There is an open set of right triangles such that for each irrational
triangle in this set (i) periodic billiards orbits are dense in the phase
space, (ii) there is a unique nonsingular perpendicular billiard orbit which is
not periodic, and (iii) the perpendicular periodic orbits fill the
corresponding invariant surface

Troubetzkoy, Serge

English

arXiv

Periodic billiard orbits are dense in the phase space of an irrational right triangle. A stronger pointwise density result is also proven

Serge Troubetzkoy

CiteSeerX

Periodic billiard orbits in right triangles II

This paper had a serious error. In fixing the error the emphasis of the paper has changed completely, thus meriting a new name: ``Periodic orbits in right triangles''. I have made a new submission to arXiv with this name

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Central symmetries of periodic billiard orbits in right triangles

Numérisation de Documents Anciens Mathématiques

Periodic billiard orbits in right triangles

Periodic billiard orbits in right triangles IISerge TroubetzkoyTo cite this version:Serge Troubetzkoy. Periodic billiard orbits in right triangles II. 2005. <hal-00005014v2>HAL Id: hal-00005014https://hal.archives-ouvertes.fr/hal-00005014v2Submitted on 22 Nov 2005HAL is a multi-disciplinary open accessarchive for the deposit and dissemination of sci-entific research documents, whether they are pub-lished or not. The documents may come fromteaching and research institutions in France orabroad, or from public or private research centers.L’archive ouverte pluridisciplinaire HAL, estdestine´e au de´poˆt et a` la diffusion de documentsscientifiques de niveau recherche, publie´s ou non,e´manant des e´tablissements d’enseignement et derecherche franc¸ais ou e´trangers, des laboratoirespublics ou prive´s.ccsd-00005014, version 2 - 22 Nov 2005PERIODIC BILLIARD ORBITSIN RIGHT TRIANGLES IISERGE TROUBETZKOYAbstract. Periodic billiard orbits are dense in the phase spaceof an irrational right triangle. A stronger pointwise density resultis also proven.1. IntroductionA billiard ball, i.e. a point mass, moves inside a polygon P ⊂ R2 withunit speed along a straight line until it reaches the boundary ∂P , theninstantaneously changes direction according to the mirror law: “theangle of incidence is equal to the angle of reflection,” and continuesalong the new line. If the trajectory hits a corner of the polygon, ingeneral it does not have a unique continuation and thus by definitionit stops there.It is an open question if there exists a periodic billiard orbit in everypolygon. None the less for certain classes of polygons one can exhibitthe existence of many periodic orbits. In particular one can ask howdense the periodic orbits are. Most of the known results are aboutrational polygons, i.e. polygons for which the angles between the sidesare rational multiples of pi. The first result in this direction was that ofH. Masur who showed that the directions of periodic orbits are densein a rational polygon [Ma]. This was strengthened by Boshernitzan etal. who showed that for a rational polygon periodic orbits are dense inthe phase space [BoGaKrTr]. In this article they also showed a point-wise density result in the configuration space: in a rational polygonP there exists a dense Gδ set G ⊂ P such that for each point p ∈ Gthe orbit of (p, θ) is periodic for a dense subset of directions θ ∈ S1.Vorobets strengthened this result to show that the set G is also of fullmeasure [Vo].Recently I showed that there is an open set O of right trianglessuch that for each irrational P ∈ O the set of periodic billiard orbitsis dense in the phase space [Tr]. The main result of this article is atwofold strengthening of this result. First of all I extend the result toall irrational right triangles.Theorem 1. Periodic orbits are dense in the phase space of any irra-tional right triangle.12 Periodic billiard orbitsRemark that this density result also holds for rational