This is a preprint of: “Pointwise periodic maps with quantized first integrals”, Anna Cima, Ar- mengol Gasull, Vı́ctor Mañosa, Francesc Mañosas, Commun. Nonlinear Sci. Numer. Simul., vol. 108, 106150:1–26, 2022. DOI: [10.1016/j.cnsns.2021.106150] Pointwise periodic maps with quantized first integrals Anna Cimap1q , Armengol Gasullp1,2q , Vı́ctor Mañosap3q and Francesc Mañosasp1q p1q Departament de Matemàtiques, Facultat de Ciències, Universitat Autònoma de Barcelona, 08193 Bellaterra, Barcelona, Spain p2q Centre de Recerca Matemàtica, Campus de Bellaterra, 08193 Bellaterra, Barcelona, Spain cima@mat.uab.cat, gasull@mat.uab.cat, manyosas@mat.uab.cat p3q Departament de Matemàtiques, Institut de Matemàtiques de la UPC-BarcelonaTech (IMTech), Universitat Politècnica de Catalunya Colom 11, 08222 Terrassa, Spain victor.manosa@upc.edu Abstract We describe the global dynamics of some pointwise periodic piecewise linear maps in the plane that exhibit interesting dynamic features. For each of these maps we find a first integral. For these integrals the set of values are discrete, thus quantized. Furthermore, the level sets are bounded sets whose interior is formed by a finite number of open tiles of certain regular or uniform tessellations. The action of the maps on each invariant set of tiles is described geometrically. Mathematics Subject Classification 2010: Primary: 37C25, 39A23. Secondary: 37C55, 37J35, 52C20. Keywords: Periodic points; pointwise periodic maps; piecewise linear maps; quantized first integrals; regular and uniform tessellations. 1 1 Introduction A pointwise periodic map is a bijective self-map in a topological space such that each point is periodic. A periodic map is a bijective self-map in a topological space such that some iterated of the map is the identity. For a periodic map F : X ÝÑ X the minimum natural number p satisfying F p “ Id is called the period of F. Notice that a pointwise periodic map satisfying that the period of the points has an upper bound is periodic and its period is the least common multiple of the periods of the elements of the space. A classical result of Montgomery establishes that any pointwise periodic homeomeor- phism in an Euclidean space is periodic, [19]. Non-periodic but pointwise periodic bijective maps do exist when the continuity assumption is relaxed, see [23] for instance. In the series of papers [7, 8] and [10], the authors introduce three explicit examples of pointwise periodic maps that are not periodic. The examples given by these authors in the above mentioned references belong to the family of piecewise affine maps with a line of discontinuity: # `1, if y ě 0; Gpx, yq “ py, ´x ´ ρy ` signpyqq , where signpyq “ (1) ´1, otherwise, for |ρ| ă 2. In particular they correspond to the cases ρ P t´1, 0, 1u. There are other values of ρ for which there exist non-periodic points, see [9]. Notice also that maps (1) correspond to the second order discontinuous difference equations xn`2 “ ´xn ´ ρxn`1 ` signpxn`1 q. As we will see in next section, each map G is linearly conjugate with the piecewise rotation map ˜ ¸˜ ¸ cospαq sinpαq x ´ signpyq F px, yq “ , (2) ´ sinpαq cospαq y where ρ “ ´2 cospαq with α P p0, πq. Observe that the maps G with ρ “ ´1, 0 and 1 are conjugate with the maps F with α “ π{3, π{2 and 2π{3, respectively. As we will see, the normal form F regularizes the shape of the invariant sets and keeps the same discontinuity line y “ 0. These maps are included in the class of symmetric maps studied in the remarkable paper [14] together with other more general piecewise rotations, see a further comment below. As noticed in [5], they exhibit complex dynamics and they belong to the type of piecewise rotations with the same rotation angle that elude the generic dichotomy that appears in most piecewise rotations of being globally attracting or globally repelling maps, see [5, Theorem 1]. Piecewise affine maps with a line of discontinuity appear as models in many fields like in the study of mechanical systems with friction, power electronics, relay control systems or economics [2, 6, 24]. In fact, as is explained in [7, 8, 10], the three maps (1) with ρ P t´1, 0, 1u appear in the study of steady states of certain cellular neural networks. 2 Despite their apparent simplicity, piecewise affine maps exhibit great dynamic richness and a variety of phenomena that are characteristic of these systems, see [3, 5, 14, 22, 24] and references therein. As we will show, the examples considered in this paper are also very rich from a dynamical viewpoint, even though each orbit is periodic. In fact, one of our motivations was to highlight the beautiful features of these examples. Recall that a first integral of a discrete dynamical system associated with a map F is a non-constant real valuated function V such that V ˝ F “ V , which means that the level sets tV “ cu, typically called the energy levels, are invariant under the action of the map. It is known that periodicity issues are related with integrability since most continuous periodic maps are completely integrable (there exist as many functionally independent first integrals as the dimension of the phase space), see [11] and [12]. In this work we consider the piecewise affine maps F with α P tπ{3, π{2, 2π{3u under the light of their properties as integrable systems. For each of these three maps, we obtain a non-trivial first integral which is defined in an open and dense set of R2 and have discrete (or quantized) energy levels. Then we describe their global features in terms of the dynamics induced by the maps on the level sets of the first integrals. These level sets are bounded, with positive measure and their interior is formed by a finite number of some prescribed tiles of certain regular or uniform tessellations forming necklaces, see Figures 1, 2 and 3. The existence of necklaces in piecewise isometries is well known. For instance, in [14, Theorem 1 and Lemma 11] it is established the existence of some invariant necklaces defined by convex polygons containing periodic islands for a family of maps that contain the ones studied in this papers. These necklaces and the set of periods associated with their periodic orbits are characterized both analytically and geometrically, and its existence is the key to prove the boundedness of the orbits of the maps considered there. We want to point out that in our maps, all the integral’s level sets are necklaces. In Remark 2 we comment the relation between both families of necklaces. In addition, as we will see, the maps F with α P tπ{3, π{2, 2π{3u have also a second continuous first integral, see Remark 23. This second first integral, however, is not useful to control the set of periods. Planar piecewise isometries appear in the study of polygonal dual billiards ([13, 17, 20]). The results in the literature indicate that some polygonal dual billiards should also have quantized integrals, see Figure 3 in [20], Figure 2 in [13], or Figures 3–5 and the results in [23]. We believe that the explicitness of the analytic expression of the quantized integrals with positive measure level sets for the maps (2) is quite novel in the context of discrete dynamical systems theory. It is interesting to notice the fact that the regular tessellations that we find in this paper also appear in the study of some polygonal dual billiards like the one introduced by Moser in [20] or those that appear in [23]. Observe, however, that these 3 dual billiards are not conjugated to any map considered in our paper, because they exhibit different sets of periods. A consequence of our results for F, when α P tπ{3, π{2, 2π{3u, is the existence of an open and dense subset U on which the dynamics of the map is strongly stable and simple. We will see that for any x P U there exists an open neighborhood of x, say Ux , and nx P N such that F nx |Ux “ Id . Moreover, varying x P U, the values nx are unbounded. We will study the three cases separately in three different sections. In a few words, the main results that we will state in detail in the next section, are: (a) We present first integrals V for each case. See Section 6 for a constructive approach for obtaining them. (b) The interior of the level sets of each first integral is described in terms of some prescribed open tiles of a regular or uniform tessellation of R2 , see Figures 1, 2 and 3. In all cases, each of them is a necklace whose beads (the open tiles) are open sets having one of the following three shapes: squares (α “ π{2q, triangles and hexagons (α P tπ{3, 2π{3u). In the three figures the beads of a necklace have the same color. In fact, the shape and the number of beads, say M, only depend on the level set k and α. Moreover, the inter-tile dynamics can be described in a very simple way: if we collapse each of the open tiles in a point, the interior of tpx, yq : V px, yq “ cu can be identified with ZM , simply following the order given by the necklace in clockwise sense, were, as usual, given q P Z, we denote by Zq the set of the residue classes induced by the congruence n ” m if and only if n ´ m is a multiple of q with n, m P Z. Then we will prove that the dynamical system generated by F, restricted to this set, is conjugated to an affine discrete dynamical system generated by a map h : ZM Ñ ZM , where hpiq “ i ` upc, αq, for some upc, αq P ZM that we also determine explicitly in this paper, see Theorems A, B and C. Notice that, geometrically, F acts as a rotation among the beads of any necklace. A similar inter-tile dynamics’ description in the context of dual polygonal billiards can be found in [17], and also in the context of piecewise linear maps [14, Theorem 1 and Lemma 11]. Due to the above conjugation, the dynamics on the interior of each level set can be completely understood, see Theorems A, B and C. Roughly speaking, for each map and for each necklace (set of tiles with the same energy level), there exists a certain number k P tM, M {2u X N, that depends (explicitly) on the energy level, so that each tile is invariant by F k . Furthermore, on each tile, F k is a rotation of order p around the center of the tile, where p P t2, 4u when α “ π{2, or p P t3, 6u when α P tπ{3, 2π{3u and it is determined explicitly by the energy level. As a consequence, on each tile there 4 is a k-periodic point (the center) and the rest of points are kp-periodic. The dynamics in the necklaces is, therefore, a discrete version of an epicyclic motion around a discrete deferent which is the locus of the centers of the tiles, [21, p. 123]. As we will see, the dynamics on the boundaries of the tiles (edges and vertices) requires a little bit more elaborated description. (c) As a consequence of the above results and the study of the dynamics on the boundary of the level sets, for each map, we easily characterize the period of every point in terms of the value of its associate first integral and obtain the global dynamics of the map. For instance in Proposition 1 we present our results in an algorithmic way when α “ π{2. In particular, the set of periods of the maps are presented. 