LONDON MATHEMATICAL SOCIETY STUDENT TEXTS Managing editor: Dr C.M. Series, Mathematics Institute University of Warwick, Coventry CV4 7AL, United Kingdom
1
Introduction to combinators and A.-calculus, J.R. HINDLEY & J.P. SELDIN
2
Building models by games, WILFRID HODGES
Local fields, J.W.S. CASSELS 4 An introduction to twistor theory, S.A. HUGGETT & K.P. TOD 5 Introduction to general relativity, L.P. HUGHSTON & K.P. TOD 6 Lectures on stochastic analysis: diffusion theory, DANIEL W. STROOCK 7 The theory of evolution and dynamical systems, J. HOFBAUER & K. SIGMUND 8 Summing and nuclear norms in Banach space theory, G.J.O. JAMESON 9 Automorphisms of surfaces after Nielsen and Thurston, A. CASSON & S. BLEILER 10 Nonstandard analysis and its applications, N. CUTLAND (ed) 3
11 Spacetime and singularities, G. NABER 12 Undergraduate algebraic geometry, MILES REID 13 An introduction to Hankel operators, J.R. PARTINGTON 14 Combinatorial group theory: a topological approach, DANIEL E. COHEN 15 Presentations of groups, D.L. JOHNSON 16 An introduction to noncommutative Noetherian rings, K.R. GOODEARL &
R.B. WARFIELD, JR. 17 Aspects of quantum field theory in curved spacetime, S.A. FULLING 18 Braids and coverings: selected topics, VAGN LUNDSGAARD HANSEN 19 Steps in commutative algebra, R.Y. SHARP 20 Communication theory, C.M. GOLDIE & R.G.E. PINCH 21 Representations of finite groups of Lie type, FRAN4;OIS DIGNE & JEAN MICHEL 22 Designs, graphs, codes, and their links, P.J. CAMERON & J.H. VAN LINT 23 Complex algebraic curves, FRANCES KIRWAN
24 Lectures on elliptic curves, J.W.S. CASSELS 25 Hyperbolic geometry, BIRGER IVERSEN 26 An Introduction to the theory of L-functions and Eisenstein series, H. HIDA 27 Hilbert Space: Compact Operators and the Trace Theorem, J.R. RETHERFORD
London Mathematical Society Student Texts 25
Hyperbolic Geometry Birger Iversen Aarhus University
AMBRIDGE
UNIVERSITY PRESS
CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, Sao Paulo, Delhi
Cambridge University Press The Edinburgh Building, Cambridge C132 8RU, UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521435086
© Cambridge University Press 1992
This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 1992 Re-issued in this digitally printed version 2008
A catalogue record for this publication is available from the British Library ISBN 978-0-521-43508-6 hardback ISBN 978-0-521-43528-4 paperback
CONTENTS
INTRODUCTION
ix
I QUADRATIC FORMS I.1
ORTHOGONALITY
1
1.2
WITT'S THEOREM
6
1.3
SYLVESTER TYPES
9
1.4
EUCLIDEAN VECTOR SPACES
12
1.5
PARABOLIC FORMS
16
1.6
LORENTZ GROUP
23
1.7
MOBIUS TRANSFORMATIONS
32
1.8
INVERSIVE PRODUCTS OF SPHERES
39
1.9
RIEMANN SPHERE
44
1. EXERCISES
50
II GEOMETRIES II.1
GEODESICS
57
11.2
EUCLIDEAN SPACE
59
11.3
SPHERES
65
II.4
HYPERBOLIC SPACE
70
11.5
KLEIN DISC
75
II.6
POINCARE DISC
76
11.7
POINCARE HALF-SPACE
78
11.8
POINCARE HALF-PLANE
80
H. EXERCISES
84
III HYPERBOLIC PLANE III.1
VECTOR CALCULUS
88
111.2
PENCILS OF GEODESICS
92
111.3
CLASSIFICATION OF ISOMETRIES
99
111.4
THE SPECIAL LINEAR GROUP
104
111.5
TRIGONOMETRY
107
111.6
ANGLE OF PARALLELISM
111
CONTENTS
vi
111.7
RIGHT ANGLED PENTAGONS
112
111.8
RIGHT ANGLED HEXAGONS
114
111.9
FROM POINCARE TO KLEIN
116
III.10
LIGHT CONE
118
III EXERCISES
121
IV FUCHSIAN GROUPS DISCRETE SUBGROUPS
125
APPENDIX: DISCRETE SUBSETS
129
IV.2
ELEMENTARY SUBGROUPS
131
IV.3
GEOMETRY OF COMMUTATORS
137
IV.4
JACOB NIELSEN'S THEOREM
141
IV.5
CUSPS
144
IV.1
IV EXERCISES
148
V FUNDAMENTAL DOMAINS V.1
THE MODULAR GROUP
150
V.2
LOCALLY FINITE DOMAINS
155
V.3
CONVEX DOMAINS
159
V.4
DIRICHLET DOMAINS
167
V EXERCISES
170
VI COVERINGS VI.1
HYPERBOLIC SURFACES
171
VI.2
HOPF-RINOW THEOREM
175
VI.3
UNIFORMIZATION
179
VI.4
MONODROMY
182
VI.5
ORBIT SPACES
186
VI.6
CLASSIFICATION
189
VI.7
INSTANT JET SERVICE
191
VI.8
HOMOTOPY
196
VI EXERCISES
198
CONTENTS
vu
VII POINCARE'S THEOREM VII.1
COMPACT POLYGONS
200
VII.2
TRIANGLE GROUPS
209
VII.3
PARABOLIC CYCLE CONDITIONS
213
VII.4
CONJUGACY CLASSES
218
VII.5
SUBGROUPS OF THE MODULAR GROUP
221
VII.6
COMBINATORIAL TOPOLOGY
226
VII.7
FRICKE-KLEIN SIGNATURES
230
VII EXERCISES
235
VIII HYPERBOLIC 3-SPACE VIII.1
EXTERIOR ALGEBRA
240
VIII.2 GRASSMANN RELATIONS
245
VIII.3 NORMAL VECTORS
248
VIII.4
PENCILS OF PLANES
251
VIII.5 BUNDLES OF GEODESICS
255
VIII.6 HERMITIAN MATRICES
260
VIII.7
DIRAC'S ALGEBRA
VIII.8 CLIFFORD GROUP VIII.9
LIGHT CONE
262 265
268
VIII.10 QUATERNIONS
271
VIII.11 FIXED POINT THEOREMS
272
VIII EXERCISES
278
APPENDIX: AXIOMS FOR PLANE GEOMETRY 284 SOURCES
291
BIBLIOGRAPHY
292
SYMBOLS
295
INDEX
296
INTRODUCTION
What is hyperbolic geometry? Let me try to give an answer by telling the
story of the parallel axiom. I shall use modern language which will ruin part of the story but highlight the basic points.
Axioms for plane geometry
A simple set of axioms for plane geometry can be presented in the framework of metric spaces. By a line in a metric space X we understand the image of a distance preserving map The three axioms of plane geometry are (the axioms are analysed in an appendix) INCIDENCE AXIOM
Through two distinct points of X there passes a unique
line. The space X has at least one point. REFLECTION AXIOM
The complement of a given line in X has two connected
components. There exists an isometry o of X which fixes the points of the line, but interchanges the two connected components of its complement. PARALLEL AXIOM
Through a given point outside a given line there passes a
unique line which does not intersect the given line.
Investigations of the parallel axiom by among others J.Bolyai (1802 1860), C.F.Gauss (1777 - 1855), N.I.Lobachevsky (1793 - 1856) show that this axiom is independent of the other axioms in the sense that there exists a plane, the so called hyperbolic Wane H2, which satisfies the first two axioms of plane geome-
try but not the parallel axiom. H2 is unique in the following sense.
CLASSIFICATION THEOREM A metric space which satisfies the three axioms of plane geometry is isometric to the Euclidean plane. A space which satisfies the first two axioms but not the parallel axiom is isometric to H2 (after rescalingl).
x
INTRODUCTION
The discovery of the hyperbolic plane began with a series of attempts to prove the parallel axiom from the other axioms. Such attempts should inevitably
lead to the discovery of the hyperbolic plane since a system of axioms for the hyperbolic plane consists in the incidence axiom, the reflection axiom and the negation of the parallel axiom. G.Sacheri (1667- 1733) was the first to penetrate deeper into this universe. He was followed by J.H.Lambert (1728 - 77) who made some important speculations on the nature of a non-Euclidean geometry. By 1850
most of the properties of a new geometry were known to Bolyai, Gauss and Lobachevsky. At that time the biggest problem was the existence of such a geometry! Twenty years later this was firmly settled through the works of E.Beltrami (1835 - 1900), A.Cayley (1821 - 95) and F.Klein (1849 - 1925).
For
more information on the history of hyperbolic geometry see [Milnor] or [Greenberg] and the references given there.
The Poincare half-plane In 1882 H.Poincare (1854
- 1912) discovered
a new model of H2, which I shall now describe. The model is the upper half-plane of the complex plane equipped with the metric given by
cosh d(z,w) = 1 + Iz - w12lm[z]-1 Im[w]-1 2
The lines in the Poincare half-plane are traced by Euclidean circles with centres on
the x-axis or Euclidean lines perpendicular to the x-axis. It is quite obvious that this model satisfies the incidence and reflection axioms but not the parallel axiom: Euclidean inversions furnish the isometries required by the reflection axiom.
The usefulness of the Poincare half-plane model lies in the fact that the group of orientation preserving isometries consists of analytic transformations of the form
z H az+b
cz + d
E S12(R)
In fact the full group of isometries of the Poincare half-plane is the group PG12(R).
In particular the group G12(Z) acts as isometries on H2, a fact which lies at the root of the applications of hyperbolic geometry to number theory.
INTRODUCTION
xi
The Poincare disc The open unit disc D is the second most commonly used model for the hyperbolic plane. We shall not write down the metric here, but
mention that the lines in this model are traced by Euclidean circles and lines orthogonal to the unit circle OD. The Poincare disc has certain aesthetic qualities, as shown below by a one of the many beautiful illustrations from [Klein, Fricke].
Discrete groups
Reflections in the of lines of the tesselation above
generate a discrete group of isometries of H2. The systematic study of such groups
was initiated by Poincare in his "Memoir stir les groupes fuchsienne" 1882. Starting from a discrete group t of isometries of H2 he used a procedure due to
Dirichlet to construct a polygon 0 with a side pairing, i.e. instructions for identification of the sides of the polygon, which allows us to reconstruct the space
H2/t. Then he took up the converse problem: given a hyperbolic polygon with a side pairing, when does the side pairing generate a discrete group with the given polygon as fundamental domain.
Poincare laid down necessary and sufficient
conditions for a given polygon with side pairings to determine a discrete group. The theorem has a bonus: from the geometry of the polygon we can read off a
INTRODUCTION
xii
complete system of generators and relations for the group. In this way the topic
has had a big influence on what today is known as combinatorial group theory. The theorem of Poincare is one of the main themes of this book. The proof of the theorem is based on some geometric ideas which I shall outline next.
New geometries
Let us for a moment go back to the discussion of the
parallel axiom and recall that the hyperbolic plane is quite unique. However, we find numerous new geometries in the class of hyperbolic surfaces: spaces locally isometric to H2. We shall see that complete hyperbolic surfaces can be classified by discrete subgroups of PG12(R).
The main point in the proof of Poincare's
theorem is the monodromy theorem: A local isometry f:X--+H2 from a complete hyperbolic surface X to the hyperbolic plane H2 is an isomorphism.
Higher dimensions
So far we have only talked about hyperbolic
geometry in two dimensions, but the hyperbolic space H" exists in all dimensions.
It is the second main theme of this book to construct these spaces and to identify
their isometry groups with the Lorentz group known from special relativity. Hyperbolic n-space is constructed as one of the two sheets of the hyperbola x02
- x12 - x32 - X42..... -xn2 = 1
It turns out that the most natural framework for this construction is the theory of
quadratic forms. This is actually where the books starts.
We offer a general
introduction to quadratic forms in Chapter I as an opener to the construction of geometries in Chapter H. It is natural to develop Euclidean and spherical geometries in a parallel fashion in order to be able to draw on analogies with these more familiar geometries. A special feature of hyperbolic geometry is the presence
of a natural boundary OHn. In terms of the hyperbola above, a boundary point can be thought of as an asymptote to the hyperbola, i.e. line in the cone X02
- x12 - - X42..... - Xn2 = 0 X32
It follows that the boundary OHn is a sphere. In the end we find that the Lorentz group acts on the sphere as the group of Mobius transformations.
INTRODUCTION
xiii
Three-dimensional geometry
A
feature
specific
of
three-
dimensional hyperbolic geometry is that the group S12(C) operates on H3 (see below) in such a way as to give an isomorphism between PG12(C) and the group of
orientation preserving isometries of 113, the special Lorentz group. This has the consequence that a Kleinian rou 3-manifold
H3/F.
,
i.e. is a discrete subgroup F of S12(C), defines a
The work of W.P.Thurston on 3-manifolds [Thurston] has
generated new interest in hyperbolic geometry.
Exterior algebra
Let us think of H3 as a sheet of the unit hyperbola in a
Minkowski space M, i.e. a four-dimensional vector space with a form of type (-3,1). Once we pick an orientation of M, the space A 2M comes equipped with the structure of a three-dimensional complex vector space with a complex quadratic form Q. The assignment of normal vector n E A 2M to an oriented geodesic line h in H3 creates a bijection between the complex quadric Q = - 1 and the set of oriented geodesics in H3.
The Hermitian model
As mentioned above, each Minkowski space M
comes equipped with its own copy of 113, a sheet of the unit hyperbola. It turns out that there exist Minkowski spaces with more algebraic structures, in particular a structure of a linear representation of S12(C). We shall be interested in the space M of 2 x 2 Hermitian matrices, i.e. complex matrices of the form
;a,dER,bEC The restriction of the determinant to M is a quadratic form of type (-3,1). The
sheet of the unit hyperbola (ad - bb = 1) consisting of positive definite forms (a > 0) is called the Hermitian model of H3. The action of S12(C) on M is given
by the formula
oX=oXo*
;oES12(C),XEM
The key fact about the Herinitian model is the following trace formula: The action of S E G12(C) on H3 can be decomposed into a,6 where a and 0 are halfturns with respect to geodesics a and b. For any such decomposition we have that
INTRODUCTION
xiv
tr2(S) = 4 <m,n>2 where m and n denote normal vectors for the geodesics a and b.
The realisation of Minkowski space through the space M of Hermitian matrices has been used in physics for a long time. We shall introduce another tool
from physics namely the Dirac algebra, which is a concrete realisation of the Clifford algebra of M. The use of Clifford algebras was suggested by the "strange formulas" from the book of [Fenchel].
Prerequisites
The general prerequisites are linear algebra and a modest
amount of point set topology, including compact subsets of a finite dimensional real vector space.
Familiarity with the concepts of group theory and, for the
understanding of Poincare's polygon theorem, a bare minimum of combinatorial
group theory ("generators and relations") is needed.
Elementary topology is
needed for VI.8, VII.5, VII.7, while VII.6 is a good introduction to combinatorial
topology. The final chapter of the book requires familiarity with the exterior algebra.
University of Aarhus, Denmark. July 1992
ACKNOWLEDGEMENT
Birger Iversen
I would like to thank Jakob Bjerregaard, Elisabeth
Husum, Claus Jungersen and Pernille Pind for comments on the various notes on
which the book is built. A particular thank you goes to Gunvor Juul and editor Caroline Series who helped the manuscript through its final stages.
I QUADRATIC FORMS
In the first two sections we lay down the general principles for quadratic
forms and the associated orthogonal groups. The third section will classify real quadratic forms into Sylvester types, while the remaining sections will deal with real forms of specific types.
Three types of real quadratic forms are of particular interest: positive definite forms, parabolic forms and hyperbolic forms. The corresponding
orthogonal groups provide us with the isometry groups of the three basis geometries: spherical geometry, Euclidean geometry and hyperbolic geometry.
For example, the orthogonal group of a hyperbolic form contains the Lorentz group as a subgroup of index 2. The Lorentz group, on the one hand, is the isometry group of hyperbolic geometry, while, on the other hand, it can be identified with the somewhat older group of Mobius transformations. We shall see that "the Mobius group is the boundary action of the Lorentz group".
1.1 ORTHOGONALITY
Let k denote a field' of characteristic # 2 and E a vector space over k of finite dimension n.
By a quadratic form on E we understand a function Q: E-+k
homogeneous of degree 2, i.e.
1.1
Q(\z)=A2Q(z)
;AEk,zEE
with the property that the symbol <x,Y> =
1.2 1
2(Q(x + Y) - Q(x) - Q(y))
;x,yEE
In this text we have applications only for the field R of real numbers,
the field C of complex numbers and occasionally the field Q of rational numbers.
QUADRATIC FORMS
2
is bilinear in x and y. The relation 1.2 between the bilinear form <x,y> and the
quadratic form Q(z) is called the formula for polarization.
Observe that the
bilinear form <x,y> is symmetrical in x and y.
Proposition 1.3
Polarization gives a one to one correspondence between
quadratic forms and symmetrical bilinear forms over a field k with char(k) # 2.
Proof A quadratic form Q(z) satisfies the formula Q(2z) = 4Q(z) as a special case of 1.1. We can now substitute z for x and y in formula 1.2 to get that
Q(z) _
1.4
;zEE
which recovers the function Q(z) from the symbol <x,y>. Conversely, if we start
with a symmetrical bilinear form <x,y> we can use formula 1.4 to define a k-
valued function Q on E which is homogeneous of degree 2.
Bilinearity and
symmetry give us the formula
<x + y, x + y> = <x,x> + + 2<x,y> which shows that Q(z) = is a quadratic form.
x,y E E
The evaluation, Q(x), of the quadratic form at a vector x E E is often referred to as the norm of x. Vectors x,y E E are called orthogonal if <x,y> = 0.
Linear subspaces U and V of E are called orthogonal if all vectors in U are orthogonal to all vectors in V. For a given linear subspace K of E we define the total orthogonal subspace K
1
by the formula
1.5
K1 = {eEEjVxEK<e,x>=0}
Definition 1.6
A quadratic form Q(z) on the finite dimensional vector
space E over k is called non-singular if the bilinear form <x,y> satisfies
1.1 ORTHOGONALITY
3
Vx E E (<x,y> = 0)
y=0
;yEE
Otherwise expressed, the form Q(z) is non-singular if and only if E
Theorem 1.7
1
= 0.
Let E be a finite dimensional vector space equipped with a
non-singular quadratic form Q.
For any linear subspace K of E we have
dim K + dim K 1- = dim E
Proof
Let us recall that a linear form on E is nothing but a k-linear map
:E--+k. Linear forms may be added and multiplied by a constant from k to form a new vector space, the dual space E*. Let us prove that
1.8
dim E* = dim E
To this end we simply pick a basis el,...,ell for E and introduce the coordinate forms xl,...,xn E E* by the formula
ei x1(e)e1 Let us verify that any linear form
;eEE
on E satisfies the formula
E E* = 1: i (e;) xi To this end we simply evaluate both sides of the formula on the basis el,...,en. This shows that xl,...,xn generates the vector space E*. To see that xl,...,xn are linearly independent, consider a relation of the form E )i x1 = 0 , A1,...,An E k.
Evaluate the relation on ej, j = 1,...,n, and conclude that Ai = 0. Let us introduce the map 1.9
q:E _ E*, q(e)=(x- <e,x>)
;eEE
Observe that the kernel for q is E 1 and use that Q is non-singular to conclude that q is injective; formula 1.8 allows us to conclude that q is surjective as well. Restriction of a form along the inclusion i:K- E defines a k-linear map
i*:E* K*
;fHoi,EE*
QUADRATIC FORMS
4
which is surjective: In the argument above pick the basis el,...,en for E such that
el,...,ek is a basis for K and observe that the coordinate forms for this basis are xt o i,...,xk o i.
We shall focus on the composite map E
q
E* -,
K*
Since this is surjective we conclude from the dimension formula of Grassmann2
dim E = dim K* + dim Ker i" q
We ask the reader to show that Ker i*q = K
The result follows from the
formula above in combination with 1.8.
Let us consider a pair (E,Q) consisting of a finite dimensional vector space
E and a quadratic form Q on E. By an isomorphism from (E,Q) to a second such pair (F,P) we understand a linear isomorphism o:E=*F such that Q = P o o.
In
particular we may talk about automorphisms of (E,Q) which may be composed to
form a group, the orthogonal group of (E,Q) which is denoted 0(Q) or just O(E) when no confusion is possible.
Let us quite generally consider a quadratic form (E,Q) and pick a basis el,...,en for the vector space E. The Gram matrix G E MM(k) is given by
1.10
Gii = <ei,ei>
Proposition 1.11
;
i,j = 1,...,n
Let Q be a non-singular quadratic form on E. An
orthogonal transformation o E O(Q) has determinant 1 or -1.
Proof
Let us pick a basis el,...,e. for E and let A denote the matrix for o, thus o(ej) = E; A;ie1 for j = 1,...,n. Direct calculation gives us = < E kAkiek, > hAhleh> = E h,k TAik<ek,eh>Ahi
which shows that the Gram matrix for Q o o is TAGA.
In particular we find
that o is an orthogonal transformation for Q if and only if the matrix A obeys 2 A linear map f.E-F between finite dimensional vector spaces satisfies
dim E = dim Im(f) + dim Ker(f)
1.1 ORTHOGONALITY
1.12
5
T
AGA=G
Let us observe that the Gram matrix G is the matrix for the map q:E _ E* with respect to the basis e1,...,en for E and x1,...,x11 for E*. Since Q is non-singular we
conclude that del G # 0. The formula 1.12 gives us del TA del G del A = del G
from which we conclude that det2A = 1, or del A = 1 or -1.
