Billiards: a Geometric and Ergodic Approach Nicolás Daniel Mart´Inez

Billiards: A Geometric and Ergodic Approach

Nicol´asDaniel Mart´ınezRamos Department of Mathematics Universidad de Los Andes

Supervisor Alexander Getmanenko

In partial fulﬁllment of the requirements for the degree of Mathematician November 27, 2019

Acknowledgements

As always, there are no possible words, either written or spoken which could show my gratitude to all people that in any way have helped me through my studies and my life. First of all, to my family: my parents, my brother, and my cat, for giving me a home in which I could grow up and become what I am today and what I will be in the future. All of the love, understanding and advice I have received from them guide the rest of my existence. Please, be assured, all that I will ever do in my life goes to you even if you do not even know about it. Now, I would like to thank my friends, from the university, from school, from life, from anywhere. You guys and gals are the reason living is worth all the trouble. I would like to mention all of you, but I am sure I would miss a lot of people, so, please, all of those who ﬁnd a friend in me, believe me, this is for you. And of course, I would not be a student if I did not take time to properly thank all of the teachers and professors I have had the plea- sure to know and learn from, each one of you, for even the tiniest amount of knowledge you have passed down to me, deserves the most sincere gratitude. I would like to make a special mention to my thesis advisor, Alexander Getmanenko, not only for being an outstanding mentor and professor, but for taking once again the hardships of being a student along with me. To being a student: Young years for smiles and tears. Present for Future. Abstract

This thesis comes from the interest to study dynamical systems which do not depend on differential equations and which allow trajectories with nondifferentiable points. At first, the main motivation was the study of laser dynamics in conics to further understand how tunneling worked in such systems, but in the process I found the rich and vast theory of Dynamical Billiards and from then on, our study focused on understanding them, as well as the tools that have been developed to study them. With this in mind, this work is divided into three parts. At first, the purpose of the document will be to properly define what a Billiard is and explore the ways in which movement can be studied. Moreover, we will study some basic properties of the movement in these systems and we will give some examples which we will explore later with more detail. Afterwards, we proceeded to explain the proof of one of the fundamen- tal tools in the study of dynamics and ergodic theory, the Oseledec’s Multiplicative Theorem. This theorem ensures us that the limit by which the Lyapunov Exponents are defined exist provided some mild conditions on a transformation which preserves the measure. More- over, we exposed it using rather simple ideas and theorems derived mostly from functional and real analysis. Finally, we used the theorem proved in the previous section and our geometric analysis of the system to derive some conclusions regarding the Billiard Map. This work we have done is not only important for its pedagogic value as a first experience writing a formal mathematical document, but also as a first stepping stone into the world of Dynamics. With this particular example, we have used some of the main concepts on this branch of mathematics, namely the stable and unstable spaces, Lya- punov Exponents and hyperbolic systems. Contents

1 Basic concepts and first examples1 1.1 Definition and Construction of Billiards ...... 1 1.2 Motion: Flow and Map ...... 3 1.2.1 The Billiard Flow ...... 3 1.2.2 The Billiard Map ...... 8 1.2.3 Families of reflected paths ...... 15 1.3 Some examples ...... 19 1.3.1 Billiard in a circle ...... 19 1.3.2 The Ellipse ...... 21 1.3.3 The Stadium ...... 22

2 Oseledec’s Multiplicative Ergodic Theorem 24 2.1 Statement of the theorem ...... 24 2.2 Proof of the ﬁrst three conclusions ...... 25 2.2.1 Some theorems about measurability ...... 28 2.3 Proof of the forth conclusion ...... 30 2.3.1 A simpler setup ...... 30 2.3.2 Skew Products ...... 36 2.3.3 Conclusion ...... 42

v CONTENTS

3 Lyapunov Exponents for the Billiard Map 43 3.1 Understanding Lyapunov Exponents of Diﬀeomorphisms . . . . . 43 3.2 The Billiard Map and its Lyapunov Exponents ...... 46 3.2.1 A simple example ...... 49 3.3 A Projective Criterion for nonzero Lyapunov Exponents . . . . . 50 3.3.1 A hyperbolic example ...... 54

4 Appendix: Some Ergodic Theorems 56 4.1 Ergodic Theorems ...... 56

References 59

vi Chapter 1

Basic concepts and ﬁrst examples

In the following chapter we will go over the basic deﬁnitions for studying billiards. We will give some examples that illustrate the main ideas when we view it as a dynamical system. Moreover, we will take a look at some examples that will appear later in the document.

1.1 Deﬁnition and Construction of Billiards

In this thesis, we will use the following setup used by Chernov and Markarian in their book [1]: Let D ⊂ R2 be a domain with a smooth or a piece-wise smooth boundary. The billiard system corresponds to the motion of a point particle inside this domain with specular reflections (angle of incidence is the angle of reflection) off the boundary δD. The definition of motion is not in itself a mathematical concept, however, to fill this gap we will shortly introduce the concepts of billiard flow and billiard map. In simple terms, the particle moves in a straight motion until the moment it collides with a point of the boundary, at that moment, the particle starts moving in the direction that makes the same angle with the normal to the wall as the original direction. Moreover, we assume that this particle moves at unit speed.

The previous ideas are too vague to properly work with them, then we will impose certain assumptions which will give us more structure on the systems [1]:

2 Assumption 1.1.1. Let D0 ⊂ R be an open, connected set. Our billiard domain will be D0 = D. Assumption 1.1.2. The boundary ∂D is a union of a ﬁnite number of Cr, r ≥ 3 compact curves. We will number these curves and denote them as Γi so that we

1 1.1 Deﬁnition and Construction of Billiards have: δD = Γ = Γ1 ∪ · · · ∪ Γn. We assume our curves to be at least thrice diﬀerentiable to be able to work with their curvature.

Assumption 1.1.3. The curves Γi can only intersect each other at their endpoints.

The endpoints of the curves must be handled with precaution, as at these points it may be impossible to find a tangent and therefore the specular reflection is not defined. Thus, we define the corners of the system as:

Γ∗ = δΓ1 ∪ · · · ∪ δΓn.

For the last assumption, we deﬁne a parametrization of each Γi:

2 fi :[ai, bi] → R . Furthermore, we take the intervals where our curves are deﬁned so that:

[ai, bi] ∩ [aj, bj] = ∅ for i 6= j.

2 Assumption 1.1.4. Let each Γi be parametrized by a map fi :[a, b] → R . Then the second derivative of the fi never vanishes or is identically zero. By the previous assumption there are three possible kinds of boundaries which we will also call walls. The first, which is a flat wall, that is, a wall which is given 00 by a line and thus, fi (x) = 0. The second possible wall is the dispersing wall, that is, a wall that is concave. The name comes from the fact that in optics these kind of walls scatters nearby 00 rays of light. As the parametrization is concave we have that the vector fi (x) 6= 0 and it points outside the domain D. On the other hand, the last possible kind of wall is the focusing wall, that is convex. As before, the name comes from the fact that rays of light reflected by these walls get closer. In terms of the derivative, we can characterize these wall 00 by fi (x) 6= 0 and it points inside the domain D. As an example of the classification of these walls, consider the following image:

2 1.2 Motion: Flow and Map

Figure 1.1: A billiard table with three types of wall.

Note that walls 1 and 3 are ﬂat walls as they are described by a line, wall 2 and 5 are dispersing and wall 4 is focusing. We will need the folowing deﬁnition:

Deﬁnition 1.1.1. The curvature K of each wall Γi is given by:  0 if Γ is ﬂat  i 00 K = −kfi k if Γi is focusing  00 kfi k if Γi is focusing.

1.2 Motion: Flow and Map

As it is often the case in dynamical systems, we can either study the system by a continuous motion (which we will denote ﬂow) or a discrete motion (called map). In the following section we will brieﬂy go over each point of view and discuss the results we can gather from them. The following section uses ideas both found in [1] and [2]

1.2.1 The Billiard Flow

We will denote the position of the particle q ∈ R2 and its velocity v ∈ S1. As we assume that the movement is continuous, both parameters will depend on t ∈ R, the time in which the movement takes place. Modeling the motion of the particle as a rectilinear movement, a change in the velocity can only be given by a specular reﬂection at the points where the trajectory of the particle intersects

3 1.2 Motion: Flow and Map a wall of our system. Moreover, the change of velocity is given by :

vf = vi − 2hn, viin, (1.1)

where vf and vi are the ﬁnal and initial velocities and n is the normal vector at the boundary position q 6∈ Γ∗ :

Figure 1.2: A billiard reﬂection

Now, as we can completely deﬁne the movement of the particle with q ∈ D and v ∈ S1, we have that the phase space is given by:

Ω = {(q, v) ∈ D × S1/ ∼}, where ∼ is the equivalence relation given by:

(q, vi) ∼ (q, vf ) for q ∈ ∂D.

Then, we deﬁne inductively the ﬂow:

Φt :Ω → Ω,

+ + for t ∈ R and (q, vi) ∈ Ω in the following way: if there is no t1 ∈ R such that t1 < t and q + t1vi ∈ ∂D, then the ﬂow is deﬁned as:

t Φ (q, vi) = (q + tvi, vi).

In the case that there is exactly one t1 such as the one previously deﬁned, once the linear trajectory intersects one of the walls, the velocity is changed from vi

4 1.2 Motion: Flow and Map

to vf as in ﬁgure 1.2. In this case, the ﬂow is explicitly:

t Φ (q, vi) = (q + t1vi + (t − t1)vf , vf ), for n ∈ N intersections with the boundary of the domain, we define inductively from the previous cases. The previous definition presents some problems for the point q∗ ∈ Γ∗, as we cannot properly define a reflection in these endpoints. Then, if for some t ∈ R+ t and some (q, vi) ∈ Ω we obtain that Φ (q, vi) = (q∗, vi), for any T > t, we define T that Φ (q, vi) = (q∗, vi), that is, we force the movement to stop. To avoid such problems, we define Ω˜ ⊂ Ω as the subset of the phase space where movement is defined for all t ∈ R. Then, we can work with the flow: Φt : Ω˜ → Ω˜.

Remark. With our identiﬁcation of the boundary, the ﬂow is continuous.

Knowing that the flow is continuous with the topology we have set in the system, a natural question to ask is weather the flow is smooth at any moment. ˜ Thus, we set a particle with arbitrary coordinates (xi, yi, θi) ∈ Ω and study the action of the flow Φt on it for t ∈ R.

If after a time t the particle has not reached a boundary, we ﬁnd that:

xf = xi + t cos(θi),

yf = yi + t sin(θi),

θf = θi. Now, assume that during a certain interval (0, t) ⊂ R there is exclusively one intersection of the path of the particle with a wall of the system. We will need to introduce the following new variables: Let (xb, yb) ∈ Γ be the point of intersection of ﬂow of the particle with a wall, T the tangent to Γ at (xb, yb) and γ the angle between T and the positive x axis. Additionally, let si the time to the intersection with the boundary and sf = t − si the time after the collision. Finally, we consider the angle ψ from T to θf .In this way, we ﬁnd the

5 1.2 Motion: Flow and Map

Figure 1.3: Flow after one collision following relations:

xi = xb − si cos θi,

yi = yb − si sin θi, θ = γ − ψ, i (1.2) xf = xb + sf cos θf ,

yi = yb + sf sin θf ,

θf = γ + ψ.

Now, let us parametrize the curve Γ by its arc-length with parameter r. In this way, we ﬁnd the following diﬀerentials on the curve Γ:

dxb = cos γdr,

dyb = sin γdr, (1.3) dγ = −Kdr.

6 1.2 Motion: Flow and Map

Diﬀerentiating the equations in (1.1):

dxi = cos γdr − cos θidsi + si sin θidθi,

dyi = sin γdr − sin θidsi − si cos θidθi, dθ = −Kdr − dψ, i (1.4) dxf = cos γdr + cos θf dsf − sf sin θf dθf ,

dyi = sin γdr + sin θf dsf + sf cos θf dθf ,

dθf = −Kdr + dψ.

The movement we have deﬁned for our particle are piece-wise linear translations on the space and its change of velocity also comes from a linear transformation, which means we can expect our ﬂow to preserve the Lebesgue measure of the set Ω,˜ which we can express in geometrical terms as the volume form dx ∧ dy ∧ dθ. The previous idea is contained in the following theorem:

Theorem 1.2.1. The ﬂow Φt : Ω˜ → Ω˜ preserves the volume form dx ∧ dy ∧ dθ.

Proof. We will prove it by factorizing the number of reflections in the flow of the particle. If there are no reflections in the trajectory, the flow is given by a translation along straight lines, which clearly preserves the volume form.

On the other hand, if there is one collision on our system, in the notation of (1.2), we obtain the following:

dxi ∧ dyi = cos γdr ∧ − sin θidsi + cos γdr ∧ −si cos θidθi

− cos θidsi ∧ sin γdr − cos θidsi ∧ −si cos θidθi

+ si sin θidθi ∧ − sin θidsi + si sin θidθi ∧ sin γdr.