2 Preliminaries and main results The families of maps F and G given in (1) and (2), respectively, are linearly conjugated because ˜ ¸ ` ´1 ` ˘˘t 1 ´ cospαq F px, yq “ Q ¨ G Q ¨ px, yqt , where Q“ . 0 sinpαq In this paper we will, therefore, work with the above normalized one-parameter family of maps F. Notice that each map F is bijective with inverse ˜ ¸˜ ¸ ˜ ¸ ´1 cospαq ´ sinpαq x signpsin pαq x ` cos pαq yq F px, yq “ ` . sinpαq cospαq y 0 Notice also that F is discontinuous in the set LC0 “ tpx, 0q : x P Ru. We will consider the critical lines LC´i “ tpx, yq such that F i px, yq P LC0 u, and also the critical set F “ Ť iPN LC´i formed by all the preimages of the critical line LC0 , where we use the notation introduced by Mira et al. in [18] (see also [1, 4]). We call the open set U “ R2 zF, the zero-free set because none of the orbits starting at point in U touches the discontinuity line LC0 , where the second coordinate of the points is zero. Regarding the above conjugation, of course it is only defined in the case sinpαq ‰ 0 which corresponds with the cases ρ ‰ ˘2. In the case α “ 0 (resp. α “ π) the map F (resp. F 2 ) has trivial dynamics of translation type, and do not correspond to the initial family G with ρ “ ˘2. For each α P tπ{2, 2π{3, π{3u we introduce some specific notations and also state our main results: Theorems A, B and C, respectively. Each one of them will be proved in a different section. The results in these theorems have the following structure: in (i) we 5 characterize the geometry of the critical set; in (ii) we state the existence of a first integral in the non-critical set and we characterize the global dynamics in this set proving that is conjugated with the composition of two rotations; in (iii) we establish the dynamics in the critical set; in (iv) we characterize the set of periods of the maps. While statements (iv) are known in the literature, the geometric description given in statements (i)–(iii) is, as far as we know, novel. 2.1 The case α “ π{2 When α “ π{2, the map F is the one in (1) with ρ “ 0 and was studied in [8]. Consider Fπ{2 the grid formed by the straight lines x “ k and y “ `, with k, ` P Z. This grid defines the square Euclidean regular tiling [15, 16], also named quadrile, see Figure 1. Each (open) tile is denoted by Tk,` “ tpx, yq; such that k ă x ă k ` 1 and ` ă y ă ` ` 1u. The centers of each of these tiles are denoted by pk,` “ pk ` 1{2, ` ` 1{2q . We also introduce the set ď Uπ{2 “ Tk,` “ R2 zFπ{2 , pk,`qPZ2 and the function Vπ{2 px, yq “ max p|Epxq ` Epyq ` 1| ´ 1, |Epxq ´ Epyq|q , (3) where Epzq “ tzu is the floor function of z P R that recall gives as output the greatest integer less than or equal to z. We also define Vk,` “ Vπ{2 ppk,` q and denote N0 “ N Y t0u. We prove: Theorem A. Consider the discrete dynamical system (DDS) generated by the map F given in (2) with α “ π{2, F px, yq “ py, ´x ` signpyqq. Then: (i) Its critical set is F “ Fπ{2 . (ii) The function V “ Vπ{2 is a first integral of F on the free-zero set U “ Uπ{2 . Each level set tV px, yq “ cu X U, with c P N0 , is a necklace formed by 4c ` 2 squares, see Figure 1. If we identify each square with a point (for instance the center), the DDS restricted to this set is conjugated with the DDS generated by the map h : Z4c`2 Ñ Z4c`2 , hpiq “ i ` c. As a consequence, when c is odd (resp. even), each square in this level set is invariant by F 4c`2 (resp. F 2c`1 ) and restricted to this square, F 4c`2 (resp. F 2c`1 ) is a rotation of order 2 (resp. 4), around the center of the tile. In particular, all points but the center in each of these tiles have period 8c ` 4. 6 (iii) All orbits with initial condition on F are p8n ` 4q-periodic for some n P N0 , see Theorem 10 for more details. (iv) The map F is pointwise periodic. Furthermore, its set of periods is PerpF q “ t4n ` 1; 8n ` 4; and 8n ` 6 for all n P N0 u . Figure 1: Level sets of the first integral V of F for α “ π{2, given in (3). In each tile Tk,` , the level Vk,` and the period of the center pk,` are indicated, between brackets. The other points in the tile have period 8Vk,` ` 4. Item pivq was already proved in [8]. All the geometric description of the dynamics of F given in the other items is new. Observe that the statement (ii) in the above result can be formalized in the following way: the dynamics of F on each necklace tV “ cu X U with c P N0 , is conjugate with the 7 dynamics of the map ϕ : Z4c`2 ˆ Zq ÝÑ Z4c`2 ˆ Zq pi, jq ÝÑ pi ` c, j ` 1q where q “ 2 if c is odd, and q “ 4 if c is even. This map can be seen as the product of two finite order rotations, its first component gives the dynamics on the discrete deferent formed by the set of centers of tiles, and the second component gives the dynamics on a epicycle. A similar situation is described in statements (ii) of Theorems B and C. Also notice that a simple check shows that the function V is not a first integral of F on the whole plane, since the relation V pF q “ V is not satisfied for some points in F “ R2 zU. As a consequence of the above theorem we can easily give a simple algorithm to know the period of each orbit in terms of its initial condition. Recall that given a point px, yq P R2 , k “ Epxq, ` “ Epyq and Vk,` “ Vπ{2 ppk,` q. For the forthcoming cases α P t2π{3, π{3u, from our results a more complicated algorithm could be obtained, but for the sake of brevity, we do not detail it. Proposition 1. Any point px, yq P R2 is a p-periodic point of F , where: (a) When x R Z and y R Z and, moreover, either x ´ k ‰ 1{2 or y ´ ` ‰ 1{2, then p “ 8Vk,` ` 2. When x ´ k “ 1{2 and y ´ ` “ 1{2, then p “ 2Vk,` ` 1 if Vk,` is even, and p “ 4Vk,` ` 2 if Vk,` is odd. (b) When x P Z and y R Z, if k is even, p “ 8Vk,` ` 4 and if k is odd, p “ 8Vk´1,` ` 4. (c) When x R Z and y P Z, if ` is even, p “ 8Vk,` ` 4 and if ` is odd, p “ 8Vk,`´1 ` 4. (d) If x P Z and y P Z, then: (i) When k is even, p “ 8Vk,` ` 4 if ` is even, and p “ 8Vk,`´1 ` 4 if ` is odd. (ii) When k is odd, p “ 8Vk´1,` ` 4 if ` is even, and p “ 8Vk´1,`´1 ` 4 if ` is odd. The statement (d) in the above result is a consequence of Theorem 10. To obtain the result we will identify some tiles, that we will call perfect squares, such that their boundaries (including their edges and all the vertices in F) avoid the discontinuity effects and, therefore, the points on the boundary of such a tile have the same periodic behavior as the interior points, except their centers. To study the periodicity in the rest of the edges (without vertices) we will associate them with an appropriate tile, so that the points in the edge follow the periodic behavior of the interior points. See Section 3.3 for more details. 8 2.2 The case α “ 2π{3 In this case, the map F in (2) is conjugate with the map G in (1) with ρ “ 1, which was studied in [10]. ? ? We define F2π{3 as the grid formed by the straight lines y “ 3px ´ 2kq, y “ 3` and ? y “ ´ 3px ´ 2m ´ 1q, with k, `, m P Z and call U2π{3 “ R2 zF2π{3 . Notice that U2π{3 is the (open) trihexagonal Euclidean uniform tiling (the tessellation 3.6.3.6 in the notation of [16]), see Figure 2. In fact, each tile in U2π{3 is defined by ? ? ? ? Tk,`,m “ px, yq, such that 3px ´ 2k ´ 2q ă y ă 3px ´ 2kq, 3` ă y ă 3p` ` 1q, ? ? ( and ´ 3px ´ 2m ` 1q ă y ă ´ 3px ´ 2m ´ 1q , with m P tk ` `, k ` ` ` 1, k ` ` ` 2u, where k “ Bpx, yq, ` “ Cpyq and m “ Dpx, yq, being ´ ? ¯ ´? ¯ ´ ? ¯ Bpx, yq “ E p3x ´ 3yq{6 , Cpyq “ E 3y{3 and Dpx, yq “ E p3x ` 3y ` 3q{6 . Moreover, • The tile Tk,`,k```1 is a regular hexagon and its geometric center (simply center, from ` ? ˘ now on) is the point pk,` “ 2k ` ` ` 3{2, 3p` ` 1{2q ; • The tiles Tk,`,k`` and Tk,`,k```2 are equilateral triangles whose respective centers are ` ? ˘ ` ? ˘ qk,` “ 2k ` ` ` 1{2, 3p` ` 1{6q and rk,` “ 2k ` ` ` 5{2, 3p` ` 5{6q ; and the adherence of the union of the three tiles is a parallelogram whose sides are y “ ? ? ? ? 3` , y “ 3p` ` 1q , y “ 3px ´ 2kq and y “ 3px ´ 2k ´ 2q, see Figure 7 and Lemma 13 for more details. Finally, we introduce the function ` V2π{3 px, yq “ max |Bpx, yq ´ Cpyq ` Dpx, yq| , ˘ |Bpx, yq ` Cpyq ` Dpx, yq ` 1| ´ 1, |´Bpx, yq ` Cpyq ` Dpx, yq| . (4) Observe that its level sets are discrete and ImagepV2π{3 q “ N0 . Clearly, V2π{3 is constant on each tile Tk,`,m and we denote its value as Vk,`,m “ max p|k ´ ` ` m|, |k ` ` ` m ` 1| ´ 1, | ´ k ` ` ` m|q . (5) Theorem B. Consider the discrete dynamical system generated by the map F given in (2) with α “ 2π{3. Then: (i) Its critical set is F “ F2π{3 . (ii) The function V “ V2π{3 is a first integral of F on the free-zero set U “ U2π{3 “ R2 zF2π{3 . 9 (a) Each level set tV px, yq “ cu, with c P 2N0 even, in U is a necklace formed by 6c ` 2 triangles, see Figure 2. If we identify each triangle with a point (the center, for instance), the DDS restricted to this set is conjugated with the DDS generated by the map h : Z6c`2 Ñ Z6c`2 , hpiq “ i ` 2c. As a consequence, each tile in this level set is invariant by F 3c`1 and restricted to this triangle, F 3c`1 is a rotation of order 3 around the center of the tile. In particular, all points but the center in each of these tiles have period 9c ` 3. (b) Each level set tV px, yq “ cu, with c P 2N0 ` 1 odd, in U is a necklace formed by 3c ` 1 hexagons, see Figure 2. If we identify each hexagon with a point, the DDS restricted to this set is conjugated with the DDS generated by the map h : Z3c`1 Ñ Z3c`1 , hpiq “ i ` c. As a consequence, each tile in this level set is invariant by F 3c`1 and restricted to this hexagon, F 3c`1 is a rotation of order 3 around the center of the tile. In particular, all points but the center in each of these tiles have period 9c ` 3. (iii) All orbits with initial condition on F are periodic with period 9n ` 3 for some n P N0 . (iv) The map F is pointwise periodic. Furthermore, its set of periods is PerpF q “ t3n ` 1 and 9n ` 3 for all n P N0 u . Similarly to Theorem A, item pivq was already known, see [10]. Again, all the geometric description of the dynamics of F given in the other items is new. From statement (ii), on each necklace, F is conjugate with the product of rotations ϕ : Z6c`2 ˆ Z3 ý given by ϕpi, jq “ pi ` 2c, j ` 1q when c is even, and ϕ : Z3c`1 ˆ Z3 ý given by ϕpi, jq “ pi ` c, j ` 1q when c is odd. 10 Figure 2: Level sets of the first integral V given in (4). In each tile the level and the period of the center are indicated respectively between brackets. 