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This article is cited in 1 scientific paper (total in 2 paper) Outer billiards outside regular polygons: tame case Ph. D. Rukhovich Moscow Institute of Physics and Technology (National Research University), Dolgoprudny, Moscow Region
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     § 1. IntroductionThis paper is an extended version of [1] and [2]. Consider Fig. 1. Let $\gamma$ be a convex subset of $\mathbb{R}^2$ and let $p$ be a point outside $\gamma$. Using the tangent line to $\gamma$ through $p$ which lies to the right of $p$, we define $Tp \equiv T(p)$ as the point symmetric to $p$ with respect to the tangency point. Definition 1. The map $T\colon \mathbb{R}^2 \setminus \gamma \to \mathbb{R}^2 \setminus \gamma$ is called the outer billiard and $\gamma$ is called the table of the outer billiard. Here the notation $f\colon X \to Y$ does not necessarily mean that $f$ is defined on the whole of $X$. However, in the majority of cases $f$ is undefined on a set of measure zero. The inverse of this map is the ‘left’ outer billiard. We denote it by $T^{-1}$. Definition 2. A point $p \in \mathbb{R}^2 \setminus \gamma$ is said to be periodic if there is a positive integer $n$ such that $T^np=p$. The smallest such $n$ is called the period of $p$ and is denoted by $\mathrm{per}(p)$. Definition 3. A point $p \in \mathbb{R}^2 \setminus \gamma$ is said to be aperiodic if it is not periodic and its trajectory is infinite in both directions. Definition 4. A point $p \in \mathbb{R}^2 \setminus \gamma$ is said to be finite if $T^np$ is undefined for some $n \in \mathbb{Z}$. We assume throughout that $\gamma$ is a convex polygon. Outer billiards were introduced by Neumann in the 1950s and became popular in the 1970s thanks to Moser [3]. Outer billiards have been studied by a number of authors (see, for example, [4]–[9] and the monograph [10]). In particular, Schwartz [5] showed that the trajectory of an initial point can be unbounded, thus answering a question of Moser and Neumann posed in [3]. We focus our attention on the following periodicity problems, which are open in the general case: 1) Is there an aperiodic point for the outer billiard outside a regular $n$-gon? 2) Is it true that the set of periodic points is a set of full measure outside the table for the outer billiard outside a regular $n$-gon? The lattice cases $n=3,4,6$ are trivial: there are no aperiodic points and, therefore, the set of periodic points is of full measure. In Schwartz’ opinion, the cases $n=5,10,8,12$ follow these in terms of complexity. The case $n=5$ was studied in detail by Tabachnikov [4]. Here aperiodic points exist, but their measure is equal to zero. In his study of pentagons, Tabachnikov pioneered the renormalization scheme, which subsequently has became the classical method for studying transformations whose behaviour reveals self-similarity. More precisely, given a transformation $f$ of a set $X$, assume that there is a subset $X' \subset X$ such that $X'$ is ‘similar’ (for example, homothetic) to $X$ and the first return map of $f$ to $X'$ coincides with $f$ on $X$ up to ‘similarity’. The regular pentagon and the related symbolic dynamics were subsequently studied in detail by Bedaride and Cassaigne [11] (see also their monograph [12]). In particular, they gave a complete description of the language as the set of finite words arising as the codes of orbits. The outer billiard outside a regular decagon was studied in detail by the present author in [13]. This case is similar to $n=5$. There is an analogous relation (see [11] and [13]) between the outer billiards outside a regular $n$-gon and a regular $2n$-gon, where $n$ is any odd number greater than or equal to three. In [13] we obtained the analogous set of all possible periods in the case $n= 10$. This was a cumbersome list of sequences of numbers. After an improvement of the exposition (resulting from a discussion with Vereshchagin and Shabanov), this set turned out to be just the set of all positive integers divisible by five. On the one hand, it is clear that all the sequences in that set consist of multiples of five. On the other hand, this set contains $5$, $10$ and the sequences $10k+15$, $20k+20$ and $20k+30$, whose union is the set of all multiples of $5$ beginning with $15$. Thus, it is true that all possible periods of points in the case of a regular decagon form the set of all positive integers divisible by $5$. In the present paper we study the ‘tame cases’ for regular $n$-gons with $n=8$ and, most importantly, $n=12$. By ‘tame’ we mean that self-similarity can be found in these cases. In accordance with Tabachnikov’s scheme, this enables us to describe the outer billiard in terms of symbolic dynamics related to permutation words. There is a consensus in the community (but no proof!) that self-similarity does not exist in the other cases (except for the trivial cases $n=3,4,6$ and the cases $5$, $10$ discussed above). The study of ‘wild’ (that is, ‘non-tame’) cases reduces to studying parametric families of piecewise Euclidean transformations, which were investigated by a number of authors (see, for example, [14] and [15], as well as the monograph [16]). We merely mention these questions, which are beyond the scope of the present paper, and express optimism about the study of ‘wild cases’. The rest of the paper is structured as follows. In § 2 we introduce the main definitions and facts about outer billiards outside convex polygons and the coding of orbits as a whole. In § 3 we consider the case of regular polygons, where the outer billiard is reduced to a piecewise isometry on some angle. We also introduce several definitions related to the combinatorics of words. In § 4 we state our main results as theorems. They are proved in §§ 5–7 and § 8. Namely, in § 5 we establish a relation between the outer billiards outside a regular $n$-gon and a regular $2n$-gon for an odd $n$ as well as a reduction of the outer billiard to a piecewise isometry of a bounded set. The main idea is that of self-similarity, which holds in the general case and enables us to assert that the structure of periodic, aperiodic and finite points outside the table is in a certain sense ‘generated’ by the structure of points in the bounded set, and it suffices to study the latter. We stress that this self-similarity exists in the general case, but gives no complete description of the outer billiard since the general study of the transformations of bounded sets is still an open problem. In § 6 we perform such a study in the particular case when $n=8$. We describe the self-similarity inside the resulting bounded set and use Tabachnikov’s renormalization scheme to solve the periodicity problem for the regular octagon. The structure of § 7, devoted to the case $n=12$, is similar. The only difference is that we use rigorous computer-assisted arguments. Finally, in § 8 we find the period codes of words in the cases $n=8, 12$. We conclude § 1 by saying a few words about the relations between outer billiards and other problems. On the sphere it is known that the outer billiard is dual to the inner one. The situation on the plane is more complicated. The question of conceptual relationship between these two kinds of billiards is important, interesting and non-trivial. On the other hand, ordinary billiards with rational angles are related to interval exchange transformations. Such billiards were studied by a number of authors (see the preprint [17] and the survey [18]). It would be great to obtain a description of classes of outer billiards related to interval exchange transformations and systems of substitutions. The methods of study of outer billiards differ substantially from the approaches developed in Teichmüller dynamics, but it is all the more important to establish their relationship. This is a very important and deep problem. Topologically different problems appear to be much closer to one another when considered from the standpoint of symbolic dynamics. This may result in a breakthrough in our understanding of both problems. We refer to [19] in this connection. The author expresses his deep gratitude to Professor A. Ya. Kanel–Belov for stating the problem and for his invaluable help, to Academician of the Russian Academy of Sciences A. L. Semenov and to Professor A. I. Bufetov for fruitful discussions of the themes of this paper and for their overall support, as well as to Professor N. K. Vereshchagin and Professor D. A. Shabanov for their interest in this work and their most valuable remarks. § 2. Outer billiards outside convex polygonsIn this paper all numerical subscripts are assumed to be integers unless otherwise stated. In this section we give the main definitions concerning outer billiards outside convex polygons and state the basic facts concerning outer billiards. Some of these facts will be stated without proof. The omitted proofs can be found in [13] (our § 2 is essentially a brief version of Chap. 2 in [13]). For simplicity, we say about a transformation $f$ that it is defined as $f\colon X \to Y$ meaning that $f$ is in fact defined only on a subset of $X$ and not necessarily on the whole of $X$. Suppose that the table $\gamma\subset \mathbb{R}^2$ is an arbitrary convex $n$-gon with $n\geqslant 3$ centred at a point $O$. Let $A_0A_1\dots A_{n-1}$ be its vertices enumerated anti-clockwise. The rays $A_1A_0,A_2A_1,\dots,A_0A_{n-1}$ divide $\mathbb{R}^2 \setminus \gamma$ into $n$ angles, whose vertices are those of $\gamma$. Let $V_i$, $0 \leqslant i<n$, be the angle with vertex $A_i$. An example of this notation for $n=5$ is shown in Fig. 2. Lemma 1 follows directly from the definition of an outer billiard. Lemma 1. Suppose that $p \in \mathbb{R}^2 \setminus \gamma$ and $i \in [0, n)$. The transformation $T$ is defined at $p$ and is the central symmetry with respect to $A_i$ if and only if $p \in \operatorname{int} (V_i)$. The next two lemmas are important from the point of view of the periodicity problem. The following obvious lemma is also important. Lemma 4. Let $X \subset \mathbb{R}^2$ be an arbitrary set of measure zero and let $F\,{=}\,\{f_1, f_2, \dots \}$ be a countable set of isometries of the plane. Then $\{f(x) \mid x \in X,\, f \in F\}$ is a set of measure zero. The following coding is ‘classical’ for outer billiards outside regular polygons. Definition 5. Let $p\,{\in}\,\mathbb{R}^2 \setminus \gamma$ be a periodic or aperiodic point. Then the code $\rho(p) \equiv \rho_{\gamma}(p)$ is a sequence $(\dots u_{-2}u_{-1}u_0u_1u_2 \dots)$ such that $T^i(p) \in \operatorname{int}(V_{u_i})$ for all $i \in \mathbb{Z}$. Definition 6. Let $p \in \mathbb{R}^2 \setminus \gamma$ be a finite point to which the outer billiard map $T$ (resp. $T^{-1}$) can be successively applied exactly $m$ times (resp. exactly $l$ times), where $0 \leqslant l, m \leqslant +\infty$. Then the code $\rho(p)=\rho_{\gamma}(p)$ is a sequence $(u_i)_{i \in [-l, m)}$ such that $u_i \in [0, n)$ and $u_i=k$ if and only if $T^i(p) \in \operatorname{int}(V_k)$. We also denote the elements $u_i$ of a code by $\rho(p)[i]$, and the sequence $u_lu_{l+1}\dots u_r$, $-\infty<l \leqslant r<+\infty$, by $\rho(p)[l, r]$). Definition 7. A component is a maximal set (with respect to inclusion) of points with the same code $\rho$. The component containing a point $p$ is denoted by $\operatorname{comp}(p)$. Lemma 5 follows directly from the definitions. Lemma 5. Suppose that the code $\rho(p)$ of $p$ is infinite in both directions and is of the form $\dots u_{-2}u_{-1}u_0u_1u_2 \dots$ . Then $\operatorname{comp}(p)=\bigcap_{i \in \mathbb{Z}} U_i$, where $U_i$ is equal to a) $\operatorname{int}(V_{u_0})$ when $i=0$; b) $T^{-i}(\operatorname{int}(V_{u_i}) \cap T^{i}(U_{i-1}))$ when $i>0$; c) $T^{-i}(\operatorname{int}(V_{u_i}) \cap T^{i}(U_{i+1}))$ when $i<0$. Note that every set $U_i$, $i \in \mathbb{Z}$, is the intersection of finitely many half-planes, the area of which is non-zero but may be infinite. When such an intersection is bounded, it is a convex polygon. Unbounded (resp. bounded) intersections of half-planes will be referred to as infinite (resp. finite) polygons. The following lemma describes the structure of components of periodic points. Lemma 6. If $p$ is a periodic point of period $m$, then the following assertions hold: a) $\operatorname{comp}(p)=U_{2m}$ (the set $U$ is defined in the statement of Lemma 5). b) All points in $\operatorname{comp}(p)$ are periodic and each of them has period $2m$ (possibly non-minimal). c) $\operatorname{comp}(p)$ is an open finite convex polygon whose sides are parallel to those of $\gamma$ and consist only of finite points. d) All points in $\operatorname{comp}(p)$, except possibly one of them, possess the same even period. Such an exceptional point exists if and only if $\operatorname{comp}(p)$ is a centrally symmetric polygon whose centre has an odd period $z$. In this case the periods of all remaining points are equal to $2z$. Here is another important property of periodic components. Lemma 7. Let $p$ be a periodic point. Then, for every $n \in \mathbb{N}$, either $T^n(\operatorname{comp}(p))=\operatorname{comp}(p)$ or $T^n(\operatorname{comp}(p)) \cap \operatorname{comp}(p)=\varnothing$. Let us introduce the period of a component. Definition 8. Let $p$ be a periodic point. Then the period $\mathrm{per}(\operatorname{comp}(p))$ of the component $\operatorname{comp}(p)$ is the smallest positive integer $k$ with $T^k(\operatorname{comp}(p))=\operatorname{comp}(p)$. The following lemma describes the relation between $\mathrm{per}(\operatorname{comp}(p))$ and the periods of the points in $\operatorname{comp}(p)$. Lemma 8. Let $p$ be a periodic point. Put $k=\mathrm{per}(\operatorname{comp}(p))$. Then the following assertions hold: – If $k$ is even, then all the points in $\operatorname{comp}(p)$ have period $k$. – If $k$ is odd, then: а) $\operatorname{comp}(p)$ is a centrally symmetric polygon with centre at some point $c$; b) the period of $c$ is equal to $k$ and the period of any other point is equal to $2k$. In view of the lemma, the component of a periodic point will be referred to as a periodic component. Lemma 8 enables us to find the set of all possible periods of points for an outer billiard outside $\gamma$, provided that we known the set of all possible periods of periodic components. The exact relation between these sets is given by the following lemma, which is a direct corollary of Lemma 8. Lemma 9. Let $B_c$ (resp. $B_p$) be the set of all possible periods of the components (resp. points) for the outer billiard outside a polygon $\gamma$. Then $B_p=B_c \cup \{2 \cdot l\mid l \in B_c,\, l \text{ is odd}\}$. The following lemma completes our investigation of periodic components. Lemma 10. Let $p$ be a periodic point. Then $T(\operatorname{comp}(p))\,{=} \operatorname{comp}(T(p))$. Proof. It is clear that $T(\operatorname{comp}(p)) \subset \operatorname{comp}(T(p))$ and Hence $\operatorname{comp}(T(p)) \subset T(\operatorname{comp}(p))$. Therefore, $T(\operatorname{comp}(p))=\operatorname{comp}(T(p))$. $\Box$
