Topological Theory of Dynamical Systems (North-Holland Mathematical Library)

TOPOLOGICAL THEORY OF DYNAMICAL SYSTEMS Recent Advances

North-Holland Mathematical Library Board of Advisory Editors: M. Artin, H. Bass, J. Eells, W. Feit, P.J. Freyd, F.W. Gehring, H. Halberstam, L.V. Hormander, J.H.B. Kemperman, H.A. Lauwerier, W.A.J. Luxemburg, F.P. Peterson, I.M. Singer and A.C. Zaanen

VOLUME 52

NORTH-HOLLAND AMSTERDAM LONDON NEW YORK TOKYO

Topological Theory of Dynamical Systems Recent Advances

N. AOKI Department of Mathematics Tokyo Metropolitan University Minami-Ohsawa 1- I , Hachioji-shi Tokyo,Japan

K. HIRAIDE Department of Mathematics Faculty of Science Ehime University Matsuyama, Japan

1994

NORTH-HOLLAND AMSTERDAM LONDON NEW YORK TOKYO

ELSEVIER SCIENCE B.V. Sara Burgerhartstraat 25 P.O. Box 211, 1000 AE Amsterdam, The Netherlands

L i b r a r y o f C o n g r e s s Cataloging-in-Publication

Data

Aoki. Nobuo. 1927Topological theory of dynamical s y s t e m s : r e c e n t a d v a n c e s / N. Aoki. K . Hiraide. p. cm. -- (North-Holland mathematical library) I n c l u d e s bibllographical r e f e r e n c e s and index. I S B N 0-444-89917-0 1. Differentiable dynamical systems. 2. T o p o l o g i c a l dynamics. I. Hiraide. K. (KEichi) 11. Title. 111. S e r i e s . QA614.8.A52 1994 514'.72--d~20 94-11678 CIP

ISBN: 0 444 89917 0 0 1994 ELSEVIER SCIENCE B.V. All rights reserved.

No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior written permission of the publisher, Elsevier Science B.V., Copyright & Permissions Department, P.O. Box 521, 1000 AM Amsterdam, The Netherlands. Special regulations for readers in the U.S.A. - This publication has been registered with the Copyright Clearance Center Inc. (CCC), Salem, Massachusetts. Information can be obtained from the CCC about conditions under which photocopies of parts of this publication may be made in the U.S.A. All other copyright questions, including photocopying outside of the U.S.A., should be referred to the publisher. No responsibility is assumed by the publisher for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions or ideas contained in the material herein. This book is printed on acid-free paper Printed in The Netherlands

PREFACE

The history of theory of dynamics began with the work of Isaac Newton who formulated the law of motion and who stated the universal law of gravitation. In Newton’s theory the motion of a dynamical system was described by a system of differential equations. The problem of applying the theory to planetary motion was challenging. People thought they would eventually succeed in explicitly integrating the differential equations. However, their hope did not come true. In the late nineteenth century, H.Poincar6 created a new branch of mathematics by publishing his famous memoir “Les mCthodes nouvelles de la mkanique ckleste” on the qualitative theory of differential equations. The point of his idea was to relate the geometry of the phase space with the analysis. As a result of PoincarB’s qualitative approach, the focus of the theory of dynamics shifted away from the differential equations. In G.D.Birkhoff’s treatise on dynamical systems, he discussed many dynamical phenomena in the context of transformation groups acting on general metric spaces. Since then, many mathematicians have considered this theory in the environment of geometry. The purpose of this book is to provide an advanced account of some aspects of dynamical systems in the framework of general topology, and is intended for use by interested graduate students and working mathematicians. Even though some of the topics discussed here are relatively new, others are not. Therefore the book is not a collection of research papers, but it is a text book to present recent developments of the theory that could be the foundations for future developments. This book does not necessarily cover all aspects of modern topological dynamics, but we intend here t o give a wider scope of various trends in recent developments of dynamical systems while avoiding overlap with other books in the concerned field. Nevertheless, the topics were selected from only the most significant results which may serve as the foundations for further developments of topological dynamics, and much effort has been done to make the subject matter accessible to a larger number of readers. This book contains a new theory developed by the authors to deal with the problems occurring in differentiable dynamics that are within the scope of general topology. To follow it, the book provides an adequate foundation for topological theory of dynamical systems, and contains tools which are sufficiently powerful through out the book. We assume that the reader has a rudimentary knowledge of algebra and analysis while a little more is presapposed from basic general topology, basic topological dynamics, and basic algebraic topology that may be obtained V

vi

PREFACE

in the graduate course, e.g., the books, Modern General Topology by J. Nagata, Topics in General Topology by K. Morita and J. Nagata, and Algebraic Topology by E. Spanier. We believe that graduate students and some undergraduate students with sufficient knowledge of basic general topology, basic topological dynamics, and basic algebraic topology will find little difficulty in reading this book. We would be especially pleased if our book could help them in the preparation of their theses. The authors wish to express their gratitude to Professor Jun-iti Nagata and Professor Jacob Palis without whose help and encouragement would have made writing this book covering the significant area of mathematics impossible. The authors also wish to acknowledge that this book was written during their stay at IMPA in Rio de Janeiro. N. Aoki and K. Hiraide

CONTENTS

INTRODUCTION

.....................................................

1

................15 CHAPTER 1 Some Properties of Anosov Systems $1.1Toral endomorphisms .......................................... 15 $1.2 Anosov differentiable systems .................................. 19

..-.-

- - - - - - - .- - - .. - . .31

* * CHAPTER 2 Dynamics of Continuous Maps ............................................. $2.1 Self-covering maps ................................................... $2.2 Expansivity $2.3 Pseudo orbit tracing property .................................. 92.4 Topological Anosov maps (TA-maps)

- - - - - - - - - - - - - -.- -.- - - - -

31 36 78 88

CHAPTER 5 Local Product Structures ........................... .................................... $5.1 Stable sets in strong Sense -. $5.2 Local product structures for TA-covering maps .............................. $5.3 Expanding factors of TA-maps $5.4 Subclasses of the class of TA-maps ............................

147 147 155 * -162 166

.................................. CHAPTER 6 TA-Covering Maps $6.1 Fundamental groups .......................................... $6.2 Universal covering spaces ..................................... 86.3 Covering transformation groups ............................... $6.4 S-injectivity of TA-covering maps .............................. $6.5 Structure groups for inverse limit systems $6.6 Lifting of local product structures $6.7 TA-covering maps of closed topological manifolds $6.8 Classification of TA-covering maps on tori .....................

168 168 174 182 189 193 -203 211 219

- - - - - - - .- - - -

----- --- ------------------- --------- ------- . - - - - - .- - . - .

vii

CONTENTS

viii

- - - - - -222 - - - - - - - - - - - - - - .222 - .- ..- - - - -. - - 231 ........................... 241 CHAPTER 8 TA-Covering Maps of Tori -.241 $8.1 Toral endomorphisms homotopic to TA-covering maps .-.- - - - - - - - - - - - - - . .. - . - 244 $8.2 Construction of semi-conjugacy maps ........................................... 250 $8.3 Nonwandering sets $8.4 Injectivity of semi-conjugacy maps - ..- - - - .- - - - - - - - - - - - - - - - - 257

CHAPTER 7 Solenoidal Groups and Self-covering Maps $7.1 Geometrical structures of solenoidal groups $7.2 Inverse limit systems of self-covering maps on tori

* *

*

*

*

*

....................................... $8.5 Proof of Theorem 6.8.1 ....................................... $8.6 Proof of Theorem 6.8.2 ................................................... $8.7 Remarks

-

268 269 ~275

.- . - - ..- -

CHAPTER 9 Perturbations of Hyperbolic Toral Endomorphisms 280 * * * 280 $9.1 TA-C"" regular maps that are not Anosov ................... 288 $9.2 One-parameter families of homeomorphisms

-

.............................. CHAPTER 10 Fixed Point Indices ............................................ $10.1 Chain complexes .......................................... $10.2 Singular homology . * $10.3 Euclidean neighborhood retracts (ENRs) .......................................... $10.4 Fixed point indices .......................................... $10.5 Lefschetz numbers ................................... $10.6 Orientability of manifolds $10.7 Orientability of generalized foliations . $10.8 Fixed point indices of expanding maps ....................... $10.9 Fixed point indices of TA-covering maps

- -304

-- -- ------- ------

- - .. - - . - - . - -.- - - - . - - - - - - - - - - - - .- - - . .

304 310 320 326 330 332 337 346 348

- - - - - - - - - - - - - - - - - .- 360

CHAPTER 11 Foundations of Ergodic Theory ............................................. $11.1Measure theory $11.2 Measure preserving transformations .......................... ........................................... $11-3 Ergodic theorems $11.4 Probability measures of compact metric spaces $11.5 Applications to topological dynamics

- - - ..- - - - - - .-

- - - - - - - - - - - - - - - .- - - - - - .................................................... 402 +

REFERENCES INDEX

360 363 366 375 387

.............................................................

411

INTRODUCTION

Take in its broad sense, Dynamical Systems involves the branches of mathematics: differential equation, differential topology, general topology and ergodic theory. This book takes up the subjects in the narrow sense, omitting differential equation from the above list and minimizing differential topology and ergodic theory. The phase space of a dynamical system will be taken to be a differentiable manifold. A continuous surjection f: M + N of metric spaces is a homeomorphism if it is injective and if the inverse map f : N --$ M is also continuous. A metric space M is called an n-dimensional topological manifold (or a manifold of class C o ) if there exist open subsets Ui of M and homeomorphisms ai of Ui onto open subsets of the n-dimensional euclidean space W", such that {Ui}covers M. It follows that the composite map ai o a;' : a j ( U i n U j ) --t a@i n Uj) is a homeomorphism whenever Ui and Uj overlap. If, in addition, all the composite maps ai o a;' are r-times continuously differentiable, then M is called a differentiable manifold of class C' (1 5 T 5 CQ). A differentiable manifold of class C" is also said to be smooth. The local charts (Ui, a ; ) of an n-dimensional differentiable manifold M enable us to use the differential calculus in the neighborhood of any point x € M. Let M and N be n-dimensional differentiable manifolds and let f : M + N be a continuous map. If ( U , a ) and ( V , p ) are local charts of M and N respectively with f(U) C V, then we say that f is diflerentiable at x E U if p o f o a-' : a ( U ) + p ( V ) is differentiable at a(.). This definition does not depend on the choice of local charts. A continuous map f : M + N of a differentiable manifold is said to be a differentiable map of class C ' i f f is r-times continuously differentiable at all points of M. If f : M --f N is a homeomorphism and if both f and f-' are r-times continuously differentiable, then the homeomorphism f : M + N is a diffeomorphism of class C'. By a curve in a manifold M we mean the image into M of an open interval of the real line under a continuously differentiable map. If y is a curve, then the tangent vector to y at x E y can be defined. For x E M all such tangent vectors form a real n-dimensional vector space T,M, called the tangent space to M at x. The set of all pairs (5, v), where x E M and v E T,M, can be vested with the structure of a 2n-dimensional differentiable manifold TM, called the tangent bundle of M. If M is of class C', then TM is of class C'-l.

-'

Dynamical Systems. We are now ready to define a dynamical system. A dynamical system with continuous time, or flow on a metric space X is a family {cpt : t E W} of homeomorphisms of X, such that the map (t, x) I+ cpt(x) 1

2

INTRODUCTION

is continuous and c p t + a ( z ) = cpt o cpa(z) for all x E X and all t , s E R. It follows that cpo is the identity map of X and cp-t is the inverse map of cpt. A family {cpt : t 2 0) of continuous maps of X is called a semi-flow if cpo is the identity map and if cpt+a(z) = cpt o cpd(z)for all x E X and all t , s 2 0, and ( t ,x) w cpt(z) is continuous. The orbit of a point x E X under the flow or semi-flow is the set of points { c p t ( x ) : t E R} or { c p t ( x ) : t 2 0). The point x is said to be a fixed point if cpt(x) = x for all t , and t o be a periodic point if cpp(z)= x for some p > 0. This implies cpt+P(x) = cpt(x) for all t . A dynamical system with discrete time on a metric space X is a family {f" : n E Z) of homeomorphisms of X ,or a family { f n : n 2 0) of continuous maps of X with fo the identity map, such that f"+'"(z) = f" o fm(z) for o f (n-times) for n > 0, all x E X and all n,m. It follows that f" = f o where f = f l , and if f is a homeomorphism then f-" = (f")-' for n > 0. Conversely, the iterations of a homeomorphism or a continuous map f:X + X form a discrete dynamical system. A point x is said to be a fixed point if f(z) = x and be a periodic point if fP(x) = z for some p > 0. A dynamical system with continuous time is obtained as solutions of differential equation of the form d -3 = F ( s ) dt where x belongs to a domain D of R" and F : D + R" is a map on the phase space D. In the theory of differential equations a fundamental theorem says that if F is continuously differentiable then, for any given point 20, the equation has a unique solution cpt(xo) which is defined for It1 sufficiently small and satisfies the initial condition cpO(z0) = 20. The time-dependence together with uniqueness ensures that cpt+a(x,-,) = cpt(cpa(zo)).The restriction to small value of It1 is essential. In many problems the phase space will be a domain of the real n-dimensional space R". However, other spaces also arise quite naturally. For example, if the differential equation can be restricted on a compact m-dimensional manifold M contained in the domain D (i.e. for every z E M the value F ( s ) is a tangent vector of M at x), for any given point xo E M the equation has then a solution cpt(x0) E M for all t E R. Therefore we shall have a flow {cpt} on the phase space M. There are two ways to relate such an equation La:= F ( x ) to a dynamical dt system with discrete time. The first method is to discretize time, i.e. for any choice of small t o > 0 this leads to the equation z ~ + Iv = to(zn) = f(sn).

Then the iterations {f"} of the map f becomes a dynamical system with discrete time on the space. The other method can be applied if the equation has a periodic solution, i.e. cpT(x0) = t o for some 20 and some T > 0. Then we consider a hypersurface S transverse to the curve t H cpt(xo), and in this hypersurface a neighborhood U of 50. Then a map g: U --+ S is induced by

INTRODUCTION

3

associating t o y E U the next intersection with S of the trajectory {cpt(y)}.

If the first such intersection occurs at y', then we define g ( y ) = y', and g is a map U --t S which is reduced from the flow. Thus we have a dynamical system with discrete time modelled on the differential equation. In this book we shall discuss mostly dynamical systems with discrete time on a compact metric space X. As a special case, we shall regard X as a differentiable manifold and f as a differentiable map of X. The general case may be called a topological dynamical system and the special case a differentiable dynamical system. We shall be primarily interested in differentiable maps, or diffeomorphisms of a differentiable manifold. However, such maps may have an invariant set which is not a manifold. The topological theory of dynamical systems is useful to investigate the behaviour of the differentiable map on the invariant set. Our book is written from this point of view. We give an important example of a discrete dynamical system with a topological structure. Let 12 2 1 be an integer and let Yf denote the set of all bi, ) where z, E Yk = (0, 1, * ,k - 1). infinite sequences z = (. ,z-1, z ~21, The set Yf becomes a compact metric space if we define the distance between two points z,y by d ( z ,y ) = 0 if z = y and d ( z ,y ) = 2-" if z # y where m is the largest integer such that z, = y , for all n with In1 < m. The set Yf is homeomorphic to the Cantor set. The shift map u defined by u ( z ) = y where z = (z,), y = ( y , ) and y , = ~ , + ~ ,E nZ, is a homeomorphism of Y f . Thus the family of homeomorphisms un,n E Z,is a discrete dynamical system, called a symbolic dynamics. The periodic points of this system are dense in Yf and there exists a point x E YE whose orbit O , ( x ) = { u " ( z ) : n E Z} is dense in Y f . If S is a closed subset of Yf and u-invariant (i.e. a ( S ) = S), then the restriction of u to S is called a subshift, denoted by u p . The subshift TI^ defines a discrete dynamical systems on S. As an important class of subshifts we give the following class. Let A = ( a i j ) be a k x k matrix of 0's and 1's and let S A denote the set of all 2 = (2,) E Yf with aznon+l= 1 for all n E Z. Then SA is closed in Yf and u(SA) = SA. A subshift of this type is called a Markov subshift. A continuous surjection f : X + X of a metric space is said to be topologically transitive if for any nonempty open sets U,V there is n > 0 such that f n ( U ) n V is nonempty. It is said to be topologically mixing if for any nonempty open sets U,V there is N > 0 such that f "(U)n V is nonempty for all n 2 N. A Markov subshift uIsA : S A --f S A is topologically transitive if and only if the matrix A is irreducible, and it is topologically mixing if and only if every element of A" is positive for some m > 0. It may seem that Markov subshifts have no relevance to physical systems. But the fact is that they are basic components of some physical systems. For example, Smale's theorem says that in a neighborhood of the orbit through

-

---

INTRODUCTION

4

any transversal homoclinic point, associated to a hyperbolic fixed point of a diffeomorphism, there is an invariant set such that some iterations of the diffeomorphism on the invariant set act a~ the shift map of Yt for some k > 0. This theorem applies to the restricted three-body problem.

1-dimensional dynamics. The iterations of maps of an interval into itself certainly present an example of nonlinear dynamical systems. The dynamics on an interval has many applications to physical systems. Let f : [0,1] + [0,1]be a continuous map of the interval. The iterations of the form z, H zn+l = f(zn) can be viewed as a discrete time version of a topological dynamical system. Here n plays the role of time variable. In natural sciences, the dynamical system often depends on quantities, called parameters. Therefore it is important to study a one-parameter family of maps which is the most basic system. For example, we consider a continuous map fr(z) = rz(1 - t) from [0,1] to itself, which is called the logistic map. Fix r with 0 5 T 5 4 and define a sequence {zn}by zn+l = rz,(l - 2,) for n 2 0. It can be shown that the behaviours of these maps, with the parameter T , where 0 5 r 5 4, has the following properties : (1) If 0 5 r < 1 then the sequence {z,} is decreasing and t, + 0 as n+m. (2) If 1 5 r 5 2 then the sequence {z,} is increasing and z, + 1- 1/r as

n -+ 00.

(3) If 2 < r 5 3 then the sequence (2,) converges to 1 - 1/r. (4) If 3 < T 5 1 & then the sequence {zn}converges to some periodic orbit with period 2. (5) If 1 & < r 5 4 then the behaviour of (2,) is very complicated, and when r varies increasingly in the interval [1+&,41 there exists a sequence of parameters r, in [1+&,41 such that fr, has 2,-periodic points. On the set of natural numbers we can define an ordering, which is called the $harkovskii ordering, as follows.

+

+

3 >5 >7>9 >2-3 > 2.5 > 2.7 > 2 - 9> >

-

-

a

...

>2"*3 > 2 5 5 > 2, - 7 > 2" - 9 > > - a *

*.*

> 2m > * * * > 8 > 4 > 2 > 1

The Sharkovskii's theorem says that if f : [0,1]---t [0,1]is continuous and has a periodic point with period p, then it has a periodic point with period q for every q < p with respect to the above ordering.

INTRODUCTION

5

From this result it follows that if the continuous map has a 3-periodic point then it has periodic points of all periods. Concerning the question whether the behaviour of the continuous map is chaotic, Li-Yorke showed that the 3-period implies chaos, i.e. if f: [0,1] + [0,1] is a unimordal map with a periodic point of period 3, then there is an uncountable set S of points and E > 0 such that for every z , y E S,z # y

liminfIf"(z)n-+w

fn(y)l = O .

It can be shown that the logistic map f, has a 3-periodic point if r 2 3.83. Thus Li-Yorke's theorem holds for such a map and its behaviour is chaotic. For the case r = 4 in particular, the logistic map f 4 is topologically conjugate to a continuous map g : [0,1] + [0,1], called the tent map, i.e. g ( z ) = 22 for 0 5 z 5 1/2 and g ( z ) = 2(1 - z) for 1/2 5 z 5 1. It can be shown that the map g is topologically semi-conjugate to a symbolic dynamics. Let X and Y be metric spaces. If continuous maps f: X -+ X and g: Y + Y satisfy the relation f o h = h o g for some homeomorphism h : Y -+ X, then we say that f is topologically conjugate t o g . When the relation f o h = h o g holds for some continuous surjection h: Y + X , we say that f is topologically semi-conjugate to g. The dynamics of maps of an interval is quite interesting. However, we do not discuss the subject in this book.

Higher dimensional dynamics. A 2-dimensional torus T2 = B2/Z2is obtained from the 2-dimensional euclidean space R2 by identifying the points ( z l , y l ) and (z2,y2) whenever 2 2 - 21 and y2 - y1 are integers, i.e. (z2 zl,y2 - yl) E Z2. The torus is a compact smooth manifold. A linear map of the plane with matrix

induces a map f = f A of the torus to itself if a, b,c,d are integers. The map f: T2t T2is a differentiable map. If the determinant ad - bc of A is non-zero, then f is called a toral endomorphism. If ad - bc = f l then f is bijective, and we call it a tom1 autornorphisrn. The toral automorphism (endomorphism) f is said t o be hyperbolic if A has no eigenvalues on the unit circle. Then the behaviour of f is quite complicated. It follows that a point p is periodic if and only if p has a coordinate consisting of rational numbers, and that the periodic points are dense in the torus. Moreover, there exists a point p whose orbit O f ( p )is dense on the torus. It may be asked whether the behaviour of f is chaotic. However, the question is somewhat vague because there is no widely accepted definition of chaotic behaviour of higher dimension. Nevertheless, the toral automorphism f is

INTRODUCTION

6

chaotic according to reasonable definitions which are not equivalent in general.

Anosov Systems. In a similar way we can define an n-dimensional torus

T" = Rn/Znfrom the n-dimensional euclidean space R". An automorphism (endomorphism) f = f A of T" is induced by an n x n matrix A of integers. If the determinant of A is non-zero and A has no eigenvalues on the unit circle, then W" is the direct sum of two subspaces E" and E", invariant under A, such that every eigenvalue of the restriction of A to E", respectively E" has an absolute value less than, respectively greater than one. Thus the behaviour o f f is similar t o that in the 2-dimensional case. To generalize the situation further, we consider a compact smooth manifold M . A diffeomorphism f : M -+ M is said to be an Anosov difeomorphism if (1) for each x E M the tangent space T,M is the direct sum of two subspaces T, M = Ei @ E," and

(2) there exist

C > 0 and 0 < X < 1 such that for all x E M and all n > 0 I l ~ z f n ( ~ )Il lCA"llvll

if v E E;t, IIDef-"(v)ll 5 CA"IIvII if w E E,". By use of (1) and (2) it can be shown that the subspaces Ei and E," vary continuously with respect to x. For a differentiable map f : M -+ M of a compact smooth manifold a point z E M is called a regular point if the differential D,f : T,M -+ Tf(,,M is surjective, and a singular point otherwise. If z is a regular point, by the inverse function theorem some open neighborhood of x is mapped diffeomorphically onto an open set of M by f . Hence the set S( f ) of all singular points of f is closed in M . If S( f ) is empty, then f is called a reguIar map. The set A( f ) = f n ( M ) is the maximal closed f-invariant set, i.e. f(A( f ) ) = A(f). If A is a closed invariant set (f (A) = A) then A is a subset of A( f ), and fla : A -+A is a local homeomorphism whenever A n S(f ) = 0. The set

nn2,,

Mf = {(xi) : xi E A(f) and f(zi) = xi+l,i E Z} is a closed subset of the product topological space M Z = nE-m Mi where each Mi is a replica of M , which can be regarded as the set of all (two sided) orbits of f . A closed invariant set A is called a hyperbolic set of f if A n S(f ) is empty and there exist C > 0 and 0 < X < 1 such that for every x = (xi)E Af = {(xi) : xi E A and f (xi)= xi+l,i E Z} there is a splitting

T,M =

UT,~M= U(E;~CB EJ: i

i

INTRODUCTION

7

so that for all i E ;Z

(1) DZif (EZi) = Eli+, where (2) for all n 2 0

0

= s,u,

If, in particular, T,M = UiE,", for all x = (xi) E A,, then flA : A + A is said to be expanding. If the entire space M is a hyperbolic set of f , then f: M + M is called an Anosov diflerentiable map. It follows that if an Anosov differentiable map is injective then it is an Anosov diffeomorphism. From the definitions of Anosov diffeomorphism and Anosov differentiable map, deduced are the following two important general properties : (A) Whenever f is an Anosov diffeomorphism, f is expansive, i.e. there exists e > 0 such that if x and y are any two distinct points of M then d( f "(x), f "(y)) > e for some integer n. Whenever f is an Anosov differentiable map, f is c-expansive, i.e. there exists e > 0 such that if (xi) and (yi) are any two distinct points of Mf then d(x,, yn) > e for some integer n. Whenever f is an expanding differentiable map, f is positively expansive, i.e. there exists e > 0 such that if x # y then d( f "(x),f "(y)) > e for some n 2 0. (B) Whenever f is an Anosov diffeomorphism, for each 6 > 0 there is a corresponding E > 0 such that if d(x,y) < E then there exists z E M such that d(f*(x), f n ( z ) ) < band d(f-n(y), f-"(z)) < 6 for all n 2 0. Whenever f is an Anosov differentiable map or an expanding differentiable map, for each 6 > 0 there is E > 0 such that for (xi),(yi) E Mf if d(zo,yo) < E then there is (zi)E Mf with the property that for all i 2 0, d(zi,z i ) < 6 and d(y-i, z-i) < 6. It follows from (A) that if 6 < e / 2 then the points z and (zi) in (B) are uniquely determined by the points x,y and (xi),(yi) respectively. For a differentiable map f : M + M it can be shown that if A is a hyperbolic set of f and if (xi) E Mf implies ( z i ) E Af whenever all xi are in some neighborhood of A, then the restriction o f f to A satisfies the same properties as in (A) and (B). The properties (A) and (B) make sense for any metric space and any continuous map, or any homeomorphism from the space to itself. We shall study more generally continuous surjections of a compact metric space with the properties (A) and (B). This approach will turn out to be a reasonably clear and coherent one. It can be shown that (A) and (B) are equivalent to (A) and (B)' f has the pseudo orbit tracing property (POTP).

INTRODUCTION

8

A sequence of points, {z,}, is said to be a 6-pseudo orbit if d( f (z,), z,+1) 5 6 for n. This is quite a natural notion since on account of rounding errors a computer will actually calculate a pseudo orbit, rather than an orbit. A 6pseudo orbit (2,) is said to be €-traced by an orbit { f ,(z)} if d( f n(z),2), 5 E for all n. Thus the pseudo orbit is uniformly approximated by a genuine orbit. We say that f has POTP if for each E > 0 there is 6 > 0 such that every &pseudo orbit can be &-tracedby some orbit. It can be deduced that if a homeomorphism f : X + X of a compact metric space has (A) and (B)' then it is topologically stable, i.e. for every E > 0 there is 6 > 0 such that if g : X t X is any homeomorphism with d ( g ( z ) ,f ( z ) )< 6 for all E X,then there is a continuous map h: X + X with h o g = f o h and d ( h ( z ) ,z)< E for all z E X. The same definition applies to a continuous surjection. From topological stability follows the structural stability theorem of Anosov diffeomorphisms of a compact smooth manifold. Let M be a compact smooth manifold and Diffl(M) the set of all diffeomorphisms endowed with the C' topology. The structural stability theorem says that there is a C' neighborhood U of an Anosov diffeomorphism f such that for every g E U,there exists a homeomorphism h: M t M for which g o h = h o f holds, i.e. g is topologically conjugate to f . The structural stability means that even if the dynamics is very complicated, the perturbed systems behave in the same way. By the work of Robbin and Robinson it was shown that if a C' diffeomorphism f: M --f M of a compact smooth manifold is Axiom A and satisfies strong transversality (see the following definitions) then f is structurally stable. Finally MaiiC (1988) has shown that the converse is true. A point z belonging to M is said to be a nonwandering point if for any neighborhood U of z there is an integer n # 0 such that U n f n ( U ) is nonempty. The set a(f)of all nonwandering points is called the nonwandering set. Clearly a(f ) is closed in M and invariant under f . A diffeomorphism f : M --f M is said to be Adorn A if a(f ) has a subset of all periodic points which is dense in a(f ) and a(f ) is a hyperbolic set of f . Topological decomposition theorem says that if f is Axiom A then a(f ) can be written as the finite disjoint U Ad of closed invariant sets Ai such that each fini is union a(f)= A1 U topologically transitive. Such a set Ai is called a basic set. For z E M the stable and unstable manifolds are defined by

---

Wa(z) = {Y E M : d( f "(z),f"(y)) 0 as n t oo}, WU(z)= ( 9 E M : d ( f F n ( z )f,- " ( y ) ) 0 as n t oo}. --f

--f

The stable manifold theorem ensures that if f is Axiom A then each of W s ( z ) and WU(z) is actually a manifold which is an injective immersion of some euclidean space, and M = U+En(f)W'(z) for B = s,u. We write Wu(Ai)= UBEAi W'(z), 0 = s,u, for basic sets Ai. A relation between the basic sets hi, 1 5 i 5 s, is defined as Ai > A, if (Wd(A,)\Ai) nwU(Aj)# 0, that is, there is

INTRODUCTION

9

a point x which comes from Ai and goes to A j by negative iterations of f. We say that f has no cycles with respect to O(f) if AiQ > Ail > . * - > Aij > AiQ cannot happen between the basic sets. An Axiom A diffeomorphism f: M -+ M is said to satisfy strong transversality if, for x E M, the stable manifold W " ( x )and the unstable manifold W"(x)are transversal, i.e. T,M = TzWs(x) T,W"(x). Fkom Smale's theorem it follows that f has no cycles with respect to n(f) if it satisfies strong transversality. The structural stability of differentiable maps is defined in a form analogous to that of diffeomorphisms. However, it is known (by Przytycki) that if the manifold M has dimension 2 2 then every Anosov differentiable map which is neither injective nor expanding does not satisfy structural stability. This indicates that the behaviour of an Anosov differentiable map is more complicated than that of an Anosov diffeomorphism. We can also give the notion of Axiom A for a differentiable map in a similar fashion. Then every differentiable map satisfying Axiom A yields the topological decomposition theorem. Thus we can take of no cycles condition for differentiable maps. However we can find in the literatures no definition of strong transversality for differentiable map. This seems likely due to the obstruction caused by singularities. In particular let M be the unit circle and R ( M ) the set of regular maps of M endowed with the C 1 topology. Then it can be shown that the set of maps in R ( M ) statisfying structural stability coincides with the set of maps in R( M ) satisfying Axiom A and no cycles condition.

+

Dynamical Systems in the General Situation. We now return to the general situation to think of a continuous surjection, or a homeomorphism of a compact metric space. A &pseudo orbit (2,) is said to be periodic if there is q > 0 such that x , + ~ = x, for all n. A point x E X is said to be chain recurrent if for 6 > 0 there is a periodic &pseudo orbit which passes through x. The set C R of all chain recurrent points is a closed set and f ( C R ) C C R (f (CR ) = C R for a homeomorphism). Chain recurrence is useful in certain aspects. It can be shown that if a continuous surjection f of a compact metric space has the properties (A) and (B)' then every chain recurrent point is a limit point of periodic points and thus f ( C R ) = C R . Moreover, the chain recurrent set C R is the union of finitely many disjoint closed subsets R1, ,R, invariant under f, and such that the restriction of f to each Ri is topologically transitive. The sets R1, ,R, are called basic sets. If, in particular, s = 1 then we have CR = X. Each set R; is the union of a finite number mi of disjoint closed subsets, which are permuted cyclically by f in such a way that the restriction of f mito each of subsets is topologically mixing. This topological decomposition theorem gives us a concept of how the continuous surjection behaves on the chain recurrent set C R . A more precise description of the behaviour on the chain recurrent set can be given. This is useful only in case f is a homeomorphism. It can be shown that for each set

---

-..

INTRODUCTION

10

Ri there is a Markov subshift (Ec,~, u ) and a continuous map ai of Ec,~onto Ri, with ai o u = f o ai, such that each point of Ri is the image under ai of at most d points of C A ~for , a certain positive integer d. This implies that the homeomorphism f restricted to Ri behaves as the shift map u restricted to Ec,~does. Even if a continuous surjection has the properties (A) and (B)', the restriction of the map to a basic set can not necessarily be realized by a Markov subshift under at most d points to one continuous map. More precise descriptions of the behaviours of dynamics having POTP and of expansive dynamics will be found in the following chapters.

Topological Anosov Maps. We return to the case of a continuous surjection of the torus. To see how the dynamics is complicated, we raise the question whether a continuous surjection with the properties (A) and (B)' of an n-dimensional torus is Co-perturbed to a certain hyperbolic toral automorphism (endomorphism). To deal with this problem, let C(T") denote the set of all covering maps of an n-dimensional torus T". A continuous surjection f : T" --+ T" is a covering map if f is a local homeomorphism. Let f = f A : T" --+ Tn be a hyperbolic differentiable map induced by an n x n matrix A. Then f is a differentiable map belonging to one of the following three types : (I) f is a toral automorphism and A has eigenvalues of modulus < 1 and modulus > 1, (11) f is a toral endomorphism and all eigenvalues of A are larger than one, when f is called an e v a d i n g toral endomorphism, (111) f is a toral endomorphism which is not injective, and A has eigenvalues of modulus < 1 and modulus > 1. We say that a continuous surjection f : W --+ T" is a topological Anosov map if f is c-expansive and has POTP. A continuous surjection f is said to be a special topological Anosov map if f satisfies the conditions : (1) f is a topological Anosov map, (2) for every ( z i ) , ( y ; )E (T"),with 10 = yo

where W u ( ( z i )= ) {yo E T" : there is ( y i ) E

(T")fsuch that d(zi,ya) + 0 as i -+ -m}

(this set is called an unstable set). There exist continuous surjections that satisfy some condition stronger than the one satisfied by special topological

INTRODUCTION

11

Anosov maps. A continuous surjection f:'Ir" + 'IP' is said to be a strongly special topological Anosov map if f satisfies the conditions : (1) f is a special topological Anosov map which is not injective, (2) for each z E X define the stable set Wa(z) = {y E T" : d( f"(z), fn(y)) --t 0 as n -, 00); then the path connected component of x in W " ( z ) is dense in T". The complicated behaviours of topological Anosov maps belonging to C(Tn) can be classified by use of topological methods. Among the main results to be obtained in this book are the following statements: (i) Every topological Anosov homeomorphism belonging to C('Ir") is topologically conjugate to a toral automorphism of type (I). (ii) Every positively expansive map belonging to C(Tn)is topologically conjugate to a toral endomorphism of type (11). (iii) Every strongly special topological Anosov map belonging to C(T") is topologically conjugate to a toral endomorphism of type (111). (iv) If a topological Anosov map which is injective is neither positively expansive nor strongly special, then its inverse limit system is topologically conjugate to a solenoidal group automorphism, in other words, to the inverse limit system of a toral endomorphism of type (111).

Ergodic Theory. This book contains arguments on measure-theoretical aspects of continuous surjections of a compact metric space. At the origins of ergodic theory were problems in statistical mechanics, related to Hamiltonian flows, one-parameter groups of diffeomorphisms preserving Liouville measure, geodesic flows etc. However, many theorems in ergodic theory are formulated in a much simpler setting, i.e. the main objects of the (ergodic) theory are measure preserving transformations of measure spaces. In the last chapter of this book we shall discuss some results (described below) of ergodic theory to apply them to topological dynamics. Let X be a set. A o-algebra of subsets of X is a collection B of subsets of X such that (1) 23 contains the entire space X, (2) X \ B E B when B E B, (3) B; E B when each Bi contains in B.

ui21

A measure space is a triple ( X , B , m ) where m is a function m : B --t R+ satisfying m(U;=, B,) = C,"==, m(B,) if { B , } is a pairwise disjoint sequence of elements of B. If m(X)= 1 then (X, 23,m)is called a probability space. We say that a map f:X+ X is a measure preserving transformation if E E B implies f -l(E) E B and m(f -'(E)) = m(E). Here m is said to be f -invariant. Let f:X--t X be a measure preserving transformation of a probability space (X, B,m). For a function ( : X -+ R such that Jx ((ldm < 00 we define ) the space mean of 6 to the time mean of ( to be limn+oo Cyzl < ( f i ( z )and

INTRODUCTION

12

sx

be ((z)dm. The first major result in ergodic theory is the ergodic theorem which was obtained in 1931 by G.D.Birkhoff. The ergodic theorem says that these means are almost everywhere equal if and only if f: X + X is ergodic, i.e. if f(E) = E and E E L? then m(E) = 0 or m(E) = 1. Such a measure m is called an ergodic measure. Let X be a compact metric space and 23 the family of Bore1 sets, i.e. the smallest a-algebra containing all open subsets of X . If C ( X ) is the B a n d space of continuous real-valued functions of X with the sup norm, then every probability measure p on X induces a non-negative linear functional on C ( X ) by the map H S t d p . Conversely, the Riesz representation theorem says that to any non-negative linear functional J on C ( X ) with J(l) = 1, there ( d p for corresponds a unique probability measure p on X such that J ( t ) =

sx

€ E C(X).

The space M ( X ) of all probability measures on X is obviously a convex set. We may define a topology in M ( X )by taking as a basis of open neighborhoods for p E M ( X ) the sets

VAG, * * *

,€k; € 1 ,- - ,a) *

with ~j > 0 and < j E C ( X ) . This topology is called the weak topology of M ( X ) . With this topology M ( X ) becomes a compact metrizable space. A sequence pn E M ( X ) converges to p if and only if one of the following equivalent conditions holds : (1) limn+, JtdPn = J ( d p for dl t E C ( X ) , (2) liminf,,, pn(U) 2 p ( U ) for all open U c X , (3) limn+- pn(A) = p(A) for all A E 23 with p(3A) = 0. Here 8A denotes the boundary of the set A. The support Supp(p) of a probability measure p is the smallest closed set C with p ( C ) = 1. Equivalently, Supp(p) is the set of all z E X with the property that p ( U ) > 0 for any open U containing x. Let f : X + X be a continuous surjection of a compact metric space. It can be shown that the set M ( f ) of all f-invariant probability measures on X is nonempty and a compact convex subset of M ( X ) . As before u : Yt + Yf is the shift map for a fixed k 2 1. A probability measure p on Yt is a-invariant if for all cyclinders [at, ,a,], p satisfies p( [at, ,a,]) = p( [ae+l, ,a,+l]). In the set of a-invariant probability measures, Bernoulli measures are important. The k-tuple T = ( p l , . . . , p k ) with pj 2: 0 and p j = 1 defines a Bernoulli measure p by

-.

-

-.

zfl

at, * *

* 9

am]) = patpac+l

* *

.Pam*

This is the simplest type of probability measures on Y f . The system (Y,", p, a) is called a Bernoulli shift if p is a Bernoulli measure.

INTRODUCTION

13

Probability spaces (X,B,p) and (X',B',p') are said to be isomorphic if there are M E 0 with p ( M ) = 1,M' E B' with p'(M') = 1 and a bijection h: M -+ M' such that h and h-' are measurable and p(h-l(A')) = $(A') for all A' in the restriction of B' to M'. If a measurable map h: M + M' is surjective and satisfies p(h-'(A')) = p(A') for all A' in the restriction of €3' to M',then we say that (X', B',p') is a factor of (X, a, p). Let f:X + X and f' : X' -+ X' be measure preserving continuous maps of compact metric spaces. Then (X,B,p, f ) and (X',B',p', f') are said to be measure theoretically conjugate if h is a measurable bijection and satisfies h o f = f' o h. When h is a measurable surjection and satisfies h o f = f' o h, ( X ' , B', p', f') is called a factor of ( X ,B,p, f ). A system (X, 23, p, f ) is called a Bernoulli shift if the system is measure theoretically conjugate to a Bernoulli shift (Y,", p, u). A probability measure p is said to be absolutely continuous with respect to u, written p 0. Then 'Xe, and so A is not ergodic then A , has 1 as an eigenvalue. Thus 2 has an Gth root of unity as an eigenvalue. Conversely, if has 1 has an L-th root of unity as an eigenvalue, then as an eigenvalue and so - I induces a singular linear map of R" where I denotes the identity matrix. Thus - I induces a many-to-one map from Z" into Z". Therefore there is 0 # m E Z" such that = 124.

-e

txe

'xe

'xe

Remark 1.1.5. An endomorphism A : T" + T" is ergodic if and only if there is 20 E T" such that the orbit OA(ZO)= {Ai(zo)}E0 is dense in T". Suppose A is ergodic. Let {uk}& be a countable base for P. Then we have that {z E T" : c l ( O ~ ( z ) # ) T"} = A-'(T" \ u k ) (see $02 and 4 of Chapter 2). Since \ uk)is A-invariant (p-a.e.), by ergodicity we have measure 0 or 1. Notice that the Lebesgue measure is a probability Bore1 measure of which gives a positive measure to every nonempty open set. Since uk is open, we have p ( n z oA-i(Tn \ uk))= 0 and so p ( { z E F : cl(OA(z)) # F}) = 0. Therefore, p ( { z E T" : cl(OA(z)) = T"}) = 1. Conversely, suppose O A ( ~ Ois)dense in Y. Since T" is connected, OAi(z0) is dense in T" for all i > 0. If y o Ae = y for some 1 > 0 and a character y, then we have y = constant. Therefore A is ergodic by Remark 1.1.3.

nzo

uEl

Remark 1.1.6 (Kroneker). Let Z[z] denote the ring of polynomials in z with integer coefficients. If all the roots X of p ( z ) E Z[z] has 1x1 = 1, then each root X is a root of unity. Remark 1.1.7. Let A : a4 --f

a4 be an automorphism

corresponding to

$1.2 Anosov differentiable systems

19

Then the eigenvalues of the matrix are

(3 - & fi4=)/4

(3 + & - i46& Thus A :

+

( both absolute value = l),

- 2)/4

(x 0.46).

T4 is ergodic. We leave the proof to the readers.

Remark 1.1.8. matrix

Let

2 : W2

+

R2 be an linear map corresponding to the

Then the eigenvalues of the matrix are (3 f 4i)/5 ( absolute values = 1). then G is an Denote as G the smallest subgroup generated by A-invariant subgroup and Z2 c G = X(G) c W2. Suppose G is endowed with the discrete topology. Then the character group of G, S 2 , is a solenoidal group (see Chapter 7 for details). Since Z2 # G, S2 is not a torus. As described in Chapter 7 an automorphism A : S2 + S2 is induced by the transposed matrix t?l. Then A preserves the probability Haar measure. By applying Remark 1.1.3 we can check that A is ergodic. However, notice that Remark 1.1.4 does not hold.

urmx'(Z2),

Though ergodic automorphisms are not necessarily hyperbolic, it was proved (Katznelson [Ka]) that every ergodic toral automorphism is a Bernoulli shift. After this the same result was checked for solenoidal group case ( [ Ao ~ ][Li], , [Mi-Tl], [Mi-TZ], [Mi-T31). Aoki's proof [A061 is much simpler than the other extant proof. $1.2 Anosov differentiable systems

The survey presented in this section is written in view of the differentiable dynamics, and t o give the reader an understanding of some of the basic properties in Anosov differentiable systems. Let M be a closed smooth manifold, i.e. a compact connected smooth manifold without boundary. A C1 diffeomorphism f : M + M is called an Anosov diffeomorphism if there is a splitting of the tangent bundle TM = E"$E" which is preserved by the differential D f off and if there are constants C > 0, 0 < p < 1 and a Riemannian metric 11 11 on TM such that for all n 2 0 (1) (2)

llDf"(v>Il I Cp"llvll

for v E E", 11Df"(v)11 2 C-lp-nllvlI for w E E".

20

CHAPTER 1

A typical example of such diffeomorphisms is the toral automorphisms of type (I). The definition of Anosov diffeomorphism does not depend on the choice of Riemannian metrics because M is compact. For 2 E M define the following subsets :

W '(z) = {y E M : d(fn(z),fn(y)) W,d(z)= {y E M : d(f"(z), f"(y))

0 as n --t oo}, I E for all n 2 0},

---t

W u ( z )= {y E M : d(f-"(z), f-"(y)) + 0 as n + oo},

W,U(z)= {y E M : d(f-"(z), f-"(y)) 5 E for all n 2 0) where d is the metric for M induced by a Riemannian metric on TM. Then the behaviour of Anosov diffeomorphisms is characterized as follows. For E > 0 small (A) W,"(z)and W,U(z)are C1disks, T,Wl(z) = E i and T,W,U(z) = E"I> (B) there exist constants c > 0, 0 < X < 1 such that for all n 2 0

d(f"(z),f"b))I

cX"d(w)

d(f-n(z),f-n(Y)) I cX"d(z,

for Y E W,d(z), for

E W,U(4,

(C) W,"(z)and W,"(z)vary continuously with respect to z. See Hirsch and Pugh [H-PI. From (B) we have W,"(z) C W'(z) for z E M (a= s , u ) and then

W " ( z )=

u f-n(~,d(f"z)), u

"20

W"(z)=

f"(W,U(f-"z)).

n10

Because W:(z) intersects W,U(z)transversally and W,"(z)n W,U(z)= {z}, we have that for E > 0 there is 6 > 0 such that WF(z)n W,"(y) consists of a single point, say [z,y], whenever d ( z , y ) < 6. Then the map

[ , ]:{(z,y) E M x M : d(z,y) < 6)

---t

M

is continuous. For the interesting properties of Anosov diffeomorphisms, the readers may refer to Anosov [Anl], Smale [Sml, Sm21, Franks [Fl, F2], Manning [ManS], and Brin and Manning [B-MI. Let C ' ( M , M) be the set of all C' maps of a closed smooth manifold M endowed with the C' topology. A map f E C ' ( M , M) is said to be ezpanding if there are constants C > 0, 0 < p < 1 and a Riemannian metric 11 11 on TM such that for all n 2 0 and allv~TM

11~f"(~~11 2 CCL-nll"II.

51.2 Anosov differentiable systems

21

A typical example of such maps is the toral endomorphisms of type (11). If f: M -+ M is expanding, then there exist constants So > 0, 0 < X < 1 such that for z,y E M

(D)d(z,y) I 60

* d(f(z),f(y)) 2 X--'d(%Y)

under some Riemannian distance. From (D) it is checked that every expanding differentiable map has fixed points (see the proof of Theorem 2.2.14 of the next chapter). However it is unknown for Anosov diffeomorphisms. For the properties of expanding differentiable maps, the readers may refer to [F2], Shub [Shl], Gromov (Gr], and Shub and Sullivan [S-S]. The above two classes of dynamical systems were extended to a wider class by Przytycki [Pr] as follows. A map f E C'(M, M ) is called an Anosov diferentiable map if f is a C' regular map and if there exist constants C > 0, 0 < p < 1 and a Riemannian metric 11 11 on TM such that for a sequence (2,) of points in M satisfying f (2,) = z,+1 for every integer n, there is a splitting

u n=-m

u E:,$E,", 00

00

T,,M=

=Ea$EU

n=--a,

which is preserved by the differential Df,and conditions (l),(2) mentioned above are satisfied. Note that EZo depends on the sequence (z,). Thus it may Such a phenomenon is happen that E:o # E:o if 20 = yo but (2,) # (y,). impossible for E& , that is, it depends only on 20. Let ( x i ) be a sequence of points in M satisfying f (xi) = zi+l for i E Z and define a set W z ( ( z i ) )= {YO E M : 3 ( y i ) s-t. d ( z - i , y - i )

5 ~ , 2i 0)

where d is the metric for M induced by a Riemannian metric on T M . The behaviour of Anosov differentiable maps was characterized in [Pr] as follows. For E > 0 small (E) W,"(z) and W,U((zi)) are C' disks, T,W,"(z)= E i and TzoW:((zi))= Ego3 (F) there exist constants c > 0, 0 < X < 1 such that for all n 2 0 d(f"(z), f%)) I cXnd(z,Y) for Y E Wcd(z), d(z-n,Y-n) 5 cX"d(zo,yo) for some ( y i ) with yo E W:((zi)), (G) W,"(z) varies continuously with respect to z,and if (z?)converges to (zi) as n -, 00 then W,U((zCq)) varies continuously to W,U((zi)). If a sequence ( z i ) satisfies 20 = z,then W,"(z) intersects W,U((zi)) transversally and W,"(z)nW,"((zi)) = {z}.Thus we have that for E > 0 there is S > 0

CHAPTER 1

22

such that W : ( z ) n W : ( ( y i ) ) is exactly one point [z, (yi)]whenever d ( z , y o ) < 6. Then the map

[9

1

{ ( z , ( ~ iE) M )

x Mf : ~ ( % , Y O< ) 6)

+

M

is continuous. Here Mf = { ( z i ) : zi E M and f ( z i ) = z i + l , i E Z} is a closed set of the product topological space M Z = { ( z i ) : Z i E M , i E Z}. A map f E C 1 (M , M ) is called a special Anosov differentiable map if f is an Anosov differentiable map and E," does not depend on the sequence (2,) with zo = z. Thus every special Anosov differentiable map has a splitting of the tangent bundle as well as diffeomorphisms. A typical example of such a map is the toral endomorphisms of type (111).The class of special differentiable Anosov maps contains Anosov diffeomorphisms, expanding differentiable maps and the products of them. For the properties of Anosov differentiable maps and special Anosov differentiable maps, the readers may refer to Maii6 and Pugh [M-P] and Przytycki PrIA diffeomorphism f: M -, M is said to be expansive if there is a constant e > 0 , called an expansive constant, such that if z , y E M and 2 # y then

for some n E Z.We say that a differentiable map f: M 4 M is positively expansive if there is e > 0 such that z # y implies d ( f n ( z ) ,f n ( y ) ) > e for some n > 0. When there is e > 0 such that for ( z i ) , ( y i ) E Mf if d ( z i , y i ) 5 e for all i E Z then (zi)= ( y i ) , the differentiable map f: M M is said to be c-expansive. A sequence of points {zi E M : a < i < b} is called a 6pseudo orbit of f if d ( f ( z i ) , z i + l )< 6 for i E ( a , b - 1). For given E > 0 a &pseudo orbit { z i } is said to be E-traced by z E M if d ( f i ( z ) , z i )< E for i E ( a ,b). Here a and b are taken as -00 5 a < b 5 00 if f is bijective, and as 0 5 a < b 5 00 if f is not bijective. We say that f has the pseudo orbit tracing property (abbreviated POTP) if for E > 0 there is 6 > 0 such that every &pseudo orbit of f can be E-traced by some point of M. For Anosov systems we have the following

Theorem 1.2.1. (1) every Anosov diffeomorphism has expansivity and POTP, (2) every expanding diflerentiable map has positive expansivity and POTP, ( 3 ) every Anosov diflerentiable map has c-ezpansivity and POTP. Proof. Expansivity of (1) and (3) follows from the facts that W,"(z)n W ; ( z ) = {z} and W,J(z) n W:((zi)) = {z} respectively, and that of (2) follows from (*). We give the proof of P O T P of (l),(2) and (3). It is clear that if N > 0 and E > 0 then there is a > 0 such that each finite a-pseudo orbit {zi : 0 5 i 5 N} satisfies d ( f i ( z o ) , z ; )< E (0 5 i 5 N ) .

$1.2 Anosov differentiable systems

23

We first show POTP of (1) by using the proof due to Bowen [Boll. It is enough to see that for p > 0 there is a > 0 such that each a-pseudo orbit { x i : 0 5 i 5 a} is &traced by some point. Indeed, let { z i : i E Z} be an a-pseudo orbit of f. For k > 0 we set x i = zi-k ( 0 5 i 5 2k). Then {xi : 0 5 i 5 212) is an a-pseudo orbit o f f , and so it is @traced by some point w k E M . It is easily checked that an accumulation point of {fk(wk)} &traces

(4. Let

E

> 0 be a number to be determined later W,"(z)n W,"(y) # 0

if

and 6 > 0 such that

d(z,y)

< 6.

Choose N so large that C X ~ E< 6/2 and then there is a > 0 such that if { y i : 0 5 i 5 N} is an a-pseudo orbit then d ( f i ( ( y o ) , y i ) < 6/2 for 0 5 i 5 N. Let r > 0 and consider an a-pseudo orbit { x i : 0 5 i 5 TN}. Put xkN = X , N and then d(fN(x(r-l)N),xiN) < 6. Take V i N E W:(fN(x(r-l)N)) n W:(xkjV) and put = f - N ( y i N ) . Then we have

~ i ~ - ~ ) ~

by the choice of a. Thus d ( f N ( z ( , _ 2 ) N ) , x ; r - l ) ~ ) < 6. By induction { x i N : 0 5 i 5 r } is constructed. Now set x = xo and for 0 5 i 5 r N choose a such that sN 5 i < (s l ) N . Then we have

+

since f N ( xrt N ) E W ~ ( Z ' ( ~ - ~On ) ~the ) . other hand, since j N ( x b N ) E W 2 ( f N ( 2 , ~ ) we )~

have d ( f i - 8 N ( ~ : N ) fi-8N ,


. Since d ( d 2 ) , zn-') < E , we have d ( z , , - l , d 2 ) ) < 6(X2 A) and obviouslyd(z(2), f ( z n - 2 ) ) < 6(X2 + X + l ) . In this fashion we have that for di)E Uqxi+...+A+') ( f ( Z n - i ) ) there is di+l)E Ue(zn-i)such that

+

Ji)

= f(#+')),

+ - + A), d(x(i+'),f(x,,-i-')) < 6(X' + + x + 1). d(x(i+l),x,,-i) < 6(X'

* *

* * *

Since n is arbitrary, the conclusion is easily obtained. 0 Theorem 1.2.2. Let A : T2 T2 be a tom1 endomorphism. Then there exists a C" regular map g: T2 + T2 near to A under Co topology such that (a) i f A is a hyperbolic automorphism then g is an expansive C" difleomorphism with POTP that fails to be Anosov, (b) i f A is a hyperbolic endomorphism of type (111) then g is a c-expansive C" regular map with POTP that fails to be Anosov.

For the proof we need some preparations and so we shall give it in Chapter 9.

Remark 1.2.3 (Lewowicz [Ll]). An example of expansive diffeomorphism that fails to be Anosov is constructed as follows. We define a diffeomorphism g:T2 + T2 by 1

+y,z - -sin2~z + y 2T Then g is non-hyperbolic at x = y = 0 because of

But g is expansive. This is followed by defining a Lyapunovfunction V((zl, y l ) , ( X 2 , Y Z ) ) by = -(Y2 - Y l ) ( Y 2 - 2 2 - ( Y l - x1)).

v

51.2 Anosov differentiable systems

25

In fact, V : T2 x T2 4 R is continuous, V(z',z')= 0 for z' E T2 and V(g(z'), g(y')) -V(z',y') > 0 whenever 0 < d(z',y') 5 a for a > 0 small enough. Then g is expansive. This is easily checked as follows. Let 0 < e < a. Then e is an expansive constant for 9. Indeed, suppose 3: # y and d(g"(z),g"(y)) 5 e for all n E %. If we have V(z,y)2 0, then V(g(z),g(y)) 2 E for some E > 0 since V(g(z),g(y)) > V(z,y).Since T2 is compact, there exists 6 > 0 such that for u,v E T2 ). < 6 V M u ) ,d v ) ) < E , from which we have d ( z , y ) 2 6. Since V(g2(z),g2((y))> V(g(z),g(y)) 2 E , we have d(g(z), g(y)) 2 6 and by induction

*

4%

L 0.

6 5 d(gn(z),gn(P)) 5 e,

Define

v = min{V(g(u),g(v)) - v ( ~ , v : (>u , ~E) T ' satisfying 6 5 d(u,v) 5 e}. Then we have

2 y + V(gn(z),gn(y)), n L 0

v(g"+'(z>,g"+'(Y)) and thus

V(g"(s1, g n ( y ) )L nv

+ V b ,Y) L nv,

n L 0However, we have max{V(u,v) : (u,v) E T2}< 00, thus contradicting. For the case V(z,y) 5 0 we obtain the conclusion by the same way. Let X and Y be compact metric spaces and let f :X + X and g:Y + Y be continuous surjections. Then f is said t o be topologically conjugate to g if there exists a homeomorphism 'p : Y 4 X such that f o 'p = cp o 9. If 'p is a continuous surjection, then f is said to be toplogically semi-conjugate to g, in other words, f is called a factor of g. When f is topologically conjugate to g, let us define a relation f g. Then the relation is an equivalence relation on the class of all continuous surjections. If f g, then dynamics of f and g are regarded as the same in topological sense because there is a homeomorphic correspondence between orbits of f and orbits of g. A homeomorphism f :X X is topologically transitive if there exists 20 E X such that the orbit { f " ( ~ 0 ) : n E Z} is dense in X. If, for all z E X, the orbit { f"(z) : n E Z} is dense in X, then f is called a minimal homeomorphism.

-

-

N

.--)

Let f : X X and g : Y ---t Y be homeomorphisms and suppose f is topologically conjugate to g. Then f is expansive if and only if so is g, and f has POTP if and only if so does g. Also, f is minimal if and only if so is g , and f is topologically transitive if and only if so is g.

Remark 1.2.4.

---f

It is known that every expansive homeomorphism g:T2 4 T2 is topologically conjugate to a hyperbolic toral automorphism. See Hiraide [Hi51 and Lewowicz [L2].

CHAPTER 1

26

Remark 1.2.5 (Coven and Reddy [C-R]). As an example of a positively expansive C" regular map on the circle that fails to be expanding, we have the following: Let 7 : R + R be a C" regular map satisfying (1) 'f(0) = 0 , m = 2, (2) T ( 0 ) = T'(1) = 1, T I ( . ) > 1 (0 < 2 < l), (3) = 2 n + f ( z - n ) ( n 5 2 5 n + l , n E Z). Let f:T' --+ T1 be a C" regular map of the torus induced by 7. Then f is positively expansive but not expanding. This is checked as follows. Since Dfn(u) = u for n 2 1 and all tangent vectors u at z = 0, f is not expanding. To show positive expansivity o f f we first show that there is 6 > 0 such that if [T(z) -fi(y)l < 6 for all i 2 0 then 2 = y, that is, f : R --+ R is positively expansive. After that we see that f is positively expansive by applying Lemma 2.2.34 of the next chapter. Choose 0 < 6 < 1/2 such that

7(z)

1 5

and X

- yI < 26

> 1 such that 1 2

-

IF(2)-P(y)I < 1/2

- n~ 2 6 ( n E Z)

Then we have the following : If infinitely many i 2 0 such that

T'(z> 2 A.

Ti(%)9 Z for all j

2 0, then there exist

If(z) - nl 2 26

for all n E Z. Indeed, if this is false and k is the largest integer with n1 2 26 for all n E Z, then there exists n E Z such that either

Without loss of generality we suppose the former. -k+l (2)- n. By induction it is easily checked that 0 Let y = f for all i 2 0. The mean value theorem derives that

).(kfI

-

< T'(y) < 26

where 0 < ti < T ( y ) . Since fi(y) < 1, we have ? ' ( t i ) > 1. Therefore {f(y)} is increasing. Since it is bounded, T ( y ) + z < 1 and then T ( z ) = E. But f ( z ) = f ' ( t ) z where 0 < t < 1, and so y ( z ) # z. This is a contradiction.

51.2 Anosov differentiable systems

i

27

We now prove that 6 is an expansive constant for 7. For x, y with x

> y if

+ j f ( z ) , 7 J ( y )E Z for some i,j 2 0, then i f ( x ) , p J ( y )E Z and so i + j

If

i + j

(z)-f

(Y)l

2 6.

Suppose 7'(x) 4 Z for all j 2 0. The sequence decreasing. This follows from the fact that

{I?(.)

- T(y)l}

is non-

for some ti with ?(y) < ti < f ( x ) . Since f'(x) @ Z for all j E Z, there are infinitely many i > 0 such that If(x) - nl 2 26 for all n E Z. For any such i and any n E Zwe have that l?(x) -

?(?I)/< 6 * [ti - nl L I?(%)

- n l - l?(x) - ?(?/)I > 6

).('+'fI

and thus T ' ( t ; ) 2 A. If If'(x) - 'ji(y)l < 6 for all i 2 0, then f (y)l 2 Al?(x)-f'(y)l for infinitely many i 2 0. Therefore { T ( x ) - F ( y ) } is unbounded, thus a contradiction. Therefore 7 is positively expansive. i + l

A map f E C1(M,M) is said to be structurally stable if there is an open neighborhood N ( f ) of f in C ' ( M , M) such that g E N ( f ) implies that f and g are topologically conjugate. Anosov [Anl] proved that every Anosov diffeomorphism is structually stable, and Shub [Shl] showed the same result for expanding differentiable maps. Przytycki proved in [Pr] that Anosov differentiable maps do not be structurally stable; more precisely this is mentioned as follows, every non-empty open subset of the set of all Anosov differentiable maps of class C', which are not diffeomorphisms nor expanding, contains an uncountable subset such that, if f # g are elements of its, then there exist no continuous surjections cp : M + M which make the commutative diagram

M

L

M

M-M 9

We denote as Diff (M) the set of all C' diffeomorphisms of a closed smooth ' topology where 1 5 T 5 00. For f E Diffl(M), manifold endowed with the C MaiiC proved recently in [Ma31 that every structurally stable diffeomorphism of a closed smooth manifold satisfies Axiom A, and Palis showed in [PI that every $%stable diffeomorphism satisfies Axiom A. A diffeomorphism satisfying the conditions (1) all periodic points of f are hyperbolic, and (2) given all the pair ( x , y ) of periodic points, W " ( z )and Wu(y) meet t ransversally,

28

CHAPTER 1

is called KupLa-Smale. It is well known that the set of all Kupka-Smale diffeomorphisms is a residual subset of Diff(M) for all T 2 1. Denote as P ' ( M ) the set of f E Diff'(M) such that all the periodic points of f are hyperbolic. Obviously P 1 ( M )is a residual subset of Diff'(M). In [A051 Aoki proved that every diffeomorphism belonging to the interior of P 1 ( M ) satisfies Axiom A and no cycles, and that every diffeomorphism belonging t o the C1interior of the set of Kupka-Smale diffeomorphisms satisfies Axiom A and strong transversality. Maiik showed in [Ma41 that every diffeomorphism belonging to the C' interior of the set of all expansive diffeomorphisms satisfies Axiom A and T,W"(z) n T,W"(z) = (0) for z E M, which is called a quasi-Anosov difleomorphism. It was proved in Sakai [S3] that each diffeomorphism belonging to the C' interior of the set of all diffeomorphisms having POTP satisfies Axiom A and strong transversality. If the manifold M is a torus, then it is well known that every Anosov diffeomorphism is topologically conjugate to a toral automorphism of type (I) (Manning [ManS]) and every expanding differentiable map is topologically conjugate to a toral endomorphism of type (11) (Shub [Shl]). However the following problem is unsolved.

Problem 1. Is every special Anosov diflerentiable map of a torus topologically conjugate to a hyperbolic toral endomorphism ? One of aims of this book is to deal with the problem in topological setting. Moreover this book will present topological developments that could be the foundation for forward dealing with the following problem for the interesting class of topological Anosov maps. Let M be a closed topological manifold, i.e. a compact connected topological manifold without boundary. If a continuous surjection f: M + M is c-expansive and has POTP, then f is called a topological Anosov map (abbreviated TA-map). By Theorem 1.2.1 every Anosov differentiable map is topological Anosov. We say that f: M + M is a special topological Anosov map (abbrev. special TA-map) i f f satisfies the conditions : (1) f is a topological Anosov map, (2) W " ( ( z i ) = ) W"((yi)) for ( z i ) ,(yi) E Mf with 20 = yo. It follows that every special Anosov differentiable map is special topological Anosov.

Problem 2. Is every special topological Anosov covering maps of an arbitrary closed topological manifold topologically conjugate to a hyperbolic infranil-endomorphism of an infm-nil-manifold ? Let L be a connected simply connected nilpotent Lie group. Let C be a compact group of automorphisms of L, and let G = L C be the Lie group obtained by considering L as acting on itself by left translation and taking the semi-direct product of L and C in Diff(L). Let I' be a uniform discrete

51.2 Anosov differentiable systems

29

subgroup of G. Then I'n L is a uniform discrete subgroup of L and I'/(rn L) is finite (see L.Auslander [Au]). If I' is torsion free, then the orbit space X = &/I' is a compact manifold. We call it an infm-nil-manifold. Let A: L -+ L be an automorphism of L such that by conjugating I' by A in Diff(L), AI'A-' c I'. Then A projects to a differetiable map of X. Such a map is called an infranil-endomorphism. Manning [Man31 proved that every Anosov diffeomorphism of an infra-nilmanifold is topologically conjugate to a hyperbolic infra-nil-automorphism. After this Brin and Manning [B-M] proved, by using a result related to growth of finitely generated groups of Gromov [Gr], that every Anosov diffeomorphism of an arbitrary closed smooth manifold satisfying a certain spectral condition is topologically conjugate to a hyperbolic infra-nil-automorphism. In [Gr] Gromov showed that every expanding map of an arbitrary closed smooth manifold is topologically conjugate to an expanding infra-nil-endomorphism. After this Hiraide proved in [Hi41 the same result for positively expansive maps. On the direction, he obtained that if a continuous map of a compact connected locally connected semilocally 1-connected metric space is open and positively expansive, then the space must be homeomorphic to an infra-nil-manifold and the map is topologically conjugate to an expanding infra-nil-endomorphism ([Hi6]). A topological space X is said to be semilocally 1-connected if for x E X there is a neighborhood U of x such that every loop contained in U with a base point x (i.e. a continuous map u : [0,1] -+ U satisfying u(0) = u(1) = x) can be deformed continuously in X to one point. We refer the reader to Smale [Sml, Sm21, Bowen [Boll, Irwin [I],Palis and de Melo [P-MI, Shub [Sh2], and MaiiC [Ma21 for background information of Anosov Systems. In Halmos and von Neuman's treatise on ergodic theory, they discussed algebraic models which indicate a characteristic of some subclass of homeomorphisms. We can find in their paper [Hal-N] the following theorem.

Theorem 1.2.6. Let f : X -+ X be a homeomorphism of a compact metric space. Iff is topologically transitive and an isometry under some metric for X , then f is topologically conjugate to a minimal rotation on a compact metric abelian group. Proof. Let d be an isometry metric for X. By assumption there exists 20 E X such that Of(x0) is dense in X. Define a multiplication * in Of(x0) by

f Yxo) * f m(xo)= f "+m(xO) for n,m E Z. Then we have

d(f"(x0) * f"(xo),fP(xo) * f9(zo)) = d(fn+m(201, P+*(xo))

5 d(f"+"(ZO), fP+"(xo))

+

+ d(fP+m(xo),f'+'(xo))

= d(fn(xO),fP(xO)) d(frn(xo),fQ(xo)),

30

CHAPTER 1

from which the map * : Of(z0) x Of(z0) + Of(z0) is uniformly continuous and can therefore be extended uniquely to a continuous map * : X x X + X. Since d ( f - ' Y 2 0 ) , f-'"(zo)) = d(f"(zo), f n ( z o ) )the , map : Of(z0) + O f ( 2 0 ) (inverse) is uniformly continuous and can be uniquely extended to a continuous map of X. Thus X is a topological group and is abelian since Of(z0) is a dense abelian group of X. Since

for all n E Z,we have f(z) = f(z0) * z for all rotation by f(z0). 0

2

E X, and therefore f is the

CHAPTER 2 Dyn a mi c s of Continuous Maps

It is sometimes said that the theory of dynamical systems is the study of the behaviours of orbits of maps. Among many ways of the study of dynamical systems, we select the path of general topology. We use expansivity and pseudo orbit tracing property, related to Anosov and Axiom A systems, as tools of study. This chapter provides these tools which are sufficiently powerful through out this book. 52.1 Self-covering maps Let X and Y be metric spaces. A continuous surjection f : X + Y is called a covering map if for y E Y there exists an open neighborhood V, of y in Y

such that

f-'(v,)

=

U U ~(i # 'i +-

~i

n vil = 0)

i

where each of Ui is open in X and flui : Ui + V, is a homeomorphism. A covering map f : X + Y is especially called a self-covering map if X = Y. We say that a continuous surjection f : X + Y is a local homeomorphism if for x E X there is an open neighborhood Ux of x in X such that f(Ux) is is a homeomorphism. It is clear that every open in Y and flu= : Ux -+ f(Ux) covering map is a local homeomorphism. Conversely, if X is compact, then a local homeomorphism f : X + Y is a covering map. This is stated precisely as follows: T h e o r e m 2.1.1 (Eilenberg [El). Let X be compact and f : X --+ Y a continuous surjection. Iff is a local homeomorphism, then them exist two positive numbers X and 8 such that each subset D of Y with diameter less than X determines a decomposition of the set f -l (D) with the following properties: (1) f-'(D)=D1UDzU.*.UDk, (2) f maps each Di homeomorphically onto D, ( 3 ) if i # j then no point of Di is closer than 28 to a point of D j , (4) for every 11 > 0 them is 0 < E < X such that if the diameter of D is less than E then each diameter of Dj is less than 7 . If, in addition, Y is connected, then there is a constant k > 0 such that f : X + Y is a 12-to-one map. The numbers (Ale) in Theorem 2.1.1 are called Eilenberg's constants for the map f : X + Y.For the proof we need the following lemma. 31

CHAPTER 2

32

Lemma 2.1.2. Let X be a compact metric space and a a finite open cover of X . Then there exists 6 = 6(a)> 0 such that for any subset A if the diameter of A is less than 6 then A c B for some B E a . Such a number 6 is called a Lebesgue number of a. Proof. If this is false, for any n > 0 there exists a subset A, such that B for all B E a. diam(A,) = sup{d(a,b) : a , b E A,} 5 l / n and A, Take x , E A,, for n 2 1. Since X is compact, there is a subsequence {x,,} such that x,, 4 x as i -+ 00. Then x E B for some B E a. Since X \ B is compact, we have a = d ( x , X \ B ) = inf{d(x,b) : b E X \ B} > 0. Take sufficiently large ni satisfying ni > 2 / a and d(x,, ,x ) < a/2. Then, for y E Ani we have d ( y , ~ ) I d ( ~ i x n i ) + d ( z n i , l~/ n ) Ii + a / 2 < a


0 such that if x # y and f ( x ) = f (y) then d ( x , y) > 48. If this is false, then for any n > 0 there exist X n , Y n E X with 2 , # Yn and f(x,) = f(y,) such that d(x,, y , ) 5 l / n . Without loss of generality we suppose that x, -+ x and yn 4 y as n + 00. Then x = y. Since f:X + Y is a local homeomorphism, there exists an open neighborhood Ux of x such that flu. : Ux4 f(Ux) is a homeomorphism. Thus x,, yn 6 U, for sufficiently large n and so f (2,) = f (y,), thus contradicting. Let k > 0 and define a set Yk by

Then we have Y&n Y&l= 0 for integers k # k'. Next, we prove that there is K > 0 such that Y = Y1 U U YK and each of Yk is open. Since is compact, by the above fact it follows that Yk = 0 for integers k larger than a certain K > 0. Thus Y = Yl U U YK.To show openness of Y& (1 5 k 5 K ) take y E Yk. Then f-'(y) = { ~ 1 , . - * , x k } . We can choose open neighborhoods Uxi of x i such that flu=,is a homeomorphism and Uxi n Uxj = 0 if i # j. Clearly

x

-.. ...

We claim that f ( U z i )contains a neighborhood U(y) of y such that U(y) C Yk. Indeed, if this is false, for n > 0 there is yn E f(Uxi) such that

$2.1 Self-covering maps

33

Iff-l(y") = {z?,..., z g , z g + l , - - - } a n d z l zi (asn + m ) f o r l 5 i 5 k+1, then we have that f ( ~ = ) y and d ( z i , z j ) > 48 for 1 5 i 5 k 1. Therefore, y E Yk+1 U * * * U YK. This is a contradiction. We are now ready to prove Theorem 2.1.1. Let 8 > 0 and K > 0 be as above. Since X is compact and f is a local homeomorphism, we can choose a finite open cover a = { U1, ,Urn) satisfying (a) f ( U i ) is open in Y, (b) f:cl(Ui) --+ f(cl(Ui)) is injective, (c) diam(cl(Ui)) < 8 for 1 5 i 5 m. Here cl(E) denotes the closure of E. Let y E Y, then y E y k for some k (1 5 k 5 K ) and so define Q(y) = y k fl ( n { f ( U i ) : y E f ( U i ) } ) . Then Q(y) is an open neighborhood of y. Since y is arbitrary in Y,we can take y1,---,ye such that U = { Q ( y l ) , - - . ,Q(ye)} is a finite open cover of Y. Let X > 0 be a Lebesgue number of U and let D be a subset satisfying diam(D) < A. Then D C Q(y) for some y E {yl,... ,ye}. If y E y k for some 12, then f-'(y) = {z1,-*,zk} and zi E Uni for some Uni E a. Notice that f-l(y) n Uni = { z i } for 1 5 i 5 k. We now define Di = Uni n f - ' ( D ) for 1 5 i 5 k. Then we have f ( D i ) = D for 1 5 i 5 k. Indeed, since Q(y) = Yk r l f(Uni)), if z E D then z = f(z) for some z E Uni. Thus z E Uni n f-'(z) c Uni n f - ' ( D > = Di and therefore z = f(z) E f ( D i ) , i.e. D c f(Di).Since /(Di)= f(Uni n f - ' ( D ) ) c f(Uni) n D = D,we have D = f ( D i ) for 1 5 i 5 k. (1)was proved. Since Di c Uni and D c f(Uni),fpi maps each Di homeomorphically onto D . Thus (2) was proved. (3) is computed as follows. If i # j , then d(Di,Dj) 2 d(Uni,Unj) 2 d ( z i , z j ) - diam(U,,,) - diam(Unj) > 48 - 8 - 8 = 28. Here d(A,B) is defined by d ( A , B )= inf{d(a,b) : a E A,b E B}. To show (4) let 1 > 0. For 1 5 i 5 m there is 0 < ~i < X such that if d(f(z),f(z)) < ~i (f(z), f(z) E f(cl(Ui))) then d(z,z ) < 7. This follows from the fact that f:cl(Ui) --f f(cl(Ui)) is injective. Put E = min{si : 1 5 i 5 m } . If diam(D) < E , then we have diam(Di) < 1 for 1 5 i 5 k. Since Y = Yk and each yk is open and closed, when Y is connected, we have Y = y k for some k, i.e. f is k-to-one. 0 --f

+

.- .

(n;=,

u,"==,

In the remainder of this section we describe well known theorems that will be used in the sequel. Let X be a topological space. A path in X is a continuous map from the unit interval [0,1] to X. If any two points in X are joined by a path, then X is said to be path connected. In general, a connected space need not be path connected. An arc in X is an injective continuous map from [0,1] to X. We say that X is arcwise connected if any two points in X are joined by an arc. It is clear that if X is arcwise connected then it is path connected.

Theorem 2.1.3. A Hausdorfl space X i s path connected if and only if X is arcwise connected.

CHAPTER 2

34

Proof. Let f : [0,1] + X be any path with f(0) # f(1). We now construct an arc h in X joining f(0) and f (1). Choose a closed subinterval I1 = [a1,bl] C [0,1] whose length 0 5 l ( I 1 ) = bl - al < 1 is maximal, satisfying the condition that f ( a 1 ) = f ( b 1 ) . The existence of such a subinterval is ensured by the fact that X is Hausdorff. Next, among all subintervals of [0,1] which are disjoint from 11 , choose an interval I2 = [az,bz] of maximal length satisfying f ( a z ) = f ( b 2 ) . Continue this process inductively, we can construct disjoint subintervals of maximal lengths e ( 1 1 ) 2 C(I2) 2 . 2 0 satisfying the condition that f is constant on the boundary of each I,.

--

arc

path

Figure 1 Let a : [0,1] + X be the unique map which is constant on each closed interval Ij and which coincides with f outside of these subintervals. Then a ( t ) coincides with f ( t ) for t E 8 I j . It is easily checked that a is continuous, and that for each point x E a([O,l]), a-'(x) c [0,1] is either one point or a closed interval. From now on we show that these conditions imply that A = a([O,11) is the image of some arc joining a(0) and a(1). To do so choose a countable dense subset { t l , t z , . - . } c [0,1]. Define a total ordering of A by a ( s ) < a ( t ) if a ( s ) # a ( t ) for s < t. Then an order preserving homeomorphism h: [0,1] -+ A can be constructed inductively as follows. Set h(0) = a(0) and h(1) = a(1). For each dyadic fraction 0 < m/2k < 1 with m odd, suppose that h((m - 1)/2k) and h ( ( m + 1 ) / 2 k ) have already been defined. Then there exists the smallest index j such that

and define h(m/2'") = a ( t j ) . It is not difficult to check that the h constructed in this manner on dyadic fractions extends uniquely to an order preserving map from [0,1] to A, which is a homeomorphism. 0 By definition a topological space X is locally connected if each point in X has an arbitrary small connected neighborhood. There exists a corresponding concept of locally path connected, or locally arcwise connected.

$2.2 Expansivity

35

Theorem 2.1.4. If a compact metric space X is locally connected, then X is locally path connected. Moreover every connected component of X is path connected. Proof. Let E > 0. Since X is locally connected, there exists a sequence of numbers 6, > 0 such that any two points with d ( z , z ' ) < 6, are contained in a connected set of diameter less than ~ / 2 , . We show that any two points z(0) and z(1) with d ( z ( O ) , z ( l ) )< 60 can be jointed by a path of diameter at most 4 ~ To . do this choose a sequence of denominators 1 = ko < k1 < kz < , eack of which divides the next. For each fraction of the form i / k n between 0 and 1 we choose an intermediate point z ( i / k , ) satisfying two conditions. (1) Any two consecutive fractions i / k n and (i l ) / k n correspond to points s ( i / k n ) and z ( ( i l)/kn) which have distance less than 6,. Thus any two such points are contained in a connected set C ( i ,k,) which has diameter less than ~ 1 2 The ~ . second condition is the following: (2) Each point of the form ~ ( j / k , + ~where ), j / k , + l lies between i / k n and (i l)/kn, belongs to the set C ( i ,k,), and thus

+

+

+

By induction on n the construction of such denominators 1, and intermediate points x ( i / k n ) is straightforward. Thus we may suppose that Z ( T ) has been defined for a dense set of rational numbers r in [0,1]. Next we prove that the correspondence T H Z ( T ) is uniformly continuous. Let r and T' be any two rational numbers such that z ( r ) and z ( r ' ) are defined. Suppose that IT - T'I < l/kn. Then we can choose i l k , as close as possible to both r and T ' . Then we have

and so d(z(r),z(r')< ) 4E/k,. This proves uniform continuity. Thus there is a unique continuous extension t H z ( t ) which is defined for all t E [0,1]. We have constructed the required path of diameter 5 4~ from z(0) to z(1). Therefore X is locally path connected. The rest of the argument is straightforward. 0

Theorem 2.1.5. Let X and Y be Hausdorff spaces and f : X --f Y a continuous surjection. If X is arcwise connected, then Y is arcwise connected. Proof. Take arbitrary points go,y1 in Y. Since f is surjective, there exist in X such that f(z0) = go, f ( z 1 )= 91. Since X is arcwise connected, we can find a path u joining 20 and 2 1 . Then f o u : I Y is a path joining yo and y1. This shows that Y is arcwise connected. 0

z 0 , q

--f

CHAPTER 2

36

52.2 Expansivity Let X be a metric space with metric d. A homeomorphism f : X --t X is ezpansive if there is a constant e > 0 such that x # y (2, y E X) implies

for some integer n. Such a constant e is called an ezpansive constant for f. This property (although not e ) is independent of the choice of metrics for X when X is compact, but not so for noncompact spaces. As one of notions that are weaker than expansivity we capture the notion which is called sensitive dependence on initial conditions. This is defined by the property that if there is 6 > 0 such that for each E X and each neighborhood U of z there exist y E U and n E Zsuch that d ( f " ( z ) , fn(y)) > 6. From definition it follows that X has no isolated points. A homeomorphism f:X + X is topologically transitive if there exists zo E X such that the orbit Of(z0) = {fn(zo): n E Z}is dense in X.

Remark 2.2.1. Let f : X + X be a homeomorphism of a compact metric space. Then f is topologically transitive if and only if for U,V nonempty open sets there is n E Zsuch that f n ( U )n V # 8. Suppose Of(z0) is dense in X and U,V are nonempty open sets. Then fn(zo) E U and f"(z0) E V for some n and m. Thus f k ( U )n V # 0 where k=n-m. To see the converse let {Ui : i > 0) be a countable base for X. For z E X we have Of(z) is not dense in X a Of(z) n Un= 0 for some n f"(z) E X \ U,,for each m E Z

n P-"(x\ un> u n ~-Yx\u,). 00

azE

for some n

m=-m 0

0

0

0

U=;,f-"(Un) is dense in X by the assumption, nz=-,(X \ f-" (Un)) is nowhere dense and thus lJ~..p=, f-"(X \ Un)is a set of first category. Since X is a compact metric space, {z E X : Of(z) is dense in X} Since

nz=-,

is dense in X.

Remark 2.2.2 ([B-S]). Let f : X -, X be a homeomorphism of a compact metric space. Suppose X is an infinite set. I f f is topologically transitive and

$2.2 Expansivity

37

Per(f) = {z E X : fn(z) = z for some n > 0) is dense in X, then f has sensitive dependence on initial conditions. This is easily checked as follows. We first show that Per(f) # X. For n > 0 let F,, = {z E X : fn(z) = z}. Then F,, is closed in X and F, = Per(f). Suppose Per(f) = X. Then there exists n > 0 such that int(F,) # 8. Since f is topologically transitive, it follows that F, = X, and so F,, is a finite set, thus contradicting. Notice that there is 6 > 0 such that for z E X there exists p E Per(f) such that d ( z , O f ( p ) )> 6. Indeed, if this is false, then for n > 0 there is z, E X such that d(z,,,Oj(q)) I l l n for all q E Per(f). Since X is compact, without loss of generality suppose 2, --+ z as n --+ 00. Then d(z, O f ( q ) )= 0 and thus 2 E O f ( q )for all q E Per(f). This can not happen. Let X = 614. For z E X let U ( z ) be an arbitrary neighborhood of z in X. Then U = U ( z )n Ux(z)is a neighborhood of z where Ux(z)= {y E X : d(z, y) < A}. Take p E Per(f) n U with f n ( p ) = p for some n > 0 and choose qz E Per(f) satisfying d(z, O j ( q z ) )> 6. Since v = f-j(Ux(fj(qz))) is a nonempty open set and f is topologically transitive, there is k E Z such that f k ( V )n U # 8. Notice that Ikl is chosen sufficiently large. Let j' > 0 be the integer such that lkl = nj' r where 0 612. In any case we have d ( f f n j ( s ) , f * " j ( y ) ) > X or d ( f * " j ( z ) , f * " j ( p ) ) > A, i.e. f has sensitive dependence on initial conditions. Let f : X -+ X be a homeomorphism of a compact metric space. A finite open cover a of X is a generator (weak generator) for f if for every bisequence {A,} of members of a, f-,,(cl(A,,)) is at most one point f-,(A,,) is at most one point). Here cl(E) denotes the closure of a subset E. These concepts are due to Keynes and Robertson [K-R].

(nr=-,

CHAPTER 2

38

If a,P are open covers of X , then their joint a V /3 is given by a V P = { A n B : A E a , B E P}. An open cover p is a refinement of an open cover a (written a 5 p ) if every member of p is a subset of some member of a. It is clear that a 5 a V p and p 5 a V p. If f : X --t X is a continuous surjection, then f-'(a) = { f - ' ( A ) : A E a} is an open cover of X. It is easily checked that

Theorem 2.2.3. Let f: X + X be a homeomorphism of a compact metric space. Then the following are equivalent. (1) f is expansive, (2) f has a generator, ( 3 ) f has a weak generator. Proof. (2) + (3) is clear (3) + (2) : Let /3 = { B 1 , . ,Bd}be a weak generator for f and let 6 > 0 be a Lebesgue number for p. Let a be a finite open cover consisting of sets Ai with diam(cl(A;)) 5 6. If {Ai,} is a bisequence in a , for every n there is j , such that cl(Aj,) c Bj,, and so f-"(cl( Aj,)) c f-"(Bjn). Therefore a is a generator. (1) + (2) : Let 6 > 0 be an expansive constant for f and let a be a 00 finite cover consisting of open balls of radius 612. Suppose z , y E f-,(cl( A,)) where A, E a. Then d(fn(x), fn(y)) 5 6 for every n, and by the assumption x = y. (3) =+ (1) : Suppose a is a weak generator. Let 6 > 0 be a Lebesgue number for a. If d(f"(z), fn(y)) < 6 for n E Z,then for n there is A, E Q such that f "(x), f "(y) E A, and so x, y E f-"(A,) which is at most one point.

-

n,"=-,

Or=-,

n,=-,

n= :-,

0

Theorem 2.2.4. Let f : X X be a homeomorphism of a compact metric space and let k # 0 be an integer. Then f is expansive if and only if f k as expansive. .--)

vf!,'

Proof. This follows from the facts that if a is a generator for f then f-i (a)= a V f - l ( a ) V . V flkl-'(a) is a generator for f k , and that if a is a generator for f k then a is also a generator for f . 0

-.

The following theorem is easily checked. Thus we leave the proof to the readers.

92.2 Expansivity

39

Theorem 2.2.5. (1) Iff:X + X is expansive and Y is a closed subset of X with f (Y) = Y, then fly :Y -+ Y is expansive. (2) If f i : Xi Xi (i = 1,2) are expansive, then the homeomorphism f1 x fa : XI x X2 -+ X1 x X2 defined by --f

fl

x

fz(x1,22)= (f1(x1),f2(x2))

(219x2) E

Xl x xz

is expansive. Every finite direct product of expansive homeomorphisms is expansive. ( 3 ) If X is compact and f:X + X is expansive, then h o f o h-' : Y + Y is expansive where h: X + Y is a homeomorphism. Let k be a fixed natural number and let Yk = (0, l , . . . ,k - 1). Put the discrete topology on Yk and define the product space Yf = n r m Y k equipped with the product topology. Then a homeomorphism u : Yf + Yf is defined by ( o ( x ) ) i = xi+l,i E Z, for x = ( x i ) . u is called the shift map. Theorem 2.2.6. Let k 2 2. Then the shift map (I :Yf

-+

Yf is expansive.

Proof.F o r O I i 5 k - l d e f i n e A i = { ( z , ) : z o = i } . Thena={Ao,...,Ak -1) is an open and closed cover of Yf. If x E u-"(Ai,) where Ain E a, then x = (... ,i-1,io,i1,-..) and thus a is a generator. 0

n= :-,

Theorem 2.2.7. Let f :X --t X be an expansive homeomorphism of a compact metric space. Then there exist k > 0 and a closed subset E of Yf such that Z is u-invariant (u(E) = C ) , and moreover there is a continuous surjection T : E + X satisfying ?r o u = f o T , i.e. the diagmm

E

A

.1x-x

E commutes.

f

Proof.Let 26 > 0 be an expansive constant for f. Choose a finite cover a! = {Ao,... ,&-I} by closed sets such that diam(Ai) < 6 for 0 5 i 5 k - 1, and such that Ai intersect only in their boundaries. Let 8 denote the union of the boundaries of Ai. Then 8, = fn(8) is a set of first category and so X\&, is dense in X. For x E X\8, we can assign uniquely a member of

u=:-,

y ,by x

G (a,)

whenever f "(x)E Aa, . Let A denote the collection of all sequences arising in this way. Then A C Yf and CP 0 o((xn)) = f

0

CP((~:,)>, (xn) E A.

CHAPTER 2

40

If cp is uniformly continuous, then cp : A -t X can be uniquely extended to a continuous map 7r : C + X, satisfying 7r 0 a ( ( z n ) )= f 0 r((xn)) for (2,) E C, where C = cl(A). It only remains to see uniform continuity of cp. Let E > 0. Since a is a generator for f , there is N > 0 such that each member of V ,! f"(a) has a diameter less than E . If this is false, then there is E > 0 such that for every j > 0 there are xj, y j E X with d(zj, yj) > E and Aj,i E a (-j 5 i 5 j) with zj,yj E f-i(Aj,i). Suppose x j -t z and yj + y. Then x # y . Since a is finite, there is A0 E {Aj,o : j E 9 ) such that z j , y j E A0 for infinitely many j. Thus, z , y E cl(A0). Similarly, for each n infinitely many elements of {Aj,, : j E Z} coincide and we have A, E a with z , y E f-"(cl(A,)). fwn(cl(An)). This is a contradiction. Therefore, z , y E O0 If ( a , ) , (b,) E A and a , = b, for In1 5 N , then cp((a,)) and cp((b,)) are in the same member of Vf, f"(a) and thus d(cp((a,,)),(p((b,)))< E . Therefore cp is uniformly continuous. 0

ni=-j

n,=-,

Let X be a metric space. A continuous surjection f :X -t X is positively expansive if there is a constant e > 0 such that if x # y then d( f ,(z),f "(y)) > e for some non-negative integer n ( e is called an expansive constant for f ) . For compact spaces, this is independent of the compatible metrics used, although not the expansive constants.