2.3 The case α “ π{3 In this last case, the map F in (2) is conjugate with the map G in (1) with ρ “ ´1, which was studied in [7]. ? ? We consider Fπ{3 the grid formed by the straight lines y “ 3px ´ 2k ´ 1q; y “ 3` ? and y “ ´ 3px ´ 2mq, with k, `, m P Z that, again, form a trihexagonal Euclidean uniform tiling which is a translation of the one that appeared in the previous case α “ 2π{3, see Figure 3. The interior of each tile is defined by ? ? ? ? Tk,`,m “ px, yq, such that 3px ´ 2k ´ 1q ă y ă 3px ´ 2k ` 1q, 3` ă y ă 3p` ` 1q, ? ? ( and ´ 3px ´ 2mq ă y ă ´ 3px ´ 2m ´ 2q . As before, we call Uπ{3 the complement of this grid. Any point px, yq P Uπ{3 belongs 11 (only) to the tile Tk,`,m with k “ Bpx, yq, ` “ Cpyq and m “ Dpx, yq, where ´ ? ¯ ´? ¯ ´ ? ¯ Bpx, yq “ E p3x ´ 3y ` 3q{6 , Cpyq “ E 3y{3 and Dpx, yq “ E p3x ` 3yq{6 and now it can be seen that m “ k ` ` ´ 1 or m “ k ` ` or m “ k ` ` ` 1. In this case, • The tile Tk,`,k`` is a regular hexagon and its center is at the point ´ ? ? ¯ pk,` “ 2k ` ` ` 1{2, 3` ` 3{2 . • The tiles Tk,`,k``´1 and Tk,`,k```1 are equilateral triangles whose centers are qk,` “ ` ? ? ˘ ` ? ? ˘ 2k ` ` ´ 1{2, 3` ` 3{6 and rk,` “ 2k ` ` ` 3{2, 3` ` 5 3{6 , respectively. We also introduce the following function ` Vπ{3 px, yq “ max |Bpx, yq ´ Cpyq ` Dpx, yq| , |Bpx, yq ` Cpyq ` Dpx, yq ` 1| ´ 1 ˘ , |´Bpx, yq ` Cpyq ` Dpx, yq ` 1| ´ 1 . (6) Observe that by construction, it is constant on each tile Tk,`,m . Hence we can associate to each point in this tile, the value Vk,`,m “ max p|k ´ ` ` m|, |k ` ` ` m ` 1| ´ 1, | ´ k ` ` ` m ` 1| ´ 1q . Our results for this case are collected in the next theorem. We remark that the proof of item piiiq will be the more complicated part of the paper. Theorem C. Consider the discrete dynamical system generated by the map F given in (2) with α “ π{3. Then: (i) Its critical set is F “ Fπ{3 . (ii) The function V “ Vπ{3 is a first integral of F on the free-zero set U “ Uπ{3 “ R2 zFπ{3 . (a) Each level set tV px, yq “ cu, with c P 2N0 even, in U is a necklace formed by 3c ` 2 hexagons, see Figure 3. If we identify each one of them with a point, the DDS restricted to this set is conjugated with the DDS generated by the map h : Z3c`2 Ñ Z3c`2 , hpiq “ i`c{2. As a consequence, when c “ 4j (resp. c “ 4j`2), each tile in this level set is invariant by F 3c{2`1 (resp. F 3c`2 ) and restricted to this hexagon, F 3c{2`1 (resp. F 3c`2 ) is a rotation of order 6 (resp. 3) around the center of the tile. In particular, all points but the center in each of these tiles have period 9c ` 6. 12 (b) Each level set tV px, yq “ cu, with c P 2N0 ` 1 odd, in U is a necklace formed by 6c ` 4 triangles, see Figure 1. If we identify each one of them with a point, the DDS restricted to this set is conjugated with the DDS generated by the map h : Z6c`4 Ñ Z6c`4 , hpiq “ i ` c. As a consequence, each tile in this level set is invariant by F 6c`4 and restricted to this triangle, F 6c`4 is a rotation of order 3 around the center of the tile. In particular, all points but the center in each of these tiles have period 18c ` 12. (iii) All orbits with initial condition on F are periodic with periods 36n`6, 18n`9, 18n`15 or 108n ` 72, for some n P N0 , for more details see Theorem 22. (iv) The map F is pointwise periodic. Furthermore, the set of periods is PerpF q “ t6n ` 1; 12n ` 8; 12n ` 10; 18n ` 9; 18n ` 15; 36n ` 6; 36n ` 24; 36n ` 30 and 108n ` 72 for all n P N0 u . Once more, although item pivq is known, see [7], all the geometric description of the dynamics of F given in the other items is new. From the above result, on each necklace, F is conjugate with the map ϕ : Z3c`2 ˆ Zq ý given by ϕpi, jq “ pi ` c{2, j ` 1q, where q “ 6 when c ” 0 mod p4q, and q “ 3 when c ” 2 mod p4q; or ϕ : Z6c`4 ˆ Z3 ý where ϕpi, jq “ pi ` c, j ` 1q, when c is odd. 13 Figure 3: Level sets of the first integral V of F for α “ π{3, given in in (6). In each tile the level and the period of the center are indicated respectively between brackets. Remark 2. As we have mentioned, in [14] an infinite number of necklaces of a family of maps that include the ones studied in this work are characterized. We want to note that Theorems A–C show that for our maps all energy levels are necklaces. In particular, in the case α “ π{2 the necklaces studied in [14] correspond to the energy levels whose centers have period 4n ` 1 (which are those with even energy level). Let us observe that from Theorem A we know that there are other necklaces whose period is 4n ` 2 (those with odd energy level). In the case α “ 2π{3, the necklaces in [14] cover all energy levels since all the necklaces have centers of period 3n ` 1. In the case α “ π{3 the necklaces in [14] are those whose centers have period 6n ` 1. Observe that Theorem C guarantees the existence of much more necklaces. 14 2.4 The address of a point We end this section with the concept of address of a point that will be used in the proof of Theorems A, B, C. Recall that every map in the considered one-parameter family F has discontinuity line LC0 “ ty “ 0u, so we introduce the sets H` “ tpx, yq P R2 : y ě 0u and H´ “ tpx, yq P R2 : y ă 0u, and we call F` and F´ the map F restricted to H` and H´ respectively. For any point px, yq P R2 we define its address Apx, yq as follows: # `, if px, yq P H` , Apx, yq “ ´, otherwise. Moreover for every n P N, we call the itinerary of length n of the point px, yq the sequence of n symbols I n px, yq “ pApx, yq, ApF px, yqq, . . . , ApF n´1 px, yqqq. Notice that if I n px, yq “ pi1 , . . . , in q then F n px, yq “ Fin ˝ Fin´1 ˝ ¨ ¨ ¨ ˝ Fi1 px, yq. For instance if px, yq P H` , F px, yq P H´ and F 2 px, yq P H´ , then the length 3 itinerary of the point is t`, ´, ´u, and F 3 px, yq “ F´ ˝ F´ ˝ F` px, yq. Lemma 3. Let J n “ pi1 , . . . , in q be a sequence of symbols of length n with ii P t`, ´u and consider the set BpJ n q “ tpx, yq P R2 such that I n px, yq “ J n u. Then BpJ n q is convex. Moreover F n restricted to BpJ n q is an affine map. Proof. The proof of the convexity follows easily by induction. If n “ 1, BpJ n q is either H` or H´ both convex sets. Assume that the result holds for sequences of length n ´ 1 and set J n´1 “ pi1 , . . . , in´1 q. Therefore we have BpJ n q “ tpx, yq P BpJ n´1 q : F n´1 px, yq P Hin u. Moreover, F n´1 restricted to BpJ n´1 q is the affine map G “ Fin´1 ˝ . . . ˝ Fi1 . So we have BpJ n q “ BpJ n´1 q X G´1 pHin q. This fact proves that BpJ n q is convex because it is the intersection of two convex sets. This ends the inductive proof of convexity. Furthermore, F n px, yq “ Fin ˝ Fin´1 ˝ ¨ ¨ ¨ ˝ Fi1 px, yq, for all px, yq P BpJ n q, showing that F n restricted to BpJ n q is an affine map. 15 3 Proof of Theorem A 3.1 Preliminaries We start by determining the set of tile centers, pk,` , such that V ppk,` q “ c for c P N0 . First, we observe that there are two tiles corresponding to the level set c “ 0. These two tiles contain the two fixed points of F which are: p0,0 “ p1{2, 1{2q that belongs to the tile T0,0 “ p0, 1q ˆ p0, 1q and p´1,´1 “ p´1{2, ´1{2q that belongs to T´1,´1 “ p´1, 0q ˆ p´1, 0q. It is easy to see that these two tiles are invariant. To describe the rest of level sets, we denote Q1 “ tpx, yq P R2 : x ă 0, y ą 0u, Q2 “ tpx, yq P R2 : x ą 0, y ą 0u, Q3 “ tpx, yq P R2 : x ą 0, y ă 0u, and Q4 “ tpx, yq P R2 : x ă 0, y ă 0u. Lemma 4. For each level set tV “ cu with c P N0 there are 4c`2 centers pk,` . Furthermore, for each natural number c ě 1 we have: (a) tpk,` : Vk,` “ cu X Q1 “ tpk ` 1{2, ` ` 1{2q : l “ k ` c, k “ ´c, ´c ` 1, . . . , ´1u. We denote by X1 , X2 , . . . , Xc these c centers for k “ ´c, ´c ` 1, . . . , ´1 respectively. Every one of them lies on the straight line y “ x ` c. (b) tpk,` : Vk,` “ cu X Q2 “ tpk ` 1{2, ` ` 1{2q : l “ ´k ` c, k “ 0, 1, . . . , cu. We denote by Xc`1 , Xc`2 , . . . , X2c`1 these c ` 1 centers for k “ 0, 1, . . . , c respectively. Every one of them lies on the straight line y “ ´x ` c ` 1. (c) tpk,` : Vk,` “ cu X Q3 “ tpk ` 1{2, ` ` 1{2q : l “ k ´ c, k “ 0, 1, . . . , c ´ 1u. We denote by X2c`2 , X2c`3 , . . . , X3c`1 these c centers for k “ c ´ 1, c ´ 2, . . . , 0 respectively. Every one of them lies on the straight line y “ x ´ c. (d) tpk,` : Vk,` “ cu X Q4 “ tpk ` 1{2, ` ` 1{2q : l “ ´k ´ c ´ 2, k “ ´1, ´2, . . . , ´c ´ 1u. We denote by X3c`2 , X3c`3 , . . . , X4c`2 these c ` 1 centers for k “ ´1, ´2, . . . , ´c ´ 1 respectively. Every one of them lies on the straight line y “ ´x ´ c ´ 1. Proof. In order to prove paq we begin by considering the points pk ` 1{2, k ` c ` 1{2q with ´c ď k ď ´1. Then Vk,k`c “ max p|2k ` c ` 1| ´ 1, cq . The inequality ´c ď k ď ´1 implies ´c ` 1 ď 2k ` c ` 1 ď c ´ 1, i.e. |2k ` c ` 1| ď c ´ 1. Therefore |2k ` c ` 1| ´ 1 ď c ´ 2 and consequently Vk,k`c “ c. Clearly, k ` 1{2 ă 0 and k ` c ` 1{2 ą 0, and hence the points belong to Q1 . To see the other inclusion take pk ` 1{2, ` ` 1{2q P Q1 with Vk,` “ c. We have to prove that ` “ k ` c and ´c ď k ď ´1. We know that k ă 0, ` ě 0 which easily implies that ` ą k. Hence |` ´ k| “ ` ´ k, and ` ` k ă ` ă ` ´ k. Then Vk,` “ max p|k ` ` ` 1| ´ 1, |k ´ `|q “ max p|k ` ` ` 1| ´ 1, ` ´ kq . 16 Consider the following two cases: (i) Assume ` ` k ` 1 ě 0. Then Vk,` “ maxp` ` k, ` ´ kq “ ` ´ k because ` ` k ă ` ă ` ´ k. It implies that ` ´ k “ c, that is ` “ k ` c. Furthermore, since ` “ k ` c and ` ě 0 we get k ě ´c. (ii) Assume ` ` k ` 1 ă 0. Then Vk,` “ maxp´` ´ k ´ 2, ` ´ kq. Since p` ´ kq ´ p´` ´ k ´ 2q “ 2` ` 2 “ 2p` ` 1q and ` ` 1 ą 0 also ` “ k ` c and the result follows as above. The proof of statements pbq, pcq and pdq follows using the same easy arguments. In Figure 4 we show the points pk,` in the levels c “ 2 and c “ 3, respectively. Figure 4: The centers in the levels c “ 2 and c “ 3. 