The Pennsylvania State University The Graduate School Department of Mathematics
REGULAR AND CHAOTIC DYNAMICS OF OUTER BILLIARDS
A Thesis in
Mathematics
by
Daniel I. Genin
c 2005 Daniel I. Genin
Submitted in Partial Fulfillment
of the Requirements
for the Degree of
Doctor of Philosophy
August 2005 The thesis of Daniel I. Genin was reviewed and approved∗ by the fol- lowing:
Sergei Tabahchnikov
Professor of Mathematics Thesis Adviser Chair of Committee
Mark Levi
Professor of Mathematics
Yakov Pesin Distinguished Professor of Mathematics
Howard Weiss Professor of Mathematics
Elena Katok Associate Professor of Business Administration
Nigel Higson Professor of Mathematics
Head of Department of Mathematics ∗Signatures are on file in the Graduate School. ABSTRACT
This thesis explores two quite different types of dynamics occurring in outer billiards. First, we construct the first example of a hyperbolic outer billiard. Thus settling the question of whether chaotic outer billiards exist. The much more regular (quasi-periodic) dynamics for a certain class of quadrilaterals is considered next. Some results on complexity growth and boundedness of orbits are obtained from sym- bolic dynamical description of the mapping. In conclusion, we present numerical studies of outer billiard systems extending the the above examples.
iii Contents
1 Introduction 1
1.1 History, known results and open questions ...... 1
1.2 Plan of exposition ...... 5
1.3 Definition and Properties ...... 7
1.4 Chaotic outer billiards ...... 12
1.5 Polygonal Outer Billiards ...... 15
1.6 Numerical results and conjectures ...... 18
2 Chaotic Outer Billiards 21
2.1 Introduction ...... 21
2.2 Area construction ...... 22
2.3 Table ...... 23
2.4 Cone field ...... 25
2.5 Hyperbolicity ...... 28
3 Polygonal Outer Billiards 43
3.1 Introduction ...... 43
iv 3.2 Polygonal outer billiards ...... 45 3.2.1 Definitions and notation ...... 45 3.2.2 Symbolic dynamics, vertex coding and complexity 51 3.3 Trapezoidal outer billiards ...... 53
3.3.1 First return map on x = 0 ...... 54
3.3.2 Symbolic dynamics of TX ...... 62 3.3.3 Dynamics for rational α ...... 70 3.3.4 Poincare section of T ...... 78 3.3.5 Boundedness of orbits ...... 87
3.3.6 Complexity ...... 91
4 Numerical Explorations 99
4.1 Introduction ...... 99 4.2 More chaotic outer billiards ...... 101 4.3 Unbounded orbits? ...... 106
References 113
v List of Figures
1.1 Outer billiard map ...... 1
1.2 Definition of outer billiard map ...... 7
1.3 Impact oscillator and outer billiards ...... 10
1.4 Sinai billiards ...... 12
1.5 Bunimovich billiards ...... 13
1.6 Defocusing ...... 13
1.7 Wojtkowski billiards ...... 14
2.1 The outer billiard table ...... 24
2.2 Cone construction ...... 27
2.3 Cone preservation ...... 31
2.4 Cone nesting ...... 32
2.5 Orbits reflecting in corners ...... 36
2.6 Order one orbits ...... 38
2.7 A point reflecting in opposite corners ...... 39
2.8 Order two orbit ...... 40
2.9 Order three orbit ...... 41
vi 2.10 Proof illustration ...... 42
3.1 L1 ...... 46 3.2 D for a regular pentagon ...... 48 3.3 Structure of T 2 ...... 49
3.4 Dual polygon P ∗ ...... 51 3.5 Trapezoid ...... 54 3.6 Diagram describing T + ...... 58 3.7 Black and white sets ...... 86
4.1 Regular pentagon a=0.1, 0.2 ...... 102
4.2 Regular pentagon a=0.3, 0.4 ...... 102 4.3 Irregular pentagon a=0.1, 0.3 ...... 103 4.4 Irregular pentagon a=0.5, 0.8 ...... 103 4.5 Regular pentagon ...... 104
4.6 Irregular pentagon ...... 105 4.7 Square ...... 105 4.8 Dynamics around (100,0) ...... 106 4.9 Orbit of a satellite domain ...... 107
4.10 Orbits of satellite domains (50,0) ...... 108 4.11 Dart quadrilateral ...... 109 4.12 2000 iterates of (1,1.001) for α = √2 ...... 110 4.13 Distance from the origin along x = 1 ...... 111
vii ACKNOWLEDGEMENTS
I am grateful to my parents and grandparents for their vigorous encour- agement to complete my doctoral work and to my friends for helping me stay sane. To my advisor for teaching me how to approach and solve mathematical problems and to all others who have taught me all the mathematics I know, and even more that I still do not. And especially to the professors of the PennState Dynamical Systems Group. Thank you.
viii Chapter 1
Introduction
1.1 History, known results and open
questions
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Figure 1.1: Outer billiard map
The outer billiard map was first introduced by B.H. Neumann [16] and later popularized by J. Moser in the article “Is the solar system
1 stable?” [15] and his book “Stable and Random Motions in Dynam- ical Systems” [14]. Since then outer billiards have been explored by many mathematicians from many different directions. One source of questions are ordinary billiards – interesting questions about ordinary billiards can be reformulated and are often still meaningful for outer billiards. On the other hand, there are some interesting questions about outer billiards which do not make sense for ordinary billiards.
For example, the original question of Moser, which to this day re- mains only partially answered, is about unbounded orbits of the outer billiard map. This question, of course, makes no sense for ordinary billiards because the phase space in the later case is compact. Moser indicated that KAM theory can be used to show that any sufficiently smooth strictly convex table has only bounded orbits (the details were worked out in the thesis of R. Douday (unpublished)), but the wider question of whether there are any tables (even not strictly convex) for which the outer billiard map has unbounded orbits remains unsolved. Kolodziej [11], and Shaidenko and Vivaldi [17] independently arrived at a result on boundedness of orbits for a certain class of polygonal outer billiard maps, which was also later reproved using a different method by Gutkin and Simanyi [8]. The exact result is discussed in Section 1.5. Tabachnikov showed that near infinity the orbits of any outer billiard are well approximated by a periodic Hamiltonian flow [21], which puts a bound on the rate of escape of unbounded orbits if such exist, and
2 suggested a non-polygonal, although still discontinuous, outer billiard for which an open set of points appears to escape to infinity.
Another fruitful area of research has been the study of dynamical structures for polygonal outer billiards. This is still a relatively new area so there are many results on interesting special cases but little in a way of general theory. It is not hard to see that polygonal outer billiards are piece-wise isometries and so this area has significant con- nections with the study of interval exchanges. The most interesting re- sults in my opinion are on renormalizability of dynamics. Tabachnikov obtained a complete symbolic description for dynamics of pentagonal outer billiard [20] and as a consequence proved that its dynamics is renormalizable about certain points and that the set of non-periodic orbits has a fractal structure with dimension log 6/ log(√5 + 2). Adler,
Kitchens and Tresser derived a similar result for a piece-wise rotation of the torus by π/4 [1]. Outer billiards about regular odd n-gons can be shown to factor over piece-wise toral rotations so the two systems are closely related. More recently Kouptsov, Lowenstein and Vivaldi
[12] used rigorous numerical computations to prove that renormaliz- ability appears for all piecewise rotations of the torus by θ if cos θ is a quadratic irrational. These results hint at a strong connection between dynamics and arithmetic properties of the polygon and seem to point in the direction of Boshernitzan’s result on renormalizability of interval exchanges over quadratic fields [2]. It is plausible that similar results
3 may be proved for piecewise toral rotations and by extension for outer billiards about regular polygons.
Complexity of piecewise isometries and of polygonal outer billiards in particular has also recently become a subject of intensive study. Al- though it has been known for some time that growth of complexity for this type of maps is sub-exponential, implying that metric and topo- logical entropies are zero, few exact results are known. For ordinary polygonal billiards results on sub-exponential growth were obtained by Katok [10], Galperin, Kruger and Troubetzkoy [6], and Gutkin and Haydn [7], also recently some more precise bounds for several cases were obtained by Cassaigne, Hubert and Troubetzkoy [5]. A first step in this direction for outer billiards was a recent paper by Gutkin and Tabachnikov [9] in which bounds on complexity growth are derived for a wide class of maps that includes outer billiards. In particular, they prove that for a general p-gon complexity grows at most as np+2 and it grows as n2 for lattice polygons. In view of the afore mentioned results connecting arithmetic properties of the polygon and dynamics of the outer billiard one would expect this to be manifested in the complexity function of the dynamical system. This question is explored here for a special case of the trapezoid.
Then there are some interesting questions which to this day re- ceived little or no attention. Among the most exciting is the Birkhoff conjecture for outer billiards, that ellipses are the only tables for which
4 the outer billiard map is completely integrable. Birkhoff conjecture for outer billiards is the counterpart of the corresponding conjecture for ordinary billiards, which is also still open. Another interesting ques- tion is : Do chaotic outer billiards exist? This question is answered positively in this thesis.
1.2 Plan of exposition
This thesis explores hyperbolic and parabolic behavior in outer billiard maps.
In Chapter 2 we explore chaotic behavior of outer billiard maps.
The main result of this chapter is Theorem 2.5.1, which shows that hyperbolic outer billiards exist. Up to now it was unknown whether hyperbolic behavior in outer billiards is possible. We construct a one parameter family of outer billiard maps with hyperbolic behavior in an invariant region near the table. The construction parallels the con- struction of Bunimovich stadium billiard in combining arcs with locally integrable behavior in a way that produces exponential instability and chaos. The proof of the result hinges on the construction of an invariant cone field for the outer billiard map.
In Chapter 3 we turn our attention to polygonal outer billiards and pursue an in-depth study of a special one parameter family – outer billiard maps about trapezoids. We show that the dynamics in this
5 case is essentially one dimensional and obtain a complete symbolic de- scription. Armed with this tool we are able to answer many interesting questions about this family. In particular, we are able to show in The- orem 3.3.29 that all orbits are bounded. This is quite remarkable since there is apparently no obstacle to escape of orbits, unlike the case of quasi-rational outer billiards. This point is discussed in more depth in Section 1.5 below. In Theorem 3.3.30 we obtain an upper bound on the complexity growth of the vertex coding. This result also shows that complexity growth of a trapezoidal outer billiard is sensitive to the arithmetic properties of the table which is in agreement with result of Kouptsov, Lowenstein and Vivaldi [12]. Another, remarkable feature of this example is largeness (in the measure theoretic sense) of the set of non-periodic orbits. Indeed, this is the first example of an outer billiard for which non-periodic orbits occupy a positive measure set.
In Chapter 4 we present results of some numerical explorations of hyperbolic as well as polygonal outer billiards and make several con- jectures about their dynamics. More specifically, we explore Moser’s question about unbounded orbits and conclude that there is good nu- merical evidence to suggest that there indeed are tables for which some orbits are unbounded. We also conjecture that hyperbolic behavior is common among outer billiard maps with polygonal invariant curves.
In the rest of the introduction we make precise definition of the outer billiard, introduce notation necessary for later discussions and
6 discuss some of the relevant results.
1.3 Definition and Properties of the
Outer Billiard Map
We begin with a definition of the outer billiard map
p
pt
T(p)
T(p) t
T 2(p)
Figure 1.2: Definition of outer billiard map
Definition 1.3.1. Let Γ be an oriented convex plane curve in R2 and Ω the domain enclosed by Γ. Suppose for the moment that Ω is strictly convex and let D = R2 Ω. The outer billiard map is the continuous \ transformation T : D D uniquely specified by →
i the oriented segment [p, T (p)] is tangent to Γ at some pt;
7 ii orientation of Γ agrees with orientation of the segment [p, T (p)]
at pt;
iii pt bisects [p, T (p)]
This definition can be extended to non-strictly convex tables by choosing D to be the set of all points in p R2 Ω such that ∈ \ T k(p) = [T k−1(p), T k(p)] Γ is an extremal point of Ω for all k Z t ∩ ∈
By analogy with ordinary billiards, which we will from now on call simply billiards, we will refer to Ω as the table of the outer billiard system and we will say that p reflects in p if [p, T (p)] Γ = p . t ∩ t We begin by describing some basic properties of T . We state them without proof and instead refer the reader to the literature. Needless to say, that most of the statements are simple enough that they are either obvious or can be easily worked out by a dedicated reader.
T is continuous iff Γ is strictly convex. •
T is Sl(2, R) equivariant. That is, if T is outer billiard about Γ • Γ and g Sl(2, R) then T (gp) = gT (p). This indicates that only ∈ gΓ Γ affine properties (properties invariant under Sl(2, R) action) of Γ matter to the dynamics of T .
T preserves Lebesgue measure. This property guarantees inter- • esting dynamics in the sense of recurrence on bounded invariant
domains of D.
8 T is a twist map. Thus Poincare’s theorem on periodic orbits of • twist maps and Aubry-Mather theory can be applied to T .
Periodic orbits of T are circumscribed polygons of critical area. •
There are two other remarkable properties of the outer billiards that are worth noting.
First, it turns out that if one defines outer billiards on the two dimensional sphere by replacing R2 in the above definition with S2 then they are isomorphic to ordinary billiards defined on the sphere. The isomorphism is given by projective duality on S2. Projective duality is an identification between oriented geodesics on S2 and points of S2. For instance, we can assign to every great circle the corresponding north pole. The space of oriented geodesics on S2 which is the quotient of
ST S2 by the geodesic flow is thus isomorphic to S2. Projective duality further allows to identify not only points in these two spaces but also curves. Given an oriented curve γ on S2 we call the dual curve γ∗ the curve traced out in the space of oriented geodesics by geodesics tangent to γ. We can now make precise the original statement about isomorphism between the ordinary and outer billiards. Given an outer billiard about a geodesically convex curve Γ on S2 and a piecewise geodesic curve γ obtained by joining consecutive points of an orbit, the piecewise geodesic dual curve γ∗ is a trajectory of an ordinary billiard inside the dual curve Γ∗. Thus in the case of S2 the two billiards are
9 dual to each other. For this reason outer billiards are also sometimes called dual billiards.
