UFDC Home  Search all Groups  UF Institutional Repository  UF Institutional Repository  UF Theses & Dissertations  Vendor Digitized Files   Help 
Material Information
Subjects
Notes
Record Information

Full Text 
AN ALGEBRAIC CHARACTERIZATION OF MINKOWSKI SPACE By RICHARD K. WHITE A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA 2001 ACKNOWLEDGMENTS I thank my advisor Dr. Stephen J. Summers for his guidance in the preparation of this dissertation. I would also like to thank all of my committee members for their support and for serving on my committee. TABLE OF CONTENTS page ACKNOW LEDGM ENTS........................................................................................ ii ABSTRACT .......................................................................................................... v CHAPTERS 1 INTRODUCTION............................................................................................. 1 2 A CONSTRUCTION OF THREEDIMENSIONAL MINKOWSKI SPACE .... 7 2.1 Preliminaries ............................................................................................ 9 2.2 Construction of 1 ...................................................................................... 11 2.3 Reflections About Exterior Points ............................................................ 18 2.4 Embedding a Hyperbolic ProjectiveM etric Plane .................................... 22 2.5 Exterior Point Reflections Generate Motions in an Affine Space ............. 23 2.6 Conclusion ................................................................................................ 24 3 A CONSTRUCTION OF FOURDIMENSIONAL MINKOWSKI SPACE ...... 26 3.1 Preliminaries and General Theorems ....................................................... 26 3.1.1 Properties of M ............................................................................ 30 3.1.2 Properties of P ............................................................................. 31 3.1.3 Properties of X ............................................................................... 32 3.1.4 General Consequences of the Axioms ............................................. 32 3.1.5 Perpendicular Plane Theorems ...................................................... 33 3.1.6 Parallel Planes ............................................................................... 34 3.1.7 Consequences of Axiom 11 and 3.1.6.18 ......................................... 37 3.2 Lines and Planes ....................................................................................... 38 3.2.1 General Theorems and Definitions .................................................. 38 3.2.2 Isotropic Lines ............................................................................... 41 3.3 A Reduction to Two Dimensions .............................................................. 44 3.4 Consequences of Section 3.3 .................................................................... 51 3.5 Construction of the Field .......................................................................... 52 3.6 Dilations and the Construction of ( ,,V,K) ............................................. 58 3.7 Subspaces and Dimensions ...................................................................... 66 3.8 Orthogonality ........................................................................................... 69 3.9 The Polarity ........................................................................................... 80 3.10 Spacelike Planes and Their Reflections ................................................... 86 iii page 4 AN EXAMPLE OF THE THREEDIMIENSIONAL MODEL .......................... 91 5 C O N C L U SIO N ................................................................................................ 96 R E F E R E N C E S ..................................................................................................... 103 BIOGRAPHICAL SKETCH ................................................................................. 106 Abstract of Dissertation Presented to the Graduate School of the University of Florida in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy AN ALGEBRAIC CHARACTERIZATION OF MINKOWSKI SPACE By Richard K. White May 2001 Chairman: Dr. Stephen J. Summers Major Department: Mathematics We give an algebraic characterization of threedimensional and four dimensional Minkowski space. We construct both spaces from a set of involution elements and the group it generates. We then identify the elements of the original generating set with spacelike lines and their corresponding reflections in the three dimensional case and with spacelike planes and their corresponding reflections in the fourdimensional case. Further, we explore the relationship between these characterizations and the condition of geometric modular action in algebraic quantum field theory. CHAPTER 1 INTRODUCTION The research program which this dissertation is a part of started with a paper by Detlev Buchholz and Stephen J. Summers entitled "An Algebraic Character ization of Vacuum States in Minkowski Space" [9]. In 1975, Bisognano and Wichmann 15] showed that for quantum theories satisfying the Wightman axioms the modular objects associated by TomitaTakesaki theory to the vacuum state and local algebras generated by field operators with support in wedgelike spacetime regions in Minkowski space have geometrical meaning. Motivated by this work, Buchholz and Summers gave an algebraic characterization of vacuum states on nets of C*algebras over Minkowski space and reconstructed the spacetime translations with the help of the modular structures associated with such states. Their result suggested that a "condition of geometric modular ;, i iii" might hold in quantum field theories on a wider class of spacetime manifolds. To explain the abstract version of this condition, first some notation is introduced and the basic setup is given. Let {A, ,},1 be a collection of C*algebras labeled by the elements of some index set I such that (Q,<) is an ordered set and the property of isotony holds. Tli it is, if i0,i2 e I such that il < i2, then A, c A,2. Let A be a C*algebra containing {AI}iE/. It is also required that the assignment i 'f A, is an orderpreserving bijection. In algebraic quantum field theory, the index set I is usually a collection of open subsets of an appropriate spacetime (M,g). In such a case, the algebra A, is interpreted as the C*algebra generated by all the observables measured in a 2 spacetime region i. Hence, to different spacetime regions should correspond different algebras. Given a state co on A, let (7R., 7tr, Q) be the corresponding GNS representation and let Ri, = no)(Ai)", i e 1, be the von Neumann algebras generated by the nr.,(A), i E L. Assume that the map i '> 7Zi is an orderpreserving bijection and that the GNS vector Q is cyclic and separating for each algebra 71i, i e I. From TomitaTakesaki theory, we thus have a collection {J}i*e of modular involutions and a collection {Ai}iEJ of modular operators directly derivable from the state and the algebras. Each Ji is an antilinear involution on RH. such that JiTiJ, = 7.i' and Ji, = Q. In addition, the set {Ji}i i generates a group J which becomes a topological group in the strong operator topology on B3(7c), the set of all bounded operators on 7(H. The modular operators {Ai}, I are positive (unbounded) invertible operators such that AJ'RTjAi" = j, j e I, t e R, i= 7 and A"iQ = Q. In algebraic quantum field theory the state o models the preparations in the laboratory and the algebras Ai model the observables in the laboratory and are therefore, viewed as idealizations of operationally determined quantities. Since TomitaTakesaki theory uniquely gives these modular objects corresponding to (7Rj,Q), it thus follows that these modular objects can be viewed as operationally determined. Motivated by the earlier work of Bisognano and Wichmann [5], Buchholz and Summers [9] proposed that physically interesting states could be selected by looking at those states which satisfied the condition of geometric modular action, CGMA. Given the structures indicated above, the pair ({A,}ii,o) satisfies the abstract version of the CGMA if {T4}iI is left invariant under the adjoint action of the 3 modular conjugations {Ji} /,; that is, if for every i,j in I there is a k in I such that adJi(Rj) Ji jJi = R"k, where JUjJi = {JAJi : A e Rj}. Thus, for each i in I, there is an orderpreserving bijection, automorphism, ti on I, (I,<), such that JiR7jJi = T,(/), forj e L.The set {T,}il, is a set of involutions which generate a group T, which is a subgroup of the group of translations on L Buchholz, Dreyer, Florig, and Summers [6] have shown that the groups Tarising in this manner satisfy certain structure properties, but for the purposes of this thesis, it is only emphasized that Tis generated by involutions and is hence, a Coxeter group. Thus there are two groups generated by involutions operating on two different levels. 1. The group Tacting on the index set L. 2. The group Jacting on the set {'1"}ie. To elaborate further the relation between the groups Tand J, consider the following. Proposition 1.1 [6] The surjective map : j+T given by (Ji ...Jim) = t i "i,,, is a group homomorphism. Its kernel S lies in the center of 7J and the adjoint action of S leaves each 1i fixed. U Thus, Jis a central extension of the group Tby S. As an immediate consequence of this proposition, J provides a projective representation of Twith coefficients in an abelian group Z in the center of J. Thus, the condition of geometric modular action induces a transformation group on the index set I and provides it with a projective representation. With this in mind, the following program was then posed. Given the operational data available from algebraic quantum field theory, can one determine the spacetime symmetries, the dimension of the spacetime, and the spacetime itself? That is to say, given a net of C*algebras and a state co satisfying the CGMA, can 4 one determine the spacetime symmetries, the dimension of the spacetime, and even the spacetime itself? Part of this has been carried out by Buchholz, Dreyer, Florig, and Summers for Minkowski space and de Sitter space [6]. However, in order to do so, they had to presume the respective spacetime as a topological manifold. But would it not be possible to completely derive the spacetime from the operationally given data without any assumption about dimension or topology? As was pointed out by Dr. Summers, a possibility to do so was opened up in this program in the following manner. As already seen, the CGMA yields an involution generated group complete with a projective representation and there is in the literature a way of deriving spacetimes from such groups going under the name of absolute geometry. In general, absolute geometry refers to a geometry that includes both Euclidean and nonEuclidean geometry as special cases. Thus, one has a system of axioms not yet implying any decision about parallelism. In our case, the axioms are given in terms of a group of motions as an extension of Klein's Erlangen Program. A group of motions is defined as a set g of involution elements closed under conjugation and the group 5 it generates. In a group of motions the representations of geometric objects and relations depend only on the given multiplication for the group elements, without reference to any additional structure. The system of axioms is formulated in terms of the involutory generators alone, so that geometric concepts like point, line, and incidence no longer are primary but are derived. The necessary means for setting up this representation are provided by the totality of reflections in points, lines, and planes (a subset of the set of motions). Points, lines, and planes are in onetoone correspondence with the reflections in them so that geometric relations among points, lines, and planes correspond to grouptheoretic equations among the reflections. This enables one to be able to 5 formulate geometric theorems about elements of the group of motions and to be able to then prove these theorems by grouptheoretic calculation. To summarize, we are to find conditions on an index set I and a corresponding net of C*algebras {4J,}i;e as well as a state o satisfying the CGMA such that the elements of I can be naturally identified with open sets of Minkowski space and such that the group T"is implemented by the Poincare group on this Minkowski space. Out of the group Twe wish to construct Minkowski space such that T's natural action on Minkowski space is that of the Poincare group. This involves two steps. First we carry out the absolute geometry program for three and fourdimensional Minkowski space. That is, characterize three and fourdimensional Minkowski space in terms of a group of motions (G, 0). Second, we must determine what additional structure on the ordered set I would yield from TomitaTakesaki theory algebraic relations among the J1 (and hence, among the ti) which coincide with the algebraic characterization found in step one. The organization of the thesis is as follows: in Chapter 2, the given pair (G, 0) is used to construct a threedimensional Minkowski space out of the plane at infinity. Then identification of the involutory elements of 9 with spacelike lines and their group action in 5 with reflections about spacelike lines is made. In Chapter 3 using the same initial data, (Q, 0), as was given in Chapter 2 but satisfying different axioms, a fourdimensional Minkowski space is constructed. The approach taken here differs from that taken in Chapter 2. This time the affine space is constructed first and then the hyperplane at infinity is used to obtain the metric. The identification of the elements of g with spacelike planes and their group action in 5 with reflections about spacelike planes is made. In Chapter 4 a concrete example of the threedimensional characterization is given. As already mentioned, Bisognano and Wichmann showed that for quantum field theories satisfying the Wightman axioms the modular objects associated by 6 TomitaTakesaki theory to the vacuum state and local algebras in wedgelike regions in threedimensional Minkowski space have geometrical interpretation [5]. In particular, the modular conjugations, {Ji}ie, act as reflections about spacelike lines. In this chapter it is shown that if one chooses the set of wedgelike regions as the index set I, the group J generated by the set {Jri}i satisfies the axiom system given in Chapter 2 for the construction of a threedimensional Minkowski space. In Chapter 5 some concluding remarks about the second step described above are made. It is noted that if one assumes the modular stability condition [6] and the halfsided modular inclusion relations given by Wiesbrock [29], then one does obtain a unitary representation of the 2+1dimensional Poincare group. CHAPTER 2 A CONSTRUCTION OF THREEDIMENSIONAL MINKOWSKI SPACE In this chapter we give an absolute geometric, that is, an algebraic, characterization of threedimensional Minkowski space. This chapter is a version of a preprint by the author entitled "A GroupTheoretic Construction Of Minkowski 3Space Out Of The Plane At Infinity" [28]. Along with the wellknown mathematical motivations [1, 2] there are also physical motivations, as we discussed in Chapter 1. Threedimensional Minkowski space is an affine space whose plane at infinity is a hyperbolic projectivemetric plane [12]. In "Absolute Geometry" [2], Bachmann, Pejas, Wolff, and Bauer (BPWB) took an abstract group 0 generated by an invariant system g of generators in which each of the generators was involutory, satisfying a set of axioms and constructed a hyperbolic projectivemetric plane in which the given group 5 was isomorphic to a subgroup of the group of congruent transformations (motions) of the projectivemetric plane. By interpreting the elements of G as line reflections in a hyperbolic plane, BPWB showed that the hyperbolic projectivemetric plane could be generated by these line reflections in such a way that these line reflections form a subgroup of the motions group of the projectivemetric plane. Coxeter showed in [13] that every motion of the hyperbolic plane is generated by a suitable product of orthogonal line reflections, where an orthogonal line reflection is defined as a harmonic homology with center exterior point and axis the given ordinary line and where the center and axis are a polepolar pair. Here we show that Coxeter's and BPWB's notions of motions coincide in the hyperbolic 8 projectivemetric plane and that the motions can be viewed as reflections about exterior points. Next we embed our projectivemetric plane into a threedimensional projective space. By singling out our original plane as the plane at infinity, we obtain an affine space whose plane at infinity is a hyperbolic projectivemetric plane, threedimensional Minkowski space. Finally, we show that the motions of our original plane induce motions in the affine space and, by a suitable identification, we show that any motion in Minkowski space can be generated by reflections about spacelike lines. Thus, to construct a threedimensional Minkowski space, one can start with a generating set g of reflections about spacelike lines in the plane at infinity. So g may be viewed as a set of reflections about exterior points in a hyperbolic projectivemetric plane. Out of the plane at infinity, one can obtain a threedimensional affine space with the Minkowski metric, which is constructed from a group generated by a set of even isometries or rotations. The approach in this chapter differs from the method used by Wolff [301 for twodimensional Minkowski space and by Klotzek and Ottenburg [19] for fourdimensional Minkowski space. The approach in these papers is to begin by constructing the affine space first. For Wolff's [30] twodimensional case, the elements of the generating set g are identified with line reflections in an affine plane. For Klotzek and Ottenburg's [19] fourdimensional case, the elements of the generating set g are identified with reflections about hyperplanes in an affine space. Thus, in each of these papers, the generating set g is identified with a set of symmetries or odd isometries. A map of affine subspaces is then obtained using the definition of orthogonality given by commuting generators. This map induces a hyperbolic polarity in the hyperplane at infinity, yielding the Minkowski metric. To briefly recap the two approaches described above, note that both approaches, ours and the one given by Klotzek and Ottenburg [19] and by Wolff 9 [30], start with a generating set g of involution elements. In our approach, one can identify the elements of g with a set of even isometries (rotations) and use the definition of orthogonality induced by the commutation relations of the generators in the hyperplane at infinity to obtain the polarity and then embed this in an affine space to get Minkowski space. In the approach of Klotzek and Ottenburg [19] and Wolff [30], one can identify the elements of G with a set of odd isometries symmetriess), construct an affine space first, and then use the definition of orthogonality induced by the commutation relations of the generators in the affine space to obtain a polarity in the hyperplane at infinity to get Minkowski space. 