Thus, multiplying by dθi:

dxi ∧ dyi ∧ dθi = − sin θi cos γdr ∧ dsi ∧ dθi + dsi ∧ − cos θi sin γdr ∧ dθi

= (− sin θi cos γ + cos θi sin γ)dr ∧ dsi ∧ (−Kdr − dψ)

= − sin ψdr ∧ dsi ∧ dψ.

In a process similar to the previous one, we ﬁnd that:

dxf ∧ dyf ∧ dθf = sin ψdr ∧ dsi ∧ dψ.

However, as we have that si + sf = t and in this case t is a constant, we see that

7 1.2 Motion: Flow and Map

dsi = −dsf which yields the conclusion we expected. In the case we have more collisions, due to our initial assumptions on the billiard system, we factorize the ﬂow so that in each component there is no more than one collision.

Corollary 1.2.1.1. The ﬂow Φt preserves the volume form dx ∧ dy ∧ dθ by precomposition. Moreover, it preserves the Lebesgue measure on Ω˜.

Now, we can conclude our study with the following theorem:

Theorem 1.2.2. The ﬂow Φt is Cl−1 smooth at the points (x, y, θ) ∈ Ω˜ if for all collisions at times ti < t the velocity before the collision and after the collision are diﬀerent.

Proof. The equations (1.2), (1.3), (1.4) show us that the derivative of the ﬂow depends on K, thus, as this is the derivative of the Cl curve that describes the boundary, it is necessary that the ﬂow is Cl−1.

1.2.2 The Billiard Map From this natural implementation of the motion of the ’ball’ on the billiard, we can now study the motion by uniquely considering the points of the boundary where the ﬂow and the boundary intersect.

Remember that each of the curves Γi is parametrized by:

2 fi :[ai, bi] → R . We characterize every collision of Φt with the boundary with two parameters, ri ∈ [ai, bi] where fi(r) is the point where the collision occurs, and the angle 0 θ ∈ [0, π] of the reﬂected velocity respecting to f (ri). The domain and range of this map is the set given by:

[˙ M = [ai, bi] × [0, π] with the product topology. 1≤i≤n

And, so, we deﬁne the map as:

T : M → M,

t t where T (r, θ) = (r1, θ1) if for t = mint∈R{Φ (r, θ) ∈ M} we have Φ (r, θ) = (r1, θ1). That is, the map T sends the parameters of a collision to the parameters of the next collision.

8 1.2 Motion: Flow and Map

We see that the properties of the map and the ﬂow are immediately linked. In particular:

Theorem 1.2.3. The form ω = sin θdθ∧dr is preserved by T . That is, T ∗(ω) = ω

Proof. As 0 ≤ θ ≤ π, we see that sin θ is positive, which implies this is an area form. Now, let us consider the euclidean distance D(r, r1) where r, r1 determine positions fi(r), fj(r1) in Γ and T (r, θ) = (r1, θ1). We write the boundary in coordinates as: fi(r) = (xi(r), yi(r)). Thus, we have that:

∂D(r, r ) ∂p(x (r) − x (r ))2 + (y (r) − y (r ))2 1 = i j 1 i j 1 ∂r ∂r (x (r) − x (r )) dxi + (y (r) − y (r )) dyi = i j 1 dr i j 1 dr . p 2 2 (xi(r) − xj(r1)) + (yi(r) − yj(r1))

We see that this is a projection of the vector that goes from fi(r) to fj(r1) into the curve at fi(r). Thus, as the angle between the curve Γ and the velocity of ∂D(r,r1) the particle is θ and ∂r is unitary, we have that the projection is given by: ∂D(r, r ) 1 = cos θ. ∂r With a similar procedure, we ﬁnd that:

∂D(r, r1) = − cos θ1. ∂r1 Thus, we have that: dD = cos θdr − cos θ1dr1, which yields: 2 0 = d D = − sin θdθ ∧ dr + sin θ1dθ1 ∧ dr1,

hence sin θ1dθ1 ∧ dr1 = sin θdθ ∧ dr.

Additionally, the map T is a homeomorphism when restricted to certain well behaved sets as we will show shortly. First we want to identify the subsets of M where the map may encounter problems. As before, we know hat the endpoints of each Γi may not be smooth, which forces us to stop the ﬂow, as we previously discussed. Moreover, we may ﬁnd

9 1.2 Motion: Flow and Map problems when for a particle at (r, θ) the velocity is parallel to the tangent at f(r), that is, if θ = 0 or θ = π. This problems come from the fact that the particle would be moving along the border without defining a reflection. Then, we define: P0 = {(r, θ) ∈ M|r = ai, bi and θ = 0, π}. Similarly to the last case, in the case that for some (r, θ) ∈ M we obtain T (r, θ) = (r1, θ1) with θ1 = 0 or θ1 = π, we encounter the problem that the movement should continue in a rectilinear path, even if there was a collision with the boundary. These intersection with the boundary Γ such that the overall trajectory of the particle is not truly affected is called a grazing collision. The points where such phenomenon appears may add some unnecessary difficulties to our study, and thus, we decided to disregard them. That is why we define:

P1 = P0 ∪ {x ∈ M|T (x) is a grazing collison}.

Finally, to further simplify our work and to be able to properly deﬁne an inverse for our billiard map, we disregard the points whose preimage intersects the pathological cases that deﬁne P0, that is:

−1 P−1 = P0 ∪ {x ∈ Int M|T {x} 6⊂ Int M}.

Lemma 1.2.1. The set M\P1 is open in M with the product topology. Further- more, the map T : M\P1 → M is a local homeomorphism.

Proof. Consider (r, θ) ∈ M\P1 where r is the parameter for a curve Γi = fi(r). Let T (r, θ) = (r1, θ1) where r1 is a parameter for a curve Γj = fj(r1). Set a system of coordinates where the x-axis the y-axis are respectively parallel and perpendicular to the derivative of f(r) at the point r. For this reason, we set:

fi(r) = (xi(r), yi(r)),

fj(r1) = (xj(r1), yj(r1)) : t Now, let t ∈ R such that Φ (fi(r), θ) = T (r, θ). Then, we ﬁnd that: x (r) + t cos θ = x (r ) =⇒ F (r, θ, t, r ) = x (r) + t cos θ − x (r ) = 0, i j 1 1 1 i j 1 (1.5) yi(r) + t sin θ = yj(r1) =⇒ F2(r, θ, t, r1) = yi(r) + t sin θ − yj(r1) = 0.

Now, as all functions are diﬀerentiable, we can apply the implicit function theorem to get the variables t, r1 in terms of r, θ. The jacobian equals:

∂F1 ∂F1 cos θ −x0 (r ) ∂t ∂r1 j 1 0 0 ∂F2 ∂F2 = 0 = xj sin θ − yj cos θ. sin θ −y (r1) ∂t ∂r1 j

10 1.2 Motion: Flow and Map

The jacobian in nonzero except at the points:

0 0 xj sin θ − yj cos θ = 0 0 0 xj sin θ = yj cos θ 0 yj tan θ = 0 . xj We see that this happens when there would be a grazing collision, which we do not need to consider as we removed the points which yield these situations when we removed P1. The implicit function theorem yields that r1 = r1(r, θ) and t = t(r, θ) are continuous in M\P1. Furthermore, as the angle of incidence α, that is, the angle formed between the initial velocity and the tangent at the point of the collision is: 0 −1 hf (r1), (t cos θ, t sin θ)i α(r, θ) = cos 0 (1.6) kf (r1)kk(t cos θ, t sin θ)k

The angle of refection θ1 depends continuously on the angle of incidence as shown in formula (1.1), which implies that:

θ1 = θ1(α(r, θ)) = θ1(r, θ), is a continuous function and thus, T is a continuous function. We see that P0 is a closed set as it is a union of ﬁnite segments. Moreover, as the points (r, θ) which yield grazing collisions under the map T are segments in M, they form a closed set, so we have that T −1({x ∈ M|T (x)is a grazing collition}) is closed. Therefore, we have that P1 is a closed set and we can conclude that M\P1 is open. Now, let us prove that T is in fact a local homeomorphism on M\P1. Consider a point (r, θ) ∈ M\P1 and a neighborhood V of (r, θ). Now, as this space is locally compact as an open subspace of a locally compact space, we can ﬁnd a compact neighborhood K such that (r, θ) ∈ V . Now, we have to prove that:

T : K → T (K) is injective , as we will have a continuous bijection between Hausdorﬀ compact spaces, yielding the desired local homeomorphism. This injectivity follows easily from (1.5), as each r determines a unique point in the boundary and cos is injective in 0 < θ < π. We can conclude as both K and T (K) are Hausdorﬀ and compact.

−1 Lemma 1.2.2. The set M\P−1 is open in M. Furthermore, the map T : M\P−1 → M is a local homeomorphism.

11 1.2 Motion: Flow and Map

−1 Proof. Consider (r, θ) ∈ M\P−1, that is (r, θ) ∈ Int M such that T {(r, θ)} ⊂ Int M. Then, as T is continuous, for all open sets where (r, θ) ∈ V we ob- −1 −1 tain T (r, θ) ⊂ T (V ) ⊂ Int M, which implies that for any (r1, θ1) ∈ V , −1 T (r1, θ1) ⊂ Int M. Then, we have that the M\P−1 is an open set. Now, to prove that T −1 is continuous, we make a similar process to the previous −1 lemma. Let (r, θ) ∈ M\P−1 such that T (r, θ) = (r1, θ1). Then, by knowing the angle of reﬂection θ , we can get the angle of incidence α with the use of formula (1.1). With this information, by changing θ for α in equations (1.5) it is possible to express r1 in terms of r and α by way of a continuous function. Furthermore, by applying (1.6) we can express θ with the use of a continuous function of r and α. Thus, we can conclude that T −1 is continuous. To show T −1 is a local homeomorphism, one uses the same argument as the proof used in the previous lemma.

Theorem 1.2.4.

T : M\P1 → M\P−1 is a homeomorphism

Proof. By the previous two lemmas, T and T −1 are continuous. It remains to prove that T (M\P1) = M\P−1. From the deﬁnitions of sets P1,P−1, we have that T (M\P1) ⊂ M\P−1. Now, consider (r, θ) ∈ M\P−1. Then, (r, θ) ∈ Int M and T −1(r, θ) ∈ Int M. Hence, T −1(r, θ) does not generate a grazing collision and −1 −1 T (r, θ) 6∈ P0. Then , we can conclude that T (M\P−1) ⊂ M\P1 from which we get M\P−1 ⊂ T (M\P1). Furthermore, we can prove that this map is a diﬀeomorphism.

12 1.2 Motion: Flow and Map

Figure 1.4: The Billiard Map seen as points of the ﬂow. (Colored in the electronic version.)

Let us explicitly describe the map T starting the motion at a point (r, θ) ∈ M\P1 where we know that T (r, θ) = (r1, θ1). Call fi(r) = (xi, yi) the initial point on the boundary and fj(r1) = (xj, yj) the point where the particle collides with the wall. We will proceed to calculate D(r,θ)T in terms of these parameters and the curvatures K, K1. Now, call ω the angle between the positive x-axis and the

13 1.2 Motion: Flow and Map trajectory of the particle after the collision. We then have:

x = x + t cos ω, j i (1.7) yj = yi + t sin ω.

From our discussion about the differential of the flow, we know that on the curve Γ we find the following differentials:

dxi = cos γdr,

dyi = sin γdr, (1.8) dγ = −Kdr.

By diﬀerentiating the relation γ = ω − θ on R2, we obtain: dω = −Kdr + dθ. (1.9)

On the other hand, if we deﬁne γ1 as the angle between the tangent vector fj(r1) and the positive x-axis, we have that ω = γ1 − θ1. Then, we get the relations (1.8) for the collision point (xj, yj) with angle γ1 and curvature K1. Moreover, we also get that:

dω = −K1dr1 − dθ1. (1.10)

By diﬀerentiating the equations in (1.7) we get that:

dx = dx − t sin ωdω + cos ωdt, j i (1.11) dyj = dyi + t cos ωdω + sin ωdt.

Multiplying the ﬁrst equality by − sin ω and the second by cos ω and replacing by the relations in (1.8):

2 − sin ω cos γ1dr1 = − sin ω cos γdr + t sin ωdω − cos ω sin ωdt, 2 cos ω sin γ1dr1 = cos ω sin γdr + t cos ωdω + sin ω cos ωdt.