T(p)
p
O
T(p)
Γ
p
Wall Mass Figure 1.3: Impact oscillator and outer billiards
Another interesting connection is between outer billiards and a physical system called the impact oscillator. An impact oscillator is a one dimensional system composed of a mass attached to a spring and a wall moving along the same axis as the mass, with which it collides elastically. The correspondence between the outer billiards and the impact oscillators was discovered and described by P. Boyland in [3]. The identification is obtained as follows : we pick a point O inside Γ and an initial point p to the right of the rightmost vertical supporting
10 line to Γ; now rotate Γ and p clockwise about O until p arrives at the rightmost vertical supporting line to Γ, then reflect p along this line through the point of tangency and repeat. Clearly this is the same as the outer billiard map. To identify the system with the impact oscil- lator we project Γ and p vertically onto the x-axis. The “shadow” of p corresponds to the mass on the spring and the rightmost point of the shadow of Γ – to the moving wall. The defining property of outer billiard p p = p T (p) guarantees elasticity of the collision. Be- | − t| | t − | cause the motion of the wall is determined by the support function p(θ) of Γ it must satisfy p00(θ) + p(θ) = r(θ) where r(θ) is an L2 function orthogonal to sin and cos. Hence not all motions of the wall can be realized.
Interestingly this correspondence allows us to restate the question about unbounded orbits in physical terms. Namely, the question be- comes : Is it possible to transfer an infinite amount of energy to the mass through collisions with the wall? From college physics everyone knows that if a harmonic oscillator is driven at its natural frequency than resonance occurs and amplitude of oscillations grows to infinity. In this case, though, such forcing functions are ruled out by the restriction on the motion of the wall so the answer is far from obvious.
Similarly, the result of Chapter 2 on chaotic behavior acquires, through this correspondence physical significance. It can now be in- terpreted to mean that for certain initial conditions and certain forcing
11 functions the impact oscillator can exhibit chaotic behavior.
1.4 Chaotic outer billiards
In Chapter 2 we explore the previously untouched question about chaotic behavior in outer billiards. Hyperbolicity turns out to be a much more complicated matter for outer billiards than for ordinary billiards be- cause there is no obvious analogue to Sinai billiards (Figure 1.4). The example constructed in Chapter 2 is inspired by work on convex chaotic billiards of Bunimovich and Wojtkowski and so we review their work here.
It has been known for a long time that billiards with concave bound-
aries, also known as Sinai billiards, have hyperbolic behavior [19]. This
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Figure 1.4: Sinai billiards is naturally expected because concave boundaries typically cause inci- dent trajectories to diverge. Billiards in convex domains on the other
12 hand were expected to exhibit at least some elliptic behavior because convex boundaries tend to focus trajectories. It was therefore a great surprise when Bunimovich showed [4] that the stadium billiard can exhibit purely hyperbolic behavior.
Figure 1.5: Bunimovich billiards
Figure 1.6: Defocusing
His construction was based on a simple and beautiful idea that if the distance between focusing pieces of the boundary is sufficiently large then trajectories, initially focused by such pieces, will eventually pass through the focusing point and travel far enough to make nearby tra-
13 jectories diverge (Figure 1.6). Thus convex pieces of the boundary can be dispersing as well and such billiards became known as defocusing. Wojtkowski [23] generalized the original construction of Bunimovich by showing that any curve satisfying d2ρ/ds2 0, where ρ(s) is the ≤ radius of curvature as a function of arc length, and not just circles and linear segments, can be used to construct convex defocusing billiards. The latest but certainly not the last turn in this story is the result of
Figure 1.7: Wojtkowski billiards
Markarian [13] that shows that billiards satisfying the opposite of the Wojtkowski condition can also be defocusing.
As mentioned above for outer billiards the situation is complicated by the fact that there is no obvious analogue of dispersing billiards. The hyperbolic outer billiard we construct in Chapter 2 is similar to the Bunimvoich stadium in that it combines pieces of the boundary that alone have integrable behavior into a table that has hyperbolic behav- ior. In our construction the role of circles of the stadium is played by
14 pieces of hyperbolas and the role of straight segments – by corners. The proof proceeds along the same lines as the proof of Wojtkowski in [23]. We construct a strictly preserved cone bundle over an invariant set which by another result of Wojtkowski guarantees non-vanishing of
Lyapunov exponents [22]. Non-vanishing of the Lyapunov exponents immediately gives such properties as a countable number of ergodic components, positive entropy, Bernoulli property etc. Ergodicity un- fortunately does not automatically follow and requires a separate ar- gument.
1.5 Polygonal outer billiards : bounded
orbits and complexity
To put the results of Chapter 3 in context we describe some of the relevant research. The two main questions in the study of polygonal outer billiards so far have been, the original question of Moser, on the boundedness of orbits and the complexity growth.
As far as the first question is concerned there is only one known result in this direction, which has been proved independently by sev- eral different authors [11], [17] [8]. The following definition is due to Kolodziej.
Definition 1.5.1. Let P be a polygon without parallel sides, with
15 vertices v0, v1,..., vn. For each side [vi, vi+1] we construct a strip Si by taking one boundary to be the line containing the side and the other – its reflection in the vertex furthest from it. Now consider the set of all parallelograms Qi obtained by intersecting pairs of strips. Since there are no parallel sides intersection of any two different strips gives a parallelogram. Let Ai be the area of parallelogram Qi. We will say that P is quasi-rational if Ai are rationally commensurable, i.e. if A /A Q for all i and j. i j ∈ A polygon P is said to be lattice or rational if coordinates of its vertices are rationally commensurable.
Clearly, rational polygons without parallel sides are quasi-rational. Regular polygons are also easily seen to be quasi-rational.
We can now state the result
Theorem 1.5.2. If P is quasi-rational then all orbits of outer billiard about P are bounded.
This has the following simple but beautiful corollary
Corollary 1.5.3. If P is a rational polygon without parallel sides all orbits of outer billiard about P are periodic.
The proof of the above result is achieved by showing that certain periodic domains form connected chains akin to invariant curves in smooth dynamics.
16 This result does not apply to trapezoidal outer billiards considered in Chapter 3. They do not satisfy the definition of quasi-rationality even when they are on a lattice because they have parallel sides. In Theorem 3.3.29 we prove that for trapezoidal outer billiards all orbits are bounded. This result is quite surprising, especially because Lemma 3.3.28 shows that there are no chains of periodic domains to contain the orbits as in the case of quasi-rational billiards. Another consequence of this result is that for lattice trapezoids all orbits are periodic. This although expected was previously unknown.
Even less is known about the growth of complexity for polygonal outer billiards. The only results in this direction are the polynomial bounds recently proved by Gutkin and Tabachnikov [9].
Theorem 1.5.4. Let P be a convex Euclidean p-gon, and let r p 1 ≤ − be the rank of the abelian group generated by translations in the sides of P . Then the complexity of the outer billiard about P is bounded by nr+2.
In the case when P is rational more can be said
Theorem 1.5.5. Let P be a rational polygon, and let p( ) be the com- · 2 2 plexity of the outer billiard about P . Then C1n < p(n) < C2n where
C1 and C2 depend only on P .
In Chapter 3 we obtain an exact symbolic description of dynamics in the case of trapezoidal outer billiards similar in spirit to the work
17 of Tabachnikov [20] and Adler, Kitchins and Tresser [1]. Using exact symbolic description we are able to obtain an upper bound on complex- ity growth p(n) of the vertex coding. In the case when the coefficients in the continued fraction expansion of the ratio of the bases of the trapezoid are bounded p(n) Cn1+τ ≤ where τ is the Diophantine exponent of the ratio and C depends only on P . This is the content of Theorem 3.3.30. Note that this improves the bound of Tabachnikov and Gutkin above for trapezoid since τ < 3. Computing a lower bound turns out to be a more complicated propo- sition. Nevertheless, there is hope that it too can be obtained from the symbolic description.
1.6 Numerical results and conjectures
In Chapter 4 we present results of some numerical experiments that extend problems of Chapters 2 and 3 and present some conjectures. In particular, we consider several outer billiard examples with polygonal invariant curves. We present numerical evidence indicating that these examples may also have non-vanishing Lyapunov exponents just as the example of Chapter 2. Based on this and many other numerical ex- periments which are not included in this thesis we conjecture that for sufficiently small area parameters hyperbolicity is typical in the class
18 of outer billiards with invariant polygonal domains. Next we present studies of two examples which are good candidates for outer billiards with unbounded orbits. The first example which was discovered by Tabachnikov is that of a half-circle. Numerical simula- tions appear to show that there are whole open domains escaping to infinity. They also indicate that the motion of these domains is quite regular which gives hope that one may be able to prove “escape to in- finity”. Interestingly, the behavior inside these domains appears to be in some sense elliptic and techniques of KAM may play a role. Another example with possibly unbounded orbits is the a dart quadrilateral, which is the next simplest quadrilateral after the trapezoid. It too has a sort of first integral, as the example of the trapezoid, and so has essentially one dimensional dynamics. Nevertheless, dynamics on the levels of the integral in this case is significantly more complicated than in the case of the trapezoid and results of Chapter 3 do not carry over trivially. Numerical studies show that orbits for the dart quadrilateral also have some regularity which again gives hope that a rigorous proof may be within reach.
19 Chapter 2
Chaotic Outer Billiards
2.1 Introduction
In this section we construct an example of an outer billiard with hyper- bolic behavior on a subset of its domain. It seems unlikely that an outer billiard can be hyperbolic on its entire domain. This appears to be a consequence of near integrability of motion near infinity. Although this statement is exactly true only for sufficiently smooth tables, for which invariant curves are known to exist arbitrarily far from the table [14], numerical experiments indicate that there are still invariant curves and elliptic islands even for non-smooth tables. In particular, this appears to be the case for the example presented in this chapter. For numerical explorations of this question see Chapter 3.
The main result of this section is Theorem 2.5.1 which states that
20 there is a one parameter family of outer billiard tables which have non-vanishing Lyapunov exponents on a subset of the invariant square. The theorem is proved by applying the result of Wojtkowski [22] which shows that it is enough to construct an eventually strictly preserved
field of cones in the tangent bundle of the invariant region to prove the non-vanishing of the Lyapunov exponents a.e. Ergodicity does not follow directly but the method of Chernov and Sinai [18] is likely to yield a proof in this case.
It is likely that the cone field construction described in this section can be extended to a much wider class of outer billiards – outer bil- liards with polygonal (rather than just square) invariant domains. The main difficulty appears to be identifying the elliptic domains inside the invariant polygons. In this respect the square is the simplest example because precise description of the elliptic domains is possible.
2.2 Area construction
We begin by describing a construction that given a convex plane curve produces an outer billiard table for which the curve is an invariant one. This construction parallels the string construction of inner billiards which does the same for caustics.
Given a plane curve γ and a parameter a satisfying 0 < a < A, where A is the area enclosed by γ, we consider the family of lines L
21 that divide the convex hull of γ into parts with area a and A a. We − have the following result
Lemma 2.2.1. The envelope of is a closed curve Γ, which is convex L if there are no cusps. Furthermore, γ is an invariant curve of outer billiard about Γ.
The proof of this result can be found in [20].
Using the area construction we can construct a table with an arbi- trary (convex) invariant region. This allows us to overcome the problem mentioned in the introduction of difficulty of attaining hyperbolic be- havior near infinity.
2.3 Table
We consider a 1-parameter family of outer billiards obtained by the area construction with γ – a square with unit sides. Fix the area parameter to be 2a and let B be the (closed) billiard table, i.e. the convex hull of Γ, and D be the complement of B in the open unit square. Let Γ be oriented in the counter-clockwise direction.
Lemma 2.3.1. Γ is composed of four arcs of hyperbolas, asymptotic to lines containing the sides of γ, meeting in corners opposite centers of sides of γ.
22
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Figure 2.1: The outer billiard table
Proof. Lines of can intersect either opposite sides or adjacent sides L of γ. If a line ` intersects opposite sides then rotating it about its midpoint does not change the area cut off by ` and so the envelope of will have a corner at the midpoint of such a line. The angle at L the corner is determined by the angle between the limiting lines, i.e. members of passing through the corners of γ and is easily computed L to be π 2 tan−1 2a. − If a line ` intersects adjacent sides of γ we show that the resulting envelope is a hyperbola. We fix a coordinate system with the origin in the corner, where the adjacent sides meet, and with the extensions of the sides being positive axes. In these coordinates the slope of ` as function of the intersection point with one of the axes is given by
23 2a/x2. This is exactly the formula for the slope of lines tangent to − the hyperbola y = a/x as a function of the intersection with the x-axis. Uniqueness of the envelope implies that the two curves are the same.
Depending on which arc of B contains pt, it is convenient to use a different coordinate system on D. Each arc naturally corresponds to a corner, which we choose to be the origin when considering points reflecting in this arc. Choose the outgoing edge (considering γ ori- ented clockwise) to be the x-axis and the incoming – the y-axis. We introduce corresponding coordinate functions x(p) and y(p). In these coordinates the hyperbolic arcs of B from the above lemma are given by the equation y(p)x(p) = a.
As a consequence of the construction D is an invariant region for the outer billiard about Γ.
2.4 Cone field
We will define an invariant cone field on a subset Dh of D defined by D = p D [T k(p), T k−1(p)] B is not a corner for some k Z . h { ∈ | ∩ ∈ }
That is Dh is the set of all points in D whose orbits touch interiors of hyperbolic arcs of B.
Lemma 2.4.1. The points in D D are periodic. \ h
24 Proof. A point in D D reflects only in corners of B. Hence as far \ h as it is concerned the table is a square. But it is well known that outer billiard about any lattice polygon has only periodic points (see for example 3.6).
The set D D is indicated in Figure 2.5 in red. It is easy to see \ h that all points in D D have the same period – four. So D D is \ h \ h fixed by T 4 and hence is an elliptic domain.