right triangles[BoGaKrTr]. Next I strengthen this density to a pointwise densitystatement.Theorem 2. Suppose that P is any irrational right triangle. Thenthere exists an at most countable set B ⊂ P such that for every p ∈P\B the orbit of (p, θ) is periodic for a dense subset of directions θ ∈S1.Billiards in right triangles are well known to be equivalent to the mo-tion of two elastic point masses on a segment (see for example [MaTa]).Theorem 2 tells us that except for an at most countable set B of ini-tial positions 0 ≤ x1 ≤ x2 ≤ 1 if (x1, x2) 6∈ B then the orbit of(x1, v1), (x2, v2) is periodic for a dense set of velocities (v1, v2).Somewhat surprisingly Theorem 2 is stronger than the result of Voro-bets for rational polygons. There is a special class of rational poly-gons known as Veech polygons which are well studied, see for example[MaTa] for the definition. Combing known results on Veech polygonsand the arguments of this article yieldsProposition 3. In P is a Veech polygon then there exists an at mostcountable set B ⊂ P such that for every p ∈ P\B the orbit of (p, θ) isperiodic for a dense subset of directions θ ∈ S1.Arithmetic or square tiled polygons form a subclass of Veech poly-gons [Zo]. Let V (P ) be the set of corners of P . By definition thereare no periodic orbits passing through p ∈ V (P ). Thus Theorem 2 andProposition 3 imply that B is not empty since V (P ) ⊂ B. A simplegeometric argument showsProposition 4. If P is arithmetic then for every p ∈ P\V (P ) theorbit of (p, θ) is periodic for a dense subset of directions θ ∈ S1.Note that there is a unique continuation of the billiard orbit throughvertices with angle pi/n. Such a vertex is called regularisable. If weconsider such orbits as defined by this continuation then Proposition 4holds for V (P ) defined as the nonregularisable vertices.2. StrategyInstead of reflection a billiard trajectory in a side of P one can reflectP in this side and unfold the trajectory to a straight line. Using thisbuild an invariant flat surface as follows. Unfold P to a rhombus R.Nest unfold R to a surface S consist of a countable union {Rn : n ∈ Z}of copies of R. The interiors of the Rn are disjoint, and parallel copies ofthe same side are identified by translation (see Figure 1 where parallelsides are labelled by an integer). Let α be the smaller angle of thetriangle P . Consider orbits which start in a direction θ in R0. Thenthe label n ∈ Z corresponds to the billiard orbits in R in the directionSERGE TROUBETZKOY 3R0R1 R−11122Figure 1. The invariant surface of an irrational triangle.θn := θ+2nα. When we fix the initial direction θ we denote the surfaceby Sθ. Sometimes we will refer to the rhombus Rn simply as level n.Consider the nonsingular orbit of a point x := (s, θ) and the sequenceof labels {l(i) : i ∈ Z} corresponding to the sequence of labelled rhombiits orbit visits. The orbit is called a forward escape orbit if limi→∞ l(i) =∞ and a backward escape orbit if limi→∞ l(i) = −∞. Extend thesenotions to singular escape orbits, i.e. “orbits” which pass through one(or more) singularity which are defined by continuity from the left orright. The first step is to extend and simplify a result of [Tr].Theorem 5. In any irrational right triangle P , for each θ which isnot parallel to the hypotenuse there is exactly one (possibly singular)forward escape orbit and exactly one (possibly singular) backward escapeorbit on the surface Sθ.Call a direction perpendicular if it is perpendicular to one of the legsof the triangle or the hypotenuse. Let L⊥ be the side of the triangle inquestion. An endpoint of L⊥ is good if the perpendicular orbit of thispoint is twice perpendicular, and the second perpendicular hit is at aninterior point of L⊥. If L⊥ is a leg is then call θ end point good if theendpoint which connects L⊥ to the hypotenuse is good. If L⊥ is thehypotenuse θ is end point good if both end points of L⊥ are good.Theorem 6. Suppose P is an irrational right triangle. Fix