Entropy of polyhedral billiard

We consider the billiard map in a convex polyhedron of $\mathbb{R}^3$, and we prove that it is of zero topological entropy.


Introduction
A billiard ball, i.e. a point mass, moves inside a polyhedron P with unit speed along a straight line until it reaches the boundary ∂P , then it instantaneously changes direction according to the mirror law, and continues along the new line.
Label the faces of P by symbols from a finite alphabet A whose cardinality equals the number of faces of P . Consider the set of all billiard orbits. After coding, the set of all the words is a language. We define the complexity of the language, p(n), by the number of words of length n that appears in this system. How complex is the game of billiard inside a polygon or a polyhedron? For the cube the computations have been done, see [B03,BH07], but there is no result for a general polyhedron. One way to answer this question is to compute the topological entropy of the billiard map.
There are three different proofs that polygonal billiard have zero topological entropy [Kat87,GKT95,GH97]. Here we consider the billiard map inside a polyhedron. We want to compute the topological entropy of the billiard map in a polyhedron. The idea is to improve the proof of Katok. Thus we must compute the metric entropy of each ergodic measure. When we follow this proof some difficulties appear. In particular a non atomic ergodic measure for the related shift can have its support included in the boundary of the definition set. Such examples were known for some piecewise isometries of R 2 since the works of Adler, Kitchens and Tresser [AKT01]; Goetz and Poggiaspalla [Goe98,GP04]. Piecewise isometries and billiard are related since the first return map of the directional billiard flow inside a rational polyhedron is a piecewise isometry.
Our main result is the following Theorem 1. 1. Let P be a convex polyhedron of R 3 and let T be the billiard map, then h top (T ) = 0.
For the standard definitions and properties of entropy we refer to Katok and Hasselblatt [HK02].

Overview of the proof
We consider the shift map associated to the billiard map, see Section 2, and compute the metric entropy for each ergodic measure of this shift. We must treat several cases depending on the support of the measure. If the ergodic measure has its support included in the definition set, then the method of Katok can be used with minor changes, see Section 3. The other case can not appear in dimension two and represent the main problem in dimension three. We treat this case by looking at the billiard orbits which pass through singularities. By a geometric argument we prove in Section 4 that the support of a such measure is the union of two sets: a countable set and a set of words whose complexity can be bounded, see Proposition 4.11 and Lemma 5.1.
If we want to generalize this result to any dimension some problems appear. Im dimension three, we treat two cases by different methods depending on the dimension of the cells. In dimension d there would be at least d − 1 different cases and actually we have no method for these cases. Moreover we must generalize Lemma 4.5 and the followings . Unfortunately this is much harder and cannot be made with computations.

Definitions
We consider the billiard map inside a convex polyhedron P . This map is defined on the set E ⊂ ∂P × PR 3 , by the following method: First we define the set E ⊂ ∂P × PR 3 . A point (m, θ) belongs to E if and only if one of the two following points is true: • The line m + R * [θ] intersects an edge of P , where [θ] is a vector of R 3 which represents θ.
Then we define E as the set Now we define the map T : Consider (m, θ) ∈ E, then we have T (m, θ) = (m , θ ) if and only if mm is colinear to [θ], and [θ ] = s[θ], where s is the linear reflection over the face which contains m .
Remark 2.1. In the following we identify PR 3 with the unit vectors of R 3 (i.e we identify θ and [θ]).
Definition 2.2. The set E is called the phase space.

Combinatorics
Definition 2.3. Let A be a finite set called the alphabet. By a language L over A we mean always a factorial extendable language: a language is a collection of sets (L n ) n≥0 where the only element of L 0 is the empty word, and each L n consists of words of the form a 1 a 2 . . . a n where a i ∈ A and such that for each v ∈ L n there exist a, b ∈ A with av, vb ∈ L n+1 , and for all v ∈ L n+1 if v = au = u b with a, b ∈ A then u, u ∈ L n . The complexity function of the language L, p : N → N is defined by p(n) = card(L n ).

Coding
We label each face of the polyhedron with a letter from the alphabet {1 . . . N }. Let E be the phase space of the billiard map and d = {d 1 . . . d N } the cover of E related to the coding. The phase space is of dimension four : two coordinates for the point on the boundary of P and two coordinates for the direction.
Let E 0 be the points of E such that T n is defined, continuous in a neighborhood for all n ∈ Z. Denote by φ the coding map, it means the map We want to compute the topological entropy of the billiard map. We define the topological entropy of the billiard map as the topological entropy of the subshift, see Definition 2. 5.
We remark that the proof of Theorem 1.1 given in [GKT95] as a corollary of their result is not complete: They do not consider the case, where the ergodic measure is supported on the boundary of φ(E 0 ).