Adding both equalities:

tdω = sin θdr + sin θ1dr1. (1.12)

From (1.9) and (1.12) we get that:

sin θ1dr1 = (−tK − sin θ)dr + tdθ. (1.13)

14 1.2 Motion: Flow and Map

From this result along with (1.10):

−K sin θ1dr + sin θ1dθ = − sin θ1K1dr1 − sin θ1dθ1

sin θ1dθ1 = K sin θ1dr − sin θ1dθ − sin θ1K1dr1

sin θ1dθ1 = K sin θ1dr − sin θ1dθ − K1((−tK − sin θ)dr + tdθ)

sin θ1dθ1 = (K sin θ1 + K1 sin θ + tK1K1)dr + (−tK1 − sin θ1)dθ. (1.14)

Therefore, we have that the derivative of T at (r, θ) equals:

1 −tK − sin θ t D(r,θ)T = . sin θ1 K sin θ1 + K1 sin θ + tK1K1 −tK1 − sin θ1 With a similar process, we also ﬁnd that: −1 1 −tK1 − sin θ1 −t D(r1,θ1)T = . sin θ −K sin θ1 − K1 sin θ − tK1K1 −tK − sin θ

Theorem 1.2.5.

l−1 T : M\P1 → M\P−1 is a C diﬀeomorphism

Proof. We obtained that the derivative DT depends on K and K1 and we know that these curvature is Cl−2, we have that T is Cl−1. Moreover, as we had that T is a homeomorphism when restricted to the previously described sets, we obtain that it is a diﬀeomorphism.

1.2.3 Families of reflected paths So far, we have studied both the billiard map and the billiard flow taking into account the movement of a single particle. However, some interesting phenomena appears when discussing the movement of a family of trajectories with similar initial conditions. Here we follow [3]. Consider the arclength parameter r ∈ R along Γ as previously defined, t ∈ R a time parameter and a smooth family of segments l(r, t) = f(r) + tv(r) where f(r) is the point on the boundary of the billiard corresponding to the parameter r and v(r) is a unit vector that shows the direction of the family of segments and depends on r. Moreover, we only consider families of paths that intersect a small arc of the billiard boundary, thus, we take r ∈ (r−, r+) for a small > 0.

Furthermore, consider g : R → R deﬁned such that l(r, g(r)) = f(r)+g(r)v(r), that is, g(r) = t is the envelope of the smooth family of segments l(r, t). Taking

15 1.2 Motion: Flow and Map

Figure 1.5: Families of reﬂected rays the derivative with respect to r, we obtain: d l(r, g(r)) = f 0(r) + g0(r)v(r) + g(r)v0(r), dr and taking the inner product with v0(r):

hl0, v0i = hf 0, v0i + g(r)hv0, v0i.

As l’ is parallel to v because l(r, g(r)) is an envelope:

hf 0, v0i g(r) = − . (1.15) hv0, v0i

Now, consider l1 the family of reﬂected rays from the family l onto the curve Γ. Consider ~τ(r), ~n(r) be respectively the unit tangent and the unit inward normal vectors at f(r). Moreover, from Figure 1.3 we see that v(r) = cos ψ~τ − sin ψ~n and v1(r) = cos ψ~τ + sin ψ~n. From this analysis, we can get the following result:

Lemma 1.2.3. Consider t = g(r) and t1 = g1(r) such that l = l(r, g(r)), l1 = l1(r, g1(r)), then: 1 1 2K − + = . g g1 sin ψ π Proof. We know that the normal vector ~n is obtained by a rotation by 2 (either counterclockwise or clockwise) of the tangent vector ~τ. Then, we will denote d~τ K ~n = R~τ where R stands for the rotation. Moreover, we know that dr = ~n which

16 1.2 Motion: Flow and Map

d~τ K implies dr = R~τ. Diﬀerentiating, we obtain v: dv dψ d~τ dψ d~n = − sin ψ~τ + cos ψ − cos ψ~n − sin ψ dr dr dr dr dr dψ dψ d~τ = − sin ψ~τ + cos ψKR~τ − cos ψ~n − sin ψR dr dr dr dψ dψ = − sin ψ~τ − sin ψKR2~τ + cos ψKR~τ − cos ψ~n dr dr dψ dψ = − + K sin ψ~τ + − + K cos ψ~n. dr dr

In a similar way, we can prove that:

dv dψ dψ 1 = − + KR sin ψ~τ + + K cos ψ~n. dr dr dr

Moreover:

dv dv dψ 2 dψ 2 , = − + K sin2 ψh~τ, ~τi + − + K cos2 ψh~n,~ni dr dr dr dr dψ 2 = − + K , dr dv dψ , ~τ = − + K sin ψ. dr dr

Now, inserting f 0 = ~τ in the previous formulas, and using (1.15), we obtain that:

1 ψ0 − K = . g sin ψ Similarly: 1 ψ0 + K = . g1 sin ψ Adding both equalities, we get our result.

Now, consider r0 such that K(r0) 6= 0. Then, consider a circle of radius −1 (2|K|) such that it is tangent to f(r0). Call d the distance of segment given by the intersection between l(r0, g(r0)) and the previously constructed disk. Lemma 1.2.4. With the previous notation: 1 1 2 − + = sgn(K) . g(r0) g1(r0) d

17 1.2 Motion: Flow and Map

Figure 1.6: A graphic representation of lemma 1.2.4

Proof. By basic trigonometry: 1 d = 2|K| 2 sin ψ 1 |K| = . d sin ψ Then, from the previous lemma, we obtain: 1 1 2K 2 − + = = sgn(K) . g(r0) g1(r0) sin ψ d

The results we have obtained so far give us some insight into the way the geometrical shape of the billiard aﬀects the paths taken by the particles in the systems. Moreover, it shows how small variations on the point of collision with the boundary locally aﬀect the orbit of the particle.

At this point, the previous ﬁndings may seem disconnected from the other results we have obtained this far. However, as this results are local, they will be of great importance once we study whether the system is hyperbolic. This deﬁnition will be presented later once the necessary framework has been developed.

18 1.3 Some examples

1.3 Some examples

1.3.1 Billiard in a circle Let D be the unit disk with boundary Γ = S1. We parametrize the circle by the arclength φ ∈ [0, 2π]/ ∼ where ∼ is such that 0 ∼ 2π. Therefore, the domain of our map T is given by M = ([0, 2π)/ ∼) × [0, π]. By the rotational symmetry, as shown in the next image, we have that our billiard map is simply a rotation given by 2θ. That is:

T (φ, θ) = (φ + 2θ, θ) =⇒ T n(φ, θ) = (φ + 2nθ, θ)

Figure 1.7: The Billiard as a rotation

Moreover, Φt is given by following the broken line in the image with unit speed. From our understanding of the billiard map as a rotation we can get several conclusions. First of all, we know that if the rotation if given by θ/π ∈ Q, then the orbit of the rotation is periodic and in the case p/π 6∈ R\Q then the orbit will be dense in the circle [2]. Thus, in case our ﬁrst point (φ, θ) in the billiard map θ T is such that π 6∈ Q, the successive images of T (φ, θ) will be dense in the line [0, 2π] × {θ}.

However, even if the projection of the image T (M) to S1 is dense, the ﬂow is

19 1.3 Some examples not dense in the disk due to the existence of a circular caustic, that is, we can find a circle centered in (0, 0) of radius 0 < r < 1 such that each segment given by a reflection in the trajectory of the particle is tangent to this disk as we explain in the following lemma: Lemma 1.3.1. Consider T the billiard map in a circular billiard with starting point (φ, θ) for 0 < θ < π. Then, the flow Φt consists of line segments tangent to the circle x2 + y2 = cos2 θ. Proof. Without loss of generality, suppose that (φ, θ) = (0, θ). Then, T (0, θ) = (2θ, θ). Thus, consider the segment with endpoints (1, 0) and (cos 2θ, sin 2θ):

(x(t), y(t)) = ((1 − t) + t cos 2θ, t sin 2θ).

Let us ﬁnd the intersection of this segment with the circle x2 + y2 = cos2 θ :

((1 − t) + t cos 2θ)2 + (t sin 2θ)2 = cos2 θ, (1 − 2t + 2t cos2 θ)2 + 4t2 cos2 θ − 4t2 cos4 θ = cos2 θ, 1 + 4t2 + 4t2 cos4 θ + 4t cos2 θ − 4t − 8t2 cos2 θ + 4t2 cos2 θ − 4t2 cos4 θ = cos2 θ, (1 − 2t)2 − cos2 θ(1 − 2t)2 = 0.

1 Then, as cos θ 6= ±1, we have that for t = 2 this segment and the circle intersect. The condition of tangency is that the equation has a double root, which is pre- 1 cisely t = 2 . By rotational symmetry, the same argument applies to a line segment connecting any two succesive boundary points.

Figure 1.8: Caustic of the ﬂow

20 1.3 Some examples

This example specially interesting as we have a simple tool to understand the motion in the system, namely, the circular rotation, which gives us insight into both the map and the ﬂow.

1.3.2 The Ellipse Let D be the domain bounded by the ellipse given by:

x2 y2 + = 1, a2 b2 for a > 0, b > 0. Furthermore, denote F1 and F2 the foci of the ellipse, which in this particular case will lie in the x-axis. Due to the lack of rotational symmetry on the system, we cannot describe a general motion for any particle, but the following result found in [1] gives us some insight into the motion of a particle in the system.

Lemma 1.3.2. Call the trajectories that intersect the segment F1F2 inner trajectories. Otherwise, we will denote them as outer trajectories. Then, for every outer trajectory there is an elliptic caustic, that is, there is an ellipse D0 with foci 0 F1,F2 such that each reflection in the trajectory is tangent to D . On the other hand, inner trajectories have a hyperbolic caustic, that is, it is possible to find a hyperbola H of foci F1,F2 such that each reflection in the trajectory is tangent to H.

The following images from [1] represent the previous lemma:

In this way, even if the explicit calculations of the elliptic billiard may get cumbersome after a certain number of reﬂections, with the initial behavior of the particle it is possible to understand the general behavior of the movement of the system.

21 1.3 Some examples

1.3.3 The Stadium The following example is a modiﬁcation of the previous example which greatly changes the behavior of the system, we will study it in detail later on.

Consider a circle centered in the origin of radius R ∈ R and cut it by the diameter which is parallel to the y-axis. Then, move each part by a length of H symmetrically respecting to the y-axis and connect the previously connected endpoints by a segment of length 2H.

This system is considerably more diﬃcult due to the lack of a rotational symmetry. Thus, we do not have tools such as the circular rotation to deepen our knowledge of the system. Moreover, as the curvature is nor constant, the movement of a particle inside this boundary heavily depends on the initial conditions we set for the particle.

Figure 1.9: The Stadium billiard

22 1.3 Some examples

To better understand the motion inside this system, we will need more elab- orate tools, which we will develop throughout this document.

23 Chapter 2

Oseledec’s Multiplicative Ergodic Theorem

In this section we will prove The Multiplicative Ergodic Theorem based on the proof by Peter Walters in [4]. However, the ﬁnal conclusion will be based on notes of [5]. We dedicate an entire section to the proof of this theorem due to its length and its result which indicates the existence of Lyapunov Exponents, which will be central to deepen our knowledge of the billiard systems.

2.1 Statement of the theorem

Let Gr(k, Rd) denote the Grassmanian of the k-dimensional subspaces in Rd. Theorem 2.1.1. Let T be a measure-preserving transformation (m.p.t) of a probability space (X, B, µ). Let f be a measurable map f : X → GL(d) where x → Ax + + −1 such that the real-valued functions x → log (kAxk) and x → log (kAxk) are integrable. Then there exists Y ⊂ X of full measure such that TY ⊂Y and the following properties hold:

1. There is a measurable function s : Y → N such that s ◦ T = s. 2. For each y ∈ Y , there are real numbers λ(s(y)) < λ(s(y)−1) < ··· < λ(2) < λ(1) where λi(T y) = λ(i)y for 1 ≤ i ≤ s(y) and y → λ(i)y is measurable.

s(x)+1 s(x) 1 d 3. For each y ∈ Y there are subspaces {0} = Vy ⊂ Vy ⊂ · · · ⊂ Vy ⊂ R . i i i Such that AyVy = VT y and y → {Vy } is a measurable map from Y to Gr(1, Rd).

i+1 i 1 (i) 4. For all y ∈ Y and all v ∈ Vy \Vy , limn→∞ n kAT n−1y ...AT yAyk = λ y

24 2.2 Proof of the ﬁrst three conclusions

2.2 Proof of the ﬁrst three conclusions

Lemma 2.2.1. Consider (an)n∈N, (bn)n∈N with an, bn ≥ 0 for all n ∈ N, then: 1 1 1 i) lim sup log(an + bn) = max lim sup log an, lim sup log bn , n→∞ n n→∞ n n→∞ n

1 1 1 ii) lim inf log(an + bn) ≥ max lim inf log an, lim inf log bn . n→∞ n n→∞ n n→∞ n Proof. First, we will prove i) by two inequalities. First, as logarithm is increasing and bn ≥ 0: 1 1 lim sup log an ≤ lim sup log(an + bn). n→∞ n n→∞ n Similarly: 1 1 lim sup log bn ≤ lim sup log(an + bn). n→∞ n n→∞ n

We now prove the reverse inequality. Let (anm ) ⊂ (an), (bnm ) ⊂ (bn) be subse- quences that converge to the limsup of their corresponding sequence. Call these limits limnm→∞ anm = A, limnm→∞ bnm = B. Without loss of generality, let A ≤ B. Then: 1 1 anm lim log(an + bn ) = lim log(bn ) + log + 1 . n →∞ m m n →∞ m m nm m nm bnm Now, A/B ≤ 1, we have that log anm + 1 is bounded and hence: bnm 1 1 = lim log bnm = lim sup log bn, nm→∞ nm n→∞ n which yields: 1 1 1 lim sup log(an + bn) ≤ lim log(anm + bnm ) = lim sup log bn. n→∞ n nm→∞ nm n→∞ n This proves i). For ii), as an, bn ≥ 0 for all n ∈ N, we have that an + bn ≥ an, bn, therefore: 1 1 lim inf log an ≤ lim inf log(an + bn), n→∞ n n→∞ n 1 1 lim inf log bn ≤ lim inf log(an + bn). n→∞ n n→∞ n

25 2.2 Proof of the ﬁrst three conclusions

This proves ii). Lemma 2.2.2. Let T : X → X a m.p.t. of (X, B, µ) and let A : X → M + (the space of d × d matrices) be such that x → log kAxk is integrable. Consider d 1 n n χ : X × R → R deﬁned as χ(x, v) = lim supn→∞ n log(k(A )xvk) where (A )x = AT n−1x ...AT xAx. Then: 1. There exists Y ⊂ X of full measure such that TY ⊂ Y and χ(x, v) ∈ R ∪ {−∞} for all x ∈ Y , v ∈ Rd. 2. χ(x, 0) = −∞ for all x ∈ X.