A cone C(p) T D will be denoted by a pair of vectors (u, v) ⊂ p h ∈ T D 2, C(p) = w T D [u, w][w, v] > 0 where [ , ] stands for the p h { ∈ p h| } · · 2 cross product induced by identifying TpDh with the ambient R . It is clear that scaling u and v in the above definition by positive constants does not change C(p) so whenever we will talk about equality of vectors defining cones we shall mean equality up-to multiplication by a positive constant.
We will first define the invariant cone field on the points in Dh for which pt is not a corner of B. The cone field can then be extended to all points in Dh which do not land on discontinuities of dT by pulling back by dT . We set C(p) = (u(p), v(p)), where for p with x(p) = x, y(p) = y
u(p) = (1, y/x) −
25 is a vector tangent to the homothetic hyperbola passing through p, and
a + a2 axy y v(p) = p pt = x − , y − − y − a + a2 axy ! p − p In the last formula we used
a + a2 axy y (x(pt), y(pt)) = − , y a + a2 axy ! p − p
dTp(C(p)) u(p) p
v(p)
C(p)
Homothetic hyperbola
Figure 2.2: Cone construction
For convenience we will define the cone field at points of γ (which are not in D) by taking u(p) to be the tangent vector to γ in the corresponding direction.
26 2.5 Hyperbolicity
Now we are ready to prove the main result.
Theorem 2.5.1. Outer billiard about B constructed in Section 2.3 has non-vanishing Lyapunov exponents on a full measure subset of Dh for a (0, 1/4(3 √5)). ∈ − We will prove this by showing that the cone field defined in the previous section is eventually strictly preserved.
Definition 2.5.2.
1. A cone field C(p) is preserved at a point p if dT C(p) C(T (p)). p ⊂
2. A cone field is strictly preserved if C(p) C ◦(T (p)). ⊂
3. A cone field is eventually strictly preserved if for almost every p there exists an n(p) such that the cone field is strictly preserved at T n(p)(p).
The result above then follows from the result of Wojtkowski intro- duced in [22].
Theorem 2.5.3. If there exists a measurable bundle of sectors which is eventually strictly preserved by Φ : M 2 M 2, where M 2 is a two → dimensional Riemannian manifold, satisfying
1. Φ preserves a probabilistic measure µ which has a non-vanishing
density with respect to the Riemann area element on M 2;
27 2. The singularities of Φ satisfy
+ 1 log DxΦ dµ(x) < + 2 k k ∞ ZM
where log+ t = max(log t, 0).
Then the Lyapunov exponent λ+ of Φ is positive µ a.e.
T satisfies the first condition above because it preserves the Lebesgue measure on Dh. It also satisfies the second condition because the dif- ferential of T is bounded in norm on Dh. Indeed, differential of an outer billiard map blows up only if the table has points of vanishing curvature, which our example does not. Thus the above result may be applied to T if the cone field is preserved. This is what we prove next.
Theorem 2.5.4. The cone field C(p) defined above is eventually strictly preserved by the outer billiard about B for a (0, 1/4(3 √5)). ∈ − This statement will be proved in a series of lemmas because the argument naturally divides into cases. The argument is essentially different for points with pt and T (p)t belonging to the interior of the same hyperbolic segment of B and for points for which they belong to interiors of different hyperbolic segments. We begin by examining the first case.
Lemma 2.5.5. C(p) is preserved at points such that pt and T (p)t belong to the interior of the same hyperbolic segment.
28 Proof. Since we are concerned with one hyperbolic arc we can for the moment forget about the rest of the table and consider outer billiard about a single branch of hyperbola.
We first show that the family of hyperbolas y = a0/x, 0 < a0 < a, is invariant under the billiard map. Using the formula for pt we have
1 √1 xy y (x0, y0) = T (x, y) = (2 − − x, 2 y) y − 1 √1 xy − − −
So
y 1 √1 xy x0y0 = 4 2x 2y − − + xy − 1 √1 xy − y − − xy = 4 + xy 2 1 1 xy + − − − 1 √1 xy − − 2 p2√1 xy = 4 + xy 2 − − − 1 √1 xy − − = xy
Thus xy is integral of T and so homothetic hyperbolas are preserved.
Since T preserves the order of points on the invariant hyperbolas a tangent vector to such a hyperbola at some point p – u(p), is mapped by dT into a vector with the same direction as the tangent vector to the same hyperbola at the image point. That is dTpu(p) = u(T (p)). Also the tangent vector v(p) = p t is mapped to minus itself dT v(p) = − p p v(p) since restricted to the line of tangency the billiard map is simply − a reflection in the tangency point. So we have the situation as in Figure
29 2.3.
y=a/2x
y=a'/2x
T(p)
pt
p
Figure 2.3: Cone preservation
Hence the image cone dT C(p) = (u(T (p)), v(p)). Noting that the p − vectors u(T (p)) and dTpu(p) are the same up-to rescaling by a positive constant it is enough to check that dT v(p) C(T (p)). We will show p ∈ that the angle of C(p) is always obtuse and the angle of dTpC(p) is always acute. Observe that v(p) always lies in the fourth quadrant while u(p) always lies in the second. So the angle between them is always obtuse (Figure 2.2). Similarly u(T (p)) is always in the second quadrant and so is v(p) so the angle between them is always acute. − Hence dT C(p) C(T (p)) for points reflecting in a single hyperbolic p ⊂ segment.
Note that dTpv(p) is strictly inside C(T (p)) although dTpu(p) = u(T (p))
30 utpo rescaling by positive constant so the inclusion is not strict.
The following consequence of the proof is useful in itself.
Lemma 2.5.6. For every point p D such that p is in the interior ∈ h t of a hyperbolic arc.
1. C(p) has an obtuse angle
2. dTpC(p) has an acute angle
Before moving on to the second case we show that cones C(p) are nested in a special way along the line through p, pt.
C(T(p)b) T(p ) b
dTpC(p)
T(p)t p t
C(p)
C(p b )
Figure 2.4: Cone nesting
0 0 Lemma 2.5.7. Let p and p be two points satisfying pt = pt and p p > p0 p then | − t| | − t|
31 1. C(p) C(p0) ⊂
0 2. dT 0 C(p ) dT C(p) p ⊂ p where the tangent spaces are identified by parallel translation to deter- mine the inclusion relationship between the cones (Figure 2.4).
Proof.
1. Let p and p0 be two points satisfying conditions of the lemma.
Notice that v(p) = v(p0) so to prove the first half it is enough to show that u(p) C(p0). We will show that 0 > m(u(p)) > ∈ m(u(p0)) where m is the slope of a vector. Then since v(p) is in the fourth quadrant and u(p), u(p0) are in the second u(p) C(p0). ∈ Indeed, for any two homothetic hyperbole xy = a and xy = a0 the slopes of tangent lines at (x, y) and (x0, y0), satisfying x0 > x and y0 < y, satisfy m < m0, since m = y/x and m0 = y0/x0. − − Because p, p0 lie along a line with negative slope x(p0) > x(p) and y(p0) < y(p). The desired conclusion follows. Notice that u(p) is
in the interior of C(p0). This is crucial to strict inclusion of the cones.
2. Similarly to prove the second part it is enough to show that
0 0 dT 0 u(p ) = u(T (p )) dT C(p). Following the same reasoning as p ⊂ p above it is enough to show that 0 > m(v(T (p))) > m(u(T (p0))) >
0 m(u(T (p))) since the two cones dTpC(p) and dTp0 C(p ) share
32 v(T (p)) = v(p) in common and all vectors determining the cones − 0 lie in the second quadrant. Since pt, p , and p satisfy the relation-
0 0 ships x(pt) > x(T (p )) > x(T (p)) and y(pt) < y(T (p )) < y(T (p)) the desired inequality follows as above. Notice that u(T (p0)) is in
the interior of C(p).
The above lemma provides an easy way of determining whether a cone field at p is preserved. Let pb be the intersection of the ray from pt through p with γ, then to check that the cone field is preserved at
T (p) it is enough to check the inclusion between cones at pb and T (p)b. Indeed, if dT C(p ) C(T (p) ) then pb b ⊂ b
dT C(p) dT C(p ) C(T (p) ) C(T (p)) p ⊂ pb b ⊂ b ⊂ so dT C(p) C(T (p)) p ⊂
Note that for p D◦ if dT v(p ) is in the interior of C(T (p) ) then ∈ h pb b b dTpC(p) ( C(T (p)), i.e. the inclusion becomes strict, because from the above lemma dTpu(p) is in the interior of dTpb C(pb) (Figure 2.4).
We now proceed to examine the points for which pt and T (p)t belong to interiors of different hyperbolic segments of B or corners.
33 Lemma 2.5.8. C(p) is strictly preserved at points such that pt and
T (p)t belong to interiors of different hyperbolic segments.
Proof. If pt and T (p)t belong to interiors of different hyperbolic seg- ments these segments must be adjacent. In this case (Figure 2.4) T (pb) and T (p)b belong to the same side of γ which means that dTpb u(pb) = u(T (p) ). Furthermore, dT v(p ) = v(T (p )) C◦(T (p) ) by Lemma b pb b b ∈ b
2.5.6. This implies that dTp(C(p)) ( C(T (p)).
Lemma 2.5.9. C(p) is strictly preserved at points such that pt is a corner point.
Proof. Considering all points such that pt is a corner point we obtain Figure 2.5 which shows points reflecting in corners and their images.
In Figure 2.5 all points that touch corners are colored : those that touch one corner between touching interiors of hyperbolic arcs are blue, those that touch two are turquoise, three – green. Any point that hits all four corners is periodic, as was proved in Lemma 2.4.1, and these points are colored red. The black lines in the diagram are the lines of discontinuity of the derivative and their preimages. We will use this diagram to reduce verification of preservation of the cone field at col- ored points to consideration of orbits of a few representative points. For this it is important to note that the diagram in Figure 2.5 is struc- turally invariant for a (0, 1/4) (which includes the parameter range ∈
34 Figure 2.5: Orbits reflecting in corners
under consideration), the interval in which the members of that pass L through corners of γ do not intersect in the interior of D.
We first argue that we only need to consider preservation inside a few of the colored polygons. This follows because the cone field and the map are both preserved under the action of the rotation subgroup of D4. Hence if the cone field is preserved inside some colored polygon of Figure 2.5 then it is preserved in every polygon of its orbit under the rotation subgroup. Therefore, it is enough to check preservation for one polygon per rotation orbit.
Next, we show that for points in the colored polygons only cones at the first and last points of the orbit segment inside the colored set need to be compared. T is a central symmetry for every point p with
35 p – a corner of B. So dT at such a point is I and dT C(p) = C(p). t p − p Hence it is enough to check inclusion of cones at successive points of the orbit that reflect in the interiors of hyperbolic arcs. Thus we ignore the points of the orbit segment that reflect in corners, and let p and p0 be
0 the first and last points respectively such that pt and pt belong to the
0 interiors of hyperbolic arcs. As before pb and pb will be intersections of the corresponding rays with γ. We further show that for points in the interior of a given polygon of the above diagram there is only one way for the boundary points
0 pb, T (p)b,..., pb to be distributed on the sides of γ. This allows us to determine preservation of cones for points in a given polygon by exam- ining a diagram like Figure 2.4 for one of its interior points. Suppose
k k there are two points p and q for which T (p)b and T (q)b lie on different sides for some k. Since the polygons are convex T k(p) and T k(q) can be joined by a line segment contained in the interior of the polygon.
By continuity there will be a point r on this line segment such that rb will be a corner point. This is a contradiction since lines in passing L through corners form the boundaries of these polygons. In what follows we will say that a colored polygon (or point) has order k if it belongs to an orbit segment that undergoes k reflections in corners between successive reflections in interiors of hyperbolic arcs. We will consider domains of each order in turn.
1. We first consider order one domains. Here there are two possi-
36 bilities corresponding to two different kinds of order one domains in the above diagram (see Figure 2.6). In this case, it is enough
p'b T(p b) p' T2(p)=p' b T2(p)=p' T(p)
T(pb ) T(p) p p't p t
p't pt
p
Figure 2.6: Order one orbits
to look at which sides the cones lie on to determine inclusion. In the first case (the left diagram in Figure 2.6), the situation
is identical to the case of a point touching different adjacent hy- perbolic segments, and so the cone field is strictly preserved for
0 these points. In the second case the cones dTpb C(pb) and C(pb) lie on opposite sides so u(p0 ) = dT u(p ). As before, because the b − pb b
angle between dTpb u(pb) and dTpb v(pb) is always acute and the
0 0 angle between u(pb) and v(pb) is always obtuse (Lemma 2.5.6), dT v(p ) C(p0 ) and we have the desired strict inclusion. pb b ∈ b
2. For points touching two corners there is only one possible config-
uration (Figure 2.8) because the restriction on a in the statement
37 of the theorem guarantees that a point can touch only adjacent corners of B. Indeed if there was a point that reflected in op- posite corners then it would have to be positioned as in Figure 2.7. In this case a would have to satisfy 4a > 1 but a < 1/4 by assumption. Hence the only possible configuration is as in
p
T(p) 2a
T 2(p)
Figure 2.7: A point reflecting in opposite corners
Figure 2.8.
0 The cone is strictly preserved if pb precedes pb (the order is given by the clockwise orientation of the square). From the diagram it is clear that this is certainly true when T (p) and p0 are on the same side of the line `, tangent to the midpoint of the hyperbolic
0 segment containing pt and pt, as B. Since T (p) reflects in a corner it has to lie between the tangents to the hyperbolic arcs that meet at that corner, call them µ and µ which are in turn lines in 1 2 L
38 T2(p)
p'b T3(p)=p' µ2
l
pb p' t p
p t µ1
T(p)
T(p b ) −1/2 (2a) 2a Figure 2.8: Order two orbit
that pass through corners of γ. The same reasoning applies to p0 for reverse time since T 2(p) also reflects in a corner. Clearly the above condition will be satisfied if the triangle bounded by µ1, µ2 and γ, containing T (p) does not intersect `. In this case T (p) is guaranteed to lie on the right side of `.