§ 3. Outer billiards outside regular polygons: main conceptsWe proceed to study outer billiards outside regular polygons. Throughout this section $\gamma$ is a regular polygon $A_0A_1\dots A_{n-1}$ with $n \geqslant 3$ whose vertices are enumerated anti-clockwise. We introduce the main concepts related to such outer billiards and to the coding of their orbits and consider a number of important lemmas. 3.1. The restricted transformationWe restrict the transformation $T$ in a way similar to that used in [11] and [4], but with a different final realization. To do this we need the following notation and conventions: – Let $R$ be the clockwise rotation by $2\pi/n$ around the centre of $\gamma$. – Note that $T$ is invariant under $R$, that is, $T(p)$ is defined as $T(R(p))=R(T(p))$ for each $p \in \mathbb{R}^2 \setminus \gamma$. We identify points that are mapped to each other under $R$. – Write $V'$ for the angle (with vertex $A_1$) centrally symmetric to $V_1$ with respect to $A_1$. – For every $p \in \mathbb{R}^2 \setminus \gamma$, let $k_p$ be the smallest $k \in \mathbb{Z}_{\geqslant 0}$ such that $R^k(p) \in V'$, and let $R'(p)$ be equal to $R^{k_p}(p)$. In other words, $R'(p)$ is a representative of the equivalence class of $p$ in $V'$ (unless $p$ lies on the extension of one of the sides). We now define the desired restriction. Definition 9. The induced transformation $T'\colon V' \to V'$ is the transformation induced by the identification of points with respect to $R$ and constructed as follows. Let $p \in \operatorname{int}(V')$. Then: a) $T'(p)$ is defined if and only if $T(p)$ is defined; b) if $T'(p)$ is defined, then $T'(p)=R'(T(p))$. For convenience, we assume that $T'$ is undefined on $\partial V'$. We regard a point $p \in V'$ as finite (periodic, aperiodic) with respect to $T'$ in exactly the same sense as $p \in \mathbb{R}^2 \setminus \gamma$ with respect to $T$ considered earlier. The following lemma is clear from the structure of the identification. Lemma 11. A point $p \in V'$ is finite (periodic, aperiodic) with respect to $T'$ if and only if $p \in V'$ is finite (periodic, aperiodic) with respect to $T$. In this case the points $R^k(p)$, $k=0, 1, \dots, n-1$, will also be finite (periodic, aperiodic) with respect to $T$. The following lemma, which is a direct corollary of Lemma 11, reduces the solution of periodicity problems on the whole plane with respect to $T$ to their solution in $\operatorname{int}(V')$ with respect to $T'$. Lemma 12. 1) There is a point aperiodic with respect to $T$ if and only if $\operatorname{int}(V')$ contains a point aperiodic with respect to $T'$. 2) The set of periodic points with respect to $T$ is a set of full measure outside $\gamma$ if and only if the set of periodic points with respect to $T'$ is of full measure in $\operatorname{int}(V')$. To reduce the problem of finding the periods to the case of the transformation $T'$, we introduce the induced code with respect to $T'$. Definition 10. Let $p \in V'$ be a periodic or aperiodic point. Then the induced code $\rho'(p) \equiv \rho'_{\gamma}(p)$ is a sequence $(\dots v_{-2}v_{-1}v_0v_1v_2\dots)$ such that $T'^i(p) \in \operatorname{int}(V_{v_i+1})$ for all $i \in \mathbb{Z}$. Note that the codes $\rho$ and $\rho'$ are related in the following way. Lemma 13. Let $p \in V'$ be a periodic or aperiodic point. Then $\rho'(p)[i]=(\rho(p)[i]-\rho(p)[i-1]) \, \operatorname{mod} n$ for all $i \in \mathbb{Z}$. Proof. The structure of $T$, $T'$ and $R'$ is such that if $T(q_1)=q_2$, then $T'(R'(q_1))= R'(q_2)$, and $R'(R(q_1))=R'(q_1)$. Note that for any non-finite point $q$, replacing $q$ by $R(q)$ reduces all the values of the code $\rho(q)$ by one modulo $n$ and does not change the values of the code $\rho'(R'(q))$. Fix an arbitrary $i \in \mathbb{Z}$, and put $k=k_{T^i(p)}$ and $p'=R^k(p)$. We have $\rho'(R'(p'))[i]=\rho'(p)[i]$ and On the other hand, $T^i(p') \in \operatorname{int}(V')$ and $T^{-1}(\operatorname{int}(V'))= \operatorname{int}(V_1)$. Hence, by the definition, $\rho(p')[i-1]=1$ and $\rho(p')[i]=1+ \rho'(R'(p'))[i]=\rho(p')[i-1]+\rho'(p)[i]$. It follows that $(\rho(p)[i]-\rho(p)[i-1]) \, \operatorname{mod} n=\rho'(p')[i]$. Since $i$ is arbitrary, this completes the proof. $\Box$ In what follows, by a ‘code’ we mean $\rho'$ unless otherwise stated. Periodic components inside $V'$ with respect to $\rho$ and $\rho'$ are clearly the same (because the code $\rho(p)$ of any non-finite point $p \in V'$ can be recovered uniquely from $\rho'(p)$ and vice versa). This enables us to define the period $\mathrm{per}'(\operatorname{comp})$ as the period of the periodic component $\operatorname{comp} \subset V'$ with respect to $T'$. All the elements of $\rho'(p)$ are integers in the interval from $1$ to $n'$, where $n'=\lceil n/2 \rceil$. We shall encode not only the orbit as a whole, but also its period. Definition 11. Let $p \in \mathbb{R}^2 \setminus \gamma$ (resp. $p\in V'$) be a periodic point with period $l$ (resp. $l'$) with respect to $T$ (resp. $T'$). Then the period code $\rho_{\mathrm{per}}(p)$ (resp. $\rho'_{\mathrm{per}}(p)$) is the word $\rho(p)[0, l-1]$ (resp. $\rho'(p)[0, l'-1]$). The following two lemmas are also simple generalizations of Lemmas 18 and 19 in [13], and their proofs are completely analogous to those in [13]. Lemma 14. Let $p \in V'$ be a periodic point with $\mathrm{per}'(p)=m$. Then $\mathrm{per}(p)=m \cdot (n/\operatorname{\textrm{GCD}}(s, n))$, where $s=\sum_{i=1}^{m} \rho'(p)[i]$. Lemma 15. Let $p \in (\subset) V'$ be a periodic point (component) with $\mathrm{per}'(p)=m$. For every $j=1, 2, \dots, n'$, let $a_j$ be the number of occurrences of $j$ in the sequence $\rho'(p)[1, m]$. Then 3.2. Definitions related to the combinatorics of wordsWe introduce some definitions based on [20]. Let $A$ and $B$ be finite sets, referred to as alphabets. Definition 12. A finite word of length $l \in \mathbb{Z}_{\geqslant 0}$ over the alphabet $A$ is a sequence $u_0u_1\dots u_{l-1}$, where $u_i \in A$ for $i=0, 1, \dots, l-1$. The set of all finite words over $A$ is denoted by $A^*$. Definition 13. A doubly infinite word over the alphabet $A$ is a sequence $(u_i)_{i \in \mathbb{Z}}=\dots u_{-2}u_{-1}u_0u_1u_2 \dots $, where $u_i \in A$ for all $i \in \mathbb{Z}$. The set of all doubly infinite words over $A$ is denoted by $A^{\mathbb{Z}}$. One can also define words that are infinite in one direction, but doubly infinite words suffice for our purposes. Note that the codes of periodic and aperiodic points are such words. Namely, the following assertions hold: – If $p \in \mathbb{R}^2 \setminus \gamma$ is a non-finite point, then $\rho(p) \in \{0,1,\dots,9\}^{\mathbb{Z}}$. – If, moreover, $p \in V'$, then $\rho'(p) \in \{0,1,\dots,4\}^{\mathbb{Z}}$. As in § 2, given any finite or infinite word $U$, we shall use the notation $U[i]$ and/or $U[l, r]$, provided that the integers $i$, $l$, $r$ do not drive us beyond $U$ and $l \leqslant r$. Definition 14. Suppose that $U,V \in A^*$, $U=u_0u_1\dots u_{l-1}$ and $V=v_0v_1\dots v_{m-1}$. Then the concatenation of $U$ and $V$ is the word $UV=u_0u_1\dots u_{l-1}v_0v_1\dots v_{m-1}$. We write $W^k$ for the concatenation of $k \in \mathbb{Z}_{\geqslant 0}$ identical words equal to $W \in A^*$. The word consisting of $k$ identical letters $a \in A$ is denoted by $a^k$. Definition 15. For every $i \in \mathbb{Z}$, let $U_i \in A^* \setminus \{\epsilon\}$ be a non-empty word of length $l_i$. Then the concatenation of the words $\dots, U_{-2}, U_{-1}, U_0, U_1, U_2, \dots$ is a doubly infinite word defined as follows. Let $m_i$, $i \in \mathbb{Z}$, be a sequence (infinite in both directions) of integers such that a) $m_0=0$; b) $m_i=m_{i-1}+l_{i-1}$ for all $i \in \mathbb{Z}_+$; c) $m_i=m_{i+1}-l_i$ for all $i \in \mathbb{Z}_-$. Then the required concatenation is a word $U \in A^{\mathbb{Z}}$ such that $U[m_i, m_{i+1}-1]=U_i$ for all $i \in \mathbb{Z}$. The following concept of substitution plays a key role in our investigation. Definition 16. Let $\sigma\colon A\to B^* \setminus \{\epsilon\}$ be an arbitrary function, where $\epsilon$ is the empty word. We extend it to a map $\sigma\colon A^* \cup A^{\mathbb{Z}} \to B^* \cup B^{\mathbb{Z}}$ by the following rules: a) $\sigma(\epsilon)=\epsilon$. b) If $W=u_0u_1\dots u_{l-1} \in A^*$, then $\sigma(W)=\sigma(u_0)\sigma(u_1)\dots \sigma(u_{l-1}) \in B^*$. c) If $W=\dots u_{-2}u_{-1}u_0u_1u_2\ldots \in A^{\mathbb{Z}}$, then
$$
\begin{equation*}
\sigma(W)= \dots \sigma(u_{-2})\sigma(u_{-1})\sigma(u_0)\sigma(u_1)\sigma(u_2) \dots\,.
\end{equation*}
\notag
$$
The resulting function $\sigma$ is called a substitution. It is sometimes convenient to say that $\sigma$ is defined on the symbols in $A$, that is, to consider $\sigma$ as a map $A \to B^* \setminus \{\epsilon\}$, having in mind its definition on $A^* \cup A^{\mathbb{Z}}$. To facilitate the search for the periods themselves we introduce two further concepts, which were used in [20]. Definition 17. Let $A=\{a_1, a_2, \dots, a_d\}$ be a finite alphabet of size $d$. Then the canonical homomorphism or Abelianization homomorphism is a map $c\colon A^* \to \mathbb{Z}^d$ such that, for every word $W \in A^*$, $c(W)$ is the column whose $i$ th coordinate is the number of occurrences of $a_i$ in $W$, where $i=1,2,\dots,d$. For example, if $A=\{1, 2, 3, 4, 5\}$ and $W=514232154141$ (or any anagram), then $c(W)=(4, 2, 1, 3, 2)$. Definition 18. Let $A=\{a_1, a_2, \dots, a_d\}$ and $B=\{b_1, b_2, \dots, b_m\}$ be finite alphabets and let $\sigma\colon A\to B \setminus \{\epsilon\}$ be an arbitrary substitution. Then the substitution matrix $M_{\sigma}$ is the matrix $\bigl(c(\sigma(a_1)), c(\sigma(a_2)),\dots, c(\sigma(a_d))\bigr)$. In other words, $M_{\sigma}=\|c_{ij}\|_{m \times d}$, where $c_{ij}$ is the number of occurrences of $b_i$ in the word $\sigma(a_j)$. For example, if $A=\{1, 2, 3\}$, $B=\{1, 2\}$ and $\sigma$ is given by $\sigma(1)=212221$, $\sigma(2)=111112$ and $\sigma(3)=22$, then $M_{\sigma}= \begin{pmatrix} 2 & 5 & 0 \\ 4 & 1 & 2 \end{pmatrix}$. Lemma 16 follows directly from the definitions. Lemma 16. Let $A=\{a_1, a_2, \dots, a_d\}$ and $B=\{b_1, b_2, \dots, b_m\}$ be finite alphabets, and let $\sigma\colon A\to B \setminus \{\epsilon\}$ be an arbitrary substitution. Then $c(\sigma(W))=M_{\sigma}(c(W))$ for all words $W \in A^*$. We conclude this section with the following useful notation. Let $Z \subseteq A^*$ be a set of words over an alphabet $A$. Then the set of all cyclic shifts of all the words in $Z$ is denoted by $\operatorname{cycle}(Z)$. For example, if $Z=\{'1234', '15151', '23', '32'\}$, then
$$
\begin{equation*}
\operatorname{cycle}(Z)=\{'1234', '2341', '3412', '4123', '15151', '51515', '23', '32'\}.
\end{equation*}
\notag
$$
§ 4. Main resultsWe state our main results in terms of the concepts introduced in § 2 and § 3. All theorems will be proved in subsequent sections. Theorem 1. For the outer billiards outside regular octagons and dodecagons, there are aperiodic points. Theorem 2. In the case of outer billiards outside regular octagons and dodecagons, the set of periodic points is a set of full measure outside the table. We introduce further notation: $\bullet$ Let $p \in (\subseteq) V'$ be a periodic point (component). Then $c(p)= c(\rho_{\mathrm{per}}(p))$ and $c'(p)=c(\rho'_{\mathrm{per}}(p))$. $\bullet$ Suppose that $Y\,{\subseteq}\, V'$. Then $P_{Y} := \{\rho'_{\mathrm{per}}(p) \mid p \subseteq Y$, $p$ is a periodic component$\}$. $\bullet$ $C_{Y}:= \{c(w) \mid w \in P_{Y} \}$. Thus, $P_{Y}$ is the set of the period codes of points in $Y$ with respect to the code $\rho'$ and $C_{Y}$ is the Abelianization of $P_{Y}$. The most important case is $Y=V'$. Then $P_{V'}$ is the set of all possible period codes with respect to $\rho'$ and $C_{V'}$ is the set of all possible Abelianizations of these codes. The following results concern the set of periodic words. Theorem 3. Let $\sigma\colon \{1, 2, 3\} \to \{1, 2, 3\}^* \setminus \{\epsilon\}$ be the substitution given by $1 \to 122323232323221$, $2 \to 122323221$ and $3 \to 111$, and let be the substitution given by $1 \to 3334$, $2 \to 334$, $3 \to 34$ and $4 \to 4$. Then, in the case of a regular octagon, 1) the set of periodic words is
$$
\begin{equation*}
P_{V'}=\operatorname{cycle}\bigl(\bigl\{\psi^l(\sigma^k(w))\mid k, l \in \mathbb{Z}_{\geqslant 0},\, w \in \{1, 2, 32\} \bigr\}\bigr) \cup \{34^l\mid l \in \mathbb{Z}_{\geqslant 0} \};
\end{equation*}
\notag
$$
2) the set of their Abelianizations is
$$
\begin{equation*}
\begin{aligned} \, C_{V'} &= \left\{\frac{1}{8}\begin{pmatrix}1+4\cdot (-3)^k+3\cdot 9^k\\-2-4\cdot (-3)^k+6\cdot 9^k\\1- 4\cdot (-3)^k+3\cdot 9^k\\0\end{pmatrix}, \frac{1}{4}\begin{pmatrix}-1+9^k\\2+2\cdot 9^k\\-1+ 9^k\\0\end{pmatrix},\right. \\ &\qquad\frac{1}{8}\begin{pmatrix}1-4\cdot (-3)^k+3\cdot 9^k\\-2+4\cdot (-3)^k+6\cdot 9^k\\1+4\cdot (-3)^k+ 3\cdot 9^k\\0\end{pmatrix}, \begin{pmatrix}0\\0\\1\\ l\end{pmatrix}, \frac{1}{2}\begin{pmatrix}0\\0\\6\cdot 9^k\\-(-3)^k+(3+6l)\cdot 9^k\end{pmatrix}, \\ &\qquad \left.\begin{pmatrix}0\\0\\2\cdot 9^k\\(2l+1)\cdot 9^k\end{pmatrix}, \frac{1}{2}\begin{pmatrix}0\\0\\6\cdot 9^k\\+(-3)^k+(3+6l)\cdot 9^k\end{pmatrix}\Biggm| k, l \in \mathbb{Z}_{\geqslant 0} \right\}; \end{aligned}
\end{equation*}
\notag
$$
