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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 Lfunctions 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 Reissued in this digitally printed version 2008
A catalogue record for this publication is available from the British Library ISBN 9780521435086 hardback ISBN 9780521435284 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 HALFSPACE
78
11.8
POINCARE HALFPLANE
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
HOPFRINOW 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
FRICKEKLEIN SIGNATURES
230
VII EXERCISES
235
VIII HYPERBOLIC 3SPACE 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 nonEuclidean 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 halfplane In 1882 H.Poincare (1854
 1912) discovered
a new model of H2, which I shall now describe. The model is the upper halfplane 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 halfplane are traced by Euclidean circles with centres on
the xaxis or Euclidean lines perpendicular to the xaxis. 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 halfplane 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 halfplane 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 nspace 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
Threedimensional 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 3manifold
H3/F.
,
i.e. is a discrete subgroup F of S12(C), defines a
The work of W.P.Thurston on 3manifolds [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 fourdimensional 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 threedimensional 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 nonsingular 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 nonsingular if and only if E
Theorem 1.7
1
= 0.
Let E be a finite dimensional vector space equipped with a
nonsingular 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 klinear 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 nonsingular 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 klinear 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 nonsingular 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 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(Alv) = 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 nonsingular. 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 nonsingular 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 nonisotropic if Q(n) # 0. A nonisotropic 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 nonisotropic vector e is a linear combination of e  f and e + f). If the vector e  f is nonisotropic, 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 nonsingular, 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 nonsingular 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 nonsingular. This will be done
by induction on dim(U). In case dim(U) = 1 let us pick a nonzero 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 nonisotropic 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 r1a: U,E to an orthogonal transformation of E. To recapi
tulate it suffices to extend v:UE 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:WF to an orthogonal transformation of F. Let us now assume that U is singular. Let there be given data consisting
of a nonzero 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 nonsingular
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 nonsingular, and then
apply the first part of the proof.
The orthogonal group 0(E) of a nonsingular form is generated by reflections along nonisotropic 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 nonsingular form of dimension n can be written as a product of no more that n reflections along nonisotropic 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 nonzero e E E. The form Q is called negative definite if Q(e) < 0
for all nonzero 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 = Ale 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)=uv 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 nonsingular 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 ni 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 nonsingular since del Ai # 0. Let us pick a nonzero 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
CauchySchwarz 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 CauchySchwartz inequality we can introduce the angle L(e,f) between two nonzero 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 n1 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 n1 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 3space 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 Clinear 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
CauchySchwarz inequality 5.1 ;e,fEF
<e,f>2 < <e,e>
Proof The CauchySchwarz 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 mdimensional
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 nonisotropic. 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:EN 2°
Im(o  t) C F 1 .
c(e+n)=e+v(e){ii
; eEE,nEN
We can arrange for v:EN to be injective: a nonzero vector e E E of Ker v is nonisotropic and fixed by o, which allows us to use the induction hypothesis on e
1 . We can now arrange for v:EN 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,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),  + 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 :EE. It follows at once that
(Qr) = TQ
; 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 nspace E. By this we understand that H is an affine hyperplane in E
and P is one of the open halfspaces 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 nontrivial subset J of 0,...,n exists such that
a,,,=0 forrEJands0J then there exists a Euclidean nsimplex 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 nonzero 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 nonzero 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 CauchySchwarz 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
00 (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>
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 n1 > 2 and must have nonzero 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
Sn1
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
= (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 nonzero 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': Sn1 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 nonzero 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 ,
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 n1 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,A1,
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 xHx 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)
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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
Ixc12
;
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 1point 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> +
2(ad + bc)
; e,f E E, a,b,c,d E R
It is easily seen that this space has Sylvester type (n1,1). 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  cl2 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 nonzero 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,Aib)
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,() 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},
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) = vr01 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
halfspaces 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*`T1
#(inT)=1 #(inf)=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) = r1 (x  f)
,
N(s) =  r1(f,a,1)
When `J is an ordinary sphere with radius s and centre g we can write
nx(S) = s '(xg) , 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 twodimensional 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  1sin 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 twodimensional 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+y2b+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
(xA)2+y2=A2 which shows that all circles pass through (0,0) and are tangent to the yaxis at this point. 3°
b < 0 . Put b =  k2 and rewrite the equation as follows
(xA)2+y2 = A2k2 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 1point 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
;adbc#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 S1 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 nonzero 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 zz 1 of C.
This transformation is the
46
QUADRATIC FORMS
composite of inversion in the unit circle and reflection z
in the xaxis as follows
from the basic formula Z1 _ZIZI2
;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 4touples 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 71a 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 nonzero 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 71 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+a1
Notice, that the value of a+ a' is unchanged if a is replaced by a1. 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 nonsingular 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 nonsingular 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 nonsingular 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=2ge 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 nonsingular 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 nonsingular 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 nonisotropic 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 halfturns in lines K and L (reflections in K and L in the notation from Exercise 1.1). 3°
Show that any two halfturns 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 halfturn. Use the next exercise actually to find a halfturn 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 halfturn with respect to a line L with the property that L and
a(L) are orthogonal. Show that aAa A is a halfturn. 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 CauchySchwarz 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) , 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 CauchySchwarz, 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)=x2<xu,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 p1 reflections has been found to transform AI,...,Ap1 into BI,...,Bp1. When p(Ap) = Bp we can take a = p. When p(Ap) $ Bp, observe that
i = 1,...,p1 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 ndimensional 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
GEOMETRIES
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 = ruar11, 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 = Tau a
1
,
a(v) = v
this may be rewritten as
e = v + (u  au)
; v E Ker(at), 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(at)®Im.(at)
Proof
Let us first show that the two subspaces Ker(at) and Im(at) 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 at 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 CauchySchwarz 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 nonzero distance. If P and Q are linearly independent, the CauchySchwarz 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(cb)
When cb is positive we deduce that a > cb which gives c < a+b. Observe that this inequality is trivial when cb 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)
GEOMETRIES
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
;sEJ7r,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(tu) + T sin e(tu)
;
tEJ
We conclude that 7 is continuously differentiable on J with y'(u) = eT , thus
3.9
y(t) = y(u) cos(tu) + y'(u) sin(tu)
;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 nonempty, 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.
GEOMETRIES
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 Grammatrix 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 =
(ch2b1)(ch2c1)(cbb chccha)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+bc) sh2(a+cb) sh.'(a+b+c) sh2(c+ba) 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
pa=2(ca)+2b>0, pb=2(cb)+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(II") inherits a quadratic form from the ambient space F. It is easily seen that the tangent space has Sylvester type (n,0)
GEOMETRIES
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(ts) = cosh Itsi 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(tu) + Tsinh E(t  u)
;tEJ
This implies that y is continuously differentiable on J with y'(u) = eT. Thus
y(t) = y(u) cosh(tu) + yo(u) sinh(tu)
;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
=22 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 halfspace. 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 halfspace 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 halfspace E+ equipped with this metric is called the Poincare halfspace.
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 halfspace. In particular, any isometry of E+ can be extended to a Mobius transformation of E.
11.7 POINCARE 11ALFSPACE
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 halfspace 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 halfspace E+ are traced by circles in E orthogonal to the boundary aE+.
GEOMETRIES
80
11.8 POINCARE HALFPLANE
The open upper halfplane 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
Iwzl
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 CR 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, 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 xaxis 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.
IIint: 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:RCyl 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(7rp) sin(pa) sin(pb) sin(pc) 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 > 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 oa(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 halfturn or symmetry with respect to A. A
halfturn 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 halfturns with respect to points of k. Conversely, the composite of two halfturns 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 halfturns 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
oXo1
; 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
13 + y or a + a + y < 7r as required.
When 7r / 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(ira) 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 (172874). 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) = +
Combine this formula with 7.3 with the specification R = N4 to get that
111.7 RIGHT ANGLED PENTAGONS
7.4
113
vol(N1,N2,N3)vol(N3,N4,N5) = +
Let us take advantage of formula 2.8 and deduce that
7.5
sinha2 sinha4 cosh a3 = + cosh a2 cosh a4
In case LPI = 2 we have = 0 , which concludes the proof.
Proposition 7.6
Let a1 > 0 and a2 > 0 be real numbers. There exists a
right angled pentagon with two consecutive sides of length a1 and a2 if and only if
sinhat sinha2 > 1
Proof
The condition is necessary as follows from the first formula in 7.1. To
see that the condition
is sufficient,
start from a right angle LP1P2P3 with
d(PI,P2) = a1 and d(P2,P3) = a2. Draw the geodesic k3 through P3 perpendicular to P2P3 and the geodesic k5 through P1 perpendicular to P1P2
Formula 5.7, used in the analysis of Lambert quadrangles, at once gives that
sinhat sinha2 =
It follows that > 1 which implies that k3 and k5 have a common perpendicular geodesic, compare 2.9.
HYPERBOLIC PLANE
114
111.8 RIGHT ANGLED HEXAGONS
Let us now turn to the right angled hexagon. The method we have used
in the analysis of the right angled pentagon applies to the hexagon as well. We
shall show that right angled hexagons are classified by the length of three alternating sides a2,a4,a6.
For geometric applications see [Fathi, Laudenbach, Poenaru], in particular, "expose 3 et appendice aux expose 8".
The point is that the basic building
blocks of hyperbolic geometry, hyperbolic pants, can be sewed out of two isometric right angled hexagons, see VII.2.6.
P4
Sine relations 8.1 sinha2 sinha6 _ sinha4 = sinha5 sinha3  sinhal P1
Proof
With a notation similar to the one applied in the previous section
we get from formula 7.2 that vol(N4,N5,N6) vol(N5,N6,N1) =
vol(N1,N2,N3) vol(N2,N3,N4) _
Using 2.8 we deduce equality between the two first ratios. The remaining formulas follow by symmetry.
HI.8 RIGHT ANGLED HEXAGONS
115
Cosine relations 8.2 sinh a2 sinh a4 cosh a3 = cosh a6 + cosh a2 cosh a4
Proof
Let us observe that formula 7.5 remains valid in this context. Thus
the result follows after the specification = cosh a6 , compare 2.8.
Classification theorem 8.3
0
Let a2,a4,a6 be three strictly positive real
numbers. Then, there exists a right angled hexagon in H2 such that three alternating sides have lengths a2,a4,a6. The hexagon is unique up to isometry.
Proof
Let the number a3 > 0 be determined by the cosine relation 8.2, and
start the construction of the hexagon on the basis of the numbers a2,a3,a4
Let N1 and N5 denote the inward directed normal vectors for h1 and h5 and deduce
from the relations 8.2 and 7.5 that = cosh a6
We conclude from this that the geodesics h1 and h5 have a common perpendicular
h6 say. We leave the remaining details to the reader.
I would like to conclude this chapter with a reference to [Fenchel] which contains a wealth of trigonometric formulas. In particular, [Fenchel] p.84 gives information on "improper" right angled hexagons.
HYPERBOLIC PLANE
116
111.9
FROM POINCARE TO KLEIN
In this section we shall describe an isometry from the Poincare halfplane H2 to the s12(R)model of the hyperbolic plane discussed in this chapter. Let us recall that the metric on the Poincare halfplane is given by
9.1
coshd(z,w) = 1 +
I
zwI2
z,wEH2
2Im[z]Im[w]
To a point z E H2 we assign the matrix F(z) E s12(R) given by _IzI2
x
9.2
;z=x+iyEH2
F(z) = 1
Y
x
1
We ask the reader to verify that F is an isometry of hyperbolic planes. We also
leave it to the reader to show that F is G12(R)equivariant in the sense that it relates the action from 11.8.4 to the action of 111.4.3 as follows
9.3
F(o(z)) = sign(o) oF(z)o"1
Proposition 9.4
; z E H2, o E G12(IR)
The transformation F from the Poincare halfplane H2
to the s12(R)model preserves oriented angles.
Proof Let z vary along a smooth curve in F(z)I
32
H2
and differentiate F(z) to get
x
I z2
x/
2(xx'+yy')
1
x
0
x/
I
Let us rewrite this in terms of the tangent map of F
_y1 TZF(u) =
1
0
2x 1
I
x
x2+y2
y2
1
x
u
v
The norm of the tangent vector TZF(u) at the point F(z) is given by
; u,v E R
111.9 FROM POINCARE TO KLEIN
9.5
= del TF(u) =y2 (u2+v2)
From this follows that TF(U) preserves angles. It is also worth making the calculation
9.6
vol (F(z), TZF(s), TZF(u)) =
2
(rvsu)
; u,v,r,s E R
y
Let us end this section with some remarks for readers familiar with differential forms. The formula 9.6 may be written
F*vol = 2 dx A dy
9.7
y
Formula 9.5 gives us the Riemannian metric on the Poincare halfplane 9.8
x+iy E C, y>O
y2(dx2+dy2)
Hyperbolic area
Let us define the hyperbolic area of an ordinary
hyperbolic polygon A on n vertices by the formula
9.9
Area A = (n2)a  E LintP P
where the sum is over all vertices P of A. We ask the reader to check that the area is additive under simple subdivision of polygons. Once this is done, it follows
from 5.3 that
Area A> 0. In the Poincare halfplane model the area can be
calculated from the areaform 9.7
9.10
Area A =
J
12 dxdy A
Y
There are hints on how to verify this in the exercises. In general, formula 9.8 may be taken as the definition of area.
HYPERBOLIC PLANE
118
111.10 LIGHT CONE
Let us investigate the space PC of isotropic lines in s12(R). This means that we are investigating lines generated by matrices of determinant 0. To a nonzero vector e E K2 we shall assign the matrix
10.1
zw Z
L(e)
w2
e
 zw
'
z
w
E R2
The matrix L(e) has determinant 0, more precisely the vector e belongs to the kernel of the linear map with matrix L(e). From this it easily follows that our construction defines a bijection
L: R + PC(s12(K)).
This map obeys the trans
formation rule (see 1.9.2 for notation)
10.2
L(o(e)) = oL(e) o'
; o E G12(K), e E K2
Proof Let us first perform the decomposition z
z
0
1
w
w
1
0
;e=l zIEK2 w
L(c) =
IT
I
Next, observe the matrix formula T
a
b
0
1
c
d
1
0
]=[
0
1
1
0
and the transformation rule follows rather easily.
To an ordered pair (Z,U) of distinct points of K we assign the vector L(Z,U) E s12(R) of determinant 1 given by
111.10 LIGIIT CONE
119
zv+wu
10.3
L(Z,U) = wu
1
zv
 2zu
Z=[z], w U=[v]
2wv  zv  w u
Alternatively, this can be described by means of the wedge product 1.5
10.4
L(Z,U) =
22 L( w) n L(v)
; D = det [W vl
It follows from 1.6 that the geodesic in H2 with normal vector L(Z,U) has ends Moreover, we conclude from 10.2 and 10.4 that
(Z,U).
10.5
o,L(Z,U)o' = L(o(Z),o(U))
;
a E G12(R)
We shall now evaluate the inner product of these vectors in terms of the cross ratio
on A. For four distinct points A,B,P,Q we shall prove that
10.6
< L(A,B),L(P,Q)> = [A,B,P,Q] 1
Proof
According to 10.5 and 1.9.7 both sides of the formula are invariant under S12(R). From the fact, 1.9.3, that the action of S12(R) on U2 is triply transitive it follows that it suffices to verify the formula in the case A = oo, B = 0,
P = 1, Q = q, q E R. Let us recall from 1.9.9 that [oo,0,1,q] = q. Thus it suffices to prove that
= 9 + l This is a straightforward verification, which we leave to the reader.
qER 0
We ask the reader to establish the following supplement to the general formula displayed in 10.6
10.7
= I
;A$B$C
HYPERBOLIC PLANE
120
Let us consider two distinct geodesics h and k in H2 with normal vectors
H and K and ends (A,B) and (P,Q). We can summarise our investigations in the following table
10.8 Relative position of h and k
< H,K >
intersecting
]1,1[
perpendicular
Sylvester type
[A,B,P,Q]
]oo,0[
0
(2,0) (2,0)
common perpendicular
R[1,1]
(1,1)
]0,1[ U ] 1,+oo[
common end
1,1
(1,1)
1
1
The column entitled Sylvester type refers to the pencil generated by h and k, see the end of 111.2.
III EXERCISES
121
III EXERCISES
EXERCISE 1.1 Determine the eigenspace decomposition for the operator X H X° on M2(R) . For notation see 1.9.2.
EXERCISE 1.2
Let E denote a twodimensional real vector space. By a complex
structure on E we understand a linear endomorphism J of E with J2 = I. 10
2°
tr J = 0 and del J = 1. Show that two complex structures on E belong to the same sheet of the Show that J E End(E) is a complex structure 4*
hyperbola del = 1 in sl(E)
R E End(E) I tr R = 0 } if and only if they define
the same orientation of E. EXERCISE 2.1 1° Show that the vertices of an equilateral triangle lie on a circle. 2°
Show that this is not true for a general triangle in H2.
EXERCISE 2.2
Let h be a geodesic in H2. For a point P E H2 let Ph denote the
perpendicular projection of P onto h.
Show that the map P , Ph decreases
distance, in the sense that d(P,Q) > d(Ph,Qh) for all points P and Q. Let OABCD be a Parallelogram, i.e. a convex quadrangle in
EXERCISE 3.1
which opposite sides have equal length.
Show that the diagonals bisect each other
and that the geodesics through opposite sides have a common perpendicular. Hint:
Show that halfturn µ with respect to the midpoint M of the diagonal AC takes the quadrangle into itself. EXERCISE 3.2
Let S be an isotropic line in s12(R). Show that the horocycles
with centre S are cut by the affine planes in s12(R) parallel to TS(H2) = S
1.
Given a triangle DABC. Let PA denote the rotation with centre A which transforms the ray AB into the ray AC. Show that PA2 = YO EXERCISE 3.3
1°
where y is reflection in the geodesic AB and ,Q is reflection in the line AC.
IIYPERBOLIC PLANE
122
2° Define rotations PB and pC as above and prove that PA2 ° PB2 o pC2 =t. 3°
If DABC is equilateral show that PA,PB,pC are conjugated in Isom+.
Show that an involution in PG12(R) can be represented by a
EXERCISE 4.1
matrix R E Gl2(ld) of trace 0 and determinant 1 or 1. Hint: Show that a matrix R E G12(R) which represents an involution in PG12(0R) is an eigenvector for the operator X H X.
Let A E s12(R) represent a point of H2.
EXERCISE 4.2
transformation X r»
EXERCISE 4.3
AXA"1 of s12(R)
1°
Show that the
is halfturn with respect to the point A.
Show that the set of vectors in s12(R) of norm 1 is
connected. 2°
Prove that S12(R) is connected. Hint: Any o E Sl2(I) can be written in the
form o = XY, where X,Y E s12(R) are vectors of norm 1.
EXERCISE 4.4
1° Show that an even Mobius transformation a of t with matrix
A=
a
_
b
b
a,bEC, Ja12lb12>0
a
leaves the unit disc D invariant. 2°
Show that any even isometry of D has this form. Hint: Find all even
Mobius transformations which commute with reflection in 8D. 3°
Find the connection between lr2A and the "type" of the isometry a.
40
Show that an even isometry of D can be written exp2aiO1+Cz
EXERCISE 5.1
;cED,0EIR
Consider a triangle DABC with LC a right angle. Show that
sin A = lim
a40 c
cos A= lim b b>o c
III EXERCISES
123
Show that a triangle LABC has an inscribed circle with radius
EXERCISE 5.2
cos2A + cos2B + cos2C + 2 cos A cos B cos C
r
EXERCISE 6.1
2 (1 + sin A)(1 + sin B)(1 + sin C) Let S be an isotropic line in s12(R). Show that the set of vectors
of norm 1 in S 1 has two connected components. Use this to replace the second half of the proof of 6.1 by a "deformation argument".
1° Show that the normal vectors of a right angled hexagon
EXERCISE 8.1 satisfy the formula
vol(N6,N1,N2) vol(N3,N4,N5) =
+ 2°
Prove the following cotangent relation for a right angled hexagon cosh a2 cosh a3 = coth a1 sinh a3 + coth a4 sinh a2
3°
Investigate an improper hexagon with five right angles in which the sixth
vertex P6 lies at infinity.
Hint: Derive a formula for the normal vectors similar to
the formula above (the term must be added to the right hand side).
EXERCISE 9.1
Let DABC be a triangle in the hyperbolic plane H2, and let
a,Q,y be translation from B to C, C to A resp. A to B. 1°
Show that /3ay is a rotation around A of angle Area A relative to the
orientation defined by the triangle. Hint: For two distinct points P and Q we let
tpQ denote the unit tangent vector at P to the geodesic from P to Q. Justify the following "mod 27r" calculation Lor(tCA, aytAB) = Lor(tCA, tCB) + Lor(tCB, aytAB) = LintA  Lor(tCB, atBA)
LintA  Lor(a 1tCB, tBA) = LintA + Lor(tBC, tBA) = LintA + LintB Let 9 denote the rotation angle in accordance with 3.7. Justify the calculation Lor(QtCA, QaytAB) = Lor(QtCA, tAB) + Lor(tAB, /3aytAB) =  LintA + 9 Finally, combine the two results. 2°
Generalise the result to an arbitrary polygon in H2.
30
Prove a similar result for the sphere S2.
HYPERBOLIC PLANE
124
EXERCISE 9.2 Show that a disc D in H2 with radius r has
Area(D) = 2a (cosh r 1) EXERCISE 9.3
Let the hyperbolic area of say a compact subset . of the
Poincare halfplane H2 be defined by formula 9.8. Show that
Area(o(.)) = Area(.) 2°
a E Isom(H2)
Given r>O and v E [0,2]. For s>r let E(s) denote the subset of H2 bounded by
the yaxis, x = rcosv, y = s and the circle with centre 0 and radius r. Show that
lims_;00 Area E(s) = 2  v 3°
Prove that the area of .ABC is 7r  LA  LB  LC. Hint: Move the triangle
such that two vertices are located on the yaxis and apply 2°.
IV FUCHSIAN GROUPS
We shall introduce the central topic of this book, discrete groups of isometries of the hyperbolic plane H2. A discrete group of even isometries is called
a Fuchsian
rg
oun.
The set of fixed points for the elliptic elements of a discrete
group r' form a discrete subset of H2.
We shall prove after Jacob Nielsen a
converse to this: A nonelementary group r with discrete elliptic fixed point set is itself discrete.
It is a common feature of the Poincare halfplane and the s12(IR)model of
H2 that these models provide natural isomorphisms between Isom(H2) and PG12(R), compare 111.9.3. The results of this chapter can be interpreted in either
of the two models.
The s12(R)model is used throughout in the more technical
parts of the proofs.
IV.1 DISCRETE SUBGROUPS
The group PG12(R), being the factor group of the topological group G12(R)
with respect to a its centre R*, carries the structure of a topological group: the open subsets of PG12(IR) correspond to open subsets of G12(R) stable under R*. We
shall use the action of PG12(R) on H2 to study discrete' subgroups of PG12(R). But first, we shall exhibit some convenient compact neighbourhoods of the identity t in PG12(R).
Theorem 1.1
Acnpoint z E H2 and an c > 0 determine a subset of PG12(UR) .N(z;() = for E PG12(R) I d(z,o'z) < c }
which is a compact neighbourhood of the identity i. in PG12(R).
'The notion of a discrete subset is discussed in an appendix to this section.
FUCIISIAN GROUPS
126
Proof
Let us introduce the norm II a II of a matrix a E G12(R) as
1.2
hall =
a2+b2+c2+d2
a
b
c
d
a
I det or I
Observe the formula II as II = II a II , a E 13*, and conclude that the norm induces a
continuous function on PG12(IR). It follows that a set of the form
1.3
{ or E PG12(13) 111 allF2
Let us show that a set of the form 1.3 is compact. Observe first that the set 1.3 is the image by a continuous function of is a neighbourhood of a in PGI2(IR). the set
{ o EG12(13) 111o1 1
IV.4 JACOB NIELSEN'S THEOREM
141
JACOB NIELSEN'S THEOREM
IV.4
We have seen that a discrete subgroup r of PG12(Il;) acts on H2 with a
discrete set of elliptic fixed points.
We shall prove the converse for a non
elementary group.
Jacob Nielsen's theorem 4.1
A nonelementary subgroup r of
PG12(R) is discrete if and only if the set P of elliptic fixed points on H2 is discrete.
Proof
It follows from 2.7 that a subgroup r of PG12(Id) is elementary if and
only if r+ is elementary. Thus we may assume that r is Fuchsian.
Let us assume that P is non empty. Pick a point z E P and observe that its orbit under r is discrete since it is a subset of the discrete set P. According to 1.5 it suffices to prove that FZ is finite. To see this, use that r is nonelementary to pick a second point w E P, w $ z. The orbit of w under FZ is finite since it is a
discrete subset of the circle S' with centre z through w. We can use that the action of F. on S1 is faithful to conclude that FZ is finite as required.
In the rest of the proof we assume that F is nonelementary and contains no rotations. It follows from 2.8 that r must contain proper translations. Let us pick a geodesic s such that the subgroup F3 made up of translations along s is nontrivial. Since r is nonelementary we have r # F3.
For r l FS we have r(s) $ s
since we can rule out halfturns. It follows from 2.5 that r(s) and s will intersect or have a common perpendicular; we shall examine the two cases separately.
When s and r(s) have a common perpendicular, let us introduce the distance d between s and r(s). For a nontrivial c E FS let us apply corollary 3.4 to v
and rar 1 and deduce the inequality
4.2
sin.h(2T0.) sinh(2d) > 1
; v E F3 , a 54 c
When s and r(s) intersect at an angle 0 E ]0,2] say, we can apply corollary 3.7 to
FUCIISIAN GROUPS
142
o and 7or1 and deduce the inequality
o E r, , o # t
sinh 2(ZTo) sin(g) > 1
4.3
Let us fix r for a moment. We can then use 4.2 and 4.3 to conclude that the numbers T. are bounded away from 0 as o runs through r,{t). It follows that IF, is discrete, in fact, r, is cyclic generated by an element o of translation length T, say. For any r E r such that s and r(s) have a common perpendicular we get from 4.2 that 4.4
sinh(2T) sinh(2 1) > 1
;
d = d(s,r(s))
For any r E I' such that s and r(s) intersect we get from 4.3 that
4.5
sin.h2(2T) sin(O) > 1
;
0 = L(s,r(s))
We shall now study the action of r on N = N(s12(68)), the space of vectors in sl2(lt)
of norm 1. Let us pick a normal vector N for s and observe that the stabiliser
for N is r,. Let us recall from 111.2.5 and 111.2.8 that for the inner product between N and rN we have respectively
(l = cosh d ,
Il = cos 0
We conclude from 4.4 that d is bounded away from 0 and from 4.5 that 0 is bounded away from 0. It follows that we can find a constant c>0 such that
II 0 ]1 c,l +c[
;
r o rN
From this we conclude that N is isolated in its orbit rN in N. Recall once again that rN is discrete and conclude from 1.14 that r is discrete.
A slight modification of the proof of Jacob Nielsen's theorem yields the following variation due to C.L. Siegel.
IV.4 JACOB NIELSEN'S THEOREM
Theorem 4.6
143
A nonelementary subgroup I' of PG12(R) is discrete if and
only if the elliptic elements of r do not accumulate at t.
Proof Let us assume that r is nonelementary and that the elliptic elements do not accumulate at t. Let us use 2.8 to pick a proper translation o E I' with axis s, say, and introduce the group rs made up of the translations in I' with axis s.
