geodesic and horocyclic trajectories cramnotes/5dterra not… · trajectories of the geodesic ﬂow...

Geodesic and Horocyclic Trajectories

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Françoise Dal’Bo

Geodesic andHorocyclicTrajectories

Françoise Dal’BoIRMARUniversité Rennes 1Rennes [email protected]

Editorial board:Sheldon Axler, San Francisco State UniversityVincenzo Capasso, Universitá degli Studi di MilanoCarles Casacuberta, Universitat de BarcelonaAngus Macintyre, Queen Mary, University of LondonKenneth Ribet, University of California, BerkeleyClaude Sabbah, CNRS, École PolytechniqueEndre Süli, University of OxfordWojbor Woyczynski, Case Western Reserve University

Translation from the French language edition:Trajectoires géodésiques et horocycliquesby Françoise Dal’BoEDP Sciences ISBN 978-2-86883-997-8Copyright c© 2007 EDP Sciences, CNRS Editions, France.http://www.edpsciences.org/http://www.cnrseditions.fr/All Rights Reserved

ISBN 978-0-85729-072-4 e-ISBN 978-0-85729-073-1DOI 10.1007/978-0-85729-073-1Springer London Dordrecht Heidelberg New York

British Library Cataloguing in Publication DataA catalogue record for this book is available from the British Library

Library of Congress Control Number: 2010937998

Mathematics Subject Classification (2010): 37Bxx, 11Jxx, 20H10, 37D40, 11A55

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To Dominique,to Alma and Romance

Preface

In this text, we present an introduction to the topological dynamics of twoclassical flows associated with surfaces of curvature −1, namely the geodesicand horocycle flows. Since the end of the nineteenth century, many texts havebeen written on this subject.

Why have we undertaken this project?In the course of several talks that we have given on this topic, notably

during some summer workshops which were organized by the University ofSavoie, we have often regretted not being able to recommend a book to thosewho wanted to find out about this area for themselves. Due to their deter-mination and their enthusiasm for the subject, we decided to go beyond thestage of regret and write up our notes.

In the past thirty years, some very strong connections have been estab-lished between dynamical systems and number theory. The intersection ofthese two fields relies on a change in point of view which, in dimension 2,essentially consists of considering a real number to be a point on the bound-ary at infinity of the Poincare half-plane and of associating this point to ageodesic, on the modular surface, pointing in its direction (Sect. VII.3). Thiscase study is still the source of inspiration for a large number of specialists.It is sometimes so present in our minds that it is absent from our texts. Oneof our motivations has been to put this idea back in the spotlight.

Who does this book address?The reader is expected to have some knowledge of differential geometry and

topological dynamics. Our goal has been to produce a text which is readableby a motivated student. Experts in other areas who are interested in thissubject have also been considered.

We have attempted to keep the reader from being overwhelmed by proofswhich are either too detailed or too succinct by punctuating the text withexercises.

In what spirit was the text written?This text has been written primarily with the idea of highlighting, in

a relatively elementary framework, the existence of gateways between somemathematical fields, and the advantages of using them.

viii Preface

We have chosen not to address the historical aspects of this field and toreserve most of the references until the end of each chapter in the Commentssection. Some of our proofs have been borrowed from the literature. In somecases, we have worked to simplify those proofs.

Since the degree of difficulty (or of simplicity) of each chapter is relativelysimilar, the unfolding of this text may not appear to reach a climax. Theapplications are the true focus of its progression.

What does the text cover?We begin with a chapter introducing the geometry of the hyperbolic plane

and the dynamics of Fuchsian groups, inspired by S. Katok’s book “FuchsianGroups” [41], and A. Katok’s and V. Climenhaga’s “Lectures on Surfaces”[39]. The action of a Fuchsian group on the Poincare half plane H is properlydiscontinuous. If it is not finite, its orbits accumulate on the boundary atinfinity H(∞) of H in a constellation of points called the limit set of thegroup. We focus on several ways in which these points are approximated,which requires us to define conical and parabolic points, and to introduce thenotion of geometrically finite groups (Sect. I.4).

In Chap. II, we study some examples of these groups with special atten-tion given to Schottky groups and the modular group. In each of these cases,we construct a method of encoding their limit sets into sequences and es-tablish a one-to-one correspondence between certain geometric properties oflimit points and other combinatorial properties of sequences. For example, thecoding introduced for the modular group allows us to interpret the continuedfraction expansion of the real numbers in terms of hyperbolic geometry. Thiscoding also allows us to identify a relationship between the golden ratio andthe length of the shortest compact geodesic on the modular surface (Sect. II.4).

In Chap. III, we study the topological dynamics of the geodesic flow gR onthe quotient T 1S of the unitary tangent bundle of H by a Fuchsian group Γ .The main idea here is to connect the dynamics of this flow to that of theaction of Γ on H(∞). We show that if Γ is not elementary, the set of periodicelements with respect to gR is dense in the non-wandering set Ωg(T 1S) ofthis flow, and that there exist some trajectories which are dense in Ωg(T 1S)(Sects. III.3 and III.4). Also, when Γ is geometrically finite, we construct acompact set which intersects every trajectory in Ωg(T 1S) (Sect. III.2).

In Chap. IV, we restrict ourselves to the case in which the Fuchsian groupis a Schottky group. Using the coding of its limit set constructed in Chap. II,we develop a symbolic approach which allows us to study the topology oftrajectories of the geodesic flow on Ωg(T 1S) and to appreciate its complexity.For example, we construct some trajectories in Ωg(T 1S) which are neithercompact nor dense, and we obtain, in the general case of non-elementaryFuchsian groups, the existence of non-periodic, minimal compact sets whichare invariant with respect to the geodesic flow (Sect. IV.3).

Chapter V is devoted to the study of the horocycle flow hR on T 1S. Themethod that we use relies on a correspondence between horocycles of H and

Preface ix

non-zero vectors in R2, modulo ± Id. This vectorial point of view allows us

to link the action of hR on T 1S to that of a linear group on a vector space,and to determine, for example, the existence of trajectories which are densein the non-wandering set Ωh(T 1S) of this flow (Sects. V.2 and V.3). Whenthe group Γ is geometrically finite, the dynamics of the horocycle flow, unlikethat of the geodesic flow, is simple since a trajectory in Ωh(T 1S) is eitherdense or periodic (Sect. V.4). Despite this very different behavior, these twoflows are intimately related in the sense that the flow hR reflects the collectivebehavior of asymptotic trajectories of the flow gR.

The last two chapters are dedicated to some applications of the study ofthese flows, one in the area of linear actions, the other in that of Diophantineapproximations.

In Chap. VI, we focus on the Lorentz space R3 equipped with a bilin-

ear form of signature (2, 1). We connect the topology of orbits of a discretegroup G of orthogonal transformations of this form to that of the trajectoriesof the geodesic and horocycle flows on the quotient of the unitary tangentbundle of H over a Fuchsian group. Translating the results proved about thehorocycle flow into this vectorial context, one obtains, for example, a completedescription of the orbits of G located in the light cone, when this group is offinite type (Sect. VI.3).

In Chap. VII, we translate the Diophantine approximation of a real num-ber by rational ones into the terms of hyperbolic geometry. Relying on thedynamics of the geodesic flow on the modular surface, we rediscover amongother things that a real number is badly approximated if and only if the coeffi-cients involved in its continued fraction expansion are bounded (Sect. VII.3).

We have chosen to make geometric simplicity a priority and to avoid dis-cussing the metric aspects of these flows. As such, we have limited the scope ofsome statements and have occasionally hidden some important ideas in thesearguments. So that the reader does not come away with the impression thatwe have said everything there is to say about these subjects, at the end ofeach chapter we have added some comments in which we recast our treatmentof the material into a general Riemannian context and introduce the readerto the vast field of ergodic geometry. We conclude these comments with someopen problems as a reminder that this general area has its share of unknownanswers and that it has a place in contemporary research.

We are indebted to Claude Sabbah for his attention to the presentation ofthis text and for his precise re-readings of it. We are equally indebted to Ray-mond Seroul for creating the figures and to Steven Broad for the translation ofthe original French text into English. We would also like to thank the refereesof the French and English versions for useful comments and suggestions.

Francoise Dal’Bo

Contents

I Dynamics of Fuchsian groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Introduction to the planar hyperbolic geometry . . . . . . . . . . . . 12 Positive isometries and Fuchsian groups . . . . . . . . . . . . . . . . . . . 143 Limit points of Fuchsian groups . . . . . . . . . . . . . . . . . . . . . . . . . . 244 Geometric finiteness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

II Examples of Fuchsian groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451 Schottky groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452 Encoding the limit set of a Schottky group . . . . . . . . . . . . . . . . 543 The modular group and two subgroups . . . . . . . . . . . . . . . . . . . . 584 Expansions of continued fractions . . . . . . . . . . . . . . . . . . . . . . . . . 665 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

III Topological dynamics of the geodesic flow . . . . . . . . . . . . . . . . 791 Preliminaries on the geodesic flow . . . . . . . . . . . . . . . . . . . . . . . . 792 Topological properties of geodesic trajectories . . . . . . . . . . . . . . 833 Periodic trajectories and their periods . . . . . . . . . . . . . . . . . . . . . 894 Dense trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 945 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

IV Schottky groups and symbolic dynamics . . . . . . . . . . . . . . . . . . 971 Coding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 982 The density of periodic and dense trajectories . . . . . . . . . . . . . . 1003 Applications to the general case . . . . . . . . . . . . . . . . . . . . . . . . . . 1034 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

V Topological dynamics of the horocycle flow . . . . . . . . . . . . . . . 1091 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1092 The horocycle flow on a quotient . . . . . . . . . . . . . . . . . . . . . . . . . 1133 Dense and periodic trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

xii Contents

4 Geometrically finite Fuchsian groups . . . . . . . . . . . . . . . . . . . . . . 1225 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

VI The Lorentzian point of view . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1271 The hyperboloid model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1282 Dynamics of the geodesic flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1343 Dynamics of the horocycle flow . . . . . . . . . . . . . . . . . . . . . . . . . . . 1394 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

VII Trajectories and Diophantine approximations . . . . . . . . . . . . . 1431 Excursions of a geodesic ray into a cusp . . . . . . . . . . . . . . . . . . . 1442 Geometrically badly approximated points . . . . . . . . . . . . . . . . . . 1493 Diophantine approximations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1524 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

A Basic concepts in topological dynamics . . . . . . . . . . . . . . . . . . . 163

B Basic concepts in Riemannian geometry . . . . . . . . . . . . . . . . . . 167

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

I

Dynamics of Fuchsian groups

This chapter is an introduction to the planar hyperbolic geometry. There aremany books which cover it. Our text is inspired by three of them: A. Beardon’s“The geometry of discrete groups” [7], A. Katok’s and V. Climenhaga’s “Lec-tures on Surfaces” [39], and S. Katok’s “Fuchsian groups” [41]. The reader willfind in these books the solutions of the exercises suggested in this chapter.

We assume that the reader has some background in complex analysis anddifferential geometry. For a short introduction to Riemannian geometry, seeAppendix B.

Sections 3 and 4 do not include many examples. Readers who prefer to seeexamples of Fuchsian groups before studying their properties are invited tobrowse through Chap. II.

1 Introduction to the planar hyperbolic geometry

We follow the conformal approach. Recall that a diffeomorphism ψ betweentwo open subsets U and V of the affine Euclidean plane R

2 is conformal if itpreserves the oriented angles. More precisely, ψ is conformal if there exists amap f from U to R

∗+ such that for any point x in U and any vectors −→u , −→v in

the Euclidean plane R2, equipped with the standard scalar product 〈 ·, · 〉, we

have:〈Txψ(−→u ), Txψ(−→v )〉 = f(x)〈−→u , −→v 〉.

When U = V , we denote by Conf(U) the group of conformal diffeomorphismson U . In terms of complex geometry, this group coincides with the group ofbiholomorphic transformations on U .

Let us describe Conf(U) when U is the open unit disk D. Observe that thegroup of Mobius transformations of the form

hα,β(z) =αz + β

βz + α,

F. Dal’Bo, Geodesic and Horocyclic Trajectories, Universitext,DOI 10.1007/978-0-85729-073-1 1, c© Springer-Verlag London Limited 2011

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2 I Dynamics of Fuchsian groups

where α and β are complex numbers with |α|2 − |β|2 = 1, is included inConf(D). Moreover we have:

Proposition 1.1. The Mobius transformations of the form hα,β , where αand β are complex numbers satisfying |α|2 − |β|2 = 1, are the only one con-formal diffeomorphisms on D.

Proof. Let ψ ∈ Conf(D). There exists hα,β such that the map φ = hα,βψ fixesthe center of D. Applying the classical Schwartz Lemma to the holomorphicmaps φ and φ−1, we obtain that |φ(z)| � |z| and |φ−1(z)| � |z| for any z ∈ D.It follows that |φ(z)| = |z|. Applying again the same Lemma, we conclude thatφ = hexp iθ,0 for some θ ∈ R, and hence that ψ is a Mobius transformation ofthe form hα,β . �

Consider now the open half-plane H = {z ∈ C | Im z > 0}. This set isconformal to D since the map

Ψ : H −→ D

z �−→ iz − i

z + i

is a holomorphic diffeomorphism. In particular Conf(H) = Ψ −1 Conf(D)Ψ .Moreover, one checks that for any hα,β in Conf(D), the map h = Ψ −1hα,βΨis a Mobius transformation of the form:

h(z) =az + b

cz + d,

where a, b, c, d are real numbers satisfying ad − bc = 1. We denote by G thegroup of such real Mobius transformations. We have:

Corollary 1.2. The group G coincides with Conf(H).

Since G contains all the transformations of the form h(z) = az + b, witha > 0 and b ∈ R, the group G acts clearly transitively on H:

Property 1.3. For any z and z′ in H, there exists g ∈ G such that z′ = g(z).

Observe that G does not acts simply transitively on H. In particular thestabilizer of i in G is the group K defined by:

K ={

r(z) =z cos θ − sin θ

z sin θ + cos θ

∣∣∣ θ ∈ R

}.

Let h ∈ G, write h(z) = (az + b)/(cz + d), where ad − bc = 1. Notice thatfor any point z in H and vectors −→u , −→v in the Euclidean plane R

2, we have(Fig. I.1):

Im h(z) =Im z

|cz + d|2 and 〈Tzh(−→u ), Tzh(−→v )〉 =1

|cz + d|4 〈 −→u , −→v 〉.

1 Introduction to the planar hyperbolic geometry 3

Fig. I.1.

It follows that:

1Im h(z)2

〈Tzh(−→u ), Tzh(−→v )〉 =1

Im z2 〈 −→u , −→v 〉.

We deduce from this formula that the group G acts by isometries on H, ifwe replace the global Euclidean scalar product 〈 , 〉 by the scalar product gz

depending on each z ∈ H and defined on each tangent plane TzH by:

gz(−→u , −→v ) =1

Im z2 〈 −→u , −→v 〉.

The family of (gz)z∈H defines a Riemannian metric on H, called the hy-perbolic metric.

Throughout this text, we consider H equipped with the metric (gz)z∈H

and call it the Poincare half-plane.By construction the angles defined by this metric are the same as the

Euclidean one, and hence the group G is included in the group of orientationpreserving isometries of H which we will simply refer to as the group of positiveisometries.

By recalling some facts from Euclidean geometry, we observe that thehyperbolic metric on H allows us to define new notions of length and area oneach tangent plane TzH. Namely, if −→u is in TzH its length is

√gz(−→u , −→u ), and

the area of a parallelogram is its Euclidean area divided by Im z2.These notions give rise to the following global definitions: the hyperbolic

length of a parametric piecewise-smooth curve c : [a, b] → H, with c(t) =x(t) + iy(t), is defined by

length(c) =∫ b

a

√x′(t)2 + y′(t)2

y(t)dt,

and the hyperbolic area of a domain B ⊂ H is defined by

A(B) =∫∫

B

dx dy

y2,

when this integral exists.


One can check that all these definitions do not depend on a particularparametrization of the curve c and of the domain B. Thus the notion ofhyperbolic length is well-defined for piecewise-smooth geometric curves (bygeometric curve, we mean the image—sometimes also called the trace—of thecurve which is a set of points in H.)

Notice that the hyperbolic length of the segment [ib, a + ib] with b > 0 is|a|/b; likewise, for the segment [i, ib] with b > 0, it is | ln b|.

Clearly, the group G preserves all these notions. Namely for any g ∈ G wehave:

length(g(c)) = length(c) and A(g(B)) = A(B).

1.1 Geodesics and distance

Let us now define the analogue in H of a basic geometric object in the Eu-clidean affine plane, a straight line. Recall that the Euclidean segment betweentwo points in the plane is the shortest curve between them.

As subgroup of Mobius transformations, the group G acts on the extendedcomplex plane C ∪ {∞}, and preserves the family of circles (we regard straightlines in C as being circles in C ∪ { ∞} which pass through ∞). Moreover thecircle R ∪ {∞} is globally invariant by G and G preserves the angles. It followsthat G preserves the subfamily of vertical half straight lines and half-circlesorthogonal to the real axis included in H.

Let z and z′ be in H, denote by S the set of piecewise-smooth parametriccurves in H having z and z′ as endpoints.

Proposition 1.4. There exists a unique piecewise-smooth geometric curve Cwith endpoints z and z′ satisfying length(C) = infc∈S length(c).

• If Re(z) = Re(z′), the curve C is the line segment with endpoints z and z′.• Otherwise, consider the half-circle in H which passes through z and z′ and

is centered on the real axis. Then C is the arc of this half-circle havingendpoints z and z′.

Proof. We will begin with the case in which z = is and z′ = is′, where s > 0and s′ > 0. Let c : [a, b] → H be a piecewise-smooth curve with endpoints zand z′. Define c(t) = x(t) + iy(t).

We have:

length(c) =∫ b

a

√x′(t)2 + y′(t)2

y(t)dt �

∣∣∣∣∫ b

a

y′(t)y(t)

dt

∣∣∣∣with equality if and only if x(t) = 0 for all t in [a, b], and y′ does not changesign. Thus length(c) � | ln(s/s′)| with equality if and only if c([a, b]) is thesegment [is, is′].

Now let z and z′ be any two points. If Re(z) = Re(z′), using a transla-tion, we deduce from the previous case that the segment [z, z′] is the unique


curve C in S such that length(C) = infc∈S length(c). Otherwise, since G actstransitivity on H, there exists g ∈ G such that g(z) = i. Moreover there existsk ∈ K such that kg(z) = i and kg(z′) is in the positive imaginary axis. It fol-lows from the first case that the shortest curve between kg(z) = i and kg(z′)is the segment [i, kg(z′)]. Since the group G acts on H by isometries, andpreserves the family of vertical half-lines and half-circles centered on the realaxis, we obtain that g−1k−1[i, kg(z′)] is the shortest curve between z and z′,and is included in a half-circle in H centered on the real axis. �Definition 1.5. Vertical half-lines and Euclidean half-circles centered on thereal axis in H are called geodesics (Fig. I.2).

Fig. I.2.

With this characterization of geodesics, we can immediately see that theEuclid’s parallel postulate fails in H; given for example the point i and thevertical geodesic C = {z ∈ H | Re(z) = 2}, there are many geodesics passingthrough i which do not intersect C.

Given two points z and z′ in H, the circular arc or line segment withendpoints z and z′ contained in the geodesic passing through these two pointsis called the hyperbolic segment and is denoted [z, z′]h (Fig. I.3). This segmentis therefore the shortest (in the sense of the hyperbolic metric) piecewise-smooth geometric curve with endpoints z and z′.

Fig. I.3.

By analogy with the Euclidean affine plane, we define the hyperbolic dis-tance between z and z′ as follow:

Proposition 1.6. The function d : H × H → R+ defined by

d(z, z′) = infS

length(c)

is a distance function.


Exercise 1.7. Prove Proposition 1.6.

Notice that, by construction, we have for any g in G:

d(g(z), g(z′)) = d(z, z′).

For some specific points z and z′, the distance d(z, z′) is easy to calculateusing Proposition 1.4. This is the case when z = it and z′ = it′ with t, t′ > 0:one has d(it, it′) = | ln(t/t′)|. Observe that, when t goes to 0 or to +∞,the points i and it are moving away from each other. On the other hand,d(it, it + 1) is less than 1/t, meaning that the points it and it + 1 are gettingcloser together as t goes to +∞.

The following exercise suggests a formula relating hyperbolic distance andEuclidean distance. The hyperbolic sine function, denoted by sinh, is definedfor real numbers x by

sinh(x) =ex − e−x

2.

Exercise 1.8. Let z and z′ be points in H. Show that the following equalityholds:

sinh(

12d(z, z′)

)=

|z − z′ |2(Im z Im z′)1/2

.

(Hint: prove that both sides are invariant under G, and reduce this problemto the case of purely imaginary z and z′ [41, Theorem 1.2.6].)

Exercise 1.9. Let z and z′ be distinct points in H. Prove that the set ofpoints z′ ′ in H satisfying d(z′ ′, z) = d(z′ ′, z′) is the geodesic passing throughthe hyperbolic midpoint of [z, z′]h, orthogonal to this hyperbolic segment.This set is called the perpendicular bisector of [z, z′]h.(Hint: reduce the problem to the case where Im z = Im z and use Exercise1.8.)

We end this subsection with the construction of a distance on the unitarytangent bundle of H defined by

T 1H = {(z, −→u ) | z ∈ H, −→u ∈ TzH and gz(−→u , −→u ) = 1}.

This distance will be useful in Chaps. III and V. Let (z, −→v ) be an ele-ment of T 1

H. Denote by (v(t))t∈R the unique geodesic through z satisfyingv(0) = z and dv/dt(0) = −→v which is parametrized by (hyperbolic) arclength(i.e., d(v(t), v(t′)) = |t − t′ |) (Fig. I.4). Such a geodesic is sometimes lessformally called a “unit speed” geodesic.

Given two elements (z, −→v ) and (z′, −→v ′) of T 1H, we introduce the func-

tion f : R → R+ defined by f(t) = d(v(t), v′(t))e− |t|. Clearly f satisfies the

following inequality

f(t) � (2|t| + d(v(0), v′(0)))e− |t|.


Fig. I.4.

This inequality implies that this function is integrable.Define

D((z, −→u ), (z′, −→v ′)) =∫ +∞

− ∞e− |t|d(v(t), v′(t)) dt.

Proposition 1.10. The function D : T 1H×T 1

H → R+ is a distance function

that is G-invariant.

Exercise 1.11. Prove Proposition 1.10.

1.2 Compactification of H

The topology induced on H by d is the same as the one induced by Euclideandistance. In this topology, H is not compact. We compactify it by taking itsclosure in the extended complex C ∪ {∞}. The set H(∞) = R ∪ { ∞} is calledthe boundary at infinity of H. The restriction to H of the topology on H∪H(∞)retains the topology induced by d. More precisely, an open set of H ∪ H(∞) iseither an open set of H ∪ R (relative to the topology induced by the Euclideandistance on R

2) or the union of the point ∞ and the complement of a compactset in H ∪ R.

Exercise 1.12. Prove that the map

Ψ : H −→ D

z �−→ iz − i

z + i

extends to a homeomorphism between H ∪ H(∞) and the closed unit disk ofthe plane.

Notation. Given a subset A of H ∪ H(∞), let◦A denote its interior and A its

closure.

Definition 1.13. The boundary at infinity of A is the set denoted by A(∞)defined by

A(∞) = A ∩ H(∞).


The boundary at infinity of a geodesic is a set containing two elements thatare called the endpoints of the geodesic. Notice that a geodesic is uniquelydetermined by its endpoints. Also a geodesic is a vertical half-line if and onlyif one of its endpoints is the point ∞.

Let x− and x+ be two distinct points of H(∞). Denote by (x−x+) thegeodesic having endpoints x−, x+, oriented from x− to x+. If z belongs to H,the geodesic ray originating at z and ending at x+ is denoted by [z, x+)(Fig. I.5).

Fig. I.5.

Since G is a subgroup of Mobius transformations, G acts onC ∪ { ∞} by homeomorphisms and hence on H(∞). More precisely, forg(z) = (az + b)/(cz + d), and x in R we have:

• if c = 0, then g(∞) = ∞ and g(x) = (ax + b)/d,• if c �= 0, then g(∞) = a/c, g(−d/c) = ∞ and if x �= −d/c, then

g(x) = (ax + b)/(cx + d).

The following exercise is a projective approach of the action of G on H(∞).Let RP

1 be the real projective line. For −→u �= −→o in R2, denote R

∗ −→u itsequivalence class in RP

1 (i.e., the non-zero real multiples of −→u ).

Exercise 1.14. Prove that the map Φ : H(∞) → RP1 defined by

Φ(x) =

{R

∗(x1

)if x ∈ R,

R∗(

10

)if x = ∞,

is a homeomorphism, and that the action of G on H(∞) is conjugate to theaction of the group { ± Id} \ SL(2, R) = PSL(2, R) on RP

1.

1.3 Hyperbolic triangles and circles

Each geodesic of H separates it into two connected components. Each of thesecomponents is called a half-plane. By definition, a hyperbolic triangle T isthe intersection of three closed half-planes whose hyperbolic area is finite andnon-zero. The following proposition is technical but classical, for its proof see[7, Chap. 7], [41, Theorem 1.4.2], [39, Lecture 30].


Proposition 1.15. Let T be a hyperbolic triangle, its boundary at infinityT (∞) contains at most three points. Furthermore, let F (T ) = T −

◦T (Fig. I.6),

we have

• if T (∞) = {x1, x2, x3}, then

F (T ) = (x1x2) ∪ (x2x3) ∪ (x3x1) and A(T ) = π,

• if T (∞) = {x1, x2}, then there is z ∈ H such that

F (T ) = (x1, z] ∪ [z, x2) ∪ (x2, x1) and A(T ) = π − α,

where α is the angle at z,• if T (∞) = {x}, then there are z1 and z2 in H such that

F (T ) = (x, z1] ∪ [z1, z2]h ∪ [z2, x) and A(T ) = π − (α1 + α2),

where αi is the angle at zi, for i = 1, 2,• if T (∞) = ∅, then there are z1, z2 and z3 in H such that

F (T ) = [z1, z2]h ∪ [z2, z3]h ∪ [z3, z1]h and A(T ) = π − (α1 + α2 + α3),

where αi is the angle at zi, for i = 1, 2, 3.

Fig. I.6.

Notice that the previous Proposition shows in particular that, contraryto the Euclidean situation, a hyperbolic triangle is not necessary a compactsubset of H, and that the sum of the angles of a compact hyperbolic triangleincluded in H is strictly lesser than π.

The hyperbolic circle (resp. hyperbolic disk) of radius r > 0 centered atz ∈ H is the set of all z′ in H such that d(z, z′) = r (resp. d(z, z′) � r). Suchsets are the same as the Euclidean one.

Exercise 1.16. Prove that the hyperbolic circle of radius r > 0 centered atz = a + ib is the Euclidean circle having the segment [a + iber, a + ibe−r] asa diameter.


Exercise 1.17. Prove that the hyperbolic circumference of any hyperboliccircle of radius r > 0 is 2π sinh r, and that the hyperbolic area of any hyper-bolic disk of radius r > 0 is 4π sinh2 r/2.

Let K(z) denote the Gauss curvature at a point z of H. By definition, K(z)measures the difference between the circumference of a Euclidean circle ofradius r centered at z, and the hyperbolic circumference c(r) of the hyperboliccircle of the same center and radius, for small r. More precisely, the followingformula holds [9, 10.5.1.3], [39, Lecture 32]:

K(z) = 3 limr→0

2πr − c(r)πr3

.

Since c(r) = 2π sinh r (Exercise 1.17), for any z ∈ H we have K(z) = −1.

1.4 Horocycles and Busemann cocycles

The family of the extended horizontal lines (i.e., with {∞}) and of circlestangent at the real line is another family of curves in H∪H(∞) invariant by G,since this group preserves the family of circles in C ∪ { ∞}, and G(H(∞)) =H(∞). There are different approaches to these curves.

One is related to the geodesics of H. Clearly, a horizontal line is orthogonalto the pencil of all vertical geodesics. Replacing this line by a circle tangentat the real line to some point x ∈ R, and using a transformation g ∈ G suchthat g(∞) = x, we obtain that this circle (without x), is orthogonal to thepencil of all geodesics (x−x+), with x+ = x.

Such curves can be also viewed as limit circles. Namely, using Exercise1.16, one checks that an extended horizontal line is the limit in H ∪ H(∞) ofhyperbolic circles passing through a fixed point z in H, with center convergingto ∞ along the geodesic ray [z, ∞). The same property holds for a circletangent at the real line, replacing the point ∞ by the point of tangency. Forthis reason, an horizontal line or a circle tangent at the real line (without itspoint of tangency) is usually called an horocycle, and its boundary at infinity,its center.

We give now a metric approach to these curves, which we will use in thenext Chapters. The idea is to sit at a point x on H(∞) and observe the pointsof H from x. To do this, we will associate to each pair of points z and z′ in H,an algebraic quantity reflecting the relative position of these two points, asseen from x.

Theorem 1.18. Let (r(t))t�0 be a geodesic ray with endpoint x, parametrizedby arclength. For any z and z′ in H, the function f(t) = d(z, r(t)) − d(z′, r(t))has a limit when t goes to +∞. This limit, called the Busemann cocycle cen-tered at x, calculated at z, z′, does not depend on the origin r(0) of the geodesicray. It is denoted by Bx(z, z′). By construction, the function Bx(z, .) is con-stant along each horocycle centered at x.


Proof. We want to show that f has a limit at +∞. Let us begin with thecase in which x = ∞, z = ib and (r(t))t�0 is the geodesic ray [z, ∞). Letz′ = a′ + ib′ and s(t) = a′ + ibet (Fig. I.7). For large t, one has

d(s(t), z′) = ln(b/b′) + t,

thusf(t) = d(s(t), z′) − d(z′, r(t)) + ln(b′/b).

Fig. I.7.

In addition, since d(s(t), r(t)) is lesser than the hyperbolic length ofthe Euclidean segment [s(t), r(t)], we have d(s(t), r(t)) � |a′ |/bet, thuslimt→+∞ f(t) = ln(b′/b).

Replace now (r(t))t�0 by a geodesic ray (r′ ′(t))t�0 = [z′ ′, ∞) with z′ ′ =a′ ′ + ib′ ′. Using the previous case, we have

limt→+∞

d(z′ ′, r′ ′(t)) − d(z, r′ ′(t)) = ln(b/b′ ′)

andlim

t→+∞d(z′ ′, r′ ′(t)) − d(z′, r′ ′(t)) = ln(b′/b′ ′).

It follows that

limt→+∞

d(z, r′ ′(t)) − d(z′, r′ ′(t)) = limt→+∞

d(z, r(t)) − d(z′, r(t)).

Moreover, notice that for z fixed, the limit of f(t) = d(z, r(t)) − d(z′, r(t))only depends on Im z′. It follows that the function B∞(z, .) is constant alongeach horizontal line. Furthermore, if Re(z′) = 0 then B∞(z, z′) = d(z, z′) forb′ � b, and B∞(z, z′) = −d(z, z′) for b′ < b.

A translation reduces the case in which x = ∞ and z is arbitrary to theoriginal case (Fig. I.7).

If x is not ∞, then the Mobius transformation h ∈ G defined by

h(z) =xz − x2 − 1

z − x

similarly recovers the original case. Thus one can conclude that f(t) has alimit at +∞ which does not depend on the origin of the geodesic ray (r(t))t�0,and that the function Bx(z, .) is constant along each horocycle centered at x(Fig. I.8). �


Fig. I.8.

Property 1.19. Let g be in G, x in H(∞), and z, z′, z′ ′ in H. One has:

(i) Bg(x)(g(z), g(z′)) = Bx(z, z′);(ii) Bx(z, z′) = Bx(z, z′ ′) + Bx(z′ ′, z′);(iii) −d(z, z′) � Bx(z, z′) � d(z, z′);(iv) Bx(z, z′) = d(z, z′) (resp. −d(z, z′)) if and only if z′ belongs to the ray

[z, x) (resp. z ∈ [z′, x)).

Exercise 1.20. Prove Property 1.19.(Hint: for (ii), (iii), (iv) reduce the problem to the case in which x = ∞.)

Notation. For any t > 0, the horocycle (resp. horodisk) centered at x definedby {z ∈ H | Bx(i, z) = ln t} (resp. {z ∈ H | Bx(i, z) � ln t}) is denoted Ht(x)(resp. H+

t (x)) (Fig. I.9).

If x = ∞, then Ht(∞) is the horizontal line defined by Im z = t, andH+

t (∞) is the closed Euclidean half-plane in H bounded by this line. Other-wise, consider g in G satisfying g(∞) = x. From Property 1.19(i) and (ii), wehave

Ht(x) = g(Ht′ (∞)) with t′ = te−Bx(i,g(i)).

Fig. I.9.


1.5 The Poincare disk

Let us come back to the open unit disk D = {z ∈ C | |z| < 1}. Re-member that this set is the image of H by the Mobius transformationΨ(z) = i(z − i)/(z + i). This transformation allows us to transport on D allthe hyperbolic notions defined on H in the previous sections.

More precisely, we consider the metric on D, already denoted (gz)z∈D,defined for any −→u and −→v in the tangent plane TzD by:

gz(−→u , −→v ) =1

| ImΨ −1(z)|2 〈TzΨ−1(−→u ), TzΨ

−1(−→v )〉.

We have:

gz(−→u , −→v ) =(

21 − |z|2

)2

〈 −→u , −→v 〉.

The disk D equipped with this metric is called the Poincare disk. By con-struction it is isometric to the Poincare half-plane. As we seen in Sect. 1.1,the group ΨGΨ −1 is the group of Mobius transformations of the form(az + b)/(bz + a), with complex coefficients a and b such that |a|2 − |b|2 = 1.Clearly this group acts by isometries on the Poincare disk.

We will retain all the notations introduced for H in our discussion of D.As such, ΨGΨ −1 is already denoted G, and d represents the distance inducedby the metric (gz)z∈D on D.

One of the advantages of this model is that its compactification corre-sponds to the Euclidean one: the boundary at infinity of D is the unit circleD(∞) = {z ∈ C | |z| = 1}.

One easy checks that geodesics are circular arcs orthogonal to D(∞) orEuclidean diameters of D. Horocycles are circles contained in D which aretangent to the unit circle (Fig. I.10).

Fig. I.10.

Another advantage of this model is that the Euclidean rotation around theorigin is an isometry. It implies for example that a hyperbolic circle of D withcenter the origin is an Euclidean circle with the same center (but differentradius!).


The Poincare disk can be also very useful to study properties of hyperbolictriangles, since using a isometry we can reduce our attention to triangleshaving one vertex at the origin. For such triangles, two of these sides areEuclidean segments and the third one is a part of an Euclidean circle, whichis convex in the Euclidean sense. Following this way, it is clear for examplethat hyperbolic triangles have angles whose sum is less than π, as we alreadyknow (see [39, Lecture 30] for another applications).

From now, we will switch back and forth between these two models de-pending on the type of symmetry for which a particular problem calls.

2 Positive isometries and Fuchsian groups

In this section, we will be interested in the dynamics of the elements of Gacting on H. We have already shown that G acts by positive isometries on H.

2.1 Decompositions of the group G

We first introduce the following three subgroups of G:

K ={

r(z) =z cos θ − sin θ

z sin θ + cos θ

∣∣∣ θ ∈ R

},

A = {h(z) = az | a > 0} ,

N = {t(z) = z + b | b ∈ R} .

Recall that K is the stabilizer of i in G. The groups A and N can be alsocharacterized by their fixed points in H = H ∪ H(∞).

Exercise 2.1. Prove that an element of G is in A if and only if it fixes 0and ∞, and that a non-identity element of G is in N if and only if the point ∞is its only fixed point in H.

Observe that elements of A leave the geodesic (0∞) invariant (Fig. I.11).Also, elements of N preserve each horocycle centered at the point ∞(Fig. I.12).

Fig. I.11. h ∈ A Fig. I.12. t ∈ N

2 Positive isometries and Fuchsian groups 15

Proposition 2.2. The group G is the group of positive isometries of H.

Proof. It is enough to prove that a positive isometry f of H is in G. Since theaction of G on H is transitive and isometric, after composing f with an elementof G, one may assume that f fixes i. Choose r ∈ K such that rf(0) = 0.Then rf maps the geodesic from 0 through i (i.e., the half imaginary axis) toitself, so that rf(∞) = ∞. Thus rf is in A and hence rf(z) = az, for somea > 0. Since rf(i) = i, we have a = 1. It follows that f is in G. �

Recall that the unit tangent bundle of H is defined by

T 1H = {(z, −→u ) | z ∈ H, −→u ∈ TzH and gz(−→u , −→u ) = 1}.

For g(z) = (az + b)/(cz + d), where a, b, c, d are real numbers satisfyingad − bc = 1, we have

g(z, −→u ) =(

az + b

cz + d,

−→u(cz + d)2

),

where multiplying −→u by a non-zero complex number means taking the imageof −→u by the linear transformation represented by this complex number.

Notice that for r(z) = (z cos θ − sin θ)/(z sin θ + cos θ), h(z) = az witha > 0, and t(z) = z + b with b ∈ R, we have

r(i, −→u ) = (i, exp −2iθ−→u ), h(z, −→u ) = (az, a−→u ), t(z, −→u ) = (z + b, −→u ).

Property 2.3. For any (z, −→v ) and (z′, −→v ′) in T 1H, there exists an unique g

in G such that g(z, −→v ) = (z′, −→v ′) (i.e., the action of G on T 1H is simply

transitive).

Proof. Since the action of G on H is transitive (Property 1.3), it is enough toprove that for any −→v and −→v ′ in Ti

1H, there exists only one g ∈ G such that

−→v ′ = Tig(−→v ). Consider the real θ such that −2θ is the measure of the orientedangle between −→v and −→v ′, and set r(z) = (z cos θ − sin θ)/(z sin θ + cos θ). Wehave r(i, −→v ) = (i, −→v ′). Now suppose that some g in G fixes (i, −→v ), then g isin K and hence g(z) = (z cos θ′ − sin θ′)/(z sin θ′ + cos θ′), for some real θ′.Since exp −2iθ′ −→v = −→v , the real θ′ is a multiple of π and hence g = id. �

The following proposition gives two decompositions of G along the groupsK, A, N . The Cartan’s decomposition will be used in the Chap. VI (Exer-cise VI.1.5).

Proposition 2.4. Let g �= id in G.

• Iwasawa’s decomposition: there exists an unique triple (n, a, k) in N ×A×Ksuch that g = nak.

• Cartan’s decomposition: there exists a (non-unique) triple (k, a, k′) in K ×A × K such that g = kak′.


Proof.Iwasawa’s decomposition: Fix −→u in TiH. Let g in G, set g(i, −→u ) = (x+ iy, −→v )and consider the transformations: a(z) = yz, n(z) = z + x and k(z) =z cos θ − sin θ/z sin θ + cos θ, where −2θ is the oriented angle between −→u and−→v (Fig. I.13). By construction nak(i, −→u ) = (x + iy, −→v ), and hence g = nak

Fig. I.13. r ∈ K

since the action of G on T 1H is simply transitive.

Suppose now that nak = n′a′k′, for n′ ∈ N , a′ ∈ A and k′ ∈ K. Thisimplies a′ −1

n′ −1na = k′k−1 and hence k = k′ since k′k−1(∞) = ∞. It follows

that n′ −1n = a′a−1. Using the fact that N ∩ A = {Id}, we obtain a = a′ and

n = n′.Cartan’s decomposition: Take g ∈ G and consider the transformation a ∈ Adefined by a(z) = zed(i,g(i)). This map sends i into the hyperbolic circle cen-tered at i passing through g(i). The action of the group K on this circle istransitive, so there is some k in K such that ka(i) = g(i). The isometry g−1kafixes i and is positive, hence there is a k′ in K such that g = kak′.

Notice that this decomposition is not unique since for example for k in K,we have k = k′ Id k′ −1k for any k′ in K. �

2.2 The dynamics of positive isometries

Now we turn to the task of classifying the positive isometries of H and under-standing what they look like geometrically. We begin by considering g in Gand look for fixed points in H = H ∪ H(∞).

Write g(z) = (az + b)/(cz + d), where a, b, c, d are real numbers withad − bc = 1. Suppose g �= Id. If c �= 0, then clearly z ∈ H is fixed by g ifand only if

z =a − d ±

√(a + d)2 − 42c

.

If c = 0, then g belongs to A or N . Let us introduce the absolute value ofthe trace of g defined by

|tr(g)| = |a + d|.

We have:


Property 2.5. Let g in G − {Id}.

• If |tr(g)| > 2, then g fixes exactly two points of H, both of which are inH(∞), and g is conjugate in G to an element of A.

• If |tr(g)| < 2, then g fixes exactly one point of H which is in H, and g isconjugate in G to an element of K.

• If |tr(g)| = 2, then g fixes exactly one point of H which is in H(∞), and gis conjugate in G to an element of N .

Exercise 2.6. Prove Property 2.5.

When |tr(g)| > 2, then g is said to be hyperbolic. Such an isometry preservesthe geodesic having its endpoints at the two fixed points. This geodesic iscalled the axis of translation of g or more simply the axis of g. Each hy-perbolic g acts on its axis by translation. For any point z in this axis, thesequences (gn(z))n�1 and (g−n(z))n�1 converge to the fixed points of g. Thelimit of the sequence (gn(z))n�1 is its attractive fixed point, and the limit of(g−n(z))n�1 is its repulsive fixed point. The attractive (resp. repulsive) fixedpoint is denoted g+ (resp. g−) (Fig. I.14).

Fig. I.14.

When |tr(g)| < 2, then g is said to be elliptic. This transformation fixes anunique point z ∈ H. Clearly, the image by g of any geodesic passing through zis a geodesic passing through z. Moreover the angle between such a geodesicand its image does not depend on the geodesic. Thus g is analogous to whatwe term rotation in the Euclidean context.When |tr(g)| = 2, then g is said to be parabolic. Such a g is actually conjugateto a translation t(z) = z + b, thus g preserves each horocycle centered at itsfixed point, on which it acts by translation (Fig. I.14, g = p).

Let us give another approach to these classes of isometries. Let g inG − {Id} and z0 ∈ H which is not fixed by g, recall (Exercise 1.9) thatthe perpendicular bisector of the hyperbolic segment [z0, g(z0)]h defined byMz0(g) = {z ∈ H | d(z, z0) = d(z, g(z0))}, is the geodesic orthogonal to thesegment [z0, g(z0)]h, passing through its middle.

This geodesic separates H into two connected components, we denoteDz0(g) the closed half-plane in H bounded by Mz0(g) containing g(z0)(Fig. I.15).


Fig. I.15.

Clearly we have

g(Mz0(g−1)) = Mz0(g) and g(

◦Dz0(g

−1)) = H − Dz0(g).

The type of the transformation g can be characterized by relative positionof Mz0(g) and Mz0(g

−1).

Property 2.7. Let g in G − {Id} and z0 ∈ H which is not fixed by g.

• The geodesics Mz0(g) and Mz0(g−1) intersect in H if and only if g is

elliptic (Fig. I.16).• The closures of the geodesics Mz0(g) and Mz0(g

−1) are disjoint in H ifand only if g is hyperbolic (Fig. I.17).

• The geodesics Mz0(g) and Mz0(g−1) have exactly one endpoint in common

if and only if g is parabolic. Furthermore, the common endpoint is the(only) fixed point of g (Fig. I.18).

Fig. I.16. g elliptic

Exercise 1. Prove Property 2.7.(Hint: check that for any g′ ∈ G, we have Mz0(g

′gg′ −1) = g′(Mg′ −1(z0)(g)),and recover the case in which g is contained in K, A or N .)