an endpoint good perpendicular direction θ. Then all orbits on Sθ except foran at most countable collection of generalized diagonals and the uniqueforward and backward escape orbits are periodic.A special case of Theorem 6 was proven in [Tr], namely when L⊥is a leg of the triangle and the smaller irrational angle of P satisfiesα ∈ (pi/6, pi/4).Next verify thatLemma 7. Every irrational right triangle has an end point good per-pendicular direction.4 Periodic billiard orbitsThe following Corollary follows immediately by combining Theorem6 with Lemma 7.Corollary 8. Suppose that P is any irrational right triangle. Then, forone at least one of the three perpendicular directions θ, the invariantsurface Sθ is foliated by periodic orbits except for an at most countablecollection of orbits/generalized diagonals Oi.Proof of Theorem 1. The theorem follows from Corollary 8 and thefact that the surface Sθ is dense in the phase space of an irrationalpolygon. Turning to the pointwise result we need the followingLemma 9. Suppose P is an irrational polygon. Suppose that thereexists a direction θ ∈ S1 such that the invariant surface Sθ is foli-ated by periodic orbits except for an at most countable collection oforbits/generalized diagonals Oi. Then there exists an at most count-able set B ⊂ P such that for every p ∈ P\B the orbit of (p, θ) isperiodic for a dense subset of directions θ ∈ S1.Proof of Theorem 2. The theorem follows immediately by combin-ing Corollary 8 and Lemma 9. 3. ProofsAn orbit segment which begins and ends at a vertex of the polygonis called a generalized diagonal. A direction θ is called simple if thereare no generalized diagonals on Sθ.Proof of Theorem 5. First consider a simple direction θ. For simpledirections the Theorem has already been proven in [Tr], however forcompleteness and to develop the proper notation I sketch the proofhere. The billiard orbits in Rn enter either Rn−1 or Rn+1. A single orbitseparates these orbits into two sets, resp. R−n and R+n which projectivelycan be thought of as intervals (see Figure 2). Call the orbit of such aninterval a strip. All the results are obtained by considering compactregions. Let KN := R+0 ∪ R−N ∪ ∪1≤n≤N−1Rn. Continue the backwardorbit of the N − 1 separatrices in KN until the first time they hit theset R+0 ∪ R−N . Projectively the set R+0 ∪ R−N consists of two interval.This pull back procedure splits it into N + 1 intervals. Map each ofthese intervals until they return to the set R0 ∪RN . Note that this setis larger than R+0 ∪ R−N . This covers KN by a union of N + 1 strips(with disjoint interiors).Next consider the centers cn of the rhombi Rn. The orbit cn has acentrally symmetric code (labels of the rhombi it visits). In particularthis implies that each of the N − 1 centers in KN\(R+0 ∪R−N ) must bein different strips. By the central symmetry each of the strips whichSERGE TROUBETZKOY 5Rn+1 Rn−1R+nRnR−nFigure 2. Projective view.includes a center must start and stop on the same level (i.e. both onR0 or both on RN )There are 2 = N+1−(N−1) exceptional strips. A simple argument(Lemma 10 of [Tr]) shows that for simple directions there must be anorbit from level 0 to level N and another from N to 0. Thus one ofthe exceptional strips, call it E+0,N , must start on level 0 and end onlevel N while the other, E−N,0, starts on level N and ends on level 0.Taking the intersection ∩N≥0E+N,0 yields (the forward orbit of) a uniqueforward escape orbit. We can replace level 0 by an arbitrary negativeinteger to see that the forward escape orbit does not depend on level0, i.e. for each n ∈ Z the image of the forward escape orbit on level nis the forward escape orbit on level n+ 1. Similarly ∩N≥0E−0,N yields aunique backward escape orbit.In [Tr] the above argument was extended to directions perpendicularto one of the legs of the triangle formally under the condition that theangle α of the triangle satisfies α ∈ (pi/6, pi/4). Infact, as mentioned inSection 4 Extensions of [Tr] “this