3) the set of periods (with respect to $T$, not $T'$) is the set of all positive integers divisible by $4$. Theorem 4. Define the following matrices: 1) $H=C_{V'}$; 2) the set of all possible periods (also with respect to $T$) of points for the outer billiard outside a regular dodecagon is the union We omit an explicit description of $P_{V'}$ since it is very cumbersome. (Some ‘basis’ words, like $1$, $2$, $32$ in the octagon case, consist of several hundreds of symbols; their lengths are the sums of components of the vectors in the sets $F$ and $G$.) We also establish the relation between the outer billiards outside a regular $n$-gon and a regular $n/2$-gon in the general case when $n$ is even and $n/2$ is odd. Namely, we prove the following theorem. Theorem 5. Suppose that $n \in \mathbb{Z}_{\geqslant 6}$, $n$ is even and $n/2$ is odd. Let $T_n$ (resp. $T_{n/2}$) be the outer billiard outside a regular $n$-gon (resp. $n/2$-gon). Then the following assertions hold: – An aperiodic point exists for $T_n$ if and only if an aperiodic point exists for $T_{n/2}$. – The set of points periodic with respect to $T_n$ is a set of full measure outside the regular $n$-gon table if and only if the set of points periodic with respect to $T_{n/2}$ is a set of full measure outside the regular $n/2$-gon table. In other words, the periodicity problems for the outer billiards outside a regular $n$-gon and a regular $n/2$-gon are equivalent to each other provided that $n$ is even and $n/2$ is odd. Another important result is Theorem 6. Theorem 6. Suppose that $n \in \mathbb{Z}_{\geqslant 4}$, $n$ is even, $\gamma$ is a regular $n$-gon and $T_n$ is the outer billiard outside $\gamma$. Then there is a bounded domain $Z \subseteq \mathbb{R}^2 \setminus \gamma$ such that $T_n(Z) \subseteq Z$ and the following assertions hold for $T_n$: – An aperiodic point $p \in \mathbb{R}^2 \setminus \gamma$ exists if and only if an aperiodic point $p' \in Z$ exists. – The set of periodic points is a set of full measure outside $\gamma$ if and only if the set of periodic points in $Z$ is of full measure. In the case when $n=5$ such a domain and the outer billiard on it were discovered and studied in detail by Tabachnikov [4]. The relation between the outer billiards for a pentagon and a dodecagon was established by Bedaride and Cassaigne [11], [12]. Theorems 6 and 5 generalize these observations. § 5. Outer billiards outside regular polygons: reduction to a bounded caseIn this section, as in § 3, we assume that $\gamma$ is a regular polygon $A_0A_1\dots A_{n-1}$ with $n \geqslant 3$ whose vertices are enumerated anti-clockwise. We shall prove that the periodicity problems for the outer billiard outside a regular $n$-gon can be reduced to the periodicity problems for a piecewise affine map of some polygon to itself. We shall also establish a relation between the outer billiards outside a regular $n$-gon and a regular $2n$-gon for odd $n$. For regular pentagons and dodecagons, this reduction and relation were investigated in [13] (see also [11] and [21]). Here we generalize this investigation to the case of arbitrary $n$ and prove Theorems 5 and 6. 5.1. $T'$ as a ‘piecewise isometry’To describe the transformation $T'$ we define $V'_i$, $0 \leqslant i<n$, as the angles symmetric to $V_i$ with respect to the vertex $A_i$. In particular, $V'=V'_1$. It follows from the definition that for every $p \in \mathbb{R}^2 \setminus \gamma$ and $j \in [0, n)$, we have $\rho(p)[0]=j \Longleftrightarrow T(p) \in \operatorname{int}(V'_j)$ (in particular, $\rho(p)[0]$ is defined if and only if $T(p)$ is defined) and, moreover, $R(V'_j)=V'_{(j-1) \operatorname{mod} n}$ for all $j \in [0, n)$. Lemma 17. Let $p \in \operatorname{int}(V')$ be such that $T(p)$ is defined. Then $T'(p)=R^{v_0}(T(p))$, where $v_0=\rho'(p)[0]$. Proof. By the definition of $\rho'$, $T$ is equal to the central symmetry with respect to $A_{v_0+1}$ for such points $p$. Hence $T(p) \in \operatorname{int}(V'_{v_0+1})$. Since (clearly) $R^{v_0}(V'_{v_0+ 1})= V'_1=V'$, we have $R^{v_0}(T(p)) \in \operatorname{int}(V')$. Therefore, $T'(p)= R^{v_0}(T(p))$. $\Box$ Put $n'=\lceil n/2 \rceil$. Let $\alpha_i \subset V'$, $i=1,2,\dots, n'$, be the set of points $p \in V'$ such that $\rho'(p)[0]=i$. We write $P_i$, $i=1, 2, \dots, n'-1$ (resp. $Q_j$, $j=2, 3, \dots, n'$) for the intersection point of the rays $A_0A_1$ and $A_{i+1}A_i$ (resp. $A_1A_2$ and $A_{j+1}A_j$). In particular, $P_1= A_1$ and $Q_2=A_2$. Examples of the transformations $T'$ and the corresponding domains and points, some of which are to be defined below, are shown in Figs. 3 and 4 (for $n=7$) and in Figs. 5 and 6 (for $n=10$). It is clear from Lemma 1 that $\alpha_i=\operatorname{int}(V') \cap \operatorname{int}(V_{i+1})$. This yields Lemma 18. Lemma 18. If $n \geqslant 5$, then 1) $\alpha_1$ is the open triangle $P_1P_2Q_2$; 2) $\alpha_i$, $i=2, 3, \dots, n'-2$, is the open quadrangle $P_iQ_iQ_{i+1}P_{i+1}$; 3) $\alpha_{n'-1}$ is the open infinite polygon bounded by the ray $A_0A_1$, the intervals $P_{n'-1}Q_{n'-1}$ and $Q_{n'-1}Q_{n'}$, and the ray $A_{n'+1}A_{n'}$; 4) $\alpha_{n'}$ is the open angle between the rays $A_1A_2$ and $A_{n'+1}A_{n'}$. But if $n$ is equal to $3$ or $4$, then 1) $\alpha_1$ is the open infinite polygon bounded by the ray $A_0A_1$, the interval $A_1A_2$, and the ray $A_{3 \operatorname{mod} n}A_2$; 2) $\alpha_2$ is the open angle between the rays $A_1A_2$ and $A_{3 \operatorname{mod} n}A_2$. An important role in our description is played by the points $O_i$, $i=1,2,\dots, n'- 1$, which are defined as the intersection points between the bisector of the angle $V'=P_2P_1Q_2$ and the bisectors of the angles $V_{i+1}=P_iA_{i+1}P_{i+1}$. Proof. Pick $i \in [1,n'-1]$ and let $O$ be the centre of the table $\gamma$. Note that the angle $V_{i+1}$ is symmetric to the angle $V'$ with respect to the line $OQ_i$. Hence, $A_1O_i=A_{i+1}O_i$. On the other hand, $T'(O_i)=R^i(T(O_i))$ by Lemma 17. Note that $T$ maps the bisector of $V_{i+1}$ isometrically onto the bisector of $V'_{i+1}$, while $R^i$ maps the latter bisector isometrically to the bisector of $V'_1=V'$. Therefore, $T'(O_i)$ lies on the bisector of $V'$ and $A_1T'(O_i)=A_{i+1}O_i=A_1O_i$, whence $O_i=T'(O_i)$. $\Box$ We use Lemma 19 to describe the transformation $T'$. Lemma 20. 1) For every $i \in \{1,2,\dots,n'-1\}$, $T'(\alpha_i)$ is the anti-clockwise rotation by $(n-2i)\pi/n$ around the point $O_i$. 2) For every $i \in \{1,2,\dots,n'-1\}$ and for all $p \in \partial(\alpha_i)$, $T'(p)$ is undefined. Proof. Part 2 obviously follows from the definition of $T'$ since the boundaries of $\alpha_i$ lie on the extensions of the sides of $\gamma$.
To prove part 1, we note that, according to Lemma 17, $T'(\alpha_i)$ is equal to $R^{i}(T(\alpha_i))$ for all $i=1,2,\dots,{n'-1}$. Since the central symmetry is the anti-clockwise rotation by $\pi$ (we choose this direction for convenience), $T'(\alpha_i)$ is the anti-clockwise rotation by $\pi-i\cdot (2\pi/n)$ around some fixed point for every $i= 1,2,\dots,n'\,{-}\,1$. This fixed point is equal to $O_i$ by Lemma 19. $\Box$ Note that Lemma 20 does not describe the structure of $T'(\alpha_{n'})$. In order to make a correct argument in this case, we introduce more notation and definitions, which are important for our investigation. Definition 19. Let $\alpha, \beta \subset \mathbb{R}^2$ be finite or infinite polygons. We say that $\beta$ is inscribed in $\alpha$ if $\beta \subset \alpha$ and every side of $\alpha$ contains a side of $\beta$. We write $\beta_i$, $i=1, 2, \dots, n'-1$, for the open convex polygon $\operatorname{comp}(O_i)$. The polygon $\beta_{n'-1}$ is also denoted by $\beta$. Examples of the domains $\beta$ for various $n$ are shown in Figs. 4 and 6. It follows from Lemma 20 that $T'(\beta_i)= \beta_i$ and $T'$ rotates $\beta_i$ anti-clockwise by $(n-2i)\pi/n$ around the centre $O_i$ for all $i \in [1, n'-1]$. We now consider the structure of $\beta=\beta_{n'-1}$. Lemma 21. Put $n_2=2n/\operatorname{\textrm{GCD}}(n, 2)$. Then $\beta_{n'-1}$ is a regular $n_2$-gon inscribed in the infinite polygon $\alpha_{n'-1}$, and the points $P_{n'-1}$ and $Q_{n'}$ are vertices of $\alpha_{n'-1}$. Proof. By definition, $O_{n'-1}$ is the intersection point of the bisectors of $V'$ and $V_{n'}$. In view of symmetry, this point lies on the line $OQ_{n'-1}$, where $O$ is the centre of the table $\gamma$. Hence $O_{n'-1}$ is equidistant from all the four sides of $\alpha_{n'-1}$. In their turn, these sides are parallel to the sides of $\gamma$. If we consider an infinitely small regular $n_2$-gon with centre $O_{n'-1}$ and sides parallel to those of $\gamma$ and ‘blow it up’ (that is, continuously increase its size without moving its centre), then the four sides parallel to those of $\alpha_{n'-1}$ and closest to them will be ‘placed’ simultaneously on the four sides of $\alpha_{n'-1}$. Hence there is a regular $n_2$-gon centred at $O_{n'-1}$ and inscribed in $\alpha_{n'-1}$. We denote this polygon by $\delta$ and assume that it is open.
By Lemma 20, the transformation $T'$ for $\alpha_{n'-1}$ is an anti-clockwise rotation by $(n\,{-}\,2n'+2)\pi/n$. We can easily see that the angle of rotation is equal to $2\pi/n_2$. Hence $\delta$ and $\beta_{n'-1}$ are invariant under the rotation by $2\pi/n_2$. On the one hand, this means that $\delta \subset \alpha_{n'-1}$ consists only of periodic points $p$ with code
$$
\begin{equation*}
\mathrm{per}'(p)=\cdots (n'-1)(n'-1)(n'-1)(n'-1) \dots=\mathrm{per}'(O_{n'-1}),
\end{equation*}
\notag
$$
hence $\delta \subseteq \beta_{n'-1}$ by definition. On the other hand, since $T'$ rotates $\delta$ anti-clockwise ‘by one side’, it follows that any side of $\delta$ will coincide after sufficiently many such rotations with a side of $\delta$ contained in a side of $\alpha_{n'-1}$. Hence the boundary of $\delta$ consists only of finite points and, therefore, $\delta=\beta_{n'-1}$.
To complete the proof notice that the angle between the rays $A_0A_1$ and $A_{n'-1}A_{n'}$ is equal to $2(n'-1)\pi/n=(n_2-2)\pi/n_2$, which coincides with the angle of a regular $n_2$-gon. Hence two vertices of $\delta$ coincide with the vertices $P_{n'-1}$ and, by symmetry, $Q_{n'}$ of $\alpha_{n'-1}$. $\Box$ Denote the vertices of $\beta_{n'-1}$ by $B_0B_1\dots B_{n_2-1}$ in such a way that the enumeration is anti-clockwise and the points $B_0$, $B_1$ lie on the ray $A_0A_1$. Then $P_{n'-1}= B_0$. Moreover, $Q_{n'}=B_{n'+1}$ for even $n$ and $Q_{n'}=B_{n+2}$ for odd $n$. Let us explain the structure of the transformation $T'$ for $\alpha_{n'}$. In the sense of equivalence with respect to $R$, the point $Q_{n'}$ is equivalent to $P_{n'-1}=R(Q_{n'})$ and, moreover, the angle of $\alpha_{n'}$ between the rays $A_1A_2$ and $A_{n'+1}A_{n'}$ is equivalent to the angle of $R(\alpha_{n'})$ between the rays $A_{n'}A_{n'-1}$ and $A_0A_1$. The part of the ray $A_0A_1$ which is a side of $R(\alpha_{n'})$ begins with the interval $B_0B_1$. Note that $R(\alpha_{n'}) \cup \beta_{n'-1} \subset V_{n'}$. We know from the proof of Lemma 21 that $T'$ rotates $\beta_{n'-1}$ anti-clockwise ‘by one side’. Under the transformation $T \circ R^{n'-1}$, which is equal to $T'$ for $V_{n'-1}$, $R(\alpha_{n'})$ is mapped to the angle between the ray $B_0B_1$ (a part of the ray $A_0A_1$) and the ray $B_1B_2$. In view of the equivalence with respect to $R$ and since this angle lies completely inside $\alpha_{n'-1}$, we obtain the following lemma, which complements Lemma 20 and completes the description of $T'$. Lemma 22. 1) $T'(\alpha_{n'})$ is the angle between the rays $B_0B_1$ and $B_1B_2$; 2) $T'$ for $\alpha_{n'}$ is a translation for even $n$ and an anti-clockwise rotation by $\pi/n$ otherwise. Part 2 can easily be deduced from a precise calculation of the angles of rotation. 5.2. The first return map as an outer billiardBy § 5.1, there is a $T'$-invariant open regular polygon $\beta=\beta_{n'-1}$ inscribed in the angle $V'$ (we temporarily omit the subscript expressing the code of this polygon as a periodic component). Lemma 23 follows obviously from this and the description of $T'$. Lemma 23. The domain $\operatorname{int}(V' \setminus \overline{\beta})$ is the union of two disjoint $T'$-invariant open domains, of which one is bounded and touches $\gamma$ while the other one is unbounded. Let $Z'$ (resp. $Z''$) be the bounded (resp. unbounded) subdomain of $\operatorname{int}(V') \setminus \beta$ described in Lemma 23. Examples of the domains $Z'$ and $Z''$ can be found in Fig. 7. Since the bounded domain $Z'$ is $T'$-invariant, the domain $Z= \operatorname{int} \bigl(\bigcup_{i=0}^{n-1} R(\overline{Z'})\bigr)$ is bounded and $T$-invariant. We shall prove in this section that $Z$ has the properties stated in Theorem 6. In this subsection we focus on the analysis of $Z''$. Let us introduce another general concept. Definition 20. Let $X$ be a set and let $f\colon X \to X$ be a transformation. Suppose that $Y \subset X$. Then the first return map of $f$ (or with respect to $f$) to $Y$ is the following transformation $f_Y\colon Y \to Y$. Suppose that $p \in Y$. Then $f_Y(p)=f^{k}(p)$, where $k=k(p)$ is the smallest positive integer such that $f^k(p) \in Y$. If there is no such integer $k(p)$, then $f_Y$ is undefined at $p$. Here are some further technical notions. We write $\operatorname{ang}_1$ (resp. $\operatorname{ang}_2$) for the open angle $T'(\alpha_{n'})$ (resp. $\alpha_{n'}$ if $n$ is even, and $T'^{-1}(\alpha_{n'})$ if $n$ is odd). Let $h\colon \operatorname{ang}_2 \to \operatorname{ang}_1$ be the translation sending $\operatorname{ang}_2$ to $\operatorname{ang}_1$. Examples of $\operatorname{ang}_1$ and $\operatorname{ang}_2$ are shown in Fig. 8. We define a transformation $T'_{\operatorname{ang}_1 \equiv T'_{\gamma, \operatorname{ang}_1}} \colon \operatorname{ang}_1 \to \operatorname{ang}_1$ as the first return map of $T'$ to $\operatorname{ang}_1$. It turns out that $T'_{\gamma, \operatorname{ang}_1}$ coincides with the induced outer billiard outside the regular polygon $\beta_{n'}$! We now formalize this statement. Let $T_{\beta}$ be the outer billiard outside the regular polygon $\beta$ and let $T'_{\beta}\colon \operatorname{ang}_1 \to \operatorname{ang}_1$ be the corresponding transformation $T'$ induced by the outer billiard and by the equivalence of points under the rotation by $2\pi/n_2$. Definition 21. Suppose that $X_1, Y_1, X_2, Y_2 \subset \mathbb{R}^2$ and we are given some transformations $f_1\colon X_1 \to Y_1$ and $f_2\colon X_2 \to Y_2$. Let $h\colon X_1 \to X_2$ be a bijective transformation. We say that $f_1$ is conjugate to $f_2$ with respect to $h$ if the following assertions hold for any point $x_1 \in X_1$ and the corresponding point $x_2=h(x_1)$: 1) $f_1(x_1)$ is defined if and only if $f_2(x_2)$ is defined. 2) If $f_1(x_1)$ is defined, then $h(f_1(x_1))=f_2(x_2)=f_2(h(x_1))$. In the case when $X_1=X_2$, $X'_1=X'_2$ and $h$ is the identity transformation, we say that $f_1$ and $f_2$ are equal, or $f_1=f_2$. Proof. By Lemmas 20 and 22, the value of $T'_{\gamma, \operatorname{ang}_1}$ at an arbitrary point $p \in \operatorname{ang}_1$ can be found in the following way. Apply to $p$ the anti-clockwise rotation $R^{-1}_{\beta}$ by $2\pi/n_2$ around the point $O_{\beta}=O_{n'-1} \equiv O_{\gamma, n'-1}$ until $p$ hits the angle $\alpha_{n'}$ or its boundary. After that, if $p$ lies on the boundary of $\alpha_{n'}$, then $T'_{\gamma, \operatorname{ang}_1}$ is undefined for the original point $p$. Otherwise the return map is given by applying $T'$ to $p$.