The classification of the positive isometries of H in terms of hyperbolic,elliptic and parabolic transformations can be also obtained using the notionof displacement (g) of an isometry g in G defined by

(g) = infz∈H

d(z, g(z)).


Fig. I.17. g hyperbolic

Fig. I.18. g parabolic

Property 2.8. Let g be in G − {Id}.

• g is elliptic if and only if (g) = 0 and this lower bound is attained.• g is hyperbolic if and only if (g) > 0, and d(z, g(z)) = (g) if and only

if z is in the axis of g.• g is parabolic if and only if (g) = 0 and this lower bound is not attained.

Proof. Since for any g′ ∈ G, we have (g′gg′ −1) = (g), we can suppose that gis in K, A or N .

If g is in K, then it fixes the point i and thus (g) = 0.If g is in A, write g(z) = λz with λ > 1. Since positive dilations and

translations are isometries, one has

(g) = infx∈R

d(i, x + iλ).

Consider the hyperbolic circle centered at i, passing through x + iλ(Fig. I.19). Recall (Exercise 1.16) that this circle is the Euclidean circle hav-ing the segment [ier, ie−r] as its diameter, where r = d(i, x+ iλ). It intersectsthe horizontal line described by the equation Im z = λ. Thus λ � er withequality if and only if x = 0. As a result, d(i, x + iλ) � ln λ with equality ifand only if x = 0. One obtains that (g) > 0 and d(z, g(z)) = (g) if and onlyif z is in the axis of g.

Finally, if g is in N write g(z) = z + b. Since the distance d(z, g(z)) issmaller than the hyperbolic length of the Euclidean segment [z, z + b], wehave d(z, g(z))) � |b|/Im z and hence (g) = 0. Moreover, since g does not fixany point in H, this lower bound is never attained.


Fig. I.19.

2.3 Fuchsian Groups and Dirichlet domains

Now that we have studied individual positive isometries of H, we will turnour focus to subgroups Γ of G.

Since our motivation is to obtain topologically regular surfaces in the quo-tient of H by Γ , we will restrict our interest to so-called Fuchsian groups,which are discrete subgroups of G with respect to the topology on G inducedby the Euclidean distance on R

4.We say that the action of a subgroup Γ of G on H is properly discontinuous

if for all compact subsets K of H, only finitely many elements γ of Γ satisfyγK ∩ K �= ∅.

Property 2.9. The action of a Fuchsian group Γ on H is properly discontin-uous.

Proof. Let K be a compact subset of H and denote by K1 the compact subsetof T 1

H composed of elements of the form (z, −→u ) where z ∈ K, and −→u ∈ Tz1H

satisfies gz(−→u , −→u ) = 1. Fix an element (z′, −→v ′) in T 1H. The action of G on

T 1H is clearly continuous. Moreover it is simply transitive (Property 2.3),

hence there is some compact subset C of G satisfying C((z′, −→v ′)) = K1. Con-sider an element γ of Γ . If γK ∩ K �= ∅, then γK1 ∩ K1 �= ∅. Thus γ is inCC−1 which is compact. The group Γ being discrete, γ is in a finite set. �

Exercise 2.10. Prove that if Γ is a Fuchsian group, then the orbits of Γ on H

are closed and discrete, and the topological quotient space Γ \H is separable.(Hint: [33, Theorem I.6.7].)

A Fuchsian group Γ tessellates H, in the sense that there is a subset Fof H satisfying the following conditions.

(i) F is a closed connected subset of H with non-empty interior;(ii)

⋃γ∈Γ γF = H;

(iii)◦F ∩ γ

◦F = ∅, for all γ ∈ Γ − {Id}.

Such a set is called a fundamental domain. One established method forobtaining such domains is to choose a point z0 of H which is not fixed by each


element of Γ − {Id} and to associate to z0 the intersection of the half-planes(Figs. I.20 and I.21)

Hz0(γ) = {z ∈ H | d(z, z0) � d(z, γ(z0))}.

The half-plane Hz0(γ) is bounded by the perpendicular bisector Mz0(γ) of thehyperbolic segment [z0, γ(z0)]h. Notice that, using the notation introducedjust before Property 2.7, we have Hz0(γ) = H −

◦Dz0(γ).

Fig. I.20. h(z) = 2z Fig. I.21. t(z) = z + 1

Since the perpendicular bisector of the segment [z0, γ(z0)]h is a geodesic,the set Hz0(γ) is convex (i.e., for any z and z′ in Hz0(γ), the hyperbolicsegment [z, z′]h is included in this set). It follows that the intersection of allHz0(γ) is also convex.

DefineDz0(Γ ) =

⋂γ∈Γγ �=Id

Hz0(γ).

This set is called the Dirichlet domain of Γ centered at z0.

Theorem 2.11. A Dirichlet domain is a convex fundamental domain of Γ .

Exercise 2.12. Prove Theorem 2.11.(Hint: [41, Theorem 3.2.2].)

Let us examine the boundary of the Dirichlet domain Dz0(Γ ) in H.

Property 2.13. The set Dz0(Γ )−◦

Dz0(Γ ) is in the union of the perpendicularbisectors of the segments [z0, γ(z0)]h, with γ in Γ − {Id}.

Proof. Take z′ in Dz0(Γ ) −◦

Dz0(Γ ). By definition, z′ is a limit of a se-quence of points in Dz0(Γ ), and of points (zn)n�1 in H satisfying d(z0, zn) >d(z0, γn(zn)) for some γn in Γ . It follows that the sequence (γ−1

n (z0))n�1 isbounded. Since Γ is discrete, the set of γn is finite and hence for some γ′ �= Idin Γ , we have d(z0, z

′) � d(z0, γ′(z′)). Since z′ is in Dz(Γ ), the point z′ is in

the perpendicular bisector of [z0, γ′(z0)]h. �


Corollary 2.14. The hyperbolic area of Dz0(Γ ) −◦

Dz0(Γ ) is zero.

Some fundamental domains of Γ may be topologically very wild (see [7,Example 9.2.5]). This is not the case with Dirichlet domains.

Property 2.15. A Dirichlet domain Dz(Γ ) is locally finite (i.e., for any com-pact K in H, the set of γ in Γ satisfying γDz(Γ ) ∩ K �= ∅ is finite).

Proof. Suppose not. Then for some compact subset K of H and some infinitesubset Γ ′ of Γ , given any γ in Γ ′ the statement γDz(Γ ) ∩ K �= ∅ holds.The set Γ ′ can be written as a sequence (γn)n�1 in Γ . For any n � 1, thereis zn ∈ Dz(Γ ) such that γn(zn) is in K. Since zn is in Dz(Γ ), one has thatd(zn, z) � d(γn(zn), z). From this inequality, one obtains that the sequence(γn(z))n�1 is bounded. The group Γ being Fuchsian, the set {γn | n � 1} isnecessarily finite, which contradicts our initial assumption. �

Examples 2.16. The following figures are examples of Dirichlet domains cen-tered at i and associated to Fuchsian groups generated by a positive isometry g(Figs. I.22 (g(z) = 2z), I.23 (g(z) = z + 1) and I.24 (g is elliptic and does notfix i)).

Fig. I.22. g(z) = 2z

Fig. I.23. g(z) = z + 1

The following result, proved in [7, Theorem 9.2.4], shows that Dirichletdomains allow us to visualize the surface S = Γ \H associated with Γ . More


Fig. I.24. g elliptic

precisely, let Γ \Dz0(Γ ) be the set of elements of Dz0(Γ ) modulo Γ , the func-tion θ : Γ \Dz0(Γ ) → S defined by

θ(Γz′ ∩ Dz0(Γ )) = Γz′,

satisfies

Proposition 2.17 ([7, Theorem 9.2.4]). The map θ : Γ \Dz0(Γ ) → S is ahomeomorphism.

As applications of this previous proposition, we can draw the surfacesassociated to discrete cyclic groups.

Examples 2.18. The Figs. I.25 show surfaces associated to discrete cyclicgroups introduced in Example 2.16. The first two surfaces are topologically

Fig. I.25.

equivalent. Both are homeomorphic to a cylinder. From the metric pointof view, however, there are some differences. To highlight these differences,choose a generator g of Γ . Define the curve c = [g−1(i), i]h ∪ [i, g(i)]h. Let c′

be the intersection of c with the Dirichlet domain Di(Γ ). Now (referring toFig. I.25) remove c′ from Di(Γ ). In the first case, the curve c is the segment[2i, 1/2i]h, thus c′ splits the domain into two subdomains of infinite area.

In the second case, the isometry g is a translation, the curve c′ which isnot a geodesic segment splits the domain into two subdomains, one with finitearea and the other of infinite area.


Observe that Proposition 2.17 implies that if some Dirichlet domainDz0(Γ ) is compact in H, then all Dirichlet domains of Γ are also compact.

Exercise 2.19. Prove that if the area of one domain Dz(Γ ) is finite, then thearea of any Dirichlet domain of Γ is finite and is equal to A(Dz(Γ )).(Hint: use the fact that the area of the boundary of a Dirichlet domain is zero(Corollary 2.14).)

Definition 2.20. A Fuchsian group Γ is called a lattice if the area of each(or of one) Dirichlet domain Dz(Γ ) is finite. Moreover, a lattice is said to beuniform if each (or one) Dirichlet domain is compact.

3 Limit points of Fuchsian groups

In this section, we will focus on the action of a Fuchsian group Γ on H(∞).When Γ is not elementary, we associate to it a constellation of points in H(∞),which will play a crucial role in the next chapters.

3.1 Limit set

Let us first analyze the Γ -orbit of a point z in H. If Γ is infinite, since itsaction on H is properly discontinuous, Γz accumulates on H(∞). In particular,there is a sequence (γn(z))n�1 of Γz converging to a point x in H(∞). Thispoint does not depend on z. To prove it, we can assume that x = 0. Setγn(z) = an + ibn. For all z′ in H, we have d(γn(z′), γn(z)) = d(z, z′), henceγn(z′) is in the hyperbolic circle centered at γn(z) with ray d(z, z′). This circleis the Euclidean circle having [an+ibned(z,z′), an+ibne−d(z,z′)] as its diameter.The Euclidean length of this diameter tends to 0 and γn(z) tends to 0, thus(γn(z′))n�1 also converges to 0.

Definition 3.1. The limit set L(Γ ) of Γ is the closed—possibly empty—subset of H(∞) defined by

L(Γ ) = Γz ∩ H(∞).

This set does not depend on our choice of z and is Γ -invariant.

Exercise 3.2. Prove that if γ is a non-elliptic isometry of Γ , then L(Γ ) con-tains the fixed point(s) of γ.

The following proposition relates the existence of hyperbolic isometriesin Γ to the cardinality of L(Γ ).

Proposition 3.3. If Γ contains at least two hyperbolic isometries that do nothave a fixed point in common, then L(Γ ) contains infinitely many elements.Otherwise,

3 Limit points of Fuchsian groups 25

• if all hyperbolic isometries of Γ have the same axis, then L(Γ ) is reducedto the endpoints of that axis,

• if Γ contains no hyperbolic isometries, then either L(Γ ) is the empty set,or L(Γ ) is reduced to a single point and Γ is generated by a parabolicisometry fixing this point.

Proof. Suppose that Γ contains two hyperbolic isometries h1, h2 that do nothave a fixed point in common. For n � 1, consider the transformation gn =hn

1h2h−n1 . Each gn belongs to G and is hyperbolic. Moreover the fixed points

of gn are the image by hn1 of the fixed points of h2. Since h1, h2 do not have

a fixed point in common, all gn are distinct. These fixed points are in L(Γ )(Exercise 3.2).

Suppose now that all hyperbolic isometries of Γ have the same axis. Let xand y denote the endpoints of this axis. These two points are in L(Γ ) andare fixed by a hyperbolic isometry h of Γ . For all γ in Γ , the isometry γhγ−1

fixes x and y thus γ preserves the geodesic (xy), which shows that these twopoints are the only elements of L(Γ ).

Finally we consider the remaining case where Γ does not contain anyhyperbolic isometries. Suppose that it contains a parabolic isometry p. Afterconjugating Γ , we can restrict our attention to the case where p(z) = z + 1.Let γ in Γ , we have γ(∞) = ∞. Actually, if γ(∞) �= ∞, then for n largeenough |tr(pnγ)| > 2 and hence pnγ is hyperbolic which is impossible. Itfollows that the group Γ is in the group of parabolic transformations of theform z + b where b is a real number. Since Γ is discrete, it is generated by atranslation and hence L(Γ ) is reduced to the point ∞.

It remains to analyze the case in which Γ only contains elliptic isometries.If all its elements are of order 2, then Γ is abelian and cyclic of order 2.Thus Γ fixes a single point of H. If Γ is not cyclic of order 2, use the Poincaredisk model instead. After conjugating Γ , one may assume that it contains anisometry of the form r(z) = eiθz with θ �= kπ. Let g ∈ Γ written as g(z) =(az + b)/(bz + a) with |a|2 − |b|2 = 1. Calculating the trace of rgr−1g−1, onefinds

tr(rgr−1g−1) = 2 + 4|b|2 sin2 θ.

Since rgr−1g−1 is elliptic (or trivial), Property 2.5 implies that b = 0 andthus g(0) = 0. Consequently Γ is a subgroup of elliptic isometries fixing 0 andhence L(Γ ) = ∅. �

Exercise 3.4. Prove that if L(Γ ) is reduced to two points, then either Γ isgenerated by a hyperbolic isometry, or Γ contains a subgroup of index 2 ofthis form.

We deduce from Proposition 3.3 and Exercise 3.4 that, if the limit set ofa Fuchsian Γ group is finite, then this set contains at most 2 points.

Definition 3.5. A Fuchsian group is said to be elementary if its limit set isfinite.


Proposition 3.6. If Γ is not elementary, then L(Γ ) is minimal, in the sensethat L(Γ ) is the smallest (ordered by inclusion) non-empty, closed subset ofH(∞) which is Γ -invariant.

Proof. Let F be a closed, non-empty, Γ -invariant subset of L(Γ ). Since Γis not elementary, it contains infinitely many hyperbolic isometries (γn)n�1

having no shared fixed points (Proposition 3.3). Since the closed set F isinvariant with respect to each γn, the fixed points γ+

n and γ−n are necessarily

contained in F . Fix a positive integer N and choose a point z on the geodesic(γ−

Nγ+N ). Let x be a point in L(Γ ) and let (gn)n�1 be a sequence in Γ such

that limn→+∞ gn(z) = x. Passing to a subsequence, one may assume that thesequences (gn(γ−

N ))n�1 and (gn(γ+N ))n�1 converge to f − and f+ in F . Since

gn(z) is in the geodesic (gn(γ−N )gn(γ+

N )), the point x is necessarily in {f −, f+}.This shows that L(Γ ) is contained in F and thus F = L(Γ ). �

Exercise 3.7. Prove that if Γ is not elementary, then none of the points inL(Γ ) is isolated.

Note that if Γ is not elementary, then L(Γ ) is uncountable since this set isclosed, non-empty and contains no isolated points (Baire Category Theorem).

Exercise 3.8. Prove that if L(Γ ) differs from H(∞), then it is totally discon-tinuous.(Hint: use the density in L(Γ ) of the orbit of any point in L(Γ ).)

3.2 Horocyclic, conical and parabolic points

Different types of points in L(Γ ) may be distinguished by the ways in whichthey are approached by sequences in Γz.

Let us start with a point in L(Γ ) which is an attractor γ+ of a hyperbolicisometry γ in Γ . After conjugating Γ , one may assume that γ+ = ∞ and thatγ(z) = λz, with λ > 1. Since Im(γn(z)) = λn Im z, for any a > 0 and for nlarge enough (Fig. I.26), we have Im(γn(z)) > ln a. It follows that the pointsγn(z), for n large enough, belong to the horodisk centered at ∞ defined byH+

a = {z | Im z � a}. Notice that this property does not depend on z. Inconclusion if x ∈ L(Γ ) is fixed by a hyperbolic isometry, then for any z ∈ H,the orbit Γz meets every horodisk centered at x.

More generally, we have the following definition.

Definition 3.9. A point x in L(Γ ) is horocyclic if for any (or for one) z in H,its orbit Γz meets every horodisk centered at x.

The set of such points is denoted by Lh(Γ ). In the course of our discus-sion we will show that, if Γ is not elementary, then Lh(Γ ) is uncountable(Lemma II.1.2 and Corollary II.1.7). Since Γ is countable, this property im-plies that most of the points in Lh(Γ ) are not fixed points of hyperbolicisometries of Γ .

Horocyclic points can be characterized in terms of Busemann cocycles.


Fig. I.26.

Proposition 3.10. A point x in L(Γ ) is horocyclic if and only if for any zin H, we have supγ∈Γ Bx(z, γ(z)) = +∞.

Proof. Let x ∈ L(Γ ), after conjugating Γ , one may assume that x = ∞. Bydefinition, ∞ is horocyclic if for any horodisk H+

a = {z | Im z � a}, with a > 0,there exists γ in Γ such that γ(i) is in H+

a (x). This assertion is equivalent tothe fact that for any n � 1, there exists γn in Γ such that Im γn(i) � n. SinceB∞(i, z) = ln(Im z), we obtain that the point ∞ is horocyclic if and onlyif there exists a sequence (γn)n�1 in Γ such that limn→+∞ B∞(i, γn(i)) =+∞. We achieve the proof using the fact that B∞(z, γn(z)) = B∞(z, i) +B∞(i, γn(i)) + B∞(γn(i), γn(z)) and B∞(γn(i), γn(z)) � d(i, z). �

Some horocyclic points x are distinguished by having a sequence in Γzwhich approaches x along a path which remains within a bounded distance ofthe geodesic ray [z, x).

Definition 3.11. A point x in L(Γ ) is conical if for some z in H, there existε > 0 and (γn)n�0 in Γ such that the sequence (γn(z))n�0 converges to x andd(γn(z), [z, x)) � ε.

Notice that the fixed point of a hyperbolic isometry γ of Γ is conical since,if z belongs to the axis of γ, then the sequence (γn(z))n�1 is in this axis.

The following exercise shows that, if x is conical, then for any z in H, thereis an infinite subsequence at a bounded distance of the ray [z, x).

Exercise 3.12. Let x in L(Γ ) and (γn)n�0 in Γ . Prove that, if for some z ∈ H

the sequence (γn(z))n�0 remains at a bounded distance of [z, x), then the sameproperty holds for any z′ in H.

The set of conical points is denoted Lc(Γ ). Clearly we have:

Lc(Γ ) ⊂ Lh(Γ ).

The terminology “conical” arises from the shape of ε-neighborhoods ofvertical geodesics.


Fig. I.27.

Exercise 3.13. Let ε > 0. Prove that there is some ε′ > 0 such that:d(z, (0∞)) � ε if and only if |Re z/Im z| � ε′ (Fig. I.27).(Hint: use Exercise 1.8.)

Conical points can be characterized in terms of the action of Γ on theproduct H(∞) × H(∞).

Proposition 3.14. A point x in L(Γ ) is conical if and only if there existsa sequence (γn)n�0 of different transformations of Γ such that, for all y inH(∞) different from x, the sequence (γn(x), γn(y))n�0 remains in a compactsubset of H(∞) × H(∞) with its diagonal removed.

Proof. Take the Poincare disk model and x �= y in D(∞). Clearly a sequence(γn(x), γn(y))n�0 remains in a compact subset of D(∞) × D(∞) with its di-agonal removed, if and only if the set of the hyperbolic distances between theorigin 0 of D and the geodesics (γn(x)γn(y))n�0 is bounded. This conditionis equivalent to the fact that the set of points γn

−1(0) is included in someε-neighborhood of the geodesic (xy). Notice that the point 0 can be replacedby any z ∈ D.

Suppose that x is a conical point in L(Γ ). For any z in D, there existε > 0 and (γn)n�0 in Γ such that the sequence (γn(z))n�0 converges to x andd(γn(z), [z, x)) � ε. Take y in D(∞) different from x, and z′ in the geodesic(xy). The sequence d(γn(z′), [z′, x)) is also bounded, and hence the point z′ re-mains at a bounded distance from the sequence of geodesics (γn

−1(y)γn−1(x)).

It follows that the sequence (γn−1(x), γn

−1(y))n�0 remains in a compact sub-set of D(∞) × D(∞) with its diagonal removed.

Suppose now that there exists a sequence (γn)n�0 of different transfor-mations of Γ such that, for all y in H(∞) different from x, the sequence(γn(x), γn(y))n�0 remains in a compact subset of H(∞) × H(∞) with its diag-onal removed. It follows that the points γn

−1(0) remain at a bounded distancefrom the geodesic (xy). Since all γn are different, the set S of points γn

−1(0)accumulates to D(∞). Translating Exercise 3.13 in D, one obtains that theclosure of S in D ∪ D(∞) is included in {x, y}, for any y �= x, and hence thatthe sequence (γ−1

n (0))n�0 converges to x. �


Later in the text, we will assume a regularity hypothesis on the group Γ(see Sect. 4), namely equality of Lh(Γ ) and Lc(Γ ). In general, this equalitydoes not hold (in an article by A. Starkov [59], the reader will find someexamples of groups having horocyclic points which are not conical).

Conical points can be characterized in terms of distance and Busemanncocycles.

Proposition 3.15. Let z be in H. A point x in L(Γ ) is conical if and only ifthere is a sequence (γn)n�1 in Γ satisfying the following two conditions:

(i) limn→+∞ Bx(z, γn(z)) = +∞;(ii) (d(z, γn(z)) − Bx(z, γn(z)))n�1 is a bounded sequence.

Proof. After conjugating Γ , one can recover the case in which z = i andx = ∞.

Suppose that the point ∞ is conical. Then there are some ε > 0 and somesequences (γn)n�1 in Γ and (tn)n�1 in R

+ such that

limn→+∞

tn = +∞ and d(itn, γn(i)) � ε.

In particular, we have limn→+∞ Bx(z, γn(z)) = +∞. Moreover, the pointγn(i) is in the Euclidean disk having [itneε, itne−ε] as a diameter, therefore thepoint z′

n = i Im γn(i) is also in this disk. For large enough integers n, we haveB∞(i, γn(i)) = d(i, z′

n); therefore |d(i, γn(i)) − B∞(i, γn(i))| � d(γn(i), z′n).

This inequality implies that |d(i, γn(i)) − B∞(i, γn(i))| is less than 2ε.Suppose now that there is a sequence (γn)n�1 in Γ satisfying the conditions

of Proposition 3.15. Then for n large enough and for z′n = i Im γn(i), we have

B∞(i, γn(i)) = d(i, z′n),

and limn→+∞ z′n = ∞.

Additionally, there is an A > 0 such that

|d(i, γn(i)) − ln(Im γn(i))| � A.

The sequence (e1/2d(i,γn(i))/√

Im γn(i))n�1 is therefore bounded above.According to Exercise 1.8, the sequence (|i − γn(i)|/Im γn(i))n�1 is like-

wise bounded above. Thus there exists A′ > 0 such that∣∣∣∣Re γn(i)Im γn(i)

∣∣∣∣ � A′.

We deduce from Exercise 3.13, that the sequence (γn(i))n�1 is in anε-neighborhood of [i, ∞). �

The last type of limit points that we introduce in this text are parabolicpoints.


Definition 3.16. A point x in L(Γ ) is parabolic if there exists a parabolictransformation γ �= Id in Γ such that γ(x) = x. The set of parabolic points isdenoted Lp(Γ ).

Recall that, if x is fixed by a parabolic isometry γ in Γ , then the sequence(γn(z))n�1 converges to x along the horocycle centered at x passing through z.

Is it possible for a parabolic point to be horocyclic? To answer this ques-tion, we will prove a stronger result.

Theorem 3.17. Let x be a point in Lp(Γ ). There exists a horodisk H+(x)centered at x such that

γH+(x) ∩ H+(x) = ∅,

for any γ in Γ which does not fix x (Fig. I.28).

Fig. I.28.

Before we prove this theorem, notice that if x is parabolic and z is chosenin a horodisk H+(x) given by Theorem 3.17, then Γz does not meet anyhorodisks centered at x which is properly contained in H+(x). Therefore xcannot be horocyclic.

Corollary 3.18. A parabolic point is not horocyclic.

It is sufficient to prove Theorem 3.17 in the case where x = ∞. In theproof, we will use the following two results on the Euclidean diameter of theimage of a horodisk centered at ∞ under some isometry which does not fixthe point ∞.

Lemma 3.19. Let g(z) = (az + b)/(cz + d) be a Mobius transformation,where a, b, c, d are real numbers satisfying c �= 0 and ad − bc = 1. For t > 0,consider the horodisk H+ centered at ∞ defined by H+ = {z ∈ H | Im z � t}.The horodisk g(H+) is the Euclidean disk tangent to the real axis at the pointa/c, with Euclidean diameter 1/c2t.


Fig. I.29.

Proof. Since c �= 0, the isometry g sends H+ onto the Euclidean disk tangentto the real axis at the point g(∞) = a/c. Moreover the point g(it) is in thehorocycle g(H) associated to g(H+) (Fig. I.29 Ht(∞) = H). Let δ denotethe Euclidean diameter of g(H). By definition of the Busemann cocycle, wehave

Ba/c (it + a/c, iδ + a/c) = ln t/δ

and thus, B∞(g−1(it + a/c), g−1(iδ + a/c)) = ln t/δ.Additionally, the point iδ + a/c is in g(H), hence g−1(iδ + a/c) is in H.

Since it is also in H, we have B∞(it, g−1(iδ + a/c)) = 0 and hence

B∞(g−1(it + a/c), g−1(iδ + a/c)) = B∞(g−1(it + a/c), it).

Therefore,

ln t/δ = lnt

Im g−1(it + a/c),

which implies the equality δ = 1/c2(g)t. �

Notation. Let g(z) = (az + b)/(cz + d) be a Mobius transformation, wherea, b, c, d are real numbers satisfying ad − bc = 1. The positive real number |c|is denoted c(g).

Notice that c(g) = 0 if and only if g fixes the point ∞. Moreover if g is atranslation, then c(gnhgm) = c(h) for all n, m and h ∈ G.

Proposition 3.20. If Γ contains a non-trivial translation, then there existsA > 0, such that c(γ) � A for all γ in Γ satisfying γ(∞) �= ∞.

Proof. Let g(z) = z+α, where α > 0, be a translation in Γ . Suppose that thereis a sequence (γn)n�1 in Γ such that γn(∞) �= ∞ and limn→+∞ c(γn) = 0.Write γn in the form γn(z) = (anz + bn)/(cnz + dn) with andn − bncn = 1and cn > 0.

Consider the integer parts, en and e′n, of an/αcn and dn/αcn respectively.

Setting gn = g−e′nγng−en , one has

gn(z) =(an − cnenα)z + b′

n

cnz + (−cne′nα + dn)

.


We know that the sequence (cn)n�1 is bounded. One also has the in-equalities 0 < an − cnenα < cnα and 0 < dn − cne′

nα < cnα. Thus thesequence (gngg−1

n )n�1 is also bounded. The group Γ being discrete, theset {gngg−1

n | n � 1} is finite. As a result, in the tail of the sequence,c(gngg−1

n ) = αc(γn)2 = 0. This contradicts the hypothesis γn(∞) �= ∞. �

Proof (of Theorem 3.17). After conjugating Γ , one may assume that x is thepoint ∞, and thus that Γ contains a non-trivial translation. Following fromProposition 3.20, there exists A > 0 such that c(γ) > A for all γ in Γ whichdo not fix the point ∞. Take t = 2/A. From Lemma 3.19, such γ sends thehorodisk H+ = {z ∈ H | Im z � t} onto an Euclidean disk having Euclideandiameter A/2c2(γ), tangent to the real axis. Since c(γ) > A, this diameter isless than 1/2A, hence gH+ does not meet H+. �

Let y be a point in H and denote by Γy the subgroup of Γ fixing y.

Exercise 3.21. Prove that Γy is cyclic.

If x is a point in Lp(Γ ), then the group Γx is generated by a parabolicisometry and hence preserves each horodisk centered at x. Fix a horodiskH+(x) given by Theorem 3.17.

Lemma 3.22. The natural projection q : Γx\H+(x) → Γ \H is injective.

Proof. Take y and y′ in H+(x) and suppose that q(Γx(y)) = q(Γx(y′)). Thereexists γ in Γ such that y′ = γ(y). Hence H+(x) ∩ γH+(x) �= ∅. It followsfrom Theorem 3.17 that γ is in Γx and hence, that Γx(y) = Γx(y′). �

Definition 3.23. The subset q(Γx\H+(x)) of Γ \H is called the cusp associ-ated with H+(x) and is denoted by C(H+(x)).

To visualize the shape of a cusp, without loss of generality we can supposex = ∞, H+(∞) = {z ∈ H | Im z � t} with t > 0, and Γx is generated by atranslation g(z) = z + a, with a > 0. Fix a point z ∈ H such that Re z = a/2and that γ(z) �= z for any γ �= Id in Γ . Applying Proposition 2.17 to theDirichlet domains Dz(Γx) and Dz(Γ ), we obtain that the cusp C(H+(∞)) ishomeomorphic to the vertical strip B = {z′ ∈ H | 0 � Re(z′) � a, Im z′ � t}(Fig. I.30) in which the vertical edges are identified by the translation g(Fig. I.31).

4 Geometric finiteness

Recall that a lattice is a Fuchsian group Γ such that the area of each (or ofone) Dirichlet domain Dz(Γ ) is finite, and that a lattice is said to be uniformif each (or one) Dirichlet domain is compact (Definition 2.20). The purposeof this section is to generalize these notions by adopting two points of view:one involving the geometry of a Dirichlet domain, the other involving theproperties of points in the limit set.

4 Geometric finiteness 33

Fig. I.30 Fig. I.31

4.1 Geometric finiteness and Dirichlet domains

Let Γ be a Fuchsian group. We associate to it the following subset of H definedby:

Ω(Γ ) = {z ∈ H | there exist x and y in L(Γ ) such that z ∈ (xy)} .

If Γ is elementary, then from Proposition 3.3, either Ω(Γ ) is empty or thisset is a single geodesic.

Definition 4.1. Let Γ be a non-elementary Fuchsian group, the Nielsen re-gion of Γ is the convex hull of Ω(Γ ), in the sense of hyperbolic segments. Itis denoted by N(Γ ).

Exercise 4.2. Let Γ be a non-elementary Fuchsian group, prove that itsNielsen region is the smallest closed, non-empty and Γ -invariant subset of H.

Proposition 4.3. We have N(Γ ) = H if and only if L(Γ ) = H(∞).

Proof. If L(Γ ) = H(∞), then N(Γ ) = H. Suppose now that L(Γ ) �= H(∞),and let us prove that N(Γ ) is properly contained in H(∞). Since L(Γ ) �=H(∞), there exists an open, non-empty interval I in R which has no inter-section with L(Γ ). Clearly, the set Ω(Γ ) is included in the closed half-planebounded by the geodesic whose endpoints are those of I, and whose boundaryat infinity is H(∞) − I. Since this half-plane is convex the Nielsen region of Γis also included in it, and hence N(Γ ) �= H. �

Notice that the interior of the Nielsen region of a non-elementary Fuchsiangroup is not empty.

Definition 4.4. A Fuchsian group Γ is said to be geometrically finite if,either Γ is elementary, or there is a Dirichlet Dz(Γ ) such that the setN(Γ ) ∩ Dz(Γ ) has finite area.

For example, lattices are geometrically finite.

Clearly, the Nielsen region N(Γ ) of a Fuchsian group Γ is tessellated bythe images by Γ of N(Γ ) ∩ Dz(Γ ), for any Dirichlet domain. Thus the action


of a geometrically finite group Γ on N(Γ ) behaves like the action of a latticeon H.

Among geometrically finite groups, those whose action on N(Γ ) is cocom-pact are especially interesting. The following definition generalizes the notionof uniform lattices.

Definition 4.5. A non-elementary Fuchsian group Γ is called convex-cocom-pact if there is a Dirichlet Dz(Γ ) such that the set N(Γ ) ∩ Dz(Γ ) is compact.

Can the geometric finiteness of a Fuchsian group Γ be checked directlyfrom the shape of one of its Dirichlet domains?

Before we answer this question, let us introduce the notion of edges andvertices of a Dirichlet domain. Take a Dirichlet domain Dz(Γ ) of Γ and a non-trivial transformation γ of Γ . When the intersection of Dz(Γ ) and γ(Dz(Γ ))is non-empty, it is contained in the perpendicular bisector Mz(γ) of [z, γ(z)]h.This intersection, is a point, a non-trivial geodesic segment, a geodesic ray ora geodesic. In the latter three cases, we say that this intersection is an edgeand denote it by C(γ):

C(γ) = Dz(Γ ) ∩ γDz(Γ ).

Notice that γ−1C(γ) is also a edge since γ−1C(γ) = γ−1Dz(Γ ) ∩ Dz(Γ ).Moreover this set is included in the perpendicular bisector of [z, γ−1(z)]h,hence γ−1C(γ) = C(γ−1).

The vertices are the endpoints of the edges. An infinite vertex is a vertexcontained in H(∞).

The group Γ being countable, the set of edges, and hence the set of vertices,of Dz(Γ ) is also countable. Let (Ci = C(γi))i∈I the (possibly finite) sequenceof edges of Dz(Γ ), where I is a subset of N and γi is in Γ .

Exercise 4.6. Prove that Dz(Γ ) is the intersection of the closed half-planesassociated to the edges C(γi)i∈I of this domain, defined by {z′ ∈ H | d(z′, z) �d(z′, γi(z))}, for all i in I.

Exercise 4.7. Suppose that I is finite and that Γ is not elementary. Provethat the area of Dz(Γ ) is infinite if and only if the boundary at infinity ofthis domain, Dz(Γ )(∞) = Dz(Γ ) ∩ H(∞), contains a closed interval whoseendpoints are two distinct infinite vertices.

Suppose now that Γ is not elementary. Let us prove that, if the domainDz(Γ ) has finitely many edges, then Γ is geometrically finite.

If the area of this domain is finite, then Γ clearly is geometrically finite(it is a lattice).

If the area of the domain Dz(Γ ) is infinite, applying Exercise 4.7, we obtaina closed interval J contained in Dz(Γ )(∞) whose endpoints are two distinctinfinite vertices of Dz(Γ ). Consider the finite sequence (Jj)1�j�k of all such


intervals. These intervals are pairwise disjoint. By construction, the polygon Pbounded by the geodesics Lj whose endpoints are those of Jj , and by theedges (Ci = C(γi))i∈I of Dz(Γ ) has finite area. Since Γ (z) ∩ Dz(Γ ) = {z},the interior of each interval Jj does not meet L(Γ ). It follows that the setΩ(Γ ), and hence N(Γ ), is included in the intersections of the closed half-planes bounded by Lj , whose boundary at infinity is H(∞) − Jj , with theclosed half-planes bounded by Ci containing z. Consequently, N(Γ ) ∩ Dz(Γ )is a subset of the polygon P (Fig. I.32), and hence Γ is geometrically finite.

Fig. I.32.

The converse is also true. Its proof, which is more technical, requires astudy of the vertices of a Dirichlet domain that we have decided not to includein the development of this text. Instead, we direct the reader to the proof byA. Beardon. These results are stated in the following theorem:

Theorem 4.8 ((i) ⇒ (ii) [7, Theorem 10.1.2]). Let Γ be a non-elemen-tary Fuchsian group. Then the following are equivalent:

(i) The area of Dz(Γ ) ∩ N(Γ ) is finite.(ii) The Dirichlet domain Dz(Γ ) has finitely many edges.

4.2 Geometric finiteness and limit points

The goal of this subsection is to give a characterization of the geometricfiniteness in terms of limit points.

Let us first analyze the intersection of the boundary at infinity of theDirichlet domain Dz(Γ ) of Γ with its limit set.

Proposition 4.9. Let Γ be a Fuchsian group and z ∈ H. A point x in H(∞)is in Dz(Γ )(∞) if and only if

supγ∈Γ

Bx(z, γ(z)) = 0.

Proof. Let z ∈ H such that γ(z) �= z for any γ �= Id in Γ and x ∈ H(∞).Denote by r : [0, +∞) → H the arclength parametrization of the geodesic ray[z, x).


Suppose that supγ∈Γ Bx(z, γ(z)) = 0. To each γ in Γ we associate a func-tion f : R

+ → R defined by f(t) = t − d(γ(z), r(t)). This is an increasingfunction since if s > t, one has d(γ(z), r(s)) � d(γ(z), r(t)) + s − t. Also,by definition of the Busemann cocycle, we have limt→+∞f(t) = Bx(z, γ(z)).Since Bx(z, γ(z)) is negative, we obtain f(t) = d(z, r(t))−d(γ(z), r(t)) � 0, forall t ∈ R

+. This shows that the ray [z, x) is included in Dz(Γ ), and thereforethat x is in Dz(Γ )(∞).

Suppose now that x is in Dz(Γ )(∞) and consider a sequence of points(zn)n�0 in Dz(Γ ) converging to x. Because Dz(Γ ) is convex, one may assumethat the sequence (zn)n�0 is in the ray [z, x). The point zn is in Dz(Γ ), thusfor all γ in Γ one has d(z, zn) − d(γ(z), zn) � 0. Passing to the limit, oneobtains Bx(z, γ(z)) � 0 for all γ in Γ , and thus supγ∈Γ Bx(z, γ(z)) = 0. �

Geometrically, the previous proposition says that a point x in Dz(Γ )(∞)is characterized by the fact that the orbit Γz does not intersect the interiorof the horodisk centered at x passing through z.

Corollary 4.10. Let Γ be a non-elementary Fuchsian group and x in H(∞).

• If x /∈ L(Γ ), then there exists γ ∈ Γ such that γ(x) is in Dz(Γ )(∞).• If x ∈ Lp(Γ ), then there exists γ ∈ Γ such that γ(x) is in Dz(Γ )(∞).

Moreover each point in Lp(Γ ) ∩ Dz(Γ )(∞) is isolated in Dz(Γ )(∞).• The set Lh(Γ ) ∩ Dz(Γ )(∞) is the empty set.

Proof. Notice that the third property is a direct consequence of Propositions4.9 and 3.10. Let us now prove the first two properties.

Let g ∈ Γ , applying Proposition 4.9 we obtain that for the point g(x) isin Dz(Γ )(∞) if and only if supγ∈Γ Bg(x)(z, γ(z)) = 0. Since Bg(x)(z, γ(z)) =Bx(g−1z, z)+Bx(z, g−1γ(z)), we obtain that g(x) is in Dz(Γ )(∞) if and onlyif the number S = supγ∈Γ Bx(z, γ(z)) is equal to Bx(z, g−1(z)).

Notice that, if the point x is not in L(Γ ), or is in Lp(Γ ), then S is finite,because under these conditions x cannot be horocyclic (Corollary 3.18).

Suppose that there is some sequence (γn)n�0 in Γ such that the sequence(Bx(z, γn(z)))n�0 is not stationary and converges to S.

For n large enough, γn(z) is in the horodisk {z′ ∈ H | Bx(z, z′) � S − 1}.The intersection of this horodisk with H(∞) is x. Therefore the sequence(γn(z))n�0 converges to x. If x does not belong to L(Γ ), we obtain a contra-diction.

If x is parabolic, after conjugating Γ one may suppose that x = ∞. Since(B∞(z, γn(z)))n�0 is not stationary and converges to 0, one can choose Aand B strictly positive such that for any n:

A � Im γn(z) � B.

Let g be a non-trivial translation in Γ . For each n, there exists kn for whichthe sequence (gknγn(z))n�0 is bounded.


Since the group Γ is Fuchsian, the set {gknγn(z) | n � 0} is finite,which is impossible since B∞(z, gknγn(z)) = B∞(z, γn(z)) and the sequence(B∞(z, γn(z)))n�0 is not stationary.

To prove the last part of the second property, take a point inLp(Γ ) ∩ Dz(Γ )(∞) and suppose x = ∞. Since the point ∞ is a parabolicpoint, the group Γ contains a non-trivial translation g, and hence Dz(Γ ) isin the vertical strip bounded by the perpendicular bisectors of the segments[z, g(z)]h and [z, g−1(z)]h. It follows that the set Dz(Γ )(∞) with the point ∞removed, is in a bounded interval of R, which shows that the point ∞ isisolated in Dz(Γ )(∞). �

Applying Corollary 4.10 to the particular case where the group Γ is alattice, we obtain

Proposition 4.11. If Γ is a lattice, then L(Γ ) = H(∞). Moreover the set ofparabolic points Lp(Γ ) is empty or is the union of finitely many Γ -orbits.

Proof. If Γ is a uniform lattice, then Proposition 4.11 is an immediate conse-quence of Corollary 4.10, since in this case Dz(Γ ) is a compact subset of H.

If Γ is a nonuniform lattice, then Dz(Γ )(∞) is not empty, let us show thatthis set is finite.

Suppose that this is not the case and consider infinitely many points(xn)n�0 in Dz(Γ )(∞). Each xn is the limit of a sequence of points in Dz(Γ ).However, this domain is convex, thus the ray [z, xn) is a subset of Dz(Γ ).Let Tn be the hyperbolic triangle with vertices z, xn, xn+1. After some re-ordering of (xn)n�0, one may assume that these triangles Tn are adjacent.Let αn denote the measure of the geometrical angle of Tn at z. The areaof Tn is

A(Tn) = π − αn.

Additionally, the triangles Tn being adjacent, one has∑+∞

n=0 αn � 2π. For allN � 0, the union of the triangles

⋃Nn=0 Tn is in Dz(Γ ) and A

( ⋃Nn=0 Tn

)�

(N + 1)π − 2π, which is impossible since the area of Dz(Γ ) is finite. In con-clusion, the set Dz(Γ )(∞) is finite. Applying Corollary 4.10, we obtain thatthe sets H(∞) − L(Γ ) and Lp(Γ ) are the finite union of Γ -orbits (or areempty). Since H(∞) − L(Γ ) is an open set, we have H(∞) − L(Γ ) = ∅, henceL(Γ ) = H(∞). �

More generally we have:

Theorem 4.12. If Γ is a non-elementary geometrically finite Fuchsiangroup, then the set L(Γ ) ∩ Dz(Γ )(∞) is finite, and is equal to the setLp(Γ ) ∩ Dz(Γ )(∞). Moreover the set Lp(Γ ) is a finite union of Γ -orbits.

Proof. For g ∈ G and z in H such that g(z) �= z, recall that the closedhalf-plane defined by Dz(g) = {z′ ∈ H | d(z′, z) � d(z′, g(z))} satisfies:g(

◦Dz(g−1)) = H − Dz(g). We choose z and denote Dz(g) = D(g).


The group Γ being geometrically finite, some Dirichlet domain Dz(Γ )has finitely many edges. We have Γ (z) ∩ Dz(Γ ) = {z}, hence x belongsto L(Γ ) ∩ Dz(Γ )(∞) and x is an infinite vertex of Dz(Γ ). It follows thatL(Γ ) ∩ Dz(Γ )(∞) is finite.

Consider now x in L(Γ ) ∩ Dz(Γ )(∞). Since Γ is not elementary, the pointsof L(Γ ) are not isolated in L(Γ ). It follows that there is a non-stationarysequence (xn)n�1 in L(Γ ) converging to x. One may assume that this sequencedoes not meet Dz(Γ )(∞), and thus that there is some γ1 in Γ satisfying: x isan endpoint of an edge C(γ−1

1 ), and the sequence (xn)n�1 is in the boundaryat infinity of the half-plane D(γ−1

1 ) (Fig. I.33).

Fig. I.33.

The isometry γ1 sends C(γ−11 ) to the edge C(γ1), thus the point γ1(x)

is an endpoint of C(γ1). Additionally, the sequence (γ1(xn))n�1 converges toγ1(x), is contained in L(Γ ), and never goes into D(γ1)(∞). It follows thatthere is some γ−1

2 distinct from γ1, such that γ1(x) is an endpoint of the edgeC(γ−1

2 ). Suppose that x is not parabolic. In this case, the isometry γ−12 is

distinct from γ−11 by Exercise 2.7.