2.1 Preliminaries The starting point for an algebraic characterization of Minkowski space is therefore far from unique. Our particular choice of algebraic characterization, in terms of reflections about spacelike lines in three dimensional Minkowki space, is motivated by physical considerations [6] which we briefly explain in the conclusion. A hyperbolic projectivemetric plane is a projective plane in which a hyperbolic polarity is singled out and used to define orthogonality in the plane. A polarity is an involutory projective correlation. A correlation is a onetoone mapping of the set of points of the projective plane onto the set of lines, and of the set of lines onto the set of points such that incidence is preserved. A projective correlation is a correlation that transforms the points Y on a line b into the lines y' through the corresponding point B'. So, in general, a correlation maps each point A of the plane into a line a of the plane and maps this line into a new point A'. When the correlation is involutory, A' always coincides with A. Thus a polarity relates A to a, and vice versa. A is called the pole of a and a is called the polar of A. Since this is a projective correlation, the polars of all the points on a form a projectively related pencil of lines through A. 10 The polarity dualizes incidences: if A lies on b, then the polar of A, a, contains the pole of b,B. In this case we say that A and B are conjugate points, and that a and b are conjugate lines. If A and a are incident, then A and a are said to be selfconjugate: A on its own polar and a through its own pole [141. A hyperbolic polarity is a polarity which admits selfconjugate points and selfconjugate lines. The set of all selfconjugate points is called a conic, which we shall call the absolute. In a projective plane in which the theorem of Pappas and the axiom of Fano hold, the polarity can be used to introduce a metric into the plane. Orthogonality is defined as follows: two lines (or two points or a line and a point) are said to be orthogonal or perpendicular to each other if they are conjugate with respect to the polarity. Congruent transformations of the plane are those collineations of the plane which preserve the absolute; that is, those collineations which leave the absolute invariant. In a projective plane with a hyperbolic polarity as absolute, the group of all collineations in the plane leaving the absolute invariant is called the hyperbolic metric group and the corresponding geometry is called the hyperbolic metric geometry in the plane 1311. The conic or absolute, separates the points of the projective plane into three disjoint classes: ordinary or interior points, points on the absolute, and exterior points. The lines of the projective plane are likewise separated into three disjoint classes. Secant lines, lines that contain interior points, exterior points, and precisely two points on the absolute. Exterior lines, lines that contain only exterior points. And tangent lines, lines that meet the absolute in precisely one point and in which every other point is an exterior point. Definition 2.1.1. Two lines containing ordinary points, two secant lines, are said to be parallel if they have a point of the absolute in common. Remark. The set of all interior points and the set of lines formed by intersecting 11 secant lines with the set of interior points, ordinary lines, is classical hyperbolic plane geometry. 2.2 Construction of I In this section we list the axioms and main results of BPWB [2] and provide a sketch of some of the arguments they used which are pertinent to this work. For detailed proofs, one is referred to the work of BPWB [2]. Definition 2.2.1. A set of elements of a group is said to be an invariant system if it is mapped into itself (and thus onto itself) by every conjugation by an element of the group. An element a of a group Q5 is called an involution if a2 = 1I, where 1e is the identity element of the group Q5. Basic assumption: A given group Q5 is generated by an invariant system g of involution elements. The elements of g are denoted by lowercase Latin letters. Those involutory elements of !5 that can be represented as ab, where a,b E= are denoted by uppercase Latin letters. If ,i e 5 and ril is an involution, we denote this by 4r. Axioms Axiom 1: For every P and Q there is a g with P, Q 1g. Axiom 2: IfP,Q ig,h then P = Q or g = h. Axiom 3: If a,b,c \P then abc = d e g. Axiom 4: Ifa,b,c Ig then abc = d e 9. Axiom 5: There exist g,h,j such that g Ih butj I g,h,gh. Axiom 6: There exist elements d,a,b e G such that d,a,b I P,c for P,c e 0. (There exist lines which have neither a line nor a point in common.) Axiom 7: For each P and for each g there exist at most two elements h,j 6 9 such that PIh,j but g,h I A,c and g,j B,d for any A,B,c,d E 0 (that is, have neither a point nor a line in common). Axiom 8: One never has P = g. We call the set of axioms just given axiom system A, denoted by as A. The initial interpretation of the elements of g is as secant or ordinary lines in a hyperbolic plane for BPWB [2]. In our approach, we view the elements of g initially as exterior points in a hyperbolic plane. After embedding our hyperbolic projectivemetric plane into an affine space, we can identify the elements of g, our generating set, with spacelike lines and their corresponding reflections in a threedimensional Minkowski space. By realizing that statements about the geometry of the plane at infinity correspond to statements about the geometry of the whole space where all lines and all planes are considered through a point, we see that the axioms also are statements about spacelike lines, the elements of g, and timelike lines, the elements P of 0, through any point in threedimensional Minkowski space. The models of the system of axioms are called groups of motions; that is, a group of motions is a pair (5, G) consisting of a group Q5 and a system g of generators of the group Q5 satisfying the basic assumption and the axioms. To give a precise form to the geometric language used here to describe grouptheoretic concepts occurring in the system of axioms, we associate with the group of motions (0,G9) the group plane (50,9), described as follows. The elements of Q are called lines of the group plane, and those involutory group elements that can be represented as the product of two elements of g are called points of the group plane. Two lines g and h of the group plane are said to be perpendicular if g 1h. Thus, the points are those elements of the group that can be written as the product of two perpendicular lines. A point P is incident with a line g in the group plane if P 1g. Two lines are said to be parallel if they satisfy Axiom 6. Thus, if P # Q, then by Axiomi and Axiom 2, the points P and Q in the group plane are joined by a unique line. If P I g then Axiom 7 says that there are at most two lines through P parallel to g. 13 Lemma 2.2.2.[21 For each a e 05, the mappings a : g * ga = aga and aa : P * Pp' aPa arc onetoone mappings of the set of lines and the set of points, each onto itself in the group plane. Proof: Let a E 0, and consider the mapping y = aya of 0 onto itself. It is easily seen that this mapping is bijective. Because G is an invariant system (ab e g for every a, b E 9) 9 is mapped onto itself, and if P is a point, so that P = gh with glh, then Pa = gaha and gajha, so that P1 is also a point. Thus, g ga, P  Pa are onetoone mappings of the set of lines and the set of points, each onto itself in the group plane. U Definition 2.2.3. A onetoone mapping a of the set of points and the set of lines each onto itself is called an orthogonal collineation if it preserves incidence and orthogonality. Since the "I" relation is preserved under the above mappings, the above mappings preserve incidence and orthogonality as defined above. Corollary 2.2.4.121 The mappings a : g * ga andaa : P P are orthogonal collineations of the group plane and are called motions of the group plane induced by a. In particular, if ac is a line a we have a reflection about the line a in the group plane, and if ac is a point A, we have a point reflection about A in the group plane. If to every a E S one assigns the motion of the group plane induced by a, one obtains a homomorphism of 5 onto the group of motions of the group plane. Bachmann [1] showed that this homomorphism is in fact an isomorphism so that points and lines in the group plane may be identified with their respective reflections. Thus, 0 is seen to be the group of orthogonal collineations of S generated by G. 14 Definition 2.2.5. Planes that are representable as an isomorphic image, with respect to incidence and orthogonality, of the group plane of a group of motions (O, g), are called metric planes. BPWB showed how one can embed a metric plane into a projectivemetric plane by constructing an ideal plane using pencils of lines [2]. We shall now outline how this is done. Definition 2.2.6. Three lines are said to lie in a pencil if their product is a line; that is, a,b,c lie in a pencil if abc = d e g. (*) Definition 2.2.7. Given two lines a,b with a # b, the set of lines satisfying (*) is called a pencil of lines and is denoted by G(ab), since it depends only on the product ab. Note that the relation (*) is symmetric, that is, it is independent of the order in which the three lines are taken, since cba = (abc)1 is a line, the invariance of g implies that cab = (abc)c is a line and that every motion of the group plane takes triples of lines lying in a pencil into triples in a pencil. The invariance of 9 also shows that (*) holds whenever at least two of the three lines coincide. Using the given axioms, BPWB 12] showed that there are three distinct classes of pencils. If a, b V then G(ab) = {c : c IV}. In this case, G(ab) is called a pencil of lines with center V and is denoted by G(V). If a,b Ic then G(ab) = {d : d 1c}. In this case, G(ab) is called a pencil of lines with axis c and is denoted by G(c). By Axiom 6, there exist lines a, b, c which do not have a common point or a common line. Recall that lines of this type are called parallel. Thus, in this case G(ab) = {c : c 11 a,b where a 11 b}, which we denote by Px. Using the above definitions of pencils of lines and the above theorems, BPWP [2] proved that an ideal projective plane, H, is constructed in the following way. An 15 ideal point is any pencil of lines G(ab) of the metric plane. The pencils G(P) correspond in a onetoone way to the points of the metric plane. An ideal line is a certain set of ideal points. There are three types: 1. A proper ideal line g(a), is the set of ideal points that have in common a line a of the metric plane. 2. The set of pencils G(x) with x IP for a fixed point P of the metric plane, which we denote by P. 3. Each set of ideal points that can be transformed by a halfrotation about a fixed point P of the metric plane into a proper ideal line, which we denote by po. The polarity is defined by the mappings G(QC) C and C G(C)Q; POO po and po > Po; G(c) * g(c) and g(c) ' G(c). Bachmann [1] showed that the resulting ideal plane is a hyperbolic projective plane in which the theorem of Pappus and the Fano axiom hold; that is, it is a hyperbolic projectivemetric plane. In this model, the ideal points of the form G(P) are the interior points of the hyperbolic projectivemetric plane. Thus the points of the metric plane correspond in a onetoone way with the interior points of the hyperbolic projectivemetric plane. The ideal points G(x), for x e g are the exterior points of the hyperbolic projectivemetric plane. Theorem 2.2.8. Each x e g corresponds in a onetoone way with the exterior points of the hyperbolic projectivemetric plane. Proof: Because each line d of the metric plane is incident with at least three points and a point is of the form ab with a I b, then each x e g is the axis of a pencil. From the uniqueness of perpendiculars each x e g corresponds in a onetoone way with the pencils G(x). Hence, each x e g corresponds in a onetoone way with the exterior points of the hyperbolic projectivemetric plane. U 16 Thus, the axioms can be viewed as axioms concerning the interior and exterior points of a hyperbolic projectivemetric plane. The ideal points of the form G(ab) where a I1 b are the points on the absolute, that is, the points at infinity in the hyperbolic projectivemetric plane. Now consider the ideal lines. A proper ideal line g(a) is a set of ideal points that have in common a line a of the metric plane. Theorem 2.2.9. A proper ideal line g(a) is a secant line of the form g(a) = {P,x,G(bc) : x, PIa and abc E Q where b 1 c}. Proof: Every two pencils of lines of the metric plane has at most one line in common. By Axiom 7, each line belongs to at most two pencils of parallels and each line g e g belongs to precisely two such pencils. Thus, a proper ideal line contains two points on the absolute, interior points, and exterior points; that is, a proper ideal line is a secant line. If we identify the points P with the pencils G(P) and the lines x with the pencils G(x), then a secant line is the set g(c) = {P,x,G(ab) : x,P Ic and abc E G where a 11 b}. U Corollary 2.2.10. The ideal line which consists of pencils G(x) with x IP for a fixed point P of the metric plane consists of only exterior points; that is, it is an exterior line. Under the identification ofx with G(x) then P = {x E g : x\P}. The last type of ideal line is a tangent line. It contains only one point, G(ab) = P., on the absolute. Denoting this line by po, then p.o = {G(ab)} u {x e g : abx e g}, where a 11 b. Recalling that each x e Q corresponds to an exterior point in the hyperbolic projectivemetric plane, we see that a tangent line consists of one point on the absolute and every other point is an exterior point. Also note that under the above identifications, each secant line g(c) corresponds to a unique "exterior point", c, c 0 g(c) since one only considers those 17 x,P Ic such that xc # 1 and Pc # 1. Each exterior line corresponds to a unique interior point P and each tangent line corresponds to a unique point on the absolute. Theorem 2.2.11. The map D given by (i) (c) = g(c), D(g(c)) = c (ii) D(P) = D(,) = P (iii) D(poo) = P., I(Po) = po is a polarity. Proof: Let P be the set of all points of H and C the set of all lines of H. From the remarks above it follows that D is a welldefined onetoone pointtoline mapping of P onto C and a welldefined onetoone linetopoint mapping of C onto P. Next we show that D is a correlation and for this it suffices to show that (D preserves incidence. Let g(c) = {P,x,G(ab) : x,Plc and abc e 9 where a 11 b} be a secant line. Let A,B,d,P. e g(c), where P, = G(ej) = {x E 9 : xef e 9 and e IIf}.Then A,B,d I c and cab e 9. <(A) = A= {x : x I A}, D(B) = B, I(d) =g(d), (D(P) = p = {P,} u {x : efx E 9} D(g(c)) = c e A n B r) g(d) r p and t(g(c)) E D(A), 0(B), <(d), F(Poo). Hence, D preserves incidence on a secant line. Now consider an exterior line P = {x : xjP} and let a, b e P. Then a, bIP and it follows that P E g(a) r g(b); that is, (D(P) e D(a) r 1(b) and (D preserves incidence on an exterior line. Finally, let po = {G(ab)} u {x : abx e 9 where a 11 b} be a tangent line. Clearly, since 1(G(ab)) = p., then P. = G(ab) e p.. Now suppose that d e po. Then abd e 9 and D(d) = g(d)= {A,x,G(ej) : A,xld and def e 9 where e 1}. Thus, d e G(ab) r G(eJ) and Po, E g(d). This implies that d e po and O(po) e D(d). Hence, D preserves incidence and is a correlation. 18 Note also that from the work above, D transforms the points Y on a line b into the lines D(Y) through the point 0(b). Thus, ( is a projective correlation. Since O2 = 10, then (D is a polarity. Moreover, since D(p.) = Po with Po. e p., then (D is a hyperbolic polarity. E Theorem 2.2.12 The definition of orthogonality given by the polarity agrees with and is induced by the definition of orthogonality in the group plane. Proof: If we define perpendicularity with respect to our polarity then the following are true (we use the notation +> to denote the phrase "if and only if "): (i) g(c) 1_ g(a) <> 1(g(c)) = c e g(a) and '(g(a)) = a e g(c) * a c. (ii) g(c) _L P> (g(c)) =c E P c JP. (iii) C _L P O ((P) = P E g(c) = (D(c) ++ P c. (iv) C _L g(x) (g(x)) =xeg(c)= D(c) x Ic. Instead of interpreting our original generators as ordinary lines in a hyperbolic plane, we now interpret them as exterior points. We can construct a hyperbolic projectivemetric plane in which the theorem of Pappus and Fano's axiom hold, which is generated by the exterior points of the hyperbolic projectivemetric plane. With the identifications above and the geometric objects above, we show in the next section that the motions of the hyperbolic projectivemetric plane above can be generated by reflections about exterior points; that is, any transformation in the hyperbolic plane which leaves the absolute invariant can be generated by a suitable product of reflections about exterior points. 2.3 Reflections About Exterior Points Definition 2.3.1. A collineation is a onetoone map of the set of points onto the set of points and a onetoone map of the set of lines onto the set of lines that preserves the incidence relation. Definition 2.3.2. A perspective collineation is a collineation which leaves a line 19 pointwise fixed, its axis, and a point linewise fixed, its center. Definition 2.3.3. A homology is a perspective collineation with center a point B and axis a line b where B is not incident with b. Definition 2.3.4. A harmonic homology with center B and axis b, where B is not incident with b, is a homology which relates each point A in the plane to its harmonic conjugate with respect to the two points B and (b,[A,B]), where [A,B] is the line joining A and B and (b, [A,B]) is the point of intersection of b and [A,B]. Definition 2.3.5. A complete quadrangle is a figure consisting of four points (the vertices), no three of which are collinear, and of the six lines joining pairs of these points. If 1 is one of these lines, called a side, then it lies on two of the vertices, and the line joining the other two vertices is called the opposite side to 1. The intersection of two opposite sides is called a diagonal point. Definition 2.3.6. A point D is the harmonic conjugate of a point C with respect to points A and B if A and B are two vertices of a complete quadrangle, C is the diagonal point on the line joining A and B, and D is the point where the line joining the other two diagonal points cuts [A,B]. One denotes this relationship by H(AB, CD). Example 2.3.7. Let A,B, and C be three collinear points. For a quick construction of the harmonic conjugate D of C with respect to A and B let Q,R,S be any points such that [Q,R],[Q,S], and [R,S] pass through A,B,C respectively. Let {P} = [A,S] r) [B,R], then {D} = [A,B] n [P,Q] ([11]). Note that if [R,S] 11 [A,C] then D is the midpoint of A and B. Coxeter [13] showed that any congruent transformation of the hyperbolic plane is a collineation which preserves the absolute and that any such transformation is a product of reflections about ordinary lines in the hyperbolic plane where a line reflection about a line m is a harmonic homology with center M and axis m, where M and m are a polepolar pair and M is an exterior point. A point 20 reflection is defined similarly, a harmonic homology with center M and axis m, where M and m are a polepolar pair, M is an interior point, and m is an exterior line. Note that in both cases, M and m are nonincident. In keeping with the notation employed at the end of 2.2, let b be an exterior point and g(b) its pole. SA Ab and d ' db Lemma 2.3.8. The map Tb : A d i g(d)b is a BPoo b pb p p collineation. Proof: This follows from the earlier observation that the motions of the group plane map pencils onto pencils preserving the "I" relation. U Lemma 2.3.9. Tb is a perspective collineation and, hence, a homology. Proof: Recall that g(b) = {A,x,P. : x,A lb and where b lies in the pencil P.