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 Title:
 An algebraic characterization of Minkowski space
 Creator:
 White, Richard K., 1960
 Publication Date:
 2001
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 English
 Physical Description:
 v, 106 leaves : ; 29 cm.
Subjects
 Subjects / Keywords:
 Algebra ( jstor )
Axioms ( jstor ) Distance functions ( jstor ) Geometric lines ( jstor ) Geometric planes ( jstor ) Lines in space ( jstor ) Minkowski space ( jstor ) Parallel lines ( jstor ) Pencils ( jstor ) Secant lines ( jstor ) Dissertations, Academic  Mathematics  UF ( lcsh ) Mathematics thesis, Ph. D ( lcsh )
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 bibliography ( marcgt )
theses ( marcgt ) nonfiction ( marcgt )
Notes
 Thesis:
 Thesis (Ph. D.)University of Florida, 2001.
 Bibliography:
 Includes bibliographical references (leaves 103105).
 General Note:
 Printout.
 General Note:
 Vita.
 Statement of Responsibility:
 by Richard K. White.
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 University of Florida
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 University of Florida
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 Copyright [name of dissertation author]. Permission granted to the University of Florida to digitize, archive and distribute this item for nonprofit research and educational purposes. Any reuse of this item in excess of fair use or other copyright exemptions requires permission of the copyright holder.
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Full Text 
AN ALGEBRAIC CHARACTERIZATION OF MINKOWSKI SPACE
By
RICHARD K. WHITE
A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL
OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE DEGREE OF
DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
2001
ACKNOWLEDGMENTS
I thank my advisor Dr. Stephen J. Summers for his guidance in the
preparation of this dissertation. I would also like to thank all of my committee
members for their support and for serving on my committee.
TABLE OF CONTENTS
page
ACKNOW LEDGM ENTS........................................................................................ ii
ABSTRACT .......................................................................................................... v
CHAPTERS
1 INTRODUCTION............................................................................................. 1
2 A CONSTRUCTION OF THREEDIMENSIONAL MINKOWSKI SPACE .... 7
2.1 Preliminaries ............................................................................................ 9
2.2 Construction of 1 ...................................................................................... 11
2.3 Reflections About Exterior Points ............................................................ 18
2.4 Embedding a Hyperbolic ProjectiveM etric Plane .................................... 22
2.5 Exterior Point Reflections Generate Motions in an Affine Space ............. 23
2.6 Conclusion ................................................................................................ 24
3 A CONSTRUCTION OF FOURDIMENSIONAL MINKOWSKI SPACE ...... 26
3.1 Preliminaries and General Theorems ....................................................... 26
3.1.1 Properties of M ............................................................................ 30
3.1.2 Properties of P ............................................................................. 31
3.1.3 Properties of X ............................................................................... 32
3.1.4 General Consequences of the Axioms ............................................. 32
3.1.5 Perpendicular Plane Theorems ...................................................... 33
3.1.6 Parallel Planes ............................................................................... 34
3.1.7 Consequences of Axiom 11 and 3.1.6.18 ......................................... 37
3.2 Lines and Planes ....................................................................................... 38
3.2.1 General Theorems and Definitions .................................................. 38
3.2.2 Isotropic Lines ............................................................................... 41
3.3 A Reduction to Two Dimensions .............................................................. 44
3.4 Consequences of Section 3.3 .................................................................... 51
3.5 Construction of the Field .......................................................................... 52
3.6 Dilations and the Construction of ( ,,V,K) ............................................. 58
3.7 Subspaces and Dimensions ...................................................................... 66
3.8 Orthogonality ........................................................................................... 69
3.9 The Polarity ........................................................................................... 80
3.10 Spacelike Planes and Their Reflections ................................................... 86
iii
page
4 AN EXAMPLE OF THE THREEDIMIENSIONAL MODEL .......................... 91
5 C O N C L U SIO N ................................................................................................ 96
R E F E R E N C E S ..................................................................................................... 103
BIOGRAPHICAL SKETCH ................................................................................. 106
Abstract of Dissertation Presented to the Graduate School
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy
AN ALGEBRAIC CHARACTERIZATION OF MINKOWSKI SPACE
By
Richard K. White
May 2001
Chairman: Dr. Stephen J. Summers
Major Department: Mathematics
We give an algebraic characterization of threedimensional and four
dimensional Minkowski space. We construct both spaces from a set of involution
elements and the group it generates. We then identify the elements of the original
generating set with spacelike lines and their corresponding reflections in the three
dimensional case and with spacelike planes and their corresponding reflections in the
fourdimensional case. Further, we explore the relationship between these
characterizations and the condition of geometric modular action in algebraic
quantum field theory.
CHAPTER 1
INTRODUCTION
The research program which this dissertation is a part of started with a paper
by Detlev Buchholz and Stephen J. Summers entitled "An Algebraic Character
ization of Vacuum States in Minkowski Space" [9]. In 1975, Bisognano and
Wichmann 15] showed that for quantum theories satisfying the Wightman axioms
the modular objects associated by TomitaTakesaki theory to the vacuum state and
local algebras generated by field operators with support in wedgelike spacetime
regions in Minkowski space have geometrical meaning. Motivated by this work,
Buchholz and Summers gave an algebraic characterization of vacuum states on nets
of C*algebras over Minkowski space and reconstructed the spacetime translations
with the help of the modular structures associated with such states. Their result
suggested that a "condition of geometric modular ;, i iii" might hold in quantum
field theories on a wider class of spacetime manifolds.
To explain the abstract version of this condition, first some notation is
introduced and the basic setup is given. Let {A, ,},1 be a collection of C*algebras
labeled by the elements of some index set I such that (Q,<) is an ordered set and the
property of isotony holds. Tli it is, if i0,i2 e I such that il < i2, then A, c A,2.
Let A be a C*algebra containing {AI}iE/. It is also required that the assignment
i 'f A, is an orderpreserving bijection.
In algebraic quantum field theory, the index set I is usually a collection of
open subsets of an appropriate spacetime (M,g). In such a case, the algebra A, is
interpreted as the C*algebra generated by all the observables measured in a
2
spacetime region i. Hence, to different spacetime regions should correspond different
algebras.
Given a state co on A, let (7R., 7tr, Q) be the corresponding GNS
representation and let Ri, = no)(Ai)", i e 1, be the von Neumann algebras generated
by the nr.,(A), i E L. Assume that the map i '> 7Zi is an orderpreserving bijection
and that the GNS vector Q is cyclic and separating for each algebra 71i, i e I. From
TomitaTakesaki theory, we thus have a collection {J}i*e of modular involutions
and a collection {Ai}iEJ of modular operators directly derivable from the state and
the algebras. Each Ji is an antilinear involution on RH. such that JiTiJ, = 7.i' and
Ji, = Q.
In addition, the set {Ji}i i generates a group J which becomes a topological
group in the strong operator topology on B3(7c), the set of all bounded operators on
7(H. The modular operators {Ai}, I are positive (unbounded) invertible operators
such that
AJ'RTjAi" = j, j e I, t e R, i= 7 and A"iQ = Q.
In algebraic quantum field theory the state o models the preparations in the
laboratory and the algebras Ai model the observables in the laboratory and are
therefore, viewed as idealizations of operationally determined quantities. Since
TomitaTakesaki theory uniquely gives these modular objects corresponding to
(7Rj,Q), it thus follows that these modular objects can be viewed as operationally
determined.
Motivated by the earlier work of Bisognano and Wichmann [5], Buchholz and
Summers [9] proposed that physically interesting states could be selected by looking
at those states which satisfied the condition of geometric modular action, CGMA.
Given the structures indicated above, the pair ({A,}ii,o) satisfies the abstract
version of the CGMA if {T4}iI is left invariant under the adjoint action of the
3
modular conjugations {Ji} /,; that is, if for every i,j in I there is a k in I such that
adJi(Rj) Ji jJi = R"k, where JUjJi = {JAJi : A e Rj}.
Thus, for each i in I, there is an orderpreserving bijection, automorphism, ti on I,
(I,<), such that JiR7jJi = T,(/), forj e L.The set {T,}il, is a set of involutions which
generate a group T, which is a subgroup of the group of translations on L Buchholz,
Dreyer, Florig, and Summers [6] have shown that the groups Tarising in this
manner satisfy certain structure properties, but for the purposes of this thesis, it is
only emphasized that Tis generated by involutions and is hence, a Coxeter group.
Thus there are two groups generated by involutions operating on two different
levels.
1. The group Tacting on the index set L.
2. The group Jacting on the set {'1"}ie.
To elaborate further the relation between the groups Tand J, consider the
following.
Proposition 1.1 [6] The surjective map : j+T given by (Ji ...Jim) = t i "i,,, is
a group homomorphism. Its kernel S lies in the center of 7J and the adjoint action of
S leaves each 1i fixed. U
Thus, Jis a central extension of the group Tby S.
As an immediate consequence of this proposition, J provides a projective
representation of Twith coefficients in an abelian group Z in the center of J. Thus,
the condition of geometric modular action induces a transformation group on the
index set I and provides it with a projective representation.
With this in mind, the following program was then posed. Given the
operational data available from algebraic quantum field theory, can one determine
the spacetime symmetries, the dimension of the spacetime, and the spacetime itself?
That is to say, given a net of C*algebras and a state co satisfying the CGMA, can
4
one determine the spacetime symmetries, the dimension of the spacetime, and even
the spacetime itself?
Part of this has been carried out by Buchholz, Dreyer, Florig, and Summers
for Minkowski space and de Sitter space [6]. However, in order to do so, they had to
presume the respective spacetime as a topological manifold. But would it not be
possible to completely derive the spacetime from the operationally given data
without any assumption about dimension or topology?
As was pointed out by Dr. Summers, a possibility to do so was opened up in
this program in the following manner. As already seen, the CGMA yields an
involution generated group complete with a projective representation and there is in
the literature a way of deriving spacetimes from such groups going under the name
of absolute geometry.
In general, absolute geometry refers to a geometry that includes both
Euclidean and nonEuclidean geometry as special cases. Thus, one has a system of
axioms not yet implying any decision about parallelism. In our case, the axioms are
given in terms of a group of motions as an extension of Klein's Erlangen Program. A
group of motions is defined as a set g of involution elements closed under
conjugation and the group 5 it generates. In a group of motions the representations
of geometric objects and relations depend only on the given multiplication for the
group elements, without reference to any additional structure. The system of axioms
is formulated in terms of the involutory generators alone, so that geometric concepts
like point, line, and incidence no longer are primary but are derived.
The necessary means for setting up this representation are provided by the
totality of reflections in points, lines, and planes (a subset of the set of motions).
Points, lines, and planes are in onetoone correspondence with the reflections in
them so that geometric relations among points, lines, and planes correspond to
grouptheoretic equations among the reflections. This enables one to be able to
5
formulate geometric theorems about elements of the group of motions and to be able
to then prove these theorems by grouptheoretic calculation.
To summarize, we are to find conditions on an index set I and a corresponding
net of C*algebras {4J,}i;e as well as a state o satisfying the CGMA such that the
elements of I can be naturally identified with open sets of Minkowski space and such
that the group T"is implemented by the Poincare group on this Minkowski space.
Out of the group Twe wish to construct Minkowski space such that T's natural
action on Minkowski space is that of the Poincare group.
This involves two steps. First we carry out the absolute geometry program for
three and fourdimensional Minkowski space. That is, characterize three and
fourdimensional Minkowski space in terms of a group of motions (G, 0). Second, we
must determine what additional structure on the ordered set I would yield from
TomitaTakesaki theory algebraic relations among the J1 (and hence, among the ti)
which coincide with the algebraic characterization found in step one.
The organization of the thesis is as follows: in Chapter 2, the given pair (G, 0)
is used to construct a threedimensional Minkowski space out of the plane at infinity.
Then identification of the involutory elements of 9 with spacelike lines and their
group action in 5 with reflections about spacelike lines is made.
In Chapter 3 using the same initial data, (Q, 0), as was given in Chapter 2 but
satisfying different axioms, a fourdimensional Minkowski space is constructed. The
approach taken here differs from that taken in Chapter 2. This time the affine space
is constructed first and then the hyperplane at infinity is used to obtain the metric.
The identification of the elements of g with spacelike planes and their group action
in 5 with reflections about spacelike planes is made.
In Chapter 4 a concrete example of the threedimensional characterization is
given. As already mentioned, Bisognano and Wichmann showed that for quantum
field theories satisfying the Wightman axioms the modular objects associated by
6
TomitaTakesaki theory to the vacuum state and local algebras in wedgelike regions
in threedimensional Minkowski space have geometrical interpretation [5]. In
particular, the modular conjugations, {Ji}ie, act as reflections about spacelike lines.
In this chapter it is shown that if one chooses the set of wedgelike regions as the
index set I, the group J generated by the set {Jri}i satisfies the axiom system given
in Chapter 2 for the construction of a threedimensional Minkowski space.
In Chapter 5 some concluding remarks about the second step described above
are made. It is noted that if one assumes the modular stability condition [6] and the
halfsided modular inclusion relations given by Wiesbrock [29], then one does obtain
a unitary representation of the 2+1dimensional Poincare group.
CHAPTER 2
A CONSTRUCTION OF THREEDIMENSIONAL MINKOWSKI SPACE
In this chapter we give an absolute geometric, that is, an algebraic,
characterization of threedimensional Minkowski space. This chapter is a version of a
preprint by the author entitled "A GroupTheoretic Construction Of Minkowski
3Space Out Of The Plane At Infinity" [28]. Along with the wellknown
mathematical motivations [1, 2] there are also physical motivations, as we discussed
in Chapter 1. Threedimensional Minkowski space is an affine space whose plane at
infinity is a hyperbolic projectivemetric plane [12]. In "Absolute Geometry" [2],
Bachmann, Pejas, Wolff, and Bauer (BPWB) took an abstract group 0 generated
by an invariant system g of generators in which each of the generators was
involutory, satisfying a set of axioms and constructed a hyperbolic projectivemetric
plane in which the given group 5 was isomorphic to a subgroup of the group of
congruent transformations (motions) of the projectivemetric plane. By interpreting
the elements of G as line reflections in a hyperbolic plane, BPWB showed that the
hyperbolic projectivemetric plane could be generated by these line reflections in
such a way that these line reflections form a subgroup of the motions group of the
projectivemetric plane.
Coxeter showed in [13] that every motion of the hyperbolic plane is generated
by a suitable product of orthogonal line reflections, where an orthogonal line
reflection is defined as a harmonic homology with center exterior point and axis the
given ordinary line and where the center and axis are a polepolar pair. Here we
show that Coxeter's and BPWB's notions of motions coincide in the hyperbolic
8
projectivemetric plane and that the motions can be viewed as reflections about
exterior points.
Next we embed our projectivemetric plane into a threedimensional projective
space. By singling out our original plane as the plane at infinity, we obtain an affine
space whose plane at infinity is a hyperbolic projectivemetric plane,
threedimensional Minkowski space. Finally, we show that the motions of our
original plane induce motions in the affine space and, by a suitable identification, we
show that any motion in Minkowski space can be generated by reflections about
spacelike lines. Thus, to construct a threedimensional Minkowski space, one can
start with a generating set g of reflections about spacelike lines in the plane at
infinity. So g may be viewed as a set of reflections about exterior points in a
hyperbolic projectivemetric plane. Out of the plane at infinity, one can obtain a
threedimensional affine space with the Minkowski metric, which is constructed from
a group generated by a set of even isometries or rotations.
The approach in this chapter differs from the method used by Wolff [301 for
twodimensional Minkowski space and by Klotzek and Ottenburg [19] for
fourdimensional Minkowski space. The approach in these papers is to begin by
constructing the affine space first. For Wolff's [30] twodimensional case, the
elements of the generating set g are identified with line reflections in an affine plane.
For Klotzek and Ottenburg's [19] fourdimensional case, the elements of the
generating set g are identified with reflections about hyperplanes in an affine space.
Thus, in each of these papers, the generating set g is identified with a set of
symmetries or odd isometries. A map of affine subspaces is then obtained using the
definition of orthogonality given by commuting generators. This map induces a
hyperbolic polarity in the hyperplane at infinity, yielding the Minkowski metric.
To briefly recap the two approaches described above, note that both
approaches, ours and the one given by Klotzek and Ottenburg [19] and by Wolff
9
[30], start with a generating set g of involution elements. In our approach, one can
identify the elements of g with a set of even isometries (rotations) and use the
definition of orthogonality induced by the commutation relations of the generators in
the hyperplane at infinity to obtain the polarity and then embed this in an affine
space to get Minkowski space. In the approach of Klotzek and Ottenburg [19] and
Wolff [30], one can identify the elements of G with a set of odd isometries
symmetriess), construct an affine space first, and then use the definition of
orthogonality induced by the commutation relations of the generators in the affine
space to obtain a polarity in the hyperplane at infinity to get Minkowski space.
2.1 Preliminaries
The starting point for an algebraic characterization of Minkowski space is
therefore far from unique. Our particular choice of algebraic characterization, in
terms of reflections about spacelike lines in three dimensional Minkowki space, is
motivated by physical considerations [6] which we briefly explain in the conclusion.
A hyperbolic projectivemetric plane is a projective plane in which a
hyperbolic polarity is singled out and used to define orthogonality in the plane. A
polarity is an involutory projective correlation. A correlation is a onetoone
mapping of the set of points of the projective plane onto the set of lines, and of the
set of lines onto the set of points such that incidence is preserved. A projective
correlation is a correlation that transforms the points Y on a line b into the lines y'
through the corresponding point B'. So, in general, a correlation maps each point A
of the plane into a line a of the plane and maps this line into a new point A'. When
the correlation is involutory, A' always coincides with A. Thus a polarity relates A to
a, and vice versa. A is called the pole of a and a is called the polar of A. Since this is
a projective correlation, the polars of all the points on a form a projectively related
pencil of lines through A.
10
The polarity dualizes incidences: if A lies on b, then the polar of A, a, contains
the pole of b,B. In this case we say that A and B are conjugate points, and that a
and b are conjugate lines. If A and a are incident, then A and a are said to be
selfconjugate: A on its own polar and a through its own pole [141. A hyperbolic
polarity is a polarity which admits selfconjugate points and selfconjugate lines. The
set of all selfconjugate points is called a conic, which we shall call the absolute.
In a projective plane in which the theorem of Pappas and the axiom of Fano
hold, the polarity can be used to introduce a metric into the plane. Orthogonality is
defined as follows: two lines (or two points or a line and a point) are said to be
orthogonal or perpendicular to each other if they are conjugate with respect to the
polarity.
Congruent transformations of the plane are those collineations of the plane
which preserve the absolute; that is, those collineations which leave the absolute
invariant. In a projective plane with a hyperbolic polarity as absolute, the group of
all collineations in the plane leaving the absolute invariant is called the hyperbolic
metric group and the corresponding geometry is called the hyperbolic metric
geometry in the plane 1311.
The conic or absolute, separates the points of the projective plane into three
disjoint classes: ordinary or interior points, points on the absolute, and exterior
points. The lines of the projective plane are likewise separated into three disjoint
classes. Secant lines, lines that contain interior points, exterior points, and precisely
two points on the absolute. Exterior lines, lines that contain only exterior points.
And tangent lines, lines that meet the absolute in precisely one point and in which
every other point is an exterior point.
Definition 2.1.1. Two lines containing ordinary points, two secant lines, are said to
be parallel if they have a point of the absolute in common.
Remark. The set of all interior points and the set of lines formed by intersecting
11
secant lines with the set of interior points, ordinary lines, is classical hyperbolic
plane geometry.
2.2 Construction of I
In this section we list the axioms and main results of BPWB [2] and provide a
sketch of some of the arguments they used which are pertinent to this work. For
detailed proofs, one is referred to the work of BPWB [2].
Definition 2.2.1. A set of elements of a group is said to be an invariant system if it
is mapped into itself (and thus onto itself) by every conjugation by an element of the
group. An element a of a group Q5 is called an involution if a2 = 1I, where 1e is the
identity element of the group Q5.
Basic assumption: A given group Q5 is generated by an invariant system g of
involution elements.
The elements of g are denoted by lowercase Latin letters. Those involutory
elements of !5 that can be represented as ab, where a,b E= are denoted by
uppercase Latin letters. If ,i e 5 and ril is an involution, we denote this by 4r.
Axioms
Axiom 1: For every P and Q there is a g with P, Q 1g.
Axiom 2: IfP,Q ig,h then P = Q or g = h.
Axiom 3: If a,b,c \P then abc = d e g.
Axiom 4: Ifa,b,c Ig then abc = d e 9.
Axiom 5: There exist g,h,j such that g Ih butj I g,h,gh.
Axiom 6: There exist elements d,a,b e G such that d,a,b I P,c for P,c e 0. (There
exist lines which have neither a line nor a point in common.)
Axiom 7: For each P and for each g there exist at most two elements h,j 6 9 such
that PIh,j but g,h I A,c and g,j B,d for any A,B,c,d E 0 (that is, have neither a
point nor a line in common).
Axiom 8: One never has P = g.
We call the set of axioms just given axiom system A, denoted by as A.
The initial interpretation of the elements of g is as secant or ordinary lines in
a hyperbolic plane for BPWB [2]. In our approach, we view the elements of g
initially as exterior points in a hyperbolic plane. After embedding our hyperbolic
projectivemetric plane into an affine space, we can identify the elements of g, our
generating set, with spacelike lines and their corresponding reflections in a
threedimensional Minkowski space. By realizing that statements about the geometry
of the plane at infinity correspond to statements about the geometry of the whole
space where all lines and all planes are considered through a point, we see that the
axioms also are statements about spacelike lines, the elements of g, and timelike
lines, the elements P of 0, through any point in threedimensional Minkowski space.
The models of the system of axioms are called groups of motions; that is, a
group of motions is a pair (5, G) consisting of a group Q5 and a system g of
generators of the group Q5 satisfying the basic assumption and the axioms.
To give a precise form to the geometric language used here to describe
grouptheoretic concepts occurring in the system of axioms, we associate with the
group of motions (0,G9) the group plane (50,9), described as follows.
The elements of Q are called lines of the group plane, and those involutory
group elements that can be represented as the product of two elements of g are
called points of the group plane. Two lines g and h of the group plane are said to be
perpendicular if g 1h. Thus, the points are those elements of the group that can be
written as the product of two perpendicular lines. A point P is incident with a line g
in the group plane if P 1g. Two lines are said to be parallel if they satisfy Axiom 6.
Thus, if P # Q, then by Axiomi and Axiom 2, the points P and Q in the group
plane are joined by a unique line. If P I g then Axiom 7 says that there are at most
two lines through P parallel to g.
13
Lemma 2.2.2.[21 For each a e 05, the mappings a : g * ga = aga and
aa : P * Pp' aPa arc onetoone mappings of the set of lines and the set of points,
each onto itself in the group plane.
Proof: Let a E 0, and consider the mapping y = aya of 0 onto itself. It is
easily seen that this mapping is bijective. Because G is an invariant system (ab e g
for every a, b E 9) 9 is mapped onto itself, and if P is a point, so that P = gh with
glh, then Pa = gaha and gajha, so that P1 is also a point. Thus, g ga, P  Pa
are onetoone mappings of the set of lines and the set of points, each onto itself in
the group plane. U
Definition 2.2.3. A onetoone mapping a of the set of points and the set of lines
each onto itself is called an orthogonal collineation if it preserves incidence and
orthogonality.
Since the "I" relation is preserved under the above mappings, the above
mappings preserve incidence and orthogonality as defined above.
Corollary 2.2.4.121 The mappings
a : g * ga andaa : P P
are orthogonal collineations of the group plane and are called motions of the group
plane induced by a.
In particular, if ac is a line a we have a reflection about the line a in the group
plane, and if ac is a point A, we have a point reflection about A in the group plane.
If to every a E S one assigns the motion of the group plane induced by a,
one obtains a homomorphism of 5 onto the group of motions of the group plane.
Bachmann [1] showed that this homomorphism is in fact an isomorphism so that
points and lines in the group plane may be identified with their respective
reflections. Thus, 0 is seen to be the group of orthogonal collineations of S
generated by G.
14
Definition 2.2.5. Planes that are representable as an isomorphic image, with respect
to incidence and orthogonality, of the group plane of a group of motions (O, g), are
called metric planes.
BPWB showed how one can embed a metric plane into a projectivemetric
plane by constructing an ideal plane using pencils of lines [2]. We shall now outline
how this is done.
Definition 2.2.6. Three lines are said to lie in a pencil if their product is a line; that
is, a,b,c lie in a pencil if
abc = d e g. (*)
Definition 2.2.7. Given two lines a,b with a # b, the set of lines satisfying (*) is
called a pencil of lines and is denoted by G(ab), since it depends only on the product
ab.
Note that the relation (*) is symmetric, that is, it is independent of the order
in which the three lines are taken, since cba = (abc)1 is a line, the invariance of g
implies that cab = (abc)c is a line and that every motion of the group plane takes
triples of lines lying in a pencil into triples in a pencil. The invariance of 9 also
shows that (*) holds whenever at least two of the three lines coincide.
Using the given axioms, BPWB 12] showed that there are three distinct classes
of pencils. If a, b V then G(ab) = {c : c IV}. In this case, G(ab) is called a pencil of
lines with center V and is denoted by G(V). If a,b Ic then G(ab) = {d : d 1c}. In this
case, G(ab) is called a pencil of lines with axis c and is denoted by G(c).
By Axiom 6, there exist lines a, b, c which do not have a common point or a
common line. Recall that lines of this type are called parallel. Thus, in this case
G(ab) = {c : c 11 a,b where a 11 b}, which we denote by Px.
Using the above definitions of pencils of lines and the above theorems, BPWP
[2] proved that an ideal projective plane, H, is constructed in the following way. An
15
ideal point is any pencil of lines G(ab) of the metric plane. The pencils G(P)
correspond in a onetoone way to the points of the metric plane. An ideal line is a
certain set of ideal points. There are three types:
1. A proper ideal line g(a), is the set of ideal points that have in common a line a
of the metric plane.
2. The set of pencils G(x) with x IP for a fixed point P of the metric plane, which
we denote by P.
3. Each set of ideal points that can be transformed by a halfrotation about a fixed
point P of the metric plane into a proper ideal line, which we denote by po.
The polarity is defined by the mappings
G(QC) C and C G(C)Q;
POO po and po > Po;
G(c) * g(c) and g(c) ' G(c).
Bachmann [1] showed that the resulting ideal plane is a hyperbolic projective plane
in which the theorem of Pappus and the Fano axiom hold; that is, it is a hyperbolic
projectivemetric plane.
In this model, the ideal points of the form G(P) are the interior points of the
hyperbolic projectivemetric plane. Thus the points of the metric plane correspond in
a onetoone way with the interior points of the hyperbolic projectivemetric plane.
The ideal points G(x), for x e g are the exterior points of the hyperbolic
projectivemetric plane.
Theorem 2.2.8. Each x e g corresponds in a onetoone way with the exterior points
of the hyperbolic projectivemetric plane.
Proof: Because each line d of the metric plane is incident with at least three points
and a point is of the form ab with a I b, then each x e g is the axis of a pencil. From
the uniqueness of perpendiculars each x e g corresponds in a onetoone way with
the pencils G(x). Hence, each x e g corresponds in a onetoone way with the
exterior points of the hyperbolic projectivemetric plane. U
16
Thus, the axioms can be viewed as axioms concerning the interior and exterior
points of a hyperbolic projectivemetric plane. The ideal points of the form G(ab)
where a I1 b are the points on the absolute, that is, the points at infinity in the
hyperbolic projectivemetric plane.
Now consider the ideal lines. A proper ideal line g(a) is a set of ideal points
that have in common a line a of the metric plane.
Theorem 2.2.9. A proper ideal line g(a) is a secant line of the form
g(a) = {P,x,G(bc) : x, PIa and abc E Q where b 1 c}.
Proof: Every two pencils of lines of the metric plane has at most one line in
common. By Axiom 7, each line belongs to at most two pencils of parallels and each
line g e g belongs to precisely two such pencils. Thus, a proper ideal line contains
two points on the absolute, interior points, and exterior points; that is, a proper
ideal line is a secant line. If we identify the points P with the pencils G(P) and the
lines x with the pencils G(x), then a secant line is the set g(c) = {P,x,G(ab) : x,P Ic
and abc E G where a 11 b}. U
Corollary 2.2.10. The ideal line which consists of pencils G(x) with x IP for a fixed
point P of the metric plane consists of only exterior points; that is, it is an exterior
line. Under the identification ofx with G(x) then P = {x E g : x\P}.
The last type of ideal line is a tangent line. It contains only one point,
G(ab) = P., on the absolute. Denoting this line by po, then
p.o = {G(ab)} u {x e g : abx e g}, where a 11 b. Recalling that each x e Q
corresponds to an exterior point in the hyperbolic projectivemetric plane, we see
that a tangent line consists of one point on the absolute and every other point is an
exterior point.
Also note that under the above identifications, each secant line g(c)
corresponds to a unique "exterior point", c, c 0 g(c) since one only considers those
17
x,P Ic such that xc # 1 and Pc # 1. Each exterior line corresponds to a unique
interior point P and each tangent line corresponds to a unique point on the absolute.
Theorem 2.2.11. The map D given by
(i) (c) = g(c), D(g(c)) = c
(ii) D(P) = D(,) = P
(iii) D(poo) = P., I(Po) = po
is a polarity.
Proof: Let P be the set of all points of H and C the set of all lines of H. From the
remarks above it follows that D is a welldefined onetoone pointtoline mapping of
P onto C and a welldefined onetoone linetopoint mapping of C onto P. Next we
show that D is a correlation and for this it suffices to show that (D preserves
incidence.
Let g(c) = {P,x,G(ab) : x,Plc and abc e 9 where a 11 b} be a secant line. Let
A,B,d,P. e g(c), where P, = G(ej) = {x E 9 : xef e 9 and e IIf}.Then A,B,d I c and
cab e 9.
<(A) = A= {x : x I A}, D(B) = B, I(d) =g(d), (D(P) = p = {P,} u {x : efx E 9}
D(g(c)) = c e A n B r) g(d) r p and t(g(c)) E D(A), 0(B), <(d), F(Poo).
Hence, D preserves incidence on a secant line. Now consider an exterior line
P = {x : xjP} and let a, b e P. Then a, bIP and it follows that P E g(a) r g(b); that
is, (D(P) e D(a) r 1(b) and (D preserves incidence on an exterior line.
Finally, let po = {G(ab)} u {x : abx e 9 where a 11 b} be a tangent line.
Clearly, since 1(G(ab)) = p., then P. = G(ab) e p.. Now suppose that d e po.
Then abd e 9 and
D(d) = g(d)= {A,x,G(ej) : A,xld and def e 9 where e 1}.
Thus, d e G(ab) r G(eJ) and Po, E g(d). This implies that d e po and O(po) e D(d).
Hence, D preserves incidence and is a correlation.
18
Note also that from the work above, D transforms the points Y on a line b into
the lines D(Y) through the point 0(b). Thus, ( is a projective correlation. Since
O2 = 10, then (D is a polarity. Moreover, since D(p.) = Po with Po. e p., then (D is
a hyperbolic polarity. E
Theorem 2.2.12 The definition of orthogonality given by the polarity agrees with
and is induced by the definition of orthogonality in the group plane.
Proof: If we define perpendicularity with respect to our polarity then the following
are true (we use the notation +> to denote the phrase "if and only if "):
(i) g(c) 1_ g(a) <> 1(g(c)) = c e g(a) and '(g(a)) = a e g(c) * a c.
(ii) g(c) _L P> (g(c)) =c E P c JP.
(iii) C _L P O ((P) = P E g(c) = (D(c) ++ P c.
(iv) C _L g(x) (g(x)) =xeg(c)= D(c) x Ic.
Instead of interpreting our original generators as ordinary lines in a hyperbolic
plane, we now interpret them as exterior points. We can construct a hyperbolic
projectivemetric plane in which the theorem of Pappus and Fano's axiom hold,
which is generated by the exterior points of the hyperbolic projectivemetric plane.
With the identifications above and the geometric objects above, we show in
the next section that the motions of the hyperbolic projectivemetric plane above can
be generated by reflections about exterior points; that is, any transformation in the
hyperbolic plane which leaves the absolute invariant can be generated by a suitable
product of reflections about exterior points.
2.3 Reflections About Exterior Points
Definition 2.3.1. A collineation is a onetoone map of the set of points onto the set
of points and a onetoone map of the set of lines onto the set of lines that preserves
the incidence relation.
Definition 2.3.2. A perspective collineation is a collineation which leaves a line
19
pointwise fixed, its axis, and a point linewise fixed, its center.
Definition 2.3.3. A homology is a perspective collineation with center a point B and
axis a line b where B is not incident with b.
Definition 2.3.4. A harmonic homology with center B and axis b, where B is not
incident with b, is a homology which relates each point A in the plane to its
harmonic conjugate with respect to the two points B and (b,[A,B]), where [A,B] is
the line joining A and B and (b, [A,B]) is the point of intersection of b and [A,B].
Definition 2.3.5. A complete quadrangle is a figure consisting of four points (the
vertices), no three of which are collinear, and of the six lines joining pairs of these
points. If 1 is one of these lines, called a side, then it lies on two of the vertices, and
the line joining the other two vertices is called the opposite side to 1. The
intersection of two opposite sides is called a diagonal point.
Definition 2.3.6. A point D is the harmonic conjugate of a point C with respect to
points A and B if A and B are two vertices of a complete quadrangle, C is the
diagonal point on the line joining A and B, and D is the point where the line joining
the other two diagonal points cuts [A,B]. One denotes this relationship by
H(AB, CD).
Example 2.3.7. Let A,B, and C be three collinear points. For a quick construction of
the harmonic conjugate D of C with respect to A and B let Q,R,S be any points such
that [Q,R],[Q,S], and [R,S] pass through A,B,C respectively. Let
{P} = [A,S] r) [B,R], then {D} = [A,B] n [P,Q] ([11]). Note that if [R,S] 11 [A,C] then
D is the midpoint of A and B.
Coxeter [13] showed that any congruent transformation of the hyperbolic
plane is a collineation which preserves the absolute and that any such
transformation is a product of reflections about ordinary lines in the hyperbolic
plane where a line reflection about a line m is a harmonic homology with center M
and axis m, where M and m are a polepolar pair and M is an exterior point. A point
20
reflection is defined similarly, a harmonic homology with center M and axis m, where
M and m are a polepolar pair, M is an interior point, and m is an exterior line. Note
that in both cases, M and m are nonincident.
In keeping with the notation employed at the end of 2.2, let b be an exterior
point and g(b) its pole.
SA Ab and d ' db
Lemma 2.3.8. The map Tb : A d i g(d)b is a
BPoo b pb p p
collineation.
Proof: This follows from the earlier observation that the motions of the group plane
map pencils onto pencils preserving the "I" relation. U
Lemma 2.3.9. Tb is a perspective collineation and, hence, a homology.
Proof: Recall that g(b) = {A,x,P. : x,A lb and where b lies in the pencil P.}. For
any A and x in g(b) we have Ab = A and xb = x since A,x lb and if A',x' 0 g(b) then
A',x' I b and Ab # A, xb # x, and Abxb I b. Thus, Abxb o g(b).
Recall also that Po = G(cd), where c and d do not have a common
perpendicular nor a common point and thus, G(cd) = {f : fcd e g}. Now g(b) is a
secant line, so that it contains two such distinct points, Po and Q., say, on the
absolute. Since the motions of the group plane map pencils onto pencils preserving
the "I" relation it follows that if c,d e P. then cb, d e Po and hence pb = Po and
Qb= Qo. Moreover, if R. o g(b) then it follows that Rb o g(b). Thus, Tb leaves
g(b) pointwise invariant.
Now let g(d), Q, and r. be a secant line, exterior line, and tangent line,
respectively, containing b. For e E g(d) we have e Id and e' I db = d since b I d,
thus eb e g(d). For A E g(d), Ab \db = d, so Ab e g(d). Similarly, it follows that if
P., e g(d) then pb e g(d), and that g(d)b = g(d). One easily sees that Qb = Q and
r = r.. Thus, Tb leaves every line through b invariant and Tb is a perspective
collineation for each b e g. 0
Theorem 2.3.10. Tb is a harmonic homology.
Proof: Since A b is again a point in the original group plane and since db is again a
line in the original group plane and from the observations above, we have, for each
b e G, Tb maps interior points to interior points, exterior points to exterior points,
points on the absolute to points on the absolute, secant lines to secant lines, exterior
lines to exterior lines, and tangent lines to tangent lines. Moreover, since (Zb)b =
for any e 5, Tb is involutory for each b E g. Now in a projective plane in which
the theorem of Pappus holds, the only collineations which are involutory are
harmonic homologies [10], thus Tb is a harmonic homology for each b E g. N
Theorem 2.3.11. Interior point reflections are generated by exterior point reflections.
Proof: A similar argument shows that for each interior point A, TA is a harmonic
homology with center A and axis A where A is the polar of A, A o A, and where TA
is defined analogously to Tb. Thus, each TA is a point reflection and since A is the
product of two exterior points, we see that point reflections about interior points are
generated by reflections about exterior points. U
Theorem 2.3.12. The reflection of an interior point about a secant line is the same
as reflecting the interior point about an exterior point. Moreover, since any motion
of the hyperbolic plane is a product of line reflections about secant lines, any motion
of the hyperbolic plane is generated by reflections about exterior points.
Proof: Consider a line reflection in the hyperbolic plane; that is, the harmonic
homology with axis g(b) and center b. Let A be an interior point and g(d) a line
through A meeting b. Since b e g(d) then b Id and g(d) is orthogonal to g(b). Let E
be the point where g(b) meets g(d). Since E E g(b) then E lb and Eb =ffor some
f e g. It follows that the reflection of A about g(b) is the same as the reflection of A
about E. Since b Id and Eld then bd = C and we have E,C Ib,d with b # d. Thus, by
Axiom 2, E = C = bd. Hence, AE = Adb = Ab. *
22
Theorem 2.3.13 Reflections of exterior points about exterior points and about
exterior lines are also motions of the projectivemetric plane; that is, the +b 's for
b e g acting on exterior points and exterior lines are motions of the hyperbolic
projectivemetric plane.
Proof: The motions of the projectivemetric plane are precisely those collineations
which leave the absolute invariant. U
We also point out that the proof that each Tb is an involutory homology also
showed that the Fano axiom holds, since in a projective plane in which the Fano
axiom does not hold no homology can be an involution [4].
2.4 Embedding a Hyperbolic ProjectiveMetric Plane
In this section we embed our hyperbolic projectivemetric plane into a three
dimensional projective space, finally obtaining an affine space whose plane at infinity
is isomorphic to our original projectivemetric plane. Any projective plane H in
which the theorem of Pappus holds can be represented as the projective coordinate
plane over a field )C. (The theorem of Pappus guarantees the commutivity of C.)
Then by means of considering quadruples of elements of /C, one can define a
projective space P3(APC) in which the coordinate plane corresponding to IH is included.
If the Fano axiom holds, then the corresponding coordinate field )C is not of
characteristic 2 [4]. By singling out the coordinate plane corresponding to H as the
plane at infinity, one obtains an affine space whose plane at infinity is a hyperbolic
projectivemetric plane: that is, threedimensional Minkowski space.
To say that a plane H is a projective coordinate plane over a field AC means
that each point of H is a triple of numbers (xo,xI,x2), not all x, = 0, together with
all multiples (0xo,kxi, xx2), for X 0 and k s /C. Similarly, each line of H is a triple
of numbers [uo,uI,u2], not all ui = 0, together with all multiples [kuo,Xu\,Xku],
X # 0. In P3(/C), all the quadruples of numbers with the last entry zero correspond
23
to H. One can now obtain an affine space A by defining the points of A to be those
of P3(kA) l; that is, those points whose last entry is nonzero; a line I of A to be a
line I' in P3(kC) 1 minus the intersection point of the line 1' with I; and by
defining a point P in A to be incident with a line 1 of A if, and only if, P is incident
with the corresponding 1'. Planes of A are obtained in a similar way [14].
Thus, each point in H7 represents the set of all lines in A parallel to a given
line, where lines and planes are said to be parallel if their first three coordinates are
the same, and each line in I represents the set of all planes parallel to a given plane.
Because parallel objects can be considered to intersect at infinity, we call H the
plane at infinity.
2.5 Exterior Point Reflections Generate Motions in an Affine Space
In this section we state and prove the main result of this chapter. Coxeter
showed that threedimensional Minkowski space is an affine space whose plane at
infinity is a hyperbolic projectivemetric plane t11]. He also classified the lines and
planes of the affine space according to their sections by the plane at infinity as
follows:
Line or Plane Section at Infinity
Timelike line Interior point
Lightlike line Point on the absolute
Spacelike line Exterior point
Characteristic plane Tangent line
Minkowski plane Secant line
Spacelike plane Exterior line
He also showed that if one starts with an affine space and introduces a
hyperbolic polarity in the plane at infinity of the affine space, then the polarity
induces a Minkowskian metric on the whole space. Under a hyperbolic polarity, a
line and a plane or a plane and a plane are perpendicular if their elements at infinity
24
correspond. Two lines are said to be perpendicular if they intersect and their
elements at infinity correspond under the polarity.
Theorem 2.5.1. Every motion in a threedimensional affine space with a hyperbolic
polarity defined on its plane at infinity, is generated by reflections about exterior
points. Moreover, because exterior points correspond to spacelike lines, then any
motion in threedimensional Minkowski space is generated by reflections about
spacelike lines.
Proof: Because any motion in threedimensional Minkowski space can be generated
by a suitable product of plane reflections, it suffices to show that reflections about
exterior points generate plane reflections.
Let a be any Minkowski plane or spacelike plane. Let P be any point in
Minkowski space. Let / be the line through P parallel to a. Let a. denote the
section of a at infinity. Applying the polarity to cc. we get a point go _L cc.. Let g
be a line through P whose section at infinity is goo, so that g is a line through P
orthogonal to a. because each line in the plane at infinity contains at least 3 points,
there exists a line / in a which is orthogonal to g as g. 1 ao0. Now let m be a line
through P not in a which intersects 1. It follows that the reflection of P about a is
the same as reflecting m about I and taking the intersection of the image of m under
the reflection with g. By the construction of the affine space and the definition of
orthogonality in the affine space it follows that 1 and m must act as their sections at
infinity act. because any point reflection in the hyperbolic projectivemetric plane
can be generated by reflections about exterior points, we have that the reflection of
P about a is generated by reflections of P about spacelike lines. M
2.6 Conclusion
As already indicated above, the geometric model for the generators of g which
lies behind the choice of algebraic characterization of threedimensional Minkowski
25
space differs significantly from those of previous absolute geometric characterizations
of Minkowski space. The model given here is the set of reflections about spacelike
lines, which is not a choice which would be made a priori by other mathematicians.
However, this is yet another example of a situation where the initial data are
imposed by physical, as opposed to purely mathematical, considerations.
In the next chapter starting with the same initial data, but satisfying different
axioms, a construction of fourdimensional Minkowski space is given. First the affine
space is constructed and then the hyperplane at infinity is used. Also given is an
explicit construction of the field, the vector space, and the metric.
CHAPTER 3
A CONSTRUCTION OF FOURDIMENSIONAL MINKOWSKI SPACE
In this chapter a construction of fourdimensional Minkowski space will be
given using the same initial data as in Chapter 2 but satisfying different axioms. The
actual construction is quite long, so to aid the reader in following, A brief outline of
the procedure shall be given here. First some general theorems and the basic
definitions will be given in the first two sections. In Sections 3.3 and 3.4 attention is
restricted to two dimensions in order to obtain the necessary machinery to construct
the field. Once the field has been obtained, then a vector space is constructed and
given a definition of orthogonality. It is then shown that the vector definition of
orthogonality is induced by and agrees with the initial definition of perpendicularity
for the geometry generated by the original set of involutions 9.
To obtain a metric vector space from the constructed vector space V, a map n
is defined on the subspaces of V. The map nt is defined to send a subspace to its
orthogonal complement. Using the work of 131 (which is given for the convenience of
the reader) the Minkowski metric is obtained and hence, Minkowski space. The last
section of this chapter identifies the elements of 9 with spacelike planes in
Minkowski space and the motions of the elements of Q with reflections about
spacelike planes, as was required in Chapter 1.
3.1 Preliminaries and General Theorems
Let there be given a nonempty set 9 of involution elements and the group S it
generates, where for any a e 9 and for any 4 E S we have ca = 'a4 E 9. If the
product of two distinct elements 41,42 e @ is an involution then we denote this by
27
writing t I2. We note that if 1 \12 then  for all ,1, 2 in 05 because
Let M = {ap3: a I 13 and a,P13 e g} and P = 9 u M. We consider the
elements of g as spacelike planes and the elements of M as Minkowskian or
Lorentzian planes. Thus, P consists of the totality of "nonsingular" planes and we
denote the elements of P by a, 13, y ....
We begin by giving the basic assumption and by making some preliminary
definitions. All the axioms are then listed for the convenience of the reader. The
geometric meaning of the axioms and the symbols used will be made clear in the
appropriate sections. The first four sections examine the incidence axioms. The order
axioms are reintroduced in Section 3.5, where we construct and order the field to
obtain a field isomorphic to the reals. In Section 3.6 we give the motivation behind
the particular choice of dilation axioms. Using these axioms we are able to define a
scalar multiplication and thereby obtain an affine vector space. The polarity axioms
are given again and examined in Section 3.8, where we define orthogonal vectors.
Basic assumption: If a e g and 13 E M then a e g and P E M for every E in g.
For the following, let a, 3 e P with at 13.
Definition 3.1. If a, 3 e g, then a13 e M by definition and we write a 1 P3 and we
say a is perpendicular to or orthogonal to 13.
Definition 3.2. Suppose that a e g, 13 e M.
(i) If ap13 e G then we write a 13 and we say a is perpendicular to 13.
(ii) If ap13 g, then we write a L 13 and we say a is absolutely perpendicular to 13.
Definition 3.3. Let X = {a13 : a 13}. We call the elements of X points and we
denote these elements by A, B, C,....
Definition 3.4. If a,P3 e M and a13 e M, then we write a 13 and we say a is
perpendicular to 13.
28
Definition 3.5. The point A and the plane a are called incident when A I a. For each
a e P, set X = {A : AIa}.se that A,BJal,a2 where A # B and ac # a2.
Suppose that A,Bla\,a2; where A # B and a1 # a2. We define the line g
containing A and B as
g = gAB [a,aX2] = {C e X : CaiO,a2}.
We say that g is the intersection of a(x and a2 (Xoa and 3O2), g c a 1,a2
(g c Xa IXa2). The point C is incident with g, C e g, if CIai,a2.
If A # B are two points such that there exist a and P3 with A, BI a, 13 and a JL 3
then we say that A and B are joinable and we write A,B e gAB [a,3,aP3]. If g is a
line which can be put into the form g = [a, P3, a3] where a 13 P then we say that g is
nonisotropic.
If A and B are two distinct points such that there do not exist a, 13 with
A, B I a,f 3 and a 1 3 then we say that A and B are unjoinable. If g is a line which
cannot be put into the form g = [a, 3, ap] with a 1 3 then we say that g is isotropic
or null.
Incidence Axioms
Axiom 1. For each P, a there exists a unique P3 E P such that P 13 and a[3 = Q.
Axiom 2. IfA,B la,[3,c andCla,P then Cls.
Axiom 3. IfP, Q I a,3 P and a 1 P then P = Q.
Axiom 4. Ifa, P3,y e g are distinct and a 1 (31 y 1 a then aP3y e 4.
Axiom 5. If a,13 e M and a I3P then a3 e AM.
Axiom 6. For all A,B; A # B, there exists a,P3 such that A,B Ia,3 anda a 13.
Axiom 7. If a 13 then there exists A, B I a, P3 such that A # B.
Axiom 8. For allMA,B,C; ABC = D e 3X.
Axiom 9. If OI a, P3,y, 5 with P3,y,6 a then 3Py6 = c.
29
Axiom 10. IfA,B,C are pairwise unjoinable points and A,B a then CIa.
Axiom 11. For alla e g, there exist distinct P,7y,8 E g such that a I P 1 y ac
but 8, ca,P,y.
Axiom 12. IfA,B Ia; a e G, then there exists P e G such that A,B I and P a.
Axiom 13. For each Pl a, Oa e M, there are distinct points A, BIa such that
P # A,B and P is unjoinable with both A and B, but A and B are joinable, and if CI a
is unjoinable with P then C is unjoinable with A or with B or C = A or C = B.
Axiom 14. If A and B arejoinable andA,BI a, then there exists I a such that
A,B1 .
Axiom 15. If a,P,y are distinct with P,7y a andA, BIa,,y; A B, then ac = Py
and ifA, B I y; y 1 a then ac = Py or P = y.
Order Axioms
Axiom F. (Formally Real Axiom) [21] Let O,E I a, a e M with 0 and E unjoinable.
Let X,6,1 e P. IfO 1X,6,,1 and X,6, L I ac then there is ay e P such that
OIy, y 1 ac, and EkO60E'k'1 = EXYXY.
Axiom L. (LUB) If A c /C, A # 0, and A is bounded above, then there exists an A
in C such that A > X, for all X in A and ifB > X for all X in A then A < B.
Dilation Axioms
Axiom T. IfO e t,g, with t # g, where t is timelike or t and g are both isotropic,
then there exists a unique a e M such that g, t c Xa.
Axiom D. (Desargues) Let g,h,k be any three distinct lines, not necessarily coplanar,
which intersect in a point 0. Let P, Q e g; R,S e h; and T, U e k. If gpT gQu and
gRT II gsu then gp II1 gQs.
Axiom R. Let 0 e g,h; P,Q e g, and R,S e h. If gpR II gQs then go,poR II go,Qos.
Polarity Axioms
Axiom U. (U' subspace axiom) Let O,A,B,T and C be four points with 0, C y6;
A, 0 ca; O, B I P; with a i y and l5. Then there exist X, e P such thatX IE ;
0,AOB1X; and O,Cs.
Axiom Si. Ifg c 3C,, a E G, h c Xp, P e g, and there exist y, e P7 such that
iL8; y78 e go h, g c Xy, and h a 6 then there exists e g such that g,h c 3 XeC.
(If g and h are two orthogonal spacelike lines then there is a spacelike plane
containing them.)
Axiom S2. Let g and h be two distinct lines such that P e g n h but there does not
exist pe P such that g 3CX0 and h c Xp. Then either there are a,0' e P such that
P = ac', g a Xa, and h c Xp, or for allA e g there exists B e h such that P and
APB are unjoinable.
This concludes the list of axioms. Note that by Axiom 5, ac3 A M for a c 13,
a, P3 e 7P and if a, P3 e M are distinct with al3, then. a3 e M. By Axiom 6, every
two points is contained in a line.
Notation. Due to the brevity of the theorems and proofs in sections 3.1
through 3.4, we shall follow the usual convention in absolute geometry [1, 2, 19, 30]
of simply numbering the results in these sections.
3.1.1. Properties of M.
3.1.1.1. The set M # 0.
Proof: By assumption 9 # 0, so let a e g. By Axiom 11, there exists y e G such
that a 1 y. Hence, ay e M by definition and M # 0. U
3.1.1.2. The elements of M are involutions.
Proof: Let y e M. Then we may write y = aC where Ca, P3 E g and ca P3. We have
yy = apap3 = aa3pp3 = 1. U
3.1.1.3. For every P e M and for every e 5, p" e M.
31
Proof: Let [3 = aU12 with a 1,42 e Q and a1 oX2. By our assumptions on 9 and
because aX Ia2 we have for all 4 e 0, aaC 2 E and cafa so that a4ax e M. *
3.1.1.4. The sets g and M are disjoint, g r) M = 0.
Proof: Let G' = \M. Then G' consists of involution elements, g' r M = 0, and
P = u M = u M. Let y E g' and 4 e 0. Now ifY E M, then by 3.1.1.3 above
y = (y) ' E M, a contradiction. Thus, g' is an invariant system of generators and
without loss of generality, we may assume that g r M = 0. U
3.1.1.5. Ifa e G, P3 E M and a 1 P3, then a3 e M.
Proof: If ap3 = ye M then a = P3y e g where 3,y e .M and 131y, which contradicts
Axiom 5. U
3.1.2. Properties of P.
For the remainder of this dissertation, let the symbol <" denote the phrase
"if, and only if'.
3.1.2.1. Ifa,f3 e P and F e 0 then a 1 P3 if, and only if, ax 1 3.
Proof: First suppose that a, 3 e g. Then a 1 13 0* a+ 2 3 because a P3 implies
that a 13 and a 113 + a:j3 If a E g and 3 E M and a 1 P3 then a3 = y e ;
aky E g, P3 E M, and a([3 = y e implies that a L p3. Conversely, if
a' 1[3 then aP[ = 8 e g and a13 = 8' E g, so that a p 3. Finally, suppose that
a,P3 E M. Then a:,3P e M and a3 e M + + *=3 M. E
3.1.2.2. If a,3 E P andF e E5, then a 13_< a 113.
Proof: Let a e g and P3 e M and F E 0. If a 1 3 then a P so that ao I 13. Thus,
a 13 or a I3P. If a 1[3 then a[3 = y So we have a13 = Y' e 6 by
the invariance of g, which contradicts a 1 3. Hence, ax 3' .
Conversely, suppose that a 13. because a' e G and E3 M by the
invariance of g and M then a = y, 83 = 8 imply that y 1 6 and y' = a p3 = 64
by the paragraph above. U
3.1.2.3. For each e (5, P7 = 7P.
Proof: because 7P = G u M, the result follows from the invariance of g and M. U
3.1.3. Properties of 3.
3.1.3.1. There exists a point; that is, T # 0.
Proof: Let a e G. By Axiom 11, there exists y,3 e g such that a, 3,y are distinct
and mutually perpendicular. By Axiom 4, a3y = a8 ?, where 5 = P3y e M and
a15 as ae 3,y. Thus, xa 1 and P = a6 is a point. U
3.1.3.2. IfO1a,P3,y; a,3,,y e ; a 1 3P 1 7 1 a; then 0 = a37y.
Proof: By the proof of 3.1.3.1 above, P = ap3y is a point and 8 = 13P7y e M because
13 l y with 3,y e g. So we have a 18, 06 because 013,y. This yields A, 01a,6
withA = a6 so that A = 0 by Axiom 3. U
3.1.3.3. IfA e XI and 6 e P, then A # 5; that is, a point does not equal a plane.
Proof: Let A = ao3 with a e g, 3 e M, and a 1 P3. Suppose that A = ae3 = 6 e P. If
5 e M then we have a = P38 e with P315 and P3,5 e M, which contradicts Axiom
5. If 6 e g, then A = a3 e g and this contradicts the definition of a point. U
3.1.3.4. The elements of X are involutory.
Proof: Let A E X, so that we may write A = ae3 with a L 13. In particular, a13 = P3a
and AA = a3ap3 = aapp313 = l. U
3.1.4. General Consequences of the Axioms
3.1.4.1. If PI a, 3 anda 13 then P = a3 and ifP = ap3 then P Ia, 3 and 1t3.
Proof: If P I a, 3 and a 13 then a3 = A for some A e X and we have A, P a, 3 with
al 3. Thus, by Axiom 3, A = P.
IfP = ap3, then by 3.1.3.4 alp. because P o P by 3.1.3.3 then al 3. Also
Pa = P3 and P3 = a imply that the products Pa and P3 are involutory so that
Pla,P3. U
3.1.4.2. IfPIa then Pa e 7P; that is, Pa is a plane P3 and P\ 3.
33
Proof: By Axiom 1 there exists P3 such that P 13 and 1 a. Thus, P[ a, 13 with a p
so by 3.1.4.1 above, P = a13 and 13 = Pa. U
3.1.4.3. For each P e T and each e Q5, P e X.
Proof: By the definition of a point we may write P = a13 with a E M 13 E G, and
al13. By 3.1.2.2, a '13, and P [Ia, 3P so that P = a134 is a point. M
3.1.4.4. IfP a,3 and y L a,3 P then a = 13.( Given a point P and a plane there exists
a unique plane a such that P I a and a L y.)
Proof: This follows immediately from Axiom 1. U
3.1.4.5. There do not exist three planes pairwise absolutely perpendicular.
Proof: Suppose that a, P3, and y are pairwise absolutely perpendicular. Then for
P = a3 we have P I a, P3 with a,3 p y so that a = 3, which contradicts a 13. U
3.1.4.6. IfA,B,CIa then ABC = DIa.
Proof: By Axiom 8, ABC is a point D and Da = ABCa = aABC = aD. U
3.1.4.7. IfA,B, CIl, 3 then ABC Ia,13.
Remark. From 3.1.4.6 and 3.1.4.7 above the product of three coplanar (and as we
shall see, collinear) points yields a point which is coplanar collinearr) with the other
three.
3.1.4.8. Ifoa1, a2,a3 i a then alC a2a3 = a4 e P and a4 i a.
Proof: Let A = aal, B = aa2, and C = aa3. Then by Axiom 8, ABC is a point D
and D = aa Iaa2aa3 = aala2a3. So DIa, by 3.1.4.6. By 3.1.4.2,
aI a2a3 = Da = 4 e 'P, D = aa4, and aI0a2a03 = 04 ia. a
3.1.5. Perpendicular Plane Theorems
3.1.5.1. If a 13 ; 13y = A andAl a then a 1 y. (A4 plane perpendicular to one of two
absolutely perpendicular planes, and passing through their point of intersection, is
perpendicular to both.)
Proof: From our assumptions above it follows that ay = a3A = f3Aa = ya, so a 1 y
34
or a I y ( note that a = y implies that ac L P3 because A = Py and A I a). Suppose that
a 17 so that B = ay. Then we have A, B I a, y with a 17, which implies that A = B by
Axiom 3. But if A = B then P3y = ay and P3 = a, which contradicts P3 1 a. Thus, ay is
a plane and a L y.
3.1.5.2. Suppose that a 1 3; AIa,p3,y,8; y7a; and 5813. Then 8 L y. (If two planes
are perpendicular, their absolutely perpendicular planes at any point of their
intersection are perpendicular.)
Proof: By 3.1.4.1, A = 7ya = 8 and 6y = ap3 E P as a I P. Thus, 6 71y.
3.1.5.3. IfO = aa =yyi = ssI with a,7,s e M and a 7 s  a then ay = s.
Proof: Because a,y,s e M, then a1,y1,E1 E g. Because a L I 1 I a and
aal = yyj = ese, then ya = y1a1; sa = sia,; sy = eily imply that
71 L ai L E I I y 7. Because O0x,ai,yic, then 0 = ay17Sc by 3.1.3.2. Because points
are involutions by 3.1.3.4, 1b = 00 = Oa171E = Oaiyi00si = ays and a7 = e. N
3.1.6. Parallel Planes.
We say that two planes a and P3 are parallel, denoted by a  3, if a = 3, or
there exists a 7 such that a, [317.
3.1.6.1. Parallelism is an equivalence relation on the set of planes P.
Proof: That the relation is reflexive and symmetric is clear. For transitivity suppose
that a P13 and P3 1 y where a, 3, and y are distinct. Then there exists 8, e PP such
that a, p3S and P3, 7yls. Let A = a5, B = 18, and C = P3. Then by Axiom 8 we have
D = ABC = aop13 = a& and aL1 so that a 11 7. 0
3.1.6.2 Ifa, ,3L6 and 3is then a1e.
3.1.6.3. If aiLy, p36, and7 6 8 then a 1 P3. (Two planes absolutely perpendicular to
two parallel planes are parallel.)
Proof: If y = 6 then we have a, 3P8 and the result follows. So assume that there is
an s such that 7,5iE. Then by 3.1.6.2 above, al6, P31y, oa,p3L, and a 1 P3. U
35
3.1.6.4. If a I 3, y 1I 8, and 13P then cdy. (Two planes parallel to two absolutely
perpendicular planes are absolutely perpendicular.)
Proof: because cx 1 P3 then a, P3LE for some c and because 835 then cI6. Similarly,
Y, 8&' for some c' and 5Lp3 imply that yip3. Hence, a, 3PL68 and P3y yield acly.
3.1.6.5. Ifa e M(9) anda II P then P3 = 9(M).
Proof: Suppose that a e 9 and a 11 P, so a, Piy for some y. Because a e 9 then
Y E M, which implies that 3 E 9. U
3.1.6.6. If a 13 P then c\\ 1 P for every 4 e 5.
Proof: Let a, Piy for some y e P. By 3.1.2.2, aP,13'iy so that ac  3. From
3.1.6.5 and 3.1.6.6 above and the invariance of 9 and M, we have that if ca 1 P3 and
ca E G(M) then P3 e 9(M) and ct 11 P3 with a4, 3P e 9(M). N
3.1.6.7. If a 113 then X, rP X p = 0 ora = P3.
Proof: Suppose that c, P3Iy and PIcc,P3. Then a = 3 by 3.1.4.4. U
3.1.6.8. IfA # B then A I B; that is, AB # BA.
Proof: By Axiom 6 there exists Ca E P such that A, BI c. By 3.1.4.1 and 3.1.4.2 we
may write A = ctci and B = aCV2. Suppose that AB = BA. Then we have
al(X2 = a2(Ci so that alia2 or ai c(2. But il  c2 because ai,a21Ca so that
either a1 = aC2 or a, r 3a2 = 0. If al I ci2 then by Axiom 7, Xa n # 0 so
UI = a2 and A = B. If al1 I2 then C = CC I2 e Xar X12 and again we have that
C I = a2. Hence, A # B. U
3.1.6.9.a. If AB = A, then B = A.
b. If A # B, then A,B, and A8 are pairwise distinct. U
3.1.6.10. IfAla; BI3; Cly; anda 1i P 13 y thenABCap3y.
Proof: Let Act = a', B3 = 3', Cy = y'. Then a', P3',y' iot, 3,y and
ABC = xac'PP313'yy' = c'P3'y' = 6'8 with a'13'y' = 6'i6 = a3P7y.
Conclusion: A I c; B1 I3; ca 1 P3, imply that B' P3.
36
3.1.6.11. IfAA' = BB'; A,A' a; BI3; a 13 thenB'P3.
Proof: Because AA' = BB' then B' = BAA'; 3  a a. From 3.1.6.10 above,
B' = BAA'Paa = P13. 0
3.1.6.12. For each P and each P3 there is a unique a such that P la anda aI 3.
Proof: By Axiom 1 there exists a unique y such that yip3. because Ply then Py = a,
aIy and a 11 P3. Now if P and 8 1 P then P8,a, so that a = 8 by 3.1.6.7. U
3.1.6.12. Let a,P3 e P and M e X. Then aip + aMi3 and thus, a 11 aM.
Proof: Suppose that aLp. By Axiom 1 there exists unique y, 1M such that aLy and
813. By 3.1.6.2 we have P3,yla and P318 which implies that y48. By 3.1.4.1,
M = y6. thus aM = aY6 = a61388 = P3. Therefore, aMl3 and aM 11 a.
Conversely, suppose that aM"'3. As above, there exists unique y,81M such
that aMly and 8l13. It follows that P3168, M = y6, and a6 = ac = aMip and
a = (a6)C103 = P3. U
If AM = B, then we say that M is the midpoint of A and B. Clearly, M is also
the midpoint of B and A.
3.1.6.14. IfPA = pB then A = B. ( Uniqueness of midpoints.)
Proof: From pA = pB we have PAB = ABP and also PAB = PBA because ABP is a
point and hence, an involution. Thus, AB = BA which implies that A = B by 3.1.6.8.
U
3.1.6.15. IfA,BIa,13 and AA = B then MIa,13.
Proof: By Axiom 6 there exists y,8 such that A,Mly,6. because B = AM = MAM then
by 3.1.4.7 we have Bly,6. Thus, A,BIa,y,6 ; May,6; and A,B13,y,86 so by Axiom2,
M113,a. U
3.1.6.16. .IfA,B,C16; a,,y1L86; Ala; BI13; CIy; a ii 3 II y then ABCIapy and
a0y 1 8.
Proof: From 3.1.6.10 we know that D = ABCI a3y = c. Because 5la,P3,y then E18 so
that c 1 8 or s18. If E = 5s then we have D,EI 8, with 68E which implies that
37
D =E by Axiom 3. Let A = aa', B = P13P, and C = yy'. Because a 1113 P then
a' P 3' 1I y'. Thus, s6 = D = ABC = apya'P3'y' = a'P3'y' which implies that
8 = a'P3'y'. Because A 18 then we may write A = aa' = 886' and by 3.1.5.1 we have
A a',a,; aLta'; and a 1 5 which implies that a' 1 8. Hence, a'8 = a'a'P3y' = p3'y'
is a plane, so P3' y'. By Axiom 7 and 3.1.6.7 we have 3P' = y' so that a' = 6. But
this yields at18 and a 1 8, which is a contradiction. Thus, aP3y 1 6. U
3.1.6.17. IfA,Bla; Ala'; a' Ia; BI3P; and P B a' then P 1 a.
Proof: By 3.1.6.16 above B = AABIa'a't3 = P3 and P3 1 a. U
3.1.6.18. IfE\ a,e; a 1 P3, and 6113 then c I a.
Proof: If El 13 then the result follows from 3.1.6.7. Suppose that E % 3 and let
M = E3. Then, E # EP and E3P I aP = a so that E, E I a. Now EM = Ec = EP, so that
Mis the midpoint of E and EO. EM = EPEla' so that E, EMI a, a and by 3.1.6.15,
Ml a,a'. In particular, MI a and we have MI a, P3, ; a P3; and P31c so that a 1 e
by 3.1.6.7. U
3.1.7. Consequences of Axiom 11 and 3.1.6.18.
3.1.7.1. IfAla; a P3; and a, 3 e g then there exists ay7 in Q such that AIy and
Y 1 P3,a.