In the preceding argument, replace x with γ1(x) and γ1 with γ2. In sodoing, if x is not parabolic one obtains an element γ3 in Γ − {γ±1

2 , Id} suchthat γ2γ1(x) is in the boundary at infinity of D(γ2). Continuing this way, oneconstructs a sequence (γn)n�1 in Γ − {Id} satisfying

γn · · · γ1(x) ∈ Dz(Γ )(∞) ∩ L(Γ ) and γn+1 �= γ±1n .

Since Dz(Γ ) has finitely many edges, the set of points in this sequence isfinite. Hence two integers n < m can be chosen satisfying γm · · · γ1(x) =γn · · · γ1(x). Because the point x is neither parabolic nor horocyclic, we haveγm · · · γn+1 = Id.

Let us examine the image of the point z, center of the Dirichlet domainDz(Γ ), by this element. The point γn+1(z) is contained in the open half-plane

◦D(γn+1). Furthermore, the half-planes

◦D(γn+1) and

◦D(γn+2) are dis-

joint. Hence γn+2γn+1(z) is in◦D(γn+2). Continuing along these lines, one ob-

tains that the point γm · · · γn+1(z) is in◦D(γm), which contradicts the equality

γm · · · γn+1(z) = z. Thus we conclude that x is parabolic. Applying Corollary4.10, we obtain that the set L(Γ ) is a finite union of Γ -orbits. �


The following theorem gives a characterization of the geometric finitenessin terms of points in the limit set.

Theorem 4.13. Let Γ be a Fuchsian group. Then the following are equivalent:

(i) The group Γ is geometrically finite.(ii) L(Γ ) = Lp(Γ ) ∪ Lh(Γ ).(iii) L(Γ ) = Lp(Γ ) ∪ Lc(Γ ).

Recall that the parabolic and conical points of the limit set of a Fuchsiangroup can be detected from the dynamics of this group on H(∞) (Proposi-tion 3.14). By Theorem 4.13, it turns out that the geometric finiteness of aFuchsian group Γ , which was originally defined by the action of Γ on H, isentirely determined by the dynamics of the group on H(∞).

Notice that (iii) ⇒ (ii) in Theorem 4.13 is clear since conical points arehorocyclic.

Proof of (ii) ⇒ (i) in Theorem 4.13.

Proposition 4.14. If Γ is a Fuchsian group which is not geometrically finite,then for any Dirichlet domain Dz(Γ ), there exists a point in L(Γ )∩ Dz(Γ )(∞)which is not parabolic.

Proof. Recall that C(γ) denotes an edge of a Dirichlet domain Dz(Γ ) whichincluded in the perpendicular bisector of [z, γ(z)]h.

Since Γ is not geometrically finite, any domain Dz(Γ ) has infinitely manyedges (C(γn))n�1. One can suppose that the sequence (γn(z))n�1 converges tosome point x. Since the edge C(γn) is included in the perpendicular bisectorsMz(γn) of segments [z, γn(z)]h, the sequence of edges (C(γn))n�1 convergesto x and hence x is in Dz(Γ )(∞) ∩ L(Γ ). One can suppose x = ∞. If xis parabolic, then the group Γx is generated by a non-trivial translation p.It follows that the domain Dz(Γ ) is in the vertical strip bounded by theperpendicular bisectors of the segments [z, g(z)]h and [z, g−1(z)]h. On theother hand, the sequence of perpendicular bisectors (Mz(γn))n�1 convergesto x and each Mz(γn) meets Dz(Γ ). Thus, for large enough n, the geodesicMz(γn) intersects Mz(p) and Mz(p−1), and the half-plane bounded by Mz(γn)containing z is a half-disk perpendicular to the real axis. This contradicts thefact that the point ∞ is in Dz(Γ )(∞). �

We deduce from this proposition and from Corollary 4.10 the followingcorollary

Corollary 4.15 (Theorem 4.13(ii) ⇒ (i)). If L(Γ ) = Lp(Γ ) ∪ Lh(Γ ),then Γ is geometrically finite.

In this part, Γ is a non-elementary geometrically finite group and Dz(Γ )is a Dirichlet domain having finitely many edges.

Before giving the proof of (i) ⇒ (iii) in Theorem 4.13, let us analyze theintersection of the Nielsen region N(Γ ) of the group Γ with Dz(Γ ).


First, we associate to each point x in the finite set Lp(Γ ) ∩ Dz(Γ )(∞) ahorodisk H+(x) centered at x satisfying the condition of Theorem 3.17

H+(x) ∩ γH+(x) = ∅,

for all γ in Γ not fixing x. One may choose the horodisks to be pairwisedisjoint. Moreover one can assume that the horodisks

(γ(H+(x)))γ∈Γx∈Lp(Γ )∩Dz(Γ )(∞)

are pairwise either disjoint or identical.To prove this, let x and y in Lp(Γ ) ∩ Dz(Γ )(∞), with y not in Γ (x). Con-

sider the set A of γ ∈ Γ such that γH+(x) ∩ H+(y) �= ∅. Let B denote the setof two-sided cosets Γy \A/Γx (i.e., equivalence classes via two relations). If Bis finite, it suffices to replace H+(x) with a smaller horodisk than H+(x), forwhich γH+(x) ∩ H+(y) = ∅ for all γ ∈ Γ . Otherwise, consider a sequence ofdistinct elements of B written as (bn = ΓyanΓx)n�1, where an ∈ A. Fix a com-pact fundamental domain K for the action of Γy on H(y). For all n � 1, thereexists pn ∈ Γy such that pnanH(x) ∩ K �= ∅. The horocycles (pnanH(x))n�1

are pairwise disjoint but also intersect K, which is impossible.The following proposition gives a decomposition of the intersection of the

Nielsen region N(Γ ) of the group Γ with Dz(Γ ).

Proposition 4.16. Let Γ be a non-elementary geometrically finite group.Then there exists a relatively compact set K ⊂ H such that

N(Γ ) ∩ Dz(Γ ) = K⋃

x∈Lp(Γ )∩Dz(Γ )(∞)

H+(x) ∩ Dz(Γ ).

Proof. Suppose that the closure of the intersection of the Nielsen region N(Γ )with the set Dz(Γ ) −

⋃x∈Lp(Γ )∩Dz(Γ )(∞) H+(x) ∩ Dz(Γ ) is not compact. Con-

sider y in the intersection of this closure with H(∞). Such a point is notin L(Γ ). Otherwise, y is in Dz(Γ )(∞) ∩ L(Γ ) and hence, since Dz(Γ ) hasfinitely many edges, y would be a parabolic point (Theorem 4.12). Let p bea generator of Γy. The domain Dz(Γ ) is included in the intersection of thehalf-planes

H(p) = {z′ ∈ H | d(z′, z) � d(z′, p(z))} and

H(p−1) = {z′ ∈ H | d(z′, z) � d(z′, p−1(z))}.

It follows that set Dz(Γ ) −⋃

x∈Lp(Γ )∩Dz(Γ )(∞) H+(x) ∩ Dz(Γ ) is included inthe set H(p) ∩ H(p−1) − H(p) ∩ H(p−1) ∩ H+(y), which is impossible since theboundary at infinity of this last set does not contain y.

In conclusion, the point y is not in L(Γ ) and hence is in an open interval Iin Dz(Γ )(∞) which does not intersect L(Γ ). Let L be the geodesic whoseendpoints are the endpoints of I. The set Ω(Γ ), and hence N(Γ ), is in the

5 Comments 41

closed half-plane bounded by L, whose boundary at infinity is (H(∞) − I).Since y is in the boundary at infinity of N(Γ ), this point is in (H(∞) − I) ∩ I,which is not possible. In conclusion, the intersection of the Nielsen regionN(Γ ) with the set Dz(Γ ) −

⋃x∈Lp(Γ )∩Dz(Γ )(∞) H+(x) ∩ Dz(Γ ) is relatively

compact in H. �

Proof of (i) ⇒ (iii) in Theorem 4.13.Suppose that Γ is geometrically finite. We are ready now to prove that

L(Γ ) = Lp(Γ ) ∪ Lc(Γ ). Since Γ is not elementary, the set Lc(Γ ) is not empty.Take any point y in L(Γ ) − Lp(Γ ), and let z′ in H be such that [z′, y) is inN(Γ ). Since Dz(Γ ) is a fundamental domain, the ray [z′, y) is contained inthe union

⋃γ∈Γ γ(Dz(Γ )) ∩ [z′, y). Suppose that there is z′ ′ in [z′, y) such that

[z′ ′, y) is contained in⋃

γ∈Γ, x∈Lp(Γ )∩Dz(Γ )(∞) γ(H+(x)). Since the horodisks(γ(H+(x)))γ∈Γ,x∈Lp(Γ )∩Dz(Γ )(∞) are either disjoint or identical, the ray [z′ ′, y)is contained in one such horodisk, which is impossible since y is not parabolic.

From Proposition 4.16, we obtain a compact K ⊂ H, a sequence (zn)n�1

in [z′, y) converging to y, and (γn)n�1 in Γ , such that zn is in γn(K). Since Kis compact, the sequence (γn(z′))n�1 converges to y while remaining in anε-neighborhood of [z′, y). This shows that y is conical.

Recall that a geometrically finite Fuchsian Γ group is said to be convex-cocompact if for some Dirichlet domain Dz(Γ ), the set Dz(Γ ) ∩ N(Γ ) is com-pact. From Theorems 4.12 and 4.13, we deduce the following characterizationof convex-cocompact groups and of lattices in terms of their points at infinity.

Corollary 4.17. Let Γ be a Fuchsian group.

• The group Γ is convex-cocompact if and only if L(Γ ) = Lc(Γ ).• The group Γ is a lattice if and only if H(∞) = Lp(Γ ) ∪ Lc(Γ ).

5 Comments

The notions and main results of this chapter can be generalized to the caseof an oriented Riemannian manifold X called a pinched Hadamard manifold ,of dimension n � 2, which is simply connected, complete and has sectionalcurvature which is bounded by two strictly negative constants [12, 6, 14, 28].Below we give a broad outline of this generalization.

In this context, geodesics are well-defined and the boundary at infinityX(∞) of this manifold is defined to be the set of equivalence classes of asymp-totic geodesic rays. While viewing the points of X as endpoints of geodesicsegments of fixed origin, one can provide X ∪ X(∞) with a natural topology,which in fact is a compactification of X in which X is an open dense sub-set. For this compactification, the set of oriented geodesics of X is identifiedwith the pairs of distinct points of X(∞). The notion of Busemann cocycles(Sect. 1.4) can also be extended to this general setting and allows us to definehorocycles (for n = 2) and horospheres (for n � 3) of X.


On the other hand, unlike the transitivity of the action of G on H, thegroup of positive isometries of X may be very poor, possibly even trivial.The proofs given in this chapter which require transitivity cannot be directlyadapted to the general case. However, the majority of them can be trans-lated into Riemannian terms. This is the case, for example, in the proof ofProperty 1.19.

As with H, if a positive isometry does not fix any point in X, then it fixesexactly one (parabolic isometry) or two (hyperbolic isometry) points in X(∞).

We call a subgroup of positive isometries whose action is properly discon-tinuous on X a Kleinian group. The existence of such non-trivial groups isnot guaranteed.

When such a group does exist, the definitions of limit set, horocyclic (horo-spheric), conical and parabolic points, given in Sect. 3, remain unchanged. Inour text, we concentrate on a category of Fuchsian groups for which conicaland horocyclic points are interchangeable. In an article by A. Starkov [59], thereader will find some examples of groups having horocyclic points which arenot conical.

The notion of geometric finiteness has arisen in the context of hyperbolicspace of curvature −1 in dimension 3, the motivation being to study the actionof discrete groups of finite type on this space. When dimX = 2, the notion ofdiscrete groups of finite type and geometrically finite groups coincide. This isnot the case in higher dimensions [10].

When the dimension of X is � 3, the stabilizer of a parabolic point xin Γ does not necessarily act cocompactly on L(Γ ) − {x}. If this action is co-compact, x is called a bounded parabolic point. This family of points emergesin an essential way in the generalization of the notion of geometric finitenessof a group to pinched Hadamard manifolds. In this setting, the definition isdelicate and can be formulated in several ways [12]. One of these formula-tions rests on the decomposition of L(Γ ) into conical and bounded parabolicpoints. Under this hypothesis, the number of Γ -orbits of parabolic points isfinite. Each of these points x has an associated horodisk (horoball) H+(x)whose image by the group Γ is either disjoint from or identical to H+(x) [54,Lemma 1.9]. The quotient of H+(x) by the stabilizer of x in Γ injects intothe manifold M = Γ \X, producing what we will continue to call a cusp. If Γis a geometrically finite Kleinian group, the set of points of X contained ingeodesics whose endpoints are in the limit set of the group, projects into theunion of a compact subset of M and a finite number of cusps. In this contextit is relatively simple to see that most of the results about the topologicaldynamics of geodesic flow and horospheric foliations proved in the followingchapters can be generalized.

We conclude these comments with an outline of some metric propertiesof the limit set of a non-elementary Fuchsian group acting on the Poincaredisk D.

5 Comments 43

One of the keys to this rich area requires the development of the Poincareseries Ps(Γ ) of Γ which is defined by

Ps(Γ ) =∑γ∈Γ

e−sd(0,γ(0)),

where 0 is the center of the disk D. Its critical exponent δ(Γ ) can also bedefined by [54]:

δ(Γ ) = limR→+∞

1R

(ln card{γ ∈ Γ | d(0, γ(0)) � R}).

This series allows us to establish a relationship between the statistical behaviorof Γ (0) and the metric properties of L(Γ ). One can show for example thatif Γ is geometrically finite, then δ(Γ ) is equal to the Hausdorff dimension ofL(Γ ) ([51], [48, Theorem 9.3.6]).

This series also allows us to construct measures m, called Pattersonmeasures, whose support is L(Γ ) and which, while not being Γ -invariant(Γ -invariant measures do not exist if Γ is not elementary), are conformal inthe sense that they satisfy the relation

∀ γ ∈ Γ,dγ−1m

dm(x) = |γ′(x)|δ(Γ ),

where |γ′(x)| represents the conformal factor at the point x of the map γ, seenas a conformal transformation [51, 48] of the disk D.

The construction of these measures is due to S. Patterson ([51], [8,Chap. 9]). If the series Ps(Γ ) diverges for s = δ(Γ ), which is the case when Γis geometrically finite [48], such a measure is obtained by taking the weaklimit, when s tends to δ(Γ ) from above, of a sequence of orbital measures ms

defined by

ms =1

Ps(Γ )

∑γ∈Γ

e−sd(0,γ(0))Dγ(0),

where Dγ(0) represents the Dirac (point mass) measure at γ(0).If Γ is a lattice, this measure is proportional to the Lebesgue measure on

D(∞). As we will show in the Comments following Chaps. III and V, oneinteresting aspect of Patterson measures is that they allow a constructionof measures on Γ \T 1

H which is invariant with respect to geodesic flow andhorocyclic foliation.

The construction of Patterson measures on L(Γ ) can be generalized toKleinian groups acting on pinched Hadamard manifolds [11, 54].

II

Examples of Fuchsian groups

In this chapter, we study concrete examples of Fuchsian groups and illustratethe results of the previous chapter.

The first family of groups that we will consider consists of geometricallyfinite free groups, called Schottky groups. Its construction is based on thedynamics of isometries.

The second family comes from number theory. It consists of three non-uniform lattices: the modular group PSL(2, Z), its congruence modulo 2 sub-group and its commutator subgroup.

We will study each of these groups according the same general outline:

• description of a fundamental domain;• shape of the associated topological surface;• properties of its isometries;• study of its limit set;• characterization of its parabolic points.

We will also construct a coding of the limit sets of Schottky groups and ofthe modular group. We will use this coding in Chap. IV to study the dynamicsof the geodesic flow, and in Chap. VII to translate the behavior of geodesicrays on the modular surface into the terms of Diophantine approximations.

1 Schottky groups

The Poincare disk model D is the ambient space in this discussion. We willfix a point 0 in this set, which is not necessary the origin of the disk.

Recall that, if g is a positive isometry of D which does not fix 0, thenthe set D0(g) = D(g) represents the closed half-plane in D bounded by theperpendicular bisector of the hyperbolic segment [0, g(0)]h, containing g(0).The sets D(g) and D(g−1) are disjoint (resp. tangent) if and only if g ishyperbolic (resp. parabolic) (Property I.2.7). Moreover we have:

g(D(g−1)) = D −◦D(g).


http://dx.doi.org/10.1007/978-0-85729-073-1_2

46 II Examples of Fuchsian groups

Fig. II.1. g hyperbolic

Fig. II.2. g parabolic

Definition 1.1. Let p be an integer � 2. A Schottky group of rank p is asubgroup of G which has a collection of non-elliptic, non-trivial generators{g1, . . . , gp} satisfying the following condition: there exists a point 0 in D

such that the closures in D = D ∪ D(∞) of the sets D0(g±1i ) = D(g±1

i ), fori = 1, . . . , p, satisfy

(D(gi) ∪ D(g−1i )) ∩ (D(gj) ∪ D(g−1

j )) = ∅,

for all i �= j in {1, . . . , p}.

Let S(g1, . . . , gp) denote such a group whose collection of generators is{g1, . . . , gp}.

In the rest of this discussion, in order to avoid notational clutter, we willrestrict ourselves to the case where p = 2.

Figure II.3 represents the four possible configurations associated withSchottky groups of rank 2.

Schottky groups are not especially difficult to find. The following lemmashows that their construction only requires two non-elliptic isometries.

1 Schottky groups 47

Fig. II.3.

Lemma 1.2. Let g and g′ be two non-elliptic isometries in G which have nocommon fixed points. Then there exists N > 0 such that gN and g′N generatea Schottky group S(gN , g′N ).

Proof. Fix a point 0 in D. The sequence (gn(0))n�1 converges to a point xwhich is fixed by g. Thus the sequence of perpendicular bisectors of [0, gn(0)]hsimilarly converges to x. Since g and g′ do not have any fixed points in com-mon, for large enough n the closed sets D(gn) ∪ D(g−n) and D(g′n) ∪ D(g′ −n)are disjoint. �

Since a non-elementary Fuchsian group contains infinitely many non-elliptic isometries which have no common fixed points, we deduce fromLemma 1.2 the following result

Corollary 1.3. A non-elementary Fuchsian group contains infinitely manySchottky groups.

1.1 Dynamics of Schottky groups

Fix a Schottky group S(g1, g2) of rank 2. The alphabet of this group is bydefinition the set A = {g±1

1 , g±12 }. A product of n letters s1 · · · sn in A is said

to be a reduced word of S(g1, g2) if n = 1, or if n > 1 and si �= s−1i+1 for all

1 � i � n − 1. The integer n is called the length of s1 · · · sn. We associate to a


reduced word s1 · · · sn the set D(s1) if n = 1, and if n > 1 the set D(s1, . . . , sn)defined by:

D(s1, . . . , sn) = s1 · · · sn−1D(sn).

Property 1.4. Let s1 · · · sn be a reduced word. The following properties hold:

(i) s1 · · · sn(D −◦D(s−1

n )) ⊂ D(s1).(ii) If n � 2, D(s1, . . . , sn) ⊂ D(s1, . . . , sn−1).(iii) If s1 · · · sn and s′

1 · · · s′n are two distinct reduced words, then the half-

planes D(s1, . . . , sn) and D(s′1, . . . , s

′n) are either tangent or disjoint.

Proof.

(i) We prove (i) by induction on n � 1. When n = 1, part (i) is a consequenceof the following relation:

∀ s ∈ A, s(D −◦D(s−1)) = D(s).

Suppose that (i) is true up to n = N , for some N � 1. Consider thereduced word s1 · · · sNsN+1. By induction hypothesis, one has

s2 · · · sN+1(D −◦D(s−1

N+1)) ⊂ D(s2).

Since s2 �= s−11 , the set D(s2) is contained in D −

◦D(s−1

1 ). Furthermore,the set s1(D −

◦D(s−1

1 )) is equal to D(s1), thus

s1 · · · sN+1(D −◦D(s−1

N+1)) ⊂ D(s1).

(ii) Since sn �= s−1n−1, one has D(sn) ⊂ D −

◦D(s−1

n−1). Thus the set sn−1D(sn)is contained in D(sn−1), which implies (ii).

(iii) Let k � 1 be the smallest integer � n such that s′k �= sk. Proving part

(iii) reduces to proving that D(s′k, . . . , s′

n) and D(sk, . . . , sn) are tangentor disjoint.By (i) and (ii), the set D(s′

k, . . . , s′n) = s′

k · · · s′n−1D(s′

n) is a subsetof D(s′

k). Likewise D(sk, . . . , sn) is a subset of D(sk). Since sk �= s′k,

the sets D(sk) and D(s′k) are tangent or disjoint, hence the half-planes

D(s′k, . . . , s′

n) and D(sk, . . . , sn) are as well. �It is of interest to note that the only hypothesis which played a role in

proving these properties was the following: if a and b are contained in A witha �= b−1, then D(a) is contained in D −

◦D(b−1). As a result, these properties

remain valid for generalized Schottky groups of rank p. This means that thegroup admits a collection of non-elliptic generators {g1, . . . , gp} satisfying thefollowing weaker condition:

(D(gi) ∪ D(g−1i )) ∩ (D(gj) ∪ D(g−1

j )) = ∅,

for all i �= j in {1, . . . , p}.We will meet such groups again in Sect. 3.


Exercise 1.5. Prove that in each of the cases a, a’, b, c represented inFig. II.3, the configuration of half-planes D(g1, g2) is as follows in Fig. II.4.

Fig. II.4.

Let g in G and z ∈ D such that g(z) �= z. Following the notation used inChap. I, we denote by Dz(g) the closed half-plane containing z, bounded by theperpendicular bisector of the segment [z, g(z)]h. We have: Dz(g) = D −

◦Dz(g).

Recall that the Dirichlet domain Dz(Γ ) of a Fuchsian group Γ centered at zis the intersection of all sets Dz(γ), with γ in Γ − {Id}.

Proposition 1.6. The group S(g1, g2) is free with respect to g1, g2, and isdiscrete. Furthermore, the set

⋂i=1,2ε=±1

D −◦D(gε

i ) is the Dirichlet domain of

this group centered at the point 0.

Proof. Let s1, . . . , sn be a reduced word. From Property 1.4(i), the points1 · · · sn(0) is contained is D(s1). The sets D(s1) and

◦D(S(g1, g2)) are dis-

joint. In particular s1 · · · sn(0) �= 0, which shows that S(g1, g2) is free.Let us prove that S(g1, g2) is discrete. Consider a sequence (γn)n�1 in

S(g1, g2) − {Id}. Each γn can be written as a reduced word sn,1 · · · sn,�n .Passing to a subsequence (γnk

)k�1, one may assume that snk,1 = s1 for allk � 1. The point γnk

(0) is contained in D(s1), thus there exists c > 0 suchthat d(γnk

(0), 0) > c for all k � 1. This shows that (γn)n�1 cannot convergeto the identity and thus S(g1, g2) is discrete.


Since S(g1, g2) is free and discrete, none of its elements is elliptic. There-fore, the Dirichlet domain D0(S(g1, g2)) centered at 0 is well defined.

It only remains to show that F =⋂

i=1,2ε=±1

D −◦D(gε

i ) and D0(S(g1, g2))

are equal. The set D0(S(g1, g2)) is certainly a subset of F , since D0(gi) =D −

◦D(gi). If it is a proper subset, there exist z in D0(S(g1, g2)) and γ in

S(g1, g2) − {Id} such that γ(z) is contained in◦F . Writing γ as a reduced word

s1 · · · sn, Property 1.4(i) implies that the point γ(z) is an element of D(s1).This is impossible since γ(z) is contained in

◦D(S(g1, g2)). �

Since the group S(g1, g2) is discrete and admits a Dirichlet domain havingfinitely many edges, one can state the following result:

Corollary 1.7. The group S(g1, g2) is a geometrically finite Fuchsian group.

Notice that the proof of Proposition 1.6 is essentially an application ofProperty 1.4(i). As such, this proposition and its corollary are also valid forgeneralized Schottky groups.

From the dynamic point of view, Schottky groups S(g1, g2), where g1

and g2 are hyperbolic, are—in some sense—the simplest non-elementary Fuch-sian groups.

The following exercise shows that the construction of these Schottkygroups can be extended to an infinite collection of generators.

Exercise 1.8. Let (gi)i�1 be an infinite sequence of non-elliptic isometriesin G satisfying the following condition for all i �= j:

(D(gi) ∪ D(g−1i )) ∩ (D(gj) ∪ D(g−1

j )) = ∅.

Prove that the group generated by this sequence is a Fuchsian free groupwhich is not geometrically finite.

We now focus on the nature of the isometries of the S(g1, g2).

Property 1.9. Let S(g1, g2) be a Schottky group.

(i) If g1 and g2 are both hyperbolic, then every element of S(g1, g2) − {Id} ishyperbolic.

(ii) If not, the non-hyperbolic isometries in S(g1, g2) − {Id} are conjugate topowers of parabolic generators in S(g1, g2).

Proof. Since the group S(g1, g2) is free, it does not contain elliptic isome-tries. Let x be a point in D(∞) fixed by a non-trivial parabolic isometry ofS(g1, g2). Applying Corollary I.4.10, we obtain γ in Γ such that y = γ(x)is in D0(S(g1, g2))(∞). Since y is in L(S(g1, g2)), this point is an endpointof an edge of D0(S(g1, g2)). Using the dynamics of the parabolic isometries,we have that any open arc in D(∞) with extremity y meets L(S(g1, g2)). Itfollows that y is the common endpoint of two edges, and hence that it is fixed


by a parabolic generator a ∈ A (see Fig. II.3). The group S(g1, g2) being dis-crete, any isometry in S(g1, g2) fixing y belongs to the cyclic group generatedby a. This implies that x is fixed by an isometry of the form γakγ−1, for somek �= 0. �

These properties cannot be extended to generalized Schottky groups. Itwill be shown in the next section that some such groups can contain parabolicisometries which are not conjugate to powers of generators.

Using Proposition I.2.17, we obtain that the surfaces S(g1, g2)\D associ-ated to each of the four cases in Fig. II.3 are of the form shown in Fig. II.5.

Fig. II.5.

1.2 Limit set

Clearly, the limit set of a Schottky group S(g1, g2) is a proper subset ofD(∞) since it does not meet the non-empty, open, circular arcs included inD0(S(g1, g2))(∞).

What is the topological structure of L(S(g1, g2))? To begin answering thisquestion, we prove the following lemma:

Lemma 1.10. Given a sequence (si)i�1 in A satisfying si+1 �= s−1i , the se-

quence of Euclidean diameters of the sets D(s1, . . . , sn) converges to zero.


Proof. By Property 1.4(ii), the half-planes D(s1, . . . , sn) are nested. If thesequence of Euclidean radii do not converge to 0, there is a compact set K in D

such that for all geodesics of the form s1 · · · sn−1(Cn), where Cn is the boundaryof D(sn), we have K ∩ s1 · · · sn−1(Cn) �= ∅. The geodesics Cn are edges of theDirichlet domain D0(S(g1, g2)), thus the image of this domain under the mapss1 · · · sn−1 intersects K. Yet this is impossible since Property I.2.15 states thatthis domain is locally finite. �

Proposition 1.11.

L(S(g1, g2)) =+∞⋂n=1

⋃reduced words

of length n

D(s1, . . . , sn).

Proof. Let y be an element of L(S(g1, g2)), and consider a sequence (γn)n�1

in S(g1, g2) such that limn→+∞ γn(0) = y. Write γn as a reduced wordγn = sn,1 · · · sn,�n , where sn,i ∈ A and sn,i �= s−1

n,i+1. Since A is finite, one canassume (by passing to a subsequence) that there exist (si)i�1 with si+1 �= s−1

i ,and a sequence of positive integers (�n)n�1 which is strictly increasing suchthat γn = s1 · · · s�n .

The point s�n(0) is an element of D(s�n), therefore γn(0) is inD(s1, . . . , s�n), for any n � 1. Since the sets D(s1, . . . , sn) are nestedand their diameter go to 0 (Lemma 1.10), we have

{y} =+∞⋂n=1

D(s1, . . . , s�n).

This shows that L(S(g1, g2)) is a subset of

+∞⋂n=1

⋃reduced words

of length n

D(s1, . . . , sn).

The reverse inclusion is a consequence of Property 1.4 and Lemma 1.10. �

Using this proposition, we obtain a construction of L(S(g1, g2)) by an iter-ative procedure analogous to the construction of Cantor sets. More precisely,consider the case in which g1 and g2 are hyperbolic and denote by I1, I2, I3, I4

the connected components of the following set:

D(∞) −⋃

a∈AD(a)(∞).

Step 1: Remove these four arcs from the set D(∞). One obtains

D(∞) −⋃

1�i�4

Ii =⋃

a∈AD(a)(∞).


Step 2: For each a, remove the four arcs a(I1), a(I2), a(I3), a(I4) from the setD(a)(∞). One obtains

∀ a ∈ A, D(a)(∞) −⋃

1�i�4

(Ii) =⋃

1�i�4

⋃b∈A

b �=a−1

D(a, b)(∞).

More generally, one does the following for n � 2:Step n: Remove from each set of the form D(s1, . . . , sn−1)(∞), wheres1 · · · sn−1 is a reduced word, the four arcs s1 · · · sn−1(I1), s1 · · · sn−1(I2),s1 · · · sn−1(I3), s1 · · · sn−1(I4). One obtains

D(s1, . . . , sn−1)(∞) −⋃

1�i�4

s1 · · · sn−1(Ii) =⋃

s∈As�=s−1

n−1

D(s1, . . . , sn−1, s)(∞).

It follows from Proposition 1.11, that the set L(S(g1, g2)) is the intersectionover the integers n � 1, of 4 × 3n−1 arcs obtained at Step n of this procedure(Fig. II.6).

Fig. II.6.

If some of the generators is parabolic, then the procedure above mustbe modified by grouping together arcs of the form D(s1, . . . , sn)(∞) andD(s′

1, . . . , s′n)(∞) having an endpoint in common.

Exercise 1.12. Prove that the set L(S(g1, g2)) is a totally discontinuous set(i.e., its connected components are points), without isolated points.(Hint: [13].)

Recall that the Nielsen region N(S(g1, g2)) of S(g1, g2) is the convex hullof the set of points in D belonging to geodesics whose endpoints are in thelimit set L(S(g1, g2)) (Sect. I.4).

Figure II.7 shows the form of the intersection of N(S(g1, g2)) with theDirichlet domain D0(S(g1, g2)) in cases a, a’, b, c associated with Fig. II.3.

If neither of its generators are parabolic, the group S(g1, g2) is geometri-cally finite and does not contain any parabolic isometries, and thus this group


Fig. II.7.

is convex-cocompact (Corollary I.4.17). This property is shown by Fig. II.7(cases a and a’) since the set N(S(g1, g2)) ∩ D0(S(g1, g2)) is compact (Defini-tion I.4.5). This is not the case if at least one of the generators is parabolic.

Thus one can state the following property.

Property 1.13. The group S(g1, g2) is convex-cocompact if and only if g1

and g2 are hyperbolic.

2 Encoding the limit set of a Schottky group

At this stage, we enter the world of symbolic dynamics which will be exploredfurther in Chap. IV.

The purpose of this section is simply to construct a dictionary betweenL(S(g1, g2)) and a specific set of sequences of elements of A. Using this dic-tionary, we establish a correspondence between some properties of these se-quences and some geometric properties of points in L(S(g1, g2)).

Since the group S(g1, g2) is free, there is a bijection between the set offinite reduced sequences s1, . . . , sn, with si �= s−1

i+1, and S(g1, g2) − {Id}. Letus extend this bijection to the set of infinite reduced sequences Σ+ defined by

Σ+ = {(si)i�1 | si ∈ A, si+1 �= s−1i }.

Let (si)i�1 be such a sequence. Define

γn = s1 · · · sn.

2 Encoding the limit set of a Schottky group 55

The point γn(0) is in the half-plane D(s1, . . . , sn). As a consequence of Prop-erty 1.4(ii), the half-planes (D(s1, . . . , sn))n�1 are nested and, by Lemma 1.10,their Euclidean diameter tends to 0. Hence the sequence (γn(0))n�1 convergesto a point in L(S(g1, g2)).

Let f : Σ+ → L(S(g1, g2)) denote the function which sends s = (si)i�1 tothe point x defined by

x(s) = limn→+∞

γn(0).

Exercise 2.1. Prove that the function f is surjective.(Hint: Use Property 1.11.)

Is this function injective? To answer this question, the cases with andwithout parabolic generators must be considered separately.

Consider the subset Σ+c of Σ+ consisting of sequences (sn)n�1 ∈ Σ+ for

which, if the term sn is parabolic, there exists m > n such that sm �= sn.Recall that, since S(g1, g2) is geometrically finite, its limit set can be de-

composed into a disjoint union of the set of its parabolic points Lp((S(g1, g2))and its conical points Lc(S(g1, g2)) (Theorem I.4.13).

Proposition 2.2. If g1 and g2 are hyperbolic, then the function f : Σ+ →L(S(g1, g2)) is a bijection. Otherwise, this function is surjective but not injec-tive and its restriction to Σ+

c is a bijection onto Lc(S(g1, g2)).

Proof.Case 1: g1 and g2 hyperbolic. In this case Σ+

c = Σ+ and for all a in A and b inA − {a}, the closure of the sets D(a) and D(b) are disjoint. Let s = (si)i�1 ands′ = (s′

i)i�1 be in Σ+. Suppose that there exists i � 1 such that si �= s′i. Let k

denote the smallest of these integers and define γ = s1 · · · sk−1 if k > 1 andγ = Id otherwise. For all n � k, the points γ−1s1 · · · sn(0) and γ−1s′

1 · · · s′n(0)

are contained in D(sk) and D(s′k) respectively. These two sets are disjoint,

thus limn→+∞ s1 · · · sn(0) �= limn→+∞ s′1 · · · s′

n(0).This shows that f is injective.

Case 2: g1 or g2 is parabolic. Suppose that g1 is parabolic. In this case, thesequences (gn

1 (0))n�1 and (g−n1 (0))n�1 converge to the same point, thus f is

not injective.Let us show that f(Σ+

c ) = Lc(S(g1, g2)). Let s = (si)i�1 be in (Σ+ − Σ+c )

and let k denote the smallest integer for which sk is parabolic and si = sk

for all i � k. Define γ = s1 · · · sk−1 if k > 1 and γ = Id otherwise. The pointγ−1f(s) is parabolic since it is fixed by sk, thus f(s) is parabolic. As a result,the set f(Σ+ − Σ+

c ) is a subset of Lp(S(g1, g2)).Let y be in Lp(S(g1, g2)). By Property 1.9(ii), there exists γ in S(g1, g2)

such that γ(y) is fixed by a parabolic generator g. Let s = (si)i�1 be a sequencein Σ+ such that f(s) = y. Since γ can be written as a finite reduced word,there exists s′ = (s′

i)i�1 in Σ+, k � 0 and k′ � 0 such that γ(y) = f(s′), andfor all i � 1, s′

k′+i = sk+i. The point γ(y) is in D(s′1)(∞). On the other hand,


γ(y) is the point of tangency between D(g) and D(g−1), hence s′1 ∈ {g, g−1}.

If we replace γ(y) with s−11 (y), following the same argument we obtain s′

1 = s′2.

Continuing this process iteratively, we find that the sequence s′ must beconstant. Hence for some k � 0, the sequence (sk+i)i�1 is constant. Thisimplies that s is not contained in Σ+

c . Hence f −1(Lp(S(g1, g2))) = (Σ − Σ+c ).

In conclusion, f(Σ − Σ+c ) = Lp(S(g1, g2)) and hence f(Σ+

c ) =Lc(S(g1, g2)).

Finally we must show that the function f restricted to Σ+c is injective. Let

s = (si)i�1 and s′ = (s′i)i�1 be elements of Σ+

c . Suppose that s and s′ aredistinct and denote by k the smallest integer i � 1 such that si �= s′

i. Defineγ = s1 · · · sk−1 if k �= 1 and γ = Id otherwise.

If s−1k �= s′

k, or if one of these two letters are hyperbolic, then the setD(sk)(∞) ∩ D(s′

k)(∞) is empty, thus γ−1f(s) �= γ−1f(s′).If s−1

k = s′k and sk is parabolic, consider the smallest i > k such that

si �= sk and define g = γsk · · · si−1. Then

g−1(f(s)) = limn→+∞

si · · · si+n(0),

g−1(f(s′)) = limn→+∞

s−1i−1 · · · s−1

k s−1k s′

k+1 · · · s′k+n(0).

Since s′k+1 �= sk, the point g−1(f(s′)) is in D(s−1

i−1)(∞). Furthermore,g−1(f(s)) is in D(si)(∞), and D(si)(∞) ∩ D(s−1

i−1)(∞) = ∅, since si isnot contained in {si−1, s

−1i−1}. Therefore g−1(f(s′)) �= g−1(f(s)) and hence

f(s) �= f(s′). �It follows that the fixed points of parabolic isometries in L(S(g1, g2)) are

encoded (non-uniquely) by the sequences (si)i�1 in Σ+ which are constantfor large i, and whose repeated term is a parabolic generator.

Let us now analyze the encoding of all the fixed points of the isometriesin S(g1, g2). Consider the shift function T : Σ+ → Σ+ defined by

T ((si)i�1) = (si+1)i�1.

A sequence s is periodic if there exists k � 1 such that T ks = s. In this case,one writes

s = (s1, . . . , sk).

More generally, if there exists n � 1 such that Tns is periodic, then thesequence s is said to be almost periodic.

Note that if s is a sequence in (Σ+ − Σ+c ), there exists k � 0 such that

T k(s) = (sk+1) with sk+1 parabolic. Hence, such a sequence is almost periodicand f(s) is the fixed point of γsk+1γ

−1, where γ = s1 · · · sk. The followingproperty generalizes this connection between almost periodic sequences andfixed points of isometries of S(g1, g2).

Property 2.3. A point y in L(S(g1, g2)) is fixed by a non-trivial isometry inS(g1, g2) if and only if there exists an almost periodic sequence s in Σ+ suchthat f(s) = y.

2 Encoding the limit set of a Schottky group 57

Proof. Let y be in L(S(g1, g2)). Suppose that there exists a non-trivial γ inS(g1, g2) such that y = limn→+∞ γn(0). Write γ as a reduced word γ =s1 · · · sn. If s1 �= s−1

n , then the periodic sequence s = (s1, . . . , sn) is containedin Σ+ and y = f(s).

Otherwise, consider the largest 1 � k < n such that sk = s−1n−k+1,

and define g = s1 · · · sk. The point g−1(y) is fixed by the reduced wordsk+1 · · · sn−k. Since sk+1 �= s−1

n−k, one has g−1(y) = f(sk+1, . . . , sn−k). Con-sider the almost periodic sequence s′ defined by s′

i = si for all 1 � i � kand T k(s′) = (sk+1, . . . , sn−k). Since sn−k �= s−1

k+1, this sequence is in Σ+ andf(s′) = y.

Conversely, consider an almost periodic sequence s in Σ+. Let k � 0 besuch that T k(s) is the periodic sequence s′ = (sk+1 · · · sn). Then

f(s′) = limp→+∞

(sk+1 · · · sn)p(0),

hence f(s′) is fixed by γ = sk+1 · · · sn. If k = 0, then f(s′) = f(s); otherwisef(s) = g(f(s′)) with g = s1 · · · sk. Thus f(s) is fixed by gγg−1. �

Since Γ is geometrically finite, we have L(S(g1, g2)) = Lp(S(g1, g2)) ∪Lc(S(g1, g2)). By definition of conical points, if x is a point in the setL(S(g1, g2)) − Lp(S(g1, g2)), then there exists a sequence (γn)n�1 in S(g1, g2)such that (γn(0))n�1 converges to x, remaining at a bounded distance fromthe geodesic ray [0, x).

How to construct such a sequence (γn)n�1? The answer is found in thecoding.

Since x is not parabolic, there exists a unique sequence s in Σ+c satisfying

f(s) = x. Consider a new sequence s′ = (s′i)i�1 constructed from s by grouping

together consecutive terms corresponding to the same parabolic generator,defined by:

• s′1 = s1 if s1 is hyperbolic and n = 1,

• s′1 = sn

1 if s1 is parabolic, with n � 1 satisfying s1 = s2 = · · · = sn andsn+1 �= s1. Such an n exists by the definition of Σ+

c .

Repeat this procedure, beginning with the sequence (sn+i)i�1, to find s′2. Step

by step, this procedure produces a sequence (s′i)i�1 satisfying the following

properties:

(i) s′i = ani

i with ai ∈ A. If ai is hyperbolic, then ni = 1 and ai+1 �= a−1i ; if

ai is parabolic, then ni ∈ N∗ and ai+1 �= a±1

i .(ii) For all i � 1, the arcs D(ai)(∞) and D(ai+1)(∞) are disjoint.(iii) limn→+∞ s′

1 · · · s′n(0) = x.

Note that, if g1 and g2 are hyperbolic, then s = s′.

Property 2.4. Let x in Lc(S(g1, g2)). There exists ε > 0 such that the se-quence (s′

1 · · · s′n(0))n�1 is contained in an ε-neighborhood of the geodesic ray

[0, x).


Proof. Write γn = s′1 · · · s′

n. Fix a point y �= x in D0(S(g1, g2))(∞).The point γ−1

n (x) is in D(s′n+1)(∞). Since y does not belong to the interior

of the arcs D(a)(∞) for all a in A, it follows from Property 1.4(i) that the pointγ−1

n (y) is in D(s′n)(∞). The construction of the sequence (s′

i)i�1 requires thearcs D(s′

i)(∞) and D(s′i+1)(∞) to be disjoint. Thus the Euclidean distance

between γ−1n (x) and γ−1

n (y) is bounded below by a positive constant whichdoes not depend on n. This property implies that there exists a compactsubset of D whose image under γn intersects the geodesic (yx). Take z in thegeodesic (yx). Since limn→+∞ γn(0) = x, the sequence (γn(0))n�1 convergesto x and remains within a bounded distance from the geodesic ray [z, x). Itfollows that there exists ε > 0 such that the sequence (γn(0))n�1 is in theε-neighborhood of the geodesic ray [0, x). �

3 The modular group and two subgroups

Let us now return to the Poincare half-plane. In this section, we study theaction on H of the modular group PSL(2, Z) composed of Mobius transforma-tions h of the form

h(z) =az + b

cz + dwith a, b, c, d ∈ Z and ad − bc = 1,

and of two of its subgroups.

3.1 The modular group

By definition, the modular group is Fuchsian. We are going to describe one ofits Dirichlet domains.

Exercise 3.1. Prove that only the trivial isometry in PSL(2, Z) fixes thepoint 2i.

Define the isometries T1(z) = z+1 and s(z) = −1/z. These two isometrieswill allow us to construct the Dirichlet domain of the modular group.

Property 3.2.

D2i(PSL(2, Z)) = {z ∈ H | |z| � 1 and − 1/2 � Re z � 1/2}.

Proof. Set E = {z ∈ H | |z| � 1 and − 1/2 � Re z � 1/2}. By definition ofthe Dirichlet domain (see Sect. I.2.3), D2i(PSL(2, Z)) is contained in the setH2i(T1) ∩ H2i(T −1

1 ) ∩ H2i(s). On the other hand,

H2i(T1) = {z ∈ H | Re z � 1/2},

H2i(T −11 ) = {z ∈ H | Re z � −1/2} and

H2i(s) = {z ∈ H | |z| � 1},

hence D2i(PSL(2, Z) is contained in E (Fig. II.8).

3 The modular group and two subgroups 59

Fig. II.8.