}. For any A and x in g(b) we have Ab = A and xb = x since A,x lb and if A',x' 0 g(b) then A',x' I b and Ab # A, xb # x, and Abxb I b. Thus, Abxb o g(b). Recall also that Po = G(cd), where c and d do not have a common perpendicular nor a common point and thus, G(cd) = {f : fcd e g}. Now g(b) is a secant line, so that it contains two such distinct points, Po and Q., say, on the absolute. Since the motions of the group plane map pencils onto pencils preserving the "I" relation it follows that if c,d e P. then cb, d e Po and hence pb = Po and Qb= Qo. Moreover, if R. o g(b) then it follows that Rb o g(b). Thus, Tb leaves g(b) pointwise invariant. Now let g(d), Q, and r. be a secant line, exterior line, and tangent line, respectively, containing b. For e E g(d) we have e Id and e' I db = d since b I d, thus eb e g(d). For A E g(d), Ab \db = d, so Ab e g(d). Similarly, it follows that if P., e g(d) then pb e g(d), and that g(d)b = g(d). One easily sees that Qb = Q and r = r.. Thus, Tb leaves every line through b invariant and Tb is a perspective collineation for each b e g. 0 Theorem 2.3.10. Tb is a harmonic homology. Proof: Since A b is again a point in the original group plane and since db is again a line in the original group plane and from the observations above, we have, for each b e G, Tb maps interior points to interior points, exterior points to exterior points, points on the absolute to points on the absolute, secant lines to secant lines, exterior lines to exterior lines, and tangent lines to tangent lines. Moreover, since (Zb)b = for any e 5, Tb is involutory for each b E g. Now in a projective plane in which the theorem of Pappus holds, the only collineations which are involutory are harmonic homologies [10], thus Tb is a harmonic homology for each b E g. N Theorem 2.3.11. Interior point reflections are generated by exterior point reflections. Proof: A similar argument shows that for each interior point A, TA is a harmonic homology with center A and axis A where A is the polar of A, A o A, and where TA is defined analogously to Tb. Thus, each TA is a point reflection and since A is the product of two exterior points, we see that point reflections about interior points are generated by reflections about exterior points. U Theorem 2.3.12. The reflection of an interior point about a secant line is the same as reflecting the interior point about an exterior point. Moreover, since any motion of the hyperbolic plane is a product of line reflections about secant lines, any motion of the hyperbolic plane is generated by reflections about exterior points. Proof: Consider a line reflection in the hyperbolic plane; that is, the harmonic homology with axis g(b) and center b. Let A be an interior point and g(d) a line through A meeting b. Since b e g(d) then b Id and g(d) is orthogonal to g(b). Let E be the point where g(b) meets g(d). Since E E g(b) then E lb and Eb =ffor some f e g. It follows that the reflection of A about g(b) is the same as the reflection of A about E. Since b Id and Eld then bd = C and we have E,C Ib,d with b # d. Thus, by Axiom 2, E = C = bd. Hence, AE = Adb = Ab. * 22 Theorem 2.3.13 Reflections of exterior points about exterior points and about exterior lines are also motions of the projectivemetric plane; that is, the +b 's for b e g acting on exterior points and exterior lines are motions of the hyperbolic projectivemetric plane. Proof: The motions of the projectivemetric plane are precisely those collineations which leave the absolute invariant. U We also point out that the proof that each Tb is an involutory homology also showed that the Fano axiom holds, since in a projective plane in which the Fano axiom does not hold no homology can be an involution [4]. 2.4 Embedding a Hyperbolic ProjectiveMetric Plane In this section we embed our hyperbolic projectivemetric plane into a three dimensional projective space, finally obtaining an affine space whose plane at infinity is isomorphic to our original projectivemetric plane. Any projective plane H in which the theorem of Pappus holds can be represented as the projective coordinate plane over a field )C. (The theorem of Pappus guarantees the commutivity of C.) Then by means of considering quadruples of elements of /C, one can define a projective space P3(APC) in which the coordinate plane corresponding to IH is included. If the Fano axiom holds, then the corresponding coordinate field )C is not of characteristic 2 [4]. By singling out the coordinate plane corresponding to H as the plane at infinity, one obtains an affine space whose plane at infinity is a hyperbolic projectivemetric plane: that is, threedimensional Minkowski space. To say that a plane H is a projective coordinate plane over a field AC means that each point of H is a triple of numbers (xo,xI,x2), not all x, = 0, together with all multiples (0xo,kxi, xx2), for X 0 and k s /C. Similarly, each line of H is a triple of numbers [uo,uI,u2], not all ui = 0, together with all multiples [kuo,Xu\,Xku], X # 0. In P3(/C), all the quadruples of numbers with the last entry zero correspond 23 to H. One can now obtain an affine space A by defining the points of A to be those of P3(kA) l; that is, those points whose last entry is nonzero; a line I of A to be a line I' in P3(kC) 1 minus the intersection point of the line 1' with I; and by defining a point P in A to be incident with a line 1 of A if, and only if, P is incident with the corresponding 1'. Planes of A are obtained in a similar way [14]. Thus, each point in H7 represents the set of all lines in A parallel to a given line, where lines and planes are said to be parallel if their first three coordinates are the same, and each line in I represents the set of all planes parallel to a given plane. Because parallel objects can be considered to intersect at infinity, we call H the plane at infinity. 2.5 Exterior Point Reflections Generate Motions in an Affine Space In this section we state and prove the main result of this chapter. Coxeter showed that threedimensional Minkowski space is an affine space whose plane at infinity is a hyperbolic projectivemetric plane t11]. He also classified the lines and planes of the affine space according to their sections by the plane at infinity as follows: Line or Plane Section at Infinity Timelike line Interior point Lightlike line Point on the absolute Spacelike line Exterior point Characteristic plane Tangent line Minkowski plane Secant line Spacelike plane Exterior line He also showed that if one starts with an affine space and introduces a hyperbolic polarity in the plane at infinity of the affine space, then the polarity induces a Minkowskian metric on the whole space. Under a hyperbolic polarity, a line and a plane or a plane and a plane are perpendicular if their elements at infinity 24 correspond. Two lines are said to be perpendicular if they intersect and their elements at infinity correspond under the polarity. Theorem 2.5.1. Every motion in a threedimensional affine space with a hyperbolic polarity defined on its plane at infinity, is generated by reflections about exterior points. Moreover, because exterior points correspond to spacelike lines, then any motion in threedimensional Minkowski space is generated by reflections about spacelike lines. Proof: Because any motion in threedimensional Minkowski space can be generated by a suitable product of plane reflections, it suffices to show that reflections about exterior points generate plane reflections. Let a be any Minkowski plane or spacelike plane. Let P be any point in Minkowski space. Let / be the line through P parallel to a. Let a. denote the section of a at infinity. Applying the polarity to cc. we get a point go _L cc.. Let g be a line through P whose section at infinity is goo, so that g is a line through P orthogonal to a. because each line in the plane at infinity contains at least 3 points, there exists a line / in a which is orthogonal to g as g. 1 ao0. Now let m be a line through P not in a which intersects 1. It follows that the reflection of P about a is the same as reflecting m about I and taking the intersection of the image of m under the reflection with g. By the construction of the affine space and the definition of orthogonality in the affine space it follows that 1 and m must act as their sections at infinity act. because any point reflection in the hyperbolic projectivemetric plane can be generated by reflections about exterior points, we have that the reflection of P about a is generated by reflections of P about spacelike lines. M 2.6 Conclusion As already indicated above, the geometric model for the generators of g which lies behind the choice of algebraic characterization of threedimensional Minkowski 25 space differs significantly from those of previous absolute geometric characterizations of Minkowski space. The model given here is the set of reflections about spacelike lines, which is not a choice which would be made a priori by other mathematicians. However, this is yet another example of a situation where the initial data are imposed by physical, as opposed to purely mathematical, considerations. In the next chapter starting with the same initial data, but satisfying different axioms, a construction of fourdimensional Minkowski space is given. First the affine space is constructed and then the hyperplane at infinity is used. Also given is an explicit construction of the field, the vector space, and the metric. CHAPTER 3 A CONSTRUCTION OF FOURDIMENSIONAL MINKOWSKI SPACE In this chapter a construction of fourdimensional Minkowski space will be given using the same initial data as in Chapter 2 but satisfying different axioms. The actual construction is quite long, so to aid the reader in following, A brief outline of the procedure shall be given here. First some general theorems and the basic definitions will be given in the first two sections. In Sections 3.3 and 3.4 attention is restricted to two dimensions in order to obtain the necessary machinery to construct the field. Once the field has been obtained, then a vector space is constructed and given a definition of orthogonality. It is then shown that the vector definition of orthogonality is induced by and agrees with the initial definition of perpendicularity for the geometry generated by the original set of involutions 9. To obtain a metric vector space from the constructed vector space V, a map n is defined on the subspaces of V. The map nt is defined to send a subspace to its orthogonal complement. Using the work of 131 (which is given for the convenience of the reader) the Minkowski metric is obtained and hence, Minkowski space. The last section of this chapter identifies the elements of 9 with spacelike planes in Minkowski space and the motions of the elements of Q with reflections about spacelike planes, as was required in Chapter 1. 3.1 Preliminaries and General Theorems Let there be given a nonempty set 9 of involution elements and the group S it generates, where for any a e 9 and for any 4 E S we have ca = 'a4 E 9. If the product of two distinct elements 41,42 e @ is an involution then we denote this by 27 writing t I2. We note that if 1 \12 then  for all ,1, 2 in 05 because Let M = {ap3: a I 13 and a,P13 e g} and P = 9 u M. We consider the elements of g as spacelike planes and the elements of M as Minkowskian or Lorentzian planes. Thus, P consists of the totality of "nonsingular" planes and we denote the elements of P by a, 13, y .... We begin by giving the basic assumption and by making some preliminary definitions. All the axioms are then listed for the convenience of the reader. The geometric meaning of the axioms and the symbols used will be made clear in the appropriate sections. The first four sections examine the incidence axioms. The order axioms are reintroduced in Section 3.5, where we construct and order the field to obtain a field isomorphic to the reals. In Section 3.6 we give the motivation behind the particular choice of dilation axioms. Using these axioms we are able to define a scalar multiplication and thereby obtain an affine vector space. The polarity axioms are given again and examined in Section 3.8, where we define orthogonal vectors. Basic assumption: If a e g and 13 E M then a e g and P E M for every E in g. For the following, let a, 3 e P with at 13. Definition 3.1. If a, 3 e g, then a13 e M by definition and we write a 1 P3 and we say a is perpendicular to or orthogonal to 13. Definition 3.2. Suppose that a e g, 13 e M. (i) If ap13 e G then we write a 13 and we say a is perpendicular to 13. (ii) If ap13 g, then we write a L 13 and we say a is absolutely perpendicular to 13. Definition 3.3. Let X = {a13 : a 13}. We call the elements of X points and we denote these elements by A, B, C,.... Definition 3.4. If a,P3 e M and a13 e M, then we write a 13 and we say a is perpendicular to 13. 28 Definition 3.5. The point A and the plane a are called incident when A I a. For each a e P, set X = {A : AIa}.se that A,BJal,a2 where A # B and ac # a2. Suppose that A,Bla\,a2; where A # B and a1 # a2. We define the line g containing A and B as g = gAB [a,aX2] = {C e X : CaiO,a2}. We say that g is the intersection of a(x and a2 (Xoa and 3O2), g c a 1,a2 (g c Xa IXa2). The point C is incident with g, C e g, if CIai,a2. If A # B are two points such that there exist a and P3 with A, BI a, 13 and a JL 3 then we say that A and B are joinable and we write A,B e gAB [a,3,aP3]. If g is a line which can be put into the form g = [a, P3, a3] where a 13 P then we say that g is nonisotropic. If A and B are two distinct points such that there do not exist a, 13 with A, B I a,f 3 and a 1 3 then we say that A and B are unjoinable. If g is a line which cannot be put into the form g = [a, 3, ap] with a 1 3 then we say that g is isotropic or null. Incidence Axioms Axiom 1. For each P, a there exists a unique P3 E P such that P 13 and a[3 = Q. Axiom 2. IfA,B la,[3,c andCla,P then Cls. Axiom 3. IfP, Q I a,3 P and a 1 P then P = Q. Axiom 4. Ifa, P3,y e g are distinct and a 1 (31 y 1 a then aP3y e 4. Axiom 5. If a,13 e M and a I3P then a3 e AM. Axiom 6. For all A,B; A # B, there exists a,P3 such that A,B Ia,3 anda a 13. Axiom 7. If a 13 then there exists A, B I a, P3 such that A # B. Axiom 8. For allMA,B,C; ABC = D e 3X. Axiom 9. If OI a, P3,y, 5 with P3,y,6 a then 3Py6 = c. 29 Axiom 10. IfA,B,C are pairwise unjoinable points and A,B a then CIa. Axiom 11. For alla e g, there exist distinct P,7y,8 E g such that a I P 1 y ac but 8, ca,P,y. Axiom 12. IfA,B Ia; a e G, then there exists P e G such that A,B I and P a. Axiom 13. For each Pl a, Oa e M, there are distinct points A, BIa such that P # A,B and P is unjoinable with both A and B, but A and B are joinable, and if CI a is unjoinable with P then C is unjoinable with A or with B or C = A or C = B. Axiom 14. If A and B arejoinable andA,BI a, then there exists I a such that A,B1 . Axiom 15. If a,P,y are distinct with P,7y a andA, BIa,,y; A B, then ac = Py and ifA, B I y; y 1 a then ac = Py or P = y. Order Axioms Axiom F. (Formally Real Axiom) [21] Let O,E I a, a e M with 0 and E unjoinable. Let X,6,1 e P. IfO 1X,6,,1 and X,6, L I ac then there is ay e P such that OIy, y 1 ac, and EkO60E'k'1 = EXYXY. Axiom L. (LUB) If A c /C, A # 0, and A is bounded above, then there exists an A in C such that A > X, for all X in A and ifB > X for all X in A then A < B. Dilation Axioms Axiom T. IfO e t,g, with t # g, where t is timelike or t and g are both isotropic, then there exists a unique a e M such that g, t c Xa. Axiom D. (Desargues) Let g,h,k be any three distinct lines, not necessarily coplanar, which intersect in a point 0. Let P, Q e g; R,S e h; and T, U e k. If gpT gQu and gRT II gsu then gp II1 gQs. Axiom R. Let 0 e g,h; P,Q e g, and R,S e h. If gpR II gQs then go,poR II go,Qos. Polarity Axioms Axiom U. (U' subspace axiom) Let O,A,B,T and C be four points with 0, C y6; A, 0 ca; O, B I P; with a i y and l5. Then there exist X, e P such thatX IE ; 0,AOB1X; and O,Cs. Axiom Si. Ifg c 3C,, a E G, h c Xp, P e g, and there exist y, e P7 such that iL8; y78 e go h, g c Xy, and h a 6 then there exists e g such that g,h c 3 XeC. (If g and h are two orthogonal spacelike lines then there is a spacelike plane containing them.) Axiom S2. Let g and h be two distinct lines such that P e g n h but there does not exist pe P such that g 3CX0 and h c Xp. Then either there are a,0' e P such that P = ac', g a Xa, and h c Xp, or for allA e g there exists B e h such that P and APB are unjoinable. This concludes the list of axioms. Note that by Axiom 5, ac3 A M for a c 13, a, P3 e 7P and if a, P3 e M are distinct with al3, then. a3 e M. By Axiom 6, every two points is contained in a line. Notation. Due to the brevity of the theorems and proofs in sections 3.1 through 3.4, we shall follow the usual convention in absolute geometry [1, 2, 19, 30] of simply numbering the results in these sections. 3.1.1. Properties of M. 3.1.1.1. The set M # 0. Proof: By assumption 9 # 0, so let a e g. By Axiom 11, there exists y e G such that a 1 y. Hence, ay e M by definition and M # 0. U 3.1.1.2. The elements of M are involutions. Proof: Let y e M. Then we may write y = aC where Ca, P3 E g and ca P3. We have yy = apap3 = aa3pp3 = 1. U 3.1.1.3. For every P e M and for every e 5, p" e M. 31 Proof: Let [3 = aU12 with a 1,42 e Q and a1 oX2. By our assumptions on 9 and because aX Ia2 we have for all 4 e 0, aaC 2 E and cafa so that a4ax e M. * 3.1.1.4. The sets g and M are disjoint, g r) M = 0. Proof: Let G' = \M. Then G' consists of involution elements, g' r M = 0, and P = u M = u M. Let y E g' and 4 e 0. Now ifY E M, then by 3.1.1.3 above y = (y) ' E M, a contradiction. Thus, g' is an invariant system of generators and without loss of generality, we may assume that g r M = 0. U 3.1.1.5. Ifa e G, P3 E M and a 1 P3, then a3 e M. Proof: If ap3 = ye M then a = P3y e g where 3,y e .M and 131y, which contradicts Axiom 5. U 3.1.2. Properties of P. For the remainder of this dissertation, let the symbol <" denote the phrase "if, and only if'. 3.1.2.1. Ifa,f3 e P and F e 0 then a 1 P3 if, and only if, ax 1 3. Proof: First suppose that a, 3 e g. Then a 1 13 0* a+ 2 3 because a P3 implies that a 13 and a 113 + a:j3 If a E g and 3 E M and a 1 P3 then a3 = y e ; aky E g, P3 E M, and a([3 = y e implies that a L p3. Conversely, if a' 1[3 then aP[ = 8 e g and a13 = 8' E g, so that a p 3. Finally, suppose that a,P3 E M. Then a:,3P e M and a3 e M + + *=3 M. E 3.1.2.2. If a,3 E P andF e E5, then a 13_< a 113. Proof: Let a e g and P3 e M and F E 0. If a 1 3 then a P so that ao I 13. Thus, a 13 or a I3P. If a 1[3 then a[3 = y So we have a13 = Y' e 6 by the invariance of g, which contradicts a 1 3. Hence, ax 3' . Conversely, suppose that a 13. because a' e G and E3 M by the invariance of g and M then a = y, 83 = 8 imply that y 1 6 and y' = a p3 = 64 by the paragraph above. U 3.1.2.3. For each e (5, P7 = 7P. Proof: because 7P = G u M, the result follows from the invariance of g and M. U 3.1.3. Properties of 3. 3.1.3.1. There exists a point; that is, T # 0. Proof: Let a e G. By Axiom 11, there exists y,3 e g such that a, 3,y are distinct and mutually perpendicular. By Axiom 4, a3y = a8 ?, where 5 = P3y e M and a15 as ae 3,y. Thus, xa 1 and P = a6 is a point. U 3.1.3.2. IfO1a,P3,y; a,3,,y e ; a 1 3P 1 7 1 a; then 0 = a37y. Proof: By the proof of 3.1.3.1 above, P = ap3y is a point and 8 = 13P7y e M because 13 l y with 3,y e g. So we have a 18, 06 because 013,y. This yields A, 01a,6 withA = a6 so that A = 0 by Axiom 3. U 3.1.3.3. IfA e XI and 6 e P, then A # 5; that is, a point does not equal a plane. Proof: Let A = ao3 with a e g, 3 e M, and a 1 P3. Suppose that A = ae3 = 6 e P. If 5 e M then we have a = P38 e with P315 and P3,5 e M, which contradicts Axiom 5. If 6 e g, then A = a3 e g and this contradicts the definition of a point. U 3.1.3.4. The elements of X are involutory. Proof: Let A E X, so that we may write A = ae3 with a L 13. In particular, a13 = P3a and AA = a3ap3 = aapp313 = l. U 3.1.4. General Consequences of the Axioms 3.1.4.1. If PI a, 3 anda 13 then P = a3 and ifP = ap3 then P Ia, 3 and 1t3. Proof: If P I a, 3 and a 13 then a3 = A for some A e X and we have A, P a, 3 with al 3. Thus, by Axiom 3, A = P. IfP = ap3, then by 3.1.3.4 alp. because P o P by 3.1.3.3 then al 3. Also Pa = P3 and P3 = a imply that the products Pa and P3 are involutory so that Pla,P3. U 3.1.4.2. IfPIa then Pa e 7P; that is, Pa is a plane P3 and P\ 3. 33 Proof: By Axiom 1 there exists P3 such that P 13 and 1 a. Thus, P[ a, 13 with a p so by 3.1.4.1 above, P = a13 and 13 = Pa. U 3.1.4.3. For each P e T and each e Q5, P e X. Proof: By the definition of a point we may write P = a13 with a E M 13 E G, and al13. By 3.1.2.2, a '13, and P [Ia, 3P so that P = a134 is a point. M 3.1.4.4. IfP a,3 and y L a,3 P then a = 13.( Given a point P and a plane there exists a unique plane a such that P I a and a L y.) Proof: This follows immediately from Axiom 1. U 3.1.4.5. There do not exist three planes pairwise absolutely perpendicular. Proof: Suppose that a, P3, and y are pairwise absolutely perpendicular. Then for P = a3 we have P I a, P3 with a,3 p y so that a = 3, which contradicts a 13. U 3.1.4.6. IfA,B,CIa then ABC = DIa. Proof: By Axiom 8, ABC is a point D and Da = ABCa = aABC = aD. U 3.1.4.7. IfA,B, CIl, 3 then ABC Ia,13. Remark. From 3.1.4.6 and 3.1.4.7 above the product of three coplanar (and as we shall see, collinear) points yields a point which is coplanar collinearr) with the other three. 