` and µ1, µ2 do not intersect inside D if in Figure 2.8 the triangles bound by ` and γ and µ2 and γ do not overlap. Since both triangles have area 2a by construction and γ has unit sides this is true if √2a+2a < 1, i.e. the lengths of the sides of triangles lying along γ add up to less than the length of the side. Solving the inequality we get a < 1/4(3 √5) which is precisely the condition − in the statement of the theorem.
39 p' b p b 3 p T (p) T3(p) T4(p)=p' T2(p)
p't T(p ) b pt
T(p) pt p't T(p) p T2(p) T4(p)=p'
T(p ) p' b b p b Figure 2.9: Order three orbit
3. Finally, for points touching three corners there are again two pos- sible configurations (Figure 2.9) which are symmetric about the diagonal. From the diagram we see that the cone field is strictly
0 preserved if pb precedes pb. Observe that reflecting a point in three consecutive corners of B gives a central symmetry about the fourth corner since the corners of B are vertices of a square. It follows that p0 and T (p) are symmetric about the one corner untouched by the orbit segment from p to p0, and so that [p0, T (p)]
contains this corner. On the other hand [p0, T (p)] must intersect
0 the interior of B because otherwise pt is a corner and then p has
order four and so is not in Dh. So the line through p and T (p) must leave p0 on the same side as B (see Figure 2.10). Noting
0 that pt and pt belong to the interior of the same hyperbolic arc
0 it follows that pt precedes pt, if the boundary of B is oriented
40 0 clockwise, and so pb precedes pb.
T(p)
pt
p't
p' p
pb p'b
Figure 2.10: Proof illustration
We have proved that the cone field is strictly preserved at every point of Dh with the exception of points that land on discontinuities of the derivative which clearly form a set of measure zero.
41 Chapter 3
Polygonal Outer Billiards
3.1 Introduction
In this section we study in detail outer billiard about trapezoids. Be- cause outer billiards are Sl(2, R) equivariant trapezoidal outer billiards form a one parameter family, parameterized by the ratio of lengths of bases, which we will denote by α. Trapezoids are distinguished among other quadrilaterals because trapezoidal outer billiards have a first in- tegral of motion in the form of the fractional part of the orthogonal projection onto the line normal to the bases. The dynamics thus re- duces to dynamics on a discrete set of lines parallel to the bases of the trapezoid. Since polygonal outer billiards are piecewise isometries the restriction to the invariant lines gives an infinite interval exchange.
Moreover, it turns out that on each line dynamics is closely related to
42 that of a circle rotation by α. This allows us to give a complete sym- bolic description of dynamics in terms of rotation sequences similar in some ways to works of Tabachnikov [20] and Adler,Kitchins and Tresser [1].
We note that trapezoidal outer billiards are not quasi-rational even when α is rational, i.e. trapezoid is lattice, because they have parallel sides, and so are not covered by results on boundedness of orbits men- tioned in Section 1.5. Thus this section gives first results on dynamics of non-quasi-rational outer billiards.
The main results of this section are Theorem 3.3.29 and Theorem 3.3.30. Theorem 3.3.29 shows that in spite of not being quasi-rational trapezoidal outer billiards have bounded orbits. This is especially sur- prising because there appear to be no obstructions to escape of orbits to infinity. Theorem 3.3.30 gives an upper bound on the growth of complexity in terms of the Diophantine exponent of α.
Trapezoidal outer billiards are not the only polygonal outer billiards with a first integral of motion. Any polygon for which projection of the vertices on to some line is a subset of a lattice has this property. We expect that some of the ideas introduced in this chapter can be applied to this wider class. It is interesting that the “next” simplest polygon with this property – a dart – appears to have unbounded orbits. We explore this question numerically in Chapter 3.
The rest of the chapter is divided into two parts. The first is a
43 brief introduction to dynamics of polygonal outer billiards in general where results important to further exposition are stated and notation is introduced. The second is devoted to the study of the trapezoidal outer billiard. We begin by exploring dynamics of the first return map on the line x = 0 in Section 3.3.1. After developing some understanding of this map we apply it to the study of lattice trapezoids in Section 3.3.3. We then return to the case of non-lattice trapezoids and explore boundedness of orbits in Section 3.3.5 and complexity growth in Section 3.3.6.
3.2 Polygonal outer billiards
3.2.1 Definitions and notation
Let P be a convex n-gon with vertices v1,v2,...,vn. We will assume that center of mass of P is at the origin since outer billiard dynamics are invariant under translation. A polygonal outer billiard is an outer billiard map T defined for table P . In this case, the outer billiard map is undefined on the set of points L1 that reflect in the the sides of the polygon (Figure 3.1). The set L1 is easily seen to be the set of rays emanating from the vertices of the polygon P and passing through preceding vertices (given by the orientation of P ). The complement of L1 is composed of cones D1, ..., Dn indexed by the corresponding vertices of P .
44 D3
D4
D2 v2 v v1 3 O
v6 v4
v5 D D1 5
D6
Figure 3.1: L1
Since we are interested in the long time behavior of the map we will be particularly interested in the set obtained by removing from the plane all preimages of the set L1. This is the set of points on which T is defined for all future times. We also define sets L with k N k ∈ { ∪ ∞} consisting of points for which T k is undefined, L = ∞ L . As well ∞ ∪k=1 k as D = R2 (L P ), which is the set of points for which all future \ ∞ ∪ iterates of T are defined.
Lemma 3.2.1. D is a union of convex polygons, segments and points.
Proof. Consider a connected component C of D. Given any two points
45 p, q in C, we have T k(p), T k(q) D for all k and i is the same ∈ ik k for both since otherwise the images are in deferent components of D and C can not be connected. Hence C T −k(D ) for all k. Since ⊂ ∩ ik the intersection of a countable collection of convex sets is convex C, is convex.
To see that C is a polygon we show that the boundary of C can lie only along a finite number of directions. T is a piecewise central symmetry and so every T k is either a piecewise central symmetry or a piecewise translation. It follows that Lk is composed of line segments parallel to sides of P . Since the limit of a sequence of parallel segments is again a segment parallel to the same direction or a point it follows that the boundary of C is composed of finitely many segments parallel to sides of P .
The set D has full measure in the complement of P and often has a surprising and beautiful geometry as, for instance, in the case of the regular pentagon, for which it was shown by S. Tabachnikov [20] to have a fractal structure (Figure 3.2).
We close this section with several results and definitions describing the general structure of T that will be used in subsequent sections.
Lemma 3.2.2. T 2 is a piecewise translation of polygonal domains (pos- sibly unbounded) D ,D ,..., where D = D T −1(D ) if the inter- 12 13 ij i ∩ j
46 Figure 3.2: D for a regular pentagon
section is non-empty. The translation vectors v12,v13,... are given by diagonals of P , where v = 2(v v ). ij j − i
Proof. From elementary geometry, composition of two central symme- tries about points v and v is a translation by the vector 2(v v ). i j j − i Hence T 2 is a piecewise translation and translation vectors are as de- scribed in the statement of the lemma. Furthermore, every point in
Dij is translated by vij by construction.
47 D35
D34 D D23 24 v2 v D41 v1 3 O D D 12 v 45 5 v4
D51 D13
D52
Figure 3.3: Structure of T 2
Lemma 3.2.3. Unbounded Dij are polygonal cones and vij is transverse to D , that is given any point p D there exists a t > 0 such that ij ∈ ij p + tv / D . ij ∈ ij
Proof. Since there are finitely many Dij some of them must be un-
−1 bounded. Every Dij is an intersection of two cones Di and T (Dj) which if unbounded is a polygonal cone.
To see that transversality holds observe that for a vector to be not transverse to Dij the corresponding ray must be contained in both cones
−1 Di and T (Dj) whose intersection forms Dij. On the other hand, vij lies is in the direction joining the vertices of the two cones and since
48 neither is contained inside the other by convexity of P , it doesn’t lie in either. Therefore, vij is transverse to Dij as claimed.
Next we state a result approximating the motion of T 2 by a periodic polygonal flow. This result was proved by Tabachnikov in [21]. Con- sider the vector field defined by V (x) = 2vij where vi, vj are vertices such that lines parallel to Ox support P at these vertices.
Lemma 3.2.4. The integral curves of the vector field V are homothetic centrally symmetric polygons.
Definition 3.2.5. We call the integral curves of V the dual polygons of P . We denote unique upto dilation dual polygon of P by P ∗ and by P ∗(x) the dual polygon passing through x.
We have the following lemma describing T 2 near infinity.
Lemma 3.2.6. If x is sufficiently far from the origin there is an -
∗ neighborhood N of P (x) such that the orbit of x is in N for at least one complete revolution, and depends only on P .
Definition 3.2.7. We will say that a point of D is regular if the lemma above can be applied to its orbit and irregular otherwise.
49 P*
P
Figure 3.4: Dual polygon P ∗
3.2.2 Symbolic dynamics, vertex coding and
complexity
Points inside a single connected component of D have virtually iden- tical orbits, more precisely they are identical upto translation (except for points at the centers of odd periodic domains). Therefore, it makes sense to identify all points belonging to the same connected compo- nent of D. This corresponds to the following symbolic coding scheme, standard in the study of both ordinary and outer billiards.
Definition 3.2.8. A vertex code of a point x in D is an infinite string
50 wx of symbols from the alphabet 1, ..., N, where N is the number of vertices of P . We define w by w (k) = j if T k(x) D . x ∈ j By construction every connected component of D is labeled by a unique vertex code.
Lemma 3.2.9. An orbit of a point x in D is periodic iff w(x) is peri- odic.
Proof. If the orbit of x is periodic it is clear that w(x) is also periodic. To prove the converse suppose w(x) is periodic but the orbit of x is not. Let t be the period of w(x) and u = T t(x) x. Without loss of generality − we may assume that t is even, otherwise we simply take 2t instead of t. Since the vertex code is periodic we will have T kt(x) = x + ku. This orbit is clearly unbounded and so there exists a large K such that for all k > K, T kt(x) D for some i, j. The number of iterates of ∈ ij 2 T necessary to traverse the cone Dij clearly grows with distance from the origin. Therefore, there is K large such that T tK (x), T tK+2(x),..., T t(K+1)(x) are in D . It follows that u = T tK(x) T t(K+1)(x) = t v ij − ij but this is impossible since by Lemma 3.2.3 vij is transverse to Dij and
kt so T (x) can not remain in Dij for all k > K.
Thus we have equivalence between symbolic dynamics on the space of all allowed sequences X = w(x) x D with the left shift and the { | ∈ } positive iterates of T .
51 Given a symbolic dynamical system we can define its complexity which in turn provides important information about the dynamical sys- tem from which symbolic dynamics was derived. We define complexity of a symbolic dynamical system.
Definition 3.2.10. A language of a symbolic dynamical system is L the set of all allowed infinite sequences. Complexity function of a sym- bolic dynamical system, usually denoted by p(n), is the rate of growth of the set which is the set of all words of length n that occur as Ln subwords of elements of . L Dynamical systems are broadly characterized based on their com- plexity into hyperbolic, parabolic and elliptic depending on whether p(n) grows exponentially, polynomially or sub-linearly. Hyperbolic sys- tems are characterized as chaotic because they typically possess strong mixing properties and exponential instability with respect to initial con- ditions. Parabolic and elliptic systems on the other hand typically lack these properties but are complicated in a different and as yet poorly understood way which is explored further in this chapter.
3.3 Trapezoidal outer billiards
Here we begin the study of the trapezoidal outer billiards. Since outer billiards are Sl(2, R) equivariant we only need to consider trapezoids P upto affine equivalence. We may assume without loss of generality that
52 the trapezoid is of the form (1/2, 0), (1/2, 1), ( 1/2, 1 α), ( 1/2, 0) − − − (Figure 3.5) with α (0, 2/3) (this assumption simplifies analysis but ∈ does not impose significant restrictions on the possible dynamics). The vertices of P will be numbered from 1 to 4 in the order in which they are listed. P is assumed to be oriented counter-clockwise (Figure 3.5).
1
1−α
−1/2 1/2
Figure 3.5: Trapezoid
3.3.1 First return map on x = 0
To get a handle on the dynamics of the outer billiard map we begin by studying the first return map to the positive y-axis more precisely to the segment X = (1 α/2, ) along the y-axis. − ∞ First, we compute the translation vectors for T 2.
53 Lemma 3.3.1. T 2 is a piecewise translation with vectors
v12 = (0, 2)
v = ( 2, 2 2α) 13 − − v = ( 2, 0) 14 − v = ( 2, 2α) 23 − − v = ( 2, 2) 24 − − v = v ij − ji and domains
D = x < 1/2, y > α(x 1/2) 1, y < 0 12 { − − } D = x < 1/2, y < α(x 1/2) 1 13 { − − } D = y > 2, y < α(x 1/2) + 1 21 { − } D = 1/2 < x < 3/2, y < α(x 1/2) + 1 23 { − } D = x > 1/2, y < 2, α(x 1/2) + 1 24 { − } D = x > 1/2, y > 2 2α, y > α(x 1/2) + 1 31 { − − − } D = x > 1/2, y > α(x 1/2) + 1, y < 2 2α 34 { − − − } D = 3/2 < x < 1/2, y > 0 41 {− − } D = x < 3/2, y > 0 42 { − }
54 Proof. As was shown in the previous section T 2 is a piecewise trans- lation. The verification that translation vectors and domains are as described is a straight forward computation.
Let TX be the first return map to X, or simply T when no ambiguity is possible. The following lemma describes TX as an infinite interval exchange.