Let z be in◦E. Suppose that there exists γ(z) = (az + b)/(cz + d) in

PSL(2, Z) − {Id} such that γ(z) is in E. Then c �= 0 necessarily since|Re(z + b)| > 1/2 for all b ∈ Z

∗. One has Im(γz) = Im z/|cz + d|2. Also,since z is contained in

◦E,

|cz + d|2 > (|c| − |d|)2 + |c| |d|.

Therefore, c �= 0 implies Im z > Im(γ(z)).If γ(z) is contained in

◦E, the same reasoning using γ−1 allows us to con-

clude that Im(γ(z)) > Im z, a contradiction.In conclusion, for all γ in PSL(2, Z) − {Id}, one has γ

◦E ∩

◦E = ∅. This

implies that◦E is contained in D2i(PSL(2, Z)). Thus D2i(PSL(2, Z)) = E. �

This proposition immediately produces the following result.

Corollary 3.3. The group PSL(2, Z) is a non-uniform lattice.

Exercise 3.4. Verify that the modular surface PSL(2, Z)\H has the form ofFig. II.9.(Hint: use Proposition I.2.17.)

Fig. II.9.

In Sect. 4, we will use another fundamental domain constructedfrom D2i(PSL(2, Z)), as defined in the following exercise.


Exercise 3.5. Prove that the set

Δ = D2i(PSL(2, Z)) ∩ {z ∈ H | Re z � 0}∩ T1(D2i(PSL(2, Z)) ∩ {z ∈ H | Re z � 0}),

is a fundamental domain of the modular group (Fig. II.10).

Fig. II.10.

The group PSL(2, Z), unlike Schottky groups, contains elliptic elements,as s(z) = −1/z and r(z) = (z − 1)/z.

Proposition 3.6. An elliptic element of PSL(2, Z) is conjugate in PSL(2, Z)to some power of either s or r.

Exercise 3.7. Prove Proposition 3.6.(Hint: use the fact that the trace of such element is −1, 0, or 1.)

Which isometries in the modular group are parabolic? As a consequence ofCorollary I.4.10, if p is such an isometry, the orbit of its fixed point intersects

D2i(PSL(2, Z))(∞). This set is reduced to the point ∞, which is parabolicsince it is fixed by T1. The stabilizer of this point in PSL(2, Z) is generatedby T1, hence p is conjugate in PSL(2, Z) to a power of T1.

From the preceding discussion, it follows that all parabolic points are ofthe form γ(∞), with γ in PSL(2, Z). If γ does not fix the point ∞, the trans-formation γ can be written as γ(z) = (az + b)/(cz + d) with c �= 0, thus γ(∞)is the rational number a/c.

Conversely, let p/q be in Q, with gcd(p, q) = 1. Consider p′, q′ in Z suchthat pq′ − qp′ = 1. Define g(z) = (pz + p′)/(qz + q′). This isometry belongsto PSL(2, Z) and g(∞) = p/q thus p/q is fixed by gT1g

−1.We have proved the following facts.

Property 3.8.

(i) The parabolic isometries of the modular group are conjugate in PSL(2, Z)to powers of T1.

(ii) Lp(PSL(2, Z)) = Q ∪ { ∞}.


Therefore, the rational numbers correspond to the parabolic points ofPSL(2, Z), distinct from ∞. Furthermore, since this group is a lattice, itslimit set is H(∞) and is the disjoint union of the set of parabolic points andconical points. It follows that the irrational numbers correspond to conicalpoints.

This geometric characterization of a rational is the key to the last sectionof Chap. VII. It allows us to relate the theory of Diophantine approximationto the behavior of geodesics on the modular surface.

3.2 Congruence modulo 2 subgroup and the commutator subgroup

One interesting aspect of these two subgroups is that they share the samefundamental domain. However, this domain is Dirichlet only in the first case.

The congruence modulo 2 subgroup. Let P be the group homomorphismof PSL(2, Z) into SL(2, Z/2Z) sending any Mobius transformation h(z) =(az + b)/(cz + d) with integer coefficients to the matrix

P (h) =(

a• b•

c• d•

),

where n• denotes the class of n in Z/2Z. The group Γ (2) = P −1(

1•

0•

0•

1•)

iscalled the congruence modulo 2 subgroup [41, Chap. V.5]. This is a normalsubgroup of PSL(2, Z) of index 6.

Let r be the Mobius transformation which sends z to r(z) = (z − 1)/z.Then

(∗) Γ (2)\ PSL(2, Z) = {Id, r, r2, T −11 , T −1

1 r, T −11 r2}.

Consider the set Δ′ (Fig. II.11) defined by

Δ′ = Δ ∪ rΔ ∪ r2Δ ∪ T −11 (Δ) ∪ T −1

1 r(Δ) ∪ T −11 r2(Δ).

Fig. II.11.


Exercise 3.9. Prove that Δ′ is a fundamental domain of Γ (2).(Hint: use Exercise 3.5 and the relation (*).)

The isometry T−1(z) = z/(z + 1) will useful in the following discussion.

Property 3.10. The domain Δ′ is the Dirichlet domain of Γ (2) at the point i.

Proof. None of the elements of Γ (2)− {Id} fixes i. Furthermore, the isometriesT ±2

1 , T ±2−1 belong to Γ (2), so one has

Hi(T 21 ) = {z ∈ H | Re z � 1},

Hi(T −21 ) = {z ∈ H | Re z � −1},

Hi(T 2−1) = {z ∈ H | |z + 1/2| � 1/2}, and

Hi(T −2−1 ) = {z ∈ H | |z − 1/2| � 1/2}.

Therefore, Δ′ = Hi(T 21 ) ∩ Hi(T −2

1 ) ∩ Hi(T 2−1) ∩ Hi(T −2

−1 ). It follows thatDi(Γ (2)) is contained in Δ′. However, since Di(Γ (2)) and Δ′ are two fun-damental domains of Γ (2), one has Δ′ = Di(Γ (2)). �Corollary 3.11. The group Γ (2) is a non-uniform lattice.

Exercise 3.12. Verify that the surface Γ (2)\H is of the form given byFig. II.12.(Hint: use Proposition I.2.17.)

Fig. II.12.

The group Γ (2) is a generalized Schottky group (see Sect. 1.1 for thedefinition). Recall that the Mobius transformation ψ(z) = i z−i

z+i (see Sect. 1.5)sends H into the Poincare disk. We have 0 = ψ(i). Set:

A = {ψT 21 ψ−1, ψT −2

1 ψ−1, ψT 21 ψ−1, ψT −2

1 ψ−1}.

For all a in A, the half-planes D(a) bounded by the perpendicular bisectors ofthe hyperbolic segments [0, a(0)]h are either tangent or disjoint. From Propo-sition 1.6, the group generated by T 2

1 and T 2−1 is therefore free (and discrete)

and it has Δ′ as its fundamental domain. Since this group is a subset of Γ (2)and has the same fundamental domain, they are equal.

Thus we have the following property:


Property 3.13. The group Γ (2) is generated by T 21 and T 2

−1, and is freerelative to these generators.

We now consider the parabolic isometries in Γ (2).

Property 3.14. The parabolic isometries of Γ (2) are conjugate to powers ofT 2

1 , T 2−1 or T −2

−1 T 21 in Γ (2).

Proof. Notice first that the point ∞ is fixed by T 21 , the point 0 is fixed by

T 2−1, the point −1 is fixed by T −2

−1 T 21 , and the point 1 is fixed by T 2

−1T −21 .

These four points are parabolic. Furthermore, −1 and 1 are in the same orbitsince T 2

1 (−1) = 1, and the sets Γ (2)(0), Γ (2)(∞), Γ (2)(1) are three disjointorbits.

Let γ be a parabolic isometry in Γ (2). From Corollary I.4.10, its fixedpoint is contained in one of the three orbits described above. Also since eachof T 2

1 , T −2−1 T 2

1 , T 2−1 generates the stabilizer in Γ (2) of its fixed point, γ is

conjugate to a power of one of these three isometries. �

Note that, unlike Schottky groups, the parabolic isometries of a generalizedSchottky group admitting a collection of non-elliptic generators {g1, . . . , gp}satisfying the condition (D(gi) ∪ D(g−1

i )) ∩ (D(gj) ∪ D(g−1j )) = ∅, for all i �= j

in {1, . . . , p} are not always conjugate to powers of the parabolic generators gi.

Exercise 3.15. Prove that the set Lp(Γ (2)) is equal to Q ∪ { ∞}.(Hint: use Property 3.8 and the fact that Γ (2) is normal in PSL(2, Z), (seealso [41, Chap. V, Example F]).)

The commutator subgroup. We now introduce another subgroup of the mod-ular group defined this time by a given collection of generators. Let α1 and α2

be two Mobius transformations defined as

α1(z) =z + 1z + 2

, α2(z) =z − 1

−z + 2,

and consider the half-planes

B(α1) = {z ∈ H | |z − 1/2| � 1/2}, B(α−11 ) = {z ∈ H | Re z � −1},

B(α2) = {z ∈ H | |z + 1/2| � 1/2}, B(α−12 ) = {z ∈ H | Re z � 1}.

For all i = 1, 2 and ε = ±1, one has

αεi (H −

◦B(α−ε

i )) = B(αεi ).

Note that we have (Fig. II.13) the following relation:

Δ′ =⋂

ε=±1i=1,2

H −◦B(α−ε

i ).


Fig. II.13.

Exercise 3.16.

(i) Prove that if the perpendicular bisector of a hyperbolic segment [z1, z2]his a vertical half-line, then Im z1 = Im z2.(Hint: use Exercise I.1.8.)

(ii) Conclude that there is no point z in H such that the half-planes B(α−11 )

and B(α−12 ) are bounded by the perpendicular bisectors of the segments

[z, α−11 (z)]h and [z, α−1

2 (z)]h respectively.

Let Γ be the group generated by α1 and α2. The group Γ being a subgroupof PSL(2, Z), it is Fuchsian. Observe that Γ is not a generalized Schottkygroup with respect to α1 and α2, since there is no z ∈ D such that B(αε

i ) =Dz(αε

i ). Even so, the arguments presented in the first part of the proof ofProposition 1.6 being purely dynamic, they still apply.

Property 3.17. The group Γ generated by α1 and α2 is free relative to thesegenerators. Moreover Γ is the commutator subgroup of PSL(2, Z) (i.e., it isgenerated by the elements [g, h] = ghg−1h−1, where g and h are in PSL(2, Z)).

Exercise 3.18. Prove Property 3.17.(Hint: For the first part of Property 3.17, see proof of Proposition 1.6. For thesecond part, use the identities [s, T −1

1 ] = α1 and [s, T1] = α2.)

Exercise 3.19. Prove that Γ is a normal subgroup of index 6 in PSL(2, Z).Moreover, give the structure of the two finite groups PSL(2, Z)/Γ (2) andPSL(2, Z)/Γ .

Let us prove that the set Δ′ is a fundamental domain of Γ . Notice that,since Γ is not a generalized Schottky group, the second part of Proposition 1.6does not apply directly.

Property 3.20. The set Δ′ is a fundamental domain of Γ .

Proof. Define A = {α±11 , α±1

2 }. Let us show that, for all z in H, there exists γin Γ such that γ(z) belongs to Δ′.

If z is not in Δ′, there exists a1 in A such that z belongs to B(a1). Definez1 = a−1

1 (z). If z1 is contained in Δ′, one may define zn = z1 for all n � 1.Otherwise there exists a2 in A − {a−1

1 } such that z1 belongs to B(a2) and one


may define z2 = a−12 (z1). Following this process, we construct the sequence

(zn)n�1.If (zn)n�1 is constant after some N , then a−1

N · · · a−11 (z) is contained in Δ′.

Otherwise, define γn = a1 · · · an. By construction, γ−1n (z) is contained in

B(an+1). Consider a subsequence (γnp)p�1 such that anp+1 = a. The point zbelongs to γnp(B(a)) for all p. For all p � 2, the half-plane γnp(B(a)) is con-tained in γnp−1(B(a)) (see the argument in Property 1.4(ii)). The point z liesin each of these half-planes, hence there is a compact K in H such that theimage by a1 · · · anp of the geodesic bounding B(a) meets K, for all p � 1. Thisgeodesic is a subset of Δ′ and Δ′ = Δ ∪ rΔ ∪ r2Δ ∪ T −1

1 Δ ∪ T −11 rΔ ∪ T −1

1 r2Δ.Therefore, there exist infinitely many elements γ in PSL(2, Z) such that γΔintersects K. This contradicts the fact that Δ is a locally finite fundamen-tal domain of PSL(2, Z) (since it is a Dirichlet domain). Thus the sequence(zn)n�1 is necessarily constant and there exists γ in Γ such that γ(z) ∈ Δ′.

Furthermore, for any non-trivial γ in Γ , the open set γ(◦Δ′) is a subset

of some open half-plane◦B(a) with a ∈ A (see the argument from Prop-

erty 1.4(i)), thus its intersection with◦Δ′ is empty. �

Exercise 3.21. Verify that the surface Γ \H is of the form given by Fig. II.14.(Hint: use the fact that since Δ′ is locally finite, and that the function from Δ′

modulo Γ to in Γ \H, sending Γz ∩ Δ′ to Γz, is a homeomorphism (see Propo-sition I.2.17).)

Fig. II.14.

Since the domain Δ′ is not a Dirichlet domain, we cannot use the resultsof Chap. I to conclude that Γ is a non-uniform lattice. This property is in facttrue, but requires proof. Our chosen proof is not especially direct, but has theadvantage of illustrating some interesting properties of the group and of usingCriterion I.4.17: Γ is a non-uniform lattice if and only if L(Γ ) = H(∞) andL(Γ ) = Lp(Γ ) ∪ Lh(Γ ), with Lp(Γ ) �= ∅.

Let us consider the non-trivial translation

[α−12 , α−1

1 ] = α−12 α−1

1 α2α1.


Exercise 3.22. Prove that [α−12 , α−1

1 ] generates the stabilizer of the point ∞in Γ .

Property 3.23. The parabolic isometries of Γ are conjugate to powers of[α−1

2 , α−11 ] in Γ .

Proof. Fix z in◦Δ′. Consider a parabolic isometry γ in Γ and write it in

the form of a reduced word s1 · · · sn, where A = {α±11 , α±1

2 }. One mayassume that s1 �= s−1

n . In this case, limk→+∞ γk(i) is contained in B(s1)and limk→+∞ γ−k(i) is contained in B(s−1

n ). These two limits are equalto the unique fixed point x of γ, thus x is an element of {−1, 0, 1, ∞}. Ifx = ∞, then γ belongs to the group generated by [α−1

2 , α−11 ]. Otherwise, since

1 = α1(∞), −1 = α2(∞), and 0 = α−12 α1(∞), the element γ is conjugate to

a power of [α−12 , α−1

1 ]. �

Exercise 3.24. Let H be a non-elementary Fuchsian group and N be a nor-mal subgroup. Prove that L(H) = L(N).(Hint: use the minimality of the limit set.)

Property 3.25. The group Γ is a non-uniform lattice and Lp(Γ ) = Q ∪ {∞}.

Proof. The group Γ is normal in PSL(2, Z), thus one has L(Γ ) = H(∞). Letus examine Lp(Γ ). We know that the point ∞ is contained in this set. Fur-thermore, for all γ in PSL(2, Z), the Mobius transformation γ[α−1

2 , α−11 ]γ−1

is a parabolic isometry in Γ which fixes γ(∞). Thus Lp(Γ ) = Q ∪ {∞} (Prop-erty 3.8).

Consider now an irrational number x. This point is horocyclic with respectto PSL(2, Z), hence there exists a sequence (γn)n�1 in the modular groupsatisfying

limn→+∞

Bx(z, γn(z)) = +∞.

Since Γ has finite index in PSL(2, Z), passing to a subsequence one has γn =gnγ, where gn is in Γ and limn→+∞ Bx(z, gn(z)) = +∞. Thus x is a horocyclicpoint with respect to Γ .

In conclusion,

L(Γ ) = H(∞), Lp(Γ ) = Q ∪ { ∞} and L(Γ ) = Lp(Γ ) ∪ Lh(Γ ). �

4 Expansions of continued fractions

We have shown in the preceding section that Q ∪ {∞} is the set of parabolicpoints associated to the modular group. In this section, we continue to weavethe relationships between number theory and hyperbolic geometry, by creatinga geometric context for representations of irrational numbers x as continuedfractions.

4 Expansions of continued fractions 67

We begin by recalling the algorithmic definition of this representation. Wedefine x0 = x and n0 = E(x0), where E(x) designates the integer part of x.For all i � 1, define xi and ni by the following recurrence relation:

xi = 1/(xi−1 − ni−1) and ni = E(xi).

Let [n0; n1, . . . , nk] denote the rational number defined by

[n0; n1, . . . , nk] = n0 +1

n1 +1

n2 + .. .+

1

nk−1 +1nk

The sequence of rational numbers ([n0; n1, . . . , nk])k�1 converges to x (see[42]). The continued fraction expansion of x is by definition the expressionof x as [n0; n1, . . .].

4.1 Geometric interpretation of continued fraction expansions

Consider the hyperbolic triangle T having infinite vertices at the points∞, 1, 0. This triangle is related to the fundamental domain Δ of PSL(2, Z)introduced in Exercise 3.5 and defined to be

Δ = {z ∈ H | 0 � Re z � 1, |z| � 1 and |z − 1| � 1}.

More precisely, if r denotes the transformation defined by r(z) = (z − 1)/z,one has (see Fig. II.15)

T = Δ ∪ rΔ ∪ r2Δ.

Fig. II.15.

It follows that⋃

γ∈PSL(2,Z)

γT = H and if γ◦T ∩

◦T �= ∅, then γ ∈ {Id, r, r2}.


This tiling of H by images of T is called the Farey tiling . Let L denote theunoriented geodesic whose endpoints are 0 and ∞. The images of L underPSL(2, Z) are called Farey lines.

Recall that

T1(z) = z + 1 and T−1(z) = z/(z + 1).

The endpoints of T1(L) are the points 1, ∞. Those of T−1(L) are 0, 1.Thus the edges of T and of γT for γ in PSL(2, Z), are Farey lines (Fig. II.16).

Property 4.1. Let L+ be the geodesic L oriented from 0 to ∞. For anyoriented Farey lines (xy), there exists an unique γ in PSL(2, Z), such that(xy) = γ(L+).

Proof. By definition, given an oriented Farey lines (xy), there exists γ inPSL(2, Z) such that γ(L) is the unoriented Farey line whose endpoints are xand y. If γ(0) = x and γ(∞) = y, then γ(L+) = (xy). Otherwise, γs(L+) =(xy), where s(z) = −1/z.

Fig. II.16.

It follows that there exists g in PSL(2, Z) satisfying g(L+) = (xy). Thiselement is unique since PSL(2, Z) does not contain any non-trivial isometryfixing the points 0 and ∞. �

Definition 4.2. Let n > 1. A collection of n oriented Farey lines L+1 , . . . , L+

n

are called consecutive, if there exists γ in PSL(2, Z) and ε in {±1} such thatfor all 1 � i � n

L+i = γT i

ε L+.

Set L+i = (xiyi). Suppose that L+

1 , . . . , L+n are consecutive. If ε = 1, then

yi = γ(∞) for all 1 � i � n; likewise if ε = −1, then xi = γ(0) for all1 � i � n.

Figures II.17, II.18, II.19 show several cases in which three oriented Fareylines are consecutive.

If x is an element of H(∞), the irrationality of x is characterized by thenumber of Farey lines which intersect the geodesic ray [i, x).

Proposition 4.3. Let x be in H(∞). The ray [i, x) intersects finitely manyFarey lines if and only if x is in Q ∪ { ∞}.


Fig. II.17.

Fig. II.18.

Fig. II.19.

Proof. Suppose that x belongs to Q ∪ { ∞}. In this case, after Property 3.8(ii),there exists γ in PSL(2, Z) such that γ(x) = ∞. Since the ray γ−1([i, x))is a vertical half-line passing through γ−1(i), there exists n in Z and z in[i, x) such that the ray T n

1 γ−1([z, x)) is in T . The domain Δ is locally finite(since it is constructed from a finite number of subsets of a Dirichlet domain(Property 3.2)), thus T n

1 γ−1([i, z]h) intersects only a finite number of imagesof T under PSL(2, Z).

It follows that T n1 γ−1([i, x))—and thus [i, x)—only intersects finitely many

Farey lines.Suppose now that [i, x) only intersects a finite number of Farey lines.

Then there exists z in [i, x) and γ in PSL(2, Z) such that [z, x) is containedin γ(T ). In other words, [γ−1(z), γ−1(x)) is a geodesic ray contained in T .Hence, γ−1(x) is in {0, 1, ∞} and thus x is in Q ∪ { ∞}. �

From now on, we focus on positive irrational numbers. Let x be such a realnumber and r : [0, +∞) → [i, x) be the arclength parametrization of [i, x).


According to the previous proposition, [i, x) crosses infinitely many Fareylines. Let (Ln)n�1 denote the sequence of Farey lines crossed by (r(t))t>0

in order by increasing t. For each n, let L+n = (xnyn) be the orientation

on Ln defined by: at the point of intersection r(tn) between [i, x) and Ln,the oriented angle from [r(tn), x) to [r(tn), yn) belongs to (0, π). If L+

n is nota vertical half-line, then L+

n = (xnyn) with xn < yn, if L+n is vertical, then

yn = ∞ (Fig. II.20). Denote by γn the unique element of PSL(2, Z) such that(Property 4.1)

γn(L+) = L+n .

Fig. II.20.

Consider the geodesic ray γ−1n ([i, x)). This ray crosses L+ at the point

γ−1n (r(tn)). By construction, at their point of intersection γ−1

n (r(tn)), theangle oriented from ([γ−1

n (r(tn)), γ−1n (x)) to [γ−1

n (r(tn)), ∞) belongs to (0, π).Thus γ−1

n ([i, x)) meets the Farey lines T−1(L+) = (01) or T1(L+) = (1∞) andhence, γ−1

n (L+n+1) is equal to T−1(L+) or T1(L+). It follows that γn+1 = γnTεn ,

where εn = ±1.Set

n0 =

{max{n � 1 | ∀ k ∈ [1, n], εk = 1} if ε1 = 1,

0 if ε1 = −1,

np = max{n > np−1 | ∀ k ∈ (np−1, n], εk = (−1)p}, if p � 1.

Note that nk is a positive integer for all k � 1.

Exercise 4.4. Prove that, for all k � 1, the positive integer nk represents thelargest integer p � 1 such that L+

np−1+1, . . . , L+np−1+p are oriented consecutive

Farey lines.


By construction of the integers nk, we have for all k � 1:

γnk= T n0

1 · · · T nk

(−1)k .

Moreover, notice that, for k � 1, the intervals of R with extremities γnk(0)

and γnk(∞) are nested. Set gk = γnk

. The next exercise relates the rationalnumber [n0; n1, . . . , nk], which was defined at the beginning of this section, tothe point gk(0).

Exercise 4.5. Let k � 2. Prove that if k is even, then gk(0) = [n0; n1, . . . , nk]and gk(∞) = [n0; n1 · · · nk−1]. Also if k is odd, then gk(0) = [n0; n1, . . . , nk−1]and gk(∞) = [n0; n1 · · · nk].(Hint: use the relations T n

1 (z) = z + n and T n−1(z) = 1/(n + 1/z).)

The following proposition shows that [n0; n1, . . .] is the continued fractionexpansion of x.

Proposition 4.6. The sequence of rational numbers ([n0; n1, . . . , nk])k�1 con-verges to x. In addition, if there exists a sequence (n′

k)k�1 satisfying n′0 ∈ N,

n′k ∈ N

∗ for all k � 1 and limk→+∞[n′0; n

′1, . . . , n

′k] = x, then nk = n′

k for allk � 0.

Proof. The geodesic gk(L) intersects the geodesic ray [i, x). By construction,the point x belongs to the interval of R having endpoints gk(0), gk(∞), andthese intervals are nested. For all k � 1, the rational numbers gk(0) and gk(∞)belong to the interval [n0, n0 +1] (Exercise 4.5), thus 0 < |gk(0) − gk(∞)| � 1.Let us show that limk→+∞ |gk(0)−gk(∞)| = 0. Suppose that there exists d > 0and a subsequence (gkp)p�1 such that |gkp(0) − gkp(∞)| > d. In this case, thegeodesic gkp(L) intersects the Euclidean segment I in H whose endpoints aren0 + id/2 and n0 + 1 + id/2. Since L is an edge of T , and since T is the finiteunion of images of Δ, there exist infinitely many isometries γ in PSL(2, Z)such that γΔ intersects the compact set I. This contradicts the fact that Δ islocally finite (since it is a finite union of subsets of a Dirichlet domain). Oneconcludes from this property that the sequence ([n0; n1, . . . , nk])k�1 convergesto x.

Let us now show uniqueness. Suppose that ([n′0; n

′1, . . . , n

′k])k�1 converges

to x. Then

limp→+∞

T n′0

1 · · · T n′2p

1 (0) = limp→+∞

T n01 · · · T n2p

1 (0).

The rational number T n′0

1 · · · T n′2p

1 (0) is in (n′0, n

′0 + 1) and T n0

1 · · · T n2p

1 (0) isin (n0, n0 + 1). Thus n′

0 = n0. It follows that,

limp→+∞

T n′1

−1 · · · T n′2p

1 (0) = limp→+∞

T n1−1 · · · T n2p

1 (0).


Applying the same reasoning to the sequences( 1[0; n′

1, . . . , n′2p]

)p�1

= ([n′1; n

′2, . . . , n

′2p])p�1 and

( 1[0; n1, . . . , n2p]p�1

)= ([n1; n2, . . . , n2p])p�1,

one obtains n1 = n′1. Iteratively, one has nk = n′

k for all k � 0. �In summary, to find the continued fraction expansion of a positive irra-

tional number x in terms of hyperbolic geometry, it suffices to identify x witha point in H(∞), to associate to the ray [i, x) the infinite sequence of orientedFarey lines (L+

i = (xiyi))i�1 in the order in which this ray crosses them per theprocedure described above, and to count the maximal number of consecutiveoriented Farey lines. Then

• n0 = E(x), and n1 is defined by: xn0+1 = · · · = xn1 = n0, xn1+1 �= n0;• for all k � 1:

– if k is even, then nk is defined by ynk−1 = ynk−1+1 = · · · = ynkand

ynk+1 �= ynk,

– if k is odd, then nk is defined by xnk−1 = xnk−1+1 = · · · = xnkand

xnk+1 �= xnk.

Examples 4.7. Verify that in the situation of Fig. II.21, n0 = 2, n1 = 2, andn2 � 2.

Fig. II.21.

Let S+ denote the set of integer sequences defined by

S+ = {(ni)i�0 | n0 ∈ N, ni ∈ N∗ for i � 1}.

Consider the function F : R+ − Q

+ → S+, which sends x to the sequence(ni)i�0 corresponding to the continued fraction expansion of x.


Property 4.8. The map F : R+ − Q

+ → S+ is bijective.

Proof. According to Proposition 4.6, it is enough to show that F is surjective.For any sequence (nk)k�0 in S+, set gk = T n0

1 ◦ · · · ◦ T nk

(−1)k . Reusing theargument from the proof of Proposition 4.6, one obtains limk→+∞ |gk(0) −gk(∞)| = 0. Moreover the intervals with extremities gk(0) and gk(∞) arenested. More precisely, if k is even, then gk+1(∞) = gk(∞), and gk+1(0) isin the segment having endpoints gk(∞), gk(0). If k is odd, then gk+1(0) =gk(0) and gk+1(∞) is in the segment having endpoints gk(∞), gk(0). Thusthe sequences (gk(0))k�1 and (gk(∞))k�1 converge to the same positive realnumber x. Exercise 4.5 implies that x = limk→+∞[n0; n1, n2, . . . , nk]. Thereal x is irrational since [i, x) intersects each gk(L) (Property 4.3). �

We have restricted ourselves to positive irrational number. However, if y isa negative irrational, one may identify it with a sequence of integers (mi)i�0,such that m0 = E(y) ∈ Z and (mi)i�1 = F (y − m0).

In this way, one obtains a bijection between the set of irrational num-bers with the set of sequences of integers whose terms are positive—with thepossible exception of the first.

Let us come back to the geometry. Since the modular group is a non-uniform lattice, its limit set is the disjoint union of parabolic points andconical points. We know that conical points correspond to irrational numbers.In the same spirit as Property 2.4 for Schottky groups, let us prove that thecontinued fraction expansion of an irrational number x is related to a sequence(γk(i))k�0 with γk in PSL(2, Z), remaining at a bounded distance from thegeodesic ray [i, x). It suffices to prove this relation when x is positive.

Property 4.9. Let x be a positive irrational number. Set F (x) = (ni)i�0 andγk = T n0

1 · · · T n2k+1−1 for all k � 0. The sequence (γk(i))k�0 remains at a

uniformly bounded distance from the geodesic ray [i, x).

Proof. Let s(z) = −1/z. Note that the point i belongs to the geodesic (s(x) x).Since −1/x < 0 and Re γk(i) > 0, proving Property 4.9 amounts to prov-ing that the sequence (γk(i))k�0 remains at a uniformly bounded distancefrom the geodesic (s(x)x), and hence that the sequence of couples of points((γ−1

k (s(x)), γ−1k (x)))k�0 is contained in a compact subset of H(∞) × H(∞)

minus its diagonal (Proposition I.3.14).The continued fraction expansion of γ−1

k (x) is [n2k+2; n2k+3, . . .]. Sincen2k+2 is non-zero, the real γ−1

k (x) is greater than 1. Furthermore, sincesT −1

1 s = T−1 and sT −1−1 s = T1, one has

sγ−1k (s(x)) = T n2k+1

1 T n2k−1 · · · T n0

−1 (x).

Hence, the continued fraction expansion of the real sγ−1k s(x) is

{[n2k+1; n2k, . . . , n0, n0, n1, n2, . . .] if n0 �= 0,

[n2k+1; n2k, . . . , n1, n1, n2, . . .] otherwise.


Since n2k+1 �= 0, in both cases one obtains that the real γ−1k (s(x)) belongs

to (−1, 0). It follows that the geodesics ((γ−1k (x′)γ−1

k (x)))k�0 remain at abounded distance from i. �

4.2 Application to the hyperbolic isometries of the modular group

We focus here on positive irrational numbers such that the sequence F (x) =(ni)i�0 is almost periodic (i.e., for some k � 0, the sequence (nk+i)i�0 isperiodic). As with the coding of the conical points of a Schottky group, weare going to show that almost periodic sequences encode the fixed points ofhyperbolic isometries of the modular group.

Property 4.10.

(i) A positive irrational number x is fixed by a hyperbolic isometry inPSL(2, Z) if and only if the sequence F (x) is almost periodic.

(ii) An isometry in PSL(2, Z) is hyperbolic if and only if it is conjugate inPSL(2, Z) to an isometry of the form T m1

1 T m2−1 · · · T mk

−1 with mi > 0 and keven.

Proof.

(i) Let x be a positive irrational number. Suppose that the sequence F (x) =(ni)i�0 is periodic, in which case n0 is non-zero. Let T denote the periodof this sequence, and define k = T − 1 if T is even, and k = 2T − 1otherwise. Then x = limp→+∞(T n0

1 · · · T nk−1 )p(0). This shows that x is

fixed by T n01 · · · T nk

−1 , which is hyperbolic since x is irrational.If F (x) is almost periodic, after q initial terms for some q � 1, it sufficesto apply the preceding reasoning to the point (T n0

1 · · · T nq

−1 )−1(x) if q isodd, and to the point (T n0

1 · · · T nq−1−1 )−1(x) if q is even.

Consider now a hyperbolic isometry γ in PSL(2, Z). Let F (γ+) = (ni)i�0

and set gk = T n01 · · · T nk

(−1)k . Recall that the geodesic ray [i, x) meetsall oriented Farey lines L+

k = (gk(0)gk(∞)). According to Exercise 4.5,the sequences (gk(0))k�0 and (gk(∞))k�0 converge to γ+. Thus for largeenough k, the Euclidean segment having endpoints gk(0), gk(∞) doesnot contain γ−, and hence there exists k′ > k such that γL+

k = L+k′ .

Applying Property 4.1, one obtains that γgk = gk′ . It follows thatγ = gk T nk+1

(−1)(k+1) · · · T n′k

(−1)k′ g−1k . If k and k′ are both odd or even, then

the sequence F (g−1k (γ+)) is periodic. Otherwise, F (T −nk+1

(−1)(k+1)g−1k (γ+)) is

periodic. In both cases, the sequence F (γ+) is almost periodic.(ii) Let γ be a hyperbolic isometry in PSL(2, Z). After conjugating γ by a

translation, one may suppose that γ+ > 0. According to the end of theproof of part (i), γ is conjugate to T nk+1

1 · · · T n′k

(−1)k′ if k and k′ are both


odd or even, and to T nk+21 · · · T n′

k

(−1)k′ , otherwise. Hence it is conjugate to

an isometry of the form T m11 · · · T mp

−1 , with mi > 0 and p even.Conversely, an isometry of the form T m1

1 T m2−1 · · · T mk

−1 with mi > 0and k even, is hyperbolic since it fixes limp→+∞(T m1

1 · · · T mk−1 )p(0) and

limp→ − ∞(T m11 · · · T mk

−1 )p(0), which are distinct real numbers. �

This property allows us to establish a relationship between the fixed pointsof hyperbolic isometries of the modular group and the quadratic real numbers,which are solutions to equations like

ax2 + bx + c = 0 with a ∈ N∗ and b, c ∈ Z.

Proposition 4.11. Let x be an irrational number. Then the following areequivalent:

(i) x is fixed by a hyperbolic isometry in PSL(2, Z);(ii) x is quadratic.

We give a proof of this well-known result (see for example [42]) using thetransformations T1 and T−1.

Proof. The implication from (i) to (ii) requires only two facts. The first oneis that a fixed point x of a Mobius transformation γ(z) = (az + b)/(cz + d)satisfies the Diophantine equation

Ax2 + Bx − C = 0,

with A = c, B = d − a and C = −b. The second one is that the integer c isnon-zero since γ is hyperbolic.

Let us now prove that (ii) implies (i). Let α and β be two distinct rootsof an equation of the form

Ax2 + Bx − C = 0,

with A �= 0 and B, C in Z.We want to show that these two real numbers are fixed by some hyperbolic

isometry of the modular group. After replacing them by g(α) and g(β), where gis in PSL(2, Z), one may assume that α > 0 and β < 0. Hence A > 0 andC > 0. Set F (α) = (ni)i�0. For all even integer k > 0, define the real numbers

xk = (T n01 · · · T nk−1

−1 )−1(α) and yk = (T n01 · · · T nk−1

−1 )−1(β),

and set x0 = α, y0 = β.We have xk = limp→+∞[nk; nk+1, . . . , nk+p], hence xk is positive. Further-

more, an induction argument shows that yk is negative and that the two realnumbers xk and yk are solutions of an equation of the form

Akx2 + Bkx − Ck = 0,


where Ak, Bk, Ck ∈ Z, Ak > 0, Ck > 0 and B2k + 4AkCk = B2 + 4AC. Thus

the coefficients Ak, Bk, Ck belong to a finite set. It follows that there exist twoeven integers k2 > k1 � 0 such that Ak1 = Ak2 , Bk1 = Bk2 , Ck1 = Ck2 . Thisimplies that xk1 = xk2 , and hence

T nk11 · · · T nk2−1

−1 (xk2) = xk2 .

We obtain that the real number xk2 is the fixed point of the hyperbolic isome-try g′ = T nk1

1 · · · T nk2−1

−1 . Since α = T n01 · · · T nk2−1

−1 (xk2), this real is fixed by aconjugate of g′. The same reasoning applied to yk2 implies that β is similarlyfixed by the same hyperbolic isometry. �

We conclude this section by focusing on the displacements of isometriesin PSL(2, Z). Recall that the displacement �(γ) of an isometry γ is defined(see I.2.2) by

�(γ) = infz∈H

d(z, γ(z)).

Exercise 4.12. Let γ(z) = (az + b)/(cz + d) be a hyperbolic isometry in G.Denote λ the eigenvalue of the matrix

(a bc d

)whose absolute value is > 1.

Prove the equality�(γ) = 2 ln |λ|.

The following property relates the fixed point of a hyperbolic isometry inPSL(2, Z) to its displacement. Let σ : (R − Q) ∩ (1, +∞) → (R − Q) ∩ (1, +∞)defined by

σ(x) =1

x − n0,

where n0 is the first term of the sequence F (x). Notice that the sequenceF (σ(x)) is the shifted sequence (ni+1)i�0.

Property 4.13. If γ is an isometry of the form γ = T m11 T m2

−1 · · · T mk−1 , with

mi in N∗ and k even, then

�(γ) = 2 ln(γ+ × σ(γ+) × · · · × σk−1(γ+)).

Proof. Set γ(z) = (az + b)/(cz + d), M =(

a bc d

)and λ = e�(γ)/2. We have

M

(γ+

1

)= λ

(γ+

1

).

Consider the matrices Dn =(

0 11 n

)and R =

(0 11 0

). These matrices satisfy the

following relations:

R2 = Id, DnR =(

1 0n 1

)and RDn =

(1 n0 1

).


Using these relations, one obtains

λ

(1

γ+

)= Dm1 · · · Dmk

(1

γ+

).

If x �= 0, then Dm

(1x

)= x

(1

m+1/x

). Therefore, Dmk

( 1γ+

)= γ+

( 1mk+1/γ+

).

However, mk + 1/γ+ = σk−1(γ+), hence

λ

(1

γ+

)= γ+Dm1 · · · Dmk−1

(1

σk−1(γ+)

).

Repeating this process, one obtains

λ

(1

γ+

)= γ+σk−1(γ+) · · · σ2(γ+)Dm1

(1

σ(γ+)

).

Furthermore, Dm1

( 1σ(γ+)

)= σ(γ+)

( 1γ+

), thus λ = γ+

∏k−1i=1 σi(γ+). �

Using this property, one obtains an interpretation—in terms of hyperbolicgeometry—of the golden ratio

N =1 +

√5

2.

Corollary 4.14. If γ is a hyperbolic isometry in PSL(2, Z), then

�(γ) � 2 ln(T1T−1)+ = 4 ln N .

Proof. Suppose γ is a hyperbolic isometry in PSL(2, Z). According to Prop-erty 4.10(ii), we can suppose that this isometry is of the form T m1

1 · · · T mk−1 ,

with mi > 0. Let us prove that �(T m11 · · · T mk

−1 ) > 4 ln N , if some mi �= 1.Let x be the attractive fixed point of this isometry, the sequence F (x) is theperiodic sequence m1, . . . , mk, m1, . . . , mk, m1, . . . . For 1 � i � k, notice thatthe real σi(x) is of the form mi+1 + 1/(mi+2 + xi), where 0 < xi < 1. There-fore, if one of the mi is equal to 2, there exist j, l with 0 � j, l � k − 1 andj �= l such that σj(x) = 2 + 1/y, where y > 1, and σl(x) = ml+1 + 1/(2 + xl),with 0 < xl < 1. From these remarks and Property 4.13, one obtains

�(γ) > 2 ln (2 × (1 + 1/3)) = 2 ln 8/3 > 4 ln N .

If one of the mi is � 3, then �(γ) � 2 ln 3 > 4 ln N .Suppose now that all of the mi are 1, then T m1

1 · · · T mk−1 is a power of

T1T−1. The attractor x of T1T−1 satisfies x = 1+1/x, hence x = N and, afterProperty 4.13, one has �(T1T−1) = 2 ln N 2.

In conclusion, �(T m11 · · · T mk

−1 ) > 4 ln N , except if k = 2 and m1 = m2 = 1.�


Corollary 4.14 has a geometric interpretation on the modular surface S =PSL(2, Z)\H. We will see in Chap. III that the set of all �(γ), where γ isa hyperbolic isometry of PSL(2, Z), is the set of the lengths of all compactgeodesics in the surface S (see Sect. III.3 and Appendix B for the notion ofa geodesic on S). In this context, Corollary 4.14 says that the real 4 ln Ncorresponds to the length of the shortest compact geodesic on the modularsurface.

5 Comments

The construction of Schottky groups and the coding of their limit sets thathave been introduced in this chapter can be generalized to pinched Hadamardmanifolds X (see the Comments in Chap. I) whose group of positive isometriescontains at least two non-elliptic elements g1, g2 having no common fixedpoints. Under this condition, for large enough n0, there exist two disjointcompact subsets K1 and K2 of X(∞) satisfying the following relation for alli = 1, 2 and |n| � n0:

gni (X(∞) − Ki) ⊂ Ki.

An application of the Ping-Pong Lemma [36] shows that the group gener-ated by gn0

1 , gn02 is a free Kleinian group. Such a group is geometrically finite

[21]. Without any other hypotheses on X, these groups—which are againcalled Schottky groups—and their variants are in general the only accessiblenon-elementary Kleinian groups.

On the other hand, if one restricts to the case in which X is a symmetricspace of rank 1, one can construct Kleinian groups (in general, lattices) usingnumber theory.

In the particular case of the Poincare half-plane, the arithmetic groups areknown [41, Chap. 5]. The modular groups and Γ (2) belong to this rich family.Let us mention (see [45]) a rather unexpected example of a lattice containedin the group of Mobius transformations having rational coefficients which isnot commensurable to the modular group but for which the set of parabolicpoints is still Q ∪ { ∞}.

The geometric construction of continued fraction expansions that was in-troduced in this chapter is essentially taken from two articles [57, 20]. Inaddition to these two references, the text of C. Series in [8] also studies thelimit set of Schottky groups and the modular group, but goes further towardthe construction of a coding of the limit set of an arbitrary geometrically finiteFuchsian group.

III

Topological dynamics of the geodesic flow

In this chapter we focus on the geodesic flow on the quotient of T 1H by a

Fuchsian group. Our motivation is to give relations between the behavior ofthis flow and the nature of the points in the limit set of the group. Generalnotions related to topological dynamics are introduced in Appendix A.

1 Preliminaries on the geodesic flow

1.1 The geodesic flow on T 1H

Recall from Proposition I.1.10 that T 1H is equipped with the G-invariant

distance D which is defined by

D((z, −→v ), (z′, −→v ′)) =∫ +∞

− ∞e− |t|d(v(t), v′(t)) dt,

where (v(t))t∈R represents the arclength parametrization of the unique ori-ented geodesic passing through z whose tangent line at z is in the directionof −→v , and which satisfies

v(0) = z anddv

dt(0) = −→v .

Let v(−∞), v(+∞) the points in H(∞) corresponding respectively to thenegative and positive endpoints of this geodesic (Fig. III.1).

Exercise 1.1. Let ((zn, −→vn))n�1 be a sequence in T 1H. Prove that

limn→+∞

D((zn, −→vn), (z, −→v )) = 0

⇐⇒ limn→+∞

vn(+∞) = v(+∞) and limn→+∞

zn = z.


http://dx.doi.org/10.1007/978-0-85729-073-1_3

80 III Topological dynamics of the geodesic flow

Fig. III.1.

We now define the geodesic flow on T 1H. Let g be the function from

R × T 1H into T 1

H defined by

g(t′, (z, −→v )) =(

v(t′),d

dtv(t′)

).

Exercise 1.2. Prove that for all real numbers t the function gt is a homeo-morphism of T 1

H into itself.

Note that for all t, t′ in R one has

D(gt(z, −→v ), gt′ (z, −→v )) = 2|t − t′ |.

It follows from this remark and Exercise 1.2 that the function g is continuous.

Exercise 1.3. Prove that gt+t′ = gt ◦ gt′ , for all t, t′ in R.

Thus the function g is a well-defined flow on T 1H. This flow is called the

geodesic flow . Its dynamics is analogous to the dynamics of the flow on R2

associated to a non-zero vector field (Example A.2(i) in Appendix A).