Theorem 2.2.8. Suppose X is compact. Let k positively expansive if and only if so is f k.

> 0 be an integer. Then f is

Proof. Let f :X -t X be positively expansive and e > 0 an expansive constant. Since f is uniformly continuous, there is 6 > 0 such that for 1 5 i 5 k d ( z , y )< 6

* d(f'(z),f'(y))< el

which implies that 6 is a n expansive constant for f k . Similarly, the converse is proved. 0 The following theorem is easily checked. Thus we leave the proof to the readers.

Theorem 2.2.9. (1) If f :X -t X is positively expansive and Y is a closed subset of X with f ( Y )= Y , then f p : Y -+ Y is positively expansive. (2) If fi : Xi -t Xi (i = 1,2) are positively expansive, then the continuous sujection f1 x f 2 : X1 x X2 -+ X1 x Xz defined by

fl

x

f2(51,22)

= (f1(51),f2(22))

(x1,22) E x1 x

x 2

i s positively expansive. Every finite direct product of positively expansive maps is positively expansive. ( 3 ) If X is compact and f : X -t X is positively expansive, then h o f o h-' : Y -t Y is positively expansive where h:X -t Y is a homeomorphism.

$2.2 Expansivity

41

Theorem 2.2.10 (Reddy [Rl]). If X is compact and f : X 4 X is a positively expansive map, then there exist a compatible metric D and constants 5 > 0 , X > 1 such that for x , y E X

D(x, Y) I 6 -*- D(f (21, f(Y)) 1 XD(x, Y).

For the proof we need the following result due to Frink [F'r]. Lemma 2.2.11. Let Vn be a decmasing sequence of open symmetric neighborhoods of the diagonal set A = { (z,x) : x € X } of X x X such that 00

v O = x x x ~,

v , = A

and Vn+lo Vn+lo Vn+l C Vn for all n 2 1. Then there is a metric D compatible with the topology of X such that for n 1 1 1 Vn c { ( z , ~ ) :D(s,Y)< -} c Vn-12n Here a subset A of X x X is said to be symmetric if A satisfies the property t h a t ( z , y ) € AH ( y , x ) E A , a n d A o B i s d e f i n e d b y A o B = { ( x , y ) : 3 z € X s.t. (x,z) E A , ( z , y ) E B}.

Proof. Define a function ( :X x X

4

R by

...

, z,, and for (z,y) E X x X denote by c(x, y) a finite sequence x = 20, x l , x,+1 = y. We write C(z,y) the collection of all such sequences c(x, y). Let us Put ~ ( xY), = i n f { C ((Si,xi+l) : ~ ( 2Y> , E c(z, Y)). i

It is easily checked that for any points x, y, z in X (i) (ii) (iii)

CHAPTER 2

42

If we hold that for n 2 1

n,"==,

then z = y whenever D ( z ,y) = 0, since (2, y) E V, = A. From (ii) and (iii) together with this fact, it follows that D is a metric compatible with the topology of To obtain (iv) it sufficies to show that for each finite sequence z = 2 0 , z~,"', zn+1 = Y

x.

For, if D(z,y) < 1/2", by (v) we have

and thus (z, y) E V,-l. The remainder of the proof is only to show that (v) is true. If R. = 0 then >.

52.2 Expansivity

45

Theorem 2.2.12. If a homeomorphism of a compact metric space is positively expansive, then the space is a set consisting of finite points. Proof. Since f is positively expansive, there are a compatible metric D, constants 6 > 0, 0 < X < 1 such that D ( x , y ) 6 implies D ( f - l ( x ) , f - l ( y ) ) 5 X D ( x ). Thus @- = { f - i : i 2 0) is uniformly equi-continuous. P u t @+ = : i 2 0). By a metric p defined by p ( f , g ) = max{D(f(z), g(x)) : x E X}, @- is totally bounded. Define a map G : $- -+ @+ by G(f -*) = f for i 2 0 Then G is pisometric. Therefore @+ is totally bounded. Since X is compact, @+ is uniformly equi-continuous and so there is v > 0 such that D ( x ,y ) < v implies D ( f i ( x ) f, i ( y ) ) 5 e for all i 2 0 ( e is an expansive constant). This shows x = y and therefore z is an isolated point. 0


0

so that there exist x1,xz E I o ( E / ~with ) z1 # 2 2 such that f ( x 1 ) = f ( x 2 ) . This cannot happen. If a = ~ / gives 2 the greatest values o f f on I o ( E / ~ then ) , the point E must be a point of the greatest value of f on I , l 2 ( ~ / 2 ) Continue . this argument, then we have that f: I -t I is a homeomorphism, which contradicts Theorem 2.2.12. 0

Theorem 2.2.14. Let f : X t X be a positively expansive map of a compact connected metric space. If f is an open map, then f has fixed points in X . Proof. Notice that f is a local homeomorphism. Since X is compact, f is a covering map. Positive expansivity ensures that there exists a compatible metric D and constants 6 > 0,X > 1 such that D ( z , y ) < 6 implies D(f (x),f ( y ) ) 2 XD(z,y). If V is an open set of diameter less than 6 and if q E f - ' ( V ) , then there is a unique open set containing q of diameter less than (1/X)6, which is mapped homeomorphically onto V. This enables us to lift chains of small open sets. Let {K : 1 5 i 5 n} be a fixed open cover of X so that the diameter of each is less than 6. Take xo E X with f ( x 0 ) # 20. Since X is connected, there is a chain of open sets chosen from among {X}from f ( x 0 ) to xo. Lift this chain to a chain of open sets from xo to x1 such that f (z1)= 20. Further, lift this chain to a chain of open sets from x1 to 2 2 such that f ( x 2 ) = 2 1 , and continue this inductively. Since the original chain had length at most n6, the

CHAPTER 2

46

length of the first lifted chain is at most (l/A)nS and in general the length of the t t h lifted chain is at most (l/A)"n6. If j < t then

D ( x j ,2 k )

I

1 '

1 - - - + (X)'}nS

{(X)J+l+

1 . nS

< ( -A) J

A-1'

Thus { z j } is a Cauchy sequence, which therefore converges to y E X. Since f ( x j ) = xj-1, we have f ( y ) = f(1imxj) = lim f ( x j ) = limzj-1 = y. 0 For f : X + X a homeomorphism of a compact metric space, define a local stable set W:(x,d) and a local unstabZe set W,"(x,d)by

W,"(z,d)= {Y E x : d(fn(z),fn(y)) I E,n 2 01, W:(x,d) = {y E X :d(f-"(x), f-"(y)) I E,n 10). The symbols

Wf(x, d ) ( 0 = s,u ) will be changed by W z ( z )later.

L e m m a 2.2.15. f : X -+ X is an expansive homeomorphism with expansive constant e if and only if there exists e > 0 such that for all y > 0 there is N > 0 such that for all x E X and all n 2 N

c W,(fp"(.)ld), f -"(WeU(x,d)) c W;(f -n(x), 4. f"(W,.(x,d))

Proof. If this is false, then we can find sequences x,, yn E X , m n > 0 such limn-,oomn = 00 and d(fmn(x,), fmn(yn)) > y. Since that yn E W:(xn,d), Yn E WeJ(x,,d),we have

d ( f i o fmn(x,), f i o f m n ( y n ) ) I e

If fmn(x,) -+ x and fmn(y,) + y as n i E Z.On the other hand, we have

-+ 00,

for all i 2 -m,. then d(fi(x), fi(y)) 5 e for all

d(x,y) = n+oo lim d(frnn(xn),frnn(y*))L 7 , thus contradicting. Conversely if d(fn(x), f n ( y ) ) 5 e for all n E Z,for any y y E Wy9(x,d). Since y is arbtrary, we have x = y. 0

> 0 we have

For x E X the stable set and unstable set of a homeomorphism f are defined bY

Wa(x,d ) = {y E X : lim d( f n ( z ) , f"(y)) = O}, n-oo

W " ( x , d ) = {y E X : lim d(f"(x), f"(y)) = 0). n+-w

$2.2 Expansivity

Lemma 2.2.16. If f:X+ X is an expansive homeomorphism, for E number less than an expansive constant

47

>0 a

y E Un,of-nW:(fn(z),d), then there is n 2 0 with fn(y) W:(fn(x), d). Thus, for y > 0 there is N 2 0 such that for all m 2 N

Proof, If

E

and so d(frn+"(y), frn+"(z))5 7 for m 2 N. Therefore, y E Wa(z,d). Conversely, if t E Wa(z,d) and E > 0, then we can find N 2 0 such that d(fn(x),fn(y)) 5 c for n >_ N. Thus d(fj o f N ( z ) , f jo fN(y)) 5 e for all j 2 0. Therefore, f N ( y ) E W:(fN(x),d) and so

For expansive homeomorphisms of a compact metric space we have the following theorem (compare to Theorem 2.2.10). Theorem 2.2.17 (Reddy [R2]). If f:X+ X is an expansive homeomorphism, then there exist a compatible metric D and constants E > 0 , c > 0 and 0 < X < 1 such that

for all n 2 0 .

We say that the metric D as in Theorems 2.2.10 and 2.2.17 is a hyperbolic metric for the map f: X + X.

Proof,Define WO= X

x X and for n 2 1

where e is an expansive constant. Then {Wn} is a decreasing sequence of open symmetric neighborhoods of the diagonal set A. Since f is expansive, if

CHAPTER 2

48

(2, y) 4 A, there is n > 0 with (5, y) 4 W,. This implies there are E > 0 and N 2 1 such that

n,, - W, = A. Thus

We now define a new sequence of open symmetric neighborhoods of A by

vo = wo, vn = Wl+(*-l)N, Obviously,

n 2 1.

V, = A. To use Lemma 2.2.11 we must prove that V,+l

o Vn+l o V,+l

for n 2 0.

c V,

It is clear that V1 o V1 o V1 c VOand by the choice of E and N , V2 o Vz o VZ C V1. Let k 2 2 and take (z,y) E vk+1Ovk+1Ovk+1. Then we prove that (z,y)E v k . Let z , w E X be points such that (z,w), (w,z),(z,y) E Vk+1. If ( p , q ) is any of these three points, then we have

from which

d(fi(z),fi(y)) < e,

lil < 1

+ (k - l)N.

Therefore, (2,y) E Vk. Let D be a compatible metric satisfying

vn c {(.,y)

:D(z,v)
0 such that D(z,y) < q implies d ( z , y ) < e, we have W,(z,D) c W,"(z,d)for z E X. Put v = min{q, 1/4} and take y E W,"(z,D).We first show that

Since (z,y) E

Let

V,, and (z,y)4 Vn+l 3 N1/2m+a(A,D)for some n, we have

L = 3N. It follows from induction that

Define X = (1/2)*. To obtain the conclusion we show that if y E W,"(z,D) then D(f"(z), f"b)) I 8XrnD(z,Y), m L 0. Put m = kL+j fork 2 0 and 0 5 j for some n 2 1, we have

< L. Since (f"(z),

fkWE W,"(f"(4,d)

f"(y))

n fkL(4W1+(n-1)N,

E Vn-Vn+l

CHAPTER 2

50

D(f kL+j Since m = bL

+ j, we have D ( f"(z),f "(Y))

1

I 4 ( j ) k D ( z , Y )= 4XkLD(z9Y)

< 8A"+jD(z, y) = 8XmD(z,y). By replacing f by f - ' , we can establish the required property for y E W,U(z,D).0 We denote as H" the half space of R",i.e. H" = { ( z 1 , z 2 , - *, *~ n E) R" : z, 2 0). Then H" 2 R"-'. A metric space M is called an n-dimensional topological manifold if for each p E M there is a neighborhood U ( p ) such that U ( p ) is homeomorphic t o an open subset V of H". Let qp denote the point in V corresponding to p . We write a M = { p E M : qp E Rn-'} and OM is called the boundary of M. If 8 M = 0, we can take R" instead of H". In this case M is called a topological manifold without boundary. We here say that a topological manifold M is closed if it has no boundary and is compact, connected. If U is an open subset of M and 1c, is a homeomorphism from U onto an open subset of R",then (U, $) is called a chart (or a coordinate neighborhood) of M . For p E U we can write 1c,(p)= ( ~ ' ( p ) , ,~ " ( p ) ) Then . zl, ,Z" is continuous functions on U. We say that (zl,-.,xn) is a local coordinate system of M at the chart (U,$), and that ( ~ ' ( p ) , . . ., z " ( p ) ) is a local coordinate of a point p in U. Let S = {(U,,$a) : a E A} be a family of charts of M. If {U, : a E A} is an open cover of M, then S is called an atlas (or a system of coordinate neighborhoods) of M . A continuous surjection f : X + X of a metric space is said to be ezpanding if f is positively expansive and it is an open map. If X is compact, then an expanding map f : X + X is a local homeomorphism, and so f is a selfcovering map by Theorem 2.1.1. In the case when X is a closed topological manifold, we have that every positively expansive map of X is expanding in our sense because it is a local homeomorphism by the following theorem (cf. PPI *

. ..

-

$2.2 Expansivity

51

Theorem 2.2.18 (Brouwer theorem). If U and V are homeomorphic subsets of R" and U is open in R", then V is open in R". The following theorem is a wider result which contains Theorem 2.2.13.

Theorem 2.2.19 (Hiraide [Hi7]). No compact connected topological manifold with boundary admits a positively expansive map. For the proof we need the following lemma.

Lemma 2.2.20. Let X be a compact connected locally connected metric space and f :X + X a positively expansive map. If a closed proper subset K satisfy the conditions :

then K = 0 . Let M be a compact connected topological manifold and 8M denote the boundary of M. If M admits a positively expansive map f , then f: M + M is locally injective. By applying Theorem 2.2.18 we have that f (M \ 8M)c M \ 8M)and flM\BM is an open map. Therefore BM = 0 by Lemma 2.2.20. Proof of Lemma 2.2.20. Since f: X + X is positively expansive, there exists a metric D satisfying all the properties of Theorem 2.2.10. For any x E X define U,(x)= {y E X : D(z,y) < E } and denote as Cc(x)the connected component of x in Uc(x).Then C,(x)is open in X. Since X is connected and locally connected, X is locally arcwise connected by Theorems 2.1.3 and 2.1.4. Let 6 > 0 be as in Theorem 2.2.10 and take E with 0 < E < 612. We first show that for x E X \ K

G(.) c x \ K

-

f (Cc(.))

3

CAL(f(2))

where X > 0 is as in Theorem 2.2.10. Suppose y E Cx,( f (x))\ f (C,(x)>. Since Cx,( f (x)) is arcwise connected, there is an arc w : [0,1] + Cx,(x)such that w ( 0 ) = f(x) and w(1) = y. Since C,(x)c X \ K and Cc(x) is open in X , f (Cc(x)) is open by (2). Since w ( 0 ) = f(x) E f(Cc(x)), there is 0 < to < 1 satisfying w([O,to))C f(Cc(x)) and w ( t 0 ) 6 /(C,(z)).Then we have

Since cl(Cc(s)) c U b p ( x ) ,by the choice of 6we have that f l C ~ ( ~ * ( + ) ) : cl(Cc(x)) + cl( f (Cc(x))) is a homeomorphism. Since w([O,to]) c cl( f (C,(x))), there is an arc G : [O,to] + cl(Cc(z)) such that f o = w. Since f o G([O,to))= w([O,to))c f(G(z)),we have G([o,to))c G ( x ) and since 4 t 0 ) $ f(Cc(Z)), obviously G(to) 6 Cc(x).

CHAPTER 2

52

On the other hand, since w ( t 0 ) E Cx,(f(z)), we have

AD(G(to),z)5 D(w(to),f(z))< S j ( t 0 ) E Uc(z).Thus G(t0) E Cc(z) since z E S j ( [ O , t o ) ) C Cc(z) C Uc(z).This is a contradiction. Therefore, Cx,(f(z)) C f(Cc(z)).

and so

Let us define X(E) = {z E X \ K : Cc(z)c X

\ K}.

Since K # X, there is 0 < EO < 6/2 such that X(Q) # 0. To obtain the conclusion of the lemma suppose K # 8. Take z E X(EO). Then we have C X C O ( f ( 2 ) ) c f(Cc,(z)).Thus, f(X(E0)) c X(AE0). It is easily checked that X(AEO)c X(pe0) c X ( E Ofor ) 1 < p < A. We show that cl(X(k0)) C X(peo). To do this let { z i } be a sequence of X ( k 0 ) and suppose x i + z as i + 00. Since CEO(.)is open in X, we have zi 6 Cco(z) for sufficiently large i. Thus Cco(z)C CPco(z:i) c Cxco(zi)which implies z E X ( ~ E O Therefore ). cl(X(Ae0)) c X ( p ~ 0 ) . We have shown that f(cl(x(e0))) = Cl(f(X(E0))) c Cl(X(AE0)) c X(PE0) c X(E0).

n:=o

Thus Y = fi(cl(X(eo))) is a nonempty closed set and f ( Y ) = Y . Since Y # X and X is connected, Y is clearly not open in X. For A c X and a > 0 define Na(A) = U a E A C Q ( a Then ). N(P-l)co(Y) c X(EO).Indeed, let z E Y and z E C(P-l)co(z). Then Cco(z)c CPCO(z). Since e E Y C X ( ~ E Owe ) , have Cco(z)C CPco(z) C X \ K and so z E X(EO). Since f ( Y )= Y , we have

nm(p-l)co(~)) cn cY 00

Yc

00

m(Eo))

i=O

i=O

nzo

and thus Y = fi(N(p-l)co(Y)). Since C ( p - l ) c o ( zC) Cclco(z) C X \ K for C x ( , - l ) c 0 ( f ( ~and ) ) thus f(N(P--l)CO

L

E Y , we have f ( C ( P - l ) c o ( z3) )

( Y ) )3 N x ( P - l ) c o ( y ) 3 ~(jA-1)co(Y).

Therefore, Y = N(P-l)co(Y) and Y is open in X. This is a contradiction. 0

Remark 2.2.21. We can construct an example of positively expansive map which is not open (an example due to Rosenholtz ([R])).Consider the subset X in the plane defined by

x = {% : I%[

3 1 = 1) u {% : 1% - -1 = -} 2 2

u {% : 1%

+ -132

1 = -}. 2

$2.2 Expansivity

53

Give X the arclength metric. Then a map is defined as follows : stretch each of the small circles onto the big circle, stretch each of the upper and under semicircles of the big circle first around a small circle, and then across the other semicircle and finally around the other smaller circle. More precisely we describe the map f by

2(2 - 3/2) 2(2 3/2) z6/2 - 3/2 z3 -2*/2 3/2

+

+

R e ( z ) 2 1, R e ( z ) 5 -1, 1/2 5 Re(z) 5 1, - 1/2 2 R e ( z ) 5 1/2, - 1 5 Re(e) 5 -1/2.

if if if if if

Then it is easily checked that f is positively expansive, but not open.

1

-11

I

Figure 2

If f : X + X is a homeomorphism of a compact metric space and x is a point in X , then a-limit set of x , denoted as a ( x ) , consists of those points y E X such that y = lim+,w f " j ( x ) for some strictly decreasing sequence of integers nj. The w-limit set of x , denoted as w ( x ) , is similarly defined for strictly increasing sequence. The set a ( x ) ( ~ ( x )is) closed, nonempty and f invariant, i.e. f ( a ( x ) )= a ( . ) for z E X . If, for some point x , a ( ~and ) W(X) each consists of a single point,then we say that x has converging semi-orbits under f . Theorem 2.2.22 (Reddy [R3]). If f : X -+ X is an expansive homeomorphism of a compact metric space, the set of points having converging semiorbits under f is a countable set. Proof. Since f is expansive, it follows that f has at most finitely many fixed points, say q l , . ,q k . Let A denote the set of points having converging semiorbits under f . Suppose A is uncountable.

.

54

CHAPTER 2

If x E A, both a(.) and ~ ( z are ) fixed points. Let A ( i , j ) be the set of points satisfying a(.) = qi and ~ ( z=) qj. Then A is the union of the finitely many sets A ( i , j ) , so that one of these, say B , is uncountable. For each N > 0 let B ( N ) be the collection of points x E B such that, for n 2 N,fn(z) is ) f-n(x) is e/a-close to a(.). Since B is the union of the e/a-close to ~ ( zand sets B ( N ) , one of them (say B ( M ) ) must be infinite. Since X is compact, there exist distinct points y and z of B ( M ) with d ( y , z ) < e/2 so small that fn(z) is e-close to fn(y) if 171.1 5 M, where e denotes an expansive constant for f . Thus, from the definition of B ( M ) , we have that d(fn(y),fn(z)) < e for all n E Z, thus contradicting the choice of e. Therefore, A is countable. 0 Theorem 2.2.23 (Jacobsen and Utz [J-U]). There ezists no expansive homeomorphism of a closed am. Proof. If f : [0,1] --f [0,1] is a homeomorphism, then f has either f (0) = 0 and f(1) = 1, or f(0) = 1 and f(1) = 0. In both cases f a induces a homeomorphism satisfying f 2 ( 0 ) = 0 and f a ( l ) = 1. P u t Fix(f2) = {z E [0,1] : f2(x ) = z}. Then Fix(f2) is closed. If Fix(f2) = [0,1], then all points of [0,1] have converging semi-orbits under f and so f is not expansive by Theorem 2.2.22.

Figure 3

If Fix(f2) # [0,1], then U = [0,1]\Fix(f2) is a nonempty open set. Thus U is the union of the countable intervals Ij where Ij’s are mutually disjoint. For x E U there is Ij such that z E Ij. Then the extremal points of Ij are or fixed points of f a . Thus we have that for x E Ij, x > fa((.) > f4(z)> z < f2(z) < f4(z)< . In any case {f2j(x) : j 2 0) converges to a fixed point of f Since Ij is uncountable, f is not expansive. 0 m e - ,

Let f:X 4 X be a homeomorphism of a compact metric space. If, for all x E X,the orbit Of(”) = { fn(x), n E Z} is dense in X ,then f is called a minimal homeomorphism. Obviously every minimal homeomorphism is topologically transitive.

$2.2 Expansivity

55

Theorem 2.2.24. A homeomorphism f : X + X is minimal if and only if E = 0 or X whenever f (E) = E and E is closed.

Proof. Suppose f is minimal. If E is closed and f ( E ) = E # 8, then E 3 Of(x) for z E E and cl(Of(x)) C E. Thus we have E = X. Conversely, for z E X,cl(Of(x)) is a closed f-invariant set and thus cl(Of(x)) = X. 0

Figure 4

A closed subset E which is f-invariant is called a minimal set with respect to f if f p is minimal.

Theorem 2.2.25. Every homeomorphism f:X

+X

has a minimal set.

Proof. Let 0 denote the collection of all closed nonempty f-invariant subsets of X. Clearly 0 # 0 because of X E 0.0 is a partially ordered set under the inclusion. Every linearly ordered set of 0 has the least element. Thus 0 has a minimum element by Zorn's lemma. This element is a minimal set with respect to f. 0

Theorem 2.2.26. There exists no expansive homeomorphism of the unit circle S'. Proof, Suppose f:S' + S' is expansive. By Theorems 2.2.22. and 2.2.23, f has no periodic points. Indeed, suppose f has distinct periodic points z and y. Then f k(x ) = z and f k ( ( y ) = y for some L > 0. For simplicity put g = f k . There exists a closed arc C in S1 such that g(C) = C , or g a ( C ) = C. If g(C) = C, then glc : C + C is expansive. This is impossible by Theorem 2.2.23. If f has only one fixed point, for any point 2 E S ' the semi-orbit {x, f(z), f'(x),...) converges to the fixed point, and also {x, f-'(x), f - 2 ( z ) , - - -converges } to the fixed point. Thus f:S' + S' is not expansive by Theorem 2.2.22. Let M be a minimal set of f in S'. Since f :M + M is expansive and minimal, we see that M is totally disconnected (see Theorem 2.2.44 below). Notice that M is an infinite set because f has no periodic points.

CHAPTER 2

56

We claim that isolated points are not in M. Indeed, if x E M is an isolated point, then Ua(z)n M = {z} for some 6 > 0 and so {z} is open in M. Since M is minimal, the closure of the orbit of x equals M and thus fn(x) E Ua(x) n M = {x} for some n E Z, a contradiction. Since S'\ M is open, S' \ M is expressed as countable disjoint union S'\ M = Uj Ij of open arcs I, in S1.Obviously diam(1j) + 0 as j -+ 00 and for fixed j,fn(lj)# Ij for all n # 0. Thus, diam(f"(1j)) t 0 as In1 + 00. Take

-

n

a, b E Ij and let fno(ab) be the arc with the longest length in { f "(ab) : n E Z}. Let E > 0. Then we can choose I, and a, b E I, such that the length between n

and fno(b) is less than E . Then, diam(f"(ab)) < E for all n E Z and therefore f : S' -+ S' is not expansive. This is a contradiction. 0 f"o(a)

The following theorem was proved in Hiraide [Hi51 and Lewowicz [L2].

Theorem 2.2.27. There exists no expansive homeomorphism of the 2-sphere

S2. We omit the proof. If the theorem is of interest, the reader should see [Hi51 and [L2]. Let X be a compact metric space with metric d and Xzdenote the product topological space X z = {(?i) : x; E X,i E Z}. Then X z is compact. We define a compatible metric d for Xz by

A homeomorphism u : X z

-+

Xz,which is defined by

.((xi)) = (yi) and yi = xi+' for all i E

Z,

is called the shift map. For f : X -+ X a continuous surjection, we let

Xf= { ( z i ) : z;E X and f(xi) = xi+l,i E Z}. Then Xf is a closed subset of X z . Moreover we have .((xi)) = (f(xi)) for all (xi) E Xf,and so Xf is u-invariant, i.e. u(Xf) = Xf. The space Xf is called the inverse limit space constructed by f and we sometimes write Xf= lim(X, f ) . The restriction u = qx, : Xf+ Xfis called the shift map t determined by f . Let i E Z and denote as p i : Xf -+ X the projection defined by (xi) H xi. Then pi o u = f o pi holds, i.e. the diagram

Xf A Xf Pi

j,

1Pi

x - x f

commutes,

52.2 Expansivity

57

and PO o ui = f o po for all i 2 0. If, in particular, f is bijective, then it is clear that each pi is bijective. In the case where X is a torus and f: X + X is a toral endomorphism, we can show that the inverse limit space Xfhas a structure of compact connected finite dimensional abelian group, which is called the solenoidal group. The properties of solenoidal groups will be discussed in Chapter 7. It is easily checked that Xf is homeomorphic to the space

X; = {(xi)?

: zi

E X and f (zi+l) = z i , i 2 0)

equipped with the product topology, and the shift map (T is topologically conjugate to a homeomorphism u' defined by (~'((zi)?)= (f (zi))? for (zi)? E X;, i.e. there exists a homeomorphism h: Xf+ X; such that the diagram

Xf 0 Xf

1.

commutes.

A continuous surjection f:X -+ X is c-expansive if there is a constant e > 0 (called an ezpansive constant) such that for (xi), (yi) E Xfif d(zi,yi) 5 e for all i E Z then (zi) = (yi). This is independent of the compatible metrics used. The notion of c-expansivity is weaker than that of positive expansivity. For homeomorphisms c-expansivity implies expansivity. Theorem 2.2.28. Let k and only if so is f k .

>0

be an integer. Then f:X + X is c-expansive if

Proof. We notice that f is uniformly continuous. Let e constant for f . Then we have d(z,y)

< 6 -----r. @(z),fi(y))

> 0 be an expansive

< e, 1 5 i 5 k

for some 6 > 0. This 6 is an expansive constant for f k . In fact, for (zi),(yi) E Xf suppose d ( z k i , yki) < 6 for all i E Z. Then we have d(zi, yi) < e for i E Z. By expansivity, zi = yi for i E from which z k i = y k i for i E Z. The converse is proved by the same way. 0

z,

For the relation between continuous surjections and their inverse limit systems we have the following Theorem 2.2.29. Let X be a compact metric space. A continuous surjection f: X 3 X as c-ezpansive if and only tf u : Xf--+ Xf is eapnsive. Proof. If (zi) # (yi) then there is k E Z such that d(zk,Y&) > e by cexpansivity. Then i(uk((zi)),uk((yi))) 2 d(zk,Yk) > e, where d((zi),(yi)) = 00 d(zi, yi)/21il. Therefore (r is expansive.

xi=-,,

CHAPTER 2

58

Conversely, let e

> 0 be an expansive constant for u and let d(si, y i ) < e / 4

(i E Z) for (xi),( y i ) E Xf.Then we have (i(d'((xi)),a"((yi))) 5 e for n E Z, and hence ( S i ) = ( y i ) by expansivity of u. Therefore e / 4 is an expansive constant for f : X t X. 0 The following theorem is easily checked. Thus we leave the proof to the readers.

Theorem 2.2.30. (1) Iff:X --t X is c-expansive and Y is a closed subset of X with f ( Y )= Y , then fly : Y + Y is c-expansive. (2) If fi : Xi + Xi (i = 1,2) are c-expansive, then the continuous surjection f1 x fz : XI x X Z + X I x X2 defined by

is c-expansive. Every finite direct product of c-expansive maps is c-expansive. ( 3 ) If X i s compact and f:X + X is c-expansive, then h o f o h-l : Y + Y is c-expansive where h : X + Y is a homeomorphism.

Theorem 2.2.31. The closed interval does not admit c-expansive maps.

Proof. Let I denote the closed interval [0,1]. For f:I + I a continuous surjection, its inverse limit system (If,u) is defined by

Suppose that u2 : If + If is expansive. It is easily checked that (If,u2)is topologically conjugate to ( I f a , ~ ) .Since f is surjective, there exist a , b E I with 0 5 a < b 5 1 such that either

(1) f(a) = 0 and f ( b ) = 1

or

(2) f ( a ) = 1 and f ( b ) = 0.

If we have (2), then we see that f2(a') = 0 and f 2 ( b ' ) = 1 for some u', b' E I with 0 5 a' < b' 5 1. Thus, to obtain the conclusion it is enough to prove (1). Take two points u,v E I ( . < v) such that (3)

f ( u ) = u,

f ( v ) = 'u,

f(x) # x for u < x

< v.

This is ensured by the facts that f has at least two fixed points and the set of fixed points is not dense in I. (3) is divided into the following cases,

( 4 ) f(x) > x for u < x We first check the case ( 4 ) .

xy > in [u,v] with

-

xo0 - x0 , f(XP) = x;+

Since x i

-+

u as n + 00, for 0

< E < x2 - xo we have

u

for some N

> 0.

i 2 1.

< 2% < U+& /2

Take 6 > 0 so small that for z,y E I

< 26 * If"(.)

Ix - yI

Write c1 = 2% - 6 and c2 = x& IfN(c')

- f"(y)l < E / 2 ,

0 5 n 5 N.

+ 6. Then

- fN(x&)I = I f N ( c i ) - 501 < &/2, i = 1,2,

and thus f N ( c ' ) < xo + E < x2. By the definition of yo we have f N + l ( c i ) 5 yo for i = 1,2. If f N + ' ( c i ) = yo for some i, then we can find a sequence {cj :j

2 O} c [u,v]

such that ci = ci and f ( c i ) = cj-, for j 2 1. Notice that {ci : j 2 0) is strictly decreasing. Define a sequence { b j } in [u,v] by

b3. -- f N - j ( ci) , bN+j

Since Ici - 20"

OljlN, j

= Ci,

> 0.

< 26 for i = 1,2, we have

If"(c')

- f n ( X g ) I = 1bN-n

0

- XN-"~

< &,

0

5 n 5 N.

Sinceu+~/2>x&>x$+~ > . . . > u a n d x & ~[u,v],wehave 21

+

&

> ci = c; > cf >

*

>u

CHAPTER 2

60

and thus lbj - z ! 1 < E for j 2 N. Therefore, lbj < E for all j 2 0. Since fN+'(ci) = f(zo) = yo, we have f j ( b 0 ) = fj(z8) for j > 0. Therefore, for j > 0, j-th components of a"(bo, bl ,* * * )

and

Q"(z:, z : ,

* *

-)

are closer than E for all n E Z. Let z3 = max{ fN+'(c') : i = 1,2}. If z3 < yo, then there exist c3, c4 with

such that f N+'(C3)

= fN+'(C4)

= c4.

By (4) we can find sequences {cj : j 2 O } , i = 3,4, in [u,v]such that ci = ci and f(c3) = for j 2 1. They are strictly decreasing sequences. By the similar argument we have that for j 2 0, j-th components of

g"(fN ( C3o ) , f N-1 ."(f N ( C4o ) , f N-1

3

(co),-**,c:,c;,---), 4

,c;,c:,4

(%I),***

are closer than E for all n E Z. Therefore (T : If + If is not expansive. The case ( 5 ) is obtained in the same argument. 0

For toral endomorphisms we obtain the following theorem from Lemmas 2.2.33 and 2.2.34 below.

Theorem 2.2.32. (1) A toml automorphism is expansive if and only i f it is an automorphism of type (0. (2) A toml endomorphism is positively ezpansive i f and only if it is an endomorphism of type (II). (3) A toml endomorphism which is not injective is c-expansive but not positively expansive if and only i f it is an endomorphism of type (III). Lemma 2.2.33. Let f be a linear map of the euclidean space B", where n 2 1, and let d be the euclidean metric for R". Then f is expansive under d i f and only i f f is hyperbolic, i.e. it ha8 no etgenvalues of modulus 1.

Proof. Suppose all eigenvalues o f f are off the unit circle. Then Rn splits into the direct sum B" = E" €3 E" of subspaces E" and E" such that f ( E d )= E" and f ( E " )= E", and such that there are c > 1,0 < X < 1 so that

$2.2 Expansivity

01

for n 2 0. Therefore, f is expansive under d. Conversely, if f has an eigenvalue of modulus 1, then Rn splits into the direct sum of subspaces as follows : (i) R" = EC@ E" @ E", (ii) f(E') = E"(0 = c , s , u ) , (iii) there are c > 0 and 0 < X < 1 such that (1) and (2) hold, and (iv) flEC is linearly conjugate to a linear map coresponding to Jordan form (in the real field)

where either

for 1 5 j 5 k. Here I is a 2 x 2 identity matrix and

0 < 8j

R(8j) =

< T.

By (iv) we can find a subspace F of Ec such that f p is an isometry under d. Therefore, f is not expansive. 0 Let

x and X be metric spaces with metrics 2 and d respectively, and let

X be a continuous surjection. Then T is called a locally isometric covering map if for each z E X there exists a neighborhood U ( z ) of x such that T - ~ ( v ( x ) )= u, ((Y# + u, n ual= 0) T :

-+

U 0

where U, is open and

TIU,

: U,

-+

U ( x ) is an isometry.

x

Lemma 2.2.34. Let f : -+ and g : X -+ X be continuous sujections and T : -+ X a locally isometric covering map. Suppose T o f = g o T and X i s compact, and suppose there exists 60 > 0 such that for each x E and 0 < 6 5 60 the open ball Ua(z)of x with radius 6 is connected and

x

x

T

is an isometry. Then

: Ua(.)

+

u6(.(.))

CHAPTER 2

62

( 1 ) f is ezpansive (under 2 ) i f and only if g is expansive, when f and g are homeomorphisms, (2) f i s positively expansive if and only if g is positively expansive, ( 3 ) f is c-expansive if g is c-expansive, and i f f is c-expansive and satisfies the following condition (C): for E > 0 there ezists N > 0 such that for (pi),(q;) E Ef one has d(p0,qo) 5 E whenever a ( p ; , q i ) 5 e for all i with lil 5 N where e > 0 i s an expansive constant for f , then g is c-expansive.

Proof. From the assumptions it follows that for each x E X and 0

< 6 5 60

and if p , q E T - ' ( x ) and p # q, then v&(p) r l u6(q)= 0. Indeed, it is clear Let a E ?r-'(u6(.)). Since qU6(,,) : that T - ' ( u 6 ( z ) ) ZI U,,,-l(z,U6(p). us(.) 4 U6(.(a)) is bijective and x E u6(7r(a)), we have a E Ua(p) for some p E T-'(x), and so (i) holds. If b E Ua(p)n Ua(q) # 8, then p = q because ?TIUs(b) : U6(b) U6(a(b)) is bijective. To obtain the lemma, we first show that f is uniformly continuous. Let 0 < E 5 60 and take 0 < 6 5 60 such that for x , y E X

Now, if &,q)

< 6, then

, implies and so n o f ( q ) E U,(g o ~ ( p ) )which

Since 6 5 60,we have that f(Ua(p))is connected. Notice that U , ( f ( p ) ) is the connected component of f ( p ) in .-'(Uc(go7r(p))). Then f (Ua(p))C Uc(f ( p ) ) , and thus f is uniformly continuous. (1) : Suppose g is an expansive homeomorphism with expansive constant e, and let 7 = min{e,60}. For p,q E if f i ( p ) , f ' ( q ) ) 5 7 for i E Z,then we have 4g"(.(P)),gi(.(q>>) = ' ( P I , f i ( q ) ) I 79 i E Z,

x, a(

4f

and so ~ ( p=) r(q).Since z(p,q) 5 60, it follows that p = q.

52.2 Expansivity

63

Conversely, suppose f is expansive. Choose 0 < 6 < 60 such that if Z ( p , q ) < 6 then a ( f ( p ) , f ( q ) )< 60 and Z(f-'(p),f-l(q)) < 60,and put 7 = min{e,S} where e is an expansive constant. For x, y E X with d(z, y) 5 7 we take p, q E such that ~ ( p =) x, ?r(q) = y and a ( p , q ) = d(z,y). If d(gi(x),gi(y)) I: 7 for i E Z, then

since a(f(p), f ( q ) ) < 60. Inductively, we have a ( f i ( p ) ,f i ( q ) ) 5 7 for all i 1 0. In the same way, a(fi(p),fi(q))5 7 for all i I: 0. By expansivity we have p = q, and therefore x = y. (2) : This is proved in the same manner as (1). (3) : Suppose g is c-expansive (with expansive constant e > 0). Let 7 = min{e,60} and ( p i ) , ( q i ) E Since A o f = g o A , it is clear that ( r ( P i ) ) , ( r ( q i ) ) E X,. If a ( p i , q i ) I: 7 for i E Z,then d(n(pi), r ( q i ) ) 5 7 for i E Z,and so pi)) = ( r ( q i ) ) , which implies (pi) = (qi). Conversely, suppose f is c-expansive and satisfies the condition (C), and suppose d(xi, yi) _< 7 (i E Z)for ( z i ) ,(yi) E X, where 7 = min{e, 6 ) and 0 < 6 5 60 is chosen such that a ( f ( p ) ,f ( q ) ) I: 60 whenever &I, q ) < 6. For each ! E A - ' ( z - n ) , q!!,, E ~-'(y-,) such that && ,!I), 5 7, n 2 0 we take ,p and define ( p ; ) , (q;) E by

xf.

xf

Then ( ~ ( p ? ) = ) (zi)and ( A ( $ ) ) = (yi), and &?,qy) 5 7 for all i 2 -n. By the condition (C) it follows that d (p t,q t) + 0 as n --+ 00. Thus zo = yo. Similarly, we have zi = yi for i E Z. 0

Remark 2.2.35. -

The condition (C) in Lemma 2.2.34 (3)is always satisfied if

X is compact. However, it remains a question of whether it is true for general

case. Let X be a compact metric space. Topological dimension of the space X is said to be less than n if for all 7 > 0 there exists a cover Q of X by open sets with diameter < 7 such that each point belongs to at most n 1 sets of a. (If the topological dimension is of interest, the reader should see Hurewicz and Wallman [H-W]).

+

Theorem 2.2.36. Let X be a compact metric space. X is 0-dimensional if and only if it is totally disconnected (i.e. the connected component of each point is a singre point). Before starting with the proof we prepare the following

CHAPTER 2

64

Lemma 2.2.37. Let 0 be the collection of open closed subsets of X. Then 0 is a base of X if and only if X is totally disconnected. Proof. Take x , y E X with x # y. Then we can find an open set U, such that x E U, and y # U,. Suppose 0 is a base of X. Then x E B c U, for some B E 0.Since X \ B is open closed and y E X \ B , the connected component of x , ~ ( x )is, a subset of B . However, since y is arbitrary, we have

and therefore X is totally disconnected. For fixed x E X let G be an open neighborhood of x . Then it suffices to find an open closed subset B satisfying x E B c G . If X is totally disconnected, then each x E X is expressed as { x } = U ( x ) where { U ( x ) } is the collection of open closed subsets containing x. Indeed, let L = n U ( x ) . To see {x} = L it suffices to prove that L is the connected component of x. If this is false, then L is expressed as L = A U B where A and B are closed and A n B = 0. Thus we have that x E A or x E B . We suppose x E A. Choose open subsets A1 and B1 such that A c A1, B C B1 and Al n B1 = 0. Then L c A1 U B 1 . Since L = U ( x ) ,there is U ( x ) such that L c U ( x ) c A1 U B1. Then it is easily checked that A: = U ( x )n A1 and Bi = U ( x )n B1 are open closed in X. Since x E A, A: is an open closed subset containing x and L n A: = 0. This contradicts the definition of L. Therefore, for y E X we can find an open closed subset H , satisfying y E H , and x # Hy.Since X \ G is compact, a finite cover { H,, , ,H,, } of X \ G exists. Notice that each HVi satisfies x @ ITvi. H = lJy HVi is open closed and x # H. Therefore x E B C G where B = X \ H. 0

n

n

--

Proof of Theorem 2.2.36. Suppose X is totally disconnected. By Lemma 2.2.37 the collection of all open closed subsets is a base of X. Let I' be an open cover of X and let x E X. Then there is U, E I' such that x E U, and then there is an open closed V, satisfying z E V, C U,. Since {Vz : x E X} covers X, choose a finite cover { W1, * ,Wk} and define

.

F1 = W i , F2 =W2\Wl,

...,

.. -

Fk=Wk\(W1U..*UWk-1).

Then { F1,. ,Fk} is a refinement of r, and a point of X belongs to only one set of {Fl, * * ,Fk}. Therefore X is zero dimensional. Suppose X is zero dimensional. Take arbitrarily a, b E X with a # b. We set U = X \ { a } a n d V = X \ { b } . Then{U,V}coversX. Let a = { F 1 , . . - , F k } be a closed cover of X such that Fi n Fj = 0 for i # j and a is a refinement of { U,V}. Without loss of generality we suppose a E 4 . Since X = F1 U F and F1 n F = 0 where F = F2 U U Fk, The connected component of a, c(a), is a subset of F1 and so c(a) c X \ {b}. Since b is arbitrary, we have that c(a) c n { X \ {b} : b E X , a # b } = { a } , i.e. X is totally disconnected.

---

52.2 Expansivity

65

We here construct the Cantor set which is of particular importance. To do as follows. First, denote the closed interval [0,1] by C1. Next, delete from C1 the open interval (1/3,2/3) which is its middle third, and denote the remaining closed set by C2. Clearly, C2 = [O, 1/31 U[2/3,11. so we proceed the argument

Next, delete from Cz the open intervals (1/9,2/9) and (7/9,8/9) which are the middle thirds of its two pieces, and denote the remaining closed set by C3. It is easy to see that

Figure 5

If we continue this process, at each stage deleting the open middle third of each closed interval remaining from the previous stage, we obtain a sequence of closed sets C,,,each of which contains all its successors. The Cantor set C is defined by m

n= 1

and it is closed. C consists of those points in the closed interval [0,1] which ultimately remain after the removal of all the open intervals (1/3,2/3),

(1/9,2/9),

(7/9,8/9),

* * * *

Clearly C contains the end points of the closed intervals which make up each set Cn ; 0, 1, 1/3, 2/3, 1/9, 2/9, 7/9, 8/9, . a

*

*

The set of these end points is clearly countable. However, the cardinal number of C itself is the cardinal number of the continuum.

CHAPTER 2

66

To prove this it suffices to exhibit an injection f of [0,1) into C. We be its construct such a map as follows. Let z E [0, 1) and let z = .blb2 binary expansion. Each b,, is either 0 or 1. Let a, = 2b,, and regard

---

f(z)= 3-lal-k 3-2a2

+---

as the ternary expansion of a real number f(z)in [0,1). The reader will easily convince himself that j(z) is in the Cantor set C. In fact, since a1 is 0 or 2, f(z)is not in [1/3,2/3). Since a2 is 0 or 2, f(z)is not in [1/9,2/9) or [7/9,8/9) ; etc. Also, it is easy to see that f: [0, 1) C is surjective. This shows that C is the set of the numbers z E [0,1] with --f

For T 2 1 we call the set C n [3+i,3+(i subinterval with mnk T if

+ l)] (0 5 i 5 3r - 1) a Cantor

We denote as I ( i , r ) the i-th Cantor subinterval with rank Obviously

T

from the left.

2'

c = UI(i,T),

I(i,r)=I(2i- l,T+l)U1(2i,T+l).

i=l

Remark 2.2.38. Let X be a compact totally disconnected metric space having no isolated points. Then X is homeomorphic to the Cantor set C. This is checked as follows. Let d and d' be metrics for X and C respectively. Since X is totally disconnected, we can choose open closed subsets X1,X2,*** ,X2k1(IZ1> 0) such that diam(Xi) < diam(X)/2

(1 5 i 5 2'')

2k1

X = UXi,

xinxj=~ (i+j)

i=l

Since each of Xi has no isolated points, as above we can choose open closed subsets X ~ J ,* * ,Xi,)ka such that

-

diam(X;,j)

< diam(X)/22 (1 5 j 5 2'l),

2ka

x i = u xi,j, j=1

Xi,j nXi,ji = 0 ( j

+j').

52.2 Expansivity

07

Continue inductively this manner. Then there is a sequence {kj}gl of positive integers and open closed subsets Xil,...,i,( 1 5 it 5 2", 1 5 I 5 m ) satisfying the conditions

< diam(X)/2"

(i>

di=(Xi, ,...,i,)

(ii)

Xil,...,i,-l,j n xil,...,im-l,jl = 0 ( j # j ' ) ,

(1 5 it 5 2",1 5 I 5 m ) .

2hm

xil,...,im-l

(iii)

=

U Xil,...

,i,-l,j-

j=1

If (i~,... , i n , . - - )E n g l { l , 2 , . - - ,2kj}, then we have {z} = for some z E X. Define k ( m ) = Cj"=, kj for rn 2 1 and

nj"=,Xi, ,...,

ij

n= :,

Then there is y E C such that {y} = I(p(rn),k(rn)),and so we define h ( z ) = y. Then h: X + C is a one-to-one continuous surjection ~ choose 6 Indeed, for E > 0 take mo 1 1 such that 3k1+*'*+kmo > l / and such that

< 6 < min{d(Xil ,...,imo-l,j,Xil,...,imo-l,j~) : 1 5 in 5 2kn, 1 5 n 5 mo - 1, 1 5 j # j ' 5 2km-0). If d ( z , y ) < 6(z,y E X), then h ( z ) and h ( y ) belong to the Cantor subinterval with the rank Cj"=ol kj. Thus we have d'(h(z),h ( y ) ) < E and therefore h is 0

continuous. That h is bijective is easily checked.

Theorem 2.2.39. Let f : X + X be an expanding map of a compact metric space X. Then the topological dimension of X is finite. Proof. To see dim(X) < 00 we consider the inverse limit system ( X f , a ) of (X, f ) . Since f is expanding, by definition f is a local homeomorphism and positively expansive. Then, by using Theorems 2.1.1 and 2.2.10, we see that there is a base of neighborhoods for Xf each of whose members is the direct product of a neighborhood of X with Cantor set. Since f is positively expansive, d : Xf + Xf is expansive. If Xf is finite dimensional, obviously the dimension of X is finite. That dim(Xf) < 00 follows from the following result. 0

Theorem 2.2.40 (Mafib [Mall). If f : X + X is an expansive homeomorphism of a compact metric space X, then the topological dimension of X is finite

.

For the proof we need the following lemmas. Let e constant and fix 0 < E < e/2.

> 0 be an expansive

CHAPTER 2

68

Lemma 2.2.41. There is a 6 > 0 such that ifd(z,y) E

for some n

< 6 (z,y E X) and

5 sup{d(fj(z),fj(y)) : o 5 j 5 n } 5 2.5

2 0 , then d(fn(z), f"(y)) 2 6.

Proof. If this is false, then there are sequences Z,,yn E X, m, > Cn > 0 such that lim d(xn,yn) = 0,

n+oo

lim

n+oo

d(fmm(xn),fmm(Yn))

2 E, :0 I m 5 m n } 5 2

= 0,

d(f'n(xn),r'"(yn))

sup{d(fm(zn), f"(Yn))

~ .

Since X is compact, we suppose that fC* (xn)+ x and f '"(y,) + y as n + 00. Then d(z,y) 2 E and d(fn(x), f"(y)) 5 28 for all n E Z.Therefore Lemma is proved. 0

Lemma 2.2.42. For all p E

> 0 there exists N

< SUP{d(f"(4, f"(Y))

= N ( p ) > 0 such that : In1

I NI

whenever d(z,y) 2 p. Proof. If this is false, then there are sequences s,,y, E X with d ( z n , yn) 2 p such that suP{d(P(zn), fj(yn>> : IjI I n ) I E If x, + x and y, + y as n + 00, then d ( z , y ) 2 p and d(fn(x),fn(y)) 5 E for all n E Z. 0 Proof of Theorem 2.2.40. By using Lemmas 2.2.41 and 2.2.42 we derive the conclusion. Let 6 be as in Lemma 2.2.41 and choose a cover { Ui : 1 5 i 5 1) of X by open sets with diameter < 6. We show that dim(X) I L2. For each n 2 0, choose 6, > 0 such that d(x,y) 5 6, implies d(fj(z),fj(y)) < E for all Ijl 5 n. We define

-

utj = fn(ui)n f-"(uj)

and x y for x,y E Ucj if there exists a finite sequence x = 20, X I , . *, x p = y such that d(z,,x,+~)< 6, for all 0 5 T 5 p - 1 and 2, E Ucj for all 0 5 T I p. Denote as nk Ui,; , 1 5 12 5 k ( i , j , n ) the 6,-components of Uej (i.e. the equevalence class of Ucj under the relation x y). It is easily checked that each U,$ is open and they cover X. We claim that N

2.2 Expansivity

69

Indeed, if this is false, then we could find p > 0 and a large number n, say N(p) given in Lemma 2.2.42, such that in some U$! there are z,y with d(z,y) > p. Let z = zo,q,. ,z p = y be a sequence in U;, such that d ( Z r , z r + l ) 5 6, for d O 5 T 5 p - I.Let US put n

> 2N(p),

.-

sr = sup{d(fm(zr), ~ " ( Z O ) ): Iml I n } . Then sr

> E by Lemma 2.2.42. 81

Choose T such that

S,I

< E,

From the choice of 6, it follows that

Isr+1

- 8.1

€7

1 IT

I E if T' < T and a, > E . d(f-,(zo),

since f-,(z~),f-"(z,) E U;.Since f"(z,)) < 6 and SO 8,

I

f-"(zr))

< P*

Then

8,

5 2~ and therefore

I6

fn(zg), fn(z,)E

U,,we have d( fn(zO),

= su~{d(f"(zo), f"(zr)) : I m l I n } I 2 ~ .

For simplicity put zb = f-"(zo) and z',= f - " ( ~ , ) . Then sup{d(fm(zb),

f"(zi)) : o 5 m 5 2n} 5 2 ~ .

Since d(z;,z',) < 6, by Lemma 2.2.41 we have

d(f2n(zb), f2"(.5))

= d(f"(zo),

fn(z?-)) 2 6,

thus contradicting d(fn(zO), f"(z,)) < 6. Therefore (1) is proved. It only remains to show that for each n, any point of X belongs to at most L2 sets of the cover

{U*$ :1 I i , j I e,1 5 k 5 k ( i , j , n ) } . Suppose that

n{uc:;.m: 1 5 m 1 s } # 0 . If ( i m , j m ) = ( i ~ , j ~then ) , we have UzT :'m = Uz!& since they are 6,components of Ui",,jn. and have nonempty intersection. This implies that to different values of m correspond different values of the couple (im,jm). Therefore, s I L2. 0

Remark 2.2.43. Theorem 2.2.40 is true for a c-expansive continuous surjection. This is checked in the way as follows. Let 3 = { Fi : 1 5 i 5 L } be a finite closed cover of X such that each Fi has sufficiently small diameter. For n > 0 we define FCj = fn(F;)n f-"(Fj) for 1 5 i,j I 1. Then 3" = {FC,} is a finite closed cover of X. Use this cover 3" in the proof of Theorem 2.2.40. Then we can show that the dimension of X is finite.

CHAPTER 2

70

Theorem 2.2.44 (Maii6 [Mall). If f:X + X is an ezpansiue homeomorphism of a compact metric space, then every minimal set for f is zerodimensional. For the proof we must investigate the structure of orbits for expansive homeomorphisms. We first prepare the following

Lemma 2.2.45. Let X be a compact connected metric space and A a proper closed subset of X . If x E A and C is the connected component of x in A, then C n BA # 0 where BA denotes the boundary of A. Proof, Let C be the collection of all open and closed subsets K under the relative topology of A such that C C K C A. Since X is compact, we have C = n { K : K E C } . To obtain the conclusion suppose C n 8A = 0. Then U{X\K : K E C} 3 BA. Since OA is compact and X\K is open and closed, BA is covered by the finite union of X\Ki. Since Ki = K is open and closed, we have K E C and K n t3A = 0. Notice that K is expressed as K = U n A for some open subset U of X because K is open in A. Since A = BA U int(A), we have K = U n A = U n (BA U int(A)) = U n int(A). Thus K is open in X. Since K is closed in X , we have X = K and so BA = 0. Therefore, A = X which is a contradiction. 0

ni

>

0. For z E X let C , ( x ) be the connected component of z in : d ( z , y ) 5 7},and write BB,(x) = {y E X : d ( z , y ) = 7). We prepare some lemmas that are used in proving Theorem 2.2.44.

Let 7

B,(z)= {y E X

Lemma 2.2.46. If, for some x E X and some m Wa(z,d ) # 0, then X contains a periodic point.

> O,fm(Wa(x,d)) n

Proof.Take y E W " ( x , d ) nfm(Ws(x,d)) and put z = f-m(y). Then fm(z) E W"(z,d)= W " ( z , d ) . Thus, lim,,,d(f" o fm(z), fn(z)) = 0. Since X is compact, a subsequence { fnj(z)} converges to w . Therefore, d ( w ,fm(w)) = limn-,a, d(fjn o fm(z), f i n ( % ) ) = 0. Define Ci(z) and X:(x) as the connected components of x in W,"(z,d)n and W:(z, d) fl &(Z) respectively. Let e > 0 be an expansive constant for f . We fix 0 < E < e / 2 . B6(Z)

Lemma 2.2.47. If dim(X) > 0, then there exists 0 < 7 < E such that for 0 < 6 < 7 there is a E X such that

Xi(.)

n a & ( U ) # 0 or C;(a) n BBa(a) # 0.

Proof. Let C , ( x ) denote the connected component of x in B,(x). Since dim(X) > 0 , X is not totally disconnected. Thus there exist x E X and 7 > 0 such that W x c )n BB,(x:)# 0 by Lemma 2.2.45, and so &(Z) n a&(z) # 0 for 0 < 6 < 7.

$2.2 Expansivity

Suppose there is 0 < 6

< -y

71

such that

x!(Y> n aB6(y> = 0 for all y E X. Then we prove that for some a E X

X:(a) n 8Ba(a)# 0.

Figure 6

To find such an a E X we construct a collection of compact connected sets A,, n 2 0, and a sequence of points xn E A, such that for a certain subsequence {m,} the following conditions hold :

If we establish the above conditions, then the conclusion is obtained a8 follows. Take a subsequence {zni}of {z,} such that tni + a as i --t 00. We fix {zni} and define

It is clear that the point a is contained in A. We show that A is connected. Indeed, suppose A is not connected. Then there exist nonempty closed sets 4, FZsuch that F1 nFZ = 0 and A = 4 LJFz. Since { z , ~ , ~is} a subsequence of the sequence {xni}, we have + a it9 j + 00. The point a belongs to F1 or F2. Without loss of generality we

CHAPTER 2

72

suppose the former, i.e. a E F1. For b E F2 let {b,i,j) be a sequence such that bni,j E Anivj and b,i,j -+ b as j -+ 00. Since F1 n F2 = 0,there exist neighborhoods U ( F i ) of Fi such that U(F1)n U(F2) = 0. But each hni,jis connected. Thus we can find yni,j E Ani,j satisfying Yni,j 6 U(F1)u U(F2).

If yni,j 4 y as j + 00, then we have y E A, but y By (iv) we have

6 F1 UF2, thus contradicting.

Since z, = fmn(z,+l)by (i), from (iv)

from which

“nij

b,.BJ. Figure 7

By induction on n we have f’(An)CBe(f’(zn)),

If n

-+ 00,

then fj(A)

05j<mo+ml+...+mn-1*

c B,(fj(a)) for 0 5 j < 00,

and thus

A c W,d(a,d).

Since zni,j+ a, there is yni,jE A,i,j n a B a ( ~ , , satisfying ~,~) 5 = !im d(Yni,j, zniqj)= d(y, a ) . 3 --)‘=a

52.2 Expansivity

73

Therefore, A n aBa(a) # 8. Since A C W i ( a , d ) and A is connected, we have E $ ( a )n a&(.) # 0. The remainder of the proof is only to construct the sequence {A,} of closed connected subsets satisfying (i), (ii), (iii) and (iv). Let x and 0 < 6 < E be as above. Put A0 = E:a(Z), then A0 n & 3 6 ( 2 ) # 8. Notice that E:(y) n 0&(y) = 8 for all y E X. To obtain the conclusion suppose Ao, 111,. -.,A,-1 are constructed. Then we have An-1

P WZ(zn-1, d).

Indeed, if A,-1 c W;(x,-l,d), then E;(zn-l) 3 An-l since E:(zn-l) connected component of z,-~ in &(&-1) n W,"(z,-,,d). Thus E;(Zn-l)

n BB6(%-1)

thus contradicting. Therefore, An-l\W,"(Zn-l,d) some m > 0 we have

n oB6(%-1) # 8,

# 8, from which choose a point y. Then, for

s u ~ { d ( f - ~ ( z f-"(zn-l)) ), (v)

3 An-1

is the

: E E &-I}

2 d ( f - m ( y ) , f-m(%-l)) > E , ~~p(d(f-'(z),f-'(x,-1)) : E E An-l,O 5 j < m } < E .

For this m we write m,-l = m and z, = f-mn-l(x,-l). Denote as An the connected component of x, in B6(3&,)n f-"'n-I(A,-l). Then (ii) holds, and A, n aB6(%,) # 8 implies (iii). By (v) we have (iv). 0

Lemma 2.2.48. For 0 < 6 < E there exists N > 0 such that for all x E X and y E W,"(z,d)with d(y,x) = 6, there is 0 < n 5 N such that 4f-n(y),

f-"(z)) > E -

Proof. If this is false, then there exist sequences x, E X,31, E W,"(Zn,d ) such that d(x,, y,) = 6 and d(f-j(x,), f-j(y,)) 5 E for 0 5 j 5 n. If x, + x and yn + y as n + 00, then x # y and d(fn(z),fn(y)) 5 E for all n E Z, thus contradicting. 0 Lemma 2.2.49. There exists 60 > 0 such that

w:(z,~)n B 6 ( 4 = w;J~,d)n &(z) for all

x E X and 0 < 6 < 60.

Proof.If this is false, for n > 0 then there exist x, E X and 0 such that W;(Zn,d) n B6,(%) # W&(%,d) n B6,(%)-

< 6, < 1/n

CHAPTER 2

74

Thus we can find yn E W&(z,,d)and m, d(Zn,Yn)

+

0 as n

-+

00,

d(fmn(zn),fmn(yn))

Since yn E W&(z,,d),for all m with m

4f" frnn(z*), f" 0

> 0 satisfying > ~ , >n 0.

> -m, 0

frnn(Yn))

from which

I

z

d(frn(z),frn(y))5 2&, m E where fmn(zn) + z and fmn(yn) + y as n --t 00. Therefore, z = y, but d ( z , y ) 2 E . This is a contradiction. 0

Figure 8

Lemma 2.2.50. Let Z = inf{d(z,y) : d(f-l(z),f-'(y)) > &,z,y E x} and let 6, > 0 be as an Lemma 2.2.49. For 0 < 6 < min{60,Z/2} there exists N = N ( 6 ) > 0 such that if z E X and A c Wi(z, d ) i s a compact connected set containing x and if A n Ba(z) # 0, then there exist 0 < m < N , a,P E f-"(a) and compact connected sets Aa,Ap such that

(4 (b) (c)

(4

P E Ap, ha n OB6(a) # 0, a E Aa,

a E W&(P,d),

n aB6(P)# 0,

inf{d(z,w) : z E Ba(a),w E B @ ) } > 6, ha c W2"c(a, d ) n B6(a), c W;c(P,d ) n B6(P)*

Proof. Let N = N ( 6 ) > 0 be as in Lemma 2.2.48. Since A f l d B & ( z )# 0 and z E A C W:(z, d ) by assumption, for y E W : ( x , d ) with d(z, y) = 6 we can find 0 5 m 5 N - 1 such that (el

(f)

s~p{d(f-~~+'~(z),f-(~+')(z)): z E A}

> - d(f-("+')(y), f-(m+"(z)) > E , sup{d(f-j(z),f-j(z)) : z E A,O I j 5 m } 5 E.

f 2.2 Expansivity

75

By (f) and the fact that A C W:(z,d)

(4

f-"(A)

Since f-"(z) E f-"(A),

we have

(h)

c W,"(f-"(z),

f-"(z) E W;(w,d)

4.

for w E f-"(A).

Figure 9 From (g) and (h) it follows that for all w E f-"(A)

f-v) c W&(w,4.

(i)

By (e), d(f-l o f-"(z), f-" (z))2 z.

f-'

o

f-"(x)) > E for some z E A, and so ~ I ( f - ~ ( z ) ,

Figure 10 Since z , x E A, we have diam(f-"(A)) 2 Z. Since 36 < Z, we can find a,@ E /-"(A) such that d ( a , @ )> 36. Obviously a 6 B6(p) and p pl &(a). Therefore, (c) holds. Let A,,Ap be the connected components of a and p in f-"(A) n &(a), f-"(A) n B,s(@)respectively. From (g) it follows that (a) holds. Since

CHAPTER 2

76

A, Ap

C f-"(A)

and Ap C f-m(A),

by (i) we have A, C W,d,(a,d)and

c W.&(p,d).Thus (d) holds. It remains to check (b).

This is followed by replacing X , A and C with f--(A), f-m(A) n &(a) and A, respectively (in fact, since fqm(A) nBa(a) is a closed ball with radius 6 of a,we have i3(f-"(A) n &(a)) = {z E f-"(A) : d ( z , a ) = a} ). 0

Proof of Theorem 2.2.44. It is enough t o verify that if f: X -+ X is minimal and expansive then dim(X) = 0. Suppose dim(X) > 0 and 7 > 0 are as in Lemma 2.2.47. For 0 < 6 < E choose N = N ( 6 ) > 0 as in Lemma 2.2.48 and let 60 > 0 be as in Lemma 2.2.49. Take Z > 0 as in Lemma 2.2.50, and for 0 < 6 < min(bo,F/2,7} define

If 71 = 0, then X is a set consisting of finite points. Indeed, for all n 2 0 there exist zn,yn E X and 0 5 in, j, 5 N ( 6 ) such that

We can suppose that n 2 0. Thus

2,

-+

z,y,

4

y as n

-+

00

and j , = j,i, = i for all

Notice that i # j. Suppose i - j > 0, then we have fj-;(z) = y E W&(z, d ) C Wa(z,d) and f + - i ( ~ a ( ~ , d ) ) n ~ ~ (d )3# 0 , since f j - i ( z ) E fj-;(Wa(z,d)). By Lemma 2.2.46, X contains a periodic point zo and thus the orbit Of(z0) equals X (since f : X + X is minimal). It remains to check the case 71 > 0. For this case we shall derive a contradiction by showing the existence of a nonempty open set U and a point p such that f n ( p ) # U for all n 2 0. If we have U and p as above, then B n U = 0 and f ( B ) c B where B is the closure of the semi-orbit {p, f ( p ) , .}. Since A = f n ( B )is nonempty and f ( A ) = A , we have X = A , thus contradicting. We first construct a collection of compact connected sets A,, a sequence of points 2, E A, and a nonempty openset U c X that satisfy the following

nz=o

-.

52.2 Expansivity

Since 6 define

77

< 7,there is u E X with X$(u)fl 6 & ( U ) # 0 by Lemma 2.2.47. We 20

= U,

A0

= E;(Q)

and show the existence of U # 0 with diam(U) < 71/2 such that U n A. # 0. Suppose U n A0 # 8 for all open sets U. Then A,, is dense in X and so A0 = X (since A0 is closed). Since diam(A0) 5 26 < e and f is expansive, X consists of one point set, thus contradicting dim(X) > 0.

/

Figure 11

If mo = 0, then (a), (b) and (d) hold for Ao. To construct A1 satisfying (c) we use Lemma 2.2.50. In fact, since A0 n 6&(xO) # 0 and Ao C W:(zo,d), by Lemma 2.2.50 we can find 0 < ml < N ( 6 ) , a , P E f - m l ( A ~ ) and compact connected sets A,, Ap satisfying (a), (b), (c) and (d) of Lemma 2.2.50. Then

Indeed, if y E Un{U,”o fj(Ap)}, then y E fj(Ap) for some 0 5 j 5 m l . Thus there is a E A p such that fj(a) = y. By assumption of (e), for z E U n fj(A,) there is w E A, such that f j ( w ) = x. Since x, y E U, (f)

d(fj(w>,P(4)= d(x,y)

and since w E A, and

t E Ap

d(w, z )

>6

< 71/2

(by (c) and (d) of Lemma 2.2.50).

By Lemma 2.2.50 (d), w E W i ( a ,d) and by Lemma 2.2.50 (a), a E W&(P,d ) . Thus, w E W & ( a , d ) . However d(fi(w),fj(z)) = d ( z , y ) 1 71 by definition, thus contradicting (f).

CHAPTER 2

78

Since p E f - m l ( A ~ ) n Ba(/3),we denote as A1 the connected component of /3 in f-ml(Ao) n B @ ) and put z1 = 0. Then we have (a') A1 n 8B6(zl) # 0 (by Lemma 2.2.50) , (by Lemma 2.2.50 (d)) , (b') A1 c Wc(zl,4

(4

Ai C f-"'(Ao),

(d')

{Uf j ( A 1 ) l n u = 0

ml

(by (el)

j=O

Replace A0 by A1 and repeat the above technique. Then there exist a compact connected set A2 and m2 > 0 satisfying (a'), (b'), (c') and (d'). Continuing inductively this technique, we have {A,},{z,} and U satisfying (a), (b), (c) and (d). By (c) we have

(Al) 3 ... 3 fml+***+mm(A,) 3 ..* and thus fml+".+mn (A,) # 0, from which take a point p. Obviously f-ml(p) E A, By (d) we have f-ml+j(p) 4 U for 0 5 j 5 ml and thus

n,

A,, 3

fml

f - ' ( p ) $2u, 0 5 i 5 m1.

Since f - ( m i + m a ) (PI E

A29

by

(4

f - ' ( p ) $ U, o 5 i 5 ml and consequently f - ' ( p ) $ U for all i 2 0. 0

+ m2

Let f: X --t X be a continuous surjection of a compact metric space. The map f is said to be minimal if, for all z E X ,the orbit O f ( z )= {fn(z) :n 2 0) is dense in X. A closed subset E is said to be a minimal set if f ( E ) = E and f p : E --t E is minimal. Remark 2.2.51. It remains a question of whether Theorem 2.2.44 is true for a c-expansive continuous surjection. We remark that the question can not be shown by the technique of the proof of Theorem 2.2.44, nor by the way through the inverse limit as in Theorem 2.2.39 (because the map is not a local homeomorphism). 82.3 Pseudo orbit tracing property

A sequence of points {zi : a < i < b} of a metric space X is called a Spseudo orbit off if d(f(zi), .;+I) < 6 for i E ( a , b - 1). Given E > O a S-pseudo orbit { z i } is said to be &-traced by a point z E X if d ( f i ( z ) , z i )< E for every i E (a,b). Here the symbols a and b are taken as -00 5 a < b 5 00 if f is bijective and as 0 5 a < b 5 00 if f is not bijective. We say that f has the pseudo orbit tracing property (abbrev. POTP) if for every E > 0 there is S > 0 such that every 6-pseudo orbit of f can be &-tracedby some point of X. For compact spaces this property is independent of the compatible metrics used.

$2.3 Pseudo orbit tracing property

79

Remark 2.3.1. In the case where X is compact, a homeomorphism f : X -+ X has P O T P if for every E > 0 there is 6 > 0 such that every (one sided) 6-pseudo orbit {zi :i 2 0) o f f can be E-traced by some point of X. Indeed, let {zi : i E Z} be a &pseudo orbit of f . For each n 2 0 define a (one sided) &pseudo orbit { z r : i 2 0) by zr = zi-n for all i 2 0, and let zn be a tracing point for {z? : i 2 0). Then it is easily checked that an accumulation point of { f n(zn)} is a tracing point for {zi : i E Z}. Theorem 2.3.2. Let X be a compact metric space and denote as id the identity map of X. Then id : X -+ X has P O T P if and only if X is totally disconnected. Proof. If X is totally disconnected, for E > 0 there exists a finite open cover {Uo,...,Un} of X such that Ui n Uj = 0 for i # j and diam(Ui) < e for a. Choose 6 with 0 < 6 < min{d(Ui,Uj) : i # j} and fix a &pseudo orbit { z k : k E Z} for id. Then there is Ui such that {zk} c Ui and so we can find in Ui an €-tracing point for {zk}. Therefore id : X --f X has POTP. Conversely, suppose dim(X) # 0. Then there is a closed connected subset F such that diam(F) > 0. Since F is compact, diam(F) = d(z0, yo) = €0 for some z0,yo E F. Let ~1 = E 0 / 3 . Since F is connected, for any 6 > 0 we can construct a &pseudo orbit from xo to yo in F which is not €1-traced in X. This is a contradiction. 0

Theorem 2.3.3. Let f : X + X be a continuous map of a compact metric space and let It > 0 be an integer. Then f has P O T P if and only if so does

fk. Proof. We first notice the following (l),(2) and (3). (1) Let E > 0. Then there is E > el > 0 such that each €1-finite pseudo orbit {zi : 0 5 i 5 12) satisfies , < ~ / 2 , 0 5 i 5 It d( f i ( ~ o ) zi)

and d(z, y)

< €1

implies max{d(fi(z), fi(y)) : 0

I i I k} < ~ / 2 .

(2) Let €1 be as in (1). Then there is 61 > 0 such that each &-pseudo orbit f k is El-traced by some point. (3) Let €1 and 61 be as in (1) and (2). Then there is 6 > 0 such that each &finite pseudo orbit { z i : 0 5 i I 12) is 61-traced by zo E X. With these properties we show that each &pseudo orbit {yi : i 2 0) for f is E-traced by some point. Write

for

zi = y k i , For fixed i, {yki+j : 0 5 j

I 12)

i 2 0.

is a &finite pseudo orbit for f . By (3) we have

d(f'(yki), Yki+j)

< 61, 0 I j 5 12.

ao

CHAPTER 2

I f j = k then d ( f k ( y k i ) , y k i + k ) = d ( f k ( z i ) , z i + l ) < 61. Thus { Z i } is a 61pseudo orbit for f k . By (2) there is y E X such that d ( f k i ( y ) , z i ) < c1 for i 2 0. Thus, d ( f k i ( y ) , y k i ) < E I for i 2 0. On the other hand, since (yki+j : 0 5 j 5 k} is an El-pseudo orbit for f, by (1) we have

Since i is arbitrary, we have d( fn(y), yn) 6-pseudo orbit {yi}. 0

< E for

n >_ 0 and y &-tracesthe

Theorem 2.3.4. Let X be a compact metric space. If f :X omorphism with POTP, then so is f - l .

4

X i s a home-

Proof. For every E > 0 let 6 > 0 be a number such that each 6-pseudo orbit { X i } is &-tracedby a point y E X. Choose 6' > 0 such that d ( z , y) < 6' implies d ( f ( z ) ,f(y)) < 6 and put g = f - ' . It is enough to see that each 6'-pseudo orbit { p i } for g is &-tracedby some point. Since d ( g ( y i ) , y i + l ) < 6' for i E Z, we have d ( y j , f ( y i + l ) ) < 6 for i E Z. Letting zi = y-i for i E Z , { z i } is a 6-pseudo orbit for f and thus there is y E X with d ( f i ( y ) , z j ) < E for i E Z. This implies d ( f - ; ( y ) , z - i ) < E for i E Z and therefore d ( g i ( y ) , y i ) < E for i E Z. 0

Theorem 2.3.5. Let X and Y be metric spaces and X x Y the product topological space with metric

where dl and dz are metrics for X and Y respectively. Let f : X g : Y + Y be continuous maps and let f x g be the map defined by

+

X and

Then f x g has POTP if and only if both f and g have POTP. Proof. Suppose f x g has POTP. For every E > 0 let {zi} and {yi} be 6-pseudo orbit for f and g respectively. Then {(z;,yi)} is a 6-pseudo orbit for f x g . Thus there is (2, y) E X x Y with d ( ( f x g ) i ( z , y), ( z i ,yi)) < E for i 2 0. Then, d l ( f ( z ) , z i ) < E and d 2 ( g i ( y ) , y i ) < E for i 2 0. Also the converse is proved. 0

$2.3 Pseudo orbit tracing property

81

Theorem 2.3.6. Let f : X -, X be a continuous map of a compact metric space X and let h : X -+ Y be a homeomorphism. Then g = h o f o h-' has POTP if and only i f so does f . Proof. For every E > 0 there is 61 > 0 such that d(z,y) < ~1 implies d'(h(z), h ( y ) )< E where d' is a metric for Y . If f has POTP, then there is b1 > 0 such that each 61-pseudo orbit { x i } for f is sl-traced by some point. Let 6 > 0 be a number such that d'(z,y) < 6 implies d(h-l(z), h-l(y)) < 6,. Then it is enough to see that each 6-pseudo orbit {yi} for g is &-traced by some point. To do this put xi = h-'(yi) for i 2 0. Since d'(g(yi),yi+l)< 6 for i 2 0, we have d(f(zi),zi+l) = d(h-' 0 g(yi),h-l (yi+l))< 61 for i 2. 0. Thus { z i }is a 61-pseudo orbit for f and so d ( f i ( y ) , z i ) < e l ( i 2. 0 ) for some y E X. Therefore, d'(h o f i ( y ) , h(zi))= d'(gi o h ( y ) ,y i ) < E for i 2. 0. 0

Theorem 2.3.7. Let X be a compact metric space. A continuous surjection f:X+ X has POTP if and only if for every E > 0 there is 6 > 0 such that for ( z i ) E Xz with d(zi,zi+1)< 6 (i E Z) there exists a point ( y i ) E X f SO that d ( y i , z i ) < E for i E Z. Proof. This is easily checked as in Remark 2.3.1. 0 By Theorem 2.3.7 we can give the definitions of &pseudo orbit and &-tracing point for continuous surjections as follows. If ( z i ) E Xz has the property that d(f(zi),zi+l)< 6 for i E Z, then ( z i ) is called a &pseudo orbit of f . If (yi) E Xf satisfies d ( y i , z i ) < 8 (i E Z), then ( y i ) is called an s-tracing point for (zi).

Theorem 2.3.8. Let X be a compact metric space. If a continuous sujection f : X + X has POTP then u :Xf + Xf obeys POTP. Proof. Let a = diam(X) and E > 0. Choose N > 0 with c ~ / 2 ~ 0 be a number such that

By POTP o f f there is 6' > 0 such that any 6'-pseudo orbit o f f is ytraced. Choose 6 > 0 with 0 < 2N6 < 6'. Let k > 0 and suppose {(zy): 0 5 n 5 k} is a finite 6-pseudo orbit of u in Xf.Then we have

where i((zi),(pi)) = CZ-, d(zi,y;)/21il, and hence {"IN : 0 5 n 5 k} is a 6'-pseudo orbit of f , from which we can find y E X such that d ( f "(y), z 2 N ) 5 7 (0

5 n 5 k)*

CHAPTER 2

82

+

3 ~ / 8 €14

+ €14 < E

becauseof d ( f n ( y i ) , x ? ) 5 ~ / 8 ( l i 5 l N) by the fact that d ( f n ( y - N ) , z E N ) 5 7 . 0

Theorem 2.3.9. If u : Xf + Xf has POTP and f:X homeomorphism, then f has POTP.

--+

X is a local

Proof. For 8 > 0 let 6 > 0 be a number such that every &pseudo orbit of u is E-traced by some point of Xf.Let a = diam(X) and choose N such that 0 0 such that if d ( z , y ) 5 7 then there are {zi : -N 5 i 5 N} and (9; : - N 5 i 5 N} such that 20 = Z, f ( z i )= zi+l and yo = y, f ( y i ) = yi+l and d(zi,yi) < 618 for lil 5 N. Let { z i :0 5 i < XI} be a 7-pseudo orbit of f. Then for i 2 0 we can find ( z i ) E Xf such that zi = zi and { ( z i , ) } is a &pseudo orbit of u. Hence there is (2,) E Xf such that i ( u i ( z n ) (zk)) , 5 E for 0 5 i < 00, and then E

2 J(ui(zn),(zf)) 2 d ( f i ( z o > , zi) = d ( f i ( z o ) , zi).o

Theorem 2.3.10. Let X be a compact metric space. A positively expansive map f: X + X has POTP i f and only if f is an open map ( i e . expanding). Proof. I f f is expanding, then f is a local homeomorphism and X has a hyperbolic metric (Theorem 2.2.10). Thus, by using the technique of Theorem 1.2.1 (2) we see that the map has POTP. Conversely, suppose f has POTP. Let U be an open set of X and for x E U choose E > 0 such that the €-neighborhood, Uc(z),is contained in U. Then there is 0 < 6 5 e such that every 6-pseudo orbit of f is E-traced, then a sequence where e is an expansive constant for f. If z E Ua(f(z)), { z , z , f ( z ) , f 2 ( z ) , . . . ,f i ( z ) , * - *is} a &pseudo orbit o f f , and so it is €-traced by some point y E X. Then d ( f ' ( f ( y ) ) , f ' ( z ) )< 6 5 e for all i 2 0, which implies f(y) = z. Since y E Uc(z), we have f ( U , ( z ) ) 3 U a ( f ( z ) ) . Therefore f ( U ) is open in X. 0

$2.3 Pseudo orbit tracing property

83

Theorem 2.3.11. Let X be a compact metric space. If X is connected and f :X --t X is an expanding map, then the set of periodic points of f is dense in X . Proof. Let D be a hyperbolic metric for f and let (60,8) be Eilenberg's con61 stants for (X,f ) . Then there are 61 > 0 and X > 1 such that D(z,y) (z,y E X) implies D(f (z),f (3)) 2 XD(z, y). Notice that 8 is chosen such that 8 < el. Then D(f(z),y) < 8 implies U6,lx(z) n f-l(y) # 0 where Ua(z)= { z E X : D ( e , z ) < a}. It is enough to see that for each v > 0, U v ( z ) contains a periodic point of f. Fix 0 < v < ~ 1 / 2 .Since f has POTP by Theorem 2.3.10, there is 6 > 0 such that every &pseudo orbit of f is v-traced by some point of X. Let { U1,..,Un} be a finite open cover of X such that the diameter of each Ui is less than 8. Then we can find T > 0 such that (n 1)8/Xr-' < 5. Take and fix zo E X. Since X is connected, there is a finite sequence


0 be as in Lemma 2.2.34. For E > 0 there is 0 < 6 < 6o such that Let { y i } be any each 6-pseudo orbit { p i } for f is E-traced by some point of such that ?T(p-k) = y-k 6-pseudo orbit of g . Fix k > 0 and choose p - k E and fix p 4 . Inductively we choose pi E such that r ( p i ) = yi for i > - -12. Since f ( P i ) E we have a 0 f ( p i ) = g 0 x ( p i ) = g ( y i ) - Since yj+l E U a ( g ( y i ) ) , there is a unique pi+l E U h ( f ( p i ) ) such that r ( p i + l ) = yi+l. By the choice of { P i : i 2 - k } , { p i } is a &pseudo orbit for f and thus there is an €-tracing Let r ( y k ) = z k . Then, for point y k of { p i } , i.e. z ( f i ( y k ) , p i ) 5 E for i 2 4.

f:x x

x

x,

x

x.

i20 d(gi(zk)7y i ) = d(gi 0 r ( y k ) ,Y i ) = d ( r 0 f i ( v k > r, ( p i ) )

-

= d ( r i ( Y k ) , p i )I E.

If k

+

: i E Z} is a 6-pseudo orbit and there is a subsequence z as j -+ 00. Thus d ( g n ( z k j ) , y n ) 5 E for n 2 -k, and for all n E Z. Therefore g has POTP.

00, then { Y i

{ z k J } such that

zkj -+

so d ( g n ( z ) , y , ) 5 E To show the converse, let 60 > 0 be as above. Then, for E > 0 we can find 0 < 6 < 60/2 such that each 6-pseudo orbit for g is E-traced by some point of X . Since f : -+ is uniformly continuous, we may suppose that &I, q ) < E implies (P),f (4)< bop. Let { p i } be a 6-pseudo orbit for f and let yi = % ( p i ) for i. Obviously { y i } is a 6-pseudo orbit for g . Indeed, since f ( p i ) , p i + l ) < 6 for all i,

af

x x

z(

d ( g ( Y i ) , y i + l ) = 4.0

f ( p i ) , r ( p i + l ) )= a ( f ( p i ) , p i + l ) < 6.

Thus there is z E X such that d ( g i ( z ) , y i ) < E for all i. Since d(z,yo) there is q E U6(pO) such that r ( q ) = z. Note that

d(.

0

< E,

f i ( q ) ,.(pi>> = d(gi 0 r(q),y i ) = d(gi(z>7y i ) < E

for all i. Since r is locally isometric, we have z ( q , p o ) < E when i = 0. If -d ( f - ' ( q ) , p i - l ) < E for i 2 1, then we have z ( f i ( q ) , p i ) < E . Indeed, since d ( f o f i - ' ( q ) 7 f ( ~ i - 1 ) )< 60/2 and a ( f ( p i - l ) i p i ) < 6,we haveZ(fi(q),pi) < 60 and -

d ( f ' ( q ) , p i ) = d ( r 0 f ' ( q ) , r ( p i ) ) = d ( g i ( z ) ,y i ) < E Thus the point q E-traces the one sided 6-pseudo orbit { p i : i 2 0). Similarly we can show that a ( f i ( q ) , p i ) < E for i I 0. Therefore the conclusion is obtained. 0

CHAPTER 2

86

Theorem 2.3.15. Let f be a linear map of the euclidean space R" and let d be the euclidean metric for W". Then f has POTP under d if and only iff is hyperbolic, i.e. it has no eigenvalues of modulus 1. Proof. If f is hyperbolic, then R" splits into the direct sum of f-invariant subspaces, W" = Es@E", such that f: Es + E d is contracting and f : EU -+ E" is expanding. See the proof of Lemma 2.2.33. Thus, in the same manner as the proof of Theorem 1.2.1 (2) we have that both flEa and fpUhave POTP. Since f is linearly conjugate to f l E a x flEY, by Theorem 2.3.5 it follows that f has POTP. Conversely, suppose f has an eigenvalue with absolute value one. Then there is an f -invariant subspace F such that f 1~ : F --f F is an isometry under d. Let E > 0. Since f is a linear map, by using the Jordan form of f (in the real field) we can choose K = K ( E )> 0 such that llp"(x)11 -+ 00 as i -+ 00 whenever 1 1 ~ 1 1< E and I(fi(x)ll > K for some i 2 0. Since f p is an isometry, for any 6 > 0 we can construct in F an &pseudo orbit which comes from the origin and goes to a set of points x with 11x11 = 2K, which implies that f does not have POTP. 0 Theorem 2.3.10. A toral endomorphism has POTP if and only if it i s hyperbolic. Proof. This is obtained from Theorems 2.3.14 and 2.3.15. 0 Let k be a natural number and Yk = { l , . . .,k}. A metric d for Yf is defined by d ( x ,y ) = 2-"' if m is the largest natural number with xn = yn for all In\< m, and d ( x , y ) = 1 if xo # yo. Such a metric is uniformly equivalent to the metric d' defined as above. If S is a closed subset of Yf with u(S) = S, then u p : S -+ S is called a subshift. We sometimes write u p as u : S -+ S. A subshift u : S -+ S is said to be of finite type if there exist some natural number N and a subset C C Yk with the property that z = (zi)E S if and only if each block ( x i , - - -, z ; + N ) in z of length N 1is one of the prescribed blocks. The smallest such natural number N is called the order of the subshift of finite type.

n,"

+

Theorem 2.3.17. Every subshift of finite type is topologically conjugate to one of order 1. Proof. We just take a new symbolic space Y z consisting of the allowable blocks of length N, and then the conjugacy cp : Yf + Y z is given by

cp(.

.. , X - 1 , 2 0 , 5 1 , " ' )

=

(...,

(q,( 7 ),(?) XN-2

The shift 5 :Y z -+ Y z is of order 1 and

'p o

XN-1

XN

o = 5 o cp holds. 0

,... ) .

$2.3 Pseudo orbit tracing property

87

Let A be a k x 12 matrix of 0's and 1'8, called a transition matrix. We define the compact set Then XA is invariant under the shift map ( ~ ( X A = ) EA). Such a shift map u p A is called a Markov subshift. Every subshift of finite type with order 1 is a Markov subshift.

Theorem 2.3.18 (Walters [W3]). Let S be a closed subset of Yf and let u : S + S be the subshift. Then u : S + S has POTP if and only if u is a subshift of finite type. Proof. If u : S + S is a Markov subshift, obviously u is of order 1. Let > 0 and take m > 0 such that 2-" < E . Then xi = yi for lil 5 m when d ( z , y ) < 2-('"+l). Let { x i : i E Z} be a 2-("+l)-pseudo orbit for up, i.e.

E

{xi}

cS

d ( u ( z i ) , x i + + ' 0 such that

5 N -1 I d ( z , y ) < 6. Denote by B the set of all blocks xi-^, - * ,x i , * * - , xi+^) in z with length 2N + 1 for all x E S and write xi

= yi, lil

S(B)={x~Y~:(zi-~,...,zi,...,xi+~E ) €ZB} ,. Obviously, S ( B ) 3 B and u ( S ( B ) )= S ( B ) . It is easy to see that S ( B ) is closed and u ~ ( Bis) of finite. Since ( y i - ~ , ,y i + ~ E ) B for each y E S ( B ) and i E Z, for i E Zthere is xi E S such that yi+j = xf for Ijl 5 N. Thus, d ( u i ( y ) , z i ) < 6 for i E Z,i.e. { x i } is a &pseudo orbit for up. Since u p is POTP, ther is z E S such that d(ui(z),zi) < &/2 for all i E Z. Then we have

--

d(u'(z), u i ( y ) ) 5 d(u"z), xi)

+ d ( x i ,ui(y))
0. We set for x E X

as before (henceforth, as the metric d is fixed, the local stable set W:(t, d ) is written by W , " ( x ) )and for x = ( x i ) E Xf

W,u(x)= {yo E M : 3(&) E Xfs.t. d(Xi,Yi) 5 E , i 5 0). Lemma 2.4.1. Let f:X + X be a c-expansive continuous suljection with expansive constant e . For 7 > 0 there exists ny > 0 such that for x = ( x i ) E Xf and x E X (1) f"(W,.(m)) c W , " ( f " ( x ) )for n 2 ny, (2) if y = ( y i ) E Xf and d ( y i ,x i ) 5 e for i 5 0 ( i.e. yo E W,"(X) ), then d ( y - , , x - , ) 5 7 for n 2 ny ( i.e. W:(X) c fn(Wy(cr-nx)) for n 2 ny ).

Proof. If (1) is false, then there are x n , y n E X and m, 2 n such that yn E W,b(xn) and d ( f m n ( x n )fmn(y")) , 2 7 . Fix i > 0. If fmm-i(xn) + xLi and fmm-i(yn) + yLi as n + 00, then d ( x L i , y L i ) 2 e since d ( f j ( x n ) , fj(y*)) 5 e for 0 5 j < m,. We have d(xb,yh) 1 7 when i = 0. Thus 96 E W t ( x ' )

$2.4 Topological Anosov maps (TA-maps)

89

where x' = (2:) because fi(zLi) = zb and fi(yLi) = yb for i 2 0. Since y* E W:(xn),we have

d(fmn+j (2, 1, fmn+j(y")) 5 e ( j 2 -m,) and hence g; E Wl(zb). Since Wc(x')n W ; ( x & )3 yk, we have z& = y& by c-expansivity, thus contradicting. If (2) is false, then we can find ( x : ) , ( y l ) E Xf and m, 2 n such that - 7 and d(z?,yl) 5 e for i 5 0. From this we obtain a d(z:m,, ~-m, > contradiction. 0

For a continuous surjection f:X + X we define the stable and unstable sets

W " ( z )= {y E X : lim d(f"(z), fn(y)) = 0) n+oo

W u ( ( x i ) )= {yo E X : 3(yi) E Xfs.t. lim d(x-i,y-;) = 0) i-rw

for z E X and (xi) E Xf. Let f : X --t X and 9: Y --f Y be continuous surjections of compact metric spaces. Suppose h o f = g o h holds for some homeomorphism h: X 4 Y . Then the following are easily checked : (1) if x E X and W s ( z )is the stable set of f , then h ( W " ( z ) )is the stable set of g at h(x), (2) if (zi)is a point in Xf and Wu((zi)) is the unstable set of f , then ( h ( x i ) ) belongs to Ygand h ( W u ( ( z i ) ) )is the unstable set of g at (h(Xi)).