3.1.4.8. Ifoa1, a2,a3 i a then alC a2a3 = a4 e P and a4 i a. Proof: Let A = aal, B = aa2, and C = aa3. Then by Axiom 8, ABC is a point D and D = aa Iaa2aa3 = aala2a3. So DIa, by 3.1.4.6. By 3.1.4.2, aI a2a3 = Da = 4 e 'P, D = aa4, and aI0a2a03 = 04 ia. a 3.1.5. Perpendicular Plane Theorems 3.1.5.1. If a 13 ; 13y = A andAl a then a 1 y. (A4 plane perpendicular to one of two absolutely perpendicular planes, and passing through their point of intersection, is perpendicular to both.) Proof: From our assumptions above it follows that ay = a3A = f3Aa = ya, so a 1 y 34 or a I y ( note that a = y implies that ac L P3 because A = Py and A I a). Suppose that a 17 so that B = ay. Then we have A, B I a, y with a 17, which implies that A = B by Axiom 3. But if A = B then P3y = ay and P3 = a, which contradicts P3 1 a. Thus, ay is a plane and a L y. 3.1.5.2. Suppose that a 1 3; AIa,p3,y,8; y7a; and 5813. Then 8 L y. (If two planes are perpendicular, their absolutely perpendicular planes at any point of their intersection are perpendicular.) Proof: By 3.1.4.1, A = 7ya = 8 and 6y = ap3 E P as a I P. Thus, 6 71y. 3.1.5.3. IfO = aa =yyi = ssI with a,7,s e M and a 7 s  a then ay = s. Proof: Because a,y,s e M, then a1,y1,E1 E g. Because a L I 1 I a and aal = yyj = ese, then ya = y1a1; sa = sia,; sy = eily imply that 71 L ai L E I I y 7. Because O0x,ai,yic, then 0 = ay17Sc by 3.1.3.2. Because points are involutions by 3.1.3.4, 1b = 00 = Oa171E = Oaiyi00si = ays and a7 = e. N 3.1.6. Parallel Planes. We say that two planes a and P3 are parallel, denoted by a  3, if a = 3, or there exists a 7 such that a, [317. 3.1.6.1. Parallelism is an equivalence relation on the set of planes P. Proof: That the relation is reflexive and symmetric is clear. For transitivity suppose that a P13 and P3 1 y where a, 3, and y are distinct. Then there exists 8, e PP such that a, p3S and P3, 7yls. Let A = a5, B = 18, and C = P3. Then by Axiom 8 we have D = ABC = aop13 = a& and aL1 so that a 11 7. 0 3.1.6.2 Ifa, ,3L6 and 3is then a1e. 3.1.6.3. If aiLy, p36, and7 6 8 then a 1 P3. (Two planes absolutely perpendicular to two parallel planes are parallel.) Proof: If y = 6 then we have a, 3P8 and the result follows. So assume that there is an s such that 7,5iE. Then by 3.1.6.2 above, al6, P31y, oa,p3L, and a 1 P3. U 35 3.1.6.4. If a I 3, y 1I 8, and 13P then cdy. (Two planes parallel to two absolutely perpendicular planes are absolutely perpendicular.) Proof: because cx 1 P3 then a, P3LE for some c and because 835 then cI6. Similarly, Y, 8&' for some c' and 5Lp3 imply that yip3. Hence, a, 3PL68 and P3y yield acly. 3.1.6.5. Ifa e M(9) anda II P then P3 = 9(M). Proof: Suppose that a e 9 and a 11 P, so a, Piy for some y. Because a e 9 then Y E M, which implies that 3 E 9. U 3.1.6.6. If a 13 P then c\\ 1 P for every 4 e 5. Proof: Let a, Piy for some y e P. By 3.1.2.2, aP,13'iy so that ac  3. From 3.1.6.5 and 3.1.6.6 above and the invariance of 9 and M, we have that if ca 1 P3 and ca E G(M) then P3 e 9(M) and ct 11 P3 with a4, 3P e 9(M). N 3.1.6.7. If a 113 then X, rP X p = 0 ora = P3. Proof: Suppose that c, P3Iy and PIcc,P3. Then a = 3 by 3.1.4.4. U 3.1.6.8. IfA # B then A I B; that is, AB # BA. Proof: By Axiom 6 there exists Ca E P such that A, BI c. By 3.1.4.1 and 3.1.4.2 we may write A = ctci and B = aCV2. Suppose that AB = BA. Then we have al(X2 = a2(Ci so that alia2 or ai c(2. But il  c2 because ai,a21Ca so that either a1 = aC2 or a, r 3a2 = 0. If al I ci2 then by Axiom 7, Xa n # 0 so UI = a2 and A = B. If al1 I2 then C = CC I2 e Xar X12 and again we have that C I = a2. Hence, A # B. U 3.1.6.9.a. If AB = A, then B = A. b. If A # B, then A,B, and A8 are pairwise distinct. U 3.1.6.10. IfAla; BI3; Cly; anda 1i P 13 y thenABCap3y. Proof: Let Act = a', B3 = 3', Cy = y'. Then a', P3',y' iot, 3,y and ABC = xac'PP313'yy' = c'P3'y' = 6'8 with a'13'y' = 6'i6 = a3P7y. Conclusion: A I c; B1 I3; ca 1 P3, imply that B' P3. 36 3.1.6.11. IfAA' = BB'; A,A' a; BI3; a 13 thenB'P3. Proof: Because AA' = BB' then B' = BAA'; 3  a a. From 3.1.6.10 above, B' = BAA'Paa = P13. 0 3.1.6.12. For each P and each P3 there is a unique a such that P la anda aI 3. Proof: By Axiom 1 there exists a unique y such that yip3. because Ply then Py = a, aIy and a 11 P3. Now if P and 8 1 P then P8,a, so that a = 8 by 3.1.6.7. U 3.1.6.12. Let a,P3 e P and M e X. Then aip + aMi3 and thus, a 11 aM. Proof: Suppose that aLp. By Axiom 1 there exists unique y, 1M such that aLy and 813. By 3.1.6.2 we have P3,yla and P318 which implies that y48. By 3.1.4.1, M = y6. thus aM = aY6 = a61388 = P3. Therefore, aMl3 and aM 11 a. Conversely, suppose that aM"'3. As above, there exists unique y,81M such that aMly and 8l13. It follows that P3168, M = y6, and a6 = ac = aMip and a = (a6)C103 = P3. U If AM = B, then we say that M is the midpoint of A and B. Clearly, M is also the midpoint of B and A. 3.1.6.14. IfPA = pB then A = B. ( Uniqueness of midpoints.) Proof: From pA = pB we have PAB = ABP and also PAB = PBA because ABP is a point and hence, an involution. Thus, AB = BA which implies that A = B by 3.1.6.8. U 3.1.6.15. IfA,BIa,13 and AA = B then MIa,13. Proof: By Axiom 6 there exists y,8 such that A,Mly,6. because B = AM = MAM then by 3.1.4.7 we have Bly,6. Thus, A,BIa,y,6 ; May,6; and A,B13,y,86 so by Axiom2, M113,a. U 3.1.6.16. .IfA,B,C16; a,,y1L86; Ala; BI13; CIy; a ii 3 II y then ABCIapy and a0y 1 8. Proof: From 3.1.6.10 we know that D = ABCI a3y = c. Because 5la,P3,y then E18 so that c 1 8 or s18. If E = 5s then we have D,EI 8, with 68E which implies that 37 D =E by Axiom 3. Let A = aa', B = P13P, and C = yy'. Because a 1113 P then a' P 3' 1I y'. Thus, s6 = D = ABC = apya'P3'y' = a'P3'y' which implies that 8 = a'P3'y'. Because A 18 then we may write A = aa' = 886' and by 3.1.5.1 we have A a',a,; aLta'; and a 1 5 which implies that a' 1 8. Hence, a'8 = a'a'P3y' = p3'y' is a plane, so P3' y'. By Axiom 7 and 3.1.6.7 we have 3P' = y' so that a' = 6. But this yields at18 and a 1 8, which is a contradiction. Thus, aP3y 1 6. U 3.1.6.17. IfA,Bla; Ala'; a' Ia; BI3P; and P B a' then P 1 a. Proof: By 3.1.6.16 above B = AABIa'a't3 = P3 and P3 1 a. U 3.1.6.18. IfE\ a,e; a 1 P3, and 6113 then c I a. Proof: If El 13 then the result follows from 3.1.6.7. Suppose that E % 3 and let M = E3. Then, E # EP and E3P I aP = a so that E, E I a. Now EM = Ec = EP, so that Mis the midpoint of E and EO. EM = EPEla' so that E, EMI a, a and by 3.1.6.15, Ml a,a'. In particular, MI a and we have MI a, P3, ; a P3; and P31c so that a 1 e by 3.1.6.7. U 3.1.7. Consequences of Axiom 11 and 3.1.6.18. 3.1.7.1. IfAla; a P3; and a, 3 e g then there exists ay7 in Q such that AIy and Y 1 P3,a. Proof: Let A 68 with 56iP. Then 6 e M and A I a, 8 with al 3, PL8 so that a 1 6 by 3.1.6.18. Because 5 e M, 6 1 a, and a e Q then c = 8a E g, A\E, and E a. We claim that 1 P3. Indeed, P3 = 36a = 6p3a = 6ap3 = 63 with 6, P3 e 9 so that c 13 or s = P3. But # 13 because then 6a = P3 and a = 813 = P, as 613. This contradicts 3.1.3.3. M 3.1.7.2. Every point may be written as a product of three mutually perpendicular planes from g. Proof: Let A be any point. Then by definition A = a3 for some a E 9 and some P3 e M with alp3. By Axiom 11 there exists 7 E 9 such that y 1 a. By 3.1.7.1 there 38 exists a S E g such that AI and 8 1 a,y. Again by 3.1.7.1 there exists an C E such that A I and E 1 8, ,a. Hence, A I a, 6; 8,a e 9, and s L 6 L a I s which yields by 3.1.3.2, A = 8as. U We note that in this case, a(3 = A = (xaS8 implies that 63 = 8; that is, if A P and P e Mthen there exists P1, 32 e g such that APi,P132 and P3 = PI3P2. 3.1.7.3. IfA I a then there is a P in 9 such thatAl P and3 P a. Proof: This follows directly from the proof of 3.1.7.2, for if a e 9 then we can find 8,e 6 G such that A 18,e and 6,s L a. If a e M then we can find a l,a2 e g such that A I ai,(X2 and a 1,a 2 I a. That is ifAIa then there exists P[1,32 e G such that A I P l, 32 and 3 1 P32. Moreover, if a E M then we can find P 1, P2 such that a = Pip31132 and if a e G then we can find P31,32 e G such that apIp31132. N 3.1.7.4. IfA a e M and 3 I a then there is ay such that A y andy 3,a. Proof: Let 8A with 6113P. Then 8 1 a by 3.1.6.18 and As = aS. Because a 1 P3 and 5i3 then E p. If dLp3 then we would have A 186, with 6,e1&3 so 5 = s and a = 1I. Thus, s&3. We note that if P3 E 9 then 8, e M andif3 P e M then e 9G. N 3.2. Lines and Planes 3.2.1. General Theorems and Definitions. 3.2.1.1. For any A,B e X and any a,b e G, ifA,B e a,b then A = B or a = b. Hence, by Axiom 6, for all A,B e 3C, there exists a unique g e 9, such that A,B e g. Proof: Let a = [a, 3], b = [y, 8], A,B Ia, P,y, 5, and suppose that A B. Let C e a, so that C Ia, 3. By Axiom 2 it follows that C y, 5 so C E b and a c b. Similarly, b c a and a = b. U 3.2.1.2. Every line contains at least three points. Proof: By definition, every line g = gAB contains at least two points A and B. Let gAB = [a,3] with A,Ba, 3. Then by Axiom 8 and 3.1.4.7, ABa,P and AB e 3. AB A by 3.1.6.9. U 3.2.1.3. Suppose that a13 = y. a. Ifaa, = P1313Pi then a I\ 13 i andac131 = 7, so7 L a,131. b. If atal = PP313I = 7, then a 13, ai l 31, y1i a1,P31, and ap3 I = a113 = . c. IfA Iaa,3 then A ly. d. The line g = [a,13] = [13,y] = [a,y ] [a,13,y] is a nonisotropic line. Thus, a, P, and y are three mutually perpendicular planes which intersect in a line. Proof: a. From cat = PP1313i we have y = 13a = 13Pial. b. If Pp1313 = yy and aal = Yyi then y = y71313i = ap131 and yi = yaal = 13al. c. Because A I a, P3 and a13 = y then Ay = Aat3 = ap13A = yA and A 1y. d. By Axiom 7 there exist points A and B such that A,BI a,13 and by 3.2.1.3.c. above, A,B]y. Thus, A,B E [a,13],[a,y 7],[13,y] and [a,13] = [a,y ] = [13p,y] [a,13,y]7 is a nonisotropic line. U 3.2.1.4. If A and B are collinear then A and B are collinear for any E 0. Proof: This follows from the fact that A, B I a,13 A,B' I a,13",. For each 4 e 0, we define a {C : C E a}. By 3.2.1.4 above, a4 is a line for every 4 in 5 and if a = [a, 13] then a4 = [a,P13]. Definition of parallel lines and planes. We say the line a is parallel to the plane a, denoted by a 11 a, if there exists a 13 such that a c Xp and 13 I a; that is, a = [13,y] and a,P 1 6 for some 6. We say that two lines a and b are parallel, a b, if there exists a,3,y,8 such that a = [a,y], b = [13,], where a I 13 and y II 6. 3.2.1.5. IfAA' = BB' # 1I; A,A'I a,a'; a # a'; B,B'1,13'; 13 13', thengAA,'  gB"6 Proof: Let BI13P* with 13"* a (3.1.6.12). Then B' P13* by 3.1.6.11. Let B130, with P 131 al. Then B'130 by 3.1.6.11 and B,B' P13,131,13*,13P which implies that [13PP13] = [13P*,13P] = g88' by 3.2.1.1. Hence, we have gAA' = [a,a ], gBB' = [13*,13] with a II 13p* and at 1 13i; thus, gA' I gBB'. * 40 3.2.1.6. Two lines, a = [a, aI] and b = [P3,I]I are parallel precisely when there exist A,A' E a and B,B' e b such that AA' = BB' 1 1o. Proof: Suppose that a I1 b. Then, a  P3 and a1 11 P1. Let A,A' e a and B e b, so that B' = AAB'Ito3ap, and 0axliPI = I,pi, by 3.1.6.10. Thus, B,B' e b with AA' =BB'. IfAA' =BB' = lo thenA =A'. U 3.2.1.7. If a I b andb I1 c then a I] c. Proof: Let a = [a,a']. From the proof of 3.2.1.5 above we may write b = [p3,3'] with a P 3 and a' i because a 1 b, and c = [7,7'], where y 1 P3 and y' II 3' because b il c. By 3.1.6.1, ot 1 y and a' ii y' so that a I1 c. U 3.2.1.8. For each line a and point A there is a unique line b such that A e b and b i1 a. Proof: Let B, C e a be distinct and if A E a we can choose B, C # A because every line contains at least three points by 3.2.1.2. Then ABC = D by Axiom 10 and BC = AD l 1I, so gAD II a by 3.2.1.6. Now suppose that A E c and c I a. By 3.2.1.7 above, c II gAD, so there exist W,X e c and Y,Z E gAD such that WX = YZ # 1 by 3.2.1.6. By 3.1.4.7, A' = AWX e c and AI = AYZ e gAD. Thus, 1 AA' = WX= YZ=AAi1 implies that A' =A1. Hence, A,A' E gAD,c; A A', because WX # 1 and gAD = c by 3.2.1.1. 0 3.2.1.9. IfA,B e g, A # B, and C e h then g 11 h if and only ifABC E h. Proof: If ABC = H e h, then 1o # AB = HC and g 11 h. Conversely, suppose that g I1 h and put D = ABC. Then because A # B, lo # AB = DC and C e gjjc, with gDC 11 g. Because g I1 h and C e h, then by 3.2.1.8, gDc = h and D e h. U 3.2.1.10. I/a 11 b then a = b or a n b = 0. Proof: Supposea 11 band anb 0. LetA E a,b andC e with C A, B E b, with A # B. Then, a = gAc, b = gAB, and therefore, D = ACB e a, b. 41 If D = A, then ACB = A, C = B, and a = b. If D # A, then it follows that a=gAD =gAc =gBD =gAB = b. U Classification of nonisotropic lines. Following the terminology of physics we make the following definitions. Let a be a nonisotropic line. If there are elements a,3,y e M such that a = P3y and a = [ct,P3,y], then we say that a is timelike. If there are elements ot, 3 E g such a (3 and a = [a, P3, ab], then we say that a is spacelike. Remark. Let A,BoI c e with A # B. Then by Axiom 14, there is a P3 such that A,B I 3 and P3 I a. Thus, a = [a, 3] is a spacelike line by definition, A,B e a, and every pair of points in a spacelike plane is joinable with a spacelike line. 3.2.2. Isotropic Lines. 3.2.2.1. IfA # B; A,B Ia,3 with P3 a, a e M; andA andB are unjoinable with P; P Ia,y; andy La,P3, then AY = B. Proof: First, AY Iaty, 3Y = a, P3. If A7 = A then A I7 and A is unjoinable with P. Similarly, By ac, P3 and By # B. Suppose that A" and P are joinable. Then there are 6 and e such that P,AYI\,c and 6 1 s. But then P,A = P',(A")"'6",s" and 8" 1 &Y, which implies that P and A are joinable. Thus, A" and P are unjoinable, so by Axiom 13, A" is unjoinable with A or A" is unjoinable with B or Ay = A or A" = B. Hence, AY=B. U 3.2.2.2. Suppose that A # B; A,Ba,P3 with a 1 P, a e M; A and B are unjoinable with Pla; Py, y 1 p3. then AY =B. Proof: First observe that because PIa,y; a 1 P3; and yip3, then a I y by 3.1.6.18. Then A It ay, PY = a, P3 and A" is joinable with B. AY # A because A and P are unjoinable as in 3.2.2.1 above. A" joinable with P implies that A = (A")" is joinable with Py = P. Hence, by Axiom 13, A" = B. U 3.2.2.3. If A and B are unjoinable then X E gAB precisely when X= A or X = B orX is unjoinable with both A and B. 42 Proof: By Axiom 6, gAB = [a, P3] for some a and P3 in P. Let X E gAB and suppose that X # A, B. If X is joinable with A then there exists y, 6 such that y 1 8 and A,XI y, 6. By Axiom 2, B I y, 6 and A and B are joinable. Hence, X is unjoinable with both A and B. U Remark. If X is unjoinable with A and B; A,B a, x3; then by Axiom 10, X1 a, 3 and X e gAB. 3.2.2.4. If gAB is isotropic; C,D e gAB; and C # D, then C and D are unjoinable. Proof: If C, DIa,3 with a I P3 then A,BIa,3 by Axiom 2; that is, gAB = gCD. 3.2.2.5. IfA,B, CI a are pairwise joinable and distinct, then each P a is joinable with at least one ofA,B, and C. Proof: If a e g then the result follows from Axiom 12, so assume that a e M. By Axiom 13 there exist D, EI a such that A,B, and C are all unjoinable with P. Then by Axiom 13, at least two of A,B, and C must lie on one of gpo and gPE. By 3.2.2.4 above, this implies that two of A,B, and C are unjoinable, which contradicts our assumption. U 3.2.2.6. If gAB is isotropic then gAB is isotropic for all 7 e 65. Proof: IfA,B4lIa,1 with a 1 P then A,BIa,] 1' and a' 1 3'. 3.2.2.7. IfP Ia E M then there are at most two isotropic lines in Xa through P. Proof: If gAP,gBP,gcp c Xa are three distinct isotropic lines through P, then P is unjoinable with A,B, and C. The points A,B, C must be pairwise distinct and are joinable by Axiom 13, in contradiction to 3.2.2.5. U 3.2.2.8. By Axiom 13 and3.2.2.7,for each P Ia, a e MA4, there are precisely two isotropic lines in Xa through P. 3.2.2.9. Let gAB,gAC c Xa, a e M, be two isotropic lines. a. IfA Ap, with p3iLa, then gA = gAc. b. If Ay, with yLa, then gAB =gAc. 43 Proof: Because gAB # gAc by assumption, then B is joinable with C by Axiom 13. The results follow from 3.2.2.1 and 3.2.2.2. M 3.2.2.10. IfA,B,CI a E M; gAB # gAC are isotropic then gBC is nonisotropic and there is a p with p3A such that I aandC = BP. Proof: Because gAB # gAC, then gBC is nonisotropic. By Axiom 14 there exists a y, 7yB,C, such that y I a. Let AI3, 131 7 (Axiom 1). Because Aa,[3; a I y; and P3 1 y, then by 3.1.6.18, a L P3. If BP=B then BI P3 and gAB is nonisotropic. Similarly, CO C and by 3.2.2.1, BO = C. U 3.2.2.11. If A,B, C I a, a E M, are pairwise unjoinable then there exist P3,y a with A113, y such thatC = BOY. Proof: First observe that by 3.2.2.3 and 3.2.2.4, gAB = gBC = gAC C is isotropic. By 3.1.7.3 and 3.1.7.4, there exist PI\A such that 3 c a. Because B and C are unjoinable with A, then B,C P3 and BO # B. By Axiom 1 and 3.1.6.18, there is a 5113 such that 6 1 3 and 6 1 a. Thus, B,BP16S,6P = 6,a and B and BP are joinable. Suppose that BO3 and C are unjoinable. By 3.2.2.3, either gBOC = gAc or A is joinable with BO. But ifA, BO Ie, & I a, then A = AP,BBe EP a1 = ax imply that A and B are joinable. If gsBC = gAc = gAB then BO3 is unjoinable with B. Hence, BO3 and C are joinable and by 3.2.2.10, there is a y with yJA such that y I a and C = BO M 3.2.2.12. Let PFa,13,y with P3,y a, a e M. IfgAB c Xa is isotropic then gY c is isotropic and gAB II AB Proof: By 3.2.2.8, there are precisely two isotropic lines in X.a through P, say gpc and gpQ. By 3.2.2.9, g)y = gpQ and g = gpc. Now PAB = D is a point by Axiom 8, so AB = PD and D is unjoinable with P as A is unjoinable with B. (Suppose that P, DIa' with a' I a. Let B\I with k 11 a'. Then because B\ a, X. 1 a by 3.1.6.17. But then by 3.1.6.16, A = PDBIa'a'k = X and A is joinable with B.) Because 44 A,B,PIa then D = PAB Ia. Hence, D e gpc or D E gpQ, and gAB II gpc or gAB II gPQ. thus gA II gQ = gPQ II gAB or g II g = gP c II gAB. 3.3. A Reduction to Two Dimensions In this section we restrict our attention to two dimensions. In this way we are able to use the work of Wolff [30] to construct our field. Let t be any element in 0; then the map a4 : X X given by a(A)= A1 is bijective and maps lines onto lines and planes onto planes by 3.2.1.4,5,6 and 3.1.2.3. Hence, it is a collineation of (0,g,9). In the following, for any element e 0, the collineation induced by it is denoted by ro. Let a e (P be fixed throughout this section. We wish to define a set Ca of maps on Xa which can be viewed as line reflections in a given plane Xa. We then show that each element of Ca is involutory and that COa forms an invariant system of generators within the group ga it generates. Finally we will show that (Ca,Ga) satisfy Wolff's axioms [30] for a two dimensional Minkowski space; that is; the Lorentz plane, for a e M. Let Ca = {g c 3CXa : g is nonisotropic}. By Axiom 14, each g E Ca may be written uniquely in the form g = [a, 3, y] where a = Py. Let A e 3a and g = [a,p3,y] e Ca. because A la and a = P3y, then A = A' = AP and AP = Ay. Hence, we define the map ag : X3a 3 Xa by ag(A) = Ag = A = AY, for A E Xa, and cyg : Ca  Ca by ag(h) = h9 = {ag(A) : A e h}. Note that by 3.2.1.4, if h = [a,6,e], then ag(h) = [a,6P,S1] = [a,8Y,Y]. 3.3.1. For each = [a,13,y] e Ca, ag : Xa X oa is bijective. Proof: If A, B I a and ag(A) = ag(B) then A7 = By and A = B, so Cg is injective. If A a then A I ay = a because a 1 y and Gg(AY) = (AI)y = A, so ag is onto. U 3.3.2. Forg = [a,p3,y] E Ca, Ocg : Ca * Ca is bijective. Proof: Let h = [a,E,rI] e Ca. Because a y and s 1 <> y 1 ly by 3.1.2.1 then 45 a = aY = EY1Y. It follows that ag(h) = [ac, cy,ly] e Ca. IfAIe,fr then AYIc7,Tr7 so ag(A) E ag(h) for all A e h and Gg is a welldefined collineation. If I = [a, .,p] e La and ag(h) = cg(o) then for every A in h we have Tg(A) = AY ag(l) = [a,,]. This implies that A = (AY)Y e [aE ,] = /;that is, h = I. Hence, Ug : Ca f a is injective. Because e r +' EY ily then it follows that h = [a,,] e + c g(h) = [a,yqY] E 4a. Hence, ag : a * a is surjective. U 3.3.3. Each ag, for g E 4a, is involutory. Proof: Let g = [a, P3,y]. Then for A a we have cgrg(A) = ag(AY) = (AY)Y = A. E Let g = [a, 8,y], h = [a,, rj] E Ca. We say that g is perpendicular to or orthogonal to h, denoted g 1 h, if one of y and 5 is absolutely perpendicular to one of E and q and g r h 0. 3.3.4. Forg and h as above, ifyls then 85_q, 8 1 e, andy 1 e. Moreover, M = ye = 5e E g,h. Proof: Let M = ye. Because y, E 1 a then MA = (ye)U = y's0 = ye = M and MI a. So we have MIota,e which implies that MIae = rq and M e h. Also, MIa,y so Mf ay = 8 and M E g. Also from M = ye it follows that Me = y = 8a = TIE so that M = 6r1 and 61lt. Applying 3.1.6.18 to MI y, r with y 1 5, 7115, and M,e with 6 1 y, ely7, we obtain y 1L 1 and 8 1 e. U 3.3.5. Ifa,b,g e 4a then a 1 b > ag bg. Proof: By 3.1.2.1 and 3.1.2.2 we have ac.Xy; yl6; bc X6 a 3cX; y+I; b c. for all 4 E . The result follows. N 46 Remark. Because AIs <> A:js' for all E e Q5, A E X, and s e P, it follows that for each g e Ca, ag maps: (i) The set X(a onto itself. (ii) The set Ca onto itself. (iii) Collinear points in X3a onto collinear points in Xa. (iv) orthogonal lines in Xa onto orthogonal lines in X3Ca. That is, ag is an orthogonal collineation of Xa. Let C( = {Tg : g e Ca} and T, = {"p : P E Xa}. 