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# An algebraic characterization of Minkowski space

## Material Information

Title:
An algebraic characterization of Minkowski space
Physical Description:
v, 106 leaves : ; 29 cm.
Language:
English
Creator:
White, Richard K., 1960-
Publication Date:

## Subjects

Subjects / Keywords:
Mathematics thesis, Ph. D   ( lcsh )
Dissertations, Academic -- Mathematics -- UF   ( lcsh )
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bibliography   ( marcgt )
theses   ( marcgt )
non-fiction   ( marcgt )

## Notes

Thesis:
Thesis (Ph. D.)--University of Florida, 2001.
Bibliography:
Includes bibliographical references (leaves 103-105).
Statement of Responsibility:
by Richard K. White.
General Note:
Printout.
General Note:
Vita.

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University of Florida
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All applicable rights reserved by the source institution and holding location.
Resource Identifier:
aleph - 027029748
oclc - 47365928
System ID:
AA00020454:00001

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.

page
ACKNOW LEDGM ENTS........................................................................................ ii

ABSTRACT .......................................................................................................... v

CHAPTERS

1 INTRODUCTION............................................................................................. 1

2 A CONSTRUCTION OF THREE-DIMENSIONAL 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 Projective-M etric Plane .................................... 22
2.5 Exterior Point Reflections Generate Motions in an Affine Space ............. 23
2.6 Conclusion ................................................................................................ 24

3 A CONSTRUCTION OF FOUR-DIMENSIONAL 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 THREE-DIMIENSIONAL 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 three-dimensional 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

four-dimensional 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 Tomita-Takesaki 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 set-up 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 order-preserving 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 order-preserving bijection

and that the GNS vector Q is cyclic and separating for each algebra 71i, i e I. From

Tomita-Takesaki 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 anti-linear 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(7-c), 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

Tomita-Takesaki 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 order-preserving 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 non-Euclidean 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 one-to-one correspondence with the reflections in

them so that geometric relations among points, lines, and planes correspond to

group-theoretic 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 group-theoretic 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 four-dimensional Minkowski space. That is, characterize three- and

four-dimensional 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

Tomita-Takesaki 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 three-dimensional Minkowski space out of the plane at infinity.

Then identification of the involutory elements of 9 with spacelike lines and their

In Chapter 3 using the same initial data, (Q, 0), as was given in Chapter 2 but

satisfying different axioms, a four-dimensional 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 Chapter 4 a concrete example of the three-dimensional 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

Tomita-Takesaki theory to the vacuum state and local algebras in wedgelike regions

in three-dimensional 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 three-dimensional 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

half-sided modular inclusion relations given by Wiesbrock [29], then one does obtain

a unitary representation of the 2+1-dimensional Poincare group.

CHAPTER 2
A CONSTRUCTION OF THREE-DIMENSIONAL MINKOWSKI SPACE

In this chapter we give an absolute geometric, that is, an algebraic,

characterization of three-dimensional Minkowski space. This chapter is a version of a

preprint by the author entitled "A Group-Theoretic Construction Of Minkowski

3-Space Out Of The Plane At Infinity" [28]. Along with the well-known

mathematical motivations [1, 2] there are also physical motivations, as we discussed

in Chapter 1. Three-dimensional Minkowski space is an affine space whose plane at

infinity is a hyperbolic projective-metric 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 projective-metric

plane in which the given group 5 was isomorphic to a subgroup of the group of

congruent transformations (motions) of the projective-metric plane. By interpreting

the elements of G as line reflections in a hyperbolic plane, BPWB showed that the

hyperbolic projective-metric 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

projective-metric 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 pole-polar pair. Here we

show that Coxeter's and BPWB's notions of motions coincide in the hyperbolic

8

projective-metric plane and that the motions can be viewed as reflections about

exterior points.

Next we embed our projective-metric plane into a three-dimensional 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 projective-metric plane,

three-dimensional 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 three-dimensional Minkowski space, one can

infinity. So g may be viewed as a set of reflections about exterior points in a

hyperbolic projective-metric plane. Out of the plane at infinity, one can obtain a

three-dimensional 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

two-dimensional Minkowski space and by Klotzek and Ottenburg [19] for

four-dimensional Minkowski space. The approach in these papers is to begin by

constructing the affine space first. For Wolff's [30] two-dimensional case, the

elements of the generating set g are identified with line reflections in an affine plane.

For Klotzek and Ottenburg's [19] four-dimensional 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 projective-metric 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 one-to-one

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

self-conjugate: A on its own polar and a through its own pole [141. A hyperbolic

polarity is a polarity which admits self-conjugate points and self-conjugate lines. The

set of all self-conjugate 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 4|r|.