The orthogonal group for the standard quadratic form on kn
1.13
n
Q(x) = x12+x22+ ... + x112
is denoted On(k). The subgroup of transformations with determinant 1 is denoted
SOn(k). When k = C or any other algebraically closed field with char(k) # 2 any non-singular quadratic form is isomorphic to the standard form 1.13.
Theorem 1.14
Let Q be a non-singular quadratic form on the vector
space E over C of dimension n. There exists a basis e1,...,e11 for E with <e;,ej> = 6ii
Proof
;
i,j = 1,...,n
Let us observe that Q can't be identically zero since this implies that
the bilinear form <x,y> is identically zero. Thus we can find a vector v E E with Q(v) # 0.
Let us write Q(v) = A2, A E C, or Q(A-lv) = 1 to conclude that we can
find e E E with Q(e) = 1. Let K denote the line spanned by e and put F = K 1
and observe that E = K ® F (meaning that any vector x E E has a unique decomposition x = f + k, where f E F and k E K). It is easily seen that the restriction of Q to F is non-singular. It is left to the reader to complete the proof by simple induction on dim(E).
QUADRATIC FORMS
6
1.2
WITT'S THEOREM
In this section we shall be concerned with a non-singular quadratic form Q
on a vector space E of dimension n over a field k with char(k)
2. The theme is
the flexibility of the orthogonal group 0(Q). The tool is reflections, an important class of orthogonal transformations, which we proceed to introduce.
We say that a vector n E E is non-isotropic if Q(n) # 0. A non-isotropic vector n E E defines a linear transformation rn of E by the formula
<x,n>
2.1
rn(x) = x - n
;xEE
_ <x,x>
;x E E
Direct computation gives us
i.e. rn is an orthogonal transformation called reflection along n. The hyperplane
orthogonal to n is fixed by rn while 7n(n) involution , i.e. rn2 = t but rn # t, and that
2.2
n.
It follows that rn is an
delrn=-1
Proposition 2.3
;nEE, #0
Let Q be a quadratic form on the vector space E. For
any A E k*, the orthogonal group 0(Q) acts transitively3 on the set
{eEEI Q(e)=A} Proof Consider vectors e,f E E with Q(e) = Q(f) = A # 0. Observe, that <e - f, e + f> = <e,e> - = 0 It follows that we can't have Q(e + f) = 0 and Q(e - f) = 0 at the same time (the non-isotropic vector e is a linear combination of e - f and e + f). If the vector e - f is non-isotropic, reflection r along this vector fixes e + f. Thus
7(e -f) = f - e
,
r(e+f) = e+f
3 We say that the action of a group G on a set X is transitive, if for all x E X and y E X there exists a E G with a(x) = y.
1.2 WITT'S THEOREM
7
Now, add these two formulas to see that r(e) = f.
along this vector interchanges e and - f.
If e + f is isotropic, reflection p
In conclusion the orthogonal trans-
formation rip maps e to f as required.
When Q is non-singular, it is also true that O(Q) acts transitively on the set of isotropic lines, i.e. lines in E generated by isotropic vectors. In fact,
Witt's theorem 2.4
Let (E,Q) be a non-singular quadratic form. Any
isomorphism a: (U,Q)-=-(V,Q) , where U and V are linear subspaces of E, can be extended to an orthogonal transformation of (E,Q).
Proof
Let us first treat the case where U is non-singular. This will be done
by induction on dim(U). In case dim(U) = 1 let us pick a non-zero u E U and put A = Q(u).
Since Q(o (u)) = A, we can apply 2.3 to extend v to an orthogonal
transformation of E.
To accomplish the induction step let us pick a non-isotropic vector e E U.
Since Q(e) = Q(v(e)) we can use 2.3 to pick r E O(E) such that o(e) = r(e). It will suffice to extend r-1a: U,E to an orthogonal transformation of E. To recapi-
tulate it suffices to extend v:U-E under the assumption that a fixes a nonisotropic vector e E U.
Let us introduce the total orthogonal space W to L = k.e
in U and the total orthogonal space F to L in E. Observe that W and F are non-
singular and that W C F and o -(W) C F.
Thus we can use the induction
hypothesis to extend the restriction a:W-F to an orthogonal transformation of F. Let us now assume that U is singular. Let there be given data consisting
of a non-zero vector e E U fl U 1 and a linear complement R to L in U. We are going to show that there exists an isotropic vector fin E which is orthogonal to R
and such that <e,f> = 1. Observe that we can find ,Q E E" with /3(e) = 1 and /i(R) = 0 ; the vector b E E with q(b) _ /3 is orthogonal to R and satisfies = 1. Let us show that the vector f = b - 2e is isotropic
= - <e,b> = 0
QUADRATIC FORMS
8
The isotropic vector f is orthogonal to R and satisfies <e,f> = 1. Let us observe
that f 0 U since e is orthogonal to U but <e,f> = 1. The space W = U + Rf is the direct sum of the two orthogonal subspaces R and ke + kf. Let us perform the same construction on the data o(U), or(e), o(R) to find an isotropic vector g E E
which is orthogonal to o(R) and which satisfies = 1. Let us extend o:U-+E to a linear map WE by the convention o(f) = g. It is easy to check that the extended map preserves the inner product: use that o(W) is the direct sum of the orthogonal subspaces o(R) and ko(e) + ko(f).
Let us finally consider the general subspace U of E. If U is non-singular
we can use the first part of the proof to extend o. If U is singular we can use the middle part of the proof a number of times until U becomes non-singular, and then
apply the first part of the proof.
The orthogonal group 0(E) of a non-singular form is generated by reflections along non-isotropic vectors as it follows from exercise 2.3. The number
of reflections needed can be bounded by the following theorem of Elie Cartan: An orthogonal transformation of a non-singular form of dimension n can be written as a product of no more that n reflections along non-isotropic vectors.
The proof of the theorem of E. Cartan is rather complicated, see [Berger) and [Deheuvels]. Moreover, the theorem is not precise enough for our purposes.
For these reasons we shall not make use of the theorem in its generality, but prove a number of more specific results along these lines: namely 4.5, 5.3, 6.11
1.3 SYLVESTER TYPES
9
SYLVESTER TYPES
1.3
In this section we shall deal exclusively with quadratic forms over the field
R of real numbers.
By a Euclidean vector space we understand a finite dimen-
sional vector space E with a quadratic form Q which is positive definite, i.e. Q(e) > 0 for all non-zero e E E. The form Q is called negative definite if Q(e) < 0
for all non-zero e E E.
Our first objective is to generalise the concept of an
orthonormal basis.
Proposition 3.1
Let (E,Q) be a quadratic form on the vector space E
over R of dimension n.
There exists an orthonormal basis for E, i.e. a basis
el,...,en with j
r
0
-1, 0, +1
;
36j
i =j
Proof Let us use induction with respect to dim(E) = n.
If the form Q is identi-
cally 0, any basis for E is orthonormal. Thus we may assume the existence of an e E E with Q(e) # 0.
Let us write Q(e) = eat with e = ± 1 and A E R*. Put
el = A-le to get Q(e1)=± 1. Let us use the induction hypothesis to find an orthonormal basis e2,...,e. for (Re1) -1' . The basis e1,...,e1, meets our requirements. 0
Sylvester's theorem 3.2
Let e1,...,en be an orthonormal basis for the
quadratic form (E,Q) over R. The numbers
p=#{iI<e;,e;>=-1}, q=#{iI <e;,e;>=1} are independent of the orthonormal basis considered.
Proof
Let us fix the basis as in the statement of the proposition, and let E_
denote the subspace of E, generated by those elements e; from our basis for which
<e;,e;> = 0, - 1. Observe that Q(e) < 0 for all e E E_ and conclude that for any Euclidean subspace F of E we have F fl E_ = 0. According to 3.4 we have that
QUADRATIC FORMS
10
dim(F) + dim(E_) = dim(F + E_) + dim(F fl E_)
This gives us dim(F) + n - q:5 n, i.e. dim(F) < q. It follows that sup { dim(F) I Euclidean F C E } = q
3.3
This shows that q is independent of the basis chosen. Let us apply this to the space (E,-Q) and conclude that p too is independent of the basis.
With the notation of the theorem we say that the quadratic form (E,Q)
has Sylvester tune (-p,q).
Observe that two quadratic forms (E,Q) and (F,R)
with dim(E) = dim(F) of equal Sylvester type are isomorphic.
Dimension formula 3.4
For subspaces U and V of the finite dimen-
sional vector space E we have that dim(U fl v) + diin(U+V) = dim(U) + dim(V)
Proof Let
us apply the Grassmann dimension formula to the linear map
f:U®V,E, f(u,v)=u-v observing that Ker f
;uEU,vEV
U fl V and Ini f = U + V.
Discriminant inequality 3.5 form of Sylvester type (-s,r).
Let (E,Q) be a non-singular quadratic
For any basis el,...,ell for E we have
sign detij <ei,ej> = (-1)s
Proof
Let us consider a second basis
related to the first basis
through the transition matrix B. This gives us = < >r B;rer, Es Bjses> From this we conclude that
rs Bir<er,es>TBsj
det j = det B detrs<e,es> del TB = det2B delii<e1,ej>
1.3 SYLVESTER TYPES
11
This shows that sign detij<ei,ej> is independent of the basis considered. We leave it to the reader to verify the formula in the case of an orthonormal basis.
Symmetric matrices
From a symmetric matrix A E MM(R) we can
construct a quadratic form on the free vector space E with basis el,...,en
Q(E xiei) = E,. arsxrxs By the Sylvester type of A we understand the Sylvester type of the form Q. The Sylvester type of A can be evaluated by means of the principal minors deiA1,..., delAn
here Ai is the i x i matrix obtained by deleting the n-i last rows and columns of A.
Lemma 3.6
The symmetrical matrix A E Mn(R) is positive definite if and
only if all principal minors deMA1,..., detAn are strictly positive.
Proof To begin let us just assume that the principal minors are different from zero and let us show how to determine the signature of A. With the notation above the subspace Ei of E generated by e1,...,ei is non-singular since del Ai # 0. Let us pick a non-zero vector fi E Ei, orthogonal to Ei_1. The Gram matrices of the two bases el,...,ei and el,.... ei_1,f1 have the same sign, thus
sign del Ai = sign sign del Ai_1
In the case at hand we deduce that > 0 for i = 1,...,n.
QUADRATIC FORMS
12
1.4
EUCLIDEAN VECTOR SPACES
Let us consider a Euclidean vector space E , i.e. a finite dimensional real vector space equipped with a positive definite quadratic form. A vector e E E has length IeI = <e,e>.
Let us start with the inequality of
Cauchy-Schwarz 4.1
I <e,f> I < I e I
; e,f E E
IfI
Addendum: The inequality is sharp when e and f are linearly independent
Proof The inequality is trivial if e and f are linearly dependent. If e and f are linearly independent, they span a Euclidean plane and we get from 3.5 that det
<e,e> <e,f>
>0 0
from which the sharp inequality follows.
Triangle inequality 4.2
I e + fI < IeI +IfI
; e,f E E
Addendum: The triangle inequality is sharp when e and f are linearly independent.
Proof
A simple direct calculation gives us le +f12 = IeI2+ IfI2+2IeIIfI +2(<e,f> - IelIfI)
0
and the result follows from 4.1.
On the basis of the Cauchy-Schwartz inequality we can introduce the angle L(e,f) between two non-zero vectors e,f by the formula
4.3
cos L(e,f) = ee,f> I
IIfI
;
L(e,f) E [0,ir]
1.4 EUCLIDEAN VECTOR SPACES
13
In the rest of this chapter we shall be concerned with the orthogonal group O(E).
The basic example of an orthogonal transformation is reflection r in a linear hyperplane H. If n E E is a unit normal vector for H we have that 4.4
r(x) = x - 2<x,n>n
Theorem 4.5
An orthogonal transformation o of the Euclidean vector
;xEE
space E of dimension n is the product of at most n reflections through hyperplanes.
Proof We shall
use induction on n = dim(E). To accomplish the induction
step we pick a vector f E E of unit length.
If o(f) = f observe that o stabilises
F = f 1 and use the induction hypothesis to decompose o into a product of at most n-1 reflections. In case o(f) # f observe that reflection r along the vector
c(f) - f interchanges f and o(f). It follows that ro fixes the vector f, so we can use induction to write ro as a product of at most n-1 reflections.
Let us recall from 1.11 that an orthogonal transformation has determinant
1 or -1. It follows that the special orthogonal 4.6
arg ouL
SO(E) = { o E O(E) I det o = 1 }
has index 2 in 0(E). We have seen that O(E) operates transitively on the sphere 4.7
S(E)=f eEEJ lei =11
We ask the reader to verify that SO(E) acts transitively on S(E), dim E > 2.
QUADRATIC FORMS
14
Euclidean plane
Let us assume that dim(E) = 2. An orthogonal transfor-
mation of E with determinant 1 is called a rotation.
In order to analyse an
orthogonal transformation o of E in more detail, let us pick an orthonormal basis i and j for E; we can express that o(i) lies on the circle S(E) by writing
o(i)=icos0+jsill 0
;OER
It follows that o(j) is one of the unit vectors orthogonal to o(i), i.e. the vector -i sinO+j cos 0 or its negative. Thus the matrix for o takes one of the forms
4.8
cos 0
-sin 0
cos 0
sin 0
sin 0
cos 0
sin. 0
-cos 0
; OER
Evaluation of the determinants show that the first matrix defines a rotation, while
It follows from the addition formulas for the
the second defines a reflection.
trigonometric functions that the first matrix in 4.8 defines an isomorphism between the circle group SO(E) and the group R/2irZ.
Euler's proposition 4.9
An orthogonal transformation o E SO(E) of
Euclidean 3-space E is a rotation: there is a line L fixed by o and the transformation induced by o in the plane L 1 is a rotation. An orthogonal transformation o E 0(E) of determinant -1 has the form pr where p is a rotation as above with axis L and r is reflection in the plane L 1 .
Proof Consider the characteristic polynomial for the transformation x(t) = det( tt - o ) Let us analyse a real root A: pick a vector e # 0 with o(e) = Ae and take norms to
see that A2 = 1, i.e. A= 1 or A = -1.
If dec(o) = 1 we have X(0) = -1. Observe that y(t) tends to +oo as t
tends to +oo and conclude that 1 is a root for X(t). Consider a line L with eigenvalue 1 and observe that o acts on the plane L
1 with determinant
1.
If dec(o) _ -1 we have x(0) = 1. Observe that X(t) tends to -oo as t
tends to -oo and conclude that -1 is a root for X(t). Consider a line L with eigenvalue -1 and observe that or acts on the plane L 1 with determinant 1. 0
1.4 EUCLIDEAN VECTOR SPACES
Theorem 4.10
15
Let o be an orthogonal transformation of the Euclidean
vector space E of dimension n.
There exists a decomposition of E into an
orthogonal sum of lines and planes stable under o.
Proof
According to lemma 4.11 below we can find a linear subspace R C E of
dimension 1 or 2 stable under o. It follows that R 1 is stable under o as well. A simple induction on dim E concludes the proof.
Lemma 4.11 Let o:E--*E be an endomorphism of a finite dimensional real vector space E. There exists a subspace R C E of dimension 1 or 2 stable under o.
Proof Let
us present a construction known as complexification of o. We ask
the reader to turn EC = E ® E into a vector space over C through the convention (x + iy) (e,f) = (xe - yf, xf + ye)
; x,y E R, e,f E E
Let us identify e E E with (e,0) E EC to get the formula
(e,f)=e+if
;
e,fEE
With this notation, we can introduce the C-linear endomorphism ac: EC-'EC by oC(e + if) = o(e) + io(f)
; e,f E E
the details are straightforward and left to the reader. From the fundamental theorem of algebra we conclude that oC has an eigenvalue A =a+ ib. A nontrivial eigenvector e+if with eigenvalue A satisfies the equation
o(e) + io(f) = (a + ib)(e + if) From this it follows that e and f generate a subspace of E stable under o-.
QUADRATIC FORMS
16
1.5
PARABOLIC FORMS
In this section we shall investigate a finite dimensional vector space F over
R equipped with a positive quadratic form Q, which of course means that Q(f) > 0
for all f E F. The point of departure is the
Cauchy-Schwarz inequality 5.1 ;e,fEF
<e,f>2 < <e,e>
Proof The Cauchy-Schwarz inequality may be rewritten in determinant form det
<e,e> <e,f>
> 0
This is trivial when e and f are linearly dependent. When e and f are linearly independent we get from 1.12 that the sign of the determinant is unchanged if we
0
replace e and f by an orthonorinal basis for the plane they span.
Corollary 5.2
Let (F,Q) be a positive quadratic form. A vector f E F is
isotropic if and only if f E F 1 , i.e. <e,f> = 0 for all e E F.
We shall investigate the orthogonal group O(F), in particular the subgroup generated by reflections. Let us observe that a reflection r fixes all isotropic vectors as follows from 5.2 and formula 2.1.
Theorem 5.3
Let Q be a positive quadratic form on the m-dimensional
real vector space F. An isometry o E O(F) which is the identity on F written as a product of at most m reflections.
1
can be
1.5 PARABOLIC FORMS
17
Proof We proceed by induction on m. Let us distinguish between two cases. 10 Im(o - t) F 1 . Choose f E F such that p = o(f) - f is non-isotropic. Reflection a along p interchanges f and o(f), i.e. 7ra(f) = f. Observe that f is non-
isotropic since o(f) # f and apply the induction hypothesis to the restriction of Tro to the hyperplane f 1
.
Let us decompose F into a direct sum F = E ® N where E is Euclidean and N = F 1 . Let us describe o in terms of a linear map v:E-N 2°
Im(o - t) C F 1 .
c(e+n)=e+v(e)-{-ii
; eEE,nEN
We can arrange for v:E-N to be injective: a non-zero vector e E E of Ker v is non-isotropic and fixed by o, which allows us to use the induction hypothesis on e
1 . We can now arrange for v:E-N to be bijective by replacing N by v(E).
From now on we may assume that
dint F=2dimF Observe that o acts as the identity on F 1 and on F/F 1 and conclude that det(o) = 1. Let us pick any reflection p in F and observe that we are back to case
1° with pa since det(po) = -1. Thus we can write
Po=pl...ps
s<m
where pl,.... ps are reflections. From the facts that m is even and det(po) _ -1, it
0
follows that s < m as required.
Let us say that a space (F,Q) is parabolic if Q is positive and the space
F 1 is of dimension 1.
The isotropic line F 1 is invariant under orthogonal
transformations. It follows that we have a morphism of groups
5.4
µ:0(F)-;R*
where the multiplier p(o) E R* denotes the eigenvalue of o E 0(F) on the isotropic
line. This allows us to introduce a subgroup of 0(F) of index 2
5.5
O.(F)_ {0' E0(F)I p(c)>0}
QUADRATIC FORMS
18
Let us observe that the antipodal map 1,
belongs to the centre of O(F)
and that µ(i) = -1. This defines a decomposition of the orthogonal group 0(F) of a parabolic space F
O(F) = 0,,(F) x 7L/(2)
5.6
The line at infinity
We shall pursue the study of parabolic spaces in a
setting which is particular useful for Euclidean geometry. Let E be a Euclidean space of dimension n and consider the following bilinear form on F = E ® R
<(e,a),(f,b)> = <e,f>
5.7
e,f E E, a,b E R
This form is easily seen to be parabolic. The isotropic line is generated by the vector (0,1) and is called the line at infinity. Let us use the inner product on F to
introduce the evaluation map ev: F x E . -> R ; x,y E E, l; E I$ ev((x,l;),y) = <x,y> + The evaluation map identifies F with the space of affine functions on E.
Similarities
Let us recall that a similarity of the Euclidean space E is an
affine transformation
of E of the form
fi(x)=aj(x)+f
; 0E 0(E),fEE, AER,.A>0
Our main objective is to identify the group Siinl(E) of similarities with the orthogonal group 0,,(E).
Proposition 5.8
For any orthogonal transformation or of parabolic space
F=E ®R, there exists a unique similarity & E Siml(E) such that ev(o(f),o(e)) = p(o) ev(f,e) The assignment o,-*o induces an isomorphism of groups
0,0(F) - Si1nl(E)
fEF ,eE E
1.5 PARABOLIC FORMS
19
The similarity )4 + v, 0 E O(E), A > 0, v E E, corresponds to 1 E 0,,(F) given by
; z E E, (E R
11(z,O = (4'(z), - <¢(z),v> + AO
Proof
Let us first observe that a point e E E defines a linear form f
ev(f,e)
on F with evaluation 1 against the isotropic vector (0,1). It is easy to see that any
linear form 0 on F with 4'(0,1) = 1 can be represented in this way and that such a representation is unique.
Let us return to 0 E O(F) and observe that p(o)-1o is a linear automorphism of F which fixes the isotropic vector (0,1). It follows that for a given e E E we can find a uniquely determined vector 0"(e) in E such that
ev(o(f),e) = µ(o) ev(f,o'(e))
;fEF
Variation of e E E defines a transformation a :E-E. It follows at once that
(Qr) = T-Q
; o,r E 0(F) We conclude that o' is invertible and that the transformation i7 = 0'-1 satisfies ,
t- = 6
the required formula. In order to show that o is a similarity, we shall work our way through three special cases. 10
A vector v E E gives rise to a transformation Ov E O(F) given by
Ov(z,O = (z,( - )
;
z E E,
E 02
Direct verification shows that Ov(e) = e + v, e E E. 2°
A real scalar A E Jr gives rise to a transformation sA E O(F) given by sA(z,() = (z,aS)
; z E E, r; E R
Direct verification gives us s'A(e) = A e, e E E. 3°
An orthogonal transformation o E O(E) defines v E 0(F) given by v(z,C) = (o(z),C)
; z E E, S E R
Another direct verification gives us o- = 0. We leave it to the reader to show
that any orthogonal transformation of F can be decomposed into a product of transformations of the three types we have just considered.
