ISSN 1064-5624, Doklady Mathematics, 2018, Vol. 98, No. 1, pp. 1–4. © Pleiades Publishing, Ltd., 2018. Original Russian Text © F.D. Rukhovich, 2018, published in Doklady Akademii Nauk, 2018, Vol. 481, No. 3, pp. 00000–00000.
MATHEMATICS
Outer Billiards outside a Regular Octagon: Periodicity of Almost All Orbits and Existence of an Aperiodic Orbit F. D. Rukhovich Presented by Academician of the RAS A.L. Semenov November 30, 2017
Received November 30, 2017
Abstract—The existence of an aperiodic orbit for an outer billiard outside a regular octagon is proved. Addi- tionally, almost all orbits of such an outer billiard are proved to be periodic. All possible periods are explicitly listed.
DOI: 10.1134/S1064562418050101
Let be a convex shape and p be a point outside it. Tabachnikov [1] has solved the periodicity prob- Consider the right (with respect to p ) tangent to . lems in the case of regular 3-, 4-, and 6-gons. Specifi- LetTp T() p be the point symmetric to p with cally, there is no aperiodic point for them, as well as for respect to the tangency point. 5-gons, for which aperiodic points exist, but have Definition 1. The mapping T is called an outer bil- measure zero. In his 2005 monograph [6], Tabach- liard, and is called an outer billiard table. nikov presented computer simulation results for an octagon, but noted that a rigor analysis is still not Definition 2. A point p outside is said to be peri- available and there are no results for other cases. Later, odic if there exists a positive integer n such that a regular pentagon and associated symbolic dynamics Tpn p. The minimum of such n is called the period were investigated in detail by Bedaride and Cassaigne of the point p and is denoted by per(p ) . [5] (see also [8]). Definition 3. A point p outside is said to be aperi- In [9, 10] Schwartz studied an outer billiard outside odic if it is not periodic and its trajectory is unbounded a regular octagon and related questions, but he did not in both directions. obtain solutions of the periodicity problems. Definition 4. A point p outside is said to be a The main result of this paper is the following theo- rems. boundary point if Tpn is not defined for some nZ . Theorem 1. For an outer billiard outside a regular In this paper, we assume that is a convex polygon. octagon, there exists an aperiodic point. Outer billiards were introduced by Bernhard Neu- Theorem 2. For an outer billiard outside a regular mann in the 1950s and became popular in the 1970s octagon, the periodic points outside the table form a set of thanks to Moser [2]. Outer billiards were studied by a full measure. series of authors (see, e.g., [1, 3, 4, 6]). For example, Schwartz [3] showed that the trajectory of an initial Theorem 3. The set of all possible periods of points for point can be unbounded, thus solving the Moser– an outer billiard outside a regular octagon is the union of Neumann problem posed in [2]. the following sets: n n n n This paper deals with the following periodicity 1. {12 * 9 – 4 * (–3) , 12 * 9 + 4 * (–3) , problems, which are open in the general case. 49|n nZn 0}; 1. Is there an aperiodic point for an outer billiard n outside a regular n -gon? 2. {8 8kk 8 9 | nkZn 0 k 2}; 2. What is the measure of periodic orbits of an outer 3. {24k * 9n – 4 * (–3)n + 12 * 9n|n, n, billiard outside a regular n -gon? kZnk02}; 4. {24k * 9n + 4 * (–3)n – 4 * 9n|n, Moscow Institute of Physics and Technology (State kZnk02}. University), Dolgoprudnyi, Moscow oblast, 141700 Russia The proof of the theorems is based on the fact that e-mail: [email protected] there are self-similar structures inside a shape invari-