Remark 2.4.2.

Lemma 2.4.3. Let f: X t X be a c-ezpansive continuous suvjection with expansive constant e and let e > E > 0 and z E X . Then WYX) =

u f-i(w:(fi(x))) i20

and for x = ( z i ) E Xf W y x )=

u

fi(w,u(o-i(x))).

i20

Proof. This follows from Lemma 2.4.1. Now we shall give Anosov property for continuous surjections of metric X spaces, in a general setting. We say that a continuous surjection f : X is a topological Anosov map (abbreviated TA-map ) if f is c-expansive and

90

CHAPTER 2

has POTP. If, in particular, f is a homeomorphism, then f is said to be a topological Anosov homeomorphism if it is expansive and has POTP. A TAmap f : X --+ X is special if f satisfies the property that W'((x;)) = W " ( ( y i ) ) for every ( x i ) , ( y i )E Xf with xo = yo. The notion of special TA-maps is a generalization of that of special Anosov differentiable maps. Let X and Y be metric spaces. We say that two continuous maps f:X X and g:Y --+ Y are topologically conjugate if there exists a homeomorphism h : Y --+ X such that f o h = h o g . In this case any orbit of g is mapped by h to a homeomorphic orbit of f . If h is a continuous surjection, then f is said to be topologically semi-conjugate to g , in other words, f is called a factor of 9. Let f : X -+ X and g : Y --+ Y be continuous surjection of compact metric spaces. If f and g are topologically conjugate, then (1) f is a TA-homeomorphism if and only if so is g, (2) f is a TA-map if and only if so is g , (3) f is a special TA-map if and only if so is g . Indeed, (1) and (2) follow from Theorem 2.3.6 together with Theorems 2.2.5 (3) and 2.2.30 (3). By Remark 2.4.2 (2) we obtain (3).

Remark 2.4.4.

In the remainder of this chapter we discuss the relation between TA-maps and the topological stability of homeomorphisms (or self-covering maps) on closed topological manifolds. Let X be a compact metric space with metric d. A homeomorphism f : X --+ X is said to be topologically stable in the class of homeomorphims if for E > 0 there is 6 > 0 such that for a homeomorphism g with d( f (x),g ( x ) ) < 6 for all x there is a continuous map h so that h o g = f o h and d ( h ( z ) , x )< E for all x. A self-covering map f is said to be topologically stable in the class of selfcovering maps i f f satisfies the conditions as above in the class of self-covering maps.

Theorem 2.4.5 (Walters [W3]). If a homeomorphism f:X -+ X of a compact metric space is topological Anosov, then f i s topologically stable in the class of homeomorphisms. Proof. Let e > 0 be an expansive constant for f and fix 0 < E < e / 3 . Let 0 < 6 < e / 3 be a number with the property of POTP. By expansivity it is checked that there is a unique x E X which &-traces a given &pseudo orbit {xi}. Indeed, let y E X be another €-tracing point of {xi}. Then we have

for all i E Z and thus x = y . Let g : X --+ X be a homeomorphism with d(g(x), f(x)) < 6 for all 2 E X. Let x E X. Since d( f o g"(x),g"+'(x)) < 6 for n, { g " ( x ) } is a &pseudo orbit of f. Thus there is a unique point h ( z ) E X whose f-orbit &-traces { g " ( z ) } .

$2.4 Topological Anosov maps (TA-maps)

91

This defines a map h : X + X with d(f" o h(x),gn(x)) < E for n and x E X. Letting n = 0, we have d(h(x), x ) < E for x E X. Since d(f" o h o g(x),g"+'(x)) < E for n E Z and

d(f"

0

f 0 h(x),g"+'(x)) = d(f"+'

0

h(x),g"+'(x))

<E

for n E Z, we have h o g(x) = f o h(x) for x E X. Finally, we show that h is continuous. Let X > 0. Then we can choose N > 0 such that d(fn(x),fn(y)) < e for In1 5 N implies d(x, y) < A. This is checked as follows. Suppose this is false. Let a be an open cover of X with diameter < e. Then there is E > 0 such that for each j > 0 there are x,, y, E X with d ( q , yj) > E and Aj,i E a (-j 5 i 5 j) with xj, yj E f-j(Aj,i). Since X is compact, we can suppose that x j -+ x and yj -+ y. Thus x # y. Consider the sets Aj,o for j. Then infinitely many sets of them coincide since a is finite. Thus xj, yj E A0 for infinitely many j and so x, y E 20.Similarly, for infinitely many Aj," they coincide and we have A,, E a with 2, y E f-"(&). Therefore x, y E 00 thus contracting the fact that f is expansive. Choose 7 > 0 such that d(x,y) < 1implies d(g"(x),gn(y)) < e/3 for In1 5 N. If d(x, y) < 7 then

ni=-j

nn=-=f-=(a,,),

d(f" 0 h(x), f" 0 W Y ) ) = d(h 0 9"(x), h 0 9"(Y)> 5 d(h 0 9"(z), 9"(x>>+ d(g"(x), 9YY)) + 49"(Y), h 0 9YY)) 5 E + e/3 + E < e, In1 5 N . Therefore, d(x, y)

< 7 implies d(h(x), h(y)) < X and continuity of h is proved.

0

-m

-1

0

m-1

rn

Figure 12 Let h:X -+ X be as above and suppose that X is a closed topological manifold. Then h : X + X is surjective. For the proof the readers may refer to Remarks 6.7.10 and 10.6.4 which will be mentioned in Chapters 6 and 10 respectively. However, in general h is not surjective. We give an example due to Walters.

Remark 2.4.6.

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CHAPTER 2

Consider the shift u : %z -t Yt where Y2 = {O,l}. Then and has POTP. Let m > 0 and define g: yz" -t y2" by

xi zi+l

x-,

Q

is expansive

lil > m, -m 5 i < m, i=m

Then d ( g ( x ) , u ( x ) )5 1/2m where d is the metric for yz" given as before. If m is sufficiently large then we have h o u = g o h for a continuous map with d(h(z),z)< 1/2,(x E Y,").Since g2m+'(z)= x for all z E Yt,we have h ( z ) = h o gam+l(x) = a2,+l o h ( x ) for all x E y,". Therefore h is not surjective.

Theorem 2.4.7. Let M be a closed topological manifold and f: M + M a self-covering map, but not injective. If f is a TA-map and has topological stability in the class of self-covering maps, then f is positively expansive. For the proof we need some properties of TA-maps on closed toplogical manifolds. Thus we shall postpone the proof until Theorem 6.7.11 in $6.7 of Chapter 6. Let f: X + X be a homeomorphism of a compact metric space. A point p E X is nonwandaring for f if, for any neighborhood U of p and any integer no > 0, there exists n with In1 > no such that fn(U) n U # 8. The set n(f) of nonwandering points is closed and f-invariant (f(fl(f) = n(f>).The limit sets w ( q ) and a(q),for q E X,are contained in n(f). Such a set fl( f ) is called the nonwandering set. Let g: Y + Y be a homeomorphism of a compact metric space and h:X + Y a continuous surjection. If g o h = h o f, then we can easily prove that h ( n ( f ) )c f l ( g ) . If, in particular, h is bijective, then h ( f l ( f ) )= f l ( g ) .

Theorem 2.4.8. Let f: M -t M be a homeomorphism of a closed topological manifold. Iff is topologically stable in the class of homeomorphisms, then the set of all periodic points of f, Per( f), is dense in n(f ). P m f . Since f is topologically stable, for E > 0 there exists 6 > 0 satisfying the definition of topological stability. We may suppose that 6 < E . Take and fix x E fl( f ) , and choose a coordinate neighborhood U of x contained in u6/2(x) (where u6/2(x) = {y E M : d ( x , y ) < 6/2}). Notice that U is chosen such that it is connected. Since x E n(f),there is 12 > 0 such that f k ( U )n U # 0 and f ' ( U ) n U = 0 for 0 < i < k. Thus f'(x') E U for some x' E U. Since U is a connected coordinate neighborhood, we can construct a homeomorphism 9: M + M satisfying 9 0 f"x') = x', g ( U ) = u, g ( y ) = y for all y $ U .

52.4 Topological Anosov maps (TA-maps)

93

Clearly, since d ( g ( z ) ,z)< 6 for z E M, we define cp = go f . Then, d(cp(z),f (z)) < 6 for E M. Therefore we can find a continuous surjection h: M t M such that h o cp = f 0 h, d ( h ( z ) , z )< E (zE M ) .

It is easy to see that cpk(z’)= 2’. Since h o cpk = f k o h, we have h(z’) = f k ( h ( z ’ ) )and thus h(z’) E Per(f). On the other hand, since d(z,h(z’)) 5 d(z,z‘) + d(z’,h(z‘)) < 2 ~ we , have h(z’) E UzL(z).Since E is arbitrary, it follows that Per( f ) is dense in GI( f ). 0 Theorem 2.4.9(Walters [W3]). Let M be a closed topological manifold with dimension 2 2. I f f : M --+ M is topologically stable in the class of homeomorphisms, then f has POTP. For the proof we need the following two lemmas.

Lemma 2.4.10. Let f : M --t M be a homeomorphism of a closed topological manifold. Let k 2 0 be an integer and let 7 , q be positive numbers. Then for any set of points {zo,z1,--* , z k } with d ( f ( z i ) , z i + l< ) ~ ( 50 i 5 k - 1 ) there exists a set of points {zh, z: , - ,z;} such that

Proof. This is proved by induction on 12. For k = 0 this is true. Suppose the lemma is true for k - 1 and we prove it for k. Let T > 0 and > 0 be given. Choose X > 0 such that d ( z , y) < X implies d( f (z),f ( 9 ) ) < 7 . We may suppose q < min{.r, A}. Let {zo,z1,. ,Z k } be a .r-pseudo orbit of f . By assumption we can choose {zb, z;, , such that d(zi,3:) < X(0 5 i 5 k - l ) , d ( f ( z { ) , z { + l 0 there is 6 > 0 such that if a finite collection 3 = {(pi,qi) E M x M : 1 5 i 5 T } satisfies the following conditions :

CHAPTER 2

94

then there exists a homeomorphism f : M -+ M such that

Proof. Let n = dimM. From the dimension theory we can take a finite open cover V of M which can be represented as the union of n 1 families V1,. ,Vn+l, such that each element of V is a connected coordinate neighborhood and for 1 L n 1 all elements of Ve are disjoint. Moreover, given E > 0 we can choose the open cover V such that the diameters of all elements of V are less than E. Let 6 > 0 be a Lebesgue number of V and assume that a finite collection F = { ( p i , q i ) E M x M : 1 5 i I T } satisfies the conditions as in the lemma. Let 3 e = {(pi,q i ) E F : 3 6 E Ve s.t. pi, qi E Vj}. Then F1 U U Fn+l= F. Since dim M = n 2 2, for each 1 5 L n 1 there is a homeomorphism f e : M + M such that (1) f & i ) = qi for (pitqi) E Fe, ( 2 ) f e ( p i ) = pi for (pi,qi) 9 Fe, and ( 3 ) f e is the identity outside elements of Ve. Let f = f i o ...o f n + l . Then f ( p i ) = qi for all ( p i , q i ) E F,and d(f(z),z) < (n 1 ) for ~ 3: E M . 0

+

-.

< < +

< +

---

+

Proof of Theorem 2.4.9. Let E > 0 be given and let 6 correspond to E as in the definition of topological Stability. We may assume that S has the property as in Lemma 2.4.10 (by taking 6 sufficiently small if necessary). Suppose {ZO, 21, ,Z k } is any pseudo orbit such that d( f (xi),zi+l)< 6 / 2 for 0 5 i < k - 1. By Lemma 2.4.10 there exists {zh,zi,*-, z i } such that d(zi,3::) < E(O < i k ) , d ( f ( z : ) , z : + , ) < 6(0 5 i k - 1),z: # 3: if i # j ( i , j < k) and f ( z : )# f ( z $ )if i # j ( i , j 5 k - 1). By Lemma 2.4.11 there is a homeomorphism cp : M + M with d(cp(z),z) < S(z E M ) and cp o f ( 3 : : ) = Z:+~(O 5 i 5 k - 1). Let g = 'p o f. Then d(g(z), f ( 3 : ) ) < S(z E M) and g(z:) = z:+,(O i 5 k - 1). By topological stability there is a continuous surjection h: M --+ M such that d ( h ( z ) ,z) < E and h o g(z) = f o h ( z ) for all z E M. Then

--
0. Since E C R ( f ) ,there is a 6/2-pseudo ,x,, d')').Therefore {x,q, ,x,, x} is an a-pseudo orbit. orbit {x(~),z~,-.. Since a is arbitrary, we have x E C R ( f ) . 0

X and

-

-

-- -- - N

-

-

-

[XI

-

-

--

96

--

$3.1 Chain recurrent sets

97

Theorem 3.1.2. Iff has POTP, then n(f)= C R ( f ) .

n(f). If z E C R ( f ) ,then for every a > 0 there is a pseudo orbit {xi} such that xo = 2, z l , . ,x k = x and d ( f i ( y ) , x i )< a(0 5 i 5 k) for some y E X . Therefore, fk(va(2))nv,(z) #0 where Ua(z)= {y E X : d(x,y) < a}. Since a is arbitrary, we have x E n(f). 0 Proof.It is enough to see that C R ( f ) C

--

Let X be a topological space and 2x the family of all nonempty closed subsets of X . The exponential topology of 2x is defined by assuming that the family of all sets

B ( G ) = { F E 2x :F

c G},

C ( H ) = { F E 2x : F n H

# 0)

is an open subbase of 2x provided G and H are open subsets of X . It is easily checked that the family of all sets

B ( G o , G 1 , * . .,Gn) = {F E 2x : F

c G o , F n G i # 0,l 5 i 5 n},

where each of Gi is an open set of X , is a base of 2x.

Remark 3.1.3. If X is compact, then so is 2x. It is enough to prove that every cover of 2x whose sets belong to an open subbase of 2x contains a finite subcover. Suppose that

2x =

(U V t

t ) )

u (UC(Ha)) 8

where Gt and H , are open in X . Put FO = X\ U, Ha. Then, for each s we have FO r l Ha = 0, i.e. FO! I$ C(Ha) and thus FO E B ( G t o )for some to, i.e. Fo c Gt, and SO X\Gt, C X\Fo = Ha.

U 8

Since X\Gt, is compact, there are si(1 5 i 5 n) such that

X\Gt, C Ha, U * * - U Ha,,. Let F E 2x. Then there exist the following two cases : (i) (ii)

F c Gt,, F $tG t , , i.e. F n (X\Gt,) #

0.

For (i) we have F E B(Gto),and for (ii) we can find j such that F n Haj # 0, i.e. F E C(Haj).In any case, F E B(Gt,)UC(H,,)U---UC(H.,). Therefore 2x is compact.

CHAPTER 3

98

Remark 3.1.4. When X is BausdorfF, X is compact if so is 2 x . Indeed, let X = U t Gt where each Gt is open. Obviously, 2 x = U t C(Gt). Since 2 x is compact, we have 2 x = C(Gt,) U * * * U C(Gt,)

for some n > 0. Let zo E X . Then {zo} E C ( G t j )for some j 5 n. This implies that xo E Gtj and therefore

X = G t , U...UGt,

.

Lek X be a metric space with bounded metric d. A metric p defined by

p(A,B ) = max{supd(A, b), sup d(a, B)}, be B

where d( A, b) = inf { d( a, b ) : a

A, B E 2 x

QEA

A } , is called the Hausdorf metric for 2 x .

Remark 3.1.5. If a metric space ( X , d ) is compact, then so is ( 2 x , p ) . To obtain this it is enough to see that 0 ( 2 x ) = 0(2x), where 0 ( 2 x ) denotes the exponential topology of 2 x and 0 ( 2 x ) , denotes the topology induced by p. We first show that 0 ( 2 x ) , c 0 ( 2 x ) , i.e. each open ball with center A , R = { F E 2x : p ( A , F ) < E } ,

is open in 2x. Since A is compact, for each k 2 1 there are a:,... that for each x E A we have d ( a t , x ) < 1/k for some i. Let us put : d(z, a t ) < E - l/k},

(1)

Gt = { x E X

(2)

G = {z E X : d ( x , A )< E } .

Then it is enough to prove that m

(3)

R

=

(JB(G, G:, . ,G&,) k=l 00

= U { F E ~ ~ : F CF G n G, f # 8 , l s i s n k } . k=l

If F E R, then p ( A , F ) < E - l/k for some k. Thus (4)

(5)

yEA y E F

*d ( y , F ) < * d ( y , A )
0. Thus we have A 3 R where

fi = { F E 2 x

:p(A,F) < E }

and so

FER&FcG. Indeed, suppose p E F\G. Then d(p, A ) 2 E and thus p(F, A ) 2 E , so F # R. Therefore, R c A. Let A = { F E 2 x : F n G # 0) where G is open in X . Let A be a compact subset such that A n G # 0. For a E A n G put d(a, X\G) = E . Since p(A,F ) < E for F E R, we have d ( a , F ) < E and so F n G # 0. Therefore, R c A. Making use of Hausdorff metric, we have the following result.

Theorem 3.1.6. Let f:X4 X be a homeomorphism of a compact metric space. Then the chain recurrent set of f ICR(f) coincides with CR(f). Proof. We give here a proof due to Robinson [Ro]. Let z E C R ( f ) and C, = {z!"'} be a periodic l/n-pseudo orbit through z. Then C , is a finite set. In the Hausdorff metric, there is a subsequence C,, that converges to some f-invariant compact set C c X . If we establish that for every y E C and E > 0 there is a periodic E-pseudo orbit {zi} through y with t; E C,then it follows that y E CR( f Ic) c CR(f ) . Since y E C,we have C c CR( f ) and 2 E C C CR(f Ic) C CR(f ICR(f)). This is true for all 2 E CR(f) and thus CR(f) c CR(f l C R ( f ) ) * It remains only to show that there is a periodic €-pseudo orbit through y with zi E C. Since f : X + X is uniformly continuous, there is ~ / 3> 6 = 6 ( ~ / 3 )> 0 such that d ( a , b ) < 6 implies d ( f ( a ) , f ( b ) )< ~ / 3 .Since C,,

100

CHAPTER 3

converges to C,we can find n = nk such that 1/n < €13 and the distance from C, to C in the Hausdorff metric is less than 6. Suppose that {z!"'} has a period j and z i;; = for all i. For each z!") take zi E C with

zin)

d(~i"),z i ) < 6, ~ i +=j Z i for all i, and zi = y for some i. Then

Therefore { z i } is a periodic €-pseudo orbit in C through y. 0

Theorem 3.1.7. Let f:X + X be a TA-homeomorphism of a compact metric space. If fpqf)is topologically tmnsitiue, then X = C R ( f ) . Proof. Suppose X # CR(f ). Then we shall derive a contradiction by showing (X\CR( f))rlPer( f ) # 0 where Per( f ) denotes the set of all periodic points of f. Since w(z0) and a(z0)are contained in CR(f ) for zo E X\CR( f ), we have cl({zo,f(zo),-.}) n Wf)#Oandcl({zo,f-l(~O),...})nc~(f) #0. 0 < E < d ( z 0 , CR(f )) and let 6 > 0 be a number with the property of POTP. , < 6 and d ( C R ( f ) ,f-"(zo>) < 6 for some large n. Then d ( C R ( f ) f"(z0)) Thus d(zn+l, f"(z0)) < 6 and d ( ~ - ~ -f-,(zo)) l, < 6 for some zn+l, %-,,-I E C R ( f1. Now construct a 6-pseudo orbit

to %,+I. Since flCR(f) is topologically transitive, we can find in from z-,-1 CR(f) a &pseudo orbit

from xn+l to z-,-1. By combining the above two &pseudo orbits, a periodic &pseudo orbit is constructed. Since f has POTP, the periodic &pseudo orbit is &-traced by a point y E X. Since f is expansive, y is periodic and since d ( z 0 , CR(f)) > E we have y $ CR(f ) . This is a contradiction. 0

Theorem 3.1.8 (Aoki [Ao~]).Let f:X + X be a homeomorphism of a compact metric space. Iff has POTP, then so does fp(f).If, in addition, f is expansive, then the set of all periodic points, Per( f ), is dense in fl( f ) . This is given in Theorem 3.4.2 for continuous surjections and so we here omit the proof of Theorem 3.1.8 since the technique of the proof is similar.

Remark 3.1.9. There exists an example of a TA-homeomorphism f of a compact metric space such that X # CR(f). This is easily constructed as follows. As before let 0 : YF + YF be the shift map, and S = { ( z i ) E Yt : (zi,zi+l) E C,iE Z} where C = { ( O , O ) , ( O , l ) , ( l , l ) } . Then 0 : S + S is

$3.1 Chain recurrent sets

101

a Markov subshift. Since 01s is expansive and has POTP, by Theorem 3.1.8 the set P e r ( a p) of periodic points is dense in C R ( a p ) . However S contains , O , O , - - . ) and y = ,1,1,...). The point only two periodic points z = z = (..- , O , O , l , l , - . . ) is in S and not periodic. Therefore, z Per(op) = ( . - a

(as-

C R ( q s )# s. Let X be a compact metric space. A closed subset E of X is isolated for the homeomorphism f : X + X if f (E) = E and if there is a compact f n ( U )= E . neighborhood U of E such that

nrm

Theorem 3.1.10. Let X be a compact metric space. Iff:X + X is expansive and f : CR(f ) + CR(f ) has POTP, then CR(f ) is isolated. Proof. Let e > 0 be an expansive constant for f . For 0 < P < e / 2 let a > 0 be a number with the property of POTP. By uniform continuity of f we can take 0 < 7 < min{a/2,e/2} such that

4%Y) < 7 (z,Y E X I

-

d(f (21,f (?I)) 0 such that U n f "(V) # 0 for all n 2 N. If f is topologically mixing, then it is topologically transitive (see Remark 2.2.1). Theorem 3.1.11 (Topological decomposition theorem). Let f : X + X be a homeomorphism of a compact metric space and let C R ( f ) be the chain recurrent set. If f p R ( f ) : C R ( f ) --t C R ( f ) is a TA-homeomorphism, then the following properties hold : ( 1 ) (Spectral decomposition theorem due to Smale) CR(f ) contains a finite sequence Bi ( 1 I i 5 L ) of f -invariant closed subsets such that (i) CR( f ) = B; (disjoint union), (ii) flBi : B; + B; is topologically transitive. ( 2 ) (Decomposition theorem due to Bowen) For each Bk there exist ak > 0 and a finite sequence Ci ( 0 5 i 5 ak - 1) of closed subsets such that (i) Ci n Cj = 0 (i # j), f(Ci) = Ci+l and fah(Cj)= Ci,

ut,

CHAPTER 3

102

(ii) B&= u;2i1ci, (iii) f p i : C; -+ C; is topologically mixing. The sets Bj and Cj are called basic sets and elementary sets respectively. For more general case (i.e. continuous surjections case) we shall give the proof of this result in Theorem 3.4.4 of 53.4. Theorem 3.1.12. Suppose f : C R ( f )-+ C R ( f ) is a TA-homeomorphism. If, in addition, f : CR(f ) + CR(f ) is topologically transitive and CR(f ) contains a fied point o f f , then f : CR(f ) + CR(f ) i s topologically mixing.

Proof. This follows from Theorem 3.1.11 (2). 0

53.2 Stable and unstable sets

For A a subset of a compact metric space X define

W'(A) = u { W u ( z d, ) : z E A } , W"(A)= u { W u ( z d, ) : z E A } and for E

>0

Here the subsets W ' ( z , d ) and W " ( z , d ) are defined as in 52.2 of Chapter 2. Theorem 3.2.1. Let C R ( f ) be the chain recurrent set for f . If f : C R ( f )+ C R ( f ) is a TA-homeomorphism and if C R ( f ) splits into the finite union CR(f ) = R1 U . U Re of basic sets Rj, then for 1 5 j 5 1

..

W'(Rj) = {z E X : lim d ( f n ( z ) , R j ) = O}, n-roo

W " ( R j ) = {z E X : lim d(f-"(z),R j ) = 0) n+oo

and for each E > 0 there exists an open set Uj 3 Rj such that

Proof. Since each Rj is open and closed in C R ( f ) f, p , has POTP. Thus, for > 0 there exists a > 0 such that every a-pseudo orbit of Rj is ~/2-tracedby some point in Rj. Let 0 < y < min{a/2, ~ / 2 }be a number such that E

d(z,Y> < Y

-

d ( f (21,f (Y))

0 with d( f"(y), R j ) < 7 for n 2 N, and so there is X , E Rj such that d(zn,f"(y)) < 7 for all n 2 N. Let zn = f " - N ( ~ ~for ) n < N. Then { x , : n E Z} is an a-pseudo orbit in Rj. This follows from the facts that if n 2 N then

+

I

I a/2 + 7 < a

d ( f ( ~ n ) , ~ n + l d(f(Zn)tf"+'(y)) ) d(fn+l(y),xn+1)

and if n

< N then

Thus, f N ( y ) E W , ' ( f N ( z )C) W 8 ( f N ( . ) )and W " ( z ) .Therefore, y E W * ( R j )and {y E

X

: lim d ( f " ( g ) ,R j ) = 0) n-w

SO

y E

f - N ( W d ( f N ( ~ ) ) )=

c W*(Rj).

To show the second statement let Uj = {y E X : d ( y , Rj) < 7) and take f-'((Vj). Since d(fk(z), R j ) < 7 for It 2 0, there is yk E Rj with d(fk((z),yk) < 7 for k 2 0. Put y-; = f-'(yo) for i > 0. Then {yk : It E Z} is an a-pseudo orbit of Rj. If x E Rj is an &/2-tracing point of {yk}, then for 1220 %

E nk>o

d(f"4, f Y.)) 5 d(f"4, and thus

E

Yk)

+ d(Yk, f k (L44 2 + 7 < E

E W,"(Z)C W , ( R j ) .0

Theorem 3.2.2. Under the assumptions of Theorem 3.2.1, given y E X there e&sts Ri ( R j ) such that lim n-w

d(fn(y), Ri) = 0

and furthermore

X =

(n+w lim d(f-"(y), R j ) = 0)

e

e

i=l

j=1

UW"(Ri)= U W"(Rj).

If i # j , then

W"(R;)n W * ( R j )= 0, W"(Ri)n W'(Rj) = 0.

CHAPTER 3

104

P m f . Since f : C R ( f )+ C R ( f ) has POTP, we have that C R ( f ) = fl(f). Thus ~ ( tC )C R ( f )= R1 U * U Re for every t E X. For fixed t E X,to show the existence of Ri with W ( X ) c Ri, it is enough to see that Ri with Ri n ~ ( t# )8 is only one in {Ri : 1 5 i 5 f}. To do so suppose 0 .

W(Z)

n Rj # 8,

n Rj' # 8

~ ( t )

# j'. Let U, be an open neighborhood of R, and suppose they satisfy U a n u b = 8 for a # b. Choose an open neighborhood V, of R, such that

for j

Since W ( Z ) n Rj # 0 and and { f " i ( t ) } such that lim

i-m

n

~ ( t R )

~ #I 0, there exist subsequences

f n i ( tE) Rj,

,lim

a-+m

fmi(t) E

{fni(t)}

R~I.

Without loss of generality we suppose that {ni} and {mi} are chosen such that n1 < ml < n2 < m2
0 sufficiently large

Since f(V,) C U,(l 5 a 5 f) and = 0 for a # b, we have fni+l(t) E Uj and fni+'(z) $! U, for all a with a # j. Thus, for i 2 I there exists ni < f i < mi such that

4 v, n %,. Since X \ (4U 4 1 is ) closed and {f'i(z) : i 2 I } c X \ (5U V,,), there is a ffd(t)

subsequence of { f e i ( t ) } for which the limit point z is contained in X\(&UQ,). Thus, z $! Rj U Rjl. However, since z E w ( t ) , there exists Rk such that z E Rk for 12 # j,j'. This is a contradiction. Therefore, ~ ( t c )R, for some j . Finally we show that d(fn(t), R j ) = 0 whenever ~ ( tc )Rj. If this is false, then there exist €0 > 0 and an infinite sequence {ni} such that d(f"*(z),Rj) 2 €0 for all i 2 0. If f n i ( t+ ) z, then d(z, R j ) 2 EO. But we have z E Rj since z E w ( t ) , thus contradicting. The proof of the rest of the statement is easy and therefore omitted. 0 Let f : X X be a homeomorphism of a compact metric space. If f is expansive and f: C R ( f ) + C R ( f ) has POTP, then C R ( f ) is an isolated set (by Theorem 3.1.10). Moreover, by Theorem 3.1.11 C R ( f ) splits into the disjoint union C R ( f )= R1 U s . . U Rc of basic sets. Then each R, is an isolated set. --$

$3.2 Stable and unstable sets

105

- - - ,Ri, such that

A cycle for the family { R; : 1 5 i 5 L} is a sequence R;, , Ri, = R,,, and (W"(Rij)\ Rij) n (W'(Rij+l)\ Rij+I)# for 1 5 j < k. The sequence Ri, , such that

0

- -.,Rib = R;, is a cycle if there are points

xj

@

21,*

- ,Zk-1 *

k

U Rim,

a(zj) C Rij,

~ ( z jC) Rij+l

m=l

15 j < 12. For this case we write Ri, < R;, < < R;, = R;,. Theorem 3.2.3. If f:X + X is eqansive and f : C R ( f )4 C R ( f ) has

for

POTP, then f has no cycles.

Proof. It only shows no existence of families {R;,} such that Ri, < Ri, < < R;, = R,, . If, in particular, R; > Ri for some i, then there are x,y E R;

-

such that E

E W u ( x )n W " ( y ) , z $ C R ( f ) .

Take and fix E > 0 sufficiently small such that d(z,CR(f))2 2 ~ .Since f : C R ( f )+ C R ( f )has POTP, we let 6 > 0 such that every &pseudo orbit o ff in C R ( f )is &-traced. Since 6/2 > d( f -"(,z), f -"x)) and 6 / 2 > d( f k ( z ) ,fh(y)) for k large and f: Ri + Rj is topologically transitive, there exist w E Ri and m > 0 such that

Thus a (212

+ m)-periodic &pseudo orbit {f--"E),

* *

,%,

* * *

,f " - ' ( E ) ,

w ,* * * ,f m - y w ) }

is constructed. If p E C R ( f )is an &-tracingpoint of the pseudo orbit, then it is easily checked that p is a (2k m)-periodic point and d ( a , p ) < E . However, p $! CR(f ), thus contradicting. In the same way we can derive a contradiction for the more general case Ril < Ri, < * * < R;, = Ri,. 0

+

Let Ri be a basic set. We say that Ri is a sink if W"(Ri)is a neighborhood of Ri in X and Ri is a source if Wu( Ri) is a neighborhood of Ri in X .

Theorem 3.2.4. Let f:X + X be an expansive homeomorphism having POTP and let R, be a basic set. Then Ri as a sink if and only if W"(Ri) = Ri, and Ri is a source if and only if W"(Ri)= R;. Proof. If Ri is asink, then we have W " ( R i )3 B,(Ri) = {x E X : d(z, Ri) 5 E } for E > 0 sufficiently small.

CHAPTER 3

106

Suppose z E W"(R;)\R;. Then f - n ( z ) E W"(R;)nB,(Ri)for some n > 0. Since Bc(R;)C W"(R;),we have f - n ( z ) E W"(y) for some y E Ri. Let z E R; be a point such that f - n ( z ) E W"(z). Since f-"(z> is a wandering point, we have Ri > R;, thus contradicting. Therefore, W"(R;)= R;. For E > 0 small enough let 6 > 0 be a number such that every 6-pseudo orbit of f is &-traced. If W U ( R ; = ) R;, then we show that Ri is a sink, i.e. W a ( R ; is ) a neighborhood of R;. If this is false, then there is y E B6/2(R;)\R; such that f n ( y ) --* Rj for some j # i . Take z E Ri with d ( y , z ) < 6. Then

is a 6-pseudo orbit. Let w E X be an &-tracing point of the 6-pseudo orbit. Since E is small, we have

f n ( w )+ Rj,

f-n(W)

--*

Ri.

However, since w is a wandering point, we have W"(&) # R;, thus contradicting. Therefore, Ri is a sink. The proof of the rest is done in the same argument and therefore we omit it. 0

Theorem 3.2.5. Under the assumptions of Theorem 3.2.4, f has sinks and sources.

Proof. Let Ri be a minimal basic set with respect to the ordering relation >. Then Ri is a sink. Indeed, if R; is not, then we can take z E W"(Ri)\ Ri. Since X = Ui W"(R;),we have z E W a ( R j )for some j . Thus there are z E R; and y E Rj such that z E W " ( z )n Wu(y). Since z is a wandering point, we have R;> Rj. However, since R; is minimal, we have a contradiction. 0

Theorem 3.2.6. Under the assumption of Theorem 3.2.3, let C be an elementary set containing in a basic set Ri. Then for each z E C, W'(x) n C i s dense in C for u = s,u. Proof. Notice that f n o ( C )= C for some no > 0. Since C is closed, for c > 0 there are w1, ,wq E C such that Bc(wi)3 C . To show that Wa(z)n C is dense in C, let z E C and take E > 0 small enough. Since is topologically mixing, we can take N > 0 such that

.

u:

f?

Let 6 > 0 be a number such that every 6-pseudo orbit of f p q f )is &-traced by some point. Since c > 0 is arbitrary, let c = 6/4. Then f n o N ( zE )B6/4(Wj) for some j and thus & / 4 ( w j ) C Ba/2(fnoNN(x)). Take w E f n O N ( B c ( z )n ) Ba/z(fnoN(z)) n C R ( f ) . Then a 6-pseudo orbit

{... ,f - y w ) , f-'(w), f n o N ( z ) , f n o N + 1 ( z ) , * . . }

$3.3 Recurrent sets and Birkhoff centers

107

is constructed in C R ( f ) . If y E C R ( f ) is a €-tracing point of the 6-pseudo orbit, then y E W i (j n o N ( z ) )and so f-"oN(y) E W,"(z). Thus f-"oN(y) E B,(z). Since f-"ON(w) E BE(%) and d ( f - " o N ( y ) , f - " o N ( w ) ) < E , we have d(f-noN(y), z ) < 2~ and thus W " ( z )f l Bz,(z) # 8. Since z is arbitrary in C, W " ( z )n C is dense in C. 0

53.3 Recurrent sets and Birkhoff centers Let X be a compact metric space. A point E X is said to be recurrent for a homeomorphism f if, for any neighborhood U of z,there exist infinitely many n with f"(z) E U (in other words, z E a(.) n ~ ( x ) ) Every . recurrent point belongs to Q(f).The closure of all recurrent points, c(f), is called the Birkhoff center. The notion of chain recurrent point is a generalization of that of recurrent point. Let F : X + R be a function. Given a neighborhood u 6 ( x ) with radius 6 centered at x we define

M ( z , 6) = sup{F(y) : y u 6 ( x ) )9 m(z,6) = inf{F(y) : y E u6(.)}. Then M ( z ) = lim6+oM(z,6) and m ( x ) = lim6+om(z,6) satisfy -MI 5 m ( z ) 5 M ( z ) 5 00. The function F : X --t R is said to be upper semicontinuous if M ( x ) = F ( x ) for x € X, and to be lower semi-continuous if m ( z )= F(z) for x E X.

Remark 3.3.1. If the set of all recurrent points is dense in Q ( f ) , then the set is a Baire set in n(f) (a Baire set means the intersection n j A j of sets Aj, j 2 0, that are open dense). Indeed, let F ( z ) = inf{d(f"(z),z) :n > 0). Then the set is contained in {x E Q ( f ) : F ( x ) = 0). Since F ( z ) is upper semi-continuous, for k 2 1,{z E Q ( f ) : F ( z ) < 1/k) is open dense in Q ( f ) and therefore the set of all recurrent points is a Baire set in Q ( f ) . Let f:X + X be a homeomorphism of a compact metric space. A closed f invariant set, A, is called a basic set for f if A is an isolated set and flA : A + A is topologically transitive. Let A be a basic set for f and let U be a compact neighborhood of A such that f n ( U ) = A. Define

nm :

w;.,,(w) =

nf - ~ ) ,WW,U) n m. =

"10

"20

Then we have

Wd(A)' =

u f-"(WLc(A,U>),

n20

WU(A>'=

u f"(W&(A, U))

n20

CHAPTER 3

108

where

Wa(A)' = {z E X : d( f "(z),A) ---t 0 as n ---t oo}, Wu(A)' = {z E X : d( f -"(z),A) + 0 as n t oo}. If f : CR(f ) + CR(f ) is a TA-homeomorphism and A is a basic set as in Theorem 3.2.1, it follows that W'(A)' = W"(A) for 0 = s,u. Let {A,, ,At} be a finite family of basic sets for a homeomorphism f of a compact metric space. We say that the family has no cycles if there is no sequence Ail, * ,Aih of distinct members such that Ail = Aih and

--. --

for 1 5 j

< k.

Theorem 3.3.2 (Malta [Mall). Let f:X --t X be a homeomorphism of a compact metric space. If the Birkhof center c( f ) is an isolated set for f and admits a decomposition c( f ) = A1 U U At into basic sets having no cycles, then c( f ) = R( f).

---

From this theorem we have the following

Theorem 3.3.3. Let f : X + X be a homeomorphism of a compact metric space. If f : X + X is expansive and f : c( f ) c( f ) has POTP, then c( f ) =

Nf 1. Proof. Replace c( f ) by S2(f ) in Theorem 3.1.8. Then we have that the set of points is dense in c( f ) and by Theorem 3.1.11, c( f ) splits into the disjoint union c( f ) = A1 U U At of basic sets A,. By applying Theorem 3.1.10 we see that c ( f ) is isolated. Replacing {I&} by {Ai} in the proof of Theorem 3.2.3, we have that cycles for the family {Ai} do not exist. Therefore the conclusion is obtained by Theorem 3.3.2. 0 all periodic

...

We first prepare some lemmas which need to prove Theorem 3.3.2. Let A be an isolated closed f-invariant set and let U be a compact neighborhood of A such that flyrnf "(U)= A. Define W:oc(A, U )as above. For a compact neighborhood V of A in WLc(A, U )with f (V) C V, the set

D

= Cl(V - f(V))

is called a fundamental domain for WP,(A,U). In a similar way we define a fundamental domain for W&(A, U).If x E Wa(A)'\A, then there exists 12 E Z such that f ' ( x ) E D. If D n A = 0,we call D a proper fundamental domain.

$3.3 Recurrent sets and Birkhoff centers

109

Lemma 3.3.4. If A is isolated, then

is a proper fundamental domain for WLC(A,U).

Proof. Since f(WLc(A, U))= f (U)n Wtoc(A,U),we have A c f ( i n W ) ) n W,.,,@, U )c f (W,.,,V, and f (int(U))nW&(A, fore, D" n A = 0.

U))

U )is an open neighborhood of A in WLc(A, U).There-

Lemma 3.3.6. Let V be a neighborhood of D" in X. Then U' = W,U,,(h,U) U O+(V) is a neighborhood of A an X. Here O+(V) = Un>Ofn(V).

-

<
0 such that fk(y) # U. Thus fk(yn) = f-mm+k(x,) # U for n sufficiently large. Since m, + 00 as n --t 00, we have 0 < m, - k 5 m, for n sufficiently large, which is a contradiction. Therefore, y E W,.b,(A,U)\ f(W&(A, U))c D" and thus Yn = f-mm(xn)E V when n is large. We have x, E U',but contradicting our assumption. 0

n,,

Lemma 3.3.6. Let

A1

and

A2

be isolated sets of X . If Al n A2 = 0 , then

Proof. Let U be a compact neighborhood of A2 such that 0:- fn(U) = A2. Denote as D" a proper fundamental domain for WLc(A2,U). To show (i) let X n E W"(A1)' be a sequence such that x, + x E cl(W"(A1)') n A2. Define F = cl(U,>, O(xn)). Then F C cl(W"(A1)'). Thus it sufficies to prove that F n D" # 6 Let V be a small neighborhood of D" such that V r l A2 = 0, and

CHAPTER 3

110

define U' = W&(Az,U) U O + ( V ) .Let k > 0 be an integer such that 2& E U'. Since x & E W"(Al)', we have X& # W"(A2)' and there is nk 2 0 such that zk E fnk(V). Thus U,,,,, Of(",,) n V # 0 and therefore F n D" # 0. To show (ii) if 2 cl(Wu(Al)') r l [WU(A2)'- Az], then 0 # a(.) C cl(W"(A1)') n Az. Thus cl(W"(A1)') n W'(A2)' # 0. From (i) we have cl(W"(Al)') n [W"(Az)' - A,] # 0. 0


n [W"(Aj,)' - Ajo] # 0, from which we choose a point z. Since X = U: W"(Ai)', there is 1 5 jl 5 I such that 2 E Wu(Ajl)'. Then we have 2

E [Wu(Aji)' - ~ j l n[W"@j,,)' l -~

j o l ,

but 2 E cl(W"(Ai)') n [W"(Aj,)' - Aj,] and so by Lemma 3.3.6 (ii) we have cl(W"(Ai)') n W'(Ajl)' # 0. NOW suppose Aj,,... ,Aj, are defined for p < I such that Aja # Ai for 1 I a < p, and

If j , # i, then we can obtain A,p+l by the same process used to get Aj1. Since 1 5 j , 5 t , by the assumption of no cycles we have j , = i for some a which proves the lemma. 0

Lemma 3.3.8. Let A be an isolated set for f and let x E X . If a ( x ) fl A # 0 and 2 # Wu(A)', then (i) (ii)

nrm

Proof. (i) : Let U be a compact neighborhood of A such that f n ( U )= A and let D" = D"(h,U)be a proper fundamental domain for W&,(A, U). To

$3.3 Recurrent sets and Birkhoff centers

111

obtain (i) it suffices to prove that O-(z) n V # 0 for a neighborhood V of D* such that V n A = 0. By Lemma 3.3.5, U' = W&(A,U) U O+V is a neighborhood of A. Since a(z)n A # 0, there exists k > 0 such that f-'(z) E U'. By the assumption z # Wu(A)', thus f-&(z)# W,U,,(A,U) and so f-'(z) E f"(V) for some n 2 0, which implies O-(z) n V # 0. (ii) : Let U be as above such that z # U. Since f(WL,(A, 17)) c WL,(A, U), we have O-(z) n WLc(A, U)= 0. Since a(.) n A # 0, there is a sequence n k with n k + 00 as k + 00 such that z k = f - n h ( z ) E U. Since z k # WLc(A,U), we have O+(zk) $Z U.For each k > 0 let mk 2 0 be such that fm(zk) E U for 0 5 m 5 mk and f m h + l ( z k ) # U.Then we have y& = f r n h ( Z & )

E U - f-'(U)

and mk - n k < 0. Thus yk E O-(z). Indeed, if not, we have 0 5 n k 5 mk and so f " r ( z k ) = z E U,which is a contradiction. Since z k converges to A, given N > 0 there is ko > 0 such that x k E n : f - " ( U ) for k 2 ko. Then f n ( z k ) E U for 0 5 n 5 N and so mk 2 N for k 2 ko. Thus mk + 00 as k + 00. Without loss of generality we suppose that yk converges to y E cl(U - f - ' ( U ) ) . To see y E Wu(A)' - A it suffices to prove that y E Wu(A)' since A c int(f-'(U)). If y # Wu(A)', then there exists j > 0 such that f-j(y) # U. Therefore, there exists ko > 0 such that for 12 2 we have mk -j > 0 and fmh-j(zk) = f - j ( y k ) # U,which is a contradiction since 0 < mk - j 5 mk.

---

Lemma 3.3.9. If {Al, ,At} is a family of disjoint isolated sets for f such that a ( z ) c Ai for all z E X , then a ( z ) C Ai for some i.

uf=,

This lemma can be derived immediately from lemma 3.3.8. Lemma 3.3.10. Let F be a compact f-invariant subset of X and let Q be a compact neighborhood of F such that n,,,f"(Q> = F . Then there i s a compact neighborhood V of F such that

V

c Q, f ( V ) c int(V).

ni=o

Proof. Define A, = fn(Q) for T 2 0. Then {A,} is a decreasing sequence of compact sets converging to F. Since Q is a compact neighborhood of F, we have A, c int( Q)for large T . For U = Q n f-' (Q)we have F c int(U)

c U C Q , f ( U )C Q

and so fix an T large enough such that A,

c int(U). Then

CHAPTER 3

112

Since A, is a neighborhood of F , there is n so large that f"(A,) c int(A,). If n = 1, then we obtain the conclusion. When n 2 2, we can take a small compact neighborhood W of A, such that A, C W C U, f " ( W )C int(A,). Define E = f"-l(W) U A,. Then f"-l(A,)

C int(f"-'(W))

f"-'(f"-'(w)) C f"-'(f"(W))C f"'2(int(Ar))

c int(E) and

C int(A,)

c int(E).

Thus we have f"-l(E) C int(E). As above choose a small compact neighborhood G of E satisfying E C G c U and f"-'(G)C int(E), and repeate inductively the above argument. Then our required set is produced. 0

Lemma 3.3.11 (Filtration lemma). If { A , , . . . , A t } as a family of disjoint isolated sets for f having no cycles such that a ( z )c Ai for all z E X, then (1) there en'st compact sets

ufZl

xt = x

B = xo

Figure 13

Proof. (1) : We define a relation Ai 5 A, on {Ai : 1 5 i 5 L} if there is a sequence Aj = A j , , * * ., h j , = Ai such that [Wu(Ajo-l)'-Aja-~]n[W"(Ajo)'Aja] # 0 for 1 5 a 5 k. By the assumption of no cycles this relation is a partial ordering. It induces a total ordering A1 < A2 < < At of the family where A; < Aj if Ai 5 A j and Ai # A,.

---

53.3 Recurrent sets and Birkhoff centers

113

We define X j by induction. Let XO= 0 and suppose that we have defined compact sets

0=xocx1c...cxj for j

< L such that

+

Let F = U{Wu(Ai)' : i 5 j 1). Then F is closed and F n A& = 0 for k > j 1. Indeed, let {zn}be a sequence in F converging to z E X. Without loss of generality we suppose that for all n > 0 we have zn E Wu(Ai)' for some fixed i 5 j 1. By Lemma 3.3.9 there is 1 5 k 5 L such that a(.) c Ah, i.e. z E W"(Ak)'. Thus z E cl(W"(Ai)') f l Wu(A,)'. By Lemma 3.3.7 we have A&5 Ai and so k 5 i 5 j 1, i.e. z E F. Thus F n A& = 0 for k > j 1. Now let Q C X be a compact neighborhood of F such that

+

+

+

+

Qn(

U Ai) =0.

i>j+l

then f-n(z) E Q for Then n:==,fn(Q) = F. Indeed, if z E n:=ofn(Q) 2 0. Thus O-(z) c Q and therefore a ( z )c Q. Since a(z)c Ai for some i 5 j 1, we have z E Ui<j+l Wu(Ai)' = F. Since f ( F ) = F,by Lemma 3.3.10 we have a compact neighborhood 5 C Q of F such that f(V'j) c int(5). Define Xj+l = Xj U 5. Then f(Xj+l) C int(Xj+l). To conclude (1)it remains to see that nFwf"(Xj+l -Xi)= Aj+l. First we observe that n

+

is closed. Indeed, if zk E S is a sequence converging to E X, then fn(z)E Xj+l for all n since Xj+l is compact. This follows from the fact that if fn(z)E Xj for some n then fn+l(z) E int(Xj), and thus f"+l(zk) E int(Xj) for large k (which contradicts zk E S). Thus a(.) C S for all z E S and so a(.) c Xj+l for all z E S, from which we have S c W"(Aj+l)'. Similarly, w ( z ) c S for all z E S. By using the dual of Lemma 3.3.8 we have S C W8(Aj+l)'. Therefore,

Lemma 3.3.9 ensures that X =

u t , Wu(Ai)' and

so

we observe Xe = X.

CHAPTER 3

114

(2) : To obtain the conclusion it suffices to show that if y E X - u t , Ai then there exist y' E O-(y) and a neighborhood V of y' such that fn(V) n V = 0 for all n > 0. Choose A, such that a ( y ) C A, c int(Xj) - Xj-1. Take y1 E O-(y) such that O-(yl) C Xj - Xj-1. Since y1 4 Aj, there exists > 0 such that ~2

= fno(yl) 4

Xj

- Xj-1.

However, O+(yl) C Xj and so fno(yl) E Xj-1. Let mo be an integer such that fFm(y2) E Xj-1 for 0 5 m 5 mo and f-(m0+')(y2) 4 Xj-1. Then

I' = f-mo(Ya) E

x j-

f(Xj-1).

Take a neighborhood W c int(Xj-1) of f(Xj-1) and a neighborhood V of y' such that W n V = 0 and f(V) c W. Then we have fn(V)c fn-l(W) c W for n > 0 and therefore V n fn(V) = 0 for n > 0. 0

L e m m a 3.3.12. If { A l , - - . ,At} is a family of disjoint isolated sets for f hawing no cycles and if z E X satisfies a(.) c U:=l W"(Ai)', then z E Uf=1 W"(Ai)'. The reader will understand the relationship between this lemma and Theorem 3.3.2.

&

Proof of Lemma 3.3.12. If z $2 W"(Ai)' and a(.) c Uf=l W"(A;)', then a cycle exists. Indeed, choose A;, ,1 5 il 5 l such that a(.) n A;, # 0. Since a(.) (Z A;,, by Lemma 3.3.8 (i) we have a ( . ) n [Wa(A;,)' - A;,] # 0. Since a(.) c W"(Ai)',there exists 1 5 i 2 5 l such that

ut,

- Ail] # [W"(Ai,)'- A;,] n[Wa(Ai1)'

0,

a ( . )

n A;, # 0.

If i 2 = il, then we have a cycle. If i 2 # i l , then we repeat the process as above. Then we have 1 5 i3 5 l such that [W"(AiS)'- Ail] n[W'(Ai,)' - La]#

0,

~ ( zn) Ais

# 0.

Continuing this, we have a cycle. 0

Lemma 3.3.13. If y E a(z), for E > 0 thew is an €-pseudo orbit y = y o , y 1 , - - - ,yk = y in a(.) (i.e. d ( f ( y i - l ) , y i ) < E for 15 i 5 k). Proof. For E > 0 let 0 < 6 < ~ / such 2 that if d ( p , q ) < 6 then d ( f ( p ) ,f ( q ) ) 5 ~ / 2 .Since f-"(z) converges to a(z)as n + 00, there is no > 0 such that d ( f - n ( ~ ) , a ( z< ) )6 for n 2 no. Since y E a(z),there are n1,k 2 0 such that n1 - A! 2 no, d(f-nl(z),y) < 6, d(f-nl+k ( X ) , Y ) < 6.

$3.3 Recurrent sets and Birkhoff centers

115

Let z' = f-"l(z). Since nl - k 2 no, when 1 5 j 5 k we have d(fj(z'),a(z)) = d(f-"l+j(z),a(z))

< 6.

Thus there is y j E a(%)such that

L e m m a 3.3.14. Let {C, : n 2 1) be a family of f -invariant compact sets such that if n 1 m then C, c C,. If z, E C, and yn E a(%,) converges to yEC= En, for E > 0 there is an &-pseudo orbit y = E,jji,. ,% = y in C ( i e . d ( f ( E ) , ~ 0 such that if z E C, then there is yl E C with d ( x , y') < 6. Take n > 0 such that y, E a(%,) C C,, and d(y,y,) < 6. By Lemma 3.3.13 there is an ~/3-pseudoorbit yn = 20, zl, ,Zk = yn in a(zn). Thus we have

-

for 1 5 j 5 k. 0

Lemma 3.3.15. If A is f-invariant and compact and if f l ( f p ) = A, then A c df). Proof. For n

> 0 define

< 1/n for some k > 0}, y) < 1/n for some k > 0).

A: = {y E A :d(fk(y), y) A, = {y E A : d( f-'(y),

Clearly A; and A$ are open dense subsets of A, and so is A, = A; n A.: Thus A = A, is a Baire set of A, i.e. cl(A) = A. Since z E a(t)n ~ ( x ) for x E A, we have A = cl(A) c c(f). 0

n,"==,

CHAPTER 3

116

Lemma 3.3.16. I f z E a(z), then z E c ( f ) .

Proof. a ( z ) is compact and fl(fl,[,,)

4.1 c c(f).

= a(.).

By Lemma 3.3.15 we have

0

Proofof Theonm 3.3.2. If we establish that a ( z ) C c(f) for all z E X ,then we have Q(f) C c ( f ) by Lemma 3.3.11, and therefore the conclusion of Theorem 3.3.2. Suppose there is z E X such that a(.) q! c ( f ) . Let 3 be the family of compact f-invariant subsets C of X such that there is z E C with a ( z ) @ c ( f ) . Introduce t o 3 an ordered relation by inclusion. If 3 possesses a minimal set E, then a(.) @ c ( f ) for some z E X. By Lemma 3.3.16 we have z @ a(.). Thus a(z)is a proper subset of E. Since f ( a ( z ) )= a(%)and a(z) is compact, a ( . ) E F,which is a contradiction. Therefore, t o obtain Theorem 3.3.2 it suffices to show the following Lemma. 0 Lemma 3.3.17. 3 possesses a minimal set C.

Proof. Let F' = {Cp : p E B} be any totally ordered subfamily of 3. Let C = n { C p : p E B},then we need to prove I: E 3. If E $! 3,then a(.) C c ( f ) for all z E C. By Lemma 3.3.11 the family Al

n E,...,At n c

ut1

admits a filtration and Q(jp)c AinC. Take z p E Cp with a(zp) @ c(f) for 3/ E B. Since a(zp) n (U: Ai) # 0, without loss of generality we suppose that a ( z p ) n A1 # 0 for all p E B. Let U be a compact neighborhood of A1 such that nZrnfn(U) = A1 and Un Ai) = 0. Since a ( z p )n A, # 0 and a(zp) q! Al, by Lemma 3.3.8 we conclude that

(ut,

a ( z p )n D" # 0 where D" = D"(A1,U) C U is a proper fundamental domain for W&,(Al). Choose yp E a ( z p ) n D" for p E B and take a subsequence yp,, such that yp, converges t o y E C n D". Recall that D" n c ( f ) = 0. Since y $! c ( f ) and C admits a filtration, there is a compact set K c C and y' E Of(y) such that

f ( K )c int(K),

y' E K - f ( K ) .

Let W c K be an open set in C such that y' @ W and f ( K ) c C. Since f ( K )is compact, there is E > 0 such that if z E E and d(z,z') < E for some z' E f ( K ) then z E W . By Lemma 3.3.14 there exists an E-pseudo orbit y' = yo,yl,... ,yk = y' in C. Then we have y j E W for 1 5 j 5 k. Indeed, since f(y0) = f(y') E f ( K ) and d(f(yo),yl) < E , we have y1 E W C K. Suppose we have proved that y j E W for j 2 1. Then f(yj) E f ( K ) and d(f(yj),yj+l) < E . Thus yj+1 E W. Consequently, y' E W ,which is a contradiction. Therefore, E E F. 0

53.4 Nonwandering sets of TA-maps

117

Remark 3.3.18. For a continuous surjection f:X + X of a compact metric space, the Birkhoff center c(f) is defined by c ( f ) = cl{z E X : x E ~ ( x ) } . As Theorem 3.3.3 we can show that iff: X + X is c-expansive and f: c ( f ) + c(f) has POTP, then c ( f ) = n(f). The proof will be given in Remark 3.5.4 of 53.5.

53.4 Nonwandering sets of TA-maps

We shall discuss the decomposition theorem of the nonwandering set for TA-maps. As before, let X be a compact metric space with metric d and let f:X -+ X be a continuous surjection. Recall that x E X is a nonwandering point if for any open neighborhood U of z there is n > 0 such that fn(U)n U # 0. Denote as n(f) the set of all nonwandering points of f ( n(f) is called the nonwandering set of f). Then n(f) is a non-empty closed subset and

fWf>) c n(f>*

Theorem 3.4.1. Let f:X + X be a continuous surjection of a compact metric space. Iff has POTP,then f(n(f))= n(f).

Proof.Suppose n(f) - f(Q(f)) # 0. Then there are z E Q(f) - f(n(f)) and E

> 0 such that

Let 6 > 0 be a number with the property as in the definition of POTP. Since x E fl(f), there is n > 0 and an n-periodic &pseudo orbit (xi) E Xz with 20 = x. Since f has POTP, there is (yi) E Xf such that d(yi,zi) < E for all i E Z (X, = {(xi) E Xz : f(x:i) = zi+l,iE Z}). Hence, { f n i ( 2 / o ) : 0 5 i < m} C B,(zo). Since X is compact, a subsequence of {fni(yo)} converges to some yt E X and, hence, yt belongs to the w-limit set of yo. Since fn(y') E B,(z)and fn(y') E f(n(f)),this can not happen. 0 Let

n(f) be the nonwandering set of a continuous surjection f. As before

let

Rf = { ( x i ) E

n(f)Z: f(q) = 2i+l,i E

Z}.

The chain recurrent set of the continuous surjection f, C R ( f ) ,is defined in the same way as given for a homeomorphism. Then f ( C R ( f ) )c C R ( f ) and C R ( f )is closed (cf. see Lemma 3.1.1). I f f has POTP, then n(f) = C R ( f ) (cf. see Theorem 3.1.2). Theorem 3.4.2. If a continuous surjection f has POTP then so does f l q j ) . If, in addition, flqf)is c-expansive, then the set of all periodic points, Per( f), as dense in n(f).

CHAPTER 3

118

Proof. For e > 0 let 6 > 0 be as in the definition of POTP. Since n(f) = C R ( f ) ,the &relation A is defined on n(f) as in 53.1. By this relation, n(f) is decomposed as a union n(f) = UxAx of distinct equivalent sets Ax. We first prove that each Ax is open in n(f). Take and fix z E Ax. For any y E Ax let zo = z,51, -. ,zp = y be a &pseudo orbit in n(f). Choose 0 < 7 < 6/3 such that f(U,(zo)) c U~(z1).To get the conclusion it is enough to see that U,(zo)r l n(f) c Ax. For any zb E U,(zo)n n(f), {zb,zl,---,zp}is a 6-pseudo orbit in Q(f). Since z,y E n(f), a &pseudo orbit 0 = {yo = y,yl,... ,yt = z} exists. If f ( Y t - 1 ) E cl(u,(zo))nn(f), then (0\ { y t } ) u {zb} = { Y O , Y l , . . . , Y t - l P ; } is 6 a 6-pseudo orbit since d(f(yl-l),z;) 5 27 < 6. Thus we have y zb. When f ( ~ t - ~g!) cl(U,(zo)) n n(f),then there is z E c1(UY(zo))n n(f) such that

-

-

d(f(Yt-l),cl(U,(zo))

n W)) = 4f(Yt--l),Z) < 6,

-

and so d(zb,z ) 5 27. Since z E n(f)= C R ( f ) ,we have z z, i.e. there is a periodic 7-pseudo orbit { Z O = z , 21, ,Zb, z } in n(f). From

---

d(f(Zb),z;)

5 d ( f ( z b ) , z ) + d ( z , z b )5 37 < 6,

the sequence (0\ {yt}) u { z o , - - * , Z b , Z & } = {yo,... ,yt--1,zo,--- ,.~b,zb} is a &pseudo orbit from yo to a$. Therefore, zb E Ax. Since n(f) is compact, {Ax} is finite and n(f) is covered by finite open sets Ax. Since each Ai is open and closed in n(f), we have d(Ai,Aj) = inf{d(a,b) : a E Ai,b E Aj} > 0 if i # j. Put 61 = min{d(Ai,Aj) : i # j}. For 0 < a < min{6,61} let (zi) be an a-pseudo orbit in n(f)'. Then we must prove that an €-tracing orbit of (xi)is chosen in Rj. If zo E Ai for some i then each point of ( z i ) belongs to Ai (since z is always a-related t o f(z)for z E n(f)). Take Za,Zb E (Zi) ( 4 < b). Then 2, % , zb, from which we can find a (L1 Itz)-periodic 6-pseudo orbit ( q )E n(f )' such that for i 2 0

+

"'za

Put It = It1

+

= za,za+1,"'

It2

9x6

= zkl,Zkl+l,"'

92,

=tkl+ka,"'

.

for simplicity. Since f has POTP, there is yavb E X such that d(fi(yavb), Z i )

0 such that ft(yavb) = yaqb,and then yayb E a(f ). If D is not discrete, then a subsequence of

53.4 Nonwandering sets of TA-maps converges to some za*bE X and d ( f j ( z a g b ) , z j )5

E

119

for 0 5 j

< k.

Thus we

have z " ' ~for any a' > 0. This follows from the fact that for any a' > 0 there is io > 0 so large that

for in 2 io. Therefore,

E C R ( f )= R(f). We put z: = f i ( z a ' b )for i 2 a and choose zL-l E f - l ( z a p b )n R(f) and z:-i-l E f - l ( . ~ : - ~ ) r l R(f) for i 2 1. Then Za,b = ( z i ) E C I ( ~ and ) ~ d ( z : , z i ) 5 E for a 5 i 5 b. Since R is compact, a subsequence of { z a ' b } converges to z = ( z i ) E R(f)' as a + --oo and b + 00. Hence d ( z i , z j ) 5 E for i E Z since d ( z : , x i ) 5 E ( a 5 i 5 b) and a,b are arbitrary. Clearly z = (zj) E Rf.

To show density of Per(f), take x E R(f). Then there are t > 0 and an t-periodic &pseudo orbit (q) E R(f)' with zo = 2. Since fp(f)has POTP, there is ( z i ) E Rf such that d ( z j , z i ) < e for i E Z. Then d(ze+j,zi) 5 2~ for i E Z. Choose e > 0 less than c-expansive constant, then we have (ze+i) = ( z ; ) and therefore f'(z0) = ze = zo E Uc(z).0

A continuous surjection f : X + X of a metric space is topologically tmnsitive if there is t o E X such that the orbit O+(ZO)= { z o , ~ ( z o ) , * -is . }dense in X.

Remark 3.4.3. Let X be a compact metric space. A continuous surjection f:X X is topologically transitive if and only if for U,V nonempty open sets there is n > 0 such that fn(U) n V # 0. For the proof suppose O+(x:o)is dense in X and U,V are nonempty open sets. Then there are n > m > 0 such that fn(xo) E U and fm(xo) E V. Thus --$

fk(u)n v # 0 for

Let {Uj : j equivalent :

= n - m.

2 1) be a countable base for X. For x

Cl(O+(Z)) #

E

X the following are

x,

w o+(z) n Un = 0 f m ( z )E

X

\ Un

for some n, for every m 2 0 and some n,

oo

f - " ( ~ \ un)

xE

for some n,

m=O

U n f - m ( ~\ un). o o w

a5 E

n=l m=O

Since U =:, f-m(Un) is dense in X by the assumptions, X = n z = , ( X \ f-m(Un)) is nowhere dense. Thus U,"==,

\U =:,

f-m(Un)

n,"==, f-"(X \ U,)

120

CHAPTER 3

= U= :l n:=,(x \ f-m(U,,)) is a set of first category. Since X is a compact metric space, {x E X :cl(O+(x)) # X } is a set of first category.

A continuous surjection f : X -+ X of a metric space is topologically mixing if for nonempty open sets U,V there exists N > 0 such that U n f " ( V )# 0 for all n 2 N. Topological mixing implies topological transitivity. Theorem 3.4.4 (Topological decomposition theorem). Let f:X -+ X be a continuous surjection of a compact metric space. If f : X + X is a TAmap, then the following properties hold: ( 1 ) (Spectml decompsition theorem due to Smale) fl( f ) contains a finite sequence Bi (1 5 i 5 L) o f f -invariant closed subsets such that (i) f l ( f ) = UfE1 Bi (disjoint union), (ii) f p i :Bi t Bi is topologically transitive, (such the subsets Bi are called basic sets.) (2) (Decomposition theorem due to Bowen) For B a basic set there exist a > 0 and a finite sequence Ci ( 0 5 i 5 a - 1) of closed subsets such that (i) Ci n C, = 0 (i # j ) , f ( C i )= Ci+l and f a ( C i )= Ci, (ii) B = Uyzj Ci, (iii) fbi : Ci + Ci is topologically mixing (such the subsets Ci are called elementary sets.) Proof. Since f : X + X has POTP, we can prove that f l ( f ) = C R ( f ) as in Theorem 3.1.2. Thus n(f)splits into the union f l ( f ) = Bx of the equivalence classes Bx under the relation which is defined in CR(f ) . Each Bx is closed and f ( B x ) = Bx. If Bx is open in fl(f), then f l ( f ) = Uf=lBi for some k > 0 (since n(f)is compact). In this case f p i is topologically transitive. For, let U and V be non-empty open sets in Bi. Since x y for x E U and y E V , we can find in Bi a tracing point for a pseudo orbit from x to y (since f p i has POTP). This shows that U n f'(V) # 0 for some I > 0. To obtain (l), it is enough to show openess of Bx. For E > 0 there is 6 > 0 such that any &pseudo orbit of f p ( f )is €-traced by some point of f l ( f ) . Then, for p E Ua(Bx)n Per(f) there is y E Bx such that d(y,p) < 6 (here Ua(Bx) denotes an open neighborhood of Bx in fl(f ) ) . By Lemma 2.4.3 we have

-

ux

-

and WU(X) =

uf'(w,u(.-i(x))) i>O

for x E Xf.Since f i n ( f )has POTP, we have W"(y) r l W"((pi))# 0 for a periodic orbit ( p i ) E flf with PO = p , and W " ( p )n Wu((yi)) # 0 for an orbit (yi) E Rf with yo = y. Since we can choose such that each yi of the orbit (yi)

53.4 Nonwandering sets of TA-maps

121

is in Bx, in our case suppose that so does (yi). Then we have y which p E Bx. And so

-

p, from

Bx 3 cl(u6(BA) n per(f)) 3 U6(BA) n cl(Per(f)) = U~(BA). To prove (2), take p E Per(f) n B with fm(p) = p and put C, = cl(W"(p) n B). Then C, is open in B. For, take q E u,5(cp)r l Per(f) n B (fn(q) = q for some n > 0), then we can find z E W"(p)n B such that d(z,q) < 6.

\

\ \ I I I

.

/ /

/

0

Figure 14 Let ( q i ) E Bf be an n-periodic orbit with qo = q. Since f l has ~ POTP, there is z' such that z' E WU((qi))nWa(z)nB,and so d(z:,qi) -+ 0 (i -+ -00) for some (xi) E Bf with zb = 2'. Since zb = z' E Wa(z) = W"(p),we have d(fi(zb),fi(p)) -+ 0 as i --+ 00. Note that x'_,,,~ + q as k -+ 00. Fix 12 > 0 and let j = mnk i for i 2 0. Then

+

d(P(zkmnk), f'(p)>

= d(fi(ziI, fi(p))

+

0

(i --t

00)

since fmnk(z'_mnk) = zb and f m n k ( p ) = p. Thus zkmnkE Wa(p) for k 2 0 and so q E C,. This shows that C, is open in B. We next show C, = C, for q E C, n Per(f). Let fm(p) = p and f n ( q ) = q. For 7 > 0 let n., > 0 be as in Lemma 2.4.1. Fix E > 0 and choose 6 > 0 as in the definition of POTP. Let z E W"(q)n B. Then there is J., > 0 such that rnnJ., 2 n., implies d(fmnJ7(z),q)< 6/2. Since q E C,, we can choose y E Ual2(q)n W 8 ( p n ) B, and so d ( f m n J ~ ( z ) , y< ) 6. Thus

n

z0

WU((PmnJvzi)) ~ " ( y )

for some orbit ( z i ) E Bf with z = zo (since f p has POTP). Therefore there is an orbit (2;) E Bf such that 20' E

W U ( ( f m n J 7 z i ) wS(y), )

CHAPTER 3

122

and so

and so t ~ , , E W s ( p ) n B .Since 7 is arbitrary, we have x E C, and therefore J,

c, c c,.

On the other hand, suppose p 4 C,. Then we have 0 < d = d ( K ,C,)where K = C, - C,. Since q E C,, there is z E W 8 ( p )n B such that d ( z , q ) < d. Clearly t E C,. But

so that f ""'(z) 4 C,,which is a contradiction. It is easily checked that if C,nC,, # 0 for q, q' E Per( f ) nB then C, = C,,. Since Cfm(,) = C,,there is the smallest integer a > 0 such that a 5 m and Cfa(,.) .= C,. Thus B = C, U Cf(,)U U Cfa-'(,) since f l is ~ topologically transitive. We finally show that fpC, is topologically mixing. Let U and V be nonempty open sets in C,. Take q E V n Per(f) with f"(q) = q. Choose E > 0 with Ue(q)c V. Then, for 1 5 j 5 n - 1 there are z j E U n Ws(faj(q)) and N j > 0 such that t 2 Nj implies

--.

Therefore, fa(nt+"-j)(U)n V # 0 for t 2 Nj and 0 5 j 5 n - 1. Put N = max{Nj : 0 5 j 5 n - 1). Then 9 2 n N implies fa8(U)n V # 0. 0

53.5 Inverse limit systems

Let f : X + X be a continuous surjection of a compact metric space. In the previous chapter, for expansivity and POTP we have discussed the relationship between the continuous map and the shift map of the inverse limit space. This section will investigate those relationships for w-limit sets, recurrent sets, nonwandering sets, chain recurrent sets and Birkhoff centers. Denote as c(f) the closure of {z E X : z 6 w(z)}. Then c(f) is called the Birkhoff center for the continuous surjection.

53.5 Inverse limit systems

123

Theorem 3.5.1. Let f : X + X be a continuous surjection of a compact metric space and let X) = l@(X, f ) be the inverse limit space with the shift map u : X) + X). Then (1) if n > 0 is an integer and F, is the set of periodic points of f with period n (i.e. F, = {z E X : f n ( z )= z}), then lii(F,, f ) is the set of periodic points of u with period n and the natural projection po : lim(F,, f ) + F, defined by ( z i ) r H zo is bijective, t (2) if z E X and x = (z,zl,z2,.--) E X), then w(x,u) = l&(w(z), f ) is the w-limit set for u, (3) i f R( f ) = {z : z E ~ ( z ) is } the recurrent set, then R(u) = l i i ( R ( f ) f, ) is the recurrent set for u, (4) c ( u ) = lii(c(f ) , f ) is the Birkhofl center for u, (5) n(u) = l i i ( n ( f )f ,) is the nonwandering set for u, (6) CR(u)= l@(CR(f), f ) is the chain recurrent set for u. Proof. For i 5 0 let pi : X) + X be the natural projection to the i-th coordinate. We first claim that if u ( A )= A is a closed set of X) then A = l@(A, f ) where A = po(A). Indeed, since p i ( A ) = pi(u'(A)) = p o ( A ) = A for all i 2 0, we have A c lii(A, f ) . If y E l&(A, f ) , then p i ( y ) E A for all i 2 0. Since A = p i ( A ) for i 2 0, there exists zi E A such that p i ( z i ) = pi(y). Then p j ( s i ) = p j ( y ) for j 5 i. Since A is compact, { z i } converges to z E A. Therefore, z = y E A. (1) : If z E F,, then (z,fn-1(z),fn-2(z),...,z,...) E l@(F,,f) is a periodic point of u with period n. Coversely, if x E X) is a periodic point of ~7with period n, then x is expressed as x = (zo,z1,--,z,,--l,zo,**-), which implies zo E F,, and so x E lim(F,, f ) . Therefore l h ( F , , f ) = {x E X) : t un(x)= x}. It is clear that po : lii(F,, f ) --t F, is bijective. (2) : Clearly p o ( w ( x , u ) ) c ~ ( z ) For . y E ~ ( z there ) is a sequence f n i ( z ) such that f n i ( z )4 y as i --+ 00. Then {uni(x)}has an accumulation point y E w(x,u). Thus p o ( y ) = y and w ( z ) = po(w(x,u)). Therefore we have 4x9 0) = 9 ( 4 z , f ), f 1. (3) : If x E R(u), then x E W ( X ) C l i i ( w ( z ) , f ) and SO p i ( x ) E W ( z i ) = ~ ( z ) Therefore, . x E l&(R(f), f ) . Let x E l i i ( R ( f )f , ) and put z = po(x). Then pi(x) = zi 6 R ( f ) for all i 2 0. Thus we have p i ( x ) E ~ ( z i=) ~ ( zand ) thus x E l&(w(z), f ) = w ( x ) by (2). Therefore, x E R(u). (4) : By (3) we have R ( o ) = l@(R(f

1, f

c l i i ( c ( f1, f)

and so c ( u ) = cl(R(u)) c l&(c(f), f ) . To obtain the conclusion take x E li$c(f), f ) . Since pi(x) = zi 6 c ( f ) for i 2 0, for E > 0 there is N > 0

CHAPTER 3

124

such that for any m 2 N there is satisfying

Ym

E {y E X : y E w(y)} (so close to )2 ,


, f). To show n(a) 3 lii(n(f),f)let M be the diameter of X. Let x = ( Z O , Z I , * - *, z r n , - - .E ) lim(n(f),f) and take m > 0 such that M/2" < ~ / 4 . t Since z , E Q(f) and f ' ( z m )= zm-; for 0 5 i 5 m,by uniform continuity we can choose ym E X satisfying

for some It that

> 0. Here we take points ym+l, ym+2, - * and yk+l, yk+,,

are in lii(X, f). Then

E

< -2 +

c - 0 small we have W,"(z)nW,U(y) # 0 when x is very near to y. If, in addition, f is expansive, then we see that W,"(z) n W,U(y) is a set consisting of single point and it is denoted as [z,y]. Throughout this section let f : X + X be a TA-homeomorphism of a compact metric space. 127

CHAPTER 4

128

Theorem 4.1.1. The homeomorphism f has a local product structure. Proof. Let e > 0 be an expansive constant for f and ~IX EO = e / 4 . Then there is 0 < 60 < EO such that W:o(z)n Wz(y) = [z,y] for z,y E X with d(z,y)

5 60.

First we show that [ , ] : A(&) --+ X is continuous. Suppose a sequence {(zn,yn)} of A(&) converges to (z,y)E A(&). P u t z, = [zn,y,]. Since X is compact, there is a subsequence {z,,} of {z,} that converges to z E X. Since zni E W,9(znj), we have d(fi(znj), fi(znj)) 5 EO for i 2: 0 and nj, and so d ( f ( z ) , p ( z ) )5 EO for i 2 0. Thus, z E W:o(z). Similarly, z E Wz(y)and z = [z,y]. This shows that {z,} converges to [z,y]. It is clear that [z,z] = z for all E X. Since [z,y] E Wc",(z), we have [[z,y],z] E W&,(z) n W t ( z ) and then [[z,y],z] = [z,e] by expansivity. Similarly, [z,[y,z]] = [z,z]. It is easily checked that f [ z , y ] = [f(z),f(y)] by uniform continuity. To conclude the theorem we must prove (B). To do so define g1 : X x A(6o) ---t by Sl(Z, (Y,

4)= 43,[? .I> I,

for z E X and (y, z ) E A(&). Then g1 is continuous and g l ( z , (2, z))= 0. By uniform continuity of g1 we can find 0 < 61 < 60/2such that diam( {z,y, z } ) < 261 implies d(z, [y,z]) < &/3. If (y,z ) E V{(z) x (z),then d(z,[y, z]) < 60/3 and thus diam(N,) < 60. To show openness of N , let w E N,. Then there are y E Vt(z) and z E V[ (z) with w = [y, z]. Since d(z,w) < 60/3,we can find maps PU

' B60/3(w)

+

Wz(z),

PS

:B60/3(W)

w.~(~)

by pU(w) = [v,z] and p 8 ( w ) = [z,w]for w E B 6 0 / 3 ( W ) . They are clearly contin= y and p 8 ( w )= z. Thus there uous. Since w = [y,z], we have p,(w) = [y,z] is a neighborhood U C BSo/3(w)of w in X such that p u ( U ) C V{(z) and p 8 ( U )c Va",(z).If w E U,then w E N, since w = [[w,21, (2,w]] by expansivity. This implies that N , is open in X. Therefore, (a) was proved. (b) is easily checked as follows. Define a map h: N, --+ V{(z) x V'(z) by 21, [z, 741, h ( 4 = ([w,

w E N,.

Obviously, h is continuous and h is the inverse map of [ , 1; i.e. h is a homeomorphism. To see (c) put 92(z,ar) = diam{z,[y,4[z,yI) for (z,y)E A(&). Then 9 2 is a continuous map and there is 0 < p < 61 such that d(z,y) < p implies ga(z,y) < 61. This shows that [y,z] E Vt(z)and [z, y] E V,.l (z)and therefore y = [[y,21, [z, y]] E N,. 0

54.1 Markov partitions and subshifts

129

for z , y E X .

Lemma 4.1.3. For z , y E X with d(z,y) o f z in Vz(z)(a= s,u), and the maps

< p, D& is an open neighborhood

are homeomorphisms.

Proof. Since Ny is open in X,Dz,y is open in Vz(z). If d(z,y) 5 p, then z E B,(y) C Ny and z E D&. Thus D& is an open neighborhood of z in V;(z). Let z E D&. Since z E Ny, we have [z,y] E V{(y). Since 2: E V { ( z ) C N , and y E I?,,(%) C N,, we have [r,y] E N,. Thus [z,y] E Dlf,,. Similarly we have [z,z] E D& for z E D;,,.That [D!& y] = D;,, follows from the properties of [ , 1. By (b) the map [ ,y] : D!& + D;,,is a homeomorphism and [ ,z] : D;,,-t D& is its inverse map. The same result is true for u = s. 0

Lemma 4.1.4. For z E X (4 fV,",(z)n V,d,(f(z))is open in V:l(f(4), (b) f-'V;(z) n V;(f-'(z)) is open in V;(f-'(z)). Proof. Take w E f ( N , ) n V,(f(z)). Then f-'(w) E N , and f-'(w) = [y,z] for some y E V$(z)and z E V;(z). Since V;(z) C W z ( z ) ,we have f-'(w) E

W.&).

On the other hand, since w E V;(f(z)),we have w E V;(f(z))c W.&(f( By expansivity we have f-'(w) = z and so w = f(z) E fV4 (z).Therefore

2)).

Since N , is open in X,we have (a) and similarly (b). 0

A subset R of X is called a rectangle if diam(R) 5 p and [z,y]E R for z,y E R. Throughout this section let R denote a rectangle of X.

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Remark 4.1.5. Since p is chosen such that p < 61,we have [z, y] = V,"1(z)r l V< (y) for z,y E cl(R). Thus cl(R) is a rectangle. For convention we write

v"(z,R) = V;(Z) n R,

vU(z,R) = V;(Z) n R

for z E X. Denote aa int V'(z, R) the interior of V"(z, R) in V 0 there exists

ufn(auR) m

%'

E Va"(.)

\

-m

such that Set(%')= Set(%)and [z, E'] # 8%. (b) If E # 8"R,for S > 0 there exists

ufn(asR) 00

E'

E Va"(E)\

-m

such that Set(%')= Set(z) and [z',x] # 8"R. Proof. Remember that i9"R = U(8"R : R E 72) where 8"R = [V"(z, R), 8Vs(z, R)], 8Va(z,R) = Vs(x, R) \ int(Vs(x, R)). Since 8Vs( z,R) is closed and nowhere dense in V:(z), then V= int(V'(z,R))\

u

RESet(z)

we see that if

Reset( I )

R

E

# 8"'R

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144

is an open neighborhood of z in V,"(z).Obviously, Set(z') = Set(%)for all z' E V and V n d " R = 0. Since 8% n V,"(z)is contained in a finite union of nowhere dense set of the form aVs(y, R), it is clear that a"R n V,"(x)is nowhere dense in Since each f-"(v,"(z)) can be covered by finitely many V:(x), we have that

v(x).

fn(aUR) n v;(B)

c pn(uZv,.(x) n a%)

is nowhere dense in v ( z ) . By Baire's theorem, U,f"(a"R)~V,"(z) is nowhere dense in V'(z). By Lemma 4.1.3, [z, ] : V,"(z) n N , -+ V,"(x)n N , is a homeomorphism. Therefore, [z, 1' $! a"R for some z' E V,"(z)\ U,fn(auR). 0 Lemma 4.3.10. Let R1, R2, R3 E R. Suppose that

Then x E P R . Proof. Lemma 4.3.1 gives

v = vU(x,fil) n vU(x,R

~3) ~-'V~(~(X),R~).

Since V"(f (x),R3) contains an open subset contained in every neighborhood of f(x) in V;(f(z)),V contains an open subset contained in every neighborhood of z in Vt(x).As the proof of Lemma 4.3.9 we can find z' E V such that x' $! 0%. Since R1 # R2, we have V c R1 n R2 c aR1. Since aR1 = a"R1 U PR1, we have x' E a"R1 and then V'(x', R 1 )c BUR1c a"R. Therefore, x E ayR. 0

Lemma 4.3.11. (a) Suppose t $! a"R. For R;E Set(f(z)) there i s a unique Rj = T,(Ri) E Set(z) such that A;j = 1. The map T, : Set(f(z)) + Set(z) is and Set(y) = Set(z), then Set(f(y)) = Set(f(z)) surjective. Ify E Va",(z)\d"R and Ty = T,. (b) Suppose z # 8%. For R; E Set(f-'(z)) there is a unique Rj = T:(R;) E Set(z) such that A;j = 1. The map Ti : Set(f-'(z)) + Set(z) is surjective. If9 E Vc(z)\ 8% and Set(y) = Set(z), then Set(f-'(y)) = Set(f-'(z)) and Tb = Ti.

Proof. By Lemma 4.3.8, for Ri E Set(f(z)) there is (wi)E .-'(f(z)) with wo E Ri and SO o-'((w;)) E ~ - ' ( z ) . Thus z E R,-,. Since (wi) E XA, we have A,-,,, - A,,,-]; = 1. By Lemma 4.3.10, w-1 is a unique since L # duR. Thus T, is well defined. If R; E Set(%),then there is ( z ; ) E .-'(z) with zo E R;. Then R,, E Set(f(z)) and Azozl = 1. Thus T,(R,,) = R,, and T, is surjective.

$4.3 Symbolic dynamics

145

Take y E V&(z) \ 8"R with Set(y) = Set(2). Then y E V"(z,Ri) for Ri E Set(z). If R j E Set(f(z)) and Ri = T,(Rj), then Lemma 4.3.1 shows that f ( ~E)fV'(z,Ri) c V"(f(Z),Rj) c RjThus Set(f(2)) C Set(f(y)). Symmetrically, Set(f(y)) c Set(f(z)). Thus Set(f(y)) = Set(f(z)). Since the definition of T, depends only on the sets Set(r) and Set(f(z)) (not z itself), we can easily check that Ty= TI.(b) is proved in the same way. 0 Let 61 > 0 be as in (B). For z E R, define

J,(z)= {V C R : 61 > V6,% E Va(z) \ 8% with Set(z) = D}, Ju(z)= {D C R : 61 > V6,3z E V,"(z) \ 8"R with Set(z) = V}, Note that we can choose 6 = 6(z) so small that Set(z) E Ja(z)if z E Vt(z) \ BUR, Set(z) E Ju(z)if z E %"(z) \ 8"R.

--

Let N > 0 and KN be the set of all n-tuples ( a l ,a2,. ,a,) of integers with 1 5 n 5 N , 1 5 ai 5 N and a1 2 a2 2 2 a,. Define a partial ordering 5 on KN by (al,... , a n ) 2 ( b , , . . . ,bm) if either n > m or N = m and bi 2 ai for all 1 5 i 5 m (note that this ordering is not the natural condition ai 2 bi). We write (al,... ,a,) > (bl,... , b m ) if (al,... ,a,) 2 (bl,..- ,bm), but ( a l , . . . ,an) # ( b ~ , . . ybm). . Let #E denote the cardinality of E and P(B)denote the family of subsets of B. Fix N 2 2211a.For L E P ( P ( R ) )we define L* E KN to be (al,... , a n ) where n = flL and al, ,a , denote the cardinalities of the n-sets in L.

...

---

Lemma 4.3.12. J , ( ~ ( z ) ) *5 J,(z)* and JU(f-l(z))* 5 Ju(z)*.

Proof. Define the map R i : J6(z) -+ Ja(f(z))by Ri(V) = Set(f(z)) for V E J,(x). Lemma 4.3.11 ensures that RZ is well defined. We show that R; is surjective. If € E J , ( f ( z ) ) ,by Lemma 4.3.9, for arbitrarily small 6 > 0 we can find w E Vt(f(z))\B"'R with Set(w) = € and w $! f(8"R).Then € = R:(Set(f-l(w))). ,D,,}, and Let J,(z)* = (al,...a,) where ai = #Vi and J6(z)= {VI,... let Ja(f(z))*= (bl,... ,bm) where bi = #Ei and J , ( f ( z ) ) = { € 1 , - - *,&}. Then RZ induces a surjection 9: [l,n] [1,m] by Ri(Vi) = € g ( i ) . Here we can consider the two cases : (al,... ,an) > (bl,... ,bm), n = m and g is bijective . 7~

> m and

Considering the second case, for a certain we have the surjection T, : €gg(i)-+ Vi and thus bg(q 2 ai. For any i there is j E [l,i] with g ( j ) 2 i. Otherwise

CHAPTER 4

146

g([l,i]) c [ l , i - 11 which contradicts that g is one-to-one. Then bi 2 b g ( j ) 2 aj 2 ai. Here we use the fact that all a; and bi are arranged in descending order. The second statement is proved in the same way. 0 We say that x E Ra is an s-branch (u-branch)point if there exist ( x i ) ,( y i ) E with xo = yo but 2 1 # yl (but 5 - 1 # y-1).

r-l(x)

Lemma 4.3.13. If 3c E Rs i s an s-branch point, then Ja(x)* > JS(f(z))*.If x E R, is a u-branch point, then J,(x)* > Ju(f-'(x))*.

Proof. Let x be an s-branch point. It is enough to show that one of T, with I E VQi,)(x) \ d"R is not bijective. Indeed, if (a1,- * - ,an)= (bl , * * ,bn), then g is a bijective and so n

n

n

n

i= 1

j=1

i=l

i=l

We here use that ai 5 bg(i). To have the equality we need ai = bg(i) for all i, i.e. T, is always a bijection. Since x is an s-branch point, we can find Ri E Set(%) and Rj,Rk E Set(f(z)) with Rj # Rk and Aij = Aik = 1. By Lemma 4.3.9 we can choose z E

{ v a R( i~) n,v ; , ) ( ~\ B"R, >)

and since

fV'(Z7Ri) c V ' ( f ( x ) , R j )n V " ( f ( z ) , R k ) (by Lemma 4.3.1), we have R,,Rk E Set(f(z)) and T,(Rj) = T,(Rk) = Ri. Therefore T, is not bijective. 0 Proof of Theowm 4.3.6. Let c = #'R and e be the length of the longest chain in the partially ordered set ( K N ,I ) .Let 0 5 n1 < n2 < < nk be all the nonnegative integers such that fnd(x) is an s-branch point. By Lemmas 4.3.12 and 4.3.13 Ja(fn*(3c))*

> Ja(fnl(x))* > ... > Ja(fnk(z))*,

which shows k 5 e. Let An be the set of all sequences ( R h o , . . .,Rk,) of elements of 'R such that there is (xi) E r - ' ( x ) with xi E Rki for all 0 5 i 5 n. By the definition of s-branch points, #An+l = #An unless fn(z)is an s-branch point. Thus UAn+l 5 c#An for every n. Since UAo 5 c, we have

#An 5 ck+l 5 ce+' for all n 2 0. Thus there exist at most ce+' possibilities for ( x i ) r with ( 2 ; )E

?r-'(x).

Similarly there exist at most ce+l possibilities for (xi)!m with (xi) E r - l ( x ) . Therefore, # r - ' ( x ) 5 ~ ? ( ~ + l0 ).