QUADRATIC FORMS
20
Normal vectors
Let there be given an oriented affine hyperplane (H,P)
in Eucliden n-space E. By this we understand that H is an affine hyperplane in E
and P is one of the open half-spaces of E bounded by H. The normal vector to (H,P) is the unit vector n E F which is zero along H and positive on P. Let us observe that reflection rn is given by
rn(z,O = (z,() - 2(n,v)
; n = (n,v)
while the corresponding transformation ?n is given by the formula
rn(x) = x - 2(<x,n> + v)n
;xEE
which is a reflection in the affine hyperplane H, compare 11.2.
Let us observe that v E OA(F) transforms the normal vector n for (H,P) into the normal vector for (5(H),5(P)) as follows from the formula
ev(v(n),5(x)) = p(v)ev(n,x)
Coxeter matrix
;
xEE
Let us consider an affine simplex D, i.e. the convex hull
of n+1 points eo,el,...,en, not contained in an affine hyperplane. Let no, nt,..., nn
be the corresponding normal vectors: the evaluation of ns against ej is zero, i # s, while the evaluation against es is positive.
From this description it follows that
the normal vectors form a basis for F = E ® R. Let us write 5.9
XOn0+)lnl
(0,1)
and observe that the A is are strictly positive which follows from evaluation of formula 5.9 on the vertices of the simplex.
The Coxeter matrix for D is
rs
Proposition 5.10
A basis no,nl...... nn of unit vectors for E ®R such that
the isotropic vector (0,1) can be written in the form
,Ono + atnt + ... + A,,n = (0,1) defines a Euclidean simplex in E.
Ai > 0 , i = 0,...,n
1.5 PARABOLIC FORMS
21
Proof For s = 0,...,n let £s denote the linear form on E (D R with Qns) = as and {$(nr) = 0 for s # r. The relation above ensures that s has evaluation 1 on
the isotropic vector (0,1). It follows from the remark made in the beginning of the
proof of 5.7 that we can find a point es E E with ev(f,es) = s(f) for all f E F. The points eo,...,en define the simplex we are looking for.
Proposition 5.11
Two simplices D and P in Euclidean space E are
similar if and only if they have the same Coxeter matrix.
Proof We have already seen that two similar simplices have the same Coxeter matrices. With the notation above let m0,mi,...,mn be the normal vectors for the simplex P enumerated such that
= <mr,ms>
;
r, s = 0,1,...,n
Let v E O(F) be such that o (nl.) = o(mr), for r = 0,...,n. It remains to prove that p(Q) > 0. To this end let us apply o- to the relation 5.9 to get that
)0m0 + Alm1 +... +.nrnn = (0,h(o,)) Let us evaluate this against an interior point p E P to get that .Dev(m0,p) +... +
µ(o,)
which shows that u(o) > 0 as required.
0
In the rest of this section we shall be concerned with the existence of simplices with given Coxeter matrices (ars).
For application to groups generated
by reflections, see [Coxeter2] and [Bourbaki2], we shall be concerned with acute angled simplices, i.e. simplices whose matrices satisfies J ars < 0 for r
QUADRATIC FORMS
22
Indecomposable matrices 5.12
Let A E M"(R) be a positive sym-
metric matrix with del A=O, which has 1 s along the diagonal and satisfies
a,.s < 0
,
r # s, r,s = 0.... ,n
; A = (ars)
If the matrix is indecomposable, in the sense that no non-trivial subset J of 0,...,n exists such that
a,,,=0 forrEJands0J then there exists a Euclidean n-simplex with A as Coxeter matrix.
Proof Let us consider a real vector space F equipped with a basis eo,...,e. and a quadratic form such that
; r,s = 0,...,n
<er,es> = a,,,
By assumption the form is positive but singular. We shall eventually prove that F is parabolic.
To this end we shall show that for a non-zero isotropic vector
n = E xsas we have as # 0 for all s = 0,1,...,n. From the inequality Q(E xses) = E r,sarsxrxs >_ E r,s arslxr1Ixsl = Q(E Ix5Ies) it follows that we can assume that all coefficients of n are positive. Let us for a moment assume that the set J = { s I xs > 0) is different from 0,...,n. Let us recall from 5.2 that the isotropic vector n is orthogonal to any vector: ; s = 0.... ,n 0 = = E r E J xr<el.,es> In particular we conclude that a1. = <er,es> = 0 for s 0 J, contradicting the
hypothesis that the matrix A is indecomposable.
Let us prove that F l has dimension 1. To this end let us introduce the space N C R"+1 representing F 1 N = {(xo,...,x") E R"+1 I Q( Exses) = 0}
By the observations made above we conclude that the intersection between N and
the hyperplane x0 = 0
is zero;
this implies
dim(N) = 1.
In conclusion if
Exses 0 is isotropic then all coefficients are non-zero of the same sign. Finally we can use an isometry of F with E ® R to construct a basis for E ® R and apply 5.10.
1.6 LORENTZ GROUP
1.6
23
LORENTZ GROUP
Let us consider a vector space F of dimension n+1 over R equipped with a
quadratic form of Sylvester type (-n,l). We shall be concerned with the action of 0(F) on the hvperboloid or pseudosphere
S(F)={fEFI =1}
6.1
Let us start our investigations by a Cauchy-Schwarz type inequality.
Lemma 6.2 Two points x and u of the hyperboloid satisfy the inequality 1 < I<x,u>I
; x,u E S(F)
Moreover, the inequality is sharp when x and u are linearly independent in F.
Proof When
x and u are linearly dependent we have x = u or x = -u and the
inequality is trivial. When x and u are linearly independent, the plane R they
span is of Sylvester type (-1,1), (-1,0) or (-2,0), compare 6.3. From the fact that R contains vectors of strictly positive norm it follows that the type is (-1,1). The discriminant lemma 6.3 gives us <x,x> < 2 as required. 0
Discriminant lemma 6.3
Let R be a plane in F generated by two
vectors e and f. The Sylvester type of R is determined by the discriminant
A = <e,e> - <e,f>2 according to the following table
0<0
A>0 (-2,0)
Proof Let
us first prove that the plane R contains a vector of norm -1. Use
any orthonormal basis for F to exhibit a hyperplane E of Sylvester type (-n,0). It follows from the dimension formula 3.4 that,
QUADRATIC FORMS
24
dim(R) + dim(E) = dim(R fl E) + dim(R + E)
Using dim(R + E) < n+1 we get that dim(R fl E) > 1. Thus R contains vectors of strictly negative norm. From this we conclude that the only possible Sylvester
types for R are (-1,1), (-1,0) and (-2,0). It follows from the discriminant inequality 3.5 that these three cases correspond to A < 0, A=O and A > 0 0
respectively.
A linear hyperplane in F must be of type (-n+1,1), (-n,0) or (-n+1,0) as it follows from the argument used in the previous proof.
Proposition 6.4
The hyperboloid S(F) has two connected components.
Points x,u E S(F) are in the same connected component if and only if <x,u> > 0.
Proof
Let us fix the point u E S(F) and let E denote the hyperplane orthogo-
nal to u. Since E has type (-n,0) we find that E separates S(F) into two parts
H={xES(F)I <x,u> > 0}, H-={xES(F)I<x,u><0} These two parts are interchanged by reflection along u. We are going to prove
that stereo ranhic projection g:H-+E with centre -u maps H homeomorphically onto the open unit disc
D={yEEI-<1} The affine line through x E H and -u is parametrised (x + u)t - u, t E R. Its inter-
section g(x) with E is given by <(x + u)t - u,u> = 0 , which gives us
6.5
g (x) =
x - <x,u>u 1 + <x,u>
Direct computation gives us 1 = 1 -+<x,u> <x,u>
1.6 LORENTZ GROUP
25
From 6.2 we conclude that <x,u> > 1, so the formula above enables us to conclude that g(x) E D as required. We ask the reader to work out that the intersection between H and the affine line through the points z E D and -u is given by
6.6
f(z) = 2
; zED
+u 1+z -u
These investigations show that f:D-+H is a homeomorphism.
Definition 6.7
By a Lorentz transformation4 of F we understand an
orthogonal transformation o E 0(F) which preserves the connected components
of the pseudosphere. The group of Lorentz transformations is called the Lorentz
rou and is denoted Lor(F). It follows from 6.4 that a Lorentz transformation can be recognised by the inequality
6.8
<x,o(x)> > 0
; o E Lor(F), X E S(F)
The antipodal map x'-+-x is in the centre of O(F) but is not a Lorentz transformation.
6.9
It follows that Lor(F) has index 2 in O(F), more precisely O(F) = Lor(F)x7L/(2)
The basic example of a Lorentz transformation is provided by reflection along a vector c E F with = -1
rc(x) = x+ 2<x,c>c
;xEF
Let us use 6.4 to check that this is a Lorentz transformation: = 1 + 2<x,c> 2
; x E S(F)
Lorentz transformations with determinant 1 are called even or special Lorentz 4 In literature with reference to physics such a transformation is called an orthochroneous Lorentz transformation.
QUADRATIC FORMS
26
transformations.
These transformations make up the special Lorentz
rou
Lor+(F), which is a subgroup of Lor(F) of index 2.
Let us make a detailed examination of a Lorentz transformation a when the space F has dimension 2. Pick an orthonormal basis e,f for F with <e,e> = 1 and x2
_ - 1. The point xe + tf lies on the hyperbola S(F) if and only if
- t2 = 1. The branch of the hyperbola containing (1,0) can be parametrised
(coshs, sinks), s E R. Thus we can write a(e) = e coshs+fsinhs t
cosh s, sinh s) x
I The vector
a(f) =
has norm -1 and is orthogonal to or(e), so we conclude that
(e sinks + f coshs). In the first case the matrix for o, is coshs
-sinks
sinks
-coslcs
It follows that det(er) = -1 and lr(o-) = 0. This implies that a has eigenvalues 1 and -1. We ask the reader to verify that the basis e,f can be picked such that the matrix for or corresponds to the case s = 0. This shows that a is a reflection along
a vector of norm -1. In the second case the matrix for o is
L(s)_
coshs
sinks
sinks
coshs
;sER
From the addition formulas for the hyperbolic trigonometric functions it follows that L(s) is a morphism of groups
6.10
L(s + t) = L(s) L(t)
; s,t E R
1.6 LORENTZ GROUP
27
This shows that the special Lorentz group is isomorphic to R when dim(F) = 2.
Theorem 6.11
Let F be a vector space of dimension n+1 equipped with
a quadratic form of Sylvester type (-ii,l). Any Lorentz transformation o of F is
the product of at most n+1 reflections of the type = -1.
rc where c E F with
Proof We shall proceed by induction on n. If n = 1 the result follows from our investigation above. If n > 2 let us pick a hyperplane E of F of type (-n,0)
and focus on the map E
F given by
eEE If this map is not injective we can find a vector e with norm -1 fixed by o. The
space D = e 1 has Sylvester type (-n+1,1) and the restriction of o to D is a Lorentz transformation as it follows from 6.8
If the map a i--+ o(e) - e, e E E, is injective, then the image space has
dimension n-1 > 2 and must have non-zero intersection with the hyperplane E. It follows that the image space contains a vector c say of norm -1.
c = o(e) - e
;
Let us write
e E F, <e,e> < 0
Since o(e) and e have the same norm we conclude from the proof of 2.3 that reflection r, along c interchanges e and o(e). It follows that rca fixes the vector e
of strictly negative norm. Apply the induction hypothesis to the space e -- to conclude the proof.
Let us turn to the action of the Lorentz group on the isotropic cone C(F), i.e. the set of vectors of norm 0. The multiplicative group IR* acts on C(F)-0; the orbit space is called the proiectivised cone of F and is denoted PC(F). We may of course think of PC(F) as the set of isotropic lines in F.
QUADRATIC FORMS
28
Proposition 6.12
Let F be a vector space of dimension n+1 equipped
with a quadratic form of Sylvester type (-n,l). The projectivised cone PC(F) is homeomorphic to the sphere
Proof Let
Sn-1
us return to the proof of 6.3 and introduce the unit sphere
Sn"1
= 8D
in the hyperplane E orthogonal to the base vector u. We shall make use of the map 4:E-*F given by
6.13
m(z) = 2z + u - u
;zEE
Direct calculation using that = 0 gives us
<5(z),c(z)> = (1 + )2
;zEE
which shows that 0 induces a map D:8U,PC(F). Observe that u and z E aU span a plane of Sylvester type (-1,1) and that q(z) and 0(-z) are two linearly independent isotropic vectors in that plane. Observe also that this plane is generated by ¢(z) and u. From these remarks it follows that 4D is injective. In order to see that I is surjective note that an isotropic line L in F together with the
vector u span a plane R of type (-1,1): pick a non-zero vector v E L and let us examine the sign of the discriminant
- 2 = - 2 < 0 The intersection E n R is a line generated by a vector z say of norm -1. It follows that L is generated by one of the two vectors O(z) and c5(-z).
Corollary 6.14
When dim(F) > 3, the
0
Lorentz group Lor(F) acts
faithfully on PC(F).
Proof
Let a be a Lorentz transformation of F which acts trivially on PC(F).
We shall analyse the problem in terms of the map': Sn-1- PC(F) introduced in the proof of 6.13. For z E
S°-1
let A (z) be the eigenvalue belonging to the isotropic
vector O(z). The function )t:Sn"1,R is continuous which follows from the formula
«¢(z),u> = \(z) = 2)(z)
; z E Sn"1
1.6 LORENTZ GROUP
29
The set of values of the function A is finite simply because it is a subset of the set of eigenvalues of o. Since dim(F) = n+1 > 3, we can use the connectedness of S' 1 to conclude that A is constant.
Let us prove that F is generated by C(F).
To this end we shall prove
that any vector f E F with non-zero norm is contained in a hyperbolic plane. If < 0, then f 1 is of type (-n+1,1), so we can select e orthogonal to f with <e,e> > 0. When > 0 the space f 1 has type (-n,0) so we will have no trouble finding e orthogonal to f with <e,e> < 0. Now, observe that a hyperbolic plane is generated by two isotropic vectors.
Let us observe that multiplication by a scalar A is not an orthogonal transformation of F unless A = 1 or A = -1. We have already observed that multiplication by -1 interchanges the two sheets of the hyperboloid S(F).
Proposition 6.15
The truncated isotropic cone C*(F) = C(F) - {0} has
two connected components provided that dim(F) > 3. The connected components are preserved by the Lorentz group Lor(F).
Proof Let us fix a point u E S(F) and observe that C*(F) is the union of C+(F)={ceFI > 0 }
,
C (F)={ceFI <0)
It follows from the proof of the previous proposition that these two sets are the connected components of C*(F).
Let us prove that the two connected components are unchanged if we replace the point of reference u with an other point v of the same sheet of the hyperboloid S(F). To this end let us focus on a point c E C*(F) and observe that the function
xH<x,c>,
omits the value zero since the hyperplane
orthogonal to c has type (-n+1,0), compare the remark made after 6.4. It follows
that and have the same sign. From the fact that o(u) and u belong to the same sheet of S(F) it follows
that and have the same sign for x E C*(F). In particular for x E C+(F) we have = <x,u> > 0, and we conclude that is positive; thus o(x) E C+(F).
QUADRATIC FORMS
30
Proposition 6.16
Let L denote an isotropic line in F, and let LorL(F)
denote the group of Lorentz transformations stabilising L. Restriction from F to
L l defines an isomorphism
LorL(F) -: 0,,(L 1 )
where 0,,(L 1) denotes the subgroup of 0(L 1) made up of orthogonal transformations with positive eigenvalue on the isotropic line L, compare 5.5.
Proof
Let OL(F) denote the subgroup of O(F) made up of transformations
which stabilises the line L.
Let us prove that restriction defines an isomorphism OL(F) =: 0(L l )
First, restriction is surjective as it follows from Witt's theorem 2.4. - Let us analyse a o E OL(F) which restricts to the identity on L 1 . To this end we introduce an n-1 dimensional subspace E of L 1 of type (-n+1,0). From the fact that o has trivial restriction to E follows that we can write o = i ® a where a is an orthogonal transformation of E 1 . The plane E 1 has type (-1,1) and can be decomposed E 1 = L ® K where K is the second isotropic line in E -L . The transformation a preserves K and L and the eigenvalues must have the form A,A-1,
A E R*, since the inner product is preserved.
Remembering that o has trivial
restriction to L 1 we conclude first that A = I and second that p = e as required. The final statement of the proposition is a consequence of 6.15 in case dim(F) > 3. In case dim(F) = 2 the result follows from the investigations a few lines above.
Proposition 6.17
The special Lorentz group Lor+(F) acts transitively on
the space N(F) of vectors of norm -1, provided dim(F) > 3.
Proof
The full orthogonal group O(F) acts transitively on N(F) as a
consequence of Witt's theorem 2.4. To see that Lor(F) acts transitively it suffices
to prove that the stabiliser in O(F) of a given N E N(F) is not contained in the Lorentz group. As an example we take the transformation which fixes N but transform xH-x on the orthogonal complement of N in F. We ask the reader to check that the stabiliser of N in Lor(F) is not contained in Lor+(F).
1.6 LORENTZ GROUP
31
The projectivised cone PC(F) is a subset of projective snare P(F), that i.e. the set of all lines in F. Let PS(F) denote the subset of P(F) consisting of lines
in F generated by vectors of strictly positive norm. In the natural topology of P(F) we have 8PS(F) = PC(F). Let Hn denote any one of the two sheets of the hyperbola S(F). We have
a natural homeomorphism Hn_PS(F) which assigns to a point X E Hn the line L in F generated by the vector X. Using this identification we can write
6.18
all- = PC(F)
QUADRATIC FORMS
32
1.7
MOBIUS TRANSFORMATIONS
Let E denote a Euclidean vector space of dimension n. A sphere C with centre c and radius r > 0 gives rise to an involution o of E-{c} given by
o(x) = c + r2 x - c
7.1
Ix-c12
;
x E E - {c}
The transformation o is called inversion in the sphere C. It is apparent from the
formula that o(x) lies on the ray through x emanating from c. The sphere C can be recovered as the fixed point set for o.
In the following we shall consider the 1-point compactification E of E, obtained from E by adding an extra point oo. Reflection o in the sphere C can be extended to an involution & of t which interchanges a_ and the centre c of C. It is easy to check directly that & is continuous.
Let us consider an affine hyperplane H in E through the point u E E. Using a unit normal vector n E E for H we can describe write Euclidean reflection in H by the formula
7.2
\(x) = x- 2<x - u , n>n
;
xEE
We leave it to the reader to extend this to a continuous transformation A of E, which preserves the point at infinity.
In order to unify the treatment let us introduce the word sphere C in E for
a Euclidean sphere in E or a subset of t of the form C = {oo} U H where H is an affine hyperplane in E. The two types of transformation of t we have introduced are called inversions in spheres in E.
The subgroup of the group of homeo-
morphisms of t generated by inversions in spheres in t is called the M6bius group of E and is denoted Mob(E).
The main purpose of this section is to identify the Mobius group with a Lorentz group. To this end we introduce the auxiliary space E ®R2 equipped with the bilinear form
1.7 MOBIUS TRANSFORMATIONS
7.3
33
<e,f> +
<(e,a,b),(f,c,d)>
2(ad + bc)
; e,f E E, a,b,c,d E R
It is easily seen that this space has Sylvester type (-n-1,1). Let us consider the transformation t: E-> C(E + R2) given by Le = (e,<e,e>,1)
7.4
;
eEE
This induces a map from E into PC(E+922) which identifies E with the complement of the point of PC(E+R2) representing the line through the point (0,1,0). Let us extend this to a bijection t: E-+PC(E + R2), where L(oo) is the line through the point (0,1,0).
Theorem 7.5 projectivised
cone
The action of the Lorentz group Lor(E (D922) on the PC(E + R2)
and
the homeomorphism
a E - + PC(E (D R2)
identifies the Mobius group M"ob(E) with the Lorentz group Lor(E+R2).
Proof
Let us return to the sphere C from 7.1 and form the vector
C = (c, -r2,1) We ask the reader to verify the formulas
) = Ix - c12 - r2
= - r2 ,
xEE
From this we can deduce a commutative square PC(E(D R2)
E
[ 7C
&
E
PC(E(D R2)
More precisely we ask the reader to verify the formula a t (x) = r2lx - cl-2 rC L (x)
Next, we return to the reflection A from 7.2 and introduce the vector
N = (n,2,0) From this we can deduce a commutative square
x E E - {c}
QUADRATIC FORMS
34
E
PC(E (D R 2)
la
1 TN
E
PC(E ® 082)
More precisely, we ask the reader to verify that t rN = A t.
According to 6.11 we can write any Lorentz transformation as a product of
transformations of the form rN where N E E ® R2 with < 0.
Let us
observe that the locus
{xEEI =0} is a sphere or an affine hyperplane: Let us write N = (e,a,b); upon multiplying N by a non-zero scalar we may assume that b = 1 or b = 0.
Case N = (e,a,1).
Let us introduce the number r = -
and
observe that a = <e,e> - r2 to get the formula 2 = - 2<x,e> + (a + <x,x>) = Ix - e12 - r2 which shows that the locus is the circle with centre e and radius r.
Case N=(e,a,0). We have = - <e,e>, in particular e # 0, so we can multiply N by a constant to obtain that e is a unit vector. Choose u E E such that = 2a, and observe that
<x,e> + 2a =
which shows that the locus is the affine hyperplane through u with normal vector e. At this point it remains only to quote 6.14.
Corollary 7.6
A Mobius transformation a of the Euclidean space of
dimension n can be written as a product of at most n+2 inversions.