3. χ(x, av) = χ(x, v) for all a ∈ R∗, x ∈ X and v ∈ Rd. d 4. χ(x, v1 + v2) = max(χ(x, v1), χ(x, v2)) for all v1, v2 ∈ R and x ∈ X. d 5. χ(T x, Axv) = χ(x, v) for all x ∈ X and v ∈ R . Proof. n Qn−1 1. We have k(A )xvk ≤ kvk i=0 kAT ixk. Applying log and dividing by n, we ﬁnd:

n−1 ! n−1 −1 Y −1 X lim n log kAT ixk = lim n log kAT ixk. n→∞ n→∞ i=0 i=0

+ + −1 Since the functions x → log kAxk and x → log kAx k are integrable, we conclude that x → log kAxk is integrable. −1 Pn−1 By Birkhoﬀ’s Ergodic Theorem [6], limn→∞ n i=0 log kAT ixk is an integrable function, which implies that for a set Y of full measure and x ∈ Y :

n−1 −1 X χ(x, v) = lim n log kAT ixk ∈ ∪ {−∞}. n→∞ R i=0

~ 1 ~ 2. χ(x, 0) = lim supn→∞ n log(k0k) = −∞. 3.

1 n χ(x, av) = lim sup log(k(A )xavk) n→∞ n

1 n = lim sup (log a + log(k(A )xvk)) n→∞ n

1 n = lim sup log(k(A )xvk) = χ(x, v). n→∞ n

26 2.2 Proof of the ﬁrst three conclusions

1 n χ(x, v1 + v2) = lim sup log(k(A )x(v1 + v2)k) n→∞ n

1 n n = lim sup log(k(A )xv1k + k(A )xv2k). n→∞ n By lemma 2.21:

≤ max(χ(x, v1), χ(x, v2)).

1 n χ(T x, Axv) = lim sup log(k(A )T xAxvk) n→∞ n 1 = lim sup log(k((AT nx ...AT 2xAT x)Axvk) n→∞ n n + 1 1 = lim sup log(k((AT nx ...AT 2xAT x)Axvk) = χ(x, v). n→∞ n n + 1

Lemma 2.2.3. Following the notation of lemma 2.2.2:

d 1. For all x ∈ Y and t ∈ R, the set Vx(t) = {v ∈ R |χ(x, v) ≤ t} is a linear d subspace of R . Moreover, we have AxVx(t) ⊂ VT x(t). And for s ≤ t, Vx(s) ⊂ Vx(t).

2. For all x ∈ Y , χ(x, ·): Rd → R ∪ {−∞} takes at most s(x) diﬀerent values λ(s(x))(x) < ··· < λ1(x) where λ(s(x))(x) could be −∞. Furthermore, s(T x) ≥ s(x) and λ(s(x))(x), . . . , λ1(x) are among the values of {λi(T x): 1 ≤ i ≤ s(T x)}.

i i 3. For x ∈ Y we denote Vx = Vx(λ (x)). Then:

s(x)+1 s(x) 1 d {0} ≡ Vx ⊂ Vx · · · ⊂ Vx = R .

i i+1 (i) Moreover, v ∈ Vx \Vx if and only if χ(x, v) = λ (x). Proof.

1. By items 3 and 4 of lemma 2.2.2 we have that Vx(t) is a linear subspace. Using item 5 of lemma 2.2.2, we see that if v ∈ Vx(t), then χ(x, v) =

27 2.2 Proof of the ﬁrst three conclusions

χ(T x, Axv) ≤ t, which implies Axv ∈ VT x(t). In the last claim, it suﬃces to see that for v ∈ Vx(s), we have that χ(x, v) ≤ s ≤ t.

2. Let x ∈ Y . From the previous item we know that s < t, implies Vx(s) ⊂ Vx(t) which in turn implies that dim Vx(s) ≤ dim Vx(t). Then χ can only take up to a finitely many different values as our linear space is finite dimensional. Denote these images λ(s(x))(x), . . . , λ1(x). Moreover, item 5 of last lemma yields our last claim.

3. We take λ(s(x))(x) < ··· < λ1(x), then by item 1 of this lemma, we ﬁnd i i+1 that the related subspaces form a ﬂag. Moreover, if v ∈ Vx \Vx , then λ(i+1)(x) < χ(x, v) ≤ λ(i)(x) and as it needs to take a value of some lambda, we have that χ(x, v) = λ(i)(x).

2.2.1 Some theorems about measurability

Let Y be a complete separable metric space and denote PK(Y ) the collection of nonempty compact subsets of Y . Then, we can endow PK(Y ) with the Hausdorﬀ metric, the following three theorems are proved in C.Casting and M.Valadier[7]:

Theorem 2.2.1. Let (X, B, µ) a complete probability space and let Y be a complete separable metric space. Consider the map Γ: X → PK(Y ). The following are equivalent:

1. Γ is measurable with respect to the Borel σ-algebra of PK(Y ). 2. If U ⊂Y is open, {x ∈ X|Γ(x) ∩ U 6= ∅} is a Borel set.

3. There is a sequence of measurable maps σn : X → Y with σn(x) ∈ Γ(x) and

Γ(x) = {σn(x)}n∈N. Theorem 2.2.2. Let (X, B, µ) is a complete probability space (in the sense of measure) and Y is a complete separable metric space. Let π : X × Y → X be the natural projection. Then for A ∈ B × B(Y ), π(A) ∈ B.

Theorem 2.2.3. Consider (X, B, µ) is a complete probability space and consider d a map x → Vx of X to Gr(1, R ). Deﬁne r : X → N by r(x) = dim Vx. We have the following equivalences:

1. x → Vx is measurable.

2. {(x, v)|x ∈ X, v ∈ Vx} ∈ B × B(Y ).

28 2.2 Proof of the ﬁrst three conclusions

d 3. {(x, y)|x ∈ X, y ∈ P (Vx)} ∈ B × B(P (R )).

d 4. x → P (Vx) is measurable from X to PK(P (R ).

5. r : X → R is measurable and for 0 ≤ k ≤ d, we have that x → Vx is measurable from r−1(k) → Gr(k, Rd).

6. r : X → R is measurable and for all k ∈ N there exist measurable maps −1 d −1 v1, . . . , vk : r (k) → R . such that for all x ∈ r (k), {v1(x), . . . , vk(x)} is an orthonormal basis for Vx.

7. r : X → R is measurable and for all k ∈ N there exist measurable maps −1 d −1 u1, . . . , uk : r (k) → R . such that for all x ∈ r (k), {u1(x), . . . , uk(x)} is a basis for Vx. Theorem 2.2.4. With the notation of lemma 2.2.2:

1. The map s : X1 → N is measurable and then there exist X2 ∈ B with µ(X2) = 1, TX2 ⊂ X2 and s ◦ T = s in X2.

(i) 2. The map λ : {x ∈ X1|s(x) ≥ i} → R ∪ {−∞} is measurable and (i) (i) λ (T x) = λ (x) for all x ∈ X2 ∩ {x|s(x) ≥ i}. Proof. Let us first prove that the functions are measurable. We have that χ : X → d (1) R is measurable, then as we know from lemma 2.2.2 λ (x) = supv χ(x, v) = d (1) maxi χ(x, ei) is measurable. Now, let Λ2 = {(x, v) ∈ X × R |χ(x, v) < λ }. −1 (1) As a preimage, Λ2 = χ (−∞, λ (x)), we have that it is measurable, that is d d Λ2 ∈ B×B(R ). By theorem 2.2.2, if we consider the projection Π1 : X×R → X, then Π1(Λ2) ∈ B. By definition, we also have that Π1(Λ2) = {x|s(x) > 1}. −1 d Consider the map Q(x) = Π1 (x) ∩ Λ2 defined as Q :Π1(Λ2) → Gr(1, R ). We −1 (1) see that Π1 (x) ∩ Λ2 = Vx(λ (x)). Then, we can apply theorem 2.2.3. As Λ2 is measurable, we have that r :Π1(Λ2) → N is measurable and u1, . . . , uk : −1 d Π1(Λ1 ∩ r (k) → R measurable maps such that u1(x), . . . , ur(x)(x) is a basis for −1 (2) Π1 (x) ∩ Λ2. Consequently, we have that λ (x) = maxi χ(x, ui(x)), from which we can conclude that it is measurable. We can continue this process defining Λn to show that λ(n)(x) is measurable for {x|s(x) ≥ n}. Furthermore, as we found (−1) that s (n, ∞) = Π1(Λn+1) ∈ B, we have that s : X → N is measurable. Now, let us prove that there is a set of full measure X2 such that for x ∈ X2, s ◦ T (x) = T (x). From lemma 2.2.3 we know that s ◦ T ≥ s. For n ∈ N, consider the sets: n Dn = {x ∈ X1|s ◦ T (x) > s(x)}. n −1 We see that Dn = (s◦T −s) (0, ∞), which implies the sets Dn are measurable. Now, suppose for contradiction that there exists some n ∈ N such that µ(Dn) > 0.

29 2.3 Proof of the forth conclusion

Then , by Poincar´erecurrence [6], we know that there is a x ∈ Dn and a sequence nk n+nk nk (nk)k∈N ⊂ N such that T (x) ∈ Dn. Then, for all k ∈ N, s◦T (x) > s◦T (x). Then, as this is an strict inequality, we have that there is some k0 ∈ N such that s ◦ T (x) > d, which is impossible, since s(x) ≤ d for x ∈ X1. Thus, µ(Dn) = 0. We then construct: [ X2 = X1\ Dn, n∈N which we can see has full measure. Moreover, we see that for x ∈ X2, s(x) = n S(T (x)) for all n, which in turn implies that TX2 ⊂ X2. To conclude, from (i) (i) lemma 2.2.3, we have that λ ◦ T = λ for x ∈ X2 ⊂ X1.

2.3 Proof of the forth conclusion

Remark. It will suﬃce for our proof to show the limit found in the fourth item of theorem 2.1.1 for ergodic T-invariant measures. In the case we have a non-ergodic T-invariant measure, we can employ its ergodic decomposition.

2.3.1 A simpler setup Our ﬁrst objective will be to ﬁnd a special case for which we can prove the theorem

Lemma 2.3.1. Let T be a measure preserving transformation of a probability 1 n space (X, B, µ) and let h : X → [0, ∞) be a measurable map. Then lim infn→∞ n h(T x) = 0 a.e.

Proof. Consider the sets: [ Ak = {x ∈ X|h(x) ≤ k}, then X = Ak. k∈N

If we have that for some k, µ(Ak) > 0, then by Poincar´eRecurrence, for almost nk all x ∈ Ak there is a sequence (which depends on x)(nk)k∈N such that T (x) ∈ nk 1 nk Ak. Therefore, h(T (x)) ≤ k and therefore, limk→∞ k h(T (x)) = 0. Hence 1 n lim infn→∞ n h(T x) = 0. Lemma 2.3.2. Let T be a measure preserving transformation on a probability space (X, B, µ) and h : X → R a measurable function. Assume (h − h ◦ T )+ ∈ 1 1 n L (µ). Then n h(T x) → 0.