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6 6 i A 3 A 2 The polygons form a “necklace” around and 6 6 12 A 4 6 A 1 7 7 Y 5 5 touch their neighbors at vertices; for example, AA73 , A 3 A 2 6 6 A 3 A 2 A 5 A 0 5 5 23 i 7 7 A 4 A 1 , etc. Thus, bound a connected domain, AA A 4 7 A 1 6 6 5 2 04 Y A 3 A 7 Y = T(Y ) 7 7 000111 777 A A 5 5 namely, the polygon Z := AAAAAA…AAA . 5 0 Z A 5 A 0 076107 765 7 7 5 5 A 6 A 7 C6 A 6 A 7 Below are several assertions that are straightfor- 0 0 C7 A A C5 4 4 ward consequences of Fig. 1. A 3 A 2 7 6 A 5 A 66 0 0 Z A 4 A 1 A A 4 4 0 0 5 A 4 4 A 7 ii(3)mod8 Y C0 Y C4 Y Lemma 1. i…T{0 1 7}: ( ) . 0 0 4 4 A 5 A 0 A0 A4 A 5 A 0 0 0 Z 4 4 1 A 6 A 7 C1 A2 A3 C3 A 6 A 7 Lemma 2. TZ() Z T () Z. A1 A1 C2 A3 A3 3 2 Z 3 2 Let us investigate points inside the polygon Z . A1 A1 3 3 4 1 1 A 4 3 A 1 Note that Z and T are invariant under the rotation Y 2 2 Y 1 1 A 3 A 2 3 3 A 5 A 0 A A k 2 2 5 0 through the angle k Z . Identifying the points of 1 1 A 4 A 1 3 3 A 6 A 7 2 A 6 A 7 4 2 Y 2 A 5 A 0 Z under this rotation, we can study only the quadrilat- 2 2 222 A 6 A 7 eral AAAA14 3 2 , on which the outer billiard T induces a map T'. For a further analysis, it is convenient to relabel the points inside 222 (Fig. 2). Specifically, is Fig. 1. Outer billiard outside a regular octagon: the shapes AAAA14 3 2 A1 2 2 2 , i, and Z. denoted by O ; A4 , by K ; A3 , by L ; A2 , by M ; C3 , by R ; and A2 , by Q . Let P be the intersection point of the segment OK and the ray AA32 and S be the intersection ant under an outer billiard outside a regular octagon. point of the segment OK and the ray RM. Schwartz conjectured the existence of self-similar structures in the case of 5-, 10-, 8-, and 12-gons [9], Consider the rays AA32 and AA43 . They divide but the proof is still not available. OKLM into three polygons: OPQ , PQRK , and LMR . Let us describe self-similar structures. For this pur- For each of them, T is a central symmetry with respect pose, we consider a regular octagon := to the vertices A2 , A3 , and A4 , respectively. These poly- gons are designated as , , and , respectively. AAAAAAAA01234567 with vertices numbered in a coun- W1 W2 W3 terclockwise direction (Fig. 4) and an outer billiard Lemma 3. The induced map T' is map T. Let l be a straight line through the vertices A i i 3 and A , i…01 7 . Let Ci[0 7] , be the (i) a rotation through around the intersection point (1)mod8i i 4 intersection point of the lines l(1)mod8i and l(1)mod8i , U of the bisector of the angle KOM and the bisector of i iiiiiiii and let AAAAAAAA01 234567 be an open polygon the segment AA12 for the open triangle int(W1) = symmetric to with respect to the point Ci . int(OPQ );
U T(U ) P Q PQ
T ' R R V T(V)
LLMM
Fig. 2. Induced map T ' on the quadrilateral OKLM.
DOKLADY MATHEMATICS Vol. 98 No. 1 2018 OUTER BILLIARDS OUTSIDE A REGULAR OCTAGON 3
O (ii) a rotation through around the intersection point 2 V of the bisectors of the angles KOM and LRQ for the M ' K ' L' open quadrilateral int(WPQRK2 ) int( ); and (iii) a rotation through around the intersection point U 4 P W of the segment bisector LM and the bisector of the angle Q KOM for the open triangle int(W3) = int(LRM ) . The map T' is not defined on the common boundar- R ies of the circumscribed polygons, i.e., on PQ and KR . V Lemma 4. A point xOKLM is a boundary (peri- odic, aperiodic) point with respect to the map T' if and W2 only if x is a boundary (periodic, aperiodic) point with M respect to the outer billiard map T . L Let us study T' on the quadrilateral OKLM . W1 Notation 1. Let U1 be a regular octagon with sides parallel to those of that is “inscribed” in the triangle OPQ. K Notation 2. Let be an affine transformation such 3 that ()OO and ()AU1 . We also introduce the Fig. 3. Trajectory of the transformation T ''. points KK'() , LL'() , and MM'() . Note that ()'''OKLM OK L M . The locations of all periodic shapes in OKLM can Let us define a transformation T'' playing a key role in the subsequent study. be described as follows. Inside OKLM, there exists a periodic shape with side h0 . Its location is uniquely Notation 3. Let xOKLM int( ' ' ') . Let nx be the min- determined: this is an octagon inscribed in the quadri- imum positive integer n such that Tx'(n ) int( OKLM ' ' ') . lateral PQRK. After deleting this octagon from OKLM, nx the latter break up into three quadrilaterals similar to Then Tx''( ) T' ( x ) . If there is no such nx , then Tx''( ) is not defined. OKLM with coefficient . By Lemma 6, they contain Lemma 5. Let xintOKLM (). Then the following three regular octagons with side h1 . Their locations can assertions hold: also be uniquely recovered. After deleting these three open polygons, the remaining set of points of 1. Tx'( ) is defined if and only if Tx''(()) is defined. OKLM breaks up into nine polygons similar