Let
us now pick a neighbourhood V of t in PG12(R) such that
< 0 and TUT 1u 1 is not a proper rotation
; o,T E V
Let us analyse a pair of elements T E Vrs. o E rs fl V, o # t. Observe that T is
not a halfturn around a point P of
a halfturn 1r around P satisfies = 1. We conclude that T(s) # s. Let us remark that s and T(s) cannot s:
have a common end as follows from 2.5 applied to o and 7017 1. If T(s) intersects
s, then we can observe that U(TU lr1)U 1(TQ 1r1)1 = (TUT 1o 1)2
is not a rotation and we conclude from 3.4 that inequality 4.3 holds. If T(s) and s
have a common perpendicular then we conclude from < 0 that c 1 and Tor 1 translate in opposite directions, compare the proof of 3.2. We conclude from
3.1 that inequality 4.2 holds. This is sufficient to conclude that rs is discrete, so
let us pick a definite generator o E rs and change the neighbourhood V of t such
that < 0 and To7'1o 1 is not a proper rotation At this point we leave the remaining details to the reader.
;TEV
FUCIISIAN GROUPS
144
IV.5
CUSPS
In this section we shall be concerned with the action of a discrete group r on the boundary of the hyperbolic plane H2.
We say that a point S E 8H2 is a
cusp for H2 if it is fixed by some proper parabolic transformation 63 E r.
The
stabiliser I'S of a cusp S E OH2 consists of horolations and reflections in geodesics
through S, 2.4.
By a horodisc
we understand a region in H2 bounded by a
horocycle.
Proposition 5.1
Given a cusp S E 8II2 for the discrete group I' of
isometries of H2. There exists a horodisc D with centre S E H2 such that
; y E t Fs
y(D) n D= O
Proof We shall work with the point oo of the Poincare upper halfplane. Moreover, we shall assume that the group 1 s+ is generated by A = [o i]. Let B = [ a] E G12(R) with det B = f 1 represent 13 E FF. According to lemma 5.3 we have jcj > 1. An elementary estimate based on 11.8.3 and 11.8.4 gives us Im[/3(x + iy)] = y Ic(x ± iy) + d 12 < y c 2y 2 < yt
Otherwise expressed
5.2
Im[z] Im[9(z)] < 1
; z c H2, ,3 E tI oo
It follows that the horodisc Im[z] > 1 will serve our purpose.
Lemma 5.3
Let us consider two matrices A and B from G12(R)
B= If A and B generate a discrete subgroup of PG12(R) then Ici > 1.
c#0, detB= ± 1
IV.5 CUSPS
Proof
145
Let us proceed under the assumption that Icl < 0. In order to derive a
contradiction we consider the sequence of matrices (Bn)n E N given by BO = B and Bn+1 = BnABn
1
;
nEN
We ask the reader to establish the recursion formula an+1
bn+1
1ancn
Cn+1
do+1
cn
2
an 2
;n>1
1+ancn
A simple induction on n E N gives us that
icnl = Ic012n
and
n + 1aol
nEN
Using co = c and Icl < 1 we conclude from this that c11+0 and ancn+0 for n+oo.
Put this information into the right hand side of the recursion formula and conclude that Bn+A for n+oo. Let us assume that the corresponding classes a
and (3 generate a discrete subgroup of PG12(R). Then we get that fl. = a for n large and deduce, by descending induction on n, that oo is fixed by Since 0 = Qo this contradicts the assumption c t 0.
With the notation of 5.1, let us observe that the orbit of a point A E 8D does not meet D: for y E rs we have y(OD) = OD and for y E FI s we find that y(8D) and D are disjoint.
Proposition 5.4
Let S E 8H2 be a cusp for the discrete group r of iso
metries of H2. For any compact subset K of H2 there exists a horodisc D with
centre S E H2 such that y(D) fl K = 0 for ally E F.
Proof Let us return to the proof of 5.1 and choose r>1 such that K lies in the vertical strip of the Poincare halfplane bounded by Im[z] = r and lm[z] = r"1. For
y E FS we have y(D) = D and consequence of the inequality 5.2.
for
y E f I s we have y(D) fl K = 0 as a
FUCHSIAN GROUPS
146
Corollary 5.5
Given cusps S,T E 49H2 in different 17orbits, for any horo
disc E with centre T there exists a horodisc D with centre S such that y(D)f1E=0for all yEr.
Proof Let
us pick a compact subset K of dE such that FTK = E. Use 5.4 to
choose a horodisc D with centre S such that D fl rK = 0. This gives rD f OE = 0. For y E r we conclude from the connectedness of yD that yD C E or yD fl E = 0. The first option can be ruled out on the grounds that 7(S) # T.
Corollary 5.6
If H2/17 is compact, then the group r contains no proper
horolations.
Proof Let
us first observe that the projection H2H2/r maps open subsets to
open subset. Next, cover H2 with open discs and apply the BorelHeine theorem
to the images of the discs in the compact space H2/r. The upshot is that we can find a finite set of discs in H2 which meet all rorbits. It follows that we can find
a compact set K in H2 which meets all rorbits. We conclude from 5.4 that F contains no horolations.
Horocyclic Topology
For a discrete group r of isometrics of H2 let us
consider the subset Y of H2 consisting of all ordinary points of H2 and the set of cusps for F. Let us introduce the horocyclic topology on Y: a subset W of Y is
open if for all points S E W there exists a disc/horodisc with centre S entirely contained in W. The horodiscs with centres at a cusp S E OH2 form a fundamental
system of neighbourhoods for S, while the ordinary discs with a given centre A E H2 form a fundamental system of neighbourhoods of A E Y.
The group r acts continuously on Y, which gives the orbit space X = Y/r a structure of topological space. This new space is a twodimensional topological manifold with boundary:
IV.5 CUSPS
147
Proposition 5.7
With the notation above, the topological space X = Y/r
is a Hausdorff space in which every point has a neighbourhood homeomorphic to a neighbourhood of 0 in R2 or a neighbourhood of 0 in R x [0,oo[.
Proof The space x = Y/r is Hausdorff as follows from 5.4 and 5.5.
In order
to construct a neighbourhood of a cusp point, it suffices to treat the point oo of the Poincare halfplane.
It follows from 5.1 that it suffices to treat the case where
r = rte. Let us assume that r is generated by zi+z+k, k>O. The space Y/r is homeomorphic to the open unit disc D in the complex plane
Y/r  + D
,
z i+ exp(21rik1z) , z E H2
In the general case we can arrange for r to be generated by zF +  i and
zF*z + k
as above. In this case we find that the orbit space X is homeomorphic to the orbit space for the action of complex conjugation on the open unit disc D in C.
In order to investigate a finite point, it suffices to treat the origin 0 in the Poincare disc D. According to 1.5 (for more details see VI.5.8) we may assume
that the full group r fixes 0. Let us assume that r+ is generated by zHBz, where 0 is a primitive nth root of unity. Using the well known fact that z*z" maps open set to open set, we can make the identification
H2/r+=, D A nontrivial element of r/r+ corresponds to reflection of D in a line through 0.
Corollary 5.8
The space X = Y/r is compact if and only if there exists a
compact subset K of H2 and a finite number of horocycles D1,...,Dr whose centres S1.... ,Sr are cusps for r such that any rorbit in H2 meets K U D1 U ... U Dr.
In the case of a Fuchsian group r, the proof of 5.7 provides a structure of
Riemann surface [Farkas, Kra] for the space X = Y/F.
When the space X is compact, it is called the horocyclic compactification of H2/r, compare VII.5.
For general information see [Lang] and [Shimura].
FUCIISIAN GROUPS
148
IV EXERCISES
EXERCISE 1.1
Let 1' be a subgroup of G12(R) and put A = 1' fl S12(R). Let P1'
and PO denote images of I' and A in PGI2(R) and assume that [PI':PD] is finite. Show that PI' is discrete in PG12(R) if and only if PA is discrete in PG12(R).
EXERCISE 1.2
The additive group U8 of real numbers acts on R/27rZ. Show that
the iorbit of 1 mod 2a is not discrete.
EXERCISE 2.1
1° Show that the group of rotations around a point A E H2
equals its normaliser in PS12(IR). 2°
Show that group of even isometries fixing a given geodesic line k equals its
own normaliser in PS12(IR). 3°
Show that the group of even isometries fixing a given end S equals its own
normaliser in PGl2(E).
EXERCISE 2.2
Let a and ,3 be translations whose axes have a common perpen
dicular. Show that the commutator a/3a1/31 is a translation.
EXERCISE 2.3
1° Show that if a subgroup F of PSI2(E) has a finite orbit on H2
of order n, then n = 1. Hint: If n > 2, show that 2°
= t for all y E G.
Show that if a subgroup r of PS12(IR) has a finite orbit of order n on OH2,
then n = 1,2. Hint: If n > 3 show that
EXERCISE 2.4 1°
yn!
yni
= t for ally E I'.
Let n. E PG12(R) be a proper translation with fixed points A,B.
Show that for any point Q 0 {A,B} the set {t:n(Q) I n E Z } accumulates
IV EXERCISES
149
1rof
at A and B. Hint: Take A=O and B=oo, in which case K = 110,
20
1
Show that a nonempty icstable closed subset Z of R contains A and B.
EXERCISE 3.1
Let S E G12(R) and put U =
an eigenvector for or and use
2(S'  S).
Show that U E s12(R) is
detS = (2 trS)2  trS2 (exercise 1.3.1) to prove that
= a det S (4  tr2 S) 2°
For a second matrix K E s12(R) prove that
= IrSK 1 IrS trK 3°
Use this to give an alternative proof of lemma 2.10.
EXERCISE 4.1
1° For A E H2 and r > 0 fixed, prove that the set below is
compact
{ N E N(s12(R)) (Il < r }
Hint: Pick a basis S,T for the tangent space of H2 at A and use that any point of F can be written in the form N = xA + yS + zT , x,y,z E R. 2°
Let `ll be a discrete subset of N(42(R)). Given a point A E H2 and r > 0,
show that only finitely many geodesics in H2 with normal vectors in 9 will meet the hyperbolic disc with centre A and radius r.
V FUNDAMENTAL DOMAINS
The most important way to get information about a discrete group r of isometries of H2 is through fundamental domains. The art of reading information
about discrete group from a fundamental domain dates from the last century. A
good example for this
is
the "modular figure" which served Gauss as a
fundamental domain for G12(Z).
The modular figure can be constructed by a
general method due to Dirichlet to be presented in V.4.
The finer analysis of a fundamental domain involves a decomposition of
its boundary into linear pieces which are paired together through side transformations. A good fundamental domain leads to a representation of the group by generators and relations, compare Poincare's theorem in chapter VII.
V.1 THE MODULAR GROUP
By a fundamental domain U for a group t of isometries of H2 we understand an open subset U of H2 such that U and 7(U) are disjoint for ally # r,
and such that each torbit meets U.
As a basic example, let us construct a
fundamental domain for the modular group PS12(7L).
= erp2ai/3
V.1
TIIE MODULAR GROUP
151
The modular figure 1.1 The three geodesics in the upper halfplane Re[z]=2, Re[z]=2, zj = 1
H2
bound an open set M which is a fundamental domain for the group PSl2(7).
Proof Let us fix a point w = yi, y > 1 and introduce the three transformations o(z) = z 1, T(Z) = z + 1
,
W(Z) = z  1
We ask the reader to verify that the three geodesics bounding M are the perpendicular bisectors for w,w(w), w,o(w) and w,T(w). In order to find a point of M in the orbit through a given point z E H2 we shall minimise the hyperbolic distance d(w,p(z)), p E PS12(7L). Precisely if v E S12(7L) is chosen such that
d(w,v(z)) < d(w,p(z))
E S12(7)
Then v(z) E M as follows from the inequality d(w,v(z)) < d(w,c,i 1(z)) = d(w(w),v(z))
and similar inequalities with w replaced by o and r.
Let us consider an orbit 0 of S12(Z) in H2 and let us show that Im[z] is a
bounded function of z E 0 which assumes its upper bound whenever z E M. In concrete terms, let us consider a z E M and p E r represented by
Let us note the inequality Icz +d F = c21zI2 + 2Re[z] cd + d2 > c2 + d2  jcdj = (Icl  Id 1)2 + IcdI > 1
and draw the conclusion from the elementary formula
f Im [ bz + d] = Im.[z] Icz + d12 del I c b
When jzi > 1 and c # 0, then the inequality above is strict w rich gives us that Im[,u(z)] < Im[z]. When c = 0, the transformation µ takes the form zHZ+b and
Im[z] is preserved. We conclude that the orbit 0 meets M in at least one point and M in at most one point as required. Taking into account that M is open, we find that an orbit which meets 8M will not meet M. We ask the reader to verify
that the figure M, which is the union of M and the part of 8M which lies in the halfplane Re[z] > 0, meets each S12(71)orbit in precisely one point. The identification of 8M can be realised by the transformations o,T introduced above.
FUNDAMENTAL DOMAINS
152
1.2
We shall see in the next, section that the modular group is generated by o and T. The transformation p = ro is a rotation of order 3. Thus we have the relations
1.3
0"
2=t
,
p3
I
i
p=To
It follows from Poincare's theorem VII.3.7 that this set of relations is complete.
The hyperbolic triangle A with vertices i, C, oo is a fundamental domain for the extended modular group PG12(7L) as follows from the fact that reflection 6
in the yaxis divides the fundamental domain for the modular group into two halves. Let us introduce reflections a,/3,ry in the sides of A as given by the figure
Let us construct a fundamental domain for some subgroups of PS12(7) using the general lemma
V.1 THE MODULAR GROUP
Lemma 1.4
153
Let I' denote a discrete group of isometries of H2 and II a
subgroup of F of finite index. If D is a fundamental domain for I' and S C F is a
full set of representatives for II\F, then the interior U of
F= UaESaD is a fundamental domain for H.
Proof
Since S is finite, we conclude that F is the closure of Ua
ES
aD.
Since this set is open, it must be contained in U (the interior of F), and it follows
that F = U. Let us show that the IIorbit of a given point z E H2 meets F. To this end choose y E F with yz E D and write
y 1 = 7r1a with a E S and 7r E II to
get that a1az E D or irz E A. Let us analyse a 7r E II and a point u E F with 7ru E U.
Pick a E S such
that u E aD and let us observe that a1F is a neighbourhood of u. It follows that
a 1F meets aD or F fl iraD 0 0. Since F is the closure of U a,OD , we conclude that 7raD meets ,6D for some Q E S. This means Ira = 0 which implies a = t.
The level 2 modular group
Let us describe a fundamental domain
for the group G(2) given by the exact sequence
1.5
G(2)  PG12(7) f G12(F2) ' 0
0
The group G12(F2) is isomorphic to the permutation group S3 through the action of G12(F2) on the three lines in F2 ®12. This group is lifted back to PG12(7L) by the dihedral group D3 generated by a and y whose matrices are 1
0
1 1
[
,
1
0
1
1
0
According to lemma 1.4 the union of the six translates of z by D3 is a fundamental domain for G(2).
It follows from the illustration above that this
union is the hyperbolic triangle 0,1,c
.
FUNDAMENTAL DOMAINS
154
The level 2 modular group I'(2) = G(2)+ has fundamental domain the quadrangle 1,0,l,oo as follows from 1.4 applied to the reflection 0. We shall see in the next section that r(2) is generated by
µ(z) = z + 2
,
v(z) =
z
2z+1
In fact we shall see as a consequence of Poincare's theorem VII.3.7 that the level 2 modular group I'(2) is free on these two generators.
I'(2) = G(2)+
1
0
1
V.2 LOCALLY FINITE DOMAINS
155
V.2 LOCALLY FINITE DOMAINS
Let I' denote a discrete group of isometries of the hyperbolic plane H2 and
let us consider a fundamental domain P C HZ which is locally finite in the sense that any point z of H2 has a neighbourhood which meets o(P) for at most finitely
many a E F. We shall relate the group theory of I' and the topology of H2/F to the geometry of P.
Proposition 2.1
Let P C H2 be a locally finite fundamental domain for
the discrete group r . The canonical projection is a homeomorphism
p:P/r=; H2/r
Proof Since
our map p is continuous, we must prove that p maps open subsets
of P/r to open subsets of H2/r. To this end it suffices to show that for any given (relative) open subset B of P the set
V=UyEry(B) is an open subset of H2. It suffices to consider a point b E B and construct an
open disc with centre b entirely contained in V. Let us use the fact that P is locally finite to see that the following subset S of r is finite
S={aEIF Ivl(b)EP} Since P is locally finite we can choose a>O so small that
D(b;c)na(P)=0
; o¢S
The disc D(b;c) is contained in the union of all IFtranslates of P, thus
D(b;) C Uo E S o(P) Let us take advantage of the fact that S is finite to decrease e > 0 further to get
D(Q 1(b);e) (l PC B
; or E S
Let us combine these two pieces of information to get
D(b;e) = Uo E S o(P) n D(b;e) = Uo
E
So(P n D(v 1(b);e)) C Ua
This concludes the construction of the required disc.
E
So(B)
0
FUNDAMENTAL DOMAINS
156
Corollary 2.2
Let P C H2 be a locally finite fundamental domain for the
discrete group IF. Then H2/I' is compact if and only if P is bounded.
Proof A closer look at the proof of 2.1 reveals that the projection P onto open subsets of H2/I'. If H2/F is compact
then we can use BorelHeine to find a finite set of open discs D1,...,Dn with centres z1,...,zn
in P such that each rorbit in H2 meets one of the sets
P fl D1,...,P fl D. It follows that P is contained in the union of these discs.
Proposition 2.3
Let P C H2 be a locally finite fundamental domain for
the discrete group r of isometries of H2. The group r is generated by
{yErI7(P)flPs0)
Proof
Let H denote the subgroup of r generated by this subset.
For z E H2
choose o E IF with z E o(P) and observe that the class [o] E 17/11 is independent of o. In other words we have defined a map
p : H2 > r/H This map is surjective since the I'translates of P cover H2. The map a is locally constant: at a given point z E H2 we can use the fact that P is locally finite to
find a finite subset S of F such that z E cr(P) for all o E S and such that N = Uo E So(P) is a neighbourhood of z in H2; it follows that p takes the value [o], o E S.
We can use the fact that H2 is connected to conclude that p is
constant. Taking into account that p is surjective, we conclude that H = F.
Let us conclude this section with some results on unbounded fundamental
domains. The problem is to relate cusp points for r to "boundary" points of a locally finite fundamental domain P. The horocvclic boundary 8hP is the set of points R E OH2 such that any horodisc with center R meets P. to compare this material with section IV.5 on cusps.
We ask the reader
V.2 LOCALLY FINITE DOMAINS
Proposition 2.4
157
Let P be a locally finite fundamental domain for 1' and
A E OH2 for which there exists a point L E P such that the geodesic segment [L,A[ is contained in P.
Proof
A transformation y E r+ with y(A) = A is parabolic.
The transformation y is elliptic, hyperbolic or parabolic. A proper
elliptic transformation has no fixed points on 8H2. Let us assume that y is proper hyperbolic with translation axis h. Upon replacing y by 71 we may assume that
y translates towards A. Let the horocycle with centre A through L intersect h in
W and put r = d(L,W). Let us contradict that P is locally finite by showing that D(W;r) n 7 n(P) i4o
; nEN,n>1
For n > 1, let Ln E [L,A[ denote the point of intersection between [L,A[ and the horocycle with centre A through yn(W). From the inequality
d(Ln,yn(W)) < d(L,W)
0
it follows that y n(Ln) E D(W;r) as required.
Example 2.5
Let I' denote the group of isometries of the Poincare half
plane generated by 7(z) = 2z, z E H2. The region P bounded by the two curves
x+i(21xl +1) , x+i(21xl+2)
xER
is a locally finite fundamental domain. Observe that y(oo) = oo and that any horodisc with centre oo meets P.
Thus oo E al,(P) but oo is not a cusp point for I.
FUNDAMENTAL DOMAINS
158
Proposition 2.6
Let A E 0112 be a cusp for the discrete group r and D a
horodisc with centre A. A locally finite fundamental domain P for IF meets (at least one but) at most finitely many Ftranslates of D.
Proof Let
us assume that the horodisc D is closed and use the remarks fol
lowing IV.5.3 to pick an open horodisc U contained in D such that there exists a
full Forbit outside U.
Let y E F be a parabolic element with fixed point A.
Moreover, we pick a geodesic It emanating from A and let C be the compact subset
of D  U bounded by h and y(h). Observe that DU is union of translates of C by powers of y. Let us enumerate the Ftranslates of P which meet C ol(P),...,on(P)
; ol...... n E F
Let us now assume that o1(D) meets P or what amounts to the same thing that o(P) meets D. Assume for a moment that o(P) C U; this implies o(P) C U which
contradicts the existence of a full Forbit outside U. In conclusion o(P) meets D  U. It follows that we can find s E Z such that yso(P) meets C. This proves that yso = of for some i = 1,...,n, and we conclude that o(D) = o1ys(D) = o1(D). It follows that P meets at most finitely Ftranslates of D.
Corollary 2.7
Let F denote a discrete group and P a locally finite funda
mental domain for F. The forbit of a cusp meets the horocyclic boundary 01iP.
Proof
Let us pick a horodisc K with its centre at a cusp A and let K, a1(K)...... on(K)
; 0"i,,0'n E IF
be a complete enumeration of the Ftranslates of K which meet P.
For any
horodisc L contained in K we find that at least one of the horodiscs L, ol(L)....... n(L)
; 01,...,on E F
will meet P since P meets any Forbit. From this it follows that at least one of the points A,o1(A),...,on(A) belongs to 01iP.
V.3 CONVEX DOMAINS
159
V.3 CONVEX DOMAINS
In this section we shall study a group
IF
of isometries of H2 and a
fundamental domain P, which is convex and locally finite. Recall that a subset R
of H2 is convex' if for any two points A,B E R the geodesic arc [A,B] C R. The main objective is to relate the geometry of 8P to r. It turns out that although no
polygonal structure is supposed from the outset, the boundary comes out as a geodesic polygon.
Let us start with a basic result on open convex subsets of H2.
HahnBanach principle 3.1
Let U be an open convex subset H2.
Through any point A E H2 outside U passes a geodesic not meeting U.
Proof Let S' denote the unit circle in the tangent space TA(H2). Let us define a map p:US' as follows: for Q E U draw the oriented geodesic from Q to A and let p(Q) E S' denote its unit tangent vector. The map p is continuous and the image V = p(U) is an open subset of S' which does not contain antipodal points.
Let .A:S'+S' denote the antipodal map, and observe that V and A(V) are nonempty open disjoint subsets of S'. Since S' is connected we can find a point s E S' not in V U A(V). It follows that neither s nor A(s) belongs to V. It follows that
the geodesic through A with tangent vector s does not meet U.
'In terms of Klein's disc model 11.5, the set R is convex in the hyperbolic sense if and only if R is a convex subset of the ambient Euclidean plane.
FUNDAMENTAL DOMAINS
160
Corollary 3.2
Let U be an open convex subset of H2. For points A E U
and B E 8U the geodesic segment [A,B[ is contained in U.
Proof
If Q E [A,B[ is not in U we can draw a geodesic h through Q not
meeting U. The closed halfplane H with 8H = h and A E H will contain U but not
B. This makes B an exterior point for U.
Corollary 3.3
Let U be a convex open subset of H2. The closure U is a
convex subset of H2.
Proof
Let us first show that we can write
U=UII,HjUH where the union is taken over the set of all closed halfplanes H in H2 containing
U. To this end consider a point w 0 U with the intention of constructing a closed
halfplane H 7 U with w 0 H.
Pick a base point u E U and a point z E ]u,w[
outside U and use the IlahnBanach principle to draw a geodesic h through z not
meeting U. The closed halfplane H with Oil = h and u E H does the job. The representation of U as a union of closed halfspaces shows immediately that U is a convex subset of 112.
Corollary 3.4
Let U be a convex open subset of H2.
The interior of the
closure U equals U.
Proof We must show that any neighbourhood of a given point z E SU meets the complement of U. To see this, use the HahnBanach principle to draw a geodesic h through z not meeting U. Since U is connected, it is contained in a closed halfplane, say, iI hounding h.
It follows that H contains U, but any
neighbourhood of z meets the complement of H.
V.3 CONVEX DOMAINS
Lemma 3.5
161
Let P be a locally finite convex fundamental domain for the
discrete group y. A set of the form
0 0t,0' Er
Pflcr(P) is either empty or a closed geodesic segment. A set of the form
; t#o#r#t, o,rEI
Pflo(P)flr(P) is empty or reduced to a point.
Proof From c(P) fl P = 0 it follows that aP fl o(P) = 0 and o i(P) fl aP = 0. Write P = P U OP and o(P) = o(P) U o(OP) to get that P fl c(P) = aP fl o(aP)
;oEr,o
We conclude that P fl c(P) is a convex subset of OP. Let us show that P fl o(P) is contained in a geodesic line. If not, we can find three points of P n o(P) not on a geodesic.
An inner point z of the triangle based on these points furnishes an
interior point of aP , contradicting 3.4.
Let us investigate an interior point z of s = P fl o(P). Pick points x,y E s
such that z E ]x,y[ and consider a point u E P. The interior On of the triangle with base [x,y] and vertex u is contained in P, 3.4. Similarly, the interior Dou of
the triangle with base [x,y] and vertex o(u) is contained in o(P). It follows that
we can find an open disc D with centre z such that D fl s is a diameter which divides D into two open halfdiscs D' and D" entirely contained in P and o(P) respectively. The halfdiscs D' and D" are disjoint from r(P). From D C D' U D" we conclude that D fl r(P) = 0. We can now conclude the proof by recalling that
0
the set P fl o(P) fl r(P) is convex.
Definition 3.6 Let P be a locally finite convex fundamental domain for the discrete group I'. By a side of P we understand a closed geodesic segment, not
reduced to a point, of the form P fl o(P) with c E r{t}.
understand a point which can o,r E r{t}, o 54 r.
be presented in
By a vertex of P we
the form P n o(P) fl r(P),
FUNDAMENTAL DOMAINS
162
Proposition 3.7
The boundary 8P of a locally finite convex fundamental
domain is the union of its sides.
Proof Let z E OP and consider the following subset S of r S = {rEr{t}IzEo(P)} Since P is locally finite we can find an open disc D with centre z such that
D C P U U or E S°(P) Observe that S:0: the alternative gives D C P which according to 3.4 gives z E P, contradicting z E 8P.
As a consequence we have that
3.8
aP C UO'E r{1}°(P)
Returning to the situation at hand, we cannot have P n a (P) = {z} for all o E S since D{z} is connected. Thus we have found a side of P through z E 8P.
0
A side s of P generates a geodesic h. The boundary points for s in h are called the the end points for the side s. The number of end points for s is 0,1 or 2.
Proposition 3.9
Let s be a side of the locally finite convex fundamental
domain P. A point z E s is a vertex for P if and only if z is an end point for s.
Proof
We have already shown during the proof of 3.5 that an interior point
of s is not a vertex of P.
a vertex of P.
Let us investigate a point z of s = P n a(P), which is not
From the previous proof we find an open disc D with centre z such
that D C P U o(P). This gives us a partition
D = DnP U Dno(P) U Dns Since D  D n s is disconnected and D n s is a geodesic arc through the centre z of D, we conclude that D n s is a diameter of D.