CHAPTER 5 Local Product Structures

In the previous chapter we have seen that every TA-homeomorphism of a compact metric space has a local product structure, and such a homeomorphism is represented as a factor of a symbolic dynamics. For a general TA-map of a compact metric space the shift map of the inverse limit system is a TA-homeomorphism (by Theorems 2.2.29 and 2.3.8) and hence it follows that every TA-map is represented as a factor of a symbolic dynamics through the inverse limit system. The purpose of this chapter is to establish a local product structure theorem for TA-maps under some assumptions. The structure will play an important role for a deep understanding of dynamics of TA-maps, as well as TA-homeomorphisms. In the final section we divide the collection of TA-maps into five classes, each of which exhibits a topological characteristic. $5.1 Stable sets in strong sense Let f: X t X be a continuous surjection of a compact metric space. As before, for x E X and E > 0 we denote as W,S(x)the local stable set { y E X : d ( f i ( x ) ,f i ( y ) ) 5 &,i 2 0). A finite sequence x = x o , x 1 , - . - ,x, = y of points in X is called an €-chain of s-direction from x to y if zi+l E W:(xi) for all 0 5 i 5 n - 1. It is clear that if 2 = 20, zl, ,x, = y is an &-chainof s-direction, then the sequence is also an &'-chain of s-direction for E' > 6. We write x y if for any E > 0 there is an €-chain of s-direction joining x and y . It is easily checked that N is an equivalence relation in X. For x E X we denote as Wa(x) the equivalence class of x and call it to be the stable set in strong sense. By definition the family

-..

N

F; = (W'(2) : 2 E X } is a decomposition of X. If f is c-expansive, from Lemma 2.4.1 it follows that w ' ( x ) is actually a subset of the stable set W'(x) = {y E X : d ( f i ( x ) ,f i ( y ) ) + 0 as i -+ 00). By the fact that f(W:(x)) c W,d(f(x)) for all x E X, clearly f(x) f(y) if x y . Thus an inclusion f (T@'(z)) C W'( f (x)) always holds.

-

-

Theorem 5.1.1. Let f : X t X be a positively expansive map of a compact metric space. Then T@'(x) = {x) for all .2: E X .

-

Proof. Let e > 0 be an expansive constant for f. By definition it follows that W,"(x)= {x} for all x E X,which implies that x = y if x y . Thus the conclusion is obtained. 0 147

CHAPTER 5

148

Theorem 5.1.2. Let f:X -+ X be a continuous surjection of a compact metric space. Iff is a local homeomorphism, then for x E X there is k 2 1 such that the restriction f: @"(x) + @"( f (x)) is a 12-to-one surjection.

Proof. Let (A,e) be Eilenberg's constants for f . For 17 > 0 take 0 < E < min{A,q} as in Theorem 2.1.1 (4). Suppose z E @"(f (x)). By definition there is an €-chain yo, y1, * * ,y, of s-direction from f (3) to z. Since diam(W,d(yo)) < A, by Theorem 2.1.1 the set W,"(yo) can be lifted by f to a set BO with the property that x E Bo and diam(B0) < 7. Then BO C W,(z). Since yl E Wi(y0) and f (Bo)= W,"(yo), there is a point z1 E W,"(x)with f (21) = y1. Next we lift by f the set W,"(yl) to a set B1 such that z1 E B1 and diam(B1) < q, and take a point 2 2 E W,"(zl) with f(z2) = y2. Continue this ,x, is constructed, so that f(xn) = z process, a sequence x = SO, xl,zz, and xi+l E W,"(ti) for 0 5 i 5 n - 1. Therefore we have an 7-chain of Since f-l(z) is finite, we can assume that x, is s-direction from x to 5., independent of the choice of 17 > 0. Hence x,, E @'(x). Since z is arbitrary, we obtain that f: w ' ( x ) + @'(f(x)) is surjective. Let {xl,--,zk} be the inverse image of f(x) by f:@"(x) @(f(z)). For a E @'(f(x)) there is an E-chain f ( x ) = yo,y1,--. ,yn = z of s-direction where 0 < E < min{A,q}. .In the . same way as above, for each 1 5 j 5 k construct an q-chain xj = xi,xi, ,xi of s-direction with f (xi)= z . Then xi E @(xi) = @(x) for 1 5 j 5 k . By the choice of 8 we have that d(zi,xjn)) 2 20 if j # j'. Therefore the cardinal number of the inverse image of z by f : *(x) + ~ "f (x)), ( say l , is not less than k. Changing roles of z by f(x), we have l < k, and thus l = k. 0

-

---

--

Theorem 5.1.3. Let f:X-+ X be a TA-covering map of a compact metric space. If X i s locally connected, then for every x E X (1) wa(x) is the path connected component of x in W " ( z ) , (2) letting F = {y E X : 3i 2 0 s.t. f ( y ) = f"(x)} one has W'(x) =

u

W(y).

v€F

We recall that if a compact metric space is locally connected, then the space is locally path connected (see Theorem 2.1.4). For the proof of Theorem 5.1.3 we prepare the following proposition.

Proposition 5.1.4. Under the assumptions of Theorem 5.1.3, there is eo > 0 such that Weo(x)C ~'(z) f o r x E X . Furthermore, for 0 < E 5 eo there is 0 < 6 < E such that each point x in X has a path connected set C satisfying the following relation : W,s(x)c c c WEd(x). For E > 0 a finite sequence x = xo,x1,. . ,z, = y of points in X is an &-chain from x to y if d ( z i , z i + l ) < E for all 0 5 i 5 n - 1. For a continuous

-

$5.1 Stable sets in strong sense

149

surjection f:X --t X let Xf denote the space { ( x i ) E Xz : f ( x i ) = x;+l, i E Z} of the inverse limit system. The following lemma is easily checked (c.f. Theorem 2.1.1). Thus we leave the proof to the readers. Lemma 5.1.5. Let f:X + X be a local homeomorphism of a compact metric space. T h e n f o r ~ > O a n d n > O t h e r e i s a > O s u c h t h a t i f x o , x l , ~ ~ ~is, x m an a-chain of points in X,then for (x:) E Xf with xg = xo there is a sequence (x:), ( x t ) , ,( x r ) of points in Xf such that

- -.

(1) x j = x j o f o r ~ < j < m , (2) d ( x i , x i + l ) < E for -n 5 i

5 0 and 0 5 j 5 m - 1.

Lemma 5.1.6. Let X be a compact locally connected metric space and f:X --t X a local homeomorphism. If u : [0,1] + X is a path, then for x = ( x i ) E Xf with xo = u ( 0 ) there is a path v : [0,1] 3 Xfsuch that v ( 0 ) = x and po ov = u where po : Xf + X is the natural projection to the zero-th coordinate.

Proof,Notice that f is a covering map (c.f. Theorem 2.1.1). By the following Theorem 5.1.7 it follows that for each i 0 there is a path ui : [0,1] -+ X such that u;(O) = xi and f o u; = u. For i > 0 let u; = f' o u : [0,1] + X, and define v : [0,1] + Xf by v ( t ) = (ui(t)). Then it is easily checked that v satisfies the desired conditions. 0


0 such that .([a - 5,a 51) C U. Take t E [a - 5,a 51 with t < a. By definition, t E J and u ~ [ ~ has , ~allift E I [ ~ by , ~ pI such that ?il[o,tl(0)= y. Since - u(t) E U , there is an open set 0 of Y such that E ~ p , q ( t E ) and p l :~U + U is a homeomorphism. Let v : [O,a 51 --t Y be defined by v ( s ) = ?i~[O,~](s)for

+

+

+

CHAPTER 5

150

+

0 I s 5 t and v(s) = p d o u(s)for t 5 s 5 a 6. Then v is a continuous map IU satisfying p o v = u on [0, a 61. Thus a 6 E J. This implies J = [0,1]. 0

+

+

,

0

1

1"

-a I

/

X

'U

Figure 18 As before, for E {yo E X : 3(yi) E

> 0 and ( z i ) E Xfdefine the local unstable set W,"((zi)) = Xf s.t. d(zi,y;) 5 ~ ,Ii0). Compare the following lemma

with Lemma 2.2.49.

Lemma 5.1.8. Let f:X + X be a continuous surjection of a compact metric space. Iff is c-expansive and e > 0 as an expansive constant for f, then for 0 < E < e there i s 6 > 0 such that for all z E X and (zi)E Xf with zo =

where &(z) = {y E X : d ( z , y ) 5 6). Proof. By Lemma 2.4.1 there is n > 0 such that f"(W,"(z))c W,6(fn(z)) for all z E X, and f"(W,"((zi-,))) 3 W,"((zi))for d (z;)E Xf.Since f is uniformly continuous, clearly there is 6 > 0 such that if d ( z , y ) 5 6 then d(fi(z), f'(y)) 5 E for 0 5 i 5 n. Then we have the above first equality. The second equality is checked as follows. By definition for yo E W,U((zi)) there is (y;) E Xf such that d(z,,yi) I e for i 5 0. Since e is an expansive constant, by c-expansivity it follows that each yi is unique for yo, which implies that for i 5 0 an inverse map of f-' is defined by yo H yi on W,U((si)).Then, by c-expansivity the continuity of the inverse map is easily obtained. Since W,U((zi)) is compact, for 6 > 0 small if yo,zo E W,U((zi)) and d(yo,zo) I 5 then there is (2;) E Xf such that d(yi,zi) 5 E for -n 5 i 5 0. Thus the second equality also holds. 0

Let e > 0 be an expansive constant for a TA-map f and put EO = ef4. Since f has POTP, there is 0 < 60 < ~ 0 / 2such that every 60-pseudo orbit

$5.1 Stable sets in strong sense

of f is ~0/2-tracedby some point in d ( z 0 , y) < 60. Then a sequence

{. * .

9

2-29 2 - 1 9

Xf.For (zi)E Xfand Y, f (I/), f 2 ( Y ) ,* *

151

y E

X suppose

1

is a 60-pseudo orbit of f , and hence it is ~ o /Zt r a c e dby some (zi) E XI.By expansivity it follows that such a tracing point is unique, which implies that WG((zi)) n Wio(y) consists of exactly one point zo. We write [(q),y]‘ = (zi) and [(zi),y] = zo. Thus the following maps are defined.

where A(&) = {((zi),y) E Xf x X : d ( z 0 , y ) < 60). Compare with the maps in $4.1 of the previous chapter.

Lemma 5.1.9. Under the above notations one has the following (1) [[(2i),Y]’,z] = [(li),z] and [ [ ( z i ) , ~ ]t’ti),^]]' , = [ ( z i ) , w ] ’ whenever the two sides of the relations are defined, (2) [ , 1’ : A(&) + Xf is continuous, ( 3 ) [ , ] : A(&) + X is continuous, (4) for E > 0 there is 0 < 6 4 60 such that d ( z 0 , y ) < 6 implies diam{zo,y,

((4, Y11
0 there is n > 0 such that

whenever d(zi,yi) 5 E~ for -n 5 i 5 0. Proof. By c-expansivity (1) is clear. (2) is checked as follows. Suppose a sequence {(Xny Yn)}r=o of A(6o) converges to (x, y) E A(&). Let 2, = [ x ~~ ,n ] ’ and let Zn = (zi,,) and X, = (z;,,). Since Xf is compact, there is a subsequence {znj) that converges to some z = ( z i ) E Xf.By the definition of [ , ]’, we have d ( ~ i , , ~Zi,n,) , 5 EO for i 5 0 and n j , and SO d(zi,zi) 5 €0 for i 5 0. Similarly we have d( f ‘(y), zi) 5 EO for i 2 0. Thus (2,) converges to [x,y]. Therefore (2) holds. Since [ , 1 = po o [ ]’, we have (3). (4) is obtained by using POTP.(5) is checked as follows. Let UO = [(zi), z] and vo = [(yi), z ] . By definition U O ,vo E W:,,(z). Also, there are (ui),( ~ i )E Xf such that d(zi, ui) 5 EO and d(yi, vi) 5 EO for i 5 0. For 7 = E take a number n > 0 as in Lemma 2.4.1. If d(si,yi). 5 EO for -n 5 i 5 0, then by the above facts it follows that d ( f j ( u - , ) , f J ( V - , ) ) 5 3 ~ < 0 e for j 2 0. Therefore V-,, E W : ( U - ~ )and , by Lemma 2.4.1 (l),vo E W:(uo). 0 Proof of Proposition 5.1.4. Let [ , ] : A(&,) + X be the map as in Lemma 5.1.9 and suppose X is locally connected. Let U ( z ) be the connected component of x in the &-open neighborhood Ua,(2).Then there is 7 > 0 such that U ( z ) 3 U,(z)for all 2 E X. For E > 0 take 0 < 6 5 60 and n > 0

CHAPTER 5

152

as in Lemma 5.1.9 (4) and (5) respectively. For €0 and n let a > 0 be as in Lemma 5.1.5. Suppose z E W:,,(z) n V(z). Since V(z) is connected, there is an a-chain z = u:,ui,--, u r = z in V(z). By Lemma 5.1.5 we can take points ( u ~ ) , ( u f ) , . ,(ur) -. E Xf such that d ( ~ { , u { + < ~ ) €0 for -n 5 i 5 0 and for 0 5 j 5 m - 1. Put y j = [ ( u ; ) , ~ ]Then . y j E Wio(z), and hence yj+l E W&,,(yj). By Lemma 5.1.9 (5) we have d(yj,yj+l) < E , and by Lemma 5.1.8, yj+l E W,S(yj) for some E' > 0. Thus the sequence z = yo, yl , ,ym = z is an €'-chain of s-direction. Since E' can be taken arbitrarily small, we obtain z E fia(z), and therefore W:? n V(z) c fia(z). Since V(z) 3 Vr(z),it follows that W:,,(z) n U,(z) c Wa(z). By Lemma 5.1.8 we have W:,,(z) c for some 0 < eo 5 € 0 . Notice that the number eo does not depend on a point 2 E X. Next we prove that there is a path connected set C(z) such that W.,(z) C C(z) c W&,,(z). To do this, let V(z) be as above. Since V(z) is path connected, for z E W:,,(z) n V(z) we can take a path u : [0,1]+ V(z) joining z and z. By Lemma 5.1.6 there is a path v : [0,1]+ Xf such that v(0) = x and PO o v = u. Define w, : [0,1]+ X by w Z ( t )= [ v ( t ) , z ] . Then w,(O) = z, w,(l) = z and w, (t) E W&,(z) for t E [0,1].Letting

- .-

w(z)

we have that W;,(z) C C(z) C Wie0(z). Let 0 < E 5 eo. Then by Lemma 2.4.1 there is n > 0 such that f"(W&,,(z)) c W:(fn(z)) for all z E X. Since f" is a local homeomorphism, it is easily checked that there is S > 0 such that f"(W:,,(z)) 3 W,b(f"(z)) for all z E X. For z E X take y E f-,,(z) and put C = f"(C(y)). Then it is clear that C c W:(z). Since C(y) 3 W:,,(y), it follows that W,"(z) C C. 0 + X be a c-expansive continuous sujection of a compact metric space and let e > 0 be an expansive constant for f . For x E Xf and z E X the intersection W&(x) n Wa(z) is at most countable.

L e m m a 5.1.10. Let f : X

n Wa(z), by Lemma 2.4.3 there is n > 0 such that fn(z) E Wed/2(fn(z)). Put F,,= fn(W&(x)). To show that for each n 2 0, F,, n W:/,( fn(z)) is finite, it suffices to prove that F,, is contained in a finite Proof.If

z E W&(X)

union of some local unstable sets W&(yi). Since f is uniformly continuous, there is E > 0 such that if d(a,b) 5 E then d(fi(a), fi(b)) 5 e/2 for 0 5 i 5 n. Then we have f"(W:(y)) C W&(u"(y)) for all y E Xf,where u : Xf + Xf is the shift map. Let 6 > 0 be as in Lemma 5.1.8. By compactness there is a finite S-dense subset {y1,y2,--- }'y, of WG2(x).Let x = (zi) and take y j = (yf) E Xf such that 3 = y j and

$5.1 Stable sets in strong sense

d(yf,zj)

153

5 e/2 for i 5 0. Then we have

and hence W,U12(x) C

W,U(yj). Therefore, F,

c U:=l

W,u/2(on(y')). 0

Proof of Theorem 5.1.3. Let eo > 0 be as in Proposition 5.1.4 and let y E W"(z). Then by Lemma 2.4.3, f n ( y ) E W:o(f n ( z ) )for some n > 0. By Proposition 5.1.4 we have f n ( y ) E W"(f"(z)), and so Wd(fn(y)) = W " ( f " ( z ) ) . Since f " : *(y) --* W d ( f n ( y ) ) is surjective by Theorem 5.1.2, there is P E @'"(y) such that fn(z) = f"(z), and hence z E F. Therefore (2) holds. To prove ( l ) , we show first that each of W"(z) is path connected. By Proposition 5.1.4 there is 6 > 0 such that for each z E X we can take a path connected set C(z) satisfying W,"(z)c C(z) c W&(z). Let z E W"(z). By definition there is a 6-chain x = x o , x 1 , ~ ,x, ~ ~= P of s-direction. Then zi+l E C ( z i ) c W:o(zj) c fi"(zi) for all 0 5 i 5 n - 1. Therefore there is a path u in X joining z and z such that u([O,11) c W " ( z ) . Let u : [0,1] + X be a path such that u(0) = z and u([O,l]) c W " ( z ) .Let [ , ] : A(&) + X be as in Lemma 5.1.9. Clearly there is 0 < t 5 1 such that u([O,t])is contained in the 60-open neighborhood of z. Take x = ( x i ) E Xf with zo = z and define v : [O,t] + WG(x) by v ( s ) = [x,u(s)]. Then v is continuous by Lemma 5.1.9. By the definition of [ , ] clearly v ( s ) E Wio(u(s)). Since u(s) E W'(z), we have v ( s ) E W"(z), and by Lemma 5.1.10 it follows that v ( s ) = z. Hence u(s) E W,"o(z).Since u is continuous, by applying Lemma 5.1.8 we can take 0 < to 5 t such that u(s) E W&(z) C W"(z) for 0 5 s 5 t o . Hence, letting I = {s E [0,1] : u ( s ) E W"(z)}, we have [ O , t o ] C I. In the same way, it is easily checked that 1 is open and closed in [0,1]. Thus u([O,11) c W"(2).0 Let f : X + X be a TA-covering map of a compact locally connected metric space. Let eo > 0 be as in Proposition 5.1.4. Then W:o(y) C for y E @"(z). By this fact together with the following general properties of local stable sets we can introduce a topology in Wa(z)such that for y E W'"(z) the collection {W,"(y) : 0 < E 5 eo) is a fundamental neighborhood system on the base of neighborhoods of y.

The topology is called the intrinsic topology of W s ( z ) .Notice that the intrinsic topology is not weaker than the restriction of the topology of X .

Lemma 5.1.11. Let eo > 0 be as in Proposition 5.1.4. Then for 0 < E 5 eo and z E X the restriction to W,"(z)of the intrinsic topology is consistent with

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the restriction to W,"(x)of the topology of X . Moreover, there is 6 > 0 such that W,"(x)nV6(x) is an open set of Wa(z) under the intrinsic topology, where u6(.) denotes the 6-open neighborhood in X . Proof. Suppose a sequence { y i } converges to a point y on W . ( z )under the topology of X. Let 7 > 0. By applying Lemma 2.4.1 we can take 6 > 0 such that W ; ( y ) 3 W,"(z) r l u 6 ( y ) (see Lemma 5.1.8). Hence yi E W ; ( y ) for large i. Conversely, if {yi} converges to y under the intrinsic topology, then the convergency under the topology of X is easily obtained. By Lemma 5.1.8 there is 6 > 0 such that W,"(x)n Ua(z)= WEb/2(z) for all x E X . Then it follows that W,"(y) C W,"(z)r l u6(Z) for sufficiently small 7 > 0 if y E W,"(z)n u6(Z). 0 Theorem 5.1.12. Let f : X + X be a TA-covering map of a compact locally connected metric space. Then for x E X the intrinsic topology of W " ( z ) satisfies the following properties: (1) locally path connected, (2) path connected, (3) first countable, (4) second countable. Proof. (1) follows from Proposition 5.1.4 and Lemma 5.1.11. (2) is easily checked from Theorem 5.1.3. (3) follows from the definition. To prove (4), let eo > 0 be as in Proposition 5.1.4 and let x E X . Since f has POTP, there is 6 > 0 such that every 6-pseudo orbit of f is eoJ2-traced by some point in Xf.Let {xl,x2, ,z'}. be a 6-dense subset of X and choose x j = (zi) E Xf (1 5 j 5 k) such that xi = x i . By the choice of 6 for y E w a ( z ) there is x j such that W,"oIz(y)nW,uO12(x3) consists of one point, say z. Then y E W,"o12(z). Therefore, by Lemma 5.1.10 it follows that wa(x)is expressed as the union of at most countable local stable sets W:o,2(z).Therefore (4) is obtained from Lemma 5.1.11. 0

-

We say that a TA-map f : X + X of a compact metric space is s-injective if the restriction f : p a ( x )+ Wa(f(x)) is injective for all z E X. If f is a self-covering map, then by Theorem 5.1.2 we have that f: Wa(x) -+ Wa(f (z)) is bijective whenever f is s-injective. For this case, if X is locally connected, -+ I@.( f (3)) is a homeomorphism under the intrinsic it follows that f : I@('.) topologies.

Remark 5.1.13. If a compact metric space X is locally connected, then clearly X splits into a finite union X = X I U X Z U . U Xt of connected components. In this case we have that f " ( X i ) = Xi (1 5 i 5 t ) for some n > 0 if f : X + X is a continuous surjection. In the next Chapter 6 we will introduce fundamental groups and universal covering spaces. It can be proved that a TA-covering map f:X+ X of a

..

$5.2 Local product structures for TA-covering maps

155

compact locally connected metric space is always s-injective in the case where X has the universal covering space (see $6.4 of Chapter 6). However, it remains a problem of whether it is true for general cases.

$6.2 Local product structures for TA-covering maps

In this section we shall establish a local product structure theorem for sinjective TA-covering maps of compact locally connected metric spaces. Let f : X -+ X be a continuous surjection of a compact metric space. As before, we denote as u : Xf + Xf the shift map of the inverse limit system and i18 po : Xf + X the natural projection to the zero-th coordinate. Let A(€) = {((zi),y) : d ( z 0 , y ) 5 E } for E > 0. We say that f has a local product structure if the following conditions (A), (B) and (C) are satisfied : (A) There is 60 > 0 and a continuous map [ , ] : A(&) --* X such that (1) [ ( z i ) , ~=~ 20 ] for all (4 E Xf, (2) f ([x,y]) = [u(x),f (y)] whenever the two sides of the relation are defined. (B) For each x = ( z i ) E Xf there is a subset N , of X with zo E N,, and a continuous section S, : N, + Xffor the projection PO, that is, PO o S, is the identity map on N,, such that

(3) Sx(z0)= x, (4) [Sx(z),YIE N x for Z,Y E N x , (5) for z,y, z E N,

(C) There is e

> 0 such that for x = (xi) E Xf,letting

the following property holds: (6) a, : D"(z0) x D"(x) + N, is a homeomorphism.

Theorem 5.2.1. Let X be a compact connected locally connected metric space. If a TA-covering map f : X + X is s-injective, then f has a local product structure satisfying the following conditions (1) { [ ( ~ i ) , ~=l }w,"((zi)) n w : ( ~for ) ( ( z i ) , E~ A(60)i ) (2) each of N, i s a connected open set of X ,

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( 3 ) there is p > 0 such that N, 2 B,(zo) for x = (xi) E Xf where B&o) = {Y E x : d(xo,y) 5 P } , (4) for x = (xi) E Xf

Furthermore, iff i s expanding then D"(z0) = (20) and D"(x) = N , for Xf,and otherwise (5) D"(ZO) (20) and D"(x) {xo} for x = (xi) E Xf.

x = (zi) E

2

2

Before starting the proof we prepare the following theorem.

Theorem 6.2.2 (Uniqueness of lifting). Let p : Y -t X be a covering map and let Z be a connected topological space. For a continuous map f : Z -t X --I suppose f , f : 2 -+ Y am lifts o f f by p . If 7(x:0)= ~ ' ( z ofor ) some point zo 4 in Z,then 7 = f .

7(x)

P m f . Let 0 = {x E 2 : = ?'(x)} and 0' = {z E Z : 7 ( x ) # T'(x)}. It is sufficient to show that 0 and 0' are both open in Z. Let x E 2. Since p is a covering map, by definition there is an open neighborhood -U of f ( s ) such that p - ' ( U ) splits into a disjoint union p-'(U) = UxGn U X of open 6 VX and T'(z) E VXt for some X,X' E A. Clearly sets of Y. Then V = f (UX) n p-'(vAt)is an open neighborhood of z in Z. It is easily checked that X = A' and V c 0 if 2 E 0, and X # A' and V c 0' if x # 0. Therefore 0 and 0' are open in 2. 0

--'

7(x)

Proof of Theorem 5.2.1. Let f : X -t X be as in Theorem 5.2.1 and choose eo > 0 as the number satisfying the property of Proposition 5.1.4. Notice that eo is an expansive constant for f . As in the previous section, we take 60 > 0 small for EO = eo/4 and define a map [ , 1' : A(&) + Xf by assigning

55.2 Local product structures for TA-covering maps

157

((zi),y) to a unique point ( z i ) with the property that d ( z i , z i ) 5 €0 for i 5 0 and d(f'(y), zi) 5 €0 for i 2 0. By Lemma 5.1.9 (2) it follows that the map is continuous. Letting [ , ] = po o [ , ]', we have a continuous map [ , ] : A(S0) + X. Since

(54

{ [ ( z i h y ] )= W,",((zi)) n W & b ) = W:((zi))nW,",b),

(1) of Theorem 5.2.1 holds. From definition the condition (A) is easily checked. To show (B) and (C),we prepare three claims. First, by Lemma 5.1.9 (4) choose 0 < €1 < 60/4 such that if d ( z 0 , y ) 5 2 ~ 1then diam{[(zi),y],zo,y) < 60/4. Next take and fix 0 < el < €1 as in Proposition 5.1.4. Then, for each z E X there is a connected set C(z) satisfying W,.,(z) c C(z) c W,.,(z) c W'(2).

Xj and

C(z0) be as above. Since f is s-injective, for each I@'(zi) + W'(z0) is bijective. Since C(z0) c fi'(zo), we define a map yi : C(z0) + fi'(zj) by the formula f-i(yi(y)) = y. Then ( y i ( y ) ) E Xj for Y E C ( ~ O ) -

Let x = ( z i ) E

i E Z the restriction

Claim 1. For each i E Zthe map yi : C(z0) + p'(zi) is continuous (under the topology of X). Indeed, since f-' : Wd(zi) + Wd(z0) is a homeomorphism under the intrinsic topology, it follows that yi : C(z0) + l@'(zi) is continuous under the intrinsic topology. Since C(z0) c We",(zo),by Lemma 5.1.11 the conclusion is obtained. For z E W,U,(x),by definition there is ( z i ) E Xf such that zo = z and d(zi,z i ) 5 el < EO for i 5 0. Since f is c-expansive, such a point ( z i ) is unique for z. So we write ( z i ( z ) ) = ( z i ) , and for i a map zi : W,U,(x)+ X is obtained. By c-expansivity it is easily checked that each the map zi is continuous. Let (y,z) E C(z0) x W,U,(x). Then d ( y , z ) < 6. Since (yi(y)) E Xj and yo(y) = y, a point [(yi(y)),z)' in Xj is defined. We write (wi(y,z)) = [(yi(y)),z]' and consider for each i a map wi : C(z0) x W,U,(x)---f X. From

Claim 1 together with the fact that [ , ] is continuous, it follows that each w ; is a continuous map. By definition wo(y,z ) = [(yi(y)), z ] , and hence by (5.1) we have for z E W,U,(x)

D ( z ) = {wo(y,z ) :y E C(z0)) c W,.(z) c W'(%).

Claim 2. For (y, z ) E C(z0) x W,U,(x)one has wi(y, z ) E I?ld(zi(z)) for all i E Z. Fix z E W,U,(x).Since 20 E C(zo), by definition wi(20,z)= z i ( z ) . Clearly f-; o wi(y, z ) = wO(y,z). Thus each wi(y, z ) is a lift of wo(y, z ) by f-' such that wi(z0, z ) = z i ( z ) .

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On the other hand, since D ( z ) c wa(z), by s-injectivity for each point wo(y,z) E D ( z ) there is a unique point w:(y,z) in Wa(zi(z)) satisfying the formula f-i(w:(y,z)) = wo(y,z). Thus a map w: : C(z0) wa(.zi(z)) was defined. Notice that the point z is fixed. Since W O is continuous, as in Claim 1 the continuity of w: is easily checked. Since f-i o w:(y, z ) r= wo(y,z ) , the map w: is also a lift of wo by f - ’ . By definition w:(zo,z> = z i ( z ) . Since C(z0) is connected, by the uniqueness theorem for liftings we have wi(y, z ) = w:(y, z ) for all y E C(z0). Thus wj(y,z) E wa(zi(z)).

Claim 3. Fix z E W,U,(x).Then the map wo : C(z0) --t D ( z ) is injective. Indeed, suppose wo = wo(y,z) = wo(y’,z) for some y,y’ E C ( Q ) and let (y)= [(yi(y)),z]‘ and ( w : ) = [(yj(y’)),z]’. By Claim 2 we have wi,wi E W ” ( z i ( z ) ) .Since wo = wb,by s-injectivity of f it follows that wi = w: for all i, and so (wi)= (w:).By the choice of €1 we have d(wo,zo)< 60. Hence by Lemma 5.1.9

Since el < ~ 1 ,by the choice of ~1 we have diam(R,) < 60. The following properties are easily checked by definition : for all y - k W:;(zo) and all z E

W,u,(4 Since a, = wo,clearly ax

: W:l

(50) x

W,u1(x)

+

Rx

is a continuous map. If a , ( y , z ) = a,($, z’) = w,by c-expansivity it follows that z = z’, and hence y = y’ by Claim 3. Therefore a, is injective, which implies that a, is a homeomorphism. For a E R,, take (y,z) E W:l(z~) x WC(x)such that a = a x ( y , z ) , and define Sx(u) = [(yi(y)),z]’(= (wi(y,z))). From the above results it follows that S, : R, + Xf is a continuous map. Since

S, is a section on R, for PO. Hence, by Lemma 5.1.9 we have that Xf satisfies the condition (B).

S, : R, +

55.2 Local product structures for TA-covering maps

159

Next, we prove that there is p' > 0 such that R, 3 BP,(z0).Since f has

POTP,there is p' > 0 such that every p'-pseudo orbit is el-traced by some point in Xf. Suppose w E BPf(zo)and let ( z j ) = [ ( z i ) , ~ ]E' Xf. Then (10) = W,U,((zi)) n W:l(w). Since W,S(w)c ~ " ( Z O ) by , s-injectivity we can take a unique (wi)E Xf such that wo = w and wj E W ' " ( z j ) for i E Z. Since d(w,zo) < p', it follows that W,U,((wi))n W,",(zo) consists of one ponit, say yo. Since yo E W,",(zo) C ~ " ( z o )we , choose (yi) E Xfsuch that yi E p " ( z i ) for all i. In the same way as the proof of Claim 2, we obtain d(w;,yi) 5 el for all i 5 0. Hence w E W,U,((yi)), and so w E W,.l ((yi)) n W:l (10). This implies w E R,. Let M, denote the interior of R, in X and suppose a , b E M,. Then there are y,y' E W:l(z~), Z,I' E W,U,(x)such that a = a , ( y , x ) and b = a x ( y ' , 2'). Letting c = a x ( y ' , z)(= wo(yi(y'), I)), we can show c E M,. Indeed, for c = S x ( c ) = (w;(y',z)) E Xf take el > E > 0 sufficiently small and in the same manner as above construct Rc(e) for E. Since diam(R,) < 60, we have diam(R, U Rc(c)) < 60 for E small. Since a E M,, by applying Lemma 5.1.9 we have [x,W,"(c)]C WeU,( x ) whenever E is small. This implies W,U(c)C a,(y', W,U,(x)).On the other hand, since b 6 M,, there is 7 > 0 such that W,d(b)c M,. Then [ S x ( w ) , c ]E R, for w E W,"(b).Since [Sc(c),b]= [c,b]= b, if E is taken small, by the continuity of [ , ] we can define a map W,"(c)+ W;(b)by v H [Sc(v),b].In the same way as the proof of Claim 2, we have [Sx[Sc(v),b],c] = v, which implies W,"(c)c a x ( W : l ( z ~ ) , zFrom ). this result together with the above fact it follows that Rc(c) c R,. Since Rc(c) contains a neighborhood of c in X ,we obtain c E M,. By the above result the inverse image of M, by ax can be written as V"(z0)x V u ( x ) . We denote as N, the connected component of 20 in M,. Then it follows that a, : D"(z0) x D"(x) + N , is a homeomorphism for some D " ( z 0 ) c V " ( z 0 )and D"(x) c V"(x). Since X is locally connected, (2) of Theorem 5.2.1 is obtained. Clearly (3) of Theorem 5.2.1 follows from definition. Restricting the section S, to N,, we have (B). (C) is clear from the construction. To prove (4) of Theorem 5.2.1, first we show that D"(z0) is the connected component of 2 0 belonging to the interior of W:l(z~) c ~ " ( z ounder ) the intrinsic topology. Since D"(z0) = W:o(z~) n N , and N, is open in X,for y E D"(z0) we have W,d!y) c D"(z0) if 7 is sufficiently small, which means that D"(z0) is open in W"(z0). The connectivity follows from the fact that N , is connected and a, : D"(z0) x D"(x) + N , is a homeomorphism. Hence D"(zo)is contained in the connected component, say D', of 20 belonging to the interior of Wi1(zo) under the intrinsic topology. Conversely, let c E D'. Then D' 3 W,"(c)if E > 0 is small. As above define Rc(&)where c = S,(c). Then it is easily checked that R c ( ~ c ) R,. Thus c is a point in M , which is the interior of R,, and so c E W:l(c~). This implies D' C D"(z0). Therefore

D' = ~ " ( z o ) .

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Since f(W,",(zo)) c W,q(zl) and f : @"(z0) + @"(z1) is a homeomorphism under the intrinsic topology, by the above result f(D"(z0))is connected and open in W,l (zl),and hence f(D"(z0)) c D"(z1). From this fact it follows that N = a,,(,)(f(D"(zo))x DU(a(x)))is a connected open subset of Nu(,).Take a lift R of N by f such that 20 E Then we have D'(z0) = Wi1(zo)fl and

w.

W 4 X ) )

c

w,-,(+))

where f ( S ( o ( x ) ) )= D"(o(x))and f(Tl(a(x)))= W,U,(o(x)). Therefore

-

N C Rx = ax(W,d,(ZO) X W.",(X)),

w

from which we have c N, because is connected and open in X. Thus D (a(x)) c D"(x),and so we obtain f(Du(x))3 D"(o(x)). To prove the last part of Theorem 5.2.1, let X" = {z E X : D " ( z ) = {z}}. Then X" is open in X. Indeed, if E X" then

11

N , = a x ( D d ( z x) DU(X)) = ax({.) x D"(x)) = D"(x) c W.",(x), and for y E N,

W Y ) n N x c W 1 ( Y ) n W.",(x) = {v}. Since D"(y)nN, is open in Dd(y)and D'(y) is connected, we have D"(y) = {y} and hence y E X",i.e. N, c X". Since B p ( z )C N, for z E X (by (3)), X" is closed and therefore X" = X unless X" = 0. If X"= X, then we have N, = Du(x) for x E Xj. Then, by using c-expansivity, we obtain that f : A4 t M is positively expansive, that is, expanding (because f is an open map). Otherwise, X" = 0. It is easily proved that

xu=

(20

E

x : D"((.i)) = ( 2 0 ) )

is open and closed in X. Supposing Xu# 0, we have Xu= X and then

x = W,",(z1)u *..u We"l(zm) ..

for some 51,. ,zm E X. It follows from Lemma 2.4.1 that fn(X) # X for some n > 0, thus contradicting. 17 Let f: X + X be an s-injective TA-covering map of a compact connected locally connected metric space. For x = (zi) E Xj let a, : D"(z0)x Du(x)-+ Nx be as in Theorem 5.2.1. Then we have the following lemmas.

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161

Lemma 6.2.3. Let S , : N , + Xf be a continuous section for po such that S,(zo) = x. For k 2 0 let p-k : Xf + X be the natural projection to the -12-th coordinate and define

Then z-k E N,(-k) and the restriction f k : N,(-k) phism.

+ N,

is a homeomor-

Proof. This is easily checked from the definition of Sx. 0

Lemma 6.2.4. Let p x = ( Z i ) E Xf

> 0 be the number in Theorem 5.2.1 (3). Then for all

Proof. Let w E W;(ZO). Since w E B,,(zo) c N,, there is (y,z) E D"(z0) x D"(x) such that a&,%) = w. Then w E Wt,(z). Since p < E O , we have z E W&,(zO)nW,U,((zi)), and z = zo by c-expansivity. Thus w = ax(y,zo)= y E D"(z0). In the same way, we have W,"(x)c D"(x). 0

Lemma 6.2.5. The set D"(z) is a connected open subset of W"(z) under the intrinsic topology. Proof. Since D"(z) = W,"(z)n N, and N, is open in X ,for y E D"(z) we have W;(y) c D"(z) if 7 is sufficiently small,which means that D " ( z )is open in I@"(%) (see Lemma 5.1.11). The connectivity follows from the fact that N, is connected and a, : D"(z) x D"(x) + N , is a homeomorphism. 0

Theorem 6.2.6 (Uniformly local splitting theorem). Under the notations and assumptions in Theorem 5.2.1, for any 7 > 0 there is m > 0 such that for each x = ( z i ) E Xf if one let K"(z,) = fm(Da(zo)) and fm(K"(x)) = D U ( o m ( x ) )then , ( ~ 1K"(zrn) ) c By(%,), (92) Ka(zm)is open in Da(zm), (93) f" : D"(z0)+ K'(z,) is a homeomorphism, (u1) KU(x)c B,(zo), (u2) K"(x) is open in D"(x), (u3) f" : K"(x) + Du(om(x))is a homeomorphism, (su) let N' = a x ( D a ( zx) K " (x )) and N 2 = a,m(x~(Ka(z,)x D"(o"(x))), then N' and N 2 are open subsets of X, and a, : D"(z0)x K " ( x ) CYp(,)

N', : K " ( z m )x D"(a"(x)) + N 2

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are both homeomorphisms and the diagmm D"(z0)x K"(x)

N'

f"xf"

K"(z,) x D"(a"(x))

1

Qm"(x)

commutes.

N2

f"

Furthermore, there is 0 < p' x = (xi) E Xf.

0 is an expansive constant for f , then (1) there is 6 > 0 such that U6(zO)c W,U((z;)) for all (z;)E Xf where U6(zO)= {y E : d ( z 0 , ?/)< 6). Indeed, since f has POTP, there is 0 < 6 < e such that every &pseudo orbit o f f is e-traced. Fix (z;)E Xf and let y E Ua(z0).Then we have a &pseudo orbit 9 2 - 2 9 2 - 1 9 Y, f ( Y), f 2 ( Y ) , * . 9 which is e-traced by ( 2 ; ) E Xf.Clearly d(f'(zO),f i ( y ) ) < e for i 2 0. Since f is positively expansive, zo = y and therefore y E W,U((z;)). Thus (1) holds. Since f is a TA-map, by Theorem 3.4.4 the nonwansering set n(f) is decompsed into basic sets, n(f) = R1 U n Re, each of which splits into elai) and emetary sets, fli = C;J U * . . U C;,ai. Here f(Ci,j) = Ci,j+l (1 j Ci,ai+l= Ci,l. The following are easily checked: (2) if (z;) E Xf and zo E Ci,j, then Ci,j C W"((zi)), (3) if U is a non-empty open set of C;,j, then there is n 2 0 such that f""i(U) = c;,j. Indeed, let e > 0 be an expansive constant for f and let 6 > 0 be as in (1). Let (z;)E Xf and suppose zo E C;,j. Then, Ua(z0)n C;,, c C;,,. Since f Q i: Ci,j + Ci,j is topologically mixing, there is n > 0 such that fnai(U6(ZO)) is 6-dense in C;,j, that is, for w E C;,, we can take z E fnQi(U6(20)) with w E u6(Z). Hence there are finite points ( z : ) , . . . ,( z f ) E Xfsuch that 2: E Ua(z0) and d ( z i , z i ) 0 as i + --oo for (1 5 j 5 k), and & u6(%lai) = Ci,j. This implies that W"((z;))n Ci,j = C;,j. Therefore (2) holds. To show (3), let U be a non-empty open set of Ci,j and take (xi) E Per(o) with 30 E U. Then, W,U((z;)) f l Ci,j C U for some e with 0 < E < e. If p is a period of (z;),then by Lemma 2.4.3 we have

Remark 5.3.3.

x

...

-

- 1 .

uj,l

--f

Ci,j = ~ " ( ( z i )n) ~

<
0 the &-open neighborhood Ue(z)is also simply connected.

-

+

Lemma 6.1.6. Let X be a simply connected space and let zo,z1 be arbitrary points of X . Then arbitrary paths u and v from zo to 2 1 am homotopic.

-

-

Proof. Since u 6 is a closed path at the base point xo and X is simply connected, we have ?r1 ( X ,20) = 0 . Thus u @ O,, and so u

- o, u

u * (a * v )

-

-

(u * @)* v

Let X and Y be topological spaces and f : X

-,o, *

---f

v

v.0

Y a continuous map. Take

zo E X and let vo = f(z0). It is clear that f u = f o u E n ( X , y o ) for u E n ( X , z : , ) .Thus we can find a map fp : n ( X , z o ) n(Y,y o ) by ---f

fa(.)

=f

21,

uE

q x ,zo).

86.1 Fundamental groups

-

173

-

If u v ( F ) for u,v E R(X,zo),then we have f u f v (f o F ) , from which a map f* : ?rl(X,SO) + m(Y,Y o )

-

is induced by fu. Since f (u * v) = (fu) (fv) for u,v E n ( X ,zo),it is easily checked that f*([u][v]) = f*([u])f*([v]). Thus f* is a homomorphism. We say that f* : T I (X, XO) + a1 (Y, yo) is a homomorphism induced f m m a continuous map f : X + Y .

Lemma 6.1.7. Let X,Y and Z be topological spaces and let 20 be an arbitmry point in X . (1) For the identity map id : X -+ X the induced homomorphism id, : T I (X, zo) -+ ?r1 ( X ,20) is the identity map. (2) If f:X + Y and g:Y -+ Z are continuous maps and f (20) = yo and g(y0) = Z O , then (g o f)* = g* o fa, that is, the diagram .Irl(X,ZO)

bf)*\

f.

?rl(Y,?/O)

commutes.

J 9*

m(Z,zo) Proof. This is clear from the definition of homomorphisms induced by continuous maps. 0 Let f,g : X -+ Y be continuous maps of topological spaces. If there is a continuous map F : X x [0,1]-+ Y such that F(z,O) = f(z) and F(z,l)= g(z), then we say that f is homotopic to g, and write f g (or f g ( F ) ) . Here F is called a homotopy from f to g. The following lemma is easily checked as the proof of Lemma 6.1.1, and so we omit the proof.

-

N

Lemma 8.1.8. Let X and Y be topological spaces. Then the relation equivalence relation in the class of all continuous maps from X to Y .

-

i s an

Let f,g : X --t Y be homotopic and F a homopoty from f to g. Then for zo E X we can define a path w E R ( Y ;f (zo),g(z0)) by

w(t)= F(zo,t)

t E I,

and the relation between homomorphisms f* : 7~1(X,zo) xl(Y,f(z0))and g* : ?rl (X, zo) -+ r1(Y, g(z0)) is expressed as follows. -+

Lemma 8.1.9. g* = ti3* of*. Proof. For each t E I we define a path wt E R ( Y ;F(zo,t),g(zo))by q ( s ) = w((1

- t)s +t)

sEI

and put ut = 2Tlt(Ft 0 u)wt

tEI

CHAPTER 6

174

-

( f u ) * w to u1 = gu, for [u]E ?rl ( X ,xo). Then ut is a homotopy from uo = and hence w*of*([u])=[w~(fu)~w]=[g~]=g*[u].

Thus

w eo f *

= g+ holds.

A continuous map f : X + Y is called a homotopy equivalence if there is a continuous map g: Y --+ X such that g o f idx and f o g i d y where idx and idy are the identity maps of X and Y respectively. Here g is called a homotopy inverse of f . It is clear that f is a homotopy equivalence if it is a homeomorphism. N

N

Proposition 6.1.10. If f:X + Y is a homotopty equivalence, then for xo E X , f * : ?rl(X,zo)+ nl(Y,f ( x 0 ) ) is an isomorphism.

-

Proof. Let g:Y + X be a homotopy inverse of f . Since g o f idx and f og i d y , by Lemma 6.1.9 there are paths w E R ( X ; q , g o f(zo)) and w’ E q y ;f (xo),f 0 g ( f ( x 0 ) ) )such that N

(9 0 f)* = a * o (idx)*,

( f 0 g)* = W: 0 ( i d y ) , .

From Lemmas 6.1.4 and 6.1.7 it follows that f * is an isomorphism. 0

$6.2 Universal covering spaces The aim of this section is to establish the existence theorem of universal covering spaces for a class of topological spaces that contains the class of topological manifolds. We recall that a continuous surjection p : Y 4 X is a covering map if for x E X there is a canonical neighborhood U of x in X , which is evenly covered by p (see 55.1 of Chapter 5).

Theorem 6.2.1 (Ho m o to p y Lifting p ro p ert y ) . Let p : Y + X be a covering map of topological spaces and let f : 2 + Y be a continuous map. If a continuous map F : 2 x [0,1] -+ X satisfies F(z,O) = p o f ( z )

z E 2,

then there is a continuous map G : 2 x [0,1] -+ Y such that PO

G = F,

G(z,O)= f ( z )

E 2.

Namely, if the square part of the following diagram commutes, then there is a continuous map G represented by the right up arrow that makes the two triangle parts commutative : 2 +f Y

7

iL 1P 2 x [0,1] + F

x

$6.2 Universal covering spaces

where i : X -+

175

X x [0,1] is an inclusion map defined by i ( z )= (z,0).

Proof. First, we show that for z E Z there are an open neighborhood N , of z and a continuous map G , : N , x [0,1] + Y such that

Since [0,1] is compact, F ( z x [0,1]) is covered by a finite number of canonical neighborhoods U1,UZ,* ,U&.We may suppose that for some finite sequence 0 = t o < tl < < t &= 1 in [O, 11, F ( z x [ti,t i + l ] )c Ui for 0 5 i 5 k - 1. Since each Ui is open, there is an open neighborhood N , of z such that F ( N , x (ti,ti+l))C Ui ( 0 5 i 5 k - 1). If continuous maps Gi : N, x [ti,ti+l]-+ Y with the following properties are constructed : (1) each of Gi is a lift of Flnzx[ti,ti+ll by p, (2) Go(z’,O)= f(z’) for z’ E N,, ( 3 ) Gi(z’,ti+l)= Gi+l(z’,ti+l)for Z’ E N, and 0 5 i 5 k - 1, then the desired map G , is gived by

---

--

I

Figure 24

To construct Gi we use induction on i. Since UOis a canonical neighborhood, by definition p-’(Uo> is decomposed into disjoint open sets of Y ,say A0 E ho, such that each restriction p : Uxo ---t UO is a homeomorphism. n N,. Then {Vx,} covers N , and Vx, n V,, = 8 if Let Vx, = f A0 # PO. Hence, since F ( N , x [to,tl])c UO, there is a unique continuous map Go : N , x [to,t l ] + Y satisfying the following conditions

uxo,

-‘(v~,)

ux0,

(a) Go(Vx, x [ t o , t i ] )C (b) G0lVA,x[to,tlj is a lift of qvx, X(t0,tlj by P.

176

CHAPTER 6

Suppose Gi-1 is constructed. Since F ( N , x [ti,ti+l])c Ui and p-'(Ui) splits into disjoint open sets UX,( X i E Ai), we let V A = ~ ( 2 E N , : Gi-'(z',ti) E U x , } . In the same way a8 above, we can choose uniquely a continuous map Gi : N, x [ti,t;+l]-+ Y such that ( c ) Gi(Vxi x [ti,ti+l])c V X ~

(4 GilVAiX[ti,ti+l]is a lift of FlV., x[ti,ti+l] by PFrom the construction it follows that the maps G; satisfy the conditions ( l ) , (2) and (3). Next, for z 1 , z ~ E Z let ( N z l , G z 1 )and (N,,,G,,) be as above. If y E NZl n NZ,?then GZl lvx[O,l]and G,,IvxIo,llare continuous maps from y x [0,1] to Y and

Thus, by Theorem 5.2.2 we have GZllarXIO,ll = G,,Ivx[o,ll.Since y is arbitrary, it follows that G,, is consistent with G,, on (N,,n N Z 2 )x [0,1].Thus we can define G : 2 x [0,1]-+ Y by G I N x[O,ll . = G , ( z E Z ) , and G satisfies all the conditions in the theorem. 0

Lemma 6.2.2. Let p : Y -+ X be a covering map. Let xo E X and bo E p-l(x0). Suppose that u, u' are paths of X starting at 20 and v,v' are paths of Y starting at bo satisfying p o v = u and p o v' = u', if u and u' are homotopic, then the terminal point of v coincides with that of v', and v is homotopic to 21'.

Proof. Let F be a homotopy from u to u'; i.e. F : I x I -+ X is a continuous map satisfying

F(t,O) = u ( t ) , F ( t , 1) = "'(t), F ( 0 , s ) = 50, F ( 1 , s ) = ~ ( 1=) ~ ' ( 1 ) . By the homotopy lifting property there exists a continuous map G : I x I * Y such that G(t,O) = v ( t ) and p o G = F . Since G(O x I) = 50, it follows that G(0,s) = v(0) = bo. Similarly, G(1,s) = v(1). Put h ( t ) = G ( t , l ) for t E I, then h: I -+ Y and

h(0) = G ( 0 , l )= bo, p o h(t) = p

o

G(t,1 ) = F ( t , 1 ) = ~ ' ( t ) .

Thus h is a path of X starting at bo and p o h = u'. By Theorem 5.2.2 we have h = v', and therefore v(1) = G(1,O) = G ( 1 , l ) = h(1) = ~ ' ( 1 ) and G is a homotopy from v to v'. 0

56.2 Universal covering spaces

177

Lemma 6.2.3. Let p : Y + X be a covering map. Then for y E Y, p+ : rl(Y, y ) + rl(X,p( y ) ) is injective. Proof. For [u],[v] E ?rl(Y,y ) supposep,([u]) =p+([v]). Thenpou is homotopic to p o v. Therefore [u] = [v] by Lemma 6.2.2. 0

Lemma 6.2.4. Let p : Y + X be a covering map. If X is connected, then the cardinal number of p-'(x) is constant. Proof. Since p : Y + X is a covering map, by definition each point x in X has a neighborhood evenly covered by p. Hence the cardinal number of p-'(x) is locally constant. Since X is connected, the conclusion is obtained. 0 For a covering map p : Y + X the cardinal number of p-'(x) covering degree of p if it is constant.

is called the

Lemma 6.2.6. Let p : Y + X be a covering map. If Y is path connected, then the index of p+(rl(Y,b)) in rl(X,p(b)) is equal to the covering degree of

P. Proof. For each y E p-'(p(b)) take a path uy in Y from b to y . Then p o uy is a closed path from p(b) to p(b), and so p+(rl(Y,b))[p o uy]is a right coset in rl(X,p(b)). Let [v] E q(Y,p(b)) and u : [0,1] + Y be a lift of v by p such that u(0) = b. Then y = u(1) E p-l(p(b)) and u * B yis a closed path from b to b. Therefore, [v] = [p o ( u O,)][p o uy] E p.(rl(Y,b))[p o uy]. On the other hand, let [v] E p.(?rl(Y,b))[pou,] npt(?rl(Y,b))[pouyr]. Then [v] = [p o u][p o uy] = [p o u'][p o uyt] for some [u],[u'] E ?rl(Y,b). Since (p o u) (p o u Y )is homotopic to (p o u') (p o u y # ) by , Lemma 6.2.2 we have u * ~ ~ (=121') * uyt(1), and SO y = ~ ~ ( =1~) ~ ' ( = 1 y'. ) 0

-

-

-

Let p : Y + X be a covering map. If Y is locally path connected and simply connected, then Y is called a universal covering space of X and p :Y + X is said to be the universal covering. Let p : Y + X and p' :Y' + X be covering maps. If there is a homeomorphism h: Y' + Y such that p o h = p', that is, the diagram

Y'

I then p :Y

+X

h

Y J

commutes.

is equivalent to p' : Y' + X.

Lemma 6.2.6. Let p :Y + X be a covering map and let X be path connected. Suppose Y' is locally path connected and simply connected. If xo E X and bo E p-'(xo) and if yo E Y', then for a continuous map h:Y' + X with h ( y 0 ) = xo there ezists only one lift : Y' + Y satisfying z ( y 0 ) = b~ and

z

CHAPTER 6

178

poIE=h.

-3. -

Y'

h

Y l P

x

Proof. We first define a map h : Y' + Y as follows. Take y E Y'. Since Y' is path connected, yo and y are joined by a path w. P u t 'u = h o w. Then u is a path in X starting at 20. Choose a path v of Y starting at bo such that p o v = u. Let b = v(1) and define h(y) = b. To show that h is well defined, let w and w' be paths in Y' joining yo and y. Since Y' is simply connected, w and w' are homotopic (w w') by Lemma 6.1.6. Thus h o w h o w'. Let v and v' be paths in Y starting at bo such that p o v = h o w and p o v1 = h o w'. By Lemma 6.2.2 we have v(1) = v'(1). Thus is well defined. From the construction it follows that Z(y0) = bo and p o h = h. Next we show that h : Y' + Y is continuous. Fix y1 E Y'. Put 2 1 = h(y1) and take a canonical neighborhood U of x1 for p. Since h is continuous, clearly V = h-'(U) is an open neighborhood of 21. Hence, since Y' is locally path connected, we can choose a neighborhood W of y1 such that W c V, and such that for any y E W there is a path in V joining y and y1.

-

-

Figure 25 Let bl = h(y1). Since p(b1) = 21 and U is a canonical neighborhood of there is an open neighborhood U(b1) of bl such that p : U(b1) + U is a homeomorphism. Now, define g: W + Y by g(y) = (plu(bl))-l o h(y). Clearly g is continuous. We claim that h = g on W. Indeed, take a path w joining yo and y1 in Y and a path w' joining y1 and y in V. If v is a path starting at bo satisfying p o v = h o w , then by definition v(1) = b1. Since h o w' is a path in U starting at 21, clearly v' is a path in U(b1) starting at bl where v' = (p~u(b~))-' o how'. By definition v'(1) = g(y), and so v v'(1) = g(y). On the other hand, since p o (v d ) = h o (w w'), we 21,

-

-

-

$6.2 Universal covering spaces

x

have v v'(1) = x(y). Therefore, = g on W , from which we obtain that continuous. The uniqueness of follows from Theorem 5.2.2. 0

179

x is

Proposition 6.2.7. Let p : Y -+ X be a covering map and let p' :Y' + X' be the universal covering. Let zo E X,bo E p-'(zo) and zh E XI, bh E p'-'(zh). If X is path connected and h: X' + X is a continuous map with h(z6) = to, then there is a unique continuous map : Y' -+ Y satisfying x(bh) = b~ and p 0 X = h op'.

lp -

Y' P'1

X'

h

h

Y

x

Proof. Since Y' is locally path connected and simply connected, by applying Lemma 6.2.6 to a map h o p' : Y' --t X we obtain the conclusion. 0 Theorem 6.2.8 (Covering theorem). Let p : Y -+ X be a covering space and let p' : Y' -+ X be the universal covering. Let zo E X,bo E p-'(zo) and bh E p'-'(zo). If Y is path connected, then there exists a unique continuous map h:Y'+Y

satisfying x(b&)= b~ and p o h = p', and

h :Y' + Y is the universal covering.

Proof. By choosing the identity map instead of h of Proposition 6.2.7 there a unique : Y' + Y satisfying h(bb)= bo and p o x = p'. We show that exists h is surjective. Indeed, for b E Y let v be a path joining bo and b and put u = p o v . Take a- path v' of Y' starting at bh such that p o v' = u and set b' = v'(1). Then h(b') = b and therefore h is surjective. For b E Y let U be a path connected neighborhood of p(b) in X which is evenly coverd by both p and p'. Then there exists a neighborhood U ( b ) of b in Y such that p : U(b) -+ U is a homeomorphism, and for each 6 E h-' ( b ) there is a neighborhood U(6)of 6 in Y' satisfying

x

(1) for each 6 E X-'(b),E : U(6) + U ( b ) is a homeomorphism, (2) X-'(U(b)) = U(U(6): 6 E X-'(b)}, ( 3 ) if 6,v E h-'(b) and 5 # then U(6)n U(v)= 0.

v,

Therefore

: Y' + Y is a covering space. 0

Theorem 6.2.9 (Uniqueness theorem). Let p : Y -+ X and p' : Y' -+ X be the universal coverings. Then p : Y + X i s equivalent to p' : Y' + X .

CHAPTER 6

180

Proof. Take xo E X. For bo

E p-'(xo)

and bh E p'-'(xo)

there exist continuous

maps

satisfying the conditions that z(bh) = 4 , g(b0) = bh, p o We consider two the continuous maps g o x : Y' + Y',

x

idyl : Y' -+

=p ' and p' o 3 = p.

Y'.

Since p ' o 3 o = p' = p' o id and 3 o x(bh) = bh, by_Theorem 6.2.8 it follows o = idyl. Similarly, o g = i d y . Therefore h is a homeomorphism. 0 that ?j

x

x

A topological space X is said to be semilocally 1-connected if for each x E X there is a neighborhood U of x such that the homomorphism i, : ~1(U,x) -t ?rl (X, x) induced by the inclusion map i : U + X is trivial. A topological space X is said to be locally Contractible if for each x E X and for a neighborhood N of x there is a neighborhood U of x with U C N such that the inclusion U + N is homotopic in N to a constant map U + {x}. Remark 6.2.10. By definition it follows that every topological manifold is locally contractible and every locally contractible space is semilocally 1connected

.

Theorem 6.2.11 (Existence theorem). Let X be a locally path connected, semilocally 1-connected and path connected space. Then there ezists the universal covering p : Y --t X.

Proof. Take and fix xo E X. We define

R(X;20,X )= { u : u is a path from [0,1] to X such that

-

-

u(0) = 20).

For u,u' E S2(X;xo,X), if u(1) = u'(1) and u is homotopic to u', then we 20,X). Denote as write u u'. Clearly is an equivalence relation in R(X; [u] the equivalence class of u for u E R(X;zo,X) and define p ( [ u ] )= u(1). Then p is a map from Y = {[u]:u E R(X; 20,X)}to X. In fact p :Y -t X is surjective. This follows from the fact that if u is a path joining xo and x then P([Ul)= 3. To introduce a topology of Y we define a fundamental neighborhood system as follows. Since X is locally path connected and semilocally 1-connected, for x E X and for any neighborhood 0 of x there exists a neighborhood U of x with U c 0 such that U is path connected and ?rl(U,x) + ?rl(X,x) is trivial. Hereafter, for x E X we choose such a neighborhood U.To define a fundamental neighborhood U ( b ) of b = [u] take a neighborhood U of u(1) = x and Put U' = {u u : u is a path in U starting at, x},

-

56.2 Universal covering spaces

181

and let

U ( b )= { [ u* u] : u u E U'}. Then {U( b) } is a fundamental neighborhood system of b. Indeed, (1) it is clear that b E U(b). (2) Two sets U ( b ) and V(b) are constructed by certain neighborhoods U and V of z = u(1) respectively. Now take a neighborhood W such that W c U n V. Then W constructs a set W ( b )satisfying

W(b)c ~ ( bn)v(b). (3) Take b' E U(b). Then b' = [u'] and u' = us u for some path in U. Let V be a neighborhood of 2' = u'(1) such that V C U. Construct a set V(b') by V. Then we have b' E V(b') c U ( b ) . We show that p : Y + X is a continuous open map. Let p ( b ) = 2 and take U a neighborhood of z. Then we have p ( U ( b ) ) = U. Indeed, clearly p ( U ( b ) ) c U. Since U is path connected, for x' E U let u be a path in U from z to 2'. Then p ( [ u u ] )= z' and thus U C p ( U ( b ) ) . Therefore, p ( U ( b ) ) = U, which shows that p is continuous and open. We prove that p : U(b) -, U is a homeomorphism. To do this it su5cies to see that p is injective. Let p(b1) = p(b2)(= z') for bl,b2 E U ( b ) . If u1 and u2 are paths of U from p ( b ) = z to z', then b = [u],bl = [u U I ] and bz = [u* uz]. Since ?rl(U,z)-+ ? r l ( X ,z)is trivial, it follows that 6 1 uz (c.f. Lemma 6.1.6). Therefore u u1 u uz and bl = [u 6 1 1 = [u u2] = bz. We check that if b, b' E p - ' ( z ) and b # b', then U(b)n U(b') = 0. Indeed, suppose U ( b ) n U(b') # 0, from which take a point c. If b = [u] and b' = [u'], then c = [u a] = [u' u'] for some paths u and u' in U starting at z. Thus we have u u u u'. Since u u', we have

-

-

-

-

0

-

N

-

N

-

-

-

N

-

-

N

(u * u) * ( u ' - u ' ) = (u * u)*

(a 2) *

N

(u

-

(0.;')).

3

N

u . i',

from which u N u',

and so b = [u] = [u'] = b', thus contradicting. It is checked that p - ' ( U ) = U { U ( b ) : b E p - l ( z ) } . Indeed, it is clear that p - ' ( U ) 3 U { U ( b ) : b E p - ' ( z ) } . Take b' = [u'] E p - ' ( U ) and put z' = p(b'). Let u be a path of U from z to z', and define u = u' a. If b = [u], then p ( b ) = z and u' 11' * (a * u) (u' a) * u = u * 6.

.

N

N

-

Thus we have b' = [u'] = [u u] E U ( b ) . We have seen that p : Y + X is a covering map. To show that p : Y + X is the universal covering, it sufficies to show that (i) Y is path connected and (ii) Y is simply connected. Proof of (i) : Let 4 = [O,,] and take any point b E Y.To show the existence of a path of Y joining b and bo we choose a path u of X joining 20 and p ( b )

CHAPTER 6

182

such that [u]= b, and define paths u, : I --t X by ud(t)= u(st) for 0 5 s 5 1. If b ( s ) is an equivalence class containing a path u,, then a map v : I + Y is defined by v ( s ) = b ( s ) , and we have that v ( 0 ) = bo and v(1) = b. Continuity of v is proved as follows. Take a neighborhood U of udo(l)in X and let U(b(s0)) be a fundamental neighborhood of b(s0) in Y.Since u : I -+ X is continuous, we choose 6 > 0 such that 1s - so1 < 6

* u(s)E u.

For s with 1s - so1 < 6 let u be a path in U from u ( s 0 ) to u(s). Then the path u, is homotopic to u,, * u. Therefore, [u,]= b(s) E U(b(s0)). Proof of (ii) : To show rl(Y, bo) = 0 let v be a closed path from bO to 4. Put u = p o v and for s with 0 5 s 5 1 define a path u, : I --$ X by u,(t) = u(st). Then a map v' : I -+ Y defined by v'(s) = [us]is continuous, that is, a path

in Y.Since

v'(0) = [uo] = bo,

p

0

v'(s) = u,(l) = u(s) = p 0 v(s),

by Theorem 5.2.2 we have v' = v , and in particular [u]= v'( 1) = v( 1) = bO = [O,,],

which implies that u and O,, are homotopic, and by Lemma 6.2.2, v is homotopic to Ob,. Thus Y is simply connected. 0

As an easy corollary we have the following proposition. Theorem 6.2.12. Let X be a locally path connected, semilocally 1-connected XO). Then and path connected space. For xo E X let p be a subgroup of 7~1(X, there ezist a covering space p : Y + X and bo E p-'(xo) such that P*'IFl(Y,bo) = p.

-

Proof. Let u,u' E R(X;zo,X). We write u u' if u(1) = u'(1) and [u.7?] E p. Put Y = R(X; xo,X)/ N and define a map p : Y + X by p ( [ u ) )= u(1). By introducing an adequate topology in Y we see that p : Y --t X is a covering map. The proof is very similar to that of Theorem 6.2.11 and so we omit the proof. 0

56.3 Covering transformation groups

-

Let X be a topological space and let be an equivalence relation in X. As before, the identifying space X/ is a family { [z]: z E X} of all equivalence classes. Let p : X 4 X/ be the natural projection defined by p(x) = [z]. Clearly p is surjective. We define a topology of X/ as the strongest topology such that p is continuous, i.e. a subset 0 of X/ is open if and only if p-'( 0) is open in X.

--

56.3 Covering transformation groups

183

-

Lemma 6.3.1. Let X be a topological space and p : X + X / the natural projection. Let Y be a topological space and suppose f :X + Y and 9: XI -+ Y satisfy a relation f = g o p. Then g is continuous i f and only i f f is continuous.

X

Pl

x/-

-< B

Y

Proof. Suppose f is continuous, To see continuity of g it suffices to show that g-'(O) is open when 0 is an open subset of Y . Since f = g o p , we have p-'(g-'(O)) = f-'(O). Since f is continuous, p-'(g-'(O)) is open if O is open, so that g-'(O) is open. Conversely, if g is continuous, then f = p o g is continuous since p is continuous. Let X be a topological space and let G be a group. We say that G acts (continuously) on X if to (9,z)E G x X there corresponds a point g(z) in X and the following conditions are satisfied: ( 1 ) e ( z ) = z for z E X where e is the identity, (2) g(g'(z)) = gg'(x) for z E X and 9,g' E G , ( 3 ) for each g E G a map z H g(z) is a homeomorphism of X . When G acts on X , for x, y E X letting

z

N

y

-

++

y = g(z) for some g E G,

an equivalence relation in X is defined. Then the identifying space X / -, denoted as X / G , is called the orbit space by G of X. It follows that for z E X, {g(z) : g E G} is the equivalence class.

Lemma 6.3.2. The natural projection p : X surjection.

--t

X / G is

Q

continuous open

Proof. Let 0 be an open subset of X. Then p - ' ( p ( 0 ) ) = U{gO : g E G}. Here g o = {g(z) : z E 0).Since go is open in X, p - ' ( p ( 0 ) ) is open in X. By the definition of topology of XIG, we have that p ( 0 ) is open in X / G , and so p is an open map. The continuity of p follows from the definition of topology. 0

Let p : Y -+ X be a covering map. A homeomorphism a : Y + Y is called a covering transformation for p if p o a = p holds. We denote as G(p)the set of all covering transformations for p . Since (1) the identity map i d y of Y belongs to G(p), (2) if a', a2 E G(p),then a1 o a2 E G(p), ( 3 ) if a E G(p),then (Y-' E G(p), it follows that G(p) is a group, which is called the covering transformation group for p .

184

CHAPTER 6

L e m m a 6.3.3. Let p : Y --$ X be the universal covering. If a continuous map h:Y -+ Y satisfies p o h = p , then h E G(p). Proof. Suppose a continuous map h:Y + Y satisfies p o h = p. Take b E Y and let c = h(b). By Lemma 6.2.6 there is a continuous map h' : Y -+ Y such that h'(c) = b and p o h' = p . Since h o h'(c) = c and p o (h o h') = p , by Theorem 5.2.2 we have h o h' = i d y . Similarly, h' o h = i d y . Thus h is a homeomorphism. 0 We say that an action of G on X is said to be properly discontinuous if for each b E X there exists a neighborhood U(b) of b such that U ( b )n g U ( b ) = 8 for all g E G with g # 1. Here gU(b) = {g(z) : z E U(b)}.

Theorem 6.3.4. If p : Y -+ X is the universal covering, then for each b E Y ( 1 ) the map a H a(b) is a bijection from G(p) onto p-'(p(b)), (2) the map $b : a H [p o ?&(a)] i s an isomorphism from G(p) onto the fundamental group r l ( X , p ( b ) ) where u a ( b ) is a path from b to a ( b ) . Furthermore, the action of G(p) on Y i s properly discontinuous and Y / G ( p ) is homeomorphic to X .

Proof. (1): If t E p-'(p(b)), then by Lemma 6.2.6 there is a unique continuous map h: Y + Y such that h(b) = z and poh = p , and by Lemma 6.3.3, h E G(p). Thus (1) holds. (2): Let u and u' be paths from b to a@). Since Y is simply connected, by Lemma 6.1.6 u and u' are homotoptic, which implies [p o u] = [p o u'] E r,(X,p(b)). Thus '$b is well defined. For a,@ E G(p) suppose &(a) = $ b ( @ ) . Since [p o u , ( b ) ] = [p o u p ( b ) ] , it follows that p o and p o u o ( b ) are homotopic, and hence u a ( b ) ( l ) = " p ( b ) ( l ) by Lemma 6.2.2. By uniqueness a = @. Therefore $b is injective. If [v]E r l ( X , p ( b ) ) ,then there is a path u starting at b such that p o u = v, and u(1) E p-'(p(b)). From this and (1)we have that $b is surjective. To show that '$b is a homomorphism, let a,@ E G(p). By definition @ o u , ( b ) is a path starting at @(b)and so

from which we have &,(a o @)= $ b ( a ) ? , ! J b ( P ) . Thus '$b is a homomorphism. (2) was proved. For b E Y let U be a canonical neighborhood of p ( b ) and take an open neighborhood U ( b ) such that p : U(b) -+ U is a homeomorphism. If a E G ( p ) and a ( U ( b ) )n U(b) # 0, then there is z , y E U ( b ) such that a ( z ) = y , and so p ( a ( z ) ) = p ( z ) = p(y). This implies z = y , and so a is the identity map. Thus the action of G(p) is properly discontinuous. To see that Y / G ( p )

56.3 Covering transformation groups

186

is homeomorphic to X, define a map f : Y / G ( p ) + X by f ( G ( p ) b ) = p ( b ) . Clearly, f is surjective. By (1)we have that f is injective. The continuity o f f follows from that of p. Since the natural projection Y + Y / G ( p )is continuous, we obtain the continuity of the inverse map f-'. 0

Remark 6.3.5. Let p :X -+ X be a covering map and let Y be a connected locally path connected semilocally 1-connected space. Suppose f:Y -+ X is a continuous map and let f(y0) = 20 and 50 E p-'(zo). A necessary and satisfying g(y0) = "0 and sufficient condition that f has a lift g:Y -+ p 0 g = f is that f * m ( Y , YO) C P * T ~ ~ , " o ) . Indeed, if there is a lift g:Y -+ X,then

x

f*+.'(Y,YO) = (pOg)*.rrl(Y,yo) c P*m(x,zo), and hence the condition is necessary. -

To see the converse, let q : Y -+ Y be the universal covering-and - take go E q - l ( y 0 ) . Then by Proposition 6.2.7 there is a continuous map f : Y -+

x

such that p o 7 = f o q and T(g0)= 50.For E P put y = q ( g ) and 5 = 7@). If Ti7 is a path in P from g to go,then w = q o Ti7 is a path in Y joining y and yo, and hence

f2r1(Y,y) = f* 0 w,?rl(Y,yo) (by Lemma 6.1.4) = (f 0 w)* 0 f*Xl(Y, Yo) c ( f 0 w)*0 p*n1p,"0) = p* 0 (7 0 E)*m(X, "0) =p*Tl(X,Z). Let Q be a covering transformation for q and take a path ii in P from to (I!@). Since q o E is a closed path from y to y, it follows that [ f o q o i i ] E f*~l(Y,y) c p * r 1 ( X , E ) . S i n c e p o T o E = f o q o E , we have ~ O ~ OEp*?rl(x,5),which E ] (see Lemma 6.2.2). Therfore, implies that 7 o ii is a closed path in - - f(g) = T ( ( I ! ( ~Since ) ) . (I! is arbitrary, by Theorem 6.3.4 we can project f :X -+ to a map g from Y to Thus the condition is sufficient.

x

x.

Let G be a group and S a subgroup of G. The largest subgroup of G containing S as a normal subgroup is called the normaliaer of S in G. We denote this subgroup as N ( S ;G). Remark 6.3.6. Let X be a connected locally path connected semilocally 1-connected space and let p : + X be the universal covering. Suppose Y is path connected and q : Y -+ X is a covering map and let T : + Y be the universal covering that satisfies p = q o T (see Theorem 6.2.8).

x

-

X - G Y v! \ P cl X

x

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186

Then a covering transformation a E G(p) can be projected to a map in G(q) by T if and only if a E N ( G ( r ) ; G ( p ) ) .Moreover G(q) is isomorphic to the ; (~))/G(T). factor group N ( G ( T )G Indeed, suppose a E G ( p ) can be projected to a map of Y by T . Let zE Then for p E G ( T )we have T o a o P ( z ) = T o a(%),which means that a o p = p' o a for some P' E G ( r ) , and hence aG(r)a-' c G ( T ) .Therefore a E N(G(?);G(p)).It is easy to check the converse. Thus we leave the proof to the readers. X".X

x.

I. Y-Y &(a)

Since every member in G(q)can be lifted to a map in G(p)by Proposition 6.2.7, a map Q : N ( G ( T )G ; ( p ) )--t G(q)defined by assigning each a E N ( G ( T )G ; (p)) to the projection by T is surjective. Obviously Q is a homomorphism. By definition it follows that the kernel Ker(Q) coincides with G ( T ) .Therefore the second statement is obtained from the homomorphism theorem described as follows: if A and B are groups and if Q :A + B is a homomorphism, then the quotient group A/Ker(Q) is isomorphic to the image Im(q5). The following theorem describes the converse of Theorem 6.3.4.

Theorem 6.3.7. Let G be a group and X a topological space. Suppose that G acts on X and the action is properly discontinuous. Then the following (1) and (2) hold : (1) the natural projection p : X -,X / G is a covering map, (2) if X is simply connected, then the fundamental group a l ( X / G ) is isomorphic to G . Proof. (1) : By Lemma 6.3.2 the projection p : X + X / G is continuous and it is an open map. Let b E X . Since the action of G is properly discontinuous, there is a neighborhood U ( b ) such that gU(b) n U(b) = 0 if g E G is not the identity. Put U = p ( U ( b ) ) . Then U is open in X . For b', b" E U ( b ) let p(b') = p(b"). Then there is g E G such that g(b') = b". This implies g = 1, and so b' = b". Thus p : U(b) + U is a homeomorphism. Similarly, we have that p :gU(b) + U is a homeomorphism for g E G . Since p-'(U) = U { g U ( b ): g E G } and gU(b) fl g'U(b) = 0 for g # g', it follows that U is evenly covered by p. Therefore p :X -+ X / G is a covering map. (2) : Take 20 E X / G and bo E p - ' ( z o ) , and define a map $ao : G -, n l ( X / G , z o )as in (2)of Theorem 6.3.4. In fact, for a E G let u be a path in X from bo to 4 4 ) .Put v = p o u . Then [v] E a l ( X / G , z o ) .Since X is simply connected, by Lemma 6.1.6 it follows that [v] is uniquely determined for a. Thus we define $ b o ( a ) = [v]. Then, in the same way as the proof of Theorem 6.3.4 (2), we obtain that ?,6bo is an isomorphism. 0

56.3 Covering transformation groups

187

Remark 6.3.8. Let T" = Rn/Z" be an n-torus and p :R" + 'P the natural projection. Then p : Rn + T" is the universal covering and T I(11'") E Z" holds. Since a map p : Z" x R" + R" is defined by p(n, z) = n + z,the group Z" acts on R". For z E R" let U ( z ) = {y E R" : IIz - yII < 1/2}. Then p(n, U ( z ) )n U ( z ) = 8. Hence the action p is properly discontinuous, and by Theorem 6.3.7 (1)the natural projection p :Rn + 11'" is a covering map. Since

R" is simply connected, it follows that R" is the universal covering space of T" and ~~(11'") Z" by Theorem 6.3.7 (2).

x

Let p : -+ X be a covering - - map and f : X + X a continuous map. We say that a continuous map f : X + is a lift of f by p if p o 7 = f o p holds.

x

x-x f

x

Theorem 6.3.9. Let p : -+ X be the universal covering and f: X + X a continuous map. - -Let zo E X , bo E p - ' ( z o ) and b6 E p-'( f (20)) Then there is a unique lift f :X + of f by p such that T ( b 0 ) = b;. If continuous maps f 1,72: + are lifts of f by p , then f, = a o T2 for some a E G ( p ) .

x

x x

- Proof. The first statement follows from Proposition 6.2.7. Suppose f ,,f : -

x

w.

X + are lifts o f f byp and let z E Since p o T l ( z ) = f o p ( z ) = p o T 2 ( z ) , there is a E G ( p )such that rl(z)= a o f 2 ( z ) .By Theorem 5.2.2 the conclusion is obtained. 0 Let - p-:55 -+ X be the universal covering and f :X + X a continuous map. For f : X + a lift of f by p we define a homomorphism 7, : G(p) + G(p) bY

x

x,

where b E c = T ( b ) , f* : r , ( X , p ( b ) )+ r l ( X , p ( c ) )is the induced homomorphism; and +b and are isomorphisms as in Theorem 6.3.4. Then we obtain the following lemma.

+=

Lemma 6.3.10. E G(P)

7,

does not depend on the choice of b. Furthermore for every -

f

(6.2)

x

oa=T*(a)oT.

Proof. For b' E and c' = f ( b ' ) define a homomorphism : G(p) + G ( p ) as in (6.1). To show that 7, = take a path w in from b to b' and let

T,,

x

CHAPTER 6

188

a E G(p). Then a o w is a path from a ( b ) to a(b’). By the definition of 7, and f, it is easily checked that 7 0a o w is a path from 7 , ( a ) ( c ) to Ti(a)(c’). On the other hand, since

po(Toaow)= f opow=po(~ow), there is p E G(p) such that 7 o a o w = /3 o (7 o w). This implies 7,(a)= p = f*(a).Thus 7, = 7;. Next, we show (6.2). Let a E G(p) and ua(b) be a path from b to .(a). Since U a ( b ) ( l ) = a ( b ) , clearly (7 o u a ( b ) ) ( l )= 7 o a(b). Since 7 o ua(b) is a lift of f o (p o U a ( b ) ) by p starting at c, by definition f,(a)(c)= (70 U,(b))(1). Therefore, 7 o a ( b ) = 7*(a) o 7(b). Since the two sides of (6.2) are both lifts of f by p , by Theorem 5.2.2 we obtain the equality of (6.2). 0

-I

Under the assumptions of Remark 6.3.6, let f:X + X be a continuous map and f‘ : Y -t Y a lift of f by q. Then a homomorphism f: : G(q) -t G(q) is defined and for every t E G(q),f’ o 1 = f:(t) o f’ holds. Indeed, let p : -+ X be the universal covering and let 7 : + be a lift of f by p. Then 7 is also a lift of f’ by T . Here T : -t Y is the universal covering such that p = q o T . Let 7, : G(p) + G(p) be as in (6.1). If a E G ( T ) , then we have

Remark 6.3.11.

x

x x

r o 7 0a = f ’ o

T

o a = f’ o r

= T 0 7*(a) 0 7, from which ?*(a)E G ( T ) . Hence ~ , ( G ( T )c) G ( T ) . From this fact it is easy to see that ~ , ( N ( G ( T ) ; G ( c ~ )N) () G ( r ) ; G ( p ) )Therefore . 7, induces a homomorphism 7: : N ( G ( T )G ; ( p ) ) / G ( r -t ) N ( G ( T )G ; ( p ) ) / G ( p ) Since . G(q) is isomorphic to N ( G ( T )G ; ( ~ ) ) / G ( Twe) ,define a homomorphism f: : G(q) -t G(q) by the following commutative diagram:

r:

G(q)

1

r:

Then, for t E G(q) we have 0T

1

-

N(G(T);G b ) ) / G ( T )

f’ 0 t

G(q)

N(G(T);G(P))/G(T)

-

= f’ 0 T 0 a = T 0 f 0 a = T 0 7*(a) 0 7 = E(t)0 T 0 7 =

f:(q 0 f’ 0 T

where a E G(p) is a covering transformation satisfying I o T = T o a. Thus f‘ 0 c = f:(e) 0 f’.

$6.4 S-injectivity of TA-covering maps

x

189

Theorem 6.3.12. Let-p :- 4 X be the universal covering and f:X + X a self-covering map. Let f : X -+ be a lift o f f by p . Then ( 1 ) 7 is a homeomorphism, (2) the cardinal number of G(p)/T,(G(p)) is equal to the covering degree of f 9 ( 3 ) f is a homeomorphism i f and only if j , ( G ( p ) ) = G(p).

x

-

Proof. Since p : X -+ X is the universal covering, by Proposition 6.2.7 it follows that f : + is a self-covering map, and by Lemma 6.2.5 it must be a homeomorphism because is simply connected. Thus (1)holds. (2)and (3) are obtained from Lemma 6.2.5 and the definition of 7,. 0

x x

x

$6.4 S-iqjectivity of TA-covering maps In the previous chapter we have seen that if a TA-covering map of a compact locally connected metric space is s-injective, then the local product structure theorem is established. In this section we may find a sufficient condition for TA-covering maps to be s-injective.

x

Theorem 6.4.1. Let X be a compact metric space with metric d and let be topological space. Let p : + X be a covering map. If X is locally connected, then there are a compatible metric f o r and a constant 60 > 0 such that (1) for 0 < 6 5 60 and z E

x

x

x

P : u6(Z>--t

v15(P(~))

x

is an isometry where u,5(Z)= {y E : a(z,y) < 6 ) and U & ( p ( z )= ) { Y E x : d(P(Z),Y) < 61, (2) all covering tmnsformations for p are isometries, (3) as a complete metric space with respect to 2.

x

Proof. Let {Uj} be a finite cover of X by connected open sets such that each Ui is evenly covered by p. Then p-'(Uj) is decomposed into the disjoint union UxEa of connected open sets in such that each restriction p :vi,x -+ Ui is a homeomorphism. Let T > 0 be a Lebesgue number of {Ui},and choose a finite cover {V,} of X by connected open sets such that the diameter of each y is less than r/2. Since 5 is contained in some Uj, the inverse image p-l(Y) also splits into disjoint connected open sets vj,x,X E A. Let 60 > 0 be a Lebesgue number of (5). To define the metric for for Z, y E put

vj,~

x

x,

d(X,Y) =

y E Vj,A for some Vj,x, otherwise.

Z,

x

CHAPTER 6

190

In fact the following is easily checked: x = y if and only if a(z, y ) = 0, and d ( z , y ) = a(y,x) holds. The triangle inequality is proved as follows. Suppose d ( a , z ) +d(z,y) < 60. Then d ( p ( z ) , p ( z ) ) d ( p ( z ) , p ( y ) ) < 60 and hence d ( p ( z ) , p ( y ) ) < 60. It suffices to show that d ( p ( z ) , p ( y ) ) = a(x,y). Since d ( x , z ) < 60, by definition there are such that x, z E Similarly, z and y are points in some vjt,At. Since -

+

vj,~

vj,~.

diam(p(Vj,A) Up(Vj8,p)) = diam(Vj U 5)) < T,

vjlAnvjf

we ,A', it follows that - have Vj UVj, c Ui for some Ui E { Ui}.Since z E Vj,x U is contained in some V;J. Therefore, 2, y E G~J.On the other hand, since d ( p ( z ) , p ( y ) )< 60, clearly p(x) and p(y) are points in some V j t t , and then V j , t c Ui. This implies z,y E Vj,,,x,t.Thus, a(x,y) = d ( p ( z ) , p ( y ) ) . By the definition of 2, the conditions (2) and (3) are easily proved. To show (I), let 0 < 6 5 60. It is clear that p(ua(z)) C Va(p(.)). Let d(p(z),y') < 6. Then p ( x ) , g ' E Vj for some Vj E {Vj}. Hence there is y E p-'(y') such that z,y E and we have a(x,y) = d(p(z),y'). Therefore, p(ua(Z:>)= Ua(p(Z)). To see that p : u6(Z) --t Ua(p(x)) is injective, let = d(p(z),y') for some z E p-'(y'). Then some v j t , A t contains x and z , from which p(z),y' E p(vJz) n p ( v j ~ , p ) Hence . p(vj,A),p(vjt,At) c Ui for some Ui E {Ui}.Since x E Vj,x n r j t , A t , there is V;,Awhich contains V,,A and Vj#,xt, and therefore 2, y, z E u i , ~This . implies y = z since p ( y ) = p ( z ) . Therefore p : u6(.) + Ua(p(Z))is bijective, hence an isometry. 0

vj,,~~

vj,~,

a(.,.)

Let X be a metric space and d a metric for X. If for every E > 0 and x E X the open ball Uc(x) of radius E centered at x is connected, we say that d is a connected metric for X . It follows that X is connected and locally connected whenever it admits a connected metric. Conversely, the following holds.

Lemma 6.4.2. Let X be a compact metric space. If X is connected and locally connected, then there exists a connected metric for X . Proof. Let d' be a metric for X. Since X is path connected by Theorem 2.1.4, for x,y we define d(z,y) = inf{diam(u([O,l])) : u is a path in X from x to y}

where diam(u([O, 11))denotes the diameter of u([O,11) with respect to d'. It is clear that (1): d(z,y) = 0 x = y, (2): d ( z , y ) = d ( y , z ) and (3): d ( z , y ) 5 d(z,z ) d ( t , y). Therefore d is a metric for X. We show that d is compatible with the topology of X. Since X is locally path connected (Theorem 2.1.4), for E > 0 we can choose 6 > 0 such that for x E X the path connected component of z in U,(z;d') contains Ua(Z;d'). Here Uc(x;d') = {y E X : d ' ( z , y ) < E } . Then, if d ' ( z , y ) < 6 then d ( z , y ) < E . Since d ' ( z , y ) 5 d ( z , y ) for z , y E X by definition, the metrics d and d' are equivalent. Connectivity of U c ( x ; d )

+

$6.4 S-injectivity of TA-covering maps

191

is checked as follows. If y E U e ( x ; d ) ,by definition there is a path u in X joining x and y such that the diameter of u([O,11) is less than E. This implies U([O,1])c Ue(z;d). 0 From the following theorem together with Theorem 5.2.1 we have that if f: X -+ X is a TA-covering map of a compact connected locally connected metric space and if X is semilocally 1-connected, then f has a local product structure.

Theorem 6.4.3. Let X be a compact connected locally connected metric space. Let f:X+ X be a TA-covering map. If X is semilocally 1-connected, then f is s-injective. For the proof we prepare the following lemma.

x

Lemma 6.4.4. Let and X be metric space with metrics 2 and d respectively. Let 7 : + and f:X+ X be continuous sujections and let p : -+ X be a covering map. Suppose p o 7 = f o p and X is compact. If there exists 60 > 0 such that for each x E and 0 < 6 5 60 the open ball U&(.) of x with radius S is connected and

x w

x

x

P : U6(.) * U 6 ( d x ) ) i s an isometry, then there is EO > 0 such that for 0 restriction P: WYP(2)) is an isometry. Here

m.1

-

w:(x) = {y E WP(.))

< E < € 0 and x

E

x the

+

x : Z ( ? ( x ) , 7 ( y ) ) 5 E , i 2 O},

= {Y E

x : d ( f i ( P ( 4 ) , Y ) IE , i

2 01.

x x

Proof. As we saw in the proof of Lemma 2.2.34, the map 7 : ---t is uniformly continuous. Hence we take 0 < €0 < 60 such that if z ( x , y ) < € 0 then 2 ( 7 ( x ) , 7 ( y ) )< 60. Let 0 < E < € 0 . If y E w z ( x ) and i 2 0, then we have E 2 2(7(4, = d ( f 0 P(Z),f 0 P(Y)) and hence p ( y ) E W:(p(x)). To see that p r ~ ; ( =is , surjective, if y' E W:(p(z)) then - there is y E such that y' = p ( y ) and d(y,x) 5 E. Since E < € 0 , we have 4f(4,7(d)< 60 and

T(Y))

--

d ( f (4,7(Y)) = d ( f 0 P(x),f

0

P(Y)> < E.