Lemma 3.3.2. TX is an orientation preserving interval exchange. Let [x] = x mod 2 := x 2 x/2 , then − b c
x + 2α if x I ∈ k,1 x 2 + 2α if x Ik,2 T (x) = − ∈ X x + 2 if x Ik,3 ∈ x if x I k,4 ∈ where
I = x I [x + 2α(k(x) + 1)] > 1 α/2, x < 2(k(x) + 1) 2α k,1 { ∈ k| − − } I = x I [x + 2α(k(x) + 1)] > 1 α/2, x > 2(k(x) + 1) 2α k,2 { ∈ k| − − } I = x I [x + 2α(k(x) + 1)] < 1 α/2, x < 2(k(x) + 1) k,3 { ∈ k| − } I = x I [x + 2α(k(x) + 1)] < 1 α/2, x > 2(k(x) + 1) k,4 { ∈ k| − }
55 and I = (2k + (1 α/2), 2(k + 1) + (1 α/2)) k − − x (1 α/2) k(x) = − − b 2 c the index of the interval Ik in which x lies.
Proof. Denote by T + the first arrival map under T 2 from the positive y axis to the negative y-axis. Consider an interval Ik along the y-axis with k > (1/α + 1/2)/2 (the case 0 k (1/a + 1/2)/2 has to d e ≤ ≤ d e be considered separately). Under T 2 this interval will be translated by v = 2(1, α 1) as long as it remains in D . Note that the endpoints 31 − 31 2m of T (Ik) have the form
(2m, 2k + 1 α/2 + 2m(α 1)) − − and (2m, 2(k + 1) + 1 α/2 + 2m(α 1)) − − so for m = k and k + 1 respectively they satisfy y = αx + (1 α/2) − which is the first discontinuity line it will encounter as is evident from the figure. It is thus not cut by this discontinuity. In the case when k (1/α + 1/2)/2 the first encountered discontinuity will be along ≤ d e the line y = 2 2α − 2 Next it enters D21 so under T the interval will be translated by v = ( 2, 0) until it will land on the discontinuity y = 2. A simple 21 −
56 X
I4
I3
7−α/2 I2
5−α/2
I1
3−α/2
2 4 6 8
−1−α/2
−3−α/2
−5−α/2
Figure 3.6: Diagram describing T +
computation shows that the interval will be split into two with the splitting point ω = [2(k + 1)α + 1 α/2] relative to the lower endpoint k − of the interval. The endpoints of the interval at this moment are (2k, 2 − ω ), (2k, 2 + 2 ω ). k − k
57 The lower half of the interval is now in D and since v = ( 1, 1) 24 24 − − after 2k more applications of T 2 it will arrive to the negative y-axis. Its image will be the interval [ 2(k 1), 2(k 1) ω )]. Similarly − − − − − k the upper half will arrive one T 2 iteration later and its image will be [ 2(k 1) ω , 2k]. The first return arrival to the negative y- − − − k − axis for each interval Ik is a composition of a two interval exchange and a translation. The two interval exchange corresponds to the circle rotation by ωk.
Finally we have
T +(x) = [[x (1 α/2)] + ω ] 2(k(x) + 1) − − k − = [x + 2(k(x) + 1)α] 2(k(x) + 1) −
In the case k(x) (1/α + 1/2)/2 the intervals will first be cut by ≤ d e the second order discontinuity segment along the line y = 2 2α but − the formula for T + is still the same (because the discontinuity segment maps onto the y-axis under T 2).
Similar considerations show that the return map from the negative to the positive y-axis is given by
T −(x) = [x 2(k˜ + 1)α] + 2(k˜ + 1) −
58 where x + (1 + α/2) k˜(x) = 1 − 2 − Composing these two maps we obtain
T (x) = T − T +(x) X ◦ = 2(k˜ + 1) + [T +(x) 2(k˜ + 1)α] − = 2(k˜ + 1) + [x + 2(k + 1)α 2(k˜ + 1)α] − where k and k˜ are evaluated at x and T +(x) respectively. Noting that
[x + 2(k + 1)α] 2(k + 1) + 1 + α/2 k˜(T +(x)) = − 1 − 2 − [x + 2(k + 1)α] + 1 + α/2 = k + 1 1 − 2 − k if [x + 2(k + 1)α] < 1 α/2 = − k 1 if [x + 2(k + 1)α] > 1 α/2 − −
59 for x I we have T (x) ∈ k X
T (x) = [x + 2(k + 1)α 2(k˜ + 1)α 2(k + 1)] + 2(k˜ + 1) X − − = [x + 2(k k˜)α 2(k + 1)] + 2(k˜ + 1) − − = x + 2(k k˜)α 2(k k˜) − − − x 2(k + 1) + 2(k k˜)α 2 − − − b 2 c x + 2α if [x + 2(k + 1)α] > 1 α/2, − x < 2(k + 1) 2α − x α if x k α > α/ , 2 + 2 [ + 2( + 1) ] 1 2 − − x > 2(k + 1) 2α = − x + 2 if [x + 2(k + 1)α] < 1 α/2, − x < 2(k + 1) x if [x + 2(k + 1)α] < 1 α/2, − x > 2(k + 1) Here in the last step we use the assumption 0 < α < 2/3.
An immediate consequence of this lemma is the following interesting result.
Lemma 3.3.3. For irrational α a point x is either a fixed point of TX or it is non-periodic.
Proof. If x is not fixed then T k(x) = x + 2i + 2jα for some i Z, ∈
60 j N 0 , and i, j are not simultaneously zero. T k(x) = x implies ∈ ∪ { } 2i + 2jα = 0 which is only possible if i = j = 0 since α is irrational.
3.3.2 Symbolic dynamics of TX
In this section we introduce a symbolic coding and symbolic dynamics for the first return map TX .
Definition 3.3.4. Symbolic code of a point x X is a sequence ω , ∈ x where ω (n) = k(T n+1(x)) k(T n (x)) for n 0. Since k changes by x X − X ≥ at most 1 or -1 under one application of TX by Lemma 3.3.2, ωx is a string of 0’s, 1’s and -1’s.
Definition 3.3.5. Let +(x) denote the positive orbit of x under some OF transformation F .
Next we show that TX is “almost” a rotation by 2α in the sense that there is a piece-wise linear surjective map (continuous on each Ik) from
1 X to S that maps orbits of TX to orbits of the rotation, with at most a
finite number of points of an orbit of TX mapping to a point of an orbit of the rotation. Let R be the map R : [0, 2] [0, 2], R(x) = [x + 2α], → i.e. rotation by 2α on a circle of length 2. We define two projection maps π (x) = [x (1 α/2)] and π (x) = [x + 2k(x)α + (1 + α/2)] −1 − − 1 that give the equivalence.
Lemma 3.3.6.
61 1. π ( +(x)) = +(π (x)) −1 OT OR −1
2. π ( +(x)) = +(π (x)) 1 OT OR 1 the correspondence is 1-to-1 and order preserving.
Proof.
1.
[π (T (x)) π (x)] = [[T (x) (1 α/2)] [x (1 α/2)]] −1 − −1 − − − − − = [T (x) x] − = 2α or 0
by Lemma 3.3.2.
2. Noting that
0 if x I ∈ k,1 1 if x Ik,2 and x < 2k + 1 α/2 2α − ∈ − − k(T (x)) k(x) = 0 if x Ik,2 and x > 2k + 1 α/2 2α − ∈ − − 1 if x I ∈ k,3 0 if x Ik,4 ∈ we have
62 [π (T (x)) π (x)] = [[T (x) + 2k(T (x))α + 1 + α/2] 1 − 1 [x + 2k(x)α + 1 + α/2]] − = [T (x) x + 2(k(T (x)) k(x))α] − − 2α if x I ∈ k,1 0 if x Ik,2 and x < 2k + 1 α/2 2α ∈ − − = 2α if x Ik,2 and x > 2k + 1 α/2 2α ∈ − − 2α if x I ∈ k,3 0 if x Ik,4 ∈
In other words orbits of TX can be projected to orbits of R in two different ways. We can also identify the fibers of the equivalence - the points of an orbit of TX that map to a single point of orbit of R.
Lemma 3.3.7.
n+1 n 1. π−1T (x)) = π−1(T (x)) iff ωx(n) = 1
2. π T n+1(x)) = π (T n(x)) iff ω (n) = 1 1 1 x −
Proof.
1. If π (T n+1(x)) = π (T n(x)) then T n+1(x) T n(x) is a multiple −1 −1 − of 2, assuming x is not fixed, it must be exactly 2 by Lemma 3.3.2.
63 From the definition of k(x) it is clear that k(x + 2) = k(x) + 1 and so ω (n) = 1. Conversely, if ω (n) = 1 then k(T n+1(x)) x x − k(T n(x)) = 1 and so T n+1(x) T n(x) = 2 by Lemma 3.3.2. −
2. If π T n+1(x)) = π (T n(x)) then [T n+1(x) T n(x)+2(k(T n+1(x)) 1 1 − − k(T n(x)))α] is a multiple of 2, so T n+1(x) T n(x) = 2m 2ω (n)α − − x
where m can be 1,0, or -1 by Lemma 3.3.2 and ωx(n) can not be 1 by the same lemma (if ω (n) = 1 T n+1(x) T n(x) = 2). x −
ωx(n) = 0 implies m = 0, because by Lemma 3.3.2 ωx(n) = 0 implies T n+1(x) T n(x) < 2. Hence ω (n) = 1. Conversely, if − x − ω (n) = 1 then T n+1(x) T n(x) = 2 + 2α and k(T n+1(x)) x − − − − k(T n(x)) = 1 so −
π (T n+1(x)) = [T n(x) 2 + 2α + 2(k(T n(x) 1)α] 1 − − = [T n(x) + 2k(T n(x))α]
n = π1(T (x))
Using the above lemmas we can now give a description of ω-coding of TX as a “rotation” sequence.
Definition 3.3.8. A rotation sequence is a semi-infinite sequence of 0’s and 1’s χ (kθ + φ mod 1) ∞ where χ is the characteristic function { J }k=0 J
64 of some subinterval J, θ is the rotation number and φ is the “phase shift”.
Lemma 3.3.9.
1 π (T n(x)) [1 + α/2 2α, 2 2α] and − −1 ∈ − − π (T n(x)) / [1 + α/2 2α, 2 2α] 1 ∈ − − ωx(n) = n 1 π1(T (x)) [1 + α/2 2α, 2 2α] ∈ − − 0 otherwise Proof. Let x0 = T n(x).
1. If π (x0) [1 + α/2 2α, 2 2α] then x0 [2(k(x0) + 1) −1 ∈ − − ∈ − 2α, 2(k(x0) + 1) + 1 α/2 2α]. Hence by Lemma 3.3.2 x0 I − − ∈ k,2 or I (assuming x0 is not fixed). If π (x0) / [1+α/2 2α, 2 2α] k,3 1 ∈ − − then [x0 + 2(k(x0) + 1)α] / [0, 1 α/2] so x0 / I . Thus if the ∈ − ∈ k,3 first conditions holds x0 I and T (x0) = x0 2 + 2α. It follows ∈ k,2 − that k(T (x0)) k(x0) = 1. − −
2. If π (x0) [1 + α/2 2α, 2 2α] then [x0 + 2(k(x0) + 1)α] 1 ∈ − − ∈ [0, 1 α/2]. Hence x0 I by Lemma 3.3.2 (assuming x0 is not − ∈ k,3 fixed). So T (x0) = x0 + 2 and k(T (x0)) k(x0) = 1. −
3. If neither of the above holds then since π (x0) / [1+α/2 2α, 2 −1 ∈ − − 2α] then x0 [2k(x0) + 1 α/2 2α, 2(k(x0) + 1) 2α] by Lemma ∈ − − − 3.3.2 x0 I or I . If x0 I then x0 > 2k(x0) + 1 α/2 and ∈ k,1 k,3 ∈ k,1 −
65 T (x0) = x0 + 2α and so k(T (x0)) k(x0) = 0. If x0 I then x0 − ∈ k,3 ∈ [2k(x0)+1 α/2 2α, 2k(x0)+1 α/2] and so k(T (x0)) k(x0) = 0. − − − −
Next we extend the result of Lemma 3.3.6 on near equivalence of
TX to a rotation to ω-coding. Namely, we prove that ω-codes of TX orbits are almost rotation sequences. We define projections on symbolic
ω ω ω sequences π1 , π−1, corresponding to π1, π−1 respectively, by π1 (ωx) =
0 0 ω 0 0 ωx where ωx is ωx with -1’s removed and π−1(ωx) = ωx where ωx is ωx with 1’s removed.
Lemma 3.3.10. We have the following diagram
+(x) π ( +(x)) OT −→ i OT l l ω πω(ω ) x −→ i x where the left vertical arrow is given by T n(x) ω (n), while the right 7→ x arrow is obtained from the coding of the rotation sequence with θ = 2α, φ = π (x) and J = [1 + α/2 2α, 2 2α], by x iχ (x) and i = 1, 1. i − − 7→ J −
Proof. The statement is obvious except for the description of the right vertical arrow. From Lemma 3.3.6 we know that the points of π ( +)(x) i OT are in one-to-one correspondence with points of +(π (x)) and fur- OR i thermore all points with ω (n) = i are projected to a single point x −
66 n0 0 πi(T (x)) with ωx(n ) = i. From Lemma 3.3.9 we know ωx(n) = i iff π (T n(x)) J. It follows that the projection of an orbit is in one-to- i ∈ one correspondence with the projection of the corresponding rotation sequence.
Thus just as TX projects to orbits of R, ω codes of orbits of TX project to rotation sequences. So ω codes of orbits of TX are comprised of two interlocking rotation sequences with different phase shifts. Fi- nally using Lemma 3.3.10 we can give an algorithm for producing ω- code of an orbit of TX as an algorithm that superimposes two rotation sequences.
Lemma 3.3.11. Code of a point x can be obtained by iterating the following algorithm :
Setup : The algorithm consists of moving two points t1 and t−1 around a circle of length 2 with a distinguished interval J = [1 + α/2 − 2α, 2 2α]. −
Input : To start – take two points t−1 = π−1(x) and t1 = π1(x) on the circle. Set the counter i to 0.