Proof: Let A 68 with 56iP. Then 6 e M and A I a, 8 with al 3, PL8 so that a 1 6
by 3.1.6.18. Because 5 e M, 6 1 a, and a e Q then c = 8a E g, A\E, and E a.
We claim that 1 P3. Indeed, P3 = 36a = 6p3a = 6ap3 = 63 with 6, P3 e 9 so that
c 13 or s = P3. But # 13 because then 6a = P3 and a = 813 = P, as 613. This
contradicts 3.1.3.3. M
3.1.7.2. Every point may be written as a product of three mutually perpendicular
planes from g.
Proof: Let A be any point. Then by definition A = a3 for some a E 9 and some
P3 e M with alp3. By Axiom 11 there exists 7 E 9 such that y 1 a. By 3.1.7.1 there
38
exists a S E g such that AI and 8 1 a,y. Again by 3.1.7.1 there exists an C E
such that A I and E 1 8, ,a. Hence, A I a, 6; 8,a e 9, and s L 6 L a I s which
yields by 3.1.3.2, A = 8as. U
We note that in this case, a(3 = A = (xaS8 implies that 63 = 8; that is, if A P
and P e Mthen there exists P1, 32 e g such that APi,P132 and P3 = PI3P2.
3.1.7.3. IfA I a then there is a P in 9 such thatAl P and3 P a.
Proof: This follows directly from the proof of 3.1.7.2, for if a e 9 then we can find
8,e 6 G such that A 18,e and 6,s L a. If a e M then we can find a l,a2 e g such
that A I ai,(X2 and a 1,a 2 I a. That is ifAIa then there exists P[1,32 e G such that
A I P l, 32 and 3 1 P32. Moreover, if a E M then we can find P 1, P2 such that
a = Pip31132 and if a e G then we can find P31,32 e G such that apIp31132. N
3.1.7.4. IfA a e M and 3 I a then there is ay such that A y andy 3,a.
Proof: Let 8A with 6113P. Then 8 1 a by 3.1.6.18 and As = aS. Because a 1 P3 and
5i3 then E p. If dLp3 then we would have A 186, with 6,e1&3 so 5 = s and a = 1I.
Thus, s&3. We note that if P3 E 9 then 8, e M andif3 P e M then e 9G. N
3.2. Lines and Planes
3.2.1. General Theorems and Definitions.
3.2.1.1. For any A,B e X and any a,b e G, ifA,B e a,b then A = B or a = b. Hence,
by Axiom 6, for all A,B e 3C, there exists a unique g e 9, such that A,B e g.
Proof: Let a = [a, 3], b = [y, 8], A,B Ia, P,y, 5, and suppose that A B. Let C e a, so
that C Ia, 3. By Axiom 2 it follows that C y, 5 so C E b and a c b. Similarly, b c a
and a = b. U
3.2.1.2. Every line contains at least three points.
Proof: By definition, every line g = gAB contains at least two points A and B. Let
gAB = [a,3] with A,Ba, 3. Then by Axiom 8 and 3.1.4.7, ABa,P and AB e 3.
AB A by 3.1.6.9. U
3.2.1.3. Suppose that a13 = y.
a. Ifaa, = P1313Pi then a I\ 13 i andac131 = 7, so7 L a,131.
b. If atal = PP313I = 7, then a 13, ai l 31, y1i a1,P31, and ap3 I = a113 = .
c. IfA Iaa,3 then A ly.
d. The line g = [a,13] = [13,y] = [a,y ] [a,13,y] is a nonisotropic line. Thus, a, P, and
y are three mutually perpendicular planes which intersect in a line.
Proof: a. From cat = PP1313i we have y = 13a = 13Pial.
b. If Pp1313 = yy and aal = Yyi then y = y71313i = ap131 and yi = yaal = 13al.
c. Because A I a, P3 and a13 = y then Ay = Aat3 = ap13A = yA and A 1y.
d. By Axiom 7 there exist points A and B such that A,BI a,13 and by 3.2.1.3.c. above,
A,B]y. Thus, A,B E [a,13],[a,y 7],[13,y] and [a,13] = [a,y ] = [13p,y] [a,13,y]7 is a
nonisotropic line. U
3.2.1.4. If A and B are collinear then A and B are collinear for any E 0.
Proof: This follows from the fact that A, B I a,13 A,B' I a,13",.
For each 4 e 0, we define a {C : C E a}. By 3.2.1.4 above, a4 is a line
for every 4 in 5 and if a = [a, 13] then a4 = [a,P13].
Definition of parallel lines and planes. We say the line a is parallel to the
plane a, denoted by a 11 a, if there exists a 13 such that a c Xp and 13 I a; that is,
a = [13,y] and a,P 1 6 for some 6.
We say that two lines a and b are parallel, a b, if there exists a,3,y,8 such
that a = [a,y], b = [13,], where a I 13 and y II 6.
3.2.1.5. IfAA' = BB' # 1I; A,A'I a,a'; a # a'; B,B'1,13'; 13 13', thengAA,'  gB"6
Proof: Let BI13P* with 13"* a (3.1.6.12). Then B' P13* by 3.1.6.11. Let B130, with
P 131 al. Then B'130 by 3.1.6.11 and B,B' P13,131,13*,13P which implies that
[13PP13] = [13P*,13P] = g88' by 3.2.1.1. Hence, we have gAA' = [a,a ], gBB' = [13*,13] with
a II 13p* and at 1 13i; thus, gA' I gBB'. *
40
3.2.1.6. Two lines, a = [a, aI] and b = [P3,I]I are parallel precisely when there exist
A,A' E a and B,B' e b such that AA' = BB' 1 1o.
Proof: Suppose that a I1 b. Then, a  P3 and a1 11 P1. Let A,A' e a and B e b, so
that B' = AAB'Ito3ap, and 0axliPI = I,pi, by 3.1.6.10. Thus, B,B' e b with
AA' =BB'. IfAA' =BB' = lo thenA =A'. U
3.2.1.7. If a I b andb I1 c then a I] c.
Proof: Let a = [a,a']. From the proof of 3.2.1.5 above we may write b = [p3,3'] with
a P 3 and a' i because a 1 b, and c = [7,7'], where y 1 P3 and y' II 3' because
b il c. By 3.1.6.1, ot 1 y and a' ii y' so that a I1 c. U
3.2.1.8. For each line a and point A there is a unique line b such that A e b and
b i1 a.
Proof: Let B, C e a be distinct and if A E a we can choose B, C # A because every
line contains at least three points by 3.2.1.2. Then ABC = D by Axiom 10 and
BC = AD l 1I, so gAD II a by 3.2.1.6. Now suppose that A E c and c I a. By 3.2.1.7
above, c II gAD, so there exist W,X e c and Y,Z E gAD such that WX = YZ # 1 by
3.2.1.6. By 3.1.4.7, A' = AWX e c and AI = AYZ e gAD. Thus,
1 AA' = WX= YZ=AAi1 implies that A' =A1. Hence, A,A' E gAD,c; A A',
because WX # 1 and gAD = c by 3.2.1.1. 0
3.2.1.9. IfA,B e g, A # B, and C e h then g 11 h if and only ifABC E h.
Proof: If ABC = H e h, then 1o # AB = HC and g 11 h. Conversely, suppose that
g I1 h and put D = ABC. Then because A # B,
lo # AB = DC and C e gjjc, with gDC 11 g.
Because g I1 h and C e h, then by 3.2.1.8, gDc = h and D e h. U
3.2.1.10. I/a 11 b then a = b or a n b = 0.
Proof: Supposea 11 band anb 0. LetA E a,b andC e with C A, B E b,
with A # B. Then, a = gAc, b = gAB, and therefore, D = ACB e a, b.
41
If D = A, then ACB = A, C = B, and a = b. If D # A, then it follows that
a=gAD =gAc =gBD =gAB = b. U
Classification of nonisotropic lines. Following the terminology of physics we
make the following definitions. Let a be a nonisotropic line. If there are elements
a,3,y e M such that a = P3y and a = [ct,P3,y], then we say that a is timelike. If there
are elements ot, 3 E g such a (3 and a = [a, P3, ab], then we say that a is spacelike.
Remark. Let A,BoI c e with A # B. Then by Axiom 14, there is a P3 such
that A,B I 3 and P3 I a. Thus, a = [a, 3] is a spacelike line by definition, A,B e a,
and every pair of points in a spacelike plane is joinable with a spacelike line.
3.2.2. Isotropic Lines.
3.2.2.1. IfA # B; A,B Ia,3 with P3 a, a e M; andA andB are unjoinable with P;
P Ia,y; andy La,P3, then AY = B.
Proof: First, AY Iaty, 3Y = a, P3. If A7 = A then A I7 and A is unjoinable with P.
Similarly, By ac, P3 and By # B. Suppose that A" and P are joinable. Then there are 6
and e such that P,AYI\,c and 6 1 s. But then P,A = P',(A")"'6",s" and 8" 1 &Y,
which implies that P and A are joinable. Thus, A" and P are unjoinable, so by Axiom
13, A" is unjoinable with A or A" is unjoinable with B or Ay = A or A" = B. Hence,
AY=B. U
3.2.2.2. Suppose that A # B; A,Ba,P3 with a 1 P, a e M; A and B are unjoinable
with Pla; Py, y 1 p3. then AY =B.
Proof: First observe that because PIa,y; a 1 P3; and yip3, then a I y by 3.1.6.18.
Then A It ay, PY = a, P3 and A" is joinable with B. AY # A because A and P are
unjoinable as in 3.2.2.1 above. A" joinable with P implies that A = (A")" is joinable
with Py = P. Hence, by Axiom 13, A" = B. U
3.2.2.3. If A and B are unjoinable then X E gAB precisely when X= A or X = B orX
is unjoinable with both A and B.
42
Proof: By Axiom 6, gAB = [a, P3] for some a and P3 in P. Let X E gAB and suppose
that X # A, B. If X is joinable with A then there exists y, 6 such that y 1 8 and
A,XI y, 6. By Axiom 2, B I y, 6 and A and B are joinable. Hence, X is unjoinable with
both A and B. U
Remark. If X is unjoinable with A and B; A,B a, x3; then by Axiom 10, X1 a, 3
and X e gAB.
3.2.2.4. If gAB is isotropic; C,D e gAB; and C # D, then C and D are unjoinable.
Proof: If C, DIa,3 with a I P3 then A,BIa,3 by Axiom 2; that is, gAB = gCD.
3.2.2.5. IfA,B, CI a are pairwise joinable and distinct, then each P a is joinable with
at least one ofA,B, and C.
Proof: If a e g then the result follows from Axiom 12, so assume that a e M. By
Axiom 13 there exist D, EI a such that A,B, and C are all unjoinable with P. Then
by Axiom 13, at least two of A,B, and C must lie on one of gpo and gPE. By 3.2.2.4
above, this implies that two of A,B, and C are unjoinable, which contradicts our
assumption. U
3.2.2.6. If gAB is isotropic then gAB is isotropic for all 7 e 65.
Proof: IfA,B4lIa,1 with a 1 P then A,BIa,] 1' and a' 1 3'.
3.2.2.7. IfP Ia E M then there are at most two isotropic lines in Xa through P.
Proof: If gAP,gBP,gcp c Xa are three distinct isotropic lines through P, then P is
unjoinable with A,B, and C. The points A,B, C must be pairwise distinct and are
joinable by Axiom 13, in contradiction to 3.2.2.5. U
3.2.2.8. By Axiom 13 and3.2.2.7,for each P Ia, a e MA4, there are precisely two
isotropic lines in Xa through P.
3.2.2.9. Let gAB,gAC c Xa, a e M, be two isotropic lines.
a. IfA Ap, with p3iLa, then gA = gAc.
b. If Ay, with yLa, then gAB =gAc.
43
Proof: Because gAB # gAc by assumption, then B is joinable with C by Axiom 13.
The results follow from 3.2.2.1 and 3.2.2.2. M
3.2.2.10. IfA,B,CI a E M; gAB # gAC are isotropic then gBC is nonisotropic and
there is a p with p3A such that I aandC = BP.
Proof: Because gAB # gAC, then gBC is nonisotropic. By Axiom 14 there exists a y,
7yB,C, such that y I a. Let AI3, 131 7 (Axiom 1). Because Aa,[3; a I y; and
P3 1 y, then by 3.1.6.18, a L P3. If BP=B then BI P3 and gAB is nonisotropic. Similarly,
CO C and by 3.2.2.1, BO = C. U
3.2.2.11. If A,B, C I a, a E M, are pairwise unjoinable then there exist P3,y a with
A113, y such thatC = BOY.
Proof: First observe that by 3.2.2.3 and 3.2.2.4,
gAB = gBC = gAC C
is isotropic. By 3.1.7.3 and 3.1.7.4, there exist PI\A such that 3 c a. Because B and
C are unjoinable with A, then B,C P3 and BO # B. By Axiom 1 and 3.1.6.18, there
is a 5113 such that 6 1 3 and 6 1 a. Thus, B,BP16S,6P = 6,a and B and BP are
joinable. Suppose that BO3 and C are unjoinable. By 3.2.2.3, either gBOC = gAc or A is
joinable with BO. But ifA, BO Ie, & I a, then A = AP,BBe EP a1 = ax imply that A
and B are joinable. If gsBC = gAc = gAB then BO3 is unjoinable with B. Hence, BO3 and
C are joinable and by 3.2.2.10, there is a y with yJA such that y I a and C = BO M
3.2.2.12. Let PFa,13,y with P3,y a, a e M. IfgAB c Xa is isotropic then gY c
is isotropic and gAB II AB
Proof: By 3.2.2.8, there are precisely two isotropic lines in X.a through P, say gpc
and gpQ. By 3.2.2.9, g)y = gpQ and g = gpc. Now PAB = D is a point by Axiom
8, so AB = PD and D is unjoinable with P as A is unjoinable with B. (Suppose that
P, DIa' with a' I a. Let B\I with k 11 a'. Then because B\ a, X. 1 a by 3.1.6.17.
But then by 3.1.6.16, A = PDBIa'a'k = X and A is joinable with B.) Because
44
A,B,PIa then D = PAB Ia. Hence, D e gpc or D E gpQ, and gAB II gpc or
gAB II gPQ. thus gA II gQ = gPQ II gAB or g II g = gP c II gAB.
3.3. A Reduction to Two Dimensions
In this section we restrict our attention to two dimensions. In this way we are
able to use the work of Wolff [30] to construct our field.
Let t be any element in 0; then the map a4 : X X given by a(A)= A1
is bijective and maps lines onto lines and planes onto planes by 3.2.1.4,5,6 and
3.1.2.3. Hence, it is a collineation of (0,g,9). In the following, for any element
e 0, the collineation induced by it is denoted by ro.
Let a e (P be fixed throughout this section. We wish to define a set Ca of
maps on Xa which can be viewed as line reflections in a given plane Xa. We then
show that each element of Ca is involutory and that COa forms an invariant system of
generators within the group ga it generates. Finally we will show that (Ca,Ga)
satisfy Wolff's axioms [30] for a two dimensional Minkowski space; that is; the
Lorentz plane, for a e M.
Let Ca = {g c 3CXa : g is nonisotropic}. By Axiom 14, each g E Ca may be
written uniquely in the form g = [a, 3, y] where a = Py. Let A e 3a and
g = [a,p3,y] e Ca. because A la and a = P3y, then A = A' = AP and AP = Ay. Hence,
we define the map ag : X3a 3 Xa by ag(A) = Ag = A = AY, for A E Xa, and
cyg : Ca  Ca by ag(h) = h9 = {ag(A) : A e h}. Note that by 3.2.1.4, if
h = [a,6,e], then ag(h) = [a,6P,S1] = [a,8Y,Y].
3.3.1. For each = [a,13,y] e Ca, ag : Xa X oa is bijective.
Proof: If A, B I a and ag(A) = ag(B) then A7 = By and A = B, so Cg is injective. If A a
then A I ay = a because a 1 y and Gg(AY) = (AI)y = A, so ag is onto. U
3.3.2. Forg = [a,p3,y] E Ca, Ocg : Ca * Ca is bijective.
Proof: Let h = [a,E,rI] e Ca. Because a y and s 1 <> y 1 ly by 3.1.2.1 then
45
a = aY = EY1Y. It follows that ag(h) = [ac, cy,ly] e Ca. IfAIe,fr then AYIc7,Tr7 so
ag(A) E ag(h) for all A e h and Gg is a welldefined collineation.
If I = [a, .,p] e La and ag(h) = cg(o) then for every A in h we have
Tg(A) = AY ag(l) = [a,,].
This implies that A = (AY)Y e [aE ,] = /;that is, h = I. Hence, Ug : Ca f a is
injective. Because e r +' EY ily then it follows that
h = [a,,] e + c g(h) = [a,yqY] E 4a.
Hence, ag : a * a is surjective. U
3.3.3. Each ag, for g E 4a, is involutory.
Proof: Let g = [a, P3,y]. Then for A a we have cgrg(A) = ag(AY) = (AY)Y = A. E
Let g = [a, 8,y], h = [a,, rj] E Ca. We say that g is perpendicular to or
orthogonal to h, denoted g 1 h, if one of y and 5 is absolutely perpendicular to one
of E and q and g r h 0.
3.3.4. Forg and h as above, ifyls then 85_q, 8 1 e, andy 1 e. Moreover,
M = ye = 5e E g,h.
Proof: Let M = ye. Because y, E 1 a then MA = (ye)U = y's0 = ye = M and MI a. So
we have MIota,e which implies that MIae = rq and M e h. Also, MIa,y so Mf ay = 8
and M E g. Also from M = ye it follows that Me = y = 8a = TIE so that M = 6r1 and
61lt. Applying 3.1.6.18 to MI y, r with y 1 5, 7115, and M,e with 6 1 y, ely7, we
obtain y 1L 1 and 8 1 e. U
3.3.5. Ifa,b,g e 4a then a 1 b > ag bg.
Proof: By 3.1.2.1 and 3.1.2.2 we have
ac.Xy; yl6; bc X6 a 3cX; y+I; b c. for all 4 E .
The result follows. N
46
Remark. Because AIs <> A:js' for all E e Q5, A E X, and s e P, it follows
that for each g e Ca, ag maps:
(i) The set X(a onto itself.
(ii) The set Ca onto itself.
(iii) Collinear points in X3a onto collinear points in Xa.
(iv) orthogonal lines in Xa onto orthogonal lines in X3Ca.
That is, ag is an orthogonal collineation of Xa.
Let C( = {Tg : g e Ca} and T, = {"p : P E Xa}.
3.3.6. The sets Ca and 3Ca are nonempty. Hence, Cao 0 and T3c 0.
Proof: If a e MA then we may write a = a'6 where a', e g and a' 1 8 by
definition of MA. By Axiom 7, there exist A,B such that A # B and A,Bl a',6. Thus,
A,BIa'8 = a and Xa # 0. Moreover, gAB = [a, a',8] e Ca and Ca 0. If a e g
then by 3.1.3.1, there exist AIa and by 3.1.7.3, there is a P3 in g such that Aj1 and
13 a. Thus, A e Xa and g = [a, P3, ap3] e Ca. U
Note that from 3.3.3, Ca consists of involutory elements and by 3.1.3.4, p3a
consists of involutory elements.
3.3.7. IfP e g,h e Ca andg I h then ap = Cgah = ahCg.
Proof: Let g = [a,y,6] and h = [a,e,ft], and without loss of generality, assume that
y s, so that 81ri. By 3.3.4, P = y = 6rI e g,h and by 3.2.1.1, {P} =gr h. Then
for A e Xa,
agFh(A) = ag(As) = A = ap(A) A" = h (A') = ahcg(A). E
3.3.8. For everyA in Xa and for every h e Ca, there is a unique g e 4a such that
A e g andg 1 h.
Proof: Let h = [a, r] e Ca and A e Xa. By Axiom 1, there exist a ylA such that
y's. Thus, Ala,y with a 1 s and yLs so a 1 y by 3.1.6.18. because a I y then
ay = 8 and Al a,y implies A 16 so that A E g = [a,7y,8] E Ca. because g c Xy, yl',
47
A E g = [a,y,6] e La. Because g c Xy, y, and h c XT, then g h. By 3.3.4
above, r1I5 and {B} = {ye} = {T1} = g h.
Now suppose that I c Ca such that A e 1 and I L h. Now we may uniquely
write I = [cx,a', P] in Ta where a = xa'p. because / L h then without loss of
generality, we may assume that a' Ie and PfLT. because yi_ and 81TI it follows that
a' 11 y and 0 11 6. By the definition of parallel lines in Section 3.2 we have 1 1 g.
because A e I nrg then by 3.2.1.10, / = g. U
3.3.9. (i) The fixed points of (g E Ca are the points in X3a incident with g.
(ii) The fixed lines of Gg E Ca( consist of g and all lines in 4,, which are
orthogonal to g.
Proof: Let g = [a,y,65]. IfA = ag(A) = Ay for Ala, then AI17. because AT = A6, VAja
then A6 = A and A16. Thus, A e g and 3.3.9.(i) follows.
Let h = [a,E,'q] e a and suppose that cg(h) = h; g # h. Then there is an
A e h such that A f g. By 3.3.8, there is a k = [a,co,9] e Ca such that A e k and
k I g. Because A E h, cg(A) = Ay e h and because A E k, k g then
A I coy,Oy = o), and AY e k. BecauseA 0 g then A I y (for if A 1y then A I a implies
that A jay = 8 which implies that A e g) and A # A7. Thus we have A,AY E h,k, so
that h = k by 3.2.1.1. Hence, h I g. U
3.3.10. (i) The only fixed point of a point reflection, ap, is the point P.
(ii) The fixed lines of ap are the lines incident with P.
Proof: If pA = P, then this implies that APA = A, or AP = lo, or A = P. Thus,
3.3.10.(i) holds. Let x = [P,u] be any line (not necessarily in Xa or nonisotropic) and
A E x. Then PA e x implies that pA p,t), so that P I pA,A = P3,u, and P E x. U
Let 9( be the group acting on Xa generated by Ca. By 3.3.5, every element of
Ga is an orthogonal collineation of Xa. Let c E a. By a transformation with T, we
mean the adjoint action of a on G : (1 e a a' f a e ?a.Not,: that every
such transformation is an inner automorphism of the group.
48
3.3.11. Let h = [a,, ri], g = [a, 'y,] e La, then a" = aTh(g).
Proof: Let A I ac and put B = rg(A) = A Then
O(ah(A)) = ahagCTh'h(A) = Ghg(0A) = oh(B).
Thus, CG9h(Ch(A)) = aah(g)(ah(A)) for all A E Xa. Because ah is injective, then
T = aoa(g) for every A in 3Xa. Similarly, ah(a) = aoh(g)(a), for every a in La.
Hence, ag' = aah(g). Analogously, we can obtain aTh = Cah(P), for every P in Xa. U
Because Ca generates 9a and every element of go, is an orthogonal collineation
of (Xa,La) then we obtain the following.
3.3.12. For every a e go and for every ag E Ca, we have ag, e Ca; that is, Ca is an
invariant system of Ga.
Consider the two mappings: g e Ca  ag E Ca and P E Xa 3  ap E T3a.
The first one is from the set of nonisotropic lines in Ca onto the set of line
reflections, and the second is from the set of points in Xa onto the set of point
reflections in Xa. These mappings are injective, because reflections in two distinct
points have distinct sets of fixed points by 3.3.10, and similarly for lines and line
reflections. Thus, it follows from 3.3.11, 3.3.12, and the preceding remark that the
next result obtains.
3.3.13. If a e a and P,Q e Xa then a(P) = Q <> a' = OQ.
Proof: Because
a(P) = Q +> ag(p) = aQ > a(p) = aa,
the result follows. U
3.3.14. If a e ga and g,h E La then a(g)= h +> *g = Ch. 
3.3.15. IfP e Xa andg E Ca then P e g <> apCFpag is involutory.
Proof: Suppose that P e g = [a,y,8] e La. Then for A a,
Gp(og(A) = op(AY) = AYP = A = agop(A). Thus, (apag)2 = Ix. Nw assume that
49
apcg is involutory. Then P = agapaggp(P) = PPP = YPY. Because yPy = PY e Xa
by 3.1.4.3, then by 3.1.6.9, P = Py. Because PY = P8 then PIy,, and hence, P e g.
U
3.3.16. Ifg,h e a, then g I h <> 9gch is in volutory.
Proof: Now g I h if and only if ag(h) = h. For g # h, by 3.3.9, ag(h) = h if and only
if CFg = h. For crg 5jh by 3.3.14, TC = Oh if and only if (agah)2 = lx,, and
Ogah l Xa. U
3.3.17. The point reflections jax are the involutory products of two line reflections
from Ca.
Proof: Let ap e Ta where P e XC. Then we may write P = y6s where a = y5 and
Y,5, ,e e by 3.1.7.3. Let g = [a,'y,5]. Because a = 78, then g e a. By 3.3.8, there
is an l in C, such that P e I and I l g. By 3.3.7, op = ar/g = agaC. U
We now show that if a e M, then the pair (Ca,Ga), acting as maps on
(XCa,a,a), satisfies Wolff's axiom system [30] for his construction of
twodimensional Minkowski space. We give Wolff's axiom system below.
Basic assumption: A given group Q5, and its generating set 9 of involution elements,
form invariant system (9, ()
The elements of G will be denoted by lowercase Latin letters. Those involutory
elements of 5 that can be represented as the product of two elements in G, ab with
a b, will be denoted by uppercase Latin letters.
Axiom 1. For each P and for each g, there is an I with P, g 1l.
Axiom 2. IfP,QIg,l then P = Q org = 1.
Axiom 3. IfP a,b,c then abc e G.
Axiom 4. Ifg a,b,c then abc e G.
Axiom 5. There exist Q,g,h such that gj h but Q f g,h,gh.
Axiom 6. There exist A,B; A # B, such that A,B f g for any g e G. (There exist
unjoinable points.)
50
Axiom 7. For each P andA,B,C such that A,B,CIg, there is a v e Q such that P,AIv
orP,Blv orP,Clv.
Geometric meaning of the axioms. The elements denoted by small Latin
letters (elements of Q) are called lines and those elements denoted by large Latin
letters (thus the element ab with al b), points. We say the point A and the line b are
incident if A Ib; the lines a and b are perpendicular orthogonall), if aI b. Further we
say two points A and B are joinable when there is a line g such that A, BIg.
Replacing Ca with Ca and Xa with T3a, the pair (Ca, gQ) satisfies the basic
assumption by 3.3.3, 3.3.6, and 3.3.12. It follows from 3.3.15, 3.3.16, and 3.3.17, that
our definition of points, our incidence relation, and our definition of orthogonality
agree with those of Wolff. Hence, we may identify LCa with Ca, Ca with g, and Qa
with 0.
Verification of Wolff's axioms. Axiom 1 follows from 3.3.8 and Axiom 2 from
3.2.1.1. For Axiom 3, let a = [a,a', 4e], b = [a, 3P, P'], c = [a,yy'] e Q. Then
a',P3,y I a, and by our Axiom 9, a'Py =8 L ac and d= [a,6,a8] e La. For Ala,
aaabac(A) = Aya' = Aa'Y = A6 = ad(A).
Hence, CaaGbaCC = d ( Ca. If we then identify "" with "e", we get Wolff's Axiom 3.
M
Axiom 4. Ifa,b, c lg, then abc e g.
Proof: Let g = [a,X,X'], a = [ac,a',], b = [a,3,3P'], and c = [a,y,'y'], with
a',P3,y I X. Then by 3.1.4.8, a'py = s I X. Let A e a, B e b, and C e c, then
Ala',a; Bla,P; and Cla,y. By 3.1.6.10 ABC = Dla'py = s and ABC = Dla by
3.1.4.6. Thus, DIE,a; c 1 X; a I X, and c 1 a. So, d = [a,E,aE] e La. And ifX1a,
then aCabac(X) = AX1a' = X' = Od(X) and caabac = ad E Ca U
Axiom 5: There exist Q,g,h such that g h but Q I g,h,gh.
51
Proof: Because a e A, we may write a = 13Py, where 3P,y E g and P3 I y. By 3.2.1.2
every line contains at least three distinct points. Let g = [a, 3,y] E CL and let
B e g. By 3.3.8, there is an h in Ca such that B in h and h J g. By 3.2.1.2, there is
anA in h such that A # B. By 3.3.8, there is an e La such that A E l and / I h.
By 3.2.1.2, there exists Q e I such that Q # A.
Now if Q e h, then A, Q E ,h implies that Q = A or I = h, so Q e h. Suppose
that Q E g. Let Q E m, m 1 1, so that aQ = 010m. Because GB = 0g
and BAQ = E e Xa, then
GE = (OBaCAaQ = 3gCThahC1/CT/ClTm = cTgCrm and g L m.
Because Q e g,m and g m then cQ = agaim = cGam and cg = a1 or g = 1. This
implies A,B e 1,m. Thus A = B or I = m, a contradiction. U
Axiom 6: There exist A, B;A # B, such that A,B f g for any g e G.
Proof: This follows from our Axiom 13. U
Axiom 7: For each P and each A, B, C such that A, B, CIg, there is a v in 9 such that
P,Afv, orP,BIv, orP,Cjv.
Proof: By our Axiom 13, there exist two isotropic lines in Xa through P, say gpQ
and gpR. If no such v in Ca satisfies the above, then two of the three points must lie
on one of the isotropic lines by our Axiom 13. But this implies that gpQ = g or
gPR = g; that is, g is isotropic; a contradiction. U
3.4. Consequences of Section 3.3
3.4.1. For every A and B, there exist a e M such that A, B Ia. (Every line lies in
some Minkowskian plane.)
Proof: By Axiom 6, there exist Xa e P such that A,BI a. If a e MA4, the result
follows. So suppose that a E g. By Axiom 12, there exist P3 e such that A,B 3
and 31 a. Then A,BI a3 = y, and y e M by the definition of M. M
52
3.4.2. For every A and B, there existsM such that AM = B; that is, every two points
has a midpoint and by 3.1.6.14, the midpoint is unique.
Proof: By 3.4.1 above, there exists a e M such that A,B Ia. From Section 3.3, there
exists Mi a such that AM = B. U
3.4.3. If AA' = BB' then A and A' are joinable precisely when B and B' are.
Proof: Suppose that A and A' are joinable and let A,A' ca,a' with a I a'. By 3.4.2,
there exists an Msuch that AM = B. Then B = AMIaM, a'M and from 3.1.2.1 and
3.1.6.13 it follows that aM I a'M, a 1 acM, and a' 11 a'M. By 3.1.6.11,
AA' = BB'; A,A'ja,a'; BlaM,alM; a  aM; and a' 11 a'M.
Thus, B' 1aM, cM' and, B and B' are joinable. U
3.4.4. If a 1 a'; a 1 P3; a' p3'; and[a,a'] 11 [p3,3'], then P3 1 3'.
Proof: Because [a,a'] 11 [13,3'], then there exist A,A'; A,A'Ia,a and there exist
B,B'; B,B' P133', such that AA' = BB'. Then for AM = B, as in the proof of 3.4.3,
[P3,3'] = gSBB' = [aM,a'] II [a, a']. That is, BIaM, M,3,3'; with 3,aM 11 a, and
3',aCM 11 a'.Thus, 3 = a' and 3' = aM by 3.1.6.12. Hence, P3 I 3' because
aM I aM. N
3.5. Construction of the Field
The basic construction. For the construction of the field we will follow the
path of Lingenberg [201. Throughout this section let a e M and 01 a be fixed.
Define the sets:
Oa = g E La : 0 e g} and Da(O) {agCh : g,h e O}.
Proposition 3.5.1 The set Da(O), acting on the points ofX3, is an abelian group.
Proof: By 3.1.7.2 we may write 0 = a3 = yrip3 with a = yTI; y,"1,3 E g; and y, l, and
P3 mutually orthogonal. Thus, g = [a,y7,1] e a; 0 e g, and Oa 0. Because each
53
ag e Ca is involutory then lx e Da(O). Now let aa,ab,ac,cd E Dca(O) where
a = [a,a',aa'], b = [a,13',ap13'], c = [a,y',ay'], and d = [a,6,a8] are nonisotropic.
Now &y'P' = E _ a with 0e by Axiom 9 and f= [a,, cM] e Oa. Thus, for AIa,
aabGcad(At ) = AkY'c'' = A;' = aaC(AA).
Hence, aabacad = aaaf e VDa(O). Because each al is involutory for 1 E L,
ca1 = al and (aCaab) = abCaa e Pa(0). From Axiom 9 and the calculation above,
the product of any three of 5,y7', 3', and x' is an involution.Thus,
A8y'P'a' = AW'Y'a' = A'a'8' and CaabOcad = CcaCaCab.That is, Da(0) is abelian.
Clearly, TEa(0) is associative so that Da(O) is indeed an abelian group. U
Lemma 3.5.2. Let g be an isotropic line in Xa with 0 e g. Then for every
CaF/h E D6 (0), alCah(g) = g.
Proof: Let I = [a,3,ap] and h = [a,y,ay] be lines in La with 0 e 1,h. By 3.2.3.12, if
01 M, with P,7y 1 a then gfY I1 g. But OPY = 0 so that ahalT(O) = O0Y = 0. Thus,
g Y = g by 3.2.1.10. Hence, ahGt(O) = g, for all ahal e Da(0). U
Lemma 3.5.3. Let g be an isotropic line in X. through 0. Then for every A,E E g,
E # 0, A # 0, there is a unique Calah e Da(O) such that Clah(E) = A.
Proof: By 3.2.3.11, there exist y,8 1 a with O0y,6 such that EY6 A. Take
I = [a,y,ay] and h = [a, 8,a6]. Note that if E = A then EP6 = E implies that Ey = E8
or El = Eh, which implies that / = h [30]. To show uniqueness, suppose that
Gaah,ak, l e Da(0), where / = [a,8,a6], k = [a,s,ae], g = [a,y,ay], and
h = [a, P3, ap3] are lines in X?(0) and E # Eha = Elk. Then Ehakl = E and EOl = E.
By Axiom 9, p3y = X with 01 X, k 1 a. Thus, m = [a, X,aX] e Oa; akaaah = am, and
E = Eml. Let E a',8' with xa' X., 8'18, and put M= ac'X and N = 6'8. Now
Ea',a,6' with a'iX, a I X, 6' L 8, and a I 5, so that a', 86' 1 a by 3.1.6.18. It
follows that MA = a'a X = a'k = Mand Na = 8'a8a = N. So that Mla,X ; NJa,8;
and m = goM and / = goN. Then because E"m' = E, we have EP5 = E; EX = E5 and
54
EM = Ea'x = E6'8 = EN and M = N by 3.1.6.14. So m = goM = gON = and al = a,,.
Therefore Gaah = (Tkal. U
Let us denote this unique map by 8A. So for all A e goE, A 0,
(i) 8A(O) = 0
(ii) 8A(E) = A
(iii) 6A E Da(0).
Translations. For every pair A,B of distinct points we can define a translation
TAB : 1 X given by TAB(A) = AAB = B. We now restrict our attention to a set of
translations defined on Ta and we note that if A,B, CaI then D = ABCIca by 3.1.4.6.
Thus, TAB : ta * Xa is a welldefined map for all A, BI a. Let C = goE be an
isotropic line in Ta and define To = {TOA : A e K)}.
Theorem 3.5.4. The set Ta is an abelian group.
Proof: Let TOA, TOB e Ta. Because C = BOA = AOB e 1C, for XI a, we calculate
(TOA o ToB)(X) = TOA(XOB) = XOBOA = XOC = To((X). Thus, TOC = TOA o TOB e Ta.
Also, Too(X) = XOO = X. Hence, Too E= Ta is the identity xa on 3Ea. To find To1,
we compute
(ToA o TOAo)(X) = XOAOA = XOOAOOA = X = XOAOA0 = (TOAo o ToA)(X).
Hence, T7oA = TOAo, A = OAO e KC, and To1 7E Ta. Clearly,
TOA o (ToB o Toc) = (TOA o TOB) o Toc for A,B,C E IC. Therefore the action of Ta is
associative and Ta is a group. Because ABC is a point and hence, an involution for
all points A,B,C, then for A,B e KC and X xa,
(TOA 0 TOB)(X) = XOBOA = XOAOB = (ToB TOA)(X).
Therefore, Ta is abelian. U
Lemma 3.5.5. For all TOA e Ta, TOA(IC) = KC.
Proof: Because OAB E 1C for A,B e 1C then TOA(K) c C for all TOA e Ta". Now let
55
C E IC and A e /C. Then OCA = D e KC so that C = ODA and TOA(D) = C. Hence,
each TOA e Ta maps KC onto KC. U
Lemma 3.5.6. Let TOA E Ta and g c Xa. Then TOA(g) = g if and only if g is
parallel to /C.
Proof: Let gHF = g be a line in Xa such that TOA(gHF) = gHF. Then
TOA(H) = HOA e gHF and gHF II IC by 3.2.1.9. Conversely, suppose that gHF 11 IC.
Then again by 3.2.1.9 it follows that for B e gHF,
ToA(B) = BOA e gHF and TOA(g) = g. U
Lemma 3.5.7. IfA,B e h, h c Xa, and h 11 AC, then there exists ToC E T"a such that
Toc(A) = B.
Proof: By 3.2.1.9 we have C = OAB e AC and it follows that B = AOC = Toc(A). U
Lemma 3.5.8. For each A e IC, there is a unique TOA E Ta such that TOA(O) = A.
Proof: Clearly, TOA(O) = OOA = A. So suppose that Toc(O) = A. Then
OOC = C = A. U
We denote this unique translation mapping of 0 into A by TA.
Lemma 3.5.9. If a e Da(O) and TA E Ta, then cOTAa = Ta(A).
Proof: Let a = agah where g = [a, P3, ac3], h = [ay, ay] E Oa. Then for any XI a,
(agChTAohaTg)(X) = (XOYOA)y = (XJPY)YPOYOAYP = XOAY = Tag(AA)A). U
By 3.2.3.8 there exist precisely two isotropic lines in 3Xa through 0 : IC = goE and
C' = gOF. We define multiplication and addition on the points of AC so that the
points of C form a field. For A,B e AC, define:
A + B = (TB o TA)(O)
A B =B A 8A)(E), where A,B # 0 and E E /C is the multiplicative identity.
A 0 O.A = 0.
Theorem 3.5.10. For every A,B e IC, TA+B = TA TB.
56
Proof: For XI Za, TA+B(X) = TAOB(X) = XOAOB = XOBOA = (TA o TB)(X). U
Theorem 3.5.11. For all A,B 0 in kC, 8A.B = 6A o 8B.
Proof: Let 8A = CJaCa' and 8B = Gbabb. Then A B = 6BA(E) = Eaab'b. Put
Cc = abCraaal, then, (CbCTc e Da(O) and C bac(E) = Ecb = Ea'abb = A B. Hence, by
3.5.1.3, 5A.B = abTc = (b(Tbf(a(aa = CaaabfabCra' = GaaaC(Fb(Tb' = 8A4B M
Hence, (/C,+) is a group isomorphic to Ta and (C\A{0},) is a group isomorphic
to Da(0). It remains to show that the distributive laws hold.
Theorem 3.5.12. Let A,B,C e IC, then (A + B) C = A C + B C.
Proof: If C = 0, then
(A+B).C=0=0 .0+0.O=A.C+B.C.
If C # 0, then we compute
(A + B) C = 6c(A + B) = 6cTA+B(O) = 6CTA+B c'(0) = 6CTATB8c'8(0) = 6CTA8' 8CTB8
= T6((A)T6(.(B)(O)= TA.CTB.C(O) = TA.C+B.C(O) = A C + B C.
Because multiplication is commutative,
C.(A+B) = (A+B)C=A C+BC=C.A+C.B. Hence, ()C,+,) is a field. U
Ordering the field and obtaining R. To order the field IC we make use of the
following 1211. Let F be a field and A 1,...,An E F. If A 1,... ,An # 0 implies that
IL A2 # 0, then ? is called formally real.
Theorem 3.5.13. (ArtinSchreier) Every formally real field can be ordered. U
To make the field AC formally real the following axiom is posited.
Axiom F. (formally real axiom). Let 0, E a, a e M with 0 and E unjoinable. Let
k, 8,11 e P. If 0 X, ,i and ,6, q a then there is a y e P such that
017y, y 1 a, and EkXk6OEkqk9 = EXyXy.
Theorem 3.5.14. If Axiom F holds on 1C, then /C is formally real.
Proof: Let I = [a,?,caXj] e 0a and let Ugah E D a(O). Then from Proposition 3.5.1,
57
Cgahat = al for some I e Oa and GgCh = aatC. Clearly, if e Oa0 with a/'at = a/at
then we have CT' = a1 and 1' = 1. Hence, for all Ogah E Da(0), there is a unique
1 e Oa, such that CgTh = IaGt. So if 0 # A E KC, then we may uniquely write 6A in
the form caaa,; that is,for every 0 # A e KC, there is a unique a, a = [a,a',Ca'J E] Oa
such that
A = aaat(E) = E1' = E^'.
Suppose that 0 A = EX'. Then A2 = Ek"'k" = 0 implies that
E = O'k'k = 0, a contradiction. Hence, if 0 # A e /C then A2 # 0. Now suppose
that A,B e KIC and A,B # 0. Let a = [a, x',aa'], b = [a,P3,4ap] e Oa such that
A = Ea' and B = E'O. Then by Axiom F there exists a y e P such that 0 171 a
and
A2 +B2 = Ek'a'0OEx44 = EkyX. (3.5.1)
Since Oy 1 ca then c = [a,y,ay] E Oa,,acat e TDa(O), and
C = acat(E) = Etc = E e C.
So equation (3.5.1) reads A2 +B2 = C2. If C2 = EYy = 0O, then E = OYky = 0.
Since E 0, it follows that if A ,...,An /C are all nonzero, then Y' A2 0 and
kC is formally real. U
To finally obtain a field isomorphic to the real numbers, R, we add the least
upper bound property to our axiom system.
Axiom L. If0 # A c KC and A is bounded above, then there exists an A E KC such
that A >X, for all X e A, and ifB e KC with B > X for all X E A then A < B.
The only ordered field up to isomorphism with the least upper bound property
is the real number field, R.
Theorem 3.5.15. The field K constructed above, along with Axiom F and Axiom L,
is isomorphic to the real number field, R.
58
In the next section an affine vector space is constructed from products of pairs
of points. The scalar multiplication is obtained by adapting and extending the
definition of multiplication of elements of KA.
3.6. Dilations and the Construction of (X,V, C)
The additive group V and dilations. First we construct a vector space V over
the field 1C. Let V = {OX : X e X }.First note that the product, AB, of any two
points A,B e X, is in V because AB = 0(OAB) = OOAB. We view the elements
OX e V as directed line segments with initial point 0 and terminal point X on the
line gox. We define an addition on V by setting OX+ OY M OXOY. The product of
three points is a point, so XOY = Z e X, OXOY = OZ E V.
Theorem 3.6.1. (V,+) is an abelian group.
Proof: Let X, Y,Z e X be distinct. Then, OX+ OY = OXOY = OYOX = OY+ OX, and
addition is abelian. The zero vector.is lo since 1I = 00 E V and
1o + OX = loOX = OX = OX+ 1o. To complete the proof we calculate
OX+ OX = OXOX = OXOOXO = OXXO = 00= 1, so OX = OX.
(OX+ OY) + OZ= (OXOY)OZ = OX(OYOZ) = OX+ (OY+ OZ).
Hence, (V ,+) is an abelian group. E
We still need to define a scalar multiplication of KC on V. To do this note that
in an affine space the group of dilations with fixed point C is isomorphic to the
multiplicative group of the field. Noting this, we geometrically construct such a
group of mappings and use these mappings to define our scalar multiplication.
A dilation of X is a mapping 5 : X '* X which is bijective and which maps
every line of X onto a parallel line. J23, p.37j
Theorem 3.6.2. [23, p.42] A dilation 5 is completely determined by the images of
two points.
59
Proof: Let 5 : X ' X be a dilation and assume that 5(X) = X' and 8(Y) = Y1 of two
points X, Y e 3X are known. We must show that the image of any Z e X is known.
Suppose that Z 0 gxry. Then clearly Z # X and Z # Y and we consider the two lines
gXZ and gyz. Observe that Z e gxz n gyz. If these lines had a point in common
besides Z, they would be equal and then gxz = gyz = gXY. But this is impossible
because Z o gxy. Therefore, {Z} = gxz r gyz. Since 5 is bijective, from settheoretic
reasons alone, 8(gxz r gyz) = 5(gXZ) n 6(gyz); or equivalently,
{8(Z)} = S(gxz) f((gyz). In other words, the lines 6(gxz) and S(gyz) have precisely
one point in common, namely, the point 5(Z) for which we are looking. The line
6(gxz) is completely known because it is the unique line which passes through X' and
is parallel to gxz. Similarly, the line 5(gyz) is the unique line which passes through
the point Y1 and is parallel to gyz. because 5(Z) is the unique point of intersection of
(gtA7) and 5(grz) (3.2.1.1), the point 5(Z) is completely determined by X, and Y1.
Conversely, assume that Z e gxy. If Z is X or Y, we are given 5(Z), so assume
that Z # Xand Z # Y. By 3.4.1, there is an a e (M such that X, Ya and there exists
a PI a such that P o gxy. Then Z 0 gxp and from the previous paragraph, 5(P) is
known. Hence, using the line gxp instead of the line gxy, we conclude from the
earlier proof that 5(2) is known. U
To define a scalar multiplication, we fix a timelike line t and use it to
geometrically define dilations. To aid in the construction, the following facts are
used to add the appropriate axioms.
In a threedimensional or fourdimensional Minkowski space, if t is any
timelike line through a point 0 and g is any other line through 0, then there is a
unique Lorentz plane containing the two lines. Two distinct isotropic lines
intersecting in a point in Minkowski space determine a unique Lorentz plane.
Desargue's axiom, D, holds in any affine space of dimension d > 3.
60
Axiom T. If O E t,g where t is timelike or t and g are both isotropic then is a unique
a e M such that g,t c Xa.
Axiom D. Let g\, g2, and g3 be any three distinct lines, not necessarily coplanar,
which intersect in a pointO. Let PI,Q\ E gj; P2,Q2 e g2; and P3,Q3 e g3. If
gPIP3 11 gQ1Q3 andgp2P3 II gQ2Q3 then gpp2 II gQIQ2"
Axiom R. LetO e g1,g2; P1,Q1 e gj; andP2,Q2 E g2. Ifgpp 2 goQQ2 then
go,P OP2 = gO.Q i OQ2.
Axiom T refers to the first statements. Axiom T is used to put an isomorphic
copy of the field on every isotropic line through 0. A scalar multiplication is then
defined in a manner similar to the definition of multiplication for the field elements.
Axiom D is Desargue's axiom, the "dilation" axiom. Axiom D ensures welldefined
dilations with the standard properties of such maps. Axiom R is used to distribute a
scalar over the sum of two vectors. The dilations are constructed next.
Let IC c Xa, a e M. By 3.1.7.2, there exist a 1,a2 e G such that O0ac1,a2
and a = aIla2. Because l0a, then Oca = P3 e and 0 = a13 = aIaC213 with aC1,2 1 3
by 3.1.6.18. Let y = a1I3 e M. Then ya = Pa3cIaIa2 = P3a2 e M, so that
t = [ca,y,a cy] c 3a is timelike and 0 E t as 0 a, ca 1,a2,13 implies that 01a,y,acty. So
let t = [a,y,6] E Oa be any timelike line through 0 and /C' c Xa the other isotropic
line through 0. Then for each AI e t, there is a unique A e IC and there is a unique
B e /C' such that A, = AOB. Because A, e t, then B = At. (From this point on
denote a line reflection ag(X) by Xg. Thus, xghr means Crrahag(X) = CgChcr(X).).
For each A, e t, there is a unique A e IC such that At = AOA'. Similarly, for
each A in IC, there is a unique AI in t such that At = AOA'. Thus, there is a
onetoone correspondence between the points of /C, the field, and the points of t.
Fix Et = EOE' as the unit point on t. For each 0 # A e IC, we use t to
construct a dilation WA of X in the following way. Let X e X.
61
IfX o t, then by Axiom T, there is a unique 71 e M such that gox,t c Xrj.
From Sections 3.5 and 3.6, X1 is an affine plane and our definition of parallel lines in
X9 is equivalent to the affine definition. So, there is a unique line h c 3eX such that
At e h and h II gxE,. Because
gox rn t = {O} 0, h ) t= {At,} 0, andh II gE,x,
then h n gox = {B} 0. In this case, set 8A(X) = B.
If Xe t and X # 0, then because t,K a Xa with a E M, there exists a
unique g c Xa such that A e g and g 11 gxE. Because gxE n t = {E} # 0 then
g r t = {B} # 0. Set 4A(X) = B.
If X = 0, set 8A(A) = 0. Note that 8A(X) e gox by construction. It is clear
that 8A : X i X is onetoone. We find it useful to make some observations.
Let 0 A e kC and recall that E e KC is the multiplicative unit point;
E .A =A, forA e C. Put At =AOAt.
Lemma 3.6.3. For the map 8A defined above, the following are true:
1 8A(E)=A.
2 8A(E) = A.
3 8A is onto.
Proof: 1. Et = EOE' and A, = AOA' imply that EEt = O' and AA, = OAt. If
EEt = OE' = l then 0 = E' and 0 = Ot = E. Similarly, because A # 0 then
AAI = OA' 1o. By 3.2.2.9, At = A' e K' as 0,A e KC, AC is isotropic, and 0y 1 a.
Similarly, E' e K'. Thus by 3.2.1.6 we have gEE, II goE' = K' II gAA' and gEE, II gAA,
by 3.2.1.7. By 3.2.1.8, h = gAA, and h r goE = gAA, r =C {A}. Hence 8A(E) = A.
2 6A(Et) = At. Again, ggA(E, = gAA, II gEE, and gAA, r t = {AI}.
3 Let P e X and P # 0. If P e t then by 3.2.1.8, there is a unique line g such that
Et e g, g II gA,P, and grgop = {Q}, for some Q e X. Then it follows that
6A(Q) =P. IfP et then 8A(Q) = P, where {Q} =g tn, E E g, and g II gAp. 0
62
Lemma 3.6.4. If C # D, then gcD II g8A()A(D); hence, for all 0 # A e KC, 8A is a
dilation of X.
Proof: Let gj = t, g2 = goc, g3 = goD, Pi = Et, Q = A2 e g, P2 = C, Q2 = 8A(C),
Q2 E g2, P3 = D, and Q3 = 8A(D). Then gcE, II g,8() and gDE, II g,,A(D)" Thus, by
Axiom D, gcD II gA(C)A(D) N
Consider now the plane Xa. Recall that by 3.5.1.3, for 0 # A e 1C, there is a
unique 8A e Dac,(O) such that SA(E) = A, where 5A is of the form Cgch for g,h e Oa.
From Axiom 9 and the proof of 3.5.1.1, for any r E Oc, Cgahor = CTw for a unique
w E Oa. That is, for every fixed r E Ox, every 0 # A e KC can be uniquely written in
the form 85A = awCar where Cw(Er) = A. (The uniqueness follows from the fact that
(At)w, = (Ar)w2 implies wl = w2 if A'r A [30].)
Therefore, for every 0 # A e AC, there is a unique a e Oa, such that E"ta = A.
That is, there is a unique a e Oa, such that GaOt = 6A. Also, every P E Xa, PI a,
can be uniquely written as P = P07IOT2, where P\ E AC and P2 E AC'.
Let P2 = 2'e C K. Then we may uniquely write P = P\OP', where P1,P2 e KC.
Define the map 6A : Xa + Xa, by A(P) = PtaO(Ptt)', for 0 # A e 1C, Pla.
Theorem 3.6.5. The map 8A : XJa * Xa is a dilation on Xa with fixed point 0 and
dilation factor A, for all 0 # A e K.
Proof: If P'laOP'a = Qa'OQ'a, then by unicity, pa = Qp. So P1 = Q0, Pft = Qa
and P2 = Q2. Hence, P = P OP2 = Q I OQ2 = Q, and 6A is injective. Let QIa with
Q = QOQ2; QI,Q2 e AC. Let PI = Qg' E AC and P2 = Q0t e C. Then P = PIOP Iaot
and 8A(P) = P1aOPfI = (f)O(f)2t = Q I0OQ = Q. Therefore, 8A is onto.
We claim that if P# Q, then gpQ I\ gp,, First we show that
8A(P) e gop. If P E IC, then P = PO0 and 6A(P) = pta00o = ptao0 = pta e 1C. If
P e AC', then putting R = P' e AC, we have P = OOR' and 6A(P) = OtaORtat =
= OOR'a' = Rtat = pat E C'.
63
So suppose that gop g is nonisotropic and write P = PIOP2, where P1 = P'.
If 0 e g, then XOY e g <> X = Yg, where X e /C and Y e /C'. Thus, 8A(P) = pla'Oa
and P,' pga = ptag which implies that 8A(P) e g.
Claim. 8A (POQ) = 8A(P)06A(Q), for all P,Q e Xa. Let P = PIOP2 and Q = Q\OQ2.
Then POQ = PIOPQIOQOQ2 = (PIOQi)O(Pz2OQ2)t, with PiOQi E KC and
(P20Q2)' e IC'. Therefore,
8A(POQ) = (P1OQ1)tao(P20Q2)tat = ptaQQa'ltQalf = (ptaoptt)oQ(Q21ltoa)
= 8A(P)OA(Q).
The claim follows.
Proof of (iii): Assume that P,0, and Q are collinear. Then there is a line g, such that
P,0,Q e g. Thus, 8A(P),8A(Q) e g and gpQ = g II g = gA(P)'A(Q)
Conversely, suppose that P, Q, and 0 are noncollinear. Let g, = gop, g2 = goQ,
and g3 = go,pOQ. Then g\,g2, and g3 are distinct lines through 0. Indeed, if g3 = g1,
say, then POQ e gop and we would have OP(POQ) = Q e gop, which contradicts
our assumption of noncollinearity. Now,
P(POQ) = OQ and 8A(P)(8A(POQ)) = 8A(8A(P)0OA(Q)) = 08A(Q)
gP,POQ II goQ = gO8A(Q) II gSA(P)8 (POQ)' and gP,POQ IIgA(P)A(POQ)
Similarly, Q(POQ) = Q(QOP) = OP; and 8A (Q)(A (POQ)) = 8A (A (Q)00A (P)) = 08A (P).
This implies that gQ,POQ II gop and gop = go6(p) II g8A(Q)6(POQ)" Hence,
gQ,POQ II gSA(Q)N(POQ)" By Axiom D, it follows that gpQ 11 gA(P)A(Q) Therefore,
8A : fa i+ 3Xa is a dilation on Xa. U
Now we show that 8A = 8A on Xa. Because 8A and 8A are dilations on Xa, a
dilation is uniquely determined by the images of two points,and 8A(O) = 0 = 8A(O),
then it suffices to show that 8A(E) = 8A(E). By definition of 8A, 8A(E) = A. By
Lemma 3.6.3, 8A(E) = A.
64
To extend the above idea to any rI e M such that t c Xq, let a # r1 e M
such that t c Xi. Let AC1 and /k2 be the isotropic lines in XrC through 0. Because t is
nonisotropic, then by Axiom 14 there exist y,6 such that il = y8 and t = [ry,S]. Thus
a, : n '* X, is welldefined. Every B e t c Xq may be uniquely written as
SB= B\OB2, where Bi e AC,. Because 0 e t and t is nonisotropic, then B2 = B'. Thus,
B = BIOBt where B1 e AC1. So, in particular, there are unique E\,A\ e IC\ such that
Et = EiOEP and A = AIOA', for AI e t.As before, there is a unique aI e O, such
that Ea' = A 1. Hence, every X e Xri can be uniquely written as X = X\OX2, where
X\,X2 (e AZ1 c X. This defines a map 6A8 : Xn * X, given by
6A(A)=
Proposition 3.6.6. Let 8A : X X3 be the map defined above. Then
1. '5A4 is a dilation on X.
2. 6A = 8A on .
3. ifP, 0, andQ are collinear points in X.r, not necessarily distinct, then
6An(POQ) = 6Aq(P)O5Aq(Q).
4. Moreover, 8A(POQ) = A(P)O0A(Q), for every P, Q\ r\ such that O,P, and Q are
collinear. U
To obtain a scalar multiplication on V, for all 0 # A e AC and all OX e V,
define
A OX = 8A (O)A (X) = 086A (X) and O. OX = 1.
We now verify the vector space properties.
Lemma 3.6.7. IfA,A' e IC and OP e V, then (A + A') OP = A OP + A' OP.
Proof: Suppose P e I and P # 0, and recall t,AC c a. Let P, EIa,P; AIy withy  P3;
A'y', y' 3 P; and O06 with 6 II P3. Then y 11 y' 1I 6; g, = [ac,y] I gEp with g, c ;
92 = [a,y']  g p with g2 c X.a; and g3 = [aX,6] I gEp with g3 c XC,. Thus,
6A(P) e gi r t and 8A(P) e 92 t. Thus, A(P)I)a,y and 8A(P)la,y'. It follows that
65
8A(P)O9A'(P)Ia,y68Y'; 8A(P)08A(P) e t; 787' = 1I P3; and AOAt'I ,875' = s. Therefore,
AOA' e [oa,] II gEP and 8A(P)06A(P) e t n [a, E]. Hence, 8AOA'(P) = 8A(P)08A'(P).
Suppose P o t. Because gop,t c X3, then replacing E with Et, A with At, A'
with A,, and c with ri in the first part of the proof and the result follows. U
Lemma 3.6.8. IfA e C and OP, OQ e V then A (OP + OQ) = A OP + A OQ.
Proof: We need to show that 8A(POQ) = 8A(P)08A(Q). Suppose P,0, and Q are
collinear. Let g = gop = goQ = goyOQ and r e M such that g,t c Xr. Then the
result follows from Proposition 3.6.6(iv).
Conversely, suppose that P, 0, and Q are not collinear. Because P, 0, and Q are
not collinear then P # Q, and from Lemma 3.6.4 it follows that gpQ 11 gsA(P)8(Q)"
Applying Axiom R we obtain goPOQ = gO8A(P)8A(Q). That is, 8A(P)O0A(Q) e goPOQ
by construction. Again by Lemma 3.6.4,
gSA(P)A(POQ) II goPOQ and 8A(P)SA(P)OSA(Q) = 08A(Q).
This implies that gA(P),8A(P)08A(Q) II go08,(Q) = goQ 11 gp,POQ, as P(POQ) = OQ. Thus,
g6A(P)..(/i'.)4(Q) = gSA(P)SA(POQ) because both lines contain 8A(P) and are parallel to
gP,POQ. Since
g8A(P).6A(P)O0A(Q) gP.POQ = {8A(P)OSA(Q)} and gsA(I, ,,POQ) n go.POQ = {8A(POQ)},
then 6A(P)06A(Q) = 6A(POQ). M
Lemma 3.6.9. IfA,A' CE a and OP e V then A (A' OP) = (A A') OP.
Proof: We need to show that 8A(6A(P)) = 6A.A'(p); that is, 86A o 6A' = 8A.A. Now
5A o 6A' and 6A.A' are dilations on X and 6 A.A'(O) = 0 by construction, so it suffices
to show that (8A o 6A,)(E) = 8.1 (E) and (6A o 8A')(0) = 8A.A,(O). Now E e /C c X,
so on Xa, 6A o 6A' = 6A o 64' and 6A.A' = 8A.A'. But on AC, by Lemma 3.5.11,
6A o 6A' = A.A'" Therefore, (8A o 8A')(E) = A (A' E E) = (A A'). E = 8AA'(E). U
Lemma 3.6.10. For E e 1C, the multiplicative unit, and OP e V, E. OP = OP.
66
Proof: We need to show that 8E = 1.. Now l1 is clearly a dilation on X and
1(0O) = 0 = 8E(O). Thus, 1.ic(E) = E= E E = E(E) = 6E(E). U
Theorem 3.6.11. The space (V,1C) constructed above is a vector space. U
The triple (X,V,KC), is an affine space.1[23,p.6] A set X along with a vector
space V over a field /C is an affine space if for every P e V and for every AX X,
there is defined a point PX e X such that the following conditions hold.
1. If v,w E V and X e 3C, then (f + ,)X = v("X).
2. If 0 denotes the zero vector, OX = X for all X e X.
3. For every ordered pair (X, Y) of points of iX, there is one and only one vector v e V
such that vX = Y.
The dimension n of the vector space V is also called the dimension of the affine space
T.
Theorem 3.6.12. (iX,V,/C) is an affine space.
Proof: If OV,OW e V and X E X, we have
(i) (OV)X= OVXe X.
(ii) (OV + OW)X = (OVOW)X = OV(OWX).
(iii) OOX= loX =X.
(iv) for Ye X., OYX = Z e iX and (OZ)X = Y.
Now if OPX = Y, then OPX = OZX, P = Z, and OP = OZ. M
3.7 Subspaces and Dimensions
In this section we show that our lines and planes have the proper dimensions.
We are then able to conclude that (V, C) and (t, V, C) are fourdimensional spaces.
Proposition 3.7.1 Let g be any line through 0 and put 0(O) = {OA : A e g}. Then
(0) is a one dimensional subspace of V.
Proof: First note that 15 = 00 e g(0), so the zero vector is in g(0). Let A,B E g.
Then C = AOB E g by 3.1.4.7 and OA + OB = OAOB = OC e g(0). From Section 3.6,
67
5A(B) e g, for all A in /C and all B in g. So A OB e ((O) for all A e /C and
OB e (O). Hence, g(O) is a subspace of V.
It must now be shown that the dimension of ,(O) is one. If g = t, then
0(O) = (OEI) because for every At e t, At = 8A(Et) by Lemma 3.6.3. So suppose that
g # t and fix B E g. Let h = gBE,. Then for all 0 # D E g, there ia a unique such
that D e d, d 11 h, and dr t 0. Put {Ft} = d t. Then for Ft = FOF' with F e /C
it follows that
6F(B) = D and OD = F. OB.
Hence, A(O) = (OB). U
Corollary 3.7.2. Following the terminology of Snapper and Troyer [23, p.11],
g = S(O,g(O)) {A = O(OA) : OA e (O0)}
is an affine subspace of dimension one.
Proposition 3.7.3. Let a e P with Oa and put ta(O) = {OA : AIa}. Then ta(0) is
a two dimensional subspace of V.
Proof: Clearly, 1e = 00 e ta(O), so the zero vector is in 1tc(O). Let C, DI a and
A,B e 1C. Then by 3.1.4.7 we have COD = FI a and by Lemma 3.6.3,
A (C) e goc c Xa and 8B(D) e gOD c Xa
so that 8A(C)O0B(D) a. It follows that OC + OD = OCOD = OF e tra(O), and
A OC + B OD = O0A(C) + 08B(D) = O(8A(C)OB(D) E 1a(0).
Hence, ta(O) is a subspace of V.
Thus, it remains to show that ta(O) is two dimensional. There are two cases:
a e M and a e g. Suppose first that a e M. We construct a basis for lt(0) using
isotropic lines. To this end, let C1 and KC2 be the isotropic lines in Xa through 0. For
any Pla we may uniquely write in Xa, P = PLOP2, with PI e IC1 and P2 E IC2.
68
From Proposition 3.7.1 above, kC1(O) = (OB) for any 0 # B e /C1 and fC2(O) = (OC)
for any 0 # C e /C2. From this it follows that P1 = 8A(B) and P2 = 8A'(C) for some
A,A' e 1C. Thus,
OP =A OB+A' *OC
and {OC,OB} span ta(O). Now ifA OB+A'. OC = Io, then O8A(B)O6A(C) = 1.
This implies that 8A(B)08A'(C) 0 and 06A'(C) = 6A(B)0. Because O,6A(C) /EC2
and 0,8A(B) E ACI, then either /C1 II AC2 or OA,(C) = 1 = 5A(B)0. But /CI 9 /C2, so
A,(0 = 8A(B) = 0. Because ca e M, then from 3.3 and 3.6 there exists t',a' e Oa
such that 8A(C) = CVa' = 0. This implies that C = 0a' = 0 or A' = 0. By
assumption C 0 and D # 0, thus it must be the case that A = A' = 0. Hence,
{OC,OB} is a linearly independent set in ta(O) and Ia(0) is two dimensional.