Exercise 1.4. Prove the following properties:

(i) the non wandering set (Definition A.11 in Appendix A) Ωg(T 1H) is empty;

(ii) all points in T 1H are divergent points.

1.2 The geodesic flow on a quotient

By analogy with the flow on the torus T2, viewed as the quotient of the

Euclidean plane by the translations group Z2, induced by a linear flow on R

2

(Examples A.2(i) and (ii) in Appendix A), the geodesic flow on T 1H induces

a flow on the quotient of this space by a Fuchsian group.More precisely, consider a Fuchsian group Γ and let π (respectively π1)

be the projection of H (respectively T 1H) onto the quotient S = Γ \H (re-

spectively T 1S = Γ \T 1H) (Figs. III.2 and III.3). Each of these quotients is

equipped with a distance defined respectively by

dΓ (π(x), π(y)) = infγ∈Γ

d(x, γ(y)),

DΓ (π1(z, −→u ), π1(z′, −→u ′)) = infγ∈Γ

D((z, −→u ), γ(z′, −→u ′)).

1 Preliminaries on the geodesic flow 81

Fig. III.2. Γ = PSL(2, Z)


Exercise 1.5. Prove that dΓ and DΓ are distance functions and that thetopologies induced by these distances on S and T 1S are the same as thoseinduced by π and π1.

The notion of convergence of a sequence can be interpreted in S and T 1Sin the following ways:

(i) A sequence (π(zn))n�1 in S converges to π(z) if and only if there existsa sequence (γn)n�1 in Γ such that (γn(zn))n�1 converges to z.

(ii) A sequence (π1((zn, −→un)))n�1 in T 1S converges to π1((z, −→u )) if and only ifthere exists a sequence (γn)n�1 in Γ such that (γn((zn, −→un)))n�1 convergesto (z, −→u ).

If Γ does not have any elliptic element, then the surface S is a differentiablemanifold whose Riemannian structure is induced by that of H, and T 1S is itsunitary tangent bundle. This is not the case if there are elliptic elements in Γ .The group PSL(2, Z) is an example of such a group. When Γ = PSL(2, Z),the π-projection of a hyperbolic disk, of sufficiently small radius, centered atj = 1/2+ i

√3/2 is homeomorphic to the cone obtained by taking the quotient

of this disk by the cyclic group of order 3 generated by r(z) = (z − 1)/z.Therefore, in a neighborhood of π(j), the modular surface does not inheritthe manifold structure of H. The same is true in a neighborhood of π(i). Inthis context one cannot talk about the Riemannian structure in the classicalsense. Therefore, despite its misleading notation, T 1S is not always the unitarytangent bundle of S. Whether or not Γ has elliptic elements, we will show thatthe flow g induces a flow on the topological space T 1S.

Let (z, −→v ) in T 1H and (v(t))t∈R be the arclength parametrization of the

oriented geodesic (γ(v(−∞))γ(v(+∞))), such that v(0) = z. For any positiveisometry γ ∈ Γ , the function γ ◦ v : R → H which sends t to γ(v(t)) is


the arclength parametrization of the oriented geodesic (γ(v(−∞))γ(v(+∞)))satisfying

γ(v(0)) = γ(z) andd

dtγ ◦ v(0) = Tzγ(−→v ).

It follows that gt(γ(z, −→v )) = γ(gt((z, −→v ))), for all t ∈ R. This last relationallows us to define the geodesic flow g : R × T 1S → T 1S (Fig. III.4) by

g(t, π1((z, −→v ))) = π1(g(t, (z, −→v ))).

The rest of this chapter is devoted to the topological dynamics of thisflow. We use—especially in Sects. 3 and 4—the following convergence criterionrelating the action of gR on Γ \T 1

H to the dual action of Γ on the set oforiented geodesics gR\T 1

H.


Proposition 1.6. Let ((zn, −→un))n�1 be a sequence in T 1H and (z, −→u ) be an

element of T 1H. The following properties are equivalent:

(i) there exists a sequence of real numbers (sn)n�1 such that

limn→+∞

gsn(π1((zn, −→un))) = π1((z, −→u );

(ii) there exists a sequence (γn)n�1 in Γ such that

limn→+∞

(γn(un(−∞)), γn(un(+∞))) = (u(−∞), u(+∞)).

Proof.(i) ⇒ (ii). By definition of the topology on T 1S, there exists a sequence

(γn)n�1 in Γ such that

limn→+∞

D(γngsn((zn, −→un)), (z, −→u )) = 0.

This convergence together with Exercise 1.1 implies that the ordered pair

(γn(un(−∞)), γn(un(+∞)))

converges to (u(−∞), u(+∞)).

2 Topological properties of geodesic trajectories 83

(ii) ⇒ (i). Consider the sequence of oriented geodesics

Ln = (γn(un(−∞))γn(un(+∞))).

This sequence converges to the geodesic L = (u(−∞)u(+∞)), hence thereexists a sequence of points z′

n in Ln converging to z. Let (z′n, −→vn) be the

element of T 1H such that −→vn is the unit vector at z′

n which is tangent to theray [z′

n, γn(un(+∞))). There exists sn ∈ R such that

(z′n, −→vn) = gsn(γn((zn, −→un))).

Since limn→+∞ z′n = z and limn→+∞ vn(+∞) = u(+∞), by Exercise 1.1 one

has limn→+∞ gsn(π1((zn, −→un))) = π1((z, −→u )). �

2 Topological properties of geodesic trajectories

We fix a non-elementary Fuchsian group Γ . The motivation of this chapter isto study the behavior of the trajectories of the geodesic flow g on T 1S. Weuse the notions introduced in Appendix A.

2.1 Characterization of the wandering and divergent points

Since Γ is not elementary group, its limit set L(Γ ) is minimal (Propo-sition I.3.6). The following theorem gives a characterization of the non-wandering set Ωg(T 1S) (Appendix A) of the geodesic flow on T 1S in termsof points in L(Γ ).

Theorem 2.1. Let (z, −→u ) be in T 1H. Then the following are equivalent:

(i) π1((z, −→u )) belongs to Ωg(T 1S);(ii) u(−∞) and u(+∞) belong to L(Γ ).

Before we prove this theorem, we will prove the following lemma.

Lemma 2.2. Let x, y be points in L(Γ ). There exists a sequence (γn)n�1 in Γsuch that limn→+∞ γn(i) = x and limn→+∞ γ−1

n (i) = y.

Proof. Fix some x in L(Γ ). Let A denote the set of x′ in L(Γ ) for whichthere exists a sequence (hn)n�1 in Γ satisfying limn→+∞ hn(i) = x andlimn→+∞ h−1

n (i) = x′. This set is non-empty and Γ -invariant. We will showthat it is also closed.

Let (x′p)p�1 be a sequence in A converging to a point x′ in H(∞). For

all p, there exists a sequence (hp,k)k�1 in Γ such that limk→+∞ hp,k(i) = xand limk→+∞ h−1

p,k(i) = x′p. Therefore, there exists a sequence (hp,kp)p�1 sat-

isfying limp→+∞ hp,kp(i) = x and limp→+∞ h−1p,kp

(i) = x′. This implies thatthe point x′ is in A and therefore that A is closed. Since A is a non-emptyclosed subset of L(Γ ), and L(Γ ) is minimal, we have A = L(Γ ). �


Proof (of Theorem 2.1).(ii) ⇒ (i). Let (γn)n�1 be in Γ such that limn→+∞ γn(i) = u(+∞) and

limn→+∞ γ−1n (i) = u(−∞). Define tn = d(z, γ−1

n (z)). The sequence (tn)n�0

converges to +∞. Consider the element (zn, −→vn) in T 1H, where zn = γ−1

n (z)and −→vn is the unit vector tangent to the segment [zn, z]h at zn (Fig. III.5).

Fig. III.5.

One has limn→+∞ vn(−∞) = u(−∞) and limn→+∞ vn(+∞) = u(+∞).Furthermore vn(tn) = z and therefore limn→+∞ gtn((zn, −→vn)) = (z, −→u ).

Consider now γn((zn, −→vn)). This element corresponds to the orderedpair composed of the point z and the unit vector tangent to the geodesicsegment [z, γn(z)]h at z. Observe that limn→+∞ γn((zn, −→vn)) = (z, −→u ). Let Vbe a neighborhood of π1((z, −→u )). For large enough n, π1((zn, −→vn)) andgtn(π1((zn, −→vn))) belong to V , thus gtnV ∩ V = ∅. This shows that π1((z, −→u ))is non-wandering.

(i) ⇒ (ii). Let (Vn)n�1 be a sequence of neighborhoods of π1((z, −→u )) suchthat

⋂+∞n=1 Vn = {π1((z, −→u ))}. Since π1((z, −→u )) is non-wandering, there exists

a sequence tn → +∞ such that gtnVn ∩ Vn = ∅. From this remark, it followsthat there exists a sequence (π1((zn, −→un)))n�1 in T 1S satisfying

limn→+∞

π1((zn, −→un)) = π1((z, −→u )) and limn→+∞

gtnπ1((zn, −→un)) = π1((z, −→u )).

Replacing (zn, −→un) by an element of Γ ((zn, −→un)), there exists a sequence(γn)n�1 in Γ satisfying

limn→+∞

(zn, −→un) = (z, −→u ) and limn→+∞

γngtn((zn, −→un)) = (z, −→u ).

Since limn→+∞ tn = +∞, one has limn→+∞ un(tn) = u(+∞). Furthermorelimn→+∞ d(un(tn), γ−1

n z) = 0, thus limn→+∞ γ−1n (i) = u(+∞). This shows

that u(+∞) belongs to L(Γ ).Replacing (tn)n�1 by (−tn)n�1 in the preceding argument, it is clear that

u(−∞) also belongs to L(Γ ). �


Corollary 2.3. The set Ωg(T 1S) is equal to T 1S if and only if L(Γ ) = H(∞).

The following proposition characterizes the fact that the set Ωg(T 1S) iscompact in terms of points in L(Γ ).

Proposition 2.4. The set Ωg(T 1S) is compact if and only if all points ofL(Γ ) are conical.

Before we prove Proposition 2.4, we introduce the subset Ωg(T 1S) ⊂ T 1H

defined by Ωg(T 1S) = (π1)−1Ωg(T 1S). It follows from Theorem 2.1 that

Ωg(T 1S) = {(z, −→u ) ∈ T 1H | u(−∞) ∈ L(Γ ), u(+∞) ∈ L(Γ )}.

Proof. The projection to H of Ωg(T 1S) is the set

Ω(Γ ) = {z ∈ H | z ∈ (xy) with x, y ∈ L(Γ )}

(this set was introduced in Sect. I.4.1). Note that Ωg(T 1S) is compact if andonly if there exists a compact K ⊂ H such that Ω(Γ ) =

⋃γ∈Γ γK.

If every point in L(Γ ) is conical, Corollary I.4.17 implies that the group Γ isconvex-cocompact. By definition the group Γ acts on the convex-hull of Ω(Γ )with a compact fundamental domain and hence that Ω(Γ ) =

⋃γ∈Γ γK, for

some compact K ⊂ H.Conversely, suppose that there exists a compact subset K ⊂ H such that

Ω(Γ ) =⋃

γ∈Γ γK. For any geodesic (xy) in Ω(Γ ), there exists a sequenceof points on this geodesic of the form (γn(kn))n�1 converging to x, whereγn ∈ Γ and kn ∈ K. Fix z on (xy). The sequence (γn(z))n�1 remains withina bounded distance of the ray [z, x). Thus x is conical. �

In the particular case where Γ is a Schottky group S(g1, g2) (see Chap. II),we obtain that the set Ωg(T 1S) is compact if and only if g1 and g2 are hyper-bolic (Figs. III.6 and III.7).

Fig. III.6. Γ = S(g1, g2), g1 and g2 hyperbolic

Recall that a point y ∈ Y is divergent (respectively positively or negativelydivergent) for a flow φ on Y , if for all unbounded sequences (tn)n�1 in R

(respectively R+ or R

−), the sequence (φtn(y))n�0 diverges (see Appendix A).


Fig. III.7. Γ = S(g1, g2), g1 hyperbolic and g2 parabolic

Let us analyze the divergent points for the geodesic flow on T 1S. No-tice that we can restrict our attention to the positively divergent points. Toprove it, we introduce the flip map on each unitary tangent plane T 1

z H, whichassociates to (z, −→u ), the point −(z, −→u ) = (z, −−→u ).

Exercise 2.5. Prove that the flip map is continuous, and that for all t ∈ R

and γ ∈ G, one has

−gt(−(z, −→u )) = g−t((z, −→u )) and γ(−(z, −→u )) = −γ((z, −→u )).

Using this exercise, we obtain

Lemma 2.6. The point π1((z, −→u )) is positively divergent if and only ifπ1(−(z, −→u )) is negatively divergent.

Suppose now that π1((z, −→u )) is a positively divergent point. There exista positive unbounded sequence (tn)n�0 and a sequence (γn)n�0 in Γ suchthat (gtnγn(z, −→u ))n�0 converges to some (z′, −→u ′) in T 1

H. Set gtnγn(z, −→u ) =(zn, −→un). The sequence (d(zn, z′))n�0 = (d(γ−1

n zn, γ−1n z′))n�0 is bounded.

Moreover the sequence (γ−1n zn)n�0 converges to u(+∞). It follows that the

sequence (γ−1n z′))n�0 converges to u(+∞), and hence that this point is in

L(Γ ). Using Lemma 2.6, we obtain the following result

Proposition 2.7. Let (z, −→u ) be in T 1H. If π1((z, −→u )) is not in Ωg(T 1S),

then it is a divergent point.

Clearly, when the set Ωg(T 1S) is compact, none of the elements of this setdiverge with respect to the geodesic flow. In the general case, let us charac-terize the divergent points in Ωg(T 1S).

Proposition 2.8. Let (z, −→u ) be in T 1H. Then the following are equivalent:

(i) π1((z, −→u )) is not positively (resp. negatively) divergent;(ii) u(+∞) (resp. u(−∞)) is a conical point in L(Γ ).


Proof.(i) ⇒ (ii). Let (tn)n�1 be an unbounded sequence in R

+ such that(gtn(π1((z, −→u ))))n�1 converges. There exists a sequence (γn)n�1 in Γ forwhich (γngtn((z, −→u )))n�1 converges to an element (z′, −→u ′) in T 1

H. We have

limn→+∞

u(tn) = u(+∞) and limn→+∞

d(u(tn), γ−1n (z′)) = 0.

The points u(tn) belong to the ray [z, u(+∞)). Furthermore

d(γ−1n (z), γ−1

n (z′)) = d(z, z′),

hence there exists ε > 0 and N > 0 such that

d(γ−1n (z), [z, u(+∞))) < ε

whenever n � N . This shows that u(+∞) is conical.(ii) ⇒ (i). Let (γn)n�1 be a sequence in Γ such that

d(γn(z), [z, u(+∞))) < ε.

It follows that there exists sn > 0 satisfying d(γn(z), u(sn)) < ε. Passing toa subsequence, one may assume that the sequence (γ−1

n gsn((z, −→u )))n�1 con-verges, which implies the convergence of the sequence (gsn(π1((z, −→u ))))n�1.

�

We deduce from this proposition and from Proposition 2.4, the followingresult

Corollary 2.9. The set Ωg(T 1S) contains divergent points of and only if itis not compact.

Using the preceding results and Exercise A.16 of Appendix A, we obtainthe following property for semi-trajectories

Property 2.10. Let (z, −→u ) be in T 1H. For some T ∈ R, the semi-trajectory

g[T,+∞)(π1(z, −→u )) (respectively g(− ∞,T ](π1(z, −→u )) is an embedding from[T, +∞) (respectively (−∞, T ])) into T 1S if and only if u(+∞) (respectivelyu(−∞)) is not conical.

2.2 Applications to geometrically finite groups

We suppose that Γ is a non-elementary, geometrically finite Fuchsian group(see Sect. I.4). Recall that there exist a Dirichlet domain Dz(Γ ) and a compactsubset K ⊂ H, such that the intersection of this domain with the Nielsenregion N(Γ ) of the group (i.e., the convex hull of the set of points in H

belonging to geodesics with endpoints in L(Γ )) satisfies (Proposition I.4.16)

N(Γ ) ∩ Dz(Γ ) = K⋃

x∈Lp(Γ )∩Dz(Γ )(∞)

H+(x) ∩ Dz(Γ ).


Moreover the set Lp(Γ ) ∩ Dz(Γ )(∞) is a finite set {x1, . . . , xn}, and(H+(xi))1�i�n is a collection of horodisks centered at xi, pairwise disjointssatisfying

γH+(xi) ∩ H+(xi) = ∅,

for all γ ∈ Γ − Γxi .Such a group Γ is also characterized by the fact that L(Γ ) is the dis-

joint union of the set of its conical points and that of its parabolic points(Theorem I.4.13).

Theorem 2.11. Let Γ be a non-elementary, geometrically finite group.

(i) There exists a compact K0 ⊂ T 1S such that, if (z, −→u ) ∈ T 1H and u(+∞)

(resp. u(−∞)) is conical, then the set of real numbers t > 0 (resp. t < 0)for which gt(π1((z, −→u ))) ∈ K0 is unbounded.

(ii) If (z, −→u ) ∈ T 1H is such that u(+∞) (resp. u(−∞)) is parabolic, then there

exists T > 0, and a cusp of S for which the projection to S of the semi-trajectory (gt(π1((z, −→u ))))t�T (resp. (gt(π1((z, −→u ))))t�−T ) is included inthe cusp.Additionally, when restricted to g[T,+∞)((z, −→u )) (resp. g(− ∞,−T ]((z, −→u ))),the projection of T 1

H to T 1S is a homeomorphism onto the semi-trajectoryg[T,+∞)(π1((z, −→u ))) (resp. g(− ∞,−T ](π1((z, −→u ))).

Proof.

(i) We begin by assuming that (z, −→u ) is such that u(+∞) is conical andu(−∞) ∈ L(Γ ), then π1((z, −→u )) is in Ωg(T 1S). To prove property (i) it isenough to prove that there exists an unbounded sequence (tn)n�1 in R

+

such that π(u(tn)) ∈ π(K). If this is not the case, there exists T > 0 suchthat, for all t � T , the point π(u(t)) is in the union of the cusps C(xi) as-sociated to H+(xi). Since the cusps C(xi) are disjoint, π([u(T ), u(+∞)))is contained in a single cusp C(xi). Hence the ray [u(T ), u(+∞)) is ina horodisk γ(H+(xi)) and thus u(+∞) = γ(xi), for some γ ∈ Γ , whichcontradicts the fact that u(+∞) is conical.Fix ε > 0. Consider now an element (z′, −→u ′) in T 1

H such that u′(+∞) isconical. Take (z, −→u ) such that u(+∞) = u′(+∞) and u(−∞) ∈ L(Γ ). Thegeodesic rays [z, u′(+∞)) and [z, u(+∞)) are asymptotic, thus there existsT > 0 such that [u′(T ), u′(+∞)) is in the ε-neighborhood of [z, u(+∞))(see Exercise 3.13). Furthermore, the preceding argument implies the ex-istence of an unbounded sequence (tn)n�1 in R

+ such that the sequence(gtn(π1((z, −→u ))))n�1 is in π(K). From these properties one deduces theexistence of an unbounded sequence (t′

n)n�1 ⊂ R+, such that the sequence

(π((z′, −→u ′)(t′n)))n�1 is in the ε-neighborhood of π(K). Such a neighbor-

hood is compact.(ii) To avoid unnecessary notation, we present an argument on S which can be

extended T 1S. Let (z, −→u ) ∈ T 1H such that u(+∞) is parabolic. One can

suppose u(+∞) = xi (Corollary I.4.10). The projection q of the quotient

3 Periodic trajectories and their periods 89

Γxi \H+(xi) to the cusp C(xi) is a homeomorphism (see Sect. I.3.22). Forlarge enough T , the ray [u(T ), xi) is contained in H+(xi). Furthermore,if one restricts the projection of H+(xi) to Γxi \H+(xi) to this ray, theresulting map p is a homeomorphism. Thus q ◦ p is a homeomorphism of[u(T ), xi) onto π([u(T ), xi)). �

Let Ei denote the set of all (z, −→u ) in T 1H such that z is in the horocycle

H(xi) associated to H+(xi). Notice that π1(Ei) is a compact subset of T 1S.Take (z′, −→u ′) in T 1

H whose positive endpoint u′(+∞) is parabolic; there existi ∈ {1, . . . , n} and γ in Γ such that γ(v(+∞)) = xi. A geodesic having xi asan endpoint intersects H(xi), hence γgR((z′, −→u ′)) ∩ Ei = ∅. It follows that theunion of the compact set K0 given by Theorem 2.11(i) with the projectionto T 1S of the Ei is a compact set intersected by all the semi-trajectoriesgR+(π1((z, −→u ))), with u(+∞) ∈ L(Γ ).

Corollary 2.12. Let Γ be a non-elementary, geometrically finite group.There exists a compact subset of T 1S intersected by every semi-trajectoryg+

R(π1((z, −→u ))), with u(+∞) ∈ L(Γ ).

Some of the trajectories of gR on Ωg(T 1S) are closed. This is the case if, forexample, u(−∞) and u(+∞) are parabolic (Theorem 2.11). Some trajectoriesare compact. In the following section it will be shown that the compactnesscorresponds to the case in which the points u(+∞) and u(−∞) are fixedby a hyperbolic isometry of Γ . Some trajectories are dense (see Sect. 4).Others are very “chaotic.” Having information about the conical and parabolicnature of u(−∞) and u(+∞) alone is not sufficient to generally describe thetopology of gR(π1((z, −→u ))). In Chap. IV, we will study the case in which Γ is aSchottky group generated by two hyperbolic isometries. Using doubly-infinitesequences as a coding of Ωg(T 1S), we will establish a correspondence betweenthe topological dynamics of gR on Ωg(T 1S) and that of the shift on the spaceof these doubly-infinite sequences. We will then see the emergence of a widevariety of topological structures for trajectories of gR. A notable example ofthis is the existence of minimal compact sets which are gR-invariant, yet arenot periodic trajectories.

When Ωg(T 1S) is not compact, this set contains unbounded trajecto-ries. For example, if u(+∞) or u(−∞) is parabolic, then (gt(π1(u))t∈R isunbounded. This condition on u(+∞) and u(−∞) is sufficient but not neces-sary. In Chap. VII, in the context of the modular surface, we will relate theboundedness of gR(π1(u)) to a property of the continued fraction expansionof u(−∞) and u(+∞) (Theorem VII.3.4).

3 Periodic trajectories and their periods

Returning to the general case of a non-elementary Fuchsian group Γ , wefocus on the periodic trajectories of the geodesic flow on T 1S. Recall that


π1((z, −→u )) ∈ T 1S is periodic for the geodesic flow if there exists some t > 0such that gt(π1((z, −→u ))) = π1((z, −→u )).

3.1 Density of periodic trajectories

The following proposition gives a characterization of the periodic elements.

Proposition 3.1. Let (z, −→u ) be in T 1H. The following are equivalent:

(i) the element π1((z, −→u )) is periodic;(ii) there exists a hyperbolic isometry γ in Γ such that u(+∞) = γ+ and

u(−∞) = γ−.

Proof.(ii) ⇒ (i). The isometry γ leaves the oriented geodesic (γ−γ+) invariant.

Furthermore, Property I.2.8 implies that given a point z on this geodesic, onehas that d(z, γ(z)) = �(γ). Therefore γ((z, −→u )) = g�(γ)((z, −→u )).

(i) ⇒ (ii). There exists t > 0 and γ in Γ such that gt((z, −→u )) = γ((z, −→u )).It follows that gnt((z, −→u )) = γn((z, −→u )). Thus γ fixes u(+∞) and u(−∞).These two points are distinct and γ is not the identity, hence this isometry ishyperbolic. �

Let π1((z, −→u )) in T 1S be a periodic point for the geodesic flow. Apply-ing Proposition 3.1, we obtain a hyperbolic isometry γ ∈ Γ fixing the pointsu(+∞) and u(−∞) such that t = d(z, γ(z)). Since Γ is discrete, the sub-group of hyperbolic isometries fixing u(+∞) and u(−∞) is generated byone primitive element γ0 = Id (i.e., there is no isometry h in Γ satisfyinghn = γ0 for n > 1). It follows that the set of real numbers t such thatgt(π1((z, −→u ))) = π1((z, −→u )), which is a closed subgroup of (R, +), is the setZ�(γ0). Since �(γ0) is the smallest t > 0 such that gt(π1((z, −→u ))) = π1((z, −→u )),it is called the period of π1((z, −→u )) and is denoted Tu.

Let Hyp(Γ ) denote the set of conjugacy classes in Γ of primitive hyperbolicisometries of Γ . According to Proposition 3.1, there is a bijection from Hyp(Γ )onto the set of periodic trajectories of gR, which sends an equivalence class [γ]to the trajectory gR(π1((z, −→u ))), where (z, −→u ) is an element of T 1

H satisfyingu(+∞) = γ+ and u(−∞) = γ−. Since the group Γ is not elementary, byCorollary II.1.3, the set Hyp(Γ ) is infinite. This allows us to state the followingproperty.

Property 3.2. The set Ωg(T 1S) contains infinitely many periodic trajecto-ries.

In Chap. IV, we will give another proof of the following Theorem using thetechniques of symbolic dynamics in the particular case of convex-cocompactSchottky groups (see Sect. IV.2.1).

Theorem 3.3. The set of periodic elements is dense in Ωg(T 1S).


Proof. Fix (z, −→u ) in Ωg(T 1S) (i.e., u(+∞) and u(−∞) belong to L(Γ )).Proposition 1.6 reduces the proof of this theorem to proving that there exists asequence of hyperbolic isometries (γn)n�1 in Γ , which satisfy limn→+∞ γ+

n =u(+∞) and limn→+∞ γ−

n = u(−∞).By Lemma 2.2, there exists (γn)n�1 in Γ such that limn→+∞ γn(i) =

u(+∞) and limn→+∞ γ−1n (i) = u(−∞).

We will now show that for large enough n, γn is hyperbolic. For thistask, we work in the Poincare disk with a center 0. Let D0(γn) be the closedhalf-plane bounded by the perpendicular bisector of the segment [0, γn(0)]h,containing γn(0). The sequence of Euclidean diameters of D0(γ±1

n ) convergesto zero since limn→+∞ γ±1

n (0) = u(+∞). The points u(−∞) and u(+∞) aredistinct, thus for large enough n, the half diks D0(γn) and D0(γ−1

n ) are dis-joint. Property I.2.7, implies that γn is hyperbolic. Moreover, γ+

n (respec-tively γ−

n ) is in the boundary at infinity of D0(γn) (respectively D0(γ−1n )),

hence limn→+∞ γ+n = u(+∞) and limn→+∞ γ−

n = u(−∞). �

3.2 Length spectrum

We define a geodesic (respectively geodesic segment) of the surface S as theimage under the canonical projection π of a geodesic (respectively geodesicsegment) of H (Fig. III.8).

Fig. III.8. Γ = S(g1, g2), g1 and g2 hyperbolic

If the group Γ does not contain any elliptic elements, S inherits its Rieman-nian structure. Geodesics on S coincide with geodesics for this Riemannianstructure.

Exercise 3.4. Prove that a geodesic of S is compact if and only if it is theprojection of a periodic trajectory of gR to S.

Let γ be a hyperbolic element of Γ . Consider a primitive element h in Γhaving the same axis as γ. If the group Γ does not contain any elliptic element,


then S is a Riemannian surface and the real number �(h) is the length, in theRiemannian sense, of the geodesic π(γ−γ+). More generally, one defines thelength of π(γ−γ+), as

lengthS(π(γ−γ+)) = �(h).

Notice that, if (z, −→u ) is such that u(+∞) = γ+ and u(−∞) = γ−, thenπ1((z, −→u )) is periodic and its period Tu is given by:

Tu = lengthS(π(γ−γ+)).

It follows that the set of lengths of compact geodesics of S is in one-to-one correspondence with the set SP(gR) of periods associated to the periodictrajectories of gR.

We are interested in the set of all nT , where n ∈ N and T ∈ SP(gR). It isin one-to-one correspondence with the set L(Γ ) = {�(γ) | γ ∈ Γ }, called thelength spectrum of Γ .

Property 3.5. Let Γ be a geometrically finite group and (γn)n�1 be a se-quence of hyperbolic isometries in Γ . If (�(γn))n�1 is a bounded sequence,then there exist k isometries g1, . . . , gk in Γ such that, for all n � 1, theisometry γn is conjugate in Γ to one of the gi.

Proof. Let (zn, −→un) be in T 1H such that un(+∞) = γ+

n and un(−∞) = γ−n .

Since the group Γ is geometrically finite, Corollary 2.11 implies that somecompact subset of T 1S is intersected by all trajectories gR(π1((zn, −→un))).It follows that after conjugating γn, replacing (zn, −→un) with an element ofgR((zn, −→un)), and passing to a subsequence, one may assume that the se-quence ((zn, −→un))n�1 converges to (z′, −→u ′). One has d(γn(zn), zn) = �(γn).Furthermore, passing to another subsequence, (�(γn))n�1 converges. Hencethere exists ε > 0 and M > 0 such that d(γn(z′), z′) � ε for all n � M . Sincethe group Γ is discrete, the set of such γn is finite. �

In terms of periods of the geodesic flow, the preceding property impliesthat, if a sequence (Tun)n�1 in SP(gR) is bounded then the elements π1(un)are contained in a finite number of periodic trajectories.

Another consequence is that for all t > 0, the subset of Hyp(Γ ) composedof the classes [γ] satisfying �(γ) � t, is finite.

In particular, there is a finite number of compact geodesics on S whoselength is less than that of all other geodesics. On the modular surface, applyingCorollary II.4.14, we obtain that there is only one such geodesic and that itslength is equal to 4 ln N , where N is the golden ratio.

Theorem 3.6. The additive group generated by the length spectrum L(Γ )(and thus by the periods SP(gR)) is dense in R.


Proof. Since the group Γ is non-elementary, Corollary II.1.3 implies that itcontains a Schottky group generated by a pair of hyperbolic isometries h and g.Let us use the Poincare disk model and fix a point 0. One may assume thatthe half-planes D0(h) = D(h), D0(h−1) = D(h−1), D0(g) = D(g), D0(g−1) =D(g−1), defined in Sect. I.2.2, are located as in Fig. III.9.

Fig. III.9.

By Proposition II.1.4, the geodesics (h−h+) and ((ghn)−(ghn)+) intersecteach other at a point zn in D for all n > 0. The sequence (zn)n�1 convergesto h−. Thus one may assume that the points zn are all distinct. One has

�(ghn) = d(zn, ghn(zn)) and �(h) = d(zn, h(zn)).

Furthermore, zn is not in the axis of ghn+1 thus from Property I.2.7(i), onehas �(ghn+1) < d(zn, ghn+1(zn)). It follows that for all n � 1

�(ghn+1) < d(h−ng−1(zn), h(zn)),

thus

(∗) �(ghn+1) < �(ghn) + �(h).

Let us return to the Poincare half-plane and choose an isometry γ(z) =(az + b)/(cz + d) in H. For such an isometry, we have

cosh(

�(γ)2

)=

|a + d|2

,

where cosh denotes the hyperbolic cosine.Using this relationship and supposing, after conjugating Γ , that h(z) = λz

with λ > 1, one obtains

limn→+∞

�(ghn+1) − �(ghn) = �(h).

If L(Γ ) generates a discrete group, then �(ghn+1) − �(ghn) = �(h) for largeenough n, which contradicts the inequality (∗). �


4 Dense trajectories

In this section, we prove the existence of trajectories of the geodesic flow thatare dense in its non-wandering set.

We use the criterion (Proposition 1.6) relating the action of gR on T 1S tothe dual action of Γ on the set of oriented geodesics gR\T 1

H, or more preciselyon the set L(Γ ) ×Δ L(Γ ) defined to be the product L(Γ ) × L(Γ ) minus itsdiagonal. Let us begin by proving the following lemma.

Lemma 4.1. Let Γ be a non-elementary Fuchsian group. For any open, non-empty subsets O and V of L(Γ ) ×Δ L(Γ ), there exists γ ∈ Γ such thatγ(O) ∩ V =∅.

Proof. One can assume that O and V are products of open, non-empty setsO = O1 × O2, V = V1 × V2. Since the set L(Γ ) is minimal, V1 contains theattractive fixed point γ+ of a hyperbolic isometry γ in Γ . It follows that, forlarge enough n, γnO1 ∩ V1 = ∅. Furthermore, from Theorem 3.3, there existsa hyperbolic isometry h in Γ such that h− is contained in γnO1 ∩ V1 and h+

is in V2. Thus for large enough k, one has hkγnO2 ∩ V2 = ∅. Moreover, sincethe point h− is in γnO1 ∩ V1, we have hkγnO1 ∩ V1 = ∅. This implies thathkγnO ∩ V = ∅. �

Theorem 4.2. There exists (z, −→u ) in Ωg(T 1S) such that gR(π1((z, −→u ))) =Ωg(T 1S).

Proof. Consider a countable family of open, non-empty subsets (On)n�1 inL(Γ ) ×Δ L(Γ ) such that every open subset of L(Γ ) ×Δ L(Γ ) contains one ofthe On. Fix an open, non-empty subset O of L(Γ ) ×Δ L(Γ ). After Lemma 4.1,there exists γ1 in Γ such that γ1O ∩ O1 = ∅. Let K1 be an open, relativelycompact subset of O such that γ1K1 is in O1. Repeating this argument andreplacing O with K1 and O1 with O2, one obtains γ2 in Γ and an open,relatively compact subset K2 such that K2 ⊂ K1 and γ2K2 ⊂ O2. Continuingthis process, one obtains a sequence (Kn)n�1 of open, relatively compact,nested subsets. The set

⋂+∞n=1 Kn is not empty. Let x in

⋂+∞n=1 Kn. For all

n � 1, the point γn(x) is in On. Consider a point x′ in L(Γ ) ×Δ L(Γ ) and aneighborhood V ′ of x′. This neighborhood contains an open set On, thus γn(x)is in V ′. The orbit Γx therefore intersects every neighborhood of x′, whichshows that x′ is in Γx. It follows that Γx = L(Γ ) ×Δ L(Γ ). It is sufficient toapply Proposition 1.6 to complete the proof. �

Theorem 4.2 will be proved again in the next chapter for the particular caseof convex-cocompact Schottky groups. This case provides a characterization ofdense trajectories gR(π1((z, −→u ))) in terms of the coding of u(−∞) and u(+∞)(see Sect. IV.2.2).

The following theorem implies Lemma 4.1. It will be proved later usingthe horocycle flow (see Sect. V.3).

5 Comments 95

Theorem 4.3 (Topological mixing). Let O and V be two open, non-emptysubsets of Ωg(T 1S). There exists T > 0 such that for all t � T , gtO ∩ V = ∅.

5 Comments

The geodesic flow is also well-defined on the unitary tangent bundle of a com-plete Riemannian manifold and in particular on the unitary tangent bundleof a pinched Hadamard manifold T 1X, and on its quotient by a torsion-freeKleinian group Γ [14, 49].

The results of this chapter, with the exception of Property 3.6 and Theo-rem 4.3 and their proofs, are inspired by two articles on the geodesic flow onΓ \T 1X by P. Eberlein ([26] and [28]).

In this general context, the length spectrum is a source of open problems.One of these problems consists of knowing whether or not Property 3.6 aboutthe density of the length spectrum always holds for Schottky groups. It is openif X is not a symmetric space and if dim(X) � 3 [11, 18, 29]. The densityproperty of the length spectrum is especially important since it is equivalentto that of the topological mixing of the geodesic flow [18].

In this book we do not study the metric properties of the geodesic flow.Let us give the reader some idea of these properties. Consider a Fuchsiangroup Γ acting on the Poincare disk D. A construction due to D. Sullivan[61, 62] allows us to obtain from a Patterson measure m on L(Γ ) (introducedin the Comments at the end of Chap. I) a measure M on L(Γ ) ×Δ L(Γ ) × R

defined as

M(dx dy ds) =m(dx)m(dy) ds

|x − y|2δ(Γ ),

where δ(Γ ) is the critical exponent of the Poincare series associated with Γ .Identifying T 1

D with triplets of points (x, y, s) such that x = y and s ∈ R,the measure M becomes a Γ -invariant measure on T 1

D. This measure is alsoflow invariant with respect to gR. Therefore it induces a measure M on Γ \T 1

D

which is again gR-flow invariant and preserves the non-wandering set of thisflow [48]. If Γ is a lattice, M is proportional to the Liouville measure. Moregenerally, if Γ is geometrically finite, M is finite and the geodesic flow isergodic and mixing [5, 54].

This construction can be generalized to the case of pinched Hadamardmanifolds [11], but the geometric finiteness of Γ does not necessarily implythe finiteness of M [23, 54].

The measure M plays a crucial role in solving counting problems, like forexample counting points in an orbit or counting closed geodesics.

IV

Schottky groups and symbolic dynamics

Throughout this chapter, the group Γ will designate a Schottky group gen-erated by two hyperbolic isometries g1, g2 (see Sect. II.1). By definition, sucha group admits a Dirichlet domain centered at a point designated to be 0 inthe Poincare disk. The possible cases are diagrammed below in Fig. IV.1. Forfurther details, the reader may refer to Sect. II.1.

Fig. IV.1.

Recall that if g is an isometry which does not fix 0, then the set D0(g)represents the closed half-plane in the disk D, containing the point g(0), andbounded by the perpendicular bisector of the segment [0, g(0)]h.

The surface S = Γ \D is homeomorphic to a sphere with three pointsremoved, or to a torus minus one point. In each of these cases, the limit set of Γis composed exclusively of conical points (Property II.1.13). Hence the non-wandering set of the geodesic flow, Ωg(T 1S), is compact (Proposition III.2.4).

The goal of this chapter is to encode the trajectories of the geodesic flowrestricted to Ωg(T 1S) into doubly-infinite sequences, and to develop this pointof view into a method of studying the dynamics of this flow. This symbolicapproach will allow us to present new proofs of Theorems III.3.3 and III.4.2.


http://dx.doi.org/10.1007/978-0-85729-073-1_4

98 IV Schottky groups and symbolic dynamics

Moreover we will complete the latter theorem by characterizing the dense tra-jectories of Ωg(T 1S) in terms of sequences. As applications, we will construct,in the general case of a non-elementary Fuchsian group Γ ′, trajectories of thegeodesic flow on Ωg(Γ ′ \T 1

D) which are neither periodic nor dense.

1 Coding

Recall that Σ+ represents the set of sequences s = (si)i�1 satisfying

si ∈ A = {g±11 , g±1

2 } and si+1 �= s−1i .

We have also defined f : Σ+ → L(Γ ) to be the map which sends a sequences = (si)i�1 to the following point

f(s) = limn→+∞

s1 · · · sn(0) (see Sects. II.1 and II.2).

Since g1 and g2 are hyperbolic, this map is a bijection (Proposition II.2.2).We equip Σ+ with the following metric δ:

δ(s, s′) =

⎧⎨⎩

0 if s = s′,1

min{i � 1 | si �= s′i} otherwise.

Exercise 1.1. Prove that δ is a metric, and that the metric space (Σ+, δ) iscompact.

Lemma 1.2. The map f : (Σ+, δ) → L(Γ ) is a homeomorphism.

Proof. It is sufficient to prove that this map is continuous. Consider a sequence(un)n�1 in Σ+ which converges to an element s of Σ+. Define

un = (un,i)i�1 and s = (si)i�1.

For all k � 2, there exists N > 1 such that, for each 1 � i � k and n � N ,one has un,i = si. Let T be the shift operator on Σ+. For all n � N , we have

f(un) = s1 · · · sk−1(f(T k−1(un))).

Since the point f(T k−1(un)) belongs to the boundary at infinity of the half-disk D0(sk), the point f(un) is in D(s1, . . . , sk)(∞) = s1 · · · sk−1D0(sk)(∞).This set also contains f(s) since

f(s) = s1 · · · sk limn→+∞

sk+1 · · · sn(0).

According to Lemma II.1.10, the sequence of Euclidean diameters of thenested sets D(s1, . . . , sk) converges to 0 as k tends to +∞. Thereforelimn→+∞ |f(un) − f(s)| = 0. �

1 Coding 99

We now consider the set Σ of doubly-infinite sequences S = (Si)i∈Z whichsatisfy the following conditions:

(Si)i�1 ∈ Σ+, (S−1−i+1)i�1 ∈ Σ+ and S1 �= S−1

0 .

We associate to S be in Σ, two sequences defined by

S+ = (Si)i�1 and S− = (S−1−i+1)i�1.

The condition S1 �= S−10 together with Property II.1.4(i) implies that the

points f(S+) and f(S−) are distinct.The distance function δ on Σ+ induces a distance function Δ on Σ defined

by

Δ(S, S′) =√

δ2(S+, S′+) + δ2(S−, S′ −).

We still denote by T : Σ → Σ the shift operator T (S) = (Si+1)i∈Z. Thisoperator is a bijection of Σ onto itself.

Exercise 1.3. Prove that the shift operator T on (Σ, Δ) is continuous.

Exercise 1.4. Prove that the metric space (Σ, Δ) is compact.

We will establish a correspondence between the topology of orbits of Ton (Σ, Δ) and the trajectories of the geodesic flow on Ωg(T 1S). To do this,recall first that L(Γ ) ×Δ L(Γ ) denotes the product of L(Γ ) with itself minusits diagonal, and denote by F : Σ → L(Γ ) ×Δ L(Γ ), the map defined by

F (S) = (x(S−), x(S+)).

This map is continuous and injective since the map f is. However, it is notsurjective since x(S−) and x(S+) respectively belong to the disjoint arcsD0(S−1

0 )(∞) and D0(S1)(∞).

Lemma 1.5. Given (x−, x+) in L(Γ ) ×Δ L(Γ ), there exist γ in Γ and S in Σsuch that γ(x−, x+) = F (S).

Proof. Let a = (ai)i�1 and b = (bi)i�1 be the elements of Σ+ such thatx− = f(a) and x+ = f(b). By hypothesis x− �= x+. Consider the smallestN � 1 for which aN �= bN . Let S denote the doubly-infinite sequence definedby

Si =

{aN+i−1 if i � 1,

b−1N −i if i � 0.

This sequence belongs to Σ. If N = 1, then F (S) = (x−, x+); otherwiseF (S) = (a−1

1 · · · a−1N −1(x−), a−1

1 · · · a−1N −1(x+)). �

This lemma shows that the map F is a surjection to a set of representativesof Γ -orbits on L(Γ ) ×Δ L(Γ ).

Recall that π1 denotes the projection of T 1H to T 1S.


Proposition 1.6. Let S, S′ be in Σ and (z, −→u ), (z′, −→u ′) in T 1D such that

(u(−∞), u(+∞)) = F (S), (u′(−∞), u′(+∞)) = F (S′).

Then the following are equivalent:

(i) S′ ∈ T Z(S);(ii) π1((z′, −→u ′)) ∈ gR(π1((z, −→u ))).

Proof. Part (ii) is equivalent to the existence of sequences (sn)n�1 in R and(γn)n�1 in Γ such that limn→+∞ γn(gsn(u)) = u′. According to Proposi-tion III.1.6, this is in turn equivalent to the existence of a sequence (γn)n�1

in Γ satisfying

(iii) limn→+∞

(γn(u(−∞)), γn(u(+∞))) = (u′(−∞), u′(+∞)).

The implication (i) ⇒ (iii) follows directly from the equality

F (Tn(S)) = (γn(u(−∞)), γn(u(+∞))),

where γn = S−1n · · · S−1

1 if n > 0 and γn = Sn+1 · · · S0 otherwise.Let us show (iii) ⇒ (i). Consider a sequence (γn)n�1 in Γ satisfying

limn→+∞

(γn(u(−∞)), γn(u(+∞))) = (u′(−∞), u′(+∞)).