Iterate : if t J 1 ∈
ωx(i) = 1 t t + 2α 1 → 1 else if t J −1 ∈
67 ω (i) = 1 x − t t + 2α −1 → −1 else t t + 2α, i = 1, 1 i → i −
ωx(i) = 0 i i + 1 → In words, we move the two points around the circle by R until one or the other enters the distinguished interval J at which point the other point stops while the point inside the interval continues to move. Once it exits the motion resumes. If both points enter the distinguished segment simultaneously t1 has ”the right of way”. 1’s and -1’s are generated when t1 or t−1 respectively passes through the distinguished segment.
Proof. The lemma follows directly from Lemma 3.3.9 and Lemma 3.3.7.
From this it follows that for irrational α the ω-coding is faithful in the sense that any two different non-periodic points have different
ω-sequences. We state this as a lemma.
Lemma 3.3.12. Let α be irrational then every pair of distinct non-fixed points have distinct ω-sequences.
Proof. The proof follows from the fact that rotation sequence of every point on the circle is unique if the angle of rotation is irrational. Indeed, if x1 and x2 are two points on the circle then because the orbit of x1
68 is dense there will be a time when the segment [x1, x2] contains one of the end points of the selected interval J. Hence x1 and x2 will have different rotation sequences.
Combining this with the above lemma we have that any two points with different t1 and t−1 will have different ω-sequences but every point of X is uniquely determined by its π1 and π−1 projections. Hence every non-fixed point of X has a unique ω-code.
3.3.3 Dynamics for rational α
In this section we apply results we have accumulated so far to the case when α is rational. The main results of this section are Corollary 3.3.14, which shows that all orbits are periodic, and Theorem 3.3.19, which counts the number of periodic orbits with distinct ω-sequences, and gives a lower bound on the number of distinct periodic orbits in a period of TX .
Through out this section α will be assumed to be rational. We begin by showing that in this case TX is periodic.
Theorem 3.3.13. If α = p , p, q N then under T , X splits into q ∈ X countably many invariant segments
[2mq + 1 α/2, 2(m + 1)q + 1 α/2] − −
69 if q 1 mod 2 and ≡
[mq + 1 α/2, (m + 1)q + 1 α/2] − − if q 1 mod 2. T restricts to the same interval exchange on each ≡ X invariant segment.
Proof. Suppose first that q 1 mod 2 (the other case is handled ≡ similarly) then k(x + 2q) = k(x) + q. Substituting this into the above formula for T we have T (x + 2q) = T (x).
Next notice that for certain values of k some of the intervals are empty. In particular for k such that 2(k + 1)α 0 mod 2, I = ≡ k,3 ∅ since x I and x < 2(k + 1) implies [x] 1 α/2. Further, the only ∈ k ≥ − way of passing from Ik to Ik+1 is through Ik,3 because points in Ik,4 and
Ik,2 move in the opposite direction and points in Ik,1 after some number of iterations always end up in Ik,2. Thus we have an upper bound on any orbit
T n(x) < inf 2(k + 1) + 1 α/2 k > k(x), 2(k + 1)α 0 mod 2 { − | ≡ }
If 2(k + 1)α 2α mod 2 then T (I ) I since in this case I = ≡ k,2 ⊂ k k,2 (2(k + 1) + 1 α/2 2α, 2(k + 1) + 1 α/2) so T (I ) = (2k + 1 − − − k,2 − α/2, 2k + 1 α/2 + 2α) I . The only way a point can jump from I − ⊂ k k to Ik−1 is through Ik,2 since points in Ik,1 and Ik,3 move in the opposite
70 direction. Hence we have a lower bound on any orbit
T n(x) < sup 2k + 1 α/2 k < k(x), 2(k + 1)α 2α mod 2 { − | ≡ }
A consequence of this boundedness is that all orbits are periodic. This result although expected because of the corresponding result for lattice polygons without parallel sides, was nevertheless unknown.
Corollary 3.3.14. For rational α all orbits are periodic.
Next we use the power of ω-coding to compute the number of dis- tinct periodic orbits in one period of TX . We begin by showing that the number of 1’s and -1’s in one period of ωx is the same.
Lemma 3.3.15. The number of 1’s and -1’s in a period of ωx is the same.
Proof. It is clear from the above result that ωx is periodic. Suppose the number of 1’s and -1’s in one period is different then after one period of
ωx the image of x is in Ik different from the one it started in. Since ωx is periodic the orbit of x will be unbounded which contradicts Lemma 3.3.13.
In the next lemma we compute the length of the period of ωx.
71 Lemma 3.3.16. Suppose α = p/q with gcd(p, q) = 1 and x is not a
fixed point then ωx is periodic with period 2n1(x) + n0(x) where
2jα + π (x) J 0 j < q if q 1 mod 2 |{ 1 ∈ | ≤ }| ≡ n1(x) = 2jα + π1(x) J 0 j < q/2 if q 0 mod 2 |{ ∈ | ≤ }| ≡ and
2jα + π (x) / J 0 j < q if q 1 mod 2 |{ 1 ∈ | ≤ }| ≡ n0(x) = 2jα + π1(x) / J 0 j < q/2 if q 0 mod 2 |{ ∈ | ≤ }| ≡ (The reason for different values is that the orbit projects to a rota- tion by 2α modulo 2.)
Proof. Let x be some point and t1, t−1 the corresponding inputs for the algorithm generating ωx. Now consider how many iterations of the algorithm (and equivalently of T ) are necessary for t1 to return to its original position. The rotation by 2α returns after n0 + n1 steps but t1 stands still for every turn that t−1 is in the distinguished interval and there are exactly n1 such points. Thus the period of t1 is 2n1 + n0. The same argument also shows that t−1 has the same period. Thus after
2n1 + n0 iterations the initial conditions of the algorithm are restored and so ωx is periodic with period 2n1 + n0.
72 Next we show that the period of x under TX is actually the same as the period of ωx.
p Lemma 3.3.17. Let α = q , gcd(p, q) = 1 then under TX every non-
fixed point has a period 2n1(x) + n0(x), where n0, n1 are as in the previous lemma.
Proof. Let x be a non-fixed periodic point of T . By Lemma 3.3.6 π1 projects the orbit of x to the orbit of circle rotation by 2α. Let τ denote the period of π T . To show that this is also a period of x under T 1 ◦ τ it is enough to show that x and T (x) belong to the same Ik because
τ τ this together with π1(x) = π1(T (x)) will imply that x = T (x). From the above lemma it is clear that one period of T composed with projection is exactly the period of ωx, but the number of 1’s and
-1’s in one period of ωx is the same by Lemma 3.3.15. Hence after one period of T composed with projection a point must be in the same Ik in which it started.
Finally, the period of ωx was shown to be 2n1(x) + n0(x) and this is also the period of x as stated.
Lemma 3.3.18. The number of distinct rotation sequences for θ = p/q with gcd(p, q) = 1 up to a shift of index is 2.
Proof. First observe that for two different phase shifts φ1 and φ2 such that φ φ = m/q the rotation sequences are identical upto a shift 1 − 2
73 of index since kθ + φ mod 1 (k + j)θ + φ mod 1 for j satisfying 1 ≡ 2 m/q jθ mod 1. Hence we only need to consider rotation sequences ≡ with φ in [0, 1/q].
Now consider what happens as φ changes continuously from 0 to 1/q. The orbit of rotation is q evenly distributed points on the circle, changing φ rigidly rotates this configuration. The rotation sequence will change every time a point of the orbit crosses into or out of the interval I. If the endpoints of the interval are not conjugate mod 1/q then as φ changes there will be two crossings, when the left and the right boundaries of the interval are crossed respectively, though not necessarily in this order. Because φ must be considered mod 1/q there are 2 distinct rotation sequences. If the ends of the interval are con- jugate mod 1/q the two crossings occur simultaneously but there are still 2 distinct rotation sequences (but one of them occurs only for one value of φ).
Finally, we compute the number of distinct ωx sequences in one period of TX which gives a lower bound on the number of distinct periodic orbits.
Theorem 3.3.19. For α Q the number of distinct ω sequences up ∈
74 to a shift of index is
q p q p + + + + 1 b2 4c d2 4e
Proof. Using the algorithmic description of ωx we see that the sequence in the first place depends on the type of the rotation sequence underly- ing ωx. As was proved above, for every α there are 2 distinct rotation sequences.
For each type of the rotation sequence we can compute the number of corresponding distinct ω sequences. Fixing a particular rotation or- bit representing the rotation sequence we can compute the number of valid configurations of t1 and t−1. The space of all valid configurations is partitioned by orbits of the algorithm corresponding to distinct se- quences ωx. Since every sequence ωx has the same period, dividing the number of all valid configurations by the period of ωx gives the number of distinct ω sequences.
We now compute the number of valid configurations for points t1, t−1 arising in the algorithm generating the ωx. We divide the set of all possible configurations into 4 subsets : S1 - t1, t−1 are both outside of the distinguished interval, S2 - t−1 is in the distinguished interval, t1 is not, S3 - t1 is in the distinguished interval, t−1 is not, and S4 - both t1 and t−1 are in the distinguished interval. We consider each element of this partition individually.
75 First, if t and t are both outside of the distinguished inter- • 1 −1
val there are no restrictions and any combination of t1 and t2 is
allowed. Since there are n0 points of the rotation orbit in the complement of the distinguished interval, S = n2 | 1| 0
If t is inside the distinguished interval but t is outside there • −1 1 are again no restrictions and so S = n n . | 2| 0 1
If t is outside but t is inside one configuration is forbidden. • −1 1
Specifically t−1 can not be right after the distinguished interval
because the only way that this can occur is if t1 and t−1 enter
the distinguished segment simultaneously and t−1 proceeds while
t1 stays put. But this contradicts the definition of the algorithm since t has the ”right of way” in this situation. Hence S = 1 | 3| (n 1)n . 0 − 1
Finally, if both t and t are inside there are n possible config- • 1 −1 1 urations. This follows because the only way both points can be inside the distinguished interval is by entering simultaneously but
then only t1 proceeds and t−1 remains stationary until t1 quits the interval.
Adding the cardinality of the four sets S1, S2, S3, S4 we have – the total number of allowed configurations of t1 and t−1 for a fixed rotation orbit is n0(2n1 + n0). This gives n0 distinct ω sequences for a
76 given rotation sequence. Since there are two distinct rotation sequences
(Lemma 3.3.18) there are n0(x1)+n0(x2) distinct ω sequences, where x1 and x2 are some points representing the rotation orbits corresponding to the two rotation sequences. A trivial computation shows that this sum is equal to q/2 + p/4 + q/2 + p/4 . Taking into account the b c d e fixed points whose ω code is a string of 0’s we obtain the desired result.
3.3.4 Poincare section of T
Although the line X is not a section of the billiard map because not all orbits intersect X it is now easy to describe a section of the entire map by looking at the strip Y = (x, y) 1/2 < x < 1/2, y > 1 α/2+αx . { |− − } We begin by showing that Y is indeed a Poincare section for T .
Lemma 3.3.20. Every orbit of the billiard map intersects Y and T (Y ) ∩ Y = φ.
Proof. That T (Y ) Y = φ is clear since T (Y ) is an image of Y under ∩ central symmetry in one of its corner points.
To show that every orbit of T intersects Y we first show that the orbit of every orbit intersects either Y or Y − = (x, y) 1/2 < x < { | − 1/2, y < 0 under iterates of T and then that every T -orbit of a point } in Y − intersects Y .
77 Without loss of generality the x coordinate satisfies 1/2 < [x] < 1/2. − If this condition is not satisfied then the x coordinate of the T -image of the point satisfies it. Because the x coordinate of a point is changed by either 2, -2, or 0 under iterates of T 2 and because the orbit of any point is not confined to the half-planes x > 1/2 and x < 1/2 by { } { − } Lemma 3.3.21 it must eventually cross either Y or Y −.
Now consider a point p Y −. Noting that T 2(Y −) is contained in ∈ x < 1/2 and that for every point in the cone x < 1/2, y < 0 { − } { − } we have v (1, 0) > 0, it follows that T 2 orbit of p passes through ij · the half-plane y > 0. This together with the contents of the previous paragraph implies that T 2 orbit of p intersects Y .
Lemma 3.3.21. The orbits of T 2 are not confined in any half-plane that does not contain P .
Proof. The statement follows for regular points of the outer billiard by Lemma 3.2.6.
It remains to consider the set of irregular points. Suppose there is a point x in the half-plane whose T 2-orbit is confined to it. We consider the angle between consecutive points of the T 2-orbit of x. Because by Lemma 3.2.6 the set of irregular points is bounded and the set of translation vectors for T 2 is finite the angle between consecutive points of the orbit, measured in the direction opposite to the orientation of the
78 polygon, is strictly between 0 and π and uniformly bounded away from both. It follows that an orbit can not be contained in the half-plane.
Notice that every point in Y returns after an even number of iter- ations of T because if x Y then 1/2 < [T 2k+1(x)] < 3/2 and so can ∈ not intersect Y . Let the first return map to Y be denoted by TY or simply T where there is no chance of confusion. Because x mod 1 is a integral of T the dynamics on Y can be decomposed into dynamics on levels sets of the integral, which are vertical lines.
Lemma 3.3.22. Every half-line x = t Y is preserved under T . { } ∩ Y
Proof. From the definition of T 2 we know that x mod 1 is preserved by T 2. Since there is only one t [ 1/2, 1/2] satisfying t = x mod 1 ∈ − each half-line x = t in Y must be preserved by TY which for each point is a power of T 2.
Let T restricted to the half-line X = (x, y) x = t, y > 1 + α(t t { | − 1/2) be denoted by T (we will drop subscript where no ambiguity is } t possible). We will show that these maps are only slight modifications of TX .
