Billiards in irrational polygonsiml.univ-mrs.fr/~troubetz/preprints/talk.zero-entropy-cirm-2017.pdf · A generalized diagonal is an orbit segment which starts and ends in a vertex

Billiards in irrational polygons

Serge Troubetzkoy

Zero Entropy System

CIRM 2017

A billiard ball, i.e., a point mass, moves inside a polygon P withunit speed along a straight line until it reaches the boundary of thepolygon, then instantaneously changes direction according to themirror law: “the angle of incidence is equal to the angle ofreflection,” and continues along the new line.

•

θ1 θ1

θ2

θ2


•

θ1 θ1

θ2

θ2


•

θ1

θ1

θ2

θ2


•

θ1 θ1

θ2

θ2


•

θ1 θ1

θ2

θ2


•

θ1 θ1

θ2

θ2

Let CP denote the set of corners of P. The billiard flow φt acts onthe phase space

YP :={

(x , ξ) ∈ (P \ CP)× S1 : ξ points into P whenever

x belongs to an edge}/ ∼,

where ∼ identifies (x , ξ) with (x , δξ) whenever x lies in the interiorof an edge and δ is the reflection in that edge.

The billiard map T : XP → XP is the first return of the flow φt tothe boundary of P

XP := YP ∩ (∂P × S1).


YP :={





XP := YP ∩ (∂P × S1).


YP :={





XP := YP ∩ (∂P × S1).

In a rectangle, if we fix the original direction ξ ∈ S1 of the particle,then the billiard flow takes only four directions {±ξ,±(ξ + π)}(for ξ = 0, π/2, π or 3π/2 this set degenerates to two directions).

This set of directions is an integral of motion. By the classicalunfolding construction we reduce the study of the billiard in arectangle to the study of the straight line flow on a flat torus.

•



•



•



•



•

The straight line flow in the torus is well understood: if the slope isrational then all orbits are periodic, while when the slope isirrational the flow is minimal and uniquely ergodic. Thus the sameis true for the billiard in a rectangle.

This construction generalizes to rational polygons, that is polygonswhere the angles between sides are rational multiples of π. It yieldsa translation surface with conical singularities. For example the(π/6, π/3, π/2) triangle unfolds to a regular octagon, withopposite sides identified.


This construction generalizes to rational polygons, that is polygonswhere the angles between sides are rational multiples of π. It yieldsa translation surface with conical singularities.

For example the(π/6, π/3, π/2) triangle unfolds to a regular octagon, withopposite sides identified.


This construction generalizes to rational polygons, that is polygonswhere the angles between sides are rational multiples of π. It yieldsa translation surface with conical singularities. For example the(π/6, π/3, π/2) triangle unfolds to a regular octagon, withopposite sides identified.

There are strong tools available to study billiards in rationalpolygons: Rauzy-Veech induction and Teichmuller flow. Manyresults have been obtained, but in this talk I will concentrate onirrational polygons.

A polygon is called irrational if it is not rational. Unlike rationalpolygons there are few techniques to prove results about billiards inirrational polygons. The main techniques are basic geometry orapproximation by rational polygons

There are strong tools available to study billiards in rationalpolygons: Rauzy-Veech induction and Teichmuller flow. Manyresults have been obtained, but in this talk I will concentrate onirrational polygons.

A polygon is called irrational if it is not rational. Unlike rationalpolygons there are few techniques to prove results about billiards inirrational polygons. The main techniques are basic geometry orapproximation by rational polygons

There are four possible kinds of results.

1) Results for all polygons.

2) Results for all irrational polygons.

3) Results for almost all polygons.

4) Results for Baire typical polygons.

Results for all polygons

, zero entropy

The metric entropy of the billiard ball map with respect toLebesgue measure is zero (1976, 1994 Sinai and 1978 Boldrighini,Keane and Marchetti)

The metric entropy of every invariant measure is zero (1987 Katok)

A topological proof that the metric entropy of every invariantmeasure is zero + topological entropy is zero via (n, ε)-separatedsets (1995 Galperin, Kruger, T.)

Zero topological entropy via generalized polygonal exchanges(1995 Gukin, Haydn)

Results for all polygons, zero entropy

























Results for all polygons, entropy and growth

Enumerate the sides of P and code billiard orbits by the sequenceof sides the hit. Consider the set of all possible “words” of lengthn, et let p(n) be the cardinality of this set.

A generalized diagonal is an orbit segment which starts and ends ina vertex. Let N(n) be the number of generalized diagonals with nlinks. The above results imply that these two quantities growsubexpontially.

