ∈ F S , it follows that also α − β γ = ( α − β √ γ )( α + β √ γ ) is infinitesimal.But the only infinitesimal element of R is , hence γ = ( αβ ) , a contradiction. Weconclude that R ( √ γ ) ⊆ M S ∪ { } and it follows that R ( √ γ ) is Archimedean. Lemma 6.10.
Let S be a Euclidean field. Then there is a smallest Euclidean subfield R of S . Moreover, R is isomorphic to an ordered subfield of R .Proof. Let Q be the rational subfield of S and let R be the Euclidean closure of Q , that is, the smallest subfield of S containing Q and such that α ∈ R + implies √ α ∈ R . Any Euclidean subfield of S contains Q and hence R , hence R is thesmallest Euclidean subfield of S . Q is isomorphic to Q and since Q can be ordered in only one way, Q is in fact anordered subfield of S that is order-isomorphic to Q equipped with its natural order.We conclude that Q is Archimedean.Furthermore, the formation of the Euclidean closure of Q is the result of a doubleinductive process, each step being a quadratic extension; cf., e.g., [Lam, Proposi-tion 2.12]. Since by Lemma 6.9 the Archimedean property is preserved in each stepand since the union of Archimedean subfields is Archimedean again, we concludethat also R is Archimedean and hence an ordered subfield of R .Let K be a Baer ordered (cid:63) -field. Then its fixed field S K is endowed with a linearorder (cid:54) with the effect that ( S K ; + , , (cid:54) ) is a totally ordered group and, for any α, β ∈ S K , α (cid:62) implies αβ (cid:62) . We will say that the fixed field of K is Euclideanif, in S K , any positive element is a square. Clearly, S K is in this case actually anordered field and this ordered field is Euclidean. Lemma 6.11.
Let K be a Baer ordered (cid:63) -field whose fixed field is Euclidean. Then K , endowed with the standard inner product, does not possess a two-valued meas-ure.Proof. By Lemma 6.10, S K possesses a subfield that is isomorphic to a Euclideansubfield of the reals. Hence the assertion follows from Lemma 6.8.We arrive at our main result. We denote by E OS the full subcategory of
LOS , andhence of
N OS , consisting of orthogonality spaces that arise from (finite-dimensional)positive definite Hermitian spaces over Baer ordered (cid:63) -field whose fixed field is Euc-lidean. 24 heorem 6.12.
Let H be a positive definite Hermitian space of finite dimension (cid:62) over a Baer ordered (cid:63) -field with a Euclidean fixed field. Let H (cid:48) be a furtherfinite-dimensional Hermitian space. Then any morphism in LOS between P ( H ) and P ( H (cid:48) ) is induced by a generalised semiunitary map.In particular, any morphism in E OS between orthogonality spaces of rank (cid:62) isinduced by a generalised semiunitary map.Proof. We verify the two conditions in Theorem 6.6.Let K be the scalar (cid:63) -field of H . Since the fixed field of K is Euclidean, condition (1)is fulfilled. Moreover, by Lemma 6.11, K does not possess two-valued measures.Hence also condition (2) holds. The objective of this paper has been to establish a categorical framework for ortho-gonality spaces. The latter structures can be identified with undirected graphs and inthe context of graph theory, categories have already been studied, e.g., in [Faw]. How-ever, the categories discussed by the graph theorists have turned out to be unsuitablein the present context. Our primary example originates from quantum physics andhence our intention has been to introduce a category whose morphisms, when appliedto linear orthogonality spaces, come close to linear mappings. We have thereforeintroduced normal orthogonality spaces, which are still more general than linear or-thogonality spaces. But normality suggests a definition of morphisms such that, whenapplied in the context of inner-product spaces, not only the orthogonality relation istaken into account but also the linear structure.We believe that the presented work is a first step into an area that offers numerousissues for further investigations. For instance, we have shown that the morphismbetween specific Hermitian spaces can be represented by generalised semiunitarymaps. It has remained open whether a similar statement is possible for a broaderclass. In fact, whereas generalised semilinear maps have been studied by several au-thors, there does not seem to exist any detailed account on maps also preserving aninner product. Moreover, we have seen that the existence of two-valued measuresplays a role in the discussion. This question as well as Gleason’s Theorem have beenstudied, with some exceptions [Dvu], in the context of classical fields, whereas thepresent context suggests to take into account further non-classical fields.To mention finally a particularly interesting issue, recall that the lattice-theoretic ap-proach has often been criticised for its inability to deal appropriately with commonconstructions of Hilbert spaces, like direct sums and tensor products. In the frame-work of orthogonality spaces, the situation is much different and a categorical frame-work might be useful for these matters.
Acknowledgement.
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