3.2 Proof of items (i) and (ii) of Theorem A: dynamics on the zero-free set Recall that in this case, F px, yq “ py, ´x ` signpyqq, F` px, yq “ py, ´x ` 1q and F´ px, yq “ py, ´x ´ 1q. We will split our proof of items piq and piiq of Theorem A in several lemmas and propositions. We start facing the dynamics of the center points of the tiles, or in other words, the dynamics among the beads of each necklace, that as we will prove will be invariant under the map F. Lemma 5. Fixed c P N, consider the centers X1 , X2 , . . . , X4c`2 which belong to the level set tV “ cu. Then F pXi q “ Xj with j ” i ` c mod p4c ` 2q. (7) 17 Proof. Consider i “ 1, 2, . . . , c, then Xi P Q1 . From Lemma 4 we know that every one of these centers is pk ` 1{2, k ` c ` 1{2q with k “ ´c, ´c ` 1, . . . , ´1. Since they belong to H` we have F pXi q “ F` pk ` 1{2, k ` c ` 1{2q “ pk ` c ` 1{2, ´k ` 1{2q . ` ˘ Denoting k̄ “ k ` c we see that pk ` c ` 1{2, ´k ` 1{2q “ k̄ ` 1{2, ´k̄ ` c ` 1{2 which satisfies the condition (b) of the Lemma 4. When k runs from ´c to ´1 (corresponding with the points X1 , X2 , . . . , Xc ), then k̄ runs from 0 to c ´ 1 (corresponding with the points Xc`1 , Xc`2 , . . . , X2c ). Hence we have proved (7) for i “ 1, 2, . . . , c. If i “ c ` 1 then Xc`1 “ p1{2, c ` 1{2q , hence F pXc`1 q “ F` pXc`1 q “ pc ` 1{2, 1{2q “ X2c`1 . The proof for i “ c ` 2, c ` 3, . . . , 2c ` 1 is done in a similar way, and also for the rest of values of i “ 2c ` 2, . . . , 4c ` 2, but taking into account that in these cases F pXi q “ F´ pXi q. As a consequence of Lemma 5 the center points of a level set form an invariant set and we can prove that the function Vπ{2 defined in (3) is a first integral of F. Proof of the first part of item piiq of Theorem A. We start proving that the function V “ Vπ{2 defined in (3) is a first integral of F on the set U “ Uπ{2 . Consider a point px, yq P U, then px, yq P Tk,` for a certain k, ` and by definition we know that V px, yq “ V ppk,` q. From Lemma 5 we know that F ppk,` q “ pk̄,`¯ with V ppk,` q “ V ppk̄,`¯q. On the other hand, since each tile is entirely contained in H` zty “ 0u or in H´ , F pTk,` q “ F` pTk,` q or F pTk,` q “ F´ pTk,` q. Since F` and F´ are rotations (thus isometries) we get that F sends tiles to tiles. In particular F pTk,` q “ Tk̄,`¯ and hence V px, yq “ V ppk,` q “ V ppk̄,`¯q “ V pF px, yqq. Now we are able to describe the dynamics of the center points and, in particular, to prove that they are periodic. Proposition 6. Every center pk,` is a periodic point of F. Furthermore, setting Vk,` “ c we have that when c is even (resp. odd), then pk,` has period 2c ` 1 (resp. 4c ` 2). Proof. Fix a level tV “ cu with c P N. From Lemma 4 we know that on tV “ cu there are 4c ` 2 different centers. From Lemma 5, we know that F sends centers to centers, that is, the set tX1 , X2 , . . . , X4c`1 u is invariant by F. Hence, given a center Xi1 of the previous set F F F we can study the sequence Xi1 ÝÑ Xi2 ÝÑ Xi3 ÝÑ ¨ ¨ ¨ . Since the orbit of every center has a finite number of elements and, since F is a bijective map and therefore the orbit of Xi1 can not be preperiodic, we get that Xip “ Xi1 for a certain p, and therefore it is periodic. Clearly the period must be less or equal to 4c ` 2. 18 From Lemma 5, the map F restricted to tX1 , X2 , . . . , X4c`1 u is conjugate to the map h : Z4c`2 ÝÑ Z4c`2 defined by hpiq “ i ` c. Then F p pXi q “ Xi ô hp piq “ i ô i ` cp ” i mod p4c ` 2q ô D n P N s.t. cp “ np4c ` 2q. Assume that c “ 2k is an even number. Then 2kp “ np8k ` 2q ô kp “ np4k ` 1q. It implies that p is a multiple of 4k ` 1 “ 2c ` 1. Since p ď 4c ` 2 we get that p “ 2c ` 1 or p “ 4c ` 2. But we observe that the orbit of Xi only contains some points Xj with j having the same parity of i. Hence, we get two different periodic orbits, each one of them of period 2c ` 1. Assume that c “ 2k ` 1 is an odd number. Then p2k ` 1qp “ np8k ` 6q. It implies that p is a multiple of 8k ` 6 “ 4c ` 2. Hence p “ 4c ` 2. We introduce now the concept of itinerary map associated with a center. Definition 7. Fix c P N and consider one of the centers of the tiles Xj for some j “ 1, 2, . . . , 4c`2, with V pXj q “ c. Since Xj is p-periodic with p “ 2c`1 or p “ 4c`2 depending on whether c is even or it is odd, if we consider its itinerary of length p: I p “ pi1 , i2 , . . . , ip q we have that F p pXj q “ Fip ˝ Fip´1 ˝ ¨ ¨ ¨ ˝ Fi1 pXj q “ Xj . We denote this composition by Ij “ Fip ˝ Fip´1 ˝ ¨ ¨ ¨ ˝ Fi1 and we call it the itinerary map associated with Xj . For instance, if c “ 2 then the center X1 is 5-periodic and its orbit is F` F` F` F´ F´ X1 ÝÑ X3 ÝÑ X5 ÝÑ X7 ÝÑ X9 ÝÑ X1 This can be easily obtained using the formula (7) in Lemma 5 (see also Figure 4). Hence I 5 pX1 q “ p`, `, `, ´, ´q and its itinerary map is I1 “ F´2 ˝ F`3 . When c “ 3, then using again formula (7) in Lemma 5 (see again Figure 4) we get: F` F` F` F´ F´ F` F` X1 ÝÑ X4 ÝÑ X7 ÝÑ X10 ÝÑ X13 ÝÑ X2 ÝÑ X5 ÝÑ X8 F´ F´ F´ F` F` F´ F´ ÝÑ X11 ÝÑ X14 ÝÑ X3 ÝÑ X6 ÝÑ X9 ÝÑ X12 ÝÑ X1 , and hence the itinerary map of X1 is I1 “ F´2 ˝ F`2 ˝ F´3 ˝ F`2 ˝ F´2 ˝ F`3 . Lemma 8. Fixed c P N, consider the centers X1 , X2 , . . . , X4c`2 lying in the level set tV “ cu. Then for all j “ 1, 2, . . . , 4c ` 2, the itinerary map Ij is a rotation centered at Xj of order 4 if c is even (angle π{2), and of order 2 if c is odd (angle π). In particular Xj is an isolated fixed point of Ij . 19 Proof. We already know that Ij pXj q “ Xj . We write ˜ ¸ ˜ ¸ x ´ signpyq 0 1 F px, yq “ A ¨ where A “ Rπ{2 “ . y ´1 0 If c “ 2k, then by using Proposition 6 we have that Xj is p2c ` 1q-periodic, hence, using also that A4 “ Id we obtain ˜ ¸ ˜ ¸ ˜ ¸ 2c`1 x 4k`1 x x Ij px, yq “ A ` vj “ A ` vj “ A ` vj y y y for a certain vj P R2 . Hence Ij is a rotation of order 4 centered at Xj . Since it has a unique fixed point (as RankpA ´ Idq “ 2) then the center of this rotation is Xj . If c “ 2k ` 1, then Xj has period 4c ` 2, hence using that A2 “ Rπ “ ´ Id, we have: ˜ ¸ ˜ ¸ ˜ ¸ 4c`2 x 8k`6 x 2 x Ij px, yq “ A ` wj “ A ` wj “ A ` wj , y y y for a certain wj P R2 . By using the same argument as before, Ij is a rotation of order 2 centered at Xj . To end the technical results we establish the next lemma which ensures the all the points in a tile have the same itinerary of arbitrary length: Lemma 9. All the points in a given tile Tk,` have the same itinerary of length n P N for every n P N. Proof. Fix n P N, and suppose that there exist two points p and q in Tk,` with different itinerary of length n and let j ď n´1 the first time that ApF j ppqq ‰ ApF j pqqq. That is p and q have the same itinerary of length j but F j ppq and F j pqq have different addresses. From Lemma 3 we have that all the points in the segment p q have the same itinerary of length j and therefore F j restricted to p q is continuous. Since F j ppq and F j pqq have different addresses it follows that there exists a point r P p q such that F j prq P LC0 . A contradiction because since Tk,` is also convex, r P Tk,` and must be zero-free. We can now prove item piq, that is, the zero-free points are exactly the points U “ Uπ{2 , which belong to the tiles. Proof of item piq of Theorem A. We have already noticed that the zero-free set is included in Uπ{2 . Now we are going to see that the boundaries of the tiles are formed by points which are not zero-free. Consider a point p “ pk, yq where ` ď y ď ` ` 1 for a certain k, ` P Z. Then p belongs to the right-boundary of Tk´1,` and to the left boundary of Tk,` . Consider 20 also the segment rs where r “ pk ´ 1{2, yq and s “ pk ` 1{2, yq. From Lemma 9 the itinerary of any length of r coincides with the itinerary of the same length of pk´1,l and the itinerary of any length of s coincides with the corresponding itinerary of pk,l . Since pk´1,l and pk,l have different infinite itineraries, there exists j such that I j prq “ I j psq but ApF j prqq ‰ ApF j psq. Now from Lemma 3 it follows that there exists t P rs such that F j ptq P LC0 . Clearly this point must be p. If we consider a point which belongs to a horizontal boundary of two consecutive tiles, then its iterate belongs to a vertical boundary of two consecutive tiles and then we can apply the above result. Continuation of the proof of item piiq of Theorem A. Consider the tile Tk,` . Let Xj be its center. By Proposition 6 it is a p-periodic point, and by Lemma 9 all the points in the tile have the same itinerary of length p, hence F p |Tk,` “ Ij . Moreover, by Lemma 8, on each tile Ij is a rotation centered at Xj of the order established in the statement. Assume that Vk,` “ c with c an even number. We have already proved that each center Xj in this level set has period 2c ` 1 (see Proposition 6). The points in the orbit of Xj are points Xi with i having the same parity of j (see Lemma 5). Hence we have two orbits, the first one formed by X1 , X3 , . . . , X4c`1 and the second one formed by X2 , X4 , . . . , X4c`2 , and therefore we know the period of the centers of the tiles. The period of all the points of the tiles Tk,` but the centers is a consequence that we have proved that F p |Tk,` “ Ipk,` , where Ipk,` is the itinerary map of pk,` , which is a rotation of order 4, see again Lemma 8. When Vk,` “ c with c an odd number the proof is similar. 3.3 Proof of item piiiq of Theorem A: dynamics on the non zero-free set Following the notation introduced in Lemma 4, for any fixed energy level c of V, there are 4c ` 2 tiles with centers X1 , X2 , . . . , X4c`2 . Let us denote Tj to the tile with center Xj . Also, for a fixed energy level c, we denote by Qj the closed square formed by the tile Tj and its boundary, that is Qj “ Tj Y BTj . For each energy level c even, we will call the squares Q1 , Q3 , . . . , Q4c`1 perfect squares because, as we will see, these closed squares evolve avoiding the discontinuity effects of F. Clearly every edge of a square is also an edge of the consecutive square. The perfect squares are positioned as Figure 5 displays, the perfect squares being the red ones. From the above figure we see that it is enough to prove the periodicity of the points on the boundary of the perfect squares and the periodicity of the points on the boundary on the squares of odd levels which are not in the boundary of the perfect squares. 21 Figure 5: Position of the perfect squares in red. The green tiles correspond to odd energy levels and the blue ones correspond to the even energy levels such that their corresponding squares are not perfect. The borders and the vertices are highlighted with the color of the tile having the same itinerary map. The level set tV “ cu is indicated in each square. Our result also will ensure that the points of the border (including the vertices) of any perfect square are periodic with the same period as the points of the tile corresponding to the perfect square (excluding the center). Observe that any vertex point in F belongs uniquely to a perfect square. Hence the result will characterize the dynamics of all the vertices. The rest of the non-zero points are periodic with the same period as the points of the adjacent tile (excluding the center) with odd energy levels. See again Figure 5. Item piiiq of Theorem A is a straightforward consequence of the next theorem. Theorem 10. Consider the level set V “ c. (a) If c is an even number, then every point on the boundary of the squares Q1 , Q3 , . . . , Q4c`1 is a p8c ` 4q-periodic point. (b) If c is an odd number, then when j is odd (resp. even) the two horizontal (resp. vertical) edges of Qj , without the vertices, are formed by p8c ` 4q-periodic points. Prior to proving the result we stress the following fact: Remark 11. On every point in H` the map F “ F` . Hence, for all j “ 1, 2, . . . , 2c ` 1, we have that F pQj q “ F` pQj q “ Qi with i ” j ` c mod 4c ` 2 (see Lemmas 5 and 9). Analogously, for i “ 2c ` 3, . . . , 4c ` 1 we have F pQj q “ F´ pQj q “ Qi with i ” j ` c mod 22 4c ` 2, since these squares are contained in H´ . Observe, however, that the situation for the squares Q2c`2 and Q2c`4 is quite different because on the top edge of these two squares F “ F` while in the rest of the square F “ F´ . In Figure 6, we display the position of the tiles corresponding with the centers X1 , X2c`1 , X2c`2 and X4c`2 with respect to the discontinuity line LC0 . Figure 6: Position of the squares Q1 , Q2c`1 , Q2c`2 and Q4c`2 , together with its centers, for any level set V “ c. Proof of Theorem 10. Consider the squares Q1 , Q3 , . . . , Q4c`1 for c even, that is, the perfect squares on this level. The only squares in this particular collection Qj with j odd which intersect y “ 0 are Q1 and Q2c`1 . But from Remark 11 we know that F pQj q “ Qk with k ” j ` c mod 4c ` 2 for all j odd, including the cases with j “ 1 and j “ 2c ` 1. In particular this implies that this set of squares is invariant. In consequence, by continuity, the points in the boundary of Qj inherit the dynamics of the points in Tj zXj and, therefore, ˇ they are periodic with period 4p2c ` 1q. Furthermore, F 2c`1 ˇ is a rotation of order 4. Qj Now assume that c is odd. We notice that the squares Q1 , Q3 , . . . , Q4c`1 (resp. Q2 , Q4 , . . . , Q4c`2 ) share every vertical (resp. horizontal) edge with an edge of a perfect square, which we already know is periodic. Hence we need to follow the dynamics of their horizontal (resp. vertical) edges or, in other words, the dynamics of Q r j “ Qj zt its vertical edges, r j “ Qj ztits horizontal edges, including the verticesu). including the verticesu (resp. Q Since now X1 , X2 , . . . , X4c`2 belong to the same periodic orbit, the set of corresponding r 2c`2 and Q squares contains Q r 4c`2 . The result will be proved if we can ensure the invariance r j . In order to do this, we must ensure that the edges we are studying of the set of squares Q are not pre-images of the top edges of the squares Q2c`2 and Q4c`2 . So, first, we study for which values of p, F p pXj q “ X2c`2 or F p pXj q “ X4c`2 . • From Lemma 5 we have F p pXj q “ X2c`2 if and only if j ` pc ” 2c ` 2 mod 4c ` 2, that is if there exists n P N such that j ` pc “ 2c ` 2 ` np4c ` 2q. Hence j ` pc is an even number and since c is odd we get that p and j have the same parity. 