technical assumption guarantees thatthe orbit starting at the endpoints of L (the leg of P ) are simple periodicloops.” The proof in [Tr] used an counting argument which showed thateven if there are generalized diagonals the number of nonexceptionalstrips, i.e. those which do not contain a point central symmetry, is stilltwo.Here I give a much simple proof which holds for all irrational righttriangles for all (nonsimple) directions except for directions parallel tothe hypotenuse. Note that if the direction is parallel to the hypotenusethe invariant surface breaks into two invariant sets, the positive levelsand the negative levels. They are separated by a generalized diagonalwhich is the hypotenuse. In particular the above picture breaks down.Fix a nonsimple direction φ which is not parallel to a side of therhombus. Suppose that there are at least two distinct forward escape6 Periodic billiard orbitsorbits on Sφ. Let x1 := (s1, φ) and x2 := (s2, φ) be a point on each ofthese orbits when it passes the last time through rhombus R0. Sup-pose s1 < s2. Since almost every point is angularly recurrent there isa nonsingular point x3 := (s3, φ) ∈ R0 with s1 < s3 < s2 whose orbitreturns to R0, i.e. there is a M3 > 0 such that the angle of TM3x3 isφ. Fix N > 0 sufficiently large. Since x1 and x2 are forward escapeorbits we can find positive integers M1,M2 such that TM1x1 ∈ RN andTM2x2 ∈ RN . Remark that if the orbit segment {x1, Tx1, . . . , TM1x1},is singular, one can replace x1 by a close point x′1 whose orbit segmenthas the same labels at the orbit segment of x1. The same holds forx2. In particular the orbit segment of x′i starts on level 0 and arrivesat time Mi to level N without revisiting level 0. This remark allowsus to assume without loss of generality that the orbits of the xi arenonsingular. Let M := max{M1,M2,M3}. Fix ε > 0 such that TMis continuous when restricted to the ε-ball around xi, i = 1, 2, 3. Bythis continuity there is a simple directions θ close to φ such that onthe surface Sθ there are at least two exception strips of the form E+0,N ,a contradiction. Thus there is a unique forward escape orbit on Sφ.Similarly there is a unique backward escape orbit. Proof of Theorem 6. Simply repeat the proof of Theorem 3 of [Tr].The fact that the (implicit) assumption of this proof is verified is ex-actly the result of Theorem 5. Proof of Lemma 7. Consider the right triangle and its related rhom-bus so that the longer leg is horizontal. Suppose that the angle α be-tween the hypotenuse and this leg satisfies 2nα < pi2and (2n+1)α > pi2for some n ≥ 1. Then the vertical orbit g starting at the left endpointof this leg hits the leg again perpendicularly at an interior point (thecase n = 1, α ∈ (pi6, pi4) is illustrated in Figure 3a). Thus this leg is endpoint good for triangles for which α ∈ ∪n≥1(pi4n+2, pi4n).Next consider the orbits perpendicular to the hypotenuse. If theangle α satisfies 2nα > pi2and (2n − 1)α < pi2for some n ≥ 2 thenthe perpendicular orbits starting at the endpoints of the hypotenusehit the hypotenuse perpendicularly again at an interior point (the casen = 2 is illustrated in Figure 3b). Thus the hypotenuse is end pointgood for right triangles for which α ∈ ∪n≥2(pi4n, pi4n−2). Proof of Lemma 9. Think of Sθ as being tiled by copies of P whichcan be enumerated Pi. Fix i and consider Pi ∩ {Oj}. This set is an atmost countable collection of parallel line segments Ui,j . Let Ui = ∪jUi,j .Consider the projection pi : Sθ → P . This is a countable to one map.Enumerate the points in pi−1(p): (p1, θ1), (p2, θ2), . . . . For each p ∈ Pthe set of directions {θk} is dense in S1. By assumption each of thesedirections is the direction of a periodic orbit through p, unless pk isSERGE TROUBETZKOY 7LLLLααααα αα001gggFigure 3. g hits perpendicularly at an interior point.in some Ui,j. If the periodic directions through p are not dense thenthere exists infinitely many Ui such that p ∈ pi(Ui). Suppose there ex-ists i, i′ such that p ∈ pi(Ui) ∩ pi(Ui′). Since the line segments in pi(Ui)and pi(Ui′) are transverse this intersection is at most countable. Onecompletes the proof by taking a union over i, i′. Proof of Proposition 3. In Veech polygons there is a dense set ofperiodic directions, and each periodic direction is completely foliatedby periodic orbits except for a finite number of generalized diagonals.Apply the argument of Lemma 9 to finish the proof. Proof of Proposition 4. Consider the square, its invariant surfaceis a torus tiled by four squares. Fix a rational direction. Every pointin this direction is periodic except for generalized diagonals. Supposethat the square unfolds to a torus. Consider the tiling of the plane bysquares representing the torus. The intersection points of two general-ized diagonals have rational coordinates. Thus we conclude that anypoint in the square for which at least one coordinate is irrational liesin at most one generalized diagonal and thus satisfies the conclusionsof the proposition.Consider a rational point (p1/q, p2/q) with gcd(p1, p2, q) = 1. Tilethe plane by (1/q, 1/q) squares. Consider the orbits of slope a/b withgcd(a, b) = 1 starting at the point (p1, p2). Such a orbit either corre-spond to a periodic generalized diagonal or a periodic billiard orbit. Itis a generalized diagonal if and only if there exists an i ∈ Z such thatq|(p1 + ia) and q|(p2 + ib) (see Figure 4). Note that pi ∈ {0, 1, . . . , q}and not being in the set V (P ) is equivalent to at least one of the pidiffering from 0 or q. If p1 6∈ {0, q} and q|a then q never divides p1 + iaand thus applying the above condition yields that the orbit is periodic.Thus the periodic directions through the given point include the set8 Periodic billiard orbitsFigure 4. Solid lines are periodic directions whiledashed lines are generalized diagonals starting at thepoint (1/3, 1/3).{qa′/b : qa′ ∧ b = 1}. Similarly applying the condition in the casep2 6∈ {0, q} and q|b yields that the periodic directions include the set{a/qb′ : a ∧ qb′ = 1}. Both of these sets of directions are dense in S1.Now if P is arithmetic then the unfolded surface is a torus cover.The periodic orbits constructed above for the square lift to periodicorbits in P . References[BoGaKrTr] M. Boshernitzan, G. Galperin, T. Kru¨ger, and S. TroubetzkoyPeriodicbilliard orbits are dense in rational polygons, Transactions AMS 350 (1998)3523-3535.[GaStVo] G. Galperin, A. Stepin and Ya. Vorobets, Periodic billiard trajectories inpolygons: generating mechanisms, Russian Math. Surveys 47 (1992) 5–80.[Ma] H. Masur, Closed trajectories of a quadratic differential with an applicationto billiards, Duke Math. Jour. 53 (1986) 307–313.[MaTa] H. Masur and S. Tabachnikov, Rational billiards and flat structures, Hand-book of dynamical systems, Vol. 1A, 1015–1089, North-Holland, Amsterdam,2002.[Tr] S. Troubetzkoy, Periodic billiard orbits in right triangles, Annales de l’Inst.Fourier 55 (2005) 29–46.[Vo] Ya. Vorobets, Periodic geodesics on translation surfaces, preprint[Zo] A. Zorich, Square tiled surfaces and Teichmu¨ller volumes of the moduli spacesof abelian differentials, in Rigidity in dynamics and geometry (Cambridge,2000), 459–471, Springer, Berlin, 2002.Centre de physique the´orique, Federation de Recherches des Unitesde Mathematique de Marseille, Institut de mathe´matiques de Luminyand, Universite´ de la Me´diterrane´e, Luminy, Case 907, F-13288 Mar-seille Cedex 9, FranceE-mail address : troubetz@iml.univ-mrs.frURL: http://iml.univ-mrs.fr/~troubetz/

International audienceThere is an open set of right triangles such that for each irrational triangle in this set (i) periodic billiards orbits are dense in the phase space, (ii) there is a unique nonsingular perpendicular billiard orbit which is not periodic, and (iii) the perpendicular periodic orbits fill the corresponding invariant surface

Periodic billiard orbits in right triangle

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