Write γn in the form of a reduced word γn = an,1 · · · an,�n . If there exists asubsequence (γnk

)k�1 satisfying

γnk= S−1

�nk· · · S−1

1 or γnk= S−�nk

+1 · · · S0,

then (γnk(u(−∞)), γnk

(u(+∞))) = F (T �nk (S)). Since F is a homeomorphismonto its image, it follows that limn→+∞ T �nk (S) = S′.

Otherwise, for large enough n, γn is distinct from both S−1�n

· · · S−11 and

S−�n+1 · · · S0. Thus Property II.1.4 implies that the points γn(u(−∞)) andγn(u(+∞)) are in the same circular arc of D(an,1)(∞). Passing to a sub-sequence, one may assume that an,1 = a1. Thus the points u′(−∞) andu′(+∞) are elements of D(a1)(∞). This is impossible since by hypothesisu′(−∞) ∈ D(S′ −1

0 )(∞), u′(+∞) ∈ D(S′1)(∞) and S′ −1

0 �= S′1. �

2 The density of periodic and dense trajectories

2.1 An alternate proof of Theorem III.3.3

We first establish relationship between the sequences in Σ which are periodicwith respect to the shift T , and the elements of Ωg(T 1S) which are periodicwith respect to the geodesic flow.

2 The density of periodic and dense trajectories 101

Let S = (Si)i�1 in Σ be T -periodic of period n. The point f(S+) is theattracting fixed point of γ = S1 · · · Sn, and f(S−) is the repulsive one. If(z, −→u ) ∈ T 1

D satisfies (u(−∞), u(+∞)) = F (S), then by Proposition III.3.1,the element π1((z, −→u )) is periodic with respect to the geodesic flow.

Conversely, if π1((z, −→u )) is periodic with respect to gR, after replacing(z, −→u ) with an element of Γ (z, −→u ), one can assume that there exists a primi-tive hyperbolic isometry γ = a1a2 · · · an with ai ∈ A, ai+1 �= a−1

i and a1 �= a−1n

satisfying u(−∞) = γ− and u(+∞) = γ+. Since a1 �= a−1n , the sequences

f −1(γ+) and f −1(γ−) are T -periodic of period n. Consider the doubly-infiniteperiodic sequence (Si)i∈Z of period n defined by S1 = a1, . . . , Sn = an. Thissequence belongs to Σ and satisfies F (S) = (γ−, γ+). In conclusion we obtainthe following property:

Property 2.1. Let (z, −→u ) be in T 1D. The element π1((z, −→u )) is gR-periodic

if and only if there exists S in Σ which is T -periodic, and an isometry γ in Γsuch that

γ(u(−∞), u(+∞)) = F (S).

Using this dictionary between periodic sequences and periodic trajectoriesfor the geodesic flow, we give another proof of the density of the set of thegR-periodic elements in Ωg(Γ \T 1

D) (Theorem III.3.3), when Γ is a Schottkygroup.

A proof of Theorem III.3.3 using symbolic dynamics

Let π1((z, −→u )) be in Ωg(Γ \T 1H). Lemma 1.5 implies that there exist γ in Γ ,

and S in Σ such that γ(u(−∞), u(+∞)) = F (S). For each n � 1, choosean+1 in A − {S−1

n , S−1−n}. Consider the sequence (Uk)k�1 of elements of Σ in

which each term Uk = (Uk,i)k�1 is a periodic sequence of period 2k + 2 suchthat

Uk,1 = S1, Uk,2 = S2, . . . , Uk,k = Sk,

Uk,k+1 = ak+1,

Uk,k+2 = S−k, Uk,k+3 = S−k+1, . . . , Uk,2k+2 = S0.

Then Δ(Uk, S) �√

2/k, which further implies that limk→+∞ Uk = S. Foreach k � 1, choose (zk, −→uk) in T 1

D such that (uk(−∞), uk(+∞)) = F (Uk).It follows from Property 2.1 that the element π1((zk, −→uk)) is periodic.Furthermore, since F is continuous,

limk→+∞

(uk(−∞), uk(+∞)) = (u(−∞), u(+∞)).

Thus there exists a sequence (sk)k�1 in R such that (gsk(π1((zk, −→uk))))k�1

converges to π1((z, −→u )).


2.2 An alternate proof of Theorem III.4.2

In order to prove the existence of geodesic trajectories which are dense inΩg(T 1S), by Proposition 1.6 it is sufficient to prove the existence of T -orbitsthat are dense on Σ. Let us characterize sequences S ∈ Σ such that T ZS = Σ.

Let V = (Vi)i∈I , with I ⊂ Z, be a finite or infinite sequence with termsin A. By definition, a block of V is a finite sequence B composed of consecutiveterms of V . Equivalently B is of the form B = (Vn+i)1�i�k, with n + i ∈ Ifor each 1 � i � k.

Property 2.2. Let S ∈ Σ. Then the following are equivalent:

(i) all reduced words a1 · · · an with ai ∈ A and ai+1 �= a−1i are a block of S;

(ii) T Z(S) = Σ.

Proof.(i) ⇒ (ii). Let S′ = (S′

i)i∈Z ∈ Σ. Consider the reduced word

S′−nS′

−n+1 · · · S′0S

′1 · · · S′

n.

For each n, by hypothesis there exists kn ∈ Z such that T kn(S) is a se-quence Un satisfying Un,i = S′

i for all −n � i � n. It follows thatΔ(S′, Un) �

√2/n, which further implies that limn→+∞ T kn(S) = S′.

(ii) ⇒ (i). Let m = a1 · · · an be a reduced word and c be in A − {a−1n , a−1

1 }.Consider the doubly-infinite periodic sequence S′ having period n+1, definedby

S′1 = a1, S′

2 = a2, S′n = an, S′

n+1 = c.

This sequence belongs to Σ. Since T Z(S) = Σ, there exists (kp)p�1 in N

such that limp→+∞ T kp(S) = S′. Define T kp(S) = Up. For large enough p,Δ(S′, Up) � 2/(n + 1). Thus Up,i = S′

i for all 1 � i � n + 1. This shows thatthe finite sequence a1, a2, . . . , an is a block of S. �

A proof of Theorem III.4.2 using symbolic dynamics

To prove Theorem III.4.2, it remains to construct a sequence satisfying part(ii) of the above property. For each n in N

∗, let En denote the set of reducedwords mn = a1 · · · an of length n, with ai ∈ A and ai �= a−1

i+1. Let n be thenumber of elements of this set. Choose an enumeration (mn,i)1�i��n of theelements of En. For all 1 � i < n, choose a letter ai in A which is not theinverse of either the last letter of the word mn,i or the first letter of the wordmn,i+1. Chosen this way, the word mn,1a1mn,2a2 · · · a�n −1mn,�n is a reducedword. Let Bn = (Bn,i)1�i�pn denote the finite sequence of letters forming thisword. Choose d0 �= B−1

1,1 in A. For n � 1, choose dn in A − {B−1n,pn

, B−1n+1,1}.

Finally, we are ready to present the doubly-infinite sequence S defined bySi = d0 for all i � 0 and whose sequence S+ is constructed from blocks

3 Applications to the general case 103

(Bn)n�1 and the sequence (dn)n�1 in the following way:

S+ = B1,1, B1,2, . . . , B1,p1︸︷︷︸B1

, d1, B2,1, . . . , B2,p2︸︷︷︸B2

, d2, . . . ,

Bn,1, . . . , Bn,pn︸︷︷︸Bn

, dn, Bn+1,1, . . .︸︷︷︸Bn+1

, . . .

The sequence S belongs to Σ. Yet by construction, the sequence S+ containsall finite reduced words. Hence, by Property 2.2, one has T Z(S) = Σ.

3 Applications to the general case

In Chap. III, we pointed out the existence of geodesic trajectories that areeither periodic or dense in the non-wandering set associated with the quotientof T 1

D by some non-elementary Fuchsian group.Having settled that question, we now focus on the complementary ques-

tion: are there any trajectories in the non-wandering set which satisfy neitherof these two properties? To answer this question, we will use the fact that anon-elementary group contains some Schottky groups S(g1, g2) generated bytwo hyperbolic isometries (Corollary II.1.3).

Proposition 3.1. Let S(g1, g2) be a Schottky group generated by two hyper-bolic isometries. The non-wandering set of the geodesic flow on S(g1, g2)\T 1

D

contains geodesic trajectories which are neither dense nor periodic.

Exercise 3.2. Prove Proposition 3.1.(Hint: construct an example of a non-periodic doubly-infinite sequence be-longing to Σ which does not use all of the letters of the alphabet A, andapply Proposition 1.6.)

Let Γ be a non-elementary Fuchsian group and S(g1, g2) be a Schottkygroup included in Γ . Set T 1S0 = S(g1, g2)\T 1

D and T 1S = Γ \T 1D. Consider

the projectionP : T 1S0 −→ T 1S.

Property 3.3. The projection P satisfies the following properties:

(i) For all t ∈ R and π1((z, −→u )) ∈ T 1S, one has P (gt(π1((z, −→u )))) =gt(P (π1((z, −→u )))).

(ii) If P (Ωg(T 1S0)) = Ωg(T 1S), then L(Γ ) = L(S(g1, g2)).


Lemma 3.5. If a geodesic trajectory in Ωg(T 1S0) is not periodic, then itsimage by P is not periodic.


Proof. Let gR(π1((z, −→u ))) be a non-periodic trajectory in Ωg(T 1S0). Sup-pose that its image by P is periodic. There exist a hyperbolic isometry γ inΓ − S(g1, g2) and a real number T �= 0 such that γ((z, −→u )) = gT ((z, −→u )). Onthe other hand, since gR(π1((z, −→u ))) is included in a compact set, there existan unbounded sequence (tn)n�1 and a sequence (γn)n�1 in S(g1, g2) for whichthe sequence γngtn((z, −→u )) converges to an element in T 1

D. Using γ, one ob-tains a unbounded sequence of integers (kn)n�1, and a bounded real sequence(sn)n�1 such that (γnγkn gsn((z, −→u )))n�1 converges in T 1

D. The group Γ be-ing discrete, the set of γnγkn is finite, this implies that γk is in S(g1, g2) forsome k �= 0, which contradicts the fact that gR(π1((z, −→u ))) is not periodic. �

Corollary 3.6. Let Γ be a non-elementary Fuchsian group. The non-wandering set of the geodesic flow on Γ \T 1

D contains geodesic trajectorieswhich are neither dense nor periodic.

Proof. We choose S(g1, g2) sufficiently “small” so that: L(Γ ) �= L(S(g1, g2)).It follows from Property 3.3, that the set P (Ωg(T 1S0)) is a proper compactsubset of Ωg(T 1S) which is invariant with respect to the geodesic flow. Takethe image by P of a geodesic trajectory on Ωg(T 1S0), which is neither periodicnor dense (Proposition 3.1). This image is a geodesic trajectory which notdense in Ωg(T 1S) and not periodic (Lemma 3.5). �

We focus now on the existence of minimal compact sets which are invariantwith respect to the geodesic flow on Ωg(T 1S).

Recall that a subset F of a topological space is minimal relative to agroup H of homeomorphisms if it is closed, non-empty, H-invariant and min-imal in the sense of inclusion for these properties (Appendix A). Such a set isnecessarily the closure of an orbit of H.

A decreasing sequence of compact sets which are invariant with respectto gR contains a smaller element. Hence all compact subsets of Ωg(T 1S) whichare invariant with respect to the geodesic flow contain a minimal subset. Thisimplies that, if the set Ωg(T 1S) is compact, then it contains minimal sets. Thisgeneral argument does not guarantee the existence of non-periodic minimalsets. We will prove that such sets do exist in Ωg(T 1S). First we consider thecase where Γ = S(g1, g2) and use the doubly-infinite sequences introduced inSect. 2.

Property 3.7. Let S = (Si)i∈Z be an element of Σ. If for any n in N∗, there

exists N(n) > 0 such that for any integer j, the sequence S−n, S−n+1, . . . ,

S0, S1, . . . , Sn is a block of the sequence Sj+1, . . . , Sj+N(n), then T Z(S) is aminimal set for T .

Proof. We are going to show that if S′ belongs to T Z(S), then S belongs toT Z(S′). This will show that T Z(S) is minimal.

Let (pk)k�1 be a sequence in Z for which limk→+∞ T pk(S) = S′. One hasT pk(S) = (Spk+i)i∈Z. Fix n in N

∗. Let N(n) be the integer associated to n.

3 Applications to the general case 105

Since limk→+∞ T pk(S) = S′, there exists kn > 0 such that Spkn+i = S′i for all

1 � i � N(n). It follows that there exists 0 � jn � N(n) − 2n − 1 such thatS′

jn+i = S−n+i−1 for all 1 � i � 2n + 1. Thus Δ(T jn+n+1(S′), S) �√

2/n.This shows that limn→+∞ T jn+n+1(S′) = S. �

It remains to construct a sequence in Σ satisfying Property 3.7. For thispurpose, consider the sequence of words (mn)n�1 defined recursively by

m1 = g1 and mn+1 = mng2mnmng2mn,

where g1, g2 are the generators of S(g1, g2). Notice that in each step, eachword mn begins and ends with the letter g1 and therefore each is a reducedword. Let n denote the length of the word mn. The n initial letters of mn+1

coincide with those of mn. The first letter of mn is also different from theinverse of its last letter. Thus there exists a (unique) doubly-infinite sequenceS = (Si)i∈Z, with Si ∈ A satisfying the conditions

S−�n+1S−�n+2 · · · S0S1 · · · S�n = mnmn.

Exercise 3.8. Prove that S belongs to Σ and is not periodic.

Lemma 3.9. Let n � 1. For any p � n + 1, all blocks of length 6n + 3 ofthe sequence S−�p+1, . . . , S0, S1, . . . , S�p contain the block S−�n , S−�n+1, . . . ,S0, S1, . . . , S�n .

Proof. Fix n � 1. Define Ap = mpmp and proceed by induction on p � n + 1.One has An+1 = mng2Ang2Ang2Ang2mn. The length of the sequences

associated to mn and An being respectively n and 2n, all blocks of length6n + 3 of the sequence associated to An+1 contain the sequence associatedto An, which proves the property for p = n + 1.

Take some p � n + 1. Assume now that the property is true for this p andlet us show that it is true up to p + 1.

Consider Ap+1 = mpg2Apg2Apg2Apg2mp. Choose a block B of length6n + 3 of the sequence associated to Ap+1. If B is a block of Ap or of mp,the induction hypothesis applies. Otherwise B is a block of the sequence as-sociated to one of the following words wi for i = 1, 2, 3 defined as follow:

w1 = mpg2Ap, w2 = Apg2Ap, w3 = Apg2mp.

Such a block must contain the letter g2 written in one of the words above.Let g

2denote this copy of g2. By construction, any wi contains

Mn = Ang2mng2mng2An.

Since mp = mp−1g2Ap−1g2mp−1 and Ap = mpmp, the sequence associatedto the word Mn is a block of length 6n + 3 of the sequence associated tothe words w1, w2, w3. Since block B has length 6n + 3 and B contains g

2,

B contains the sequence associated to An. �


Corollary 3.10. The closed set T Z(S) is minimal and non-periodic.

Proof. We show that Property 3.7(i) is satisfied. Let us fix n ∈ N∗. For

all j ∈ N∗, there exists k � n + 1 such that the finite sequence B =

Sj+1, . . . , Sj+6�n+3 is a block of S−�k+1, . . . , S0, S1, . . . , S�k. Lemma 3.9 im-

plies that the block B contains S−�n , . . . , S0, . . . , S�n , thus in particular itcontains S−n, . . . , S0, . . . , Sn. Furthermore, by Exercise 3.8, S is not periodic.

�

Let us return to the geodesic flow on Ωg(S(g1, g2)\T 1D). Let (z, −→u ) ∈ T 1

D

be such that (u(−∞), u(+∞)) = F (S), where S is given by Corollary 3.10.Notice that, since S is not periodic, neither is π1((z, −→u )).

Exercise 3.11. Let Γ be a Fuchsian group. Prove that a compact trajectoryfor the geodesic flow on Ωg(Γ \T 1

D) is periodic.

Since Ωg(S(g1, g2)\T 1D) is compact and π1((z, −→u )) is not periodic, it fol-

lows from Exercise 3.11 that gR(π1((z, −→u ))) is not closed.

Theorem 3.12. The set gR(π1((z, −→u ))) ⊂ Ωg(S(g1, g2)\T 1D) is a compact

minimal set relative to gR, which is not periodic.

Proof. Let us show that gR(π1((z, −→u )u)) is minimal relative to gR. Takeπ1((z′, −→u ′)) ∈ gR(π1((z, −→u ))). By Lemma 1.5, after replacing (z′, −→u ′) with anelement of Γ (z′, −→u ′), one can assume that there exists a sequence S′ in Σ suchthat (u′(−∞), u′(+∞)) = F (S′). By Proposition 1.6, one has S′ ∈ T Z(S). Theset T Z(S) is minimal and S belongs to T Z(S′), therefore π1((z, −→u )) belongsto gR(π1((z′, −→u ′))). This shows that gR(π1((z, −→u ))) is minimal. �

Consider now a non-elementary Fuchsian group Γ .

Corollary 3.13. The set Ωg(Γ \T 1D) contains compact minimal sets which

are non-periodic for the geodesic flow.

Proof. Choose a Schottky subgroup S(g1, g2) of Γ . Consider a compact min-imal non-periodic set K ⊂ Ωg(S(g1, g2)\T 1

D) given by Theorem 3.12. Theprojection P sends K to a compact subset K ′ of Ωg(Γ \T 1

D) which is invari-ant with respect to the geodesic flow (Property 3.3). It is minimal. To seethis, suppose that K ′ properly contains a compact non-empty K0 which isinvariant with respect to gR. The set P −1(K0) ∩ K would therefore be a com-pact non-empty gR-invariant proper subset of K. This is impossible since Kis minimal. Moreover, K ′ is not minimal, according to Lemma 3.5. �

4 Comments 107

4 Comments

This approach to geodesic flow by way of symbolic dynamics was originallydeveloped for the modular surface [2, 3, 20, 57]. It extends to quotients of thePoincare half-plane by geometrically finite Fuchsian groups [55].

This point of view can be generalized to quotients of pinched Hadamardmanifolds by some cocompact Kleinian groups (method of Markov parti-tions [58]) and by some Schottky groups [21, 44].

The coding relates the ergodic theory of geodesic flow to that of sub-shiftsof finite type and Ruelle-Perron-Frobenius operators. For example, it allowsone to recover the Gauss measure dx/(1 + x) on [0, 1] which is invariant withrespect to the Gauss function t(x) = 1/x − [1/x], from the Liouville measureon PSL(2, Z)\T 1

H which is invariant with respect to the geodesic flow [20, 57].This symbolic method is especially effective for calculating the entropy of

the geodesic flow or enumerating the closed geodesics of a manifold [44, 21].The book by T. Bedford, M. Keane & C. Series [8] is a good reference for thisapproach and its applications. As we saw in Sect. 3, this coding also allows usto construct some minimal sets for the geodesic flow. This construction is dueto Morse and is described in the book by W. Gottschalk and G. Hedlund [34].

The sets that we have constructed are compact. In [22], we present aconstruction of non-trivial non-compact minimal sets, when the surface hascusps.

V

Topological dynamics of the horocycle flow

In this chapter, we analyze the topology of the trajectories of another classicalexample of flow on the quotient of T 1

H by a Fuchsian group: the horocycleflow. Our method is based on a correspondence between the set of horocyclesof H and the space of non-zero vectors in R

2 modulo {± Id}. This vectorialpoint of view allows one to relate the topological dynamics of the linear actionon R

2 of a discrete subgroup Γ of SL(2, R) to that of the horocycle flowon the quotient of T 1

H by the Fuchsian group corresponding to Γ . In thegeometrically finite case, we show that the horocycle flow is less topologicallyturbulent than the geodesic flow (Sect. 4).

Throughout this chapter, we use the definitions and notations associatedwith the dynamics of a flow as originally introduced in Appendix A.

1 Preliminaries

1.1 The horocycle flow on T 1H

To each element (z, −→u ) ∈ T 1H, we associate the horocycle H passing

through z centered at u(+∞). Let β : R → H, be the arclength parametriza-tion of H such that: β(0) = z, and the pair of vectors (dβ/ds(0), −→u ) formsa positive basis for TzH (Fig. V.1). The image of β is called the orientedhorocycle associated to (z, −→u ).

Exercise 1.1. Prove that if z = a+ib and u(+∞) = ∞, then β(s) = a+sb+ibfor all s ∈ R (Fig. V.2).

For fixed t, we introduce the function ht : T 1H → T 1

H defined by

ht((z, −→u )) =(β(t), −→v (t)

),

where −→v (t) is the unit vector in T 1β(t)H for which the pair (dβ/ds(t), −→v (t)) is

a positive orthonormal basis (Fig. V.3).


http://dx.doi.org/10.1007/978-0-85729-073-1_5

110 V Topological dynamics of the horocycle flow

Fig. V.1.

Fig. V.2.

Fig. V.3.

Exercise 1.2. Prove that for all positive isometries g ∈ G and for all realnumbers t, the following equality holds:

g ◦ ht = ht ◦ g.

Exercise 1.3. Prove that ht is a homeomorphism of T 1H equipped with the

metric D.(Hint: using Property I.2.3, Exercises III.1.1 and 1.2, prove that if a sequence((zn, −→un))n�1 in T 1

H converges to (z, −→u ), then (ht((zn, −→un)))n�1 converges toht((z, −→u )).)

For all t and t′ in R, one has

ht+t′ = ht ◦ ht′ .

1 Preliminaries 111

To see this, take (i, −→u ) in T 1H such that u(+∞) = ∞. By Exercises 1.1 and

1.2 one has, for all positive isometries g,

ht+t′ (g((i, −→u ))) = g((z′, −→u ′)) with z′ = i + t + t′ and u′(+∞) = ∞.

Yet (z′, −→u ′) = ht(ht′ ((i, −→u ))). Therefore,

ht+t′ (g((i, −→u ))) = ht(ht′ (g((i, −→u )))).

Since the group of positive isometries of H acts transitively on T 1H (Prop-

erty I.2.3), the above statement is satisfied by all elements of T 1H.

It follows that the map from (R, +) into the group of homeomorphisms ofT 1

H which sends t to ht is a group homomorphism.

Exercise 1.4. Prove that for all t, t′ ∈ R and (z, −→u ) in T 1H, one has

D(ht((z, −→u )), ht′ ((z, −→u ))) = 4 ln(

12(

|t′ − t| +√

|t′ − t|2 + 4))

.

Exercises 1.3 and 1.4 imply that the map h : R × T 1H → T 1

H defined by

h(t, (z, −→u )) = ht((z, −→u ))

is continuous. This map defines a flow (Appendix A) on T 1H, that we call the

horocycle flow .As in the case of the geodesic flow (see Exercise III.1.4), the dynamics of

this flow on T 1H are fairly straightforward.

Exercise 1.5. Prove the following properties:

(i) the set Ωh(T 1H) is empty;

(ii) for all (z, −→u ) in T 1H, the map from R into T 1

H which sends t to ht((z, −→u ))is an embedding.

1.2 A vectorial point of view on the space of trajectories of hR

Let E be the quotient of R2 − {(0, 0)} over { ± Id}. Consider the map (Fig. V.4)

vect : T 1H −→ E

defined by

(z, −→u ) �→ vect((z, −→u ))

=

⎧⎨⎩

±eBu(+∞)(i,z)/2/√

1 + u2(+∞)( u(+∞)

1

)if u(+∞) �= ∞,

±eBu(+∞)(i,z)/2(

10

)if u(+∞) = ∞.

Note that if u(+∞) = ∞ and z = i, then vect((z, −→u )) = ±(

10

). Clearly, we

have


Fig. V.4.

Property 1.6. The map vect is surjective, constant on the trajectories of hR.

The map vect induces an action on E of the group G of positive isometriesof H. The following proposition clarifies the nature of this action. Let usintroduce some notation.

For all g ∈ G, set g(z) = (az + b)/(cz + d), with ad − bc = 1, and let Mg

denote the linear transformation acting on E defined by

∀ ±(

xy

)∈ E = { ± Id} \(R∗)2, Mg

(±

(xy

))= ±

(a bc d

) (xy

).

Proposition 1.7. For all g in G and (z, −→u ) in T 1H, one has

vect(g((z, −→u ))) = Mg(vect((z, −→u ))).

Proof. Consider (i, −→u1) in T 1H defined by u1(+∞) = ∞. Recall that

vect((i, −→u1)) = ±e1, where e1 =(

10

). We first show that for all g in G,

vect(g((i, −→u1))) = Mg(±e1). Let us decompose Mg = ±(

a bc d

)as Mg =

±KAN , where

K =(

cos θ − sin θsin θ cos θ

), A =

(λ 00 λ−1

), λ > 0, N =

(1 t0 1

)

(Proposition I.2.4). Let k and a be the Mobius transformations associatedwith K and A. We have

B∞(i, g−1(i)) = B∞(i, a−1(i)) and B∞(i, a−1(i)) = B∞(i, λ−2i).

Hence B∞(i, g−1(i)) = ln λ−2. Furthermore, ‖M(e1)‖ = λ, which implies that‖vect(g((i, −→u1)))‖ = ‖Mg(±e1)‖. Additionally, g(u1(+∞)) = k(∞) and M(e1)is colinear with K(e1). These two facts prove the desired equality in the casewhere (z, −→u ) = (i, −→u1).

We now show that, for all (z, −→u ) in T 1H,

vect(g((z, −→u ))) = Mg vect((z, −→u )).

2 The horocycle flow on a quotient 113

Since the action of G on T 1H is simply transitive (Property I.2.3), there

exists g′ in G such that g′((i, −→u1)) = (z, −→u ). Therefore, vect(g((z, −→u ))) =vect(gg′((i, −→u1))), and hence

vect(g((z, −→u ))) = Mgg′ (vect((i, −→u1))).

At this point, it is sufficient to observe that

Mgg′ (vect((i, −→u1))) = Mg(Mg′ (vect((i, −→u1))))

and Mg′ (vect((i, −→u1))) = vect(g′((i, −→u1))). �

Exercise 1.8. Prove that the map vect is continuous.(Hint: Use Property I.2.2 and Proposition 1.7.)

The following property provides a further characterization of the horocyclicand parabolic points of the limit set of a Fuchsian group (Sect. I.3.2).

Property 1.9. Let (z, −→u ) be in T 1H and (gn)n�1 be a sequence in G.

(i) The sequence (Bu(+∞)(i, g−1n (i)))n�1 tends to +∞ if and only if

(‖Mgn(vect((z, −→u )))‖)n�1 converges to 0.(ii) The point u(+∞) is fixed by a parabolic isometry g ∈ G − {Id} if and only

if Mg(vect((z, −→u ))) = vect((z, −→u )).


2 The horocycle flow on a quotient

Let us consider a Fuchsian group Γ . We will retain the notation introduced inSect. III.1.2. The commutativity of hR and G proved in Exercise 1.2, allowsone to define the horocycle flow hR on the quotient T 1S = Γ \T 1

H (Fig. V.5):for all (z, −→u ) in T 1

H, we set

ht(π1((z, −→u ))) = π1(ht((z, −→u ))).

By definition of the topology of T 1S, a sequence (π1((zn, −→un)))n�1 con-verges to π1((z, −→u )) if and only if there exists a sequence (γn)n�1 in Γ suchthat limn→+∞ γn(zn) = z and limn→+∞ γn(un(+∞)) = u(+∞).

Thus (htn(π1((z, −→u ))))n�1 converges to π1((z′, −→u ′)) if and only if thereexists (γn)n�1 in Γ such that

limn→+∞

γnzn = z′ and limn→+∞

γn(u(+∞)) = u′(+∞),

where zn is the projection on H of htn((z, −→u )).


Fig. V.5. Γ = PSL(2, Z)

2.1 A vectorial point of view on hR

Let MΓ denote the subgroup of { ± Id} \ SL(2, R) consisting of all Mγ with γin Γ . This group of linear transformations is isomorphic to Γ . The followingproposition relates its dynamics on E to those of hR on T 1S.

Proposition 2.1. Let (z, −→u ) and (z′, −→u ′) be in T 1H. There exists a

sequence (tn)n�1 in R such that (htn(π1((z, −→u ))))n�1 converges toπ1((z′, −→u ′)) if and only if there exists a sequence (Mγn)n�1 in MΓ suchthat (Mγn(vect((z, −→u ))))n�1 converges to vect((z′, −→u ′)).

Proof. Assume that there exists (γn)n�1 in Γ for which (γnhtn((z, −→u )))n�1

converges to (z′, −→u ′). Its image by the map vect converges to vect((z′, −→u ′)),since vect is continuous. We have

vect(γnhtn((z, −→u ))) = Mγn(vect((z, −→u ))),

hence (Mγn(vect((z, −→u ))))n�1 converges to vect((z′, −→u ′)).Conversely, assume that (Mγn(vect((z, −→u ))))n�1 converges to

vect((z′, −→u ′)). By definition of the map vect, the sequences (γn(u(+∞)))n�1

and (Bγn(u(+∞))(i, γn(z)))n�1 converge to u′(+∞) and Bu′(+∞)(i, z′) re-spectively. Consider the real number tn such that htn(γn((z, −→u ))) is tangentto the geodesic passing through z′ having γn(u(+∞)) as an endpoint.Set htn(γn((z, −→u ))) = (zn, −→un). Since un(+∞) = γn(u(+∞)), the se-quence (un(+∞))n�1 converges to u′(+∞). Moreover Bγn(u(+∞))(i, γn(z)) =Bun(+∞)(i, zn), hence (zn)n�1 converges to z′. One thus obtains

limn→+∞

htn(π1((z, −→u ))) = π1((z′, −→u ′)). �

Corollary 2.2. Let (z, −→u ) be in T 1H. The trajectory hR(π1((z, −→u ))) is closed

in T 1S if and only if the orbit of the vector vect((z, −→u )) with respect to thegroup MΓ is closed in E.

Exercise 2.3. Prove Corollary 2.2.

2 The horocycle flow on a quotient 115

2.2 Characterization of the non-wandering set

Let us introduce the subset of E defined by

E(Γ ) = {vect((z, −→u )) | (z, −→u ) ∈ T 1H, u(+∞) ∈ L(Γ )}.

Exercise 2.4. Prove that the set E(Γ ) is closed in E, and invariant withrespect to the action of the group MΓ .

If L(Γ ) = H(∞), in particular if Γ is the modular group, then E(Γ ) = E.Otherwise, E(Γ ) is a proper subset of E. For example, if Γ is a Schottkygroup, then E(Γ ) is homeomorphic to the product of R

∗+ with a Cantor set.

Let PE (Γ ) denote the image of E(Γ ) by the projection from E to theprojective line RP

1.

Exercise 2.5. Prove that the set PE (Γ ) ⊂ RP1 is closed and invariant with

respect to the projective action of MΓ on RP1.

Prove that this action, when conjugated by a homeomorphism, is identicalto the action of Γ on L(Γ ).(Hint: use Exercise I.1.14.)

It follows that, if Γ is not elementary, then every projective orbits of MΓ

on PE (Γ ) is dense; in other words, PE (Γ ) is a minimal set for this action.Is this property also satisfied by the action of MΓ on E(Γ )?At least, in the case where Γ is the modular group, the answer is “No”

since E(Γ ) = {± Id} \R2 and the quotient of Z

2 − {(0, 0)} over {± Id} is aclosed subset of E, invariant under PSL(2, Z).

In the next section, we will specify a necessary and sufficient conditionon Γ to allow the answer to this question to be “Yes” (Proposition 4.3(ii)).In general, only the existence of dense orbits is guaranteed.

Proposition 2.6. Let Γ be a non-elementary Fuchsian group. There exists±w ∈ E(Γ ) such that MΓ (w) = E(Γ ).

To prove this proposition, we use the vector space R2 and consider the

subgroup MΓ of SL(2, R) which is the pre-image of MΓ with respect to theprojection of SL(2, R) onto PSL(2, R). This group acts on the set E(Γ ) ⊂R

2 − {(0, 0)} which is the pre-image of E(Γ ) with respect to the projectionof R

2 − {(0, 0)} onto E.Let us begin by proving the following lemma:

Lemma 2.7. Let B1 and B2 be two open disks in R2− {(0, 0)} which have non-

empty intersection with E(Γ ). There exists M in MΓ such that MB1∩B2 �= ∅.

Proof. For i = 1, 2, we let Ci denote the positive open cone generated by Bi.The disk B1 intersects E(Γ ), and the projective action of MΓ on PE (Γ ) isminimal (since Γ is non-elementary and according to Exercise 2.5), hence there


exists an eigenvector u+1 in B1 associated to M1 ∈ MΓ which is associated

with a hyperbolic isometry γ1 in Γ . One can assume that M1(u+1 ) = λ1u

+1 ,

where λ1 > 1. Let γ be a hyperbolic isometry in Γ not having any fixedpoints in common with those of γ1. Choose M ∈ MΓ which projects to thematrix associated with γ, and two eigenvectors u+ and u− of M such thatM(u+) = λu+, M(u+) = λ−1u−, where |λ| > 1. After replacing M1 by Mn

1

and u+, u− by some vectors in R∗u+, R∗u−, one can assume that M1(u+)

and M1(u−) belong to B1. Let M2 be an element of MΓ which projects tothe matrix associated with a hyperbolic isometry in Γ whose eigen directionsare distinct from those of M1 and whose attractive eigenvectors belong to C2.After replacing M2 by ±Mn

2 , one can assume that M2(u+) belongs to C2. Thesegment [M2M

n(u−), M2Mn(u+)] converges to the open ray Δ beginning at 0

in the direction of M2(u+), as n tends to +∞. This ray is also the limit of theimages of the maps M2M

nM −11 restricted to the segment [M1(u−), M1(u+)].

This limit is contained in B1. Since M2(u+) belongs to C2, the ray Δ intersectsthe open set B2 in a segment of non-zero length (i.e., not a point). Therefore,there exist w in B1 and n such that M2M

nM −11 (w) is in B2. �

Proof (of Proposition 2.6). We shall prove that there exists x in R2 − {(0, 0)}

such that MΓ (x) = E(Γ ). The idea is the same as that used to prove Theo-rem III.4.2.

Let (Bn)n�1 be a sequence of open disks of R2 − {(0, 0)} each having non-

empty intersection with E(Γ ), and such that any open subset of R2 − {(0, 0)}

intersecting E(Γ ) contains a disk in this sequence. Fix such an open set O. ByLemma 2.7, there exists M1 in MΓ such that M1(O) ∩ B1 �= ∅. Let K1 be anopen, pre-compact set intersecting E(Γ ), contained in O, such that M1(K1)is in B1. In the preceding argument, replacing O with K1 and B1 by B2, oneobtains M2 ∈ MΓ and an open relatively compact set K2 ⊂ K1 intersectingE(Γ ) such that M2K2 ⊂ B2. In this way, one obtains two sequences (Mn)n�1

in MΓ and (Kn)n�1 of nested, pre-compact open sets intersecting E(Γ ) suchthat Mn(Kn) ⊂ Bn. Let x be in

⋂+∞n=1 Kn ∩ E(Γ ). For all n � 1, the point

Mn(x) is in Bn. Consider an element x′ in E(Γ ) and a sequence (Dn)n�1 ofdisks centered at x′ whose radius converges to 0. Each Dn contains a disk Bin

thus Min(x) is in Dn. This shows that x′ belongs to MΓ (x) and thus thatMΓ (x) is dense in E(Γ ). �

Let us return to the horocycle flow and introduce the subset F (Γ ) in T 1H,

defined byF (Γ ) = vect−1(E(Γ )).

This set is also defined by

F (Γ ) = {(z, −→u ) ∈ T 1H | u(+∞) ∈ L(Γ )}.

3 Dense and periodic trajectories 117

Exercise 2.8. Prove that F (Γ ) is closed and invariant with respect to Γ

and hR.

Let F (Γ ) denote the image of F (Γ ) by the projection π1 to T 1S. This setis closed and invariant with respect to the flow hR. Furthermore, it is a propersubset of T 1S if and only if L(Γ ) �= H(∞).

Exercise 2.9. Prove that F (Γ ) is compact if and only if S is compact.

Notice that Propositions 2.1 and 2.6 together imply the following propo-sition.

Proposition 2.10. There exists (z, −→u ) ∈ F (Γ ) such that hR(π1((z, −→u ))) =F (Γ ).

Can F (Γ ) be characterized by some property of the horocycle flow? Thisquestion is answered by the following proposition.

Proposition 2.11. The set F (Γ ) is the non-wandering set, Ωh(T 1S), of thehorocycle flow on T 1S.

Proof. Assume that π1((z, −→u )) is non-wandering (see Appendix A). There isa sequence (Vn)n�1 of nested neighborhoods of π1((z, −→u )) and a sequenceof positive real numbers (tn)n�1 ⊂ R

+ such that limn→+∞ tn = +∞ andhtn(Vn) ∩ Vn �= ∅. Consider a sequence ((zn, −→un))n�1 such that

π1((zn, −→un)) ∈ Vn, limn→+∞

π1((zn, −→un)) = π1((z, −→u )),

limn→+∞

(zn, −→un) = (z, −→u ) and

limn→+∞

htn(π1((zn, −→un))) = π1((z, −→u )).

The last of these limits when considered on T 1H implies the existence of

a sequence (γn)n�1 in Γ such that limn→+∞ htnγn((zn, −→un)) = (z, −→u ). Sethtn((zn, −→un)) = (z′

n,−→u′

n). We have

limn→+∞

z′n = u(+∞) and lim

n→+∞d(z′

n, γ−1n (z)) = 0.

This implies that limn→+∞ γ−1n (z) = u(+∞) and hence that u(+∞) is in

L(Γ ).Consider an element in F (Γ ) whose horocyclic trajectory is dense in F (Γ ).

By Proposition 2.10, such a point exists. Moreover such point is non-wanderingwith respect to the horocycle flow. Since the non-wandering set is closed andinvariant with respect to the horocycle flow, it contains F (Γ ). �

3 Dense and periodic trajectories

Assume that Γ is a non-elementary Fuchsian group. Our motivation is tocharacterize the topology of a trajectory hR(π1((z, −→u ))) included in Ωh(T 1S)in terms of the properties of the point u(+∞).


3.1 Dense horocyclic trajectories

If Γ is not elementary, Propositions 2.11 and 2.10 together imply that the setΩh(T 1S) contains some dense horocyclic trajectories.

Theorem 3.1. Let (z, −→u ) be in T 1H. The following are equivalent:

(i) the point u(+∞) is a horocyclic point in L(Γ );(ii) the orbit hR(π1((z, −→u ))) is dense in Ωh(T 1S).

To prove this theorem, we use the vectorial point of view and the fol-lowing lemma, which follows from the definition of a horocyclic point andProperty 1.9(i).

Lemma 3.2. Let (z, −→u ) ∈ T 1H. The point u(+∞) is horocyclic

if and only if there exists a sequence (Mn)n�1 in MΓ such thatlimn→+∞ ‖Mn(vect((z, −→u )))‖ = 0.

Proof (of Theorem 3.1). To prove Theorem 3.1, it is sufficient (by Proposi-tion 2.1 and Lemma 3.2) to prove that the following statements are equivalent:

(i′) There exists a sequence (Mn)n�1 in MΓ such that

limn→+∞

‖Mn(vect((z, −→u )))‖ = 0.

(ii′) MΓ vect((z, −→u )) = E(Γ ).

It is clear that (ii′) implies (i′).Let us prove (i′) ⇒ (ii′). We begin with the case in which u(+∞) is

fixed by a hyperbolic isometry γ ∈ Γ . Let M be the element of MΓ as-sociated with γ. We can assume that M(vect((z, −→u ))) = ±λ vect((z, −→u )),with 0 < λ < 1. By Proposition 2.6, there exists (z′, −→u ′) in T 1

H such thatMΓ vect((z′, −→u ′)) = E(Γ ). If we prove that MΓ vect((z, −→u )) contains an ele-ment of ±R

∗ vect((z′, −→u ′)), then MΓ vect((z, −→u )) = E(Γ ).The point u′(+∞) belongs to L(Γ ), which is minimal, thus there ex-

ists (γn)n�1 in Γ such that limn→+∞ γn(u(+∞)) = u′(+∞). Let Mn de-note the element of MΓ associated with γn and (pn)n�1 be a sequence in Z

such that (λpn ‖Mn(vect((z, −→u )))‖)n�1 converges to a real number α �= 0.Since MnMpn(vect((z, −→u ))) = ±λpnMn(vect((z, −→u ))) and γnγpnu(+∞) =γnu(+∞), one has

limn→+∞

MnMpn vect((z, −→u )) = ±αvect((z′, −→u ′))

‖ vect((z′, −→u ′))‖ .

Assume now that u(+∞) is horocyclic and is not fixed by any hyperbolicisometry in Γ . Let us show that MΓ vect((z, −→u )) contains an eigenvector,modulo ±1, of an element of MΓ associated to a hyperbolic isometry in Γ .


This will prove that MΓ vect((z, −→u )) is dense in E(Γ ). Let γ be a hyper-bolic isometry in Γ . Denote M the element of MΓ associated with γ. Thesequence (γ−nu(+∞))n�1 converges to γ−. Consider a sequence (Mn)n�1 inMΓ such that limn→+∞ ‖Mn(vect((z, −→u )))‖ = 0. Take a vector w in (R∗)2

which projects to vect((z, −→u )) and M, Mn in SL(2, R) which project to M

and Mn respectively. Let w+, w− be eigenvectors of M such that

Mw+ = λw+ and Mw− =1λ

w−,

with |λ| > 1. Set Mn(w) = anw+ + bnw−. Then

limn→+∞

(a2n + b2

n) = 0.

Furthermore,M −nMn(w) = (an/λn)w+ + bnλnw−,

hence, after passing to a subsequence, one can assume that (M −nMn(w))n�1

converges to βw− with β �= 0. It follows that (M −nMn(vect((z, −→u ))))n�1

converges to an eigenvector, modulo ±1, of a matrix associated to a hyperbolicisometry of Γ . �

As an application of this theorem, we prove Theorem III.4.3 about thetopological mixing property of the geodesic flow on Ωg(T 1S).

3.2 Relationship between hR and gR, and application

The horocycle flow is closely related to geodesic flow. Namely we have

Property 3.3. Let (z, −→u ) be in T 1H and let s, t be in R. One has

gt ◦ hs ◦ g−t((z, −→u )) = hse−t((z, −→u )).

Proof. Because the action of the group G of positive isometries of H commuteswith those of gR and hR, and because this group acts transitively on T 1

H, itis sufficient to prove this relationship when z = i and u(+∞) = ∞.

In this case, the image by g−t of (i, −→u ) is (e−ti, −→v ), with v(+∞) = ∞. Itfollows that hsg−t(i, −→u ) = (e−ti + se−t, −→u ′), with u′(+∞) = ∞ (Fig. V.6).Hence

gt(hs(g−t((i, −→u )))) = (i + e−ts,−→u′ ′),

with u′ ′(+∞) = ∞.We conclude using the fact that hse−t((i, −→u )) = (i + e−ts,

−→u′ ′). �

As application of this relationship, we use properties of the horocyclic flowon T 1S to obtain Theorem III.4.3 relative to the behavior of the geodesic flow.


Fig. V.6.

Proof of Theorem III.4.3

Recall the statement of this theorem:Let O and V be open and non-empty subsets of Ωg(T 1S). There exists

T > 0 such that for all t � T , gt(O) ∩ V �= ∅.