Proof
Any Lorentz transformation of the space E ® 082 can be written as a
product of at most n+2 reflections, compare 6.11.
Let us say that a Mobius transformation a of E is odd or even depending
on the sign of the determinant of the corresponding Lorentz transformation. An
1.7 MOBIUS TRANSFORMATIONS
35
even Mobius transformation can be expressed as a product of an even number of inversions, while an odd Mobius transformations is a product of an odd number of inversions.
Corollary 7.7
The restriction to E of the group of Mobius transformations
of k which fixes oo is the full group Siml(E) of Euclidean similarities of E.
Proof
Let us first check through three special cases. First observe that a A>0
gives rise to a Lorentz transformation (x,a,b)'--' (x,Aa,A-ib)
which induces the dilatation
of E. Next observe that a o E O(E) is induced
on E by the Lorentz transformation (x,a,b) '-4 (o(x),a,b) Let us finally observe that a vector v E E gives rise to the Lorentz transformation
(x,a,b) '- (x + by, a + b + 2<x,v>, b) which induces translation x F x + v on E. This shows that any similarity of E is a Mobius transformation. In order to prove the converse, we must show that these
three types of transformation generate the group LorL(E ® 082) consisting of Lorentz transformations which stabilise the line L generated by (0,1,0).
To this end we shall use the map (z,O -- (-z,2(,0) to identify E ®08 with the subspace L -L of E ® R2. This allows us to express the evaluation map from 1.5 in terms of the inner product on E ® 082.
7.8
ev((z,O,x) = <(-z,2(,0) , ex>
; (z,() E E (D 08, x E E
We ask the reader to complete the proof by means of 6.16 and 5.8.
0
DISCS Recall also that N = N(E ® 082) denotes the set of vectors in E ® 082 of norm -1. A vector U E N defines a sphere f = {x E E I = 0} and decomposes the complement of Y into the union of two connected open subsets
QUADRATIC FORMS
36
`b ={xEEI<0}
`.b={xEEI > 0},
which are called the discs bounding the sphere 1. The set 5 is called the disc with normal vector U.
Let us make a slight refinement of our earlier concepts by focusing on the connected component of the truncated isotropic cone C(E (D 022) made up of points whose inner product with (0,1,1) is positive:
7.9
C+(E e R2) = { (A,a,b) I - + ab = 0, a + b > 0 }
Observe that the map t: M
C factors through C+. It is useful to reinterpret the
projectivised cone as
PC(E ® R2) = C+(E e R2)/ R+
Corollary 7.10
; R+= ]0,+o0[
A Mobius transformation v E M'ob(E) maps a sphere Y
into a sphere v(Y). Inversion in the sphere v(1) is given by vov"1 where o is inversion in the sphere f.
Proof
Let us agree to denote by v the Lorentz transformation of E ® R2
corresponding to v by theorem 7.5. Let U be the normal vector for one of the discs ll bounding the sphere 1, we ask the reader to identify the set v(`.D) with the
disc with normal vector v(U). Observe that v = rU and use the general formula
rp(U) = vr0-1 to conclude that voi 1 is reflection in the sphere v(s).
Corollary 7.11
The Mobius group
acts transitively on the set of
discs in E.
Proof We can identify the action of M'ob(E) on the set of discs in E with the action of the Lorentz group Lor(E (DR2) on the space N(E (D R2) of vectors of norm
-1. Once this is done, the result follows from 6.16.
1.7 MOBIUS TRANSFORMATIONS
Corollary 7.12 Let 9G be
37
a sphere in E. A Mobius transformation o which
fixes 9G pointwise is either the identity or inversion in JG (dim(E) > 2) .
Proof Let K be
a normal vector for one of the discs `D bounding 9G and let k
denote inversion in the sphere 9G. From o(9G) = 9G we conclude that o(`D) = `D or
o(S) = tc(s). Upon replacing o by tco we may assume that o(`D) = `D or otherwise
expressed o(K) = K. Let us observe that restriction of v to F = K 1 acts trivially on the PC(F); it follows from 6.14 that the restriction of a to F is t, the identity. This gives O = t, which implies that o = t.
Definition 7.13
We say that two spheres I and T are orthogonal if the
normal vectors for the bounding discs are orthogonal5.
Proposition 7.14
Let o denote inversion in the sphere Y and r inversion
in the sphere 4T. If Y $ T then the following conditions are equivalent 10
1 and T are orthogonal
20
or = ro
30
r(Y) _ 1
40
o(C)_CT
Proof
It is a consequence of 7.10 that the last three conditions are equivalent.
To proceed, introduce normal vectors H and K for two bounding spheres. The correspondance 7.7 identifies the Mobius inversion r with the Lorentz reflection rK along the vector K.
From the formula
7K(H) = H + 2K
we conclude that rKK(H) = ± H if and only if = 0.
This shows that
condition 1° is equivalent to condition 3°.
5 The geometric meaning of this concept will be given in next section.
QUADRATIC FORMS
38
Proposition 7.15
Let A and B be points of t conjugated with respect to
the sphere I (by this we mean that A and B are points outside Y interchanged by inversion in Y). Any sphere through A and B is orthogonal to Y.
Proof
Let a generate the isotropic line in E ® R2 represented by A and let n
denote a normal vector for T. The vector b = a+ 2n generates the isotropic
Since A 0 Y we have $ 0 and the formula for b
line represented by B.
shows that n belongs to the plane spanned by a and b. It follows that the normal vector for any sphere through A and B is orthogonal to n.
Proposition 7.16
Let there be given an open disc `h in E. The group
Mob(9) of Mobius transformations of E which leaves `D invariant is generated by inversions in spheres orthogonal to Y = c99.
Proof
Let N denote the normal vector for `D.
A Mobius transformation
leaves `D invariant if and only if the corresponding Lorentz transformation
fixes
N. It follows that we may identify M"ob(`D) with the Lorentz group of N 1 . This
group is generated by reflections rI< where = -1 and = 0.
Corollary 7.17
Let L denote a linear hyperplane in E and `ll one of the
half-spaces in E bounded by H. Restriction from k to L defines an isomorphism Mob(9)
Proof
M"ob(L)
Specialise the proof of 7.16 to this case.
The composite of the inverse of 7.17 and the inclusion Mob(`D)-+M'ob(E) is known as Poincare extension : M'ob(L)- M'ob(E).
1.8 INVERSIVE PRODUCT OF SPHERES
1.8
39
INVERSIVE PRODUCT OF SPHERES
Let E denote a Euclidean vector space of dimension n > 2. We shall make
a closer study of the intersection of two spheres I and T in E. This will be done through a numerical character Y* J . Let us represent the spheres by their standard equations
8.1
Y:
- 2<x,f> + b<x,x> + a = 0
`T:
- 2<x,g> + d<x,x> + c = 0
; - + ab < 0 ; - + cd < 0
The inversive product of I and `T is given by
I -lad - !be - ab - cd
8.2
From the investigations made in the previous section it follows that the inversive product of I and T is preserved under Mobius transformations:
8.3
o(Y)*o(T) = oj*gj
;
o E M"ob(E)
The relative positions of two distinct spheres S and T can be read off T*`J
8.4
:
Y*T>1
Y*`T<1
Y*1f = 1
#(inT)>1
#(inT)=1 #(in-f)=o
Proof Let H and K be normal vectors for discs bounding Y and T. These two vectors span a plane R, and we can make the identification
InGT=PC(R1) The geometry is ruled by the discriminant
0 = - 2 = 1 - (Y* r)2 From the discriminant lemma 6.3 we deduce that
t*T < 1
:
0 < 0 , the plane R has Sylvester type (-2,0) which implies that
QUADRATIC FORMS
40
R
1 has Sylvester type (-n+l,l). Thus PC(R 1)
Y*°J = 1
R
:
consists of at least two points.
A = 1 , the plane R has Sylvester type (-1,0) which implies that
1 has Sylvester type (-n,0). Thus #PC(R 1) = 1.
Y*°J > 1
:
A < 0 , the plane R has Sylvester type (-1,0) which implies that
R 1 has Sylvester type (-n,0). Thus PC(R 1)
Proposition 8.5
= 0.
The Mobius group M''b(E) acts transitively on pairs of
spheres (Y,r) with given inversive products.
Proof This follows easily from Witt's theorem 2.4.
Let us investigate an ordinary point x E E of intersection of the spheres Y
and IT, x E Y fl of. To this end we pick a disc 9 bounding Y and let nx(`D) E E denote the outward pointing unit normal to Y at x. Similarly, let S be a disc boun-
ding T and let n.(S) E E denote the outward pointing unit normal to `J at x. We have that
8.6
Y*T = jj
We shall prove more precisely that the inner product of the normal vectors N('D),N(g) E N(E (D R2) is given by
8.7
Proof
Let us suppose that Y is an ordinary sphere in E and that `D is its
interior. In the equation 8.1 for Y we can take b = 1 which gives us that f is the
centre for Y, while the radius r is given by r2 = - a. Thus nx(`D) = r-1 (x - f)
,
N(s) = - r-1(f,a,1)
When `J is an ordinary sphere with radius s and centre g we can write
nx(S) = s '(x-g) , N(9) = -s't(g,c,1)
1.8 INVERSIVE PRODUCT OF SPHERES
41
where c = - s2. Direct calculation gives us
= r 1s1(- + za +11c) Then using 8.1 with b = 1 and d = 1 to eliminate a and c we get that
= r 1s 1(<x,f> - + <x,g> - <x,x>) = - When Gl is an affine hyperplane through x with nx(g) = n we have that nx(g) = n
,
N(S) = (n,2<x,n>,O)
Maintaining that I is an ordinary sphere as above, we get that
= r l( - + <x,n>) = - The case where both Y and T are affine hyperplanes is now obvious.
In the rest of this section we shall limit ourselves to the two-dimensional case.
First a rather amusing example due to Jacob Steiner.
Steiner's alternatives 8.8
Let A and % be circles in the Euclidean
plane of which °.B contains A in its interior. It is possible to find a kissing chain of
n circles each touching A on the outside and `.B on the inside, if and only if
A*% = 1 + 2 tan 2 n
Proof According to 8.5 we may assume that the circles A and `.B are concentric with radii a and b, a < b. The condition for the existence of a chain of kissing circles can easily be worked out by Euclidean trigonometry. The result is
sin 11= 1+B
0
'
-b
QUADRATIC FORMS
42
From this formula we deduce without difficulty that
a - 1-sin il 1+sin n
b
On the other hand we find by direct computation
.A*`B=2(b+a) A simple straightforward calculation concludes the proof.
Definition 8.9
By a
0
of circles in the Euclidean plane E we under-
stand a two-dimensional subspace R of E
R2. The actual circles in the pencil are
the circles in k with normal vector in R.
Proposition 8.10
Let R be a pencil of circles in E. The set of circles in
the pencil can be described as follows, depending on the Sylvester type
(-2,0)
:
circles through two given distinct points P and Q
(-1,0) : circles through a given point P orthogonal to a given circle N through P. (-1,1) circles which conjugate two given distinct points P and Q.
Proof Let us check through the three possible Sylvester types. Case (-2,0). The orthogonal space R 1 has type (-1,1) and contains precisely two isotropic lines P and Q say. We can recover R as the set of vectors orthogonal to P and Q. In geometric terms, a circle Y belongs to the pencil if and only if it passes through P and Q, compare the proof of 7.15. Case (-1,1).
The space R contains exactly two isotropic lines. Let us
observe that a circle I with normal vector N conjugates P and Q if and only if N E R. Alternatively, let us treat the case P = oo and Q = 0 directly. The circles in k which conjugate these two points are the ordinary circles in E with centre 0. Notice that through each point of t - {0,oo} passes a unique circle of this pencil.
1.8 INVERSIVE PRODUCT OF SPHERES
43
Case (-1,0). Let us observe that the dual pencil R 1 has the same type
and that R fl R 1 = P is an isotropic line. Pick a circle N in the pencil R
1
and
observe that the circles in R are precisely the circles through P, orthogonal to N.
Example 8.11
Consider the following families of circles in R2
x2+y2-b+2Ax=0 10
b > 0 . All circles pass through (0,4) and (0,-4g)
2°
b = 0 . The equation for the circle can be written
;AER
(x-A)2+y2=A2 which shows that all circles pass through (0,0) and are tangent to the y-axis at this point. 3°
b < 0 . Put b = - k2 and rewrite the equation as follows
(x-A)2+y2 = A2-k2 For A = k , the point (k,0) is the only solution, and similarly A = -k gives (-k,0). The system is, in fact, orthogonal to the pencil of circles through (k,0) and (-k,0).
QUADRATIC FORMS
44
1.9 RIEMANN SPHERE
Let us introduce the Riemann sphere t as the 1-point compactification of
of the complex plane C. This compactification can be realised in a number of ways, but we shall use a construction from projective geometry which displays the underlying algebra in a natural way.
Consider the scalar action of the complex multiplicative group C* on the complex vector space C x C, and let t denote the orbit space for the induced action
on the complement of the origin.
The equivalence class of a pair of complex
numbers (z,w) $ (0,0) is denoted [ w J. Thus za
wa
; aEC* 1=
The group G12(C) acts on the Riemann sphere in virtue of the formula
9.1
a
b
z
az+bw
c
d
w
cz+dw
;ad-bc#0
Let us remark that scalar matrices act trivially on C, thus we are dealing with an action of PG12(C) = G12(C)/C* on C.
In order to calculate the action of the
inverse matrix S-1 we can use the cofactor matrix S' given by
9.2
a
b
c
d
-
d
-b
-c
a
This is justified by the well known formula S S" = S" S = t del S.
Proposition 9.3
The action of the group PG12(C) on t is triply transi-
tive: for any two triples A,B,C and P,Q,R, each consisting of three distinct points, there is a unique v E PG12(C) which transforms the first triple into the second.
1.9 RIEMANN SPHERE
45
Proof Let the points A,B,C be represented by the vectors a, b, c in C x C. By assumption, the first two vectors are linearly independent so we can write
c=aA+bit
; A,pEC Since A and p are different from zero we can change a and b such that c = a + b. Quite similarly, we can represent the points P,Q,R by vectors p, q, p + q. The
with o(a) = p and
linear automorphism o of C x C
o(b) = q
induces the
required Mobius transformation.
It remains to investigate a linear automorphism r of C x C which induces
the identity on C. By assumption, all non-zero vectors of C x C are eigenvectors
0
for r. It follows easily that r is multiplication by a scalar it E C*.
Let us introduce the point at infinity oo = [ ] and observe that we can o
identify the complement C - {oo} with C once we identify the complex number z
Let us investigate the transformation o from 9.1 in this
with the point [ i l of C.
terminology. When o(oo) = oo we have c = 0. If c t_ 0 we find that o(oo) =
o l(oo) _ -d
To recapitulate 0 - ( _A )
9.4
Theorem 9.5 mations.
_
u(00)
00
0.(L) =
,
cz + b
and
zEC,z#-a
The group G12(C) acts on t as even Mobius transfor-
More precisely, this action induces an isomorphism of groups PG12(C)
Proof
c
M1ob+(C)
Let us first observe that the group of upper triangular matrices acts on
C fixing oo and induces the group of even similarities of C. In general, the matrix
identity (c # 0) a C
c
d
0
1
leads us to consider the transformation z-z 1 of C.
This transformation is the
46
QUADRATIC FORMS
composite of inversion in the unit circle and reflection z-
in the x-axis as follows
from the basic formula Z1 _ZIZI-2
;zEC*
This concludes the proof of the fact that G12(C) acts through even Mobius transformations. In order to conclude the proof, observe that PG12(C) acts transitively
on t and that the group of even Mobius transformations fixing oo induces the 0
group of even similarities on C, compare 7.7.
Cross ratio
We shall introduce a fundamental invariant of four distinct
points (P,Q,R,S) of the Riemann sphere C. To this end let us represent each of the four points by 2 x 1 matrices p, q, r, s and define the cross ratio as the point of C given by 9.6
del[p,r] del[q,s] 1 [P,Q,R,S] - rl det[p,s] det[q,r] ]
where the symbol [p,q] denotes the 2 x 2 matrix with first column p and second column q. For a Mobius transformation v we have
del [o(p),o(q)] = del(a) del [p,q] From this it follows that the cross ratio is invariant under Mobius transformations
9.7
[o-(P),a(Q),a(R),o(S)] = [P,Q,R,S]
; a E G12(C)
At this point let us remark that the cross ratio [P,Q,R,S] is well defined as long as
P,Q,R,S represent at least three distinct points of C. This allows us to fix three
distinct points P,Q,R of C and make a free variation of the fourth point S. The result is a Mobius transformation:
Proposition 9.8
Let P,Q,R denote three distinct points of C. The map S H [P,Q,R,S]
is an even Mobius transformation, which transforms P,Q,R into 00,0,1.
;
SEC
1.9 RIEMANN SPHERE
Proof
47
Let us rewrite formula 9.6 as follows [P,Q,R,S] =
-g2det[p,r] -p2det[q,r]
gldel[p,r]
pl det[q,r]
st I
S2
I
Observe that the 2 x 2 matrix has determinant det[p,r] det[q,r] det[p,q] $ 0. Direct
calculation reveals that this transformation maps P,Q,R into oo,0,1.
Proposition 9.8 in connection with 9.3 offers an implicit definition of the
cross product as the even Mobius transformation which maps P,Q,R into 00,0,1. In particular we find that
9.9
[00,0,1,S]=S
Proposition 9.10
; SEC
Two 4-touples of distinct points of the Riemann
sphere P,Q,R,S and X,Y,Z,W can be transformed into another by an even Mobius transformation if and only if [P,Q,R,S] = [X,Y,Z,W]
Proof It follows from 9.7 that the condition is necessary. Conversely, if the two cross ratios are identical, let o and r be the Mobius transformations given by
o(P,Q,R) = (oo,0,1) , According to formulas 9.7 and 9.9 we get that
r(X,Y,Z) = (c ,0,1)
[P,Q,R,S] = [oo,0,l,o(S)] = o(S)
and similarly, [X,Y,Z,W] = r(W). Thus the Mobius transformation 7-1a maps S to W as required.
Proposition 9.11
An odd Mobius transformation Q E Mob(C) satisfies
[o(P),o(Q),o(R),o(S)] = [P,Q,R,S] whenever P,Q,R,S represent four distinct points of C.
QUADRATIC FORMS
48
Proof
In order to generate the full Mobius group we need PG12(C) and one
single odd Mobius transformation, for example complex conjugation
[wz ,
F'[ w]
The formula is easily verified for complex conjugation.
Let us record two symmetry properties of the cross ratio. The verification is immediate and is left to the reader.
9.12
[P,Q,R,S] = [R,S,P,Q]
Proposition 9.13
,
[P,Q,R,S] = [Q,P,S,R]
Any a E PG12(C) is conjugated to a transformation
given by a matrix of the following form 1
1
0
1
;
Proof
a E C*
I
Let a be given by the matrix S E G12(C). Note that a non-zero vector
V E C X C is an eigenvector for S if and only if [v] E C is a fixed point for a. From
this we conclude that a $ e has one or two fixed points on C.
When a has two distinct fixed points A and B, let r be an even Mobius transformation with r(oo) = A and r(0) = B. The transformation 7--1 ar will fix 0 and oo. It follows that a representing matrix must be diagonal. After multipli-
cation by a complex number we get a matrix of the first type. When a has only one fixed point A, choose an auxiliary point C # A and let r be the even Mobius transformation given by
r(0) = C , r(1) = o(C) Observe that the transformation rlar fixes oo and maps 0 to 1. It follows that r(oo) = A ,
this transformation can be represented by a matrix of the form
1.9 RIEMANN SPHERE
49
a
1
0
1
; aEC* 0
But the matrix can only have one eigenvalue so we must have a = 1.
Let us introduce an important invariant of a matrix S E G12(C)
+2c-
9.14
(trS)2
; S E G12(C)
(let S
Observe that lr2 S depends only on the conjugacy class of S in PG12(C). In particular, we may introduce the invariant 1r2o of an even Mobius transformation o.
Proposition 9.15
Let o and r be even Mobius transformations, both
different from t. Then o is conjugated to r in PG12(C) if and only if tr2o = tr2r.
Proof
With the notation of 9.14 we have the following explicit formulas
trz
1
1
0
1
]=4
a
0
0
1
tr2
,
aEC-{0,1}
1=2+a+a-1
Notice, that the value of a+ a-' is unchanged if a is replaced by a-1. On the other hand we have the matrix formulas 0
-1
a
0
0
1
1
0
0
1
-1
0
The remaining details are left to the reader.
J=[
1
0
0
a
]
_
[
a-
0
0
1
0
QUADRATIC FORMS
50
I
EXERCISES
EXERCISE 1.1
Let E be a finite dimensional vector space equipped with a non-
singular quadratic form. 10
Let F be a non-singular subspace of E, show that E = F ® F 1 and that
(x,y)-(x,-y)
;xEF,yEF1
defines an isometry a E 0(E) with a2 = 1.
Conversely, let a E 0(E) be given with a2 = c. Show that E = E+ ®E_
2°
where E+ is the eigenspace for 1 and E_ is the eigenspace for -1. Hint: Use that
x = z(x - a(x)) + z(x + a(x)) ;xEE Given an isometry a E 0(E) with a2 = t. Show that the fixed point space
3°
E. for a is a non-singular subspace of E and that a(x) = -x for x E E01
1.
1° Let (E,R) and (F,S) be quadratic spaces. Show that the
EXERCISE 1.2 function
Q(e,f) = R(e) + S(f)
defines a quadratic form on the direct sum E Fg F of E and F.
; (e,f) E E ® F
The form Q on
E ®F is denoted (E,R) 1 (F,S)
Let (D,Q) be a quadratic form and E and F orthogonal subspaces of D such that D = E + F and E fl F = 0. Show that (D,Q) is isomorphic to the form 2°
(E,R)
I
(F,S) where R and S denote the restrictions of Q to E and F respectively.