23 • Analogously, F p pXj q “ X4c`2 if and only if j ` pc ” 0 mod 4c ` 2, which means that there exists n P N such that j ` pc “ np4c ` 2q. As before p and j have the same parity. Assume that j is odd, and let the p-iterate of Qj be the first one that reaches Q2c`2 (or Q4c`2 ). Since it is the first time that the images of Qj intersect ty “ 0u, we can still apply the arguments in the proof of Lemma 8 and therefore F p |Qj is an even-order rotation. Thus, since p is also odd, the horizontal edges of Qj are mapped via F p to the vertical edges of Q2c`2 (or Q4c`2 ). Therefore F p pQ rj q “ Q r 2c`2 (or Q r 4c`2 ). It implies that for all j odd, rj q “ Q F pQ r j`c . The same arguments work when j is even: p is even too and F p sends the vertical edges rj q “ Q of Qj to the vertical edges of Q2c`2 (or Q4c`2 ). So also F pQ r j`c for every j even. The condition F pQrj q “ Q r j`c implies that, by continuity, the points in the edges under study of Qj inherit the dynamics of the points in Tj zXj , which are periodic with period 2p4c ` 1q. Consider the square Qk,` “ Tk,` Y BTk,` , and set c “ Vk,` ; then, we will say that the square has odd label (resp. even label) if Qk,` “ Qj for an odd value j (resp. even) in the order introduced in Lemma 4. Observe that, in particular, with the proof of Theorem 10 we also have proved the following result that gives the dynamics of F on all the points in F. Corollary 12. Consider the square Qk,` “ Tk,` Y BTk,` , and set c “ Vk,` , then: (a) If c is even, and the square has odd label then Qk,` is invariant under the action of the ˇ map F 2c`1 , and F 2c`1 ˇQ is a rotation of order 4 centered at pk,` ; as a consequence k,` the edges of these Qk,` are formed by 4p2c ` 1q´periodic points. (b) If c is odd, and the square has odd label (resp. even label) then the horizontal (resp. vertical) edges (excluding the vertices) are invariant under the action of the map F 4c`2 which is also a rotation of order 2 centered at pk,` on that edges (excluding the vertices); as a consequence the edges of these squares which are not edges of a perfect square are formed by 2p4c ` 2q´periodic points. Remember that if the square has even energy level and even label we treat their bound- aries as being part of the boundary of the adjacent odd-energy level tile. 3.4 Proof of item pivq of Theorem A The proof simply follows by collecting the results of the previous items. 24 4 Proof of Theorem B 4.1 Preliminaries For each tile Tk,`,m , we start determining m in terms of k and `. Lemma 13. Any point px, yq P U “ U2π{3 belongs to a tile Tk,`,m where either, m “ k ` ` or m “ k ` ` ` 1 or m “ k ` ` ` 2. ? ? Proof. From the inequalities k ă x{2 ´ 3y{6 ă k ` 1 and ` ă 3y{3 ă ` ` 1 it follows ? ? that 1{2 ` k ` ` ă x{2 ` 3y{6 ` 1{2 ă k ` ` ` 5{2. Hence m “ Epx{2 ` 3y{6 ` 1{2q is either m “ k ` `, m “ k ` ` ` 1 or m “ k ` ` ` 2. ? ? In fact, the set of points satisfying k ă x{2 ´ 3y{6 ă k ` 1 and ` ă 3y{6 ă ` ` 1 ? ? ? form a parallelogram whose sides are y “ 3` , y “ 3p` ` 1q , y “ 3px ´ 2kq and ? y “ 3px ´ 2k ´ 2q. From the proof of the above lemma we see that ? ? (a) m “ k ` ` ô k ` ` ` 1{2 ă x{2 ` 3y{6 ` 1{2 ă k ` ` ` 1 ô 3p´x ` 2k ` 2`q ă y ă ? 3p´x ` 2k ` 2` ` 1q ? ? (b) m “ k ` ` ` 1 ô k ` ` ` 1 ă x{2 ` 3y{6 ` 1{2 ă k ` ` ` 2 ô 3p´x ` 2k ` 2` ` 1q ă ? y ă 3p´x ` 2k ` 2` ` 3q ? ? (c) m “ k ` ` ` 2 ô k ` ` ` 2 ă x{2 ` 3y{6 ` 1{2 ă k ` ` ` 52 ô 3p´x ` 2k ` 2` ` 3q ă ? y ă 3p´x ` 2k ` 2` ` 4q ? ? Denoting by L1 “ ty “ 3p´x ` 2k ` 2`qu, L2 “ ty “ 3p´x ` 2k ` 2` ` 1qu, L3 “ ty “ ? ? 3p´x ` 2k ` 2` ` 3qu and L4 “ ty “ 3p´x ` 2k ` 2` ` 4qu, we can draw its graphics in Figure 7. Figure 7: The tiles Tk,`,k`` Y Tk,`,k```1 Y Tk,`,k```2 when α “ 2π{3. 25 In Figure 7, we can see that the parallelogram has been partitioned into three sets: two triangles and one hexagon. Thus, we have proved the following lemma, where we use the notation introduced in Section 2.2. Lemma 14. Let Tk,`,m be one tile of U2π{3 “ U. (a) If m “ k ` ` or m “ k ` ` ` 2, then Tk,`,m is a triangle whose center is either qk,` or rk,` , respectively. (b) If m “ k ` ` ` 1 then Tk,`,m is a hexagon whose center is pk,` . 4.2 Proof of items piq and piiq: dynamics on the zero-free set As in the case α “ π{2 we split the proof of these two items into several lemmas and propositions. Here there is an added difficulty, there are tiles with hexagonal shape and others with triangular shape. We study them separately. 4.2.1 Dynamics on the hexagonal tiles. The case m “ k ` ` ` 1 From Lemma 14 the tile Tk,`,k```1 is a regular hexagon. Set Vk,` for the value of V on Tk,`,k```1 . Then Vk,` “ Vk,`,k```1 “ max p|2k ` 1|, 2|k ` ` ` 1| ´ 1, |2` ` 1|q . The next two results characterize the set of centroids of the hexagons, that is their number and geometric locus, for this case. ` ? ˘ Lemma 15. Let m “ k ` ` ` 1 and pk,` “ 2k ` ` ` 3{2, 3 p` ` 1{2q . Then (a) V ppk,` q “ c is an odd number. (b) The set tpk,` : Vk,` “ cu has 3c ` 1 points. In particular there are 3c ` 1 hexagons Tk,`,k```1 in this energy level. (c) The points pk,` with Vk,` “ c lie in the irregular hexagon determined by the intersection ? ? ? of the straight lines y “ 3px ˘ cq, y “ ˘ 3c{2 and y “ 3p´x ˘ p1 ` cqq, see Figure 8. 26 Figure 8: The centers points pk,` in the level c “ 3. Proof. Since Vk,` depends on the signs of 2k ` 1, 2` ` 1 and k ` ` ` 1, we are going to consider the different cases. (1) Assume that k ă 0, l ě 0 and k ` ` ` 1 ď 0. From (5) we have that Vk,` “ max p´2k ´ 1, 2` ` 1, ´2k ´ 2` ´ 3q , and from the inequalities p2` ` 1q ´ p´2k ´ 1q ď 0 and p´2k ´ 2` ´ 3q ´ p´2k ´ 1q ă 0 we get that Vk,` “ ´2k ´ 1 “ c. Hence c is odd, k “ ´p1 ` cq{2, 0 ď ` ď ´k ´ 1 “ pc ´ 1q{2, and the points pk,l can be written as ` ? ˘ pk,l “ ´c ` ` ` 1{2, 3 p` ` 1{2q with 0 ď ` ď pc ´ 1q{2. Hence, every pk,l lies on the ? straight line y “ 3px ` cq and that there are pc ` 1q{2 such points. (2) Assume that k ă 0, l ě 0 and k```1 ą 0. Then, proceeding analogously to the previous case we get Vk,` “ 2` ` 1 “ c. Hence c is odd, ` “ pc ´ 1q{2 and ´pc ` 1q{2 ă k ă 0. We ` ? ˘ get the points pk,l “ 2k ` pc ` 2q{2, 3c{2 with p1 ´ cq{2 ď k ď ´1. These points lie ? on the straight line y “ 3c{2 and there are pc ´ 1q{2 such points. (3) Assume that k ě 0 and ` ě 0. Then Vk,` “ 2k ` 2` ` 1 “ c ą 0. Hence c is odd, k “ ` ? ˘ pc ´ 1 ´ 2`q{2 and 0 ď ` ď pc ´ 1q{2. We get the points pk,l “ c ´ ` ` 1{2, 3 p` ` 1{2q ? with 0 ď ` ď pc ´ 1q{2. These points lie on y “ 3p´x ` 1 ` cq and there are pc ` 1q{2 such points. (4) Assume that ` ă 0, k ě 0 and k ` ` ` 1 ě 0. Then Vk,` “ 2k ` 1 “ c. Hence c is odd, k “ ` ? ˘ pc ´ 1q{2 and ´pc ` 1q{2 ď ` ă 0. We obtain the points pk,l “ c ` ` ` 1{2, 3 p` ` 1{2q 27 ? with ´pc ` 1q{2 ď ` ď ´1, which lie on y “ 3px ´ cq and there are pc ` 1q{2 such points. (5) Assume that ` ă 0, k ě 0 and k ```1 ă 0. Then Vk,` “ ´2`´1 “ c. Hence c is odd, ` “ ` ? ˘ ´pc ` 1q{2 and 0 ď k ă pc ´ 1q{2. We get the points pk,l “ 2k ` p2 ´ cq{2, ´ 3c{2 ? with 0 ď k ď pc ´ 3q{2, which lie on y “ ´ 3c{2 and there are pc ´ 1q{2 such points. (6) Assume k ă 0 and ` ă 0. Then k ` ` ` 1 ă 0 and Vk,` “ ´2k ´ 2` ´ 3 “ c. Hence c is odd, k “ ´` ´ pc ` 3q{2 with ´pc ` 3q{2 ă ` ă 0. We get the points pk,l “ ` ? ˘ ? ´c ´ ` ´ 3{2, 3 p` ` 1{2q with ´pc ` 1q{2 ď ` ď ´1, which lie on y “ ´ 3px`c`1q and there are pc ` 1q{2 such points. The Lemma follows from the above case-by-case study. Consider an odd energy level Vk,` “ c; we will label the center points pk,` analogously as in the case α “ π{2: we denote by X1 the point on the corresponding irregular hexagon defined by the lines in the above lemma, which belongs to H` and its first component is ? the smallest one; that is, X1 “ p´c ` 1{2, 3{2q. After we denote by X2 , X3 , . . . X3c`1 the consecutive points on the hexagon turning clockwise (see Figure 8 for instance). The set of center points in such a level set is invariant under the action of F and its dynamics is given in the next result: Proposition 16. Assume m “ k ` ` ` 1. Fixed Vk,` “ c an odd number, consider the points X1 , X2 , . . . , X3c`1 introduced above. Then (a) For all i “ 1, 2, . . . , 3c ` 1 , F pXi q “ Xj with j ” i ` c mod p3c ` 1q. (b) The set tX1 , X2 , . . . , X3c`1 u is a periodic orbit of period 3c ` 1. Proof. To prove statement paq we are going to consider the points that are on each of the six sides of the irregular hexagon delimited by the straight lines in Lemma 15. ? • Consider the points X1 , X2 , . . . , Xpc`1q{2 which lie on y “ 3px ` cq. The map F` ? sends this straight line to y “ 3p´x ` 1 ` cq and ´ ? ¯ ´ ? ¯ F pX1 q “ F` ´c ` 1{2, 3{2 “ pc ` 2q{2, 3c{2 “ Xc`1 . Since the distance between two consecutive points is constant and F` is an isometry we get that X2 , X3 , . . . , Xpc`1q{2 are mapped to Xc`2 , Xc`3 , . . . , Xp3c`1q{2 respectively. In particular F pXi q “ Xi`c (8) for all i “ 1, 2, . . . , Xpc`1q{2 . 28 ? • Now consider Xpc`1q{2 , Xpc`3q{2 , . . . , Xc`1 which lie on y “ 3c{2. F` sends y “ ? ? 