Suppose now that a e g. We construct an orthogonall" basis. By 3.1.7.2, there
exist P3,y e g such that 0 = c0y and a 1 P31 y 1 a. Let x = [a, 3,a c] and
y = [a,y,ay]. Note that ifx =y then by Axiom 14, a(3 = y and 0 = a13y = 1I, so
x y. Let Pl a and let 13',y'IP with 13' 13 and 7y' y. Put P1 = Pp1313' and P2 = 7Y'.
Now PIa,13'; a I 13; 13' 1 13, so that a I 13' by 3.1.6.18. Because PIa,y'; a 7 y;
y' 1 y, then a 1 7y'. Thus,
P' = (13p,)c = 131t, = 1313' = Pi and P( = (n')r = 7' = P2,
so Pi,P2 a. This implies that PI Ia,P3 and P2Ia,y so PI e x and P2 e y. If
Q = PiOP2 then Q = POP2 = 13'PP313ayyy' = 13'ay'.Because 13' 1 a, let 8 = a13' E P.
Then Q = 86y' and 6 1'. Thus PI P13',ac implies that P 13'a = 8. Thus we have
P,QjS,y' with 6 1 y', and P = Q by Axiom 3. That is, P = POP2 with P1 e x and
P2 E Y.
Let 0 X e x and 0 Y e y. Then from Proposition 3.7.1 above we have
x = (OX) andy = (OY) and there exist A,B e IC such that OP = A OX+B OY, where
69
P1 = 8A(X) and P2 = 5B(Y). Hence, {OX,OY} spans a(O). IfA OX+B.OY= 1I,
then we obtain 0OA(X) = 8B(Y)O. Because x Jf y, it follows that 8A(X) = 0 = 5(Y).
Let Xin be the unique plane containing t and x and /CI and /C2 the isotropic lines in
X3 through 0 Then we may write 8A(A) = X1a(OX'la', where X1i e /Ci, and X E K/2.
As above it follows that A = 0 and similarly, B = 0. Therefore, 1a(0) is two
dimensional. M
Corollary 3.7.4. X. = S(O,t_(O)) is an affine subspace of dimension two for all
a E P, 0La.
Theorem 3.7.5. (V,/) is a fourdimensional vector space and hence, (X, V,C) is a four
dimensional affine space.
Proof: Let OP e V and let 0 = a43 with a e M and P3 e g. Let PIa',3P' with a' I a
and P3' 1 P3. Put Y' = xa'c and P" = P3'P3. Then Q = P'OP" = a'aacPP313P3' = a'3'. Thus,
P,Q Icc', 3' with a' 1 3'. Therefore P = Q = P'OP" with P'Ia and P" I3. Since ta(O)
and .tp(O) are twodimensional then there exist bases {OZ, OT} c .tc(O) and
{OX,OY} c tp(O0) such that
OP' = A OZ+A' OT and OP" = B OX+ B' OY
for some A,A',B,B' e /C. Thus, P = A. OZ+A' OT+B OX+ B' OYand
{OX, OY, OZ, OT} span V. If A *OZ+A' *OT+ B OX+ B' OY= lo, then in
particular, OP = OP' + OP" = lo. This implies that OP" = P'O. So either
gOP" II gp'o or P" = P' = 0. But gop" 1 gop' and therefore,
A OZ+A' OT = OP' = =OP" = B OX+B' OY. As was shown in Proposition
3.7.2, we obtain A = A' = B B' = 0 and the result follows. U
3.8 Orthogonality
In this section we extend the definition of orthogonality to include lines and
then use this definition to define orthogonal vectors.
70
Definition 3.8.1. Let g and h be two lines. We say that g is perpendicular to or
orthogonal to h, denoted by g L h, if there exist Oa, P3 E P such that g c Xa, h c Xp,
ax 1 and P = (a e g,h. In this case, P is the point of intersection of g and h.
Lemma 3.8.2. If g and h are isotropic then g is not orthogonal to h.
Proof: This follows directly from our definition above and from our definition of
a 1 p. For if a 1 P3 then one of a and P3 must be in G and by Axiom 12, all lines in a
plane 5Xp for P3 e G are nonisotropic. U
In Minkowski space, if g and h are two isotropic lines then g L h <> g 1 h.
Thus we extend the above definition in the following way.
Definition 3.8.3. If g and h are isotropic lines, then g I h <> g 11 h .
Definition 3.8.4. If g is a line and a e 7) then we say that g is orthogonal to or
perpendicular to ta, g I Xa, if there exists a P e P such that g c 1p, P3 1 a, and
P = ap g. In this case, P = ap is the point of intersection of g and Xa.
Definition 3.8.5. For every 1I OA,OB e V, we say that OA is orthogonal to OB,
OA OB, if and only if goA L gOB; that is, there exist (x, a3 e P such that goA c X3,
goB c Xp, and 0 = ap. For the zero vector 1Q = 00, we define 1I OA, for all
OA E V.
Lemma 3.8.6. From 3.1.2.1 and 3.1.2.2 it follows that fort e (5:
i g 1 h <h g L h:.
(hi) g I T X~ <> g~ I X.
(iii) OA I OB <> (OA) I (OB),. U
Lemma 3.8.7. If g is a nonisotropic line then g is not orthogonal to g.
Proof: Ifg I g then there exist a,P e P such that g c X, Xp; a I 3, and
P = ap e g. But for every point Q e g, Q I(xa,P with a 1 P3 which implies that P = Q
by Axiom 3; that is, g is a line which contains only one point, which contradicts the
definition of a line. U
71
Additional axioms and their immediate consequences. To complete our
preparations for defining our polarity and thus obtaining the Minkowski metric, we
recall our final three axioms.
Axiom U. (UL subspace axiom) Let O,A,B, and C be any four, not necessarily
distinct, points with A, 0 a; O,BI ; O,C y,6 and a 17 y and 1 6. Then there exists
k,& e P such that X 1 ; O,AOBIX; and O,CCe.
Axiom S1. Ifg c Xa, a e G, h c X p, P e g, and there exists y,8 E P such that
y 16 5; y6 E grn h; g c Xy; and h c X 6 then there exists 8 e 9 such that g,h c .. (If
g and h are two orthogonal spacelike lines then there is a spacelike plane containing
them.)
Axiom S2. Let g and h be two distinct lines such that P e g r) h but there does not
exist P e P such that g,h c Xp. Then either there exists a,y e P such that a 1 y,
g c Xa, andh c Xy or for allA E g, there exists B e h such that P andAPB are
unjoinable.
Lemma 3.8.8. Ifg,h c Xa are nonisotropic for a E P, then g _L h in the sense of
Section 3.3 if and only if g L h in the sense of Definition 3.8.1.
Proof: Let g = [a,P,5], h = [a,y,,.] c Xa with a = 36 = y.. Recall that g I h in the
sense of 3.3 if, without loss of generality, P3 y and A 7= (= g gr h. So, in
particular, A = P3y with A E gr h and g c Xp and h c Ty. So g _ h by 3.8.1.
Now suppose that there exist rl,s e P such that B = TE e grn h and g c
and h c XE, but it is not the case that 13 1 y, nor that P3 1 ., nor that 6 1 y, nor
that 6 1 .. Then by 3.3.8 there exists a unique e CLO such that B E 1 and
I = [a,u, p] for some u, p E P with a = ug, u 1 13, and g 1 6. If / # h then la(B) and
ha(B) span ta(B). Thus if C E g there exists L e I and H e h such that BLBH = BC.
By Axiom U, BL 1 BC and BH 1 BC imply that BC = BLBH 1 BC. That is, g I g,
which cannot happen for nonisotropic g. Hence, I = h and the result follows. U
72
Lemma 3.8.9. If a e P then for each P a and for each nonisotropic line g c X,
there is a unique nonisotropic line h c Xa such that P e h and h L g.
Proof: This follows directly from 3.3.8 and Lemma 3.8.8. U
Lemma 3.8.10. Ifg,h c Xca for X e P and g,h are nonisotropic, then g L h if and
only if Ogah = GChg # 1Xa
Proof: This follows from 3.3.16 and Lemma 3.8.8. U
Lemma 3.8.11. Suppose that 0 e c,d; c,d c Xa, a E P with c and d nonisotropic
and c L d. Let C e c and D e d, then go,DOC is not orthogonal to c if C,D # 0.
Proof: First note that go.DOC # c or d because then we would have DOC = D1 E d,
say, so that C = ODD1 e d and 0 e d implies that C = 0 or c = d. Now if
go.DOC L c, then in X3a we have (O = c(d = (gODO *(Yc which implies that
d = gO,DOC. 0
Lemma 3.8.12. Let g,x c 3Ea with g isotropic, ca e 7P, andg n x = {O}. Then g is
not orthogonal to x if g # x.
Proof: If x is isotropic and x g then because x # g, there exists y, 6 e P such that
g c Xy, x c X6, and y 1 6. But then one of y and 6 must lie in g. But no element in
g can contain an isotropic line so x is not orthogonal to g.
Suppose that x is nonisotropic and let 0 e g fn x. Because g is isotropic then
cc e M and by 3.3.8, there is a unique nonisotropic line h c Xa such that 0 e h and
h I x. Suppose that g I x and let 0 # K e g, 0 # H E h, and 0 X( e x. Then OH
I OX and OK I OX so that by Axiom U, O(HOK) = OHOK I OX. If go.HOK is
nonisotropic then goHOK = h. And because go,HOK C 3a, then HOK = HI e h. Then
we have K = OHH1 e h and g = h, a contradiction.
Suppose that go,HOK is isotropic. If go,HoK = g then we may write HOK = K]
for some KI e g. It follows that H = K1KO e g and g = h. If goHOK # g then go,HOK
is the other isotropic line through 0 in Xa. Because g and go.HOK span Xa and
73
g,go,HOK I x then by Axiom U, x is orthogonal to every line in Xa through 0. So in
particular, x I h' which implies that x = h. U
Corollary 3.8.13. If g is isotropic and x I g then either x = g or x is nonisotropic and
x and g are noncoplanar; that is, there does not exist 6 e P such that x,g c X6.
Lemma 3.8.14. If g is isotropic, x is nonisotropic, x I g with x # g, and {P} = x fl g,
then for all P # A e g and for all P # B e x, g and gAPB are not coplanar.
Proof: From Lemma 3.8.12, x and g are noncoplanar. Suppose that A,O,AOB 17 for
some y e P. Then B = OAAOB\y, which implies that x = goB c Xy and
g = gOA c X'y. a contradiction to Lemma 3.8.11. U
Lemma 3.8.15. If OC e V is isotropic and OB e V is nonisotropic with OC I OB then
OCOB I OC.
Proof: By Lemma 3.8.13, goc and goB are not coplanar and by Lemma 3.8.14, goc
and go.coB are not coplanar.By Axiom S2, either goc I gOCOB or there exists
D e gocoB such that 0 and COD are unjoinable. So if goc is not orthogonal to gocoB
then go,coD is isotropic and by Axiom T, there exists a unique 6 E 'P such that
goc,go.coD c 6. This implies that O,C,CODI6, D = OCCOD 1,
gOD = go.COD c X6, and therefore, goc c X6, a contradiction. U
Lemma 3.8.16. The zero vector, lo = 00, is the only vector orthogonal to every
vector in V.
Proof: This follows immediately from Lemma 3.8.11 above. U
Theorem 3.8.17. IfU is a subspace of V, then UL = {OA e V : OA I OB, VOB U}
is a subspace of V.
Proof: because the zero vector 1 is orthogonal to every vector in V by definition,
then 1, e U'. Let OA e U't and R E K.C because R OA e< OA >, the subspace
generated by OA, and go.4A I goB for every OB e U by the definitions of U' and
orthogonal vectors, then R OA e UL.
74
Let OA, OB e U and OC e U. If OC is nonisotropic then OC OA, OB and
there exist a,P3,y,6 e P such that
A,O\La; 0,B13; 0,CIy,6; a_ 1y; and 3 16 8.
By Axiom U, there exists k,E e P such that 0 = kE, AOB Ik, and 0,C I&. Thus,
OA + OB = 0(AOB) L OC.
Now suppose that OC is isotropic. The following possibilities exist.
(i) If OC O A, OB, then as in (a) above, (OA + OB) 1 OC.
(ii) If OC = OA = OB, then because OC is isotropic, AOB = COC e goc and
(OA + OB) I OC.
(iii) If OC = OA # OB, then OB I OC implies that OB is nonisotropic and
(OC+ OB) = OCOB OC by Lemma 3.8.14.
Hence, if OA, OB e U then OA + OB e UL and UL is a subspace of V. U
Theorem 3.8.18. If O = a3 then t,(0)' = tp(0) and.tp(0) = ta(O0).
Proof: From the definition of orthogonal vectors we clearly have 1ta(O) c !p(0O)
and lp(0)' c ,a(0). Now suppose that OA I ta(O); that is, goA L a. Then, there
exists y e P such that OA e 1t(0) and y 1 ac. But then we obtain 0 = ca3 = ay so
3 = y. Hence, ta(0)' = tp(0) and ta(0) = t1(0)L. M
An immediate consequence of Theorem 3.8.18 is the following. For each a e P
with 01 a, there exists a unique P3 e P such that03, ta(O)1 = .tp(0), and
((0H =t(0).
Theorem 3.8.19. If d(0) is nonisotropic then there is a unique hyperplane A(O) such
that d(0)' = A(0) and (d(0)')1 = d(0).
Proof: Let a e P such that a c X, where a is the nonisotropic line associated with
d(0) = {OA : A e a}. Then for 0 = a13 we have a L P3 and a 1 g for every g c Xp
with 0 e g. Thus, tp(0O) c d(O). Now a is nonisotropic, so by 3.3.8, there exists a
unique h c X.a such that 0 E h and a 1 h. By Axiom U, (/(0),ptp(0)) c d(O).
75
Suppose 1 OB e d(O) r (<(O), tp(O)).Then B e a c Xa and there exists
OC e /(O) and there .exists OD e tp1(O) such that OB = OCOD; that is, there exists
C e h and D 13 such that B = COD. because h c Xa, CIa and C = cx 1. Because
Di3, D = PPI33 for some P3i e P. Now a l 13 and P31 P31, so a 1I 13. But B = COD
and COD = cxaaoppp3131I = ai3PI so that B3p1,at with a 11 13. This implies that a = PI
so D = 0, B = C, and a = h, a contradiction. Hence, d(O) r (/(O), Ip(0)) = {1 }
and V = d(O) (h(O),!tp1(O)). If 0 e d and d I1 a then d(O) c (h/(O),tp(O)). For
otherwise we would have
V = d(0) d(O) (4(O),.tp(O)>,
which is not possible. Hence, 6(0)L = (/(O),.tp(0)).
On the other hand, from tp(O)' = ta(O), a IL h, and a I 3, then,
d(O) c (/i(O),t (O))'. If 0 g 1 P3, then by definition g c X,. If 0 e g 1 h, then
g = a. Hence, d(O) = (/h(O), Ip(0)).
Claim. d(O)' is independent of the plane containing a. Suppose a c Xy .
Then for 0 = y76, we have y # a, P13, and a L X6. Again there exists a unique
I c Xy such that 0 e I and 1 L a, so that d(O)' = (l(O),.t6(0)) as above. Now any
point P e X may be written as P = P\OP2 with P\ a and P2 I 3. Since h L a, then we
may write P\ = HOA with H e h and A e a. Then
((O), tp(O)) d(O) = V = ((0), ty(O)) d(O0),
and it follows that (
Theorem 3.8.20. If d(O) is isotropic then there is an unique hyperplane A(O) such
that d(O)' = A(O),d(O) c A(O), and (J(O)')' = d(0).
Proof: Let d(0) c ta(O) and a the isotropic line in Xa corresponding to d(0).
because a is isotropic then a e M. Let 0 = ap3, 13 e 9, and put
A(O) = (d(O),t1(O)). Since a c Xa, a 13, and 0 = a13 e a, then a 1 X1. Because
76
a is isotropic then a I a. By Axiom U and Lemma 3.8.14 it follows that
(d(O),p(O)) = A(O) c d(O)'. If k(O) c (d(O),tp(O)) is isotropic and g # a, then
there exists a unique y e M such that g,a c Xy. Because g(O) c A(O) c d(O)',
then g 1 a, which contradicts Lemma 3.8.11. Thus, A(O) contains no other isotropic
line.
Let h I a with 0 e h. For each H e h we may write H = H'OB, with HNc\ and
BI3. If 0 # A e= a, it follows that OB OA; OH OA, so that OB I OA and
OH' = OHOB0 1 OA. This implies that gOH' = a so H/ e a and h(0) e (d(O),p(O)).
Thus, d(O)' = (d(O),t(O)).
If h(O) c (d(O),tp3(O)), then
/h(O) c 13(O)' = ta(O) and /h(O) I i(O).
Thus, h(O) = d(O). Hence, (d(O),Tp(O))> = d(O) and (d(O)')i = d(O).
The subspace di(O)I is independent of the plane containing a. Suppose a c X.,
y e M, y # ac. Put 0 = 7y6. Let h(O) c (d(O),t6(O)>. Then h 1 a and
h(O) c (d(O),Jp(O)). If
4(0) C
then k 1 a and k(O) c (d(O), t(0)). Therefore,
(d(O),.t6(O)) = (d(O),tp(0)). U
Remark. If g is a line then g is either nonisotropic or isotropic. From Theorem
3.8.19 and Theorem 3.8.20, if U_< V is a onedimensional subspace then UW is a
uniquely determined hyperplane.
In an affine space a plane is uniquely determined by two distinct intersecting
lines. Let g and h be two distinct intersecting lines and let Kg,h) denote the unique
plane determined by g and h.
77
Definition 3.8.21. If a e P such that g,h c Xa then we say that (g,h) = Xa is a
nonsingular plane. If there does not exist a e P such that g, h c Xa then we say that
(g, h) is singular or (g, h) is a singular plane.
It is clear that every plane in (X, V, IC) is either singular or nonsingular. Note
that by Theorem 3.8.18, if U < V is a nonsingular twodimensional subspace of V,
then U1 is a uniquely determined nonsingular twodimensional subspace of V. Now
consider the following cases for two distinct intersecting lines g and h.
Suppose that g is isotropic and h is isotropic. Then by Axiom T, there exists a
unique a e MA such that g,h c Xa. If one of g or h is timelike, then by Axiom T,
there exists a unique a e M such that g,h c a.
Thus, if (g,h) is singular then either g is isotropic and h is spacelike, or g and h
are both spacelike.
Proposition 3.8.22. Let (g, h) be a singular plane with g isotropic and h spacelike.
Then g I h.
Proof: Let {P} = g n h. By Axiom S2, either g I h or for each A in g, there exists B
in h such that P and APB are unjoinable. Suppose that A e g, B e h, and P and APB
are unjoinable. It must be the case that A is joinable with APB; that is, gAAPB is
nonisotropic. Let gpA = [6,&], so A,P\6,E. If A is unjoinable with APB then A, P, and
APB are pairwise unjoinable points, so by Axiom 10, APBs6,6 and APB E g. Thus,
gPAPB = gPA = g and B = PAAPB e g, so g = h.
Thus, gpA4pB t g and gP,4PB is isotropic, so by Axiom T, there exists a unique
y e M such that g,gpAPB c Xy. This means that P,A,APBIy so B = PAAPBI7, and
h c Xiy, which contradicts our initial assumption. Therefore, g IL h. 0
Theorem 3.8.23. Let (g(O),h(O)) be a two dimensional singular subspace of V where
9(0) is isotropic and h(O) is spa celike. Then (g(O), h(O))I is a two dimensional
singular subspace of V which contains g(0). Moreover, ((kg(O), fh(O))1)1 = (gO), h(O)).
Proof: First we observe that if W is a subspace of V generated by subspaces U and U'
78
then W' = (UU') = U N', v e (U,)' <4 v e U and
v e U"' <> v eU' r) U'. Consider (g(O),/f(O))' = g(0)' rn /(0)1. Because g I h by
Proposition 3.8.22 then there exists y,6 e P such that 0 = y6 with g c Xy and
h c X6. From Theorem 3.8.20, g(0)' = (2(0),t6(0)) and by Theorem 3.8.19,
/(O) = (1(0), Oty(0)) where I c 3E is the unique line through 0 in X orthogonal to
h. Hence,
(XO), fi(o)> = 2(0 ) ) )
= (AO), 1t6(0)> < )(0), t(0)>
= <(O),(o9)>.
Moreover,
((0), (0)>)L = g(O) n(o)1
= (g(O),.t (0)> r) <(0), t(0)>
= g(o),h(0)>.
Now if g,l c X, for some E e P then by Theorem 3.8.18, for 0 = E,
(k(O),i(O)) = t;(O) and (0(),/h(0)) = (2(O),i(0))' = .t,(0)' = .t10(0). This says
that g,h c Xp; that is, (g,h) is nonsingular. Thus, (g(0),l(0)) is nonsingular.
If (g, h) is a singular plane, 0 e g r h, and g and h are spacelike, then by
Axiom S1, g is not orthogonal to h. So by Axiom 2, for all A e g, there is a B e h
such that 0 and AOB are unjoinable. U
Theorem 3.8.24. For the above setup:
1. gOOB E (g,h}.
2. goAoB I g, h.
3. if C e g and D e h such that goCOD is isotropic then go.coD = goAOB.
4. (g(O),fih(O))' = (gooB(0),g(O))'.
Proof: For 1., if there exists y E P such that go0oB,g c Xy, then 0,A,AOBI\;. implies
that B = OAAOIy and h c Xy. This yields (g,h) = Xy is nonsingular. For 2.
79
go,AOB I g,h by Axiom S2 because by 1. above go,COD c (g,h) and (g,h) is singular.
If C e g and D e h such that go,coD is isotropic then by 1,.go,coD,go,AOB c (g,h). If
go.COD goAOB then by Axiom T there exists a unique ri E P such that
gO,COD, gO0AOB C XY.
Because two intersecting lines uniquely determine a plane, then (g,h) = Xy. Hence,
go,COD = gOAOB. The last conclusion follows from (g,h) = (g,goAOB).
If (g,h) is a singular plane then from the proof of Proposition 3.8.22 and from
Theorem 3.8.24, it follows that for each P e (g,h) there exists a unique isotropic line
I c (g,h) such that P e 1; that is, (k(O),/h(O)) contains one isotropic line.
To complete the classification of orthogonal subspaces of V, it remains to
consider hyperplanes. Now if (X,V ,C) is any four dimensional affine space then a line
and a plane which intersect in a point uniquely determine a hyperplane and any
hyperplane in the space can be characterized as the subspace generated by a
corresponding line and plane.
Let A = (h, p) be the hyperplane generated by a line h and a plane p which
intersect in a point 0. Then
A(o) fi(o), p(0))1 fi(o) (0).
From Theorem 3.8.19 and Theorem 3.8.20, we know that the dimension of h(O)' is
three. By Theorems 3.8.18, 3.8.23, and 3.8.24 it follows that the dimension of
(0(0)L) = 2. Consider the following possibilities. If h(0) r) 0(0) = {lo}, then
h(O)' 0(O)' c V has dimension five whereas the dimension of V is four. If
i(O) r 0(9)' = p(O)', then it follows that p(O)_ c hf(O); h(0) I 0(0)L and
h(o) c (p(0)=) p(0).
Which contradicts the assumption that h and p intersect only in 0.
80
Therefore, h(O) n ((O)( = k(O) for some line g containing 0. Because g must
be either isotropic or nonisotropic, then the following is true.
Theorem 3.8.25. If A(O) is a hyperplane the there is a unique line g such that
A(O) =(0) and (A(0))' = A(O). E
3.9 The Polarity
In this section a polarity is defined so that we can obtain the metric through a
process given by Baer [3]. For the convenience of the reader, the pertinent definitions
and theorems [3] are given below.
Definition 3.9.1. An autoduality 7t of the vector space V over the field IC is a
correspondence with the following properties:
1. Every subspace U of V is mapped onto a uniquely determined subspace nr(M) of V.
2. To every subspace U of V there exists one and only one subspace W of V such that
7=() = U.
3. For subspaces U and W of V, U < W if, and only if, n(W) < t(7.
In other words, an autoduality is a onetoone monotone decreasing mapping of
the totality of the subspaces of V onto the totality of the subspaces of V.
Definition 3.9.2. An autoduality 7t of the vector space (V,K) of dimension not less
than two is called a polarity, if 7t2 = 1, the identity.
Definition 3.9.3. A semibilinear form over (V,/C) is a pair consisting of an
antiautomorphism a of the field K and a function (fix, y) with the following
properties:
(i) J(x, y) is, for every x, y e V, a uniquely determined number in KC.
(ii) f(a + b, c) = (a, c) +(b, c) and J(a, b + c) = (a, b) +fa, c), for a, b, c E V.
(iii) J(tx, y) = /fx, y) and J(x, ty) = f(x, y)a(t) for x, y e V and t e 1C.
If a = 1 then f is called a bilinear form.
Definition 3.9.4. Iff is a semibilinear form over (V,/C) and if LU is a subset of V, then
81
{x e V : fix, u) = 0 for every e U} and {x e V :fJ(u, x) = 0 for every u e U} are
subspaces of V. We say that the autoduality 7t of (V, K) upon itself is represented by
the semibilinear form Ax, y) if
n(U) = {x e V :fix,U) = 0} {x e V :Afix, u) = 0 for every u e U1}.
Theorem 3.9.5.[3] Autodualities of vector spaces of dimension not less than 3 are
represented by semibilinear forms. U
Theorem 3.9.6.[3] If the semibilinear formsf and g over (V, )C) represent the same
autoduality of V, and if dim(V) > 2, then there exists a 0 # d e kC such that
g(x,y)= fJ(x,y)d for every x,y E1V. U
Definition 3.9.7. If 7t is an autoduality of the vector space (V,/C), then a subspace W
of V is called an Nsubspace of V with respect to 7r, if (v)
In this case, 7t is said to be a null system on the subspace W of V.
Theorem 3.9.8.[3] Suppose that (f, ) represents the autoduality 7 of(V, C). Then t is
a null system on the subspace W +> J(w, w) = 0 for every w e WV. U
Definition 3.9.9. A line (v) is called isotropic if (v) < t((v)). (So an isotropic line is
an Nline.)
Theorem 3.9.10.1[3] If the semibilinear form (fa) represents a polarity 7r, and if
J(w,w) = 1 for some wEV, then a2 = 1 and
a(f(x,y)) =J(y,x) for every x,y cV.
In this case we say that f is a symmetrical or just symmetrical. U
Theorem 3.9.11.[3] Suppose that 7 is an autoduality of the vector space (V, C) and
that dim() > 3. Then n is a polarity if, and only if, 7r is either a null system or else 7t
may be represented by a symmetrical semibilinear form (f, ac) with involutorial a. U
Theorem 3.9.12.[3] Suppose that the polarity 7r of the vector space (V, KC) possess
82
isotropic lines and the dim(V) > 3. Then 7 may be represented by bilinear forms if,
and only if,
(a) planes containing more than two isotropic lines are Nplanes and
(b) K: is commutative. U
Theorem 3.9.13. Suppose that t is a polarity of the vector space (V,)IQ such that the
conditions of Theorem 9.12 are met. Then from Theorem 3.9.11 it follows that if t is
not a null system then 7t may be represented by symmetrical bilinear forms. U
Defining the polarity 7t and obtaining the metric g. Consider (V,/C)
constructed in this work. If U is any subspace of V then by Theorem 3.8.17, U' is also
a subspace of V. From the end of section 3.8, if U is a subspace of V, U # {1}, and
U # V, then U' is a uniquely determined subspace of V. From Lemma 3.8.11 and
Definition 3.8.5, {1Q}' = V and V1 = {1}. So we define the mapping it on the
subspaces of V as follows:
Definition 3.9.14. If U is a subspace of V then i(bL) = UL.
From the remarks above and from Section 3.8 it is clear that to every subspace
U of V there exists one and only one subspace WV of V such that it(W) = U.
Theorem 3.9.15. Let U and WV be subspaces of V, then U
74(W) it(4).
Proof: Suppose that U < W. If OV e W41 then OV 1 OW for every OW e W, U c W,
so OV I OU for every OU e UL and OV UL'. Thus, U < W/V implies that
7t(V) = W1 < U' = 7i(4). Conversely, suppose that WV' < U. By Lemma 3.8.11 and
Definition 3.8.5 we have ({1o}I)' = {1} and (V')1 = V. From the previous section
it follows that, if U is a nontrivial proper subspace of V then (UL')' = U. because /V1
and U' are subspaces, then from the first part of the proof we have W' < UL' implies
that U = (UL')' < (W1)' = W. U
Corollary 3.9.16. From Definition 3.9.14 and Theorem 3.9.15 above it follows that
E : U < V 'r U' < V is an autoduality. Moreover, because (UL')' = UL for every
subspace U of V, then T is a polarity.
Lemma 3.9.17. Let e E 5 and OA e V. Then 7r(l,) = 7r(U) for every e 0 and for
every subspace U < V. That is, 7T is invariant under a for every 4 e S.
Proof: We have O(OA) = O09A' = B e X, so (OA) = OB e V. Hence, from Lemma
3.8.6.the result follows. U
Theorem 3.9.18. The polarity 7 is not a null system.
Proof: Let a be any nonisotropic line with 0 e a. By Lemma 3.8.7, a is not
orthogonal to itself so that d(0) : d(0)' and hence, d(0) % n(d(0)). By Definition
3.9.7, 7r is not a null system. U
Theorem 3.9.19. The polarity 7t defined above may be represented by bilinear forms.
Proof:. Let g be any isotropic line through 0. By Definition 3.8.5, g(0) < g(O)' so
that (O) < n(7rg(0)) and k(O) is an isotropic line in the sense of Definition 3.9.9.
Conversely, if h is a line such that h(0) < nt(h(O)) = h(O)' then h I h and by Lemma
3.8.7, h must be isotropic in the sense of Section 3.2. Thus, (V,/C) possesses isotropic
lines. By Proposition 3.7.3, dimni(V) = 4 > 3. From Definition 3.8.21 any plane in
(V,/C) is singular or nonsingular. Now a singular plane contains only one isotropic line
by Theorem 3.8.24. If U < V is a nonsingular plane then U = .ty(O) for some y e P. If
y e M then U has precisely two isotropic lines, as was shown in 3.2.2.8. If y e,
then by Axiom 12 and Section 3.2, U does not have isotropic lines. Thus, no plane of
V contains more than two isotropic lines. Since the field AC = R is commutative then
by Theorem 3.8.12 the result follows. U
Theorem 3.9.20. The polarity 7 may be represented by symmetrical bilinear forms;
that is, there is a symmetrical bilinear form g such that for any subspace U of V,
UL = r(M) = {OA e V : g(OA,U) = 0} or
g(OA,OB) = 0 = g(OB,OA)if, and only if, OA 1 OB for OA,OB e V.
Proof: This follows directly from Theorems 3.9.18, 3.9.19. and 3.9.13. U
84
Thus, we have a metric g, a symmetric bilinear form, induced by our polarity,
which agrees with our definition of orthogonal vectors, which in turn is induced by
and is defined in terms of the commutation relations of the elements of our
generating set g.
Lemma 3.9.21. Let g be a symmetric bilinear form representing 71. Then g is
nondegenerate.
Proof: This follows from the fact that ic(V) = V1' = {1}. U
Theorem 3.9.22. Let g be a symmetric bilinear form representing T. Then there is a
basis of V such that the matrix of g with respect to this basis has the form
SC 0 0 0
0 C 0 0
r #, 0 Ce IC;
0 0 C 0
0 0 0 C
that is, there is an orthogonal basis of V such that the matrix of g with respect to
this basis has the above form.
Proof: Let a e M such that kC c Xa and let 0 = ac3. Let Oy7,5 e g such that
a = y6 and y,6 13. Let s = yp e M and 11 = 63 e AM.Then 8Tl = y7pp3138 = y = a so
6 1 1. Put x = [ay,5], y = [3,y, ], z = [13,r,], and t = [a, r1].
Claim. x,y,z,t are four mutually orthogonal nonisotropic lines through 0. We
have 0 = ap3, 0 = x,t c Xa and 0 e y,z c Xp so that x,t I y,z. Also,
0 = a3 = y6p3 = yr1 = 6c so that y I il and 6 1 E. Then
0 e x c Xy,0 e t c X^,O e Xy, and 0 e z c 3C,
implies that x I t and y I z.
The construction of the basis. Let E E /C, the multiplicative identity,
considered as a point in X. Put T = EOEt E t and X = EOEX x. Note that because
85
0 e x,t c X3C and x I t then axlt = cro = CaT(x in X3a. Now t c 3E and e M, so
there exist precisely two isotropic lines, /Ci, and /C2, through 0 in XC. Thus, there is
an unique E E /Ci1 such that T = EgOE. Because y c XC then Y EOEy E y and
ryCat = ao = atoy in XC. Similarly, because t c X1; M1 E A4, there are precisely two
isotropic lines, ACIn and /C2, through 0 in Xi and there is Eq in /CIn such that
T = EnOEt. because z c X then Z = EnOEt e z and aCzat = To = ctraz in Xn. We
calculate XOT = (EOEX)O(EOE') = (EOE)O(EX OE) = (EOE)O(E0tO'Et)
= (EOE)0(EOE)' = (EOE)0(OEOOE)t = (EOE)O0t
= (EOE)OO = EOE e 1C.
YOT = (EgOE0)O(E9OEt) = (EcOE,)O(EOEE)t = EgOEg e KC1.
ZOT = (E OE)O(EqOEt) = EqOE9 e K l. Thus, if we put E1 = OX, E2 = OY,
E3 = OZ, and R4 = OT it follows that the set {E1,R2,ER3,E4} consists of four mutually
orthogonal vectors such that Ei + 4 is isotropic for i = 1,2,3. So if g is a symmetric
bilinear form representing 7t then from Ei + E4 L Ei + R4 for i = 1,2,3 and Ei IL Ej for
i j it follows that for i = 1,2,3,
O = g(Ei +R4, Ei + E4) = g(Efi, Ei) + 2g(Ei, 4) + g(!4, 4) = g(Ri) + g(4,/4).
So that g(E,,Ei) = g(04,94) 0, because E4 is nonisotropic. Thus, it remains to
show that {V],/2,E3,/4} is a basis for V. But this follows from Section 3.7. U
Theorem 3.9.23. (V,/C,g) is a fourdimensional Minkowski vector space. Moreover,
(Iy(O),g) is a fourdimensional Minkowski space for every y in P0.
. . . "4
Proof: Put g(E4,E4) = 1 and g(Ei,, E) = 1, for i = 1,2,3. Minkowski space is the only
nonsingular real fourdimensional vector space with metric
1 00 00
S0 1 0 0 (3.24)
0 001 0
0 0 0 1
86
In the last section of Chapter 3 we show that each X in G can be identified with
a spacelike plane and each &k e g* with a reflection about a spacelike plane.
3.10 Spacelike Planes and Their Relections
For each X E Q with 0\1 define a5X V V by a(OA) = (OA) OAX, for
OA e V. To extend this definition to any X e g, we note that if X f 0 and OA e V,
then there exists a unique D in X such that O0XAX = D; that is, there is a unique D
in X such that (OA)X = OXAX = OD. Thus, we define 6(OA) = OD, where
O0XAx = D. We note that if k 10 then D = AX.
First we show that &k is a semilinear automorphism of V for each X, E V. [23]
Now, a function f : V ', V is a semilinear automorphism if, and only if, it has the
following properties:
1. f is an automorphism of the additive group of V onto itself.
2. f sends one dimensional subspaces of V onto one dimensional subspaces (that is, f
is a collineation).
3. If A and B are linearly independent vectors of V, the vectors J(A) and J(B) are
also linearly independent.
Theorem 3.10.1. The map & is an automorphism of the additive group of V onto
itself
Proof: Suppose that 6(OA) = 6&(OB). Then 6&(OA) = OD where
OD = 0'A = OXBX. This implies that AX = BX or A = B. Thus, OA = OB and 6& is
injective. To show that 6X is onto, let OB e V and A = 00B. Then
&(OA) = OA' = O(OOBX)k = OkOX OB = OB. We show &x is additive. Let
OA,OB e V. Let D = OOXAk, F = OOBX. Then
6&(OA + OB) = (OAOB) = OAXOBX0 = ODOF = 6&(OA) + 6X(OB). U
Lemma 3.10.2. The map &X sends one dimensional subspaces of V onto one
dimensional subspaces of V.
87
Proof: Let < OA > be a one dimensional subspace of V. Then there exists a line
g = [a,P,7y] such that OB e< OA > if, and only if, O,B e g. Since
O,B I a,P3,y7 < OA,BX  aI W, y7 then &x(< OA >) =< &x(OA) > is a one
dimensional subspace of V. N
Lemma 3.10.3. The transformation, &k, maps linearly independent vectors of V to
linearly independent vectors.
Proof: The vectors OA, OB e V are linearly independent if, and only if, goA # gOB if,
and only if, gO9Ax # g9OB;. Hence, &X is a semilinear automorphism of V for each
2 e G. 0
Theorem 3.10.4. The map Csk : V '* V is a linear automorphism of V onto V.
Proof: Consider the definition of a semilinear automorphism [23,defn.73.1]. Let (V,K)
and (V',/C') be vector spaces over the division rings )C and KC', respectively. Suppose
that pt : /C /C' is an isomorphism from /C onto KC'. A map X : V 'k V' is called
semilinear with respect to p. if
1. (A + B) = X(A) + X(B) for all A,B e V.
2. X(tA) = p(t)X(A) for all A e V and t e /C.
The only isomorphism p. : P '+ JR is the identity. Thus, &X is a linear
automorphism of V onto V. N
Let (V, KC,g) be a metric vector space. A similarity y of V is a linear
automorphism of V for which there exists a nonzero r e K; such that
g(yA4,yA)Thsalrri
g(yA,yB) = rg(A,B), for all A,B E V. If A is nonisotropic, r = yAA). The scalar r is
called the square ratio of the similarity.
Lemma 3.10.5. [23] A linear automorphism of a metric vector space V is a similarity
if, and only if, it preserves orthogonality.
Proof: For all lines h and I in our space, h L / <> hx I lk, hence,
J(OA) 1 &(OB) <> OA OB. U
88
Theorem 3.10.6. The linear automrnorphism %z is an isominetry.
Proof: Let y : V i V be a similarity with square ratio r # 0. Then for A,B e V we
have g(yA,yB) = rg(A,B) = rg(y1 (yA),y 1 (yB)) and g( (yA),y (yB)) g(YAYB).
That is, yI has square ratio r1. Now 6i is an involution, so that &k = ftl. Hence,
if r 0 is the square ratio of 6x then r = orr2 = 1, so r = 1. [231 Because the
field C is isomorphic to R, then V is not an Artinian space. Thus, for each similarity
& of V there is an unique r > 0 and there is an unique isometry a such that
6( = M(0,r); where M(0,r)(A) = rA for all A in V. The map 6. is thus a similarity
with square ratio r2. Because r > 0, then from above, r = 1 and it follows that 6& is
an isometry. U
Proposition 3.10.7. The isometry 6& is a 180 rotation; that is, a reflection about a
plane (a twodimensional subspace of V).
Proof: Let ; E g. Suppose that 0\2k and put 0 = ka with c e M. Then for all AIk
and for all Ba, 6(OA) = (OA)k = O6AA = OA and
6(OB) = (OB) = O= B = OB"I = OB = OB.
Since 1,a(0) = .tx(0)' and V = J(O) I(O)' then ey = 1tx(o) ljt(o). Thus,
&z is a reflection about the plane .tX(O).
Suppose that 0 X. Let 6 I 0 such that c I1 k and let P I X be arbitrary.
Then if R I ,, POR= Q I X and OR = PQ. Thus,
&a(OR) = (OR); = (PQ)k = PQ = OR and x 1(o) = lt(o).
Now let 310 such that P3 1 c, so 0 = c3. Let B I P3. Now P3 1 X because 11 X. and
POB = DZy where P = ky and y 113. We calculate,
6(OB) = (OB) = (PD)^ = (PD)Y = (PD)p = DP = BO = OB = OB.
Hence, t = lt (o) ~1t(o)0 and &k is a reflection about a plane. U
89
To show that 6. is a reflection about a spacelike plane (Euclidean plane) for
every X e G, we use the following theorem from Snapper and Troyer [231.
Theorem 3.10.8. [231 Let (V, C5,g) be an ndimensional metric vector space over a
field k1 with metric g, IC = R, and n = 2. Every nonsingular real plane has a
coordinate system such that the matrix of its metric is one of the following.
S0 the Euclidean plane, ( the Lorentz plane, and
( the negative Euclidean plane. U
Hence, the Euclidean plane, the Lorentz plane, and the negative Euclidean
plane are the only nonisometric, nonsingular real planes. This also follows from
Sylvester's Theorem [23] (It states that there are precisely n+1 nonisometric,
nonsingular spaces of dimension n).
Lemma 3.10.9. Let kC = R, n = 4, and V be Minkowski space. Then the orthogonal
complement of a Lorentz plane is a Euclidean plane.
Proof: Let {eI},i, ..4 be a basis for V such that the metric of V with respect to this
basis has matrix of the form (3.24). Let m = e3 + e4 and m = e3 e4 so that
< m >,< n > are the unique isotropic lines in the plane < e3,e4 >. Let a be a Lorentz
plane with isotropic basis m I and n I. Then there is an isometry a :< e3,e4 >* a
such that o(m) = m 1 and 0(n) = n1. By the Witt Theorem [23] c can be extended to
an isometry of V, which we also denote by a. This implies that
Ca : V =< e3,e4 > G < ei,e2 >F+ V = oc D< a(ei),a(e2) >;
that is, {a(ei),a(e2)} is a basis for a1. Now a is an isometry and
g(el,el) = g(e2,e2) = I, so the metric g with respect to a1 has matrix a2, the 2x2
identity matrix, and a1 is a Euclidean plane. M
90
Theorem 3.10.10. The map 6& is a reflection about a spacelike (Euclidean) plane for
every X e G.
Proof: From the proof of Proposition 3.10.7 it suffices to consider e g with 0[X.
Let 0 = OX with 0 e M and let Q(.) denote the quadratic form associated to the
metric g obtained in Section 3.9. Because 0 e M there are precisely two isotropic
lines /C,01, KC02 c XO. Let 0 A E /C0, and 0 B e KC02. Then OA and OB are
isotropic vectors which form a basis for t0(O) and Q(OA) = 0 = Q(OB). Therefore the
metric of 10o(0) with respect to OA and OB has matrix
9 = ( A OA OB) 0 r *
S OB r 0(
where the products of the matrix elements are the inner products defined by the
metric g and g(OA,OB) = r # 0. Since  OA E KCo,, and OB e C02 then k0,, K02
also form an isotropic basis for .o0(O). Hence we may assume that r = 1. Let
OTI (OA OB) and OXI =  (OA + OB).
Then g'(OXI, OTI) = 0, Q'(OXI) = I, and Q'(OTi) = 1. Because OX1 I OTI and
10(0) is nonsingular then OXI and OT, are linearly independent and hence, form a
basis for 10(0). Moreover, the metric of e(0) with respect to OXIand OT, has the
matrix Therefore, by Theorem 3.10.8, l0(O) is a Lorentz plane and by
Lemma 3.10.9, t() = t() is a Euclidean plane. 1
Lemma 3.10.9, I.(O) = to(O)L is a Euclidean plane. N
CHAPTER 4
AN EXAMPLE OF THE THREEDIMENSIONAL MODEL
This chapter begins by considering a net of von Neumann algebras, {TZ(0)}o I,
and a state o), coming from a finite component Wightman quantum field theory in
threedimensional Minkowski space. There are various senses to the phrase "coming
from a Wightman quantum field theory". The assumption here is the version given by
Bisognano and Wichmann [5]. That is, given a finite component Wightman quantum
field, ((x), assume that the quantum field operator, 0(t), is essentially selfadjoint and
its closure is affiliated with the algebra R(0) (in the sense of von Neumann algebras)
for every test function f whose support lies in the spacetime region 0. Driessler,
Summers, and Wichmann show these conditions can be weakened [15]. But free boson
field theories satisfy these conditions in threedimensional Minkowski space [5].
For such theories the modular involutions, Jo, associated by TomitaTakesaki
theory to the vacuum state and local algebras of wedgelike regions, 0, in three
dimensional Minkowski space, act like reflections about the spacelike edge of the
wedge [5]. Since the modular involutions have that action upon the net, the
hypotheses of Buchholz, Dreyer, Florig, and Summers (BDFS) are satisfied [6].
Therefore the Condition of Geometric Modular Action, CGMA, obtains for the set of
wedgelike regions [271 in Minkowski space. The precise wording of this version of the
CGMA is given below.
Let 1i and 12 be two lightlike linearly independent vectors belonging to the
forward light cone in threedimensional Minkowski space. The wedges are defined as
the subsets W[l1,12] {a/cij + f12 +1 R1'2 : C > 0, p3 < 0, (1L, li) = 0, i = 1,2},
where (, ) denotes the Minkowski inner product.
91
92
Let II = (1,1,0) and 12 = (1,1,0) be lightlike vectors in R1,2, threedimensional
Minkowski space, and let P' be the Poincar6 group, the isometry group of this space.
Then the set of wedges, W, is given by W = {XKW[l1,12] : X e P}, where
X .W[l,12] = {.(x) :x E W4[112]}.
The CGMA for Minkowski space is defined as follows. Let {7Z(W)} ww be a net
of von Neiiiniiiii algebras acting on a Hilbert space 'H7 with common cyclic and
separating vector Q 7E satisfying the abstract version of the CGMA and where the
index set I is chosen to be the collection of wedgelike regions W in R1'2 defined as
above. Recall from Chapter 1, with ({1Z(W)} weW,',R,) there is the following.
1. A collection of modular involutions {Jw} wew.
2. The group J generated by {Jw} w&w.
3. A collection of involutory transformations on W, {fw} ww.
4. The group Tgenerated by {Tw}wEw.
Assume also that:
5. The group Tacts transitively upon the set W, that is, for every W1, W2 e W there
is a W3 e W such that Tw3(WI) = W2.
Note that this assumption is implied by the algebraic condition that the set
{adJw} WW acts transitively upon the net {R(W)} wEw. At this point the following
two assumptions are added (127] which have been verified for general Wightman
fields).
4.6 For W1, W2 E W, if W\ n WK2 # 0, then Q is cyclic and separating for
Z(W) n 7IZ(W2).
4.7 For W1, W2 E W, if Q is cyclic and separating for 1Z(W,) r 7Z(W2), then
W, r W2 0.
The CGMA for Minkowski space is the abstract version of the CGMA with the
choice of A' for the index set I, together with assumptions 4.6 and 4.7 above less the
transitivity assumption [6].
93
Buchholz, Dreyer, Florig, and Summers [6] showed that with the above
assumptions one can construct a subgroup T3 of the Poincare group P, which is
isomorphic to Tand related to the group Tin the following way. For each T E Tthere
exists an element g, e T such that r(W) = gW = {g,(x) : x e W}. To each of the
defining involutions Tw e T, W e W, there is a unique corresponding gw e q3c P [6].
Moreover, BDFS obtained the following (suitably modified for three dimensions and
abbreviated for our purposes).
Theorem 4.1 [6] Let the group Tact transitively upon the set W of wedges in R1,2,
and let 3 be the corresponding subgroup of P. Moreover, let gw be the corresponding
involutive element of P corresponding to the involution 'rw e T Then gw is a
reflection about the spacelike orthogonal line which forms the edge of the wedge W. In
particular, one has gwW = W,' the causal complement of W, for every W e W. In
addition, T exactly equals the proper Poincare group P+. U
Recall from Chapter 2 that the initial model of (9, 0) is as a group plane. This
means that each g e g is viewed as a line in a plane and each P = gh, gjh, as a point
in a plane. Let us call the axiom system given in Chapter 2 as A. Thus as A is a
set of axioms about "points" and "lines" in a "plane".
Let P denote the collection of points P e 0. For each a E 0 define the map
ra : Pug +P u by
ra(P) P aPalI for P e P and ca(g) = g a agaI for g e g.
Since (Q, 0) is an invariant system then each cra is a bijective mapping of the set of
points and the set of lines, each onto itself, which preserves the incidence and
orthogonality relations, defined by "I", of the plane. We say that aF is a motion of the
group plane. Since g generates 0 then the set of line reflections g, = {ag : g e g}
generates the group of motions 0O = {ra : a E 5}. Let D : 0 +> Oa be the map
defined by D(a) = ac,, for a E 6. Then 0 is in fact a group isomorphism [2]. This
94
means that (9, 0) is isomorphic to (G9, 0a) (in the sense that g is equivalent to G9 as
sets and ((G) = G, (D(O) = 0o, where D is a group isomorphism). This implies that
X(as A), which we denote by as D(A4), is an axiom system concerning the group of
motions; line reflections of a plane, the group it generates, and point reflections of a
plane. A plane whose points and lines satisfy as A.
As was shown in Chapter 2, given (9, 0) satisfying as A, one obtains IR 12,
threedimensional Minkowski space. Under the identifications given in Chapter 2, we
find that each g e g corresponds to a spacelike line in R1,2. Thus, as (D(A) is a set
of true statements concerning reflections about spacelike lines and the motions such
reflections generate in threedimensional Minkowski space. Moreover, since such
motions are in fact isometrics in R1'2 then D(O) is isomorphic to a subgroup of the
threedimensional Poincare group.
Theorem 4.2 Under the same conditions as in Theorem 4.1 it follows that
({tw} wew,T) acting on W satisfies as D(A).
Proof: From Theorem 4.1 we have ({gw}w)w, W3) satisfies as D(A) since 3 is the
subgroup of T generated by reflections about spacelike lines. Also from Theorem 4.1,
({tw} wew,T) is isomorphic to ({gw} ww,T3) so ({tw} ww,7) satisfies as0(A). N
The net continuity condition assumed by BDFS [6] for the next theorem was
later shown to be superfluous [81 for this theorem and the remaining theorems.
Theorem 4.3 [6] Assume the CGMA with the spacetimne R 1,2 and W the described set
of wedges. If J acts transitively upon the set {f(R W)} ww then there exists a strongly
(anti) continuous unitary representation U(P +) of the proper Poincare group which
acts geometrically correctly upon the net {1(W)} wew and which satisfies U(gw) = Jw,
for every W e W. Moreover, U(Pt) equals the subgroup of j consisting of all products
of even numbers ofJw 's and J = U(Pt) u Jw U(Pt), where
WR = {x E R12 : xI > Ixo}.
95
Theorem 4.4 Under the same hypotheses as Theorem 4.3, the group J is isomorphic
to P + = T, which is generated by the set of involutions {gw I W e W}. Moreover,
(.{Jw } ww, J ) satisfies as I(A4).
Proof: By Proposition 1.1 there is surjective homomorphism : J T, where the
kernel of 4, ker(4), is contained in the center of J, Z(J). By Theorem 4.3 there is a
faithful representation U(P+) such that U(gw) = Jw, for every W e W. Since the
center of PT is trivial, U(.) is a faithful representation of P+ and hence an injective
map preserving the algebraic relations, Z(J) is trivial. This implies that ker(4)= {1}
and 4 is an isomorphism. If T : T> 3 denotes the isomorphism of T and 3 given by
BDFS from Theorem 4.1 then T o 4 : J + 3 = P+ is an isomorphism. It therefore
follows that the pair ({Jw} wewV, J) satisfies as ((A). E
We can now give the main result of this chapter.
Theorem 4.6 Any state and net of von Neumann algebras, coming from a (finite
component) Wightman quantum field in threedimensional Minkowski space, which
satisfies the Wightman axioms, provides a set of modular involutions i,,ti.v'iug
as S(A). U
As a final remark we note that since free boson field theories satisfy the
Wightman axioms and therefore the CGMA for Minkowski space holds, then these
theories give a concrete example of the threedimensional case of this dissertation.

Full Text 
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 (0, Q), are
called metric planes.
BPWB showed how one can embed a metric plane into a projectivemetric
plane by constructing an ideal plane using pencils of lines [2], We shall now outline
how this is done.
Definition 2.2.6. Three lines are said to lie in a pencil if their product is a line; that
is, a,b,c lie in a pencil if
abe = d e Q. (*)
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)~l is a line, the invariance of Q
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 Q also
shows that (*) holds whenever at least two of the three lines coincide.
Using the given axioms, BPWB [2] showed that there are three distinct classes
of pencils. If a,b\V then G{ab) = {c : c \ V}. In this case, G(ab) is called a pencil of
lines with center V and is denoted by G(V). If a,b c then G(ab) = {d : d c}. 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  a,b where a  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, II, is constructed in the following way. An
53
Gg e Ca is involutory then l.ta e Ua(0). Now let Ga,Gb,Gc, a(0) where
a = [a,a',aa'], b = [a,p',ap'], c = [a,y',aY'], and d = [a,8,a5] are nonisotropic.
Now 8y'p' = 8 X a with 0\e by Axiom 9 and/ = [a,e,as] e Oa. Thus, for A a,
GaGftGcG^X) = XÂ§Y P a = Aea = GaG/(X).
Hence, OaO^OcG^ = ao/ e Va(0). Because each G/ is involutory for / e Ca,
071 = 0/ and (g0g,)1 = GGa e T>a(0). From Axiom 9 and the calculation above,
the product of any three of 8,Y,,pi, and a' is an involution.Thus,
^Sy'p'a ^p'y'Sa' ^p'a'Sy' an(j aaabGcad = GcGd
Clearly, Va(0) is associative so that Va(0) is indeed an abelian group.
Lemma 3.5.2. Let g be an isotropic line in Xa with O e g. Then for every
G,Gh e Va(0), a,Gh(g) = g.
Proof: Let / = [a,p,aP] and h = [a,y,ay] be lines in Ca with O e /,/?. By 3.2.3.12, if
0\ p,y with p,Y T a then gPy  g. But = O so that G/,G/(0) = C>PY = O. Thus,
gPv = g by 3.2.1.10. Hence, G/,g/(C) = g, for all 0/,0/ e Va(0).
Lemma 3.5.3. Let g be an isotropic line in Xa through O. Then for every A, E e g,
E O, A O, there is a unique o/o/, e Va(0) such that g/G/,(Â£) = A.
Proof: By 3.2.3.11, there exist y,8 X a with 0y,8 such that X78 = A. Take
/ = [a,y,ay] and h = [a,8,a5]. Note that if E = A then X78 = E implies that EA = X8
or E1 = Eh, which implies that l = h [30]. To show uniqueness, suppose that
Gac/GkGi e T>a(0), where / = [a,8,aS], k = [a,8,as], g = [a,y,ay], and
h = [a,p.ap] are lines in Xa(0) and E Eha = Elk. Then Ehakl = E and E^ = E.
By Axiom 9, Pys = X with 0\X, X X a. Thus, m = [a,X,a>.] e Oa;GkGaGh = Vm, and
E = Eml. Let Xa',8' with a'XA., S;X8, and put M = a'X and N = 8'8. Now
EI a',a,8' with a'iA, a X X, 8' X 8, and a X 8, so that a',81 X a by 3.1.6.18. It
follows that AT = a'aXa = a'X = Man N = 8'8a = N So that M\a,X ; Aa,S;
and m go.\i and / = goN Then because Emi = E, we have EK& = E; El = Xs and
32
3.1.2.3. For each Â£ e , = V.
Proof: because V = Q u M, the result follows from the invariance of Q and M..
3.1.3. Properties of X.
3.1.3.1. There exists a point; that is, X 0.
Proof: Let a e Q. By Axiom 11, there exists y,p e Q such that a,p,y are distinct
and mutually perpendicular. By Axiom 4, aPy = a8 Â£ G, where 8 = Py e A4 and
a8 as ctp,y. Thus, a 8 and P = a8 is a point.
3.1.3.2. IfOa,p,y; a,p,y e Q\ a 1 P 1 y 1 a; then O = aPy.
Proof: By the proof of 3.1.3.1 above, P = aPy is a point and 5 = Py e M. because
P _L y with p,y e Q. So we have a 8, 08 because 0p,y. This yields A,0\a,8
with A a8 so that A = O by Axiom 3.
3.1.3.3. If A e X and 8 e V, then A 8; that is, a point does not equal a plane.
Proof: Let A ap with a e Q, p e M, and a ip. Suppose that A aP = 8 e V. If
8 e M. then we have a = PS e Q with P18 and p,8 e A4, which contradicts Axiom
5. If 8 e Q, then A = ap e Q and this contradicts the definition of a point.
3.1.3.4. The elements ofX are involutory.
Proof: Let A e X, so that we may write A aÂ¡5 with aip. In particular, ap = pa
and AA = apap = aapp = 1.
3.1.4. General Consequences of the Axioms
3.1.4.1. IfP a,p anda A. P then P = aP and ifP = ap then Pa,p and a L p.
Proof: If Pa,p and a_Lp then aP A for some A e X and we have A,Pa,p with
ai p. Thus, by Axiom 3, A = P.
If P ap, then by 3.1.3.4 ap. because P V by 3.1.3.3 then aip. Also
Pa. P and PP = a imply that the products Pa and Pp are involutory so that
Pa,p.
3.1.4.2. IfP  a then Pa e V; that is, Pa is a plane P and P\p.
34
or a L y ( note that a = y implies that a X P because A py and A a). Suppose that
a Xy so that B = ay. Then we have A,B a,y with a Xy, which implies that A = B by
Axiom 3. But if A = B then Py = ay and p = a, which contradicts pla. Thus, ay is
a plane and aXy.
3.1.5.2. Suppose that a T P; ^a,p,y,5; yXa; and 8Xp. Then 5 1 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 ya = 8p and 8y = ap e V as a JL p. Thus, 5 1 y.
3.1.5.3. If O = aaj =yyi = esj with a,y,s e M anda lylsla then ay = s.
Proof: Because a,y,e e M, then ai,yi,8 e Q. Because alyltia and
aa i = yy i = eei, then ya = y i a i; sa = sjai; ey = s i y i imply that
yi 1 ai 1 E i yi. Because 0ai,yi,8i, then O = aiyjS] by 3.1.3.2. Because points
are involutions by 3.1.3.4, l
3.1.6. Parallel Planes.
We say that two planes a and P are paralleldenoted by a  p, if a = p, or
there exists a y such that a, pXy.
3.1.6.1. Parallelism is an equivalence relation on the set of planes V.
Proof: That the relation is reflexive and symmetric is clear. For transitivity suppose
that a  P and P  y where a.p, and y are distinct. Then there exists 8,e e V such
that a, pX8 and p,yXe. Let A = a8, B P8, and C = pe. Then by Axiom 8 we have
D = ABC = a88ppe = ae and aXs so that a  y.
3.1.6.2 //a, pX8 and pXe then aXe.
3.1.6.3. If aXy, pX8, and y  8 then a  p. (Two planes absolutely perpendicular to
two parallel planes are parallel.)
Proof: If y = 8 then we have a. pX8 and the result follows. So assume that there is
an e such that y,8Xe. Then by 3.1.6.2 above, aX8, pXy, a,pXs, and a  p.
CHAPTER 2
A CONSTRUCTION OF THREEDIMENSIONAL MINKOWSKI SPACE
In this chapter we give an absolute geometric, that is, an algebraic,
characterization of threedimensional Minkowski space. This chapter is a version of a
preprint by the author entitled A GroupTheoretic Construction Of Minkowski
3Space Out Of The Plane At Infinity [28]. Along with the wellknown
mathematical motivations [1, 2] there are also physical motivations, as we discussed
in Chapter 1. Threedimensional Minkowski space is an affine space whose plane at
infinity is a hyperbolic projectivemetric plane [12]. In Absolute Geometry [2],
Bachmann, Pejas, Wolff, and Bauer (BPWB) took an abstract group 0 generated
by an invariant system Q of generators in which each of the generators was
involutory, satisfying a set of axioms and constructed a hyperbolic projectivemetric
plane in which the given group 0 was isomorphic to a subgroup of the group of
congruent transformations (motions) of the projectivemetric plane. By interpreting
the elements of Q as line reflections in a hyperbolic plane, BPWB showed that the
hyperbolic projectivemetric plane could be generated by these line reflections in
such a way that these line reflections form a subgroup of the motions group of the
projectivemetric plane.
Coxeter showed in [13] that every motion of the hyperbolic plane is generated
by a suitable product of orthogonal line reflections, where an orthogonal line
reflection is defined as a harmonic homology with center exterior point and axis the
given ordinary line and where the center and axis are a polepolar pair. Here we
show that Coxeters and BPWBs notions of motions coincide in the hyperbolic
7
72
Lemma 3.8.9. If a e V then for each P a and for each nonisotropic line g c Xn
there is a unique nonisotropic line h c Xa such that P e h and h T g.
Proof: This follows directly from 3.3.8 and Lemma 3.8.8.
Lemma 3.8.10. Ifg,h c Xa fora e V and g,h are nonisotropic, then g h if and
only if GgGh = GhGg lXa
Proof: This follows from 3.3.16 and Lemma 3.8.8.
Lemma 3.8.11. Suppose that O e c,d\ c,d
and c 1 d. Let C e c and D e d, then go,DOC Is not orthogonal to c if C\D ^ O.
Proof: First note that go.DOC c ov d because then we would have DOC = D\ e d,
say, so that C = ODD\ e d and O e d implies that C = O or c = d. Now if
go.DOC L c, then in 3Ea we have Go = Gcd = ^gojxx ^c, which implies that
d = go,doc
Lemma 3.8.12. Let g,x c with g isotropic, a e V, andg D x {O}. Then g is
not orthogonal to x if g jc.
Proof: If x is isotropic and x T g then because x g, there exists y,5 6 V such that
g a Xy, x c Tg, and y!8. But then one of y and 8 must lie in Q. But no element in
Q can contain an isotropic line so jc is not orthogonal to g.
Suppose that x is nonisotropic and let O e g n jc. Because g is isotropic then
a e M. and by 3.3.8, there is a unique nonisotropic line h c Xa such that O e h and
h Lx. Suppose that glr and let O^Keg, O^Heh, and O X e jc. Then OH
1 OX and OK 1 OX so that by Axiom U, O(HOK) = OHOK 1 OX. If go.HOK is
nonisotropic then go,HOK = h. And because go.HOK <= Toe, then HOK = H\ eh. Then
we have K = OHH] 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 K\ e g. It follows that H = K\KO e g and g = h. If go.HOK g then go,HOK
is the other isotropic line through O in Because g and go.HOK span Xa and
the result of Bisognano and Wichmann [5], the modular conjugations of the wedge
algebras act as reflections. These properties are encoded in condition II.2 and II.3.
A physical framework will now be given as a description of quantum field
theories in terms of local nets of algebras [17]. The basic assumptions are the
following. Let {A(0)}oeV <= B(H) be a net of von Neumann algebras indexed by the
closed double cones T> in R1,2 which satisfy the following properties:
1. (Isotony) If 0\ a O2, then A(0\) a A(02).
2. (Locality) If 0\ 2)'
Where A(Q)' denotes the commutant of A(O) in B(H) and O' is the causal
complement of O c M12.
3. (Poincar covariance) There is a unitary representation
U : SO^(2,1) > M12 > UifH) of the Poincar group with positive energy.
4. (Vacuum vector) There is a unique U invariant vector Q e H.
The algebra A(0), the inductive limit of the net, is called the local algebra of
observables localized in O c R1,2.
As was mentioned in Chapter 4, if the local net is generated by Wightman
fields then the modular groups associated with algebras of observables localized in
wedges act as Lorentz boosts in the directions of different wedges and the modular
conjugations act as reflections [5].
In particular, the adjoint action of the modular conjugations on the net act as
reflections about the spacelike edge of the wedge. For what follows we shall call the
properties in the previous paragraph the Bisognano and Wichmann property.
Theorem 5.6 [29] Let A(0), O c: R12, be a local net fulfilling the Bisognano and
Wichmann property for wedges. Let M = A(flT/ ,/2]), M = A(W[li,/3]),
C s AMhM), andM= A{W[lul2l\\), where l\di, and13 are three linearly
independent light rays. Then this set of algebras together with the vacuum vector Q
fulfill the assumptions of Theorem 5.4. Conversely, let hi, M, C, and if be a set of A
99
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 /C.
3.6. Dilations and the Construction of (T, V,/C)
The additive group V and dilations. First we construct a vector space V over
the field /C. 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 = O(OAB) = OOAB. We view the elements
OX e V as directed line segments with initial point O and terminal point X on the
line gox We define an addition on V by setting OX + OY = 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 1 since 1 = OO e V and
1 + OX = 1 & OX = OX = OX+ 1. To complete the proof we calculate
OX+ OX = OXOX = OXOOXO = OXXO OO = 1, so OX = OX.
(OX+ OY) + OZ = (OXOY)OZ = OX(OYOZ) = OX + (OY+ OZ).
Hence, (V ,+) is an abelian group.
We still need to define a scalar multiplication of /C 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. [23, p.37]
Theorem 3.6.2. [23, p.42] A dilation 5 is completely determined by the images of
two points.