For any polygon we have

p(n) =n−1∑j=0

N(j)

(2002 Cassaigne, Hubert, T., 2003 Bedaride)





p(n) =n−1∑j=0

N(j)






p(n) =n−1∑j=0

N(j)






p(n) =n−1∑j=0

N(j)


For each rational polygon, there exists a constant C > 0 such thatp(n) ≤ Cn3 (CHT,B).

(After Sinai). Consider an arbitrary polygon and a billiard word oflength n. The associated n-cell is the set of points whose orbit oflength n has this code.Then for any δ > 0 and γ > 0 there is a set G of Lebesguemeasure at least 1− γ and there exists K > 0 such that for anyn ≥ 0 there is a collection of n-cells which cover G havingcardinality at most Kn7+2δ.

Any invariant measure ν which satisfies ν(B(∂XP , ε)) ≤ Cεa forsome a > 0 will have a similar result, only the degree will dependon a.







Results for all polygons, distribution of orbits

The orbit of every non-periodic point accumulates at a vertex(1995 GKT).

Every polygon contains a minimal set (2017 Kruger, T.)Since the billiard map is not a topological dynamical system wemust give an ad hoc definition of minimality.







Let X+P ⊂ XP be the set of points with infinite forward orbit

and X−P ⊂ XP the set of points with infinite backward orbit.

DefinitionWe say that a non-empty, closed set M ⊂ XP is a billiard minimalset if

1. either M is a periodic orbit orM ∩ X±P are uncountable and dense in M,M \ (X+

P ∪ X−P ) is at most countable,

2. for each y ∈ M ∩ X±P we have O(y) = M, and

3. for each y ∈ M ∩ X+P and each ε > 0, the set

{n ≥ 0 : T ny ∈ B(y , ε)} has bounded gaps(and similarly for X−P ).

























Galperin has given examples of (aperiodic) minimal sets in somespecial irrational polygons (1983 Galperin). Consider the projectionto the angular component π2 : XP → S1. In Galperin’s examplesπ2(M) is finite.

A direction θ ∈ S1 is called exceptional if the invariant surface Sθcontains a a generalized diagonal. Note that there are onlycountably many exceptional directions.In Galperin’s examples all directions in the set π2(M) areexceptional.

We do not know if all billiard minimal sets are of this type.







Results for all irrational polygons

We call a point non-exceptional if its direction is non-exceptionaland it has infinite forward orbit.

The forward orbit of any non-exceptional point takes an infinitenumber of directions (2011, Bobok, T.).

If a forward orbit takes a finite number of directions then its orbitclosure is a minimal set (2011, Bobok, T.).

In a irrational right triangle the directions of the forward orbit ofany non-exceptional point is dense in S1.





















Results for almost every (irrational) polygon

The billiard in isometric polygons, or rescaled polygons areconjugate. Up to scaling and isometry, a triangle is determined bytwo of its angles, thus the set of triangles can be identified with anopen subset of R2, almost every refers to Lebesgue measurerestricted to this subset.

For almost every triangle, for every ε > 0, there exists a constantC > 0 such that p(n) ≤ Cen

ε(Scheglov, arXiv 2012).

Best published result

p(n) ≤ Cen√3−1+ε

(Scheglov 2013)

Scheglov’s proofs use a diophantine approximation result ofKaloshin and Rodnianski.






p(n) ≤ Cen√3−1+ε

(Scheglov 2013)







p(n) ≤ Cen√3−1+ε

(Scheglov 2013)







p(n) ≤ Cen√3−1+ε

(Scheglov 2013)







p(n) ≤ Cen√3−1+ε

(Scheglov 2013)


Results for Baire typical polygons

Consider the set XN of all N-gons. We number the angles{α1, α2, . . . , αN} and the set of lengths of sides {l1, l2, . . . , lN}clockwise. Since the billiard flow does not depend on scaling wecan normalize by setting l1 = 1. The angle α1 is determined by theother angles and the lengths l2, l3 are determined by the remainingangles and lengths. Thus we see that XN is an open subset ofR2N−4.

The billiard flow/map in the Baire typical polygon is topologicallytransitive (1975 Katok, Zemlyakov).

The billiard flow/map in the Baire typical polygon is ergodic (1986Kerchoff, Masur, Smillie).

The billiard map in the Baire typical polygon is totally ergodic(2004, T.).

The billiard map in the Baire typical polygon is topologicallyweakly mixing (2017, Bobok, T.).































The three old results are based on the fact that we know that the“goal” property holds in rational polygons for the billiard flow/maprestricted to most invariant surfaces.