0
V.3 CONVEX DOMAINS
Corollary 3.10
Proof
163
A vertex A of P is contained in exactly two sides of P.
Consider a open disc D with centre A which meets only those sides of
P which have vertex Az. Let these sides be enumerated cyclically $1,s2,.... sr. Note
also that D fl P is connected since it is a convex set. Let us change the enumeration above such that D fl P C L(sl,s2). Observe that s fl D C P for i = 1,...,r and conclude that r = 2. 0
A side s of P can be put in the form s = P fl o(P) where o E r{t}. According to 3.6 the transformation o is unique and will be called the side transformation os generated by s. os(*s) = s.
Let us put *s = P fl o *'(P) and observe that
The assignment sN*s is an involution on the set of sides of P which
is called the side pairina of P.
By an edge of P we understand an oriented side. The operator * extends
in a natural way to an operator on the set of S of edges of P. We shall in particular be interested in the subset S+ of edges whose initial vertex is finite. For an edge s with initial vertex A we let is denote the second edge with initial vertex A, compare 3.8. Let us combine these two operators to obtain a new operator %Fs=J*s
The operator %k is a bijection since it is composed of two involutions.
; sES+
FUNDAMENTAL DOMAINS
164
Lemma 3.11 The operator W: g+>g+ has finite orbits.
Proof
Let us consider a sequence of iterates of ' on s E g si=%P's,sE7L
S17...'Sri...
Observe that the side transformation of = on maps the initial vertex Ai of si onto the initial vertex A1_1 of
si_1.
It follows that the sequence A = A1,...,Ar,... of
0
initial points all belong to the finite set P fl TA.
Key lemma 3.12
Let s1 be an edge of P with initial vertex A and let st,...,Sri...
ol,...,Orr,...
be the sequence of *iterates of st and corresponding side transformations. There t and such that the first of the two unions
exists n > 2 with
n
n
0"1o2...o1(P) 0102...oi(P) U U i=1 i=1 is neighbourhood of A while the second union is a disjoint union.
Proof side of is1.
Let us make use of the orientation of H2 which has P on the positive Let us define the sign of an edge s of P by sign(s) = +1 when P is on
the positive side of s and sign(s) = 1 otherwise. Observe that sign(*s) =  det(os) sign(s) Using sign( is)
;sEg
sign(s) we get for s = s1_1 that sign([si) = det(o11) sign(Tsi_1)
;
Let us investigate the angle Lor(jsi,sl) with vertex A1. We have that mod 29r Lor(lsi,si) = sign (Jsi)L1ncA1 Observe that o1...oi_1P has vertex A = o1o2...o1_1(A1) and angle Lor(0*1 ...a1_1 f 1,01...oi_lsi) = det(o1...0i1)Lor(jsi,si)
From the above formulas and the normalisation sign(js1) = 1 we get that Lor(ot...oi_1 jsi,ot...oi_tsi) = Lint Al
mod 27r
Let us also observe that ot...oi_t Jsi = ol...oi_1*si_1 = Q1...oi_25i_1
It is now clear that the integer n for which
21r does the job.
i = 2,...
V.3 CONVEX DOMAINS
165
Let sl,...,s,, be a full edge cycle, i.e. a full orbit for $ on C. The corresponding sequence Al,...,Ar of initial points is called a vertex cycle. The second vertex cycle which starts out with Al is A1,Ar.,Ar_1,...,A2, as follows by observing that is1,1sl.,Jsi._1,...,Js2 form a full edge cycle.
Theorem 3.13
An interior point z of P is requivalent to no other point
of P. An interior point z of a side s of P is Fequivalent to the point O's '(Z) of P,
but to no other point of P. Two vertices of P are equivalent under r if and only if they belong to the same vertex cycle.
Proof
Let us investigate an interior point z of a side s=P fl o(P). A point
w E P equivalent to z can be written in the form w=r(z), r E I'. This implies z E r"1(P) fl P fl o(P) and we conclude that r = t or r = 0"1 = os 1 Let us now turn to the case where A is a vertex of P and let us consider y E r such that y(A) E P. This means that A belongs to the closure of 7 1(P) and we conclude from 3.12 that we can find i = 1,...,n such that 'V
This gives y I = 0102...0j
1(P)fo1...o1(P)
0
Evaluation of this formula at the vertex Al+1 gives
1(A;+1) = A or Ai = y(A). y
Corollary 3.14 Let P be a locally finite convex fundamental domain for F. The group IF is generated by the side transformations us, as s runs through the sides of P.
Proof According to 2.3 it suffices to show that the subgroup of r generated by the side transformations contains any y E r for which there exists a z E P with y(z) E P. But this problem has already been dealt with during the proof of 3.13.
FUNDAMENTAL DOMAINS
166
Corollary 3.15
Let sl,...,sr denote a full edge cycle for the operator 'I,
and let ol,...,or denote the corresponding side transformations. The cycle man
o = ol...or is a rotation around the initial vertex Al of s1. The order of o is given by 2a = ord(o) (LintA1 + ... + LintAr) where Al,...,Ar are the initial vertices of the edge cycle.
Proof Let us return to the proof of 3.12. Evaluation of the formula al.... on = on the edge sn+1 gives us sl = sn+1 Taking into account that r is the length of the cycle, we find that q = n/r is an integer. From the proof of 3.12 we get that sign1(s,.+1) = del(o1...o,.) sign(sl)
Since sr+1 = s1, we get that del or = 1 which makes or a rotation. Observe that 27r = E i "1 LintAl = cl E i r l LintAl
and the result follows.
Proposition 3.16 Let P be
a locally finite convex fundamental domain for
F and A E 0hP a cusp for r. The number of sides of P with end A is two.
Proof Let us first observe that for any point Q E P and any point A E 8hP the geodesic segment [Q,A[ is contained in P as follows from 3.1.
Let us assume that A E 8hP is a cusp and let us pick a horocycle D with center D. Recall from 2.6 that D meets at most finitely many translates of P and conclude that D meets at most finitely many sides of P. This allows us to shrink
D such that all sides of P which meets D have ends at A. Let us observe that D fl P # 0 and D T= P and conclude from the connectedness of D that D f OP # 0.
It follows from 3.7 that D f OP is traced on D by a number of geodesics hl,...,hr through A.
It follows from this that D fl P is fibred by traces of geodesics.
Observe also P fl yP = 0 for a generator y of r A. The remaining details are left of the reader.
V.4 DIRICIILET DOMAINS
167
V.4 DIRICHLET DOMAINS
We shall present a method due to Dirichlet for construction of fundamen
tal domains for a discrete group r of isometrics of H2. Let us recall from IV.1.9 that for a compact subset K of H2 we have that o(K) meets K for finitely many o only.
This allows us to fix a point w E H2 with r,, = {t}. For a nontrivial o E IF let Lo(w) denote the perpendicular bisector for w,
o(w) and let Ho(w) denote the open halfplane of the complement of Lo(w) containing w.
Lo(w)
4.1 w
The Dirichlet domain with centre w is the set P(w) given by
fl
P(w) =
4.2
Ho(w)
CEr,o#t
Proposition 4.3
The Dirichlet domain P(w) is a fundamental domain for
IF in H2. Its closure P(w) is given by
P(w) =
I
I
Ho(w)
or E r,C
Proof Let
us verify that the intersection P(w) fl D(w;r) is open for all r > 0.
Observe that the set S = {o E r{t} I o(w) E D(w;2r)} is finite and that
P(w) fl D(w;r) = n a E S Ho(w)
FUNDAMENTAL DOMAINS
168
We find that P(w) is open, since P(w) fl D(w;r) is open for all r>O. Next, we ask the reader to verify the formula
Er 7(P(w)) = P(7(w)) From this it follows immediately that P(w) and y(P(w)) are disjoint for 7 54 t. It is now time to verify the formula for P(w). Observe that the left hand side is included in the right hand side since the intersection of a family of closed
sets is closed. To verify the opposite inclusion, consider a point z E H2 which belongs to Ho(w) for all o E r{t). Notice that [w,z] belongs to Ho(w) for all o and conclude that [w,z[ belongs to P(w); it follows that z E P(w).
Let us show that any orbit A meets P(w). To this end pick a z E A with the shortest possible distance from w. This gives d(z,w) < d(o 1(z),w)
or d(z,w) < d(z,o(w)) for all o
; o E IF, o 0 t
t, which shows that z E H. for all o E F 1t). It
follows that z E P(w), since P(w) is the intersection of such sets.
Proposition 4.4 Let r be a discrete group of isometries of H2 and let w be a point of H2 such that rN, = {t}. The Dirichlet domain P with centre w is a locally finite convex fundamental domain.
Proof
Let us fix a number r > 0 and show that the open disc D(w;r) with centre w and radius r > 0 meets o(P) for at most finitely many o E 17. If o(P) meets P, we can find z E P with o(z) E D(w;r); this gives us
d(w,o(w)) d(f(u),f(v))
; u,v E X
Let us fix a point x E X and let W be an open neighbourhood of x which is mapped by f homeomorphically onto a hyperbolic disc D(o(x);r) in Y (r > 0). For
points u and v in W let ) be the curve in W such that fA is the geodesic arc from f(u) to f(v). This gives d(u,v) < 1(fa) = d(f(u),f(v)) and we conclude that d(u,v) = d(f(u),f(v))
;
u,v E W
Let us show that d(x,z) > 0 for any point z # x. If z E W, this follows from the formula above. For z 0 W, let y be any continuous curve in X with fy rectifiable
which joins x and z. Pick P E ]0,r[ and let us show that fy meets the circle S in Y
with centre y = o(x) and radius p. To see this, let us prove that y meets the inverse image S' of S by f:W*Y (we shall give an outline only since a similar argument has been used in the proof of 1.3). Observe that the complement of S' in X is partitioned into two open sets: the inverse image D' of D(y;p) by f:W>Y
and the complement in X of the inverse image K' of D(y,p) by f:W;Y. Finally use that the image of 7 is connected to conclude that y meets S'. It follows that 1(fy) > p. Let us make a variation of y and conclude that d(x,z) > p. We claim that a continuous curve y:[a,b],X is rectifiable in X if and only
if fy is rectifiable in Y. To see this, use the BorelHeine theorem to introduce a subdivision a = ao < al... < an = b of [a,b] such that the restriction y:[a1l,ai],X takes values in a hyperbolic disc W as considered above. It follows that
1(7) = E; 1(yi) = Ei 1(fy;) = 1(fy) This reveals that X has the shortest length property.
D
VI.2 IIOPFRINOW THEOREM
VI.2
175
HOPFRINOW THEOREM
In this section we shall be concerned with the geodesics on a hyperbolic surface X. The basic problem is to join two points x and y of X by a geodesic
curve of arc length equal to the distance between x and y. Substantial results will only be obtained when X is complete in the sense that any Cauchy sequence on X is convergent.
To begin let us start with a geodesic curve y:J*X defined on the interval J, compare 11.1, and let us show that the following inequality is satisfied d(7(s),7(t)) X.
Proof Let
us first remark that the result is true for X = H2 as follows from the
proof of 11.4.8. In the general case let us assume that 0 E J and try to extend y to
the right of J. To this end put j = { x E ]0,+co[ 1 y extends to ]0,x[ }
The set
is a subinterval of ]0,+oo[. Let us prove that a E I implies a+e E j for
some a>O. From the inequality 2.1 we conclude that y:[O,a[+ X is uniformly conti
nuous. From the completeness of X it follows1 that y can be extended to a continuous curve on [O,a]. In order to extend y to the right of the point a consider a
hyperbolic disc in X with centre y(a) and use the result for H2. It follows that I = ]0,+oo[. Extension of y to the left of J can be done in a similar way.
Let us now consider two fixed points x and y of a hyperbolic surface X.
The distance d = d(x,y) can almost (1.2) be realized as a the length of a curve connecting x to y, but not quite in general. Consider for example the punctured Poincare disc D{0} and the points x = 1/2 and y = 1/2. On a complete surface, however, we can realize the distance as the length of a geodesic connecting x to y:
Hopf Rinow theorem 2.4
Let X be a complete hyperbolic surface.
For any two points x and y on X of distance d = d(x,y), there exists a geodesic curve o:I2+X with a(0) = x and o(d) = y.
1 We are referring to the following general fact: let X be a complete metric space, any uniformly continuous curve y:[0,a[+X can be extended to a continuous curve on [0,a]. A proof can be based on the observation that a uniformly continuous map y preserves Cauchy sequences.
VI.2 IIOPFRINOW THEOREM
177
Proof Let us introduce an abbreviation which will facilitate the exposition and underline the strategy of the proof %%(x'z'y) :
x and z satisfy the HopfRinow theorem and d(x,z) + d(z,y) = d(x,y)
J
Let us first show that for any two distinct points x and y of X, we can find z # x such that %%(x,z,y). To this end choose an r E ]O,d[ such that the disc D(x;p) is hyperbolic for some p > r. The circle S(x;r) with centre x and radius r is compact as follows from 1.3. Let the continuous function z ,, d(z,y)
;
z E S(x;r)
take its minimum at the point z E S(x;r). Consider a rectifiable curve y:[a,c], X with y(a) = x and y(c) = y. Choose a point b E ]a,c[ with d(x,y(b)) = r. We have 1(y) _ k(Y(a,h)) + 1(y[b,e]) >_ d(x,7(b)) +d(7(b),Y) > r+d(z,Y)
Take infimum over all rectifiable curves y from x to y to get d(x,y) > d(x,z) + d(z,y)
The opposite inequality is the triangle inequality and we have verified ]b,(x,z,y). Let us establish the basic relation
2.5
J
R (x,Z,Y) & Jt%(Z,w,Y)
N%(x,w,Y)
; x,Y,z,w E X
Simple manipulations of the two identities and the triangle inequality yield d(x,w) + d(w,y) < d(x,z) + d(z,w) + d(w,y) = d(x,z) + d(z,y) = d(x,y) Another application of the triangle inequality gives us d(x,w) + d(w,y) = d(x,y)
Put this information back into the first inequality and deduce that in fact d(x,z) + d(z,w) = d(x,w)
It follows from this and lemma 2.6 that x and w satisfy the HopfRinow theorem.
To conclude the proof, let x and y be points with d = d(x,y). Pick a point
v $ x satisfying N (x,v,y), put e = d(x,v) and use 2.3 to construct a geodesic curve o:R X with o(0) = x and o(e) = v. We shall study the set J = { s E [O,d] I d(x,o(s)) = s & d(o(s),y) = d  s }
It is easily seen that the set J is a closed subinterval of [O,d] containing 0. Let us
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178
show that s E ]O,d[ fl J implies t E J for some t > s. Observe that z = a(s) satisfies 3G%(x,z,y). Pick a point w $ z such that 3G%(z,w,y) and conclude from 2.8 that
36,(z,w,y). It follows from 2.6 and 2.3 that w belongs to the image of o. The parameter t E R with o(t) = w satisfies our requirements. It follows from this that
0
J=[O,d]. In particular d(o(d),y) = 0, i.e. o(d) = y.
Lemma 2.6 Let
r :[a,c]+X be a curve in the metric space X and let there be
given b E [a,c] such that d(7(a),7(b)) + d(7(b),7(c)) = d(7(a),7(c))
If the restrictions y:[a,b]+X and y:[b,c]X are distance preserving, then the curve y:[a,c]>X is distance preserving.
Proof
Let us first prove the following simple inequality
d(y(s),y(t)) < t  s
; s,t E [a,c], s < t
It suffices to treat the case s E [a,b] and t E [b,c]; the triangle inequality yields
d(y(s),y(t)) < d(7(s),y(b)) + d(y(b),7(t)) = b  s + t  b = t  s which completes the proof of the inequality. inequality as well.
We need to prove the opposite
To this end let us observe that
c  a = (c  b) + (b  a) = d(y(a),y(c)) < d(y(a),y(s)) + d(y(s),y(t)) + d(y(t),y(c)) Let us combine this with the first inequality to get that
ca<sa+d(y(s),y(t))+ct which may be rewritten t  s < d(y(s),y(t)).
0
VI.3 UNIFORMIZATION
VI.3
179
UNIFORMIZATION
In this section we shall show that a complete hyperbolic surface X admits a local isometry of the form f.H2+X , where H2 is the hyperbolic plane. But first
a result to the effect that plane figures cannot be isometric without being congruent. Such figures can't be deformed: they are rigid.
Rigidity 3.1
Let D be a subset of H2 not contained in a hyperbolic line.
Any isometry o:DJH2 has a unique extension to an isometry of H2.
Proof
Let us pick three points A,B,C of D not on a geodesic. We can use
lemma 11.4.8 to pick an isometry r:H2_H2 which agrees with o at the three points A,B,C. It follows from the proof of 11.4.9 that the restriction of r to D is actually
0
equal to a.
Uniformization theorem 3.2
Let D be an open disc in the
¢:D'X of D into a complete hyperbolic surface X has a unique extension to a local isometry H2 X. hyperbolic plane H2. Any isometry
Proof
Let 0 denote the centre of D and let us use 2.3 to extend the isometry D+X to a map 0:H2.X whose restriction to each hyperbolic line through 0 is a geodesic. Let us fix a point A of H2 and proceed to find an open disc with centre A which is mapped isometrically into X by 0. To this end use the
BorelHeine theorem to find an r > 0 such that for all C E [0,A] the metric disc D(O(C);2r) is a hyperbolic disc.
Let us first prove that
d(,A(A),O(B)) = d(A,B)
;
B E D(A;r)
To this end consider a subdivision of the segment [0,A]
0 = A0, A1,.... Ai_1, A = A
d(Ai_1)A;) < r
,
i = 1,...,n
and let B0,B1,...,Bn be the perpendicular projections of the sequence of As onto
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180
the line through 0 and B. Let us prove2 that for i = 1,...,n ,,,
LAi1Bi1A1
(Ai1)4(Bi1)0(Ai)
Odi(Bi1)0(Ai)0(Bi)
OB11A1Bi
where
z
means that the triangles are congruent in the sense that distances between
pairs of corresponding vertices are equal.
This is done by induction on i. To accomplish the induction step, let us focus on the hyperbolic discs Di = D(Ai,2r) and Si = (O(Ai);2r). Observe at first that the restriction 0:[Ai,Ai+1]'X has image in `11i and conclude that d(Ai,Ai+1) = d(0(Ai),0(Ai+1))
From the induction hypothesis we get that d(0(Ai),0(Bi)) = d(Ai,Bi) < d(An,Bn) < r
Let us combine this with the inequality
d(0(Bi),q(T)) < d(Bi,T) < r
; T E [B1,Bi+1]
to conclude that the restriction m:[Bi,Bi+1],X has image in `Di. This gives d(0(Bi),0(Bi+1)) = d(B1.Bi+1)
Let us now focus on the following configuration in Di
and the image configuration in `.bi. From the induction hypothesis we conclude that LAi1A;B11
LBiiA;Bi
= L0(Ai1)0(Ai)0(Bi1) = L0(Bi1)c5(Ai)c5(B;)
with the conclusion that
2The details of the proof are elementary and can be administered in a number of ways. The reader is advised to work out his own version.
VI.3 UNIFORMIZATION
181
LA;_1A;B;
L0(Ai1)0(Ai)0(Bi) From this we draw the same conclusion for the complementary angle LAi+1A;Bi
=
AAi+IAiBi
=
A0(Ai+1)0(Ai)0(Bi)
LBi1BiAi
=
Ld'(Bi1)0(Bi)0(Ai)
LA,B;AJ+1
=
LAi+1BiBi+1
=
L0(Ai+1)0(Ai)0(Bi) We can now use elementary geometry to conclude that
In a similar way we can first deduce that
with the conclusion that Another reference to elementary geometry gives us AAi+1B;Bi+1
A0(Ai+1)0(Bi)0(Bi+1)
which concludes the induction. Let us now prove that AAnB11B
OO(An)O(Bn)O(B)
Observe that LBn=LO(Bn) is a right angle, and conclude as above that the corresponding sides have equal length. In conclusion d(A,B) = d(O(A),0(B)). Let
us remark that at the same time we have proved that the following two angles are equal LAn_1AB = L0(Ai1)0(A)0(B).
It remains to prove that the distance between any two points B and C of D(A;r) is preserved.
To see this, observe that the proof above shows that the
angle LOAB is preserved by 0. A similar remark applies to the point C and the result follows by hyperbolic trigonometry, 111.5.1.
The proof of the uniformization theorem is adapted from the Euclidean case given in [Nikulin, Shafarevich] II, §10. The reader is advised to consult this source for further comments.
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VI.4
MONODROMY
In this section we shall be concerned with general properties of a local isometry f.X+Y between hyperbolic surfaces. It follows from rigidity 3.1 that a
local isometry is an open map in the sense that any open subset U of X is mapped by f onto an open subset f(U) of Y. Let us establish an important inequality 4.1
d(f(x),f(y)) < d(x,y)
; x,y E X
Proof Consider a point x E X and an r > 0 such that the metric disc D(x;r) is a hyperbolic disc and such that the restriction f:D(x;r)Y is an isometry. From the
fact that f is open it follows that D(x;r) is mapped homeomorphically onto an open neighbourhood of f(x) in Y. We conclude from lemma 1.3 that f(D(x;r)) = D(f(x);r)
Let there be given points x,y E X and a continuous curve y:[a,b]X with y(a) = x
and y(b) = y. We ask the reader to use the BorelHeine theorem to find an r > 0
such that for all s E [a,b] the point y(s) is the centre of a hyperbolic disc with
radius r such that f maps D(y(s);r) isometrically onto D(fy(s);r).
A simple
subdivision argument shows that y is rectifiable in X if and only if fy is rectifiable
in Y. For a rectifiable curve y in X we find that 1(fy) = 1(y)
The formula 4.1 follows from variation of y.
Geodesic lifting property 4.2
0
Let f: X+Y be a local isometry
between complete hyperbolic surfaces and consider points x E X and y E Y with
f(x) = y. For any geodesic curve 71:R:Y with ij(0) = y, there exists a unique geodesic curve £: RX with f{ = q and (0) = x.
Proof
Choose r>0 such that f induces an isometry D(x;r) D(y;r). Next,
choose an open interval J around 0 in R such that 71(J) C D(y;r). Let f:JD(x;r)
VI.4 MONODROMY
be given by ff =,9.
183
Let us use 2.3 to extend
that the identity ff _ 17 is generally valid.
to all of O and use 2.2 to conclude
0
Proposition 4.3 A local isometry f:X*Y between complete hyperbolic surfaces is surjective. Moreover, let there be given a point y E Y and a p > 0 such
that D(y;p) is hyperbolic, then for any x E X with f(x) = y the disc D(x;p) is mapped isometrically onto D(y;p) by f.
Proof
We have already observed that f(X) is open. Let us show that f(X) is
closed. To this end consider a point w in the closure of f(X). Pick a hyperbolic disc D with centre w and a point y E D fl f(X). Let us join the centre w of D with y along the radius of D and extend this to a geodesic ij:l >Y. According to 4.2 we
From this it follows that all points of y can find a geodesic x:R,X with p = belong to f(X). Since f(X) is open and closed we can use the fact that Y is connected to conclude that f(X) = Y.
Let us now investigate the point x. For small values of r > 0, the map f induces an isometry D(x;r)+D(y;r).
The inverse of this induces an isometry
s:D(y;r)+X with fs = t and s(y) = x. Let us use the uniformization theorem 3.2 to
extend s to a local isometry s:D(y;p)X. The extension will still satisfy s(y) = x and fs = c as follows from 2.2 applied to the various radii in the disc D(y;p). In
fact s is an isometry: for u,v E D(y;p) we get from the inequality 4.1 applied twice
d(u,v) = d(f(s(u)),f(s(v)) < d(s(u),s(v)) < d(u,v) which shows that d(u,v) = d(s(u),s(v)). We conclude from the proof of 4.1 that s(D(y;p)) = D(x;p). Let us recall the relation fs = t and it follows that D(x;p) is mapped by f isometrically onto D(y,p).
We shall see that a local isometry f:X>Y between complete hyperbolic surfaces X and Y is a covering projection in the sense of homotopy theory. For more topological results and references see VI.8.
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Corollary 4.4
Let f:X Y denote a local isometry between complete
hyperbolic surfaces and let y E Y and p > 0 be given such that the metric disc D(y;p) is hyperbolic. For distinct points x and z of the fibre f'({y}) ; x,z E f1{y} , x # z
D(x;p) fl D(z;p) = 0
Moreover, any point w E X with f(w) E D(y;p) is contained in the disc D(x;p) for some x E f1{y}.
Proof Let
us put D = D(y;p) and make a general remark on two continuous
sections r,s of f:X}Y over D C Y. Let us show that
r=s
r(D)fls(D)54 0
We may write a common point u of r(D) and s(D) in the form u=r(vl)=s(v1), where v1,v2 E D. This, however, gives f(u) = v1 = v2. Thus we have found a v E D with r(v) = s(v). Observe that the connected set D is partitioned into the two sets
{wEDIr(w)0s(w)}, {vEDIr(v)=s(v)) The first set is open since r and s are continuous and X is Hausdorff. To see that
the second set is open consider a point v E D with r(v) = s(v); choose a neighbourhood U of r(v) = s(v) such that the restriction of f:X'Y to U is injective and choose a neighbourhood V of v in D with r(V) C U and s(V) C U. We have that
w= f(r(w)) = f(s(w)) which gives
r(w) = s(w) for all w E V.
;wEV
In conclusion we have partitioned the
connected set D into two open subsets of which the second is nonempty.
It
follows that r = s.
To conclude the proof of the first statement, observe that at a point of the
we can use the restriction of f to the disc D(x;p) to get an ,: D(x;p) D(y;p). The inverse s,, = f,' defines a section s, of
fibre x E f1{y} isometry
f:X*Y over D with s,(D) = D(x;p).
Consider a point w E X with d(f(w),y) = r < p. geodesic with q(O) = f(w) and il(r) = y.
Let q:R4Y denote a
Let us use 4.2 to find a geodesic e:R+X
with f. = p and (O) = w. This gives f(i(r)) = zl(r) = y. In conclusion we have found a geodesic {: RX with (0) = w and fi(r) = x E f1 {y} where r E ]0,p[. It follows that w E D(x;p) as required.
VI.4 MONODROMY
Proposition 4.5
185
Let f:XY be a local isometry of complete hyperbolic
surfaces. The metric on Y can be recovered from that of X by the formula
d(y,w) = inf { d(x,v) I x,v E X, f(x) = y, f(v) = w }
Proof Use the HopfRhiniw theorem 2.4 to pick a geodesic curve il:[O,d]>Y with 9(0) = y and q(d) = w, where d = d(y,w). Let :[O,d]*X be a lifting of rl to
X, compare 4.2.
This gives points x = (O) and v = (d) with f(x) = y and
f(v) = w. It follows from the inequality 2.1 that d(x,v) < d. The required identity is now a consequence of the general inequality 4.1.
A refinement of the argument above gives us
4.6
f(D(x;r)) = D(f(x);r)
Proposition 4.7
;xEX, r>0
Let Z be a complete hyperbolic surface. Any closed and
bounded subset of Z is compact.
Proof
By 3.2 we can find a local isometry from
f
that all closed metric discs in Z are compact.
Monodromy theorem 4.8 Any local isometry
Let X be a complete hyperbolic surface.
is a bijection.
Proof Let us first show that any local isometry o:H2+H2 is an isometry. To this end choose an open disc D C H2 such that the restriction o': D+H2 is an isometry.
