%THIS IS A FILE WRITTEN IN AMSTEX 2.1 %DEFINITIONS %here I redifine "\enddemo" \predefine\enddimost{\enddemo} \redefine\enddemo {\penalty 5000\hskip15pt plus1pt minus5pt\penalty1000 \qed\enddimost} \catcode@=11 \font\s@=cmss10 \predefine\reall{\Re} \def\Id{{\hbox{\s@ 1\kern-.8mm l}}} %this is the identity for matrices \redefine\Re{\Bbb R} \def\N{\Bbb N} \def\Z{\Bbb Z} \def\To{\Bbb T} %here are some usefull def \define\RT {\widetilde{\Cal S}} \define\RB {\Cal S_\delta} \define\TB {\widetilde{\Cal S}_\delta} \define\len{\text{\rm length\,}} \define\lenq{\text{{\rm length}$\!{}_{\Cal Q}\,$}} \define\lvp{\left\langle} \define\rvp{\right\rangle} %here are large angle braket \define\pp{\delta p} \define\qq{\delta q} \define\ph{\varphi} \define\Q{\Cal Q} \define\G{\Cal R} %collection of squares \define\C{\Cal C} \define\R{\Cal S} \define\U{\Cal U} %this is the name of the neigh. of x \define\M{\Cal M} \define\T{\Cal T} \define\Su{\Cal S} %the right part of the L^1 norm \define\Lon{\biggr\|_{\lower 2pt\hbox{$\scriptstyle 1$}}} \define\lon{\bigr\|_{\lower 2pt\hbox{$\scriptstyle 1$}}} %\define\subh #1{\vskip2cm\penalty-1000\subhead % #1 \endsubhead\penalty 2000\vskip1cm\penalty 2000} %This file is the "master" file (although you can call it as you %wish) %It contains the various definitions and it uses the files that %contain the different sections as input. %you may have to modify the next line (%\input amstex) according to %your local system. % HERE WE GO ! %\input amstex \documentstyle {amsppt} \magnification \magstep1 \openup3\jot \NoBlackBoxes \pageno=1 \hsize 6 truein %this file must be called "abstract.tex" \topmatter \title ERGODICITY IN\\ HAMILTONIAN SYSTEMS. \endtitle \author Carlangelo Liverani,\ Maciej Wojtkowski \endauthor \address Liverani Carlangelo, Mathematics Department, University of Rome II, Tor Vergata, Rome, Italy. \endaddress \email liverani@vaxtvm.infn.it \endemail \address Maciej Wojtkowski, Department of Mathematics, University of Arizona, Tucson, AZ 85721, USA. \endaddress \email maciejw@math.arizona.edu \endemail \date 28-7-92 \enddate \abstract We discuss the Sinai method of proving ergodicity of a discontinuous Hamiltonian system with (non-uniform) hyperbolic behavior. \endabstract \thanks \bf We would like to thank N. Chernov, L. Chierchia, V. Donnay, A. Katok, N. Sim\'anyi, D. Sz\'asz and L.-S. Young for helpful and enlightening discussions. The first author wishes to thank the Mathematics Department of the University of Arizona, Tucson and the Center for Applied Mathematics at Cornell University, in particular its director J. Guckenheimer, where he was visiting during part of this work, he also acknowledges the partial support received by CNR, grant n. 203.01.52, and by the GNFM. The second author gratefully acknowledges the hospitality of Forschungsinstitut f\"ur Mathematik at ETH Z\"urich, where the first draft of this paper was written. He also acknowledges the partial support from NSF Grant DMS--9017993. \endthanks \endtopmatter \document %this file must be called "table.tex" \vskip.5cm \centerline{\bf CONTENT} \vskip.5cm \roster \item"0." Introduction\dotfill p. \ 2 \item"1." A Model Problem\dotfill p. \ 3 \item"2." The Sinai Method\dotfill p. \ 9 \item"3." Proof of Sinai Theorem (for linear discontinuous maps of the $2$-torus)\dotfill p. 14 \item"4." Sectors in linear symplectic spaces\dotfill p. 19 \item"5." The space of Lagrangian subspaces contained in a sector\dotfill p. 22 \item"6." Unbounded Sequences of Monotone Maps\dotfill p. 26 \item"7." Properties of the system and the formulation of the results \dotfill p. 34 \item"8." Construction of the neighborhood and the coordinate system \dotfill p. 43 \item"9." Unstable manifolds in the neighborhood $\U$\dotfill p. 46 \item"10." Local ergodicity in the smooth case\dotfill p. 51 \item"11." Local ergodicity in the discontinuous case\dotfill p. 54 \item"12." Proof of Sinai Theorem\dotfill p. 58 \item"13." Tail Bound'\dotfill p. 62 \item"14." Applications\dotfill p. 67 \item" " References\dotfill p. 76 \endroster %this file must be called "fs0.tex" \vskip.7cm \subhead \S 0. Introduction \endsubhead \vskip.4cm The notion of ergodicity was introduced by Boltzman as a property satisfied by a Hamiltonian flow on its energy manifold. The emergence of the KAM (Kolmogorov-Arnold-Moser) theory of quasiperiodic motions made it clear that very few Hamiltonian systems are actually ergodic. Moreover, those systems which seem to be ergodic do not lend themselves easily to rigorous methods. Ergodicity is a rather weak property in the hierarchy of stochastic behavior of a dynamical system. The study of strong properties (mixing, K-property and Bernoulliness) in smooth dynamical systems began from the geodesic flows on surfaces of negative curvature. In particular, Hopf \cite {H} invented a method of proving ergodicity, using horocycles, which turned out to be so versatile that it endured a lot of generalizations. It was developed by Anosov and Sinai \cite {AS} and applied to Anosov systems with a smooth invariant measure. With the advances of the theory of Kolmogorov - Sinai entropy the Hopf method turned out to be also a basis for proving the K-property of Anosov systems. The key role in this approach is played by the hyperbolic behavior in a dynamical system. By the hyperbolic behavior we mean the property of exponential divergence of nearby orbits. In the strongest form it is present in Anosov systems and Smale systems. It leads there to a rigid topological behavior. In weaker forms it seems to be a common phenomenon. In his pioneering work on billiard systems Sinai \cite{S} showed that already weak hyperbolic properties are sufficient to establish the strong mixing properties. Even the discontinuity of the system can be accommodated. The Multiplicative Ergodic Theorem of Oseledets \cite{O} makes Lyapunov exponents a natural tool to describe the hyperbolic behavior of a dynamical system with a smooth invariant measure. Pesin \cite {P} made the nonvanishing of Lyapunov exponents the starting point for the study of hyperbolic behavior. He showed that, if a diffeomorphism preserving a smooth measure has only nonvanishing Lyapunov exponents, then it has at most countably many ergodic components and (roughly speaking) on each component it has the Bernoulli property. Pesin's work raised the question of sufficient conditions for ergodicity or, more modestly, for the openness (modulo sets of measure zero) of the ergodic components. In his work, spanning two decades, on the system of colliding balls (gas of hard balls) Sinai developed a method of proving (local) ergodicity in discontinuous systems with nonuniform hyperbolic behavior. We will refer to it as the Sinai method. It was improved by Sinai and Chernov \cite {CS} and by A.Kr\' amli, N.Sim\' anyi and D.Sz\' asz \cite {KSS}. In both papers the discussion is confined to the realm of semidispersing billiards. The purpose of the present paper is to recover the Sinai method as a part of the theory of hyperbolic dynamical systems. In the process we have simplified some of the aspects of the method, and we have revealed its logical structure and limitations. We rely on two developments. The first is the work of Katok and Strelcyn \cite {KS} in which they generalized Pesin Theory to discontinuous systems. The other is the development of criteria for nonvanishing of Lyapunov exponents in Hamiltonian systems in papers \cite {W1}, \cite {W2} and \cite{W3}. In the language of these criteria Burns and Gerber \cite {BG} found a sufficient condition for (local) ergodicity in the smooth case of lowest dimension (3 for flows preserving a smooth measure). It was later generalized by Katok \cite {K1} to arbitrary dimension. As a byproduct of our general approach, which includes discontinuous systems, we obtain a similar theorem (Main Theorem in the smooth case) and a new proof. Let us give some advice to the reader on how to use our paper. The first three Sections demonstrate what the Sinai method is and how it works. The discussion is conducted in the simplest possible environment of a linear discontinuous system on the two dimensional torus. It is reasonable to stop here, especially if the reader is only interested in two dimensional uniformly hyperbolic systems. But we do not recommend trying to read the heart of the paper without going through the first three Sections. In Sections 4,5 and 6 we develop the linear symplectic language in which we formulate our results. We suggest that the reader skips these sections and goes straight to Section 7 where we formulate the multitude of hypotheses and the two Main Theorems on local ergodicity, one for smooth systems and the other (much harder) for discontinuous systems. The reading of Section 7, and the following Sections, will require numerous trips back to Sections 4-6 for the necessary definitions and theorems. If the reader does not care about the discontinuous case, she needs to read only Sections 8, 9 and 10 with significant leaps (since everything is simpler in the smooth case). Sections 11 and 12 contain almost the whole proof of the Main Theorem in the discontinuous case (it also relies on the results of Sections 8-10). The remaining part of the proof is contained in Section 13. It stands out by the level of technical complications. Section 14 contains some classes of examples where all the hard work can be put to use, and one class where it cannot. The interest in this last example comes from the fact that it is multidimensional and all the Lyapunov exponents are different from zero. Unfortunately, it does not satisfy an important property (proper alignment of singularity sets). It points towards the need for a more flexible scheme. %this file should be called s1.tex \vskip.7cm \subhead \S 1. A MODEL PROBLEM \endsubhead \vskip.4cm We will discuss here a very simple model problem in which the important features of the Sinai's method are not obscured by technical details. Our discussion will be very careful so that in the future when the technical details will cloud the horizon we will be able to refer the reader to these basic clarifications. We consider a family of linear maps of the plane defined by \aligned x_1' &= x_1 + a x_2 \\ x_2' &= x_2 , \endaligned where $a$ is a real parameter. We use these linear maps to define (discontinuous) maps of the torus by restricting the formulas to the strip $\{ 0 \leq x_2 \leq 1 \}$ and further taking them modulo 1. In this way we define a mapping $T_1$ of the torus $\To ^2 = \Re^2/\Z^2$ which is discontinuous on the circle $\{ x_2 \in \Z \}$ (except when $a$ is equal to an integer) and preserves the Lebesgue measure $\mu$. Similarly we define another family of maps depending on the same parameter $a$ by restricting the formulas \aligned x_1' &= x_1 \\ x_2' &= a x_1 + x_2 \endaligned to the strip $\{ 0 \leq x_1 \leq 1 \}$ and then taking them modulo 1. Thus for each $a$ we get a mapping $T_2$ of the torus which is discontinuous on the circle $\{ x_1 \in \Z \}$ (except when $a$ is equal to an integer) and preserves the Lebesgue measure $\mu$. Finally we introduce the composition of these maps $T = T_2 T_1$ which depends on one real parameter $a$. An alternative way of describing the map $T$ is by introducing two fundamental domains for the torus $\M^+ = \{ 0 \leq x_1 + a x_2 \leq 1,\, 0 \leq x_2 \leq 1 \}$ and $\M^- = \{ 0 \leq x_1 \leq 1, \,0 \leq - a x_1 + x_2 \leq 1,\, \}$ (see Fig.1). \topinsert \vskip 3in \hsize=4.5in \raggedright \noindent{\bf Figure 1} The map. \endinsert The linear map defined by the matrix $$\left(\matrix 1 & a \\ a & 1 + a^2\endmatrix\right)= \left(\matrix 1 & 0 \\ a& 1 \endmatrix\right) \left(\matrix 1 & a \\ 0 & 1 \endmatrix\right)$$ takes $\M^+$ onto $\M^-$ thus defining a map of the torus which is discontinuous at most on the boundary of $\M^+$ and preserves the Lebesgue measure. This is our map $T$. Let $\Cal S^{\pm}= \partial\M^{\pm}$ be the boundary of $\M^{\pm}$. Except for integer values of $a$ the mapping $T$ is discontinuous on $\Cal S^+$ and its inverse $T^{-1}$ is discontinuous on $\Cal S^-$. Let us stress that the map $T$ is well defined in the closed domain $\M^+$ but two different points on the boundary $\Cal S^+$ which correspond to the {\it same} point on the torus will be mapped onto two different points on the boundary $\Cal S^-$ which correspond to two {\it different} points on the torus (except for the corner). We adopt the convention that the image under $T$ of a point from $\Cal S^+$ is the pair of image points in $\Cal S^-$. With this convention we can apply $T$ or any of its powers to any subset in the torus. For integer values of $a \neq 0$ we have a hyperbolic algebraic automorphism of the torus, a prime example of an Anosov system. It is thus a Bernoulli system and has a nice Markov partition \cite{AW}. We restrict ourselves to the study of ergodicity and we repeat the proof of ergodicity by the Hopf method, since the Sinai method is built upon it. Let $f:\To ^2 \to \Re$ be a continuous function. We want to prove that for almost every $x\in \To ^2$ the time averages $$\frac{f(x) + f(T x) + \dots + f(T^{n-1}x)}{n}$$ converge as $n \rightarrow +\infty$ to the average value of $f$, i.e., $\int f d\mu$. Once this is established one can obtain the same property for all integrable functions by an approximation argument. >From Birkhoff Ergodic Theorem (BET) we know that the time averages converge almost everywhere to a function $f^+ \in L^1(\To ^2,\,\mu)$ which is invariant on the orbits of $T$, i.e., $f^+\circ T = f^+$, and has the same average value as $f$, i.e., $\int f^+ d\mu = \int f d\mu$. Further applying BET to $f$ and $T^{-1}$ we obtain that the time averages in the past $$\frac{f(x) + f(T^{-1} x) + \dots + f(T^{-n+1}x)}{n}$$ converge almost everywhere as $n \rightarrow +\infty$ to $f^- \in L^1(\To ^2,\,\mu)$ for which $f^-\circ T = f^-$ and $\int f^- d\mu = \int f d\mu$. It is the usual magic of the ergodic theory which forces the functions $f^+$ and $f^-$ to coincide almost everywhere. (Let us recall the argument: let $$\Cal A_+=\{x\in\To^2\;|\;f_+(x)> f_-(x)\};$$ by definition $\Cal A_+$ is an invariant set, hence $$\int_{\Cal A_+}\left [f_+(x)-f_-(x)\right ]d\mu(x)= \int_{\Cal A_+}f(x)d\mu(x)-\int_{\Cal A_+}f(x)d\mu(x)=0$$ which implies $\mu(\Cal A_+)=0$ and $f_+\leq f_-$ $\mu$-almost everywhere. The same argument, this time applied to the set $\Cal A_-=\{x\in\To^2\;|\;f_-(x)> f_+(x)\}$, implies the converse inequality.) For $a \neq 0$ the matrix $$\left(\matrix 1 & a \\ a & 1 + a^2\endmatrix\right)$$ is a hyperbolic matrix with eigenvalues $\lambda = \lambda(a) > 1$ and $\frac1\lambda < 1$. For $x\in \To^2$ let us denote by $W^u(x)$ $(W^s(x))$ the line in $\To^2$ passing through $x$ and having the direction of the unstable eigenvector (the stable eigenvector), i.e., the eigenvector with eigenvalue $\lambda$ ($\frac1\lambda$). We call $W^u(x)$ $(W^s(x))$ the unstable (stable) leaf of $x$. The leaves of $x$ have the following property. If $y\in W^{u}(x)$ $(y\in W^s(x))$ then the distance $$d(T^ny,\,T^nx) = \lambda^{-|n|} d(y,\,x) \rightarrow 0 \ \ \text{as} \ \ n \rightarrow -\infty (+\infty).$$ Hence for $y,z \in W^{u(s)}(x)$ $$|f(T^ny) - f(T^nz) | \rightarrow 0 \ \ \text{as} \ \ n \rightarrow -\infty (+\infty).$$ It follows that for $y,z \in W^{u(s)}(x)$ either $f^{\pm}(y)$ and $f^{\pm}(z)$ are both defined and equal or they are both undefined. Lifting the functions $f^+$ and $f^-$ to $\Re^2$ and using the directions of the eigenvalues as coordinate directions we can say that $f^+$ is a function of one coordinate alone and $f^-$ is a function of only the other coordinate. Since the two functions coincide almost everywhere they must be constant. \par Let us examine what can be saved of this argument when $a$ is not an integer. In such a case, we still have the stable and unstable directions but a line parallel to, say, the unstable direction is cut by $\Cal S^-$ into pieces and if $y$ and $z$ belong to two different pieces the distance $d(T^ny,\,T^nz)$ does not decrease to zero as $n \rightarrow -\infty$. Since this last property is of crucial importance in the Hopf method, the unstable (and stable) leaves have to be much shorter than before. Here is how we construct them. For simplicity of notation we will formulate everything for the unstable leaves alone. \par We proceed inductively. Thus, for $x \in int \M^-$, we define $W_1^u(x)$ as the open segment of the line through $x$ with the direction of the unstable eigenvector which contains $x$ and has both endpoints on $\Cal S^-$. The preimage $T^{-1}W_1^u(x)$ is by a factor of $\lambda$ shorter than $\Cal W_1^u(x)$ and, in general, is cut into two or three pieces by $\Cal S^-$. We pick the piece which contains $T^{-1}x$ and take its image under $T$; this is our second approximate unstable leaf $W_2^u(x)$, i.e., $$W_2^u(x) = T\left(T^{-1}W_1^u(x) \cap W_1^u(T^{-1}(x))\right).$$ Unless $T^{-1}x \in \Cal S^-$ the second approximate unstable leaf $W_2^u(x)$ is again an open segment containing $x$ with endpoints on $\Cal S^- \cup T \Cal S^-$ and naturally $W_2^u(x) \subset W_1^u(x)$. Given $W_n^u(x)$, $n=1,2,\dots ,$ we define the $n+1$ approximate unstable leaf of $x$ $W_{n+1}^u(x)$ by $$W_{n+1}^u(x) = T^n\left(T^{-n}W_n^u(x) \cap W_1^u(T^{-n}(x))\right).$$ If $x \notin \bigcup_{i=0}^{+\infty} T^i\Cal S^-$ then this inductive procedure will yield a nested sequence of open segments containing $x$ $$W_1^u(x) \supset W_2^u(x) \supset \dots$$ with endpoints on $$\bigcup_{i=0}^{+\infty} T^i \Cal S^-\ .$$ We can also describe this construction in the following way. First we consider a fairly long segment $W_1^u(x)$. Then we look at $T \Cal S^-$, if it does not intersect $W_1^u(x)$ then we do not change it, if it splits $W_1^u(x)$ into several segments, then we keep the segment which contains $x$. We repeat it with $T^2 \Cal S^-$ and further images of $\Cal S^-$, so that the segment may be cut shorter infinitely many times. The property $x \notin \bigcup_{i=0}^{+\infty} T^i\Cal S^-$ ensures that $x$ stays always strictly inside the segment. It is quite remarkable that, for almost every $x$, this inductive process shortens the segment only finitely many times. More precisely we have \proclaim{Proposition 1.1} For almost all $x \in \M^- \setminus \bigcup_{i=0}^{+\infty} T^i\Cal S^-$ the sequence of approximate unstable leaves of $x$ stabilizes, i.e., there is a natural $N = N(x)$ such that $$\bigcap_{i=1}^{+\infty} W_i^u(x) = \bigcap_{i=1}^N W_i^u(x).$$ \endproclaim \demo{Proof} For $t >0$, let $$X_t = \{ x \in \M^- \;|\; d(x,\,\Cal S^-) \leq t\}$$ where $d(\cdot,\,\cdot)$ is the distance of a point form a set. Because $\Cal S^-$ is a finite union of segments we have $$\mu \left(X_t \right) \leq \text{const} \ t.$$ Choosing $t_n = \frac 1{n^2}$ we get $$\sum_{n=1}^{+\infty} \mu\left( X_{t_n} \right) < +\infty,$$ hence also $$\sum_{n=1}^{+\infty} \mu\left(T^n X_{t_n}\right) < +\infty.$$ It follows by the Borel-Cantelli Lemma that almost every $x$ belongs to only finitely many of the sets $$T X_{t_1},T^2 X_{t_2},\dots ,$$ which means that except for finitely many values of $n$ $$d(T^{-n}x,\,\Cal S^-) > \frac 1{n^2}.$$ Choosing $c(x) > 0$ sufficiently small we can take care of the finite number of exceptional values of $n$ so that $$d(T^{-n}x,\,\Cal S^-) > \frac {c(x)}{n^2}$$ for each $n = 1,2,\dots \ .$ Each time $W_{n+1}^u(x)$ is shorter than $W_{n}^u(x)$ we must have $$d(T^{-n} x,\,\Cal S^-) < \frac{\text{length}\left(\Cal W_{n}^u(x)\right)} {\lambda^{n}}.$$ But then $$\frac {c(x)}{n^2} < \frac{\text{length}\left(\Cal W_{n}^u(x)\right)}{\lambda^{n}} \leq \frac{\text{length}\left(W_1^u(x)\right)}{\lambda^{n}},$$ which can hold for at most finitely many values of $n$. \enddemo We define the unstable leaf only for points $x$ in the set of full measure described in Proposition 1.1, by taking the intersection $$W^u(x) = \bigcap_{i=1}^{+\infty} W_i^u(x).$$ In view of Proposition 1.1, for each $W^u(x)$, there are natural numbers $n_l(x)$ and $n_r(x)$ such that $T^{n_l(x)}W^u(x)$ has the left endpoint on $\Cal S^-$ and $T^{n_r(x)}W^u(x)$ has the right endpoint on $\Cal S^-$. Most importantly we have the exponential contraction of $W^u(x)$, i.e., for $y \in W^u(x)$ the distance $$d(T^{-n}y,\,T^{-n}x) = \frac{d( y,\, x)}{\lambda^n} \rightarrow 0 \ \ \text{as} \ \ n \rightarrow +\infty .