Lemma 3.3.23. Tt is an orientation preserving interval exchange
79 x + 2α if x I ∈ k,1 x 2 + 2α if x Ik,2 T (x) = − ∈ x + 2 if x Ik,3 ∈ x if x I k,4 ∈ where
I = x I [x + 2α(k (x) + 1)] > 1 + α(t 1/2), x < 2(k (x) + 1) 2α k,1 { ∈ k| t − t − } I = x I [x + 2α(k (x) + 1)] > 1 + α(t 1/2), x > 2(k (x) + 1) 2α k,2 { ∈ k| t − t − } I = x I [x + 2α(k (x) + 1)] < 1 + α(t 1/2), x < 2(k (x) + 1) k,3 { ∈ k| t − t } I = x I [x + 2α(k (x) + 1)] < 1 + α(t 1/2), x > 2(k (x) + 1) k,4 { ∈ k| t − t }
and
I = (2k + (1 + α(t 1/2)), 2(k + 1) + (1 + α(t 1/2))) k − −
x (1 + α(t 1/2)) k (x) = − − t b 2 c
the index of the interval Ik in which x lies.
Proof. Following the same steps as in the proof of theorem 3.3.2 we can
80 + − + show that Tt is a composition of Tt and Tt , where Tt maps the ray (x, y) Y x = t into its complement and T − maps the complement { ∈ | } t back into the strip. Furthermore,
T +(x) = [x + 2(k (x) + 1)α] 2(k (x) + 1) t t − t T −(x) = [x 2(k˜ (x) + 1)α] + 2(k˜ (x) + 1) t − t t where
x 1 α(t 1/2) k (x) = − − − t 2 x + 1 α(t 1/2) k˜ (x) = − − 1 t − 2 −
Composing these transformations to obtain Tt we get
T (x) = 2 k˜(T +(x)) + 1 + T +(x) 2 k˜(T +(x)) + 1 α t − h i = 2 k˜(T +(x)) + 1 + x + 2(k + 1)α 2 k˜(T +(x)) + 1 α 2(k + 1) − − h i
81 Noting that
[x + 2(k + 1)α] 2(k + 1) + 1 α(t 1/2) k˜(T +(x)) = − − − 1 − 2 − [x + 2(k + 1)α] + 1 α(t 1/2) = k + 1 − − 1 − 2 − k if [x + 2(k + 1)α] < 1 + α(t 1/2) − = k 1 if [x + 2(k + 1)α] > 1 + α(t 1/2) − − for x I we have ∈ k
T (x) = [x + 2(k + 1)α 2(k˜ + 1)α 2(k + 1)] + 2(k˜ + 1) t − − = [x + 2(k k˜)α 2(k + 1)] + 2(k˜ + 1) − − x 2(k + 1) + 2(k k˜)α = x + 2(k k˜)α 2(k k˜) 2 − − − − − − $ 2 % x + 2α if [x + 2(k + 1)α] > 1 + α(t 1/2), − x < 2(k + 1) 2α − x 2 + 2α if [x + 2(k + 1)α] > 1 + α(t 1/2), − − x > 2(k + 1) 2α = − x + 2 if [x + 2(k + 1)α] < 1 + α(t 1/2), − x < 2(k + 1) x if [x + 2(k + 1)α] < 1 + α(t 1/2), − x > 2(k + 1)
82 Note that Tt(x) is a function of two variables and defines a polygonal exchange on Y . With this we have a formula for the section map since
TY (p) = Tu(v) if p = (u, v)
We define a symbolic coding for orbits of Tt just as we did with
TX . Namely, we assign to every point x of Xt a symbolic sequence ωt (n) = k (T n+1(x)) k (T n(x)). The arguments and constructions of x t t − t t the previous section apply with trivial modifications to the dynamics of Tt. We have the following nearly identical lemmas.
Lemma 3.3.24.
1. π ( + (x)) = +(π (x)) −1 OTt OR −1
2. π ( + (x)) = +(π (x)) 1 OTt OR 1 the correspondence is 1-to-1 and order preserving.
Lemma 3.3.25. We have the following diagram
+ (x) π ( + (x)) OTt −→ i OTt l l ωt πω(ωt ) x −→ i x where the left vertical arrow is given by T n(x) ωt (n), while the right t 7→ x arrow is obtained from the coding of the rotation sequence with θ = 2α, φ = π (x) and J = [1 α(t 1/2) 2α, 2 2α], by x iχ (x) and i t − − − − 7→ J i = 1, 1. −
83 Lemma 3.3.26. Code of a point x can be obtained by iterating the following algorithm :
Setup : The algorithm consists of moving two points t1 and t−1 around a circle of length 2.
Input : To start – take two points t−1 = π−1(x) and t1 = π1(x) on the circle. Set the counter i to 0.
Iterate :
if t J 1 ∈ t
ωx(i) = 1 t t + 2α 1 → 1 else if t J −1 ∈ t ω (i) = 1 x − t t + 2α −1 → −1 else
t t + 2α, i = 1, 1 i → i −
ωx(i) = 0 i i + 1 → In words, we move the two points by a circle rotation until one or the other enters the distinguished interval Jt at which point the other point stops while the point inside the interval continues to move. Once it exits the motion resumes. If both points enter the distinguished segment simultaneously t1 has ”the right of way”. 1’s and -1’s are generated when t1 or t−1 respectively passes through the distinguished segment.
84 If α is rational we have
Lemma 3.3.27. The number of distinct ω sequences up to a shift of index is q p t q p t + + + + 1 b2 4 − 2c d2 4 − 2e
Figure 3.7: Black and white sets
With this we have a picture of the action of T in the strip Y . The
85 entire strip will be divided into domains filled with either periodic or non-periodic points. If one were to plot all preimages of discontinuity lines in black on white background the regions corresponding to non- periodic points would appear black, since the lines of discontinuity are dense in this set, and so we will call the closure of the set of non-periodic points in Y and more generally in the entire plane the “black” set. The closure of the set of periodic points will be called the “white” set for the same reasons (Figure 3.7).
3.3.5 Boundedness of orbits
In this section we prove that all orbits of TY and hence all orbits of T are bounded for all α. This is one of the main results of this chapter. As mentioned in the introduction it is known that orbits of outer billiards about quasi-rational polygons are bounded. Obstruction to escape of orbits for this class of polygons are “necklaces” formed around the table by orbits of periodic domains. Here we begin by showing that for trapezoids such necklaces do not exist.
Lemma 3.3.28. Closures of orbits of periodic domains do not form continuous chains.
Proof. We show that the intersection of an orbit of a periodic domain with Y Y + (1, 0) is disconnected. Let Z Y be a periodic domain. ∪ ⊂ Because the first return map to Y fixes Z, the first return map to
86 Y + (1, 0) fixes the image of Z in this strip. Hence there is exactly one image of Z in the strip Y + (1, 0).
To find the image of Z in Y + (1, 0) we take the image of T (T −(Z))
− − where T stands for Tt taken collectively to define a map from Y to Y −. Clearly this image lies in the desired strip. From the definition of T − we have T −(Z) = [Z] + (0, [2(k + 1)α] 2(k + 1)) where [z] = − [(z , z )] = (z , [z ]) for z Z, and Z = 1/2 I . So T (T −(Z)) = 1 2 1 2 ∈ ∪t=−1/2 k,4 (1, 2(k + 1) [2(k + 1)α]) [Z]. Taking into account that [Z] = Z − − − (0, 2(k + 1)), we have T (T −(Z)) is a central symmetry of Z about the point (1/2, 2(k + 1) [2(k + 1)α]/2). Since Z lies above y = 2(k + 1) − by definition and we just showed that T (T −(Z)) lies below the line y = 2(k + 1) [2(k + 1)α]/2] the closures of the two do not intersect. −
It is now interesting to ask whether non-periodic orbits of TY are bounded. The answer surprisingly turns out to be – yes.
N Let S(x, N) = i=0 ωx(i). Clearly the orbit of x is bounded if and only if sup S(x, NP) < . Note that S(x, N) is always bounded below N ∞ because the orbit lives on a half-line.
Theorem 3.3.29. Every orbit of TY is bounded.
Proof. We fix x and t and proceed by induction on k. Using the algo-
t rithmic description of ωx from Lemma 3.3.26 we have that for x such that kt(x) = 0 initial separation between the points t1 and t−1 used in
87 the algorithm is 0. It is now easy to see that because the only way one of the points can get ahead of the other is by traveling within the dis- tinguished interval, J, and because the other point has to stand still at the same time, one point can not separate from the other by more than a constant C equal to the length of the longest block of 1’s for the corre- sponding rotation sequence. Therefore, a 1 is never separated by more than C places from the corresponding -1 in ωx and so sup S(x, N) < C for all x such that kt(x) = 0
Now suppose that sup S(x, N) is finite for x + 2K for which kt(x + 2K) = K and consider what happens for x + 2(K + 1). Incrementing x by 2 shifts t2 by one step along its orbit. This follows from lemma 3.3.26.
t Consider now how ωx is transformed. Shifting t2 forward along the orbit corresponds to shifting all 1’s left by one position. There are two special situations.
First when a block of -1’s is separated from a block of 1’s by a single
t 0. From the above description of ωx it is easy to see that a block of
1’s can never follow a block a block of -1’s directly because t1 has ”the right of way” when traversing the distinguished segment. Hence in this case a block of 1’s can not simply be shifted left because the new code will not correspond to an orbit. It is easy to see that this situation happens when t1 enters the segment one step after t−1 exits it. Shifting t1 forward along its orbit will make both points enter the distinguished
88 segment simultaneously which will mean that in the resulting code the two blocks will switch places. That is if a block of -1’s is directly followed is separated from a block of 1’s by a single 0 then in the new code the two blocks will be adjacent and will have switched places, the single 0 will have moved to the end of the block of 1’s.
The second special situation arises when a block of -1’s directly follows a block of 1’s. This situation corresponds to the two points entering the distinguished segment simultaneously. Shifting t1 along its orbit will cause it to enter the segment first if it was ahead of t−1 in which case the first 0 after the block of -1’s will move between the two blocks as the block of 1’s is shifted to the left. If t−1 is ahead when the two points enter the distinguished segment then after the shift the two points still enter the segment simultaneously and so the blocks of 1’s and -1’s remain adjacent and both are shifted to the left by one step.
From this it is clear that any 1 or -1 doesn’t change it’s position by more than C places again. Hence S(x+2(K+1), N) S(x+2K, N)+ | | ≤ C < . It follows that for a fixed x and t every point (t, x + 2k) has ∞ a bounded orbit under TY . Since choices of x and t were arbitrary the same is true for all points of Y . Combining this with Lemma 3.2.6 it follows that every orbit whose restriction to Y is bounded is globally bounded and so we have proved the result.
89 3.3.6 Complexity
Here we compute the complexity of TY with respect to the ω-coding and obtain a bound on the growth of complexity of the vertex coding defined in the introduction. We will denote the complexity function of the ω-coding, as defined in the introduction, by p(n). For trapezoidal outer billiards we have :
Theorem 3.3.30. For trapezoidal outer billiard with α R Q ∈ \
2n2α p(n) ≤ δ(n) where δ(n) = min 2kα mod 1 . k≤n{ }
Proof. We first compute the complexity for Tt for a fixed t. Clearly, ωt (1...n) = ωt (1...n) iff πω(ωt )(1...n) = πω(ωt )(1...n) for i = 1, 1. x y i x i y − The later by Lemma 3.3.24 are rotation sequences. By Lemma 3.3.31 there are 2n distinct rotation sequences of length n for a fixed θ and
I, which is the case for Tt. From Lemma 3.3.26 we know that each
ω t of the rotation sequences πi (ωx)(1...n) is determined by πi(x). Since the orbits of R2α are dense every rotation sequence of length n can be realized as a πω(ωt )(1...n) for some x X . Hence there are (2n)2 i x ∈ t t possible ωx(1...n). Next we compute the number of distinct sets of length n rotation sequences as the distinguished interval J = [1 α(t 1/2) 2α, 2 2α] t − − − − varies for 1/2 < t < 1/2. By Lemma 3.3.32 we have that this number −
90 is at most ( J J )/δ(n)+1. So there are at most α/2δ(n) such | 1/2|−| −1/2| sets.
Combining the two computations we have the desired result.
Lemma 3.3.31. The complexity of a rotation sequences with θ R Q ∈ \ and θ and J rationally independent is 2n. | |
Proof. We proceed by induction on n. Clearly the statement is true for n = 1.
Suppose it is true for n. The set J = x J ω (1...n) = ω is ω { ∈ | x } n −i C an interval since J = R (J x ) where J = I and J = I . For ω ∩i=1 θ ω (i) 1 0 n + 1 every J such that J R−1(I) = J or J R−1(IC ) = J will ω ω ∩ θ ω ω ∩ θ ω have a unique extension and so will contribute 1 to p(n+1). There will be exactly 2 J such that J R−1(I) = J each of which will have two ω ω ∩ θ 6 ω extensions one by a 1 and one by a 0. These correspond to Jω which receive the endpoints of I under R−1 since by rational independence of θ and I all preimages of the endpoints are distinct. Hence the | | complexity for n + 1 will be 2n + 2 = 2(n + 1)
Lemma 3.3.32. Two sets of length n rotation sequences for θ, J and θ, J 0 are indistinguishable if
kδ < J , J 0 < (k + 1)δ | | | |
91 where δ = min kθ mod 1 . k≤n{ }
Proof. Pick a rotation sequence from the first set and choose a realiza- tion K, i.e. a length n orbit of rotation by θ of a point φ. Suppose without loss of generality that J 0 < J . We can translate J 0 inside | | | | 0 J so that every point of K inside J will be inside J . Indeed, let d1 and d2 be the distances between the endpoints of J and the leftmost and rightmost points of K J respectively, then d + d δ since δ ∩ 1 2 ≥ is the smallest distance between any two points of K. On the other hand, J J 0 < δ and so the distance between the extreme points of | | − | | K J is less than J 0 . Therefore, J 0 can be positioned inside J so that ∩ | | every point of K J is also in J 0 and so the corresponding rotation ∩ sequences will be identical. Since the set of length n rotation sequences is invariant under translations of J 0 we are done.
Next we will prove a lemma that will relate complexity of the ω- coding to the complexity of the vertex coding described in the intro- duction. Let ρ(n) be the complexity of the trapezoidal outer billiard with respect to the vertex coding.