3c{2 to the straight line y “ 3px ´ cq and we already check that Xpc`1q{2 is mapped to Xp3c`1q{2 . Being F` an isometry we get that Xpc`3q{2 , . . . , Xc`1 are mapped to Xp3c`3q{2 , . . . , X2c`1 respectively. Finally, equation (8) also holds for every i “ pc ` 1q{2, . . . , c ` 1. • Following the same argument it is seen that Xc`1 , Xc`2 , . . . , Xp3c`1q{2 are mapped to X2c`1 , X2c`2 , . . . , Xp5c`1q{2 respectively. ? • The points Xp3c`3q{2 , . . . , X2c`1 lie on y “ 3px ´ cq and on H´ . Then we have to ? ? take into account F´ which maps y “ 3px ´ cq to y “ ´ 3px ` c ` 1q and ´ ? ¯ ´ ? ¯ F pXp3c`3q{2 q “ F´ c ´ 1{2, ´ 32 “ ´pc ` 2q{2, ´ 3c{2 “ Xp5c`3q{2 . Hence F´ sends Xp3c`5q{2 , . . . , X2c`1 to Xp5c`5q{2 , . . . , X3c`1 respectively and (8) holds for every i “ p3c ` 3q{2, p3c ` 5q{2, . . . , 2c ` 1. • Now consider the points of the irregular hexagon which lie on the straight line y “ ? ? ´ 3c{2, that is, X2c`1 , . . . , Xp5c`3q{2 . The map F´ sends y “ ´ 3c{2 to y “ ? 3px ` cq and we verify that X2c`1 is sent to X3c`1 . Then X2c`2 , . . . , X 5c`3 are 2 sent to X1 , . . . , Xpc`1q{2 . So F pXi q “ Xj with j ” i ` c mod p3c ` 1q for i “ 2c ` 1, . . . , p5c ` 3q{2. ? • Finally, the points Xp5c`5q{2 , . . . , X3c`1 are on H´ and also lie on y “ ´ 3px ` ? c ` 1q. The map F´ sends this straight line to y “ 3c{2 and since we already know that that F´ pXp5c`5q{2 q “ Xpc`1q{2 we get that Xp5c`5q{2 , . . . , X3c`1 are sent to Xpc`3q{2 , . . . , Xc respectively. Again F pXi q “ Xj with j ” i ` c mod p3c ` 1q for every i “ p5c ` 5q{2, . . . , X3c`1 . In order to prove pbq we proceed as in the proof of Proposition 6. We use that the map F restricted to tX1 , X2 , . . . , X3c`1 u is conjugated to h : Z3c`1 ÝÑ Z3c`1 defined by hpiq “ i`c. Then F p pXi q “ Xi ô hp piq “ i ô i ` cp ” i mod p3c ` 1q ô Dn P N : cp “ np3c ` 1q. This implies that p must be a multiple of 3c ` 1, and since p ď 3c ` 1 we get that the minimal period is p “ 3c ` 1 as we wanted to see. 4.2.2 Dynamics on the triangular tiles. The cases m “ k ` ` and m “ k ` ` ` 2 For the triangular tiles, a result analogous to Lemma 15 is the following. We omit all the details. 29 ` ? ˘ Lemma 17. (i) Let m “ k ` ` and the points qk,` “ 2k ` ` ` 1{2, 3p` ` 1{6q . Then (a) V pqk,` q “ c is an even number. (b) The set tqk,` : V pqk,` q “ cu has 3c ` 1 elements. In particular there are 3c ` 1 triangles Tk,`,k`` in this energy level. (c) The points qk,` : V pqk,` q “ c lie in the irregular hexagon determined by the in- ? ? tersection of the six straight lines y “ 3px ` c ´ 1{3q, y “ 3p3c ` 1q{6, y “ ? ? ? ? ´ 3px´c´2{3q, y “ 3px´c´1{3q, y “ ´ 3p3c´1q{6 and y “ ´ 3px`c`4{3q. ` ? ˘ (ii) Let m “ k ` ` ` 2 and the points rk,` “ 2k ` ` ` 5{2, 3p` ` 5{6q . Then (a) V prk,` q “ c is an even number. (b) The set trk,` : V prk,` q “ cu has 3c ` 1 elements. In particular there are 3c ` 1 triangles Tk,`,k`` in this energy level. (c) The points rk,` : V prk,` q “ c lie in the irregular hexagon determined by the in- ? ? tersection of the six straight lines y “ 3px ´ c ` 1{3q, y “ 3p3c ´ 1q{6, y “ ? ? ? ? ´ 3px`c`4{3q, y “ 3px´c`1{3q, y “ ´ 3p3c`1q{6 and y “ ´ 3px`c´4{3q. Figure 9: Position of the centers qk,` and rk,` in the level c “ 4. For a fixed even number c ě 2 consider tqk,` : V pqk,` q “ cu (resp. trk,` : V prk,` q “ cu). We denote by Y1 (resp. Z1 ) the point qk,` (resp. rk,` ) in the corresponding irregular hexagon defined by the lines in the above lemma, which belongs to H` and its first component is the ` ? ˘ ` ? ˘ smallest one, that is Y1 “ ´c ` 1{2, 3{6 (resp. Z1 “ ´c ` 1{2, 5 3{6 ). We denote 30 by Y2 , Y3 , . . . , Y3c`1 (resp. Z2 , Z3 , . . . , Z3c`1 ) the consecutive points on the corresponding hexagon turning clockwise. See Figure 9 where the center points of the energy level c “ 4 are shown. Proposition 18. Consider a fixed even energy level Vk,` “ c and the points Y1 , Y2 , . . . , Y3c`1 and Z1 , Z2 , . . . , Z3c`1 defined above. Then (a) For all i “ 1, 2, . . . , 3c ` 1, F pYi q “ Yj with j ” i ` c mod p3c ` 1q and F pZi q “ Zj with j ” i ` c mod p3c ` 1q. (b) The set tY1 , Y2 , . . . , Y3c`1 u is a periodic orbit of period 3c`1 and the set Z1 , Z2 , . . . , Z3c`1 also is a periodic orbit of period 3c ` 1. The proof follows exactly by the same arguments involved in the proofs of Lemma 5 and Proposition 6. The next corollary simply consists of gluing (in a suitable way) the two sets given in the previous proposition, to form a single necklace with 6c ` 2 triangular beads. Corollary 19. Consider a fixed even energy level Vk,` “ c and denote the set of 6c ` 2 ordered points Y1 , Z1 , Y2 , Z2 . . . , Y3c`1 , Z3c`1 , that we will denote Wi , i “ 1, 2, . . . , 6c ` 2. Then for all i “ 1, 2, . . . , 6c ` 2, F pWi q “ Wj with j ” i ` 2c mod p6c ` 2q. Proof of item piq of Theorem B. Following the spirit of Definition 7, we can introduce the concept of itinerary map for the centers pk,` , rk,` and qk,` in an analogous way. Then, the proof is exactly the same proof as for item piq of Theorem A. It is based on the fact that all the points in the same tile have the same itineraries of arbitrary length (a result analogous to Lemma 9) and also on Lemma 3. Proof of item piiq of Theorem B. We start proving that V “ V3π{2 is a first integral. As in the proof of item piiq of Theorem A, we notice that since the tiles are completely contained in H` zty “ 0u or H´ and the maps F˘ are rotations, then F sends tiles to tiles. Remember that by its definition V is constant on each tile, and in particular takes the value attained at the center point. The result follows now from the fact that in each level set, the set of centers is invariant, see Propositions 16 and 18. Similarly that in the proof of Theorem A we consider the tile Tk,`,m . We know that all the points in the tile have the same itinerary than its center which, by Propositions 16, 18 and Corollary 19 give the discrete dynamical systems generated by the functions h given in the statement of Theorem B between the corresponding ZM . Moreover, we know that the centers are periodic with period 3c ` 1. Hence, if I is the itinerary map associated ˇ with the center point, that is I “ F 3c`1 ˇT , we have that IpTk,`,m q “ Tk,`,m . Writing k,`,m F px, yq “ A ¨ px ´ signpyq, yqt where A “ R2π{3 , we have I “ A3c`1 ` v “ A ` v for a 31 certain v P R2 , v ‰ 0, which implies that I is a rotation with a unique fixed point, hence it is the center point. Furthermore I 3 “ Id because I 3 “ A3 ` pA2 ` A ` Idqv “ Id, since A2 ` A ` Id “ 0. In summary, I is a rotation of angle α “ 2π{3 which implies that the points px, yq P Tk,`,m which are not centers are 3-periodic for F 3c`1 and consequently, they are p9c ` 3q-periodic. 4.3 Proof of item piiiq of Theorem B: dynamics in the non zero-free set From the previous results, we know that the non zero-free set F is formed by the borders of the tiles, both hexagons and triangles. Consider an energy level c P N0 . Assume that c is an odd number, then the level set tV “ cu is formed by 3c`1 hexagonal tiles, whose centers X1 , X2 , . . . , X3c`1 form a periodic orbit. Denoting by Hi the closure of this hexagon we also know that H1 and H2c´1 intersect y “ 0 at the bottom edge while H2c and H3c`1 intersect y “ 0 at the top edge. When c is even, we have the points Y1 , Y2 , . . . , Y3c`1 (respectively, Z1 , Z2 , . . . , Z3c`1 ). Each Yi (resp. Zi ) is the center of an upward (resp. downward) facing triangle; its closure intersects y “ 0 only when i “ 1 and i “ p3c ` 2q{2 (resp. i “ p3c ` 2q{2 and i “ 3c ` 1), see Figure 10. Figure 10: Position of the tiles which intersect y “ 0. We are going to call perfect triangles the ones corresponding to Y1 , Y2 , . . . , Y3c`1 . As for perfect squares, we will prove that these figures will evolve avoiding the discontinuity of F. They are positioned as the Figure 11 shows, the perfect triangles being the red ones, which are precisely the ones pointing upwards. The blue ones correspond to Z1 , Z2 , . . . , Z3c`1 . 32 Figure 11: Position of the perfect triangles in red. The borders and the vertices are high- lighted with the color of the tile having the same itinerary map. The level set tV “ cu is indicated in each tile. Proof of item piiiq of Theorem B. First, we observe that the borders of the perfect triangles (including the vertices) have the same period as the interior points which are not centers. Indeed, set an even number c P N0 and denote by Ti the closed triangle (i.e. including the boundary with vertices) which contain the point Yi . For i ‰ 1, 3c{2 ` 1, the triangle Ti does not intersect y “ 0, hence F pTi q “ Ti`c . For i “ 1 or i “ 3c{2 ` 1, F pTi q “ F` pTi q which also is Ti`c . Therefore, by continuity, the points in the boundary of Ti are periodic with period 3p3c ` 1q “ 9c ` 3, as for the points in the interior of Ti . Now take c odd and let Hi be the closed hexagon which contains Xi in its interior. Looking at Figure 11 we see that Hi has three edges which also are edges of a perfect r i “ Hi zperfect edges. (the triangle; if we call these edges the perfect edges, we consider H motivation for this name is similar that the ones of perfect triangles, or squares, and will be apparent later). Then, H r i contains three alternate edges, say l1 , l2 , l3 , such that the ? ? slopes of the straight lines which contain them are 3, ´ 3 and 0 respectively. Observe that l3 is always at the bottom of the hexagon, hence H r i is always fully contained in H` and H´ , and therefore F pH r i q “ F` pH r i q or F pH r i q “ F´ pH r i q. In any case the three edges included in F pH r i q are three alternate edges with the edge of slope 0 in the bottom of the hexagon F pHr i q. That is, F pH riq “ Hr i`c for all i “ 1, 2, . . . , 3c ` 1. As in the previous case, r i are periodic with period 9c ` 3, as for the by continuity, the points in the boundary of H points in the interior of Hi . Hence we have proved that all the points in the edges of the 33 hexagons are periodic points. It remains to consider the edges of the triangles which are not perfect. But, as can be seen in Figure 11, all these edges are also the edges of the contiguous hexagons, which we have already proved that all of them are periodic. Observe also that all the vertices belong to perfect triangles. 4.4 Proof of item pivq of Theorem B As in the case α “ π{2, the proof follows by replacing the value of V by 2n or 2n ` 1, for n P N0 , in the results of the previous items. We re-obtain the results of [10]. 5 Proof of Theorem C 5.1 Preliminaries As in the case studied in the previous section, for each tile Tk,`,m , the values k, `, m are not independent. Here, either m “ k ` ` ´ 1 or m “ k ` ` or m “ k ` ` ` 1. Lemma 20. Let Tk,`,m be one tile of Uπ{3 “ U. ` ? ? ˘ (a) If m “ k`` then Tk,`,m is a hexagon whose center is pk,` “ 2k ` ` ` 1{2, 3` ` 3{2 . (b) If m “ k ` ` ´ 1 or m “ k ` ` ` 1, then Tk,`,m is a triangle whose center is either ` ? ? ˘ ` ? ? ˘ qk,` “ 2k ` ` ´ 1{2, 3` ` 3{6 or rk,` “ 2k ` ` ` 3{2, 3` ` 5 3{6 , respectively. 