$$ \par Everything that we have done to construct the unstable leaves can be repeated for the stable leaves and they have analogous properties. Once we have the stable and unstable leaves we are ready to do the Hopf argument. For any continuous function $f:\To ^2 \to \Re$ the forward ergodic average $f^+$ is constant on the stable leaves and the backward ergodic average $f^-$ is constant on the unstable leaves. Let us call a point $x\in \To ^2$ $f$-typical, if $f^+(x)$, $f^-(x)$, $W^u(x)$ and $W^s(x)$ are well defined and $f^+(x) = f^-(x)$. The set of $f$-typical points has full measure, so a stable (or an unstable) leaf contains a set of $f$-typical points of full arc-length, except for a family of leaves of total measure zero. If $W^s(x)$ is not one of those exceptional leaves, then the set $$C_1 = \bigcup_{\Sb y \in W^s(x)\\y\ \text{is}\ f-\text{typical} \endSb} W^u(y)$$ has positive measure and $f^- = f^+ = const$ on $C_1$. We can proceed by adding all the stable leaves through $f$-typical points in $C_1$ to obtain $C_2$, etc., but a priori there is no reason to expect that we will be able to cover all of the torus in this way. (Indeed one can imagine that there is a dividing line between two ergodic components of our system and that all the stable and unstable leaves stop short of crossing this line.) That is where the Hopf method breaks down. It can only tell us that the ergodic components have positive measure and, therefore, that there are at most countably many of them. (To be more precise, we cannot really claim that $C_1$ belongs to one ergodic component. To argue this we have to modify our argument by taking a sequence of continuous functions dense in $L^1$ and considering the set of points which are $f$-typical for all the functions $f$ in the sequence. This set, as the intersection of countably many sets of full measure, has full measure. We can then use it in the definition of $C_1$ and claim that $f^- = f^+ = const$ on $C_1$ for all the functions in our dense sequence. This implies that such $C_1$ does belong to one ergodic component. It follows easily that every invariant subset of positive measure contains an ergodic component of positive measure. Hence all ergodic components have positive measure.) %this file must be called "s2x3.tex" \vskip.7cm \subhead \S 2. THE SINAI METHOD \endsubhead \vskip.4cm We have seen, in the previous section, that the Hopf method is not sufficient to prove the ergodicity of a discontinuous map because the stable and unstable leaves may be short. The Sinai method amounts to establishing that most of the stable and unstable leaves are, in a certain sense, sufficiently long. The first (highly nontrivial) step in this method is to formulate precisely what is meant by sufficiently long". As before, we do it only for the unstable leaves; the changes necessary in the case of stable leaves are automatic. Let $\U \subset \To^2$ be a (small) square with the sides parallel to unstable and stable directions respectively (to make the geometry simpler let us think that the unstable direction is horizontal and the stable direction vertical). For any $0 < c <1$ we construct a sequence $\Cal G_n(c), n=1,2,\dots ,$ of coverings of $\U$ in the following way. Without loss of generality we can let $$\U =\{(u,v)\;|\;-b < u < b,\, -b < v < b\}.$$ We consider the net $\Cal N(n,c)$ defined by $$\Cal N(n,c)= \{\frac cn (m,\,k)\in \U\;|\; m,k\in\Z\} .$$ Now the covering $\Cal G_n(c)$ is the collection of squares having centers at points from $\Cal N(n,c)$ and sides, of length $\frac 1n$, parallel to the sides of $\U$. If $c< \frac 12$ then $\Cal G_n(c)$ is a covering of $\U$ (otherwise $\Cal G_n(c)$ may cover only a smaller square). The parameter $c$ will be chosen later to be very small, so that many squares in $\Cal G_n(c)$ overlap. However, once $c$ is fixed, a point in $\U$ may belong to at most $k(c)$ squares in $\Cal G_n(c)$, where $k(c)$ is a fixed number independent of $n=1,2,\dots$ ; (one can easily establish that $k(c) \leq (\frac 1{2c} + 1)^2$, but we will not use any explicit estimate). \topinsert \vskip 3in \hsize=4.5in \raggedright \noindent{\bf Figure 2} The covering. \endinsert We call two squares, in $\Cal G_n(c)$, immediate neighbors if the distance between their centers is $\frac cn$.Two immediate neighbors overlap on $1-c$ part of their areas. One can naturally define a column of squares and a row of squares as special collections of squares in $\Cal G_n(c)$ (see Figure 2). For example, a sequence $\{R_i\}_{i=1}^l$ of squares from $\Cal G_n(c)$ is called a column of squares if, for every $i = 1,\dots,l-1$, $R_i$ and $R_{i+1}$ are immediate neighbors, $R_{i+1}$ is above $R_i$, and there is no square in $\Cal G_n(c)$ below $R_1$ or above $R_l$. For each square $R \in \Cal G_n$ we introduce the stable, $\partial _s R$, and unstable, $\partial _u R$, boundaries of $R$; $\partial _s R$ is the union of the two boundary segments of $R$ which have the stable (vertical) direction and $\partial _u R$ is the union of the two boundary segments of $R$ which have the unstable (horizontal) direction. Given a point $x \in R$, the unstable leaf $W^u(x)$ may intersect both segments in $\partial _s R$ or it may be too short to reach one of them (or both). In the first case we say that $W^u(x)$ is long in $R$, or that it is connecting in $R$ , in the second that it is short in $R$ or that it is not connecting in $R$. \proclaim{Definition 2.1} Given $\alpha,\, 0 < \alpha < 1,$ we call a square $R \in \Cal G_n(c)$ $\alpha$-connecting if the measure of the set of points $x \in R$ whose unstable leaf $W^u(x)$ is long in $R$ is at least $\alpha$ part of the total area of $R$. \endproclaim Sinai formulates the property that most of unstable leaves are sufficiently long in the following way. \proclaim{Sinai Theorem 2.2} There is $\alpha_0 <1$ such that for any $\alpha,\, 0 < \alpha \leq \alpha_0$ and any $c,\, 0 0$. We choose and fix $M= M(\epsilon)$ so large that the last series does not exceed $\frac \epsilon {2n}$, i.e., $$\frac {k(c)}{\alpha} \sum_{i=M+1}^{+\infty} \frac {const} {\lambda^i} < \frac \epsilon 2 .$$ Given $M$, we can still choose $n_0 = n_0(\epsilon,\,M)$ so large that, for any $n \geq n_0$, the measure of the union of $M$-bad squares in $\Cal G_n(c)$ is less than $\frac \epsilon {2n}$. To estimate the measure of the union of squares in $\Cal G_n(c)$, for $n \geq n_0$, which are not $\alpha$-connecting we split them into those which are $M$-bad and those which are not. For both families of squares the measure of their union is less than $\frac \epsilon {2n}$. This proves our claim. \qed \proclaim {Remark 3.6} \endproclaim Let us point out that the property that the sector $\Cal C$, defined by the directions of the segments in $\Cal S^-$, is sufficiently narrow ($\kappa(a)<1$) can be relaxed. For a general hyperbolic piecewise linear map it is sufficient that the segments in $\Cal S^-$ are not parallel to the stable direction. In such a case we can find a natural $N$ such that all the segments in $\bigcup_{i=N+1}^{+\infty} T^i \Cal S^-$ have directions contained in a chosen narrow sector $\Cal C$ ( $N$ is the number of iterates of $T$ which do not put the singularity lines $\Cal S^-$ into the chosen sector $\Cal C$). Then the argument above applies to any square neighborhood $\U$ which does not intersect $$\Cal S^-_N = \bigcup_{i=0}^{N} T^i \Cal S^-\ .$$ Similarly, in the version of the Sinai Theorem for the stable leaves we would have arrived at a natural $N'$ such that the claim holds for any square $\U$ which does not intersect $$\Cal S^+_{N'} = \bigcup_{i=0}^{N'} T^{-i} \Cal S^+\ .$$ Hence, it follows from Proposition 2.3 that any open square, with horizontal and vertical sides, which does not intersect $\Cal S^-_N \cup \Cal S^+_{N'}$ belongs to one ergodic component. This implies that the partition of $\Bbb T^2$ into ergodic components is coarser than the partition into (open) connected components of $$\Bbb T^2 \setminus \left( \Cal S^-_N \cup \Cal S^+_{N'} \right)\ .$$ Since $\Cal S^-_N \cup \Cal S^+_{N'}$ is a finite collection of segments, we obtain that there are at most finitely many ergodic components. To argue that there is only one component let us note that $\Cal S^-_{N-1} \cup \Cal S^+_{N'}$ and $T^{N} \Cal S^-$ intersect in at most finitely many points which split the segments in $T^{N} \Cal S^-$ into finitely many segments $\{ I_k\}_{k=1}^{K_N}$ so that the interior of every $I_k$ lies in the boundary of at most two connected components of $\Bbb T^2 \setminus \left( \Cal S^-_N \cup \Cal S^+_{N'} \right)$, i.e., it has only one connected component on each side. Suppose that for such a segment $I_k$ is in the boundary of two different ergodic components. Then $T I_k$ is also in the boundary of two different ergodic components. But $T I_k$ and $\Cal S^-_N \cup \Cal S^+_{N'}$ have only finitely many points of intersection, so that whole open sub-intervals of $T I_k$ must end up inside one connected component of $\Bbb T^2 \setminus \left( \Cal S^-_N \cup \Cal S^+_{N'} \right)$ and thus it must have the same ergodic component on both sides. This contradiction implies that $I_k$ does not take part in the splitting of $\Bbb T^2$ into ergodic components so we can drop it. In this way we can drop all of $T^{N} \Cal S^-$ and claim that the partition into ergodic components is coarser than the partition into connected components of $$\Bbb T^2 \setminus \left( \Cal S^-_{N-1} \cup \Cal S^+_{N'} \right) \ .$$ It is now clear that we can proceed by dropping $T^{N-1} \Cal S^-$ and $T^{-N'} \Cal S^+$ as possible boundaries for the ergodic components and arriving eventually at $\Cal S^+ \cup \Cal S^-$ as the only possible boundaries we see that even these can be dropped. Hence there is only one ergodic component. Let us spell out the property of $T$ which is basic in this argument: {\it Although some points of $\Cal S^-$ return to $\Cal S^-$ under iterates of $T$, no interval in $\Cal S^-$ can do it.} %this file has to be called s4x6.tex \vskip.7cm \subhead \S 4. SECTORS IN A LINEAR SYMPLECTIC SPACE \endsubhead \vskip.4cm For the convenience of the reader we will repeat here some of the material from \cite {W3} and \cite {LW}. Let $\Cal W$ be a linear symplectic space of dimension $2d$ with the symplectic form $\omega$. For instance we call $\Cal W = \Bbb{R}^{d}\times\Bbb{R}^{d}$ the standard linear symplectic space if $$\omega (w_1,w_2) = \langle\xi^1,\eta^2\rangle-\langle\xi^2,\eta^1\rangle,$$ $$\text{where} \ w_i =(\xi^i,\eta^i),\ i =1,2, \ \ \text{and} \ \ \langle\xi,\eta\rangle = \xi_1\eta_1 + \dots + \xi_d\eta_d .$$ The symplectic group $Sp\left(d,\Bbb{R}\right)$ is the group of linear maps of $\Cal W$ ($2d\times 2d$ matrices if $\Cal W = \Bbb{R}^{d}\times \Bbb{R}^{d}$) preserving the symplectic form i.e., $L\in Sp\left(d,\Bbb{R}\right)$ if $$\omega(Lw_1,Lw_2)=\omega(w_1,w_2)$$ for every $w_1,w_2 \in \Cal W$. By definition a Lagrangian subspace of a linear symplectic space $\Cal W$ is a $d$-dimensional subspace on which the restriction of $\omega$ is zero (equivalently it is a maximal subspace on which $\omega$ vanishes). \proclaim{Definition 4.1} Given two transversal Lagrangian subspaces $V_1$ and $V_2$ we define the sector between $V_1$ and $V_2$ by $$\Cal{C} = \Cal{C}\left(V_1,V_2\right) = \{w\in \Cal W\ |\ \omega(v_1,v_2)\geq 0 \ \text{for}\ w=v_1+v_2, v_i\in V_i, i=1,\, 2\}$$ \endproclaim Equivalently, we define first the quadratic form $\Cal Q$ associated with an ordered pair of transversal Lagrangian subspaces, $$\Cal{Q}(w) = \omega (v_1,v_2)$$ where $w =v_1 + v_2$, $v_i\in V_i, i=1,2,$ is the unique decomposition of $w$.We have $$\Cal{C} =\{w\in \Cal W\ |\ \Cal{Q}(w)\geq 0\}.$$ In the case of the standard symplectic space, $V_1=\Bbb{R}^d\times\{0\}$ and $V_2=\{0\}\times\Bbb{R}^d$ we get $$\Cal{Q}\left((\xi,\eta)\right)=\langle\xi,\eta\rangle$$ and $$\Cal{C} =\{(\xi,\eta)\in \Bbb {R}^{d}\times \Bbb{R}^{d}\ |\ \langle\xi,\eta\rangle\geq 0\}.$$ We will refer to this $\C$ as the standard sector. Since any two pairs of transversal Lagrangian subspaces are symplectically equivalent we may consider only this case without any loss of generality. In the following we will alternate between the coordinate free geometric formulations and this special case. On the one hand, coordinate free formulations are important because we need to apply these concepts to the case of the derivative map which in general acts between two different tangent spaces, each one with its own sector. On the other hand, it turns out that many arguments are greatly simplified by resorting to these special coordinates. It is natural to ask if a sector determines uniquely its sides. It is not a vacuous question since, for $d>1$, there are many Lagrangian subspaces in the boundary of a sector. The answer is positive. \proclaim{Proposition 4.2} For two pairs of transversal Lagrangian subspaces $V_1,V_2$ and $V_1',V_2'$ if $$\Cal{C}\left(V_1,V_2\right) = \Cal{C}\left(V_1',V_2'\right)$$ then $$V_1 = V_1' \ \ \text{and} \ \ V_2 = V_2'.$$ Moreover $V_1$ and $V_2$ are the only isolated Lagrangian subspaces contained in the boundary of the sector $\Cal{C}\left(V_1,V_2\right)$. \endproclaim The proof of this Proposition can be found in \cite {W3}. Based on the notion of the sector between two transversal Lagrangian subspaces (or the quadratic form $\Cal{Q}$) we define two monotonicity properties of a linear symplectic map. By $int\C$ we denote the interior of the sector, i.e., $$int\C =\{w \in \Cal W | \Q(w) > 0\}.$$ \proclaim{Definition 4.3} Given the sector $\Cal C$ between two transversal Lagrangian subspaces we call a linear symplectic map $L$ monotone if $$L\Cal{C}\subset\Cal{C}$$ and strictly monotone if $$L\Cal{C} \subset int\Cal{C} \cup \{0\}.$$ \endproclaim A very useful characterization of monotonicity is given in the following \proclaim{Theorem 4.4} $L$ is (strictly) monotone if and only if $\Cal{Q}\left(Lw\right)\geq \Cal{Q}\left(w\right)$ for every $w\in \Cal W$ ($\Cal{Q}\left(Lw\right) > \Cal{Q}\left(w\right)$ for every $w\in \Cal W ,\ w \neq 0$). \endproclaim The fact that monotonicity implies the increase of the quadratic form defining the sector is a manifestation of a very special geometric structure of a sector and does not hold for cones defined by general quadratic forms. Before proving this theorem we will discuss several useful facts. For a pair of transversal Lagrangian subspaces $V_1$ and $V_2$ and a linear map $L:\Cal W\to \Cal W$, we can define the following block' operators: \aligned A:V_1\to V_1 ,\ & B:V_2\to V_1 \\ C:V_1\to V_2 ,\ & D:V_2\to V_2. \endaligned They are uniquely defined by the requirement that for any $v_1\in V_1, v_2\in V_2$ $$L\left(v_1+v_2\right)= Av_1 + Bv_2 + Cv_1 + Dv_2 .$$ We will need the following Lemma. \proclaim{Lemma 4.5} If $L$ is monotone with respect to the sector between $V_1$ and $V_2$ then $LV_1$ is transversal to $V_2$ and $LV_2$ is transversal to $V_1$. \endproclaim \demo{Proof} Suppose that there is $0\neq \bar{v}_{1} \in V_1$ such that $L\bar{v}_{1} \in V_2 .$ We choose $\bar{v}_{2} \in V_2$ so that $$\Cal{Q}\left(\bar{v}_{1} + \bar{v}_{2}\right) = \omega\left(\bar{v}_{1},\bar{v}_{2}\right) > 0.$$ We have also $$\omega\left(\bar{v}_{1},\bar{v}_{2}\right)= \omega\left(L\bar{v}_{1},L\bar{v}_{2}\right)= \omega\left(L\bar{v}_{1},B\bar{v}_{2}+ D\bar{v}_{2}\right)= \omega\left(L\bar{v}_{1},B\bar{v}_{2}\right).$$ Let $v_{\epsilon} = \bar{v}_{1}+\epsilon \bar{v}_{2}.$ We have that for $\epsilon >0$\ $v_{\epsilon}$belongs to $\text{int}\Cal{C}.$ Hence also $\Cal{Q}\left(Lv_{\epsilon}\right)\geq 0$ for $\epsilon >0.$ On the other hand $$\Cal{Q}\left(Lv_{\epsilon}\right) = \epsilon ^2 \omega\left(B\bar{v}_{2},D\bar{v}_{2}\right) - \epsilon\omega\left(L\bar{v}_{1},B\bar{v}_{2}\right)$$ which is negative for sufficiently small positive $\epsilon.$ This contradiction proves the Lemma. \enddemo It follows from Lemma 4.5 that the operators $A:V_1\to V_1$ and $D:V_2\to V_2$ are invertible. We switch now to the coordinate language. Let $$L = \left(\matrix A& B\\ C& D\endmatrix\right)$$ be a symplectic map of the standard symplectic space $\Bbb{R}^{d}\times\Bbb{R}^{d}$ monotone with respect to the standard sector. $A,B,C,D$ are now just $d\times d$ matrices. Let us describe those symplectic matrices which are monotone in the weakest sense, namely they preserve the quadratic form $\Cal{Q}.$ We will call such matrices $\Cal{Q}$-isometries. Obviously a $\Cal{Q}$-isometry maps the sector onto itself. The converse is also true. \proclaim {Proposition 4.6} If $L$ is a linear symplectic map and $$L\Cal{C} = \Cal{C}$$ then $$L = \left(\matrix A& 0\\ 0& A^{*-1}\endmatrix\right) .$$ In particular it preserves the quadratic form $\Cal{Q}$ $$\Cal{Q}\circ L = \Cal{Q} .$$ \endproclaim \demo{Proof} If $L\Cal{C} = \Cal{C}$ then $L$ maps also the boundary of the sector $\Cal{C}$ onto itself. It follows from Proposition 4.2 that both sides of the sector stay put under $L$. Hence $B=C=0.$ By symplecticity $D = A^{*-1}.$ \enddemo By Lemma 4.5 given a monotone $L$ we can always factor out the following $\Cal{Q}$-isometry on the left $$L = \left(\matrix A& B\\ C& D\endmatrix\right) = \left(\matrix A& 0\\ 0& A^{*-1}\endmatrix\right)\left(\matrix I& P \\ R &\cdot \endmatrix\right)$$ Symplecticity of $L$ forces $P$ and $R$ to be symmetric and allows the further unique factorization $$L = \left(\matrix A& 0\\ 0& A^{*-1}\endmatrix\right)\left(\matrix I& 0 \\ P &I \endmatrix\right)\left(\matrix I& R \\ 0 &I \endmatrix\right). \tag {4.7}$$ Moreover monotonicity forces $P$ and $R$ to be positive semidefinite ($P\ge 0,\,R\ge 0$). Strict monotonicity means that $P$ and $R$ are positive definite ($P> 0,\, R> 0$). These claims follow from the following \demo{Proof of Theorem 4.4} Using the above factorization we get for $w = (\xi,\,\eta)$ $$\Cal{Q}(Lw) = \langle\xi,\,\eta\rangle + \langle R\eta,\,\eta\rangle + \langle P(\xi+R\eta),\,\xi+R\eta\rangle.$$ Putting $\eta = 0$ we obtain that $P\geq 0$. To show that also $R \geq 0$ let us consider an eigenvector $\eta_0$ of $R$ with eigenvalue $\lambda$ and let $\xi = a\eta_0$. We get that if $a\geq 0$ then $w = (\xi,\,\eta_0) \in \Cal C$ so that $\Cal{Q}(Lw) \geq 0$. It follows that $$(a+\lambda)\langle\eta,\,\eta\rangle + (a+\lambda)^2\langle P\eta,\,\eta\rangle\geq 0 .$$ This implies immediately that $\lambda \geq 0$. This proves the monotone version of the Theorem. The strictly monotone version is obtained in a similar way. \enddemo As a byproduct of the proof we get the following useful observation \proclaim{Proposition 4.8} A monotone map $L$ is strictly monotone if and only if $$LV_i \subset \text{int }\Cal{C} \cup \{ 0\} ,\ i=1,2.$$ \endproclaim \qed The following Proposition simplifies computations with monotone maps. \proclaim {Proposition 4.9} If $$L = \left(\matrix A& B\\ C& D\endmatrix\right)$$ is a strictly monotone map then by multiplying it by $\Cal{Q}$-isometries on the left and on the right we can bring it to the form $$\left(\matrix I& I\\ T& I+T\endmatrix\right)$$ where $T$ is diagonal and has the same eigenvalues as $C^*B$. \endproclaim \demo{Proof} The factorization of the monotone map $L$ yields $$\left(\matrix A& 0\\0& A^{*-1}\endmatrix\right)L = \left(\matrix I& R\\ P& I+PR\endmatrix\right)$$ where $P > 0$, $R > 0$ and $PR = C^*B$. We have further $$\left(\matrix R^{-\frac{1}{2}}& 0\\ 0&R^{\frac{1}{2}}\endmatrix\right) \left(\matrix I& R\\ P& I+PR\endmatrix\right) \left(\matrix R^{\frac{1}{2}}& 0\\ 0& R^{-\frac{1}{2}}\endmatrix\right) = \left(\matrix I& I\\ K& I+K\endmatrix\right)$$ where $K= R^{\frac{1}{2}}PR^{\frac{1}{2}}$ has the same eigenvalues as $C^*B= PR.$ Finally if $F$ is the orthogonal matrix which diagonalizes $K$, i.e., $F^{-1}KF$ is diagonal, then $$\left(\matrix F^{-1}& 0\\ 0& F^{-1}\endmatrix\right)\left(\matrix I& I\\K& I+K\endmatrix\right)\left(\matrix F& 0\\ 0& F\endmatrix\right) = \left(\matrix I& I\\ T& I+T\endmatrix\right)$$ has the desired form with $T = F^{-1}KF$ having the same eigenvalues as $C^*B$. \enddemo Let us note that in the last Proposition we can ask for the diagonal entries of $T$ to be ordered because any permutation of the entries can be accomplished by an appropriate $\Cal Q$-isometry. \vskip.7cm \subhead \S 5. THE SPACE OF LAGRANGIAN SUBSPACES CONTAINED IN A SECTOR \endsubhead \vskip.4cm Let us fix a sector $\C = \C(V_1,\,V_2)$ between two transversal Lagrangian subspaces $V_1$ and $V_2$. We say that a Lagrangian subspace $E$ is strictly contained in $\C$ if $$E \subset int\,\C \cup \{0\}.$$ We denote by $Lag(\C)$ the manifold of all such Lagrangian subspaces and by $\widehat{Lag}(\C)$ its closure in the Lagrangian Grassmanian, i.e., $\widehat{Lag}(\C)$ is the set of all Lagrangian subspaces contained in $\C$. We will introduce a metric and a partial order into $Lag(\C)$. Let $$\pi_i : \Cal W \to V_i, \ i = 1,2,$$ be the natural projections, i.e., $$w = \pi_1w +\pi_2w\ \ \text{ for every}\ \ w \in \Cal W.$$ If a Lagrangian subspace $E$ is strictly contained in $\C$ then $\pi_iE = V_i, \ i = 1,2$, so that $\pi_i|_E$ (the restriction of $\pi_i$ to the subspace $E$) is a one to one map of $E$ onto $V_i$. With every subspace $E \in Lag(\C)$ we can associate a positive definite quadratic form on $V_1$ obtained by the formula $$\Q \circ \left(\pi_1|_E\right)^{-1}.$$ It will turn out that this is actually a one-to-one correspondence between positive definite quadratic forms on $V_1$ and Lagrangian subspaces contained strictly in $\C$. \proclaim{Definition 5.1} For two Lagrangian subspaces $E_1,E_2 \in Lag(\C)$ we define the relation $E_1 \leq E_2$ ($E_1 < E_2$) by the inequality of the corresponding quadratic forms $$\Q \circ \left(\pi_1|_{E_1}\right)^{-1} \leq (<) \Q \circ \left(\pi_1|_{E_2}\right)^{-1}.$$ We define the distance of two Lagrangian subspaces $E_1,E_2 \in Lag(\C)$ by $$d(E_1,\,E_2) = \frac 12 \sup_{0\neq v \in V_1} |\ln\Q \circ \left(\pi_1|_{E_1}\right)^{-1}(v) - \ln\Q \circ \left(\pi_1|_{E_2}\right)^{-1}(v)|.$$ \endproclaim It is easy to see that $d(\cdot,\,\cdot)$ is indeed a metric. There are other ways to introduce the partial order and the metric. The coordinate free definitions simplify some of the arguments in the following. For equivalent definitions of the metric see \cite {LW}, \cite {Ve}. Theses definitions are justified by the following theorem. \proclaim{Theorem 5.2} For two transversal Lagrangian subspaces $E_1,E_2 \in Lag(\C)$, $$E_1 < E_2 \ \ \text{ if and only if } \ \ \C(E_1,\,E_2) \subset \C(V_1,\,V_2).$$ Further, if $E_1 < E_2$ then for a Lagrangian subspace $E \in Lag(\C)$, $$E \subset \C(E_1,\,E_2) \ \ \text{ if and only if } \ \ E_1 \leq E \leq E_2.$$ \endproclaim \proclaim{Corollary 5.3} If $E_1,E_2 \in Lag(\C)$ and $E_1 < E_2$ then the diameter of the set $\widehat{Lag}\left(\C(E_1,\,E_2)\right)$ in $Lag(\C)$ is equal to the distance of $E_1$ and $E_2$. \endproclaim \qed We will prove Theorem 5.2 at the end of this Section. Let us introduce a convenient parametrization of $Lag(\C)$ by symmetric positive definite matrices. We consider the standard sector $\C$ in $\Bbb{R}^{d}\times \Bbb{R}^{d}$ with $V_1 = \Bbb R^d \times \{0\}$ and $V_2 = \{0\} \times \Bbb R^d$. Let $U : \Bbb{R}^d \to \Bbb{R}^d$ be a linear map and $$gU =\{\left(\xi,\eta\right)\in \Bbb{R}^{d}\times\Bbb{R}^{d}\;|\;\eta = U\xi\}$$ be its graph. The linear subspace $gU$ is a Lagrangian subspace if and only if $U$ is symmetric and further for a symmetric $U$ its graph $gU \subset \Cal{C}$ if and only if $U \ge 0.