3.3.6. The sets Ca and 3Ca are nonempty. Hence, Cao 0 and T3c 0. Proof: If a e MA then we may write a = a'6 where a', e g and a' 1 8 by definition of MA. By Axiom 7, there exist A,B such that A # B and A,Bl a',6. Thus, A,BIa'8 = a and Xa # 0. Moreover, gAB = [a, a',8] e Ca and Ca 0. If a e g then by 3.1.3.1, there exist AIa and by 3.1.7.3, there is a P3 in g such that Aj1 and 13 a. Thus, A e Xa and g = [a, P3, ap3] e Ca. U Note that from 3.3.3, Ca consists of involutory elements and by 3.1.3.4, p3a consists of involutory elements. 3.3.7. IfP e g,h e Ca andg I h then ap = Cgah = ahCg. Proof: Let g = [a,y,6] and h = [a,e,ft], and without loss of generality, assume that y s, so that 81ri. By 3.3.4, P = y = 6rI e g,h and by 3.2.1.1, {P} =gr h. Then for A e Xa, agFh(A) = ag(As) = A = ap(A) A" = h (A') = ahcg(A). E 3.3.8. For everyA in Xa and for every h e Ca, there is a unique g e 4a such that A e g andg 1 h. Proof: Let h = [a, r] e Ca and A e Xa. By Axiom 1, there exist a ylA such that y's. Thus, Ala,y with a 1 s and yLs so a 1 y by 3.1.6.18. because a I y then ay = 8 and Al a,y implies A 16 so that A E g = [a,7y,8] E Ca. because g c Xy, yl', 47 A E g = [a,y,6] e La. Because g c Xy, y, and h c XT, then g h. By 3.3.4 above, r1I5 and {B} = {ye} = {T1} = g h. Now suppose that I c Ca such that A e 1 and I L h. Now we may uniquely write I = [cx,a', P] in Ta where a = xa'p. because / L h then without loss of generality, we may assume that a' Ie and PfLT. because yi_ and 81TI it follows that a' 11 y and 0 11 6. By the definition of parallel lines in Section 3.2 we have 1 1 g. because A e I nrg then by 3.2.1.10, / = g. U 3.3.9. (i) The fixed points of (g E Ca are the points in X3a incident with g. (ii) The fixed lines of Gg E Ca( consist of g and all lines in 4,, which are orthogonal to g. Proof: Let g = [a,y,65]. IfA = ag(A) = Ay for Ala, then AI17. because AT = A6, VAja then A6 = A and A16. Thus, A e g and 3.3.9.(i) follows. Let h = [a,E,'q] e a and suppose that cg(h) = h; g # h. Then there is an A e h such that A f g. By 3.3.8, there is a k = [a,co,9] e Ca such that A e k and k I g. Because A E h, cg(A) = Ay e h and because A E k, k g then A I coy,Oy = o), and AY e k. BecauseA 0 g then A I y (for if A 1y then A I a implies that A jay = 8 which implies that A e g) and A # A7. Thus we have A,AY E h,k, so that h = k by 3.2.1.1. Hence, h I g. U 3.3.10. (i) The only fixed point of a point reflection, ap, is the point P. (ii) The fixed lines of ap are the lines incident with P. Proof: If pA = P, then this implies that APA = A, or AP = lo, or A = P. Thus, 3.3.10.(i) holds. Let x = [P,u] be any line (not necessarily in Xa or nonisotropic) and A E x. Then PA e x implies that pA p,t), so that P I pA,A = P3,u, and P E x. U Let 9( be the group acting on Xa generated by Ca. By 3.3.5, every element of Ga is an orthogonal collineation of Xa. Let c E a. By a transformation with T, we mean the adjoint action of a on G : (1 e a a' f a e ?a.Not,: that every such transformation is an inner automorphism of the group. 48 3.3.11. Let h = [a,, ri], g = [a, 'y,] e La, then a" = aTh(g). Proof: Let A I ac and put B = rg(A) = A Then O(ah(A)) = ahagCTh'h(A) = Ghg(0A) = oh(B). Thus, CG9h(Ch(A)) = aah(g)(ah(A)) for all A E Xa. Because ah is injective, then T = aoa(g) for every A in 3Xa. Similarly, ah(a) = aoh(g)(a), for every a in La. Hence, ag' = aah(g). Analogously, we can obtain aTh = Cah(P), for every P in Xa. U Because Ca generates 9a and every element of go, is an orthogonal collineation of (Xa,La) then we obtain the following. 3.3.12. For every a e go and for every ag E Ca, we have ag, e Ca; that is, Ca is an invariant system of Ga. Consider the two mappings: g e Ca  ag E Ca and P E Xa 3  ap E T3a. The first one is from the set of nonisotropic lines in Ca onto the set of line reflections, and the second is from the set of points in Xa onto the set of point reflections in Xa. These mappings are injective, because reflections in two distinct points have distinct sets of fixed points by 3.3.10, and similarly for lines and line reflections. Thus, it follows from 3.3.11, 3.3.12, and the preceding remark that the next result obtains. 3.3.13. If a e a and P,Q e Xa then a(P) = Q <> a' = OQ. Proof: Because a(P) = Q +> ag(p) = aQ > a(p) = aa, the result follows. U 3.3.14. If a e ga and g,h E La then a(g)= h +> *g = Ch.  3.3.15. IfP e Xa andg E Ca then P e g <> apCFpag is involutory. Proof: Suppose that P e g = [a,y,8] e La. Then for A a, Gp(og(A) = op(AY) = AYP = A = agop(A). Thus, (apag)2 = Ix. Nw assume that 49 apcg is involutory. Then P = agapaggp(P) = PPP = YPY. Because yPy = PY e Xa by 3.1.4.3, then by 3.1.6.9, P = Py. Because PY = P8 then PIy,, and hence, P e g. U 3.3.16. Ifg,h e a, then g I h <> 9gch is in volutory. Proof: Now g I h if and only if ag(h) = h. For g # h, by 3.3.9, ag(h) = h if and only if CFg = h. For crg 5jh by 3.3.14, TC = Oh if and only if (agah)2 = lx,, and Ogah l Xa. U 3.3.17. The point reflections jax are the involutory products of two line reflections from Ca. Proof: Let ap e Ta where P e XC. Then we may write P = y6s where a = y5 and Y,5, ,e e by 3.1.7.3. Let g = [a,'y,5]. Because a = 78, then g e a. By 3.3.8, there is an l in C, such that P e I and I l g. By 3.3.7, op = ar/g = agaC. U We now show that if a e M, then the pair (Ca,Ga), acting as maps on (XCa,a,a), satisfies Wolff's axiom system [30] for his construction of twodimensional Minkowski space. We give Wolff's axiom system below. Basic assumption: A given group Q5, and its generating set 9 of involution elements, form invariant system (9, () The elements of G will be denoted by lowercase Latin letters. Those involutory elements of 5 that can be represented as the product of two elements in G, ab with a b, will be denoted by uppercase Latin letters. Axiom 1. For each P and for each g, there is an I with P, g 1l. Axiom 2. IfP,QIg,l then P = Q org = 1. Axiom 3. IfP a,b,c then abc e G. Axiom 4. Ifg a,b,c then abc e G. Axiom 5. There exist Q,g,h such that gj h but Q f g,h,gh. Axiom 6. There exist A,B; A # B, such that A,B f g for any g e G. (There exist unjoinable points.) 50 Axiom 7. For each P andA,B,C such that A,B,CIg, there is a v e Q such that P,AIv orP,Blv orP,Clv. Geometric meaning of the axioms. The elements denoted by small Latin letters (elements of Q) are called lines and those elements denoted by large Latin letters (thus the element ab with al b), points. We say the point A and the line b are incident if A Ib; the lines a and b are perpendicular orthogonall), if aI b. Further we say two points A and B are joinable when there is a line g such that A, BIg. Replacing Ca with Ca and Xa with T3a, the pair (Ca, gQ) satisfies the basic assumption by 3.3.3, 3.3.6, and 3.3.12. It follows from 3.3.15, 3.3.16, and 3.3.17, that our definition of points, our incidence relation, and our definition of orthogonality agree with those of Wolff. Hence, we may identify LCa with Ca, Ca with g, and Qa with 0. Verification of Wolff's axioms. Axiom 1 follows from 3.3.8 and Axiom 2 from 3.2.1.1. For Axiom 3, let a = [a,a', 4e], b = [a, 3P, P'], c = [a,yy'] e Q. Then a',P3,y I a, and by our Axiom 9, a'Py =8 L ac and d= [a,6,a8] e La. For Ala, aaabac(A) = Aya' = Aa'Y = A6 = ad(A). Hence, CaaGbaCC = d ( Ca. If we then identify "" with "e", we get Wolff's Axiom 3. M Axiom 4. Ifa,b, c lg, then abc e g. Proof: Let g = [a,X,X'], a = [ac,a',], b = [a,3,3P'], and c = [a,y,'y'], with a',P3,y I X. Then by 3.1.4.8, a'py = s I X. Let A e a, B e b, and C e c, then Ala',a; Bla,P; and Cla,y. By 3.1.6.10 ABC = Dla'py = s and ABC = Dla by 3.1.4.6. Thus, DIE,a; c 1 X; a I X, and c 1 a. So, d = [a,E,aE] e La. And ifX1a, then aCabac(X) = AX1a' = X' = Od(X) and caabac = ad E Ca U Axiom 5: There exist Q,g,h such that g h but Q I g,h,gh. 51 Proof: Because a e A, we may write a = 13Py, where 3P,y E g and P3 I y. By 3.2.1.2 every line contains at least three distinct points. Let g = [a, 3,y] E CL and let B e g. By 3.3.8, there is an h in Ca such that B in h and h J g. By 3.2.1.2, there is anA in h such that A # B. By 3.3.8, there is an e La such that A E l and / I h. By 3.2.1.2, there exists Q e I such that Q # A. Now if Q e h, then A, Q E ,h implies that Q = A or I = h, so Q e h. Suppose that Q E g. Let Q E m, m 1 1, so that aQ = 010m. Because GB = 0g and BAQ = E e Xa, then GE = (OBaCAaQ = 3gCThahC1/CT/ClTm = cTgCrm and g L m. Because Q e g,m and g m then cQ = agaim = cGam and cg = a1 or g = 1. This implies A,B e 1,m. Thus A = B or I = m, a contradiction. U Axiom 6: There exist A, B;A # B, such that A,B f g for any g e G. Proof: This follows from our Axiom 13. U Axiom 7: For each P and each A, B, C such that A, B, CIg, there is a v in 9 such that P,Afv, orP,BIv, orP,Cjv. Proof: By our Axiom 13, there exist two isotropic lines in Xa through P, say gpQ and gpR. If no such v in Ca satisfies the above, then two of the three points must lie on one of the isotropic lines by our Axiom 13. But this implies that gpQ = g or gPR = g; that is, g is isotropic; a contradiction. U 3.4. Consequences of Section 3.3 3.4.1. For every A and B, there exist a e M such that A, B Ia. (Every line lies in some Minkowskian plane.) Proof: By Axiom 6, there exist Xa e P such that A,BI a. If a e MA4, the result follows. So suppose that a E g. By Axiom 12, there exist P3 e such that A,B 3 and 31 a. Then A,BI a3 = y, and y e M by the definition of M. M 52 3.4.2. For every A and B, there existsM such that AM = B; that is, every two points has a midpoint and by 3.1.6.14, the midpoint is unique. Proof: By 3.4.1 above, there exists a e M such that A,B Ia. From Section 3.3, there exists Mi a such that AM = B. U 3.4.3. If AA' = BB' then A and A' are joinable precisely when B and B' are. Proof: Suppose that A and A' are joinable and let A,A' ca,a' with a I a'. By 3.4.2, there exists an Msuch that AM = B. Then B = AMIaM, a'M and from 3.1.2.1 and 3.1.6.13 it follows that aM I a'M, a 1 acM, and a' 11 a'M. By 3.1.6.11, AA' = BB'; A,A'ja,a'; BlaM,alM; a  aM; and a' 11 a'M. Thus, B' 1aM, cM' and, B and B' are joinable. U 3.4.4. If a 1 a'; a 1 P3; a' p3'; and[a,a'] 11 [p3,3'], then P3 1 3'. Proof: Because [a,a'] 11 [13,3'], then there exist A,A'; A,A'Ia,a and there exist B,B'; B,B' P133', such that AA' = BB'. Then for AM = B, as in the proof of 3.4.3, [P3,3'] = gSBB' = [aM,a'] II [a, a']. That is, BIaM, M,3,3'; with 3,aM 11 a, and 3',aCM 11 a'.Thus, 3 = a' and 3' = aM by 3.1.6.12. Hence, P3 I 3' because aM I aM. N 3.5. Construction of the Field The basic construction. For the construction of the field we will follow the path of Lingenberg [201. Throughout this section let a e M and 01 a be fixed. Define the sets: Oa = g E La : 0 e g} and Da(O) {agCh : g,h e O}. Proposition 3.5.1 The set Da(O), acting on the points ofX3, is an abelian group. Proof: By 3.1.7.2 we may write 0 = a3 = yrip3 with a = yTI; y,"1,3 E g; and y, l, and P3 mutually orthogonal. Thus, g = [a,y7,1] e a; 0 e g, and Oa 0. Because each 53 ag e Ca is involutory then lx e Da(O). Now let aa,ab,ac,cd E Dca(O) where a = [a,a',aa'], b = [a,13',ap13'], c = [a,y',ay'], and d = [a,6,a8] are nonisotropic. Now &y'P' = E _ a with 0e by Axiom 9 and f= [a,, cM] e Oa. Thus, for AIa, aabGcad(At ) = AkY'c'' = A;' = aaC(AA). Hence, aabacad = aaaf e VDa(O). Because each al is involutory for 1 E L, ca1 = al and (aCaab) = abCaa e Pa(0). From Axiom 9 and the calculation above, the product of any three of 5,y7', 3', and x' is an involution.Thus, A8y'P'a' = AW'Y'a' = A'a'8' and CaabOcad = CcaCaCab.That is, Da(0) is abelian. Clearly, TEa(0) is associative so that Da(O) is indeed an abelian group. U Lemma 3.5.2. Let g be an isotropic line in Xa with 0 e g. Then for every CaF/h E D6 (0), alCah(g) = g. Proof: Let I = [a,3,ap] and h = [a,y,ay] be lines in La with 0 e 1,h. By 3.2.3.12, if 01 M, with P,7y 1 a then gfY I1 g. But OPY = 0 so that ahalT(O) = O0Y = 0. Thus, g Y = g by 3.2.1.10. Hence, ahGt(O) = g, for all ahal e Da(0). U Lemma 3.5.3. Let g be an isotropic line in X. through 0. Then for every A,E E g, E # 0, A # 0, there is a unique Calah e Da(O) such that Clah(E) = A. Proof: By 3.2.3.11, there exist y,8 1 a with O0y,6 such that EY6 A. Take I = [a,y,ay] and h = [a, 8,a6]. Note that if E = A then EP6 = E implies that Ey = E8 or El = Eh, which implies that / = h [30]. To show uniqueness, suppose that Gaah,ak, l e Da(0), where / = [a,8,a6], k = [a,s,ae], g = [a,y,ay], and h = [a, P3, ap3] are lines in X?(0) and E # Eha = Elk. Then Ehakl = E and EOl = E. By Axiom 9, p3y = X with 01 X, k 1 a. Thus, m = [a, X,aX] e Oa; akaaah = am, and E = Eml. Let E a',8' with xa' X., 8'18, and put M= ac'X and N = 6'8. Now Ea',a,6' with a'iX, a I X, 6' L 8, and a I 5, so that a', 86' 1 a by 3.1.6.18. It follows that MA = a'a X = a'k = Mand Na = 8'a8a = N. So that Mla,X ; NJa,8; and m = goM and / = goN. Then because E"m' = E, we have EP5 = E; EX = E5 and 54 EM = Ea'x = E6'8 = EN and M = N by 3.1.6.14. So m = goM = gON = and al = a,,. Therefore Gaah = (Tkal. U Let us denote this unique map by 8A. So for all A e goE, A 0, (i) 8A(O) = 0 (ii) 8A(E) = A (iii) 6A E Da(0). Translations. For every pair A,B of distinct points we can define a translation TAB : 1 X given by TAB(A) = AAB = B. We now restrict our attention to a set of translations defined on Ta and we note that if A,B, CaI then D = ABCIca by 3.1.4.6. Thus, TAB : ta * Xa is a welldefined map for all A, BI a. Let C = goE be an isotropic line in Ta and define To = {TOA : A e K)}. Theorem 3.5.4. The set Ta is an abelian group. Proof: Let TOA, TOB e Ta. Because C = BOA = AOB e 1C, for XI a, we calculate (TOA o ToB)(X) = TOA(XOB) = XOBOA = XOC = To((X). Thus, TOC = TOA o TOB e Ta. Also, Too(X) = XOO = X. Hence, Too E= Ta is the identity xa on 3Ea. To find To1, we compute (ToA o TOAo)(X) = XOAOA = XOOAOOA = X = XOAOA0 = (TOAo o ToA)(X). Hence, T7oA = TOAo, A = OAO e KC, and To1 7E Ta. Clearly, TOA o (ToB o Toc) = (TOA o TOB) o Toc for A,B,C E IC. Therefore the action of Ta is associative and Ta is a group. Because ABC is a point and hence, an involution for all points A,B,C, then for A,B e KC and X xa, (TOA 0 TOB)(X) = XOBOA = XOAOB = (ToB TOA)(X). Therefore, Ta is abelian. U Lemma 3.5.5. For all TOA e Ta, TOA(IC) = KC. Proof: Because OAB E 1C for A,B e 1C then TOA(K) c C for all TOA e Ta". Now let 55 C E IC and A e /C. Then OCA = D e KC so that C = ODA and TOA(D) = C. Hence, each TOA e Ta maps KC onto KC. U Lemma 3.5.6. Let TOA E Ta and g c Xa. Then TOA(g) = g if and only if g is parallel to /C. Proof: Let gHF = g be a line in Xa such that TOA(gHF) = gHF. Then TOA(H) = HOA e gHF and gHF II IC by 3.2.1.9. Conversely, suppose that gHF 11 IC. Then again by 3.2.1.9 it follows that for B e gHF, ToA(B) = BOA e gHF and TOA(g) = g. U Lemma 3.5.7. IfA,B e h, h c Xa, and h 11 AC, then there exists ToC E T"a such that Toc(A) = B. Proof: By 3.2.1.9 we have C = OAB e AC and it follows that B = AOC = Toc(A). U Lemma 3.5.8. For each A e IC, there is a unique TOA E Ta such that TOA(O) = A. Proof: Clearly, TOA(O) = OOA = A. So suppose that Toc(O) = A. Then OOC = C = A. U We denote this unique translation mapping of 0 into A by TA. Lemma 3.5.9. If a e Da(O) and TA E Ta, then cOTAa = Ta(A). Proof: Let a = agah where g = [a, P3, ac3], h = [ay, ay] E Oa. Then for any XI a, (agChTAohaTg)(X) = (XOYOA)y = (XJPY)YPOYOAYP = XOAY = Tag(AA)A). U By 3.2.3.8 there exist precisely two isotropic lines in 3Xa through 0 : IC = goE and C' = gOF. We define multiplication and addition on the points of AC so that the points of C form a field. For A,B e AC, define: A + B = (TB o TA)(O) A B =B A 8A)(E), where A,B # 0 and E E /C is the multiplicative identity. A 0 O.A = 0. Theorem 3.5.10. For every A,B e IC, TA+B = TA TB. 56 Proof: For XI Za, TA+B(X) = TAOB(X) = XOAOB = XOBOA = (TA o TB)(X). U Theorem 3.5.11. For all A,B 0 in kC, 8A.B = 6A o 8B. Proof: Let 8A = CJaCa' and 8B = Gbabb. Then A B = 6BA(E) = Eaab'b. Put Cc = abCraaal, then, (CbCTc e Da(O) and C bac(E) = Ecb = Ea'abb = A B. Hence, by 3.5.1.3, 5A.B = abTc = (b(Tbf(a(aa = CaaabfabCra' = GaaaC(Fb(Tb' = 8A4B M Hence, (/C,+) is a group isomorphic to Ta and (C\A{0},) is a group isomorphic to Da(0). It remains to show that the distributive laws hold. Theorem 3.5.12. Let A,B,C e IC, then (A + B) C = A C + B C. Proof: If C = 0, then (A+B).C=0=0 .0+0.O=A.C+B.C. If C # 0, then we compute (A + B) C = 6c(A + B) = 6cTA+B(O) = 6CTA+B c'(0) = 6CTATB8c'8(0) = 6CTA8' 8CTB8 = T6((A)T6(.(B)(O)= TA.CTB.C(O) = TA.C+B.C(O) = A C + B C. Because multiplication is commutative, C.(A+B) = (A+B)C=A C+BC=C.A+C.B. Hence, ()C,+,) is a field. U Ordering the field and obtaining R. To order the field IC we make use of the following 1211. Let F be a field and A 1,...,An E F. If A 1,... ,An # 0 implies that IL A2 # 0, then ? is called formally real. Theorem 3.5.13. (ArtinSchreier) Every formally real field can be ordered. U To make the field AC formally real the following axiom is posited. Axiom F. (formally real axiom). Let 0, E a, a e M with 0 and E unjoinable. Let k, 8,11 e P. If 0 X, ,i and ,6, q a then there is a y e P such that 017y, y 1 a, and EkXk6OEkqk9 = EXyXy. Theorem 3.5.14. If Axiom F holds on 1C, then /C is formally real. Proof: Let I = [a,?,caXj] e 0a and let Ugah E D a(O). Then from Proposition 3.5.1, 57 Cgahat = al for some I e Oa and GgCh = aatC. Clearly, if e Oa0 with a/'at = a/at then we have CT' = a1 and 1' = 1. Hence, for all Ogah E Da(0), there is a unique 1 e Oa, such that CgTh = IaGt. So if 0 # A E KC, then we may uniquely write 6A in the form caaa,; that is,for every 0 # A e KC, there is a unique a, a = [a,a',Ca'J E] Oa such that A = aaat(E) = E1' = E^'. Suppose that 0 A = EX'. Then A2 = Ek"'k" = 0 implies that E = O'k'k = 0, a contradiction. Hence, if 0 # A e /C then A2 # 0. Now suppose that A,B e KIC and A,B # 0. Let a = [a, x',aa'], b = [a,P3,4ap] e Oa such that A = Ea' and B = E'O. Then by Axiom F there exists a y e P such that 0 171 a and A2 +B2 = Ek'a'0OEx44 = EkyX. (3.5.1) Since Oy 1 ca then c = [a,y,ay] E Oa,,acat e TDa(O), and C = acat(E) = Etc = E e C. So equation (3.5.1) reads A2 +B2 = C2. If C2 = EYy = 0O, then E = OYky = 0. Since E 0, it follows that if A ,...,An /C are all nonzero, then Y' A2 0 and kC is formally real. U To finally obtain a field isomorphic to the real numbers, R, we add the least upper bound property to our axiom system. Axiom L. If0 # A c KC and A is bounded above, then there exists an A E KC such that A >X, for all X e A, and ifB e KC with B > X for all X E A then A < B. The only ordered field up to isomorphism with the least upper bound property is the real number field, R. Theorem 3.5.15. The field K constructed above, along with Axiom F and Axiom L, is isomorphic to the real number field, R. 58 In the next section an affine vector space is constructed from products of pairs of points. The scalar multiplication is obtained by adapting and extending the definition of multiplication of elements of KA. 3.6. Dilations and the Construction of (X,V, C) The additive group V and dilations. First we construct a vector space V over the field 1C. Let V = {OX : X e X }.First note that the product, AB, of any two points A,B e X, is in V because AB = 0(OAB) = OOAB. We view the elements OX e V as directed line segments with initial point 0 and terminal point X on the line gox. We define an addition on V by setting OX+ OY M OXOY. The product of three points is a point, so XOY = Z e X, OXOY = OZ E V. Theorem 3.6.1. (V,+) is an abelian group. Proof: Let X, Y,Z e X be distinct. Then, OX+ OY = OXOY = OYOX = OY+ OX, and addition is abelian. The zero vector.is lo since 1I = 00 E V and 1o + OX = loOX = OX = OX+ 1o. To complete the proof we calculate OX+ OX = OXOX = OXOOXO = OXXO = 00= 1, so OX = OX. (OX+ OY) + OZ= (OXOY)OZ = OX(OYOZ) = OX+ (OY+ OZ). Hence, (V ,+) is an abelian group. E We still need to define a scalar multiplication of KC on V. To do this note that in an affine space the group of dilations with fixed point C is isomorphic to the multiplicative group of the field. Noting this, we geometrically construct such a group of mappings and use these mappings to define our scalar multiplication. A dilation of X is a mapping 5 : X '* X which is bijective and which maps every line of X onto a parallel line. J23, p.37j Theorem 3.6.2. [23, p.42] A dilation 5 is completely determined by the images of two points. 