To unify our analysis in the cases of even and odd $n$ we note that $T'_{\gamma, \operatorname{ang}_1}$ can be calculated in the following way: – Let $k=k(p) \in \mathbb{Z}_+$ be the smallest number such that $q=T'^k(p) \in \overline{\operatorname{ang}_2}$. – If $q \in \partial \operatorname{ang}_2$, then $T'_{\gamma, \operatorname{ang}_1}(p)$ is undefined. – Otherwise, $T'_{\gamma, \operatorname{ang}_1}(p)=h(q)$. The last step yields the desired result since $h(q)=T'(p)$ for even $n$ and $T'^2(q)$ for odd $n$. Examples of $T'_{\gamma, \operatorname{ang}_1}$ and the trajectories of the first return map of domains in $\operatorname{ang}_1$ with respect to $T'$ are shown in Figs. 9 and 10. It follows from this algorithm that $T'_{\gamma, \operatorname{ang}_1}$ (as well as $T'_{\beta}$) is defined at those and only those points of $\operatorname{ang}_1$ which do not lie on the rays $B_2B_1, B_3B_2, \dots, B_{n'_2+1}B_{n'_2}$, where $n'_2=n_2/2$. These rays divide $\operatorname{ang}_1$ into open domains $\alpha_{\beta, 1}$, $\alpha_{\beta, 2}$, …, $\alpha_{\beta, n'_2}$ enumerated according to increasing distance to the vertex $B_1$ of the angle $\operatorname{ang}_1$. On each of these domains $T'_{\gamma, \operatorname{ang}_1}$ and $T'_{\beta}$ are isometries, and for $T'_{\beta}$ the isometries can be described similarly to Lemmas 20 and 22. Consider both transformations on the domain $\alpha_{\beta, n'_2}$. In this domain $T'_{\beta}$ is, by definition, equal to the composite of the central symmetry with respect to the vertex $B_{n'_2+1}$ and the anti-clockwise rotation by $2\pi n'_2/n_2=\pi$ around the point $O_{n'_2-1}$, that is, the central symmetry with respect to $O_{n'_2-1}$. By high school geometry, this composite of two central symmetries is equal to the translation along the vector $2 \cdot \overrightarrow{B_{n'_2+1}O_{n'_2-1}}= \overrightarrow{B_{n'_2+ 1}{B_1}}$. Since $h$ is also equal to this translation, we conclude that $T'_{\gamma, \operatorname{ang}_1}=T'_{\beta}$ on $\alpha_{\beta, n'_2}$. We now fix an arbitrary integer $i \in [1, n'_2)$ and prove that these transformations coincide on $\alpha_{\beta, i}$. By Lemma 20, $T'_{\beta}(\alpha_{\beta, i})$ is an anti-clockwise rotation by $(n_2- 2i)\pi/n_2$. On the other hand, $T'_{\gamma, \operatorname{ang}_1}(\alpha_{\beta, i})$ is the composite of $n'_2-i$ anti-clockwise rotations by $2\pi/n_2$ (this is the number of rotations needed for the image of $\alpha_{\beta, i}$ to lie inside $\operatorname{ang}_2$) and a translation. This composite is an anti-clockwise rotation by $2(n'_2-i)\pi/n_2= (n_2-2i)\pi/n_2$ (because $n_2$ is obviously even). Therefore it suffices to prove that $T'_{\beta}(\alpha_{\beta, i})$ and $T'_{\gamma, \operatorname{ang}_1}(\alpha_{\beta, i})$ have a common fixed point. By Lemma 20, $T'_{\beta}(\alpha_{\beta, i})$ has a fixed point $O_{\beta, i}$, the intersection point of the bisectors of $\operatorname{ang}_1$ and the angle with vertex $B_{i+1}$ formed by the rays $B_{i+1}B_i$ and $B_{i+2}B_{i+1}$. By symmetry, $|B_1O_{\beta, i}|=|B_{i+1}O_{\beta, i}|=l_i$ for some $l_i \in \mathbb{R}_+$. To calculate the value of $T'_{\gamma, \operatorname{ang}_1}$ at $O_{\beta, i}$ by the procedure described above, we apply $R^{-1}_{\beta}$ to $O_{\beta, i}$ $n'_2-i$ times. This transformation maps the half-open interval $B_{i+1}O_{\beta, i}$ (excluding $B_{i+1}$), which is the beginning of the bisector of the angle between the rays $B_{i+1}B_i$ and $B_{i+2}{i+1}$, to an interval of the same length, which is the beginning of the bisector of the angle between the rays $B_{n'_2+1}B_{n'_2}$ and $B_{n'_2+2}B_{n'_2+1}$, that is, the angle $\operatorname{ang}_2$. It follows from this and the subsequent steps of the procedure that $T'_{\gamma, \operatorname{ang}_1}$ maps $O_{\beta, i}$ to the point lying on the bisector of $\operatorname{ang}_1$ at a distance $l_i$, that is, it maps it to $O_{\beta, i}$. Thus $O_{\beta, i}$ is a fixed point for $T'_{\beta}(\alpha_{\beta, i})$ and for $T'_{\gamma, \operatorname{ang}_1}$. $\Box$ Here is another simple but important property of the first return map $T'_{\beta}$. Lemma 25. Let $q \in \operatorname{int}(Z'')$ be a non-finite point. Then one can find a point $p \in \operatorname{ang}_1$ and a non-negative integer $k$ such that $q=T'^k(p)$. Proof. When $q \in \operatorname{ang}_1$, the lemma is obvious. Otherwise, $T'^{-1}(q)= R_{\beta}(q)$. We successively apply $T^{-1}$ to $q$ until $q$ hits either $\operatorname{ang}_1$ or the ray $B_1B_2$. However, in view of the absence of a non-finite preimage under $T'$, the ray $B_1B_2$ consists only of finite points. Therefore $q$ does hit $\operatorname{ang}_1$, and the lemma follows. $\Box$ Combining Lemma 25 and Lemma 4, we obtain the following fact, which is important for the theorems we are proving. Lemma 26. 1) An aperiodic point inside $Z''$ with respect to $T'$ exists if and only if there is an aperiodic point inside $\operatorname{ang}_1$ with respect to $T'_{\gamma, \operatorname{ang}_1}$. 2) The set of points inside $Z''$ periodic with respect to $T'$ is a set of full measure outside the table if and only if the set of points inside $\operatorname{ang}_1$ periodic with respect to $T'_{\gamma, \operatorname{ang}_1}$ is of full measure. 5.3. Odd $n/2$: proof of Theorem 6We assume that $n$ is even and $n/2$ is odd. Our aim is to prove Theorem 6, that is, establish the equivalence of the periodicity problems for the outer billiards outside a regular $n$-gon and a regular $n/2$-gon. To make our analysis precise, we also assume that $n \geqslant 10$. An analogue for $n=6$ also exists, but in a somewhat more degenerate form. It was studied in [12]. In any case, it follows from § 1 that the periodicity problems for $n=3,6$ have been solved and possess the required equivalence. Lemma 27 follows obviously from Lemma 24. Lemma 27. 1) If there is an aperiodic point for the outer billiard outside a regular $n$-gon, then there is also an aperiodic point for the outer billiard outside a regular $n/2$-gon. 2) If the set of periodic points for the outer billiard outside a regular $n/2$-gon is a set of full measure outside the table, then the set of periodic points for the outer billiard outside a regular $n$-gon is also a set of full measure outside the table. Since $n$ is even, we have $n'=n/2$ and $n_2=n$. Hence $\beta=\beta_{n/2- 1}$ is a regular $n$-gon equal to $\gamma$ (because $\gamma$ and $\beta$ are ‘sandwiched’ between the parallel lines $A_0A_1$ and $A_{n'}A_{n'+1}$). Moreover, $\gamma$ and $\beta$ are centrally symmetric with respect to $Q_{n'-1}$, the intersection point of the rays $A_1A_2$ and $A_{n'-1}A_{n'}$. Let $\delta$ be the open intersection of the angles $V'$, $V_{n'-1}$ as well as the angle between the rays $B_1B_0$ and $B_0B_{n-1}$ and the angle between the rays $B_{n'+1}B_{n'+2}$ and $B_{n'+2}B_{n'+3}$ (note that the last two angles are symmetric to $V'$ and $V_{n'-1}$ with respect to the line passing through $Q_{n'-1}$ and perpendicular to the line $A_0A_1$). This intersection is the pentagon $P_{n'-2}P'_{n'-2}Q'_{n'-2}Q_{n'-1}Q_{n'-2}$, where $P'_{n'-2}$ and $Q'_{n'-2}$ are the intersection points of the ray $B_{n'+2}B_{n'+3}$ and the rays $A_0A_1$ (which contains the points $B_0$ and $B_1$) and $A_{n'}A_{n'-1}$ (which contains $B_0$ and $B_{n-1}$), respectively. Note that in the case $n=10$, and only in this case, we have $\beta'=\delta$ in view of appropriate angles inside $\delta$. Note that the outer angles at the vertices $P_{n'-2}$, $Q_{n'-2}$ and $Q_{n'-1}$ of the pentagon $\delta$ are equal to $4\pi/n=2\pi/n'$ and, therefore, the outer angles at the vertices $Q_{n'-2}$ and $Q'_{n'-2}$ of $\delta$ are equal to
$$
\begin{equation*}
\frac{2\pi-3\cdot 4\pi/n}{2}= \frac{(n-6)\pi}{n}=\frac{(n'-3)\pi}{n'},
\end{equation*}
\notag
$$
which is divisible by $2\pi/n'$ since $n'$ is odd. Moreover, $O_{n'-2}$, the intersection point of the bisectors of the four angles whose intersection is $\delta$, is equidistant from all the five sides of $\delta$ by symmetry. By combining these facts with the inclusion $\delta \subseteq \alpha_{n'-2}$ and taking into account that $T'$ rotates $\alpha_{n'-2}$ by the angle $(n-2(n'-2)/n)\pi= 4\pi/n=2\pi/n'$, we obtain the following lemma. Lemma 28. 1) $\beta_{n'-2} \equiv \operatorname{comp}(O_{n'-2})$ is an open regular $n'$-gon inscribed in $\delta$. 2) The points $P_{n'-2}$ and $Q_{n'-2}$, $Q_{n'-1}$ are vertices of $\operatorname{comp}(O_{n'-2})$. We denote the $n'$-gon $\operatorname{comp}(O_{n'-2})$ by $\beta'$. The vertices of $\beta'$ are denoted by $B'_0B'_1\dots B'_{n'-1}$, where the enumeration is anti-clockwise and the points $B'_0$ and $B'_1$ lie on the ray $A_0A_1$. Thus, $B'_0=P_{n'-2}$, $B'_1=P'_{n'-2}$ and $B'_{(n'+1)/2}=Q_{n'-1}$. Note that, as in the case of $V' \setminus \beta$, the set $Z' \setminus \beta'$ is a union of two disjoint polygonal domains. We denote these domains by $Z'_1$ and $Z'_2$, where $Z'_1$ is closer to the vertex $A_1$ (moreover, $A_1$ is a vertex of $Z'_1$). By construction, the points $B'_1$, $B'_2$, $B_{n'+3}$ and $B_{n'+2}$ lie on a straight line. This line divides $Z'_2$ into the polygonal domains $B'_0B_0B_{n-1}B_{n-2}\dots B_{n'+3}$ and $B_{n'+2}B'_2B'_3\dots B'_{(n'+1)/2}$. We denote these domains by $Z'_{21}$ and $Z'_{22}$, respectively, and regard them as open domains. Examples of the domains $Z'_1$, $Z'_{21}$ and $Z'_{22}$ are shown in Fig. 11. We introduce the following notation: a) $T'_{Z'_{21}}$ and $T'_{Z'_{22}}$ are the first return maps of $T'$ to $Z'_{21}$ and $Z'_{22}$, respectively; b) $V'_{\beta'}$ is the open angle between the rays $B'_0B'_1$ and $B'_1B'_2$; c) $T_{\beta'}$ is the outer billiard outside $\beta'$, and $T'_{\beta'}\colon V'_{\beta'} \to V'_{\beta'}$ is the transformation induced by $T_{\beta'}$ and the equivalence under the rotation by $2\pi/n'$ around the centre of $\beta'$. Since $O_{n'-1}$ is the intersection point of the bisector of $V'_{\beta'}$ and the bisector of the angle formed by the rays $B'_{(n'+1)/2}B'_{(n'-1)/2}$ and $B'_{(n'+3)/2}B'_{(n'+1)/2}$ (the latter bisector is parallel to the lines $A_0A_1$ and $A_{n'}A_{n'+1}$ and equidistant from these lines), Lemma 29 follows from the description of $\beta$. This lemma means that the domain $Z'_{21}$ is invariant under $T'_{\beta'}$. The following lemma establishes the most important relation between outer billiards. Lemma 30. 1) If we regard $T'_{\beta'}$ as a transformation of $Z'_{21}$, then $T'_{\beta'}\,{=}\,T'_{Z'_{21}}$. 2) If we regard $T'$ as a transformation of $Z'_1$, then $T'_{Z'_{22}}$ is conjugate to $T'$ with respect to the anti-clockwise rotation by $(n'-1)\pi/n'$ around $O'_{n'-2}$. This rotation maps $Z'_{22}$ to $Z'_1$. 3) Let $p$ be a non-finite point or component (periodic or aperiodic) lying in $Z'_2$. Then one can find a non-finite point or component $q$ lying in $Z'_{21}$ and a non-negative integer $k=k(p)$ such that $T'^k(q)=p$. Lemma 30 can be proved by the method used in the proof of Lemma 24. The first return trajectories for $T'_{Z'_{21}}$ and $T'_{Z'_{22}}$ in the case $n=14$ are shown in Fig. 12. Lemma 31 is a direct corollary of Lemma 30 and the equivalence of the periodic structures on the angles $\operatorname{ang}_1$ generating the periodic structures in $Z''$ for the regular $n$-gon $\gamma$ and the regular $n/2$-gon $\beta'$ (see Lemma 26). Lemma 31. 1) If there is an aperiodic point for the outer billiard outside a regular $n/2$-gon, then such a point also exists for the outer billiard outside a regular $n$-gon. 2) If the set of periodic points for the outer billiard outside a regular $n$-gon is a set of full measure, then the set of periodic points for the outer billiard outside a regular $n/2$-gon is also a set of full measure. Combining this with Lemma 27, we obtain the following result, which is a restatement of Theorem 6. Lemma 32. 1) An aperiodic point for the outer billiard outside a regular $n/2$-gon exists if and only if such a point also exists for the outer billiard outside a regular $n$-gon. 2) The set of periodic points for the outer billiard outside a regular $n$-gon is a set of full measure if and only if the set of periodic points for the outer billiard outside a regular $n/2$-gon is a set of full measure. Note that these lemmas yield not only the equivalence of the periodicity problems, but also a certain identification of the periodic structures arising in the outer billiards outside regular $n$-gons and $n/2$-gons. This property was indicated for the regular pentagon and the regular dodecagon by a number of authors (see, for example, [11], [22], as well as [13]). We were able to prove that a similar phenomenon holds for an arbitrary $n=4k+2$, where $k \in \mathbb{Z}$ and $k \geqslant 2$. When $k=1$ and $n=6$, the domain $Z'$ degenerates to a triangle, which is also an analogue of the domain $\beta'$. In the case of a triangle, the domain $Z'$ degenerates to an empty set. 5.4. Even $n$: proof of Theorem 5As in § 5.3, we assume that $n$ is even. But now we need not assume that $n/2$ is odd. The argument works for any even $n \geqslant 4$. Our aim is to prove Theorem 5, that is, to reduce the study of periodicity problems for the map $T'$ on the angle $V'$ to the investigation of these problems for the same map $T'$ on the domain $Z'$ (more precisely, on $Z' \cup \beta$). For $n=10$, this reduction was performed in [13]. Here we generalize it to the case of an arbitrary even $n \geqslant 4$. Let $H\colon V' \to \alpha_{n'}=\operatorname{ang}_2$ (resp. $H'\colon V' \to \operatorname{ang}_1$) be the translation along the vector $A_1Q_{n'+1}$ (resp. $A_1B_1$). Note that the extension of $H'$ to the whole of $\mathbb{R}^2$ maps $\gamma$ to $\beta$. Lemma 33 follows obviously from Lemma 24, the definition of $T'_{\gamma, \operatorname{ang}_1}$ and the equality $T'(\operatorname{ang}_2)=\operatorname{ang}_1$. Lemma 33. Suppose that $p \in \operatorname{int}(V')$ and $q=H'(p)$ or $q=H(p)$. Then $p$ is finite (periodic, aperiodic) if and only if $q$ is finite (periodic, aperiodic). The following lemma turns out to be of utmost importance for us. Lemma 34. Let $q \in V'$ be a periodic (aperiodic) point. Then one can find a periodic (aperiodic) point $p \in Z' \cup \beta$, a number $m \in \mathbb{Z}_{\geqslant 0}$ and a sequence of transformations $f_1,f_2,\dots,f_m$ such that $q=f_m(f_{m-1}(\dots f_2(f_1(p))\dots ))$ and every transformation $f_j$, $j=1,2,\dots,m$, is either $T'$ or $H$. Proof. It follows automatically from Lemma 33 that the set of periodic (aperiodic) points is invariant under $T'$ and $H$. Hence it suffices to obtain from $q$ a point in $Z'$ by means of the transformations $T'^{-1}$ and $H^{-1}$.