52
3.4.2. For every A and B. there exists M 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,Ba. From Section 3.3, there
exists M a such that AM = B.
3.4.3. If A A1 = BB1 then A and A' are joinable precisely when B and t are.
Proof: Suppose that A and A' are joinable and let A,A1 a,a/ with a J. a'. By 3.4.2,
there exists an Msuch that AM = B. Then B = AM \aM,a'M and from 3.1.2.1 and
3.1.6.13 it follows that aM T a'M, a  aM, and a'  a'u. By 3.1.6.11,
AA1 BB'; A,A1 a,a;; B\aM,a'M; a  aw; and a  a'M.
Thus, t \ and, B and Bf are joinable.
3.4.4. If a 1 a'; a  p; a'  p'; and [ a,a']  [p,p'], then p 1 p'.
Proof: Because [a,a ]  [p,p ], then there exist A,A1; A,A1 a,a/ and there exist
B.B1; B,B' pp,) such that AA' = BB'. Then for AM = B. as in the proof of 3.4.3,
[p,p ] = gBB' = [au,a'u]  [a,a']. That is, 5aM,a,A/,p,p'; with P,aM  a, and
P ,a w  a'.Thus, P = aM and P' = a'M by 3.1.6.12. Hence, P T P' because
aMla'M. U
3.5. Construction of the Field
The basic construction. For the construction of the field we will follow the
path of Lingenberg [20]. Throughout this section let a e M and 0a be fixed.
Define the sets:
Oa = {g e Ca : O e g} and Va(0) s : g,h e Oa).
Proposition 3.5.1 The set T>a(0), acting on the points ofXa, is an abelian group.
Proof: By 3.1.7.2 we may write O = ap = yrp with a = yr; y,rpP e Q\ and y,ri, and
P mutually orthogonal. Thus, g = [a,y,r] e Ca; O e g, and Oa 0. Because each
45
a = aY = eyty. It follows that ag(h) = [a,Â£Y,rY] e Â£a. If^s,r) then ^4Y  sY,rY so
ag(A) e og(h) for all A eh and ag is a welldefined collineation.
If l = [a,A.,p] e Â£a and cg(h) = ag(l) then for every A in h we have
ag(A) = Ay e og(t) = [a,X.Y,pY].
This implies that A = (Ay)y e [a,A,,p] = /;that is, h = l. Hence, ag : Ca Â£a is
injective. Because Â£_Lr<sY_LrY then it follows that
h = [a,s,r}] e Ca <> csg(b) = [a,EY,riY] e Â£
Hence, og : Â£a 1 Â£a is surjective.
3.3.3. Each <5g, for g e Â£a, is involutory.
Proof: Let g = [a, p,y]. Then for A  a we have ogog(A) = og(Ay) = (Ay)y = A.
Let g = [a,8,y], h = [a,E,r] e Â£a. We say that g is perpendicular to or
orthogonal to h, denoted g _L h, if one of y and 5 is absolutely perpendicular to one
of e and r and g n h 0.
3.3.4. For g and h as above, if y.z then 8r), Sis, andy T Â£. Moreover,
A/ = yB = 8r) eg,/?.
Proof: Let M = ys. Because y,Eia then hf1 = (ye) = yaÂ£a = ys M and M\ a. So
we have Ma,Â£ which implies that Mas = r and M e h. Also, Ma,y so May = 8
and Meg. Also from M = ye it follows that Me = y = 8a = 8r)s so that M = 6r\ and
8lr. Applying 3.1.6.18 to M\y,x\ with y 1 8, r8, and M8,e with 8 1 y, Ely, we
obtain y A. rj and 8 1e.
3.3.5. If a,b,g e Â£a then a J_ b a8 X b8.
Proof: By 3.1.2.1 and 3.1.2.2 we have
o c Xy; yi.8; c ^ ^ c Xy; y^Â£8?; b*= c 3fg, for all ^ e .
The result follows.
88
Theorem 3.10.6. The linear automorphism ox is an isometry.
Proof: Let y : V < V be a similarity with square ratio r 0. Then for A,B e V we
have g(yA,yB) = rg(A,B) = rgfy~x(yA),y~x(yB)) and gfy~1(yA),fl(yB)) = Â¡rg(yA,yB).
That is, y_l has square ratio r~K Now ox is an involution, so that Ox = d^1. Hence,
if r 0 is the square ratio of ox then r  or r1 1, so r 1. [23] Because the
field /C is isomorphic to R, then V is not an Artinian space. Thus, for each similarity
dx of V there is an unique r > 0 and there is an unique isometry a such that
ox = AY(0,r); where M(t),r)(A) = rA for all A in V. The map ox is thus a similarity
with square ratio r2. Because r > 0, then from above, r 1 and it follows that dx. is
an isometry.
Proposition 3.10.7. The isometry ox is a 180 rotation; that is, a reflection about a
plane (a twodimensional subspace ofV).
Proof: Let X e Q. Suppose that 0\X and put O Xa with a e M.. Then for all A\X
and for all Z?a, dx(OA) = (OA)* = 0*A* = OA and
6x(OB) = (OB)* = 0*B* = OBa* = OB = OB.
Since .ta(0) = .tx(O)1 and V = .tk(0) .tx(O)1 then dx = lfx(0) 1 $x{0) Thus,
ox is a reflection about the plane &x(0).
Suppose that O / X. Let s  O such that s  X and let P  X be arbitrary.
Then if/?  e, POR = Q \ X and OR = PQ. Thus,
6x(OR) = (OR)* = (PQ)* = PQ = OR and d ]ke{0) = 1e(G).
Now let PO such that p 1 s, so O = sp. Let B  p. Now p 1 X because s  X and
POB = D\y where P = Xy and y  p. We calculate,
x(OB) = (OB)* = (PD)* = (PDy* = (PDf = DP = BO = OB = OB.
Hence, ox l e(0) and &k is a reflection about a plane.
47
A s g = [a,y,8] e Â£ Because g c 3Cy, yis, and h c XE, then g J_ h. By 3.3.4
above, ri_8 and {B} = {ye} = {r)5} = g n h.
Now suppose that / c Ca such that A s l and / 1 h. Now we may uniquely
write / = [a,a',p] in .Ta where a = a'p. because / T h then without loss of
generality, we may assume that as and P_Lt]. because yTs and 8i_r it follows that
a  y and P  8. By the definition of parallel lines in Section 3.2 we have /  g.
because A s l ng then by 3.2.1.10, / = g.
3.3.9. (i) The fixed points ofag e Ca are the points in Xa incident with g.
(ii) The fixed lines of ag e Ca consist of g and all lines in Ca which are
orthogonal to g.
Proof: Let g = [a,y,8]. If A = ag(/l) = A'1 for A \ a, then A \ y. because Ay = A5, VA  a
then As = A and ^418. Thus, A e g and 3.3.9.(i) follows.
Let h = [a,s, q] e Ca and suppose that ag(h) = h; g h. Then there is an
A e h such that A <Â£ g. By 3.3.8, there is a k [a,co,0] e Ca such that Ask and
k J g. Because Ash, ag(A) = Ay s h and because A s k, k J. g then
AY  coy,9Y = to,0 and AY s k. Because A <Â£ g then A / y (for if A \ y then A \ a implies
that A  ay = 8 which implies that A s g) and A AY. Thus we have A,Ar s h,k, so
that h = k by 3.2.1.1. Hence, h X g.
3.3.10. (i) The only fixed point of a point reflection, op, is the point P.
(ii) The fixed lines of dp are the lines incident with P.
Proof: If PA = P, then this implies that APA = A, or AP = 1, or A = P. Thus,
3.3.10.(i) holds. Let x = [p,u] be any line (not necessarily in 3Â£a or nonisotropic) and
A s x. Then PA s x implies that PA \ p,u, so that /*! P'4,v>'4 = p,o, and P s x.
Let Qa be the group acting on Xa generated by Ca. By 3.3.5, every element of
Qa is an orthogonal collineation of Ta Let cr e Qa. By a transformation with cr, we
mean the adjoint action ofaonia : C s Qa > cjj7 = ckti
such transformation is an inner automorphism of the group.
18
Note also that from the work above, O transforms the points Y on a line b into
the lines (y) through the point <&(b). Thus, C> is a projective correlation. Since
is
a hyperbolic polarity.
Theorem 2.2.12 The definition of orthogonality given by the polarity agrees with
and is induced by the defnition 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) <> (g(c)) = c e g(a) and a c.
(ii) g(c) iff) = c e P <> c \P.
(iii) ClPo (P) P e g(c) = Q>(c) < P c.
(iv) Clg(i)o (g(*)) = x e g(c) = cD(c) <> x c.
Instead of interpreting our original generators as ordinary lines in a hyperbolic
plane, we now interpret them as exterior points. We can construct a hyperbolic
projectivemetric plane in which the theorem of Pappus and Fanos axiom hold,
which is generated by the exterior points of the hyperbolic projectivemetric plane.
With the identifications above and the geometric objects above, we show in
the next section that the motions of the hyperbolic projectivemetric plane above can
be generated by reflections about exterior points; that is, any transformation in the
hyperbolic plane which leaves the absolute invariant can be generated by a suitable
product of reflections about exterior points.
2.3 Reflections About Exterior Points
Definition 2.3.1. A collineation is a onetoone map of the set of points onto the set
of points and a onetoone map of the set of lines onto the set of lines that preserves
the incidence relation.
Definition 2.3.2. A perspective collineation is a collineation which leaves a line
2
spacetime region /'. Hence, to different spacetime regions should correspond different
algebras.
Given a state to on A, let (7fco,7ia>,G) be the corresponding GNS
representation and let 7ZÂ¡ = n(Ai)", i e /, be the von Neumann algebras generated
by the 7tco(4./), / e /. Assume that the map i 7ZÂ¡ is an orderpreserving bijection
and that the GNS vector Q is cyclic and separating for each algebra 7ZÂ¡, i e I. From
TomitaTakesaki theory, we thus have a collection of modular involutions
and a collection {A,}(<=/ of modular operators directly derivable from the state and
the algebras. Each JÂ¡ is an antilinear involution on 7such that J/72./J, = IZf and
JÂ¡ Q = Q.
In addition, the set {J,}/e/ generates a group J which becomes a topological
group in the strong operator topology on B(7iw), the set of all bounded operators on
7t0). The modular operators {A,},e/ are positive (unbounded) invertible operators
such that
AfUjAf = 7Zj, j e I, t e R, i = fA and A"Q = Q.
In algebraic quantum field theory the state co models the preparations in the
laboratory and the algebras AÂ¡ model the observables in the laboratory and are
therefore, viewed as idealizations of operationally determined quantities. Since
TomitaTakesaki theory uniquely gives these modular objects corresponding to
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/} ,e/, co) satisfies the abstract
version of the CGMA if {7Â£,},e/ is left invariant under the adjoint action of the
59
Proof: Let 5 : X >* X be a dilation and assume that 8(dQ = ^ and 8(h) = Y of two
points X,Y e X are known. We must show that the image of any Z e X is known.
Suppose that Z Â£ gxY Then clearly Z X and Z Y and we consider the two lines
gxz and gyz Observe that Z e gxz^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 g gxY Therefore, {Z} = gxz n gYZ Since 8 is bijective, from settheoretic
reasons alone, 8(gxz n gYZ) = 8(gxz) n 8(gyz); or equivalently,
{8(Z)} = 8(gxz) ^ 8(gYz) In other words, the lines 8(gxz) and 8(gYz) have precisely
one point in common, namely, the point 8(Z) for which we are looking. The line
8(gxz) is completely known because it is the unique line which passes through X! and
is parallel to gxz Similarly, the line 8(gYz) is the unique line which passes through
the point Y1 and is parallel to gYZ because 8(Z) is the unique point of intersection of
8(gxz) and 8(gYz) (3.2.1.1), the point 8(Z) is completely determined by X1 and Y.
Conversely, assume that Z e gxY If Z is X or Y, we axe given 8(Z), so assume
that Z X and Z Y. By 3.4.1, there is an a e M. such that X, T a and there exists
a P  a such that P gxY Then Z gxp and from the previous paragraph, 8(P) is
known. Hence, using the line gxp instead of the line gxY, we conclude from the
earlier proof that 8(Z) is known.
To define a scalar multiplication, we fix a timelike line t and use it to
geometrically define dilations. To aid in the construction, the following facts are
used to add the appropriate axioms.
In a threedimensional or fourdimensional Minkowski space, if t is any
timelike line through a point O and g is any other line through O, 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.
Desargues axiom, D, holds in any affine space of dimension d > 3.
83
subspace U ofV, then n is a polarity.
Lemma 3.9.17. Let Â£, e and OA e V. Then n(ff) n(Lf)^ for every \ e and for
every subspace U < V. That is, n is invariant under a= for every Â£, e .
Proof: We have 0(0A= OO^A^ = B e X, so (OA= OB e V. Hence, from Lemma
3.8.6,the result follows.
Theorem 3.9.18. The polarity n is not a null system.
Proof: Let a be any nonisotropic line with O e a. By Lemma 3.8.7, a is not
orthogonal to itself so that a(0) $ a(0)L and hence, (O) $ n(d(0)). By Definition
3.9.7, n is not a null system.
Theorem 3.9.19. The polarity n defined above may be represented by bilinear forms.
Proof:. Let g be any isotropic line through O. By Definition 3.8.5, g(0) < (O)1 so
that g(0) < 7i((0)) and g(0) is an isotropic line in the sense of Definition 3.9.9.
Conversely, if h is a line such that h(O) < n(h(0)) = /ifO)1 then h _L 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, dim(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 V. If
y e M then U has precisely two isotropic lines, as was shown in 3.2.2.8. If y e Q,
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 K. = R is commutative then
by Theorem 3.8.12 the result follows.
Theorem 3.9.20. The polarity 7t may be represented by symmetrical bilinear forms;
that is, there is a symmetrical bilinear form g such that for any subspace U ofV,
UL = n (70 = {OA e V: g(OA,U) = 0} or
g(OA,OB) = O = g(OB, OA)if and only if, OA _L OB for OA,OB e V.
Proof: This follows directly from Theorems 3.9.18, 3.9.19, and 3.9.13.
87
Proof: Let < OA > be a one dimensional subspace of V. Then there exists a line
g = [a, p,y] such that OB e< OA > if, and only if, 0,B eg. Since
0,B  a,p,y <> Ox,Bk  a\p\y* then ^(< OA >) =< x(OA) > is a one
dimensional subspace of V.
Lemma 3.10.3. The transformation, G\, maps linearly independent vectors ofV to
linearly independent vectors.
Proof: The vectors OA,OB e V are linearly independent if, and only if, goA gOB if,
and only if, gQiji ^ goxBx Hence, is a semilinear automorphism of V for each
XeQ. m
Theorem 3.10.4. The map : V > V is a linear automorphism ofV onto V.
Proof: Consider the definition of a semilinear automorphism (23,defn.73.1). Let (V,IC)
and be vector spaces over the division rings 1C and 1C1, respectively. Suppose
that p : K. 1C' is an isomorphism from /C onto K.'. A map X : V i V1 is called
semilinear with respect to p if
1 .\{A + B) = k(A) + X(B) for all A,B e V.
2. X(tA) = p(/)X(A) for all A e V and / e 1C.
The only isomorphism p : R > R is the identity. Thus, K is a linear
automorphism of V onto V.
Let (V,/C,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 1C such that
giyA,yB) = rg{A,B), for all A. B e V. If A is nonisotropic, r = 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 / in our space, h _L / < hx _L /\ hence,
i(OA) 1 &x(OB) <> OA 1 OB.
TABLE OF CONTENTS
page
ACKNOWLEDGMENTS ii
ABSTRACT v
CHAPTERS
1 INTRODUCTION 1
2 A CONSTRUCTION OF THREEDIMENSIONAL MINKOWSKI SPACE .... 7
2.1 Preliminaries 9
2.2 Construction of n 11
2.3 Reflections About Exterior Points 18
2.4 Embedding a Hyperbolic ProjectiveMetric Plane 22
2.5 Exterior Point Reflections Generate Motions in an Affine Space 23
2.6 Conclusion 24
3 A CONSTRUCTION OF FOURDIMENSIONAL MINKOWSKI SPACE 26
3.1 Preliminaries and General Theorems 26
3.1.1 Properties of M. 30
3.1.2 Properties of V 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 (T,V,/C) 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
9
[30], start with a generating set Q of involution elements. In our approach, one can
identify the elements of Q 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 Q with a set of odd isometries
(symmetries), 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 Minkowski space, is
motivated by physical considerations [6] which we briefly explain in the conclusion.
A hyperbolic projectivemetric plane is a projective plane in which a
hyperbolic polarity is singled out and used to define orthogonality in the plane. A
polarity is an involutory projective correlation. A correlation is a onetoone
mapping of the set of points of the projective plane onto the set of lines, and of the
set of lines onto the set of points such that incidence is preserved. A projective
correlation is a correlation that transforms the points Y on a line b into the lines y'
through the corresponding point B'. So, in general, a correlation maps each point A
of the plane into a line a of the plane and maps this line into a new point A1. 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.
46
Remark. Because A\z <> A^\e^ for all Â£, e 0, A e X, and s e V, it follows
that for each g e Â£a, Gg maps:
(i) The set Xa onto itself.
(ii) The set Â£a onto itself.
(iii) Collinear points in Xa onto collinear points in Xa.
(iv) orthogonal lines in Xa onto orthogonal lines in 3fa
That is, ag is an orthogonal collineation of Xa
Let Ca = {dg : g e Â£a} and = {a/> : P e Ta}.
3.3.6. The sets Â£a and Xa are nonempty. Hence, Ca ^ 0 and 0.
Proof: If a e M. then we may write a = a/8 where a\8 e Q and a' T 5 by
definition of M.. By Axiom 7, there exist A,B such that A B and A,B\
A,Pa'8 = a and Xa 0 Moreover, gB [a, a', 8] e Â£a and Â£a 0. If a e Q
then by 3.1.3.1, there exist A  a and by 3.1.7.3, there is a p in Q such that A  P and
p 1 a. Thus, A e Xa and g [a,p,aP] e Â£a
Note that from 3.3.3, Ca consists of involutory elements and by 3.1.3.4, *}3a
consists of involutory elements.
3.3.7. IfP e g,h e Â£a and g X h then gp = aga/, = G^Gg.
Proof: Let g = [a,y,8] and h = [a,s,r], and without loss of generality, assume that
yXs, so that 8Xri. By 3.3.4, P = ye = 8r) e g,h and by 3.2.1.1, {P} = gr\h. Then
for A e Xa,
GgG/}(A) = ag(Ae) = A^ = ap(A) = Aye = G,My) =
3.3.8. For every A in Xa and for every h e Â£, there is a unique g e Â£a such that
A e g and g X h.
Proof: Let h = [a,s,r] e Â£a and A e Xa. By Axiom 1, there exist a yA such that
yis. Thus, A a,y with a X s and yXs so a X y by 3.1.6.18. because a X y then
ay = 8 and A  a,y implies A  8 so that A e g = [a,y,8] e Â£a. because g c= Xy, yle,
17
x,P c 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 O given by
(i)
(ii)
(Hi) O(poo) Pco, 0(Poo) Pco
is a polarity.
Proof: Let V be the set of all points of n and C the set of all lines of IT From the
remarks above it follows that O is a welldefined onetoone pointtoline mapping of
V onto C and a welldefined onetoone linetopoint mapping of C onto V. Next we
show that is a correlation and for this it suffices to show that preserves
incidence.
Let g(c) = {P,x,G(ab) : x,P\c and abe e Q where a  b) be a secant line. Let
A,B,d,P0o e g(c), where Pa0 = G(ef) = {x e Q : xef e Q and e /}.Then A,B,d \ c and
cab e Q.
0(g(c)) = c e A n B n g(d) n px and 0(^(c)) 6 0>(d), (D(5), cb(/), DfPoo).
Hence, d) preserves incidence on a secant line. Now consider an exterior line
P = {x : xP} and let a, b e P. Then a, b\P and it follows that P e g(a) n g(b); that
is, d>(P) e d>(/) n d>() and d> preserves incidence on an exterior line.
Finally, let p> = {G(ab)} u {x : abx e Q where a  b} be a tangent line.
Clearly, since (G(ab)) = p&, then P = G(ab) e p&. Now suppose that d e p.
Then abd e Q and
^(^0 = g(d) = {A,x,G(ef) : A,x\d and def e Q where e  /}.
Thus, d e G(ab) n G(ef) and P e g(d). This implies that d e /> and d>(/?) e O(iZ).
Hence, d> preserves incidence and is a correlation.
55
C e 1C and A e 1C. Then OCA = D e K so that C = ODA and T0a(P) = C. Hence,
each Toa e Ta maps 1C onto 1C.
Lemma 3.5.6. Let T0a e Ta and g c Xa Then TOA(g) = g if and only if g is
parallel to 1C.
Proof: Let gHF = gbea line in 3Â£a such that ToAigHF) = gHF Then
Toa(H) = HOA e gHF and gHF II 1C by 3.2.1.9. Conversely, suppose that gHF II 1C.
Then again by 3.2.1.9 it follows that for B e gHF,
Toa{B) = BOA g gHF and T0A(g) = g.
Lemma 3.5.7. If A. B e h, h a Xa, and h  /C, then there exists Toe e Ta such that
Toc(A) = B.
Proof: By 3.2.1.9 we have C = OAB e 1C and it follows that B = AOC = Toc(4).
Lemma 3.5.8. For each A e 1C, there is a unique Toa e Ta such that Toa(0) = A.
Proof: Clearly, Toa(0) = OOA = A. So suppose that TociP) = A. Then
ooc = c = a. m
We denote this unique translation mapping of O into A by TA.
Lemma 3.5.9. 7/
Proof: Let a = CgCSh where g = [a,p,ap], h = [ay,ay] e Oa. Then for any X\a,
(agchTAahGg)(X) = QfrOAyrt = = XOA# = TagaiMU0
By 3.2.3.8 there exist precisely two isotropic lines in Xa through O : 1C = goE and
1C = goF We define multiplication and addition on the points of 1C so that the
points of 1C form a field. For A,B e 1C, define:
A + B = (Tb o Ta)(0)
A B = (8b 8a)(Â£), where A,B O and E e 1C is the multiplicative identity.
A O = O A = O.
Theorem 3.5.10. For every A, B e 1C, Ta+b = TA Tb
63
So suppose that gop = g is nonisotropic and write P = P\OP2, where P\ = P2.
If O e g, then XOY e g < X = Yg, where X e 1C and Y e X.'. Thus, 8A(P) = PfOPf1
and P2 = Pf" = Pfs, which implies that 8A(P) e g.
Claim. 8A(POQ) = 8A(P)08A(Q), for all P,Q e Â£. Let P = P\OP\ and Q = Q\OQ2.
Then POQ = PiOP2OQ]OQl2 = (P\0Q\)0(P20Q2)', with P\OQ\ e K and
(PlOQi) e 1C'. Therefore,
8A(P0Q) = (P\OQ\),aO(P2OQ2yal = P\aOQ\aOPfOQf = (PfOPfyJiQfOQf)
= 8A(P)08A(Q).
The claim follows.
Proof of (iii): Assume that P, O, and Q are collinear. Then there is a line g, such that
P,0,Q e g. Thus, 8A(P),8A(0) e g and gPQ = g  g = g^P^Qy
Conversely, suppose that P, Q, and O are noncollinear. Let g\ = gop, gi = gOQ>
and g3 = go.POQ Then g\,g2, and g3 are distinct lines through O. Indeed, if g2 = g\,
say, then POQ e gop and we would have OPQPOQ) = Q e gop, which contradicts
our assumption of noncollinearity. Now,
P(POQ) = OQ and 8A(P)(8A(POQ)) = 8A(8A(P)08A(Q)) = 08A(Q)
gP.POQ II gOQ = gobA(Q) II &8A(P)8A(POQY an<^ gP.POQ II gbA(P)8A(POQy
Similarly, Q(POQ) = Q(QOP) = OP; and 8A(Q)(8A(POQ)) = 8A(8A(Q)08A(P)) = 08A(P).
This implies that gQ^poQ II gop and gop go8A(P) II &8A(Q)8A(POQ)m Hence,
gQ.POQ II g8A(Q)8A(POQy BY Axiom D, it follows that gPQ  g^P)8A(Qy Therefore,
8a : Xa ** Ta is a dilation on JÂ£a.
Now we show that 8A = 8A on Xa. Because 8A and 8 A are dilations on Ta, a
dilation is uniquely determined by the images of two points,and 8A(0) = O 8A(0),
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.
65
5,4(/5)054'(/>)a,y8Y,; 8A(P)08A'{P) e t\ y8y' = Â£  p; and AOA1  a,y5y' = e. Therefore,
AOA1 e [a,e]  gEp and 5A{P)08A{P) e tn [a,e]. Hence, 8AOA{P) = 8A{P)08A'{P).
Suppose Pit. Because gopd c jÂ£t), then replacing E with Et, A with At, A1
with A\, and a with rj in the first part of the proof and the result follows.
Lemma 3.6.8. If A e 1C 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 = gojOQ and r e M such that g,t c 3^. Then the
result follows from Proposition 3.6.6(iv).
Conversely, suppose that P, O, and Q are not collinear. Because P, O, and Q are
not collinear then P ^ Q, and from Lemma 3.6.4 it follows that gpQ  SlA(P)hA(Q)
Applying Axiom R we obtain go,poo = Zo8a(P)8a(Q) That is> 8A(P)08A(Q) e go.POQ
by construction. Again by Lemma 3.6.4,
S8a(P)8a(POQ) II So.poq and 8A(P)8A(P)()8A(Q) = 08A{Q).
This implies that gEA(P).8A(P)08A(Q) II So8a(Q) = Soq II gp.POQ, as P(POQ) = OQ. Thus,
gÂ§A(P),8A(P)O8A(0 = gA(P)8A(POQ) because both lines contain 8A(P) and are parallel to
gp.POQ Since
gsA(P).8A(P)OSA(g) ngP.POQ = {8.4(P)O5,4(0} and gs^p^gpoQ) ^go.POQ = {8A(POQ)},
then 8a(P)08a(Q) = 5A(POQ).
Lemma 3.6.9. If A, A e 1C and OP e V then A (A1 OP) = (A A') OP.
Proof: We need to show that 8A(8A>(P)) = 8A.A'(P); that is, 8^ 8A> = 8a.ai. Now
8a o 8Ai and 8A.A are dilations on X and 8A.A>(0) = O by construction, so it suffices
to show that (8A 8A )(E) = 8A.A(E) and (8A 8A')(0) = 8A.A'(0). Now E e K e Xa,
so on Xa, 8A o 8A' = 8A o SA, and 8A.A> 8a.ai. But on /C, by Lemma 3.5.11,
8a o 8a, = 8a.ai. Therefore, (8^ 8A')(E) = A (A1 E) = (A A') E = 8A.A'{E).
Lemma 3.6.10. For E e 1C, the multiplicative unit, and OP e V, E OP = OP.
22
Theorem 2.3.13 Reflections of exterior points about exterior points and about
exterior lines are also motions of the projectivemetric plane; that is, the T>s for
b e Q acting on exterior points and exterior lines are motions of the hyperbolic
projectivemetric plane.
Proof: The motions of the projectivemetric plane are precisely those collineations
which leave the absolute invariant.
We also point out that the proof that each is an involutory homology also
showed that the Fano axiom holds, since in a projective plane in which the Fano
axiom does not hold no homology can be an involution [4],
2.4 Embedding a Hyperbolic ProjectiveMetric Plane
In this section we embed our hyperbolic projectivemetric plane into a three
dimensional projective space, finally obtaining an affine space whose plane at infinity
is isomorphic to our original projectivemetric plane. Any projective plane FI in
which the theorem of Pappus holds can be represented as the projective coordinate
plane over a field 1C. (The theorem of Pappus guarantees the commutivity of 1C.)
Then by means of considering quadruples of elements of 1C, one can define a
projective space P3(/C) in which the coordinate plane corresponding to FI is included.
If the Fano axiom holds, then the corresponding coordinate field 1C is not of
characteristic 2 [4], By singling out the coordinate plane corresponding to FI as the
plane at infinity, one obtains an affine space whose plane at infinity is a hyperbolic
projectivemetric plane: that is, threedimensional Minkowski space.
To say that a plane IF is a projective coordinate plane over a field K. means
that each point of II is a triple of numbers (xo,xi,X2), not xi ~ 0, together with
all multiples (Axo, Ax i,Ax2), for A. ^ 0 and A. e 1C. Similarly, each line of FI is a triple
of numbers [uq,u\,U2\, not all uÂ¡ = 0, together with all multiples [Xuq,Xu\,Xu2\,
X 0. In P3(/C), all the quadruples of numbers with the last entry zero correspond
67
8A(B) e g, for all A in /C and all B in g. So A OB e g(0) for all ^4 e 1C and
OB e g(0). Hence, g(0) is a subspace of V.
It must now be shown that the dimension of g(O) is one. If g t, then
g(0) = (OEt) because for every At e t, A, = 8A(E,) by Lemma 3.6.3. So suppose that
g t and fix Beg. Let h = gBEr Then for all O D e g, there ia a unique dsuch
that D e d. d  h, and d n t 0. Put {F,} = dnt. Then for Ft = FOF' with F e 1C
it follows that
5 f(B) = D and OD = F OB.
Hence, g(0) = (OB).
Corollary 3.7.2. Following the terminology of Snapper and Troyer [23, p. 11],
g S(0,g(0)) S {A = 0(0A) : OA e g(0)}
is an affine subspace of dimension one.
Proposition 3.7.3. Let a e V with 0\a and put $La(0) = {OA : A a}. Then ta(C) is
a two dimensional subspace ofV.
Proof: Clearly, 1 = OO e '.ta(0), so the zero vector is in ta(C). Let C,D a and
A,B e 1C. Then by 3.1.4.7 we have COD Fa and by Lemma 3.6.3,
8^(0 e goc <= Ta and 85(D) e gOD c= Xa
so that MQOSgfDila. It follows that OC + OD = OCOD OF e Â£a(0), and
A OC + B OD = 084(C) + 08b(D) = 0(8A(Q08B(D) e ta(0).
Hence, ta(C) is a subspace of V.
Thus, it remains to show that Â£a(C) is two dimensional. There are two cases:
a e M and a e Q. Suppose first that a e M. We construct a basis for ia(0) using
isotropic lines. To this end, let 1C] and 1C2 be the isotropic lines in Xa through O. For
any Fa we may uniquely write in Xa, P = P\OPi, with P\ e 1C\ and P2 e /C2
BIOGRAPHICAL SKETCH
Richard K. White was born in Richmond, Virginia on August 19, 1960. He
graduated summa cum laude from the University of North Florida in 1991 with a
Bachelor of Science degree in Mathematics. He graduated from the University of
Florida in 1994 with a Master of Science degree in Mathematics. In May 2001 he
graduated with a Ph.D. in Mathematics from the University of Florida. He is the
proud parent of a sixyearold angel, Jackie.
106
37
D E by Axiom 3. Let A aa', B = p(3r, and C = yy\ Because a  p  y then
a'  p'  y\ Thus, e5 = D = ABC = aPya'pY = ea'pY which implies that
8 = a'PY Because A 5 then we may write A = aa' = 88/ and by 3.1.5.1 we have
A led,a,8; aXar; and a X 8 which implies that a' X 8. Hence, od8 = a^a'pY = PY
is a plane, so pr X y'. By Axiom 7 and 3.1.6.7 we have p/ = y1 so that a1 = 8. But
this yields aX8 and a X 8, which is a contradiction. Thus, aPy _L 8.
3.1.6.17. IfA,B\a; A\a'\ a' Xa; B\ p; and p  a' then P X a.
Proof: By 3.1.6.16 above B = AAB\a'a'fi = p and p X a.
3.1.6.18. IfEa,c; a X p, andsXP then s X a.
Proof: If E\ P then the result follows from 3.1.6.7. Suppose that E / P and let
M = sp. Then, E E$ and E$ \ = a so that E,E$ \ a. Now EM = Â£eP = so that
M is the midpoint of E and E$. EM = Â£'Peae so that Â£,Â£'/a,ae and by 3.1.6.15,
Ma,ae. In particular, Ma and we have M\a,p,e; a X p; and PXe so that a X Â£
by 3.1.6.7.
3.1.7. Consequences of Axiom 11 and 3.1.6.18.
3.1.7.1. If A  a; a X P; and a, P e Q then there exists ay in Q such that A \ y and
y X p,a.
Proof: Let A \ 8 with 8Xp. Then 8 e M. and A \ a, 8 with a X p, pX8 so that a X 8
by 3.1.6.18. Because 8 e M, 8 X a, and a e Q then Â£ = 8a e Q, A\e, and Â£ X a.
We claim that Â£ X p. Indeed, pe = P8a = Spa = 8aP = ep with Â£,p e Q so that
Â£ X P or Â£ = p. But Â£ p because then 8a = P and a = 8p = P, as 8Xp. This
contradicts 3.1.3.3.
3.1.7.2. Every point may be written as a product of three mutually perpendicular
planes from Q.
Proof: Let A be any point. Then by definition A = ap for some a e Q and some
P e M with aXp. By Axiom 11 there exists y e Q such that y X a. By 3.1.7.1 there
51
Proof: Because a e M, we may write a = Py, where P,y e G and p L y. By 3.2.1.2
every line contains at least three distinct points. Let g = [a,p,y] e Â£a and let
Beg. By 3.3.8, there is an h in Ca such that B in h and h _L g. By 3.2.1.2, there is
an A in h such that A B. By 3.3.8, there is an / e Ca such that A e l and / J. h.
By 3.2.1.2, there exists Q e l such that Q A.
Now if Q e h, then A,Q e l,h implies that Q = A or / = h, so Q Â£ h. Suppose
that Q e g. Let Q e m. m 1 /, so that gq = GÂ¡Gm. Because gb = QgGh, Ga = G^G/,
and BAQ = E e 3Ca, then
GE = GbG.4Q = &g&hhÂ¡GÂ¡Gm = GgGm and g L ITl.
Because Q e g,m and g _L m then gq = GgGm GÂ¡Gm and Gg = gÂ¡ or g = l. This
implies A,B e Lm. Thus A = B or / = m, a contradiction.
Axiom 6: There exist A, B; A B, such that A. B / g for any g e Q.
Proof: This follows from our Axiom 13.
Axiom 7: For each P and each A, B.C such that A,B,C\g, there is a v in Q such that
P,A  v, orP,B v, or P,C v.
Proof: By our Axiom 13, there exist two isotropic lines in .Ta 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.
3.4. Consequences of Section 3.3
3.4.1. For every A and B. there exist a e M such that A.B\a. {Every line lies in
some Minkowskian plane.)
Proof: By Axiom 6, there exist a e V such that A,B a. If a e A4, the result
follows. So suppose that a e Q. By Axiom 12, there exist P e Q such that A,B\$
and P i a. Then A,B ap = y, and y e M. by the definition of M..
89
To show that d>. is a reflection about a spacelike plane (Euclidean plane) for
every X e Q, we use the following theorem from Snapper and Troyer [23].
Theorem 3.10.8. [23] Let (V,/C,g) be an ndimensional metric vector space over a
field K. with metric g, K. = R, and n 2. Every nonsingular real plane has a
coordinate system such that the matrix of its metric is one of the following.
the Euclidean plane,
1 0
0 1
 the Lorentz plane, and
\
 the negative Euclidean plane.
Hence, the Euclidean plane, the Lorentz plane, and the negative Euclidean
plane are the only nonisometric, nonsingular real planes. This also follows from
Sylvesters Theorem [23) (It states that there are precisely n+1 nonisometric,
nonsingular spaces of dimension n).
Lemma 3.10.9. Let K. = R, n = 4, andV be Minkowski space. Then the orthogonal
complement of a Lorentz plane is a Euclidean plane.
Proof: Let {e,},=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 mi and nj. Then there is an isometry a :< e3,e4 >> a
such that cr(m) = mi and a(n) = n,. By the Witt Theorem [23] a can be extended to
an isometry of V, which we also denote by a. This implies that
a : V =< e3,e4 > < ei,e2 >* V = a < a(ei),a(e2) >;
that is, (a(ei).a(e2)} is a basis for a1. Now a is an isometry and
g(ei,ei) = g(e2,e2) = 1, so the metric g with respect to a1 has matrix I2, the 2x2
identity matrix, and a1 is a Euclidean plane.
66
Proof: We need to show that 6e = l.t Now l j is clearly a dilation on X and
l.r(<9) = O = SffO). Thus, l.r(Â£) = E = E E = 8f(Â£) = Â§Â£(Â£).
Theorem 3.6.11. The space (V. AC) constructed above is a vector space.
The triple (3l,V,/C), is an affine space. [23,p.6] A set X along with a vector
space V over a field K, is an affine space if for every v e V and for every X g X,
there is defined a point vX e X such that the following conditions hold.
1. If v,w g V and X e X, then (v + w)X = v(wA).
2. If denotes the zero vector, OX = X for all X e X.
3. For every ordered pair (X, Y) of points of T. there is one and only one vector v g V
such that vX = Y.
The dimension n of the vector space V is also called the dimension of the affine space
X.
Theorem 3.6.12. (T. V, K) is an affine space.
Proof: If OV,OW e V and X e X, we have
(i) (OV)X = OVX g X.
(ii) (OV+ OW)X (OVOW)X = OV(OWX).
(iii) OOX= 1
(iv) for Y g X, OYX = Z e X and (OZ)X = Y.
Now if OPX = Y, then OPX = OZX, P = Z, and OP OZ. U
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, 1C) and (.T, V,/C) are fourdimensional spaces.
Proposition 3.7.1 Let g be any line through O and put g{0) = {OA : A e g}. Then
g(0) is a one dimensional subspace ofV.
Proof: First note that 1 = OO g 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,
75
Suppose 1 OB e (O) n (/(0),.tp(0)).Then B e a cz Xa and there exists
OC e h(0) and there .exists OD e .tp(O) such that OB = OCOD\ that is, there exists
C e h and D P such that B = COD. because h c Xa, Ca and C = aai. Because
Z)p, D = ppi for some Pi e P. Now a 1 P and P 1 Pi, so a  Pi. But B = COD
and COD = aiaapPPi = a.Pi so that Ppi,a with a  Pi. This implies that a = Pi
so D = O, B C, and a = h, a contradiction. Hence, () n (h(O), ip(0)) = {1}
and V = (0) (/(), p(0)). If O e d and d T a then d(O) c:
otherwise we would have
V = d(0) d(0)
which is not possible. Hence, d(O)1 = .
On the other hand, from p(O)1 = Â£a(0), a T h, and a _L p, then,
d(O) c (h(0),Â£a(0))J. If O e g _L p. then by definition g a Xa If O e g h, then
g = a. Hence. (O) = (hCO^tpiO))1.
Claim. (O)1 is independent of the plane containing a. Suppose a
Then for O yd, we have y ^ a, 6 ^ p, and a _L jCÂ§. Again there exists a unique
/ c Xy such that O e l and /1 a, so that d(O)1 = (l(0),X8(0)) as above. Now any
point P e X may be written as P P\OP2 with Pi a and P2 \ P Since h 1 a, then we
may write P\ = HOA with H e h and A e a. Then
(h(0),Xp(0)) (O) = V= (l(0),Xy(0)) d(O),
and it follows that (h(0),Xp(0)) = (~/(0),Xy(0)) = A(O).
Theorem 3.8.20. If (Q) is isotropic then there is an unique hyperplane A{()) such
that d(0)x = A(0), d(0) c= A(O), and (d(O)1)1 = d(O).
Proof: Let d(O)
because a is isotropic then a e M. Let O = ap, p e Q, and put
A(0) = (d(0),Xp(0)). Since a c Xa, alp, and O = ap e a, then a 1 Xp. Because
73
g,go.HOK L then by Axiom U, x is orthogonal to every line in Aa through O. So in
particular, x A. hx which implies that x = h.
Corollary 3.8.13. If g is isotropic andx _!_ g then either x = g or x is nonisotropic and
x and g are noncoplanar; that is, there does not exist 5 e V such that x,g cz jÂ£Â§.
Lemma 3.8.14. If g is isotropic, x is nonisotropic, rig with x g, and {P} = x fl g,
then for all P A eg and for all P B e x, g and gAPB are not coplanar.
Proof: From Lemma 3.8.12, jc and g are noncoplanar. Suppose that A,0,A0B y for
some y e V. Then B = OAAOB\y, which implies that x = goB c: Â£y and
g Soa c: Xy, a contradiction to Lemma 3.8.11.
Lemma 3.8.15. If OC e V is isotropic and OB e V is nonisotropic with OC T OB then
OCOB 1 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 1 gocoB or there exists
D e gocoB such that O 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 8 6 V such that
goc, go.coD c Ag. This implies that O, C, COD \ 8, D OCCOD18,
gOD = go.coD c: ^8, and therefore, goc ^ Xd, a contradiction.
Lemma 3.8.16. The zero vector, 1 = OO, is the only vector orthogonal to every
vector in V.
Proof: This follows immediately from Lemma 3.8.11 above.
Theorem 3.8.17. IfU is a subspace ofV, then UL = {OA e V : OA _L OB, dOB e U)
is a subspace ofV.
Proof: because the zero vector 1 is orthogonal to every vector in V by definition,
then 1 e Y1. Let OA e U1 and R e 1C. because R OA e< OA >, the subspace
generated by OA, and goA L goB for every OB e U by the definitions of and
orthogonal vectors, then R OA e U1.
6
TomitaTakesaki theory to the vacuum state and local algebras in wedgelike regions
in threedimensional Minkowski space have geometrical interpretation [5]. In
particular, the modular conjugations, (J,}/6/, 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 /, the group J generated by the set {7/}/e/ satisfies the axiom system given
in Chapter 2 for the construction of a threedimensional Minkowski space.
In Chapter 5 some concluding remarks about the second step described above
are made. It is noted that if one assumes the modular stability condition [6] and the
halfsided modular inclusion relations given by Wiesbrock [29], then one does obtain
a unitary representation of the 2+1dimensional Poincar group.
56
Proof: For X\a, TA+B(X) = TAOB(X) = XOAOB = XOBOA = (TA TB)(X). U
Theorem 3.5.11. For all A,B 0 in tC, 8A.B = 8A 8B.
Proof: Let 8,1 = GaGa' and 8B = GbGb'. Then A B = 8B8A(E) = Eaabb. Put
gc = Gb'GaGa', then, Gbac e Va(0) and GbGc(E) = Ecb = Ea ab b = A B. Hence, by
3.5.1.3, 8A.B = CTbGc = G bG b'GaGa' = GaGb'GbGai = GaGa'GbGb' = 8A8B.
Hence, (/C,+) is a group isomorphic to Ta and (JC\{0},) is a group isomorphic
to T>a{Q). It remains to show that the distributive laws hold.
Theorem 3.5.12. Let A,B,C e K, then (A + B)'C = A'C+B*C.
Proof: If C O, then
(A + B)C=0=00 + 0 0 = A C+ BC.
If C O, then we compute
(A + B) C = 8C(A + B) = 8cTa+b(0) = 8cTA+B8~c\0) = 8cTATB8~c\0) = 8cTA8~8cTB8
= Tr(A)Tbr(B)(0) = Ta.cTb.c(0) = Ta.c+b.c{0) = A C + B C.
Because multiplication is commutative,
C(A + B) = (A + B)C = AC + BC = CA + CB. Hence, (Â£,+, ) is a field.
Ordering the field and obtaining M. To order the field K. we make use of the
following [21]. Let F be a field and A\,...,A e F. If A\,...,A 0 implies that
Xit 0 then F is called formally real.
Theorem 3.5.13. (ArtinSchreier) Every formally real field can be ordered.
To make the field K, formally real the following axiom is posited.
Axiom F. (formally real axiom). Let O, E\ a, a e M with O and E unjoinable. Let
A., 5, rj e V. If 0\f,8,r\ andL,8,r\ A. a then there is ay e V such that
0y, y 1 a, and E^OE^i = E^xL
Theorem 3.5.14. If Axiom F holds on 1C, then /C is formally real.
Proof: Let t [a,A.,aX] e Oa and let gsgb e Va(0). Then from Proposition 3.5.1,
30
Polarity Axioms
Axiom U. (Ux subspace axiom) Let 0,A,B,T and C be four points with 0,Cy6;
A,01 a; 0,5P; with a i. y and pi 6. Then there exist A,,e e V such that A, is;
0,AOB\X; and 0,C\z.
Axiom SI. Ifg c Xa, a e Q, h cz Xp, P e Q, and there exist y,8 e P sucA that
>,8; y8 e gn A, g c Xy, and h a Xg then there exists e e Q such that g,h c= Xe.
(Ifg 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 p e V such that g c Xp and h <= Xp. Then either there are a,P' e V such that
P = apr, g a Xa, and h c: Xp' or for all A eg there exists B e h such that P and
APB are unjoinable.
This concludes the list of axioms. Note that by Axiom 5, aP M. for a 1 P,
a, P e V and if a, P e A4 are distinct with ap, then.ap 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 Q 0, so let a e Q. By Axiom 11, there exists y e Q such
that aly. Hence, ay e M. by definition and M. 0.
3.1.1.2. The elements of M. are involutions.
Proof: Let y e M. Then we may write y = aP where a,P e Q and alp. We have
yy = apap = aapp 1.
3.1.1.3. For every P e M. and for every Â£, e 0, p^ e M..
10
The polarity dualizes incidences: if A lies on b, then the polar of A, a, contains
the pole of b,B. In this case we say that A and B are conjugate points, and that a
and b are conjugate lines. If A and a are incident, then A and a are said to be
selfconjugate: A on its own polar and a through its own pole [14]. A hyperbolic
polarity is a polarity which admits selfconjugate points and selfconjugate lines. The
set of all selfconjugate points is called a conic, which we shall call the absolute.
In a projective plane in which the theorem of Pappas and the axiom of Fano
hold, the polarity can be used to introduce a metric into the plane. Orthogonality is
defined as follows: two lines (or two points or a line and a point) are said to be
orthogonal or perpendicular to each other if they are conjugate with respect to the
polarity.
Congruent transformations of the plane are those collineations of the plane
which preserve the absolute; that is, those collineations which leave the absolute
invariant. In a projective plane with a hyperbolic polarity as absolute, the group of
all collineations in the plane leaving the absolute invariant is called the hyperbolic
metric group and the corresponding geometry is called the hyperbolic metric
geometry in the plane [31].
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
29
Axiom 10. If A. B, C are pairwise unjoinable points and A,B  a then C\a.
Axiom 11. For all a e Q, there exist distinct p,y,8 e Q such that a 1 (5 1 y 1 a
but 8 / a, p,y.
Axiom 12. If A,B  a; a e Q, then there exists p e G such that A, B P and P J. a.
Axiom 13. For each P a, a e M, there are distinct points A, B\ a such that
P A,B and P is unjoinable with both A and B, but A and B are joinable, and if Ca
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 are joinable and A, B\ a, then there exists P 1 a such that
A,B\fi.
Axiom 15. If a, p,y are distinct with P,y J. a and A,B\a, fi,y; A B, then a = Py
and if A, Ry; y _L a then a = Py or p = y.
Order Axioms
Axiom F. (Formally Real Axiom) [21 j Let 0,E  a, a e M. with O and E unjoinable.
Let X, 8, r e V. If O \ X,8, r andL,6,r\ _L a then there is ay e V such that
0y, y 1 a, and EX5k5OE1^ = Ex^.
Axiom L. (LUB) If A c /C, A 0, and A is bounded above, then there exists an A
in K such that A > X, for all X in A and if B > X for all X in A then A < B.
Dilation Axioms
Axiom T. If O 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 O. Let P,Q e g; R,S e h; and T,U e k. Ifgpj  gqu and
gRT II gsu then gpR  ggS.
Axiom R. Let O e g,h; P,Q e g, and R,S e h. If gPR  gQS then g0,p0R II go.QOS
85
O e x,t c Xa and x _L t then GxGt = go = tGx in Xa Now clÂ£ and s e A4, so
there exist precisely two isotropic lines, /CÂ¡z and Kze, through O in Xe. Thus, there is
an unique Â£e e IC\Z such that T = EzOEz. Because y c Xe then Y = EzOEz e y and
OyC7; = Go in XE Similarly, because c Xti; q e M., there are precisely two
isotropic lines, and through O in X^ and there is Ex\ in /Cir, such that
T = Er\OE^. because z a X^ then Z = E^OE^ e z and G:Gt = Go = GtGz in Xn We
calculate XOT = (E0Ex)0(E0El) = (EOE)()(ExOEl) = (E0E)0(E0,0E')
= (EOE)()(E0OE)t = (E0E)0{0E00E)t = (E0E)00
= (E0E)00 = EOE e X.
YOT = (Ez0E{)0(Ez0E'e) = (Â£E0Â£e)0(Â£OÂ£E)' = EzOEz e Â£le.
ZOT = (E^OE^O^E^OE^) = E^OE^ e ATin. Thus, if we put Â£\ = OX, E2 OY,
Â£3 = OZ, and Â£4 = OT it follows that the set {Â£^Â£2, Â£3, Â£4} consists of four mutually
orthogonal vectors such that Â£, + Â£4 is isotropic for / = 1,2,3. So if g is a symmetric
bilinear form representing n then from Â£, + Â£4 1 Â£, + Â£4 for / = 1,2,3 and E, 1 Â£, for
i j it follows that for / = 1,2,3,
O = g(Â¡ + Â£4,Â£, + Â£4) = g(,Ei,EÂ¡) + 2g(Â£ Â£4) + g(Â£4, Â£4) = g(Â£Â£,) + giÂ£4,Â£4).
So that g(Ej,Ei) = ^(Â£4, Â£4) O, because Â£4 is nonisotropic. Thus, it remains to
show that {Â£4,Â£2,Â£3,Â£4} is a basis for V. But this follows from Section 3.7.
Theorem 3.9.23. (V,/C,g) is a fourdimensional Minkowski vector space. Moreover,
(Xy(0),g) is a fourdimensional Minkowski space for every y in Vo
>
Proof: Put #(Â£4,Â£4) = 1 and g(Â£Â£,) = 1, for / = 1,2,3. Minkowski space is the only
nonsingular real fourdimensional vector space with metric
r 1 0 0 0 ^
0 10 0
0 0 1 0
^ 0 0 0 1^
(3.24)
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.
li
TABLE OF CONTENTS
page
ACKNOWLEDGMENTS ii
ABSTRACT v
CHAPTERS
1 INTRODUCTION 1
2 A CONSTRUCTION OF THREEDIMENSIONAL MINKOWSKI SPACE .... 7
2.1 Preliminaries 9
2.2 Construction of n 11
2.3 Reflections About Exterior Points 18
2.4 Embedding a Hyperbolic ProjectiveMetric Plane 22
2.5 Exterior Point Reflections Generate Motions in an Affine Space 23
2.6 Conclusion 24
3 A CONSTRUCTION OF FOURDIMENSIONAL MINKOWSKI SPACE 26
3.1 Preliminaries and General Theorems 26
3.1.1 Properties of M. 30
3.1.2 Properties of V 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 (T,V,/C) 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 THREEDIMENSIONAL MODEL 91
5 CONCLUSION 96
REFERENCES 103
BIOGRAPHICAL SKETCH 106
IV
Abstract of Dissertation Presented to the Graduate School
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy
AN ALGEBRAIC CHARACTERIZATION OF MINKOWSKI SPACE
By
Richard K. White
May 2001
Chairman: Dr. Stephen J. Summers
Major Department: Mathematics
We give an algebraic characterization of threedimensional and four
dimensional Minkowski space. We construct both spaces from a set of involution
elements and the group it generates. We then identify the elements of the original
generating set with spacelike lines and their corresponding reflections in the three
dimensional case and with spacelike planes and their corresponding reflections in the
fourdimensional case. Further, we explore the relationship between these
characterizations and the condition of geometric modular action in algebraic
quantum field theory.
v
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 [5] showed that for quantum theories satisfying the Wightman axioms
the modular objects associated by TomitaTakesaki theory to the vacuum state and
local algebras generated by field operators with support in wedgelike spacetime
regions in Minkowski space have geometrical meaning. Motivated by this work,
Buchholz and Summers gave an algebraic characterization of vacuum states on nets
of C*algebras over Minkowski space and reconstructed the spacetime translations
with the help of the modular structures associated with such states. Their result
suggested that a condition of geometric modular action might hold in quantum
field theories on a wider class of spacetime manifolds.
To explain the abstract version of this condition, first some notation is
introduced and the basic setup is given. Let {A };e/ be a collection of C*algebras
labeled by the elements of some index set / such that (/,<) is an ordered set and the
property of isotony holds. That is, if i\,ii e I such that i\ < Â¡2, then AÂ¡x c A,2.
Let A be a C*algebra containing {A,},e/. It is also required that the assignment
i A, is an orderpreserving bijection.
In algebraic quantum field theory, the index set I is usually a collection of
open subsets of an appropriate spacetime (M,g). In such a case, the algebra A, is
interpreted as the C*algebra generated by all the observables measured in a
1
2
spacetime region /'. Hence, to different spacetime regions should correspond different
algebras.
Given a state to on A, let (7fco,7ia>,G) be the corresponding GNS
representation and let 7ZÂ¡ = n(Ai)", i e /, be the von Neumann algebras generated
by the 7tco(4./), / e /. Assume that the map i 7ZÂ¡ is an orderpreserving bijection
and that the GNS vector Q is cyclic and separating for each algebra 7ZÂ¡, i e I. From
TomitaTakesaki theory, we thus have a collection of modular involutions
and a collection {A,}(<=/ of modular operators directly derivable from the state and
the algebras. Each JÂ¡ is an antilinear involution on 7such that J/72./J, = IZf and
JÂ¡ Q = Q.
In addition, the set {J,}/e/ generates a group J which becomes a topological
group in the strong operator topology on B(7iw), the set of all bounded operators on
7t0). The modular operators {A,},e/ are positive (unbounded) invertible operators
such that
AfUjAf = 7Zj, j e I, t e R, i = fA and A"Q = Q.
In algebraic quantum field theory the state co models the preparations in the
laboratory and the algebras AÂ¡ model the observables in the laboratory and are
therefore, viewed as idealizations of operationally determined quantities. Since
TomitaTakesaki theory uniquely gives these modular objects corresponding to
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/} ,e/, co) satisfies the abstract
version of the CGMA if {7Â£,},e/ is left invariant under the adjoint action of the
3
modular conjugations {J,},e/; that is, if for every i,j in I there is a k in I such that
adJ,(TZj) s JjTZjJi TZk, where JÂ¡TZjJÂ¡ = {JÂ¡AJÂ¡ : A e 1lj}.
Thus, for each i in I, there is an orderpreserving bijection, automorphism, xÂ¡ on I,
(/,<), such that JÂ¡HjJÂ¡ = for j e /.The set is a set of involutions which
generate a group % which is a subgroup of the group of translations on I. 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 I.
2. The group J acting on the set {/Â£, }/Â£/
To elaborate further the relation between the groups Tand J, consider the
following.
Proposition 1.1 [6] The surjective map ^ given by J,m) = T,yT,m, is
a group homomorphism. Its kernel S lies in the center of J and the adjoint action of
S leaves each 7ZÂ¡ fixed.
Thus, J is a central extension of the group Tby S.
As an immediate consequence of this proposition, J provides a projective
representation of Twith coefficients in an abelian group Z in the center of J. Thus,
the condition of geometric modular action induces a transformation group on the
index set I and provides it with a projective representation.
With this in mind, the following program was then posed. Given the
operational data available from algebraic quantum field theory, can one determine
the spacetime symmetries, the dimension of the spacetime, and the spacetime itself?
That is to say, given a net of C*algebras and a state co satisfying the CGMA, can
4
one determine the spacetime symmetries, the dimension of the spacetime, and even
the spacetime itself?
Part of this has been carried out by Buchholz, Dreyer, Florig, and Summers
for Minkowski space and de Sitter space [6]. However, in order to do so, they had to
presume the respective spacetime as a topological manifold. But would it not be
possible to completely derive the spacetime from the operationally given data
without any assumption about dimension or topology?
As was pointed out by Dr. Summers, a possibility to do so was opened up in
this program in the following manner. As already seen, the CGMA yields an
involution generated group complete with a projective representation and there is in
the literature a way of deriving spacetimes from such groups going under the name
of absolute geometry.
In general, absolute geometry refers to a geometry that includes both
Euclidean and nonEuclidean geometry as special cases. Thus, one has a system of
axioms not yet implying any decision about parallelism. In our case, the axioms are
given in terms of a group of motions as an extension of Kleins Erlangen Program. A
group of motions is defined as a set Q of involution elements closed under
conjugation and the group 0 it generates. In a group of motions the representations
of geometric objects and relations depend only on the given multiplication for the
group elements, without reference to any additional structure. The system of axioms
is formulated in terms of the involutory generators alone, so that geometric concepts
like point, line, and incidence no longer are primary but are derived.
The necessary means for setting up this representation are provided by the
totality of reflections in points, lines, and planes (a subset of the set of motions).
Points, lines, and planes are in onetoone correspondence with the reflections in
them so that geometric relations among points, lines, and planes correspond to
grouptheoretic equations among the reflections. This enables one to be able to
5
formulate geometric theorems about elements of the group of motions and to be able
to then prove these theorems by grouptheoretic calculation.
To summarize, we are to find conditions on an index set I and a corresponding
net of C*algebras {AÂ¡}Â¡ei as well as a state co satisfying the CGMA such that the
elements of I can be naturally identified with open sets of Minkowski space and such
that the group Tis implemented by the Poincar group on this Minkowski space.
Out of the group Twe wish to construct Minkowski space such that Ts natural
action on Minkowski space is that of the Poincar group.
This involves two steps. First we carry out the absolute geometry program for
three and fourdimensional Minkowski space. That is, characterize three and
fourdimensional Minkowski space in terms of a group of motions (Q, 0). Second, we
must determine what additional structure on the ordered set I would yield from
TomitaTakesaki theory algebraic relations among the J, (and hence, among the x,)
which coincide with the algebraic characterization found in step one.
The organization of the thesis is as follows: in Chapter 2, the given pair (Q, 0)
is used to construct a threedimensional Minkowski space out of the plane at infinity.
Then identification of the involutory elements of Q with spacelike lines and their
group action in with reflections about spacelike lines is made.
In Chapter 3 using the same initial data, (Q, ), as was given in Chapter 2 but
satisfying different axioms, a fourdimensional Minkowski space is constructed. The
approach taken here differs from that taken in Chapter 2. This time the affine space
is constructed first and then the hyperplane at infinity is used to obtain the metric.
The identification of the elements of Q with spacelike planes and their group action
in 0 with reflections about spacelike planes is made.
In Chapter 4 a concrete example of the threedimensional characterization is
given. As already mentioned, Bisognano and Wichmann showed that for quantum
field theories satisfying the Wightman axioms the modular objects associated by
6
TomitaTakesaki theory to the vacuum state and local algebras in wedgelike regions
in threedimensional Minkowski space have geometrical interpretation [5]. In
particular, the modular conjugations, (J,}/6/, 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 /, the group J generated by the set {7/}/e/ satisfies the axiom system given
in Chapter 2 for the construction of a threedimensional Minkowski space.
In Chapter 5 some concluding remarks about the second step described above
are made. It is noted that if one assumes the modular stability condition [6] and the
halfsided modular inclusion relations given by Wiesbrock [29], then one does obtain
a unitary representation of the 2+1dimensional Poincar group.
CHAPTER 2
A CONSTRUCTION OF THREEDIMENSIONAL MINKOWSKI SPACE
In this chapter we give an absolute geometric, that is, an algebraic,
characterization of threedimensional Minkowski space. This chapter is a version of a
preprint by the author entitled A GroupTheoretic Construction Of Minkowski
3Space Out Of The Plane At Infinity [28]. Along with the wellknown
mathematical motivations [1, 2] there are also physical motivations, as we discussed
in Chapter 1. Threedimensional Minkowski space is an affine space whose plane at
infinity is a hyperbolic projectivemetric plane [12]. In Absolute Geometry [2],
Bachmann, Pejas, Wolff, and Bauer (BPWB) took an abstract group 0 generated
by an invariant system Q of generators in which each of the generators was
involutory, satisfying a set of axioms and constructed a hyperbolic projectivemetric
plane in which the given group 0 was isomorphic to a subgroup of the group of
congruent transformations (motions) of the projectivemetric plane. By interpreting
the elements of Q as line reflections in a hyperbolic plane, BPWB showed that the
hyperbolic projectivemetric plane could be generated by these line reflections in
such a way that these line reflections form a subgroup of the motions group of the
projectivemetric plane.
Coxeter showed in [13] that every motion of the hyperbolic plane is generated
by a suitable product of orthogonal line reflections, where an orthogonal line
reflection is defined as a harmonic homology with center exterior point and axis the
given ordinary line and where the center and axis are a polepolar pair. Here we
show that Coxeters and BPWBs notions of motions coincide in the hyperbolic
7
8
projectivemetric plane and that the motions can be viewed as reflections about
exterior points.
Next we embed our projectivemetric plane into a threedimensional projective
space. By singling out our original plane as the plane at infinity, we obtain an affine
space whose plane at infinity is a hyperbolic projectivemetric plane,
threedimensional Minkowski space. Finally, we show that the motions of our
original plane induce motions in the affine space and, by a suitable identification, we
show that any motion in Minkowski space can be generated by reflections about
spacelike lines. Thus, to construct a threedimensional Minkowski space, one can
start with a generating set Q of reflections about spacelike lines in the plane at
infinity. So Q may be viewed as a set of reflections about exterior points in a
hyperbolic projectivemetric plane. Out of the plane at infinity, one can obtain a
threedimensional affine space with the Minkowski metric, which is constructed from
a group generated by a set of even isometries or rotations.
The approach in this chapter differs from the method used by Wolff [30] for
twodimensional Minkowski space and by Klotzek and Ottenburg [19] for
fourdimensional Minkowski space. The approach in these papers is to begin by
constructing the affine space first. For Wolffs [30] twodimensional case, the
elements of the generating set Q are identified with line reflections in an affine plane.
For Klotzek and Ottenburgs [19] fourdimensional case, the elements of the
generating set Q are identified with reflections about hyperplanes in an affine space.
Thus, in each of these papers, the generating set Q 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 Q of involution elements. In our approach, one can
identify the elements of Q 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 Q with a set of odd isometries
(symmetries), 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 Minkowski space, is
motivated by physical considerations [6] which we briefly explain in the conclusion.
A hyperbolic projectivemetric plane is a projective plane in which a
hyperbolic polarity is singled out and used to define orthogonality in the plane. A
polarity is an involutory projective correlation. A correlation is a onetoone
mapping of the set of points of the projective plane onto the set of lines, and of the
set of lines onto the set of points such that incidence is preserved. A projective
correlation is a correlation that transforms the points Y on a line b into the lines y'
through the corresponding point B'. So, in general, a correlation maps each point A
of the plane into a line a of the plane and maps this line into a new point A1. When
the correlation is involutory, A' always coincides with A. Thus a polarity relates A to
a, and vice versa. A is called the pole of a and a is called the polar of A. Since this is
a projective correlation, the polars of all the points on a form a projectively related
pencil of lines through A.
10
The polarity dualizes incidences: if A lies on b, then the polar of A, a, contains
the pole of b,B. In this case we say that A and B are conjugate points, and that a
and b are conjugate lines. If A and a are incident, then A and a are said to be
selfconjugate: A on its own polar and a through its own pole [14]. A hyperbolic
polarity is a polarity which admits selfconjugate points and selfconjugate lines. The
set of all selfconjugate points is called a conic, which we shall call the absolute.
In a projective plane in which the theorem of Pappas and the axiom of Fano
hold, the polarity can be used to introduce a metric into the plane. Orthogonality is
defined as follows: two lines (or two points or a line and a point) are said to be
orthogonal or perpendicular to each other if they are conjugate with respect to the
polarity.
Congruent transformations of the plane are those collineations of the plane
which preserve the absolute; that is, those collineations which leave the absolute
invariant. In a projective plane with a hyperbolic polarity as absolute, the group of
all collineations in the plane leaving the absolute invariant is called the hyperbolic
metric group and the corresponding geometry is called the hyperbolic metric
geometry in the plane [31].
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 n
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 is called an involution if a2 = 1, where 1 is the
identity element of the group .
Basic assumption: A given group is generated by an invariant system Q of
involution elements.
The elements of Q are denoted by lowercase Latin letters. Those involutory
elements of that can be represented as ab, where a,b e Q, are denoted by
uppercase Latin letters. If r e and fyr\ is an involution, we denote this by E,r.
Axioms
Axiom 1: For every P and Q there is a g with P, Q g.
Axiom 2: IfP,Q \g,h then P = Q or g = h.
Axiom 3: If a,b,c \P then abe d e Q.
Axiom 4: If a,b,c g then abe = d e Q.
Axiom 5: There exist g,h,j such that g \h butj / g,h,gh.