59 Proof: Let 5 : X ' X be a dilation and assume that 5(X) = X' and 8(Y) = Y1 of two points X, Y e 3X are known. We must show that the image of any Z e X is known. Suppose that Z 0 gxry. Then clearly Z # X and Z # Y and we consider the two lines gXZ and gyz. Observe that Z e gxz n gyz. If these lines had a point in common besides Z, they would be equal and then gxz = gyz = gXY. But this is impossible because Z o gxy. Therefore, {Z} = gxz r gyz. Since 5 is bijective, from settheoretic reasons alone, 8(gxz r gyz) = 5(gXZ) n 6(gyz); or equivalently, {8(Z)} = S(gxz) f((gyz). In other words, the lines 6(gxz) and S(gyz) have precisely one point in common, namely, the point 5(Z) for which we are looking. The line 6(gxz) is completely known because it is the unique line which passes through X' and is parallel to gxz. Similarly, the line 5(gyz) is the unique line which passes through the point Y1 and is parallel to gyz. because 5(Z) is the unique point of intersection of (gtA7) and 5(grz) (3.2.1.1), the point 5(Z) is completely determined by X, and Y1. Conversely, assume that Z e gxy. If Z is X or Y, we are given 5(Z), so assume that Z # Xand Z # Y. By 3.4.1, there is an a e (M such that X, Ya and there exists a PI a such that P o gxy. Then Z 0 gxp and from the previous paragraph, 5(P) is known. Hence, using the line gxp instead of the line gxy, we conclude from the earlier proof that 5(2) is known. U To define a scalar multiplication, we fix a timelike line t and use it to geometrically define dilations. To aid in the construction, the following facts are used to add the appropriate axioms. In a threedimensional or fourdimensional Minkowski space, if t is any timelike line through a point 0 and g is any other line through 0, then there is a unique Lorentz plane containing the two lines. Two distinct isotropic lines intersecting in a point in Minkowski space determine a unique Lorentz plane. Desargue's axiom, D, holds in any affine space of dimension d > 3. 60 Axiom T. If O E t,g where t is timelike or t and g are both isotropic then is a unique a e M such that g,t c Xa. Axiom D. Let g\, g2, and g3 be any three distinct lines, not necessarily coplanar, which intersect in a pointO. Let PI,Q\ E gj; P2,Q2 e g2; and P3,Q3 e g3. If gPIP3 11 gQ1Q3 andgp2P3 II gQ2Q3 then gpp2 II gQIQ2" Axiom R. LetO e g1,g2; P1,Q1 e gj; andP2,Q2 E g2. Ifgpp 2 goQQ2 then go,P OP2 = gO.Q i OQ2. Axiom T refers to the first statements. Axiom T is used to put an isomorphic copy of the field on every isotropic line through 0. A scalar multiplication is then defined in a manner similar to the definition of multiplication for the field elements. Axiom D is Desargue's axiom, the "dilation" axiom. Axiom D ensures welldefined dilations with the standard properties of such maps. Axiom R is used to distribute a scalar over the sum of two vectors. The dilations are constructed next. Let IC c Xa, a e M. By 3.1.7.2, there exist a 1,a2 e G such that O0ac1,a2 and a = aIla2. Because l0a, then Oca = P3 e and 0 = a13 = aIaC213 with aC1,2 1 3 by 3.1.6.18. Let y = a1I3 e M. Then ya = Pa3cIaIa2 = P3a2 e M, so that t = [ca,y,a cy] c 3a is timelike and 0 E t as 0 a, ca 1,a2,13 implies that 01a,y,acty. So let t = [a,y,6] E Oa be any timelike line through 0 and /C' c Xa the other isotropic line through 0. Then for each AI e t, there is a unique A e IC and there is a unique B e /C' such that A, = AOB. Because A, e t, then B = At. (From this point on denote a line reflection ag(X) by Xg. Thus, xghr means Crrahag(X) = CgChcr(X).). For each A, e t, there is a unique A e IC such that At = AOA'. Similarly, for each A in IC, there is a unique AI in t such that At = AOA'. Thus, there is a onetoone correspondence between the points of /C, the field, and the points of t. Fix Et = EOE' as the unit point on t. For each 0 # A e IC, we use t to construct a dilation WA of X in the following way. Let X e X. 61 IfX o t, then by Axiom T, there is a unique 71 e M such that gox,t c Xrj. From Sections 3.5 and 3.6, X1 is an affine plane and our definition of parallel lines in X9 is equivalent to the affine definition. So, there is a unique line h c 3eX such that At e h and h II gxE,. Because gox rn t = {O} 0, h ) t= {At,} 0, andh II gE,x, then h n gox = {B} 0. In this case, set 8A(X) = B. If Xe t and X # 0, then because t,K a Xa with a E M, there exists a unique g c Xa such that A e g and g 11 gxE. Because gxE n t = {E} # 0 then g r t = {B} # 0. Set 4A(X) = B. If X = 0, set 8A(A) = 0. Note that 8A(X) e gox by construction. It is clear that 8A : X i X is onetoone. We find it useful to make some observations. Let 0 A e kC and recall that E e KC is the multiplicative unit point; E .A =A, forA e C. Put At =AOAt. Lemma 3.6.3. For the map 8A defined above, the following are true: 1 8A(E)=A. 2 8A(E) = A. 3 8A is onto. Proof: 1. Et = EOE' and A, = AOA' imply that EEt = O' and AA, = OAt. If EEt = OE' = l then 0 = E' and 0 = Ot = E. Similarly, because A # 0 then AAI = OA' 1o. By 3.2.2.9, At = A' e K' as 0,A e KC, AC is isotropic, and 0y 1 a. Similarly, E' e K'. Thus by 3.2.1.6 we have gEE, II goE' = K' II gAA' and gEE, II gAA, by 3.2.1.7. By 3.2.1.8, h = gAA, and h r goE = gAA, r =C {A}. Hence 8A(E) = A. 2 6A(Et) = At. Again, ggA(E, = gAA, II gEE, and gAA, r t = {AI}. 3 Let P e X and P # 0. If P e t then by 3.2.1.8, there is a unique line g such that Et e g, g II gA,P, and grgop = {Q}, for some Q e X. Then it follows that 6A(Q) =P. IfP et then 8A(Q) = P, where {Q} =g tn, E E g, and g II gAp. 0 62 Lemma 3.6.4. If C # D, then gcD II g8A()A(D); hence, for all 0 # A e KC, 8A is a dilation of X. Proof: Let gj = t, g2 = goc, g3 = goD, Pi = Et, Q = A2 e g, P2 = C, Q2 = 8A(C), Q2 E g2, P3 = D, and Q3 = 8A(D). Then gcE, II g,8() and gDE, II g,,A(D)" Thus, by Axiom D, gcD II gA(C)A(D) N Consider now the plane Xa. Recall that by 3.5.1.3, for 0 # A e 1C, there is a unique 8A e Dac,(O) such that SA(E) = A, where 5A is of the form Cgch for g,h e Oa. From Axiom 9 and the proof of 3.5.1.1, for any r E Oc, Cgahor = CTw for a unique w E Oa. That is, for every fixed r E Ox, every 0 # A e KC can be uniquely written in the form 85A = awCar where Cw(Er) = A. (The uniqueness follows from the fact that (At)w, = (Ar)w2 implies wl = w2 if A'r A [30].) Therefore, for every 0 # A e AC, there is a unique a e Oa, such that E"ta = A. That is, there is a unique a e Oa, such that GaOt = 6A. Also, every P E Xa, PI a, can be uniquely written as P = P07IOT2, where P\ E AC and P2 E AC'. Let P2 = 2'e C K. Then we may uniquely write P = P\OP', where P1,P2 e KC. Define the map 6A : Xa + Xa, by A(P) = PtaO(Ptt)', for 0 # A e 1C, Pla. Theorem 3.6.5. The map 8A : XJa * Xa is a dilation on Xa with fixed point 0 and dilation factor A, for all 0 # A e K. Proof: If P'laOP'a = Qa'OQ'a, then by unicity, pa = Qp. So P1 = Q0, Pft = Qa and P2 = Q2. Hence, P = P OP2 = Q I OQ2 = Q, and 6A is injective. Let QIa with Q = QOQ2; QI,Q2 e AC. Let PI = Qg' E AC and P2 = Q0t e C. Then P = PIOP Iaot and 8A(P) = P1aOPfI = (f)O(f)2t = Q I0OQ = Q. Therefore, 8A is onto. We claim that if P# Q, then gpQ I\ gp,, First we show that 8A(P) e gop. If P E IC, then P = PO0 and 6A(P) = pta00o = ptao0 = pta e 1C. If P e AC', then putting R = P' e AC, we have P = OOR' and 6A(P) = OtaORtat = = OOR'a' = Rtat = pat E C'. 63 So suppose that gop g is nonisotropic and write P = PIOP2, where P1 = P'. If 0 e g, then XOY e g <> X = Yg, where X e /C and Y e /C'. Thus, 8A(P) = pla'Oa and P,' pga = ptag which implies that 8A(P) e g. Claim. 8A (POQ) = 8A(P)06A(Q), for all P,Q e Xa. Let P = PIOP2 and Q = Q\OQ2. Then POQ = PIOPQIOQOQ2 = (PIOQi)O(Pz2OQ2)t, with PiOQi E KC and (P20Q2)' e IC'. Therefore, 8A(POQ) = (P1OQ1)tao(P20Q2)tat = ptaQQa'ltQalf = (ptaoptt)oQ(Q21ltoa) = 8A(P)OA(Q). The claim follows. Proof of (iii): Assume that P,0, and Q are collinear. Then there is a line g, such that P,0,Q e g. Thus, 8A(P),8A(Q) e g and gpQ = g II g = gA(P)'A(Q) Conversely, suppose that P, Q, and 0 are noncollinear. Let g, = gop, g2 = goQ, and g3 = go,pOQ. Then g\,g2, and g3 are distinct lines through 0. Indeed, if g3 = g1, say, then POQ e gop and we would have OP(POQ) = Q e gop, which contradicts our assumption of noncollinearity. Now, P(POQ) = OQ and 8A(P)(8A(POQ)) = 8A(8A(P)0OA(Q)) = 08A(Q) gP,POQ II goQ = gO8A(Q) II gSA(P)8 (POQ)' and gP,POQ IIgA(P)A(POQ) Similarly, Q(POQ) = Q(QOP) = OP; and 8A (Q)(A (POQ)) = 8A (A (Q)00A (P)) = 08A (P). This implies that gQ,POQ II gop and gop = go6(p) II g8A(Q)6(POQ)" Hence, gQ,POQ II gSA(Q)N(POQ)" By Axiom D, it follows that gpQ 11 gA(P)A(Q) Therefore, 8A : fa i+ 3Xa is a dilation on Xa. U Now we show that 8A = 8A on Xa. Because 8A and 8A are dilations on Xa, a dilation is uniquely determined by the images of two points,and 8A(O) = 0 = 8A(O), then it suffices to show that 8A(E) = 8A(E). By definition of 8A, 8A(E) = A. By Lemma 3.6.3, 8A(E) = A. 64 To extend the above idea to any rI e M such that t c Xq, let a # r1 e M such that t c Xi. Let AC1 and /k2 be the isotropic lines in XrC through 0. Because t is nonisotropic, then by Axiom 14 there exist y,6 such that il = y8 and t = [ry,S]. Thus a, : n '* X, is welldefined. Every B e t c Xq may be uniquely written as SB= B\OB2, where Bi e AC,. Because 0 e t and t is nonisotropic, then B2 = B'. Thus, B = BIOBt where B1 e AC1. So, in particular, there are unique E\,A\ e IC\ such that Et = EiOEP and A = AIOA', for AI e t.As before, there is a unique aI e O, such that Ea' = A 1. Hence, every X e Xri can be uniquely written as X = X\OX2, where X\,X2 (e AZ1 c X. This defines a map 6A8 : Xn * X, given by 6A(A)= Proposition 3.6.6. Let 8A : X X3 be the map defined above. Then 1. '5A4 is a dilation on X. 2. 6A = 8A on . 3. ifP, 0, andQ are collinear points in X.r, not necessarily distinct, then 6An(POQ) = 6Aq(P)O5Aq(Q). 4. Moreover, 8A(POQ) = A(P)O0A(Q), for every P, Q\ r\ such that O,P, and Q are collinear. U To obtain a scalar multiplication on V, for all 0 # A e AC and all OX e V, define A OX = 8A (O)A (X) = 086A (X) and O. OX = 1. We now verify the vector space properties. Lemma 3.6.7. IfA,A' e IC and OP e V, then (A + A') OP = A OP + A' OP. Proof: Suppose P e I and P # 0, and recall t,AC c a. Let P, EIa,P; AIy withy  P3; A'y', y' 3 P; and O06 with 6 II P3. Then y 11 y' 1I 6; g, = [ac,y] I gEp with g, c ; 92 = [a,y']  g p with g2 c X.a; and g3 = [aX,6] I gEp with g3 c XC,. Thus, 6A(P) e gi r t and 8A(P) e 92 t. Thus, A(P)I)a,y and 8A(P)la,y'. It follows that 65 8A(P)O9A'(P)Ia,y68Y'; 8A(P)08A(P) e t; 787' = 1I P3; and AOAt'I ,875' = s. Therefore, AOA' e [oa,] II gEP and 8A(P)06A(P) e t n [a, E]. Hence, 8AOA'(P) = 8A(P)08A'(P). Suppose P o t. Because gop,t c X3, then replacing E with Et, A with At, A' with A,, and c with ri in the first part of the proof and the result follows. U Lemma 3.6.8. IfA e C and OP, OQ e V then A (OP + OQ) = A OP + A OQ. Proof: We need to show that 8A(POQ) = 8A(P)08A(Q). Suppose P,0, and Q are collinear. Let g = gop = goQ = goyOQ and r e M such that g,t c Xr. Then the result follows from Proposition 3.6.6(iv). Conversely, suppose that P, 0, and Q are not collinear. Because P, 0, and Q are not collinear then P # Q, and from Lemma 3.6.4 it follows that gpQ 11 gsA(P)8(Q)" Applying Axiom R we obtain goPOQ = gO8A(P)8A(Q). That is, 8A(P)O0A(Q) e goPOQ by construction. Again by Lemma 3.6.4, gSA(P)A(POQ) II goPOQ and 8A(P)SA(P)OSA(Q) = 08A(Q). This implies that gA(P),8A(P)08A(Q) II go08,(Q) = goQ 11 gp,POQ, as P(POQ) = OQ. Thus, g6A(P)..(/i'.)4(Q) = gSA(P)SA(POQ) because both lines contain 8A(P) and are parallel to gP,POQ. Since g8A(P).6A(P)O0A(Q) gP.POQ = {8A(P)OSA(Q)} and gsA(I, ,,POQ) n go.POQ = {8A(POQ)}, then 6A(P)06A(Q) = 6A(POQ). M Lemma 3.6.9. IfA,A' CE a and OP e V then A (A' OP) = (A A') OP. Proof: We need to show that 8A(6A(P)) = 6A.A'(p); that is, 86A o 6A' = 8A.A. Now 5A o 6A' and 6A.A' are dilations on X and 6 A.A'(O) = 0 by construction, so it suffices to show that (8A o 6A,)(E) = 8.1 (E) and (6A o 8A')(0) = 8A.A,(O). Now E e /C c X, so on Xa, 6A o 6A' = 6A o 64' and 6A.A' = 8A.A'. But on AC, by Lemma 3.5.11, 6A o 6A' = A.A'" Therefore, (8A o 8A')(E) = A (A' E E) = (A A'). E = 8AA'(E). U Lemma 3.6.10. For E e 1C, the multiplicative unit, and OP e V, E. OP = OP. 66 Proof: We need to show that 8E = 1.. Now l1 is clearly a dilation on X and 1(0O) = 0 = 8E(O). Thus, 1.ic(E) = E= E E = E(E) = 6E(E). U Theorem 3.6.11. The space (V,1C) constructed above is a vector space. U The triple (X,V,KC), is an affine space.1[23,p.6] A set X along with a vector space V over a field /C is an affine space if for every P e V and for every AX X, there is defined a point PX e X such that the following conditions hold. 1. If v,w E V and X e 3C, then (f + ,)X = v("X). 2. If 0 denotes the zero vector, OX = X for all X e X. 3. For every ordered pair (X, Y) of points of iX, there is one and only one vector v e V such that vX = Y. The dimension n of the vector space V is also called the dimension of the affine space T. Theorem 3.6.12. (iX,V,/C) is an affine space. Proof: If OV,OW e V and X E X, we have (i) (OV)X= OVXe X. (ii) (OV + OW)X = (OVOW)X = OV(OWX). (iii) OOX= loX =X. (iv) for Ye X., OYX = Z e iX and (OZ)X = Y. Now if OPX = Y, then OPX = OZX, P = Z, and OP = OZ. M 3.7 Subspaces and Dimensions In this section we show that our lines and planes have the proper dimensions. We are then able to conclude that (V, C) and (t, V, C) are fourdimensional spaces. Proposition 3.7.1 Let g be any line through 0 and put 0(O) = {OA : A e g}. Then (0) is a one dimensional subspace of V. Proof: First note that 15 = 00 e g(0), so the zero vector is in g(0). Let A,B E g. Then C = AOB E g by 3.1.4.7 and OA + OB = OAOB = OC e g(0). From Section 3.6, 67 5A(B) e g, for all A in /C and all B in g. So A OB e ((O) for all A e /C and OB e (O). Hence, g(O) is a subspace of V. It must now be shown that the dimension of ,(O) is one. If g = t, then 0(O) = (OEI) because for every At e t, At = 8A(Et) by Lemma 3.6.3. So suppose that g # t and fix B E g. Let h = gBE,. Then for all 0 # D E g, there ia a unique such that D e d, d 11 h, and dr t 0. Put {Ft} = d t. Then for Ft = FOF' with F e /C it follows that 6F(B) = D and OD = F. OB. Hence, A(O) = (OB). U Corollary 3.7.2. Following the terminology of Snapper and Troyer [23, p.11], g = S(O,g(O)) {A = O(OA) : OA e (O0)} is an affine subspace of dimension one. Proposition 3.7.3. Let a e P with Oa and put ta(O) = {OA : AIa}. Then ta(0) is a two dimensional subspace of V. Proof: Clearly, 1e = 00 e ta(O), so the zero vector is in 1tc(O). Let C, DI a and A,B e 1C. Then by 3.1.4.7 we have COD = FI a and by Lemma 3.6.3, A (C) e goc c Xa and 8B(D) e gOD c Xa so that 8A(C)O0B(D) a. It follows that OC + OD = OCOD = OF e tra(O), and A OC + B OD = O0A(C) + 08B(D) = O(8A(C)OB(D) E 1a(0). Hence, ta(O) is a subspace of V. Thus, it remains to show that ta(O) is two dimensional. There are two cases: a e M and a e g. Suppose first that a e M. We construct a basis for lt(0) using isotropic lines. To this end, let C1 and KC2 be the isotropic lines in Xa through 0. For any Pla we may uniquely write in Xa, P = PLOP2, with PI e IC1 and P2 E IC2. 68 From Proposition 3.7.1 above, kC1(O) = (OB) for any 0 # B e /C1 and fC2(O) = (OC) for any 0 # C e /C2. From this it follows that P1 = 8A(B) and P2 = 8A'(C) for some A,A' e 1C. Thus, OP =A OB+A' *OC and {OC,OB} span ta(O). Now ifA OB+A'. OC = Io, then O8A(B)O6A(C) = 1. This implies that 8A(B)08A'(C) 0 and 06A'(C) = 6A(B)0. Because O,6A(C) /EC2 and 0,8A(B) E ACI, then either /C1 II AC2 or OA,(C) = 1 = 5A(B)0. But /CI 9 /C2, so A,(0 = 8A(B) = 0. Because ca e M, then from 3.3 and 3.6 there exists t',a' e Oa such that 8A(C) = CVa' = 0. This implies that C = 0a' = 0 or A' = 0. By assumption C 0 and D # 0, thus it must be the case that A = A' = 0. Hence, {OC,OB} is a linearly independent set in ta(O) and Ia(0) is two dimensional. Suppose now that a e g. We construct an orthogonall" basis. By 3.1.7.2, there exist P3,y e g such that 0 = c0y and a 1 P31 y 1 a. Let x = [a, 3,a c] and y = [a,y,ay]. Note that ifx =y then by Axiom 14, a(3 = y and 0 = a13y = 1I, so x y. Let Pl a and let 13',y'IP with 13' 13 and 7y' y. Put P1 = Pp1313' and P2 = 7Y'. Now PIa,13'; a I 13; 13' 1 13, so that a I 13' by 3.1.6.18. Because PIa,y'; a 7 y; y' 1 y, then a 1 7y'. Thus, P' = (13p,)c = 131t, = 1313' = Pi and P( = (n')r = 7' = P2, so Pi,P2 a. This implies that PI Ia,P3 and P2Ia,y so PI e x and P2 e y. If Q = PiOP2 then Q = POP2 = 13'PP313ayyy' = 13'ay'.Because 13' 1 a, let 8 = a13' E P. Then Q = 86y' and 6 1'. Thus PI P13',ac implies that P 13'a = 8. Thus we have P,QjS,y' with 6 1 y', and P = Q by Axiom 3. That is, P = POP2 with P1 e x and P2 E Y. Let 0 X e x and 0 Y e y. Then from Proposition 3.7.1 above we have x = (OX) andy = (OY) and there exist A,B e IC such that OP = A OX+B OY, where 69 P1 = 8A(X) and P2 = 5B(Y). Hence, {OX,OY} spans a(O). IfA OX+B.OY= 1I, then we obtain 0OA(X) = 8B(Y)O. Because x Jf y, it follows that 8A(X) = 0 = 5(Y). Let Xin be the unique plane containing t and x and /CI and /C2 the isotropic lines in X3 through 0 Then we may write 8A(A) = X1a(OX'la', where X1i e /Ci, and X E K/2. As above it follows that A = 0 and similarly, B = 0. Therefore, 1a(0) is two dimensional. M Corollary 3.7.4. X. = S(O,t_(O)) is an affine subspace of dimension two for all a E P, 0La. Theorem 3.7.5. (V,/) is a fourdimensional vector space and hence, (X, V,C) is a four dimensional affine space. Proof: Let OP e V and let 0 = a43 with a e M and P3 e g. Let PIa',3P' with a' I a and P3' 1 P3. Put Y' = xa'c and P" = P3'P3. Then Q = P'OP" = a'aacPP313P3' = a'3'. Thus, P,Q Icc', 3' with a' 1 3'. Therefore P = Q = P'OP" with P'Ia and P" I3. Since ta(O) and .tp(O) are twodimensional then there exist bases {OZ, OT} c .tc(O) and {OX,OY} c tp(O0) such that OP' = A OZ+A' OT and OP" = B OX+ B' OY for some A,A',B,B' e /C. Thus, P = A. OZ+A' OT+B OX+ B' OYand {OX, OY, OZ, OT} span V. If A *OZ+A' *OT+ B OX+ B' OY= lo, then in particular, OP = OP' + OP" = lo. This implies that OP" = P'O. So either gOP" II gp'o or P" = P' = 0. But gop" 1 gop' and therefore, A OZ+A' OT = OP' = =OP" = B OX+B' OY. As was shown in Proposition 3.7.2, we obtain A = A' = B B' = 0 and the result follows. U 3.8 Orthogonality In this section we extend the definition of orthogonality to include lines and then use this definition to define orthogonal vectors. 