$ Every Lagrangian subspace in $Lag(\C)$ is transversal to $V_2$ so that it is a graph of a linear map as above. We will find the following Lemma useful. \proclaim {Lemma 5.4} If a Langrangian subspace $E \subset \Cal C(V_1,\,V_2)$ is transversal to both $V_1$ and $V_2$ then it is strictly contained in the sector. \endproclaim \demo{Proof} We use the coordinate description of the standard sector. Thus the Lagrangian subspace $E$ being transversal to $V_2$ is the graph of a symmetric positive semidefinite matrix. Since $E$ is also transversal to $V_1$ the matrix is nondegenerate and hence positive definite. It follows immediately that $E$ is strictly contained in the sector. \enddemo We have obtained a one-to-one correspondence between Lagrangian subspaces in $Lag(\C)$ and symmetric positive definite matrices. The quadratic form on $V_1$ introduced in Definition 5.1 becomes the form defined by the positive definite matrix. The partial order becomes the familiar partial order between symmetric matrices. The image of a Lagrangian subspace under a symplectic linear map is again a Lagrangian subspace. Moreover monotone maps take Lagrangian subspaces strictly contained in $\C$ into Lagrangian subspaces strictly contained in $\C$. Hence a monotone map $L$ defines a map of $Lag(\C)$ into itself. We will denote it again by $L: Lag(\C) \to Lag(\C)$. To simplify notation we will also write $U$ instead of $gU$. We have that $$L = \left(\matrix A& B\\ C& D\endmatrix\right)$$ acts on Lagrangian subspaces by the following M\"obius transformation $$LU=\left(C+DU\right)\left(A+BU\right)^{-1}.$$ In particular the action of a $\Q$-isometry $$L = \left(\matrix A& 0\\ 0& A^{*-1}\endmatrix\right)$$ is given by $$LU = A^{*-1}UA^{-1}.$$ By putting $A = U^{\frac 12}$ we see that any $U > 0$ can be mapped onto identity matrix $I$. Thus $\Q$-isometries act transitively on $Lag(\C)$. Moreover it is not hard to see that \proclaim{Proposition 5.5} The action of a $\Q$-isometry on $Lag(\C)$ preserves the partial order and the metric. \endproclaim \qed Let $E_0 = \{(\xi,\,\eta)\ |\ \xi = \eta\}$. By straightforward computations we find that \aligned \C(V_1,\,E_0) = \{(\xi,\,\eta)\ |\ \langle\xi,\,\eta\rangle - \langle\eta,\,\eta\rangle \geq 0\},\\ \C(E_0,\,V_2) = \{(\xi,\,\eta)\ |\ \langle\xi,\,\eta\rangle - \langle\xi,\,\xi\rangle \geq 0\}. \endaligned \tag{5.6} We get that \aligned \C(V_1,\,E_0) \subset \C(V_1,\,V_2), \\ \C(E_0,\,V_2)\subset \C(V_1,\,V_2), \\ \C(V_1,\,E_0) \cap \C(E_0,\,V_2) = E_0. \endaligned \tag 5.7 Because the group of $\Q$-isometries acts transitively on $Lag(\C)$ \thetag {5.7} holds not just for the special Lagrangian subspace $E_0$ from \thetag {5.6} but for any Lagrangian subspace from $Lag(\C)$. (It just happens that the easiest way to establish \thetag{5.7} is to do the calculation in the standard sector.) \proclaim{Proposition 5.8} For two Lagrangian subspaces $E_1,E_2 \in Lag(\C)$ the following are equivalent \aligned &(1) \ \ E_1 \leq E_2,\\ &(2) \ \ E_2 \subset \C(E_1,\,V_2),\\ &(3) \ \ E_1 \subset \C(V_1,\,E_2). \endaligned \endproclaim \demo{Proof} We will be using the coordinate description of the standard sector. Since the group of $\Q$-isometries acts transitively on $Lag(\C)$ we can assume that $E_1$ is equal to $E_0$ from \thetag{5.6}. Let $U_2$ be the positive definite matrix defining $E_2$. We get from \thetag{5.6} that $E_2 \subset \C(E_0,\,V_2)$ if and only if $U_2 \geq I$. Hence (1) is equivalent to (2). Similarly let $E_2$ be equal to $E_0$ and $U_1$ be the positive definite matrix defining $E_1$. Using \thetag{5.6} again we get that $E_1 \subset \C(V_1,\,E_0)$ if and only if $U_1 -U_1^2 \geq 0$ which is equivalent to $U_1 \leq I$. This proves the equivalence of (1) and (3). \enddemo \demo{Proof of Theorem 5.2} If $E_1 < E_2$ then, by Proposition 5.8 and Lemma 5.4, $E_2$ is strictly contained in $\C(E_1,\,V_2)$. Using \thetag{5.7} we get $$\C(E_1,\,E_2) \subset \C(E_1,\,V_2) \subset \C(V_1,\,V_2).$$ Suppose now that $\C(E_1,\,E_2) \subset \C(V_1,\,V_2)$. By Proposition 5.8, it suffices to show that $E_2 \subset \C(E_1,\,V_2)$. If it is not so then there is $e_2 \in E_2$ which does not belong to $\C(E_1,\,V_2)$. Let us consider $v_1 = \pi_1e_2$ where $\pi_1 : \Cal W \to V_1$ is the projection onto $V_1$ in the direction of $V_2$. Let further $e_1$ be the unique element in $E_1$ such that $\pi_1e_1 = v_1$ (i.e., $e_1 = \left(\pi_1|_{E_1}\right)^{-1}v_1$). Clearly the difference between the two vectors $v_2 = e_2 - e_1$ belongs to $V_2$. Because $e_2 = e_1 + v_2$ and $e_2 \notin \C(E_1,\,V_2)$ we have $\omega(e_1,\,v_2) < 0$ so that $\omega(-e_1,\,e_2) > 0$. It follows that $v_2 =e_2 - e_1 \in int\,\C(E_1,\,E_2) \subset int\,\C(V_1,\,V_2)$. We have thus reached a contradiction, that $v_2$ belongs to $V_2$ and also to $int\,\C(V_1,\,V_2)$. This contradiction proves that indeed $E_2 \subset \C(E_1,\,V_2)$ which, by Proposition 5.8, implies that $E_1 < E_2$ (remember that $E_1$ and $E_2$ are assumed to be transversal). The first part of the Theorem is proven. To prove the second part, let $E_1 < E_2$ and $E \subset \C(E_1,\,E_2)$. By Proposition 5.8, we get $E_2 \subset \C(E_1,\,V_2)$. It follows, in view of \thetag{5.7}, that $\C(E_1,\,E_2) \subset \C(E_1,\,V_2)$, and hence $E \subset \C(E_1,\,V_2)$ which is equivalent (again by Proposition 5.8) to $E_1 \leq E$. Similarly, we get $E \leq E_2$. In the opposite direction, if $E_1 \leq E < E_2$ then, by Proposition 5.8, $E_1$ and $E$ are strictly contained in $\C(V_1,\,E_2)$ and $E_1 \subset \C(V_1,\,E)$. Applying now the equivalence of (2) and (3) in Proposition 5.8 to the case of $E_1,E \in Lag(\C(V_1,\,E_2))$, we get immediately $E \subset \C(E_1,\,E_2)$. The case of $E_1 \leq E \leq E_2$ can be now treated by continuity. \enddemo Let us consider a special family of Lagrangian subspaces in the standard sector: the graphs of multiples of the identity matrix, i.e., for a real number $u$ let $$Z_u = \{(\xi,\,\eta)\ |\ \eta = e^u \xi\}.$$ We have that $$d(Z_{u_1},\,Z_{u_2}) = \frac 12 | u_1 - u_2|.$$ \proclaim{Lemma 5.9} Let us fix two Lagrangian subspaces, $Z_{u_1}$ and $Z_{u_2}$, $u_1 < u_2$. If for a Lagrangian subspace $E \in Lag(\C)$ $$d(Z_{u_1},\,E) \leq \frac 12 (u_2 - u_1)$$ then $$E \leq Z_{u_2}.$$ \endproclaim \demo{Proof} Let the Lagrangian subspace $E$ be the graph of a positive definite matrix $U$. For every nonzero $\xi \in \Bbb R^d$, we have $$\ln\langle\xi,\,U\xi\rangle -\ln\langle\xi,\,e^{u_1}\xi\rangle \leq u_2 -u_1.$$ It follows that, for every nonzero $\xi \in \Bbb R^d$, $$\ln\frac{\langle\xi,\,U\xi\rangle}{\langle\xi,\,\xi\rangle} \leq u_2 .$$ We conclude that $U \leq e^{u_2}I$. \enddemo We will use the following consequence of the last Lemma. \proclaim{Proposition 5.10} Let $E_1 < E_2$ be two Lagrangian subspaces contained strictly in $\C(V_1,\,V_2)$. There is a symplectic map which maps the sector $\C(V_1,\,V_2)$ {\bf onto} the standard sector $\C$ and the sector $\C(E_1,\,E_2)$ {\bf into} the sector $\C(Z_{-u},\,Z_u)$ if and only if $d(E_1,\,E_2) \leq u$. \endproclaim \demo{Proof} By a symplectic map we can map the subspace $V_1$ onto $\Bbb R^d \times \{0\}$, the subspace $V_2$ onto $\{0\} \times \Bbb R^d$ and $E_1$ onto $Z_{-u}$ (because $\Q$-isometries act transitively on $Lag(\C)$). It follows from Lemma 5.9 that the sector $\C(E_1,\,E_2)$ will be then automatically mapped into $\C(Z_{-u},\,Z_u)$. The converse follows from Corollary 5.3. \enddemo For aesthetical reasons we will be using Proposition 5.10 in a different coordinate system. Let us introduce the family of sectors $$\C_{\rho} = \{(\xi,\,\eta)\ |\ \|\eta\| \leq \rho\|\xi\| \}$$ for any real $\rho > 0$. \proclaim{Proposition 5.11} Let $E_1 < E_2$ be two Lagrangian subspaces contained strictly in $\C(V_1,\,V_2)$. There is a symplectic map which maps the sector $\C(V_1,\,V_2)$ {\bf onto} the sector $\C_{\rho^{-1}}$ and the sector $\C(E_1,\,E_2)$ {\bf into} the sector $\C_\rho$ if and only if $$d(E_1,\,E_2) \leq \ln \frac {1+\rho^2}{1-\rho^2},$$ with $0<\rho<1$. \endproclaim \demo{Proof} Let us define a symplectic map $L$ by \aligned \xi' &= \frac 1{\sqrt 2}(\rho^{-\frac 12}\xi -\rho^{\frac 12}\eta),\\ \eta' &= \frac 1{\sqrt 2}(\rho^{-\frac 12}\xi +\rho^{\frac 12}\eta). \endaligned A direct computation shows that, if $\rho<1$, $L\C_{\rho^{-1}}=\C$ and $L\C_{\rho}=\C(Z_{-u},\,Z_{u})$, with $u=\log\frac{1+\rho^2}{1-\rho^2}$. The result follows then from Proposition 5.10. \enddemo \vskip.7cm \subhead \S 6. UNBOUNDED SEQUENCES OF LINEAR MONOTONE MAPS \endsubhead \vskip.4cm In this section we fix a sector $\C = \C(V_1,\,V_2)$ between two Lagrangian subspaces. One can think that $\C$ is the standard sector. We start by computing the coefficient of expansion of $\Q$ under the action of a monotone symplectic map. For a linear symplectic map $L$, monotone with respect to the sector $\Cal{C}$, we define the coefficient of expansion at $w \in int\Cal{C}$ by $$\beta\left(w,L\right) = \sqrt{\frac{\Cal{Q}\left(Lw\right)}{\Cal{Q}\left(w\right)}} .$$ Further, we define the least coefficient of expansion by $$\sigma_{\Cal{C}} \left(L\right)= \inf_{w\in int\Cal{C}} \beta\left(w,L\right).$$ Let us note that, for two monotone maps $L_1$ and $L_2$, $$\sigma_{\Cal{C}} \left(L_2L_1\right) \geq \sigma_{\Cal{C}} \left(L_2\right)\sigma_{\Cal{C}} \left(L_1\right),$$ i.e., the coefficient of expansion $\sigma_{\Cal{C}}$ is supermultiplicative. We will omit the index $\Cal{C}$ in $\sigma_{\C}(L)$ when it is clear which sector we have in mind. We want to find the value of the expansion coefficient in coordinates. We will use the fact that this infimum does not change if $L$ is multiplied on the left or on the right by $\Cal{Q}$-isometries. So let $$L = \left(\matrix A& B\\ C& D\endmatrix\right)$$ be a monotone matrix. By the factorization \thetag{4.7}, $C^*B = PR$ is equal to the product of two positive semidefinite matrices, and so it has only real non-negative eigenvalues. Let us denote them by $0\leq t_1 \leq \dots \leq t_d$. The monotone map $L$ is strictly monotone if and only if $t_1 > 0$. \proclaim {Proposition 6.1} For a monotone map $L$ $$\sigma \left(L\right) = \sqrt{1+t_1}+\sqrt{t_1}= \exp \sinh^{-1}\sqrt{t_1}.$$ Moreover, if $L$ is strictly monotone $$\sigma \left(L\right) = \beta\left(w,L\right)$$ for some $w\in \text{int }\Cal{C}$. \endproclaim \demo{Proof} Let us put $$m \left(L\right) = \sqrt{1+t_1}+\sqrt{t_1} = \min_{1\le i \le d}\left(\sqrt{1+t_i}+\sqrt{t_i}\right) .$$ First we prove the inequality $\beta\left(w,L\right) \ge m \left(L\right)$ for $w\in\text{int}\Cal{C}$. Since both $\beta\left(w,L\right)$ and $m \left(L\right)$ are continuous functions of $L$ it is sufficient to prove the inequality for strictly monotone maps only. In view of Proposition 4.9 we can take $$L = \left(\matrix I&I\\ T& I+T\endmatrix\right)$$ with diagonal $T$ and $t_1,\dots ,t_d$ on the diagonal. We compute directly, for $w=\left(\xi ,\eta\right)$ such that $\Cal{Q}\left(w\right)=1$, \aligned \left(\beta\left(w,L\right) \right)^2 & = \sum_{i=1}^{d}\left(t_i \xi_i^2 + \left(1+2t_i\right)\xi_i\eta_i + \left(1+t_i\right)\eta_i^2\right)\\ &=\sum_{i:\xi_i\eta_i \ge 0}\left(\left(\sqrt{t_i}\xi_i - \sqrt{1+t_i}\eta_i\right)^2 +\left(\sqrt{1+t_i} + \sqrt{t_i}\right)^2\xi_i\eta_i \right)\\ &+ \sum_{i:\xi_i\eta_i < 0}\left(\left(\sqrt{t_i}\xi_i + \sqrt{1+t_i}\eta_i\right)^2 +\left(\sqrt{1+t_i} - \sqrt{t_i}\right)^2\xi_i\eta_i \right) \ge\\ &\ge \sum_{i:\xi_i\eta_i \ge 0}\left(\sqrt{1+t_i} + \sqrt{t_i}\right)^2\xi_i\eta_i + \sum_{i:\xi_i\eta_i < 0}\left(\sqrt{1+t_i} + \sqrt{t_i}\right)^{-2}\xi_i\eta_i\ge\\ &\ge\left(1+\delta\right) m\left(L\right)^2- \delta m\left(L\right)^{-2} \ge m\left(L\right)^2 \endaligned where $$\delta =\left(\sum_{i:\xi_i\eta_i \ge 0}\xi_i\eta_i\right) -1 = - \sum_{i:\xi_i\eta_i<0} \xi_i\eta_i\ge 0$$ and all the inequalities become equalities for $$\xi_1 = \left(\frac{1+t_1}{t_1}\right)^{\frac 14}, \eta_1 = \left(\frac{t_1}{1+t_1}\right)^{\frac 14}, \ \xi_i = 0, \eta_i =0,\ i= 2,\dots,d.$$ Thus the Proposition is proven for strictly monotone matrices, and for all monotone matrices we get the inequality $\sigma(L) \geq m(L)$. To get the equality $\sigma(L) = m(L)$ for all monotone matrices, we proceed as follows. For any $\epsilon > 0$, we choose a strictly monotone matrix $L_{\epsilon}$, so close to the identity that $m\left(L_{\epsilon}L\right) < m\left(L\right) +{\epsilon}.$ Since $L_{\epsilon}L$ is strictly monotone and our Proposition has been proven for strictly monotone matrices, there is $w_{\epsilon}\in \text{int}\Cal{C}$ such that $$\beta\left(w_{\epsilon},L_{\epsilon}L\right) = m(L_{\epsilon}L)= \sigma\left(L_{\epsilon}L\right).$$ But $\beta\left(w,L_{\epsilon}L\right) > \beta\left(w,L\right)$, for any $w \in \text{int}\Cal{C}.$ Hence $$m(L) \leq \sigma\left(L\right) \le \beta\left(w_{\epsilon},L\right) < \beta\left(w_{\epsilon},L_{\epsilon}L\right) = m\left(L_{\epsilon}L\right) < m\left(L\right) +{\epsilon},$$ which ends the proof. \enddemo For a given sector $\Cal{C} = \Cal{C}(V_1,V_2)$ let $\Cal{C'} = \Cal{C}(V_2,V_1)$ be the complementary sector. We have \proclaim{Proposition 6.2} If $L$ is (strictly) monotone with respect to $\Cal{C}$ then $L^{-1}$ is (strictly) monotone with respect to $\Cal{C'}$ and $\sigma_{\Cal{C}} (L) = \sigma_{\Cal{C'}} (L^{-1})$. \endproclaim \demo{Proof} We have that the union $$\Cal{C}\left(V_1,V_2\right) \cup \text{int}\Cal{C}\left(V_2,V_1\right)$$ is equal to the whole linear symplectic space $\Cal W$. Hence if $$L \Cal{C}\left(V_1,V_2\right) \subset \Cal{C}\left(V_1,V_2\right)$$ then $$\Cal{C}\left(V_1,V_2\right) \subset L^{-1} \Cal{C}\left(V_1,V_2\right)$$ and finally $$L^{-1}\text{int}\Cal{C}\left(V_2,V_1\right) \subset \text{int}\Cal{C}\left(V_2,V_1\right).$$ The last property is easily seen to be equivalent to the monotonicity of $L^{-1}$. To obtain the equality of the coefficients of least expansion we will use the standard sector and the block description of $L$. Let (see \thetag{4.7}) $$L = \left(\matrix A& 0\\ 0& A^{*-1}\endmatrix\right)\left(\matrix I& 0 \\ P &I \endmatrix\right)\left(\matrix I& R \\ 0 &I \endmatrix\right).$$ The linear symplectic map $\left(\matrix 0 & I \\ -I & 0 \endmatrix\right)$ takes the standard sector $\Cal{C}$ onto $\Cal{C'}$ and further $$L_1 = \left(\matrix 0 & -I \\ I & 0 \endmatrix\right)L^{-1} \left(\matrix 0 & I \\ -I & 0 \endmatrix\right)$$ has the same least coefficient of expansion with respect to $\Cal{C}$ as $L^{-1}$ with respect to $\Cal{C'}$. Since $$L^{-1} = \left(\matrix I& -R \\ 0 &I \endmatrix\right) \left(\matrix I& 0 \\ -P &I \endmatrix\right) \left(\matrix A^{-1}& 0\\ 0& A^*\endmatrix\right)$$ we get $$L_1 = \left(\matrix I & P \\ R & I+RP \endmatrix\right) \left(\matrix A^*& 0\\ 0& A^{-1}\endmatrix\right).$$ Our claim follows now from the formula in Proposition 6.1 and the fact that $PR$ has the same eigenvalues as $RP$. \enddemo The next Proposition is a useful addition to the Corollary 5.3. \proclaim {Proposition 6.3} For a strictly monotone map $L$ $$d(LV_1,\,LV_2) =\ln\frac{\sigma(L)^2 +1}{\sigma(L)^2 -1} .$$ \endproclaim \demo{Proof} Since $\Cal Q-\text{isometries}$ preserve the distance between Lagrangian subspaces it follows from Proposition 4.9 that we can restrict our calculations to $$L = \left(\matrix I& I\\ T& I+T\endmatrix\right)$$ with diagonal $T$. By the Definition 5.1 we have \aligned d(LV_1,\,LV_2)&=\frac 12 \sup_{0\neq \xi \in \Bbb R^d} \ |\ \ln\langle \xi,\,T\xi \rangle - \ln\langle \xi,\,(T+I)\xi \rangle|\\ &= \frac 12\sup_{0\neq \xi \in \Bbb R^d}\ln\frac{\langle \xi,\,(I+T^{-1})\xi \rangle} {\langle \xi,\,\xi \rangle} = \max_{i}\frac{\ln \left(1+t_i^{-1}\right)}{2}= \frac{\ln \left(1+t_1^{-1}\right)}{2} \endaligned where $t_1\leq t_2 \leq \dots \leq t_d$ are the eigenvalues of $T$. The desired formula is now obtained by a straightforward calculation. \enddemo We introduce now an important property of a sequence of monotone maps. Let us consider a sequence of linear symplectic monotone maps $\{L_i\}_{i=1}^{+\infty}$. To simplify notation let us put $L^n = L_n\dots L_1.$ \proclaim{Definition 6.4} A sequence $\{L_1,L_2,\dots\}$ of monotone maps is called unbounded if, for all $w \in \text{int}\Cal{C}$, $$\Cal{Q}(L^nw) \rightarrow +\infty \ \ \text{as} \ \ n \rightarrow +\infty .$$ It is called strictly unbounded if, for all $w \in \Cal{C}, w \neq 0,$, $$\Cal{Q}(L^nw) \rightarrow +\infty \ \ \text{as} \ \ n \rightarrow +\infty .$$ \endproclaim \proclaim{Theorem 6.5} A sequence $\{L_1,L_2,\dots\}$ of maps, monotone with respect to $\Cal{C}$, is unbounded if and only if $$\bigcap_{n = 1}^{+\infty}L_1^{-1}L_2^{-1}\dots L_n^{-1} \C' = \text{one Lagrangian subspace},$$ where $\C'$ is the complementary sector. \endproclaim \proclaim{Corollary 6.6} If a sequence of monotone maps $\{L_1,L_2,\dots\}$ is unbounded then the sequence $\{L_2,L_3,\dots\}$ is also unbounded. \endproclaim \qed We were not able to find a proof of Corollary 6.6 independent of Theorem 6.5. \demo{Proof of Theorem 6.5} We note that $\{L_1,L_2,\dots\}$ is unbounded if and only if for any strictly monotone $L$ the sequence $\{L,L_1,L_2,\dots\}$ is unbounded. The next step is to prove that $\{L_1,L_2,\dots\}$ is unbounded if and only if for every strictly monotone $L$ $$\sigma_{\Cal C}\left(L^nL\right) \rightarrow +\infty \ \ \text{as} \ \ n \rightarrow +\infty .\tag{6.7}$$ Indeed the last property implies immediately that $\{L,L_1,L_2,\dots\}$ is unbounded and so, if it holds for all strictly monotone $L$, then also $\{L_1,L_2,\dots\}$ is unbounded. To prove the converse we will need the following fact from point set topology: \proclaim{Lemma} Let $f_1 \leq f_2\leq \dots ,$ be a nondecreasing sequence of real-valued continuous functions defined on a compact Hausdorff space $X$. If for every $x\in X$ $$\lim_{n\to +\infty}f_n(x) = +\infty,$$ then $$\lim_{n\to +\infty}\inf_{x\in X}f_n(x) = +\infty .$$ \endproclaim If $\{L_1,L_2,\dots \}$ is unbounded and $L$ is strictly monotone then we have $$\sigma_{\C}\left(L^nL\right) = \inf_{w\in int\C} {{\sqrt{\Q(L^nLw)}} \over {\sqrt{\Q(w)}}} \ge \inf_{0\neq w\in \C} {{\sqrt{\Q(L^nLw)}} \over {\sqrt{\Q(Lw)}}}\ \sigma_{\C}\left(L\right) .$$ Applying the Lemma to $$f_n(w)= {{\sqrt{\Q(L^nLw)}} \over {\sqrt{\Q(Lw)}}},\ n=1,2,\dots ,$$ which can be considered as a sequence of functions on the compact space of rays in $\Cal C$ we obtain \thetag{6.7}. Now we will be proving that \thetag{6.7} is equivalent to $$\bigcap_{n=1}^{+\infty}L^{-1}L_1^{-1}L_2^{-1}\dots L_n^{-1}\C' = \text{one Lagrangian subspace},$$ where $\C'= \C(V_2,\,V_1)$ is the complementary sector. The sectors $$\C'_n = L^{-1}L_1^{-1}L_2^{-1}\dots L_n^{-1}\C' = L^{-1}\left(L^n\right)^{-1}\C' = \C(L^{-1}\left(L^n\right)^{-1}V_2,\,L^{-1}\left(L^n\right)^{-1}V_1)$$ $n=1,2,\dots,$ form a nested sequence. We consider the space $Lag(\C')$ of all Lagrangian subspaces contained strictly in $\C'$ with the metric defined in Section 5. The sequence of subsets $\widehat{Lag}(\C'_n) \subset Lag(\C'), n= 1,2,\dots,\dots,$ is a nested sequence of compact subsets. Hence its intersection contains exactly one point (= Lagrangian subspace) if and only if their diameters converge to zero. By Corollary 5.3 the diameter of $\widehat{Lag}(\C'_n)$ is equal to the distance of the Lagrangian subspaces $L^{-1}\left(L^n\right)^{-1}V_2$ and $L^{-1}\left(L^n\right)^{-1}V_1$. By Proposition 6.3 this distance is equal to $$\ln \frac{s_n^2+1}{s_n^2-1},$$ where $s_n = \sigma_{\C'}\left(L^{-1}(L^n)^{-1}\right).$ But, by Proposition 6.2, $$\sigma_{\C'}\left(L^{-1}(L^n)^{-1}\right) =\sigma_{\C}(L^nL).$$ This shows that indeed the set $$\bigcap_{n=1}^{+\infty}\widehat{Lag}(C'_n)$$ contains exactly one point if and only if \thetag{6.7} holds. \enddemo We will use the following characterization of strict unboundedness. \proclaim{Theorem 6.8} Let $\{L_i\}_{i=1}^{+\infty}$ be a sequence of linear symplectic monotone maps. The following are equivalent. \aligned (1)& \ \text { The sequence } \ \{L_i\}_{i=1}^{+\infty} \ \text { is strictly unbounded, } \\ (2)& \ \inf_{0\neq w\in \C} {{\sqrt{\Q(L^nw)}} \over {\|w\|}} \rightarrow +\infty \ \text{as} \ n \rightarrow +\infty ,\\ (3)& \ \sigma(L^n) \rightarrow +\infty \ \text{as} \ n \rightarrow +\infty ,\\ (4)& \text {the sequence } \ \{L_i\}_{i=1}^{+\infty} \ \text { is unbounded and}\ L^{n_0}\ \text{is strictly monotone for some}\\ &\ n_0 \geq 1. \endaligned \endproclaim \demo{Proof} The Lemma used in the proof of Theorem 7.5 applied to the sequence of functions $$f_n(w)= {{\sqrt{\Q(L^nw)}} \over {\|w\|}},\ n=1,2,\dots,$$ shows that (1) $\Rightarrow$ (2). Further (2) $\Rightarrow$ (3) because $$\sigma(L^n) = \inf_{w\in int\C} {{\sqrt{\Q(L^nw)}} \over {\sqrt{\Q(w)}}} \geq \inf_{0\neq w \in \C}{{\sqrt{\Q(L^nw)}} \over {\|w\|}} \inf_{ w \in int\C}{{\|w\|}\over {\sqrt{\Q(w)}}}.$$ The implication (3) $\Rightarrow$ (4) is obvious ($\sigma(L^n) > 1$ if and only if $L^n$ is strictly monotone, cf. Proposition 6.1). Finally, let the sequence $\{L_i\}_{i=1}^{+\infty}$ be unbounded, and let $L^{n_0}$ be strictly monotone. By Corollary 6.6, the sequence $\{L_{n_0+1},L_{n_0+2},\dots\}$ is also unbounded. It follows that $\{L_i\}_{i=1}^{+\infty}$ is strictly unbounded. \enddemo The following example plays a role in the study of special Hamiltonian systems. \vskip.5cm \centerline{\bf Example.} \vskip.2cm Let $$L_n = \left(\matrix A_n& 0\\ 0& A_n^{*-1}\endmatrix\right) \left(\matrix I& 0 \\ P_n &I \endmatrix\right) \left(\matrix I& R_n \\ 0 &I \endmatrix\right),$$ $n=1,2,\dots ,$ be a sequence of monotone symplectic matrices with nonexpanding $A_n$, i.e., $\|A_n\xi\| \leq \|\xi\|$ for all $\xi$. We assume further that the symmetric matrices $R_n$ satisfy $$\tau_n' I \geq R_n \geq \tau_n I \ \ \text{and}\ \ \frac{\tau_n'}{\tau_n} \leq C,$$ for some positive constants $C$ and $\tau_n,\tau_n',\ n= 1,2,\dots .