Let $S$ be the infinite strip bounded by the interval $A_1B_1$ and the rays $A_1A_2$ and $B_1B_2$. Note that, up to the finite points of the ray $B_0B_1$, we have $V' \setminus S= T'(\alpha_{n'})=T'(\operatorname{ang}_2)=\operatorname{ang}_1$ and, for any point $r \in \operatorname{ang}_1$, the point $T'^{-1}(r)$ is the translation of $r$ along the vector $\overrightarrow{B_1B_{n'+1}}=\overrightarrow{A_1A_{n'+1}}$. We apply $T'^{-1}$ to $q$ until $q$ hits $S$. Then we apply $H^{-1}$ to $q$ as many times as possible until it leaves $V'$. Now, by construction, $q$ lies either in $Z' \cup \beta$ or in the image of $Z'$ under the central symmetry with respect to $O_{n'-1}$. In the former case the lemma is proved. In the latter case the proof can be completed by applying the transformation $T'^{-1}$ at most $n'-2$ times to $q$ in such a way that $q$ hits $\operatorname{ang}_1$, and then applying $T'^{-1}$ once more. The resulting point is easily seen to be inside $H(Z')$, and an application of $H^{-1}$ sends it inside $Z'$. $\Box$ Lemma 34 can be strengthened. Lemma 35. Let $q \in V'$ be a periodic (aperiodic) point. Then one can find a periodic (aperiodic) point $p \in Z' \cup \beta$ and numbers $k, l \in \mathbb{Z}_{\geqslant 0}$ such that $q=T'^l(H^k(p))$. Proof. By Lemma 24 and since $T'(\operatorname{ang}_2)= \operatorname{ang}_1$ (and $T'(\operatorname{ang}_2)$ is a translation), we have $H(T'(p'))=T'_{\operatorname{ang}_2}(H(p'))= T'^m(H(p'))$ for every non-finite point $p' \in V'$, where $m=m(p')$. Applying this equality sufficiently many times to the chain of transformations obtained in Lemma 34, we arrive at the desired result. $\Box$ Lemma 35 enables us to reduce (restrict) the domain of the periodicity problem from $V'$ to $Z'$, that is, to prove Theorem 5 (strictly speaking, in the case of even $n$). This will be done right now. Lemma 36. 1) An aperiodic point $q \in V'$ exists if and only if there is an aperiodic point $p \in Z'$. 2) The set of periodic points inside $V'$ is a set of full measure if and only if so is the set of periodic points inside $Z'$. Proof. In view of Lemma 35, the only non-obvious point to prove that if the set of periodic points in $Z'$ of full measure, then so is the set of such points in $V'$. We now prove this with the aid of Lemma 4.
By Lemma 35, every aperiodic point $q \in V'$ can be represented as $q=f(p)$, where $f=H^k \circ T'^l$ and $k, l \in \mathbb{Z}_{\geqslant 0}$. The set of all possible such $f$’s is countable. Note that $T'$ is not an isometry, it is merely a piecewise isometry, but the number of pieces in $T'$ is equal to ${n'}$. Thus, applying $T'$ to a point $p' \in V'$ is equivalent to applying to $p'$ one of the $n'$ isometries, which are fixed in advance and described in Lemmas 20 and 22. Replacing every transformation $T'$ in all possible compositions $f$ by one of these $n'$ isometries in all possible ways (independently of the other transformations $T'$), we also obtain a countable set $F'$ of isometries such that for every aperiodic point $q \in V'$ one can find an aperiodic point $p \in V'$ and an isometry $f' \in F'$ with $q=f'(p)$. Therefore, if the set of aperiodic points in $Z'$ is of measure zero, then so is the set of aperiodic points in $V'$ (formally speaking; see also Lemma 3). $\Box$ Thus, it suffices to study the periodicity problems for outer billiards outside regular polygons only in the invariant domains $Z'$. Strictly speaking, Lemma 36 provides this conclusion (as well as the conclusion of Theorem 5) only in the case when $n$ is even. However, if $n/2$ is odd, then it follows directly from Lemma 30 that the periodicity problems in $Z'$ outside an $n$-gon are equivalent to those in $Z'$ outside the $n/2$-gon. Combining this with Lemma 36, we obtain the complete proof of Theorem 5. Moreover, from the standpoint of periodicity problems (and many other problems related to the tilings generated by the outer billiards outside regular polygons) for an arbitrary $n$, we can see that it suffices to investigate the case when $n$ is even. § 6. The outer billiard outside a regular octagon: Tabachnikov’s renormalization schemeIn this section we study the periodicity problems for the outer billiard outside a regular octagon. We apply a method of analysis similar to that in [13]. It is based on the renormalization scheme used by Tabachnikov [4]. 6.1. Main definitions and componentsIn this subsection we put $n=8$. The domain $\beta=\beta_3$ is a regular octagon equal to $\gamma$, the domain $Z'$ is the quadrangle $A_1B_0B_7B_6$, the domain $\alpha_1$ is an open triangle $A_1P_2Q_2$, and $\alpha_2$ is the quadrangle $P_2P_3Q_3Q_2$. The transformation $T'$ leaves the domain $Z'$ invariant and splits it into the domains $\alpha_1$, $\alpha_2$ and the triangle $Q_3B_7B_6$, which we denote by $\alpha'_3$ (Fig. 13). By Lemma 20, $T'$ rotates $\alpha_1$ (resp. $\alpha_2$ and $\alpha'_3$) anti-clockwise by $3\pi/4$ around $O_1$ (resp. by $\pi/2$ around $O_2$ and by $\pi/4$ around $O_3$, the centre of $\beta_3$). See § 5.1 for the definitions of the points $O_i$ and other notation. What do $\beta_1=\operatorname{comp}(O_1)$ and $\beta_2=\operatorname{comp}(O_2)$ look like? Lemma 37. The domains $\beta_1$ and $\beta_2$ are regular octagons inscribed in $\alpha_1$ and $\alpha_2$ and centred at $O_1$ and $O_2$, respectively. These polygons are shown in Fig. 14. To prove the lemma it suffices to verify that these octagons are mapped by $T'$ to themselves and every boundary interval is mapped by some iteration of $T'$ to a subset of one of the rays $A_0A_1$, $A_2A_1$, $A_3A_2$ and $A_4A_3$. We denote the vertices of $\beta_i$, $i=1, 2, 3$, by $B^i_0$, $B^i_1$, $\dots$, $B^i_7$ in such a way that the enumeration is anti-clockwise and the points $B^i_0$, $B^i_1$ lie on the ray $A_0A_1$. In particular, the points $B^3_0$, $B^3_1$, $\dots$, $B^3_7$ coincide with $B_0$, $B_1$, $\dots$, $B_7$, respectively. We introduce another important pair of points. Let $O_{32}$ be the intersection point of the bisectors of all the angles of the quadrangle $B^2_3B^2_4B^2_5B^3_6$ (this point exists since the quadrangle is symmetric) and let $O_{23}$ be the point of intersection of the bisectors of the quadrangle $B^2_1B^2_2B^2_3B^3_0$. All assertions in the following lemma are easily verifiable. Lemma 38. 1) We have $T'(O_{32})=O_{23}$ and $T'(O_{23})=O_{32}$. 2) $\operatorname{comp}(O_{32})$ is the open regular octagon centred at $O_{32}$ and inscribed in the angle $B^2_3B_6B^2_5$. 3) $\operatorname{comp}(O_{23})$ is the open regular octagon centred at $O_{23}$ and inscribed in the angle $B^2_1B_0B^2_3$. 4) $B^2_4$ is a vertex of $\operatorname{comp}(O_{32})$ and $B^2_2$ is a vertex of $\operatorname{comp}(O_{23})$. 5) $T'(\operatorname{comp}(O_{32}))=\operatorname{comp}(O_{23})$ and $T'(\operatorname{comp}(O_{23}))=\operatorname{comp}(O_{32})$. We denote $\operatorname{comp}(O_{32})$ (resp. $\operatorname{comp}(O_{23})$) by $\beta_{32}$ (resp. $\beta_{23}$). 6.2. Self-similarity and its propertiesA key element of our analysis is the property of self-similarity enjoyed by this outer billiard. It can be established in the following way. Let $\Gamma$ be the homothety with centre $A_1$ and coefficient $\lambda=|A_1O_1|/|A_1O_3|$. Then $\Gamma$ sends $O_3$ to $O_1$, and $B^3_j$ to $B^1_j$ for $j=0, 1, \dots, 7$. We also put $Z'_\Gamma=\Gamma(Z')$ and let $T'_{Z'_\Gamma}$: $Z'_\Gamma \to Z'_\Gamma$ be the first return map on $Z'_\Gamma$ with respect to $T'$. Lemma 39. Consider $T'$ as a transformation $T'\colon Z' \to Z'$. Then $T'$ is conjugate to $T'_{Z'_\Gamma}$ with respect to $\Gamma$. Proof. Consider Fig. 15. It shows the domains $\Gamma(\alpha_1)$, $\Gamma(\alpha_2)$, $\Gamma(\alpha'_3)$ (marked by $0$) and their trajectories under successive application of $T'$. The successive application of $T'$ to each of these three open domains is an isometry till we return to $Z'_{\Gamma}$. The number of each domain is the number of iterations of $T'$ needed to transform $\Gamma(\alpha_1)$, $\Gamma(\alpha_2)$, $\Gamma(\alpha'_3)$ into the current state.
We draw the following conclusions from Fig. 15: 1) For every domain $U \in \{\Gamma(\alpha_1), \Gamma(\alpha_2),\Gamma(\alpha'_3) \}$, we have where $k(U)$ is the number of iterations of $T'$ till the first return of the corresponding domain.2) The first return map $T'_{Z'_{\Gamma}}$ is undefined at the points of the open boundary intervals between $\Gamma(\alpha_1)$ and $\Gamma(\alpha_2)$ as well as between $\Gamma(\alpha_2)$ and $\Gamma(\alpha'_3)$ (the first interval is mapped to the ray $A_2A_1$ after one iteration of $T'$, and the second one is mapped to the ray $A_4A_3$ after seven iterations of $T'$). These observations prove the lemma. $\Box$ The following properties of self-similarity hold in view of Fig. 15 and the proof just completed. All further notions and lemmas in this subsection are introduced by analogy with § 3.9 in [13]. For completeness, we give their full proofs. We introduce the notion of rank of a point. Definition 22. Suppose that $p \in Z'$. Then the rank of $p$, or $\operatorname{rk}(p)$, is the largest non-negative integer $k=k(p)$ such that $\Gamma^{-k}(p) \in Z'$. Definition 23. Let $p\,{\in}Z'$ be a periodic point with period $\mathrm{per}'(p)\,{=}\,k$. Then the rank of the orbit of $p$, or $\operatorname{rko}(p)$, is the number $\max\{\operatorname{rk}(T'^j(p)) \mid j\,{\in}\,\mathbb{Z}, j \in [0, k)\}$. Lemma 41. Let $q \in Z'$ be a periodic point with $\operatorname{rk}(q)=\operatorname{rko}(q)=k>0$. Then $\operatorname{rk}(p)=\operatorname{rko}(p)=k-1$, where $p=\Gamma^{-1}(q)$. Proof. The point $p$ is also periodic in view of the conjugacy in Lemma 39. Put $l= \mathrm{per}'(p)$. Then Lemma 39 yields $T^{\prime j}_{Z'_{\Gamma}}(q)= \Gamma({T'}^j(p))$ for all $j$ such that $0 \leqslant j \leqslant l$, and we have ${T^{\prime l}_{Z'_{\Gamma}}}(q)=q$. By the definitions of $T'_{Z'_{\Gamma}}$ and of the rank, the only points with non-zero rank in the orbit of $q$ under $T'$ are the points $ \{ T_{Z'_{\Gamma}}^{\prime j}(q), 0 \leqslant j<l \} $, and we have $\operatorname{rk}(T_{Z'_{\Gamma}}^{\prime j}(q))=1+\operatorname{rk}({T'}^j(p)) $ for all $j \in [0, l)$. Hence $\operatorname{rko}(p)=\operatorname{rk}(p)=k-1$. $\Box$ Lemma 42 follows from Lemma 41 and is analogous to Lemma 35. Lemma 42. Every periodic point (component) $q \in Z'$ ($q \subset Z'$) can be represented as $q=T'^l(\Gamma^k(p))$, where $p \in Z' \setminus Z'_\Gamma$ ($p \subset Z' \setminus Z'_\Gamma$), $p$ is a periodic point of rank $0$ ($p \in \{\beta_1, \beta_2, \beta_{32}\}$) and $k, l \in \mathbb{Z}_{\geqslant 0}$. Proof. The ‘point’ part follows obviously from Lemma 41. The ‘component’ part of Lemma 42 follows since the boundary $(\partial \Gamma(Z')) \cap Z'$ consists only of finite points. This fact also implies that the notions of rank and rank of the orbit can be generalized naturally to periodic components inside $Z'$. It follows from Fig. 15 that the periodic components of rank $0$ are the regular polygons $\beta_1$, $\beta_2$, $\beta_{32}$ and $\beta_{23}=T'(\beta_{23})$ because all periodic points of $Z'$ not belonging to $\beta_1$, $\beta_2$, $\beta_{32}$ or $\beta_{23}$ lie on the return trajectories of $T'$ in $Z'_\Gamma$. $\Box$ 6.3. Proof of Theorem 1Here we prove that the outer billiard outside a regular octagon admits an aperiodic point. Put $f=\Gamma^2 \circ {T'}^2$ and $X=f(Z')$. Let $T'_X\colon X \to X$ be the first return map of $T'$ to $X$. Note that $X$ is a quadrangle similar to $Z'$ with coefficient of similarity $\lambda^2$ (see Fig. 16). Lemma 43 follows obviously from Lemma 39. Lemma 43. The transformation $T'\colon Z' \to Z'$ is conjugate to $T_X\colon X \to X$ with respect to $f$. We consider a sequence of periodic components $C_0, C_1, C_2, \dots$ by putting $C_0\,{=}\, \beta_2$, $C_1=\beta_{23}$, $C_2=T'^2(\Gamma(\beta_1))$ ($C_2=T'^2(\Gamma^2(\beta))$) and $C_{i+3}=f(C_i)$ for $i=0, 1, 2, \dots$ . The domains $C_0$, $C_1$, $C_2$ and $X$ are shown in Fig. 16. Finally, let $c_{\mathrm{inf}} \in X$ be the limit of the sequence $C=\{C_i\}_{i= 0}^{\infty}$. We claim that $c_{\mathrm{inf}}$ (a fixed point of $f$) is the desired aperiodic point. To this end we note that the following lemma holds. Lemma 44. The transformation $f$ is the composite of the contraction with coefficient $\lambda^2$ centred at $c_{\mathrm{inf}}$ and the clockwise rotation by $\pi/4$ around $c_{\mathrm{inf}}$. We use this representation of $f$ to prove the following two lemmas. Proof. Suppose that $c_{\mathrm{inf}}$ is periodic. Then the sequence $C$ of distinct periodic components tends to a point lying inside the periodic component of $c_{\mathrm{inf}}$. This gives rise to a contradiction. $\Box$ Proof. Suppose that $c_{\mathrm{inf}}$ is finite. Then, by Lemma 2, there is an open interval of length $2l$, $l \in \mathbb{R}_+$, with midpoint $c_{\mathrm{inf}}$ consisting only of finite points. On the other hand, the components $C_0$, $C_1$, $C_2$ are such that any line through $c_{\mathrm{inf}}$ passes through at least one of them. This imposes an upper bound on the length $l$. However, by Lemma 44, this assertion also holds for the triples of components $(C_3, C_4, C_5)$, $(C_6, C_7, C_8)$, $\dots$, and the upper bounds for $l$ imposed by these triples tend to zero (because $C$ converges to $c_{\mathrm{inf}}$). Therefore, $l$ cannot be greater than $0$. This gives rise to a contradiction. $\Box$ Thus, the point $c_{\mathrm{inf}}$ is neither finite nor periodic. It follows that $c_{\mathrm{inf}}$ is aperiodic and Theorem 1 is proved. 6.4. Proof of Theorem 2To prove Theorem 2 we introduce two definitions. Definition 24. Let $W \subset \mathbb{R}^2$ be a polygon. A partition of $W$ is a finite set $\mathcal{W}=\{W_1, W_2, \dots, W_k\}$, $k \in \mathbb{Z}_+$, of polygons such that a) $W=\bigcup_{j=1}^{k} W_j$; b) $\operatorname{int}(W_i) \cap \operatorname{int}(W_j)=\varnothing$ for all $i,j$ such that $1 \leqslant i<j \leqslant k$. Definition 25. Let $\mathcal{W}^1$, $\mathcal{W}^2$ be partitions of a polygon $W \subset \mathbb{R}^2$. We say that $\mathcal{W}^2$ is a refinement of $\mathcal{W}^1$ if, for every $Q \in \mathcal{W}^2$, there is a $P \in \mathcal{W}^1$ such that $Q \subset P$. Thus, a refinement of $\mathcal{W}^1$ is a union of partitions of the polygons occurring in $\mathcal{W}^1$. The following lemma is related to partitions. Lemma 47. Let $W$ be an arbitrary polygon. Consider a sequence $(\mathcal{W}^l)_{l \in \mathbb{Z}_+}$ of partitions of $W$ into finitely many polygons with the following properties: 1) $\mathcal{W}^{l+1}$ is a refinement of $\mathcal{W}^l$ for every $l \in \mathbb{Z}_+$. 2) Every polygon occurring in at least one partition of the sequence $(\mathcal{W}^l)_{l \in \mathbb{Z}_+}$ is coloured green or red. The colour may depend on the partition. 3) For every $l \in \mathbb{Z}_+$ and every $A \in \mathcal{W}^l$ which is green in $\mathcal{W}^l$, we have $A \in \mathcal{W}^{l+1}$ and $A$ is green in $\mathcal{W}^{l+1}$. 4) There are numbers $\epsilon \in \mathbb{R}_+$ and $k \in \mathbb{Z}_+$ such that, for every $l \in \mathbb{Z}_+$ and every red polygon $U \in \mathcal(W^l)$, it is guaranteed that the total area of green polygons of the $(l+k)$th partition that lie inside $U$ is greater than or equal to $\epsilon A$, where $A$ is the area of $U$. Then the union of all green polygons occurring in these partitions is a set of full measure in $W$. Proof. Let $S_l$, $l \in \mathbb{Z}_+$, be the total area of red polygons in $\mathcal(W^l)$. It suffices to prove that $S_l$ tends to zero. But this follows since for every $l \in \mathbb{Z}_+$ we have $S_l \geqslant S_{l+1}$ and $S_{l+k} \leqslant (1-\epsilon)\cdot S_l$, where $1-\epsilon$ is a positive constant strictly smaller than 1. $\Box$ We use Lemma 47 to prove the following lemma, which yields, in combination with Lemma 36, а proof of Theorem 2. Proof. In order to use Lemma 47 we introduce the following sequence $(\mathcal{Z}^l)_{l \in \mathbb{Z}_+}$ of partitions (coloured red and green) of the polygon $\overline{Z'}$:
a) $(\mathcal{Z}^1)$ consists of three red polygons $\overline{\alpha_1}$, $\overline{\alpha_2}$ and $\overline{\alpha'_3}$. b) For every $l\,{\in}\,\mathbb{Z}_+$, we have $\mathcal{Z}^{l+1}{=}\bigl\{\bigcup\{\overline{T^{\prime j}(\Gamma(\operatorname{int}(P)))}\} \,|\, P{\kern1pt}{\in}{\kern1pt}\mathcal{Z}^l,\, 0\,{\leqslant}\, j\,{<}\,k(P)\bigr\} \cup \{\overline{\beta_2}, \overline{\beta_3}, \overline{\beta_{32}}, \overline{\beta_{23}}\}$, where $k(P)$ is the smallest $t \in \mathbb{Z}_+$ such that $(T^{\prime k}(\operatorname{int}(P))) \subset Z'_\Gamma$. c) In the notation of part b) the polygon $\overline{T^{\prime j}(\Gamma(\operatorname{int}(P)))}$ is green in $(\mathcal{Z})^{l+1}$ if and only if $P$ is green in $(\mathcal{Z})^l$. d) The polygons $\overline{\beta_2}$, $\overline{\beta_3}$, $\overline{\beta_{32}}$, $\overline{\beta_{23}}$ are green in all partitions where they occur. The first four partitions of the sequence $\mathcal{Z}^l$ are shown in Fig. 17. It follows from Fig. 15 that $\mathcal{Z}^2$ is a partition of $Z'$. Using induction and Lemma 39, we can easily see that $\mathcal{X}^{l+1}$ is a refinement of $\mathcal{X}^l$ for every $l \in \mathbb{Z}_+$. Moreover, it can be proved by induction using the same lemma that all green polygons in any partition $\mathcal{Z}^l$, $l \in \mathbb{Z}_+$, are periodic components (more precisely, their closures), while all red polygons are the trajectories of first return of the open triangles $\Gamma_{l-1}(\alpha_1)$, $\Gamma_{l-1}(\alpha_2)$, $\Gamma_{l-1}(\alpha'_3)$ to $\Gamma^{l-1}(Z')$ with respect to $T'$. Thus, conditions 1)–3) of Lemma 47 hold. On the other hand, when we pass from $\mathcal{Z}^l$ to $\mathcal{Z}^{l+1}$, the domain $\Gamma^{l-1}(\alpha_1)$ is subdivided into several domains, one of which is the green domain $\Gamma^{l-1}(\beta_1)$. Along with it, all the domains of the trajectory of first return of $\Gamma^{l-1}(\alpha_1)$ to $\Gamma^{l-1}(Z')$ with respect to $T'$ are subdivided conjugately with respect to iterations of $T'$. Hence at least an $\epsilon_1$th fraction of the area in all the domains of this trajectory has become green, where $\epsilon_1= \operatorname{Area}(\beta_1)/\operatorname{Area}(\alpha_1)$. A similar situation holds for the domains generated by $\Gamma^{l-1}(\alpha_2)$ and $\Gamma^{l-1}(\alpha'_3)$. In these cases we have
$$
\begin{equation*}
\epsilon_2= (\operatorname{Area}(\beta_1)+ \operatorname{Area}(\beta_{23}))/\operatorname{Area}(\alpha_2)\quad\text{and}\quad \epsilon_3= \operatorname{Area}(\beta_{32})/\operatorname{Area}(\alpha'_3).
\end{equation*}
\notag
$$
Hence, condition 4) of Lemma 47 holds with $\epsilon= \min(\epsilon_1, \epsilon_2, \epsilon_3)$ and $k=1$. Thus, all the hypotheses of Lemma 47 hold. It follows that the union of green domains, which are periodic components, is a set of full measure in $Z'$. $\Box$ § 7. The outer billiard outside a regular dodecagon: Tabachnikov’s renormalization schemeIn § 6 we analyzed the outer billiard outside a regular octagon. We were able to describe completely all its periodic components in the domain $Z'$ and, as a corollary, in the angle $V'$. We now consider the regular dodecagon. Our investigation uses Tabachnikov’s renormalization scheme, as in the case of the regular octagon, but some assertions will be verified by means of computer-assisted proofs (using an algorithm realized in Python). 7.1. Computer calculationsIn this technical section we discuss the main points enabling us to perform absolutely precise calculations. We assume that the system of basis vectors is right-oriented. The main decision is to create a data type $Absqrt3$, where the required numbers will be stored. Definition 26. A number $x \in \mathbb{R}$ is said to be computable if $x \in \mathbb{Q}[\sqrt{3}\,]$. In other words, there are rational numbers $a$, $b$ such that $x=a+b\sqrt{3}$. We create a class $Absqrt3$ for storing the fields $a$ and $b$ of type $Fraction$. This data type, supported by the fractions module, stores a rational number as an irreducible fraction and, therefore, absolutely precisely. The numerator and denominator can be arbitrarily large (unlike many other languages, Python imposes no restrictions on the size of a number stored as a variable of integer type $int$). Python supports all arithmetic operators for data types $int$ and $Fraction$. The following operations over computable numbers can be performed algorithmically: – addition: $(a+b\sqrt{3})+(c+d\sqrt{3})=(a+c)+(b+d)\sqrt{3} \in \mathbb{Q}[\sqrt{3}\,]$; – subtraction: $(a+b\sqrt{3})-(c+d\sqrt{3})=(a-c)+(b-d)\sqrt{3} \in \mathbb{Q}[\sqrt{3}\,]$; – multiplication: $(a+b\sqrt{3}) \cdot (c+d\sqrt{3})=(ac+3bd)+(bc+ad)\sqrt{3} \in \mathbb{Q}[\sqrt{3}\,]$; – division: $(a+b\sqrt{3})/(c+d\sqrt{3})= (a+b\sqrt{3})(c- d\sqrt{3})/(c^2-3d^2) \in \mathbb{Q}[\sqrt{3}\,]$; – comparison with $0$: $a+b\sqrt{3}>0$ if and only if one of the following conditions holds:
– comparison: if $q1, q2 \in \mathbb{Q}[\sqrt{3}\,]$, then $q1>q2 \Longleftrightarrow q1-q2>0$. Definition 27. An arbitrary point $p$ (an arbitrary vector $v$) is said to be computable if both its coordinates are computable numbers. Points and vectors are very similar from the point of view of computer calculations (both are encoded by the coordinates $x, y$). Moreover, every point can be interpreted as the radius-vector with this endpoint. Therefore we create a class $MyPoint$ for storing fields $x$ and $y$ of type $Absqrt3$ and realize the following operations over such vectors: – addition: $v1+v2=MyPoint(v1.x+v2.x, v1.y+v2.y)$; – subtraction: $v1-v2=MyPoint(v1.x-v2.x, v1.y-v2.y)$; – multiplication by a computable number: $v \cdot d=d \cdot v=MyPoint(d \cdot v.x, d \cdot v.y)$; – division by a computable number: $v / d=MyPoint(v.x / d, v.y / d)$; – scalar product: $(v1, v2)=v1.x \cdot v2.x+v1.y \cdot v2.y$; – vector (skew, or pseudo-vector) product: $[v1, v2]=v1.x \cdot v2.y-v2.x \cdot v1.y$; – anti-clockwise rotation by $\pi/6$: if then
$$
\begin{equation*}
\begin{pmatrix} v'.x \\ v'.y \end{pmatrix}=\begin{pmatrix} \dfrac{\sqrt{3}}{2} & -\dfrac{1}{2} \\ \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \end{pmatrix} \cdot \begin{pmatrix} v.x \\ v.y \end{pmatrix}.
\end{equation*}
\notag
$$
Intersection points of lines play an important role in the study of outer billiards. Definition 28. A straight line is said to be computable if it contains at least two computable points. We introduce a class $MyLine$, which specifies straight lines. It stores fields $p1$ and $p2$ of type $MyPoint$ (two points lying on the line). The following lemma is a key fact enabling us to perform computer calculations. Proof. Let $p1$, $p2$ be distinct points on the plane. The line $l$ passing through them can be described in the following two ways:
a) $l=\{p1+(p2-p1) \cdot t\mid t \in \mathbb{R} \}$; b) $l=\{p \in \mathbb{R}^2\mid [p-p1,\, p2-p1]=0 \}$. Let $l1$ and $l2$ be variables of type $Line$ specifying non-parallel straight lines, and let $p$ be their intersection point. Then we have the following system of equations for the point $p$ and the variable $t \in \mathbb{R}$:
$$
\begin{equation*}
\begin{cases} p=l1.p1+(l1.p2-l1.p1) \cdot t, \\ [p-l2.p1,\, l2.p2-l2.p1]=0. \end{cases}
\end{equation*}
\notag
$$
Substituting the first equation into the second one, we obtain the equation $[l1.p1+ (l1.p2-l1.p1) \cdot t-l2.p1,\, l2.p2-l2.p1]=0$, which is linear in $t$. Solving it, we obtain $t= [l2.p1- l1.p1,\, l2.p2-l2.p1]/[l1.p2-l1.p1,\, l2.p2-l2.p1]$ and then find $p$ from the first equation. Since $t$ turns out to be computable (the denominator is non-zero since the lines are non-parallel), so is $p$. $\Box$ We specify a polygon as a list (array) of points. To perform calculations we need to realize $\gamma$ as a computable polygon. This is possible because if $A_0$ and $A_1$ are computable, then the other vertices can be found by using the formulae $A_{i+1}-A_i=(A_i-A_{i- 1}).rotate30Counterclockwise()$, $i=1, 2, \dots, 10$. It follows from Lemma 49 that all the points occurring in our notation are computable. Strictly speaking, to prove the computability of $O_i$ for $i=1, 2, \dots$ we need to show that the bisector of an angle of magnitude $\pi/6$ formed by two computable rays (which are encoded similarly to computable lines) is also computable. In view of this computability, the calculations in our computer-assisted proof will be absolutely precise. 7.2. The main notation, points and componentsWe proceed directly to the analysis. Throughout this section we assume that $\gamma$ is a computable regular dodecagon $A_0A_1\dots A_{11}$ whose vertices are enumerated anti-clockwise. The periodic component $\beta= \beta_{n'-1}$ is a regular dodecagon $B_0B_1\dots B_{11}$ with ${\overrightarrow {B_0B_1}}={\overrightarrow{A_0A_1}}$ and $B_0=P_5$. We study the transformation $T'\colon Z' \to Z'$, where $Z'$ is the open polygon $A_1B_0B_{11}B_{10}B_9B_8$. This transformation is shown in Fig. 18. It splits $Z'$ into the domains $\alpha_1$, $\alpha_2$, $\alpha_3$, $\alpha_4$ defined above and the domain $\alpha'_5=\alpha_5 \cap Z'$. Lemma 21 enables us to find $\beta$. But what are the periodic components $\beta_i$, $i=1, 2, 3, 4$? We answer this question with the aid of computer calculations. To do this we describe an algorithm for finding the periodic component of any given periodic point $startPoint \in Z'$. Lemma 50. Algorithm 1 terminates and returns the correct component. Proof. It follows from the construction of the algorithm that, at every moment of time, the point $p$ lies on the orbit of $startPoint$ under $T'$ and we have $\operatorname{comp}(p) \subseteq U$. On the other hand, in the terminology of Lemma 5, $U=T'^{k}(U_k)$ after $k$ iterations of the cycle. Hence the first part of Lemma 6 implies that $U$ will be equal to $\operatorname{comp}(p)$ after at most $2m$ iterations, where $m=\mathrm{per}(p)=\mathrm{per}(startPoint)$ (not to be confused with $\mathrm{per}'(p)$). From this moment on, the area of $U$ will not decrease and $U$ will reappear after at most $m$ iterations. An earlier version of $U$ may reappear first. It follows again from the construction that if a polygon $U$ reappears, then $U \subseteq \operatorname{comp}(p)$ and, therefore, $U= \operatorname{comp}(p)$.
Thus, when the ‘for’ loop terminates, we have $U=\operatorname{comp}(p)$. The subsequent ‘while’ loop transforms $\operatorname{comp}(p)$ to $\operatorname{comp}(startPoint)$. Thus the polygon returned is $\operatorname{comp}(startPoint)$. $\Box$ We now calculate the domains $\beta_i=\operatorname{comp}(O_i)$ by using a slight modification of Algorithm 1. The modification consists in noticing that since $O_i= T'(O_i)$, it suffices to simply perform the ‘for’ loop until $U$ becomes equal to $T'(U)$. It is realized in the function getStableZoneFrom12Gon at https://github.com/dprpavlin/OuterBilliards/blob/master/workWith12Gon.py. The result is stated in the following lemma. Lemma 51. 1) $\beta_1$ and $\beta_4$ are regular dodecagons inscribed in $\alpha_1$ and $\alpha_4$, respectively. 2) $\beta_2$ is an equilateral hexagon with successive angles
$$
\begin{equation*}
\frac{5\pi}{6},\quad \frac{\pi}{2},\quad \frac{5\pi}{6},\quad \frac{\pi}{2},\quad \frac{5\pi}{6},\quad \frac{\pi}{2}.
\end{equation*}
\notag
$$
It is inscribed in $\alpha_2$ in such a way that $P_3$ and $Q_2$ are vertices of $\beta_2$. 3) $\beta_3$ is an equilateral octagon with successive angles It is inscribed in $\alpha_3$ in such a way that $Q_3$ is a vertex of $\beta_3$.The polygons $\beta_i$, $i=1, 2, 3, 4, 5$, are shown in Fig. 19. This result may seem counterintuitive because all periodic components were regular octagons in the previous case $n=8$ and regular pentagons and decagons for $n=5, 10$ (see [4] and [13]). Moreover, only regular polygons occur as periodic components in the lattice cases $n=3, 4, 6$ when the tables are affine equivalent to polygons with integer coordinates (folklore). However, in the case under investigation we encounter ‘irregular’ components at a very early stage. We denote by $B^1_0, B^1_1, \dots, B^1_{11}$ the vertices of $\beta_1$ in such a way that the enumeration is anti-clockwise and the vertices $B^1_0$, $B^1_1$ lie on the ray $A_0A_1$. We use similar notation for the vertices of $\beta_2$, $\beta_3$ and $\beta_4$. Then we have, for example, $A_2=B^2_4$ and $Q_3=B^3_6$, and the points $B^1_{10}$, $B^1_4$, $B^2_5$, $B^2_2= B^3_7$, $B^3_3=B^4_{10}$ and $B^4_4$ lie on the bisector of the angle $V'=P_2P_1Q_2$. 7.3. First return maps and nice self-returns: search algorithmTo follow our plan we need to be able to find the first return map for some polygonal domains. We shall search the first return map for domains $S$ of special form. Definition 30. Let $S$ be a domain lying inside the angle $P_2P_1Q_2$. We say that $S$ has a nice self-return with respect to $T'$ if the following conditions hold: 1) Suppose that $x$ and $y$ are non-finite points in $S$ and $k$ is a positive integer. Assume that $T'^k(x)$ and $T'^k(y)$ are defined, we have $T'^i(x), T'^i(y) \notin S$ for all $i \in [1, k)$, and the sequences $\rho'(x)[0, k-1]$ and $\rho'(y)[0, k-1]$ coincide. Then either both points $T'^k(x)$ and $T'^k(y)$ lie inside $S$, or none of them lies inside $S$. 2) The domain $S$ is open and the boundary of $S$ consists of finite points. Lemma 52. Let $S \subseteq Z'$ be an open polygon whose boundary consists only of finite points, each of which lies either on the boundary of $Z'$ or on the boundary of a periodic component (depending on the point) whose trajectory under $T'$ is disjoint from $S$. Then $S$ has a nice self-return with respect to $T'$. Proof. Assume the opposite. Then there are $x$, $y$ and $k$ contradicting the definition of nice self-return. By the properties of $T'$ and the definition of a code, $T'^k$ sends $xy$ to the interval $T'^k(x)T'^k(y)$. Since exactly one of the points $x$ or $y$ lies inside $S$, the closed interval $xy$ contains a point $z$ such that $T'^k(z)$ lies on the boundary of $S$ and $z$ is adjacent to a periodic component $\zeta$ whose trajectory is disjoint from $S$.