70 Definition 3.8.1. Let g and h be two lines. We say that g is perpendicular to or orthogonal to h, denoted by g L h, if there exist Oa, P3 E P such that g c Xa, h c Xp, ax 1 and P = (a e g,h. In this case, P is the point of intersection of g and h. Lemma 3.8.2. If g and h are isotropic then g is not orthogonal to h. Proof: This follows directly from our definition above and from our definition of a 1 p. For if a 1 P3 then one of a and P3 must be in G and by Axiom 12, all lines in a plane 5Xp for P3 e G are nonisotropic. U In Minkowski space, if g and h are two isotropic lines then g L h <> g 1 h. Thus we extend the above definition in the following way. Definition 3.8.3. If g and h are isotropic lines, then g I h <> g 11 h . Definition 3.8.4. If g is a line and a e 7) then we say that g is orthogonal to or perpendicular to ta, g I Xa, if there exists a P e P such that g c 1p, P3 1 a, and P = ap g. In this case, P = ap is the point of intersection of g and Xa. Definition 3.8.5. For every 1I OA,OB e V, we say that OA is orthogonal to OB, OA OB, if and only if goA L gOB; that is, there exist (x, a3 e P such that goA c X3, goB c Xp, and 0 = ap. For the zero vector 1Q = 00, we define 1I OA, for all OA E V. Lemma 3.8.6. From 3.1.2.1 and 3.1.2.2 it follows that fort e (5: i g 1 h <h g L h:. (hi) g I T X~ <> g~ I X. (iii) OA I OB <> (OA) I (OB),. U Lemma 3.8.7. If g is a nonisotropic line then g is not orthogonal to g. Proof: Ifg I g then there exist a,P e P such that g c X, Xp; a I 3, and P = ap e g. But for every point Q e g, Q I(xa,P with a 1 P3 which implies that P = Q by Axiom 3; that is, g is a line which contains only one point, which contradicts the definition of a line. U 71 Additional axioms and their immediate consequences. To complete our preparations for defining our polarity and thus obtaining the Minkowski metric, we recall our final three axioms. Axiom U. (UL subspace axiom) Let O,A,B, and C be any four, not necessarily distinct, points with A, 0 a; O,BI ; O,C y,6 and a 17 y and 1 6. Then there exists k,& e P such that X 1 ; O,AOBIX; and O,CCe. Axiom S1. Ifg c Xa, a e G, h c X p, P e g, and there exists y,8 E P such that y 16 5; y6 E grn h; g c Xy; and h c X 6 then there exists 8 e 9 such that g,h c .. (If g and h are two orthogonal spacelike lines then there is a spacelike plane containing them.) Axiom S2. Let g and h be two distinct lines such that P e g r) h but there does not exist P e P such that g,h c Xp. Then either there exists a,y e P such that a 1 y, g c Xa, andh c Xy or for allA E g, there exists B e h such that P andAPB are unjoinable. Lemma 3.8.8. Ifg,h c Xa are nonisotropic for a E P, then g _L h in the sense of Section 3.3 if and only if g L h in the sense of Definition 3.8.1. Proof: Let g = [a,P,5], h = [a,y,,.] c Xa with a = 36 = y.. Recall that g I h in the sense of 3.3 if, without loss of generality, P3 y and A 7= (= g gr h. So, in particular, A = P3y with A E gr h and g c Xp and h c Ty. So g _ h by 3.8.1. Now suppose that there exist rl,s e P such that B = TE e grn h and g c and h c XE, but it is not the case that 13 1 y, nor that P3 1 ., nor that 6 1 y, nor that 6 1 .. Then by 3.3.8 there exists a unique e CLO such that B E 1 and I = [a,u, p] for some u, p E P with a = ug, u 1 13, and g 1 6. If / # h then la(B) and ha(B) span ta(B). Thus if C E g there exists L e I and H e h such that BLBH = BC. By Axiom U, BL 1 BC and BH 1 BC imply that BC = BLBH 1 BC. That is, g I g, which cannot happen for nonisotropic g. Hence, I = h and the result follows. U 72 Lemma 3.8.9. If a e P then for each P a and for each nonisotropic line g c X, there is a unique nonisotropic line h c Xa such that P e h and h L g. Proof: This follows directly from 3.3.8 and Lemma 3.8.8. U Lemma 3.8.10. Ifg,h c Xca for X e P and g,h are nonisotropic, then g L h if and only if Ogah = GChg # 1Xa Proof: This follows from 3.3.16 and Lemma 3.8.8. U Lemma 3.8.11. Suppose that 0 e c,d; c,d c Xa, a E P with c and d nonisotropic and c L d. Let C e c and D e d, then go,DOC is not orthogonal to c if C,D # 0. Proof: First note that go.DOC # c or d because then we would have DOC = D1 E d, say, so that C = ODD1 e d and 0 e d implies that C = 0 or c = d. Now if go.DOC L c, then in X3a we have (O = c(d = (gODO *(Yc which implies that d = gO,DOC. 0 Lemma 3.8.12. Let g,x c 3Ea with g isotropic, ca e 7P, andg n x = {O}. Then g is not orthogonal to x if g # x. Proof: If x is isotropic and x g then because x # g, there exists y, 6 e P such that g c Xy, x c X6, and y 1 6. But then one of y and 6 must lie in g. But no element in g can contain an isotropic line so x is not orthogonal to g. Suppose that x is nonisotropic and let 0 e g fn x. Because g is isotropic then cc e M and by 3.3.8, there is a unique nonisotropic line h c Xa such that 0 e h and h I x. Suppose that g I x and let 0 # K e g, 0 # H E h, and 0 X( e x. Then OH I OX and OK I OX so that by Axiom U, O(HOK) = OHOK I OX. If go.HOK is nonisotropic then goHOK = h. And because go,HOK C 3a, then HOK = HI e h. Then we have K = OHH1 e h and g = h, a contradiction. Suppose that go,HOK is isotropic. If go,HoK = g then we may write HOK = K] for some KI e g. It follows that H = K1KO e g and g = h. If goHOK # g then go,HOK is the other isotropic line through 0 in Xa. Because g and go.HOK span Xa and 73 g,go,HOK I x then by Axiom U, x is orthogonal to every line in Xa through 0. So in particular, x I h' which implies that x = h. U Corollary 3.8.13. If g is isotropic and x I g then either x = g or x is nonisotropic and x and g are noncoplanar; that is, there does not exist 6 e P such that x,g c X6. Lemma 3.8.14. If g is isotropic, x is nonisotropic, x I g with x # g, and {P} = x fl g, then for all P # A e g and for all P # B e x, g and gAPB are not coplanar. Proof: From Lemma 3.8.12, x and g are noncoplanar. Suppose that A,O,AOB 17 for some y e P. Then B = OAAOB\y, which implies that x = goB c Xy and g = gOA c X'y. a contradiction to Lemma 3.8.11. U Lemma 3.8.15. If OC e V is isotropic and OB e V is nonisotropic with OC I OB then OCOB I OC. Proof: By Lemma 3.8.13, goc and goB are not coplanar and by Lemma 3.8.14, goc and go.coB are not coplanar.By Axiom S2, either goc I gOCOB or there exists D e gocoB such that 0 and COD are unjoinable. So if goc is not orthogonal to gocoB then go,coD is isotropic and by Axiom T, there exists a unique 6 E 'P such that goc,go.coD c 6. This implies that O,C,CODI6, D = OCCOD 1, gOD = go.COD c X6, and therefore, goc c X6, a contradiction. U Lemma 3.8.16. The zero vector, lo = 00, is the only vector orthogonal to every vector in V. Proof: This follows immediately from Lemma 3.8.11 above. U Theorem 3.8.17. IfU is a subspace of V, then UL = {OA e V : OA I OB, VOB U} is a subspace of V. Proof: because the zero vector 1 is orthogonal to every vector in V by definition, then 1, e U'. Let OA e U't and R E K.C because R OA e< OA >, the subspace generated by OA, and go.4A I goB for every OB e U by the definitions of U' and orthogonal vectors, then R OA e UL. 74 Let OA, OB e U and OC e U. If OC is nonisotropic then OC OA, OB and there exist a,P3,y,6 e P such that A,O\La; 0,B13; 0,CIy,6; a_ 1y; and 3 16 8. By Axiom U, there exists k,E e P such that 0 = kE, AOB Ik, and 0,C I&. Thus, OA + OB = 0(AOB) L OC. Now suppose that OC is isotropic. The following possibilities exist. (i) If OC O A, OB, then as in (a) above, (OA + OB) 1 OC. (ii) If OC = OA = OB, then because OC is isotropic, AOB = COC e goc and (OA + OB) I OC. (iii) If OC = OA # OB, then OB I OC implies that OB is nonisotropic and (OC+ OB) = OCOB OC by Lemma 3.8.14. Hence, if OA, OB e U then OA + OB e UL and UL is a subspace of V. U Theorem 3.8.18. If O = a3 then t,(0)' = tp(0) and.tp(0) = ta(O0). Proof: From the definition of orthogonal vectors we clearly have 1ta(O) c !p(0O) and lp(0)' c ,a(0). Now suppose that OA I ta(O); that is, goA L a. Then, there exists y e P such that OA e 1t(0) and y 1 ac. But then we obtain 0 = ca3 = ay so 3 = y. Hence, ta(0)' = tp(0) and ta(0) = t1(0)L. M An immediate consequence of Theorem 3.8.18 is the following. For each a e P with 01 a, there exists a unique P3 e P such that03, ta(O)1 = .tp(0), and ((0H =t(0). Theorem 3.8.19. If d(0) is nonisotropic then there is a unique hyperplane A(O) such that d(0)' = A(0) and (d(0)')1 = d(0). Proof: Let a e P such that a c X, where a is the nonisotropic line associated with d(0) = {OA : A e a}. Then for 0 = a13 we have a L P3 and a 1 g for every g c Xp with 0 e g. Thus, tp(0O) c d(O). Now a is nonisotropic, so by 3.3.8, there exists a unique h c X.a such that 0 E h and a 1 h. By Axiom U, (/(0),ptp(0)) c d(O). 75 Suppose 1 OB e d(O) r (<(O), tp(O)).Then B e a c Xa and there exists OC e /(O) and there .exists OD e tp1(O) such that OB = OCOD; that is, there exists C e h and D 13 such that B = COD. because h c Xa, CIa and C = cx 1. Because Di3, D = PPI33 for some P3i e P. Now a l 13 and P31 P31, so a 1I 13. But B = COD and COD = cxaaoppp3131I = ai3PI so that B3p1,at with a 11 13. This implies that a = PI so D = 0, B = C, and a = h, a contradiction. Hence, d(O) r (/(O), Ip(0)) = {1 } and V = d(O) (h(O),!tp1(O)). If 0 e d and d I1 a then d(O) c (h/(O),tp(O)). For otherwise we would have V = d(0) d(O) (4(O),.tp(O)>, which is not possible. Hence, 6(0)L = (/(O),.tp(0)). On the other hand, from tp(O)' = ta(O), a IL h, and a I 3, then, d(O) c (/i(O),t (O))'. If 0 g 1 P3, then by definition g c X,. If 0 e g 1 h, then g = a. Hence, d(O) = (/h(O), Ip(0)). Claim. d(O)' is independent of the plane containing a. Suppose a c Xy . Then for 0 = y76, we have y # a, P13, and a L X6. Again there exists a unique I c Xy such that 0 e I and 1 L a, so that d(O)' = (l(O),.t6(0)) as above. Now any point P e X may be written as P = P\OP2 with P\ a and P2 I 3. Since h L a, then we may write P\ = HOA with H e h and A e a. Then ((O), tp(O)) d(O) = V = ((0), ty(O)) d(O0), and it follows that ( Theorem 3.8.20. If d(O) is isotropic then there is an unique hyperplane A(O) such that d(O)' = A(O),d(O) c A(O), and (J(O)')' = d(0). Proof: Let d(0) c ta(O) and a the isotropic line in Xa corresponding to d(0). because a is isotropic then a e M. Let 0 = ap3, 13 e 9, and put A(O) = (d(O),t1(O)). Since a c Xa, a 13, and 0 = a13 e a, then a 1 X1. Because 76 a is isotropic then a I a. By Axiom U and Lemma 3.8.14 it follows that (d(O),p(O)) = A(O) c d(O)'. If k(O) c (d(O),tp(O)) is isotropic and g # a, then there exists a unique y e M such that g,a c Xy. Because g(O) c A(O) c d(O)', then g 1 a, which contradicts Lemma 3.8.11. Thus, A(O) contains no other isotropic line. Let h I a with 0 e h. For each H e h we may write H = H'OB, with HNc\ and BI3. If 0 # A e= a, it follows that OB OA; OH OA, so that OB I OA and OH' = OHOB0 1 OA. This implies that gOH' = a so H/ e a and h(0) e (d(O),p(O)). Thus, d(O)' = (d(O),t(O)). If h(O) c (d(O),tp3(O)), then /h(O) c 13(O)' = ta(O) and /h(O) I i(O). Thus, h(O) = d(O). Hence, (d(O),Tp(O))> = d(O) and (d(O)')i = d(O). The subspace di(O)I is independent of the plane containing a. Suppose a c X., y e M, y # ac. Put 0 = 7y6. Let h(O) c (d(O),t6(O)>. Then h 1 a and h(O) c (d(O),Jp(O)). If 4(0) C then k 1 a and k(O) c (d(O), t(0)). Therefore, (d(O),.t6(O)) = (d(O),tp(0)). U Remark. If g is a line then g is either nonisotropic or isotropic. From Theorem 3.8.19 and Theorem 3.8.20, if U_< V is a onedimensional subspace then UW is a uniquely determined hyperplane. In an affine space a plane is uniquely determined by two distinct intersecting lines. Let g and h be two distinct intersecting lines and let Kg,h) denote the unique plane determined by g and h. 77 Definition 3.8.21. If a e P such that g,h c Xa then we say that (g,h) = Xa is a nonsingular plane. If there does not exist a e P such that g, h c Xa then we say that (g, h) is singular or (g, h) is a singular plane. It is clear that every plane in (X, V, IC) is either singular or nonsingular. Note that by Theorem 3.8.18, if U < V is a nonsingular twodimensional subspace of V, then U1 is a uniquely determined nonsingular twodimensional subspace of V. Now consider the following cases for two distinct intersecting lines g and h. Suppose that g is isotropic and h is isotropic. Then by Axiom T, there exists a unique a e MA such that g,h c Xa. If one of g or h is timelike, then by Axiom T, there exists a unique a e M such that g,h c a. Thus, if (g,h) is singular then either g is isotropic and h is spacelike, or g and h are both spacelike. Proposition 3.8.22. Let (g, h) be a singular plane with g isotropic and h spacelike. Then g I h. Proof: Let {P} = g n h. By Axiom S2, either g I h or for each A in g, there exists B in h such that P and APB are unjoinable. Suppose that A e g, B e h, and P and APB are unjoinable. It must be the case that A is joinable with APB; that is, gAAPB is nonisotropic. Let gpA = [6,&], so A,P\6,E. If A is unjoinable with APB then A, P, and APB are pairwise unjoinable points, so by Axiom 10, APBs6,6 and APB E g. Thus, gPAPB = gPA = g and B = PAAPB e g, so g = h. Thus, gpA4pB t g and gP,4PB is isotropic, so by Axiom T, there exists a unique y e M such that g,gpAPB c Xy. This means that P,A,APBIy so B = PAAPBI7, and h c Xiy, which contradicts our initial assumption. Therefore, g IL h. 0 Theorem 3.8.23. Let (g(O),h(O)) be a two dimensional singular subspace of V where 9(0) is isotropic and h(O) is spa celike. Then (g(O), h(O))I is a two dimensional singular subspace of V which contains g(0). Moreover, ((kg(O), fh(O))1)1 = (gO), h(O)). Proof: First we observe that if W is a subspace of V generated by subspaces U and U' 78 then W' = (UU') = U N', v e (U,)' <4 v e U and v e U"' <> v eU' r) U'. Consider (g(O),/f(O))' = g(0)' rn /(0)1. Because g I h by Proposition 3.8.22 then there exists y,6 e P such that 0 = y6 with g c Xy and h c X6. From Theorem 3.8.20, g(0)' = (2(0),t6(0)) and by Theorem 3.8.19, /(O) = (1(0), Oty(0)) where I c 3E is the unique line through 0 in X orthogonal to h. Hence, (XO), fi(o)> = 2(0 ) ) ) = (AO), 1t6(0)> < )(0), t(0)> = <(O),(o9)>. Moreover, ((0), (0)>)L = g(O) n(o)1 = (g(O),.t (0)> r) <(0), t(0)> = g(o),h(0)>. Now if g,l c X, for some E e P then by Theorem 3.8.18, for 0 = E, (k(O),i(O)) = t;(O) and (0(),/h(0)) = (2(O),i(0))' = .t,(0)' = .t10(0). This says that g,h c Xp; that is, (g,h) is nonsingular. Thus, (g(0),l(0)) is nonsingular. If (g, h) is a singular plane, 0 e g r h, and g and h are spacelike, then by Axiom S1, g is not orthogonal to h. So by Axiom 2, for all A e g, there is a B e h such that 0 and AOB are unjoinable. U Theorem 3.8.24. For the above setup: 1. gOOB E (g,h}. 2. goAoB I g, h. 3. if C e g and D e h such that goCOD is isotropic then go.coD = goAOB. 4. (g(O),fih(O))' = (gooB(0),g(O))'. Proof: For 1., if there exists y E P such that go0oB,g c Xy, then 0,A,AOBI\;. implies that B = OAAOIy and h c Xy. This yields (g,h) = Xy is nonsingular. For 2. 79 go,AOB I g,h by Axiom S2 because by 1. above go,COD c (g,h) and (g,h) is singular. If C e g and D e h such that go,coD is isotropic then by 1,.go,coD,go,AOB c (g,h). If go.COD goAOB then by Axiom T there exists a unique ri E P such that gO,COD, gO0AOB C XY. Because two intersecting lines uniquely determine a plane, then (g,h) = Xy. Hence, go,COD = gOAOB. The last conclusion follows from (g,h) = (g,goAOB). If (g,h) is a singular plane then from the proof of Proposition 3.8.22 and from Theorem 3.8.24, it follows that for each P e (g,h) there exists a unique isotropic line I c (g,h) such that P e 1; that is, (k(O),/h(O)) contains one isotropic line. To complete the classification of orthogonal subspaces of V, it remains to consider hyperplanes. Now if (X,V ,C) is any four dimensional affine space then a line and a plane which intersect in a point uniquely determine a hyperplane and any hyperplane in the space can be characterized as the subspace generated by a corresponding line and plane. Let A = (h, p) be the hyperplane generated by a line h and a plane p which intersect in a point 0. Then A(o) fi(o), p(0))1 fi(o) (0). From Theorem 3.8.19 and Theorem 3.8.20, we know that the dimension of h(O)' is three. By Theorems 3.8.18, 3.8.23, and 3.8.24 it follows that the dimension of (0(0)L) = 2. Consider the following possibilities. If h(0) r) 0(0) = {lo}, then h(O)' 0(O)' c V has dimension five whereas the dimension of V is four. If i(O) r 0(9)' = p(O)', then it follows that p(O)_ c hf(O); h(0) I 0(0)L and h(o) c (p(0)=) p(0). Which contradicts the assumption that h and p intersect only in 0. 80 Therefore, h(O) n ((O)( = k(O) for some line g containing 0. Because g must be either isotropic or nonisotropic, then the following is true. Theorem 3.8.25. If A(O) is a hyperplane the there is a unique line g such that A(O) =(0) and (A(0))' = A(O). E 3.9 The Polarity In this section a polarity is defined so that we can obtain the metric through a process given by Baer [3]. For the convenience of the reader, the pertinent definitions and theorems [3] are given below. Definition 3.9.1. An autoduality 7t of the vector space V over the field IC is a correspondence with the following properties: 1. Every subspace U of V is mapped onto a uniquely determined subspace nr(M) of V. 2. To every subspace U of V there exists one and only one subspace W of V such that 7=() = U. 3. For subspaces U and W of V, U < W if, and only if, n(W) < t(7. In other words, an autoduality is a onetoone monotone decreasing mapping of the totality of the subspaces of V onto the totality of the subspaces of V. Definition 3.9.2. An autoduality 7t of the vector space (V,K) of dimension not less than two is called a polarity, if 7t2 = 1, the identity. Definition 3.9.3. A semibilinear form over (V,/C) is a pair consisting of an antiautomorphism a of the field K and a function (fix, y) with the following properties: (i) J(x, y) is, for every x, y e V, a uniquely determined number in KC. (ii) f(a + b, c) = (a, c) +(b, c) and J(a, b + c) = (a, b) +fa, c), for a, b, c E V. (iii) J(tx, y) = /fx, y) and J(x, ty) = f(x, y)a(t) for x, y e V and t e 1C. If a = 1 then f is called a bilinear form. Definition 3.9.4. Iff is a semibilinear form over (V,/C) and if LU is a subset of V, then 81 {x e V : fix, u) = 0 for every e U} and {x e V :fJ(u, x) = 0 for every u e U} are subspaces of V. We say that the autoduality 7t of (V, K) upon itself is represented by the semibilinear form Ax, y) if n(U) = {x e V :fix,U) = 0} {x e V :Afix, u) = 0 for every u e U1}. Theorem 3.9.5.[3] Autodualities of vector spaces of dimension not less than 3 are represented by semibilinear forms. U Theorem 3.9.6.[3] If the semibilinear formsf and g over (V, )C) represent the same autoduality of V, and if dim(V) > 2, then there exists a 0 # d e kC such that g(x,y)= fJ(x,y)d for every x,y E1V. U Definition 3.9.7. If 7t is an autoduality of the vector space (V,/C), then a subspace W of V is called an Nsubspace of V with respect to 7r, if (v) In this case, 7t is said to be a null system on the subspace W of V. Theorem 3.9.8.[3] Suppose that (f, ) represents the autoduality 7 of(V, C). Then t is a null system on the subspace W +> J(w, w) = 0 for every w e WV. U Definition 3.9.9. A line (v) is called isotropic if (v) < t((v)). (So an isotropic line is an Nline.) Theorem 3.9.10.1[3] If the semibilinear form (fa) represents a polarity 7r, and if J(w,w) = 1 for some wEV, then a2 = 1 and a(f(x,y)) =J(y,x) for every x,y cV. In this case we say that f is a symmetrical or just symmetrical. U Theorem 3.9.11.[3] Suppose that 7 is an autoduality of the vector space (V, C) and that dim() > 3. Then n is a polarity if, and only if, 7r is either a null system or else 7t may be represented by a symmetrical semibilinear form (f, ac) with involutorial a. U Theorem 3.9.12.