Next use 3.1 to extend o' to a global isometry o": H2*H2. Observe
that or = o" as follows by restriction to hyperbolic lines through the centre of D.
In the general case, let us use the uniformization theorem 3.2 to construct
a local isometry ir:H2+X. consequently a bijection.
The composite Oir:H2H2 is a local isometry and We conclude from this that it is injective and from 4.3
that rr is surjective. This shows that 0 is a bijection as well.
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VI.5 ORBIT SPACES
Let us consider a hyperbolic surface X and a group G of isometries of X
which acts discontinuously on X. By this we mean that any point x E X has an open neighbourhood U such that
o(x) 0 U for all but finitely many o E G
5.1
For the Gorbit A through x we conclude the existence of r > 0 such that
D(x;r) fl A = {x} For two distinct points p,q of A we can choose o E G with o(p) = x and conclude
that d(o(p),o(q)) > r. Since o is an isometry we can conclude that
;p,gEA,p
d(p,q) ? r
5.2
Proposition 5.3
q
A discontinuous group G of isometries of the hyperbolic
surface X acts with closed orbits on X.
Proof
Let A be a Gorbit on X and z a point of X outside A. With the
notation of 5.2 the disc D(z; r/2) intersects A in at most one point. From this it is
0
easy to find a disc with centre z not meeting A.
We intend to put a metric on the space X/G of orbits. To get started let us introduce the distance between two Gorbits A and `.9 by the formula
5.4
d(A,%) = inf{d(x,y) I xEA,yE`.B}
Let us fix points x E A and Y E S. We can rewrite the definition above as
d(A,`$) = inf { d(o(x),r(y))
I
o,r E G }
Since we can write d(o(x),r(y)) = d(x,oir(y)), we conclude that
VI.5 ORBIT SPACES
5.5
187
d(A,`.B) = inf { d(x,y) I y E `
Proposition 5.6
;xEA
}
The distance function defines a structure of metric space
on the orbit space X/G. This metric has the shortest length property.
Proof
The distance between two orbits is symmetrical by 5.4. In order to
prove the triangle inequality for orbits .A,`B,C let us fix a point y E S. This gives
d(A,C) < d(x,z) < d(x,y) + d(y,z)
;
x E .A, z E C
Fix x E .A and take infimum over all z E C and conclude from 5.5 that
d(A,C) < d(x,y) + d(%,C)
;
XEA
Take infimum over all x E A and apply 5.5 to get
d(A,C) < d(A,B) + d(B,C) If A 54 `9 we conclude from 5.5 and 5.3 that d(A,`.B) > 0. Let f:XX/G denote
the projection. Observe that f decreases distances to conclude that a rectifiable curve y in X joining x to v maps to a rectifiable curve f7 in X/G with 1(y) > 1(fy) > d(f(x),f(v))
For points y,w E Y choose x,v E X with f(x) = y, f(v) = w and d(x,v) = d(y,w). For c > 0 given, choose a rectifiable curve 'y in X joining x to v which satisfies 1(y) < d(x,z) + e. From the inequality above we get that 1(f7) < d(y,w) + e
0
which shows that X/G has the shortest length property.
It follows from 5.1 that the canonical projection f:X+X/G satisfies
5.7
f(D(x;r)) = D(f(x);r)
; x E X , r>0
Let us make a closer analysis of this in terms of the stabiliser of x given by
Gx={uEGIv(x)=x} Let us notice that Gx acts on any disc D(x;r) with centre x and radius r > 0. Thus we can form the orbit space D(x;r)/G.
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Proposition 5.8 tinuously on X.
Let G be a group of isometries of X acting discon
For any point x of X there exists r > 0 such that f induces an
isometry of D(x;r)/Gx onto D(y;r).
Proof Let us use 5.1 to choose r > 0 such that d(x,o(x)) > 4r
;
o E G  Gx
Using the triangle inequality we get for y E D(x;r) and a E GGx that
4r < d(x,o(x)) < d(x,o(y)) + d(o(x),o(y)) < d(x,a(y)) + r which allows us to conclude that d(x,o(y)) > 3r for y E D(x;r) and a E G  G. From this it follows that for z,y E D(x;r) we have that
inf{ d(z,o(y)) I o E G}= inf{ d(z,o(y)) I a E GX }
0
which is the required result.
Proposition 5.9 Let G be a discontinuous group of isometries of the hyperbolic surface X. The orbit space X/G is complete if and only if X is complete.
Proof When X is complete, we get from 5.7 that any closed metric disc in X is compact. It follows from 5.4 that any closed metric disc in X/G is compact. To
prove that a Cauchy sequence {z11) in X/G is convergent we can refer to the elementary fact that a Cauchy sequence is bounded, i.e. contained in some closed
metric disc in X/G. Let us finally recall the fact that a Cauchy sequence in a compact metric space is convergent.
Suppose conversely that X/G is complete and let us investigate a Cauchy sequence {x17} in X.
The sequence {p(x11)} is a Cauchy sequence in X/G and
converges towards a point y E X/G, say. Let 9J be the corresponding Gorbit in X
and choose r>0 such that the discs D(x;3r)x E c are disjoint. Let us choose an n E N such that d(p(xn),y) < r and such that d(xn,xp) < r for all p > n. Let the point x E qJ be given by the condition d(x1l,x) < r. It follows that
xp E D(x;2r)
;
p>n
Upon shrinking r we can obtain that the closure of D(x;2r) is compact, compare 1.3. It follows that the sequence {x11} is convergent in X.
0
VI6 CLASSIFICATION
189
VI.6 CLASSIFICATION
In this section we bring our theory of complete hyperbolic surfaces to culmination by classifying all such surfaces in terms of torsion free3 discrete subgroups of PG12(R).
Let us observe that a torsion free discrete subgroup I' of
PG12(R) acts discontinuously on H2 with trivial stabilisers.
It follows from 5.8
that the orbit space H2/I is a complete hyperbolic surface.
It is useful to observe that the canonical projection a:H2+H2/I' has the following mapping property: a map f:H2/F+Y into a hyperbolic surface Y is a local isometry if and only if fir:H2+Y is a local isometry.
Theorem 6.1
Any complete hyperbolic surface X is isometric to a surface
of the form H2/F where r is a torsion free discrete subgroup of PGl2(03). Two such subgroups r and E define isometric surfaces H2/F and H2/E if and only if r and E are conjugated subgroups of PG1.2(R).
Proof Given
a hyperbolic surface X. Use the uniformization theorem 3.2 to
pick a local isometry a:H2+X. Let us introduce the subgroup r of PG12(IR) by
r = { o E PGL2(R) I ro = it }
Let us investigate the action of r in a neighbourhood of a point z E H2. To this
end pick an open disc D with centre z such that the restriction of a:D>X is injective. Let us show that
o(z)0D for all oEF  {t} Assume for a moment that o E f satisfies o(z) E D. From a(o(z)) = a(z) we conclude that o(z) = z, which implies o(D) = D. This gives ir(o(x)) = a(x) for all
x E D and therefore o(x) = x for all x E D; thus 3.1 gives us or = t. This shows that r acts discontinuously on H2 without fixed points. We can now conclude that F is a discrete subgroup of PGl2(Od) which is torsion free: elements of finite order
of PG12(R) are rotations and reflections, but both types have fixed points. 3
A group r is torsion free if all nontrivial elements have infinite order.
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Consider points z,w E 112 with x = a(z) = a(w). Let us show that we can
find a transformation y E r with y(z) = w. The restrictions it
and
?r,,:D(w;r)>D(x;r) are isornetries for small values of r > 0. Introduce the isometry 0 = aw 11r : D(z;r) ' D(w;r)
and use rigidity 3.1 to extend 0 to a global isometry o of H2 into itself. The identity ao = n holds in D(z;r), but it extends to all of H2 by rigidity 3.1. We can now use 7r:H2>X to produce a bijection 0: H2/r
X. But 4.5
allows us to conclude that 0 is an isometry.
Let us now study local isometries from H2/r to H2/E.
Let us first show
that a o E PG12(U8) with arc r1 C E induces a map
6.2
H2/r
H2/E
;
; orf 1 C E
ll H or (h)
In other words, we must verify that " x  y mod r
o(x)
u(y) mod E ".
To this end we write y = 7(x) with y E r to get o(y) = oy(x)) = (uyo1)o(x) and
observe that oyo1 E E to get o(x) = o(y) mod E as required. It follows from the remarks from beginning of this section that the map 6.2 is a local isometry.
Let us finally verify that the transformation 6.2 is an isometry if and only
if oro' = E. To see this make the factorization of 6.2
H2/r
H2/oro 1 , H2/E
where the first map is the isomorphism induced by o and the second map is simple
projection. This second map is bijective if and only if oro 1 = E.
Corollary 6.3
0
Let r c PG12(R) denote a discrete torsion free subgroup.
The automorphism group of H2/r is isomorphic to Nr/r where the normaliser AT of r is given by
AT ={oEPGl2(R)Ioro1=r}
VI.7 INSTANT JET SERVICE
191
VU INSTANT JET SERVICE
Let us consider a group I' of isometrics of a hyperbolic surface X and a closed subset Z of X with the property that the family (QZ)a E I. forms a locally finite covering of X. This assumption implies that r acts discontinuously on X.
The action of r induces an equivalence relation Q on Z and the projection
f:X X/I' induces a homeomorphism Z/gj
X/1', as follows from the proof of
V.2.1. We are interested in transporting the metric from X/1' to Z/9. Let us illustrate this by the analogous example of a cylinder realized as the
orbit space of the Euclidean plane X under the group r generated by a single parallel translation r; the set Z is a plane strip perpendicular to the translation direction and of width equal to the length of the translation. The corresponding examples in hyperbolic geometry are treated in 7.7. These examples are basic for
the proof given in VII.3 of the Poincare theorem in the case of unbounded polygons.
Definition 7.1
Let fj be an equivalence relation on the metric space Z.
The distance between two equivalence classes X; and W is defined by n
d(%,4V) = 111f E d(zl1,wl) 1=1
where the infimum is taken over finite sequences in Z of the form ZQ,...,Zn1, W I ,..., Wn
where
;zoE%,wnE`W
zi = wi mod 9 for all i = 1,...,n1.
This "instant jet service" distance on Z/gj is symmetrical and satisfies the triangle inequality. Two distinct equivalence classes may have distance 0:
Example 7.2
Consider the closed interval Z=[0,1] with the identifications
t = 1 t t E ]0,1[. The "instant jet service" distance between {0} and {1} is 0.
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Lemma 7.3
Let Z be a metric space equipped with an equivalence relation
(j, and let f:Z+W be a distance decreasing map into a second metric space. If f is constant on the equivalence classes modulo g, then f induces a distance decreasing map f:Z/g+W.
Proof Let us first observe that a sum of the type which occurs in 7.1 gives rise to the following inequality, where we have put z11 = wn n
n
n
E d(zi1,wi) ? E d(f(zi1),f(wi)) ? E d(f(zi1),f(zi)) ? d(f(zo),f(wn))
i=1
i=1
i=1
Take infimum over all sequences of the type considered and conclude that the map f:X/g>W is distance decreasing.
Proposition 7.4
Let r be a group of isometries of a hyperbolic surface
X and Z a closed subset of X whose I'translates form a locally finite covering of X.
The projection f:X X/I' induces an isornetry
f:Z/gj =+ X/1'
where gj is the
equivalence relation on Z induced by the action of r on x and the metric on Z/fj is instant jet service.
Proof A simple application of lemma 7.3 shows that f:Z/gj
X/I decreases
distances. Let us prove that the inverse map f1 decreases distances. To this end
we consider points z,w E Z with equivalence classes Z,W E Z/gj and show that
d(z,u(w)) > d(%,°W)
;
o E 1'
Let us fix c E I' and consider a rectifiable curve y:[a,b]X with y(a) = z and y(b) = c(w). We shall show below that there exists a subdivision of [a,b] a = to < t1... < to = b and elements i = c1,...,cn = c of r such that
7.5
7(ti) E ci(Z) noi+1(z)
Let us use 7.5 to pick a sequence of points z = y(ti) = ki(wi) = i+1(zi)
;
i = 1,..,n1
w of Z with ;
i = 1,..,n1
VI.7 INSTANT JET SERVICE
193
This gives us the estimate n
n
1(7) ? . d(y(ti1),7(ti)) _
d(o;(zi_1),oi(wi)) _ . d(z;_1,wi) ? d(`%,`W)
Finally, make a variation of y and deduce that d(x,o(w)) > d(Z,` V) as required.
Let us start the construction of a subdivision of the interval [a,b] subject
to 7.5 by observing that y([a,b]) meets only finitely many I'translates of Z as follows from BorelHeine. When y(b) E Z we can use n = 2 and ti = b. Let us assume that o(w) 0 Z and let tl E [a,b] be the last time y leaves Z: t1 = sup{t E [a,b] I y(t) E Z)
Let us observe that any union of I'translates of Z is a closed set as follows from
the fact that the covering of X with Ftranslates of Z is locally finite. We can now conclude that 1(t1) E U ,
54 ,
o(Z)
since the alternative would make y(t1) an interior point of Z, contradicting its definition. Choose 02
of = c such that y(tl) E o2(Z). When y(b) E o2(Z) we
can use n = 3 with t2 = b. If y(b) ¢ o2(Z), let t2 E [a,b] be the last time y leaves o2(Z) and observe that
y(t2) E U o 0 al p2
o(Z)
since the alternative makes y(t2) an interior point of ol(Z) U o2(Z) contradicting
the definition of t2 and y(b) 0 o2(Z), that is t2 < b. This procedure terminates since y meets only finitely many Ftranslates of Z.
We are going to stydy some examples which are usefull for the understanding of Poincares theorem in VII.3 in case of unbounded polygones.
Example 7.6
Let h and k be the geodesics in the Poincare halfplane H2
given by Re[z] = 1 and Re[z] = a where a > 1. Observe that z+az transforms h into k.
Let `R, denote the equivalence relation in the strip Z in H2 bounded by h
and k based on the identification z " az , z E h.
We are going to show that Z/`J
is not complete. (For a different approach see exercise 5.2).
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194
h
k
zi=awi
wi=zi 1a+1
zi1
Consider the sequence {zi}i E N in Z given by the recursion formulas above starting from zo=a+ib, b > 0. It will be useful to observe that Im[zi] = a'b, i E N. Let us recall from 11.8.2
d(z,w) < 2 sinh 2 (z, w) = 1w  zi
(Ir[w]Im[z])1/2
Let us use this to estimate the distance in Z/% d([z]i,[zi1]) = d([wi],[z1_1]) < d(wi,zi1) < (a 
1)b1a1i
;
i>1
Recall that a > 1 to see that the infinite series Ed([zi_1],[zi]) is convergent and conclude that [zi]i E N is a Cauchy sequence in Z/R. In order to show that our sequence is not convergent, consider the function f:ZHIS given by f(z) = y/x. The function f induces a continuous function on Z/% (see exercise 5.2). If our sequence was convergent, then f would be bounded on the sequence, which is not the case.
Proposition 7.7
Let Z denote the closed subset of H2 bounded by two
nonintersecting geodesics h and k with ends A,B and C,D in the relative positions
shown below. Let o be an even isometry with o(A) = D and o(B) = C. The equivalence relation given by the identification z  o(z), z E h, is denoted %. If the ends of h and k are distinct, then Z/% is complete. If h and k have a common end A = D, then Z/% is complete if and only if o is a horolation.
VI.7 INSTANT JET SERVICE
Proof
195
Suppose that the ends of h and k are distinct. Consider a sequence
(wn)n E N of points of Z representing a Cauchy sequence in Z/`3G. We shall prove
that the sequence is bounded in H2. {wn}n
EN
Once this is done, we conclude that
is contained in a compact subset of Z. It follows that our Cauchy
sequence is contained in a compact subset of Z/%, which makes it convergent. Let us introduce the geodesic s through A and D and consider the function
f,:H2+R which is zero on the halfplane containing B and C, while its value at a
point z on the other halfplane is the hyperbolic distance from z to s. We ask the
reader to show that f,:X;R is distance decreasing and we conclude from lemma 7.3 that the sequence f,(zn)n E N is a Cauchy sequence. This implies that we can find a constant c such that
fs(z1l) < c
;nEN
The same conclusion is available for the geodesic through B and C.
In order to show that our initial sequence is bounded away from the points
A and D, pick a horocycle A with centre A and define the polar function pA:H2+R which is zero outside the horodisc bounded by A while the value at a
point z of the horodisc is the distance from z to A measured along the geodesic through z and A.
Let us pick a second horocycle 9 with centre D disjoint from A
and such that c(A fl h) = 9 fl k.
The function
p = 2(pA + pct) is distance
decreasing and constant on the equivalence classes modulo `J,. It follows from
lemma 7.3 applied to p:Z*R that p is bounded on our sequence.
The same
conclusion is available for the points B and C. Thus we have proved that our sequence is bounded in H2.
Let us now assume that A = D and that a is a horolation with centre A. Pick a horocycle A with centre A and observe that the function PA is constant on the equivalence classes of !. As above we conclude that Z/`R is complete.
When A = D and a is a translation we can use example 7.6 to conclude that Z/% is not complete.
0
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V1.8 HOMOTOPY
In this section we shall study homotopy classes of paths on a complete hyperbolic surface X. The key fact of topological nature is 4.4 according to which
a local isometry f:X+Y between complete hyperbolic surfaces is a covering projection in the sense of 8.1 below. For proofs of 8.2 and 8.3 see [Massey].
Definition 8.1 A continuous map f:X=Y is called a covering projection if every y E Y has an open neighbourhood D such that for each x E f1{y} there exists a continuous section sX: D+X with s`(y) = x inducing a homeomorphism
f1(D)
f 1{y} x D
Unique lifting lemma 8.2
;
(x,v)'+sx(v)
Let f:X+Y denote a covering projection
and r,s:W+X continuous maps with fs = fr. If W is connected and X is Hausdorff,
then s(w) = r(w) for some w E W implies that s = r.
Homotopy lifting lemma 8.3
Let f:X+Y be a covering projection.
For any topological space T, any solid diagram of continuous maps T x [0]
+0
I T x [0,1]
X
I
,
f
Y
can be completed with a dashed arrow to a commutative diagram in the category of continuous maps.
Theorem 8.4
Let x and y be distinct points of the complete hyperbolic
surface X. Any homotopy class of continuous curves joining x to y can in a unique way be represented by a geodesic curve joining x to y.
VI.8 IIOMOTOPY
Proof
197
Let us use the uniformization theorem 3.2 to find a local isometry
f:H2+X and let us use 4.3 to pick a base point z E H2 with f(z) = x. For a curve
7:[a,b]+X with y(a) = x and y(b) = y there is a unique curve ,6:[a,b]+H2 with
,6(a) = z and f,6 = y. It follows from 8.2 and 8.3 that the homotopy type of ,6 is determined by the point ,3(b) E f1 {y}.
The image in Y of the geodesic line
segment [,6(a) 6(b)] represents the homotopy class of ,6. The uniqueness in the statement follows from the geodesic lifting property 4.2.
Definition 8.5
By a closed geodesic we understand a geodesic curve
y:R+X such that there exists a > 0 with 7(x + a = y(x) for all x E R.
Let us recall that two continuous curves a,/3:S1X are freely homotopic if there exists a continuous map W:S1 x [0,1]X such that
'(s,0) = a(s) , 'Y(s,1) = Q(s)
Theorem 8.6
s E S'
Let X be a compact hyperbolic surface. A closed curve in X
not homotopic to zero is freely homotopic to a unique closed geodesic in X.
Proof
With the notation from the proof of 8.1, let us consider a curve
g:[0,1]X with g(0) = g(1) = x. Let h:[0,1]i[2 be a lifting of g with h(0) = z and let y E IF be determined by h(1) = yz. Since g is not homotopic to zero we have
y # t. We conclude from IV.5.6 that y is a translation with translation axis say k.
We intend to show that the restriction of the projection f:H2X to k defines a closed geodesic L+X which is freely homotopic to g. A free homotopy is formed by
the restriction of f:II2*X to the hyperbolic quadrangle given below
7z
The horizontal coordinate lines in the quadrangle are formed by yhypercycles.
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VI EXERCISES
EXERCISE 1.1 10
Let y:[a,c]*X be a continuous curve in a metric space X.
Show that for any point b between a and c the restrictions yl:[a,b]*X and
y2:[b,c]aX are rectifiable if and only if y:[a,c]*X is rectifiable. 2°
With the notation above assume that y is rectifiable and show that 1(7) = 1(71) + 1(72)
30
Show that a rectifiable curve y:[a,c]YX which satisfies
1(7) = d(7(a),y(b)) is a geodesic curve. Hint: Show that y is distance preserving.
EXERCISE 1.2 Let us fix a point u+iv of the Poincare halfplane and consider
the real function V)(x,y) in two variables given by 1G(x,Y) = d(x+iy,u+iv)2 10
Show that the Hessian of this function at the point u+iv is
= v(dx2 + dy2) Show that a continuously differentiable curve y:[a,b]+H2 is rectifiable d2.0(x,Y)
2°
with curve length
b
(x'(t)2 + Y'(t)2)/Y(t) dt
1(7) _
where 7(t) = (x(t),y(t)) , t E [a,b].
EXERCISE 1.3
a
Show that a circle in H2 with radius r has arc length 27r sink r,
compare VI.1. Show for example first that the length k11 of the edge of an inscribed
ngon is given by the formula sinh(k,i) = sinh(r) 2 sin(7f/n)
and use this to calculate the length of the circle as
EXERCISE 1.4
lien
n;00
nkn
Consider a fixed real number y > 0 and let Clf denote the
Euclidean line in the upper halfplane through iy and parallel to the xaxis.
VI EXERCISES 10
199
Show that 'J is a horocycle, and that the are length is given by
length (iy,x+iy) = 2°
Verify the formula
length (iy,x+iy) = 2 sinh 12d(iy,x+iy) EXERCISE 2.1
;
x>0
;
x>0
Y
Let X be a metric space and 7:[a,b]#X a geodesic curve. Show
that the curve y is rectifiable with arc length 1(y) = b  a. EXERCISE 5.1
Let X be a hyperbolic surface and G a group of isometries of X
which acts discontinuously on X and let f:X X/G denote the canonical projection.
Show that a subset U of X/G is open (in the sense of the metric on X/G) if and only if f"1(U) is an open subset of X. EXERCISE 5.2
Let a be the isometry of the Poincare halfplane given by zp az,
where a > 1 is a fixed real number. 10
Show that the group
IF
generated by a acts discontinuously on
X = {zEH2IRe[z]>0). 2° Let f:XQ be the function given by f(x+iy) = y/x, x+iy E X. Show that f induces a continuous function on the surface X/I'. 3°
Show that the surface X/I' is not complete. Hint: Use 5.9.
EXERCISE 7.1
With the notations and assumptions of proposition 7.7. Show
that the set X = U
EZ
onZ
complete if and only if X = H2.
EXERCISE 8.1
is open and connected.
Prove that Z/' is
Hint: Apply propositions 5.9 and 7.4.
Let I' be a discrete group of isometries of H2 and y E r a
horolation. Explain how to construct from y a closed curve in H2/17 unique up to free homotopy. Hint: Model the proof of 8.6.
VII POINCARE`S THEOREM
We have arrived at the central theme of the book, Poincare's theorem, which allows us to construct a discrete group from a given hyperbolic polygon with
side pairing and to write down generators and relations for the group.
The proof of the theorem is a beautiful application of the monodromy theorem VI.4.8. In order to display the geometric ideas of the proof, we have first
treated the case where the polygon is compact. The general case requires a more
technical argument based on completeness and is given in section 3.
Let me
mention that Poincare's original proof [Poincare] is incomplete in the noncompact case, compare [de Rham].
VII.1 COMPACT POLYGONS
Let us consider an ordinary polygon A in the hyperbolic plane H2 with a side airin
.
By this we understand an involution si+*s on the set of edges g of A
such that s and *s have the same hyperbolic length and such that 1.1
;sEg
1
Recall that an edge s is an oriented side and that s1 denotes the result of changing
the orientation. For s E 6 we let ors denote the isometry of H2 which maps *s to s
in such a way, that locally, the halfplane bounded by *s containing A is mapped
to the halfplane bounded by s but opposite A. The isometry as is called the it transformation determined by s. Let us at once observe that we have the
Reflection relations 1.2 1
or *s = as
;sEg
VII.! COMPACT POLYGONS
201
Given an edge s of A, we let Is denote the edge with the same initial vertex as s but different final vertex. This defines for us a second involution on 6, su s. The combination of these two operators defines the edge operator T:99 given by
;sES
IQs= f*s
1.3
The edge operator I on g is bijective and g is finite so we conclude that W has finite order. The cycles for ' on g are called edge cycles. The sequence of initial
points of an edge cycle is called a vertex cycle (a vertex cycle may contain repetitions).
Lemma 1.4
Let sl,...,s, be an edge cycle with vertex cycle P1,...,Pr and side
transformations
(oi = oS , i = 1,..,r).
The cycle ma o = 03
r
is a
rotation around P1 of angle (measured in the direction determined by A and sl) congruent mod 27r to the sum of the interior angles
LintPi + LintP2 + ... LintPr
Proof Let
us choose the orientation of 112 which places A on the positive side
of s1 near P1. With the notation of 111.2.4 we have
Lor(si,1si) = sign(si) LintP1 where sign(si) = ei equals +1 (resp.1) in case negative) side of si.
mod 27r z
;
i = 1,..,r
lies on the positive(resp.
From the very construction of the side transformations it
follows that
m = 1,...,r
sign(sm) = sign(s,u+1)
Let us multiply these equations together for m = 1,...,r to get that sign(s1) = det(ol...or) sign(sr+1)
Since sr+1 = s1 we conclude that the cycle map is even.
Let us proceed to
calculate the rotation angle of the cycle rrta.p. Let us prove a general formula
Ei=m...r.int Pi = emLor (sm, om...or(sl)) To this end we recall that si = oi(*si) , °i+1 = *si
,
(Si = 0i(si+1)
m = 1,...,r
POINCARE'S THEOREM
202
This will be done by decreasing induction on in. The case m = 1 runs as follows LintPr = Er' or(Sr'Tsr) = ErLor(Sr,0r(Sl)) The induction step m+1 follows from the formulas we have just derived Ei=m.... rLint Pi = emLor(sm'lsm) + em+1Lor(Sm+1, 47m+1"'0r(Sl))
emLor(Sm,lsm)+Em+l det(om)Lor(lsnl, am"'Qr(Si)) _ emLor(Sm,lsm) +EmLor(1Sm,Om...Or(S1)) = EmLor(sm, om...or(sl))
This concludes the proof of our formula.
Finally, put m = 1 in our formula and
0
recall that el = 1 to get the announced rotation angle.
Examples Consider
an 8gon with side pairing as indicated in the drawing.
P3
a=0a, 0 =0b,'Y=0c1 b=od. Edge cycle:
a*b"1*a 1b c
*d"1 *c1
d
Vertex cycle: P1P4P3P2P5P8P7P6 Cycle map: a/3 1a'113yb"ly"1b
The identification space is a sphere with 2 handles (oriented surface of genus 2)
Our next example is an isoceles triangle with the side pairing indicated on the drawing, the base edge c is paired with itself *c=c. The side transformation a=oa is a rotation with centre C, while y=ac is a reflection in the geodesic through c. C
Edge cycles
a
*a 1 ,
Vertex cycles
C
C
Cycle maps
a,
a
a c *a c ,
1
,
The identification space is a disc.