By induction -4

d(f (4,fi(Y)) =d(f

0

P(.),

f 0 P(Y>>I& (i 2 0)

CHAPTER 6

192

and so y E w : ( x ) . Since w : ( x ) c Ua,(z),it follows that p , i p ( = l is injective. The proof is complete. 0

Proof of Theorem 6.4.3. By Theorem 6.2.8 there is the universal covering p : --f X . Let d be a connected metric for X . Apply Theorem 6.4.1 to the and a constant covering map p : -+ X . Then there exist a metric for 60 > 0 such that the properties of Theorem 6.4.1 hold. Let 7 : + be a lift o f f by p (see Theorem 6.3.9). Since f is a self-covering map, by Theorem 6.3.12 it follows that 7 is a homeomorphism. Take 0 < EO < SO as in Lemma 6.4.4 and let 0 < eo < EO be a number with the property of Proposition 5.1.4. Notice that eo is an expansive constant for f. For x E X let y E w d ( x ) . Since p d ( x )is path connected by Theorem 5.1.3, there is a path w : [0,1]+ m d ( x )from x to y. Then by Lemma 5.1.11 for some < tk = 1 we have w ( [ t i , t j + l ] )C Wc',(w(ti)) finite sequence to = 0 < t l < for 0 5 i 5 k - 1. Take b E p-'(x) and choose a path u in starting at b such that p o u = w . Then, by Lemma 6.4.4 it follows that for 0 5 i 5 k - 1 the restriction p : W:o(u(ti))+ W,",(w(ti))is a homeomorphism, and so u([ti,ti+11)c K 0 ( U ( t i ) ) . Let E > 0 be a number such that ke < SO. By Lemma 2.4.1 there is n > 0 -n -d such that fn(W:o(z))C W:(fn(z)) for all z E X . Then, f (Weo(z'))C 3 - d d 1 1 w:(fn(zr)> for all zr E X. In particular, f W e o ( u ( t i ) c ) W e ( f ( u ( t i ) ) )for 0 5 i 5 12 - 1. Hence

x

x

x

x x

x

-*

from which we have d ( f " ( x ) , j " ( v ) ) = d ( f (u(O)),F(u(l))). On the other hand, since 7 is a homeomorphism, F ( u ( 0 ) )# F ( u ( 1 ) ) . Therefore, fn(z)# fn(y) and so f ( x ) # f(y). Since y is arbitrary in w d ( x ) ,it follows that the restriction f: m d ( x )+ m d ( f ( x ) )is injective. Thus f is s-injective.

Theorem 6.4.5. Let X be a compact connected locally connected metric space. Suppose X is semilocally 1-connected. Then n l ( X ) is finitely generated. Proof. Let { Ui}be a finite open cover of X such that each Ui is path connected and if Ui n Uj # 0 for i and j then the inclusion Ui U Uj --t X induces the zero map on the fundamental groups. W e and fix p E X . Let xi E Ui for i and denote by ui a path from p to xi. If Ui n Uj # 0, then we can choose a path ui,j in U;U Uj from xi to x j . Then it is not difficult to show that the finite family of closed paths vi,, = ui ui,j i i j at a base point p generates the fundamental group. 0

-

-

$6.5 Structure groups for inverse limit systems

193

$8.5 Structure groups for inverse limit systems

In this section we will introduce a transformation group for the inverse limit system of a self-covering map, and show that the transformation group possesses similar properties as covering transformation groups. For the case where the space is an n-torus, in the next chapter (Chapter 7) we shall discuss in more detail the geometric structure of the inverse limit systems of the selfcovering maps, by making use of the algebraic structure of solenoidal groups. We say that (Y,X , F , p ) is a fiber bundle if the following conditions are satisfied: (1) Y , X and F are topological spaces, ( 2 ) p :Y -+ X is a continuous surjection, (3) there exists an open cover {Ux : X E A} of X such that for X E A there is a homeomorphism

satisfying p o cpx(x,y ) = z. Here Y , X ,F and p are called a total space, a base space, a fiber and a bundle projection respectively. And each homeomorphism c p is ~ called a coordinate function. For x E X we say that p - ' ( x ) is the fiber over x , and write F, = p-'(x). Note that F, is homeomorphic to F . Indeed, let x E Ux,define cpx,,(y) = cpx(x, y ) for y E F . Then (PA,= : F + F, is a homeomorphism. If a fiber F has the discrete topology, then it follows that a fiber bundle (Y,X , F , p ) means a covering map p : Y -+ X with the covering degree equal to the cardinal number of F . Let f : X + X be a continuous surjection of a compact metric space and let n : Xf-+ Xfbe the inverse limit system. As before, we denote as po : Xf-+ X the natural projection to the zero-th coordinate.

Theorem 6.5.1. Let X be a compact connected locally connected metric space. Suppose X is semilocally l-connected. I f f :X -+ X is a self-covering map and the covering degree is greater than one, then (Xf, X , C , p o ) is a fiber bundle where C denotes the Cantor set. Proof. Let p : -+ X be the universal covering. Given x E X , we choose a canonical neighborhood U of x for p such that U is connected. Then for each n 2 0 the neighborhood U is evenly covered by f " . Indeed, let E E f - " ( x ) and take b,c E si? such that p(b) = t and p ( c ) = x . Since U is evenly covered by p , we can find an open neighborhood V ( c ) of c and a homeomorphism h: U -+ V ( c ) such that p o h is the identity map idu on U. By Theorem 6.3.9 there is a lift jj : -+ of f " by p and g(b) = c. Since 3 is a homeomorphism by Theorem 6.3.12,we put 1, = peg-' o h. Then f " 0 k, = id^, i.e. k, is a lift of idu by f " . Hence, letting U ( z ) = k,(U) we have that U ( z ) is an open neighborhood of z and k, : U + U ( z ) is a

x x

194

CHAPTER 6

homeomorphism. If U ( z )n U ( z ' ) # 0,then by Theorem 5.2.2 k, = k,,, and so z = 2. This implies that U is evenly covered by f". For each point x = (q)E Xf with 20 = x, by the above result we can , : U + Xf such that Sx(x)= x and po o S, = take a continuous map S idu. It follows that Sx(U)n S,(U) = 0 if x # y. Now, we define a map $ : U x p , ' ( z ) t p;'(U) by $(y,x) = S&). It is clear that $ is a bijection. If pi : XI + X is the projection to the i-th coordinate, then pi o $ is a map to k,,(y) if i 5 0, where k,, sending (y,x) to fi(y) if i 2 0, and sends (y,(q)) is a lift of idu by f - i such that kEi(z) = xi. This implies the continuity of $. If the closure cl(U) of U is also evenly covered by p : + X, then in the same way, we obtain a continuous bijection from cl(U) x p i ' ( . ) onto pO'(cl(U)), which is a homeomorphism because cl(U) x p,'(z) is compact. Therefore $ is a homeomorphism whenever U is taken small. Since the covering degree of f is greater than one, it follows that p,'(x) is compact and totally disconnected, and has no isolated points. Combining this fact with Remark 2.2.38 we have that pi'(.) is homeomorphic to the Cantor set. The proof is complete.

x

Figure 26 Let X be a topological space and let 0 be a family of subsets of X. If each point of X has a neighborhood which intersects at most finite members belonging to 0, then 0 is said to be locally finite. We say that X is paracompact if for any open cover V of X there is a locally finite open cover U of X such that U is a refinement of V.

Remark 6.5.2. It is known that for a fiber bundle (Y, X,F , p ) if a base space X is a paracompact Hausdorff space then the bundle projection p : Y + X is a fibration, that is, p : Y + X possesses the homotopy lifting property stated in Theorem 6.2.1. See Spanier [Sp]. Let X be a topological space and f: X + X a self-covering map. As usual, define the space of the inverse limit system as a subspace Xf = {(zi) E X z : f(q)= zi+l,i E Z} of the product topological space Xz. Then the

$6.5 Structure groups for inverse limit systems

195

natural projection po : Xf + X satisfies the following properties (compare to Theorems 5.1.7, 5.2.2 and 6.2.1.): (1) Path lifting property : If u : [0,1] + X is a path, then for x E Xf with po(x) = u(0) there is a path v : [0,1] + Xf such that v(0) = x and po o v = u. (2) Uniqueness of lifting : Let 2 be a connected topological space and let f : 2 + X be a continuous map. If continuous map g,gl : 2 + Xf satisfy po o g = po o g1 = f and there is a point EO E 2 such that g ( z 0 ) = g'(zo), then g = 9'. (3) Homotopy lifting property : Let 2 be a topological space and f : 2 + Xf a continuous map. If a continuous map F : 2 x [0,1] -, X satisfies F ( E ,0) = po o f ( z ) for z E 2, then there is a continuous map G : 2 x [0,1] + Xf such that po o G = F and G(z,0) = f ( z ) for E E 2. Indeed, given a path u : [0,1] -, X and a point x = ( x i ) E Xf with p o ( x ) = u(O), by Theorem 5.1.7, for each i 5 0 there is a path vi : [0,1] + X starting at the point xi such that f - i o vi = u. Letting vi = f i o u for i > 0, we define v : [0,1] + Xf by v ( t ) = ( v i ( t ) ) . Then v satisfies the conditions in (1). Similary, (2) and (3) are easily checked from Theorems 5.2.2 and 6.2.1 respectively. Let f : X + X be a self-covering map of a compact connected locally connected metric space. For x = (xi) e Xf we define a subgroup H,(x) of the fundametal group ? r l ( X , . ~ )as follows: [u] E H,(x) if and only if for each i < 0 the lift of u by f --I starting at zi is a closed path from xi to xi. From Lemma 6.2.3 it follows that if f * : ? r l ( X , z o )+ ? r l ( X , z l )is the induced homomorphism, then the restriction f* : H,(x) + H,(u(x)) is an isomorphism where u : Xf + Xf is the shift map. A homeomorphism a : Xf + Xf is called a transformation for the natural projection po : Xf + X if po o a = po holds. We denote as F(p0) the set of all transformations for PO. Then (1) the identity map of Xf belongs to F(po), (2) ifQl,Q2 E F(Po), then a1 0 a 2 E F(Po), (3) if a E F(po), then a-l E F ( p 0 ) . Hence F ( p 0 ) is a group. In fact, F ( p 0 ) is a topological group equipped with the Co topology. F ( p 0 ) is called the structure group for the inverse limit system

u:xf+ X f .

Theorem 6.5.3. Let X be a compact connected locally connected metric space and let f : X + X be a self-covering map with the covering degree greater than one. Suppose X is semilocally 1-connected and H,((ci)) is nornaal in ?rl ( X ,CO) for some ( c i ) E Xf. If the factor group n l ( X , c o ) / H , ( ( c i ) ) i s abelian, then the structure group F ( p 0 ) for the inverse limit system u : Xf + Xf possesses the following properties: (1) for x = (xi) E Xf the map a H a ( x ) is a homeomorphism from F ( p 0 ) onto the fiber p i 1 (xo), (2) the orbit space Xf/F(po) is homeomorphic to X ,

CHAPTER 6

196

(3) there is a (homeomorphic) automorphism u, : F(p0 that for every Q E F(p0) u 0 Q = u,(a)0 u,

(4) let p : X

-+ X be the universal covering, G(p) the covering tmnsformation group for p , and for a lift 7 : -+ of f by p let f , : G(p) -+ G(p) be the induced homomorphism, then there is a homomorphism r, : G(p) + F(p0) such that r,(G(p)) is dense in F(p0) and u* o r, = r, o 7, holds, i.e. the following diagmm commutes :

x

G(P)

-5

G(P)

1 Moreover for every (zi) E Xf the subgroup H , ( ( z i ) ) possesses the following properties: ( 5 ) if 2 0 = co then H , ( ( Z i ) ) = L ( ( C i ) ) , (6) if w is a path in X from ~0 to zo then w * ( H , ( ( z i ) ) ) = Hdo((ci)) where w, : ?rl(X,zo)+ nl(X, CO) denotes the induced isomorphism in Lemma 6.1.4. For the proof we need some lemmas. Hereafter, let X be a compact connected locally connected metric space with -be a self-covering map. Suppose X is semilocally metric d and let f : X --t X 1-connected and let p : X + X be the universal covering. We take and fix a metric 2 for and a constant 60 satisfying the properties of Theorem 6.4.1. Also, we denote as G(p)the covering transformation group for p and as O(f ) the family of all lifts of f by p. Notice that every element in O(f ) is a homeomorphism of (Theorem 6.3.12). A compact subset C of is called a compact covering domain for p if p(int(C)) = X where int(C) denotes the interior of C in Since X is compact, it follows that a compact covering domain for p always exists.

x.

L e m m a 6.5.4. For all 2,y E

x

-

4%ar) < 6

E

> 0 there is 6 > 0 such that for

*

m=@(3(z),

all 7j E O ( f ) and for

3(Y>>, &T-l(2), 3-1(Y))) < E .

Proof. Let m be the covering degree of f . Fix 7 an element in O(f).By Theorem 6.3.12 (2) there is a finite sequence cq, * ,am-l in G(p) such that

-

G(P)= 7*(G(P))u a1 0 7*(G(P))u * - . u am-1

0

7,WP)).

$6.5 Structure groups for inverse limit systems

197

If C is acompact covering domain forp, then CO= C U a ~ l ( C ) u - - - U a ; ; l ' _ l ( C ) is compact, and hence there is 6 > 0 such that for z , y E CO

d(z,y)

< 6 * max{a(3'(z),7(y)),a(7-1(z),7-1(y))} < E-

-

For z,y E X suppose a(z,y) < 6. Since p(int(C)) = X ,if 6 is small then a ( z ) ,a ( y ) E C for some a E G(p). Since a = a;of,(/?) for some p E G(p), we have T*(p)(z),7*(P)(y) E CO.Hence, by Lemma 6.3.10 it follows that

-d(f(z)93'(y)) =

W * ( a >3'(4,7*(a> 7(ar)) -0

0

= d(f

0

a(4,7 0 4Y)) < E

and

For any 3 E O ( f )there is 7 E G ( p )such that 3 = 703' (Theorem 6.3.9). Thus, d(3( z),3(y)) = f (y )) < E and a(3-l (z),3-' (9)) = 2(7-'07-' (z),7-'0 ~ - ' ( y ) ) < E because 7 is an isometry. -

-2

X

a@(%),

x2

Define a product set = { ( u i ) : ui E as usual by a ( ( u ; ) )= (u;+1). Write

Xf = { ( Z i ) E

(6.3)

x2:

x,i E Z} and a shift map T :xz

--f

f(P(Zi))

= p(z;+1),iE Z}.

xf.

It is = XIholds. Let u = ( u i ) E For each i E Z denote - clear that F(xf) f u i , u i + l the element f in O(f) such that ?(u;) = u;+l and define fUi--l,Ui

(6.4)

f, i

=

O..*ofuo,tLl

('fui,ui+l)-l

{

idx

0

0

(3'u-1,uo)-1

ifi>O

if i < o if i = 0.

x

Then for each i E Z, 7" is a homeomorphism from onto itself and by Lemma 6.5.4 it is bi-uniformly continuous with respect to the metric 2. Given 3 E O(f) we have i f, = a;o 2 for some a; E G(p) if i 2 0. If, in particular, f is a homeomorphism, then 7; = a;o 3 for all i E Z where a; E G(p). We Xf by define a map r, : X --f

CHAPTER 6

198

By definition the following properties are easily checked:

{

(6.6)

TU('110)

pi

= (P(Ui)), = P 0 T;(z),

0 ~"(2)

Crn 0 Tll(2:) = %~(U,Cr",(.))

( n E Z).

where p i : Xf -, X denotes the natural projection to the i-th coordinate. From now on, let us suppose that the covering degree of f is greater than one. By Theorem 6.5.1 it follows that ( X f , C , X , p o )is a fiber bundle where C is the Cantor set. We note that a coordinate function cp : U x C -, p,'(U) for ( X f ,C, X , p o ) exists whenever U is a connected open set of X with small diameter.

xf

Lemma 6.5.5. For u = (ui)E the following properties hold: (1) T, : X + Xf is continuous, (2) T,(X) is dense in Xf, ( 3 ) T,(X) is the path connected component of ~,,(uo)in X f , (4) if U is a connected open set of X and cp : U x C + p,'(U) as a coordinate function for ( X f , C , X , p o ) ,then there is a dense subset C' of C such that cp(U x C') = p;'(U) n

~,(x).

Proof. (1) is clear. To show (2), let (xi) E Xf and let n > 0. Take a point y in p - ' ( ~ - ~and ) choose 2: E such that Tin(z)= y. Then pi o T,,(z) = 5 , for i 2 -n. Since n is arbitrary, the point (xi)must belong to the closure of T,(X) in Xf.Thus (2) holds. is path connected. Since is path connected, by (1) it follows that ) Then po o u is a path Conversely, let u be a path starting at ~ ~ ( ' 1in~ 0Xf. in X starting at p(u0). Lift po o u by p to a path v with the initial point uo. Then 7, o v is a path in Xf starting at T,(uo). Since po o 7, o v = po o u, by o v = u. This means that T,@) contains the path Theorem 5.2.2 we have connected component of T,(UO) in Xf.Therefore (3) holds. Since U is path connected and C is totally disconnected, by (3) we have n p,'(U) = cp(U x C') for some subset C' of C, and by (2) it follows that C' is dense in C. Thus (4) holds. 0

x

~,(x)

x

~,(x)

xf.

Let u E We define a family T, of subsets of T,(X) as follows: V E T, if and only if there is a connected open set U of X such that V is expressed as V = cp(U x { a } ) by a coordinate function cp : U x C -, p;'(U) for ( X f , C , X , p o ) , where a is a point in C. It is easily checked that (1) any point in T,(X) belongs to some V E T,, (2) if V', V2 E T, and x E V1 n V2, then there is V, E T, such that x E V, c vl n v2. We call this topology Therefore, the family T, generates a topology of the intrinsic topology of

~,(x).

~,(x).

$6.5 Structure groups for inverse limit systems

-

a,

.,(a)

199

Lemma 6.5.6. For u E the map T, : X + and the restriction po : + X are both covering maps under the intrinsic topology of and the following diagram commutes.

~,,(x)

-

x

I P

3

~,,(x),

Tu(x)

J PO

X

Proof. By defintion it is clear that the diagram commutes. Let x E X and let U be a connected canonical neighborhood of x in X for p. By Lemma 6.5.5 (4) it follows that the U is evenly covered by po : -+ X if diam(U) is small, from which po : T,,(X)+ X is a covering map. Let V be an open set of such that po : V + U is a homeomorphism. Since U is evenly covered by PO, by the above commutative diagram we see that V is evenly covered by T,,, and so T,, is a covering map. 0

~,,(x)

~,,(x)

Lemma 6.5.7. If po+ : T ~ ( T , , ( ~ ) , T , , ( z-+) ) nl(X,(x)) is the induced homo= H=(T,,(Z)). morphism of fundamental groups, then po,(nl(.r,,(X), ~~(2)))

Proof.If [u] E H o o ( ~ , , ( 2 ) )then , by definition for each i 5 0 there is a closed path u; from ~ ; ( T , , ( z )to) pi(^,,(^)) such that f -;o u; = u. Letting ui = f' o u for i > 0, we have a closed path v = ( u ; ) in from T,(z) to T,,(z). Then [v] E T ~ ( T , , ( ~ ) , T , , ( z ) )Since . po o v = u, it follows that p o + ( [ w ] )= [u]. Conversely, if [v] E nl(~,,(X), T,,(z)),then for each i 5 0, pi o w is a closed path in X from pi(^,,(^)) to P ; ( T , ( Z ) ) and f - i o p; o v = po o v, and so p o * ( [ v ] )= [Po 0 I. E ~ o c ( T u ( 4 ) .0 Lemma 6.5.8. Let h,h' E F(p0). If h ( x ) = h'(x) for some x E Xf,then h = h'.

~,,(x)

Proof. Choose points u = (u;) and v = ( w ; ) in such that (p(u;))= x and ( p ( v i ) )= h(x). Then x E and h ( x ) E Since is the path connected component of x in Xfby Lemma 6.5.5 (3), we have that h ( ~ , ( x ) ) is path connected, and so h ( ~ , ( x )c) Since h is a homeomorphism, it follows that h ( ~ , ( x )=) Next, we show that h : ~ , , ( x+ ) is a homeomorphism under the intrinsic topologies. Let U be a connected open set of X and suppose diam(U) is small. Since h E F ( p o ) , it follows that h(p,'(U)) = p,'(U). Notice that p,'(U) is expressed i18 p,'(U) = cp(U x C) by a coordinate function cp. If V = cp(U x { a } ) c then we have h ( V ) = cp(U x { a ' } ) for some a' E C. This means that h: + is a homeomorphism. ) Since h ( x ) = h'(x), in the same way as above we have that h ' ( ~ , ( x ) = and h' : -+ is a homeomorphism. Since pooh = po = pooh', we obtain h = h' (see Remark 6.5.2 (2)). 0

~,,(x)

~,,(x). ~ , ( x ) ~,,(x). ~,,(x). ~,,(x)

~,,(x), ~,,(x)~,,(x) ~,,(x) ~,,(x)~ " ( x )

CHAPTER 6

200

a,,

~=(,,)(x)

Lemma 6.5.9. For u E u ( T , , ( ~ )= ) and the restriction u : + ~z(,,)(x) is a homeomorphism under the intrinsic topologies. Furthermore the following diagram commutes:

~,,(x)

x -

f

x

Proof. Since u is a homeomorphism, in the same way as the first part in the proof of Lemma 6.5.8 we have u ( T , , ( ~ )=) ~z(,,)(x).It is clear that the diagram commutes. Let U be a connected open set of X with small diameter. Since poou = f opo, clearly u(p;'(U)) c p,'(f(U)). If cp : U x C + p,'(U) is a coordinate function and V = cp(U x { a } ) for a E C, then u ( V ) = cp'(f(U) x {a'}) for some coordinate function 'p' : f ( U ) x C + p o ( f ( U ) ) and some a' E C, because po o u ( V ) = f o po(V) = f (U).This implies the continuity of u : -t T!(,,)(X).By the commutative diagram together with Lemma 6.5.6 the conclusion is obtained.

~,,(x)

Now, we suppose that H,((ci)) is normal in .~(X,CO) for some ( c i ) E Xf, and choose b = (a;) E such that ( p ( b i ) ) = ( c i ) . Since q, : + is a covering map by Lemma 6.5.6, we denote as G(q,) the covering transformation group for q,. By Lemma 6.5.6 it is clear that G(q,) is a subgroup of G(p). Apply Theorem 6.3.4 (2) to p : -+ X and q, : + FYom Lemma 6.5.7 together with the assumption that H,((ci)) is normal in .1(X,cg), it follows that G(q,) is normal in G(p). Hence we can consider G(p)/G(q,) as the covering transformation group for po : -+ X (see Remark 6.3.6).

x q,(x)

x q,(x).

x

q,(x)

Lemma 8.5.10. Let z E X . Then G(p)/G(q,) is transitive on

q,(x)n

PO'(X:)*

P m f . This follows from Theorem 6.3.4 (1). 0 Moreover we suppose that ? r l ( X , ~ ) / H ~ ( ( cisi an ) ) abelian group. Then G(p)/G(q,) is abelian. Lemma 8.5.11. G(p)/G(q,) is u uniformly equi-continuous family with respect to the compatible metrics for Xf.

Proof. We recall that Xf has a metric d' defined by

56.5 Structure groups for inverse limit systems

201

q,(x).

Fix 6 > 0 small and suppose J((zi),(yi)) < 6 for (zi),(yi) E Then d ( z 0 , yo) < 6. Since 6 is small, there is a connected canonical neighborhood U for p such that 20 and yo belong to U. Notice that U is evenly covered by f -i for i 5 0 (see the proof of Theorem 6.5.1). Hence for each i 5 0 we can take such that zi E U ( z i ) and f - i : U ( z i ) + U a connected open set U ( z i ) of is a homeomorphism. Since d ( ( z i ) , ( y i ) ) < 6, it follows that for some large n(6) > 0 if i 2 -n(6) then yi E U(zi). Take z,y E such that ( z i ) = q,(z),(yi) = q,(y). We claim that for each a E G(p),if -n(6) 5 i 5 0 then p o x ( a ( z ) ) and p o z ( a ( y ) )are both points in the same component of f ’ ( U ) . Indeed, let u be a path in from z to a(.). Then p o u is a closed path from 10 to 20. If v denotes a path form z to y, then p o v is a path from zo to yo. Put w = p 6 v p o u * p o v. Then w is a closed path from yo to yo and its lift starting at y by p has a(y)as the terminal point. Let z: = Ti(a(z)) and y: = Ti(?(y)). Then p ( z { ) is the terminal point of the lift of p o u starting at zi by f-’ and p(y:) is that of the lift of w starting at yi by f - i . Since G ( p ) / G ( q , )is abelian by assumption, this implies that p ( z { ) and p(y:) are both points in same component of f ’ ( U ) whenever yi E U(zj). Therefore the above claim holds. Let E > 0. By the above result it follows that d ( p o f ’ , ( a ( z ) ) , p o x ( a ( y ) ) )< E for -.(a) 5 i 5 n(6) if d(zo,yo) = d ( p ( z ) , p ( y ) )is small, and then d(q,(a(z)), % ( a ( y ) ) ) < E because n(6) is taken large. Hence J ( a ’ ( ( z i ) ) , d ( ( Y i ) ) )= d ( q , ( a ( z ) )q,(a(y))) , < E for a E G(p) whenever d((zj),(yi)) < 6 and 6 is small. Thus G(p)/G(q,) is uniformly equi-continuous. 0

x

x

x

-

q,(x)

Since is dense in Xj by Lemma 6.5.5 (2), we have by Lemma 6.5.11 that each a’ E G ( p ) / G ( q , )is extended to a continuous map of Xj,written as %,(a). From the fact that G(p)/G(q,) is a group, it follows that %+(a)is a homeomorphism. Since po o a’ = po holds, clearly %+(a)is a member in the structure group F(po).

Lemma 6.5.12. q,+ : G ( p ) ---t F ( p 0 ) i s a homomorphism. Proof. This is clear from definition. 0

We denote as %(Xf)the set of all homeomorphisms of Xf.Then a metric d for ‘H(Xf)is defined by

d(h, h’) = m a { sup{d(h(z), h‘(z)) : z E X ~ I , sUp{d(h-l(z),h’-l(z))

:z E

Xf}}.

‘H(Xf) is a complete metric space. Since q,+(G(p))is a uniformly equicontinuous subgroup by Lemma 6.5.11, it follows that q,*(G(p))is relatively compact in ‘H(Xf),that is, the closure F of q,+(G(p))is compact. Notice that each element in F belongs to F ( p 0 ) .

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202

Proof of Theorem 6.5.3. Let x = (q)E X j . Since q,*(G(p))xis dense in p,'(so) by Lemmas 6.5.5 (4)and 6.5.10, we have Fx = p r ' ( z 0 ) . Here Fx = { h ( x ) : h E F}. Hence by Lemma 6.5.7, F = F ( p 0 ) and therefore (1) holds. To show (2), define g : X j / F ( p o )-+ X by g(F(p0)x)= po(x). By Lemma 6.3.1 it is clear that g is continuous. Since g is bijective, (2) holds. From (1) together with Lemma 6.5.7 we obtain (5). To see (6),let w be a path in X from co to 20. We can take a path iij in X j starting at the point c, such that PO O E = w (see Remark 6.5.2 (1)). Since c = q,(bo) E the path Ti7 is in from which an isomorphism E* : ~~(q,(x),E(l)) -+ 7rl(q,(X),c) is induced as in Lemma 6.1.4. Project this isomorphism by PO and use Lemma 6.5.7. Then w*(Hm(Ti7(l))) = H,(c). Since po(E(1)) = 20, by (1) and Lemma 6.5.7 we-have - H,(W(l)) = H , ( ( z o ) ) . Therefore (6)holds. To show (3) and (4),let f :X --t X be a lift off by p . Since q(X,zo)/H,( (zi)) is abelian for all - ( z i ) E X j by (6), we may suppose 7; = (W E Z). Then q, = T q b ) on X ,and hence q,* = T q b ) * : G(p) + G ( p ) / G ( q , ) . Since the shift map u satisfies PO o u = f o PO, we have the restriction u : p , ' ( ~ ) + p,'(f(~)). We calculate for a E G(p)

q,(x),

q,(x),

7

0

%bz(a)(Tb(bO)) = g o '

b ( a ) ( ( P0 X ( b 0 ) ) )

= 4 ( P 0 X(aP0)))) = (P

z(b)(fb(a(bO)))))

= (P &b)(T*(a)(?(bO)))) = TZ(b)*(?*(a))

= Tb*(?*(a))

TZ(b)(?(bO)) 'J(Tb(b0)).

Since Po

0 00

q,*(a)(n(bo)) = f 0 P ( 4 ) = PO

TZ(b)*(?*(a)) O(%(bO)),

by uniquenss of lifting (Remark 6.5.2 (2)) it follows that u 0 T b + ( a ) = q,+(f, (a))o u for all a E G(p). Since a map from q,*(G(p))to F ( p 0 ) defined by **(a)

7b*(lf*b))

is uniformly continuous, we have its extension u* : F(p0) -+ F ( p 0 ) . Therefore (4)holds. Let ?,bC : F ( p 0 ) -+ p r ' ( c 0 ) be defined by &(h) = h(c). By (l), $.J~ is a homeomorphism. Similary, define )$,(, : F ( p 0 ) -+ p;'(f(co)). Then we have on pi' (CO) 0 = & ( c ) 0 u* 0

$2,

$6.6 Lifting of local product structures

203

from which cr* is a homeomorphism. Therefore (3) holds. 0

36.6

Lifting of local product structures

Let X be a compact connected locally connected metric space with metric d . In $85.2 and 6.4 we have shown that a TA-covering map f:X+ X has a local product structure if X is semilocally 1-connected (in other words, X has the universal covering space). This section will discuss that the local product structure can be lifted on the universal covering space, and investigates some properties of the stable and unstable sets on the space. Furthermore, a characterization of special TA-covering maps may be found. Throughout this section we suppose that X is semilocally 1-connected. -+ X the universal covering and as G(p) the As before, denote as p : covering transformation group for p , and take a metric 2 for and a constant 60 > 0 satisfying the properties of Theorem 6.4.1. Let f:X --+ X be a self-covering map. For x E and u E we define a local stable set w:(z; u) and a local unstable set w r ( x ;u) by

x

x

xf

x

7

If we fix an element 7 in O ( f ) , then there is p; E G ( p ) such that 7; = /3; o (i 2 0). Hence = z ( y ( x ) , T ( y ) )for i 2 0, from which the local stable set w:(x;u) does not depend on the choice of u. For simplicity we write W : ( z )= V:(x;u) (x E and u E

z(T“(x),Ti(y))

x

xf).

wr(x;

If 7, : G(p) + G(p)is not isomorphic, then the local unstable set u) is not necessarily independent of u. Let EO > 0 be a number with € 0 < 60 such that the implication in Lemma 6.5.4 holds for E = 60 and d = E O . In the case where f is c-expansive, we take € 0 as an expansive constant for f because EO is small. We note that for 0 < E < € 0 the sets m : ( x ) and m r ( x ;u) are projected isometrically on X by p, because p : u6,,(2)+ U&,,(p(a))is an isometry for x E

x.

Lemma 6.6.1. Let x E

x and u E xf.Then for 0


for all n 2 ny. Proof. This follows from Lemmas 2.4.1 and 6.6.1. 0

xf.

Let x E Fix u E set wy (2;u) by

X

and define a stable set

-

x : ~(Fi(x),A ( y ) ) w (x;u) = {y E x :a(T;(z),f"l(y)) ~ ' ( xu); = {y E

--f

Y

--+

ws(x;u) and an unstable o as i

o as i

--+

--+

oo},

-..I.

56.6 Lifting of local product structures

205

Since wd(z; u) is independent of u, we write

-

Wd(z) = W"(z;u).

It follows that the families

-

*

f = {Wd(z): z E X}

are decompositions of

7, = {W(z;u) : 2 E X}

xland for i E Z

Lemma 6.6.4. Suppose f: X

+X

is c-expansive. For 0 < e < € 0 and

EX

Proof. This is clear by Lemma 6.6.3. 0 Theorem 6.6.6 (Lifting of local product structure). If f : X -+ X is a TA-covering map and 0 < e < € 0 is a small number, then for E and u E there is a connected open neighborhood r ( x ; u ) of z in and a continuous map E,,: N ( z ;u) x r ( z ; u) + r ( z ; u) such that

x

a,

(1)

{~'U(Yl.))

=-T(Y;

(2) for-Yl 21 w E

"3;

u>n m4 for Y, E

4

w; 4,

au(Y1Y) = Y9 -

QU(YlEU(.Z, 'UI)) = -%&41.1) = W G l ( Y 1 el, w), (3) the restriction E,,: Dd(z)x P ( z ; u) + u) is a homeomorphism where = w:(z) n r ( z ; u ) and s ( z ; u) = w,"(z;u) n E(z;u), (4) there-is -a constant p > 0 such that m(z;u) 3 B,(z) where Z,(z) = {Y E x : 4 Z l Y ) I PI, ( 5 ) 7,(Dd(.)) c Dd(T,(4)and 7,(Wz; u>>3 s ( 7 u ( 4 ; q u ) ) . Furthermore, i f f is expanding then ?;j"(z) = {z} and P ( z ; u) = N(z;u), and otherwise (6) Is"(.) {z} and s ( z ; u) {z}.

x(z;

nd(z)

2

2

CHAPTER 6

206

Proof. Since X is semilocally 1-connected, by Theorem 6.4.3 it follows that f is s-injective, and hence by Theorem 5.2.1, f has a local product structure satisfying the conditions stated in Theorem 5.2.1. Fix 0 < e < € 0 . For x E and u E let NTU(=) be as in Theorem 5.2.1. Since diarn(NTu(*))< €0 < SO, we define T(x;u) as the inverse image of NTu(z) by the isometry p : Ua,(x)+ Va,(p(s)). Then, by Theorem 5.2.1 (2) we have that x(x;u) is a connected open set. (4) follows from Theorem 5.2.1 (3). To show the existence of the map E,,,let STu(z): NTU(=) + Xf be the continuous section satisfying the condition (B) of the local product structure. We claim that STu(z)(p(y))= T,,(Y) for all y E x ( x ;u). Indeed, consider a map T,, o T : NTu(x) -+ Xf where T denotes the inverse Since S,(,)(p(x)) = T,(z) by definition, by map of p : z ( x ; u ) + NTU(=). Theorem 5.2.2 we have STu(z)= T,, o T. This implies the above claim. Since for a, b E N r , ( p ( z ) )

x

xf

[STu(z)(a), bl = ~,u(S,c=,(a))n W ( b ) by Theorem 5.2.1 (l), from Lemma 6.6.2 it follows that v ( y ; u ) n w:(z) consists - of only one point, say &(y, z), for y, z E N(z;u). Therefore a map E,, : N ( z ;u) xx(z; u) -+ z ( x ; u) is obtained, so that (1) holds. The continuity of E,, follows from those of [ , ] and STU(=).By the condition (B) we have (2). (3) is obtained from (2) and the continuity of E,,. Notice that and ----u D (x;u) are mapped by p to Da(x)and D"(T,,(z))respectively. Hence (5) follows from Theorem 5.2.1 (4). The last statement is clear from the second part in Theorem 5.2.1. 0 Suppose f : X 4 X is a TA-covering map and fix 0 < e < € 0 small. Under the notations of Theorem 6.6.5 we have the following Lemmas 6.6.6-6.6.12.

n"(x)

Lemma 8.6.6. Let p > 0 be as in Theorem 6.6.5(4). Then for x E UEXf

-

W i ( x )c D"(x)

x and

u

WJx; u) c r ( z ;u).

Proof. This is easily checked from Lemma 5.2.4. 0

a,.

x

Lemma 6.6.7. Let x E and u E Then w " ( x ) and w"(x;u) are path connected Proof. Combining Lemmas 6.6.6 and 6.6.4, we have

-

W " ( 4 = (J7 & ) ( m U ( x ) ) ) ,

-

W"(Z;U>=

i20

UZ-i(,,)(D( f , , ( s ) ; ~ - i ( u ) ) ) . --u

--i

i20

On the other hand, by Theorem 6.6.5 (3) it follows that are path connected. Thus the conclusion is obtained. 0

Da(x)and g ( x ;u)

56.6 Lifting of local product structures

Lemma 6.6.8. For z E

207

x and u E xf,

(1) P ( m z ;4) = WU(TU(Z)), (2) p ( w 8 ( z )= ) F b ( p ( z ) ) where t?l'"(p(z)) is the stab2e set in strong sense, and p : w"(z)+ w " ( p ( z )i)s bijective.

a(E(z),Ti(y))

Proof. Let y E T ( z ; u ) . Since -+ 0 as i -+ -00, we have d(p; o ~ ~ ( p2; o) T,,(Y)) --.t 0 (i + -00). Thus p(y) E W u ( ~ u ( ~and ) ) so p ( V ( z ;u)) c WU(7u(z)). If y E WU(~,,(z)),by definition we can find (y;) E Xfsuch that yo = y and d(y;,pi o ~ ~ ( 5-+) )0 (i -+ -00). Let p > 0 be as in Theorem 6.6.5 (4). Then by Lemma 6.6.6 we have --u -;

Ir-i

W p ( f u ( 4 ; ~ ( u )c) D

and so W;(oi o ~

c

(fu(4;m),

~ ( 2 ) Du(oi ) o ~ ~ ( 5for ) )

i 5 0. Hence there is i < 0 such

U-i

that yi E Du(oi o ~ ~ ( 3 )Since ). p ( D (f,(z);$(u))) = D u ( do ~ ~ ( 2 we ) ) )can 11 -; find jji E D (fu(z);$(u)) such that p(jji) = y;, and then

-_

-u

tT:(u)(gi) E

-i

G : ( ~ ) ( D(fu(z>;ai(u>>)

c ~i:~u)(~(Tu(z);~(u)))

c T ( z ; u). Since poTi:(,,)(jj;) = f - i o p ( j j i ) = f - i ( y i ) (note i 5 0) and y = yo = f-i(yi), we have y E p ( w ( z ;u)), i.e. W"(T,,(Z)) c p ( W ( z ;u)). Therefore (1) holds. Let y E w"(z). Since Z(Ti(z),T"(y))+ 0 as i -+ 00, it follows that d(fi(p(z)),fi(p(y))) + 0 as i -t 00. Hence p ( w " ( z )c ) W 8 ( p ( z ) ) .Since Wd(z) is path connected by Lemma 6.6.7, so is p ( w " ( z ) )F'rom . these facts together with Theorem 5.1.3 (1) we have p ( w " ( z )c ) P8(p(z)). Conversely, let z E @ " ( p ( z ) ) . Since r?l"(p(z))is path connected, there is a path u from p ( z ) t o z. By Lemm 5.1.11 we have that for some sequence 0 = t o < tl < < t k = 1, u([tj,tj+l]) c W,d(u(tj)) (0 I j 5 k - 1) where 0 < E < € 0 . If v is a path in starting at z such that p o v = u, then by Lemma 6.6.1, v([tj,tj+l]) c w:(v(tj)) for 0 5 j 5 k - 1. Hence ~ ( T i ( ~ ( 0 ) ) ~ T ~ ( v-+ ( l 0) ) ) (i -t 00) by Lemma 6.6.3, and so v(1) E w"(z). Since p(v(1)) = z , we have p(m"(z)) 3 w " ( p ( z ) )Therefore, . p ( w s ( z )= ) w " ( p ( z ) )Let . y,z E and y # z. Then there is i > 0 such that

...

x

ma(.)

60

2 a(Ti',(y),Tu(~,)= d ( f i 0 P(Y),f i

0

P(Z))

from which p , ~ . (is~injective. ) Thus (2) is obtained. 0

> 0,

CHAPTER 6

208

a,.

x

Lemma 6.6.9. Let u E For x,y E there are at most countable sets A c nd(y) and B C s ( y ; u ) such that (1) W(v;U) n W"(Z)= E,@"(Y) x B ) , (2) W(y; U) n W ( ZU); = Eu(A x r ( y ; u ) ) , (3) the connected component of z in X(x;u)nW'(z) coincides with zi'(z) and that of x in W(x;u )n W(z;u ) coincides with r ( z ; u). Proof. Letting B = r ( y ; U)nw"(z),from Lemma 5.1.10 together with Lemmas 6.6.1 and 6.6.8 (2) we have that B is at most countable. On the other hand, if w:(a)n wd(b) # 0 for a, b E then w:(a)c wd(b) (by Lemma 6.6.3). Hence, by Theorem 6.6.5 (1) and (2) we have

w,

Therefore (1)holds. To show (2), let A = W(z;u)n nd(y). Then A is at most countable. is uniformly This is checked by following the proof of Lemma 5.1.10 since continuous for each i 5 0 by Lemma 6.5.4. (The detail is left to the readers.) In the similar way as above, we obtain (2). (3) follows from (1) and (2). 0

--

Lemma 6.6.10. For x E X , W"(z) is the path connected component of z in P-' ( J + " ( P ( W Proof. Let u be a path starting at z in p-'(@"(p(x))). Then p o u is a path ) p o u(0) = p ( z ) . This implies u(1) E w d ( x )(see the proof of in W " ( p ( z )with Lemma 6.6.8 (2)). Thus the conclusion is obtained. 0 Lemma 6.6.11. Let G ( p ) be the group of all cowering tmnsformations for p . Then (1) a(W'(z))= w"(a(x)) for a E G(p) and x E (2) i f f : X X is a special TA-covering map, for a E G(p), u E and x E X the following properties hold: (2.a) T ( z ; u ) i s the path connected component of x in p - ' ( W " ( ~ , ( x ) ) ) , (2.b) a ( W ( z ;u)) = W ( a ( x )u). ;

x,

--f

Proof. (1) is clear. By Lemma 6.6.8 (1) we have

xf

$6.6 Lifting of local product structures

209

then W " ( a ( z ) ; u )is the path connected component of a(x)in p - l (W"(T,, (x))). This is checked as follows: since the fundamental group ?rl (X) is at most countable by Theorem 6.4.5, from Theorem 6.3.4 it follows that p - l ( p ( z ) ) is at most countable. Hence, by Lemma 6.6.9 (2) for z E there is at most a countable set A, c nd(z)such that

x

nqz;

p-1(wu(TU(z)))

u) = Eu(Az x ~ ( zu)). ;

x

Let u be a path in starting at x such that u([O,1)) is contained in p-'(W"( T,,(z))). Then u([O,tl])C x(x;u) for some 0 < t1 5 1. Since A, is at most countable, it follows that u([O,tl])C g ( z ; u ) c r ( z ; u ) . Hence, letting J = {t E [0,1]: u ( t ) E W(z;u)), we have [O,t,] c J. In the same way as above, it is easily checked that J is open in [O,l]. Since each w ( z ;u) contains the pneighborhood of z (Theorem 6.6.5 (4)), J is closed in [0,1].Thus J = [0,1].Combining this fact with Lemma 6.6.7, we obtain that w ' ( y ; u) is the path connected component of y in p-'(W"(~,,(z))). On the other hand, since a(W"(z;u)) is path connected, we have a(W"(.; u)) c WU(a(z);u)

and therefore a ( W " ( x ; u)) = r ( a ( x ) u). ; We must clear away the above assumption to complete the proof of (2). Notice that po o T,,(z) = po o T , , ( ~ ( Z >for ) xE uE and a E G(p). Since f: X X is a special TA-covering map, we have W " ( T , , ( ~ ( Z=) )W) u ( ~ , , ( 2 ) ) . Since p ( r ( a ( z ) ;u)) = W"(T,,(CY(Z))) for a E G(p) (by Lemma 6.6.8 (l)), we and hence have p ( W " ( a ( z ) ;u)) = W"(T,,(Z))

x, xf

P-'(Wu(Tu(4)) 3

u

W - ( a ( z ) ;u).

UEG(P)

Let y E p - ' ( W " ( ~ , ( x ) ) ) .Then p(y) E W"(T,,(Z)) and so W"(T,,(Z)) = W " ( T , , ( ~ ) )By . Lemma 6.6.8 (l), p(W(y;u)) = W"(T,,(Y)), from which p ( v u ( y ;u)) = W"(T,,(Z)). Hence we have P-l(W"(Tu(4)) =

(J

a(W"(y;

4)

amp)

and so a(%)E w ' ( y ; u) for some a E G(p). Therefore, y E r ( a ( z ) ;u). 0

Lemma 6.6.12. Suppose f is a special TA-covering map. Let u,u' E Then for x E u w (2; u) = W(z;u').

w

xf.

Proof.This follows from Lemma 6.6.11 (2.a). 0 As above, let X be a compact connected locally connected semilocally 1connected metric space. As a characterization of special TA-maps we have then the following Theorem 6.6.13.

CHAPTER 6

210

Theorem 6.6.13. Let f : X -+ X be a TA-covering map. Then f : X + X is special if and only if for E > 0 there is 6 > 0 such that for x , y E Xf with zo = yo = 2 Wz(x) fl B6(z)= w:(Y) B6(z).

where Bs(z)= { y E X : d ( s , y ) 5 6). Proof. e): This follows from Lemma 2.4.3. +): Letting e = E , we take p > 0 as in Lemma 6.6.6. By Lemma 6.6.3 there is n > 0 such that 7 ; n-( W , ( a ) )c u --n (a),Z-"(v)) for all v E and a E Let z E p - ' ( z ) and choose u,u' E such that ~ ~ ( = 2 x) and T,,,(z) = y . Since f is special, by Lemma 6.6.12 it follows that V ( z ; u ) = u W (z;u'). Hence

m,,(f,, wf

x.

- --'

wf

n

By the choice of n we have g(Du@-'(z);(~"(u)))c w]r(z; u) where 3 = 7, , and so g ( D ( g (z);an(u))) c X(z;u') nm"(z;u'). Since g ( s ( g - ' ( z > ; ( ~ " (u))) is connected, we have g ( D ( g ( z ) ; a * ( u ) ) )c Z ( Z ; U ' ) by Lemma 6.6.9 (3), and g(s(7j-'(z);an(u))) c v:(z;u'). Notice that T ( z ; u ) C --1 g ( D ( g ( z ) ; an(u))) for some 7 = -y(e,n) > 0. H e n c e T ( z ; u ) C r ( z ; u ' ) , and by Lemma 6.6.1 we obtain W,U(x)C W,U(y).Similarly, W,U(y)C W,U(x). Therefore the conclusion is obtained from Lemma 5.1.8. 0

- --'

-

Theorem 8.6.14. Let f : X -+ X be a TA-covering map. For x E Xf let Q, : D"(z0) x D"(x) N , be as in Theorem 5.2.1. Then for 2 E X there is at most countable subset B c Du(x) such that -+

Furthermore, i f f is special, then there is at most countable set A C D"(z0) such that W u ( y )n Nx = a,(A x D " ( x ) ) .

Proof.The first part follows from Lemma 5.1.10. Combining Lemma 6.6.9 (2) with Lemma 6.6.11 (2.a), the second part is easily checked. 0 Theorem 6.6.15. Let Y and X be both semilocally l-connected and let q : Y -+ X be a covering map. Let g:Y -+ Y and f : X -+ X be self-covering maps. Suppose f o q = q o g. Then (1) f is a TA-covering map i f and only i f so is g ,

$6.7 TA-covering maps of closed topological manifolds

211

(2) f is a special TA-covering map if and only if so is g , (3) i f f is a TA-covering map then for y E Y the stable set r?..(y) in strong sense i s the path connected component of y in q-'(t?l'"(q(y))). Furthermore, iff is special, then for (yi) € Y, the unstable set W"((yi)) is the path connected component of yo in q-'(W"((q(yi)))).

Proof. From Lemma 2.2.34 (3) and Theorem 2.3.14 together with Remark 2.2.35 we have (1) (use Theorem 6.4.1 and Lemma 6.4.2). (2) follows from Theorem 6.6.13. As Lemma 6.6.10 we have the first part of (3). The second part is easily checked by Lemma 6.6.11 (2.a). 0

$6.7 TA-covering maps of closed topological manifolds In the previous sections we have seen the behaviour of orbits of TA-covering maps. In this section we restrict ourselves to closed topological manifolds. Let M be a connected topological manifold without boundary and let 3be a family of subsets of M. We say that 3 is a generalized foliation on M if the following holds; (1)F is a decomposition of M, (2) each L E 3,called a leaf, is path connected, (3) if x E M then there exist non-trivial connected subsets D,,K , with {x} = D, n K,, a connected open neighborhood N , of x, and a homeomorphism cp, : D, x K , N,, called a local coordinate at x, such that (4 ( P = ( X , X ) = x, (b) cpz(Y, x) = Y(Y E Dx) and V d Z , z ) = 4 2 E G), (c) for each L E 3 there is an at most countable set B C K , such that N, n L = cp,(D, x B ) . --f

Figure 27 Let 3 be a generalized foliation on M. If D, and K, are manifolds for all x E M, then 3 is a foliation of class Co in the usual sense.

CHAPTER 6

212

Remark 8.7.1. If M is two or three dimentional manifold without boundary then 3is a foliation of class Coin the usual sense. (cf. Bing [Bin] and Wilder [Wll).

For the case of dimM = 2, this is checked as follows. Fix x E M and let cp : D x K --t N be a local coordinate at x. It suffices to show that D and K are one dimensional manifolds. We show that this is true for D. The conclusion for K will be obtained in the same way. For simplicity, denote by I, and If the intervals [0,1], ( 0 , l ) and (0,l) respectively. Since D x K is a connected manifold, it is checked that D is arcwise connected. Indeed, let a, b E D. Since D x K is connected, there is a continuous map a : I --.t D such that a(0) = a and a(1)= b. By Theorem 2.1.5, a ( I ) is arcwise connected. Hence we can find an arc from a to b in a ( I ) . Let AD (resp. A K ) be defined as the set of all injective continuous maps from I to D (resp. K). For a E AD, a : --t a ( p )is a homeomorphism since a is injective. Moreover a ( p )is open in D. Indeed, take and fix p E A K . If (U, cp) is a chart of D x K, then the map cp 0 (ax P )

P

P

n(.

P ) - ~ (ax/? u ) u 2 ~2

is injective and continuous. By Theorem 2.2.18 the image of cp o (a x p ) is open in W2,and therefore a ( I o )is open in D. Put VD = UaEAD a ( I o ) .If VD = D, it follows that D is 1-manifold without boundary. For the case where VD # D, we show that D is homeomorphic to I or I'. Take and fix x # VD and define

A', = {(I! E AD :a(0)= x}. First we claim that there is the implication between al(1) and az(I) for C Y ~ , C YE~ A',. Indeed, assume that al(1)@ az(1).Then we have that az(I)C al(1).For, let J = al(I)\az(I).Since al(0)= az(0)= x, J al(I)and so a,'(J) I. Since J # 0, there is a boundary point c of a,'(J). Since J is open in al(I), a ; ' ( J ) is open in I. Hence c # a ; ' ( J ) and so a~(c) # J. This implies that a l ( c ) E az(1). But az(I0)is open in D. Since c is a boundary point we have that a1(c)= a z ( 0 ) or al(c)= az(1). If al(c)= az(O), of aI1(J), then we must have c = 0 (since a1,az E A',). In this case there is d E (0,1] such that al([O,d]) n az([O,11) = {x}. This shows that there is in D an arc of q ( d ) to az(1)through the point x. By definition we have x E VD,thus a contradiction. Therefore a1(c) = az(1). Since c is arbitrary in the boundary set of a F 1 ( J ) ,the boundary set of a ; ' ( J ) is a single ponit. Therefore a;l(J) is an open interval in I and a,'(J) 3 1, from which we see that a F 1 ( J )= (c, 11 and hence a1([0,c ] ) c a 2 ( [ 0 , 1 ] )and therefore az(1)c al(I).We proved that {a(I): a E AL} is a totally ordered set under inclusion.

5

$6.7TA-covering maps of closed topological manifolds

213

On the other hand, since D is arcwise connected, we have

D=

u

a(I).

aEA’D

These two facts implies that D is homeomorphic to either I or I1. For the case of dim M = 3, the proof is done by making use of singular homology. For details, see Wilder [Wi]. If dim M 2 4, then leaves of generalized foliations are not necessarily locally euclidean. Indeed, there is a topological space X without manifold structure such that X x R is homeomorphic to R4. See Bing [Bin]. Let 3 be a generalized folitaion on M. If x E L for some L E 3, then we have D, c L by (b) and (c). For fixed L E 3 let O L be a family of subsets D of L such that there is an open subset 0 of M such that D is a connected component in 0 n L. Then the topology generated by OL is called the leaf topology of L. If D, is as in (3) and x E L, then D, is open in L with respect to OL.The restriction of the topology on the leaf L to D, is consistent with that of the topology on M t o D,. Hence, D is open in L if and only if for x E L, D n D, is open in D,. Since M is locally contractible and has a countable base, the topology on the leaf L has the following properties; (1) path connected, (2) locally contractible, and (3) second countabdity. Let 3 and 3’ be generalized foliations on M and let f: M --+ M be a homeomorphism satisfying f (3)= 3‘.Then, for L E 3 we have that f : L 3 f ( L ) is a homeomorphism with respect to OL and O ~ ( L ) .

IL

Remark 6.7.2. Let 3 be a generalized foliation on M and let U be an open subset of M. Denote by L ( x ) the leaf through x of 3 and put V = { X E M : L ( x )n U

# 8).

Then V is open in M. Indeed, for x E V we have L ( x ) n U # 0. Take w E L ( x ) n U. Then there is an arc in L ( x ) with end points x and w. By (3) there is a finite sequence { x i : 1 Ii 5 1 ) C L ( x ) such that DXi contains the arc, D,, c U , DXin Dxi+l # 8 (1 5 i 5 1 - l), and x E Dxt. Let wi E DXin Dxi+lfor 1 5 i 5 1 - 1 and let KXi(1 5 i 5 1) be a connected subset as in (3). Since ( P + , ( w ~ , z ~=) w1 by (b), we have w1 E U and hence there is an open set KL2 c K,, such that (pxo(wl,KL,) c U. Put NLa = (pxa(Dxa x K;,). Then NL, c N,, and L ( z ) n u # 8 for z E N&. Replace NA, by U and use the above technique. Then we have that there is N i 3 c NX3bsuch that L ( z ) n N;, # 8 for z E NL3. Using induction, we have

ut,

CHAPTER 6

214

L ( B )n NL,-l # 0 for z E NL,. In fact, L ( z ) n U # 8 for E D,, C NLt c V, V is open in M.

8

E NLt. Since

Let 3and 3' be generalized foliations on M. We say that 3is tmnsverse to 3' if, for z E M, there exist non-trivial connected subsets D,, DL with {z} = D, n D:, a connected open neighborhood N, of x in M (such a neighborhood N, is called a coordinate domain at z), and a homeomorphism $, :D, x DL --$ N, (in particular called a canonical coordinate at z)such that

(4 4 X ( W ) = 2, (b) $ X ( V , ~ ) = Y ( Y E D X ) and $ z ( z , z ) = 4 2 E DL), (c) for any L E 3 there is an at most countable set B' C DL such that N, n L = $,(Dx x B'), (d) for any L' E 3' there is an at most countable set B c D, such that N, n L' = +,(I3 x 0:). It is clear that if 3 is transverse to 3' then 3' is transverse to 3.

Figure 28 Let 3 and 3' be transverse generalized foliations on M and denote as L ( z ) and L'(z) the leaves through a point z in M respectively. Given N , a coordinate domain, for y E N , we define D(y) and D'(y) as the connected components of y in N , fl L (y ) and N , r l L'(y) respectively. Then, for y , z E N , the intersection D'(y) n D ( z ) is a single point. Hence we can define a map (6.8)

7 N ,

: NXX

NX--t NX

by ( y , z ) H D'(y) n D (z), and have then TN. 7 N a ( Y , 7 N s ('9

(y, Y) = Y9

w))= 7 N a (Y,w)9

7,.

(7Na

(Y9 '), w, = Y N . ( Y , w)'

If Nx and Nv are coordinate domains and if N , is a coordinate domain such that N , C N, n N,, then it follows that (6.9)

- 7N,IN,xN, - 7NYIN.xN.' -

7 N z

$6.7 TA-covering maps of closed topological manifolds

215

Let p : N + M be a covering map and 3a generalized foliation on M . We define a family 7 of subsets of N by

3 = {i/ : 3 L E 3 s.t. i/ is a path connected component of p - ' ( L ) } . Then this family is a generalized foliation on N. For ?; E F the image p ( z ) is an element in 3 and the restriction p : --f p ( z ) is a covering map under the leaf topologies. We say that 7 is the lift of 3 by p . If 3' is another generalized foliation on M and it is transverse to 3,then it follows that the lift 7' by p is transverse to 7. The following is obtained from Theorems 5.1.3 and 6.6.14 together with Lemmas 6.6.7 and 6.6.8 (1).

z

Theorem 6.7.3. Let f : M + M be a TA-covering map of a closed topological manifold. Suppose f is not eqanding. Then the family 3;= { W 5 ( x ) : x E M } of all stable sets in strong sense is a generalized foliation on M . I f f is special, then the family F; = {W"(x): x E Mf} of all unstable sets is also a generalized foliation on M and the families 3;and 7 are transverse. Notice that the intrinsic topology for stable sets in strong sense is consistent with the leaf topology. From Lemmas 6.6.7 and 6.6.9 together with Lemmas 6.6.10 and 6.6.11 (2.a), we have the following theorem.

z

Theorem 6.7.4. Under the assumptions in Theorem 6.7.3, let p : +M be as in (6.3). For u E the families be the universal covering and let -5 F = {W5(z) : z E z} and T", = u) : x E are tmnsverse The family F5 is the lift of 3; by p and if f is generalized foliations on special, then is the lift of by p . In Chapter 11 we will introduce the concept of orientablility for generalized foliations by making use of singular homology. And we will establish in Chapter 12 a fixed point index theorem for TA-covering maps of closed topological manifolds, which says that any iteration of a TA-covering map has the same fixed point index 1 or -1 at each fixed point if the family of the stable sets in strong sense and the closed topological manifold are both orientable. This will be useful in counting the number of fixed points of TA-covering maps. Remark 6.7.5. For f : M M a TA-covering map the homeomorphism ax : D"(x0)x D U ( x )-+ N, in Theorem 5.2.1 is a local coodinate (or a canonical coordinate ) at xo for F; ( and 3;). Hence

mf

3:

a.

{w(z;

mf

z}

--f

ax

= rN,

ID'(Z0)XD'(X)

where -yN, is defined as above (see (6.8)). Similarly, the homeomorphism E,, : xDu(z;u) 3 N(x;u) of Theorem 6.6.5 is a canonical neighborhood at x for F5 and 7;.

a"(,)

CHAPTER 6

216

We denote subsets of

R" by

I" = { ( t l ,...,tn) : 0 6 t; 5 l } ,

n n

81" = {(tl,.

1

e .

tn) :

t i ( 1 - t i ) = 01,

i= 1

and let X be a topological space. Take and fix 20 E X . In the family of maps f : I n --f X with f ( a I n ) = xo we introduce the operation of product f g by

(f * g ) ( t l ,-.- ,tn) =

{

f(tl,*--,~n--l,2tn) g(tl,...,L-1,2tn-1)

5 tn I 1 / 2 ) ( 1 / 2 I t n 5 1). (0

In + X denote a homotopy satisfying f t ( 8 I " ) = 20 between f o and gt : (1",8In) --f (X,ZO) denote that between go and 91. Then f t gt gives a homotopy between f o go and f 1 91. Thus the product of homotopy classes { f } and { g } is defined by { f } {g} = { f g } , and so the set of homotopy classes, n;,(X,ZO), becomes a group under the operation. If X is path connected, then 7rn(X,zo) is isomorphic to xn(X,Yo) for 20,yo E X. We say that 7rn(X,xo) is the n-homotopy group, and write x n ( X ) = ~,(X,xo) when X is path connected. If in particular n = 1, then 7rl(X)is the fundamental group. Let and

-

ft : f1,

-

-

-

-

Proposition 6.7.6. Let p : Y X be a covering map of topological spaces and let zo E X and yo E p-'(zO), If n 2 2, then the induced homomorphism p* : nn(Y, yo) + a,(X, 20) is an isomorphism. --f

For the proof, see Spanier (Sp]. Under the assumptions in Theorem 6.7.4, let x E % and

Remark 6.7.7. uE

mf. Then we have

7rk(W"(Z))

= 7 r k ( w ( z ;u)) = 0

(12 2 1).

This is easily proved as follows. Choose p > 0 as in Theorem 6.6.5 (4). Then we can find a finite open cover {U}of M such that each U is homeomorphic to R" and diam(l7) < p. Then {V} = {V : V is a connected component of p-'(U)} is a n open cover of In fact, each V is homeomorphic to R" and diam(V) < p. Let 6 > 0 be a Lebesgue number of {V}. For 2 E ii? and u E %if we have

z.

-

W " ( z )=

(J7;:(u)(Ds(7:(x))) iy0

--i

-&q

4

and fai(,,)(D (f,(x))) is open in W"(x)under the leaf topology. If 7 :'I -+ W"(x)is a continuous map with 7 ( a I k )= 2, then there is i 2 0 such that

56.7 TA-covering maps of closed topological manifolds

and so

TU

0

$Ik)

217

c S(?L(z)).

By Lemma 6.6.3

TU for sufficiently large i, and so 6.6.5 (4) we have

0

7 V k )c

Bs(TU(.))

o ~(1') C

V for some V E {V}. By Theorem

v c x(7;(z);a'(u)). We identify V with R" and define a homotopy Ft : ( I k ,ark)+ (wa(z), z)by

for all 0 5 t 5 1 and z E Ik where

denotes the projection. Then F l ( z ) = ~ ( z and ) Fo(z)is a constant. Thus nk(wa(z))= 0. Similarly we have n k ( r ( z ; u)) = 0. Let n be a group. We say that a path connected topological space X is of type K ( n ,1) if n l ( X ) 2 n and n k ( X ) = 0 for k # 1.

Proposition 6.7.8. Let N be a topological space of type K ( n , l ) and let M be a compact connected topological manifold. Let 20 E M and yo E N . Then, given a homomorphism C#J : x l ( M , z o ) + nl(N,yo), there exists a continuous map f: M + N with f(z0) = yo such that f* = 4. Conversely, i f f ,g : M + N are continuous maps with f(zo) = g ( z 0 ) = yo and if f + , g + : q ( M , z o ) + m ( N , y o ) satisfies f,(a) = pg*(a)p-',Va E x l ( M , z o ) for some P E ~ ( N , Y o ) , then f and g are homotopic.

For the proof, see Spanier [Sp] and Kirby-Siebenman [Ki-S].

Remark 6.7.9. Let f , g : T" + T" be continuous maps of an n-torus and let f(z0) = g(z0) for some zo E T". Then f and g are homotopic if and only if f* = g+ : nl(Tn,zo) + nl('IP,f(zo)).From this fact we have that if f:T* + T" is a continuous map, then there is a unique toral endmorphism A : 'IP + T" homotopic to f . Indeed, since the natural projection R" + T" is a covering map and nk(Rn) = 0 for all 12, by Proposition 6.7.6 the torus T" is of type K(n,l). Here n = Z" (Remark 6.3.8). Hence by Proposition 6.7.8 we obtain the conclusion.

CHAPTER 6

218

Remark 6.7.10. Let M be a compact topological manifold and let d be a metric for M. Then there is 6 > 0 such that for continuous maps f , g of M if d ( f ( z ) , g ( z ) )< 6 for all z E M then f and g are homotopic. Moreover, one can choose a homotopy F : M x [0,1] -+ M from f to g satisfying the property that d ( f ( z ) ,F ( z ,t)) is small for all z E M and t E [0,1] whenever so is 6. In Chapter 10 we will show that M is an euclidean neighborhood retract, i.e. there exist an open set 0 of an euclidean space R", and continuous maps r : 0 -t M and i : M + 0 such that r o i is the identity map of M. By making use of this fact, the above statement is easily checked as follows. Define a subset U of M x M by

U = {(z,y)

: (1

- t ) i ( z )+ ti(y) E 0 for all t E [O,l]}.

Then U is an open neighborhood of the diagonal in M x M. Since M is compact, we can find 6 > 0 such that {(z,y) : d ( z , y ) < 6) c U,and define F : M x [0,1] -t M by

F ( z , t ) = r ( ( 1 - t ) i o f ( z )+ t i o g ( z ) ) . It is easy t o see that F satisfies all the desired properties. In the remainder of this section we give the proof of the following statement mentioned in Theorem 2.4.7 of Chapter 2.

Theorem 6.7.11. Let M be a closed topological manifold and f : M + M a self-covering map, but not injective. If f is a TA-covering map and has topological stability in the class of self-covering maps, then f is expanding. Proof. Suppose that f : M M fails to be expanding. Since f is a TAcovering map, there exists a periodic point po E M with period n > 0. Let /3 be a number such that z # y and f ( z ) = f(y) implies d ( z , y ) 2 p. Let e > 0 be a n expansive constant for f and choose 6 > 0 in the definition of topological stability. For an n-cyclic sequence ( p i ) E Mf there exist pLl E f - l ( p 0 ) - {p-1) and a neighborhood U(pLl) such that diam(U(pL,)) < p / 2 and diam( f ( U ( p L l ) ) )< 6, and such that ( p i ) n U(p'_,)= 0 and f p q g L , ) is a homeomorphism. Since f is not expanding, we have W,"(pLl) # { p L 1 ) by Theorem 5.2.1, and so Q-1 E w ? Y P ' _ l ) - {PLl). Now construct a self-covering map g such that f = g on M - U(pLl) and g(q-1) = po. Obviously d( f , g ) 5 6 . Thus there exists a continuous surjection h: M M such that h o g = f o h and d(h,i d ) e by topological stability in the class of self-covering maps. Since (pi) n U(p'_,)= 0, we have --f


0 such that nx = 0), then the rank of G is zero. If X is a compact metric abelian group, then its character group G is countable and the topological dimension of X is equal to the rank of G ([Po]). We say that S is a solenoidal group if S is compact, connected, finitedimensional and abelian. Thus, every finite-dimensional torus is clearly solenoidal. A compact metric space which is homeomorphic to a solenoidal group is called a classical solenoid. Denote as G the dual group (or the character group) of S. If S is n-dimensional then the rank of G is n. Thus there is 8 c G such that B = { e l , . , e n } is linearly independent, and so the factor group G/gp8 is a torsion group (the notation gpE means the subgroup generated by E). Hence every 0 # g E G is expressed as ag=alel +-..+anen

+ -+

--

.

..

.

.

for some a # 0 and some a l l . ..,a, with ( a l l . . ,a,) # (0,.. ,0). Since the existence of ( a l / a , ,a , / a ) is unique, we can define an injection 'p : G + Q" bY v(g)= (a1/a, * * ,& / a ) . To simplify the notations, we identify g with ( a l / a , .,a,/a) under 'p. Then we can consider 8 is the canonical base of Z"(i.e. el = (1101 ,O), .. .,en = (0,.. ,0, l)),and so gpe = iz" c G c ~n c R". For t = ( t l , . , t n ) E R" define

...

-

..

.

..

...

+ + t,a,/a

(addition mod 1) for all g = (a1/a1...,u,,/a) E G, i.e. $ ( t ) g = ( t , g ) where ( , ) denotes the inner product. Then we obtain $ ( t ) E S. In fact $ : R" + S is a continuous homomorphism (Pontrjagin [Po]). The following lemmas will be used subsequently. Lemma 7.1.1. $(Rn) is dense in S . I f S is the torus then $(Rn) = S . Proof. From the definition of $ it follows that if g E G and $(t)g = 0 for all t E R" then g = 0. This shows the first statement of the lemma. If S is an n-torus, then G/gp8 is finite (since gp8 = Z").Let F = ann(S,gp8). Then F is finite since so is G/gp8. Thus mF = {mx : x E F } = (0) $(t)g = t l a l / a

for some m > 0. Therefore, by the following Lemma 7.1.2 (2) we have S = +(W") F = m[$(R") F ] = $(R") mF = $(R").O

+

+

+

CHAPTER 7

224

Lemma 7.1.2. Let F be the annihilator of gp6 in S. Then (1) F is totally disconnected and $-l{$@?') n F } = Z", (2) S = $(R") F, ( 3 ) thew is a small open neighborhood V of 0 in R" such that $ ( V ) n F = { 0 } and the direct product V x F is homeomorphic to $ ( V ) F , (4) Let V be as in (3). Then $ ( V )+ F is an open neighborhood of 0 in S , ( 5 ) $(R") is the path connected component of 0 in S .

+

+

Proof. The dual group G/gp6 of F is a torsion group (i.e. for any x E G/gp6 there exists n > 0 such that n5 = 0). Hence F is totally disconnected. The equality $-l{$(R") n F} = Z" follows from the definition of $. (1) was proved. Let 7 r :~S S / F be the natural projection. From (1) we have ---f

7rF 0

$(R") = 7rF 0 $ { ( t l , . .., t n ) E R" :0 5 ti 5 1 , l 5 i 5 n } ,

which implies that n~o$(W") is compact. Since $(Rn)is dense in S by Lemma 7.1.1, T F o $(R") = S / F and so

S = $(It")

+ F.

(2) was shown. To show (3), we take a bounded closed set

1

1

u = { ( t1 , ...,t n ) E R" : - -3 -< t ". 0 such that u([O,to])C $ ( V ) . Hence [&to] c J. Let x E S and denote as cpx : S + S the translation defined by cpz(z) = x z. Then cp,($(V) F) is homeomorphic to V x F , which implies that $(an) f l p z ( $ ( V ) F ) is expressed as cp,($(V) C) for some C c F . From this fact it follows that J is open and closed in [0,1].Therefore J = [0,1]. (5) was proved. 0

+

+

+

+

+

+

s7.1 Geometrical structures of solenoidal groups

225

Lemma 7.1.3. $(Z")is a closed subgroup in $(Rn). Proof. By Lemma 7.1.2 (1) it is clear that $(Zn) cl($(Z")) n $(Exn)

c F n $(It")

c F. By Lemma 7.1.2 (2) = $(Z")

n $(R") = $(Zn). Here cl(E) denotes the closure of E. Lemma 7.1.4. $(Z")is dense in F . and so cl($(Z"))

+

+

0

Proof. Write B = cl($(Z")). Then S I B = {($(R") B ) / B } { F / B } . Since $(Rn)/$(Zn) is a factor group of Rn/Zn, it is a torus and so ($(a") B)/B is also a torus because of

+

($(R")+ B ) / B N $ ( R " ) / W " ) by the following lemma. On the other hand, S / B is connected, from which we have F = B. 0

Lemma 7.1.5. Let H be a normal subgroup of a topological group G and let M be any subgroup of G . Let cp(mH) = m ( M n H ) for m E M . Then cp : M H / H M / ( M n H ) is an open map. --f

Proof. Let A H be an open subset of M H / H . That means A H = cp(AH),A c M , and A H is relatively open in M H c G. But (p(AH) = { x ( M n H): x E A } . Since A ( M n H ) = AH n M , A ( M n H ) is relatively open in M. Therefore, by the definition of the quotient topology of M / ( M nH ) , (p(AH)= { x ( M n H): x E A } is open in M / ( M n H).0

Lemma 7.1.6. If S contains no torus subgroups, then 11, : Rn +

$(an)i s

bijective.

Proof. If K is the kernel of$, then K is discrete and $ induces the continuous bijective homomorphism $' : Rn/K + $(EXn). If K # {0} then Rn/K contains a torus T and so (0) # I)'(") c S. Since S has no torus, we arrived at a contradiction. 0

Lemma 7.1.7. The family of all open subgroups of F is a base of open neighborhoods of the identity 0 of F . Proof. Since F is totally disconnected, obviously F has a base of open closed neighborhoods of 0 in F. To obtain the conclusion it suffices to see that if U is an arbitrary open neighborhood of the identity 0 of F, then there is an open closed subgroup H such that H C U. Since 0 is the connected component of F, there is an open closed subset P such that 0 6 P C U. Define Q = { q E F : P q C P}. Then H = Q n (-Q) is an open closed subgroup satisfying H c U. To see that Q is open in F take and fix q E Q. Let z E P be arbitrary. Since x + p E P and P is open, there are open neighborhoods U, and V, such that

+

CHAPTER 7

226

x E U,,q E V, and U, + V, C P. Since {Uz: x E P} is an open cover of P, we ,U,,} in the cover. Set V = V,, n * r l V,, . can find a finite cover {U,, Then P + V c P and q E V c P. This means that Q is open in F. To prove that Q is closed in F we show that F \ Q is open. Take T E F \ Q. Since P + r p P, there is p E P such that p + r E F \ P. Since F \ P is open, we can find an open neighborhood W of r such that p + W c F \ P. This implies that W C F \ Q. Thus F \ Q is open and so Q is closed. Since 0 E P, for y E Q we have y = y + 0 E P and thus Q c P. Since P + 0 = P c P, clearly 0 E Q. Therefore H = Q n (-Q) is an open closed

-

-

subset containing the identity 0. It remains to see that H is a subgroup. Take hl,h2 E H, then hl E Q and -h2 E Q and so P (hl - h2) = ( P h l ) - h2 c P - h2 C P . This shows that hl - h2 E Q . Similarly we have -(hl - h2) = h2 - hl E Q. Therefore, hl - h2 E H and H is a subgroup of F. 0

+

+

+

Lemma 7.1.8. S = $(Itn) FO for every open subgroup FO of F. Proof. There is a subgroup Zt with finite index such that Fon$(R") = $(Z;). Thus we have R" = C Z Zt for some compact subset C of R", and so

+

+

$@") = $(C)

c $(C)+ Fo c

Since $(R") is dense in S, we have S = +(R")

+ Fo.

+ Fo.0

Let d denote a translation invariant metric for S and d p denote the euclidean metric for R". For 6 > 0 put

Ua(O)= {V

E R" : dgn(v,O)

< 6},

F6(O)= { Z E F : d(z,O) < 6). Since F is totally disconnected and Fa(0) is symmetric (i.e. if 2 E Fa(O) then -x belongs to Fb(O)),Fb(O) contains an open subgroup of F. To avoid complication we may assume that Fs(O) itself is an open subgroup of F. Let a0 > 0 be small enough. By Lemmas 7.1.2 (3) and 7.1.7 (7.1)

W = $(uao (0)) + Fa0 (0)

is an open neighborhood of 0 in S. Throughout this chapter, a0 is fixed. Define a function

fors-ye W for z - y $ W.

tc

by

$7.1 Geometrical structures of solenoidal groups

227

Then do is equivalent to the original metric d for S and a translation invariant metric for S. If, in particular, v, v' E Rn and d p (v, v') < a0 then we have do($(v),$(v'))= dan(v,v'). We sometimes adapt the metric do in stead of the metric d. For 0 the open neighborhood Wa(O) with respect to do is expressed as

< 6 < a0

+

w6(0) = $(u6(0)) F6(0) where Va(O) and F6(O) are open neighborhoods of R" and F respectively. Clearly Wa(O) is symmetric, and Wa(O) n F is a symmetric open neighborhood of 0 in F. Remark that $(Ua(O))is not open in $(R"). If F1 = Fa(O)is a i'C Z"such that subgroup of F,by Lemma 7.1.2 (1) there is a subgroup Z Fl

n

= $(Z;)

and then

+

+

w6(0) n $@") = $(u6(0)) $(Z?)= $(u6(0) Z?)(7.3) If F1 is open in F,then Zr is of finite index. Let G (C 0") be a discrete countable abelian group and let 71 : G + G be an automorphism. Then G is said t o be finitely generated under if G contains a finite set (91,. . . , g n } such that G = gp{xi(gj) : i E Z, 1 5 j 5 n}. Especially G is said to be finitely generated if there is a finite set E such that G = gp(E). Let S,,be an n-solenoidal group and A : S,,+ S, an automorphism. As before, let ( G , X )be the dual of (S,,,A) (i.e. G is the character group of S,, and ( x g ) ( z ) = g ( A z ) for all g E G and z E Sn). Since S,,is an n-solenoidal group, there is a continuous homomorphism $J : R" + S,,such that

x

Sn

+F

= $(Rn)

where F is a totally disconnected subgroup of S,. Obviously A o $J(Rn) = $(Rn).As seen above, since $(s)g = ( s , g ) for s E R" and g E G, we have

( A 0 $(s)>g = $ ( s ) ( & 7 )= (s,&) = ( % , g ) (where denotes the transposed matrix of A)

tx

(s E R n , g E G ) .

= ($ 0 "(s))g Here

- 1.

< , > denotes the inner product. R"

'A

R"

+I $(an)A

This shows that the diagram

NRn>

commutes.

CHAPTER 7

228

Theorem 7.1.9. A : Sn -t Sn is ezpansive if and only if all eigenvalues of are 08the unit circle and G is finitely generated under

tx

x.

-

Proof. First we prove that A : S, -+ S, is expansive when G = gp{A (9) : -m < n < oo} and all the eigenvalues of 2 are off the unit circle. Since Sn is n-dimensional, we have rank(G) = n. Thus {g,?1(g), , *-l A ( 9 ) ) is linearly independent in G. By Lemma 7.1.2 (3) there is a small

-..

open neighborhood V of 0 in Wn such that $ ( V ) r l F = (0) and the direct 'a-"(V) product V x F is homeomorphic to $(V) F. Define V1 = and FI = n;=-, A-"(F). Then $(V1) F1 is an open neighborhood of the identity in S,. Indeed, V1 is an open neighborhood of 0 in W". Since the character group of F/F1 is a finite group

+

nf=-,

+

n=-1

+

we have that F/F1 is finite (see [Po]). Thus F1 is open in F. Since $ ( V ) F is open in S, (by Lemma 7.1.2 (4)), we have that I/J(Vl) F1 is open in S,. Since all the eigenvalues of are off the unit circle and 2 is similar to %, the linear map : R" + W" is expansive by Lemma 2.2.33, from which we have m

+

x

tx

n=-m

--l

i

Since X Z - - A gp{g,x(g),... , A ( 9 ) ) = G and F is the annihilator of ---n-1 gP{g,X(g), , A (9)) in sn, we have

---

n A = ( F ~c) n m

n=-m

00

A ~ ( F=) (0).

n=-m

Since $ ( V ) n F = (0) by Lemma 7.1.2 (3), it is easy to see that

n=-j

n= -j

n=-j

nrrn

for all j 2 0. Thus Aj(I/J(Vl)+Fl)= { 0 } , which implies that A : Sn -+ S, is expansive. From now on we prove the implication e).Since G is finitely generated under A , by definition G contains a finite set {g1,... ,gk} such that G = E!=l Gi where

Gi = gp{Aj(gi)

: -00

< j < OO},

1 5 i 5 k.

$7.1 Geometrical structures of solenoidal groups

229

Let Ki denote the annihilator of Gi in S, for 1 5 i 5 k. Then each of Gi is the character group of Sn/Ki. Thus the factor automorphism A : Sn/Ki --t Sn/Ki is expansive. Denote as pi the projection from S,,onto S,/Ki for 1 5 i 5 k, and define u = p,'(u1)n n pil(uk) where each of Ui is an open neighborhood of the identity of Sn/Ki. Then we have m

.

n

A ~ ( u= ) (0).

-00

z:=l

ntZl

This follows from the fact that Ki = (0) because of Gi = G. Therefore A : S,,+ S,, is expansive. +) : To see that all the eigenvalues of t X are off the unit circle choose a small neighborhood U of the identity in S,. Then there is a neighborhood V of 0 in R" such that + ( V ) c U and : V + U is injective (by Lemma 7.1.2). Thus m

+

(0) =

fi

AW

3

-m

fi

AWV)) =

-m

+(n

*T(VN

--oo

nzm

*x

and so * a " ( V )= ( 0 ) . This implies that : R" + R" is expansive. By Lemma 2.2.33 all the eigenvalues of 'X are off the unit circle. To prove that G is finitely generated under it suffices to see that G/gp( UTmX i @ " ) ) is finite. Since F = ann(X, Z") by Lemma 7.1.1, we have nTm

A j ( F ) = ann(X, gp(Urm X'(Z"))). If H = nyrnA j ( F ) is finite, then 00 - j G/gp(U-,A (Z")) is finite since it is the character group of H. To prove that H is finite take a small open neighborhood V of 0 in R". As above define Vl = "'(V) and Fl = A j ( F ) . Then $J(V~)+ F1is open in S,. Since A : S,,+ S, is expansive, we have

nyl

+ n+@(v1)) + nA ~ ( F ' ) , = nrm A j ( F l ) = nrm A j ( F ) = H. Therefore G is finitely gener-

n m

(01 =

m

-m

-W

Aj(+(Vl) F') =

-m

and so (0) ated under

00

A.

In the remaining part of this section, we give a simple example of a solenoid which admits a TA-homeomorphism (due to Smale). A 1-solenoidal group S1 is constructed by the map u S' = ( z E C : IzI = 1). Obviously u : S' + S' is expanding. Let D 2= (w E C : lwl 5 1) and define a map f of the solid torus N = S' x D 2into N by 1 1 f(z,w) = ( E 2 - 2 -.I). '2 4

Remark 7.1.10. (u : E

H

z 2 ) of the circle

+

CHAPTER 7

230

n,"==,

Set A = f n ( N ) , then A is l-dimensional. Define the natural projection p :N + S ' by p ( a , w ) = a and a homeomorphism h: A --+ (S1)N by h ( z ) = ( p ( f - ' ~ ) ) Then ~ ~ . we have h(A) = S1 where S1 = { ( a ; ) E (S')N : a ( a i + l ) = a i } , and the following diagram

A - Af

hl 4

hofoh-1

1.

commutes.

s1

It is easily checked that ho f oh-' = u. Therefore, f is a TA- homeomorphism.

Figure 29

Remark 7.1.11. Solenoids introduced in Williams [Wi11,2,3] is defined as follows. We proceed in defining a l-manifold K such that the following three types of coordinate neighborhoods are allowed: R, H = (3 E R : 2 2 0) and Y = ((3,y) E R : y = 0 or y = 'p(z)} where 'p : R + R is a fixed C" function such that ~ ( z=)0 for 2 5 0 and ~ ( z>) 0 for z > 0. The boundary aK is defined as usual to be points of K corresponding to 0 E H. The branch set B of K is the set of all points of K corresponding to (0,O) E Y. Obviously aK and B are finite if K is compact. A Cr-structure for a branched l-manifold is defind as usual. Since two the submanifolds containing a branch point b have the same tangent at b, the branched l-manifold K has a tangent bundle TK. A differentiable map f: K + K induces a map D f : TK TK. Then f is an immersion if D f is a monomorphism on the tangent space at each point. Branched manifolds occur naturally in topological dynamics and they can be defined for all dimensions.

57.2 Inverse limit systems of self-covering maps on tori

231

If g: K + K is an immersion of a branched manifold, then g is an eqansion relative to a Riemannian metric on a tangent bundle TK if there are c > 0 and X > 1 such that JIDg"vlJ2 cA"1Jv11for v E TK and n 2 0. Let K be a compact branched 1-manifold with branch set B and let g:K + K be an immersion such that g is an expansion, all points of K are nonwandering under g and each point of K has a neighborhood whose image under g is an arc. Let K, be the inverse limit system of (K,g). Then K, is called a solenoid. For a point a = (ao, al, ) E Kf define h-'(a) = ( a l , a2, a3,. ) as before. Then h:Kf -t Kf is a homeomorphism. The dynamics of h:Kf -+ Kf is discussed in [Will, 21 and the dynamical systems on solenoids for all dimensions were developed in [Wil3].

...

..

$7.2 Inverse limit systems of self-covering maps on tori

Let T" be an n-torus and let A : R" + T" be the natural projection. Let denote the euclidean metric for R". Then dRn induces a metric d for ll"'

dRn

bY

d(z+Z",y+Z")=inf(dp(z+d,y+d'):d,d'

EZ"}

for z,y E R". We denote aa G(A) the covering transformation group for A. Then G(A) is isomorphic t o Z"(see Remark 6.3.8). Throughout this section, we suppose that f : T" -t 11'" is a self-covering map. Let 7 : R" + R" be a lift o ff by A. Then f :R" -t W" is a homeomorphism by the homomorphism 7* : G(A) -+ G(A) Theorem 6.3.12 (1). Since G(A) Z Z", defined as (6.1) can be considered as a group homomorphism of Z", in which case 7, : Z"-+ Z"can be written as an n by n matrix 2, all the entries of which are integers. - The matrix can be considered as a linear map of R" such that Z1p = f , . Then by Lemma 6.3.10 we have

(7.4)

f(v

+ 1) = A(1)+ f ( v )

for L E

Z"and v E R".

Note that f:T -+ ll"' a homeomorphism if and only if ?I@") = Z".Since f is a self-covering map in our case, we have Z(Zn) c Z"and 2 ( Z n ) is of finite index (see Theorem 6.3.12 (2)). A (Z").Then x(Zp")= Zp" and every element k of Zp" Put Zp" = 00 -j satisfies the property that there is a non-zero polynomial p ( z ) E Z[z], where p(s) is monic and its constant term is 1 or -1, such that p ( Z ) k = 0. In this y c Z"such that case there is a subgroup Z

no

A@;) c

z;,

Z;n Z;= o

and Z; @ Zt is of finite index. Indeed, let Q" denote the n-vector space over Q and let QF be the smallest subspace containing Z ;.Then Q" is 2-invariant

CHAPTER 7

232

and since 2(Zp”) = Zp”, Qp” is 2-invariant. If p ( z ) denotes the characteristic polynomial of then p ( z ) is divided by the characteristic polynomial p l ( z ) of ZQ,and so p ( z)is expressed as p ( z)= p l ( z ) p z(z) for some pa (z)E Q[z] 0bviously Q; = ( V E Q” :p l ( Z ) = ~ 0).

x,

.

We write

Qr=

{V

E Q” :p a ( 2 ) = ~ 0).

Then Q” = QF @ Qlf and X(Qlf) = Qlf. Therefore x(Zlf) Qtn Z”, and Zp” @ Zlf is of finite index.

c Zlf

where Zlf =

nrzJ(Zy)

Remark that = (0). Let ‘b’ and t q denote the tori with Zp” and Z,: respectively, as the groups of all covering transformations. Then 9” = @ x @ is a finite covering space of T”. We denote as T’ the finite covering map from ’k” onto P.Since a(ZF) = Z; and x(Zlf) c Zlf, 2 induces an automorphism Ap : @ + @ and an endomorphism A , : ‘b -, fq. The following cases are considered: (i) gn = ~ p (z; ” = (0))

(7.5)

(z;

I P=

@,

= 10))

T”= Tq, (iii) Z” 3 Zp” @ Zlf (Zp” # (0) and Zlf # (0)) * en is a finite covering space of P with covering map T’ : ’k” 4 P.

(ii) Z” = Z;

In the case (i) we have that f : P+ P is a homeomorphism. Hence, for the discussion about the inverse limit system of (P, f) it suffices to consider the cases (ii) and (iii). From now on we deal with the cases (ii) and (iii) of (7.5), i.e. Zlf # (0). - Since ’k” has Zp” x Z t as the group of all covering transformations and Alz; x is the group homomorphism with

&,

f: T”4 T”induces a self-covering map f’ : ’k” -t %“’ such that the diagram ex?,

4T”

f ’ @ x j p

f

1.’ T”

commutes.

$7.2 Inverse limit systems of self-covering maps on tori

233

As above denote as '2the transposed matrix of 2.Then det(3) = det('3). (Z:) Z : , we have 'X(Z:) Zt and write Since x

5

5

Obviously 'Z(G)= G and G is a subgroup in Q'. Impose the discrete topology in G and denote as the dual of (G,'Z) (g(dx) = ('Xg)(z), g E G and x E S,). Then S, is a q-solenoidal group with metric do defined as in (7.2) and d : S, -+ S, is an automorphism. Thus we can find a continuous homomorphism $1 :Rq + Sq such that S, is expressed as

(&,A)

(74

sq

+

= $1 (aq) Fq

where Fq is the annihilator of Zt in S, (by Lemma 7.1.2 (2)). Then F, is the closure of $1 (Zt) by Lemma 7.1.4. It is clear that d o $1 (Rq) = $1 (W), and so

for s E Rq and g E G. Thus we have

That S, contains no tori is checked as follows: Suppose that S, contains a torus and let T be the maximal torus of S,. Then d(T) = T and the character group of T is the factor group GIG' where G' is the annihilator of T (G' = ann(G,T)). Thus GIG' is finitely generated. Then there exist 91,.. ,gk E G such that G = K @ G' (direct sum) where K = gp(g1,. , ,gk}, from which Sq splits into the direct sum S, = T @ Si of the maximal torus T and a solenoidal group Si where T = ann(S,,G') and Si = ann(S,,K). Obviously K is the character group of T and G' is that of S;. Since F, is totally disconnected, T n F, is finite and so T n LF, = (0) for some L > 0. Thus we have

.

.

If Z1 = ann(G,LFq), then Zt C Z1 and Z,/Ztis algebraically isomorphic to F,/CF,, from which mZ1 c Zi for some m > 0. Remark that Z1 is the character group of Sq/CF,. Let Zl,land Z I , ~denote the character groups of (T @ CF,)/LF, and (Si+ LF,)/LFq respectively (i.e. Z ~ isJthe annihilator of

CHAPTER 7

234

(Si+LF,)/LF, and Z1,zis that of (T@LF,)/LF,). From the splitting of S,/LF, it follows that

Z1 = Zl,l63 Z1,2. Since the automorphism 2 : S, + S, induces an automorphism from (T@ LF,)/LF, onto itself, we have z(Z1,~) = Z1,2and so Z(rnZ1,z)= rnZ1,~ c Zi. QO -i This shows mZl,z c 0,A (Z;) = {0}, thus contradicting. Therefore the homomorphism $1 : Rq + S, is injective by Lemma 7.1.6. We define the homomorphism $ : RP @ Rq + RP @ $1 (Rq) by

(7.8)

= ( U P , $1

$ ( U P , v9)

for up E RP and v, E

Rq,

and put

--I

(7.9)

('u,))

and A ' = $ o ? I o $ - l ,

f =$oTo$-'

i.e. the diagrams

-

R"

f

-

A

R"

1.

commute.