$ We do not make any assumptions about $P_n$ (beyond $P_n \geq 0$ which is forced by the monotonicity of $L_n$). Note that if a symmetric matrix $R$ satisfies $\tau I \leq R \leq \tau ' I$ then $\tau \|\eta \| \leq \|R \eta \| \leq \tau '\|\eta \|$. Indeed $$\left = \frac{\left} {\left}\ \left,$$ which yields the estimate. \proclaim{Proposition 6.9} If $\sum_{n=1}^{+\infty}\tau_n = +\infty$ then the sequence $\{L_1,L_2,\dots\}$ is unbounded. \endproclaim \demo{Proof} Let $w_1 = (\xi_1,\,\eta_1) \in int\C$ and $w_{n+1} = (\xi_{n+1},\,\eta_{n+1}) = L_nw_n, n=1,2,\dots$. Our goal is to show that $$q_n = \Q (w_n) \rightarrow +\infty\ \ \text{as}\ \ n \rightarrow +\infty.$$ We have $\xi_{n+1} = A_n\left( \xi_n + R_n \eta_n \right)$, so that $$\|\xi_{n+1}\| \leq \|\xi_n\| + \|R_n \eta_n \| \leq \|\xi_n\| + \tau_n'\|\eta_n \| \leq \|\xi_1\| + \sum_{i=1}^{n}\tau_i'\|\eta_i \|. \tag{6.10}$$ At the same time, $q_n = \left<\xi_n,\,\eta_n\right> \leq \|\xi_n\|\|\eta_n \|,$ so that $$\|\eta_n \| \geq \frac{q_n}{\|\xi_n\|}, \tag{6.11}$$ and hence (see also the proof of Theorem 4.4) $$q_{n+1} \geq q_n + \left \geq q_n + \tau_n \|\eta_n \|^2 \geq q_n + \tau_n \|\eta_n \| \frac{q_n}{\|\xi_n\|}.$$ Using \thetag{6.10}, we obtain from the last inequality $$\frac{q_{n+1}}{q_n} \geq 1 + \frac{\tau_n \|\eta_n \|} {\|\xi_1\| + \sum_{i=1}^{n-1}\tau_i'\|\eta_i \|} \geq 1 + \frac1C\frac{\tau_n' \|\eta_n \|} {\|\xi_1\| + \sum_{i=1}^{n-1}\tau_i'\|\eta_i \|}. \tag{6.12}$$ If $\sum_{i=1}^{+\infty}\tau_i'\|\eta_i \| < +\infty$ then by \thetag{6.10} the sequence $\|\xi_n\|$ is bounded from above, and hence by \thetag{6.11} the sequence $\|\eta_n \|$ is bounded away from zero which is a contradiction (in view of $\sum_{i=1}^{+\infty}\tau_i' = +\infty$). Hence $$\sum_{i=1}^{+\infty}\tau_i'\|\eta_i \| = +\infty.$$ Now the claim follows from \thetag{6.12} and the following \proclaim{Lemma 6.13} For a sequence of positive numbers $a_0,a_1,\dots ,$ if $$\sum_{n=1}^{+\infty}a_n = +\infty \ \ \text{then } \ \ \sum_{n=1}^{+\infty}\frac{a_n} {\sum_{i=0}^{n-1}a_i} = +\infty.$$ \endproclaim \demo{Proof of the Lemma} We have for $1 \leq k \leq l$ $$\sum_{n=k}^{l}\frac{a_n}{\sum_{i=0}^{n-1}a_i} \geq \frac{\sum_{n=k}^{l}a_n} {\sum_{n=0}^{l}a_n} \rightarrow 1\ \ \text{as} \ \ l \rightarrow +\infty .$$ \enddemo \enddemo %this file must be called "s7.tex" \vskip.7cm \subhead \S 7. PROPERTIES OF THE SYSTEM AND THE FORMULATION OF THE RESULTS \endsubhead \vskip.4cm In this section we describe in detail the class of systems to which the present paper applies. We divide the conditions that the systems must satisfy into several groups. The multitude of conditions is justified by the fact that we want to include discontinuous systems (there is only one way to be continuous but many ways to be discontinuous !). In the case of a symplectomorphism of a compact symplectic manifold most of these conditions are vacuous. Because of that we will single out this case and we will refer to it as the smooth case. The bulk of our effort is devoted to the discontinuous case. \medskip {\bf A. The phase space.}\par In the smooth case the phase space $\Cal M$ is a smooth compact symplectic manifold. In the discontinuous case it is a disjoint union of nice subsets of the linear symplectic space. More precisely, let us consider the standard linear symplectic space $\Cal W = \Re^d \times \Re^d$ equipped with a Riemannian metric uniformly equivalent to the standard Euclidean scalar product. For simplicity we assume that the Riemannian volume element is equal to the symplectic volume element (measure) $\mu$. By a submanifold of $\Cal W$ we mean an embedded submanifold of $\Cal W$. Further, we define a piece of a submanifold $\Cal N$ to be a compact subset of $\Cal N$ which is the closure of its interior (in the relative topology of the submanifold $\Cal N$). A piece $X$ of a submanifold has a well defined boundary which we will denote by $\partial X$ (it is the set of boundary points with respect to the relative topology of the submanifold). Notice that at every point of a piece of a submanifold, including a boundary point, we have a well defined tangent subspace. A submanifold carries the measure defined by the Riemannian volume element. The measure of the boundary of a piece of a submanifold is not necessarily zero. The phase space is made up of pieces of $\Cal W$ which have regular boundaries in the sense of the following definition. \proclaim {Definition 7.1} A compact subset $X \subset \Cal W$ is called regular if it is a finite union of pieces $X_i,i=1,\dots,k,$ of $2d-1$-dimensional submanifolds $$X = X_1 \cup \dots \cup X_k.$$ The pieces overlap at most on their boundaries, i.e., $$X_i \cap X_j \subset \partial X_i \cup \partial X_j,\ i,j = 1,\dots k;$$ and the boundary $\partial X_i$ of each piece $X_i,\ i=1,\dots k,$ is a finite union of compact subsets of $2d-2$-dimensional submanifolds. \endproclaim To picture such sets one can think of the boundary of a $2d$-dimensional cube. The faces are pieces of $2d-1$-dimensional submanifolds and they clearly overlap only at their boundaries. The boundary of each face is a union of pieces of $2d-2$ dimensional submanifolds (actually it is a union of $2d-2$ dimensional cubes). Let us stress that in the definition of a regular set we do not impose any requirements on the $2d-2$ dimensional subsets in the boundary. Due to the generality of the definition one cannot even claim that the union of two regular sets is regular. As a consequence of Definition 7.1 the natural measures on the pieces $X_i,i=1,\dots,k,$ of any regular subset $X$ can be concocted to give a well defined measure $\mu_X$ on $X$ (the $2d-1$ dimensional Riemannian volume). It is so because the boundaries of the pieces being themselves finite unions of subsets of submanifolds of lower dimension have zero measure. Hence if we put $$\partial X = \bigcup_{i=1}^k\partial X_i,$$ then $$\mu_X\left(\partial X\right) = 0. \tag {7.2}$$ Moreover, by the regularity of the measure $\mu_{X}$, it follows from \thetag{7.2} that, if we denote by $(\partial X)^\delta$ the $\delta$-neighborhood of $\partial X$ in $X$, then $$\lim_{\delta \to 0} \mu_X\left((\partial X)^\delta\right) = 0. \tag {7.3}$$ Further we have the following Proposition. \proclaim{Proposition 7.4} For a subset $Y$ of $X\subset \Cal W$ let the $\delta$-neighborhood of $Y$ in $\Cal W$ be denoted by $Y^\delta$, i.e., $$Y^\delta = \{x \in \Cal W\,|\, \text{d}(x,Y) \leq \delta \}.$$ If $X$ is a regular ($2d-1$-dimensional) subset of $\Cal W$ and $Y \subset X$ is {\bf closed} then $$\lim_{\delta \rightarrow 0} {{\mu(Y^\delta)}\over 2\delta } = \mu_X(Y).$$ \endproclaim We will use in the following only the weaker property $$\limsup_{\delta \rightarrow 0} {{\mu(Y^\delta)}\over \delta } \leq const\mu_X(Y). \tag 7.5$$ We leave the proof of Proposition \thetag {7.4} or of the weaker property \thetag{7.5} to the reader. \proclaim {Definition 7.6} A compact subset $\Cal M \subset \Cal W$ is called a symplectic box if the boundary $\partial \Cal M$ of $\Cal M$ is a regular subset of $\Cal W$ and the interior $int\Cal M$ of $\Cal M$ is connected and dense in $\Cal M$. \endproclaim We can now formulate the requirements on the phase space of a discontinuous system. {\it The phase space of our system is a finite disjoint union of symplectic boxes.} To simplify notation we assume that the phase space is just one symplectic box $\Cal M$. It will be quite obvious how to generalize the subsequent formulations to the case of several symplectic boxes. \medskip {\bf B. The map $T$ (the dynamical system).} In the smooth case the map $T$ is a symplectomorphism $T: \Cal M \to \Cal M$. In the discontinuous case we assume that the symplectic box $\Cal M$ is partitioned in two ways into unions of equal number of symplectic boxes $$\Cal M = \Cal M_1^+ \cup \dots \cup \Cal M_m^+ = \Cal M_1^- \cup \dots \cup \Cal M_m^-.$$ Two boxes of one partition can overlap at most on their boundaries, i.e., $$\Cal M_i^{\pm}\cap \Cal M_j^{\pm} \subset \partial \Cal M_i^{\pm}\cap \partial \Cal M_j^{\pm} ,\ \ i,j = 1,\dots ,m.$$ The map $T$ is defined separately on each of the symplectic boxes $\Cal M_i^+, \ i =1,\dots,m$. It is a symplectomorphism of the interior of each $\Cal M_i^+$ onto the interior $\Cal M_i^-, \ i =1,\dots,m$ and a homomorphism of $\Cal M_i^+$ onto $\Cal M_i^-, \ i =1,\dots,m$. We assume that the derivative $DT$ is well behaved near the boundaries of the symplectic boxes. Namely, we assume that it satisfies the Katok-Strelcyn conditions so that we can apply their results \cite {K-S} on the existence of (un)stable manifolds and the property of absolute continuity. We will say that $T$ is a (discontinuous) symplectic map of $\Cal M$. Formally $T$ is not well defined on the set of points which belong to the boundaries of several plus-boxes: it has several values. We adopt the convention that the image of a subset of $\Cal M$ under $T$ contains all such values. Let us introduce the singularity sets $\R^{+}$ and $\R^-$. $$\R^{\pm} = \{ p\in \Cal M \;|\; p \text { belongs to at least two of the boxes }\Cal M_i^{\pm}, i =1,\dots,m\}.$$ The plus-singularity set $\R^+$ is a closed subset and $T$ is continuous on its complement. Similarly $T^{-1}$ is continuous on the complement of $\R^-$. We have that $\R^+ \cup \partial \Cal M$ is the union of all the boundaries of the plus-boxes and $\R^- \cup \partial \Cal M$ is the union of all the boundaries of the minus-boxes, i.e., $$\R^{\pm} \cup \partial \Cal M = \bigcup_{i=1}^m\partial \Cal M_i^{\pm}.$$ Note that most of the points in the boundary $\partial \Cal M$ of $\Cal M$ do not belong to $\R^-$ or $\R^+$. We assume that the singularity sets $\R^{\pm}$ and the union of boundaries $\bigcup_{i=1}^m\partial \Cal M_i^{\pm}$ are regular sets. An important role in our discussion will be played by the singularity sets of the higher iterates of $T$. We define for $n\geq 1$ $$\R^+_n = \R^+ \cup T^{-1}\R^+ \cup \dots \cup T^{-n+1}\R^+.$$ and $$\R^-_n = \R^- \cup T\R^- \cup \dots \cup T^{n-1}\R^-.$$ We have that $T^n$ is continuous on the complement of $\R^+_n$ and $T^{-n}$ is continuous on the complement of $\R^-_n$. \proclaim{Regularity of singularity sets} We assume that for every $n \geq 1$ the singularity sets $\R^+_n$ and $\R^-_n$ are regular. \endproclaim We will formulate, in Lemma 7.7, an abstract condition on the first power of $T$ alone that guarantees the regularity of the singularity sets but it requires that the map is a diffeomorphism on every symplectic box up to and including its boundary i.e., it can be extended to a diffeomorphism of an open neighborhood of $\Cal M_i^+$ onto an open neighborhood of $\Cal M_i^-,\ i=1,\dots,m$. Hence it is very appealing to restrict the discussion to such maps. Unfortunately, such a restriction would leave out important examples: billiard systems where the derivative may blow up at the boundary. The conditions in the work of Katok and Strelcyn \cite{K-S} were tailored for such systems. Nevertheless the reader is invited to be generous with the restrictions on the regularity of $T$, this will make it easier to follow the main line of the argument. \medskip {\bf C. Monotonicity of $T$.} \par In the smooth case we assume that two continuous bundles of transversal Lagrangian subspaces are chosen in an open subset $U \subset \Cal M$ ($U$ is not necessarily dense). We denote them by $\{V_1(p)\}_{p\in \U}$ and $\{V_2(p)\}_{p\in \U}$ respectively. In the discontinuous case we assume that two continuous bundles of transversal Lagrangian subspaces are chosen in the interior of the symplectic box $\Cal M$. Their limits (if they exist at all) at the boundary $\partial\M$ are allowed to be nontransversal (to have nonzero intersection). We consider the bundle of sectors (see Definition 4.1) defined by these Lagrangian subspaces $$\Cal C(p)=\Cal C(V_1(p),\,V_2(p)).$$ Let $$\Cal C'(p)=\Cal C(V_2(p),\,V_1(p))$$ be the complementary sector. We require that the derivative of the map and its iterates, where defined, is monotone, if only monotonicity is well defined (cf. Definition 4.3). More precisely, in the smooth case we require that, if $p\in U$ and $T^kp\in U$ for $k\geq 1$, then $$D_pT^k\C(p)\subset\C(T^kp).$$ In the discontinuous case we assume that $$D_pT \Cal C(p) \subset \Cal C(Tp)$$ for points $p$ in the interior of every symplectic box $\Cal M_i^+, i = 1,\dots,m$ . We call a point $p \in int\Cal M$ ($p\in U$ in the smooth case) {\it strictly monotone in the future} if there is $n \geq 1$ such that $D_pT^{n}$ is defined and it is strictly monotone (in the smooth case we require, naturally, that $T^np\in U$), i.e., $$D_pT^{n} \Cal C(p) \subset int\Cal C(T^{n}p) \cup \{0\}.$$ Similarly a point $p$ is called {\it strictly monotone in the past} if there is $n \geq 1$ such that $D_pT^{-n}$ is strictly monotone with respect to the complementary sectors, i.e., $$D_pT^{-n} \C'(p) \subset int\C'(T^{-n}p) \cup \{0\}.$$ It is clear that if $p$ is strictly monotone in the future then its preimages are also strictly monotone in the future. By Proposition 6.2 we also have that if $p$ is strictly monotone in the future then there is $n \geq 1$ such that $T^np$ is strictly monotone in the past. \proclaim{Strict monotonicity almost everywhere} We assume that almost all points in $\Cal M$ ($U$ in the smooth case) are strictly monotone. \endproclaim This property implies that all Lyapunov exponents are non-zero almost everywhere in $\Cal M$ (in $U$ in the smooth case). The proof of this fact is quite simple and can be found in \cite {W1}. It will also follow easily from our Proposition 8.4. Thus by the work of Pesin \cite {P} in the smooth case and of Katok and Strelcyn \cite {K-S} in the discontinuous case, through almost every point there are local stable and unstable manifolds of dimension $d$ and the foliations into these manifolds are absolutely continuous. The sectors $\C(p)$ contain the unstable Lagrangian subspaces (tangent to the unstable manifolds) and the complementary sectors $\C'(p)$ contain the stable Lagrangian subspaces (tangent to the stable manifolds). The sectors can be viewed as a priori approximations to the unstable and stable subspaces. We will refer to the sectors as unstable sector and stable sector respectively. This ends the list of required properties for the smooth case. The last three properties of our system are introduced only for the discontinuous case. \medskip {\bf D. Alignment of Singularity sets} \par For a codimension one subspace in a linear symplectic space its characteristic line is, by definition, the skeworthogonal complement (which is a one dimensional subspace). \proclaim {Proper alignment of $\R^-$ and $\R^+$} We assume that the tangent subspace of $\R^-$ at any $p \in \R^-$ has the characteristic line contained strictly in the sector $\C(p)$ and that the tangent subspace of $\R^+$ at any $p \in \R^+$ has the characteristic line contained strictly in the complementary sector $\C'(p)$. We say that the singularity sets $\R^-$ and $\R^+$ are properly aligned. \endproclaim Let us note that if a point in $\R^\pm$ belongs to several pieces of submanifolds then we require that the tangent subspaces to {\it all} of these pieces have characteristic lines in the interior of the sector. \par It will be clear from the way in which the proper alignment of singularity sets is used in Section 12 that it is sufficient to assume that there is $N$ such that $T^N\R^-$ and $T^{-N}\R^+$ are properly aligned. We will show, in section 14, that for the system of falling balls even this weaker property fails. Hence the study of ergodicity of this system would require some further relaxation of this property. \par Let us note that it is helpful in establishing the regularity of singularity sets $\R^{\pm}_n$ if the boundaries of $\Cal M$ have tangent subspaces with characteristic lines contained in the boundary of the sector. It is so in some examples. More precisely we have the following lemma. \proclaim {LEMMA 7.7} If the map $T$ is a diffeomorphism up to and including the boundaries of the symplectic boxes $\Cal M^+_1,\dots,\Cal M^+_m$, satisfies properties C, D and the boundary $\partial\Cal M$ of $\Cal M$ has all the tangent subspaces with characteristic lines contained in the boundary of the sectors then the sets $\R^{\pm}_n, n\geq 1,$ are regular (i.e. the property {\bf B} is automatically verified). \endproclaim \demo{Proof} Let us recall that, by assumption, $\R^-$ and $\bigcup_{i=1}^m\partial\Cal M^+_i$ are regular subsets. Further the intersection of any properly aligned regular subset $X$ (the characteristic lines of its tangent subspaces are contained strictly in the unstable sector $\C$) with any of the symplectic boxes $\Cal M^+_1,\dots,\Cal M^+_m$ is a regular subset. Indeed let $X_1,\dots,X_p$ be the pieces of $2d-1$ dimensional manifolds which make up $X$ ($X=\bigcup_{i=1}^p X_i$) and $Y_1,\dots,Y_q$ be the pieces of $2d-1$ dimensional manifolds which make up the boundary of say $\Cal M^+_1$ ($\partial\Cal M^+_1=\bigcup_{j=1}^q Y_j$). By the proper alignment of the pieces we can assume that any $X_i$ and $Y_j$ are pieces of transversal submanifolds. Hence, the intersection of the submanifolds is a submanifold of dimension $2d-2$, and therefore $X_i\cap Y_j$ is a compact subset of the $2d-2$-dimensional manifold. It follows that the intersection of $X_i$ with $\Cal M^+_1$ is a finite union of pieces of the $2d-1$ dimensional manifold and also a regular subset. The same can be repeated for the other symplectic boxes $\Cal M^+_2,\dots,\Cal M^+_m$. Moreover we have that any $(X_i\cap \Cal M^+_1)\cup\partial \Cal M^+_1,\ i = 1,\dots,p,$ is a regular subset and further $(X\cap \Cal M^+_1)\cup\partial \Cal M^+_1$ is a regular subset. It follows that $T\left((X\cap\M^+_1)\cup\partial\M^+_1\right)$ $=(TX\cap \Cal M^-_1)\cup\partial \Cal M^-_1$ is a regular subset and, after repeating the argument for the other symplectic boxes, we get that for any regular and properly aligned subset $X$ the sets $TX\cup\bigcup_{i=1}^m\partial \Cal M^-_i$ and $TX\cup \R^-$ are regular. Moreover $TX\cup \R^-$ is properly aligned. Now the proof can clearly be completed by induction since $$\R^-_{n+1} = T\R^-_n\cup \R^-.$$ The argument for $\R^+$ is completely analogous. \enddemo \par The last two properties are rather technical. They are used only in Section 13 in the proof of the tail bound'. It remains an open question if one can do without them. \medskip {\bf E. Noncontraction property.} \par There is a constant $a, \ 0 < a \leq 1,$ such that, for any $n \geq 1$ and every $p \in \Cal M \setminus \R^+_n$, $$\|D_pT^nv\| \geq a \|v\|,$$ for every vector $v$ in the sector $\C(p)$. This condition is hard (or impossible) to check unless we know of a special norm (or any other homogeneous function on the tangent vectors in the sector uniformly equivalent to the Riemannian metric) which does not decrease under $DT$. In the theory of Anosov systems a metric with such a property can be constructed, and it is called a Lyapunov metric. In the case of semi-dispersing billiards the noncontraction property does not hold in general. Chernov and Sinai \cite{CS} were using a semi-norm (the configuration norm) which does not decrease on vectors in the unstable sector, but it is not equivalent to the Riemannian metric. The reason why it suffices is quite intricate. It turns out that the ($d$-dimensional) subspace of null vectors of the semi-norm (on which the seminorm vanishes) is contained in the boundary of the sectors and in the subspaces tangent to $\partial \M$ . Moreover, all of $\Cal S^-$($\Cal S^+$) is in the image (preimage) of $\partial \M$. Because of these properties, Chernov and Sinai were able to establish that the measure of a $\delta$-neighborhood of the singularity set in this semi-norm is $\Cal O(\delta)$. It is possible to formulate such properties axiomatically, but we found it too cumbersome. \par Finally, let us note that the noncontraction property is automatically satisfied, if $T$ is differentiable up to and including the boundaries of symplectic boxes, the bundle of sectors is defined in all of $\M$ (including the boundaries), and $DT$ is strictly monotone everywhere. Indeed, in such a case we can measure vectors in $\C$ by the form $\Q$ defined by the bundle of sectors $DT^{-1}\C$ which are uniformly larger than $\C$ by compactness. The form $\Q$ does not decrease under $DT$, and, by a compactness argument, it is equivalent to the Riemannian metric in the sectors $\C$. \medskip {\bf F. Sinai - Chernov Ansatz.} \par This is a property of the derivatives of the iterates of $T$ on the singularity set itself, of $T^{-1}$ on $\R^+$ and of $T$ on $\R^{-}$. Namely, we require that, for almost every point in $\R^-$ with respect to the measure $\mu_{\R}$ ($\mu_{\R}$ is the $2d-1$ dimensional Riemannian volume on $\R^- \cup \R^+$), all iterates of $T$ are differentiable and for almost every point in $\R^+$ all iterates of $T^{-1}$ are differentiable. Note that the last requirement holds automatically under the assumptions of Lemma 7.7. Moreover, \medbreak{\sl we assume that for almost every point $p \in \R^{-}$ with respect to the measure $\mu_{\R}$, the sequence of derivatives $\{D_{T^np}T\}_{n\geq 0}$ is strictly unbounded (cf. Definition 6.4). Analogous property must hold for $\R^+$ and $T^{-1}$.