BAAB
ayay
VII.! COMPACT POLYGONS
203
Let us consider a different side pairing on the same triangle given below: again
The transformation y = o,
*c = c.
is
a reflection as before, while the
transformation a = oa is a glide reflection. Edge cycle:
a a 1 c *a 1 *a c'1
Cycle map:
a2ya 2y
Vertex cycle:
BCAACB
The identification space is a Mobius band.
Remark 1.5
Let us consider an edge cycle SI S2,...Sr. From sj+1 = 1*sj,
i = 1,...,r, we conclude that Jsj+1 = *si. This gives us ; i = 1,... I'(lsi+1) = ls, It follows that lsl,Jsr,lsr_1,...,1s2 form an edge cycle. Notice that the corresponding
vertex cycle is P1,P,,Pi_1,...,P1, which is nothing but the reverse cycle of the original vertex cycle.
It follows that a vertex cycle makes up a full equivalence
class of A under the equivalence relation generated by the side pairing. Let us observe that 1S111Sr,1sr_11 ...,152 = *Sr,*Sr_1,...,*S1
and conclude that the cycle map is
o,.
10,, 2 1...01 1, which is the inverse of the
cycle map of the original cycle.
Cycle condition 1.6
We say that an edge cycle s11 ...,sr satisfies the
cycle condition if the angle sum along the corresponding vertex cycle P1,..,Pr has the form
i=1...r
Li.,t P =
2'r nc
;
nc = 1,2,3...
The cycle map o = alor of an edge cycle which satisfies the cycle condition is a rotation about P1 of order no.1 'Cycles c with nc = 1 are sometimes called accidental cycles
POINCARE'S THEOREM
204
Poineare's Theorem 1.7
Let A be a compact convex polygon A with
a side pairing which satisfies the cycle conditions 1.6. Then, the group G genera
ted by the side transformations as, s E g, is a discrete group with A as a fundamental domain. A complete set of relations for these generators is the reflection relations 1.2 and the cycle relations o.cnc = d
one for each edge cycle c.
Proof Let I' denote the universal group on the generators as , s E g, subject to the reflection relations 1.2 and the cycle relations. The canonical map F+G is denoted
On the basis of these data we shall construct a topological space X
and a local homeornorphism f: X H2, compare the introduction to chapter VI. The set underlying X is the quotient space for the equivalence relation on r x A generated by relations of the form
1.8
(7QS,h)'.,('Y,vs(h))
;
'YEr,sEfi,hE*s
The equivalence class of the point (y,k) is denoted y. k E X. Let us equip I' x A
with the product topology: an open subset has the form U = U7 E r {0.) x U0. where U. is an open subset of A, u E r. The topology on X is the quotient topology: open subsets of X correspond to open,  stable subsets of IF x A. The evaluation map (y,h)i (h), I' x A+H2 is continuous and compatible with our equivalence relation, so it will induce a continuous map f.X*H2 y(h)
; 7EI, heA
The natural action of r on r x A induces an action of r on X through the formula a(13. s) = (a13). s
I',sEA
Let us make the observation that f:X H2 is Fequivariant in the sense that
f(rx)=rf(x)
;rEI ,xEX
Let us prove that X is connected. To this end let us note that sr+t.s induces a continuous section of f:XH2 over A and conclude that t.A is a connected subset
of X. For x E X given, let us construct a connected subset of X containing x and meeting t.A. It is clear from the construction that X is the union of all 17translates
VII.1 COMPACT POLYGONS
205
of t.0. So let us pick r E r with x E r.4 and let us write r = al...a. where the cis are side transformations with respect to the edges s1,...,sn of A. For i = 1,...,n we
get from 1.8 that t.0 meets oi.A, which implies that Oi...Oi_i.0 meets The chain
t.0 U o 1.0 U QiQ2.0 U ... 0,
10'2...On_1.0 U T.0
is a connected set of the required sort.
We proceed to construct some special neighbourhoods of a given point x E X.
For convenience we introduce the following terminology concerning the
equivalence class of r x A represented by x. But first let us agree that for z E A
and c > 0 we shall put A(z;c) = {x E i d(x,z) < (}.
Local tesselation property 1.9
We say that an equivalence class
C of F x A has the local tesselation property if the following three conditions are satisfied. 1° C is a finite subset of r x A C : (i 1,k1),...,(ilm,kn1)
t)i E r , ki E A
;
2° For e>0 sufficiently small, the set {,71) x A(k1;e) U ... U {77m) x A(km;f)
is a  stable subset of r x A and 3° the corresponding open subset of X 171.0(kl;r) U ... U is mapped by f bijectively onto the open disc D(z;c), z = 1(k1) _
'
We are going to show that all equivalence classes of r x A have the local tesselation property. Observe that if the equivalence class C has the local
tesselation property, then the same is true for the class yC for all y E F. Thus it suffices to verify that for each z E A, the equivalence class generated by (t,z) has the local tesselation property. Let us investigate a number of special cases.
The full equivalence class of a point of the form (t,z), z E Into, consists of
that point alone. Any subset of {t} x A is
stable in r x, and the open subset
t.IntA of X is mapped by f:X+H2 onto InI A.
The full equivalence class in r x A generated by a point of the form (t,a), where a is an interior point of an edge s, consists of two points (t,a) , (os,b)
; b = vs 1(a)
POINCARE'S THEOREM
206
For e > 0 sufficiently small, the following subset of I' x A
{t} x A(a;e) U {as} x O(b;()
is open and stable. The image in X of this set
t.A(a;e) U vs.O(b;c) is mapped by f:X'H2 onto D(a;e).
Let us now consider a vertex P1 of A. Pick an edge sl with initial point Pl and iterate the edge operator T to get a sequence of edges sl,...,sr,...,sd, where r is the length of the edge cycle and d = nr where n is defines in 1.6 such that
n1_1 ,...,r
Lint Pi = 2ir
We let P1,...,Pd denote the corresponding sequence of initial vertices and
the sequence of side transformations.
Let me stress that both sequences have
period r. Observe that P1}1 is the initial vertex of s1+1 and (si+1 = *si, this gives
us vi(Pi+1) = Pi for i = 1,...,r. We conclude that the full equivalence of class of the point (1,P1) in I' x A is (t'P1) , (o1,P2) , (C1O21P3) ,..., (Q1O2...Ud1'Pd)
; d = rn
Let us prove that this equivalence class of r x L satisfies the local tesselation property 1.9.
The stability condition follows basically by observing that the
boundary of O(Pi;c) is located on the edges si and f si and that Ili10Si) = oi1(*si1) = Si1
'
Si = o1(*Si) = o1(Ts1+1)
from which we deduce that 0'i1.1si = t. si_1
,
Si = 0'i 1si+1
It is now time to calculate the image of
in H2, that is
vlo2...vi_1O(Pi;e)
The sector A(Pi;e) is bounded by L(si,f s1). The image is bounded by the circle with centre P1 and radius a and the two edges 0"1"',7i1(Si)
,
.71...o11(jSi) = 01...0i_2(si1)
Let us use the notation from the proof of lemma 1.4 to calculate the oriented angle between these two edges L0r(ol...oi1(si)'Q1...Q11(lsi)) = det(o1)...det(ai1)Lor(sPlsi)
det(ol)...det(v1_1)sign(si)LintP1 = sign(s1) Lint Pi
Observe that if these angles are added from i = 1 to i = d = nr we get sign(sl)21r. From this it follows that our class satisfies the local tesselation property 1.9.
VII.! COMPACT POLYGONS
207
Let us show that the open subsets of X exhibited in 1.9 form a basis for the topology on X: with the notation above, an open subset U of X which contains
the point x representing the equivalence class C can be thought of as a  stable open subset V of r x A containing
(rj1,k1),...,(r11,,,k1 ).
This implies that V contains
{r1j} x A(k;;e) for i = 1,...,m and e > 0 sufficiently small as required. As a consequence of the local tesselation property, we find that f:X*H2 maps open subsets of X to open subsets of H2 . Indeed f is a local homeomorphism.
It is time to show that X is a Hausdorff space: let 10 C be equivalence classes where C is given by 1.9 and `D is given by
; j E I' , hl E A Let us consider a fixed pair ij and show that we can find (ij > 0, such that {t1i} x 0(k,;E) n
x A(hj;E) = 0
;
E E }O,Ejj]
If rh # tj any eij > 0 will do. If T1, = j then k, # hj and it suffices to make sure that A(k1;e) and A(hj;e) are disjoint.
As another consequence of the local tesselation property, we conclude that
the rtranslates of t.0 form a locally finite covering of X: let us consider the point x E X given by the equivalence class C from 1.9. If we let U denote an open
neighbourhood of x in X of the form displayed in 1.9, we have that
o(t.A) n u = 0
a7
711,...,17m
In particular, we have ax E U for finitely many o E I only; that is the action of r on X is discontinuous.
Let us prove that X/I' is complete. The continuous map AX, hp+t.h, induces a surjective map ._A>X/1' which implies that X/I is compact and therefore
complete. We conclude from VI.5.9 that X is complete. We can now use the monodromy theorem VI.4.8 to show that f:X?H2 is
bijective. Surjectivity of f means that H2
=UgEG g(A)
From the very definition of the equivalence relation
o.0nt.In(L =0
follows that
;0El,a}tt
The injective map f:X+H2 transforms disjoint sets into disjoint sets, thus we get
; a E r{t} 0(A) n InlZ = 0 This shows that the kernel of rG is trivial. It follows that the canonical map
POINCARE'S THEOREM
208
from the universal group r to the group G is a bijection and that A is a fundamental domain for the action of G on H2. 0
Parallelogram groups 1.10
Let us consider a parallelogram ABCD
in H2. By this we mean a convex quadrangle ABCD where the opposite sides
have equal length. The elementary geometry of the parallelogram tells us that a
halfturn with respect to the midpoint of AC transforms DABC into OCDA. Otherwise expressed, the diagonals have a common midpoint M and a halfturn with respect to M transforms the parallelogram into itself. The angle sum of the quadrangle is less that 27r, 111.9.7. Let us assume that this has the form
LA+LB+LC+LD = 2n
; n= 2,3,...
We can apply Poincare's theorem to the side pairing *AD = BC and *AB = DC. The corresponding side transformations are denoted or and 0.
Edge cycle :
a b *a 1*b1
Cycle map :
a/3a 101
Vertex cycle :
ABCD
The cycle transformation a,3a101 is a rotation around A of order n. We find that the parallelogram group G is the universal group on a and )3 subject to
(a)3a1,31)11=c
VII.2 TRIANGLE GROUPS
209
VII.2 TRIANGLE GROUPS
Let us consider DABC in the hyperbolic plane H2. Reflection in the geodesic BC is denoted a, reflection in CA is denoted ,3, while reflection in AB is denoted y. We shall assume that the angles have measures as follows
LA=p, LB=E, LC=r
2.1
p,q,rEN
Since the angle sum of ABC is < 7r we must of course assume that
2.2
;
p,q,r E N
Conversely, for any three integers p,q,r satisfying this condition, we can find DABC with angles of these measures; this triangle is unique up to congruence, compare 111.5.
The subgroup of Isom(H2) generated by the reflections a,(3,y is
denoted G(p,q,r). The isometry yQ is a rotation around A of angle 2a/p which makes y/3 a
rotation of order p, similar remarks apply to the vertices B and C. We can now
write down three reflection relations and three cycle relations for our three generators
2.3
a2 = t, l32 = i, y2 = t, (7/)p = t, (ay)q =
,
(a)3)r = t
Let us apply Poincare's theorem to L ABC with the trivial pairing *a = a, *b = b, *c = c. The edge cycles are
c b1
,
a c1,
a1 b
This realises the relations 2.3 as reflection relations and cycle relations.
A picture of the tesselation of the Poincare disc by the fundamental domain for the triangle group D(2,3,7) is given in the introduction.
POINCARE'S THEOREM
210
Let us work out generators and relations for the even part G+(p,q,r) of the
triangle group. It is quite clear that G+(p,q,r) is generated by
P2A=N7, P2B=rya
P2C=a13 From 2.3 we derive the following relations for the ps
2.4
,
P2A P2B P2C=' , P2AP=1, P2Bq=t , P2Cr=,
Let us show that the list 2.4 is a complete set of relations for G+(p,q,r). To this end we apply Poincare's theorem to the quadrangle ABCD where the point D is reflection of B in the geodesic through A and C.
Edge cycles: a*c 1, c 1*a, a, *a 1, c, *c.
The side transformations are oa=P2C
and o*c = P2A as follows from
P2C(D) = aa(D) = a(B) = B , P2A(B) = /37(B) =,3(B) = D Thus Poincare's theorem gives us the following complete set of relations (P2CP2A)q = P2C' = P2AP =
which is the same as 2.4.
We shall continue to give some applications of Poincare's theorem to compact polygons. The following notation is useful and will be exploited in VII.6.
Definition 2.5
By a boundary cycle of a polygon A we understand a
cyclic enumeration of its edges a1,...,a1,
equals the initial vertex of a1+1 for i = 1,...,n.
a1) where the final vertex of ai
Once the hyperbolic plane is
oriented, the positive boundary cycle is defined by requiring that locally A is situated on the positive side of the edges of the boundary.
VII.2 TRIANGLE GROUPS
Hyperbolic pants
211
Consider a rectangular hexagon `A in H2 with boun
dary cycle a1,...,a6 and side pairing *ai = ai, i=1,...,6. Poincare's theorem tells us
that reflections p1,...,p6 in the sides of `3v generate a discrete group H with fundamental domain %. A complete set of twelve relations is given below
pie ;
=t
,
P,Pi+l = Pi+1Pi
i = 1,...,6 (P7 = Pl)
Consider a second copy of the same hexagon and sew the two hexagons together
along the seems a2,a4,a6. The result is a pair of hyperbolic pants. In order to
realise this as an orbit space, consider the rectangular octagon A =
U p4(A)
with the side pairing given below (*a = a, *c = c, *e = e, *f = f).
c
The boundary cycle is a b c
*b1
d e f *e1.
The two side transformations are
ob = p2p4 and oe = p4p6. We leave it to the reader to apply Poincare's theorem and conclude that the group I' generated by P11 P3, P5, P64' P42' P26
Pij = PiPj
is a discrete subgroup of index 2 in II with fundamental domain A. The cycle and reflection relations may be rewritten in the following form
2.6
P12 =
P32 = t, P52 = t, P64 42P26 = t
P1P26 = P26P1 , P3P42 = P42P3 , P5P64 = P605
POINCARE'S THEOREM
212
Torus with a hole
Let us start with a right angled hexagon with
boundary cycle a1a2a3a,1a5a6 as before. This time we require a2 and a6 to have
the same length; formula III.8.1 implies that a3 and a5 have the same length as well.
Reflect this in the side a1 to get an octagon with the side pairing as
indicated
Edge cycles:
ab*a1*c, a*c1f1c1
The group is generated by the side transformations a (3 y r) 0 subject to
2.7
a,3a"1 =7,712=t, 2=I,'Y?11'y1=0
It follows that the group is generated by a, 3 q subject to the sole relation q2 = 1.
Example 2.8
Let us fix a pentagon A with angle sum inn and consider the
side pairing with boundary cycle a*b"1*a1 be
in particular the edge c is reflected in itself. The edge cycle is
ab*a1*b"1cb'1a1*b*ac1 With the convention a = oa... we find a complete set of relations (a,3a"1)3ly fa,6la 1y)"
It is convenient to introduce e =
[a,3]"1
= c, y2 = t
and C1y( = 6 to get
[a,3]e,6=c1'Yc,(6Y)n=e,62=y2=t The orbit space is a torus with a hole and a marked point on the boundary.
VII.3 PARABOLIC CYCLE CONDITION
213
VII.3 PARABOLIC CYCLE CONDITION
Let us consider a polygon A in H2 with some of its vertices on OH2. To be sure the sides of the polygon are geodesic arcs in H2 or arcs of OH2. The edges
supported by arcs on the boundary are called free edges.
The set of edges
including the free ones is denoted 9. Let there be given a side pairing that is an involution si *s on g with
3.1
=(*s)1
;sE9
The free edges are supposed to be fixed points under * while the nonfree edges are subject to the following conditions: if the initial, resp. final vertex of the edge s is
finite we require the initial, resp. final vertex of *s to be finite as well
;
if both
ends of s are finite, we require s and *s to have the same length.
We shall assume that for each edge s E 9 there is given an isometry as of H2 subject to the conditions: for a nonfree edge s the map as maps the edge *s onto s in such a way that locally the halfplane determined by *s and A is mapped to the halfplane bounded by s but opposite A. For a free edge s we require as = t.
Observe that unless both ends of the nonfree edge s E S are infinite then
the isometry as is uniquely determined by these requirements.
In general the
precise determination of as is part of the structure.
The isometry as is called the side transformation determined by s. It is required that these data satisfy the
Reflection relations 3.2 als = as
, aas *s =
;sE9
For an edge s of A we let is denote edge with the same initial vertex as s but different final vertex. This defines for us a second involution on g, s I
f s.
The
POINCARE'S THEOREM
214
composite of these two operators defines the edge operator 41:
3.3
given by
;sE9
1Ys=1*s
Let us introduce the set 1900 of edges with initial point on OH2. Observe that goo is stable under * and 1 and conclude that goo is stable under the edge operator W.
The cycles sl,.... sr of 'Y on go, are of two sorts: free cycles which contain free
edges and parabolic cycles which do not contain free edges.
a=
The cycle man
a parabolic cycle is even, as follows from the proof of 1.4.
Let us
remark that o is a parabolic transformation if the side transformations generate a discrete group with fundamental domain .A, compare V.2.4. The cycle map of a free cycle is a reflection or a as follows from the general
Lemma 3.4
Let sl,...,sr be an edge cycle with side transformations o1,...,0r
and cycle map o = 0102... or. If none of the edges of the cycle are fixed by * then
the vertex cycle P1,...,Pr has no repetitions.
Conversely, if the vertex cycle
P1,...,Pr contains a repetition, P1 = P11 say, then the edge cycle s1,...,Sr contains exactly two edges fixed by *, and the "half cycle maps" 01...0n1 , 0n...or
are conjugates of the side transformations relative to the two edges fixed by *.
Proof From P1 = Pn and s1 $ sn we get s1 = Isn, and we conclude that s1 = 11*sn_1 = *sn_1, S2 = 1*s1 = 1Sn.1 = 11*sn2 = *Sn2
A simple induction on
gives us
Let us assume for a second that n is odd, say n = 21n+1. Put i = n1 in the formula
above to get that sin = *snl+1 or *sm = Sm+1 = !*s., which is absurd. conclusion the number in must be even, n = 2m.
From the boxed formula with
n = 2m and i = m we get that sm = *sm. From the same source we get that sl,...,sm,...,s2m1 = 511 ...1sii1'sm'*sin1,...,*Sl
and we conclude that
In
VII.3 PARABOLIC CYCLE CONDITION
215
0'1 ...on_1 
1
which shows that a,...on_1 is a conjugate of o,u Let us proceed to apply this result to the opposite edge cycle From
sl = sn we get Isl = f sn or *Sr = *sii_1. Thus we can conclude that the
sequence sn,...,sr consists of an odd number of edges and that the middle edge sf is fixed, *sf = sf. Moreover, we find that on...or is a conjugate of of.
Conversely, let us suppose that our cycle sl'...'sr contains and edge c with *c = c. Let us consider the full infinite sequence of edges defined by
ci='Y'c
;iEZL
Let us make a simple observation that *c = c = j*c_1 which gives c1=*c_1.
By
continuation of this process we obtain the formula
*cm=cm
;mEZ
or cl_m = J,cm. Thus we can rewrite our sequence as Ic4, Jc3, Jc2, Icl, c1, c2, C3 ...
It follows that the sequence P1,...,Pr. of initial points of sl,...,sr contains a
repetition, PI = P. say. With the notation above we must have c = sm or c = sf.
Example 3.5
Let us consider a pentagon with all vertices on OH2 A
Boundary cycle : a*b"l*al
be
Edge cycle :
a b *al*blc bla l*b
*a c1
Cycle transformation :
aQa la1y )3a,6'al'Y C
The vertex cycle is ADCBEEBCDA.
The "halfcycles" relative to a point P are
reflections, providing a decomposition of the cycle map into a product of two reflections in geodesics through P. It follows that the cycle map is parabolic.
POINCARE'S THEOREM
216
Example 3.6
Let us consider a quadrangle with a free edge cycle.
Edge cycle : a c *a b
Vertex cycle: ACCA
Cycle map: reflection fi in b
Theorem 3.7
The conclusion of Poincare's theorem 1.7 remains valid for
polygons with vertices at infinity, provided that each parabolic cycle s1,...,sr generates a parabolic cycle map
Proof Let us return to the proof of 1.7 and observe that it suffices to prove that X/r is complete. According to VI.7.4 this space is isometric to the space t.0/g where g is the equivalence relation on t.0 induced by the action of I' on X. Projection t.0*
decreases distances2 and induces a distance decreasing bijection
t.0/g
A/%
where % denotes the equivalence relation on A generated by the side pairing. On
the other hand, the presence of the continuous section, zHt.z , 0>t.A shows that
the bijection above has a continuous inverse. Thus it suffices to prove that A/96 is complete.
Let us consider a sequence (wn)n E
N
of points of A representing a Cauchy
sequence in A/%. We intend to show that this sequence is bounded in H2. Once
this is done, we conclude that our Cauchy sequence is contained in a compact subset of A/`3U, which makes it convergent in A/%.
Consider first a free edge AB of A. Let us pick a hypercycle C with ends A and B, which does not meet OA. The function f f:H2*R introduced in the proof
of VI.7.7 is distance decreasing and induces a distance decreasing function on 2 When A is convex, the projection t.A= A is an isometry, a priori.
VIL3 PARABOLIC CYCLE CONDITION
217
It follows that fe is bounded on the sequence (wn)n E N. As a consequence we can find a hypercycle C as above such that none of the points of A/%, VI.7.3.
our sequence belongs to the region bounded by C and the free edge AB. Let us consider a parabolic cycle s1,...,sr. Pick a horocycle A at the initial
point P1 of s1 and let Al,...,Ar be defined recursively by Ai = a1Ai+t , Ar _ o,.Al
Notice that .A = Al since o = Q1.... Or is either a horolation or t. If repetitions occur in the sequence P1,...,Pr, the same repetitions occur in the sequence Al,...,Ar as follows from 3.4 : "halfcycle maps" are reflections.
With the notation from the proof of VI.7.7 let pi denote the radial function defined by A1. Recall that pi is zero outside the horodisc bounded by Ai
and that pi is distance decreasing. Let us form the distance decreasing function p:A*R
p = r (P1+... + and observe that this function is constant on the equivalence classes modulo `Jb when the horocycle A is sufficiently small. We conclude from VI.7.3 that p defines
a distance decreasing function on A/%. It follows that the sequence (p(wn))n E N is convergent and we conclude that p is bounded on (wn)n E N' In conclusion we can choose the horocycle A such that the sequence (wn)n E N does not meet any of the horodiscs bounded by 411...)A,.
Let us consider a free cycle s1,...,sr. It follows from 3.4 that the cycle contains one or two free edges. In the case of two free edges we get from 3.4 that
all "halfcycle maps" are t. If the number of free edges is 1 we get from 3.4 that the "halfcycle maps" are i and reflection in a geodesic. It follows that the cycle map is a reflection in a geodesic with end P1. Thus a free cycle can be treated just like a parabolic cycle with parabolic cycle map. Our journey has ended !
POINCARE'S THEOREM
218
VII.4 CONJUGACY CLASSES
In this section we shall show how to use a fundamental domain to determine the conjugacy classes of parabolic and elliptic elements in a discrete group r.
More precisely we are given a polygon A with a side pairing satisfying the hypothesis of Poincare's theorem 3.7 and we look for conjugacy classes in the group r generated by the side transformations of A.
Let us start by the observation that an elliptic element is conjugated to an
an element which stabilises a point of 8
while a parabolic element is conjugated
to an element which stabilises a point of (9hA, V.2.7. This makes it relevant to determine the stabiliser IPp of point of an edge s of A.
Proposition 4.1
Let P be an interior point of the edge s. The stabiliser rp
is trivial except in the following two cases *s = s : rp is generated by the reflection os.
*s = s1 and P is the midpoint of s : rp is generated by os.
Proof Let
us assume that P E 112 is an inner point of the edge s. Observe that
A U arA is a neighbourhood of P and conclude that a y E I'p must equal t or os. When y = os, we conclude that P is a fixed point for os and we get that *s = s or *s =s1. In the first case as is a reflection in s, in the second case os is a halfturn in P.
Let us assume that P E 0h0. Observe first that an even y E rp is parabolic as follows from V.2.4. It follows that an odd y E rp must be a reflection
in a geodesic through P since y2 can't be proper hyperbolic. It also follows that
horocycles with centre P are stable under r. When P is an interior point of a free
side s of A we conclude that A contains a full horocycle A with centre P. For y E rp we conclude that hat A meets y hat A, thus y = t.
11
VILA CONJUGACY CLASSES
Theorem 4.2
219
Let P be a vertex of the polygon A and let sl,.,.,sr be a full
edge cycle with vertex cycle P = P11 ...,Pr. If the vertex cycle is without repetitions
then F is generated by the cycle map a,...a,. If the vertex cycle has a repetition, P1 = Pn say, then 1'p is generated by the two "halfcycle maps" Q1...ani
Proof
an...61.
+
Let us first consider the case where P is a point of H2.
Let us inves
tigate the following union of translates of I'
3=AU0_1AUa10_2AU ...Ua1a2...ar1L From the formula ai(Pi+1) = Pi it follows that each of these translates passes through the point P1; in fact P1 = °1...oi(Pi+1) Observe that
Q1...ai10 fl ul...oi(A) D o1...vi1(si) as follows from si = vi(*si). Let us consider a small circle A with centre Pt and
observe that 3, = j fl A is a connected with J, fl vJ, # 0. More precise information is available: the arc 2. subtends an angle 27r/n where n = ord(o ). It follows that
translates of J of the form Qsj, s E Z, cover an open neighbourhood of P.
For
'Y E 1'p we conclude that y Int A meets one of these translates ; in fact y must have the form
y=as or yasQ1...Qn.1
;n=2,...,r,sEZ
In the second case, we must have PI = u1...an1(P1), i.e. P1 = P. It follows that 0"
is a "halfcycle map".
Let us consider a parabolic cycle sl,...,sr. The main point is to show that the cycle map Q1...an is nontrivial. Pick a 1lorocycle A at the initial point P1 of sl and let A1,...,A,. be defined recursively by
Ai = °iAi+1
,
A,. = O A
1,...,n
Notice that A = Al since o = al.... ar is either a horolation or L. If repetitions occur in the sequence P1,...,Pr, the same repetitions occur in the sequence Al,...,Ar as follows from 3.4: "half cycle maps" are reflections.
We can now measure the
the distance from st to a(s1) along the horocycle A and show that this equals the
sum E i=1
r 0i where ©i is the length of the horocyclic arc cut by (si,1si) in A.
The detailed argument is like the proof of 1.4
POINCARE'S THEOREM
220
We can introduce the horocycle arc L = .A n J and observe that I. n o3. 0 0 implies
that .A is covered by translate o5L, s E Z. We can proceed as above taking the second half of the proof of 4.2 into account.