5.2 Proof of items piq and piiq of Theorem C: dynamics on the zero-free set These results can be proved by the same arguments that we have used in the proofs of Theorems A and B, in Sections 3 and 4. Although we will not give all the details of their proofs, we want to highlight the main features and results that allow to give the dynamics in this case. Consider an even number c. Then, by Lemma 20, the tiles on the level set V “ c are hexagons whose centers are some of the points pk,` for some k, `. It can be proved that there are 3c ` 2 centers in this level. This centers lie in certain hexagons. We denote them by tX1 , X2 , . . . , X3c`2 u labeling them as in the case α “ 2π{3, see the Figure 12. 34 Figure 12: Position of the centers in the level c “ 4. Observe that the points do not belong to the same orbit. In this case there are two different orbits. The lines linking the centers are plotted because their expressions play a role in order to obtain the expression of the first integrals V , as explained in Section 6. In this case F restricted to tX1 , X2 , . . . , X3c`2 u is conjugated to h : Z3c`2 ÝÑ Z3c`2 where hpiq “ i ` c{2. (9) From this equality, and using that 3c ” ´2 mod 3c ` 2, one easily gets that when c{2 is even then the minimal period is 3c{2 ` 1, and that when c{2 is odd, then the minimal period is 3c ` 2. In the first case we get two periodic orbits tX1 , X3 , . . . , X3c`1 u and tX2 , X4 , . . . , X3c`2 u while in the second one all the points Xi , i “ 1, 2, . . . , 3c ` 1 belong to the same periodic orbit. To study the periodicity of the points in the hexagonal tile different from its center, for each Xj we consider its itinerary map Ij . • When c{2 is even, Ij has the form Ij “ A3c{2`1 ` v (where v “ Xj ´ A3c{2`1 Xj ) and 3c{2 ` 1 “ 3 ¨ 2n ` 1 “ 6n ` 1 for some n P N0 . Hence, since A6 “ Id, it holds that Ij “ A ` v. Therefore Ij restricted to the hexagon which contains Xj , is a rotation of angle π{3 centered at Xj and every point in the hexagon is a 6-periodic point for Ij . It implies that these points are 6p3c{2 ` 1q “ 9c ` 6 periodic points for F. • When c{2 is odd, Ij “ A3c`2 ` v and 3c ` 2 “ 3 ¨ 2p2n ` 1q ` 2 “ 6p2n ` 1q ` 2 for some n P N0 . Thus Ij “ A2 ` v, using again that A6 “ Id . This implies that every 35 point in the hexagon is a 3p3c ` 2q “ 9c ` 6 periodic point for F. Now let c be an odd number. Then the tiles in tV “ cu are triangles whose centers are either the points qk,` or the points rk,` introduced in Lemma 20, for some k, ` P Z. The points qk,` lie in some lines that define a hexagon, as do the points rk,` . But now all these centers belong to the same periodic orbit. To prove this, as usual, we label these points in a clockwise direction, as Figure 13 shows for the case c “ 3: the red points are the points qk,` P tV “ 3u while the blue ones are rk,` P tV “ 3u. Figure 13: Position of the centers in the level c “ 3. Observe that all the points belong to the same orbit. As shown in Section 6, the lines joining the centers play a role in the determination of the first integral V and do not represent two different orbits. With this labeling it can be proved that F pXi q “ Xj with j ” i ` c (mod 6c ` 4) which implies that the minimal period is p “ 6c ` 4. To see the periodicity of the points in the triangles different from its center we consider the itinerary function of the center Xj which has the form Ij “ A4 ` v. Then Ij3 “ A12 ` pA4 ` A2 ` Idqv “ Id . Arguing as before we get that each point in the triangle different from its center is a 3p6c ` 4q “ 18c ` 12 periodic point. 5.3 Proof of item (iii) of Theorem C: dynamics on the non zero-free set In this case, the dynamics of the points on the edges and vertices of the tiles is more complicated than the ones found in the cases α “ π{2 and α “ 2π{3, so we are going to give the details. 36 5.3.1 Perfect edges and vertices We begin by considering the levels c “ 4k, k P N. We already know that in these levels there are 3c ` 2 “ 12k ` 2 centers, X1 , X2 , . . . , X12k`2 and F pXi q “ Xj where j “ i ` 2k mod p12k ` 2q. Also X1 , X3 , . . . , X12k`1 form a periodic orbit of period 6k ` 1, as does X2 , X4 , . . . X12k`2 . Let Hj be the hexagon such that Xj P Hj including its boundary (hence also its vertices). Then the hexagons that meet y “ 0 are H1 , H6k`1 (its bottom edge is contained in y “ 0) and H6k`2 , H12k`2 (its top edge is contained in y “ 0). See Figure 14. Figure 14: Position of the hexagonal tiles of a level c “ 4k which intersect y “ 0. The red ones belong to the set of perfect hexagons. Clearly for all j “ 1, 2, . . . , 6k ` 1, F pHj q “ F` pHj q “ Hj`2k . while for j “ 6k ` 3, 6k ` 4, . . . , 12k ` 1 also F pHj q “ F´ pHj q “ Hj`2k mod p12k`2q . But the hexagons H6k`2 , H12k`2 do not satisfy this property because on the top edge of these hexagons F “ F ` . Then we easily get: Lemma 21. Assume that c “ 4k and consider the (closed) hexagons H1 , H3 , . . . , H12k`1 . Then for all j “ 1, 3, . . . , 12k ` 1 every point in Hj different from its center is periodic of period 36k ` 6. Proof. For j “ 1, 3, . . . , 12k ` 1, the hexagons satisfy F pHj q “ Hj`2k mod p12k`2q , hence it is easy to observe that their images are never the hexagons H6k`2 and H12k`2 . Then, by continuity, every point on the boundary of Hj has the same periodicity as the points inside the hexagon (except the center). In particular, H1 , H3 , . . . , H12k`1 form an invariant set. As in the above sections we call H1 , H3 , . . . , H12k`1 perfect hexagons and their edges and vertices behave as the corresponding interior points, apart from the centers, that is they are p36k ` 6q´ periodic. Also we will call non-perfect edges or vertices those which do not collide with a perfect hexagon. 37 Figure 15: Position of the perfect hexagons in red. The level tV “ cu is indicated in each set. 5.3.2 Non-perfect edges We continue the study considering the even levels of the form c “ 4k ` 2. We know that in these level sets there are 3c ` 2 “ 12k ` 8 centers and that all of them belong to the same periodic orbit. The hexagons which meet y “ 0 are H1 , H6k`4 , H6k`5 and H12k`8 , see Figure 16. Figure 16: Position of the hexagonal tiles of a level c “ 4k ` 2, which intersect y “ 0. We are going to follow the dynamics of the interior of the bottom edge of H1 , that we will denote as L (for simplicity we will use the term edge although the two boundary points are not included). This dynamics is, by far, the most complex of those we have studied in this paper. Since the argument is long, we first briefly summarize it: we will show that every point in L is p108k ` 72q-periodic. The edge is rigidly mapped by by iterating F into 38 the edges of the hexagons in the level 4k ` 2, but also into the edges of the triangles in the levels 4k ` 1 and 4k ` 3. Indeed, after the first iteration, F pLq is the edge of H2k`2 obtained after rotating L an angle equal to π{3, because F pX1 q “ X2k`2 (remember that from (9), F pXi q “ Xi`2k`1 mod p12k`8q .q We continue iterating until we find the hexagon H6k`5 . To com- pute how many iterations we need for X1 to reach X6k`5 we ask for the minimal pos- itive number p such that F p pX1 q “ X6k`5 . That is, F p pX1 q “ X6k`5 , or equivalently, 1 ` pp2k ` 1q ” 6k ` 5 mod p12k ` 8q. Thus, p ” p2k ` 1q´1 p6k ` 4q “ p6k ` 1qp6k ` 4q “ 36k 2 ` 30k ` 4 ” 6k ` 4 mod p12k ` 8q. That is F 6k`4 pX1 q “ X6k`5 . Now F 6k`4 pLq is the edge of H6k`5 after rotating L 4π an angle equal to 3 . Hence we follow iterating until we arrive to X12k`8 , that is, three iterates more: F 3 pX6k`5 q “ X12k`8 . This implies that F 6k`7 pLq is the edge of H12k`8 obtained after rotating L an angle equal to π{3. Now we ask for the minimal p such that F p pX12k`8 q “ X6k`5 . The computation gives that p “ 12k ` 5. Hence we can write F 6k`4 F3 F 12k`5 F3 X1 ÝÑ X6k`5 ÝÑ X12k`8 ÝÑ X6k`5 ÝÑ X12k`8 , and following the edge L we have that after 18k ` 15 iterates the initial edge L of H1 is the top edge of H12k`8 , which is the bottom edge of the triangle T1 in the level 4k ` 3. Next, we follow the orbit of the centers of the triangles in the level c “ 4k ` 3. Let Y1 , Y2 , . . . , Y24k`22 be the centers of the triangles T1 , T2 , . . . , T24k`22 . All of them form a unique periodic orbit and F pYi q “ Yj where j “ i ` 4k ` 3 mod p24k ` 22q, remember that in the triangles F pYi q “ Yi`c mod p6c`4q . The triangles with edges in the critical line are displayed in the Figure 17. Figure 17: Position of the triangular tiles which intersect y “ 0. Taking into account that p4k `3q´1 “ 18k `15 in Z24k`22 and solving the corresponding congruences we find that: F 6k`7 F 6k`4 F 18k`18 F 6k`4 Y1 ÝÑ Y24k`22 ÝÑ Y12k`12 ÝÑ Y24k`22 ÝÑ X12k`12 . 39 Then we see that the bottom edge of T1 is transformed into the top edge of T12k`12 after 36k ` 33 iterates. This one also is the bottom edge of the hexagon H6k`4 in the level c “ 4k ` 2, see again Figure 15, and also Figure 3. Following the same procedure it can be seen that F 6k`1 pX6k`4 q “ X6k`5 and using the calculations made before we obtain F 6k`1 F3 F 12k`5 X6k`4 ÝÑ X6k`5 ÝÑ X12k`8 ÝÑ X6k`5 . Hence, the bottom edge of H6k`4 is transformed into the top edge of H6k`5 after 18k ` 9 iterates. The top edge of H6k`5 is also the bottom edge of one triangle whose center belongs to the level set 4k ` 1. In this level set there are 24k ` 10 centers of triangles, that we denote by Z1 , Z2 , . . . , Z24k`10 , and we know that F pZi q “ Zj with j “ i ` 4k ` 1 mod p24k ` 10q. We call T1 , T2 , . . . , T24k`10 these triangles. Specifically, the top edge of H6k`5 is the bottom edge of T12k`5 , see Figures 15 and 18. Figure 18: Position of some of the triangular the tiles which intersect y “ 0. Using that in Z24k`10 , p4k ` 1q´1 “ 6k ` 1 and solving the corresponding congruences we have that F 6k`1 F 6k`4 F 18k`6 F 6k`4 Z12k`5 ÝÑ Z12k`6 ÝÑ Z24k`10 ÝÑ Z12k`6 ÝÑ Z24k`10 . It follows that after 36k ` 15 iterates, the bottom edge of T12k`5 is transformed in the top edge of T24k`10 . But this top edge of T24k`10 is exactly the edge L. Hence summing up the involved iterates we have that every point in L is a 108k ` 72 periodic point. Also the same holds for all the points belonging to the 108k ` 72 edges obtained iterating L. In other words, we get a periodic orbit of edges of period 108k ` 72 and, of course, the points of L are mapped to themselves after these iterations. 