Notations
Let Σ be the closure of φ(E 0 ), and consider the cover The cover d, when restricted to E 0 , is a partition. The sets of this cover are called n-cells. If v ∈ Σ we denote It is the closure of the set of points of E 0 such that the orbit is coded by v.
Definition 2.4. Let ξ = {c 1 , . . . , c k } be the partition of Σ given by Finally we can define the topological entropy Definition 2. 5. Consider a polyhedron of R 3 , and T the billiard map, then we define where p(n) is the number of n-cells.
This definition is made with the help of the following lemma which links it to the topological entropy of the shift.
Remark 2.7. The number of cells, p(n), is equal to the complexity of the language Σ. There are several other possible definitions (Bowen definition . . . ) but we use this one since we are interested in the complexity function of the billiard map.

Cell
We denote by π the following map: Consider an infinite word v ∈ φ(E 0 ).
Definition 2.8. We consider the elements (m, θ) of ∂P × PR 3 as vectors θ with base point m.
We say that X ⊂ ∂P × PR 3 is a strip if all x ∈ X are parallel vectors whose base points form an interval. We say that X ⊂ ∂P × PR 3 is a tube if all x ∈ X are parallel vectors whose base points form an open polygon or an open ellipse.
Now we recall the theorem of Galperin, Kruger and Troubetzkoy [GKT95], which describe the shape of σ − v : Lemma 2.9. Let v ∈ φ(E 0 ) be an infinite word, then there are three cases: Remark 2. 10. The preceding lemma shows that φ is not bijective on E 0 .
By the preceding lemma for each infinite word v the set π(σ − v ) is unique. If the base points form an interval we say that σ − v is of dimension one, and of dimension two if the base points form a polygon or an ellipse.

Geometry
First we define the rational polyhedron. Let P be a polyhedron of R 3 , consider the linear reflections s i over the faces of P .
Definition 2.12. We denote by G(P ) the group generated by the s i , and we say that P is rational if G(P ) is finite.
In R 2 a polygon is rational if and only if all the angles are rational multiples of π. Thus the rational polygons with k edges are dense in the set of polygons with k edges. In higher dimension, there is no simple characterization of rational polyhedrons, moreover their set is not dense in the set of polyhedrons with fixed combinatorial type (number of edges, vertices, faces).
An useful tool in the billiard study is the unfolding. When a trajectory passes through a face, there is reflection of the line. The unfolding consists in following the same line and in reflecting the polyhedron over the face. For example for the billiard in the square/cube, we obtain the usual square/cube tiling. In the following we will use this tool, and an edge means an edge of an unfolded polyhedron.

Related results
If P is a rational polyhedron, then we can define the first return map of the directional flow in a fixed direction ω. This map T ω is a polygon exchange (generalization of interval exchange). Gutkin Moreover if µ is any invariant measure then Buzzi [Buz01], has generalized this result. He proves that each piecewise isometrie of R n have zero topological entropy. Remark that a polygonal exchange is a piecewise isometry.

Variational principle
We use the variational principle to compute the entropy Remark that we cannot apply it to the map T since it is not continuous on a compact metric space. The knowledge of h µ (T ) does not allow to compute h top (T ). We are not interested in the atomic measures because the associated system is periodic, thus their entropy is equal to zero. We split into two cases supp(µ) ⊂ φ(E 0 ) or not. We begin by treating the first case which is in the same spirit as the argument in Katok [Kat87].
By Lemma 2.9 the dimension of σ − v can take three values.
The ergodicity of µ implies that this set either has zero measure or full measure.
Assume it is of full measure, then d − is a partition of points, and same thing for ξ − . Then ξ − is a refinement of Sξ − , this implies that those two sets are equal.
Assume it is of zero measure. Then by ergodicity there are two cases : σ − v is an interval or of dimension two for a set of full measure.
Consider a line included in the plane of the strip and orthogonal to the axis Rθ, and denote L(σ − v ) the length of the set at the intersection of the line and the strip, see Figure 2.
we conclude that the function L is a sub-invariant of S. Since µ is ergodic the function L is constant µ a.e. Thus for µ a.e v we obtain two intervals of same length, one included in the other. They are equal. We deduce • If σ − v is of dimension 2 for a positive measure set of v, by ergodicity it is of the same dimension for µ a.e v. It implies that v is a periodic word µ a.e, thus Sξ − = ξ − µ a.e.