} \medbreak By Theorem 6.8 the forward part of Sinai - Chernov Ansatz is equivalent to the following property. For almost every point $p \in \R^{-}$ with respect to the measure $\mu_{\R}$ $$\lim_{n\to +\infty} \sigma(D_pT^n) = +\infty,$$ where the coefficient $\sigma$ is defined at the beginning of Section 6. \par In several examples unboundedness holds for all orbits by virtue of Proposition 6.9, but strict monotonicity is hard to establish. We have completed the formulation of the conditions. Under these conditions we will prove the following two theorems. \proclaim{Main Theorem (Smooth case)} For any $n\geq 1$ and any $p \in U$ such that $T^np\in U$ and $\sigma(D_pT^n) > 1$ (i.e., $p$ is strictly monotone) there is a neighborhood of $p$ which is contained in one ergodic component of $T$. \endproclaim It follows from this theorem that if $U$ is connected and every point in it is strictly monotone then $\bigcup_{i=-\infty}^{+\infty}T^iU$ belongs to one ergodic component. Such a theorem was first proven by Burns and Gerber \cite {BG} for flows in dimension 3. It was later generalized by Katok \cite {K} to arbitrary dimension, and recently also to a non-symplectic framework \cite{K1}. Our proof is a byproduct of the preparatory steps in the proof of the following \proclaim{Main Theorem (Discontinuous case)} For any $n\geq 1$ and for any $p \in \Cal M \setminus \R^+_n$ such that $\sigma(D_pT^n) > 3$ there is a neighborhood of $p$ which is contained in one ergodic component of $T$. \endproclaim Let us note that the conditions of the last theorem are satisfied for almost all points $p\in \Cal M$. Indeed let $$\Cal M_{n,\epsilon} = \{p\in\M\;|\;\sigma(D_pT^n) > 1+ \epsilon \}.$$ Since almost all points are strictly monotone, then $$\bigcup_{n=1}^{+\infty}\bigcup_{\epsilon>0} \Cal M_{n,\epsilon}$$ has full measure. By the Poincare Recurrence Theorem and the super-multiplicativity of the coefficient $\sigma$, we conclude that $$\bigcup_{n=1}^{+\infty} \Cal M_{n,2}$$ has also full measure. Hence the theorem implies, in particular, that all ergodic components are essentially open. The theorem allows also to go further since we assume that only finitely many iterates of $T$ are differentiable at $p$, so that we can apply it to orbits that end up on the singularity sets both in the future and in the past (e.g. $p\in \R^-$ and $T^np \in \R^+$). We need though a specific amount of hyperbolicity on this finite orbit ($\sigma(D_pT^n) > 3$); note that in the smooth case any amount of hyperbolicity ($\sigma(D_pT^n) > 1$) is sufficient. This theorem gives a fairly explicit description of points which can lie in the boundary of an ergodic component. By checking that there are only few such points (e.g. that they form a set of codimension 2) one can conclude that a given system is ergodic. In all the examples that we know, any point with an infinite orbit (in the future or in the past) has the unbounded sequence of derivatives (in the sense of Definition 6.4). In such a case, it follows from Theorem 6.8 that for any strictly monotone point with the infinite orbit in the future the condition $\sigma(D_pT^n) > 3$ is satisfied automatically, if only n is sufficciently large. There is no need to formulate the Main Theorem separately for a point $p$ which has only the backward orbit ($p \in \Cal S^+$). We can simply apply the theorem to $T^-np$ (one can appreciate now the convenience of Proposition 6.2). \topinsert \vskip 3in \hsize=4.5in \noindent{\bf Figure 6} The Baker Map and the Modified Baker Map. \endinsert \topinsert \vskip 3in \hsize=4.5in \noindent{\bf Figure 7} The discontinuity lines of the Modified Baker Map. \endinsert Let us finish this Section with an example where the role of the proper alignment of singularity sets is exposed. The well known Baker's Transformation maps the unit square as shown in Fig.6a and it is ergodic. Let us consider a variation of this construction where the square is stretched and squeezed as before but now the middle one half is left at the bottom and the quarters on the left and right are translated to the top as shown in Fig.6b. This time the map $T$ is not ergodic. The ergodic components are separated by the vertical segment in the middle although for any point $p$ on this segment we have that $$\sigma(D_pT^2) = 4.$$ Of all the conditions formulated above only the proper alignment of singularity sets is violated. Namely, part of $\R^-$ has stable (vertical) direction (all of $\R^+$ has stable direction, which is fine), see Fig.7 where $\R^\pm$ are indicated by bold lines. For the standard Baker's transformation the condition of the proper alignment is clearly satisfied. %this file should be called s8.tex \vskip.7cm \subhead \S 8. CONSTRUCTION OF THE NEIGHBORHOOD AND THE COORDINATE SYSTEM \endsubhead \vskip.4cm We will construct a convenient coordinate system in a neighborhood of a strictly monotone point $p\in\Cal M$. There are two cases: strict monotonicity in the past and strict monotonicity in the future but they are completely symmetric. Therefore, we will discuss only one of them. Namely, we assume that there is $N \ge 1$ such that \aligned i)&\qquad T^{-N} \text{ \ is differentiable at \ } p\;:\; p \not\in \R^-_N\cup\partial \M, \text{\it\ (discontinuous case)}\\ &\qquad T^{-N}p \in U, \text{\it\ (smooth case)}\\ ii)&\qquad D_pT^{-N} \text{ \ is strictly monotone.} \endaligned \tag 8.1 We will find a neighborhood $\Cal U(p)$ in which there is an abundance of long'' stable and unstable manifolds. Let us emphasize that we have assumed only that $p$ (and its $N$ preimages) does not belong to $\R^-$ but it may very well belong to $\R^+$. Such a level of generality is crucial in obtaining local ergodicity also for points in the singularity sets $\R^\pm$. \par Our first requirement on the neighborhood is that $T^{-N}$ is a diffeomorphism of $\Cal U(p)$ onto a neighborhood of $\bar p = T^{-N}p$ (and in the smooth case both neighborhoods are contained in $U$). By the Darboux theorem a symplectic manifold looks locally like a piece of the standard linear symplectic space. Hence reducing $\Cal U(p)$ further, if necessary, we can identify it with a neighborhood $\Cal U$ of the standard linear symplectic space $\Re^{d}\times\Re^{d}$ $$\Cal U = \Cal U_a = \Cal V_a \times \Cal V_a,$$ where $$\Cal V_a = \{x = (x^1,\dots,x^d)\in\Re^{d}\;|\; |x^i| < a, i = 1,\dots,d \}.$$ (In the discontinuous case we have assumed from the very beginning that a symplectic box is a subset in $\Re^d \times \Re^d$). We assume that the point $p$ becomes the zero point and the symplectic structure is the standard one. In particular all the tangent spaces in $\Cal U(p)$ can be identified with $\Re^{d}\times\Re^{d}$. The choice of a cube for the shape of the neighborhood is important only for some of the arguments in Section 11, otherwise we want to stress that our neighborhood $\Cal U$ is the cartesian product of neighborhoods $\Cal V_a$ in the $d$-dimensional linear space and we will not use any special directions there. Let us further introduce for any positive $\rho$ the following sectors in the tangent space of $\Cal U$. $$\C_\rho = \{(\xi,\,\eta)\in\Re^{d}\times\Re^{d}\;|\; \|\eta\| \leq \rho\|\xi\| \}$$ and the complementary sector $$\C'_\rho = \{(\xi,\,\eta)\in\Re^{d}\times\Re^{d}\;|\; \|\xi\| \leq \rho^{-1}\|\eta\| \}.$$ By the assumption \thetag {8.1} the sector $D_{\bar p}T^{N}\C(\bar p)$ is strictly inside the sector $\C(p)$. We change coordinates in $\Cal U$ in such a way that for some $\tilde{\rho} < 1$ $$\C'(p) = \C'_{{\tilde{\rho}}^{-1}}$$ and $$D_{\bar p}T^{N}\C(\bar p) \subset \C_{\tilde{\rho}}.$$ By Propositions 5.11 and 6.3 this can be done with $\tilde{\rho} = (\sigma(D_{\bar p}T^{N}))^{-1}$ . We pick $\rho,\ \tilde{\rho} < \rho <1$. By the continuity of the sector bundle $\C(z), z\in \Cal U$, and of the derivative $D_{y}T^{N}, y \in T^{-N}\Cal U$, if we reduce the size of $\Cal U$ appropriately, we can achieve that for any $z \in \Cal U$ $$\C'(z) \subset \C'_{\rho^{-1}} \tag {8.2}$$ and for any $y \in T^{-N}\Cal U$ $$D_{y}T^{N}\C(y) \subset \C_\rho . \tag {8.3}$$ The properties \thetag {8.2} and \thetag {8.3} can be viewed as a uniform separation of sectors $D_{y}T^{N}\C(y)$ and $\C(T^{N}y)$ over $\U$ ( $y \in T^{-N}\U$). This property seems to be asymmetric in time, i.e., $T$ plays here a different role than $T^{-1}$. Nevertheless we can obtain from \thetag {8.2} and \thetag {8.3} the following fundamental Proposition which is perfectly symmetric in time. We will say that a point $z \in \Cal U$ has $k$ spaced returns in a given time interval if there are $k$ moments of time in this interval $$i_1 < i_2 < \dots < i_k$$ at which $z$ visits $\Cal U$, i.e., $$T^{i_j}z \in \Cal U \text {\ for \ }j=1,\dots,k,$$ and the visits are spaced by at least time $N$, i.e., $$i_{j+1} - i_j \geq N \text {\ for \ }j=1,\dots,k-1.$$ \proclaim{Proposition 8.4} If $T^n$ is differentiable at $z\in \Cal U$ for $n \geq N$ and $z' = T^nz\in \Cal U$ then $$D_zT^n \C_{\rho^{-1}} \subset \C_\rho \tag 8.5u$$ and $$D_{z'}T^{-n} \C'_\rho \subset \C'_{\rho^{-1}}. \tag 8.5s$$ Moreover for $(\xi',\,\eta') = D_zT^n(\xi,\,\eta)$ if $(\xi,\,\eta) \in \C_\rho$ then $$\|\xi'\| \geq b\rho^{-k}\|\xi\|, \tag 8.6u$$ and if $(\xi',\,\eta') \in \C'_{\rho^{-1}}$ then $$\|\eta\| \geq b\rho^{-k}\|\eta'\|, \tag 8.6s$$ where $k$ is the maximal number of spaced returns of $z$ in the time interval from $N$ to $n$ and $$b = \sqrt{1-\rho^4}.$$ \endproclaim \demo{Proof} It follows from \thetag {8.2} that for any $x \in \Cal U$ $$\C_\rho \subset \C_{\rho^{-1}} \subset \C(x).$$ Hence $$D_zT^{n-N}\C_{\rho^{-1}} \subset \C(T^{n-N}z).$$ Now \thetag {8.5u} follows from \thetag {8.3}. Let us further note that \thetag {8.3} implies that for any $x \in \Cal U$ $$D_xT^{-N}\C'_{\rho} \subset \C'(T^{-N}x).$$ We obtain \thetag {8.5s} by applying first $D_{z'}T^{-N}$, then $D_{T^{-N}z'}T^{-n+N}$ and using \thetag {8.2} again. The properties \thetag {8.6u} and \thetag {8.6s} follow from \thetag {8.5u} and \thetag {8.5s} respectively in exactly the same way. We will prove only the unstable version. To measure vectors in $\C_\rho$ we use the form $\Q$ associated with the sector $\C_{\rho^{-1}}$. It is equal to $$\rho^{-1}\|\xi\|^2 - \rho\|\eta\|^2$$ and on every spaced return to $\Cal U$ the value of this form on vectors from $\C_{\rho^{-1}}$ gets increased by at least the factor $\rho^{-2}$, cf. Propositions 5.11 and 6.3 . It remains to compare the value of this form at $(\xi,\,\eta) \in \C_\rho$ with $\|\xi\|^2$. We have $$\rho^{-1}\|\xi\|^2 \geq \rho^{-1}\|\xi\|^2 - \rho\|\eta\|^2 \geq (\rho^{-1}-\rho^3)\|\xi\|^2$$ which immediately yields \thetag {8.6u}. \enddemo Having achieved the symmetry with respect to the direction of time we will restrict the discussion in the next section to the case of unstable manifolds using the unstable version of Proposition 8.4. It can be then repeated for the stable manifolds with the use of the stable version. \proclaim{Remark 8.7} If $p$ is not a periodic point then by reducing the neighborhood $\Cal U$ we can guarantee that any successive visits to $\Cal U$ are spaced by at least the time $N$. In such a case the number of spaced returns becomes simply the number of returns to $\Cal U$. It is so also if $N=1$. \endproclaim % this file should be called s9x11.tex \vskip.7cm \subhead \S 9. UNSTABLE MANIFOLDS IN THE NEIGHBORHOOD $\Cal U$ \endsubhead \vskip.4cm Let us repeat the properties of $T$ and $\Cal U$ established in the previous section which we will rely upon. Note that the original point $p$ does not appear explicitly. There is a positive number $\rho < 1$ such that for any $z \in \Cal U$ $$\C_\rho \subset \C_{\rho^{-1}} \subset \C(z) \tag 9.1$$ and for any $y \in T^{-N}\Cal U$ $$D_yT^N \C(y) \subset \C_\rho . \tag 9.2$$ It follows that if $z\in \Cal U$ and $T^nz\in \Cal U$ for $n \geq N$ then $$D_zT^n \C_{\rho^{-1}} \subset \C_\rho. \tag 9.3$$ Moreover if $$(\xi,\,\eta) \in \C_\rho \text{ and } (\xi',\,\eta') = D_zT^n (\xi,\,\eta)$$ then $$\|\xi'\| \geq b\rho^{-k}\|\xi\| \tag 9.4$$ where $k$ is the maximal number of spaced returns to $\Cal U$ between the times $N$ and $n$ and $b= \sqrt{1-\rho^{4}}$. By the Pesin theory \cite {P} in the smooth case and the Katok-Strelcyn theory \cite {K-S} in the general case, for almost all $z \in \Cal U$, we have a local unstable manifold $W_{loc}^u(z)$ through $z$. Further, the tangent spaces of $W_{loc}^u(z) \cap \Cal U$ are Lagrangian subspaces contained in $\C_\rho$. But the general theory does not give us a good hold on their size. Let $\pi_i : \Cal V \times \Cal V \to \Cal V,\ i=1,2,$ be the projection on the first and second component respectively. We denote by $\Cal B(c;r)$ the open ball with the center at $c$ and the radius $r$. \proclaim{Definition 9.5 } We say that an unstable manifold in $\Cal U$ of a point $z=(z_1,\,z_2) \in \Cal U$ has size $\varepsilon$ if it contains the graph of a smooth mapping from $\Cal B(z_1;\varepsilon)$ to $\Cal V$. We denote such a graph by $W_{\varepsilon}^u(z)$ and we will call it the unstable manifold of size $\varepsilon$. \endproclaim By the definition of an unstable manifold $W_{\varepsilon}^u(z)$ of size $\varepsilon$ its projection onto the first component is the open ball with the center at $\pi_1z$ and radius $\varepsilon$. \proclaim{Lemma 9.6} The projection onto the second component of an unstable manifold through $z =(z_1,\,z_2)\in \Cal U$ of size $\varepsilon$ lies in the open ball with the center at $z_2$ and the radius $\rho\varepsilon$, i.e., $$\pi_2 \left(W_{\varepsilon}^u(z)\right) \subset \Cal B(z_2; \rho\varepsilon).$$ \endproclaim \demo{Proof} Let $W_{\varepsilon}^u(z)$ be the graph of $$\psi : \Cal B(z_1;\varepsilon) \to \Cal V.$$ The subspace $\{(\xi,\,D\psi\xi)| \xi\in \Re^{d}\}$ is tangent to $W_{\varepsilon}^u(z)$ and hence is contained in $\C_\rho$. It follows that $$\|D\psi\| \leq \rho.$$ By the mean value theorem if $z'=(z_1',\,z_2') \in W_{\varepsilon}^u(z)$ then $$\|z_2' - z_2\| = \|\psi(z_1') - \psi(z_1)\| \leq \sup \|D\psi\| \|z_1' - z_1\| < \rho \varepsilon .$$ \enddemo In contrast to the model problem at the beginning where we had fairly long initial unstable leaves and then we cut them because of the discontinuity of our system, we start here with small unstable manifolds and grow'' them until they are large or until they hit the singularity, whichever comes first. This is done in the proof of the following Theorem. \proclaim{Theorem 9.7} For any $\delta >0$ almost every point $z$ in $\Cal U^1_\delta$, $$\Cal U^1_\delta = \Cal U_{a_1(\delta)}$$ where $a_1(\delta) =a-b^{-1}\delta$ ($\U_a$ is defined in \S 8 and $b= \sqrt{1-\rho^4}$), either has an unstable manifold of size $\delta$ or it has an unstable manifold of size $\delta' < \delta$ such that the closure of $W^u_{\delta'}(z)$ intersects $\bigcup_{j> N}T^j\R^-$. \endproclaim \demo{Proof} Let $\Cal A(\varepsilon) \subset \Cal U^1_\delta$ be the set of points which have unstable manifolds of size $\varepsilon$. By the Katok-Strelcyn theory almost all points in $\Cal U^1 _\delta$ belong to $\bigcup_{\varepsilon > 0}\Cal A(\varepsilon)$. Let us fix $\Cal A(\varepsilon)$ of positive measure and let $k$ be the smallest natural number such that $$b\rho^{-k}\varepsilon \geq \delta.$$ Almost all points in $\Cal A(\varepsilon)$ have $k$ spaced returns to $\Cal A(\varepsilon)$ in the past. Let $z$ be such a point and let $$-N \geq -i_1 > \dots > -i_k = -n$$ be the $k$ times of spaced returns of this point, i.e., $$T^{-i_j}z \in \Cal A(\varepsilon),\ j =1,\dots,k.$$ The geometric idea for growing unstable manifolds is to take the unstable manifold of size $\varepsilon$ through the point $T^{-n}z$ and map it forward under $T^n$. The expansion property \thetag {9.4} guarantees then that the image contains the unstable manifold of size $\delta$. There are two complications in this argument. First it may happen that $T^{n}$ is not continuous on the unstable manifold $W^u_\varepsilon(T^{-n}z)$, that is $$W^u_{\varepsilon}(T^{-n}z)\cap \R^+_n \neq\emptyset .$$ The other problem occurs when parts of the images of the unstable manifold are outside of $\Cal U$ where the expansion property \thetag {9.4} may fail. To present clearly the core of the argument, we ignore for the time being these two difficulties and assume that $T^n$ is differentiable on $W^u_\varepsilon(T^{-n}z)$ and that $$T^{n-i_j}W^u_\varepsilon(T^{-n}z) \subset \Cal U,\ j = 0,\dots,k,$$ here we set $i_0=0$. We can prove then that $z$ has an unstable manifold of size $\delta$. Indeed let $W^u_\varepsilon(T^{-n}z)$ be the graph of $$\psi : \Cal B(\pi_1(T^{-n}z);\varepsilon) \to \Cal V$$ and let us consider the map $$\varphi : \Cal B(\pi_1(T^{-n}z);\varepsilon) \to \Cal V$$ defined by $\varphi (x) = \pi_1\left( T^n(x,\psi x)\right)$. By \thetag {9.4} this map is an expanding map with the coefficient of expansion not less than $b\rho^{-k}$, i.e., $$\|D\varphi\xi\| \geq b\rho^{-k}\|\xi\|.$$ Hence the image of $\Cal B(\pi_1(T^{-n}z);\varepsilon)$ by $\varphi$ contains the ball $\Cal B(\pi_1z;\delta)$. Additional complication is caused by the fact that $\varphi$ is not necessarily one-to-one. But since $\varphi$ is a local diffeomorphism, we can define $\varphi^{-1}$ on $\Cal B(\pi_1z;\delta)$ as the branch of the inverse for which $\varphi^{-1}\pi_1z = \pi_1(T^{-n}z)$. Therefore, $T^nW^u_{\varepsilon}(T^{-n}z)$ contains the graph of the map $$\pi_2\circ T^n\circ(id\times\psi)\circ\varphi^{-1}$$ which defines $W^u_{\delta}(z)$. Let us now address the general case. We will construct the maximal subset of $W^u_\varepsilon(T^{-n}z)$ on which $T^n$ is differentiable and its images at the return times to $\Cal U$ are contained in $\Cal U$. Our first step is to consider the connected component of $$W^u_\varepsilon(T^{-n}z) \setminus \R^+_n$$ which contains $T^{-n}z$ and denote it by $\widetilde{\widetilde {W^u_\varepsilon}}(T^{-n}z)$. Further, the connected component of $$\bigcap_{j=0}^k T^{i_j-n} \left(T^{n-i_j}\widetilde{\widetilde {W^u_\varepsilon}}(T^{-n}z) \cap\Cal U\right)$$ which contains $T^{-n}z$ will be denoted it by $\widetilde{W^u_\varepsilon}(T^{-n}z)$. It is the part of the unstable manifold which has the desired properties. Now we consider the image $$T^{n}\widetilde {W^u_\varepsilon}(T^{-n}z)$$ and we let $\delta'$ be the largest positive number such that $W^u_{\delta'}(z)$ is well defined and contained in $T^{n}\widetilde {W^u_\varepsilon}(T^{-n}z)$. If $\delta' \geq \delta$ then we are done. Let us hence assume that $\delta' < \delta$. It follows from the maximality of $\delta'$, that the boundary of $W^u_{\delta'}(z)$ must contain a point from the boundary of $T^{n}\widetilde{ W^u_\varepsilon}(T^{-n}z)$. Let $z'$ be such a point. If $z'$ belongs to $\bigcup_{i \geq N}^{n-1}T^{i}\R^-$, then we are again done. If not then $T^{-n}$ is differentiable at $z'$, and hence $T^{-n}z'$ belongs to the boundary of $\widetilde {W^u_\varepsilon}(T^{-n}z)$ and does not belong to $\R^+_n$. It follows now from the construction of $\widetilde {W^u_\varepsilon}(T^{-n}z)$ that $T^{-n}z'$ must belong to the boundary of $W^u_\varepsilon(T^{-n}z)$ or for some $j, 0\leq j \leq k,$ $T^{-i_j}z'$ belongs to the boundary of $\Cal U$. We will obtain now a contradiction by using the expansion property \thetag {9.4} . Let $W_{\delta'}^u(z)$ be the graph of $$\chi : \Cal B(\pi_1z;\delta') \to \Cal V$$ and let $$\gamma_0 : [0,1) \to \Cal B(\pi_1z;\delta')$$ be the segment connecting $\pi_1z$ and $\pi_1z'$. We consider the preimages of the curve $$\{(\gamma_0(t),\,\chi\gamma_0(t))\;|\;0\leq t< 1\}\subset W_{\delta'}^u(z),$$ and obtain $\gamma_j : [0,1) \to \Cal V,\ j=0,\dots,k$ by the formula $$\gamma_j(t) = \pi_1\left( T^{-i_j}(\gamma_0(t),\,\chi\gamma_0(t))\right).$$ It follows from \thetag {9.4} that the length of $\gamma_0$ is not smaller than the length of $\gamma_j$ times $b\rho^{-j}$. If $T^{-n}z'$ belongs to the boundary of $W^u_\varepsilon(T^{-n}z)$ then the length of $\gamma_k$ is at least $\varepsilon$ and we get the contradiction $$\delta' \geq b\rho^{-k}\varepsilon \geq \delta .$$ Finally if $T^{-i_j}z'$ belongs to the boundary of $\Cal U$ for some $j, 0\leq j \leq k,$ then $\gamma_j$ which connects $\pi_1(T^{-i_j}z) \in \Cal U^1_\delta$ and $\pi_1(T^{-i_j}z')$ must have the length at least $b^{-1}\delta$. We get again the contradiction $$\delta' \geq b\rho^{-j} b^{-1} \delta \geq \delta.$$ \enddemo \proclaim{Definition 9.8} We say that the unstable manifold of size $\delta$ $\Cal W^u_{\delta}(z)$ is cut by $T^i\R^-, i\geq 0,$ if its boundary contains a point from $T^i\R^-$. \endproclaim By Theorem 9.7 to guarantee that at least some points (and in the case of a smooth map almost all points) have unstable manifolds of size $\delta$ we need to step away from the boundary of $\Cal U$ by at least $b^{-1}\delta$. In the following we fix a sufficiently small $\delta_0$ and restrict our discussions to $\Cal U^1 = \Cal U^1_{\delta_0}$. We can then claim that in $\Cal U^1$ almost every point has a uniformly large unstable manifold (of size $\delta_0$) or a smaller unstable manifold cut by some image of the singularity set $\R^-$. By $\bar{\Cal B}(c;r)$ we denote the closed ball with the center at $c$ and the radius $r$. We define a rectangle $R(z;\delta)$ with the center at $z= (z_1,\,z_2)$ and the size $\delta$ as the Cartesian product of closed balls $$R(z;\delta) = \bar{\Cal B}(z_1;\frac \delta 2) \times\bar{\Cal B}(z_2;\frac \delta 2).$$ \proclaim{Definition 9.10} We say that the unstable manifold $W^u_{\delta'}(z')$ of $z'=(z_1',\,z_2')$ of size $\delta'$ is connecting in the rectangle $R(z;\delta)$ with the center at $z= (z_1,\,z_2)$ and size $\delta$ if $$\bar{\Cal B}(z_1;\frac \delta 2) \subset \Cal B(z_1';\delta')$$ and $$\pi_2\left(W^u_{\delta'}(z')\cap R(z;\delta)\right) \subset \Cal B(z_2;\frac \delta 2).