Also, it follows from the properties of outer billiards that Moreover, we can easily see that this equality holds for an arbitrary point $p$ in an $\epsilon$-neighbourhood $U_{\epsilon}(z)$ of $z$, where $\epsilon>0$ and $U_{\epsilon}(z) \subset S$. Since $T'^k(U_{\epsilon}(z))= U_{\epsilon}(T^k(z))$, we can choose $p$ in such a way that $T'^k(p) \in \zeta$. But this contradicts the defining property of $\zeta$ since $T'^{-k}(p)$ lies in $S$. $\Box$We use Algorithm 2 to calculate the first return map $T'_S$. This algorithm emulates the first return map almost directly (of course, not for every point, but for the pieces with the same return code $\rho'$; the restriction of the first return map to every such piece is an isometry). It is realized by the functions tryMakeFirstReturnMapOnlyPolygons and reverseAndTestZonesOnlyPolygons (see https://github.com/dprpavlin/OuterBilliards/blob/master/workWith12Gon.py). The first function emulates the first return map and finds the first return images, and the second one recovers the preimages from the images (an analogue of ‘foreach’ in Algorithm 2). Lemma 53. If Algorithm 2 terminates, then the result is correct. Proof. Using induction, we can easily see that the following assertions hold after $k$ iterations, where $k=0, 1, \dots$ :
– Every polygon $pol$ in the array $pols$ is $T'^k(basePol)$, where $basePol \subseteq S$ is the maximal set (with respect to inclusion) of points $p \in S$ with the same part $\rho'(p)[0,k-1]$ of the code. – Every polygon $s_2$ in the array $S_2$ is $T'^l(s_1)$, where $l \in (0, k]$ is an integer and $s_1 \subseteq S$ is the maximal set (with respect to inclusion) of points with the same code $\rho'(p)[0, l-1]$ such that $T'^l(s_1) \subseteq S$ and $T'^l(s_1)$ is disjoint from $S$ for all $l' \in (0, l)$. (Nice self-return was introduced in order to guarantee such an invariant.) – All the polygons $basePol$ and $s_1$ described in the preceding two points form a partition. – $T'_S$ is undefined at the points lying on the boundaries of these polygons. Thus, if the ‘repeat’ loop terminates, we find a partition of $S$ into a family $S_1$ of open polygons $s_1$, for each of which $T'_S$ is an isometry and $T'_S(s_1)$ lies in $S_2$. To complete the algorithm we only need to find all such $s_1$ explicitly, which is provided by the last loop. $\Box$ Note that if the number of iterations up to the first return to the given polygon $S$ is unbounded, then the algorithm will work for an infinitely long time. Such a situation is theoretically possible. Indeed, $T'\colon Z' \to Z'$ is a piecewise isometry, and an example of a piecewise isometry with infinite time of first return was given by Goetz and Poggiaspalla [23]. But if the return time to $S$ is bounded, then $T'_S$ induces a partition into domains lying in the array $S_1$. Definition 31. We say that $T'_S$ splits $S$ into domains in the set $W$ if the result of running Algorithm 2 is that the array $S_1$ consists precisely of domains in $W$. In this case it follows from the proof of Lemma 53 that $W$ is a partition of $S$. 7.4. Self-similarityWe follow the plan in § 6. Let $\Gamma_1$ be the contraction centred at the point $A_1=P_1$ and sending $O_5$ to $O_1$. We put $Z'_1=\Gamma_1(Z')$. The following lemma holds because the polygon $Z'_1$ is bounded by the sides of the angle $V'$ and by the periodic component $\beta_1$. One might expect by analogy with Lemma 39 that $T'_{Z'_1}$ is similar to $T'$ for $Z'$. However, if we run Algorithm 2 on a computer, it terminates and informs us that Lemma 55 holds. Lemma 55. The transformation $T'_{Z'_1}$ splits $Z'_1$ into ten polygons, of which four are triangles, five are quadrangles, and one is a non-equilateral hexagon. Thus, our plan cannot be realized in its original form. We are nevertheless able to find self-similarity, albeit in a more complicated way. Let $\Gamma_4$ be the contraction centred at the same point $A_1$, but sending $O_5$ to $O_4$. We put $Z'_4=\Gamma_4(Z')$ and $Z'_{14}= \Gamma_1(Z'_4)$. Proof. The polygon $Z'_4$ has a nice self-return since it is bounded by the sides of the angle $V'$ and by the periodic component $\beta_4$. So does $Z'_{14}$ because the domain $\Gamma_1(\beta_4)$ is the periodic component of the point $O_{14}=\Gamma_1(O_4)$ (this fact was verified with the aid of Algorithm 1). $\Box$ Put $\beta_{14}=\Gamma_1(\beta_4)$. Let $T'_4$ and $T'_{14}$ be the first return maps of $T'$ to $Z'_4$ and $Z'_{14}$, respectively. By running Algorithm 2 for the domains $Z'_4$ and $Z'_{14}$, we find their first return maps $T'_4$ and $T'_{14}$. The result is now positive now and can be captured by the following lemmas. Thus, self-similarity exists in the case of a regular dodecagon. The structure of $T'_4$ is shown in Fig. 20. 7.5. Proof of Theorem 1The self-similarity found in Lemma 58 can be used to prove the existence of an aperiodic trajectory, that is, to prove Theorem 1 in the case of a regular dodecagon. Let $X$ be the open polygon $T'^{-2}(Z'_{14})$. We put $\Gamma_X=\Gamma_1 \circ T^{-2}$ and write $T'_X$ for the first return map with respect to $T'$ on $X$. Clearly, $X=\Gamma_X(Z'_4)$. We can see from Fig. 21 that $Z'_{14}$ is not split by $T'^{-2}$. Moreover, $T'^{-2}$ does not separate $Z'_{14}$ from $\beta_{14}$. Therefore $X$ turns out to be ‘sandwiched’ between the domains $\beta_2$, $\beta_3$ and $T'^{-2}(\beta_{14})$. This yields Lemma 59 (without further computer verification). Lemma 59. 1) The polygon $X$ has a nice self-return. 2) The transformation $T'_4$ is conjugate to $T'_X$ with respect to $\Gamma_X$. Let $c_{\mathrm{inf}} \in X$ be a fixed point of the map $\Gamma_X\colon Z'_4 \to Z'_4$. The following Lemma 60 completes the proof of Theorem 1 in the case of a regular dodecagon. Proof. Consider the following sequence $C_0, C_1, C_2, \dots$ of periodic components: $C_0=\beta_3$, $C_1=\beta_2$, $C_2=T'^{-2}(\beta_{14})$ and $C_{i+3}=\Gamma_X(C_i)$ for $i=0, 1,\dots$ .
The components $C_0$, $C_1$ and $C_2$ are shown in Fig. 21. Since $\Gamma_X$ is a contraction, the sequence $(C_i)$ converges to $c_{\mathrm{inf}}$. As in the case of a regular octagon, one can show that the point $c_{\mathrm{inf}}$ is neither finite nor periodic. Hence $c_{\mathrm{inf}}$ is aperiodic. $\Box$ 7.6. Proof of Theorem 2Note that a complete description of all periodic domains was not needed in the proof of the existence of an aperiodic point. However, we need it in the proof of Theorem 2. It does not follow from the previous analysis that the set of points in $Z'_4$ whose trajectories are disjoint from $Z'_{14}$ splits into a finite union of periodic domains. Moreover, we need to verify a similar fact for $Z'$ and $Z'_4$. To do this we consider Algorithm 3, which uses the ‘Queue’ data structure. This structure supports the operations ‘add an element to the queue’ and ‘retrieve the element added before the other elements of the queue’. This algorithm is realized by the function findPeriodicComponents (see https://github.com/dprpavlin/OuterBilliards/blob/master/workWith12Gon.py). As in the case of Algorithm 2, it is possible that Algorithm 3 will not terminate. However, the following lemma holds. Lemma 61. If Algorithm 3 terminates, then the answer found is correct. Proof. It is easy to prove by induction that the boundary of every polygon in our algorithm consists of finite points only.
Consider an arbitrary non-finite point $p \in Z' \setminus S$ and define a sequence of polygons $F_0, F_1, F_2, \dots$ in the following way: a) $F_0=Z'$; b) for every $i \in \mathbb{Z}_+$, $F_i=T'(F_{i-1} \cap \alpha_j)$, where $j=j(i)$ is such that $T^{i-1}(p)\,{\in}\,\alpha_j$. We actually have $F_i=Z' \cap \{T'^i(q) | \rho'(q)[0,i-1]=\rho'(p)[0,i-1]\}$. The boundaries of all the polygons $F_i$ clearly consist of finite points. It is clear from the structure of the algorithm that if some polygon $F_i$, $i \in \mathbb{Z}_{\geqslant 0}$, falls within the scope of the algorithm (that is, occurs in $performedPolygons$), then so does $F_{i+1}$, as long as $F_{i+1} \not\subseteq S$. Consider two cases: – There is an $n \in \mathbb{Z}_+$ such that $T'^n(p) \in S$. Since Algorithm 2 terminates for $S$ and the point $p$ is non-finite, we can see that $p$ lies on the trajectory of first return to $S$, that is, inside one of the domains of a first return trajectory to $S$. (Indeed, if the points $p, T^{-1}(p), T^{-2}(p),\dots$ do not belong to $S$, then $p$ is aperiodic, hence the algorithm performs at least $j$ operations $ncp=T'(ncp)$ for every point $T^{-j}(p)$, contradicting termination in finite time.) – There is no such $n$. Then all the polygons in the sequence $F$ fall within the scope of the algorithm. However, the algorithm terminates. This means that $F$ eventually becomes a cyclic sequence. For example, let $k, l \in \mathbb{Z}_+$ be such that $k<l$ and $F_k=F_l$. Then $F_k$ is the periodic component containing $T'^k(p)$. (All the points of $F_k$ are clearly periodic, while the boundary of $F_k$ consists of finite points only.) In other words, $\operatorname{comp}(T'^k(p))$ falls within the scope of the algorithm. Hence $\operatorname{comp}(p)$ and the whole periodic trajectory of $\operatorname{comp}(p)$ fall within the scope of the algorithm, and at least one of the polygons in this trajectory occurs also in $ans$. We also note that no polygons in the periodic trajectory intersect $S$ (otherwise such a polygon would have a non-empty intersection with $\partial S$ and, therefore, would contain finite points). Thus, every non-finite point in $Z'$ occurs either on the trajectory of first return to $S$ (including $S$ itself) or on the trajectory of a periodic component whose trajectory is disjoint from $S$. Any such periodic component will necessarily be considered (since $p$ can belong to all of them). Hence the number of such components is finite (because the algorithm terminates). Since the closure of the set of all non-finite points in $Z'$ gives the closure of $Z'$ (see Lemma 3), we can conclude that the union of periodic components whose trajectories are disjoint from $S$ and first return trajectories of $S_1$ to $S$ is indeed a partition of $Z'$. Thus, we have shown that $ans$ contains a good subset of periodic domains from the standpoint of the answer. We now consider all the operations related to $performedPolygons$ and prove that there are no ‘redundant’ polygons in $ans$. Indeed, for any polygon occurring in $performedPolygons$, any operation other than verification of its occurrence in $performedPolygons$ will be performed at most once. Moreover, the structure of the internal $while$ is such that if a periodic component $\operatorname{comp}(p)$ whose trajectory is disjoint from $S$ occurs in $performedPolygons$, then all the polygons in the trajectory of $\operatorname{comp}(p)$ occur simultaneously in $performedPolygons$ and, therefore, exactly one of these polygons will occur in $ans$. It is also clear that for every non-finite point $p$ whose trajectory is disjoint from $S$, only a polygon from the trajectory of $\operatorname{comp}(p)$ can be added to $ans$ (since these polygons are not separated by $T'$ while any superset of any of them must be separated by some iteration). Thus, we need only to prove that no non-finite point whose trajectory passes through $S$ will be covered by a polygon in $ans$. Indeed, let $p$ be such a point and let $W_p$ be its polygon. It follows from what has been said above that $W_p= \operatorname{comp}(p)$. Hence there is a $k \in \mathbb{Z}_+$ such that $T'^k(\operatorname{comp}(p))$ is not disjoint from $S$. Since all the points of $\partial S$ are finite, $T'^k(\operatorname{comp}(p)) \subseteq S$. Hence the algorithm stops processing the trajectory of $\operatorname{comp}(p)$ at this point without completing the loop, and $W_p$ does not occur in $ans$. $\Box$ We run Algorithm 3 with $S=Z'_4$ as input data. The result is described in the following lemma. Lemma 62. Algorithm 3 terminates for $S= Z'_4$, with the returned array $ans$ containing seven polygons. It follows from Lemmas 62 and 61 that the periodicity problems can be reduced to problems on $Z'_4$. This idea is stated more formally in the following lemma. Lemma 63. 1) An aperiodic point with respect to $T'$ inside $Z'$ exists if and only if there is an aperiodic point with respect to $T'$ or $T'_4$ inside $Z'_4$. 2) The set of points periodic with respect to $T'$ is a set of full measure in $Z'$ if and only if the set of points periodic with respect to $T'$ or $T'_{4}$ is a set of full measure inside $Z'_4$. In view of the results in § 7.5, the main point of the first part of Lemma 63 is the absence of aperiodic points whose trajectories lie strictly outside $Z'_4$. Now we run Algorithm 3 for $S=Z'_{14}$ (applying $T'$, not $T'_4$, to polygons). The result is described in the following lemma. Lemma 64. Algorithm 3 terminates for $S= Z'_{14}$, with the returned array $ans$ containing 20 polygons. Thus, there are $20-7=13$ periodic components with respect to $T'_{Z'_4}$ in $Z'_4$ such that the trajectories of these components are disjoint from $Z'_{14}$ and from one another. Using these results, we need only to prove the following lemma. Combined with Lemmas 63 and 36, it yields Theorem 2 for a regular dodecagon. Proof. The proof follows the same scheme as Lemma 48 and uses Lemma 47. Namely, we introduce the following sequence of partitions $(\mathcal{Z}^l)_{l \in \mathbb{Z}_+}$ (coloured red and green) of the polygon $\overline{Z'_4}$:
a) $(\mathcal{Z}^1)$ consists of the eight domains into which $T'_4$ splits $Z'_4$. b) We have $\mathcal{Z}^{l+1}= \bigl\{\bigcup\{\overline{T_{4}^{\prime j}(\Gamma(\operatorname{int}(P)))}\} \bigm| P \in \mathcal{Z}^l, 0 \leqslant j<t(P)\bigr\} \cup \mathcal(Q)$ for every $l \in \mathbb{Z}_+$. Here $t(P)$ is the smallest $t \in \mathbb{Z}_+$ such that $(T_{4}^{\prime t}(\operatorname{int}(P))) \subset Z'_\Gamma$, and $\mathcal(Q)$ is the set of periodic domains inside $Z'_{4}$ whose trajectories are disjoint from $Z'_{14}$. (The set $\mathcal(Q)$ is generated by thirteen domains and by the transformation $T'_4$.) c) $\overline{T_4^{\prime j}(\Gamma(\operatorname{int}(P)))}$ is green in the partition $(\mathcal{Z})^{l+1}$ if and only if $P$ is green in $(\mathcal{Z})^l$. d) The domains in $\mathcal(Q)$ are green in all partitions where they occur. To use Lemma 47 it suffices to show that the partitions of every polygon in $(\mathcal{Z}^1)$ eventually contain a green domain. By using a computer program, one can verify that this holds for the partition $(\mathcal{Z}^3)$ (but not $(\mathcal{Z}^2)$, as in the octagon case). Hence the hypotheses of Lemma 47 hold with $k=2$ and with $\epsilon$ equal to the smallest ratio of the area of green domains in $\mathcal{Z}^3$ lying inside a certain element of $\mathcal{Z}^1$ to the area of this element. It follows that the union of green periodic domains is a set of full measure inside $Z'_4$. $\Box$ Thus, the fullness of measure of the set of periodic points for the outer billiard outside a regular dodecagon is established by a computer-assisted proof. In § 8 we use some simple additional calculations to describe all the period codes and the set of periods itself. § 8. Application: computation of periods and period codes of periodic pointsTo study the possible values of the periods for outer billiards outside regular polygons we first introduce the main definitions, then state and prove some general facts and, finally, use these facts to obtain the sets of periods for the outer billiards outside regular octagons and dodecagons. 8.1. Outer billiards: reduction to the bounded caseWe reconsider outer billiards. Let $\gamma$ be a regular $n$-gon. Then all the definitions introduced in §§ 3–5 hold for $\gamma$. We use Lemma 9 to seek the periods and period codes of periodic components. We begin with some notation extending the definitions in § 4. Let $S$ be an arbitrary subset of $V'$. Then we put a) $P_S := \{\rho'_{\mathrm{per}}(p) \mid p \subseteq S,\, p\text{ is a periodic component}\}$; b) $C_S := \{c(w) \mid w \in C_{S} \}$. We shall work with the sets $S=V'$ and $Z' \cup \beta$, $Z'$. Note that $P_{Z' \cup \beta}=P_{Z'} \cup \{n'-1\}$ and $C_{Z' \cup \beta}=C_{Z'} \cup \{(0, 0, \dots, 0, 1, 0)^\top\}$ (because $\rho'_{\mathrm{per}}(\beta)=n'-1$). Lemma 15 can be restated in the following way in terms of the Abelianization homomorphism and the incidence matrix (we recall that $n'\,{=}\,\lceil n/2 \rceil$). We introduce a substitution $\psi$ induced by the conjugacy in Lemma 24. Let $A=\{1, 2, \dots, n'\}$ be an alphabet. If $n$ is even, we define $\psi\colon A\to A^* \setminus \{\epsilon \}$ by the following rules:
$$
\begin{equation*}
i \to \underbrace{(n'-1)(n'-1)\dots (n'-1)}_{n'-i \text{ times} }n',\qquad i=1, 2, \dots, n'.