Axiom 6; There exist elements d, a, b e Q such that d, a, b / P,c for P,c e . (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 hj e Q such
that P\h,j but g, h f A,c and g,j / B,d for anyA,B,c,d e (that is, have neither a
point nor a line in common).
12
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 Q is as secant or ordinary lines in
a hyperbolic plane for BPWB [2], In our approach, we view the elements of Q
initially as exterior points in a hyperbolic plane. After embedding our hyperbolic
projectivemetric plane into an affine space, we can identify the elements of Q. our
generating set, with spacelike lines and their corresponding reflections in a
threedimensional Minkowski space. By realizing that statements about the geometry
of the plane at infinity correspond to statements about the geometry of the whole
space where all lines and all planes are considered through a point, we see that the
axioms also are statements about spacelike lines, the elements of Q, and timelike
lines, the elements P of , through any point in threedimensional Minkowski space.
The models of the system of axioms are called groups of motions; that is, a
group of motions is a pair (, Q) consisting of a group and a system Q of
generators of the group satisfying the basic assumption and the axioms.
To give a precise form to the geometric language used here to describe
grouptheoretic concepts occurring in the system of axioms, we associate with the
group of motions (,Â£) the group plane (,Â£), 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 Q are
called points of the group plane. Two lines g and h of the group plane are said to be
perpendicular if g \h. 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 g. Two lines are said to be parallel if they satisfy Axiom 6.
Thus, if P Q, then by Axioml and Axiom 2, the points P and Q in the group
plane are joined by a unique line. If P / g then Axiom 7 says that there are at most
two lines through P parallel to g.
13
Lemma 2.2.2.[2) For each a e 0, the mappings aa : g >* ga = ago. and
<5 a : P '> Pa = a Pa are onetoone mappings of the set of lines and the set of points,
each onto itself in the group plane.
Proof: Let a e 0, and consider the mapping y t* ya = aya of 0 onto itself. It is
easily seen that this mapping is bijective. Because Q is an invariant system (ah e Q
for every a,b e Q) Q is mapped onto itself, and if P is a point, so that P = gh with
g\h, then Pa = gaha and ga\ha, so that Pa is also a point. Thus, g ga, P Pa
are onetoone mappings of the set of lines and the set of points, each onto itself in
the group plane.
Definition 2.2.3. A onetoone mapping a of the set of points and the set of lines
each onto itself is called an orthogonal collineation if it preserves incidence and
orthogonality.
Since the  relation is preserved under the above mappings, the above
mappings preserve incidence and orthogonality as defined above.
Corollary 2.2.4. [2] The mappings
Ga g >* ga and eta P '* Pa
are orthogonal collineations of the group plane and are called motions of the group
plane induced by a.
In particular, if a is a line a we have a reflection about the line a in the group
plane, and if a is a point A, we have a point reflection about A in the group plane.
If to every a e 0 one assigns the motion of the group plane induced by a,
one obtains a homomorphism of 0 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 0
generated by Q.
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 (0, Q), are
called metric planes.
BPWB showed how one can embed a metric plane into a projectivemetric
plane by constructing an ideal plane using pencils of lines [2], We shall now outline
how this is done.
Definition 2.2.6. Three lines are said to lie in a pencil if their product is a line; that
is, a,b,c lie in a pencil if
abe = d e Q. (*)
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)~l is a line, the invariance of Q
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 Q also
shows that (*) holds whenever at least two of the three lines coincide.
Using the given axioms, BPWB [2] showed that there are three distinct classes
of pencils. If a,b\V then G{ab) = {c : c \ V}. In this case, G(ab) is called a pencil of
lines with center V and is denoted by G(V). If a,b c then G(ab) = {d : d c}. 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  a,b where a  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, II, is constructed in the following way. An
15
ideal point is any pencil of lines G(ab) of the metric plane. The pencils G(JP)
correspond in a onetoone way to the points of the metric plane. An ideal line is a
certain set of ideal points. There are three types:
1. A proper ideal line g(a), is the set of ideal points that have in common a line a
of the metric plane.
2. The set of pencils G(x) with x \P 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 px.
The polarity is defined by the mappings
G(C) C and C G(C);
Poo >< poo and px > P<;
G(c) i* gjc) and g(c) G(c).
Bachmann [1] showed that the resulting ideal plane is a hyperbolic projective plane
in which the theorem of Pappus and the Fano axiom hold; that is, it is a hyperbolic
projectivemetric plane.
In this model, the ideal points of the form G(P) are the interior points of the
hyperbolic projectivemetric plane. Thus the points of the metric plane correspond in
a onetoone way with the interior points of the hyperbolic projectivemetric plane.
The ideal points G(x), for x e Q are the exterior points of the hyperbolic
projectivemetric plane.
Theorem 2.2.8. Each x e Q corresponds in a onetoone way with the exterior points
of the hyperbolic projectivemetric plane.
Proof: Because each line d of the metric plane is incident with at least three points
and a point is of the form ab with a \ b, then each x e Q is the axis of a pencil. From
the uniqueness of perpendiculars each x e Q corresponds in a onetoone way with
the pencils G(x). Hence, each x e Q corresponds in a onetoone way with the
exterior points of the hyperbolic projectivemetric plane.
16
Thus, the axioms can be viewed as axioms concerning the interior and exterior
points of a hyperbolic projectivemetric plane. The ideal points of the form G(ab)
where a  b are the points on the absolute, that is, the points at infinity in the
hyperbolic projectivemetric plane.
Now consider the ideal lines. A proper ideal line g(o) 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,P\a and abe e Q where b  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 Q 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 \c
and abe e G where a \\ b}.
Corollary 2.2.10. The ideal line which consists of pencils G(x) with x \P 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 Q : xP}.
The last type of ideal line is a tangent line. It contains only one point,
G(ab) = Poo, on the absolute. Denoting this line by /?>, then
poo = {G(a)} u {x e Q : abx e Qj, where a  b. Recalling that each x e Q
corresponds to an exterior point in the hyperbolic projectivemetric plane, we see
that a tangent line consists of one point on the absolute and every other point is an
exterior point.
Also note that under the above identifications, each secant line g(c)
corresponds to a unique exterior point, c, c g{c) since one only considers those
17
x,P c 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 O given by
(i)
(ii)
(Hi) O(poo) Pco, 0(Poo) Pco
is a polarity.
Proof: Let V be the set of all points of n and C the set of all lines of IT From the
remarks above it follows that O is a welldefined onetoone pointtoline mapping of
V onto C and a welldefined onetoone linetopoint mapping of C onto V. Next we
show that is a correlation and for this it suffices to show that preserves
incidence.
Let g(c) = {P,x,G(ab) : x,P\c and abe e Q where a  b) be a secant line. Let
A,B,d,P0o e g(c), where Pa0 = G(ef) = {x e Q : xef e Q and e /}.Then A,B,d \ c and
cab e Q.
0(g(c)) = c e A n B n g(d) n px and 0(^(c)) 6 0>(d), (D(5), cb(/), DfPoo).
Hence, d) preserves incidence on a secant line. Now consider an exterior line
P = {x : xP} and let a, b e P. Then a, b\P and it follows that P e g(a) n g(b); that
is, d>(P) e d>(/) n d>() and d> preserves incidence on an exterior line.
Finally, let p> = {G(ab)} u {x : abx e Q where a  b} be a tangent line.
Clearly, since (G(ab)) = p&, then P = G(ab) e p&. Now suppose that d e p.
Then abd e Q and
^(^0 = g(d) = {A,x,G(ef) : A,x\d and def e Q where e  /}.
Thus, d e G(ab) n G(ef) and P e g(d). This implies that d e /> and d>(/?) e O(iZ).
Hence, d> preserves incidence and is a correlation.
18
Note also that from the work above, O transforms the points Y on a line b into
the lines (y) through the point <&(b). Thus, C> is a projective correlation. Since
is
a hyperbolic polarity.
Theorem 2.2.12 The definition of orthogonality given by the polarity agrees with
and is induced by the defnition 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) <> (g(c)) = c e g(a) and a c.
(ii) g(c) iff) = c e P <> c \P.
(iii) ClPo (P) P e g(c) = Q>(c) < P c.
(iv) Clg(i)o (g(*)) = x e g(c) = cD(c) <> x c.
Instead of interpreting our original generators as ordinary lines in a hyperbolic
plane, we now interpret them as exterior points. We can construct a hyperbolic
projectivemetric plane in which the theorem of Pappus and Fanos axiom hold,
which is generated by the exterior points of the hyperbolic projectivemetric plane.
With the identifications above and the geometric objects above, we show in
the next section that the motions of the hyperbolic projectivemetric plane above can
be generated by reflections about exterior points; that is, any transformation in the
hyperbolic plane which leaves the absolute invariant can be generated by a suitable
product of reflections about exterior points.
2.3 Reflections About Exterior Points
Definition 2.3.1. A collineation is a onetoone map of the set of points onto the set
of points and a onetoone map of the set of lines onto the set of lines that preserves
the incidence relation.
Definition 2.3.2. A perspective collineation is a collineation which leaves a line
19
pointwise fixed, its axis, and a point linewise fixed, its center.
Definition 2.3.3. A homology is a perspective collineation with center a point B and
axis a line b where B is not incident with b.
Definition 2.3.4. A harmonic homology with center B and axis b, where B is not
incident with b, is a homology which relates each point A in the plane to its
harmonic conjugate with respect to the two points B and (b, [A, 5]), 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 / 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 /. 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 pass through A, B, C respectively. Let
{P} = [A,S] n [5,/?], then {D} = [A,B] n [P,Q\ ([11]). Note that if [R,S]  [A,C] then
D is the midpoint of A and B.
Coxeter [13] showed that any congruent transformation of the hyperbolic
plane is a collineation which preserves the absolute and that any such
transformation is a product of reflections about ordinary lines in the hyperbolic
plane where a line reflection about a line m is a harmonic homology with center M
and axis m, where M and m are a polepolar pair and M is an exterior point. A point
20
reflection is defined similarly, a harmonic homology with center M and axis m, where
M and m are a polepolar pair, M is an interior point, and m is an exterior line. Note
that in both cases, M and m are nonincident.
In keeping with the notation employed at the end of Â§2.2, let b be an exterior
point and g{b) its pole.
Lemma 2.3.8. The map : <
A
A
Ab and d
Ab d
P$0 Poo
db
g{d)b
~N
> is a
nb
J
collineation.
Proof: This follows from the earlier observation that the motions of the group plane
map pencils onto pencils preserving the  relation.
Lemma 2.3.9. *P, is a perspective collineation and, hence, a homology.
Proof: Recall that g(b) = {A,x,P*> : x,A \b and where b lies in the pencil Poo}. For
any A and x in g(b) we have Ab A and xh = x since A,x \b and if A1,x' g gfb) then
A',x' / b and Ab A, xb x, and Ab,xb / b. Thus, Ab,xb g g(b).
Recall also that Poo = G(cd), where c and d do not have a common
perpendicular nor a common point and thus, G(cd) = [f: fed e Q}. Now g(b) is a
secant line, so that it contains two such distinct points, Pec and Qoo, 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 Poo then cb,db e Pec and hence, P* = Poo and
Q = Qec. Moreover, if Poo Â£ g(b) then it follows that Rb
g{b) pointwise invariant.
Now let g(d), Q, and roo be a secant line, exterior line, and tangent line,
respectively, containing b. For e e g(d) we have e \d and eb \ db = d since b  d,
thus eb Â£ g[d). For A e g{d), Ab \db d, so Ab e g{d). Similarly, it follows that if
Poo e gid) then P* e g(d), and that g(d)b = g(d). One easily sees that Qb = Q and
r* = roo. Thus, leaves every line through b invariant and T* is a perspective
21
collineation for each b e Q.
Theorem 2.3.10. T is a harmonic homology.
Proof: Since Ab 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 Q, T/, 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 (^b)b = Â£,
for any h, e 0, T/, is involutory for each b e Q. Now in a projective plane in which
the theorem of Pappus holds, the only collineations which are involutory are
harmonic homologies [10], thus T, is a harmonic homology for each b e Q.
Theorem 2.3.11. Interior point reections are generated by exterior point reflections.
Proof: A similar argument shows that for each interior point A, is a harmonic
homology with center A and axis A where A is the polar of A, A Â£ A, and where
is defined analogously to T. Thus, each 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.
Theorem 2.3.12. The reection 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 \d and g(d) is orthogonal to g(b). Let E
be the point where g(b) meets g(d). Since E e gfb) then E \b and Eb =f for 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 \d and E\d then bd = C and we have E,C \b,d with b d. Thus, by
Axiom 2, E = C = bd. Hence, AE Adb Ab.
22
Theorem 2.3.13 Reflections of exterior points about exterior points and about
exterior lines are also motions of the projectivemetric plane; that is, the T>s for
b e Q acting on exterior points and exterior lines are motions of the hyperbolic
projectivemetric plane.
Proof: The motions of the projectivemetric plane are precisely those collineations
which leave the absolute invariant.
We also point out that the proof that each is an involutory homology also
showed that the Fano axiom holds, since in a projective plane in which the Fano
axiom does not hold no homology can be an involution [4],
2.4 Embedding a Hyperbolic ProjectiveMetric Plane
In this section we embed our hyperbolic projectivemetric plane into a three
dimensional projective space, finally obtaining an affine space whose plane at infinity
is isomorphic to our original projectivemetric plane. Any projective plane FI in
which the theorem of Pappus holds can be represented as the projective coordinate
plane over a field 1C. (The theorem of Pappus guarantees the commutivity of 1C.)
Then by means of considering quadruples of elements of 1C, one can define a
projective space P3(/C) in which the coordinate plane corresponding to FI is included.
If the Fano axiom holds, then the corresponding coordinate field 1C is not of
characteristic 2 [4], By singling out the coordinate plane corresponding to FI as the
plane at infinity, one obtains an affine space whose plane at infinity is a hyperbolic
projectivemetric plane: that is, threedimensional Minkowski space.
To say that a plane IF is a projective coordinate plane over a field K. means
that each point of II is a triple of numbers (xo,xi,X2), not xi ~ 0, together with
all multiples (Axo, Ax i,Ax2), for A. ^ 0 and A. e 1C. Similarly, each line of FI is a triple
of numbers [uq,u\,U2\, not all uÂ¡ = 0, together with all multiples [Xuq,Xu\,Xu2\,
X 0. In P3(/C), all the quadruples of numbers with the last entry zero correspond
23
to n. One can now obtain an affine space A by defining the points of A to be those
of P3(/C) FI; that is, those points whose last entry is nonzero; a line / of A to be a
line /' in P3(/C) Id minus the intersection point of the line /' with FI; and by
defining a point P in A to be incident with a line / of A if, and only if, P is incident
with the corresponding /'. Planes of A are obtained in a similar way [14].
Thus, each point in Id 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 FI represents the set of all planes parallel to a given plane.
Because parallel objects can be considered to intersect at infinity, we call FI the
plane at infinity.
2.5 Exterior Point Reflections Generate Motions in an Affine Space
In this section we state and prove the main result of this chapter. Coxeter
showed that threedimensional Minkowski space is an affine space whose plane at
infinity is a hyperbolic projectivemetric plane [11]. 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
Timelike line
Lightlike line
Spacelike line
Characteristic plane
Minkowski plane
Spacelike plane
Section at Infinity
Interior point
Point on the absolute
Exterior point
Tangent line
Secant line
Exterior line
He also showed that if one starts with an affine space and introduces a
hyperbolic polarity in the plane at infinity of the affine space, then the polarity
induces a Minkowskian metric on the whole space. Under a hyperbolic polarity, a
line and a plane or a plane and a plane are perpendicular if their elements at infinity
24
correspond. Two lines are said to be perpendicular if they intersect and their
elements at infinity correspond under the polarity.
Theorem 2.5.1. Every motion in a threedimensional affine space with a hyperbolic
polarity defned on its plane at infnity, is generated by reflections about exterior
points. Moreover, because exterior points correspond to spacelike lines, then any
motion in threedimensional Minkowski space is generated by reflections about
spacelike lines.
Proof: Because any motion in threedimensional Minkowski space can be generated
by a suitable product of plane reflections, it suffices to show that reflections about
exterior points generate plane reflections.
Let a be any Minkowski plane or spacelike plane. Let P be any point in
Minkowski space. Let / be the line through P parallel to a. Let a denote the
section of a at infinity. Applying the polarity to ctoo we get a point gao X a 00 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 T ctoo. Now let m be a line
through P not in a which intersects /. It follows that the reflection of P about a is
the same as reflecting m about / 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 / and m must act as their sections at
infinity act. because any point reflection in the hyperbolic projectivemetric plane
can be generated by reflections about exterior points, we have that the reflection of
P about a is generated by reflections of P about spacelike lines.
2.6 Conclusion
As already indicated above, the geometric model for the generators of Q which
lies behind the choice of algebraic characterization of threedimensional Minkowski
25
space differs significantly from those of previous absolute geometric characterizations
of Minkowski space. The model given here is the set of reflections about spacelike
lines, which is not a choice which would be made a priori by other mathematicians.
However, this is yet another example of a situation where the initial data are
imposed by physical, as opposed to purely mathematical, considerations.
In the next chapter starting with the same initial data, but satisfying different
axioms, a construction of fourdimensional Minkowski space is given. First the affine
space is constructed and then the hyperplane at infinity is used. Also given is an
explicit construction of the field, the vector space, and the metric.
CHAPTER 3
A CONSTRUCTION OF FOURDIMENSIONAL MINKOWSKI SPACE
In this chapter a construction of fourdimensional Minkowski space will be
given using the same initial data as in Chapter 2 but satisfying different axioms. The
actual construction is quite long, so to aid the reader in following, A brief outline of
the procedure shall be given here. First some general theorems and the basic
definitions will be given in the first two sections. In Sections 3.3 and 3.4 attention is
restricted to two dimensions in order to obtain the necessary machinery to construct
the field. Once the field has been obtained, then a vector space is constructed and
given a definition of orthogonality. It is then shown that the vector definition of
orthogonality is induced by and agrees with the initial definition of perpendicularity
for the geometry generated by the original set of involutions Q.
To obtain a metric vector space from the constructed vector space V, a map n
is defined on the subspaces of V. The map n is defined to send a subspace to its
orthogonal complement. Using the work of [3] (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 Q 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 Q of involution elements and the group it
generates, where for any a e Q and for any Â£ we have oU = Â£'a% e Q. If the
product of two distinct elements e is an involution then we denote this by
26
27
writing Â£11^2 We note that if then ^ for all in because
$ = = r1^ = r1^.^ = 
Let M = {aP: a  P and a,p e <3} and V = Q u AL We consider the
elements of Â£/ as spacelike planes and the elements of M. as Minkowskian or
Lorentzian planes. Thus, V consists of the totality of nonsingular planes and we
denote the elements of V by a, P, 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 Q and P e M then a e G and ft e M for every E, in Q.
For the following, let a, P e V with a p.
Definition 3.1. If a, p e Q, then aP e M by definition and we write alp and we
say a is perpendicular to or orthogonal to p.
Definition 3.2. Suppose that a e Q, P e AL
(i) If aP e Q, then we write a 1 P and we say a is perpendicular to p.
(ii) If aP g Q, then we write a J_ P and we say a is absolutely perpendicular to p.
Definition 3.3. Let X = {aP : a T P}. We call the elements of X points and we
denote these elements by A, B, C,....
Definition 3.4. If a, p e M. and a.p e A4, then we write a P and we say a is
perpendicular to p.
28
Definition 3.5. The point A and the plane a are called incident when A a. For each
a e V, set = {A : ^a}.se that A,B\a\,ct2 where A B and ai a2.
Suppose that A,B\a\,ap, where A B and ai a2. We define the line g
containing A and B as
g = gAB = [ai,a2] = {C e X : Cai,a2}.
We say that g is the intersection of a.) and a2 (Tai and X
(g a Xai,Xa2) The point C is incident with g, C e g, if Cai,a2.
If A B are two points such that there exist a and P with A,B a, P and alp
then we say that A and B are joinable and we write A,B e gAB = [a, p,aP]. If g is a
line which can be put into the form g = [a, P,ap] where a 1 P then we say that g is
nonisotropic.
If A and B are two distinct points such that there do not exist a, p with
A,B a, P and a 1 P then we say that A and B are unjoinable. If g is a line which
cannot be put into the form g = [a, P,aP] with a _L P then we say that g is isotropic
or null.
Incidence Axioms
Axiom 1. For each P.a there exists a unique p e V such that P\ P and a,p = Q.
Axiom 2. IfA,B a,p,s and C\a, P then C\s.
Axiom 3. IfP,Q a,P and a L p then P = Q.
Axiom 4. If a,p,y e Q are distinct anda ipiyla then aPy g Q.
Axiom 5. Ifa,P e M andaP then ap e M.
Axiom 6. For all A, B: A B, there exists a,P such that A,B\a,fi anda p.
Axiom 7. If a _L P then there exists A,B a,p such that A B.
Axiom 8. For all A,B,C\ ABC = D e X.
Axiom 9. If 0a,p,y,8 with P,y,5 1 a then Py5 = s.
29
Axiom 10. If A. B, C are pairwise unjoinable points and A,B  a then C\a.
Axiom 11. For all a e Q, there exist distinct p,y,8 e Q such that a 1 (5 1 y 1 a
but 8 / a, p,y.
Axiom 12. If A,B  a; a e Q, then there exists p e G such that A, B P and P J. a.
Axiom 13. For each P a, a e M, there are distinct points A, B\ a such that
P A,B and P is unjoinable with both A and B, but A and B are joinable, and if Ca
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 are joinable and A, B\ a, then there exists P 1 a such that
A,B\fi.
Axiom 15. If a, p,y are distinct with P,y J. a and A,B\a, fi,y; A B, then a = Py
and if A, Ry; y _L a then a = Py or p = y.
Order Axioms
Axiom F. (Formally Real Axiom) [21 j Let 0,E  a, a e M. with O and E unjoinable.
Let X, 8, r e V. If O \ X,8, r andL,6,r\ _L a then there is ay e V such that
0y, y 1 a, and EX5k5OE1^ = Ex^.
Axiom L. (LUB) If A c /C, A 0, and A is bounded above, then there exists an A
in K such that A > X, for all X in A and if B > X for all X in A then A < B.
Dilation Axioms
Axiom T. If O 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 O. Let P,Q e g; R,S e h; and T,U e k. Ifgpj  gqu and
gRT II gsu then gpR  ggS.
Axiom R. Let O e g,h; P,Q e g, and R,S e h. If gPR  gQS then g0,p0R II go.QOS
30
Polarity Axioms
Axiom U. (Ux subspace axiom) Let 0,A,B,T and C be four points with 0,Cy6;
A,01 a; 0,5P; with a i. y and pi 6. Then there exist A,,e e V such that A, is;
0,AOB\X; and 0,C\z.
Axiom SI. Ifg c Xa, a e Q, h cz Xp, P e Q, and there exist y,8 e P sucA that
>,8; y8 e gn A, g c Xy, and h a Xg then there exists e e Q such that g,h c= Xe.
(Ifg 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 p e V such that g c Xp and h <= Xp. Then either there are a,P' e V such that
P = apr, g a Xa, and h c: Xp' or for all A eg there exists B e h such that P and
APB are unjoinable.
This concludes the list of axioms. Note that by Axiom 5, aP M. for a 1 P,
a, P e V and if a, P e A4 are distinct with ap, then.ap 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 Q 0, so let a e Q. By Axiom 11, there exists y e Q such
that aly. Hence, ay e M. by definition and M. 0.
3.1.1.2. The elements of M. are involutions.
Proof: Let y e M. Then we may write y = aP where a,P e Q and alp. We have
yy = apap = aapp 1.
3.1.1.3. For every P e M. and for every Â£, e 0, p^ e M..
31
Proof: Let p = aja2 with a 1,012 e Q and ai a2 By our assumptions on Q and
because aj (X2 we have for all ^ e 0, a^a e Q and afla^ so that aa e M.
3.1.1.4. The sets Q and M. are disjoint, Q n M = 0.
Proof: Let Q1 = Q\M. Then Q1 consists of involution elements, Q' n M. = 0, and
V Q1 u M. = Qu M. Let y e Q' and ^6 0. Now if e M., then by 3.1.1.3 above
y = (y^)4 1 e M, a contradiction. Thus, Q' is an invariant system of generators and
without loss of generality, we may assume that Q r\ M. 0. M
3.1.1.5. If a e Q, P e M anda 1 p, then ap Â£ M.
Proof: If a.p = y e M. then a = py e Q where p,y e M and Py, which contradicts
Axiom 5.
3.1.2. Properties of V.
For the remainder of this dissertation, let the symbol < denote the phrase
if, and only if.
3.1.2.1. Zfa,p e V and Â£, e 0 then a T P if, and only if, T pi
Proof: First suppose that a,p e Q. Then a 1 P f) 1 p^ because a P implies
that a I P and a  P < a>  pi If a e Q and P e M and al P then ap = y e Q;
a^,y^ Q, p^ e M, and a^P^ = y^ e Q implies that a^ T pi Conversely, if
a; J. p^ then a^p^ = 8 e Q and ap = 8^ 1 e Q, so that a 1 p. Finally, suppose that
a,p e M. Then aip^ e M and ap e M <> a^p^ e M.
3.1.2.2. If a, p e V and ^6 0, then a i P < a^ i pi
Proof: Let a e Q and P e M and Â£ e 0. If a ip then ap so that a^pi Thus,
a^ T Ps or a^ T pi If a^ i p^ then a^P^ = y e Q. So we have ap = y^ 1 e Q by
the invariance of Q, which contradicts a ip. Hence, a^ i pi
Conversely, suppose that a^ i pi because a^ e Q and P^ e M by the
invariance of Q and M then a^ = y, p^ = 5 imply that y 15 and = ai P = 8^
by the paragraph above.
32
3.1.2.3. For each Â£ e , = V.
Proof: because V = Q u M, the result follows from the invariance of Q and M..
3.1.3. Properties of X.
3.1.3.1. There exists a point; that is, X 0.
Proof: Let a e Q. By Axiom 11, there exists y,p e Q such that a,p,y are distinct
and mutually perpendicular. By Axiom 4, aPy = a8 Â£ G, where 8 = Py e A4 and
a8 as ctp,y. Thus, a 8 and P = a8 is a point.
3.1.3.2. IfOa,p,y; a,p,y e Q\ a 1 P 1 y 1 a; then O = aPy.
Proof: By the proof of 3.1.3.1 above, P = aPy is a point and 5 = Py e M. because
P _L y with p,y e Q. So we have a 8, 08 because 0p,y. This yields A,0\a,8
with A a8 so that A = O by Axiom 3.
3.1.3.3. If A e X and 8 e V, then A 8; that is, a point does not equal a plane.
Proof: Let A ap with a e Q, p e M, and a ip. Suppose that A aP = 8 e V. If
8 e M. then we have a = PS e Q with P18 and p,8 e A4, which contradicts Axiom
5. If 8 e Q, then A = ap e Q and this contradicts the definition of a point.
3.1.3.4. The elements ofX are involutory.
Proof: Let A e X, so that we may write A aÂ¡5 with aip. In particular, ap = pa
and AA = apap = aapp = 1.
3.1.4. General Consequences of the Axioms
3.1.4.1. IfP a,p anda A. P then P = aP and ifP = ap then Pa,p and a L p.
Proof: If Pa,p and a_Lp then aP A for some A e X and we have A,Pa,p with
ai p. Thus, by Axiom 3, A = P.
If P ap, then by 3.1.3.4 ap. because P V by 3.1.3.3 then aip. Also
Pa. P and PP = a imply that the products Pa and Pp are involutory so that
Pa,p.
3.1.4.2. IfP  a then Pa e V; that is, Pa is a plane P and P\p.
33
Proof: By Axiom 1 there exists (3 such that P  p and P X a. Thus, P  a, P with a X P
so by 3.1.4.1 above, P ap and P = Pa. U
3.1.4.3. For each P e X and each Â£ e , P^ e X.
Proof: By the definition of a point we may write P = ap with a e M P e Q, and
a X p. By 3.1.2.2, a^Xp^ and P^aXp^ so that P% = a^P^ is a point.
3.1.4.4. IfP a,p and yXa,p then a = p. ( Given a point P and a plane y there exists
a unique plane a such that P  a and a X y.)
Proof: This follows immediately from Axiom 1.
3.1.4.5. There do not exist three planes pairwise absolutely perpendicular.
Proof: Suppose that a, p, and y are pairwise absolutely perpendicular. Then for
P aP we have P  a, P with a, p X y so that a = p, which contradicts a X p.
3.1.4.6. If A,B,C\a then ABC = D\a.
Proof: By Axiom 8, ABC is a point D and Da = ABCa = aABC aD.
3.1.4.7. IfA,B,C\a,$ thenABC\a,$.
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 (collinear) with the other
three.
3.1.4.8. If a 1,012,013X0. then axajai, = 014 e V and 014X01.
Proof: Let A = aai, B = aa.2, and C = aa.3. Then by Axiom 8, ABC is a point D
and D = aaiaa20UX3 = aaia2(X3. So Da, by 3.1.4.6. By 3.1.4.2,
aia2(X3 = Da = (X4 e V, D = aa4, and aia.2(X3 = c^Xa.
3.1.5. Perpendicular Plane Theorems
3.1.5.1. If a 1 P; Py = A and A \ a then aly. (A 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 = ap^ = fiAa = ya, so a Xy
34
or a L y ( note that a = y implies that a X P because A py and A a). Suppose that
a Xy so that B = ay. Then we have A,B a,y with a Xy, which implies that A = B by
Axiom 3. But if A = B then Py = ay and p = a, which contradicts pla. Thus, ay is
a plane and aXy.
3.1.5.2. Suppose that a T P; ^a,p,y,5; yXa; and 8Xp. Then 5 1 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 ya = 8p and 8y = ap e V as a JL p. Thus, 5 1 y.
3.1.5.3. If O = aaj =yyi = esj with a,y,s e M anda lylsla then ay = s.
Proof: Because a,y,e e M, then ai,yi,8 e Q. Because alyltia and
aa i = yy i = eei, then ya = y i a i; sa = sjai; ey = s i y i imply that
yi 1 ai 1 E i yi. Because 0ai,yi,8i, then O = aiyjS] by 3.1.3.2. Because points
are involutions by 3.1.3.4, l
3.1.6. Parallel Planes.
We say that two planes a and P are paralleldenoted by a  p, if a = p, or
there exists a y such that a, pXy.
3.1.6.1. Parallelism is an equivalence relation on the set of planes V.
Proof: That the relation is reflexive and symmetric is clear. For transitivity suppose
that a  P and P  y where a.p, and y are distinct. Then there exists 8,e e V such
that a, pX8 and p,yXe. Let A = a8, B P8, and C = pe. Then by Axiom 8 we have
D = ABC = a88ppe = ae and aXs so that a  y.
3.1.6.2 //a, pX8 and pXe then aXe.
3.1.6.3. If aXy, pX8, and y  8 then a  p. (Two planes absolutely perpendicular to
two parallel planes are parallel.)
Proof: If y = 8 then we have a. pX8 and the result follows. So assume that there is
an e such that y,8Xe. Then by 3.1.6.2 above, aX8, pXy, a,pXs, and a  p.
35
3.1.6.4. If a  p, y  5, and PT8 then aiy. (Two planes parallel to two absolutely
perpendicular planes are absolutely perpendicular.)
Proof: because a  p then a, PJ_e for some e and because Pthen aT5. Similarly,
y^Ls' for some s' and SJLp imply that yiP. Hence, a, P8 and pj_y yield aXy.
3.1.6.5. If a e M(G) anda  p then P e G(M).
Proof: Suppose that a e Q and a  P, so a, pXy for some y. Because a e Q then
y e A4, which implies that P e Q.
3.1.6.6. If a  P then a*=  P^ for every Â£, e 0.
Proof: Let a, pTy for some y e V. By 3.1.2.2, a^P^Ly^ so that  pX From
3.1.6.5 and 3.1.6.6 above and the invariance of Q and A4, we have that if a  P and
a e G(M) then P e Q(A4) and a^  P^ with a^P^ e G(M).
3.1.6.7. If a  p then Xa n jÂ£p = 0 ora = p.
Proof: Suppose that a, pXy and Pa,p. Then a = P by 3.1.4.4.
3.1.6.8. If A B then A / B; that is, AB ^ BA.
Proof: By Axiom 6 there exists a e V such that A,B a. By 3.1.4.1 and 3.1.4.2 we
may write A = aaj and B = acc2. Suppose that AB = BA. Then we have
aia.2 = a2(X so that aiT(X2 or ai 1 a2. But ai  CX2 because a.,(X2Ta so that
either ai = a2 or iÂ£a, n Xa2 = 0 If oti T a2 then by Axiom 7, Xa n 3fp 0 so
a\ = a2 and A = B. If a\(La2 then C a\a2 e Tai n Xa2 and again we have that
ai = a2. Hence, A B.
3.1.6.9.a. If Ab = A, then B A.
b. If A B. then A, B, and AB are pairwise distinct.
3.1.6.10. If A  a; 5p; Cy; anda  P  y then ABC\afy.
Proof: Let Aa = a', 5p = p', Cy = y'. Then a',p',y'la,p,y and
ABC aa'pp'yy' = a'pY = d'd with a/pY = 5'T8 aPy.
Conclusion: A  a; B  P; a  p, imply that B11 p.
36
3.1.6.11. IfAA' =BB'; A,A'\a; 5P; a  p then B'  p.
Proof: Because AA1 = BB1 then B1 = BAA1; p  a  a. From 3.1.6.10 above,
B1 = BAA' paa = p.
3.1.6.12. For each P and each P there is a unique a such that P  a and a  p.
Proof: By Axiom 1 there exists a unique y such that yJLp. because Py then Py = a,
aJLy and a  p. Now if P8 and 6  p then P\8,a, so that a = 5 by 3.1.6.7.
3.1.6.12.Let a,p e V and M e X. Then aip aMP and thus, a  aw.
Proof: Suppose that a_Lp. By Axiom 1 there exists unique y, 51M such that a_Ly and
8p. By 3.1.6.2 we have p,yXa and p8 which implies that yT8. By 3.1.4.1,
M = y8. thus
Conversely, suppose that a wi.p. As above, there exists unique y, 81M such
that aA/Xy and S_Lp. It follows that pi8, M y8, and a8 = aYÂ§ = awTP and
a = (a8)8P8 = p.
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.
3.1.6.15. IfA,Ba,p andAM = B then Ma,p.
Proof: By Axiom 6 there exists y,S such that A,A/y,8. because B = AM = MAM then
by 3.1.4.7 we have By,8. Thus, A,Ba,y,8 ; My,8; and A,B\p,y,8 so by Axiom2,
M\ p,a.
3.1.6.16. .IfA,B,C18; a,p,yT8; A a; 5P; Cy; a  p  y then ABC\aPy and
aPy T 8.
Proof: From 3.1.6.10 we know that D = ABC\ct$y = s. Because 8a,p,y then e8 so
that s 1 8 or eT8. If E = 8s then we have D,Â£8,s with 8Te which implies that
37
D E by Axiom 3. Let A aa', B = p(3r, and C = yy\ Because a  p  y then
a'  p'  y\ Thus, e5 = D = ABC = aPya'pY = ea'pY which implies that
8 = a'PY Because A 5 then we may write A = aa' = 88/ and by 3.1.5.1 we have
A led,a,8; aXar; and a X 8 which implies that a' X 8. Hence, od8 = a^a'pY = PY
is a plane, so pr X y'. By Axiom 7 and 3.1.6.7 we have p/ = y1 so that a1 = 8. But
this yields aX8 and a X 8, which is a contradiction. Thus, aPy _L 8.
3.1.6.17. IfA,B\a; A\a'\ a' Xa; B\ p; and p  a' then P X a.
Proof: By 3.1.6.16 above B = AAB\a'a'fi = p and p X a.
3.1.6.18. IfEa,c; a X p, andsXP then s X a.
Proof: If E\ P then the result follows from 3.1.6.7. Suppose that E / P and let
M = sp. Then, E E$ and E$ \ = a so that E,E$ \ a. Now EM = Â£eP = so that
M is the midpoint of E and E$. EM = Â£'Peae so that Â£,Â£'/a,ae and by 3.1.6.15,
Ma,ae. In particular, Ma and we have M\a,p,e; a X p; and PXe so that a X Â£
by 3.1.6.7.
3.1.7. Consequences of Axiom 11 and 3.1.6.18.
3.1.7.1. If A  a; a X P; and a, P e Q then there exists ay in Q such that A \ y and
y X p,a.
Proof: Let A \ 8 with 8Xp. Then 8 e M. and A \ a, 8 with a X p, pX8 so that a X 8
by 3.1.6.18. Because 8 e M, 8 X a, and a e Q then Â£ = 8a e Q, A\e, and Â£ X a.
We claim that Â£ X p. Indeed, pe = P8a = Spa = 8aP = ep with Â£,p e Q so that
Â£ X P or Â£ = p. But Â£ p because then 8a = P and a = 8p = P, as 8Xp. This
contradicts 3.1.3.3.
3.1.7.2. Every point may be written as a product of three mutually perpendicular
planes from Q.
Proof: Let A be any point. Then by definition A = ap for some a e Q and some
P e M with aXp. By Axiom 11 there exists y e Q such that y X a. By 3.1.7.1 there
38
exists a 5 e Q such that d8 and 8 _L a,y. Again by 3.1.7.1 there exists an s e Q
such that A s and 8 1 8, a. Hence, de,a,8; s,8,a e Q, and s 1 8 1 a 1 8 which
yields by 3.1.3.2, A = 8as.
We note that in this case, ap = A = a8s implies that p = 8s; that is, if A  P
and P e Ad then there exists Pi,p2 e Q such that dPi,p2 and p = piP2
3.1.7.3. If A I a then there is a P in Q such that A \ P and P J_ a.
Proof: This follows directly from the proof of 3.1.7.2, for if a e Q then we can find
8,s e Q such that A 8,e and 8,s _L a. If a e Ad then we can find
that A ai,(X2 and a 1,012 L a. That is if A a then there exists P1, P2 e Q such that
dPi,p2 and pi JL p2 Moreover, if a e Ad then we can find Pi,p2 such that
a = P1P2 and if a e Q then we can find p 1, P2 e Q such that (XJP1P2.
3.1.7.4. If A a e Ad and P a then there is ay such that A\y andy p,a.
Proof: Let 8d with 8p. Then 8 _L a by 3.1.6.18 and A s = a8. Because a J_ P and
A A A
8P then s p. If e_LP then we would have A \ 8,s with S,8P so 8 = 8 and a = 1.
Thus, sip. We note that if p e Q then 8,s e M and if P e Ad then s e Q.
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 Q, ifA,B e a,b then A = B ora = b. Hence,
by Axiom 6, for all A,B e X, there exists a unique g e Q, such that A, B e g.
Proof: Let a = [a. P], b = [y,8], A,B\a, P,y,8, and suppose that A B. Let Cea, so
that Ca,p. By Axiom 2 it follows that Cy,8 so C e b and a c b. Similarly, b c a
and a = b.
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,p] with d,5a,p. Then by Axiom 8 and 3.1.4.7, A8a,p and A8 e X.
A8 A by 3.1.6.9.
39
3.2.1.3. Suppose that ap = y.
a. Ifaai = PPi then ai J. Pi andaiPi = y, soy _L ai,pi.
b. If aai = PPi = YYi then a T Pi, ai _L p, yi _L ai,Pi, anda$\ = aiP = y.
c. If A  a, P then A \ y.
d. The line g = [a, P] = [P,y] = [a,y] = [a, p,y] is a nonisotropic line. Thus, a, p, and
y are three mutually perpendicular planes which intersect in a line.
Proof: a. From aai = PPi we have y = Pa = P i a i.
b. If ppi = yyi and aai = TYi then yi yPPi aPi and yi yaai = Paj.
c. Because A  a, p and ap = y then Ay = A aP = aPA = yA and A \ y.
d. By Axiom 7 there exist points A and B such that A,Ba,p and by 3.2.1.3.c. above,
A,B\y. Thus, A,B e [a,P],[a,y],[P,y] and [a,P] = [a,y] = [p,y] s [a,p,y] is a
nonisotropic line.
3.2.1.4. If A and B are collinear then A'~ and Bf are collinear for anyE, e 0.
Proof: This follows from the fact that A,B\a, P < A^,B^ a;,p;.
For each Â£ e 0, we define ciS = {C^ : C e a}. By 3.2.1.4 above, is a line
for every Â£ in 0 and if a = [a, P] then aS = [a^,p^].
Definition of parallel lines and planes. We say the line a is parallel to the
planea, denoted by a  a, if there exists a P such that a c 3Â£p and P  a; that is,
a = [P,y] and a,p 1 5 for some 5.
We say that two lines a and b are parallel, a  b, if there exists a, P,y,8 such
that a = [a,y], b = [p, 8], where a  P and y  5.
3.2.1.5. If AA1 = BB' 1; A,A'\a,a'; a a'; 5,B'p,p'; p ^ p', then gAA>  gBB>.
Proof: Let 5p* with p*  a (3.1.6.12). Then B' P* by 3.1.6.11. Let Bp*, with
Pi  aj. Then B1 \ P^ by 3.1.6.11 and B,B' \ p,Pi,p*,pÂ¡ which implies that
[p,pi] = [p*,pt] = gBB' by 3.2.1.1. Hence, we have gAA< = [a,ai], gBB< = [P*,pÂ¡] with
a  p* and ai  pi; thus, gAA<  gBBi. U
40
3.2.1.6. Two lines, a = [a,ai] and b = [P,Pi] are parallel precisely when there exist
A, A' e a and B,B' e b such that AA' = BB1 1.
Proof: Suppose that a  b. Then, a  P and ai  Pi. Let A,A1 e a and B e b, so
that B1 = AA' B\aa$, and aictipi = P,pi, by 3.1.6.10. Thus, B,B' e b with
AA' = BB'. IAA' = BB1 = 1 then A = A1. U
3.2.1.7. If a  b and b  c then a  c.
Proof: Let a = [a,a/]. From the proof of 3.2.1.5 above we may write b = [p, p'] with
a  P and a'  p' because a  b, and c [y,y;], where y  p and y'  p; because
b  c. By 3.1.6.1, a  y and a'  y' so that a  c.
3.2.1.8. For each line a and point A there is a unique line b such that A e b and
b  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 1, so gAD II by 3.2.1.6. Now suppose that A e c and c  a. By 3.2.1.7
above, c  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, A1 = AWX e c and A, = AYZ e gAD Thus,
1 AA1 = WX = YZ AAi implies that A1 = A\. Hence, A,A1 e gAD,c; A ^ A1,
because WX ^ 1 and gAD = c by 3.2.1.1. I
3.2.1.9. IfA,B e g, A B, and C e h then g  h if and only if ABC e h.
Proof: If ABC = H e h, then 1 AB = HC and g  h. Conversely, suppose that
g  h and put D = ABC. Then because A B,
1 AB DC and C e goc, with goc II g
Because g  h and C e h, then by 3.2.1.8, gDc = h and D e h.
3.2.1.10. If a  b then a b ora r\b = 0.
Proof: Suppose a  b and a n b 0. Let A e a,b and Cea with C A, B e b,
with A B. Then, a = gAc, b = gABi 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.
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, p,y e M. such that a = py and a = [a, p,y], then we say that a is timelike. If there
are elements a, P e Q such a p and a = [a, p,a6], then we say that a is spacelike.
Remark. Let A,Ba e Q with A B. Then by Axiom 14, there is a P such
that A,B  P and P 1 a. Thus, a = [a, p] 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. If A B\ A,B a,p with P 1 a, a e M.\ and A and B are unjoinable with P;
P\a,y; andy T a,p, then Ay = B.
Proof: First, ^4Y  aY, pY = a, p. If Ay = A then A y and A is unjoinable with P.
Similarly, By\a, P and By B. Suppose that Ay and P are joinable. Then there are 5
and e such that P,4y5,e and 5 s. But then P,A = />Y,(AY)Y 8Y,sY and 5Y 1 sY,
which implies that P and A are joinable. Thus, Ay and P are unjoinable, so by Axiom
13, Ay is unjoinable with A or Ay is unjoinable with B or Ay = A or Ay = B. Hence,
Ay = B.
3.2.2.2. Suppose that A B: A,B a,p with a i p, a e M; A and B are unjoinable
with P\a, P\y, y 1 p. then Ay = B.
Proof: First observe that because Pa,y; alp; and yip, then a 1 y by 3.1.6.18.
Then TYaY, pY = a,p and Ay is joinable with B. Ay A because A and P are
unjoinable as in 3.2.2.1 above. Ay joinable with P implies that A = (Ay)y is joinable
with Py = P. Hence, by Axiom 13, Ay = B.
3.2.2.3. If A and B are unjoinable then X e gAB precisely when X = A or X = B or X
is unjoinable with both A and B.
42
Proof: By Axiom 6, gAB = [a, P] for some a and P in V. Let X e gAB and suppose
that X A, B. If A is joinable with A then there exists y,5 such that y A. 8 and
A,Xy,5. By Axiom 2, B y,5 and A and B are joinable. Hence, X is unjoinable with
both A and B.
Remark. If X is unjoinable with A and B\ A,B a, P; then by Axiom 10, Aa, P
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,Z)a, P with a 1 P then A,B a, p by Axiom 2; that is, gAB = gcD
3.2.2.5. If A ,B,C\a are pairwise joinable and distinct, then each P \ a is joinable with
at least one of A, B, and C.
Proof: If a e Q then the result follows from Axiom 12, so assume that a e AT By
Axiom 13 there exist D,Ea 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 gpB. By 3.2.2.4
above, this implies that two of A, B, and C are unjoinable, which contradicts our
assumption.
3.2.2.6. If gAB is isotropic then gB is isotropic for all Â£, e .
Proof: If A^,B^ a,p with a 1 P then A,Ba^ ',p^ 1 and a^' ip^'.
3.2.2.7. If P\ a e M then there are at most two isotropic lines in Xa through P.
Proof: If gAp,gBp,gcp ^ 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.
3.2.2.8. By Axiom 13 and 3.2.2.7, for each P \ a, a e M., there are precisely two
isotropic lines in Xa through P.
3.2.2.9. Let gAB,gAc Ta, a e M, be two isotropic lines.
a. If A  (5, with pla, then g^4B = gAC.
b. If A  y, with yia, then g\B =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.
3.2.2.10. IfA,B,C a, a e M; gAB gAC are isotropic then gBc is nonisotropic and
there is a p with p A such that (3 1a and C = 2?P.
Proof: Because gAB gAc> then gBc is nonisotropic. By Axiom 14 there exists a y,
y\B,C, such that y 1 a. Let A  p, ply (Axiom 1). Because A  a, P; a _L y; and
ply, then by 3.1.6.18, a 1 p. If B$=B then B\P and gAB is nonisotropic. Similarly,
CP C and by 3.2.2.1, flP = C.
3.2.2.11. If A .B,C\ a, a e A4, are pairwise unjoinable then there exist p,y _L a with
A  p,y such that C = #Py.
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 p A such that P a. Because B and
C are unjoinable with A, then B,C / p and BP B. By Axiom 1 and 3.1.6.18, there
is a 81 p such that 6 1 P and 8 1a. Thus, B,B$ 18, Â§P = 6, a and B and i?P are
joinable. Suppose that BP and C are unjoinable. By 3.2.2.3, either ggPc = gAC or A is
joinable with Z?P. But if A,B^\z, e a, then A = A$,B\z$ _L aP = a imply that A
and B are joinable. If = gAc = gAB then fiP is unjoinable with B. Hence, BP and
C are joinable and by 3.2.2.10, there is a y with y A such that y _L a and C = B^
3.2.2.12. Let P a,p,y with p,y J. a, a e M. If gAB c: Xa is isotropic then g^g c Xa
is isotropic and gAB  g%.
Proof: By 3.2.2.8, there are precisely two isotropic lines in Xa through P, say gpc
and gpQ. By 3.2.2.9, g^PC = gpQ and g^pC = 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,D\a' with a1 _L a. Let B\X with X  a'. Then because 5a, X a by 3.1.6.17.
But then by 3.1.6.16, A = PDB\a'a'X = X and A is joinable with B.) Because
44
A,B,P  a then D = PAB a. Hence, D e gPc or D e gpQ, and gAB II gPC or
gAB II gPQ thus g% II gpQ = gPQ II gAB or g% II gpc = gPC 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 B, be any element in 0; then the map : X ^ 3Â£ given by az(A) = B/1B,~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 (<3,Q,X). In the following, for any element
B, e 0, the collineation induced by it is denoted by a^.
Let a e V be fixed throughout this section. We wish to define a set Ca of
maps on 3Â£a which can be viewed as line reflections in a given plane 3fa. We then
show that each element of Ca is involutory and that Ca forms an invariant system of
generators within the group Qa it generates. Finally we will show that (Ca,Ga)
satisfy Wolffs axioms [30] for a two dimensional Minkowski space; that is; the
Lorentz plane, for a 6 M..
Let Â£a = {g c Xa : g is nonisotropic}. By Axiom 14, each g e Ca may be
written uniquely in the form g = [a, p,y] where a = Py. Let A e Xa and
g = [obP>Y] e Ca. because A  a and a = py, then A Aa = A^ and A$ = Ar. Hence,
we define the map ag : Xa >* Xa by ag(A) = A8 = A$ = Ay, for A e Xa, and
dg : Ca ^ Ca by Gg{h) = hg = {og(H) : A e h}. Note that by 3.2.1.4, if
h = [a,8,s], then ag(h) = [a,5P,sP] = [a,8y,sY].
3.3.1. For each g = [a,p,y] e Â£a, : Xa La is bijective.
Proof: If A,B \ a and og(A) ag(B) then Ay = By and A = B, so ag is injective. If A \ a
then Ay \ ay a because aly and cg(Ay) = (Ay)y = A, so ag is onto.
3.3.2. For g = [a, P,y] e Â£a,
Proof: Let h = [a,s,r] e Â£a. Because a 1 y and e 1 r\ <> eY 1 r\y by 3.1.2.1 then
45
a = aY = eyty. It follows that ag(h) = [a,Â£Y,rY] e Â£a. If^s,r) then ^4Y  sY,rY so
ag(A) e og(h) for all A eh and ag is a welldefined collineation.
If l = [a,A.,p] e Â£a and cg(h) = ag(l) then for every A in h we have
ag(A) = Ay e og(t) = [a,X.Y,pY].
This implies that A = (Ay)y e [a,A,,p] = /;that is, h = l. Hence, ag : Ca Â£a is
injective. Because Â£_Lr<sY_LrY then it follows that
h = [a,s,r}] e Ca <> csg(b) = [a,EY,riY] e Â£
Hence, og : Â£a 1 Â£a is surjective.
3.3.3. Each <5g, for g e Â£a, is involutory.
Proof: Let g = [a, p,y]. Then for A  a we have ogog(A) = og(Ay) = (Ay)y = A.
Let g = [a,8,y], h = [a,E,r] e Â£a. We say that g is perpendicular to or
orthogonal to h, denoted g _L h, if one of y and 5 is absolutely perpendicular to one
of e and r and g n h 0.
3.3.4. For g and h as above, if y.z then 8r), Sis, andy T Â£. Moreover,
A/ = yB = 8r) eg,/?.
Proof: Let M = ys. Because y,Eia then hf1 = (ye) = yaÂ£a = ys M and M\ a. So
we have Ma,Â£ which implies that Mas = r and M e h. Also, Ma,y so May = 8
and Meg. Also from M = ye it follows that Me = y = 8a = 8r)s so that M = 6r\ and
8lr. Applying 3.1.6.18 to M\y,x\ with y 1 8, r8, and M8,e with 8 1 y, Ely, we
obtain y A. rj and 8 1e.
3.3.5. If a,b,g e Â£a then a J_ b a8 X b8.
Proof: By 3.1.2.1 and 3.1.2.2 we have
o c Xy; yi.8; c ^ ^ c Xy; y^Â£8?; b*= c 3fg, for all ^ e .
The result follows.
46
Remark. Because A\z <> A^\e^ for all Â£, e 0, A e X, and s e V, it follows
that for each g e Â£a, Gg maps:
(i) The set Xa onto itself.
(ii) The set Â£a onto itself.
(iii) Collinear points in Xa onto collinear points in Xa.
(iv) orthogonal lines in Xa onto orthogonal lines in 3fa
That is, ag is an orthogonal collineation of Xa
Let Ca = {dg : g e Â£a} and = {a/> : P e Ta}.
3.3.6. The sets Â£a and Xa are nonempty. Hence, Ca ^ 0 and 0.
Proof: If a e M. then we may write a = a/8 where a\8 e Q and a' T 5 by
definition of M.. By Axiom 7, there exist A,B such that A B and A,B\
A,Pa'8 = a and Xa 0 Moreover, gB [a, a', 8] e Â£a and Â£a 0. If a e Q
then by 3.1.3.1, there exist A  a and by 3.1.7.3, there is a p in Q such that A  P and
p 1 a. Thus, A e Xa and g [a,p,aP] e Â£a
Note that from 3.3.3, Ca consists of involutory elements and by 3.1.3.4, *}3a
consists of involutory elements.
3.3.7. IfP e g,h e Â£a and g X h then gp = aga/, = G^Gg.
Proof: Let g = [a,y,8] and h = [a,s,r], and without loss of generality, assume that
yXs, so that 8Xri. By 3.3.4, P = ye = 8r) e g,h and by 3.2.1.1, {P} = gr\h. Then
for A e Xa,
GgG/}(A) = ag(Ae) = A^ = ap(A) = Aye = G,My) =
3.3.8. For every A in Xa and for every h e Â£, there is a unique g e Â£a such that
A e g and g X h.
Proof: Let h = [a,s,r] e Â£a and A e Xa. By Axiom 1, there exist a yA such that
yis. Thus, A a,y with a X s and yXs so a X y by 3.1.6.18. because a X y then
ay = 8 and A  a,y implies A  8 so that A e g = [a,y,8] e Â£a. because g c= Xy, yle,
47
A s g = [a,y,8] e Â£ Because g c 3Cy, yis, and h c XE, then g J_ h. By 3.3.4
above, ri_8 and {B} = {ye} = {r)5} = g n h.
Now suppose that / c Ca such that A s l and / 1 h. Now we may uniquely
write / = [a,a',p] in .Ta where a = a'p. because / T h then without loss of
generality, we may assume that as and P_Lt]. because yTs and 8i_r it follows that
a  y and P  8. By the definition of parallel lines in Section 3.2 we have /  g.
because A s l ng then by 3.2.1.10, / = g.
3.3.9. (i) The fixed points ofag e Ca are the points in Xa incident with g.
(ii) The fixed lines of ag e Ca consist of g and all lines in Ca which are
orthogonal to g.
Proof: Let g = [a,y,8]. If A = ag(/l) = A'1 for A \ a, then A \ y. because Ay = A5, VA  a
then As = A and ^418. Thus, A e g and 3.3.9.(i) follows.
Let h = [a,s, q] e Ca and suppose that ag(h) = h; g h. Then there is an
A e h such that A <Â£ g. By 3.3.8, there is a k [a,co,0] e Ca such that Ask and
k J g. Because Ash, ag(A) = Ay s h and because A s k, k J. g then
AY  coy,9Y = to,0 and AY s k. Because A <Â£ g then A / y (for if A \ y then A \ a implies
that A  ay = 8 which implies that A s g) and A AY. Thus we have A,Ar s h,k, so
that h = k by 3.2.1.1. Hence, h X g.
3.3.10. (i) The only fixed point of a point reflection, op, is the point P.
(ii) The fixed lines of dp are the lines incident with P.
Proof: If PA = P, then this implies that APA = A, or AP = 1, or A = P. Thus,
3.3.10.(i) holds. Let x = [p,u] be any line (not necessarily in 3Â£a or nonisotropic) and
A s x. Then PA s x implies that PA \ p,u, so that /*! P'4,v>'4 = p,o, and P s x.
Let Qa be the group acting on Xa generated by Ca. By 3.3.5, every element of
Qa is an orthogonal collineation of Ta Let cr e Qa. By a transformation with cr, we
mean the adjoint action ofaonia : C s Qa > cjj7 = ckti
such transformation is an inner automorphism of the group.
48
3.3.11. Let h = [a,Â£,r], g = [a,7,5] e Â£a, then Ggh = aa,,(g)
Proof: Let A  a and put B = ag(/l) = Ay. Then
opiVhCA)) = g hGgG^Gh(A) = <*h<* g() = oh(B).
Thus, Og''(o/,(A)) = Oa/,(g)(^/;(d)) for all A e XaBecause G/, is injective, then
Ggh = aCT/i(g) for every A in .Ta Similarly, Ogh(a) = Ga/i(g)(a), for every a in Â£a.
Hence, Ggh = Goh(g) Analogously, we can obtain G^h = G0h(P), for every P in Xa B
Because Ca generates Qa and every element of Qa is an orthogonal collineation
of (Xa,Â£a) then we obtain the following.
3.3.12. For every g e Qa and for every gg e Ca, we have a e Ca; that is, Ca is an
invariant system ofQa B
Consider the two mappings: g e Â£a 1 ag e Ca and P Ta *+ ap e iPa
The first one is from the set of nonisotropic lines in Â£a onto the set of line
reflections, and the second is from the set of points in Xa onto the set of point
reflections in 3Ca 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. Ig e Qa and P,Q e Xa then g(P) = Q <> Gp = gq.
Proof: Because
G(P) = Q <> Ga(P) = Gq <> Ga(P) = Gp,
the result follows. B
3.3.14. Ig e Qa and g,h e Â£a then G(g) = h < ag = a/,. B
3.3.15. IfP e Xa andg e Â£a then P e g GpGg is involutory.
Proof: Suppose that P e g = [a,y,8] e Â£a. Then for A\a,
GpGg(A) = Gp(Ay) = Ayl = A11 = GgGp(A). Thus, (GpGg)2 = lxa Now assume that
49
GpGg is involutory. Then P = GgGpGgGp(P) = P= Py/>y. Because yPy Py e 3f
by 3.1.4.3, then by 3.1.6.9, P = Py. Because Py = P8 then Py,8, and hence, Peg.
3.3.16. Ifg,h e Â£a, then g 1 h *> GgG/, is involutory.
Proof: Nowg h if and only if Gg(h) = h. For g h, by 3.3.9, Gg(h) h if and only
if = Gh For Gg Gh by 3.3.14, Gg = O/, if and only if (GgG/,)2 = ljra and
^ lia
3.3.17. The point reflections *pa are the involutory products of two line reflections
from Ca
Proof: Let Gp e ^3a where P e jÂ£a. Then we may write P = y8s where a = y8 and
y,8,s e Q by 3.1.7.3. Let g = [a,y,S], Because a = y8, then g e Â£ By 3.3.8, there
is an / in Â£a such that Pel and Id. g. By 3.3.7, Gp G/Gg GgGÂ¡.
We now show that if a e M., then the pair (Ca,Ga), acting as maps on
(Â£,Â£,), satisfies Wolffs axiom system [30] for his construction of
twodimensional Minkowski space. We give Wolffs axiom system below.
Basic assumption: A given group 0, and its generating set G of involution elements,
form invariant system (Â£/, 0)
The elements of G will be denoted by lowercase Latin letters. Those involutory
elements of 0 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 l with P,g\ l.
Axiom 2. IfP,Q\g,l then P = Q or g l.
Axiom 3. IfP\a,b,c then abe e G.
Axiom 4. If g\a,b,c then abe e G
Axiom 5. There exist Q,g,h such that g\h but Q / g.h.gh.
Axiom 6. There exist A, B; A B, such that A, B l g for any g e G (There exist
unjoinable points.)
50
Axiom 7. For each P and A,B,C such that A,B,C\g, there is a v e Q such that P,A  v
or P,B\v orP,Cv.
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 a\b), points. We say the point A and the line b are
incident if A \ b; the lines a and b are perpendicular (orthogonal), if a\ b. Further we
say two points A and B are joinable when there is a line g such that A,B\g.
Replacing Ca with Ca and Xa with the pair (Ca,Ga) 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 Ca with Ca, Ca with Q, and Qa
with
Verification of Wolffs axioms. Axiom 1 follows from 3.3.8 and Axiom 2 from
3.2.1.1. For Axiom 3, let a = [a,a',s], b = [a,p,p'], c = [a,y,y'] e Q. Then
a',p,y T a, and by our Axiom 9, a'Py = 8 i a and d = [a,8,a8] e Â£ For A a,
Hence, aaobGc = cd e Ca. If we then identify  with e, we get Wolffs Axiom 3.
Axiom 4. Ifa,b,c\g, then abe e Q.
Proof: Let g = [a, a = [a, a', 8], b = [a,p,p'], and c = [a,y,y'], with
a',p,y 1 X. Then by 3.1.4.8, a'Py = s 1 X. Let A e a, B e b, and C e c, then
A\a',a Ra,p; and Ca,y. By 3.1.6.10 ABC = Z)a'Py = s and ABC = Da by
3.I.4.6. Thus, Ds,a; s i X; a 1 X, and s 1 a. So, d = [a,8,as] Ca. And if X\a,
then
Axiom 5: There exist Q,g,h such thatg\h but Q / g,h,gh.
51
Proof: Because a e M, we may write a = Py, where P,y e G and p L y. By 3.2.1.2
every line contains at least three distinct points. Let g = [a,p,y] e Â£a and let
Beg. By 3.3.8, there is an h in Ca such that B in h and h _L g. By 3.2.1.2, there is
an A in h such that A B. By 3.3.8, there is an / e Ca such that A e l and / J. h.
By 3.2.1.2, there exists Q e l such that Q A.
Now if Q e h, then A,Q e l,h implies that Q = A or / = h, so Q Â£ h. Suppose
that Q e g. Let Q e m. m 1 /, so that gq = GÂ¡Gm. Because gb = QgGh, Ga = G^G/,
and BAQ = E e 3Ca, then
GE = GbG.4Q = &g&hhÂ¡GÂ¡Gm = GgGm and g L ITl.
Because Q e g,m and g _L m then gq = GgGm GÂ¡Gm and Gg = gÂ¡ or g = l. This
implies A,B e Lm. Thus A = B or / = m, a contradiction.
Axiom 6: There exist A, B; A B, such that A. B / g for any g e Q.
Proof: This follows from our Axiom 13.
Axiom 7: For each P and each A, B.C such that A,B,C\g, there is a v in Q such that
P,A  v, orP,B v, or P,C v.
Proof: By our Axiom 13, there exist two isotropic lines in .Ta 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.
3.4. Consequences of Section 3.3
3.4.1. For every A and B. there exist a e M such that A.B\a. {Every line lies in
some Minkowskian plane.)
Proof: By Axiom 6, there exist a e V such that A,B a. If a e A4, the result
follows. So suppose that a e Q. By Axiom 12, there exist P e Q such that A,B\$
and P i a. Then A,B ap = y, and y e M. by the definition of M..
52
3.4.2. For every A and B. there exists M 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,Ba. From Section 3.3, there
exists M a such that AM = B.
3.4.3. If A A1 = BB1 then A and A' are joinable precisely when B and t are.
Proof: Suppose that A and A' are joinable and let A,A1 a,a/ with a J. a'. By 3.4.2,
there exists an Msuch that AM = B. Then B = AM \aM,a'M and from 3.1.2.1 and
3.1.6.13 it follows that aM T a'M, a  aM, and a'  a'u. By 3.1.6.11,
AA1 BB'; A,A1 a,a;; B\aM,a'M; a  aw; and a  a'M.
Thus, t \ and, B and Bf are joinable.
3.4.4. If a 1 a'; a  p; a'  p'; and [ a,a']  [p,p'], then p 1 p'.
Proof: Because [a,a ]  [p,p ], then there exist A,A1; A,A1 a,a/ and there exist
B.B1; B,B' pp,) such that AA' = BB'. Then for AM = B. as in the proof of 3.4.3,
[p,p ] = gBB' = [au,a'u]  [a,a']. That is, 5aM,a,A/,p,p'; with P,aM  a, and
P ,a w  a'.Thus, P = aM and P' = a'M by 3.1.6.12. Hence, P T P' because
aMla'M. U
3.5. Construction of the Field
The basic construction. For the construction of the field we will follow the
path of Lingenberg [20]. Throughout this section let a e M and 0a be fixed.
Define the sets:
Oa = {g e Ca : O e g} and Va(0) s : g,h e Oa).
Proposition 3.5.1 The set T>a(0), acting on the points ofXa, is an abelian group.
Proof: By 3.1.7.2 we may write O = ap = yrp with a = yr; y,rpP e Q\ and y,ri, and
P mutually orthogonal. Thus, g = [a,y,r] e Ca; O e g, and Oa 0. Because each
53
Gg e Ca is involutory then l.ta e Ua(0). Now let Ga,Gb,Gc,a(0) where
a = [a,a',aa'], b = [a,p',ap'], c = [a,y',aY'], and d = [a,8,a5] are nonisotropic.