Proof. We prove this statement by contradiction. Assume that there are twonon-empty open subsets O and V of Ωg(T 1S) and an unbounded sequence(tn)n�1 such that O ∩ gtn(V ) = ∅. We can suppose that limn→+∞ tn = −∞.By Theorem III.3.3, there exists π1((z, −→u )) in V which is periodic with respectto gR. Let T denote its period and define tn = rnT + sn where −rn ∈ N

and −T < sn � 0. After passing to a subsequence, one can assume that(sn)n�1 converges to a real number s. The point u(+∞) is horocyclic. Thusby Theorem 3.1, one has hR(π1((z, −→u ))) = Ωh(T 1S). By Proposition 2.11and Theorem III.2.1, we have Ωg(T 1S) ⊂ Ωh(T 1S). Hence there exists tin R such that ht(π1((z, −→u ))) belongs to g−s(O). Consider the hyperbolicisometry γ in Γ such that u(+∞) = γ+ and �(γ) = T . One has γn((z, −→u )) =gnT ((z, −→u )). Using this relationship and Property 3.3, one obtains the equalityγ−rn g−rnT (ht((z, −→u ))) = hternT ◦ g−2rnT ((z, −→u )). This equality implies that(g−rnT (ht(π1((z, −→u ))))n�1 converges to π1((z, −→u )) and thus that for largeenough n, gs+rnT V ∩ O �= ∅. Since limn→+∞(sn − s) = 0, it follows that forlarge enough n, we have gtnV ∩O �= ∅. This contradicts our initial hypothesis.

�

3.3 Periodic horocyclic trajectories and their periods

We now focus on the existence of periodic elements of the flow hR on T 1S.By definition, if π1((z, −→u )) is periodic, there exist T > 0 and γ ∈ Γ such that

hT ((z, −→u )) = γ((z, −→u )).

Proposition 3.4. Let (z, −→u ) be in T 1H and let γ be a positive isometry of H.

The following are equivalent:

(i) there exists T > 0 such that hT ((z, −→u )) = γ((z, −→u ));(ii) the isometry γ is parabolic and fixes u(+∞).


Proof.(i) ⇒ (ii). Suppose that there exist T > 0 and γ such that hT ((z, −→u )) =

γ((z, −→u )). For all n in Z∗, one has hnT ((z, −→u )) = γn((z, −→u )). Hence γ fixes

u(+∞). Furthermore γ �= Id and γ is not hyperbolic since limn→+∞ γ−n(z) =limn→+∞ γn(z). Therefore γ is parabolic.

(ii) ⇒ (i). Let γ be a parabolic isometry and let (z, −→u ) be an elementof T 1

H satisfying γ(u(+∞)) = u(+∞). This isometry preserves the orientedhorocycle associated with (z, −→u ) (see Sect. I.2.2). Hence there exists T > 0such that γ((z, −→u )) = hT ((z, −→u )) (Fig. V.7). �

Fig. V.7.

Clearly, one obtains:

Corollary 3.5.

(i) The element π1((z, −→u )) is periodic with respect to the horocycle flow onΩh(T 1S) if and only if u(+∞) is fixed by a parabolic isometry of Γ .

(ii) The set Ωh(T 1S) contains periodic trajectories if and only if the group Γcontains parabolic isometries.

For example, if Γ is a Schottky group generated by two hyperbolic isome-tries, then Ωh(T 1S) does not contain any periodic trajectory. On the otherhand, such trajectories do exist if Γ is the modular group.

Note that, if π1((z, −→u )) is periodic with respect to hR, then π1(gt((z, −→u )))is likewise periodic for all t ∈ R. It follows that, if it is not empty, the set ofperiodic trajectories of hR contains a subset which is in one-to-one correspon-dence with R.

As in the case of the geodesic flow, if π1((z, −→u )) is a periodic element withrespect to the flow hR, we write Tu to denote its period. Let γ be the parabolicisometry of Γ satisfying

hTu((z, −→u )) = γ((z, −→u )).

Does γ determine the period Tu? Unlike the case of the geodesic flow (seeSect. III.2), the answer is “No.” To see this, suppose that u(+∞) = ∞. Under


this hypothesis, γ is necessarily a translation t(z) = z + a and therefore

Tu =|a|

Im z.

This equality shows that Tu does not depend only on the translation t, andthat the set of all periods of periodic elements of Ωh(T 1S) is R

∗+.

Proposition 3.6. If the set of periodic elements of the horocycle flow is non-empty, then it is dense in Ωh(T 1S).

Proof. Suppose that there exists a periodic element π1((z0,−→u0)) in

T 1S. Let π1((z, −→u )) be in Ωh(T 1S). Since L(Γ ) is a minimal setwith respect to the action of Γ , there exists (γn)n�1 in Γ such thatlimn→+∞ γn(u0(+∞)) = u(+∞). Consider the sequence ((z, −→un))n�1 in T 1

H

defined by un(+∞) = γn(u0(+∞)). One has limn→+∞(z, −→un) = (z, −→u ), hencelimn→+∞ π1((z, −→un)) = π1((z, −→u )). Furthermore, the point γn(u(+∞)) isparabolic, thus π1((z, −→un)) is periodic. �

4 Characterization of geometrically finite Fuchsiangroups

Recall that a Fuchsian group is geometrically finite if and only if every pointin its limit set is either horocyclic or parabolic (see Theorem I.4.13).

The geometric finiteness of a group can be characterized using the topo-logical dynamics of hR. Actually, the following theorem stems directly fromTheorem 3.1 and from Proposition 3.5.

Theorem 4.1. A non-elementary Fuchsian group is geometrically finite ifand only if the trajectories of hR restricted to Ωh(T 1S) are dense in Ωh(T 1S)or periodic.

For example, if Γ is a Schottky group generated by two hyperbolic isome-tries, all trajectories of hR are dense in Ωh(T 1S). If Γ is the modular group,the horocyclic trajectories are dense in T 1S or they are periodic.

Among the geometrically finite Fuchsian groups, recall that convex-cocompact groups are those whose limit sets are entirely composed ofhorocyclic points (Corollary I.4.17). If Γ is such a group, all trajectories of hR

are dense in Ωh(T 1S); in other words, Ωh(T 1S) is a minimal set with respectto the flow hR.

Conversely, if all trajectories of hR are dense in Ωh(T 1S), then all pointsof L(Γ ) are horocyclic and therefore Γ is convex-cocompact.

One can state the following properties:

5 Comments 123

Proposition 4.2. Let Γ be a non-elementary Fuchsian group.

(i) The group Γ is convex-cocompact if and only if Ωh(T 1S) is a minimal setwith respect to the flow hR.

(ii) The group Γ is a uniform lattice if and only if T 1S is a minimal set withrespect to the horocycle flow on T 1S.

Using Proposition 2.1, one can translate Propositions 4.1 and 4.2 intovectorial terms. We thus obtain the following characterization of geometricfiniteness and of convex-cocompactness in terms of the dynamics of the groupMΓ on E(Γ ).

Proposition 4.3. Let Γ be a non-elementary Fuchsian group.

(i) The group Γ is geometrically finite if and only if for all vector v ∈ E(Γ ),either MΓ (v) = E(Γ ) or there exists M ∈ MΓ − {Id} such that Mv = v.

(ii) The group Γ is convex-cocompact if and only if MΓ (v) = E(Γ ) for allv ∈ E(Γ ).

(iii) The group Γ is a uniform lattice if and only if MΓ (v) = E for all v ∈ E.

In the case of a Schottky group generated by two hyperbolic isometries,every orbit of MΓ has dense restriction to E(Γ ). If Γ is the modular group,one deduces from Proposition 4.3 the following result:

Property 4.4. Let( x

y

)∈ R

2. If y �= 0 and if x/y /∈ Q, then SL(2, Z)( x

y

)=R

2.


In the case of the modular group, if a vector w in E is fixed by a non-trivial element of MΓ then w = λ

( pq

)for some (p, q) ∈ Z × Z. Thus MΓ (v) is

a discrete set in E. This is a general phenomenon as described in the followingexercise.

Exercise 4.6. Let Γ be a non-elementary Fuchsian group and let w in E(Γ )be fixed by a non-trivial element of MΓ . Prove that the orbit MΓ (v) is adiscrete subset of E(Γ ).(Hint: use Corollary 2.2.)

5 Comments

The horocycle flow is closely related to the geodesic flow. The collective behav-ior of the geodesic trajectories is reflected by the horocycle flow in the sense ofProperty 3.3: two elements π1((z, −→u )) and π1((z′, −→u ′)) in the quotient Γ \T 1

H

belong to the same horocyclic trajectory if and only if the distance betweengt(π1((z, −→u ))) and gt(π1((z′, −→u ′))) converges to 0 when t tends to +∞.

Although the behavior of individual geodesics is very unpredictable asshown in Chaps. III and IV, when Γ is geometrically finite, the collective


behavior of the geodesic flow is regular (Proposition 4.2). If the hypothesis ofgeometric finiteness is not satisfied, the horocyclic flow can be very irregular,as illustrated in examples constructed by M. Kulikov [43]. In these examples,a group Γ is devised for which the horocycle flow on Γ \T 1

H does not admitany minimal set.

The relationship between these two flows allows one, for example, to provethe topological mixing of gR (Theorem III.4.3). This relationship is especiallyuseful when tackling metric questions, the historical example being the proofdue to G. Hedlund [32] in the setting of lattices, of the ergodicity of hR withrespect to the Liouville measure.

In the general case where X is a pinched Hadamard manifold and Γ is anon-elementary Kleinian group acting on X, the notion of horocycle flow onT 1X only makes sense if the dimension of X is equal to 2. If X is not a surface,this notion is replaced by that of the strong stable foliation on T 1X whichgeneralizes the foliation of T 1

H by the trajectories of hR [5, 27, 54]. The pro-jection of the leaves of this foliation to X are the horospheres of X. The partof the non-wandering set of the horocycle flow on Γ \T 1

H is then played byΩh(Γ \T 1X), the set obtained by projecting to Γ \T 1X the leaves of T 1X cor-responding to horospheres centered at points of L(Γ ). In this general setting,the existence of a leaf which is dense in Ωh(Γ \T 1X) is an open question. Thisquestion is equivalent to that of the topological mixing of the geodesic flowon Ωg(Γ \T 1X) as well as that of the density of the length spectrum [19]. Inthe context of the Poincare half plane, we have proved this existence (Corol-lary 2.10) using a vectorial point of view. This approach is found in [16].Adding the assumption that there exists a dense leaf in Ωh(Γ \T 1X), most ofthe results of this chapter can be generalized [5, 19, 27, 54].

As with the geodesic flow, we have not addressed the metric aspect of thehorocycle flow. The texts of E. Ghys [32] and S. Starkov [59] provide a goodintroduction. Let us outline the basics.

Consider the horocycle flow hR on T 1D where D is the Poincare disk. Let 0

be the center of this disk. Each trajectory is identified with a pair (x, s) wherex ∈ D(∞), s ∈ R and s = Bx(0, z), for points z on the horocycle associatedwith this trajectory.

Let Γ be a non-elementary Fuchsian group. This identification allows oneto construct a measure N on T 1

H which is invariant with respect to this flowand with respect to Γ , defined by

N(dx ds) = esδ(Γ )m(dx) ds,

where m is a Patterson measure on L(Γ ) and δ(Γ ) is the critical exponent ofthe Poincare series associated with Γ (see the Comments of Chap. I).

The measure N induces another measure N on Γ \T 1H which is invariant

with respect to the horocycle flow and whose support is Ωh(Γ \T 1H) [5, 54].

If Γ is a lattice, this measure is finite. Otherwise it is infinite. Under theassumption that Γ is geometrically finite, the horocycle flow is ergodic withrespect to this measure [32, 53, 54].

5 Comments 125

In the general case of a pinched Hadamard manifold of arbitrary dimen-sion, the construction of the measure N and its resulting ergodicity can begeneralized, provided that one again assumes the existence of a leaf which isdense in Ωh(Γ \T 1X) [54]. Adding a further assumption of finiteness on thePatterson-Sullivan measures on Γ \T 1X (see the Comments from Chap. III),one obtains a classification of ergodic measures which are invariant with re-spect to the strong stable foliation ([47, 53] and [54, Theorem 6.5]).

VI

The Lorentzian point of view

In the previous chapter (Sect. V.2), we established a correspondence betweenthe dynamics of the horocycle flow on Γ \T 1

H and the dynamics of the lineargroup associated with Γ on { ± Id} \R

2 − {0}.Our motivation in this chapter, is to construct a linear representation of Γ

taking into account simultaneously the dynamics of the horocycle and of thegeodesic flows. Many proofs are reformulations of proofs given in the previouschapters. In this case, they are left to the reader. Appendix B can be usefulin this chapter.

To this end, we work in the space R3 equipped with the Lorentz bilinear

formb(x, x′) = x1x

′1 + x2x

′2 − x3x

′3.

Each real number t is associated with a surface

Ht = {x ∈ R3 | b(x, x) = t}.

If t is strictly negative, Ht is a hyperboloid of two sheets (Fig. VI.1). In this

Fig. VI.1.


http://dx.doi.org/10.1007/978-0-85729-073-1_6

128 VI The Lorentzian point of view

case, one sets

H+t = Ht ∩ (R2 × R

+) and H −t = Ht ∩ (R2 × R

−).

If t is strictly positive, Ht is connected and is a hyperboloid of one sheet(Fig. VI.1).

Finally, H0 is the light cone, an object associated with special relativity(Fig. VI.1). This cone minus the point (0, 0, 0) has two connected components

H+∗0 = H0 ∩ R

2 × R∗+ and H − ∗

0 = H0 ∩ R2 × R

∗−.

The group O(2, 1) of orthogonal transformations of b acts on each sur-face Ht. This group is not connected. Let O0(2, 1) denote the connected com-ponent of O0(2, 1) containing the identity.

Throughout this chapter, ‖x‖ will indicate the Euclidean norm of x in R3.

1 The hyperboloid model

In our present context, the Poincare disk is considered to be the subset of R3

defined byD = {x ∈ R

2 × {0} | ‖x‖ < 1},

equipped with the metric g as defined in Sect. I.1.5.The purpose of this section is to construct a Riemannian structure on the

sheet H+−1, for which H+

−1 is isometric to the Poincare disk, and to understandits geometry from a Lorentzian point of view.

1.1 Construction of the metric and compactification

Consider the stereographic projection P : H+−1 → D, which sends x to the

point P (x) which is the intersection of D with the line passing through x andthrough the point (0, 0, −1) (Fig. VI.2). This map is a diffeomorphism whoseanalytic expression is

P (x1, x2, t) =(

x1

1 + t,

x2

1 + t, 0

).

Let gL denote the metric on H+−1 obtained by pulling back the hyperbolic

metric g on D by P −1. By definition, for x ∈ H+−1 and −→v , −→v ′ ∈ T 1

x H+−1, we

havegL

x (−→v , −→v ′) =4

(1 − ‖P (x)‖2)2〈TxP (−→v ), TxP (−→v ′)〉.

Exercise 1.1. Prove the equality

gLx (−→v , −→v ′) = b(−→v , −→v ′).

1 The hyperboloid model 129

Fig. VI.2.

We deduce from Exercise 1.1, that on each tangent plane of H+−1, the

metric gL corresponds to the restriction of b to that tangent plane. Equippedwith this metric, H+

−1 is isometric to the Poincare disk and therefore to thePoincare half-plane.

Exercise 1.2. Recall that G is the group of positive isometries of (D, g).

(i) Prove that H+−1 is invariant with respect to the group O0(2, 1).

(ii) Prove the equalityP O0(2, 1)P −1 = G.

It follows from Exercise 1.2 that the group O0(2, 1) is the group oforientation-preserving isometries of (H+

−1, gL). The action of this group is

transitive on H+−1 and simply transitive on the unitary tangent bundle T 1H+

−1

of H+−1 (Property I.2.3).

Exercise 1.3. Prove that the geodesics of (H+−1, g

L) correspond to the inter-sections of H+

−1 with planes passing through the point (0, 0, 0) and through apoint of H+

−1 (Fig. VI.3).

Fig. VI.3.

We compactify (H+−1, g

L) by applying P −1 to the compactification of(D, g). To accomplish this, consider the space H+

−1(∞) of lines in the cone H0.The bijection P −1 extends to a bijection, again denoted P −1, of D ∪ D(∞) onH+

−1 ∪ H+−1(∞) defined on D(∞) by

P −1((cos σ, sin σ, 0)) = {(t cosσ, t sin σ, t) | t ∈ R}.


In other words, if x ∈ D(∞), then the line P −1(x) contains the point (0, 0, 0)and is parallel to the line passing through x and (0, 0, −1) (Fig. VI.4).

Fig. VI.4.

The set H+−1(∞) is called the boundary at infinity of H+

−1. Let us equipH+

−1 ∪ H+−1(∞) with the topology induced by P in which a neighborhood

of a point y is the pre-image of a neighborhood of P (y). In this topology,H+

−1 ∪ H+−1(∞) is compact and P is a homeomorphism. More explicitly, the

convergence of an unbounded sequence (yn)n�1 in H+−1 to a line D in H+

−1(∞)corresponds to the Euclidean convergence to D of the sequence of lines passingthrough the origin and yn.

The endpoints of a geodesic in H+−1 correspond to two lines in the cone H0

contained in the plane passing through this geodesic and the point (0, 0, 0)(Fig. VI.5).

Fig. VI.5.

Since the action of G on D extends to D(∞) via an action of homeomor-phisms, the action of O0(2, 1) extended similarly to H+

−1 ∪ H+−1(∞).

Exercise 1.4. Prove that the action of O0(2, 1) on H+−1(∞), which is

P -conjugate to the action of G on D(∞), corresponds to the projective actionof O0(2, 1) on the space of lines in H0.


1.2 Classification of positive isometries and Busemann cocycles

Let f be a non-trivial transformation of O0(2, 1). One says that f is respec-tively elliptic, parabolic or hyperbolic if PfP −1 is (see Sect. I.2). Translatedin terms of isotropic eigenvectors (i.e., b(−→v , −→v )) = 0 and −→v = −→

0 ), this clas-sification boils down to the following property:

• either f does not admit any isotropic eigenvector (f is elliptic);• or f admits exactly one isotropic eigendirection (f is parabolic);• or f admits two distinct isotropic eigendirections (f is hyperbolic).

Let A denote the subgroup of elements of O0(2, 1) globally fixing the linesD0 = R(1, 0, 1) and D1 = R(−1, 0, 1). Let N denote the subgroup of elementsof O0(2, 1) globally fixing D0. Let K denote the subgroup of elements fixingthe point x0 = (0, 0, 1).

For all t ∈ R, set

at =

⎛⎝cosh t 0 sinh t

0 1 0sinh t 0 cosh t

⎞⎠ , nt =

⎛⎝1 − t2/2 t t2/2

−t 1 t−t2/2 t t2/2 + 1

⎞⎠ ,

kt =

⎛⎝cos t − sin t 0

sin t cos t 00 0 1

⎞⎠ .

Exercise 1.5. Prove the following properties

(i) A = {at/t ∈ R}, N = {nt | t ∈ R}, K = {kt | t ∈ [0, 2π)}.(ii) For all f in O0(2, 1), there exist t, t′, t′ ′ and s, s′, s′ ′ such that

f = ktat′ nt′ ′ (see Proposition I.2.4),f = ksas′ ks′ ′ (see Proposition I.2.4).

The notion of the Busemann cocycle makes sense on H+−1. To see this,

let D be an element of H+−1(∞), and let x, y be two points in H+

−1. Considerthe arclength parametrization (R(t))t�0 of the geodesic ray beginning at xand ending at D and set

F (t) = dL(x, R(t)) − dL(y, R(t)),

where dL is the distance function induced on H+−1 by gL (see Sect. I.1).

By construction of the metric gL, one has

F (t) = d(P (x), r(t)) − d(P (y), r(t)),

where (r(t))t�0 is the arclength parametrization of the geodesic ray[P (x), P (D)). It follows that the limit of F as t goes to +∞ exists andone has (see Sect. I.1)

BP (D)(P (x), P (y)) = limt→+∞

F (t).


Thus the Busemann cocycle centered at D, calculated at x and y, is definedas follows:

BD(x, y) = BP (D)(P (x), P (y)).

A horocycle of H+−1 centered at D ∈ H+

−1(∞) is by definition a level setof the function

H+−1 −→ R

x �−→ BD((0, 0, 1), x).

Clearly, the image by P of such a horocycle is a horocycle of D. Moreovera horocycle of H+

−1 centered at D is invariant with respect to the group ofparabolic isometries of O0(2, 1) fixing D.

Exercise 1.6. Prove that a horocycle of H+−1 centered at D passing through

x ∈ H+−1 is the intersection of H+

−1 with the plane passing through x which isparallel to the tangent plane of the cone H0 containing the line D (Fig. VI.6).

Fig. VI.6.

1.3 Lorentz groups and limit sets

By definition, a Lorentz group is a subgroup of O0(2, 1) which is P -conjugateto a Fuchsian group; equivalently, such a group is a discrete subgroup ofO0(2, 1).

Let ΓL be such a group and ΓF its associated Fuchsian group. One has

ΓF = PΓLP −1.

The action of a Lorentz group on H+−1 is properly discontinuous (Prop-

erty I.2.9).When ΓF is infinite, we define its limit set, L(ΓF), as the intersection of

H+−1(∞) with the closure of any orbit ΓF(x), with x ∈ H+

−1. We have

L(ΓF) = P (L(ΓL)).

One says that ΓL is elementary if ΓF is.


As with L(ΓF), the notions of horocyclic, conical and parabolic points canbe defined (see Sect. I.3). The following proposition gives a characterizationof these points in terms of the linear action of ΓL. For f ∈ O0(2, 1), we definethe norm ‖f ‖ = supx �=0 ‖f(x)‖/‖x‖.

Proposition 1.7. Let ΓL be a Lorentz group. Take D ∈ H+−1(∞) and let y be

a direction vector for D.

(i) The line D is horocyclic with respect to ΓL if and only if there exists asequence (γn)n�1 in ΓL such that limn→+∞ ‖γny‖ = 0.

(ii) The line D is conical with respect to ΓL if and only if there exists a se-quence (γn)n�1 in ΓL such that limn→+∞ ‖γ−1

n ‖ = +∞ and such that thesequence (‖γ−1

n ‖ ‖γny‖)n�1 is bounded.

Proof. Set D0 = R( 1

01

), y0 =

( 101

)and x0 =

( 001

).

(i) Let us begin with the case in which D = D0. By Exercise 1.5(ii),a transformation f in O0(2, 1) can be decomposed into ktat′ nt′ ′ with kt ∈ K,at′ ∈ A and nt′ ′ ∈ N . One has BD0(x0, f

−1(x0)) = BD0(x0, a−1t′ (x0)), hence

BD0(x0, f−1(x0)) = −t′. Furthermore, ‖f(y0)‖ =

√2et′

, hence ‖f(y0)‖ =‖y0‖e−BD0 (x0,f −1(x0)).

Therefore, limn→+∞ ‖γn(y0)‖ = 0 if and only if

limn→+∞

BD0(x0, γ−1n (x0)) = +∞.

This proves the equivalence in (i) when y = y0.Suppose now that D is arbitrary in H+

−1(∞). Notice that the groupO0(2, 1) acts transitively on H+∗

0 . Let f be in O0(2, 1) such that y = f(y0) is adirection of D. One has ‖γn(y)‖ = ‖γnf(y0)‖, hence limn→+∞ ‖γn(y)‖ = 0 ifand only if limn→+∞ BD0 (x0, f

−1γ−1n (x0)) = +∞. The equivalence in part (i)

can then be deduced from the relation

(∗) BD0(x0, f−1γ−1

n (x0)) = BD(f(x0), x0) + BD(x0, γ−1n (x0)).

(ii) Again let us begin with the case where D = D0. By Exercise 1.5(iii), atransformation f in O0(2, 1) can be decomposed into ktat′ kt′ ′ with kt, kt′ ′ ∈ Kand at′ ∈ A. One has dL(x0, f(x0)) = dL(x0, at′ (x0)) and dL(x0, at′ (x0))= |t′ |.Furthermore, ‖f −1‖ = e|t′ |, hence ‖f −1‖ = edL(x0,f(x0)). From this remarkand from the proof of (i), one obtains

‖γny0‖ ‖γ−1n ‖ = ‖y0‖e−BD0 (x0,γ−1

n (x0))+dL(x0,γ−1n (x0)).

It follows that the conditions

limn→+∞

‖γ−1n ‖ = +∞ and (‖γny0‖ ‖γ−1

n ‖)n�1 is bounded


are equivalent to the conditions

limn→+∞

dL(x0, γ−1n (x0)) = +∞ and

(−BD0(x0, γ−1n (x0)) + dL(x0, γ

−1n (x0)))n�1 is bounded.

The last two conditions are equivalent to the fact that the point D is conical(Proposition I.3.15).

The case in which D is arbitrary is treated like it was in the proof ofpart (i), replacing D with f(D0) and using the relation (∗). �

2 Lorentzian interpretation of the dynamics of thegeodesic flow

We write gR to denote the geodesic flow on T 1H+−1. By definition, if (x, −→v ) ∈

T 1H+−1 and (v(t))t∈R is the arclength parametrization of the oriented geodesic

associated with (z, −→v ) (see Sect. III.1), one has (Fig. VI.7)

gt′ ((z, −→v )) =(

v(t′),dv

dt(t′)

).

Fig. VI.7.

2.1 Lorentzian point of view on the set of trajectories of gR

Let (z, −→v ) in T 1H+−1 and (v(t))t∈R be the oriented geodesic associated to

(z, −→v ) such that v(0) = z. We denote by D−(v) and D+(v) respectively, thenegative and positive endpoints of (v(t))t∈R.

Exercise 2.1. Prove that if (z, −→v ) = ((0, 0, 1), (1, 0, 0), then D−(v) =R(−1, 0, 1) and D+(v) = R(1, 0, 1).

Let u−(v), u+(v) be the direction vectors of D−(v) and D+(v) satisfying(Fig. VI.8)

‖u−(v)‖ = ‖u+(v)‖ = 1 and u−(v) ∈ H+0 , u+(v) ∈ H+

0 .

2 Dynamics of the geodesic flow 135

Fig. VI.8.

The vector w(v) ∈ H1 is defined as satisfying

b(w(v), w(v)) = 1, b(w(v), u−(v)) = 0, b(w(v), u+(v)) = 0 and

(w(v), u−(v), u+(v)) is a positive basis.

For example if (z, −→v ) = ((0, 0, 1), (1, 0, 0)), then

w(v) = (0, 1, 0).

Exercise 2.2. For all f ∈ O0(2, 1), prove that

w(f(v)) = f(w(v)),

where f(v) represents the geodesic (f(v(t))t∈R, and f(w(v)) represents theimage of w(v) by the linear map f .

Let W denote the map from T 1H+−1 → H1 defined by (Fig. VI.9):

W ((z, −→v )) = w(v).

Fig. VI.9.

The map W satisfies the following properties.


Property 2.3.

(i) The map W : T 1H+−1 → H1 is continuous with respect to the metric DL

on T 1H+−1 (see Exercise I.1.10).

(ii) The map W : T 1H+−1 → H1 is surjective and

W −1(W ((z, −→v ))) = gR((z, −→v )).

Exercise 2.4. Prove Property 2.3.(Hint: for (i), use the fact that the action of O0(2, 1) is simply transitive onT 1H+

−1.)

Therefore, the map W induces a bijection between the set of trajectoriesof gR and the hyperboloid of one sheet H1.

In this model, the action of O0(2, 1) on the set of geodesic trajectoriescorresponds to the linear action of this group on H1.

2.2 Linear action of a Lorentz group on H1 and the dynamics ofthe geodesic flow

Consider now a non-elementary Lorentz group ΓL. Let ΓF denote the Fuchsiangroup associated with ΓL.

We use notations introduced in Chap. III. In our present context, π denotesthe projection from H+

−1 to the surface S = ΓL\H+−1, and π1 denotes the

projection from T 1H+−1 to T 1S = ΓL\T 1H+

−1. Recall that the map P is thestereographic projection from H+

−1 to D.Take f ∈ ΓL, and set PfP −1 = γ. For all (z, −→v ) ∈ T 1H+

−1, we have

f((z, −→v )) = γ((P (z),TPz(−→v ))),

where TPz is the tangent map of P at z. This map induces a homeomor-phism ϕ from T 1S → ΓF\T 1

D defined by

ϕ(ΓL((z, −→v ))) = ΓF((P (z),TPz(−→v ))).

The action of O0(2, 1) on T 1H+−1 commutes with that of gR, thus this flow

induces a flow, denoted gR, on T 1S, called the geodesic flow .By construction, for all (z, −→v ) in T 1S, one has

ϕ(gt((z, −→v )) = gt(ϕ((z, −→v )).

Our purpose now is to use results proved in Chap. III about the geodesic flowto obtain properties of the orbits of ΓL on H1.

The following proposition, which is analogous to Proposition III.1.6, con-nects these two worlds.

2 Dynamics of the geodesic flow 137

Proposition 2.5. Let u1 and u2 be in H1, and let (z1,−→v1) and (z2,

−→v2) be el-ements of T 1H+

−1 such that W ((z1,−→v1)) = u1 and W ((z2,

−→v2)) = u2. Considera sequence (γn)n�1 in ΓL. The following are equivalent:

(i) the sequence (γn(u1))n�1 converges to u2;(ii) there exists a sequence (sn)n�1 in R such that (gsn(γn((z1,

−→v1))))n�1 con-verges to (z2,

−→v2).

Exercise 2.6. Prove Proposition 2.5.(Hint: use Exercises 2.2 and 2.3, and reuse the arguments from the proof ofProposition III.1.6.)

Let us now focus on the closed orbits of ΓL in H1.

Exercise 2.7. Let u ∈ H1. Prove that ΓL(u) is closed if and only if everysequence in ΓL(u) converging in H1 is constant after some initial terms.

Let u1 ∈ H1 and (z1,−→v1) ∈ W −1(u1). In Chap. III, we proved that the

existence of a convergent sequence (gsn(π1((z, −→v ))))n�1, with (sn)n�1 un-bounded, is equivalent to the fact that v(+∞) or v(−∞) is conical (Proposi-tion III.2.8). This result, added to Proposition 2.5 and Exercise 2.7, impliesthat if ΓL(u1) is not closed then D−(v1) or D+(v1) is conical.

The converse is not true, because if π1((z1,−→v1)) is periodic, in other words

if there exists γ ∈ ΓL − {Id} fixing u1, then the trajectory of π1((z1,−→v1)) is

compact, and hence, by Proposition 2.5, the orbit ΓL(u1) is closed.However, if π1((z1,

−→v1)) is not periodic and if either D−(v1) or D+(v1)is conical, then ΓL(u1) is not closed. Indeed, under these hypotheses, thereexist an unbounded sequence (sn)n�1 and a sequence (γn)n�1 in ΓL such that(gsn(γn((z1,

−→v1))))n�1 converges, and such that the sequence of trajectories(gR(γn((z1,

−→v1))))n�1 is not stationary. This implies, by Proposition 2.5 andExercise 2.7, that ΓL(u1) is not closed.

Fig. VI.10.


Let u+1 (respectively u−

1 ) denote the element of H+0 of Euclidean norm 1

satisfying (Fig. VI.10)

D+(v1) = Ru+1 (respectively D−(v1) = Ru−

1 ).

The preceding argument, together with the Lorentzian characterization ofconical points (Proposition 1.7), implies the following result:

Theorem 2.8. Let u1 ∈ H1. The orbit ΓL(u1) is closed if and only if one ofthe following conditions is satisfied:

(i) there exists γ in ΓL − {Id} such that γ(u1) = u1;(ii) for every sequence (γn)n�1 in ΓL satisfying limn→+∞ ‖γ−1

n ‖ = +∞, thesequences (‖γ−1

n ‖ ‖γn(u−1 )‖)n�1 and (‖γ−1

n ‖ ‖γn(u+1 )‖)n�1 are not bounded.

One can deduce from this theorem and from Corollary I.4.17 the followingcharacterization of Lorentz lattices in terms of their action on H1.

Corollary 2.9. A Lorentz group ΓL is a lattice if and only if the only closedorbits of ΓL in H1 are either orbits of vectors which are fixed by a hyperbolicisometry of ΓL, or are orbits of the form ΓL(u), where Ru− and Ru+ areeigenlines of a parabolic isometry in ΓL.

Note that, if L(ΓL) = H+−1(∞), which is the case for example if ΓF is a

Schottky group (see Chap. II) then, by Proposition 1.7 and Theorem 2.8, ifu ∈ H1 and if Ru− and Ru+ do not belong to L(ΓL), then the orbit of u underΓL is closed.

A Lorentz group is geometrically finite if L(ΓL) consists entirely of conicalor parabolic points (see Theorem I.4.13). Corollary 2.9 can be generalized tothis family of groups in the following way.

Corollary 2.10. A Lorentz group ΓL is geometrically finite if and only if theonly closed orbits of ΓL on H1 are either the orbits of vectors fixed by anhyperbolic isometry of ΓL, or are orbits of the form ΓL(u), where Ru− andRu+ are in the set Lp(ΓL) ∪ (H+

−1(∞) − L(ΓL)).

Let us introduce the set H1(ΓL) defined by

H1(ΓL) = {u ∈ H1 | Ru− and Ru+ are in L(ΓL)}.

Exercise 2.11. Prove that H1(ΓL) is closed and invariant with respect to ΓL.

The set H1(ΓL) is related to the non-wandering set of the geodesic flow(Theorem III.2.1) in the following way:

π1(W −1(H1(Γ ))) = Ωg(T 1S).

As we have shown in Chaps. III and IV, the topological nature of a non-closed trajectory of the geodesic flow on Ωg(T 1S) may be very complex. It

3 Dynamics of the horocycle flow 139

follows that, if u ∈ H1(Γ ) and ΓL(u) is not closed, no information about theclosure of ΓL(u) is available without additional hypotheses on u.

However, one can state the following properties which follow directly fromTheorems III.3.3, III.4.2 and from the continuity of the map W :

Property 2.12. Let ΓL be a non-elementary Lorentz group.

(i) The set of vectors in H1 fixed by hyperbolic isometries of ΓL is dense inH1(ΓL).

(ii) Some orbits of ΓL are dense in H1(ΓL).

3 Lorentzian interpretation of the dynamics of thehorocycle flow

We continue to use notations introduced in the preceding section. In Chap. Vwe established a correspondence between the set of trajectories of the horocy-cle flow on T 1

H and { ± Id} \R2 − {0}. The Lorentzian model that we propose

below brings the methods that were used in Chap. V into play. For this reason,many of the proofs are left as exercises.

We will again write hR to denote the horocycle flow on T 1H+−1 defined by

ht′ ((z, −→v )) = (β(t′),−→v′ ),

where (β(t))t∈R is the arclength parametrization of the horocycle centered atD+(v) passing through z for which one has β(0) = z, and for which the orderedpair (dβ/dt(t′),

−→v′ ) is a positive orthonormal basis for Tz H+

−1 (Fig. VI.11).

Fig. VI.11.

3.1 Lorentzian point of view on the set of trajectories of hR

Consider the map V , from T 1H+−1 into the positive half cone H+∗

0 , defined by

V ((z, −→v )) = eBD+(v)(x0,z)/2u+(v),


where x0 = (0, 0, 1) and u+(v) is the unit vector (in the Euclidean norm)belonging to H+

0 and D+(v).For example, if z = (0, 0, 1) and −→v = (1, 0, 0), then V ((z, −→v )) =

(1/√

2)(1, 0, 1) (Fig. VI.12).

Fig. VI.12.

Property 3.1. Let f ∈ O0(2, 1) and (z, −→v ) ∈ T H+−1. The following properties

are satisfied

(i) V (f((z, −→v ))) = f(V ((z, −→v )));(ii) the map V : T 1H+

−1 → H+∗0 is surjective and V −1(V ((z, −→v ))) =

hR((z, −→v ));(iii) the map V : T 1H+

−1 → H+∗0 is continuous.

Exercise 3.2. Prove Property 3.1.(Hint: see Exercises V.1.6, V.1.8 and Proposition V.1.7.)

3.2 Linear action of a Lorentz group on H+∗0 and dynamics of the

horocycle flow

As in Sect. 2.2 of this chapter, we consider a non-elementary Lorentz group ΓL.Since the action of O0(2, 1) commutes with that of hR, this flow induces an-other flow, denoted hR, on T 1S = ΓL\T 1H+

−1. The following proposition al-lows us to establish a relationship between the topological behavior of theorbits of ΓL on H+∗

0 and that of the trajectories of hR.

Proposition 3.3. Let u1, u2 ∈ H+∗0 and (z1,

−→v1), (z2,−→v2) ∈ T 1H+

−1 such thatD+(vi) = Rui for i = 1, 2. Consider a sequence (γn)n�1 in ΓL. The followingare equivalent:

(i) the sequence (γn(u1))n�1 converges to u2;(ii) there exists a sequence (sn)n�1 in R such that limn→+∞ hsn(γn((z1,

−→v1))) =(z2,

−→v2).

Exercise 3.4. Prove Proposition 3.3.(Hint: Rewrite the proof of Proposition V.2.1.)

3 Dynamics of the horocycle flow 141

Consider the set H+0 (ΓL) defined as follow

H+0 (ΓL) = {u ∈ H+∗

0 | Ru ∈ L(ΓL)}.

Exercise 3.5. Prove that H+0 (ΓL) is closed in H+∗

0 and invariant with respectto ΓL.

This set is related to the non-wandering set of the horocycle flow (Propo-sition V.2.11) by the following equality:

π1(V −1(H+0 (ΓL))) = Ωh(T 1S).

In particular, H0(ΓL) contains the isotropic eigenvectors of the parabolicand hyperbolic isometries of ΓL.

Exercise 3.6. Prove that, if u ∈ H+∗0 is fixed by a parabolic isometry of ΓL

or if u does not belong to H+0 (ΓL), then ΓL(u) is closed in H+∗

0 .

In Chap. V, we proved that the trajectory of π1((z, −→v )) is dense inΩh(T 1S) if and only if v(+∞) is horocyclic. This result, together with Propo-sition 1.7 and Lemma 3.3, allows us to state the following proposition.

Proposition 3.7. Let u ∈ H+∗0 . The ΓL-orbit of u is dense in H0(ΓL) if and

only if there exists (γn)n�1 in ΓL such that limn→+∞ γn(u) = 0.

Exercise 3.8. Let u ∈ H+∗0 .

(i) Prove that if ΓL(u) is closed in H+∗0 , then ΓL(u) is closed in H0.

(ii) Prove that ΓL(u) is closed in H+∗0 if and only if every sequence ΓL(u)

converging in H0 is stationary.

If ΓL is a lattice, then H0(ΓL) = H+∗0 . Moreover, if ΓL is uniform, then

H+−1(∞) consists entirely of horocyclic points, and Proposition 3.7 implies

that every orbit of ΓL on H+0 is dense.

More generally, when ΓL is geometrically finite, Proposition V.4.3 trans-lated into the Lorentzian context, becomes the following proposition.

Proposition 3.9.

(i) The group ΓL is geometrically finite and non-elementary if and only iffor all u ∈ H0(ΓL), either ΓL(u) is dense in H+

0 (ΓL) or u is fixed by aparabolic isometry of ΓL.

(ii) The group ΓL is a uniform lattice if and only if every orbit of ΓL on H+0

is dense in H+0 .


4 Comments

Following the work of G. Hedlund [37] and L. Greenberg [4], the study of orbitsof groups acting linearly on a vector space has become a research area in itsown right. One of the first results in this area was the equivalence, for latticesin SL(n, R), between the presence of the zero vector in the closure of a non-trivial orbit and the density of that orbit in R

n [4]. This result has sincebeen extended by J.-P. Conze and Y. Guivarc’h to some discrete subgroups ofSL(n, R) [16]. For n = 2, it has also been proved in Chap. V.

In this chapter (and Sect. V.2), we have introduced a relationship betweenthe linear action of a discrete subgroup ΓL of SO0(2, 1) on R

3 and the dynamicsof the geodesic or horocycle flow on ΓL\T 1H+

−1 (or ΓF\T 1H). This relationship

is based on a change in point of view which consists of interpreting the linearaction of ΓL as an action on the set of trajectories of a flow on T 1H+

−1 (orT 1

H). We have also obtained a correspondence between the topology of linearorbits and that of the trajectories of these flows, which has been used forexample to prove the existence of dense trajectories in the non-wandering setof the horocycle flow (Corollary V.2.10).

In the metric context, there are many applications of this new point ofview ([5, 32] and [60, Chap. II]). One example is the study of the asymptoticbehavior of the number of vectors in H−1 ∩ Z

3 having Euclidean norm � T ,discussed in M. Babillot’s text [5, Sect. 3.2], which reduces to counting thepoints of an orbit of the group SL(3, Z) ∩ SO0(2, 1) in a disk on the surfaceH+

−1 equipped with the metric gL.

VII

Trajectories and Diophantine approximations

In this chapter, our setting is the Poincare half-plane. Consider a non-elementary, geometrically finite Fuchsian group Γ (see Chap. I for the defini-tions) which contains a non-trivial translation. With these hypotheses, the sur-face S = Γ \H admits finitely many cusps (see Sects. I.3 and I.4) (Fig. VII.1).

Fig. VII.1.

As in the previous chapters, we let π denote the projection from H to S.In the first step, we study the excursions of a geodesic ray π([z, x)) into thecusp corresponding to the image of the restriction of π to a horodisk centeredat the point ∞. Our purpose is to relate the frequency of these excursions tothe way in which the real number x is approximated by the Γ -orbit of thepoint ∞.

In the second step, we restrict our attention to the modular group andrediscover, in the spirit of Chap. III, some classical results of the theory ofDiophantine approximations.


http://dx.doi.org/10.1007/978-0-85729-073-1_7

144 VII Trajectories and Diophantine approximations

1 Excursions of a geodesic ray into a cusp

Let us begin by noting that the point ∞ is a parabolic point of the limit setof Γ since this group contains a translation.

For all t > 0, set

Ht = {z ∈ H | Im z = t} and H+t = {z ∈ H | Im z � t}.

The set Ht is the horocycle centered at ∞, which is the level curve for thevalue ln t of the function f(z) = B∞(i, z). The set H+

t is its correspondinghorodisk.

Recall that there exists t0 > 0 such that for all γ in Γ − Γ∞, one has

γHt0 ∩ Ht0 = ∅ (Theorem I.3.17).

With this condition, the projection from Γ∞ \H+t0 to the cusp π(H+

t0) is injec-tive (see the end of Sect. I.3) (Fig. VII.2).

Fig. VII.2.

Let us now fix such a horodisk H+t0 and a point z ∈ H. Take x ∈ H(∞)

and denote (r(s))s�0 the arclength parametrization of the ray [z, x).Does the geodesic ray π([z, x)) visit the cusp π(H+

t0)? As we will see, theanswer depends on properties of the point x.

Suppose first that, for some T � 0, the ray π([r(T ), x)) is in π(H+t0). This

implies that the ray [r(T ), x) is covered by the family γ(H+t0) with γ ∈ Γ −Γ∞.

Since these horodisks are disjoint, this ray is contained in only one of thesehorodisks, which implies that x = γ(∞) for some γ ∈ Γ .

Conversely if x = γ(∞), then there exists T � 0 such that π([r(T ), x)) isin π(H+

t0), by Theorem III.2.11.Thus one obtains:

Proposition 1.1. There exists T � 0 such that π([r(T ), x)) is contained inπ(H+

t0) if and only if x belongs to the Γ -orbit of the point ∞.

Furthermore, by Theorem III.2.11, if π([r(T ), x)) is in π(H+t0), then the

map from [T, ∞) to π(H+t0) which sends a real number s to π(r(s)) is an

embedding (Fig. VII.3).

1 Excursions of a geodesic ray into a cusp 145

Fig. VII.3. x ∈ Γ (∞)

Suppose now that the point x is not in Γ (∞). According to Proposition 1.1,the ray π([z, x)) is not contained in the horodisk π(H+

t0). As consequence, weobtain

Corollary 1.2. Suppose that x ∈ H(∞) does not belong to Γ (∞). Let t > 0be such that γ(Ht) ∩ Ht = ∅ for all γ ∈ Γ − Γ∞. If there exists s � 0 suchthat π(r(s)) ∈ π(H+

t ), then there exists s′ > s such that π(r(s′)) ∈ π(Ht)(Fig. VII.4).