30 2.3 Proof of the forth conclusion

Proof. By considering a telescopic series, we have that:

n−1 1 1 X (h(T n(x)) − h(x)) = − (h − h ◦ T )(T ix). n n i=0

Then, as (h − h ◦ T )+ ∈ L1(µ), we can apply Kingman’s ergodic theorem [8] 1 n and we have that limn→∞ n h(T (x)) exists almost everywhere. Moreover, by the previous lemma: 1 1 1 lim h(T n(x)) = lim (h ◦ T n)+(x) − lim (h ◦ T n)−(x) n→∞ n n→∞ n n→∞ n 1 1 = lim inf (h ◦ T n)+(x) − lim sup (h ◦ T n)−(x) = 0. n→∞ n n→∞ n

Lemma 2.3.3. Let T be a ergodic measure-preserving transformation of (X, B, µ) and let x → Ax be a measurable map from X to M. Furthermore, let x → + log kAxk be integrable. Suppose that {Ux}x∈X is a measurable subbundle of X × d d R , that is, a collection of measurable sets such that {Ux}x∈X ⊂ X × R such that there exists a projection Π which allows that Π(Ux) = x. Moreover, we set the property that AxUx ⊂ UT x. As we know that by Kingsman’s ergodic theorem 1 n limn→∞ n log kA |Ux k exists (and could possible be −∞) and is constant almost everywhere. Suppose that the value of this limit L is such that L ≤ ρ for some ρ ∈ R. For > 0, we deﬁne:

n −n(ρ+) a(x) = sup kA |Ux ke . n≥0

Then, we have that: 1 log a (T nx) → 0 a.e. n Proof. By our choise of ρ, we have that:

1 ≤ a(x) < ∞.

31 2.3 Proof of the forth conclusion

With that information, we get the following inequality:

n −n(ρ+) n −n(ρ+) a(x) supn≥0 kA |Ux ke kA |Ux ke = m −m(ρ+) ≤ sup n −n(ρ+) a(T x) supm≥0 kA |UT x ke n≥0 kA |UT x ke

≤ max(1, kA|Ux k).

Applying logarithm to the previous inequality, we obtain:

+ log(a(x)) − log(a(T x)) ≤ max(log kAUx k, 0).

+ Thus, we have that x → (log(a(x))−log(a(T x)) is integrable and we can apply the previous lemma.

Theorem 2.3.1. Let T be an ergodic measure preserving transformation on the + probability space (X, B, µ) and let A : X → M(d) be such that x → log kAxk is d integrable. Let {Ux}x ∈ X, {Vx}x ∈ X measurable subbundles of X × R such that AxUx ⊂ UT x,AxVx ⊂ VT x. Let Y be a set of full measure such that TY ⊂ Y and let ρ, λ real numbers such that:

1 n lim sup log k(A )xuk ≤ ρ for u ∈ Ux\{0}, x ∈ Y, n→∞ n

1 n lim sup log k(A )xvk = λ for v ∈ Vx\Ux, x ∈ Y. n→∞ n L⊥ Deﬁne the measurable subbundle {Wx}x∈X such that Vx = Ux Wx using the inner product of the euclidean space. Let Bx : Wx → UT x and Cx : Wx → WT x be L⊥ maps induced by A|Vx : Vx → VT x, by considering that Ax(w) = Bx(w) Cx(w). Then, there exists Y 0 ∈ B of full measure such that TY 0 ⊂ Y 0 where:

1 n 1 n 0 lim sup log k(C )xwk = lim sup log k(A )x(u ⊕ w)k for w ∈ Wx\{0}, u ∈ Ux, x ∈ Y . n→∞ n n→∞ n

1 n 0 Furthermore, if limn→∞ n log k(C )xwk exists for some w ∈ Wx\{0}, x ∈ Y , then 1 n limn→∞ n log k(A )x(u ⊕ w)k exists for all u ∈ Ux. L⊥ Proof. Let u ∈ Ux, w ∈ Wx. By our deﬁnition of Ax on Ux Wx:

n−1 ! n n X n−i−1 i n (A )x(u ⊕ w) = ((A )xu + (A )T i+1xBT ix(C )x w) ⊕ (C )xw i=0

32 2.3 Proof of the forth conclusion

(n) Denote the large parenthesis by Dx . Then the above formula reads:

n n (n) n (A )x(u ⊕ w) = ((A )xu + Dx w) ⊕ (C )xw.

By lemma 2.2.1, we obtain: 1 lim sup log k(An) (u ⊕ w)k n x n→∞ (2.1) 1 n (n) 1 n = max(lim sup log k(A )xu + Dx wk, lim sup log k(C )xwk). n→∞ n n→∞ n Thus, by setting u = 0: 1 lim sup log k(An) wk n x n→∞ (2.2) 1 (n) 1 n = max(lim sup log kDx wk, lim sup log k(C )xwk). n→∞ n n→∞ n

∞ For > 0, we deﬁne the a as in lemma 2.3.3 on page 29. Now, consider a {p}p=1 a decreasing sequence which converges to 0. By the previous two lemmas, we 1 n 1 n n have that n log ap (T x) → 0 and n log kAT xk → 0 for x in a set of full measure Yp. Our objective is to show that for x ∈ ∩p∈NYp, the conclusion holds. For contradiction, assume that for some x ∈ ∩p∈NYp:

1 n 1 n lim sup log k(C )xwk = τ < λ = lim sup log k(A )xwk. n→∞ n n→∞ n

Consider p such that max(τ, ρ) + p < λ. Note that taking exponential in the previous inequality we see that there exists N(x, w, p) such that for n ≥ N we n n(τ+p) have that k(C )xwk < e . We also write Lx = Ax . We ﬁnd that: Ux

n−1 (n) X n−i−1 i kDx wk ≤ k(L )T i+1xk · kBT ixk · k(C )xwk i=0 n−i−1 i ≤ n max k(L )T i+1xk · kBT ixk · k(C )xwk 0≤i≤n−1

n−in−1 in = nk(L )T in+1xk · kBT in xk · k(C )xwk, where in(x, w) is the index where the maximum is attained. Now, fix x, w and consider the sequence {in}n≥1. We see that it is increasing. We will show that we can find a contradiction both when {in}n≥1 is bounded and when it is unbounded. When {in} is unbounded, then for every N,M ∈ N, we can find a value of the

33 2.3 Proof of the forth conclusion

index n ≥ M, in > N, so we have that:

1 (n) 1 1 n−in−1 1 1 in log kD wk = log n + log(k(L ) in+1 k) + log(kB in k) + log(k(C ) wk) n x n n T x n T x n x

1 1 1+in (n−in−1)(ρ+p) 1 ≤ log n + log(a (T x)e ) + log(kB in k) n n p n T x 1 + log(k(Cin ) wk) n x 1 1 n − i − 1 ≤ log n + log(a (T 1+in x)) + n (ρ + )) n n p n p

1 + 1 in(τ+p) + log (kB in k) + log(e ) n T x n 1 1 n − i − 1 ≤ log n + log(a (T 1+in x)) + n (ρ + )) n n p n p 1 + in + log (kA in k) + (τ + ) n T x n p 1 1 n − i − 1 ≤ log n + log(a (T 1+in x)) + n (max(ρ, τ) + )) n n p n p 1 + in + log (kA in k) + (max(ρ, τ) + ). n T x n p

Applying lim sup, we see that this inequality yields:

1 (n) lim sup log kDx wk ≤ max(ρ, τ) + p < λ. n→∞ n This contradicts the equality (2.2). On the other hand, if {in} ≤ M for some constant, then:

1 (n) 1 1 n−i−1 1 log kDx wk = log n + max log(k(L )T i+1xk) + max log(kBT ixk) n n n i≤M n i≤M 1 i + max log(k(C )xwk) n i≤M 1 1 1+i n − 1 1 ≤ log n + max log(ap (T x)) + (ρ + p) + max log(kBT ixk) n n i≤M n n i≤M 1 i + max log(k(C )xwk). n i≤M

34 2.3 Proof of the forth conclusion

Once again, applying lim sup, we ﬁnd that:

1 (n) lim sup log kDx wk ≤ ρ + p < λ. n→∞ n which is once again a contradiction. In summary, we have shown:

1 n 1 n lim sup log k(C )xwk = lim sup log k(A )xwk, n→∞ n n→∞ n for all x ∈ T Y 0 and for w ∈ W , w 6= 0. Furthermore, we have that: p∈N x

1 n (n) 1 n (n) lim sup log k(A )xu + Dx wk ≤ lim sup log(k(A )xuk + kDx wk) n→∞ n n→∞ n 1 n 1 (n) = max lim sup log k(A )xuk, lim sup log kDx wk n→∞ n n→∞ n (2.3) 1 ≤ max(ρ, lim sup log k(Cn)wk) n 1 = lim sup log k(Cn)wk, n and then, from (2.1) we ﬁnd that for u ∈ Ux:

1 n 1 n lim sup log k(A )x(u ⊕ w)k = lim sup log k(C )wk. (2.4) n→∞ n n

1 n 0 Moreover, if lim n log k(C )wk exists for some w ∈ Wx and x ∈ Y , we have that:

1 n 1 n lim sup log k(A )x(u ⊕ w)k = lim log k(C )wk. n→∞ n n And on the other hand, taking lim inf, by lemma 2.2.1:

1 n 1 n lim inf log k(A )x(u ⊕ w)k ≥ lim log k(C )wk, n→∞ n n

1 n which implies the limit exists for n log k(A )x(u ⊕ w)k. Corollary 2.3.1.1. Let T be ergodic. It suﬃces to prove Osceledec’s theorem for + the case when Ax ∈ GL(d) a.e. and x → log kAxk in the case that Ax ∈ M.

Proof. Let Ax 6∈ GL(d), then there exists some nonzero v ∈ V such that Axv = 0. Then, for this x, λ(s(x))x = −∞. If this occurs on a set H ⊂ Y 0 such that

35 2.3 Proof of the forth conclusion

µ(H) > 0, we have that: ∞ [ L = T −nH. n=0 Has full measure as T is ergodic. Now, if we take l ∈ L. Then, there is some m ∈ N such that l ∈ T −m(H), thus, consider h ∈ H such that T ml = h. Now, by theorem 2.2.4, we have that λs(x)(l) = λs(x)(h) = −∞. Then, λs(x)(x) = −∞ a.e. s(x) ⊥ Consider the subbundle {Wx}x∈X by Wx = (Vx ) and consider the linear map Cx : Wx → WT x deﬁned as the orthogonal projection of Axw onto WT x. Then + x → log kCxk is integrable and each Cx is invertible, as, by the previous theorem 1 n lim supn→∞ n log k(C )xwk= 6 −∞. Furthermore, by the same theorem, we see that the Lyapunov exponents for (s(x)−1) (1) x → Cx are λ , . . . , λ and we have a ﬁltration:

(s(x)−1) (1) Vx ∩ Wx,...,Vx ∩ Wx = Wx.

Now, suppose that we have Osceledec’s theorem for x → Cx. Therefore, we have 0 j (j+1) that for almost all x ∈ Y and for w ∈ (Vx ∩ Wx)\(Vx ∩ Wx):

1 n (j) lim log k(C )xwk = λ (x). n→∞ n Then, by the previous theorem, the limit: .

1 n (j) (j) (j+1) lim log k(A )xvk = λ (x) exists a.e. for v ∈ Vx \Vx n→∞ n

Thus, the Osceledec’s theorem holds for x → Ax. Then, we only have to take into account invertible matrices. However, as it is explained in [4], from now on we also have to take the extra hypothesis that + −1 x → log kAx k is also integrable, as there seems to be no other way to bridge −1 some gaps that come from the necessity to integrate x → log kAx k. This, of course, weakens the theorem, however, as our purpose here is to use this theorem as a way to investigate the properties of the Billiard Map, which we have proved is a diﬀeomorphism, this setback will not aﬀect our overall work.

2.3.2 Skew Products For our last part, we will use skew products, which we will deﬁne shortly, to show that we can change the limsup we found in lemma 2.2.3 to a uniform limit. Once we have this statement, we will have completed our proof of the theorem. The following arguments are used in Bochi [5].

36 2.3 Proof of the forth conclusion

Deﬁnition 2.3.1. Let C be a compact metric space and (X, µ, T ) a measure preserving system where T is ergodic. We say that S : X × C → X × C is a skew product over T if:

• ΠX ◦ S = T where ΠX is the natural projection of X × C into X. • The map S(x, ·): C → C is continuous for almost all x.

Deﬁnition 2.3.2. Let φ : X × C → R and S : X × C → X × C a skew product (n) Pn−1 i over T. Deﬁne φ = i=0 φ ◦ S .

Moreover, we will denote F the measurable functions φ : X × C → R such 1 that φ(x, ·): C → R is continuous and supc∈C |φ(x, c)| ∈ L (µ).

Lemma 2.3.4. Let M be the set of probability measures on X × C such that ∗ (ΠX )∗ν = µ.Then we can ﬁnd a space of functionals S to some Banach space S, a topology τ of S∗ and a compact K such that M ⊂ K ⊂ S∗ and a sequence {νn}n≥0 ⊂ M converges if and only if for every φ ∈ F: Z Z φdνn → φdν. X×C X×C

Proof. First, we see that F is a vector space. Then consider the completion F¯ of R F endowed with the norm kφk = supc∈C |φ|dµ. Each ν ∈ M determines a linear functional on F by: Z φdν for φ ∈ F.