$(Rn) f $@">

z

Since ?I : RP@RQ+ RP@RQis expressed as the direct sum 2 = of the linear maps : RP + RP and X p q : Rq + Rq,we have

z;;ilwP

X-I =

XlWP

@ ($1

0

zpp 6 3 z i R q

z1w* $,'), 0

z'

which shows that : +(Rn) -+ $(Rn) is continuous, because &,l(Rq) = $1 0 A p 0 $1' by (7.7). Let dKP and dRq be the euclidean metrics for Rp and RQ respectively and define for (vp,vq),( v k , ~ ; ) E RP x RQ

(7.10)

d((vp,v,), (v;,~;)) = m a x ( d w p ( v p , v ~ ) , d a q ( ~ q , ~ ~ ) } .

Then, 2 is uniformly equivalent to d p . A metric 2' for RP @ S, is defined by (7.11)

--I

d ( ( v p , ~ ) , ( v k , ~ - Im) a) x ( d ~ p ( v p , 2 1 k ) , d o ( ~ , y - I ) )

for (vp,y), (vk, y') E RP x S, where do is the adapted metric defined as in (7.2) for S,. Since ?,hl(Rq) c S,, we have

dO($l(v,),$l(v;))= dWq(vq,v;)

$7.2 Inverse limit systems of self-covering maps on tori

235

if dwq('uq,'uL) < (YO ((YO is a number chosen for S, as in (7.2)), and so (7.11) equals max(d~~('u*,'U~),dw~('u,,'ub)} whenever dwq('u,,$) < 1y0. Thus we have --I

d (+('u),$(w)) = z(w,w)

if z('u,w)

< (YO

(v,w E R").

Note that a base of neighborhoods of the identity of S, is $1(U)

+ F,J

:

U is an open neighborhood of 0 in Rq and Fq,x is an open subgroup of F,

(see Lemma 7.1.2 (4)). From the existence of a subgroup Zx,; index such that

of Z : with finite

the relative topology of +l(Rq) is given by the following base of neighborhoods of O in +l(Rq):

$1

(U

+

U is an open neighborhood of 0 in RQand Z;,-+ : )Z;,x is a subgroup such that +I(Z;,~) = F,,x n$(Re)for some open subgroup F,,x of F

For Z;,A a subgroup of follows that

Zt with finite index let Fq,x = ~l(+l(Z;,~)).Then it F,,X

n +l(R,)

= dl (Z;,,).

Lemma 7.2.1. (1) f'(+(v P)) = ~ ' ( + ( P ) ) f'(+(v)) for 'u E R" and P E Z", (2) 7' and 2' are d-biunifownly continuous.

+

+

Proof. (1) follows from (7.4). Since j I $ l ( w q ) = $1 o Alp o +11 by (7.7) and : S, + S, is an automorphism, we have 2'-biuniform continuity of 71'. Since 7 : R" + Rn is 2-uniformly continuous, for every open sets U c R p and U' c IWq there are open sets V c Rp and V' c RQsuch that ?(V@V') c U@U', and so 7'(V @ +l(V')) c u @ +l(U').

A

: +l(RQ) + $1(RQ) is continuous, there is a subgroup Z:,2 of Since iZ: with finite index such that

and

CHAPTER 7

236

Therefore we have -I

f (v@$I(v'+Z;,,))c u@$l(u'+Z;,l)

from which f' is d-uniformly continuous. The same result holds for the inverse of 7'. 0

x'

7

From Lemma 7.2.1 it follows that the maps and induce homeomorphisms of WP@S, onto itself respectively. We denote them as the same symbols. Let p : W P @ S, + @' CB S, be the natural projection defined by (v

+ y)

-

(v

+ Z,") +y

(v E W" and y

E

S,).

Then we can define a homeomorphism : 'b @ S, + @ @ S, and a group automorphism A : 'b' @ S, + @' @ S, by jop=po7'

(7.13)

and A o p = p o x '

respectively. The following diagram illustrates the relationship among maps described above.

Lemma 7.2.2. For z E @' @ S, and y E F,, j ( z + y) = A(y) + f(z). Proof. This follows from Lemma 7.2.1 (1) and continuity. I3

x'

Since o $1(Zi)C $1 (Zy), we have A(F,) c Fq. Thus a continuous map : (%P @ Sq)/Fq+ (@ @ S,)/F, is defined by

p(z + Fq) = j ( z ) + Fq

(zE @ @ Sq).

$7.2 Inverse limit systems of self-covering maps on tori

Since S, = $(Rq)

+ F,,

237

it follows that

Thus a map ( : (@ @ Sq)/Fq+ 'b' @ ?q is defined by

Then ( is continuous and bijective.

Proposition 7.2.3. The following diagram

Proof. The commutativity is calculated as follows: for x E

'b and v E Rq,

Theorem 7.2.4. Let f:T" + T" be a self-covering map of an n-torus and A : If" -+ T" denote the endornorphisrn homotopic to f . Then the following properties hold :

CHAPTER 7

238

(1) There are an n-dimensional solenoidal group S and an automorphism : S + S such that ( S , A ) is topologically (algebraically) conjugate to the inverse limit system of (T", A). ( 2 ) There is a homeomorphism f of the solenoidal group S such that (S,f ) is topologically conjugate to the inverse limit system of (Tn,f).

d

Proof. We define the inverse limit system (X, p') of ((@' @ S,)/F,,

x=

{(f-n(2)

and

f"((f-n(~)

+ Fq)n20 : 2 E T p @ s,

71

p ) by

2 0)

+ Fq)n>O) = ( F n + ' ( z ) + Fqih)n>o.

Then X is a closed subset of an infinite product topological space (('k@ S q / F , ) N ,and : X + X is a homeomorphism. To see the relation between the systems (@'@ S,, f ) and (X, p'),we let ((2) = (f"(2)

+ F,),>o

for 2 E

PPCB S,.

Obviously ( : @'@ S, + X is continuous and surjective. The injectivity of C is checked as follows. If ((2) = ( ( y ) then f - n ( z ) F, = f - " ( y ) F, for n 2 0, and hence f"(f-,(z) F,) = P ( f n ( y ) Fp),from which z - y E A"(F,) for n 2 0 (by using Lemma 7.2.2). 00 -i Since x(Zy)C Zt and A (ZF) = {0}, we have x'($~(Zy)) c $1(Zy) 00 --ri A ( $ l ( i L ; ) ) = (0). Since F, is the closure of $l(Z;), obviously and cl(x'($l(Zlf))) = A(F,) c F, and the following is easily checked:

+

+

no

no

n

n

00

Q)

(7.14)

cl(xij(~q)) = A ~ F , )= (0). 0

0

Therefore,

+

+

2

= y . The following commutative diagram was obtained : 7b$Sq

c1 X

T ('ifp @ S,)/F,

€1 fP@f,

i

TP@S,

1c:

j,'

X

-

T

i'

(%p

homeomorphism

inverse limit

@ S,)/F,

1

.$: homeomorphism

f'

___)

VP@@

57.2 Inverse limit systems of self-covering maps on tori

239

From the diagram it follows that (@ @ S,,f) is the inverse limit system of (p",f ' ) . Therefore, (2) is concluded by the following Lemma 7.2.5. Letting f = A, we obtain (1) as a special case of (2). 0

Lemma 7.2.5. Let f:T" + T" be a self-covering map which is not a homeomorphism. Let R' : p" t T" be a finite covering map and denote as G(n') the covering tmnsformation group for R'. Then a self-covering map f ' : TP t TP satisfying f o R' = R' o f ' induces a homomorphism f : : G(?r') --t G(n') (see Remark 6.3.11). If j ' i ( G ( ~ ' )= ) {id}, then the inverse limit system of (T", f ) i s topologically conjugate to the inverse limit system of (p, f ' ) , i.e. letting

njz0

Xf = {(zi);>o: xi E T" and f ( z i + l )= 2i (i 2 0 ) } , Xp = {(fi);>o : 2i E 9" and

P (&) one has that

f'(&+l)

= 2; (i >O)},

H( T I ( & ) ) ,

P :Xff 4 Xf

as a homeomorphism and the diagmm

Xf'

d Xfl

4

-

Xf

0

commutes.

Xf

Here u' and u denote the shift maps induced by f ' and f respectively. Proof. Since f o R' = R' o f ' , by definition we have the above commutative diagram. Thus it suffices to show only that P is a homeomorphism. The continuity of R' implies that of P. To show injectivity of P, for (i;), ($;) E X f f let (~'(5;)) = ( R ' ( & ) ) . Then for i 2 0 there is a; E G(R') such that f i = cri(5.i) (see Remark 6.3.6). Thus we have Oi

= .?(&+I) = f'(ai+l(j.i+l)) = f:(ai+l) 0 f ' ( S i + l ) = f:(ai+l)(zi)

n,

and so ai = f:(ai+l),which implies ai E f ' i ( G ( d ) ) = {id}, i.e. a;= id. Therefore, (&) = ( g i ) . This shows that P is injective. It is easy to see that /3(X;) is dense in Xf. Since X; is compact, it follows that P is surjective. Therefore P : Xf, -+ Xf is a homeomorphism. 0 Let C(Tn)be the set of all self-covering maps of 'IP'. Then C(Tn)becomes a metric space with respect to d defined by d ( f , g ) = m a x { d ( f ( x ) , g ( z ) :) z E T"} for f , g E C(T") where d(z,y) denotes the metric for 'IP' induced by the euclidean metric 2 for an.

CHAPTER 7

240

As above, let f be a self-covering map in C('JP) and denote as 7 : B" + R" a homeomorphism which is a lift of f by ?r. Let { fi} be a sequence of self-covering maps in C(Tn)and suppose fi converges uniformly to f under d. If i is sufficiently large, then each of fi is homotopic t o f (see Remark 6.7.10). Hence, if Ti : Rn + Rn is a lift of fi by ?r, then the following property holds:

fi(w

+ l ) = x ( l )+ T i ( w )

for I E

Zn and w E B".

Since d( fi, f) + 0 as i + 00, we may suppose that Ti converges uniformly 7 under 2. Indeed, let F : T" x [0,1]+ 'JP be a homotopy from f to fi. By applying the homotopy lifting property (Theorem 6.2.1) we can find a homotopy : R" x [0,1] + R" starting at 7. Define Ti : In + B" as the terminal map of F. Obviously Ti is a lift of fi by ?r. Since ?r is a locally isometric covering map, it is easy to check that z ( 7 ( z ) , f i ( z ) = ) d(f(?r(z)), - - fi(?r(x)))for all z 6 R" if i is large (see Remark 6.7.10). Thus d ( f i , f ) = sup{z(Ti(z),f(x)) : z E a"}+ 0 as i + 00. Then As in (7.9)define :R*@$l(RQ) + RP@$l(Rq) by = $ofio+-'. Lemma 7.2.1 holds for 7;. Therefore each 7: induces a homeomorphism - - - from B P @ S, onto RP@S,, which is denoted as the same symbol. Since d( f i , f ) -+ 0 as i + 00, it follows that the sequence of homeomorphisms of RP @ S, -I converges uniformly to f under b, where 7' denotes the homeomorphism of RP @ S, induced by 7 and d is a metric for R* @ S, as in (7.11). Thus we have the following lemma.

to

3

{Ti}

Lemma 7.2.6. Let f:'ll"' + 'JP be a self-cowering map and suppose { fi} is a sequence of self-covering maps in C(T"). If d ( f i , f ) --+ 0 as i + 00, then there exists a sequence of homeomorphisms of RP @ S, such that 7; converges uniformly to under 2'.

f

{ri}

Theorem 7.2.7. Let f and {fi} be as in Lemma 7.2.6. Then there exist a homeomorphism f of 'b @ S, and a sequence {fi} of homeomorphisms of ?b'@ S, such that f i converges uniformly to f.

Pro05 As in (7.13) define homeomorphisms Lemma 7.2.6 the conclusion is obtained. 0

f

and

fi

of

'b @ S,.

Then by

CHAPTER 8 TA-Covering Maps of Tori

In this chapter we shall restrict ourselves to tori and deal with the class

7d of all TA-covering maps. We recall that each map f in I d belongs to the subclass 7d3-1if it is injective, and ortherwise f is a member in one of the following three subclasses

PEM, SS7d, 7d \ (7d3-1U PEM U SS7d). The purpose of this chapter is to show that for each subclass, a map in it is characterized, in a topological sense, as an algebraic endomorphism which indicates a characteristic of the subclass.

$8.1 Toral endomorphisms homotopic to TA-covering maps In this section we investigate some properties of TA-coveringmaps which are invariant under homotopy. To show the theorems stated below we need some results on fixed point indices of TA-covering maps, which will be established in Chapters 11 and 12.

Theorem 8.1.1. Let f:T" + 'Iln be a self-covering map of an n-torus and A : T" + 'Iln denote the tom1 endomorphism homotopic to f . Iff is a TAcovering map, then A is hyperbolic.

Proof.When f is not expanding (i.e. f is a map in 7d \ PEM), from Theorem 6.7.3 it follows that the family F; of all stable sets in strong sense is a generalized foliation on T".In this case we can say whether or not the family F; is orientable (this will be discussed in Chapter 10). If F; is non-orientable, we use Proposition 10.7.2 and take a double covering space of 'Iln. Then a lift of f by the double covering map belongs to I d \ PEM (see Lemma 2.2.34) and the lift of 3;is orientable. Hence we may suppose that 3;is orientable. Since T" is orientable, we can use Theorem 10.9.1. When f belongs to PEM, we apply Theorem 10.8.1. In any case the following holds; there is P > 0 such that for each m 2 L all fixed points of f" have the same fixed point index 1 or -1. Choose a positive integer mo with mo 2 P such that fmo is topologically mixing on each elementary set, and write g = f " 0 . Obviously 9:T" +Pis a TA-covering map and g is homotopic to A"o. It is enough to show that A"O is hyperbolic. 241

CHAPTER 8

242

By the result stated above together with the property of fixed point index (see Property 10.4.4 of Chapter 10) we have for m 2 1

where N ( g m ) is the number of fixed points of g", I ( g m , x ) is the fixed point index of gm at x ; and I ( g m ) is the fixed point index of g". Let X i (1 i n) denote the eigenvalues of the linear map of Bn which is a lift of A"". Then from Lefschetz fixed point formula it follows that

<
0 such that any &-pseudo orbit of g , ( x i ) , is e/3-traced by some point in (lP')g.Since g is topologically mixing on an elementary set B, there is k > 0 such that g k ( K ) n K' # 0 for any K , K' of a finite cover consisting of ~/P-ballsin B. Let x E B be a fixed point of gm and choose y E B such that d ( x , y) < E and d ( z , g k ( y ) )< E . Then we construct a one sided (m k)-periodic &-pseudoorbit

+

+

which coincides with the one sided sequence ( z i ) r of a two sided (m k)Hence there is (yi) E (Tn)gsuch that periodic €-pseudo orbit ( z ; ) in (Im)'. d ( y i , zi) < e/3 for all i E Z. By c-expansivity we have gm+k((yo) = yo. Therefore N(gm) < N(grn+&)for m 1. Note that each Xi is not a root of unity. Indeed, this follows from the fact that Per(g) # 0 and N ( g " ) = 11 - AT/. To see 1A;I # 1 for 1 5 i 5 n, suppose [ A i l = l ( 1 5 i s), I X i ( < 1 (s+l 5 i 5 t) and IXiI > 1 ( t + l i n). Since N ( g m ) N(gm+') for m 1, we have

>


0 such that for m 1 e each fixed point of f" : T" + Tn has the same fixed point index 1 or -1. Fix the natural number rn here. Let 3 : R" + R" denote the linear map which covers the toral endomorphism A homotopic to f . Since 71" : R" + R" is hyperbolic by Theorem 8.1.1, it follows that '7i" has the single fixed point 0 and the fixed point index, I ( T , O ) , equals to f l (see Remark 10.4.8 of Chapter 10). Since f" and A" are homotopic, the relation between 7 and 2" is expressed as

from which we have M = sup{llr(x)-f"(x)I( : x E R"} < 00. S i n c e r - i d is an isomorphism of R" by hyperbolicity of Z", we let p = inf{Il(?lm id)(x)ll : 11x11 = 1) > 0. Here id is the identity map and 11 11 denotes the euclidean norm of R". Choose r such that p r > M. If 11x11 > r , then

which implies that all fixed points of 7 are contained in the ball of radius T . Since 7 is expansive, the fixed points must be isolated. Therefore F i x ( 7 ) is finite. Let us define bt : R" + R",O 5 t 5 1, by k t ( x ) = tA"(2)

+ (1 - t > T ( x ) .

Then kt is a homotopy from 7" to 2" and if 11x11 > T then

Hence, for any t, bt has no fixed points outside the ball of radius T . This shows = I ( 7 ) = CZEFix(f=) I ( 7 , x ) (see Properties 10.4.4 and 10.4.5 that I@") of Chapter 10).

CHAPTER 8

244

7,

By the fact that f o 7r = 7r o it is clear that x(Fix(7)) c Fix(f "). Let x E F i x ( 7 ) and choose an open neighborhood U, of x such that the restrictions 7rl(1= and nlTm(Vs) are both injective. Since the diagram

commutes, 4 U Z )

f"- 4 7 Y U x ) )

we have I ( 7 , x ) = I(fm,7r(x)) (see Property 10.4.7 of Chapter lo), from which each 2 E F i x ( 7 ) has the same index by Theorems 10.8.1 and 10.9.1 of Chapter 10. Hence #Fix(F) = I

I ( 7 , z ) l = I I ( 7 ) l = l I ( r ) [= 1. zEFix(f".)

Therefore, 7 :R" + R" has exactly one fixed point and so does 7. 0

$8.2 Construction of semi-coqjugacy maps

In the previous section we have seen that toral endomorphisms homotopic to TA-covering maps are hyperbolic. In this section we discuss dynamics of hyperbolic linear automorphisms of the euclidean space R", and show the following Theorem 8.2.1. Let 2 be a metric for R" induced from a norm of R". For continuous maps f and g of R" we write -

d ( f , g ) = sup{~(f(x),g(x)): 2 E R"). Notice that

a(f , g ) is not necessarily finite.

Theorem 8.2.1. Let f:"" + 11'" be a self-cowering map and A : 'lr" + 'IP denote the tom1 endomorphism homotopic to f . Let 3: : R" + R" be a lift of f by the natuml projection x : Rn + 'IT" and let 2 : R" + R" be the linear automorphism which i s a lift of A by 7 ~ . If A is hyperbolic, then there as a unique continuous surjection : R" R" such that the following properties hold: (1) Z o h = E o ? , (2) a(%,i d ) i s finite, ( 3 ) E is uniformly continuous under& where i d denotes the identity map of R". First we establish the following proposition to obtain Theorem 8.2.1.

98.2 Construction of semi-conjugacy maps

245

Proposition 8.2.2. Let L : R" + R" be a hyperbolic linear automorphism and let T : R" + Rn be a homoemorphism. If J( L, T) is finite, then there is a unique map 4 :R" + R" such that (1) L o 4 = 4 o T, (2) 2(4,id) is finite. Furthermore, for K > 0 there i s a constant SK > 0 such that if a(L,T) < K , then the above map Cp has the following properties : (3) a('$,id) < S K , (4) Cp is a continuous sujeetion, ( 5 ) Cp i s uniformly continuous under 2 if so is T . For the proof we need the following Lemmas 8.2.3 and 8.2.4.

Lemma 8.2.3. Let L :Rn + R" be a hyperbolic linear automorphism. Then the following properties hold : ( 1 ) for K > 0 and E > 0 there exists N > 0 such that if a(Li(z), Li(y)) 5 K for all i with lil 5 N , then z(z,y) 5 E , (2) for given K > 0, if a(Li(z), L'(y)) 5 K for all i E Z, then z = y, Proof. Since L : R" ---t R" is hyperbolic, there are L-invariant subspaces E m , E" and constants 0 < A < 1 , c 2 1 such that R" = E" @ E mand (8.2)

IILiE.II 5 cX"

(Vn 1 O),

IILiE.II I cX-"

(Vn 50).

For simplicity, let L, = LIE.and L, = LIE.. Obviously L is linearly conjugate to L, x L,. - (1): Let K > 0 and E > 0. Choose N > 0 satisfying cKXN < E. If d ( L r ( z ) ,L$(y)) 5 K , then we have

-

d(z,y ) = ;i(L,N

0 L:(z),

L,N

5 cXNa(L:(z), L:(y))

0 L,N(y))

< cXNK < E .

Similarly, if a ( L ; N ( z ) , L ; N ( y ) ) 5 K then d ( z , y ) < E . Hence, L , x L, possesses the property of (1) with respect to a metric induced from a norm of E" x E m .Since all norms of Rn are mutually equivalent, from the fact that L is linearly conjugate to L , x L, we obtain the conclusion of (1). (2) is clear from (1). 0

Lemma 8.2.4. Under the assumptions of Lemma 8.2.3, for K > 0 there exists 6~ > 0 such that for any K-pseudo orbit { z i : i E Z} of L there is a unique z E R" so that a(Li(z), z i ) 5 SK for i E Z.

Proof. Let L, and L, be as in the proof of Lemma 8.2.3. Since L is linearly conjugate to L, x L,, it is enough to show the lemma for L, because the analogous result holds for L,.

CHAPTER 8

246

Let c and X be numbers satisfying (8.2), and put 6~ = c K / ( l - A). Let {zi : ( ( a k (< ( K where ark = z k + l - Lu(2k)

i 2 0) be a K-pseudo orbit of L,. Then for k 2 0, and we have zk

= Lt(Z0)

+ Lt-'(CUo) + + a&-1

Write g = L i l for simplicity. Then ()gkl((5 cXk and = g(a0)

Forp

for k 2 1.

- a *

+ *. - +

zk

= Lk(z0

+ a h ) where

gk(a!g-l).

> Fz > 0 we have

cKXk++'

I- 1 - X ' and so

{vk}

converges to some p in E". If y = zo + p then k

k

L,(Y) - z k = Lt(50 + p

- 20 - -&"(ai4)) 1

m

M

for k 2 0. Applying this fact, we can easily prove the existence of a SK-tracing point for any K-pseudo orbit {zi :i E Z}.By Lemma 8.2.3 (2) its uniqueness is clear. 0

Proof of Proposition 8.2.2. Since z ( L , T ) is finite, we can take K > 0 such ) K for all z E R". Let SK > 0 be as in Lemma 8.2.4. For that d ( L ( z ) , T ( z )< any z E R" the sequence { T j ( z ): j E Z} is a K-pseudo orbit of L. Hence there is a unique y E R" such that

We define a map q5 : Rn + R" by +(z) = y. Since z(z,y) I 6 ~ obviously , sup{z(q5(z),z) : 2 E R"} I S K . Hence (2) holds. Since ( T j ( T ( z ) ): j E Z} is SK-traced by a point L(y), we have L ( $ ( z ) )= L(y) = q5(T(z)),from which (1)is obtained.

$8.2 Construction of semi-conjugacy maps

The uniqueness of 4 is easily checked as follows. If a map satisfies (1) and (2), then for v E Rn and i E Z

-

d(Li 0 4(v),Li 0 c#J'(v)) 5 sup{;i(L'

L'

0 Q ( 2 ) ): 2

fp'

0

0 $(2),

= sup{&b 0 Ti(.),

247

4' : R" + R" E Rn}

T'(2)):2 E R"}

= sup{Z(fp(2),4'(2)) :2 E a"}< 00. Thus +(v) = #(v) by Lemma 8.2.3 (2). Next, we show the second statement. For K > 0 let 6~ be as in Lemma 8.2.4. If a(L,T) < K, then Z(4,id) < 6~ by the construction of 4. Hence (3) holds. To see (4), suppose 4 is not continuous at some 20 E Rn. Then there is a sequence 2, + zo ( n + 00) such that yn = 4(zn)does not tend to yo = +(Q) as n + 00. Since 12,) is &bounded and 2(4(xn),zn)5 6~ for n 2. 0, the set {+(zn)}is also &bounded. Hence we can suppose, by taking a subsequence if necessary, yn + 9; # yo as n + 00. By Lemma 8.2.3 (2) we can find k E Z satisfying

-

k

I

d ( L (yo),

Lk (YO)) > K/4 + 36K.

Fix k and choose no > 0 such that a(Lk((yn),Lk((y&)) < K/4 for n 2 no. Since 2, + 20 as n + 00, we have ~(Tk(zn),Tk(zO)) < 6~ for n 2 no if no is sufficiently large. Then we have -

d ( L k ( y n ) ,L'((Y~))= Z ( L ~0 4(2n)r L~ 0 fp(20))

-

=~

( ~ o T ~ ( ~ ~ ) , ~ o T ~ ( ~ O ) )

5 Z(4 0 Tk(On),Tk(2n))f;i(Tk(Zn),Tk(Z0)) +~ < 36K

( ~ ~ ( ~ 0 ) , 4 0 ~ ~ ( ~ 0 ) )

and hence

which is a contradiction. By (2) the map 4 is extended to a continuous map on Sn = R" U {m} by &z) = 4(z)for x E R" and &oo) = 00, and a homotopy ht between and the identity map is defined by

4

ht(z)= t+(x)

6

+ (1 - t ) z (2 E R")

and

4

ht(00) = 00.

Hence : S" + S" is surjective (see Remark 10.6.4 of Chapter 10) and so 4 :Rn -t R" is surjective. Therefore (4) was proved.

CHAPTER 8

248

To show (5), let 6~

>0

be as above. For given

E

> 0, by Lemma 8.2.3

(1) there is N > 0 such that if a(Li(z),Li(y))< 36K for i with (il 5 N , then d ( z , y ) < E . Since T is uniformly continuous, we can take y > 0 satisfying the property that a(Ti(z),Ti(y)) < 6~ (-N 5 i 5 N) whenever ;i 1 and a compatible metric for R" such that (1) D is complete, ( 2 ) all covering transformations for n are isometries under 0, (3) m4,7(Y))1 X%Y) for Z,Y E R".

n

Proof. Since f is positively expansive, by Theorem 2.2.10 there exists a compatible metric D for T" and constants 6' > 0 and X > 1such that D(z,y) 5 6' implies D(f(z),f(y)) _> XD(z,y). Since n : R" -+ Tn is a covering map, there exist a metric D for R" and a constant 60 > 0 satisfying the properties in Theorem 6.4.1. For 6 = min(S',So}, Lemma 6.5.4 ensures the existence of 0 < 61 < 6 such that D(f(z),T(y>) < 61 implies D(z,y) < 6. Note that B(z,i) = D ( z ,y ) since 6 5 60. F'rom these facts we have that d(z,y) < & / A if d(f(51,?(I/)) < 61. For z,y E R" let {xi : 0 5 i 5 L 1) be a 61-chain from z to y (i.e. D(zi,s;+l) < 61 for 0 5 i 5 t ) and define by

+

$8.3 Nonwandering sets

249

where the infimum is taken over all finite &-chains from x to y. By the triangle inequality of D we have D(x,y) 2 B(x,y), from which D is a metric for R". Clearly B(z,y) = B( x , y ) if D(z,y) 5 61. Thus 'ZT is compatible and by Theorem 6.4.1 (3), (1) holds. (2) is clear from the construction of together with Theorem 6.4.1 (2). It remains to show only (3). Let {xi : 0 5 i 5 m } be a finite sequence from f(x) to f ( y ) with B(xi,xi+l) -- 1 < 61 for 0 5 i 5 m - 1. Then {f-l(xo), ... ,f (2,) is a finite sequence from x to y such that w-1(xi),7-1(xi+l))

for 0 5 i


0 there exists 6, > 0 such that for any K-pseudo orbit {xi :i 2 0) o f 7 there is a unique x E R" so that D(T(x),xi) 5 6~ for i 2 0. Proof. This is proved in the same technique as the proof of Lemma 8.2.4. In fact, put x: = 7 - ' ( x i ) for i 2 0. By Proposition 8.2.5 we have

-

D ( 4 - , , 2 iO ) -

a(f-- i 1--

0

7(x&l),7-i(zi)) K

< +(f(xi-l),xi) - A

L

3

(i 2 0).

Thus 1x4) is a Cauchy sequence and so there is a point z in R" such that x: -+x as i -+ 00. Fix i > 0 and let 0 5 j < i, then we have . . D ( x j J j ( x ; ) ) = D(xj,fJ-*(xi))

-

-< D ( x j , 7 - 1 ( x j + l ) ) + + Z(T++l 5 K(X-l + .. . + A-(i-j) 1< s * * *

(xi-l),7j-i(Zi))

where 6~ = K/(X - 1). Therefore D(xj,T'(x))< 6~ for j 2 0. 0

Proposition 8.2.7. Under the assumptions and notations of Proposition 8.2.5, there exists a unique continuous surjection :R" + Iw" such that (1) f O L-%-O X , (2) sup{D(k(x), x) : x E R"} is finite, ( 3 ) % is uniformly continuous under

n.

CHAPTER 8

250

--

Proof. The finiteness of sup{D(f(z),z(z)) : z E R"} is followed from the fact for L E Z" and z E R". The existence of that 7 ( x + 1 ) = a(1) satisfying (1) and (2) is easily checked from Lemma 8.2.6. The proof of (3) and the uniqueness of K is similar to that of Proposition 8.2.2. 0

+ 7(z)

$8.3 Nonwandering sets In Chapter 7 we have seen that the inverse limit system of a self-covering map of 'P is topologically conjugate to a homeomorphism of an n-dimensional solenoidal group S (see Theorem 7.2.4). In this section we shall establish the following Theorem 8.3.1, and show that for every map belonging to ?A the nonwandering set coincides with the entire space T". Throughout this section, let f : ' P + 'P be a self-covering map of an ntorus and A : 'P + T" denote the toral endomorphism homotopic to f. Let f be a lift of f by the natural projection ?r : Rn + T" and denote the linear map which is a lift of A by T. As seen in $7.2, these lifts induce a homeomorphism : 'b @ S, + 'b @ S, of an n-solenoidal group and an automorphism 2 : 'b @ S, + 'b' @ S, respectively (see (7.13) of $7.2).

Theorem 8.3.1. If A is hyperbolic, then there is a continuous surjection

i: 'b' @ S, -+ 'b' @ S,

such that

(1) A L o k = i o j , (2) each path connected component in 'b @ S, is k-invariant.

To show the above theorem, suppose A is hyperbolic. By Theorem 8.2.1 there is a semi-conjugacy map 'il :R" + R" between and 7.

Lemma 8.3.2. Let Zp" and Z t be as in (7.5) of 57.2. Then (1) for v E R" and 1 E Zp"

-

h(v

+ 1) = P + 'il(v),

(2) for any X > 0 there exists a subgroup Zt,l c Zt, which is of finite index, and a continuous map i@: R" x Zt,l + Bx(O), where Bx(0) i s a closed ball at 0 of mdius A, such that

-

h(v

+ 1) = P + 'il(v) + q ( w , L )

for v E Rn and P E Zi,l.

Proof, (1) : Since f and A are homotopic, there is K > 0 such that a(?(v), A(v)) < K for v E R". Hence, for L E Z; the sequence {?'(v 1) : j E Z} is a K-pseudo orbit of 51. Let 6~ be a number as in Lemma 8.2.4. Then there

-

+

$8.3 Nonwandering sets

251

+

is a unique w E R” such that a(fj(v C),Z’(w))5 6~ for j E Z,and by the construction of h we have x(v 1) = w. Also {rj(v) : j E Z}is a K-pseudo - -j -orbit of Thus d(f (v),Z’(w‘))5 SK ( j E Z)for some w’ E W”,and hence h(v) = w’. Let 6 = w - 1. Since z(Zp”)= Zp” by definition, we have rj(v + Zp”) =

+

x.

Zp”

+ TJ(v) for j E Z.Thus ~(~j(v),Zj(~))=~(T’(v+l),Z’(65+~)) < 6K

for all j E Z and so w‘ = 65 by Lemma 8.2.3 (2). Therefore

-

h(v

+ 1) = w = w‘ + e = -h ( v )+ e.

(2) : Let A be a positive number. Since 2 is hyperbolic, by Lemma 8.2.3 (1) we can find N > 0 such that for w,w E R”

-(8.3)

d(A’(V),Z’(W))5 26K ( V j 2 - N )

a ( ~W ,) < A.

Choose then a subgroup ZX1 of finite index such that 2’(Z7,1)

c 257

for

j 2 -N. Obviously

fj(v

+ q l )= x’(Z;,l) + ?’(v)

(v E R”)

+

for j 2 -N. Let v E R” and t? E Z L,; Then {Tj(v) : j E Z} and {r’(v e) : j E Z}are both K-pseudo orbits of A. Thus, there are w’,w E R” such that -d(f’(v),Z’(w‘))5 6~ and l),Z’(w))5 6~ for j E Z,and h(v) = W’ - -j and h(v+l)= w hold. Put 6 = w -e. Then we have d ( A (6),xJ(w’)) 5 26K for j 2 -N, and hence a(6,w’)< A by (8.3). Since 6 - w’ depends on v and 1, letting F(v,!) = 6 - w’,a map T7j : R” x Zt,l --+ Bx(0) is defined, and so

a(?’(~+

-

h(v

+ e) = w = 6 + e = w’+ F(v,l)+ e -

= h(v) from which the continuity of Let 11, : RP @ Rp

--i

+ e + a ( v ,q,

is obtained. 0

RP @ $1(Rg) be as in (7.8) of $7.2 and define

h =$oho$-’, F’($(u),$(l)) = $ oF(u,!) --I

for u E R” and l E Z;,l

where (p : R” x Z:,l --f Bx(0) is a continuous map of Lemma 8.3.2 (2). Since A oh = ho 7 on R”, we have on

$(R~>=R”@$~(IwQ)

(8.4)

~‘oh‘=Zor’

where 2’and

f denote maps defined as in (7.9).

CHAPTER 8

252

Lemma 8.3.3. Let 2’ be a metric for

@ S, as in

$l(Rq) Rp @ $1(Rq) and p’ : $(Itn)x

$l(Z;,,) --$

--$

(7.11). Then

x’: RP @

$(Bx(O)) aTe both 2’-

uniformly continuous.

Proof. Let 6~ > 0 be a number such that 2(E(v),v) < 6~ for v E R”. Since A :RP @ Rq 4 RP @ RQ is hyperbolic, by Lemma 8.2.3 (1) for any sufficiently small q > 0 there is N 2 0 such that

-

I

4

By Lemma 7.2.1 (2) we have that f is d -biuniformly continuous. Let 0 q’ < q be a number to be determined later and take a > 0 such that


+ $(c, y)

$(It") and y E $(Zl;,l).

Proof of Theorem 8.3.1. Since $1(Rq) is dense in S, by Lemma 7.1.1, by Lemma 8.3.3 it follows that is extended to a continuous map of R P @ S, onto itself, which is denoted as the same symbol. Then x'(z+Z;) = Z;+K'(e) for c E RJ'GI S, by Lemma 8.3.2 ( l ) , and hence Et induces a map of @' GI S, onto itself. Let us denote the induced map as k, then by (8.4) we have

xt

Aoh=koj

on

PGIS,.

Therefore (1) holds. To show (2), let Fi be the closure of $l(Zl;,l). Then by Lemma 8.3.3 the map ij9 is extended to a continuous map, say pt, of (RP@ S,) x Fi into $(Bx(O))where X > 0 is the number appeared in Lemma 8.3.2. By Lemma 8.3.2 we have Et(z

+ y) = y + E 1 ( e ) +$(z,y)

for z E R p @ S ,and y E Fi,

~

CHAPTER 8

254

from which there exists a continuous map satisfying

9 : (@' @ S,) X Fg' 4p 0 $(Bx(O))

k(z + y) = y + k(z)+ @(z,y)

(8.6)

for z E

@'

@ S, and y E Fi

where p : R* @ S, --t T P @ S, is the natural projection. By Lemma 7.1.2 (5), +l(Rq) is the path connected component of 0 in SQ and hence @ @ $1(Rq) is that of 0 in @ @ S,. Let L be a path connected component in @ s,. Since

TP

+

there is y E Fg' such that L = 'b' @ +1(R*) y. By (8.6) we have k('b @ +l(R,) y) = @ @ 1(W) y, and so k ( L ) = L. Thus (2) was proved. 0

+

+

+

The remainder of this section is devoted to the proof of the following theorem. Theorem 8.3.4. I f f : T" + T" is a TA-cowering map, then the nonwandering set n(f) coincides with the entire space S .

E o m the above theorem together with Theorem 7.2.4 (2) we have the following theorem. Theorem 8.3.5. If f:'P4 T" is a TA-cowering map, then

n(f) = 'Ir".

To show Theorem 8.3.4, suppose f: T" -+ T" is a TA-covering map. Then by Theorem 8.1.1 it follows that A is hyperbolic. Hence, by Theorem 8.2.1 we R" between 2 and 7. have a semi-conjugacy map % : R" Lemma 8.3.6. is compact).

h is proper (i.e.

the inverse image by h of any compact subset

Proof. This is clear by Theorem 8.2.1 (2). 0

- --I

Since 2o h = h o 7 on R", we have f(h

-

Lemma 8.3.7. f : h

(0)) = h-l(O).

( 0 ) --+ h-'(O) has POTP.

Proof. Since 7 has POTP, for E > 0 there is 6 > 0 such that every 6-pseudo orbit of 7 is &-tracedby some point of R". If {vi} c h-'(O) is a &pseudo orbit of 7, then an €-tracing point v for {wi} exists in R". Since x(v;) = 0 for all i, each of o x(v) is near to 0 in R", and hence x(v) = 0 by expansivity of 2, i.e. v E K-'(o).

2

$8.3 Nonwandering sets

Lemma 8.3.8.

L-'(O)

255

is the set of one point.

Proof. 7 : E-'(O) + E-'(O) has POTP by Lemma 8.3.7 and is expansive. Let Sl denote the nonwandering set of 71K-1(o). Then the set of all periodic points of T , x - ~ ( ois) dense in Sl by Theorem 3.4.2. Since 7 : R" + Rn has exactly one fixed point by Theorem 8.1.3, we have that R consists of one point. This implies E-'(O) = by Theorems 3.1.2 and 3.1.7. 0 Since A is hyperbolic, by Theorem 8.3.1 there is a continuous surjection

k : @@ S, + '@' @ S, such that d o k = i o j. Lemma 8.3.0. If 9:@'@Sq-+ 'b'@S, is a continuous surjection and satisfies g(0) = 0, then g(@ CB +~(RQ)) c ?P CD +~(wP). Proof. This follows from the fact that component of o in TP @ s,.

'kp

@

+l(Rq) is the path connected

Lemma 8.3.10. If 9:fP@S, + @' @S, i s a local homeomorphism and g(0) = 0, then g(?p CB $1 (w)) = @J@ +1(a,>. Proof. Let x be a point in 'b'@ +l(Rq).Since @@ +l(Rq) is path connected, we can take a path u joining 0 and x in 'b@ S,. Since g is a covering map and g(0) = 0, by Theorem 5.1.7 there is a lift v of u by g starting at 0. Then the image of v is contained in @P @ +l(Rq), which implies that g(@' @ +1(W)) 3 T P @ +1(Rq), and hence the conclusion is obtained from Lemma 8.3.9. 0

Proposition 8.3.11. Ifxo E Per(A), then k - l ( x ~ )is the set of one point. Proof. Without loss of generality we may suppose A ( x 0 ) = $0. Since d o i= k o j , we have j(k-'(xo)) = i-'(xo). Thus j : k-l(x0) + k-l(x0) has POTP. The proof is similar to that of Lemma 8.3.7. Since flk-l(zo) is expansive, clearly k-l(x0) contains a periodic point yo of j. To avoid complication, suppose f(y0) = yo. We denote as cpzo and 'pgo translations of 'kP @ S, + 'b'@ S, defined by 'pzo( z ) = xo z and 'pgo( z ) = yo z respectively. Write

+

Then ij(0) = k(0) = 0 and x~@@S,andz~F,

(8.7)

ij(x

+

+

Aok

=

o ij.

By Lemma 7.2.2 we have for

+ j(y0 + + z ) = -Yo + + + x) = A ( z ) + ij(x).

= -yo

5

4 2 )

j(y0

CHAPTER 8

256

Since k : 'b' @ S, --+ @ '@ S, is a continuous surjection, by Lemma 8.3.9

i(@

@ $,(Ex,)

c qp @ +l(R,),

and since i j is a homeomorphism, by Lemma 8.3.10

a(* t

CI3 +l(R,)) =

*

G3 $l(RP).

Define a continuous bijection $I : 'ib' @ Rq for t v E ?P @ Rq,and put

+ $I(.)

+

3' = $)I-' -I

0 ij 0 $I,

--I

A = $I-'

od o

+

@' CI3 $ l ( R q ) by -I

+I,

-I

$'(t

+ v) =

ole$)'.

12 =

--I

Then, A o k = %I 03'. It is easy to see that 3' and A are uniformly continuous bijections, and that %I is uniformly continuous and El(@' @Rq) C @'@RQ. By (8.7) we have $(v

+ C)

=

x(L)+ ~ ' ( v )

for v E

@' @ Rq

and C E

2

Z : .

and E denote lifts of 3', 2' and by the natural projection IWP@Rq @'@ R'J respectively. Then the following diagram Let 3,

+

-

Rp@Rq 9 RP@R'I

4 1% k(RpCI3Rq)-k(RP@R*) -

-

A

commutes and

+ P) = a(P)+ 3(v) for v E Rp@ R4 and P E ZF @ Zt. Since x(Zn)c Z"and 3 :R" + R" is bijective, by (8.8)the map 3 induces

(8.8)

Tj(v

a self-covering map g:W + p,and we have then the commutative diagram

n,"

2 ( F q ) = (0) where [ is a homeomorphism as in Proposition 7.2.3. Since by (7.14), it follows that ( @ @ S , , j ) is theinverse limit system of ('kp@@,gl).

$8.4 Injectivity of semi-conjugacy maps

257

Thus, g' is a TA-covering map because j is a TA-homeomorphism. Since = K-l(O) and therefore by Lemma 8.3.8,

A o % = % o 3, we have J(%-'(O)) -- 1 k (0) is the set of one point.

On the other hand, by Theorem 8.3.1 (2) we have i - l ( L ) = L for each path component L in 'lb @ S,. Since L o 'pw = k and k(0) = 0, it follows that i-'(fp

'Ib @ +1(Rq), and so i - l ( O ) = %-'(O). Since k ' ( 0 ) = by the above result we conclude that k-'(zo) consists of one

@ +1(Rq)) =

'p;:(i-'(zo)),

point. 0

Lemma 8.3.12.

L(n(f))= @ @ S,.

Proof. The periodic points of d are dense in @@S,. If h(n(f))# @@S, then k(n(f))is a proper compact subset of 'b@ S,. Hence we can find z E @' @ S, such that * ( z ) = z for some T and e 4 L(Q(f)). Then UL, L - ' ( ~ ( z ) ) is a non-empty compact f-invariant subset of @ @ S, that disjoints from n(i), which is impossible. 0

Lemma 8.3.13. There exists a basic set

n1 such that L(n1)= f@ S,.

Proof. Let z be a point in f p @ S, such that the orbit { & ( z ) : i E Z} is dense in f p @ S,. By Lemma 8.3.12 there is z E a(?) such that i ( z ) = z. If R1 is the basic set in which z belongs, then i(n1) contains the closure of {di(z) : i E Z}, which is f p @ S,. 0 Proof of Theorem 8.3.4. By Proposition 8.3.11, h : k-'(Per(d)) -+ Per(d) is bijective and so n(j)itself is a basic set. Thus f,*(i) is topologically transitive, in which case we have n(f) = f p @ S, by Theorems 3.1.2 and 3.1.7 because is a TA-homeomorphism. 0

58.4 Injectivity of semi-conjugacy maps

x

The purpose of this section is to show that the semi-conjugacy map of Theorem 8.2.1 is actually a homeomorphism whenever f is a TA-covering map. For the case when f is expanding, we have the following Proposition 8.4.2.

Proposition 8.4.1. Let f : Tn + W be expanding and let % : R" -+ R" be the semi-conjugacy map obtained in Theorem 8.2.1. Then is 0 homeomorphism and satisfies x ( 1 + x ) = 1 x(z) for 2 E B" and 1 E Z".

x

+

a

Proof. We already know that there exists a metric for R" such that 'f has the property of Proposition 8.2.5 and, further, a tracing property in Lemma 8.2.6. Let % : R" --f R" be a semi-conjugacy map as in Proposition 8.2.7. In the similar way as the proof of Lemma 8.3.2 (l),we have %(a: 1) = 1 %(z)

+

+

CHAPTER 8

258

for z E R" and I E Z". Thus, sup{2(Z(z),z) : z E R"} is finite, i.e. there is B > 0 such that @(z),z) < B for z E R". From Theorem 8.2.1 (1) and Proposition 8.2.7 (1) it follows that

(SE 0 Z) 0 2 = 51 0 (SE 0 Z), (Z 0 h)0 3: = 7 0 (Z 0 K). Since a(h(z),z)< 6~ for z E R", we have for all z E R"

--

d ( h o Z ( z ) , z ) < c,

--

d ( k o h ( z ) , z )< c

- -j

where c = B + ~ KProposition . 8.2.5 implies D ( f o (%oh)(z),7'(z))+ 00 as j + 00 when K O X(z) # z. But 2(7' o (zoh)(z),r J ( z )< ) c for-j 2 - 0. This is impossible - - since and are uniformly equivalent. Therefore, k o h ( z )= z, and so k o h is the identity map. Similarly, SE o E is the identity map. 0 For TA-covering maps in 7d \ PEM we have the following proposition.

Proposition 8.4.2. Let f:T" --+ T" be a TA-covering map and let : R" t R" be the semi-conjugacy map obtained in Theorem 8.21. Then SE is a homeomorphism and h-' is 2-uniformly continuous where 2 is a metric for R" induced from a norm of R". For the proof we need some lemmas. By Theorem 8.1.3 we have T(b0) = bo for some bo E R". Let b denote the two sided sequence (... ,bo,bo,bo,...) of boys and for i E Z define : R" --$ R" as in (6.4) of Chapter 6. Then = for all i E Z. Since f is a map in I d \ P & M , from Proposition 6.6.5 we have the following Lemma.

2

f

Lemma 8.4.3. Let 2 E R" and 0 < e < EO be as i n Theorem 6.6.5. Then there is an open neighborhood R(z;b ) of 2 in R" such that (1) d i a m ( r ( z ;b ) )< E O , (2) E b : D"(z) x r ( z ;b) --+ r ( z ;b ) is a homeomorphism where (i) Bs(z)= n F(z;b), (ii) r ( z ; b) = E ( z ;b) n r ( z ;b) (iii) {Eb(y,z)} = T ( y ; b) r l for (y, 2 ) E x Z ( z ; b) ( 3 ) there is a constant p > 0 such that T(z; b ) 3 EJz), (4) D"(z)# {z} and r ( z ; b) # {z}.

v:(z)

w",(z)

a"(,)

For z E R" define the stable and unstable sets w"(z)and w " ( z ; b) as in 56.6. Then by Theorem 6.7.4 the families 7" = : z E R"} and 7; = { r ( z ; b ) : z E R"} are generalized foliations on R" and they are x B"(z; b) is a canonical coordinate at z transverse. Notice that E b : a"(z) for 7' and 7;.

{v"(z)

$8.4 Injectivity of semi-conjugacy maps

259

-

Lemma 8.4.4. For a ,y E R", W"(a) n r ( y ; b) is at most one point.

Proof. Let a , b E w"(a) n T ( y ; b) and suppose a # b. Then there is m > 0 - -such that d(f "(u),T-"(b)) < p where p is as in Lemma 8.4.3 (3). Put m a' = 7- (u) and b' = j-m(b), and let e > 0 be as in Lemma 8.4.3. For sufficiently large m we have b'

(8.9)

4 Wz(u').

It is clear that b' E w " ( u ' ) since a , b E ws(a),and that b' E Z p ( a ' ) c W ( d ;b) since a(a',b') < p. Hence there is (b1,bz) E b " ( u ' ) x nu(u';b) such that b' = Zb(b1,bZ) E wz(b2).Then we obtain b2 # a'. For, if b2 = a' then W:(a') = w:(bz)3 b', which is inconsistent with (8.9). Let Vat and u b , be open neighborhoods of u' and bz in g ( u ' ; b), respectively, such that Vat n Ub, = 0, and put N,I = &-,(b"(,') X vat),

Nbt

= ?jib(b"(a')X

ub,).

Obviously N,I and Nbt are open neighborhoods of a' and b' in R" respectively. Since N,, fl Nbt = 0, by Lemma 6.6.2 we have (8.10)

-

W:(v) r l w:(w) = 0

for v E N,t and w E

Nbt.

Figure 30

-

If V, = { z E R" : W g ( z fl ) Nbt # 0}, then V, is open in R" since 7" is a generalized foliation on R" (see Remark 6.7.2), and u' E V, since b' E w " ( u ' ) . Since Per(f) is dense in T" by Theorem 8.3.5, there is p E V, n N,, such ) let u = (ui) E (EX")? that ~ ( p E) Per(f). Let k be a period of ~ ( p and be a k-periodic sequence with p = uo. Define

7:

:

R"

--t

R" as in (6.4)

CHAPTER 8

260

-k

and write ij = f, for simplicity. Then g ( p ) = p. Since p E V,, we can choose w E wd(p)n N w . Since p,w E z(a';u), by Theorem 6.6.5 (1) we have w,'(p; u) n w:(w) = { q } for some q E W(a';u). Hence lim # ( q ) = p r+m

since w"(w) = wd(p), and .lim $(q) = p . Using (8.10), we have p

-

#

q

14--00

because p E N,,, q E Wz(w) and w E Nb,. Let E = mi.{& q ) , e'}/4 where el is an expansive constant for ij. Then there is 0 < 6 < 2~ such that every 6-pseudo orbit of i j is &-traced by some point of R". Choose t > 0 such that d(ge+'(q),p) < 6/2 and a(B-'(q),p) < 6/2. Then the sequence

+

is a (2t 1)-periodic &pseudo orbit of ij. By using POTP and expansivity we can find qo E R" such that Ijze+'(qo) = qo and a(q,qO) < E . It is checked that Be+'(qo) # qo. Indeed, if ge+'(q0) = qo then

-

d(p,$+'(qo)) I a p , i j e + ' ( q ) )

+ 2(ije+'(q),ije+'(qo))

< 6/2 + E < 2 E .

Thus we have &, q ) < 3.5 which is impossible since 4~ I d(p, q ) . Therefore ij2'+l has at least two distinct fixed points, which contradicts Theorem 8.1.3. 0 Since ?I is a hyperbolic linear map,

Rn splits into the direct sum R"

=

zd(0) @ r ( 0 ) of Ah v arian t subspaces z"(0) and c ( 0 ) where z"(0)is the subspaces corresponding to eigenvalues with absolute d u e smaller than one and r ( 0 ) is that of eigenvalues with absolute value greater than one. For 2 E R" let ( 0 = s,u) denote the translation of r ( 0 ) to z. For 17= s,u the family 7; = {Z"(,) : 2 E R"}

z"(z)

is a (linear) foliation on R" and the following holds:

%(w"(z)) c Z"(h(z)),%(w"(z;b)) c Zu(%(z)) for all s E R" -- -since Aoh = ho f holds and : R" --f R" is a 2-uniformly continuous surjection (see Theorem 8.2.1). (8.11)

Lemma 8.4.5 (Franks [Fl]). Let

2, y

E Rn. Then ?(z; b) w d ( y ) i s the

set of one point.

Proof,Let yo E R" and put s = wd(y0). It is enough to show that w ' ( z ;b)n s # 0 for all 2 E R". Let us put

Q = {z E Rn : V ( z ; b) n s # 0},

88.4 Injectivity of semi-conjugacy maps

then we have

--u

Q = {z E R" :W -

(2; b)

261

n U ( s )# 0)

where U ( s ) = UzEaN ( z ;b). Indeed, choose x from the right hand set of the above equality. Then z E v W (x;b) n U ( s ) and hence z E r ( x ;b) n w ( z ' ;b) for some z' E s. Since b) = Eb(D'(z') x r ( z ' ; b)), there is (y1,y2) E B"(z')x g ( z ' ; b) such that z = Eb(yl,y2) E q ( y 1 ; b ) . Hence y1 E R ( z ; b ) c r ( z ; b ) and on the other hand, y1 E D"(z') c s. Therefore, T ( z ; b) n s # 0 which implies x E Q. Hence Q is open in R" (see Remark 6.7.2). If Q = R" then the lemma holds. Thus we suppose Q R" and then derive a contradiction. Let w E Q. If %(w; b) (Z Q, then Q does not contain D"(w). For, take z E r ( w ; b), then there is ( x ' , ~ ' ' )E D"(w)x z ( w ; b )

r(z';

5

such that x = Eb(Z',Z'') E C ( z ' ; b). If D'"(w)C Q then W(x'; b) n s # since x' E D"(w) C Q. Since T ( x ;b) = T ( x ' ; b), we have r ( x ; b ) n s # and therefore x E Q, i.e. %(w; b) c Q, thus contradicting.

0 0

Figure 31 Choose and fix a E D"(w) \ Q. Let y : [0,1] -+ D"(w)be a path such that $0) = w and y(1) = a, and let p : [0,1] -t W(w;b) be a path such that p ( 0 ) = w and p(1) E r ( w ; b) f l s.

If we set

then R is non-empty since ([0,1] x {O})U({O} x [0,1]) c R and by transversdity of 7" and R is open in [0,1] x [0,1]. Note that R [0,1] x [0,1]. Since

x,

5

CHAPTER 8

262

1 1

W (7(~); b) n W"(p(t))is a single point for B : R + R n by

(T,

t) E R,we can define a map

Then B is continuous. By (8.11) we have

-h(W

( Y ( T ) ; b))

Define by p' : R" that

c 1;"(xo T ( T ) ) and

+ z"(0)the

x ( w ' ( p ( t ) )c Z"(x o p ( t ) ) .

natural projection (u = s,u). Then it follows

Figure 32 Notice that the last part of the above relation is compact. Since by Lemma 8.3.6, we obtain that B(R) is bounded. Let us put to = sup{t^: p ( y ( ~ )b) ; nr " ( p ( t ) ) # TO

Then

= SUP{+

to)

(TO,

0, 0 5 T 5 1 and : W(T(T);b)n W " ( p ( t 0 ) )# 0, 0 5 T 5 f } .

x is proper

0 _< t 5

4 R. Since B(R) is bounded, we can choose a sequence

t^},

88.4 Injectivity of semi-conjugacy maps

263

converging t o ( T ~t ,o ) ,such that 8 ( T n , t n ) converges in R". Let lim8(Tn, t n ) = v. Take a compact neighborhoods C" and C" of v in fs"(v) and g ( v ; b ) respectively, and let C = &(c" x C"). Then C is a compact neighborhood t n ) E C for n 2 1. of v in R". Since lim 8 ( T n , t n ) = v, we may assume "-boo

Then for n 2 1 there is (u,,,~,,) E C" x C" such that 8(Tn,tn) = Eb(%,%)

and hence Y

w

('Y(Tn);

b) n m d ( p ( t n )= ) {8(Tn,tn)} = { b ( U n , %)} C F ( u , ; b) n W"(w,),

from which u ( ' Y ( T n ) ;b)

w

-

= V ( u , ; b), W"(p(t*>) = W"(.n>.

Thus we have

nm"(%)

{8(Tlltn)} = m"(y(Ti);b)nW"(p(t,)) = r ( u i ; b ) 3 Eb(ul,vn), and so 8(rl, t n ) = Eb(U1, v,) E C . In the similar way, 8(T,, t l ) = Eb(Un1 v1) E C. Since 8 is continuous, we have 8(rl ,tn) --t 8(rl, t o ) and B(T,, t l ) -+ B(r0,t l ) ( n + m). Thus 8(rl, to), 8(ro,t l ) E C , from which there are (w, z), (w',2 ' ) E C" x C" such that

= &(wl In the same fashion we have 8 ( T l , to)

%),

-

8(TO, t l )

= Eb(w',

%')a

-

b) = W"(w'; b) W"(p(t0))= W"(Z),W"(T(TO);

and hence u ( r ( T 0 ) ; b)

w

n W'(p(t0)) = W(w';b) n ma(%) 3 Eb(w', %).

Therefore (r0,to) E R, thus contradicting. 0 By Lemma 8.4.5 we can define 9 : R" x R" {Tb(2,

y)} =

mu(%; b) fl W"(y)

-+

for

R" by (2,y) E

Rn X R".

Then ib is a continuous map such that for x E R" =ab 2b,P(.)x i P ( . ; b ) where Eb is a map as in Lemma 8.4.3 (2). It is easily checked that the following properties ; for 2,y, z E R" zb(z,z)

= 2,

ib(xc,zb(y,z)) = i b ( z , z ) ,

ib

satisfies

Fb(Zb(Z,y)i%)=Zb(Z,%).

x

Proof of Proposition 8.4.2. First we show that is a homeomorphism. Since h is surjective by Theorem 8.2.1 (3), it is enough to prove that %-'(z) is the set of one point for all 2 E R". To do this, let 2 E R" and define for y E x-'(z)

I ; , ~= ~

n

' ( 2 )~ " ( y ) ,

I ; , ~= ~ ' ( z n) W - ( y ; b).

CHAPTER 8

264

Claim 1. ib(I:,,, x I:,,,) = %-'(z). Indeed, for V , w E %-'(z) -

- 3

( V ;b) n w " ( W ) )

h 0 Zb(V, W ) = h(W

- ( since X o SE = IL o f )

c Z"(SE(v)) n Z"(X(w)) = (31 and so Zb(V,w) E

I:,,,)

c x-'(z).

il-'(z>. Since I ; , ~c S E - ' ( ~ )for u = s,u, we have

x

Conversely, let y E %-'(x). Then for any z E x-'(z) ib(z,!/)

from which &,(Z, ib(Y, z ) ) E ib(I:,y

E

E-'(z),

?b(z,!/)

w"(!/)

y) E I:,,,. Similarly, zb(y, z ) E I:,y. Therefore, z = ib(Zb(z, y), x IZ",,,).

As seen in Theorem 8.2.1, we have a(~(z),z)5 6 (z E Bn)for some 6 > 0 and so diam(L-'(z)) 5 26, i.e. X-l(z) c B26(Y) for y E %-'(z) where B6(3!)= { z E B" : d ( z , y ) 5 6). We define for y E %-'(z) --d

(8.12)

In,,,

= Zb(B26(Y), y),

--u

Iz,y

= Zb(Y,B26(3')).

--o

Claim 2. I:,,, c Iz,,,for u = s,u. Indeed, since ib(fE-'(Z),y) = ?.b(ib(I:,,, x ~:,,,),y) = Zb(I,",,,,Y) = I : , ~ ,we have I:,,, c ib(B26(Y)r y) = Tt,y because X-'(z)

c B26(v).

Also we obtain the same result for u = u.

Let us put for (y E S E - ' ( ~ ) )

- -s R x , ~= 2b(Iz,y

--u

Iz,y)*

Claim 3. There is N > 0 such that Rz,y y E SE-'(z). Indeed, by (8.12) it follows that

c B N ( ~for) all x

E

Bn and

58.4 Injectivity of semi-conjugacy maps

and hence

Rz,,c

u

265

n B26(za(w))

B26(r(v))

V,W

from which there exists N y E F'(2). Let e

> 0 such that R,,,

> 0 be as in Lemma 8.4.3

and let

-

ci(v,w;W'(v)) = min{m 2

'u

C

B N ( ~for) all 2 E Rn and all

E R". We define for w E w a ( v )

a * o :F ( w ) E We(f

(v))},

and for w E w ' ( v ; b)

-

d(v,w;w'(v;b)) = min{m 2 0 :

7-m(w)

u --m

E W e ( f (v);b)}.

Note that these are well defined by Lemma 6.6.4.

Claim 4. There exists KO > 0 such that for z E R" and y E %-'(z) (1)if v E R,,, and w E R,,,nWd(v),then J(v,w;w'(v)) 5 KO, (2) if v E R,,,and w E R2,, r l r ( v ; b), then ~ ( v , w ; ~ ( b)) v ; 5 KO. Indeed, let p be as in Lemma 8.4.3 and N as in Claim 3. Then there are P > 0 and a sequence (21,...,zl}c R" such that z ~ ( y c) U: BJzi). Hence R,,,C fl(zi;b) by Claim 3. Let v E R,,,and define

uf

D = R,,,n W'(V). Then we have D = &,(f:,,, v) and hence D is connected. C W"(v) then t = ib(%I,v) for some z1 E T:,,. Indeed, if z E ?b(T:,,,v) Since v E R,,,,there is (211, v2) E T:,, x such that v = ib(V1, v2). Hence

z,,

z =Zb(zl,Tb(vl,v2))

=ib(zl,v2)

E ib(T:,,

x

z,,) Rz,, =

and so z E D. Conversely, let z E D. Then z = ib(wl,w2) for some (w1,w2)E I,,, X T:,,. Since z = zb(%,'u), we have

-8

z =Tb(Tb(wl,w2),'")

E zb(T:,,,v)

and therefore D C Zb(T:,v, v). Since R,,, C U: N ( q ;b), we have D = U: x(zi;b) n D. To avoid complication, we may suppose that each T ( q b) ; n D is non-empty. Choose y i E D f l ?s(zi;b) for 1 5 i 5 e. Then

CHAPTER 8

266

This is checked as follows. Since yi E r ( z i ; b), there is zi E z ( z i ;b) such that y i E w:(zi).If y' E D n b) then we have also yl E W:(z)for some z E B'(zi;b). Since yi,y' E D C W"(v), clearly zi,z E w s ( v ) and so

x(zi;

zi, z E Z(Z~; b) n W ( v ) c T ( z i ; b) n Ws(v), which shows zi = z by Lemma 8.4.4. Therefore yl E W;e(yi), from which

By Lemma 6.6.3 there is KO

> 0 such that

for z E Rn. Hence we have

e f K 0 ( ~ c)

e

U f K 0 ( K e ( y i ) )c 1

UW:/~~(?~O(Y~))* 1

Since D is connected, for i l , i 2 with 1 5 i l , i 2 5 t? we can find a sequence = il,jl,j2,. ,j , = i 2 such that

..

j1

w;e(yji) n G e ( y j i + l )

+0

(1 L i Im - 1).

By using this fact we have

f K O ( Dc) K / 2 ( f K 0 ( Y i ) ) and therefore a(., w;Ws(v)) 5 KOfor any w E D . The analogous result holds for b). Claim 4 was proved.

mu(.;

If v # w then there is no > 0 such that f-"(v) (w)) for n 2 no (since 7 is expansive) and hence

Let v,w E I,&. -s

---n

We(f

- --

s

--7%

4f -n(~),7--n(w);W (f (.))I

2 n - no.

Let KObe as in Claim 4 and write nl =no+Ko +1,

--?I1

f

Then we have a(v', w';wa(v')) 2 KO

--nl

(v) and w'= f

+ 1.

(w).

4

$8.4 Injectivity of semi-conjugacy maps - Since A o h = E o 7 on

267

Rn,it follows that

n1

--n1

= 2- (z)and y' = f (y). Therefore, v',w' E Iil,ylC c R,r,y~nW"(y'). Using Claim 4, we have Z(vr,w';m8(v')) 5 KO,thus contradicting. This shows that I&, is a set of one point. In the same fashion we

where

2 '

have that I&, is a single point set. By Claim 1 we obtain that X-'(z) is the set of one point. Next, we show that fl-' : R" 3 R" is 2-uniformly continuous. Since h : R" + R" is bijective, using the fact that o ;i:= f: o and ( x ) , z ) < 6~ (x E a"), we have that for j E Z d(h

z-'

--'

+ ( 2 ( y ) - Z'(x))E Bza,(O). R" is expansive (under a), if we establish the following:

(8.13)

(7' 0 z-'(z)

Since 7 : R"

4

-

7'

0 E-'(y))

---N

for E

(8.14)

z-'

> 0 there is N > 0 such that if d(f

- -N 5 3 6 ~ a n dd(f

--N

(z),f

(y))

(z),~~(y5 ) ) 36K, then ~ ( x , y < > E.

Then by uniform continuity of 2,we can find 6 > 0 such that z(x,y) < 6 implies 2(Z'(z),x'(y)) < 6~ for j = N and j = - N . From (8.13) we o E-'(y) E E S ~ ~ (for O j) = N and j = - N and so have 7' o x - ' ( x ) - --1 (x),E-'(y)) < E . Therefore E-' is 2-uniformly continuous. d(h

f:'

To prove (8.14) write

BZ

=Zb(B36K(z),B36K(2))

for x E R". Then we can find K > 0 such that B, C B K ( ~for) x E R" (see the proof of Claim 3). Let e > 0 be as in Lemma 8.4.3. As above, define d(v,w;W"(v)) for w E and Z(v,w;m"(v;b)) for w E W"(v;b). To obtain that there exists KO> 0 such that for all x E R"

ma(.)

if v E B, and w E B, nW"(v), d(w,w;W"(w)) 5 KO d(w,w ;W(v;b)) 5 KO if v E B, and w E B, n b)

{I

we can use the proof of Claim 4.

mu(.;

268

CHAPTER 8

Since 7 is expansive (under there exists rn > 0 such that

a), by Lemma 6.6.3 it follows that for E

- -j

To see that N = m+Ko is our requirement, suppose d(f (z),T'(y)) for j = N and j = -N. For the case j = - N we have

>0

I 36K

and thus

This implies that

from which

Therefore a(&(Z, y), z)< &/2. For the case j = N we have a ( i b ( Z , y), y) in the same argument. 0

< &/2

38.5 Proof of Theorem 6.8.1

By Theorem 7.2.4 it suffices to show that the semi-conjugacy map h: 'b @ S, --t @' @ S, obtained in Theorem 8.3.1 is a homeomorphism if f is a TAcovering map. To do this, let X : R" --t R" be the semi-conjugacy map obtained in Theorem 8.2.1 and let E' : RP @ +1(Rq) + RP @ +l(RQ)be defined as in (8.4). Then E' is bijective by Proposition 8.4.2. We already know that E' is -I d -uniformly continuous (see Lemma 8.3.3). The uniform continuity of El-' is easily checked by the following facts: (1) 7 : R" + R" satisfies the property of (8.14), (2) the relative topology of +(RP @ Rq) = RP @ +1(Rq) is of the form mentioned in (7.12).

$8.6 Proof of Theorem 6.8.2

269

x'

Hence fE' is 2'-biuniformly continuous and so induces a homeomorphism of E P @ S, onto R P @ S, which is denoted as the same symbol. On the other hand, by Lemma 7.2.1 (2) the maps and induce homeomorphisms of RP @ S, onto RP @ S, respectively. We denote them by the same o = E' o 7' on RP @ S,. Since symbols respectively. Then we have

7

2

E'(vp

x'

+ ZP + z,) = q v p + z,) + ZP

for up E E P and zq E S,, a homeomorphism i: 'b' @ S, + 'b @ S, is induced by Z' : R P @ S, + RP @ S,. Also homeomorphisms d and f of 'b' @ S, onto 'b' @ S, are induced by and 7' respectively, and A o i= k o f holds.

x'

$8.6 Proof of Theorem 6.8.2

Let f:T" + Tn be a special TA-covering map, i.e. f is a map in S7d. By Propositions 8.2.1 and 8.4.2 we have a constant 6~ > 0 and a homeomorphism

h : R n + R n s u c h t h a t ~ o ~ = ~ 0 7 a n d ~ ( x ( z ) , z(ZER"). ) 0 such that for w,w E if a(w,w) > E then d(L ( v ) , ~ " ( w ) ) > E'. Take n > 0 with ne' 2 2 6 and ~ let 6' > 0 be a number such that n6' 5 6 ~ By . Lemma 8.6.2 we can find 6 > 0 such that a(z,y) < 6 implies -8 (y) C u61(ma(2)). Now, suppose that z,y E satisfy (8.16), and put

QY

w

I, = ( ~ ; ~ ~ ~ ) - ' ( and z ) I, = (p!!lz,,)-l(y). f by the definition of p!.

f

Then I, E mB(z)and L, E Wa(y)

Thus

Since p;(L,) E za(l,)and p!!(L,) E

za(L,),it follows that

-d(L -a ( I , ) , ~ " ( t , ) )> E l . Letting t = L ,

- I,,

we have

and by induction

Figure 33 On the other hand, since t, E

wB(z)and I, E m a ( y ) , it is clear that

$8.6 Proof of Theorem 6.8.2

and hence L, = L,

273

+ 1 E up(w"(&>). This implies that

-

w"(& + e) = & + w8(!) c U & ( w " ( e z )= ) u6,(v8(o)) + 1,

from which

r"(. c! Ua,(w"(O)). ) Then we have -

w"(24!) c u6~(Wd(c))

and by induction

-

w"(d)c Uij,(w"((n- 1)L)).

Thus w"(nL)c Unp(W8(0)).By (8.15) it follows that w " ( n 1 ) c U,,a,(W'(O)) C Uza,(z"(O)). Therefore we have

& + ne E

c

un6f(w"(e~)) u6~+n6~(z"(12))c

U26~(z"(&))

and arrived at a contradiction. 0 Let

2

E Q?. Then p f ( L ) = 2 for some e f

f

( P ; , ~ . , ) - ~ ( Z )=

E Z" and hence

w"(e;o) n Z"

= {.!

+ w " ( 0 ; 0 ) ) n Z"

(by Lemma 6.6.11 (2b))

=e+rj (by Lemma 8.6.3 (2))

Ce+r,

= I;"(e)

n zn,

from which p L o (p!Iz.,)-l(z) = p:(L). Therefore, we can define a map A t Qf -, Qf by sending x to

e(L).

x" :

Lemma 8.6.7. ( ~ ) Z O ~ " = R Of ~ O ~ Q ~ , (2) K" i s Z-uniformly continuous. Proof. This is similar to that of Lemma 8.6.6. 0

By Lemmas 8.6.6 (2) and 8.6.7 (2) the maps continuous maps

i" : w " ( 0 ; 0 ) + z"(0)

and

2 and x8 are extended to

id : w'(0)+ z"(0)

respectively. From Lemmas 8.6.6 (1) and 8.6.7 (1) we have (8.17)

{

!oh"=I'oy

onW"(0)

A o jtu = jtu o 7

on W"(O; 0).

CHAPTER 8

274

Let a, b E R", then ~ ' " ( a ) n ~ " is ( bexactly ) one point, say P(a, b). It follows that

p : R" x R" -t Rn is z-uniformly continuous. Thus a &uniformly continuous map i: R" -t W n is defined by i ( z ) = p o (iu o p;(z),

jld o p ; ( z ) )

for z E

w".

Then by (8.17)we have

-

A O ~ = ~ OonWn ?

and by definition for L E Z"

i(L)

= P ( P ; ( C ) , P y ) ) = p.

- It is easy t o check that sup{J(h(z),z) : z E R"} is bounded. For, since iis d-uniformly continuous, for m

d(z,y) < m (2, y

> 0 there is K > 0 such that

* Z(i(z) - z,i ( y ) - y) I K .

E

_Let - z E R", then we can find L E Z" such that a(z,1) < m, and so z(i(z), z)= d ( h ( s ) - s,O) = Z(i(z) - z,i ( L ) - L ) < K . Therefore,

h = iby Proposition 8.2.2

and x ( L ) = L for all P E Z".

Figure 34 From the following Proposition 8.6.8 we obtain (3) of Theorem 6.8.2.

E(L + v) = L + E(v) for all L E Z" and all v E R". -u Proof. Let v E UeEznW (L; 0 ) . Then there exists L, E Zn such that v E

Proposition 8.6.8. u

-

W ( L , ; O ) . Since h(F,,) = F x and x ( L v ) = L,, we have x(v) E ?;"(P,;O) and for L E Z" h(v) e E q e , 1;0 ) . -¶I

+

+

$8.7 Remarks

275

Since v+P E ~ ( P , , + 1 0; ) by Lemma 6.6.11 (2), clearly x ( v + l ) E r ( l , + P ; O ) and therefore there exists c > 0 such that

-

d ( 2 o ii(v + ~ ) , 2 ( i i ( v + ) 1))< c

(i 5 01. - On the other hand, since A o h = h o f and d(x(z),z)< 6~ for z E R",we have z ( x i o E ( z ) , T ( z ) )< 6~ for i E Zand z E R". Note that f(z+k') = f(z)+x(t) for 2 E R". Then, for i 2 0 we have -4- -i d ( A (h(v)+ L),T(v+ P)) = d ( A ox(v) + 2(P),T(v)+ $ ( l ) ) -

- -

=

and so

d(2

-4-

+

d ( A ( h ( v ) l),20 K(V

0

x ( v ) , T ( v )< ) 6K

+ l ) )< 26K

(i 2 0).

+

+

Therefore, by Lemma 8.2.3 (2) it follows that &(v l ) = h(v) l . Since UtEz,,V ( l0;) is dense in R" by Lemma 8.6.1 (2), the conclusion is obtained. 0

58.7 Remarks

In this section we mention some remarks about TA-covering maps of ton.

Remark 8.7.1. In Theorem 6.8.2 we required T" to be an n-torus. However the result is true for expanding maps of an n-solenoidal group. An outline of the proof is stated as follows. Let S be solenoidal, then S is abelian. Thus S splits into the direct sum S = S,,@ T" of the maximum torus 1"" and a solenoidal group S,, without tori. If S, = {0}, then S is a torus and Theorem 6.8.1 holds. For the case S,, # (0) let n = dim(&). As saw in Chapter 7, S,, is expressed as S,, = $(R") F where F is a totally disconnected compact subgroup. Let f: S --+ S be an expanding map. Then f is a local homeomorphism; hence, f has a fixed point. We may assume that the identity 0 is a fixed point. Then we obtain f($(Rn) @ T") = $(Rn)@ T" and a homeomorphism f : R" x R" + R" x R" which is a lift of f: $(R") @ T" --+ $(Rn)@ T". It is checked that for X > 0 small there are a subgroup Z' (C Z" x Zm)of finite index and a continuous map F : Z' x (R"x R") + BA(O) and an injective homomorphism F : Z'--+ Z"x Z" such that

+

-

f(v

+ l ) = F ( l ) + F(l,v) + f(v)

for l E Z'and v E R" x R" (here Bx(0)is a closed ball of radius A). Moreover, we can prove that F : R" x R" --+ R" x R" is expanding. This fact leads us to the end goal (see Aoki [A021 for details).

276

CHAPTER 8

For general TA-covering maps it remains a problem of whether Theorem 6.8.2 is true in the case of solenoidals. In 57.2 of Chapter 7 we have seen that a toral endomorphism splits into two subsystems throughout a method of lifting by some finite covering map (see (7.5)). In the case when a toral endomorphism A : 'IP + I[n is hyperbolic, if we have the case (i) of (7.5) then R P = R" and 71 : R" + R" is hyperbolic. For the case (ii) of (7.5) we have that WQ = R" and 71 :R" + R" is hyperbolic (or expanding). The case (iii) of (7.5) implies that R" splits into the direct = R P @ WQ of z-invariant non-trivial subspaces R P and RQ,and 2 : sum In R" + R" is hyperbolic. Since is hyperbolic (or expanding), RQ splits into the direct sum RQ = E" @ E" of subspaces En and E" where E" is a subspace corresponding to the eigenvalues of with modulus < 1 and E" is that with modulus > 1 (or WQ = E"). Put Z' = E" n Z t and define a subspace K = span&'. Since x(Z')c Z', we have x(V1) = V1 and Tipl is expanding. Obviously induces a toral endomorphism of K /Z' onto V1 /Z'. Then VI is the maximum subspace of RQsuch that the induced endomorphism is expanding. Thus, it is easily checked that there is an Ti-invariant subspace V2 such that

xi^^

x

Z" = vzn Z;

IWq = Vl CB V2,

xp2

satisfies Ti@") c Z"; and spanpZ" = Va. induces a hyperbolic toral endomorphism of type (111). Since x(Zp")= Zp", it is clear that Tipp induces a hyperbolic toral automorphism. Thus the following remark is obtained as a result.

Remark 8.7.2. If A : T" + 1" is a hyperbolic endomorphism, then there exist a finite covering space 'k" (n-torus) of 1" and a hyperbolic toral endomorphism A' : 'k" + 2" such that the diagram

T"

A

T"

commutes, and such that there are the maximum torus subgroups 1,2), the torus subgroup 'k"3 and (a) a hyperbolic automorphism A1 : 'b + el, (b) an expanding endomorphism A2 : + @a, (c) a hyperbolic endomorphism A3 : 'k"3 + 'k"3,

'k"i (i =

$8.7 Remarks

277

so that the diagram

I

I commutes homeomorphically and algebraically.

Remark 8.7.3. (1) If f: T" --t II'" is a TA-covering map and if ~?.'.(z,~) is dense in 7"' for some zo E P, then so is w(z)for all z E T". (2) Let f : T" + 11" be a special TA-covering map which is not injective nor expanding and let A : 'I"' + 11"" be the endomorphism homotopic to f. Let

9.2

be a torus subgroup as in Remark 8.7.2. Then, f is strongly special if and only if the space p a is trivial, i.e. p a = (0). These are checked as follows. As before let 7 :Rn --t R" be a lift o f f by the natural projection T :R" + T" and suppose 7(0) = 0 for simplicity. If 2 : R" --t R" denotes the linear map which covers A, then 2 is hyperbolic (by Theorem 8.1.1). Thus there exists a map E : R" --t R" such that 2 o = E o 7 and z(Z(z),z) < 6~ for all z E R" by Theorem 8.2.1. From Theorem 8.2.1 and Proposition 8.4.2 it follows that fE is a homeomorphism and it is 2-biuniformly continuous. To show (1) let e > 0 be as in Lemma 8.4.3 and choose ~1 > 0 and e > ~2 > 0 satisfying

x

d(v,w) 5 El -

-7 ~(~-l(v),E-'(w)) 5 e,

d ( v , w ) 5 E 2 -7 a(x(v),x(w)) 5 El.

E > 0 denote w:(v) and respectively. Then we have

For v E R" and f and

Choose E such that e 2 E

and for m 2 0

w:(v;a)as the local stable sets of

> 0 and for w E wa(v)write

CHAPTER 8

278

Then we have

Since n(w6(v)) = fi'(?r(v)), clearly n(Dm(v,&)) c f i ' ( n ( v ) ) . To obtain the conclusion it is enough to see that for every ,d > 0 there exists P 2 0 such that if ~ ( v E) 'IP then n(Dc(v,e))is P-dense in 'IP (i.e. x(De(v,Ez))n Up(z)# 0 for any z E 'P and any @neighborhood Up(z)). Since is 2-biuniformly continuous, we can find 7 > 0 and 6 > 0 such that d(v,w) < 27andd(v,w) < 6imply@-1(v),x-1(w)) < PandZ(x(v),E(w)) < 7 respectively. Take and fix vo E ~ ~ ( 2 0 )Since . fi'(z0) is dense in T", there exists k 2 0 such that a ( D k ( v 0 ; ~ ~ )is) &dense in 'Il"'. Thus, for a 6neighborhood U = Ua(Dk(v0;~~)) of D ~ ( v o , we E ~have ) ?r(U)= 'IT" and

x

-

(by the choice of 6) h(U) C U,(h(D~,(vo;~z))) (by the choice of €1) C U7(Bk(z(vo);~1))

where &(v;

€1)

= {w E Z8(v);2(v, w ;Z"(V),

Here Z"(v) denotes the leaf of the linear foliation

€1)

5 k}.

F; through 5 and as above

define ~ ( v , w ; Z ' ( V ) ,= E ~min{i ) 2 0 ; 2 ( w ) E ~ ~ l ( $ ( v ) ; ~ ) } Then . it is easily checked that there exists P > 0 such that for all u E U,(Bk(x(v0);~1))

and then for all u E U,(Bk(x(wo); €1)

Therefore, for v E U we have Up(De(v;e)) 2 U and therefore n(De(v;e)) is P-dense in T". Next, we show (2). Since T(0) = 0, we have flzn = A l p and thus for P E Z"

2

0

X ( P ) = z 0 f'(P)

= 7i 0 $(t)

(i 2 0).

58.7 Remarks -4

279

-

Sincez(x(z),z) < 6~ for z E R", clearly d(A oh(I),;I'(P)) < 6, for i 2 0 and --1 -a -a --1 so x(t)E Since I E h ( L ( I ) ) = we have W (h ( I ) ) =

w"(e)

ma(%-'(!)),

za(I).

u W8(I) u wa(K-l(P)) u Z"(I)). =

earn

=%-I(

eEz= If wa(0)is dense in 'll"', then Uazn W a ( I is ) dense in R" and so is UeEZnza (I). This implies that ' k " 2 is trivial ( p a = (0)). Conversely, suppose = (0). Then UeEznLa(!)is dense in R" and so is UeEZ,,W d ( I ) .Therefore Wa(z) eEz*

is dense in 'll"' for all z E T".

Remark 8.7.4. Let fi : Iln 3 Tn converge to a self-covering map f:Tn -+ T" as i -+ 00, and let A denote the toral endomorphism homotopic to f . If each f i is a strongly special TA-covering map, then there exists a continuous surjection h:T" + Iln such that the diagram

Iln-Iln f

T"

-

T"

A

commutes. This is easily checked as follows. - As above, let A : "" + 'P be the toral endomorphism homotopic to f, and A : R" -+ R" the linear map induced by A. Then A is hyperbolic since f i is a TA-covering map. Use Theorem 8.2.1 for fi, then there is a continuous surjection Ei : Rn -+ R" such that z(&(v),v) < 6~ for v E R". Hence, for fixedjEZ

-

A' o xi(.) - Ti(v) = x i o T:(v) - Ti(.) E Ba,(O)

(i 2 1)

and there is N ( j ) > 0 such that

ff(v)

-Ti(v)

E Ba,(O)

(v E R" and i , q 1 N ( j ) ) .

Thus, we have

-

hi(v) - E,(w) E F ( B j a , ( O ) )

(v E R" and i , q 2 N ( j ) )

x

which implies that x i converges uniformly to some by Lemma 8.2.3 (1). Obviously 6 : 1"+ Rn satisfies all the properties of Theorem 8.2.1. If each f i is a strongly special TA-covering map, then satisfies the condition that & ( I + z )= I + x i ( z ) for z E R" and I E Z", and so does 51. Therefore, the conclusion is obtained.

CHAPTER 9

Perturbations of Hy p erb o lic Toral Endomorphisms

In the previous chapter we have shown that every TA-covering map of an ntorus is characterized by a hyperbolic toral endomorphism which is homotopic to the map. To advance our discussion of the dynamics, we shall investigate some of characteristics of hyperbolic toral endomorphisms. The first section of this chapter is devoted to discuss Co-perturbations of hyperbolic toral endomorphisms in the class of self-covering maps, which is a result mentioned in 51.2 of Chapter 1. 59.1 TA-C" regular maps that are not Anosov

In this section we restrict ourselves a 2-torus and deal with hyperbolic toral endomorphisms of a 2-torus which are not expanding. We notice that such a toral endomorphism belongs to the subclass 7 d H U SS7d. The purpose of this section is to show the following statement mentioned in Theorem 1.2.2 of 51.2.

Theorem 9.1.1. Let A : T2 -, T2 be a hyperbolic toral endomorphism but not expanding and let a > 0 be an arbitmry number. (1) If A is injective (2.e. A belongs to 7dH), then there exists a TAdiffeomorphism of class CO",9: T2--f T2, with d(g, A ) < a that fails to be an Anosov diffeomorphism. (2) If A is not injective (i.e. A belongs to SS7d), then there is a TAregular map of class C", 9 : T2--f T2,with d(g,A ) < a that fails to be a special TA-map or an Anosov differentiable map. where d(g,A ) = m a x { d ( g ( z ) A(%)) , : 2 E a'}. R e m a r k 9.1.2. It seems likely that Theorem 9.1.1 can be carried over to tori with dimension greater than 2. However, we do not settle here the case. Let A : T2--t T2be as in Theorem 9.1.1 and 2 : R2 + R2 denote the linear automorphism which is a lift of A by the natural projection T : R2 -, T2.Let p1 and p2 be the eigenvalues of 2.Since A is not expanding, we may suppose that 0 < 1p1( < 1 < Ip2l. Denote as z"(0)the eigenspace of 2 for p1 and write L"(z)= z"(0) z for z E R2.Then 7; = {z"(z): z E R2} is a linear foliation on R2. It follows that 31 = ~ ( 7 % is the ) family of stable sets in -0 strong sense and for each z E R", T : L (z) + L " ( T ( z )is ) bijective. Here L"(z') denotes the leaf of 3; through z' E T2.We already know that each

+

280

g9.l TA-C" regular maps that are not Anosov

281

F$ is dense in T2. Let 2' = n ( x ) E T2 and define a metric d" for L"(z')

leaf of bY

for y, z E z"(z)

d"(?r(y),?r(z)) = a(y, z )

where 2 is a metric for W2 induced by a norm of W2. For the proof of Theorem 9.1.1 we need the following lemma.

Lemma 9.1.3. Let cp : T2 --t T2 be a homeomorphism satisfying the following conditions (i), (ii) and (iii); (i) cp is homotopic to the identity map i d , (ii) each leaf of Fi are cp-invariant, i.e. cp(L"(z'))= L"(z')for z' E T2, (iii) for y', z' E L"(d)there esists e(y', 2)> 0 such that (4 d'(cp(y'), (P(z')) I Z')d"(Y', 4, (b) 0 < .)' I P-l where P = 1P1I? (c) for e > 0 there is a constant 6 > 0 such that if d"(y',z') > E then

w,

w,

((y',z')

5 p-1 - 6.

Then the following properties hold ; (1) if A is a hyperbolic automorphism, then g = A o cp is a TA-homeomorphism, (2) if A is a hyperbolic endomorphism but not injective nor expanding, then g = A o cp is a TA-covering map. Proof. Since cp(L"(0))= L"(0)by (ii), we can choose a lift p : W2 -+ R2 of cp by ?r satisfying p(Za(0))= z"(0).Then, for L E Z2

@7q)

+ v ~ e=) Z"(O)+ 1 = ~ " ( q

+

= p(z"(0) 1)= ~ ( z " ( o ) )

and hence for all z E W2 (9.1)

p(Z"(z))= Z " ( x )

because there is always a sequence {li} c Z2 such that Z"(&) + Z"(z) as i + 00. Let j j = 2 o p. Then 3 is a lift of g by ?r, and by (9.1) we have for

x E R2

-5 g(L

(3))=

Z"(X(z)).

To obtain the conclusion of the lemma we prepare the following three claims. Claim 1. As before let w"(z)denote the stable set of 3 at z. Then Z"(z) = ~ " ( z for ) 2ER ~ . -s i Indeed, by (9.2) we have @(z) E L (A (z)) for i 2 0. If y E then - -" -i -s i d(L (A (z)),L (A (y))) converges to 0 as i + 00 because a(?(z),$(y)) -+ 0

w"(z),

CHAPTER 9

282

as i + 00. Since -2 is hyperbolic, it follows that z"(z) = zs(y), and so Conversely, let y E Then $(z), y E i.e. W " ( z ) c -s i $(y) E L (A (z)) for i 2 0 by (9.2), from which we have

z"(z),

z"(z).

z"(z).

Hence, t o show that W"(z) 3 z"(z),it is enough to see that d"(gi o r ( z ) , g i o ~ ( y ) )-+ 0 as i + 00. If this is false, then we can find E > 0 satisfying the property that for j > 0 there is ij > j such that d"(gij o x ( z ) , g * j o n(y)) > E , and then

which shows that ds(gij OA(Z), gij OA(Y)) + 0 as j 1, thus contradicting.

+ 00

because of p ( p - l - 6 )
0 we have &j,x) < K and hence there-exist - a constant SK > 0 and a continuous surjection ?E : B2 + B2 such that Ao h = Eo7j and @,id) < SK (see Theorem 8.2.1). - - Since x i s-s&uniformly continuous by Theorem 8.2.1 (3), it follows that h ( W s ( z ) )= L ( h ( z ) )for - -s -s z E EX2, and hence by Claim 1 we have h ( L (z)) = L ( h ( z ) )for all z E B2. -8

-

Claim 2. L ( h ( z ) )= z"(z)for z E B2. Indeed, let r ( 0 ) be the eigenspace of for p2 and define a map : -U L (0) z'(0) by assigning to each z E z"(0)the point in r ( 0 ) n z"(h(z)). ---f

$9.1 TA-C" regular maps that are not Anosov

283

Then we have

-8 A , ~ o~Xu(,) ( ~ )= Z ( T ( o > n L (I+))) = T(o) n X(Z"(E(z)))

= Z'(o) n 7i o h(Z"(z)) = T(o)n h o ~(Z"(Z)) = z"(o) n h(Z"(sl(~))) = T(o)n Z"(h o

(by (9.2))

a(,))

= 2 0 +(o)("). From the first part of Proposition 8.2.2 together with the fact that &distance between 2 and the identity map of zu(0)is bounded, it follows that h" -8 coincides with the identity map, and therefore L ( h ( z ) )= z"(z).