EXERCISE 2.1
By a hyperbolic plane over k, we understand a quadratic form
(F,P) which has a basis consisting of two isotropic vectors e and f with <e,f> # 0. 1°
Show that all hyperbolic planes over k are isomorphic.
2°
Let (E,Q) be a non-singular quadratic form which contains an isotropic vector
e 36 0. Show that e is contained in a hyperbolic plane V C E.
Hint: Choose g E E with <e,g> = 1 and observe that f=2g-e is isotropic with <e,f> = 2. 3°
Under the assumptions of 2°, show that the equation Q(x) = a has a solution
xin EforallaEE.
I EXERCISES
51
EXERCISE 2.2
Let us consider a non-singular quadratic form (E,Q). We say
that a subspace F of E is isotropic if the restriction of Q to F is identically zero.
Given maximal isotropic subspaces C and D of E. Show there exists a o E 0(E) such that o(C) = D. Hint: Use Witt's theorem. 1°
2°
Show that an isotropic subspace F of E is contained in a maximal isotropic
subspace. 3°
Hint: Use Witt's theorem.
Let q denote the dimension of a maximal isotropic subspace of E. Show that
(E,Q) is isomorphic to a quadratic form of the type
Hi 1...1Hg1F with 111,...,Hq hyperbolic planes and F non-singular without isotropic vectors. EXERCISE 2.3
Let Q be a quadratic form on the vector space E, and let
OP(E) = {o E 0(E) I o(e) = e; e E E 1°
11
Show that OP(E) is a normal subgroup of O(E) and that it contains all
reflections along non-isotropic vectors. 2°
Given A E k*, show that OP(E) acts transitively on the set
{eEEI Q(e)=a} Hint:
Use the proof of proposition 2.3.
3°
Show that any o E 0P(E) can be
written as a product of at most 2 dimE orthogonal reflections. 4°
When dim E = 2 show that any or E OP(E) can be written as a product of
one or two orthogonal reflections.
EXERCISE 2.4
G=
1°
Consider the quadratic form on R 3 with Gram matrix G (below)
A=
R=
0
0
-1/s
-r
1
-r/s
-s
0
0
r,s E R
s#0
Show that any orthogonal reflection of R3 has a matrix of the form R
above. 20
Show that the matrix A represents an orthogonal transformation of R3
which can't be written as a product of two orthogonal reflections.
QUADRATIC FORMS
52
EXERCISE 3.1
1° Show that the determinant det is a quadratic form on M2(R)
and that its polarisation is given by
= 2 it AB`
; A,B E M2(R)
where the symbol B' denotes the cofactor matrix of B given by a
b
c
d
J=[
d
-b
-c
a
2°
Show that the Sylvester type of det on M2(OI) is (-2,2).
3°
Show that
=
2(
it A tr B - Ir AB )
EXERCISE 4.1 Let E denote a Euclidean space of dimension 3. 1°
Given two planes P and Q, show that there exists a plane R whose normal
vector is orthogonal to the normal vectors of P and Q. 2°
Show that an even isometry a E SO(E) can be written or = k,\ , where k
and ) are half-turns in lines K and L (reflections in K and L in the notation from Exercise 1.1). 3°
Show that any two half-turns are conjugated in SO(E).
EXERCISE 4.2
Show that S03(UR) is a simple group, in the sense that any
normal subgroup N 54 {t} equals S03(R).
Hint: Use the previous exercise to see
that it suffices to show that N contains a half-turn. Use the next exercise actually to find a half-turn in N. EXERCISE 4.3
Let E be a Euclidean vector space of dimension 3 and let
a E 50(E) be a rotation with angle > 7r/2. 1°
Show that there exists a line L through 0 such that L and a(L) are
orthogonal. Hint:
Observe that the angle between L and a(L) is a continuous
function of L. 2°
Let A be half-turn with respect to a line L with the property that L and
a(L) are orthogonal. Show that aAa A is a half-turn. Hint: Show that a(L) is
stable under a)o . with eigenvalue -1.
I EXERCISES EXERCISE 4.4
53
1°
Show that the Killing form <X,Y> = ir(X Ty)
; X,Y E MM(R)
is a symmetrical bilinear form on Mn(R) and that the corresponding quadratic form satisfies the Cauchy-Schwarz inequality
<X,Y>I < 1XI IYI 2°
; X,Y E
Show that the Killing form is invariant under 0,,,(R) in the sense that ; X E MM(OZ), g E OO(R) = <X,X>
EXERCISE 6.1
1° Show that Lor+(F) acts transitively on each of the sheets Hn
of the hyperboloid S(F).
2° Show that the special Lorentz group Lor+(F) is connected.
EXERCISE 6.2
Show that the orthogonal group 0(F) has four connected
components.
EXERCISE 7.1
Justify the classical construction by ruler and compass of
inversion B of the point A in the sphere with centre 0 and radius r
Hint: Note that SOAP is similar to AOPB and deduce that d(O,A)d(O,B) = r2. Alternatively, observe that the circle with diameter BP is orthogonal to the initial circle, and passes through the point A; compare 1.7.14 and 1.8.7.
QUADRATIC FORMS
54
EXERCISE 7.2 1°
Show that dilatation with centre 0 and ratio r>0 can be
realised as the composite of reflections in the Euclidean spheres with centre 0 and radius 1 and q-r-. 2°
Show that parallel translation in E can be written as a composite of
Euclidean reflections in two parallel affine planes.
EXERCISE 7.3 (Ptolemy's theorem) Let E be an Euclidean vector space of dimension n.
Show that n+2 points x1,...,x,,
of E lie on a sphere if and only if
2
detij d(x1,xj)2 = 0
where d denotes the Euclidean distance, compare 11.2.1. Hint: With the notation
of 1.7.4 show that = 2 d(x,y)2 for x,y E E.
EXERCISE 8.1
Let us consider Euclidean discs 1.t and 11 with centres u and v
and radii r and s. Show that v12
in -
niar -
-
r2
- s2
2rs
EXERCISE 8.2 Let us consider a square matrix with n+2 rows and n+2 columns
diagonal entries: -1
-1
1
1
1
1
-1
1
1
1
1
-1
1
1
1
1
-1
entries elsewhere: 1
1°
Show that this matrix defines a quadratic form of type (-n-1,1)
2°
Let E be a Euclidean n-space. Show that there exists n+2 spheres in E
which are two and two tangent to each other.
the above matrix as Gram matrix.
Hint: Pick a basis for E ® R2 with
I EXERCISES
55
EXERCISE 8.3 Let R denote a pencil of circles in the Euclidean plane E, and let G denote the group generated by reflections in the circles of the pencil. 1°
Show that the action of G on R identifies G with a subgroup of O(R), and
identify this subgroup depending on the type of the pencil. 2°
Show that a product a/y of three reflections in circles of the pencil is itself
a reflection in a circle from the pencil.
EXERCISE 9.110
Show that A E G12(C) with tr A = 0 defines a Mobius
transformation on t, which is an involution. 2°
Show that all involutions in Mob(C) can be represented in this way.
EXERCISE 9.2 Let P,Q,R denote three distinct points of C. Show that the point S E C belongs to the circle through P,Q,R if and only if [P,Q,R,S] E R.
EXERCISE 9.3 Let P,Q,R,S be four distinct points on a circle 9G. Show that [ P,Q,R,S] < 0
if and only if the pair {P,Q} separates the pair {R,S}; meaning that P and Q belongs to different connected components of 9G - {R,S}.
EXERCISE 9.4
By a Half-turn we understand an even Mobius transformation
o#t with o,2=t. 10
Show that a half-turn has precisely two fixed points on C.
2°
Show that there exists a unique half-turn fixing two given points of C.
3°
Let there be given four distinct points P,Q,R,S of C. Show that there
exists a unique o E PG12(C) which transposes P and Q and transposes R and S. In
the following this half-turn will be denoted (P,Q)(R,S). Observe that this symbol is still meaningful in case R = S. 4°
Show that an even Mobius transformation of C can be written as a
product of two half-turns; otherwise expressed, show that a given triple A,B,C of
points of t can be transformed into another given triple P,Q,R by a product of two half-turns. Hint: In case A 54 Q and B # P let D denote the image of C by (A,Q) o (B,P) and consider the transformation
QUADRATIC FORMS
56
(R,D)(P,Q) ° (A,Q)(B,P)
When A = P and B = Q, use the transformation (A,B)(R,R) o (A,B)(C,R). EXERCISE 9.5 1° Show that the cross ratio of four distinct points is a complex
number distinct from 0 and 1 which obeys the rules [P,Q,R,S] = [Q,P,R,S]-1 = [P,Q,S,R]-1
2°
Five distinct points P,Q,R,S,T of t satisfy the cancellation rule [P,Q,R,S] [P,Q,S,T] = [P,Q,R,T]
EXERCISE 9.6
Fix three distinct points P,Q,R of C and consider the rational
function A on C given by A(S) = [P,Q,R,S]. Show that permutations of P,Q,R A- 1 A 1- A 1 1 A, give the six rational functions
A-1
1,
EXERCISE 9.7
Let 0,7 E PG12(C) be transformations represented by matrices
S,T E G12(C). Show that tr[o,r] = lr STS-1T-1 is independent of S and T. 2°
Show that lr[o,r] = 2 if and only if a and r have a common fixed point
on the Riemann sphere C. Hint: Use the normal forms for S from exercise 9.1. 3°
Give an example of transformations o and r with lr[o,r] = - 2.
EXERCISE 9.8 For o E PG12(C) let Fix(o) denote the set of fixed points for o on C. Show that if r E PG12(C) commutes with or then o(Fix(r)) = r(Fix(o)). 2°
Show the converse of 1°. In particular, show that Fix(o) = Fix(r) implies
that o and r commute. 3°
Find the condition for two half-turns to commute.
EXERCISE 9.9
Show that the evaluation map PG12(C)
Cxtxt
; o ,--, (-(00),o(0),o(1))
is a homeomorphisrn of PG12(C) onto the open subset W of C x C x C consisting of
triples of distinct points of C. Hint. With the notation of 11.9.6 show that the following Mobius transformation maps (00,0,1) onto (P,Q,R) pl del[q,r]
-q1 det[Rr]
p2det[q,r]
-g2det[p,r]
II GEOMETRIES
We are now ready to introduce the three basic geometries: Euclidean, spherical and hyperbolic geometry. Using the algebra from the previous chapter
we shall work our way through the three geometries and identify their isometrics and geodesic' curves. In the case of hyperbolic geometry we present three "non-
linear models" : the Klein disc model, the Poincare disc model and the Poincare half-space model.
11.1 GEODESICS
Let us recall that a metric space is a set X equipped with a symmetrical function d: X x X - IR with the property that d(A,B) = 0 if and only if A = B and which satisfies the
Triangle inequality 1.1
d(A,C) < d(A,B) + d(B,C)
;
A,B,C E X
A map o,: X--+Y from one metric space to another is distance preserving if
1.2
d(o(A),o(B)) = d(A,B)
; A,B E X
It is clear from the definition that a distance preserving map is injective.
A
distance preserving map o: X--*Y which is surjective is called an isometry. Observe
that the inverse map of an isometry is itself an isometry. It follows that the set of
isometrics of a given metric space form a group. We shall say that metric spaces X and Y are isometric if there exists an isometry of X onto Y.
'The geometries we are going to study all carry the structure of metric space. The notion of a geodesic curve in a metric space, as introduced in section 11.1, is consistent with the notion of a geodesic curve in Riemannian geometry.
GEOMETRIES
58
The basic example of a metric space is the set R of real numbers equipped
with the standard metric d(x,y)
y - x 1, x,y E R. It will be important for us to
know all isometries of R.
Proposition 1.3 An isometry o:R - R is either a translation x
x+a or
a reflection x i-+ b-x.
Proof For a given isometry o of R, let r be the reflection or translation which agrees with o at 0 and 1. We are going to prove that o = r. Suppose for a moment that c E R is a point where r(c) # a(c): but this implies that the point a = 0(0) = r(0) is the midpoint of o(c) and r(c) since
d(o(c),a) = d(o(c),o(0)) = d(c,0) = d(r(c),r(0)) = d(r(c),a) Similarly, b = u(1) = r(l) is the midpoint of o(c) and r(c), contradicting a # b.
Definition 1.4
Let J C R be an interval and X a metric space. A curve
y:J-+X is said to be a geodesic curve if each point c E J has a neighbourhood U C J such that the restriction of y:U-'X is distance preserving.
Lemma 1.5
Let u be a point of the open interval J of R and y:J->R a geodesic curve with y(u)=0. We can find e = ± 1 such that
y(t)=e(t-u)
Proof
;tEJ
Let us choose an open neighbourhood U C J of a given point u such
that the restriction of y to U is distance preserving.
We conclude from the proof
of 1.3 that the restriction of y to U is an affine function. It follows that y is differentiable on U with constant derivative. Using that R is connected we conclude that y' is globally constant on J. At this point the result follows from elementary calculus.
0
11.2 EUCLIDEAN SPACE
59
11.2 EUCLIDEAN SPACE
Let E denote an n-dimensional Euclidean space. The scalar product of vectors e and f will be denoted < e,f > , while the length of a vector c is given by Icl = . The distance between two points A and B of E is defined by
d(A,B)=I A - BI
2.1
; A,BEE
This turns E into a metric space as follows from Cauchy-Schwarz, I.4.1.
Let us observe that a point v E E and a vector e of norm 1 defines a geodesic curve t '-, to + v, t E R, whose image is an affine line. Conversely
Proposition 2.2
Any geodesic curve y : R - E has the form
y(t) = et + v
;tEIR
where v is a point of E and e is a vector of norm 1.
Proof
Let us consider a point a E R and pick an open interval J around a such
that the restriction of y to J is distance preserving. It follows from the sharp triangle inequality 1.4.2 that for any three distinct parameters r,s,t E J the points y(r), y(s),y(t) lie on an affine line. If we fix two distinct points c,d E J we find that y(J) is contained in the affine line through y(c) and y(d). It follows from 1.5 that we can find a vector e of norm 1 such that
y(t) = e(t - a) + ry(a)
;tEJ
This implies that y is a differentiable curve whose derivative y'(t) is locally constant. Since IR is connected, we conclude that ye(t) is independent of t E R. It follows from elementary calculus that y has the required form.
0
Let us find all isometries of the Euclidean space E. A basic example is reflection r in an affine hyperplane H of E. In terms of a unit normal vector n to
GEOMETRIES
60
H and a point u E H, the action of r is given by
r(x)=x-2<x-u,n>n
2.3
;xEE
The isometry r fixes H while the image r(x) of a point x 0 H can be described by observing that the affine line through x and r(x) is perpendicular to H and that H intersects this line in the midpoint of x and r(x).
Lemma 2.4
Let there be given two sequences A1,...,Ap and B1,...,Bp of
points of E such that d(Ai,Ai) = d(Bi,B3)
;
i,j = 1,...,p
Then there exists an isometry a of E composed of at most p reflections such that cT(Ai) = Bi
Proof
;
i = 1,...,p
Let us recall that the bisecting hvnernlane for two distinct points A
and B of E is given by { X E E I d(X,A) = d(X,B) } It is easily seen that reflection in the bisecting hyperplane interchanges A and B.
We shall prove the lemma by induction on p. Observe that we have just
taken care of the case p = 1. To accomplish the induction step, suppose that an
isometry p composed of at most p-1 reflections has been found to transform AI,...,Ap-1 into BI,...,Bp-1. When p(Ap) = Bp we can take a = p. When p(Ap) $ Bp, observe that
i = 1,...,p-1 which means that the points BI,...,Bp_1 all lie on the hyperplane bisecting p(Ap) d(p(Ap),Bi) = d(p(Ap),p(Ai)) = d(Ap,Ai) = d(Bp,Bi)
;
and B. If r denotes reflection in this hyperplane then the isometry o = rp serves our purpose. 0
By a Euclidean simplex in the n-dimensional space E we understand a sequence A0,A1,...,An of points of E not contained in an affine hyperplane.
11.2 EUCLIDEAN SPACE
Lemma 2.5
61
Let there be given a Euclidean simplex A0,A1,...,An of E. If
two isometries a and ,3 of E agree on A0,...,A,, then a= 0.
Proof Put o = /34a and assume that P is a point of E where o(P)
P. From
; i = o,...,n d(o(P),Ai) = d(o(P),o(A1)) = d(P,Ai) it follows that the points A0,...,A. all belong to the affine hyperplane bisecting P
and o(P). This contradicts that the points A0,...,An form a simplex.
Corollary 2.6
Let o be an isometry of E which fixes an affine hyperplane K
pointwise. Then, either o is reflection in K or o = t.
Proof Let us assume that o $ t and pick P E E with o(P) # P. It follows from the proof of 2.5 that K equals the perpendicular bisector for o(P) and P.
Let us
form the composite of r, reflection in K, and o to get an isometry ro which fixes P and the points of K. We conclude from 2.5 that ro = t.
Theorem 2.7
Any isometry /3 of a Euclidean space of dimension n can be
written as the product of at most n+1 reflections.
Proof Let
us pick a Euclidean simplex A0,...,A. in E. We can use lemma 2.4
to pick an isometry o which is the composite of at most n+1 reflections with o(Ai) = /3(Ai)
;
i = 0,...,n
It follows from lemma 2.5 that o = 0.
We are now in a position to describe all isometries of Euclidean space. First of all a vector e E E defines translation re given by the formula
2.8
re(x)=e+x
;xEE
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62
The translations form a subgroup 7(E) of the group Isom(E) of all isometrics.
Lemma 2.9
Any isometry p of Euclidean space E can be written
;eEE,oEO(E)
p=Teo
The orthogonal transformation a is called the linearisation of p and is denoted p.
Proof
Let Isom'(E) denote the set of isometrics of E which can be written in
the form TeU as above. We ask the reader to verify the formula
;e EE, 0EO(E)
are °1=To.(e) We can now verify that Isom'(E) is closed under composition TV a Te P = TV TQ(e)
P = TV+O-(e) vP
Observe also that Isom'(E) contains all reflections in hyperplanes in E as follows
0
by inspection of the formula 2.3. Thus we get Isonz(E) = Isom'(E) from 2.7.
The linearisation pi-+p of an isometry plays an important role in Euclidean geometry. Let us notice some of the basic features. First of all
2.10
; 04 E Isom(E)
(0 i/i) = $ o i'
Observe that for 0 E Isom(E) we have i = c if and only if 0 is a translation. This fact can be recaptured by saying that the following sequence of groups is exact
2.11
0
-+
T(E)
Isom(E)
0(E)
-+
0
As a final general remark on the structure of Isom(E), we ask the reader to verify
2.12
. TV O 1 =
T.i(V)
;oEIsom(E),vEE
We shall classify isometrics of Euclidean space of dimension 2 and 3. discussion is based on the following innocent looking theorem.
The
11.2 EUCLIDEAN SPACE
63
Classification theorem 2.13 An isometry ik of Euclidean space E can be decomposed into a product of two commuting isometries
jG=TVOO =00Tv where TV is a translation along a vector v E E and 0 is an isometry with a fixed point. Such a decomposition of 0 is unique.
Proof
Let us decompose Vi = Tea as in 2.9.
The transformation 0, we are
looking for have a fixed point u, say, in E. This means that we are looking for a
From 2.12 we find that the condition for 0 to commute with a translation r, v E E , is that q5(v) = v or transformation of the form .0 = ruar-11, u E E.
a(v) = v. Thus we are looking for two vectors u and v in E such that Tea = T,,r crr Using orU 1 = T-au a
1
,
a(v) = v
this may be rewritten as
e = v + (u - au)
; v E Ker(a-t), u E E
Such a decomposition of e E E is always possible by the following lemma. Uniqueness of v comes from the same source.
Lemma 2.14
A transformation a E O(E) gives rise to an orthogonal
decomposition of E
E = Ker(a-t)®Im.(a-t)
Proof
Let us first show that the two subspaces Ker(a-t) and Im(a-t) of E
are orthogonal to each other
<x,a(y)-y> = <x,a(y)> - <x,y> = - <x,y> = 0 This gives us Ker(a - t) n Im (a - t) = 0 . From Grassmann's dimension formula applied to the linear map a-t we get that
dim Ker(o - t) + dim Im(a - t) = n The result follows from the dimension formula 1.3.4.
Let us return to the classification theorem 2.13. The fact that rv and 0 commute can be written in the form
GEOMETRIES
64
O(x+v)=O(x)+v
; xEE
This shows that the fixed point set for 0 is stabilised by the translation rv.
Let us apply the classification theorem to dimensions 2 and 3. As a consequence of our investigations of O(E) made in 1.4.8 and 1.4.9 we get
Euclidean plane 2.15 An even isometry of the plane is a rotation or a translation. An odd isometry is a fide reflection, i.e. an isometry rlc composed of
a reflection is in a line k and a translation r along k. The special case where r = t is a reflection.
Euclidean space 2.16
An even isometry is a screw composed of a
rotation and a translation along the axis of rotation. An odd isometry is either a
rotary reflection, composed of a reflection in a plane and a rotation with axis perpendicular to the same plane, or a fide reflection , composed of a reflection in a plane and a translation along that plane.
11.3 SPHERES
65
11.3 SPHERES
Let us consider a Euclidean space F of dimension n+l. We shall introduce a metric on its unit sphere Sn = S(F). To begin, observe that the Cauchy-Schwarz inequality 1.4.1 gives us ; P,Q E Sn E [-1,1] This allows us to define the spherical distance d(P,Q) between two points P and Q
of Sn through the formula
3.1
cos d(P,Q) =
; d(P,Q) E [0,a]
It is clear that the spherical distance d(P,Q) is symmetrical in P and Q. Let us show that two distinct points P and Q have non-zero distance. If P and Q are linearly independent, the Cauchy-Schwarz inequality 1.4.1 is sharp and we find
that E ]-1,1[. If P and Q are linearly dependent, we have P = - Q which gives = -1 and d(P,Q) = 7r. The triangle inequality will be proved in 3.6 after an introduction to spherical trigonometry.