5.3.3 Non-perfect vertices And what about the vertices? As we will see in the proof of Theorem 22, we only need to prove the periodicity of the vertices in y “ 0. Observe that if such a vertex belongs to 40 a perfect hexagon, then we already know that it is periodic with the same period as the interior points. If it is non-perfect, then either paq it is mapped to a vertex colliding from the top with a triangle of level V “ 4k ` 1 and a hexagon of level V “ 4k ` 2, both in H` , as the solid-circle point in Figure 19; or (b) it collides from the top with a hexagon of level V “ 4k ` 2 and triangle of level V “ 4k ` 3, both in H` , as the box-shaped point in Figure 19. Figure 19: Position of the non-perfect vertices in y “ 0. If H1 is the left non-perfect hexagon with level V “ 4k ` 2 then the vertex V1 pH1 q is the box-shaped point and the vertex V6 pH1 q is the solid-circle point. The perfect hexagons are the red ones. To study the dynamics of the non-perfect vertices in y “ 0, we will use the following notation: given a hexagon H, we label its vertices as Vi pHq with i “ 1, ¨ ¨ ¨ , 6 starting from the left-bottom vertex and in clockwise sense, see Figure 20. Figure 20: Labeling the vertices of a hexagon H Let H1 be the non-perfect hexagon at level V “ 4k ` 2 in Q1 , whose intersection with y “ 0 is its bottom edge. Then: paq We will follow the orbit of the point V6 pH1 q (the blue point in Figure 19) by us- ing the results found in Section 5.3.2. In particular, we know that F 3 pH1 q “ H6k`4 , F 6k`1 pH6k`4 q “ H6k`5 , F 3 pH6k`5 q “ H12k`8 and F 6k`1 pH12k`8 q “ H1 . Therefore, we 41 easily find F3 F 6k`1 F 6k`1 V6 pH1 q ÝÑ V3 pH6k`4 q ÝÑ V4 pH6k`5 q “ V1 pH6k`4 q ÝÑ F3 F 6k`1 V2 pH6k`5 q ÝÑ V5 pH12k`8 q ÝÑ V6 pH1 q, hence the point V6 pH1 q is p18k ` 9q-periodic. (b) We will pursue the orbit of the point V1 pH1 q (the box-shaped point in Figure 19). F3 F 6k`1 F3 F 6k`1 F3 V1 pH1 q ÝÑ V4 pH6k`4 q ÝÑ V5 pH6k`5 q ÝÑ V2 pH12k`8 q ÝÑ V3 pH1 q ÝÑ F 6k`1 F3 V6 pH6k`4 q ÝÑ V1 pH6k`5 q ÝÑ V4 pH12k`8 q “ V1 pH1 q, hence the point V1 pH1 q is p18k ` 15q-periodic. Now we have all the ingredients to prove the main result of this section, that clearly implies item piiiq of Theorem C. Theorem 22. Every non zero-free point of F is periodic. Furthermore: (a) If px, yq is a point in the edge of a perfect hexagon (which has energy level V “ 4k), then it is periodic with period 36k ` 6. (b) If px, yq is a point in a non-perfect (open) edge of a tile then it is periodic with period 108k ` 72 for some k P N0 . (c) If px, yq is a non-perfect vertex then it is periodic with period 18k ` 9 or 18k ` 15 for some k P N0 . Observe that any non zero-free point belongs to one of the above cases. Proof. We already know that the set of the non zero-free points is formed by the edges and the vertices of the hexagons and triangles introduced before. Consider the points in one edge. Then after a finite number of iterates this edge is transformed in one edge contained in y “ 0. If this edge correspond to an edge of a perfect hexagon with center belonging to the level 4k, every point will be 6p6k ` 1q´periodic. If not, it will be an edge of a polygon with center belonging to the level either 4k ` 1, 4k ` 2 or 4k ` 3. From the discussion above we know that every point will be p108k ` 72q´periodic. With respect to the vertices, observe that since any vertex belongs to F, after some iterates it will be mapped to a vertex point in y “ 0. Hence there are three possibilities: it is mapped to a perfect vertex of a perfect hexagon in H` , which has energy level V “ 4k (and in this case it is periodic of period 36k ` 6); or it is mapped to a vertex colliding from the top with a triangle of level V “ 4k ` 1 and a hexagon of level V “ 4k ` 2 (and in this case it is periodic with period 18k ` 9); or it is mapped to a vertex colliding from the top with a hexagon of level V “ 4k ` 2 and a triangle of level V “ 4k ` 3. In this case it is periodic with period 18k ` 15. 42 The proof of item pivq is a straightforward consequence of all the previous results. 6 Obtaining the first integrals We have intuited the expressions of the first integrals after several simulations. For com- pleteness we present in detail a three-step constructive procedure that allows to obtain the first integrals of F given in (3) corresponding to α “ π{2. For the other two cases the line of argument is the same, and the details are analogous, and we only give some comments. Step 1. By displaying some preimages of the critical line LC´i we realize that the zero-free set is formed by open tiles of a regular or uniform tessellation of R2 . This fact is trivial in this case where α “ π{2 but, a priori, it was not so obvious in the cases α “ 2π{3 and α “ π{3 studied in Sections 4 and 5. The normal form of F given in (2) regularizes the tesselation. Step 2. From preliminary numerical explorations we also realize that the centers of some tiles form an invariant set under the dynamics of F. In the case α “ π{2 these centers were located in the lines y “ x ` c, y “ ´x ` c ` 1, y “ x ´ c and y “ ´x ´ c ´ 1 for a certain fixed value c P N0 , depending on the quadrant where the center points are located (see Lemma 4 and Figure 4, and see Lemmas 15 and 17 for the case α “ 2π{3). Step 3. Isolating the value in the expression of the lines linking the centers we obtain that c “ y ´ x for px, yq P Q1 , c “ y ` x ´ 1 for px, yq P Q2 , c “ x ´ y for px, yq P Q3 and c “ ´x ´ y ´ 1 for px, yq P Q4 , where recall that Qj , j “ 1, 2, 3, 4, are the four quadrants of R2 . From these expressions and taking into account that c P N0 and that given a zero-free point px, yq the center point of its associated tile is pEpxq ` 1{2, Epyq ` 1{2q, we arrive to the expression of the first integral Vπ{2 px, yq “ max p|Epxq ` Epyq ` 1| ´ 1, |Epxq ´ Epyq|q . 7 Final comments. We have proved that for α P tπ{3, π{2, 2π{3u, the corresponding zero-free sets U are the union of a countable number of open sets (the tiles), hence the associated critical sets F “ R2 zU are closed sets. In consequence, for any point px, yq P R2 , the distance dist ppx, yq, Fq is well defined. Since F is also invariant, we have: Remark 23. Any map (2) with α P tπ{3, π{2, 2π{3u has the non-quantized continuous first integral W px, yq “ dist ppx, yq, Fq. We believe that the only pointwise periodic cases for the maps F with α P p0, 2πq, are the ones studied in this work as well the cases α P t4π{3, 3π{2, 5π{3u (recall that we where 43 motivated by the study of the maps G in (1) with |ρ| ă 2, which are conjugated with the maps F in (2) with α P p0, πq). In these later cases we have observed that the quantized first integrals given in this paper are also integrals of these maps. In particular: Vπ{2 , V2π{3 and Vπ{3 are first integrals of F when α “ 3π{2, 5π{3 and 4π{3 respectively. However notice that none of these last maps are conjugated to the maps considered in this work: for example, note that the maps F with α P t4π{3, 3π{2, 5π{3u do not have fixed points since the centers of the rotations are virtual. The maps F belong to the class of symmetric maps studied in the relevant paper [14]. We refer the reader to this reference to learn about the general properties of the maps F with α a general value in r0, 2πqztπ{3, π{2, 2π{3, 4π{3, 3π{2, 5π{3u. For instance, in that paper it is proved that for any α ‰ ˘π being a rational multiple of π there exists a sequence of open invariant nested necklaces, that tend to infinity, each one of them being similar to the level sets of our quantized first integrals, whose beads are polygons, and where the dynamics of F is given by a product of two rotations. Remarkably, although the adherence of the union of all these invariant necklaces does not fill the full plane, it allows to prove that all orbits of F are bounded. Acknowledgements We want to thank the anonymous reviewers for their comments that have allowed us to better contextualize the problem, improve our article and have provided us with very interesting references. The authors are supported by Ministry of Science and Innovation– State Research Agency of the Spanish Government through grants PID2019-104658GB- I00 (first and second authors), DPI2016-77407-P (AEI/FEDER, UE, third author) and MTM2017-86795-C3-1-P (fourth autor). The first, second and fourth authors are also supported by the grant 2017-SGR-1617 from AGAUR, Generalitat de Catalunya. The third author acknowledges the group’s research recognition 2017-SGR-388 from AGAUR, Generalitat de Catalunya. References [1] V. Avrutin, L. Gardini, I. Sushko, F. Tramontana. Continuous and discontinuous piecewise-smooth one-dimensional maps. Invariant sets and bifurcation structures. World Scientific, Hackensack (NJ) 2019. [2] S. Banerjee, G.C. Verghese (eds). Nonlinear Phenomena in Power Electronics. IEEE Press, New York 2001. 44 [3] M. di Bernardo, C.J. Budd, A.R. Champneys, P. Kowalczyk. Piecewise-Smooth Dy- namical Systems: Theory and Applications. Springer, London 2008. [4] G.I. Bischi, L. Gardini, C. Mira. Maps with denominator. Part 1: some generic prop- erties. Int. J. Bifurcation & Chaos 9 (1999), 119–153. [5] M. Boshernitzan, A. Goetz. A dichotomy for a two-parameter piecewise rotation. Er- god. Th. & Dynam. Sys. 23 (2003), 759–770. [6] B. Brogliato. Nonsmooth Mechanics: Models, Dynamics and Control. Springer-Verlag, New York 1999. [7] Y. C. Chang, S. S. Cheng. Complete periodicity analysis for a discontinuous recurrence equation. Int. J. Bifurcations and Chaos 23 (2013), 1330012 (34 pages). [8] Y. C. Chang, S. S. Cheng. Complete periodic behaviours of real and complex bang bang dynamical systems. J. Difference Equations and Appl. 20 (2014), 765–810. [9] Y. C. Chang, S. S. Cheng, Y. C. Yeh. Abundant periodic and aperiodic solutions of a discontinuous three-term recurrence relation. J. Difference Equations and Appl. 25 (2019), 1082–1106. [10] Y. C. Chang, G. Q. Wang, S. S. Cheng. Complete set of periodic solutions of a discon- tinuous recurrence equation. J. Difference Equations and Appl. 18 (2012), 1133–1162. [11] A. Cima, A. Gasull, V. Mañosa. Global periodicity and complete integrability of dis- crete dynamical systems. J. Difference Equations and Appl. 12 (2006), 697–716. [12] A. Cima, A. Gasull, V. Mañosa, F. Mañosas. Different approaches to the global pe- riodicity problem. In Ll. Alsedà et al. eds. Difference Equations, Discrete Dynamical Systems and Applications (ICDEA 2012). Springer Proceedings in Mathematics & Statistics Vol. 180. pp. 85–106. Springer, Berlin 2016. [13] F. Dogru, S. Tabachnikov. Dual billiards. Math. Intelligencer 27 (2005), 18–25. [14] A. Goetz, A. Quas. Global properties of a family of piecewise isometries. Ergod. Th. & Dynam. Sys. 29 (2009), 545–568. [15] B. Grünbaum, G.C. Shephard. Tilings by regular polygons. Math. Mag. 50 (1977), 227–247. [16] B. Grünbaum, G.C. Shephard. Tilings and Patterns. W.H. Freeman and Co., New York 1987. 45 [17] E. Gutkin, N. Simanyi. Dual polygonal billiards and necklace dynamics. Commun. Math. Phys. 143 (1992), 431–449. [18] C. Mira, L. Gardini, A. Barugola, J.C. Cathala. Chaotic Dynamics in Two-Dimensional Noninvertible Maps. World Scientific, Singapore 1996. [19] D. Montgomery. Pointwise periodic homeomorphisms Amer. J. Math. 59 (1937), 118– 120. [20] J. Moser. Is the solar system stable? Math. Intelligencer 1 (1978), 65–71. [21] O. Neugebauer. The Exact Sciences in Antiquity. Dover, New York 1969. [22] D.J.W. Simpson, J.D. Meiss. Neimark-Sacker bifurcations in planar, piecewise-smooth, continuous maps. SIAM J. Appl. Dyn. Syst. 7 (2008), 795–824. [23] F. Vivaldi, A.V. Shaidenko Global stability of a class of discontinuous dual billiards. Commun. Math. Phys. 110 (1987), 625–640. [24] Z.T. Zhusubaliyev, E. Mosekilde. Bifurcations and Chaos in Piecewise-Smooth Dynam- ical Systems. World Scientific. Singapore 2003. 46