Let us now turn to a free cycle. Let us treat the case where *sr = Is1 is a free edge. The full cycle can be written, compare the proof of 3.4 s1...s11_1sn*sii_1...*sls1
;
r = 2n
with *sn = s, . In the case where the edge sn is nonfree we have found that sn is a
reflection and it follows that o = 01...oi_1 is a reflection.
Observe that the
horocyclic arc L _ .A n J contains .A n A and .A n o0. Since L is connected we find that 3, = .A and we can proceed as before.
In the case where the edge sn is
free, let us observe that Int 0 and o1...on1Int 0 are disjoint. It follows that A and
cover the two ends of A and we can conclude that k = A. We can finally conclude the proof as above.
0
VII.5 SUBGROUPS OF THE MODULAR GROUP
VII.5
221
SUBGROUPS OF THE MODULAR GROUP
In this section we shall apply our theory to some important subgroups of the modular group introduced in V.1. Let us take our point of departure at the
fact that the hyperbolic triangle 0 with vertices i, (, oo is a fundamental domain for the extended modular group G = PGI2(7L). For unexplained notation see V.1.
Straightforward application of Poincare's theorem to
z
gives us the
generators a,#,y for G and the complete set of relations
5.1
1/2
1/2
Commutator group From the relations for G it follows by pure group theory that its commutator subgroup [G,G] has index 4 in G with its factor group
generated by Q and y (observe that a = y mod [G,G]). It follows from V.1.4 applied to that we can use the hyperbolic quadrangle (, 0, C, oo as the fundamental domain for [G,G].
of=p=TQ= 3a7O
Let us apply Poincare's theorem to get:
f
0'r=ay=a)30y=op10" Boundary cycle :
r*r 1*flf
Elliptic edge cycles :
r'1
Parabolic edge cycle :
r *f1, *r fl
,
f
C
POINCARE'S THEOREM
222
It follows that [G,G] is generated by the third order rotations p and opo and that there is no relation between these two generators. For further reference let us note that the parabolic cycle transformations are (oP 1.Y) P 1 = r2
,
(op 10) 1P = OT2o
It follows from our investigations that G has three subgroups of index 2.
The modular group r is generated by o and p with the complete set of
relations p3 = t and a2 = t
follows by applying Poincare's theorem to the modular figure from V.I. From these relations it follows that [r,r] has index 6 in it
F with a factor group generated by ap = T. In particular, we conclude that r has a unique subgroup F2 of index 2, in fact r2 = [G,G].
From our investigations above it follows that r has a unique normal sub
group r3 of index 3. The group r3 contains or = 03 and its conjugates ror1 and
r for.
Let us show that r3 is generated by these three elements of order two
subject to no further conditions.
From p = ro it follows that r/r3 is cyclic of
order 3 generated by T. We can use V.1.4 to conclude that r 14 U 4 U T.d represents a fundamental domain.
7
3
Side transformations : r3, 7 _1C 7 ,
U, 7?r
1
Cycle relation :
(r1oT) o (ror1)T3=t T
1cr
a
TOT1
The cycle relation, which may be read 72p37 2 = t , eliminates the generator
T3.
The remaining cycle conditions simply express the fact that r1or, 0, TUT I all have order 2. We may express this by saying that r3 is the free product of three cyclic groups of order 2.
VII.5 SUBGROUPS OF THE MODULAR GROUP
Proposition 5.2
223
The commutator group [r,r] has index 6 in r and the
group mr,r] is cyclic. The group [F,I'] is a free group on the two generators ?T IOT
Proof The group
QrUT 1
,
has index 2 in r3 with a factor group generated by o.
It follows from the discussion above that [r,r] has the fundamental domain U
*k
*n
m
n
1 R fk
Two of the side pairings can at once be written down p = 0T 1QT
,
V = 0ro.r 1
taking into account that 11(r"1(()) = ( and v(T(O) _ (. Observe that µ(oo) = 1, µ(1) = 0, v(oo) = 1, v(1) = 0 and conclude that the remaining side transformations are A = vp and is = µv. The counter clockwise boundary cycle, starting at oo, can now be written as *1 *m n 1"1k m1*n 1 *k1
The edge cycles are
n1 m *11 and r n1 n *k1. The cycle relations are
vµ.1'=t
,
µvtil=t
For later reference let us notice that the parabolic cycle *k 1 gives us the parabolic V1/11 I vp=r6
It follows that the stabiliser [F,r],, has index 6 in r,,.
POINCARE'S THEOREM
224
Let us conclude this section with a general formula for the genus of the compact Riemann surface, IV.5.8, associated with a normal subgroup II of finite index in the modular group F. The important invariants are the indices [Fx:IIx] of the stabiliser IIx of the point x, where x = i, (, oo (( = exp(27ri/3)).
Genus formula 5.3
Let II denote a normal subgroup of finite index of
the modular group F. The horocyclic compactification X of H2/H is a Riemann surface with Euler characteristic
X(X) = 2[r:II]  [r:II]
E
(1  [rx:llx]1)
x=i,(,eo
Proof
Let us consider the triangulation `r of the space X induced by the
triangulation of H2 formed by the hyperbolic triangle A with vertices i, (, oo and its translates by the extended modular group. The Euler characteristic is given by
X(X)=FE+V where the number of vertices in `r is V, the number of unoriented edges is E and
the number of unoriented faces is F. We have F = 2[F:II] and E = 2 where e is the number of edges. For a vertex v of T let ev denote the number of edges with
an initial point at v to get
X(X)=2[r:II] Ev(2 v1) The number e, can also be calculated as the number of triangles from `r with a vertex at v. From the fact that H is a normal subgroup follows that F acts on X and T. It also follows that we can separate the vertices of 'r into the three orbits of v = [x], x = i, C, oo. This identifies the orbit of [x] and the set r/fl.Fx and
gives us the formula e, = 2(Fx:IIj. The standard exact sequence of groups
o  rx/IIx
r/fi
r/II.F, , o
gives us the numerical formula [r:II] = [r:II.rx][rx:flx]
Simple elimination between the three formulas completes the proof.
VII.5 SUBGROUPS OF THE MODULAR GROUP
225
The genus g of the compact Riemann surface X, given by the formula X = 2  2g , can be computed from the branch scheme
([rt,j ], [r(,nkl, In particular we find that the group [r,r] has a branch scheme (2,3,6) and the genus formula gives g = 1. Observe also that the branch scheme (2,3,6) gives genus 1 independent of the index
The group r2 has a branch scheme (2,1,2)
while the group r3 has a branch scheme (1,3,3).
Let us conclude this section with the introduction of a series of important
normal subgroups of the modular group r. Let us consider a positive integer n and introduce F(n), modular rou of level n, by the exact sequence 5.4
0
r(n)
PS12(7)
PS12(7L/n)
0
We leave it to the reader to prove that the reduction map S12(7L)+S12(7L/n) is surjective. This is a nontrivial exercise in elementary arithmetic. It is also left to
the reader to show that the branch scheme of r(n) is (2,3,n).
For more information on this topic refer the reader to the books of [Magnus] and [Newman] and the references given there. In particular [Newman] p.143: apart from the three exceptions r, r2 and r3, every normal subgroup of r
is free.
POINCARE'S THEOREM
226
VII.6 COMBINATORIAL TOPOLOGY
In this section we shall consider a discrete group of isometries I' of H2 for
which H2/1' is compact. From the Dirichlet construction V.4 it follows that r has
a compact fundamental polygon A, compare V.2.2. We shall introduce a simple
cut and paste technique to modify A in such a way that the topological type of H2/F can be read off the boundary cycle of A.
In order to justify the cut and paste technique we must generalise the concept of the fundamental polygon to allow for the sides of the polygon to be composed of finitely many geodesic arcs. This generalisation does not destroy the validity of Poincare's theorem as follows from its proof.
The edges of the polygon are paired by the symbol
ai
*a and we insist
on the condition *a # a'. Let us enumerate one half of the edges (one edge from each side of the polygon)
el,...,eel,*el,...*e,,,fl,...,fm
its understood that f. _ *f1, i = 1,...,m. The f1s are referred to as the unpaired letters. We can write the boundary cycle as a word of these 2n + m symbols and their inverses.
Let us assume that the boundary cycle of A has the form ApQB*p"1W (see the illustration on the next page), where some of the words A,B,Q,W may be
empty. Let us introduce a curve p in the interior of A with the same initial point as p and the same final point as Q. Let me stress that a curve is oriented, without
double points, and composed of finitely many hyperbolic arcs.
Let the side transformation o'1 = oP 1 map the polygon pQp 1 into *pQ*p 1 and let us construct a new fundamental polygon
A,= (APQp1) U *PQ*p1 The boundary cycle of A' is ApBQ*p1W. This justifies the first of the four rules given below. The remaining rules are left to the reader to justify.
VII.6 COMBINATORIAL TOPOLOGY
6.1 I ApQB*p'1W i. ApBQ*p1W
6.2 [A p QB*pW H ApB*pQ1W
6.3
IAQPB*p'1W ti ApB*p'1QW
6.4
AQpB*pW H ApBQ1*pW
227
POINCARE'S THEOREM
228
Theorem 6.5 The fundamental polygon A for r can be chosen in such a way that the boundary cycle for A has the form
(albl*al'1*bl I)
fI ldk*dk1) A k
1=1
fI (ci*ci) (fI ehUh*eh 1) V i=1
h=1
where V and U1,...,U1. are words of the unpaired letters.
Proof
Let us depart from the boundary cycle of any fundamental polygon and
apply the rules 6.1 through 6.4. Let us first observe that if the boundary cycle contains p and *p, then we can use 6.2 with B = 0 to bring these symbols together
in such a way that the boundary cycle contains the syllable p*p. This syllable can be moved freely around as follows from 6.4 and 6.2 with B = 0 Qp*p
PQ1*P
'+
'+ P*PQ
Let us now assume that the boundary cycle doesn't contain pairs of the form p,*p.
Observe that a syllable of the form a*a 1 can be brought up front in virtue of 6.3 with B = 0 Qa*a 1
.
a*a 1Q
Let us suppose that the boundary cycle contains the letters a,*a 1,b,*b1.
The
heart of the matter is the relative position of these four symbols. To get the general idea we ask the reader to work out the two alternative reduction steps
aXbY*a1
aZYXb*a1*b1Hab*a1*b"1ZYX
aXbY*b"1Z*a1F+ aZXbY*b"1*a1
aZX*a"1 by*b"1
We trust the reader to complete the proof.
Remark 6.6
The standard form of the boundary cycle as given in 1.5 is
not unique. The following reductions can easily be deduced from our four rules
ab*a1*b"1c*c u+ ab*a1c*b*c H abc*b*a*c ac1b*b*a*c i+ a*c"1c1b*b*a I *c"1c1b*ba*a
u+
VII.6 COMBINATORIAL TOPOLOGY
229
The number 2g + s is unique, being the epos of the surface H2/I'.
Let us also
mention that the surface is orientable if s = 0 and unorientable if s > 0. The number of nonempty words on the list U1,...,Ur,V equals the number of boundary components of H2/17. These facts follow from Reidemeister's theory of surfaces: a
boundary cycle for the surface H2/1' is obtained by deleting the symbol * in the boundary cycle 6.5 for A, see for example [Zieschang, Vogt, Coldewey] p.114.
Let us conclude this section with some standard notations, see [Griffiths].
Sphere
a*a1b*1)1
Torus
a *b1 *a1 b
Mobius band
a *a c
Projective plane
a *a b *b1
Klein bottle
a *a b* b
,
a b *a1 *b
The first presentation of the Klein bottle is the result of gluing two Mobius bands
together along their boundaries.
The second presentation alludes to a cylinder
whose boundary circles have been given different orientations and glued together accordingly.
POINCARE'S THEOREM
230
VII.7 FRICKEKLEIN SIGNATURES
In this section we consider a Fuchsian group r of H2 with compact orbit
We try to gather enough geometric information about r to be able to
space.
reconstruct r as an abstract group. The signature itself is a collection of geometric data sufficient to provide a reconstruction of the group by generators and relations.
As a general principle, two discrete groups should have the same signature if and only if they are isomorphic as abstract groups.
The basic part of the signature is the genus g of the compact Riemann surface H2/r given through the Euler characteristic
7.1
x(H2/r)=222g
Moreover, we let nl,...,nr, denote the order of the conjugacy classes of elliptic elements of r. The fact that the number of conjugacy classes is finite follows from
the presence of a compact Dirichlet polygon. The signature of f is
7.2
sign r = ( g; n1,...,ni,)
Thus the signature is collection of integers with g > 0 and n1 > 2 , i = 1,...,r. In order to investigate which data can occur as the signature we consider a Dirichlet
polygon A for r. We shall show that
7.3
Proof

i=1
Area(O) = 2a[ x(H2/r) + E (1Let the polygon A have v vertices. By the definition of area, 111.9.9
Area(s) = 7r (v  2)  E p L;ncP
where the sum is over all vertices P of A.
Let us recall from VII.3.5 that the
vertex cycles of a Fuchsian group are without repetitions.
This allows us to
partition the sum above into vertex cycles and use V.3.15 to rewrite the formula
VII.7 FRJCKEKLEIN SIGNATURES
231
Area(A) = 7r(v  2)  r 21r =1 '
The polygon A modulo identification of sides defines a complex on H2/F with
x(H2/F)=VE+F where V = r , E = 2v and F = 1. Elimination of v and r gives us
Area(A)=27r(E1)V+27rE(1n) i=1
which is the required result, once its recalled that F = 1.
'
0
As an interesting application of the area formula 7.3, we find a lower bound for the area of a Dirichlet polygon A for a Fuchsian group 7.4
Proof
Area(A) >
21
According to 7.3, it amounts to the same to prove the estimate r 2g2+n.)?42
Let us first remark that this is clear for g > 1. For g = 1 we must have r > 1 and the result follows since (1 
j
) > 2. When g = 0 we have r > 3 since the area is
positive. When r > 5, the estimate follows by observing that each term in the sum is at least 2. When r = 4, the minimum is taken for signature (0;2,2,2,3), but
02+2+2+2+13G It remains to treat the case g = 0 and r = 3. We must prove the inequality
;2 4 use hyperbolic geometry, but when (p  2)(q  2) use Euclidean geometry. EXERCISE 2.4
Let A be an rgon in the hyperbolic plane whose vertices has
angles of measures
LPi = 7r/n; 10
;
n, E N, n, > 2, i=1,...,r
Show that reflections in the sides of A generate a discrete group D(0)
with A as fundamental domain. 2°
30
Show that "Dyck's group" D+(0) can be generated by pl,...,pr subject to nl = n1, ... =P =P1P2...p1.=c pl Show that the genus of the Riemann surface H2/D+(0) is zero.
EXERCISE 2.5
Let n > 2 be an integer and let A be a regular 2ngon with
angles of measure a/n.
Let us consider a side pairing on A with boundary cycle al*al a2 *a2 ... an *an
For i = 1,...,n we let Mi resp. *M1 denote the midpoint of the edge a; resp. *Mi. Show that the side transformation ai is glide transformation along M;*M4. 2°
Show that al,...,an generate a discrete group G with fundamental domain
A and that a complete set of relations is 2 2 anon1 ... al2 =t
3°
Identify the group G with a subgroup of index 4n in the triangle group
G(2,2n,2n).
EXERCISE 2.6
Let g > 2 be an integer and let A be a regular 4ggon with
angles of measure 7r/2g.
Let us consider the side pairing on .A which identifies
each side with the opposite side of the polygon by means of parallel translation along the midpoints of the two sides. Let us enumerate the sides of A such that the boundary cycle reads
1° Show that
al bl a2 b2 ... ag bg *al1*bl... *ag 1*bg 1 a discrete group G with fundamental
domain A and that a complete set of relations is
VII EXERCISES
237 _1
1
_1
_1
2p2 ... an/3n_1:_ l(3» = Q» a» ... 0111 x101 a2)322°
Show that the Riemann surface H2/G has genus g.
3°
Identify the group G with a subgroup of index 8g in the triangle group
G(2,4g,4g).
Let 0 be the regular 8gon with angles of measure it/4.
EXERCISE 2.7
Consider the side pairing with boundary cycle a b *a 1*c 1*d1 c *b"1d
Show that the corresponding Riemann surface has genus 2 and a fundamental group generated by a, 13, y, 6 subject to the relation
6 Y60a'
01 y1
a=t
Show that the commutator subgroup [r,r] of the modular
EXERCISE 5.1
group r has the fundamental domain A U r/, U r20 U r30 U r40 U r50 with side transformations r6, r3o, r4or 1, r5or 2.
Work out the edge cycles and apply
Poincare's theorem. EXERCISE 5.2
1°
Show that the subgroup r2 of index 2 in the modular group
1' has a branch scheme (2,1,2), and that this is the only normal subgroup of I' with a branch scheme of the form (2,1,n). 2°
Show that the normal subgroup r3 of r of index 3 has a branch scheme
(1,3,3) and that this is the only normal subgroup of finite index with a branch scheme of the form (1,3,n). 30
Show that no normal subgroup of finite index of r has a branch scheme of
the form (1,1,n), except for 1' itself. EXERCISE 5.3
Let i denote the hyperbolic triangle oo,0, l with side trans
formations o, p1 ap, pop 1 where p = ro.
Show that Poincare's theorem applies
and deduce that r3 is the free product of the three cyclic subgroups generated by 0',p 1l7P, P7p1
POINCARE'S THEOREM
238
EXERCISE 5.4 I 'Show that 1.3 is the smallest normal subgroup of r containing p. 2°
Show that the subgroup I'2 of I' of index 2 is the smallest normal subgroup
of IF containing o. 3°
Show that 1'2 and 1'3 are the only nontrivial normal subgroups of r which
contains elements of finite orders.
EXERCISE 5.5
Let 11 denote a normal subgroup of finite index p of I' with a
branch scheme of the the form (2,3,n). 10
Show that the corresponding compact Riemann surface has genus 0 for the
following values of (µ,n) only (6,2)
2°
(12,3) (24,4) (60,5)
Show that the table above is realised by 1'(2), I'(3), r(4), 1'(5)
EXERCISE 5.6
Verify the following calculation of genus of F(n)
n
1
2
3
4
5
6
7
genus
0
0
0
0
0
1
3
8 5
9 10
10
11
13 26
12 25
VIII HYPERBOLIC 3SPACE
In the first two sections of this chapter we return to the generality of chapter I and study the exterior algebra over an oriented ndimensional real vector space E equipped with a nonsingular quadratic form. The main issue is parametrisation of subspaces of E of given Sylvester type by exterior vectors.
This construction is particularly satisfactory where our space is an oriented
Minkowski space M, i.e. is a fourdimensional space of Sylvester type (3,1). In
this case the space A 2M has a natural complex structure and a complex inner product familiar from special relativity.
We find that a 2vector z E A 2M
is
decomposable (primitive) if and only if Im[] = 0. A decomposable 2vector
z # 0 determines a plane in M whose Sylvester type is determined by the sign of
the norm .
In particular we find a one to one correspondence between
oriented geodesics in H3 and 2vectors z E A 2M with
1.
Our general
discussion of the geometry of H3 is based on this fact.
So far, our discussion has been based on a general Minkowski space. In
section VIII.6 we introduce a specific Minkowski space, namely the space M of 2 x 2 Hermitian matrices. This space is, in a natural way, a linear representation of the group S12(C) and we deduce an isomorphism PS12(C)
Lor+(M)
where Lor+(M) is the group of special Lorentz transformations of M. We make a definite choice of a sheet of the hyperbola in M and call this the Hermitian model of H3, compare [Shafarevich p.149] and [Mania].
Beyond its own intrinsic interest, the Hermitian model has a trace formula not available in general: Let S E S12(C) act on H3 as the product of halfturns a and 0 in two geodesics h. and k with normal vectors H and K. Then lr2(S) = 4 2
This leads to a classification theorem: the conjugacy classes in the special Lorentz group are in one to one correspondence with classes of pairs of geodesics in H3.
The Hernmitian model has its origin in physics. We shall make use of another tool with its origin in physics, the Dira.c algebra. For readers familiar
with the formalism of Clifford algebra we can add that Dirac's algebra is a
HYPERBOLIC 3SPACE
240
concrete realisation of the Clifford algebra of M.
The trace formula above makes it clear that geodesics have an important role to play in the study of even isometries of H3. Geodesics occur in bundles as we shall see in section VIII.5. In the final section of the book we shall illustrate
how the geometric notion of bundles of geodesics in H3 applies to fixed point theorems.
VIII.1 EXTERIOR ALGEBRA
In this section we consider a vector space E of dimension n over R equipped with a nonsingular quadratic form.
We shall be concerned with the
extra structure induced by the form on the exterior algebra A E. For p E N we can introduce a symmetric bilinear form on APE by the formula
1.1
= det1j
A vector e E E gives rise to an operator Le on A E given by 1.2
u1 A ... A up Le = E
.(1)i1 u1 A
... A uj... Au P
It is easily verified that our operation satisfies the general rule 1.3
(u A v)Le = (uLe) A v+(1)ru A Me)
;u E A rE,v E ASE, e E E
Observe that the operator La on A E has square 0. It follows that we can extend the operator L to all of A E making A E into a r ht module over the ring A E
1.4
u L(v A w) = (u Iv) Lw
; u,v,w E A E
VIII.! EXTERIOR ALGEBRA
241
It is now time to introduce the key formula
1.5
u L v =
; u,v E APE
Proof Let us proceed by induction on
p Consider two decomposable vectors u = ul A ... A uP and v = vl A ... A vp. Using formulas 1.2 and 1.3 we get that u L v = (E .(1)i+l ul A ... A iii... A up) L V2 A ... A vp =
(1)1+1 _ where we have used expansion of the determinant after the first row.
0
Let us combine the two previous formulas to get that
1.6
=
; u E A P+qE, v E APE, w E n qE
Orientations Let us recall the algebra underlying the concept of orientation By a frame in E we understand an ordered basis el,...,en for E. The frames make up a subset Fr(E) of En on which the group G1(E) acts transitively. The group GC+(E) of transformations with of our ndimensional real vector space E.
positive determinant acts with two orbits on Fr(E), the orientations of E.
In the rest of this section we shall assume that an orientation of E has been singled out. A frame el,...,en is said to be positively oriented if it represents the preferred orientation. In particular, if we depart from a positively oriented orthonormal basis el,...,en for E, the orientation vector or = el A ... A en is
It is now time to introduce the Node star operator * in the exterior algebra of the oriented space E of Sylvester type (s,t). To a pvector v we shall associate the (np)independent of the positively oriented orthonormal basis considered.
vector *v given by
HYPERBOLIC 3SPACE
242
*v=(1)5or[v
1.7
; vEAPE
By definition *1 = (1)sor. Simple evaluation gives us
*or= 1
1.8
The * operator satisfies the formula
1.9
** V =
Proof
; vE APE
(l)P(I1P)+sv
Let us observe that APE
is spanned by pvectors of the form el h ... A ep where e1,...,ep are part of a positively oriented orthonormal basis el,...,ep,...,e. for E. Simple direct calculation yields
1.10
(l)'* el h ... h ep = 7P 11 <ei,ei> ep+1 A... A en i=1
Let us remark that the operator * changes sign if the orientation is changed; the orientation must be changed (n  p)p times to make eP+j.... oriented.
jen,el,...ep positively
Thus we get from 1.7 that (1)n(np)
(1)s* eP+1 A ... hen =
It
fi
<ei,ei> el h ... A ep
i=P+1
Apply 1.7 once more and observe that
n
r[ <ei,ei> = (1)5.
11
i=1
Consider an (np)vector w and a pvector u and let us write w h u = A or with A E R. The constant A can be determined by formula 1.6 :
= = (1)s Taking into account that = (1)5 we get that A = . Thus w h u = or Let us apply the Hodge operator and use 1.8 to get that
VIII.1 EXTERIOR ALGEBRA
243
; w E AnPE, u E APE
*(w A u) =
In particular take w = *v , v E ARE, and use 1.9 to get that
1.12
; u,v E APE
= (1)5*(u A *v)
A combination of 1.11 and 1.12 gives that
1.13
= (l)5
Complex structure
;
u,v E APE
Let M be an oriented Minkowski space, i.e. a space
of Sylvester type (3,1). The operator * on
A 2M satisfies *2 = t and is self
adjoint as follows from 1.13
=
; u,v E A 2M
This allows us to introduce a complex structure on A 2E through the formula
1.14
(x+iy)v=xv+y*v
;x,yER, vE A2M
and a symmetrical Cvalued bilinear form
1.15
c =  i
;
u,v E A 2M
The subscript c will be omitted when no confusion is possible. Observe that
c = c = + i = i(  i) = ic the structure of a threedimensional vector
In conclusion we have given A 2E
space over C with a complex quadratic form. Let us look at this in terms of a positively oriented orthonormal basis aO,o1,a2,o3 for our Minkowski space with eH(2) A ... A eH(p)
which gives aH = 0 whenever 11(1) = 1. Thus Ev C e 1 as required.
HYPERBOLIC 3SPACE
246
Proposition 2.3
If the vector w E APE is pure, then the vector
*w E A 'PE is pure and E*,, = EN, 1
.
Proof From 1.3 it follows that for u E A rE and v E ASE we have the formula
2.4
(u A v)Le1 A ... A ep = uL(e1 A ... A ep) A v
; ej E E, vLe1=0, i=1,...,p
Let us consider a pure vector w = el A ... A ep and let E, denote the space spanned by el,...,ep. Pick a basis f1,...,fn for E with fp+l,...,fn E Eµ, 1 and apply 2.5 to get (f1 A ... A fn)Le1 A... A ep = fp+l A ... A fn
Observe that we can arrange for or = f1 A ... A fn and the result follows.
Corollary 2.5 Any nonzero vector w E An1E is pure.
Theorem 2.6 A nonzero z E APE is pure if and only if the composite map AP1E
zL
E
_zA Ap+1E
is zero. When z is pure, the sequence is exact.
Proof
Let us write down the general formula
2.7
*(z A e) = or L(z A e) = (or Lz)Le = *zle
;eEE
to see that z A e = 0 if and only if e c Ker(*zL). We conclude from 2.2 that the image of the first map is Mz while the kernel of the second map is M*Z 1 . Thus
(z A ) o (zL) = 0 a Ez is orthogonal to E*Z If z is pure, we have seen in 2.3 that EZ is orthogonal to E*z. Conversely, if EZ is orthogonal to E*,L we conclude from 1.1.7 that dim Ez + dim E*Z < n
Since in general dim EZ > p and dim. E*z > n  p we get dim EZ = p. This means
that z is a pure pvector.
VIII.2 GRASSMANN RELATIONS
247
Let us now specify our theory to a Minkowski space M. Our main concern will be the classification of twodimensional subspaces of M by 2vectors.
Theorem 2.8 A
nonzero 2vector w E A 2M is pure if and only if
<w,w>c E R. A pure 2vector w gives rise to an exact sequence
awl M +*wl M
M
For a pure 2vector w the plane Mw = Ker(w A: M> A 3M) has Sylvester type (1,1), (1,0),(2,0) depending on whether <w,w> is negative, zero or positive.
Proof The composite of the two maps we have just written is zero if and only if w is pure as follows from 2.6 and 2.7. The first statement will follow once we have established the formula
2.9
e = Im[<w,w>j e
*wl(wle) 2
;eEM
2
We shall be content with a particular case, namely, with the notation of 1.16
w_(x+iy)Q1Av2=xo'1Aa2+Ya3AQo, *w=iw= yo1Aa2+xc3Aco observe that Im[<w,w>,] = Im[(x + iy)2] = 2xy. Finally check through the formula as e varies through ao,al,a2,a3. An alternative proof can be given using the technique from VIII.7.