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

projective-metric plane into an affine space, we can identify the elements of g, our

generating set, with spacelike lines and their corresponding reflections in a

three-dimensional 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 three-dimensional 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

group-theoretic 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 one-to-one 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 one-to-one mappings of the set of lines and the set of points, each onto itself in

the group plane. U

Definition 2.2.3. A one-to-one 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 projective-metric

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 one-to-one 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;
PO-O 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

projective-metric plane.

In this model, the ideal points of the form G(P) are the interior points of the

hyperbolic projective-metric plane. Thus the points of the metric plane correspond in

a one-to-one way with the interior points of the hyperbolic projective-metric plane.

The ideal points G(x), for x e g are the exterior points of the hyperbolic

projective-metric plane.

Theorem 2.2.8. Each x e g corresponds in a one-to-one way with the exterior points

of the hyperbolic projective-metric 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 one-to-one way with

the pencils G(x). Hence, each x e g corresponds in a one-to-one way with the

exterior points of the hyperbolic projective-metric plane. U

16

Thus, the axioms can be viewed as axioms concerning the interior and exterior

points of a hyperbolic projective-metric 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 projective-metric 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 projective-metric 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 well-defined one-to-one point-to-line mapping of

P onto C and a well-defined one-to-one line-to-point 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

projective-metric plane in which the theorem of Pappus and Fano's axiom hold,

which is generated by the exterior points of the hyperbolic projective-metric plane.

With the identifications above and the geometric objects above, we show in

the next section that the motions of the hyperbolic projective-metric 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.

Definition 2.3.1. A collineation is a one-to-one map of the set of points onto the set

of points and a one-to-one 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 line-wise 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 pole-polar 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 pole-polar 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

exterior lines are also motions of the projective-metric plane; that is, the +b 's for

b e g acting on exterior points and exterior lines are motions of the hyperbolic

projective-metric plane.

Proof: The motions of the projective-metric 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 Projective-Metric Plane

In this section we embed our hyperbolic projective-metric plane into a three

dimensional projective space, finally obtaining an affine space whose plane at infinity

is isomorphic to our original projective-metric 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

projective-metric plane: that is, three-dimensional 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 three-dimensional Minkowski space is an affine space whose plane at

infinity is a hyperbolic projective-metric 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 three-dimensional 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 three-dimensional Minkowski space is generated by reflections about

spacelike lines.

Proof: Because any motion in three-dimensional 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 projective-metric 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 three-dimensional 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 four-dimensional 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 FOUR-DIMENSIONAL MINKOWSKI SPACE

In this chapter a construction of four-dimensional 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 I|2. 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 "non-singular" 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

i-L8; 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:j|3 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, 0|6 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 O0|x,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, 3P-8 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, 3P-L68 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, P-iy 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 I-2 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; BI|3; 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 P|8,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 = a6-1388 = 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, P-L8 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

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 a-pIp31132. 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 8|A 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,B|a, 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 B|130, 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,B|a,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 A|a,[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)= A-1

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 well-defined 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 y-Ls 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 X-a 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

two-dimensional 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 0|e 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

E|a',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 well-defined 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+B-C=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. (Artin-Schreier) 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 O|y 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 set-theoretic

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 three-dimensional or four-dimensional 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 well-defined

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

one-to-one 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 one-to-one. 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 0|y 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(Q21-ltoa)
= 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 X-i. 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 well-defined. 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' = 8-A.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 = 8A-A'(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 four-dimensional 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 four-dimensional 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 two-dimensional 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. (U-L 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 that0|3, 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 X-a, 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 one-dimensional 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 two-dimensional subspace of V,

then U1 is a uniquely determined nonsingular two-dimensional 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 one-to-one 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

anti-automorphism 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 N-subspace 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 N-line.)

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 c-V.

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 N-planes 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 four-dimensional Minkowski vector space. Moreover,

(Iy(O),g) is a four-dimensional 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 four-dimensional 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(y-1 (yA),y -1 (yB)) and g(- (yA),y- (yB)) -g(YAYB).

That is, y-I has square ratio r-1. 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 two-dimensional 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 B|a, 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 3|10 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 n-dimensional 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 THREE-DIMENSIONAL 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

three-dimensional 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 self-adjoint 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 three-dimensional Minkowski space [5].

For such theories the modular involutions, Jo, associated by Tomita-Takesaki

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 three-dimensional 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, three-dimensional

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 7-E 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} w-w.

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)} wE-w. 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 aPa-lI for P e P and ca(g) =- g a aga-I 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,

three-dimensional 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 three-dimensional Minkowski space. Moreover, since such

motions are in fact isometrics in R1'2 then D(O) is isomorphic to a subgroup of the

three-dimensional 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 as-0(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 three-dimensional 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 three-dimensional case of this dissertation.