Definition 3.2
By a tan ent vector to a point A E S" we understand a
vector T E F with = 0. tangent vector.
A tangent vector of norm 1 is called a unit
The space of tangent vectors to S" at A forms a linear hyperplane
TA(Sn) of F called the tangent space to the sphere Sn at A.
Lemma 3.3
Let A and B be points of the sphere Sn. A unit tangent vector
U to S" at A can be chosen such that B = A cos d(A,B) + U sin d(A,B)
Proof
Let us at first suppose that A and B are linearly independent.
Let U
be a unit tangent vector at A in the plane spanned by A and B, and let us write
GEOMETRIES
66
B =xA+yU
;x,yER
Using = 0 we get x2 + y2 = 1. Thus we can determine s such that B = A cos s + U sin s
;
sE[-7r,7r]
Notice that the formula is unchanged under the substitution (s,U)H(-s,-U), and conclude that we can write B as above with s E [0,7r]. Observe that
cos d(A,B) = = cos s and conclude that s = d(A,B). If A and B are linearly dependent we have B = A
or A = -B. It follows that sin d(A,B) = 0 and that any unit tangent vector U to Sn at A will serve our purpose.
Let us consider a spherical triangle AABC. By this we understand three points A,B,C of Sn linearly independent in the ambient space F. The following notation is standard
a = d(B,C)
b = d(A,C)
c = d(A,B)
Let us use lemma 3.3 to pick tangent vectors U and V at A such that
B=Acosc+Usinc
,
C=Acosb+Vsinb
We can use this to introduce LA = a as the angle between U and V
3.4
cos a =
; a E ]0,7r[
We are now in a position to prove the basic formula of spherical trigonometry.
111.3 SPHERES
67
Cosine relation 3.5 cos a = cos b cos c + sin b sin c cos a
1
Proof
We ask the reader to verify the determinant formula
= sin c sin b
Hint: Observe that the left hand side is zero if B is replaced by A or if C is
replaced by A. Evaluation of the determinants using 3.4 gives us 3.5.
Triangle inequality 3.6
Three points A,B,C in Sn satisfy
d(A,B) < d(A,C) + d(C,B) This inequality is sharp when A,B,C are linearly independent in F.
Proof When A,B,C are linearly independent we can use the cosine formula with cosa < 1 to get the inequality cosa < cos(c-b)
When c-b is positive we deduce that a > c-b which gives c < a+b. Observe that this inequality is trivial when c-b is negative. This concludes the proof of the sharp triangle inequality.
When A,B,C lie in a plane we can modify the
trigonometry to maintain the cosine formula 3.5 allowing for a = O,ir.
We can
deduce the triangle inequality from 3.5 using the inequality cosa < 1. An alternative proof of the triangle inequality including the sharp triangle inequality can be found in exercise 111.5
Let us observe that the metric we have introduced on the sphere Sn defines the topology on Sn induced from the ambient Euclidean space F as follows from
3.7
1 P- Q I =
2- 2 =
2- 2 cosd(P,Q)
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68
Proposition 3.8
can be written in the form
Any geodesic curve
; tER
y(t)=Acost+Tsint where A E S° and T is a unit tangent vector to Sn at A.
Proof
Let us investigate y in a neighbourhood of a point u E R. To this end
we pick an open interval J around u of length less than a and such that the restriction of y to J is distance preserving. It follows from 3.6 that for any three
distinct parameters r,s,t E J, the points y(r),y(s),y(t) are linearly dependent.
In
particular if we fix two distinct points c,d E J such that 7(c) and y(d) are linearly
independent, we find that q'(J) is contained in the plane R spanned by y(c) and 7(d). Put A = y(u) and pick a unit tangent vector T E TA(S") contained in R
and consider the curve
B(s)=Acoss+Tsins
;sEJ-7r,7r[
It follows by a simple direct calculation using 3.3 that 9 induces an isometry between ]-ir,a[ and the set of points X E R n Sn with d(A,X) < 7r. According to 1.5 we can find e = f 1 such that
-y(t) = A cos e(t-u) + T sin e(t-u)
;
tEJ
We conclude that 7 is continuously differentiable on J with y'(u) = eT , thus
3.9
y(t) = y(u) cos(t-u) + y'(u) sin(t-u)
;tEJ
Let us show that two geodesic curves o,,-f: R- Sn which coincide in a neighbourhood of 0 are identical.
To this end we introduce the set { t E R I o(t) = -y(t)
,
o'(t) = -y(t)
}
Observe that the set is closed since 7,0,7',o' are continuous. The set is open as well, as follows from formula 3.9. Since the set is non-empty, by assumption, we
get from the connectedness of R that y = o. Apply this to the given geodesic curve
7 and the geodesic curve o given by the right hand side of 3.9 with u = 0, and the result follows.
11.3 SPHERES
69
Isometries of the sphere
An orthogonal transformation a E O(F)
induces an isometry of the sphere S. We intend to show that any isometry of the sphere has this form; but first a lemma.
Lemma 3.10
Let there be given two sequences A1,...,Ap and B1,...,BP of
points of S" such that d(A1,Aj) = d(Bj,Bj)
;
i,j = 1,...,p
Then there exists an isometry v composed of at most p hyperplane reflections with o,(Ai) = B.
;
i = 1,...,p
Proof Consider two distinct points A and B of S", let us analyse the set { P E S" I d(A,P) = d(B,P) }
From the definition of the spherical distance it follows that this is the intersection
between S" and the linear hyperplane K orthogonal to the vector A - B. Orthogonal reflection in K will interchange A and B. We can now conclude the proof by the method used in the Euclidean case, see the proof of 2.4.
Theorem
3.11
Any isometry 0 of the sphere S" is induced by an
orthogonal transformation of the ambient Euclidean space.
Proof
Let us pick a spherical simplex in S", i.e. a sequence A0,...,An of points
of S" linearly independent in F. We can use lemma 3.10 to pick an isometry o which is the composite of at most n+1 reflections in hyperplanes with o(A;) = /3(Ai)
;
i = 0,...,n
We can conclude the proof by the method used in the Euclidean case, see 2.5.
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70
11.4 HYPERBOLIC SPACE
Let us consider a real vector space F of dimension n+1 equipped with a quadratic form of Sylvester type (-n,1). The hyperboloid S(F) consisting of all vectors of norm 1 has two sheets as we have seen in 1.6.3. There is no way in general to distinguish the two sheets of the hyperboloid, but let Hn denote any one
of them. We are going to introduce a metric on H".
The point of departure is the inequality > 1 obeyed by any two points P and Q of H", compare 1.6.2 and 1.6.4. We define the hyperbolic distance
d(P,Q) between P and Q to be the positive real number d(P,Q) such that
4.1
cosh d(P,Q) =
; P,Q E H"
It is clear that the hyperbolic distance d(P,Q) is symmetrical in P and Q. Let us
verify that d(P,Q) > 0 for P $ Q : observe that the vectors P and Q are linearly independent and conclude from 1.6.2 that > 1. Let us now turn to the triangle inequality.
Triangle inequality 4.2 Any three points A,B,C in H" satisfy d(A,B) < d(A,C)+ d(C,B) The triangle inequality is sharp in case A,B,C are linearly independent in F.
Proof
Consider points A,B,C on H" and put a = d(B,C), b = d(C,A) and
We shall calculate the determinant of the Gram-matrix 1.1.10 of A,B,C using the standard abbreviations cha = cosha and sha = sinha. c = d(A,B).
4.3
A = del
1
cha
chb
cha
1
chc
chb
chc
1
H.4 HYPERBOLIC SPACE
71
Direct expansion after the first row of -A gives us
A = (1 - ch2c) - cha(ch a - chb chc) + ch b(ch a chc - chb) _ 1 - ch2a - ch2b - ch2c + 2 cha chb chc =
(ch2b-1)(ch2c-1)-(cbb chc-cha)2sh2b sh2c- (chb chc
- cha)2 =
(cha- chbchc+shb shc)(chb chc+shb shc- cha) _ [cha - ch(b - c)] [ch(c + b) - ch(a)] =
4sh2(a+b-c) sh2(a+c-b) sh.'(a+b+c) sh2(c+b-a) This gives us the following factorisation of the discriminant
O = 4 sh p sh(p - a) sh.(p - b) sh(p - c)
4.4
; p=2(a+b+c)
Let us assume that the vectors A,B,C are linearly independent in the ambient space F.
It follows that they generate a subspace of F of type (-2,1) and we
conclude from 1.3.5 that A > 0. If c > a and c > b we have that
p-a=2(c-a)+2b>0, p-b=2(c-b)+2a>0 Using A > 0 and p > 0 we conclude from the factorisation 4.4 that p - c > 0. It follows that a + b - c = 2(p - c) = 0. If the vectors A,B,C are linearly dependent in the ambient space F, then A = 0. If A,B,C do not represent the same point then p $ 0 and we conclude from
the factorisation 4.4 that p - a = 0 , p - b = 0 or p - c = 0. The remaining details are left to the reader.
Definition 4.5
By a tangent vector T to a point A E H" we understand a
T E F with = 0. A tangent vector of norm -1 is called a unit tangent vector. The space of tangent vectors to H" at A forms a hyperplane of F of type (-n,0) which is called the tangent space to H" at A and is denoted TA(H").
The tangent space TA(I-I") inherits a quadratic form from the ambient space F. It is easily seen that the tangent space has Sylvester type (-n,0)
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72
Lemma 4.6 Let A and B be points of H". It is possible to choose a unit tangent vector U to H" at A such that B = A cosh d(A,B) + U sinh d(A,B)
Proof If A = B we can use any unit tangent vector U at A.
If A # B, the two
vectors are linearly independent and the plan R they span has Sylvester type (-1,1): according to the discriminant lemma 1.6.3 only three types are possible
(-1,1), (-1,0), (-2,0), but only the first type contains vectors of norm 1.
It
follows that we can pick a unit tangent vector U in R and write
B=xA+yU
;x,yER
Using = 0 we conclude that x2 - y2 = 1. Thus we can find s such that
B=Acosh s+Usin.hs
;sER
Notice, that the formula is unchanged under the substitution (U,s) H (-U,-s) and conclude that we can arrange for the formula to hold with s > 0. Observe that cosh d(A,B) = = cosh s and conclude that s = d(A,B) as required.
Geodesics
Starting from a point A E H" and a unit tangent vector T to H"
at A we can construct a curve 0: R - II" by the explicit formula
5(s) = A cosh s+ T sinh s
;sER
Let us calculate the hyperbolic distance between y(s) and y(t), s,t E R
cosh d(y(t),y(s)) = = cosh(t-s) = cosh It-si From this we conclude that
;s,tER d(y(t),y(s)) = I t - s This shows that 0: R - H" is a geodesic curve. The image of the curve is contained in the plane R spanned by A and T; in fact it follows from lemma 4.6 that the image of y is H" fl R.
11.4 HYPERBOLIC SPACE
Proposition 4.7
73
Any geodesic curve 7:R--+H" can be written in the form
y(t) = A cosh t + T sinh t
tER
;
where A E H" and T is a unit tangent vector to H" at A.
Proof
Let us fix a point u E R and pick an open interval J around u such that
the restriction of y to J is distance preserving. It follows from the sharp triangle inequality 4.2 that for any three distinct parameters r,s,t E J, the points
y(r),y(s),y(t) are linearly dependent. In particular if we fix two distinct points
c,d E J such that 7(c) and y(d) are linearly independent we find that O(J) is contained in the plane R spanned by y(c) and y(d). Put A = y(u) and pick a unit
tangent vector T E TA(H') contained
in
R. It
follows from the discussion
preceding 4.7 that we can use lemma 1.5 to find e = ± 1 such that
y(t) = A cosh c(t-u) + Tsinh E(t - u)
;tEJ
This implies that y is continuously differentiable on J with y'(u) = eT. Thus
y(t) = y(u) cosh(t-u) + yo(u) sinh(t-u)
;tEJ
The merits of this formula is to reconstruct y in a neighbourhood of u E R from
the values of 7(u) and yo(u). We ask the reader to conclude the proof by the method used in the proof of 3.8.
In recapitulation, a geodesic curve y : R - H" is distance preserving and not merely locally distance preserving. By a geodesic line or hyperbolic line in H"
we understand the image of a geodesic curve y: R -+ H".
We have seen that
through two distinct points of H" there passes a unique geodesic line.
Isometries A Lorentz transformation o E Lor(F) preserves the two sheets of the unit hyperbola S(F), in fact it induces an isometry of hyperbolic space H".
We intend to show that this procedure induces an isomorphism between the Lorentz group and the group of isometrics of H".
GEOMETRIES
74
Lemma 4.8
Let there be given two sequences A1,...,Ap and B1,...,Bp of
points of Hn such that d(A;,Ai) = d(Bi,Bj)
; i,j = 1,...,p
Then there exists an isometry a composed of at most p Lorentz reflections with o-(Ai) = B;
Proof
; i = 1,...,p
Consider two distinct points A and B of Hn and let us analyse the
perpendicular bisector for A and B { P E HIn I d(A,P) = d(B,P) }
From the definition of the hyperbolic distance it follows that this is the intersection between H' and the linear hyperplane orthogonal to the vector N = A - B. We can use 1.6.2 and 1.6.4 to conclude that
=2-2 <0 Reflection along N will interchange A and B. We can now conclude the proof by the method used in the proof of 2.4.
Theorem 4.9
0
A. isometry 13 of the hyperbolic space Hn is induced by a
Lorentz transformation of the ambient space F.
Proof
Let us pick a hyperbolic simplex, i.e. a sequence A0,...,An of points of
H" linearly independent in F. We can use lemina 4.8 to pick an isometry a which is the composite of at most n+1 Lorentz reflections such that a(A1) = 13(A;)
It follows that a = /3, compare the proof of 2.5.
Definition 4.10
; i = 0,...,n
0
A. isometry a of the hyperbolic space Hn is called even
or odd if the corresponding Lorentz transformation is even or odd.
11.5 KLEIN DISC
75
11.5 KLEIN DISC
In this section we shall construct a model of hyperbolic n-space based on the open unit disc D" of a Euclidean space E of dimension n. To begin we observe
that the auxiliary space E ® R carries a quite natural quadratic form of Sylvester type (-n,l): the corresponding bilinear form is given by 5.1
<(A,a),(B,b)> = - + ab
; A,B E E, a,b E R
For hyperbolic space H" we take the sheet of the unit hyperbola in E ® R consisting of points (A,a) with - +a 2 = 1 and a > 0. We shall make use of the parametrisation p:D"-+H" given by the formula
5.2
P(A)
-
(A,1)
;AED"
1 -
Geometrically, the point p(A) is the intersection of H" and the line through (A,1)
and 0. The Klein model of hyperbolic n-space is D" equipped with the metric obtained by transporting the hyperbolic metric along p: D" - H". It follows from 5.2 and 4.1 that the metric in the Klein model is given by
5.3
coshd(A,B) =
I - 4 (1 - )(l - )
;A,BED"
The geodesic lines in the Klein model are precisely traces on D of ordinary affine lines in E. This can be seen from the description of the geodesic lines H" given in 4.7 and the geometric origin of the parametrisation p:D"
H".
The boundary 8H" (in the sense of 6.18) can be parametrised by the Euclidean boundary of OD" by assigning to a point A E 8D" the isotropic line in E ®R through the point (A,1).
GEOMETRIES
76
11.6 POINCARE DISC
Let us once again start with an n-dimensional Euclidean space E and its open unit disc D. We shall make use of the auxiliary space E ® R equipped with the hyperbolic form given in 5.1 and the sheet Hn of the unit hyperboloid in E ® R
passing through (0,1). This time we shall consider the parametrisation f: Dn - Hn given by the formula
6.1
f(P) =
(2P,1 + )
;PEDn
1-
From the geometric point of view, this is the stereographic projection with centre
(0,-1) already considered in
1.6.6.
The result of transporting the hyperbolic
metric along f is given by
6.2
cosh d(P,Q) = 1 + 2
,P,QED"
1P2 Q12
(1-IP))(1-1Q7 2
The open unit disc Dn equipped with this metric is called the Poincare disc.
Theorem 6.3
The group M'o6(Dn) of A18bius transformations of E which
leaves Dn invariant acts as the group of isometries of the Poincare disc Dn. In particular, any isometry of Dn extends to a Nl6bius transformation of E.
Proof Let us first record from 1.6.5 that f-1:Hn-*Dn is given by the formula f-1(A,a) = A(l+a)-1
; (A,a) E Hn
Let r denote reflection along a vector (N,n) in E 9 R of norm -1. Let us transform r to Dn. We ask the reader to verify the formula
f-'7-f(P) - P + (n + n - 2) N I N-nP 12
; PEDn
When n = 0, we have = 1 and f-1rf is Euclidean reflection along N.
II.6 POINCARE DISC
When n 10, we have
f1 ' rf(P) _
1
77
+n2 = and f-1rf is given by the formula
P+(InP-N12-1)Nii1 nP-N 12
Nri +ii 2 P-Nn1 1
=
1 P-Nn-1 12
; P E Dn
This shows that f-1rf is induced by inversion in the Euclidean sphere N with centre Nn-1 and radius n-1. Let us observe that (n-1 )2+ 1 =
which shows that X is orthogonal to Sr-1 =
DDn.
The result follows by combining
1.7.16 with 11.4.9 and 1.6.11.
Lemma 6.4
An isometry 0 of the Poincare disc Dn C E is induced by an
orthogonal transformation of E if and only if 6(0) = 0.
Proof
An isometry of D can be represented by a Mobius transformation 0 of
E which commutes with reflection o in the unit sphere, 1.7.15. When ¢(0) = 0, let
us evaluate the formula oO = 0o at oo to get o(O(oo)) = 0(0) = 0 or 0(oo) = oo. It follows from 1.7.7 that 0 is a Euclidean similarity of E. Using once again that 0(0) = 0 and ¢(D) = D, we conclude that 0 is orthogonal.
Proposition 6.5 An isometry
0 of the Poincare disc D° C E can be written
pp, where p E 0(E) and p is reflection in a sphere orthogonal to
Proof
Sn"1
= DD'.
Let us focus on the point P = 0(0) E Dn. If P = 0, the result follows
from 6.4. If P $ 0, put R = o(P) where a is inversion in the sphere Sn"1. Let
be a circle with centre R orthogonal to Sn"1. We ask the reader to verify that * Sn 1 = 2i. I p(0) - o(R) I
Observe that o(R) = P and conclude that p(O) = P or p(P) = 0. It follows that po fixes 0 and we conclude from the previous lemma that pO E 0(E).
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78
11.7 POINCARE HALF-SPACE
Let us start from a Euclidean vector space L of dimension n-I and form the Euclidean space E = L ® R. We shall be concerned with the inner half-space E+ consisting of points (p,h) of E with h > 0. Inversion o in a sphere with radius and centre in the south pole S = (0,-1) will map E+ onto the unit disc Dn
S
Observe that o induces a stereographic projection of 8Dn onto the boundary L of the upper half-space. The explicit formula 1.7.1 for o:E+ _a Dn may be written
7.1
(p , I+h) o(p,h) = (0,-1) + 2 IP12 + (I +h)2
;pEL,heR
Let us use o to transport the metric of the Poincare disc onto the upper half-space E+. Explicit calculation based on 6.2 and 7.1 gives us 7.2
coshd(P,Q) = 1+
1P
-Q
ph
2
P = (p,h), Q = (q,k)
The open half-space E+ equipped with this metric is called the Poincare half-space.
Theorem 7.3
The group Mob(E+) of Mobius transformations of E which
leaves E+ invariant acts as the group of isometries of the Poincare half-space. In particular, any isometry of E+ can be extended to a Mobius transformation of E.
11.7 POINCARE 11ALF-SPACE
Proof
79
Let v denote the inversion considered in 7.1. Conjugation by o in
Mbb(E) transforms the subgroup Mob(Dn) into the subgroup Mbb(E+). This allows
us to transform theorem 6.3 into 7.3.
Corollary 7.4
0
The group of isometries of the Poincare half-space E+ is
isomorphic to the full Mobius group of L = 8E+.
Proof This results from a simple combination of 1.7.16 and 11.7.3.
The geodesics in the Poincare half-space E+ are traced by circles in E orthogonal to the boundary aE+.
GEOMETRIES
80
11.8 POINCARE HALF-PLANE
The open upper half-plane H2 of C is a model of for the hyperbolic plane when we specify the metric according to 7.2
_
8.1
12
; z,wEH2
coshd(z,w) = 1 + 2 Im[w] II [z]
For the convenience of the reader we shall give two different formulas
8.2
sinhd(z,w) _ z
Iw-zl
Im[w] Im[z] '
2
cosh d(z , w) _ 2
I w -Y I Im[w] Im[z]
In order to describe the isometries of H2 recall from 1.9 that G12(R) acts on C-R through the formula a
b
c
d
az+b cz + d
We ask the reader to work out the following useful expression for the imaginary part of this fraction. 8.3
IM[ az + b
_
Im[z]
ra d] bl
cz+d] -Icz+d12 det Lc
c a] E G12(R)
This leads us to define the following action of G12(R) on H2
8.4
az + b cz + d
deter>0
cz + d
deto, <0
a+b
0- = Lc
db]
Proposition 8.5 The action of G12(R) on H2 identifies PG12(R) and the group Isom(H2) of isometries of H2.
11.8 POINCARIs IIALF-PLANE
81
Proof
The formula 8.4 in fact defines an action of G12(R) on t, which preserves H2. By 7.3 it suffices to prove that any Mobius transformation of C which preserves H2 has this form.