Now 8y'p' = 8 X a with 0\e by Axiom 9 and/ = [a,e,as] e Oa. Thus, for A a,
GaGftGcG^X) = XÂ§Y P a = Aea = GaG/(X).
Hence, OaO^OcG^ = ao/ e Va(0). Because each G/ is involutory for / e Ca,
071 = 0/ and (g0g,)1 = GGa e T>a(0). From Axiom 9 and the calculation above,
the product of any three of 8,Y,,pi, and a' is an involution.Thus,
^Sy'p'a ^p'y'Sa' ^p'a'Sy' an(j aaabGcad = GcGd
Clearly, Va(0) is associative so that Va(0) is indeed an abelian group.
Lemma 3.5.2. Let g be an isotropic line in Xa with O e g. Then for every
G,Gh e Va(0), a,Gh(g) = g.
Proof: Let / = [a,p,aP] and h = [a,y,ay] be lines in Ca with O e /,/?. By 3.2.3.12, if
0\ p,y with p,Y T a then gPy  g. But = O so that G/,G/(0) = C>PY = O. Thus,
gPv = g by 3.2.1.10. Hence, G/,g/(C) = g, for all 0/,0/ e Va(0).
Lemma 3.5.3. Let g be an isotropic line in Xa through O. Then for every A, E e g,
E O, A O, there is a unique o/o/, e Va(0) such that g/G/,(Â£) = A.
Proof: By 3.2.3.11, there exist y,8 X a with 0y,8 such that X78 = A. Take
/ = [a,y,ay] and h = [a,8,a5]. Note that if E = A then X78 = E implies that EA = X8
or E1 = Eh, which implies that l = h [30]. To show uniqueness, suppose that
Gac/GkGi e T>a(0), where / = [a,8,aS], k = [a,8,as], g = [a,y,ay], and
h = [a,p.ap] are lines in Xa(0) and E Eha = Elk. Then Ehakl = E and E^ = E.
By Axiom 9, Pys = X with 0\X, X X a. Thus, m = [a,X,a>.] e Oa;GkGaGh = Vm, and
E = Eml. Let Xa',8' with a'XA., S;X8, and put M = a'X and N = 8'8. Now
EI a',a,8' with a'iA, a X X, 8' X 8, and a X 8, so that a',81 X a by 3.1.6.18. It
follows that AT = a'aXa = a'X = Man N = 8'8a = N So that M\a,X ; Aa,S;
and m go.\i and / = goN Then because Emi = E, we have EK& = E; El = Xs and
54
Em = Ea X = Â£8 8 = En and M = N by 3.1.6.14. So m = goM = gON = / and ct/ = om.
Therefore aa<7/, = ct*ct/.
Let us denote this unique map by 8a So for all A e goE A O,
(i) 8^(0) = 0
(ii) 8(Â£) = A
(iii) 5A e X>a(0).
Translations. For every pair A, 5 of distinct points we can define a translation
Tab : 3E > X given by Tab{A) = AAB = fi. We now restrict our attention to a set of
translations defined on Xa and we note that if A,B,Ca. then D = ABCa by 3.1.4.6.
Thus, Tab ^a 1 Xa is a welldefined map for all A,Ba. Let 1C = goE be an
isotropic line in Ta and define Ta = {Toa 'A e 1C}.
Theorem 3.5.4. The set Ta is an abelian group.
Proof: Let Toa^ob e Ta. Because C = BOA = AOB e 1C, for X\a, we calculate
{Toa o Tob){X) = TO{XOB) = XOBOA = XOC = TOC(X). Thus, Toc = TOA TOB e Ta.
Also, Too(X) = XOO = X. Hence, Too 6 Ta is the identity l.ra on Xa. To find TlqA,
we compute
(Toa TOo){X) = XOAOA = XOOAOOA = X= XOAOA = {TOAo TOA){X).
Hence, ToA = TOAo, A = OAO e 1C, and TqA e Ta. Clearly,
Toa {Tqb Toe) = {Toa Tob) 0 Toc for A,B,C e 1C. 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 1C and X\a,
{Toa Tob){X) = XOBOA XOAOB = {TOB TOA){X).
Therefore, Ta is abelian.
Lemma 3.5.5. For all T0a e Ta, Toa{1C) = 1C.
Proof: Because OAB e K for A,B e Â¡C then Toa{1C) c K. for all TOA Ta. Now let
55
C e 1C and A e 1C. Then OCA = D e K so that C = ODA and T0a(P) = C. Hence,
each Toa e Ta maps 1C onto 1C.
Lemma 3.5.6. Let T0a e Ta and g c Xa Then TOA(g) = g if and only if g is
parallel to 1C.
Proof: Let gHF = gbea line in 3Â£a such that ToAigHF) = gHF Then
Toa(H) = HOA e gHF and gHF II 1C by 3.2.1.9. Conversely, suppose that gHF II 1C.
Then again by 3.2.1.9 it follows that for B e gHF,
Toa{B) = BOA g gHF and T0A(g) = g.
Lemma 3.5.7. If A. B e h, h a Xa, and h  /C, then there exists Toe e Ta such that
Toc(A) = B.
Proof: By 3.2.1.9 we have C = OAB e 1C and it follows that B = AOC = Toc(4).
Lemma 3.5.8. For each A e 1C, there is a unique Toa e Ta such that Toa(0) = A.
Proof: Clearly, Toa(0) = OOA = A. So suppose that TociP) = A. Then
ooc = c = a. m
We denote this unique translation mapping of O into A by TA.
Lemma 3.5.9. 7/
Proof: Let a = CgCSh where g = [a,p,ap], h = [ay,ay] e Oa. Then for any X\a,
(agchTAahGg)(X) = QfrOAyrt = = XOA# = TagaiMU0
By 3.2.3.8 there exist precisely two isotropic lines in Xa through O : 1C = goE and
1C = goF We define multiplication and addition on the points of 1C so that the
points of 1C form a field. For A,B e 1C, define:
A + B = (Tb o Ta)(0)
A B = (8b 8a)(Â£), where A,B O and E e 1C is the multiplicative identity.
A O = O A = O.
Theorem 3.5.10. For every A, B e 1C, Ta+b = TA Tb
56
Proof: For X\a, TA+B(X) = TAOB(X) = XOAOB = XOBOA = (TA TB)(X). U
Theorem 3.5.11. For all A,B 0 in tC, 8A.B = 8A 8B.
Proof: Let 8,1 = GaGa' and 8B = GbGb'. Then A B = 8B8A(E) = Eaabb. Put
gc = Gb'GaGa', then, Gbac e Va(0) and GbGc(E) = Ecb = Ea ab b = A B. Hence, by
3.5.1.3, 8A.B = CTbGc = G bG b'GaGa' = GaGb'GbGai = GaGa'GbGb' = 8A8B.
Hence, (/C,+) is a group isomorphic to Ta and (JC\{0},) is a group isomorphic
to T>a{Q). It remains to show that the distributive laws hold.
Theorem 3.5.12. Let A,B,C e K, then (A + B)'C = A'C+B*C.
Proof: If C O, then
(A + B)C=0=00 + 0 0 = A C+ BC.
If C O, then we compute
(A + B) C = 8C(A + B) = 8cTa+b(0) = 8cTA+B8~c\0) = 8cTATB8~c\0) = 8cTA8~8cTB8
= Tr(A)Tbr(B)(0) = Ta.cTb.c(0) = Ta.c+b.c{0) = A C + B C.
Because multiplication is commutative,
C(A + B) = (A + B)C = AC + BC = CA + CB. Hence, (Â£,+, ) is a field.
Ordering the field and obtaining M. To order the field K. we make use of the
following [21]. Let F be a field and A\,...,A e F. If A\,...,A 0 implies that
Xit 0 then F is called formally real.
Theorem 3.5.13. (ArtinSchreier) Every formally real field can be ordered.
To make the field K, formally real the following axiom is posited.
Axiom F. (formally real axiom). Let O, E\ a, a e M with O and E unjoinable. Let
A., 5, rj e V. If 0\f,8,r\ andL,8,r\ A. a then there is ay e V such that
0y, y 1 a, and E^OE^i = E^xL
Theorem 3.5.14. If Axiom F holds on 1C, then /C is formally real.
Proof: Let t [a,A.,aX] e Oa and let gsgb e Va(0). Then from Proposition 3.5.1,
57
GgG/,Gt = &i for some l e Oa and OgGh = CJ/CT/. Clearly, if / e Oa with G/'Gi = G/GÂ¡
then we have cry = gÂ¡ and l1 = l. Hence, for all GgGÂ¡, e T>a(0), there is a unique
/ e Oa, such that GgG/, = G/Gt. So if O A e 1C, then we may uniquely write 5a in
the form GaG,, that is,for every O A e 1C, there is a unique a, a = [a,a',aa] e Oa
such that
A = GaGt(E) = E,a = EXa'.
Suppose that O A = EXa Then A2 = EXa Xa = O implies that
E = Oa xa x = O, a contradiction. Hence, if O A e 1C then A2 O. Now suppose
that A,B e 1C and A,B O. Let a = [a,a\aa'], b = [a, p,aP] e Oa such that
A = EXa and B = Ex$. Then by Axiom F there exists a y e V such that 0y JL a
and
A2 + B2 = EXa'Xa'OExVW = Ex^. (3.5.1)
Since 0y X a then c = [a,y,ay] e Oa,GcG, e Va(0), and
C = gcg,(E) = E,c = Exy e 1C.
So equation (3.5.1) reads A2 + B2 = C2. If C2 = EXyXy = O, then E = OyX^x = O.
Since E O, it follows that if A\,...,A e 1C are all nonzero, then A2 O and
1C is formally real.
To finally obtain a field isomorphic to the real numbers, R, we add the least
upper bound property to our axiom system.
Axiom L. 7/0 A c K. and A is bounded above, then there exists an A e 1C such
that A > X, for all X e A, and if B e 1C 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 Held 1C 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 /C.
3.6. Dilations and the Construction of (T, V,/C)
The additive group V and dilations. First we construct a vector space V over
the field /C. 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 = O(OAB) = OOAB. We view the elements
OX e V as directed line segments with initial point O and terminal point X on the
line gox We define an addition on V by setting OX + OY = 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 1 since 1 = OO e V and
1 + OX = 1 & OX = OX = OX+ 1. To complete the proof we calculate
OX+ OX = OXOX = OXOOXO = OXXO OO = 1, so OX = OX.
(OX+ OY) + OZ = (OXOY)OZ = OX(OYOZ) = OX + (OY+ OZ).
Hence, (V ,+) is an abelian group.
We still need to define a scalar multiplication of /C 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. [23, p.37]
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 8(dQ = ^ and 8(h) = Y of two
points X,Y e X are known. We must show that the image of any Z e X is known.
Suppose that Z Â£ gxY Then clearly Z X and Z Y and we consider the two lines
gxz and gyz Observe that Z e gxz^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 g gxY Therefore, {Z} = gxz n gYZ Since 8 is bijective, from settheoretic
reasons alone, 8(gxz n gYZ) = 8(gxz) n 8(gyz); or equivalently,
{8(Z)} = 8(gxz) ^ 8(gYz) In other words, the lines 8(gxz) and 8(gYz) have precisely
one point in common, namely, the point 8(Z) for which we are looking. The line
8(gxz) is completely known because it is the unique line which passes through X! and
is parallel to gxz Similarly, the line 8(gYz) is the unique line which passes through
the point Y1 and is parallel to gYZ because 8(Z) is the unique point of intersection of
8(gxz) and 8(gYz) (3.2.1.1), the point 8(Z) is completely determined by X1 and Y.
Conversely, assume that Z e gxY If Z is X or Y, we axe given 8(Z), so assume
that Z X and Z Y. By 3.4.1, there is an a e M. such that X, T a and there exists
a P  a such that P gxY Then Z gxp and from the previous paragraph, 8(P) is
known. Hence, using the line gxp instead of the line gxY, we conclude from the
earlier proof that 8(Z) is known.
To define a scalar multiplication, we fix a timelike line t and use it to
geometrically define dilations. To aid in the construction, the following facts are
used to add the appropriate axioms.
In a threedimensional or fourdimensional Minkowski space, if t is any
timelike line through a point O and g is any other line through O, 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.
Desargues 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 gi, g2, andgj be any three distinct lines, not necessarily coplanar,
which intersect in a point O. LetP\,Q\ e gi; P2,Q2 e g2i and P2,Q2 e g3. If
gP\Py II gQtQi andgp2p3 II gQ2Q3 then gP]p2  gQ{Q2.
Axiom R. LetO e g\,g2\ P\,Q\ e gi; andP2,Q2 e g2. Ifgp{p2  gQxQ2 then
gO,P\OP2 = gO,Q\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 O. A scalar multiplication is then
defined in a manner similar to the definition of multiplication for the field elements.
Axiom D is Desargues axiom, the dilation axiom. Axiom D ensures welldefined
dilations with the standard properies of such maps. Axiom R is used to distribute a
scalar over the sum of two vectors. The dilations are constructed next.
Let /C c= 3fa, a e M. By 3.1.7.2, there exist ai,(X2 e Q such that 0ai,oi2
and a = ai(X2. Because 0a, then Oa = p e Q and O = ap = ai
by 3.1.6.18. Let y = aiP e M. Then ya = Paiai(X2 = Poo e M., so that
t = [a,y,ay] c Xa is timelike and O e t as 0a,ai,a2,P implies that 0a,y,ay. So
let t = [a,y,5] Oa be any timelike line through O and K.' cz Xa the other isotropic
line through O. Then for each At e t, there is a unique A e 1C and there is a unique
B e K! such that At = AOB. Because At e t, then B = A'. (From this point on
denote a line reflection ag(A) by Xs. Thus, X%hr means arai,cg(X) = aga/,ar(A).).
For each At e /, there is a unique A e JC such that At AO A1. Similarly, for
each A in 1C, there is a unique At in t such that At = AOA. Thus, there is a
onetoone correspondence between the points of /C, the field, and the points of t.
Fix Et = EOE1 as the unit point on t. For each O A e 1C, we use t to
construct a dilation 5 a of X in the following way. Let X e X.
61
If A g t, then by Axiom T, there is a unique r e M. such that goxA c Xri
From Sections 3.5 and 3.6, JCtj is an affine plane and our definition of parallel lines in
An is equivalent to the affine definition. So, there is a unique line h c An such that
At h and h  gxE, Because
goxi^t = {O} 0, hnt = {At} 0, and h  gE,x,
then hr^gox = {B} 0. In this case, set = B.
If A e t and A O, then because t,lC a Xa with a e M., there exists a
unique g c Xa such that A e g and g  gxE Because gxE ^ t = {E} ^0 then
gnt= {B} 0. Set 8a(X) = B.
If X = O, set 8a(X) = O. Note that e gox by construction. It is clear
that 8a : X >> X is onetoone. We find it useful to make some observations.
Let O A e /C and recall that E e 1C is the multiplicative unit point;
E A = A, for A e K.. Put At = AOA.
Lemma 3.6.3. For the map 5,4 defined above, the following are true:
1 8a(E)=A.
2 8A(Et)=At.
3 5,4 is onto.
Proof: 1. Et = EOE1 and At = AO A1 imply that EEt OE and AAt = OA. If
EEt = OE = 1 then O E and O = O' = E. Similarly, because A * O then
AAt OAl 1. By 3.2.2.9, A' = A^ e 1C' as 0,A e 1C, K, is isotropic, and 0y _L a.
Similarly, E e A'. Thus by 3.2.1.6 we have gEEl II gOE< = K. II gAA> and gEE, II gAA,
by 3.2.1.7. By 3.2.1.8, h = gAA, and h n goE = gAA, n A = {A}. Hence 8A(E) = A.
2 8a(E,) = A,. Again, gsAE)A, = gAA, II gEE, and gAA, n / = {At}.
3 Let P e X and P O. If P g t then by 3.2.1.8, there is a unique line g such that
Et e g, g II gA,p, and gng0p = {Q}, for some Q e X. Then it follows that
8a(Q) = P. If P e t then 8A(Q) = P, where {Q} = gr\t, E eg, and g  gAP.
62
Lemma 3.6.4. IfC D, then gCD  hence, for all O A e 1C, 8A is a
dilation ofX.
Proof: Let g\ = t, gi = goc, g3 = gOD, Pl = Et, Q\ = A2 e g\, P2 = C, Qi = 6A(C),
Qi e g2, Pz = D, and Q3 = 5A(D). Then gCE, II gA,8A(Q and Sde, II gAl8A(D) Thus> by
Axiom D, gcD II gdA(c$A(Dy
Consider now the plane Xa Recall that by 3.5.1.3, for O A e 1C, there is a
unique 8A e Va(0) such that 8A(E) A, where 8A is of the form GgGj, for g,h e Oa
From Axiom 9 and the proof of 3.5.1.1, for any r e Oa, GgG/,Gr = for a unique
w e Oa That is, for every fixed r e Oa, every O A e /C can be uniquely writen in
the form 8A = GwGr where Gw(Er) A. (The uniqueness follows from the fact that
(Ar)w' = (Ar)w2 implies wj = u>2 if Ar # A [30].)
Therefore, for every O A e 1C, there is a unique a e Oa, such that Ela = A.
That is, there is a unique a e Oa, such that GaGÂ¡ = 8A. Also, every P e Xa, Pa,
can be uniquely writen as P = P\OPi, where P\ e 1C and Pi e 1C'.
Let Pi = Pi e 1C. Then we may uniquely write P = P\OP\, where P\,Pi e 1C.
Define the map 8A : Xa Xa, by 8A(P) = P\aO(Pfy, for O A e 1C, P\a.
Theorem 3.6.5. The map 8A : Xa Xa is a dilation on Xa with fixed point O and
dilation factor A, for all O A e 1C.
Proof: If PyOP'f1 = Q'fOQf, then by unicity, Pf = Qf. So P\ = Q\, P'f1 Qi,
and Pi = Qi Hence, P = P\OP\ = Q\OQl2 = Q, and 4 is injective. Let Q\ct with
Q = Q\OQ\; Q\,Qi e 1C. Let P, = Qf e 1C and P2 = Qf e 1C. Then P = PxOP2 a
and SA(P) = PfOPf = (&TO(Q?)at = Q\OQ\ = Q. Therefore, SA is onto.
We claim that if P ^ Q, then gpQ  gsA(P)8A(Qy First we show that
&a(P) e gop. If P e 1C, then P = POO1 and 8A(P) = P,aOO,a = P,aOO = P,a e 1C. If
P e K\ then putting R = P' e 1C, we have P = OOR and 8A(P) = OtaORtat =
= OORlal = R,at = Pal e 1C'.
63
So suppose that gop = g is nonisotropic and write P = P\OP2, where P\ = P2.
If O e g, then XOY e g < X = Yg, where X e 1C and Y e X.'. Thus, 8A(P) = PfOPf1
and P2 = Pf" = Pfs, which implies that 8A(P) e g.
Claim. 8A(POQ) = 8A(P)08A(Q), for all P,Q e Â£. Let P = P\OP\ and Q = Q\OQ2.
Then POQ = PiOP2OQ]OQl2 = (P\0Q\)0(P20Q2)', with P\OQ\ e K and
(PlOQi) e 1C'. Therefore,
8A(P0Q) = (P\OQ\),aO(P2OQ2yal = P\aOQ\aOPfOQf = (PfOPfyJiQfOQf)
= 8A(P)08A(Q).
The claim follows.
Proof of (iii): Assume that P, O, and Q are collinear. Then there is a line g, such that
P,0,Q e g. Thus, 8A(P),8A(0) e g and gPQ = g  g = g^P^Qy
Conversely, suppose that P, Q, and O are noncollinear. Let g\ = gop, gi = gOQ>
and g3 = go.POQ Then g\,g2, and g3 are distinct lines through O. Indeed, if g2 = g\,
say, then POQ e gop and we would have OPQPOQ) = Q e gop, which contradicts
our assumption of noncollinearity. Now,
P(POQ) = OQ and 8A(P)(8A(POQ)) = 8A(8A(P)08A(Q)) = 08A(Q)
gP.POQ II gOQ = gobA(Q) II &8A(P)8A(POQY an<^ gP.POQ II gbA(P)8A(POQy
Similarly, Q(POQ) = Q(QOP) = OP; and 8A(Q)(8A(POQ)) = 8A(8A(Q)08A(P)) = 08A(P).
This implies that gQ^poQ II gop and gop go8A(P) II &8A(Q)8A(POQ)m Hence,
gQ.POQ II g8A(Q)8A(POQy BY Axiom D, it follows that gPQ  g^P)8A(Qy Therefore,
8a : Xa ** Ta is a dilation on JÂ£a.
Now we show that 8A = 8A on Xa. Because 8A and 8 A are dilations on Ta, a
dilation is uniquely determined by the images of two points,and 8A(0) = O 8A(0),
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 r e M. such that t c 3Â£n, let a r e M.
such that t a .in Let K.\ and 1C2 be the isotropic lines in 3^ through O. Because t is
nonisotropic, then by Axiom 14 there exist y, 8 such that r = y8 and t = [r),y,8]. Thus
a, : 3En > Tr) is welldefined. Every B e t c 3Â£r may be uniquely written as
B = B\OB2, where BÂ¡ e JCÂ¡. Because O e t and t is nonisotropic, then B2 = B\. Thus,
B = B\OB\ where B\ e 1C\. So, in particular, there are unique E\,A\ e 1C\ such that
Et = E\OE\ and At = A\OA\, for AÂ¡ e t. As before, there is a unique aÂ¡ e Or such
that Ej = A\. Hence, every X e Tr can be uniquely written as X X\OXt2, where
X\,X2 e IC\ c This defines a map 8,^ : Tn Tr, given by
8^n(A) = A?1 OX'j'1 for A e
Proposition 3.6.6. Let 8^n : be the map defined above. Then
1. S^n is a dilation on 3^.
2. 8^n = 5a on Tn.
3. if P, O, and Q are collinear points in 3fn, not necessarily distinct, then
5 a^POQ) = 5Ax](P)05Al](Q).
4. Moreover, 8A(POQ) = 8A(P)O8A(0, for every P, Q\ q such that (),P. and Q are
collinear.
To obtain a scalar multiplication on V, for all O A e 1C and all OX e V,
define
A OX = 8a(0)8a(X) = 05AX) and O OX = 1.
We now verify the vector space properties.
Lemma 3.6.7. If A,A' e 1C and OP e V, then (A + A1) OP = A OP + A1 OP.
Proof: Suppose P e t and P O, and recall t,K. c Xa. Let P,Â£a,p; Ay with y  p;
A' ly'< Y' II P: and 08 with 8  p. Then y  y'  8; g, = [a,y]  gEP with g, c Xa;
g2 = [ot^y ] II gEP with g2 c Ta; and g2 = [a,8]  gEp with #3
h(P) e g\ r\t and 5y(P) e g2 n t. Thus, 8^(P)a,y and 8y(Â£)ot,y'. It follows that
65
5,4(/5)054'(/>)a,y8Y,; 8A(P)08A'{P) e t\ y8y' = Â£  p; and AOA1  a,y5y' = e. Therefore,
AOA1 e [a,e]  gEp and 5A{P)08A{P) e tn [a,e]. Hence, 8AOA{P) = 8A{P)08A'{P).
Suppose Pit. Because gopd c jÂ£t), then replacing E with Et, A with At, A1
with A\, and a with rj in the first part of the proof and the result follows.
Lemma 3.6.8. If A e 1C 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 = gojOQ and r e M such that g,t c 3^. Then the
result follows from Proposition 3.6.6(iv).
Conversely, suppose that P, O, and Q are not collinear. Because P, O, and Q are
not collinear then P ^ Q, and from Lemma 3.6.4 it follows that gpQ  SlA(P)hA(Q)
Applying Axiom R we obtain go,poo = Zo8a(P)8a(Q) That is> 8A(P)08A(Q) e go.POQ
by construction. Again by Lemma 3.6.4,
S8a(P)8a(POQ) II So.poq and 8A(P)8A(P)()8A(Q) = 08A{Q).
This implies that gEA(P).8A(P)08A(Q) II So8a(Q) = Soq II gp.POQ, as P(POQ) = OQ. Thus,
gÂ§A(P),8A(P)O8A(0 = gA(P)8A(POQ) because both lines contain 8A(P) and are parallel to
gp.POQ Since
gsA(P).8A(P)OSA(g) ngP.POQ = {8.4(P)O5,4(0} and gs^p^gpoQ) ^go.POQ = {8A(POQ)},
then 8a(P)08a(Q) = 5A(POQ).
Lemma 3.6.9. If A, A e 1C and OP e V then A (A1 OP) = (A A') OP.
Proof: We need to show that 8A(8A>(P)) = 8A.A'(P); that is, 8^ 8A> = 8a.ai. Now
8a o 8Ai and 8A.A are dilations on X and 8A.A>(0) = O by construction, so it suffices
to show that (8A 8A )(E) = 8A.A(E) and (8A 8A')(0) = 8A.A'(0). Now E e K e Xa,
so on Xa, 8A o 8A' = 8A o SA, and 8A.A> 8a.ai. But on /C, by Lemma 3.5.11,
8a o 8a, = 8a.ai. Therefore, (8^ 8A')(E) = A (A1 E) = (A A') E = 8A.A'{E).
Lemma 3.6.10. For E e 1C, the multiplicative unit, and OP e V, E OP = OP.
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Proof: We need to show that 6e = l.t Now l j is clearly a dilation on X and
l.r(<9) = O = SffO). Thus, l.r(Â£) = E = E E = 8f(Â£) = Â§Â£(Â£).
Theorem 3.6.11. The space (V. AC) constructed above is a vector space.
The triple (3l,V,/C), is an affine space. [23,p.6] A set X along with a vector
space V over a field K, is an affine space if for every v e V and for every X g X,
there is defined a point vX e X such that the following conditions hold.
1. If v,w g V and X e X, then (v + w)X = v(wA).
2. If denotes the zero vector, OX = X for all X e X.
3. For every ordered pair (X, Y) of points of T. there is one and only one vector v g V
such that vX = Y.
The dimension n of the vector space V is also called the dimension of the affine space
X.
Theorem 3.6.12. (T. V, K) is an affine space.
Proof: If OV,OW e V and X e X, we have
(i) (OV)X = OVX g X.
(ii) (OV+ OW)X (OVOW)X = OV(OWX).
(iii) OOX= 1
(iv) for Y g X, OYX = Z e X and (OZ)X = Y.
Now if OPX = Y, then OPX = OZX, P = Z, and OP OZ. U
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, 1C) and (.T, V,/C) are fourdimensional spaces.
Proposition 3.7.1 Let g be any line through O and put g{0) = {OA : A e g}. Then
g(0) is a one dimensional subspace ofV.
Proof: First note that 1 = OO g 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,
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8A(B) e g, for all A in /C and all B in g. So A OB e g(0) for all ^4 e 1C and
OB e g(0). Hence, g(0) is a subspace of V.
It must now be shown that the dimension of g(O) is one. If g t, then
g(0) = (OEt) because for every At e t, A, = 8A(E,) by Lemma 3.6.3. So suppose that
g t and fix Beg. Let h = gBEr Then for all O D e g, there ia a unique dsuch
that D e d. d  h, and d n t 0. Put {F,} = dnt. Then for Ft = FOF' with F e 1C
it follows that
5 f(B) = D and OD = F OB.
Hence, g(0) = (OB).
Corollary 3.7.2. Following the terminology of Snapper and Troyer [23, p. 11],
g S(0,g(0)) S {A = 0(0A) : OA e g(0)}
is an affine subspace of dimension one.
Proposition 3.7.3. Let a e V with 0\a and put $La(0) = {OA : A a}. Then ta(C) is
a two dimensional subspace ofV.
Proof: Clearly, 1 = OO e '.ta(0), so the zero vector is in ta(C). Let C,D a and
A,B e 1C. Then by 3.1.4.7 we have COD Fa and by Lemma 3.6.3,
8^(0 e goc <= Ta and 85(D) e gOD c= Xa
so that MQOSgfDila. It follows that OC + OD = OCOD OF e Â£a(0), and
A OC + B OD = 084(C) + 08b(D) = 0(8A(Q08B(D) e ta(0).
Hence, ta(C) is a subspace of V.
Thus, it remains to show that Â£a(C) is two dimensional. There are two cases:
a e M and a e Q. Suppose first that a e M. We construct a basis for ia(0) using
isotropic lines. To this end, let 1C] and 1C2 be the isotropic lines in Xa through O. For
any Fa we may uniquely write in Xa, P = P\OPi, with P\ e 1C\ and P2 e /C2
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From Proposition 3.7.1 above, K,\(O) = (OB) for any O B e IC\ and K2(0) = (OC)
for any O C e 1C2. From this it follows that P\ = 6A(B) and P2 = 8A'(C) for some
A,A' e 1C. Thus,
OP = A OB + A' OC
and {OC,OB} span ta(0). Now if A OB + A1 OC = 1, then 08A(B)08A'(C) = 1.
This implies that 8A(B)08 y(C) = O and 08A>(C) = 8A(B)0. Because 0,8A(C) e 1C2
and 0,8A(B) e K\, then either 1C\  K2 or 08A'(C) 1 = 8A(B)0. But IC\ Jf 1C2, so
8A'(C) 8a(B) = O. Because a e Ai, then from 3.3 and 3.6 there exists t',a e Oa
such that 8A'(C) = C,a = O. This implies that C Oa 1 = O or A1 = O. By
assumption C O and D O, thus it must be the case that A A1 O. Hence,
{OC,OB} is a linearly independent set in $.a(O) and Xa(0) is two dimensional.
Suppose now that a e Q. We construct an orthogonal basis. By 3.1.7.2, there
exist p,y e 5 such that O = aPy and a 1 p 1 y 1 a. Let x = [a, P,aP] and
y = [a,y,ay]. Note that if x = y then by Axiom 14, aP = y and O = apy = 1, so
x y. Let P\a and let p',y'P with P' T p and y' T y. Put P\ = PP' and P2 = yy'.
Now P\ a, p'; al p; P' T P, so that a 1 pf by 3.1.6.18. Because P\a,y'; aly;
y' T y, then aly1. Thus,
p = (pp')a = pp'a = pp' = P] and p = (yy') = yy' = P2,
so P \,P 2 I ct. This implies that P\ a,p and P2ot,y so Pi ex and P2 e y. If
Q = P\OP2 then Q = P1OP2 = p^Pay/y* = p'ay'.Because pr T a, let 8 = aP* e V.
Then Q = 8y' and 6 T y'. Thus P\ P',a implies that P P'a = 5. Thus we have
P,Q\8,y' with 8 A y', and P = Q by Axiom 3. That is, P = P1OP2 with P\ e x and
P2 e y.
Let O X e x and O Y e y. Then from Proposition 3.7.1 above we have
x = (OX) and y = (OY) and there exist A,B e K. such that OP A OX + B OY, where
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Px = lA(X) and P2 = 8B(Y). Hence, {OX,OY} spans If A 0X+ B OY = 1,
then we obtain 08A(X) = 8b(Y)0. Because x Jf y, it follows that 8.4(A) = O = Sb(Y).
Let Xr be the unique plane containing t and x and IC\ and K2 the isotropic lines in
3En through O Then we may write Â§A(X) = XfOXf', where X\ e 1Ci, and XÂ¡ e JC2
As above it follows that A O and similarly, B = O. Therefore, Â£(0) is two
dimensional.
Corollary 3.7.4. Ta = B((), Aa((7)) is an affine subspace of dimension two for all
a e V, 0\a.
Theorem 3.7.5. (V./C) is a fourdimensional vector space and hence, (T, V,/C) is a four
dimensional affine space.
Proof: Let OP e V and let O = ap with a e M. and P e Q. Let Pla^p' with cd T a
and P' 1 p. Put P1 = cl'cl and P" = p'p. Then Q P OP" = a,aapppi = a'pd Thus,
P,Q\a',pi with a' X P'. Therefore P = Q = P'OP" with P' a and P"  p. Since Â£a(0)
and tp((9) are twodimensional then there exist bases {OZ,OT) c .ta(C) and
{OX,OY} a p(O) such that
OP' = A OZ + A' OT and OP" = B OX+ B' OY
for some A,A\B,B' e K. Thus, P A OZ + A' OT + B OX + B' OY and
{OX,OY,OZ,OT} span V. If A OZ + A1 OT+ B OX+ B1 OY = 1, then in
particular, OP = OP1 + OP" = 1. This implies that OP'" = P'O. So either
Sop" II Sp'o or p" = P' = O. But gOP" f gOP' and therefore,
A OZ + A1 OT OP1 = 1 = OP" = B OX + B1 OY. As was shown in Proposition
3.7.2, we obtain A=A' = B = B= 0 and the result follows.
3.8 Orthogonality
In this section we extend the definition of orthogonality to include lines and
then use this definition to define orthogonal vectors.
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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 a,(l e P such that g c Xa, h c 3fp,
alp, and P = ap 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
alp. For if a 1 P then one of a and P must be in Q and by Axiom 12, all lines in a
plane Tp for P e Q are nonisotropic.
In Minkowski space, if g and h are two isotropic lines then g _L h <> g  h.
Thus we extend the above definition in the following way.
Definition 3.8.3. If g and h are isotropic lines, then g _L h < g  h .
Definition 3.8.4. If g is a line and a e V then we say that g is orthogonal to or
perpendicular to Xa, g _L Xa, if there exists a P e V such that g c Jfp, P 1 a, and
P = ap 6 g. In this case, P ap is the point of intersection of g and Xa
Definition 3.8.5. For every 1 OA.OB e V, we say that OA is orthogonal to OB,
OA 1 OB, if and only if goA 1 goBthat is, there exist a,p e V such that goA c Xa,
gOB c 3fp, and O = ap. For the zero vector 1 = OO, we define 1 _L 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 T h <> g^ _L h^.
(ii) g 1 Xa <> g*> 1 Xaz.
(iii) OA A. OB <> (OA)S 1 (05)1
Lemma 3.8.7. If g is a nonisotropic line then g is not orthogonal to g.
Proof: If g 1 g then there exist a,p e V such that g c Ta,Tp; alp, and
P = P e g. But for every point Q e g, Q a,p with alp 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.
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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. (U1 subspace axiom) Let 0,A,B, and C be any four, not necessarily
distinct, points with A, 0\a; 0,B  [3; O, C  y, 8 and a 1 y and P 1 8. Then there exists
X,e e V such that X 1 s; 0,A0B\X; and 0,C e.
Axiom Si. If g c.ia, a e Q, h cz Tp, P e Q. and there exists y,8 e V such that
y 1 5; y5 eg h\ g c and h cz then there exists E Â£ Q such that g,h cz (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 P e V such that g,h c= Tp. Then either there exists a,y e V such that a 1 y,
g c Aa, and h cz Xy or for all A e g, there exists B e h such that P and APB are
unjoinable.
Lemma 3.8.8. If g. h c Xa are nonisotropic for a e V, then g _L h in the sense of
Section 3.3 if and only if g 1 h in the sense of Definition 3.8.1.
Proof: Let g = [a,p,8], h = [a,y,X] c 3Â£a with a = p8 = yX. Recall that g 1 h in the
sense of 3.3 if without loss of generality, ply and A = Py e gn h. So, in
particular, A = py with A e gn h and g cz 3fp and h cz Xy. So g _L h by 3.8.1.
Now suppose that there exist rpe e V such that B = qe e gn h and g cz 3^
and h cz XE, but it is not the case that ply, nor that p 1 X, nor that 8 1 y, nor
that 8 1 X. Then by 3.3.8 there exists a unique / e Ca such that B e l and
/ = [a,u,p] for some u,p e V with a = op, u 1 p, and p 1 8. If / h then la(B) and
ha(B) span ta(R). Thus if C e g there exists Lei 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, gig,
which cannot happen for nonisotropic g. Hence, / = h and the result follows.
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Lemma 3.8.9. If a e V then for each P a and for each nonisotropic line g c Xn
there is a unique nonisotropic line h c Xa such that P e h and h T g.
Proof: This follows directly from 3.3.8 and Lemma 3.8.8.
Lemma 3.8.10. Ifg,h c Xa fora e V and g,h are nonisotropic, then g h if and
only if GgGh = GhGg lXa
Proof: This follows from 3.3.16 and Lemma 3.8.8.
Lemma 3.8.11. Suppose that O e c,d\ c,d
and c 1 d. Let C e c and D e d, then go,DOC Is not orthogonal to c if C\D ^ O.
Proof: First note that go.DOC c ov d because then we would have DOC = D\ e d,
say, so that C = ODD\ e d and O e d implies that C = O or c = d. Now if
go.DOC L c, then in 3Ea we have Go = Gcd = ^gojxx ^c, which implies that
d = go,doc
Lemma 3.8.12. Let g,x c with g isotropic, a e V, andg D x {O}. Then g is
not orthogonal to x if g jc.
Proof: If x is isotropic and x T g then because x g, there exists y,5 6 V such that
g a Xy, x c Tg, and y!8. But then one of y and 8 must lie in Q. But no element in
Q can contain an isotropic line so jc is not orthogonal to g.
Suppose that x is nonisotropic and let O e g n jc. Because g is isotropic then
a e M. and by 3.3.8, there is a unique nonisotropic line h c Xa such that O e h and
h Lx. Suppose that glr and let O^Keg, O^Heh, and O X e jc. Then OH
1 OX and OK 1 OX so that by Axiom U, O(HOK) = OHOK 1 OX. If go.HOK is
nonisotropic then go,HOK = h. And because go.HOK <= Toe, then HOK = H\ eh. Then
we have K = OHH] 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 K\ e g. It follows that H = K\KO e g and g = h. If go.HOK g then go,HOK
is the other isotropic line through O in Because g and go.HOK span Xa and
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g,go.HOK L then by Axiom U, x is orthogonal to every line in Aa through O. So in
particular, x A. hx which implies that x = h.
Corollary 3.8.13. If g is isotropic andx _!_ g then either x = g or x is nonisotropic and
x and g are noncoplanar; that is, there does not exist 5 e V such that x,g cz jÂ£Â§.
Lemma 3.8.14. If g is isotropic, x is nonisotropic, rig with x g, and {P} = x fl g,
then for all P A eg and for all P B e x, g and gAPB are not coplanar.
Proof: From Lemma 3.8.12, jc and g are noncoplanar. Suppose that A,0,A0B y for
some y e V. Then B = OAAOB\y, which implies that x = goB c: Â£y and
g Soa c: Xy, a contradiction to Lemma 3.8.11.
Lemma 3.8.15. If OC e V is isotropic and OB e V is nonisotropic with OC T OB then
OCOB 1 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 1 gocoB or there exists
D e gocoB such that O 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 8 6 V such that
goc, go.coD c Ag. This implies that O, C, COD \ 8, D OCCOD18,
gOD = go.coD c: ^8, and therefore, goc ^ Xd, a contradiction.
Lemma 3.8.16. The zero vector, 1 = OO, is the only vector orthogonal to every
vector in V.
Proof: This follows immediately from Lemma 3.8.11 above.
Theorem 3.8.17. IfU is a subspace ofV, then UL = {OA e V : OA _L OB, dOB e U)
is a subspace ofV.
Proof: because the zero vector 1 is orthogonal to every vector in V by definition,
then 1 e Y1. Let OA e U1 and R e 1C. because R OA e< OA >, the subspace
generated by OA, and goA L goB for every OB e U by the definitions of and
orthogonal vectors, then R OA e U1.
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Let OA,OB e U1 and OC e U. If OC is nonisotropic then OC OA,OB and
there exist a,p,y,5 e V such that
A,0\a\ 0,BP; 0,Cy,8; a 1 y; and P 1 6.
By Axiom U, there exists X,8 e V such that O = Xz, AOB\X, and 0,C e. Thus,
OA + OB = O(AOB) 1 OC.
Now suppose that OC is isotropic. The following possibilities exist.
(i) If OC OA,OB, then as in (a) above, (OA + OB) A. OC.
(ii) If OC = OA = OB, then because OC is isotropic, AOB = COC e goc and
(OA + OB) 1 OC.
(iii) If OC OA OB, then OB _L OC implies that OB is nonisotropic and
(OC + OB) = OCOB T OC by Lemma 3.8.14.
Hence, if OA,OB e U1 then OA + OB e UL and UL is a subspace of V.
Theorem 3.8.18. HO = ap then .ta(O)1 = .tp(O) and tp(O)1 =
Proof: From the definition of orthogonal vectors we clearly have Â£(0) cr ipiO)1
and tp(O)1 c= ia(0) Now suppose that OA _L .ta(0); that is, goA h a Then, there
exists y e V such that OA e Â£y(0) and y 1 a. But then we obtain O = ap = ay so
P = y. Hence, .ta(O)1 = .tp(O) and Â£a(0) .tp(O)1.
An immediate consequence of Theorem 3.8.18 is the following. For each a e V
with O  a, there exists a unique P e V such t,hatOp, .ta(O)1 = Â£p(0), and
(ta (O)1)1 = a (O).
Theorem 3.8.19. Ifa(O) is nonisotropic then there is a unique hyperplane A(0) such
that (0)L = A(0) and (d(O)1)1 = d(O).
Proof: Let a e V such that a c: Ta, where a is the nonisotropic line associated with
0(0) {OA : A e a}. Then for O = ap we have a _L p and a T g for every g a 3fp
with O e g. Thus, .tp(6>) c (0)L. Now a is nonisotropic, so by 3.3.8, there exists a
unique h c Xa such that O e h and a 1 h. By Axiom U, (h(0),Â£$(0)) c (0)L.
75
Suppose 1 OB e (O) n (/(0),.tp(0)).Then B e a cz Xa and there exists
OC e h(0) and there .exists OD e .tp(O) such that OB = OCOD\ that is, there exists
C e h and D P such that B = COD. because h c Xa, Ca and C = aai. Because
Z)p, D = ppi for some Pi e P. Now a 1 P and P 1 Pi, so a  Pi. But B = COD
and COD = aiaapPPi = a.Pi so that Ppi,a with a  Pi. This implies that a = Pi
so D = O, B C, and a = h, a contradiction. Hence, () n (h(O), ip(0)) = {1}
and V = (0) (/(), p(0)). If O e d and d T a then d(O) c:
otherwise we would have
V = d(0) d(0)
which is not possible. Hence, d(O)1 = .
On the other hand, from p(O)1 = Â£a(0), a T h, and a _L p, then,
d(O) c (h(0),Â£a(0))J. If O e g _L p. then by definition g a Xa If O e g h, then
g = a. Hence. (O) = (hCO^tpiO))1.
Claim. (O)1 is independent of the plane containing a. Suppose a
Then for O yd, we have y ^ a, 6 ^ p, and a _L jCÂ§. Again there exists a unique
/ c Xy such that O e l and /1 a, so that d(O)1 = (l(0),X8(0)) as above. Now any
point P e X may be written as P P\OP2 with Pi a and P2 \ P Since h 1 a, then we
may write P\ = HOA with H e h and A e a. Then
(h(0),Xp(0)) (O) = V= (l(0),Xy(0)) d(O),
and it follows that (h(0),Xp(0)) = (~/(0),Xy(0)) = A(O).
Theorem 3.8.20. If (Q) is isotropic then there is an unique hyperplane A{()) such
that d(0)x = A(0), d(0) c= A(O), and (d(O)1)1 = d(O).
Proof: Let d(O)
because a is isotropic then a e M. Let O = ap, p e Q, and put
A(0) = (d(0),Xp(0)). Since a c Xa, alp, and O = ap e a, then a 1 Xp. Because
76
a is isotropic then ala. By Axiom U and Lemma 3.8.14 it follows that
is isotropic and g a, then
there exists a unique y e M such that g,a a Xy. Because g(0) a A(0) a {0)L,
then g 1 a, which contradicts Lemma 3.8.11. Thus, A(0) contains no other isotropic
line.
Let h 1 a with O e h. For each H e h we may write H = H1 OB, with H'\a and
Hp. If O A e a, it follows that OB T OA\ OH 1 OA, so that OB T OA and
OH' = OHOB T OA. This implies that gOH< = a so H* e a and h(O) e (d(0),_tp(0)).
Thus, (O)1 = .
If h(0) c {(Oi^tpfO))1, then
h(O) c tp(O)1 = .ta(O) and h(O) 1 a(O).
Thus, h(P) = a(0). Hence, ((0),Xp(O))1 = a(0) and (dfO)1)1 = a(O).
The subspace (O)1 is independent of the plane containing a. Suppose a c: Xy,
y e M, y a. Put O = y5. Let h(0) c (a(0),Â£5(0)). Then h 1 a and
h(0) c . If
kO) c <(0),tp(0)>
then k 1 a and k(O)
.
Remark. If g is a line then g is either nonisotropic or isotropic. From Theorem
3.8.19 and Theorem 3.8.20, if U < V is a onedimensional subspace then UL 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 (g, h) denote the unique
plane determined by g and h.
77
Definition 3.8.21. If a e V such that g,h a Ta then we say that (g,h> = Xa is a
nonsingular plane. If there does not exist a e V such that g,h cz Xa. then we say that
(g, h) is singular or (g. h) is a singular plane.
It is clear that every plane in (3f,V,/C) is either singular or nonsingular. Note
that by Theorem 3.8.18, if U < V is a nonsingular twodimensional subspace of V,
then UL is a uniquely determined nonsingular twodimensional subspace of V. Now
consider the following cases for two distinct intersecting lines g and h.
Suppose that g is isotropic and h is isotropic. Then by Axiom T, there exists a
unique a e M 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 a Xa
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 T h.
Proof: Let {P} = gn h. By Axiom S2, either g 1 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, gAjPB is
nonisotropic. Let gPA = [8,s], so A,P\b,z. If A is unjoinable with APB then A, P, and
APB are pairwise unjoinable points, so by Axiom 10, APB\h,z and APB e g. Thus,
gPjPB = gPA = g and B = PAAPB eg, so g = h.
Thus, gpjPB g and gpjps is isotropic, so by Axiom T, there exists a unique
y e M such that g,gpjps c This means that P,A,APB\y so B PAAPB\y, and
h c Xy, which contradicts our initial assumption. Therefore, g T h.
Theorem 3.8.23. Let (g(0),h(0)) be a two dimensional singular subspace ofV where
g(O) is isotropic and h(()) is spacelike. Then (g(0), h(O))1 is a two dimensional
singular subspace of V which contains g(O). Moreover, ((g(()),h{0))L)L = (g(0),h(0)).
Proof: First we observe that if W is a subspace of V generated by subspaces U and Id!
78
then W1 = {UM')l = W1 n Z/1, v e (U,U')L < v e UL, and
v e U'1 < v eZ/1 n ZY1. Consider (g((9),/ifO))1 = g(0)x n /i(O)1. Because g J. /? by
Proposition 3.8.22 then there exists y,8 e V such that O = y5 with g e Xy and
h c .Tg. From Theorem 3.8.20, gfO)1 = (g(0), Â£g(0)> and by Theorem 3.8.19,
O)1 = (j(0),Xy(0)) where / c 3Â£g is the unique line through O in 3Â£g orthogonal to
h. Hence,
((0),h(0))L =g(0)^f(0)
= n
=
Moreover,
1=g(0)1n/(0)1
= n
 (M0),h(0)).
Now if g,/ cz 3Â£e for some 8 e V then by Theorem 3.8.18, for O = sp,
{giO)Jm = XE(0) and (g(0),h(0)) = (g(0), /(O))1 = .te(0)X = .tp(O). This says
that g,h c .Tp; that is,
If (g,h) is a singular plane, O e g n h, and g and h are spacelike, then by
Axiom Si, g is not orthogonal to h. So by Axiom 2, for all A e g, there is a B e h
such that O and A OB are unjoinable.
Theorem 3.8.24. For the above setup:
1 go job e (g,h).
2. go job T g,/z.
3. ifC e g and D e h such that go,cod is isotropic then go.coo = go job
4. <&P),HP))1 =
Proof: For 1., if there exists ye? such that gojOB,g c Xy, then 0,A,A0B\\y implies
that B = OAAO y and h = Xy is nonsingular. For 2.
79
go job L g,h by Axiom S2 because by 1. above go,cod <= (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,gOJOB <= (g,H If
go.coD go job then by Axiom T there exists a unique r\ e V such that
go,COD, gOJOB <= 3Cy.
Because two intersecting lines uniquely determine a plane, then
go,COD = gOJOB The last conclusion follows from (g,h) = (g,go job)
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
/ c (g,h) such that Pel; that is,
To complete the classification of orthogonal subspaces of V, it remains to
consider hyperplanes. Now if (X, V, 1C) 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 O. Then
A(O)1 = {h{Q)MO))L = KO)1 n IKO)1.
From Theorem 3.8.19 and Theorem 3.8.20, we know that the dimension of /i(O)1 is
three. By Theorems 3.8.18, 3.8.23, and 3.8.24 it follows that the dimension of
(f)(0)1) = 2. Consider the following possibilities. If h(O)1 n fHO)1 = {1}, then
h(0)L fi(O)1 e V has dimension five whereas the dimension of V is four. If
h(0)L n fKO)1 = fKO)1, then it follows that p(O)1 c h(0)x; h(0) 1 p(O)1 and
HO) c (^(O)1)1 = P(0).
Which contradicts the assumption that h and p intersect only in O.
80
Therefore, h^O)1 n p(O)1 = g(0) for some line g containing O. Because g must
be either isotropic or nonisotropic, then the following is true.
Theorem 3.8.25. If A(0) is a hyperplane the there is a unique line g such that
A(0)L = g(0) andiAiO)1)1 = A{0). U
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 n of the vector space V over the field 1C is a
correspondence with the following properties:
1. Every subspace U of V is mapped onto a uniquely determined subspace n(M) of V.
2. To every subspace U of V there exists one and only one subspace W of V such that
n(W) = U.
3. For subspaces U and W of V, U < W if, and only if, n(W) < n(U).
In other words, an autoduality is a onetoone monotone decreasing mapping of
the totality of the subspaces of V onto the totality of the subspaces of V.
Definition 3.9.2. An autoduality n of the vector space (V,/C) of dimension not less
than two is called a polarity, if n2 = 1, the identity.
Definition 3.9.3. A semi bilinear form over (V,/C) is a pair consisting of an
antiautomorphism a of the field K, and a function /(x, y) with the following
properties:
(i) ^(x, y) is, for every x, y e V, a uniquely determined number in /C.
(ii) J(a + b, c) =y(a, c) +./fb, c) and J[a, b + c) = ./(a, b) + /(a, c), for a, b, c e V.
(iii) J(tx, y) = tf(x, y) and /(x, ty) =/x, y)a(/) for x, y e V and t e 1C.
If a = 1 then /is called a bilinear form.
Definition 3.9.4. If/ is a semibilinear form over (V,/C) and if U is a subset of V, then
81
{x e V : fx, u) = 0 for everyu e U} and {x e V : /(u, x) = 0 for every u e U) are
subspaces of V. We say that the autoduality n of (V,/C) upon itself is represented by
the semibilinear form fix, y) if
n (U) = {x e V : fx,U) = 0} = {x e V : fx, u) = 0 for every u e U}.
Theorem 3.9.5. [3] Autodualities of vector spaces of dimension not less than 3 are
represented by semibilinear forms. M
Theorem 3.9.6. [3] If the semibilinear forms f and g over (V, 1C) represent the same
autoduality ofV, and if dim(V) > 2, then there exists a 0 ^ d e /C such that
&(x, y) = fx, y )d for every x, y e V.
Definition 3.9.7. If n is an autoduality of the vector space (V,/C), then a subspace W
of V is called an Nsubspace of V with respect to 7T, if <7iv)) for every v e W.
In this case, n is said to be a null system on the subspace W of V.
Theorem 3.9.8. [3] Suppose that if, a) represents the autoduality it of (V,fC). Then 7t is
a null system on the subspace W
Definition 3.9.9. A line (v) is called isotropic if (v) < 7iv)). (So an isotropic line is
an Nline.)
Theorem 3.9.10. [3] If the semibilinear form if, a) represents a polarity n, and if
fw,w) = 1 for some we V, then a2 1 and
affix,y)) fy,x) for everyx,y eV.
In this case we say that f is a symmetrical or just symmetrical.
Theorem 3.9.11. [3] Suppose that n is an autoduality of the vector space (V,/C) and
that dimfV) > 3. Then n is a polarity if, and only if, n is either a null system or else n
may be represented by a symmetrical semibilinear form if, a) with involutorial a.
Theorem 3.9.12. [3] Suppose that the polarity n of the vector space (V,/C) possess
82
isotropic lines and the dimfV) > 3. Then n may be represented by bilinear forms if,
and only if,
(a) planes containing more than two isotropic lines are Nplanes and
(b) 1C is commutative.
Theorem 3.9.13. Suppose that n is a polarity of the vector space (V,/C) such that the
conditions of Theorem 9.12 are met. Then from Theorem 3.9.11 it follows that if n is
not a null system then n may be represented by symmetrical bilinear forms. M
Defining the polarity n 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, U1 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 71 is a uniquely determined subspace of V. From Lemma 3.8.11 and
Definition 3.8.5, {l}1 = V and V1 = {1}. So we define the mapping n on the
subspaces of V as follows:
Definition 3.9.14. If U is a subspace of V then n(JJ) = U1.
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 W of V such that 7t(14) = U.
Theorem 3.9.15. Let U and W be subspaces ofV, then U < W if, and only if,
71(14) < K (U).
Proof: Suppose that U < W. If OV e W1 then OV _L OfV for every OfV e W, U c W,
so OV OU for every OU e U and OV e U1. Thus, U < W implies that
n(W) VV1 < U1 n(lf). Conversely, suppose that W1 < U1. By Lemma 3.8.11 and
Definition 3.8.5 we have ({l}1)1 = {1} and (V1)1 = V. From the previous section
it follws that, if U is a nontrivial proper subspace of V then (71)1 = U. because W1
and UL are subspaces, then from the first part of the proof we have W1 < UL implies
that U = (U1)1 < (W1)1 = W.
Corollary 3.9.16. From Definition 3.9.14 and Theorem 3.9.15 above it follows that
n : U < V Id1 < V is an autoduality. Moreover, because (Y1)1 = U for every
83
subspace U ofV, then n is a polarity.
Lemma 3.9.17. Let Â£, e and OA e V. Then n(ff) n(Lf)^ for every \ e and for
every subspace U < V. That is, n is invariant under a= for every Â£, e .
Proof: We have 0(0A= OO^A^ = B e X, so (OA= OB e V. Hence, from Lemma
3.8.6,the result follows.
Theorem 3.9.18. The polarity n is not a null system.
Proof: Let a be any nonisotropic line with O e a. By Lemma 3.8.7, a is not
orthogonal to itself so that a(0) $ a(0)L and hence, (O) $ n(d(0)). By Definition
3.9.7, n is not a null system.
Theorem 3.9.19. The polarity n defined above may be represented by bilinear forms.
Proof:. Let g be any isotropic line through O. By Definition 3.8.5, g(0) < (O)1 so
that g(0) < 7i((0)) and g(0) is an isotropic line in the sense of Definition 3.9.9.
Conversely, if h is a line such that h(O) < n(h(0)) = /ifO)1 then h _L 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, dim(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 V. If
y e M then U has precisely two isotropic lines, as was shown in 3.2.2.8. If y e Q,
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 K. = R is commutative then
by Theorem 3.8.12 the result follows.
Theorem 3.9.20. The polarity 7t may be represented by symmetrical bilinear forms;
that is, there is a symmetrical bilinear form g such that for any subspace U ofV,
UL = n (70 = {OA e V: g(OA,U) = 0} or
g(OA,OB) = O = g(OB, OA)if and only if, OA _L OB for OA,OB e V.
Proof: This follows directly from Theorems 3.9.18, 3.9.19, and 3.9.13.
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 Q.
Lemma 3.9.21. Let g be a symmetric bilinear form representing 7t. Then g is
nondegenerate.
Proof: This follows from the fact that 7i(V) = V1 = {1}.
Theorem 3.9.22. Let g be a symmetric bilinear form representing n. Then there is a
basis ofV such that the matrix of g with respect to this basis has the form
r C 0 0
0 C 0
0 0 c
^000
that is, there is an orthogonal basis ofV such that the matrix of g with respect to
this basis has the above form.
Proof: Let a e M. such that 1C a Xa and let O = ap. Let 0y,8 e Q such that
a = y8 and y,5 _L p. Let 8 = yP e M and r = 8p e ALThen sr = yPP8 = y8 = a so
sir). Put = [a,y, 8], y = [P,y,s], z = [p,r,8], and t = [a,s,r]].
Claim. x,y,z,t are four mutually orthogonal nonisotropic lines through O. We
have O = ap, O e x,t c 3Â£a and O e y,z c Tp so that x,t 1 y,z. Also,
O = ap = y8p = yr = 8s so that y 1 q and 8 1s. Then
O e x a Xy,0 eta X^,0 e Xy, and O e z a 3Â£n
implies that jc _L t and y L z.
The construction of the basis. Let E e 1C, the multiplicative identity,
considered as a point in Xa. Put T = EOE e t and X = EOEx e x. Note that because
0 ^
0
0
C J
0 e 1C;
85
O e x,t c Xa and x _L t then GxGt = go = tGx in Xa Now clÂ£ and s e A4, so
there exist precisely two isotropic lines, /CÂ¡z and Kze, through O in Xe. Thus, there is
an unique Â£e e IC\Z such that T = EzOEz. Because y c Xe then Y = EzOEz e y and
OyC7; = Go in XE Similarly, because c Xti; q e M., there are precisely two
isotropic lines, and through O in X^ and there is Ex\ in /Cir, such that
T = Er\OE^. because z a X^ then Z = E^OE^ e z and G:Gt = Go = GtGz in Xn We
calculate XOT = (E0Ex)0(E0El) = (EOE)()(ExOEl) = (E0E)0(E0,0E')
= (EOE)()(E0OE)t = (E0E)0{0E00E)t = (E0E)00
= (E0E)00 = EOE e X.
YOT = (Ez0E{)0(Ez0E'e) = (Â£E0Â£e)0(Â£OÂ£E)' = EzOEz e Â£le.
ZOT = (E^OE^O^E^OE^) = E^OE^ e ATin. Thus, if we put Â£\ = OX, E2 OY,
Â£3 = OZ, and Â£4 = OT it follows that the set {Â£^Â£2, Â£3, Â£4} consists of four mutually
orthogonal vectors such that Â£, + Â£4 is isotropic for / = 1,2,3. So if g is a symmetric
bilinear form representing n then from Â£, + Â£4 1 Â£, + Â£4 for / = 1,2,3 and E, 1 Â£, for
i j it follows that for / = 1,2,3,
O = g(Â¡ + Â£4,Â£, + Â£4) = g(,Ei,EÂ¡) + 2g(Â£ Â£4) + g(Â£4, Â£4) = g(Â£Â£,) + giÂ£4,Â£4).
So that g(Ej,Ei) = ^(Â£4, Â£4) O, because Â£4 is nonisotropic. Thus, it remains to
show that {Â£4,Â£2,Â£3,Â£4} is a basis for V. But this follows from Section 3.7.
Theorem 3.9.23. (V,/C,g) is a fourdimensional Minkowski vector space. Moreover,
(Xy(0),g) is a fourdimensional Minkowski space for every y in Vo
>
Proof: Put #(Â£4,Â£4) = 1 and g(Â£Â£,) = 1, for / = 1,2,3. Minkowski space is the only
nonsingular real fourdimensional vector space with metric
r 1 0 0 0 ^
0 10 0
0 0 1 0
^ 0 0 0 1^
(3.24)
86
In the last section of Chapter 3 we show that each X in Q can be identified with
a spacelike plane and each \ e Q* with a reflection about a spacelike plane.
3.10 Spacelike Planes and Their Relections
For each X e Q with 01 A., define a\ : V V by
0,1 e V. To extend this definition to any X e Q, we note that if X / O and OA e V,
then there exists a unique D in X such that OOxAx = D; that is, there is a unique D
in X such that (OA)1 = OxAx = OD. Thus, we define x(OA) = OD, where
OOxAx = D. We note that if X \ O then D = Ax.
First we show that is a semilinear automorphism of V for each X e V. [23]
Now, a function /: V > V is a semilinear automorphism if, and only if, it has the
following properties:
1. /is an automorphism of the additive group of V onto itself.
2. / sends one dimensional subspaces of V onto one dimensional subspaces (that is, /
is a collineation).
3. If A and B are linearly independent vectors of V, the vectors/(A) and/(B) are
also linearly independent.
Theorem 3.10.1. The map cr>_ is an automorphism of the additive group ofV onto
itself
Proof: Suppose that 6\(OA) = dx(OB). Then x(OA) = OD where
OD = OkAx = OlBx. This implies that Ax Bx or A B. Thus, OA = OB and G\ is
injective. To show that g\ is onto, let OB e V and A = OOxBx. Then
ox(OA) = OxAx = Ox(OOxBx)x = OxOxOB = OB. We show is additive. Let
OA,OB e V. Let D = OOxAx, F = OOxBx. Then
6i(OA + OB) = (OAOB)x = OxAxOxBx = ODOF = x(OA) + dk(OB).
Lemma 3.10.2. The map dk sends one dimensional subspaces ofV onto one
dimensional subspaces ofV.
87
Proof: Let < OA > be a one dimensional subspace of V. Then there exists a line
g = [a, p,y] such that OB e< OA > if, and only if, 0,B eg. Since
0,B  a,p,y <> Ox,Bk  a\p\y* then ^(< OA >) =< x(OA) > is a one
dimensional subspace of V.
Lemma 3.10.3. The transformation, G\, maps linearly independent vectors ofV to
linearly independent vectors.
Proof: The vectors OA,OB e V are linearly independent if, and only if, goA gOB if,
and only if, gQiji ^ goxBx Hence, is a semilinear automorphism of V for each
XeQ. m
Theorem 3.10.4. The map : V > V is a linear automorphism ofV onto V.
Proof: Consider the definition of a semilinear automorphism (23,defn.73.1). Let (V,IC)
and be vector spaces over the division rings 1C and 1C1, respectively. Suppose
that p : K. 1C' is an isomorphism from /C onto K.'. A map X : V i V1 is called
semilinear with respect to p if
1 .\{A + B) = k(A) + X(B) for all A,B e V.
2. X(tA) = p(/)X(A) for all A e V and / e 1C.
The only isomorphism p : R > R is the identity. Thus, K is a linear
automorphism of V onto V.
Let (V,/C,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 1C such that
giyA,yB) = rg{A,B), for all A. B e V. If A is nonisotropic, r = 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 / in our space, h _L / < hx _L /\ hence,
i(OA) 1 &x(OB) <> OA 1 OB.
88
Theorem 3.10.6. The linear automorphism ox is an isometry.
Proof: Let y : V < V be a similarity with square ratio r 0. Then for A,B e V we
have g(yA,yB) = rg(A,B) = rgfy~x(yA),y~x(yB)) and gfy~1(yA),fl(yB)) = Â¡rg(yA,yB).
That is, y_l has square ratio r~K Now ox is an involution, so that Ox = d^1. Hence,
if r 0 is the square ratio of ox then r  or r1 1, so r 1. [23] Because the
field /C is isomorphic to R, then V is not an Artinian space. Thus, for each similarity
dx of V there is an unique r > 0 and there is an unique isometry a such that
ox = AY(0,r); where M(t),r)(A) = rA for all A in V. The map ox is thus a similarity
with square ratio r2. Because r > 0, then from above, r 1 and it follows that dx. is
an isometry.
Proposition 3.10.7. The isometry ox is a 180 rotation; that is, a reflection about a
plane (a twodimensional subspace ofV).
Proof: Let X e Q. Suppose that 0\X and put O Xa with a e M.. Then for all A\X
and for all Z?a, dx(OA) = (OA)* = 0*A* = OA and
6x(OB) = (OB)* = 0*B* = OBa* = OB = OB.
Since .ta(0) = .tx(O)1 and V = .tk(0) .tx(O)1 then dx = lfx(0) 1 $x{0) Thus,
ox is a reflection about the plane &x(0).
Suppose that O / X. Let s  O such that s  X and let P  X be arbitrary.
Then if/?  e, POR = Q \ X and OR = PQ. Thus,
6x(OR) = (OR)* = (PQ)* = PQ = OR and d ]ke{0) = 1e(G).
Now let PO such that p 1 s, so O = sp. Let B  p. Now p 1 X because s  X and
POB = D\y where P = Xy and y  p. We calculate,
x(OB) = (OB)* = (PD)* = (PDy* = (PDf = DP = BO = OB = OB.
Hence, ox l e(0) and &k is a reflection about a plane.
89
To show that d>. is a reflection about a spacelike plane (Euclidean plane) for
every X e Q, we use the following theorem from Snapper and Troyer [23].
Theorem 3.10.8. [23] Let (V,/C,g) be an ndimensional metric vector space over a
field K. with metric g, K. = R, and n 2. Every nonsingular real plane has a
coordinate system such that the matrix of its metric is one of the following.
the Euclidean plane,
1 0
0 1
 the Lorentz plane, and
\
 the negative Euclidean plane.