$$ \endproclaim (See Figure 4). \par We can say equivalently, that an unstable manifold $W^u_{\delta'}(z')$ is connecting in the rectangle $R(z;\delta)$, if the intersection of $W^u_{\delta'}(z')$ with the rectangle is the graph of a smooth mapping from the closed ball $\bar{\Cal B}(\pi_1z;\frac \delta 2)$ to the open ball $\Cal B(\pi_2z;\frac \delta 2)$. Clearly it is necessary that $\delta' > \frac \delta 2$. \proclaim{Definition 9.11} For a given rectangle $R(z;\delta)$ with the center at $z= (z_1,\,z_2)$ and size $\delta$ we define its unstable core as the subset of those points $z'=(z_1',\,z_2') \in R(z;\delta)$ for which $$\rho\|z_1' - z_1\| +\|z_2' - z_2\| < (1-\rho)\frac \delta 2.$$ \endproclaim The role of an unstable core is revealed in the following Lemma. \topinsert \vskip 3in \hsize=4.5in \raggedright \noindent{\bf Figure 7.} The core of a rectangle. \endinsert \proclaim{Lemma 9.12} If an unstable manifold $W^u_{\delta'}(z')$ of size $\delta' > \|\pi_1z' -\pi_2z\| +\frac \delta 2$ intersects the unstable core of a rectangle $R(z;\delta)$, then it is connecting in the rectangle. \endproclaim \demo{Proof} Let $z=(z_1,\,z_2)$ and $z'=(z_1',\,z_2')$, let $W^u_{\delta'}(z')$ be the graph of $\psi : \Cal B(z_1';\delta') \to \Cal V$, and let $(x_1',\,\psi x_1')$ be a point in the unstable core of the rectangle. By the condition on $\delta'$ $$\bar {\Cal B}(z_1;\frac \delta 2) \subset \Cal B(z_1;\delta').$$ We have to check only that if $x_1 \in \bar{\Cal B}(z_1;\frac \delta 2)$ then $$\|\psi x_1 -z_2\| < \frac \delta 2 .$$ We have \aligned \|\psi x_1 -z_2\| &\leq \|\psi x_1 -\psi x_1'\| + \|\psi x_1' -z_2\|\\ &\leq \sup\|D\psi\|\|x_1- x_1'\| + \|\psi x_1' -z_2\|\\ &\leq \rho\|x_1-z_1\| + \rho\|x_1'-z_1\| +\|\psi x_1' -z_2\|\\ & < \rho \frac \delta 2 + (1-\rho)\frac \delta 2 = \frac \delta 2. \endaligned \enddemo The point of the above lemma is that even a large unstable manifold may fail to be connecting in a rectangle, if it intersects the rectangle too close to the boundary. \vskip.7cm \subhead \S 10. LOCAL ERGODICITY IN THE SMOOTH CASE \endsubhead \vskip.4cm Contrary to the title of this section, we will consider here several propositions valid in the general case. Incidentally, they will suffice to obtain local ergodicity in the smooth case (Main Theorem). It is important to remember that all of Section 9 can be repeated for stable manifolds. In this section we will be using both stable and unstable manifolds. \proclaim{Lemma 10.1} If an unstable manifold and a stable manifold are connecting in a rectangle, then there is a unique point of intersection of these manifolds in the rectangle, and it belongs to the interior of the rectangle. \endproclaim \demo{Proof} Let the rectangle have the center at $z=(z_1,\,z_2)$ and size $\delta$. The intersections of the unstable and stable manifolds with the rectangle $R(z;\delta)$ are the graphs of smooth mappings $$\psi^u : \bar{\Cal B}(z_1;\frac \delta 2) \to \Cal B(z_2;\frac \delta 2)$$ and $$\psi^s : \bar{\Cal B}(z_2;\frac \delta 2) \to \Cal B(z_1;\frac \delta 2),$$ respectively. Since both $\psi^u$ and $\psi^s$ are contractions, so is their composition $$\psi^s\psi^u : \bar{\Cal B}(z_1;\frac \delta 2) \to \Cal B(z_1;\frac \delta 2).$$ Hence, the composition has a unique fixed point $x_1 \in \Cal B(z_1;\frac \delta 2)$. The point $$\left(x_1,\,\psi^u x_1\right) = \left(\psi^s\psi^u x_1,\,\psi^u x_1 \right)$$ is the desired intersection point. \enddemo For a rectangle $R$ we denote by $W^{(u)s}(R)$ the union of the intersections with $R$ of all (un)stable manifolds connecting in $R$, i.e., $$W^{(u)s}(R) = \bigcup\{R\cap W^{(u)s}_{\delta'}(z')\;|\; \ W^{(u)s}_{\delta'}(z') \ \text { is connecting in } \ R \}.$$ The union of the unstable core and the stable core of a rectangle will be in the following called simply the core of the rectangle. \proclaim {Proposition 10.2} For any rectangle $R \subset \Cal U^1$, if the sets $W^s(R)$ and $W^u(R)$ have positive measure, then $W^s(R) \cup W^u(R)$ belongs to one ergodic component of $T$. \endproclaim \demo{Proof} The proof is done by the Hopf method as described in Sections 1 and 2. Let us fix a continuous function defined on our phase space. For all points in one (un)stable manifold the (backward) forward time averages are the same. As shown in Section 1 the forward and backward time averages have to coincide almost everywhere. Our goal is to show that they are constant almost everywhere in $W^s(R) \cup W^u(R)$. There is a technical difficulty stemming from the fact that the foliations into stable and unstable manifolds are not smooth in general. One has to use the absolute continuity of the foliations which was proven in \cite {KS} under the conditions which fit our scheme. (It is by far the hardest fact to prove in their theory.) It follows from absolute continuity of the foliation into unstable manifolds that except for the union of unstable manifolds from $W^u(R)$ of total measure zero almost every point (with respect to the Remannian volume in the manifold) in an unstable manifold from $W^u(R)$ has equal forward and backward time averages. Let us take such a typical unstable manifold. Again by the property of absolute continuity the union of stable manifolds in $W^s(R)$ which intersect this chosen unstable manifold at points where the forward and backward time averages exist and are equal differs from $W^s(R)$ by a set of zero measure. Hence the time average of our function is constant almost everywhere in $W^s(R)$. Similarly, the time average of our function is constant almost everywhere in $W^u(R)$. Finally, using the property of absolute continuity for the third time, we can claim that $W^u(R)$ and $W^s(R)$ intersect on a subset of positive measure. Hence the time average of our function is constant almost everywhere in $W^s(R) \cup W^u(R)$. To prove that $W^s(R) \cup W^u(R)$ belongs to one ergodic component, we proceed in the same way as at the end of Section 2. \enddemo We are ready to prove the local ergodicity in the smooth case \demo{Proof of Main Theorem (smooth case)} All the constructions started in Section 9 apply to our point $p$. We will prove that a neighborhood $\Cal U^2$, only slightly smaller than $\Cal U^1$, belongs to one ergodic component. Indeed, according to Lemma 9.12 all the points in the (un)stable core of a rectangle $R \subset \Cal U^1$ which have an (un)stable manifold of sufficiently large size belong to $W^{(u)s}(R)$. By Theorem 9.7, in the smooth case almost every point in $\Cal U^1$ has both the unstable manifold and the stable manifold of size $\delta_0$. Hence by Lemma 9.12, for any rectangle $R \subset \Cal U^1$ of size $\delta < \delta_0$, the set $W^s(R)$ contains at least the stable core of $R$ and $W^u(R)$ contains at least the unstable core of $R$. Clearly then the sets $W^s(R)$ and $W^u(R)$ have positive measure, and we can apply Proposition 10.2. To end the proof, we consider a family of rectangles of size $\delta \leq \delta_0$ contained in $\Cal U^1$ whose cores cover a slightly shrunk neighborhood $\Cal U^2 \subset \Cal U^1$. By Proposition 10.2, we can claim that each core belongs to one ergodic component. Since the cores form an open cover of the connected set $\Cal U^2$, we can conclude that $\Cal U^2$ belongs to one ergodic component. \enddemo Actually we can claim that, under the assumptions of the Main Theorem, the whole neighborhood $\Cal U$ constructed in Section 8 belongs to one ergodic component. Indeed, by taking $\delta \to 0$ the above argument applies to $\Cal U^2 \to \Cal U^1$ so that actually $\Cal U^1$ belongs to one ergodic component. If we now recall that the $\delta_0$ in the definition of $\U^1$ can be chosen arbitrarily small, we can see that also the whole neighborhood $\Cal U$ belongs to one ergodic component. This does not strengthen the theorem, but it demonstrates the usefulness of coverings with rectangles of size $\delta \to 0$. It will be crucial in the treatment of the discontinuous case. Let us outline the plan for proving local ergodicity in the general case. We cover the neighborhood $\Cal U^2$ with rectangles of size $\delta$. At least for some rectangles $R$ the sets $W^s(R)$ and $W^u(R)$ will have positive measure. We will be actually interested in the property that these sets cover certain fixed (but otherwise arbitrarily small) portion of the core of the rectangle, and we will call such rectangles connecting. One may then expect to have more and more connecting rectangles as $\delta \to 0$. The precise formulation of such a property is the subject of Sinai Theorem. The method of the proof requires that the size $rho$ of the sector is less than $\frac 1 3$. In applying Sinai Theorem it is convenient to work with more structured coverings, namely the centers of the rectangles will belong to a lattice with vertices so close that the cores of nearest neighbors will overlap almost completely. Consequently, if both nearest neighbors $R_1$ and $R_2$ are connecting then the union of $W^s(R_1) \cup W^u(R_1)$ and $W^s(R_2) \cup W^u(R_2)$ belongs to one ergodic component (as in the proof of Proposition 2.3). It will follow from Sinai Theorem that the network of connecting rectangles becomes more and more dense as $\delta \to 0$, so that we will be able to claim that one ergodic component reaches from any place in the neighborhood $\Cal U^1$ to any other place. We will conclude by using the Lebesgue Density Theorem to show that $\Cal U^2$ belongs to one ergodic component. \vskip.7cm \subhead \S 11. LOCAL ERGODICITY IN THE DISCONTINUOUS CASE \endsubhead \vskip.4cm Given $\delta > 0$, we consider a shrunk neighborhood $\Cal U^2_\delta$ defined by the requirement that a rectangle with the center in $\Cal U^2_\delta$ and size $\delta$ lies completely in $\Cal U^1$. (One can easily see that $\Cal U^2_\delta = \Cal U_{a_2(\delta)}$ where $a_2(\delta) = a_1(\delta_0) - \frac \delta 2$ .) Let us note that $\Cal U^2_\delta \to \Cal U^1$ as $\delta \to 0$. Let $\Cal N(\delta,c)$ be the net defined by $$\Cal N(\delta,c)= \{c\delta (m,\,k)\in \Cal U^2_\delta\;|\; m,\,k\in\Z^d\}.$$ We consider the family $\Cal G_\delta$ of all rectangles with the centers in $\Cal N(\delta,c)$ and size $\delta$ $$\Cal G_\delta = \{R(z;\delta)\;|\; z \in \Cal N(\delta,c)\}.$$ If $c$ is sufficiently small the family $\Cal G_\delta$ is a covering of $\Cal U^2_\delta$. The parameter $c$ will be chosen later to be very small, so that many rectangles in $\Cal R_\delta$ overlap. But once $c$ is fixed, a point may belong to at most a fixed number of rectangles, which we denote by $k(c)$ (it does not depend on $\delta$). \proclaim{Definition 11.1} Given $\alpha,\, 0 < \alpha < 1,$ we call a rectangle $R \in \Cal R_\delta$ $\alpha$-connecting in the (un)stable direction (or briefly connecting) if at least the $\alpha$ part of the measure of the (un)stable core of $R$ is covered by $W^{(u)s}(R)$. \endproclaim \proclaim{Sinai Theorem 11.2} If $\rho < \frac 13$ then there is $\alpha, 0 < \alpha < 1,$ such that for any $c$ $$\lim_{\delta\to 0}\delta^{-1} \mu\left (\bigcup\{R\in\Cal G_\delta\;|\;R \text{ is not } \alpha\text{-connecting }\}\right )=0,$$ i.e., the union of rectangles which are not $\alpha$-connecting in either the stable or the unstable direction has measure $o(\delta)$ \endproclaim It is very important for the application of this theorem that, given $\rho < \frac 13$, we get a certain $\alpha$ (which may be very small if $\rho$ is close to $\frac 13$), and we are free to choose $c$ (which determines the overlap of the rectangles in $\Cal G_\delta$) as small as we may need. We will prove Sinai Theorem in Sections 12 and 13. In the remainder of this Section we will show how to obtain the Main Theorem in the discontinuous case from Sinai Theorem. We start with some auxiliary abstract facts. The first one is in Measure Theory. For any finite subset $A$, we will denote by $|A|$ the number of elements in $A$. \proclaim{Lemma 11.3} Let $\{Y_a\;|\; a\in A\}$ be a finite family of measurable subsets of equal measure $m$ in the measure space $\left(X,\,\nu\right)$, such that no point in $X$ belongs to more than $k$ elements of the family. For any subfamily $\{Y_a\;|\; a\in A_1\}, A_1 \subset A,$ we have $$\frac m k |A_1| \leq \nu\left(\bigcup_{a\in A_1}Y_a \right) \leq m |A_1|.$$ Further, if for a measurable subset $Y \subset X$ and some $\alpha, 0 <\alpha < 1,$ $$\nu(Y_a\cap Y) \geq \alpha \nu(Y_a) \ \text { for } \ a \in A_1,$$ then $$\nu\left(\bigcup_{a\in A_1}Y_a \cap Y \right) \geq \frac \alpha k \nu\left(\bigcup_{a\in A_1}Y_a \right).$$ \endproclaim \qed The second fact is in Combinatorics. Let us consider the lattice $\Z^d$ and its finite pieces $$L_n =L_n(d) = \{0,1,\dots ,n-1\}^d \subset \Z^d.$$ Let $K \subset L_n$ be an arbitrary subset, which we call a configuration. We think of elements of $K$ as occupied sites and elements of $L_n \setminus K$ as empty sites. For a given configuration $K \subset L_n$, we consider the graph obtained by connecting by straight segments all pairs of occupied sites which are nearest neighbors. Let $gK \subset K$ be the family of sites in the largest connected component of the graph. \proclaim{Proposition 11.4} Let $K_n \subset L_n(d), n= 1,2,\dots ,$ be a sequence of configurations. If $$n\frac {|L_n\setminus K_n|} {|L_n|} \rightarrow 0 \ \ \text {as} \ \ n \rightarrow +\infty,$$ then $$\frac {|gK_n|} {|L_n|} \rightarrow 1 \ \ \text {as} \ \ n \rightarrow +\infty.$$ \endproclaim \demo{Proof} The Proposition will follow immediately from the following combinatorial Lemma. \proclaim{Lemma 11.5} Let $K \subset L_n(d)$ be an arbitrary configuration. If $$\frac {|L_n\setminus K|} {n^{d-1}} < a <1,$$ then $$\frac {|gK|} {n^d} \geq 1-(d-1)a.$$ \endproclaim \demo{Proof} The proof is by induction on $d$. For $d=1$ the statement is obvious. Suppose it is true for some $d$. We will establish it for $d+1$. We partition $L_n(d+1)$ into subsets $L_n(d)\times\{ i \}, i = 0,\dots,n-1$ and we call them floors. We pick the floor with the fewest number of empty sites. Clearly the number of empty sites there does not exceed $an^{d-1}$, so that we can apply to it the inductive assumption. We obtain in this floor a connected graph with at least $(1-(d-1)a)n^d$ elements. Now we partition $L_n(d+1)$ into subsets $\{ z \} \times \{0,\dots,n-1\}, z \in L_n(d)$, and we call them columns. A column is called an elevator, if all of its elements are occupied. The number of elevators is at least $(1-a)n^d$. Hence the number of elevators which intersect the connected graph in the floor considered above is at least $(1-da)n^d$. Adding these elevators to the graph, we obtain a connected graph with at least $(1-da)n^{d+1}$ elements, which ends the proof of the inductive step. \enddemo \enddemo \demo{Proof of Main Theorem (Discontinuous case)} All the constructions of Sections 8 through 10 apply with some $\rho < \frac 13$. We will be proving that the neighborhood $\Cal U^1$ belongs to one ergodic component. The Sinai Theorem gives us $\alpha < 1$ which depends only on $\rho$, and may have to be very small if $\rho$ is very close to $\frac 13$. Let us consider the lattice $\Cal N(\delta,c)$ and the covering $\Cal G_\delta$. We choose $c$ so small that, if the centers of two rectangles in $\Cal G_\delta$ are nearest neighbors in $\Cal N(\delta,c)$, then their unstable cores (and then automatically also stable cores) overlap on more than $1-\alpha$ part of their measure. Note that such a property depends on $c$, but is independent of the value of $\delta$. This choice of $c$ has the following consequence. If two rectangles $R_1$ and $R_2$ with centers at nearest neighbors in $\Cal N(\delta,c)$ are $\alpha$-connecting in the unstable direction, then $W^u(R_1)$ and $W^u(R_2)$ intersect on a subset of positive measure. If in addition we also know that $W^s(R_1)$ and $W^s(R_2)$ have positive measure, then using Proposition 10.2 we obtain that $$W^u(R_1) \cup W^u(R_2) \cup W^s(R_1) \cup W^s(R_2)$$ belongs to one ergodic component. We consider the configuration $\Cal K(\delta)$ in the lattice $\Cal N(\delta,c)$ which consists of the centers of all rectangles in $\Cal G_\delta$ which are $\alpha$-connecting both in the stable and unstable directions. As in the discussion proceeding Proposition 11.4, we consider the graph obtained by connecting with straight segments all pairs of nearest neighbors in $\Cal K(\delta)$. Let, as before, $g\Cal K(\delta)$ be the collection of vertices in the largest connected component of this graph. By our construction, the set $$Y(\delta) = \bigcup \{W^u(R(z;\delta))\cup W^s(R(z;\delta)) \;|\; z \in g\Cal K(\delta)\}$$ belongs to one ergodic component. This set is crucial in our proof that $\Cal U^1$ belongs to one ergodic component. It may be very small in measure (if $\alpha$ is small), but it covers at least certain fixed $\alpha'$ part of the measure of each of the rectangles with centers in $g\Cal K(\delta)$, i.e., $$\mu \left( R(z;\delta) \cap Y(\delta)\right) \geq \alpha' \mu \left( R(z;\delta)\right) \tag{11.6}$$ for any $z \in g\Cal K(\delta)$ ($\alpha'$ is smaller than $\alpha$, since $\alpha$ is only the part of the measure of the (un)stable core covered by the connecting (un)stable manifolds). It remains to show that the points in $g\Cal K(\delta)$ reach into all parts of $\Cal U^1$. It will follow from Sinai Theorem. By Sinai Theorem, the total measure covered by rectangles which are not $\alpha$-connecting is $o(\delta)$. Using Lemma 11.3, we can translate this estimate as $$k(c)^{-1} |\Cal N(\delta,c) \setminus \Cal K(\delta)| \delta^{2d} = o(\delta).$$ Since in addition $$\frac {|\Cal N(\delta,c)|} {(c\delta)^{2d}} = O(1),$$ we see that the assumptions of Proposition 11.4 are satisfied and we can claim that $$\frac {|g\Cal K(\delta)|}{|\Cal N(\delta,c)|} \to 1 \ \text { as } \ \delta \to 0. \tag {11.7}$$ We are ready to finish the proof by a contradiction. Suppose there are two $T$ invariant disjoint subsets $E_1$ and $E_2$ which have intersections with $\Cal U^1$ of positive measure. Let us pick two Lebesgue density points $p_1$ and $p_2$ for $E_1 \cap \Cal U^1$ and $E_2 \cap \Cal U^1$, respectively. Next, we fix cubes $C_1$ and $C_2$ with centers at $p_1$ and $p_2$ so small that $$\mu(C_i \cap E_i) \geq \left(1 - \frac {\alpha'}{2k(c)}\right)\mu(C_i), \ i = 1,2.$$ It follows from \thetag{11.7} that $$\frac {|\left(\Cal N(\delta,c) \setminus g\Cal K(\delta)\right)\cap C_i|}{|\Cal N(\delta,c)|} \to 0 \ \text { as } \ \delta \to 0, \ i=1,2.$$ Since $$\frac {|\Cal N(\delta,c)|}{|\Cal N(\delta,c)\cap C_i|} = O(1), \ i=1,2,$$ we conclude that $$\frac {|\left(\Cal N(\delta,c)\cap C_i \right) \setminus g\Cal K(\delta)|} {|\Cal N(\delta,c)\cap C_i|} \to 0 \ \text { as } \ \delta \to 0, \ i=1,2.$$ Now we get immediately that $$\mu \left(\left(\bigcup\{R(z;\delta) | z\in g\Cal K(\delta)\cap C_i\}\right) \triangle C_i\right) \to 0 \ \text { as } \ \delta \to 0, \ i=1,2, \tag{11.8}$$ where $\triangle$ denotes the symmetric difference, i.e., for any two sets $A$ and $B$ $$A\triangle B = (A\setminus B)\cup (B\setminus A).$$ By \thetag{11.6} and Lemma 11.3 $$\mu \left(\bigcup\{R(z;\delta) | z\in g\Cal K(\delta)\cap C_i\} \cap Y(\delta) \right) \geq \frac {\alpha'}{k(c)}\mu \left(\bigcup\{R(z;\delta) | z\in g\Cal K(\delta)\cap C_i\}\right),$$ $i =1,2$. Comparing this with \thetag{11.8} and remembering how dense $E_i$ is in $C_i, \ i=1,2,$ we conclude that, for sufficiently small $\delta$, the set $Y(\delta)$ must intersect both $E_1$ and $E_2$ over subsets of positive measure, which contradicts the fact that it belongs to one ergodic component. \enddemo %this file should be called s12.tex \vskip.7cm \subhead \S 12. PROOF OF SINAI THEOREM \endsubhead \vskip.4cm We will be proving only the unstable version of the theorem, i.e., we will estimate the measure of the union of rectangles which are not $\alpha$-connecting in the unstable direction. Everything can be then repeated for the stable manifolds. For a point $y$ in the core of a rectangle $R(z;\delta)$ there are two possibilities: \roster \item the point $y$ has an unstable manifold of size $\delta' > \|\pi_1y -\pi_1z\| + \frac\delta 2$ (which is connecting in $R(z;\delta)$ by Lemma 9.12), \item the point $y$ has an unstable manifold of size $\delta' \leq \|\pi_1y -\pi_1z\| + \frac\delta 2$ cut by $\bigcup_{i\geq 0}T^i\Cal S^-$. \endroster If a rectangle $R(z;\delta)$ is not connecting, then the second possibility must occur for at least $1-\alpha$ part of its core. The neighborhood $\Cal U$ was chosen so small that $\Cal S^-_N = \bigcup_{i=0}^{N-1}T^i\Cal S^-$ is disjoint from $\Cal U$. It follows that, for points in $\Cal U^1$, the unstable manifolds of size $\delta' < \delta_0$ cannot be cut by this singularity set. For any $M \geq N$ let us introduce the following special case of the second property: \vskip.2cm $(2_M)$ \vtop{\hsize=4.5in\noindent\it the point $y$ has an unstable manifold of size $\delta' \leq \|\pi_1y -\pi_1z\| + \frac\delta 2$ cut by $\bigcup_{i = N}^MT^i\Cal S^-$.} \vskip.2cm Further, we introduce the auxiliary notion of a $M$-nonconnecting rectangle. Roughly speaking, it is a rectangle which is not connecting because of the singularity set $\bigcup_{i=N}^M T^i\Cal S^-$. \proclaim{DEFINITION 12.1} Given $\alpha < \frac 12$, we say that a rectangle $R$ of size $\delta$ is $M$-nonconnecting, if at least $1- 2 \alpha$ part of the measure of the unstable core of $R$ consists of points which satisfy the property $(2_M)$. \endproclaim The plan of the proof is the following. We divide the argument into two parts. We will prove that, for every $\varepsilon >0$ there is $M = M(\varepsilon)$ and $\delta_\varepsilon$ such that, for all $\delta < \delta_\varepsilon$, the total measure of all rectangles in $\Cal G_\delta$ which are not $\alpha$-connecting and are not $M$-nonconnecting is less than $\delta\varepsilon$. This is the subject of the tail bound' (section 13), and it is by far the hardest part of the proof. It will require global considerations (i.e., outside of $\Cal U$). The particular value of $\alpha$ is immaterial there. We will start with the easier part proving that, for a given $\rho < \frac 13$ there is $\alpha$ such that, for any $\varepsilon >0$ and any $M \geq N$ there is $\delta_\varepsilon$ such that, for all $\delta < \delta_\varepsilon$, the total measure of all $M$-nonconnecting rectangles of size $\delta$ is less than $\delta \varepsilon$. Let us formulate it in a separate Proposition. Its proof will be completely confined to the neighborhood $\Cal U$. \proclaim{PROPOSITION 12.2} For any $\rho < \frac 13$, there is $\alpha, 0 < \alpha < 1$, such that, for any $M \geq N$, $$\lim_{\delta\to 0}\delta^{-1} \mu\left (\bigcup\{R \;|\;R \text{ has size } \delta \text{ and is } M\text{-nonconnecting } \}\right )=0.