If y, z E Z"(z) and h(y) = h ( z ) ,then we have for i E Z

h 0 -ig (y) = 2 0 h(y) = 2 0 Z(Z) = h 0 @ ( z )

z.

and so a($(y),@(z)) < 26K for all i E Combining this fact with the following Claim 3, we obtain that h is injective.

Claim 3. If y # z for y,z E Z"(z), then ~(@(y),3'(z)) + 00 as i + -00. Indeed, let E > 0 be a number such that a(y, z ) > E and let 6 > 0 be as in (c) of the condition (iii). Then we have

-

d(g-l(y),g-l(z)) = 2(v-1 o x - l ( y ) , p - l

0

a-'(z))

= dy7roqj-l o x - 1 ( y ) , 7 r o p o P ( z ) ) = d"(q-1

((q-1

((7r

( y ) , q-1

0 7r 0 F ( Z ) )

d"(7ro~-'(y),no~-1(r))

L

from which and so

0 7r 0 P

o g-'(y>,?r o g-'(z))

0 7r 0 P

5

( y ) , 'p-1

,u-l

- 6 by

-

43-YYWM) 2

P-l

0 7r 0

P(z))

(c) of the condition (iii),

Z(Y,.>,

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284

and by induction

which shows Claim 3 because p-l/(p-l - 6)

> 1.

By (9.3) we can show (8.14) of 88.4 (replace 7 by 3). Hence, E-' is 2uniformly continuous. This is proved by the same way as the proof of Proposition 8.4.2. Since o 5 = 51 o T j , it follows that 3 : W2 + R2 is a TAFrom this fact together with Lemma 2.2.34 and homeomorphism under Theorem 2.3.14 we conclude that g is expansive (or c-expansive) and has POTP (notice that 3 satisfies the condition (C) of Lemma 2.2.34 (3) by (8.14)). Therefore, (1) and (2) hold. 0

a.

Proof of Theorem 9.1.1. Since FA and 3 (= ( T ( ~ ( X ) ) : x E Rn}) are transverse linear foliations on T2,we can find a coordinate domain D at 0 such that D = D"@ D"

where D" = {x E L"(0): d"(0,x)5 a } and D" = ?r{y E z"(0) : a(0,y) 5 a}. Here a > 0 is a small number. To avoid complication we may consider D" and D" as D" = [-a,a] and D" = [-a,a] respectively.

I?"

D'

Figure 35 Let c > 0 be a small number to be determined later and choose C" functions ~1 : D" --t D" and 61 : D" + W,as illustrated in Figure 35, satisfying

and &(O) = c, 0 5 &(x) 5 c. We define a function g1 : D" CB 13" g1(x8,xU)= x 8

+

61(xU)K1(xS)

--+

R by

59.1 TA-C" regular maps that are not Anosov

285

for (xa,z,) E D ' @ D". Then we have ag1

-(xa, OX"

2,)

=1

+ 61(Xu)-(2,)dK1 dxa

> 0.

Since g1( ,xu) is the identity on a neighborhood of the end points of D", we have g1 (D" @ 0")= D" and so write

PI(^^,^^) = (g1(z8,zu),xu)

for (xd,zu)E D" @ D".

Obviously cp1 : D -+ D is a diffeomorphism which is the identity map on a neighborhood of the boundary OD,and d(cp1,id) is small if so is c. Let us put

(4, 4)d"((% 4, (4,4).

=t((%4,

It is easy to see that for given E > 0, if d8((xs,zU),(x:,xU)) > E then there is 6 > 0 satisfying €((%%), ( 4 , x u ) ) 5 p-l - 6. In the case when A is an automorphism, we define a diffeomorphism cp : T2+ T2bv

Then cp satisfies all the conditions of Lemma 9.1.3. Therefore g = A o cp is a TA-diffeomorphism. Moreover we have

288

CHAPTER 9

and hence g is not Anosov. (1) was proved. To see (2)) choose a sequence

satisfying p-1 # 0. Here ( T 2 )denotes ~ the space of the inverse limit system for ("',A). Then there is X > 0 such that Bx(p-1) f l Bx(0) = 0 and Bx(pi) n (Bx(p-1) U Bx(p0)) = 0 for i 5 -2 where Bx(pi) is the closed disc with radius A. We may suppose that D c Bx(0) (take a > 0 small if necessary).

Figure 36 Construct C" functions n2 : D s + D' and 62 : Du + R, as Figure 36,such t h a t 1 g ( x ) 1 < l a n d - c < 6 2 ( x ) < c . Anddefineafunctiong2:Da@DU+R bY g2(%GI) = 3" 6 2 ( 4 ~ 2 ( % ) .

+

Since g 2 ( ,xu) is the identity on a neighborhood of the end points of D",it follows that g2(Ds @ 0 " ) = D", and so put

Then cp2 : D + D is a diffeomorphism which is the identity map on a neighborhood of 8D, and d(cp2, id) is small if so is c. Moreover we have

(4,

dd(cp2(3s,4, c p 2 ( 4 , %)) = t ( ( % , 4 zu))d"((%

It is easily checked that for fixed xu E Du

xu), (&zu))

$9.1 TA-Coo regular maps that are not Anosov

and for that

E

> 0, if

da((za,z,,),(zb, 2,))

> E,

€ ( ( z a , zu),( 4 , z u ) )

as above we can find 6

287

> 0 such

I P-' - 6.

Now, let us define a diffeomorphism cp :T2+ T2by ifz€D

i f z c D+p-1 if z 4 DU ( D + p l )

Figure 37 Then cp satisfies all the conditions of Lemma 9.1.3. Since A is an endomorphism but not injective, the composite g = A o cp is a TA-covering map by Lemma 9.1.3 (2) and clearly it is a C" regular map.

DU

Figure 38

It only remains to see that g is not special and fail to be Anosov. Since = (p1IDu = id, we have = A l p and hence D" = WZ(0) where 0 denotes the two sided sequence consisting of 0's. Since 'pip

Bx(pi) n {D U (D +p-1)} = 8

(i L -2),

CHAPTER 9

208

we have g l B r ( p i ) = A I B ~ for ( ~ i~5) -2 and hence

where

(r

: (T)A + (P)A denotes the shift map. Thus W,U(p) = g(D"

+

+

n BIZ(())* Since g)D'+p-l = A (PID"+P-~and p ( x ) = 9 2 ( x - P - 1 ) p-1 for x E D" + p - l , the sets W,U(O)and W,U(p) are illustrated as Figure 38. Therefore, g is not special by Theorem 6.6.13. The derivative Dog of g at 0 is calculated as above. Hence, Dog is non-hyperbolic, i.e. g is not Anosov. (2) was proved. 0

P-1)

$9.2 One-parameter families of homeomorphisms

Let X be a compact metric space with a metric d and let C ( X )denote the set of all continuous maps of X equipped with the Co-topology, i.e. the topology induced by the metric d ( f , g ) = m a x { d ( f ( x ) , g ( z ) ) ) .For h E C ( X ) the path connected component of h in C ( X ) is the homotopy class of h, written as €Iom(h). In the case when X is a compact topological manifold, from Remark 6.7.10 it follows that C ( X ) is locally contractible, which implies that each homotopy class is open and closed in C ( X ) . Let %(X)be the set of all homeomorphisms of X onto itself. Then % ( X ) is a topological group. Given f E % ( X ) , the isotopy class of f is defined as the path connected component of f in % ( X ) . We denote it as Iso(f). By the following Theorem 9.2.1 we have that each isotopy class is also open and closed in % ( X ) if X is a compact topological manifold.

Theorem 9.2.1 (Cernavskii [eel, Edwards-Kirby [E-K],). The homeomorphism group H ( M ) of a compact topological manifold M is locally contractible. In particular, %(M ) is locally path connected. We omit the proof. If it is of interest, the reader should see Cernavskii [Ce] and Edwards-Kirby [EK]. In this section we first discuss one-parameter families of homeomorphisms in a general setting, and show the following theorem.

Theorem 9.2.2. Let A : T + T" be a hyperbolic tom1 automorphism of an n-torus. Then there is a unique continuous map H : Iso(A) + Hom(id) with H(A) = id such that A o H(f ) = H ( f ) o f for all f 6 Iso(A), where id denotes the identity map. A path w in % ( X ) is called an isotopy or a path of homeomorphisms. It follows that the map F : X x I + X defined by F ( z , t ) = w ( t ) ( x ) is continuous, i.e. F is homotopy such that for each t € I, F( , t ) : X + X is a homeomorphism. Here I = [0,1]. Conversely, if F : X x I + X is a homotopy

59.2 One-parameter families of homeomorphisms

289

and each of F( , t ) is a homeomorphism, then the map t H F( , t ) is a path of homeomorphisms of X. As before, let 2 x denote the family of non-empty closed subsets of X. Then 2 x is a compact metric space under the Hausdo& metric (see Remark 3.1.5). Let Y be a metric space with a metric d'. A map f: Y + 2 x is said to be upper semi-continuous (resp. lower semi-continuous) if for y E Y and E > 0 there is 6 > 0 such that f(z) c U,(f(y)) (resp. f(y) C U,(f(z))) whenever b(y,.z) < 6 (compare with the definition of semi-continuity in 53.3 of Chapter 3). Here U,(f(y)) denotes the &-openneighborhood of f ( y ) in X. It is easy to see that f: Y + 2 x is upper semi-continuous if and only if {(z,y) : z E f(y), y E Y}is closed in X x Y.

Remark 9.2.3. If f:Y-+ 2 x is upper or lower semi-continuous, then the inverse images of open sets are F,-sets, i.e. a countable union of closed subsets. Indeed, if G is open in X, then B(G) = {F E 2x : F c G} and C ( G ) = {F E 2 x : F n G # 0) are elements of a subbase of 2 x (see 53.1 of Chapter 3). Thus it sufficies to see that f - ' ( B ( G ) ) and f - ' ( C ( G ) ) are F,-sets. For the case when f is upper semi-continuous, f - l ( B ( G ) ) = {y E Y : f ( y ) C G } = {y E Y : f(y) n G" = 0) is open in Y. Since Y is a metric space, we see that any open subset 0 is expressed as 0 = U z l A i for an increasing sequence of closed subsets A;. Applying this fact, then we have that f - l ( B ( G ) ) is a Fu-set. Since X is a metric space, we have also that G= K; for closed subsets Ki. Thus

UFl

where Fi = {y E Y : f(y) n Ki # 8) for i 2 1. Since each F; is closed, f-'( C (G)) is a F,-set. For the case when f is lower semi-continuous,the open set G is expressed as a union of closed sets K; such that Ki c int(Ki+l) for i 2 1. If y E f - ' ( B ( G ) ) , then we have f(y) c G C Uz2int(K;). Since f(y) is compact in X,there is n > 0 such that f(y) C K,, and therefore f - l ( B ( G ) ) = U,",l{y E Y : f(y) c K,,}. Since f is lower semi-continuous, {y E Y : f(y) c K , } = {y E Y : f(y) n KE = 8) is closed in Y. Therefore, #-'(B(G)) is a F,-set. Since f - ' ( C ( G ) ) = {y E Y : f ( y ) n G # 8) is open in Y ,f-'(C(G)) is also a F@-set. Let X and Y be metric spaces and let f : X + Y be a map. Denote as D(f) the set of discontinuous points of f.

CHAPTER 9

290

Remark 9.2.4.

u

D(f) =

(1)

[ C W

\ f-'(cl(f(A)))l

ACX

U [f-'(int(B))

=

(2)

\ int(f-'(~))]

BCY

=

(3)

u

[W-V))

\ f-'(B)1

BCY

This is checked as follows. (1): Take z # D ( f ) . Then, for a subset A C X we have x # cl(A) \ f-'(cl(f(A))) when x # cl(A). If z E cl(A), for {zn} c A satisfying z, + z we have that f(zn) + f(z)since f is continuous at x. Thus f(sn) E f(A) c cl(f(A)), and so x E f-'(cl(f(A))). Therefore, x # cl(A) \ f-'(cl(f(A))). In any case we have

c

DC

n u

MA)

\ f - l ( c ~ ( f ( ~ ) ) )and ~~

ACX

D 3

[cl(A) \ f-'(cw4)))1.

ACX

If z E D , then there exists a neighborhood Vf(,)of f(z)such that f(Uz) p V'(,) for any neighborhood U, of x. Put A = {y E X : f ( y ) # Vf(,)}, then x E cl(A). Since x # A, we have x # f-l(cl(f(A))). Thus x E cl(A)\f-'(cl(f(A))) and therefore D c UACX[cl(A)\ f-'(cl(f(A)))]. (1) was proved. (2): Let x E D and take a neighborhood B of f(x) such that f(Uz) 6B for any neighborhood U, of x. It is clear that z E f-'(int(B)) and x # int(f-'(B)). Thus x E [f-'(int(B))

\ int(f-'(~))l c

U [f-'(int(B))

\ int(f-'(~))].

BCY

Conversely, take a point z from the set of the above right side. Then z E f-'(int(B]) \ int(f-'(B)) for some B c Y. Thus f ( U 2 ) c int(B) for any neighborhood U, of x. This shows that f is not continuous at x, i.e. x E D. (2) was proved. It remains to show (3). Since int(Y \ B) = Y \ cl(B) and f-'(Y \ B) = f - ' ( Y )\ f - ' ( B ) = X \ f - ' ( B ) , by (2) we have the conclusion.

Remark 9.2.5. (4)

D ( f >=

u

G: open

(5)

[f--'(G)

\ int(f-'(G))I

59.2 One-parameter families of homeomorphisms

291

By the above Remark 9.2.4 we have

Since int(B) is open and int(f-'(B)) 3 int(f-'(int(B))), we have (4). By (3) the equality (5) is concluded.

Remark 9.2.6. Suppose Y is separable and let { O n } be an open base of Y. Then

where Sn = Y \ On for n. For the proof it sufficies to see that

To do so take a point x from the set of the left side. Then we have x E f - ' ( G ) and x $! int(f-'(G)) for some open set G. Since G = IJOj for Oj E {On}, there is no such that x E f-'(On,). But, x @ int(f-'(O,,)) since x 6 int(f-'(G)) = int(Unf-'(On)). Therefore, x E f-'(On,) \ int(f-'(O,,)) c Un[f-'(On) \ int(f-'(On))]-

Remark 9.2.7. Let Y be a metric space and X a compact metric space. If f :Y --t 2x is upper or lower semi-continuous, then Y \ D ( f ) is a Baire set. This may be shown by using the above remarks. Indeed, by Remark 9.2.6 we have D ( f ) = Un[cl(f-'(On)) \ f - ' ( O n ) ] . Denote as Pn the complement of cl(f-'(On)) \ f - ' ( O n ) for n. Then each Pn is open and dense. Thus Y \ D ( f )= Pn is a Baire set.

nn

Let {At : t E I}be an upper semi-continuous family of points in 2x, i.e. the map from I t o 2x defined by t H At is upper semi-continuous. For t E I let f t : At --t At be a homeomorphism. If the map ( x , t ) H f t ( x ) is a continuous map from { ( z , t ): x E A t , t E I}to X, then we say that the family { f t : t E I} is a path of homeomorphisms on {At}.

Remark 9.2.8. For a homeomorphism f : X --t X let C R ( f )denote the chain recurrent set of f . If { f t : t E I} is a path of homeomorphisms of X, then (1) { CR(f t ) : t E I}is an upper semi-continuous family, (2) { f t p R ( f t ): t E I}is a path of homeomorphisms on { C R ( f t ) } .

CHAPTER 9

292

Indeed, suppose (1) is false. Then we can find t o E I and €0 > 0 such that for every n > 0 there is t , E I with [to - tnl < 1/n so that CR(ft,) vcO(CR(ft0)). Take 2, E C R ( f t , ) such that z n $ vco(CR(ft0)). We may suppose that x , converges to some 20 E X. Then 20 $ C R ( f t , ) . Since 2 , x,, for 6 > 0 there is a periodic &pseudo orbit (2, = yo, y1, * *. ,y k = 2,) of ft,. Since d ( f t o ,ft,,) < 6 for large n, it follows that the sequence is a 20 for f r o , thus periodic 26-pseudo orbit of f t , , from which we have 20 contradicting. Therefore, (1) holds. (2) is clear from the fact that { f t } is a path of homeomorphisms of X.


0 there is t E I such that ht # hi and d(ht, hi) < E , which contradicts the following Lemma 9.2.10. 0

Lemma 9.2.10. Let (2, dz) and (W,dw) be metric spaces and let f : 2 -t 2 and g: W -+ W be homeomorphisms. Let h, h‘ : W --+ 2 are continuous maps satisfying f o h = h o g and f o h’ = h’ o g respectively. Suppose f is expansive and e > 0 is an expansive constant for f . If

then h = h’. Proof. If h # h’, then we have

Therefore, the conclusion

in

obtained. 0

$9.2 One-parameter families of homeomorphisms

293

Remark 9.2.11. In Lemma 9.2.10 the assumption that f is expansive can be changed for the weaker assumption that f has sensitive dependence on initial conditions. Let {ft : t E I} and {fi : t E I} be paths of homeomorphisms on upper semi-continuous families {At} and {A:} respectively. An inverse path of { f t } is defined by {ft} = {fl-t : t E I} which is a path of homoeomorphisms on {Al-t : t E I}. If A1 = Ah and f1 = f;, then we can define aproduct of { f t } and {fi} by {ft}.{fi} = {ft-fi : t E I} where 0I t I 112 f t . f; = f2t tI 1. 1/2 I Then { f t } { fi} is a path of homeomorphisms on an upper semi-continuous family {At A: : t E I}where

{

-

At

*

A: =

05t I 1/2 1/2 I t 51.

{

Let Ao = Ah, A1 = A: and fo = f;, f1 = fi. We say that { f t } and {ti} is homotopic if there are an upper semi-continuous family {A,,, : t,s E I} and a family { ftra : t, s E I}of homeomorphisms ft,a : At,a --+ At,,, with the property that the map from {(z,t,s) : 2 E At,,,t,s E I} to X defined by (2, t , 5) I+ ft,,(z) is continuous, such that At,o = A t , f t , o = f t (vt E I) A o , ~= Ao, fo,a = fo Ai,, = A i , fi,s = f i At,, = A;, ft,l = fl (Vt E I), Here ({At,,}, {ft+}) is called a homotopy from { f t } to {fi}.

{

{

(VS E I) (Va E I).

Remark 9.2.12. Let {ft} and {fi} be homotopic and let ( { A t , a } , { f t , 8 } ) denote a homotopy from {ft} to {fl}. Let f : A 4 A be a homeomorphism and suppose there is a continuous family {ht,,} of semi-conjugacy maps between f and {ft,a}. I f f is expansive, then ho,, and hl,, do not depend on s. This is easily checked by Lemma 9.2.10. Let f : X + X be a homeomorphism and denote as C(f) the set of homeomorphisms h of X commuting with f, i.e. f o h = h o f . It is clear that C(f) is a subgroup of X(X).We call this subgroup the centralizer of f . If f is expansive, then by Lemma 9.2.10 it follows that C ( f ) is discrete in X ( X ) .

Remark 9.2.13. Let A4 be a compact topological manifold. Let {A, :t E I} be an upper semi-continuous family of points in 2M and let { f t : t E I} be a path of homeomorphisms on {At}. Suppose A0 = A1 and fo = fi. Then, it seems likely that the following (MC) is true whenever fo is expansive. (MC) If {hi :t E I}is a continuous family of semi-conjugacy maps between fo and {ft} and if ho is the identity map of Ao, then hl is a homeomorphism, i.e. hl belongs to C(foIho).

CHAPTER 9

294

It is easy to see that (MC) is equivalent to the following assertion. If for some ho E C(f o lAo) there is a continuous family of semi-conjugacy maps {ht :t E I } between f o and { f t } , then hl also belongs to C(fOIAo). A sequence of groups and homomorphisms

O-G1-+

h

h'

Gz-G3-0

is called a short ezact sequence if h is injective, Im(h) = Ker(h') and h' is

surjective. If, in addition, h has a left inverse, the short exact sequence is said to split. In this case Gz is isomorphic to the direct sum of G1 and G3.

Lemma 9.2.14. Let X be a pamcompact Hausdorff space and let p : X + S1 be a fiber bundle over a circle with connected fiber. Let bo E X and denote 0s X O the fiber over p(b0). Then the following sequence of fundamental groups and induced homomorphisms 0

-

r1(Xo,bo) 1 ;r1(X,bo)

-

rl(S1,p(bo))

0

is a short ezact sequence where i : XO + X denotes the inclusion map.

Proof. This is easily checked by the fact that p : X -+ S1 is a fibration (see Remark 6.5.2) and rz(S1)= 0 (see Remark 6.7.9). The detail is left to the readers. 0

Theorem 9.2.15. Let S' be a circle R/Z and e : R --t S' denote the natuml projection. Let X be a compact connected locally connected metric space and let p : X + S' be a fiber bundle with connected fiber. Suppose X is semilocally 1-connected. Let f : X -+ X be a homeomorphism which preserves each fiber for p , and for t E R let Xe(t)= p-'(e(t)) and f e ( q = fl+,. Suppose A : T" --t T" is a hyperbolic tom1 automorphism and for some to E R there i s a homomorphism q5 : r ~ ( X ~ ( ~ ~+l ,r .l (bT~" ,)c ~ ) where , bo and co are base points, such that the following daagmm commutes: rl(Xe(to),bo) 41 .l(Y,Q)

f*(*O)*

A.

r1(Xe(to),f ( b 0 ) )

1

V.O+OU.

r1(T",A(co))

where u, (resp. v , ) denotes the induced isomorphism by a path u (resp. v) in X (resp. T") from bo (resp. A(co)) to f (bo) (resp. co) as in Lemma 6.1.4. Then the following properties holds: (1) there exists a unique continuous family ht : Xe(t) + T", t E R, of continuous maps with hto(bo)= co such that hto* = q5 and A o ht = ht o fe(q for all t E R,

59.2 One-parameter families of homeomorphisms

295

(2) there exists a homeomorphism T : Iln -+ Iln homotopic to the identity map such that T E C ( A ) and T o ht = ht+l holds for all t E R, ( 3 ) if the following short exact sequence splits:

then the map T is the identity map. From the above theorem we have the following corollary.

Corollary 9.2.16. Let A : Iln + Iln be a hyperbolic tom1 automorphism and let f t : Iln + Iln, t E I , be an isotopy starting at A. Then there i s a unique continuous family {ht : t E I } of semi-conjugacy maps between A and { ft} such that ho is the identity map. Furthermore, i f fo = f1 then hl is also the identity map.

Proof. Let {gt : t E I } = { ft}.m and define g: 'IP' xS' --+ 'IP' x S 1 by g ( x ,t ) = (gt(z),t).Here S' = [O,l]/{O, 1). Since go = g1 = A, by Theorem 9.2.15 there is a unique continuous map H : T" x S' -, Iln such that HIptx{o)(x,O) = x and A o H = H o g , which shows the first part of the corollary. When fo = f1, let us define g:Iln x S' T" x S' by g ( x , t ) = (ft(z),t).Then we have a semi-conjugacy map H' from g to A such that HITnx{ol is the identity map. Since n l ( P x S') E nl(T") @ nl(S'), by Theorem 9.2.15 (3) it follows that ho = h i . 0 -+

Proof of Theorem 9.2.2. For f E Iso(A) let { ft} be a n isotopy from A to f. Then, by the first part of Corollary 9.2.16 there is a continuous family { h t } of semi-conjugacy maps between A and {ft}, and so we define H : Iso(A) --+ Hom(id) by H ( f ) = h l . This is well-defined by the last part of Corollary 9.2.16. By this fact together with Theorem 9.2.1 we have that H is continuous. 0

Remark 9.2.17. The statement of Theorem 9.2.15 can be extended to the more general case of hyperbolic infra-nil-automorphisms. This implies that Theorem 9.2.2 and Corollary 9.2.16 also hold in such a general setting. For the proof the readers may refer to Hiraide [Hi9]. For the proof of Theorem 9.2.15 we prepare the following lemma.

x x

Lemma 9.2.18. Let X be a compact metric - space with metric d and let be a connected topological space. Let p : X -+ X be a covering map. If X is connected and locally connected, then there is a compatible metric p for such that (1) all covering tmnsfonnations for p are isometries under p, (2) ( X , p ) i s a complete metric space) ( 3 ) for all z E and T > 0 the closed ball {y E : p ( z , y ) 5 T } is compact.

x

x

CHAPTER 9

296

x

x

Proof. Since p : + X is a covering map, there exist a metric ;ifor and a constant 60 > 0 satisfying the properties in Theorem 6.4.1. Fix 0 < 6 < 6012. For z,y E let {zi : 0 5 i 5 f+1} be a 6-chain from x to y, i.e. a(zi,z i + l ) 5 6 for 0 5 i 5 I , and define p by

x

t

Y) = i n f { C a(zi,zi+l 1)

~(2,

i=O

where the infimum is taken over all finite &chains from z to y. By the triangle inequality of 2 we have p ( z , y ) 2 a(z,y), from which p is a metric for Clearly y) = p ( z , y) if z(z,y) 5 6. Thus p is a compatible metric and by Theorem 6.4.1 (3) we obtain (2). From the construction of p together with Theorem 6.4.1 (2) it is easy to see (1). To show (3), let K be a compact subset of and write N6(K) = {y E : p(y,K) 5 6) where p ( y , K ) = min{p(y,z) : E K}. If p ( y , z ) 5 6, then by the definition of-p we - have a(y, z)5 6, and hence a(y, z)= p(y, z),from which N6(K) = {y E : d(y,K) 5 6). Let {yi) be a sequence of points in Na(K). Then we can find a sequence { z i ) of points in K such that yi E B6(zi)for all i. Here B6(zi) = {z E : a(zi, z ) 5 6). Since K is compact, clearly there is a subsequence {zj} converging to some ,2 E K. Hence y j E Bza(zm) if j is sufficiently large. Since 26 < 60, by Theorem 6.4.1 (1) we have that BZa(zm) is compact, and so {yj} contains a subsequence { y k } converging to some y., Since ;i(yk,zk) 5 6, it follows that y, E Ns(K). Therefore, N 6 ( K ) is compact. From the fact that a(z,y) = p ( z , y ) if p ( z , y ) 5 6, it is easy to see that Nza(K) = N 6 ( N & ( K ) which ), shows that & ( K ) is also compact. By and T > 0, then we can induction, so is N,6(K) for every n 2 1. Let z E find n 2 1 such that {y E : p(z, y) 5 T } C N,6(K). Therefore, (3) holds. 0

x.

a(.,

x

x

x

x

x

x

Proof of Theorem 9.2.15. Since X is semilocally l-connected, by Theorem 6.2.11 there is the universal covering 7r : + X. By Proposition 6.2.7 we can find a continuous map p : + R such that the diagram

x

x

X

d

x

-

L

R

- 1. S'

P

-

commutes. Since R is contractible, it follows that j5 : M + R is a trivial such that fiber bundle, i.e. there is a homeomorphism cp : p-'(O) x P j j o cp(z, t) = t for all (2, t). For t E R denote as Xt the fiber over t. For simplicity let t o = 0. Then for i E Z we have = Xe(o) and each restriction 7r : xi t Xe(o)is the universal covering. --$

.(xi)

x

59.2 One-parameter families of homeomorphisms

297

Choose 80 E W Osuch that ~ ( 8 0 = ) bo. Since u is a path in X starting at bo, we can find a lift Ti of u by A satisfying Ti(0) = 80. Then Ti( 1) E XO -and - x(B(1)) = f ( b 0 ) . Hence there is a lift 7 : X + of f by -T -such that f ( b 0 ) = B(1). Since f preserves each fiber for p, it follows that f ( X , ) = for all t E R. P ut 7o= f,x,. Let A' : Iw" + Tn be the natural projection and choose 30 E R" such that T'(Eo) = Q. Then there is a lift 2 : R" + W" of A such that 2(&)= g(0) where 5 is a lift of v by d with E(1) = Zoo. Since A is a hyperbolic toral automorphism, it is clear that is conjugate by some translation to the linear map which covers A. Hence by Lemmas 8.2.3 and 8.2.4, has the following properties:

x

x,

-4

(Pl) for K > 0 and E > 0 there exists J > 0 such that if d(A ( z ) , 2 ( y ) ) 5 K for all i with lil 5 J, then a(z,y) 5 E , (P2) for given K > 0, if 2(2(z),z(y)) 5 K for all i E Z, then z = y. (P3) for K > 0 there exist8 6~ > 0 such that for any K-pseudo orbit -4 { z i : i E Z} of there is a unique z E Rn so that d(A (z),zi) 5 6~ for i E Z. where 2 denotes the euclidean metric. Let G ( T ) denote the covering transformation group for A. Then G ( A )is isomorphic t o A ~ ( Xbo), by Theorem 6.3.4. Denote as F the subgroup of G ( A ) which corresponds to nl(X,(,),bo). Since X,(o)is the fiber over e(0) for the bo)) = 0, fiber bundle p : X + S', G ( A ) / Fis isomorphic to Z. Since p+(?rl(Xo, it follows that a ( X t )= X t for all a E F and t E R. - Let 7, : G ( A )--$ G ( A )denote the induced homomorphism as 7. Then fop = f,(p) o 7 holds for all p E G ( A )(see Lemma 6.3.10). Since f + ( ~ ~ ( X ~ ( o ) , b o ) ) = ?rl(X,(o),f(bo)), we have 7+(F)= F and the restriction 7, : F + F is consistent with the induced homomorphism To,. Denote as G the covering transformation group for A' and as A, : G -+ G the induced homomorphism. As usual, we consider 4 : n(X,(o),bo)+ A~('IP,Q) as a homomorphism from F to G . Since A, o 4 = (v+ o 4 o u,) o fe(o)+ by the assumption, from the choice of 7 and we have 2 , 0 4 = 407, on F. Hence, letting F(b) = {a(b): a E F} (b E and G(c) = {L(c) : L E G } ( c E R"), we have the following commutative diagram :

x

x w)

G(4)

A

GP(G))

- where 4' and 4'' are defined by d'(a(b0)) = 4(a)(Zo) and 4"(a(fo(bo))) 4(a)(2(-Eo)) for all a E F ,respectively.

=

CHAPTER 9

298

x

Let d be a metric for X and choose a metric p for as in Lemma 9.2.18. Then the following property holds: (P4) for t E R there is Kt > 0 such that xt C U K * ( ~where O ) U K , ( ~=O ) {x E : p(.,Xo) < K t } . For, let D be a compact covering domain for ?r : Xt + X,(t) and choose K > 0 such that D c UK($O).Here UK($O)= {y E : p(y,$o) < K}. Then U a E F a ( D= ) Xt. Hence, by Lemma 9.2.18 (1) we obtain (P4). -Since f ( X t ) = si?t, the orbit Let x E and choose t E R such that x E {?(I) : i E Z} is a subset of xt.By (P4) there is Kt > 0 such that for i E Z there is ai E F such that ai(b0) E U~,(f(x)). Notice that ai(&) E for all i E Z. Since 7 is uniformly continuous under p, we can find KI > Kt such that p(T(x),7(y)) < K ; whenever p(x,y) < K t . Put Bt = { z E : p($,z) Kt Ki } . Then Bt is compact by Lemma 9.2.18 (3). Hence

x

xt.

xo

x

-+

X = {a E F : a(Bt U T ( B t ) )n (Bt U T ( B t ) )#

0)

is finite. Let L = Lt = max,Ex p(+(a)(x(i?o)),&). Then by (P3) there exists GL > 0 such that any L-pseudo orbit of 2 is GL-traced by a unique point in R". Notice that L depends on t.

x

Claim 1. For z E let {ai : i E Z} be as above. Then { + l o a i ( ~ o ):)i E Z} is an L-pseudo orbit of A. i + l Indeed, since T*(ai)(T(&))= T(ai(80))E U,:(f (z)),we have p(T* (ai)(T($o)),ai+1(&)) < Kt t Ki, and hence o 7 * ( a i ) E X, from which there is -yi E E such that T*(ai)= ai+l o -yi. Then we have

azl

A o 4' o ai(60)= +I1

o

7o c~i(Z0)

= +I1

0 7*(.i)(7($0))

= 4''

0

lo

ri(T(G)) I1

--

= +(ai+l)0 +(ri) 0 4 (f(b0))

= +(ai+l)0 + ( r i ) ( x ( G ) )

and so o(~o+lo~i(~oO),+l~a;+~(~o))

= D(+(ai+l)0 +(~i)(x(z~)), +(ai+l)(~)) = D ( + ( r i ) m O ) ) , ~ o5)L

for all i E Z. Therefore, Claim 1 holds.

Let

t,

be a 6L-tracing point of

(4' o ai(60): i E Z}.

$9.2 One-parameter families of homeomorphisms

299

Claim 2. z, is independent of the choice of {a,}. Indeed, for i E Zlet a:($) E U~,(fi(x)) for some a: E F. Then {q5’oa:(&)) is an L-pseudo orbit of and there is a unique : z E R” which is 6L-tracing {4’0a:(bo)}.Since p ( a i @ o ) , a ~ ( & ) )< Kt+Kt < K t + K i , we have a: = a i o 7 i for some -yi E El and hence for i E Z

Therefore z, = zk by (P2).

- -

Define a map H : X -+ R” by x I+ z,. This is well-defined by Claim 2. Let x = 50. Then, from the choice of 3: and 2 it is easily checked that D(4’ o ~ , ( C O ) , $ ( C O ) ) < L = Lo for all i. This shows = ZO. For x E let {a,}be as above. Then we have for i E Z

z(&)

x

- -

x.

and hence A o H = 1? o 7 holds on Let x E and {a;}be as above. Since a,(&)E UK,(?(S)) for all i E Z, for given J > 0 there is a neighborhood U ( x ) of x in such that a,(&,-,) E U ~ , ( f i ( y ) ) for all y E U ( x ) and all i with lil 5 J . Then D ( 2 oS(y),$ o H ( x ) ) 5 2 6 for ~ i with lil 5 J. Choose J large enough, then D ( z ( x ) , a ( y ) ) is small by (P l), which shows continuity of p.

x

x

-

w

Claim 3. H o a = 4(a)o for a E F. Indeed, let x E and {a,}be as above. Since a i ( 6 0 ) E U K , ( ~ , ( ? ) it ), i andso D(A o H o a ( z ) , q 5 ’ o ~ ( a ) o follows that T:(a)oa&,) E U~~(f(a(x)), ai(50))< 6L for all i E Z.Since

x

CHAPTER 9

300

we have

from which ff o a(.)

= d(a)o ff(z). Therefore, the conclusion is obtained.

- -

By Claim 3 for each t E R the restriction H-: Xt -+ R" can be projected to a continuous map ht : X e ( t ) + T". Since A o H = ff o 7, we have that A o ht = ht o f e ( t ) for all t E R. Since = 50, clearly ho(b0) = CO, and by Claim 3, hoe = 4 holds. Since H is continuous, it follows that {ht : t E R} is a continuous family. - hh(b0) To show uniqueness, let { h i } be another continuous family such that = 9, and A o hi = hi o fe(q for all t E R. Since w : X X = Q, is the universal covering, there is a lift ff' : X + R" of the family { h i } --I such that H ( b o ) = 50. Then o = o 7. Since hh, = 9, it follows that p o a = 4(a)o ff for all a E F. By this and Claim 3 we have that D ( z , z ' ) = sup{D(ff(z),rr'(z)) : z E is finite, and hence ff = ff' by Lemma 9.2.10 and (P2). (1)of Theorem 9.2.15 was proved. Next we show (3). Let P denote the subgroup of F corresponding to Ker(4). = Since the sequence of (3) splits, we can choose p E G(w) with @(TO) such that f o p = t c o p o 7 for some tc E P. Take a lift : X + R" of the continuous family { h t } such that ff'(p(60)) = ZO. Then p' o ,f?= ff. Hence

a(&,)

x rr' rr' x}

z'

Let a E F. Then o a = K' o a o /3 for some IC' E P. This shows ff' o a = 4(a)o Hence = 1? by Lemma 9.2.10 and (P2). Since P(X0) = XI,it is clear that p(xt)= xt+lfor all t E R, and therefore ht = ht+l for all t E R. (3) of Theorem 9.2.15 was proved.

rr'.

89.2 One-parameter families of homeomorphisms

301

To show (2), take p E G(A) such that p(X0) = 1 5 1 . Then there is a lift H : X + R" of the family { h t } such that z'(p(60)) = &, i.e. r?'o p = Let a = T i ' ( p ) o p-'. Then a E F and 2o = ??' o (70a).Hence we have

o.

--I

(9.4)

Z'o(7oa) = y o o f o f

where y = 4 o f , ( a ) E G. Notice that y is a translation on R" by some !E Z". We choose c E R" satisfying the equation c ) - c = !, and define T :R" --t R" by T(z) = G c. Then

a(

+

--1

T

OAOF=-Y-'OA.

Combining this and (9.4) we have

-

Ao(Too')=(Too')oT.

z')

Since (?; o p')o a = 4(a)o ( po for a E F, by Lemma 9.2.10 and (P2)we obtain that Top' = (2) was proved. 0

z.

Most of the above discussion will be applied for one-parameter families of self-covering maps (more generally, also for those of continuous maps). In the remainder of this section we discuss the case of hyperbolic toral endomorphisms of types (11) and (111). Let M be a compact topological manifold and denote as C ( M ) the set of self-covering maps of M equipped with the Co topology. For given f E C ( M ) we denote as Iso(f) the path connected component of f in C ( M ) . &om the following theorem we have that Iso( f ) is open and closed in C(M).

Theorem 9.2.19. C ( M ) is locally contractible. Proof. Let f E C ( M ) and let d be a metric for M. By Theorem 9.2.1 together with the fact that the map from X ( M ) to C ( M ) defined by h I+ f o h is continuous, it suffices to show that for g E C ( M ) ,if d ( f , g ) = max{d(f(t),g(z)) : G E M } is small enough, then there is h E X ( M ) with d(h,id) small such that f oh=g. Let n : + M be the universal covering and 7 : + a lift of f by A. Let g E C ( M ) and suppose d(f,g) is sufficiently small. By Remark 6.7.10 there is a homotopy ft : M + M (0 5 t 5 1) from f to g. Choose a homotopy _f t- : + such that 7o= f and ft o 7~ = A 0 7 , for all t. Put 3 = f l . Then d(f,g) = d(f,g). Here Z(f,S) = sup{z(f(z),jj(z)) : z E ?i?} and 2 is a metric as in Theorem 6.4.1. for Let G(n) denote the covering transformation group for n. By Lemma 6.3.10 there are homomorphisms Tt, : G(n) + G ( T ) such that

z z

z

ft 0a

= f t * b >0 7*

for all a E G(n),

CHAPTER 9

302

and by continuity it follows that Tt* does not depend on t. Hence, 7, = g*. - --I Let a E Im(?*), then f o f , ( a ) ( z )= a o f ( z )for x E from which we have -- 1 f o a ( y ) = 7C1(a)o ?-'(y) where 7(z) = y. Hence

z,

a.

for all a E G ( T ) and x E Therefore, 7-l o can be projected to a homeomorphism of M, say h: M -t M . Then f o h = g. Since d( f , g ) = a(7,g), we obtain that d(h,i d ) is small if so is d( f ,9). 0

For expanding toral endomorphisms the following theorems are easily checked by the same way as the proof of Theorem 9.2.2. The proofs are left to the readers Theorem 9.2.20. Let A : 1In -+ "IF be an expanding endomorphism of an n-torus. Then there is a unique continuous map H : Iso(A) -t Hom(id) with H(A) = i d such that A o H ( f ) = H(f ) o f for all f E Iso(A). By Theorem 6.7.11 together with Theorem 9.2.19 we have that a hyperbolic toral endomorphism of type (111)does not possess the property as in Theorems 9.2.2 and 9.2.20. However, the following theorem holds.

Theorem 9.2.21. Let A : Iln -t T" be a hyperbolic endomorphism of an n-torus. Suppose {ft : t E I} is a path of self-covering maps f t : "IF + T" starting at A. Here I = [0,1]. Then there ezists a unique homotopy kt : S 4 S , t E I, such that (1) ho =_ id, (2) do ht = i t o it for all t E I

where it: S t S (resp. A : S -t S ) is the homeomorphism of the solenoidal group (resp. the automorphism) associated with f t (resp. A) as in Theorem 7.2.4. Proof. Let T : R" -t T" be the natural projection. Denote as 71 : R" + R" the linear map which covers A, and as 7, : R" + R",t E I , a unique path of homeomorphisms satisfying

-

f o = zand

T O ~ , = ~ ~ Of T oralltEI.

From the proof of Theorem 9.2.2 we see that there exists a unique path of continuous surjections K t , t E I, such that (1) =_id, (2) A o ht = Et o (3) Z(Et(v),v) < 6K for all v E R".

5

Tt,

59.2 One-parameter families of homeomorphisms

303

By (7.8) of 57.2 we can find a homomorphism 11, : Rn + Rn such that $(R") is dense in a certain solenoidal group S. Thus, as in (7.9) define

and then we have a commutative diagram

xi

{Ti

where = 11, o x t o $-' for t E I. Since : t E I} and 2 'are biuniformly continuous by Lemma 7.2.1,7; and A' induce continuous bijections ft and A of the solenoidal group S. Moreover, : t E I}induces a homotopy {it: t E I} of continuous surjections i t : S S such that i, =, id and A o i t = ito jt for follows from Lemma 9.2.9. 0 all t E I. The uniqueness of { i t }

{xi

--f

CHAPTER 10 Fixed Point Indices

In Chapter 6 we have seen that the family of stable sets in strong sense of a TA-covering map has a structure which is called a generalized foliation. The purpose of this chapter is to discuss orientability of generalized foliations on topological manifolds without boundary, and to establish the fixed point index theorem for TA-covering maps.

510.1 Chain complexes In this section we shall introduce some algebraic preliminaries to discuss singular homology. Let C = {a, : C , --+ Cn-l} be a chaan complez, i.e. C is a sequence

of abelian groups C, and homomorphisms a,,, called boundary operators, such that an o a,+, = 0 for all n E Z. Since 0, o a,+, = 0, we have Im(a,+l) C Ker(8,). The factor group H,(C) = K e r ( ~ , ) / I m ( ~ , + ~is) called the n-th homology group of C , and an element in Ker(8,) is a cyc2e. We say that an element in Hn(C)is a homology class, and denote as [c] a homology class represented by a cycle c. Let C = {an : C, + Cn-l} and C' = {a; : C:, + CL-l} be chain complexes. The family cp = {cp,} of homomorphisms cp, : C, + C:, is said to be a chain map if a:, o cp, = cpn-l o 8, for all n E Z, i.e. the diagrams Cn

-1 an

Cn-1

qn-

1

commute.

A chain map cp = { cp}, is an isomorphism if each of cp, is an isomorphism. The family of identities id = { i d , } : C C is a chain map. If cp = {cp,} :C 3 C' and cp' = {cp;} : C' -, C" are chain maps, then cp'ocp = {cp',ocp,} : C -+ C" is also a chain map. Let cp = {qn} : C + C' be a chain map. Then cp,(Ker(O,)) c Ker(8;) and cpn(Im(&+l)) C Im(8;+,). Thus, for each n E Z we can define a homomorphism y* : Hn(C) --t Hn(C') by P*([c])= [(P~(c)]304

$10.1 Chain complexes

305

Two chain maps cp,cp' : C + C' are said to be chain homotopic if there is a family 9 = (9,) of homomorphisms an : C, -+ C:,+l such that for n E Z = v n - 9.; In this case, the family 8 is called a chain homotopy from cp to cp' and we write cp N 9'. It is easily checked that the relation N is an equivalence relation. Proposition 10.1.1. If chain maps cp, cp' : C + C' are chain homotopic, then cp+ = cp: : H,,(C) -+ Hn(C') for all n E Z. Proof. This follows from the fact that a',+l

o*n +*,-I

oan

+

~ + ( [ c=] )[ ~ n ( c = ) ] [V',(C) aL+10 +n(c) = [cp;(c)] ( since an(.) = 0)

+ an-1

0

an(c)]

= cp:([cl).n

A chain map cp : C -+ C' is said to be a chain homotopy equivalence if there exists a chain map 11, : C' -+ C such that 11, o cp N id and cp o 11, N id. Here 11, is called a chain homotopy inverse of cp. If cp : C -+ C' is a chain homotopy equivalence, then the induced homomorphism cp+ : H,(C) --+ H,(C') is isomorphic for all n E Z. A sequence of abelian groups and homomorphisms hn+l

hn

+ An+l + A, + An-l

-

*..

is called an exact sequence if Im(h,+l) = Ker(h,) for all n. If, in particular, a sequence A" + 0 0 --t A 2 A' is exact, then it is a short exact sequence. A sequence of chain complexes and chain maps 0

-

c 5 c' -Lc" +0

is called a short exact sequence if all rows of the following commutative diagram are exact :

1

1

1

1

1

1

CHAPTER 10

306

Given a short exact sequence of chain complexes and chain maps 0 +c

5 c'

2 c" ----+ 0,

homomorphisms cp+ : Hn(C) + Hn(C') and cp: : Hn(C') + Hn(C") are induced as stated above. Moreover, a homomorphism & : Hn(C") + Hn-l(C), which is called the connecting homomorphism, is defined as follows. Let a E Hn(C"). Then a = [c"] for some c" E Ker(8:). Since cpk : Ck + C: is surjective, we have cp',(c') = c" for some c' E CL,and cp',-l

0 a;(Cl)

=

a;

0

cp:,(c') = a;(c") = 0.

Thus ak(c') E Ker(cpk-l) = Im(cp,-l). Since cpn-l is injective, there is a unique c E Cn-l such that a;(c') = cpn-l(c) and then pn-2 0 an--l(C) = 8L-10 vn-l(C)

= ~ n - 1 06 ; ( ~ ' = ) O

which shows Bn-l(c) = 0 since cpn-2 is injective. Thus, c E Ker(&-l), from which [c]E Hn-l(C). The element [c] is independent of the choice of c' and c". Therefore, we can define a map a, : Hn(C") + H n ( C ) by = [c].It is easy to see that a, is a homomorphism.

a,(.)

Proposition 10.1.2. (1) For a short exact sequence of chain complexes and chain maps 0 + c 3 C' -5c" + 0, the sequence

is exact. (2) Given the diagram of chain complexes and chain maps

'pl C"

O - C L C c '

0

--D

D"

D'

rl,

__t

0

0

rl,!

if the first row and the second mw are both exact, then the following diagram commutes :

510.1 Chain complexes

307

Proof. (1)is checked as follows. Im(cp) = Ker(cp'): It is clear that cp: o cp, = 0, i.e. Im(cp') c Ker(cp). Conversely, let y E Ker(ak) such that cpL(y) E Im(8:). Then we can find z' E C:+l with S:+l(z') = cp',(y), and y' E CL+l with ~pL+~(y') = z'. Since cpL(y - a;+l(y')) = 0, there is x E Cn such that cpn(z)= y - 8n+l(y'). Then x E Ker(8,). Therefore, [y] = cp,([z]) E Im(cp,). Im(cp') = Ker(6,): If y E Ker(a'), by the definition of 8, we have 8, o cp:([y]) = 0 , and so Im(cp:) C Ker(8,). Let z E Ker(8:). If 8,([2])= 0, then there exist y E Ck and E Cn such that cpL(y) = and an(y) = 8 n 0 cpn(z). Thus, y - cpn(x) E Ker(Bn) and cpL(y - cpn(x)) = z, from which [z] E Im(cp:). Im(8*) = Ker(cp,): By the definition of 8, we have cp, o 8, = 0. For x E Ker(Bn-l) let cpn-l(x) E Im(i3;). Then there is y E Ck such that 8L(y) = cpn-l(z), and so a:(%) = cp;-l o cpn-l(z)= 0 where z = cpL(y). Therefore, z E Ker(8;) and [XI = a,([z]). (2) is clear by definition. 0 The following lemma is easily checked and so the proof is left to the readers. L e m m a 10.1.3 (Five lemma). Given the commutative diagmm of abelian groups and homomorphisms with exact rows

if XI,

X 2 , X4

and X5 a= isomorphisms, then so is

X3.

Let C = (8, : Cn --t Cn-l} be a chain complex. Then C' = {an : C; + Ck-l} is said to be a subcomplex of C if for each n E Z (1) Ck is a subgroup of Cn, (2) a n = anlc,, (3) an(cL) c CL-1It is clear that a subcomplex C' is a chain complex. If C' = {an : CL + Ck-l} and C" = {an : C: --t C:-l} are subcomplexes of the complex C, then C' C" = {an : CL C: --+ ChF1 C:-l} is a subcomplex of C which is called the subcomplex generated by C' and C". Given chain complexes C = {an : Cn --t Cn-l} and C' = {a; : C; --+ Ck-l}, the direct sum of C and C',

+

+

+

is a chain complex. Here each 8, is defined by B,(c,c') = (an(c),a;(c')).

CHAPTER 10

308

Lemma 10.1.4. For chain complexes C and C'

for all n E Z.

Proof.This is easily checked and so we omit it. 0 Let C = {an : C,, --t Cn-l} be a chain complex and let C' = {an : t CA-l) be a subcomplex of C. Then for each n E Z the boundary operator 8, induces a homomorphism a,, : C,,/CA + Cn-l/CA-l defined by an(c CA) = a,,(c) CA-l, and the sequence

CA

+

+

is a chain complex, which is called the quotient complex. We denote it as

C/C'. Remark 10.1.5.

Let a E H,,(C/C'). Then the element a is expressed as a = [c CL] for some c E C,,with an(c) E Thus, a = 0 if and only if c = a,,+l(d) c' for some d E Cn+l and some c' E C;.

+

+

Let C/C' and D/D' be quotient complexes and let cp = (9,) : C + D be a chain map. If cp,(CA) c DL for n E Z, then the chain map 'p induces a chain map cp : C/C'+ D/D' and moreover 'p+ : H,(C/C') + H,(D/D'),n E Z, is defined. Let A and B be abelian groups. Denote as F ( A , B ) the free abelian group generated by the set {(.,a) : a E A , b E B } , i.e. F ( A , B) is the direct sum @(o,b)Z. Let F'(A, B) be the free abelian group generated by elements of the form

where a,ai E A and b, b; E B,i = 1,2. Since F'(A, B) is a subgroup of F ( A , B), an abelian factor group F ( A ,B ) / F ' ( A ,B) is defined. We denote as A 8 B the abelian factor group, which is called the tensor product of A and B. For (a,b) E F ( A , B ) let a 8 b denote the element of A 8 B represented by ( a ,a). It follows that every element in A 8 B is expressed as a finite sum of such a 8 b with integer coefficients, and that the map (a, b) H a 8 b from A @ B to A 8 B is bilinear. Hence, we have the following properties :

$10.2 Singular homology

309

Let cp : A + A' and 11, : B + B' be homomorphisms of abelian groups. Then a homomorphism YJ QD 11, : A €4 B + A' €4 B' is defined by

For chain complexes C = {an : C,,--t Cn-l} and D = {a, : D , + & - I } we define C €4 D = {an : (C8 D ) , -+ ( C €4 D),,-1}, which is called the tensor product of C and D , where

(C €4 D ) , =

c

C, €4 D ,

(c

denotes direct sum),

p+q=n

& ( c a d ) = a,(c) @ d + ( - l ) P c @ ~ q ( d )( c E C,,d E D,,p+q = n). Since 8,08,,+1 = 0 for n E Z, it follows that C @ Dis actually a chain complex. For n E Zwe can define a homomorphism

by n([c]€4 [dl) = [c€4 4. Let cp : C -+ C' and $ : D define a chain map cp €4 $ : C €4 D -+ C' 8 D' by

-+

D' be chain maps and

Then (cp 8 $)* o n = n o (cp, €4 11,*) holds. Next, let us introduce the torsion product for abelian groups. To do this we need the following lemma.

Lemma 10.1.6. Let A be an abelian group. Then there exists free abelian groups FO and F1 such that FO 3 F1 and A g Fo/F1. Given an abelian group B , let i @ i d : FI @ B+ FomB be the tensor product of the inclusion i : F1 -+ Fo and the identity id : B + B . Then Ker(i €4 id) i s independent of the choice of the groups FO and F1, that is, they are mutually isomorphic. For the proof we refer to Dold [D2]or Spanier [Sp]. Let A and B be abelian groups and let Ker(i €4 id) be as in Lemma 10.1.6. We call Ker(i 18 id) the torsion product of A and B , and denote it as A * B. It is checked that (1) A * B Z B * A , ( 2 ) if B is torsion free (i.e. nb # 0 for all b E B , n E Zwith b # 0, n # 0 ) then A * B = 0, ( 3 ) A * B is torsion (i.e. for d E A * B there is n E Zwith n # 0 such that nd = 0).

CHAPTER 10

310

Theorem 10.1.7(Kunneth formula). Let C = {an : Cn ---t Cn-l} and C’= {an: CA + C;-l} be chain complexes. Suppose each of Cn (n E Z) is a free abelian group. Then one has the following short exact sequence : 0+

c

H P ( C )Q H , ( C ’ )

c

5 Hn(C Q C‘)+

H P ( C )* H,(C’)+0

p+q=n-1

p+q=n

and this sequence splits.

For the proof, see Dold [D2] or Spanier [Sp]. $10.2 Singular homology

The purpose of this section is to recall some properties of singular homology groups of topological spaces, which will be used frequently in later sections. Let { e o , . . . , e n } be the standard base of an euclidean space IWn+’, i.e. eo = (1,0, * , O ) , * . ,e , = (0, , O , 1). We denote as An = [eo, ,en] the smallest convex set containing {eo, ,en}, which is called a standard n simplex. Let X be a topological space. A continuous map u : An + X is said to be a singular n-simplez of X . Denote as Sn(X)the collection of all singular n-simplexes of X and as Sn(X) the free abelian group generated by Sn(X). For n < 0 we let Sn(X)= 0. Let 0 5 i 5 n and define an affine map EL : An-l A,, by

--

.

-..

For D E Sn(X) the composites u o E:, we let n

an(.)

-

- ,u o E:

=C(-l)iu

0

belong to Sn-l(X), and so

Ei,

i=O

by which a homomorphism 8, : Sn(X) -+ Sn-l(X) is induced. This called the boundary homomorphism. From the following lemma it follows that the sequence

is a chain complex, which is denoted as S ( X ) .

Lemma 10.2.1. 8, o an+,= 0 for all n E Z. Proof. For u E Sn(X) n

an-l 0 an(.) = x ( - l ) i a n - l ( u

0 EL)

i=O

=

n

n-I

i=O

j=O

C(-l)iE ( - l ) j u

0

€i o&.

an is

510.2 Singular homology

.

Since E; o ~

8,-1

311

.

3 , -=~ ~i o &:Ifor ll i > j,it follows that 0 &(a)

=

c + c c

(-1)i+ja0

(€i 0 €:l)

O<jli l),

H@')

z

from which the conclusion is obtained by induction. Theorem 10.2.10 (Excision Theorem). If { Y , Z } is an excisive couple of X , then the inclusion i : (Y,Y n 2) + ( X ,2) induces an isomorphism i, : Hn(Y,Yn 2) -+ H n ( X ,2) for all n E Z.

Proof. Let us consider the following commutative diagram of inclusions and quotient maps : 0

- -

S(Y)+S(Z)

S(2)

-

(S(Y )+S(Z))/S(Z)

-I - I - + l - I 0

S(X)

S(2)

I S(X)/S(Z)

-

0

-

0.

It is clear that the rows of the above diagram are short exact sequences. Hence, by Proposition 10.1.2 we have the following commutative diagram : Hn(Z)

Hn(z)

Hn(S(Y)

S(Z))

Hn(X)

Hn((s(Y)

+S ( z ) ) / S ( z ) )

Hn-l(Z)

I

I

Hn(X, 2)

Hn-l(Z)

CHAPTER 10

316

+

Since {Y,Z} is an excisive couple, the inclusion e : S(Y) S ( Z ) + S(X) induces an isomorphism e, : Hn(S(Y) S ( Z ) ) Hn(X). From this together with Lemma 10.1.3 we have

+

On the other hand, by the homomorphism theorem we have S(Y)/(S(Y) n S ( Z ) ) 2 ( V ) + S(Z>>/S(Z>. Since S(Y) n S ( Z ) = S(Y n Z),we have Hn(Y,Y r l 2)E Hn(X,2)under the inclusion. Therefore, the conclusion is obtained. 0 Theorem 10.2.11(Excision isomorphism theorem). Let (X, A) be a topological pair and let U be an open subset of X . Suppose cl(U) c int(A). Then the inclusion i : ( X - U,A - U) + (X,A) induces an isomorphism i, : Hn(X- U,A - U )+ Hn(X, A) for all n.

Proof. Put Y = X - U and Z = A. Since cl(U) c int(A), we have X = int(Y)Uint(Z), and by Theorem 10.2.7, {Y, Z} is an excisive couple. Therefore, Theorem 10.2.10 derives the conclusion. 0 Lemma 10.2.12. (1) Hp(Rn,R" - 0) 2 0 ( p # n) and Hn(R",R" - 0 ) r Z. ( 2 ) Given T > 0 let 0: = {z E R" : 11x11 < T } whew 11 11 is the euclidean n o m . Then the inclusion i : (Rn,Rn - 0:) H (Rn,Rn - 0 ) induces an isomorphism

- 0)

i, : Hp(Rn,R" - 0:) E H,(R",R"

( p 2 0).

Proof. Since i : R" - 0: R" - 0 is a homotopy equivalence, we have that i, : Hj(R" - 0:) + Hj(Rn - 0) is an isomorphism for j. Theorem 10.2.5 --f

yields the following comutative diagram whose rows are exact. + Hj(R")

1

4

Hj(lR")

-

Hj(B",R"

-

a.

1

- 0:) 8 ,Hj-l(R" - 0:)

Hj(lR",P

- 0)

8.

a.

1-

Hj-l(B"

0)

-

Hj-l(R")

-

4

I

Hj-l(lR") +

Since there is a homotopy equivalence from S"-' to R" - OF, we have that Hj(R" - 0:)Z Hj(Sn-l)r Z ( j = 0 , n - l ) , S 0 (j# 0 , n - 1). Note that Hj(R") "L O(j # 0) and H0(Rn)2 Z. Therefore, (1) and (2) are obtained from the above diagram together with Lemma 10.1.3. 0 For topological pairs (X, A) and (Y, B) we write

(X, A) x (Y, B) = ( X x Y, A x Y U X x B).

510.2 Singular homology

Define affine maps

E;

: A p -+

An and q;1“: A,

and let Pl: X x Y t X and P2 : X x Y u E Sn(X x Y )define 0 u 0 Ep”,

I q)

(0 I k

q;(ek) = ek+n-q

u:,= PI

An by

+

5p)

(0 I k

Ep”(ek)= ek

+

317

Y be the natural projections. For

a; = Pz

0 u 0 q;

where Cm

A,-SA,&XXY-%X,

5

5 x x Y %Y. Then a homomorphism p : Sn(X x Y )+ ( S ( X )8 S(Y)),is defined by A,

p+n=n

We call the family p = { p } : S ( X x Y )+ S ( X ) @ S ( Y )the Alexander-Whitney map. For u E Sn(X)and r E &(Y)write ( a x r ) ( z ) = (o(I), ~ ( 2 ) where ) I E An. Then u x r E Sn(X x Y).For 0 5 i 5 n - 1 affine maps qk : An + An-l is defined by

-

1 Ik Let I = {il, i2, * ,i, : i k < A map q I : An + An-, is defined by

< p} be a subset of

{0,1,.

,n - 1).

il qr =qn-*+l o...oq2.

If, in particular, I = 0 then we let 71 = i d . Let 8 be the collection of all subsets consisting of p elements of (0, 1, ,n1 ) and for I E 8 we write J = { 0 , 1 , - . . ,n- 1 ) - I . For I E 8 denote as € ( I ) the cardinal number of the set { ( i , j ) : i E I , j E J,i > j}. Let p + q = n and take u E S , ( X ) and r E S,(Y). Then

---

uvr =

E ( - I ) ~o( ’ )x( ~o 773)

(T

qI)

IEe

is an element of Sn(X x Y).Thus a homomorphism

V : ( S ( X )8 S(Y))n Sn(X x Y ) is defined by u 8 T H uVT. We call the family V = {V} : S ( X ) C3 S ( Y ) --+ S(X x Y ) the Eilenberg-MacLane map, and write xVy = V(x @ y) for x E -+

S p ( X )and y E S,(Y).

CHAPTER 10

318

Theorem 10.2.13. The Alezander-Whitney map p = { p } : S ( X x Y ) + S ( X ) 8 S ( Y ) and the Eilenberg-MacLane map V = {V} : S ( X ) 8 S ( Y ) -+ S ( X x Y ) are chain maps with the following properties : (1) p and V are natural with respect t o continuous maps, i.e. i f f : X -+ X ' and 9:Y -+ Y' are continuous maps, then p o ( f x g)n = ( f n @I ga) o p and v 0 ( f n 8 gn) = ( f x 9111 0 v. (2) p and V are chain homotopy equivalences and these are a chain homotopy inverse in each other. Moreover, one can choose the chain homotopy between p o V (resp. V o p ) and the identity (resp. the identity) to be natural with respect to continuous maps. For the proof we refer to Eilenberg-Maclane [E-MI.

Remark 10.2.14. Let ( X , A ) and ( Y , B ) be topological pairs. If { A x Y , X x B } is an excisive couple, then the Alexander-Whitney map p = { p } : S ( X x Y ) + S ( X ) 8 S ( Y ) induces an isomorphism P* : H n ( ( X ,A ) x (Y,B ) ) -+ H n ( S ( X ) / S ( A 8 ) S(Y)/S(B)) for all n. Indeed, since p : S ( X x Y ) -+ S ( X )8 S ( Y ) maps S ( A x Y ) S ( X x B ) to S ( A )8 S ( Y ) S ( X ) 8 S ( B ) , by Theorem 10.2.13 (2)

+

+

S(X x Y ) : S ( A x Y ) S ( X x B ) -+

+

S ( X )8 S ( Y ) S ( A ) 8 S ( Y ) + S ( X )8 S ( B )

is a chain homotopy equivalence. The right hand of the above relation is isomorphic t o S ( X ) / S ( A )8 S ( Y ) / S ( B ) .Since { A x Y , X x B } is an excisive couple, we have for all n

Therefore, the conclusion is obtained.

For a = [z]E H p ( X ,A ) and b = [y] E Hq(Y,B ) we define a homomorphism K

:

C

~ p ( xA, ) 8 Hq (Y,B )

+

Hn(S(X)/S(A8 ) s(Y)/s(B))

p+q=n

by I([.

8 [y]) = [Z8 y] as stated in 510.1, and define a

X

b bY

p*(a x b) = n(a 8 b), which is called the cross product of a and b. By Theorem 10.2.13 the EilenbergMacLane map V : S ( X ) 8 S ( Y ) -, S ( X x Y) is a chain homotopy inverse of the Alexander-Whitney map p. Thus, we have the following corollary.

f 10.2 Singular homology

319

Corollary 10.2.16. Let a E H p ( X , A ) and b E H,(Y,B). If a = [z]and b = [y] where E S p ( X ) and y E Sq(Y),then the cross product a x b is expressed as a x b = [zVy]. Remark 10.2.16. The cross product has the following properties : (1) x : H p ( X , A )@ H q ( Y , B )+ H,,((X,A) x ( Y , B ) )is bilinear, (2) (f xg),(axb) = f,(a)xg,(b)forcontinuousmaps f : ( X , A )-+ ( X ' , A ' ) , 9: (Y, B ) (Y', B'), ( 3 ) & + ( a ) = a x [ y ] andi,,(b) = [ ~ I x b f o r i n c l u s i o n s i (~X: , A )+ ( X , A ) x Y ,i, : ( Y , B )4 X x ( Y , B )defined by i Y ( z )= ( x , y ) , i,(y) = (z,y), (4) ( u x b) x c = a x (b x c ) , ( 5 ) if T : X x Y + Y x X is a continuous map defined by T ( z ,y) = (y, z), then T,(a x b) = (-l)P% x u for a E H p ( X , A )and b E H,(Y,B). Indeed, (1) and (3) is easily checked from definition. (2) follows from Theorem 10.2.13 (1). For the proof of (4) and ( 5 ) , see Dold [D2] or Spanier [Sp]. +

&om Theorem 10.1.7 we have the following theorem. Theorem 10.2.17 (Kiinneth formula). Let ( X ,A ) and (Y,B ) be topological pairs. If { A x Y,X x B } i s an excisive couple, then the cross product yields the following split exact sequence

0

-c

HP(X,A ) 8 HqW,B ) -Hn((X,

p+q=n

c

p+q=n-1

A ) x (Y, B))

-

Hp(& A ) * Hq(Y,B )

-

0.

Remark 10.2.18. Let 'IT" = R"/Z" be an n-torus. Since 'IP is homeomorphic to S' x x S' (the direct product of n times) and

...

by Theorem 10.2.17 we have that Hp('IP)is a free abelian group of rank In particular, Hl(T")"L Z" holds. Let u1,u2, ,u,, : [0,1] + R" be paths defined by

(3.

- -.

ul(t)=(t,O,"',o), u 2 ( t ) = ( O , t , . . . ,O),

... , u n ( t ) = ( O , O , . * -, t )

and put wi = a o u i for 1 5 i 5 n where a : Rn + Tn is the natural projection. Then each wi is a closed path from 0 to 0 in T", and {[w1],... ,[wn]}is a 2 Z". system of generators of the fundamental group m('ITn,0) On the other hand, each wi can be considered as a singular 1-simplex and then they are cycles since &(wi) = 0. It is easy to see that the set { [wl], ,[wn]}of homology classes is a system of generators of H1(Tn).

-.

320

CHAPTER 10

Let Z : ?rl(Tn,O) + Hl(’P)denote an isomorphism defined by 2 ( [ w i ] ) = 5 i 5 n). Then, for a continuous map f : Tn --t ’P we have the following commutative diagram [wi] (1

where w is a path from 0 t o f ( 0 ) and w* : ?r~(’ll’”, f ( 0 ) )--t m(Tn,0) denotes the induced isomorphism in Lemma 6.1.4.

510.3 Euclidean neighborhood retracts (ENRs)

Let X be a topological space and A a subset of X. The subset A is said to be a retract of X if there is a continuous map T : X + A such that the restriction of T to A is the identity map, i.e. T I A = id. In this case, r is called a retraction. We say that A is a neighborhood retract (abbreviated NR) of X if there exists an open neighborhood U of A in X such that A is a retract of

U. Remark 10.3.1. (1)If A is open in X ,then A is an NR of X. (2) If A is an NR of X and B is an NR of A, then B is an NR of X. (1) is clear. (2) is easily checked as follows. Since A is an NR of X, there is an open subset U of X and a continuous map T I : U + A such that T I is a retraction. Since B is an NR of A, an open subset V of A and a retraction T Z : V + B exist. Put W = r,’(V) and define a continuous map T = ~2 o ( T ~ I W:) W + B. Then W is open in X and T : W + B is a retraction.

A topological space X is said to be locally compact if for any x E X and any neighborhood V of x there is a compact neighborhood of x containing in V. A subset A of X is called a locally closed set if there exist an open subset 0 of X and a closed subset C of X such that A = 0 n C. Lemma 10.3.2. Let X be a subset of Wn. Then (1) if X is an NR then X i s locally closed set, ( 2 ) X is locally closed set if and only if X is locally compact, ( 3 ) if X is a locally closed set then X is homeomorphic to some closed set of W+l.

510.3 Euclidean neighborhood retracts (EN%)

321

Proof. (1) : Since X is an NR of R",there exist an open set 0 of R" and a continuous map T such that T : 0 + X is a retraction. Since T I X is the identity map, X is closed in 0. Thus, X = 0 n C for some closed set C of Itn. (2) : If X is a locally closed set, there exist an open set 0 and a closed set C of R" such that X = 0 n C. Let x E X . Then there is a compact neighborhood K of x in R" such that K c 0. Since K n X = K n C is a compact neighborhood of x in X , X is locally compact. Conversely, suppose X is locally compact. For x E X there is a compact neighborhood K , of x in X. Thus, K, is expressed as K , = X n V, where V, is a compact neighborhood V, of x in R". Put 0, = int(V,). Since

X n O , = K , n O , =cl(X )n O ,, we have

and therefore X is a locally closed set. (3) : Since X = 0 n C for some open set 0 and some closed set C, we can define a continuous map j : 0 + R"+' by

where d denotes the euclidean metric. Then j ( 0 ) = { ( x , t ) c R" x I[$ : t d ( x , R n - 0 ) = 1) is closed in R"+'. Obviously j : 0 + j ( 0 ) is a homeomorphism, and so j(X) is closed in j ( 0 ) . Therefore, j ( X ) is closed in I[$"+'. 0

Theorem 10.3.3. Let X be an NR of R" and let Y be a subset of Rm. If Y is homeomorphic to X, then Y is an NR of R". Proof. By Lemma 10.3.2, X is locally compact and so Y is locally compact. Thus Y is a locally closed set. This ensures the existence of an open set V of R" such that Y is closed in V. Let h : Y + X be a homeomorphism. Since V is a normal space, there is a continuous map g:V -+ R" such that gly = h. Since X is a NR of R", an open set 0 of X in R" and a retraction T : 0 + X exist. The set W = g-'(0) is an open set of R" such that W 3 Y, and h-' o T o g : W + Y is a retraction. 0

A set X is said to be an euclidean neighborhood retmct (ENR) if X is homeomorphic to an NR of some euclidean space R". By Theorem 10.3.3, ENR is topologically invariant, i.e. if X is an ENR and X is homeomorphic to some subset Y of R"',then Y is an NR of R".

CHAPTER 10

322

Lemma 10.3.4. Let X be a topological space and A a subset of X . Suppose Y is an ENR. Iff and f' are continuous maps from X to Y with fp = f/At then there ecist an open neighborhood W of A in X and a homotopy F from fp to such that F : W x [0,1] -+ Y and F ( a , t ) = f ( a ) = f'(a) for all a E A and all t E [0,1].

flW

Proof. Since Y is an ENR, there are an open set 0 of an euclidean space R" and continuous maps T : 0 + Y and i : Y -+ 0 such that T o i = id. Define a subset of X by

w = {X E x : (I - t ) i o f (2) + t i o f'(z) E 0 for all t E [O, 11) and a homotopy F : W x [0,1] -+ Y by

+

F ( z , t ) = ~ ( ( 1 -t ) i o f (z) ti o f'(z)). Then we obtain the conclusion. 0

Theorem 10.3.5. Let X be a Hausdorfl space and suppose { X I , - - - ,X p } is a finite open cover of X . If each of Xi as an ENR, then X is an ENR. Proof. By the assumption X is homeomorphic to some closed set of a euclidean space R". By Lemma 10.3.2 (3) for each X i there is a homomorphism gi : Xi -+ R"i such that g i ( X i ) is a closed subset of Rni. As one point compactification of R"i we consider the sphere S"i = Rni U {m}. Define a map hi : X --t Sni by

Then hi is continuous. Indeed, let C be a closed subset of S"'. If m $2 C , then hf'(C) is homeomorphic to C n g;(Xi) which is compact. Thus, hf'(C) is closed. When 00 E C,hf'(S"i - C ) = gi'(Sni - C ) is an open subset of X i , which implies that hf (S"i - C ) is open in X , and so hf' (C) is closed in X. To obtain the conclusion it sufficies to show the case when { X l , X z } covers X . We put N = nl +nz 2 and define a continuous map h: X -+ S"1 x S"' c RN.Then h is injective and h : X + h ( X ) is a homeomorphism. By Lemma 10.3.2 each Xi is locally compact, and so is X = X1 U X Z . Thus, h ( X ) is locally compact. Lemma 10.3.2 (3) ensures that h ( X ) is homeomorphic to some closed subset of a euclidean space RN. It remains to see that X is an ENR. We identify h ( X ) with X and then X is a closed subset of RN. Since each Xi is an ENR, by Theorem 10.3.3 an open set Oi of RN with Oi 3 Xi and a retraction ri : Oi + Xi exist. Put

'

+

0 = r ; l ( x l n x 2 )n TF1(x1 n xz).

310.3 Euclidean neighborhood retracts (ENRs)

323

Obviously r l p : 0 + X1 n X2 and r2p: 0 + X1 n Xz are retractions. Since X1 n Xz is an ENR, by Lemma 10.3.4 there exist an open neighborhood Wo of X I n X Z and a homotopy H from r l p 0 to r21wo such that WOc 0 and H : Wo x [0,1] + X 1 n X z with H ( a , t ) = a (a E X 1 n X z , t E [0,11). Note that X - X1 and X - X2 are closed in R" and

(X-Xl)n(X-X2)=0,

x - x 1c02, x - x 2 c 0 1 .

Choose open sets W1 and WZsuch that

x-x1 c W2 c 0 2 , x - x z

c Wl c 0

1

cl(W1) n cl(W2) = 0. Then W = WOU W1 U W2 is an open neighborhood of X in R". Let T : R" -, [0,1] be a continuous map such that .(W1) = 0, and T ( W Z= ) 1. Define a continuous map p : W -, X by

Since p is a retraction, X is an ENR. 0

Theorem 10.3.6. Let X be a Hausdorflspace and suppose { X i : 1 5 i < m} i s a countable open cover of X . If (1) each X i is an ENR, (2) for suficiently large n, each Xi is embedded in R", i.e. there is a homeomorphism from Xi onto a subset of R", then X is also an ENR. Proof. By Lemma 10.3.2 each Xi is locally compact, and so X is locally compact. Since each Xi has a countable base, so does X and thus X is paracompact. By the assumption (2) we have dim(X) 5 n. From a fact of the dimension theory we can find a refinement {Yi} of { X i } such that {Y,} is an opencoverofXandforY,,,...,Yjt E { x } , Y j , n . . . n Y , ,# Qimpliest 5 n + l . Take subfamilies C 1 , - - * C n + l of {yi} such that C1 U . * . U Cn+1 = {yi}, and all elements of Ce are mutually disjoint for 1 5 t! 5 n 1. Then for 1 ~ . ! < n + l , & = U C ~ i s a n E N R . I n d e e d , l e t C ~ = { Y ~ , Y , ' , . . .E } .a c h T 'is an open subset of some set X i . Thus, yj' is an ENR. By the assumption (2), Y,'is embedded in R". Since R" is homeomorphic to R"-l x ( j- f ,j f ) , we have that Yi is homeomorphic to some NR of R"-' x (j- f ,j f). Therefore, Zt is an NR of R". Since { & , . . . , Z n + l } is a finite open cover of X , by Theorem 10.3.5 the conclusion is obtained. 0

+

+

+

CHAPTER 10

324

Theorem 10.3.7. Every topological manifold is an ENR. Proof. Each point of X has a neighborhood which is homeomorphic to R" or [O,oo). It is clear that R" and B" x ( 0 , ~ are ) EN&. Since X has a countable base, by Theorem 10.3.6 it follows that X is an ENR. 0

R" x

Fix n > 0 large enough. Let

vo, q ,

-

* *

,wp

be points in R". If

implies A0 = A1 = = Ap = 0, then v o , v 1 , - - - ,up are said to be linearly ,wp] the smallest convex set independent. In this case, we denote as [vo, v1, containing {vo, q , ,up}, which is called a p-simplex. Let up = [vo,v1, ,wp] be a psimplex and let {io, * * ,ip} c {0,1,. , p } . A q-simplex [vi,, ,vi.1 is called a face of up. We say that a collection K = {u} of simplexes is a simplicial complex if (1) any face of a simplex in K belongs to K, (2) if U,T E K and u n T # 0 then u n T E K, (3) for u E K and x E u there is a neighborhood V in R" such that g { E~ K :T n V # 0) is finite. Let K be a simplicial complex. The dimension of K, denoted as dim(K), is defined to equal n if K contains an n-simplex but no (n 1)-simplex. We write IKI = UoEKu. It follows that IKI is a compact subset of R" if and only if K is finite. A subset L of K is called a subcomplex of K if L is a simplicial complex. A topolological space X is a polyhedron if there are a simplicial complex K and a homoemorphism h: IKI + X. Here K is called a simplicial decomposition of X . A sequence of abelian groups, A = {A,, : n E Z}, is said to be finitely generated if each A, is finitely generated and A,, is trivial except abelian groups of finite numbers.

---

- a -

- - .--

-

.-

+

Theorem 10.3.8. If X is a compact ENR, then the homology group H , ( X ) = { H , ( X ) } is finitely genemted. Proof. We may suppose that X is a subset of some euclidean space R". Since X is an ENR, an open set 0 of R" and a retraction T : 0 + X exist. We here define a simplicial decomposition K of R" as follows. Let { 211,. ,wn} be a base of B"and let ?r be a permutation of (1,2, ,n). * a,v, for ( a l , * ,a,) E Z" and consider n-simplexes We write v = alvl

-.

+- - +

--

---

Denote as K the set of all faces of these simplexes. Then K is a simplicial complex and IK( = R". If the length of w i , l 5 i 5 n, is small, every u E K with u n X # 0 is a subset of 0 (i.e. u C 0) since X is compact. Let L be a collection of all faces of u E K with u n X # 0. Clearly L is a subcomplex of

$10.3 Euclidean neighborhood retracts (ENRs)

325

K. Since X is compact, L is finite and X C ILI. Thus, TO = ~ l p:lILI + X is a retraction, from which PO* : Hn(ILI) + Hn(X),n E Z,is surjective. To obtain the conclusion it sufficies to show that for every finite simplicial complex L, H,((LI),n E Z,is finitely generated and Hn(ILI) = 0 for n > dim(L). To do this we use induction on the dimension of L. If dim(L) = 0, then ILI is a finite set, and so

-

Thus, the case of dim( L) = 0 holds. Let dim( L) = n and let { a1, ,a,} c L be the collection of all n-simplexes. Ta,ke zi E int(ai) for 1 5 i 5 k and define

Obviously {Ul,U2} is an open cover of ILI. Hence, we have the following Mayer-Vietoris exact sequence :

Let L"-' = {a E L : dim(o) < n}. Then L"-' is a subcomplex of L, dim(L"-') = n - 1 and (L"-'( c U1. It is easy to see that the inclusion i : (L"-'l + U1 is a homotopy equivalence. By the assumption of induction Hp(U1) is finitely generated and Hp(U1) = 0 for p 2 n. It is clear that Hp(U2)= 0 for p # 0. Notice that U1 n U2 = &(int(ai) - z i ) . Since each of int(ai) - zi is homotopy equivalent to Sn-', we have that Hp(U1 n U2) E 0 for p # 0 , n .- 1 and Hp(U1 n U2) (p = 0,n - 1) is isomorphic to Z@ * @ Z (the direct sum of k-times). Therefore, by the above exact sequence it follows that Hp(JLI)is finitely generated for 0 5 p 5 n, and Hp(ILI) = 0 for p > n. 0

--

Let X be a topological space. If the homology group H , ( X ) = {Hn(X)} is finitely generated, then we define the Eder characteristic of X by

where each bn(X)is the rank of Hn(X). Theorem 10.3.8 ensures that the Euler characteristic can be defined for every compact ENR.

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510.4 Fixed point indices Let X be an ENR and let W be an open subset of X. For a continuous map f: W -+ X we will define a characteristic of f, called the fixed point index I ( f ) , in the case when Fix(f) = {z E W : f(z) = z} is compact. If X = R",then we have Hn(Rn,Rn-O) S Z by Lemma 10.2.12 (1). Choose a generator 0 of Hn(Rn,Rn - 0) and let K be a compact subset of R". We - K) as follows. define a fundamental homology class OK of Hn(Rn,Rn For T > 0 put 0: = {z E Rn : 11z11 < T } where 11 11 is the euclidean norm. Take T such that K c Om. Then, the inclusion j : (R",R" - 0:) + (R",R" - K) induces a homomorphism j , : Hn(Bn,Rn - OF) + Hn(R",Rn - K ) .

By Lemma 10.2.12 an isomorphism

i* : H"(R", R" - 0;)+ H"(R", R" - 0) is obtained. We define OK

= j, o i;'(O).

It follows that ( - O ) K = -OK holds and the definition of OK is independent of the choice of T . For an open set V with K c V c R" we have by the excision isomorphism theorem that H"(V, - K )E H"(R",IIB" - K ) .

v

Let 0; (or simply OK) denote the image of OK mapped by the isomorphism. Let L be a compact subset with L c K and let V,U be open subsets with L c K c V c U c R". If i, : Hn(V,V- K ) -+ Hn(U,U - L) is a homomorphism induced by the inclusion i : V + U , then we have (10.3)

i,(og)= 0;.

Let g:V -+ R" be a continuous map and suppose F = Fix(g) is compact. Let (id - 9 ) : (V,V- F ) + (R",Rn - 0) be a continuous map defined by (id - g)(z) = z - g(z).We define the fixed point index I(g) of g by

(id - g)*(OF) = I(g)O where (id - g)* : Hn(V,V- F ) + Hn(Rn,Rn- 0). Since ( - O F ) = -OF, the index I(g) is independent of the choice of 0. If F c K c V , then (id - g)*(OK)= I(g)o by (10.3). For the fixed point index I(g) we have the following the properties. Property 10.4.1. IfI(g) # 0 then F is nonempty.

Proof. Suppose F = 0, then Hn(V,V - F ) 2 0 and thus I ( g ) = 0. 0

$10.4 Fixed point indices

Property 10.4.2. For an open subset V' with F I(9lVd I(9) agree.

327

c V' c

V , the indices

Proof. This follows from (10.3). 0 Property 10.4.3. If g(V) = p , then I(g) = 1 when p E V and I(g) = 0 when P$V. Proof. If p $! V, then F = 0 and by Property 10.4.1, I(g) = 0. When p E V we have F = {p}. Since (id - g)(x) = x - p , we define a translation T : R" 4 Rn by T ( x )= x - p . Then it sufficies to show that T,(O,) = 0 where

T; : Hn(Rn,Rn- p ) + Hn(Rn,R" - 0). Take T > 0 such that p E 0: = {x E R" : )1x))< T } and define a homotopy ht : (Rn,Rn - 0:) -t (Rn,Rn - O),O 5 t 5 1, by ht(x) = 2 - tp. Obviously ho is the inclusion and hl = T . Thus we have a commutative diagram :

Therefore, T,(O,) = 0. 0 Property 10.4.4. Let V be expressed as the union of two open subsets V1 and V2 (V = Vi U V2). If Fix(gp,) and Fix(glv,) are disjoint and compact, then I(9) = I b l V , ) I(91V2).

+

Proof. We put F1 = Fix(glv,) and FZ = Fix(glv,). Since FI n F2 = 0, we can take open subsets U1 and U2 such that Fi c Ui c x , i = 1,2, and Ui n U2 = 8. Since I(gpi) = I ( g 1 ~by ) Property 10.4.2, we have I ( g p ) = I(g) where U = U1 U U2. On the other hand, by Lemma 10.1.4

where F = Fl U F2. The isomorphism used here maps (OF,, 0,) to OF.Thus we have

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328

Property 10.4.5. Let gt : V -+ R"(0 5 t 5 1 ) be a homotopy and put Ft = Fix(gt). If K = UoStSl Ft is compact, then I ( g 0 ) = I(g1).

Proof. Since Ft c K C V for 0 5 t 5 1, clearly (id - gt) : (Rn,R" - K) -+ (Rn,Rn - 0 ) , 0 5 t 5 1, is a homotopy, and hence ( i d - go), = ( i d - gl),. Thus, qgo) = I(g1). 0 Property 10.4.6. Let U be an open subset of Rm and let f:U-+ R" be a continuous map. Let V be open in R", and g: V + R" continuous. If Fx(f) is compact, then the continuous map f x g : U x V -+ Rm+" has the property I(f x 9) = I(f)I(g). Proof. Let 0 1 be a generator of Hm(Rm,R" - 0) and let 0 2 be that of Hn(Rn,R"-0). The cross product 0 = 01 x 0 2 is a generator of Hm+n(Rm+n, I[$"+" - 0) E Z. The inclusion i induces an isomorphism

i, : Hm+n(R"+",R"+" - 0,-x 0,") + Hm+n(R"+",R"+"

- 0,-+").

It is easily checked that O K ~ = L OK x OL for compact sets K c R" and L c R". Thus we have O ~ i ~ (xf )OF^^(^) = O ~ i ~ ( f ) ~ ~Since j ~ (Fix(f) ~ ) . x Fix(g) = Fix(f x g) and (id - f) x ( i d - g ) = ( i d - f x g), we have

I(f g)o= (id- f

x

g)+oFix(fxg)

= ( ( i d - f) x (id - g))+(oFix(f) x OFix(g)) = (id- f)*(OFix(f)) (id - g)*(OFix(g)) = W O l x I(g)Oz = I(f>I(gP and therefore

I(f

x g) = I(f)l(g). 0

Property 10.4.7. For open sets V1 c R"1 and VZ C Rna let fl : Vl -+ Rna and fz : VZ -+ Rnl be continuous maps. Put U1 = f;'(Vz) and U2 = fF1(V1). Let f1 0 f2 : UZ +R"1. fz 0 f l : Ul R"',

-

Then Fix(f2 o f 1 ) and Fix(f1 o fz) are homeomorphic. If two the sets are compact, then I(fz o f l ) = l(flo fz). Proof. Since fllFix(fiofl) : Fix(fz 0 f i ) -+ Fix(f1 o fz) is the inverse map of fZIFix(flofa): F k ( f i 0 fz) + Fix(f2 0 f i ) , it follows that Fix(f2 o f1) is homeomorphic to Fix(fl o fz). Define a homotopy gt : U1 x UZ+ I[$"l+"a by gt(21,22)

= (tf2

f1(%1)

+ (l - t)fz(2z),f1(z1)),

(0 5 t 5 1).

Then = {(21,22) E

u1 x VZ :2 2 = f 1 ( 2 1 ) , 2 1

= fZ(z2))

$10.4 Fixed point indices

329

is homeomorphic to Fix(fiof1) and so Fix(gt) is compact. Thus I ( g 0 ) = I(gl) by Property 10.4.5. Let ht : U1 x R"a -+ Rn1+"a (0 5 t 5 1) be a homotopy defined by ht(51,22) = O f 1 0 f2(21),(1- t)fl(Zl)). Since g1 = hol(rlxua, by Property 10.4.2 we have I(g1) = I(h0). The image Im($) of a continuous map $ : Fix(f2 o f 1 ) x [0,1] t U1 x R"a, which sends (z1,t)to (21, ( l - t ) f l ( q ) ) , coincides with Uoltll Fix(ht). Thus it is compact and by Property 10.4.5, I ( h 0 ) = I(h1). Since h1(21,x2) = (f2 of1(x1),0), we have I(h1) = I(f2 of1) by Properties 10.4.1 and 10.4.6, and consequently I ( f 2 o f 1 ) = I(g0). To obtain the conclusion we define g: : U1 x U2 + by g:(xl,x2) = (f2(22),tfl 0 f 2 ( z 2 ) (1 - t)fl(.l)) and hi : Rnl x U2 + Rnl+"a by

+

h:(zl,z2) = ((1 - t)f2(22),fl 0 f2(.2)). In the above way we have I(gk) = I(f1 o f 2 ) . Since go = g;, the conclusion is obtained. 0 Suppose X is an ENR and W is an open subset of X. We now give the definition of fixed point index I(f) for a continuous map f: W + X. Since X is an ENR, X is homeomorphic to an NR of R" for a certain n. Thus, there exist an open subset 0 of R" and a continuous map i : X + 0 and r : 0 + X so that T o i = id. The set Fix(i o f o r ) of fixed points of the composite r-'(W) r,w f -5 0

-

x

is homeomorphic to Fix(f). If Fix(f) is compact, then the fixed point index I(i o f O T ) of the composite is defined as above. We define the fixed point index I(f) of f by I(f) = I(i o f o r). The index I(f) is independent of the choice of 0,iand r . Indeed, let 0','i and r' correspond to 0,i and r respectively. Apply Property 10.4.7 for i o r' : 0' + R" and 'i o f o T : r-'(W) + R"'.Then we have I(i o T' o 'i o f o T ) = I(i' o f o T o i o r') and so I(i 0 f 0 r ) = I(i' 0 f 0 r'). It follows that the fixed point index of the continuous map f :W + X possesses also the above Properties 10.4.1-10.4.7. If a fixed point x of f is isolated, then I(fl(r) is defined by taking a small neighborhood U of x, and it is independent of the choice of U by Property 10.4.2. Thus, we write I ( f , z )= I ( f p ) , and I ( f , x ) is called the fixed point index off at 2. Let S" be an n-dimensional sphere and let f: S" + S" be a continuous map. Since Hn(Sn) Z Z, there is m E Z such that the induced homomorphism f* : Hn(S") + Hn(Sn)satisfies the following relation (VU E Hn(Sn))f * ( ~= ) VZU We call such an m the mupping degree of f and denote it as deg( f).