Hence, the Euclidean plane, the Lorentz plane, and the negative Euclidean
plane are the only nonisometric, nonsingular real planes. This also follows from
Sylvesters Theorem [23) (It states that there are precisely n+1 nonisometric,
nonsingular spaces of dimension n).
Lemma 3.10.9. Let K. = R, n = 4, andV be Minkowski space. Then the orthogonal
complement of a Lorentz plane is a Euclidean plane.
Proof: Let {e,},=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 mi and nj. Then there is an isometry a :< e3,e4 >> a
such that cr(m) = mi and a(n) = n,. By the Witt Theorem [23] a can be extended to
an isometry of V, which we also denote by a. This implies that
a : V =< e3,e4 > < ei,e2 >* V = a < a(ei),a(e2) >;
that is, (a(ei).a(e2)} is a basis for a1. Now a is an isometry and
g(ei,ei) = g(e2,e2) = 1, so the metric g with respect to a1 has matrix I2, the 2x2
identity matrix, and a1 is a Euclidean plane.
90
Theorem 3.10.10. The map G\ is a refection about a spacelike (Euclidean) plane for
every X e Q.
Proof: From the proof of Proposition 3.10.7 it suffices to consider X e Q with 0\X.
Let O 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 ALe,, JCq2 c Tq. Let O ^ A e Afq, and O ^ B e Kq2. Then OA and OB are
isotropic vectors which form a basis for ie(O) and Q(OA) = 0 = Q(OB). Therefore the
metric of Â£q(0) with respect to OA and OB has matrix
!
g
OA
OB
(OA OB) =
Or''
r 0 j
where the products of the matrix elements are the inner products defined by the
metric g and g(OA,OB) = r 0. Since jr OA e /C,, and OB e /Cq2 then /C,, JCq2
also form an isotropic basis for .te(O). Hence we may assume that r = 1. Let
or,
i
n
(OA OB)
and OX,
(OA + OB).
Then g'(OX,,OT,) 0, Q'(OX,) = 1, and Q'(OT,) = 1. Because OX, 1 OT, and
Â£e(0) is nonsingular then OX, and OT, are linearly independent and hence, form a
basis for ie(O). Moreover, the metric of .te(0) with respect to OX,and OT, has the
matrix
. Therefore, by Theorem 3.10.8, te(O) is a Lorentz plane and by
Lemma 3.10.9, i/.(0) = ie(O)1 is a Euclidean plane. I
CHAPTER 4
AN EXAMPLE OF THE THREEDIMENSIONAL MODEL
This chapter begins by considering a net of von Neumann algebras, {11(0)} Oeh
and a state (0, coming from a finite component Wightman quantum field theory in
threedimensional Minkowski space. There are various senses to the phrase coming
from a Wightman quantum field theory. The assumption here is the version given by
Bisognano and Wichmann [5]. That is, given a finite component Wightman quantum
field, <})(jc), assume that the quantum field operator,
its closure is affiliated with the algebra TZ(0) (in the sense of von Neumann algebras)
for every test function / whose support lies in the spacetime region O. Driessler,
Summers, and Wichmann show these conditions can be weakened [15]. But free boson
field theories satisfy these conditions in threedimensional Minkowski space [5].
For such theories the modular involutions, Jo, associated by TomitaTakesaki
theory to the vacuum state and local algebras of wedgelike regions, O, 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 [27 in Minkowski space. The precise wording of this version of the
CGMA is given below.
Let /1 and l2 be two lightlike linearly independent vectors belonging to the
forward light cone in threedimensional Minkowski space. The wedges are defined as
the subsets W[l\,l2\ = {a/i + p/2 + lL e K12 : a > 0, p < 0, (/*,//) = 0, i = 1,2},
where ( ) denotes the Minkowski inner product.
91
92
Let l\ =(1,1,0) and /2 = (1, 1,0) be lightlike vectors in R12, threedimensional
Minkowski space, and let V be the Poincar group, the isometry group of this space.
Then the set of wedges, >V, is given by W = {^W[l\,l2\ ' ^ e T5}, where
= {Hx) : X e W[hh]}.
The CGMA for Minkowski space is defined as follows. Let {R(W)} ^vv be a net
of von Neumann algebras acting on a Hilbert space H with common cyclic and
separating vector Q e H, satisfying the abstract version of the CGMA and where the
index set I is chosen to be the collection of wedgelike regions W in R12 defined as
above. Recall from Chapter 1, with ({7Z(W)} iyew,7i.,Q) there is the following.
1. A collection of modular involutions {Jw}w<w
2. The group J generated by {Jw) wew
3. A collection of involutory transformations on W, {x^}^evv
4. The group Tgenerated by {x>r}frevv
Assume also that:
5. The group Tacts transitively upon the set W, that is, for every W\,W2 e W there
is a W3 e W such that Tiv}(fVÂ¡) = W2.
Note that this assumption is implied by the algebraic condition that the set
{adJw}weW acts transitively upon the net {V,(W)} pyeW At this point the following
two assumptions are added ([27] which have been verified for general Wightman
fields).
4.6 For W\,W2 e W, if W\ o W2 9* 0, then Q is cyclic and separating for
TZ(Wi) n n{W2).
4.7 For W\,W2 e W, if Q is cyclic and separating for 7l(W\) n 7Z(IV2), then
W, nW2 0.
The CGMA for Minkowski space is the abstract version of the CGMA with the
choice of >V for the index set /, 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 of the Poincar group V, which is
isomorphic to Tand related to the group Tin the following way. For each x e Tthere
exists an element gT e ^3 such that x(W) = gjW = {gx(x) : x e Wj. To each of the
defining involutions xw e T W e W, there is a unique corresponding gw e V [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 M12,
and let be the corresponding subgroup ofV. Moreover, let gw be the corresponding
involutive element ofV corresponding to the involution xw e T Then gw is a
reection about the spacelike orthogonal line which forms the edge of the wedge W. In
particular, one hasgwW = W,' the causal complement ofW, for every W e W. In
addition, exactly equals the proper Poincare group V+.
Recall from Chapter 2 that the initial model of (Q, 0) is as a group plane. This
means that each g e Q is viewed as a line in a plane and each P = gh, g\h, 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
CTa :Â¥uQ >P u Q by
tfa(P) = Pa = aPcT1 for P e V and ca(g) = ga = agcT1 for g e Q.
Since (Q, 0) is an invariant system then each aa 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 of the plane. We say that aa is a motion of the
group plane. Since Q generates 0 then the set of line reflections Qa = {<*g g e Q}
generates the group of motions 0CT = {ct : a e 0}. Let O : 0 > 0CT be the map
defined by (a) = cra, for a e 0. Then O is in fact a group isomorphism [2], This
94
means that (Â£,0) is isomorphic to (Ga,o) (in the sense that G is equivalent to Ga as
sets and
d>(Â£W A), which we denote by as
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 ((/, 0) satisfying as A, one obtains R1'2,
threedimensional Minkowski space. Under the identifications given in Chapter 2, we
find that each g e G corresponds to a spacelike line in R1,2. Thus, as C>(A) is a set
of true statements concerning reflections about spacelike lines and the motions such
reflections generate in threedimensional Minkowski space. Moreover, since such
motions are in fact isometries in R12 then 0(0) is isomorphic to a subgroup of the
threedimensional Poincar group.
Theorem 4.2 Under the same conditions as in Theorem 4.1 it follows that
({xw} weW,T) acting on Wsatisfies as O(M).
Proof: From Theorem 4.1 we have ({gw} WeWty) satisfies as O(M) since is the
subgroup of V generated by reflections about spacelike lines. Also from Theorem 4.1,
({xw}weW*'T) is isomorphic to ({gw}wew,Â¥) so ({?w} WeW,T) satisfies as O(A).
The net continuity condition assumed by BDFS [6] for the next theorem was
later shown to be superfluous [8] for this theorem and the remaining theorems.
Theorem 4.3 [6] Assume the CGMA with the spacetime R1,2 and W the described set
of wedges. If J acts transitively upon the set {1Z(IV)} weW then there exists a strongly
(anti) continuous unitary representation U(V +) of the proper Poincar group which
acts geometrically correctly upon the net {7Z(fV)} we\v and which satisfies U(gw) = Jw,
for every W e W. Moreover, U(V\) equals the subgroup of J consisting of all products
of even numbers ofJw s and J = U(V\) u JwRU(V\), where
WR = {x e R12 : jc, > *0}.
95
Theorem 4.4 Under the same hypotheses as Theorem 4.3, the group J is isomorphic
toV + = ty, which is generated by the set of involutions {gw \ W e W}. Moreover,
({Jw} we w. J ) satisfies as O(M).
Proof: By Proposition 1.1 there is surjective homomorphism Â£ : J T, where the
kernel of ker(^), is contained in the center of J, Z( J). By Theorem 4.3 there is a
faithful representation U(V+) such that U(gw) = Jw, for every W e W. Since the
center of V\ is trivial, Â£/() is a faithful representation of V\ and hence an injective
map preserving the algebraic relations, Z(J) is trivial. This implies that ker(^)= {1}
and Â£, is an isomorphism. If T : T> ^ denotes the isomorphism of T and ^3 given by
BDFS from Theorem 4.1 then T E, : J ^ = V+ is an isomorphism. It therefore
follows that the pair ({Jw} w&v\>, J) satisfies as 0(^4).
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 Held in threedimensional Minkowski space, which
satisfies the Wightman axioms, provides a set of modular involutions satisfying
as
As a final remark we note that since free boson field theories satisfy the
Wightman axioms and therefore the CGMA for Minkowski space holds, then these
theories give a concrete example of the threedimensional case of this dissertation.
CHAPTER 5
CONCLUSION
At this point it is useful to briefly recall the starting point of this thesis, to
restate the problem, and to summarize the results obtained. It is assumed that there
is a net of C*algebras {AÂ¡}Â¡^Â¡, each of which is a subalgebra of a C*algebra A, and a
state
CGMA, is it possible to determine the spacetime symmetries (the isometry group),
the dimension of the spacetime, and even the spacetime itself, without any assumption
about the dimension or the topology of the underlying spacetime?
Recall also that the resolution of these questions involved two steps. First,
given a set of involution elements Q and the group 0 it generates, find necessary
conditions on the pair (Q, 0) that will allow a construction of three and
fourdimensional Minkowski space. Moreover, this should be done in such a way that
the generating involutions can be identified with spacelike lines or spacelike planes
and their respective reflections. This was done for threedimensional Minkowski space
in Chapter 2 and for fourdimensional Minkowski space in Chapter 3.
The second step of this process is to determine what additional structure on the
index set I would yield algebraic relations among the modular involutions, {JÂ¡}Â¡e/,
such that the pair ({*//}/g/*7) satisfies the axiom systems given in Chapters 2 and 3.
Using the work of Buchholz, Dreyer, Florig, and Summers [6] and the work of
Wiesbrock [29], we are able to obtain a result concerning the above step in three
dimensions. First we briefly recall the abstract version of the Condition of Geometric
Modular Action, CGMA, described in Chapter 1. We assume there is a net, {V,Â¡}Â¡eÂ¡,
96
of von Neumann algebras acting on a Hilbert space 7i, where the index set I is a
partiallyordered set. There is a vector Cl e H which is cyclic and separating for each
IZj, i e I. From the modular theory of TomitaTakesaki we then obtain a collection,
{J/},<=/, of modular involutions which generates a group Jand a collection, {A,}/e/, of
modular operators. The assignment /' * 7ZÂ¡ is an orderpreserving bijection and each
ad Jj leaves the set {72./},e/ invariant. The last two assumptions imply that for each
i e I, there is an orderpreserving bijection x, on 1 such that JÂ¡TZjJÂ¡ = 7ZxÂ¡q), j e I.
The group generated by {x,}/6/ is denoted by Tand forms a subgroup of the
transformations on the index set I. Assume also that the two intersection conditions
for wedges given in Chapter 4 also hold as part of the CGMA for what follows.
To help explain the additional assumptions used in our result we give the
following definitions and theorems.
Definition 5.1 [6] The Modular Stability Condition (MSC). Let {7Z(lÂ¥)} be a net
of von Neumann algebras satisfying the CGMA where the index set I is the set of
wedgelike regions W in R12 described in Chapter 4. Then the modular stability
condition is satisfied if the modular unitaries are contained in the group J generated
by the modular involutions; that is,
A '{y e J for all t e R and W e W.
Theorem 5.2 [6] Assume the CGMA for threedimensional Minkowski space with (4.6)
and (4.7), where the index set I = W, the collection of wedgelike regions in Z?12.
Assume also the transitivity of the adjoint action of J on the net {7Z(W)} weW Let
U(RU2) be the representation of the translation group. If e J, for all t R and
some W e W, that is, if the modular stability condition obtains, then sp{U)
sp(U) d V. Moreover, for every futuredirected light like vector t such that
W+ a W, there holds the relation = U(e~alt), for all t e R, where
a = 271.
97
Definition 5.3 [29] Let A/, M be von Neumann algebras acting on a Hilbert Space
Li. Let Q be a common cyclic and separating vector in H. If A/Ajif
t > 0, we call (AAc A4,f2) a +halfsided modular inclusion (+hsm). If A% M
Ac A/, for all f < 0, we call (A/"c Ad,Q) a halfsided modular inclusion (hsm).
Definition 5.4 [29] Let A/", A4 be von Neumann algebras acting on a Hilbert space Li
with fi e Li a common cyclic and separating vector for A/, Ad, and Ain M.
1. If ((Ain M) a AA,Q) and ((Ain M) c Ad,Q) are hsm inclusions.
2. And if Jy{s lim/^+oo A^AaVV = s lim,^ao A^ A
Then we say that such a pair ((A/,A4), Q) has () modular intersection, mis.
Theorem 5.5 [29] Let M, M, C, and U be von Neumann algebras acting on a Hilbert
space Li with a common cyclic and separating vector fl e Li. Assume the following.
I. 1. (J\f,M,Q) is mis,
2. (Â£, A4,Q) is + mis,
3. (Â£,AAf2) is mis,
II. 1. ( c Af. Q) is hsm,
2. Ad Jy (.1y*J J]\[J
3. [Ad J(JÂ¡Jjd),JjjJ'atJ= 0,
III. 1. Ad (Ad Jc(JyJM)A) CAd c= K
with to = ^ In 2.
Then the modular groups
A 'if, A A ,and A'fr, for t, r, s, v eR,
generate a unitary representation of the 2+1dimensional Poincar group.
Remarks [29] The conditions in I. give a unitary representation of the
2+1dimensional homogenous Lorentz group. The hsm inclusion of condition II. equips
us with a representation of the translations along some light ray. The product of the
two modular conjugations is then a finite translation of this kind. Moreover, due to
98
the result of Bisognano and Wichmann [5], the modular conjugations of the wedge
algebras act as reflections. These properties are encoded in condition II.2 and II.3.
A physical framework will now be given as a description of quantum field
theories in terms of local nets of algebras [17]. The basic assumptions are the
following. Let {A(0)}oeV <= B(H) be a net of von Neumann algebras indexed by the
closed double cones T> in R1,2 which satisfy the following properties:
1. (Isotony) If 0\ a O2, then A(0\) a A(02).
2. (Locality) If 0\ 2)'
Where A(Q)' denotes the commutant of A(O) in B(H) and O' is the causal
complement of O c M12.
3. (Poincar covariance) There is a unitary representation
U : SO^(2,1) > M12 > UifH) of the Poincar group with positive energy.
4. (Vacuum vector) There is a unique U invariant vector Q e H.
The algebra A(0), the inductive limit of the net, is called the local algebra of
observables localized in O c R1,2.
As was mentioned in Chapter 4, if the local net is generated by Wightman
fields then the modular groups associated with algebras of observables localized in
wedges act as Lorentz boosts in the directions of different wedges and the modular
conjugations act as reflections [5].
In particular, the adjoint action of the modular conjugations on the net act as
reflections about the spacelike edge of the wedge. For what follows we shall call the
properties in the previous paragraph the Bisognano and Wichmann property.
Theorem 5.6 [29] Let A(0), O c: R12, be a local net fulfilling the Bisognano and
Wichmann property for wedges. Let M = A(flT/ ,/2]), M = A(W[li,/3]),
C s AMhM), andM= A{W[lul2l\\), where l\di, and13 are three linearly
independent light rays. Then this set of algebras together with the vacuum vector Q
fulfill the assumptions of Theorem 5.4. Conversely, let hi, M, C, and if be a set of A
99
von Neumann algebras acting on a Hilbert space H together with a common cyclic
and separating vector Q e Ti. which fulfill conditions I.III. of Theorem 5.4. Then
these data determine a local (BisognanoWichmann) net A(O) c B(H), O c M12,
such that the incident algebras become the wedge algebras of the constructed net as
in the first part.
Theorem 5.7 Let {7ZÂ¡} Â¡eÂ¡ be a net of von Neumann algebras acting on a Hilbert space
7i, together with a common cyclic and separating vector Q e H satisfying the CGMA
with the MSC. Assume also that there exist Â¡NjM,k\rJc e 1 such that
lZiX.lZjM,lZkandlZic satisfy the hypotheses of Theorem 5.5. Then the following are
true.
1. There is an injective map F : W /, such that for each W e W, the modular
objects Jf(W), Nj(W) have the BisognanoWichmann Property when acting upon
1Zf(W)
2. The modular unitarios A''v, AjrM, A_, and Nfc generate a continuous unitary
representation of 7$+ which acts covariantly upon {1Zf(W)} w^w
If the additional intersection assumptions, (4.6) and (4.7), are made on the subnet
{'RF(W)} WeWi then:
3. The CGMA as stated for Minkowski space holds for the subnet {Hf(W)} wew, as
does the MSC.
4. The modular conjugations {adJF(W)} w<=w, and thus ({ipw,} T), satisfy the
axioms of Chapter 2.
5. There is a continuous {anti) unitary representation of P+ acting covariantly upon
{'R.F(W)}wnW.
Proof: For the proof of Teorem 5.7.1, we give the construction Wiesbrock gave in the
proof of Theorem 5.6 [29]. Let /j = (1,1,0), l2 = (1, 1,0), /3 = (1,0,1) e R1*2. The
local algebra of observables to wedges is defined by
KWHuhV = Kix, mmiuh]) *KjM, n(W[h,h\) = H,c.
For arbitrary linearly independent light rays lhlj e M12 pointing to the future, let
100
A Â¡"i e SO^( 1,2) with /, = A///i, and lj = This element in 50^(1,2) is
uniquely defined up to a multiplication by a boost of type A/h/2(/), e 1 with the
given asymptotics l\,h Let U denote the unitary representation of the Poincar group
according to Theorem 5.4, that is, let
U{A/lA(0) sAjg, ZY(A//2(0) =A^, ZV(A/2,/3(0) sAg; t e R.
Now define the observable algebra associated with arbitrary wedges by
TZ(W[lÂ¡,lj]) = adll(AiljjKR'iAr) ( e by the MSC and the CGMA).
For translated wedges, define for a e R12
n(W[lÂ¡,lj,a]) S ad W(a)W(A/.//)(7e,>) ( e {7Z,}ieI by the MSC and CGMA).
In this way, for any wedge region W in R1,2 there is a unique von Neumann algebra
1ZÂ¡W in {7Zj}je/. Taking F : W > / to be the map = /w, for W e W and i\y e /
as obtained above and the result follows.
Conclusion 2 follows from Theorem 5.6 and Theorem 5.5. Conclusions 3 and 4
also follow from Theorem 5.6. The last conclusion follows from Theorem 4.3. I
We conclude this chapter with a few remarks. Given a net {4./}/e/ and a state
to satisfying purely algebraic conditions, one derives threedimensional Minkowski
space, and an identification between elements of I and the wedges in
threedimensional Minkowski space in such a way that the adJÂ¡ act like a reflection
through the spacelike edge of the wedge. Therefore, solely with assumptions on the
algebras of observables {4,}/e/ and the preparation oo, we are able to derive the
physical spacetime and its symmetries. We can also derive an interpretation of
suitable elements of {M/},<=/ as local algebras associated with wedge regions, as well as
derive a prescription of how the spacetime symmetries act upon the observables. In
addition, we can get a time orientation of the spacetime from the MSC.
101
A similar process can be done for fourdimensional Minkowski space using the
work of Wiesbrock and Khler [18]. However, we refrain from giving the details here.
102
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BIOGRAPHICAL SKETCH
Richard K. White was born in Richmond, Virginia on August 19, 1960. He
graduated summa cum laude from the University of North Florida in 1991 with a
Bachelor of Science degree in Mathematics. He graduated from the University of
Florida in 1994 with a Master of Science degree in Mathematics. In May 2001 he
graduated with a Ph.D. in Mathematics from the University of Florida. He is the
proud parent of a sixyearold angel, Jackie.
106
I certify that I have read this study and that in my opinion it conforms to acceptable
standards of scholarly presentation and is fully adequate, in scope and quality, as a
dissertation for the degree of Doctor of Philosophy.
n J. Summer's, Chairmt
Stephen J. Summer's, Chairman
Professor of Mathematics
I certify that 1 have read this study and that in my opinion it conforms to acceptable
standards of scholarly presentation and is fully adequate, in scope and quality, as a
dissertation for the degree of Doctor of Philosophy.
Gerard Emch
Professor of Mathematics
I certify that 1 have read this study and that in my opinion it conforms to acceptable
standards of scholarly presentation and is fully adequate, in scope and quality, as a
dissertation for the degree of Doctor of Philosophy.
/Q .
Johft7R. Klauder *
Joint Professor of Mathematics and Physics
I certify that I have read this study and that in my opinion it conforms to acceptable
standards of scholarly presentation and is fully adequate, in scope and quality, as a
dissertation for the degree of Doctor of Philosophy.
David Groisser
Associate Professor of Mathematics
I certify that I have read this study and that in my opinion it conforms to acceptable
standards of scholarly presentation and is fully adequate, in scope and quality, as a
dissertation for the degree of Doctor of Philosophy.
X C
Khandkera Muttalib
Professor of Physics
This dissertation was submitted to the Graduate Faculty of the Department of
Mathematics in the College of Liberal Arts and Sciences and to the Graduate School and was
accepted as partial fulfillment of the requirements for the degree of Doctor of Philosophy.
May 2001
Dean, Graduate School
>
.UiSS^
5
formulate geometric theorems about elements of the group of motions and to be able
to then prove these theorems by grouptheoretic calculation.
To summarize, we are to find conditions on an index set I and a corresponding
net of C*algebras {AÂ¡}Â¡ei as well as a state co satisfying the CGMA such that the
elements of I can be naturally identified with open sets of Minkowski space and such
that the group Tis implemented by the Poincar group on this Minkowski space.
Out of the group Twe wish to construct Minkowski space such that Ts natural
action on Minkowski space is that of the Poincar group.
This involves two steps. First we carry out the absolute geometry program for
three and fourdimensional Minkowski space. That is, characterize three and
fourdimensional Minkowski space in terms of a group of motions (Q, 0). Second, we
must determine what additional structure on the ordered set I would yield from
TomitaTakesaki theory algebraic relations among the J, (and hence, among the x,)
which coincide with the algebraic characterization found in step one.
The organization of the thesis is as follows: in Chapter 2, the given pair (Q, 0)
is used to construct a threedimensional Minkowski space out of the plane at infinity.
Then identification of the involutory elements of Q with spacelike lines and their
group action in with reflections about spacelike lines is made.
In Chapter 3 using the same initial data, (Q, ), as was given in Chapter 2 but
satisfying different axioms, a fourdimensional Minkowski space is constructed. The
approach taken here differs from that taken in Chapter 2. This time the affine space
is constructed first and then the hyperplane at infinity is used to obtain the metric.
The identification of the elements of Q with spacelike planes and their group action
in 0 with reflections about spacelike planes is made.
In Chapter 4 a concrete example of the threedimensional characterization is
given. As already mentioned, Bisognano and Wichmann showed that for quantum
field theories satisfying the Wightman axioms the modular objects associated by
11
secant lines with the set of interior points, ordinary lines, is classical hyperbolic
plane geometry.
2.2 Construction of n
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 is called an involution if a2 = 1, where 1 is the
identity element of the group .
Basic assumption: A given group is generated by an invariant system Q of
involution elements.
The elements of Q are denoted by lowercase Latin letters. Those involutory
elements of that can be represented as ab, where a,b e Q, are denoted by
uppercase Latin letters. If r e and fyr\ is an involution, we denote this by E,r.
Axioms
Axiom 1: For every P and Q there is a g with P, Q g.
Axiom 2: IfP,Q \g,h then P = Q or g = h.
Axiom 3: If a,b,c \P then abe d e Q.
Axiom 4: If a,b,c g then abe = d e Q.
Axiom 5: There exist g,h,j such that g \h butj / g,h,gh.
Axiom 6; There exist elements d, a, b e Q such that d, a, b / P,c for P,c e . (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 hj e Q such
that P\h,j but g, h f A,c and g,j / B,d for anyA,B,c,d e (that is, have neither a
point nor a line in common).
74
Let OA,OB e U1 and OC e U. If OC is nonisotropic then OC OA,OB and
there exist a,p,y,5 e V such that
A,0\a\ 0,BP; 0,Cy,8; a 1 y; and P 1 6.
By Axiom U, there exists X,8 e V such that O = Xz, AOB\X, and 0,C e. Thus,
OA + OB = O(AOB) 1 OC.
Now suppose that OC is isotropic. The following possibilities exist.
(i) If OC OA,OB, then as in (a) above, (OA + OB) A. OC.
(ii) If OC = OA = OB, then because OC is isotropic, AOB = COC e goc and
(OA + OB) 1 OC.
(iii) If OC OA OB, then OB _L OC implies that OB is nonisotropic and
(OC + OB) = OCOB T OC by Lemma 3.8.14.
Hence, if OA,OB e U1 then OA + OB e UL and UL is a subspace of V.
Theorem 3.8.18. HO = ap then .ta(O)1 = .tp(O) and tp(O)1 =
Proof: From the definition of orthogonal vectors we clearly have Â£(0) cr ipiO)1
and tp(O)1 c= ia(0) Now suppose that OA _L .ta(0); that is, goA h a Then, there
exists y e V such that OA e Â£y(0) and y 1 a. But then we obtain O = ap = ay so
P = y. Hence, .ta(O)1 = .tp(O) and Â£a(0) .tp(O)1.
An immediate consequence of Theorem 3.8.18 is the following. For each a e V
with O  a, there exists a unique P e V such t,hatOp, .ta(O)1 = Â£p(0), and
(ta (O)1)1 = a (O).
Theorem 3.8.19. Ifa(O) is nonisotropic then there is a unique hyperplane A(0) such
that (0)L = A(0) and (d(O)1)1 = d(O).
Proof: Let a e V such that a c: Ta, where a is the nonisotropic line associated with
0(0) {OA : A e a}. Then for O = ap we have a _L p and a T g for every g a 3fp
with O e g. Thus, .tp(6>) c (0)L. Now a is nonisotropic, so by 3.3.8, there exists a
unique h c Xa such that O e h and a 1 h. By Axiom U, (h(0),Â£$(0)) c (0)L.
79
go job L g,h by Axiom S2 because by 1. above go,cod <= (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,gOJOB <= (g,H If
go.coD go job then by Axiom T there exists a unique r\ e V such that
go,COD, gOJOB <= 3Cy.
Because two intersecting lines uniquely determine a plane, then
go,COD = gOJOB The last conclusion follows from (g,h) = (g,go job)
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
/ c (g,h) such that Pel; that is,
To complete the classification of orthogonal subspaces of V, it remains to
consider hyperplanes. Now if (X, V, 1C) 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 O. Then
A(O)1 = {h{Q)MO))L = KO)1 n IKO)1.
From Theorem 3.8.19 and Theorem 3.8.20, we know that the dimension of /i(O)1 is
three. By Theorems 3.8.18, 3.8.23, and 3.8.24 it follows that the dimension of
(f)(0)1) = 2. Consider the following possibilities. If h(O)1 n fHO)1 = {1}, then
h(0)L fi(O)1 e V has dimension five whereas the dimension of V is four. If
h(0)L n fKO)1 = fKO)1, then it follows that p(O)1 c h(0)x; h(0) 1 p(O)1 and
HO) c (^(O)1)1 = P(0).
Which contradicts the assumption that h and p intersect only in O.
104
[14] H.S.M. Coxeter, Projective Geometry, 2nd ed., SpringerVerlag, New York,
N.Y., 1987.
[15] W. Driessler, S. J. Summers, and E. Wichmann, On the Connection Between
Quantum Fields and von Neumann Algebras of Local Operators, Commun.
Math. Phys. Vol. 105, 1986, pp.4984.
[16] M. Florig, Geometric Modular Action, Ph.D. Dissertation, University of
Florida, 1999.
117] R. Haag, Local Quantum Physics, SpringerVerlag, Berlin, 1992.
[18] R. Khler and H.W. Wiesbrock, Modular theory and the reconstruction of
fourdimensional quantum Held theories, J. Math. Phys., Vol. 42, 2001,
pp.7486.
[19] B. Klotzek and R. Ottenberg, Pseudoeuklidische Raume im Aufbau der
Geometrie aus dem Spiegelungsbegriff, Zeitschr. f. math. Logik und Grundlagen
d. Math., Vol. 26, 1980, pp.145164.
[20] R. Lingenberg, Metric Planes and Metric Vector Spaces, John Wiley and Sons,
New York, N.Y., Chichester, Brisbane, Toronto, 1979.
[21] H. Noack and H. Wolff, Zorns Lemma and the High Chain Principle,
Fundamentals of Mathematics, Vol. 1, MIT Press, Cambridge, Mass., London,
1986, pp.527529.
[22] A. Seidenberg, Lectures in projective geometry, Van Nostrand, Princeton, N.J.,
1962.
[23] E. Snapper and R. Troyer, Metric Affine Geometry, Academic Press, New
York, N.Y., London, 1971.
[24] R. F. Streater and A.S. Wightman, PCT, Spin and Statistics, and All That,
Benjamin, Reading, Mass., 1964.
[25] S. J. Summers, Geometric modular action and transformation groups, Ann. Inst.
Henri Poincar, Vol. 64, 1996, pp.409432.
[26] M. Takesaki, Tomitas Theory of Modular Hilbert Algebras and Its
Applications, Lecture Notes in Mathematics, Vol. 128 Springer, 1970.
[27] L. Thomas and E. Wichmann, Standard forms of local nets in quantum field
theory, J. Math. Phys., Vol. 39, 1998, pp.26432681.
[28] R. White, A GroupTheoretic Construction Of Minkowski 3Space Out Of The
Plane At Infinity, preprint.
[29] H.W. Wiesbrock, Modular Intersections of vonNeumannAlgebras in Quantum
Field Theory, Comm. Math. Phys., Vol. 193, 1998, pp.269285.
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.
3.2.2.10. IfA,B,C a, a e M; gAB gAC are isotropic then gBc is nonisotropic and
there is a p with p A such that (3 1a and C = 2?P.
Proof: Because gAB gAc> then gBc is nonisotropic. By Axiom 14 there exists a y,
y\B,C, such that y 1 a. Let A  p, ply (Axiom 1). Because A  a, P; a _L y; and
ply, then by 3.1.6.18, a 1 p. If B$=B then B\P and gAB is nonisotropic. Similarly,
CP C and by 3.2.2.1, flP = C.
3.2.2.11. If A .B,C\ a, a e A4, are pairwise unjoinable then there exist p,y _L a with
A  p,y such that C = #Py.
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 p A such that P a. Because B and
C are unjoinable with A, then B,C / p and BP B. By Axiom 1 and 3.1.6.18, there
is a 81 p such that 6 1 P and 8 1a. Thus, B,B$ 18, Â§P = 6, a and B and i?P are
joinable. Suppose that BP and C are unjoinable. By 3.2.2.3, either ggPc = gAC or A is
joinable with Z?P. But if A,B^\z, e a, then A = A$,B\z$ _L aP = a imply that A
and B are joinable. If = gAc = gAB then fiP is unjoinable with B. Hence, BP and
C are joinable and by 3.2.2.10, there is a y with y A such that y _L a and C = B^
3.2.2.12. Let P a,p,y with p,y J. a, a e M. If gAB c: Xa is isotropic then g^g c Xa
is isotropic and gAB  g%.
Proof: By 3.2.2.8, there are precisely two isotropic lines in Xa through P, say gpc
and gpQ. By 3.2.2.9, g^PC = gpQ and g^pC = 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,D\a' with a1 _L a. Let B\X with X  a'. Then because 5a, X a by 3.1.6.17.
But then by 3.1.6.16, A = PDB\a'a'X = X and A is joinable with B.) Because
3
modular conjugations {J,},e/; that is, if for every i,j in I there is a k in I such that
adJ,(TZj) s JjTZjJi TZk, where JÂ¡TZjJÂ¡ = {JÂ¡AJÂ¡ : A e 1lj}.
Thus, for each i in I, there is an orderpreserving bijection, automorphism, xÂ¡ on I,
(/,<), such that JÂ¡HjJÂ¡ = for j e /.The set is a set of involutions which
generate a group % which is a subgroup of the group of translations on I. 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 I.
2. The group J acting on the set {/Â£, }/Â£/
To elaborate further the relation between the groups Tand J, consider the
following.
Proposition 1.1 [6] The surjective map ^ given by J,m) = T,yT,m, is
a group homomorphism. Its kernel S lies in the center of J and the adjoint action of
S leaves each 7ZÂ¡ fixed.
Thus, J is 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
REFERENCES
[1] F. Bachmann, Aufbau der Geometrie aus deni Spiegelungsbegriff, 2nd ed.,
Springer, Berlin, New York, N.Y., 1973.
[2] F. Bachmann, A. Bauer, W. Pejas, and H. Wolff, Absolute Geometry,
Fundamentals of Mathematics, Vol. 2, MIT Press, Cambridge, Mass., London,
1986, pp. 129174.
[3] R. Baer, Linear algebra and projective geometry, Academic Press, New York,
N.Y., 1952.
[4] A. Bauer and R. Lingenberg, Affine and projective planes, Fundamentals of
Mathematics, Vol. 2, MIT Press, Cambridge, Mass., London, 1986, pp.64111.
[5] J. Bisognano and E. Wichmann, On the duality condition for a hermitian
scalar Held, J. Math. Phys., Vol.16, 1975, pp.9851007.
[6] D. Buchholz, 0. Dreyer, M.Florig,, and S. J. Summers, Geometric modular
action and spacetime symmetry groups, Rev. Math. Phys., Vol. 12, 2000,
pp.475560.
[7] D. Buchholz, M. Florig, and S. J. Summers, An algebraic characterization of
vacuum states in Minkowski space, //, Continuity aspects, Lett. Math. Phys.,
Vol. 49, 1999, pp.337350.
[8] D. Buchholz, M. Florig, and S. J. Summers, Geometric modular action and the
Lorentz, Poincar, and de Sitter groups, preprint in preparation.
[9] D. Buchholz, and S. J. Summers, An algebraic charactrization of vacuum states
in Minkowski space, Commun. Math. Phys., Vol. 155, 1993, pp.449458.
[10] H. Busemann and P. Kelly, Projective geometry and projective metrics,
Academic Press, New York, N.Y., 1953.
111] H.S.M. Coxeter, A Geometrical Background For De Sitters World,
Amer.Math.Mon., Vol. 50, 1943, pp.217228.
(12] H.S.M. Coxeter, Introduction To Geometry, Wiley and Sons, Inc., New York,
N.Y., London, 1961.
[13] H.S.M. Coxeter, NonEuclidean Geometry, 4th ed., University of Toronto
Press, Toronto, 1961.
103
68
From Proposition 3.7.1 above, K,\(O) = (OB) for any O B e IC\ and K2(0) = (OC)
for any O C e 1C2. From this it follows that P\ = 6A(B) and P2 = 8A'(C) for some
A,A' e 1C. Thus,
OP = A OB + A' OC
and {OC,OB} span ta(0). Now if A OB + A1 OC = 1, then 08A(B)08A'(C) = 1.
This implies that 8A(B)08 y(C) = O and 08A>(C) = 8A(B)0. Because 0,8A(C) e 1C2
and 0,8A(B) e K\, then either 1C\  K2 or 08A'(C) 1 = 8A(B)0. But IC\ Jf 1C2, so
8A'(C) 8a(B) = O. Because a e Ai, then from 3.3 and 3.6 there exists t',a e Oa
such that 8A'(C) = C,a = O. This implies that C Oa 1 = O or A1 = O. By
assumption C O and D O, thus it must be the case that A A1 O. Hence,
{OC,OB} is a linearly independent set in $.a(O) and Xa(0) is two dimensional.
Suppose now that a e Q. We construct an orthogonal basis. By 3.1.7.2, there
exist p,y e 5 such that O = aPy and a 1 p 1 y 1 a. Let x = [a, P,aP] and
y = [a,y,ay]. Note that if x = y then by Axiom 14, aP = y and O = apy = 1, so
x y. Let P\a and let p',y'P with P' T p and y' T y. Put P\ = PP' and P2 = yy'.
Now P\ a, p'; al p; P' T P, so that a 1 pf by 3.1.6.18. Because P\a,y'; aly;
y' T y, then aly1. Thus,
p = (pp')a = pp'a = pp' = P] and p = (yy') = yy' = P2,
so P \,P 2 I ct. This implies that P\ a,p and P2ot,y so Pi ex and P2 e y. If
Q = P\OP2 then Q = P1OP2 = p^Pay/y* = p'ay'.Because pr T a, let 8 = aP* e V.
Then Q = 8y' and 6 T y'. Thus P\ P',a implies that P P'a = 5. Thus we have
P,Q\8,y' with 8 A y', and P = Q by Axiom 3. That is, P = P1OP2 with P\ e x and
P2 e y.
Let O X e x and O Y e y. Then from Proposition 3.7.1 above we have
x = (OX) and y = (OY) and there exist A,B e K. such that OP A OX + B OY, where
27
writing Â£11^2 We note that if then ^ for all in because
$ = = r1^ = r1^.^ = 
Let M = {aP: a  P and a,p e <3} and V = Q u AL We consider the
elements of Â£/ as spacelike planes and the elements of M. as Minkowskian or
Lorentzian planes. Thus, V consists of the totality of nonsingular planes and we
denote the elements of V by a, P, 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 Q and P e M then a e G and ft e M for every E, in Q.
For the following, let a, P e V with a p.
Definition 3.1. If a, p e Q, then aP e M by definition and we write alp and we
say a is perpendicular to or orthogonal to p.
Definition 3.2. Suppose that a e Q, P e AL
(i) If aP e Q, then we write a 1 P and we say a is perpendicular to p.
(ii) If aP g Q, then we write a J_ P and we say a is absolutely perpendicular to p.
Definition 3.3. Let X = {aP : a T P}. We call the elements of X points and we
denote these elements by A, B, C,....
Definition 3.4. If a, p e M. and a.p e A4, then we write a P and we say a is
perpendicular to p.
I certify that I have read this study and that in my opinion it conforms to acceptable
standards of scholarly presentation and is fully adequate, in scope and quality, as a
dissertation for the degree of Doctor of Philosophy.
n J. Summer's, Chairmt
Stephen J. Summer's, Chairman
Professor of Mathematics
I certify that 1 have read this study and that in my opinion it conforms to acceptable
standards of scholarly presentation and is fully adequate, in scope and quality, as a
dissertation for the degree of Doctor of Philosophy.
Gerard Emch
Professor of Mathematics
I certify that 1 have read this study and that in my opinion it conforms to acceptable
standards of scholarly presentation and is fully adequate, in scope and quality, as a
dissertation for the degree of Doctor of Philosophy.
/Q .
Johft7R. Klauder *
Joint Professor of Mathematics and Physics
I certify that I have read this study and that in my opinion it conforms to acceptable
standards of scholarly presentation and is fully adequate, in scope and quality, as a
dissertation for the degree of Doctor of Philosophy.
David Groisser
Associate Professor of Mathematics
I certify that I have read this study and that in my opinion it conforms to acceptable
standards of scholarly presentation and is fully adequate, in scope and quality, as a
dissertation for the degree of Doctor of Philosophy.
X C
Khandkera Muttalib
Professor of Physics
This dissertation was submitted to the Graduate Faculty of the Department of
Mathematics in the College of Liberal Arts and Sciences and to the Graduate School and was
accepted as partial fulfillment of the requirements for the degree of Doctor of Philosophy.
May 2001
Dean, Graduate School
92
Let l\ =(1,1,0) and /2 = (1, 1,0) be lightlike vectors in R12, threedimensional
Minkowski space, and let V be the Poincar group, the isometry group of this space.
Then the set of wedges, >V, is given by W = {^W[l\,l2\ ' ^ e T5}, where
= {Hx) : X e W[hh]}.
The CGMA for Minkowski space is defined as follows. Let {R(W)} ^vv be a net
of von Neumann algebras acting on a Hilbert space H with common cyclic and
separating vector Q e H, satisfying the abstract version of the CGMA and where the
index set I is chosen to be the collection of wedgelike regions W in R12 defined as
above. Recall from Chapter 1, with ({7Z(W)} iyew,7i.,Q) there is the following.
1. A collection of modular involutions {Jw}w<w
2. The group J generated by {Jw) wew
3. A collection of involutory transformations on W, {x^}^evv
4. The group Tgenerated by {x>r}frevv
Assume also that:
5. The group Tacts transitively upon the set W, that is, for every W\,W2 e W there
is a W3 e W such that Tiv}(fVÂ¡) = W2.
Note that this assumption is implied by the algebraic condition that the set
{adJw}weW acts transitively upon the net {V,(W)} pyeW At this point the following
two assumptions are added ([27] which have been verified for general Wightman
fields).
4.6 For W\,W2 e W, if W\ o W2 9* 0, then Q is cyclic and separating for
TZ(Wi) n n{W2).
4.7 For W\,W2 e W, if Q is cyclic and separating for 7l(W\) n 7Z(IV2), then
W, nW2 0.
The CGMA for Minkowski space is the abstract version of the CGMA with the
choice of >V for the index set /, together with assumptions 4.6 and 4.7 above less the
transitivity assumption [6].
76
a is isotropic then ala. By Axiom U and Lemma 3.8.14 it follows that
is isotropic and g a, then
there exists a unique y e M such that g,a a Xy. Because g(0) a A(0) a {0)L,
then g 1 a, which contradicts Lemma 3.8.11. Thus, A(0) contains no other isotropic
line.
Let h 1 a with O e h. For each H e h we may write H = H1 OB, with H'\a and
Hp. If O A e a, it follows that OB T OA\ OH 1 OA, so that OB T OA and
OH' = OHOB T OA. This implies that gOH< = a so H* e a and h(O) e (d(0),_tp(0)).
Thus, (O)1 = .
If h(0) c {(Oi^tpfO))1, then
h(O) c tp(O)1 = .ta(O) and h(O) 1 a(O).
Thus, h(P) = a(0). Hence, ((0),Xp(O))1 = a(0) and (dfO)1)1 = a(O).
The subspace (O)1 is independent of the plane containing a. Suppose a c: Xy,
y e M, y a. Put O = y5. Let h(0) c (a(0),Â£5(0)). Then h 1 a and
h(0) c . If
kO) c <(0),tp(0)>
then k 1 a and k(O)
.
Remark. If g is a line then g is either nonisotropic or isotropic. From Theorem
3.8.19 and Theorem 3.8.20, if U < V is a onedimensional subspace then UL 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 (g, h) denote the unique
plane determined by g and h.
95
Theorem 4.4 Under the same hypotheses as Theorem 4.3, the group J is isomorphic
toV + = ty, which is generated by the set of involutions {gw \ W e W}. Moreover,
({Jw} we w. J ) satisfies as O(M).
Proof: By Proposition 1.1 there is surjective homomorphism Â£ : J T, where the
kernel of ker(^), is contained in the center of J, Z( J). By Theorem 4.3 there is a
faithful representation U(V+) such that U(gw) = Jw, for every W e W. Since the
center of V\ is trivial, Â£/() is a faithful representation of V\ and hence an injective
map preserving the algebraic relations, Z(J) is trivial. This implies that ker(^)= {1}
and Â£, is an isomorphism. If T : T> ^ denotes the isomorphism of T and ^3 given by
BDFS from Theorem 4.1 then T E, : J ^ = V+ is an isomorphism. It therefore
follows that the pair ({Jw} w&v\>, J) satisfies as 0(^4).
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 Held in threedimensional Minkowski space, which
satisfies the Wightman axioms, provides a set of modular involutions satisfying
as
As a final remark we note that since free boson field theories satisfy the
Wightman axioms and therefore the CGMA for Minkowski space holds, then these
theories give a concrete example of the threedimensional case of this dissertation.
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 Q.
Lemma 3.9.21. Let g be a symmetric bilinear form representing 7t. Then g is
nondegenerate.
Proof: This follows from the fact that 7i(V) = V1 = {1}.
Theorem 3.9.22. Let g be a symmetric bilinear form representing n. Then there is a
basis ofV such that the matrix of g with respect to this basis has the form
r C 0 0
0 C 0
0 0 c
^000
that is, there is an orthogonal basis ofV such that the matrix of g with respect to
this basis has the above form.
Proof: Let a e M. such that 1C a Xa and let O = ap. Let 0y,8 e Q such that
a = y8 and y,5 _L p. Let 8 = yP e M and r = 8p e ALThen sr = yPP8 = y8 = a so
sir). Put = [a,y, 8], y = [P,y,s], z = [p,r,8], and t = [a,s,r]].
Claim. x,y,z,t are four mutually orthogonal nonisotropic lines through O. We
have O = ap, O e x,t c 3Â£a and O e y,z c Tp so that x,t 1 y,z. Also,
O = ap = y8p = yr = 8s so that y 1 q and 8 1s. Then
O e x a Xy,0 eta X^,0 e Xy, and O e z a 3Â£n
implies that jc _L t and y L z.
The construction of the basis. Let E e 1C, the multiplicative identity,
considered as a point in Xa. Put T = EOE e t and X = EOEx e x. Note that because
0 ^
0
0
C J
0 e 1C;
CHAPTER 5
CONCLUSION
At this point it is useful to briefly recall the starting point of this thesis, to
restate the problem, and to summarize the results obtained. It is assumed that there
is a net of C*algebras {AÂ¡}Â¡^Â¡, each of which is a subalgebra of a C*algebra A, and a
state
CGMA, is it possible to determine the spacetime symmetries (the isometry group),
the dimension of the spacetime, and even the spacetime itself, without any assumption
about the dimension or the topology of the underlying spacetime?
Recall also that the resolution of these questions involved two steps. First,
given a set of involution elements Q and the group 0 it generates, find necessary
conditions on the pair (Q, 0) that will allow a construction of three and
fourdimensional Minkowski space. Moreover, this should be done in such a way that
the generating involutions can be identified with spacelike lines or spacelike planes
and their respective reflections. This was done for threedimensional Minkowski space
in Chapter 2 and for fourdimensional Minkowski space in Chapter 3.
The second step of this process is to determine what additional structure on the
index set I would yield algebraic relations among the modular involutions, {JÂ¡}Â¡e/,
such that the pair ({*//}/g/*7) satisfies the axiom systems given in Chapters 2 and 3.
Using the work of Buchholz, Dreyer, Florig, and Summers [6] and the work of
Wiesbrock [29], we are able to obtain a result concerning the above step in three
dimensions. First we briefly recall the abstract version of the Condition of Geometric
Modular Action, CGMA, described in Chapter 1. We assume there is a net, {V,Â¡}Â¡eÂ¡,
96
54
Em = Ea X = Â£8 8 = En and M = N by 3.1.6.14. So m = goM = gON = / and ct/ = om.
Therefore aa<7/, = ct*ct/.
Let us denote this unique map by 8a So for all A e goE A O,
(i) 8^(0) = 0
(ii) 8(Â£) = A
(iii) 5A e X>a(0).
Translations. For every pair A, 5 of distinct points we can define a translation
Tab : 3E > X given by Tab{A) = AAB = fi. We now restrict our attention to a set of
translations defined on Xa and we note that if A,B,Ca. then D = ABCa by 3.1.4.6.
Thus, Tab ^a 1 Xa is a welldefined map for all A,Ba. Let 1C = goE be an
isotropic line in Ta and define Ta = {Toa 'A e 1C}.
Theorem 3.5.4. The set Ta is an abelian group.
Proof: Let Toa^ob e Ta. Because C = BOA = AOB e 1C, for X\a, we calculate
{Toa o Tob){X) = TO{XOB) = XOBOA = XOC = TOC(X). Thus, Toc = TOA TOB e Ta.
Also, Too(X) = XOO = X. Hence, Too 6 Ta is the identity l.ra on Xa. To find TlqA,
we compute
(Toa TOo){X) = XOAOA = XOOAOOA = X= XOAOA = {TOAo TOA){X).
Hence, ToA = TOAo, A = OAO e 1C, and TqA e Ta. Clearly,
Toa {Tqb Toe) = {Toa Tob) 0 Toc for A,B,C e 1C. 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 1C and X\a,
{Toa Tob){X) = XOBOA XOAOB = {TOB TOA){X).
Therefore, Ta is abelian.
Lemma 3.5.5. For all T0a e Ta, Toa{1C) = 1C.
Proof: Because OAB e K for A,B e Â¡C then Toa{1C) c K. for all TOA Ta. Now let
81
{x e V : fx, u) = 0 for everyu e U} and {x e V : /(u, x) = 0 for every u e U) are
subspaces of V. We say that the autoduality n of (V,/C) upon itself is represented by
the semibilinear form fix, y) if
n (U) = {x e V : fx,U) = 0} = {x e V : fx, u) = 0 for every u e U}.
Theorem 3.9.5. [3] Autodualities of vector spaces of dimension not less than 3 are
represented by semibilinear forms. M
Theorem 3.9.6. [3] If the semibilinear forms f and g over (V, 1C) represent the same
autoduality ofV, and if dim(V) > 2, then there exists a 0 ^ d e /C such that
&(x, y) = fx, y )d for every x, y e V.
Definition 3.9.7. If n is an autoduality of the vector space (V,/C), then a subspace W
of V is called an Nsubspace of V with respect to 7T, if <7iv)) for every v e W.
In this case, n is said to be a null system on the subspace W of V.
Theorem 3.9.8. [3] Suppose that if, a) represents the autoduality it of (V,fC). Then 7t is
a null system on the subspace W
Definition 3.9.9. A line (v) is called isotropic if (v) < 7iv)). (So an isotropic line is
an Nline.)
Theorem 3.9.10. [3] If the semibilinear form if, a) represents a polarity n, and if
fw,w) = 1 for some we V, then a2 1 and
affix,y)) fy,x) for everyx,y eV.
In this case we say that f is a symmetrical or just symmetrical.
Theorem 3.9.11. [3] Suppose that n is an autoduality of the vector space (V,/C) and
that dimfV) > 3. Then n is a polarity if, and only if, n is either a null system or else n
may be represented by a symmetrical semibilinear form if, a) with involutorial a.
Theorem 3.9.12. [3] Suppose that the polarity n of the vector space (V,/C) possess
24
correspond. Two lines are said to be perpendicular if they intersect and their
elements at infinity correspond under the polarity.
Theorem 2.5.1. Every motion in a threedimensional affine space with a hyperbolic
polarity defned on its plane at infnity, is generated by reflections about exterior
points. Moreover, because exterior points correspond to spacelike lines, then any
motion in threedimensional Minkowski space is generated by reflections about
spacelike lines.
Proof: Because any motion in threedimensional Minkowski space can be generated
by a suitable product of plane reflections, it suffices to show that reflections about
exterior points generate plane reflections.
Let a be any Minkowski plane or spacelike plane. Let P be any point in
Minkowski space. Let / be the line through P parallel to a. Let a denote the
section of a at infinity. Applying the polarity to ctoo we get a point gao X a 00 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 T ctoo. Now let m be a line
through P not in a which intersects /. It follows that the reflection of P about a is
the same as reflecting m about / 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 / and m must act as their sections at
infinity act. because any point reflection in the hyperbolic projectivemetric plane
can be generated by reflections about exterior points, we have that the reflection of
P about a is generated by reflections of P about spacelike lines.
2.6 Conclusion
As already indicated above, the geometric model for the generators of Q which
lies behind the choice of algebraic characterization of threedimensional Minkowski
21
collineation for each b e Q.
Theorem 2.3.10. T is a harmonic homology.
Proof: Since Ab 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 Q, T/, 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 (^b)b = Â£,
for any h, e 0, T/, is involutory for each b e Q. Now in a projective plane in which
the theorem of Pappus holds, the only collineations which are involutory are
harmonic homologies [10], thus T, is a harmonic homology for each b e Q.
Theorem 2.3.11. Interior point reections are generated by exterior point reflections.
Proof: A similar argument shows that for each interior point A, is a harmonic
homology with center A and axis A where A is the polar of A, A Â£ A, and where
is defined analogously to T. Thus, each 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.
Theorem 2.3.12. The reection 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 \d and g(d) is orthogonal to g(b). Let E
be the point where g(b) meets g(d). Since E e gfb) then E \b and Eb =f for 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 \d and E\d then bd = C and we have E,C \b,d with b d. Thus, by
Axiom 2, E = C = bd. Hence, AE Adb Ab.
61
If A g t, then by Axiom T, there is a unique r e M. such that goxA c Xri
From Sections 3.5 and 3.6, JCtj is an affine plane and our definition of parallel lines in
An is equivalent to the affine definition. So, there is a unique line h c An such that
At h and h  gxE, Because
goxi^t = {O} 0, hnt = {At} 0, and h  gE,x,
then hr^gox = {B} 0. In this case, set = B.
If A e t and A O, then because t,lC a Xa with a e M., there exists a
unique g c Xa such that A e g and g  gxE Because gxE ^ t = {E} ^0 then
gnt= {B} 0. Set 8a(X) = B.
If X = O, set 8a(X) = O. Note that e gox by construction. It is clear
that 8a : X >> X is onetoone. We find it useful to make some observations.
Let O A e /C and recall that E e 1C is the multiplicative unit point;
E A = A, for A e K.. Put At = AOA.
Lemma 3.6.3. For the map 5,4 defined above, the following are true:
1 8a(E)=A.
2 8A(Et)=At.
3 5,4 is onto.
Proof: 1. Et = EOE1 and At = AO A1 imply that EEt OE and AAt = OA. If
EEt = OE = 1 then O E and O = O' = E. Similarly, because A * O then
AAt OAl 1. By 3.2.2.9, A' = A^ e 1C' as 0,A e 1C, K, is isotropic, and 0y _L a.
Similarly, E e A'. Thus by 3.2.1.6 we have gEEl II gOE< = K. II gAA> and gEE, II gAA,
by 3.2.1.7. By 3.2.1.8, h = gAA, and h n goE = gAA, n A = {A}. Hence 8A(E) = A.
2 8a(E,) = A,. Again, gsAE)A, = gAA, II gEE, and gAA, n / = {At}.
3 Let P e X and P O. If P g t then by 3.2.1.8, there is a unique line g such that
Et e g, g II gA,p, and gng0p = {Q}, for some Q e X. Then it follows that
8a(Q) = P. If P e t then 8A(Q) = P, where {Q} = gr\t, E eg, and g  gAP.
19
pointwise fixed, its axis, and a point linewise fixed, its center.
Definition 2.3.3. A homology is a perspective collineation with center a point B and
axis a line b where B is not incident with b.
Definition 2.3.4. A harmonic homology with center B and axis b, where B is not
incident with b, is a homology which relates each point A in the plane to its
harmonic conjugate with respect to the two points B and (b, [A, 5]), 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 / 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 /. 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 pass through A, B, C respectively. Let
{P} = [A,S] n [5,/?], then {D} = [A,B] n [P,Q\ ([11]). Note that if [R,S]  [A,C] then
D is the midpoint of A and B.
Coxeter [13] showed that any congruent transformation of the hyperbolic
plane is a collineation which preserves the absolute and that any such
transformation is a product of reflections about ordinary lines in the hyperbolic
plane where a line reflection about a line m is a harmonic homology with center M
and axis m, where M and m are a polepolar pair and M is an exterior point. A point
page
4 AN EXAMPLE OF THE THREEDIMENSIONAL MODEL 91
5 CONCLUSION 96
REFERENCES 103
BIOGRAPHICAL SKETCH 106
IV
>
.UiSS^
77
Definition 3.8.21. If a e V such that g,h a Ta then we say that (g,h> = Xa is a
nonsingular plane. If there does not exist a e V such that g,h cz Xa. then we say that
(g, h) is singular or (g. h) is a singular plane.
It is clear that every plane in (3f,V,/C) is either singular or nonsingular. Note
that by Theorem 3.8.18, if U < V is a nonsingular twodimensional subspace of V,
then UL is a uniquely determined nonsingular twodimensional subspace of V. Now
consider the following cases for two distinct intersecting lines g and h.
Suppose that g is isotropic and h is isotropic. Then by Axiom T, there exists a
unique a e M 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 a Xa
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 T h.
Proof: Let {P} = gn h. By Axiom S2, either g 1 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, gAjPB is
nonisotropic. Let gPA = [8,s], so A,P\b,z. If A is unjoinable with APB then A, P, and
APB are pairwise unjoinable points, so by Axiom 10, APB\h,z and APB e g. Thus,
gPjPB = gPA = g and B = PAAPB eg, so g = h.
Thus, gpjPB g and gpjps is isotropic, so by Axiom T, there exists a unique
y e M such that g,gpjps c This means that P,A,APB\y so B PAAPB\y, and
h c Xy, which contradicts our initial assumption. Therefore, g T h.
Theorem 3.8.23. Let (g(0),h(0)) be a two dimensional singular subspace ofV where
g(O) is isotropic and h(()) is spacelike. Then (g(0), h(O))1 is a two dimensional
singular subspace of V which contains g(O). Moreover, ((g(()),h{0))L)L = (g(0),h(0)).
Proof: First we observe that if W is a subspace of V generated by subspaces U and Id!
48
3.3.11. Let h = [a,Â£,r], g = [a,7,5] e Â£a, then Ggh = aa,,(g)
Proof: Let A  a and put B = ag(/l) = Ay. Then
opiVhCA)) = g hGgG^Gh(A) = <*h<* g() = oh(B).
Thus, Og''(o/,(A)) = Oa/,(g)(^/;(d)) for all A e XaBecause G/, is injective, then
Ggh = aCT/i(g) for every A in .Ta Similarly, Ogh(a) = Ga/i(g)(a), for every a in Â£a.
Hence, Ggh = Goh(g) Analogously, we can obtain G^h = G0h(P), for every P in Xa B
Because Ca generates Qa and every element of Qa is an orthogonal collineation
of (Xa,Â£a) then we obtain the following.
3.3.12. For every g e Qa and for every gg e Ca, we have a e Ca; that is, Ca is an
invariant system ofQa B
Consider the two mappings: g e Â£a 1 ag e Ca and P Ta *+ ap e iPa
The first one is from the set of nonisotropic lines in Â£a onto the set of line
reflections, and the second is from the set of points in Xa onto the set of point
reflections in 3Ca 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. Ig e Qa and P,Q e Xa then g(P) = Q <> Gp = gq.
Proof: Because
G(P) = Q <> Ga(P) = Gq <> Ga(P) = Gp,
the result follows. B
3.3.14. Ig e Qa and g,h e Â£a then G(g) = h < ag = a/,. B
3.3.15. IfP e Xa andg e Â£a then P e g GpGg is involutory.
Proof: Suppose that P e g = [a,y,8] e Â£a. Then for A\a,
GpGg(A) = Gp(Ay) = Ayl = A11 = GgGp(A). Thus, (GpGg)2 = lxa Now assume that
86
In the last section of Chapter 3 we show that each X in Q can be identified with
a spacelike plane and each \ e Q* with a reflection about a spacelike plane.
3.10 Spacelike Planes and Their Relections
For each X e Q with 01 A., define a\ : V V by
0,1 e V. To extend this definition to any X e Q, we note that if X / O and OA e V,
then there exists a unique D in X such that OOxAx = D; that is, there is a unique D
in X such that (OA)1 = OxAx = OD. Thus, we define x(OA) = OD, where
OOxAx = D. We note that if X \ O then D = Ax.
First we show that is a semilinear automorphism of V for each X e V. [23]
Now, a function /: V > V is a semilinear automorphism if, and only if, it has the
following properties:
1. /is an automorphism of the additive group of V onto itself.
2. / sends one dimensional subspaces of V onto one dimensional subspaces (that is, /
is a collineation).
3. If A and B are linearly independent vectors of V, the vectors/(A) and/(B) are
also linearly independent.
Theorem 3.10.1. The map cr>_ is an automorphism of the additive group ofV onto
itself
Proof: Suppose that 6\(OA) = dx(OB). Then x(OA) = OD where
OD = OkAx = OlBx. This implies that Ax Bx or A B. Thus, OA = OB and G\ is
injective. To show that g\ is onto, let OB e V and A = OOxBx. Then
ox(OA) = OxAx = Ox(OOxBx)x = OxOxOB = OB. We show is additive. Let
OA,OB e V. Let D = OOxAx, F = OOxBx. Then
6i(OA + OB) = (OAOB)x = OxAxOxBx = ODOF = x(OA) + dk(OB).
Lemma 3.10.2. The map dk sends one dimensional subspaces ofV onto one
dimensional subspaces ofV.
8
projectivemetric plane and that the motions can be viewed as reflections about
exterior points.
Next we embed our projectivemetric plane into a threedimensional projective
space. By singling out our original plane as the plane at infinity, we obtain an affine
space whose plane at infinity is a hyperbolic projectivemetric plane,
threedimensional Minkowski space. Finally, we show that the motions of our
original plane induce motions in the affine space and, by a suitable identification, we
show that any motion in Minkowski space can be generated by reflections about
spacelike lines. Thus, to construct a threedimensional Minkowski space, one can
start with a generating set Q of reflections about spacelike lines in the plane at
infinity. So Q may be viewed as a set of reflections about exterior points in a
hyperbolic projectivemetric plane. Out of the plane at infinity, one can obtain a
threedimensional affine space with the Minkowski metric, which is constructed from
a group generated by a set of even isometries or rotations.
The approach in this chapter differs from the method used by Wolff [30] for
twodimensional Minkowski space and by Klotzek and Ottenburg [19] for
fourdimensional Minkowski space. The approach in these papers is to begin by
constructing the affine space first. For Wolffs [30] twodimensional case, the
elements of the generating set Q are identified with line reflections in an affine plane.
For Klotzek and Ottenburgs [19] fourdimensional case, the elements of the
generating set Q are identified with reflections about hyperplanes in an affine space.
Thus, in each of these papers, the generating set Q 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
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
62
Lemma 3.6.4. IfC D, then gCD  hence, for all O A e 1C, 8A is a
dilation ofX.
Proof: Let g\ = t, gi = goc, g3 = gOD, Pl = Et, Q\ = A2 e g\, P2 = C, Qi = 6A(C),
Qi e g2, Pz = D, and Q3 = 5A(D). Then gCE, II gA,8A(Q and Sde, II gAl8A(D) Thus> by
Axiom D, gcD II gdA(c$A(Dy
Consider now the plane Xa Recall that by 3.5.1.3, for O A e 1C, there is a
unique 8A e Va(0) such that 8A(E) A, where 8A is of the form GgGj, for g,h e Oa
From Axiom 9 and the proof of 3.5.1.1, for any r e Oa, GgG/,Gr = for a unique
w e Oa That is, for every fixed r e Oa, every O A e /C can be uniquely writen in
the form 8A = GwGr where Gw(Er) A. (The uniqueness follows from the fact that
(Ar)w' = (Ar)w2 implies wj = u>2 if Ar # A [30].)
Therefore, for every O A e 1C, there is a unique a e Oa, such that Ela = A.
That is, there is a unique a e Oa, such that GaGÂ¡ = 8A. Also, every P e Xa, Pa,
can be uniquely writen as P = P\OPi, where P\ e 1C and Pi e 1C'.
Let Pi = Pi e 1C. Then we may uniquely write P = P\OP\, where P\,Pi e 1C.
Define the map 8A : Xa Xa, by 8A(P) = P\aO(Pfy, for O A e 1C, P\a.
Theorem 3.6.5. The map 8A : Xa Xa is a dilation on Xa with fixed point O and
dilation factor A, for all O A e 1C.
Proof: If PyOP'f1 = Q'fOQf, then by unicity, Pf = Qf. So P\ = Q\, P'f1 Qi,
and Pi = Qi Hence, P = P\OP\ = Q\OQl2 = Q, and 4 is injective. Let Q\ct with
Q = Q\OQ\; Q\,Qi e 1C. Let P, = Qf e 1C and P2 = Qf e 1C. Then P = PxOP2 a
and SA(P) = PfOPf = (&TO(Q?)at = Q\OQ\ = Q. Therefore, SA is onto.
We claim that if P ^ Q, then gpQ  gsA(P)8A(Qy First we show that
&a(P) e gop. If P e 1C, then P = POO1 and 8A(P) = P,aOO,a = P,aOO = P,a e 1C. If
P e K\ then putting R = P' e 1C, we have P = OOR and 8A(P) = OtaORtat =
= OORlal = R,at = Pal e 1C'.