[3] Suppose that the polarity 7r of the vector space (V, KC) possess 82 isotropic lines and the dim(V) > 3. Then 7 may be represented by bilinear forms if, and only if, (a) planes containing more than two isotropic lines are Nplanes and (b) K: is commutative. U Theorem 3.9.13. Suppose that t is a polarity of the vector space (V,)IQ such that the conditions of Theorem 9.12 are met. Then from Theorem 3.9.11 it follows that if t is not a null system then 7t may be represented by symmetrical bilinear forms. U Defining the polarity 7t and obtaining the metric g. Consider (V,/C) constructed in this work. If U is any subspace of V then by Theorem 3.8.17, U' is also a subspace of V. From the end of section 3.8, if U is a subspace of V, U # {1}, and U # V, then U' is a uniquely determined subspace of V. From Lemma 3.8.11 and Definition 3.8.5, {1Q}' = V and V1 = {1}. So we define the mapping it on the subspaces of V as follows: Definition 3.9.14. If U is a subspace of V then i(bL) = UL. From the remarks above and from Section 3.8 it is clear that to every subspace U of V there exists one and only one subspace WV of V such that it(W) = U. Theorem 3.9.15. Let U and WV be subspaces of V, then U 74(W) it(4). Proof: Suppose that U < W. If OV e W41 then OV 1 OW for every OW e W, U c W, so OV I OU for every OU e UL and OV UL'. Thus, U < W/V implies that 7t(V) = W1 < U' = 7i(4). Conversely, suppose that WV' < U. By Lemma 3.8.11 and Definition 3.8.5 we have ({1o}I)' = {1} and (V')1 = V. From the previous section it follows that, if U is a nontrivial proper subspace of V then (UL')' = U. because /V1 and U' are subspaces, then from the first part of the proof we have W' < UL' implies that U = (UL')' < (W1)' = W. U Corollary 3.9.16. From Definition 3.9.14 and Theorem 3.9.15 above it follows that E : U < V 'r U' < V is an autoduality. Moreover, because (UL')' = UL for every subspace U of V, then T is a polarity. Lemma 3.9.17. Let e E 5 and OA e V. Then 7r(l,) = 7r(U) for every e 0 and for every subspace U < V. That is, 7T is invariant under a for every 4 e S. Proof: We have O(OA) = O09A' = B e X, so (OA) = OB e V. Hence, from Lemma 3.8.6.the result follows. U Theorem 3.9.18. The polarity 7 is not a null system. Proof: Let a be any nonisotropic line with 0 e a. By Lemma 3.8.7, a is not orthogonal to itself so that d(0) : d(0)' and hence, d(0) % n(d(0)). By Definition 3.9.7, 7r is not a null system. U Theorem 3.9.19. The polarity 7t defined above may be represented by bilinear forms. Proof:. Let g be any isotropic line through 0. By Definition 3.8.5, g(0) < g(O)' so that (O) < n(7rg(0)) and k(O) is an isotropic line in the sense of Definition 3.9.9. Conversely, if h is a line such that h(0) < nt(h(O)) = h(O)' then h I h and by Lemma 3.8.7, h must be isotropic in the sense of Section 3.2. Thus, (V,/C) possesses isotropic lines. By Proposition 3.7.3, dimni(V) = 4 > 3. From Definition 3.8.21 any plane in (V,/C) is singular or nonsingular. Now a singular plane contains only one isotropic line by Theorem 3.8.24. If U < V is a nonsingular plane then U = .ty(O) for some y e P. If y e M then U has precisely two isotropic lines, as was shown in 3.2.2.8. If y e, then by Axiom 12 and Section 3.2, U does not have isotropic lines. Thus, no plane of V contains more than two isotropic lines. Since the field AC = R is commutative then by Theorem 3.8.12 the result follows. U Theorem 3.9.20. The polarity 7 may be represented by symmetrical bilinear forms; that is, there is a symmetrical bilinear form g such that for any subspace U of V, UL = r(M) = {OA e V : g(OA,U) = 0} or g(OA,OB) = 0 = g(OB,OA)if, and only if, OA 1 OB for OA,OB e V. Proof: This follows directly from Theorems 3.9.18, 3.9.19. and 3.9.13. U 84 Thus, we have a metric g, a symmetric bilinear form, induced by our polarity, which agrees with our definition of orthogonal vectors, which in turn is induced by and is defined in terms of the commutation relations of the elements of our generating set g. Lemma 3.9.21. Let g be a symmetric bilinear form representing 71. Then g is nondegenerate. Proof: This follows from the fact that ic(V) = V1' = {1}. U Theorem 3.9.22. Let g be a symmetric bilinear form representing T. Then there is a basis of V such that the matrix of g with respect to this basis has the form SC 0 0 0 0 C 0 0 r #, 0 Ce IC; 0 0 C 0 0 0 0 C that is, there is an orthogonal basis of V such that the matrix of g with respect to this basis has the above form. Proof: Let a e M such that kC c Xa and let 0 = ac3. Let Oy7,5 e g such that a = y6 and y,6 13. Let s = yp e M and 11 = 63 e AM.Then 8Tl = y7pp3138 = y = a so 6 1 1. Put x = [ay,5], y = [3,y, ], z = [13,r,], and t = [a, r1]. Claim. x,y,z,t are four mutually orthogonal nonisotropic lines through 0. We have 0 = ap3, 0 = x,t c Xa and 0 e y,z c Xp so that x,t I y,z. Also, 0 = a3 = y6p3 = yr1 = 6c so that y I il and 6 1 E. Then 0 e x c Xy,0 e t c X^,O e Xy, and 0 e z c 3C, implies that x I t and y I z. The construction of the basis. Let E E /C, the multiplicative identity, considered as a point in X. Put T = EOEt E t and X = EOEX x. Note that because 85 0 e x,t c X3C and x I t then axlt = cro = CaT(x in X3a. Now t c 3E and e M, so there exist precisely two isotropic lines, /Ci, and /C2, through 0 in XC. Thus, there is an unique E E /Ci1 such that T = EgOE. Because y c XC then Y EOEy E y and ryCat = ao = atoy in XC. Similarly, because t c X1; M1 E A4, there are precisely two isotropic lines, ACIn and /C2, through 0 in Xi and there is Eq in /CIn such that T = EnOEt. because z c X then Z = EnOEt e z and aCzat = To = ctraz in Xn. We calculate XOT = (EOEX)O(EOE') = (EOE)O(EX OE) = (EOE)O(E0tO'Et) = (EOE)0(EOE)' = (EOE)0(OEOOE)t = (EOE)O0t = (EOE)OO = EOE e 1C. YOT = (EgOE0)O(E9OEt) = (EcOE,)O(EOEE)t = EgOEg e KC1. ZOT = (E OE)O(EqOEt) = EqOE9 e K l. Thus, if we put E1 = OX, E2 = OY, E3 = OZ, and R4 = OT it follows that the set {E1,R2,ER3,E4} consists of four mutually orthogonal vectors such that Ei + 4 is isotropic for i = 1,2,3. So if g is a symmetric bilinear form representing 7t then from Ei + E4 L Ei + R4 for i = 1,2,3 and Ei IL Ej for i j it follows that for i = 1,2,3, O = g(Ei +R4, Ei + E4) = g(Efi, Ei) + 2g(Ei, 4) + g(!4, 4) = g(Ri) + g(4,/4). So that g(E,,Ei) = g(04,94) 0, because E4 is nonisotropic. Thus, it remains to show that {V],/2,E3,/4} is a basis for V. But this follows from Section 3.7. U Theorem 3.9.23. (V,/C,g) is a fourdimensional Minkowski vector space. Moreover, (Iy(O),g) is a fourdimensional Minkowski space for every y in P0. . . . "4 Proof: Put g(E4,E4) = 1 and g(Ei,, E) = 1, for i = 1,2,3. Minkowski space is the only nonsingular real fourdimensional vector space with metric 1 00 00 S0 1 0 0 (3.24) 0 001 0 0 0 0 1 86 In the last section of Chapter 3 we show that each X in G can be identified with a spacelike plane and each &k e g* with a reflection about a spacelike plane. 3.10 Spacelike Planes and Their Relections For each X E Q with 0\1 define a5X V V by a(OA) = (OA) OAX, for OA e V. To extend this definition to any X e g, we note that if X f 0 and OA e V, then there exists a unique D in X such that O0XAX = D; that is, there is a unique D in X such that (OA)X = OXAX = OD. Thus, we define 6(OA) = OD, where O0XAx = D. We note that if k 10 then D = AX. First we show that &k is a semilinear automorphism of V for each X, E V. [23] Now, a function f : V ', V is a semilinear automorphism if, and only if, it has the following properties: 1. f is an automorphism of the additive group of V onto itself. 2. f sends one dimensional subspaces of V onto one dimensional subspaces (that is, f is a collineation). 3. If A and B are linearly independent vectors of V, the vectors J(A) and J(B) are also linearly independent. Theorem 3.10.1. The map & is an automorphism of the additive group of V onto itself Proof: Suppose that 6(OA) = 6&(OB). Then 6&(OA) = OD where OD = 0'A = OXBX. This implies that AX = BX or A = B. Thus, OA = OB and 6& is injective. To show that 6X is onto, let OB e V and A = 00B. Then &(OA) = OA' = O(OOBX)k = OkOX OB = OB. We show &x is additive. Let OA,OB e V. Let D = OOXAk, F = OOBX. Then 6&(OA + OB) = (OAOB) = OAXOBX0 = ODOF = 6&(OA) + 6X(OB). U Lemma 3.10.2. The map &X sends one dimensional subspaces of V onto one dimensional subspaces of V. 87 Proof: Let < OA > be a one dimensional subspace of V. Then there exists a line g = [a,P,7y] such that OB e< OA > if, and only if, O,B e g. Since O,B I a,P3,y7 < OA,BX  aI W, y7 then &x(< OA >) =< &x(OA) > is a one dimensional subspace of V. N Lemma 3.10.3. The transformation, &k, maps linearly independent vectors of V to linearly independent vectors. Proof: The vectors OA, OB e V are linearly independent if, and only if, goA # gOB if, and only if, gO9Ax # g9OB;. Hence, &X is a semilinear automorphism of V for each 2 e G. 0 Theorem 3.10.4. The map Csk : V '* V is a linear automorphism of V onto V. Proof: Consider the definition of a semilinear automorphism [23,defn.73.1]. Let (V,K) and (V',/C') be vector spaces over the division rings )C and KC', respectively. Suppose that pt : /C /C' is an isomorphism from /C onto KC'. A map X : V 'k V' is called semilinear with respect to p. if 1. (A + B) = X(A) + X(B) for all A,B e V. 2. X(tA) = p(t)X(A) for all A e V and t e /C. The only isomorphism p. : P '+ JR is the identity. Thus, &X is a linear automorphism of V onto V. N Let (V, KC,g) be a metric vector space. A similarity y of V is a linear automorphism of V for which there exists a nonzero r e K; such that g(yA4,yA)Thsalrri g(yA,yB) = rg(A,B), for all A,B E V. If A is nonisotropic, r = yAA). The scalar r is called the square ratio of the similarity. Lemma 3.10.5. [23] A linear automorphism of a metric vector space V is a similarity if, and only if, it preserves orthogonality. Proof: For all lines h and I in our space, h L / <> hx I lk, hence, J(OA) 1 &(OB) <> OA OB. U 88 Theorem 3.10.6. The linear automrnorphism %z is an isominetry. Proof: Let y : V i V be a similarity with square ratio r # 0. Then for A,B e V we have g(yA,yB) = rg(A,B) = rg(y1 (yA),y 1 (yB)) and g( (yA),y (yB)) g(YAYB). That is, yI has square ratio r1. Now 6i is an involution, so that &k = ftl. Hence, if r 0 is the square ratio of 6x then r = orr2 = 1, so r = 1. [231 Because the field C is isomorphic to R, then V is not an Artinian space. Thus, for each similarity & of V there is an unique r > 0 and there is an unique isometry a such that 6( = M(0,r); where M(0,r)(A) = rA for all A in V. The map 6. is thus a similarity with square ratio r2. Because r > 0, then from above, r = 1 and it follows that 6& is an isometry. U Proposition 3.10.7. The isometry 6& is a 180 rotation; that is, a reflection about a plane (a twodimensional subspace of V). Proof: Let ; E g. Suppose that 0\2k and put 0 = ka with c e M. Then for all AIk and for all Ba, 6(OA) = (OA)k = O6AA = OA and 6(OB) = (OB) = O= B = OB"I = OB = OB. Since 1,a(0) = .tx(0)' and V = J(O) I(O)' then ey = 1tx(o) ljt(o). Thus, &z is a reflection about the plane .tX(O). Suppose that 0 X. Let 6 I 0 such that c I1 k and let P I X be arbitrary. Then if R I ,, POR= Q I X and OR = PQ. Thus, &a(OR) = (OR); = (PQ)k = PQ = OR and x 1(o) = lt(o). Now let 310 such that P3 1 c, so 0 = c3. Let B I P3. Now P3 1 X because 11 X. and POB = DZy where P = ky and y 113. We calculate, 6(OB) = (OB) = (PD)^ = (PD)Y = (PD)p = DP = BO = OB = OB. Hence, t = lt (o) ~1t(o)0 and &k is a reflection about a plane. U 89 To show that 6. is a reflection about a spacelike plane (Euclidean plane) for every X e G, we use the following theorem from Snapper and Troyer [231. Theorem 3.10.8. [231 Let (V, C5,g) be an ndimensional metric vector space over a field k1 with metric g, IC = R, and n = 2. Every nonsingular real plane has a coordinate system such that the matrix of its metric is one of the following. S0 the Euclidean plane, ( the Lorentz plane, and ( the negative Euclidean plane. U Hence, the Euclidean plane, the Lorentz plane, and the negative Euclidean plane are the only nonisometric, nonsingular real planes. This also follows from Sylvester's Theorem [23] (It states that there are precisely n+1 nonisometric, nonsingular spaces of dimension n). Lemma 3.10.9. Let kC = R, n = 4, and V be Minkowski space. Then the orthogonal complement of a Lorentz plane is a Euclidean plane. Proof: Let {eI},i, ..4 be a basis for V such that the metric of V with respect to this basis has matrix of the form (3.24). Let m = e3 + e4 and m = e3 e4 so that < m >,< n > are the unique isotropic lines in the plane < e3,e4 >. Let a be a Lorentz plane with isotropic basis m I and n I. Then there is an isometry a :< e3,e4 >* a such that o(m) = m 1 and 0(n) = n1. By the Witt Theorem [23] c can be extended to an isometry of V, which we also denote by a. This implies that Ca : V =< e3,e4 > G < ei,e2 >F+ V = oc D< a(ei),a(e2) >; that is, {a(ei),a(e2)} is a basis for a1. Now a is an isometry and g(el,el) = g(e2,e2) = I, so the metric g with respect to a1 has matrix a2, the 2x2 identity matrix, and a1 is a Euclidean plane. M 90 Theorem 3.10.10. The map 6& is a reflection about a spacelike (Euclidean) plane for every X e G. Proof: From the proof of Proposition 3.10.7 it suffices to consider e g with 0[X. Let 0 = OX with 0 e M and let Q(.) denote the quadratic form associated to the metric g obtained in Section 3.9. Because 0 e M there are precisely two isotropic lines /C,01, KC02 c XO. Let 0 A E /C0, and 0 B e KC02. Then OA and OB are isotropic vectors which form a basis for t0(O) and Q(OA) = 0 = Q(OB). Therefore the metric of 10o(0) with respect to OA and OB has matrix 9 = ( A OA OB) 0 r * S OB r 0( where the products of the matrix elements are the inner products defined by the metric g and g(OA,OB) = r # 0. Since  OA E KCo,, and OB e C02 then k0,, K02 also form an isotropic basis for .o0(O). Hence we may assume that r = 1. Let OTI (OA OB) and OXI =  (OA + OB). Then g'(OXI, OTI) = 0, Q'(OXI) = I, and Q'(OTi) = 1. Because OX1 I OTI and 10(0) is nonsingular then OXI and OT, are linearly independent and hence, form a basis for 10(0). Moreover, the metric of e(0) with respect to OXIand OT, has the matrix Therefore, by Theorem 3.10.8, l0(O) is a Lorentz plane and by Lemma 3.10.9, t() = t() is a Euclidean plane. 1 Lemma 3.10.9, I.(O) = to(O)L is a Euclidean plane. N CHAPTER 4 AN EXAMPLE OF THE THREEDIMENSIONAL MODEL This chapter begins by considering a net of von Neumann algebras, {TZ(0)}o I, and a state o), coming from a finite component Wightman quantum field theory in threedimensional Minkowski space. There are various senses to the phrase "coming from a Wightman quantum field theory". The assumption here is the version given by Bisognano and Wichmann [5]. That is, given a finite component Wightman quantum field, ((x), assume that the quantum field operator, 0(t), is essentially selfadjoint and its closure is affiliated with the algebra R(0) (in the sense of von Neumann algebras) for every test function f whose support lies in the spacetime region 0. Driessler, Summers, and Wichmann show these conditions can be weakened [15]. But free boson field theories satisfy these conditions in threedimensional Minkowski space [5]. For such theories the modular involutions, Jo, associated by TomitaTakesaki theory to the vacuum state and local algebras of wedgelike regions, 0, in three dimensional Minkowski space, act like reflections about the spacelike edge of the wedge [5]. Since the modular involutions have that action upon the net, the hypotheses of Buchholz, Dreyer, Florig, and Summers (BDFS) are satisfied [6]. Therefore the Condition of Geometric Modular Action, CGMA, obtains for the set of wedgelike regions [271 in Minkowski space. The precise wording of this version of the CGMA is given below. Let 1i and 12 be two lightlike linearly independent vectors belonging to the forward light cone in threedimensional Minkowski space. The wedges are defined as the subsets W[l1,12] {a/cij + f12 +1 R1'2 : C > 0, p3 < 0, (1L, li) = 0, i = 1,2}, where (, ) denotes the Minkowski inner product. 91 92 Let II = (1,1,0) and 12 = (1,1,0) be lightlike vectors in R1,2, threedimensional Minkowski space, and let P' be the Poincar6 group, the isometry group of this space. Then the set of wedges, W, is given by W = {XKW[l1,12] : X e P}, where X .W[l,12] = {.(x) :x E W4[112]}. The CGMA for Minkowski space is defined as follows. Let {7Z(W)} ww be a net of von Neiiiniiiii algebras acting on a Hilbert space 'H7 with common cyclic and separating vector Q 7E satisfying the abstract version of the CGMA and where the index set I is chosen to be the collection of wedgelike regions W in R1'2 defined as above. Recall from Chapter 1, with ({1Z(W)} weW,',R,) there is the following. 1. A collection of modular involutions {Jw} wew. 2. The group J generated by {Jw} w&w. 3. A collection of involutory transformations on W, {fw} ww. 4. The group Tgenerated by {Tw}wEw. Assume also that: 5. The group Tacts transitively upon the set W, that is, for every W1, W2 e W there is a W3 e W such that Tw3(WI) = W2. Note that this assumption is implied by the algebraic condition that the set {adJw} WW acts transitively upon the net {R(W)} wEw. At this point the following two assumptions are added (127] which have been verified for general Wightman fields). 4.6 For W1, W2 E W, if W\ n WK2 # 0, then Q is cyclic and separating for Z(W) n 7IZ(W2). 4.7 For W1, W2 E W, if Q is cyclic and separating for 1Z(W,) r 7Z(W2), then W, r W2 0. The CGMA for Minkowski space is the abstract version of the CGMA with the choice of A' for the index set I, together with assumptions 4.6 and 4.7 above less the transitivity assumption [6]. 93 Buchholz, Dreyer, Florig, and Summers [6] showed that with the above assumptions one can construct a subgroup T3 of the Poincare group P, which is isomorphic to Tand related to the group Tin the following way. For each T E Tthere exists an element g, e T such that r(W) = gW = {g,(x) : x e W}. To each of the defining involutions Tw e T, W e W, there is a unique corresponding gw e q3c P [6]. Moreover, BDFS obtained the following (suitably modified for three dimensions and abbreviated for our purposes). Theorem 4.1 [6] Let the group Tact transitively upon the set W of wedges in R1,2, and let 3 be the corresponding subgroup of P. Moreover, let gw be the corresponding involutive element of P corresponding to the involution 'rw e T Then gw is a reflection about the spacelike orthogonal line which forms the edge of the wedge W. In particular, one has gwW = W,' the causal complement of W, for every W e W. In addition, T exactly equals the proper Poincare group P+. U Recall from Chapter 2 that the initial model of (9, 0) is as a group plane. This means that each g e g is viewed as a line in a plane and each P = gh, gjh, as a point in a plane. Let us call the axiom system given in Chapter 2 as A. Thus as A is a set of axioms about "points" and "lines" in a "plane". Let P denote the collection of points P e 0. For each a E 0 define the map ra : Pug +P u by ra(P) P aPalI for P e P and ca(g) = g a agaI for g e g. Since (Q, 0) is an invariant system then each cra is a bijective mapping of the set of points and the set of lines, each onto itself, which preserves the incidence and orthogonality relations, defined by "I", of the plane. We say that aF is a motion of the group plane. Since g generates 0 then the set of line reflections g, = {ag : g e g} generates the group of motions 0O = {ra : a E 5}. Let D : 0 +> Oa be the map defined by D(a) = ac,, for a E 6. Then 0 is in fact a group isomorphism [2]. This 94 means that (9, 0) is isomorphic to (G9, 0a) (in the sense that g is equivalent to G9 as sets and ((G) = G, (D(O) = 0o, where D is a group isomorphism). This implies that X(as A), which we denote by as D(A4), is an axiom system concerning the group of motions; line reflections of a plane, the group it generates, and point reflections of a plane. A plane whose points and lines satisfy as A. As was shown in Chapter 2, given (9, 0) satisfying as A, one obtains IR 12, threedimensional Minkowski space. Under the identifications given in Chapter 2, we find that each g e g corresponds to a spacelike line in R1,2. Thus, as (D(A) is a set of true statements concerning reflections about spacelike lines and the motions such reflections generate in threedimensional Minkowski space. Moreover, since such motions are in fact isometrics in R1'2 then D(O) is isomorphic to a subgroup of the threedimensional Poincare group. Theorem 4.2 Under the same conditions as in Theorem 4.1 it follows that ({tw} wew,T) acting on W satisfies as D(A). Proof: From Theorem 4.1 we have ({gw}w)w, W3) satisfies as D(A) since 3 is the subgroup of T generated by reflections about spacelike lines. Also from Theorem 4.1, ({tw} wew,T) is isomorphic to ({gw} ww,T3) so ({tw} ww,7) satisfies as0(A). N The net continuity condition assumed by BDFS [6] for the next theorem was later shown to be superfluous [81 for this theorem and the remaining theorems. Theorem 4.3 [6] Assume the CGMA with the spacetimne R 1,2 and W the described set of wedges. If J acts transitively upon the set {f(R W)} ww then there exists a strongly (anti) continuous unitary representation U(P +) of the proper Poincare group which acts geometrically correctly upon the net {1(W)} wew and which satisfies U(gw) = Jw, for every W e W. Moreover, U(Pt) equals the subgroup of j consisting of all products of even numbers ofJw 's and J = U(Pt) u Jw U(Pt), where WR = {x E R12 : xI > Ixo}. 95 Theorem 4.4 Under the same hypotheses as Theorem 4.3, the group J is isomorphic to P + = T, which is generated by the set of involutions {gw I W e W}. Moreover, (.{Jw } ww, J ) satisfies as I(A4). Proof: By Proposition 1.1 there is surjective homomorphism : J T, where the kernel of 4, ker(4), is contained in the center of J, Z(J). By Theorem 4.3 there is a faithful representation U(P+) such that U(gw) = Jw, for every W e W. Since the center of PT is trivial, U(.) is a faithful representation of P+ and hence an injective map preserving the algebraic relations, Z(J) is trivial. This implies that ker(4)= {1} and 4 is an isomorphism. If T : T> 3 denotes the isomorphism of T and 3 given by BDFS from Theorem 4.1 then T o 4 : J + 3 = P+ is an isomorphism. It therefore follows that the pair ({Jw} wewV, J) satisfies as ((A). E We can now give the main result of this chapter. Theorem 4.6 Any state and net of von Neumann algebras, coming from a (finite component) Wightman quantum field in threedimensional Minkowski space, which satisfies the Wightman axioms, provides a set of modular involutions i,,ti.v'iug as S(A). U As a final remark we note that since free boson field theories satisfy the Wightman axioms and therefore the CGMA for Minkowski space holds, then these theories give a concrete example of the threedimensional case of this dissertation. 