Moreover, we see that this is continuous as for > 0, if φ1, φ2 ∈ F are such that R R kφ1 − φ2k < then (φ1 − φ2)dν ≤ supc∈C |φ1 − φ2| dµ < . This implies that M ⊂ F∗. By Hahn-Banach, we can extend these functionals to the completion, which shows that M ⊂ F¯ ∗.

Now let V be the unit ball in F¯ and let φ ∈ V . Then, for any ν ∈ M: Z Z

φdν ≤ sup |φ| dµ = 1. c∈C

Thus, M ⊂ V endowed with the weak-∗ topology, which is compact acording to the Banach-Alaoglu theorem [9]. Then, we endow M with the subspace topology. Finally, take the topology on the M given by the functionals of F. This topology is weaker than the standard weak∗ topology, which implies that the compact ball is still compact and the convergence of sequences of measures {νn}n≥0 depends

37 2.3 Proof of the forth conclusion only on the convergence of : Z Z φdνn → φdν, for all φ ∈ F.

Corollary 2.3.1.2. It is possible to endow M with a topology such that it is compact and a sequence {νn}n≥0 ⊂ M converges if and only if for every φ ∈ F: Z Z φdνn → φdν. X×C X×C

Proof. We have that M ⊂ (Measures(X) × Measures(C)) ∩ F¯ ∗. Then, as both of these spaces are Banach spaces, their intersection is a Banach Space and thus it is closed in both spaces. Moreover, as ΠX∗ : Measures(X) × Measures(C) → Measures(X) is continuous, we conclude that M is closed as Π−1{µ} = . Then, X∗ M this set is also closed in the space (Measures(X) × Measures(C)) ∩ F¯ ∗, which in turn implies that it is closed in F¯ ∗. Additionally, we see that for λ ∈ [0, 1], ν1, ν2 ∈ M, λν1 + (1 − λ)ν2 is a probability measure and ΠX∗ (λν1 + (1 − λ)ν2) = λµ + (1 − λ)µ = µ. This implies that M is convex in the topology of the dual space. Then, as convex closed sets in the dual topology are closed in the weak∗ topology, M is closed in the weaker topology. From the proof of the last lemma, we conclude it is compact in the topology given by the functionals of F. The following lemma is a translation to our case on arguments about selectors which can be found in the book Aliprantis and Border [10].

(m) Lemma 2.3.5. Let φ ∈ F. We denote Im(x) = infc∈C φ (x, c) as in deﬁnition 2.3.2. Then, we have that Im(x) is measurable and there exists um : X → C a (m) measurable function such that φ (x, um(x)) = Im(x). Proof. As C is a compact metric space, it is separable. Let L ⊂ C be a countable dense subset. Then: (m) Im(x) = inf φ (x, c) c∈L Which makes it measurable. Now, take d a compatible metric on C such that diam(C) < 1. Our objective is to construct a sequence of measurable functions

{fi}i∈N from X to C such that: 1 1 d(f (x), f (x)) < and d(f (x),D(x)) < , (2.5) n n+1 2n n 2n

38 2.3 Proof of the forth conclusion

(n) where D(x) = {c ∈ C|In(x) = φ (x, c)}. Note that D(x) is a closed set (and, therefore, compact) as it is the preimage of the minimal value φ(n)(x, ·) obtains. Then, as the sequence of functions converges to some f as it is a Cauchy sequence and D(x) is a closed set, f(x) ∈ D(x). Additionally, as f is the limit of measurable functions, it is measurable. The construction of the functions will be an inductive process. First, let L = {l0, l1, l2,... } ⊂ C be a dense and countable subset. Then, let be f0(c) = l0 be a constant function and therefore measurable. Now, let us describe the inductive step. Let fn be measurable and such that it satisﬁes relations (2.5). Then there 1 is some c ∈ D(x) such that d(c, fn(x)) < 2n . Now, as L is dense, there exists 1 1 some lk ∈ L such that d(c, lk) + d(c, fn(x)) < 2n which implies d(D(x), lk) < 2n+1 . −1 1 This construction shows us that x ∈ fn (B(lk, 2n )). Moreover, it shows that 1 −1 1 x ∈ {x ∈ X|d(lk,D(x)) < 2n }. As fn is measurable, fn (B(lk, 2n )) is measurable. 1 For {x ∈ X|d(lk,D(x)) < 2n }, consider the function:

Gk(x) = min d(lk, c). c∈D(x)

As D(x) is compact, this function is measurable. This implies that:

1 1 G−1 0, = x ∈ X|d(l ,D(x)) < is measurable . k 2n k 2n

−1 1 1 Name Ak = fn (B(lk, 2n ))∩ x ∈ X|d(lk,D(x)) < 2n . Let kn(x) be the smallest k such that x ∈ Ak. Then we set fn+1(x) = lkn(s). We see that this construction ensures that (2.5) holds true. To ﬁnish this lemma, we have to show that in fact fn+1 is measurable. Take E a measurable subset of C. As the image of each fn+1 is a subset of L:

−1 [ −1 fn+1(E) = fn+1({lk}),

lk∈E∩L where: k−1 −1 [ fn+1({lk}) = {x ∈ X|kn(x) = k} = Ak\ Ai, i=0 which is a measurable set.

Theorem 2.3.2. Let φ ∈ F and k ∈ R such that for µ almost all x ∈ X: 1 lim sup φ(n)(x, c) ≥ k for all c ∈ C. n→∞ n Then, we can replace lim sup with lim inf uniform over C. That is, for µ-almost

39 2.3 Proof of the forth conclusion

all x ∈ X, for all > 0, there exists n0 ∈ N such that: 1 φ(n)(x, c) > k − for all c ∈ C and n > n . n 0

Proof. Let In(x) be as in lemma 2.3.5. As φ ∈ F, I1 is integrable. Additionally, φ(m+n) − φ(m) = φ ◦ Sm + ··· + φ ◦ Sm+n. Then, applying inf, we ﬁnd that

m m+n Im+n − Im(x) ≥ inf φ ◦ S (x, c) + ··· + φ ◦ S (x, c) c∈C m n m m ≥ inf φ(T x, c) + ··· + φ ◦ S (T x, c) = In ◦ T . c∈C

Then, we can apply Kingman’s Subadditive Ergodic Theorem to −In, so we find −1 that n In converges almost everywhere to some constant b. Now, define un as in the last lemma, from which we know that we can take it as a measurable function and define the measures: Z 0 νn = δ(x,un(x))dµ,

where δ(x,un(x)) is the Kronecker delta,

n−1 1 X ν = = Sjν0. n n ∗ n j=0

We see that this measures are of probability and their projection to X is µ. Then, by corollary 2.3.1.2, we know that under the topology given by the functionals of F, there is a subsequence {νmi } ⊂ {νn} that converges to some ν which is a probability measure whose projection to X is µ. Furthermore, for any f ∈ F and νm ∈ {νm }: i Z

fd(S∗ν − ν)

Z Z Z

≤ fd(S∗(ν − νm)) + fd(S∗νm − νm) + fd(νm − ν)

Z Z Z Z 1 m 0 1 0 ≤ fd(S∗(ν − νm)) + fdS ν + fdν + fd(νm − ν) m ∗ m m m Z Z Z Z 1 m 0 1 0 ≤ fd(S∗(ν − νm)) + fdS ν + fdν + fd(νm − ν) m ∗ m m m Z Z 2kfkF ≤ fd(S∗(ν − νm)) + + fd(νm − ν) . m

40 2.3 Proof of the forth conclusion

Taking m → ∞, we conclude that the final member of the previous inequalities tend to 0, from which we find that ν is S−invariant. Additionally, we find that: Z Z 1 Z φdν = lim φdνn = lim In(x)dµ = b. n→∞ n→∞ n

Thus, by Birkhoﬀ’s ergodic theorem, we obtain that for ν almost all (x, c): 1 lim φn(x, c) = φ∗(x, c). n→∞ n

Furthermore: Z Z φ∗dν = φdν = b, which implies that for a set of ν-full measure: Z 1 Z 1 lim φn(x, u)dν = lim sup φn(x, u)dν ≥ k, n→∞ n n→∞ n from which we obtain that: b ≥ k. −1 Now, since b = limn→∞ n infc∈C φ(n)(x, c), we obtain our desired conclusion.

Corollary 2.3.2.1. Let φ ∈ F and k ∈ R be such that for µ almost all x ∈ X: 1 lim sup φ(n)(x, c) = k for all c ∈ C. n→∞ n Then, we can replace lim sup with lim uniform over C.

Proof. From last theorem, we have that: 1 1 lim inf φ(n)(x, c) ≥ k = lim sup φ(n)(x, c), n→∞ n n→∞ n which implies the limit exists and: 1 lim φ(n)(x, c) = k. n→∞ n

41 2.3 Proof of the forth conclusion

2.3.3 Conclusion We now can ﬁnally prove the existence of the limits instead of lim sup by using the tools we proved in the last section. For the remainder of this chapter, consider a ﬂag: s(x)+1 s(x) 1 d {0} ≡ Vx ⊂ Vx · · · ⊂ Vx = R . From what we have proved so far, we know that under the assuptions of Osceledec’s i i+1 1 n (i) theorem v ∈ Vx \Vx if and only if lim supn→∞ n log(k(A )xvk) = λ (x). Our work on the theorem will conclude with the following theorem:

Theorem 2.3.3. Under the hypothesis of Osceledec’s theorem, for almost all x i i+1 and for v ∈ Vx \Vx , the limit

1 n (i) lim log(k(A )xvk) = λ (x), n→∞ n exists.

k k+1 Proof. Consider x ∈ X and 1 ≤ k ≤ s(x). Let m = dim(Vx /Vx ). Thus, k k+1 m modulo a composition with some isomorphism, we can identify Vx /Vx = R . Then, consider the projective space C = P Rm. Now, consider the skew product S : X × C → X × C over T deﬁned as S(x, cˆ) = (T (x),Axcˆ) wherec ˆ = c is the representative for c ∈ Rm in the projective space. kAxck Deﬁne φ(x, cˆ) = log kck . As matrix multiplication and taking logarithms are + + −1 continuous, the map x → Ax is measurable and both log kAxk and log k(Ax) k are integrable, we see that φ ∈ F. Moreover:

n−1 j+1 1 1 X k(A )xck φ(n)(x, cˆ) = log n n k(Aj) ck j=0 x 1 k(An) ck = log x , n kck which implies: 1 lim sup φ(n)(x, cˆ) = λ(k)(x) for almost all x. n Then, from corollary 2.3.2.1, we get that we can replace limsup with limit.

42 Chapter 3

Lyapunov Exponents for the Billiard Map

In this chapter, we will show how Oseledec’s Theorem from Chapter 2 can be applied to billiard systems. In particular, ﬁrst we will discuss the meaning of the Lyapunov Exponents of the system and their relation with the motion of the particle. Afterwards, we will proceed to understand the Lyapunov exponents of the Billiard map and show a simple example. We will base our ideas on [2], [1], [11]

3.1 Understanding Lyapunov Exponents of Dif- feomorphisms

Consider a Riemanninan Manifold M, an open dense subset N ⊂ M and a Cr, with r > 2, map F : N → M such that F is a diﬀeomorphism to F(N). Fur- T∞ n thermore, consider the set R = −∞ F (N) where all iterations are deﬁned. . Assume a measure µ on M such that µ(R) = 1 and is preserved by F.

Now, we have that the map given by the derivative of F at point x ∈ M is continuous and therefore measurable, then to apply Oseledec’s theorem we only + + −1 need to ask that both log kDxFk and log kDxF k are µ-integrable. Thus, from this result we get that for µ almost all x ∈ R:

(1) (s(x)) TxM = Vx ⊃ · · · ⊃ Vx ⊃ {0}.

k (k) k+1 Thus, by deﬁning Wx = Vx /Vx we ﬁnd a decomposition:

(1) (s(x)) TxM ' Wx ⊕ · · · ⊕ Wx ,

43 3.1 Understanding Lyapunov Exponents of Diﬀeomorphisms

(k) where, for all nonzero w ∈ Wx :

1 n (k) lim log kDxF wk = λx . (3.1) n→∞ n

(k) (k) The values λx are called the Lyapunov exponents of the system and dim Wx is the multiplicity of each exponent. Let us understand what the limit (3.1) implies. Consider > 0. Then, there is some N ∈ N such that for n ≥ N:

1 n (k) log kDxF wk − λ < , n x which implies: (k) (k) n(−+λx ) n n(+λx ) e < kDxF wk < e . We have three general cases:

(k) n • Case λx < 0: By taking n → ∞ we get that kDxF wk → 0, which implies that under iterations of F the tangent vectors contract exponentially towards the future.