In a rational polygonthe billiard flow/map is topologically transitive on an invariantsurface for all but countably many directions (1978, Boldrighni,Keane, Marchetti),they are (uniquely ergodic) on almost all invariant surfaces (1987Kerchoff, Masur, Smillie), andthe billiard map is totally ergodic on almost every invariant surface(2004 T.).

The proof of topological weak mixing is somewhat different, we donot know that in a rational polygon the billiard map istopologically weakly mixing for for almost every direction.


In a rational polygonthe billiard flow/map is topologically transitive on an invariantsurface for all but countably many directions (1978, Boldrighni,Keane, Marchetti),

they are (uniquely ergodic) on almost all invariant surfaces (1987Kerchoff, Masur, Smillie), andthe billiard map is totally ergodic on almost every invariant surface(2004 T.).



In a rational polygonthe billiard flow/map is topologically transitive on an invariantsurface for all but countably many directions (1978, Boldrighni,Keane, Marchetti),they are (uniquely ergodic) on almost all invariant surfaces (1987Kerchoff, Masur, Smillie), and

the billiard map is totally ergodic on almost every invariant surface(2004 T.).








Definition.The billiard map T is topologically transitive if T has a dense orbit.

The billiard map T is topologically weakly-mixing if T × T has adense orbit.The billiard map T is totally transitive if for each k ≥ 1 the mapT k has a dense orbit.

Definition.The billiard map T is topologically transitive if T has a dense orbit.The billiard map T is topologically weakly-mixing if T × T has adense orbit.

The billiard map T is totally transitive if for each k ≥ 1 the mapT k has a dense orbit.

Definition.The billiard map T is topologically transitive if T has a dense orbit.The billiard map T is topologically weakly-mixing if T × T has adense orbit.The billiard map T is totally transitive if for each k ≥ 1 the mapT k has a dense orbit.

Definition.The billiard map T is topologically transitive if T has a dense orbit.The billiard map T is topologically weakly-mixing if T × T has adense orbit.The billiard map T is totally transitive if for each k ≥ 1 the mapT k has a dense orbit.

Topological dynamics results.

A topological dynamical system is a pair (X ,S) where X is atopological space X and S : X 7→ X is a continuous map.

Let (X ,S) be a totally transitive topological dynamical systemswith a periodic point, then (X , S) is Li-Yorke chaotic (2000Huang, Ye).

Let (X ,S) be a totally transitve topological dynamical systemssuch that periodic points are dense in X . Then S is topologicallyweakly mixing (1997 Banks).

Let (X ,S) be a topological dynamical systems. S is topologicallyweakly mixing if and only if for all open U,V there exists n suchthat T nU ∩ U 6= ∅ and T nU ∩ V 6= ∅ (1970 Petersen).









Let (X ,S) be a totally transitve topological dynamical systemssuch that periodic points are dense in X . Then S is topologicallyweakly mixing

(1997 Banks).












Proof of Banks theorem.

Let U,V ,G ,H be non-empty open sets.

We need to show that there exists n ≥ 1 such that(S × S)n(U × V ) ∩ (G ∩ H) 6= ∅.S is transitive, so there is a k ≥ 1 such that SkU ∩G is non-emptyand thus W := U ∩ S−k(G ) is open and non-empty.Hence W contains a periodic point x , let m be its period.For j ≥ 1 one has S jm+k(x) = Sk(x) ∈ G , thus the setSmj+k(U) ∩ G contains the point x and so it is non-empty.Since S−kH is open and non-empty, and Sm is transitive, there is aj ≥ 1 such that Smj(V ) ∩ S−k(H) 6= ∅, and thusSmj+k(V ) ∩ H 6= ∅.Let n := mj + k , the two red equations combined give the desiredresult:

(S × S)n(U × V ) ∩ (G × H) 6= ∅.


Let U,V ,G ,H be non-empty open sets.We need to show that there exists n ≥ 1 such that(S × S)n(U × V ) ∩ (G ∩ H) 6= ∅.

S is transitive, so there is a k ≥ 1 such that SkU ∩G is non-emptyand thus W := U ∩ S−k(G ) is open and non-empty.Hence W contains a periodic point x , let m be its period.For j ≥ 1 one has S jm+k(x) = Sk(x) ∈ G , thus the setSmj+k(U) ∩ G contains the point x and so it is non-empty.Since S−kH is open and non-empty, and Sm is transitive, there is aj ≥ 1 such that Smj(V ) ∩ S−k(H) 6= ∅, and thusSmj+k(V ) ∩ H 6= ∅.Let n := mj + k , the two red equations combined give the desiredresult:

(S × S)n(U × V ) ∩ (G × H) 6= ∅.