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 [5] showed that for quantum theories satisfying the Wightman axioms
the modular objects associated by TomitaTakesaki theory to the vacuum state and
local algebras generated by field operators with support in wedgelike spacetime
regions in Minkowski space have geometrical meaning. Motivated by this work,
Buchholz and Summers gave an algebraic characterization of vacuum states on nets
of C*algebras over Minkowski space and reconstructed the spacetime translations
with the help of the modular structures associated with such states. Their result
suggested that a condition of geometric modular action might hold in quantum
field theories on a wider class of spacetime manifolds.
To explain the abstract version of this condition, first some notation is
introduced and the basic setup is given. Let {A };e/ be a collection of C*algebras
labeled by the elements of some index set / such that (/,<) is an ordered set and the
property of isotony holds. That is, if i\,ii e I such that i\ < Â¡2, then AÂ¡x c A,2.
Let A be a C*algebra containing {A,},e/. It is also required that the assignment
i A, is an orderpreserving bijection.
In algebraic quantum field theory, the index set I is usually a collection of
open subsets of an appropriate spacetime (M,g). In such a case, the algebra A, is
interpreted as the C*algebra generated by all the observables measured in a
1
42
Proof: By Axiom 6, gAB = [a, P] for some a and P in V. Let X e gAB and suppose
that X A, B. If A is joinable with A then there exists y,5 such that y A. 8 and
A,Xy,5. By Axiom 2, B y,5 and A and B are joinable. Hence, X is unjoinable with
both A and B.
Remark. If X is unjoinable with A and B\ A,B a, P; then by Axiom 10, Aa, P
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,Z)a, P with a 1 P then A,B a, p by Axiom 2; that is, gAB = gcD
3.2.2.5. If A ,B,C\a are pairwise joinable and distinct, then each P \ a is joinable with
at least one of A, B, and C.
Proof: If a e Q then the result follows from Axiom 12, so assume that a e AT By
Axiom 13 there exist D,Ea 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 gpB. By 3.2.2.4
above, this implies that two of A, B, and C are unjoinable, which contradicts our
assumption.
3.2.2.6. If gAB is isotropic then gB is isotropic for all Â£, e .
Proof: If A^,B^ a,p with a 1 P then A,Ba^ ',p^ 1 and a^' ip^'.
3.2.2.7. If P\ a e M then there are at most two isotropic lines in Xa through P.
Proof: If gAp,gBp,gcp ^ 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.
3.2.2.8. By Axiom 13 and 3.2.2.7, for each P \ a, a e M., there are precisely two
isotropic lines in Xa through P.
3.2.2.9. Let gAB,gAc Ta, a e M, be two isotropic lines.
a. If A  (5, with pla, then g^4B = gAC.
b. If A  y, with yia, then g\B =gAC.
39
3.2.1.3. Suppose that ap = y.
a. Ifaai = PPi then ai J. Pi andaiPi = y, soy _L ai,pi.
b. If aai = PPi = YYi then a T Pi, ai _L p, yi _L ai,Pi, anda$\ = aiP = y.
c. If A  a, P then A \ y.
d. The line g = [a, P] = [P,y] = [a,y] = [a, p,y] is a nonisotropic line. Thus, a, p, and
y are three mutually perpendicular planes which intersect in a line.
Proof: a. From aai = PPi we have y = Pa = P i a i.
b. If ppi = yyi and aai = TYi then yi yPPi aPi and yi yaai = Paj.
c. Because A  a, p and ap = y then Ay = A aP = aPA = yA and A \ y.
d. By Axiom 7 there exist points A and B such that A,Ba,p and by 3.2.1.3.c. above,
A,B\y. Thus, A,B e [a,P],[a,y],[P,y] and [a,P] = [a,y] = [p,y] s [a,p,y] is a
nonisotropic line.
3.2.1.4. If A and B are collinear then A'~ and Bf are collinear for anyE, e 0.
Proof: This follows from the fact that A,B\a, P < A^,B^ a;,p;.
For each Â£ e 0, we define ciS = {C^ : C e a}. By 3.2.1.4 above, is a line
for every Â£ in 0 and if a = [a, P] then aS = [a^,p^].
Definition of parallel lines and planes. We say the line a is parallel to the
planea, denoted by a  a, if there exists a P such that a c 3Â£p and P  a; that is,
a = [P,y] and a,p 1 5 for some 5.
We say that two lines a and b are parallel, a  b, if there exists a, P,y,8 such
that a = [a,y], b = [p, 8], where a  P and y  5.
3.2.1.5. If AA1 = BB' 1; A,A'\a,a'; a a'; 5,B'p,p'; p ^ p', then gAA>  gBB>.
Proof: Let 5p* with p*  a (3.1.6.12). Then B' P* by 3.1.6.11. Let Bp*, with
Pi  aj. Then B1 \ P^ by 3.1.6.11 and B,B' \ p,Pi,p*,pÂ¡ which implies that
[p,pi] = [p*,pt] = gBB' by 3.2.1.1. Hence, we have gAA< = [a,ai], gBB< = [P*,pÂ¡] with
a  p* and ai  pi; thus, gAA<  gBBi. U
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 a,(l e P such that g c Xa, h c 3fp,
alp, and P = ap 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
alp. For if a 1 P then one of a and P must be in Q and by Axiom 12, all lines in a
plane Tp for P e Q are nonisotropic.
In Minkowski space, if g and h are two isotropic lines then g _L h <> g  h.
Thus we extend the above definition in the following way.
Definition 3.8.3. If g and h are isotropic lines, then g _L h < g  h .
Definition 3.8.4. If g is a line and a e V then we say that g is orthogonal to or
perpendicular to Xa, g _L Xa, if there exists a P e V such that g c Jfp, P 1 a, and
P = ap 6 g. In this case, P ap is the point of intersection of g and Xa
Definition 3.8.5. For every 1 OA.OB e V, we say that OA is orthogonal to OB,
OA 1 OB, if and only if goA 1 goBthat is, there exist a,p e V such that goA c Xa,
gOB c 3fp, and O = ap. For the zero vector 1 = OO, we define 1 _L 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 T h <> g^ _L h^.
(ii) g 1 Xa <> g*> 1 Xaz.
(iii) OA A. OB <> (OA)S 1 (05)1
Lemma 3.8.7. If g is a nonisotropic line then g is not orthogonal to g.
Proof: If g 1 g then there exist a,p e V such that g c Ta,Tp; alp, and
P = P e g. But for every point Q e g, Q a,p with alp 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.
105
[30] H. Wolff., Minkowskische und absolute Geometrie I, Math. Annalen, Vol. 171,
1967, pp. 144164.
[31] J.W. Young, Projective geometry, MAA, Chicago, Ill., 1930.
Definition 5.3 [29] Let A/, M be von Neumann algebras acting on a Hilbert Space
Li. Let Q be a common cyclic and separating vector in H. If A/Ajif
t > 0, we call (AAc A4,f2) a +halfsided modular inclusion (+hsm). If A% M
Ac A/, for all f < 0, we call (A/"c Ad,Q) a halfsided modular inclusion (hsm).
Definition 5.4 [29] Let A/", A4 be von Neumann algebras acting on a Hilbert space Li
with fi e Li a common cyclic and separating vector for A/, Ad, and Ain M.
1. If ((Ain M) a AA,Q) and ((Ain M) c Ad,Q) are hsm inclusions.
2. And if Jy{s lim/^+oo A^AaVV = s lim,^ao A^ A
Then we say that such a pair ((A/,A4), Q) has () modular intersection, mis.
Theorem 5.5 [29] Let M, M, C, and U be von Neumann algebras acting on a Hilbert
space Li with a common cyclic and separating vector fl e Li. Assume the following.
I. 1. (J\f,M,Q) is mis,
2. (Â£, A4,Q) is + mis,
3. (Â£,AAf2) is mis,
II. 1. ( c Af. Q) is hsm,
2. Ad Jy (.1y*J J]\[J
3. [Ad J(JÂ¡Jjd),JjjJ'atJ= 0,
III. 1. Ad (Ad Jc(JyJM)A) CAd c= K
with to = ^ In 2.
Then the modular groups
A 'if, A A ,and A'fr, for t, r, s, v eR,
generate a unitary representation of the 2+1dimensional Poincar group.
Remarks [29] The conditions in I. give a unitary representation of the
2+1dimensional homogenous Lorentz group. The hsm inclusion of condition II. equips
us with a representation of the translations along some light ray. The product of the
two modular conjugations is then a finite translation of this kind. Moreover, due to
98
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.
li
94
means that (Â£,0) is isomorphic to (Ga,o) (in the sense that G is equivalent to Ga as
sets and
d>(Â£W A), which we denote by as
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 ((/, 0) satisfying as A, one obtains R1'2,
threedimensional Minkowski space. Under the identifications given in Chapter 2, we
find that each g e G corresponds to a spacelike line in R1,2. Thus, as C>(A) is a set
of true statements concerning reflections about spacelike lines and the motions such
reflections generate in threedimensional Minkowski space. Moreover, since such
motions are in fact isometries in R12 then 0(0) is isomorphic to a subgroup of the
threedimensional Poincar group.
Theorem 4.2 Under the same conditions as in Theorem 4.1 it follows that
({xw} weW,T) acting on Wsatisfies as O(M).
Proof: From Theorem 4.1 we have ({gw} WeWty) satisfies as O(M) since is the
subgroup of V generated by reflections about spacelike lines. Also from Theorem 4.1,
({xw}weW*'T) is isomorphic to ({gw}wew,Â¥) so ({?w} WeW,T) satisfies as O(A).
The net continuity condition assumed by BDFS [6] for the next theorem was
later shown to be superfluous [8] for this theorem and the remaining theorems.
Theorem 4.3 [6] Assume the CGMA with the spacetime R1,2 and W the described set
of wedges. If J acts transitively upon the set {1Z(IV)} weW then there exists a strongly
(anti) continuous unitary representation U(V +) of the proper Poincar group which
acts geometrically correctly upon the net {7Z(fV)} we\v and which satisfies U(gw) = Jw,
for every W e W. Moreover, U(V\) equals the subgroup of J consisting of all products
of even numbers ofJw s and J = U(V\) u JwRU(V\), where
WR = {x e R12 : jc, > *0}.
40
3.2.1.6. Two lines, a = [a,ai] and b = [P,Pi] are parallel precisely when there exist
A, A' e a and B,B' e b such that AA' = BB1 1.
Proof: Suppose that a  b. Then, a  P and ai  Pi. Let A,A1 e a and B e b, so
that B1 = AA' B\aa$, and aictipi = P,pi, by 3.1.6.10. Thus, B,B' e b with
AA' = BB'. IAA' = BB1 = 1 then A = A1. U
3.2.1.7. If a  b and b  c then a  c.
Proof: Let a = [a,a/]. From the proof of 3.2.1.5 above we may write b = [p, p'] with
a  P and a'  p' because a  b, and c [y,y;], where y  p and y'  p; because
b  c. By 3.1.6.1, a  y and a'  y' so that a  c.
3.2.1.8. For each line a and point A there is a unique line b such that A e b and
b  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 1, so gAD II by 3.2.1.6. Now suppose that A e c and c  a. By 3.2.1.7
above, c  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, A1 = AWX e c and A, = AYZ e gAD Thus,
1 AA1 = WX = YZ AAi implies that A1 = A\. Hence, A,A1 e gAD,c; A ^ A1,
because WX ^ 1 and gAD = c by 3.2.1.1. I
3.2.1.9. IfA,B e g, A B, and C e h then g  h if and only if ABC e h.
Proof: If ABC = H e h, then 1 AB = HC and g  h. Conversely, suppose that
g  h and put D = ABC. Then because A B,
1 AB DC and C e goc, with goc II g
Because g  h and C e h, then by 3.2.1.8, gDc = h and D e h.
3.2.1.10. If a  b then a b ora r\b = 0.
Proof: Suppose a  b and a n b 0. Let A e a,b and Cea with C A, B e b,
with A B. Then, a = gAc, b = gABi and therefore, D = ACB e a,b.
CHAPTER 3
A CONSTRUCTION OF FOURDIMENSIONAL MINKOWSKI SPACE
In this chapter a construction of fourdimensional Minkowski space will be
given using the same initial data as in Chapter 2 but satisfying different axioms. The
actual construction is quite long, so to aid the reader in following, A brief outline of
the procedure shall be given here. First some general theorems and the basic
definitions will be given in the first two sections. In Sections 3.3 and 3.4 attention is
restricted to two dimensions in order to obtain the necessary machinery to construct
the field. Once the field has been obtained, then a vector space is constructed and
given a definition of orthogonality. It is then shown that the vector definition of
orthogonality is induced by and agrees with the initial definition of perpendicularity
for the geometry generated by the original set of involutions Q.
To obtain a metric vector space from the constructed vector space V, a map n
is defined on the subspaces of V. The map n is defined to send a subspace to its
orthogonal complement. Using the work of [3] (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 Q 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 Q of involution elements and the group it
generates, where for any a e Q and for any Â£ we have oU = Â£'a% e Q. If the
product of two distinct elements e is an involution then we denote this by
26
44
A,B,P  a then D = PAB a. Hence, D e gPc or D e gpQ, and gAB II gPC or
gAB II gPQ thus g% II gpQ = gPQ II gAB or g% II gpc = gPC 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 B, be any element in 0; then the map : X ^ 3Â£ given by az(A) = B/1B,~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 (<3,Q,X). In the following, for any element
B, e 0, the collineation induced by it is denoted by a^.
Let a e V be fixed throughout this section. We wish to define a set Ca of
maps on 3Â£a which can be viewed as line reflections in a given plane 3fa. We then
show that each element of Ca is involutory and that Ca forms an invariant system of
generators within the group Qa it generates. Finally we will show that (Ca,Ga)
satisfy Wolffs axioms [30] for a two dimensional Minkowski space; that is; the
Lorentz plane, for a 6 M..
Let Â£a = {g c Xa : g is nonisotropic}. By Axiom 14, each g e Ca may be
written uniquely in the form g = [a, p,y] where a = Py. Let A e Xa and
g = [obP>Y] e Ca. because A  a and a = py, then A Aa = A^ and A$ = Ar. Hence,
we define the map ag : Xa >* Xa by ag(A) = A8 = A$ = Ay, for A e Xa, and
dg : Ca ^ Ca by Gg{h) = hg = {og(H) : A e h}. Note that by 3.2.1.4, if
h = [a,8,s], then ag(h) = [a,5P,sP] = [a,8y,sY].
3.3.1. For each g = [a,p,y] e Â£a, : Xa La is bijective.
Proof: If A,B \ a and og(A) ag(B) then Ay = By and A = B, so ag is injective. If A \ a
then Ay \ ay a because aly and cg(Ay) = (Ay)y = A, so ag is onto.
3.3.2. For g = [a, P,y] e Â£a,
Proof: Let h = [a,s,r] e Â£a. Because a 1 y and e 1 r\ <> eY 1 r\y by 3.1.2.1 then
23
to n. One can now obtain an affine space A by defining the points of A to be those
of P3(/C) FI; that is, those points whose last entry is nonzero; a line / of A to be a
line /' in P3(/C) Id minus the intersection point of the line /' with FI; and by
defining a point P in A to be incident with a line / of A if, and only if, P is incident
with the corresponding /'. Planes of A are obtained in a similar way [14].
Thus, each point in Id 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 FI represents the set of all planes parallel to a given plane.
Because parallel objects can be considered to intersect at infinity, we call FI the
plane at infinity.
2.5 Exterior Point Reflections Generate Motions in an Affine Space
In this section we state and prove the main result of this chapter. Coxeter
showed that threedimensional Minkowski space is an affine space whose plane at
infinity is a hyperbolic projectivemetric plane [11]. 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
Timelike line
Lightlike line
Spacelike line
Characteristic plane
Minkowski plane
Spacelike plane
Section at Infinity
Interior point
Point on the absolute
Exterior point
Tangent line
Secant line
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
36
3.1.6.11. IfAA' =BB'; A,A'\a; 5P; a  p then B'  p.
Proof: Because AA1 = BB1 then B1 = BAA1; p  a  a. From 3.1.6.10 above,
B1 = BAA' paa = p.
3.1.6.12. For each P and each P there is a unique a such that P  a and a  p.
Proof: By Axiom 1 there exists a unique y such that yJLp. because Py then Py = a,
aJLy and a  p. Now if P8 and 6  p then P\8,a, so that a = 5 by 3.1.6.7.
3.1.6.12.Let a,p e V and M e X. Then aip aMP and thus, a  aw.
Proof: Suppose that a_Lp. By Axiom 1 there exists unique y, 51M such that a_Ly and
8p. By 3.1.6.2 we have p,yXa and p8 which implies that yT8. By 3.1.4.1,
M = y8. thus
Conversely, suppose that a wi.p. As above, there exists unique y, 81M such
that aA/Xy and S_Lp. It follows that pi8, M y8, and a8 = aYÂ§ = awTP and
a = (a8)8P8 = p.
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.
3.1.6.15. IfA,Ba,p andAM = B then Ma,p.
Proof: By Axiom 6 there exists y,S such that A,A/y,8. because B = AM = MAM then
by 3.1.4.7 we have By,8. Thus, A,Ba,y,8 ; My,8; and A,B\p,y,8 so by Axiom2,
M\ p,a.
3.1.6.16. .IfA,B,C18; a,p,yT8; A a; 5P; Cy; a  p  y then ABC\aPy and
aPy T 8.
Proof: From 3.1.6.10 we know that D = ABC\ct$y = s. Because 8a,p,y then e8 so
that s 1 8 or eT8. If E = 8s then we have D,Â£8,s with 8Te which implies that
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 gi, g2, andgj be any three distinct lines, not necessarily coplanar,
which intersect in a point O. LetP\,Q\ e gi; P2,Q2 e g2i and P2,Q2 e g3. If
gP\Py II gQtQi andgp2p3 II gQ2Q3 then gP]p2  gQ{Q2.
Axiom R. LetO e g\,g2\ P\,Q\ e gi; andP2,Q2 e g2. Ifgp{p2  gQxQ2 then
gO,P\OP2 = gO,Q\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 O. A scalar multiplication is then
defined in a manner similar to the definition of multiplication for the field elements.
Axiom D is Desargues axiom, the dilation axiom. Axiom D ensures welldefined
dilations with the standard properies of such maps. Axiom R is used to distribute a
scalar over the sum of two vectors. The dilations are constructed next.
Let /C c= 3fa, a e M. By 3.1.7.2, there exist ai,(X2 e Q such that 0ai,oi2
and a = ai(X2. Because 0a, then Oa = p e Q and O = ap = ai
by 3.1.6.18. Let y = aiP e M. Then ya = Paiai(X2 = Poo e M., so that
t = [a,y,ay] c Xa is timelike and O e t as 0a,ai,a2,P implies that 0a,y,ay. So
let t = [a,y,5] Oa be any timelike line through O and K.' cz Xa the other isotropic
line through O. Then for each At e t, there is a unique A e 1C and there is a unique
B e K! such that At = AOB. Because At e t, then B = A'. (From this point on
denote a line reflection ag(A) by Xs. Thus, X%hr means arai,cg(X) = aga/,ar(A).).
For each At e /, there is a unique A e JC such that At AO A1. Similarly, for
each A in 1C, there is a unique At in t such that At = AOA. Thus, there is a
onetoone correspondence between the points of /C, the field, and the points of t.
Fix Et = EOE1 as the unit point on t. For each O A e 1C, we use t to
construct a dilation 5 a of X in the following way. Let X e X.
33
Proof: By Axiom 1 there exists (3 such that P  p and P X a. Thus, P  a, P with a X P
so by 3.1.4.1 above, P ap and P = Pa. U
3.1.4.3. For each P e X and each Â£ e , P^ e X.
Proof: By the definition of a point we may write P = ap with a e M P e Q, and
a X p. By 3.1.2.2, a^Xp^ and P^aXp^ so that P% = a^P^ is a point.
3.1.4.4. IfP a,p and yXa,p then a = p. ( Given a point P and a plane y there exists
a unique plane a such that P  a and a X y.)
Proof: This follows immediately from Axiom 1.
3.1.4.5. There do not exist three planes pairwise absolutely perpendicular.
Proof: Suppose that a, p, and y are pairwise absolutely perpendicular. Then for
P aP we have P  a, P with a, p X y so that a = p, which contradicts a X p.
3.1.4.6. If A,B,C\a then ABC = D\a.
Proof: By Axiom 8, ABC is a point D and Da = ABCa = aABC aD.
3.1.4.7. IfA,B,C\a,$ thenABC\a,$.
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 (collinear) with the other
three.
3.1.4.8. If a 1,012,013X0. then axajai, = 014 e V and 014X01.
Proof: Let A = aai, B = aa.2, and C = aa.3. Then by Axiom 8, ABC is a point D
and D = aaiaa20UX3 = aaia2(X3. So Da, by 3.1.4.6. By 3.1.4.2,
aia2(X3 = Da = (X4 e V, D = aa4, and aia.2(X3 = c^Xa.
3.1.5. Perpendicular Plane Theorems
3.1.5.1. If a 1 P; Py = A and A \ a then aly. (A 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 = ap^ = fiAa = ya, so a Xy
49
GpGg is involutory. Then P = GgGpGgGp(P) = P= Py/>y. Because yPy Py e 3f
by 3.1.4.3, then by 3.1.6.9, P = Py. Because Py = P8 then Py,8, and hence, Peg.
3.3.16. Ifg,h e Â£a, then g 1 h *> GgG/, is involutory.
Proof: Nowg h if and only if Gg(h) = h. For g h, by 3.3.9, Gg(h) h if and only
if = Gh For Gg Gh by 3.3.14, Gg = O/, if and only if (GgG/,)2 = ljra and
^ lia
3.3.17. The point reflections *pa are the involutory products of two line reflections
from Ca
Proof: Let Gp e ^3a where P e jÂ£a. Then we may write P = y8s where a = y8 and
y,8,s e Q by 3.1.7.3. Let g = [a,y,S], Because a = y8, then g e Â£ By 3.3.8, there
is an / in Â£a such that Pel and Id. g. By 3.3.7, Gp G/Gg GgGÂ¡.
We now show that if a e M., then the pair (Ca,Ga), acting as maps on
(Â£,Â£,), satisfies Wolffs axiom system [30] for his construction of
twodimensional Minkowski space. We give Wolffs axiom system below.
Basic assumption: A given group 0, and its generating set G of involution elements,
form invariant system (Â£/, 0)
The elements of G will be denoted by lowercase Latin letters. Those involutory
elements of 0 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 l with P,g\ l.
Axiom 2. IfP,Q\g,l then P = Q or g l.
Axiom 3. IfP\a,b,c then abe e G.
Axiom 4. If g\a,b,c then abe e G
Axiom 5. There exist Q,g,h such that g\h but Q / g.h.gh.
Axiom 6. There exist A, B; A B, such that A, B l g for any g e G (There exist
unjoinable points.)
16
Thus, the axioms can be viewed as axioms concerning the interior and exterior
points of a hyperbolic projectivemetric plane. The ideal points of the form G(ab)
where a  b are the points on the absolute, that is, the points at infinity in the
hyperbolic projectivemetric plane.
Now consider the ideal lines. A proper ideal line g(o) 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,P\a and abe e Q where b  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 Q 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 \c
and abe e G where a \\ b}.
Corollary 2.2.10. The ideal line which consists of pencils G(x) with x \P 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 Q : xP}.
The last type of ideal line is a tangent line. It contains only one point,
G(ab) = Poo, on the absolute. Denoting this line by /?>, then
poo = {G(a)} u {x e Q : abx e Qj, where a  b. Recalling that each x e Q
corresponds to an exterior point in the hyperbolic projectivemetric plane, we see
that a tangent line consists of one point on the absolute and every other point is an
exterior point.
Also note that under the above identifications, each secant line g(c)
corresponds to a unique exterior point, c, c g{c) since one only considers those
25
space differs significantly from those of previous absolute geometric characterizations
of Minkowski space. The model given here is the set of reflections about spacelike
lines, which is not a choice which would be made a priori by other mathematicians.
However, this is yet another example of a situation where the initial data are
imposed by physical, as opposed to purely mathematical, considerations.
In the next chapter starting with the same initial data, but satisfying different
axioms, a construction of fourdimensional Minkowski space is given. First the affine
space is constructed and then the hyperplane at infinity is used. Also given is an
explicit construction of the field, the vector space, and the metric.
4
one determine the spacetime symmetries, the dimension of the spacetime, and even
the spacetime itself?
Part of this has been carried out by Buchholz, Dreyer, Florig, and Summers
for Minkowski space and de Sitter space [6]. However, in order to do so, they had to
presume the respective spacetime as a topological manifold. But would it not be
possible to completely derive the spacetime from the operationally given data
without any assumption about dimension or topology?
As was pointed out by Dr. Summers, a possibility to do so was opened up in
this program in the following manner. As already seen, the CGMA yields an
involution generated group complete with a projective representation and there is in
the literature a way of deriving spacetimes from such groups going under the name
of absolute geometry.
In general, absolute geometry refers to a geometry that includes both
Euclidean and nonEuclidean geometry as special cases. Thus, one has a system of
axioms not yet implying any decision about parallelism. In our case, the axioms are
given in terms of a group of motions as an extension of Kleins Erlangen Program. A
group of motions is defined as a set Q of involution elements closed under
conjugation and the group 0 it generates. In a group of motions the representations
of geometric objects and relations depend only on the given multiplication for the
group elements, without reference to any additional structure. The system of axioms
is formulated in terms of the involutory generators alone, so that geometric concepts
like point, line, and incidence no longer are primary but are derived.
The necessary means for setting up this representation are provided by the
totality of reflections in points, lines, and planes (a subset of the set of motions).
Points, lines, and planes are in onetoone correspondence with the reflections in
them so that geometric relations among points, lines, and planes correspond to
grouptheoretic equations among the reflections. This enables one to be able to
28
Definition 3.5. The point A and the plane a are called incident when A a. For each
a e V, set = {A : ^a}.se that A,B\a\,ct2 where A B and ai a2.
Suppose that A,B\a\,ap, where A B and ai a2. We define the line g
containing A and B as
g = gAB = [ai,a2] = {C e X : Cai,a2}.
We say that g is the intersection of a.) and a2 (Tai and X
(g a Xai,Xa2) The point C is incident with g, C e g, if Cai,a2.
If A B are two points such that there exist a and P with A,B a, P and alp
then we say that A and B are joinable and we write A,B e gAB = [a, p,aP]. If g is a
line which can be put into the form g = [a, P,ap] where a 1 P then we say that g is
nonisotropic.
If A and B are two distinct points such that there do not exist a, p with
A,B a, P and a 1 P then we say that A and B are unjoinable. If g is a line which
cannot be put into the form g = [a, P,aP] with a _L P then we say that g is isotropic
or null.
Incidence Axioms
Axiom 1. For each P.a there exists a unique p e V such that P\ P and a,p = Q.
Axiom 2. IfA,B a,p,s and C\a, P then C\s.
Axiom 3. IfP,Q a,P and a L p then P = Q.
Axiom 4. If a,p,y e Q are distinct anda ipiyla then aPy g Q.
Axiom 5. Ifa,P e M andaP then ap e M.
Axiom 6. For all A, B: A B, there exists a,P such that A,B\a,fi anda p.
Axiom 7. If a _L P then there exists A,B a,p such that A B.
Axiom 8. For all A,B,C\ ABC = D e X.
Axiom 9. If 0a,p,y,8 with P,y,5 1 a then Py5 = s.
78
then W1 = {UM')l = W1 n Z/1, v e (U,U')L < v e UL, and
v e U'1 < v eZ/1 n ZY1. Consider (g((9),/ifO))1 = g(0)x n /i(O)1. Because g J. /? by
Proposition 3.8.22 then there exists y,8 e V such that O = y5 with g e Xy and
h c .Tg. From Theorem 3.8.20, gfO)1 = (g(0), Â£g(0)> and by Theorem 3.8.19,
O)1 = (j(0),Xy(0)) where / c 3Â£g is the unique line through O in 3Â£g orthogonal to
h. Hence,
((0),h(0))L =g(0)^f(0)
= n
=
Moreover,
1=g(0)1n/(0)1
= n
 (M0),h(0)).
Now if g,/ cz 3Â£e for some 8 e V then by Theorem 3.8.18, for O = sp,
{giO)Jm = XE(0) and (g(0),h(0)) = (g(0), /(O))1 = .te(0)X = .tp(O). This says
that g,h c .Tp; that is,
If (g,h) is a singular plane, O e g n h, and g and h are spacelike, then by
Axiom Si, g is not orthogonal to h. So by Axiom 2, for all A e g, there is a B e h
such that O and A OB are unjoinable.
Theorem 3.8.24. For the above setup:
1 go job e (g,h).
2. go job T g,/z.
3. ifC e g and D e h such that go,cod is isotropic then go.coo = go job
4. <&P),HP))1 =
Proof: For 1., if there exists ye? such that gojOB,g c Xy, then 0,A,A0B\\y implies
that B = OAAO y and h = Xy is nonsingular. For 2.
50
Axiom 7. For each P and A,B,C such that A,B,C\g, there is a v e Q such that P,A  v
or P,B\v orP,Cv.
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 a\b), points. We say the point A and the line b are
incident if A \ b; the lines a and b are perpendicular (orthogonal), if a\ b. Further we
say two points A and B are joinable when there is a line g such that A,B\g.
Replacing Ca with Ca and Xa with the pair (Ca,Ga) 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 Ca with Ca, Ca with Q, and Qa
with
Verification of Wolffs axioms. Axiom 1 follows from 3.3.8 and Axiom 2 from
3.2.1.1. For Axiom 3, let a = [a,a',s], b = [a,p,p'], c = [a,y,y'] e Q. Then
a',p,y T a, and by our Axiom 9, a'Py = 8 i a and d = [a,8,a8] e Â£ For A a,
Hence, aaobGc = cd e Ca. If we then identify  with e, we get Wolffs Axiom 3.
Axiom 4. Ifa,b,c\g, then abe e Q.
Proof: Let g = [a, a = [a, a', 8], b = [a,p,p'], and c = [a,y,y'], with
a',p,y 1 X. Then by 3.1.4.8, a'Py = s 1 X. Let A e a, B e b, and C e c, then
A\a',a Ra,p; and Ca,y. By 3.1.6.10 ABC = Z)a'Py = s and ABC = Da by
3.I.4.6. Thus, Ds,a; s i X; a 1 X, and s 1 a. So, d = [a,8,as] Ca. And if X\a,
then
Axiom 5: There exist Q,g,h such thatg\h but Q / g,h,gh.
69
Px = lA(X) and P2 = 8B(Y). Hence, {OX,OY} spans If A 0X+ B OY = 1,
then we obtain 08A(X) = 8b(Y)0. Because x Jf y, it follows that 8.4(A) = O = Sb(Y).
Let Xr be the unique plane containing t and x and IC\ and K2 the isotropic lines in
3En through O Then we may write Â§A(X) = XfOXf', where X\ e 1Ci, and XÂ¡ e JC2
As above it follows that A O and similarly, B = O. Therefore, Â£(0) is two
dimensional.
Corollary 3.7.4. Ta = B((), Aa((7)) is an affine subspace of dimension two for all
a e V, 0\a.
Theorem 3.7.5. (V./C) is a fourdimensional vector space and hence, (T, V,/C) is a four
dimensional affine space.
Proof: Let OP e V and let O = ap with a e M. and P e Q. Let Pla^p' with cd T a
and P' 1 p. Put P1 = cl'cl and P" = p'p. Then Q P OP" = a,aapppi = a'pd Thus,
P,Q\a',pi with a' X P'. Therefore P = Q = P'OP" with P' a and P"  p. Since Â£a(0)
and tp((9) are twodimensional then there exist bases {OZ,OT) c .ta(C) and
{OX,OY} a p(O) such that
OP' = A OZ + A' OT and OP" = B OX+ B' OY
for some A,A\B,B' e K. Thus, P A OZ + A' OT + B OX + B' OY and
{OX,OY,OZ,OT} span V. If A OZ + A1 OT+ B OX+ B1 OY = 1, then in
particular, OP = OP1 + OP" = 1. This implies that OP'" = P'O. So either
Sop" II Sp'o or p" = P' = O. But gOP" f gOP' and therefore,
A OZ + A1 OT OP1 = 1 = OP" = B OX + B1 OY. As was shown in Proposition
3.7.2, we obtain A=A' = B = B= 0 and the result follows.
3.8 Orthogonality
In this section we extend the definition of orthogonality to include lines and
then use this definition to define orthogonal vectors.
38
exists a 5 e Q such that d8 and 8 _L a,y. Again by 3.1.7.1 there exists an s e Q
such that A s and 8 1 8, a. Hence, de,a,8; s,8,a e Q, and s 1 8 1 a 1 8 which
yields by 3.1.3.2, A = 8as.
We note that in this case, ap = A = a8s implies that p = 8s; that is, if A  P
and P e Ad then there exists Pi,p2 e Q such that dPi,p2 and p = piP2
3.1.7.3. If A I a then there is a P in Q such that A \ P and P J_ a.
Proof: This follows directly from the proof of 3.1.7.2, for if a e Q then we can find
8,s e Q such that A 8,e and 8,s _L a. If a e Ad then we can find
that A ai,(X2 and a 1,012 L a. That is if A a then there exists P1, P2 e Q such that
dPi,p2 and pi JL p2 Moreover, if a e Ad then we can find Pi,p2 such that
a = P1P2 and if a e Q then we can find p 1, P2 e Q such that (XJP1P2.
3.1.7.4. If A a e Ad and P a then there is ay such that A\y andy p,a.
Proof: Let 8d with 8p. Then 8 _L a by 3.1.6.18 and A s = a8. Because a J_ P and
A A A
8P then s p. If e_LP then we would have A \ 8,s with S,8P so 8 = 8 and a = 1.
Thus, sip. We note that if p e Q then 8,s e M and if P e Ad then s e Q.
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 Q, ifA,B e a,b then A = B ora = b. Hence,
by Axiom 6, for all A,B e X, there exists a unique g e Q, such that A, B e g.
Proof: Let a = [a. P], b = [y,8], A,B\a, P,y,8, and suppose that A B. Let Cea, so
that Ca,p. By Axiom 2 it follows that Cy,8 so C e b and a c b. Similarly, b c a
and a = b.
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,p] with d,5a,p. Then by Axiom 8 and 3.1.4.7, A8a,p and A8 e X.
A8 A by 3.1.6.9.
Abstract of Dissertation Presented to the Graduate School
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy
AN ALGEBRAIC CHARACTERIZATION OF MINKOWSKI SPACE
By
Richard K. White
May 2001
Chairman: Dr. Stephen J. Summers
Major Department: Mathematics
We give an algebraic characterization of threedimensional and four
dimensional Minkowski space. We construct both spaces from a set of involution
elements and the group it generates. We then identify the elements of the original
generating set with spacelike lines and their corresponding reflections in the three
dimensional case and with spacelike planes and their corresponding reflections in the
fourdimensional case. Further, we explore the relationship between these
characterizations and the condition of geometric modular action in algebraic
quantum field theory.
v
20
reflection is defined similarly, a harmonic homology with center M and axis m, where
M and m are a polepolar pair, M is an interior point, and m is an exterior line. Note
that in both cases, M and m are nonincident.
In keeping with the notation employed at the end of Â§2.2, let b be an exterior
point and g{b) its pole.
Lemma 2.3.8. The map : <
A
A
Ab and d
Ab d
P$0 Poo
db
g{d)b
~N
> is a
nb
J
collineation.
Proof: This follows from the earlier observation that the motions of the group plane
map pencils onto pencils preserving the  relation.
Lemma 2.3.9. *P, is a perspective collineation and, hence, a homology.
Proof: Recall that g(b) = {A,x,P*> : x,A \b and where b lies in the pencil Poo}. For
any A and x in g(b) we have Ab A and xh = x since A,x \b and if A1,x' g gfb) then
A',x' / b and Ab A, xb x, and Ab,xb / b. Thus, Ab,xb g g(b).
Recall also that Poo = G(cd), where c and d do not have a common
perpendicular nor a common point and thus, G(cd) = [f: fed e Q}. Now g(b) is a
secant line, so that it contains two such distinct points, Pec and Qoo, say, on the
absolute. Since the motions of the group plane map pencils onto pencils preserving
the I relation it follows that if c,d e Poo then cb,db e Pec and hence, P* = Poo and
Q = Qec. Moreover, if Poo Â£ g(b) then it follows that Rb
g{b) pointwise invariant.
Now let g(d), Q, and roo be a secant line, exterior line, and tangent line,
respectively, containing b. For e e g(d) we have e \d and eb \ db = d since b  d,
thus eb Â£ g[d). For A e g{d), Ab \db d, so Ab e g{d). Similarly, it follows that if
Poo e gid) then P* e g(d), and that g(d)b = g(d). One easily sees that Qb = Q and
r* = roo. Thus, leaves every line through b invariant and T* is a perspective
57
GgG/,Gt = &i for some l e Oa and OgGh = CJ/CT/. Clearly, if / e Oa with G/'Gi = G/GÂ¡
then we have cry = gÂ¡ and l1 = l. Hence, for all GgGÂ¡, e T>a(0), there is a unique
/ e Oa, such that GgG/, = G/Gt. So if O A e 1C, then we may uniquely write 5a in
the form GaG,, that is,for every O A e 1C, there is a unique a, a = [a,a',aa] e Oa
such that
A = GaGt(E) = E,a = EXa'.
Suppose that O A = EXa Then A2 = EXa Xa = O implies that
E = Oa xa x = O, a contradiction. Hence, if O A e 1C then A2 O. Now suppose
that A,B e 1C and A,B O. Let a = [a,a\aa'], b = [a, p,aP] e Oa such that
A = EXa and B = Ex$. Then by Axiom F there exists a y e V such that 0y JL a
and
A2 + B2 = EXa'Xa'OExVW = Ex^. (3.5.1)
Since 0y X a then c = [a,y,ay] e Oa,GcG, e Va(0), and
C = gcg,(E) = E,c = Exy e 1C.
So equation (3.5.1) reads A2 + B2 = C2. If C2 = EXyXy = O, then E = OyX^x = O.
Since E O, it follows that if A\,...,A e 1C are all nonzero, then A2 O and
1C is formally real.
To finally obtain a field isomorphic to the real numbers, R, we add the least
upper bound property to our axiom system.
Axiom L. 7/0 A c K. and A is bounded above, then there exists an A e 1C such
that A > X, for all X e A, and if B e 1C 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 Held 1C constructed above, along with Axiom F and Axiom L,
is isomorphic to the real number field, R.
31
Proof: Let p = aja2 with a 1,012 e Q and ai a2 By our assumptions on Q and
because aj (X2 we have for all ^ e 0, a^a e Q and afla^ so that aa e M.
3.1.1.4. The sets Q and M. are disjoint, Q n M = 0.
Proof: Let Q1 = Q\M. Then Q1 consists of involution elements, Q' n M. = 0, and
V Q1 u M. = Qu M. Let y e Q' and ^6 0. Now if e M., then by 3.1.1.3 above
y = (y^)4 1 e M, a contradiction. Thus, Q' is an invariant system of generators and
without loss of generality, we may assume that Q r\ M. 0. M
3.1.1.5. If a e Q, P e M anda 1 p, then ap Â£ M.
Proof: If a.p = y e M. then a = py e Q where p,y e M and Py, which contradicts
Axiom 5.
3.1.2. Properties of V.
For the remainder of this dissertation, let the symbol < denote the phrase
if, and only if.
3.1.2.1. Zfa,p e V and Â£, e 0 then a T P if, and only if, T pi
Proof: First suppose that a,p e Q. Then a 1 P f) 1 p^ because a P implies
that a I P and a  P < a>  pi If a e Q and P e M and al P then ap = y e Q;
a^,y^ Q, p^ e M, and a^P^ = y^ e Q implies that a^ T pi Conversely, if
a; J. p^ then a^p^ = 8 e Q and ap = 8^ 1 e Q, so that a 1 p. Finally, suppose that
a,p e M. Then aip^ e M and ap e M <> a^p^ e M.
3.1.2.2. If a, p e V and ^6 0, then a i P < a^ i pi
Proof: Let a e Q and P e M and Â£ e 0. If a ip then ap so that a^pi Thus,
a^ T Ps or a^ T pi If a^ i p^ then a^P^ = y e Q. So we have ap = y^ 1 e Q by
the invariance of Q, which contradicts a ip. Hence, a^ i pi
Conversely, suppose that a^ i pi because a^ e Q and P^ e M by the
invariance of Q and M then a^ = y, p^ = 5 imply that y 15 and = ai P = 8^
by the paragraph above.
64
To extend the above idea to any r e M. such that t c 3Â£n, let a r e M.
such that t a .in Let K.\ and 1C2 be the isotropic lines in 3^ through O. Because t is
nonisotropic, then by Axiom 14 there exist y, 8 such that r = y8 and t = [r),y,8]. Thus
a, : 3En > Tr) is welldefined. Every B e t c 3Â£r may be uniquely written as
B = B\OB2, where BÂ¡ e JCÂ¡. Because O e t and t is nonisotropic, then B2 = B\. Thus,
B = B\OB\ where B\ e 1C\. So, in particular, there are unique E\,A\ e 1C\ such that
Et = E\OE\ and At = A\OA\, for AÂ¡ e t. As before, there is a unique aÂ¡ e Or such
that Ej = A\. Hence, every X e Tr can be uniquely written as X X\OXt2, where
X\,X2 e IC\ c This defines a map 8,^ : Tn Tr, given by
8^n(A) = A?1 OX'j'1 for A e
Proposition 3.6.6. Let 8^n : be the map defined above. Then
1. S^n is a dilation on 3^.
2. 8^n = 5a on Tn.
3. if P, O, and Q are collinear points in 3fn, not necessarily distinct, then
5 a^POQ) = 5Ax](P)05Al](Q).
4. Moreover, 8A(POQ) = 8A(P)O8A(0, for every P, Q\ q such that (),P. and Q are
collinear.
To obtain a scalar multiplication on V, for all O A e 1C and all OX e V,
define
A OX = 8a(0)8a(X) = 05AX) and O OX = 1.
We now verify the vector space properties.
Lemma 3.6.7. If A,A' e 1C and OP e V, then (A + A1) OP = A OP + A1 OP.
Proof: Suppose P e t and P O, and recall t,K. c Xa. Let P,Â£a,p; Ay with y  p;
A' ly'< Y' II P: and 08 with 8  p. Then y  y'  8; g, = [a,y]  gEP with g, c Xa;
g2 = [ot^y ] II gEP with g2 c Ta; and g2 = [a,8]  gEp with #3
h(P) e g\ r\t and 5y(P) e g2 n t. Thus, 8^(P)a,y and 8y(Â£)ot,y'. It follows that
of von Neumann algebras acting on a Hilbert space 7i, where the index set I is a
partiallyordered set. There is a vector Cl e H which is cyclic and separating for each
IZj, i e I. From the modular theory of TomitaTakesaki we then obtain a collection,
{J/},<=/, of modular involutions which generates a group Jand a collection, {A,}/e/, of
modular operators. The assignment /' * 7ZÂ¡ is an orderpreserving bijection and each
ad Jj leaves the set {72./},e/ invariant. The last two assumptions imply that for each
i e I, there is an orderpreserving bijection x, on 1 such that JÂ¡TZjJÂ¡ = 7ZxÂ¡q), j e I.
The group generated by {x,}/6/ is denoted by Tand forms a subgroup of the
transformations on the index set I. Assume also that the two intersection conditions
for wedges given in Chapter 4 also hold as part of the CGMA for what follows.
To help explain the additional assumptions used in our result we give the
following definitions and theorems.
Definition 5.1 [6] The Modular Stability Condition (MSC). Let {7Z(lÂ¥)} be a net
of von Neumann algebras satisfying the CGMA where the index set I is the set of
wedgelike regions W in R12 described in Chapter 4. Then the modular stability
condition is satisfied if the modular unitaries are contained in the group J generated
by the modular involutions; that is,
A '{y e J for all t e R and W e W.
Theorem 5.2 [6] Assume the CGMA for threedimensional Minkowski space with (4.6)
and (4.7), where the index set I = W, the collection of wedgelike regions in Z?12.
Assume also the transitivity of the adjoint action of J on the net {7Z(W)} weW Let
U(RU2) be the representation of the translation group. If e J, for all t R and
some W e W, that is, if the modular stability condition obtains, then sp{U)
sp(U) d V. Moreover, for every futuredirected light like vector t such that
W+ a W, there holds the relation = U(e~alt), for all t e R, where
a = 271.
97
A Â¡"i e SO^( 1,2) with /, = A///i, and lj = This element in 50^(1,2) is
uniquely defined up to a multiplication by a boost of type A/h/2(/), e 1 with the
given asymptotics l\,h Let U denote the unitary representation of the Poincar group
according to Theorem 5.4, that is, let
U{A/lA(0) sAjg, ZY(A//2(0) =A^, ZV(A/2,/3(0) sAg; t e R.
Now define the observable algebra associated with arbitrary wedges by
TZ(W[lÂ¡,lj]) = adll(AiljjKR'iAr) ( e by the MSC and the CGMA).
For translated wedges, define for a e R12
n(W[lÂ¡,lj,a]) S ad W(a)W(A/.//)(7e,>) ( e {7Z,}ieI by the MSC and CGMA).
In this way, for any wedge region W in R1,2 there is a unique von Neumann algebra
1ZÂ¡W in {7Zj}je/. Taking F : W > / to be the map = /w, for W e W and i\y e /
as obtained above and the result follows.
Conclusion 2 follows from Theorem 5.6 and Theorem 5.5. Conclusions 3 and 4
also follow from Theorem 5.6. The last conclusion follows from Theorem 4.3. I
We conclude this chapter with a few remarks. Given a net {4./}/e/ and a state
to satisfying purely algebraic conditions, one derives threedimensional Minkowski
space, and an identification between elements of I and the wedges in
threedimensional Minkowski space in such a way that the adJÂ¡ act like a reflection
through the spacelike edge of the wedge. Therefore, solely with assumptions on the
algebras of observables {4,}/e/ and the preparation oo, we are able to derive the
physical spacetime and its symmetries. We can also derive an interpretation of
suitable elements of {M/},<=/ as local algebras associated with wedge regions, as well as
derive a prescription of how the spacetime symmetries act upon the observables. In
addition, we can get a time orientation of the spacetime from the MSC.
101
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. (U1 subspace axiom) Let 0,A,B, and C be any four, not necessarily
distinct, points with A, 0\a; 0,B  [3; O, C  y, 8 and a 1 y and P 1 8. Then there exists
X,e e V such that X 1 s; 0,A0B\X; and 0,C e.
Axiom Si. If g c.ia, a e Q, h cz Tp, P e Q. and there exists y,8 e V such that
y 1 5; y5 eg h\ g c and h cz then there exists E Â£ Q such that g,h cz (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 P e V such that g,h c= Tp. Then either there exists a,y e V such that a 1 y,
g c Aa, and h cz Xy or for all A e g, there exists B e h such that P and APB are
unjoinable.
Lemma 3.8.8. If g. h c Xa are nonisotropic for a e V, then g _L h in the sense of
Section 3.3 if and only if g 1 h in the sense of Definition 3.8.1.
Proof: Let g = [a,p,8], h = [a,y,X] c 3Â£a with a = p8 = yX. Recall that g 1 h in the
sense of 3.3 if without loss of generality, ply and A = Py e gn h. So, in
particular, A = py with A e gn h and g cz 3fp and h cz Xy. So g _L h by 3.8.1.
Now suppose that there exist rpe e V such that B = qe e gn h and g cz 3^
and h cz XE, but it is not the case that ply, nor that p 1 X, nor that 8 1 y, nor
that 8 1 X. Then by 3.3.8 there exists a unique / e Ca such that B e l and
/ = [a,u,p] for some u,p e V with a = op, u 1 p, and p 1 8. If / h then la(B) and
ha(B) span ta(R). Thus if C e g there exists Lei 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, gig,
which cannot happen for nonisotropic g. Hence, / = h and the result follows.
35
3.1.6.4. If a  p, y  5, and PT8 then aiy. (Two planes parallel to two absolutely
perpendicular planes are absolutely perpendicular.)
Proof: because a  p then a, PJ_e for some e and because Pthen aT5. Similarly,
y^Ls' for some s' and SJLp imply that yiP. Hence, a, P8 and pj_y yield aXy.
3.1.6.5. If a e M(G) anda  p then P e G(M).
Proof: Suppose that a e Q and a  P, so a, pXy for some y. Because a e Q then
y e A4, which implies that P e Q.
3.1.6.6. If a  P then a*=  P^ for every Â£, e 0.
Proof: Let a, pTy for some y e V. By 3.1.2.2, a^P^Ly^ so that  pX From
3.1.6.5 and 3.1.6.6 above and the invariance of Q and A4, we have that if a  P and
a e G(M) then P e Q(A4) and a^  P^ with a^P^ e G(M).
3.1.6.7. If a  p then Xa n jÂ£p = 0 ora = p.
Proof: Suppose that a, pXy and Pa,p. Then a = P by 3.1.4.4.
3.1.6.8. If A B then A / B; that is, AB ^ BA.
Proof: By Axiom 6 there exists a e V such that A,B a. By 3.1.4.1 and 3.1.4.2 we
may write A = aaj and B = acc2. Suppose that AB = BA. Then we have
aia.2 = a2(X so that aiT(X2 or ai 1 a2. But ai  CX2 because a.,(X2Ta so that
either ai = a2 or iÂ£a, n Xa2 = 0 If oti T a2 then by Axiom 7, Xa n 3fp 0 so
a\ = a2 and A = B. If a\(La2 then C a\a2 e Tai n Xa2 and again we have that
ai = a2. Hence, A B.
3.1.6.9.a. If Ab = A, then B A.
b. If A B. then A, B, and AB are pairwise distinct.
3.1.6.10. If A  a; 5p; Cy; anda  P  y then ABC\afy.
Proof: Let Aa = a', 5p = p', Cy = y'. Then a',p',y'la,p,y and
ABC aa'pp'yy' = a'pY = d'd with a/pY = 5'T8 aPy.
Conclusion: A  a; B  P; a  p, imply that B11 p.
93
Buchholz, Dreyer, Florig, and Summers [6] showed that with the above
assumptions one can construct a subgroup of the Poincar group V, which is
isomorphic to Tand related to the group Tin the following way. For each x e Tthere
exists an element gT e ^3 such that x(W) = gjW = {gx(x) : x e Wj. To each of the
defining involutions xw e T W e W, there is a unique corresponding gw e V [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 M12,
and let be the corresponding subgroup ofV. Moreover, let gw be the corresponding
involutive element ofV corresponding to the involution xw e T Then gw is a
reection about the spacelike orthogonal line which forms the edge of the wedge W. In
particular, one hasgwW = W,' the causal complement ofW, for every W e W. In
addition, exactly equals the proper Poincare group V+.
Recall from Chapter 2 that the initial model of (Q, 0) is as a group plane. This
means that each g e Q is viewed as a line in a plane and each P = gh, g\h, 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
CTa :Â¥uQ >P u Q by
tfa(P) = Pa = aPcT1 for P e V and ca(g) = ga = agcT1 for g e Q.
Since (Q, 0) is an invariant system then each aa 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 of the plane. We say that aa is a motion of the
group plane. Since Q generates 0 then the set of line reflections Qa = {<*g g e Q}
generates the group of motions 0CT = {ct : a e 0}. Let O : 0 > 0CT be the map
defined by (a) = cra, for a e 0. Then O is in fact a group isomorphism [2], This
CHAPTER 4
AN EXAMPLE OF THE THREEDIMENSIONAL MODEL
This chapter begins by considering a net of von Neumann algebras, {11(0)} Oeh
and a state (0, coming from a finite component Wightman quantum field theory in
threedimensional Minkowski space. There are various senses to the phrase coming
from a Wightman quantum field theory. The assumption here is the version given by
Bisognano and Wichmann [5]. That is, given a finite component Wightman quantum
field, <})(jc), assume that the quantum field operator,
its closure is affiliated with the algebra TZ(0) (in the sense of von Neumann algebras)
for every test function / whose support lies in the spacetime region O. Driessler,
Summers, and Wichmann show these conditions can be weakened [15]. But free boson
field theories satisfy these conditions in threedimensional Minkowski space [5].
For such theories the modular involutions, Jo, associated by TomitaTakesaki
theory to the vacuum state and local algebras of wedgelike regions, O, 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 [27 in Minkowski space. The precise wording of this version of the
CGMA is given below.
Let /1 and l2 be two lightlike linearly independent vectors belonging to the
forward light cone in threedimensional Minkowski space. The wedges are defined as
the subsets W[l\,l2\ = {a/i + p/2 + lL e K12 : a > 0, p < 0, (/*,//) = 0, i = 1,2},
where ( ) denotes the Minkowski inner product.
91
13
Lemma 2.2.2.[2) For each a e 0, the mappings aa : g >* ga = ago. and
<5 a : P '> Pa = a Pa are onetoone mappings of the set of lines and the set of points,
each onto itself in the group plane.
Proof: Let a e 0, and consider the mapping y t* ya = aya of 0 onto itself. It is
easily seen that this mapping is bijective. Because Q is an invariant system (ah e Q
for every a,b e Q) Q is mapped onto itself, and if P is a point, so that P = gh with
g\h, then Pa = gaha and ga\ha, so that Pa is also a point. Thus, g ga, P Pa
are onetoone mappings of the set of lines and the set of points, each onto itself in
the group plane.
Definition 2.2.3. A onetoone mapping a of the set of points and the set of lines
each onto itself is called an orthogonal collineation if it preserves incidence and
orthogonality.
Since the  relation is preserved under the above mappings, the above
mappings preserve incidence and orthogonality as defined above.
Corollary 2.2.4. [2] The mappings
Ga g >* ga and eta P '* Pa
are orthogonal collineations of the group plane and are called motions of the group
plane induced by a.
In particular, if a is a line a we have a reflection about the line a in the group
plane, and if a is a point A, we have a point reflection about A in the group plane.
If to every a e 0 one assigns the motion of the group plane induced by a,
one obtains a homomorphism of 0 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 0
generated by Q.
82
isotropic lines and the dimfV) > 3. Then n may be represented by bilinear forms if,
and only if,
(a) planes containing more than two isotropic lines are Nplanes and
(b) 1C is commutative.
Theorem 3.9.13. Suppose that n is a polarity of the vector space (V,/C) such that the
conditions of Theorem 9.12 are met. Then from Theorem 3.9.11 it follows that if n is
not a null system then n may be represented by symmetrical bilinear forms. M
Defining the polarity n 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, U1 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 71 is a uniquely determined subspace of V. From Lemma 3.8.11 and
Definition 3.8.5, {l}1 = V and V1 = {1}. So we define the mapping n on the
subspaces of V as follows:
Definition 3.9.14. If U is a subspace of V then n(JJ) = U1.
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 W of V such that 7t(14) = U.
Theorem 3.9.15. Let U and W be subspaces ofV, then U < W if, and only if,
71(14) < K (U).
Proof: Suppose that U < W. If OV e W1 then OV _L OfV for every OfV e W, U c W,
so OV OU for every OU e U and OV e U1. Thus, U < W implies that
n(W) VV1 < U1 n(lf). Conversely, suppose that W1 < U1. By Lemma 3.8.11 and
Definition 3.8.5 we have ({l}1)1 = {1} and (V1)1 = V. From the previous section
it follws that, if U is a nontrivial proper subspace of V then (71)1 = U. because W1
and UL are subspaces, then from the first part of the proof we have W1 < UL implies
that U = (U1)1 < (W1)1 = W.
Corollary 3.9.16. From Definition 3.9.14 and Theorem 3.9.15 above it follows that
n : U < V Id1 < V is an autoduality. Moreover, because (Y1)1 = U for every
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.
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, p,y e M. such that a = py and a = [a, p,y], then we say that a is timelike. If there
are elements a, P e Q such a p and a = [a, p,a6], then we say that a is spacelike.
Remark. Let A,Ba e Q with A B. Then by Axiom 14, there is a P such
that A,B  P and P 1 a. Thus, a = [a, p] 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. If A B\ A,B a,p with P 1 a, a e M.\ and A and B are unjoinable with P;
P\a,y; andy T a,p, then Ay = B.
Proof: First, ^4Y  aY, pY = a, p. If Ay = A then A y and A is unjoinable with P.
Similarly, By\a, P and By B. Suppose that Ay and P are joinable. Then there are 5
and e such that P,4y5,e and 5 s. But then P,A = />Y,(AY)Y 8Y,sY and 5Y 1 sY,
which implies that P and A are joinable. Thus, Ay and P are unjoinable, so by Axiom
13, Ay is unjoinable with A or Ay is unjoinable with B or Ay = A or Ay = B. Hence,
Ay = B.
3.2.2.2. Suppose that A B: A,B a,p with a i p, a e M; A and B are unjoinable
with P\a, P\y, y 1 p. then Ay = B.
Proof: First observe that because Pa,y; alp; and yip, then a 1 y by 3.1.6.18.
Then TYaY, pY = a,p and Ay is joinable with B. Ay A because A and P are
unjoinable as in 3.2.2.1 above. Ay joinable with P implies that A = (Ay)y is joinable
with Py = P. Hence, by Axiom 13, Ay = B.
3.2.2.3. If A and B are unjoinable then X e gAB precisely when X = A or X = B or X
is unjoinable with both A and B.
A similar process can be done for fourdimensional Minkowski space using the
work of Wiesbrock and Khler [18]. However, we refrain from giving the details here.
102
12
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 Q is as secant or ordinary lines in
a hyperbolic plane for BPWB [2], In our approach, we view the elements of Q
initially as exterior points in a hyperbolic plane. After embedding our hyperbolic
projectivemetric plane into an affine space, we can identify the elements of Q. our
generating set, with spacelike lines and their corresponding reflections in a
threedimensional Minkowski space. By realizing that statements about the geometry
of the plane at infinity correspond to statements about the geometry of the whole
space where all lines and all planes are considered through a point, we see that the
axioms also are statements about spacelike lines, the elements of Q, and timelike
lines, the elements P of , through any point in threedimensional Minkowski space.
The models of the system of axioms are called groups of motions; that is, a
group of motions is a pair (, Q) consisting of a group and a system Q of
generators of the group satisfying the basic assumption and the axioms.
To give a precise form to the geometric language used here to describe
grouptheoretic concepts occurring in the system of axioms, we associate with the
group of motions (,Â£) the group plane (,Â£), 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 Q are
called points of the group plane. Two lines g and h of the group plane are said to be
perpendicular if g \h. 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 g. Two lines are said to be parallel if they satisfy Axiom 6.
Thus, if P Q, then by Axioml and Axiom 2, the points P and Q in the group
plane are joined by a unique line. If P / g then Axiom 7 says that there are at most
two lines through P parallel to g.
15
ideal point is any pencil of lines G(ab) of the metric plane. The pencils G(JP)
correspond in a onetoone way to the points of the metric plane. An ideal line is a
certain set of ideal points. There are three types:
1. A proper ideal line g(a), is the set of ideal points that have in common a line a
of the metric plane.
2. The set of pencils G(x) with x \P 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 px.
The polarity is defined by the mappings
G(C) C and C G(C);
Poo >< poo and px > P<;
G(c) i* gjc) and g(c) G(c).
Bachmann [1] showed that the resulting ideal plane is a hyperbolic projective plane
in which the theorem of Pappus and the Fano axiom hold; that is, it is a hyperbolic
projectivemetric plane.
In this model, the ideal points of the form G(P) are the interior points of the
hyperbolic projectivemetric plane. Thus the points of the metric plane correspond in
a onetoone way with the interior points of the hyperbolic projectivemetric plane.
The ideal points G(x), for x e Q are the exterior points of the hyperbolic
projectivemetric plane.
Theorem 2.2.8. Each x e Q corresponds in a onetoone way with the exterior points
of the hyperbolic projectivemetric plane.
Proof: Because each line d of the metric plane is incident with at least three points
and a point is of the form ab with a \ b, then each x e Q is the axis of a pencil. From
the uniqueness of perpendiculars each x e Q corresponds in a onetoone way with
the pencils G(x). Hence, each x e Q corresponds in a onetoone way with the
exterior points of the hyperbolic projectivemetric plane.
80
Therefore, h^O)1 n p(O)1 = g(0) for some line g containing O. Because g must
be either isotropic or nonisotropic, then the following is true.
Theorem 3.8.25. If A(0) is a hyperplane the there is a unique line g such that
A(0)L = g(0) andiAiO)1)1 = A{0). U
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 n of the vector space V over the field 1C is a
correspondence with the following properties:
1. Every subspace U of V is mapped onto a uniquely determined subspace n(M) of V.
2. To every subspace U of V there exists one and only one subspace W of V such that
n(W) = U.
3. For subspaces U and W of V, U < W if, and only if, n(W) < n(U).
In other words, an autoduality is a onetoone monotone decreasing mapping of
the totality of the subspaces of V onto the totality of the subspaces of V.
Definition 3.9.2. An autoduality n of the vector space (V,/C) of dimension not less
than two is called a polarity, if n2 = 1, the identity.
Definition 3.9.3. A semi bilinear form over (V,/C) is a pair consisting of an
antiautomorphism a of the field K, and a function /(x, y) with the following
properties:
(i) ^(x, y) is, for every x, y e V, a uniquely determined number in /C.
(ii) J(a + b, c) =y(a, c) +./fb, c) and J[a, b + c) = ./(a, b) + /(a, c), for a, b, c e V.
(iii) J(tx, y) = tf(x, y) and /(x, ty) =/x, y)a(/) for x, y e V and t e 1C.
If a = 1 then /is called a bilinear form.
Definition 3.9.4. If/ is a semibilinear form over (V,/C) and if U is a subset of V, then
von Neumann algebras acting on a Hilbert space H together with a common cyclic
and separating vector Q e Ti. which fulfill conditions I.III. of Theorem 5.4. Then
these data determine a local (BisognanoWichmann) net A(O) c B(H), O c M12,
such that the incident algebras become the wedge algebras of the constructed net as
in the first part.
Theorem 5.7 Let {7ZÂ¡} Â¡eÂ¡ be a net of von Neumann algebras acting on a Hilbert space
7i, together with a common cyclic and separating vector Q e H satisfying the CGMA
with the MSC. Assume also that there exist Â¡NjM,k\rJc e 1 such that
lZiX.lZjM,lZkandlZic satisfy the hypotheses of Theorem 5.5. Then the following are
true.
1. There is an injective map F : W /, such that for each W e W, the modular
objects Jf(W), Nj(W) have the BisognanoWichmann Property when acting upon
1Zf(W)
2. The modular unitarios A''v, AjrM, A_, and Nfc generate a continuous unitary
representation of 7$+ which acts covariantly upon {1Zf(W)} w^w
If the additional intersection assumptions, (4.6) and (4.7), are made on the subnet
{'RF(W)} WeWi then:
3. The CGMA as stated for Minkowski space holds for the subnet {Hf(W)} wew, as
does the MSC.
4. The modular conjugations {adJF(W)} w<=w, and thus ({ipw,} T), satisfy the
axioms of Chapter 2.
5. There is a continuous {anti) unitary representation of P+ acting covariantly upon
{'R.F(W)}wnW.
Proof: For the proof of Teorem 5.7.1, we give the construction Wiesbrock gave in the
proof of Theorem 5.6 [29]. Let /j = (1,1,0), l2 = (1, 1,0), /3 = (1,0,1) e R1*2. The
local algebra of observables to wedges is defined by
KWHuhV = Kix, mmiuh]) *KjM, n(W[h,h\) = H,c.
For arbitrary linearly independent light rays lhlj e M12 pointing to the future, let
100
90
Theorem 3.10.10. The map G\ is a refection about a spacelike (Euclidean) plane for
every X e Q.
Proof: From the proof of Proposition 3.10.7 it suffices to consider X e Q with 0\X.
Let O 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 ALe,, JCq2 c Tq. Let O ^ A e Afq, and O ^ B e Kq2. Then OA and OB are
isotropic vectors which form a basis for ie(O) and Q(OA) = 0 = Q(OB). Therefore the
metric of Â£q(0) with respect to OA and OB has matrix
!
g
OA
OB
(OA OB) =
Or''
r 0 j
where the products of the matrix elements are the inner products defined by the
metric g and g(OA,OB) = r 0. Since jr OA e /C,, and OB e /Cq2 then /C,, JCq2
also form an isotropic basis for .te(O). Hence we may assume that r = 1. Let
or,
i
n
(OA OB)
and OX,
(OA + OB).
Then g'(OX,,OT,) 0, Q'(OX,) = 1, and Q'(OT,) = 1. Because OX, 1 OT, and
Â£e(0) is nonsingular then OX, and OT, are linearly independent and hence, form a
basis for ie(O). Moreover, the metric of .te(0) with respect to OX,and OT, has the
matrix
. Therefore, by Theorem 3.10.8, te(O) is a Lorentz plane and by
Lemma 3.10.9, i/.(0) = ie(O)1 is a Euclidean plane. I