Measures on the boundary
We will treat the cases of ergodic measures, satisfying First we generalize Lemma 2.9: Lemma 4.1. For a convex polyhedron, for any word v ∈ Σ \ φ(E 0 ) the set σ − v is connected and is a strip.
We remark that Lemma 4.1 is the only place where we use the convexity of P .
Proof. First the word v is a limit of words v n in φ(E 0 ). Each of these words v n have a unique direction θ n by Lemma 2.9. The directions θ n converge to θ, this shows that the direction of σ − v is unique. Now by convexity of P the set σ − v is convex as intersection of convex sets. By definition the projection of σ − v on ∂P is included inside an edge, thus it is of dimension less than or equal to one. This implies that the set is an interval or a point.
A priori there are several cases as dimσ − v can be equal to 0 or 1. We see here a difference with the polygonal case. In this case the dimension was always equal to zero.

Orbits passing through several edges
In this paragraph an edge means the edge which appears in the unfolding of P corresponding to v. We represent an edge by a point and a vector. The point is a vertex of a copy of P in the unfolding and the vector is the direction of the edge. We consider two edges A, B in the unfolding. Consider m ∈ A and a direction θ such that the orbit of (m, θ) passes through an edge. We identify the point m with the distance d(m, a) if a is one endpoint of the edge A. Moreover we denote by u an unit vector colinear to the edge A.  Proof. Consider the affine subspace generated by the edge A 0 and the line m + Rθ. There are two cases : • A 1 ∈ Af f (A 0 , m + Rθ). Assume A 0 , A 1 are not colinear, then the affine space generated by A 0 , A 1 is of dimension two (or one), and several points m can be associated to the same direction θ. In the case it is of dimension 2, θ is in the plane which contains A 0 , A 1 . Then there exists an affine map f which gives the equation of the plane and we obtain f (θ) = 0.
• A 1 / ∈ Af f (A 0 , m + Rθ), then the space Af f (A 0 , A 1 ) is of dimension three. If the direction is not associated to a single point then the edges A 0 , A 1 are coplanar. Thus in our case the direction is associated to a single point m. There exists a real number λ such that m + λθ ∈ A 1 . Since A 1 is an edge, it is the intersection of two planes (we take the planes of the two faces of the polyhedron). We denote the two planes by the equations h = 0; g = 0 where h, g R 3 → R. We obtain the system h(m + λθ) = 0, g(m + λθ) = 0.
Here h(x) =< v h , x > +b h where v h is a vector and < ·, · > is the scalar product and similarly for g. Then we write h(m) =< v h , mu > +b h = m < v h , u > +b h , we do the same thing for g. Since A 0 , A 1 are not coplanar the terms < v g , θ >, < v h , θ > are non null, thus we obtain the expression for λ : For a fixed θ, there can be only one point m ∈ A 0 which solves this equation, otherwise we would be in case (i). Thus we find m = F (θ) where F is the quotient of two linear polynomials : Note that F does not depend on the concrete choices of the planes h, g, but only on the edges A 0 , A 1 . We prove the last point by contradiction. If we have the same equation for two edges, it means that all the lines which pass through two edges pass through the third. We claim it implies that the three edges A 0 , A 1 , A 2 are coplanar : the first case is when A 1 , A 2 are coplanar. Then the assumption implies that the third is coplanar, contradiction. Now assume that the three edges are pairwise not coplanar. Indeed consider a first line which passes through the three edges. Call m the point on A 0 , and u the direction. Now consider a line which contains m and passes through A 1 with a different direction. Those two lines intersect A 1 , thus m and the two lines are coplanar. Since A 2 is not coplanar with A 0 , both lines can not intersect A 2 , contradiction. To finish consider the case when two edges are colinear but the third one is not colinear with either of the other two. This case can be reduced to the first case by looking at the first and third edges.
where u is an unit vector colinear to the edge A 0 .
Proof. By Lemma 4.2 each F i is the quotient of two polynomials. Consider the denominator of F i as function of θ ( we use the notations of the preceding proof). By equation ( * ) we obtain: We remark for the map F i that is orthogonal to u and to x i . Thus this vector is colinear to u ∧ x i : and : We claim that C i = C i = 1. We can choose the vectors v g i , v h i such that they are orthogonal and of norm 1. Then x i is colinear to v g i ∧ v h i and is of norm one, thus if we choose the proper orientation of x i they are equal. Then we can have Thus we deduce K i = 1. Now we compute the norm of the vector of the numerator