Let U,V ,G ,H be non-empty open sets.We need to show that there exists n ≥ 1 such that(S × S)n(U × V ) ∩ (G ∩ H) 6= ∅.S is transitive, so there is a k ≥ 1 such that SkU ∩G is non-emptyand thus W := U ∩ S−k(G ) is open and non-empty.

Hence W contains a periodic point x , let m be its period.For j ≥ 1 one has S jm+k(x) = Sk(x) ∈ G , thus the setSmj+k(U) ∩ G contains the point x and so it is non-empty.Since S−kH is open and non-empty, and Sm is transitive, there is aj ≥ 1 such that Smj(V ) ∩ S−k(H) 6= ∅, and thusSmj+k(V ) ∩ H 6= ∅.Let n := mj + k , the two red equations combined give the desiredresult:

(S × S)n(U × V ) ∩ (G × H) 6= ∅.


Let U,V ,G ,H be non-empty open sets.We need to show that there exists n ≥ 1 such that(S × S)n(U × V ) ∩ (G ∩ H) 6= ∅.S is transitive, so there is a k ≥ 1 such that SkU ∩G is non-emptyand thus W := U ∩ S−k(G ) is open and non-empty.Hence W contains a periodic point x , let m be its period.

For j ≥ 1 one has S jm+k(x) = Sk(x) ∈ G , thus the setSmj+k(U) ∩ G contains the point x and so it is non-empty.Since S−kH is open and non-empty, and Sm is transitive, there is aj ≥ 1 such that Smj(V ) ∩ S−k(H) 6= ∅, and thusSmj+k(V ) ∩ H 6= ∅.Let n := mj + k , the two red equations combined give the desiredresult:

(S × S)n(U × V ) ∩ (G × H) 6= ∅.


Let U,V ,G ,H be non-empty open sets.We need to show that there exists n ≥ 1 such that(S × S)n(U × V ) ∩ (G ∩ H) 6= ∅.S is transitive, so there is a k ≥ 1 such that SkU ∩G is non-emptyand thus W := U ∩ S−k(G ) is open and non-empty.Hence W contains a periodic point x , let m be its period.For j ≥ 1 one has S jm+k(x) = Sk(x) ∈ G , thus the setSmj+k(U) ∩ G contains the point x and so it is non-empty.

Since S−kH is open and non-empty, and Sm is transitive, there is aj ≥ 1 such that Smj(V ) ∩ S−k(H) 6= ∅, and thusSmj+k(V ) ∩ H 6= ∅.Let n := mj + k , the two red equations combined give the desiredresult:

(S × S)n(U × V ) ∩ (G × H) 6= ∅.


Let U,V ,G ,H be non-empty open sets.We need to show that there exists n ≥ 1 such that(S × S)n(U × V ) ∩ (G ∩ H) 6= ∅.S is transitive, so there is a k ≥ 1 such that SkU ∩G is non-emptyand thus W := U ∩ S−k(G ) is open and non-empty.Hence W contains a periodic point x , let m be its period.For j ≥ 1 one has S jm+k(x) = Sk(x) ∈ G , thus the setSmj+k(U) ∩ G contains the point x and so it is non-empty.Since S−kH is open and non-empty, and Sm is transitive, there is aj ≥ 1 such that Smj(V ) ∩ S−k(H) 6= ∅, and thusSmj+k(V ) ∩ H 6= ∅.

Let n := mj + k , the two red equations combined give the desiredresult:

(S × S)n(U × V ) ∩ (G × H) 6= ∅.


Let U,V ,G ,H be non-empty open sets.We need to show that there exists n ≥ 1 such that(S × S)n(U × V ) ∩ (G ∩ H) 6= ∅.S is transitive, so there is a k ≥ 1 such that SkU ∩G is non-emptyand thus W := U ∩ S−k(G ) is open and non-empty.Hence W contains a periodic point x , let m be its period.For j ≥ 1 one has S jm+k(x) = Sk(x) ∈ G , thus the setSmj+k(U) ∩ G contains the point x and so it is non-empty.Since S−kH is open and non-empty, and Sm is transitive, there is aj ≥ 1 such that Smj(V ) ∩ S−k(H) 6= ∅, and thusSmj+k(V ) ∩ H 6= ∅.Let n := mj + k , the two red equations combined give the desiredresult:

(S × S)n(U × V ) ∩ (G × H) 6= ∅.

Topological weak mixing in generic polygonsElements of the proof:We claim that in a rational polygon, the map T × T satisfies theBanks condition for open sets which are not too small in thevertical direction.