(k) n • Case λx > 0: Again taking n → ∞ we obtain that kDxF wk → ∞ at an exponential rate, that is, the tangent vectors grow exponentially. Another way to look at this case, as F is a diﬀeomorphism in R, is to take n → −∞. Therefore, in this case we get an exponential contraction to the past.

(k) • Case λx = 0: In this case, we do not get either a contraction or an expan- sion at an exponential rate. With this motivation, we deﬁne both the stable and unstable subspaces at the tangent space of x ∈ R as follows:

s M (k) u M (k) Ex = Wx and Ex = Wx , (k) (k) λx <0 λx >0 where the superscripts s and u denote stable and unstable subspaces.

Deﬁnition 3.1.1. We say that a point x ∈ M is hyperbolic if all of its Lyapunov exponents are nonzero, that is, if:

s u TxM = Ex ⊕ Ex .

Moreover, this motivates a deﬁnition for the complete system:

44 3.1 Understanding Lyapunov Exponents of Diﬀeomorphisms

Deﬁnition 3.1.2. We say that M is hyperbolic if µ almost every point is hyperbolic. Lastly we will proceed to study the angle between the stable and unstable subsets in each point where these sets are deﬁned. We will use the following fact found in Bochi [5]: Let L : Rd → Rd an invertible linear map, then for any v, w ∈ Rd nonzero, we have the following inequality:

1 sin θ(Lu, Lw) ≤ ≤ kLkkL−1k. kLkkL−1k sin θ(u, w)

Moreover, we will use the following lemma, which is a consequence of Birkhoﬀ’s theorem:

Lemma 3.1.1. Let (R, F, µ) a measure preserving system. Then, for f : R → R a measurable function such that f ◦ F − f is integrable we obtain that: 1 lim f(Fnx) = 0 for almost all x ∈ R n→∞ n Proof. We see that:

n−1 1 f 1 X f ◦ Fn = + (f ◦ F − f). n n n j=0

Thus, by Birkhoﬀ’s ergodic theorem, we get that: 1 f ◦ Fnx = ϕ∗x. n Where ϕ∗ is the ergodic average. Now, consider A = {x ∈ R|ϕ∗(x) 6= 0}. For x ∈ A we get that |f ◦ T nx| → ∞ when n → ∞. However, we have that {x ∈ X||f(x)| < ∞} is a set of full measure and thus, B = {x ∈ R||f ◦ Fnx| < ∞ for n ∈ N} is also a set of full measure by Poincar´eRecurrence theorem, which implies that µ(A) = 0. We then conclude that ϕ = 0 almost everywhere.

(1) (2) Theorem 3.1.1. Under the hypotheses of this section, consider Ex ,Ex ⊂ TxR (1) (2) two diﬀerent subspaces of the tangent space at x. Let γ(Ex ,Ex ) be the angle formed between the subspaces. Then for µ-almost all x ∈ X:

1 (1) (2) lim log(sin(γ(E n ,E n ))) = 0, n→∞ n F x F x

(1) (2) (1) (2) n Where EFnx,EFnx are the images of Ex ,Ex under the derivative DxF .

45 3.2 The Billiard Map and its Lyapunov Exponents

Proof. Consider g : X → R deﬁned as:

(1) (2) g(x) = γ(Ex ,Ex ) = inf arccos hw, vi. kwk=kvk=1 (1) (2) w∈Ex ,v∈Ex Then, as arccos is continuous and the sphere of both spaces is compact, we get that g is measurable. Then, f : X → R deﬁned as f(x) = log(sin(g(x))) is measurable. From the fact we mentioned earlier, we get that:

−1 −1 − log kDxFk − log kDxF k ≤ f ◦ F − f ≤ log kDxFk + log kDxF k.

Which implies that |f ◦ F − f| is integrable. Then, by lemma 3.1.1 we get that 1 n n f ◦ F → 0 as n → ∞. As with the zero Lyapunov exponents, the last result implies that the rate at which the angle tends to zero is not exponential, which will be useful in the next section.

3.2 The Billiard Map and its Lyapunov Expo- nents

Now, we want to apply the Oseledec’s theorem to the billiard map. The following lemmas show that the billiard Map satisﬁes the hypotheses of the theorem:

Lemma 3.2.1. The billiard map T preserves the measure µ = sin θdrdθ on M.

Proof. From Theorem 1.2.5 we ﬁnd: sin θ det(D(r,θ)T ) = . sin θ1 Then, changing variables, for A ∈ B: Z Z µ(T (A)) = sin θdr1dθ1 = sin θdrdθ = µ(A). T (A) A

Remark. Note that the previous measure is not necessarily a probability measure. However, if the length of the curve that bounds the billiard is ﬁnite, we can normalize by the factor: Z π Z sinθdθdr = 2|Γ|. 0 Γ

46 3.2 The Billiard Map and its Lyapunov Exponents

1 Then, from now on, our measure µ = 2|Γ| sin θdθdr. Now, we need to check that the positive part of the logarithm of the norm of the derivative of T and its inverse are integrable. We do this under mild restrictions in the following lemma: Lemma 3.2.2. Assume that the curvature K of the boundary Γ of the billiard and the length of the curve that bounds the system are ﬁnite. Then both log kDr,θT k −1 and log kDr,θT k are µ-integrable. Proof. If we have that the curvature of the boundary Γ of the billiard is bounded at every point, we can bound the derivative of the billiard map with some constant C > 1 such that: C kDr,θT k ≤ . sin θ1 Then, Z Z Z + C 1 log kDr,θT kdµ ≤ log dµ ≤ log(C) + | log(sin θ1) sin θ|dθdr M M sin θ1 2|Γ| M 1 Z = log(C) + | log(sin θ1) sin θ1|dθ1dr1 2|Γ| M π Z 2 = log(C) + | log(sin θ1) sin θ1|dθ1. 0

The only singularity of log(sin θ1) sin θ1 may occur at θ1 = 0, but notice that:

log(sin θ ) cos θ1 lim 1 = lim sin θ1 = 0, 1 cos θ1 θ1→0 θ1→0 2 sin θ1 sin θ1 the function log(sin θ1) sin θ1 is bounded and therefore it is integrable. In a similar + −1 way, it is possible to prove the same for log kDr,θT k, completing the proof. Combining the last two results, we get the following theorem. Theorem 3.2.1. If the curvature K is bounded and the length of the curve is bounded, then we can apply Osceledec’s Multiplicative Theorem to the billiard map. Then, for µ-almost all points in M, the Lyapunov exponents are well-defined. Moreover, as dim M = 2, we know that there are at most two Lyapunov exponents. We can find the following relation between those exponents: Theorem 3.2.2. At µ-almost all points, if the Lyapunov exponents are different (1) (2) at TxM, then λx + λx = 0. On the other hand, if we only have a unique Lyapunov exponent on TxM, λx = 0.

47 3.2 The Billiard Map and its Lyapunov Exponents

(1) (2) Proof. Let x = (r, θ) ∈ M and let λx ≥ λx (possibly equal). Consider P a ~ (1) ~ parallelogram in TxM of sides ~a and b such that ~a is parallel to Ex and b is parallel (2) (1) (2) (1) (2) Ex and Ex ∩ Ex = {0}. Then if γ is the angle between Ex and Ex , it is the angle between the sides of P and therefore the area of the parallelogram is ab sin γ ~ n where a, b are the lengths of the vectors ~a, b respectively. Additionally, DxT P ~ n ~ n~ is a parallelogram in TT nxM with sides ~an, bn where ~an = DxT ~a bn = DxT b. (1) ~ By Oseledec’s Multiplicative Theorem, ~an is parallel to ET nx and bn is parallel to (2) ET nx. Call γn the angle between the two sides. Then: sin (γ )a b kD T nk = n n n . x sin (γ)ab

However, from theorem 1.2.5 we have that:

n sin θ kDxT k = . sin θn Applying logarithm and dividing over n:

1 sin γ a b 1 log n + log n + log n = (log(sin θ) + log(sin θ )), n sin γ a b n n which is well deﬁned as θ, θn ∈ (0, π). Now, taking the limit when n → ∞, from theorem 3.1.1 and lemma 3.1.1: 1 1 log(sin γ ) → 0 and log(sin θ ) → 0. n n n n Moreover: 1 a 1 b log n → λ(1) and log n → λ(2). n a x n b x Combining the last two results, we obtain that: 1 sin γn an bn (1) (2) lim log + log + log = λx + λx = 0. n→∞ n sin γ a b

Remark. Note that for µ-almost all x ∈ M, either the point is hyperbolic or both Lyapunov exponents are zero.

48 3.2 The Billiard Map and its Lyapunov Exponents

3.2.1 A simple example Let us remember the billiard in a circle given in section 1.3. We know that the domain of the map T are the points (φ, θ) ∈ ([0, 2π]/ ∼)×[0, π]. Take (φ, θ) ∈ M and consider its orbit under T . Then, from our analysis in section 1.3, we know that T (φ, θ) = (φ + 2θ, θ). Interpreting (φ, θ) as a column vector, we write:

φ 1 2 φ T = . θ 0 1 θ

Which in turn yields that:

1 2n φ T n(φ, θ) = . 0 1 θ

Thus, as it is a linear, the derivative at (φ, θ) is:

n n D(φ,θ)T = T .

Now, consider v = (v1, v2) ∈ T(φ,θ)M, let us ﬁnd the Lyapunov Exponent of v:

1 n 1 2 2 lim log kD(φ,θ)T vk = lim log((v1 + 2nv2) + v2) n→∞ n n→∞ 2n 1 2 2 = lim log((v1 + 2xv2) + v2) x→∞ 2n 1 4v2(v1 + 2xv2) (Using L’Hopital’s rule:) = lim 2 2 = 0. x→∞ 2 (v1 + 2xv2) + v2

(1) Then, one of the Lyapunov Exponents for T(φ,θ)M is λ(φ,θ) = 0. By the last the- (2) orem, we know that λ(φ,θ) is also zero. This example presents us an important idea: It is not necessary for a dense orbit on the system to be hyperbolic. As we discussed previously, if we take θ/π 6∈ Q, then the projection of the map onto the circle will be dense and the Lyapunov exponents are always zero. This implies that it is not necessary to have exponential growth to move all along the system.

49 3.3 A Projective Criterion for nonzero Lyapunov Exponents

3.3 A Projective Criterion for nonzero Lyapunov Exponents

In the last section we saw that it suﬃces to prove that a Lyapunov Exponent is nonzero in the Billiard map to show that it is hyperbolic. Then, if we can show that the greatest Lyapunov exponent is greater than zero, we will have proved that the system is hyperbolic. In that way, our objective in this section will be to obtain a lower bound for this exponent. This section is based on the proofs and examples of [3] and [12].

First, we need some deﬁnitions which we will ground into our discussion over the billiard map and which are necessary for the result we will give. As usual, consider M, and T : M → M the domain for the billiard map and the billiard map respectively.

Deﬁnition 3.3.1. The set C(r, θ) = {(x, y) ∈ T(r,θ)M|xy ≥ 0} is called a sector S ˙ of (r, θ) ∈ M. Moreover, we denote C = (r,θ)∈MC(r, θ) ⊂ T M and call it the bundle of sectors.

Deﬁnition 3.3.2. We say that C is preserved by T : M → M if D(r,θ)T (C(r, θ)) ⊂ C(T (r, θ)) almost everywhere. As the title of this section suggests, we will make use of some ideas of projective spaces for our criterion. Namely, we can use a projective parameter −∞ < t ≤ ∞ to determine the lines in each T(r,θ)M, as it is a two dimensional space. Now, as each C(r, θ) is closed and connected, they represent a closed interval in the projective space, which we will denote I(r, θ) = {t|L(r, θ) ≤ t ≤ R(r, θ)}. Now, if C is preserved by T , then the sector D(r,θ)T (C(r, θ)) is represented by a subinterval of the interval I(T (r, θ)) which we will denote I1(r, θ) = {t|L1(r, θ) ≤ t ≤ R1(r, θ)}. Consider the set N ⊂ M where L(T (r, θ)) < L1(r, θ) ≤ R1(r, θ) < R(T (r, θ)). Now, we deﬁne ξ(r, θ) as the cross ratio of the lines L(T (r, θ)),L1(r, θ),R1(r, θ),R(T (r, θ)), that is: R(T (r, θ)) − L (r, θ) R (r, θ) − L(T (r, θ)) ξ(r, θ) = 1 1 . R(T (r, θ)) − R1(r, θ) L1(r, θ) − L(T (r, θ)) Now, with this knowledge, we will enunciate the main result of this section, which we will prove later on: Theorem 3.3.1. If C is preserved by T , then: Z Z p + ξ(r, θ) + 1 λ dµ ≥ log p dµ, M N ξ(r, θ) − 1

50 3.3 A Projective Criterion for nonzero Lyapunov Exponents where λ+(r, θ) is the greatest Lyapunov exponent at (r, θ).