$$ \endproclaim \demo{Proof} We rely on our assumption that $\Cal S^-$ and its images are sufficiently nice'. More precisely we have required that the singularity set $\Cal S^-_{M+1} = \bigcup_{i=0}^MT^i \Cal S^-$ is regular. The definition of regularity was tailored to the needs of this proof. In particular, the singularity set $\Cal S^-_{M+1}$ is a finite union of pieces of submanifolds $I_k$ of codimension one, with boundaries $\partial I_k, k=1,\dots,p$. The boundaries $\partial I_k, k=1,\dots,p$ are themselves finite unions of compact subsets of submanifolds of codimension $2$ . What is more $$I_k\cap I_l \subset \partial I_k \cup \partial I_l\ \text{ for any }\ k, l.$$ In each of the pieces of submanifolds $I_k, k=1,\dots,p$, we consider the open neighborhood of the boundary of radius $r$, and we denote by $J_r$ the union of these neighborhoods, i.e., $$J_r = \bigcup_{k=1}^p \{ p\in I_k \;|\; d (p,\, \partial I_k) < r \}.$$ For each $\delta$ let $r(\delta)$ be the smallest $r$ such that, for any $k \neq l$, the distance of $I_k \setminus J_r$ and $I_l \setminus J_r$ is not less than $2\delta$. (In particular, for any $k \neq l$, the sets $I_k \setminus J_r$ and $I_l \setminus J_r$ are disjoint compact subsets, and their distance is at least $2\delta$.) Clearly $$\lim_{\delta \to 0} r(\delta) = 0.$$ Hence, by the property \thetag{7.3} $$\lim_{\delta \to 0}\mu_{\Cal S}(J_{r(\delta)}) = 0 \tag {12.3}$$ where $\mu_{\Cal S}$ is the natural volume element on $\Cal S_{M+1}^-$. Let us note that, if a rectangle $R = R(z;\delta)$ contains a point with the unstable manifold of size $\delta' < \delta$ cut by $S^-_{M+1}$, then it intersects the $2\delta$-neighborhood of $S^-_{M+1}$, but it does not necessarily intersect the singularity set itself. For technical reasons we prefer to blow up every rectangle, so that the blown up rectangle must intersect $S^-_{M+1}$ itself, and not only its neighborhood. For a fixed $b_0 < \frac 13$, to be chosen later, and for any rectangle $R = R(z;\delta)$, we introduce the blown up rectangle $$\widetilde R= \Cal B(\pi_1 z,\,(1+2b_0)\frac\delta 2) \times \Cal B(\pi_2 z,\, \frac\delta 2).$$ The diameter of $\widetilde R$ is less than $2\delta$, since we assume that $b_0 < \frac 13$. Let $y$ belong to the core of $R = R(z;\delta)$, satisfy the property $(2_M)$, and $$\|\pi_1y - \pi_1 z\| \leq b_0\frac\delta 2.$$ This implies that the unstable manifold $W^u_{\delta'}(y)$ is contained in $\widetilde R$, so that $\widetilde R$ intersects $\bigcup_{i=N}^M T^i\Cal S^-$. For a fixed $b_0$, if $\alpha$ is sufficiently small, then any $M$-nonconnecting rectangle must contain such a point $y$. We conclude that, for $\alpha$ sufficiently small, if a rectangle $R$ of size $\delta$ is $M$-nonconnecting, then $\widetilde R$ intersects at least one of the pieces of submanifolds $I_k, k=1,\dots,p.$ If for a rectangle $R$ of size $\delta$ the blown up rectangle $\widetilde R$ intersects two pieces of submanifolds $I_k$ and $I_l, k \neq l$ then, by definition of $r(\delta)$, it must intersect $J_{r(\delta)}$, and so it must be contained in the neighborhood of $J_{r(\delta)}$ of radius $2\delta$. By \thetag {12.3} and Proposition 7.4 (or \thetag {7.5}) the measure of the neighborhood of $J_{r(\delta)}$ of radius $2\delta$ is $o(\delta)$ (i.e., when divided by $\delta$, it tends to zero as $\delta$ tends to zero). It remains to consider those blown up rectangles which intersect only one of the submanifolds $I_k, k=1,\dots,p$. The proof will be finished when we prove that there is $\alpha$ so small that, for any sufficiently small $\delta$, if a blown up rectangle $\widetilde R$ of size $\delta$ intersects only one of the submanifolds $I_k, k=1,\dots,p,$, then the rectangle $R$ is {\bf not} $M$-nonconnecting. Our first observation is that there is a constant $K$ depending only on the manifolds $I_k, k=1,\dots,p,$ such that for any $x,x'\in I_k$ there is $v$ in the tangent space to $I_k$ at $x$ ($v \in \Cal T_xI_k$) for which $$\| x' - x - v \| \leq K \|x' - x \|^2 \tag 12.4$$ Here we consider the tangent space $\Cal T_xI_k$ of $I_k$ at $x$ as a subspace in $\Re^{d} \times \Re^{d}$. This property is a formulation of the fact that smooth submanifolds are locally close to their tangent subspaces and follows easily from the Taylor expansion. Further, in view of the proper alignment of the singularity manifolds, the tangent subspaces $\Cal T_xI_k, x \in I_k \cap \Cal U^1$ must have their characteristic lines in $\C_\rho$. Let us now take a rectangle $R = R(z;\delta)$ such that the blown up rectangle $\widetilde R$ intersects $I_k$. We will show that $\pi_2(I_k \cap \widetilde R)$ is contained in a fairly narrow layer. To show this, let $x = (x_1,\,x_2),x' = (x'_1,\,x'_2) \in I_k \cap \widetilde R$ and let $v = (\xi,\,\eta) \in T_xI_k$ be the vector for which \thetag {12.4} holds. We pick a nonzero vector $v_0 = (\xi_0,\,\eta_0) \in T_xI_k$ with the direction of the characteristic line. For convenience, we scale it so that $\|\xi_0\| = 1$. We have, by the definition of a characteristic line, $$\omega (v,\,v_0) = \langle \xi,\,\eta_0 \rangle - \langle \eta,\,\xi_0 \rangle = 0.$$ It follows that $$\left | \langle \eta,\,\xi_0 \rangle \right | = \left | \langle \xi,\,\eta_0 \rangle \right | \leq \rho \|\xi_0\| \|\xi\| = \rho \|\xi\|.$$ Replacing $v$ by $x - x'$ in the last inequality and using \thetag{12.4}, we get $$\left | \langle \xi_0 ,\,x_2' - x_2 \rangle \right | \leq \rho (\| x_1' - x_1 \| + K \|x' - x \|^2) + K \|x' - x \|^2.$$ Since both $x$ and $x'$ are in $\widetilde R$, we have that $$\| x_1' - x_1 \| < (1+2b_0)\delta$$ and $$\|x' - x \| < 2 \delta.$$ Therefore, for any $x,x' \in I_k\cap \widetilde R$, we obtain the inequality $$\left | \langle \xi_0,\,x_2' - x_2 \rangle \right | \leq \rho (1+2b_0) \delta + const\ \delta^2, \tag {12.5}$$ where the constant depends only on $\rho$ and $K$. The inequality (12.5) shows that $\pi_2(I_k\cap\widetilde R)$ is contained in a layer perpendicular to $\xi_0$ of the width $\rho (1+2b_0) \delta + const\ \delta^2$. Hence, there is $\bar x_2$ (in the center' of the layer) such that every $x = (x_1,\,x_2) \in I_k\cap \widetilde R$ must belong to the layer defined by the inequality $$\left | \langle \xi_0,\,x_2 -\bar x_2 \rangle \right | \leq \rho (1+2b_0)\frac \delta 2 + const\ \delta^2. \tag {12.6}$$ We want to estimate the width of the layer where all the points from the core of the rectangle with short' unstable manifolds, cut by $I_k$, must lie. To that end let us take a point $y = (y_1,\,y_2)$ in the core of the rectangle $R(z;\delta)$ and such that $\|y_1 - \pi_1z\| \leq b_0\frac \delta 2$. If $y$ satisfies the property $(2_M)$ then, by Lemma 9.6, the projection $\pi_2W^u_{\delta'}(y)$ of the unstable manifold lies in the ball $$\Cal B(y_2;\rho\delta') \subset \Cal B(y_2;\rho(1+b_0)\frac \delta 2).$$ Assuming that $W^u_{\delta'}(y)$ is cut by $I_k$, there is $x = (x_1,\,x_2)\in I_k \cap \widetilde R$, for which $$\left | \langle \xi_0,\,y_2 - x_2 \rangle \right | \leq \rho(1+b_0)\frac \delta 2.$$ Hence, by \thetag{12.6}, the point $y$ must belong to the layer defined by the inequality $$\left | \langle \xi_0,\,y_2 -\bar x_2 \rangle \right | \leq \rho(1+b_0)\frac \delta 2+\rho(1+2b_0)\frac \delta 2 +const\ \delta^2 \tag {12.7}$$ %\topinsert %\vskip 3in %\hsize=4.5in %\raggedright %\noindent{\bf FIGURE 8.} Layer intersecting the core. %\endinsert The last step is to choose $b_0$ so small that this layer cannot cover all of the core. We prefer, for convenience, to fit a Cartesian product into the unstable core, and to prove that, if only $b_0$ is chosen sufficiently small, a fixed part of the measure of this set is not covered by the layer \thetag{12.7}. We choose such set to be $$X(b_0) = \Cal B(\pi_1z;b_0\frac \delta 2) \times \Cal B(\pi_2z; s(b_0)\frac \delta 2)$$ where $s(b_0) = 1-\rho -\rho b_0$. By the definition of a core, the set $X(b_0)$ is contained in the core of $R(z;\delta)$, and its measure is not less than a certain fixed part of the measure of the core, depending on $b_0$ (and the dimension d) but independent of $\delta$. If the layer \thetag {12.7} is sufficiently narrow, it cannot cover all of $X(b_0)$. The precise inequality, which guarantees that, is easily transformed into $$3\rho + const\ \delta < 1 - 4\rho b_0. \tag {12.8}$$ After a moment of reflection the reader will realize that if only $\rho < \frac 13$ we can choose $b_0$ so small that not only \thetag {12.8} is satisfied, but also a certain fixed part of $X(b_0)$ (depending on $b_0$ but independent of $\delta$) is not covered by the layer \thetag{12.7}. Thus, there is $\alpha$, depending on $\rho$ and $b_0$, such that more than $2\alpha$ part of the measure of the core is free of points satisfying the property $(2_M)$. Hence the rectangle $R$ is {\bf not} $M$-nonconnecting. \enddemo If the reader finds it hard to follow the above argument, it is because we strived to use as little hyperbolicity as possible on our finite orbit. The amount of hyperbolicity is measured by the size $\rho$ of the sector . We have managed to relax the condition on $\rho$ up to $\rho < \frac 13$. It is not hard to see that, if the last condition is relaxed further, Proposition 12.2 will not hold in general. \vskip.7cm \subhead \S 13. TAIL BOUND' \endsubhead \vskip.4cm We will be proving that, for every $\varepsilon > 0$, there is $M$ such that the measure of points $z \in \Cal U^1$ with the unstable manifold of size $\delta' < \delta$ cut by $\bigcup_{i\geq M+1}T^i\Cal S^-$ does not exceed $\varepsilon\delta$. Comparing this set with the union of rectangles in $\Cal G_\delta$ which are not $\alpha$-connecting and not $M$-nonconnecting, we establish immediately that the measure of the union can be bigger by at most an absolute (=independent of $\delta$) factor, made up of $\rho, \alpha$ and the overlap coefficient $k(c)$ (introduced prior to Definition 11.1). To arrive at this conclusion it is important that we consider only the rectangles from the covering $\Cal G_\delta$ (and not all possible rectangles of size $\delta$). We start by exploring some of the consequences of the Sinai - Chernov Ansatz. No reference to the neighborhood $\Cal U$ will be made at this stage. So we have assumed that almost all points in $\Cal S^-$ (with respect to the measure $\mu_{\Cal S}$) are strictly unbounded in the future. It follows from Theorem 6.8 that, for almost every point $p \in \Cal S^-$, $$\lim_{n\to +\infty}\inf_{0 \neq v\in \C(p)}\frac{\sqrt{\Q(D_pT^nv)}} {\|v\|} = +\infty.$$ For a linear monotone map, let us put $$\sigma_*(L) = \inf_{0 \neq v\in \C(p)}\frac{\sqrt{\Q(Lv)}}{\|v\|}.$$ Consequently, for any (arbitrarily small) $h > 0$ and any (arbitrarily large) $t >0$, there is $M = M(h,t)$ so large that the subset $$\widetilde E_t = \{p \in \Cal S^-\;|\; \sigma_*(D_pT^M) \leq t + 1 \}$$ has measure $$\mu_{\Cal S}(\widetilde E_t) \leq h.$$ The map $T^M$ is, in general, not even continuous in all of $\Cal S^-$. The coefficient $\sigma_*(D_pT^M)$ is defined only for almost every point $p \in \Cal S^-$. Hence, so far, the subset $\widetilde E_t$ is defined modulo subsets of measure zero. We need a closed subset, since we plan to use Proposition 7.4. The map $T^M$ is discontinuous olny on $\Cal S^+_M,$ which was assumed to be a regular set. Using the proper alignment of singularity sets and the monotonicity of the system, we conclude that $\Cal S^+_M$ is transversal to $\Cal S^-$ (in the natural sense). It follows that the set $B_M = \left( \Cal S^+_M \cup \partial\M\right) \cap \Cal S^-$ is a finite union of compact subsets of submanifolds of dimension $2d-2$. Further, $\Cal S^-$ is decomposed into (possibly very large) finite number of pieces of submanifolds of dimension $2d-1$ such that $T^M$ is differentiable in the interior of every piece, and their boundaries are subsets of $B_M$. It follows that the coefficient $\sigma_*(D_pT^M)$ is continuous in the interior of every piece. Let us choose $\zeta$ so small that the closure of the $\zeta$-neighborhood of $B_M$ in $\Cal S^-$ $$B_M^\zeta = \{p\in \Cal S^-\;|\; d(p,\, B_M) < \zeta\}$$ has small measure $$\mu_{\Cal S}\left(\overline{B_M^\zeta}\right) \leq h.$$ Now the set $E_t$ defined by $$E_t = \widetilde E_t \setminus B_M^\zeta = \{p \in \Cal S^-\setminus B_M^\zeta \;|\; \sigma_*(D_pT^M) \leq t + 1 \}$$ is {\bf closed}, and we have $$\mu_{\Cal S}\left(E_t\cup\overline{B_M^\zeta}\right) \leq 2h.$$ Let $$\Cal S_t = \{p \in \Cal S^-\setminus B_M^\zeta \;|\; \sigma_*(D_pT^M) \geq t + 1 \}.$$ $\Cal S_t$ is a compact set and the coefficient $\sigma_*(D_pT^M)$ is continuous in a neighborhood of $\Cal S_t$ in $\M$. Hence, there is $r >0$ such that $$\sigma_*(D_pT^M) > t,$$ for every point $p$ in the $r$-neighborhood of $\Cal S_t$ in $\M$, i.e., for every point in $$\Cal S_t^r = \{p \in \M \;|\; d(p,\,\Cal S_t) < r\}.$$ Now we look at our neighborhood $\Cal U$. Our goal is to estimate, for given $\delta$, the measure of the set $Y(\delta, M)$ of points in $\Cal U^1$ which have the unstable manifold of size $\delta' < \delta$ cut by $\bigcup_{i\geq M+1}T^i\Cal S^-$. We will achieve this by splitting $Y(\delta, M)$ into convenient pieces and showing that their preimages must end up in extremely small neighborhoods of $\Cal S^-$. For $z \in Y(\delta, M)$ the unstable manifold $\Cal W^u_{\delta'}(z)$ may be cut by several (possibly infinitely many) of the singularity sets $T^i\Cal S^-$, $i = M+1,\dots .$ Let $m(z)$ be the smallest $i \geq M+1$ such that $\Cal W^u_{\delta'}(z)$ is cut by $T^i\Cal S^-$. Let further $$k(z) = \#\{i\;|\;1 \leq i \leq m(z) - M,\ T^{-i}z \in \Cal U^1\}.$$ Roughly speaking $k(z)$ is the number of times the point $z$ visits in $\Cal U^1$ in the past in the time frame bounded by $m(z)$. We put for $k =0,1,\dots, \ m = M+1, \dots ,$ $$Y^k_m = \{z \in Y(\delta, M)\;|\; m(z) = m , k(z) = k \}.$$ We will now fix $k$ and estimate the measure of $$\bigcup_{m \geq M+1} Y^k_m.$$ \proclaim{Lemma 13.1} For $m \neq m'$ $$T^{-m}Y^k_m \cap T^{-m'}Y^k_{m'} = \emptyset.$$ \endproclaim \demo{Proof} Let $m < m'$. If $y \in T^{-m}Y^k_m \cap T^{-m'}Y^k_{m'}$ then for $z = T^my$ and $z' = T^{m'}y$ we have $$k(z') \geq k(z) + 1.$$ It contradicts the fact that $z \in Y^k_m$ and $z' \in Y^k_{m'}$. \enddemo By Lemma 13.1 we have $$\mu(\bigcup_{m \geq M+1} Y^k_m) \leq \sum_{m \geq M+1}\mu(Y^k_m) = \sum_{m \geq M+1}\mu(T^{-m}Y^k_m) = \mu(\bigcup_{m \geq M+1} T^{-m}Y^k_m).$$ Let $z \in Y^k_m$ and $z' \in T^m \Cal S^-$ be a point in the boundary of $\Cal W^u_{\delta'} (z)$. We connect $z$ and $z'$ by the curve $\gamma$ in $\Cal W^u_{\delta'} (z)$ which projects under $\pi_1$ onto the linear segment from $\pi_1z$ to $\pi_1z'$. In the neighborhood $\Cal U$ we have three ways of measuring the length of $\gamma$. We can use the quadratic form $\Q$, or the length of the projection onto the first component, or finally, we can use the Riemannian metric. All these metrics are equivalent in $\Cal U$, and we will use the following coefficients defined by their ratios $$\sup\left\{\frac {\| v \|}{\|\xi\|}\;|\;0 \neq v =(\xi,\,\eta)\in \C_\rho \right\} = \sqrt{1 + \rho^2}$$ and $$q= \sup\left\{\frac {\sqrt{\Q(v)}}{\|\xi\|}\;|\;0 \neq v =(\xi,\,\eta)\in \C_\rho \right\},$$ where the last supremum is taken also over all of $\Cal U$. Our goal is to estimate the distance of $T^{-m}z$ and $T^{-m}z'$ in the Riemannian metric, such a distance clearly does not exceed the length of the curve $T^{-m}\gamma$. To that end, let $n \leq m-M$, be the time of the $k$-th visit in the past by $z$ to $\Cal U^1$, i.e., $T^{-n}z \in \Cal U^1$. By Proposition 8.4, on every spaced return to $\Cal U$ the projection of the preimage of $\gamma$ is contracted by at least the factor $\rho$. In the $k$ visits there must be at least $\frac kN -1$ spaced returns. Hence, the projection of $T^{-n}\gamma$ has the length which, by \thetag{8.6u}, does not exceed $$c_1\lambda^k\delta,$$ where $$\lambda = \rho^{\frac 1N} \ \text{ and } \ c_1 = \frac 1{\rho b} = \frac 1 {\rho\sqrt{1-\rho^4}}.$$ It follows that the Riemannian length of $T^{-n}\gamma$ does not exceed $$c_2\lambda^k\delta,$$ where $$c_2 = \frac 1 {\rho\sqrt{1-\rho^2}},$$ and its length in the metric $\Q$ does not exceed $$c_3\lambda^k\delta,$$ where $$c_3 = \frac q {\rho\sqrt{1-\rho^2}}.$$ Now we apply $T^{-(m-n)}$ to $T^{-n}\gamma$, and we use the fact that $m-n \geq M$. There are two different cases. {\it Case 1.} $$T^{-m}z' \in E_t\cup\overline{B_M^\zeta}$$ We use the noncontraction property. Under the action of $T^{-(m-n)}$ the Riemannian length of $\gamma$ can expand at most by the factor $\frac 1a$. We conclude that the Riemannian length of $T^{-m}\gamma$ does not exceed $$\frac {c_2}a \lambda^k \delta.$$ Thus $T^{-m}z$ belongs to the neighborhood of $E_t\cup\overline{B_M^\zeta}$ in $\M$ of this radius. By Proposition 7.4 its measure does not exceed $$3h\frac{2c_2}a \lambda^k \delta, \tag 13.2$$ if only $\delta$ is small enough ($\delta \leq \delta_0$ and $\delta_0$ does not depend on $k$ or $m$). {\it Case 2.} $$T^{-m}z' \in \Cal S_t$$ We claim that, for sufficiently small $\delta$, the length of $T^{-m}\gamma$ does not exceed $$\frac 1t c_3\lambda^k\delta.$$ Indeed, it is so if $T^{-m}\gamma$ is contained in $\Cal S_t^r$ (the $r$-neighborhood of $\Cal S_t$ in $\M$). Since $m-n \geq M$, we have $$\sigma_*(D_pT^{m-n}) > t,$$ for every point $p\in \Cal S_t^r$ (if only the map $T^{m-n}$ is differentiable at $p$). Hence, the length in the metric $\Q$ of $T^{-n}\gamma$ is longer than the Riemannian length of $T^{-m}\gamma$ by at least the factor $t$. If $T^{-m}\gamma$ is not contained in $\Cal S_t^r$, then there must be a segment of this curve in $\Cal S_t^r$ which has at least length $r$. It follows that the image of this segment under $T^{m-n}$ has the length in the metric $\Q$ not less than $tr$, which is more than the total length in the metric $\Q$ of $T^{-n}\gamma$, for sufficiently small $\delta$. This contradiction shows that, for sufficiently small $\delta$, $T^{-m}\gamma \subset \Cal S_t^r$. We have proven our claim. It follows that $T^{-m}z$ belongs to the neighborhood of $\Cal S^-$ of radius $\frac 1t c_3\lambda^k\delta$. Using again Proposition 7.4, we can estimate the measure of this neighborhood by $$2\mu_{\Cal S}(\Cal S^-)\frac {2c_3}{t} \lambda^k\delta, \tag 13.3$$ if only $\delta$ is sufficiently small ($\delta \leq \delta_0$ and $\delta_0$ does not depend on $k$ or $m$). Combining the estimates \thetag {13.2} and \thetag {13.3} we obtain that for any $k=0,1,\dots ,$ $$\mu(\bigcup_{m \geq M+1} T^{-m}Y^k_m) \leq \left(h \frac {6c_2}a + \frac 1t 2c_3\mu_{\Cal S}(\Cal S^-)\right) \ \lambda^k \ \delta.$$ It follows that $$\mu(Y(\delta,M)) \leq \left(h \frac {6c_2}a + \frac 1t 4c_3\mu_{\Cal S}(\Cal S^-)\right)\frac 1{1-\lambda} \ \delta.$$ The last inequality tells us how we should choose a small $h$ and a large $t$ at the beginning of our argument to guarantee that $$\mu(Y(\delta,M)) \leq \varepsilon\delta.$$ The tail bound' is proven. %this file must be called "s14.tex" \vskip.7cm \subhead \S 14. Applications\endsubhead \vskip.4cm {\bf A. Billiard systems in convex scattering domains. } \bigskip We assume that the reader is familiar with billiard systems. If it is not the case we recommend \cite {W4} for a quick introduction into the subject. We will rely on the results of that paper. Let us consider a domain in the plane bounded by a locally convex closed curve given by the natural equation $r = r(s), 0 \leq s \leq l$ describing the radius of curvature $r$ as a function of the arc length $s$. We assume that the radius of curvature satisfies the condition $$\frac {d^2r}{ds^2} < 0, \ \text { for all } \ s, 0\leq s \leq l. \tag {14.1}$$ Curves satisfying this condition were called in \cite {W4} strictly convex scattering. \proclaim{Examples} \endproclaim 1. Perturbation of a circle. 2. Cardioid. \bigskip Such a domain cannot be convex and there is a singular point in the boundary where the curve intersects itself. If you do not like playing billiards on a table which is not convex you may take the convex hull of our domain and everything below still applies. The following theorem follows fairly easily from the Main Theorem. \proclaim {Theorem 14.2} The billiard system in a domain bounded by a strictly convex scattering curve (i.e., satisfying \thetag{14.1}) is ergodic. \endproclaim Let us consider the map $T$ describing the first return map to the boundary. $T$ is defined on the set $\Cal M$ of unit tangent vectors pointing inwards. We parametrize $\Cal M$ by the arc length parameter of the foot point $s, 0 \leq s \leq l$ and the angle $\varphi, 0 \leq \varphi \leq \pi$ the unit vector makes with the boundary (oriented counterclockwise). $\Cal M$ becomes the rectangle $[0,l] \times [0,\pi]$. The symplectic form (the invariant area element) is given by $$\sin \varphi \ ds \wedge d\varphi.