to OKLM with 2. If Tx'( ) is defined, then ('())Tx T ''(()) x . coefficient 2 . By Lemma 6, in them there are nine Transformations of type T'' are known in the litera- 2 ture as first return maps. For example, T'' is a first regular octagons with sides hh20 . Uniquely return map of OK'' L M ' with respect to T' . First return recovering their locations and deleting them yields 27 map trajectories can be seen in Fig. 3. Note that the set quadrilaterals similar to OKLM . Repeating this oper- of points whose T' -trajectories do not cross OK'' L M ' ation an infinite number of times, at the kth step can easily be described as four regular octagons. All of k 0, we obtain 3k quadrilaterals similar to OKLM them are periodic shapes of T' . with coefficient k . By Lemma 6, they contain 3k peri- Let us determine the locations of periodic compo- k nents in OKLM . odic regular octagons with sides of length hhk 0 parallel to . The locations of these polygons are Notation 4. Let h1 be the side length of an octagon uniquely recovered. Thus, the periodic components 2 (equal to the octagon ) and h0 be the side length of look like shown in Fig. 4. h0 The following result is a straightforward conse- the octagon U 2 . Let . Consider the sequence of h1 quence of this argument and the configuration of the n N periodic components. numbers hhn 0 , n . The following result can be proved using Lemma 5. Lemma 7. The periodic points inside the shape Z form a set of full measure. Lemma 6. For any nZ, the quadrilateral k OKLM contains exactly 3 periodic shapes that are reg- Consider three periodic components C0 , C1 , and C2 ular octagons with side hn. with respective side lengths h0 , h2 , and h3 (see Fig. 4).
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O 2. If Tx'( ) is defined, then HT( '()) x T4 ( H ()) x . Let Y be the union of the quadrilateral OKLM and the octagon 2 . Lemma 10. All periodic components inside KOM can be obtained from the periodic components lying in Y P Q by using a sequence of the transformations H and T '. More formally, let C be a periodic component. Then mn R CT '( HC (0 )), where C0 is a periodic component C0 C1 lying in Y, n and m are nonnegative integers, and m 3. C2 The results of this paper were reported at the Com- binatorics on Words, Calculability, and Automata research school (CIRM, Marseille, France, February L M 3, 2017) and at the Tiling Dynamical System research school (CIRM, Marseille, France, November 21, 2017).
ACKNOWLEDGMENTS K This work was supported by the Russian Science Foundation, project no. 17-11-01337. Fig. 4. Outer billiard outside a regular octagon: periodic shapes. REFERENCES 1. S. Tabachnikov, Russ. Math. Surv. 48 (6), 81–109 They bound a quadrilateral G0 similar to OKLM . Let (1993). g be an affine transformation that maps OKLM to G0 . 2. J. Moser, Math. Intell. 1, 65–71 (1978). Let CgCii3 () and GgGiii1 () 0 . Then Ci is a 3. R. E. Schwartz, Outer Billiards on Kites (Princeton sequence of periodic components with periods tend- Univ. Press, Princeton, NJ, 2009). ing to infinity and each of the quadrilaterals Gi is 4. D. Dolgopyat and B. Fayad, Ann. Henri Poincaré 10, 357–375 (2009). bounded by three periodic components C3i , C31i , C . Let the point c be the limit of the sequence C . 5. J. Bedaride and J. Cassaigne, J. London Math. Soc. 83 32i n (2), 301–323 (2011). Lemma 8. c is an aperiodic point. 6. S. Tabachnikov, Geometry and Billiards (Am. Math. The following fact allows us to prove Theorems 2 Soc., Providence, RI, 2005). and 3 not only for the shape Z, but also for the whole 7. S. Tabachnikov, Adv. Math. 115 (2), 221–249 (1995). plane. Let T4 be a first return map T' for the angle 8. J. Bedaride and J. Cassaigne, Outer Billiards outside Regular Polygons, eprint (2011). arXiv:0912.0563. AAA223. Let H be a parallel translation such that 01 6 9. R. E. Schwartz, Outer Billiards, Arithmetic Graph, and 2 HA()11 A. the Octagon, eprint (2010). arXiv:1006.2782. Lemma 9. Let xKOMint( ). Then the following 10. R. E. Schwartz, The Octagonal PETs (Am. Math. Soc., assertions hold: Providence, RI, 2014).
1. Tx'( ) is defined if and only if THx4(()) is defined. Translated by I. Ruzanova
SPELL OK
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