Lemma 3.3.33. Let ∆(k) = max min 2jα 2iα and δ(k) = i≤k j=6 i | − | mini≤k2iα then
n p(k) ρ(n) Cn2 ≤ (k 1)2 δ−1 ∆(k) Xk=2 − ◦ 92 where C is a constant that depends only on α
Proof. Fix n and consider the orbit segments contributing to ρ(n). Since every orbit of T 2 intersects Y every orbit (p) has a minimal O extension ˜(p) to an orbit of T 2 beginning and ending in Y . We par- O tition the orbits contributing to ρ(n) according to the winding number of ˜(p). Notice that this winding number is always less than n. O Observe that for a fixed n it is enough to consider orbit segments originating within some bounded distance R from the origin because very distant orbits have vertex codes composed of long blocks of alter- nating vertices corresponding to long traversals of sectors Dij. Consider the orbits that have winding number 1. Let l be such that
∗ the shortest side, measured in units of the corresponding vij, of P (p) is l Op . Then the most distant point of Y whose winding number 1 orbit | | segment can contribute to ρ(n) must satisfy l Op < n/2. So Op < | | | | n/2l The shortest orbit segment with winding number 1 contributing to ρ(n) will satisfy ˜(p) = (p) and so will be itself of length n. Hence O O the nearest orbit segment contributing to ρ(n) will satisfy L Op > n/2, | | where L is such that perimeter of P ∗(p), measured in units of the corresponding v ’s, is L Op . So Op > n/2L. ij | | | |
Thus only orbit segments originating in Y between r1 = n/2l and
0 r1 = n/2L (with winding number 1) will contribute to ρ(n). Each such orbit segment will contribute at most 2Lr1 distinct words to ρ(n). It remains to compute the number of such distinct orbit segments, which
93 is equal to the number of domains of continuity of TY . Clearly each ω 1-word corresponds to a different orbit segment but they may occur
0 with some multiplicity in the strip between r1 and r1. From Lemma
3.3.23 we have that there are at most 2 domains of continuity of TY with a given symbol per Ik so the total number of domains of continuity of T between r and r0 is (r r0 )p(1). So the total contribution from Y 1 1 1 − 1 orbit segments with winding number 1 will be
n n n p(1)(r r0 )2Lr = p(1)( )2L 1 − 1 1 2l − 2L 2l n2(L l) = p(1) − 2l2
We can now repeat the same analysis for ω k-words for every k < n since clearly ω-words longer than n can not correspond to any vertex code word of length n. As above we get two limits r = n/2L(k 1) k − 0 and rk = n/2Lk corresponding to the furthest and the nearest orbits contributing to ρ(n). To compute the multiplicity of ω k-words between these limits we look back to Lemma 3.3.26. For a fixed t for two points x and x0 to generate the same ω k-word it must be that π (x) π (x0) < | i − i | where i = 1, 1 and < max min 2jα 2iα , i.e. less than the − i≤k j=6 i | − | largest distance between neighboring points in the orbit of R2α of length k. Let ∆(k) = max min 2jα 2iα and δ(k) = min 2iα then i≤k j=6 i | − | i≤k k the distance between two domains of continuity of TY with the same ω-code is at least 2δ−1 ∆(k). Thus the contribution to ρ(n) from ◦
94 k-long ω-sequences will be
r r0 p(k) n nk p(k) k − k 2kLr = 2δ−1 ∆(k) k 2δ−1 ∆(k) 2Lk(k 1) k 1 ◦ ◦ − − 1 p(k)n2 = 2L (k 1)2 δ−1 ∆(k) − ◦
Finally, summing over all ω-words of length less than n we have
n2(L l) n2 n p(k) ρ(n) p(1) − + ≤ 2l2 2L (k 1)2 δ−1 ∆(k) Xk=2 − ◦ n p(k) = Cn2 (k 1)2 δ−1 ∆(k) Xk=2 − ◦
Thus the vertex coding complexity of the outer billiard depends on the arithmetic properties of α. We have the following corollary.
Corollary 3.3.34. Suppose the continued fraction coefficients of α are bounded then
ρ(n) Cn1+τ ≤ where C depends only on α and τ is the Diophantine type of α, i.e. α p/q > γq−τ . | − |
Proof. If continued fraction coefficients of α are bounded the denomi- nators of continued fraction approximations pn/qn satisfy qn/qn+1 > for some > 0. This follows from the relation qn+1 = qn−1 + an−2qn
95 where an is the n-th continued fraction coefficient of α, since
q q n = n qn+1 qn−1 + an−2qn q > n qn + an−2qn 1 = 1 + an−2
If qn/qn+1 is strictly bounded away from zero then it is easy to see that δ−1 ∆(k) > Ck for some k. Indeed, δ(k) = 2α max q k > q and ◦ { n{ n| n}} ∆(k) = 2α min q k < q so if q < k < q then δ−1 ∆(k) = { n{ n| n}} n−1 n ◦ q . Since q /q > we have δ−1 ∆(k) > k + const. Substituting n−1 n n−1 ◦ expression for δ(k) into the bound in Lemma 3.3.30 and substituting this and the previous bound into the bound in Lemma 3.3.33 we obtain
n k ρ(n) C n2 ≤ 0 k−τ+1(k 1)2 2 − Xn C n2 kτ−2 ≤ 1 2 X C n1+τ ≤ 2
where Ci are constants that depend only on α.
96 Chapter 4
Numerical Explorations
4.1 Introduction
In this chapter we present several numerical studies of outer billiard dynamics. We begin with numerical exploration of chaotic behavior in outer billiards of the type constructed in Chapter 2. In particular we look at outer billiards made using the area construction out of regu- lar and non-regular pentagons. Numerical simulations seem to suggest that the hyperbolic behavior inside the invariant polygonal domain is common for this type of billiards. These studies also appear to sug- gest that such examples are typically ergodic. We make the following plausible conjectures
Conjecture 4.1.1. A typical outer billiard made using the area con- struction, with sufficiently small area parameter, from a polygon has a
97 positive Lebesgue measure hyperbolic set inside the polygonal invariant region.
Conjecture 4.1.2. Dynamics on the hyperbolic set is ergodic with re- spect to Lebesgue measure.
We also look briefly at behavior near infinity for these examples and note that simulations indicate presence of invariant curves and elliptic islands in the neighborhood of infinity. This is amazing considering that near the boundary the dynamics appear to be hyperbolic. This situation is also in stark contrast to the ordinary billiard map which typically has similar behavior near both boundaries. Further, we explore some examples which appear to have unbounded orbits. We begin with an example of a semi-circle, originally con- structed by S. Tabachnikov. Numerical simulations show open domains that move to infinity in a regular fashion yet the proof that unbounded orbits really exist has proved to be elusive. At the moment this ex- ample appears to be the best candidate to answer Moser’s question. Again we state this as a conjecture with the hope that it will someday be resolved
Conjecture 4.1.3. Outer billiard about a semi-circle has unbounded orbits.
This is not the only example which seems to have unbounded orbits.
We also present a very simple quadrilateral for which numerical studies
98 show unbounded orbits following a regular pattern. We conjecture that there are polygons which have unbounded orbits of which the above mentioned quadrilateral may be a representative.
Conjecture 4.1.4. There are polygons for which outer billiards have unbounded orbits.
All simulation software was written by the author and the source code is available on-line at www.math.psu.edu/genin/thesis. Simulator for chaotic outer billiards was written in C++ using double precision arithmetic and was compiled using the GNU C compiler. The simula- tor for polygonal outer billiards was written in Mathematica 5.0 and is capable of executing exact computations by preserving numbers in the form algebraic expressions through out the computation. All simula- tions were executed on a 1.4 GHz PC running Debian Linux.
4.2 More chaotic outer billiards
We begin by looking at outer billiard obtained by an area construction from a regular pentagon. Below are the plots of 20,000 iterates of the outer billiard map for a=0.1, 0.2, 0.3, and 0.4 (Figures 4.1, 4.2) Notice that as in the case of the square there are elliptic islands near the corners of the pentagon. What makes this outer billiard much more complicated to analyze even for small a’s is the irregularity of the elliptic regions which makes it difficult to identify the hyperbolic set.
99 Figure 4.1: Regular pentagon a=0.1, 0.2
Figure 4.2: Regular pentagon a=0.3, 0.4
Another interesting observation is that as a increases more and more elliptic islands emerge and eventually other invariant curves appear. This also appears to be typical for this class of outer billiards. It seems possible to eliminate elliptic domains altogether by breaking symmetry.
For example, for an irregular pentagon with vertices (0,0), (2,0), (3,2),
100 (1,3), (0,2) the same plots look like Figures 4.3, 4.4.
Figure 4.3: Irregular pentagon a=0.1, 0.3
Figure 4.4: Irregular pentagon a=0.5, 0.8
The behavior in both cases does appear to be hyperbolic. To sup- port this statement we note that if there is indeed a positive measure hyperbolic set then one should be able to observe exponential diver-
101 gence of orbits with close initial conditions. To gauge the amount of stretching we take 100 randomly chosen pairs of points about 10−12 units apart and record distances between 20 successive iterates of the points. We then average the obtained data sets and plot the resulting function on the logarithmic scale. The resulting plots for the regular and irregular pentagons from above are in Figures 4.5 and 4.6. The plots appear to show that separation between adjacent points initially grows exponentially supporting the conjecture that there is a positive measure hyperbolic set. These plots also have striking similarity to the corresponding plot (Figure 4.7) for the ”square” outer billiard consid- ered in Chapter 2.
Figure 4.5: Regular pentagon
So the numerical simulations appear to indicate presence of a hy- perbolic set of positive measure. Moreover, numerical studies indicate that this is typical for outer billiards constructed from polygons by
102 Figure 4.6: Irregular pentagon
Figure 4.7: Square
the are construction. All together this makes Conjecture 4.1.1 rather plausible.
Surprisingly dynamics far from the table appears to be close to integrable with many invariant curves and elliptic islands as in Figure
4.8. The figure shows several orbits of the outer billiard produced from
103 the irregular pentagon around the point (100,0).
Figure 4.8: Dynamics around (100,0)
4.3 Unbounded orbits?
Numerical experiments also provide some evidence that outer billiards with unbounded orbits exist.
The first example of the half-circle was discovered through computer experiments by S. Tabachnikov. The half-circle outer billiard is discon- tinuous because of the flat segment and the discontinuity set, shown in Figures 4.9 and 4.10, has a remarkable structure. The elliptic do- mains (large white ovals) form a sort of regular lattice. Asymptotically they lie along confocal parabolas with foci at the origin. Because of
104 symmetry they are also all of odd period. It is not hard to see that or- bits of points inside the large elliptic domains are bounded. Something more interesting happens to the small satellite domains interspersed between the large ones. Each of the small domains follows a large peri- odic domain for one revolution and then ”attaches” to the next periodic domain and repeats this on and on, moving further and further away from the origin. This motion is shown in Figure 4.9 where segments of the orbit following different periodic domains are indicated by different colors.
Figure 4.9: Orbit of a satellite domain
This motion appears to persist at arbitrarily large distances from
105 the table. Figure 4.10 shows orbits of two satellite domains near the point (50,0) : the red domains are moving away from the origin while the green ones are moving toward it.
Figure 4.10: Orbits of satellite domains (50,0)
Because the motion of the satellite domains is so regular there is hope that it is possible to prove that their orbits are indeed unbounded. Regularity not withstanding the proof so far proved to be very elusive.
The regularity of motion improves as the distance from the origin in- creases and the key difficulty seems to be in obtaining estimates on dynamics that decay sufficiently quickly with distance. Another candidate for an outer billiard with unbounded orbits is a quadrilateral table with vertices (0,0), (1,1), (0,α), (-1,1), for some irra- tional α > 1 (Figure 4.11). This table is the next simplest quadrilateral
106 α
1
−1 1 Figure 4.11: Dart quadrilateral after the trapezoid considered in Chapter 3; it too almost preserves the fractional part of the x-coordinate, i.e. it takes on only two values x { 0} and 1 x for an orbit originating at (x , y ). Thus dynamics for this { − 0} 0 0 quadrilateral again reduces to an infinite interval exchange on vertical lines x = t for t satisfying t = x or 1 x for some fixed x . { } { 0} { − 0} 0 Numerical studies indicate that points of the form (1,1+) for suffi- ciently small have unbounded orbits. A plot of 2000 iterations of the point (1, 1.001) for α = √2 is shown in Figure 4.12.
One can clearly see that the orbit of the point lies along a discrete set of vertical lines and that there is a clear spiral structure to the orbit. Another surprising observation is that the orbit appears to be discrete in the vertical direction as well. Figure 4.13 shows how the distance from the origin of the restriction of the orbit to x = 1 grows with respect to the number of returns to x = 1. The graph shows
107 Figure 4.12: 2000 iterates of (1,1.001) for α = √2 clearly that the distance from the origin grows quite quickly although not monotonically which is sure to complicate any attempt at a rigorous argument.
108 Figure 4.13: Distance from the origin along x = 1
109
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113 Daniel I. Genin Curriculum Vitae [email protected] Education August 2005 Ph.D. in Mathematics, Minor in High Performance Computing Pennsylvania State University
May 1998 B.Sc. in Mathematics Pennsylvania State University Work and Research Experience 2005 Laboratory for Economic Management and Auctions Research Assistant Studied strategy evolution and equilibrium selection in the context of the • market entry game
2004 Applied Research Laboratory Research Assistant Collaborated with mechanical and electrical engineers on designing • an algorithm for stochastic modelling of sensor data
2003-04 Web Administrator for MASS program (www.psu.edu/mass)
1998-03 Teaching Assistant Prepared and delivered lectures on differential equations, calculus, probability • and algebra Used software such as Mathematica to illustrate mathematical concepts • TAed for the Mathematics Advanced Study Semester - an advanced mathematics • program for undergraduates Conference Talks ”Chaotic Outer Billiards” – Dynamical Systems Workshop (University of Maryland 2005) Honors and achievements
1999 Huskel B. Curry Fellowship 1998 Winner of Undergraduate Research Poster Competition 1997 Undergraduate Scholarship Achievement Award
114