\end{equation*}
\notag
$$
Lemma 67 follows from the description of the first return map $T'$ on $\operatorname{ang}_1$ in Lemma 24 and its proof. Lemma 67. Let $p \in V'$ be a non-finite point. Then a) $\rho'(H(p))=\psi(\rho'(p))$; b) $\rho'_{\mathrm{per}}(H(p))= \psi(\rho'_{\mathrm{per}}(p))$ if $p$ is periodic. The following lemma is a direct consequence of Lemmas 67 and 35. It reduces the computation of period codes and periods to the computation of $P_{Z' \cup \beta}$ and $C_{Z' \cup \beta}$. Lemma 68. Let $n$ be even. Then a) $P_{V'}=\operatorname{cycle}(\{\psi^l w\mid w \in P_{Z'\cup \beta},\, l \in \mathbb{Z}_{\geqslant 0} \})$; b) $C_{V'}=\{M_{\psi}^l w\mid w \in C_{Z'\cup \beta},\, l \in \mathbb{Z}_{\geqslant 0} \}$. We note that one can similarly obtain the following lemma, which makes it possible to find the Abelianizations of the period codes and, by Lemma 66, the periods of points for the outer billiard outside a regular $n'$-gon, where $n'$ is odd. Given a $k \times l$-matrix $M$ and a set $S$ of $l$-component vectors, we put $MS=\{Ms\mid s \in S\}$. Lemma 69. Suppose that $n$ is even and $n'=n/2$ is odd. Let $P(C)^{n'}_{V'}$ (resp. $P(C)^{n}_{V'}$) be the analogue of $P(C)_{V'}$ for the outer billiard outside a regular $n'$-gon (resp. $n$-gon) and let $P(C)^{n'}_{Z' \cup \beta}$ be the corresponding analogue of $P(C)_{Z' \cup \beta}$. Define a substitution $\psi'$ by the rules Then a) $P^{n'}_{V'}=P^{n'}_{Z' \cup \beta} \cup \operatorname{cycle}(\psi'(P^{n}_{V'}))$; b) $C^{n'}_{V'}=C^{n'}_{Z' \cup \beta} \cup M_{\psi'}(C^{n}_{V'})$. 8.2. Computation of periods for $n= 8$In the case of a regular octagon, we calculate the sets $P_{Z'}$ and $C_{Z'}$ and then use Lemmas 68, 14 and 9 in order to find all possible periods as well as period codes and their Abelianizations, that is, we prove Theorem 3. Since $n=8$, all the columns $C_{Z'}$ have $n'=4$ coordinates. However, since 4 does not occur in the codes of the points in $Z'$, the fourth coordinate of the columns $C_{Z'}$ is always zero. Therefore it is convenient to seek the 3-component columns $C^3_{Z'}$ obtained from $C_{Z'}$ by deleting the last coordinate. We now introduce the following substitution induced by the self-similarity in Lemma 39. Definition 32. Let $\sigma\colon \{1, 2, 3\} \to \{1, 2, 3\}^* \setminus \{\epsilon\}$ be the substitution given by the following rules: $1 \to 122323232323221$, $2 \to 122323221$ and $3 \to 111$. Note that this substitution $\sigma$ coincides with that denoted by $\sigma$ in Theorem 3. It follows that
$$
\begin{equation*}
M_{\sigma} = \begin{pmatrix} 2 & 2 & 3 \\ 8 & 5 & 0 \\ 5 & 2 & 0 \end{pmatrix}.
\end{equation*}
\notag
$$
We state an advanced version of Lemma 40 (with a similar proof). Lemma 70. Let $p \in Z'$ be a non-finite point. Then a) $\rho'(\Gamma(p))=\sigma(\rho'(p))$; b) $\rho'_{\mathrm{per}}(\Gamma(p))= \sigma(\rho'_{\mathrm{per}}(p))$ and $c(\Gamma(p))=M_{\sigma}(c(p))$ if $p$ is periodic. Lemma 71 is a direct corollary of Lemmas 70 and 42. Lemma 71. 1) $P^3_{Z'}=\operatorname{cycle}(\{\sigma^k(w) \mid k \in \mathbb{Z}_{\geqslant 0},\, w \in \{\beta_1, \beta_2, \beta_{32}\})$; 2) $C^3_{Z'}=\{M_{\sigma}^k(w) \mid k \in \mathbb{Z}_{\geqslant 0},\, w \in C_{\operatorname{rk}0}\}$ and $C_{\operatorname{rk}0}=\{c(\beta_1), c(\beta_2), c(\beta_{32})\}=\{(1, 0, 0)^\top, (0, 1, 0)^\top, (0, 1, 1)^\top\}$. The linear transformation $g\colon \mathbb{R}^3 \to \mathbb{R}^3$ with matrix $M_{\sigma}$ has a basis of eigenvectors $v_1=(1, -2, 1)^\top$ with eigenvalue $1$, $v_2=(1, -1, -1)^\top$ with eigenvalue $-3$ and $v_3=(1, 2, 1)^\top$ with eigenvalue $9$. This means that if $v=a_1v_1+a_2v_2+a_3v_3$, where $a_1,a_2,a_3 \in \mathbb{R}$, then $M_{\sigma}^k v=a_1\cdot v_1+a_2 \cdot (-3)^k \cdot v_2+a_3 \cdot 9^k \cdot v_3$ for every $k \in \mathbb{Z}_{\geqslant 0}$. By finding suitable $a_1$, $a_2$ and $a_3$ for each of the following vectors, we can rewrite Lemma 71 in the following form. Lemma 72. We also note that $P^3_{Z' \cup \beta}=P^3_{Z'} \cup \{3\}$ and $C^3_{Z' \cup \beta}= C^3_{Z'} \cup {(0, 0, 1)^\top}$. Combining this with Lemma 68, we can easily obtain the first part of Theorem 3. To calculate $C_{V'}$ we note that if $n=8$, then the substitution $\psi$ in § 8.1 is of the form $1 \to 3334$, $2 \to 334$, $3 \to 34$, $4 \to 4$. It follows that
$$
\begin{equation*}
M_{\psi}=\begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 3 & 2 & 1 & 0\\ 1 & 1 & 1 & 1 \end{pmatrix}.
\end{equation*}
\notag
$$
We use Lemma 68. Note that if $a, b, c \in \mathbb{R}$, $M_{\psi}((a, b, c, 0)^\top)=(0, 0, c', d')$ and $l \in \mathbb{Z}_+$, then $M^{l+1}_{\psi}((a, b, c, 0)^\top)=(0, 0, c', d'+c'\cdot l)^\top$. Combining this with Lemma 68, we obtain the second part of Theorem 3. To find the sets $B_c$ and $B_p$ (defined in Lemma 9) in the case of a regular octagon we use Lemma 9 and the following lemma, which is a particular case of Lemma 66. We now inspect all the seven series $C_{V'}$. Let $p \subseteq V'$ be a periodic component. We consider the following cases: 1) $c'(p)=\dfrac{1}{8}\begin{pmatrix}1+4\cdot (-3)^k+3\cdot 9^k\\-2-4\cdot (-3)^k+6\cdot 9^k\\1- 4\cdot (-3)^k+ 3\cdot 9^k\\0\end{pmatrix}$, $k \in \mathbb{Z}_{\geqslant 0}$. Then
2) $c'(p)=\dfrac{1}{4}\begin{pmatrix}-1+9^k\\2+2\cdot 9^k\\-1+9^k\\0\end{pmatrix}$, $k \in \mathbb{Z}_{\geqslant 0}$. Then
3) $c'(p)=\dfrac{1}{8}\begin{pmatrix}1-4\cdot (-3)^k+3\cdot 9^k\\-2+4\cdot (-3)^k+6\cdot 9^k\\1+4\cdot (-3)^k+ 3\cdot 9^k\\0\end{pmatrix}$, $k \in \mathbb{Z}_{\geqslant 0}$. Then
4) $c'(p)=\begin{pmatrix}0\\0\\1\\l\end{pmatrix}$, $l \in \mathbb{Z}_{\geqslant 0}$. Then
5) $c'(p)=\dfrac{1}{2}\begin{pmatrix}0\\0\\6\cdot 9^k\\-(-3)^k+(3+6l)\cdot 9^k\end{pmatrix}$, $k,l \in \mathbb{Z}_{\geqslant 0}$. Then
6) $c'(p)=\begin{pmatrix}0\\0\\2\cdot 9^k\\(2l+1)\cdot 9^k\end{pmatrix}$, $k,l \in \mathbb{Z}_{\geqslant 0}$. Then
7) $c'(p)=\dfrac{1}{2}\begin{pmatrix}0\\0\\6\cdot 9^k\\+(-3)^k+(3+6l)\cdot 9^k\end{pmatrix}$, $k,l \in \mathbb{Z}_{\geqslant 0}$. Then
Combining the resulting seven series of answers, we obtain the sets of periods $B_p$ and $B_c$ in the case when $n=8$. To complete the proof of Theorem 3, notice that in the octagon case we have $B_p=B_c$, all the periods are divisible by $4$, and the series $4\cdot 9^k$, $8l+8$ and $(8l+12)\cdot 9^k$ contain all such numbers. 8.3. Computation of periods for $n=12$To conclude § 8 we assume that $n=12$ and prove Theorem 4. Since the calculations are cumbersome, we do not present all the periods explicitly as in the octagon case. The scheme of calculations is the same as in the octagon case, but it is modified since self-similarity was found for $Z'_4$ instead of $Z'$. Definition 33. Let $\omega_1,\dots,\omega_8$ be the open domains into which $T'_4$ splits $Z'_4$, enumerated in some order. The choice of order is irrelevant, in principle. For definiteness, we use the order returned by the realization of Algorithm 2 for $Z'_4$. Definition 34. Let ${p \in Z'_4}$ be a non-finite point (with respect to $T'$, hence $T'_4$). Then the $T'_4$-code $\rho'_4(p)$ is defined as the doubly infinite sequence $\dots w_{-2}w_{-1}w_0w_1w_2 \dots$ such that every $w_j$, $j \in \mathbb{Z}$, is the integer in the range from $1$ to $8$ satisfying $T_4^{\prime j}(p) \in \omega_{w_j}$. One can naturally define the period code $\rho'_{4, \mathrm{per}}(p)$ with respect to $T'_4$ as well as the Abelianization homomorphism $c'_4(p)=c(\rho'_{4, \mathrm{per}}(p))$ for the period code of any periodic point $p$. Definition 35. We define a substitution $\sigma\colon \{1,\dots,8\} \to \{1,\dots,8\}^* \setminus \{\epsilon\}$ in the following way by using a computer. Suppose that $i \in \{1,\dots,8\}$. Choose a computable point $p \in \Gamma_1(\omega_i)$ (for example, the barycentre of the vertices of $\Gamma_1(\omega_i)$) and let $k=k(i)$ be the smallest positive integer such that $T_4^{\prime k}(p) \in Z'_{14}$. Then $\sigma(i)=\rho'_4[0,k(i)-1]$. Note that the substitution $\sigma$ in the octagon case was obtained in exactly the same way. Using a computer once again, we can easily obtain the matrix $M_{\sigma}\colon 8 \times 8$ of the substitution $\sigma$. As a result of running the program, we can see that Lemma 74 holds. Lemma 74. We have the following equality between matrices: $M_{\sigma}=M_{88}$ (the latter is the matrix defined in the statement of Theorem 4). Lemma 75 holds as an automatic corollary, Lemma 75. Let $p \in \Gamma_4$ be a non-finite point. Then a) $\rho'_4(\Gamma_1(p))=\sigma(\rho'_4(p))$; b) $\rho'_{4, \mathrm{per}}(\Gamma_1(p))\,{=}\,\sigma(\rho'_{4, \mathrm{per}}(p))$ and $c'_4(\Gamma(p))\,{=}\,M_{88}(c'_4(p))$ if $p$ is periodic. By Lemmas 62 and 64, there are 13 periodic components inside $Z'_4$ whose periodic trajectories cover all non-finite points not lying on the trajectories of first return to $Z'_{14}$. These components form the set of ‘components of rank $0$’ (by analogy with $\beta_1$, $\beta_2$, $\beta_{32}$ in the regular octagon case). Having found these components with the aid of Algorithm 3, we can easily find the set of Abelianizations of the period codes of these components (by using a computer once again) and verify that this is precisely the set $F$ in Theorem 4. The proof of the following lemma is completely analogous to that of Lemma 71. Lemma 76. Put $C_4=C_{Z'_4}=\{ c'_4(p)\mid p \in Z'_4,\, p\text{ is a periodic point} \}$. Then $C_4=\{M^k_{88}f\mid f \in F\}$. The following steps are needed to pass from $C_4$ to $C_{Z'}$ and $C_{Z' \cup \beta}$: – Describe the periodic words for the trajectories in $C_{Z' \cup \beta}$ passing through $C_4$. To do this it suffices to introduce a substitution $\phi\colon \{1,\dots,8\} \to \{1,\dots,6\}^* \setminus \{\epsilon\}$ such that, for every $i \in \{1,\dots,8\}$, $\phi(i)$ is the return code $\rho'$ of the domain $\omega_i$ to $Z'_4$ (a formal definition can be given in a similar way as that of $\sigma$) and to apply this substitution to the period codes in $Z'_4$. It is easily verifiable by computer that $M_{\phi}=M_{68}$ (the matrix defined in the statement of Theorem 4). – Describe the periodic words for the trajectories in $C_{Z'}$ that do not pass through $Z'_4$. By Lemma 62, there are exactly seven periodic components whose trajectories (taken together) cover all non-finite points in $Z'$ not lying on the trajectories of first return to $Z'_4$. Having found these components with the aid of Algorithm 3, we can easily find the set of Abelianizations of the period codes of these components (using a computer once again), add $c'(\beta)$ to them, and verify that the result is exactly the set $G$ in Theorem 4. By combining these two steps, we obtain the following lemma. When $n=12$, the substitution $\psi$ takes the form $i \to 5^{6-i}6$ for $i=1, 2, \dots, 6$. Therefore $M_{\psi}=M_{66}$. Then Lemma 68 enables us to prove easily that $C_{V'}= H$. The rest of the proof of Theorem 4 can easily be obtained by using Lemmas 66 and 9. In terms of the latter lemma and Theorem 4, we have $B=B_c$ and $B_2= B_p$. Note that odd periods occur in the dodecagon case (for example, $\mathrm{per}(\beta_4)=\mathrm{per}(O_4)=3$) in contrast to the octagon case.
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\Bibitem{Ruk22} |
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