Let us investigate the general subspace E of M of dimension 2. Pick an orthonormal basis e,f for E and calculate the norm of the 2vector e A f
<e A f,e A f> = <e,e>  <e,f>2
The discriminant lemma 1.6.3 shows that the norm is negative, zero, positive depending on the Sylvester type as stated.
0
Let us end this section by the observation that an operator on Minkowski space M of the form zl, z E A 2M, is skew adioint
= zlelf =  zlfle =  <e,zlf>
HYPERBOLIC SPACE
248
VIII.3
NORMAL VECTORS
Let H3 be one of the sheets of the unit hyperbola in Minkowski space M.
We shall study the geometry of H3 by assigning to a geodesic in H3 a normal vector N E A 2M.
The relative position of two geodesics will be studied through
the complex inner product of their normal vectors.
Proposition 3.1
The points of a geodesic curve y:R+H3 span a plane in M
of Sylvester type (1,1). The normal vector
;tEIR y(t)Ay'(t)E A2M has norm 1 and is independent of t E R. This establishes a one to one correspondence between oriented geodesics in H3 and 2vectors in A 2M of norm 1.
Proof A geodesic y:R,H3 can be written in standard form y(t) = A cosh t+ T sink t
;
tER
where A and T are vectors from M of norm 1 and 1. Let us differentiate this to get yl(t) = A sink t + T cosh t. A simple calculation gives y(t) A y'(t) = A A T (cosh2t  sinh2t) = A A T
It follows from 3.6 that the norm of N = A A T is 1. Conversely, let N E A 2M be a vector of norm 1. It follows from 2.8 that we can find vectors A and T in M with A A T = N. The discriminant
1 = = =  2
shows us that A and T span a plane of type (1,1). We may assume that A and
T are orthogonal with A E H3 and = 1.
We can use the standard
formula above to write down a geodesic curve with normal vector N = A A T.
By a hyperbolic
la aU we understand a subset of H3 traced by a linear
subspace of M of Sylvester type (2,1).
Let us observe that the linear span of a
hyperbolic plane in H3 is a linear subspace of M of type (2,1)
VIH.3 NORMAL VECTORS
Proposition 3.2
249
Two geodesics h and k in H3 with normal vectors H and
K are coplanar, i.e. contained in a hyperbolic plane if and only if c E R.
Proof Pick points A E h and B E k and unit tangent vectors S at A and T at B such that H=A A S and K=BAT. We get from 1.18 Im[c] =  vo1(A,S,B,T)
Thus c E R if and only if A,S,B,T are linearly dependent. Let us observe
that a threedimensional subspace E of M which meets H3 has type (2,1), since
the two alternative types (2,0) and (1,0) are excluded on the grounds that they do not contain vectors of norm 1, compare 1.6.3.
In conclusion, if c E R,
then the linear hyperplane E spanned by h. and k cuts H3 in a hyperbolic plane which contains h and k.
Let us say that two geodesics h and k are perpendicular if they meet in a point A and their tangent vectors at A are orthogonal.
Corollary 3.3
Two geodesics h and k are perpendicular if and only if
their normal vectors H and K are orthogonal c = 0.
Proof If c = 0 then h and k are coplanar by 3.2. Observe also that if h and k are perpendicular then they are contained in a hyperbolic plane.
Thus the
result follows from 111.2.
Theorem 3.4
The action of the special Lorentz group on A 2M defines an
isomorphism of groups
Lor+(M)  S03(C)
Proof The group S03(C) = SO( A 2M) acts transitively on orthonormal pairs H,K in A 2M, while the group Lor+(M) acts transitively on pairs h,k of oriented perpendicular geodesics.
HYPERBOLIC SPACE
250
Let OH3 denote the set of isotropic lines in M (on ints at infinity). We say that P E OH3 is an end for a geodesic h in 113 if the line S is contained in the linear
span of h. It is clear that a geodesic h has two ends, and conversely, given two distinct points A and B of 8H3, there exists a unique geodesic in H3 with ends A and B. Observe also that two geodesics h and k with a common end are coplanar.
Theorem 3.5
Let h and k be two geodesics in H3. There exists a geode
sic g perpendicular to h and k if and only if h and k don't have a common end.
Proof
Let H and K be normal vectors for h and k respectively. The vectors H
and K are linearly independent over C:
from H = AK, A E C, we get, taking
norms, that A2 = 1 and therefore H = K or H = K, which means that H and K represent the same unoriented geodesic. Let us introduce the complex vector space
E C A 2M spanned by H and K and let us form the discriminant
A = cc  c2 = 1  c2 The complex line E
1
is isotropic if and only if A = 0.
If A # 0, we can find a 2vector G with norm 1 spanning E 1 . The geodesic g with normal vector G is perpendicular to both h and k as follows from 3.3. Conversely, if the geodesic g is perpendicular to h and k, then the normal vector G
for g spans the complex line E 1 and it follows that A # 0
If A = 0, then we have ,, = ± 1, and we conclude that h and k coplanar.
Thus the result follows from the discussion in 111.2.
VIII.4 PENCILS OF PLANES
251
VIII.4 PENCILS OF PLANES
Hyperbolic planes occur in certain families called pencils. In this section
we shall find all such pencils and demonstrate their significance for simple geometric problems.
Let us recall that a hyperbolic plane P in H3 is the trace on H3 of a linear
hyperplane E of M of Sylvester type (2,1): P = E n H3.
Let there be given an
orientation of the supporting hyperplane E. A positively oriented orthonormal basis K,L,M for E defines the normal vector for P
4.1
S=KALAME A3M
The hyperplane E can be recovered from S as the kernel of S A : M  A 4M. Alternatively, the space E is the kernel of the composite map
M sn A4M * A°M By 2.7 this map is (*SL): M
R , so we find that E = (*S) 1 . In conclusion
4.2
P=H3 n (*S)1
Proposition 4.3
The normal vector S E A 3M of an oriented hyperbolic
plane P in H3 has norm
1.
This construction establishes a one to one
correspondence between 3vectors S E A 3M of norm 1 and oriented hyperbolic planes P in II3.
Proof
With the notation of 4.1, pick a positively oriented orthonormal basis
el,e2,e3 with norms 1,1,+1. This gives <S,S> = det1j<e1,ei> = 1
The main body of the proposition follows from VIII.2.
We shall investigate a number of incidence relations in terms of normal vectors.
0
HYPERBOLIC SPACE
252
Proposition 4.4 Let P be
a hyperbolic plane with normal vector S E A 3M.
A geodesic k with normal vector K E A 2M is contained in P if and only if (*K) A (*S) = 0
Proof
Let us observe that MK is contained in P if and only if MK is
orthogonal to *S as follows from 4.2. This means *S E MK follows from 2.3.
1
or *S E M*K as
0
But *S E M*K means *S A *K = 0 by 2.8.
Definition 4.5
A geodesic n is said to be perpendicular to a hyperbolic
plane P in H3 if n intersects P in a point A and all geodesics h in P through A are perpendicular to n..
Proposition 4.6 A geodesic
it with normal vector N E A 2M is perpendicu
lar to the hyperbolic plane P with normal vector S E A M3 if and only if
NA*S=0. Proof Suppose that the geodesic n is perpendicular to P. Let us observe that a unit tangent vector T to it at the point A of intersection between P and It is orthogonal to any unit tangent S to P at A. Observe that = 0 to conclude that T is orthogonal to the linear span E of P. But the vector *S is orthogonal to E too as follows from 4.2. Let us recall from 1.13 that
4.7
=  <S,S>
;
S E A3M
It follows that *S has norm 1. Thus T = ± S and
NA*S=AATA*S=0 Conversely, let us assume that the normal vector N to
it
satisfies N A *S = 0.
According to 2.8 this means that *S belongs to the linear span of n. From this we
can conclude the existence of a point A E n. with = 0. The point A is
common to n and P and the vector *S is tangent to it at A. It follows from 3.3 that the geodesic n is perpendicular to P.
0
VHI.4 PENCILS OF PLANES
Proposition 4.8
253
Let P and Q be two distinct hyperbolic planes with
normal vectors S and T. The plane MI< associated with the pure 2vector K = *(*S A *T)
is the intersection of the hyperplanes E and F in M spanned by P and Q.
Proof From the assumption that P $ Q it follows that *S and *T are linearly independent which implies that *S A *T # 0. Observe that *K A *S=0 to conclude
that *S belongs to M.,I > 1
Sylvester type
(1,1)
<S,T> I < 1
(0,2)
I<S,T>l = 1
(0,1)
Intersection P n Q
#(aP n aQ)
0, common perp.l.
0
geodesic
0, common end
2 1
Proof With the notation of 4.8, let us calculate the norm of K =  =  (  2) = <S,T>2 1 If > 0 we find that the discriminant A of 9 is negative and that 1 has type (1,i) and defines a E n F = Mx has type (0,2). The plane G = geodesic perpendicular to P and Q.
If < 0, we have A > 0 and the plane E n F has type (1,1). It
HYPERBOLIC SPACE
254
follows that P fl Q is a geodesic.
If < K,K> = 0, we have A = 0, and the plane E fl F has type (1,0). It
0
follows that P and Q have precisely one common end.
Corollary 4.10
Let n be a geodesic in H3. Through any point A E H3
there passes a unique geodesic perpendicular to n. There is a unique hyperbolic plane perpendicular to n with a given end Q E 8H3.
Proof
Let us introduce the following subset of A3M
A={SEA3MI=0} This is a threedimensional subspace of A 3M of type (0,3).
The pencil X of
hyperbolic planes perpendicular to n has type (1,1), and it follows that NIA has dimension 1, spanned by a vector of norm 1. Introduce the set
0 ={SEA3MI=0} This is a threedimensional subspace of A 3M of type (0,2) and it follows that N n Q has dimension 1, spanned by a vector of norm 1.
Proposition 4.11
Let 9 be a pencil of hyperbolic planes. Through each
point A of II3 there passes at least one plane from the pencil. For pencils of types (1,1) or (0,1) there will pass precisely one plane from the pencil through A.
Proof
Let us consider a pencil 9 of type (0,1). The condition for the hyperbolic plane P with normal vector S to pass through the point A is = 0. Thus we are lead to consider the intersection A 1 n *9. The plane
*9 in M has type ( 1,0) while A 1 has type (  3,0). It follows that the intersection is onedimensional, generated by a vector S of norm  I. We leave it to the reader to treat the remaining two types of pencil.
VIII.5 BUNDLES OF GEODESICS
VIII.5
By a
255
BUNDLES OF GEODESICS
en
ncil 9 of geodesics in H3 we understand a twodimensional linear
subspace of A 2M consisting of 2vectors with real norm and spanned by 2vectors
of norm 1. The actual geodesics in the pencil are those geodesics whose normal vectors lie in 9.
Proposition 5.1 The geodesics in a pencil 9 are contained in a hyperbolic plane. Conversely, the normal vectors of two geodesics span a pencil if and only if the two geodesics are coplanar, i.e. is contained in a hyperbolic plane.
Proof Let us pick two distinct geodesics h and k from the pencil and let H and K be normal vectors for these. Observe that the formula for polarisation 1.1.2 shows that E R since 11, K and H + K all have real norms; it follows from
3.2 that h and k are contained in the same linear hyperplane N of M of type (2,1). It follows that 9 C A 2N.
Definition 5.2
By a bundle `9 of geodesics in H3 we understand a three
dimensional linear subspace of A 2M consisting of 2vectors with real norm, and
spanned by 2vectors of norm 1.
The actual geodesics in the bundle are the
geodesics with normal vectors in `.B.
Geodesics through a point
Let us fix a point A of H3 and introduce
the following set of 2vectors
A={XEA2MIAAX=O} It is easy to check that exterior multiplication A A defines an exact sequence
5.3
0 a A °M * A 1 M
A 2M > A 3M  A 4M _ 0
HYPERBOLIC 3SPACE
256
From this it follows that .A = Im( A A: A 1M> A 2M).
This description reveals
that A has dimension 3 and consists of 2vectors with real norm, 1.18. From ; X,Y E TA(H)
= <X,Y>
it follows that A is isometric to the tangent space TA(H3), in particular the Sylvester type is (3,0).
An oriented geodesic curve k has normal vector K E A if
and only if k passes through A, compare 2.8. In resume, we have realised the set of geodesics through A as the geodesics in the bundle A.
Geodesics with a given end
Let us fix an isotropic line in M
represented by the isotropic vector Q and introduce the set Q of 2vectors
Q= {XEA2MI XAQ=0} From the exact sequence 5.3 it follows that Q = Im( A Q: A 1M* A 2M).
This
description reveals that A has dimension 3 and consists of 2vectors with real norm, 1.18.
Let us pick three linearly independent vectors A,B,C in M with
= = 0 and = 1. From _ 
; X,Y E M3
we find that Q A A, Q A B, Q A C form an orthonormal basis with norms 0, 0, 1.
An oriented geodesic k has normal vector K E Q if and only if Q belongs to the
plane spanned by k, compare 2.8.
To summarise, we have realised the set of
geodesics with end RQ as the set of geodesics in the bundle Q.
Geodesics perpendicular to a plane
Let us consider a hyperbolic
plane P in H3 with normal vector S E A 3M and let us introduce the following set of 2vectors
N={XEA2MI XA*S=0} A look at the exact sequence 5.3 reveals that N consists of vectors of real norm and that its dimension is 3. In order to calculate the Sylvester type , observe that =  1 and that X r+ X A *S defines an antiisometry between (*S) 1 and N. In particular, the Sylvester type of X is (1,2). It follows from 4.6 that a
geodesic n has normal vector N in N if and only if n is perpendicular to P.
V111I.5 BUNDLES OF GEODESICS
Geodesics in a plane
257
Let us consider a hyperbolic plane P with given
normal vector S E A 3M and form the following set of 2vectors
9_{XEA2MI*XA*S=0} From the previous case we find that 9 consists of vectors of real norm and that its
dimension is 3 of Sylvester type (2,1). A geodesic k has normal vector in 9 if and only if k C P as follows from 4.4.
Let us summarise our findings in table 5.4 below. The heading "Pencils" indicates the possible Sylvester types of pencil in the bundle.
5.2
Bundle
Sylvester type
Pencils
Geodesics through a point Geodesics with a given end
Geodesics perpendicular to a plane Geodesics in a plane
Let us notice that the Hodge * transforms spaces of type (3,0) or (1,0) into spaces of type (0,3) or (0,1); however, such spaces will not contain 2vectors of norm 1. We have, in fact, exhausted the possible Sylvester types:
Lemma 5.5
A threedimensional subspace B of A 2M consisting of 2
vectors with real norm has Sylvester type from the following list
(0,3) (3,0) (0,1) (1,0) (1,2) (1,2)
Proof
We must rule out the four types (0,2), (2,0), (1,1) and (0,0).
In
the first three cases it is easily checked that an orthonormal real basis forms a complex basis as well. These three types can now be ruled out on the grounds that the complex inner product on A 2M is nonsingular.
HYPERBOLIC 3SPACE
258
In order to rule out the fourth case, observe that a hypothetical threedimensional subspace consisting of 2vectors of norm 0 would span a complex subspace E isotropic for the complex inner product, i.e E C E 1 . Since the inner
product on A M2 is nonsingular we get that dimCE = 1,
0
,
which is a
contradiction.
Theorem 5.6
Table 5.4 is a complete list of bundles of geodesics in
hyperbolic 3space H3.
Proof
By theorem 3.4 it suffices to prove that SO3 =SO( A 2M) acts
transitively on the bundles of given type.
Let us treat type (2,1) separately. A bundle is specified by a sequence el,e2,e3 of orthogonal 2vectors of norms 1,1, 1. It is clear that 03 = O( A 2M)
acts transitively on such sequences. An orthogonal transformation has del = ± 1,
but the sign can be changed by permuting the first two vectors leaving the third invariant. Types (3,0) and (1,2) can be treated in a similar fashion. Let us consider a bundle E of type (1,0) and let us prove that E fl iE # 0.
Pick an orthogonal basis el,e2,e3 with norms 1, 0, 0 to see that E n iE = 0 is equivalent to E + iE = M, which implies that el,e2,e3 are linearly independent over
C; in particular we find that (Ce2 + Ce3) 1 = Ce2 + Ce3 contradicting that our complex form is nonsingular. Let us show that E fl iE C E
<e,z>_i
1 : starting with
;zEE,e E EniE
and the observation ie E E we conclude that <e,z> = 0 as required.
This shows
that we can construct a real basis for E as follows: pick any vector e E E with norm 1 and a nonzero vector f E E fl iE to get the basis
e, f, if.
This shows
how to recover E from a pair of orthogonal vectors e,f of norms 1, 0. It follows from Witt's theorem 1.2.4 that 03 acts transitively on such data. To see that S03
acts transitively, let us consider an orthonormal basis el, e2, e3 and put e = el, f = e2 + ie3 ; observe that we can permute e2 and e3 keeping el fixed.
VIH.5 BUNDLES OF GEODESICS
259
Let us emphasise that the fact that geometrically different pencils have different Sylvester types amplifies statements like "through a given point there passes
at most one geodesic perpendicular to a given geodesic". As an illustration let us
apply this to the theorem of Whitehead, Smid and Van der Waerden, see [Coxeterl) § 4.
Whitehead's theorem 5.7
If there are five distinct hyperbolic
planes containing pairs of four given geodesics, the remaining pair also consists of two coplanar geodesics (a more precise statement is given in the proof).
Proof
Let the four geodesics be a,b,c,d.
In the graph below we have
represented each of the five hyperbolic planes by a bar. The problem is to show that the geodesics b and d are coplanar.
F ad
b
The assumptions may be restated: the triples a, b, c and a, c, d generate bundles
`.B and `D of Sylvester type # (2,1). generated by a,c in common.
Lemma 5.8
Observe that `9 and `.D have the pencil
It follows from 5.8 below that `B = 5.
Let `.B and `.b be bundles of geodesics both of type # (2,1). If
`B and 9 have a common pencil, then `B = `D.
Proof
Inspection of 5.4 reveals that `B and `D must be of the same type in
order to have a common pencil. We ask the reader to go through the three cases.
HYPERBOLIC 3SPACE
260
VIII.6 HERMITIAN MATRICES
In this section we shall introduce the space M of 2 x 2 Hermitian matrices
and show that the determinant is a quadratic form on M of type (3,1). In fact, this realisation of Minkowski space has been used in physics for a long time.
Let us fix the notation with respect to some basic operators on the algebra M2(C) of 2 x 2 complex matrices. The cofactor matrix of A is denoted A` d
6.1 c
b I
; a,b,c,d E C
a
The basic formula for manipulations with this symbol is
6.2
AA'= AA = t det A
; A E M2(C)
Our symbol is an antiautornorphism in the sense that
6.3
(AB)"=BA"
;A,BEM2(C)
On the space M2(C) we shall consider the symmetric bilinear form
6.4
=
tr(A B")
; A, B E M2(C)
2
From the fact that A A" = t det A it follows at once that
6.5
= del A
; A E M2(C)
Polarisation of formula 6.2 gives us the fundamental formula
6.6
AB'+BA_ 2t
;
A,B E M2(C)
VIII.6 HERMITIAN MATRICES
261
The second operator of importance is the star A* = TA or complex transposed. A matrix which is invariant under the star operator is called Hermitian. The space of Hermitian matrices is denoted
M= {XEM2(C)I X"=X} Quite explicitly, a Iermitian matrix looks like this
; a,dER,bEC From this description it follows that the inner product takes real values on M. In
particular, the determinant defines a real quadratic form on M. An orthonormal basis for M is formed by the
6.7 Pauli matrices 1
0
0 0.1
0
1
1
0
1
0
'
0'2
i
i 0
'73 =
[
1
0
0
1
]
which have norms 1,1,1,1 respectively. This shows that M is a Minkowski space. The sheet of the unit hyperbola in M consisting of positive definite forms (all > 0) will be our Hermitian model of H3.
6.8
The group S12(C) acts on M:
0.X= crXCT`
; aES12(C),XEM
These transformations are orthogonal and we conclude from the fact that S12(C) is
connected that a transformation of the form 6.8 preserves orientation and the connected components of the unit hyperbola in M. Thus we deduce a morphisml S12(C)  Lor+(M) which we shall demonstrate to be surjective, see 8.5. It is this fine action of S12(C) that makes the Hermitian model of H3 extremely useful.
'A different proof of this will be presented during the proof of 8.5. As it happens, it follows from the same source that S12(C) is connected. See also 8.9.
IIYPERBOLIC 3SPACE
262
VII1.7
DIRAC'S ALGEBRA
In this section we shall introduce an algebra which has been used in physics by P.M.Dirac2. In mathematical terms it is a concrete realisation of the Clifford algebra of the Minkowski space M, [Deheuvels, X.6]. However, the reader
is not assumed to be familiar with the formalism of Clifford algebras.
The Dirac algebra D is built over the real vector space M2(C) ® M2(C) with multiplication given by means of the automorphism A i A = TA of M2(C) (A,B) (C,D) = (AC + BD,AD + BC)
7.1
; A,B,C,D E M2(C)
We ask the reader to verify that the Dirac algebra D is associative with multipli
cative unit (i,0). Let us introduce the element %P = (O,t) and let us identify a matrix A E M2(C) with the element (A,0) of the Dirac algebra. Then we can write ; A,B E M2(C)
(A,B) = A + BAY
The multiplication is ruled by the formulas
q,2=c
7.2
q'A=AP
;AEM2(C)
is the autornorphism of D given by
The principal involution
7.3
,
(A + BWY)# = A  BAY
A,B E M2(C)
;
Its eigenspaces relative to 1 and 1 are denoted D+ and D_ respectively. The principal antiinvolution X X" of the Dirac algebra is given by (A + BW) = A+ WB`
7.4
We ask the reader to check that indeed 2See
A.Messiah,
Company, Amsterdam 1961.
Quantum
; A,B E M2(C)
(XY)' = YX , X,Y E D.
Mechanics,
NorthHolland
Our next Publishing
VIII.7 DIRAC'S ALGEBRA
263
formula is a simple transcription of formula 6.6
7.5
RW SW + SAY RW = 2 t
;R,SEM
The Minkowski space M is oriented by means of the Pauli matrices 6.7. This has the effect of making A 2M into a complex vector space with a complex quadratic form, 1.15. Let us introduce the Dvalued alternating bilinear form on M
7.6
R x S= 2(RIYSW  SWRW)
; R,S E M
Observe that this symbol is invariant under the principal autornorphism but anti
invariant under the principal antiautomorphism. As a consequence we conclude that the symbol takes values in the following subspace of D+ s12(C) = {X E M2(C) I Ir X = 0 }
The space s12(C) has a natural complex structure and the determinant defines a complex quadratic form on s12(C).
Theorem 7.7
The assignment A A B H A x B defines an isometry
A 2M  s12(C) from the complex vector space A 2M equipped with the complex quadratic form from 1.14 onto s12(C) equipped with the determinant form.
Proof The Pauli matrices Qo,v1,Q2,o3 form a positively oriented orthonormal basis for M. It follows from 6.6 that
; r$s
o,.W osW + osT orW = 0
7.8
(o0'h)2 =I
,
(osW)2 =
t
It is also worth recording that the scalar matrix i = it can be written
7.9
i = o0W0r1WYo2WY0'3W = ouo10203
s=1,2,3
HYPERBOLIC 3SPACE
264
Let us prove that the map R A S f+ R x S is Clinear. According to the formulas 1.16 this is equivalent to the formulas
i(o1XO2)=0.3Xo0, i(a2X03)=al xa0' t(o3xol)=0,2X00 These can be verified by means of 7.7 and 7.10.
Let us be content with the first
00 471*aAo3" (0,1' 2W) =  00"P02P03' A = 03*00%p
It remains to verify that of X o2 , a2 X 03, 03 x a1 form an orthogonal basis for s12(C). Let us,/ verify that the first two vectors are orthogonal: 21T(o1%Yo3'Y) = 0
2t'1(a1 X a2)(a2 X 03)) The remaining details are left to the reader.
Let us end this section with some information for readers familiar with the general formalism of Clifford algebras [Deheuvels].
Corollary 7.10
The Dirac algebra D is isomorphic to the Clifford algebra
of the quadratic space M of Hermitian matrices.
Proof
Let us observe that we have the direct sum decompositions
D+=R sl2(C)®iR
,
D_=MW(B iMW
It follows from this that the Ralgebra D is generated by MW. On the other hand the real dimension of D is 16 and the dimension of the Clifford algebra is 24.
For the benefit of the reader we should mention that the Dirac matrices refer to the following eightdimensional real representation of Dirac's algebra
A+BW
It is known that D
A
B
B
A
M2(H), compare execise 7.1.
dimensional representation above is simple.
; D ' M4(C)
It follows that the eight
VIII.8 CLIFFORD GROUP
265
VIII.8 CLIFFORD GROUP
In this section we see how a certain subgroup of the multiplicative group U(D) of Dirac's algebra D acts on M.
Since we have already identified A 2M with
a subspace of D, this will bring, for example, normal vectors of geodesics in H3 to
act on M. We shall apply this to the finer study of even Lorentz transformations, and conclude the section with the main theorem of this chapter.
The map Ri+R* identifies the space M with a linear subspace of the Dirac algebra D. The Clifford group r is the subgroup of those G E U(D) which have the following stability property relative to the subspace MAY of D
8.1
G# MW G1 = MW
Thus the group r acts on M through the formula
8.2
(G.X)41 =G#XWG1
;XEM,GEF
Let us observe that G E r acts on M as an orthogonal transformation: t = (G.x)W(G.X)W =  (G.X)WY((G.X)W)# = <X,X>
As an example, let us consider a nonisotropic vector S E M and the corresponding unit St E I. The action of S%P on M is reflection TS along S:
8.3
Proof
(SW).x = rs(x)
;XEM
Let us recall from 7.5 the formulas SWSW = <S,S> t
,
SWXW + XWSW = 2 <X,S> t
;XEM
Let us multiply the second formula from the right by (SW)1 = S4/<S,S>1 to get  S1YXW(SW)1 = XW  2<X,S><S,S>1SW = rs(X)W
HYPERBOLIC 3SPACE
266
Let us identify G12(C) with the group of multiplicative units in D+. For a
matrix R E G12(C) with real determinant we have i0 = R*(det R)'1 , thus
RXIR'1 =
RXR1 F
;XEM
= RXR*,P(det R)1
which shows that R E r+, where I'+ = D+ fl r. We ask the reader to show that
r+={REG12(C) I MR ER)
8.4
;REr+,XEM
R.X=RXR''=RXR*(detR)'
Theorem 8.5 The action of S12(C) on M defines an isomorphism of groups PS12(C) = Lor+(M)
Proof
Let us show that R E r+ acts as a Lorentz transformation on M if and
only if del R is positive. To this end we ask the reader to verify the identity a
b
c
d
it
1
0
][ a
c ][
1
0
] =as+bb+cc+dd
Let us also remark that R E r+ acts trivially on M if and only if R is a real scalar matrix: this follows from the proof of 7.10. Consider RE r+ with del R > 0 and
decompose the action of R on M into a product of reflections T1T2...Td along vectors Sl,...,Sd of norm 1. It follows that we can find k E R* such that
R = k SOP S2I'...Sd Let us make use of the principal antiinvolution 7.4 to get that R R = k2 S1S1 S2S2"... SdSd. =
k2(1)d
Using that delR = RR' is positive we conclude that d is even. An even Lorentz transformation of Al can be written TATBTCTD, where
A,B,C,D E M, each of determinant 1. According to 7.5 this is the action of ABCD
This allows us to conclude that S12(C)Lor+(Af) is surjective.