By Poincare extension 1.7.16 it suffices to
identify PG12(R) and the Mobius group of the subspace R of C. This can be done
by modifying the construction in 1.9 to bring PG12(I8) to act on I8 in a triply transitive manner.
Proposition 8.6
The geodesics in H2 are traced by circles in t orthogonal
to A. Expressed in another way, the geodesics are traced by Euclidean circles with centres on the x-axis R and Euclidean lines perpendicular to the x-axis.
Proof The curve y: R- H2 given by y(s) = ies, s E R, cosh (s - t) =2t(e8-t + et-sl =
e2t,+e2s
(e
2etes
SZ
e) 2etes
t
_1+
is distance preserving: =cosh d(ies,iet)
It follows that dJ, the positive y-axis, is a geodesic. Let us observe that PG12(R) acts transitively on the set of circles in C orthogonal to A as follows from the fact that PG12(R) acts transitively on R. The action of PG12(R) on the set of geodesics in H2 is transitive as well as follows from 8.5 and 4.9.
Let us amplify the fact that the isometries of H2 are induced by Mobius transformations by asking the reader to verify that the distance can be written 8.7
cosh d(z,w) = I + 2
1
[z,`-V
',z]
I
; z,w E H2, z 0 w
Alternatively, let us introduce the circle L through z and w orthogonal to h and let z* and w* denote the points of intersection between the circle 3. and h, defined such that z,w and z*,w* give the same orientation to the geodesic Hfl1..
GEOMETRIES
82
w
z
w*
z*
For the distance between the points z and w we have
8.8
d(z,w) = log [z,w,w*,z*]
; z,w E H2, z $ w
Proof Upon performing a Mobius transformation we may assume that z and w are points of tlJ with Im[z] < Im[w]. This gives z* = 0 and w* = oo and
[z,w,x,0]
- zw
Let us write z = ies and w = iet (s < t) and conclude from the proof of 8.6 that d(z,w) = t - s. Let us put this together log [z,w,oo,0] = log z = log et'S = t - s = d(z,w)
0
and the formula follows.
Let us take a closer look at the action of a given o E S12(R) on H2 using
the theory of pencils of circles developed in I.B.
We shall make use of the
invariant tr2o = (lro)2. The discussion will be resumed in III.4 and IV.2.
Following Felix Klein we shall divide these transformations into three classes depending on the invariant lr2o.
Elliptic (rotation): lr2o < 4. The transformation o has a fixed point A E H2 and fixes the pencil of circles orthogonal to the pencil of geodesics through A. A circle from this pencil which is contained in H2 is called a hvnerbolic circle with centre A. Hyperbolic circles with centre A can be characterised as the orbits for the group of elliptic isometries with fixed point A.
11.8 POINCARE HALF-PLANE
Hyperbolic (translation) tr2o
83
The isometry a has two distinct fixed points A and B on OH2 and leaves the geodesic h with ends A and B invariant. > 4.
The pencil of circles through A and B is left invariant by o. The trace on H2 of a circle from this pencil is called a hvnercycle for the geodesic h through A and B. The hypercycles can be identified with the orbits of the group of even isometries which leave A and B invariant.
A
Parabolic (horolation) ir2v = 4. The transformation a has a unique fixed point P on OH2. The pencil of circles through P and tangent to the x-axis is
invariant under c. A circle from this pencil which is contained in H2 is called a horocycle with centre Q. The horocycles can be characterised as the orbits for the group of parabolic transformations with fixed point P E 8H2.
GEOMETRIES
84
II EXERCISES
EXERCISE 1.1 Let X and Y be metric spaces. For points P = (x,y) and Q = ({,rl)
on X x Y let us define the distance by d(yrl)2
d(P,Q) = Show that this defines a metric on X x Y.
EXERCISE 2.1 Give a direct proof of the fact that an isometry f of Euclidean space E with f(O) = 0 is linear. Hint: Use polarisation to show that
= <x,y>
; x,y E E
Conclude from this that vectors of the form below have norm zero
f(x+y) - f(x) - f(y) EXERCISE 2.2
,
f(rx) - rf(x)
; x,y E E, r E R
Let o and p be rotations of the Euclidean plane with distinct
centres A and B. 10
Justify the following procedure for the calculation of pa-:
Write p = 07
and o = ya where y is reflection in the line through A and B and Q is reflection in a line b through A and a is reflection in a line a through B. If a and b are parallel
we find that pa. = 6a is a translation. If a and b intersect in C we find that po = ,3a is a rotation with centre C. 2°
Suppose that p and a are rotations with the same angle -27r/3 but
distinct centres A and B.
Show that pa is a rotation with angle 27r/3 and that its
centre C forms an equilateral triangle with A and B. 30
Let DABC be a triangle in the Euclidean plane.
in the plane such that
Let A* denote the point
ABC is equilateral and such A and A* lie on opposite
sides of the line through B and C.
We let BA denote the rotation about the
circumcentre OA of DA*BC which transforms B to C.
mations BC and BB and show that
8C BB BA = c.
Define similar transfor-
Hint: Show that 0C BB BA is
rotation with centre B and angle 0. 4°
With the notation above, show that the circumcentres OA, OB and Oc
form an equilateral triangle. Hint: Use the method from 1° to calculate 0B BA.
II EXERCISES
85
EXERCISE 3.1
Given two lines k and I in F through the origin. Let us define
the acute angle L(k,l) E [0,2] between k and I by the formula cos L(k,l) = II where k and I are unit vectors generating k and I. 10
Show that the acute angle defines a metric on projective space P(F), the
set of lines in F.
Hint: Interpret P(F) as the orbit space for the action of the
antipodal map on the sphere S(F). Show that through two distinct points of P(F) passes a unique geodesic.
2°
EXERCISE 3.2
Let S2 denote the unit sphere in R3 and let the standard
spherical coordinates o': l 2
S2 be given by
o(9,0) = (silt 0 Cos 0, Sill 0 sill 0, Cos 0)
Show that a continuously differentiable curve y: [a,b]
1°
S2 which does not
pass through the north and the south pole can be written y = a(0,0) 0,0:[a,b] 2°
,
where
R are continuously differentiable functions.
With the notation above show that the length of the curve y is given by
l(7)= Jab 4 02 + siln2 0 ¢2 dt and deduce from this that 1(y) ? 10(b) - 9(a) I 3°
Let us consider the colatitude 0 as a continuous function
0:S2-*[O,a].
Show that a continuously differentiable curve y:[a,b]_S2 satisfies the inequality
1(y) ? I0(y(b)) - 0(7(a))I Hint: If 0 < 0(y(a)) < 9(y(b)) < it, let c E [a,b] be the last time 0(7(t)) takes the
value 0(y(a)) and d E [a,b] the first time 0(y(t)) takes the value 0(y(b)) and apply the inequality from 2° to the restriction of y to the subinterval [c,d]. 4°
Let P, Q be points on the sphere S2. Show that the spherical distance is
d(P,Q) = infy 1(7) where y runs through the set of piecewise continuously differentiable curves y on S2 connecting P with Q.
I-Iint: Arrange for P and Q to have the same longitude.
GEOMETRIES
86
EXERCISE 3.3
In R3 let the standard cylinder be given by Cy1 = { (x,y,z)
I
2 2 x +y =1}
For p,q E Cyl define the distance d(p,q) to be d(p,q) = infy 1(y)
where y runs through the set of piecewise continuously differentiable curves on the cylinder Cyl. 10
Show that the distance defined above is a metric on Cyl.
2°
Recall that the standard cylinder coordinates tc:R2-> Cyl are given by rc(O,r) = (cos 0, sin 0, r)
Show that a continuously differentiable curve y:[a,b]-+Cyl can be factored y=4cp where µ:0t-*R2 is a continuously differentiable curve. 3°
With the notation above show that the length of the curve y on Cyl is
4°
Show that the distance between points p and q on Cyl is given by
1(y) = 1(µ)
inf
d(p,q) =
d(P,Q)
P,Q; p=ti(P),q= c(Q) 5°
Show that
X:022_Cy1
is a local isometry, or more precisely that there
exists a constant k>0 such that for every P E R2 the disc with centre P and radius k is mapped bijectively onto the metric disc with centre p = ic(P) and radius k. 6°
Show that an affine line in R2 is mapped by n: onto a geodesic in Cyl and
that all geodesics on Cyl have this form. 7°
Let there be given points p and q on Cyl with d(p,q) = 1. Show that we can
find a geodesic curve y:R-Cyl with y(0) = p and 7(l) = q. EXERCISE 3.4
We shall be concerned with the torus Si x S1 considered as
metric space as described in exercise 1.1. The canonical map X:R--+S1 induces a map v: R2 1°
S1 X S1,
; x,Y E R
For points p and q on the torus S1 x S1, show that
d(p,q) = 2°
v(x,Y) = (1(x),X(Y))
inf d(P,Q) P,Q; p=v(P), q=v(Q)
Show that v:R2_S1 x S1 is a local isometry, or more precisely that there
II EXERCISES
87
exists a constant k > 0 such that for all P E R 2 the disc with centre P and radius k
is mapped bijectively onto the metric disc with centre p = v(P) and radius k. 3°
Show that an affine line in
R2
is mapped by v onto a geodesic in S1 x S1,
and that all geodesics on S1 x S1 have this form. 4°
Let there be given points p and q on S1 x S' with d(p,q) = d. Show that
we can find a geodesic curve y:R->S' x S' with y(0) = p and y(d) = q. For real numbers a,b,c consider the determinant g given by
EXERCISE 3.5
g = det
1°
1
cosa
cosb
cosa
1
cosc
cosb
cosc
1
Show that
g = 4 sin(7r-p) sin(p-a) sin(p-b) sin(p-c) 2°
p = 2(a+b+c)
For real numbers a, b, c E ]O,ir[ show that g > 0 if and only if a < b + c,
b
Show that the inequality g > 0 occurs precisely when the remaining four
inequalities all are sharp. 4°
Use this to give an alternative proof of the triangle inequality 11.3.6
EXERCISE 5.1
For d E D', d # 0, let d* denote the mirror image of d in the
unit sphere Si-1 and let
'cd
denote reflection in the circle with centre d*
orthogonal to Sr"1. Let Pd denote Euclidean reflection in the hyperplane orthogonal
to d and put rd = t'd Pd1°
Extend the definition of the symbol rd to all of Dn by the convention
ro = t , and show that any µ E Mob(D") can in a unique way be written
p=rdo 20
;dEDn,aEO(E)
Use the previous result to establish a homeomorphism Mob(Dn) -Z D' x O(E)
The topology on Mob(Dn) is uniform convergence on Sn-1. (See [Beardon] §3.7.)
EXERCISE 8.1
Show that a E G12(R) acts on H2 as reflection in a geodesic if
and onlyifdeia
III HYPERBOLIC PLANE
In this chapter we shall make a detailed study of the hyperbolic plane H2: classification of isometries and trigonometry. The even isometries of H2 can be
divided into three classes: rotations, translations and horolations. This division follows the natural occurrence of geodesics in three types of pencils.
We shall
keep track of the pencils by assigning a normal vector to an oriented geodesic. This will be dealt with in a specific model for H2 referred to as the s12(R)-model. The big advantage of the s12(R)-model over say the Poincare models is a very rich vector calculus 1.
111.1 VECTOR CALCULUS
In this section we shall present a very useful three-dimensional real vector
space with an inner product of Sylvester type (-2,1) with a good deal more The underlying real vector space is s12(R), the space of real
algebraic structure.
2 x 2 matrices with trace 0 s12(R) = { R E M2(Id) I it R = 0 }
The square of a matrix R E s12(P) is a scalar matrix, in fact
R2=-cdetR
1.1
;
REs12(R)
In view of the importance we offer direct verification of 1.1
a
b
C
-a
11
a
b
c
-a
0
_ (a2+ be) 0
1
1 An interpretation of the vector calculus on s12(R) can be given in terms of Clifford algebra. See the remarks at the end of section 111.1.
111.1 VECTOR CALCULUS
89
We shall be concerned with the symmetrical bilinear form
1.2
= - 2 tr(RS)
R,S E s12(R)
;
The corresponding quadratic form is the determinant as follows from 1.1
del R =
1.3
; RE s12(R)
Otherwise expressed, the inner product 1.2 is obtained by polarisation of the determinant, compare 1.1.2. Let us use this to "polarise" the identity 1.1
(R + S)2 - R2 - S2 = - 2 t An elementary calculation gives us the formula
RS + SR = -2 t
1.4
; R,S E s12(R)
It is time to show that the Sylvester type of our inner product is (-2,1).
To this
end we ask the reader to verify that the following three matrices form an orthonormal basis for s12(R)
I The determinants of these matrices are -1, -1, +1.
It is now time to introduce an important tri-linear form vol on s12(R). The evaluation of the form vol on three vectors K,L,M is given by
1.5
vol(K,L,M) = -1 tr(KLM)
;
K,L,M E s12(R)
This is actually an alternating form, in the sense that the form annihilates any triple where two of the vectors are equal: for example if K = M we get from
K2 = - t del K tr(KLK) = tr(LK2) = - tr(L det K) = - del K lr L = 0
HYPERBOLIC PLANE
90
The alternating form 1.5 is a volume form for s12(R) in the sense that it takes the value 1 or -1 whenever K,L,M form an orthonormal basis. To see this, we ask the reader to verify that vol has evaluation 1 against the basis above.
Let us introduce a third opeation on s12(R), the wedge product of two vectors K and L from s12(IR), given by the formula
(KL - LK)
KAL=
1.6
Z
;
K,L E S12(R)
The three elements of structure are related through the formula = vol(K,L,M)
1.7
; K,L,M E s12(R)
as follows from straightforward computation:
!r(KLM - LKM) = 2(vol(K,L,M) - vol(L,K,M)) = vol(K,L,M)
4
We shall need one more formula relating the inner product and the wedge product
1.8
A A (B A C) = B - C
; A,B,C E S12(R)
This formula can be deduced directly from the definitions using formula 1.4:
4(B - C) = - (AC + CA) B - B (AC + CA) + (AB + BA) C + C (AB + BA)
_
ABC - ACB + CBA - BCA = 4 A A (B A C)
The inner product of two wedge products can be evaluated by the formula
1.9
= - ; A,B,C,D E sl2(Id)
Let us verify this formula. From formula 1.8 we deduce that
= vol(A,B,CAD) = vol(B,CAD,A) = We find from 1.8 that B A (CAD) = C - D and 1.9 follows.
III.1 VECTOR CALCULUS
91
Let us present a very useful volume formula.
For two triples of vectors
A1,A2,A3 and B1,B2,B3 in s12(R) we have
1.10
vol(A1,A2,A3) vol(B1,B2,B3) = detij
Proof
Let us first fix A1,A2,A3. Both sides of the formula are alternating in
B11B21B3 so we may assume that B1,B2,B3 is a fixed positively oriented orthonormal basis.
We can make a variation of A1,A2,A3 and observe, that both
sides are alternating in the A1,A2,A3. Thus it suffices to treat the case where the A1,A2,A3 and B1,B2,B3 form the same orthonormal basis.
In this case the result
follows from the fact that s12(R) has Sylvester type (-2,1).
Clifford algebra 1.11
Let us add some remarks for readers familiar
with the formalism of Clifford algebra.
For reference see [Deheuvels] or the
opening chapter of [Gilbert,Murray], but it is our personal opinion that the arguments of Clifford algebra are best learned through concrete examples.
The real algebra M2(C) is easily seen to be generated by the linear subspace isl2(R). The square of an element X of this subspace is a real scalar. In fact X2 is a real quadratic form of X E isl2(OR) as follows from 1.1:
(iS)2 = det S
;
S E s12(R)
Observe that dimR M2(C) = 23 and conclude from general principles that M2(C) is
a concrete realisation of the Clifford algebra of the quadratic space s12(l) of Sylvester type (-2,1). A good many of the definitions and arguments of this section conform with the general notions of Clifford algebra. This is particularly
true for the definition of the wedge product 1.6 and the arguments presented in section 111.4.
IIYPERBOLIC PLANE
92
111.2 PENCILS OF GEODESICS
Let us introduce hyperbolic 2-space H2 as a definite sheet of the unit hyperboloid in the space s12(R). A point of the unit hyperboloid is represented by a matrix of the form
A=
; a,b,c E R, detA = 1
Observe that the entry c $ 0 which follows from - a2 - be = 1. Thus the sign of entry c distinguishes between the sheets of the hyperboloid. We let H2 denote the sheet consisting of matrices A as above with c>0
Normal vectors
Let us recall from 11.4.7 that a geodesic curve y:R-H2
has the explicit parametrisation
7(t) = A cosh t + T sinh t
;tER
where A = 7(0) and T = y'(0) is a vector orthogonal to A with
1.
With this notation the normal vector to y is
2.1
N= y'(t) A y(t)
;tER
The normal vector N is independent of time t as follows from the parametrisation of y given above. Observe that N is orthogonal to A and T and and that the
norm of N is -1 :
= - = -1 The tangent vector y'(t) can be recovered from N by the formula 2.2
y'(t) = y(t) A N
This follows from 2.1 using the general formula 1.8.
;
tER
Let us remark that the
normal vector N depends on the orientation of the geodesic curve y. Thus the normal vector of an unoriented geodesic is determined up to a sign only.
111.2 PENCILS OF GEODESICS
Proposition 2.3 connected components.
93
The complement of a geodesic h in H2 has two Reflection r along a normal vector N for h will fix h
pointwise but interchange the two connected components of its complement.
Proof
Let N be a normal vector for h.
The function X
<X,N>, X E H2,
is zero along h but non-zero elsewhere. This divides the complement of h in H2 into the open sets U and V where this function is positive, resp. negative. The sets
U and V are interchanged by r since <X,N> = - as follows from
r(X)=X+2<X,N>N
; XEH2
It remains to prove that U is connected. Consider two distinct points A and B of U. We can use 11.4.6 to pick a unit tangent T to H2 at A such that
B = A coshd+T sinhd
; d = d(A,B)
Let us form the scalar product of N and y(t) = A cosh t + T sinh t, t E ld, to get
= cosht ( + tanht)
When > 0 we find that y(t) E U for all t E R. When < 0 we can use the standard properties of the function tanht to find a number r E R such that
< 0 for all t > r and < 0 for all t < r. It follows that y(t) E U
for all t < r and y(t) E V for all t > r. This shows that the geodesic are [A,B] is entirely contained in U.
0
The two connected components of the complement of h are called the sides of h. Once the geodesic h is oriented by a normal vector N, we can distinguish
between the two sides of h: a geodesic curve through a point A E h with tangent vector N at A passes from the negative side of h to the positive side of h (warning: the function X F-+ <X,N> takes negative values on the positive side of h).
HYPERBOLIC PLANE
94
Oriented angles
Let us recall that the tangent space of H2 at a point A is
the space TA(H2) of vectors in s12(R) orthogonal to the vector A. We say that a pair of tangent vectors X,Y are positively oriented if vol(A,X,Y) > 0. Given a unit tangent vector S E TA(H2),
then S and S A A form a positively
oriented
orthonormal basis for TA(H2) as follows from 1.7 and 1.9. Given two unit tangent
vectors S and T to H2 at the point A E H2, the oriented angle Lor(S,T) between S and T is defined by
2.4
T = S cos Lor(S,T) + S A A sin Lor(S,T)
; Lor(S,T) E 02/2irl
In accordance with Euclidean geometry we shall make use of a second kind of angle, namely the interior angle LitA E ]0,2a[ of a polygon A at the vertex A.
Intersecting geodesics We shall investigate the intersection between two oriented geodesics h and k through the point A E H2. Let us introduce the directional angle a from h to k as indicated below
or = Lor(V,-U)
-U=Vcosa+VAAsina
Using 1.7 and 1.8 we deduce that = cosa and U A V = A sina. For the corresponding normal vectors H = U A A and K = V A A we deduce from 1.8 that
2.5
= cos a
,
H A K = A sin a
In particular, we find that II < 1. Conversely
111.2 PENCILS OF GEODESICS
Proposition 2.6
95
Two distinct geodesics h and k in H2 intersect if and
only if their normal vectors H and K satisfy II < 1.
Proof
Suppose that the normal vectors H and K satisfy Il < 1.
Consider the discriminant A for the plane E spanned by H and K
A = - 2 = 1- 2 Observe that A>O and conclude from the discriminant lemma 1.6.3 that E has Sylvester type (-2,0). It follows that the line E L has type (0,1) and that the
point A of intersection between H2 and E L lies on h and k.
The converse
implication follows from 2.5.
We say that geodesics h and k are perpendicular if they meet at a point A of H2 at an angle a.
Corollary 2.7 Geodesics h and k in H2 are perpendicular if and only if their normal vectors H and K satisfy = 0.
Proof
Two geodesics which meet at a right angle will have = 0 as it
follows from 2.5. Conversely, if = 0 the two geodesics will meet according
to 2.6. The angle at which they meet must be a right angle, 2.5.
Geodesics with a common perpendicular
Let us investigate
two geodesics h and k both perpendicular to the same geodesic 1. The points of
intersection with 1 will be denoted A and B respectively. Let L be the normal vector for I corresponding to the orientation of 1 from A towards B.
We shall
make use of the normal vector H = A A L for h and the normal vector K = -B A L for k, compare 2.2.
HYPERBOLIC PLANE
96
H=AAL
K= -BAL
1
Let us evaluate the inner product between the normal vectors H and K
= - = This gives us the first of the following two formulas
2.8
= cosh d(A,B)
,
vol(H,L,K) = - sinh d(A,B)
In order to prove the second formula, observe first that L A K = B as follows from
1.8 using K = L A B. Next, observe that vol(H,L,K) = to get that B = A cosh d(A,B) + H sinh d(A,B)
This gives = sinh d(A,B) = - sinh d(A,B) as required.