Thus we have |b
where a j = a j,1 and b j = b 1,j .
Now the equation (F 1 − F 2 )(θ) = 0 can be written as where a j = a j,2 and b j = b 2,j .
With the equation θ 3 = P 3 P 2 θ 2 we obtain an equation of the following form.
Thus we obtain the value of θ 1 θ 2 .
If the coefficient of θ 1 θ 2 is null we obtain an equation of the form P 2 = KP 3 . This implies that P is on a plane. It is impossible since the lines A i are non coplanar. Thus we can obtain the value of θ 1 θ 2 . Then the first line of the system gives an equation of the form where f is a homogeneous rational map of twp variables.
• Second case P 2 = 0. We obtain Corollary 4.6. Consider four edges A 0 , A 1 , A 2 , A 3 two by two non coplanar such that A 3 / ∈ S(A 0 , A 1 , A 2 ). Then the maps F 1 − F 2 , F 1 − F 3 are linearly independent.
Proof. We make the proof by contradiction. If the maps F 1 − F 2 , F 1 − F 3 are linearly dependent, it means that F 3 is a linear combination of F 1 , F 2 .
It implies that the system Thus each line which passes through A 0 , A 1 , A 2 must passes through A 3 . By preceding Lemma it implies that A 3 is in S(A 0 , A 1 , A 2 ), contradiction.

Key point
Lemma 4.7. Consider a point (m, θ) ∈ E 0 ; then the set of words v such that (m, θ) ∈ σ − v is at most countable.
For the proof we refer to [Kat87]. This proof does not depend on the dimension.

Definitions
For a fixed word v ∈ Σ \ φ(E 0 ), the set σ − v is of dimension 0 or 1 and the direction θ is unique, see Lemma 4.1. Fix a word v ∈ Σ \ φ(E 0 ), we will consider several cases: • First σ − v is an interval with endpoints a, b. For any m ∈]a, b[ we consider the set of discontinuities met in the unfolding of (m, θ). This set is independent of m ∈]a, b[ since σ − v is an interval. We denote it Disc(v, int). If the endpoint a (resp. b) is included in the interval then the orbit of (m, θ) can meet other discontinuities. We call Disc(v, a) (resp. Disc(v, b)) the set of those discontinuities.
• If σ − v is a point it is the same method as Disc(v, int), we denote the set of discontinuities by Disc(v, int).
Here there are two sorts of discontinuities. First the singularity is a point of the boundary of a face whose code contributes to v. Then the orbit is not transverse to the edge. Secondly they meet in the transversal sense. If the orbit is included in an edge, then the discontinuities met are the boundary points of that edge (and similarly if the orbit is in a face).
Definition 4.8. Let V = Σ \ φ(E 0 ) and X ⊂ V be the set of v ∈ V such that the union of the elements A i of Disc(v, int), Disc(v, a), Disc(v, b) are contained in a finite union of hyperplanes and of surfaces S(A 0 , A 1 , A 2 ).
In the following Lemma the function L refers to the width of the strip of singular orbits as it does in the proof of Lemma 3.1.
, thus the lemma follows since S is ergodic.
Let D stand for Disc(v, int), Disc(v, a), or Disc(v, b).
Remark 4. 10. For two sets A i , A j ∈ D the relation dimAf f (A i , A j ) = 2 is a transitive relation. Indeed consider three sets A i , A j , A k such that A i ∼ A j , and A j ∼ A k . Since the line m + Rθ passes through A i , A j , A k , we deduce Then we can show Proposition 4.11. The set V \ X is at most countable. Proof. By Lemma 4.9 we can assume there is a constant L ≥ 0 such that L(σ − v ) = L. Suppose first that L > 0. Suppose v ∈ support(µ). This implies that Disc(v, int) is contained in a single plane. If w ∈ support(µ) satisfies w i = v i for i ≥ 0 then Disc(w, int) is contained in the same plane. Each trajectory in φ(E 0 ) which approximates the future of v cuts this plane in a single point. Consider these sequence of approximating trajectories which converges to (m, θ). The limit of these trajectories cuts the surface at one (or zero) points. The point where it cuts the surface determines the backwards unfolding, and thus the backwards code. Thus if we ignore for the moment the boundary discontinuities the knowing the future v 0 , v 1 , v 2 , . . . determines O(n) choices of the past v −n , . . . , v −1 .
The boundary discontinuities and the case L(σ − v ) = 0 are treated analogously. Let (m, θ) = σ − v (or one of the boundary points of σ − v in the case above). By Lemma 4.9 we can assume that Disc(v, m) is contained in N planes, and that if w ∈ support(µ) satisfies w i = v i for i ≥ 0 then Disc(w, int) is contained in the same planes. Arguing as above, the point where an approximating orbit cuts these planes determines the past. Thus the future v 0 , v 1 , v 2 , . .