0
Vffl.8 CLIFFORD GROUP
Definition 8.6
267
By a halfturn in a geodesic k in H3 we understand the
orthogonal reflection in the linear span of k in the ambient Minkowski space.
Main theorem 8.7
Let S E S12(C) act on H3 as the product aQ of half
turns a and i in geodesics h and k with normal vectors H and K. Then tr2 S = 4 2
Any even isometry of H3 can be decomposed into a product of two halfturns.
Proof Pick a point A E h and a unit tangent vector T to h at A such that H = A X T = 2(AWTW  TWYAW) = A'YTWY
We conclude from formula 8.3 that H E I'+ acts on M as a. Observe that the scalar matrix i = it acts as the antipodal map x x, x E M, to get that
8.8
(i H). X = a(X)
; XEM
It follows that (iH)(iK) E S12(C) acts as a/3 on M which gives S = f iH iK. Let us recall that K" = K, since K E s12(C). This gives us
lr(ill iK) =  tr(HK) = tr(HK) = 2 The required formula follows by squaring this relation. Conclusion by 8.9.
0
Lemma 8.9 Any S E S12(C) can be written S = HK where H,K E s12(C) with detH = de1K = 1.
Proof We ask the reader to use the theory of quadratic forms to find vectors H,K E s12(C) with = = 1 and 2 = tr(S). Observe that 1r2S = 1r2HK and conclude from 1.9.15 that S and IIK are conjugated in PSI2(C).
Upon changing the sign of H we may assume that S and HK are conjugated in S12(C).
0
HYPERBOLIC 3SPACE
268
VIII.9 LIGHT CONE
In this section we shall work with the Hermitian model for H3 and identify
its boundary OH3 with t keeping track of the action of S12(C). Once this is done
we write down the normal vector L(A,B) for the oriented geodesic with ends A,B E C and evaluate in terms of the cross ratio [A,B,P,Q]. To begin let us assign to e E C2 the following Hermitian matrix T
9.1
Zz
L(e)
_
:]=[:]
w
;e=[]
Notice that L(Ae) = ).L(e) , A E C, and observe the pleasing transformation rule 9.2
L(v(e)) = o L(e)o *
; o E S12 (C), e E C2
which follows immediately from the decomposition T
Proposition 9.3
zZ
Zw
z
Z
wz
ww
w
w
;z,wEC
The transformation L induces a S12(C) equivariant
bijection from t onto the space PC(M) of isotropic lines in M.
Proof The equivariant part of the statement is formula 9.2. The verification of bijectivity is straightforward. Let me take the opportunity to point out that the
construction performed here is a special case of the general treatment in 1.7. It suffices to identify C ® R2 with M through the assignment
; zEC,x,yER Specification of 1.7.5 gives another proof of the isomorphism Mob(C) Lor(M).
0
VIII.9 LIGHT CONE
Circles
269
Let us consider a vector N E M of norm 1, quite explicitly
;bEC,a,dEl,adbb=1 This determines a hyperbolic plane P = H3 fl N 1' . By the boundary 91? of P we
understand the set of isotropic lines contained in the linear hyperplane N 1 spanned by P. If we take advantage of the isomorphism t +PC(M) we get the following equation for OP
=0 an equation which may be written T Y
9.4
d
b
z
a
w
=0
which is the general equation of a circle in C.
Pair of ends
An oriented geodesic k in H3 can be specified through its ends
which we may consider as an ordered pair (P,Q) of distinct points of C.
Let us
write P = w and Q = V ] and show that the normal vector N of k is given by
9.5
; D = det I
L(
zu
wv
J
Let us transform this through the correspondence A 2Af  s12(C) to get
9.6
L ( P ,Q)
= zv lwu
zv}wu
2zu
2wv
zvwu
;P=[z1,Q=[°I
This matrix is easily seen to have determinant 1. Let us recall from 8.11 that
HYPERBOLIC 3SPACE
270
the matrix iL(P,Q) represents a Mobius transformation which represents a halfturn with respect to the geodesic with ends P and Q. It follows from 9.2 and 9.5 that the symbol L(P,Q) obeys the transformation rule
9.7
aL(P,Q)a' = L(o(P),o(Q))
;
O 'E S12(C)
We shall evaluate the complex inner products of 2vectors of the type 9.5 in terms
of the cross ratio on C. Recall from 1.9.8 that for four distinct points A,B,P,Q of C the cross ratio [A,B,C,D] $ 0, 1, oo.
9.8
1
=
[A,B,P,Q]+1
Proof According to 9.7 and 1.9.7 both sides of the formula are invariant under S12(C). From the fact, 1.9.3, that the action of S12(C) on C is triply transitive it
follows that it suffices to verify the formula in the case where A = oo, B = 0, P = 1, Q = q with q E C, q
Recall from 1.9.9 that [oo,0,1,Q] = q. We find
0, 1.
that L(00,0)
1
0
L(1,Q) = 1_' q
tr L(oo,0)L(1,Q) = q + 1 as required.
which gives
q1
z
2 +q q 2
1q
0
Corollary 9.9 Let A,B,P,Q be four distinct points of the Riemann sphere C. The geodesics h and k in h3 with ends A,B and P,Q are coplanar if and only if [A,B,P,Q] E R. The two geodesics h and k will intersect if and only if [A,B,P,Q] is
negative; h and k are perpendicular if and only if [A,B,P,Q] =  1.
Proof The transformation
q ,, q + 1
the remaining details see 111.10.8.
q1
is an involution which preserves k For
VIII.10 QUATERNIONS
VIII.10
271
QUATERNIONS
Poincare's halfspace model of hyperbolic 3space is the open subspace C x ]0,+oo[ of C x R equipped with the global metric p given by the formula 11.7.2 2
10.1
R = (z,r), S = (w,s)
cosh p(R,S) = 1 + IR2rs I
The group S12(C) acts as isometries on this model: if we agree to view the point R
above as a quaternion R = x + iy +j r, then the action is given by the formula
10.2
;a=
a(R) = (aR + b)(cR + d)1
see, for example, [Beardon] or [Fenchel].
We can give an S12(C)equivariant isometry from the Poincare model to the Hermitian model of 113 by assigning to the point R from Poincare's halfspace the complex 2 x 2 matrix
10.3
z + r2
z
z
1
; R=z+jr
F(R) =
Direct computation gives us 2
trF(R)°F(S)=1+JS,R1 rs
;R=z+jr,S=w+js
which shows that F is an isometry as required, compare 11.4.1, 7.5. and 10.1. The relevant formula for S12(C)equivariance
10.4 is left to the reader.
F(o(R)) = a F(R) a"
; a E S12(C)
HYPERBOLIC 3SPACE
272
VIII.11
FIXED POINT THEOREMS
In this section we shall apply geometry to prove some fixed point theorems for subgroups of PS12(C).
Let us first recall that or E PS12(C), o # t, has one or
two fixed points on OH3. In the case of two fixed points, the geodesic through
these is stable under o and is called the axis of o. We shall analyse o in terms of theorem 8.7 on halfturns. Details of some of the arguments can be found in the exercises.
Elliptic transformation
1r2o E [0,4[
.
The transformation o has two
fixed points on OH3 and all points of the axis are fixed points. The transformation a can be realised as a product of halfturns in two geodesics through a given point of the axis, both perpendicular to the axis. All planes perpendicular to the axis are
invariant under o. Elliptic transformations with a fixed axis h can be realised as the product of two reflections in planes belonging to the pencil of planes through h.
Parabolic transformation
tr2o = 4. The transformation o has preci
sely one fixed point oo on OH3 and is the product of two halfturns in geodesics through oo. The hyperbolic plane P spanned by these two geodesics is an invariant plane.
The transformation o can be realised as the product of two reflections in
planes belonging to the pencil of planes through oo perpendicular to P.
The parabolic transformations (including t) fixing a given point oo E OH3 make up the commutator group of the subgroup of PS12(C) fixing oo. This can be seen by explicit computations, compare the proof of 1.9.13.
Hyperbolic transformation
tr2o E ]4,oo[.
The transformation has
two fixed points on OH3 but no fixed points on H3. The transformation leaves any
plane containing the axis invariant. The transformation can be realised as the
VIII.11 FIXED POINT THEOREMS
273
product of halfturns in two geodesics contained in a fixed invariant plane perpendicular to the axis. A hyperbolic transformation with fixed axis h can be realised as a product of two reflections in planes belonging to the pencil of planes perpendicular to h.
Strictly loxodromic transformation
tr2Q 0 [O,oo].
The trans
formation has two fixed points on 8H3 but none on H3. The transformation can be
realised as a product of halfturns in geodesics perpendicular to the axis but not
contained in a plane. A transformation which is either strictly loxodromic or hyperbolic is called loxodromic.
The group of transformations fixing two given points A and B of OH3 is
commutative, being the product of the group of rotations with axis AB and the group of parallel translations along the geodesic AB. This group has index 2 in the group of transformations stabilising the geodesic with ends A and B.
Elliptic fixed point theorem 11.1 A subgroup G of PS12(C) consisting of elliptic transformations has a fixed point on H3.
Proof Let us consider two distinct nontrivial members o and r of G and show that their axes h and k intersect. To this end we shall rule out two alternatives: common end and common perpendicular.
When h and k have a common end, the commutator oro 1r 1 is parabolic
and we conclude that o and r commute. The fixed point set h for c on H3 is stable under r as follows from the "normaliser principle" IV.2.6. It follows that h=k or ( when r is a halfturn) It is perpendicular to k.
When It and k have a common perpendicular, let A E It and B E k be points such that the geodesic I through A and B is perpendicular to h and k. Pick
geodesics a through A and b through B such that a = a\ and r = A,3 where a,3,\ are halfturns in the geodesics a,b,l. Observe that or = a# and conclude from 8.7
that a and b intersect. This implies that the plane P spanned by a and b is perpen
HYPERBOLIC 3SPACE
274
dicular to the axes h and k for a and r. We conclude from the proof of IV.2.8 that A = B and therefore h = k.
Since any two axes of rotation are coplanar, we conclude that the axes of
rotation are members of some bundle X.
Since any two axes intersect, we 0
conclude from 5.4 that all axes pass through the same point of H3.
Definition 11.2 A subgroup G of PS12(C) is called elementary if G leaves a geodesic invariant or fixes a point of H3 or a point of the boundary 8H3.
Corollary 11.3 A subgroup G of PS12(C) with a finite orbit on H3 or 8H3 is elementary.
Proof
Suppose that G has a finite orbit on H3 of order n > 1. Any non
trivial y E G acts as a permutation on the set 0 on n elements and we find that y"! has a fixed point on H3.
From this it follows that y" is a elliptic (or t) and
we conclude that y is a elliptic: the class of loxodromic (resp. parabolic elements) is stable under n!powers. It follows from 11.1 that G has a fixed point on H3.
Suppose that G has an orbit on OOH3 consisting of n > 1 elements. In case
n = 1, the group has a fixed point on 8H3. When n = 2, the geodesic through the
two point orbit is stable. When n > 3, we find that all elements of G have finite order and the elliptic fixed point theorem tells us that G has a fixed point on H3.
Corollary
11.4
A nonelementary subgroup G of PS12(C) contains a
loxodromic transformation.
Proof Let us assume that G does not contain a loxodromic element. According to the elliptic fixed point theorem, the group G contains a parabolic element y, fixing oo say. For any element a E G we have that 1r2(y"c)E[0,4]
;nEN
VIII.11 FIXED POINT THEOREMS
275
In order to exploit this we put 1
1
a
b
0
1
c
d
; a,b,c,d E C, adbc = 1
Direct calculation gives us
tr2(y°o) = (a + d + nc)2 E [0,4]
;nEN
But this is only possible if c = 0, which means that o(oo) = oo. This being true for all o E G makes G elementary contrary to the hypothesis.
0
Invariant disc theorem 11.5 A nonelementary subgroup G of PS12(C) fixes an oriented hyperbolic plane if and only if G doesn't contain strictly loxodromic elements.
Proof Suppose first that G doesn't contain strictly loxodromic elements. Let us show that any two hyperbolic transformations u,v E G have coplanar axes. To
this end we may assume that the axes m and it have a common perpendicular c.
Let us write p = ay and v = y,3 where y is a halfturn with respect to c, a is a halfturn with respect to a geodesic a perpendicular to m and a is a halfturn with
respect to a geodesic b perpendicular to transformations ay,
'y,(3
,
a,6
it.
From the fact that none of the
are strictly loxodromic it follows that a,b,c are
pairwise coplanar, compare 8.7. It follows from 5.4 that we can find a plane P
such that either a,b,c are perpendicular to P or all lie in the same plane P. In both cases we conclude that m and n lie in P.
Let us remark that G contains a hyperbolic transformation p with axis m say, 11.4.
For o E G we find that opo l is hyperbolic with axis o(m). It follows
that we can find another hyperbolic transformation v E G with axis n # m. At this point we ask the reader to use 5.4 to find a plane R containing the axes of all hyperbolic elements of G. Observe that R is stable under G.
It remains to prove that G preserves the orientation of R. To this end we
must prove that the group I' C Isom(R) induced by G consists of even transformations only.
Assume for a moment that this is not the case. Taking into
HYPERBOLIC 3SPACE
276
account that r is nonelementary, it follows from IV.2.9 that we can find a glide reflection y E r with axes g, say. Let us write y = rRKR where r is restriction to R of a translation r in H3 along g and KR is restriction to R of reflection 1s in the hyperbolic plane K perpendicular to R through g. Let p denote reflection in R and conclude that rtzp E 1'. This is a contradiction since rKp is strictly loxodromic.
Conversely, let R be an oriented plane fixed by G. An element y E G induces a transformation of the plane R which is even. It follows that y is the
0
product of two reflections in planes perpendicular to R.
Proposition 11.6
Any solvable subgroup G of PS12(C) is elementary.
More generally, a subgroup G of PS12(C) with a composition series
{t} =Go CG1 with G1 nontrivial abelian, is elementary.
Proof
Let us start by assuming that G1 contains a parabolic element
with fixed point S E OH3.
say,
Since S is the only fixed point for o on OH3, we
conclude from the "normaliser principle", IV.2.6, that the fixed point set for G1 on 8H3 is {S}. Multiple applications of the normaliser principle show that G fixes S.
Let us assume that G1 contains an element K # t with two distinct fixed points A and B on OH3 and let us prove that the geodesic k with ends A and B is
fixed by G1. When a2 = t (halfturn) we find that the set of fixed points for a on H3 is k and we conclude from the "normaliser principle" that k is stabilised by G1. When tie # t we find that k is the only geodesic fixed by n;:
if is fixes the geodesic
h with ends C and D, we find that tit has fixed points A,B,C,D on 8H3 ; it follows
that h = k. We can apply the "normaliser principle" to the action of G1 on the set of geodesics in H3 and conclude that G1 fixes k.
If k is the only geodesic fixed by G1, it follows from the "normaliser principle" that G fixes k. Thus we may assume that G1 fixes two geodesics k and
It follows from the argument above that G1 consists of halfturns. By inspection of a single halfturn, we conclude that k and l have a common 1.
perpendicular h. It follows that G1 fixes the two perpendicular geodesics h and k.
VHI.11 FIXED POINT THEOREMS
277
The geodesic m perpendicular to h and k is fixed as well and we conclude that G1
is a subgroup of the group (of order 4) generated by halfturns in h and k.
If
#G1 = 4 we find that G1 has precisely one fixed point on H3 and it follows from the "normalises principle" that G has a fixed point on H3. We have reduced the problem to the case where #G1 = 2. In this case GI
is generated by a halfturn in a geodesic it which is stabilised by G2 by the "normaliser principle". We can now repeat the arguments of the last paragraph
and conclude that G is elementary or #G2 = 2. We leave it to the reader to conclude the proof by a simple induction on n.
Theorem 11.7
The group PS12(C) is simple, i.e. the group contains no
nontrivial normal subgroup.
Proof A nontrivial element o of PS12(C) can be written as a product of halfturns in geodesics a and b respectively.
o=a,0
Let us choose an end A for a and
an end B 54 A for b. We can write a13 = (ay)(y(3) where y is a halfturn in the geodesic AB.
This provides a representation of v as a product of two parabolic
transformations. It is important to recall that any two parabolic transformations are conjugated in PS12(C), compare 1.9.13.
Let us show that a normal subgroup N # {t} contains a parabolic transformation. If not, we can consider an element v with two fixed points A and B on 8H3.
Pick a parabolic transformation a with fixed point A and observe that the
commutator aIvav1 # 1 is parabolic and belongs to N.
At this point we refer the reader to [Beardon] for more information, in particular Jorgensen's inequality.
HYPERBOLIC 3SPACE
278
VIII EXERCISES
EXERCISE 2.1
1°
Suppose that dim E = 4.
show that dim EH, = 2 or dim E,,, = 4.
Hint:
For a nonzero vector w E A 2E
Rule out the case dim EW = 3 by
means of proposition 2.5. 2°
Let w be a nonpure 2vector as above. Show that E,,, = E.W.
EXERCISE 2.2 Let Z:MM be a skew adjoint operator on Minkowski space M.
Show that there exists a unique 2vector z E A 2M such that zL = Z. Hint: A skew
adjoint operator Z defines an alternating 2form , x,y E M. Show that there is a unique z E A 2M with = , x,y E M. EXERCISE 4.1
Let P and Q be distinct hyperbolic planes in H3. Show that
there exists at most one geodesic h perpendicular to both P and Q. Hint: Use proposition 4.6. EXERCISE 6.1
1°
Prove the general formulas
U+U =t irU
;UEM2(C)
tr(UV) + tr(U"V) = trU trV 2°
;
U,V E M2(C)
;
U,V E M2(C)
Apply these formulas twice and deduce that
trUV + trUV" = 2 trU trV  2 tr(UV) EXERCISE 6.2 Consider the complex multilinear form on M2(C)
vol(A,B,C,D) = 4 tr(AB"CDA'BC'D) 1°
; A,B,C,D E M2(C)
Show that the form changes its sign under the permutation
(A,B,C,D)(D,A,B,C). 2°
Show that vol(A,B,C,D) changes sign under the permutation of A and B.
3°
Show that the form vol is alternating. Hint: Show that the group of per
mutations of the letters A,B,C,D is generated by the permutations from 2° and 3°. 40
Show that vol takes the value 1 on the Pauli matrices o0,q1,2,3
VIII EXERCISES
279
For matrices A,B E M2(C) put A x B =
EXERCISE 6.3
2(AB
BA'). With the
notation of exercise 6.2. show that for all A,B,C,D E M2(C).
=   i vol(A,B,C,D) Hint:
First observe that the formula is valid for A,B,C,D E M. Second, observe
that the formula is Clinear in all variables.
Third, observe that the complex
vector space M2(C) is spanned by M.
Show that Dirac's algebra D is isomorphic to M2(H).
EXERCISE 7.1
Hint:
Identify the field of quaternions oI with the ring of invariants under the involution
Xr+X of M2(C). Observe that the scalar matrices 1,i form a basis for M2(C) as a vector space over H and define a morphism of rings F:M2(C) M2(H) by
; X=H+iK, H,KEH
F(X) =
Extend the definition of F to Dirac's algebra D by the convention F(AY) = r o] EXERCISE 7.2 1° Given D E s12(C). View D as a 2vector on M and show that
D[X = 2(DX  XD)
;XEM
2°
Verify the identity VHI.3.2
D[iD[X = Im[] X
;XEM
3°
Verify the identity
DLDLX + iDLiDLX =  DXD
;XEM
4°
Consider R E s12(C) whose determinant is real and different from zero.
Show that
DXD' = (del D)( D[D[X + D[iD[X)
;XEM
Consider the following alternating Cvalued 3form on M
EXERCISE 7.3
A1,A2,A3 E M
sEo sign(or) t
where the sum is over the group of permutations E3. 1°
Show that this form identifies A 3M and N'Y where N is the space of
antiHermitian matrices, X* = X. A 3M with multiplication by i:M+N.
2°
Identify the lodge star *: A'M
3°
Identify the exterior product A 2M x A 1 M : A 3M with the pairing
2(WXXW*)
; WEs12(C),XEM
HYPERBOLIC 3SPACE
280
EXERCISE 7.4 Let lb denote the set of matrices X E G12(C) with X =  X.
; a,dER,bEC 10
Show that the determinant makes this into a Minkowski space. Hint: Use
the transformation T.AM given by b
iI
a
2°
I
Let us consider the following action of the group S12(C) on 13
;aESl2(C),XEA
o'.X=17 Xv' Show that S12(C) acts on A through Lorentz transformations. 30
Verify the formula
;XEM,cES12(C)
T(crxa:')=0 T(x)&I 4°
Define a multiplication on M2(C) (D M2(C) by the formula 7.1 replacing A
by A and show that the resulting algebra F is the Clifford algebra of (A,del). 5°
Show that the representation
D  M4(C)
;
(A,B)
i>
can be decomposed into two simple representations. Hint: Show that vectors of the form (z1,z2,z1,z2), z1,z2 E C make up a Dstable subspace of C4. 6°
Conclude that the Clifford algebra of a form of type (1,3) is M4(R).
EXERCISE 7.5
1° Define a multiplication on M2(C) ® M2(C) by the formula 7.1
replacing A by A in the definition 7.1 and show that the resulting Calgebra F is the Clifford algebra of (s12(C),del). 2°
Repeat 1° with C replaced by R.
EXERCISE 8.1
1°
Show that the only involutions in Lor+(M) are the half
turns. Hint: See exercise I.1.1. 2°
Find all involutions in the group Lor(M).
VIII EXERCISES
281
EXERCISE 8.2 Show that an even Lorentz transformation o of M is a product of two plane reflections if and only if tr2v E [0,+00[.
EXERCISE 8.3
For U E r show that UU' E ode and
UU#' E Ei.
Verify that the
spinor norms N:rR and S:r*R defined by N(U)i = UU'
,
S(U)t = UU#'c
;UEF
are morphisms of groups. 20
Explain how the spinor norm N was used in the proof of Theorem 8.5.
3°
Show that U E r acts on M as a Lorentz transformation if and only if the
spinor norm is S(U) is positive. EXERCISE 8.4
Let r denote an oriented geodesic with normal vector R. Show
that a halfturn p with respect to r acts on M through the formula p = RL RL + *RL *RL
Hint: Consider a positively oriented orthonormal basis A,T,U,V for M where A E r
and T is a positively oriented unit tangent vector to r at A. EXERCISE 8.5
Verify the following table of involutions.
Lorentz transformation
r
S E G12(C)
point reflection
iSW
SE M, detS=1
halfturn
is
S E s12(C), det
plane reflection
S4
SEM, detS=1
S = 1
1° Show that the Clifford group r+ acts on M as orientation preserving transformations and the induced map r+ SO(M) is surjective. EXERCISE 8.6
Hint: Show that r+ is generated by S12(C), i and R*. 2°
Show that the action of r on M defines a surjective map 17>0(M).
Hint: Observe that r = r+ U r+1P
EXERCISE 9.1
Show that
= 1
;
A o B i6C
HYPERBOLIC 3SPACE
282
EXERCISE 9.2 An even isometry a of H3 interchanges two distinct points A and
B of OH3. Show that a is a halfturn with respect a geodesic perpendicular to the geodesic with ends A and B.
Consider two nonintersecting discs in the complex plane and let
EXERCISE 9.3
the Euclidean line through their centres intersect the bounding circles in the points
A,B,C,D in this order. 10
Show that the hyperbolic distance distance d between the corresponding
hyperbolic planes is given by tanh.2 2cl
20
BC =AAD C BD
In the case where the two circles are concentric, show that d is the
logarithm of the ratio of their radii. EXERCISE 11.1
1°
1 tr 2
Let z = x + iy be a complex number. Verify the formula eZ
0
0
ez
= cosh x cos y+ i sinh x sin. y
Discuss the action of the matrix on H3 in the cases x = 0 and y = 0 separately. In general observe that eZ = ece'y. 2°
Use this to discuss the general loxodromic transformation.
EXERCISE 11.2
1°
Let a 54 1 be a parabolic transformation of H3. Show that
the set of hyperbolic planes in H3 invariant under a form a pencil. Investigate the special case a = Lo 2°
1],
Hint:
compare 1.9.15.
Let P be an invariant plane for a.
Show that or preserves the connected
components of H3P.
EXERCISE 11.3
Let 9 denote a pencil of hyperbolic planes. Show that the
product of three reflections in planes from 9 is itself a reflection in a plane from 9. Hint:
Show that there exists a hyperbolic plane Q whose normal vector is
orthogonal to the normal vector of any plane from 9.
VIII EXERCISES
EXERCISE 11.4
283
1° Show that an odd isometry ? of H3 can be decomposed as
v = ap where ti is a halfturn and p is a reflection in a plane. Hint: When c.2 # c,
pick a point A E OH3 (distinct from the fixed points of o2) and let p denote reflection in the plane through B = o(A) perpendicular to the geodesic with ends A
and C = a(B). Observe that ap interchanges the points B and C and conclude that te = up is a halfturn. 2°
Show that an odd isometry o of H3 can be decomposed into a product of
three plane reflections u = a/3y where the mirror plane of y is perpendicular to the mirror planes of a and ,13. Hint: Consider a decomposition u = icp a in 1° and use
that the pencil of planes containing the axis k of K. has type (0,2).
APPENDIX: AXIOMS FOR PLANE GEOMETRY The aim of this appendix is to prove that a metric space which satisfies the incidence axiom and the reflection axiom is isometric to the Euclidean plane or
(after rescaling) the hyperbolic plane. For a precise statement see the first page of the introduction. The need for a proof comes from the fact that our axioms are
somewhat untraditional.
The proof is presented as a systematic series of
deductions from the axioms. This will be carried out to the point where we have proved the "Pasch's axiom", see the discussion at the end of the appendix.
Let us agree that a straight geodesic curve in a metric space X is a distance preserving map y: RX and recall that a line in X is the image of a straight geodesic curve, compare 11.1. In the following we let X be a plane i.e. a metric space X which satisfies the following two axioms 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 or of X which fixes the points of the line, but interchanges the two connected components of its complement.
Sharp triangle inequality A.1
For a point B of the plane not on the
line through two distinct points A and C we have that d(A,C) < d(A,B) + d(B,C)
Proof
We shall turn the problem upside down and assume that we are given
three distinct points A,B,C with d(A,C) = d(A,B) + d(B,C) and proceed to
construct a line through all three points. Pick a,b,c E R with b  a = d(A,B) and c  b = d(B,C) and choose three straight geodesic curves 0, A, It in X with
0(a) = A , q(c) = C, A(a) = A, .\(b) = B, µ(b) = B, p(c) = C
Let o:RX denote the curve with restriction 0 to ]oo,a], restriction A to [a,b], restriction it to [b,c] and restriction 0 to [c,+oo[. We shall prove that o:IR+X is a
APPENDIX: AXIOMS FOR PLANE GEOMETRY
285
straight geodesic curve. The triangle inequality gives us (compare VI.2.6)
d(c (x)),a (y)) < y  x
;
x,y E R ,
x