Fig. VII.4. x /∈ Γ (∞) and π(r(s)) ∈ π(H+t )

Corollary 1.3. Let t0 > 0, if x does not belong to Γ (∞) and is not conical,then there exists T � 0 such that π([r(T ), x)) ∩ π(H+

t0) = ∅.

Proof. Since x does not belong to Γ (∞), then, by Proposition 1.1, two casesarise:

(i) either there exists an unbounded sequence (sn)n�1 such that π(r(sn)) ∈π(H+

t0);(ii) or there exists T � 0 such that π([r(T ), x)) ∩ π(H+

t0) = ∅.


Let us prove that the case (i) implies that x is conical. Actually, by Corol-lary 1.2, there exists an unbounded sequence (s′

n)n�1 such that π(r(s′n)) ∈

π(Ht0). Since the set π(Ht0) is compact, then, by Proposition III.2.8 (on Sinstead of Γ \T 1

H), x is conical. �

Let us analyze the topology of a ray π([z, x)). Since the group Γ is geomet-rically finite, if x is not conical then either x is parabolic or x does not belongto L(Γ ). In the first case, by the preceding argument, there exists T � 0 suchthat the map from [T, ∞) into a cusp π(H+

t (x)) associated with x, whichmaps the real number s to π(r(s)), is an embedding (Fig. VII.5).

Fig. VII.5. x /∈ Γ (∞) and x is parabolic

The second case is the subject of the following exercise.

Exercise 1.4. Prove that if x does not belong to L(Γ ), then there existsT � 0 such that the map from [T, +∞) into S which sends a real number sto π(r(s)) is an embedding (Fig. VII.6).(Hint: see Property III.2.10.)

Fig. VII.6. x /∈ L(Γ )

1 Excursions of a geodesic ray into a cusp 147

It now remains to address the case where x is conical. Note that if x isthe fixed point of a hyperbolic isometry in Γ and if z is on the axis of thatisometry, then π([z, x)) is a compact geodesic in S and thus, for sufficientlylarge t0, this ray does not intersect the cusp π(H+

t0) (Fig. VII.7).

Fig. VII.7. π([z, x)) compact geodesic

On the other hand, we will see for example when Γ is the modular group,that there exist irrational numbers x such that π([z, x)) is not bounded. In thiscase, there exists an unbounded sequence (sn)n�0 for which π(r(sn)) belongsto π(H+

t0) (Fig. VII.8).

Fig. VII.8. Γ = PSL(2, Z)

Thus the fact that x ∈ H(∞) is conical cannot be characterized solely interms of excursions of the ray π([z, x)) into π(H+

t0). It is necessary to considerthe family of horodisks π(H+

t ) with t ∈ R∗+.

We associate to each x ∈ H the set E([z, x)) ⊂ R∗+, of t > 0 for which

there exists an unbounded sequence (sn)n�1 satisfying

π(r(sn)) ∈ π(Ht).


If x ∈ Γ (∞) or if x /∈ Γ (∞) and is not conical, then E([z, x)) = ∅ (Proposi-tion 1.3). This is not the case if x is conical.

Proposition 1.5.

(i) There exists t1 > 0 such that for every conical point x in L(Γ ), one hast1 ∈ E([z, x)).

(ii) If x is a conical point in L(Γ ), the upper bound in R+ ∪ {+∞} of the set

E([z, x)) is independent of z.

Proof.

(i) We suppose that x is conical. The group Γ is geometrically finite and non-elementary, therefore by Proposition III.2.11 on S (rather than T 1S), thereexists a compact subset K1 ⊂ S (independent of x) and an unboundedsequence (sn)n�1 such that π(r(sn)) belongs to K1. Let us lift K1 to acompact set K1 included in a horodisk H+

t1 . By Property 1.2, there existss′

n � sn satisfying π(r(s′n)) ∈ π(Ht1). This shows that t1 belongs to

E([z, x)).(ii) Let x be a conical point in L(Γ ). Take z′ �= z in H, and denote by (r′(s))s�0

the arclength parametrization of the ray [z′, x). Notice that for all ε > 0,there exists T � 0 such that, [r′(T ), x) is in the ε-neighborhood of the ray[z, x).

Fix a real number t in E([z, x)). There exists a sequence (γn)n�1 in Γ andan unbounded sequence (sn)n�1 such that γn(r(sn)) belongs to Ht, in otherwords

B∞(i, γn(r(sn))) = t.

Let ε > 0 and let (s′n)n�1 be a sequence satisfying d(r′(s′

n), r(sn)) � ε. UsingProperty I.1.19 of the Busemann cocycle, one obtains the following statement:

t − ε � B∞(i, γn(r′(sn))) � t + ε.

It follows that π(r′(sn)) belongs to π(H+t−ε). Thus there exists s′ ′

n � s′n such

that π(r′(s′ ′n)) belongs to π(Ht−ε), which shows that t−ε belongs to E([z′, x)).

In conclusion, the upper bound of this set is at least that of E([z, x)). Changingthe roles of z and z′ completes the proof. �

Definition 1.6. Let x ∈ L(Γ ) be a conical point. The upper bound of the setE([z, x)) is called the height of the ray π([z, x)) and is written h(x).Moreover, x is said to be geometrically badly approximated if its height h(x)is finite.

If π([z, x)) is bounded, clearly x is geometrically badly approximated. Isthe converse true? In the following section, we give an answer to this question.

2 Geometrically badly approximated points 149

2 Geometrically badly approximated points

Consider a conical point x ∈ L(Γ ) which is geometrically badly approximated,and a real number t > h(x) satisfying

γ(H+t ) ∩ H+

t = ∅ for all γ ∈ Γ − Γ∞.

By definition of h(x), there exists T > 0 such that π([r(T ), x)) does notintersect π(Ht). For t large enough, the horocycle π(Ht) separates S into twoconnected components, therefore two cases arise:

(i) π([r(T ), x)) ⊂ π(H+t );

(ii) π([r(T ), x)) ⊂ S − π(H+t ).

The first case is excluded by Property 1.2. It remains to consider case (ii).Since the set π([z, r(T )]) is compact, there exists t′ � t such that

π([z, x)) ∩ π(H+t′ ) = ∅.

Thus one obtains the following characterization:

Proposition 2.1. A conical point x in L(Γ ) is geometrically badly approxi-mated if and only if there exists t > 0 such that

π([z, x)) ∩ π(H+t ) = ∅.

Notice that this proposition does not imply that if x is geometrically badlyapproximated, then π([z, x)) is bounded. Actually, in the case where L(Γ )contains a parabolic point y which is not in Γ (∞), then the surface S admitsat least two disjoint cusps (see for example the group Γ (2) from Chap. II),thus the ray π([z, x)) can be unbounded without meeting a cusp C(H+

t )(Fig. VII.9).

Fig. VII.9. Γ = Γ (2)

On the other hand, if the set of parabolic points of L(Γ ) is reduced to theΓ -orbit of the point ∞, then the projection to S of the Nielsen region N(Γ )is the union of a compact set and π(H+

t ) (for large t) (Proposition I.4.16).Therefore, if x is geometrically badly approximated, then π([z, x)) is bounded.


Corollary 2.2. If the set of parabolic points of L(Γ ) is equal to Γ (∞), thena conical point x in L(Γ ) is geometrically badly approximated if and only ifthe ray π([z, x)) is bounded.

Now consider the particular case where the group Γ is a Schottky groupS(p, h) generated by a translation p and a hyperbolic isometry h (see Sect. II.1)(Fig. VII.10).

Fig. VII.10. Γ = S(p, h)

As we saw in Chap. II (Property II.1.9), under these hypotheses, the setof parabolic points of S(p, h) is equal to the orbit of the point ∞, and henceCorollary 2.2 applies. Recall from Proposition II.2.2 that a conical point ofL(S(p, h)) is uniquely represented by a sequence f(x) = (si)i�1 satisfying

si ∈ {h±1, p±1}, si+1 �= s−1i

and if si ∈ {p±1}, then there exists j > i such that sj ∈ {h±1}.

The following proposition characterizes the geometrically badly approximatedpoints in coding terms.

Proposition 2.3. Let x be a conical point in L(S(p, h)). Define f(x) =(si)i�1. The following statements are equivalent:

(i) the point x is geometrically badly approximated;(ii) there exists an integer r > 0 such that if si ∈ {p±1}, then there exists

1 � j � r such that si+j ∈ {h±1}.

Proof.Not (ii) ⇒ not (i). Consider the sequence (ai)i�1 constructed from f(x) by

grouping together the consecutive p and p−1 terms. Such a sequence satisfies

ai ∈ {pn, h±1 | n ∈ Z∗ }, ai �= a−1

i+1,

if ai = pn then ai+1 ∈ {h±1},

and x = limn→+∞

a1 · · · an(z0),

where z0 is a point in H−(D0(h)∪D0(h−1)∪D0(p)∪D0(p−1)). By hypothesis,there exists a subsequence (aik

)k�1 such that aik= pnk with limk→+∞ |nk | =

+∞. Define γk = a1 · · · aik. We have,

limk→+∞

γ−1k (z0) = ∞ and γ−1

k (x) ∈ D0(h) ∪ D0(h−1)

(see Property II.1.4).

2 Geometrically badly approximated points 151

The points γ−1k (x) belong to a compact subset of R hence, passing to a subse-

quence, one can assume that the sequence of rays [γ−1k (z0), γ−1

k (x)) convergesto a geodesic (∞y). It follows that for all t > 0, there exists k � 1 such that

[γ−1k (i), γ−1

k (x)) ∩ H+t �= ∅.

This property implies that the ray π([z0, x)) is not bounded, and thereforethat x is not geometrically badly approximated.

Not (i) ⇒ not (ii). Suppose that x is not geometrically badly approximatedand choose z from the geodesic (∞x). By assumption, there exists a sequence(tn)n�1 which converges to +∞ and a sequence (γn)n�1 in S(p, h) such that

γn([z, x)) ∩ Htn �= ∅.

Passing to a subsequence, one can assume that γn can be written in the formof a reduced word c1 · · · c�n satisfying

c1 ∈ {h±1}, ci ∈ {p±1, h±1} for i � 2, and limn→+∞

�n = +∞.

The point γn(∞) belongs to D0(h) ∪ D0(h−1) ∩ R, which is a compact sub-set of R, and the sequence of radii of Euclidean circular arcs (γn(∞)γn(x))converges to +∞ thus

limn→+∞

γn(x) = ∞.

This property implies the existence of N1 > 0 such that

∀ n � N1, c1 = s−1�n

, . . . , cn = s−11 .

To see this, note that x = limn→+∞ s1 · · · sn(z0). If the preceding condition isnot satisfied, there exists a subsequence (γnp)p�1 such that the first letter ofthe reduced word corresponding to γnps1 · · · s�np

is the letter c1. In this case,by Property II.1.4(i), γnp(x) belongs to the compact set D0(h) ∪ D0(h−1) ∩ R,which is not allowed.

It follows that for n � N1, the point γn(x) belongs to D(s�n+1)(∞). Sincelimn→+∞ γn(x) = ∞, there exists N2 � N1 such that

∀ n � N2, s�n+1 ∈ {p±1}.

Let n � N2. The point s−1�n+1γn(x) belongs to D(s�n+2)(∞) and one has

limn→+∞ s−1�n+1γn(x) = ∞ thus, after reusing the same argument, there exists

N3 � N2 such that

∀ n � N3, s�n+1 ∈ {p±1} and s�n+2 = s�n+1.

Continuing to apply this argument, one obtains an increasing sequence(Nk)k�2 satisfying the following condition:

∀ n � Nk, s�n+1 ∈ {p±1}, s�n+1 = · · · = s�n+k−1,

which contradicts part (i). �


3 Applications to the theory of Diophantineapproximations

We consider now the case where Γ is the modular group PSL(2, Z). Theset of parabolic points associated with this group is simply the orbit of thepoint ∞ and is equal to Q ∪ { ∞} (Property II.3.8). The modular surfaceS = PSL(2, Z)\H therefore admits a single type of cusp π(H+

t ), where H+t is

a horodisk centered at the point ∞ (Fig. VII.11).

Fig. VII.11. Γ = PSL(2, Z)

The purpose of this section is to draw a parallel between the excursions intoπ(H+

t ) of a geodesic ray π([z, x)), where x is irrational, and an approximationof x by a sequence of rational numbers.

We begin with a discussion of three well-known results from number theory.We will prove them in this section using a hyperbolic point of view.

3.1 Three classical theorems

The idea of continued fractions emerged very early [25, Chap. V]. In Sect. II.4,we gave a geometric interpretation of this idea using the Farey tiling of H. Onebranch of the theory of Diophantine approximations consists of constructing aone-to-one correspondence between some algebraic properties of an irrationalnumber and those of the sequence of integers (ni)i�0 associated with its con-tinued fraction expansion. One example is Proposition II.4.11 which relatesthe quadratic real numbers to almost periodic sequences.

Another branch focuses on the speed of convergence of the sequence ofrational numbers associated with a continued fraction expansion. For exam-ple, one of these problems is to find the “best” (in the sense of asymptoticbehavior) function ψ : N → R

∗+ decreasing to 0 such that for all x ∈ R − Q,

there exists a sequence of rational numbers (pn/qn)n�1 satisfying

|x − pn/qn| � ψ(|qn|) and limn→+∞

|qn| = +∞.

Note that if pn is the integer part of nx, the sequence (pn/n)n�1 satisfies

|x − pn/n| � 1/n.

3 Diophantine approximations 153

It follow that the function ψ(n) that we are looking for is thus less than 1/n.The following theorem is one of the first results related to this question.

One of its classical proofs relies on some properties of the continued fractionexpansion [52, Chap. 6, Theorem 6.24].

Theorem 3.1. For all x ∈ R − Q, there is a sequence (pn/qn)n�1 of rationalnumbers satisfying

|x − pn/qn| � 1/(2q2n) and lim

n→+∞|qn| = +∞.

Can one find a function ψ(n) converging to zero faster than 1/n2? Theanswer is “No” as shown in the following exercise:

Exercise 3.2. Prove that for all p ∈ Z and q in N∗, the following inequality

is satisfied:|

√2 − p/q| � 1/(4q2).

The function ψ that we are looking for, therefore satisfies

1/4 � n2ψ(n) � 1/2.

This naturally leads us to define for each irrational number x the quantity

ν(x) = inf{

c > 0 | ∃ (pn/qn)n�1 ∈ Q,

|x − pn/qn| � c/q2n and lim

n→+∞|qn| = +∞

}.

By Theorem 3.1, this quantity is less than 1/2 for all x. The following theoremis more precise. It can be proved, for example by associating a sequence ofcircles to the sequence of rational numbers given by the continued fractionexpansion, and by studying their relative positions [52, Chap. 6, Theorem6.25] (see also [30]).

Theorem 3.3. For every irrational number x, one has

ν(x) � 1/√

5.

Also ν(x) = 1/√

5 if and only if there exist a, b, c, d in Z such that

ac − bd = 1 and x =aN + b

cN + d, where N = (1 +

√5)/2 is the golden ratio.

Among the rational numbers, the badly approximated real numbers x forwhich ν(x) is strictly positive are of special interest. For example, this is thecase for

√2.

The following theorem relates this property to a property of the sequenceof integers (ni)i�0 associated with the continuous fraction expansion of x.A proof of this theorem is given, for example, in [24, Theorem 2.20]. Ourproof is not very different from the cited example.

Theorem 3.4. Let x be an irrational number. The following are equivalent:

(i) the sequence (ni)i�0 is bounded;(ii) the real number x is badly approximated.


3.2 Hyperbolic proofs of Theorems 3.1, 3.3 and 3.4

The proofs of Theorems 3.1 and 3.3 that we propose are not more elementarythan the originals! Our purpose here is not to gain simplicity but to illustratethe fact that the mathematical world is not compartmentalized.

In the rest of this section, Γ = PSL(2, Z) and p(z) = z+1. This translationgenerates the stabilizer Γ∞ of the point ∞ in Γ .

Recall (from Lemma I.3.19) that the Euclidean diameter of the image byan isometry γ ∈ Γ − Γ∞ of the horocycle Ht centered at the point γ(∞) is1/(c2(γ)t), where c(γ) is equal to the absolute value of the coefficient c in γ(z)written in the form γ(z) = (az + b)/(cz + d), with ad − bc = 1.

Let x ∈ R, we have (Fig. VII.12):

(∗) (∞x) ∩ γ(Ht) �= ∅ =⇒ |x − a/c(γ)| � 1/(2tc2(γ)).

Fig. VII.12.

The following lemma provides a gateway between approximation theoryand the study of geodesic rays on the surface S = Γ \H. Recall that π is theprojection from H to S.

Lemma 3.5. Let x ∈ R and let (r(s))s∈R be an arclength parametrization ofthe oriented geodesic (∞x). The following are equivalent:

(i) there exists a sequence (sn)n�1 of positive real numbers such thatlimn→+∞ sn = +∞ and π(r(sn)) belongs to the horocycle π(Ht);

(ii) there exists a sequence (γn)n�1 in Γ − Γ∞ such that

|x − γn(∞)| � 1/(2tc2(γn)) and limn→+∞

c(γn) = +∞.

Proof.(i) ⇒ (ii). We will essentially recycle the arguments used in the proof of

Proposition I.3.20. The fact that π(r(sn)) belongs to π(Ht) means that thereexists γn ∈ Γ such that

r(sn) ∈ γn(Ht).


The geodesic (∞x) intersects γn(Ht), hence |x − γn(∞)| � 1/(2tc2(γn)). Sincesn goes to +∞ and γn(Ht) does not meet Ht, then the sequence (c(γn))n�1

is not bounded.(ii) ⇒ (i). By hypothesis, the geodesic (∞x) intersects each circle γn(Ht).

If we let sn denote the largest real number s such that r(s) ∈ γn(Ht), thenr(sn) = x + ibe−sn , where b is a fixed real number > 0. Let Rn denotethe Euclidean ray of γn(Ht). The center of this Euclidean circle is the pointγn(∞) + iRn, thus the following equation holds:

(x − γn(∞))2 + (be−sn − Rn)2 = R2n.

The sequences (x − γn(∞))n�1 and (Rn)n�1 converge to 0. It follows thatlimn→+∞ sn = +∞. �

We are now ready to prove Theorems 3.1 and 3.3.

Proof (of Theorem 3.1). Let x ∈ R − Q. We will show that there exists asequence (sn)n�1 of positive real numbers such that

π(r(sn)) ∈ π(H1) and limn→+∞

sn = +∞.

We argue by contradiction. Suppose that there exists z ∈ (∞x) such thatπ([z, x)) does not intersect π(H1). Lift this situation up to H and considerthe elliptic isometry r in Γ , of order 3, defined by r(z) = (z − 1)/z. Theisometries r and r2 map H1 to circles of Euclidean diameter 1, tangent to thepoints 1 and 0 respectively, and tangent to each other at the point 1/2 + i/2(Fig. VII.13).

Fig. VII.13.

As shown in Chap. II (Exercise II.3.8), the set Δ defined by

Δ = {z ∈ H | 0 � Re z � 1, |z| � 1 and |z − 1| � 1},

is a fundamental domain of Γ (Fig. VII.14). It follows that there exists γ ∈ Γsuch that γ(z) belongs to Δ. The point x does not belong to the Γ -orbit of thepoint ∞ and the ray [γ(z), γ(x)), which is a circular arc, does not intersectH1. Hence this ray is in H − H+

1 . For the same reasons, it is also in H − r(H+1 )


Fig. VII.14.

and H − r2(H+1 ). Since γ(z) belongs to Δ, this ray is in a compact set, which

is impossible.To achieve the proof, it suffices to apply statement (∗), Lemma 3.5 and

to note that if γ does not belong to Γ∞, then γ(∞) is rational of the forma/c(γ), where a and c(γ) are relatively prime. �

Proof (of Theorem 3.3). Recall that R − Q is the set of conical points of Γ(Property II.3.8). By Lemma 3.5, the quantities ν(x) and the height h(x) of x(Definition 1.6) are related by the following relation:

ν(x) = 1/(2h(x)).

Theorem 3.3 can therefore be restated in terms of the height of geodesic raysin the form

infx∈R−Q

h(x) =√

5/2.

Let us prove this statement. Consider the case where x is the fixed point ofa hyperbolic isometry γ in Γ . Choose z on the axis of this isometry, with thecondition that π([z, x)) = π((γ−γ+)), where π((γ−γ+)) is a compact geodesicon S. Thus h(x) is the largest t > 0 such that

π((γ−γ+)) ∩ π(Ht) �= ∅.

In other words, on H one has the expression (Fig. VII.15)

h(x) = maxg∈Γ

|g(γ−) − g(γ+)|/2.

Fig. VII.15.


We write γ(z) = (az + b)/(cz + d) with ad − bc = 1. Since γ is hyperbolic,we have c �= 0. The following expression is obtained from a simple calculation:

(∗∗) |g(γ−) − g(γ+)| =√

(a + d)2 − 4/c(gγg−1).

At the end of Sect. II.4 (Corollary 4.14), we introduced the particular casewhere x is the golden ratio N = (1 +

√5)/2. As we saw in that section, this

point is fixed by the hyperbolic isometry γ = T1T−1 in Γ , which can also bewritten in the form

γ(z) = (2z + 1)/(z + 1).

The following equality can be deduced from expression (∗∗):

h(N ) = (√

5/2)ming∈Γ

c(gγg−1).

Since for all g ∈ Γ the isometry gγg−1 is hyperbolic, we have c(gγg−1) �= 0.Furthermore, this quantity is an integer thus c(gγg−1) � 1. In the particularcase where γ = T1T−1, one has c(γ) = 1. It follows that

h(N ) =√

5/2,

and that for all x ∈ R − Q

h(x) �√

5/2.

We prove now the reverse inequality. Let x ∈ R − Q. The definition of theheight h(x) implies

h(γx) = h(x),

for all γ ∈ Γ . Thus without loss of generality one can assume that x is in[0, 1]. Denoted by ([0; n1, . . . , nk])k�1 the continued fraction expansion of x.As in Sect. II.4, we will use the isometries T1(z) = z + 1, T−1(z) = z/(z + 1)and s(z) = −1/z. These three isometries are interrelated by the equalitysT−1s = T −1

1 .Let k � 2 and set γk = T n1

−1 · · · T nk

(−1)k . Recall the following facts fromExercise II.4.5 and Proposition II.4.6:

• if k is even, then γk(0) = [0; n1, . . . , nk] and γk(∞) = [0; n1, . . . , nk−1];• if k is odd, then γk(0) = [0; n1, . . . , nk−1] and γk(∞) = [0; n1, . . . , nk];• limk→+∞[0; n1, . . . , nk] = x.

Our proof requires the following lemma which relates the integers (ni)i�1

to the height h(x).

Lemma 3.6. There exists a sequence (gi)i�2 in Γ such that

g−1i (∞x) ∩ H(ni+1/(ni+1+1))/2 �= ∅ and lim

i→+∞c(g−1

i ) = +∞.


Proof. If i is even, define gi = γi−1. One has

g−1i = s ◦ T ni−1

1 ◦ T ni−2−1 ◦ · · · ◦ T n1

1 ◦ s.

Thus g−1i (∞) is the rational −[0; ni−1, ni−2, . . . , n1], and the continued frac-

tion expansion of g−1i (x) is ([ni; ni+1, . . . , nk])k�1.

If i is odd, define gi = γi−1 ◦ s. One has:

g−1i = T ni−1

−1 ◦ T ni−21 ◦ · · · ◦ T n1

1 ◦ s.

Thus g−1i (∞) is the rational [0; ni−1, ni−2, . . . , n1] and g−1

i (x) has the sequence(−[ni; ni+1, . . . , nk])k�1 as its continued fraction expansion.

In these two cases one has:

|g−1i (∞) − g−1

i (x)| � ni + 1/(ni+1 + 1),

which proves the first part of the lemma.The second part of this result arises from the fact that the sequences

(c(g2i))i�1 and (c(g2i+1))i�1 are sequences of positive integers which arestrictly increasing. �

Now we return to the proof of our inequality. Consider an arclengthparametrization (r(s))s∈R of the oriented geodesic (∞x). If the sequence(ni)i�1 contains infinitely many terms � 3, by Lemmas 3.5 and 3.6, thereexists an unbounded sequence (sn)n�1 in R

+ such that

π(r(sn)) ∈ π(H3/2).

Thus h(x) � 3/2 which implies h(x) >√

5/2.Otherwise, after replacing x with γ−1

i (x) with i � 1, two cases arise:

• either the sequence (ni)i�1 contains infinitely many terms equal to 2 and1 � ni � 2 for all i � 1,

• or every term of the sequence (ni)i�1 is equal to 1.

In the first case, by Lemmas 3.5 and 3.6, there exists an unbounded se-quence (sn)n�1 in R

+ such that

π(r(sn)) ∈ π(H7/6).

Thus h(x) � 7/6, and in particular, h(x) >√

5/2.In the second case, x is related to the golden ratio. More precisely, one has

T1(x) = N , thus h(x) = h(N ) and, by the first part of the proof, h(x) =√

5/2.�

Notice that, as we saw in Corollary II.4.14, the projection to S of theaxis of the hyperbolic isometry T1T−1 is the shortest compact geodesic in S.It follows from the proof of Theorem 3.3 that it is also the geodesic thatachieves the least height in the cusp π(H+

1 ).It now remains only to prove Theorem 3.4.


Proof (of Theorem 3.4). Let x be an irrational number. We know that ν(x)and h(x) are related by the following equation:

ν(x) = 1/(2h(x)).

Thus x is badly approximately if and only if x is geometrically badly approx-imated. Recall that, by Corollary 2.2, this property is equivalent to the factthat the ray π([z, x)) is bounded.

Not (i) ⇒ not (ii). Suppose that the sequence (ni)i�1 is not bounded. Forall t > 0, there exists a subsequence (nik

)k�1 whose terms are all � 2t. Let(r(s))s∈R be an arclength parametrization of the oriented geodesic (∞x). ByLemmas 3.5 and 3.6, there exists a sequence (sn)n�1 in R

+ converging to +∞such that

π(r(sn)) ∈ π(Ht).

One can conclude that h(x) is greater than t for all t > 0 and hence that x isnot geometrically badly approximated.

Not (ii) ⇒ not (i). Suppose that π([i, x)) is not bounded. For all integersk � 2 the surface S minus the cusp π(H+

k ) is bounded. Hence there existsK � 2 such that for all k � K (Fig. VII.16)

π([i, x)) ∩ π(Hk) �= ∅.

Fig. VII.16.

On H, this property translates to the existence of a sequence (gk)k�K in Γsatisfying

g−1k ([i, x)) ∩ Hk �= ∅.

Since the coefficients of the Mobius transformation gk are integers, we haveIm g−1

k (i) � 1. Therefore, the ray g−1k ([i, x)) crosses Hk at two points, which

implies the inequality

| Re(g−1k (i)) − g−1

k (x)| > k.


Fig. VII.17.

It follows that this ray intersects at least k vertical geodesics of the form(Fig. VII.17)

((mk + 1)∞), ((mk + 2)∞), . . . , ((mk + k)∞).

Hence the ray [i, x) intersects k consecutive Farey lines

gk T mk+11 (0∞), . . . , gk T mk+k

1 (0∞).

Returning to the geometric construction of the continued fraction expansiondiscussed in Sect. II.3, one obtains that for all k � K, there exists ni � k. �

4 Comments

Artin [3] was one of the first mathematicians who developed this geometricapproach to numbers. Some years after, the key idea to built a relationshipbetween fractions and circles appeared in an elementary and informative paperof Ford [30].

The geometric approach to numbers, as presented in Sect. VII.3, allowsone to rediscover other classical results. For example, one can obtain proper-ties of the Markov spectrum by relating this object to the lengths of simple(without self intersections), compact geodesics in the quotient of H over thegroup Γ (2), which was introduced in Chap. II [15, 35, 56]. It also allows somequestions of number theory to be formulated in terms of dynamics. Considerfor example the open question about characterizing badly approximated alge-braic numbers x of degree n � 3, which reduces, on the modular surface, toasking if the associated rays π([i, x)) are bounded.

The S. Patterson’s thesis [50] is one of the founding texts which gener-alize this approach to Fuchsian groups. More generally, this approach canbe adapted for geometrically finite Kleinian groups Γ acting on a pinchedHadamard manifold [38]. Thus it widens the field of the theory of Diophan-tine approximation, replacing R with L(Γ ), and Q with the orbit of a parabolicpoint of Γ .

The metric theory of approximations can likewise be approached from thisgeometric angle, and can be generalized by allowing the role of the Lebesgue

4 Comments 161

measure to be played by a Patterson measure [51, 62]. Using this point ofview one can obtain a general version of Khintchine’s theorem, whose classi-cal statement establishes a link between the nature of the series

∑n�1 Ψ(n),

where Ψ is a strictly positive decreasing function, and the Lebesgue measureof the set of real numbers x approximated by a sequence of rational numbers(pn/qn)n�1 satisfying |x − pn/qn| � Ψ(qn)/qn.

Due to the initiative of M.S. Raghunathan, this approach has been sim-ilarly developed by G. Margulis, S. Dani and many other mathematicians tosolve some problems in Diophantine approximation in R

n ([60, Chap. IV]and [46]). The dynamical system in play in this context is the action of aclosed subgroup of SL(n, R) on the symmetric space Mn = SL(n, Z)\ SL(n, R).One good illustration of the effectiveness of this altered point of view is theproof of the Oppenheim conjecture on the non-degenerate, indefinite, irra-tional quadratic forms on R

n (n � 3). This conjecture asserts that if Q issuch a form, then for all ε> 0, there exists v ∈ Z

n − {0} such that |Q(v)| � ε.Its proof, due to G. Margulis, requires the topological description of orbitsof the group SO0(p, q) on Mn, where (p, q) is the signature of Q ([5, Ap-pendix by E. Breuillard] and [32, 47, 60, 17]). In the same spirit, G. Margulisalso showed that the Hardy-Littlewood conjecture which stated that for everypair (x, y) in R

2, there exists a sequence of integers (qn)n�1, (pn)n�1, (rn)n�1,with qn > 0, such that limn→+∞ q2

n|x − pn/qn| |y − rn/qn| = 0, is related to theorbits of the group D of diagonal 3 × 3 matrices of the form (et1+t2 , e−t1 , e−t2)on M3 [60, Sect. 30]. At the time of writing of this text, this method has not(yet) produced an answer to this conjecture but has enriched the area of dy-namical systems with an open question, namely: Are the bounded orbits of Don M3 compact?

Appendix A

Basic concepts in topological dynamics

This short introduction to abstract topological dynamics is inspired byM. Alongi’s and G.S. Nelson’s “Recurrence and topology” [1]. This book con-tains the solutions of the exercises of this appendix. We encourage a readerwho wants to know more on this field to read A. Katok’s and B. Hasselblatt’s“Introduction to the Modern Theory of Dynamical Systems” [40], andW.H. Gottschalk’s and G.A. Hedlund’s “Topological dynamics” [34].

Let Y be a topological space. By definition, a flow on Y is a map

φ : R × Y −→ Y

satisfying the following conditions:

(i) φ is continuous;(ii) φ(t, .) : Y → Y is a homeomorphism for each t ∈ R;(iii) φ(s, φ(t, y)) = φ(s + t, y), for all y in Y and any real numbers s, t.

For each real number t, we denote by φt : Y → Y the map defined for ally ∈ Y by φt(y) = φ(t, y).

Exercise A.1. Prove that φ0 = Id, and that φt = φ−1t , for each t ∈ R.

Many examples arise from smooth vector fields f on smooth manifolds,and are determined by a differential equations of the form

f(y) =dy

dt,

where dy/dt denotes the derivative of a function y with respect to a singleindependent variable.

In most cases, there exists a unique smooth function φ : R × Y → Ysatisfying

dφ(t, y)dt

(0) = f(y),

such that

F. Dal’Bo, Geodesic and Horocyclic Trajectories, Universitext,DOI 10.1007/978-0-85729-073-1, c© Springer-Verlag London Limited 2011

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164 A Basic concepts in topological dynamics

(i) φ(t, .) : Y → Y is a diffeomorphism for each t ∈ R;(ii) φ(s, φ(t, y)) = φ(s + t, y), for all y in Y and any real numbers s, t.

Examples A.2.

(i) If Y = R2 and f(y) is the constant vector field −→v �= −→

0 , then φ(t, y) =y + t−→v .

(ii) Notice that the flow φ(t, y) = y + t−→v induces a flow Φ on the torusT

2 = R2/Z

2 given by

Φ(t, y mod Z2) = y + t−→v mod Z

2.

More generally, when Y is a compact smooth manifold, classical theoremsfor ordinary differential equations guarantee the existence (and uniqueness)of a flow associated to a smooth vector field on Y .

Definition A.3. If φ : R × Y → Y is a flow, then the trajectory (respectivelythe positive or negative semi-trajectory) from a point y in Y is the set of pointsφt(y), where t is in R (respectively R

+ or R−).

In Example A.2(i), the trajectory from y ∈ R2 is the straight line passing

through y with direction −→v .In Example A.2(ii), the trajectories of Φ on T

2, are the projection on T2

of the trajectories of Example A.2(i).

Proposition A.4. Let φ : R × Y → Y be a flow, if two trajectories have anonempty intersection, then they are equal.

Exercise A.5. Prove Proposition A.4.

It follows from Proposition A.4 that the family of all trajectories is apartition of the space Y .

Definition A.6. Let φ : R × Y → Y be a flow. A point y is a periodic pointif there exits T > 0 such that φT (y) = y. The period of y is the infimum ofsuch T .

A flow φ associated to a non-zero constant vector field on R2 does not

admit periodic points. In contrary, if −→v ∈ Q × Q − {(0, 0)}, then all points inthe torus T

2 are periodic for the flow induced by φ.

Exercise A.7. Prove that a flow on T2 induced by a non-zero constant vector

field −→v on R2 has periodic point if and only if −→v ∈ Q × Q − {(0, 0)}.

Proposition A.8. Let φ : R × Y → Y be a flow. If y is a periodic point, thenits trajectory is compact.


A Basic concepts in topological dynamics 165

Exercise A.10. Let Φ be a flow on T2 induced by a non-zero constant vector

field −→v . Prove that, if −→v /∈ Q × Q − {(0, 0)}, then each trajectory is densein T

2.

Notice that, if y is a periodic point for a flow φ, or if its trajectory is dense,then there exists an unbounded sequence of real numbers (tn)n�0 such thatlimtn →∞ φtn(y) = y.

More generally we introduce the following definition

Definition A.11. Let φ : R × Y → Y be a flow. A point y is non-wanderingif for any neighborhood V of y, there exists an unbounded sequence of realnumbers (tn)n�0 such that

φtnV ∩ V �= ∅.

We denote by Ωφ(Y ) the set of non-wandering points of φ.

Notice that in examples (i) and (ii) we have: Ωφ(R2) = ∅ andΩΦ(T2) = T

2. In general the situation is more complicated.

Exercise A.12. Let φ be the flow on the closed unit disk D= {z ∈ C | |z| � 1}associated to the vector field defined in polar coordinates (r, θ) by:

dr

dt= r(r − 1) and

dθ

dt= θ.

Prove that the set of periodic points is S1 ∪ {0} and that Ωφ(D) = S

1 ∪ {0}.(Hint: see [1, Exercises 2.3.8 and 2.5.12].)

Proposition A.13. Let φ : R × Y → Y be a flow. The non-wandering setΩφ(Y ) is a closed set, invariant with respect to the flow.


In Example A.2(i), no trajectory has accumulation points. More generallywe define the notion of divergent points:

Definition A.15. Let φ : R × Y → Y be a flow. A point y is said tobe divergent (respectively positively or negatively divergent) if for all un-bounded sequences (tn)n�1 in R (respectively R

+ or R−), the sequence of

points (φtn(y))n�1 diverges.

Notice that the notion of divergent points makes sense only for non-compact manifolds.

Exercise A.16. Prove that a point y is positively divergent (respectively neg-atively divergent) if and only if for some T ∈ R the function from [T, +∞)(respectively (−∞, T ]) into Y , which sends t to φt(y) is a homeomorphismonto its image (i.e., a topological embedding).

166 A Basic concepts in topological dynamics

There is no general relations between divergent points and wanderingpoints. Exercise A.12 gives an example without divergent points, whereΩφ(Y ) �= Y . It is shown in Chap. III that, when Y is the quotient of T 1

H

by the modular group PSL(2, Z) and φ is the geodesic flow, then Ωφ(Y ) = Yand there are divergent points.

Definition A.17. A set M ⊂ Y is minimal with respect to the flow φ if Mis a nonempty closed subset in Y such that for each m ∈ M its trajectoryφR(m) is dense in M .

Equivalently, a nonempty subset in Y is minimal if it does not containproper nonempty closed subset, invariant with respect to the flow φ.

For example, if y is periodic or is positively and negatively divergent, thenits trajectory is minimal.

Appendix B

Basic concepts in Riemannian geometry

This appendix outlines some results proved in [31], which are useful in Chaps. Iand VI.

Let M be a connected smooth manifold. A Riemannian metric on M is afamily of scalar products (gm)m∈M defined on each tangent space TmM anddepending smoothly on m. For example, the Euclidean space R

n is canonicallyequipped with a Riemannian structure (gm)m∈M , where gm is the ambientscalar product. More generally, if M is a submanifold of R

n, then the restric-tion of gm to each tangent space TmM induces a Riemannian metric on M .This is the case for example for the torus T

2 viewed as revolution surfacein R

3 induced by the map ψ : R2 → R

3 defined by

ψ(θ, φ) = ((2 + cos θ) cos φ, (2 + cos θ) sin φ, sin θ).

Given a Riemannian metric (gm)m∈M on M , we are lead to define a canon-ical measure vg on M . More precisely, let (Uk, φk) be a chart and consider thelocal expression of the metric in this chart

∑1�i,j�Dim(M)

gkijdxidxj .

The volume of the parallelotope generate by the vectors ∂/∂xi is√

det (gkij).

We define the measure vg as corresponding to the density which is given inthe atlas (Uk, φk) by √

det (gkij)L,

where L is he Lebesgue measure on Rn.

By definition, the volume of a subset B ⊂ M is given by

vol(B) =∫

B

vg,

when this integral exists.


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168 B Basic concepts in Riemannian geometry

For the torus T2 viewed as revolution surface in R

3, the area of a subsetB = ψ(A) ⊂ T

2 associated with the induced metric is given by

vol(B) =∫

A

√∣∣∣∣∂dψ

dθ∧ ∂dψ

dφ

∣∣∣∣ dθ dφ,

where ∧ is the vector product in R3.

The notion of length of a piecewise C1 curve c : [0, a] → M is also welldefined and is given by

length(c) =∫ a

0

√gc(t)(c′(t), c′(t)) dt.

This notion does not depend on the choice of a regular parametrization.Using the notion of length, we define a distance on M associated to the Rie-

mannian metric (gm)m∈M . The following proposition is proved in [31, Propo-sition 2.91].

Proposition B.1. Let d : M × M → R+ be the map defined for m and m′

as the infimum of the lengths of all piecewise C1 curves from m to m′. Thismap is a distance on M , which gives back the topology of M .

In the Euclidean space, straight lines are length minimizing. The curveswhich (locally) minimize length in a Riemannian manifold are the geodesics .Namely we have [31, Corollary 2.94].

Definition B.2. A curve c : I ⊂ R → M , parametrized proportional toarclength, is a geodesic if and only if for any t ∈ I there exists ε > 0 such thatd(c(t), c(t + ε)) = length(c|[t,t+ε]).

For the metric on T2 viewed as revolution surface in R

3 meridian lines andparallels (θ = constant) parameterized proportional to length are geodesics[31, Exercise 2.83].

A diffeomorphism f between a Riemannian manifold (M, (gm)m∈M ) anda smooth manifold M ′ induces a metric on M ′ defined for m′ ∈ M ′ and−→u ′, −→v ′ ∈ T ′

mM ′ by

g′m′ (−→u ′, −→v ′) = gf −1(m′)(Tm′ f −1(−→u ′), Tm′ f −1(−→v ′)).

The Riemannian manifolds (M, (gm)m∈M ) and (M ′, (g′m′ )m′ ∈M ′ ) are iso-

metric in the following sense

Definition B.3. Let (M, (gm)m∈M ) and (M ′, (g′m′ )m′ ∈M ′ ) be two Rieman-

nian manifolds. A map f : M → M ′ is an isometry (resp. local isometry)if f is a diffeomorphism (resp. local diffeomorphism), satisfying the followingrelation for any m ∈ M and −→u , −→v ∈ TmM

g′f(m)(Tmf(−→u ), Tmf(−→v )) = gm(−→u , −→v ).

B Basic concepts in Riemannian geometry 169

When M ′ = M , the set of isometries f : M → M is a group. Let Γ bea discrete group of isometries of M . We suppose that Γ acts on M freely(i.e., for m ∈ M and g ∈ Γ − {Id}, g(m) �= m), and properly (i.e., for anym, m′ ∈ M , if m′ /∈ Γm, then there exist two neighborhoods V (m) andV (m′) such that gV ∩ V ′ = ∅, for any g ∈ Γ ). Under these conditions,there exists an unique Riemannian metric on Γ \M such that the canonicalprojection of M onto Γ \M is a smooth covering map and a local isometry[31, Proposition 2.20]. For example, if Γ is a group of translations associatedto a basis of R

2, we obtain a Riemannian metric on the torus T2, which is said

to be flat [31, Exercise 2.25]. For a flat Riemannian metric on T2, the geodesics

are the projections of the straight lines of R2 parameterized proportional to

length.More generally, we have [31, Proposition 2.81].

Proposition B.4. If Γ is a discrete group of isometries of (M, (gm)m∈M )acting freely and properly on M , then the geodesics of Γ \M are the projectionsof the geodesics of M , and the geodesics of M are the lifting of those of Γ \M .

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Index

BBadly approximated real number, 153Busemann cocycle, 132

CCongruence modulo 2 subgroup, 61Conical point, 27Convex-cocompact Fuchsian group, 34Cusp, 32

DDirichlet domain, 21Divergent point, 165

EElementary Fuchsian group, 25Elliptic isometry, 17, 131Expansion of continued fraction, 66

FFarey lines, 68Farey tiling, 68Flow, 163Fuchsian group, 20

GGeodesic, 5, 91, 129, 168Geodesic flow, 80, 82, 134, 136Geometric curve, 4Geometrically badly approximated

point, 148Geometrically finite Fuchsian group, 33Golden ratio, 77, 153

HHardy-Littlewood conjecture, 161Horocycle, 10, 132Horocycle flow, 111, 113, 139, 140Horocyclic point, 26Horodisk, 12Hyperbolic area, 3Hyperbolic isometry, 17, 131

IIsometry, 168

KKhintchine’s theorem, 161Kleinian group, 42

LLength, 168Length spectrum, 92Limit set, 24Lorentz bilinear form, 127Lorentz group, 132

MMarkov spectrum, 161Modular group, 58

NNielsen region, 33

OOppenheim conjecture, 161

PParabolic isometry, 17, 131Parabolic point, 29


http://dx.doi.org/10.1007/978-0-85729-073-1

176 Index

Patterson measure, 43Periodic point, 164Pinched Hadamard manifold, 41Ping-Pong Lemma, 78Poincare series, 43Positive isometry, 15

QQuadratic real number, 75

RRiemannian metric, 167

SSchottky group, 46Semi-trajectory, 164Strong stable foliation, 124

TTopological mixing, 95Trajectory, 164

VVolume, 167

geodesic and horocyclic trajectories cramnotes/5dterra not… · trajectories of the geodesic ﬂow...

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