$$ When we derive the formula for the derivative of $T$ we will be able to check immediately that $T$ preserves this area element. The map $T$ is discontinuous at those billiard orbits which hit the singular point of the boundary. They form a curve $\Cal S^+$ in $\Cal M$ which is a graph of a strictly decreasing function, decreasing curve for short. This curve divides the rectangle $\Cal M$ into two curvilinear triangles, $\Cal M^+_b$ with a side at the bottom and $\Cal M^+_t$ with a side at the top. To find the images of $\Cal M^+_b$ and $\Cal M^+_t$ we use the reversibility of our system. Namely, let $S : \Cal M \to \Cal M$ be defined by $S(s,\varphi) = (s,\pi-\varphi)$. We have $$T \circ S = S \circ T^{-1}.$$ We can now claim that $T^{-1}$ is continuous except on $\Cal S^- = S\Cal S^+$ which is an increasing curve (the graph of a strictly increasing function). $\Cal S^+$ divides the rectangle $\Cal M$ into two curvilinear triangles $\Cal M^-_b = S\Cal M^+_t$ and $\Cal M^-_t = S\Cal M^+_b$. We have constructed our symplectic boxes. $T$ is a diffeomorphism on their interiors and a homeomorphism on the closure. The derivative of $T$ does blow up at least at one point of the boundary $\Cal S^+$ (different for $\Cal M^+_b$ and for $\Cal M^+_t$) corresponding to the two billiard orbits tangent to one of the branches of the boundary at the singular point. In the case of the cardioid the derivative blows up at any point of $\Cal S^+$ and also at the vertical boundaries because the curvature at the cusp is infinite (see the formula for the derivative of $T$ below). It is very handy that we did not have to require in Section 8 that our map is a diffeomorphism on the closed symplectic boxes. The derivative of $DT$ at $(s_0,\,\varphi_0)$ has the form $$\left(\matrix \frac{\tau-d_0}{r_0\sin \varphi_1}& \frac{\tau}{\sin \varphi_1}\\ \frac{\tau-d_0-d_1}{r_0r_1\sin \varphi_1}&\frac{\tau-d_1}{r_1\sin \varphi_1} \endmatrix\right) \tag {14.2}$$ where $T(s_0,\,\varphi_0) = (s_1,\,\varphi_1)$, $\tau$ is the time between consecutive collisions (i.e., the length of the billiard orbit segment) and $d_i = r_i\sin \varphi_i,\ i=1,2.$ This derivative can be obtained by straightforward implicit differentiation but we do not recommend it. There is a more geometric (and safer) way to obtain the derivative by resorting to the description of billiard orbit variations by Jacobi fields. In our two dimensional situation it amounts to introducing coordinates $(J,\,J')$ in the tangent planes of $\Cal M$ \aligned J = & \sin \varphi ds,\\ J'= & -\frac 1r ds - d\varphi. \endaligned \tag {14.3} The evolution of $(J,\,J')$ between collisions is given by the matrix $$\left(\matrix 1 & \tau \\ 0 & 1 \endmatrix\right). \tag {14.4}$$ At the collision $(J,\,J')$ is changed by $$\left(\matrix -1 & 0 \\ \frac 2{d_1} & -1 \endmatrix\right). \tag {14.5}$$ Now the derivative \thetag{14.2} is obtained by multiplying the matrices \thetag{14.4} and \thetag{14.5} and taking into account \thetag{14.3}. The geometric meaning of $d_0,d_1$ and the inequality $$\tau > d_0 +d_1 \tag {14.6}$$ is explained at length in \cite {W4}. It was proven there that \thetag {14.6} holds for any billiard orbit segment if the boundary curve is strictly convex scattering (actually these two properties are essentially equivalent). It follows from \thetag {14.6} that for a strictly convex scattering curve all elements in \thetag {14.2} are positive. We choose as our family of sectors the constant sector between the horizontal line $\{d\varphi =0\}$ and the vertical line $\{ds =0\}$. We see immediately that the derivative $DT$ is strictly monotone. We are now ready to argue that the singularity sets $\Cal S^-_n = \bigcup_{i=0}^n T^i\Cal S^-$ are regular. We claim that $\Cal S^-_n$ is a finite union of increasing curves which intersect each other only at the endpoints. It can be proven by induction. Indeed $\Cal S^-$ is an increasing curve and so it is also properly aligned. The singularity set $\Cal S^+$ is a decreasing curve and as such it may intersect each of the increasing curves of $\Cal S^-_n$ in at most one point. Hence both $\Cal M^+_b \cap \Cal S^-_n$ and $\Cal M^+_t \cap \Cal S^-_n$ are finite unions of increasing curves with intersections only at the endpoints. Hence in view of the monotonicity of our system the images under $T$ are also finite unions of increasing curves in $\Cal M^-_b$ and $\Cal M^-_t$ respectively. It is clear that we can safely add $\Cal S^-$ to these images. We have thus checked that $\Cal S^-_{n+1} = \Cal S^- \cup T\Cal S^-_n$ is also a finite union of increasing curves which intersect only at the endpoints. Note that the assumptions of Lemma 7.6 are too restrictive to allow its application in this case. One can easily compute (and it was done explicitly in \cite {W4}) that $$\sigma(DT) = \sqrt{1+\omega} +\sqrt{\omega}, \ \text { where } \ \omega = \frac {(\tau -d_0 -d_1)\tau}{d_0d_1}. \tag {14.7}$$ It follows from \thetag {14.7} and from the supermultiplicativity of the coefficient of expansion $\sigma$ that the only way in which an orbit can fail to be strictly unbounded is when the lengths of the segments of the orbit go to zero. It was shown by Halpern \cite {Ha} that there are no such billiard orbits if the boundary is sufficiently smooth (if $r(s)$ is a $C^1$ function). Hence all orbits for which arbitrary power of $T$ is differentiable are strictly unbounded. This observation takes care of the Sinai - Chernov Ansatz. We can apply Main Theorem to all points except the double points of singularity sets. There are only countably many such points, so that their complement is connected. Hence, local ergodicity implies ergodicity. It remains to check the noncontraction property. It was pointed out to us by Donnay \cite{D1} that the derivative of $T$ increases $|J'|^2$ on nonzero vectors from the sector. Indeed the interior of the sector is defined by $$\frac {J'}{J} < - \frac 1d$$ so that we have $$\frac {|J'|}{|J|} > \frac 1d.$$ If $DT (J_0,\,J_0') = (J_1,\,J_1')$ then we have from \thetag {14.4} and \thetag {14.5} that $$J_1 = -J_0 - \tau J_0'.$$ It follows that $$|J_1'| \geq \frac 1{d_1} |J_1| = \frac 1{d_1} | J_0 + \tau J_0'| \geq \frac {\tau}{d_1}|J_0'| - \frac 1{d_1} |J_0| \geq \frac {\tau - d_0}{d_1}|J_0'|.$$ In view of \thetag {14.6} $\frac {\tau - d_0}{d_1} > 1$. So indeed $|J'|^2$ gets increased. Moreover, for all vectors in the sector we have the following estimates $$2(\frac 1{r^2}ds^2 + d\varphi^2) \geq |J'|^2 = |\frac 1r ds + d\varphi|^2 \geq \frac 1{r^2}ds^2 + d\varphi^2.$$ The metric $\frac 1{r^2}ds^2 + d\varphi^2$ is equivalent to the standard Riemannian metric in the $(s,\,\varphi)$ coordinates ($ds^2 + d\varphi^2$) if only $r$ is bounded away from zero. Thus noncontraction is established under this additional assumption (which excludes the cardioid). To cover the case of the cardioid we observe that the noncontraction property is used only in the proof of the 'tail bound', where subsets of the neighborhood $\U$ are transported back to the neighborhhood of the singularity set $\Cal S^-$. We need the property that vectors in $\C$ are not contracted too much, along the orbits from the vicinity of the singularity set to the neighborhood $\U$, even if the orbit is very long. We obtain readily this property from the observation that although $|J'|^2$ is, in general, only bigger than the scaled standard Riemannian metric, it is clearly equivalent to one, locally in the neighborhood $\U$. The reader may be worried that the standard Riemannian metric in the $(s,\,\varphi)$ coordinates does not generate the invariant area element. However, the Riemannian area is not smaller than the symplectic area. This is sufficient for the proof of Sinai Theorem. We could also handle this complication by introducing from the very beginning coordinates in $\M$ in which the symplectic form is standard. We can conclude that $T$ is ergodic, and so Theorem 14.2 is proven. It follows from the results of Katok and Strelcyn \cite {KS} that $T$ is a Bernoulli system. The framework of this paper allows to cover also the class of billiard systems in domains with more than one smooth piece in the boundary, which are not necessarily convex scattering. In the recent paper \cite {D2} Donnay introduced a natural condition (focusing arc) on the convex pieces of the boundary of a billiard table. He proves that if two focusing arcs are connected by sufficiently long (extremely long may be required) straight segments, then the billiard system in such a (stadium like) domain has nonvanishing Lyapunov exponents. This work puts the original stadium of Bunimovich \cite {B}, which had arcs of circles in the boundary, into a large class of billiard systems with nonuniform hyperbolic behavior, larger than the class with convex scattering pieces introduced in \cite {W4}. All the properties listed in Section 7 are satisfied for the billiards of Donnay in a straightforward fashion, with the notable exception of the noncontraction property. The problem is that the construction of the bundle of sectors depends heavily on the dynamics and the chances of finding a geometrically defined Lyapunov metric (like $|J'|^2$ for the convex scattering curves) are slim. Instead we use the following two ideas. We have remarked in Section 7 that if the map $T$ is differentiable up to and including the boundary of symplectic boxes, and $DT$ is strictly monotone, then the noncontraction property holds automatically. In the billiards of Donnay the sectors are pushed strictly inside at the time of crossing from one convex piece to the other. Hence, we can use this observation on the compact part of the phase space made up of orbits which cross over from one convex piece to the other. We have the noncontraction property for the return map to this set, where we measure vectors in $\C$ using the form $\Q$ defined by the bundle of sectors uniformly larger than $\C$. The construction of the bundle of sectors $\C$ by Donnay and his condition on the separation of convex pieces allows to introduce immediately these larger sectors with respect to which the derivative of the return map is monotone. It remains to check the noncontraction property along grazing' orbits which reflect many times in one convex piece. This is essentially done in \cite {D2}, where Lazutkin coordinates are used to put tha map $T$ in the vicinity of the boundary into a normal form. These two observations, put together, give us the unconditional noncontraction property, and thus our Main Theorem applies. \bigskip {\bf B. Piecewise linear standard map. } \bigskip Let $T:\Bbb T^2 \to \Bbb T^2$ be defined by $$T(x_1,\,x_2) = (x_1+x_2 + Af(x_1),x_2+Af(x_1))$$ where $(x_1,\,x_2)$ are taken modulo 1, $f$ is a periodic function $$f(t) = |t| -\frac 12, \ \ \text { for } -\frac12 \leq t \leq \frac 12,$$ and $A$ is a real parameter. The mapping $T$ preserves the Lebesgue measure. For $A=1$ there is a simple invariant domain $\Cal D$ in the torus shown in Figure \thetag{14.8}. It was proven in \cite {W5} that the Lyapunov exponents are different from zero almost everywhere in $\Cal D$. \proclaim{Theorem 14.9} $T$ is ergodic in $\Cal D$. \endproclaim As in the previous application it follows that $T$ is a Bernoulli system in $\Cal D$. All the conditions of Section 7 are satisfied here in a very simple fashion. The reader can find all the necessary details in \cite {W5} and \cite {W6}. In this piecewise linear case one does not have to rely on the general results of Katok and Strelcyn. The existence of stable and unstable leaves can be obtained by the straightforward approach of Sections 1-3. There are many other values of $A$ for which nonvanishing of Lyapunov exponents was established for $T$ in some domains in the torus, \cite {W5},\cite{W6}. The most interesting is the sequence of $A$'s (roughly speaking) going to zero for which there is an invariant domain, with similar geometry as $\Cal D$, where $T$ has nonvanishing Lyapunov exponents. It is a piecewise linear model for the unstable layer containing the separatrices of the saddle fixed point $(0,\,\frac 14)$. One can apply Main Theorem to all these special domains , so that in each case the map $T$ is ergodic and hence Bernoulli. The reader should not have any difficulties in recovering the details based on the two papers cited above (incidentally even the noncontraction property was considered there). \bigskip {\bf C. The system of falling balls. } \bigskip One of the original motivations for our work was to prove ergodicity of the system of falling balls. This is a monotone system (\cite {W7}, \cite {W8}, \cite {W1}) and all smooth orbits are strictly unbounded. (The unboundedness of all orbits is obtained, under mild assumptions, by the application of Proposition 6.) It follows that all Lyapunov exponents are different from zero and it looks like a prime candidate for the application of Main Theorem. It turns out, however, that in this example the singularity sets are not properly aligned, if the number of balls is greater than two. We will show this, and briefly discuss the case of two balls. The system of falling balls is the system of point particles moving on a vertical line, which also interact by elastic collisions, and are subjected to a potential external field which forces the particles to fall down. To prevent the particles from falling into an abyss we introduce the hard floor, and assume that the bottom particle bounces back upon collision with it. The masses of the particles are in general different (the system of equal masses is completely integrable, since the elastic collision of equal masses in one dimension amounts to exchanging of the momenta). The Hamiltonian of the system is $$H = \sum_{i=1}^N\left(\frac {p_i^2}{2m_i} + m_iU\left(q_i\right)\right)$$ where $q_i$ are the positions and $p_i=m_iv_i$ the momenta of the particles, $q_i,p_i\in \Bbb{R}, i=1,\dots ,N$, and $U\left(q\right)$ is the potential of the external field . The differential equations of the system are \aligned \dot{q_i}&= \frac {p_i}{m_i}\\ \dot {p_i}&= -m_iU'\left(q_i\right), \endaligned $i=1,\dots ,N$. Additionally we assume that the particles are impenetrable and collide elastically with each other and with the floor ${q =0}$. We choose the following Lagrangian subspaces $$V_1 = \{ dp_1=\dots =dp_N =0\} \ \ \ \text{and} \ \ \ V_2 = \{ dh_1=\dots =dh_N =0\},$$ where $h_i=\frac {p_i^2}{2m_i} + m_iU\left(q_i\right), i=1,\dots ,N$, are individual energies of the particles. We have $$dh_i =\frac {p_idp_i}{m_i} + m_iU'\left(q_i\right)dq_i,$$ $i=1,\dots ,N,$ so that $V_1$ and $V_2$ are indeed transversal if only $U' \neq 0$, i.e., if the external field is actually present. The form $\Cal{Q}$ is equal to $$\Cal{Q}\left(\left(\delta q, \delta p\right)\right) = \sum_{i=1}^{N} \left(\delta q_i \delta p_i + \frac{p_i}{m_i^2U'\left(q_i\right)}\left(\delta p_i\right)^2\right).$$ It was shown in the papers cited above that the system is strictly monotone provided that $$U'\left(q\right) > 0 \ \ \text { and } \ \ U''\left(q\right) < 0 .$$ and $$m_1 > m_2 >\dots > m_N.$$ The symplectic map $T$ that naturally arises in this system is the map `from collision to collision''. Our dynamical system is a suspension of the map. So that the system is ergodic if and only if the map $T$ is ergodic. As usual, the actual computations are easier done in the whole phase space of the flow. Singularity set $\Cal S^-$ corresponds to triple collisions: simultaneous collisions of three particles and the collision of two particles with the floor. Part of the first singularity set are not properly aligned. The second set is. So the methods of this paper apply only to the system of two particles. Let us show that indeed the triple collision of three particles produces the singularity set which is not properly aligned. We consider the manifold $$\{(q,\,p) | q_1 = q_2 = q_3 \}.$$ Its tangent subspace is $$\{(\delta q,\,\delta p) | \delta q_1 = \delta q_2 = \delta q_3 \}.$$ Its skew orthogonal complement is the two dimensional subspace given by equations \aligned & \delta q = 0,\\ &\delta p_1 + \delta p_2 +\delta p_3 =0,\\ & \delta p_i = 0 \ \ \text { for } \ \ i \ge 4. \endaligned \tag {14.10} Restricting the form $\Q$ to this plane we get $$\sum_{i=1}^3 \frac {p_i}{m_i^2 U'}(\delta p_i)^2. \tag {14.11}$$ We should assume that the particles emerge from collisions which means that $$\frac {p_1} {m_1} < \frac {p_2} {m_2} < \frac {p_3} {m_3}.$$ But they may, as well, be all negative which makes the quadratic form \thetag {14.11} negative definite. The actual characteristic line is obtained by intersecting the plane \thetag {14.10} by the tangent to the constant energy manifold. If all the momenta are negative it is guaranteed to be outside of the sector. It is not hard to compute that the precise condition for the characteristic line to be contained in the sector is $$\frac {v_1}{m_1} \left( v_2 -v_3 \right)^2 + \frac {v_2}{m_2} \left( v_3 -v_1 \right)^2 + \frac {v_3}{m_3} \left( v_1 -v_2 \right)^2 \geq 0$$ where $v_i = \frac {p_i}{m_i}, i \geq 1$ are the velocities. We close with the discussion of the system of two balls. For clarity, we restrict ourselves to the case of constant acceleration, $U(q) = q$. It was established in \cite {W7}, that also in this case all orbits are strictly monotone, if there are only two or three balls and their masses decrease. (For more than three balls technical problems arise, and it is an open problem to prove strict monotonicity.) Let us fix the value of the total energy of the system, $H=\frac 12$. In this manifold we consider the two dimensional section $\M$ of the flow, corresponding to the bottom particle emerging from the collision with the floor; the surface $\M$ is given by $\{H=\frac 12,\, q_1 = 0,\, v_1 \geq 0\}$. The state of the system in $M$ is completely described by the velocities of the particles $(v_1,\,v_2)$; and we use the velocities as coordinates in $\M$. Hence, our phase space $\M$ is the domain bounded by the half-ellipse $$m_1v_1^2 + m_2v_2^2 \leq 1,\ \ , v_1 \geq 0.$$ Let us calculate the symplectic form in these coordinates. We have $$\omega = dp_1\wedge dq_1 + dp_2\wedge dq_2.$$ On the surface of section $\M$ $$dq_1 \equiv 0 \ \ \text{ and } \ \ dq_2 = -\frac {m_1}{m_2}v_1dv_1 - v_2dv_2.$$ Hence, we get $$\omega = m_1 v_1 dv_1\wedge dv_2.$$ The map $T: \M \to \M$ is defined by the first return of the flow to $\M$. Our symplectic box $\M$ is split into two symplectic boxes by $\Cal S^+$, which is the arc of the ellipse $\{ m_1v_1^2 + m_2(v_2-2v_1)^2 = 1\}$ contained in $\M$. The symplectic box $\M^+_f$, above $\Cal S^+$, contains all the initial states for which the bottom particle returns to the floor without colliding with the top particle. The map $T$ in $\M^+_f$ is linear $$T(v_1,\,v_2) =(v_1,\,v_2-2v_1).$$ The symplectic box $\M^+_c$, below $\Cal S^+$, contains all the initial states for which there is a collision of the two particles before the bottom particle returns to the floor. The map $T$ in $\M^+_c$ is nonlinear and is best described in a coordinate system $(h,\,z)$ where \aligned h =& \frac 12 m_1 v_1^2\\ z= & v_2 -v_1. \endaligned (This coordinate system is derived from the canonical system of coordinates in the full phase space furnished by the individual energies and velocities of the particles. The exceptional role of these coordinates is well documented in \cite {W7}.) Note that both the energy of the bottom particle and the difference of velocities change only in collisions. Now $T = F_2\circ F_1$, where $$F_1(h,\,z) = ( - h - a z^2 + b,\, -z), \ \ a = \frac{m_1m_2(m_1-m_2)}{(m_1+m_2)^2} \ \ \text { and } \ \ b = \frac {m_1}{m_1+m_2} ,$$ describes the collision of the two particles, and $$F_2(h,\,z) = (h,\, z + c\sqrt h), \ \ c = \sqrt{\frac {8}{m_1}},$$ describes the collision of the bottom particle with the floor. To find the image symplectic boxes $\M^-_f$ and $\M^-_c$ we can use the reversibility of our system. Namely, if we put $S(v_1,\,v_2)= (v_1,\,-v_2)$ then $T \circ S = S \circ T^{-1}$, and so $\M^-_f = S \M^+_f,\, \M^-_c = S \M^+_c$. Our bundle of unstable sectors is constant in the coordinates $(h,\,z)$ and equal to the positive (and negative) quadrant; the form $\Q$ is given by $\Q(\delta h,\, \delta z ) = \delta h \delta z.$ It is immediate that $\Cal S^+$ and $\Cal S^- = S \Cal S^+$ are properly aligned. We can now check that $T$ is monotone in $\M^+_f$ and strictly monotone in $\M^+_c$ (both $F_1$ and $F_2$ are monotone). Indeed, in the $(h,\,z)$ coordinates we have $$DF_1 = \left(\matrix -1& -2a z \\ 0 & -1 \endmatrix\right) \ \text { and } DF_2 = \left(\matrix 1& 0 \\ \frac c{2\sqrt h} & 1 \endmatrix\right).$$ Moreover the map $T$ in $\M^+_f$ is equal in the coordinates $(h,\,z)$ to $F_2$. Since the collision of the two particles must eventually occur we obtain strict monotonicity of all nondegenerate orbits. Unboundedness of all nondegenerate orbits follows from Proposition 6.10. So the Sinai-Chernov Ansatz holds. To check the noncontraction property, we observe that the standard Riemannian metric in the coordinates $(h,\,z)$ does not decrease on vectors from the sector, if we apply one of the above matrices. Finally, we are guaranteed that the coefficient $\sigma (DT^n)$ can be made arbitrarily large by increasing $n$, except for points which end up on singularity set $\Cal S^+$ in the future and the singularity set $\Cal S^-$ in the past. There are only countably many such points in view of the proper alignment of singularity sets, and Main Theorem applies to all other points. It follows that $T$ is ergodic and consequently, by the results of Katok and Strelcyn, it is a Bernoulli system. The case of variable acceleration ($U'' < 0$) can be treated in a similar fashion. It is not possible to write down the formulas for the return map $T$ but its derivative in the coordinates \aligned \delta h &= \frac {p_1}{m_1} \delta p_1 \\ \delta z &= \frac {1}{m_2U'(q_2)} \delta p_2 - \frac {1}{m_1U'(q_1)} \delta p_1, \endaligned was essentially calculated in \cite {W8}. 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