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330

Remark 10.4.8. (1) Let V be a neighborhood of the origin 0 in R" and let g: V 4 R" be a continuous map with 0 E Fix(g). For E > 0 small let 0," = {Z E R" : 1 1 ~ 1 15 E } and suppose D, f l Fix(g) = (0). Let g' : Sn-' + S"-' is a continuous map defined by EZ gl(Z)

=

)IEZ

- g( E Z ) - g(EZ)11

(z E S".-l).

Then I(g, 0) = deg(g') holds. (2) Let g: V + R" be as above and suppose g is a C' map. Then (i) if det(id - Dog) is positive, then I(g,O) = 1, (ii) if det(id - Dog) is negative, then I(g,O) = -1 where Dog is the derivative of g at 0 and det(id - Dog) is the determinant of id - Dog These are checked as follows. To see (l),choose T > 0 such that D, c 0: C V, and consider the following diagram

where 3 = id-g , i'(5) = EZ (Z E S n - l ) and i is the inclusion. It is easy to see that a,, i: and i, are isomorphisms and the diagram commutes. Therefore, the conclusion is obtained. To show (2), take E > 0 small enough and define a homotopy H :0: + Rn by H ( z , t ) = g(tz)/t for 0 < t 5 1 and H(z,O) = Dog(z). By Property 10.4.5 we have I(g,O) = I(Dog,O). From this together with (1) the conclusion is obtained

.

$10.5 Lefschetz numbers

In this section we discuss Lefschetz numbers of continuous maps on compact ENRs. If an abelian group A is finitely generated, then A is expressed as the direct sum of cyclic groups of finite number. Let { e l , * ,eg, eg+l , ,e n } be a system of generators of A where el , ,eg have infinite order and eg+l ,* ,en have finite order. A homomorphism cp : A + A has the representation

---

--

--

510.5 Lefschetz numbers

33 1

6 for some integers aij(1 5 i, j 5 n). Here we write tr(cp) = Ci=l aii, which is called the trace of cp. Let X be a topological space and suppose the homology group H , ( X ) = { H n ( X ) }is finitely generated. For continuous map f:X+ X we define the Lefschett number L( f ) of f by

n

where each f + , , :H n ( X ) + H n ( X ) denotes the induced homomorphism. If, in particular, f is the identity id, then L(id) means the Euler characteristic of

X. We have easily the following theorem.

Theorem 10.6.1. If continuous maps f : X topic, then L( f ) = L(g).

+

X and g : X + X am homo-

Whenever X is a compact ENR, by Theorem 10.3.8 the homology group H , ( X ) is finitely generated. Thus, the Lefschetz number is defined for every continuous map of X . For the relation between Lefschtez number and fixed point index we have the following theorem.

Theorem 10.5.2 (Lefschetz Axed point formula). Suppose X is a compact ENR and f : X --+ X is a continuous map. Then I ( f ) = L(f ) . For the proof, see Dold [ D l , D2]. From Theorem 10.5.2 and Property 10.4.1 it follows that if f : X + X is a continuous map of a compact ENR then

L ( f )# 0

Remark 10.5.3. Let f : S" sphere. Then we have

-+

* W f #) 0.

S" be a continuous map of an n-dimensional

L ( f )= 1

+ (-l)"deg(f).

where deg( f ) is the mapping degree of f .

Remark 10.5.4. Let T" be an n-torus. Then Hl(T")E Z @ *..@ Z (the direct sum of n-times). Let { e l , , e n } be a system of generators of HI(""). For a continuous map f : T " -t T" we have f*(ei) = C;=, aijej ( 1 5 i 5 n). If A l , . ,An are eigenvalues of the matrix ( a i j ) , then the Lefschetz number L( f ) is expressed as

-

n

i= I

This is checked as follows. For 1 5 i 5 n let wi be a cycle i19 in Remark 10.2.18 and let ei = [wi].Then, we can consider the set {ei, x ei, x x ei, :

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332

..-

1 5 il < i2 < < i k 5 n } as a system of generators of Hk(1In) (see Remark 10.2.16), and the following relation is easily checked:

f*(eil x ei, x

.- - x e i , ) = ai1,il

ai1,ia

* *

ei, Xei, x - - . x e i , + . . . aij,,il

aih,ig

* * *

aik,ib

Therefore, we have

= det( I - (ai,j)) n

= n(l- X i ) . i=l

where I denotes the identity matrix.

$10.6 Orientability of manifolds

Let X be a connected locally path connected Hausdorff space and suppose the following properties hold: for all z E X

and there is a connected open neighborhood V, of z in X , called a canonical neighborhood, such that for some 0, E H,(X,X - Vz), i z * ( O z )is a generator of H,(X,X - z ) for z E V, where i, : ( X , X - V,) -+ ( X , X - z ) is the inclusion. Here 0, is a fundamental homology class of H n ( X ,X - V,). Since X is Hausdorff, it follows that if 0: E H n ( X , X - V,) is another fundamental homology class, then we have either

i,+(Oh)= i z + ( O x )(Vz E V,)

or

i,+(O:) = - i z + ( O x ) (Vz E Vz).

Let w : [0,1] + X be a path. Then there is a sequence SO = 0 < s1 < * < s, = 1 such that w([si,si+l])c Vw,,i, for 0 5 i 5 n - 1 where QSi) is a

510.6 Orientability of manifolds

333

canonical neighborhood of w(si). Let Ow(ai)be a fundametal homology class of H ( X , X - V,(,,)) and define an isomorphism

wit : H n ( X , X - ~ ( s i ) -+) Hn(X,X - w(si+l)) by wit

0

iW(8i)*(0W(8i))

wu = wn-1p

0

= iW(ai++(0W(8i)).

Then the composite

***w18o W O ~ : H n ( X , X - ~ ( 0 ) + ) Hn(X,X - ~ ( 1 ) )

does not depend on the choice of {Vw(8i)}and {Ow(,,)}. It is easy to see that ( P l ) if w is a constant path then wp is the identity map id, (P2) if two paths w1 and wz satisfy wl(0) = wz(l), then (w1 W Z ) ~= Wap 0 w11. (P3) if w1 is homotopic t o WZ , then wlp = wzp. We say that X is orientable if wl = id for all closed path w in X . In this case, there exists the family of generators

{1=E H n ( X , X - z):z E X } , called an orientation of X , such that wp(lw(o))= lw(l) for all paths w in X . Let X and Y be topological spaces as above, and let f : X ---t Y be a covering map. Since f is a local homeomorphism, the following (P4) is easily checked from defintion : (P4) f , o wn = ( f o w)p o f * for any path w in X where f , denotes the induced isomorphism. From the following Lemma 10.6.1 we can say whether or not a connected toplogical manifold without boundary is orientable.

Lemma 10.6.1. Let M be a topological n-manifold and let U be a connected open subset of M such that there is a homeomorphism h: U --t Wn. Let V = h-'({z E W n : 1 1 ~ 1 1 < 1)) where 11 11 denotes the euclidean norm. If z E V , then the inclusion i : (M, M - V) t ( M ,M - z) induces an isomorphism

i, : H.j(M, M - V ) + H j ( M , M - 3)

(j 2 0).

Proof.This is clear from Lemma 10.2.12 (2) together with the excision isomorphism theorem. 0 Remark 10.6.2. Let M be a connected topological manifold without boundary. If M is non-orientable, then there is a double covering map p : --+ M such that is orientable. Indeed, let A = {[w] E n l ( M ) : w. = i d } . Then A is a subgroup of ?rl(M) with index 2. By Theorem 6.2.12 we can find a double covering map p : &t ---t M such that p*(?rl(&f)) = A. Then by (P4) ~

p*owp = ( P o ~ ) p o P * for any path w with [w] E ?rl(M). Since [pow] E A, we have p, o wp = p , , and so w8 = id. Thus, M is orientable.

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334

Remark 10.6.3. Let M be an orientable closed topological n-manifold. Then, H n ( M ) S Z and if OM is a generator of H n ( M ) then OM) is a Zfor all 2 E M where i* : H n ( M ) H,(M, Mgenerator of H n ( M ,M - 2 ) x) denotes the homomorphism induced by the inclusion. Such an element OM, called a fundamental homology class of M, corresponds to an orientation of M. For the details, see Dold [D2] or Spanier [Sp]. For a continuous map f: M + M the mapping degree, deg(f ) E Z,is defined bY f* : Hn(M) Hn(M). f*(Olcri)= deg(f)OM -+

+

If, in addition, f is a self-covering map, then the absolute valute I deg(f)l is consistent with the covering degree of f. Indeed, let b E M and put F = f - ' ( b ) . Then the inclusion i : M + M induces a homomorphism i, : H,(M) + H n ( M , M - F). Let {lZE H n ( M , M - x) : x E X} be an orientation of M corresponding to OM. Since H,(M,M - F ) S! $cEFHn(M,M - C), we have i * ( o = ~ $ccF1c. ) On the other hand, from (P4) it follows that there is a constant k = f l such that f*(IZ) = k l f ( O! for all 2 E M where f* : H n ( M , M - x) -+ H n ( M , M - f(x)) is the induced isomorphism. Hence f * ( e c ~=~ kUFlb. l c ) Since the diagram H,(M) f*

LH n ( M , M - F )

1

1

commutes,

f*

H n ( M ) --L H n ( M , M - b) we have deg(f)lb = kdF1b. Therefore, 1 deg(f)l = UF.

Remark 10.8.4. Let M be a closed topological manifold. If a continuous map f : M + M is homotopic to a homeomorphism (or more generally, a self-covering map), then f is surjective. Indeed, suppose M is orientable and let OM be a fundamental homology class of M. If there is x E M such that x $ f ( M ) , then we have a commutative diagram

Hn(M) f*

1

i . Hn(M,M-x)

lf*

Hn(M) i . H,(M,M-2) and so f* o i * ( 0 ~ = )deg(f)i,(OU). Since f ( M ) c M - x, it follows that f* o & ( O M )= 0, thus contradicting. In the case M is non-orientable, by Remark 10.6.2 we take a double covering ) p*(nl(fi)), map p : A? -+ M such that fi is orientable. Then f * ( p * ( r l ( f i ) )= and so there is a lift f of f by p (see Remark 6.3.5) and f is homotopic to a

$10.6 Orientability of manifolds

homeomorphism (or a self-covering map). Therefore,

335

is surjective, and so is

f. In the remainder of this section we shall discuss the (homological) dimension of generalized foliations on connected topological manifolds without boundary. Proposition 10.6.5. Let M be a connected topological manifold without boundary and suppose 3 is a generalized foliation on M . Then each leaf of 3 is an ENR under the leaf topology.

Proof. Let L be a leaf of 3 and let x E L. By definition there is a local coordinate cp : D x K + N at x. Since N is open in M, N is an ENR (by Theorem 10.3.7) and so is D. Since D is open in L and the leaf topology of L has a countable base, from Theorem 10.3.6 it follows that L is an ENR. 0

Lemma 10.6.6 (cf. Bredon [Br]). Let X and Y be non-trivial topological spaces. If the product topological space X x Y is a connected topological nmanifold without boundary, then (1) there are integers p , q > 0 with p q = n such that for z E X and

+

Y E Y

( 2 ) each point of X has a canonical neighborhood in X and the same fact holds for Y .

Proof. First we claim that if A and B are abelian groups and the tensor product A 8 B is isomorphic to Z, then both A and B are isomorphic to Z. Indeed, let cp : A 8 B + Z be an isomorphism. Then we can take m

i=l

cp(xEl

such that ai 8 bi) = 1. Let $"A denote the direct sum of m-times and define a homomorphism c p :~$"A -+ Z by

Then cp~($"A) = cp(A 8 B), and so $"A E Ker(cpA) $ Z. Similarly, e m B g Ker(cpB) @ Z. Thus it follows that

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336

On the other hand, since A @ B E Z,we have

($"A) 8 (emB)r e m a Z from which Ker(cpA) and Ker(cpB) are finitely generated. Therefore, A and B are isomorphic to Z. To show (1) let (z,y)E X x Y . Since X x Y is a manifold, X and Y are Hausdorff, and hence { X x (Y - y), ( X - z)x Y} is an excisive couple in X x Y. By Kunneth formula we have the following split exact sequence:

0 4

c

H~(x,x-z)@Hj(Y,Y-y)+H~((x,x-z)x(Y,Y-y))

i+j= k --t

c

H i ( X , X - z)* Hj(Y,Y - y)

--+

0

i+j=k-1 Since dim(X x Y) = n and X x Y has no boundary, we have

Hk((X,X - ). x (Y,Y - y)) From this together with the fact that is a torsion group, it follows that

Z 0

(k=n) (k#n).

Ci+j=k-l H i ( X , X - z)* Hj(Y,Y - y)

+

and so by the first claim we can find p 2 0 and q 2 0 with p q = n such that (10.4) holds. Then, p # 0 and q # 0 since X and Y are non-trivial. Since X x Y is connected, it follows that p and q are independent of the choice of points in X x Y. It remains t o show (2). Let p and q be as in (1) and let z E X and y E Y. As representatives of generators of H p ( X , X - z) and H,(Y,Y - y) take cycles c, E A , ( X ) and d, E A , ( Y ) respectively. If a is a boundary homomorphism, then ac, E A p - l ( X - x) and so there exists a neighborhood D, of 2 in X such that ac, E A p - ~ ( X- z ) for all z E D,. Similarly, there exists a neighborhood D, of y in Y such that ad, E Aq-l(Y - z ) for all z E D,. Choose a neighborhood V of (z,y)in X x Y as in Lemma 10.6.1. Then Hn(X x Y , X x Y - V) 2 Z. As a representative of its generator we take a cycle uv E A , ( X x Y ) . Then by Lemma 10.6.1, [ U V ] is a generator of Hn(X x Y , X x Y - z ) for all z E V. Kunneth formula shows that [c,] x [d,] is a generator of H n ( ( X , X - z)x (Y,Y - y))

= z.

On the otherhand, by Corollary 10.2.15 we have that [cz] x [d,] is represented by a cycle c,Vd, where V is the Eilenberg-MacLane map. Thus, there exist

510.7 Orientability of generalized foliations

337

+

a E An+l(X x Y ) and b E A,(X x Y - (2, y)) satisfying c,Vd, - EU = €la b where E is either 1 or -1. Let V' be a neighborhood of ( x , y ) in X x Y with V' c V such that b E A,(X x Y - z ) for all z E V', and choose neighborhoods DL and D&of x in X and y in Y respectively, such that DL c D,, D&C D , and DL x D&c V'. Then, DL is a canonical neighborhood with [c,] a fundamental homology class. The similar fact holds for D&. 0

Proposition 10.6.7. Let M be a connected topological manifold without bounda y and let 3 be a genemlized foliation on M . Then there exists an integer p with 0 < p < dim(M) such that for every leaf L E 3 and for e v e y x E L

Proof. Let 'p : D x K ---t N be a local coordinate. By Lemma 10.6.6 we can find 0 < p < dim( M) such that for y E D

Take z E N and let L E 3 be the leaf through z. Then for some y E D

(i 2 0). Indeed, if D' is the connected component of z in N f l L,then there is z' E K such that cp( ,z') : D + D' is a homeomorphism, from which Hj(L, L - Z ) 2 Hi(D, D - y)

Hi(D', D'

- Z) E Hi(D, D - y)

(i 2 0)

where t = 'p(y, z'). Since D' is open in L, by applying the excision isomorphism theorem the above relation are obtained. Since N and z E N are arbitrary and M is connected, we obtain the proposition. 0 Let 3and p > 0 be as in Proposition 10.6.7. We say that the dimension of 3is p , and write dim(3) = p. If 3and 3' are transverse generalized foliations on a manifold M, then the following equality holds: dim(3)

+ dim(3') = dim(M).

$10.7 Orientability of generalized foliations The purpose of this section is to discuss orientability for generalized foliations. Let M be a connected topological manifold without boundary. Suppose 3 is a generalized foliation on M. We prepare seven claims to mention how to introduce orientability for 3. Let dim(3) = p and denote as L ( x ) the leaf through z of 3.

338

CHAPTER 10

Claim 1. Take and flx a E M. If cp : D x K 4 N is a local coordinate at a, then N is expressed as the disjoint union N = UxEKD , of subsets D , where D , = cp(D,z). By Lemma 10.6.6 (2) there is a Canonical neighborhood of a in D , written as V. Then R = cp(V x K) is connected and open in M. It is clear that R c N and R = UxEKV,. Here V, = cp(V,z) for 2 E K.

R

Figure 39

Let z E K . T hen cp (a,s)=zan d cp ( ,z):(D,D-V)-t(D,,D,-V,)is a homeomorphism. Hence, V, is a canonical neighborhood of z in D,. Take a fundamental homology class 0 E H,(D, D - V) and put

We write

K = (0,: 2 E K} and fix K. For q , x 2 E K define a homeomorphism diagram commutes :

Then we have

where Ox,, Ox,E K.

$+1,+2

such that the following

810.7 Orientability of generalized foliations

339

Claim 2. For q E R take z E K with q E V,. Then a homomorphism iq* : Hp(D,,

D, - K ) --t Hp(Dz, D, - q )

is induced by the inclusion iq

: (Dz, D, - Vz) --t

(DZ, D, - 4).

Since 0, E H,(D,, D, - V,) r l IK is a fundamental homology class, the image iq,(O,) is a generator of H,(D,,D, - q). Note that D, is the connected component of q in N n L(q). Since D, is open in L(q), the inclusion j , : ( D , , D , - q ) --t ( L ( q ) L(q) , - q ) induces an isomorphism jq*

- q)

: HP(%

+

Hp(L(q),L(q) - q).

Therefore, jq* o iq*(O,)is a generator of H,(L(q), L(q) - q).

Claim 3. qz E

Let

q1,

qz E R, then there are

: H P ( L ( q l ), L(ql) - q 1 )

by

z1,22

E K such that

q1

E V,, and

VZ2.We now define an isomorphism

+g,R(jql*

0

Hp(L(q2) ?L ( q 2 ) - 4 2 )

iq,+(OZl= ) ) jqa+o iqa+(Oza). Then the following (10.5) holds:

{

(10.5)

qR = id, '::,R

= 'ii,R

(aqa qi,R I-'

@Z,RV

= aql qp,R'

Claim 4. Let a E M be as above and take b E M. Let cp' : D' x K' --t N' be a local coordinate at b and choose a canonical neighborhood V' of b in D'. As seen above, N' = cp'(D' x K') and R' = cp'(V' x K) are expressed as the disjoint unions N'= D& R' =

U

U Vi

UEK'

U€K

where D& = cp'(D',y) and V,' = cp'(V',y). fundamental homology class and put

0;= cp'( ,y')*(0')

E

Let 0' E Hp(D',D' - V ' ) be a

Hp(D&,D&- V,').

As above we write

K'= {0&: y E K ' } and fix

K'.For y 1 , yz E K' define a homeomorphism

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340

Claim 6. Suppose R n R' # and let ql,92 E R f l R'. If there exists a path y : [0,1] + R n R' such that ~ ( 0 = ) q1 and y(1) = q2, then (10.6)

@ E , R = @::,Rf

On

Hp(L(ql),L(ql)

- q1).

Figure 40 We need the following seven steps to check (10.6). We first show the existence of t2 > 0 such that

Step 1.

Let cp and 9' be as above and let p~ : D x K + K and p ~ :l E [0,1] define

D' x K' + K' be the natural projections. For t a ( t )= PK

0

'p-'

0

')'(t),

d ( t ) = PKI

0

0 y(t).

Then a ( t ) E K and a'(t) E K' for t E [0,1]. Since V C N and V,(t) = cp(V,a(t)),we have y ( t ) E V,(q. For, since cp : D x K N is a local coordinate at a, clearly a ( t ) = cp(a,a(t)) E cp(V,a(t))= V,(t). Note that y ( t ) E R = cp(V x K) for t E [0,1]. If p v : V x K + V is the natural ) ) hence projection, then cp-' o y ( t ) = ( p v o cp-' o y ( t ) , p o~ 9-l o ~ ( t and -)

cp(V,4 t ) )3 4 P V

0

9-l 0 d t ) ,4 t ) ) = y ( t )

because p v o 'p-l o ~ ( tE)V. Therefore y ( t ) E Va(q. Similarly, y ( t ) E VLIct)holds for t E [0,1].

510.7 Orientability of generalized foliations

341

Step 2. For simplicity, we denote as t and t' the symbols a(t) and a'(t) respectively, when they are in subscripts. Then, by the above step we have for t E [O, 11

Since Dt and D:,are open in L ( r ( t ) ) ,it is clear that Dt nD:, is an open neighborhood of T ( t ) in L ( y ( t ) ) . In particular, DOn D;,is an open neighborhood of $0) = q1 in L(q1). Since the leaf L(q1) is locally compact, we can find a connected open neighborhood C of q1 in DOn D;, such that the closure of C,cl(C), is compact d(C) c Do n DA,. Then cl(C) C N f l N'. Notice that N n N' is open in M. Since +o,o : Do + DOis the identity map and +o,t = +o,t : DO-+ N is continuous, there is to > 0 such that for all t E [0, t o ]

Since +o,t(cl(C)) is connected in L(-y(t)),for some x' E K' we have +o,t (cl(C)) C D;, because N' = UzlEKl DL,. Then y ( t ) E DL, = D;,, from which +o,t (cl(C)) c D;,. Hence Pt(cl(C)) C q!(l,ol(D~l) = DAl

for all t E [0, to]

where Pt

Since

Pt

: cl(C)

---t

= +;y 0 $o,t.

D;, varies continuously with respect to t E [O,to] and

PO :cl(C) + DOn D& is the inclusion, from the fact that DOn D;, is an open neighborhood of cl(C) in D;,it follows that for some 0 < tl 5 t o (10.7)

Pt(C)

c DOr l DkI

for all t E

[O,tl].

Step 3. Since VOn V;, and C are open neighborhoods of q1 in DOn Do', there exists a canonical neighborhood W of q1 in DOn D;, satisfying

(10.8)

cl(W) is compact cl(W) c c cl(w ) c von vd, .

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342

By continuity we can find 0 < t z

{

(10.9)

5 t l such that for all t E [ O , t z ]

7 ( t ) E +o,t(W) q1 = ~ ( 0 E) Pt(w>c

von Vd,.

By (10.7) and (10.9)we have a continuous map

Pt : (C,C - W ) (Don Dhf,D~n ~k~ - ql) -+

for all t E [0, t z ] . Since Pt is homotopic to Po, clearly

Pt* = Po* : Hp(C,C - W )+ Hp(Don Dhf,D~ n D;,- ql). Hence, for each t E [O,to] we have the following commutative diagram : HP(C, c - W )

Po. J

\ dJ0.t.

where i, is a homomorphism induced by the inclusion.

Step 4. Since W C VO by (10.8), by the inclusions we obtain the following relation :

Hp(Do,Do - VO)-+ Hp(Do,Do - W )

= HP(C,c - W ) .

To avoid complication, we consider 00E H,(DO,Do - v o ) n K as a fundamental homology class of Hp(C,C - W). From the relation

HP(D;,,aV - v o r )

+

4md%,Pt(W)>= Hp(Pt(C),Pt(C) Pt(W)), -

-

the element O;, E Hp(D;,,D;,- VOI)n F is also considered as a fundamental - p t ( W ) ) . Similarly, for t E [ O , t z ] we homology class of HP(Pt(C),Pt(C) identify fundamental homology classes as follows : Hp(Dt,Dt - Vt) W

Ot

-

-

Hp(Dt,Dt - +o,t(W)) Ot

TI

m Hp(40,t(C),+o,t(C- W)) W

o:,

H

o:,

11

rn Hp(D;,,D:t -

--+

Hp(D:t,D:, -dJo,t(W)).

$10.7 Orientability of generalized foliations

Step 5. mutes :

By the definition of

Pt

*o,tp

it is clear that the following diagram comPi +

(C,c - W )

(tJt(C),tJt(C- W ) )

\

J *htpi(c)

(*o,t(C),*o,t(C - W ) )

Hence, letting + ~ , t= +O,tlC

and

+0,t*(00) = Ot

= +bf,tf,p and

i(

)

we have

+&,t,*(Obt)= Oi,.

Denote inclusions as follows:

and

Step 6. We now are in a position to prove that for t E [O,t,] @dt)

pl,R

- @Y(t)

-

pl,R’

On HP(L(ql),L(ql)

- q1)-

To do this it is enough to show that if

(10.10)

j,(t)*

0

i,(t)*(Ot>= j,(t)*

then jql* 0 &ll*(0O)

0

i,(t)*(W

=jilt i;l*(obJ, 0

and otherwise, that is, if one has

( 10.11)

j,(t)*

0

343

i,(t,*(Ot) = -&(t)*

0

i,(t)*(O3

CHAPTER 10

344

then j q l * 0 iql*(Oo) = In the case of (10.10) we have

-&*

0

i;l*(O;,).

i,(t)*(Ot) = i,(t)*(O:,) since jr(+ is an isomorphism. Notice that Ot and O:, are fundamental hoSince T ( t ) E $o,t(W), it follows mology classes of H,($o,t(C),$O,t(C that for z E & & V ) iz*(Ot) = iz*(o:,) where i,, denotes the homomorphism induced by the inclusion

w)).

iz : ($o,t(C),+o,t(C - W ) ) Hence, letting z1 = $;,,t,(ql) we have

+

($O,t(C),$O,t(C) - .).

iz1*(Ot) = i*,*(o:t). Take qi E W such that z1 = $o,t(qi), and write

4 = $O,t(C : (C,C- Q 3 If iq;. : (C, C - W ) + relation is obtained :

($o,t(C),+o,t(C) - .1). (C, C - q ; ) denotes the inclusion, then the following +

GI* 0 +o,t*(Oo) = 4*0 iq;*(Oo). Similarly, let -1

$ = +;~,t~lp,(c) : (Pt(C),Pt(C>- Q1) (40,t(C),$o,t(C) - 211, then we have i,,* 0 +;,,t,(o;,) = 0 i;l*(O;,)* Therefore lj* 0 i q ; * ( 0 o )= $6: 0 i;,*(o;l). --$

4:

On the other hand, letting; : (Pt(C),Pt(C)-ql) -+ (DonD;,,DonD;, -ql) denote the inclusion, we have

- $-:-I

Pt* = i*

0

0

4*

0

iq;*

from which

a*

4*

Pt*(Oo) = 0 &-I 0 0 i q ; * ( 0 o )= Since Pt* = PO* on H,(C,C - W), it follows that PO*(OO) =

a*

0

ibl*(O;,).

a* i;,*(o;,), 0

and therefore j p l * 0 iq,*(OO)= j;,*0 i;l*(O;t). In the same way as above, we obtain the similar result for the case of (10.11). Since t is arbitrary in [0,t z ] , we proved that for all t E [0, t 2 ] &t) =p ( t )

q1,R

q1,R”

510.7 Orientability of generalized foliations

345

Step 7. Let 4 be the set of s E [0,1] such that

for t E [O,s]. It is easily checked that 4 is open and closed, which shows 4 = [0,1]. Therefore, (10.6) holds.

Claim 7. Let w : [0,1] + M be any path. Then we can find a sequence 80

=0

< 81 < *..< 8 ,

=1

and a sequence 21,. ..,x, of points in M such that

w([si-1,

ail)

C Rq

where Rxi = cpxi(Vxi x Kxi), cpxi : Dxi x Kxi -+ Nni is a local coordinate at zi, and Vxi is a canonical neighborhood of zi in Dxi. Define an isomorphism

w* : ~ * ( J W O ) ) L(W(0)) , - w(0))

+

H P ( ~ ( W )L(W(1)) , - w(1))

bY By (10.5) and (10.6), wt is well defined. Furthermore, the following properties are easily checked : (Ll) if w is a constant path then W. = id, (L2) if two paths w1 and w2 satisfy w1(0) = w2(1), then (wl w2)* = w2* 0 Wl., (L3) if w1 is homotopic to w2, then w1, = w2,.

.

By making use of the family {we} of the isomorphisms w, defined as above, we can introduce orientability for generalized foliations as follows. A generalized foliation 3 on M is said to be orientable if w* is the identity map for all closed paths w. Let dim(3) = p . We call an orientation of 3 the family of generators {lX E HP(L(x), L ( z ) - X ) : x E M}

if ~ * ( l ~ = (lw(l) ~ )for ) any path exactly two orientations for 3.

20

in M . If 3 is orientable, then there are

L e m m a 10.7.1. Let Mi (i = 1,2) be a connected topological manifold without boundary and let 3i be a generalized foliation on Mi. If p : MI + M2 i s a covering map and 3 1 is the lift of 3 2 by p , then for any path w of M I p* 0 w* = ( P 0 w)* 0 p*. Proof. This is clear. 0

CHAPTER 10

346

Proposition 10.7.2. If F is non-orientable, then there exists Q double covering map p : M + M such that ( 1 ) if? is the lift of 3 by p then ? is orientable, (2) i f f : M t M is a self-covering map and 3 is the lifl of 3: by f then t h e r e i s a l i f t f : 2 + M off byp. Proof. Let A = {[w]E q ( M ) : w, = id}. Obviously A is a subgroup of ?rl(M) with index 2 , and then there exist a topological manifold and a double = A. Since covering map p : M + M such that p,(?rl(U)) P*

0

w* = ( p 0 w), o p ,

for [w]E T I ( $ ) , we have p* o w, = p* because [p o w ] E A, and so w, = id. Thus, ? is orientable. For [w]E A, we have

f , = f*

0 W*

= ( f 0 w). 0 f*

and so ( f ow), = id, from which [ f ow]E A. Hence, we have f,(A) C A. This implies the existence of a lift f of f by p. 0

510.8 Fixed point indices of expanding maps The purpose of this section is to show the following fked point index theorem for TA-covering maps belonging to P&M.

Theorem 10.8.1 (Fixed point index theorem). Let M be a closed topological manifold and suppose f : M 4 M i s expanding. If M is orientable, then there is C > 0 such that for evey m 2 C all jized points of f have the same index 1 or -1. Proof. As before we denote as Mf the space of the inverse limit system. Let d be a metric for M and let e > 0 be an expansive constant for f . From Remark 5.3.3 (1) we can find 6 > 0 such that for x = (xi) E Mf, Uh(z0)C W,ul,(x) where Uh(Z0)= {y E M : d ( z o , y ) < 6). Since M is a closed manifold, there exists an atlas { ( U , h ) }of M with the property that {U}is finite, each diameter of U is less than 5, and h: U + R" is a homeomorphism where n = dim( M). Let 6 > p > 0 be a Lebesgue number of {U}. By Lemma 2.4.1 (2) we can take C > 0 such that for m 2 C and (q), (yi) E Mf if d ( x i , y i ) 5 e for all i 5 0 then d ( z i - m , y i - - m ) 5 p for all i 5 0. Fix m 2 C and let Q E F i x ( f m ) . Then the sequence

a = ( ...,a , f ( a ),..., f m - l ( a ) , a ,... ) E M f

$10.8 Fixed point indices of expanding maps

347

is an m-periodic orbit of f . Choose ( U , h ) E { ( U , h ) } such that B p ( a )c U where B J a ) = {y E M : d ( a , y ) 5 p } . Since diam(U) < 6, by the choice of 6 we have U c Ua(a)c W&(a). Let V denote the set of points y in M such that there is (yj) E Mf satisfying y-, = y, yo E U and d ( y i , a i ) 5 e/4 for all i 5 0. Then a = a_, E V and V c B J a ) since m 2 1. Write Fa = for simplicity. Since f is expanding, Fa : V + U is bijective, and hence it is a homeomorphism. Notice that V is open in M. Since a is exactly one fixed point of Fa, to obtain the conclusion it is enough to show that the fixed point index of Fa is 1 or - 1 and it does not depend on the choice of a E Fix(fm). Since M is orientable, we fix an orientation of M and denote it as { l s E H , ( M , M - z): z E M}. Since f is a self-covering map, from (P4) of $10.6 it follows that there is a constant E, = f l such that

fp

Fa, o iG:(la) = €,i(l;(la) for all a E Fix(fm) where iv : (V,V - a ) -+ ( M ,M - a ) and iu : (U, U -a) + ( M ,M - a ) are inclusions. Choose an orientation ( 1 , E H,(R",R" - z ) : z E R"} of R" such that h: U + R" is orientation preserving, and define a map

G I = i d - h o Fa o h-' : ( h ( V ) , h ( V )- h ( a ) ) + (R",R" - 0). Since each 1 , is a fundamental homology class of z, by the definition of fixed point index G I *o i , ' ( l h ( a ) ) = I ( F a ) h where i : ( h ( V ) , h ( V )- h(a)) -+ (R",Rn - 0) is the inclusion. On the other hand, letting

G2 = i d - h p o Fa-'

o

h-' : (R", R" - h ( a ) )+ (R",R" - 0 ) ,

we have

G2*(lh(a))= ~ ( F ~ ' ) h and

G1=CoG20hoFaoh$ where C : R"

+

R" is a map defined by ( ( z ) = - z . Hence

GI*0 it'( lh ( a ) )= C* 0 G2* 0 h* 0 Fa* 0 h$*

0

i r ' (l h(a ))

= C* o G2* o h, o Fa, o i i ; ( l a )

= C* 0 G2* 0 h*(EmiEt(la)) = 6, G2(Emlh(a)) = C* ( ~m~( F, - ') lO ) =c~mI(F~')lo

CHAPTER 10

348

where ( + ( l o ) = clo and c = f l . Thus, I ( F o ) = e m I ( F L 1 ) . Since Frl : U + V and V C B p ( a ) c U ,we can define a homotopy Ht : U + U by H t ( z ) = h-'(th o Fr1(z)).Then HI = FT1 and HO is a constant map. Since V c B,(a), a subset {z E

U : H&) = z for some t E [ O , l ] } c B p ( z )

is compact, and so I ( F L 1 ) = 1 by Properties 10.4.3 and 10.4.5. Therefore, I(F0) = CEm. 0

510.9 Fixed point indices of TA-covering maps In this section we show the following fixed point index theorem for TAcovering maps belonging to 7d \ PEM.

Theorem 10.9.1 (Fixed point index theorem). Let M be a closed topological manifold. Suppose f: M + M is a self-covering map belonging to I d \ P & M . If M is orientable and the family 3; of stable sets in strong sense is orientable, then there is 1 > 0 such that for every m 2 C all fixed points of f has the same index 1 or - 1. For the proof we need some preparations.

Lemma 10.9.2. Let U be a neighborhood of 0 in R" and denote as A the diagonal of R" x R". Then for N e d a E R" the map

i"

: ( U , U - 0 ) + (R" x

R",R" x R n \ A ) ,

i ( z )= ( z + a , a )

is a homotopy equivalence. Proof. Consider maps

;" : (R",R"

- 0) + (R" x

d : (R" x R",R" x R"

\

R",R" x R" \ A ) , A) + (R",R" - 0),

-0

i

(2)

= (z

d(z,w) = z

+ a,a), - W.

z"

Clearly d o io = id. A homotopy from id to o d is given by H t ( z ,w) = ( z t ( a - w ) ,w t ( a - w ) ) ( 0 5 t 5 1). Hence a" is a homotopy equivalence. U - 0) + (R", R" - 0) is a homotopy equivalence, we Since the inclusion (U, obtain the conclusion. 0

+

+

mf a.

Let 7r : + M be the universal covering and define as in (6.3) of 56.5. By Theorem 6.7.4 for each u E the families 7" and 7::of stable sets and unstable sets are transverse generalized foliations on Fix an orientation of M as { l Z E H n ( M , M - z) : z E M} where n = dim(M). Then we can choose an orientation of so that the covering map ?r

af

510.9 Fixed point indices of TA-covering maps

--

349

z}.

is orientation preserving, and denote it as { 1, E H n ( M ,M - x ) : x E Let {lZE H,(R",R" - z ) : z E a"}be an orientaion of R". When V is an open set of R" and z belongs to V, we denote as

i$ : (V,V - z ) -, (Rn,Rn - z) the inclusion map. It is clear that

i;*

: Hn(V,V - 2) + Hn(Rn,Rn - z)

is an isomorphism and i;i1(lz) is a fundamental homology class of z. Let U be a coordinate domain at a point in with respect to and 7",, and define a continuous map ru : U x U 4 U as in (6.8) of $6.7. Choose a continuous injection : U -, R" which preserves the orientations of and 1". We note that +(U)is open in R" and : U 4 +(U)is a homeomorphism. For fixed x E U define a continuous map TE,4 : +( U )+ Rn by

z

+

(10.12)

z

+

TE,4(4 = 4 0 -t,u(+-lM,4 - 4 0 rdz, +--l(Z))

for z E +(U).Then T ~ , 4 ( z = ) 0 if and only if z = +(z). Thus, we have a map

TE,4 : (+(U>, W) - 4b))

+

(R*,R" - 0).

From the following lemma we can find a constant c E Z, called the intersection number of Faand Fz, which depends only on the family Fa.

Figure 41 and let : N --$ Rn Lemma 10.9.3. Let N be a connected open subset of be a continuous injection which preserves the orientations. Then there exists a constant c E Z such that for any coordinate domain U in N and for any X E U

(10.13)

*4(2)--1 G , 4 * 0 24(U)* (14(0,> = clo.

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350

Moreover, the number c does not depend on the choice of N and 4, nor on the choice of u E

mf.

Proof. To show the first statement, let x E U and choose 6 > 0 such that 4(U) 3 where B6 = {z E R" : IIq5(x) - zll 5 a}. Let u 6 / 2 be the open ball of $(x) with radius 6 / 2 . Then there is E > 0 such that if z E B,5 - u 6 / 2 then ~~T~,+ 2 (8.z Since ) ~ ~ T((I, q5 and 4-l are continuous, we can find 0 < < 6 / 2 such that for y E U if 4(y) E Bp then llTC,d(z) - T;,+(z)II I &/2 for z E Ba. Let y E U with 4(y) E Bp and define Ht : B,5 + R" by = (1 - W C & )

&(z)

+ tTt?j,+(").

Then we have for z E Ba - u 6 / 2

lHt(411= IITC&> + t(T&&) - TC,&))ll 2 IIT&dz>II - tllTr?,&> - Tr7,+)(4ll 2 & - t & / 2> 0 , and hence T& and Ti,+are homotopic continuous maps from (Ba,Ba - ua/,) to (Rn,Rn - 0), from which TC,d,Bs+ = Ti,+lB6* on H,,(B6, Ba - u6/2).Therefore T;,+* = Ti,+*: Hn(d(u),4(u)- u6/2) Hn(R",IW" - 0)+

Since b(z),$(y) E Val,, by the fact that follows that

u6/2 is

a canonical neighborhood it

Hence we can find c E Z such that (10.13) holds for all y E u6/2. Since U is connected, the integer c does not depend on the point x E U. Let U and U' be coordinate domains in N. If U nu' # 4, then a coordinate = ~ ( ( I ~ ( ( I I I ~ ( (= I I I ~ ( ( I ~ I ( ( I I I ~ holds. U I I Hence domain U" in U n U' exists and for x E U" d(z)-1 *4(2)-1 T&+*Oi+(,)* Pd(3)) = TC',+*O Z + ( u ' ) * (1d(2))9 which shows that the number c does not depend on U ,because N is connected. We proved the first part of the lemma. and let 4,4' : N + R" be Next, let N be a connected open set of orientation preserving continuous injections. Let U be a coordinate domain in N. For fixed x E U denote as D" and D" the path connected components of x in U n m " ( x ) and U n u) respectively. Then 7~ : D" x D" + U is a homeomorphism. Hence the composite maps ~ ( ( I I I

mu(.;

11, : 4 ( U )

-

U

% D" x D" 94(D")x 4(D") -1

+1-1

11,' : $'(U)

U

-

D" x D" d' X 4' 4'(D")x q5'(D")

810.9 Fixed point indices of TA-covering maps

351

are homeomorphisms. For a E R" let T,, denote the translation on R" defined by T,,(z)= z - a. Consider the following diagram

where j and j' denote inclusions, and i+(") and i+'(") are the maps defined as in Lemma 10.9.2. It is easy to see that the two squares of the above diagram commute. Since the restrictions of the map d : (R" x Rn,Rn x R"

\ A)

-+

are homotopy inverses of i+(") and definition we have

(R",R" - 0),

~ ( . z , w= ) z -w

is'(") (see the proof of Lemma 10.9.2), by

Hence, from the fact that 4' o 4-l : 4(V)-+ #(V) is orientation preserving it follows that

which shows that the number c does not depend on the choice of 4. Let N and N' be connected open sets of and let : N + R" and 4' : N' -+ R" be orientation preserving continuous injections. Since is a connected manifold, there are a connected open set N" and an orientation preserving continuous injection 4" : N" -+ R" such that N n N" # 0 and N' n N" # 0. Applying the above result to 4, 4'' : N n N" -+ R" and I$',4" : N' n N" -+ R", we obtain that the number c does not depend on the choice of N and 4. To see that the number c does not depend on a point u in Mf, take and fix x E and choose a chart (V,h) at ~ ( zsuch ) that h:V 4 R" is an orientation preserving homeomorphism. Then, for some T > 0, B p ( T ( x ) ) c V where B r ( r ( z ) )= {W E A4 : d ( r ( z ) , w ) 5 T } and d is a metric for M. For y E and u E Mf let E,, : B8(y)x r ( y ; u ) -+ N(y;u) be a local product structure as in Theorem 6.6.5. Then we may suppose that r ( y ; u ) c zr(y)

m,

a

CHAPTER 10

352

for all y E li? and u E mj. Here z r ( y ) = { w E : j ( y , w ) 5 T } and is a metric for li? as in Theorem 6.4.1. By Theorem 6.6.5 (4) there is a constant p > 0 such that C m(y;u). Choose 6 > 0 such that h(B,(z)) 3 B6 where B6 = { z E R" : llh(x) - zll 5 6). Since V is evenly covered by x , we can take an open neighborhood U of x in li? such that the restriction x : U + V is a homeomorphism. Put 4 = h o x . Then 4 : U + R" is orientation preserving, because so is x : M --t M. Let u E &if. Since x is locally isometric, r ( z ; u ) c U. We note that r ( z ; u )is a coordinate domain with respect to and Fz (see Remark 6.7.5). Hence, TX is defined as in (10.13), and

zp(y)

N(z;u),9

+ ( r ( z u;)) = h 0 x ( p ( z ; u ) )= ~ ( N T , , (3= )h(Bp(x(2))) ) 3 B6 -

where T,, : M --t Mfdenotes the map defined in (6.5) of 56.5 and NTu(z) is a neighborhood of x ( z ) as in Theorem 5.2.1. Let STu(z): NTu(z) --t Mj be a continuous section satisfying the condition (B) of 55.2 and let aTu(z) : D"(x(x))x D"(T,,(Z))+ NTu(z) be a local product structure on M (see Theorem 5.2.1). Then we have for z E h(N,,(z))

TZ N ( z ; u ) , &= 4

m(=;u)(4-'(z), 1' - 4

4-'('>>

= h 0 % ( z ) ( h - ' ( 4 , ~ ( 4 )- h 0 %(,)(+),

=

.(')]

[sTu(Z)(h-'(z))9

-

[TU(z)9

h-'(z))

'-'(')]'

Let U,5p be the open ball of 4(z) with radius 6/2. Since B6 C 4(p(z;u)), we can choose E > 0 such that llTZ (.)I[ 2 E for all z E Ba - u,5/,. N(z;u),O

Let u' E mj and let k > 0 be an integer. Then we can find a covering transformation p k :li? + li? for x such that pi o ~ ~ (=2pi) o T,f(pk(z)) for i with lil I k, where pi : Mj t M is the projection to the i-th coordinate. Put yk = pk(z). Since p k is orientation preserving, clearly 40~;' : P ~ ( t u )IW" is an orientation preserving homeomorphism. By the fact that P k is an isometry is under 2, we have r(yk;u') C B r ( y k ) C Pk(U). Hence, TZk N(ur;uf),&p;l

defined and for z E h(NTuf(pk(z)))

T_" N(vh;uf),+op;l

=4

pl'

- 4 0 Pb' =

4-'('),Yk)

%(uk;Uf)(pk

aTuf(uh)(h-'(z)9

0

%(uk;uf)(Yk9Pk x(yk))

-

0

4-'(4)

'aru#(uk)(x(9k)? '

'-'('))

= [sTuf(uk)(h-'(z)),x(x)] [Tuf(Y&),h-'(z)]. Since B p ( x ( z ) c ) N z)flNTu,(uk), by the choice of 6 we have B6 c h(NT+,))n T ! h(NTu,(uh)). From thls and the fact that pi o T,,(z) = pi o Tuf(yk) for z with (i( 5 k, it follows that if z E B6 then for i with ti1 5 k pi

sTu(Z)(h-'(z))

= Pi

sTuf(,h)(h-'(z))'

810.9 Fixed point indices of TA-covering maps

353

Hence, by Lemma 5.1.9 (4) if k is large enough then for z E B6

we have

from which

Therefore, the conclusion was obtained. 0 We take an atlas { (U,g ) } of M such that { U;} is finite and each g: U + R" is a homeomorphism. Let SO > 0 be a Lebesgue number of { U } . By Theorem 5.2.1 we can construct a local product structure a, : D"(z0) x D"(x) + N, for each x = (z;) E Mf with the property that each diameter of N x is less C N , for than SO. By (3) of Theorem 5.2.1 there is po > 0 such that BpO(zo) all (zi) = x E Xf. Choose a finite open cover {V} of M such that each diameter of V is less than po and for any V there are an open set V' and a homeomorphism h: V' + R" so that V c V' and h ( V ) = { z E R" : 1 1 ~ 1 1< 1). We recall that such a V is canonical neighborhood (see Lemma 10.6.1). Let po > 61 > 0 be a Lebesgue number of {V}. Then we can find 0 < p1 < 61 such that for (z;) = x E Mf if y E Bpl(zO)then d ( p , 0 aX(y),zo) < 61 where p , : D " ( z ) x D"(x) + D"(x) is the projection. This is ensured by compactness of M. Next, we choose an atlas {(W, h)} of M with the property that each diameter of W is less than p l , {W} is finite, and each h: W + R" is a homeomorphism. Let p1 > y > 0 be a Lebesgue number of {W}. Then for 3 E M there is W E {W} such that z E B y ( z )C W C B P 1 ( z ) .

Choose P > 0 as in Theorem 5.2.6 and fix m 2 P. Let z E Fix(fm). Then there exists x = (zi)E Mf such that 20 = 3 and P ( x ) = x where

CHAPTER 10

354

u : M f + M f denote the shift map. As in Theorem 5.2.6 we define D"(z), D"(x), K " ( s )and K"(x), and write

I

D" = D"(z) = D"(x),

{ D"

K" = K " ( z ) K" = K"(x)

if there is no confusion. From Theorem 5.2.6 it follows that F,,, = f,'& :D" --f K" and F,,, =

fG.,: K"

+ D"

are both homeomorphisms. Since

D"

c Wedo(%)and K" c D"

C

W,uO(x)

where eo is an expansive constant for f, by Lemma 2.4.1 the point 2 is only one fixed point of Fa,, and Fu+. By Theorem 5.2.1 it is clear that D dand D" are EN%. See also Lemma 5.2.5 and Proposition 10.6.5. By Theorem 5.2.6 (s2) and (u2), K" is open in Do where u = s,u.

Lemma 10.9.4. I ( f m , z )= I(Fa,,) - I(F,,,,). Proof. From Theorem 5.2.6 (su) together with Property 10.4.6 we obtain the conclusion. 0

Lemma 10.9.5. For every 2 E Fix(f "), (1) W , X ) = qq:) = 1, (2) I(FU,,) i s either 1 or -1 and so is I ( F $ ) . Proof. (1): Let { ( W , h ) }be the atlas choosen above. Since -y is a Lebesgue number of {W}, there is (W,h)E { ( W , h ) }such that By(z)c W c Bpl(z)c N,. By Theorem 5.2.6 (sl) we have K" c B,(s). Define a homotopy Ht : D" + D", 0 5 t L: 1 , by

H&/)= p" 0 a,l

0

h--l(th 0 Fs+(y))

where p" : D" x D" + D" denotes the projection. Then HOis a constant map and H I = FS+. Since {y E D" : Ht(y) = y for some t E [0,1)} is contained in a compact set pa o a;'(Bpl(~)),by Property 10.4.5 we have I ( F " , , , z ) = 1. Also I(FQ5,z)= 1 because Fa,, and FQ; are mutually in dual relation. Since 2 is arbitrary in Fix(f"), (1) was obtained. (2): Let { (U, g)} and 60 > 0 be as above. Since 60 is a Lebesgue number of {U}and diam(N,) < 60, there is ( U , g ) E { ( U , g ) }such that N , c U.Write

11, = g o a,,

K = $(K" x K") and a = 11,(z,z).

Since 11, : D" x D" + R" is injective, it follows that K is open in definition of fixed point index we have

G,o ik;'(la) = .I(F:: x Fu,,)l0

R". By the

510.9 Fixed point indices of TA-covering maps

355

where

G = id - 11, o (F$ x Fu,z) 0 1,-' We consider a map GI= id - 11, o (Fa,= x

FG-) o 11,-'

: ( K ,K

- U ) -t (Wn,R" - 0).

: (g(N,),g(N,)

- a ) + (W",Wn - 0).

Fu,z) o 11,-l is a homeomorphism, we have G = ( o G1 o 11, o (3':; x FU+)0 4-l where C : R" -t W" is a map defined by ( ( z ) = - z . Hence Since 11, o (F;: x

= f1(F8,Z F 0 . Then all points of E return infinitely often to E under positive iteration by f.

u;="=,

Proof. For N > 0 let EN = f - " ( E ) . Clearly, E E EN) = EN+^, EN C EN-' C C EO and m ( E ~ + 1= ) m ( f - ' ( E ~ )= ) E EN). Therefore, EN) = m ( E o )for each N , and m(n;=, E N )= ~ ( E o )The . set EN = U:=N f - " ( E ) is the set of all points entering E infinitely often under positive iteration by f. Moreover, F = E n E N )consists of all points of E which enter E infinitely often under positive iterates o f f . Since EN C EO and both sets have the same measure,

n=;,

n= ;,

(n;=,

n= ;,

nE ~ ) ) Q)

m ( ~=)m ( n~(

= m ( n~E ~=)m ( ~ ) .

N=O

To see that a point of F returns to F infinitely often, let such that f " i ( x ) E is a sequence 0 < nl < n2 < show that fnl(z)E F. Since fni-"1(fnl(z))= f n i ( z E) enters E infinitely often under iterates of f f n3-"l, that f a , ( , ) E F. Similarly we can show f"'(z) E F for i.

z E F. Then there E for i. We first E for i > 0, f n l ( z ) - - . This implies 0

-

A transformation f: (X, B,m ) + (X, B,m) is ergodic if for B E B,f-'(B) = B then m ( B )= 0 or m ( B ) = 1. Theorem 11.2.3. Let f : X -+ X be a measure preservang transformation. Then the following are equivalent. (1) f is ergodic, ( 2 ) m(f-l(I3)A.B) = 0 for B E B + m ( B ) = 0 or 1, (3) For A , B E B with m(A),m(B)> 0 there is n > 0 such that m(f-"(B) A B ) > 0. Here AAB = ( A \ B)U ( B \ A ) .

Proof. (1) + (2) : Suppose m ( f - ' ( B ) A B ) = 0 and let B , = f - i ( B ) E B. Since

n,"==,u&

B, = ( Buf - l ( ~ u ) ...) n ( f - l ( ~ )u f - 2 ( ~ ) u - ..) n. .. , we have f-'(B,) = B, and m(B,) = m(B). Thus m(B,) = 0 or 1 and therefore m ( B ) = 0 or 1. (2) + (3) : Let m(A) > O,m(B) > 0 and suppose (3) is false. Since m ( f - " ( A ) n B) = 0 for all n > 0, we have m((U;==, !-"(A)) n B ) = 0. Let A' = U r = l f - n ( A ) . Then f-I(A') C A' and m(A') = m ( f - l ( A ' ) ) . Thus m(f-'(A')AA') = 0. By (2) we have m(A') = 0 or 1. But f - ' ( A ) c A' and f is measure preserving. Thus m(A') = 1. This contradicts the fact that m(A' n B) = 0.

$11.2 Measure preserving transformations

385

(3) + (1) : Suppose (1) is false, i.e. f - ' ( B ) = B and 0 < m ( B ) < 1 for some B E t?. Then m ( f - " ( B )n(X\B))= 0 for n > 0, which contradicts (3). 0

A characterization of ergodicity is given as follows. Theorem 11.2.4. Let f: ( X ,13, m ) (X, B,m ) be measure preserving. Then the following statements are equivalent. (1) f is ergodic. ( 2 ) Let ( be a measumble function of X. If ( o f(z)= ((2) for z E X, then ( is a constant a.e.. (3) Whenever ( i s measumble and ( 0 f(z)= f(z)a.e. then ( is a constant a. e.. (4) Whenevert E L 2 ( m )a n d ( o f ( z ) = ((3) f o r z E X then( is a constant a. e.. ( 5 ) Whenever f E L2(m) and ( o f(z)= ((3) a.e. then ( is a constant a.e.. Here L 2 ( m ) is a collection of square integmble functions. --f

Proof. (1)

=$

Then

(3) : Suppose ( is measurable and ( o f = ( 8.e.. Define

f-'(x(k, n))AX(k, ). c {. : € 0 f(z)# €(.)I

= 0. By Theorem 11.2.3(2) we have and thus m(f-'(X(k,n))AX(k,n)) m(X(k,n)) = 0 or 1. Fix n. Then UkEZX(k,n) = X which is a disjoint union. For each n there is a unique k, such that m ( X ( k , , n ) ) = 1. Let W = X(k,,n).Then m(Y)= 1 and ( is a constant on Y. Thus f is a constant 8.e.. It is clear that (3) =$ (2) =+ (4), (5) =$ (4) and (3) + (5). Thus it only remains to show that (4) + (1). Suppose f - ' ( E ) = E for E E 13. Then 1~ E L 2 ( m )and 1~ o f(z)= 1 ~ ( z ) p 0 or 1. 0 for z E X. By (4), 1~ is a constant a.e.. Thus m ( E ) = l ~ d =

Y

nn=,

Let X be a compact metric space and let m be a Bore1 probability measure of X which give positive measure to every nonempty open set. If f:X+ X is continuous, measure preserving and ergodic, then m ( { z : O f ( z )is dense in X})= 1 where O f ( z )= {fn(z) :n 2 0). This is checked as follows. Let {Un : 1 5 n < oo} be a base for the topology of X. O f ( z )is dense in X if and only if z E UEof-'((V,). Since f-'((Vn)) c UFOf - ' ( ( V n ) and f is mewure preserving and is nonempty ergodic, we have m(UEof-'(Un)) = 0 or 1. Since UFOf-'((V,,) f-'(Un)) = 1. open set, we have m(Ur=o

Remark 11.2.5.

f-'(Ur=o

CHAPTER 11

366

Remark 11.2.6. Let G be a compact metric abelian group and let A : G + G be an endomorphism. Then A is ergodic with respect to the Haar probability measure if and only if 7 o A n = 7 for some character 7 then 7 = 1. Remark 11.2.7. Let T" be the n-torus. Then an endomorphism A : T" + T" is ergodic with respect to the Haar probability measure if and only if a lift A : R" --t R" of A has no roots of unity as eigenvalues. The proof was given in Remark 1.1.4 of Chapter 1. Remark 11.2.8. Let m be the measure of Yf constructed as ( 4 ) of Remark 11.2.1. Then the shift map a : Yf + Yf is ergodic with respect to m. Let A be the a-algebra generated by finite unions of measurable rectangles. Suppose a-'(E) = E for E E B. Let E > 0 and choose A E A such that m ( E A A ) < E . Then

Im(E) - m(A)I = Im(E n A )

+ m ( E \ A ) - m ( A n E ) - m ( A \ E)I

< m ( E \ A ) + m ( A \ E ) < E.

Choose n so large that B = a-"(A) depends on different coordinates from A. Then m ( A r l B ) = m ( A ) m ( B )= T ~ ( A Since ) ~ . m(EAB)= ~ ( o - ~ ( E ) A o - " ( A ) ) = m ( E A A ) < E and since E A ( A n B ) c EAAUEAB, we have m ( E A ( A n B ) ) < 2 ~ Thus . Im(E) - m ( A n B)I < 2~ and

since m ( A ) 5 1 and m ( B ) 5 1. Since E is arbitrary, we have m ( E ) = m(E)2 and so m ( E ) = 0 or 1.

$11.3 Ergodic theorems

The purpose of this section is to introduce the major result proved in 1931 by G.D.Birkhoff [Birl]. Theorem 11.3.1 (Birkhoff ergodic theorem). Let f :(X, B, m) + (X, B, m ) be measure preserwing (we allow ( X , B , m ) to be a-finite) and ( E L'(m) where L'(m) i s the collection of all integmble functions. Then l / n C r i i ((f'z) converges a.e. t o ( * E L'(m), a n d [ * o f =(* a.e. a n d S ( * d m = S ( d m if m(X)< 00.

$11.3 Ergodic theorems

367

Remark 11.3.2. I f f is ergodic, then (* is a constant a.e. and if m(X) < 00

&

then (* = J(dm a.e.. We define the time mean of ( to be

. n-1 and the space mean of ( to be

If f is ergodic, then the ergodic theorem implies these means are equal.

As one of applications of Birkhoff ergodic theorem we have the following result. Let f: [0, 1) --t ( 0 , l ) be defined by f(z) = 22 mod 1. Then f preserves Lebesgue measure and is ergodic (by Remark 11.2.1(2) and 11.2.7). Theorem 11.3.3 (Borel’s theorem). Let m be Lebesgue measure of [O,l).

’.( n

the number of 1 ’ s in the first n digits of the binary expansion of x E [0, 1)

Proof.Suppose x f(x) =

+3

+ . e m .

=

}

--f

21 a.e..

9 + 8 + ... has Let

((2) =

a unique binary expansion. Then l p p l l ( x ) . Then

Thus the number of 1’s in the first n digits of the dyadic expansion of x is Cyit ((fx). Dividing both sides of this equality by n and applying Birkhoff ergodic theorem, we have

C

1 n-l lim ((fix)= n+m n i=O

J

1 l[l/z,l)dm= - ax.. 0 2

Let f : ( X ,U,m) + (X, f3,rn) be measure preserving and define an operator Ufon P ( m ) by

(UfO(4 = ( ( f x ) . Then Uf(LP(m)) = P ( m ) and, since f is measure preserving, IIUf(llp = I I ( I I p where II(IIp = CJ, ItIPdm)l/P. To show Birkhoff ergodic theorem we need the following lemma.

CHAPTER 11

368

Lemma 11.3.4 (Maximal ergodic theorem). Let U : L'(m) t L'(m) be a positive linear operator (i.e. 4 2 0 j U( 2 0 ) which has a norm 5 1. For N > 0 an integer define 40 (n

= 0, =4

+ U( + + Un-' 0) OsnsN

= FN(z).

Thus 4 2 FN - UFN on A = {x : FN(s) > 0 ) and

LEdm 2

1

FNdm -

A

J,UFNdm

=

FNdm -

UFNdm (since FN = 0 on x \ A )

2

FNdm -

UFNdm (since UFN 2 0 when FN 2 0)

2 0 ( since

IIuI~= SUP{IIU~II

t

: E ~ ' ( m I)ItII,1 = 1)

5 1). 0

Lemma 11.3.5. Let f: (X, B , m ) -+ ( X , B , m ) be measure preserving. If L 1 ( m ) and 1 n-l B, = {z E X : sup E(f"x) > a } , rill

4E

"=O

then

if f - ' ( A ) = A and m ( A ) < 00. Proof. We first prove this result under the assumptions m ( X ) < 00 and A = X. Ifq = 4-a, then we have B, = u= ;{,z : FN(z) > 0). Thus qdm 2 0

,,s

§ 11.3 Ergodic theorems

sB,

369

by Lemma 11.3.4. Therefore (dm 2 am(B,). In the general case, use in the place of f . Then we have that qdm 2 am(A n 23,). 0

fiA

SAnB

Proof of Theorem 11.3.1. For ( E L'(m) we set

Then we have (* o f = (* and (* o f = &. This follows from the fact that

-.f (T)-X((fi2!)n+ 1 1 -C((fi+'Z)= 1 n For p

n-l

n-l

i=o

i=o

(XI

< a let E,,p = {Z E X : &(z)

Then f-'(E,,p)

< /3 < Q < (*(z)}.

= E,,p and n-1

To use Lemma 11.3.5 we show that m(E,,p) < 00. Suppose a > 0 and let C C E,,p with m(C) < integrable and by Lemma 11.3.4

00.

Then q = ( - alc is

where HN is defined analogously to the function FN in Lemma 11.3.4. Since C U;=,{Z : H N ( z ) > 0}, we have Itldm 2 am(C). Therefore m ( C ) 5 a [([dm for every subset C of E,,p with finite measure and thus m(E,,p)
l

By Lemma 11.3.5

-p

CHAPTER 11

370

Replacing f,a and

/3 by -f,-/3 and -a respectively, we have that

(-(I* = --L

(-o*

= -t*,

J

tdm 5 pm(E,,p).

E- ,@

However, since a > p, we have m(EQ,p)= 0, and since

{. E

u

x : e k and n # J , then n # Jk+l and Ian1 < Therefore, limn+oo,ng~ lan[ = 0. ( 2 ) +- (1) : Suppose lan/ 5 K for n. Let E > 0. Then there is N, > 0 such that n 1 N , and n $! J imply lanl < E , and Me > 0 such that n 2 Me implies aJ(n) n < E . For n 2 max{N,, M e } ,

&.

. n-1

K

< -aJ(n)

(1) e

+

+

E < (K 1)~. n (3) : This follows from the fact that

Let f : (X,B,m ) + (X, B,m) be measure preserving on a probability space. Define a transformation f x f : X x X + X x X by (f x f )(c, y) = (f (c),f (y)). Then f x f is a measure preserving transformation on (X x X, l? x B,m x m).

CHAPTER 11

374

Theorem 11.3.11. I f f is measure preserving on m), then the following are equivalent. (1) f is weakly mixing. (2) f x f is ergodic. ( 3 ) f x f is weakly mixing. Proof. (1) + (3) : Let A, B E density zero such that

Q

probability space (X,B,

B and C, D E B. Then there are J1 and JZ of

lim m(f-"(A) n B) = m ( A ) m ( B ) ,

n-oo

n4J1

lim m(f-n(C) n D) = m(C)m(D).

n+m

n4 Ja

Then we have lim

n+oo n4J1U J a

=

m x m((f x f ) - " ( A x C ) n (Bx 0))

lim ;n ;

m ( f - " ( A )n B ) m ( f - " ( C )n 0)

%a

= m ( A ) m ( B ) m ( C ) m ( D= ) m x m(A x

C)m x m(C x 0)

The above relation holds for finite disjoint unions of rectangles, which form an algebra 3. Notice that 3 generates the a-algebra B x B. By Lemma 11.3.9 we have

. n-1

for A , B E 3,and the result holds for A , B E (3) =+ (2) : This is clear. (2) + (1) : Let A, B E B. Then we have 1 n-l

-

B x B.

1 n-l

C m ( f - i ( ~n)B ) = -n C m x m((f x / I - ~ ( A x x)n (Bx xi) . i=o

-

a=o

m x m ( A x X)m x m(B x X ) = m(A)m(B) and also 1 n-l 1 n-l C(m(f-'((A) n B ) ) 2= - C m x m((f n n i=O

x f ) - i ( A x A ) n ( B x B))

i=O

+ m x m ( A x A)" x m ( B x B )

=~ ( A ) ' W Z ( B ) ~ .

$1 1.4 Probability measures of compact metric spaces

. -+

375

n--1

2 n ~ ( A ) ~ m (-B2)m~( A ) ' ~ n . ( B = ) ~0 .

Therefore f is weakly mixing by Theorem 11.3.9. 511.4 Probability measures of compact metric spaces Let X be a compact metric space. A Borel probability measure p is said to be regular if for E > 0 and a Borel set B there exist a closed set C, and an open set U, such that C, c B C U, and p(U,\C,) < E . Theorem 11.4.1. Every Borel probability measure is regular. Proof. Let p be a Borel probability measure of X and define for E > 0 there are a closed set C, and an open set U, M= AED: such that C, c A c U, and p(LJ,\C,) < E

1

{

If all closed subsets of X are members of M and if M is a a-algebra, then we have M = t?,and therefore p is regular. We first show that M is a a-algebra. Since X is open and closed, we have that X E M. If A E M, for E > 0 we have C, C A C U, and p(U,\C,) < E as in definition. Thus, X\Ce 3 X\A 3 X\U, and (X\C,)\(X\U,) = U,\C,, and so p((X\C,)\(X\U,)) < E . This shows that X\A E M. It remains to show that if A l , A2, * * E M then An E M. Let E > 0. Then C,,n C An C U,,, and p(U,,n\C,,n) < ~ 1 for 2 certain ~ closed set C,,, and open set U,+. Let U, = U,,+, clearly U, is open. But C: = C,,,, is not closed. Choose k > 0 such that p(CL\ C,,,,) I €12 (by the following Lemma 11.4.2) and k define C, = C,,n. Then C, is closed and

Un

un

un

U:=,

Unzl

00

Ce c

U An C uc, k=l

and moreover

+ P(CL\G) I C p(uc,n\Cc,n) + P ( c ~ \ c ~ )

P(U€\CE>I P w C \ c : ) 00

n= 1 00

n= 1

CHAPTER 11

376

We have shown that M is a a-algebra. Finally we show that M contains all closed subsets of X. Let C be a closed subset of X and define a sequence of open sets Un by

Un = {X E X : ~ ( C , X 0 fix k > 0 such that Clearly, Ul 3 Uz 3 p(Uk\C) 5 E and put U, = u k and C, = C. Then C, C C C U, and so C E M. Therefore M contains all closed subsets. 0

L e m m a 11.4.2. Let p be a Bore1 probability measure of X and let All Az, E B. Then

-

* *

W

( 1 ) A1 C A2

c

===+ p (

U An) = lim p(An), n+m

n=l W

(2) A1 3 Az 3

p( n=l

An) = n+w lim p(An)-

Proof. (I) : Define a sequence of measurable sets Bn by B1 = Ai, Bn = An\An-l,n = 2 , 3 , - * * Then . An = Uy=1 Bi, U,"==,An = Ur=p=, Bn and {Bn) are pairwise disjoint. Thus W

n

m

W

u n

= lim p ( n+w

i=l

Bi) = lim p(A,). n+w

(2) : Put Bn = Al\A,,n = 1 , 2 , . . . . Then B1, Bz,... E Band A n n B n = 8 for 2 1. Since A1 = An U Bn, we have 1 2 p(An) p(Bn). Since A1 = An) U C U L Bn)i

(nr=P=,

+

511.4 Probability measures of compact metric spaces

377

Remark 11.4.3. Let p be a Borel probability measure of X. For any B E 23 we have

p ( B ) = sup{p(C) : C C B and C is closed }, p ( B ) = inf{p(U) : B C U and U is open }. This follows from Theorem 11.4.1. Theorem 11.4.4. Let p and u be Borel probability measures of X . If

for all ( E C ( X ) , then p ( B ) = u ( B ) for all B E B. Proof. If p ( C ) = u ( C ) for any closed subset C, by Remark 11.4.3 we have p ( B ) = u ( B ) for every B E B. Thus it sufficies to show that p ( C ) = v ( C ) for closed sets C. Since p is regular, for e > 0 there is an open set U such that C c U and p(U\C) < e. Define a continuous function ( : X + R by

Clearly, ((2) = 0 for z E X\U, t(z) = 1 for z E C and 0 I ((z) 5 1 for z E X. Since p ( U ) 5 p ( C ) E , we have

+

Since E is arbitrary, u ( C ) I p ( C ) . By replacing p by u we have p ( C ) I u ( C ) , and therefore p ( C ) = u ( C ) . 0

-.

Theorem 11.4.5. Let X be a compact metric space and let p1 ,pz, ,p be a sequence of Borel probability measures. Then the following conditions are equivalent :

(2)limsuppn(C) 5 p ( C ) ( closed set C c X ) , n+oo

(3)liminf p n ( U ) 2 p ( U ) ( open set U n+oo

cX),

(4) n+oo lim p n ( A )= p ( A ) ( p ( B A )= 0 where a A is the boundary of A ) .

Proof. (2)

(3) is clear.

CHAPTER 11

378

(1)

+ (2) : Let C be a closed set. For k 2 1 define = (2 E

u k

x : d ( z , C )< l/k}.

nk

Then u k 3 u k + 1 for k 2 1 and u k = c. Thus p ( u k ) --+ p ( c ) by Lemma 11.4.2(2). Since X is compact, we can find continuous functions (k satisfying Ek(2)

=

{ 01

Thus limsuppn(C)

I limsup n+m

n-m

Since k 2 1 is arbitrary, we have

ifzEC if 2 @ u k .

L L &dpn =

&dp I p(vk).

limsuppn(C) I p ( C ) . n+oo

(2), (3) + (4) : Let A E B and denote as int(A) the interior of A. If p(dA) = 0, then we have p(int(A)) = p(A) = p(cl(A)) and thus limsup pn(A) I lim sup pn(C1(A)) I p(cl(A)) n+m

n+m

pu(A),

1

liminf pn(A) 2 liminf p,(int(A)) 2 p(int(A)) = p(A). n+m

n+m

Therefore, limn+- pn(A) = p(A). (4) =+ (1) : For 5 E C ( X ) define m = min{t(z) - 1 : 2 E X } ,

M = max{[(z)

+ 1 :2 E X }

and v ( E ) = p ( ( - ' ( E ) ) for Borel sets E of the interval [m, MI. Notice that v is a Borel probability measure of [m, M] and the set A = { t E [m,M ] : v ( { t } )= 0) is dense in [m, MI. For E > 0 choose a finite sequence t o , t l , . ,t k such that each t j belong to A and

..

< t l < ... < t k = M , - t j - 1 < E (1 I j I 12).

m=to t j

For 1 0 by the assumption and so p(E:) ergodicity ensures that p ( { z E X : cl(Of(z)) = X}) = 1. 0

> 0 . The

388

CHAPTER 11

Remark 11.5.2. Let n(f) be the nonwandering set of a homeomorphism f of a compact metric space X. If f has POTP, then there exists a probability measure p E M f ( X )such that p has positive measures for all nonempty open sets. Let { U1,U2,* * * } be a basis of the relative topology of n(f). For Uj let V be a closed subset with the interior points such that Uj C V. For E such that where Uc(z)= {y E n(f) : d ( z , y ) < E } , take a sufficiently small V 3 Uc(z) 6(0 < 6 < E ) satisfying the property in the definition of POTP. Since n(f) is a nonwandering set, there exist z E n(f) and no > 0 such that z,f"O(z) E u6/2

Now, define an no-periodic 6-pseudo orbit {zj} by %nok+i = f'(z) for k E Z and 0 5 i 5 no - 1. Since fp(f): n(f) + a(f)has POTP, there is y E n(f) satisfying d(fj(y), zj) c E for all j E Z. For all m 2 1 define

where 6, denotes a Dirac measure, i.e. for a Borel set B

Then each p m is a Borel probability measure on n(f) such that pm(V) 2 pm(Uc(z)) 2 1/2no. Since M ( X ) is compact by Theorem 11.4.7, a suitable subsequence {pmi} converges to a Borel probability measure p ( j ) . By Theorem 11.4.5

and so p(j)(Uj) 2 p(j)(V) > 0. Next we show that p ( j ) is f-invariant. For any Borel set B we have

and similarly (11.3)

511.5 Applications to topological dynamics

389

Let ( E C(O(f)). If ( 2 0 then ( is uniformly approximated by an increasing sequence of simple functions n2"

where i+l

i

(0 Bn,i = {X E X : - < ((z)< F } 2n Bn,n2" = {z E : ((z) 2 n} (n 2 1).

5 i 5 n2" - I),

x

Thus we have n2"

Since 0 5 ((z)- &(z) 5 1/2n (z E O(f)) for sufficiently large n, by using (11.2) and (11.3)

Fix n and take the limit of the above inequality. Then

Since n is arbitrary, we have J'( o f d p ' j ) = S ( d p ( j ) . Similarly, J'( o f-ldp(j) = J' ( d p ( j ) . These two equality hold for general continuous function ( since it is expressed as ((z) = (+(z) - t-(z) where (+(z) = max{((z),O} and (-(z) = max{-((z),O}. Therefore we have p ( j ) ( f - ' ( B ) ) = p ( j ) ( B )= p ( j ) ( f ( B )for ) all Borel sets B. The measure p ( j ) is depend on the open set U j , and so we put 0 0 .

for Borel sets B. Then p is f-invariant and p ( V j ) > 0 for all Uj.

390

CHAPTER 11

Theorem 11.5.3. Let f:X + X be a homeomorphism of a compact metric space. Suppose X is connected and not one point. I f f is minimal, then f does not have POTP.

Proof. Let L be the diameter of X, i.e. L = max{d(x, y ) : x, y E X} and put E = L/3. I f f : X + X has POTP, by the following lemma, for x E X there exist y E X and k > 0 such that Cl(ofh(Y)) C Uc(x). Thus, X = d(Op(fj(y))). From connectedness and minimality we have cl(ofh(y)) = X. For instance, if k = 3, then Aj = cl(013(fj(y))),j = 0,1,2,are not pairwise disjoint because X is connected. Thus A0 n A1 # 0, from which we take a point z. Then z , f - ' ( z ) E A0 and so we have BOU f2(B0)= X where Bo = cl(Op(z)). Since BOn f 2 ( B o )# 0, we take w E BOn f a ( & ) and then U J , ~ - ~ ( U J E) Bo. Since O,S(UJ) c [Bo n f"B0)l u {r-1(UJ),f-4(w),f-7(w),...} c P o n f2(B0)lu f 2 ( B 0 )= f 2 ( B 0 ) ,we have d ( O p ( w ) ) = X. Since w E Bo = cl(Op(z)) C A. = c1(Of3(y)), consequently c1(Of3(y)) = X. But l 5 2.5 since Of3(y) c Uc(x), thus contradicting.

'::u

Lemma 11.5.4. If a homeomorphism f has POTP, given E > 0 and x E fl( f) there ezist y E X and k = k ( x , ~ )> 0 such that C l ( o f h ( y ) ) C Vc(x).

Proof. For ~ / > 2 0 let 6 > 0 be a number satisfying the definition of POTP. Take sufficiently small 6 such that 0 < 6 < E . Since x E fl( f ) , there exist z and k > 0 such that z , f k ( z )E U612(x). Construct a 12-periodic6-pseudo orbit { Z i : i E Z} by n E Z and 0 5 i 5 k. z,k+i = fi(z), Then d ( f i ( y ) , z i )< ~ / 3 , Ei Z for some point y E X. Since d ( f n k ( y ) , z n k ) < &/3, we have d(fnk(y),x) < 5 ~ / 6for n E Z, i.e. Cl(Ofh(Y)) C Uc(x). 0 Let f : X + X be a homeomorphism of a compact metric space. The homeomorphism f is said to be distal if for x , y E X inf{d(f"(x), fn(y)) : n E Z}= 0 then x = y. Obviously an isometry is distal. Let A be a subset of Z. Then A is syndetic if there is a finite set K of Z with Z = K A. Let x E X, then x is an almost periodic point if { n E Z : f"(x) E U} is a syndetic set for all neighborhood U of x.

+

Theorem 11.5.5 (Aoki[Ao8]). Let f:X+ X be a homeomorphism of a compact metric space. Suppose X is not one point. Iff i s distal, then f does not have POTP.

Proof. Suppose f has POTP and let diam(X) = L. Let E = L/9. By Lemma 11.5.4,for yo E f l ( f ) there exist y E X and k > 0 with Cl(ofh(y)) c UC(yo). We write g = f for simplicity. Note that g: X + X has POTP and is distal. Let 0 < 6 < E be a number in the definition of P O T P for 9 . Since X is

511.5 Applications to topological dynamics

391

--

compact and connected, there is a finite sequence pl = y, p2, , p such ~ that Ua(pi). By the next Theorem d(Pi,Pi+l) < 6/2(1 5 i 5 N - 1) and = 11.5.7 each point z E X is almost periodic. Thus there exist c(i), 1 5 i 5 N, with d(pi,g"("(pi)) < 6/2, and SO we define a sequence {zi : i E Z} by

x uEl

Obviously {xi} is a 6-pseudo orbit and is &dense in X. Since g has POTP, there is E E X with d(gi(z>,z;) < E for all i E Z, and so for w E X there is w' E Og(z) with d(w,w') < 2 ~ For . simplicity write

c = c(1)

+2

c

N-1

c(i)

+ c(N).

i= 2

Then we have

Thus

Letting 0; = {gi(z) : i 5 0) and 0: = {gi(z) : i 2 0}, we have g(0;) c U,(cl(Og(y))) and gc(O$) C Ue(Cl(Og(y))). Since cl({gi(z) : i 6 Z}) = 0; U O$, by Baire theorem one of O,,O$ has interior points in 0, = cl({gi(z) : i E Z}). Since g:O, + 0, is a homeomorphism, there is a g-invariant Bore1 probability measure p on 0,.Since 0, is a minimal set for g by the next

CHAPTER 11

392

Theorem 11.5.8, we have that p ( U ) > 0 for any nonempty open set U. This g'(U) = 0, and p ( 0 , ) 5 p(U). follows from the facts that If 0; contains interior points of 0,, then 0; = g ( 0 ; ) . Indeed, suppose g-'(O;) # 0;. For the case when V = r)k>lg-k(O;) has interior points of 0, we have g - i ( z ) E V for some j 2 0. Thus 0; C V since g ( V ) c V . Therefore, g-'(O;) = O;, a contradiction. For the case when V has no interior points we have that p(O;\V) > 0 since O,\V is nonempty and open. Since 0; = U k >-O g-'((W) u V , we have p ( ~ >) 0, Put w = O;\g-'(O;). thus contradicting p ( 0 ; ) 5 1. If O$ contains interior points of 0,, we have also g(Of) = 0:. In any case, 0, = 0: or 0, = 0;.Thus 0, c Uc(cl(Og(y))). Since cl(O,(y)) C Uc(y0),

zrrn

urm

0,c UC(Cl(O,(Y))) c

Uzc(Y0).

Since {xi} is &dense in X and d(g'(a), z i ) < e(i E Z), we have max{d(O,,z) :t E X} 5 2~ and thus X = U2,(0,) = U4c(g~). Therefore, diam(X) = I dicting. 0

0 there is a syndetic set A such that fn(z) E Uc(t)for all z E X and n E A.

Theorem 11.5.6 (Ellis[El]). Iff:X f is distal.

--f

X is uniformly almost periodic, then

Proof. We first prove that i f f is uniformly almost periodic then {f" : n E Z} is equi-continuous. Let A be a syndetic set such that fn(z) E UCls(z) for all z E X and n E A. Then Z = A K for some finite set K = {n1,--,nt}.

Take 0

< 6 < €13 such that d(z,y) < 6

+

* max{d(f"'(z),

f"i(y)) : 1 5 i 5 1 )
>

E -. 3

< 6 then

5 4 f " 0 f Y Z ) , f m(z)+ > 4 f "(z),f "b)) + 4 f "(Y), f 0 f "(Y)) < €13 + €13 + €13 = E

and so { f " : n E Z} is equi-continuous. Suppose inf{d(fn(z),fn(y)) : n E Z} = 0. For j > 0 there is nj E Z with d(fnj(z), f"j(y)) < l/j. Thus d(f"j(z), fnj(y)) < 6 for all j large than a certain positive integer J. Thus we have

f "(Y)) : n E Z} sup{d(f n(zc>,

< E,

$11.5 Applications to topological dynamics

from which d ( x , y ) z=y. 0

< E . Since

E

393

is arbitrary, we have d ( z , y ) = 0 and then

Let g: X --t X be a continuous map of a compact metric space. The collection { g ( x ) : z E X } will be considered to be a point of the product topological space X x = n z E x X z where X, = X for x E X . Obviously X x is a compact Hausdorff space. Let f: X + X be a homeomorphism. The closure of {f" : n E Z} in X x is said to be an Ellis group and denotes it as E ( X ) . A nonempty subset I of E ( X ) is called a right ideal if g o h E I for all g E E ( X ) and all h E I .

Theorem 11.6.7 (Ellis[El]). If a homeomorphism f : X each x E X is an almost periodic point.

--t

X is distal, then

Proof. Before starting with the proof we prepare claims. Claim 1. Let I C E ( X ) and I # 8. If 1 is a right ideal of E ( X ) , then there is u E I such that u2 = u. Indeed, a map Rg : E ( X ) + E ( X ) defined by R g ( h )= h o g is continuous. This follows from the fact that if a discrete sequence { h a } of E ( X ) converges to h E E ( X ) , then h , ( g x ) = h ( g z ) for all x E X and thus Rg(ha)--t Rg(h). Let M denote the family of closed nonempty subsets S c I with S o S c S. Then M is the partially ordered set under the inclusion and is inductive. Thus a minimal element S exists in M by Zorn's lemma. If g E S, then S o g = Rg(S)# 8 and ( S o g ) o ( S o g )C S o S o S o g C S o g . Since S is minimal, we have S o g = S and then there is h E S with hog = g . Obviously R;l (s)ns # 8. Sinc e f l og= f i o g = g f o r f l , f 2 E R;1(S)nS,wehave(flof2)og= flog=g and so f10 f 2 E R ; l ( g ) n S . Therefore, ( R , l ( g ) n s ) o ( R , ' ( g > n s ) c R ; l ( g ) n S . Minimality ensures that R;'(g) n S = S. This shows that g2 = g . Claim 2. Let M be a closed subset of E(X).If f" o g E M for g E M and n E Z, then M = E ( X ) . Indeed, E ( X ) o M C M , by Claim 1 there is g E M with g 2 = g . P u t y = g ( x ) . Then g ( x ) = g ( y ) . Since g E E(X), we have that f" --t g for some sequence {f"} and

Since g ( x ) = g ( y ) , obviously d ( f n ( x ) , f n ( y ) )+ 0 as n + 00. Thus inf{d(f"( x ) , fm(y)) : m E Z} = 0. Since f : X + X is distal, we have x = y and then g is the identity since x is arbitrary. Therefore, f " o i d = f " E M for all n E Z. This shows that M = E ( X ) . Now we are ready to prove the theorem. Note that the identity i d is contained in E ( X ) . Let U be an open neighborhood of id in E ( X ) . We first show that there is a syndetic set A c Z such that f" E U for all n E A. Fix h E E ( X ) and define a map L h : E ( X ) + E ( X ) by L h ( g ) = h o g . It is easily checked that if h is continuous then so is L h . Thus L p is continuous for n E Z,and L f mo Lf-n = Lf-" o Lfm = i d , from which Lfn is injective

CHAPTER 11

394

and bicontinuous. Thus L p ( U ) = f " o U is open in E ( X ) and so W = E(X)\(U:f" o U )is closed in E ( X ) . If W # 0, then we have f " o g E W for g E W and n E Z. Thus W = E ( X ) by Claim 2, which contradicts W # E ( X ) . Therefore, W = 0, from which E(X)c U a: f" o U. Since E ( X ) is compact, we have E ( X ) C fni o U for some finite set K = {nl, ,nt}. Let A = {n E Z: f" E U}.Then there is ni E K with f" E f " i o U for all n E Z. Thus, f"-"; E U,i.e. n - n; E A and so n E A K. Since n is arbitrary, we have Z= K A and then A is a syndetic set. To show that each x E X is an almost periodic point define a continuous map 0, : E(X)+ X by 0 , ( g ) = g(z). Let V be a neighborhood of z in X. Then there is a neighborhood U of id in E ( X ) such that 0,(U)c V. Replacing this U by the above U,for n E A

ut,

+

+

and therefore x is an almost periodic point of f. 0

Theorem 11.5.8. I f f :X is a minimal set.

+X

is distal, for each x E X , the closure of Of(.)

Proof. By Theorem 11.5.7 each x E X is an almost periodic point. For U an open neighborhood of x define A = {n E Z: f " ( x ) E U}.Thus Zis expressed as Z= A K for some finite set K = {nl, ,nk}. Thus we have

+

k

Of (x)= u { f n + " i ( x ) : n E A } i=O

from which cl(Of(y)) n cl(U) # 0 for all y E cl(Of(x)). Here the integer no implies no = 0. This follows from the facts that f - " ; ( y ) E cl({fn(z) : n E A}) c cl(U) since y E cl{f"+"i(x) : n E A} for some integer ni. Since U is arbitrary, we have z E cl(Of(y)) and so cl(Of(x)) = cl(Of(y)), i.e. cl(Of(x)) is a minimal set. 0 Let f:X + X be a homeomorphism of a compact metric space. The homeomorphism f is said to be semisimple if there exists a sequence {En}of closed subsets such that f ( E a ) = E,,X = U E , and each E, is a minimal set. Obviously, E , f l Ep = 0 if a # p.

Theorem 11.5.9. If f : X + X is distal, then f is semisimple. Proof. {cl(Of(x)) : x E X} is a cover of X. By Theorem 11.5.8, each cl(Of(x)) is minimal. Thus we have the conclusion. 0 Let f:X X be a homeomorphism of a compact metric space. The homeomorphism f is said to be uniquely ergodic if M f ( X )is one point set. Obviously uniquely ergodicity implies ergodicity. --f

$11.5 Applications to topological dynamics

395

Theorem 11.5.10. Let f : X + X be a uniquely ergodic homeomorphism. Then f is minimal i f and only i f the f-invariant Borel probability measure p has positive measure for all nonempty open sets. Proof.

+ ) : Suppose f is minimal and p ( U ) = 0 for some nonempty open set

= Ura fn(U), we have p ( X ) = 0, thus contradicting. : Suppose f is not minimal, i.e. f ( F ) = F # X for some closed

U. Since X

-+ )

set F # 0. By Krylov-Bogolioubov theorem, f IF-invariant Borel probability measure p~ exists. Define p'(B) = ~ F ( nFB ) for Borel sets B of X . Then p' is f-invariant Borel probability measure and p'(X\F) = 0 holds. Thus p # p'. Since f is uniquely ergodic, we have a contradiction. 0

Theorem 11.5.11 (Oxtoby[Ox]). Let f : X X be a homeomorphism of a compact metric space. Then the following conditions are equivalent : (I) f is uniquely ergodic. ( 2 ) For each ( E C ( X ) the sequence ( o f j converges uniformly on X to a constant. ( 3 ) For any x E X there ezists p E M f ( X ) such that x is generic with respect to p .

cy=i'

Proof. (1) =$ (2) : Suppose that (2) is false, i.e. there exist ( E C ( X ) and > 0 such that for any N > 0 there are n 2 N and xn E X so that

E

By ( l ) , p is only one f-invariant. Fix the point xn and define a linear map Jn : C ( X ) + R by

-

n-1

By Ftiesz Representation Theorem Jn(q) =

J,qdpn

(7 E

c(x))

for some pn E M ( X ) . Since M ( X ) is compact, there is a subsequence { p n i } and v E M ( X ) such that

For any

E C(X)

CHAPTER 11

396

Thus we have

I)

I)

Y o f-'= u by Theorem 11.4.4. Similarly, u o f = u and therefore v E M f ( X ) . However,

from which

which shows that p # u. This is a contradiction. (2) + (3) : Since the constant referred to the statement of (2) is J t d p for some p E M f ( X ) ,(3) is concluded.

--- ./' /

/

/'

Figure 43

(3) + (1) : By the assumption, for 2 E X there is p E M f ( X )such that

for all

< E C ( X ) . Thus, for any u E M f ( X )

By Lebesgue Convergence Theorem we have

for all ( E C ( X ) ,and therefore u = p. This shows that M f ( X )is one point set. 0

$11.5 Applications to topological dynamics

397

Let f : X + X be a homeomorphism of a compact metric space. The homeomorphism f is said to have specification if for any E > 0 there exists M = M ( E )> 0 such that for any finite sequence of points 21,22,"'

,xk

E x,

and for j with 2 5 j 5 It, choosing any sequence of integers a1

I bl < a2 5 b2 c ..-< ak L b k ,

such that aj - bj-1 2 M ( 2 2 j 5 I t ) and an integer p with p 2 M there exists a point c E X with f P ( z ) = z so that

+(bk - al),

d ( f i ( c ) ,f i ( x j ) )< B

for a, 5 i 5 bj and 1 5 j 5 k. If a point c E X is a periodic point with period p, then pE= Czi: 6p(.) is an f-invariant Bore1 probability measure and f is ergodic with respect to pE. We denote as E( f) the set of all measures pz constructed as above. Obviously every element of E ( f ) is an extremal point in M j ( X ) by Theorem 11.4.11.

Theorem 11.5.12 (Sigmund[Sig2]). Let f : X + X be a homeomorphism of a compact metric space. Iff is a topologically mixing TA-homeomorphism, then E ( f ) is dense in M j ( X ) . Proof. Let p E M f ( X )be given. Any neighborhood of p contains a set of the form

where F is some finite subset of C ( X ) . We may assume 11(11 5 1 for all ( E F. Thus we must show that W ( p )contains an element of E ( f ) . If N is large enough, for E F and e E Q(f) (the set of quasiregular points) we have