Consider e1(r, θ), e2(r, θ) a basis for T(r,θ)M such that e1, e2 depend measurably on (r, θ). Now, as we have a basis, we know that we can consider D(r,θ)T a matrix whose entries will be denoted: a(r, θ) b(r, θ) D T = . (r,θ) c(r, θ) d(r, θ)

As C is preserved by T , then, for (x, y) ∈ T(r,θ)M such that xy ≥ 0:

(ax + by)(cx + dy) ≥ 0.

Then, as we can vary x and y over all nonnegative real numbers, we can conclude that either all entries on the matrix are nonnegative or all of them are nonpositive. Furthermore, as the measure is preserved by T , then this matrix has determinant ±1. Now, we know that in the coordinates e1(T (r, θ)), e2(T (r, θ)), L(T (r, θ)) = 0 and R(T (r, θ)) = ∞ as we have the vectors (0, 1), (1, 0) ∈ T(r,θ)M. If the c d det(D(r,θ)T ) = 1, then L1(T (r, θ)) = (a, c) = a and R1(T (r, θ)) = (b, d) = b in projective coordinates. In the case the determinant is -1, we obtain that d a L1(T (r, θ)) = b and R1(T (r, θ)) = c . Thus, we have that the cross ratio is gven by: ( ad cb , for det(D(r,θ)T ) = 1 ξ(r, θ) = cb ad , for det(D(r,θ)T ) = −1. Then, to ﬁnd our main result it will suﬃce to study the matrix that comes from the derivative of the billiard map, thus, let us discuss some properties of 2 × 2 matrices which will be useful later on.

First, we set the following notation: Z = {(x, y) ∈ R2|xy ≥ 0}, F : Z → R such that F (x, y) = xy, P = {A ∈ GL(2, R)|AZ ⊂ Z} and SP = {A ∈ 1 F (Au) 2 P ||det(A)| = 1}. Moreover, deﬁne the map: ρ : P → R by ρ(A) = infu∈int(Z) F (u) . Now, we present some lemmas which will help us prove theorem 3.2.1:

Lemma 3.3.1. Let A, B ∈ P , then ρ(AB) ≥ ρ(A)ρ(B).

Proof. As matrices are continuous and B preserves Z, then B(Int Z) ⊂ Int Z. Thus: 1 1 F (ABu) 2 F (Bu) 2 ρ(AB) = inf ≥ ρ(A)ρ(B). u∈int(Z) F (Bu) F (u)

51 3.3 A Projective Criterion for nonzero Lyapunov Exponents

Lemma 3.3.2. For A ∈ P , kAk ≥ ρ(A).

Proof. Let u1 = (1, 1). Then:

1 kAu k F (Au ) 2 kAk ≥ 1 ≥ 1 ≥ ρ(A). ku1k F (u1)

a b √ √ Lemma 3.3.3. Consider A = ∈ P . Then ρ(A) = ad + bc. c d

Proof. Consider u = (x, y) ∈ Int(Z). Then, inserting u1 in the deﬁnition of the norm of A: F (A)u (ax + by)(cx + dy) = F (u) xy x y = ad + bc + ac + bd . y x Now, minimizing this function with respect to x and y, we solve:

d F (A)u 1 y = ac − bd = 0, dx F (u) y x2 d F (A)u x 1 = −ac + bd = 0. dy F (u) y2 x √ √ The critical point is x = bd, y = ac. Moreover, using the Hessian, we can see that this point yields an the minimum. Then:

√ 1 √ √ ρ(A) = (ad + bc + 2 acbd) 2 = ( ad + bc).

From the deﬁnition of the greatest Lyapunov exponent and from Lemmas 3.2.1 and 3.2.2, we get the following inequalities:

+ 1 n λ (r, θ) = lim log kD(r,θ)T k n→∞ n 1 ≥ lim sup log ρ(A(T n(r, θ)) ...A(r, θ)) n→∞ n 1 ≥ lim sup log ρ(A(T n(r, θ))) + ··· + log(ρ(A(r, θ))), n→∞ n

52 3.3 A Projective Criterion for nonzero Lyapunov Exponents

where A(r, θ) = D(r,θ)T . Take now the function h(r, θ) = log(ρ(A(r, θ))) and a 1 cutoﬀ function hm(r, θ) = min(h(r, θ), m) for m > 1. We can see that hm ∈ L (µ) and therefore we can apply Birkhoﬀ’s ergodic theorem, which yields the existance of an a.e. limit: ∗ 1 n hm(r, θ) = lim (hm ◦ T + ··· + h), n→∞ n ∗ 1 where hm ∈ L (µ). Then, from the previous inequality, we obtain that: Z Z Z + ∗ λ dµ ≥ hmdµ = hmdµ. M M M However, as this happens for all m > 1, we can take the limit and we obtain that: Z Z λ+dµ ≥ hdµ. M M Now, we have the necessary tools to prove theorem.

Proof of theorem 3.2.1: Proof. Let (r, θ) ∈ N. In these points, we know that the projective lines of D(r,θ)T (C(r, θ)) are strictly contained in the lines of C(T (r, θ)), which implies the matrix A = D√(r,θ)T does√ not have zero entries. Then, from Lemma 3.2.3, we have ad that ρ(A) = ad + bc. In the case ξ = cb : √ √ √ ρ(A) = ad + bc = bc(pξ + 1) r bc = (pξ + 1) , ad − bc which yields: √ ξ + 1 ρ(A) = √ . ξ − 1 bc Similarly, in the case ξ = ad , we get the same equality.

Now, for (r, θ) ∈ M\N, we have that at least one entry on A is 0, which from lemma 3.2.3 implies that ρ(A) = 1 and therefore log(ρ(A)) = 0. Thus, we have that: √ Z Z Z ξ + 1 λ+dµ ≥ hdµ = log √ dµ. M M N ξ − 1

53 3.3 A Projective Criterion for nonzero Lyapunov Exponents

3.3.1 A hyperbolic example Consider again the stadium example presented in section 1.3.2. From [13] and [3] we know two important results about these systems. The ﬁrst, is that the billiard map is ergodic. This tool will be of the utmost importance; as the Lyapunov exponents are T -invariant, λ+(r, θ) is constant µ-almost everywhere and it is equal to: Z λ+ = λ+dµ. M

On the other hand, Wojtkowski proved the following result: Consider Γ1, Γ2 the semicircles used to construct the Stadium. Then, the bundle of sectors C is preserved by T and furthermore, for the set Sc = {(r, θ) ∈ M|r ∈ Γ1 and ΠΓ(T (r, θ)) ∈ Γ2 or r ∈ Γ2 and ΠF (T (r, θ)) ∈ Γ1} that is, points with successive collisions in diﬀerent semicircles. Then, C is preserved by T and additionally we have that N = Sc. Now, consider the euclidean distance from r ∈ ΠΓ(Sc) to ΠΓ(T (r, θ)) to be L. From lemma 1.2.4 we obtain that for g(r) = g0 − L. Then: 1 1 2 − + = g0 − L g1 d1 1 2 1 = + g1 d1 g0 − L d1g0 − d1L g1 = . 2g0 − 2L + d1

Figure 3.1: Paths in Sc

Then, the image of the sector C(r, θ) = {g0|0 ≤ g0 ≤ d0} is a proper subset of C(T (r, θ)) = {g1|0 ≤ g1 ≤ d1} by the derivative of T . With this information,

54 3.3 A Projective Criterion for nonzero Lyapunov Exponents we can now calculate ξ:

−d1L d1d0−d1L d1 − ξ(r, θ) = −2L+d1 2d0−2L+d1 d1d0−d1L −d1L d1 − 2d0−2L+d1 −2L+d1 d (d − L) d (d − L) = 1 1 1 0 d1(d0 + d1 − L) −d1L (L − d )(L − d ) = 1 0 , L(L − d0 − d1) from which we can calculate that: √ ξ + 1 p(L − d )(L − d ) + pL(L − d − d ) √ = 1 √0 0 1 ξ − 1 d0d1 s s L(L − d − d ) L(L − d − d ) = 1 + 0 1 + 0 1 . d0d1 d0d1

Then, we ﬁnally obtain that: √ Z Z ξ + 1 λ+ = λ+dµ ≥ log √ dµ > 0, M Sc ξ − 1 which shows us that this system is hyperbolic.

While it may seem that we had to construct too many tools to prove that what appears to be a simple example is actually a hyperbolic system, we have found an immensely useful criterion to determine the existence of hyperbolic points. Moreover, this example shows us that it is not necessary to construct artiﬁcial domains to ﬁnd nonlinear growth in movement, which demonstrates the importance of having methods to study these dynamics.

55 Chapter 4

Appendix: Some Ergodic Theorems

In this section we will state the theorems from ergodic theory that we used in this thesis. The proofs of these theorems can be found in [6] and [8].

4.1 Ergodic Theorems

First, we deﬁne ergodicity as stated in page 23 of [6]: Deﬁnition 4.1.1. A measure preserving transformation T : X → X of a probability space (X, B, µ) is called ergodic if for any B ∈ B:

T −1(B) = B =⇒ µ(B) = 0 or µ(B) = 1.

The following theorems can be found on pages 21 and 44 of [6] respectively: Theorem 4.1.1. (PoincaréRecurrence) Let T be a measure preserving transformation on a probability space (X, B, µ) and let E ⊂ X be a measurable set. Then almost every point x ∈ E returns to E infinitely often. Theorem 4.1.2. (Birkhoff’s Ergodic Theorem) Let (X, B, µ, T ) a measure pre- 1 serving system. If f ∈ Lµ, then:

n−1 1 X lim f(T ix) = f ∗(x), n→∞ n i=0

1 ∗ 1 converges almost everywhere and in Lµ to a T -invariant function f ∈ Lµ and, Z Z fdµ = f ∗dµ.

56 4.1 Ergodic Theorems

If T is ergodic, then f ∗ is constant and equals: Z fdµ almost everywhere.

Our ﬁnal theorem can be found in [8]. It is a generalization of the theorem by Birkhoﬀ:

Theorem 4.1.3. (Kingmans Subadditive Ergodic Theorem) Let (X, B, µ, T ) a ¯ measure preserving system. Let fn : X → R a sequence of measurable functions + such that f1 is integrable and:

m fn+m ≤ fm + fn ◦ T for all n, m ≥ 1 .

fn ¯ + 1 Then limn→∞ n converges a.e. to a function f : X → R. Moreover, f ∈ Lµ and: Z 1 Z 1 Z fdµ = lim fndµ = inf fndµ. n→∞ n n∈N n

57 References

[1] N. Chernov and R. Markarian, Introduction to the ergodic theory of chaotic billiards. Publica¸cõesMatemáticasdo IMPA. [IMPA Mathematical Publi- cations], Instituto de MatemáticaPura e Aplicada (IMPA), Rio de Janeiro, second ed., 2003. 24o ColóquioBrasileiro de Matemática.[24th Brazilian Mathematics Colloquium].1,3, 21, 43 [2] S. Tabachnikov, Geometry and billiards, vol. 30 of Student Mathematical Library. American Mathematical Society, Providence, RI; Mathematics Ad- vanced Study Semesters, University Park, PA, 2005.3, 19, 43 [3] M. Wojtkowski, “Principles for the design of billiards with nonvanishing lyapunov exponents,” Communications in Mathematical Physics, vol. 105, pp. 391–414, Sept. 1986. 15, 50, 54 [4] P. Walters, “A dynamical proof of the multiplicative ergodic theorem,” Transactions of the American Mathematical Society, vol. 335, no. 1, pp. 245– 257, 1993. 24, 36 [5] Bochi, “The multiplicative ergodic theorem of osceledets,” p. 12, 5 2008. 24, 36, 45 [6] M. Einsiedler and T. Ward, Ergodic Theory. Springer London, 2011. 26, 30, 56 [7] C. Castaing and M. Valadier, Convex analysis and measurable multifunc- tions. Berlin New York: Springer-Verlag, 1977. 28 [8] J. B. A Avila, “On the subbadditive ergodic theorem,” Russian Mathematical Surveys, vol. 25, p. 5, 2009. 31, 56, 57 [9] W. Rudin, Functional Analysis. International series in pure and applied mathematics, McGraw-Hill, 1991. 37 [10] C. Aliprantis, Infinite dimensional analysis : a hitchhiker’s guide. Berlin New York: Springer, 2006. 38

58 REFERENCES

[11] Y. G Sinai, “Dynamical systems with elastic reﬂections,” Russian Mathe- matical Surveys, vol. 25, p. 137, 10 2007. 43

[12] V. J. Donnay, “Using integrability to produce chaos: Billiards with positive entropy,” Communications in Mathematical Physics, vol. 141, pp. 225–257, Oct. 1991. 50

[13] L. A. Bunimovich, “On the ergodic properties of nowhere dispersing billiards,” Communications in Mathematical Physics, vol. 65, pp. 295–312, Oct. 1979. 54