Theorem 2.9
Two distinct geodesics h and k are perpendicular to the same
geodesic if and only if their normal vectors H and K satisfy Il > 1. The common perpendicular geodesic is unique whenever it exists.
Proof Observe that H and K span a plane E, and form the discriminant A = - 2 = 1 - 2 It follows that Al > 1. If A < 0 we conclude from the
discriminant lemma I.6.3 that E has Sylvester type (-1,1). It follows that E -1contains a vector L of norm -1. The geodesic 1 with normal vector L is perpendicular to h and k. Conversely, consider a geodesic I perpendicular to both h and k. It follows
IH.2 PENCILS OF GEODESICS
97
from 3.3 that a normal vector L for 1 is orthogonal to both H and K, i.e. L E E 1 This makes L unique up to a sign.
Lines with a common end
A geodesic h in H2 spans a linear plane
of type (-1,1). The two isotropic lines in this plane are called the ends of h. Let us observe that a normal vector H for h is orthogonal to the the two ends of h.
Proposition 2.10
Two distinct geodesics h and k in H2 have a common
end if and only if their normal vectors H and K satisfy Il = 1.
Proof Observe that H and K span a plane E, and form the discriminant A = - 2 = 1 - 2
If h and k have a common end S we must have S = E 1 . In particular, we conclude that the line E 1- is isotropic. It follows from the discriminant lemma 1.6.3
that the plane E has type (-1,0) and that A = 0,
i.e. Il = 1.
Conversely, if Il = 1 we get A = 0 and we get from 1.6.3 that E 1 is an
isotropic line, in fact a common end for h and
Pencils of geodesics
k.
A set `.P of geodesics in H2 is called a
en
ncil if
there exists a linear plane P in s12(R) such that `.P equals the set of geodesics with
normal vectors in P. Let us recall from 1.6.4 that a linear plane P in s12(Ot) is of
Sylvester type (-2,0), (-1,1) or (-1,0).
It follows by inspection that P is
generated by its vectors of norm -1. In other words, the plane P is generated by the normal vectors of geodesics in the pencil it determines. Let us examine the three Sylvester types separately.
Type (-2,0). The line P 1 has type +1 and intersects H2 at a point A, say. The pencil 9 is the set of geodesics through the point A.
Type (-1,1). The line P 1 has type -1 and is generated by the normal vector of a geodesic h. The pencil 9 is the set of all geodesics perpendicular to h.
IIYPERBOLIC PLANE
98
Type (-1,0). The line S = P 1 is isotropic. The pencil 9 is the set of geodesics with end S.
Let us examine the intersection of two pencils. It follows from the discussion above that two distinct pencils can't have more than one geodesic in common.
The following is a list of true statements about the intersection of
pencils.
2.11
Through two given points A and B passes a unique geodesic h.
2.12
Through a given point A passes a unique geodesic perpendicular to a given geodesic 1.
2.13
Through a point A passes a unique geodesic with a given end.
2.14
There is a unique geodesic with a given end S perpendicular to a given geodesic h provided the ends of h are different from S.
2.15
There is a unique geodesic with given ends R and S.
Proof
Let A be a point of H2 and A the pencil of geodesics through A. Observe that the plane A 1 has type (-2,0) and conclude that the intersection between A 1- and any other plane P is a line of type -1. This proves the first three statements.
To prove 2.14, pick a normal vector N to h and observe that N 1 has type
(-1,1) while S -L has type (-1,0). It follows that the intersection of these two
planes has type -1 or 0. Type 0 can be ruled out since S is not an end of h.
To prove 2.15, observe that the planes R 1- and S 1- have type (-1,0).
The intersection is a line of type -1 or 0. singular to rule out 0.
Use the fact that our form is non-
III.3 CLASSIFICATION OF ISOMETRIES 111.3
99
CLASSIFICATION OF ISOMETRIES
An isometry of the hyperbolic plane H2 is the product of at most three reflections in geodesics in H2 as we have seen in 11.4. In particular we can write an
even reflection as a product a,6 of reflections in geodesics a and b. A finer classification will depend on the relative position of the two geodesics a and b. In
other words, the classification of even isometries is tied to the three types of pencils of geodesics.
The classification of odd isometries turns out to be simpler.
An odd isometry is a glide reflection, i.e. the composite of a translation along a geodesic and a reflection in the same geodesic.
Theorem on three reflections 3.1
Let
be reflections in
three geodesics a,b,c belonging to the same pencil 9. The product a/3y is itself a reflection in a geodesic from the pencil 9.
Proof Let S # 0 be a vector orthogonal to the plane generated by the normal vectors of the geodesics from 9. Let us show that a Lorentz transformation o of s12(R) which fixes S is the product of one or two reflections in geodesics from 9.
Note that a induces an orthogonal transformation of the plane P = S 1 . When <S,S> > 0 , the plane P has type (-2,0) and the result follows from 1.4.5. When <S,S> = 0 , the plane P is parabolic and the result follows from 1.5.3 and 1.6.16. When <S,S> < 0 , the plane P is hyperbolic and the result follows from 1.6.11.
Observe that the vector S is fixed by the three reflections a, Q and y and
conclude that a$y can be written as a product of at most two reflections. On the basis of parity II.4.10, we conclude that a13y is a reflection.
0
The theorem on three reflections has an application to the geometry of a
triangle AABC. Let rna, mb, in,, be the perpendicular bisectors for AB, BC and CA, compare the proof of 11.4.8.
HYPERBOLIC PLANE
100
Corollary 3.2
The perpendicular bisectors ma, mb, m, for triangle DABC
lie in a pencil.
Proof
Let 9 denote the pencil containing mb and ma and let a E 9 be the
geodesic through A. Reflections in the geodesics ma, mb, a are denoted µa, Pb and
a. It follows from 3.1 that µaµba is a reflection in a geodesic h E 9. Observe that µaµba transforms A into B and conclude that h = mc. Thus me E 9.
Horolations
Let S be an isotropic line in s12(R). A horolation with centre S
is an isometry of the form a,Q where a and 6 are reflections in geodesics a and b with end S. The set of horolations with centre S form a group, 3.1.
Corollary 3.3 The group of horolations with centre S is abelian.
Proof
For three reflections p,a,r in geodesics with end S we have par = rap,
as follows from par = (par)-1 =
7_ 1
0`
1P- 1 .
For four such reflections, we get that
(a1)(7) = (aQy)b = (7)3a)b = y(3ab) = y(ba,Q) = (yb)(a/3) and the result follows.
A closer look at the proof of 3.1 reveals that the group of horolations with centre S is isomorphic to the additive group of R, see also IV.2 The orbits on H2 for the group of horolations with centre S are called horocycles with centre S.
111.3 CLASSIFICATION OF ISOMETRIES
Corollary 3.4
101
Two distinct points A and B belong to the same horocycle
with centre S if and only if the perpendicular bisector m for A and B has end S.
Proof Suppose that the perpendicular bisector in for A and B has end S. Let It be a reflection in m and let a be a reflection in the geodesic through A with end
S. With this notation, the horolation µa transforms A into B. Conversely, let a be a horolation with
B.
It follows from the theorem on three reflections
that as is a reflection in a geodesic it with end S. Note that o-a(A) = B and conclude that n is the perpendicular bisector for A and B.
Proposition 3.5
The group of horolations with centre S acts on the set of
geodesics with end S in a simply transitive manner.
Proof
Let a and b be geodesics with end S and let M and N be normal
vectors for a and 6 respectively. The linear automorphism a of the plane P = S
with o,(S) = S and a(M) = N is orthogonal.
1
It follows from 1.5.3 that o can be
written as a product of one or two reflections. Upon replacing N by -N we can assume that a is the product of two reflections. It is immediate to extend a to a horolation with centre S.
Let us assume that the horolation cr fixes the geodesic a. For A E a we must have a(A) = A: otherwise, the perpendicular bisector for A and
has end
S, contradicting the fact that two perpendicular lines can't have the same end.
Corollary 3.6
A horocycle 3G with centre S and a geodesic a with end S
intersect in precisely one point.
Proof
If X and a have two distinct points A and B in common, then the
perpendicular bisector m for A,B has end S, contradicting the fact that two perpen-
dicular geodesics can't have a common end.
Let us choose a point B E )G and a
horolation or which transforms the geodesic b through B with end S into the geodesic a. It follows that a(B) is a point of intersection between 7G and a.
HYPERBOLIC PLANE
102
We shall present the remaining two pencils leaving the detailed treatment to the reader. This can be done by algebra as above, but elementary geometry may be applied as well. See also 11.8.
Rotations By
a rotation around a point A of H2 we understand an isometry
of the form a13 where a and /3 are reflections in geodesics a and 6 passing through
A. The rotations form a group as it follows from the theorem on three reflections. The orbits for the action of the rotation group on H2 are the circles with centre A.
The group of rotations around A is isomorphic to R/2irZ: The rotation angle B of a rotation o is given by
3.7
0
= Lor(t,o(t))
; t E TA(H2)
,
= -1
The rotation with angle a is called a half-turn or symmetry with respect to A. A
half-turn can be expressed K A where r and A are reflections in a pair of perpendicular geodesics through A.
Translations
Let k denote a geodesic in H2. By a translation along k we
understand an isometry of the form a/0 where a and /Q are reflections in geodesics
a and b perpendicular to k. The translations along k form a group as follows from
the theorem on three reflections 3.1. The group of translations along k acts on k in a simply transitive manner. The orbits for the group of translations along k are called hypercycles .
For a translation o along k the geodesic k is called the translation axis of
o. The translation length T of o is given by T = d(A,o(A)), A E k. It is worth noticing that the translation axis has a natural orientation.
If we let w denote reflection in k then we can decompose the translation above as a# = (alc)(K#). In other words, a translation along k can be written as
the composite of two half-turns with respect to points of k. Conversely, the composite of two half-turns around points of k is a translation along k.
111.3 CLASSIFICATION OF ISOMETRIES
103
Odd isometries The odd isometries of H2 are easy to classify by virtue of the following proposition.
Proposition 3.8 An odd isometry
0 of H2 is a lg ide reflection , i.e. of the
form 0 = roc where r is translation along a geodesic k and K is a reflection in k.
Proof Let
us fix A E H2 and focus on the point O(A). If A = O(A), we find
that 0 is a reflection in a geodesic through A. If A # O(A), let in denote the geodesic through A and q5(A) and let l denote the perpendicular bisector for A and O(A).
The point of intersection between in and I is denoted M while µ and A
denote reflections in these geodesics. Observe that \q is even and fixes the point A.
Thus a4 is a rotation around A, and we can use the theorem on three
reflections to write a¢ = ACV where v is a reflection in a geodesic n through A. Let
us introduce reflection ti in the geodesic k through M perpendicular to n and write
Note that \p and inc are half-turns around points of k. It follows that the isometry r = () p)(vic) is a translation along k.
HYPERBOLIC PLANE
104
111.4
THE SPECIAL LINEAR GROUP
The group of isometries of the hyperbolic plane H2 is the Lorentz group of
the ambient space s12(R). We shall use this to identify the group of isometries of H2 with the group PG12(R) given by
PG12(R) = G12(R)/R*
4.1
Let us first bring the whole group G12(R) to act on sl2(R). To this end we shall make use of the sign of the determinant
4.2
sign: G12(R) - {-1,1}
With this notation, the group G12(R) acts on s12(R) through the formula o X = sign(o)
4.3
oXo-1
; o E G12(R), X E s12(R)
It is clear that o acts as an orthogonal transformation of s12(R). We proceed to verify that this is a Lorentz transformation. To this end we ask the reader to verify the following formula
01
],y-', [ ° 1]> =
2(deto)-1(a2+b2+c2+d2)
where a,b,c,d are the entries of o. It follows from this and 1.6.4 that conjugation
with o interchanges the two sheets of the hyperbola if and only if sign(o) < 0. It follows that the formula 4.3 defines a Lorentz transformation of s12(UR).
Theorem 4.4 The action of G12(R)
on sl2(Od) induces an isomorphism
PG12(R) - Lor(s12(R))
111.4 THE SPECIAL LINEAR GROUP
105
Proof The main point is to realise any Lorentz transformation of s12(R) by an element of Gl2(R).
Let us start with reflection rI< along a vector K E s12(R) of
norm -1. We shall involve the "strange formula" 1.4
KX + XK = -2 t
; X E s12(R)
Multiply this identity by the matrix K = K-1 and deduce the formula
4.5
-KXK-1 = X + 2K
; X E s12(R)
From the fact 1.6.11, that Lor(sl2(R)) is generated by reflections, it follows that the map G12(R) -> Lor(s12(R)) is surjective.
Let us show that the kernel of the map G12(R) -; Lor(sl2(R)) consists of
scalar matrices. Consider a o E G12(R) with dec(o) > 0 which acts trivially on s12(R).
Observe that the vector space M2(R) is generated by t and s12(R) and
conclude that o commutes with all matrices X E M2(R). From this it follows that
o is a scalar matrix. Next, consider a a E G12(R) with dec(o) < 0 which acts trivially on s12(R).
This means that o anti-commutes with s12(R). We ask the
reader to verify the general fact that a matrix o E M2(R) which anti-commutes with s12(R) is the zero matrix.
Corollary 4.6
The action of S12(R) on s12(R) identifies the group PS12(R)
with the special Lorentz group Lor+(s12(R)).
Proof We shall show that the action of a o E G12(R) defined by 4.3 is an even Lorentz transformation on s12(R) if and only if dec(o) > 0.
To this end, we pick a
decomposition of the action of o on s12(R) into a product r1...rs of reflections along vectors K1,...,Ks of norm -1. It follows from 4.5 that the matrices o and K1...Kg E G12(R) have the same action on s12(R). Thus we can use 4.4 to pick a
scalar r E R* such that
o = rK1..K5 Calculate the determinant of both sides to get that dec(o) = r2(-1)$. This shows that dec(o) > 0 if and only if s is even.
HYPERBOLIC PLANE
106
Theorem 4.7
Let the action of o E S12(U8) on H2 be decomposed as a,3
where a and ,6 are reflections in geodesics h and k. If H and K denote normal vectors for h and k, then
lr2(o) = 42
Proof From the end of the proof of the previous proposition follows that
we
can write o = eHK, where e = ± 1. This gives us
ir o = e lr(HK) = - 2e
0
The result follows by taking the square of this relation.
Let us summarise the discussion in a small table, compare also 11.8.
4.8
Position of h and k
tr2(0.)
Klein notation geometric notation
intersecting
elliptic
rotation
common perpendicular
hyperbolic
translation
common end
parabolic
horolation
For a general a E G12(R) we shall make use of the invariant tr2(o) =
(det tr o)2 )
introduced in 1.9.14.
Proposition 4.9
Let o E G12(R) be a non-scalar matrix. The conjugacy
class of a in PG12(R) is entirely determined by tr2(a) and sign deco.
Proof
It is known from linear algebra that any conjugacy class in G12(R)
contains a matrix from the following list A
0
a
-,3
A
1
0
µ
(3
a
0
A
The remaining details are left to the reader.
0
III.5 TRIGONOMETRY
107
111.5 TRIGONOMETRY
In the next four sections we shall study the trigonometry of the hyperbolic plane and apply this to a number of existence and classification problems.
Our treatment is based on the vector calculus from section III.1 and the concept of normal vectors from 111.2. Let us use the convention that the natural
orientation of a closed polygon is such that the polygon lies on the positive side of its edges. In other words, the corresponding normal vectors are inward directed, see
the remarks made after 2.3. Let us start with the hyperbolic triangle AABC.
Cosine and sine relations 5.1 cosh a = cosh b cosh c - sinh b sink c cos a
sinh a
- sinh b
sin a - sin Q
sinh c
sin y
c
Proof Let us first prove the cosine relation. Recall from 2.5 that B = A cosh c + V sinh c ,
C=Acoshb - Usinkb
Let us use this and = cosa to calculate the inner product of B and C = cosh b cosh c - sink b sink c cos a Observe that = cosh a and the cosine relation follows.
In order to prove the sine realations we observe that 1vol(A,B,C)I symmetrical in A,B,C. Let us evaluate this in an asymmetrical manner:
is
with the
notation from 2.5 we have vol(A,B,C) = vol(A,U,V) sink b sinh c
A few lines before 2.5 we found that U A V = A sin a, thus
vol(A,B,C) = sin a sinh b sink c = sin a sinh a sinh b sinh c sinh a and the result follows since I vol(A,B,C) I is symmetrical in A,B,C.
0
HYPERBOLIC PLANE
108
Alternative cosine relation 5.2 cosh a =
Proof
cos ,3 cos y + cos a sin ,Q sin y
Let H,K,L be the inward directed normal vectors of the edges of the
triangle
Formula 1.9 in combination with formula 2.5 gives us that
= - = cos /3 cosy +
cos a
From the second formula in 2.5 we get that
HAK=Csiny ,
LAH=Bsin,O
Combine this with the definition of hyperbolic distance to get that
= sin y sin 0 = cosh a sin y sin ,Q and the alternative cosine relation follows.
Corollary 5.3
The sum of the angles in
0
ABC satisfies
a+Q+y< 7r
Proof
Let us assume that a > ,3 and
> y.
We ask the reader to use
formula 5.2 to rewrite the inequality cosh a > 1 in the form cos(7r - a) < cos(,3 +'Y)
When ,8 + y < it we conclude that 7r - a > 13 + y or a + a + y < 7r as required.
When 7r <,3+y we can rewrite the inequality above as cos(a+7r) < cos(,13+y); this gives a + 7r < 0 + y, which contradicts a > / and a > y.
111.5 TRIGONOMETRY
Theorem 5.4
109
Let a,Q,y E ]0,lr[ be real numbers with a +,3 + y < a, then
there exists a triangle DABC in H2 with LA = a, LB =,3, LC = y.
Proof
Let us rewrite the basic assumption 0 + y < Tr - a and conclude that
cos (f3+y)>cos(ir-a) Using the trigonometric addition formulas we find that
sin 0 sin y < cos,3 cos y + cos a It follows that we can determine a E ]0,+Oo[ such that
sin / sin y cosh a = cos 0 cos y + cos a Let us now pick two points B and C with d(B,C) = a , and draw a geodesic I through B forming an angle ,3 with BC and a geodesic k through C forming an
angle y with the geodesic h through BC. We let H,K,L denote inward directed normal vectors for these geodesics
Proceeding as in the proof of 5.2 we find that 5.5
cosh a sin ,3 sin y = cos ,3 cos y +
This gives us cos a = , which shows that II < 1. We can use 2.6 to conclude that the geodesics I and k intersect. It follows from 5.2 that the angle of intersection is a.
0
HYPERBOLIC PLANE
110
Let us investigate a quadrangle OABCD with three right angles. Such a
quadrangle is called a Lambert quadrangle after J.H.Lambert (1728-74). It is
known since the early days of hyperbolic geometry that the fourth angle of a Lambert quadrangle is acute. We can deduce this from the second of the two trigonometric formulas to follow.
Lambert quadrangle 5.6
cosh d(A,D) = cosh d(B,C) sin y sinh d(A,B) sinh d(D,A) = cos y
Proof Let H,K,L,M be inward directed normal vectors as indicated in the illustration below
Let us substitute /3 = 2 in formula 5.5 and deduce that
cosh d(B,C) sin y = Formula 2.8 gives us = cosk d(A,D), which takes care of the first formula. To prove the second formula, use 1.10 to get that
0
5.7
vol(K,M,L) vol(M,L,H) = det
-1
0
0
-1
and the result follows from formulas 2.8 and 2.5.
<M,H> = 0
0
III.6 ANGLE OF PARALLELISM
111
ANGLE OF PARALLELISM
111.6
In this section we shall study the trigonometry of a triangle DABC in which the vertices B and C are ordinary points of H2 while A is a point at infinity
(an end).
For the history of this problem and its relations to astronomy, the
reader is referred to "Hyperbolic geometry : The first 150 years" [Milnor].
We are going to prove a relation for AABC which formally looks like the alternative cosine relation 5.2 of an ordinary triangle with LA = a = 0.
Proof Let
us first observe that formula 5.5 remains valid in this context. It
follows from 2.10 that = ± 1.
In order to decide between +1 and -1,
observe that the left hand side of formula 5.5 is positive.
.
0
In particular, if LC is a right angle we find that
cosh a sin /3 = 1
6.2
sinh a tan,6=I lanha sec(3=1
The last two formulas are elementary consequences of the first. formula, we are using the old convention sec,3 = 1/cos /3.
In the third
HYPERBOLIC PLANE
112
111.7 RIGHT ANGLED PENTAGONS
In this section we shall investigate a number of pentagons on which there are no analogues in Euclidean geometry. Let us start with the trigonometry of the right angled pentagon. The existence of such a thing will be demonstrated in 7.6. P4
7.1 cosh as = sinha2 sinha3
cosha3 = cotha2 cotha4
Proof We shall assume that the enumeration of the vertices is consistent with the natural orientation of the sides of the polygon, 11.5. For i = 1,...,5 we let Ni be the inward directed normal vector for the edge P;Pi+1. Let us at once record that = 0
;
i = 1,...,5 , N6 = Ni
From the general formula 1.10 (or the more specific formula 5.7), we get that
7.2
vol(N1,N2,N3) vol(N2,N3,N4) =
Evaluate this by means of 2.8 to get the first of the two formulas in 7.1.
Let us now turn to the second formula, which we, by the way, intend to generalise beyond the case where LP1 is a right angle. We shall first establish the following formula
7.3
vol(N1,N3,R) = vol(N1,N2,N3)
Observe that this formula is linear in R and check the formula directly in the case where R=N1,N2,N3. We ask the reader to use 1.10 to get that vol(N1,N3,N4) vol(N3,N4,N5) = + 