YP :={



XP := YP ∩ (∂P × S1).

∂P

S1

0

2π


YP :={



XP := YP ∩ (∂P × S1).

∂P

S1

0

2π


YP :={



XP := YP ∩ (∂P × S1).

∂P

S1

0

2π


YP :={



XP := YP ∩ (∂P × S1).

∂P

S1

0

2π


YP :={



XP := YP ∩ (∂P × S1).

∂P

S1

0

2π


YP :={



XP := YP ∩ (∂P × S1).

∂P

S1

0

2π


YP :={



XP := YP ∩ (∂P × S1).

∂P

S1

0

2π

Let N be the greatest common denominator of the angles betweenthe sides expressed as p

qπ. But our polygon is rational. In the

interval [0, 2πN ) ⊂ S1 there is a single θ ∈ [0, 2πN ) for each invariantsurface. We draw the phase space XP with the followingrepresentation

N copies of ∂P0

2πN

Each horizontal line traversing this representation of the phasespace is an “invariant surface” of the billiard map, more preciselythe billiard map restricted to such a horizontal line is theassociated interval exchange map.

Let N be the greatest common denominator of the angles betweenthe sides expressed as p

qπ. But our polygon is rational. In the

interval [0, 2πN ) ⊂ S1 there is a single θ ∈ [0, 2πN ) for each invariantsurface. We draw the phase space XP with the followingrepresentation

N copies of ∂P0

2πN

Each horizontal line traversing this representation of the phasespace is an “invariant surface” of the billiard map, more preciselythe billiard map restricted to such a horizontal line is theassociated interval exchange map.

2) In a rational polygon periodic points are dense (1998Boshernitzan, Galperin, Kruger, T.)

3) In a rational polygon, in any direction without a saddleconnection the billiard map T is totally minimal.The argument of Banks works perfectly for any open rectanglesU,V ,G ,H which stretch from the bottom to the top.

N copies of ∂P0

2πN

2) In a rational polygon periodic points are dense (1998Boshernitzan, Galperin, Kruger, T.)3) In a rational polygon, in any direction without a saddleconnection the billiard map T is totally minimal.The argument of Banks works perfectly for any open rectanglesU,V ,G ,H which stretch from the bottom to the top.

N copies of ∂P0

2πN

We do this for a finite collection of such rectangles of the form( i2N ,

i+22N )× (0, 2πN ) where i = 0, 1, . . . , (2N − 1)× length of ∂P.

∂P

S1

0

2π

The phase space XP varies continuously as we vary P, in particularthis finite collection varies continuously.

By continuity, for all polygons sufficiently close to P theBanks/Petersen condition holds for this finite collection ofrectangles.

We denote this neighborhood by B(P, εP).

Let P denote a countable dense collection of of rational polygons,and let PN consist of the polygons in P whose gcd of angles is N.

Then the set

∞⋂M=1

∞⋃N=M

⋃P∈PN

B(P, εP)

is a dense Gδ set and for any Q in this set, and the billiard mapsatisfies the Banks condition.





Then the set

∞⋂M=1

∞⋃N=M

⋃P∈PN

B(P, εP)






Then the set

∞⋂M=1

∞⋃N=M

⋃P∈PN

B(P, εP)






Then the set

∞⋂M=1

∞⋃N=M

⋃P∈PN

B(P, εP)


Open questions

Is the billiard flow topologically weakly mixing in Baire genericpolygons?

Is the billiard map/flow weakly mixing in Baire generic polygons?

Remark: despite the results of Avila and Forni on weak mixing ofinterval exchanges and translation surface, it is unknown if thebilliard flow/map is weakly-mixing in almost every direction inrational polygons. Some partial results in this direction are know,Avila-Delecroix showed that this is the case in Veech polygonsother than genus 2, and Malaga-Sabogal, T. showed that this isthe case in Baire generic VH-polygons.

Can the billiard flow/map be mixing in an irrational polygon?

Remark: the billiard flow/map restricted to an invariant surface isnever mixing in a rational polygon (1980 Katok)

Open questions






Open questions






Open questions






Open questions






Open questions






How about topological mixing?

Remark: In 2012 Chaika gave an example of a rational polygon forwhich the billiard map is topologically mixing in certain directions.

Does every irrational polygon have a periodic orbit?

Is the billiard flow ergodic in almost every polygon?

(Forni has been giving talks announcing partial progress in thisdirection).





















Documents

Billiards in irrational polygonsiml.univ-mrs.fr/~troubetz/preprints/talk.zero-entropy-cirm-2017.pdf · A generalized diagonal is an orbit segment which starts and ends in a vertex